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AP Statistics Curriculum 2007 IntroTools
(Difference between revisions)
Revision as of 15:37, 3 April 2013 (view source)IvoDinov (Talk | contribs)m (→Approach & Model Validation)← Older edit Current revision as of 15:48, 5 April 2013 (view source)IvoDinov (Talk | contribs) m (→Data Plots) (2 intermediate revisions not shown) Line 21: Line 21: |} |} - We can easily compute the average weight (66 kg) and height (167 cm) using the [http://en.wikipedia.org/wiki/Sample_mean sample mean-formula]. We can also compute these averages using the [http://www.socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts], or any other statistical package, as shown in the image below. + We can easily compute the average weight (66 kg) and height (167 cm) using the [http://en.wikipedia.org/wiki/Sample_mean sample mean-formula]. We can also compute these averages using the [http://www.socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts (BarChart3DDemo1)], or any other statistical package, as shown in the image below. -
[[Image:SOCR_EBook_Dinov_IntroTools_061707_Fig1.png|400px]]
+
[[Image:SOCR_EBook_Dinov_IntroTools_061707_Fig1.png|300px]] + [[Image:SOCR_EBook_Dinov_IntroTools_Fig2.png|300px]] +
You can see a [[SOCR_Data_Dinov_020108_HeightsWeights | large dataset of human weights and heights here]]. You can see a [[SOCR_Data_Dinov_020108_HeightsWeights | large dataset of human weights and heights here]]. Line 38: Line 40: ===Data Plots=== ===Data Plots=== - Let's first try to plot some of these data. Suppose we take a smaller fraction of the entire dataset. [[AP_Statistics_Curriculum_2007_IntroTools_Data1 | You can find a fragment of 21 rows and 3 columns of measurements here]]. This number is large enough to require computer software to graph the data. In column 1, this data subset includes an index of the region (blob) and in column 2, a pair of MEAN & Standard Deviation for the intensities over the blob (within the Left Occipital lobe). Now go to [http://www.socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts] and select the ''StatisticalBarChardDemo1'' Chart (under ''BarCharts'' --> ''CategoryPlot''), see figure below). '''Clear''' the default data and '''Paste''' in [[AP_Statistics_Curriculum_2007_IntroTools_Data1 | this data segment]]. '''Map''' the first column (C1) to ''Series'' and the second column (C2) to ''Categories'' and click '''UPDATE''' to redraw the graph with the new data. This plot shows the relations between the means and standard deviations of the intensities in the 21 regions (blobs, rows in table). We see that there is variation in both means and standard deviations (error bars on the box plots). + Let's first try to plot some of these data. Suppose we take a smaller fraction of the entire dataset. [[AP_Statistics_Curriculum_2007_IntroTools_Data1 | You can find a fragment of 21 rows and 3 columns of measurements here]]. This number is large enough to require computer software to graph the data. In column 1, this data subset includes an index of the region (blob) and in column 2, a pair of MEAN & Standard Deviation for the intensities over the blob (within the Left Occipital lobe). Now go to [http://www.socr.ucla.edu/htmls/SOCR_Charts.html SOCR Charts] and select the ''StatisticalBarChartDemo1'' Chart (under ''BarCharts'' --> ''CategoryPlot''), see figure below). '''Clear''' the default data and '''Paste''' in [[AP_Statistics_Curriculum_2007_IntroTools_Data1 | this data segment]]. '''Map''' the first column (C1) to ''Series'' and the second column (C2) to ''Categories'' and click '''UPDATE''' to redraw the graph with the new data. This plot shows the relations between the means and standard deviations of the intensities in the 21 regions (blobs, rows in table). We see that there is variation in both means and standard deviations (error bars on the box plots).
[[Image:SOCR_EBook_Dinov_IntroTools_061707_Fig2.png|400px]]
[[Image:SOCR_EBook_Dinov_IntroTools_061707_Fig2.png|400px]]
Current revision as of 15:48, 5 April 2013
General Advance-Placement (AP) Statistics Curriculum - Statistics with Tools
Statistics with Tools (Calculators and Computers)
A critical component in any data analysis or process understanding approach is the development of a model. Models have compact analytical representations (e.g., formulas, symbolic equations, etc.) The model is frequently used to study the process theoretically. Empirical validation of the model is carried by plugging in data and actually testing the model. This validation step may be done manually, by computing the model prediction or model inference from recorded measurements. This typically may be done by hand only for small number of observations (<10). In practice, most of the time, we write (or use existent) algorithms and computer programs that automate these calculations for greater efficiency, accuracy and consistency in applying the model to larger datasets.
There are a number of statistical software tools (programs) that one can employ for data analysis and statistical processing. Some of these are: SAS, SYSTAT, SPSS, R, SOCR, etc.
Approach & Model Validation
Before any statistical analysis tool is employed to analyze a dataset, one needs to carefully review the prerequisites and assumptions that this model demands about the data and study design.
For example, if we measure the weight and height of students and want to study gender, age or race differences or association between weight and height, we will need to make sure our sample size is large enough. These weight and height measurements are random (i.e., we do not have repeated measurements of the same student or twin-measurements) and that the students we measure are actually a representative sample of the population that we are making inference about (e.g., 8th-grade students). You can also find a real and large weight and height dataset here.
In this example, suppose we record the following 6 pairs of {weight (kg), height (cm)}:
Student_Index 1 2 3 4 5 6 Weight 60 75 58 67 56 80 Height 167 175 152 172 166 175
We can easily compute the average weight (66 kg) and height (167 cm) using the sample mean-formula. We can also compute these averages using the SOCR Charts (BarChart3DDemo1), or any other statistical package, as shown in the image below.
You can see a large dataset of human weights and heights here.
Computational Resources: Internet-based SOCR Tools
Several of the SOCR tools and resources will be shown later, and will be useful in a various situations. Here is just a list of tools with one example of each:
Hands-on Examples & Activities
Data and Study Design
As part of a brain imaging study of Alzheimer's disease *, the investigators collected the following data. We will now demonstrate how computer programs, software tools and resources, like SOCR, can help in statistically analyzing larger datasets (certainly data size over 10 is difficult to calculate by hand correctly). In this case we'll work with 240 measurements derived from data acquired by this study.
Data Plots
Let's first try to plot some of these data. Suppose we take a smaller fraction of the entire dataset. You can find a fragment of 21 rows and 3 columns of measurements here. This number is large enough to require computer software to graph the data. In column 1, this data subset includes an index of the region (blob) and in column 2, a pair of MEAN & Standard Deviation for the intensities over the blob (within the Left Occipital lobe). Now go to SOCR Charts and select the StatisticalBarChartDemo1 Chart (under BarCharts --> CategoryPlot), see figure below). Clear the default data and Paste in this data segment. Map the first column (C1) to Series and the second column (C2) to Categories and click UPDATE to redraw the graph with the new data. This plot shows the relations between the means and standard deviations of the intensities in the 21 regions (blobs, rows in table). We see that there is variation in both means and standard deviations (error bars on the box plots).
Statistical Analysis
Now we can demonstrate the use of SOCR Analyses to look for Left-Right hemispheric (HEMISPHERE) effects of the average MRI intensities (MEAN) in one Region of Interest (Occipital lobe, ROI=2). For this, we can apply simple Paired T-test. This analysis is justified as the average intensities that will follow Normal Distribution by the Central Limit Theorem because the left and right hemispheric observations are naturally paired.
• Copy in your mouse buffer the 6th (MEAN), 8th (HEMISPHERE) and 9th (ROI) columns of the following data table. You can paste these three columns in Excel, or any other spreadsheet program, and reorder the rows first by ROI and then by HEMISPHERE. This will give you an exert of 240 rows of measurements (MEAN) for ROI=2 (Occipital lobe) for each of the two hemispheres. The breakdown of this number of observations is as follows 240 = 2(hemispheres) * 3 (3D spatial locations, blobs) * 40 (Patients).
• Copy these 240 Rows and paste them in the Paired T-test Analysis under SOCR Analyses. Map the MEAN and HEMISPHERE columns to Dependent and Independent variables and then click Calculate. The results indicate that there are significant differences between the Left and Right Occipital mean intensities for these 40 subjects.
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Decimal Worksheet for Class 5
# Decimal Worksheet for Class 5
## Decimal Worksheet for Class 5
Maths worksheets play a vital role in strengthening the maths concepts. After studying the concepts of mathematics, revision of questions related to that particular topic is very important for the students. This assesses the preparation of the students. Students can assess themselves by its own by solving the worksheets problems. Parents and teachers can also assess the students by giving them these worksheets to the students to solve.
Maths-formula brings such worksheets on Decimals Class 5 for you. By solving these worksheets, you can increase the understanding of the concepts learnt. These worksheets are prepared by experienced teachers and subject matter experts.
Thus, let us try to attempt these questions to solve and assess yourself. Parents and teachers can also give these questions to the students to test them. Unit tests can also be prepared by the teachers using these questions.
## Decimal Worksheet for Class 5
Worksheet 1
1. Write in words.
a. 0.5 = ______________
b. 2.15 = _______________
c. 0.681 = ______________
d. 3.25 = ______________
2. Read and write the number names of the following:
a. 25.12 b. 138.123
c. 781.001 d. 212.51
3. Write in decimals.
a. One Point Seven Three: __________
b. Fifteen Point Eight Five Nine: __________
c. Nine Point Four Five: __________
d. Twenty-Five Point Zero Seven Nine: __________
4. Write the expanded form of the following decimal numbers.
a. 23.7 b. 528.13
c. 428.85 d. 509.09
5. Change the following decimals into like decimals.
a. 0.5, 0.42 b. 1.42, 1.5
c. 5.9, 5.875 d. 6.72, 6.719
6. Convert the following fractions into decimals.
a. 23/10 b. 71/10
c. 103/100 d. 251/25
7. Arrange the following decimals in ascending order.
a. 0.42, 0.5, 0.382
b. 11.1, 11.21, 11.001
c. 20.3, 30.2, 23.25
d. 9.82, 9.9, 9.795
8. Convert the following decimals into fractions.
a. 7.03 b. 8.1
c. 100.01 d. 32.17
a. Which is lighter — 2.56 kg or 2.8 kg? ________
b. Which scooter has more fuel filled in it:
A — 12.82 litres or B — 12.84 litres? ________
c. Raju has 120.08, whereas Nilu has 120.80. Who has more money? ________
10. Convert the following into like decimals.
a. 0.8, 0.81, 0.719
b. 2.3, 2.03, 2.003
c. 71.23, 68.13, 5.731
## Decimal Worksheet for Class 5
Worksheet 2
1. Arrange the following in ascending order.
a. 7.29, 7.2, 7.21, 7.3
b. 121.3, 120.8, 120.79, 121.29
c. 37.61, 37.5, 42.9, 45.6
d. 1.04, 1.14, 0.04, 1.004
2. Expand the following decimal numbers.
a. 0.693 = ____________________
b. 1.584 = ____________________
c. 195.43 = ____________________
d. 532.789 = ____________________
3. Arrange the following in descending order.
a. 13.45, 12.71, 17.21, 14.35
b. 81.32, 81.23, 81.03, 81.3
c. 48.5, 38.5, 17.5, 27.5
d. 12.121, 12.1, 12.21, 12.22
4. Convert the expanded forms to decimal numbers.
a. 30 + 4 + 0.1 + 0.08 = ____________
b. 500 + 40 + 3 + 0.6 + 0.007 = __________
c. 800 + 20 + 7 + 0.5 + 0.001 = __________
d. 400 + 30 + 1 + 0.02 + 0.007 = __________
5. Ritesh, Amit and Arjun took part in a long jump event on the sports day. Ritesh covered 3.8 m, Amit covered 3.81 m and Arjun covered 3.18 m. Arrange them for the first, second and third prizes based on the distances covered by them.
6. Fill in the blanks.
a. If 1 is divided into 10 equal parts, each part equals __________ of one.
b. The name of the place 10 times less than the ones place is __________.
c. The name of the place 100 times less than the ones place is __________.
d. The __________ point separates the whole numbers from the decimal fractions.
7. The height of Ravi, Ansh, Neetu and Shikha are 90.28 cm, 91.82 cm, 90.82 cm and 92.18 cm, respectively. Arrange them in increasing order of their heights.
8. Find the place value of the underlined digit.
a. 278.352 Place value = _________
b. 5.814 Place value = __________
c. 7.218 Place value = __________
d. 8.965 Place value = __________
a. 48.712 + 32.415 + 12.213
b. 72.18 + 45.54 + 54
c. 121.84 + 72.1 + 55.13
d. 39.16 + 42.47 + 81.12
10. Convert into decimals.
a. ¾ = _____________
b. 2/5 = ______________
c. 8/10 = _____________
d. 4/5 = _______________
Worksheet 1
1. a. Zero point five
b. Two point one five
c. Zero point six eight one
d. Three point two five
2. a. Twenty-five point one two
b. One hundred thirty-eight point one two three
c. Seven hundred eighty-one point zero zero one
d. Two hundred twelve point five one
3. a. 1.73 b. 15.859
c. 9.45 d. 25.079
4. a. 20 + 3 + 0.7
b. 500 + 20 + 8 + 0.1 + 0.03
c. 400 + 20 + 8 + 0.8 + 0.05
d. 500 + 9 + 0.09
5. a. 0.50, 0.42
b. 1.42, 1.50
c. 5.900, 5.875
d. 6.720, 6.719
6. a. 2.3 b. 7.1
c. 10.03 d. 25.04
7. a. 0.382, 0.42, 0.5
b. 11.001, 11.1, 11.21
c. 20.3, 23.25, 30.2
d. 9.795, 9.82, 9.9
8. a. 73/100 b. 81/10
c. 1001/100 d. 3217/100
9. a. 2.56 kg b. B c. Nilu
10. a. 0.800, 0.810, 0.719
b. 2.300, 2.030, 2.003
c. 71.230, 68.130, 5.731
Worksheet 2
1. a. 7.2, 7.21, 7.29, 7.3
b. 120.79, 120.8, 121.29, 121.3
c. 37.5, 37.61, 42.9, 45.6
d. 0.04, 1.004, 1.04, 1.14
2. a. 0.6 + 0.09 + 0.003
b. 1 + 0.5 + 0.08 + 0.004
c. 100 + 90 + 5 + 0.4 + 0.03
d. 500 + 30 + 2 + 0.7 + 0.08 + 0.009
3. a. 17.21, 14.35, 13.45, 12.71
b. 81.32, 81.3, 81.23, 81.03
c. 48.5, 38.5, 27.5, 17.5
d. 12.22, 12.21, 12.121, 12.1
4. a. 34.18 b. 543.607
c. 827.501 d. 431.027
5. Amit, Ritesh and Arjun
6. a. one-tenth
b. one-tenth
c. one-hundredth
d. decimal
7. Ravi, Neetu, Ansh and Shikha
8. a. 0.002 b. 0.8
c. 7 d. 0.005
9. a. 93.34 b. 171.72
c. 249.07 d. 162.75
10. a. 0.75 b. 0.4
c. 0.8 d. 0.8
Maths Worksheets for Other Chapters of Class 5
Maths Worksheets for Other Classes
Maths Worksheets for Class 1
Maths Worksheets for Class 2
Maths Worksheets for Class 3
Maths Worksheets for Class 4
Maths Worksheets for Class 6
Maths Worksheets for Class 7
Maths Worksheets for Class 8
Maths Worksheets for Class 9
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# Is 6000 a prime numbers?
## Is 6000 a prime numbers?
Prime Numbers Before Calculator The number 6000 is not a prime number because it is possible to express it as a product of prime factors. In other words, 6000 can be divided by 1, by itself and at least by 2, 3 and 5. So, 6000 is a ‘composite number’.
How many prime numbers are there between 1 and 199?
Prime Numbers between 1 and 1,000
2 13
71 73 101
113 127 151
173 179 199
229 233 263
### How many prime numbers are there between 1 and 10000?
There are 1229 prime numbers between 1 and 10,000.
How do you find prime numbers from 1 to 1000?
The first few prime numbers are as follows: 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97, 101, 103, 107, 109, 113, 127, 131, 137, 139, 149, 151, 157, 163, 167, 173, 179, 181, 191, 193, 197, and 199, etc.
#### What is the prime factorization of 1296?
There are total 25 factors of 1296, of which 2, 3 are its prime factors. The Prime Factorization of 1296 is 24 × 34.
What is the prime factorization of 2250?
The prime factorization of 2,250 is 2 × 32 × 53.
## What is the prime numbers between 1 to 100?
List of prime numbers to 100. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
Is there a formula for finding prime numbers?
Methods to Find Prime Numbers Method 1: Two consecutive numbers which are natural numbers and prime numbers are 2 and 3. Apart from 2 and 3, every prime number can be written in the form of 6n + 1 or 6n – 1, where n is a natural number. Note: These both are the general formula to find the prime numbers.
### What is the GCF of 16 and 36?
4
The GCF of 16 and 36 is 4. To calculate the greatest common factor (GCF) of 16 and 36, we need to factor each number (factors of 16 = 1, 2, 4, 8, 16; factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36) and choose the greatest factor that exactly divides both 16 and 36, i.e., 4.
How to make a list of prime numbers from 1 to 60?
List of Prime Numbers from 1 to 60 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59 About List of Prime Numbers This prime numbers generator is used to generate the list of prime numbers from 1 to a number you specify.
#### Are there any prime numbers from 1 to 1000?
List of Prime Numbers from 1 to 1000. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47,
How many prime numbers are there in the number system?
As we know, the prime numbers are the numbers which have only two factors which are 1 and the number itself. There are a number of primes in the number system. Let us provide here the list of prime numbers that are present between 1 and 100, along with their factors and prime factorisation. Also, get the list of prime numbers from 1 to 1000 here.
## Are there any prime numbers that have only two factors?
List of Prime Numbers 1 to 100. As we know, the prime numbers are the numbers which have only two factors which are 1 and the number itself. There are a number of primes in the number system.
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# Remembering trigonometric functions
A number of students have a tough time while remembering trigonometric functions. To make the trigonometric functions unforgettable for you, we have come up with a mnemonic “SOHCAHTOA”. It will make sure remember to figure out the sine, cosine, tangent of an angle from the sides of a right triangle and vice versa. If you remember this mnemonic, then the questions will be completely on your tips where you are expected to use the functions and determine the length of a missing side.
# Cross multiply to find the greater fraction
If you are required to find the greater of two fractions and you are not sure about the right fraction, then Cross multiply to find the greater fraction technique discussed here will simplify the confusion.
1. Simply multiply the numerator of each fraction with the denominator of the other fraction.
2. Write the answers to the corresponding numerator.
3. The fraction which will show a greater value will be bigger.
For Example
If you are given two fractions
5/7 and 2/3
5 multiply by 3 will give 15 and 2 multiply by 7 will give 14
15 5/7 2/3 14
So,
5/7 is the greater equation.
# Multiplying larger numbers quickly
One of the common problem in doing mental multiplication is many students forget that multiplying larger numbers quickly requires breaking those larger numbers into more practical pieces. For example, if we see
17 x 6
you might think that 17 is a huge number and difficult to multiply.
To simplify the problem, you can break 17 into two manageable numbers that you can both multiply times 6. To get the right answer, add the result of two numbers. You might take 9 and 8, (9×6)+(9×8). But there is an easier way i.e. to choose a number which ends in a zero as these numbers are easiest to multiply.
Let’s take the example of 17. It can be broken into
7 + 10= 17
Now multiplying the 6 by 10 is quite instantaneous.
10 x 6= 60
That leaves the 7 thus giving us
7 x 6 = 42
42 + 60 = 102
An important point to consider is that not to think of this as a writing exercise. Simply make these equations in your mind and breaking these numbers in this way will make things quite easier for you. While doing this, you multiply 10×6 and then keep 60 on hold, then you multiply 6×7 and finally add both 60 and 42.
Let’s solve another equation with slightly larger number
Suppose we have
32 x 8
One way you can follow is to multiply 10×8 three times. But a faster way is to multiply
30 x 8
We have selected 30 as it ends in zero and closest to 32. This gives us 30×8 which is same as 3×8 with a zero at the end.
3 x 8= 24
Now add zero and it will give
240
Now we will put this number on hold and multiply the 2 left over as our original number was 32 time the 8. So,
2 x 8=16
The final step is to add this number to the on-hold number i.e. 240
240+16=256
Have a great test day with making the complex multiplications way too simpler for you.
# Solving Algebraic Simultaneous Equations
For solving algebraic simultaneous equations, the first step you should follow is to add or subtract the given equation. In case you are not able to find the required result, multiple one of the equation by a number. Make sure to select a number that eliminates one of the variables when you add or subtract the equation.
# Solving Algebraic Equations
To keep the students, stay on the right track with their exam preparation, here we have listed some of the key tips for solving algebraic equations more quickly without facing any error.
1. While solving algebraic equations, treat both equations equally. What you do on one side, you must do on the other side of equations to keep it balanced. For example, if you apply square on one side, you must apply square on the other side too. Remember the algebraic equation with a Seesaw. As balancing a seesaw requires equal weight on both sides.
2. While solving an expression, keep in mind to quickly manipulate the equation to get the desired expression.
3. If two expressions are required to set equal to each other, cross multiply them or multiply by using a common denominator.
4. While multiplying or dividing an expression by a negative number, remember to switch the direction of the inequality sign or else your minor mistake will ultimately give wrong results.
# Greater than vs Less than
Greater than vs Less than makes us confused several times while operating the equations because of common looking signs.
Here we have listed two ways that will make you remember them throughout your life.
1. It’s a very common example that we have learned in our childhood that an alligator always eats the bigger number. It will completely make sense if you turn the inequality sign into an alligator.
2. Another trick is that you can simply grow up by learning “less than three” i.e. <3. In this way, you will never forget the less than sign which will definitely make you remember the greater than sign.
# Multiplication of a two-digit number with 11
Multiplication of a two-digit number with 11 is a challenging and time consuming task. Therefore, keeping in mind the following tricks and practice a few times will effectively help you in saving your precious time and solve equation within seconds.
It’s not a difficult task to multiply two-digit numbers with 11 once you know the trick. Let’s solve the following problem
For example Lets multiply 13 by 11
13 x 11
To solve this, simply add the two digits
1+3=4
Put the 4 between 1 and 3
Now consider another example where we multiply 34 by 11
34 x 11
3 + 4= 7
Now putting 7 between 3 and 4 will give us
374, and that is a correct answer
Let’s move towards a bit complex problem and make it simple than ever before. Here we will multiply 78 with 11
78 x 11
Adding both 7 and 8 will give
7 + 8 = 15
Now put 15 between 7 and will give us
7158, now add 7 and 1 which will result 8
Now the final outcome will be
858, which is the correct answer
Practice of this trick will surely increase the speed of solving complex equations and expressions enabling students to focus more on the complex parts of calculations.
# How to calculate the square of number ending with 5
Students in their exam can save enormous amount of time solving complex problems if they know how to calculate the square of number ending with 5. This quick technique is very easy to learn and handy to apply Lets say you have a Number n5 (Where n can be any digit) then the square of the number can be calculated in these simple four steps
1. Extract the trailing 5 and consider the remaining digit as n. E.g You have the number 45 then n=4.
2. To calculate the square put n(n+1) E.g n=4(4+1) =20
3. Put 25 for trailing 5
4. Concatenating them will give the correct square E.g square of 45=2025
For example calculating square for 35
• Step 1: Here n=3
• Step 2: n(n+1) = 3(3+1) =3*4 =12
• Step 3 : 25 for trailing 5
• Step 4: Concatenating 12 and 25 will give us 1225 which is the correct answer.
To understand the trick more lets take another example of calculate square for 155
• Step 1: Here n=15
• Step 2: n(n+1) = 15(15+1) =15*16 =240
• Step 3 : 25 for trailing 5
• Step 4: Concatenating 240 and 25 will give us 24025 which is the correct answer.
However we can use the technique described here where we calculate the square of any number in mind but the trick given above is easier for the case where the number’s ending digit is 5.
For example square of 35 can be calculated as x = 35
• Step 1 : y=30
• Step 2 : z= 5
• Step 3 : (35+5) *(35-5) + 52 = 40*30+25 = 40 *10 *3 +25
• breaking up 20=2*10 for easier and quick calculation
400*3+25 = 1200+25 = 1225 is the correct answer
# How to calculate the square of any number in your mind
How to calculate the square of any number in your mind is the most interesting question by the students who are planning to attempt any prep test, the technique discuss in this post help those students while saving their time in solving complex problems.
This technique contains following steps which can be followed in mind to calculate the square of any number quickly.
Lets consider x is the number which square is to be calculated.
1. Find the absolute value of number to the nearest multiple of 10. Lets consider y is that absolute value of the number x from the nearest multiple of 10.
2. Calculate the difference of the absolute value of the number with the number itself. E.g z is the difference between the number from its absolute value than z=|y-x|
3. Now the solution expression will be (x+z) * (x-z) + z2
Lets take an example,
We need to find the square of 22
so x=22
Step 1 : y=20
Step 2 : z= 2
Step 3 : (22+2) *(22-2) + 22
= 24*20+4
= 24 *10 * 2 +4 breaking up 20=2*10 for easier and quick calculation
= 240*2+4
= 480+4
= 484 is the correct answer
Lets take another example,
We need to find the square of 96
so x=96
Step 1 : y=100
Step 2 : z= 4
Step 3 : (96-4) *(96+4) + 42
= 92*100+4
= 9200+16
= 9216 is the correct answer
However there is a quicker technique for calculating square of number ending with 5 which is discussed in this post
Refer to the table below for some more examples, and examples given above
x y z (x+z) (x-z) z2 (x+z)*(x-z) +z2 1 22 20 2 24 20 4 484 2 36 40 4 40 32 16 1296 3 43 40 3 46 40 9 1849 4 57 60 3 60 54 9 3249 5 96 100 4 100 92 16 9216
# Dealing with alternative information
Deal with alternative with extra care containing more than one bit of data – ensure both or all bits are right.
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# How do you find vertical, horizontal and oblique asymptotes for (x^3-8)/(x^2-5x+6)?
Jun 13, 2018
the oblique asymptote is $y = x + 5$ and the horizontal asymptote is $x = 3$
#### Explanation:
$\frac{{x}^{3} - 8}{{x}^{2} - 5 x + 6} = \frac{\cancel{x - 2} \left({x}^{2} + 2 x + 4\right)}{\left(x - 3\right) \cancel{x - 2}}$
=$\frac{{x}^{2} + 2 x + 4}{x - 3} = \frac{\left(x - 3\right) \left(x + 5\right) + 19}{x - 3} = x + 5 + \frac{19}{x - 3}$
Therefore, the oblique asymptote is $y = x + 5$ and the horizontal asymptote is $x = 3$
The horizontal asymptote is found by letting the denominator equal to 0.
The oblique asymptote is found by long division
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# wikiHow to Do Conic Sections
Conic sections are an interesting branch of mathematics involving the cutting of a double-napped cone. By cutting the cone in different ways, you can create a shape as simple as a point or as complex as a hyperbola.
### Method 1 General
1. 1
Understand what is special about a conic section. Unlike regular coordinate equations, conic sections are general equations and don't necessarily have to be functions. For instance, x=5, while an equation, is not a function.
2. 2
Know the difference between a degenerate case and a conic section. The degenerate cases are those where the cutting plane passes through the intersection, or apex of the double-napped cone. Some examples of degenerates are lines, intersecting lines, and points. The four conic sections are circles, parabolas, ellipses and hyperbolas.[1]
3. 3
Realize the idea that conic sections rely on. A conic section on a coordinate plane is just a collection of points which follow a certain rule which relates them all to the direction and focal points of the conic.
### Method 2 Finding the Equation of a Circle
1. 1
Know what part of the cone you are looking at. A circle is defined when the cone is cut by a plane which is parallel to the top and bottom faces of the cone.
2. 2
Find the coordinates of the center of the circle. For formula sake, we will call the center (h,k) as is custom when writing the general equation of a conic section.
3. 3
Find the radius of the circle. The circle is defined as a collection of points which are the same distance away from a set center point (h,k). That distance is the radius.
4. 4
Plug them into the equation of a circle. The equation of a circle is one of the easiest to remember of all the conic sections. Given a center of (h,k) and a radius of length r, a circle is defined by (x-h)2+(y-k)2. Be sure to realize that this isn't a function. If you are trying to graph a circle on your graphing calculator, you will have to do some algebra to separate it into two equations which can be graphed using a calculator or use the "draw" feature.
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## What is the formula for a linear equation?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
## What is linear equation Give 5 example?
Solve (i) 3x − 1=5 (ii) 4x +2=9 (iii) 6 − 2x = 1 (iv) 5x +1= −9 (v) 3x +1= x − 5 (vi) 6x − 1=3 − 2x (vii) 2 − 3(x − 2) = x + 4 (viii) 5(x − 1) + 1 = 2(x − 2). In general, a linear equation in one variable has just one solution, i.e., a unique solution.
## How do you solve linear equations examples?
Solving Linear EquationsSolve: (2x + 5)/(x + 4) = 1. ⇒ 2x – x = 4 – 5 (Transferring positive x to the left hand side changes to negative x and again, positive 5 changes to negative 5) Solution: ⇒ 3x/3 = 9/3 (Dividing both sides by 3) ⇒ 5 – 2x + 2 = 12 – 4x – 2x (Removing the brackets and then simplify) x/2 + x/3 = x – 7.
## What is linear equation used for?
Linear equations are an important tool in science and many everyday applications. They allow scientist to describe relationships between two variables in the physical world, make predictions, calculate rates, and make conversions, among other things. Graphing linear equations helps make trends visible.
## What are the 3 types of equations?
There are three major forms of linear equations: point-slope form, standard form, and slope-intercept form.
## How do you plot a linear equation?
To graph a linear equation, we can use the slope and y-intercept.Locate the y-intercept on the graph and plot the point.From this point, use the slope to find a second point and plot it.Draw the line that connects the two points.
You might be interested: Equation for respiration
## What is linear equation explain with example?
The definition of a linear equation is an algebraic equation in which each term has an exponent of one and the graphing of the equation results in a straight line. An example of linear equation is y=mx + b.
## What is linear function example?
Linear functions are those whose graph is a straight line. A linear function has one independent variable and one dependent variable. The independent variable is x and the dependent variable is y. a is the constant term or the y intercept.Linear Functions.
outputtotal cost
30 unitsC = 7,000 + 30(600) = 25,000
## What is an equation in math?
An equation says that two things are equal. It will have an equals sign “=” like this: 7 + 2 = 10 − 1. That equation says: what is on the left (7 + 2) is equal to what is on the right (10 − 1) So an equation is like a statement “this equals that”
## What is a linear equation for dummies?
Linear Equations Represent Lines To make a line you need two points. Then you can draw a line through those two points. The x and y variables in the linear equation represent the x and y coordinates on a graph.
## What are the four steps for solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal. If we subtract the same number from both sides of an equation, both sides will remain equal.
## What are the 3 types of system of linear equation?
There are three types of systems of linear equations in two variables, and three types of solutions.An independent system has exactly one solution pair. The point where the two lines intersect is the only solution.An inconsistent system has no solution. A dependent system has infinitely many solutions.
### Releated
#### Navier-stokes equation
Is the Navier Stokes equation solved? In particular, solutions of the Navier–Stokes equations often include turbulence, which remains one of the greatest unsolved problems in physics, despite its immense importance in science and engineering. Even more basic properties of the solutions to Navier–Stokes have never been proven. Who Solved Navier Stokes? Russian mathematician Grigori Perelman […]
#### Equation for moment of inertia
How is moment of inertia calculated? Basically, for any rotating object, the moment of inertia can be calculated by taking the distance of each particle from the axis of rotation (r in the equation), squaring that value (that’s the r2 term), and multiplying it times the mass of that particle. What is moment of inertia […]
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Wierzycaz
2021-05-01
The average height of a 2 year old boy is 38 inches; an 8 year old averages 56 inches. Use this information to write a linear equation that models the height (in inches), y, in terms of the age (in years), x. Use the linear equation to predict the average height of a 5 year-old boy.
i1ziZ
Represent the given data as points: $\left(x1,y1\right)=$$\left(2,38\right)\to 2$ year old boy is 38 inches year old boy is 38 inches
$\left(x2,y2\right)=$$\left(8,56\right)\to 8$ year old boy is 56 inches year old boy is 56 inches
Find the slope $m=\frac{y2-y1}{x2-x1}=\frac{56-38}{8-2}=\frac{18}{6}=3$
Use the slope-intercept form of a line: $y=mx+b$
Substitute any point, say (2,38) and $m=3$ to find bb:
$38=3\left(2\right)+b$
$38=6+b$
$32=b$
So, the linear equation is: $y=3x+32$
To predict the average height of a 5 year-old boy, substitute $x=5$:
$y=3\left(5\right)+32$
$y=15+32$
$y=47\to 47$ inches
Jeffrey Jordon
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Rd Sharma Xi 2020 2021 _volume 1 Solutions for Class 11 Science Maths Chapter 3 Functions are provided here with simple step-by-step explanations. These solutions for Functions are extremely popular among Class 11 Science students for Maths Functions Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rd Sharma Xi 2020 2021 _volume 1 Book of Class 11 Science Maths Chapter 3 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rd Sharma Xi 2020 2021 _volume 1 Solutions. All Rd Sharma Xi 2020 2021 _volume 1 Solutions for class Class 11 Science Maths are prepared by experts and are 100% accurate.
#### Question 1:
If f(x) = x2 − 3x + 4, then find the values of x satisfying the equation f(x) = f(2x + 1).
#### Answer:
Given:
f (x) = x2 – 3x + 4
Therefore,
f (2x + 1) = (2x + 1)2 – 3(2x + 1) + 4
= 4x2 + 1 + 4x – 6x – 3 + 4
= 4x2 – 2x + 2
Now,
f (x) = f (2x + 1)
x2 – 3x + 4 = 4x2 – 2x + 2
⇒ 4x2x2 – 2x + 3x + 2 – 4 = 0
⇒ 3x2 + x – 2 = 0
⇒ 3x2 + 3x – 2x – 2 = 0
⇒ 3x(x + 1) – 2(x +1) = 0
⇒ (3x – 2)(x +1) = 0
⇒ (x + 1) = 0 or ( 3x – 2) = 0
Hence, $x=-1,\frac{2}{3}$.
#### Question 2:
If f(x) = (xa)2 (xb)2, find f(a + b).
#### Answer:
Given:
f (x) = (xa)2(xb)2
Thus,
f (a + b) = (a + ba)2(a + bb)2
= b2a2
Hence, f (a + b) = a2b2 .
#### Question 3:
If $y=f\left(x\right)=\frac{ax-b}{bx-a}$, show that x = f(y).
#### Answer:
Given:
$f\left(x\right)=\frac{ax-b}{bx-a}$
Let y = f (x) .
y( bx $-$ a) = ax b
xybay = axb
xybax = ayb
x(bya) = ayb
$⇒x=\frac{ay-b}{by-a}$
x = f (y)
Hence proved.
#### Question 4:
If $f\left(x\right)=\frac{1}{1-x}$, show that f[f[f(x)]] = x.
#### Answer:
Given:
$f\left(x\right)=\frac{1}{1-x}\phantom{\rule{0ex}{0ex}}$
Thus,
$f\left\{f\left(x\right)\right\}=f\left\{\frac{1}{1-x}\right\}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{1-\frac{1}{1-x}}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{\frac{1-x-1}{1-x}}\phantom{\rule{0ex}{0ex}}=\frac{1-x}{-x}\phantom{\rule{0ex}{0ex}}=\frac{x-1}{x}$
Again,
$f\left[f\left\{f\left(x\right)\right\}\right]=f\left[\frac{x-1}{x}\right]\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{1-\left(\frac{x-1}{x}\right)}\phantom{\rule{0ex}{0ex}}=\frac{1}{\frac{x-x+1}{x}}\phantom{\rule{0ex}{0ex}}=\frac{x}{1}\phantom{\rule{0ex}{0ex}}=x$
Therefore, f[f{f(x)}] = x.
Hence proved.
#### Question 5:
If $f\left(x\right)=\frac{x+1}{x-1}$, show that f[f[(x)]] = x.
#### Answer:
Given:
$f\left(x\right)=\frac{x+1}{x-1}$
Therefore,
$f\left[f\left\{\left(x\right)\right\}\right]=f\left(\frac{x+1}{x-1}\right)\phantom{\rule{0ex}{0ex}}$
$=\frac{\left(\frac{x+1}{x-1}\right)+1}{\left(\frac{x+1}{x-1}\right)-1}\phantom{\rule{0ex}{0ex}}$
$=\frac{\frac{x+1+x-1}{x-1}}{\frac{x+1-x+1}{x-1}}=\frac{\frac{2x}{x-1}}{\frac{2}{x-1}}=\frac{2x}{2}=x$
Thus,
f [ f {(x)}] = x
Hence proved.
#### Question 6:
If
find: (a) f(1/2), (b) f(−2), (c) f(1), (d) $f\left(\sqrt{3}\right)$ and (e) $f\left(\sqrt{-3}\right)$.
#### Answer:
Given:
Now,
(a) $f\left(\frac{1}{2}\right)=\frac{1}{2}$ [ Using f (x) = x, 0 ≤ x < 1]
(b) f ($-$2) = ( $-$ 2)2 = 4
(c) $f\left(1\right)=\frac{1}{1}=1$
(d) $f\left(\sqrt{3}\right)=\frac{1}{\sqrt{3}}$
(e) $f\left(\sqrt{-3}\right)$
Since x is not defined in R, $f\left(\sqrt{-3}\right)$ does not exist.
#### Question 7:
If $f\left(x\right)={x}^{3}-\frac{1}{{x}^{3}}$, show that $f\left(x\right)+f\left(\frac{1}{x}\right)=0.$
#### Answer:
Given:
$f\left(x\right)={x}^{3}-\frac{1}{{x}^{3}}\phantom{\rule{0ex}{0ex}}$ ...(i)
Thus,
$f\left(\frac{1}{x}\right)={\left(\frac{1}{x}\right)}^{3}-\frac{1}{{\left(\frac{1}{x}\right)}^{3}}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{x}^{3}}-\frac{1}{\frac{1}{{x}^{3}}}$
$\therefore f\left(\frac{1}{x}\right)=\frac{1}{{x}^{3}}-{x}^{3}$ ...(ii)
$f\left(x\right)+f\left(\frac{1}{x}\right)=\left({x}^{3}-\frac{1}{{x}^{3}}\right)+\left(\frac{1}{{x}^{3}}-{x}^{3}\right)\phantom{\rule{0ex}{0ex}}$
$={x}^{3}-\frac{1}{{x}^{3}}+\frac{1}{{x}^{3}}-{x}^{3}=0$
Hence, $f\left(x\right)+f\left(\frac{1}{x}\right)=0$ .
#### Question 8:
If $f\left(x\right)=\frac{2x}{1+{x}^{2}}$, show that f(tan θ) = sin 2θ.
#### Answer:
Given:
$f\left(x\right)=\frac{2x}{1+{x}^{2}}$
Thus,
$f\left(\mathrm{tan}\theta \right)=\frac{2\left(\mathrm{tan}\theta \right)}{1+{\mathrm{tan}}^{2}\theta }\phantom{\rule{0ex}{0ex}}$
Hence, f (tan θ) = sin 2θ.
#### Question 9:
If $f\left(x\right)=\frac{x-1}{x+1}$, then show that
(i) $f\left(\frac{1}{x}\right)=-f\left(x\right)$ (ii) $f\left(-\frac{1}{x}\right)=-\frac{1}{f\left(x\right)}$
#### Answer:
Given: $f\left(x\right)=\frac{x-1}{x+1}$ .....(1)
(i) Replacing x by $\frac{1}{x}$ in (1), we get
(ii) Replacing x by $-\frac{1}{x}$ in (1), we get
#### Question 10:
If f(x) = (axn)1/n, a > 0 and n ∈ N, then prove that f(f(x)) = x for all x.
#### Answer:
Given:
f(x) = (axn)1/n, a > 0
Now,
f{ f (x)} = f (axn)1/n
= [a – {(a xn)1/n}n]1/n
= [ a – (axn)]1/n
= [ a a + xn)]1/n = (xn)1/n = x(n × 1/n) = x
Thus, f(f(x)) = x.
Hence proved.
#### Question 11:
If for non-zero x, af(x) + bf $\left(\frac{1}{x}\right)=\frac{1}{x}-5$, where ab, then find f(x).
#### Answer:
Given:
$af\left(x\right)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5\phantom{\rule{0ex}{0ex}}$ ...(i)
$⇒af\left(\frac{1}{x}\right)+bf\left(x\right)=\frac{1}{\frac{1}{x}}-5\phantom{\rule{0ex}{0ex}}$
$⇒af\left(\frac{1}{x}\right)+bf\left(x\right)=x-5$ ...(ii)
On adding equations (i) and (ii), we get:
$af\left(x\right)+bf\left(x\right)+bf\left(\frac{1}{x}\right)+af\left(\frac{1}{x}\right)=\frac{1}{x}-5+x-5\phantom{\rule{0ex}{0ex}}$
$⇒\left(a+b\right)f\left(x\right)+\left(a+b\right)f\left(\frac{1}{x}\right)=\frac{1}{x}+x-10\phantom{\rule{0ex}{0ex}}$
$⇒f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{\left(a+b\right)}\left[\frac{1}{x}+x-10\right]$ ...(iii)
On subtracting (ii) from (i), we get:
$af\left(x\right)-bf\left(x\right)+bf\left(\frac{1}{x}\right)-af\left(\frac{1}{x}\right)=\frac{1}{x}-5-x+5\phantom{\rule{0ex}{0ex}}$
$⇒\left(a-b\right)f\left(x\right)-f\left(\frac{1}{x}\right)\left(a-b\right)=\frac{1}{x}-x\phantom{\rule{0ex}{0ex}}$
$⇒f\left(x\right)-f\left(\frac{1}{x}\right)=\frac{1}{\left(a-b\right)}\left[\frac{1}{x}-x\right]$ ...(iv)
On adding equations (iii) and (iv), we get:
$2f\left(x\right)=\frac{1}{a+b}\left[\frac{1}{x}+x-10\right]+\frac{1}{a-b}\left[\frac{1}{x}-x\right]\phantom{\rule{0ex}{0ex}}$
$⇒2f\left(x\right)=\frac{\left(a-b\right)\left[\frac{1}{x}+x-10\right]+\left(a+b\right)\left[\frac{1}{x}-x\right]}{\left(a+b\right)\left(a-b\right)}\phantom{\rule{0ex}{0ex}}$
$⇒2f\left(x\right)=\frac{\frac{a}{x}+ax-10a-\frac{b}{x}-bx+10b+\frac{a}{x}-ax+\frac{b}{x}-bx}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}$
$⇒2f\left(x\right)=\frac{\frac{2a}{x}-10a+10b-2bx}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}$
$⇒f\left(x\right)=\frac{1}{{a}^{2}-{b}^{2}}×\frac{1}{2}\left[\frac{2a}{x}-10a+10b-2bx\right]\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-5a+5b-bx\right]$
Therefore,
$f\left(x\right)=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx-5a+5b\right]\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx\right]-\frac{5\left(a-b\right)}{{a}^{2}-{b}^{2}}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx\right]-\frac{5\left(a-b\right)}{\left(a-b\right)\left(a+b\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx\right]-\frac{5}{\left(a+b\right)}$
Hence,
$f\left(x\right)=\frac{1}{{a}^{2}-{b}^{2}}\left[\frac{a}{x}-bx\right]-\frac{5}{\left(a+b\right)}$
#### Question 1:
Find the domain of each of the following real valued functions of real variable:
(i) $f\left(x\right)=\frac{1}{x}$
(ii) $f\left(x\right)=\frac{1}{x-7}$
(iii) $f\left(x\right)=\frac{3x-2}{x+1}$
(iv) $f\left(x\right)=\frac{2x+1}{{x}^{2}-9}$
(v) $f\left(x\right)=\frac{{x}^{2}+2x+1}{{x}^{2}-8x+12}$
#### Answer:
(i) Given: $f\left(x\right)=\frac{1}{x}$
Domain of f :
We observe that f (x) is defined for all x except at x = 0.
At x = 0, f (x) takes the intermediate form $\frac{1}{0}.$
Hence, domain ( f ) = R $-${ 0 }
(ii) Given: $f\left(x\right)=\frac{1}{\left(x-7\right)}$
Domain of f :
Clearly, f (x) is not defined for all (x $-$ 7) = 0 i.e. x = 7.
At x = 7, f (x) takes the intermediate form $\frac{1}{0}.$
Hence, domain ( f ) = R $-$ { 7 }.
(iii) Given: $f\left(x\right)=\frac{3x-2}{\left(x+1\right)}$
Domain of f :
Clearly, f (x) is not defined for all (x + 1) = 0, i.e. x$-$ 1.
At x = $-$1, f (x) takes the intermediate form $\frac{1}{0}.$
Hence, domain ( f ) = R $-$$-$1 }.
(iv) Given: $f\left(x\right)=\frac{2x+1}{{x}^{2}-9}$
Domain of f :
Clearly, f (x) is defined for all xR except for x2 $-$ 9 ≠ 0, i.e. x = ± 3.
At x = $-$3, 3, f (x) takes the intermediate form $\frac{1}{0}.$
Hence, domain ( f ) = R $-$$-$ 3, 3 }.
(v) Given:
$=\frac{{x}^{2}+2x+1}{{x}^{2}-6x-2x+12}\phantom{\rule{0ex}{0ex}}$
$=\frac{{x}^{2}+2x+1}{x\left(x-6\right)-2\left(x-6\right)}\phantom{\rule{0ex}{0ex}}$
$=\frac{{x}^{2}+2x+1}{\left(x-6\right)\left(x-2\right)}$
Domain of f : Clearly, f (x) is a rational function of x as $\frac{{x}^{2}+2x+1}{{x}^{2}-8x+12}$ is a rational expression.
Clearly, f (x) assumes real values for all x except for all those values of x for which x2 $-$ 8x + 12 = 0, i.e. x = 2, 6.
Hence, domain ( f ) = R $-$ {2,6}.
#### Question 2:
Find the domain of each of the following real valued functions of real variable:
(i) $f\left(x\right)=\sqrt{x-2}$
(ii) $f\left(x\right)=\frac{1}{\sqrt{{x}^{2}-1}}$
(iii) $f\left(x\right)=\sqrt{9-{x}^{2}}$
(iv) $f\left(x\right)=\frac{\sqrt{x-2}}{3-x}$
#### Answer:
(i) Given: $f\left(x\right)=\sqrt{x-2}$
Clearly, f (x) assumes real values if x $-$ 2 ≥ 0.
x ≥ 2
x ∈ [2, ∞)
Hence, domain (f) = [2, ∞) .
(ii) Given: $f\left(x\right)=\frac{1}{\sqrt{{x}^{2}-1}}$
Clearly, f (x) is defined for x2 $-$ 1 > 0 .
(x + 1)(x $-$ 1) > 0 [ Since a2 $-$ b2 = ( a + b)(a - b)]
x$-$1 and x > 1
x ∈ ($-$∞ , $-$ 1) ∪ (1, ∞)
Hence, domain (f) = ($-$ ∞ , $-$ 1) ∪ (1, ∞)
(iii) Given: $f\left(x\right)=\sqrt{9-{x}^{2}}$
We observe that f (x) is defined for all satisfying
$-$ x2 ≥ 0 .
x2 $-$ 9 ≤ 0
⇒ (x + 3)(x $-$ 3) ≤ 0
$-$3 ≤ x ≤ 3
x ∈ [ $-$ 3, 3]
Hence, domain ( f ) = [ $-$3, 3]
(iv) Given: $f\left(x\right)=\sqrt{\frac{x-2}{3-x}}$
Clearly, f (x) assumes real values if
x $-$ 2 ≥ 0 and 3 $-$ x > 0
x ≥ 2 and 3 > x
x ∈ [2, 3)
Hence, domain ( f ) = [2, 3) .
#### Question 3:
Find the domain and range of each of the following real valued functions:
(i) $f\left(x\right)=\frac{ax+b}{bx-a}$
(ii) $f\left(x\right)=\frac{ax-b}{cx-d}$
(iii) $f\left(x\right)=\sqrt{x-1}$
(iv) $f\left(x\right)=\sqrt{x-3}$
(v) $f\left(x\right)=\frac{x-2}{2-x}$
(vi) $f\left(x\right)=\left|x-1\right|$
(vii) $f\left(x\right)=-\left|x\right|$
(viii) $f\left(x\right)=\sqrt{9-{x}^{2}}$
(ix) $f\left(x\right)=\frac{1}{\sqrt{16-{x}^{2}}}$
(x) $f\left(x\right)=\sqrt{{x}^{2}-16}$
#### Answer:
(i)
Given:
$f\left(x\right)=\frac{ax+b}{bx-a}$
Domain of f : Clearly, f (x) is a rational function of x as $\frac{ax+b}{bx-a}$is a rational expression.
Clearly, f (x) assumes real values for all x except for all those values of x for which ( bx $-$ a) = 0, i.e. bx = a.
$⇒x=\frac{a}{b}$
Hence, domain ( f ) = $R-\left\{\frac{a}{b}\right\}$
Range of f :
Let f (x) = y
$⇒\frac{ax+b}{bx-a}=y\phantom{\rule{0ex}{0ex}}$
⇒ (ax + b) = y (bx $-$ a)
⇒ (ax + b) = (bxy $-$ ay)
b + ay = bxy $-$ ax
b + ay = x(by $-$ a)
$⇒x=\frac{b+ay}{by-a}$
Clearly, f (x) assumes real values for all x except for all those values of x for which ( by $-$ a) = 0, i.e. by = a.
$⇒y=\frac{a}{b}$.
Hence, range ( f ) = $R-\left\{\frac{a}{b}\right\}$
(ii)
Given:
$f\left(x\right)=\frac{ax-b}{cx-d}$
Domain of f : Clearly, f (x) is a rational function of x as $\frac{ax-b}{cx-d}$is a rational expression.
Clearly, f (x) assumes real values for all x except for all those values of x for which ( cx $-$ d) = 0, i.e. cx = d.
$⇒x=\frac{d}{c}$.
Hence, domain ( f ) = $R-\left\{\frac{d}{c}\right\}$
Range of f :
Let f (x) = y
$⇒\frac{ax-b}{cx-d}=y\phantom{\rule{0ex}{0ex}}$
⇒ (ax $-$ b) = y( cx $-$ d)
⇒ (ax $-$ b) = (cxy $-$ dy)
dy $-$ b = cxy $-$ ax
dy $-$ b = x(cy $-$ a)
$⇒x=\frac{dy-b}{cy-a}$
Clearly, f (x) assumes real values for all x except for all those values of x for which ( cy $-$ a) = 0, i.e. cy = a.
$⇒y=\frac{a}{c}$.
Hence, range ( f ) = $R-\left\{\frac{a}{c}\right\}$ .
(iii)
Given: $f\left(x\right)=\sqrt{x-1}$
Domain ( f ) : Clearly, f (x) assumes real values if x $-$ 1 ≥ 0 ⇒ x ≥ 1 ⇒ x ∈ [1, ∞) .
Hence, domain (f) = [1, ∞)
Range of f : For x ≥ 1, we have:
x $-$ 1 ≥ 0
$⇒\sqrt{x-1}\ge 0$
f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .
(iv)
Given: $f\left(x\right)=\sqrt{x-3}$
Domain ( f ) : Clearly, f (x) assumes real values if x $-$ 3 ≥ 0 ⇒ x ≥ 3 ⇒ x ∈ [3, ∞) .
Hence, domain ( f ) = [3, ∞)
Range of f : For x ≥ 3, we have:
x $-$ 3 ≥ 0
$⇒\sqrt{x-3}\ge 0$
f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞) .
(v)
Given:
$f\left(x\right)=\frac{x-2}{2-x}$
Domain ( f ) :
Clearly, f (x) is defined for all x satisfying: if 2 $-$ x ≠ 0 ⇒ ≠ 2.
Hence, domain ( f ) = R $-$ {2}.
Range of f :
Let f (x) = y
$\frac{x-2}{2-x}=y$
x $-$ 2 = y (2 $-$ x)
x $-$ 2 = $-$ y (x $-$ 2)
y$-$ 1
Hence, range ( f ) = {$-$ 1}.
(vi)
The given real function is f (x) = |x – 1|.
It is clear that |x – 1| is defined for all real numbers.
Hence, domain of f = R.
Also, for xR, (x – 1) assumes all real numbers.
Thus, the range of f is the set of all non-negative real numbers.
Hence, range of f = [0, ∞) .
(vii)
f (x) = – | x |, xR
We know that
$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x& x<0\end{array}\right\\phantom{\rule{0ex}{0ex}}$
$\therefore f\left(x\right)=-\left|x\right|=\left\{\begin{array}{ll}-x,& x\ge 0\\ x,& x<0\end{array}\right\$
Since f(x) is defined for xR, domain of f = R.
It can be observed that the range of f (x) = – | x | is all real numbers except positive real numbers.
∴ The range of f is (– ∞, 0).
(viii) Given:
$f\left(x\right)=\sqrt{9-{x}^{2}}$
$\sqrt{9-{x}^{2}}$ is defined for all real numbers that are greater than or equal to – 3 and less than or equal to 3.
Thus, domain of f (x) is {x : – 3 ≤ x ≤ 3} or [– 3, 3].
For any value of x such that – 3 ≤ x ≤ 3, the value of f (x) will lie between 0 and 3.
Hence, the range of f (x) is {x: 0 ≤ x ≤ 3} or [0, 3].
(ix) Given:
$f\left(x\right)=\frac{1}{\sqrt{16-{x}^{2}}}$
$\frac{1}{\sqrt{16-{x}^{2}}}$ is defined for all real numbers that are greater than – 4 and less than 4.
Thus, domain of f (x) is {x : – 4 < x < 4} or (– 4, 4).
Range of f :
Let f (x) = y
Hence, range ( f ) = [).
(x) Given:
$f\left(x\right)=\sqrt{{x}^{2}-16}$
$\sqrt{{x}^{2}-16}$ is defined for all real numbers that are greater than or equal to 4 and less than or equal to –4.
Thus, domain of f (x) is {x : x ≤ – 4 or x ≥ 4} or (–∞, –4] ∪ [4, ∞).
Range of f :
For
x ≥ 4, we have:
x2 $-$ 16 ≥ 0
$⇒\sqrt{{x}^{2}-16}\ge 0$
f (x) ≥ 0
For x ≤ – 4, we have:
x2 $-$ 16 ≥ 0
$⇒\sqrt{{x}^{2}-16}\ge 0$
f (x) ≥ 0
Thus, f (x) takes all real values greater than zero.
Hence, range (f) = [0, ∞).
#### Question 1:
Find f + g, fg, cf (c ∈ R, c ≠ 0), fg, in each of the following:
(a) If f(x) = x3 + 1 and g(x) = x + 1
(b) If $f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\sqrt{x+1}$
#### Answer:
(a) Given:
f (x) = x3 + 1 and g (x) = x + 1
Thus,
(f + g) (x) : RR is given by (f + g) (x) = f (x) + g (x) = x3 + 1 + x + 1 = x3 + x + 2.
($-$ g) (x) : RR is given by (f $-$ g) (x) = f (x$-$ g (x) = (x3 + 1) $-$ (x + 1 ) = x3 + 1 $-$ x $-$ 1 = x3 $-$ x.
cf : RR is given by (cf) (x) = c(x3 + 1).
(fg) (x) : RR is given by (fg) (x) = f(x).g(x) = (x3 + 1) (x + 1) = (x + 1) (x2 $-$ x + 1) (x + 1) = (x + 1)2 (x2 $-$ x + 1).
Note that : (x3 + 1) = (x + 1) (x2 $-$ x + 1)]
(b) Given:
$f\left(x\right)=\sqrt{x-1}$ and $g\left(x\right)=\sqrt{x+1}$
Thus,
(f + g) ) : [1, ∞) → R is defined by (f + g) (x) = f (x) + g (x) = $\sqrt{x-1}+\sqrt{x+1}$.
(f $-$ g) ) : [1, ∞) → R is defined by (f $-$ g) (x) = f (x$-$ g (x) = $\sqrt{x-1}-\sqrt{x+1}$ .
cf : [1, ∞) → R is defined by (cf) (x) = $c\sqrt{x-1}$ .
(fg) : [1, ∞) → R is defined by (fg) (x) = f(x).g(x) = $\sqrt{x-1}×\sqrt{x+1}=\sqrt{{x}^{2}-1}$ .
#### Question 2:
Let f(x) = 2x + 5 and g(x) = x2 + x. Describe (i) f + g (ii) fg (iii) fg (iv) f/g. Find the domain in each case.
#### Answer:
Given:
f(x) = 2x + 5 and g(x) = x2 + x
Clearly, f (x) and g (x) assume real values for all x.
Hence,
domain (f) = R and domain (g) = R.
.
Now,
(i) (f + g) : RR is given by (f + g) (x) = f (x) + g (x) = 2x + 5 + x2 + x = x2 + 3x + 5.
Hence, domain ( f + g) = R .
(ii) ($-$ g) : RR is given by (f $-$ g) (x) = f (x$-$ g (x) = (2x + 5) $-$ (x2 + x) = 5 + x $-$ x2
Hence, domain ( f $-$ g) = R.
(iii) (fg) : RR is given by (fg) (x) = f(x).g(x) = (2x + 5)(x2 + x)
= 2x3 + 2x2 + 5x2 + 5x
= 2x3 + 7x2 + 5x
Hence, domain ( f.g) = R .
(iv) Given:
g(x) = x2 + x
g(x) = 0 ⇒ x2 + x = 0 = x(x+ 1) = 0
x = 0 or (x + 1) = 0
x = 0 or x$-$ 1
Now,
.
Hence, $\text{d}\mathrm{omain}\left(\frac{\mathrm{f}}{\mathrm{g}}\right)=\mathrm{R}-\left\{-1,0\right\}$.
#### Question 3:
If f(x) be defined on [−2, 2] and is given by and g(x) $=f\left(\left|x\right|\right)+\left|f\left(x\right)\right|$, find g(x).
#### Answer:
Given:
$f\left(x\right)=\left\{\begin{array}{ll}-1,& -2⩽x⩽0\\ x-1,& 0
Thus,
$g\left(x\right)=f\left(\left|x\right|\right)+\left|f\left(x\right)\right|$
#### Question 4:
Let f and g be two real functions defined by $f\left(x\right)=\sqrt{x+1}$and $g\left(x\right)=\sqrt{9-{x}^{2}}$. Then, describe each of the following functions:
(i) f + g
(ii) gf
(iii) f g
(iv) $\frac{f}{g}$
(v) $\frac{g}{f}$
(vi)
(vii) f2 + 7f
(viii) $\frac{5}{8}$
#### Answer:
Given:
Clearly, $f\left(x\right)=\sqrt{x+1}$ is defined for all x$-$1.
Thus, domain (f) = [ $-$1, ∞]
Again,
$g\left(x\right)=\sqrt{9-{x}^{2}}$ is defined for
9 $-$x2 ≥ 0 ⇒ x2 $-$ 9 ≤ 0
x2 $-$ 32 ≤ 0
⇒ (x + 3)(x $-$ 3) ≤ 0
$x\in \left[-3,3\right]$
Thus, domain (g) = [$-$ 3, 3]
Now,
domain ( f ) ∩ domain( g ) = [ $-$1, ∞] ∩ [$-$ 3, 3]
= [ $-$1, 3]
(i) ( f + g ) : [ $-$ 1 , 3] → R is given by ( f + g ) (x) = f (x) + g (x) = $\sqrt{x+1}+\sqrt{9-{x}^{2}}$ .
(ii) ( $-$ f ) : [ $-$1 , 3] → R is given by ( g $-$ f ) (x) = g (x)$-$ f (x) = $\sqrt{9-{x}^{2}}-\sqrt{x+1}$ .
(iii) (fg) : [ $-$1 , 3] → R is given by (fg) (x) = f(x).g(x) = .
(iv) .
(v) .
(vi)
$=2\sqrt{x+1}-\sqrt{45-5{x}^{2}}$ .
(vii) {Since domain(f) = [$-$ 1, ∞]}
$={\left(\sqrt{x+1}\right)}^{2}+7\left(\sqrt{x+1}\right)=x+1+7\sqrt{x+1}\phantom{\rule{0ex}{0ex}}$
(viii) {Since domain(g) = [$-$ 3, 3]}
#### Question 5:
If f(x) = loge (1 − x) and g(x) = [x], then determine each of the following functions:
(i) f + g
(ii) fg
(iii) $\frac{f}{g}$
(iv) $\frac{g}{f}$
Also, find (f + g) (−1), (fg) (0), .
#### Answer:
Given:
f(x) = loge (1 − x) and g(x) = [x]
Clearly, f(x) = loge (1 − x) is defined for all ( 1 $-$ x) > 0.
⇒ 1 > x
x < 1
x ∈ ( $-$∞, 1)
Thus, domain (f ) = ( $-$ ∞, 1)
Again,
g(x) = [x] is defined for all x ∈ R.
Thus, domain (g) = R
∴ Domain (f) ∩ Domain (g) = ( $-$ ∞, 1) ∩ R
= ( $-$ ∞, 1)
Hence,
(i ) ( f + g ) : ( $-$∞, 1) → R is given by ( f + g ) (x) = f (x) + g (x) = loge (1 − x) + [ x ].
(ii) (fg) : ( $-$ ∞, 1) → R is given by (fg) (x) = f(x).g(x) = loge (1 − x)[ x ] = [ x ]loge (1 − x).
(iii) Given:
g(x) = [ x ]
If [ x ] = 0,
x ∈ (0, 1)
Thus,
$\mathrm{domain}\left(\frac{f}{g}\right)=\mathrm{domain}\left(f\right)\cap \mathrm{domain}\left(g\right)-\left\{x:g\left(x\right)=0\right\}$
(iv) Given:
f(x) = loge (1 − x)
$⇒\frac{1}{f\left(x\right)}=\frac{1}{{\mathrm{log}}_{e}\left(1-x\right)}$
$\frac{1}{f\left(x\right)}$ is defined if loge( 1$-$x) is defined and loge(1 – x) ≠ 0.
⇒ (1$-$ x) > 0 and (1 $-$ x) ≠ 0
x < 1 and x ≠ 0
x ∈ ( $-$ ∞, 0)∪ (0, 1)
Thus, $\text{d}\mathrm{omain}\left(\frac{g}{f}\right)=\left(-\infty ,0\right)\cup \left(0,1\right)$ = ( $-$ ∞, 1) .
(f + g)( $-$1) = f( $-$1) + g( $-$1)
= loge{1 – ($-$ 1)}+ [ $-$1]
= loge 2 – 1
Hence, (f + g)( $-$ 1) = loge 2 – 1
(fg)(0) = loge ( 1 – 0) × [0] = 0
$\left(\frac{g}{f}\right)\left(\frac{1}{2}\right)=\frac{\left[\frac{1}{2}\right]}{{\mathrm{log}}_{\mathrm{e}}\left(1-\frac{1}{2}\right)}=0$
#### Question 6:
If f, g and h are real functions defined by and h(x) = 2x2 − 3, find the values of (2f + gh) (1) and (2f + gh) (0).
#### Answer:
Given:
Clearly, f (x) is defined for x + 1 ≥ 0 .
x$-$ 1
x ∈ [ $-$1, ∞]
Thus, domain ( f ) = [ $-$1, ∞] .
Clearly, g (x) is defined for x ≠ 0 .
x ∈ R – { 0} and h(x) is defined for all x such that x ∈ R .
Thus,
domain ( f ) ∩ domain (g) ∩ domain (h) = [ $-$ 1, ∞] – { 0}.
Hence,
(2f + g h) : [ $-$ 1, ∞] – { 0} → R is given by:
(2f + gh)(x) = 2f (x) + g (x$-$ h (x)
$=2\sqrt{x+1}+\frac{1}{x}-2{x}^{2}+3\phantom{\rule{0ex}{0ex}}$
(2f + gh) (0) does not exist because 0 does not lie in the domain x [ - 1, ∞] – {0}.
#### Question 7:
The function f is defined by . Draw the graph of f(x).
#### Answer:
Here,
f (x) = 1 – x for x < 0. So,
f ($-$ 4) = 1 – ( $-$ 4) = 5
f ($-$ 3) = 1 – ( $-$ 3) = 4
f ($-$2) = 1 – ( $-$ 2) = 3
f ($-$1) = 1 – ($-$ 1) = 2 etc.
Also, f(x) = 1 for x = 0.
Lastly, f (x) = x + 1 for, x > 0.
and f (1) = 2, f (2) = 3, f (3) = 4, f (4) = 5 and so on.
Thus, the graph of f is as shown below:
#### Question 8:
Let f, g : R → R be defined, respectively by f(x) = x + 1 and g(x) = 2x − 3. Find f + g, fg and $\frac{f}{g}$.
#### Answer:
f, g : R → R is defined, respectively, by f(x) = x + 1 and g(x) = 2x − 3.
(f + g) (x) = f(x) + g(x)
= (x + 1) + (2x – 3)
= 3x $-$ 2
∴ (f + g) (x) = 3x – 2
(f $-$ g)(x) = f(x$-$ g(x)
= (x + 1) $-$ (2x – 3)
= x + 1 – 2x + 3
= $-$x + 4
∴ (f - g) (x) = $-$ x + 4
$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)},g\left(x\right)\ne 0,x\in \mathbit{R}\phantom{\rule{0ex}{0ex}}$
$⇒\left(\frac{f}{g}\right)\left(x\right)=\frac{x+1}{2x-3},x\ne \frac{3}{2}$
#### Question 9:
Let f : [0, ∞) → R and g : R → R be defined by $f\left(x\right)=\sqrt{x}$and g(x) = x. Find f + g, fg, fg and $\frac{f}{g}$.
#### Answer:
It is given that f : [0, ∞) → R and g : RR such that $f\left(x\right)=\sqrt{x}$ and g(x) = x .
$D\left(f+g\right)=\left[0,\infty \right)\cap R=\left[0,\infty \right)$
So, f + g : [0, ∞) → R is given by
$\left(f+g\right)\left(x\right)=f\left(x\right)+g\left(x\right)=\sqrt{x}+x$
$D\left(f-g\right)=D\left(f\right)\cap D\left(g\right)=\left[0,\infty \right)\cap R=\left[0,\infty \right)$
So, f - g : [0, ∞) → R is given by
$\left(f-g\right)\left(x\right)=f\left(x\right)-g\left(x\right)=\sqrt{x}-x$
$D\left(fg\right)=D\left(f\right)\cap D\left(g\right)=\left[0,\infty \right)\cap R=\left[0,\infty \right)$
So, fg : [0, ∞) → R is given by
$\left(fg\right)\left(x\right)=f\left(x\right)g\left(x\right)=\sqrt{x}.x={x}^{3}{2}}$
$D\left(\frac{f}{g}\right)=\left[D\left(f\right)\cap D\left(g\right)-\left\{x:g\left(x\right)=0\right\}\right]=\left(0,\infty \right)$
So,$\frac{f}{g}:\left(0,\infty \right)\to R$ is given by
$\left(\frac{f}{g}\right)\left(x\right)=\frac{f\left(x\right)}{g\left(x\right)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}$
#### Question 10:
Let f(x) = x2 and g(x) = 2x+ 1 be two real functions. Find (f + g) (x), (fg) (x), (fg) (x) and .
#### Answer:
Given:
f (x) = x2 and g (x) = 2x + 1
Clearly, D (f) = R and D (g) = R
$D\left(fg\right)=D\left(f\right)\cap D\left(g\right)=R\cap R=R\phantom{\rule{0ex}{0ex}}$
$D\left(\frac{f}{g}\right)=D\left(f\right)\cap D\left(g\right)-\left\{x:g\left(x\right)=0\right\}=R\cap R-\left\{-\frac{1}{2}\right\}=R-\left\{-\frac{1}{2}\right\}$
Thus,
(f + g) (x) : RR is given by (f + g) (x) = f (x) + g (x) = x2 + 2x + 1= (x + 1)2 .
($-$ g) (x) : RR is given by (f $-$ g) (x) = f (x$-$ g (x) = x2 $-$ 2x $-$1.
(fg) (x) : RR is given by (fg) (x) = f(x).g(x) = x2(2x + 1) = 2x3 + x2 .
.
#### Question 1:
Let A = {1, 2, 3} and B = {2, 3, 4}. Then which of the following is a function from A to B?
(a) {(1, 2), (1, 3), (2, 3), (3, 3)}
(b) [(1, 3), (2, 4)]
(c) {(1, 3), (2, 2), (3, 3)}
(d) {(1, 2), (2, 3), (3, 2), (3, 4)}
#### Answer:
(c) {(1, 3), (2, 2), (3, 3)}
We have
R = {(1, 3), (2, 2), (3, 3)}
We observe that each element of the given set has appeared as first component in one and only one ordered pair of R.
So, R = {(1, 3), (2, 2), (3, 3)} is a function.
#### Question 2:
If f : Q → Q is defined as f(x) = x2, then f−1 (9) is equal to
(a) 3
(b) −3
(c) {−3, 3}
(d) Ď•
#### Answer:
(c) {−3, 3}
If f : AB, such that yB, then ${f}^{-1}${ y }={xA: f (x) = y}.
In other words, ${f}^{-1}${ y} is the set of pre-images of y.
Let ${f}^{-1}${9} = x
Then, f (x) = 9
x2 = 9
x = ± 3
${f}^{-1}${9} = {- 3, 3}.
#### Question 3:
Which one of the following is not a function?
(a) {(x, y) : x, y ∈ R, x2 = y}
(b) {(x, y) : x, y ∈, R, y2 = x}
(c) {(x, y) : x, y ∈ R, x2 = y3}
(d) {(x, y) : x, y ∈, R, y = x3}
#### Answer:
(b) {(x, y) : x, y ∈, R, y2 = x}
#### Question 4:
If f(x) = cos (log x), then the value of f(x2) f(y2) −$\frac{1}{2}\left\{f\left(\frac{{x}^{2}}{{y}^{2}}\right)+f\left({x}^{2}{y}^{2}\right)\right\}$is
(a) −2
(b) −1
(c) 1/2
(d) None of these
#### Answer:
(d) None of these
Given:
$f\left(x\right)=\mathrm{cos}\left(\mathrm{log}x\right)$
$⇒f\left({x}^{2}\right)=\mathrm{cos}\left(\mathrm{log}\left({x}^{2}\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left({x}^{2}\right)=\mathrm{cos}\left(2\mathrm{log}\left(x\right)\right)$
Similarly,
$f\left({y}^{2}\right)=\mathrm{cos}\left(2\mathrm{log}\left(y\right)\right)$
Now,
$f\left(\frac{{x}^{2}}{{y}^{2}}\right)=\mathrm{cos}\left(\mathrm{log}\left(\frac{{x}^{2}}{{y}^{2}}\right)\right)=\mathrm{cos}\left(\mathrm{log}{x}^{2}-\mathrm{log}{y}^{2}\right)$
and
$f\left({x}^{2}{y}^{2}\right)=\mathrm{cos}\left(\mathrm{log}{x}^{2}{y}^{2}\right)=\mathrm{cos}\left(\mathrm{log}{x}^{2}+\mathrm{log}{y}^{2}\right)$
$⇒f\left(\frac{{x}^{2}}{{y}^{2}}\right)+f\left({x}^{2}{y}^{2}\right)=\mathrm{cos}\left(\left(2\mathrm{log}x-2\mathrm{log}y\right)\right)+\mathrm{cos}\left(\left(2\mathrm{log}x+2\mathrm{log}y\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{{x}^{2}}{{y}^{2}}\right)+f\left({x}^{2}{y}^{2}\right)=2\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left[f\left(\frac{{x}^{2}}{{y}^{2}}\right)+f\left({x}^{2}{y}^{2}\right)\right]=\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)$
$⇒f\left({x}^{2}\right)f\left({y}^{2}\right)-\frac{1}{2}\left\{f\left({x}^{2}{y}^{2}\right)+f\left(\frac{{x}^{2}}{{y}^{2}}\right)\right\}=\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)-\mathrm{cos}\left(2\mathrm{log}x\right)\mathrm{cos}\left(2\mathrm{log}y\right)=0$
#### Question 5:
If f(x) = cos (log x), then the value of f(x) f(y) −$\frac{1}{2}\left\{f\left(\frac{x}{y}\right)+f\left(xy\right)\right\}$ is
(a) −1
(b) 1/2
(c) −2
(d) None of these
#### Answer:
(d) None of these
Given:
$f\left(x\right)=\mathrm{cos}\left(\mathrm{log}x\right)$
$f\left(y\right)=\mathrm{cos}\left(\mathrm{log}y\right)$
Now,
$f\left(\frac{x}{y}\right)=\mathrm{cos}\left(\mathrm{log}\left(\frac{x}{y}\right)\right)=\mathrm{cos}\left(\mathrm{log}x-\mathrm{log}y\right)$
and
$f\left(xy\right)=\mathrm{cos}\left(\mathrm{log}xy\right)=\mathrm{cos}\left(\mathrm{log}x+\mathrm{log}y\right)$
$⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=\mathrm{cos}\left(\mathrm{log}x-\mathrm{log}y\right)+\mathrm{cos}\left(\mathrm{log}x+\mathrm{log}y\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=2\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]=\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)$
$⇒f\left(x\right)f\left(y\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}=\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)-\mathrm{cos}\left(\mathrm{log}x\right)\mathrm{cos}\left(\mathrm{log}y\right)=0$
#### Question 6:
Let f(x) = |x − 1|. Then,
(a) f(x2) = [f(x)]2
(b) f(x + y) = f(x) f(y)
(c) f(|x| = |f(x)|
(d) None of these
#### Answer:
(d) None of these
#### Question 7:
The range of f(x) = cos [x], for π/2 < x < π/2 is
(a) {−1, 1, 0}
(b) {cos 1, cos 2, 1}
(c) {cos 1, −cos 1, 1}
(d) [−1, 1]
#### Answer:
(b) {cos 1, cos 2, 1}
Since, f(x) = cos [x], where $\frac{-\mathrm{\pi }}{2},
#### Question 8:
Which of the following are functions?
(a) {(x, y) : y2 = x, x, y ∈ R}
(b) {(x, y) : y = |x|, x, y ∈ R}
(c) {(x, y) : x2 + y2 = 1, x, y ∈ R}
(d) {(x, y) : x2y2 = 1, x, y ∈ R}
#### Answer:
(b) {(x, y) : y = |x|, x, y ∈ R}
For every value of x ∈ R, there is a unique value y∈ R.
i.e. there is a unique image for all values of x ∈ R.
Also, values of x occur only once in the ordered pairs.
Thus, it is a function.
#### Question 9:
If , then f(g(x)) is equal to
(a) f(3x)
(b) {f(x)}3
(c) 3f(x)
(d) −f(x)
(c) 3f(x)
#### Question 10:
If A = {1, 2, 3} and B = {x, y}, then the number of functions that can be defined from A into B is
(a) 12
(b) 8
(c) 6
(d) 3
#### Answer:
(b) 8
Given:
Number of elements in set A = 3
Number of elements in set B = 2
Therefore, the number of functions that can be defined from A into B is = 23 = 8.
#### Question 11:
If , then $f\left(\frac{2x}{1+{x}^{2}}\right)$ is equal to
(a) {f(x)}2
(b) {f(x)}3
(c) 2f(x)
(d) 3f(x)
(c) 2f(x)
#### Question 12:
If f(x) = cos (log x), then value ofis
(a) 1
(b) −1
(c) 0
(d) ±1
#### Answer:
(c) 0
Given : f(x) = cos (log x)
Then,
#### Question 13:
If $f\left(x\right)=\frac{{2}^{x}+{2}^{-x}}{2}$, then f(x + y) f(xy) is equal to
(a) $\frac{1}{2}\left[f\left(2x\right)+f\left(2y\right)\right]$
(b) $\frac{1}{2}\left[f\left(2x\right)-f\left(2y\right)\right]$
(c) $\frac{1}{4}\left[f\left(2x\right)+f\left(2y\right)\right]$
(d) $\frac{1}{4}\left[f\left(2x\right)-f\left(2y\right)\right]$
#### Answer:
(a) $\frac{1}{2}\left[f\left(2x\right)+f\left(2y\right)\right]$
Given:
$f\left(x\right)=\frac{{2}^{x}+{2}^{-x}}{2}$
Now,
f(x + y) f(xy) = $\left(\frac{{2}^{x+y}+{2}^{-x-y}}{2}\right)\left(\frac{{2}^{x-y}+{2}^{-x+y}}{2}\right)$
⇒ f(x + y) f(xy) = $\frac{1}{4}\left({2}^{2x}+{2}^{-2y}+{2}^{2y}+{2}^{-2x}\right)$
⇒ f(x + y) f(xy) = $\frac{1}{2}\left(\frac{{2}^{2x}+{2}^{-2x}}{2}+\frac{{2}^{2y}+{2}^{-2y}}{2}\right)$
⇒ f(x + y) f(xy) = $\frac{1}{2}\left[f\left(2x\right)+f\left(2y\right)\right]$
#### Question 14:
If 2f (x) − $3f\left(\frac{1}{x}\right)={x}^{2}$ (x ≠ 0), then f(2) is equal to
(a) $-\frac{7}{4}$
(b) $\frac{5}{2}$
(c) −1
(d) None of these
#### Answer:
(a) $-\frac{7}{4}$
2f (x) − $3f\left(\frac{1}{x}\right)={x}^{2}$ (x ≠ 0) ....(1)
#### Question 15:
Let f : R → R be defined by f(x) = 2x + |x|. Then f(2x) + f(−x) − f(x) =
(a) 2x
(b) 2|x|
(c) −2x
(d) −2|x|
#### Answer:
(b) 2|x|
f(x) = 2x + |x|
Then, f(2x) + f(−x) − f(x)
#### Question 16:
The range of the function $f\left(x\right)=\frac{{x}^{2}-x}{{x}^{2}+2x}$ is
(a) R
(b) R − {1}
(c) R − {−1/2, 1}
(d) None of these
#### Answer:
(c) R − {-1/2,1}
$f\left(x\right)=\frac{{x}^{2}-x}{{x}^{2}+2x}$
#### Question 17:
If x ≠ 1 and $f\left(x\right)=\frac{x+1}{x-1}$is a real function, then f(f(f(2))) is
(a) 1
(b) 2
(c) 3
(d) 4
#### Answer:
(c) 3
$f\left(x\right)=\frac{x+1}{x-1}$
#### Question 18:
If f(x) = cos (loge x), then $f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}$is equal to
(a) cos (xy)
(b) log (cos (xy))
(c) 1
(d) cos (x + y)
#### Answer:
Given:
$f\left(x\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x\right)$
$⇒f\left(\frac{1}{x}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}\left(\frac{1}{x}\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{1}{x}\right)=\mathrm{cos}\left(-{\mathrm{log}}_{e}\left(x\right)\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{1}{x}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}\left(x\right)\right)$
Similarly,
$f\left(\frac{1}{y}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}y\right)$
Now,
$f\left(xy\right)=\mathrm{cos}\left({\mathrm{log}}_{e}xy\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x+{\mathrm{log}}_{e}y\right)$
and
$f\left(\frac{x}{y}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}\frac{x}{y}\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x-{\mathrm{log}}_{e}y\right)$
$⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=\mathrm{cos}\left({\mathrm{log}}_{e}x-{\mathrm{log}}_{e}y\right)+\mathrm{cos}\left({\mathrm{log}}_{e}x+{\mathrm{log}}_{e}y\right)\phantom{\rule{0ex}{0ex}}⇒f\left(\frac{x}{y}\right)+f\left(xy\right)=2\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)\phantom{\rule{0ex}{0ex}}⇒\frac{1}{2}\left[f\left(\frac{x}{y}\right)+f\left(xy\right)\right]=\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)$
$⇒f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f\left(xy\right)+f\left(\frac{x}{y}\right)\right\}=\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)-\mathrm{cos}\left({\mathrm{log}}_{e}x\right)\mathrm{cos}\left({\mathrm{log}}_{e}y\right)=0$
Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.
#### Question 19:
Let f(x) = x, $g\left(x\right)=\frac{1}{x}$ and h(x) = f(x) g(x). Then, h(x) = 1
(a) x ∈ R
(b) x ∈ Q
(c) x ∈ R − Q
(d) x ∈ R, x ≠ 0
#### Answer:
(d) x ∈ R, x ≠ 0
Given:
f(x) = x, $g\left(x\right)=\frac{1}{x}$ and h(x) = f(x) g(x)
Now,
$h\left(x\right)=x×\frac{1}{x}=1$
We observe that the domain of f is $\mathrm{ℝ}$ and the domain of g is $\mathrm{ℝ}-\left\{0\right\}$.
∴ Domain of h = Domain of f â‹‚ Domain of g = $\mathrm{ℝ}\cap \left[\mathrm{ℝ}-\left\{0\right\}\right]=\mathrm{ℝ}-\left\{0\right\}$
$⇒$x ∈ R, x ≠ 0
#### Question 20:
If for x ∈ R, then f (2002) =
(a) 1
(b) 2
(c) 3
(d) 4
#### Answer:
(a) 1
Given:
On dividing the numerator and denominator by , we get
(For every x ∈ R)
For x = 2002, we have
f (2002) = 1
#### Question 21:
The function f : R → R is defined by f(x) = cos2 x + sin4 x. Then, f(R) =
(a) [3/4, 1)
(b) (3/4, 1]
(c) [3/4, 1]
(d) (3/4, 1)
#### Answer:
(c) (3/4, 1)
Given:
f(x) = cos2x + sin4x
$⇒f\left(x\right)=1-{\mathrm{sin}}^{2}x+{\mathrm{sin}}^{4}x$
$⇒f\left(x\right)={\left({\mathrm{sin}}^{2}x-\frac{1}{2}\right)}^{2}+\frac{3}{4}$
The minimum value of $f\left(x\right)$ is $\frac{3}{4}$.
Also,
${\mathrm{sin}}^{2}x\le 1\phantom{\rule{0ex}{0ex}}⇒{\mathrm{sin}}^{2}x-\frac{1}{2}\le \frac{1}{2}\phantom{\rule{0ex}{0ex}}⇒{\left({\mathrm{sin}}^{2}x-\frac{1}{2}\right)}^{2}\le \frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒{\left({\mathrm{sin}}^{2}x-\frac{1}{2}\right)}^{2}+\frac{3}{4}\le \frac{1}{4}+\frac{3}{4}\phantom{\rule{0ex}{0ex}}⇒f\left(x\right)\le 1$
The maximum value of $f\left(x\right)$ is 1.
∴ f(R) = (3/4, 1)
#### Question 22:
Let A = {x ∈ R : x ≠ 0, −4 ≤ x ≤ 4} and f : A ∈ R be defined by $f\left(x\right)=\frac{\left|x\right|}{x}$for x ∈ A.
Then th (is
(a) [1, −1]
(b) [x : 0 ≤ x ≤ 4]
(c) {1}
(d) {x : −4 ≤ x ≤ 0}
#### Answer:
Disclaimer: The question in the book has some error. The solution is created according to the question given in the book.
#### Question 23:
If f : R → R and g : R → R are defined by f(x) = 2x + 3 and g(x) = x2 + 7, then the values of x such that g(f(x)) = 8 are
(a) 1, 2
(b) −1, 2
(c) −1, −2
(d) 1, −2
#### Answer:
(c) −1, −2
f(x) = 2x + 3 and g(x) = x2 + 7
#### Question 24:
If f : [−2, 2] → R is defined by , then
{x ∈ [−2, 2] : x ≤ 0 and f (|x|) = x} =
(a) {−1}
(b) {0}
(c) $\left\{-\frac{1}{2}\right\}$
(d) Ď•
#### Answer:
(c) $\left\{-\frac{1}{2}\right\}$
Given:
We know,
$\left|x\right|\ge 0$
$f\left(\left|x\right|\right)=\left|x\right|-1$ ...(1)
Also,
If $x\le 0$, then $\left|x\right|=-x$ ...(2)
∴ {x ∈ [−2, 2]: x ≤ 0 and f (|x|) = x}
=
=
=
=
= $\left\{\frac{-1}{2}\right\}$
#### Question 25:
If ${e}^{f\left(x\right)}=\frac{10+x}{10-x}$, x ∈ (−10, 10) and , then k =
(a) 0.5
(b) 0.6
(c) 0.7
(d) 0.8
#### Answer:
(a) 0.5
${e}^{f\left(x\right)}=\frac{10+x}{10-x}$
...(1)
#### Question 26:
f is a real valued function given by $f\left(x\right)=27{x}^{3}+\frac{1}{{x}^{3}}$and α, β are roots of $3x+\frac{1}{x}=12$. Then,
(a) f(α) ≠ f(β)
(b) f(α) = 10
(c) f(β) = −10
(d) None of these
#### Answer:
(d) None of these
Given:
$f\left(x\right)=27{x}^{3}+\frac{1}{{x}^{3}}$
$⇒f\left(x\right)=\left(3x+\frac{1}{x}\right)\left(9{x}^{2}+\frac{1}{{x}^{2}}-3\right)$
$⇒f\left(x\right)=\left(3x+\frac{1}{x}\right)\left({\left(3x+\frac{1}{x}\right)}^{2}-9\right)$
$⇒f\left(\alpha \right)=\left(3\alpha +\frac{1}{\alpha }\right)\left({\left(3\alpha +\frac{1}{\alpha }\right)}^{2}-9\right)$
Since α and β are the roots of $3x+\frac{1}{x}=12$,
$⇒f\left(\alpha \right)=12\left({\left(12\right)}^{2}-9\right)$ and $f\left(\beta \right)=12\left({\left(12\right)}^{2}-9\right)$
$⇒f\left(\alpha \right)=f\left(\beta \right)=12\left({\left(12\right)}^{2}-9\right)$
Disclaimer: The question in the book has some error, so none of the options are matching with the solution. The solution is created according to the question given in the book.
#### Question 27:
If $f\left(x\right)=64{x}^{3}+\frac{1}{{x}^{3}}$and α, β are the roots of $4x+\frac{1}{x}=3$. Then,
(a) f(α) = f(β) = −9
(b) f(α) = f(β) = 63
(c) f(α) ≠ f(β)
(d) none of these
#### Answer:
(a) $f\left(\alpha \right)=f\left(\beta \right)=-9$
Given:
$f\left(x\right)=64{x}^{3}+\frac{1}{{x}^{3}}$
$⇒f\left(x\right)=\left(4x+\frac{1}{x}\right)\left(16{x}^{2}+\frac{1}{{x}^{2}}-4\right)$
$⇒f\left(x\right)=\left(4x+\frac{1}{x}\right)\left({\left(4x+\frac{1}{x}\right)}^{2}-12\right)$
Since α and β are the roots of $4x+\frac{1}{x}=3$,
$⇒f\left(\alpha \right)=3\left({\left(3\right)}^{2}-12\right)=-9$ and $f\left(\beta \right)=3\left({\left(3\right)}^{2}-12\right)=-9$
$⇒f\left(\alpha \right)=f\left(\beta \right)=-9$
#### Question 28:
If $3f\left(x\right)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3$ for all non-zero x, then f(x) =
(a) $\frac{1}{14}\left(\frac{3}{x}+5x-6\right)$
(b) $\frac{1}{14}\left(-\frac{3}{x}+5x-6\right)$
(c) $\frac{1}{14}\left(-\frac{3}{x}+5x+6\right)$
(d) None of these
#### Answer:
(d) None of these
$3f\left(x\right)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3$ ...(1)
Disclaimer: The question in the book has some error, so, none of the options are matching with the solution. The solution is created according to the question given in the book.
#### Question 29:
If f : R → R be given by for all $f\left(x\right)=\frac{{4}^{x}}{{4}^{x}+2}$ x ∈ R, then
(a) f(x) = f(1 − x)
(b) f(x) + f(1 − x) = 0
(c) f(x) + f(1 − x) = 1
(d) f(x) + f(x − 1) = 1
#### Answer:
(c) f(x) + f(1 − x) = 1
$f\left(x\right)=\frac{{4}^{x}}{{4}^{x}+2}$x ∈ R
#### Question 30:
If f(x) = sin [π2] x + sin [−π]2 x, where [x] denotes the greatest integer less than or equal to x, then
(a) f(π/2) = 1
(b) f(π) = 2
(c) f(π/4) = −1
(d) None of these
#### Answer:
(a) f(π/2) = 1
f(x) = sin [π2] x + sin [−π2]x
#### Question 31:
The domain of the function $f\left(x\right)=\sqrt{2-2x-{x}^{2}}$ is
(a)
(b)
(c) [−2, 2]
(d)
#### Answer:
(b)
$f\left(x\right)=\sqrt{2-2x-{x}^{2}}$
#### Question 32:
The domain of definition of is
(a) (−∞, −3] ∪ (2, 5)
(b) (−∞, −3) ∪ (2, 5)
(c) (−∞, −3) ∪ [2, 5]
(d) None of these
#### Answer:
(a) (−∞, −3] ∪ (2, 5)
#### Question 33:
The domain of the function is
(a) [−1, 2) ∪ [3, ∞)
(b) (−1, 2) ∪ [3, ∞)
(c) [−1, 2] ∪ [3, ∞)
(d) None of these
#### Answer:
(a) [−1, 2) ∪ [3, ∞)
#### Question 34:
The domain of definition of the function $f\left(x\right)=\sqrt{x-1}+\sqrt{3-x}$ is
(a) [1, ∞)
(b) (−∞, 3)
(c) (1, 3)
(d) [1, 3]
#### Answer:
(d) [1, 3]
$f\left(x\right)=\sqrt{x-1}+\sqrt{3-x}$
#### Question 35:
The domain of definition of the function $f\left(x\right)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}$is
(a) (−∞, −2] ∪ [2, ∞)
(b) [−1, 1]
(c) Ď•
(d) None of these
#### Answer:
(c) Ď•
$f\left(x\right)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}$
#### Question 36:
The domain of definition of the function f(x) = log |x| is
(a) R
(b) (−∞, 0)
(c) (0, ∞)
(d) R − {0}
(d) R − {0}
f(x) = log |x|
#### Question 37:
The domain of definition of $f\left(x\right)=\sqrt{4x-{x}^{2}}$ is
(a) R − [0, 4]
(b) R − (0, 4)
(c) (0, 4)
(d) [0, 4]
#### Answer:
(d) [0, 4]
Given:
$f\left(x\right)=\sqrt{4x-{x}^{2}}$
Clearly, f (x) assumes real values if
4x $-$ x2 ≥ 0
x(4 $-$ x) ≥ 0
⇒ $-$ x($-$ 4) ≥ 0
x($-$ 4) ≤ 0
x ∈ [0, 4]
Hence, domain (f )= [0, 4].
#### Question 38:
The domain of definition of $f\left(x\right)=\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-3+2\sqrt{x-4}}$is
(a) [4, ∞)
(b) (−∞, 4]
(c) (4, ∞)
(d) (−∞, 4)
#### Answer:
(a) [4, ∞)
$f\left(x\right)=\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-3+2\sqrt{x-4}}$
#### Question 39:
The domain of the function is
(a) (−3, − 2) ∪ (2, 3)
(b) [−3, − 2) ∪ [2, 3)
(c) [−3, − 2] ∪ [2, 3]
(d) None of these
#### Answer:
(c) [−3, − 2] ∪ [2, 3]
#### Question 40:
The range of the function $f\left(x\right)=\frac{x}{\left|x\right|}$is
(a) R − {0}
(b) R − {−1, 1}
(c) {−1, 1}
(d) None of these
#### Answer:
(c) {−1, 1}
$f\left(x\right)=\frac{x}{\left|x\right|}$
#### Question 41:
The range of the function $f\left(x\right)=\frac{x+2}{\left|x+2\right|}$, x ≠ −2 is
(a) {−1, 1}
(b) {−1, 0, 1}
(c) {1}
(d) (0, ∞)
#### Answer:
(a) {−1, 1}
$f\left(x\right)=\frac{x+2}{\left|x+2\right|}$ , x ≠ −2
#### Question 42:
The range of the function f(x) = |x − 1| is
(a) (−∞, 0)
(b) [0, ∞)
(c) (0, ∞)
(d) R
(b) [0, ∞)
#### Question 43:
Let $f\left(x\right)=\sqrt{{x}^{2}+1}$. Then, which of the following is correct?
(a) $f\left(xy\right)=f\left(x\right)f\left(y\right)$ (b) $f\left(xy\right)\ge f\left(x\right)f\left(y\right)$ (c) $f\left(xy\right)\le f\left(x\right)f\left(y\right)$ (d) none of these
#### Answer:
Given: $f\left(x\right)=\sqrt{{x}^{2}+1}$ .....(1)
Replacing x by y in (1), we get
$f\left(y\right)=\sqrt{{y}^{2}+1}$
Also, replacing x by xy in (1), we get
$f\left(xy\right)=\sqrt{{x}^{2}{y}^{2}+1}$
Now,
${x}^{2}{y}^{2}+1\le {x}^{2}{y}^{2}+{x}^{2}+{y}^{2}+1\phantom{\rule{0ex}{0ex}}⇒\sqrt{{x}^{2}{y}^{2}+1}\le \sqrt{{x}^{2}{y}^{2}+{x}^{2}+{y}^{2}+1}\phantom{\rule{0ex}{0ex}}⇒f\left(xy\right)\le f\left(x\right)f\left(y\right)$
Hence, the correct answer is option (c).
#### Question 44:
If ${\left[x\right]}^{2}-5\left[x\right]+6=0$, where [.] denotes the greatest integer function, then
(a) x ∈ [3, 4] (b) x ∈ (2, 3] (c) x ∈ [2, 3] (d) x ∈ [2, 4)
#### Answer:
The given equation is ${\left[x\right]}^{2}-5\left[x\right]+6=0$.
${\left[x\right]}^{2}-5\left[x\right]+6=0\phantom{\rule{0ex}{0ex}}⇒{\left[x\right]}^{2}-3\left[x\right]-2\left[x\right]+6=0\phantom{\rule{0ex}{0ex}}⇒\left[x\right]\left(\left[x\right]-3\right)-2\left(\left[x\right]-3\right)=0\phantom{\rule{0ex}{0ex}}⇒\left(\left[x\right]-2\right)\left(\left[x\right]-3\right)=0$
x ∈ [2, 3) or x ∈ [3, 4)
x ∈ [2, 4)
Hence, the correct answer is option (d).
#### Question 45:
The range of $f\left(x\right)=\frac{1}{1-2\mathrm{cos}x}$ is
(a) [1/3, 1] (b) [−1, 1/3] (c) (−∞, −1) ∪ [1/3, ∞) (d) [−1/3, 1]
#### Answer:
We know that −1 ≤ cosx ≤ 1 for all x ∈ R.
Now,
But,
$\mathrm{cos}x\ne \frac{1}{2}$
$⇒1-2\mathrm{cos}x\in \left[-1,3\right]-\left\{0\right\}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{1-2\mathrm{cos}x}\in \left(-\infty ,-1\right]\cup \left[\frac{1}{3},\infty \right)$
∴ Range of f(x) = (−∞, −1] ∪[$\frac{1}{3}$, ∞)
Disclaimer: The range of the function does not matches with either of the given options. The range matches with option (c) if it is given as "(−∞, −1] ∪ [1/3, ∞)".
#### Question 46:
The domain of the function $f\left(x\right)\sqrt{4-x}+\frac{1}{\sqrt{{x}^{2}-1}}$ is equal to
(a) (−∞, −1) ∪ (1, 4)
(b) (−∞, −1] ∪ (1, 4]
(c) (−∞, −1) ∪ [1, 4]
(d) (−∞, −1) ∪ [1, 4)
#### Answer:
for $\sqrt{4-x}$
4 − x ≥ 0
i.e 4 ≥ x
i.e x ≤ 4
and $\frac{1}{\sqrt{{x}^{2}-1}}$
Since x2 − 1 > 0
i.e x2 > 1
i.e x < − 1 or x > 1
for Domain is (–∞, –1) ∪ (1, 4].
Domain of is
(a) (−a, a)
(b) [−a, a]
(c) [0, a]
(d) (−a, 0]
#### Answer:
f(x) = $\sqrt{{a}^{2}-{x}^{2}}$ ; > 0
Since a2 − x2 ≥ 0
i.e −x2 ≥ −a2
i.e x2 ≤ a2
i.e |x| ≤ a
i.e −a ≤ x ≤ a
i.e x∈ [−aa]
$\therefore$ Domain of (x) is [−aa]
Hence, the correct answer is option B.
#### Question 48:
If f(x) = ax + b, where, a and b are integers, f(−1) = −5 and f(x) = 3, then a and b are equal
(a) a = −3, b = −1
(b) a = 2, b = −3
(c) a = 0, b = 2
(d) a = 2, b = 3
#### Answer:
f(x) = ax + b
Given f(−1) = −5 and f(x) = 3
(correction)
#### Question 49:
The domain and range of the real function of defined by $f\left(x\right)=\frac{4-x}{x-4}$ is given by
(a) Domain = R, Range ={−1, 1}
(b) Domain = R −{1}, Range = R
(c) Domain = R −{4}, Range = {−1}
(d) Domain = R −{−4}, Range = {−1, 1}
#### Answer:
f(x) = $\frac{4-x}{x-4}$
Domain :- for f(x) = $\frac{4-x}{x-4}$
− 4$\ne$0
i.e x$\ne$4
∴ Domain is R −{4}
Range:- Since f(x) = $\frac{-\left(x-4\right)}{x-4}=-1$
∴ Range is {−1}
Hence, the correct answer is option C.
#### Question 50:
The domain and range of real function f defined by $f\left(x\right)=\sqrt{x-1}$ is given by
(a) Domain = (1, ∞), Range = (0, ∞)
(b) Domain = [1, ∞), Range =(0, ∞)
(c) Domain = [1, ∞), Range = [0, ∞)
(d) Domain = [1, ∞), Range = [0, ∞)
#### Answer:
f(x) = $\sqrt{x-1}$
Since x −1 ≥ 0
i.e x ≥ 1
$\therefore$ Domain of f(x) is [1, ∞)
and for x∈ [1, ∞)
f(x) ≥ 0
⇒ Range of f(x) is [0, ∞)
Hence, the correct answer is option C.
#### Question 51:
The domain of the function f given by $f\left(x\right)=\frac{{x}^{2}+2x+1}{{x}^{2}-x-6}$
(a) R −{−2, 3}
(b) R −{−3, 2}
(c) R −[−2, 3]
(d) R −(−2, 3)
#### Answer:
f(x) = $\frac{{x}^{2}+2x+1}{{x}^{2}-x-6}$
f(x) is not defined for x2 − x − 6 = 0
i.e x2 − 3x + 2x – 6 = 0
i.e x(− 3) + 2 (− 3) = 0
i.e (− 3) (+ 2) = 0
i.e x = 3 or x = −2
$\therefore$ Domain for (x) is R −{−2, 3}
Hence the correct answer is option A.
#### Question 52:
The domain and range of the function f given by f(x) = 2 − |x − 5|, is
(a) Domain = R+, Range = (−∞, 1]
(b) Domain = R, Range = (−∞, 2]
(c) Domain =R, Range = (−∞, 2)
(d) Domain = R+, Range = (−∞, 2]
#### Answer:
f(x) = 2 − |− 5|
Since f(x) is defined for every xR
$\therefore$ Domain is R
also |− 5| ≥ 0
i.e −|− 5| ≤ 0
i.e 2 − |− 5| ≤ 2
$\therefore$ f(x) ≤ 2
i.e Range for (x) is (−∞, 2]
Hence, the correct answer is option B.
#### Question 53:
If $f\left(x\right)={x}^{3}-\frac{1}{{x}^{3}},$ then $f\left(x\right)+f\left(\frac{1}{x}\right)$ is equal to
(a) 2x3
(b) $\frac{2}{{x}^{3}}$
(c) 0
(d) 1
#### Answer:
Hence, the correct answer is option C.
#### Question 54:
The domain of the function f defined by $f\left(x\right)=\frac{1}{\sqrt{x-\left|x\right|}}$ is
(a) R0
(b) R+
(c) R
(d) none of these
#### Answer:
f(x) = $f\left(x\right)=\frac{1}{\sqrt{x-\left|x\right|}}$
f(x) is defined if − |x| > 0
i.e x > |x
i.e |x| < x
Since no such real number x exist such that |x| < x
$\therefore$ Domain of f(x) is empty set.
Hence, the correct answer is option D.
#### Question 1:
Let A and B be any two sets such that n(A) = p and n(B) = q, then the total functions from A to B is equal to __________ .
#### Answer:
n(A) = p, n(B) = q.
here any element of set A, can be connected with elements of B in q ways.
and there are p such elements in A
$\therefore$ Total function possible is
i.e qp
$\therefore$ Total functions from A to B qp i.e n(Bn(A.
If __________ .
#### Answer:
f(x) = $\frac{x}{x-1}$
given $\frac{x}{x-1}=\frac{1}{y}$
xy = x − 1 i.e y $\frac{x-1}{x}$
If __________ .
#### Question 4:
The domain of the function $f\left(x\right)=\frac{1}{\sqrt{\left|x\right|-x}}$ is _____________.
#### Answer:
f(x) = $\frac{1}{\sqrt{\left|x\right|-x}}$
f(x) is defined if |x| − x > 0
i.e |x| > x
i.e x < |x
which is possible for negative real numbers
∴ Domain for f(x) is |R​ ~ {0}
#### Question 5:
The range of the function f(x) = [x] − x is __________ .
#### Answer:
f(x) = [x] − x
Since x ≥ [x]
Every number is greater than or equal to its greatest integral value
i.e x – [x] = {x} fractional part of x
∴ [x] − x = −{x} fraction part only.
also [x] = x for integral value of x
hence, for non-integral values f(x) = −{x$\in$ (−1, 0)
and for integral values f(x) = 0
Hence, Range of f(x) is (−1, 0].
#### Question 6:
The range of the function $f\left(x\right)=\frac{x+2}{\left|x+2\right|}$ is __________ .
#### Answer:
∴ Range of function f(x) is {−1, 1}
#### Question 7:
The range of the function f(x) = logax, a > 0 is __________ .
#### Answer:
f(x) = loga x ; a > 0
f(x) = $\frac{{\mathrm{log}}_{10}\left(x\right)}{{\mathrm{log}}_{10}\left(a\right)}$
Since a > 0 ⇒ f(x) is defined for R+ −{0}
∴ Range set of f(x) is R
#### Question 8:
Let f and g be two functions given by f = {(2, 4), (5, 6), (8, −1), (10, −3)} and g = {(2, 5), (7, 1) (8, 4), (10, 13), (11, −5). Then, domain of f + g is __________ .
#### Answer:
f = {(2, 4), (5, 6), (8, –1), (10, –3)}
g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, –5)}
Domain of f is {2, 5, 8, 10} and Domain of g = {2, 7, 8, 10, 11}
Since (f + g) (x) = f(x) + g(x)
∴ x should lie in Both Domain of f and Domain of g
∴ x ∈ Domain ( f) ∩ Domain (g
i.e Domain of f + g is {2, 8, 10}
#### Question 9:
Let f and g be two real functions given by f = {(10, 1), (2, 0), (3, −4), (4, 2), (5, 1)} and g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}. Then the domain fg is given by __________ .
#### Answer:
f = {(10, 1),(2, 0), (3, –4), (4, 2), (5, 1)}
g = {(1, 0), (2, 2), (3, –1), (4, 4), (5, 3)}
Since (fg) (x) = f(x) g(x
∴ Domain of fg is (Domain of f) ∩ (Domain of g
Since Domain of f = {10, 2, 3, 4, 5}
Domain of g = {1, 2, 3, 4, 5}
⇒ Domain of f g is {2, 3, 4, 5}
#### Question 10:
The domain for which the functions f(x) = 3x2 −1 and g(x) = 3 + x are equal is __________ .
#### Answer:
f(x) = 3x2 − 1
g(x) = 3 + x
Given:- f(x) = g(x
i.e 3x2 − 1 = 3 + x
3x2 − x − 1 − 3 = 0
3x2 − x − 4 = 0
3x+ 3x − 4x − 4 = 0
3x(x + 1) − 4 (x + 1) = 0
(3− 4) (x + 1) = 0
i.e x$\frac{4}{3}$ or x = −1
∴ Domain for which f(x) and g(x) are equal is $\left\{-1,\frac{4}{3}\right\}$
#### Question 11:
The domain of the function $f\left(x\right)=\frac{{x}^{2}+1}{{x}^{2}-3x+2}$ is __________ .
#### Answer:
f(x) = $\frac{{x}^{2}+1}{{x}^{2}-3x+2}$
f(x) is defined if x2 − 3x + 2 ≠ 0
i.e f(x) is not defined
for x2 − 3x + 2 = 0
i.e x2 − 2x − x + 2 = 0
(x2 − 2x) −1 (x − 2) = 0
x (x − 2) −1 (x − 2) = 0
i.e (x − 1) (x − 2) = 0
i.e x = 1 or x = 2
i.e for x = 1 or x = 2, f (x) is not defined
∴ Domain of f(x) = R ~ {1,2}
#### Question 12:
If $f\left(x\right)=\frac{x-1}{x+1},$ then $f\left(\frac{1}{x}\right)+f\left(x\right)$ is equal to __________ .
#### Question 13:
If $f\left(x\right)=\frac{x-1}{x+1},$ then is equal to __________ .
#### Question 14:
If f(x) = [x]2 − 5 [x] + 6, then the set of values of x satisfying f(x) = 0 is __________ .
#### Answer:
f(x) = [x]2 − 5 [x] + 6
Given f(x) = 0
i.e [x]− 5 [x] + 6 = 0
Let [x] = y
i.e y2 − 5y + 6 = 0
i.e y2 − 3y – 2y + 6 = 0
y(y − 3) −2 (y − 3) = 0
i.e y = 2 or y = 3
i.e [x] = 2 or [x] = 3
here, [x] = 2 if x∈ [2, 3) and [x] = 3 if x∈ [3, 4)
∴ value of x for which f(x) = 0 are [2, 3) ∪ [3, 4)
i.e x∈ [2, 4)
#### Question 15:
The domain of the function $f\left(x\right)=\sqrt{9-x}+\frac{1}{\sqrt{{x}^{2}-16}}$ is equal to __________ .
#### Answer:
here f(x) is defined for 9 − x ≥ 0
i.e 9 ≥ x
i.e x ≤ 9
and x2 − 16 > 0
i.e x2 > 16
i.e x > 4 or x < −4
f(x) is defined for common region is (−∞, −4) ∪ (4, 9]
∴ Domain of f(x) is (−∞, −4) ∪ (4, 9]
#### Question 16:
The domain and range of the function $f\left(x\right)=\frac{2-x}{x-2}$ are _________ and _________ respectively.
#### Answer:
f(x) is defined for x − 2 ≠ 0
i.e ≠ 2
∴ Domain of f(x) is R −{2}
Since
f(x) = – $\frac{\left(x-2\right)}{x-2}$
f(x) = −1
∴ Range of f(x) is {−1}
#### Question 17:
The domain of the function $f\left(x\right)=\frac{1}{\sqrt{{\left[x\right]}^{2}-3\left[x\right]+2}}$ is __________ .
#### Answer:
Since [x]2 − 3[x] + 2 = 0
if [x]2 − 2[x] − [x] + 2 = 0
[x] ([x] − 2) −1 ([x] − 2) = 0
i.e [x] = 2 or [x] = 1
f(x) is defined only if [x]2 − 3[x] + 2 > 0
i.e ([x] −2) ([x]−1) > 0
[x] > 2 or [x] < 1
i.e [x] ≥ 3 or [x] ≤ 0
i.e x ∈[3, ∞) or x∈(−∞, 1)
i.e Domain of f(x) is [3, ∞) ∪ (−∞, 1)
i.e (−∞, 1) ∪ [3, ∞)
#### Question 18:
The range of the function $f\left(x\right)=\frac{\left|x-4\right|}{x-4}$ is __________ .
#### Answer:
Range of f(x) is {−1, 1}
#### Question 19:
The domain of the function f(x) = x + [x] is __________ .
#### Answer:
f(x) = x + [x
Since Domain of [x] = R
and Domain of x = R
∴ Domain of x + [x] = R ∩ R = R
#### Question 20:
The range of the function $f\left(x\right)=\sqrt{1-{x}^{2}}$ is __________ .
#### Answer:
f(x) = $\sqrt{1-{x}^{2}}$
Since f(x) ≥ 0 ( Being a square root)
∴ Range of f(x) = [0, ∞)
#### Question 21:
The domain of the function $f\left(x\right)=\sum _{n=1}^{10}\frac{1}{\left|2x-n\right|}$ is __________ .
#### Answer:
f(x) =
f(x) is not defined if 2x n = 0
∴ Domain of f(x) is
#### Question 22:
The domain of the function $f\left(x\right)=\frac{\left|x\right|-2}{\left|x\right|-3}$ is __________ .
#### Answer:
here f(x) is not define if |x| − 3 = 0
i.e |x| = 3
i.e x = 3 or −3
∴ Domain of f(x) is R –{−3, 3}
#### Question 23:
If f(2x + 3) = 4x2 + 12x +15, then the value of f(3x + 2) is __________ .
#### Answer:
Given f(2+ 3) = 4x2 + 12x + 15
= (2x)2 + 2(2x) (3) + (3)2 + 6
f(2+ 3) = (2x + 3)2 + 6
i.e f(y) = y2 + 6 where y = 2x + 3
∴ f (3x + 2) = (3x + 2)2 + 6
= 9x2 + 4 + 12x + 6
Hence f(3x + 2) = 9x2 + 12x + 10
#### Question 24:
The number of elements of an identity function defined on a set containing four elements is __________ .
#### Answer:
The number of elements of an identity function defined on a set of containing.
Four elements is four only
Since identity function maps same element to same element.
i.e x → x
y → y
z → z
and w → w
There are four such elements.
#### Question 1:
Write the range of the real function f(x) = |x|.
#### Answer:
Given:
f (x) = | x |, xR
We know that
$\left|x\right|=\left\{\begin{array}{ll}x,& x\ge 0\\ -x& x<0\end{array}\right\\phantom{\rule{0ex}{0ex}}$
It can be observed that the range of f (x) = | x | is all real numbers except negative real numbers.
∴ The range of f is [0, ∞) .
#### Question 2:
If f is a real function satisfying $f\left(x+\frac{1}{x}\right)={x}^{2}+\frac{1}{{x}^{2}}$for all x ∈ R − {0}, then write the expression for f(x).
#### Answer:
Given:
$f\left(x+\frac{1}{x}\right)={x}^{2}+\frac{1}{{x}^{2}}$
$={x}^{2}+\frac{1}{{x}^{2}}+2-2\phantom{\rule{0ex}{0ex}}$
$={\left(x+\frac{1}{x}\right)}^{2}-2$
Thus,
$f\left(x+\frac{1}{x}\right)={\left(x+\frac{1}{x}\right)}^{2}-2$
Hence,
f (x) = x2 $-$ 2 , where | x | ≥ 2.
#### Question 3:
Write the range of the function f(x) = sin [x], where $\frac{-\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{4}$.
#### Answer:
Given : f(x) = sin [x], where $\frac{-\mathrm{\pi }}{4}\le x\le \frac{\mathrm{\pi }}{4}$.
#### Question 4:
If f(x) = cos [π2]x + cos [−π2] x, where [x] denotes the greatest integer less than or equal to x, then write the value of f(π).
#### Answer:
f(x) = cos [π2]x + cos [−π2] x
#### Question 5:
Write the range of the function f(x) = cos [x], where $\frac{-\mathrm{\pi }}{2}.
#### Answer:
Since f(x) = cos [x], where $\frac{-\mathrm{\pi }}{2},
#### Question 6:
Write the range of the function f(x) = ex[x], x ∈ R.
#### Answer:
f(x) = ex[x], x ∈ R
#### Question 7:
Let . Then write the value of α satisfying f(f(x)) = x for all x ≠ −1.
Given:
#### Question 8:
If $f\left(x\right)=1-\frac{1}{x}$, then write the value of $f\left(f\left(\frac{1}{x}\right)\right)$.
#### Answer:
Given:
$f\left(x\right)=1-\frac{1}{x}\phantom{\rule{0ex}{0ex}}$
Now,
$f\left(\frac{1}{x}\right)=1-\frac{1}{\frac{1}{x}}=1-x$
$⇒f\left(f\left(\frac{1}{x}\right)\right)=f\left(1-x\right)$
Again,
If $f\left(x\right)=1-\frac{1}{x}\phantom{\rule{0ex}{0ex}}$
Thus,
$f\left(1-x\right)=1-\frac{1}{1-x}\phantom{\rule{0ex}{0ex}}$
$=\frac{1-x-1}{1-x}\phantom{\rule{0ex}{0ex}}=\frac{-x}{1-x}\phantom{\rule{0ex}{0ex}}=\frac{-x}{-\left(x-1\right)}\phantom{\rule{0ex}{0ex}}=\frac{x}{x-1}$
#### Question 9:
Write the domain and range of the function $f\left(x\right)=\frac{x-2}{2-x}$.
#### Answer:
Given:
$f\left(x\right)=\frac{x-2}{2-x}$
Domain ( f ) :
Clearly, f (x) is defined for all x satisfying: if 2 $-$x ≠ 0 ⇒ ≠ 2.
Hence, domain ( f ) = R $-$ {2}
Range of f :
Let f (x) = y
$\frac{x-2}{2-x}=y$
x $-$ 2 = y (2 $-$ x)
x $-$ 2 = $-$ y (x $-$ 2)
y$-$ 1
Hence, range ( f ) = {$-$ 1}.
#### Question 10:
If f(x) = 4xx2, x ∈ R, then write the value of f(a + 1) −f(a − 1).
#### Answer:
Given:
f(x) = 4xx2, x ∈ R
Now,
f(a + 1) = 4(a + 1) $-$ (a + 1)2
= 4a + 4 $-$ (a2 + 1 + 2a)
= 4a + 4 $-$ a2 $-$$-$ 2a
= 2a $-$ a2 + 3
f(a $-$ 1) = 4(a $-$ 1) $-$ (a $-$ 1)2
= 4a $-$$-$ (a2 + 1 $-$ 2a)
= 4a $-$$-$ a2 $-$ 1 + 2a
= 6a $-$ a2 $-$ 5
Thus,
f(a + 1) − f(a − 1) = ( 2a $-$ a2 + 3) $-$ (6a $-$ a2 $-$ 5)
= 2a $-$ a2 + 3 $-$ 6a + a2 + 5
= 8 $-$ 4a
= 4(2 $-$ a)
#### Question 11:
If f, g, h are real functions given by f(x) = x2, g(x) = tan x and h(x) = loge x, then write the value of (hogof) $\left(\sqrt{\frac{\mathrm{\pi }}{4}}\right)$.
#### Answer:
Given : f(x) = x2, g(x) = tan x and h(x) = loge x.
(hogof) $\left(\sqrt{\frac{\mathrm{\pi }}{4}}\right)$ =
#### Question 12:
Write the domain and range of function f(x) given by $f\left(x\right)=\frac{1}{\sqrt{x-\left|x\right|}}$.
#### Answer:
Given:
$f\left(x\right)=\frac{1}{\sqrt{x-\left|x\right|}}$
We know that
x $-$ | x| ≤ 0 for all x.
$⇒\frac{1}{\sqrt{x-\left|x\right|}}$ does not take any real values for any x ∈ R.
f (x) is not defined for any x ∈ R.
Hence,
domain ( f ) = Φ and range ( f ) = Φ .
#### Question 13:
Write the domain and range of $f\left(x\right)=\sqrt{x-\left[x\right]}$.
#### Answer:
$f\left(x\right)=\sqrt{x-\left[x\right]}$
#### Question 14:
Write the domain and range of function f(x) given by $f\left(x\right)=\sqrt{\left[x\right]-x}$.
#### Answer:
$f\left(x\right)=\sqrt{\left[x\right]-x}$
#### Question 15:
Let A and B be two sets such that n(A) = p and n(B) = q, write the number of functions from A to B.
#### Answer:
It is given that A and B are two sets such that n(A) = p and n(B) = q.
Now, any element of set A, say ai (1 ≤ i ≤ p), is related with an element of set B in q ways. Similarly, other elements of set A are related with an element of set B in q ways.
Thus, every element of set A is related with every element of set B in q ways.
∴ Total number of functions from A to B = q × × q × ... × q (p times) = qp
#### Question 16:
Let f and g be two functions given by
f = {(2, 4), (5, 6), (8, −1), (10, −3)} and g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, −5)}.
Find the domain of f + g.
#### Answer:
It is given that f and g are two functions such that
f = {(2, 4), (5, 6), (8, −1), (10, −3)}
and g = {(2, 5), (7, 1), (8, 4), (10, 13), (11, −5)}
Now,
Domain of f = Df = {2, 5, 8, 10}
Domain of g = Dg = {2, 7, 8, 10, 11}
∴ Domain of f + g = Df Dg = {2, 8, 10}
#### Question 17:
Find the set of values of x for which the functions f(x) = 3x2 − 1 and g(x) = 3 + x are equal.
#### Answer:
It is given that the functions f(x) = 3x2 − 1 and g(x) = 3 + x are equal.
$\therefore f\left(x\right)=g\left(x\right)\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-1=3+x\phantom{\rule{0ex}{0ex}}⇒3{x}^{2}-x-4=0\phantom{\rule{0ex}{0ex}}⇒\left(x+1\right)\left(3x-4\right)=0$
Hence, the set of values of x for which the given functions are equal is $\left\{-1,\frac{4}{3}\right\}$.
#### Question 18:
Let f and g be two real functions given by
f = {(0, 1), (2, 0), (3, −4), (4, 2), (5, 1)} and g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}
Find the domain of fg.
#### Answer:
It is given that f and g are two real functions such that
f = {(0, 1), (2, 0), (3, −4), (4, 2), (5, 1)}
and g = {(1, 0), (2, 2), (3, −1), (4, 4), (5, 3)}
Now,
Domain of f = Df = {0, 2, 3, 4, 5}
Domain of g = Dg = {1, 2, 3, 4, 5}
∴ Domain of fg = Df Dg = {2, 3, 4, 5}
#### Question 1:
Define a function as a set of ordered pairs.
#### Answer:
A function is a set of ordered pairs with the property that no two ordered pairs have the same first component and different second components.
Sometimes we say that a function is a rule (correspondence) that assigns to each element of one set, X, only one element of another set, Y.
The elements of set X are often called inputs and the elements of set Y are called outputs.
The domain of a function is the set of all first components, x, in the ordered pairs.
The range of a function is the set of all second components, y, in the ordered pairs.
A function can be defined by a set of ordered pairs.
Example: {(1,a), (2, b), (3, c), (4,a)} is a function, since there are no two pairs with the same first component.
The domain is then the set {1,2,3,4} and the range is the set {a, b, c}.
#### Question 2:
Define a function as a correspondence between two sets.
#### Answer:
A function is a correspondence between two sets of elements, such that for each element in the first set there is only one corresponding element in the second set.
The first set is called the domain and the set of all corresponding elements in the second set is called the range.
Let A = {1, 2, 3} and B = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}
Let f : A B be the correspondence which assigns to each element in A its square.
Hence,
f (1) = 12 = 1
f (2) = 22 = 4
f (3) = 32 = 9
Since for each element (1 or 2 or 3) of A, there is exactly one element of B, so f is a function.
In this case, every element of B is not an image of some element of A.
#### Question 3:
What is the fundamental difference between a relation and a function? Is every relation a function?
#### Answer:
Differences between relation and function
1. If R is a relation from A to B, then domain of R may be a subset of A. But if f is a function from A to B, then domain f is equal to A.
2. In a relation from A to B, an element of A may be related to more than one element in B. But in a function from A to B, each element of A must be associated to one and only one element of B.
Thus, every function is a relation, but every relation is not necessarily a function.
#### Question 4:
Let A = {−2, −1, 0, 1, 2} and f : A → Z be a function defined by f(x) = x2 − 2x − 3. Find:
(a) range of f, i.e. f(A).
(b) pre-images of 6, −3 and 5.
#### Answer:
(a) Given:
f (x) = x2 − 2x − 3
f (−2) = (− 2)2 − 2(− 2) − 3
= 4 + 4 – 3
= 8 − 3 = 5
f (−1) = (−1)2 − 2(−1) − 3
= 1+ 2 − 3
= 3 − 3 = 0
f (0) = (0)2 − 2(0) − 3
= 0 − 0 − 3
= − 3
f (1) = (1)2 − 2(1) − 3
= 1 − 2 − 3
=1 − 5 = − 4
f (2) = (2)2 – 2(2) − 3
= 4 − 4 – 3
= 4 – 7 = − 3
Thus, range of f(A) = (− 4, − 3, 0, 5).
(b) Let x be the pre-image of 6.
Then,
f(6) = x2 − 2x − 3 = 6
x2 − 2x − 9 = 0
$x=1±\sqrt{10}$
Since $x=1±\sqrt{10}\notin A$, there is no pre-image of 6.
Let x be the pre-image of $-$3. Then,
f(− 3) ⇒ x2 − 2x − 3 = − 3
x2 − 2x = 0
x = 0, 2
Clearly $0,2\in A$. So, 0 and 2 are pre-images of −3.
Let x be the pre-image of 5. Then,
f(5) ⇒ x2 − 2x − 3 = 5
x2 − 2x − 8 = 0
⇒ (x − 4) (x + 2) = 0 ⇒ x = 4, − 2
Since $-2\in A$, − 2 is the pre-image of 5.
Hence,
pre-images of 6, − 3 and 5 are $\varphi ,\left\{0,2,\right\},-2$ respectively.
#### Question 5:
If a function f : R → R be defined by
$f\left(x\right)=\left\{\begin{array}{cc}3x-2,& x<0;\\ 1,& x=0;\\ 4x+1,& x>0.\end{array}\right\$
find: f(1), f(−1), f(0) and f(2).
#### Answer:
f (1) = 4 × 1 + 1 = 5 [By using f (x) = 4x + 1, x > 0]
f ($-$ 1) = 3 × ($-$1) $-$ 2 [By using f (x) = 3x $-$ 2, x < 0]
= $-$$-$ 2 = $-$ 5
f (0) = 1 [By using f (x) = 1, x = 0]
f (2) = 4 × 2 + 1 [By using f (x) = 4x + 1, x > 0]
= 9
Hence,
f (1) = 5, f ($-$1) = $-$ 5, f (0) = 1 and f (2) = 9.
#### Question 6:
A function f : R → R is defined by f(x) = x2. Determine (a) range of f, (b) {x : f(x) = 4}, (c) [y : f(y) = −1].
#### Answer:
(a) Given:
f (x) = x2
Range of f = R+ (Set of all real numbers greater than or equal to zero)
(b) Given:
f (x) = x2
⇒ x2 = 4
x = ± 2
∴ {x : f (x) = 4 } = { $-$ 2, 2}.
(c) { y : f (y) = $-$1}
f (y) = $-$1
It is clear that x2$-$1 but x2 ≥ 0 .
f (y) ≠ $-$1
∴ {y : f (y) = $-$1} = Φ
#### Question 7:
Let f : R+ → R, where R+ is the set of all positive real numbers, such that f(x) = loge x. Determine
(a) the image set of the domain of f
(b) {x : f(x) = −2}
(c) whether f(xy) = f(x) : f(y) holds
#### Answer:
Given:
f : R+ → R
and f (x) = logex .............(i)
(a) f : R+ → R
Thus, the image set of the domain f = R .
(b) {x : f (x) = $-$2
f (x ) = $-$2 .....(ii)
From equations (i) and (ii), we get :
logex = $-$2
⇒ x = ${e}^{-2}$
Hence, { x : f (x) = - 2} = { e – 2} . [Since logab = c ⇒ b = ac]
(c) f (xy) = loge(xy) {From(i)}
= logex + logey [Since logemn = loge m + logen]
= f (x) + f (y)
Thus, f (xy) = f (x) + f (y)
Hence, it is clear that f (xy) = f (x) + f (y) holds.
#### Question 8:
Write the following relations as sets of ordered pairs and find which of them are functions:
(a) {(x, y) : y = 3x, x ∈ {1, 2, 3}, y ∈ [3,6, 9, 12]}
(b) {(x, y) : y > x + 1, x = 1, 2 and y = 2, 4, 6}
(c) {(x, y) : x + y = 3, x, y, ∈ [0, 1, 2, 3]}
#### Answer:
(a) Given:
{(x, y) : y = 3x, x ∈ {1, 2, 3}, y ∈ [3,6, 9, 12]}
On substituting x = 1, 2, 3 in x, we get :
y = 3, 6, 9, respectively.
∴ R = {(1, 3) , (2, 6), (3, 9)}
Hence, we observe that each element of the given set has appeared as the first component in one and only one ordered pair in R . So, R is a function in the given set.
(b) Given:
{(x, y) : y > x + 1, x = 1, 2 and y = 2, 4, 6}
On substituting x = 1, 2 in y > x + 1, we get :
y > 2 and y > 3, respectively.
R = {(1, 4), (1, 6), (2, 4), (2, 6)}
We observe that 1 and 2 have appeared more than once as the first component of the ordered pairs. So, it is not a function.
(c) Given:
{(x, y) : x + y = 3, x, y, ∈ [0, 1, 2, 3]}
x + y = 3
y = 3 – x
On substituting x = 0,1, 2, 3 in y, we get:
y = 3, 2, 1, 0, respectively.
∴ R = {(0, 3), (1, 2), (2, 1), (3, 0)}
Hence, we observe that each element of the given set has appeared as the first component in one and only one ordered pair in R . So, R is a function in the given set.
#### Question 9:
Let f : R → R and g : C → C be two functions defined as f(x) = x2 and g(x) = x2. Are they equal functions?
#### Answer:
It is given that
f : R → R and g : C → C are two function defined as f (x) = x2 and g (x) = x2 .
Thus,
domain ( f ) = R and domain ( g ) = C .
Since, domain ( f ) ≠ domain ( g ),
f (x) and g (x) are not equal functions.
#### Question 10:
f, g, h are three function defined from R to R as follows:
(i) f(x) = x2
(ii) g(x) = sin x
(iii) h(x) = x2 + 1
Find the range of each function.
#### Answer:
(i) Given:
f (x) = x2
Range of f(x) = R+ (set of all positive integers)
= {y ∈ R| y ≥ 0}
(ii) Given:
g(x) = sin x
Range of g(x) = {y ∈ R : $-$1 ≤ y ≤ 1}
(iii) Given:
h (x) = x2 + 1
Range of h (x) = {y ∈ R : y ≥ 1}
#### Question 11:
Let X = {1, 2, 3, 4} and Y = {1, 5, 9, 11, 15, 16}
Determine which of the following sets are functions from X to Y.
(a) f1 = {(1, 1), (2, 11), (3, 1), (4, 15)}
(b) f2 = {(1, 1), (2, 7), (3, 5)}
(c) f3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
#### Answer:
(a) Given:
f1 = {(1, 1), (2, 11), (3, 1), (4, 15)}
f1 is a function from X to Y.
(b) Given:
f2 = {(1, 1), (2, 7), (3, 5)}
f2 is not a function from X to Y because 2 ∈ X has no image in Y.
(c) Given:
f3 = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}
f3 is not a function from X to Y because 2 ∈ X has two images, 9 and 11, in Y.
#### Question 12:
Let A = (12, 13, 14, 15, 16, 17) and f : A → Z be a function given by
f(x) = highest prime factor of x.
Find range of f.
#### Answer:
Given:
A ={12, 13, 14, 15, 16, 17}
f : A be defined by f (x) = the highest prime factor of x.
f (12) = the highest prime factor of 12 = 3
f (13) = the highest prime factor of 13 = 13
f (14) = the highest prime factor of 14 = 7
f (15) = the highest prime factor of 15 = 5
f (16) = the highest prime factor of 16 = 2
f (17) = the highest prime factor of 17 = 17
The range of f is the set of all f (x), where $x\in A$.nA
Therefore,
range of f = {2, 3, 5, 7, 13, 17}.
#### Question 13:
If f : R → R be defined by f(x) = x2 + 1, then find f−1 [17] and f−1 [−3].
#### Answer:
If f : AB is such that yB, then ${f}^{-1}${ y }={xA: f (x) = y}.
In other words, f -1{ y} is the set of pre - images of y.
Let ${f}^{-1}${17} = x .
Then, f (x) =17 .
x2 +1 = 17
x2 = 17 $-$ 1 = 16
x = ± 4
${f}^{-1}${17} = {$-$ 4,4}
Again,
let ${f}^{-1}${$-$3} = x .
Then, f (x) = $-$ 3
x2 + 1 = $-$ 3
x2$-$$-$ 1 = $-$ 4
$x=\sqrt{-4}$
Clearly, no solution is available in R.
So ${f}^{-1}${- 3} = Φ .
#### Question 14:
Let A = [p, q, r, s] and B = [1, 2, 3]. Which of the following relations from A to B is not a function?
(a) R1 = [(p, 1), (q, 2), (r, 1), (s, 2)]
(b) R2 = [(p, 1), (q, 1), (r, 1), (s, 1)]
(c) R3 = [(p, 1), (q, 2), (p, 2), (s, 3)
(d) R4 = [(p, 2), (q, 3), (r, 2), (s, 2)].
#### Answer:
(c) R3 = [(p, 1), (q, 2), (p, 2), (s, 3)
All the relations in (a), (b) and (d) have a unique image in B for all the elements in A.
R3 is not a function from A to B because p ∈ A has two images, 1 and 2, in B.
Hence, option (c) is not a function.
#### Question 15:
Let A = [9, 10, 11, 12, 13] and let f : A → N be defined by f(n) = the highest prime factor of n. Find the range of f.
#### Answer:
Given:
A ={9, 10, 11, 12, 13}
f : AN be defined by f (n) = the highest prime factor of n.
f (9) = the highest prime factor of 9 = 3
f (10) = the highest prime factor of 10 = 5
f (11) = the highest prime factor of 11 = 11
f (12) = the highest prime factor of 12 = 3
f (13) = the highest prime factor of 13 = 13
The range of f is the set of all f (n), where $n\in A$.
Therefore,
range of f = {3, 5, 11, 13}
#### Question 16:
The function f is defined by$f\left(x\right)=\left\{\begin{array}{cc}{x}^{2},& 0\le x\le 3\\ 3x,& 3\le x\le 10\end{array}\right\$
The relation g is defined by $g\left(x\right)=\left\{\begin{array}{ll}{x}^{2},& 0\le x\le 2\\ 3x,& 2\le x\le 10\end{array}\right\$
Show that f is a function and g is not a function.
#### Answer:
The function f is defined by $f\left(x\right)=\left\{\begin{array}{ll}{x}^{2}& 0⩽x⩽3\\ 3x& 3⩽x⩽10\end{array}\right\$
It is observed that for 0 ≤ x < 3, f (x) = x2 .
3 < x ≤ 10, f (x) = 3x
Also, at x = 3, f(x) = 32 = 9. And
f (x) = 3 × 3 = 9.
That is, at x = 3, f (x) = 9.
Therefore, for 0 ≤ x ≤ 10, the images of f (x) are unique.
Thus, the given relation is a function.
Again,
the relation g is defined as $g\left(x\right)=\left\{\begin{array}{ll}{x}^{2},& 0⩽x⩽2\\ 3x,& 2⩽x⩽10\end{array}\right\$
It can be observed that for x = 2, g(x) = 22 = 4 and also,
g(x) = 3 × 2 = 6.
Hence, 2 in the domain of the relation g corresponds to two different images, i.e. 4 and 6.
Hence, this relation is not a function.
Hence proved.
#### Question 17:
If f(x) = x2, find $\frac{f\left(1.1\right)-f\left(1\right)}{\left(1.1\right)-1}$.
#### Answer:
Given:
f(x) = x2
Therefore,
$\frac{f\left(1.1\right)-f\left(1\right)}{\left(1.1\right)-1}=\frac{{\left(1.1\right)}^{2}-{\left(1\right)}^{2}}{\left(1.1-1\right)}=\frac{1.21-1}{0.1}=\frac{0.21}{0.1}=2.1$
#### Question 18:
Express the function f : XR given by f(x) = x3 + 1 as set of ordered pairs, where X = {−1, 0, 3, 9, 7}. [NCERT EXEMPLAR]
#### Answer:
The function f : XR is defined by f(x) = x3 + 1, where X = {−1, 0, 3, 9, 7}.
Now,
$f\left(-1\right)={\left(-1\right)}^{3}+1=0\phantom{\rule{0ex}{0ex}}f\left(0\right)={0}^{3}+1=1\phantom{\rule{0ex}{0ex}}f\left(3\right)={3}^{3}+1=28\phantom{\rule{0ex}{0ex}}f\left(7\right)={7}^{3}+1=344\phantom{\rule{0ex}{0ex}}f\left(9\right)={9}^{3}+1=730$
So, $f=\left\{\left(x,f\left(x\right)\right):x\in X\right\}=\left\{\left(-1,0\right),\left(0,1\right),\left(3,28\right),\left(7,344\right),\left(9,730\right)\right\}$
View NCERT Solutions for all chapters of Class 11
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Introduction
• In general, the ratio of a number x to a number y is defined as the quotient of the numbers x and y.
• Ratios can be expressed as percentages. To express the value of a ratio as a percentage, we multiply the ratio by 100.
Thus, 4/5 = 0.8 = 80%
Some Important Properties
• If we multiply the numerator and the denominator of a ratio by the same number, the ratio remains unchanged.
That is, a/b = ma/mb
• If we divide the numerator and the denominator of a ratio by the same number, the ratio remains unchanged. Thus,
a/b= (a/m)/(b/m)
• Denominator equation method: The magnitudes of two ratios can be compared by equating the denominators of the two ratios and then checking for the value of the numerator.
• eg. Which is larger: 8/3 or 11/4
• If either or both the terms of a ratio are a surd quantity, then the ratio will never evolve into integral numbers unless the surd quantities are equal. use this principle to spot options in questions having surds.
• If a/b = c/d = e/f = g/h = k then
K = (a+c+e+g) / (b + d + f + h)
Mathematical use of ratio and Proportions
• As a bridge between 3 or more quantities:
If A: B and B:C is given A:C can be found out
• Ratio as a Multiplier
If A: B is 3:1, then the value of B has to be multiplied by 3 to get the value of A.
Calculations
Calculation methods for Ratio comparisons
The Cross Multiplication Method
Two ratios can be compared using the cross-multiplication method as follows:
• Suppose you have to compare 12/17 with 15/19. Then, to test which ratio is higher, cross multiply and compare 12 * 19 and 15 * 17.
If 12 * 19 is bigger the Ratio 12/17 will be bigger. If 15 * 17 is higher, the ratio 15/19 will be higher.
In this case, 15 * 17 being higher, the Ratio 15/19 is higher.
Proportions
When two ratios are equal, then the quantities composing them are said to be proportional. Thus, if a/b = c/d, then a, b, c, d is proportional. This is expressed as
a: b: c: d or a: b = c: d
Variations
1. Direct Proportion (A∝ B)
When it is said that A varies directly as B, you should understand the following implications:
2. Logical implication: When A increases B increases
3. Calculation implication: If A increases by 10%, B will also increase by 10%
4. Graphical implications: See following graph
5. Equation implication: The ratio A/B is constant
1. Inverse Proportion (A∝1/B):
When A varies inversely as B, the following implication arises:
a. Logical implication: When A increases B decreases
b. Calculation implication: If A decreases by 9.09%, B will increase by 10%.
c. Graphical implications: see graph above
d. Equation implication: The product A * B is constant.
Practice Questions:
1. Rs. 5783 is divided among Aman, Balan, and Chintan in such a way that if Rs. 28, Rs. 37 and Rs. 18 are deducted from their respective shares, they have money in the ratio 4:6:9. Find Aman’s share.
a.1228
b. 1245.5
c. 1328
d. 1356
2. If 10 persons can clean 10 floors by 10 mops in 10 days, in how many days can 8 persons clean 8 floors by 8 mops
a. 12 days
b. 10 days
c. 11 days
d. 8 days
3. Rs. 3650 is divided among 4 engineers, 3 MBAs and 5 CAs such that 3 CAs get as much as 2 MBAs and 3 Engineers as much as 2 CAs. Find the share of an MBA.
a. 300
b. 450
c. 475
d. None of these
4. Divide Rs. 1400 into three parts in such a way that half of the first part, one-fourth of the second part and one-eighth of the third part are equal.
a. 300, 600, 500
b. 200, 400, 800
c.100, 400, 1000
d. None of these
5. The ratio of water and milk in a 30-litre mixture is 7: 3. Find the quantity of water to be added to the mixture in order to make this ratio 6:1
a. 30
b. 32
c. 33
d. 35
6. A student appeared in 6 papers. The maximum marks are the same for each paper. His marks in these papers are in the proportion of 5: 6: 7: 8: 9: 10. overall he scored 60%. In how many numbers of papers did he score less than 60% of the maximum marks? (UPSC 2021)
a. 2
b. 3
c. 4
d. 5
7. Out of 120 applications for a post, 70 are males and 80 have a driver license. What is the ratio of between the minimum to maximum number of males having driver's license.
(UPSC 2013)
a.1:2
b.2:3
c.3:7
d.5:7
8. In a rare coin collection, there is one gold coin for every three non-gold coins. 10 more gold coins are added to the collection and the ratio of gold coins to non-gold coins would be 1 : 2. Based on the information; the total number of coins in the collection now becomes.
(UPSC 2013)
a. 90
b. 80
c. 60
d. 50
9. The ratio of a two-digit natural number to a number formed by reversing its digits is 4:7. The number of such pairs is (UPSC 2019)
a. 5
b. 4
c. 3
d. 2
10. The monthly incomes of X and Y are in the ratio of 4 : 3 and their monthly expenses are in the ratio of 3: 2. However, each saves Rs. 6,000 per month. What is their total monthly income?
(UPSC 2017)
a. 28,000
b. 42,000
c. 56,000
d. 84,000
11. The monthly incomes of Peter and Paul are in the ratio of 4:3. Their expenses are in the ratio of 3:2. If each saves Rs. 6,000 at the end of the month, their monthly incomes respectively are (in Rs.) (UPSC 2015)
a. 24,000 and 18,000
b. 28000 and 21000
c. 32,000 and 24,000
d. 34,000 and 26,000
12. A sum of Rs. 2,500 is distributed among X, Y and Z in two ratio 1/2 ∶ 3/4 ∶ 5/6. What is the difference between the maximum share and the minimum share? (UPSC 2020)
a. Rs 300
b. Rs. 350
c. Rs. 400
d. Rs. 450
13. A 6 student appeared in 6 papers. The maximum marks are the same for each paper. His marks in these are in the proportion of 5:6:7: 8:9:10 overall he scored 60%. In how many numbers of papers did he score less than 60% of the maximum marks?
(UPSC 2021)
a. 2
b. 3
c. 4
d. 5
14. Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio of increased seats?
a. 2 : 3 : 4
b. 6 : 7 : 8
c. 6 : 8 : 9
d. None of these
15. The ratio of the number of boys and girls in a college is 7: 8. If the percentage increase in the number of boys and girls be 20% and 10% respectively, what will be the new ratio?
a. 8: 9
b. 17: 18
c. 21: 22
d. Cannot be determined
16. If x varies as y, and y = 4 when x =12, find x when y = 15.
a. 45
b. 54
c. 70
d. 15
17. The ratio of water and milk in a 30-litre mixture is 7:3. Find the quantity of water to be added to the mixture in order to make this ratio 6:1.
a. 6
b. 4
c. 5
d. 7
18. Rahul has coins of the denomination of ₹ 1, 50 p and 25 p in the ratio of 12:10:7. The total worth of the coins he has is ₹ 112.5. Find the number of 25 p coins that Rahul has
a. 48
b. 72
c. 60
d. 42
19. Four numbers in the ratio 1:2:4:8 add up to give a sum of 120. Find the value of the biggest number.
a. 40
b. 30
c. 64
d. 60
20. The ratio between two numbers is 7:11 and their LCM is 154. The first number is:
a. 14
b. 7
c. 22
d. 32
21. A fort has provisions for 60 days. If after 15 days 500 men strengthen them and the food lasts 40 days longer, how many men are there in the fort?
a. 3500
b. 4000
c. 6000
d. None of these
22. The ratio of the age of a man and his wife is 4 : 3. After 4 years, this ratio will be 9 : 7. If at the time of the marriage, the ratio was 5:3, then how many years ago they were married?
a. 8 years
b. 12 years
c. 10 years
d. 15 years
23. 60 litres of diesel is required to travel 600 km using a 800 cc engine. If the volume of diesel required to cover a distance varies directly as the capacity of the engine, then how many litres of diesel is required to travel 800 km using 1200 cc engine?
a. 80 litres
b. 90 litres
c. 120 litres
d. 170 litres
24. If 20 men or 24 women or 40 boys can do a task in 12 days working for 8 hours a day, how many men working with 6 women and 2 boys take to do a task four times as big working for 5 hours a day for 12 days?
a. 8 men
b. 2 men
c. 122 men
d. 24 men
25. If ₹ 232 is divided among 150 children such that each girl and each boy gets ₹1 and ₹2 respectively. Then hoe many girls are there?
a. 52
b. 54
c. 68
d. 62
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# Fraction (mathematics)
File:Cake-quarters.jpg
A cake divided into four equal quarters. Each fraction of the cake is represented numerically as 14. It can be seen that two quarters (2 x 14 = 24) is equivalent to half (12) the cake.
In mathematics, a fraction is a way of expressing a quantity based on an amount that is divided into a number of equal-sized parts. For example, each part of a cake split into four equal parts is called a quarter (and represented numerically as 14); two quarters is half the cake, and eight quarters would make two cakes.
Mathematically, a fraction is a quotient of numbers, like 34, or more generally, an element of a quotient field.
In our cake example above, where a quarter is represented numerically as 14, the bottom number, called the denominator, is the total number of equal parts making up the cake as a whole, and the top number, called the numerator, is the number of these parts we have. For example, the fraction 34 represents three quarters. The numerator and denominator may be separated by a slanting line, or may be written above and below a horizontal line.
The numerator and denominator are the "terms" of the fraction. The word "numerator" is related to the word "enumerate," meaning to "tell how many"; thus the numerator tells us how many parts we have in the indicated fraction. To denominate means to "give a name" or "tell what kind"; thus the denominator tells us what kind of parts we have (halves, thirds, fourths, etc.). Note that because it is impossible to divide something into zero equal parts, zero can never be the denominator of a fraction.
The word is also used in related expressions, like continued fraction and algebraic fraction, see Special cases below.
## Forms of fractions
### Proper and improper fractions
If the numerator and denominator of a fraction are both positive, then the fraction is a proper fraction if the numerator is less than the denominator, but an improper fraction otherwise. If either the numerator or denominator (or both) are negative, their absolute values should be compared to determine whether the fraction is proper or improper.
### Mixed numbers
A mixed number is the sum of a whole number and a proper fraction. For instance, you could have two entire cakes and three quarters of another cake. The whole and fractional parts of the number are written right next to each other: 2 + 34 = 234.
An improper fraction can be thought of as another way to write a mixed number; in the "234" example above, imagine that the two entire cakes are each divided into quarters. Each entire cake contributes 44 to the total, so 44 + 44 + 34 = 114 is another way of writing 234.
A mixed number can be converted to an improper fraction in three steps:
1. Multiply the whole part times the denominator of the fractional part.
2. Add the numerator of the fractional part to that product.
3. The resulting sum is the numerator of the new (improper) fraction, and the new denominator is the same as that of the mixed number.
Similarly, an improper fraction can be converted to a mixed number:
1. Divide the numerator by the denominator.
2. The quotient (without remainder) becomes the whole part and the remainder becomes the numerator of the fractional part.
3. The new denominator is the same as that of the original improper fraction.
### Equivalent fractions
Multiplying the numerator and denominator of a fraction by the same (non-zero) number results in a new fraction that is said to be equivalent to the original fraction.1 The word equivalent means that the two fractions have the same value. For instance, consider the fraction 12. When the numerator and denominator are both multiplied by 2, the result is 24, which has the same value as 12. To see this, imagine cutting the example cake into four pieces; two of the pieces together (24) make up half the cake (12).
We can say, for example, that 13, 26, 39, and 100300 are all equivalent fractions.
Dividing the numerator and denominator of a fraction by the same non-zero number will also yield an equivalent fraction. We call this reducing the fraction. A fraction in which the numerator and denominator have no factors in common (other than 1) is said to be irreducible or in lowest terms. For instance, 39 is not in lowest terms because both 3 and 9 can be evenly divided by 3. In contrast, 38 is in lowest terms — the only number that's a factor of both 3 and 8 is 1.
### Reciprocals and the "invisible denominator"
The reciprocal of a fraction is another fraction with the numerator and denominator swapped. The reciprocal of 37, for instance, is 73.
Because any number divided by 1 results in the same number, it is possible to write any whole number as a fraction by using 1 as the denominator: 17 = 171. (We sometimes call the number 1 the "invisible denominator.") Therefore, we can say that, except for zero, every fraction or whole number has a reciprocal. The reciprocal of 17 would be 117.
## Arithmetic with fractions
Fractions, like whole numbers, obey the commutative, associative, and distributive laws, and the rule against division by zero.
Adding fractions can be a little tricky, since you cannot simply add the numerators and denominators. For example, if we had a cake divided into three pieces, each piece would be 1/3. Then, if we try to add one piece, 1/4, from the cake divided into four pieces, and one piece, 1/3, from the cake divided into three pieces, what would we have? Well, would we have 1/4 + 1/3 = ??? You can see this is NOT equal to 1/7 or 2/7 !!
To add fractions together, they must be changed to equivalent values having the same fractional unit -- the same denominator -- in this case 1/12. How do we do this? By multiplying each fraction by 1. By one? Yes. 1 = 3/3 and 1 = 4/4. Now watch: 1/4 = 1/4 x 1 = 1/4 x 3/3 = 3/12. And 1/3 = 1/3 x 1 = 1/3 x 4/4 = 4/12. So now 1/4 + 1/3 = 3/12 + 4/12 = 7/12 and we have the correct result.
Notice that we only add the numerators together. The denominator does not change, since we are working with the same fractional unit. Another way to see this is: 1/4 + 1/3 = 3/12 + 4/12 = 1/12 x (3 + 4) = 1/12 x 7 = 7/12.
Lets take another example. If you add a half dollar to a quarter, what will you get? You know it's 75 cents, right? When we say 75 cents we have automatically, in our mind, converted each coin into cents (pennies): One half dollar = 50 cents; one quarter = 25 cents; so 1/2 + 1/4 = 50/100 + 25/100 = 75/100 or 75 cents. Of course, we could use a smaller denominator since we know one half dollar equals two quarters. I.e., 1/2 + 1/4 = 2/4 + 1/4 = 3/4. In words, one half plus one quarter equals two quarters plus one quarter equals three quarters, or 75 cents.
So the trick is to find a common fractional unit -- a common denominator -- that will let us simply add the numerators together. Let's take one more example. Find 2/3 + 1/2. We see that the denominators are 3 and 2. We need to find a value that each denominator can be multiplied by to give a common value. Well, it's easy to see that we can multiply 3 by 2, and 2 by 3, to give a common denominator of 6. But remember, you cannot change the value of each fraction, so we must multiply both numerator and denominator by the same number. We now have: 2/3 + 1/2 = 2/2 x 2/3 + 3/3 x 1/2 = 4/6 + 3/6 = 7/6 or 1 + 1/6.
When doing arithmetic with fractions, results should usually be expressed in lowest terms. For instance,
16 + 13 = 16 + 26 = 36 = 12
Note that 36 is not an incorrect answer, because 36 and 12 are equivalent, but the reduced form is preferred, and classroom exercises will nearly always require that final fractional answers to problems be reduced.
#### Subtracting fractions
The process for subtracting fractions is, in essence, the same as that of adding them: find a common denominator, and change each fraction to an equivalent fraction with the chosen common denominator. The resulting fraction will have that denominator, and its numerator will be the result of subtracting the numerators of the original fractions. For instance, 2/3 - 1/2 = 2/2 × 2/3 - 3/3 × 1/2 = 4/6 - 3/6 = 1/6.
### Multiplication and division
#### Multiplication
##### By whole numbers
If you consider the cake example above, if you have a quarter of the cake, and you multiple the amount by three, then you end up with three quarters. We can write this numerically as follows:
$3\times {1 \over 4}={3 \over 4}$
As another example, suppose that five people work for three hours out of a seven hour day (ie. for three seventh of the work day). In total, they will have worked for 15 hours (5 x 3 hours each), or 15 sevenths of a day. Since 7 seventh of a day is a whole day, 14 sevenths is two days, then in total, they will have worked for 2 days and a seventh of day. Numerically:
$5\times {3 \over 7}={15 \over 7}=2{1 \over 7}$
##### By fractions
If you consider the cake example above, if you have a quarter of the cake, and you multiple the amount by a third, then you end up with a twelfth of the cake. In other words, a third of a quarter (or a third times a quarter), is a twelfth. Why? Because we are splitting each quarter into three pieces, and four quarters times three makes 12 parts (or twelfths). We can write this numerically as follows:
${1 \over 3}\times {1 \over 4}={1 \over 12}$
As another example, suppose that five people do an equal amount work that totals three hours out of a seven hour day. Each person will have done a fifth of the work, so they will have worked for a fifth of three sevenths of a day. Numerically:
${1 \over 5}\times {3 \over 7}={3 \over 35}$
##### General rule
You may have noticed that when we multiply fractions, we simply multiply the two numerators (the top numbers), and multiply the two denominators) (the bottom numbers). For example:
${5 \over 6}\times {7 \over 8}={5\times 7 \over 6\times 8}={35 \over 48}$
##### By mixed numbers
When multiplying mixed numbers, it's best to convert the whole part of the mixed number into a fraction. For example:
$3\times 2{3 \over 4}=3\times \left({{8 \over 4}+{3 \over 4}}\right)=3\times {11 \over 4}={33 \over 4}=8{1 \over 4}$
In other words, $2{3 \over 4}$ is the same as $\left({{8 \over 4}+{3 \over 4}}\right)$, making 11 quarters in total (because 2 cakes, each split into quarters makes 8 quarters total). And 33 quarters is $8{1 \over 4}$ since 8 cakes, each made of quarters, is 32 quarters in total.
#### Division
To divide by a fraction, simply multiply by the reciprocal of that fraction.
$5\div {1 \over 2}=5\times {2 \over 1}=5\times 2=10$
${2 \over 3}\div {2 \over 5}={2 \over 3}\times {5 \over 2}={10 \over 6}={5 \over 3}$
About 4,000 years ago Egyptians divided with fractions using slightly different methods. Egyptians used the least common multiple technique to divide unit fractions. Examples can be found at
## Special cases
A vulgar fraction (or common fraction) is a rational number written as one integer (the numerator) divided by a non-zero integer (the denominator), for example 4/3 as opposed to 11/3. The line that separates the numerator and the denominator is called the vinculum if it is horizontal, a solidus if it is slanting. A unit fraction is a vulgar fraction with a numerator of 1 (1/7). An Egyptian fraction is the sum of distinct unit fractions (1/3+1/5). A decimal fraction is a vulgar fraction where the denominator is a power of 10 (4/100). A dyadic fraction is a vulgar fraction in which the denominator is a power of two (1/8). A compound fraction is a fraction where the numerator or denominator (or both) contain fractions, ${\frac {2}{3}}{\Bigg /}{\frac {1}{5}}$, these can be simplified to give vulgar fractions.
An expression that has the form of a fraction, but actually represents division by or into an irrational number might be called an "irrational fraction" (an oxymoron). A common example is π/2, the radian measure of a right angle.
Rational numbers are the quotient field of integers. Rational functions are functions evaluated in the form of a fraction, where the numerator and denominator are polynomials. These rational expressions are the quotient field of the polynomials (over some integral domain).
A continued fraction is an expression such as $a_{0}+{\frac {1}{a_{1}+{\frac {1}{a_{2}+...}}}}$, where the ai are integers. This is not an element of a quotient field.
The term partial fraction is used in algebra, when decomposing rational functions. The goal of the method of partial fractions is to write rational functions as sums of other rational functions with denominators of lesser degree.
## Pedagogical tools
In Primary Schools, fractions have been demonstrated through Cuisenaire rods.
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# How Many Plants Do You Need for Your Space?
## How Many Plants Do You Need?
Photo by Thinkstock
How many flowers or plants do you need for your garden bed or space?
Avoid the nursery and find out that you bought too many flowers or plants—or, discover that your number fell short of the look you want in the garden space. We consulted the New York Botanical Garden in the Bronx. This is his formula for estimating the number of plants you need. Obviously, it depends on the type of plant but this gives you a good guidelines for garden design as a starting place.
### Steps for Calculating the Number of Plants
1. Determine the number of square feet in the area to be planted:
• Rectangle: Multiply length by width
• Circle: Multiply the radius by itself, and that result by pi (3.14)
• Oval: Multiply the average radius by itself, and that result by pi (3.14)
• Triangle: Multiply ½ the height by the base
2. Determine the number of square inches in the area to be planted by multiplying the number of square feet by 144 (which is the number of square inches in one square foot).
3. Determine the number of square inches a mature plant will cover by multiplying the number of inches of suggested spacing between plants by itself.
4. Divide the number of square inches in the plot by the number of square inches required for one plant. This result is the total number of plants needed for that plot.
#### Example A:
Step 1. The area to be planted is a 6x8-foot rectangle. 6 feet x 8 feet = 48 square feet.
Step 2. 48 square feet x 144 square inches = 6,912 square inches.
Step 3. The suggested spacing for wax begonia (Begonia semperflorens-cultorum) is 8 to 10 inches.
• At 8-inch centers, 8 inches x 8 inches = 64 square inches for each plant.
• At 9-inch centers, 9 inches x 9 inches = 81 square inches for each plant.
• At 10-inch centers, 10 inches x 10 inches = 100 square inches for each plant.
Note: The middle of the suggested range (9 inches) is usually recommended, but for quicker coverage, the low range (8 inches) can be used. The higher range (10 inches) is usually not recommended. If all plants do not grow, coverage will not be complete.
Step 4.
• At 8-inch centers, 6,912 square inches ÷ 64 square inches = 108 plants.
• At 9-inch centers, 6,912 square inches ÷ 81 square inches = 85.3 or 86 plants.
• At 10-inch centers, 6,912 square inches ÷ 100 square inches = 69.1 or 70 plants.
#### Example B:
Step 1. The area to be planted is a 4x4-foot square. 4 feet x 4 feet = 16 square feet.
Step 2. 16 square feet x 144 square inches = 2,304 square inches.
Step 3. The suggested spacing for French marigold (Tagetes patula) is 12 to 15 inches.
• At 12-inch centers, 12 inches x 12 inches = 144 square inches for each plant.
• At 13-inch centers, 13 inches x 13 inches = 169 square inches for each plant.
• At 14-inch centers, 14 inches x 14 inches = 196 square inches for each plant.
• At 15-inch centers, 15 inches x 15 inches = 225 square inches for each plant.
Step 4.
• At 12-inch centers, 2,304 square inches ÷ 144 square inches = 16 plants.
• At 13-inch centers, 2,304 square inches ÷ 169 square inches = 13.6 or 14 plants.
• At 14-inch centers, 2,304 square inches ÷ 196 square inches = 11.8 or 12 plants.At 15-inch centers, 2,304 square inches ÷ 225 square inches = 10.2 or 11 plants.
Note: If your calculations come out to a fraction of a plant, such as 40.4 or 80.7, always use the next whole number. It doesn’t sound like a big difference, but that one extra plant, not noticeable when first planted, may make the difference as to whether or not your design fills out.
If you are growing a vegetable garden, our online Garden Planner will automatically calculate the spacing by vegetable plant for you! Technology is indeed amazing. Explore the Almanac Garden Planner.
## Source:
2001 Early Spring Garden Guide. This article was originally published in 2011 and has been updated.
### Hi, I took up gardening this
Hi, I took up gardening this year and successor planted a few seeds. I build a 4x4 raised bed but I'm afraid I overcrowded my veggies and they are growing bigger everyday day. I am building a bigger raised bed and moving some of the plants to it. I have tomato plants, eggplant and string bean plants all about 8-9 inches in heights. Is is safe for me to move them to a different bed? I don't think I have much of a choice because my zuchinni and cucumber leaves are blocking the sun from the plants mentioned above. Thank you.
### A 14foot triangular bed - I
A 14foot triangular bed - I want to plant bush roses which mature at 4ft spread - how any roses to plant?
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Combinatorics is an area of discrete mathematics that studies collections of distinct objects and the ways that they can be counted or ordered, or used to satisfy some optimality criterion.
The most basic ideas in combinatorics include:
factorials
The number of possible arrangements of ${\displaystyle n}$ distinct items is n-factorial, written ${\displaystyle n!}$ , which equals
${\displaystyle n\times(n-1)\times(n-2)\times\cdots\times2\times1}$
• Example: Three items, A, B, and C, can be arranged in ${\displaystyle 3! = 3 \times 2 \times 1 = 6}$ different ways: ABC, ACB, BAC, BCA, CAB, and CBA.
permutations
The number of arrangements that are possible when a subset of ${\displaystyle k}$ items is taken from a set of ${\displaystyle n}$ distinct items is a "permutation of ${\displaystyle n}$ objects taken ${\displaystyle k}$ at a time", which can be written as ${\displaystyle P^n_k}$ or ${\displaystyle {}_nP_k}$ , and is equal to ${\displaystyle \frac{n!}{(n-k)!}}$ .
• Example: The number of possible arrangements of the four letters A, B, C, D, taken two at a time, is ${\displaystyle P^4_3=\frac{4!}{(4-2)!}=\frac{24}{2}=12}$: AB, BA, AC, CA, AD, DA, BC, CB, BD, DB, CD, and DC.
combinations
The number of possible subsets of ${\displaystyle k}$ items taken from a set of ${\displaystyle n}$ items, where the order of the items doesn't matter (e.g., the sets ABC and BCA are considered equivalent), is a "combination of ${\displaystyle n}$ objects taken ${\displaystyle k}$ at a time", which is written ${\displaystyle C^n_k}$ or ${\displaystyle {}_nC_k}$ or ${\displaystyle \binom{n}{k}}$ , and is equal to ${\displaystyle \frac{n!}{k!(n-k)!}}$ .
• Example: The number of subsets of two letters chosen from the four letters A, B, C, and D, is ${\displaystyle C^4_2=\frac{4!}{2!(4-2)!}=\frac{24}{2\times2}=6}$ : AB, AC, AD, BC, BD, and CD.
distributions
partitions of integers or of sets
recurrence relations
inclusions
inversions
inclusion/exclusion principle
derangements and subfactorials
repetitions and replacements
various restrictions placed on problems
fundamental counting principle
circular permutations
generating functions
free and fixed permutations, rotational symmetry and reflective symmetry
cyclic permutations
multisets
Pascal's Triangle
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A Key to Understanding Math
I've tutored in math, science, history and English; of these, I love math the most. One of my nephews recently needed help in working a "word problem" on his math homework. As we discussed the problem, it became apparent to me that he has been doing most of his math mechanically without truly understanding the core concepts underlying the problems that he is working, let alone correctly associating the meaning of words in the context of applying the principles of math. In my experience, this is far too common a problem, and one that is easily addressed with some remedial instruction.
One thing that I learned early on is that the primary problem in learning how to solve "word problems" isn't the math—the calculations—itself; it's understanding the language, the words, that's used to express it. Word problems are just another way of describing facts and the relationships between them. With a solid primary school education, the arithmetic calculations are not the challenge; the challenge is knowing how to take ordinary, everyday situations and express them in the language necessary to describe them in a way that will allow the person to use the math that they know. The key to success in math is understating that math is a language that describes relationships, a language with it's own unique terminology and contexts. It's when we understand the concepts as applied to concrete examples that we can extend those concepts to abstract ideas.
For example, consider the following word problem:
Bob has four apples. Carol has six apples. Ted has twice as many apples as do Bob and Carol combined. How many apples does Ted have?
Most of us can easily perform the calculations necessary to answer this question. Let's step through this.
In this problem, there are several key terms that indicate the relationships between the facts:
• twice — two times
• as many as — multiplication
If we were to exchange these words for their respective operations, we'd have:
Combined: Bob's 4 apples plus Carol's 6 apples, which equals 10 apples
Twice, as many as, combined: 2 times 10 apples, which equals 20 apples
Expressed in an sentence, the calculation is: "Twice as many apples as the sum of four apples and six apples," or, "Two times the sum of four apples and six apples." As an equation, which many of us can solve, this is 2 x (4 apples + 6 apples). Using the correct Order of Operations (PEMDAS or BEDMAS, whichever is easier for you to remember), we get:
$$2\times(4\text{ apples}+6\text{ apples})\\= 2\times10\text{ apples}\\= 20\text{ apples}$$
Ted has 20 apples, which is twice as many as apples as Bob and Carol have combined.
In explaining this problem to my nephew, I had to break it down into the smallest pieces, explaining the mathematical meaning of the key terms and the relationships described by them, translating them into operations and calculations. I had to use several different approaches with examples until I saw the light come on in his eyes; I was finally successful in conveying the knowledge through using wintergreen peppermints as counters as we talked through the problem, giving him a visual, concrete example of what the terms and amounts mean in the "real" world. His challenge wasn't understanding how to perform the underlying mathematical calculations, it was his lack of understanding the relationships described in concrete terms and translating these relationships into mathematical calculations that confused him.
My nephew also had a word problem in which he had to calculate the area of two differently sized rectangles that shared a side. Helping him thought it, I thought that rather than duplicating it here, we could make this problem more real and applicable to our lives. Here's an example of a situation that many of us have faced:
Alice wants to refresh the look of her dining room. With the help of a friend, she's going to paint the old, faded 1960's paneling antique white, visually opening up the space. She needs to know how much it will cost to do this, but she doesn't know how much paint she needs. She'd like our help determining how much this remodeling will cost.
I'm glad that Alice came to us. There are several facts that she needs to gather and mathematical formulas—and remember, that means understanding the relationships between the facts—that she needs to know to determine the total cost. First, she needs to find out how much paint will be required to cover the room. In order to do so, she needs to measure the room to calculate the total surface area to be painted. She needs to know how much surface area that a given amount of paint will cover. With this and the cost of paint per gallon, she can calculate the cost of the paint. Finally, she needs to add the cost of the drop cloth, as well as the brushes and trays necessary for her and her friend.
To find the total surface area, Alice needs to calculate the surface area of each surface to be painted—four walls and the ceiling—knowing that the area of a flat surface is its length times its width. If there are any windows that she's not intending on painting, she needs to find their surface area and subtract that from the surface area of the walls. If the windows are small and few in number and she doesn't want to do the extra math, she probably doesn't need to consider them: the effect on the required amount of paint will be negligible.
Since the room is a regular rectangle, she can either add the five surface areas:
\begin{align}\text{Ceiling area}\\+\:\text{Long wall area}\\+\:\text{Long wall area}\\+\:\text{Short wall area}\\+\:\text{Short wall area}\end{align}
or she can add the one surface area of the ceiling to twice the surface area of each of the long and short walls:
\begin{align}\text{Ceiling area}\\+\:2\times(\text{Long wall area})\\+\:2\times(\text{Short wall area})\end{align}
To ease the display of the calculations, we'll use the slightly more complex later form.
Upon measuring, Alice finds that her dining room is 20 feet long, 16 feet deep and 12 feet high. The room has six windows but they are relatively small, so she isn't including them in her calculations.
Let's break this down. The ceiling, covering the length and width of the room, is 20' long by 16' wide. The long walls are 20' across by 12' high. The short walls are 16' across by 12' high. This is:
\begin{align}20\text{ ft}\times16\text{ ft}\\+\:2\times(20\text{ ft}\times12\text{ ft})\\+\:2\times(16\text{ ft}\times12\text{ ft})\end{align}
To find the necessary number of gallons of paint needed, Alice needs to know that the paint will cover so many square feet per gallon, dividing the surface area to be painted by the coverage rate of the paint.
Alice calls the local hardware store and is told that her desired brand of paint costs 12 per gallon with a coverage rate of 350 sq ft per gallon. As previously noted, Alice will need to divide the surface area of the walls by the coverage rate of the paint to find how many gallons that she will need: \begin{align}((20\text{ ft}\times16\text{ ft})\\+2\times(20\text{ ft}\times12\text{ ft})\\+2\times(16\text{ ft}\times12\text{ ft}))\\\div350\text{ ft}^2\text{/gal}\end{align} We can also display this as a fraction: $$\frac{20\text{ ft}\times16\text{ ft}+2\times(20\text{ ft}\times12\text{ ft})+2\times(16\text{ ft}\times12\text{ ft})}{350\text{ ft}^2\text{/gal}}$$ Now that Alice can determine exactly how much paint this will require, she has to multiply the number of gallons of paint required by the cost per gallon to find the cost of the paint. Finally, Alice needs to add the cost of the required tools: a paint tray with a disposable liner; two roller brushes, one with an extension handle (only one because she wants to save as much money as possible); two hand brushes; a paint tray; and a drop cloth to protect the floor. Alice prices the tools as follows: • Paint tray —6
• Disposable tray liner — $3 • Roller brush —$8
• Extension handle — $12 • Hand brush —$7
• Drop cloth — $4 To recap, Alice's dining room is 20' x 16' x 12'. She has to paint every surface except the floor. The brand of paint that she wants costs$12 per gallon and covers about 350 sq ft per gallon. She needs the above listed tools.
The formula that represents this is: the sum of the surface areas of the roof and four walls, divided by the coverage rate of the paint, times the cost of the paint per gallon, plus the cost of the tools.
Being careful in the use of parentheses, the complete formula to determine the cost of painting Alice's dining room is:
$$\frac{( \text{Ceiling area})+2\times(\text{Long wall area})+2\times(\text{Short wall area)}}{(\text{Coverage rate})}\\\times(\text{Cost per gallon})\\+(\text{Cost of tools})$$
With this information, Alice can calculate the cost of painting painting her dining room. Plugging in the values, we get:
$$\frac{(20\text{ ft}\times16\text{ ft})+2(20\text{ ft}\times12\text{ ft})+2(16\text{ ft}\times12\text{ ft})}{350\text{ ft}^2\text{/gal}}\\\times(12\text{/gal})\\+(6+3+2\times8+12+2\times7+4)$$
Remembering the Order of Operations, now it's time to perform the calculations. (From this point on, I'll only list the major steps, with the assumption that the basic arithmetic—the addition, subtraction, multiplication and division of whole numbers, fractions and decimals—is understood).
$$= \frac{320\text{ ft}^2+2\times240\text{ ft}^2+2\times192\text{ ft}^2}{350\text{ ft}^2\text{/gal}}\\\times(12\text{/gal})\\+(6+3+16+12+14+4)$$
Recall that
$$1\text{ ft}\times1\text{ ft} = 1\text{ sq ft} = 1\text{ ft}^2$$
We don't want to make any mistakes on the units!
$$= \frac{320\text{ ft}^2+480\text{ ft}^2+384\text{ ft}^2}{350\text{ ft}^2\text{/gal}}\times(12\text{/gal})+55\\= \frac{1184\text{ ft}^2}{350\text{ ft}^2\text{/gal}}\times(12\text{/gal})+55$$
Since
$$\frac{1184\text{ ft}^2}{350\text{ ft}^2\text{/gal}}=1184\text{ ft}^2\times\frac{1\text{ gal}}{350\text{ ft}^2} \approx 3.38\text{ gal}$$
and the smallest quantity in which her chosen brand of paint is sold is a gallon, 4 gallons of paint are required. We input 4 into the equation in the place of 3.38.
$$= 4\text{ gal}\times(12\text{/gal})+55\\= 48+55\\= 103$$
It will cost Alice approximately $103 to paint her dining room, not including sales tax. Alice is grateful to know how much painting her dining room will cost. After thinking about it, she wonders how much of a difference that not painting the windows will make in the amount of paint required. Since she really needs only 3.38 gallons, she wants to know how much paint that subtracting the surface area of the windows reduces from the required amount. She'd love to save the$12 on that fourth gallon of paint if at all possible. She measures the windows at 6 feet by 3 feet.
We already know how to calculate the area of the windows. Remembering from earlier that Alice's dining room has six windows and plugging the values into the formula that we used earlier, we find that
$$6\times(6\text{ ft}\times3\text{ ft}) = 108\text{ ft}^2$$
which, when divided by the coverage rate of the paint:
$$108\text{ ft}^2\div(350\text{ ft}^2\text{/gal})\approx0.31\text{ gal}$$
reveals that she needs just over 3 gallons of paint (3.07 to be exact).
Given the fact that coverage rates are estimates that depend on how heavily the paint is applied in each coat, she might be able to get away with buying only 3 gallons, saving that \$12; however, she just might run short. If this was a custom color of paint, she'd have a tougher decision to make because of the difficultly of matching the original mix, but she did choose an "off the shelf" color, so picking up another gallon if necessary is an option. While I'd recommend getting the fourth gallon up front and returning it if it's not needed, the choice is hers.
If you've stayed with me through this entire post, thanks! I've demonstrated that a problem with Word Problems isn't understanding the "math"; rather, it's understanding the words—the language and concepts—as describing relationships between the relevant facts.
In summary, if we're struggling with math, the first step is to know and understand how to apply the basic mathematical concepts to our situation, and then how to translate those relationships into calculations.
If you're in need of a tutor in math or any subject, let me know. I can teach math from basic arithmetic through algebra, geometry and trigonometry. I'm available in the evenings and on the weekends. My price depends on the market and the distance driven, and I'll work with you according to your situation.
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# Multiplying fractions
Home -> Tutorials -> Multiplying fractions
A fraction represents a part of a whole. It is made up of an integer numerator (the top number) and a non-zero integer denominator (the bottom number).
## To multiply two fractions:
1. Multiply the numerators.
2. Multiply the denominators.
3. Reduce the resultant fraction to its lowest terms.
For all real numbers a, b, c, d (b≠0,d≠0)
### Example1:
Find: $\frac{1}{2} \times \frac{2}{5}$
### Solution:
Step1:
Multiply the numerators and denominators
$\frac{1}{2} \times \frac{2}{5} = \frac{1 \times 2}{2 \times 5} = \frac{2}{10}$
Step2:
Simplify the fraction
$\frac{2}{10} = \frac{1}{5}$
Answer: $\frac{1}{5}$
### Example2:
Find: $\frac{8}{9} \times \frac{3}{4}$
### Solution:
Step1:
Multiply the numerators and denominators
$\frac{8}{9} \times \frac{3}{4} = \frac{8 \times 3}{9 \times 4}$
Step2:
Divide out common factors
$\frac{8 \times 3}{9 \times 4} = \frac{ 2 \times 1}{3 \times 1}$
Step3:
Simplify the fraction
$\frac{ 2 \times 1}{3 \times 1} = \frac{2}{3}$
Answer: $\frac{2}{3}$
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# Dot Product – Definition, Types, Properties, and Solved Examples
Finding solutions to certain problems is not always with a simple multiplication of two or more terms. The difference arises when a term represents more than just a number.
Vectors are such terms that represent two factors, and not just one numerical. They are magnitude and direction.
So, to multiple vectors, there are broadly two methods to go about. The first is the Dot Product that we will discuss below, and the second is the Cross Product.
## What is a Dot Product?
The name Dot Product is because of the ‘.’ present between two terms instead of the usual ‘x’ sign. They often represent the vector lengths as their magnitudes and the angles between those magnitudes.
## Definition
It can be defined under two separate headings, applicable in 2 different ways: Algebraic and Geometric. The method that follows is significantly different
## Algebraic Dot Product
It is the simpler one of the two. An Algebraic dot product can be defined as the addition of the two terms. Each term can itself be a solution to an algebraic operation.
## Geometric Dot Product
A Geometric Dot Product is the product of two vector magnitudes and the angle between them, cosine.
## Properties
Property 1: Algebraic Dot Product = Geometric Dot Product in the final answer you get.
Property 2: If the angle between the two terms is 0°, then the cosine value is 1. This implies that the terms are parallel to each other.
On the other hand, if the angle between the two terms is 90°, the cosine value is 1/2. This implies that the terms are perpendicular to each other.
Property 3: Since dot product is a scalar product, therefore, (pa).(qb)=(pb).(qa)=pq a.b. Here p and q represent the length of the two terms a and b from the point of origin.
Property 4: If the dot product between one vector and with itself, for example, a with a or b with b is calculated, then a.a = a.a cos 0 = a2 or b.b = b.b cos0
Property 5: Both Algebraic and Geometric abide by the distributive law – a.(b + c) = a.b + a.c
The following two properties are based on coordinates of the x-axis, y-axis, and z-axis, represented as i, j, and k.
## Formula
Since they are defined on different principles, their formulae are also different. While one is a summation, the other is a mixed expression that involves incorporating the angle.
## Algebraic Dot Product
The Algebraic Dot Product is represented as –
a · b = ax × bx + ay × by
Where a is the first term
b is the second term
Here, ax and ay are the two different values of the same term, but in different dimensions of the x-axis and y-axis, respectively. For bx and by, too, the expressions represent the values in the two different dimensions.
## Geometric Dot Product
The Geometric Dot Product is represented as –
a · b = |a| × |b| × cos(θ)
Where a and b are the two terms.
θ is the angle between the two terms.
Unlike the Algebraic Dot Product, the Geometric Dot Product does not have a two-dimensional representation; only one value for one term. However, the addition here is the inclusion of the cosine angle between the two terms.
## Solved Examples
### Algebraic Dot Product
Example: The coordinates for the x and y for the vectors terms a and b are given as – (-2,6) and (3,12)
Solution: Substituting the given values in the formula of a · b = ax × bx + ay × by
We get –
a . b = (-2 × 3) + (6 × 12)
a . b = (-6) + (72)
a . b = -6 + 96
a . b = 90
So, the dot product of the terms a and b by the Algebraic Dot Product Method is 90.
### Geometric Dot Product
Example: The angle is 90°. Along with it, the given a and b values are 19 and 16.
Solution: Substitute the given values in the formula of a . b = |a| x |b| x cos (θ)
We get –
a . b = |19| x |16| x cos (90)
a . b = |19| x |16| x 1/2
a . b = 19 × 16 × 1/2
a . b = 152
So, the dot product of the terms a and b by the Geometric Dot Product Method is 304.
## Conclusion
It is essential to understand the difference between what is a dot product, its properties, and its formulae to differentiate it from other product-finding methods. Interchanging of dot and cross products is not possible as they are very different in their characteristics.
That’s it for this post. If you like this article, share it if you like, like it if you share it. You can also find us on Mix, Twitter, Pinterest, and Facebook. Hey man, If you have come this far, do give us feedback in the comment section. It would make my day. You can also make a donation. Your donations will help us to run our website and serve you BETTER. Cheers!!!
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• Graphing Linear Equations and Inequalities
At its most basic level, a linear equation is a function with variables and constants. These variables include x, y, or often times both.
There are a handful of ways to write linear equations:
• Point-slope form
• Slope-intercept form
• Standard form
The type of form determines how you graph the linear equation and determine its x and y-intercepts.
However, we can graph more than linear equations. We can also graph linear inequalities.
A linear inequality is simply a linear equation that doesn’t have an equal sign. Instead, it has one of the following symbols:
• Greater than (>)
• Less than (<)
• Greater than or equal to (≥)
• Less than or equal to (≤) symbol
Here, we will be explaining the many forms of linear equations and inequalities, and showing you how to graph them. Graphing a linear function allows you to find various combinations of points that, when plugged in, serve as a solution to the equation.
Let’s start by learning how to graph a linear equation in point-slope form…
Point-Slope Form
The point-slope form of a linear equation is written as y − y1 = m(x − x1). The values of x1 and y1 represent a specific point on the graph, and m represents the slope of the equation. This form is used when we only know one set of points and want to determine additional points by using the slope of the equation.
Here is an example of a linear equation in point-slope form:
y+2 = 3/2 (x−5)
Note that anytime a set of points is in point-slope form, any positive sign in front of the values of x1 and y1 must be reversed to negative and any negative sign must be reversed to positive. So for this equation, x1= -2 and y1=5. Let’s graph this point accordingly:
Now let’s use the slope of this equation to determine the next point. Because the slope is 3/2, let’s move the y-value in the positive direction (upward) by three points and shift the x-value in the positive direction (to the right) two points:
By moving point (5, -2) up three spaces and right two, we’ve determined that the next point on the graph is (7,1.)
Now that we’ve learned Point-Slope Form, let’s review the slope-intercept form of a linear equation…
Slope-Intercept Form
Slope-intercept form is the most common way to write a linear equation. It uses the formula
y =
mx + b.
In this form, m represents the slope of the line and b represents the y-intercept. This form makes it really easy to plug in x-values to determine their correlating y-values, and vice versa. Here is an example of a linear equation in slope-intercept form:
y = 4/3x + 2
Let’s determine the x-intercept by plugging in zero for y:
y = 4/3x + 2
(0) = 4/3x + 2
-2 = 4/3x
3/4(-2) = x
64 = x
x − intercept = -1.5
We’ve now determined that the x-intercept for this equation is (-1.5, 0.)
Now that we’ve reviewed point-slope form, let’s learn how to write a linear equation in standard form…
Standard Form
Standard form is a way of displaying linear functions with the x and y values on the same side of the equation:
Ax + By = C.
The rules of standard form are that the values of A, B, and C must all be whole numbers with no fractions or decimals. Additionally, the value of A must always be positive.
Here is an example of an equation written in standard form:
2x − 3y = 6
This form makes it easy to write find the x and y-intercepts, all you have to do is plug in zero:
2x − 3(0 = 6
2x = 6
x = 6/2
x − intercept = (3,0)
2(0) − 3y = 6
-3y = 6
y = –6/3
y − intercept = (0, -2)
How to Graph Linear Inequalities
Graphing linear inequalities uses all the same steps as graphing a linear equation. The big difference is that linear inequalities use greater than (>), less than (<), greater than or equal to (≥), and less than or equal to (≤) symbols instead of an equal sign. Because of this, additional steps must be taken when graphing a linear inequality.
When graphing an inequality, you must decide whether to use a dotted or solid line for the linear equation. You must also shade either the section left or right of the graphed line depending on each given inequality statement.
Let’s explain how to take these additional steps for graphing a linear inequality…
Step 1: Arrange Inequality into Slope-Intercept Form
When you’re working with a linear inequality, you ought to have y on its own side of the equation, separate from other numbers and variables. This means that we must convert linear inequalities into slope-intercept form. Here’s an example:
y – 3 > 2x
y > 2x +3
Now that the equation is in slope-intercept form, graph it just like a regular linear equation:
Step 2: Determining Strict Inequalities
Now that we have the equation graphed, we must decide whether it’s a strict inequality or not. Determining this will tell us if we should use a solid or dashed boundary line. A boundary line separates the graph into two sections and shows us which points can serve as possible solutions to the inequality.
A strict linear inequality has a “>” or “<” sign. In these inequalities you’ll use a dashed line to indicate that any points on this line cannot be possible solutions to the inequality.
A non-strict inequality has a “≥” or “≤” sign. For these inequalities, you’ll use a solid line to indicate that any set of points along this line can be a possible solution to the inequality equation. When you plug in any x or y values on this line, the linear inequality equation will still be true.
The equation y > 2x + 3 is a strict inequality so let’s use a dotted line to graph it:
Step 3: Deciding the Shaded Region
The last step for graphing linear inequalities is choosing which side of the boundary line you need to shade. This shaded area will incorporate every point that satisfies the inequality.
The best way to determine the shaded side is to select a testing point like (0,0) and plug it into the inequality. If plugging in this point makes the inequality true, shade the side of the graph that point lies on. If the testing point makes the inequality untrue, shade the other side of the graph.
Lets plug (0,0) into the equation y > 2x + 3:
(0) > 2(0) + 3
0 > 3
This equation is clearly untrue as 0 is not greater than 3. Because of this, we will shade the left side of the dotted line on our graph:
Still struggling to understand linear equations and inequalities? We don’t blame you! Algebra is a tough subject and many of us need extra help to master it.
That’s where Tutor Portland comes in. We provide students with tutors who specialize in algebra along with numerous other math subjects. Our tutors give you the one-on-one support you need to keep up with homework and score high on exams. Whether you want in-home or virtual sessions, Tutor Portland will find the perfect tutor to help you with concepts like Linear Equations and Inequalities. Sign up today for a free intro session!
|
# Integers
## Integers include positive whole numbers, negative whole numbers, and zero.
The “set of all integers” is often shown like this:
Integers = {… -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, …}
The dots at each end of the set mean that you can keep counting in either direction. The set can also be shown as a number line:
The arrows on each end of the number line mean that you can keep counting in either direction.
### Is It an Integer?
Integers are whole numbers and their negative opposites. Therefore, these numbers can never be integers:
• fractions
• decimals
• percents
Whether you are adding or subtracting two integers, start by using the number line to find the first number. Put your finger on it. Let's say the first number is 3.
• Then, if you are adding a positive number, move your finger to the right as many places as the value of that number. For example, if you are adding 4, move your finger 4 places to the right.
3 + 4 = 7
• If you are adding a negative number, move your finger to the left as many places as the value of that number. For example, if you are adding -4, move your finger 4 places to the left.
3 + -4 = -1
• If you are subtracting a positive number, move your finger to the left as many places as the value of that number. For example, if you are subtracting 4, move your finger 4 places to the left.
3 - 4 = -1
• If you are subtracting a negative number, move your finger to the right as many places as the value of that number. For example, if you are subtracting -4, move your finger 4 places to the right.
3 - -4 = 7
Here are two rules to remember:
• Adding a negative number is just like subtracting a positive number.
3 + -4 = 3 - 4
• Subtracting a negative number is just like adding a positive number. The two negatives cancel out each other.
3 + 4 = 3 - -4
### Multiplying and Dividing Integers
If you multiply or divide two positive numbers, the result will be positive.
6 x 2 = 12
6 / 2 = 3
If you multiply or divide a positive number with a negative number, the result will be negative.
6 x -2 = -12
6 / -2 = -3
If you multiply or divide two negative numbers, the result will be positive—the two negatives will cancel out each other.
-6 x -2 = 12
-6 / -2 = 3
### Integer Rules: A Video
Watch this video to better understand the correct procedure for adding, subtracting, multiplying, and dividing positive and negative whole numbers.
Numbers Even and Odd Numbers
|
In this post we will learn how to add numbers with 6. After the concept we will solve some practice questions which will clear your doubts and reinforce the lessons for your examinations.
All the study material prepared by us are according to USA and Canada Math Curriculum. All we have tries to make the course little entertaining and interactive for kids so they can enjoy the learning process.
On adding number with 6, we have to make six right jumps on the number line.
Let us understand the concept with the help of examples.
Example 01
7 + 6 = ?
In order to solve the above addition, do the following steps:
Step 02: Make six jumps to the right
Step 03: Stop at number 13
All the steps are explained in the below number line illustration
Hence, 7 + 6 = 13 is the solution
Example 02
5 + 6 = ?
Step 1 : Start at number 5
Step 2: Make six jumps towards right
Step 3: Stop at 11
Hence, 5 + 6 = 11 is the right answer
Example 03
12 + 6 = ?
Step 1 : Start at number 12
Step 2: Make six jumps towards right
Step 3: Stop at 18
Hence, 12 + 6 = 18 is the right answer
Conclusion: On adding 6 to any number we have to jump 6 times towards the right on the number line.
All the numbers below are added with 5. Do the calculation and find the right answer
(a) 21 + 6 = ?
Solution
27
(b) 12 + 6 = ?
Solution
18
(c) 5 + 6 = ?
Solution
11
(d) 19 + 6 = ?
Solution
25
(e) 0 + 6 = ?
Solution
6
(f) 50 + 6 = ?
Solution
56
(g) 56 + 6 = ?
Solution
62
(h) 11 + 6 = ?
Solution
17
(i) 23 + 6 = ?
Solution
29
(j) 79 + 6 = ?
Solution
85
(k) 1 + 6 = ?
Solution
7
(l) 37 + 6 = ?
Solution
43
(m) 6 + 6 = ?
Solution
12
(n) 99 + 6 = ?
Solution
105
(o) 109 + 6 = ?
Solution
115
(p) 38 + 6 = ?
Solution
34
(q) 84 + 6 = ?
Solution
90
(r) 16 + 6 = ?
Solution
22
(s) 112 + 6 = ?
Solution
118
(t) 77 + 6 = ?
Solution
83
(01) Count the number of triangles after addition and select the right answer
(a) 6
(b) 9
(c) 4
(d) 3
Solution
The above image can be expressed as:
⟹ 3 + 6 = 9
Option (b) is the right answer
(02) Count the number of triangles and select the right answer
(a) 7
(b) 9
(c) 8
(d) 10
Solution
The above image can be expressed as:
1 + 6 = 7
Option (a) is the right answer
(a) 7
(b) 9
(c) 8
(d) 10
Solution
The above image can be expressed as :
4 + 6 = 10
Option (d) is the right answer
(04) Count the number of images addition and select the right answer
(a) 7
(b) 9
(c) 8
(d) 10
Solution
The above image can be expressed as:
2 + 6 = 8
Option (c) is the right answer
(05) Do the addition and count the number of blocks and select the right answer
(a) 7
(b) 9
(c) 8
(d) 10
Solution
The above image can be illustrated as:
4 + 6 = 10
Option (d) is the right answer
(06) Count the number of leaves after addition and select the right answer
(a) 12
(b) 13
(c) 14
(d) 10
Solution
The above image can be expressed as:
6+ 6 = 12
Option (a) is the right answer
(07) Count the number of glasses after addition and select the right answer
(a) 12
(b) 13
(c) 14
(d) 10
Solution
The above image can be expressed as:
7 + 6 = 13
Option (b) is the right answer
(08) Do the addition and find the number of apples after addition
(a) 6
(b) 7
(c) 9
(d) 8
Solution
The above image can be expressed as:
2 + 6 = 8
Option (d) is the right answer
(09) Count the number of flower after addition and select the right answer
(a) 16
(b) 14
(c) 11
(d) 13
Solution
The above image can be expressed as:
5 + 6 = 11
Option (c) is the right answer
(10) Do the addition and select the right number of trees
(a) 10
(b) 7
(c) 5
(d) 4
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# Triangle Area Problem | AMC-10A, 2009 | Problem 10
Try this beautiful problem from Geometry based on Area of Triangle.
## Area of Triangle - AMC-10A, 2009- Problem 10
Triangle $A B C$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B, A D=3$, and $D C=4 .$ What is the area of $\triangle A B C$ ?
• $4 \sqrt{3}$
• $7 \sqrt{3}$
• $14 \sqrt{3}$
• $21$
• $42$
### Key Concepts
Triangle
Similarity
Geometry
Answer: $7 \sqrt{3}$
AMC-10A (2009) Problem 10
Pre College Mathematics
## Try with Hints
We have to find out the area of the Triangle ABC where $\angle B=90^{\circ}$ and $BD \perp AC$
Area of a Triangle = $\frac{1}{2}\times$ Base $\times$ Height.But we don know the value of $AB$ & $BC$. But we know $AC=7$. So if we can find out the value of $BD$ then we can find out the are of $\triangle ABC$ by $\frac{1}{2}\times AC \times BD$
Can you now finish the problem ..........
Let $\angle C=\theta$, then $\angle A=(90-\theta)$ (as $\angle B=90^{\circ}$, Sum of the angles in a triangle is $180^{\circ}$)
In $\triangle ABD$, $\angle ABD=\theta$ $\Rightarrow \angle A=(90-\theta$)
Again In $\triangle DBC$, $\angle DBC$=($90-\theta$) $\Rightarrow \angle C=\theta$
From the above condition we say that , $\triangle ABD \sim \triangle BDC$
Therefore , $\frac{BD}{CD}=\frac{AD}{BD}$ $\Rightarrow {BD}^2=AD.CD=4\times 3$
$\Rightarrow BD=\sqrt {12}$
can you finish the problem........
Therefore area of the $\triangle ABC=\frac{1}{2}\times AC \times BD=\frac{1}{2}\times 7 \times \sqrt{12}=7 \sqrt{3}$
## Subscribe to Cheenta at Youtube
Try this beautiful problem from Geometry based on Area of Triangle.
## Area of Triangle - AMC-10A, 2009- Problem 10
Triangle $A B C$ has a right angle at $B$. Point $D$ is the foot of the altitude from $B, A D=3$, and $D C=4 .$ What is the area of $\triangle A B C$ ?
• $4 \sqrt{3}$
• $7 \sqrt{3}$
• $14 \sqrt{3}$
• $21$
• $42$
### Key Concepts
Triangle
Similarity
Geometry
Answer: $7 \sqrt{3}$
AMC-10A (2009) Problem 10
Pre College Mathematics
## Try with Hints
We have to find out the area of the Triangle ABC where $\angle B=90^{\circ}$ and $BD \perp AC$
Area of a Triangle = $\frac{1}{2}\times$ Base $\times$ Height.But we don know the value of $AB$ & $BC$. But we know $AC=7$. So if we can find out the value of $BD$ then we can find out the are of $\triangle ABC$ by $\frac{1}{2}\times AC \times BD$
Can you now finish the problem ..........
Let $\angle C=\theta$, then $\angle A=(90-\theta)$ (as $\angle B=90^{\circ}$, Sum of the angles in a triangle is $180^{\circ}$)
In $\triangle ABD$, $\angle ABD=\theta$ $\Rightarrow \angle A=(90-\theta$)
Again In $\triangle DBC$, $\angle DBC$=($90-\theta$) $\Rightarrow \angle C=\theta$
From the above condition we say that , $\triangle ABD \sim \triangle BDC$
Therefore , $\frac{BD}{CD}=\frac{AD}{BD}$ $\Rightarrow {BD}^2=AD.CD=4\times 3$
$\Rightarrow BD=\sqrt {12}$
can you finish the problem........
Therefore area of the $\triangle ABC=\frac{1}{2}\times AC \times BD=\frac{1}{2}\times 7 \times \sqrt{12}=7 \sqrt{3}$
## Subscribe to Cheenta at Youtube
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# What Are The Prime Factors Of 17?
• Prime factors of 17: 1 * 17
## Is 17 A Prime Number?
• Yes the number 17 is a prime number.
• It's a prime because seventeen has no positive divisors other than 1 and itself.
## How To Calculate Prime Number Factors
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 1 7
• About Number 1. The number 1 is not a prime number, but a divider for every natural number. It is often taken as the smallest natural number (however, some authors include the natural numbers from zero). Your prime factorization is the empty product with 0 factors, which is defined as having a value of 1. The one is often referred to as one of the five most important constants of analysis (besides 0, p, e, and i). Number one is also used in other meanings in mathematics, such as a neutral element for multiplication in a ring, called the identity element. In these systems, other rules can apply, so does 1 + 1 different meanings and can give different results. With 1 are in linear algebra and vectors and one Einsmatrizen whose elements are all equal to the identity element, and refers to the identity map.
• About Number 7. Seven is a prime number. It is the lowest natural number that cannot be represented as the sum of the squares of three integers. The corresponding cyclic number is 142857. You can use this feature to calculate the result of the division of natural numbers by 7 without a calculator quickly. A seven-sided shape is a heptagon. One rule for divisibility by 7 leads to a simple algorithm to test the rest loose divisibility of a natural number by 7: Take away the last digit, double it and subtract them from the rest of the digits. If the difference is negative, then you're leaving the minus sign. If the result has more than one digit, so you repeat steps 1 through fourth. Eventually results are 7 or 0, then the number is divisible by 7 and not otherwise.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What Are Prime Factors?
• In number theory, the prime factors of a positive integer are the prime numbers that divide that integer exactly. The prime factorization of a positive integer is a list of the integer's prime factors, together with their multiplicities. The process of determining these factors is called integer factorization. The fundamental theorem of arithmetic says that every positive integer has a single unique prime factorization.
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# elementaryalgebra
```Review of
Elementary
Algebra
Content
0
1
Fractions ...........................................................................................................................................1
Integers .............................................................................................................................................5
Order of Operations .........................................................................................................................9
Exponents.......................................................................................................................................11
Polynomials....................................................................................................................................18
Factoring ........................................................................................................................................22
Solving Linear Equations ...............................................................................................................31
Solving Linear Inequalities ............................................................................................................34
Graphing Linear Equations ............................................................................................................38
Finding the Equation of a Line ......................................................................................................43
Application of Linear Equations ....................................................................................................48
Functions ........................................................................................................................................50
Systems of Linear Equations .........................................................................................................53
Application of a System of Linear Equations ................................................................................62
Solving Linear Inequalities in Two Variables ...............................................................................64
Solving Systems of Linear Inequalities .........................................................................................67
Solving Absolute Value Equations ................................................................................................72
Solving Absolute Value Inequalities .............................................................................................76
Vocabulary Used in Application Problems....................................................................................79
Solving Application Problems .......................................................................................................81
Comprehensive Review of Elementary Algebra............................................................................88
Revised May 2013
2i
3
FRACTIONS
Lowest Terms
Example:
1
Simplify:
12 4 ⋅ 3
=
15 5 ⋅ 3
1
=
4
5
Rule: To add and subtract fractions you need a Common Denominator.
Example:
2 3 2+3
+ =
7 7
7
5
=
7
Example:
1 2 1 3 2 4
+ = ⋅ + ⋅
4 3 4 3 3 4
3 8
=
+
12 12
11
=
12
Example:
7 1 7 1 3
− =
− ⋅
12 4 12 4 3
7 3
=
−
12 12
4
=
12
1
=
3
For this example, we already have the
common denominator.
This example, we need to find a common
denominator, or a denominator which is
DIVISIBLE by 3 and 4. One such denominator
is 12.
Note: Any number over itself is equal to 1. When you multiply a number by 1, you are not changing the number,
just renaming it.
4
3
9
= 1= 1 = 1
4
3
9
1
FRACTIONS
Multiplication/Division
Rule:
Multiplication and division of fractions do not require a common denominator.
Example:
3
7 12 7 12
⋅ = ⋅
8 5 8 5
Cancel a common factor of 4
2
7 ⋅3
=
2⋅5
21
=
10
Example:
Multiply across
2 5 2 6
÷ = ⋅
3 6 3 5
5
, multiply by the
6
6
reciprocal which is
5
2
To divide by
2 6
=
⋅
3 5
1
2⋅2
1⋅ 5
4
=
5
=
2
FRACTIONS
Problems
Perform the following operations.
1.
7 5
+
12 12
1
2.
2 1
+
3 5
13
15
3.
3 1
−
4 3
5
12
4.
5 5
+
6 12
5
4
5.
5 1
−
7 6
23
42
6.
7 1
−
12 2
1
12
7.
3
+2
4
11
4
8.
5−
4
3
11
3
9.
16 13
−
3 5
41
15
10.
16 12
−
3 7
76
21
3
FRACTIONS
Problems (continued)
Perform the following operations.
11.
25
+6
4
49
4
12.
1 3
⋅
2 4
3
8
13.
3 2
⋅
7 5
6
35
14.
2⋅
2
15
4
15
15.
3 2
÷
5 3
9
10
16.
3
÷2
8
3
16
17.
5 3
÷
6 14
35
9
18.
9 5
÷
4 3
27
20
4
INTEGERS
Definition:
The absolute value of a number is the distance between 0 and the number on the
number line. The symbol for “The absolute value of a” is a . The absolute value
of a number can never be negative.
Examples:
4
1) −4 =
2) 3 = 3
•
Like signs: Add their absolute values and use the common sign.
Examples:
1) −3 + ( −2 ) =− ( 3 + 2 )
= −5
2) 4 + 8 =
( 4 + 8)
= 12
•
Unlike signs: Subtract their absolute values and use the sign of the number with the
larger absolute value.
Examples:
1) −3 + 4 =+ ( 4 − 3)
=1
2) 2 + ( −6 ) =− ( 6 − 2 )
= −4
5
INTEGERS
Subtraction of Signed Numbers
1) Change the subtraction symbol to addition.
2) Change the sign of the number being subtracted.
3) Add the numbers using like signs or unlike signs rules for addition.
Examples:
1) −3 − 4 =−3 + ( −4 )
= −7
2) 6 − ( −2 ) = 6 + ( 2 )
=8
3) −2 − ( −4 ) =−2 + 4
=2
Multiplication and Division of Two Signed Numbers:
•
Like signs:
The product or quotient of two numbers with like signs is positive.
Examples:
•
Unlike signs:
12
( −3)( −4 ) =
2)
8
=2
4
3)
−16
=4
−4
=2
The product or quotient of two numbers with unlike signs is negative.
Examples:
•
1)
1)
−12
( −3)( 4 ) =
2)
−15
= −5
3
Division by 0 is undefined.
Example:
3
is undefined.
0
6
INTEGERS
Addition and Subtraction of Signed Numbers
Problems
Perform the following operations.
1.
5 + ( −3)
2
2.
−6 + ( −2 )
–8
3.
7 − 12
−5
4.
−11 − 4
−15
5.
6 + 2 + ( −13)
−5
6.
8 − ( −5 )
13
7.
−3 − ( 4 − 11)
4
8.
5 1
− −
6 2
−
9.
( −9 ) + ( −14 ) + 12
−11
10.
−4.4 − 8.6
−13
11.
2 + ( −4 − 8 )
−10
12.
9 3
+−
10 5
3
10
13.
−8 − ( −4 − 1) − ( 9 − 2 )
4
14.
−8 + ( −3) + −7 + ( −6 )
–24
15.
−4 + ( −12 + 1) − ( −1 − 9 )
−5
7
4
3
INTEGERS
Multiplication and Division of Signed Numbers
Problems
Perform the following operations.
1.
( 3)( −4 )
–12
2.
24
−6
−4
3.
( −10 )( −12 )
120
4.
0
−2
0
5.
3 10
− −
8 9
5
12
6.
( −9.8) ÷ ( −7 )
1.4
7.
−30
2−8
8.
( −5.1)(.02 )
9.
−40
8 − ( −2 )
−4
( −9 − 1)( −2 ) − ( −6 )
26
10.
5
–.102
8
ORDER OF OPERATIONS
Grouping Symbols
Excuse
( )
[]
Parentheses
Brackets
My Dear
Exponents
Aunt Sally
Multiply
Divide
"as you see it, left to right"
Subtract
"as you see it, left to right"
Square Roots
Absolute Value
Examples:
Note: When one pair of grouping symbols is inside another pair, perform the operations within
the innermost pair of grouping symbols first.
36 ÷ ( 5 − 2 ) + 6 ⋅ 7
54 ÷ 3 − ( 2 ⋅ 3) − 12 + 3
36 ÷ ( 3) + 6 ⋅ 7
54 ÷ 3 − ( 6 ) − 12 + 3
36 ÷ 9 + 6 ⋅ 7
54 ÷ [ −3] − 12 + 3
2
2
4 + 42
− 18 − 12 + 3
−30 + 3
46
− 27
Examples:
Note: For a problem with a fraction bar, perform the operations in the numerator and
denominator separately.
4 9 − 7 + −7 4 ⋅ 2 + 7 8 + 7 15
=
=
= = 3
32 − 22
9−4
5
5
( 6 − 5) − ( −21=) 14 − ( −21) = 1 − ( −21) =
( −9 )( −3) − 42 ( −9 )( −3) − 42 ( −9 )( −3) − 16
4
9
1 + 21 22
= = 2
27 − 16 11
ORDER OF OPERATIONS
Problems
Perform the following operations.
1.
5 + 3 5 − 2
18
2.
12 ÷ 2 6 + 1
37
3.
4 − 32 + 6
1
4.
62 −8÷ 2 + 4
12
5.
−12 − ( 4 + 3)
−19
6.
5 − 2 ( 43) − 5
−24
7.
9 2 ÷ 6 − 5 ( 2)
8.
9.
10.
−17
2
2−4
( −3)
2
−
+1
4 + 2 ( 2 − 5 )2 − 3
3 2 1
−
4 3 2
1
5
19
7
12
2
10
EXPONENTS
Laws of Exponents
a n = a a a ... a
Definition:
n times
a = the base
n = the exponent
Properties:
1.
Product rule: a m a n = a m + n
2 3
2+3
a=
a a
a a a=
a a=
a5
Example:
2 times
2.
am
Quotient rule: n = a m − n
a
3.
Power rule:
4.
(a )
n m
= a nm
1
1
3 times
1
1
1
Example:
a 7 aaaaaaa
7 −5
=
a=
a2
1 1 1 1 1=
5
a
aaaaa
Example:
(a=
)
3 2
3 2
a=
a6
Raising a product to a power:
Raising a quotient to a power:
( ab )
an
a
= n
b
b
n
n
=a b
n n
Zero power rule: a 0 = 1
( x=
)
0
Example:
1;
( 3x=
)
0
0
30 x=
1=
1 1
Examples:
(12ab =
)( 3a b c )
2
3 5
36
=
a1+3b 2+5c 36a 4b 7 c
4
2 x 2 24 x 2 4 16 x8
=
=
3
4 3 4
81 y12
3y 3 y
3 x 2 y 2 ( 2 xy + 5 y 2 ) =
( 3x y ) ( 2 xy ) + ( 3x y )( 5 y ) =
2
2
2
2
2
11
6 x 2+1 y 2+1 + 15 x 2 y 2+ 2 = 6 x 3 y 3 + 15 x 2 y 4
EXPONENTS
Problems
Perform the following operations and simplify.
1.
( x y )( x y )
2.
( 5a b )( −7a b )
2
5
8
6
x10 y 8
3
−35a 9b9
3 8
3.
x10 y 4
x 6 yz
x4 y3
z
4.
10a 3b 7
−2a 2b 4
−5ab3
5.
( ab c )
a 3b12 c 6
6.
( −4x
7.
( −m n )
8.
( 3n x )
9.
( 5x y ) ( −3xy )
10.
4 2 3
2
y5 z3 )
3
0
2
2
16x 4 y10 z 6
− m9 n3
3
2
9x 2
3 0
−4 x3 y
2 xyz
−27x3 y12
4 3
−8x 6
z3
3
12
EXPONENTS
Negative Exponents
Definitions:
a−m =
a−m
1
m
1
a
1
am
= am
a−m
1
a−m
b−n
Note: The act of moving a factor from the
numerator to the denominator (or from the
denominator to the numerator) changes a
negative exponent to a positive exponent.
bn
am
Examples:
10−4
=
−2
=
ab
10−4
1
1
=
4
1
10 10, 000
ab −2 a1b −2
a1 a
=
=
1
1
b2 b2
Since b −2 has a negative exponent, move
it to the denominator to become b 2
x −5 y 2
x −5 y 2 y 2 z 3
=
=
z −3
z −3
x5
Only the factors with negative exponents
are moved.
13
EXPONENTS
Example:
−2
−12 x −2 y 3
−12 y 3
=
2 4 2
4 2
3x y
3x x y
−4 y
= 6
x
( −4 )
=
=
Within the parenthesis, move the factors with
negative exponents.
−2
y −2
6 −2
x
( )
−2 −2
( −4 ) y
=
x −12
=
−2
Simplify within the parenthesis.
−2
Use the power rule for exponents.
x12
( −4 )
x12
16 y 2
2
y2
Move all factors with negative exponents.
Simplify.
14
EXPONENTS
Problems
Perform the following operations and simply.
1.
( −3)
9
2
2. −32
–9
3. 30 ⋅ 3−12 ⋅ 38
1
1
=
4
3 81
4 ⋅ 5−1
4. −3
2
−2
52
25
=
2 6
4 2 1024
5.
( 3a b )( 3ab c )
6.
( −3ebc ) ( 2ab )
18ac 4 e 2
b11
7.
(x y )
y6
x9
8.
0 4
2 2
3
9ab 7 c 2
3 2
−13
−2 −3
6ab 0 c −2
( −3a )
−1
−18a 2 c
b
bc −3
15
EXPONENTS
Problems (continued)
9.
x −8
1
x8
10.
2−3
1
8
11.
a −4 a10 a −8
1
a2
12.
a −6 ( a 2 a −3 )
1
a7
13.
x8
x −4
x12
14.
b −3
b5
1
b8
15.
a −4
a −6
a2
16.
p
p −5
p6
17.
4−5
4−2
1
64
18.
28
25
8
19.
2−1
22
1
8
20.
ab −4
a 5b −3
1
a 4b
21.
x −5 y 2
x −2 y −3
y5
x3
16
EXPONENTS
Problems (continued)
22.
b −2 c −8
b 4 c −2
1
bc
6 6
23.
3−4
9−3
9
24.
2−5 a −3
2a −6
a3
64
25.
( 2a b )
26.
( 3x )
27.
3a −4 ( 2a −5b )
28.
( 2a b ) ( a b )
29.
( −4a b )
30.
(a
4 −3 5
32a 20
b15
−4 −3
x12
27
−2
3 −4 −1
−6
3a 6
4b 2
−2
a 9b 2
2
5 −2 −2
b4
16a10
)
−4 5 −3
b
a19
b17
a −7b 2
31.
32.
−5a −1 ( a 3b −4 )
−6
−5b 24
a19
( 2a b )
( 2a b )
3a ( 2a )
( 3a )
32a
b2
5 −4 2
−3 2 −3
33.
−2
−5 −4
27 a 22
16
2 −2
17
POLYNOMIALS
Definition: A sum of a finite number of terms of the form: an x n where an is a real number and n
is a non-negative integer. (No negative exponents, no fractional exponents.)
Examples:
3 x 4 + 2 x 3 − 8 x + 1 is a polynomial
2 x −5 + 3 x 2 is not a polynomial
Types of Polynomials:
Monomial: A polynomial with 1 term
Example: 3x 4
Binomials: A polynomial with 2 terms
Example: 2 x3 + 4
Trinomial: A polynomial with 3 terms
Example: 3x 2 + 4 x − 2
Degree:
•
Degree of a term – the sum of the exponents on the variables.
Example: 3x 4 y 3 has degree 4 + 3 =
7
•
Degree of a polynomial – the largest degree of any of the terms. In a polynomial
with one variable it is the largest exponent.
Example:
5 x 4 − 3 x 3 + 2 x 2 − 8 has degree 4
3
2 x 2 y + 3 xy − 4 has degree 3 since 2x 2 y has degree 2 + 1 =
18
POLYNOMIALS
Example:
( 3x
2
+ 4 x − 5) + ( 2 x 2 − 8 x + 2 ) = 3x 2 + 4 x − 5 + 2 x 2 − 8 x + 2 = 5 x 2 − 4 x − 3
Subtraction of Polynomials
Subtracting Polynomials: Distribute the negative sign and combine like terms.
Example:
( 3x
2
+ 4 x − 5 ) − ( 2 x 2 − 8 x + 2 ) = 3 x 2 + 4 x − 5 − 2 x 2 + 8 x − 2 = x 2 + 12 x − 7
Multiplication of Polynomials
Multiplying Polynomials: Use the distributive property.
Example:
( 3x + 2 ) ( x 2 − 3x + 4=)
3x ( x 2 − 3x + 4 ) + 2 ( x 2 − 3x + 4 )
= 3 x 3 − 9 x 2 + 12 x + 2 x 2 − 6 x + 8
= 3x3 − 7 x 2 + 6 x + 8
Example:
( 2 x + 3)( 3x − 4 )
( 2 x )( 3x ) + ( 2 x )( −4 ) + 3 ( 3x ) + 3 ( −4 )
=
= 6 x 2 − 8 x + 9 x − 12
= 6 x 2 + x − 12
Example:
( 3 y + 4 )( 3 y − 4 ) The multiplication of the sum and difference of two terms.
= 3 y ( 3 y ) + 3 y ( −4 ) + 4 ( 3 y ) + 4 ( −4 )
= 9 y 2 − 12 y + 12 y − 16
9 y2 +
=
= 9 y − 16
2
0
− 16
The answer is “the difference of two squares.”
19
POLYNOMIALS
Example:
( 4 y + 3) The square of a binomial.
( 4 y + 3)( 4 y + 3) = 4 y ( 4 y ) + 4 y ( 3) + 3 ( 4 y ) + 3 ( 3)
2
= 16 y 2 + 12 y + 12 y + 9
16 y 2 + 2 (12 y ) + 9
=
= 16 y 2 + 24 y + 9
The answer is a “perfect square trinomial.”
Example:
( a − b ) The cube of a binomial.
( a − b )( a − b ) ( a − b )
( a − b )( a − b )( a − b ) =
= a a + a ( −b ) − b ( a ) − b ( −b ) ( a − b )
= a 2 − ab − ab + b 2 ( a − b )
= a 2 − 2ab + b 2 ( a − b )
= a 2 ( a − b ) − 2ab ( a − b ) + b 2 ( a − b )
3
=a − a b − 2a b + 2ab + ab − b
3
2
2
2
2
3
=
a 3 − 3a 2b + 3ab 2 − b3
Division of Polynomials
Dividing a polynomial by a monomial.
Example:
9 x 4 y 3 − 24 x 2 y 5 + xy 4
3x 2 y 2
9 x 4 y 3 24 x 2 y 5
xy 4
= 2 2−
+
3x y
3x 2 y 2 3x 2 y 2
1
= 3 x 4 − 2 y 3− 2 − 8 x 2 − 2 y 5− 2 + x1− 2 y 4 − 2
3
1
=3 x 2 y − 8 x 0 y 3 + x −1 y 2
3
2
y
= 3x 2 y − 8 y 3 +
3x
20
Use the distributive property.
POLYNOMIALS
Problems
Find the degree of the following and determine what type of
polynomial is given.
1.
3 x 3 y 2 + 2 xy − 8
Trinomial, degree 5
2.
5 − 2x 3
Binomial, degree 3
3.
x5
Monomial, degree 5
4.
8
Monomial, degree 0
5.
( 3x
3
+ 2 x 2 − 5 x + 4 ) + ( 6 x3 − 2 x − 8)
9 x3 + 2 x 2 − 7 x − 4
Subtract:
6.
( 3x
3
+ 2 x 2 − 5 x + 4 ) − ( 6 x3 − 2 x − 8)
−3 x 3 + 2 x 2 − 3 x + 12
Multiply:
7.
( x + 2 ) ( 2 x 2 − 5 x + 3)
2 x3 − x 2 − 7 x + 6
8.
( 3x + 2 )( 2 x − 3)
6 x2 − 5x − 6
9.
( 4 x + 3)( 4 x − 3)
16 x 2 − 9
10.
(5x + 7 )
25 x 2 + 70 x + 49
11.
(9 y − 2)
12.
( 3x − 4 )
Divide:
13.
2
2
81 y 2 − 36 y + 4
3
27 x3 − 108 x 2 + 144 x − 64
12 x15 − 48 x12 + x10 − 18 x8
6 x10
1 3
2 x5 − 8 x 2 + − 2
6 x
21
FACTORING
Factoring Out the Greatest Common Factor
1.
2.
3.
4.
5.
Identify the TERMS of the polynomial.
Factor each term to its prime factors.
Look for common factors in all terms.
Factor out the common factor.
Check by multiplying.
Example: Factor 6 x 2 + 8 xy 2 − 4 x
Terms: 6 x 2 ,
8 xy 2 ,
and
2 ⋅3⋅ x ⋅ x + 2 ⋅ 2 ⋅ 2 ⋅ x ⋅ y ⋅ y − 2 ⋅ 2 ⋅ x
Factor into primes
2 ⋅3⋅ x ⋅ x + 2 ⋅ 2 ⋅ 2 ⋅ x ⋅ y ⋅ y − 2 ⋅ 2 ⋅ x
Look for common factors
2 x ( 3 x + 4 y2 − 2)
Factor out the common factor
Check:
2 x ( 3x ) + 2 x ( 4 y 2 ) − 2 x ( 2 )
Check by multiplying
6 x 2 + 8 xy 2 − 4 x
Factoring Trinomials of the Form: x 2 + bx + c
To factor x 2 + 5 x + 6, look for two numbers whose product = 6, and whose sum = 5
List factors of 6
Choose the pair which adds to 5
and multiplies to 6.
16
2 + 3
= 5
−1−6
2 • 3
= 6
2•3
−2−3
Since the numbers 2 and 3 work for both, we will use them.
Substitute 2 x + 3x in for 5x.
So, x 2 + 5 x + 6 = x 2 + 2 x + 3 x + 6
= x ( x + 2) + 3( x + 2)
Factor by grouping.
=
( x + 2 )( x + 3)
( x + 2 )( x + 3 )
Check by multiplying:
( x + 2 )( x + 3) = x 2 + 2 x + 3x + 6
= x2 + 5x + 6
22
− 4x
FACTORING
Factoring Trinomials of the Form: ax 2 + bx + c
Example: Factor 6 x 2 + 14 x + 4
6 x 2 + 14 x + 4
2 ( 3x 2 + 7 x + 2 )
1.
Factor out the GCF
2.
Identify the values of
a, b, and c for the
expression 3 x 2 + 7 x + 2
3.
Find the product of a c
=
a c 2=
3 6
4.
List all possible factor pairs
that equal a ⋅ c and identify
the pair whose sum is b .
1 6
23
5.
Substitute boxed values
from Step 4 in for the bx
term.
3 x 2 +7 x + 2
3 x 2 +1x + 6 x + 2
6.
Group the two pairs
3 x 2 + 1x + 6 x + 2
First step in factor by
grouping.
7.
Factor the GCF out of each
pair.
x ( 3 x + 1) + 2 ( 3 x + 1)
The resulting common
factor is ( 3 x + 1)
8.
Factor out ( 3 x + 1)
( 3x + 1)( x + 2 )
These are the two factors
of 3 x 2 + 7 x + 2
9.
Recall the original GCF
from Step 1.
2 ( 3 x 2 + 7 x + 2 )= 2 ( 3 x + 1)( x + 2 ) There are 3 factors of
6 x 2 + 14 x + 4
2 ( 3 x + 1)( x + 2 )
Check by multiplying.
2 ( 3 x + 1)( x + 2 ) = 6 x 2 + 14 x + 4
10.
GCF is 2
=
a 3;=
b 7; and c = 2
1+6=7
2+3=
5
23
1 & 6 and 2 & 3 are the
only factor pairs.
1 & 6 is the pair that adds
up to 7 and multiplies up to
6.
Substitute
+1x + 6 x for + 7 x
FACTORING
Example: Factor 3 y 2 − 4 y − 4
Note: If a ⋅ c is negative, the factor pairs have opposite signs.
1.
Factor out the GCF.
There is no GCF.
2.
Identify the values of a,
b, and c.
a=
3; b =
−4; c =
−4
3.
Find the product of ac .
ac =
3 ( −4 ) =
−12
4.
List all possible factor
pairs that equal ac and
identify the pair whose
sum is b.
+1 ( −12 )
+ 1 + ( −12 ) =−11
−1 ( +12 )
− 1 + ( +12 ) =+11
+2 ( −6 )
+ 2 + ( −6 ) =−4
−2 ( +6 )
− 2 + ( +6 ) =+4
+3 ( −4 )
+ 3 + ( −4 ) =−1
−3 ( +4 )
− 3 + ( +4 ) =+1
3 y 2 − 4 y − 4= 3 y 2 + 2 y − 6 y − 4
5.
Substitute the boxed
values from Step 4 in for
the bx term.
6.
Group the two pairs.
7.
Factor the GCF out of
each pair
8.
Factor out ( 3 y + 2 )
( 3 y + 2 )( y − 2 )
( 3 y + 2 )( y − 2 )
Check:
( 3 y + 2 )( y − 2 ) =
9.
= 3 y 2 + 2 y −6 y − 4
= y (3 y + 2) − 2 (3 y + 2)
24
Include all possible factor
combinations, including
the + / − combinations
since a c is negative.
Substitute
2 y − 6 y in for − 4 y
First step in factor by
grouping.
The resulting common
factor is ( 3 y + 2 ) .
These are the factors.
3y2 − 4 y − 4
FACTORING
Factoring Perfect Square Trinomials
Example: Factor x 2 + 18 x + 81
STEP
1. Determine if the trinomial is a Perfect Square.
• Is the first term a perfect square?
• Is the third term a perfect square?
• Is the second term twice the product of
the square roots of the first and third
terms?
x2
+81
+18x
2. To factor a Perfect Square Trinomial:
• Find the square root of the first term
• Identify the sign of the second term
• Find the square root of the third term
• Write the factored form
x
+
9
•
NOTES
Yes x x = ( x )
2
Yes =
99 (=
9 ) 81
Yes 2 ( x9 ) = 18 x
2
x2 = x
81 = 9
2
x + 18 x + 81 = ( x + 9 )( x + 9 ) = ( x + 9 )
2
( x + 9)
( x + 9 )( x + 9 ) = x 2 + 18 x + 81
2
Example: Factor 25 x 2 + 20 xy + 4 y 2
STEP
1. Is the trinomial a Perfect Square Trinomial?
• Is the first term a perfect square?
NOTES
25x 2
Yes ( 5 x ) = 25 x 2
2
•
Is the third term a perfect square?
4 y2
Yes ( 2 y ) = 4 y 2
•
Is the second term twice the product of
the square roots of the first and third
terms?
20xy
Yes 2 ( 5 x2 y ) = 20 xy
2. Factor the trinomial
• Square root of the first term
•
•
Sign of the second term
Square root of the third term
•
Factored form
•
2
25 x 2 = 5 x
+20xy → +
4 y2 = 2 y
25 x 2 + 20 xy + 4 y 2 = ( 5 x + 2 y )( 5 x + 2 y ) = ( 5 x + 2 y )
(5 x + 2 y )
( 5 x + 2 y )( 5 x + 2 y ) =
2
25
25 x 2 + 20 xy + 4 y 2
2
FACTORING
Factoring the Difference of Two Squares
Example: Factor x 2 − 36
STEP
NOTES
1. Are there any common factors?
No
2. Determine if the binomial is the Difference of Two Squares
• Is the first term a perfect square?
x2
Yes x x = x 2
•
Is the second term a perfect square?
36
Yes 66 = 36
•
Is this the difference of the two terms?
x 2 − 36
Yes, “–” means the
difference of the two
terms.
3. To factor the Difference of Two Squares:
•
Find the square root of the first term.
x
x2 = x
•
Find the square root of the second term.
6
36 = 6
•
Write the factored form.
x 2 − 36 = ( x + 6 )( x − 6 )
( x + 6 )( x − 6 )
( x + 6 )( x − 6 ) = x 2 − 36
•
The factors are the
product of the sum
(+) and difference (–)
of the terms’ square
roots.
Example: Factor 36a 2 − 121b 2
STEP
1. Are there any common factors?
2. Is the binomial a Difference of Two Squares?
• Is the first term a perfect square?
NOTES
No
36a 2
Yes 6a 6a = 36a 2
•
Is the second term a perfect square?
121b 2
Yes 11b11b = 121b 2
•
Is this the difference of the two terms?
36a 2 − 121b 2
Yes, “–” means the difference
of the two terms.
3. To factor the Difference of Two Squares:
• Find the square root of the first term.
6a
36a 2 = 6a
121b 2 = 11b
•
Find the square root of the second term.
11b
•
Write the factored form.
36a 2121b 2 =+
( 6a 11b )( 6a − 11b )
•
( 6a + 11b )( 6a − 11b )
( 6a + 11b )( 6a − 11b ) =
26
36a 2 − 121b 2
The factors are the product of
the sum (+) and difference (–)
of the terms’ square roots.
FACTORING
FACTORING STRATEGY
1. Is there a common factor? If so, factor out the GCF, or the opposite of the GCF so that the
2. How many terms does the polynomial have?
If it has two terms, look for the following problem type:
a. The difference of two squares
If it has three terms, look for the following problem types:
a. A perfect-square trinomial
b. If the trinomial is not a perfect square, use the grouping method.
If it has four or more terms, try to factor by grouping.
3. Can any factors be factored further? If so, factor them completely.
4. Does the factorization check? Check by multiplying.
27
FACTORING STRATEGY
Always check for
Greatest Common Factor First
Then
2 – Terms
3 – Terms
Difference of
Two Squares
x2 − 4
2
2
( x ) − ( 2)
( x + 2 )( x − 2 )
Form: x2 + bx + c
x2 − 6 x + 5
4 – Terms
Form: ax2 + bx + c
5x2 − 7 x − 6
List factors of 5
5 ⋅1
−5 ⋅ −1
5 ⋅ −6 = −30
List factors of –30
Choose the pair which
−5 + −1 = − 6
( x − 5)( x − 1)
−1 ⋅ 30
−2 ⋅ 15
−3 ⋅ 10
−5 ⋅ 6
1 ⋅ −30
2 ⋅ −15
3 ⋅ −10
5 ⋅ −6
Choose the pair which adds to –7
Perfect Square Trinomials
3 x + −10 x = −7 x
4 x 2 + 20 x + 25
Substitute this pair in for the
middle term.
( 2 x ) + 2 ( 2 x ) ( 5)
( 2x + 5)( 2x + 5)
2
2
x
+
5
( )
2
continue:
+ 52
5 x 2 + 3 x − 10 x − 6
Factor by Grouping.
x ( 5 x + 3) − 2 ( 5 x + 3)
( 5 x + 3)( x − 2 )
28
Factor by
Grouping
7 x 2 + 14 x − 6 x − 12
7 x ( x + 2) − 6 ( x + 2)
( x + 2 )( 7 x − 6 )
FACTORING
Problems
Factor completely.
1. x 2 + 6 x + 8
( x + 2 )( x + 4 )
2. u 2 + 15u + 56
( u + 8)( u + 7 )
3. x 2 + 8 x − 20
( x + 10 )( x − 2 )
4. y 2 − 6 y − 40
( y − 10 )( y + 4 )
5. m 2 − 15m + 54
( m − 9 )( m − 6 )
6. x 2 + 5 xy − 14 y 2
( x + 7 y )( x − 2 y )
7. a 2 + 16ab + 28b 2
( a + 14b )( a + 2b )
8. x 3 − 3 x 2 − 18 x
x ( x − 6 )( x + 3)
9. 14 x 2 − x − 3
( 7 x + 3)( 2 x − 1)
10. 3 x 2 − 19 x + 20
( x − 5)( 3x − 4 )
29
FACTORING
Problems (continued)
2 ( x − 12 )( x + 4 )
11. 2 x 2 − 16 x − 96
54 x3 ( x − 2 )( x + 1)
12. 54 x5 − 54 x 4 − 108 x3
2 ( 3x 2 + x + 6 )
13. 6 x 2 + 2 x + 12
18 ( 2 x + 3)( x − 1)
14. 36 x 2 + 18 x − 54
( x − 1)( x + 1)
15. x 2 − 1
( 4x
16. 16 x 4 − 81
2
+ 9 ) ( 2 x + 3)( 2 x − 3)
3 x 2 ( 3 x + 4 )( 3 x − 4 )
17. 27 x 4 − 48 x 2
3 ( 2 x + 5 )( x + 3)
18. 6 x 2 + 33 x + 45
19. x 2 − 18 x + 81
( x − 9)
20. 9 x 2 + 24 x + 16
( 3x + 4 )
21. 20 x 2 + 60 x + 45
5 ( 2 x + 3)
30
2
2
2
SOLVING LINEAR EQUATIONS
To Solve a Linear Equation
1.
2.
3.
4.
5.
6.
Remove grouping symbols by using the distributive property.
Combine like terms to simplify each side.
Clear fractions by multiplying both sides of the equation by the Least Common Denominator.
Move the variable terms to one side and the constants to the other side. Do this by adding or
subtracting terms.
Solve for the variable by multiplying by the inverse or dividing by the coefficient of x.
Check by substituting the result into the original equation.
Example
Solve 3 ( x − 4 ) − ( 2 x + 8 ) = 4 x + 4
3 x − 12 − 2 x − 8= 4 x + 4
x − 20 = 4 x + 4
− 3x =
24
Use the distributive property.
Combine terms
Move variable to one side, numbers to the other
−3 x 24
=
−3 −3
x = −8
Solution:
Divide to solve for the variable.
3 ( −8 − 4 ) − ( 2 ( −8 ) + 8 ) = 4 ( −8 ) + 4
Check:
3 ( −12 ) − ( −16 + 8 ) =
−32 + 4
− 36 − ( −8 ) =−28
− 28 =
−28
Example: Solve
2
1 3
x− =
3
2 4
12 ( 23 x ) − 12 ( 12 ) =
12 ( 34 )
8x − 6 =
9
8 x 15
=
8
8
Solution:
Check:
x=
Clear fractions by multiplying by the LCD
15
8
2 15 1 3
− =
3 8 2 4
2 1 155 1 3
− =
31 84 2 4
5 2 3
− =
4 4 4
3 3
=
4 4
31
SOLVING LINEAR EQUATIONS
Special Cases:
1. When the variable terms drop out and the result is a true statement, (i.e., 2 = 2 or 0 = 0 ),
there are an infinite number of solutions. The equation is called an identity.
Example:
Solve
2 + 9=
x 3 ( 3 x + 1) − 1
2 + 9=
x 3 ( 3 x + 1) − 1
2 + 9x = 9x + 3 −1
2 + 9x = 9x + 2
0=0
Infinite number of solutions
Solution:
All real numbers
2. When the variable terms drop out and the result is a false statement (i.e., 10 = 2 or 8 = 0),
there is no solution. The equation is called a contradiction.
Example:
Solve 3 x − 5= 3 ( x − 2 ) + 4
3x − 5 = 3x − 6 + 4
3x − 5 = 3x − 2
− 5 =−2
Solution:
No solution
32
SOLVING LINEAR EQUATIONS
Problems
Solve.
1.
3 − ( 2 x + 5 ) =x − 3 ( 2 x − 5 )
x=
2.
( 12 ) ( x − 5=) ( 13 ) ( x + 2 )
x = 19
3.
2 x + 3 = 3x + 6 − x + 4
No solution
4.
2 x + 3 = 3x + 6 − x − 3
Infinite number of solutions:
All real numbers
5.
3 x + 4 − 5 x + 2= 2 ( 3 + x )
x=0
33
17
3
SOLVING LINEAR INEQUALITIES
Solving Linear Inequalities
Linear inequalities are solved almost exactly like linear equations. There is only one exception:
if it is necessary to divide or multiply by a NEGATIVE number, the inequality sign must be
reversed.
The solution can be written as a statement of inequality or in interval notation. It can also be
shown as a graph. In interval notation and graphing:
• Use a bracket, [ ], or closed circle, if the endpoint is included in the solution
• Use parenthesis, ( ), or open circle, if the endpoint is not included in the
solution
Example:
Interval Notation: ( −∞,1)
x <1
or
0
Example:
Solution:
1
2
0
1
−3 ( x − 2 ) ≤ x − 5
−3 x + 6 ≤ x − 5
−4 x ≤ − 11
−4 x
−4
≥
−11
−4
x
≥
11
4
11
x ≥ 4
2
Distribute
Move variable term to one side,
constants to the other
Divide by − 4 and reverse the
inequality because of division − 4
Simplify
11
Interval Notation: , ∞
4
[
0
1
2
11
3
4
4
5
4
5
or
0
1
2
11
3
4
34
SOLVING LINEAR INEQUALITIES
Problems
Solve.
a < −10
or
( ∞, −10 )
2. 2 ( 5 x + 3) − 3 ≤ 6 ( 2 x − 3) + 15
x≥3
or
[3, ∞ )
3. 3 x + 4 − 5 x + 2 < 2 ( 3 + x )
x>0
or
( 0,∞ )
4. 10 − 4 x + 8 ≥ 6 − 2 x + 10
x ≤1
or
( −∞,1]
5. 5 ( x + 3) − 6 x ≤ 3 ( 2 x + 1) − 4 x
x≥4
or
[ 4,∞ )
x < −11
or
( −∞, −11)
1. 4a − 7 > 3 ( 2a + 5 ) − 2
6. 2 ( x − 5 ) + 3 x < 4 ( x − 6 ) + 3
35
SOLVING LINEAR INEQUALITIES
Compound Inequalities
Compound inequalities are solved the same way as simple inequalities. Any operation (addition,
subtraction, multiplication, division) must be performed on all three pieces of the inequality.
Never remove the variable from the middle piece of the inequality. Always remember that when
dividing or multiplying by a NEGATIVE number, the inequality sign must be reversed.
Example:
Solve − 8 < 3 x − 2 ≤ 4
+2
+2 +2
−6 <
−6
<
3
−2 <
3x ≤ 6
3x
6
≤
3
3
x ≤2
Add 2 to all three parts of the inequality
Divide all three parts of the inequality by 3
Simplify
In interval notation, this is ( −2, 2] , because x is between –2 and 2, including the endpoint 2, but
not including the endpoint –2.
Interval Notation: ( −2, 2]
−2 < x ≤ 2
Solution:
-3
(-2
-1
0
1
2
○
or
3
-3 -2 -1
Example:
0
Solve −5 < −3x < 12
−
3
5
> x > −4
3
Keep the variable in the middle piece of the
inequality.
−4 < x <
5
3
Arrange the inequality so that the lesser
value is on the left and the greater value on
the right, and change the inequality symbols
to preserve the relationship.
5
3
5
Interval Notation: −4,
3
(
-4
2
When dividing an inequality by a negative
number, reverse the inequality sign.
5
−4,
3
-5
∙
5 −3 x 12
>
>
−3 −3 −3
−4 < x <
Solution:
1
)
0
or
2
36
-5
-4
0
2
SOLVING LINEAR INEQUALITIES
Problems
Solve
1.
−2 < 3 x + 1 < 4
−1 < x < 1
or
( −1,1)
2.
−5 ≤ 4 x + 5 ≤ 5
− 52 ≤ x ≤ 0
or
[ − 52 ,0]
3.
−4 < 4 x + 2 ≤ 8
− 23 < x ≤
3
2
or
( − 32 , 32 ]
4.
3 < 5 − 2x < 7
−1 < x < 1
or
(1,1)
5.
2 ≤ 1− x < 6
−5 < x ≤ −1
or
( −5, −1]
6.
−2 < 4 − 3 x < 0
4
3
<x<2
or
( 34 , 2 )
7.
0 < 1 − 4x < 7
− 32 < x <
or
( − 32 , 14 )
8.
−4 ≤ 3 − 2 x ≤ −1
2≤ x≤
or
[ 2, 72 ]
9.
−7 < 4 − 2 x < −4
4 < x < 112
or
( 4, 112 )
4 < 4 − 3x < 6
− 23 < x < 0
or
( − 23 ,0 )
10.
37
1
4
7
2
GRAPHING LINEAR EQUATIONS
Standard Form: ax + by = c
A line is made up of an infinite number of points in the form (x,y). The coordinates for each of these
points will satisfy the equation for the line (make it true).
Two special points on the line are its x and y intercepts. These are the points on the line where the line
crosses the x axis and the y axis. The x intercept is in the form (x,0), and the y intercept is in the form
(0,y).
Find the x intercept of the line by substituting 0 for y, and solving for x, (x,0).
Find the y intercept of the line by substituting 0 for x, and solving for y, (0,y).
Example: Graph the line using the intercept method 3 x − 4 y =
−12
( 4,6 )
Find the x intercept:
y
Substitute 0 for y and solve for x:
− 12
3x − 4 y =
− 12
3x − 4 ( 0 ) =
− 12
3x − 0 =
3 x = − 12
−12
3x =
3
x = −4
The x intercept is (– 4,0).
x
0
Find the y intercept:
Substitute 0 for x and solve for y:
3x − 4 y =
−12
3( 0) − 4 y =
−12
0 − 4y =
−12
− 4y =
−12
−4 y −12
=
−4
−4
y =3
The y intercept is (0,3).
To graph the line, graph these two points, and connect them.
As a check, find one more point on the line. Let x = 4 and find y = 6 . The point (4,6) is on the line and
collinear with ( − 4, 0 ) and ( 0,3) .
38
GRAPHING LINEAR EQUATIONS
Slope—Intercept Form: y = mx + b
If the equation of the line is in slope-intercept form, we use the slope m, and the y intercept,
(0,b), to graph the line.
The slope of a line passing through points ( x1 , y1 ) and ( x2 , y2 ) is
vertical change
rise change in y y2 − y1
=
m
= =
=
if x2 ≠ x1
horizontal change run change in x x2 − x1
Example: Graph the line: =
y 2x + 3 .
The slope calculation is called a
Rate of Change. The rate of change
describes how one quantity (the
numerator) changes with respect to
another (the denominator).
Identify the slope as a fraction:
In the equation =
y mx + b , m is the slope.
2
1
In the equation =
y 2 x + 3 , 2 is the slope, so m = .
Identify the y intercept as a point:
In the equation, =
y mx + b, ( 0, b ) is the y intercept.
In the equation, =
y 2 x + 3 , (0,3) is the y intercept
Graph the y intercept:
Graph the slope:
(0,3) is the first point to be graphed
m=
2 rise of 2
=
1 run of 1
y
2 is positive; move up 2 (rise) from the y
intercept, (0,3), to the point (0,5).
8
6
(1,5)
4
1 is positive; then move 1 to the right (run)
from (0,5) to (1,5).
2 (0,3)
0
2
4
6
8
x
(1,5) is the second point of the line.
Graph the line through the two points:
Graph the line through the points (0,3) and
(1,5).
39
GRAPHING LINEAR EQUATIONS
Special Cases:
The graph of the linear equation y = k , where k is a real number, is the horizontal line going through
the ( 0, k )
y
Example: y = 2
4
x
4
The graph of the linear equation x = k , where k is a real number, is the vertical line going through
the point ( k ,0 ) .
y
Example: x = 4
3
3
40
x
GRAPHING LINEAR EQUATIONS
Problems
Graph the following equations:
1. x − 5 y =
5
2. x + 3 y =
5
y
y
x
3. y= x + 4
x
4. x = 2 y
y
y
x
x
5. 6 y = 12
6. x = −2
y
y
x
x
41
GRAPHING LINEAR EQUATIONS
1. x − 5 y =
5
2. x + 3 y =
5
y
4
3
3
3. y= x + 4
x
x
2
4. x = 2 y
y
y
4
4
3
5. 6 y = 12
y
4
x
6. x = −2
y
x
y
5
2
3
2
x
42
x
FINDING THE EQUATION OF A LINE
The standard form of the equation of a line is written as
ax + by =
c
where a and b are not both 0.
Definition of slope: Let L be a line passing through the points ( x1 , y1 ) and ( x2 , y2 ) . Then the slope
of L is given by the formula:
m=
y2 − y1
x2 − x1
The slope-intercept form of the equation of a line is written as
=
y mx + b
where m is the slope and the point ( 0,b ) is the y-intercept.
Point-Slope Form: The equation of the straight line passing through ( x1 , y1 ) and having slope m
is given by y − y1= m ( x − x1 ) .
Parallel Property: Parallel lines have the same slope.
Perpendicular Property: When two lines are perpendicular, their slopes are negative (opposite)
reciprocals of one another. The product of their slopes is –1.
43
FINDING THE EQUATION OF A LINE
To find the equation of a line, use: Slope-Intercept form: =
y mx + b , slope m, y intercept ( 0,b )
or Point-Slope form: y − y1= m ( x − x1 ) , slope m, point ( x1 , y1 )
Slope-Intercept Form
When given the slope and y intercept: use the slope-intercept form =
y mx + b
Example: Find the equation of the line with slope
3
and y intercept ( 0, −2 )
4
Use the slope-intercept form, =
y mx + b , and fill in the values:
Solution: The equation of the line is:
3
=
y
x−2
4
Make the substitutions of
3
3
m = slope is and
4
4
b=
−2 ( y intercept is ( 0, −2 ) )
Point-Slope Form
When given the slope and a point on the line: use the point-slope form y − y1= m ( x − x1 )
Example: Find the equation of the line with slope
2
, through the point ( 6,1)
3
Use the point-slope form, y − y1= m ( x − x1 ) , and fill in the values.
y −=
1
y − 1=
y − 1=
y − 1 + 1=
y
=
2
( x − 6)
3
2
2 6
x− ⋅
3
3 1
2
x−4
3
2
x − 4 +1
3
2
x−3
3
2
2
Make the substitutions of m = slope is and
3
3
x1 = 6 and y1 = 1 ( point ( 6,1) ) .
=
y
Solution: The equation of the line is
44
2
x −3
3
FINDING THE EQUATION OF A LINE
When given two points on the line, first find the slope m, then use the slope and either one of the
points to find the equation using the point-slope form y − y1= m ( x − x1 )
Example: Find the equation of the line through the points ( 2,3) and (1,1) .
Slope m=
3 −1 2
= = 2
2 −1 1
y − y1= m ( x − x1 )
y − 3= 2 ( x − 2 )
y − 3 = 2x − 4
=
y 2x −1
Make the substitutions of
=
m 2 ( slope is 2 )=
and x1 2=
and y1 3
( point ( 2,3) ) .
Solution: The equation of the line is =
y 2x −1
45
FINDING THE EQUATION OF A LINE
Problems
1. Put the following in slope-intercept form.
a. 8 x − 4 y =
4
=
y 2x −1
b. 2 x + 3 y =
3
2
y=
− x +1
3
2. Find the slope.
a. ( 3, 4 ) and ( 7,9 )
5
4
b. ( −2,1) and ( 3, −3)
−
4
5
3. Find the equation of the line through ( 5,3)
and parallel to y =
−2 x + 1 .
y=
−2 x + 13
4. Find the equation of the line through ( −1, −3)
1
13
y=
− x−
4
4
and perpendicular to =
y 4x − 2 .
46
FINDING THE EQUATION OF A LINE
Problems
Find the equation of the line that satisfies the given conditions.
Write the equation in both standard form and slope-intercept form.
Standard Form
5. slope =
1
2
6. slope = −
5
6
7. slope = 1
x − 2y =
−5
line passes through ( 0,0 )
5x + 6 y =
0
5
y= − x
6
y-intercept –3
x− y=
3
y= x − 3
y=4
9. slope is undefined and passing through ( −5,6 )
x = −5
3
2
1
5
x+
2
2
line passes through ( 3, 4 )
8. horizontal line through (1, 4 )
10. slope = −
Slope-Intercept Form
−15
3x + 2 y =
x-intercept –5
=
y
=
y
−3
15
x−
2
2
1
3
x+
2
2
11. horizontal line through ( 5, −3)
y = −3
12. vertical line passing through ( 5, −4 )
x=5
13. line passing through (1, 2 ) and ( 5, 4 )
x − 2y =
−3
=
y
14. line passing through ( −3, 4 ) and ( 5, −1)
5x + 8 y =
17
=
y
15. line passing through ( 4, −3) and ( 4, −7 )
x=4
16. line parallel to 3 x + y =
6 and passing through
3x + y =
5
−5
17
x+
8
8
y=
−3 x + 5
(1, 2 )
17. line passing through ( 5, −3) and parallel to
y = −3
y−2=
0
5x − 2 y =
−9
18. line perpendicular to 2 x + 5 y =
3 and
passing through (1,7 )
47
=
y
5
9
x+
2
2
APPLICATION OF LINEAR EQUATIONS
Example: Determine Rate of Change
A credit union offers a checking account with a service charge for each check written. The
relationship between the monthly charge for each check “y” and the number of checks written
“x” is graphed below. At what rate does the monthly charge for the checking account change?
Also, find the unit cost, another way to express the rate of change.
y
22
Monthly Charge for
checking account (\$)
20
18
16
(75, 16)
14
(50, 14)
12
10
8
x
10
20
30
40
50
60
70
80
90
100
Number of checks written during the month
Find the rate of change (slope of the line). The units will be dollars per number of checks.
From the graph, we see that two points on the line are (50, 14) and (75, 16). If we let
50,14 ) and ( x 2 , y2 ) ( 75,16 ) , we have
(=
=
( x1 , y1 )
Rate of
=
change
y2 − y1 ) dollars (16 − 14 ) dollars
(=
=
( x2 − x1 ) checks ( 75 − 50 ) checks
2dollars
25checks
Rate of change can also be expressed as \$0.08 per check
The rate of change can be expressed
as \$2 for every 25 checks.
or
8¢ per check.
The monthly cost of the checking account increases \$2 for every 25 checks written. To find the
cost of 1 check (unit cost), take the fraction which represents the rate of change, 252 , and divide
both the numerator and denominator by 25.
2 dollars
2 ÷ 25
.08 dollars
= =
= \$.08 / check. The unit cost = \$.08 per check.
25 checks 25 ÷ 25
1 check
48
APPLICATION OF LINEAR EQUATIONS
Problems
The graph models the number
of members in an organization
from 2001 to 2010.
.
..
150
(2007,125)
100
2010
2005
50
2001
Number of Members
(2001,200)
200
Year
How many members did the
organization have in 2009?
100 members
Find the rate of change.
(slope of the line)
loss of 25 members every
two years
What is the rate of change per year?
loss of 12.5 members per
year
49
FUNCTIONS
Recall that relation is a set of ordered pairs and that a function is a special type of relation.
A function is a set of ordered pairs (a relation) in which to each first component, there
corresponds exactly one second component.
The set of first components is called the domain of the function and the set of second
components is called the range of the function.
Examples: Determine if the following are functions.
Domain
x
−4
−8
−16
Range
y
0
1
2
Domain Range
x
y
2
1
7
3
5
This is a function, since for each
first component, there is exactly
one second component.
This is not a function since for
the first component, 7, there
are two different second
components, 3 and 5.
Vertical Line Tests: A graph in the plane represents a function if no vertical line intersects the
graph at more than one point.
Examples:
y
(3,2)
y
x
This is a graph of
a function. It
passes the vertical
line test.
x
(3,–2)
This is not a graph of a
function. For example,
the x-value of 3 is
assigned two different y
values, 2 and −2. (for the
first component, 3, there
are two different second
components, 2 and −2.)
Since we will often work with sets of ordered pairs of the form ( x, y ) , it is helpful to define a
function using the variables x and y.
Given a relation in x and y, if to each value of x in the domain there corresponds exactly one
value of y in the range, then y is said to be a function of x.
50
FUNCTIONS
Notation: To denote that y is a function of x, we write y = f ( x ) . The expression “ f ( x ) ” is read
f of x. It does not mean f times x. Since y and f ( x ) are equal, they can be used
interchangeably. This means we can write y = x 2 , or we can write f ( x ) = x 2 .
Evaluate: To evaluate or calculate a function, replace the x in the function rule by the given x
value from the domain and then compute according to the rule. For example:
Examples:
1. Given: f ( x=) 6 x + 5
Find:
5 17
f ( 2=
) 6 ( 2 ) +=
5 5
f ( 0=
) 6 ( 0) +=
f ( −1) =6 ( −1) + 5 =−1
2. Given: g ( x ) = 3x 2 − 5 x + 8
Find: g (=
1) 3 (1) − 5 (1) +=
8 6
2
g (=
0 ) 3 ( 0 ) − 5 ( 0 ) +=
8 8
2
g ( −1) = 3 ( −1) − 5 ( −1) + 8 = 16
2
51
FUNCTIONS
Problems
1. Determine whether or not each relation defines a function. If no, explain why not.
a.
Domain
Range
x
11
21
31
y
4
8
10
b. No, for each first component, 3,
there corresponds two different
second components, 2 and 7.
Domain Range
b.
x
3
5
3
c.
a. Yes, for each first component,
there corresponds exactly one
second component.
y
2
6
7
{(1,3) , ( 2, −4 ) , (1,0 )}
c. No, for the first component, 1, there
corresponds two different second
components 3 and 0.
d. Yes, for each first component there
corresponds exactly one second
component.
{( 4,3) , ( −2,3) , (1,3)}
d.
2. Determine the domain and range of each of the following:
a.
{( −3, 4 ) , ( 4, 2 ) , ( 0,0 ) , ( −2,7 )}
a. Domain: {−3, −2,0, 4}
Range:
b.
x
–5
–7
–9
c. x
b. Domain: {−5, −7, −9}
Range: {2, 4,6}
y
2
4
6
c. Domain: {1,5,9}
Range: {7,13, 21}
y
1
5
9
7
13
21
3. Given f ( x ) =
x + 2 and g ( x ) =
x − 3, find the following:
a.
b.
c.
d.
{0, 2, 4,7}
f ( −2 )
a. 0
b. –2
f ( −4 )
g ( 0)
c. –3
g ( 2)
d. –1
52
SYSTEMS OF LINEAR EQUATIONS
The following summary compares the graphing, elimination (addition), and substitution methods
for solving linear systems of equations.
Method
Graphing
Example
2x + y =
6
x − 2y =
8
Notes
You can see the solution is where the two
lines intersect but if the solution does not
involve integers it’s impossible to tell
exactly what the solution is.
y
5
2x+y=6
The solution for this system of linear
equations is ( 4, −2 ) .
2
-5-4-3-2-1-
-2
x-2y=8
1 2 34 5
(4,-2)
-4
-
Substitution
y 3x − 1
=
−4
3 x − 2 y =
Substitute 3 x − 1 for y :
3 x − 2 ( 3 x − 1) =
−4
Solve for x:
x=2
Back-substitute:
=
y 3 ( 2 ) −=
1 5
Solution:
Elimination
Gives exact solutions.
The solution for this system of equations
is ( 2,5 ) .
( 2,5)
−8
2 x + 3 y =
−34
5 x + 4 y =
Multiply the top equation by 5 and
the bottom equation by −2. This will
result in opposite coefficients of x.
− 40
10 x + 15 y =
−10 x − 8 y =
68
Add the two equations: 7 y = 28
y=4
To find x, substitute y = 4 into either
original equation.
2 x + 3( 4) =
−8
x = −10
Solution: ( −10, 4 )
53
Gives exact solution.
Easy to use if a variable is on one
side by itself. The solution for this system
of equations is ( −10, 4 ) .
SYSTEMS OF LINEAR EQUATIONS
Three Possible Solution Types
y
No Solution (Parallel Lines)
Inconsistent System
Independent Equations
x
0
y
.
0
Exactly One Solution (point of
intersection)
Consistent System
Independent Equations
x
Solution
Infinitely Many Solutions
(Lines Coincide-any point on the
line is a solution)
Consistent System
Dependent Equations
# = same # (identity)
y
0
x
54
SYSTEMS OF LINEAR EQUATIONS
Solving Linear Systems By Substitution Method
To solve a linear system of two equations in two variables by substitution:
1. Solve one of the equations for one of the variables.
2. Substitute the expression obtained in step 1 for that variable in the other equation, and solve the
resulting equation in one variable.
3. Substitute the value for that variable into one of the original equations and solve for the other
variable.
a. if we get 0 = nonzero number or # = different # (contradiction), the system is inconsistent,
the lines are parallel, the equations are independent, and there is no solution.
b. if we get 0 = 0 or # = same # (identity), the system is consistent, the lines are the same
(coincide), the equations are dependent, and there are infinitely many solutions.
Example:
Solve the system of linear equations using the substitution method.
3
3 x + 5 y =
x= 8 − 4 y
This equation is solved for x.
3(8 − 4 y ) + 5 y =
3
Substitute 8 − 4 y in for x in
the first equation.
24 − 12 y + 5 y =
3
24 − 7y =
3
− 7y =
−21
y = 3
Solve for y.
x=
8 − 4 ( 3) =
−4
Substitute 3 in for y to find the value of x.
The solution is ( − 4, 3 ) .
Check in both original equations:
3x + 5y =
3
3 ( −4 ) + 5 ( 3) =
3
− 12 + 15 =
3
3=3
x= 8 − 4 y
−4 = 8 − 4 ( 3)
−4 = 8 − 12
−4 =−4
55
SYSTEMS OF LINEAR EQUATIONS
Example:
Solve the system of linear equations using the substitution method.
4
4 x + 12 y =
11
−5 x + y =
Since y has a coefficient of 1, it will be easy to solve this
second equation for y.
4
4 x + 12 y =
y 5 x + 11
=
4 x + 12 ( 5 x + 11) =
4
Substitute 5 x + 11 in for y
4 x + 60 x + 132 =
4
64 x = −128
x = −2
Solve for x.
y = 5 ( −2 ) + 11
Substitute – 2 in for x to find the value of y.
y =1
The solution is
( −2,1) .
Check in both original equations:
4 x + 12 y =
4
− 5x + y =
11
4 ( −2 ) + 12 (1) =
4
−5 ( −2 ) + 1 =
11
− 8 + 12 =
4
4=4
10 + 1 =
11
11 = 11
56
SYSTEMS OF LINEAR EQUATIONS
Solving Linear Systems By Elimination (Addition) Method
To solve linear systems of two equations in two variables by elimination:
1. Write the system so that each equation is in standard form.
ax + by =
c
2. Multiply one equation (or both equations if necessary), by a number to obtain additive inverse
(opposite) coefficients of one of the variables.
3. Add the resulting equations and solve the new equation in one variable.
4. Substitute the value for that variable into one of the original equations and solve for the other
variable.
a. If we get 0 = nonzero number or # = different # (contradiction), the system
is inconsistent; there are no solutions, the equations are independent, and
the lines are parallel.
b. If we get 0 = 0 or # = same # (identity), the system is consistent; there are
infinitely many solutions, the equations are dependent and the lines are the
same (coincide).
Example:
Solve the system of linear equations using the elimination (addition) method.
9
x + y =
−3
2 x − y =
The coefficients of y are opposites (additive
3x = 6
x=2
Solve for x.
x+ y =
9
2+ y =
9
y=7
Substitute 2 in for x and solve for y.
The solution is (2,7).
Check in both original equations:
x+ y=
9
2+7=
9
9=9
−3
2x − y =
2 ( 2 ) − 7 =−3
4 − 7 =−3
− 3 =−3
57
SYSTEMS OF LINEAR EQUATIONS
Example:
Solve this system of linear equations using the elimination (addition) method.
11
3 x + 2 y =
−2
2 x − 8 y =
Multiplying the first equation by 4 yields 8y.This will result in
opposite coefficients (additive inverses) of y.
44
12 x + 8 y =
−2
2x − 8 y =
14 x = 42
x=3
3 ( 3) + 2 y =
11
Solve for x.
Substitute 3 in for x and solve for y.
9 + 2y =
11
2y = 2
y =1
The solution is (3,1).
Check in both original equations:
3x + 2 y =
11
3 ( 3) + 2 (1) =
11
9+2=
11
11 = 11
2x − 8 y =
−2
2 ( 3) − 8 (1) =
−2
6 − 8 =−2
− 2 =−2
58
SYSTEMS OF LINEAR EQUATIONS
Problem
Solve this System of Linear Equations using all three methods:
Graphing Method
Substitution Method
−1
x + y =
1
x − y =
Graphing Method:
y
0
x
Substitution Method:
Answer is on the following page.
59
SYSTEMS OF LINEAR EQUATIONS
y
2
–2
0
–2
Solution:
60
2
x
SYSTEMS OF LINEAR EQUATIONS
Problems
Solve these systems of linear equations using either the substitution or
1.
2.
11
x+ y =
1
x− y =
=
x 6=
and y 5
2x + y =
9
=
x 4=
and y 1
( 6,5)
( 4,1)
3x − y =
11
3.
4.
5.
2x + y =
5
5x − 2 y =
8
=
x 2=
and y 1
3x + 4 y =
7
8x − y =
7
=
x 1=
and y 1
( 2,1)
(1,1)
1
1
=
and y
2
3
1 1
,
2 3
=
x
2x + 3y =
2
10 x − 6 y =
3
6.
7.
8.
2x + 7 y =
23
x − 4y =
−11
=
x 1=
and y 3
x+ y =
3
2x + 3y =
8
=
x 1=
and y 2
5x + y =
5
2x + 3y =
2
=
x 1=
and y 0
(1,3)
(1, 2 )
(1,0 )
9.
4x − 4 y =
8
x− y =
1
Inconsistent system with no
solution.
10.
2x − y =
4
6x − 3y =
12
Dependent system with
infinitely many solutions.
61
APPLICATION OF A SYSTEM OF LINEAR EQUATIONS
The following steps are helpful when solving problems involving two unknown quantities.
1. Analyze the problem by reading it carefully to understand the given facts. Often a
diagram or table will help you visualize the facts of the problem.
2. Define variables to represent the two unknown quantities.
3. Translate the words of the problem to form two equations involving each of the two
variables.
4. Solve the system of equations using graphing, substitution, or elimination (addition)
method.
5. State the conclusion.
6. Check the results in the words of the problem.
Example:
Determine the cost of a quart of pineapple and a container of frozen raspberries for the punch Diane is
making. Three (3) quarts of pineapple juice and 4 containers of raspberries will cost \$10. Five (5)
quarts of pineapple juice and 2 containers of raspberries will cost \$12.
Let:
p = cost of a quart of pineapple juice
r = cost of a container of raspberries
Define the variables
10
3 p + 4r =
12
5 p + 2r =
Solve the system for “p” using
10
3 p + 4r =
−24
−10 p − 4r =
−7p =
−14
p=2
3 ( 2 ) + 4r =
10
Substitute 2 in for p and solve for r
r =1
Answer: A quart of pineapples costs \$2 and a container of frozen raspberries costs \$1.
Check:
Using \$2 as the cost for a quart of pineapple
juice, 3 quarts costs \$6. Using \$1 as the cost
of a container of raspberries, 4 containers cost
\$4. \$6 for pineapple juice and \$4 for
raspberries is \$10 total cost. Do the same
technique for checking the \$12 cost.
Check the results in the words
of the problem.
62
APPLICATION OF A SYSTEM OF LINEAR EQUATIONS
Problem
Use the steps for solving a system of linear equations.
People have begun purchasing tickets to a production
of a musical at a regional theatre. A purchase of 9
adult tickets and 7 tickets for the children costs \$116.
Another purchase of 5 adult tickets and 8 tickets for
the children costs \$85. Find the cost of an adult ticket
and a child’s ticket.
63
A child’s ticket costs \$5
and an adult ticket costs \$9.
SOLVING LINEAR INEQUALITIES IN TWO VARIABLES
Graphing Linear Inequalities In Two Variables
1. Replace the inequality symbol with an equal symbol = and graph the boundary line of the
region. If the original inequality allows the possibility of equality (the symbol is either ≤ or ≥ ),
draw the boundary line as a solid line. If equality is not allowed (< or >), draw the boundary line
as a dashed line.
2. Pick a test point that is on one side of the boundary line. (Use the origin if possible; it is
easier). Replace x and y in the inequality with the coordinates of that point. If a true statement
results, shade the side (half-plane) that contains that point. If a false statement results, shade the
other side (other half-plane).
Example:
Solve 4 x + y ≥ 2
2
Step 1: 4 x + y =
≥ sign, draw the solid boundary line
y=
− 4x + 2
y
2
0
-2
x
2
-2
Step 2: pick (0,0)
4 ( 0) + 0 ≥ 2
Pick test point.
Shade the side that contains (0,0).
4 ≥ 2is true
y
.
2
0
x
2
( 0,0 )
64
SOLVING LINEAR INEQUALITIES IN TWO VARIABLES
Problems
1
2
1. Solve y ≤ x − 1
y
x
2. Solve 2 y + 3x < 6
y
x
3. Solve y ≥ 5
y
x
65
SOLVING LINEAR INEQUALITIES IN TWO VARIABLES
y
1.
4
2
x
4
y
2.
5
5
3.
x
y
2
2
x
66
SOLVING SYSTEMS OF LINEAR INEQUALITIES
1. Graph each inequality on the same rectangular coordinate system.
2. Use shading to highlight the intersection of the graphs (the region where the graphs overlap).
The points in this region are the solutions of the system.
3. As an informal check, pick a point from the region where the graphs intersect and verity that
its coordinates satisfy each inequality of the original system.
x + y ≥ 4
Example: Solve this system of inequalities.
4 x − 3 y > 9
Look at the graph of each inequality separately.
x+ y ≥4
y
≥ sign, draw solid boundary line.
Use (0, 0) as a test point.
0 ≥ 4 is false.
Shade the half-plane that does not include (0, 0).
4
4
x
4x − 3y > 9
y
> sign, draw dashed boundary line.
Use (0, 0) as a test point.
0 > 9 is false.
Shade the half-plane that does not include (0, 0).
3
x
−3
67
SOLVING SYSTEMS OF LINEAR INEQUALITIES
x + y ≥ 4
4 x − 3 y > 9
y
Graph each inequality on the same
coordinate system.
Notice the region where the graphs
overlap.
The points in this region are the
solutions of the system.
0
Solution of the System:
y
4
0
x
3
68
SOLVING SYSTEMS OF LINEAR INEQUALITIES
y
Informal Check:
The point (5,2) is in the region
of overlap. Substitute (5,2) in to both
inequalities.
4
x+ y≥4
5+ 2 4
7 ≥ 4 is true
4x − 3y > 9
• (5,2)
4 ( 5) − 3( 2 ) 9
20 − 6 9
14 > 9 is true
0
(5,2) is one of the solutions.
69
3
x
SOLVING SYSTEMS OF LINEAR INEQUALITIES
Problems
1. Solve this system of linear inequalities.
4
y ≥ − 3 x + 2
y > 5 x +1
6
y
x
2. Solve this system of linear inequalities.
3 x + y ≥ −5
−3 x − y ≥ −5
y
x
70
SOLVING SYSTEMS OF LINEAR INEQUALITIES
y
1.
5
2
x
3
-2
y
2.
7
4
x
71
SOLVING ABSOLUTE VALUE EQUATIONS
Absolute Value:
The absolute value of a number is its distance from 0 on the number line
For any positive number k and any algebraic expression X:
To solve X = k , solve the equivalent compound equation
X= k
X = −k
or
Example: x = 7
Solution: x = 7 or x = -7
–7
0
7
Example: x − 2 =
3
x − 2 =3 or
Solution: x = 5
x − 2 =−3
x = -1
–1
Check:
0
x=5
5
Check:
x = −1
x−2 =
3
x−2 =
3
5−2 =
3
−1 − 2 =3
3 =3
−3 =
3
3=3
3=3
72
SOLVING ABSOLUTE VALUE EQUATIONS
Example: Solve the following:
4 x − 64
= 32
4
Solve the equation by
rewriting it as two separate
equations.
4 x − 64
4 x − 64
= 32 or
= −32
4
4
When X = k , then X = k
and X = − k
Solve each equation for x.
4 x − 64 =
128
4 x − 64 =
−128
Multiply both sides by 4.
4 x = 192
4 x = −64
Solutions:
x = 48
Check:
or
Divide both sides by 4.
x = -16
Check x = 48
Check x = −16
4 ( 48 ) − 64
4 ( −16 ) − 64
4
= 32
4
= 32
192 − 64
= 32
4
−64 − 64
= 32
4
128
= 32
4
−128
= 32
4
32 = 32
32
−32 =
32 = 32
32 = 32
73
The two solutions check.
SOLVING ABSOLUTE VALUE EQUATIONS
Example: Solve the following: 5 x − 7= 4 ( x + 1)
Solve the equation by
rewriting it as two separate
equations.
Solve each equation for x.
5 x − 7 =4 ( x + 1) or 5 x − 7 =− 4 ( x + 1)
5 x − 7 =− [ 4 x + 4]
5 x − 7 =4 x + 4
x − 7 =4
x=
11
5 x − 7 =−4 x − 4
9 x − 7 =−4
9x = 3
1
=
x 11
=
x
or
Solutions:
3
Check:
Check x = 11
5 (11) − 7= 4 (11 + 1)
Check x =
When X = k , then
X = k and X = − k
Use the Distributive
Property.
1
3
1
1
5 − 7= 4 + 1
3
3
55 − 7 =
4 (12 )
5
4
−7 =
4
3
3
48 = 48
−16 16
=
3
3
48 = 48
16 16
=
3
3
74
The two solutions
check.
SOLVING ABSOLUTE VALUE EQUATIONS
Problems
1.
3x − 2 =
5
2.
10 x − x =
−40
No solution. You can’t solve an
absolute value equation when the
absolute value is equal to a negative
quantity.
3.
2
x+3 +4 =
10
3
x=
4.
3
5.
5 x + 3 = 3 x + 25
6.
2 x + 1= 3 ( x + 1)
x=
7
3
9
2
or x = −1
or x = −
27
2
x = 10
1
x − 5 − 4 =−4
2
x = 11 or x = −
7
2
4
x=
−2 or x =
−
5
75
SOLVING ABSOLUTE VALUE INEQUALITIES
To solve inequalities with absolute value signs, there are two cases:
Case 1:
For any positive number k and any algebraic expression X
To solve X < k , solve the equivalent compound inequality − k < X < k .
To solve X ≤ k , solve the equivalent compound inequality − k ≤ X ≤ k .
Example:
Solution:
Case 2:
x <7
Solve:
Interval Notation: (-7,7)
-7 < x < 7
-7
0
or
7
-7
0
7
For any positive number k and any algebraic expression X
To solve X > k , solve the equivalent compound inequality X < −k or X > k .
To solve X ≥ k , solve the equivalent compound inequality X ≤ −k or X ≥ k .
Example:
x ≥7
Solve:
Solution:
x ≤ −7
-7
or
Interval Notation: ﴾–∞,–7] ∪ [7,∞﴿
x≥7
7
0
or
-7
7
0
76
SOLVING ABSOLUTE VALUE INEQUALITIES
Example:
x−2 ≤3
−3 ≤ x − 2 ≤ 3
−1 ≤ x ≤ 5
Solve:
Solution:
Case 1
Equivalent compound inequality
−1 ≤ x ≤ 5
Interval Notation: [–1,5]
5
-1 0
or
5
-1 0
Example:
Solve:
x−2 >3
x − 2 < −3 or x − 2 > 3
x < −1
x>5
Solution:
Case 2
Equivalent inequalities
x < −1 or x > 5
-2 -1
Interval Notation:
0
5
or
-2 -1
0
5
77
( ∞, −1) ∪ ( 5, ∞ )
SOLVING ABSOLUTE VALUE INEQUALITIES
Problems
1.
x −5 < 3
2< x<8
( 2,8)
2.
x − 7 > 10
x < −3 or x > 17
( −∞, −3) ∪ (17, ∞ )
3.
5 − 3 x ≤ 14
−3 ≤ x ≤
19
3
19
−3, 3
4.
2x − 3 < 9
−3 < x < 6
( −3, 6 )
5.
3− x
≥6
5
x ≤ −27 or x ≥ 33
6.
3x − 8 − 4 ≥ 0
( −∞, −27] ∪ 33, ∞ )
4
or x ≥ 4
3
4
−∞, ∪ [ 4, ∞ )
3
x≤
78
VOCABULARY USED IN APPLICATION PROBLEMS
English
Algebra
sum
The sum of a number and 4.
x+4
total
Seven more than a number.
x+7
plus
Six increased by a number.
6+ x
in all
8+ x
more than
A number plus 4.
x+4
together
increased by
all together
combined
Subtraction
subtracted from
English
Algebra
difference
The difference of a number and 3.
x −3
take away
The difference of 3 and a number.
3− x
less than
Five less than a number.
x −5
minus
A number decreased by 3.
x −3
remain
A number subtracted from 8.
8− x
decreased by
Eight subtracted from a number.
x −8
have left
Two minus a number.
2− x
are left
more
fewer
Be careful with subtraction. The order is important. Three less than a
number is x−3.
79
VOCABULARY USED IN APPLICATION PROBLEMS
Multiplication
English
Algebra
product of
The product of 3 and a number.
3x
multiplied by
Three-fourths of a number.
times
Four times a number.
3
x
4
4x
of
A number multiplied by 6.
6x
Double a number.
2x
Twice a number.
2x
English
Algebra
Division
divided by
quotient of
The quotient of a number and 3.
The quotient of 3 and a number.
separated into equal
parts
A number divided by 6.
shared equally
Six divided by a number.
x
3
3
3 ÷ x or
x
x
x ÷ 6 or
6
6
6 ÷ x or
x
x ÷ 3 or
Be careful with division. The order is important. A number divided
by 6 is x
6
80
SOLVING APPLICATION PROBLEMS
Strategy for solving word problems:
1. Analyze the problem: Read the problem carefully.
2. Visualize the facts of the problem (if needed): Use diagrams and/or tables.
3. Define the variable/s: Identify the unknown quantity (or quantities) and label them,
i. e., let x = something.
4. Write an equation: Use the defined variable/s.
5. Solve the equation: Make sure you have answered the question that was asked.
Examples:
1. The sum of three times a number and 11 is –13. Find the number.
Let x = the number
Define the variable.
−13
3 x + 11 =
Write an equation..
−13 − 11
3 x + 11 − 11 =
3 x = −24
x = −8
Solve the equation.
Solution: The number is – 8.
− 24
Check: 3 ( −8 ) =
− 24 + 11 =
−13
2. Together, a lot and a house cost \$40,000. The house costs seven times more than the lot.
How much does the lot cost? The house?
Let x = the cost of the lot
7x = the cost of the house
x + 7x =
40,000
Write an equation
8 x = 40,000
x = 5,000
7x = 7•5000 = 35,000
Solution:
Define the variable.
The lot costs \$5,000.
The house costs \$35,000.
Check: 5,000 + 35,000 = 40,000
7•5000 = 35,000
81
Solve the equation.
SOLVING APPLICATION PROBLEMS
Problems:
1.
Five plus three more than a number is nineteen. What is the number?
2.
When 18 is subtracted from six times a certain number, the result is 96. What is the
number?
3.
If you double a number and then add 85, you get three-fourths of the original number.
What is the original number?
4.
A 180-m rope is cut into three pieces. The second piece is twice as long as the first. The
third piece is three times as long as the second. How long is each piece of rope?
5.
Donna and Melissa purchased rollerblades for a total of \$107. Donna paid \$17 more for
her rollerblades than Melissa did. What did Melissa pay?
6.
A student pays \$278 for a calculator and a keyboard. If the calculator costs \$64 less than
the keyboard, how much did each cost?
82
SOLVING APPLICATION PROBLEMS
1.
5 + ( x + 3) =
19
The number is 11.
x = 11
2.
3.
4.
6 x − 18 =
96
x = 19
3
2 x + 85 =x
4
x = −68
x + 2 x + 3( 2 x ) =
180
The number is 19.
The number is –68.
The lengths are 20 m, 40 m, and 120 m.
x = 20
5.
6.
107
x + x + 17 =
x = 45
278
x + ( x − 64 ) =
Melissa paid \$45.
The keyboard costs \$171, and the calculator costs \$107.
x = 171
83
SOLVING APPLICATION PROBLEMS
Problems
Solve the following word problems using one variable or two variables in a system.
Integer Problems
1. The sum of three consecutive integers is 144. Find the integers.
2. The sum of three consecutive even integers is 84. Find the integers.
3. The sum of three consecutive odd integers is 111. Find the integers.
Perimeter Problems
4. The length of a rectangle is seven more than the width. The perimeter is 34 inches. Find the
length and width.
5. The length of a rectangle is 3 inches less than twice the width. The perimeter is 18 inches.
Find the length and width.
6. The length of one side of a triangle is 4 inches more than the shortest side. The longest side
is two inches more than twice the length of the shortest side. The perimeter is 26 inches.
Find the length of all three sides.
84
SOLVING APPLICATION PROBLEMS
Mixture Problems
7. For Valentine’s Day, Candy, a candy store owner, wants a mixture of candy hearts and foilwrapped chocolates. If she has 10 pounds of candy hearts, which sell for \$2 per pound, how
many pounds of the chocolates, which sell for \$6 per pound, should be mixed to get a
mixture selling at \$5 per pound?
8. The same candy store owner has 24 pounds of chocolate creams which sell for \$12 per
pound. How many pounds of chocolate caramel nut clusters, which sell for \$9 per pound,
should he mix to get a mixture selling at \$10 per pound?
9. Sharon wants to make 100 pounds of holiday mix for her baskets. She purchases cashews at
\$8.75 per pound and walnuts at \$3.75 per pound. She feels she can afford a mixture which
costs \$6.35 per pound. How much of each type of nut should she purchase to make the mix?
Distance Problems
10. Two cars leave Chicago at 11 a.m. headed in opposite directions. At 2 p.m., the two cars are
375 miles apart. If one car is traveling 5 mph faster than the other, what are their speeds?
11. A plane leaves Chicago headed due west at 10 a.m. At 11 a.m., another plane leaves
Chicago headed due east. At 1 p.m., the two planes are 2950 miles apart. If the first plane is
flying 100 mph slower than the second plane, find their rates.
12. A car leaves Milwaukee at noon headed north. Five hours later it arrives at its destination. A
second car traveling south at a rate 10 mph slower leaves Milwaukee at 3:00 and arrives at
its destination two hours later. When they arrive at their destination, they are 400 miles
apart. How fast is each of the cars traveling?
13. A bicyclist can ride 24 miles with the wind in 2 hours. Against the wind, the return trip takes
him 3 hours. Find the speed of the wind. (Hint: Solve using systems of equations.)
85
SOLVING APPLICATION PROBLEMS
Investment Problems
14. A couple wants to invest \$12,000 in two retirement accounts, one earning 6% and the other
9%. How much should be invested in each account for them to earn an annual interest of
\$945?
15. An investor has put \$4,500 in a credit union account earning 4% annual interest. How much
should he invest in an account which pays 10% annual interest to receive total annual
interest of \$1,000 from the two accounts.
16. Three accounts generate a total annual interest of \$1,249.50. The investor deposited an equal
amount of money in each account. The accounts paid an annual rate of return of 7%, 8%,
and 10.5%. How much was invested in each account?
Number-Value Problems
17. The admission prices for a movie theater in Crystal Lake are \$9 for adults, \$8 for seniors,
and \$5 for children. A family purchased twice as many children’s tickets as adults and the
same number of senior tickets as adults. The total cost of the tickets was \$54. How many of
each type of ticket was purchased?
18. Marie has \$2.20 worth of quarters, dimes, and nickels. She has 3 times as many nickels as
quarters and 3 fewer dimes than quarters. How many of each type of coin does she have?
19. Deb went shopping at the school bookstore and purchased \$53 of computer items. The CD’s
cost \$2 each, the DVD’s cost \$3 each, and the flash drives were \$15 each. The total cost was
\$53. He purchased one more DVD than CD’s and half as many flash drives as CD’s. How
many of each did he purchase?
86
SOLVING APPLICATION PROBLEMS
1.
The 3 consecutive integers are 47, 48, and 49.
2.
The 3 consecutive even integers are 26, 28, and 30.
3.
The 3 consecutive odd integers are 35, 37, and 39.
4.
The length is 12 inches, and the width is 5 inches.
5.
The length is 5 inches, and the width is 4 inches.
6.
The lengths are 5inches, 9 inches, and 12 inches.
7.
The store owner should add 30 pounds of the chocolates worth \$6 per pound.
8.
The store owner should mix 48 pounds of chocolate caramel nut clusters worth \$9 per
pound.
9.
52 pounds of cashews and 48 pounds of walnuts.
10.
One car is traveling 60 mph; the other is traveling 65 mph.
11.
The first plane is traveling 550 mph. The second plane is traveling 650 mph.
12.
One car is traveling 60 mph; the other is traveling 50 mph.
13.
The wind speed is 2 miles per hour.
14.
\$4,500 invested at 6% and \$7,500 at 9%.
15.
\$8,200 at 10%.
16.
\$4,900 in each account.
17.
2 adult tickets, 4 children’s, and 2 senior tickets.
18.
5 quarters, 2 dimes, and 15 nickels.
19.
4 CD’s, 5 DVD’s, and 2 flash drives.
87
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
Problems:
1. Evaluate the following:
a. −32 + 42 − 52
b. ( 4 − 5 )
c.
20
−3 − ( −7 )
22 − 3
d. 12 − 2 1 − ( −8 + 2 )
2. If x = 2 and y = -4, evaluate:
a. x − xy
b.
x2 − y 2
3x + y
3. Solve the following equations for x:
2x
a. 3 ( x − 5 ) + 2 =
b.
x −5
−5 =
7
3
c.
2
1
x +1 = + x
5
3
d.
−5
x = 15
8
88
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
4. Solve the following for the indicated variable:
a. Solve: P= 2l + 2 w for w.
1
b. Solve: A = bh for h.
2
c. Solve: 3 x + 2 y =
5 for y.
a b c
d. Solve: + = for a.
2 3 4
a. 3 − ( 2 x + 4 ) < −2 ( x + 3) + x
b. −5 x + 4 ≥ 6
6. Simplify the following:
a. ( −3x 2 y 2 )
b. ( 2 y )
2
−4
c. ( x 3 x 4 )
2
d. ( ab −3c 4 )( ab 4 c −2 )
e.
a 4b 0
a −3
4t −3t 4t −5
f.
2 −6
3t t
3
89
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
7. Perform the indicated operations:
a. ( 4c 2 + 3c − 2 ) + ( 3c 2 + 4c + 2 )
b. 3 x ( 2 x + 3)
2
c. ( 2t + 3s )( 3t − s )
d. (10 x 2 + 11x + 3) ÷ ( 5 x + 3)
e. ( 5 x − 8 y ) − ( −2 x + 5 y )
f. ( −3 x + y ) ( x 2 − 8 xy + 16 y 2 )
g. ( 2 x − 32 ) ÷ 16 x
h. ( x + y )( x − y ) + x ( x + y )
8. Graph the following:
a. 4 x − 3 y =
12
b. x = 4
c. 3 x − 8 y =
0
d. y = −2
9. Find the x- and y-intercept of the following:
a. 3 x − 2 y =
5
b. 2 x − 3 y =
0
c. x = 4
d. y = −2
90
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
10. Find the equation of the following lines:
a. A line passing through (-2,4) and (6, 10)
2
passing through (0, 5)
3
1
c. A line through (1, 2) with a slope of
2
b. A line with a slope of
d. A horizontal line through (-3, 5)
e. A vertical line through (-3, 5)
f. A line with an undefined slope through (3, -2)
11. Find the slope of the following:
a. A line through (-3, 7) and (2, -5)
b. The line 3 x − 8 y =
4
c. The line y = 7
d. The line x = -4
12. Solve the following systems:
−2
x + 4 y =
a.
y =− x − 5
2
3 x − y =
b.
8
2 x + y =
0
3 x − 5 y − 16 =
c. x 5
1
2 − 6 y =
3
−3 y + 6
x =
d.
12
2 x + 6 y =
y
x − 4 =
e.
−2 y =4 − 2 x
91
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
a. 3 x − 2 < 4 and −3 x < 3
b. −2 ≤ 2 x − 5 < 6
c. −2 x + 4 > 8 or x − 1 ≥ 2
d. 3 x − 2 < 4
e. 2 x + 7 ≥ 4
f. −2 3 x − 1 > 6
g. 2 x − 4 − 3 < 2
14.
Factor.
a. 14 xyz − 16 x 2 y 2 z
b. −r 2 − 14r − 45
c. 6s 5 − 26s 4 − 20s 3
d. 9 z 2 − 1
15.
The length of a rectangle is one foot longer than twice the width. The perimeter is 20 inches.
Find the length and width.
16.
The longest side of a triangle is two centimeters longer than twice the shortest side. The
medium length side is three centimeters longer than the shortest side. The perimeter is 25
centimeters. Find the lengths of the three sides.
17.
One train leaves Chicago at 7 p.m. At 9 p.m., another train leaves Chicago heading in the
opposite direction, traveling 10 mph faster. At 11 p.m., the two trains are 530 miles apart.
How fast was each train moving?
18.
Scotty inherited \$20,000 which he invested in two different accounts. The first account paid
out an interest rate of 6%. The second account paid out an interest rate of 3%. After 1 year,
Scotty had earned \$750 interest. How much did he invest in each account
92
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
1.
a. 0
b. 1
c. 4
d. –2
2.
a. 10
b. –6
3.
a. 13
b. 41
c.
10
9
d. –24
4.
a. w=
or w =
h=
c.
3
5
y=
− x+
2
2
a.
P − 2l
2
2A
b
b.
d. =
a
5.
P
−l
2
1
2
3c − 4b
c − b or a =
2
3
6
( 5, ∞ )
2
b. −∞, −
5
93
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
6.
a. 9x 4 y 4
b.
1
16 y 4
c. x14
d. a 2bc 2
e. a 7
f.
7.
64
27
a. 7c 2 + 7c
b. 12 x3 + 36 x 2 + 27 x
c. 6t 2 + 7 st − 3s 2
d. 2x + 1
e. 7x – 13y
f. −3 x3 + 25 x 2 y − 56 xy 2 + 16 y 3
g.
1 2
−
8 x
h. 2x 2 + xy − y 2
94
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
8.
a.
4x – 3y = 12
y
x
3
−4
8. b.
x=4
y
5
5
x
95
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
8. c.
3x -8y = 0
y
4
2
8. d.
7
x
y = −2
y
2
2
x
96
COMPREHENSIVE REVIEW
OF ELEMENTARY ALGEBRA
y = -2
9.
a. x-intercept =
5
,0
3
5
y-intercept = 0, −
2
b. x-intercept = (0, 0)
y-intercept = (0, 0)
c. x-intercept = (4, 0)
No y-intercept
d. No x-intercept
y-intercept (0, -2)
10. a. =
y
3
11
x+
4
2
b. =
y
2
x+5
3
c. =
y
1
3
x+
2
2
d. y = 5
e. x = -3
f. x = 3
11. a. m = −
b. m =
12
5
3
8
c. m = 0
d. Slope undefined
97
COMPREHENSIVE REVIEW OF ELEMENTARY ALGEBRA
12. a. ( -6, 1 )
b. ( 2, 4 )
c. No solution
d. Infinite solutions
e. No solution
13. a. (-1,2)
3 11
b. ,
2 2
c. ( −∞, −2 ) ∪ [3, ∞ )
2
d. − , 2
3
11 3
e. −∞, − ∪ , ∞
2 2
f. no solution
1 9
g. − ,
2 2
14. a. 2 xyz ( 7 − 8 xy )
b. − ( r + 9 )( r + 5 )
c. 2 s 3 ( 3s + 2 )( s − 5 )
d.
( 3z + 1)( 3z − 1)
15. The length is 7 feet and the width is 3 feet.
16. Short side is 5 cm, medium side is 8 cm, and the longest side is 12 cm.
17. The train that left at 7:00 p.m., travelled at 85 mph.
The train that left at 9:00 p.m., travelled at 95 mph.
18. \$15, 000 at 3% and \$5, 000 at 6 %.
98
```
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Derive the formula to find areas underneath curves
In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…
In this diagram what you will discover is that:
• A length a exists, which starts at the origin O and ends at a;
• A length x exists, which starts at the origin O and ends at x;
• A length x+𝛿x exists, which starts at the origin O and ends at x+𝛿x;
• A length 𝛿x exists, which starts at x and ends at x+𝛿x;
• A height y exists, which starts at the origin O and ends at y;
• A height y+𝛿y exists, which starts at the origin O and ends at y+𝛿y;
• A height 𝛿y exists, which starts at y and ends at y+𝛿y;
• There is a curve called y=f(x);
• There is an area underneath the curve called A which commences at a and ends at x;
• There is an area underneath the curve called 𝛿A which commences at x and ends at x+𝛿(Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called 𝛿A);
• There is a rectangle that exists called QRUT. It has an area which is y𝛿x;
• There is a rectangle that exists called PRUS. It has an area which is (y+𝛿y)𝛿x;
• 𝛿A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.
Producing the formula with the information we’ve discovered…
Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…
Let’s start off by saying that:
Area QRUT < 𝛿A < Area PRUS
Which is something we already discovered.
If this is the case, we can say that:
y𝛿x < 𝛿A < (y+𝛿y)𝛿x
Now, check out what happens when we divide all the elements of this expression by 𝛿x:
What we end up with is…
Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:
This is because as 𝛿x approaches 0, 𝛿y approaches 0 leaving (𝛿A)/(𝛿x) sandwiched between y and y+0.000000000000000001 which is virtually y.
And, also…
As a consequence, this ultimately means that:
$y=\frac { dA }{ dx }$
This is incredibly significant, because if we then integrate both sides of this equation, we get:
$\int { ydx=\int { \frac { dA }{ dx } } } dx\quad \Rightarrow \quad A=\int { ydx }$
And…
$A=\int { ydx } =F\left( x \right) +C$
Now, this equation can actually be used to find the area A underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.
Finalising the formula…
Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:
$A=\int { ydx } =F\left( x \right) +C$
However, it is not complete. We need to know what the constant C is. So…
If we say that at x=a the area A underneath the curve is 0, watch what happens… Look at what we get…
$O=F\left( a \right) +C$
Which means that:
$C=-F\left( a \right)$
Hence, we can conclude that:
$A=\int { ydx=F\left( x \right) } -F\left( a \right)$
And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b
This is probably the formula you’re most familiar with…
$A=\int _{ a }^{ b }{ ydx=F\left( b \right) } -F\left( a \right) ={ \left[ F\left( x \right) \right] }_{ a }^{ b }$
Which is the formula which can be used to find areas underneath curves.
If you are still confused and would like to go through this proof once again, please watch my video below…
You can also leave your comments below.
Related:
Trapezium Rule Formula – Derivation
You can now find out how to derive the 3 main cosine rule formulas through a new document that I’ve created called “Cosine Rule Mastery“.
This document can be downloaded free of charge along with “Sine Rule Mastery” which is another document that explains in detail how to come up with the sine rule formula.
I’d also like to talk about a new video I’ve created (posted below). It’s related to a 4 dimensional hypercube and learning how to train your mind to see things from different mathematical perspectives.
Prior to posting up the video above, I did create another similar video. In the video below, you will see me split a prism into 3 equal parts. This video will interest those who’d like to find the volume of a square based pyramid.
Trapezium Rule Formula – Derivation
Find out how to come up with the Trapezium Rule formula from scratch.
1) Derive the formula for the area of trapeziums:
2) Use the area of trapeziums formula to come up with the Trapezium Rule formula:
Related:
Derive the formula to find areas underneath curves
Areas Of Triangles & The Sine Rule
Learn how to come up with the formula for areas of triangles, then use this formula to derive the sine rule formula.
Related:
The quickest Sine Rule proof
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NCERT Solutions for Class 6 Maths Chapter 11 Exercise 11.4 in Hindi and English Medium updated for CBSE 2022-2023.
• 6th Maths Exercise 11.4 Solutions in Hindi and English Medium
Class 6 Maths Chapter 11 Exercise 11.4 Solution
Class VI Mathematics NCERT (https://ncert.nic.in/) book Ex. 11.4 solution of chapter 11 Algebra all question answers with proper explanation updated for CBSE current session 2022-23. Hindi Medium as well as English Medium contents are given here with videos solution. Class 6 math exercise 11.4 deals with the questions related to word problems or application based questions based on algebraic expression. NCERT Questions are done in simple manner so that every student can understand easily.
Class: 6 Mathematics Chapter: 11 Exercise: 11.4 Chapter Name: Algebra Content: NCERT Textbook’s Solution Medium: Hindi and English Medium
Free App for Class 6
Properties of Multiplication of Literals
Like numerals, multiplication of literal numbers has the following properties:
For all literal numbers a, b and c:
• (i) ab = ba
• (ii) a (bc) = (ab)c
• (iii) 1a = a = a1
• (iv) 0a = 0 = a0
• (v) a (b + c) = ab + ac; (b + c) a = ba + ca
• (vi) a (b – c) = ab – ac; (b – c) a = ba – ca
Class 6 Maths Chapter 11 Exercise 11.4 Solution in Videos
Division of literal numbers
The division (÷) sign between two numbers means that the number on the left of it is to be divided by the number on the right.
In the case of literal numbers also, a ÷ b means that the literal number a is to be divided by the literal number b and is commonly written as a/b (b ≠0). For example, we write:
(i) a divided by 3 as a/3
(ii) one-fourth of a + b as (a + b)/4
(iii) one-half of the product of a and b as ab/2
(iv) 12 divided by x (x ≠0) as 12/x
Variables and Constants
A symbol having a fixed numerical value is called a constant whereas a symbol which takes on various numerical values is called a variable.
For example, the perimeter P of a square is given by the formula P = 4s, where s is the length of side of a square. Here 4 is a fixed number and hence a constant but the literal numbers P and s depend on different sizes of the square and hence they are variables.
Similarly, the volume of a sphere is given by V = 4/3 πr3
Here 4/3 and π are fixed numbers, i.e., constants but the literal numbers V and r depend on different sizes of the sphere and hence they are variables.
The score of Rashmi in English is 15 more than the three-fourth of her score in Hindi. Determine her score in English if she scored x marks in Hindi.
We have, Score in Hindi = x
Three-fourth of score in Hindi = (¾)x
So, 15 more than three-fourth of score in Hindi = 3x/4 + 15
Hence, Rashmi’s score in English = (¾)x + 15
What are mathematical literals?
In mathematical logic, a literal is an atomic formula (atom) or its negation. The definition mostly appears in proof theory (of classical logic), e.g. in conjunctive normal form and the method of resolution. Literals can be divided into two types: A positive literal is just an atom.
Mayank spends Rs. x daily and saves Rs y per week. What is his income in 5 weeks?
We have, 1 week = 7 days
So, 5 weeks = (7 × 5) days = 35 days
As Mayank spends Rs. x daily.
So, Total amount spent by him in 5 weeks (i.e., 35 days) = Rs (35 times x) = Rs 35x
He saves Rs. y per week.
So, Total amount saved by him in 5 weeks = Rs (5 times y) = Rs 5y
Now, income = expenditure + savings
So, His income in 5 weeks = Rs. (sum of 35x and 5y) = Rs (35x + 5y)
What is difference between literal and variable?
A literal is notation for representing a fixed (constant) value. A variable is storage location associated with a symbolic name (pointed to, if you’d like)
Algebraic Expression
A combination of constants and variables, connected by the symbols +, –, × and ÷ is called an algebraic expression.
The various parts of the algebraic expression separated by the sign + or – are called the terms of the expression.
Thus, (i) the algebraic expression 3x – 4y + 2xy has three terms, namely 3x, –4y and 2xy.
(ii) The algebraic expression 5a + 2b – 6 ab + 7 has four terms namely 5a, 2b, –6ab and 7.
Is exercise 11.4 of class 6th Maths complex ot solve?
Yes, exercise 11.4 of class 6th Maths is complex. In this exercise, all questions are word problems. Most of the students find word problems complicated. In this exercise, students have to use expressions practically.
How many questions are there in exercise 11.4 of class 6th Maths?
Exercise 11.4 of 6th class Maths has three questions. Question 1 has five parts, question 2 has five parts, and question 3 has three parts. This exercise doesn’t have any examples.
Which questions of exercise 11.4 of class 6th Maths can students expect in the school exams?
Exercise 11.4 of 6th class Maths has three questions. Questions of exercise 11.4 of class 6th Maths that students can expect in the school exams are questions 1 (a), (c), (d), (e), and question 2.
Is exercise 11.4 of 6th standard Maths lengthy?
Exercise 11.4 of 6th standard Maths is not very lengthy and not very short. It is moderate. Students need a maximum of 2 days to complete exercise 11.4 of class 6th Maths if they give 1 hour per day to this exercise. This time can increase or decrease according to the student’s working speed.
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Banking Exam Preparation (IBPS Exam) – “MULTIPLICATION” Part II
Banking Exam Preparation (IBPS Exam) – “MULTIPLICATION” Part II
Multiply by 2
Method – 1
Direct multiply
Double each number and carry forward unit digit to next digit
Method – 2
Minus 5
Double each digit less than 5 (a) and double each digit more than 4 after minus 5 (b).
Add 1 to each digit before each (b).
Banking Exam Preparation (IBPS Exam) – “MULTIPLICATION” Part II
Multiply by 3
Step
1. Divide first digit by 2 and remove all digits after point.
2. Minus all digits from 9 but unit digit from 10.
3. We find a new number.
4. Double each digit of new number and add half of next main number’s digit (Ignore Digit after point).
5. Add extra 5 if main digit is odd and 0 if even.
6. Carry forward answer’s digit before unit digit to next digit.
7. Minus 2 from first digit of finding.
E.g. 46357 x 3
Step
1. 4/2 = 2
2. 9 – 4, 9 – 6, 9 – 3, 9 – 5, 10 – 7 = 53643
3. 53643
4. (5 x 2) + 6/2, (3 x 2) + 3/2. (6 x 2) + 5/2, (4 x 2) + 7/2, (3 x 2) + 0/2 = 13, 7, 14, 11, 6
5. 13 + 0, 7 + 0, 14 + 5, 11 + 5, 6 + 5
6. 13, 07, 19, 16, 11 = First digit 2 = 339071
7. Finding 339071 minus 2 from first digit = 1339071
1 2 3 4 5 6 Main Number Minus 9/10 Double 2 Odd/Even ½ of next digit 3+4+5 0 0 0 0 2 2 4 5 10 0 3 13 6 3 6 0 1 7 3 6 12 5 2 19 5 4 8 5 3 16 7 3 6 5 0 11
339071
Now Minus 2 from first digit 3 – 2 = 1
Banking Exam Preparation (IBPS Exam) – “MULTIPLICATION” Part II
Multiply by 4
Same as multiply by 3 but after minus 9/10 not doubled numbers direct add half of next digit.
After finding minus 1 from first digit in place of 2.
E.g. 37485 x 4
1 2 3 4 5 Main Number Minus 9/10 Odd/Even ½ of next digit 2+3+4 0 0 0 1 1 3 6 5 3 14 7 2 5 2 9 4 5 0 4 9 8 1 0 2 3 5 5 5 0 10
Now minus 1 from first digit 2 – 1 = 1
Banking Exam Preparation (IBPS Exam) – “MULTIPLICATION” Part II
Multiply by 5
1. Take 5 for each odd digit and 0 for each even digit. Now add half of next digit (ignore digits after point)
E.g. 78439 x 5
1 2 3 4 Main Number Odd/Even ½ of next digit 2+3 0 0 3 3 7 5 4 9 8 0 2 2 4 0 1 1 3 5 4 9 9 5 0 5
2. Multiply by 10 and divide by 2.
E.g. 78439 x 5
78439 x 10 = 784390
784390/2 = 392195
Banking Exam Preparation (IBPS Exam) – “MULTIPLICATION” Part II
Multiply by 6
Write main each digit of main number and add 5 for each odd digit and 0 for each even digit also add half of next digit.
E.g. 7987 x 6
1 2 3 4 Main Number Odd/Even ½ of next digit 1+2+3 0 0 3 3 7 5 4 16 9 5 4 18 8 0 3 11 7 5 0 12
Banking Exam Preparation (IBPS Exam) – “MULTIPLICATION” Part II
You can also find
|
xWhat is x if lg(2x+2)=lg(8+3x)/(x-2)?
justaguide | Certified Educator
We have to solve lg(2x+2)=lg[(8+3x)/(x-2)]
Equate 2x + 2 = (8+3x)/(x-2)
=> (2x + 2)(x - 2) = 8 + 3x
=> 2x^2 + 2x - 4x - 4 = 8 + 3x
=> 2x^2 - 5x - 12 = 0
=> 2x^2 - 8x + 3x - 12 = 0
=> 2x(x - 4) + 3(x - 4) = 0
=> (2x + 3) = 0 and (x - 4) = 0
=> x = -3/2 and x = 4.
For x = -3/2, the given logarithms become logs of negative numbers and are undefined.
The solution of the equation is x = 4
giorgiana1976 | Student
We'll impose the constraints of existence of logarithms:
3x+8>0
x>-8/3
2x+2>0
x>-1
x-2>0
x>2
The interval of admissible values for x is (2, +infinite).
Now, we'll solve the equation:
lg(2x+2)=lg [(8+3x)/(x-2)]
Since the logarithms have the matching bases, we'll apply the one to one rule:
2x + 2 = [(8+3x)/(x-2)]
We'll cross multiply and we'll get:
[(2x+2)*(x-2)] = 8 + 3x
We'll remove the brackets from the left side:
2x^2 - 4x + 2x - 4 = 8 + 3x
We'll move all terms to the left side and we'll use symmetric property:
2x^2 - 4x + 2x - 4 - 3x - 8 = 0
2x^2 - 5x - 12 = 0
x1 = [5 + sqrt(25 + 96)]/4
x1 = (5+11)/4
x1 = 16/4
x1 = 4
x2 = -6/4
x2 = -3/2
Since the second solution does not belong to the range of admissible values, we'll reject it.
The only admissible solution is x = 4.
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# CHAPTER 1– NUMBER BASES
LEARNING AREA Number Bases LEARNING OBJECTIVES LEARNING OUTCOMES Students will be able to: (i) State zero, one, two, three,…, as a number in base: a) two b) eight c) five (ii) State the value of a digit of a number in base: a) two b) eight c) five (iii) Write a number in base: a) two b) eight c) five in expanded notation. SUGGESTED TEACHING & LEARNING ACTIVITIES Use models such as a clock face or a counter which uses a particular number base. Number base blocks of twos, eights and fives can be used to demonstrate the value of a number in the respective number bases. For example: 2435
MATHEMATICS 5
MORAL VALUES Systematic Rational Accurate Emphasise the ways to read numbers in various bases. Examples : • 1012 is read as “one zero one base two” • 72058 is read as “seven two zero five base eight” • 43255 is read as “ four three two five base five” Numbers in base two are also known as binary numbers. Examples of numbers in expanded notation : • 101102= 1×24 + 0×23 + 1×22 + 1×21 + 0×20 • 3258 = 3×82 +2×81 + 5×80 • 30415 = 3×53 + 0×52 + 4×51 + 1×50 Expanded notation POINTS TO NOTE / VOCABULARY
WEEKS
GENERICS ICT Contextual learning Cooperative learning
CCTS
Students will be taught to: 1. Understand and use the concept of number in base two, eight and five.
Conceptual Compare and contrast
2
4
3
Discuss • digits used • place values in the number system with a particular number base.
1
CHAPTER 1– NUMBER BASES
LEARNING AREA LEARNING OBJECTIVES Students will be taught to: LEARNING OUTCOMES Students will be able to: iv) convert a number in base : a) two b) eight c) five to a number in base ten and vice versa. v) convert a number in a certain base to a number in another base. SUGGESTED TEACHING & LEARNING ACTIVITIES Number base blocks of twos, eights and fives can also be used here. For example, to convert 1010 to a number in base two, use the concept of least number of blocks (23), tiles (22), rectangles (21) and squares (20). In this case, the least number of objects needed here are one block, zero tiles, one rectangle and zero squares. So, 1010 = 10102. Discuss the special case of converting a number in base two directly to a number in base eight and vice versa. For example, convert a number in base two directly to a number in base eight through grouping of three consecutive digits. Perform addition and subtraction in the conventional manner. For example : 1010 + 110 Contextual Learning Communicat ion Method of Learning Evaluation Arrange sequentially Using algorithm and relationship
MATHEMATICS 5
MORAL VALUES Systematic Consistent POINTS TO NOTE / VOCABULARY Perform repeated division to convert a number in base ten to a number in other bases. For example, convert 71410 to a number in base five : 5)714 5)142---4 5) 28---2 5) 5---3 5) 1---0 0---1 ∴ 71410 = 103245 Limit conversion of numbers to base two, eight and five only.
WEEKS
GENERICS ICT Contextual learning Cooperative learning
CCTS
Identify patterns Identify relations Arrange sequentially
Students will be taught to:
Students will be able to: (vi) Perform computations involving : a) addition b) subtraction of two numbers in base two
Appreciatio n of technology Cooperation Prudence
2
CHAPTER 2 – GRAPHS OF FUNCTIONS II
LEARNING AREA
Graphs of functions
MATHEMATICS 5
SUGGESTED TEACHING & LEARNING ACTIVITIES
Explore graphs of functions using graphing calculator or the Geometer’s Sketchpad. Compare the characteristics of graphs of functions with different values of constants. For example :
WEEKS
LEARNING OBJECTIVES
Students will be taught to: 2.1 Understand and use the concept of graphs of functions.
LEARNING OUTCOMES
Students will be able to: (i) Draw the graph of a ; a) linear function; y = ax + b , where a and b are constants b) quadratic function; 2 y = ax + bx + c , where a, b and c are constants, a ≠ 0 c) cubic function : 3 2 y = ax + bx + cx + d , where a,b,c and d are constants, a≠0 d) reciprocal function :
, where a is a x constants, a≠0. y= a
GENERICS
Constructivism Mastery learning Self-access learning
CCTS
Concept constructivis m Compare and contrast Analising Mental visualization Relationship
MORAL VALUES
Punctuality Awareness Systematic Neatness
POINTS TO NOTE / VOCABULARY
Limit cubic functions to the following forms: 3 y = ax 3 y = ax + b 3 y = ax + bx + c
A
(ii) Find from a graph : a) the value of y , given a value of x b) the value(s) of x , given a value of y.
B Graph B is broader than graph A and intersects the vertical axis above the horizontal axis.
Students will be taught to:
Students will be able to: iii) Identify: a) the shape of graph given a type of function b) the type of function given a graph c) the graph given a function and vice versa. LEARNING OUTCOMES
WEEKS
LEARNING AREA
LEARNING OBJECTIVES
As reinforcement, let students play a game; for example, matching card of graphs with their respective function. When the students have their matching partners, ask them to group themselves into four groups of types of functions. Finally, ask each group to name the type of function that is depicted on the SUGGESTED TEACHING & cards. LEARNING ACTIVITIES
- Mastery learning Cooperative learning. - Contextual learning GENERICS
- Comparing - Accuracy & - Systematic differentiatin g - Classifying - Identifying patterns MORAL VALUES
CCTS
For graph of cubic function, limit to y = ax3 and y = ax3 + b. For graph of quadratic function limit to y = ax2 + b and quadratic function which can be factorise to ( mx + n) ( px + q) where m .n.p POINTS TO NOTE / and q are integers VOCABULARY
3
Graphs of reciprocal function For graph of cubic function. Students will be taught to: Students will be able to: 2.Identifying relation . (ii)Identifyin g relations.Mental visualization (i)Identifyin g patterns.cubic or reciprocal function.2 Understand and use the concept of the solution of an equation by graphical methods. Involve everyday problems. (iv)Represe nting and interpret ing data.Graph of linear function . i) Find the point(s) of intersection of two graphs.Systematic .Graph of cubic function .To draw a graph - - Coopera tive learning Constru ctivisme Use the traditional graph plotting exercise if the graphing calculator or the Sketchpad is unavailable. Explore using graphing calculator or the Geometer’s Sketchpad to relate the xcoordinate of a point of intersection of two appropriate graph to the solution of a given equation.CHAPTER 2 – GRAPHS OF FUNCTIONS II MATHEMATICS 5 . . . Make generalization about the point(s) of intersection of the two graphs.quadratic.Graph of quadratic function .Neatness .To sketch a graph . (iii)Recogniz ing and representi ng.Precise . limit to y = ax3 and y = ax3 + b iv) Sketch the graph of a given linear.Graphs of function .Self access learning . -Rationale -Diligence -Systematic -Accuracy WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS CCTS MORAL VALUES POINTS TO NOTE / VOCABULARY 4 . (iii) Solve problems involving solution of an equation by graphical method. (ii) Obtain the solution of an equation by finding the point(s) of intersection of two graphs.
5 . include situations involving y > ax + b or y < ax + b ii) Determine the position of a given point relative to the equation y < ax + b .For the region representing y ≥ ax + b or y ≤ ax + b . the line y = ax + b is drawn as a solid line to indicate that all points on the line y = ax + b are in the region.For the region representing y > ax + b or y < ax + b .3. x ≥ a . x<a region dashed line y = ax + b iii) Identify the region satisfying y > ax + b or y < ax + b iv) Shade the regions representing the inequalities Use the Sketchpad or graphing calculator to explore points relative to a graph to make generalization about regions satisfying the given inequalities.3 Understand and use the concept of the region representing in inequalities in two variables Students will be able to: i) Determine whether a given point satisfies : y = ax + b or Discuss that if one point in a region satisfies y > ax + b or Enquirydiscovery Constructivis m Identifying patterns Systematic Determinati on Making inferences MATHEMATICS 5 For learning Objectives 2.CHAPTER 2 – GRAPHS OF FUNCTIONS II Students will be taught to: 2. x = a. Solid line > ax + b or y < ax + b b) y ≥ ax + b or y ≤ ax + b a) y v) Determine the region which satisfies two or more simultaneous linear inequalities. Emphasise that: . . then all point in the region satisfies the same inequalities. x > a. x ≤ a.the line y = ax + b is drawn as a dashed line to indicate that all points on the line are not in the region.
two enlargements. Identifying Relation Investigated the characteristics of and object and its image under combined transformation Contextual Learning Multiple Intelligence theory Drawing Diagrams Identifying Relation Systematic iv. followed by a line and a object ii. or the overhead projector and transparencies.1 Understand and use the concept of combination of two transformations. an enlargement and an isometric transformation. Explore combined transformation using the graphing calculator. TRANSFORM ATIONS III LEARNING OBJECTIVES Students will be taught to: 3. Draw the image of and object under combination of two transformations. SUGGESTED TEACHING & LEARNING ACTIVITIES Relate to transformations in real life situations such as tessellation patterns on walls. Determine the image of an object under combination of two isometric transformations. b. the geometer’s Sketchpad. Constructivi sm Contextual Learning Identifying Relation Arranging Sequentially Diligence Accuracy Consistent Combined transformation. Determine the image of an object under combination of a. ceilings or floors GENERICS Constructivi sm Contextual Learning CCTS Identifying relations Characterizi ng MORAL VALUES Systematic Determinati on Accuracy POINTS TO NOTE / VOCABULARY Begin with a point. Mastery Learning Comparing and Differentiati ng Interpreting Rules and Regulations Self Confidence Neatness Limit isometric transformations to translations. State the coordinates of the image of a point under combined transformation.CHAPTER 3 – TRANSFORMATIONS III MATHEMATICS 5 WEEKS LEARNING AREA 3. 6 . iii. LEARNING OUTCOMES Students will be able to: i. reflections and rotations.
Specify two successive transformation in a combined transformation given the object and the image Carry out projects to design patterns using combined transformations that can be used as decorative purposes. Mastery Learning ICT Using Analogies Working Out Mentally Honesty Cooperation Limit the equivalent 7 . Specify a transformation which is equivalent to the combination of two isometric transformations. Mastery Learning Identifying Patterns Identifying Relation Logical Reasoning Representin g and Interpreting Data Systematic Hardworkin g Specify vii. Use the Sketchpad to prove the single transformation which is equivalent to the combination of two isometric transformations.CHAPTER 3 – TRANSFORMATIONS III LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES v. Determine whether combined transformation AB is equivalent to combined transformation BA SUGGESTED TEACHING & LEARNING ACTIVITIES MATHEMATICS 5 MORAL VALUES Rational Cautious POINTS TO NOTE / VOCABULARY Equivalent WEEKS GENERICS Multiple Intelligence CCTS Comparing and Differentiati ng Identifying Relation vi. These projects can then be presented in classroom with the students describing or specifying the transformations involved.
Constructing a kaleidoscope.CHAPTER 3 – TRANSFORMATIONS III LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES viii. SUGGESTED TEACHING & LEARNING ACTIVITIES a. b. How to make a frieze or strip pattern. MATHEMATICS 5 MORAL VALUES Sharing Rational Diligence POINTS TO NOTE / VOCABULARY WEEKS GENERICS Mastery Learning ICT CCTS Find all possible solution Using Analogies Drawing Diagram Working out Mentally 8 . Solve problems involving transformation.
2 Understand and use the concept of equal matrices. row matrix. Matrix. LEARNING OUTCOMES Students will be able to: i) form a matrix from given information. 4. Accurate Mastery learning Use students sitting positions in the classroom by rows and columns to identify a student who is sitting in a particular row and in a particular column as a concrete example. Discuss equal matrices in terms of : a) the order b) the corresponding elements Mastery learning Using algorithm and relationship Comparing and differentiatin g Systematic Accurate Equal matrices Including finding values of unknown elements. column matrix. in table form Constructivis m and then in matrix form. the price of food on a menu.CHAPTER 4 – MATRICES LEARNING AREA Matrices LEARNING OBJECTIVES Students will be taught to: 4. 9 . MATHEMATICS 5 MORAL VALUES Neatness and systematic POINTS TO NOTE / VOCABULARY Emphasize that matrices are written in bracket. ii) Solve problems involving equal matrices. square matrix WEEKS CCTS Arranging sequentially Collecting and handling data ii) Determine : a) the number of rows b) the number of columns c) the order of a matrix iii) Identify a specific element in a matrix. i) Determine whether two matrices are equal. SUGGESTED TEACHING & GENERICS LEARNING ACTIVITIES Contextual Represent data in real life learning situations.1 understand and use the concept of matrix. Identifying patterns Identifying patterns Emphasize that a matrix of order m x n is read as ‘an m by n matrix’ Systematic Use row number and column number to specify the position of an element. for example.
ii) Express a given matrix as a multiplication of another matrix by a number.CHAPTER 4 – MATRICES LEARNING AREA LEARNING OBJECTIVES 4. MATHEMATICS 5 MORAL VALUES Cooperation Rationale Confidence Using algorithm and relationship Problem solving Systematic Limit to matrices with not more than three rows and three columns. iv) Solve matrix equation involving addition and subtraction Contextual learning Multiple intelligences Mastery learning Future studies SUGGESTED TEACHING & LEARNING ACTIVITIES Relate to real life situations such as keeping scores of metals.3 Perform addition and subtraction on matrices. iii) Perform addition and subtraction on a few matrices. iii) Perform calculation on matrices involving addition. LEARNING OUTCOMES i) Determine whether addition or subtraction can be performed on two given matrices. Relate to real life situations such as in industrial productions Mastery learning Constructivis m Contextual learning Self-access learning Multiplying a matrix by a number is known as scalar multiplication systematic 10 . POINTS TO NOTE / VOCABULARY WEEKS GENERICS Self-access learning Constructivi sm Mastery learning Communicat ion method of learning CCTS Comparing and differentiati ng Using algorithm and relationship Analyzing Making inferences Problem solving Evaluating Using algorithm and relationship Conceptuali ze and finding all possible solutions Include finding values of unknown elements/matrix equation 4. subtraction and scalar multiplication. i) Multiply a matrix by a number. ii) Find the sum or the difference of two matrices. tally or points in sport.4 perform multiplication of a matrix by a number.
Solve matrix equations involving multiplication of two matrices. Limit to matrices with not more than three rows and three columns.CHAPTER 4 – MATRICES MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES iv) Solve matrix equations involving addition. SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS Self-access learning Constructivis m Self-access learning • Constructi vism • ICT • Cooperati ve • Learning CCTS Evaluating and problems solving MORAL VALUES POINTS TO NOTE / VOCABULARY Include finding the values of unknown elements 4.5 Perform multiplication of two matrices i. subtraction and scalar multiplication. 11 . Limit to two unknown elements. iii. Find the product of two matrices For matrices A and B . The order of the matrices : (m x n) x (n x s) = (m x s) ii. Determine whether two matrices can be multiplied and state the order of the product when the two matrices can be multiplied. discuss the relationship between AB and BA. • Identifyin g patterns • Arranging sequentiall y • Recognizi ng and representin g • Making generalizati on • classifying • Determin ation • Systemat ic • Consiste nt • Diligence • Neatness The number of columns of first matrix must be same with the number of rows of second matrix. Relate to real life situations such as finding the cost of a meal in a restaurant.
iii) Perform calculation involving identity matrices.matrix for each order. For 12 . WEEKS GENERICS • Contextual learning CCTS • Making generalizati on Discuss: . SUGGESTED TEACHING & LEARNING ACTIVITIES Begin with discussing the property of the number 1 as an identity for multiplication of numbers. Identity matrix unit matrix. but not all square matrices have inverse matrices. ii) Write identity matrix of any order. Relate to the property of multiplicative inverse of numbers. an identity matrix is a square matrix . AB = BA = I • Inverse matrices can only exist for square matrices. −1 a) (ii) Find the inverse Use the method of solving simultaneous linear equations to show that not all square matrices have inverse matrices. IA = A • Constructi vism • Identifyi ng patterns • Systemati c Limit to matrices with no more than three rows and three columns. 2 is the multiplicative inverse of 2 and vice versa. LEARNING OUTCOMES i) Determine whether a given matrix is an identity matrix by multiplying it to another matrix. Emphasize that: • If matrix B is the inverse of matrix A. −1 x2 = 1 In the example.7 Understand and use the concept of inverse matrix. For example : 2x2 −1 =2 −1 • Cooperativ e learning • Constructi vism • Mastery learning • Solving problems • Comparin g • Identifyin g patterns and relations • Neatness • Cooperati on • Neatness • Systemati c The inverse of matrix A is denoted by A . 4. (i) Determine whether a 2 x 2 matrix is the inverse matrix of another 2 x 2 matrix.6 Understand and use the concept of identity matrix. MATHEMATICS 5 MORAL VALUES • Rational POINTS TO NOTE / VOCABULARY Identity matrix is usually denoted by I and is also known as unit matrix. then matrix A is also the inverse of matrix B.CHAPTER 4 – MATRICES LEARNING AREA LEARNING OBJECTIVES 4. there is only one identity . AI = A . Discuss the properties: .
q 1 0 = s 0 1 3 p + 4r = 0 3q + 4 s = 1 p q where r s is the inverse matrix. • Using formula a b For A = c d . q + 2 s = 0.CHAPTER 4 – MATRICES MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES matrix of a 2 x 2 matrix using : b) the method of solving simultaneous linear equations a formula. Express each inverse matrix as a multiplication to the original matrix and discuss how the determinant is obtained MORAL VALUES • Cooperati on • Neatness • Systemati c POINTS TO NOTE / VOCABULARY Steps to find the inverse matrix : • Solving simultaneous linear equations 1 2 p 3 4 r p + 2r = 1. SUGGESTED TEACHING & LEARNING ACTIVITIES example. Using matrices and their respective inverse matrices in the previous method to relate to the formula. −b d −1 A = ad − bc ad − bc a −c ad − bc ad − bc or 13 . ask student to try to find the inverse matrix of GENERICS • Constructi vism • Mastery learning • Communi cation method of learning • CCTS • Comparin g • Identifyin g patterns and relations • 3 2 6 4 .
carry out operations leading to the formula. • Multiple Intelligence s • Constructi vism • Identifyin g Relations • Systemati c • Neatness a b p −1 h A −1 c d q =A k 14 .CHAPTER 4 – MATRICES LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES SUGGESTED TEACHING & LEARNING ACTIVITIES MATHEMATICS 5 MORAL VALUES • POINTS TO NOTE / VOCABULARY 1 d − b A−1 = ad − bc − c a when ad − bc ≠ 0. d. Simultaneous linear equations ap + bq = h cp + dq = k in matrix form is a b p h c d = q k Where a. p ad q are constants. For example: Write 2 x + 3y = 13 • Mastery Learning • Constructi vism • Identifyin g Patterns • Rational p (ii) Find the matrix q in a b p h c d = q k Using the inverse matrix. 4x − y = 5 As equal matrices: 2 x + 3y 13 4x − y = 5 which is then expressed as: 2 3 x 13 4 − 1 = y 5 Discuss why: • The use of inverse matrix is necessary. b. WEEKS GENERICS • • CCTS 4. A -1 does not exist if the determinant is zero. Limit to two unknowns. Relate to equal matrices by writing down the simultaneous equations as equal matrices first. p and q are unknowns.8 Solve simultaneous linear equations by using matrices (i) Write simultaneous linear equations in matrix form. c. ad − bc is known as the determinant of the matrix A. Prior to use the formula. h and k are constants. Relate to solving linear equations of type ax = b • It is important to place the inverse matrix at the right place on both sides of the equation.
(iv) Solve problems involving matrices. SUGGESTED TEACHING & LEARNING ACTIVITIES Relate the use of matrices to other areas such as in business or economy. Carry out projects involving matrices using the electronic spreadsheet. Matrix method 15 . GENERICS • Cooperati ve Learning CCTS • Identifyin g Patterns • Identifyin g Relations MORAL VALUES • Rational • Systemati c • Neatness POINTS TO NOTE / VOCABULARY a b Where A = c d . science etc.CHAPTER 4 – MATRICES MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (iii) Solve simultaneous linear equations by the matrix method. • Selfaccess Learning • Mastery Learning • ICT • Represent ing & Interpreting • Data • Rational • Systemati c • Neatness • The matrix method uses inverse matrix to solve simultaneous linear equations.
the Charles’ Law or the mation of the simple pendulum. (v)Solve problems involving direct variations for the following cases: y ∝ x. 3. VOCABULARY: Direct variation Quantity Constant of variation Variable.or y1 y 2 = x1 x 2 variation. 2 1 x n .1 Understand and use the concept of direct variation SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS Contextual Learning Self. y ∝ xn . 2 discuss the y ∝ x . to get the solutions. the relation is written as y ∝ x.CHAPTER 5 – VARIATIONS LEARNING AREA VARIATIONS MATHEMATICS 5 LEARNING OUTCOMES Students will be able to: (i)State the changes in a quantity with respect to the changes in another quantity. y ∝ x2 16 . 1 . (iv)Find the value of a variable in a direct variation when sufficient information is given. 2 If limit n = 2. (iii)Express a direct variation in the form of equation involving two variables. Discuss the characteristic of the graph of y against x when y ∝ x . For the cases y ∝ xn . Relate mathematical variation to other area such as science and technology. If y varies directly as x .access Learning Communicati on Method of Leaning CCTS Identifying relations Making generalization Estimating MORAL VALUES Rationale Systematic Tolerance Hardworking POINTS TO NOTE / VOCABULARY Y varies directly as only if x if and y x is a constant. Using characteristics of against the graph of y y∝x . For example. (ii)Determine from given information whether a quantity varies directly as another quantity. y ∝ x3 . n = 2. then y = kx where k is constant of y = kx . 3. For the cases WEEKS LEARNING OBJECTIVES Students will be taught to: 5. in everyday life situations involving direct variation. 1 .
y∝ 2 . against WEEKS SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS Contructivism Communicati on method of learning CCTS Making inferences Representing and interpreting data Identifying relations MORAL VALUES Rational Systematic POINTS TO NOTE / VOCABULARY y varies inversely as x if and only if xy is a constant. For example. For the cases Relate to other areas like science and technology. in everyday life situations involving inverse variation. 2 y∝ 1 k . limit n to 2. discuss 2 Problem Solving where k is the constant of variation. Discuss the form of the graph of y ii) Determine from given information whether a quantity varies inversely as another quantity. x x 1 1 y∝ 3 . 1 1 y∝ . Using: • • the characteristics of the graph of y against 1 xn . v) Solve problems involving inverse variation for the following cases: y∝ 1 xn .2 Understand and use the concept of inverse variations MATHEMATICS 5 LEARNING OUTCOMES i) State the changes in a quantity with respect to changes in another quantity. then y = x x For the cases iv) Find the value of a variable in an inverse variation when sufficient information is given.CHAPTER 5 – VARIATIONS LEARNING AREA LEARNING OBJECTIVES 5. x Cooperative learning 1 x . If y varies inversely as x.3 and 1 . n = 2.3 and 1 . iii) Express a inverse variation in the form of equation involving two variables.y∝ 1 x x2 k or x x1 y1 = x 2 y 2 y= to get the solution. Boyle’s Law. VOCABULARY: Inverse variation 17 . 1 xn . the relation is written as y∝ y Rational Systematic Accuracy If 1 x when 1 y∝ .
3 Understand and use the concept of joint variation. c) a direct variation and an inverse variation. For example: GENERICS Constructivism Cooperative learning Multiple intelligences Self –access learning CCTS Identifying relations comparing and differentiating collecting and handling data using analogies finding all possible solutions MORAL VALUES Cooperation Punctuality Systematic Rational POINTS TO NOTE / VOCABULARY For the cases y ∝ xn zn . MATHEMATICS 5 LEARNING OUTCOMES i) Represent a joint variation by using the symbol ∝ for the following cases: a) two direct variations. z . 1 2 1 x zn n V I∝ R xn y∝ n .CHAPTER 5 – VARIATIONS LEARNING AREA LEARNING OBJECTIVES 5. ii) Express a joint variation in the form of equation. iii) Find the value of a variable in a joint variation when sufficient information is given. y ∝ and 3. means the current I varies directly as the voltage V and varies inversely as the resistance R. Relate to other areas like science and technology. WEEKS SUGGESTED TEACHING & LEARNING ACTIVITIES Discuss joint variation for the three cases in everyday life situations. b) two inverse variations. Joint variation Mastery learning 18 . limit n to 2. iv) Solve problems involving joint variation.
b) a relationship between distance and time. Comparing and differentiati ng Interpreting data The gradient of a graph represents the rate of change of a quantity on the vertical axis with respect to the change of another quantity on the horizontal axis. SUGGESTED TEACHING & LEARNING ACTIVITIES Use examples in various areas such as technology and social science. (iii) Find and interpret the gradient of a distance-time graph. MATHEMATICS 5 MORAL VALUES Rationality Respect POINTS TO NOTE / VOCABULARY Limit to graph a straight line.CHAPTER 6 – GRADIENT LEARNING AREA Gradient and area under a graph LEARNING OBJECTIVES Students will be taught to: 6. Compare and differentiate between distance-time graph and speed-time graph. The rate of change may have a specific name for example “speed” for a distance time graph. given . a) a table of distance-time values. WEEKS GENERICS Contextual learning CCTS Recognizin g and representin g (ii) Draw the distancetime graph. Emphasis that: Gradient change of distance change of time =speed = Distance-time graph Speed-time graph 19 .1 Understand and use the concept of quantity represented by the gradient of a graph. LEARNING OUTCOMES Students will be able to: (i) State the quantity represented by the gradient of a graph.
t (v) Draw a graph to show the relationship between two variables representing certain measurements and state the meaning of its gradient. v represents speed. Discuss the formula for finding the area under a graph involving.CHAPTER 6 – GRADIENT LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (iv) Find the speed for a period of time from a distance-time graph. 6. For example: distance. (i) State the quantity represented the area under a graph. POINTS TO NOTE / h and k are constants. (iii) Determine the distance by finding the area under the following types of speed-time graph: LEARNING (a) v = k WEEKS LEARNING LEARNING Discuss that in certain cases. t represents time. SUGGESTED TEACHING & LEARNING ACTIVITIES Use real life situation such as traveling from one place to another by train or by bus. Use examples in social science and economy. (ii) Find the area under a graph. the Constructivis area under a graph may not m represent any meaningful quantity. MATHEMATICS 5 MORAL VALUES POINTS TO NOTE / VOCABULARY Include graphs which consist of a combination of a few straight lines.2 Understand the concept of quantity represented by the area under a graph. For example: The area under the distance-time graph. • a straight line which is parallel to the x-axis • a straight lineTEACHING in the form of SUGGESTED & GENERICS Recognisin g and representin g Respect Include speed-time and acceleration-time graphs. Limit to graph of a straight line of a combination of a few straight lines. s WEEKS GENERICS CCTS time. CCTS MORAL 20 .
MATHEMATICS 5 VALUES VOCABULARY For example: Speed. LEARNING ACTIVITIES y= kx + h a combination of the above. v time.CHAPTER 6 – GRADIENT AREA OBJECTIVES OUTCOMES (uniform speed) (b) v = kt (c) v = kt + h (d) a combination of the above. (v) Solve problems involving gradient and area under a graph. t area under a graph acceleration-time graph uniform speed 21 .
The probability of an outcome A. A sample space in which each outcome is equally likely is called equiprobable sample space. MORAL VALUES Determinati on Cooperation Rational MATHEMATICS 5 POINTS TO NOTE / VOCABULARY Limit to sample space with equally likely outcomes. Use tree diagrams to obtain sample space for tossing a fair coin or tossing a fair die activities. with equiprobable sample space S. Drilling exercise. Find the probability. toss a fair coin b. (b) P(A) = 0 Climbing up the twin tower. 22 . Include everyday problems and making predictions. give a TRUE or FALSE question.1Understand and use the concept of probability of an event. LEARNING OUTCOMES Students will be able to: i) Determine the sample space of an experiment with equally likely outcomes SUGGESTED TEACHING & LEARNING ACTIVITIES Discuss equiprobable sample through concrete activities and begin with simple cases such as a. Tossing a fair coin . iii) Solve problems involving probability of an event. The graphing calculator may also be used to simulate these activities. Equally likely WEEKS GENERICS Contextual Learning Mastery Learning CCTS Making inference Working out mentally Finding all possible solutions. ii) Determine the probability of an event with equiprobable sample space. Discuss event that produce (a) P(A) = 1. P( Head) + P(Tail) = 1. is P(A)= n(A) n(S) Use tree diagram where appropriate. Finding all possible solutions.CHAPTER 7 – PROBABILITY II LEARNING AREA Probability II LEARNING OBJECTIVES Students will be taught to: 7.
PP} b) R = An event to get the picture at the second toss or both times showing the number.3 Understand and use the concept of probability of combined event i) List the outcomes for events : a) A or B as elements of set A ∪B ii) Find the probability by listing the outcomes of the combined event : a) A or B Example i: A coin is tossed twice consecutively. NN. NN. R= {NP. PN. MORAL VALUES Cooperation Equity Rationale Precise Making inferences Drawing diagrams Estimating Identifying Patterns Identifying Relations Finding all possible solutions MATHEMATICS 5 POINTS TO NOTE / VOCABULARY The complement of an event A is the set of all outcomes in the sample space that are not included in the outcomes of event A.CHAPTER 7 – PROBABILITY II LEARNING AREA LEARNING OBJECTIVES 7. WEEKS GENERICS CCTS Constructivism Identifying relations Contextual Learning Finding all Possible solutions 7. Mastery Learning Enquiry Discovery Tolerance Determination Event Combined event Consecutively Toss Consistent 23 . NN. LEARNING OUTCOMES (i)State the complement of an event in : a) words b) set notation (ii) Find the probability of the complement of an event SUGGESTED TEACHING & LEARNING ACTIVITIES Discuss equiprobable sample space through activities such as finding the consonants and vowels from the word given. PP} n(Q) = 3.2 Understand and use the concept of probability of the complement of an event. Include events in real life situations such as winning or losing a game and passing or failing an exam. NN } Example ii: Find the probability by listing the outcomes of the combined event a) S = { NP. PP . List the probability for each combined event a) Q = An event to get the numbers at the first go or both times showing the pictures Q = { NP. PP} n(S) = 4 Q = { NP.
CHAPTER 7 – PROBABILITY II MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES SUGGESTED TEACHING & LEARNING ACTIVITIES P(Q) = GENERICS CCTS MORAL VALUES POINTS TO NOTE / VOCABULARY n(Q) n( S ) = 3 4 Contextual Learning Identifying relations Finding all possible solution Drawing diagram Cooperation Systematic (i) list the outcomes for events A and B as elements of set A ∩ B 1. Combined event Rational A A G {A. Ask D1one student to toss 2 coins at the same time.A} {G.G} 3. List the outcomes for different event A and A = {(A. The total number of the event n(A ∩ A) = 1 n(A ∩ G) = 2 n(G ∩ G) = 1 24 .A} G {A.G). A and A = A ∩ A A and G = A ∩ G G and G = G ∩ G 5.A)} A and G = {(A. Fill in the outcomes. (G.G} {G. D2 2. State the relationship between and & ∩.A)} G and G = { (G.G)} 4.
1). 3.6) (G. Each group will be given one coin and one dice. Based on tree diagram. Find the probability of getting a ‘1’ when rolling a coin is ‘A’. The probability to getting ‘A’ and ‘1’ can be written as P(A ∩ 1) = P(A) × P(1) 1 1 × 2 6 1 = 12 = 25 . (G.6)} 4. (A. (A.CHAPTER 7 – PROBABILITY II LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (ii) Find the probability by listing the outcomes of the combined event A and B SUGGESTED TEACHING & LEARNING ACTIVITIES 1.4).1). (A.5). (A. Introduce a tree diagram 6. (G.3). (G. Split the class into the group 2.2).3).4).2). (A. (G. find the probability of :(a) getting ‘A’ (b) getting ‘1’ 1 2 1 P(1) = 6 P(A) = 7. (G.5). P(A ∩ 1) = MORAL VALUES Cooperation Systematic Finding all possible solution Drawing diagram Rational MATHEMATICS 5 POINTS TO NOTE / VOCABULARY Combined event WEEKS GENERICS Contextual Learning CCTS Identifying relations 1 12 5. List out all the possible combination when toss the coin and dice at the same time { (A.
Men Wo men Discuss: • situations where decision have to made based on probability.CHAPTER 7 – PROBABILITY II MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (iii) Solve problems involving probability of combined event. Example of a two-way classification table: MEANS OF GOING TO WORK Offic Car Bus Oth ers GENERICS ICT Mastery Learning Self-access Learning CCTS Identifying Relations Making Generalizations MORAL VALUES Systematic Neatness Responsibility Making Inferences and hypothesis POINTS TO NOTE / VOCABULARY Emphasis that: • knowledge about probability making decisions • predictions as based on probability is not definite or absolute. Ask students to create tree diagrams from these tables. 56 50 25 42 ers 83 37 26 . for example in business. such as determining the value for specific insurance policy and time the slot for TV advertisements • the statement “probability is the underlying language of statistics”. SUGGESTED TEACHING & LEARNING ACTIVITIES Use two-way classification tables of events from newspaper articles or statistical data to find probability of combined events.
northwest. east. They are measured in a clockwise direction from north. west (b) north-east. south-west Carry out activities or games involving finding directions using a compass. For cases involving degrees and minutes.1 Understand and use the concept of bearing LEARNING OUTCOMES Students will be able to : SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS CCTS MORAL VALUES POINTS TO NOTE / VOCABULARY (i) Draw and label the eight main compass directions: (a) north.CHAPTER 8 – BEARING MATHEMATICS 5 WEEKS LEARNING AREA 8 BEARING LEARNING OBJECTIVES Students will be taught to : 8. south-east. Due north is considered as bearing 000o. 000o to 360o. (ii) State the compass angle of any compass direction Making connections Visualize mentally (iii) Draw a diagram of a point which shows the direction of B relative to another point A given the bearing of B from A Comparing and differentiatin g 27 . south. state in degrees up to one decimal point. such as treasure hunt or scavenger hunt. It can also be about locating several points on a map Constructivi sm Cooperative Multiple intelligence Making connections Visualize mentally Cooperation Accuracy Neatness Carefulness North–east South–east North-west South-west Compass angle bearing Compass angle and bearing are written in three-digit form.
in map reading and navigation Contextual Constructivi sm Self-access learning (Mathematic al-logical Verballinguistic) Communicat ion Interpret Draw diagrams Recognizing relationship Problem solving Accuracy Rational Responsibili ty Appreciatio n 28 .CHAPTER 8 – BEARING MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (iv) State the bearing of point A from point b based on given information SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS Mastery learning Contextual Constructivi sm Self-access learning (Mathematic al-logical Verballinguistic) CCTS Making connections Visualize mentally MORAL VALUES Rational Accuracy Systematic Carefulness POINTS TO NOTE / VOCABULARY Begin with the case where bearing of point B from point A is given (v) Solve problems involving bearing Discuss the use of bearing in real life situations. For example.
Using any computer software to sketch a circle parallel to the equator. learning patterns Introduce the meridian through Greenwich in England as the Greenwich Meridian with longitude 00 Discuss that: (a) all points on a meridian have the same longitude. Constructivi sm Self-access learning Drawing diagrams Rational Equator Latitude Emphasize that * the latitude of the equator is 0° * latitude ranges from 0° to 90°N(or S) 29 .1 Understand and use the concept of longitude.2 Understand and use the concept of latitude (i) Sketch a circle parallel to the equator. (iii) Sketch and label a meridian with the longitude given.CHAPTER 10 –PLANS AND ELEVATION MATHEMATICS 5 WEEKS LEARNING AREA EARTH AS A SPHERE LEARNING OBJECTIVES Students will be taught to: 9. (ii) State the longitude of a given point. (c) Meridians with longitudes xoE (or W) and ( 1800 . SUGGESTED TEACHING & GENERICS CCTS LEARNING ACTIVITIES Models such as globes should be Contextual Identifying used. (ii) State the latitude of a given point. (iv) Find the difference between two longitudes. (b) There two meridians on a great circle through both poles. Identifying relations MORAL VALUES Understandin g POINTS TO NOTE / VOCABULARY Great circle Meridian Longitude Constructivis m 9.x0 )W (or E) form a great circle through both poles. LEARNING OUTCOMES Students will be able to: (i) Sketch a great circle through the north and south poles.
(iii)Sketch and label the latitude and longitude of a given place. 9. Logical Reasoning.4 Understand and use the concept of distance on the surface of the earth to solve problems Systematic Rational 30 . SUGGESTED TEACHING & LEARNING ACTIVITIES Discuss that all points on a parallel of latitude have the same latitude Carry out group activity such as station game. given the subtended angle at the centre of the earth and vice versa Use a globe or a map to find locations of cities around the world. Public Spiritedness. Contextual Learning. (i) State the latitude and longitude of a given place. GENERICS CCTS MORAL VALUES POINTS TO NOTE / VOCABULARY Parallel of latitude Cooperative learning Enquirydiscover y Communicati on method of learning Finding all possible solutions Logical reasoning Recognizing & interpreting data Cooperation Sharing Systematic Tolerance Involve actual places on the earth Express the difference between two latitudes with an angle in the range of 0°≤x≤180°. Each station will have different diagram and the student will be ask to find the difference between two latitudes for each diagram. Neatness. The location of a place A at latitude x◦N and longitude y◦E is written as A(x◦N. 9. (iv) Find the difference between two latitudes. y◦E). (ii) Mark the location of a place. (i) find the length of an arc of a great circle in nautical mile. Use a globe or a map to name a place given its location. Recognizing and Representin g. Identifying relations Systematic. Communicati on Method of Learning. Identifying Relation. Constructivis m. A place on the surface of the earth is represented by a point.CHAPTER 10 –PLANS AND ELEVATION MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (iii)Sketch and label a parallel of latitude.3 Understand he concept of location of a place.
given the latitudes of both points. • Cooperativ e learning Enquiry discovery Use models such as the globe. Discuss how to find discover the value of this angle y Representin g and interpreting data Drawing diagrams (v) find the longitudes of a point given the longitude of another point and the distance between the two points along the equator. (vii) state the relation between the length of an arc on the equator between two meridians and the length of the corresponding arc on a parallel of latitude. given the longitudes of both points. (iv) find the distance between two points measured along the equator.CHAPTER 10 –PLANS AND ELEVATION LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (ii) find the distance between two points measured along a meridian. MORAL VALUES MATHEMATICS 5 POINTS TO NOTE / VOCABULARY WEEKS GENERICS • Contextu al Learning • Enquiry discovery CCTS Sketch the angle at centre of the • Constructi earth that is subtended by the arc vism between two given points along • Enquiry the equator. (iii) find the latitude of a point given the latitude of another point and the distance between the two points along the same meridian. (vi) state the relations between the radius of the earth and the radius of a parallel of latitude. SUGGESTED TEACHING & LEARNING ACTIVITIES Use the globe to find the distance between two cities or town on the same meridian. to find relationships between the radius of the earth and radii parallel of latitudes • Constructi vism • Communic ation Method of Learning Identifying relations Neatness Systematic Rational Identifying relations 31 .
• Contextual Learning • Enquiry discovery • Self access Learning • Cooperati ve Learning • Self access Learning • Mastery Learning • Thinking skills Drawing diagrams Comparing & differentiatin g Making inferences Cooperation Sharing Tolerance Rational 32 . (x) Find the shortest distance between two points on the surface of the earth. (xi) Solve problems involving :(a) distance between two points (b) traveling on a surface of the earth.CHAPTER 10 –PLANS AND ELEVATION MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES (viii) find the distance between two points measured along a parallel of a latitude. • Cooperati ve Learning • Multiple Learning Use the globe and a few pieces of string to show how to determine the shortest distance between two points on the surface of the earth. GENERICS • Mastery Learning CCTS MORAL VALUES Cooperation Tolerance Sharing POINTS TO NOTE / VOCABULARY (ix) find the latitude of a point given the longitude of another point and the distance between the points along a parallel of latitude. SUGGESTED TEACHING & LEARNING ACTIVITIES Find the distance between two cities or towns on the same parallel of latitudes as a group project.
2 Draw orthogonal projection .2 Understand and use the Concept of plan and elevation Carry out activities in groups where students combine two or more different shapes of simple solid objects into interesting models and draw plans and elevations for these models Mastery Learning Self access learning Analyzing Synthesizi ng Accuracy Creative thinking Systemati c Self Confident Neatness Limit to full scale drawings only Include drawing plan and elevation in one diagram showing projections lines 33 .2 Draw a) the front elevation b) side elevation of a solid object Use models. cylinder.1 Draw the plan of a solid Object 10. cuboids. cone.given an object and a plane 10.1 Identify orthogonal projection 10.1.Plans and Elevations Students will be taught to : 10.1.CHAPTER 10 –PLANS AND ELEVATION MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS CCTS MORAL VALUES POINTS TO NOTE / VOCABULARY 10.2. blocks or plan and elevation kit Comparin Contextual g and learning Differenti ating Mastery Learning Accuracy Creative thinking Emphasize the different uses of dashed lines and solid lines Begin with simple solid objects such as cubic.2. prism and right pyramid Vocab Orthogonal projection Visualizati on Systemati c Identifyin g relationshi p 10.1.1 Understand and use the concept of orthogonal projection Students will be able to 10.3 Determine the difference between an object and 10.
2.2.3 Solve problems involving plans and elevation 10. for example students or teachers dream home and construct a scale model based on the drawings.CHAPTER 10 –PLANS AND ELEVATION MATHEMATICS 5 WEEKS LEARNING AREA LEARNING OBJECTIVES LEARNING OUTCOMES SUGGESTED TEACHING & LEARNING ACTIVITIES GENERICS CCTS MORAL VALUES POINTS TO NOTE / VOCABULARY Draw a) the plan b) the front elevation c) the side elevation of a solid object to scale 10. Carry out group project: Draw plan and elevation of buildings or structures.2.3 Use models to show that it is important to have a plan and at least two side elevations to construct a solid object. Involve real life situations such as in building prototypes and using actual home plans Constructi vism Identifyin g Relationsh ip Dedicatio n Determina tion Vocab Plan Front elevation Side Elevation 34 .4 10.
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Math Practice Online > free > lessons > Nevada > 9th grade > Applied Linear Equations 1
These sample problems below for Applied Linear Equations 1 were generated by the MathScore.com engine.
## Sample Problems For Applied Linear Equations 1
### Complexity=1, Mode=m-pt
1. In slope-intercept form, write the equation of a line that goes through point (-4, -4) with slope 1/4. Equation: 2. In slope-intercept form, write the equation of a line that goes through point (0, 5) with slope 5. Equation:
### Complexity=1, Mode=pt-pt
1. In slope-intercept form, write the equation of a line that goes through point (3, 2) and (0, -3). Equation: 2. In slope-intercept form, write the equation of a line that goes through point (0, -5) and (-5, 4). Equation:
### Complexity=1, Mode=m-pt
1In slope-intercept form, write the equation of a line that goes through point (-4, -4) with slope 1/4.
Equation:
Solution
Plug what you know into y=mx+b to get this:
Now solve for b:
b = - 3
Plug m and b back into y=mx+b to get the answer
Alternate Solution
Start by plugging the slope and point into point-slope form
Now solve for y
2In slope-intercept form, write the equation of a line that goes through point (0, 5) with slope 5.
Equation:
Solution
Plug what you know into y=mx+b to get this:
5 = 5(0) + b
Now solve for b:
b = 5
Plug m and b back into y=mx+b to get the answer
y = 5x + 5
Alternate Solution
Start by plugging the slope and point into point-slope form
y - 5 = 5x
Now solve for y
y = 5x + 5
### Complexity=1, Mode=pt-pt
1In slope-intercept form, write the equation of a line that goes through point (3, 2) and (0, -3).
Equation:
Solution
Calculate the slope:
(y2 - y1) / (x2 - x1) =
(-3 - 2) / (0 - 3) =
53
Plug (3, 2) into y=mx+b to get this:
Now solve for b:
b = - 3
Plug m and b back into y=mx+b to get the answer
Alternate Solution
Plug the slope and (3, 2) into point-slope form
Now solve for y
2In slope-intercept form, write the equation of a line that goes through point (0, -5) and (-5, 4).
Equation:
Solution
Calculate the slope:
(y2 - y1) / (x2 - x1) =
(4 - -5) / (-5 - 0) =
- 95
Plug (0, -5) into y=mx+b to get this:
Now solve for b:
b = - 5
Plug m and b back into y=mx+b to get the answer
Alternate Solution
Plug the slope and (0, -5) into point-slope form
Now solve for y
MathScore.com
Copyright Accurate Learning Systems Corporation 2008.
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# Responses to the Farmer MacDonald Problem
By Hartweg, Kim | Teaching Children Mathematics, May 2004 | Go to article overview
# Responses to the Farmer MacDonald Problem
Hartweg, Kim, Teaching Children Mathematics
The problem appearing in the May 2003 "Problem Solvers" section was stated as follows:
Farmer MacDonald decided to make a graph to show the number of animals on his farm. He must do some mathematical calculations before he can prepare his graph. Here is what Farmer MacDonald knows about his animals:
* He has 12 cows.
* He has 1/2 as many pigs as cows.
* He has 4 more chickens than pigs.
* He has 4 more sheep than chickens.
Organize and display the data by creating a graph to represent Farmer MacDonald's animals. Explain how you organized the data.
[ILLUSTRATION OMITTED]
The purpose of constructing a graph for the Farmer MacDonald problem is to give students the opportunity to become invested in the data and learn how graphs can display information in a variety of ways. Students in Erin Krieg's, Sue Campbell's, and Marlys Heisler's third-grade classes at Hamilton Elementary School in Hamilton, Illinois, were not given much guidance about how to construct the graph but were encouraged to create a graph that made sense to them so that they could explain the information to the class. After working with a partner to solve the problem and create a graph, students presented their work to the rest of the class and participated in a discussion about which graphs were easiest to read and understand.
Heisler's students represented the number of animals on Farmer MacDonald's farm in a variety of ways. Connor and Logan drew a grid to color squares (see fig. 1a), Jon and Scott drew squares and shaded them in (see fig. 1b), Erica and Shelby drew stars (see fig. 1c), Jacey and Kelly wrote out animal sounds (see fig. 1d), and Delaney and Jimmy numbered circles (see fig. 1e). After presenting their graphs to the class, the students decided that Connor and Logan's graph using the grid was the easiest for seeing the amount of each animal on Farmer MacDonald's farm because "they all started in the same place." Erica and Shelby's graph was the next graph of choice, but students decided that the number of stars in the first row should be consistent throughout the graph. When asked why Delaney and Jimmy drew their circles vertically for the pigs and horizontally for the cows, sheep, and chickens, Delaney stated, "We knew we would run out of room on our paper, so we decided to go across. …
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How To Put A Worm On A Treble Hook, Creamy Chicken With Mixed Vegetables, Cajun Eggplant Recipes, Fallout 4 Katana, How To Draw A Palm Tree Realistic, " /> How To Put A Worm On A Treble Hook, Creamy Chicken With Mixed Vegetables, Cajun Eggplant Recipes, Fallout 4 Katana, How To Draw A Palm Tree Realistic, "/>
# differentiable vs continuous derivative
That is, the graph of a differentiable function must have a (non-vertical) tangent line at each point in its domain, be relatively "smooth" (but not necessarily mathematically smooth), and cannot contain any breaks, corners, or cusps. How do you find the differentiable points for a graph? To explain why this is true, we are going to use the following definition of the derivative f ′ … ? The first examples of functions continuous on the entire real line but having no finite derivative at any point were constructed by B. Bolzano in 1830 (published in 1930) and by K. Weierstrass in 1860 (published in 1872). Using the mean value theorem. Your IP: 68.66.216.17 The class C1 consists of all differentiable functions whose derivative is continuous; such functions are called continuously differentiable." Because when a function is differentiable we can use all the power of calculus when working with it. Learn why this is so, and how to make sure the theorem can be applied in the context of a problem. If a function is differentiable, then it has a slope at all points of its graph. Consider a function which is continuous on a closed interval [a,b] and differentiable on the open interval (a,b). we found the derivative, 2x), 2. The differentiability theorem states that continuous partial derivatives are sufficient for a function to be differentiable . Thank you very much for your response. • For example, the function 1. f ( x ) = { x 2 sin ( 1 x ) if x ≠ 0 0 if x = 0 {\displaystyle f(x)={\begin{cases}x^{2}\sin \left({\tfrac {1}{x}}\right)&{\text{if }}x\neq 0\\0&{\text{if }}x=0\end{cases}}} is differentiable at 0, since 1. f ′ ( 0 ) = li… When this limit exist, it is called derivative of #f# at #a# and denoted #f'(a)# or #(df)/dx (a)#. Since is not continuous at , it cannot be differentiable at . and continuous derivative means analytic, but later they show that if a function is analytic it is infinitely differentiable. Pick some values for the independent variable . What did you learn to do when you were first taught about functions? The absolute value function is not differentiable at 0. which means that f(x) is continuous at x 0.Thus there is a link between continuity and differentiability: If a function is differentiable at a point, it is also continuous there. For a function to be differentiable, it must be continuous. As seen in the graphs above, a function is only differentiable at a point when the slope of the tangent line from the left and right of a point are approaching the same value, as Khan Academy also states. )For one of the example non-differentiable functions, let's see if we can visualize that indeed these partial derivatives were the problem. The Weierstrass function has historically served the role of a pathological function, being the first published example (1872) specifically concocted to challenge the notion that every continuous function is differentiable except on a set of isolated points. For each , find the corresponding (unique!) What is the derivative of a unit vector? 3. Since the one sided derivatives f ′ (2−) and f ′ (2+) are not equal, f ′ (2) does not exist. From Wikipedia's Smooth Functions: "The class C0 consists of all continuous functions. value of the dependent variable . Think about it for a moment. Consequently, there is no need to investigate for differentiability at a point, if the function fails to be continuous at that point. It is called the derivative of f with respect to x. Derivative vs Differential In differential calculus, derivative and differential of a function are closely related but have very different meanings, and used to represent two important mathematical objects related to differentiable functions. A function f {\displaystyle f} is said to be continuously differentiable if the derivative f ′ ( x ) {\displaystyle f'(x)} exists and is itself a continuous function. The reason why the derivative of the ReLU function is not defined at x=0 is that, in colloquial terms, the function is not “smooth” at x=0. up vote 0 down vote favorite Suppose I have two branches, develop and release_v1, and I want to merge the release_v1 branch into develop. I leave it to you to figure out what path this is. However, continuity and Differentiability of functional parameters are very difficult. Continuity of a function is the characteristic of a function by virtue of which, the graphical form of that function is a continuous wave. Note that the fact that all differentiable functions are continuous does not imply that every continuous function is differentiable. Cloudflare Ray ID: 6095b3035d007e49 The natural procedure to graph is: 1. See, for example, Munkres or Spivak (for RN) or Cheney (for any normed vector space). is Gateaux differentiable at (0, 0), with its derivative there being g(a, b) = 0 for all (a, b), which is a linear operator. Proof. Differentiable ⇒ Continuous. The first examples of functions continuous on the entire real line but having no finite derivative at any point were constructed by B. Bolzano in 1830 (published in 1930) and by K. Weierstrass in 1860 (published in 1872). Note: Every differentiable function is continuous but every continuous function is not differentiable. Here, we will learn everything about Continuity and Differentiability of … On what interval is the function #ln((4x^2)+9) ... Can a function be continuous and non-differentiable on a given domain? Please enable Cookies and reload the page. In handling continuity and differentiability of f, we treat the point x = 0 separately from all other points because f changes its formula at that point. However, f is not continuous at (0, 0) (one can see by approaching the origin along the curve (t, t 3)) and therefore f cannot be Fréchet … It is possible to have a function defined for real numbers such that is a differentiable function everywhere on its domain but the derivative is not a continuous function. Generally the most common forms of non-differentiable behavior involve a function going to infinity at x, or having a jump or cusp at x. Now, for a function to be considered differentiable, its derivative must exist at each point in its domain, in this case Give an example of a function which is continuous but not differentiable at exactly three points. I have found a path where the limit of this function is 1/2, which is enough to show that the function is not continuous at (0, 0). It follows that f is not differentiable at x = 0.. Differentiability is when we are able to find the slope of a function at a given point. In other words, a function is differentiable when the slope of the tangent line equals the limit of the function at a given point. It's important to recognize, however, that the differentiability theorem does not allow you to make any conclusions just from the fact that a function has discontinuous partial derivatives. There is a difference between Definition 87 and Theorem 105, though: it is possible for a function $$f$$ to be differentiable yet $$f_x$$ and/or $$f_y$$ is not continuous. One example is the function f(x) = x 2 sin(1/x). The colored line segments around the movable blue point illustrate the partial derivatives. A differentiable function is a function whose derivative exists at each point in its domain. if near any point c in the domain of f(x), it is true that . Example of a function that does not have a continuous derivative: Not all continuous functions have continuous derivatives. A function is differentiable on an interval if f ' ( a) exists for every value of a in the interval. However, a differentiable function and a continuous derivative do not necessarily go hand in hand: it’s possible to have a continuous function with a non-continuous derivative. That is, C 1 (U) is the set of functions with first order derivatives that are continuous. It's important to recognize, however, that the differentiability theorem does not allow you to make any conclusions just from the fact that a function has discontinuous partial derivatives. Equivalently, a differentiable function on the real numbers need not be a continuously differentiable function. geometrically, the function #f# is differentiable at #a# if it has a non-vertical tangent at the corresponding point on the graph, that is, at #(a,f(a))#.That means that the limit #lim_{x\to a} (f(x)-f(a))/(x-a)# exists (i.e, is a finite number, which is the slope of this tangent line). Differentiable ⇒ Continuous. The theorems assure us that essentially all functions that we see in the course of our studies here are differentiable (and hence continuous) on their natural domains. Point, then it has a discontinuous derivative if near any point where f ( =.... what is the set of functions say u & in ; 1. C1 consists of all, to continuous functions of x at x = a of functional are!: 1 the requirements for a continuous derivative: 1 we can find the derivative where. Analytic it is possible for the derivative of f ( x = 0 ( not. 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# Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3
You can Download Samacheer Kalvi 6th Maths Book Solutions Guide Pdf, Tamilnadu State Board help you to revise the complete Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 6th Maths Solutions Term 1 Chapter 6 Information Processing Ex 6.3
Question 1.
How many Triangles are there in each of the following figures?
Solution:
(i) The single portions A, B, C, D, E, F, G and H are 8 triangles.
Taking the combination of 2 E & F, F & G & H and E & 11 are 4 triangles.
There are 8 + 4 = 12 triangles.
(ii) The smaller portions A, B, C, D, E, F, G and H are 8 triangles.
Combining 2 at a time B & C, D & F, F & G, A & H are 4 triangles.
Taking 4 at a time
Combination of B, C, D and E; D, E, F and G; F, G, H and A; H, A, B and C are 4 triangles.
Total triangles = 8 + 4 + 4 = 16.
(iii) Every part A, B, C, D, E, F, G, H, I, J, K, L, M, N, O and P are 16 triangles.
Combining two at a time. A & B, B & C, C & D, D & A, J & K, L & M, N & O, P & I are 8 triangles.
Taking the combination of 4, we have I, J, E & A; K, L, F and B, M, N, G and C, P, O, H and D are 4 triangles.
Taking 8 parts together we have QRS, RST, STQ, and TRQ are 4 triangles.
Total triangles = 16 + 8 + 4 + 4 = 32 triangles.
(iv)
Taking smaller triangles we have AFB, BFG, BGC, CGH, HCD, DHI, DIE, EIJ, EJAand AJF ⇒ 10 triangles.
Taking the combination of 2.
CGD, CID, DJE, DHE, EIA, EFA, AGB, AJB, BHC and BFC are 10 triangles.
Combining 3 at a time we have CED, DAE, EBA, ACB, BDC, CEF, DBJ, GDA, EBH, ACI ⇒ 10 triangles.
Other triangles are BCE, CAD, DBE, ADB and ACE are 5 triangles.
Total 10 + 10 + 10 + 5 = 35 triangles.
Question 2.
Find the number of dots in the tenth figure of the following pattern.
Solution:
(i) The number of dots given are 1, 3, 6, 10, 15,…
1st number = 1
2nd number = 1 + 2 = 3
3rd number = 3 + 3 = 6
4th number = 6 + 4 = 10
5th number = 10 + 5 = 15
6th number will be 15 + 6 = 21
7th number will be 21 + 7 = 28
8th number will be 28 + 8 = 36
9th number will be 36 + 9 = 45
10th number will be 45 + 10 = 55
Number of dots in the 10th figure = 55
(ii) The number of dots given are 1, 4, 9, 11, 16, 25,…
The no. of dots are in the pattern 1 × 1, 2 × 2, 3 × 3, 4 × 4, 5 × 5, …… 10 × 10
The number of dots in the 10th figure = 100.
Question 3.
(i) Draw the next pattern.
(ii) Prepare a table for the number of dots used for each pattern.
(iii) Explain the pattern.
(iv) Find the number of dots in the 25th pattern.
Solution:
First number is 2
2nd number is 2 + 3 = 5
3rd number is 5 + 4 = 9
4th number is 9 + 5 = 14
5th number is 14 + 6 = 20
(iv) Number of dots in the 25th pattern is
The number of dots in the 25th pattern = 350.
Question 4.
Count the number of squares in each of the following figures?
Solution:
Smaller squares 4 × 4 = 16 (As numbered).
As a whole bigger = 4
Total squares = 20
As we see in the figure in the middle 1, 2, 3, and 4 are 4 small squares.
Also, we have 9 and 10 = 2 big squares.
Outer squares 5, 6, 7, & 8 = 4.
Total = 4 + 4 + 1 + 1 = 10 squares.
Question 5.
How many circles are there in the following figure?
Solution:
There are 7 Circles.
Question 6.
Find the minimum number of straight lines used in forming the following figures.
Solution:
There are 5 horizontal lines and 5 vertical lines.
Total of 10 lines minimum needed.
There are 12 straight lines used which is a minimum.
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# group socle
The socle of a group is the subgroup generated by all minimal normal subgroups. Because the product of normal subgroups is a subgroup, it follows we can remove the word “generated” and replace it by “product.” So the socle of a group is now the product of its minimal normal subgroups. This description can be further refined with a few observations.
###### Proposition 1.
If $M$ and $N$ are minimal normal subgroups then $M$ and $N$ centralize each other.
###### Proof.
Given two distinct minimal normal subgroup $M$ and $N$, $[M,N]$ is contained in $N$ and $M$ as both are normal. Thus $[M,N]\leq M\cap N$. But $M$ and $N$ are distinct minimal normal subgroups and $M\cap N$ is normal so $M\cap N=1$ thus $[M,N]=1$. ∎
###### Proposition 2.
The socle of a finite group is a direct product of minimal normal subgroups.
###### Proof.
Let $S$ be the socle of $G$. We already know $S$ is the product of its minimal normal subgroups, so let us assume $S=N_{1}\cdots N_{k}$ where each $N_{i}$ is a distinct minimal normal subgroup of $G$. Thus $N_{1}\cap N_{2}=1$ and $N_{1}N_{2}$ clearly contains $N_{1}$ and $N_{2}$. Now suppose we extend this to a subsquence $N_{i_{1}}=N_{1},N_{i_{2}}=N_{2},N_{i_{3}},\dots,N_{i_{j}}$ where
$N_{i_{k}}\cap(N_{i_{1}}\cdots N_{i_{k-1}})=1$
for $1\leq k and $N_{i}\leq N_{i_{1}}\cdots N_{i_{j}}$ for all $1\leq i\leq i_{j}$. Then consider $N_{i_{j}+1}$.
As $N_{i_{j}+1}$ is a minimal normal subgroup and $N_{i_{1}}\cdots N_{i_{j}}$ is a normal subgroup, $N_{i_{j}+1}$ is either contained in $N_{i_{1}}\cdots N_{i_{j}}$ or intersects trivially. If $N_{i_{j}+1}$ is contained in $N_{i_{1}}\cdots N_{i_{j}}$ then skip to the next $N_{i}$, otherwise set it to be $N_{i_{j+1}}$. The result is a squence $N_{i_{1}},\dots,N_{i_{j}}$ of minimal normal subgroups where $S=N_{i_{1}}\cdots N_{i_{s}}$ and
$N_{i_{j}}\cap(N_{i_{1}}\cdots N_{i_{j-1}})=1,\quad 1\leq j\leq s.$
As we have already seen distinct minimal normal subgroups centralize each other we conclude that $S=N_{i_{1}}\times\cdots\times N_{i_{s}}$. ∎
###### Proposition 3.
A minimal normal subgroup is characteristically simple, so if it is finite then it is a product of isomorphic simple groups.
###### Proof.
If $M$ is a minimal normal subgroup of $G$ and $1 is characteristic in $M$, then $C$ is normal in $G$ which contradicts the minimality of $M$. Thus $M$ is characteristically simple. ∎
###### Corollary 4.
The socle of a finite group is a direct product of simple groups.
###### Proof.
As each $N_{i_{j}}$ is characteristically simple each $N_{i_{j}}$ is a direct product of isomorphic simple groups, thus $S$ is a direct product simple groups. ∎
Title group socle GroupSocle 2013-03-22 15:55:12 2013-03-22 15:55:12 Algeboy (12884) Algeboy (12884) 14 Algeboy (12884) Definition msc 20E34 socle socle
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# Difference between revisions of "2021 AIME II Problems/Problem 12"
## Problem
A convex quadrilateral has area $30$ and side lengths $5, 6, 9,$ and $7,$ in that order. Denote by $\theta$ the measure of the acute angle formed by the diagonals of the quadrilateral. Then $\tan \theta$ can be written in the form $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m + n$.
## Diagram
~MRENTHUSIASM (by Geometry Expressions)
## Solution 1
We denote by $A$, $B$, $C$ and $D$ four vertices of this quadrilateral, such that $AB = 5$, $BC = 6$, $CD = 9$, $DA = 7$. We denote by $E$ the point that two diagonals $AC$ and $BD$ meet at. To simplify the notation, we denote $a = AE$, $b = BE$, $c = CE$, $d = DE$.
We denote $\theta = \angle AED$. Hence, $\angle AEB = \angle CED = 180^\circ - \theta$ and $\angle BEC = \theta$.
First, we use the triangle area formula with sines to write down an equation of the area of the quadrilateral $ABCD$.
We have \begin{align*} {\rm Area} \ ABCD & = {\rm Area} \ \triangle ABE + {\rm Area} \ \triangle BCE + {\rm Area} \ \triangle CDE + {\rm Area} \ \triangle DAE \\ & = \frac{1}{2} ab \sin \angle AEB + \frac{1}{2} bc \sin \angle BEC + \frac{1}{2} cd \sin \angle CED + \frac{1}{2} da \sin \angle DEA \\ & = \frac{1}{2} ab \sin \theta + \frac{1}{2} bc \sin \left( 180^\circ - \theta \right) + \frac{1}{2} cd \sin \theta + \frac{1}{2} da \sin \left( 180^\circ - \theta \right) \\ & = \frac{1}{2} ab \sin \theta + \frac{1}{2} bc \sin \theta + \frac{1}{2} cd \sin \theta + \frac{1}{2} da \sin \theta \\ & = \frac{1}{2} \left( ab + bc + cd + da \right) \sin \theta , \end{align*} where the second equality follows from the formula to use the sine function to compute a triangle area, the the fourth equality follows from the property that $\sin \left( 180^\circ - \theta \right) = \sin \theta$.
Because ${\rm Area} \ ABCD = 30$, we have $$\left( ab + bc + cd + da \right) \sin \theta = 60 . \ \ \ (1)$$.
Second, we use the law of cosines to establish four equations for four sides of the quadrilateral $ABCD$.
In $\triangle AEB$, following from the law of cosines, we have $$a^2 + b^2 - 2 a b \cos \angle AEB = AB^2 .$$
Because $\cos \angle AEB = \cos \left( 180^\circ - \theta \right) = \cos \theta$ and $AB = 5$, we have $$a^2 + b^2 + 2 a b \cos \theta = 5^2 . \ \ \ (2)$$
In $\triangle BEC$, following from the law of cosines, we have $$b^2 + c^2 - 2 b c \cos \angle BEC = BC^2 .$$
Because $\cos \angle AEB = \cos \theta$ and $BC = 6$, we have $$b^2 + c^2 - 2 b c \cos \theta = 6^2 . \ \ \ (3)$$
In $\triangle CED$, following from the law of cosines, we have $$c^2 + d^2 - 2 c d \cos \angle CED = CD^2 .$$
Because $\cos \angle CED = \cos \left( 180^\circ - \theta \right) = \cos \theta$ and $CD = 9$, we have $$c^2 + d^2 + 2 c d \cos \theta = 9^2 . \ \ \ (4)$$
In $\triangle DEA$, following from the law of cosines, we have $$d^2 + a^2 - 2 d a \cos \angle DEA = DA^2 .$$
Because $\cos \angle DEC = \cos \theta$ and $DA = 7$, we have $$d^2 + a^2 - 2 d a \cos \theta = 7^2 . \ \ \ (5)$$
By taking $\frac{1}{2} \left( {\rm Eq} \ (2) - {\rm Eq} \ (3) + {\rm Eq} \ (4) - {\rm Eq} \ (5) \right)$, we get
$$\left( ab + bc + cd + da \right) \cos \theta = \frac{21}{2} . \ \ \ (6)$$
By taking $\frac{{\rm Eq} \ (1)}{{\rm Eq} \ (6)}$, we get $$\tan \theta = \frac{60}{21/2} = \frac{40}{7} .$$
Therefore, by writing this answer in the form of $\frac{m}{n}$, we have $m = 40$ and $n = 7$. Therefore, the answer to this question is $m + n = 40 + 7 = \boxed{047}$.
~ Steven Chen (www.professorchenedu.com)
## Solution 2
Since we are asked to find $\tan \theta$, we can find $\sin \theta$ and $\cos \theta$ separately and use their values to get $\tan \theta$. We can start by drawing a diagram. Let the vertices of the quadrilateral be $A$, $B$, $C$, and $D$. Let $AB = 5$, $BC = 6$, $CD = 9$, and $DA = 7$. Let $AX = a$, $BX = b$, $CX = c$, and $DX = d$. We know that $\theta$ is the acute angle formed between the intersection of the diagonals $AC$ and $BD$.
$[asy] unitsize(4cm); pair A,B,C,D,X; A = (0,0); B = (1,0); C = (1.25,-1); D = (-0.75,-0.75); draw(A--B--C--D--cycle,black+1bp); X = intersectionpoint(A--C,B--D); draw(A--C); draw(B--D); label("A",A,NW); label("B",B,NE); label("C",C,SE); label("D",D,SW); dot(X); label("X",X,S); label("5",(A+B)/2,N); label("6",(B+C)/2,E); label("9",(C+D)/2,S); label("7",(D+A)/2,W); label("\theta",X,2.5E); label("a",(A+X)/2,NE); label("b",(B+X)/2,NW); label("c",(C+X)/2,SW); label("d",(D+X)/2,SE); [/asy]$
We are given that the area of quadrilateral $ABCD$ is $30$. We can express this area using the areas of triangles $AXB$, $BXC$, $CXD$, and $DXA$. Since we want to find $\sin \theta$ and $\cos \theta$, we can represent these areas using $\sin \theta$ as follows:
\begin{align*} 30 &=[ABCD] \\ &=[AXB] + [BXC] + [CXD] + [DXA] \\ &=\frac{1}{2} ab \sin (\angle AXB) + \frac{1}{2} bc \sin (\angle BXC) + \frac{1}{2} cd \sin (\angle CXD) + \frac{1}{2} da \sin (\angle AXD) \\ &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \end{align*}
We know that $\sin (180^\circ - \theta) = \sin \theta$. Therefore it follows that:
\begin{align*} 30 &=\frac{1}{2} ab \sin (180^\circ - \theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (180^\circ - \theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2} ab \sin (\theta) + \frac{1}{2} bc \sin (\theta) + \frac{1}{2} cd \sin (\theta) + \frac{1}{2} da \sin (\theta) \\ &=\frac{1}{2}\sin\theta (ab + bc + cd + da) \end{align*}
From here we see that $\sin \theta = \frac{60}{ab + bc + cd + da}$. Now we need to find $\cos \theta$. Using the Law of Cosines on each of the four smaller triangles, we get following equations:
\begin{align*} 5^2 &= a^2 + b^2 - 2ab\cos(180^\circ-\theta) \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta \\ 9^2 &= c^2 + d^2 - 2cd\cos(180^\circ-\theta) \\ 7^2 &= d^2 + a^2 - 2da\cos \theta \end{align*}
We know that $\cos (180^\circ - \theta) = -\cos \theta$. We can substitute this value into our equations to get:
\begin{align*} 5^2 &= a^2 + b^2 + 2ab\cos \theta \\ 6^2 &= b^2 + c^2 - 2bc\cos \theta \\ 9^2 &= c^2 + d^2 + 2cd\cos \theta \\ 7^2 &= d^2 + a^2 - 2da\cos \theta \end{align*}
If we subtract the sum of the first and third equation from the sum of the second and fourth equation, the squared terms cancel, leaving us with: $$5^2 + 9^2 - 6^2 - 7^2 = 2ab \cos \theta + 2bc \cos \theta + 2cd \cos \theta + 2da \cos \theta$$ $$21 = 2\cos \theta (ab + bc + cd + da)$$
From here we see that $\cos \theta = \frac{21/2}{ab + bc + cd + da}$.
Since we have figured out $\sin \theta$ and $\cos \theta$, we can calculate $\tan \theta$:
$$\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{\frac{60}{ab + bc + cd + da}}{\frac{21/2}{ab + bc + cd + da}} = \frac{60}{21/2} = \frac{120}{21} = \frac{40}{7}$$
Therefore our answer is $40 + 7 = \boxed{047}$.
~ my_aops_lessons
## Solution 3 (Pythagorean Theorem and Right Triangle Trigonometry)
This solution refers to the Diagram section.
In convex quadrilateral $ABCD,$ let $AB=5,BC=6,CD=9,$ and $DA=7.$ Let $A'$ and $C'$ be the feet of the perpendiculars from $A$ and $C$, respectively, to $\overline{BD}.$ We obtain the following diagram:
WILL BE BACK SOON.
~MRENTHUSIASM
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Derivatives of Trigonometric Functions
# Derivatives of Sine, Cosine, and Tangent
Some common functions that appear in equations are the basic trigonometric functions. The following three theorems will establish their derivatives.
Theorem 1: If $f(x) = \sin x$ then $\frac{d}{dx} f(x) = \cos x$.
• Proof: Let $f(x) = \sin x$. From the definition of a derivative it follows that:
(1)
\begin{align} f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ f'(x) = \lim_{h \to 0} \frac{\sin(x + h) - \sin(x)}{h} \\ f'(x) = \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h} \\ f'(x) = \lim_{h \to 0} \left ( \sin x \frac{(\cos h - 1)}{h} + \cos x \frac{\sin h}{h} \right ) \end{align}
• Now recall that $\lim_{h \to 0} \frac{(\cos h - 1)}{h} = 0$, and $\lim_{h \to 0} \frac{\sin h}{h} = 1$, so therefore, $f'(x) = 0 \cdot \sin x + 1 \cdot \cos x = \cos x$. $\blacksquare$
Theorem 2: If $f(x) = \cos x$ then $\frac{d}{dx} f(x) = -\sin x$.
• Proof of Property (b): Let $f(x) = \cos x$. From the definition of a derivative it follows that:
(2)
\begin{align} f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \\ f'(x) = \lim_{h \to 0} \frac{\cos(x + h) - \cos x}{h} \\ f'(x) = \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h} \\ f'(x) = \lim_{h \to 0} \cos x \frac{(\cos h - 1)}{h} - \lim_{h \to 0} \sin x \frac{\sin h}{h} \\ f'(x) = 0 \cdot \cos x - 1 \cdot \sin x \\ f'(x) = -\sin x \quad \blacksquare \end{align}
Theorem 3: If $f(x) = \tan x$ then $\frac{d}{dx} f(x) = \sec ^2 x$.
We will now prove theorem 3 at this point and time.
## Example 1
Differentiate $f(x) = 3 \cos x + 2 \sin x$.
Applying the theorems above and we get that $f'(x) = -3 \sin x + 2 \cos x$.
## Example 2
Differentiate $f(x) = 2\tan x - 2 \sin x$.
Applying the theorems above and we get that $f'(x) = 2 \sec ^2 x - 2 \cos x$.
# Cyclic Pattern of Higher Order Derivatives of Sine and Cosine
Notice that the derivative of $\sin x$ is $\cos x$… the derivative of $\cos x$ is $-\sin x$, etc…. Hopefully you can see that there is a cycle here as:
(3)
\begin{align} \frac{d}{dx} \sin x = \cos x \end{align}
(4)
\begin{align} \frac{d}{dx} \cos x = -\sin x \end{align}
(5)
\begin{align} \frac{d}{dx} -\sin x = -\cos x \end{align}
(6)
\begin{align} \frac{d}{dx} -\cos x = \sin x \end{align}
We can thus make a general rule for higher order derivatives of $\sin x$ and $\cos x$. If $f(x) = \sin x$ or $f(x) = \cos x$ and $n = 4k + 1$, then for all integers $k ≥ 0$:
(7)
\begin{align} \frac{d^n}{dx^n} f(x) = f(x) \end{align}
## Example 3
What function do you get if you differentiate $f(x) = \sin x$ one-hundred times?
Notice that every four differentiations, we will get back to $\sin x$. Therefore, if we differentiate $100$ times, we will still have $\sin x$.
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Home » Greatest Common Factor » GCF of 67 and 40
# GCF of 67 and 40
The gcf of 67 and 40 is the largest positive integer that divides the numbers 67 and 40 without a remainder. Spelled out, it is the greatest common factor of 67 and 40. Here you can find the gcf of 67 and 40, along with a total of three methods for computing it.
1
In addition, we have a calculator you should check out. Not only can it determine the gcf of 67 and 40, but also that of three or more integers including sixty-seven and forty for example. Keep reading to learn everything about the gcf (67,40) and the terms related to it.
## What is the GCF of 67 and 40
If you just want to know what is the greatest common factor of 67 and 40, it is 1. Usually, this is written as
gcf(67,40) = 1
The gcf of 67 and 40 can be obtained like this:
• The factors of 67 are 67, 1.
• The factors of 40 are 40, 20, 10, 8, 5, 4, 2, 1.
• The common factors of 67 and 40 are 1, intersecting the two sets above.
• In the intersection factors of 67 ∩ factors of 40 the greatest element is 1.
• Therefore, the greatest common factor of 67 and 40 is 1.
Taking the above into account you also know how to find all the common factors of 67 and 40, not just the greatest. In the next section we show you how to calculate the gcf of sixty-seven and forty by means of two more methods.
## How to find the GCF of 67 and 40
The greatest common factor of 67 and 40 can be computed by using the least common multiple aka lcm of 67 and 40. This is the easiest approach:
gcf (67,40) = = 1
Alternatively, the gcf of 67 and 40 can be found using the prime factorization of 67 and 40:
• The prime factorization of 67 is: 67
• The prime factorization of 40 is: 2 x 2 x 2 x 5
• The prime factors and multiplicities 67 and 40 have in common are: 1
• 1 is the gcf of 67 and 40
• gcf(67,40) = 1
In any case, the easiest way to compute the gcf of two numbers like 67 and 40 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 67,40. The calculation is conducted automatically.
Similar searched terms on our site also include:
## Use of GCF of 67 and 40
What is the greatest common factor of 67 and 40 used for? Answer: It is helpful for reducing fractions like 67 / 40. Just divide the nominator as well as the denominator by the gcf (67,40) to reduce the fraction to lowest terms.
.
## Properties of GCF of 67 and 40
The most important properties of the gcf(67,40) are:
• Commutative property: gcf(67,40) = gcf(40,67)
• Associative property: gcf(67,40,n) = gcf(gcf(40,67),n)
The associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.
To sum up, the gcf of 67 and 40 is 1. In common notation: gcf (67,40) = 1.
If you have been searching for gcf 67 and 40 or gcf 67 40 then you have come to the correct page, too. The same is the true if you typed gcf for 67 and 40 in your favorite search engine.
Note that you can find the greatest common factor of many integer pairs including sixty-seven / forty by using the search form in the sidebar of this page.
Questions and comments related to the gcf of 67 and 40 are really appreciated. Use the form below or send us a mail to get in touch.
Please hit the sharing buttons if our article about the greatest common factor of 67 and 40 has been useful to you, and make sure to bookmark our site.
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# How to find corresponding coordinates?
• Sep 21st 2012, 10:42 AM
misiaizeska
How to find corresponding coordinates?
The point (2,3) is one the graph of y=f(x). Determine the corresponding coordinates of this point on the graph of y = -2(f(2(x+5))-4.
Not sure how to do this - if anyone could help I would greatly appreciate it.
I know that there are multiple points, and the answers are : (2,3)/(1,3)/(1,6)/(1,-6)/(-4,-6)/(-4,-10). I'm not sure how to find these - thanks in advance!
• Sep 21st 2012, 02:07 PM
sakonpure6
Re: How to find corresponding coordinates?
Okay,actually there is only one answer for the image.
Here is how to find it:
1- Point $(2,3)$represents $x=2$ and $y=3$ on $y=f(x)$ So when we sub those points in we get $3=f(2)$.
2- To find the image of point $(3,2)$ of $y=-2(f(2(x+5))-4$ , first you got to realize that $y=f(x)$ is the parent function of $y=-2(f(2(x+5))-4$ which has just been moved around. So for x co-ordinate for the image, we got to find the number for x in the transformed function that would make the transformed function have $f(2)$ and that would be the image co-ordinates for x value. Your transformed function would look like this $y=-2(f(2))-4$
$(2(x+5))=2$
$(x+5)=1$
$(x)=-4$
Therefore when x= -4we get 2. x= -4 is the x-co-ordinate of the new image point.
Now to solve for y point, we do some substitution.
Since $f(2)=3$ we sub in $3$ for $f(2)$ in the transformed function and then solve for y.
$y=-2(f(2))-4$
$y=-2(3)-4$
$y=-6-4$
$y=-10$
Therefore the image of point (3,2) on the function $f(x)=y$ is (-4,-10) on the function of $y=-2(f(2(x+5))-4$
Hope this helped if you have more questions just ask :)
EDIT: Another way of solving this problem would be by applying the transformations to (2,3) in the transformed function
• Oct 2nd 2012, 03:04 PM
misiaizeska
Re: How to find corresponding coordinates?
Thank you so much!
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## What Are Inverse Trigonometric Functions?
Inverse trigonometric functions are the inverse of normal trigonometric functions. Alternatively denoted as cyclometric or arcus functions, these inverse trigonometric functions exist to counter the basic trigonometric functions, such as sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (cosec). When trigonometric ratios are calculated, the angular values can be calculated with the help of the inverse trigonometric functions.
## How Do Inverse Trigonometric Functions Work?
The term Arcus functions, or Arc functions, is also used to denote inverse trigonometric functions. If a normal trigonometric function is being considered, it has a value. Using the inverse trigonometric function, we can calculate the arc length that is used to get that specific value. Whatever operation the basic trigonometric function performs, the inverse trigonometric function does exactly the opposite.
When considering right-angled triangles, the concept of trigonometry comes into play. Using the trigonometric functions, students can measure the angles created in the triangle by the base, height, or hypotenuse. Using the inverse trigonometric functions, the exact value of the created angle can be measured.
## How Many Types of Inverse Trigonometric Functions Are There?
As stated before, the inverse trigonometric functions are the exact opposites of the basic trig functions. There are six basic functions in trigonometry. Every trigonometric ratio can be expressed with the help of these functions. As a result, there are also six inverse trigonometric functions, each acting as an inverse for the six trigonometric functions.
The inverse trigonometric functions are as follows-
• Arcsine
• Arccosine
• Arctangent
• Arccotangent
• Arcsecant
• Arccosecant
All of these six inverse trigonometric functions have been discussed in detail below, along with their ranges and domains.
### What is Arcsine Function?
The arcsine function, or arcsin, is the first of the six inverse trigonometric functions. It is the inverse function that corresponds to the sine function. As a result, it is denoted by sin-1 x. The arcsine function has a range that starts from -π/2 to π/2, and its domain lies from -1 to 1.
### What is Arccosine Function?
The arccosine function, or arcos, is the inverse trigonometric function corresponding to the cosine or cos function in trigonometry. Hence, it is also denoted as cos-1 x. The range of the arccosine function lies from 0 to π, and its domain starts at -1 and ends at 1.
### What is Arctangent Function?
The arctangent function, or arctan, is the inverse trigonometric function corresponding to the tangent or tan function in trigonometry. In other words, it is the inverse of the tangent trig function. Therefore, it can also be denoted as tan-1 x. Its range lies between -π/2 and π/2, and its domain lies between negative infinity (-∞) and positive infinity (∞).
### What is Arccotangent Function?
The arccotangent function, or arccot, is the inverse of the cotangent or cot function. It can be represented as cot-1 x. The range of the arccotangent function lies between 0 and π, and its domain lies between negative infinity and positive infinity.
### What is Arcsecant Function?
The arcsecant function, or arcsec, is the inverse of the secant or sec function. Hence, it can be represented as sec-1 x. The range of the arcsecant function lies between -π/2 and π/2, excluding zero. The domain either lies from negative infinity to -1, or from 1 to positive infinity. The values near and at zero are excluded.
### What is the Arccosecant Function?
The arccosecant function, or arccos, acts as the inverse of the cosecant or the cosec function. As a result, it can be denoted as cosec-1 x. The domain and range of the arccosecant function are the same as the arcsecant function. Hence, its range lies between -π/2 and π/2, excluding zero, and its domain lies either from negative infinity to -1, or from 1 to positive infinity. Here also, the values at and close to zero are excluded.
## Formulas
All of the inverse trigonometric functions that exist come with their own set of formulae. Students need to be familiar with these formulas to solve complex inverse trigonometric problems quickly.
Some of the most basic inverse trigonometric function formulas are given below.
### Arcsine Function
For any arcsine function x,
Here, x lies from -1 to 1.
### Arccosine Function
For any arcos function x,
Here also, x lies from -1 to 1.
### Arctangent Function
For any arctan function x,
Here, x lies in the real number set R.
### Arccotangent Function
For any arc cot function x,
Here also, x lies in the real number set R.
### Arcsecant Function
For any arcsec function x,
Here, the mod x, or |x| is always greater than or equal to 1.
### Arccosecant Function
For any arccosec function x,
Here also, |x| is always greater than or equal to 1.
These are some of the basic inverse trig function formulas that come in very handy during operations.
## What Are the Derivatives of Inverse Trigonometric Functions?
Just like normal trig functions, inverse trigonometric functions can also be differentiated. By differentiation, the first-order derivatives of the inverse trigonometric functions can be found.
All of the six inverse trigonometric functions have their first-order derivatives. They are given below.
### Arcsine Function
For y = sin-1 x,
$\frac{dy}{dx}=\frac{1}{\sqrt{1-{x}^{2}}}$
### Arccosine Function
For y = cos-1 x,
$\frac{dy}{dx}=\frac{-1}{\sqrt{1-{x}^{2}}}$
### Arctangent Function
For y = tan-1 x,
$\frac{dy}{dx}=\frac{1}{1+{x}^{2}}$
### Arccotangent Function
For y = cot-1 x,
$\frac{dy}{dx}=\frac{-1}{1+{x}^{2}}$
### Arcsecant Function
For y = sec-1 x,
$\frac{dy}{dx}=\frac{1}{\left[\left|x\right|\sqrt{\left({x}^{2}-1\right)}\right]}$
### Arccosecant Function
For y = cosec-1 x,
$\frac{dy}{dx}=\frac{-1}{\left[\left|x\right|\sqrt{\left({x}^{2}-1\right)}\right]}$
## Practice Problem
Find the value of sin (cos-1 4/5)
⇒ For assumption, let:
Therefore, sin (cos-1 4/5) = 3/5.
## Context and Applications
Inverse trigonometric functions have some major real-life applications. Examples include operations in navigating, processes in geometry, describing terms in physics, and applications in engineering work.
This topic is significant in the professional exams for both undergraduate and graduate courses, especially for:
• Bachelors of Science in Mathematics
• Masters of Science in Mathematics
### Want more help with your trigonometry homework?
We've got you covered with step-by-step solutions to millions of textbook problems, subject matter experts on standby 24/7 when you're stumped, and more.
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# How to measure length correctly
This term we began our new Maths topic – Measurement.
The skills we will be learning in measurement are:
• What is length and how can we measure it?
• Appropriate measuring tools and how to use them
• Units of measurement: millimeters., centimeters, metres and kilometres.
• Perimeter
• Area
To begin with we brainstormed what we thought units of measurement were and what units of measurement we would use to measure certain objects. Here are some examples:
Small units of length are called millimetres.
A millimetre is about the thickness of a plastic id card (or credit card).
Or about the thickness of 10 sheets of paper on top of each other.
This is a very small measurement!
When you have 10 millimetres, it can be called a centimetre.
1 centimetre = 10 millimetres
A fingernail is about one centimetre wide.
metre is equal to 100 centimetres.
1 meter = 100 centimetres
The length of this guitar is about 1 metre
kilometre is equal to 1000 meters.
When you need to get from one place to another, you measure the distance using kilometres.
The distance from one city to another or how far a plane travels would be measured using kilometeres.
We discussed as a class how to measure accurately. We need to make sure that we begin at the 0 and that the edge of object we are measuring is next to it. Its also very important to hold the ruler straight. We also discussed what the different units of measurement are on a ruler. Check out this video for more information!
Next we took part in a measuring competition as a class.
Our task was to work in groups of 5 and use our rulers to measuring different stops of paper in centimetres. The only hints we had was that no strip in all 6 coloured bags were the same length and the answers would either be whole centimetres like 18cm or 5cm or they were half centimetres such as 21.5cm or 14.5cm.
It was really important to work as a team and make sure we used our measuring skills that we discussed as a class in the previous lesson. These skills included; lining the 0 up with the end of the paper to measure, holding the ruler straight, reading a ruler correctly and using the correct units of measurement.
Watch our video below of us working together as teams by communicating and helping each other and using mathematical language.
IMG_0376 from skye baldwin on Vimeo.
IMG_0374 from skye baldwin on Vimeo.
For the next few lessons our class undertook a mini investigation involving measuring our bodies.
Our first task was to draw a table and record our estimates of the following body parts:
• Around the base of our thumb
• Around our wrist
• Around our neck
• Around our waist
• A body part of our choice.
We then worked with a partner and used measuring tapes to measure each body part. The units of measurement we used were centimetres. We are now getting better and better at estimating measurements each week with some students even estimating the exact measurement!
Next we were given an old saying to read:
Once around the waist, twice around the neck;
Once around the neck, twice around the wrist.
We had to use the information we had gathered to decide whether this saying was true for us.
We found out that whilst both measurements were not exactly the same as each other they were very very close. We decided that the saying was close to being true because of this observation.
Lastly we had to investigate if we could find any other pairs of body measurements that are double or half relationships. Some of us found that twice around the base of our thumb is the same or nearly the same measurement as once around our wrist. Check out some other statements we created that were true to our body measurements!
## 6 thoughts on “All about measurement”
1. Laksmon
Dear Mrs Baldwin,
The interesting thing about the task was the body that we measured twice around the wrist and once around the waist my did not equal the same I found that hard.
You could use measment in building and cutting and more
2. Jasmin O
Dear Mrs Baldwin what I found interesting was how much smaller or bigger my waist and my neck was compered to what I thought it was. The examples that I would use measuring in would be if you were getting new furniture so it would fit in the area or when you are doing art and you needed to measure how big the pieces of paper needed to be. From Jasmin.
3. Harrison
Dear mrs Baldwin,
what I enjoyed the most about measuring is the body measurement because you actually were able to get your full measurement of certain body parts.
I’s probably be seeing builders measuring timber,bricks and doors or something like that.
from Harrison
4. kyra
Dear Mrs Baldwin what I found most interesting about the maths topic was measuring the length of the whole class
The measurement I see in every day life is clothes designers and shop owners measuring clothes and mothers and fathers measuring their children as they grow up.
5. Chloe
Dear Mrs Baldwin,
I like that it was a group activity so we had to work together and I found that interesting because we could see who could get along together and help each other.
If you are a tyre fitter and you may measure the tyres to find the right size to fit the car or a tailor/dressmaker or a Landscaper etc.
Regards Chloe.
6. Constantina🐬
Dear Miss Baldwin,
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Kirchoff's Laws, Ohm's Law, and parallel and series resistors
More Tools for analyzing circuits
Outline: Kirchoff's Current Law Kirchoff's Voltage Law Ohm's Law Independent Sources Reducing Circuits Voltage Division Current Division
Kirchoff's Current Law
### Kirchoff's Current Law (KCL) states: The Sum of the currents entering a node is always zero. Or we can say the sum of the currents entering a node = sum of currents leaving the same node.
Before elaborating on this law, let's first define what a node is:
A node is a connection point of two or more elements.
The table below shows some circuits and their nodes:
This circuit has four nodes: This circuit has six nodes: This circuit has three nodes:
Now let's apply KCL in the following two problems:
Σientering = Σileaving
In this subcircuit, we have one node shown. At this node we have three currents entering the node and one current leaving the node. Applying KCL: i1 + i2 + i3 = i4 3 + -2 + 1 = 2 2 = 2 thus KCL holds true. In this subcircuit, we will use KCL to find iy Appying KCL at the left node: 3A + 4A = 2A + ix ix = 5A Now applying KCL at the right node: ix + iy = 1A 5A + iy = 1A iy = -4A
KCL can be written three ways:
Σientering = Σileaving
OR
Σientering - Σileaving = 0
OR
- Σientering + Σileaving = 0
Top of Page
Kirchoff's Voltage Law
### Kirchoff's Voltage Law (KVL) states: The Sum of the voltage drops equals the sum of the voltage rises around any closed path.
ΣVdrops = ΣVrises
OR
ΣVdrops - ΣVrises = 0
Now lets apply KVL in the following problems:
In this circuit, we have one closed path. Typically we refer to a closed path as a loop. In this problem we will find Vx using KVL. Starting at the top-left corner and going clockwise we apply KVL: V1 - V2 + (-3V) + Vx = 0 1V - 6V + -3V + Vx = 0 -8V = -Vx Vx = 8V In this circuit, we will use KVL to find Vx, Vy, & Vz Note: We will also write KVL assuming a clockwise direction. The right most loop has only one unknown, so we will apply KVL there first: +3V - 9V - (-2V) + Vx = 0 -4V + Vx = 0 Vx = 4V Now applying KVL at the middle and left-most loops: -Vx + Vy = 0 Vx = Vy = 4V -Vy - Vz = 0 -Vy = Vz = -4V
Top of Page
Ohm's Law
#### Before defining Ohm's Law, let's elaborate on what a resistor is: A resistor is a circuit element which is neither an insulator or conductor. Current can pass through a resistor, however there is some resistance. A resistor resists the flow of current. This resistance means that some work must be done to "push" current through the resistor. Whenever work is done on charge, we have voltage. Thus, when current flows through a resistor, there is some voltage across the resistor.
The symbol for the resistor is:
The unit for resistance is ohms and the symbol is:
Ω
Ohm's Law states:
The voltage across a resistor is proportional to the current flowing in the resistor.
The larger the current in the resistor the larger the voltage across the resistor.
The resistance of the resistor is the constant of proportionlality.
So Ohm's Law is:
V = IR
We will apply Ohm's Law and Kirchoff's Laws to solve the following problems:
In this problem we will find V using Ohm's Law: V = IR V = 2A(10 ohms) = 20 Volts Note here that current is going into the negative side of the resistor: Note: Current can only flow from a higher voltage potential to a lower voltage potential (like water flowing down stream), Therefore in this problem you must use this form of Ohm's Law: V = -IR Now find I: Since V = -IR in this problem then I = -V/R I = -(20 V)/(1 ohm) I = -20 Amps
### Summary: Use V = IR if current is shown flowing into the positive side of the resistor, and use V = - IR if the current is shown flowing into the negative side of the resistor.
Top of Page
Independent Sources
### In order to build a useful circuit, there needs to be a power source. Without a power source or Voltage, current will not flow.What is an independent source? An independent source supplies a fixed voltage or current, independent of anything else.
Below are two independent soureces:
These are independent voltage sources. The first is a power supply that provides a constant voltage. The second is a battery that also provides a constant voltage. These is an independent current sources. Independent current sources are rare in the real world so consider this to be a theoretical device. You will see later that a current source can be constructed from a voltage source.
## Study Problems
After clicking on the following link enter 1-1 for the problem and 1 for the step:
Study Problem 1-1
After clicking on the following link enter 1-2 for the problem and 1 for the step:
Study Problem 1-2
After clicking on the following link enter 1-3 for the problem and 1 for the step:
Study Problem 1-3
After clicking on the following link enter 1-4 for the problem and 1 for the step:
Study Problem 1-4
Top of Page
Reducing Circuits
### When analyzing a circuit, it is often easier to do so if the circuit can be simplified or reduced. In this section we will consider how a colletion of resistor can be reduced down to fewer resitors.
Consider the following two circuits:
Write a KVL equation for this circuit. Recall that V10=IR Start at the bottom left corner and go clockwise: -30V + 10I = 0 30 = 10I Write a KVL equation for this circuit. Recall that V5=IR Start at the bottom left corner and go clockwise: -30V + 5I + 5I = 0 -30V + 10I = 0 30 = 10I Both circuits are described by the same equation, therefore they are electrically the same circuit. This means that you can replace one circuit with the other. We can simplify the second circuit by replacing it with the first circuit.
### Series Resistors:
Consider the following two circuits:
Write a KVL equation for this circuit. -V + R1I + R2I + R3I = 0 -V + I(R1 + R2 + R3) = 0 -V + IReqv = 0 where Reqv = R1 + R2 + R3 Since Reqv = R1 + R2 + R3 then we can replace the three resistors with one resistor. Resistors which carry the same current are in series. In this case R1, R2, and R3 are all in series. They all carry current I Resistors in series can be added and then replaced with one resistor equivalent to to their sum. ΣRi = Reqv
## Study Problem
After clicking on the following link enter 1-5 for the problem and 1 for the step:
Study Problem 1-5
### Parallel Resistors:
Consider the following two circuits:
In this circuit R1 and R2 are said to be in parallel because they have the same two terminal nodes, therefore they have to have the same volrage across them. V1 = VS and V2 = VS Using Ohm's Law we know that I=V/R Therefore: I1 = VS/R1 and I2 = VS/R2 Applying KCL at the top node we generate the following equation: IS = I1 + I2 IS = VS/R1 + VS/R2 Factoring out VS we get: IS = VS(1/R1 + 1/R2) Now let's obtain the resistance which is equivalent to the two resistors in parallel: Recall that VS/IS = Requivalent and IS/VS = 1/Requivalent 1/Requivalent = IS/VS = 1/R1 + 1/R2 1/Requivalent = (R2 + R1)/{R1*R2) Threrefore two resistors in parallel can be reduced with the following expression: Requivalent = {R1*R2)/(R1 + R2)
Using this information we will reduce the following circuits:
The two circuits shown here are equivalent:
Req. = {R1*R2)/(R1 + R2)
Req. = {12*4)/(12 + 4) = 48/16 = 3 Ohms
Req. = 3 Ohms
The three circuits shown here are also equivalent:
Rx = {R1*R2)/(R1 + R2)
Rx = {12*12)/(12 + 12) = 144/24 = 6 Ohms
Req. = {R3*Rx)/(R3 + Rx)
Req. = (3*6)/(3+6) = 18/9 = 2 Ohms
### Combining Resistors in Parallel:
If we always combine resistors in parallel "two-at-a-time"
then we can use the equation:
Requivalent = {R1*R2)/(R1 + R2)
Otherwise we must use the more general form for 'n' resistors:
1/Requivalent = 1/R1 + 1/R2 + .... + 1/Rn
## Study Problem
After clicking on the following link enter 1-6 for the problem and 1 for the step:
Study Problem 1-6
Top of Page
Voltage Division
Continuing with analysis:
Req. = R1 + R2 + R3
IS = VS/Req.
V1 = IS * R1
V2 = IS * R2
V3 = IS * R3
Now plug in IS:
V1 = (VS/Req.) * R1 = (R1/Req.)*VS
V2 = (VS/Req.) * R2 = (R2/Req.)*VS
V3 = (VS/Req.) * R3 = (R3/Req.)*VS
In plain english what do these mathematical equations say?
The voltage across a series resistor is some percentage of the total voltage, VS
This percentage is equal to the individual resistor's resistance divided by the equivalent resistance, Req.
The voltage is being divided among the series resistors, hence this is called Voltage Division
## Study Problem
After clicking on the following link enter 1-10 for the problem and 1 for the step:
Study Problem 1-10
Top of Page
Current Division
Continuing with analysis:
From the circuit on the left we know that I1 = Vs/R1
and I2 = Vs/R2
From the reduced circuit on the right we know that
Vs = Is*Req.
Now we will plug the second equation into the first equation:
I1 = (Is*Req.)/R1
I2 = (Is*Req.)/R2
From the section on parallel resistors we know:
Req. = R1*R2/(R1+R2)
If we plug this into the last set of equations we get:
I1 = (Is*[R1*R2/(R1+R2)])/R1
I2 = (Is*[R1*R2/(R1+R2)])/R2
Note that R1 factors out of the first equation
and R2 factors out the the second equation giving the following:
I1 = Is*[R2/(R1+R2)]
I2 = Is*[R1/(R1+R2)]
In plain english what do these mathematical equations say?
The current through a parallel resistor is some percentage of the total source current, IS
This percentage is equal to the OTHER resistor's resistance divided by the sum of the two resistors, R1 and R2
The currrent is being divided among the parallel resistors, hence this is called Current Division
## Study Problem
After clicking on the following link enter 1-11 for the problem and 1 for the step:
Study Problem 1-11
Top of Page
Back To Index
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#### Need Solution for R.D.Sharma Maths Class 12 Chapter 18 Indefinite Integrals Exercise 18.11 Question 5 Maths Textbook Solution.
Answer:$\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec \times| C$
Hint:- Use substitution method to solve this integral.
Given:$\int \tan ^{5} x d x$
Solution: Let,$\mathrm{I}=\int \tan ^{5} x d x$
Re-Write,$\mathrm{I}=\int \tan ^{3} x \cdot \tan ^{2} x \cdot d x$
$I=\int \tan ^{3} x\left(\sec ^{2} x-1\right) d x$ $\text { (if, } \left.\tan ^{2} x=\sec ^{2} x-1\right)$
$\mathrm{I}=\int\left(\tan ^{3} x \cdot \sec ^{2} x-\tan ^{3}\right) d x$
$I=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{3} x d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \tan ^{2} x \cdot \tan x d x$
$\inline \left.=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x-1\right) \cdot \tan x d x \quad \text { (if, } \sec ^{2} x-\tan ^{2} x=1\right)$
$\inline =\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int\left(\sec ^{2} x \tan x-\tan x\right) d x$
$=\int \tan ^{3} x \cdot \sec ^{2} x \cdot d x-\int \sec ^{2} x \tan x d x+\int \tan x d x$
Put $tan x = t \Rightarrow \sec^{2}x d x = d t , then$
$\mathrm{I}=\int \mathrm{t}^{3} d t-\int t d t+\int \tan x d x$
$=\frac{t^{8+1}}{3+1}+\frac{t^{1+1}}{1+1}+\log |\sec \times| C$ $\text { (if, } \left.\int x^{n} d x=\frac{x^{n+1}}{n+1}+C \& \int \tan x d x=\log |\sec x|+c\right)$
$=\frac{t^{4}}{4}-\frac{t^{2}}{2}+\log |\sec x| C$
$\left.=\frac{\tan ^{4} x}{4}-\frac{\tan ^{2} x}{2}+\log |\sec x| C\: \: \: \: \: \: \quad \text { (if, } t=\tan x\right)$
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### Algebra I Keystone Exam Sample Questions
```Algebra I
Keystone Exam
Sample Questions
Problems taken from: Keystone Exams: Algebra I Assessment
Anchors and Eligible Content
http://static.pdesas.org/Content/Documents/KYST_ALG1_AAEC_S
mpl_Qstns_Glssry_April-2012.pdf
A1.1.1.1.1
A.1.1.1.1.2
A.1.1.1.1.2
A1.1.1.2.1
A1.1.1.3.1
A1.1.1.4.1
A1.1.1.5.1
A1.1.1.5.2
A1.1.1.5.3
A1.1.1
= ℎ2 + 4ℎ
A1.1.1
= ℎ2 + 16ℎ + 60
A1.1.1
The width of the new frame is 1 inch. To do this problem I factored ℎ2 +
8ℎ + 12 into (ℎ + 6)(ℎ + 2) to find the length and height of the canvas and
frame. The new height, h + 2, is 2 more than the height of the canvas, h, so
the new frame must add a total of 2 inches, 1 on each side.
A1.1.1
3
4
0.003, 5−3 , , , 5
8 7
A1.1.1
17
A1.1.1
Step 1 is incorrect. 7 425 should be broken down into the product of its factors,
7 25 17 .
A1.1.1
The solution using the corrected step would be:
7 425
7 25 17
7 5 17
35 17
A1.1.2.1.1
A1.1.2.1.2
A1.1.2.1.3
A1.1.2.2.1
A1.1.2.2.2
A1.1.2
6 hours
A1.1.2
3 hours
A1.1.2
ℎ > 5 From 0 hours up to 5 hours, Pat saves more money than Alex. At 5 hours they have
saved the same and at all points after 5 hours, Alex will have saved more money.
A1.1.2
= 0.75 + 1.25
A1.1.2
x-variable: the number of bowls
y-variable: the height of the stack of bowls
A1.1.2
8.75 inches
A1.1.3.1.1
A1.1.3.1.2
A1.1.3.1.3
A1.1.3.2.1
A1.1.3.2.2
A1.1.3
≤ 180
5
+ ≤ 300
4
A1.1.3
Dark purple
represents
solution set
A1.1.3
5
The maximum profit is represented by the intersection of = 180 and 4 + = 300.
The point (96, 180) is the intersection of these two equations.
A1.1.3
4 + 3 < 7 − 9
12 < 3
4<
or
>4
A1.1.3
1
+ 7.5
4
A1.1.3
<
3
−
5
+3
≥1
A1.2.1.1.1
A1.2.1.1.2
A1.2.1.1.3
A1.2.1.2.1
A1.2.1.2.2
A1.2.1
= 62ℎ + 84
A1.2.1
11
6
1
A1.2.1
A1.2.1
As the distance driven (d) increases, the amount of gasoline (g) must decline.
A1.2.1
= 24 − 220
A1.2.1
\$220.00
A1.2.1
10
A1.2.1
-220 represents the cost of materials. As the values of the range increase from
–220, Ben is losing less money. When the range becomes positive, Ben is
making a profit.
A1.2.2.1.1
A1.2.2.1.2
A1.2.2.1.3
A1.2.2.1.4
A1.2.2.2.1
A1.2.2
.40 + .25 = 10
A1.2.2
A1.2.2
8
−
5
A1.2.2
The slope represents the fact that for every 8 granola bars
that Georgia purchases, she could by 5 fruit bars.
A1.2.2
14 hours
A1.2.2
560 mph
A1.2.2
= .00032
A1.2.2
2
1166
3
miles
A1.2.3.1.1
A1.2.3.2.1
A1.2.3.2.2
A1.2.3.2.3
A1.2.3.3.1
A1.2.3
40
A1.2.3
25%
A1.2.3
Half of the students scored greater than 68. The median marks
the point where half of the students scored greater than. In
this case, half of the students scored greater than 68, not 75.
A1.2.3
A1.2.3
146
A1.2.3
153
A1.2.3
148 and 158
```
|
# Problems on Ages
Working methodology: In these problems, two persons initial ages will be given. and before or after several years, their ratio of the ages will be given. Multiply the ratio of their initial age by x or some variable and take them as their initial age. Now if final ratio has been given, equate this ratio with that ratio and find x. Or proceed according to the problem.
Solved Examples:
1. One year ago Jaya was four times as old as her daughter Nikitha. Six years hence, Mrs.Jaya's age will exceed her daughter's age by 9 years. The ratio of the present ages of Jaya and her daughter is :
a. 9 : 2
b. 11: 3
c. 12: 5
d. 13: 4
Correct Option: D
Explanation:
Let Nikitha's age 1 year ago = x
Then Jaya's age 1 year ago = 4x
After 6 years their ages are 4x+7, x+7
(4x+7) - (x+7) = 9 or x = 3
Present age of Jaya = (12+1) years = 13 years
Present age of Nikitha = (3+1) years = 4 years
Ratio of their ages = 13 : 4
2. Five years ago, the total of the ages of a father and his son was 40 years. The ratio of their present ages is 4 : 1. What is the present age of the father ?
a. 30 years
b. 20 years
c. 25 years
d. None of these
Correct Option: D
Explanation:
Let son's age = x. Then father's age = 4x
(x-5)+(4x-5)=40 $⇒$ x = 10
Present age of father = 40 years
3. A father is twice as old as his son. 20 years ago the age of the father was 12 times the age of the son. The present age of the father is :
a. 44 years
b. 32 years
c. 22 years
d. 45 years
Correct Option: A
Explanation:
Let son's age = x. Then, father's age = 2x.
12(x-20)=(2x-20)$⇒$x=22
Father's present age = 44 years
4. The ratio of the ages of Jaya and Ravi is 2:5. After 8 years, their ages will be in the ratio of 1:2. The difference in their present ages is : ( in years )
a. 24
b. 26
c. 29
d. 32
Correct Option: A
Explanation:
Let Jaya's age = 2x or Ravi's age = 5x.
$\frac{2x+8}{5x+8}=\frac{1}{2}⇒2\left(2x+8\right)$= (5x+8)$⇒$ x = 8
Jaya's age = 16 years
Ravi's age = 40 years
Difference of their ages = 24 years
5. The ages of Ravi Rani are in the ratio of 3:5. After 9 years, the ratio of their ages will become 3:4. The present age of Rani is : (in years )
a. 9
b. 15
c. 18
d. 24
Correct Option: B
Explanation:
Let Ravi's age = 3x or Rani's age = 5x
$\frac{3x+9}{5x+9}=\frac{3}{4}⇒4\left(3x+9\right)=3\left(5x+9\right)⇒x=3$
Rani's age = 15 years
6. The ratio of the ages of Meena and Meera is 4:3. The sum of their ages is 28 years. The ratio of their ages after 8 years will be :
a. 4:3
b. 12:11
c. 7:4
d. 6:5
Correct Option: D
Explanation:
Let Meena's age = 4x or Meera's age = 3x. Then,
4x+3x=28$⇒$ x = 4
Meena's age = 16 years or Meera's age=12 years
After 8 years their ages are 24 and 20. So ratio is 6 : 5
7. Ten years ago A was half of B in age. If the ratio of their present ages is 3:4 , what will be the total of their present ages ?
a. 8 years
b. 20 years
c. 35 years
d. 45 years
Correct Option: C
Explanation:
Let A's age 10 years ago = x years
Then B's age 10 years ago = 2x years
$\frac{x+10}{2x+10}=\frac{3}{4}⇒4\left(x+10\right)$ = 3(2x+10) $⇒$ x = 5
Total of heir present ages =(x+10+2x+10)=(3x+20)=35 years
8. The ratio of Vimal's age and Aruna's age is 3:5 and sum of their ages is 80 years. The ratio of their ages after 10 years will be :
a. 2 : 3
b. 1 : 2
c. 3 : 2
d. 3 : 5
Correct Option: A
Explanation:
3x+5x=80$⇒$x = 10
Ratio of their ages after 10 years =(3x+10 : 5x+10) = 40:60 = 2:3
9. Jayesh is as much younger to Anil as he is older to Prashant. If the sum of the ages of Anil and Prashant is 48 years, what is the age of Jayesh?
a. 20 years
b. 24 years
c. 30 years
d. Cannot be determined
Correct Option: B
Explanation:
The given question says, the difference between Anil and Jayesh is same as Jayesh and Prashant.
So A - J = J - P $⇒$ 2J = A + P
Given A + P = 48.
So 2J = 48 $⇒$ J = 24
Alternate Method:
Let Anil's age = x years. Then, Prasant's age = (48-x) years. Let the age of Jayesh be P years. Then
P - (48 - x) = x-P $⇒$ 2P = 48 or P = 24
10. Three years ago the average age of A and B was 18 years. With C joining them, the average becomes 22 years. How old is C now ?
a. 24 years
b. 27 years
c. 28 years
d. 30 years
Correct Option: A
Explanation:
Sum of ages of A and B, 3 years ago = $\left(18×2\right)=36$ years. Sum of ages of A,B and C, now = $\left(22×3\right)=66$ years.
Sum of ages of A and B, now =(36+6) years=42 years
C's age = (66-42)years = 24 years
11.One year ago the ratio between Samir and Ashok's age was 4:3. One year hence the ratio of their age will be 5:4. What is the sum of their present ages in years?
a. 12 years
b. 15 years
c. 16 years
d. Cannot be determined.
Correct Option: C
Explanation:
$\frac{4x+2}{3x+2}=\frac{5}{4}⇒4\left(4x+2\right)=5\left(3x+2\right)⇒x=2$
Sum of their present ages = (4x + 1 + 3x + 1)=(7x + 2) = 16 years
12.The ratio between the ages of A and B at present is 2: 3 Five years hence the ratio of their ages will be 3:4. What is the present age of A ?
a. 10 years
b. 15 years
c. 25 years
Correct Option: A
Explanation:
Let the ages of A and B be 2x and 3x years.
$\frac{2x+5}{3x+5}=\frac{3}{4}⇒4\left(2x+5\right)=3\left(3x+5\right)⇒x=5$
A's present age = 2x=10 years
13. The ratio of the ages of father and son at present is 6:1. After 5 years the ratio of will become 7:2. The present age of the son is :
a. 5 years
b. 6 years
c. 9 years
d. 10 years
Correct Option: A
Explanation:
$\frac{6x+5}{x+5}=\frac{7}{2}⇒2\left(6x+5\right)=7\left(x+5\right)⇒x=5$
Son's present age = 5 years
14.Ratio of Ashok's age to Pradeep's age is equal to 4:3 Ashok will be 26 years old after 6 years. How old is Pradeep now ?
a. 12 years
b. 15 years
c. $19\frac{1}{2}$ years
d. 21 years
Correct Option: B
Explanation:
Let Ashok's age = 4x and Pradeep's age is 3x years
4x+6=26 $⇒$ x = 5
15. Kamala got married 6 years ago. Today her age is $1\frac{1}{4}$ times her age at the time of marriage. Her son's age is $\left(\frac{1}{10}\right)$times her age. Her son's age is :
a. 2 years
b. 3 years
c. 4 years
d. 5 years
Correct Option: B
Explanation:
Let Kamala's age 6 years ago be x years, then Kamala's present age = (x+6) years
$\left(x+6\right)=\frac{5}{4}x$ or 4x+24 = 5x or x = 24
So Kamala's present age = 30 years, Son's present age = $\left[\frac{1}{10}×30\right]=3$ years
16. 10 years ago, Chandravati's mother was 4 times older than her daughter. After 10 years, the mother will be twice older than the daughter. The present age of Chandravati is:
a. 5 years
b. 10 years
c. 20 years
d. 30 years
Correct Option: C
Explanation: Let Chandravati's age 10 years ago be x years.
Mother's age 10 years ago =(4x) years
2(x+20)=(4x+20)$⇒$ x = 10
Present age of Chandravathi = (x+10)=20 years
17. The age of Arvind's father is 4 times his age. If 5 years ago, father's age was 7 times the age of his son at the time, what is Arvind's father's present age ?
a. 35 years
b. 40 years
c. 70 years
d. 84 years
Correct Option: B
Explanation:
Let Arvind's age be x years
Then his father's age = 4 x years
(4x-5)=7(x-5) or 3x = 30 or x = 10
Hence Arvind's father's age is 40 years
18. Pushpa is twice as old as Rita was two years ago. If the difference between their ages is 2 years, how old is Pushpa today ?
a. 6 years
b. 8 years
c. 10 years
d. 12 years
Correct Option: B
Explanation:
Let Rita's age 2 years ago be x years. Pushpa's present age = (2x) years.
2x-(x+2)=2 $⇒$ x = 4
Pushpa's present age = 8 years
19. Five years ago Vinay's age was one-third of the age of Vikas and now Vinay's age is 17 years. What is the present age of Vikas ?
a. 9 years
b. 36 years
c. 41 years
d. 51 years
Correct Option: C
Explanation:
Let the present age of Vikas be x years
Then, 17-5 = $\frac{1}{3}$(x-5)
or x-5=36 or x = 41
20. The sum of the ages of a father and son is 45 years. Five years ago the product of their ages was 4 times the father's age at that time. The present ages of the father and son, respectively are,
a. 25 years, 10 years
b. 36 years, 9 years
c. 39 years, 5 years
d. none of these
Correct Option: B
Explanation:
Let the present ages of father and son be x years and (45-x) years
Then, (x-5)(45-x-5)=4(x-5)
45-x-5=4 or x = 36
The present ages of father and son are 36 years and 9 years respectively.
21. Rajan's age is 3 times that of Ashok. In 12 years, Rajan's age will be double the age of Ashok. Rajan's present age is :
a. 27 years
b. 32 years
c. 36 years
d. 40 years
Correct Option: C
Explanation:
Let Ashok's age be x years
Then, Rajan's age = 3 x years
2(x+12)=(3x+12) or x = 12
Hence, Rajan's age = 3x years
2 (x+12)=(3x+12) or x = 12
Hence Rajan's present age is 36 years
22. In 10 years, A will be twice as old as B was 10 years ago. If A is now 9 years older than B, the present age of B is :
a. 19 years
b. 29 years
c. 39 years
d. 49 years
Correct Option: C
Explanation:
Let the present ages of B and A be x years and (x+9) years respectively. Then 10 years ago B's age is (x - 10) and in the next 10 years A's age is (x + 19)
Given that A's age is twice of B
So (x + 19) = 2 x (x - 10) $⇒$ x + 19 = 2x - 20 $⇒$ x = 39
23. Mr.Sohanlal is 4 times as old as his son. Four years hence the sum of their ages will be 43 years. The present age of son is :
a. 5 years
b. 7 years
c. 8 years
d. 10 years
Correct Option: B
Explanation:
Let the son's age be x years
Then, (x+4)+(4x+4)=43
or 5x = 35 or x = 7
24. The sum of the ages of a son and father is 56 years. After 4 years, the age of the father will be three times that of the son. Their ages respectively are:
a. 12 years, 44 years
b. 16 years, 42 years
c. 16 years, 48 years
d. 18 years, 36 years
Correct Option: A
Explanation:
Let the present ages of son and father be x years
and (56-x) years respectively.
Then (56-x+4)=3(x+4)
or 4x =48 or x = 12
So their ages are 12 years, 44 years respectively.
25. The sum of the ages of a mother and daughter is 50 years. Also, 5 years ago, the mother's age was 7 times the age of the daughter. The present ages of the mother and daughter respectively are :
a. 35 years, 15 years
b. 38 years, 12 years
c. 40 years, 10 years
d. 42 years, 8 years
Correct Option: C
Explanation:
Let the daughter's present age be x years. Then, mother's present age = (50-x) years.
Now, 7(x-5)=(50-x-5) or x = 10
So, their present ages are 40 years and 10 years.
|
Why can't I divide by zero?
We tell students they can't do it, but do they really understand why?
Objective: Students will use various methods to demonstrate why division by zero can't be done.
Reformatted: January 2010
Method 1: Word problems with diagrams
What does division mean it terms of every day experiences?
Write simple word problems to demonstrate what division means. Build from easy to understand examples. Be sure to show examples in both orders so students understand the meaning of the order. Have the students try to draw a diagram for each.
a: You have 2 pizzas and you want to split them between 6 people. How much does each person get? 2 / 6 = 1/3 : each person gets 1/3 pizzas. (pictured here ->) b: You have 6 pizzas and you split them between 2 people. How much does each person get? 6 / 2 = 3 : each person gets 3 c: You have 0 pizzas and you split them between 2 people. How much does each person get? 0 / 2 = 0 : each person gets nothing d: You have 2 pizzas and you split them between 0 people. How much does each person get? 2 / 0 = undefined : where are the people to eat the pizza?
Example 2:
You have a 3 meter long board and you want to cut it into pieces that are X cm long. How many pieces will you have? 3 / 0 = ?? How many 0 cm pieces will be needed to make 3 meters. What will those pieces look like? You have a 3 meter long board and you want to cut it into X equal pieces. How long will each piece be? 3 / 0 = ?? How can you take an existing item and reduce it to zero pieces? If these questions are unanswerable, then the formula 3 / 0 will also be unanswerable or undefined.
Related Pages at this site
Method 2: The limit using a simple pattern
Vocabulary: Limit is a term used in calculus meaning to approach - get real close to - with out going the whole way to that value.
What does a pattern imply if we approach division by zero?
Use a simple pattern where the numbers do not confuse the students. Let them determine the answers and then try to interpret where that pattern will take them as they approach division by zero.
100 / 100 = 1 100 / 10 = 10 100 / 1 = 100 100 / 0.1 = 1000 100 / 0.01 = 10000 100 / 0.001 = 100000 In this pattern the numerator is staying the same while the denominator is getting closer to 0. The answer is increasing rapidly. Where will this result go and we approach division by zero?
Method 3: The limit using a graph
If your students are familiar with graphs, generate the graph of y=24/x. Ask them to determine where does the y-value go as x approaches 0? How is the approach from the negative side different from the approach on the positive side? Are the negative and positive answers the same? Are they finite?
Method 4: Inverse Operations
Division is the inverse of multiplication. If we can multiply, then we can divide by reversing the numbers. What does this relationship imply for us? Show easy relationships to set up the pattern first:
2 * 7 = 14 14 / 7 = 2 14 / 2 = 7 10*52 = 520 520 / 52 = 10 520 / 10 = 52 0 * 5 = 0 0 / 0 = ?? 0 / 5 = 0 8 * X = 32 32 / 8 = X = 4 X can only be 4 0 * X = 0 0 / 0 = X X can be any answer
Special Case Challenge:
1. Pick two numbers and perform an impossible operation with the result being a variable.
2. Show the inverse operation
3. Define an X to make your statements consistent
Examples:
Creating Negative Numbers 5 - 8 = X you were told you can't subtract a large from a small. X + 8 = 5 show the inverse -3 + 8 = 5 define into existence negative numbers to make these operations work
Creating Imaginary Numbers X = sqrt(-1) you were told you can't take the square root of a negative X2 = -1 show the inverse i2 = -1 define imaginary numbers to make this work
Creating Division by Zero 5 / 0 = X show division by zero 0 * X = 5 show the inverse can you define a system of numbers that will make these equations make sense? How? or Why not? Readings: Dual Numbers chapter of A Mathematical Mosaic
|
# Proposition 6
If in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.
Let ABC be a triangle having the angle ABC equal to the angle ACB.
I say that the side AB also equals the side AC.
C.N
If AB does not equal AC, then one of them is greater.
Let AB be greater. Cut off DB from AB the greater equal to AC the less, and join DC.
Since DB equals AC, and BC is common, therefore the two sides DB and BC equal the two sides AC and CB respectively, and the angle DBC equals the angle ACB. Therefore the base DC equals the base AB, and the triangle DBC equals the triangle ACB, the less equals the greater, which is absurd. Therefore AB is not unequal to AC, it therefore equals it.
Therefore if in a triangle two angles equal one another, then the sides opposite the equal angles also equal one another.
Q.E.D.
## Guide
#### Converses of propositions
This is the converse of (part of) the previous proposition I.5. Proposition I.6 says that if angle B equals angle C, then side AB equals side AC. Proposition I.5 says that if side AB equals side AC, then angle B equals angle C. In general, the converse of a proposition of the form “If P, then Q” is the proposition “If Q, then P.” When both a proposition and its converse are valid, Euclid tends to prove the converse soon after the proposition, a practice that has continued to this day.
A proposition and its converse are not logically equivalent. There are examples where “If P, then Q” is valid, but “If Q, then P” is not valid. An example from the Elements is proposition III.5 which states “If two circles cut one another, then they do not have the same center.” The converse would be “If two circles do not have the same center, then they cut one another” which is certainly not valid since if one circle lies entirely outside the other, then they don’t cut one another.
This is the first proof by contradiction, also called reductio ad absurdum, in the Elements. In this proof, in order to prove AB equals AC, Euclid assumes they are unequal and derives a contradiction, namely, that the triangle ACB equals a part of itself, triangle DBC, which contradicts C.N.5, the whole is greater than the part. The contradiction is that triangle ACB both equals and does not equal triangle DBC.
In general, to prove a statement of the form “P” with a proof by contradiction, begin with an assumption “not P” and derive some contradiction “Q and not Q,” and finally conclude “P.”
Euclid often uses proofs by contradiction, but he does not use them to conclude the existence of geometric objects. That is, he does not use them in constructions. But he does use them to show what has been constructed is correct.
In modern mathematics nonconstructive proofs by contradiction do occur. Famous examples are Brouwer’s fixed point theorems published in 1912. One of these states that any continuous transformation f of a circle (circular disk) to itself has a fixed point x, that is, a point such that f(x) = x. In his proof, he assumed that such a point did not exist and derived a contradiction. Although his proof is logically correct, he was not satisfied since the proof does not help in constructing a fixed point. Brouwer was an adherent of a philosophy of mathematics called intuitionism that holds, among other things, that mathematical objects have not been shown to exist until constructions have been given for them.
#### The law of trichotomy in practice
The proof uses the law of trichotomy for lines: “If AB does not equal AC, then one of them is greater.” There are three cases: AB < AC, AB = AC, or AB > AC. If the middle possibility is excluded, then only the two others remain, so one of the lines is greater. The law of trichotomy is not explicitly stated as a Common Notion, but it is the sort of property of magnitudes listed as Common Notions.
Proposition I.3 can be read as a construction to determine whether one line is less than, equal to, or greater than another. Using I.3, one line is laid along another, and it will fall short, fall equal, or extend beyond the other. For this proposition I.6, the construction simplifies since the two lines AB and AC already have one end in common.
The other part of the law of trichotomy is also used in the proof, the part that says only one of the three cases can occur: “... the triangle DBC equals the triangle ACB, the less equals the greater, which is absurd.” C.N.5, the whole is greater than the part, allows the conclusion that triangle DBC (the part) is less than triangle ACB (the whole). But the contradiction arises because only one of the two cases DBC = ACB and DBC < ACB can occur.
#### Use of Proposition 6
This proposition is not used in the proofs of any of the later propositions in Book I, but it is used in Books II, III, IV, VI, and XIII.
|
Subsections
Derivatives
Purpose
The purpose of this lab is to teach you how to use Maple commands for computing derivatives.
Background
Maple has several commands for computing derivatives. The most common ones are the diff command for differentiating expressions and the D operator for differentiating functions. These are the commands we will consider here.
Functions and Expressions in Maple
Before describing the commands for differentiation, we need to review the concepts of expression and function in Maple. An expression is a collection of mathematical symbols, for example and are both expressions. A function, on the other hand, is a rule for associating an output value with an input value.
The connection is that expressions are often used to define functions. That is, we could let , which defines a function . The rule for this function is to substitute a value for into the expression to obtain the output value. Not all expressions can be used to define functions, however, and not all functions are defined by expressions so these really are distinct mathematical objects.
Maple mimics this mathematical distinction between expression and function. You can define expressions in Maple and even label them for later use with commands like the one below.
> p := x^2+sin(3*x);
This is an expression not a function, which means there is no rule associated with it. Thus evaluating the expression at a specific value of requires the subs command, as in the following example.
> subs(x=2,p);
The syntax for defining a function in Maple uses an arrow to make the idea of a function as a process explicit. For example, we can define a function in Maple using the expression with the following command.
> f := x -> x^2+sin(3*x);
Evaluating our function at a specific value of is now easy.
> f(2);
One final thing to note is that Maple will use f to denote the function we have defined, but will use f(x) to denote the expression used to define the function.
Something you should never do
Students often want to define a function with a command like the following.
> F(x) := x^2+3;
Since Maple doesn't complain, students often think that what they've done is correct. The output from the following commands, however, shows that the object we've defined doesn't behave like a function.
> F(x);
> F(1);
> F(t);
> F(2*x);
What is happening here is that we've defined something called a table in Maple. For the input it gives the output , but it doesn't behave like a function and doesn't give the value we wanted for any other input.
The Maple D and diff commands
These commands can be summarized as follows.
• The D operator acts on a function to produce the derivative of that function.
• The diff command acts on an expression and differentiates that expression with respect to a variable specified by the user.
When you use the D operator to compute the derivative of a function, the result is also a function, as shown below.
> D(f);
If you provide a label, then you get a function you can use later in the session,
> df := D(f);
However, this is usually not necessary. See the examples below.
If you want to evaluate the derivative at a specific value of or just get the expression for the derivative, you can use the following forms of the D operator.
> D(f)(2);
> D(f)(x);
This last form is the one to use for plotting, as shown below.
> plot(D(f)(x),x=-2..2);
Suppose you want to find the equaton of the line tangent to the graph of at the point . This can be done in Maple using the point slope form of a line as shown below.
> tanline := D(f)(5)*(x-5)+f(5);
The D operator cannot be used on expressions, for example trying to use it to differentiate the expresssion we defined above results in an error.
> D(p);
If you recall that Maple uses f(x) to refer to the expresssion that is used to define , then the following error shouldn't surprise you.
> D(f(x));
To differentiate expressions, you need to use the diff command. Here is an example.
> diff(p,x);
The diff command can also be applied to functions as shown below.
> diff(f(x),x);
Note, however, that the result of the diff command is an expression, not a function. This means that computing the value of the derivative at a specific value of requires you to use the subs command.
Exercises
1. Compute the derivatives of the following functions using both the diff command and D operator for each function.
2. Find the equation of the line tangent to the graph of the function at . Include a plot of the function and the tangent line on the same graph over the interval .
3. For the function in exercise 2, find all points on the graph of where the tangent line is horizontal. Remember that a point has an and a value.
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What are different ways to show division?
What are different ways to show division?
There are a number of signs that people may use to indicate division. The most common one is ÷, but the backslash / is also used. Sometimes people will write one number on top of another with a line between them. This is also called a fraction.
What are the two ways of dividing functions?
To multiply or divide functions, just multiply or divide the values at each point where it makes sense. If the functions are given by formulas, you can just multiply or divide the formulas (it doesn’t matter whether you plug in values before or after).
What are 2 ways to solve a division problem?
As we learned, a dividend is the total number we are dividing, a divisor is the number dividing into the whole number, and a whole number is a number that is not a fraction. Then we learned that to solve division problems you can make equal groups, create equal groups with place value, and use multiplication.
What is the division of function?
When you divide two such functions together, you get what is called a rational expression. A rational expression is the division of two polynomials. If they divide evenly, your answer will become a polynomial.
Can you divide using lattice method?
The lattice strategy partitions numbers into smaller parts and it may not be an efficient strategy for students to use if they do not understand how division works. To use this strategy, students should have a solid understanding of place value and dividing large quantities in equal groups.
What is graph and its types?
In discrete mathematics, a graph is a collection of points, called vertices, and lines between those points, called edges. There are many different types of graphs, such as connected and disconnected graphs, bipartite graphs, weighted graphs, directed and undirected graphs, and simple graphs.
What are the different ways to do Division?
There are a variety of ways to do division. You can divide decimals, fractions, or even exponents, and you can do long division or short division.
What are the different types of graph?
1 Line Graphs. 2 Bar Graphs. 3 Combo Chart. 4 Scatterplot. 5 Waterfall Chart. 6 Pie Graph. 7 Histogram. 8 Gauge Chart. 9 Area Graph. 10 Spider chart / radar graph.
How do you divide in Excel using long division?
Place the divisor, the number you’ll be dividing, outside the long division bar, and the dividend, the number that you’ll be dividing into, inside the long division bar. Sample problem #1 (beginner): 65 ÷ 5.
What is Division in math?
Last Updated: June 8, 2021 Division is one of the 4 major operations in arithmetic, alongside addition, subtraction, and multiplication. In addition to whole numbers, you can divide decimals, fractions, or exponents. You can do long division or, if one of the numbers is a single digit, short division.
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# Completing the square
Some quadratic expressions can be factored as perfect squares. For example, x²+6x+9=(x+3)². However, even if an expression isn't a perfect square, we can turn it into one by adding a constant number. For example, x²+6x+5 isn't a perfect square, but if we add 4 we get (x+3)². This, in essence, is the method of *completing the square*. Created by Sal Khan and CK-12 Foundation.
## Want to join the conversation?
• That wasn't very clear, may you please do another video about that?
• You don´t need another video because I´m about to explain it to you! Say you have the equation 3x^2-6x+8=23. To complete the square, first, you want to get the constant (c) on one side of the equation, and the variable(s) on the other side. To do this, you will subtract 8 from both sides to get 3x^2-6x=15. Next, you want to get rid of the coefficient before x^2 (a) because it won´t always be a perfect square. Because there is a 3 in front of x^2, you will divide both sides by 3 to get x^2-2x=5. Next, you want to add a value to the variable side so that when you factor that side, you will have a perfect square. In this case, you will add 1 because it perfectly factors out into (x-1)^2. Because you´re taking this value away from the constant, you will add it to the other side of the equation (this might not make sense at first, but if the constant were on the variable side, you would be subtracting). This will all give you the equation (x-1)^2=16. Next, you want to take the square root from both sides so that x-1 is equal to the positive or negative square root of 16 (positive or negative 4). Finally, you add 1 to both sides, taking into account that 4 could be positive OR negative. Therefore, x = -3 or 5. Situations could vary, but this is the basic idea behind the procedure. I hope this helps! :)
• Can someone please post the link to the "Last video" of which Sal speaks?
• I don't know about you guys, but I'm more confused now than I was when I started the video! Could someone please explain to me in detail how all this works?
• 5 years later...
I don't know about others, but I'm more confused now than when I started the video as well! Could someone at least show all the steps or work that is missing for understanding this kind of high-level education problems?
• Between to , May i understand why a=-2. i thought it would be 2 instead since were equating -4x to -2ax.
• I think Sal made a mistake. But 2^2 and (-2)^2 are both 4, so the result was correct.
• At , I am confused about where the (x-a)^2 came from and why that is equal to x^2-4x=5
Could you explain this?
Now x^2 - 4x = 5
x^2 - 4x + (something) = 5 + something
I want to factorise the left side of the equal sign, so I have to find a value for (something) which would allow me to factorise the left-hand side of the equation.
If (something)=4
x^2 - 4x + 4 = 5 + 4
Notice that now I could factorise the left hand side into (x - 2)^2
(x-2)^2 = 9
x-2 = root of 9 = + or - 3
x = +3+2 or -3+2
x = +5 or -1.
You want x^2 - 4x + (something) to be equal to (x-a)^2.
where a is half the coefficient of x(the number before the x), as long a there is no coefficient of x^2. Here a = 2.
A shorter way to do this is:
x^2 - 4x = 5
(x - (4/2))^2 - (4/2)^2 = 5
(x-2)^2 - 4 = 5 and so on. But remember, you only halve the coefficient of x and put it into the brackets only if there is no number before x^2 (coefficient of x^2). If it is there then you have to divide the whole equation first by the coefficient and then halve the coefficient of x and put it into the brackets.
Sal uses (x - a)^2 simply to tell this is the format of the factorised form of :
x^2 -4x +? = (x - a)^2, where a = 2 (4 divided by two).
Sorry for the large number of words used to answer your question.
• He is missing a lot of steps in his work.
• Why is it called 'completing the square?'
• It is called completing the square because once you have to "complete" a perfect square to solve it, as in all of the steps are for you to end up with a perfect square to apply a square root on it.
• Why did he say that we needed a a number times 2 to equal -3 (it was around )?
• isn't a perfect square trinomial
(ax-b)^2 = (ax)^2+2abx+b^2?
I get a bit confused as to why, when using the completing the square to derive the quadratic formula we only divide by 2, whit out also dividing by a.
but we do, right? we do it at the beginning.
ax^2+bx+c = 0
1/a*(ax^2+bx+c) = (0)1/a
x^2+bx/a + (b/2a)^2 = -x/a - (bx/2a)^2
In that step, we divide the second term by 2a to isolate b, and raise it to a second power.
I'm just trying to confirm things, not sure if I’m wrong
• I was following the quadratic formula Proof video, some expressions in my question were written wrong, so here's the correction.
ax^2+bx+c = 0
We divide by ' a ' both sides
(a/a)x^2 + (b/a)x + (c/a) = 0/a
We move the c element to the right hand side
and we add the missing expression to both sides to form a perfect square.
x^2 + (b/a)x + (b/(2a))^2 = -c/a + (b/(2a))^2
At this point we complete the square.
I was trying to ask if my thinking behind how we come up whit (b/(2a))^2 was correct.
A perfect square trinomial is
(cx+d)^2 = (cx)^2+2cdx+d^2
(I’m not using ‘a’ and ‘b’ to not mix up the variables whit the first equation of this comment)
So if we want to find the third term, we just need to take the second term, ignore the factor x, and divide by '2c', so:
(2cd/(2c))^2 = d^2 This is what I meant by "isolating b": find the missing third term using the second.
So, the relationship between deriving the quadratic formula, and completing the square to find the third element is this:
b = 2cd <- I didn't get this relationship at first
b/(2a) = (2cd)/(2c)
And finally:
( b/(2a) )^2 = d^2
But I think it's all clear now, correct me if there's something wrong :)
• Around , Sal divides the quadratic equation by 5. This process makes the coefficient of x^2 equal to 1. My question is does the coefficient of x^2 need to be 1 to complete the square.
• no it doesn't have to.
for example; 2x^2+18x+16
one can factor this by..
(x+8)(2x+2)
but if you divide everthing by 2,
you can make 2x^2+18x+16 to x^2+9x+8 then you can factor this to
(x+8)(x+1)
you see, this is the same as (x+8)(2x+1) but simpler.
so to answer your question; it doesn't matter, but is's the matter of which one is simpler
hope this helps :)
(1 vote)
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# Help with proof of triangle inequality
I don't understand one step in this proof of the triangle inequality (number 1 below).
I.e. we have the equality $|x+y|^2 =|x|^2+2xy+|y|^2$, but how can we then conclude $|x+y|^2 \leq|x|^2+2|xy|+|y|^2$?
Let $x,y \in \mathbb{R}$ and let $|x|$ be the absolute value of $x$. Then: $$|x+y|\leq|x|+|y|$$ Proof: \begin{align} |x+y|^2 &= (x+y)^2 \\ &=x^2+2xy+y^2 \\ &= |x|^2+2xy +|y|^2 \\ &\leq |x|^2+2|xy|+|y|^2 \tag{1}\\ &=|x|^2+2|x||y|+|y|^2 \\ &= (|x|+|y|)^2 \end{align}
Thanks!
• This comes from the fact that absolute value of a real number is greater or equal that number: $\forall t\in \Bbb R \quad x\le |x|$. This can be easily shown by the definition of absolute value. – TZakrevskiy Feb 17 '17 at 11:53
It's because for every number $a\in\mathbb R$, you have $a\leq |a|.$ In particular, here they set $a=xy$.
Longer explanation:
First of all, if $a\in\mathbb R$, then you have two options:
1. If $a\geq 0$, then $a=|a|$, which means that $a\leq |a|$
2. If $a<0$, then $|a|>0$ which means that $a<0<|a|$, so $a<|a|$, which again means that $a\leq |a|$.
in total, this means that $a\leq |a|$ is true for any real number $a$.
In particular, this measn that $xy\leq |xy|$ for each value of $x,y\in\mathbb R$.
We also know that if $a\leq b$, then $ac \leq bc$ if $c>0$. In particular, this means that $2xy\leq 2|xy|$.
Now, there is also a rule that if $a\leq b$, then $b+a\leq c+a$ for all $a,b,c\in\mathbb R$. In particular, this means that, because $2xy\leq 2|xy|$, we also know that $$2xy + (|x|^2+|y|^2) \leq 2|xy| + (|x|^2+|y|^2),$$ which is exactly what you need.
• In point 1, should it not just be "If $a\ge0$, then $a=\left|a\right|$"? – lioness99a Feb 17 '17 at 14:04
• @lioness99a Yes, of course. Thank you. – 5xum Feb 17 '17 at 14:08
• You also don't need the "which means that..." part of point 1 - we can't conclude that $a\leq\left|a\right|$ from the rest of the statement – lioness99a Feb 17 '17 at 14:10
• @lioness99a Of course we can. If $x=y$, then we also know that $x\leq y$. – 5xum Feb 17 '17 at 14:11
• I suppose, but in my head that's an odd step. Point 1 proves that $a=\left|a\right|$ and then point 2 further proves that $a<\left|a\right|$. We then combine the two to prove that $a\leq\left|a\right|$. But technically, yes, what you have written is correct – lioness99a Feb 17 '17 at 14:13
Because we can write $2xy\leq|2xy|=2|xy|$.
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# Rational Number Review: Integers and Number Line
Contributor: Erika Wargo. Lesson ID: 12703
You are already familiar with opposites: stop and go, light and dark, and on and off. Did you know that numbers have opposites, too? Watch and play and learn about negative numbers and number lines!
categories
## Integers/Rational Numbers and Operations
subject
Math
learning style
Visual
personality style
Beaver
Middle School (6-8)
Lesson Type
Quick Query
## Lesson Plan - Get It!
Audio:
What integer could be used to represent the following real-world situations?
Katie flew a kite 15 feet above the ground.
The shark swam 15 feet below the water.
Numbers are an important part of daily life.
Integers and rational numbers are numbers that are used to express quantities in real-world situations. The price of a book or new video game is represented by a rational number. The temperature or amount of money owed to a friend is represented by an integer.
To learn more about integers, you will watch Math Antics - Negative Numbers. As you watch the video, answer the following in your math journal, then click on the items to see the answers (no peeking!):
Integers and their negatives can be seen on a number line.
Every integer on a number line has an opposite, or a number that is the same distance from zero (0), but in the other direction. When graphing a number on a number line, the distance from zero is represented by absolute value.
Review what absolute value is and how to find the absolute value of a number as you watch Absolute Value by Shmoop. As you watch the video, answer the following in your math journal, then click on the items to see the answers (no peeking!):
Think back to the question that was presented to you at the beginning of the lesson:
What integer could be used to represent the following real-world situations?
• Katie flew a kite 15 feet above the ground.
• The shark swam 15 feet below the water.
If a kite is flying 15 feet above the ground, the ground would represent the point 0 on a number line. Since the kite is above the ground, it would represent a positive integer, or +15.
The shark is swimming below the water. Since the water is below the ground, it would be represented by a negative integer, or –15.
What other situations could be represented by integers in the world around you?
1. Write at least two examples in your math journal.
2. Then, sketch a number line showing integers from -5 to +5.
3. Identify the location of -3 and determine its absolute value.
In the Got It? section, you will practice graphing rational numbers on a number line and applying them to real-world situations as you play interactive games and practice.
Interactive Video
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# HOW TO CALCULATE MARGINAL RELATIVE FREQUENCY
Marginal relative frequency is one of the types of relative frequency that we can obtain from a two-way frequency table.
A marginal relative frequency can be calculated by dividing a row total or a column total by the Grand total.
Let us look at some examples to understand how to calculate marginal relative frequency.
Example 1 :
A survey is conducted among school students. 50 students are randomly selected and they are asked, whether they prefer dogs, cats or other pets. The table given below shows the results of the survey.
Use the above table to find each marginal relative frequency.
(i) Find the marginal relative frequency of the students who prefer cats as pets.
(ii) Find the marginal relative frequency of boys.
(iii) Find the marginal relative frequency of girls.
(iv) Find the marginal relative frequency of the students who prefer dogs as pets.
Solution (i) :
Divide the total number of students who prefer cats as pets by the grand total. Express your answer as a decimal and as a percent.
15/50 = 0.30 = 30%
Solution (ii) :
Divide the total number of boys by the grand total. Express your answer as a decimal and as a percent.
22/50 = 0.44 = 44%
Solution (iii) :
Divide the total number of girls by the grand total. Express your answer as a decimal and as a percent.
28/50 = 0.56 = 56%
Solution (iv) :
Divide the total number of students who prefer dogs as pets by the grand total. Express your answer as a decimal and as a percent.
22/50 = 0.44 = 44%
Example 2 :
A survey is made among 100 students in a middle school. They are asked, how they travel to school. The table given below shows the results of the survey.
Use the above table to find each conditional relative frequency.
(i) Find the marginal relative frequency of the students who prefer car.
(ii) Find the marginal relative frequency of boys.
(iii) Find the marginal relative frequency of girls.
(iv) Find the marginal relative frequency of the students who prefer bus.
Solution (i) :
Divide the total number of students who prefer car by the grand total. Express your answer as a decimal and as a percent.
47/200 0.24 = 24%
Solution (ii) :
Divide the total number of boys by the grand total. Express your answer as a decimal and as a percent.
100/200 = 0.50 = 50%
Solution (iii) :
Divide the total number of girls by the grand total. Express your answer as a decimal and as a percent.
100/200 = 0.50 = 50%
Solution (iv) :
Divide the total number of students who prefer bus by the grand total. Express your answer as a decimal and as a percent.
72/200 = 0.36 = 36%
Kindly mail your feedback to [email protected]
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# Square Root Functions
## Definition and Graph of the Square Root Function
The square root of a nonnegative real number $x$ is a number $y$ such $x = y^2$. For example, $3$ and $- 3$ are the square roots of $9$ because $3^2 = 9$ and $(-3)^2 = 9$
Of the two roots, the nonnegative root of a nonnegative number $x$ is written as function with the following notation $f(x) = \sqrt x$ where the symbol $\sqrt { \; }$ is called the radical and $x$ is called the radicand and must be nonnegative so that $f(x)$ is real.
The square root function defined above is evaluated for some nonnegative values of $x$ in the table below.
$x$ $f(x) = \sqrt x$
$0$ $f(x) = \sqrt 0 = 0$
$1$ $f(x) = \sqrt 1 = 1$
$2$ $f(x) = \sqrt 2 \approx 1.414$
$4$ $f(x) = \sqrt 4 = 2$
$7$ $f(x) = \sqrt 7 \approx 2.645$
$9$ $f(x) = \sqrt 9 = 3$
$16$ $f(x) = \sqrt 16 = 4$
The graph of the square root function is shown below with some points from the above table.
## Properties of the Square Root Function
Some of the properties of the square root function may be deduced from its graph
1. The domain of the square root function $f(x) = \sqrt x$ is given in interval form by: $[0, + \infty)$
2. The range of the square root function $f(x) = \sqrt x$ is given in interval form by: $[0, + \infty)$
3. The x and y intercepts are both at $(0,0)$
4. The square root function is an increasing function
5. The square root function is a one-to-one function and has an inverse.
## Common Mistakes to Avoid when Working with Square Root Functions
1. It is wrong to write $\sqrt{25} = \pm 5$. The radicand is the symbol of the square root function and a function has only one output which as defined above is equal to the positive root.
Correct answer: $\sqrt{25} = 5$
2. It is wrong to write $\sqrt{x^2} = x$. The output of the square root is nonnenegative and $x$ in the given expression may be negative, zero or postive.
Correct answer: $\sqrt{x^2} = |x|$
## Exploring Interactively the Square Root Function
Square root functions of the general form
$f(x) = a \sqrt{x - c} + d$ and the characteristics of their graphs such as domain, range, x intercept, y intercept are explored interactively. Square root equations are also explored graphically. There is also another tutorial on graphing square root functions in this site.
The exploration is carried out by changing the parameters $a, c$ and $d$ included in the expression of the square root function defined above. The Answers to the questions in the tutorial are included in this page.
click on the button above "draw" and start exploring.
a = 1 -10+10 c = 0 -10+10 d = 0 1+10
>
1. Use the sliders to set parameters $a$ and $c$ to some constant values and change $d$. What happens to the the graph when the value of parameter $d$ changes? Give an analytical explanation.
2. Use the sliders to set parameters $a$ and $d$ to some constant values and change $c$. What happens to the the graph when the value of parameter $c$ changes? Give an analytical explanation.
3. Use the sliders to set parameters $c$ and $d$ to some constant values and change parameters $a$. What happens to the graph when the value of parameter $a$ changes? Give an analytical explanation.
4. Use the sliders to set parameters $a$, $c$ and $d$ to different values and determine which parameters affect the domain of the square root function $f$ defined above? Find the domain analytically and compare it to the domain obtained graphically.
5. Use the sliders to set parameters $a$, $c$ and $d$ to different values and determine which parameters affect the range of the square root function $f$ defined above? Find the range analytically and compare it to the range obtained graphically.
6. How many x intercept the graph of $f$ has?
7. How many solutions an equation of the form
$a \sqrt{x - c} + d = 0$
has? (parameter $a$ not equal to zero).
Find the solution to this equation in terms of $a$, $c$ and $d$ and compare it to the x intercept given graphically.
8. Find the y intercept analytically and compare it to the one given by the app.
## Answers to the Above Questions
1. Changes in the parameter $d$ affect the y coordinates of all points on the graph hence the vertical translation or shifting. When $d$ increases, the graph is translated upward and when d decreases the graph is translated downward.
2. When $c$ increases, the graph is translated to the right and when $c$ decreases, the graph is translated to the left. This is also called horizontal shifting.
3. Parameter $a$ is a multiplicative factor for the y coordinates of all points on the graph of function $f$. Let $a$ be greater than zero. If $a$ gets larger than 1, the graph stretches (or expands) vertically. If $a$ gets smaller than 1, the graph shrinks vertically. If $a$ changes sign, a reflection of the graph on the x axis occurs.
4. Only parameter $c$ affects the domain. The domain of
$f(x) = a \sqrt{x - c} + d = 0$
may found by solving the inequality $x - c \ge 0$ hence the domain is the set of all values in the interval $[c , + \infty)$
5. Only parameters $a$ and $d$ affect the range. The range of function $f$ given above may found as follows: With x in the domain defined by interval $[c , + \infty)$ , the $\sqrt{x - c}$ is always positive or equal to zero hence
$\sqrt{x - c} \ge 0$
If parameter $a$ is positive then
$a \sqrt{x - c} \ge 0$
Add $d$ to both sides to obtain
$a \sqrt{x - c} + d \ge d$
Hence the range of the square root function defined above is the set of all values in the interval $[d , + \infty)$
If parameter $a$ is negative then
$a \sqrt{x - c} \le 0$
Add $d$ to both sides to obtain
$a \sqrt{x - c} + d \le d$
Hence the range of the square root function defined above is the set of all values in the interval $(- \infty , d]$
6. one solution or no solution.
7. Solve the equation
$a \sqrt{x - c} + d = 0$
add $-d$ to both sides of the equation
$a \sqrt{x - c} = - d$
If $d$ is positive, $- d$ is negative and the above equation has no solution. If $d$ is negative, we square both sides and solve to obtain
$x = (-d/a)^2 + c$
The above equation may have one solution or no solution.
8. If $c$ is positive, $\sqrt{-c}$ is not a real number and therefore the graph has no y intercept. If $c$ is negative or equal to zero then the y intercept is given by
$y = a \sqrt{-c} + d$
Square Root
functions
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Every Question Helps You Learn
The surface area of a cube (such as a dice) with sides of 5cm is 150cm2.
# Perimeter and Area (Year 6)
In KS2 Maths, children learn about perimeter and area. Year Six students figure out the perimeter of shapes like squares, triangles, and rectangles. They also tackle the area of a square, a right-angled triangle, a rectangle, or an L-shape. Converting between mm2, cm2, and m2 becomes a breeze. Now, they explore 3D shapes and how to find their area.
Area is the size of a 2D or 3D shape's surface. For a 3D shape, add the area of each face together. Calculating each face's area is similar to 2D shapes. This Perimeter and Area quiz is all about 2D shapes' perimeter and area, and 3D shapes' surface area.
Take this quiz if you're 10-11 years old and discover what you know about area and perimeter in KS2 Maths.
1.
How would you find the area of an L-shape?
Divide into rectangles; work out the areas of each and then divide one by the other
Divide into rectangles; work out the areas of each and then multiply one by the other
Divide into rectangles; work out the areas of each and then add together
Divide into rectangles; work out the areas of each and then subtract one from the other
Always simplify shapes if you can
2.
1cm2 is equivalent to how many mm2?
1mm2
10mm2
100mm2
1,000mm2
There are 10mm in a cm so 10 x 10 = 100
3.
What is the area of a right angle triangle with width 12cm and length 6cm?
18cm2
36cm2
72cm2
148cm2
12 x 6 = 72
72 ÷ 2 = 36
4.
How do we calculate the area of a right angled triangle?
Length x width then halve
Length + width then double
Length x width then double
Length x width then quarter
Right angled triangles are rectangles cut in half so the formula is the same as for a rectangle but then halved
5.
What is the perimeter of a regular octagon with sides of 12cm?
20cm
56cm
96cm
160cm
Octagons have 8 sides so 12 x 8 = 96
6.
What is the perimeter of a regular decagon with sides of 23cm?
46cm
115cm
200cm
230cm
A regular decagon has 10 equal sides, so we multiply the length of the side by 10 to calculate the perimeter
7.
What is the surface area of a cube with sides of 5cm?
25cm2
75cm2
125cm2
150cm2
The area of each face is 25cm2. There are 6 faces, so multiply 25cm2 by 6 to give 150cm2
8.
What is the formula for calculating the area of a rectangle?
2 x length x 2 x width
2 x length + 2 x width
Length + width
Length x width
A rectangle measuring 60cm by 45cm would have an area of 2,700cm2
9.
A regular heptagon has a perimeter of 63cm. What length is each side?
6.3cm
9cm
10cm
12cm
A regular heptagon has 7 equal sides. To find the length of each side divide the perimeter by 7
10.
What is the surface area of a cuboid with measurements of 4cm x 5cm x 6cm?
74cm2
100cm2
148cm2
174cm2
2 of the faces will be 4cm x 5cm = 20cm2
2 of the faces will be 4cm x 6cm = 24cm2
2 of the faces will be 5cm x 6cm = 30 cm2
20 + 20 + 24 + 24 + 30 + 30 = 148
You can find more about this topic by visiting BBC Bitesize - How to work out an area
Author: Amanda Swift
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# vector help, straight lines
• May 7th 2010, 02:11 PM
Tweety
vector help, straight lines
Quote:
Relative to a fixed origin, the points A, B and C have position vectors (2i − j + 6k),
(5i − 4j) and (7i − 6j − 4k) respectively.
(a) Show that A, B and C all lie on a single straight line.
completely lost here, any help appreciated.
• May 7th 2010, 05:02 PM
Soroban
Hello, Tweety!
There are several ways to do this . . .
Quote:
Relative to a fixed origin, the points $A, B, C$ have position vectors:
. . (2i − j + 6k), (5i − 4j), (7i − 6j − 4k) respectively.
(a) Show that $A, B, C$ all lie on a single straight line.
We have: . $\begin{Bmatrix}A\!: & (2,\text{-}1,6) \\ B\!: & (5,\text{-}4,0) \\ C\!: & (7.\text{-}6.\text{-}4) \end{Bmatrix}$
Then: . $\begin{array}{cccccc}\overrightarrow{AB} &=& \langle 3,\text{-}3,\text{-}6\rangle &=& 3\langle 1,\text{-}1,\text{-}2\rangle
\\ \overrightarrow{BC} &=& \langle 2,\text{-}2,\text{-}4 \rangle &=& 2\langle1,\text{-}1,\text{-}2\rangle \end{array}$
The vectors are parallel and contain point $B.$
Therefore, the points are collinear.
$\begin{array}{ccccccccc}
|AB| &=& \sqrt{9 + 9 + 36} &=& \sqrt{54} &=& 3\sqrt{6} \\
|BC| &=& \sqrt{4 + 4 + 16} &=& \sqrt{24} &=& 2\sqrt{6} \\
|AC| &=& \sqrt{25 + 25 + 100} &=& \sqrt{150} &=& 5\sqrt{6} \end{array}$
Since $|AB| + |BC| \:=\:|AC|$, points $A,B,C$ are collinear.
We have: . $\begin{array}{ccc}\overrightarrow{AB} &=& \langle 3,\text{-}3,\text{-}6\rangle \\ \overrightarrow{BC} &=& \langle 2,\text{-}2,\text{-}4\rangle \end{array}$
$\text{The angle }\theta \text{ between }\overrightarrow{AB}\text{ and }\overrightarrow{BC}\text{ is given by: }\; \cos\theta \;=\;\frac{\overrightarrow{AB}\cdot \overrightarrow{BC}}{|\overrightarrow{AB}|\,|\over rightarrow{BC}|}$
We have: . $\cos\theta \;=\;\frac{\langle 3,\text{-}3,\text{-}6\rangle\cdot\langle 2,\text{-}2,\text{-}4\rangle}{\sqrt{9+9+36}\,\sqrt{4+4+16}} \;=\;\frac{6+6+24}{\sqrt{54}\,\sqrt{24}} \;=\;\frac{36}{36}\;=\;1$
Since $\cos\theta \:=\:1$, then: . $\theta \:=\:0 \quad\Rightarrow\quad \overrightarrow{AB} \:\parallel\: \overrightarrow{BC}$
Therefore, points $A,B,C$ are collinear.
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# Sarin’s family and Ramadan Part 5 (Math Question)
Standard
The blog postings are about the Singapore Math. The readers can learn from the postings about Solving Singapore Primary School Mathematics. The blog presenting the Math Concept, Math Questions with solutions that teaches in Singapore Primary Schools. You or the kids can learn to deal with Math Modeling, Problem Sum from Lower Primary School to Upper Primary School level. This posting is an upper primary school math question on Fraction, Ratio, Percentage and Problem Sum.
You should read the posting on Sarin learns Average Number, Percentage and Charting in school (Math Concept) to understand about Average Number.
To read also the postings on Sarin learns the concept of Fraction (mathematics concept) and Sarin learns concept of ratio in school to understand the concept of Fraction and Ratio.
Try to solve the question before look out for the solution‼
Upper primary school mathematics question UPQ243
Sarin, Hairu and Fatimah decided to buy a present for their friend. Sarin agreed to contribute 3/10 of the cost of the present while Hairu agreed to pay 60% of the remaining amount. The balance will be paid by Fatimah. When they purchased the present, the price of the item had increased to 20%. As a result, Sarin paid \$27.90 for his share.
a) What was the original price of the present?
b) How much did Fatimah have to pay in the end?
Solution:
Hairu is paying 60% of the remaining amount
Based on the fraction model,
Hairu is paying 7 × 0.6/10 = 4.2/10 fraction of the total amount
Fatimah is paying 7/10 – 4.2/10 = 2.8/10 fraction of the total amount
The faction model,
The ratio of the amount they need to pay,
Sarin : Hairu : Fatimah = 30 : 42 : 28 = 15 : 21 : 14
The actual amount paid by Sarin was \$27.90 after 20% increased in price
The amount he should paid before increase = 27.90 ÷ 1.2 = \$23.25
Based on the payment ratio,
15 units = 23.25
1 unit = 1.55
The original price of the gift = 50 × 1.55 = \$77.50
The amount Fatimah need to pay = 1.2 × 14 × 1.55 = \$26.04
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# Every integer is a ……………………………………………A. Real numberB. Rational numberC. Irrational numberD. Natural number
Last updated date: 13th Jun 2024
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Hint: Here, we will first define all the different types of numbers. Then we will compare these numbers with the definition of an integer number. We will then select the option that comes under the integers. Integers lie between $- \infty$ to $\infty$ and integers are denoted by a symbol ‘Z’.
Complete Step by Step Solution:
We know that an integer is a number that can be positive, negative or zero but it cannot be a fraction. Whole numbers are the numbers starting from zero and with all positive numbers. We know that integers do not include decimal numbers.
We know that a rational number is defined as a number that can be expressed in the form of fractions. A rational number can also be defined as the ratio of two integers or the Quotient of two integers
An irrational number is defined as a number that cannot be expressed in the form of fractions. An irrational number can also be defined which cannot be expressed as the ratio of two integers.
We know that Real numbers are defined as a set of numbers containing both the rational number and an irrational number.
A Natural number is a number starting from one and with all the Positive numbers.
Every integer is a Real number and Rational number but not Irrational numbers and natural numbers.
Therefore, every integer is a real number and a rational number.
Thus, option (A) and (B) is the correct answer.
Note:
We know that Integers are the set of all Positive numbers and Negative numbers but cannot be zero. The major properties of integers are the Closure Property, Commutative Property, Associative Property, Distributive Property, Additive Inverse, Multiplicative Inverse and Identity Property. The additive inverse of every natural number is the set of all Integers.
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# 1 Finding Limits Graphically & Numerically Section 1.2.
## Presentation on theme: "1 Finding Limits Graphically & Numerically Section 1.2."— Presentation transcript:
1 Finding Limits Graphically & Numerically Section 1.2
2 After this lesson, you should be able to: estimate a limit using a numerical or graphical approach determine the existence of a limit
3 Introduction to Limits The functionis a rational function. If I asked you the value of the function when x = 4, you would say ______ What about when x = 2 ? Well, if you look at the function and determine its domain, you’ll see that. Look at the table and you’ll notice ERROR in the y column for –2 and 2. On your calculator, hit then then . You’ll see that no y value corresponds to x = 2.
4 Continued… Since we know that x can’t be 2, or –2, let’s see what’s happening near 2 and -2… Let’s start with x = 2. We’ll need to know what is happening to the right and to the left of 2. The notation we use is: read as: “the limit of the function as x approaches 2”. In order for this limit to exist, the limit from the right of 2 and the limit from the left of 2 has to equal the same real number.
5 Right and Left Limits To take the right limit, we’ll use the notation, The + symbol to the right of the number refers to taking the limit from values larger than 2. To take the left limit, we’ll use the notation, The - symbol to the right of the number refers to taking the limit from values smaller than 2.
6 Right Limit Numerically The right limit: Look at the table of this function. You will probably want to go to TBLSET and change the TBL to be.1 and start the table at 1.7 or so. We can also set Indpnt to Ask and enter our own values close to 2. As x approaches 2 from the right (larger values than 2), what value is y approaching? x2.22.12.012.0012 f(x)
7 Left Limit Numerically The left limit: Again, look at the table. As x approaches 2 from the left (smaller values than 2), what value is y approaching? x1.81.91.991.9992 f(x)
8 The Limit Numerically Both the left and the right limits are the same real number, therefore the limit exists. We can then conclude, To find the limit graphically, trace the graph and see what happens to the function as x approaches 2 from both the right and the left. You can ZOOM IN to see x values very close to 2.
9 Another Example: Finding Limit Numerically On your calculator, graph Where is f(x) undefined? Use the table on your calculator to estimate the limit as x approaches 0. Take the limit from the right and from the left:
10 One more Example EXAMPLE: Find the numerically and verify graphically. Answer: *Use Radian mode.
11 Methods for finding limits Limits can be estimated three ways: Numerically… looking at a table of values Graphically…. using a graph Analytically… using algebra OR calculus (covered next section) Notation: roughly translates into: The value of f(x), as x approaches c from either side, becomes close to L.
12 Limits Graphically: Example 1 c L2L2 L1L1 Discontinuity at x = c There’s a break in the graph at x = c Although it is unclear what is happening at x = c since x cannot equal c, we can at least get closer and closer to c and get a better idea of what is happening near c. In order to do this we need to approach c from the right and from the left. Right Limit Left Limit
13 Limits Graphically: Example 2 c Hole at x = c Discontinuity at x = c Right Limit Left Limit Note: The limit exists but L f(c) L The existence or nonexistence of f(x) at x = c has no bearing on the existence of the limit of f(x) as x approaches c.
14 Limits Graphically: Example 3 c Continuous Function f(c) No hole or break at x = c Right Limit Left Limit In this case, the limit exists and the limit equals the value of f(c).
15 Limit Differs From the Right and Left- Case 1 The limits from the right and the left do not equal the same number, therefore the limit ____________________________________. To graph this piecewise function, this is the menu
16 Unbounded Behavior- Case 2 Since f(x) is not approaching a real number L as x approaches 0, the limit does not exist.
17 Oscillating Behavior- Case 3 Since f(x) is oscillates between –1 and 1 as x approaches 0, the limit does not exist. Look at the graph of this function (in radian mode).
18 A limit does not exist when: 1. f(x) approaches a different number from the right side of c than it approaches from the left side. (case 1 example) 2. f(x) increases or decreases without bound as x approaches c. (The function goes to +/- infinity as x c : case 2 example) 3. f(x) oscillates between two fixed values as x approaches c. (case 3 example)
19 Homework Section 1.2: page 54 #3, 7-15 odd, 19, 49, 53, 63, 65
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# Fractions
A fraction is a part of a whole
### Slice a pizza, and you will have fractions:
1/2 1/4 3/8 (One-Half) (One-Quarter) (Three-Eighths) The top number tells how many slices you have The bottom number tells how many slices the pizza was cut into.
## Equivalent Fractions
Some fractions may look different, but are really the same, for example:
4/8 = 2/4 = 1/2 (Four-Eighths) Two-Quarters) (One-Half) = =
It is usually best to show an answer using the simplest fraction ( 1/2 in this case ). That is called Simplifying, or Reducing the Fraction
## Numerator / Denominator
We call the top number the Numerator, it is the number of parts you have.
We call the bottom number the Denominator, it is the number of parts the whole is divided into.
Numerator Denominator
You just have to remember those names! (If you forget just think "Down"-ominator)
You can add fractions easily if the bottom number (the denominator) is the same:
1/4 + 1/4 = 2/4 = 1/2 (One-Quarter) (One-Quarter) (Two-Quarters) (One-Half) + = =
Another example:
5/8 + 1/8 = 6/8 = 3/4 + = =
## Adding Fractions with Different Denominators
But what if the denominators (the bottom numbers) are not the same? As in this example:
3/8 + 1/4 = ? + =
You must somehow make the denominators the same.
In this case it is easy, because we know that 1/4 is the same as 2/8 :
3/8 + 2/8 = 5/8 + =
But it can be harder to make the denominators the same, so you may need to use one of these methods (they both work, use which one you prefer):
## Other Things You Can Do With Fractions
You can also:
And you can visit the Fractions Index to find out even more.
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# Introduction to the fourth dimension
## Last revised 2003-12-23
While giving someone a definition of the fourth dimension is relatively easy, giving someone an intuitive understanding of the fourth dimension can be quite difficult. A definition of the fourth dimension could go like this: The fourth dimension is all space that one can get to by travelling in a direction perpendicular to three-dimensional space. Whenever an uninitiated person hears this, they start pointing their finger around in the air, trying to figure out how it's possible for such a direction to exist. Such a short explanation gives them no intuitive feeling of the fourth dimension.
In order to give you a better understanding the fourth dimension, I will begin with a method that follows a sequence of n-hypercubes that starts with the zeroth dimension and progresses up to the fourth dimension. An n-hypercube is the generalization of the cube within n dimensions, with a 3-hypercube just being the traditional cube. By seeing each n-hypercube build up from the previous one, you should have a better understanding of the final step, from the third dimension to the fourth dimension.
### Step 1: Zeroth Dimension
Imagine a point in space. It is a 0-hypercube. A point is zero dimensional because it has no width, length, or height, and is infinitely small. Every point is exactly the same and has the same measurements, because it has no dimension. Below is a picture of a point, representing the zeroth dimension.
### Step 2: First Dimension
Take the zero-dimensional point and extrude it in any direction, creating a line segment, which is a 1-hypercube. All line segments are one-dimensional because they differ in size by only one measurement, length. They all have the same width and height, which is infinitely small. If you expanded the line infinitely, it would cover one-dimensional space.
### Step 3: Second Dimension
Now take the line segment and extrude it in any direction that is perpendicular to the first direction, creating a square, which is a 2-hypercube. All squares are two dimensional because they differ with each other in size by two measurements, width and length. They all have the same height, which is infinitely small. All of the edges are the same length, and all of the angles are right angles. If you expanded the square infinitely, it would cover two-dimensional space.
### Step 4 - Third Dimension
Take the non-infinite square and extrude it in a third direction, perpendicular to both of the first two directions, creating a cube, which is a 3-hypercube. All cubes are three dimensional because they differ with each other in size by all of the three measurements that we know of - width, length, and height. Just like the square, all of the edges within a single cube are the same length, and all of the angles are right angles. If you expanded the cube infinitely in all directions, it would cover three-dimensional space.
### Step 5 - Fourth Dimension
Now, the final step. Take the non-infinite cube and extrude it in yet another direction perpendicular to the first three. But how can we do this? It is impossible to do within the restrictions of the third dimension (which will I refer to as realmspace in this webpage). However, within the fourth dimension (which I call tetraspace), it is possible. The shape that results from this extrusion of a cube into tetraspace is called a tesseract, which is a 4-hypercube. All tesseracts differ from other tesseracts in size by four measurements (equal to each other within a single tesseract) - width, length, height, and a fourth measurement, which I call trength. Looking back to the previous n-dimensional cubes, they all have the same trength, which is infinitely small. Just like the cube and square, all of the edges within a single tesseract are the same length, and all of the angles are right angles. If you expanded the tesseract infinitely, it would cover four-dimensional space.
There are several ways to view the tesseract, and I will show three of them here. The first one is called an inner projection, and it is formed from a projecting the tesseract into realmspace with a perspective projection. The parts of the original tesseract that are farther away appear smaller in the inner projection. The original cube cell that existed before the extrusion into a tesseract is in gray, the paths of the vertices are in teal, and the stopping point of the extruded cube cell is in blue. The real tesseract isn't shaped like the inner projection shown below - the inner projection is a very distorted "image" of the original tesseract. All of the edges you see in the image are actually the same length as each other, and all angles between edges are right angles.
The second way to view a tesseract isn't actually a normal tesseract, but a parallel projection of a skewed tesseract. To make this shape, first you make a tesseract, then shift the top cube cell a short distance in a diagonal direction, parallel to realmspace. Since this shift is parallel to realmspace, it can actually be in any direction that you can point to. After the shift, you trace the shadow of the skewed tesseract's edges. The result is a shape that has two cubes with their vertices connected together. In the orignal shape, all of the edges within the cube cells are the same length and have right angles with each other. However, they don't have right angles with the teal connection edges, and the teal connection edges are slightly longer than the cube cells' edges.
The third way to view a tesseract is a parallel projection into realmspace. It is the same as a skewed tesseract, but without the top cube cell shifted. Since the edges of the tesseract were extruded in a direction perpendicular to realmspace, when the shape is projected back into realmspace, the edges of the blue cube cell are projected straight back onto the gray cube cell's edges. The resulting projection is a simple cube. This didn't happen with the inner projection, because that projection was a perspective projection.
This last step of trying to view a tesseract shows the difficulties in portraying objects from tetraspace within the limitations of realmspace - there is an extra perpendicular direction that we can't depict within our own space without distorting the original object. Because of these problems, it takes many examples in order to begin understanding the nature of the fourth dimension.
You have now seen a glimpse of the fourth dimension. This is only the beginning - there are many more aspects of the fourth dimension to explore. In the rest of these pages, I will discuss many properties of the fourth dimension - rotation, flatness, levitation, shapes, water, and many others. By the time you are finished, you should have learned many interesting things about the fourth dimension, and maybe you will have even made some discoveries of your own.
The text will frequently refer to technical terms, which are in bold when they first appear. You can find these terms in the glossary if you need further explanation. If you can't figure out the meaning of a word and the glossary doesn't help, post a message on the fourth dimension discussion forum and someone will answer your question.
Good luck on learning about the fourth dimension! I hope you have as much fun as I do.
For other short introductions to the fourth dimension, check out my page of links to short introductions.
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# Solve determinant
It’s important to keep them in mind when trying to figure out how to Solve determinant. We can solve math problems for you.
## Solving determinant
There are a variety of methods that can be used to Solve determinant. Intermediate algebra involves solving a variety of problems involving addition, subtraction, multiplication, and division. This type of problem typically involves the application of previous knowledge to solve new problems. Examples of intermediate algebra include finding the area of a rectangular plot using a right triangle with unknown lengths and finding the value of an unknown quantity in a series by summing certain terms. Intermediate algebra problems tend to be more complex than elementary algebra problems because they require more advanced math skills such as addition and subtraction. Consequently, students may need additional time to practice these skills before they are ready to tackle these types of problems. In order to effectively solve an intermediate algebra problem, students must understand the process of adding and subtracting integers. In addition, they need to understand the concept of multiplication and division. To practice intermediate algebra, students can use online resources such as Khan Academy or their math textbooks. By practicing intermediate algebra problems at home, students will be able to develop better math skills that will help them tackle any future algebra problem that comes their way.
Hard problems are those that are difficult to solve. The best way to describe a hard problem is as a challenge, or an obstacle that must be overcome. A hard problem can be something as simple as learning how to ski for the first time, or as complex as curing cancer. Hard problems are typically more difficult than they have any right to be. Sometimes it’s even impossible to solve them. But if you stick with it, eventually you will find a solution. There are two types of hard problems: those that can be solved and those that cannot be solved. Seemingly impossible problems often turn out to have solutions after all. The trick is finding them. It doesn’t matter whether your problem is big or small, complicated or simple. If you’re willing to put in the work, you can solve almost any problem you encounter.
The best x intercept solver is one that solves for the unknown value of x, the variable you are trying to estimate. This means that the solution should include both the mean and standard deviation of the variable in order to calculate an accurate estimate of your target value. One way to think about this is that a solver should be able to answer questions like “what is my target score?” or “what is my target GPA?” One reason why you might want a x intercept solver over a slope-intercept solver is that slope-intercept solvers tend to overfit data, which means they tend to give unrealistic estimates. A x intercept solver, on the other hand, can be trained on any dataset (as long as it has a mean and standard deviation), which means it can give accurate estimates regardless of how well your data represents your target value. Another reason why you might want a x intercept solver over a slope-intercept solver is that slope-intercept solvers require more computational effort than x intercept solvers, which could lead you to use more resources (CPU power, memory, etc) in order to process your data.
By solving math problems, you will build up your mathematical skills. You can also practice by solving math problems for free on websites. Solving math problems is an easy and fun way to practice math. Solving math problems is also good for improving concentration and focus. You can solve math problems with steps by following the steps below: 1) Choose a problem from the list of free math problems 2) Solve the problem by solving the equation 3) Check your answer 4) Repeat Steps 1 through 3 as many times as necessary
## Instant assistance with all types of math
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## How do you solve a system of linear equations by substitution?
1. Step 1 : First, solve one linear equation for y in terms of x .
2. Step 2 : Then substitute that expression for y in the other linear equation.
3. Step 3 : Solve this, and you have the x -coordinate of the intersection.
4. Step 4 : Then plug in x to either equation to find the corresponding y -coordinate.
How do you teach systems by substitution?
Step 1: Solve one equation for one of the variables. Step 2: Substitute the resulting expression into the other equation to replace the variable. Then solve the equation. Step 3: Substitute to solve for the other variable.
### How do you make a system of equations fun?
Remember that you can use any of these activities to review systems of equations later in the year.
1. System of Equations Mazes.
2. Put It Into Slope-Intercept Form Practice FREEBIE.
3. Coloring Activity.
4. Task Cards and the Target Game.
5. Think, Pick, Flip, Check.
6. Mystery Boxes.
7. Oreo Activity.
8. Quick Check.
What’s the substitution method?
One method to look at is called the substitution method. You solve one equation for one variable and then substitute this expression into the other equation. In the substitution method you solve for one variable, and then substitute that expression into the other equation.
#### How do you create a system of equations?
Here’s how it goes:
1. Step 1: Solve one of the equations for one of the variables. Let’s solve the first equation for y:
2. Step 2: Substitute that equation into the other equation, and solve for x.
3. Step 3: Substitute x = 4 x = 4 x=4 into one of the original equations, and solve for y.
What is the formula of substitution method?
The substitution method is a simple way to solve linear equations algebraically and find the solutions of the variables. As the name suggests, it involves finding the value of x-variable in terms of y-variable and then substituting or replacing the value of x-variable in the second equation.
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Chapter 8
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Chapter 8
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Presentation Transcript
1. Chapter 8 Sampling Variability and Sampling Distributions
2. Suppose we are interested in finding the true mean (m) fat content of quarter-pound hamburgers marketed by a national fast food chain. To learn something about m, we could obtain a sample of n = 50 hamburgers and determine the fat content of each one. To answer these questions, we will examine the sampling distribution, which describes the long-run behavior of sample statistic. Would the sample mean be a good estimate of m? How close is the sample mean to m? Recall that the sample mean is a statistic Will other samples of n = 50 have the same sample mean?
3. Statistic • A number that that can be computed from sample data • Some statistics we will use include x – sample mean s – standard deviation p – sample proportion • The observed value of the statistic depends on the particular sample selected from the population and it will vary from sample to sample. This variability is called sampling variability
4. We caught fish with lengths 6.3 inches, 2.2 inches, and 13.3 inches. x = 7.27 inches 2nd sample - 8.5, 4.6, and 5.6 inches. x = 6.23 inches 3rd sample – 10.3, 8.9, and 13.4 inches. x = 10.87 inches The campus of Wolf City College has a fish pond. Suppose there are 20 fish in the pond. The lengths of the fish (in inches) are given below: This is a statistic! The true mean m = 8. Notice that some sample means are closer and some farther away; some above and some below the mean. Let’s catch two more samples and look at the sample means. Suppose we randomly catch a sample of 3 fish from this pond and measure their length. What would the mean length of the sample be? This is an example of sampling variability
5. There are 1140 (20C3) different possible samples of size 3 from this population. If we were to catch all those different samples and calculate the mean length of each sample, we would have a distribution of all possible x. This would be the sampling distribution of x. Fish Pond Continued . . .
6. In this case, the sample statistic is the sample mean x. Sampling Distributions of x • The distribution that would be formed by considering the value of asample statisticforeverypossible differentsample of agiven sizefrom apopulation.
7. Fish Pond Revisited . . . Suppose there are only 5 fish in the pond. The lengths of the fish (in inches) are given below: We will keep the population size small so that we can find ALL the possible samples. What is the mean and standard deviation of this population? mx = 7.84 sx = 3.262
8. These values determine the sampling distributionof x for samples of size 2. mx = 7.84 sx = 1.998 Fish Pond Revisited . . . mx = 7.84 and sx = 3.262 Let’s find all the samples of size 2. How many samples of size 2 are possible? How do these values compare to the population mean and standard deviation? What is the mean and standard deviation of these sample means?
9. These values determine the sampling distributionof x for samples of size 3. mx = 7.84 sx = 1.332 Fish Pond Revisited . . . mx = 7.84 and sx = 3.262 Now let’s find all the samples of size 3. How many samples of size 3 are possible? What is the mean and standard deviation of these sample means? How do these values compare to the population mean and standard deviation?
10. mx=m sx What do you notice? • The mean of the sampling distribution EQUALS the mean of the population. • As the sample size increases, the standard deviation of the sampling distribution decreases. as n
11. General Properties of Sampling Distributions of x Rule 1: Rule 2: This rule is exact if the population is infinite, and is approximately correct if the population is finite and no more than 10% of the population is included in the sample Note that in the previous fish pond examples this standard deviation formula was not correct because the sample sizes were more than 10% of the population.
12. The paper “Mean Platelet Volume in Patients with Metabolic Syndrome and Its Relationship with Coronary Artery Disease” (Thrombosis Research, 2007) includes data that suggests that the distribution of platelet volume of patients who do not have metabolic syndrome is approximately normal with mean m = 8.25 and standard deviation s = 0.75. We can use Minitab to generate random samples from this population. We will generate 500 random samples of n = 5 and compute the sample mean for each.
13. Platelets Continued . . . Similarly, we will generate 500 random samples of n = 10, n = 20, and n = 30. The density histograms below display the resulting 500 x for each of the given sample sizes. What do you notice about the standard deviation of these histograms? What do you notice about the shape of these histograms? What do you notice about the means of these histograms?
14. General Properties Continued . . . Rule 3: When the population distribution is normal, the sampling distribution of x is also normal for any sample size n.
15. The paper “Is the Overtime Period in an NHL Game Long Enough?” (American Statistician, 2008) gave data on the time (in minutes) from the start of the game to the first goal scored for the 281 regular season games from the 2005-2006 season that went into overtime. The density histogram for the data is shown below. Let’s consider these 281 values as a population. The distribution is strongly positively skewed with mean m = 13 minutes and with a median of 10 minutes. Using Minitab, we will generate 500 samples of the following sample sizes from this distribution: n = 5, n = 10, n = 20, n = 30.
16. What do you notice about the standard deviations of these histograms? These are the density histograms for the 500 samples Are these histograms centered at approximately m = 13? What do you notice about the shape of these histograms?
17. General Properties Continued . . . Rule 4:Central Limit Theorem When n is sufficiently large, the sampling distribution of x is well approximated by a normal curve, even when the population distribution is not itself normal. How large is “sufficiently large” anyway? CLT can safely be applied if n exceeds 30.
18. A soft-drink bottler claims that, on average, cans contain 12 oz of soda. Let x denote the actual volume of soda in a randomly selected can. Suppose that x is normally distributed with s = .16 oz. Sixteen cans are randomly selected, and the soda volume is determined for each one. Let x = the resulting sample mean soda. If the bottler’s claim is correct, then the sampling distribution of x is normally distributed with:
19. To standardized these endpoints, use P(11.96 < x < 12.08) = Soda Problem Continued . . . Look these up in the table and subtract the probabilities. What is the probability that the sample mean soda volume is between 11.96 ounces and 12.08 ounces? .9772 - .1587 = .8185
20. So the distribution of x is approximately normal with A hot dog manufacturer asserts that one of its brands of hot dogs has a average fat content of 18 grams per hot dog with standard deviation of 1 gram. Consumers of this brand would probably not be disturbed if the mean was less than 18 grams, but would be unhappy if it exceeded 18 grams. An independent testing organization is asked to analyze a random sample of 36 hot dogs. Suppose the resulting sample mean is 18.4 grams. Does this result suggest that the manufacturer’s claim is incorrect? Since the sample size is greater than 30, the Central Limit Theorem applies.
21. Values of x at least as large as 18.4 would be observed only about .82% of the time. The sample mean of 18.4 is large enough to cause us to doubt that the manufacturer’s claim is correct. P(x> 18.4) = Hot Dogs Continued . . . Suppose the resulting sample mean is 18.4 grams. Does this result suggest that the manufacturer’s claim is incorrect? 1 - .9918 = .0082
22. Let’s explore what happens with in distributions of sample proportions (p). Have students perform the following experiment. • Toss a penny 20 times and record the number of heads. • Calculate the proportion of heads and mark it on the dot plot on the board. What shape do you think the dot plot will have? This is a statistic! The dotplot is a partial graph of the sampling distribution of all sample proportions of sample size 20. What would happen to the dotplot if we flipped the penny 50 times and recorded the proportion of heads?
23. We will use: p for the population proportion and p for the sample proportion In this case, we will use Sampling Distribution of p The distribution that would be formed by considering the value of asample statisticforeverypossible differentsample of agiven sizefrom apopulation.
24. Suppose we have a population of six students: Alice, Ben, Charles, Denise, Edward, & Frank We are interested in the proportion of females.This is called What is the proportion of females? Let’s select samples of two from this population. How many different samples are possible? We will keep the population small so that we can find ALL the possible samples of a given size. the parameter of interest 1/3 6C2 =15
25. Alice & Ben .5 Alice & Charles .5 Alice & Denise 1 Alice & Edward .5 Alice & Frank .5 Ben & Charles 0 Ben & Denise .5 Ben & Edward 0 Ben & Frank 0 Charles & Denise .5 Charles & Edward 0 Charles & Frank 0 Denise & Edward .5 Denise & Frank .5 Edward & Frank 0 Find the 15 different samples that are possible and find the sample proportion of the number of females in each sample. How does the mean of the sampling distribution compare to the population parameter (p)? Find the mean and standard deviation of these sample proportions.
26. General Properties for Sampling Distributions of p Rule 1: Rule 2: This rule is exact if the population is infinite, and is approximately correct if the population is finite and no more than 10% of the population is included in the sample Note that in the previous student example this standard deviation formula was not correct because the sample size was more than 10% of the population.
27. In the fall of 2008, there were 18,516 students enrolled at California Polytechnic State University, San Luis Obispo. Of these students, 8091 (43.7%) were female. We will use a statistical software package to simulate sampling from this Cal Poly population. We will generate 500 samples of each of the following sample sizes: n = 10, n = 25, n = 50, n = 100 and compute the proportion of females for each sample. The following histograms display the distributions of the sample proportions for the 500 samples of each sample size.
28. What do you notice about the standard deviation of these distributions? What do you notice about the shape of these distributions? Are these histograms centered around the true proportion p = .437?
29. The development of viral hepatitis after a blood transfusion can cause serious complications for a patient. The article “Lack of Awareness Results in Poor Autologous Blood Transfusions” (Health Care Management, May 15, 2003) reported that hepatitis occurs in 7% of patients who receive blood transfusions during heart surgery. We will simulate sampling from a population of blood recipients. We will generate 500 samples of each of the following sample sizes: n = 10, n = 25, n = 50, n = 100 and compute the proportion of people who contract hepatitis for each sample. The following histograms display the distributions of the sample proportions for the 500 samples of each sample size.
30. Are these histograms centered around the true proportion p = .07? What happens to the shape of these histograms as the sample size increases?
31. A conservative rule of thumb: Ifnp> 10andn (1 – p) > 10, then a normal distribution provides a reasonable approximation to the sampling distribution of p. General Properties Continued . . . Rule 3:When n is large and p is nottoo near 0 or 1, the sampling distribution of p is approximately normal. The farther the value of p is from 0.5, the largern must be for the sampling distribution of p to be approximately normal.
32. To answer this question, we must consider the sampling distribution of p. Blood Transfusions Revisited . . . Let p = proportion of patients who contract hepatitis after a blood transfusion p = .07 Suppose a new blood screening procedure is believed to reduce the incident rate of hepatitis. Blood screened using this procedure is given to n = 200 blood recipients. Only 6 of the 200 patients contract hepatitis. Does this result indicate that the true proportion of patients who contract hepatitis when the new screening is used is less than 7%?
33. Blood Transfusions Revisited . . . Let p = .07 p = 6/200 = .03 Is the sampling distribution approximately normal? np = 200(.07) = 14 > 10 n(1-p) = 200(.93) = 186 > 10 What is the mean and standard deviation of the sampling distribution? Yes, we can use a normal approximation.
34. P(p < .03) = Blood Transfusions Revisited . . . Let p = .07 p = 6/200 = .03 Does this result indicate that the true proportion of patients who contract hepatitis when the new screening is used is less than 7%? This small probability tells us that it is unlikely that a sample proportion of .03 or smaller would be observed if the screening procedure was ineffective. This new screening procedure appears to yield a smaller incidence rate for hepatitis. .0132 Assume the screening procedure is not effective and p = .07.
|
### Combined Gas LawProblems #1 - 10
The form of the Combined Gas Law most often used is this:
(P1V1) / T1 = (P2V2) / T2
Most commonly V2 is being solved for. The rearrangement looks like this:
V2 = (P1V1T2) / (T1P2)
A reminder: all these problems use Kelvin for the temperature. I will not usually comment on the change from °C to K. I will use 273 but be aware that your teacher (or computer lesson) may insist on using 273.15.
When you use the combined gas law paired with Dalton's Law, remember that a gas collected over water is always considered to be saturated with water vapor. The vapor pressure of water varies with temperature and must be looked up in a reference source.
Problem #1: A gas has a volume of 800.0 mL at minus 23.0 °C and 300.0 torr. What would the volume of the gas be at 227.0 °C and 600.0 torr of pressure?
Solution:
1) The combined gas law is rearranged to isolate V2:
P1V1T2 V2 = –––––– P2T1
2) Values are inserted into the proper places:
(300.0 torr) (800.0 mL) (500.0 K) V2 = ––––––––––––––––––––––––––– (250.0 K) (600.0 torr)
V2 = 800.0 mL
Problem #2: 500.0 liters of a gas in a flexible-walled container are prepared at 700.0 mmHg and 200.0 °C. The gas is placed into a tank under high pressure. When the tank cools to 20.0 °C, the pressure of the gas is 30.0 atm. What is the volume of the gas?
Solution:
1) The combined gas law is rearranged to isolate V2:
V2 = (P1V1T2) / (T1P2)
2) Values are inserted into the proper places:
V2 = [(0.92105) (500) (293)] / [(473) (30)]
V2 = 9.51 L
Note the use of square brackets to communicate the correct order of operations.
Note that the problem provides different pressure units for the starting and ending values. I used 700/760 to convert from mmHg to atm.
Note that I paid scant attention to setting up the problem with correct sig figs in the problem. This happens often in gas law problems. Note also I omitted all the units.
Problem #3: 690.0 mL of oxygen are collected over water at 26.0 °C and a total pressure of 725.0 mm of mercury. What is the volume of dry oxygen at 52.0 °C and 800.0 mm pressure?
Solution:
1) Use Dalton's Law to remove the pressure of the water vapor:
Ptotal = PO2 + PH2O
PO2 = Ptotal - PH2O
PO2 = 725.0 mmHg - 25.2 mmHg = 699.8 mmHg
The 25.2 value came from here. I looked up the value associated with 26.0 °C and converted it from kPa to mmHg following the instructions given.
2) Here are the values in a solution matrix:
P1 = 699.8 mmHg P2 = 800.0 mmHg V1 = 690.0 mL V2 = x T1 = 299.0 K T2 = 325.0 K
A common student error is to use Dalton's Law, but then use the total pressure value in the combined gas law instead of using the correct value.
The correct pressure to use for P1 is the 699.8 value, not the 725 value. The 725 is the pressure of an oxygen/water mixture and we want ONLY the oxygen (which is the 699.8 value).
3) Use the combined gas law:
x = [(699.8) (690.0) (325)] / [(299) (800.0)]
x = 656 mL (to three sig figs)
Problem #4: What is the volume of gas at 2.00 atm and 200.0 K if its original volume was 300.0 L at 0.250 atm and 400.0 K.
Solution:
1) Here are the values in a solution matrix:
P1 = 0.250 atm P2 = 2.00 atm V1 = 300.0 L V2 = x T1 = 400.0 K T2 = 200.0 K
Note how the problem statement is worded so as to give the starting values last.
2) The combined gas law rearranged to isolate V2:
V2 = (P1V1T2) / (T1P2)
x = [(0.25) (300) (200)] / [(400) (2)] <--- note lack of units
x = 18.75 L
To three sig figs, this is 18.8 L
Problem #5: At conditions of 785.0 torr of pressure and 15.0 °C temperature, a gas occupies a volume of 45.5 mL. What will be the volume of the same gas at 745.0 torr and 30.0 °C?
Solution:
V2 = [(785 mmHg) (45.5 mL) (303 K)] / [(288 K) (745 mmHg)]
V2 = 50.3757 mL
To three sig figs, the answer is 50.4 mL
Problem #6: What is the final volume of a 400.0 mL gas sample that is subjected to a temperature change from 22.0 °C to 30.0 °C and a pressure change from standard pressure to 360.0 mmHg?
Solution:
We are looking to determine V2 in this problem. Here's the set up:
P1 = 760.0 mmHg P2 = 360.0 mmHg V1 = 400.0 L V2 = x T1 = 295 K T2 = 303 K
V2 = [(760 mmHg) (400 mL) (303 K)] / [(295 K) (360 mmHg)]
V2 = 867 mL (to three sig figs)
Problem #7: 400.0 mL of hydrogen are collected over water at 18.0 °C and a total pressure of 740.0 mm of mercury.
(a) What is the partial pressure of H2?
(b) What is the partial pressure of H2O?
(c) What is the volume of DRY hydrogen at STP?
Solution:
1) We will use Dalton's Law to determine the partial pressure of the dry hydrogen gas. We look up the vapor pressure of water in a reference source.
Ptotal = PH2 + PH2O
PH2 = Ptotal - PH2O
740.0 - 15.5 = 724.5 mmHg
I used a different reference source than previously used for the vapor pressure of water. There are many available on the Internet.
2) The partial pressure of the water is its vapor pressure of 15.5 mmHg.
3) Combined gas law rearranged to show V2 isolated:
P1V1T2 V2 = –––––– P2T1
(724.5 mmHg) (400.0 mL) (273 K) V2 = –––––––––––––––––––––––––––– (291 K) (760.0 mmHg)
V2 = 358 mL (to three sig figs)
Problem #8: The pressure of a gas is reduced to 75% of its initial value and the volume is increased by 40% of its initial value. Find the final temperature, given that the initial temperature was -10 °C. This is a combined gas law problem.
Solution:
Let us assign P1 = 1, therefore P2 = 0.75
Let us assign V1 = 1, therefore V2 = 1.4
I won't bother with units on P or V. Your teacher may want the units added in, so I'll do that below.
T1 = -10 C = 263 K
P1V1/T1 = P2V2/T2
[(1 atm) (1 L)] / 263 K = [(0.75 atm) (1.4 L)] / x
(1 atm) (1 L) (x) = (263 K) (0.75 atm) (1.4 L)
x = 276.15 K = 3.15 °C
Problem #9: The pressure of 8.06 L of an ideal gas in a flexible container is decreased to one-third of its original pressure, and its absolute temperature is decreased by one-half. What is the final volume of the gas?
Solution:
1) Assign values as follows:
P1 = 3.00 atm P2 = 1.00 atm V1 = 8.06 L V2 = x T1 = 2.00 K T2 = 1.00 K
Note the made up values for P and T.
2) Insert values into the combined gas law equation and solve for x:
P1V1 / T1 = P2V2 / T2
[(3.00 atm) (8.06 L)] / 2.00 K = [(1.00 atm) (x)] / 1.00 K
x = 12.1 L (to three sig figs)
Problem #10: A balloon of air now occupies 10.0 L at 25.0 °C and 1.00 atm. What temperature was it initially, if it occupied 9.40 L and was in a freezer with a pressure of 0.939 atm?
Solution:
1) Assign values as follows:
P1 = 0.939 atm P2 = 1.00 atm V1 = 9.40 L V2 = 10.0 L T1 = x T2 = 298 K
Note how the problem gives you the ending conditions regarding PVT and asks you for a starting condition. Also, note that temperature is asked for. Compare this to the usual case of asking for the final volume.
2) Let's rearrange the combined gas law equation for T1:
P1V1 P2V2 ––––– = ––––– T1 T2
T1P2V2 = P1V1T2
P1V1T2 T1 = –––––– P2V2
3) Put values in and solve:
(0.939 atm) (9.40 L) (298 K) T1 = –––––––––––––––––––––––– (1.00 atm) (10.0 L)
T1 = 263 K
The form of the final temperature was not specificed, but it is usually given in Celsius, so:
T1 = -10. °C
Problem #11: A gas occupies an 8.00 mL flexible-walled container. The pressure is doubled, the absolute temperature is quadrupled, and 15% of the gas leaks out. What is the new volume?
Solution:
1) This problem involves the combined gas law that has all four variables in it:
P1V1 P2V2 ––––– = ––––– n1T1 n2T2
2) The changes are all expressed in a relative way, so I will assume 1.00 atm and 1.00 K:
P1 = 1.00 atm P2 = 2.00 atm V1 = 8.00 mL V2 = x n1 = 1.00 mol n2 = 0.85 mol T1 = 1.00 K T2 = 4.00 K
By the way, having a gas at 1.00 K is a very silly thing. The point, of course, is that the absolute temperature quadruples. We could use 100 K and 400 K and get the same answer. Or use 200 K and 800 K. The key point is that the temperature quadruples. And, note that it is the absolute temperature (in K) that quadruples, not the temperature in degrees Celsius.
Notice how I interpreted the 15% to mean 15% of the moles of gas escaped. If I decided that 15% of the mass escaped, the problem answer would be the same. I will leave it to you to figure out, if you so desire.
3) I won't bother to isolate V2 this time:
(1.00 atm) (8.00 mL) (2.00 atm) (x) ––––––––––––––––– = ––––––––––––––– (1.00 mol) (1.00 K) (0.85 mol) (4.00 K)
4) Cross-multiply:
(1.00 atm) (8.00 mL) (0.85 mol) (4.00 K) = (2.00 atm) (x) (1.00 mol) (1.00 K)
4) Divide:
(1.00 atm) (8.00 mL) (0.85 mol) (4.00 K) x = ––––––––––––––––––––––––––––––––– (2.00 atm) (1.00 mol) (1.00 K)
x = 13.6 mL
Bonus Problem: A sample of neon has a volume of 0.730 dm3 at a temperature of 21.0 °C. and pressure of 102.5 kPa. If the density of neon is 0.900 g/dm3 at 0 °C and 101.3 kPa, what is the mass of the sample?
Solution:
1) Convert gas conditions to STP:
Here's the cross-multiplied form of the combined gas law:
P1V1T2 = P2V2T1
(102.5 kPa) (0.730 dm3) (273 K) = (101.3 kPa) (V2) (294 K)
V2 = 0.685887 dm3
2) Determine mass:
0.685887 dm3 times 0.900 g/dm3 = 0.617 g
|
## Precalculus (6th Edition) Blitzer
Let us consider the statement ${{S}_{n}}:6\text{ is a factor of }n\left( n+1 \right)\left( n+2 \right)$. Firstly, show that ${{S}_{1}}$ is true. And write ${{S}_{1}}$ by taking the first term on the left and replacing n with $1$ on the right. \begin{align} & n\left( n+1 \right)\left( n+2 \right)=1\left( 1+1 \right)\left( 1+2 \right) \\ & =6 \end{align} Therefore, the above statement is true for $n=1$ as 6 is a factor of 6. So, the statement ${{S}_{1}}$ is true. Next, show that, if ${{S}_{k}}$ is true, then ${{S}_{k+1}}$ is true. Then write ${{S}_{k}}$ and ${{S}_{k+1}}$ by taking the sum of the first k and $\left( k+1 \right)$ terms on the left and replacing n with k and $\left( k+1 \right)$ on the right. \begin{align} & \text{ }k\left( k+1 \right)\left( k+2 \right)=k\left( {{k}^{2}}+3k+2 \right) \\ & ={{k}^{3}}+3{{k}^{2}}+2k \end{align} And, ${{S}_{k+1}}:6\text{ is a factor of }\left( k+1 \right)\left( k+1+1 \right)\left( k+1+2 \right)$ That is, ${{S}_{k+1}}:6\text{ is a factor of }\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)={{k}^{3}}+6{{k}^{2}}+11k+6$ Here, we assume that ${{S}_{k}}$ is true, so add the next consecutive integer after $k$, specifically $2k$ to both sides of the equation. $6\text{ is a factor of }k\left( k+1 \right)\left( k+2 \right)$. And add $3{{k}^{2}}+9k+6$ on both sides, ${{S}_{k+1}}:6\text{ is a factor of }\left( k+1 \right)\left( k+2 \right)\left( k+3 \right)={{k}^{3}}+6{{k}^{2}}+11k+6$ So the final statement is ${{S}_{k+1}}$. Now, if we assume that ${{S}_{k}}$ is true, and add the next consecutive integer after k, specifically $2k$ to both sides of ${{S}_{k}}$, then ${{S}_{k+1}}$ is also true. Thus, by the principal of mathematical induction, the statement ${{S}_{n}}:6\text{ is a factor of }n\left( n+1 \right)\left( n+2 \right)$ is true for every positive integer $n$.
|
# OPERATIONS ON SETS
When two or more sets are combined together to form another set under some given conditions, then operations on sets are carried out.
The important operations on sets are.
1. Union
2. Intersection
3. Set difference
4. Symmetric difference
5. Complement
6. Disjoint sets
Let us discuss the above operations in detail one by one.
## Union
Let X and Y be two sets.
Now, we can define the following new set.
X u Y = {z | z ∈ X or z ∈ Y}
(That is, z may be in X or in Y or in both X and Y)
X u Y is read as "X union Y"
Now that X u Y contains all the elements of X and all the elements of Y and the figure given below illustrates this.
It is clear that X ⊆ X u Y and also Y ⊆ X u Y.
## Intersection
Let X and Y be two sets.
Now, we can define the following new set.
X n Y = {z | z ∈ X and z ∈ Y}
(That is z must be in both X and Y)
X n Y is read as "X intersection Y"
Now that X n Y contains only those elements which belong to both X and Y and the figure given below illustrates this.
It is trivial that that X n Y ⊆ X and also X n Y ⊆ Y.
## Set Difference
Let X and Y be two sets.
Now, we can define the following new set.
X\Y = {z | z ∈ X but z ∉ Y}
(That is z must be in X and must not be in Y)
X\Y is read as "X difference Y"
Now that X \ Y contains only elements of X which are not in Y and the figure given below illustrates this.
Some authors use A - B for A\B. We shall use the notation A\B which is widely used in mathematics for set difference.
## Symmetric Difference
Let X and Y be two sets.
Now, we can define the following new set.
X Δ Y = (X\Y) u (Y\X)
X Δ Y is read as "X symmetric difference Y"
Now that X Δ Y contains all elements in X u Y which are not in X n Y and the figure given below illustrates this. .
## Complement of a Set
If X ⊆ U, where U is a universal set, then U \ X is called the compliment of X with respect to U. If underlying universal set is fixed, then we denote U \ X by X' and it is called compliment of X.
X' = U\X
or
Xc = U\X
The difference set set A \ B can also be viewed as the compliment of B with respect to A.
## Disjoint Sets
Two sets X and Y are said to be disjoint if they do not have any common element. That is, X and Y are disjoint if
X n Y =
It is clear that n(A u B) = n(A) + n(B), if A and B are disjoint finite set.
## Using Operations on Sets
Consider the following sets.
U = {1, 2, 3, 4, ......., 100}
A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
B = {1, 2, 4, 6, 7, 8, 12, 15}
C = {-2, -1, 0, 1, 3, 5, 7}
Now, let us find the following.
(i) AuB :
AuB = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 15}
(ii) BnC :
BnC = {1, 7}
(iii) A\C :
A/C = {2, 4, 6, 8, 9, 10, 11}
(iv) A Δ C :
Δ C = (A\C) u (C\A)
Δ C = {2, 4, 6, 8, 9, 10, 11} u {-2, -1, 0}
Δ C = {-2, -1, 0, 2, 4, 6, 8, 9, 10, 11}
(v) Ac :
Ac = U\A
Ac = U - A
Ac = {13, 14, 15, ......., 100}
Let us list out some useful results.
Let U be a universal set and A, B are subsets of U.
Then the following hold :
(i) A\B = AnBc
(ii) B\A = BnAc
(iii) A\B = A ----> AnB =
(iv) (A\B)uB = AuB
(v) (A\B)nB =
(vi) (A\B)u(B\A) = (AuB)\(AnB)
Kindly mail your feedback to [email protected]
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# A Fundamentals Of Maths Quiz!
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Are you preparing for a math quiz and are now looking for some revision material. The exam below is designed to help you get the highest grade you can get. Give it a try and get to know which problems you need to work more on. All the best of luck!
• 1.
### What is an object measured in three dimensions of length, width, and height?
• A.
One-Dimensional.
• B.
Five-Dimensional.
• C.
Three-Dimensional.
• D.
Two-Dimensional.
C. Three-Dimensional.
Explanation
An object measured in three dimensions of length, width, and height is referred to as three-dimensional. This means that the object has a physical presence and occupies space in three different directions. The dimensions of length, width, and height provide a comprehensive description of the object's size and shape, allowing us to visualize it in a three-dimensional space.
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• 2.
### What is a polygon with four sides called?
• A.
Triangle.
• B.
Circle.
• C.
• D.
Pentagon.
Explanation
A polygon with four sides is called a quadrilateral. A quadrilateral is a polygon that has four straight sides and four angles. The term "quad" refers to the number four, and "lateral" refers to sides. Therefore, a quadrilateral is a geometric shape that consists of four sides.
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• 3.
### What is the amount left over when a number can't be divided equally?
• A.
Divisor.
• B.
Whole Number.
• C.
Remainder.
• D.
Odd Number.
C. Remainder.
Explanation
When a number cannot be divided equally, the amount left over is called the remainder. This means that there is a remainder or leftover amount after dividing the number by another number. For example, if we divide 10 by 3, the quotient is 3 with a remainder of 1. So, in this case, the correct answer is "Remainder".
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• 4.
### How many sides of the same length does a rhombus have?
• A.
Three.
• B.
Eighteen.
• C.
One.
• D.
Four.
D. Four.
Explanation
A rhombus is a quadrilateral with four sides of equal length. Therefore, it has four sides of the same length.
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• 5.
### What is the part of a line that has an endpoint and runs in one direction?
• A.
Center.
• B.
Ray.
• C.
Mid-line.
• D.
B. Ray.
Explanation
A ray is a part of a line that has one endpoint and extends infinitely in one direction. It can be thought of as a beam of light emanating from a point and continuing indefinitely. In contrast, a line segment has two endpoints and a definite length, whereas a ray has no specific length. Therefore, the correct answer is ray.
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• 6.
### What kind of clock has hands moving on it for showing hours and minutes?
• A.
Stopwatch.
• B.
Sundial.
• C.
Digital.
• D.
Analog.
D. Analog.
Explanation
An analog clock is a type of clock that has hands moving on it to show hours and minutes. This type of clock typically has a circular face with numbers or markers around the edge to indicate the hours, and the hands rotate around the center of the clock to show the current time. Stopwatch, sundial, and digital clocks do not have hands moving on them to show hours and minutes, making analog the correct answer.
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• 7.
### What is 10 - 4 equal to?
• A.
2
• B.
10
• C.
19
• D.
6
D. 6
Explanation
The correct answer is 6 because when you subtract 4 from 10, you are left with 6.
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• 8.
### What is the line segment where two faces of a solid figure meet?
• A.
Surface.
• B.
Vertex.
• C.
Diameter.
• D.
Edge.
D. Edge.
Explanation
An edge is the line segment where two faces of a solid figure meet. It is the boundary between two adjacent faces and represents the intersection of these faces. In other words, it is the line that connects two vertices of a solid figure. In this context, the other options are not relevant. A surface refers to the outer layer of a solid figure, a vertex represents a point where multiple edges meet, and a diameter is a line segment that passes through the center of a circle or sphere.
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• 9.
### In which direction(s) does a plane extend?
• A.
Right To Left Only.
• B.
In All Directions.
• C.
Left To Right Only.
• D.
Up And Down Only.
B. In All Directions.
Explanation
A plane extends in all directions because it is a two-dimensional surface that has length and width but no depth. It can be thought of as an infinitely large flat surface that continues indefinitely in all directions. Therefore, it extends both horizontally (left to right and right to left) and vertically (up and down).
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• 10.
### What is a "point"?
• A.
An Estimation.
• B.
Another Word For Sum.
• C.
A Factor In A Sentence.
• D.
An Exact Location In Space.
D. An Exact Location In Space.
Explanation
The term "point" refers to an exact location in space. In geometry, a point is a fundamental concept that has no size or dimension, but represents a specific position. It is often denoted by a dot and can be used to describe the location of objects or define geometric shapes. In other contexts, "point" can also refer to a particular spot or position in various fields such as navigation, mapping, or even in a conversation.
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• 11.
### Which unit is used to measure angles or temperatures?
• A.
Inch.
• B.
Decimeter.
• C.
Degree.
• D.
Point.
C. Degree.
Explanation
The unit used to measure angles or temperatures is called a degree. This unit is commonly used in mathematics, physics, and meteorology to quantify the size or amount of rotation or inclination. It is denoted by the symbol ° and is divided into smaller units such as minutes (') and seconds ("). The other options, inch, decimeter, and point, are not used to measure angles or temperatures.
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• 12.
### If two objects have the same shape, but are different in size, then they are what?
• A.
Even.
• B.
Prime.
• C.
Similar.
• D.
Odd.
C. Similar.
Explanation
If two objects have the same shape but are different in size, they are considered similar. Similarity refers to when two or more objects have the same shape, but their sizes may vary. In this case, the objects may have different dimensions or proportions, but their overall shape remains the same. The term "similar" is commonly used in geometry to describe objects that have the same shape but are not necessarily identical in size.
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• 13.
### How many right angles does a square have?
• A.
4
• B.
1
• C.
24
• D.
26
A. 4
Explanation
A square has four right angles because all four of its sides are equal in length and each corner forms a 90-degree angle. This is a defining characteristic of a square, as it distinguishes it from other shapes that may have different angles. Therefore, the correct answer is 4.
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• 14.
### In an expression, what is missing?
• A.
Minus Signs.
• B.
An Equal Sign.
• C.
• D.
The Number Three.
B. An Equal Sign.
Explanation
The given correct answer is "An Equal Sign." In an expression, an equal sign is used to indicate that both sides of the equation are equal. Without an equal sign, the expression would not represent an equation but rather a statement or a mathematical operation. The presence of an equal sign is crucial in expressing mathematical relationships and solving equations.
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• 15.
### Which triangle has three congruent sides?
• A.
Uneven.
• B.
Double.
• C.
Equilateral.
• D.
Right.
C. Equilateral.
Explanation
An equilateral triangle is the only type of triangle that has three congruent sides. Uneven, double, and right triangles do not have all sides congruent.
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# Solve algebra question
The solver will provide step-by-step instructions on how to Solve algebra question. So let's get started!
## Solving algebra question
We can do your math homework for you, and we'll make sure that you understand how to Solve algebra question. How to solve radicals can be a tricky process, but there are a few steps that can help. First, rationalize the denominator by multiplying by an accessory root. This will eliminate any fractions in the denominator. Next, extract any perfect square roots from the radical. For example, if the radical is 4√5, you would take out the 2√5. Finally, simplify the radical by using absolute value signs and grouping like terms. How to solve radicals may seem complicated at first, but with some practice it can become second nature.
Algebra is the branch of mathematics that deals with the equations and rules governing the manipulation of algebraic expressions. Algebra is used in solving mathematical problems and in discovering new mathematical truths. Algebra is based on the concept of variables, which are symbols that represent unknown numbers or quantities. Algebra is used to solve equations, which are mathematical statements that state that two expressions are equal. The process of solving an equation for a variable is called solving for x. To solve for x, one must first identify the equation's variables and then use algebraic methods to solve for the variable. Algebraic methods include using addition, subtraction, multiplication, and division to solve for a variable. In some cases, algebraic equations can be solve by using exponential or logarithmic functions. Algebra is a powerful tool that can be used to solve mathematical problems and discover new mathematical truths.
This may seem like a lot of work, but the FOIL method can be a very helpful tool for solving trinomials. In fact, many algebra textbooks recommend using the FOIL method when solving trinomials. So next time you're stuck on a trinomial, give the FOIL method a try. You might be surprised at how helpful it can be.
How to solve for domain: There are many ways to solve for the domain of a function. In algebra, the domain is often defined as the set of all values for which a function produces a real output. However, this definition can be difficult to work with, so it is often useful to think about the domain in terms of graphing. For instance, if a function produces imaginary results for certain input values, then those input values will not be included in the function's domain. Similarly, if a function is undefined for certain input values, those values will also be excluded from the domain. In general, the graphing method is the easiest way to determine the domain of a function. However, it is sometimes necessary to use other methods, such as solving inequalities or using set notation. With practice, you will be able to solve for domain quickly and easily.
Let's say you're a cashier and need to figure out how much change to give someone from a \$20 bill. You would take the bill and subtract it from 20, which would give you the amount of change owed. So, if someone gave you a \$20 bill, you would give them back \$16 in change since 20-4 equals 16. You can use this same method to solve problems with larger numbers as well. For example, if someone gave you a \$50 bill, you would take the bill and subtract it from 50, which would give you the amount of change owed. So, if someone gave you a \$50 bill, you would give them back \$40 in change since 50-10 equals 40. As you can see, this method is simple yet effective when trying to figure out how much change to give someone. Give it a try next time you're stuck on a math problem!
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# 1.10: Perimeter and Area
Difficulty Level: At Grade Created by: CK-12
Pacing
Day 1 Day 2 Day 3 Day 4 Day 5
Triangles and Parallelograms Trapezoids, Rhombi, and Kites
More Trapezoids, Rhombi, and Kites
Start Area of Similar Polygons
Finish Area of Similar Polygons
Quiz 1
Start Circumference and Arc Length
Day 6 Day 7 Day 8 Day 9 Day 10
Finish Circumference and Arc Length
Investigation 10-1
Area of Circles and Sectors
Quiz 2
Start Review of Chapter 10
Review of Chapter 10 Chapter 10 Test
## Triangles and Parallelograms
Goal
This lesson introduces students to the area and perimeter formulas for triangles, parallelograms and rectangles.
Relevant Review
Most of this lesson should be review for students. They have learned about area and perimeter of triangles and rectangles in a previous math class (Math 6, Pre-Algebra, or equivalent).
Notation Note
In this chapter, students need to use square units. If no specific units are given, students can write \begin{align*}\text{units}^2\end{align*} or \begin{align*}u^2\end{align*}.
Teaching Strategies
If students are having a hard time with the formulas for area and perimeter of a rectangle, place Example 3 on a piece of graph paper or transparency. Then, students can could the squares for the area and perimeter and you can generate the formula together.
If you count all the squares, there are 36 squares in the area, or square centimeters (red numbers). Counting around the rectangle (blue numbers), we see there are 26 squares. Therefore, the perimeter of this square is 26 cm.
This technique will also work for squares.
An important note, each problem will have some sort of units. Remind students that the shapes might not always be drawn to scale.
Example 5 is a counterexample for the converse of the Congruent Areas Postulate. Therefore, the converse is false. An additional counterexample would to have them draw all the possible rectangles with an area of \begin{align*}20 \ in^2\end{align*}. Use graph paper so students will see that each rectangle has 20 squares. Possible answers are: \begin{align*}20 \times 1, 10 \times 2\end{align*}, and \begin{align*}5 \times 4\end{align*}.
The Area Addition Postulate encourages students to separate a figure into smaller shapes. Always divide the larger shape into smaller shapes that students know how to find the area of.
To show students the area of a parallelogram, cut out the picture (or draw a similar picture to cut out) of the parallelogram and then cut the side off and move it over so that the parallelogram is transformed into a rectangle. Explain to students that the line that you cut is the height of the parallelogram, which is not a side of the parallelogram. Then, cut this parallelogram along a diagonal to create a triangle. Here, students will see that the area of a triangle is half the area of a parallelogram. You may need to rotate the halves (triangles) so that they overlap perfectly. This will show the students that the triangles are congruent and each is exactly half of the parallelogram.
Create another set of flashcards for the area formulas in this chapter. These flashcards should be double-sided. The blank side should be a sketch of the figure and its name. The flip side should have the formula for its area and the formula for its perimeter. Students should create flashcards as the chapter progresses.
## Trapezoids, Rhombi, and Kites
Goal
This lesson further expands upon area formulas to include trapezoids, rhombi, and kites.
Relevant Review
Students might need a quick review of the definitions of trapezoids, rhombi, and kites. Go over their properties (especially that the diagonals of rhombi and kites are perpendicular) and theorems. Students may know the area formula of a trapezoid from a previous math class.
Review the Pythagorean Theorem and special right triangles. There are several examples and review questions that will use these properties. If students do not remember the special right triangle ratios, they can use the Pythagorean Theorem.
Teaching Strategies
Use the same technique discussed in the previous lesson for the area of a parallelogram and triangle. Cut out two congruent trapezoids and demonstrate the explanation at the beginning of the lesson explaining the area of a trapezoid. Going over this with students (rather than just giving them a handout or reading it) will enable them to understand the formula better. These activities are done best on an overhead projector.
Conveniently, the area formula of the rhombus and kite are the same. Again, you can cut out a rhombus and kite, then cut them on the diagonals and piece each together to form a rectangle. Generate the formula with students. Another way to write the formula of a rhombus is to say that it has 4 congruent triangles, with area \begin{align*}\frac{1}{2} \left (\frac{1}{2} d_1 \right ) \left (\frac{1}{2} d_2 \right ) = \frac{1}{8} d_1d_2\end{align*}. Multiplying this by 4, we get \begin{align*}\frac{4}{8}d_1d_2 = \frac{1}{2}d_1d_2\end{align*}. This process is not as easily done with a kite because one of the diagonals is not bisected.
Additional Example: Find two different rhombi that have an area of \begin{align*}48 \ \text{units}^2\end{align*}.
Solution: The diagonals are used to find the area, so when solving this problem, we are going to be finding the diagonals’ lengths. \begin{align*}\frac{1}{2} d_1d_2 = 48\end{align*}, so \begin{align*}d_1d_2 = 96\end{align*}. This means that the product of the diagonals is double the area.
The diagonals can be: 1 and 96, 2 and 48, 3 and 32, 4 and 24, 6 and 16, 8 and 12.
As an extension, you can students draw the rhombi. The diagonals bisect each other, so have the diagonals cut each other in half and then connect the endpoints of the diagonals to form the rhombus. Three examples are below.
## Areas of Similar Polygons
Goal
Students will learn about the relationship between the scale factor of similar polygons and their areas. Students should also be able to apply area ratios to solving problems.
Relevant Review
Review the properties of similar shapes, primarily triangles and quadrilaterals, from Chapter 7. Remind students that the perimeter, sides, diagonals, etc. have the same ratio as the scale factor. The Review Queue reviews similar squares. As an additional question, ask students to find the perimeter of both squares and then reduce the ratio (smaller square = 40, larger square = 100, ratio is 2:5, the same as the ratio of the sides). Ask students why they think the ratio of the sides is the same as the ratio of the perimeters.
Teaching Strategies
Examples 1 and 2 lead students towards the Area of Similar Polygons Theorem. As an additional example (before introducing the Area of Similar Polygons Theorem), ask students to find the area of two more similar shapes. Having students repeat problems like Example 2, they should see a pattern and arrive at the theorem on their own.
Additional Example: Two similar triangles are below. Find their areas and the ratio of the areas. How does the ratio of the areas relate to the scale factor?
Solution: Each half of the isosceles triangles are 3-4-5 triangles. The smaller triangle has a height of 3 and the larger triangle has a height of 9 (because 12 is 3 time 4, so this triangle is three time larger than the smaller triangle). The areas are: \begin{align*}A_{larger \ \Delta} = \frac{1}{2} \cdot 24 \cdot 9 = 108\end{align*} and \begin{align*}A_{smaller \ \Delta} = \frac{1}{2} \cdot 8 \cdot 3 = 12\end{align*}. The ratio of the area is \begin{align*}\frac{12}{108} = \frac{1}{9}\end{align*}. The ratios of the scale factor and areas relate by squaring the scale factor, \begin{align*}\frac{1}{9} = \left (\frac{1}{3} \right )^2\end{align*}.
## Circumference and Arc Length
Goal
The purpose of this lesson is to review the circumference formula and then derive a formula for arc length.
Relevant Review
The Review Queue is a necessary review of circles. Students need to be able to apply central angles, find intercepted arcs and inscribed angles. They also need to know that there are \begin{align*}360^\circ\end{align*} in a circle.
Teaching Strategies
Students may already know the formula for circumference, but probably do not remember where \begin{align*}\pi\end{align*} comes from. Investigation 10-1 is a useful activity so that students can see how \begin{align*}\pi\end{align*} was developed and why it is necessary to find the circumference and area of circles. You can decide to make this investigation teacher-led or allow students to work in pairs or groups. From this investigation, we see that the circumference is dependent upon \begin{align*}\pi\end{align*}.
When introducing arc length, first have students find the circumference of a circle with radius of \begin{align*}6 (12 \pi)\end{align*}. Then, see what the length of the arc of a semicircle \begin{align*}(6 \pi)\end{align*}. Students should make the connection that the arc length of the semicircle will be half of the circumference. Then ask students what the arc length of half of the semicircle is \begin{align*}(3 \pi)\end{align*}. Ask what the corresponding angle measure for this arc length would be \begin{align*}(90^\circ)\end{align*}. See if students can reduce \begin{align*}\frac{90}{360}\end{align*} and if they make the correlation that the measure of this arc is a quarter of the total circumference, just like \begin{align*}90^\circ\end{align*} is a quarter of \begin{align*}360^\circ\end{align*}. Using this same circle, see if students can find the arc length of a \begin{align*}30^\circ\end{align*} portion of the circle \begin{align*}\left (\frac{3 \pi}{3} = \pi \right )\end{align*}. Then, as students what portion of the total circumference \begin{align*}\pi\end{align*} is. \begin{align*}\pi\end{align*} is \begin{align*}\frac{1}{12}\end{align*} of \begin{align*}12 \pi\end{align*}, just like \begin{align*}30^\circ\end{align*} is \begin{align*}\frac{1}{12}\end{align*} of \begin{align*}360^\circ\end{align*}. This should lead students towards the Arc Length Formula.
Students may wonder why it is necessary to leave answer in exact value, in terms of \begin{align*}\pi\end{align*}, instead of approximate (multiplying by 3.14). This is usually a teacher preference. By using the approximate value for \begin{align*}\pi\end{align*}, the answer automatically has a rounding error. Rounding the decimal too short will cause a much larger error than using the decimal to the hundred-thousandths place. Whatever your preference, be sure to explain both methods to your students. The review questions request that answers be left in terms of \begin{align*}\pi\end{align*}, but this can be easily changed, depending on your decision.
## Area of Circles and Sectors
Goal
This lesson reviews the formula for the area of a circle and introduces the formula for the area of a sector and segment of a circle.
Teaching Strategies
If you have access to an LCD display or a computer lab, show students the animation of the area of a circle formula (link is in the FlexBook).
The formula for the area of a sector is very similar to the formula for arc length. Ask students to compare the two formulas. Stress to students that the angle fraction in the sector formula is the same as it is for the arc length formula. Therefore, students do not need to memorize a new formula; they just need to remember the angle fraction for both.
To find the area of the shaded regions (like Example 8), students will need to add or subtract areas of circles, triangles, rectangles, or squares in order to find the correct area. Encourage students to identify the shapes in these types of problems before they begin to solve it. At the end of this lesson, quickly go over problems 23-25, so that students know how to solve the problems that evening. Remind students to use the examples in the lesson to help them with homework problems.
Finding the area of a segment can be quite challenging for students. This text keeps the angles fairly simple, using special right triangle ratios. Depending on your level of student, you may decide to omit this portion of this lesson. If so, skip Example 9 and review questions 26-31.
Solution: The triangle that is inscribed in the circle is a 45-45-90 triangle and its hypotenuse is on the diameter of the circle. Therefore, the hypotenuse is \begin{align*} 24 \sqrt{2}\end{align*} and the radius is \begin{align*}12 \sqrt{2}\end{align*}. The area of the shaded region is the area of the circle minus the area of the triangle.
\begin{align*}A_{\bigodot} &= \pi (12\sqrt{2})^2 = \pi \cdot 144 \cdot 2 = 288 \pi\\ A_{\Delta} &= \frac{1}{2} \cdot 24 \cdot 24 = 288\end{align*}
The area of the shaded region is \begin{align*}288 \pi - 288 \approx 616.78 \ \text{units}^2\end{align*}
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# HOW TO SIMPLIFY RADICAL EXPRESSIONS WITH VARIABLES AND EXPONENTS
How to simplify radical expressions with variables and exponents :
To simplify radical terms or radical expressions first we have to find the factors.
• According to the index of the given radical, we have to take one common term from the radical.
• For example if we have square root, then we have to take one common term for every two same terms.
• If we have cube root then we have to take one common term for every three same terms.
• In case we have algebraic expression in the radical sign, we have to find factors using algebraic identities.
Here we can find some frequently used tables while finding the square root of a number and also you can find frequently used algebraic formulas.
Let us see some example problems based on the above concept.
## How to simplify radical expressions with variables and exponents - Examples
Example 1 :
Find the square root of 196 a⁶ b⁸ c¹⁰
Solution :
= √(196 a⁶ b⁸ c¹⁰ )
= √(14 x 14 a³ x a³ x b⁴ x b⁴ x c⁵ x c⁵)
= 14 |a³b⁴ c⁵|
Example 2 :
Find the square root of 289 (a - b)⁴ (b - c)
Solution :
= √289 (a - b)⁴ (b - c)⁶
= √(17 x 17 (a - b)² (a - b)² (b - c)³(b - c)³
= 17 |(a - b)² (b - c)³|
Example 3 :
Find the square root of (x + 11)² - 44 x
Solution :
(a + b)² = a² + 2 a b + b²
= √(x² + 2 x (11) + 11² - 44 x)
= √(x² + 22 x + 121 - 44 x)
= (x² - 22 x + 121 )
= √(x - 11)²
= |x - 11|
Example 4 :
Find the square root of (x - y)² + 4 x y
Solution :
(a - b)² = a² - 2 a b + b²
= √[(x - y)² + 4 x y]
= √[x² - 2 x y + y² + 4 x y]
= √(x² + 2 x y + y²)
=
√((x + y)²
= (x + y)
Example 5 :
Find the square root of 121 x⁸ y⁶ ÷ 81 x⁴ y
Solution:
= √(121 x⁸ y⁶/ 81 x⁴ y⁸)
= √(11 x 11 x x⁴ x x⁴ y³y³/ 9 x x² x x² y⁴ y⁴)
= (11/9) ( x⁴ y³/ x² y)
= (11/9) ( x² /y)
Example 6 :
Find the square root of [64 (a + b)⁴(x - y)⁸(b - c)⁶]/[25 (x+ y)⁴ (a - b)⁶(b + c)¹⁰]
Solution:
= √[64 (a + b)⁴(x - y)⁸(b - c)⁶]/[25 (x+ y)⁴ (a - b)⁶(b + c)¹⁰]
= (8/5)|[(a + b)² (x - y)(b - c)³/(x+ y) ² (a - b)³(b + c)⁵]|
Example 7 :
Find the square root of 16 x² - 24 x + 9
Solution :
= √(16 x² - 24 x + 9)
= √(4 x)² - 2 (4x) (3) + 3²
= √(4 x - 3)²
= |(4 x - 3)|
Example 8 :
Find the square root of (x² - 25)(x² + 8 x + 15)(x² - 2 x - 15)
Solution :
= √(x² - 25)(x² + 8 x + 15)(x² - 2 x - 15)
= √(x + 5)(x - 5) (x + 3) (x + 5) (x - 5)(x + 3)
= |(x - 5) (x + 5) (x + 3)|
Example 9 :
Find the square root of 4x² + 9y² + 25z² - 12xy + 30 yz - 20 zx How to simplify radical expressions with variables and exponents
Solution :
= √(4x² + 9y² + 25z² - 12xy + 30 yz - 20 zx)
= √(2x)²+(-3y)²+(-5z)²+2(2x)(-3y)+2(-3y)(-5z)+2(2x)(-5z)
= √(2x - 3y - 5z)²
= |(2x - 3y - 5z)|
Example 10 :
Find the square root of x⁴ + (1/x⁴) + 2
Solution :
(a + b)² = a² + 2 a b + b²
= (x²)² + (1/x²)² + 2 x² (1/x²)]
= √(x² + (1/x²))²
= |(x² + (1/x²))|
After having gone through the stuff given above, we hope that the students would have understood "How to simplify radical expressions with variables and exponents".
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Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
|
## Brickwork Quantity Calculation
To find out the Brickwork Quantity Calculation above the plinth level we need to first find out the length, width, and height of the wall in the given drawing. For this, we have to see the drawing. For the length and width of the wall, we need to see the top view of the drawing
In this top view, we can see the room of dimensions 4m x3.5m inner to inner.
So to find out the total length of the wall we need to add all the length of walls from column to column,
= 4+3.5+4+3.5
=15m
The total length of the wall = 15m
Width of the wall =230mm or say 0.23m
### The find out the height of the wall we have to see the front elevation of the wall in the drawing.
Here in this front view, the height of the wall is 3m.
Here the sill height of the window is 0.7m
(Sill height is basically the distance between the plinth level and the window bottom)
First, we find the area of the wall so that we can deduct the door and window area easily.
Length= 15m
Height=3m
Area = 45m2
Now we have to deduct the door and window space for that we have to see joinery details
Now in this joinery detail, there is one door and two windows
1. Door
Length = 0.8m
Height = 2m
Area = 0.8 x 2
= 1.6m2
2. Window
Length = 1.5m
Height = 1.2m
Area = 1.5 x 1.2
= 1.8m2 x 2
= 3.6m2
Total deduction = 1.6 + 3.6
= 5.2m2
Total area after deductions = 45-5.2
= 39.8m2
To find out the Brickwork Quantity Calculation we have to multiply the width of the wall to the area
i.e, = 39.8 x0.23
= 9.154m3is the required volume of brickwork
As discussed earlier (as per BIS) that 1m3 brickwork = 500 numbers of brick
Therefore, 9.154 x 500 = 4577 numbers of bricks are used.
Total numbers of bricks are 4577
If you Want to Know How to Calculate Brickwork quantity in the foundation of the building then click on this link below to watch
CALCULATION OF BRICKWORK QUANTITY IN THE FOUNDATION OF THE BUILDING
FOR MORE DETAIL OF COURSE VISIT: RED BRICKS ACADEMY OF CIVIL ENGINEERING
|
# Sine
## Definition
The ratio of lengths of opposite side to hypotenuse at the respective angle of a right angled triangle is called sine.
Sine is a representative to disclose the relation between angle and the ratio of length of opposite side to length of hypotenuse in the right angled triangle. Hence, it is also called as trigonometric function.
### Expression
It is expressed as a ratio of lengths of opposite side to hypotenuse to express sine in mathematical form.
$$\frac{Length \, of \, Opposite \, Side}{Length \, of \, Hypotenuse}$$
But the ratio is calculated at the angle of the right angled triangle. So, the relation of ratio with angle of the right angled triangle is expressed as sine of angle. Mathematically, sine of angle is written by writing $\sin$ first and then angle of the right angled triangle.
According to this theory, the relation of ratio of lengths of opposite side to hypotenuse with angle is expressed in mathematical form as follows.
$$\sin \, (angle) = \frac{Length \, of \, Opposite \, side}{Length \, of \, Hypotenuse}$$
Sine is defined in trigonometry from ratio. So, it is also called as a trigonometric ratio.
For example, in $\Delta BAC$, the angle of the right angled triangle is $\theta$. So, sine of angle is written as $\sin \theta$. Lengths of opposite side and hypotenuse are $BC$ and $AC$ respectively.
$$\therefore \,\, \sin \theta = \frac{BC}{AC}$$
#### Example
$\Delta DEF$ is another right angle triangle.
1. Length of the opposite side $(\overline{DF})$ is $DF = 3$ meters
2. Length of the hypotenuse $(\overline{DE})$ is $DE = 5$ meters
3. Angle of this right angled triangle is $36.87^°$
Express sine in mathematical form for this triangle.
$$\sin 36.87^\circ = \frac{DF}{DE}$$
$$\implies \sin 36.87^\circ = \frac{3}{5} = 0.6$$
It is read as the value of sine of $36.87^\circ$ is $0.6$.
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# Applied Mathematics
The subject of mathematics is related to almost all the other subjects. The advancement in the fields of engineering, science, economics statistics etc. are facilitated by use of mathematics. In other words, the application of mathematics helps in development and easier understanding of topics in other subjects. The branch of mathematics that is used for such a purpose is called as applied mathematics. It may be noted that in this branch of math, the important terms and constants used in the other topics also form as parts.
Let us give a simple introduction to applied mathematics with interesting examples from other fields.
A car leaves an airport at 8 am and runs at an average velocity 45 mph. At 9 am another car leaves the same airport in the same direction and it has to meet the first car before noon. What should be the minimum average velocity with which the second car should run to achieve the necessity?
The situation described in the above physics problem is not very unusual. A person traveling in the first car might have left out something and his friend at the airport may like to reach that article. And by noon, the first person might reach his destination and may not be reachable thereafter. Let us see how applied mathematics helps us to solve. The velocity refers to the rate of change of distance with respect to time. Hence the distance traveled by the first car in the 4 hours (from 8 am to noon) is given by the mathematical equation d1 = v1* t1 = 45*4 = 180 miles, since we know that v1 = 45 mph and t1 = 4 hours. The concept of applied math is same for the second car but now the equation is d2 = v2* t2. But in this case the known values are d2 = 180 miles, since at the point of interception both cars must have traveled the same distance and out of necessity the maximum value of t2 can only be 3 hours (from 9 am to noon). Therefore, 180 = v2*3, which gives the solution as v2 = 60 mph.
Let us study another problem related to physics but which can be considered as an applied mathematics. Two wires of ½ in. diameter are anchored at a ceiling roof as shown in the diagram. These wires are riveted to a hook which is used to hold heavy weights. The wires make angles of 45o and 60o with the ceiling. The wires have an ultimate tensile strength of 16T per sq.in. What could be the maximum weight that can be loaded on the hook?
The concept of this problem is used in material lifting equipment. Ultimate tensile strength of 16T/ sq.in. means the wire can take a load of only 16T for a cross section area of 1 in. Since the diameter of the wires are ½ in. each wire can take only a load of 16(π/4)(1/2)2 = π tons ≈ 3.14 tons. Now mathematically we can draw a vector diagram and find the solution. The same is drawn below.
The wires on the left and right take the loads that are the projection of the main load W in the direction of wires. Let those components be P and Q respectively. As per vector algebra, P = (√2)W/2 and Q = (√3)W/2. Obviously the magnitude of Q is greater and therefore it must be equal to 3.14T. Hence W can be equal to a maximum of 3.14/0.866 = 3.63T approximately.
The concept of matrices is widely used in statistical fields. We will give an introduction to matrices in our next topic.
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Comment by Dorothy Hastings on February 4, 2014 at 12:46pm
Unless and until we apply our book knowledge into practical world we found it useless and uninteresting. If we are taught to learn things not just for reading books and solving mathematical problems but also by incorporating these ideas in to real life situations, the interest will develop.
Comment by John Marsh on February 3, 2014 at 3:00am
Yes you are absolutely right Marin, you can use same concept for stress and strain calculation.
Comment by Oscar Marin on February 2, 2014 at 4:42am
I never analyzed this concept of applied mathematics, It seems really great. Can that be applied to trigonometric problem solving too. I think this has much of application on work and energy concept of physics where a stress and strain calculation for a particular object or a bridge needs to be calculated. Hope I am right!!
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> Newton's Third Law | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Physical Science Concepts For Middle School Go to the latest version.
# 4.26: Newton's Third Law
Difficulty Level: Basic Created by: CK-12
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This is a sketch of Jerod on his skateboard. He’s on his way to Newton’s Skate Park. When he pushes his foot against the ground, what happens next? He moves on his skateboard in the opposite direction. How does this happen?
### Action and Reaction
Newton’s third law of motion explains how Jerod starts his skateboard moving. This law states that every action has an equal and opposite reaction. This means that forces always act in pairs. First an action occurs—Jerod pushes against the ground with his foot. Then a reaction occurs—Jerod moves forward on his skateboard. The reaction is always equal in strength to the action but in the opposite direction.
Q : If Jerod pushes against the ground with greater force, how will this affect his forward motion?
A : His action force will be greater, so the reaction force will be greater as well. Jerod will be pushed forward with more force, and this will make him go faster and farther.
### Equal and Opposite Forces
The forces involved in actions and reactions can be represented with arrows. The way an arrow points shows the direction of the force, and the size of the arrow represents the strength of the force. Look at the skateboarders in the Figure below . In the top row, the arrows represent the forces with which the skateboarders push against each other. This is the action. In the bottom row, the arrows represent the forces with which the skateboarders move apart. This is the reaction. Compare the top and bottom arrows. They point in different directions, but they are the same size. This shows that the reaction forces are equal and opposite to the action forces.
### Equal and Opposite but Not Balanced
Because action and reaction forces are equal and opposite, you might think they would cancel out, as balanced forces do. But you would be wrong. Balanced forces are equal and opposite forces that act on the same object. That’s why they cancel out. Action-reaction forces are equal and opposite forces that act on different objects, so they don’t cancel out. In fact, they often result in motion. Think about Jerod again. He applies force with his foot to the ground, whereas the ground applies force to Jerod and the skateboard, causing them to move forward.
Q : Actions and reactions occur all the time. Can you think of an example in your daily life?
A : Here’s one example. If you lean on something like a wall or your locker, you are applying force to it. The wall or locker applies an equal and opposite force to you. If it didn’t, you would go right through it or else it would tip over.
You can watch a video about actions and reactions at this URL:
### Summary
• Newton’s third law of motion states that every action has an equal and opposite reaction. This means that forces always act in pairs.
• Action and reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don’t cancel out.
### Vocabulary
• Newton’s third law of motion : Law stating that every action has an equal and opposite reaction.
### Practice
Watch this video about Newton’s third law of motion, and then answer the questions below.
1. Outline the action and reaction demonstrated by the astronauts in the video. Why does wearing the battery pack affect the motion of the astronaut named Alexander?
2. Describe an example of Newton’s cradle.
3. How do space vehicles apply action and reaction forces to blast off?
### Review
1. State Newton’s third law of motion.
2. Describe an example of an action and reaction. Identify the forces and their directions.
3. Explain why action and reaction forces are not balanced forces.
Basic
7 , 8
## Date Created:
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# Fractions and Decimals
### 2.1 Introduction
You have learnt fractions and decimals in earlier classes. The study of fractions included proper, improper and mixed fractions as well as their addition and subtraction. We also studied comparison of fractions, equivalent fractions, representation of fractions on the number line and ordering of fractions.
Our study of decimals included, their comparison, their representation on the number line and their addition and subtraction.
We shall now learn multiplication and division of fractions as well as of decimals.
### 2.2 How Well have You Learnt about Fractions?
A proper fraction is a fraction that represents a part of a whole. Is a proper fraction? Which is bigger, the numerator or the denominator?
An improper fraction is a combination of whole and a proper fraction. Is an improper fraction? Which is bigger here, the numerator or the denominator?
The improper fraction can be written as . This is a mixed fraction.
Can you write five examples each of proper, improper and mixed fractions?
Example 1 Write five equivalent fractions of .
Solution One of the equivalent fractions of is
. Find the other four.
Example 2 Ramesh solved part of an exercise while Seema solved of it. Who solved lesser part?
Solution In order to find who solved lesser part of the exercise, let us compare and .
Converting them to like fractions we have, , .
Since 10 < 28 , so .
Thus, .
Ramesh solved lesser part than Seema.
Example 3 Sameera purchased kg apples and kg oranges. What is the total weight of fruits purchased by her?
Solution The total weight of the fruits
=
=
Example 4 Suman studies for hours daily. She devotes hours of her time for Science and Mathematics. How much time does she devote for other subjects?
Solution Total time of Suman’s study = h = h
Time devoted by her for Science and Mathematics = = h
Thus, time devoted by her for other subjects = h
= = = h = h
### Exercise 2.1
1. Solve:
(i) (ii) (iii) (iv)
(v) (vi) (vii)
2. Arrange the following in descending order:
(i) (ii) .
3. In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square?
4. A rectangular sheet of paper is cm long and cm wide. Find its perimeter.
5. Find the perimeters of (i) ABE (ii) the rectangle BCDE in this figure. Whose perimeter is greater?
6. Salil wants to put a picture in a frame. The picture is cm wide. To fit in the frame the picture cannot be more than cm wide. How much should the picture be trimmed?
7. Ritu ate part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat? Who had the larger share? By how much?
8. Michael finished colouring a picture in hour. Vaibhav finished colouring the same picture in hour. Who worked longer? By what fraction was it longer?
### 2.3 Multiplication of Fractions
You know how to find the area of a rectangle. It is equal to length × breadth. If the length and breadth of a rectangle are 7 cm and 4 cm respectively, then what will be its area? Its area would be 7 × 4 = 28 cm2.
What will be the area of the rectangle if its length and breadth are cm and
cm respectively? You will say it will be × = × cm
2. The numbers and are fractions. To calculate the area of the given rectangle, we need to know how to multiply fractions. We shall learn that now.
#### 2.3.1 Multiplication of a Fraction by a Whole Number
Observe the pictures at the left (Fig 2.1). Each shaded part is part of a circle. How much will the two shaded parts represent together? They will represent = .
Fig 2.1
Combining the two shaded parts, we get Fig 2.2 . What part of a circle does the shaded part in Fig 2.2 represent? It represents part of a circle .
Fig 2.2
The shaded portions in Fig 2.1 taken together are the same as the shaded portion in Fig 2.2, i.e., we get Fig 2.3.
Fig 2.3
or = .
Can you now tell what this picture will represent? (Fig 2.4)
Fig 2.4
And this? (Fig 2.5)
Fig 2.5
Let us now find .
We have =
We also have
So = =
Similarly = = ?
Can you tell = ?
The fractions that we considered till now, i.e., and were proper fractions.
For improper fractions also we have,
= =
Try, = ? = ?
Thus, to multiply a whole number with a proper or an improper fraction, we multiply the whole number with the numerator of the fraction, keeping the denominator same.
### Try These
1. Find: (a) (b) (c) (d)
If the product is an improper fraction express it as a mixed fraction.
2. Represent pictorially :
To multiply a mixed fraction to a whole number, first convert the mixed fraction to an improper fraction and then multiply.
Therefore, = = = .
Similarly, = = ?
### Try These
Find: (i)
(ii)
Fraction as an operator ‘of
Observe these figures (Fig 2.6)
The two squares are exactly similar.
Each shaded portion represents of 1.
Fig 2.6
So, both the shaded portions together will represent of 2.
Combine the 2 shaded parts. It represents 1.
So, we say of 2 is 1. We can also get it as × 2 = 1.
Thus, of 2 = × 2 = 1
Also, look at these similar squares (Fig 2.7).
Fig 2.7
Each shaded portion represents of 1.
So, the three shaded portions represent of 3.
It represents 1 i.e., .
So, of 3 is . Also, × 3 = .
Thus, of 3 = × 3 = .
So we see that ‘of’ represents multiplication.
Farida has 20 marbles. Reshma has of the number of marbles what Farida has. How many marbles Reshma has? As, ‘of’ indicates multiplication, so, Reshma has = 4 marbles.
Similarly, we have of 16 is = = 8.
### Try These
Can you tell, what is (i) of 10?, (ii) of 16?, (iii) of 25?
Example 5 In a class of 40 students of the total number of studetns like to study English, of the total number like to study Mathematics and the remaining students like to study Science.
(i) How many students like to study English?
(ii) How many students like to study Mathematics?
(iii) What fraction of the total number of students like to study Science?
Solution Total number of students in the class = 40.
(i) Of these of the total number of students like to study English.
Thus, the number of students who like to study English = of 40 = = 8.
(ii) Try yourself.
(iii) The number of students who like English and Mathematics = 8 + 16 = 24. Thus, the number of students who like Science = 40 – 24 = 16.
Thus, the required fraction is .
### Exercise 2.2
1. Which of the drawings (a) to (d) show :
(i) (ii) (iii) (iv)
(a) (b)
(c) (d)
2. Some pictures (a) to (c) are given below. Tell which of them show:
(i) (ii) (iii)
3. Multiply and reduce to lowest form and convert into a mixed fraction:
(i) (ii) (iii) (iv) (v)
(vi) (vii) (viii) (ix) (x)
4. Shade: (i) of the circles in box (a) (ii) of the triangles in box (b)
(iii) of the squares in box (c).
5. Find:
(a) of (i) 24 (ii) 46 (b) of (i) 18 (ii) 27
(c) of (i) 16 (ii) 36 (d) of (i) 20 (ii) 35
6. Multiply and express as a mixed fraction :
(a) (b) (c)
(d) (e) (f)
7. Find: (a) of (i) (ii) (b) of (i) (ii)
8. Vidya and Pratap went for a picnic. Their mother gave them a water bottle that contained 5 litres of water. Vidya consumed of the water. Pratap consumed the remaining water.
(i) How much water did Vidya drink?
(ii) What fraction of the total quantity of water did Pratap drink?
#### 2.3.2 Multiplication of a Fraction by a Fraction
Farida had a 9 cm long strip of ribbon. She cut this strip into four equal parts. How did she do it? She folded the strip twice. What fraction of the total length will each part represent? Each part will be of the strip. She took one part and divided it in two equal parts by folding the part once. What will one of the pieces represent? It will represent of or × .
Let us now see how to find the product of two fractions like × .
To do this we first learn to find the products like × .
(a) How do we find of a whole? We divide the whole in three equal parts. Each of the three parts represents of the whole. Take one part of these three parts, and shade it as shown in Fig 2.8.
Fig 2.8
(b) How will you find of this shaded part? Divide this one-third () shaded part into two equal parts. Each of these two parts represents of i.e., × (Fig 2.9).
Take out 1 part of these two and name it ‘A’. ‘A’ represents × .
A
Fig 2.9
(c) What fraction is ‘A’ of the whole? For this, divide each of the remaining parts also in two equal parts. How many such equal parts do you have now?
There are six such equal parts. ‘A’ is one of these parts.
So, ‘A’ is of the whole. Thus, × = .
How did we decide that ‘A’ was of the whole? The whole was divided in 6 = 2 × 3 parts and 1 = 1 × 1 part was taken out of it.
Thus, × = =
or × =
The value of × can be found in a similar way. Divide the whole into two equal parts and then divide one of these parts in three equal parts. Take one of these parts. This will represent × i.e., .
Therefore × = = as discussed earlier.
Hence × ×
Find × and × ; × and × and check whether you get
× × × ×
### Try These
Fill in these boxes:
(i) × = = (ii) × =
(iii) × = (iv) × =
Example 6 Sushant reads part of a book in 1 hour. How much part of the book will he read in hours?
Solution The part of the book read by Sushant in 1 hour = .
So, the part of the book read by him in hours = ×
= × =
Let us now find ×. We know that × 5 .
So, × × × 5 = =
Also, . Thus, × = .
This is also shown by the figures drawn below. Each of these five equal shapes (Fig 2.10) are parts of five similar circles. Take one such shape. To obtain this shape we first divide a circle in three equal parts. Further divide each of these three parts in two equal parts. One part out of it is the shape we considered. What will it represent? It will represent × = . The total of such parts would be 5 × = .
Fig 2.10
Similarly × = = .
We can thus find × as × = = .
So, we find that we multiply two fractions as .
Value of the Products
You have seen that the product of two whole numbers is bigger than each of the two whole numbers. For example, 3 × 4 = 12 and 12 > 4, 12 > 3. What happens to the value of the product when we multiply two fractions?
Let us first consider the product of two proper fractions.
We have,
### Try These
Find: × ; ×
You will find that when two proper fractions are multiplied, the product is less than each of the fractions. Or, we say the value of the product of two proper fractions is smaller than each of the two fractions.
Check this by constructing five more examples.
Let us now multiply two improper fractions.
We find that the product of two improper fractions is greater than each of the two fractions.
### Try These
Find: × ; × .
Or, the value of the product of two improper fractions is more than each of the two fractions.
Construct five more examples for yourself and verify the above statement.
Let us now multiply a proper and an improper fraction, say and .
We have × = . Here, < and >
The product obtained is less than the improper fraction and greater than the proper fraction involved in the multiplication.
Check it for × × .
### Exercise 2.3
1. Find:
(i) of (a) (b) (c)
(ii) of (a) (b) (c)
2. Multiply and reduce to lowest form (if possible) :
(i) (ii) (iii) (iv)
(v) (vi) (vii)
3. Multiply the following fractions:
(i) (ii) (iii) (iv)
(v) (vi) (vii)
4. Which is greater:
(i) of or of (ii) of or of
5. Saili plants 4 saplings, in a row, in her garden. The distance between two adjacent saplings is m. Find the distance between the first and the last sapling.
6. Lipika reads a book for hours everyday. She reads the entire book in 6 days. How many hours in all were required by her to read the book?
7. A car runs 16 km using 1 litre of petrol. How much distance will it cover using litres of petrol.
8. (a) (i) Provide the number in the box , such that .
(ii) The simplest form of the number obtained in is _____.
(b) (i) Provide the number in the box , such that .
(ii) The simplest form of the number obtained in is _____.
### 2.4 Division of Fractions
John has a paper strip of length 6 cm. He cuts this strip in smaller strips of length 2 cm each. You know that he would get 6 ÷ 2 =3 strips.
John cuts another strip of length 6 cm into smaller strips of length cm each. How many strips will he get now? He will get 6 ÷ strips.
A paper strip of length cm can be cut into smaller strips of length cm each to give ÷ pieces.
So, we are required to divide a whole number by a fraction or a fraction by another fraction. Let us see how to do that.
#### 2.4.1 Division of Whole Number by a Fraction
Let us find 1÷.
We divide a whole into a number of equal parts such that each part is half of the whole. The number of such half () parts would be 1÷. Observe the figure (Fig 2.11). How many half parts do you see?
Fig 2.11
There are two half parts.
So, 1 ÷ = 2. Also, = 1 × 2 = 2. Thus, 1 ÷ = 1 ×
Similarly, 3 ÷ = number of parts obtained when each of the 3 whole, are divided into equal parts = 12 (From Fig 2.12)
Fig 2.12
Observe also that,
Find in a similar way, 3 ÷ and .
Reciprocal of a fraction
The number can be obtained by interchanging the numerator and denominator of or by inverting . Similarly, is obtained by inverting .
Let us first see about the inverting of such numbers.
Observe these products and fill in the blanks :
Multiply five more such pairs.
The non-zero numbers whose product with each other is 1, are called the reciprocals of each other. So reciprocal of is and the reciprocal of is . What is the receiprocal of ?
You will see that the reciprocal of is obtained by inverting it. You get .
### Think, Discuss and Write
(i) Will the reciprocal of a proper fraction be again a proper fraction?
(ii) Will the reciprocal of an improper fraction be again an improper fraction?
Therefore, we can say that
1 ÷ = = 1× reciprocal of .
3 ÷ = = 3× reciprocal of .
3 ÷ = ------ = ----------------------.
So, 2 ÷ = 2 × reciprocal of = .
5 ÷ = 5 × ------------------- = 5 × -------------
Thus, to divide a whole number by any fraction, multiply that whole number by the reciprocal of that fraction.
### Try These
Find : (i) 7 ÷ (ii) 6 ÷ (iii) 2 ÷
• While dividing a whole number by a mixed fraction, first convert the mixed fraction into improper fraction and then solve it.
Thus, 4 ÷ = 4 ÷ = ? Also, 5 ÷ 3 = 3 ÷ = ?
### Try These
Find: (i) 6 ÷
(ii) 7 ÷
#### 2.4.2 Division of a Fraction by a Whole Number
What will be ÷ 3?
Based on our earlier observations we have: ÷ 3 = ÷ × = =
So, ÷ 7 = × = ? What is ÷ 6 , ÷ 8 ?
While dividing mixed fractions by whole numbers, convert the mixed fractions into improper fractions. That is,
#### 2.4.3 Division of a Fraction by Another Fraction
We can now find ÷.
÷ = × reciprocal of = × = .
Similarly, × reciprocal of = ? and, ÷ = ?
### Try These
Find: (i) (ii) (iii) (iv)
### Exercise 2.4
1. Find:
(i) (ii) (iii) (iv) (v) (vi)
2. Find the reciprocal of each of the following fractions. Classify the reciprocals as proper fractions, improper fractions and whole numbers.
(i) (ii) (iii) (iv) (v) (vi) (vii)
3. Find:
(i) (ii) (iii) (iv) (v) (vi)
4. Find:
(i) (ii) (iii) (iv) (v) (vi) (vii) (viii)
### 2.5 How Well have you Learnt about Decimal Numbers
You have learnt about decimal numbers in the earlier classes. Let us briefly recall them here. Look at the following table and fill up the blank spaces.
In the table, you wrote the decimal number, given its place-value expansion. You can do the reverse, too. That is, given the number you can write its expanded form. For example,
John has15.50 and Salma has ₹ 15.75. Who has more money? To find this we need to compare the decimal numbers 15.50 and 15.75. To do this, we first compare the digits on the left of the decimal point, starting from the leftmost digit. Here both the digits 1 and 5, to the left of the decimal point, are same. So we compare the digits on the right of the decimal point starting from the tenths place. We find that 5 < 7, so we say
15.50 < 15.75. Thus, Salma has more money than John.
If the digits at the tenths place are also same then compare the digits at the hundredths place and so on.
Now compare quickly, 35.63 and 35.67; 20.1 and 20.01; 19.36 and 29.36.
While converting lower units of money, length and weight, to their higher units, we are required to use decimals. For example,
= 0.005 kg, 7 cm = 0.07 m.
Write 75 paise = ₹ ______, 250 g = _____ kg, 85 cm = _____m.
We also know how to add and subtract decimals. Thus, 21.36 + 37.35 is
What is the value of 0.19 + 2.3 ?
The difference 29.35 4.56 is
Tell the value of 39.87 21.98.
### Exercise 2.5
1. Which is greater?
(i) 0.5 or 0.05 (ii) 0.7 or 0.5 (iii) 7 or 0.7
(iv) 1.37 or 1.49 (v) 2.03 or 2.30 (vi) 0.8 or 0.88.
2. Express as rupees using decimals :
(i) 7 paise (ii) 7 rupees 7 paise (iii) 77 rupees 77 paise
(iv) 50 paise (v) 235 paise.
3. (i) Express 5 cm in metre and kilometre (ii) Express 35 mm in cm, m and km
4. Express in kg:
(i) 200 g (ii) 3470 g (iii) 4 kg 8 g
5. Write the following decimal numbers in the expanded form:
(i) 20.03 (ii) 2.03 (iii) 200.03 (iv) 2.034
6. Write the place value of 2 in the following decimal numbers:
(i) 2.56 (ii) 21.37 (iii) 10.25 (iv) 9.42 (v) 63.352.
7. Dinesh went from place A to place B and from there to place C. A is 7.5 km from B and B is 12.7 km from C. Ayub went from place A to place D and from there to place C. D is 9.3 km from A and C is 11.8 km from D. Who travelled more and by how much?
8. Shyama bought 5 kg 300 g apples and 3 kg 250 g mangoes. Sarala bought 4 kg 800 g oranges and 4 kg 150 g bananas. Who bought more fruits?
9. How much less is 28 km than 42.6 km?
### 2.6 Multiplication of Decimal Numbers
Reshma purchased 1.5kg vegetable at the rate of ` 8.50 per kg. How much money should she pay? Certainly it would be ` (8.50 × 1.50). Both 8.5 and 1.5 are decimal numbers. So, we have come across a situation where we need to know how to multiply two decimals. Let us now learn the multiplication of two decimal numbers.
First we find 0.1 × 0.1.
Let us see it’s pictorial representation (Fig 2.13)
The fraction represents 1 part out of 10 equal parts.
The shaded part in the picture represents .
We know that,
means of . So, divide this th part into 10 equal parts and take one part out of it.
Thus, we have, (Fig 2.14).
Fig 2.14
The dotted square is one part out of 10 of the th part. That is, it represents or 0.1 × 0.1.
Can the dotted square be represented in some other way?
How many small squares do you find in Fig 2.14?
There are 100 small squares. So the dotted square represents one out of 100 or 0.01.
Hence, 0.1 × 0.1 = 0.01.
Note that 0.1 occurs two times in the product. In 0.1 there is one digit to the right of the decimal point. In 0.01 there are two digits (i.e., 1 + 1) to the right of the decimal point.
Let us now find 0.2 × 0.3.
We have, 0.2 × 0.3 =
As we did for , let us divide the square into 10 equal parts and take three parts out of it, to get . Again divide each of these three equal parts into 10 equal parts and take two from each. We get .
The dotted squares represent or 0.2 × 0.3. (Fig 2.15)
Since there are 6 dotted squares out of 100, so they also reprsent 0.06.
Thus, 0.2 × 0.3 = 0.06.
Observe that 2 × 3 = 6 and the number of digits to the right of the decimal point in 0.06 is 2 (= 1 + 1).
Check whether this applies to 0.1 × 0.1 also.
Find 0.2 × 0.4 by applying these observations.
While finding 0.1 × 0.1 and 0.2 × 0.3, you might have noticed that first we multiplied them as whole numbers ignoring the decimal point. In 0.1 × 0.1, we found 01 × 01 or 1 × 1. Similarly in 0.2 × 0.3 we found 02 × 03 or 2 × 3.
Then, we counted the number of digits starting from the rightmost digit and moved towards left. We then put the decimal point there. The number of digits to be counted is obtained by adding the number of digits to the right of the decimal point in the decimal numbers that are being multiplied.
Let us now find 1.2 × 2.5.
Multiply 12 and 25. We get 300. Both, in 1.2 and 2.5, there is 1 digit to the right of the decimal point. So, count 1 + 1 = 2 digits from the rightmost digit (i.e., 0) in 300 and move towards left. We get 3.00 or 3.
Find in a similar way 1.5 × 1.6, 2.4 × 4.2.
While multiplying 2.5 and 1.25, you will first multiply 25 and 125. For placing the decimal in the product obtained, you will count 1 + 2 = 3 (Why?) digits starting from the rightmost digit. Thus, 2.5 × 1.25 = 3.225
### Try These
Find 2.7 × 1.35.
1. Find: (i) 2.7 × 4 (ii) 1.8 × 1.2 (iii) 2.3 × 4.35
2. Arrange the products obtained in (1) in descending order.
Example 7 The side of an equilateral triangle is 3.5 cm. Find its perimeter.
Solution All the sides of an equilateral triangle are equal.
So, length of each side = 3.5 cm
Thus, perimeter = 3 × 3.5 cm = 10.5 cm
Example 8 The length of a rectangle is 7.1 cm and its breadth is 2.5 cm. What is the area of the rectangle?
Solution Length of the rectangle = 7.1 cm
Breadth of the rectangle = 2.5 cm
Therefore, area of the rectangle = 7.1 × 2.5 cm2 = 17.75 cm2
#### 2.6.1 Multiplication ofDecimal Numbers by 10, 100 and 1000
Reshma observed that 2.3 = whereas 2.35 = . Thus, she found that depending on the position of the decimal point the decimal number can be converted to a fraction with denominator 10 or 100. She wondered what would happen if a decimal number is multiplied by 10 or 100 or 1000.
Let us see if we can find a pattern of multiplying numbers by 10 or 100 or 1000.
Have a look at the table given below and fill in the blanks:
Observe the shift of the decimal point of the products in the table. Here the numbers are multiplied by 10,100 and 1000. In 1.76 × 10 = 17.6, the digits are same i.e., 1, 7 and 6. Do you observe this in other products also? Observe 1.76 and 17.6. To which side has the decimal point shifted, right or left? The decimal point has shifted to the right by one place. Note that 10 has one zero over 1.
In 1.76×100 = 176.0, observe 1.76 and 176.0. To which side and by how many digits has the decimal point shifted? The decimal point has shifted to the right by two places.
Note that 100 has two zeros over one.
Do you observe similar shifting of decimal point in other products also?
So we say, when a decimal number is multiplied by 10, 100 or 1000, the digits in the product are same as in the decimal number but the decimal point in the product is shifted to the right by as, many of places as there are zeros over one.
Based on these observations we can now say
0.07 × 10 = 0.7, 0.07 × 100 = 7 and 0.07 × 1000 = 70.
Can you now tell 2.97 × 10 = ? 2.97 × 100 = ? 2.97 × 1000 = ?
Can you now help Reshma to find the total amount i.e., ₹ 8.50 × 150, that she has to pay?
### Exercise 2.6
1. Find:
(i) 0.2 × 6 (ii) 8 × 4.6 (iii) 2.71 × 5 (iv) 20.1 × 4
(v) 0.05 × 7 (vi) 211.02 × 4 (vii) 2 × 0.86
2. Find the area of rectangle whose length is 5.7cm and breadth is 3 cm.
3. Find:
(i) 1.3 × 10 (ii) 36.8 × 10 (iii) 153.7 × 10 (iv) 168.07 × 10 (v) 31.1 × 100 (vi) 156.1 × 100
(vii) 3.62 × 100 (viii) 43.07 × 100
(ix) 0.5 × 10 (x) 0.08 × 10 (xi) 0.9 × 100 (xii) 0.03 × 1000
4. A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover in 10 litres of petrol?
5. Find:
(i) 2.5 × 0.3 (ii) 0.1 × 51.7 (iii) 0.2 × 316.8 (iv) 1.3 × 3.1
(v) 0.5 × 0.05 (vi) 11.2 × 0.15 (vii) 1.07 × 0.02
(viii) 10.05 × 1.05 (ix) 101.01 × 0.01 (x) 100.01 × 1.1
### 2.7 Division of Decimal Numbers
Savita was preparing a design to decorate her classroom. She needed a few coloured strips of paper of length 1.9 cm each. She had a strip of coloured paper of length 9.5 cm. How many pieces of the required length will she get out of this strip? She thought it would be cm. Is she correct?
Both 9.5 and 1.9 are decimal numbers. So we need to know the division of decimal numbers too!
#### 2.7.1 Division by 10, 100 and 1000
Let us find the division of a decimal number by 10, 100 and 1000.
Consider 31.5 ÷ 10.
Let us see if we can find a pattern for dividing numbers by 10, 100 or 1000. This may help us in dividing numbers by 10, 100 or 1000 in a shorter way.
Take 31.5 ÷ 10 = 3.15. In 31.5 and 3.15, the digits are same i.e., 3, 1, and 5 but the decimal point has shifted in the quotient. To which side and by how many digits? The decimal point has shifted to the left by one place. Note that 10 has one zero over 1.
Consider now 31.5 ÷ 100 = 0.315. In 31.5 and 0.315 the digits are same, but what about the decimal point in the quotient? It has shifted to the left by two places. Note that 100 has two zeros over1. Using this observation let us now quickly find:
### Try These
Find: (i) 235.4 ÷ 10
(ii) 235.4 ÷100
(iii) 235.4 ÷ 1000
So we can say that, while dividing a number by 10, 100 or 1000, the digits of the number and the quotient are same but the decimal point in the quotient shifts to the left by as many places as there are zeros over 1. Using this observation let us now quickly find: 2.38 ÷ 10 = 0.238, 2.38 ÷ 100 = 0.0238, 2.38 ÷ 1000 = 0.00238
#### 2.7.2 Division of a Decimal Number by a Whole Number
Let us find. Remember we also write it as 6.4 ÷ 2.
### Try These
(i) 35.7 ÷ 3 = ?;
(ii) 25.5 ÷ 3 = ?
Or, let us first divide 64 by 2. We get 32. There is one digit to the right of the decimal point in 6.4. Place the decimal in 32 such that there would be one digit to its right. We get 3.2 again.
To find 19.5 ÷ 5, first find 195 ÷5. We get 39. There is one digit to the right of the decimal point in 19.5. Place the decimal point in 39 such that there would be one digit to its right. You will get 3.9.
### Try These
(i) 43.15 ÷ 5 = ?;
(ii) 82.44 ÷ 6 = ?
Or, divide 1296 by 4. You get 324. There are two digits to the right of the decimal in 12.96. Making similar placement of the decimal in 324, you will get 3.24.
Note that here and in the next section, we have considered only those divisions in which, ignoring the decimal, the number would be completely divisible by another number to give remainder zero. Like, in 19.5 ÷ 5, the number 195 when divided by 5, leaves remainder zero.
### Try These
Find: (i) 15.5 ÷ 5
(ii) 126.35 ÷ 7
However, there are situations in which the number may not be completely divisible by another number, i.e., we may not get remainder zero. For example, 195 ÷ 7. We deal with such situations in later classes.
Example 9 Find the average of 4.2, 3.8 and 7.6.
Solution The average of 4.2, 3.8 and 7.6 is = 5.2.
#### 2.7.3 Division of a Decimal Number by another Decimal Number
Let us find i.e., 25.5 ÷ 0.5.
We have Thus, 25.5 ÷ 0.5 = 51
What do you observe? For , we find that there is one digit to the right of the decimal in 0.5. This could be converted to whole number by dividing by 10. Accordingly 25.5 was also converted to a fraction by dividing by 10.
Or, we say the decimal point was shifted by one place to the right in 0.5 to make it 5. So, there was a shift of one decimal point to the right in 25.5 also to make it 255.
Thus, 22.5 ÷ 1.5 = = = 15
Find and in a similar way.
### Try These
Find: (i) (ii) (iii)
Let us now find 20.55 ÷ 1.5.
We can write it is as 205.5 ÷ 15, as discussed above. We get 13.7. Find .
Consider now, . We can write it as (How?) and we get the quotient as 134.9. How will you find ? We know that 27 can be written as 27.00.
So, = 900
Example 10 Each side of a regular polygon is 2.5 cm in length. The perimeter of the polygon is 12.5cm. How many sides does the polygon have?
Solution The perimeter of a regular polygon is the sum of the lengths of all its equal sides = 12.5 cm.
Length of each side = 2.5 cm. Thus, the number of sides = = 5
The polygon has 5 sides.
Example 1 A car covers a distance of 89.1 km in 2.2 hours. What is the average distance covered by it in 1 hour?
Solution Distance covered by the car = 89.1 km.
Time required to cover this distance = 2.2 hours.
So distance covered by it in 1 hour = = = 40.5 km.
### Exercise 2.7
1. Find:
(i) 0.4 ÷ 2 (ii) 0.35 ÷ 5 (iii) 2.48 ÷ 4 (iv) 65.4 ÷ 6
(v) 651.2 ÷ 4 (vi) 14.49 ÷ 7 (vii) 3.96 ÷ 4 (viii) 0.80 ÷ 5
2. Find:
(i) 4.8 ÷ 10 (ii) 52.5 ÷ 10 (iii) 0.7 ÷ 10
(iv) 33.1 ÷ 10 (v) 272.23 ÷ 10 (vi) 0.56 ÷ 10 (vii) 3.97 ÷10
3. Find:
(i) 2.7 ÷ 100 (ii) 0.3 ÷ 100 (iii) 0.78 ÷ 100
(iv) 432.6 ÷ 100 (v) 23.6 ÷100 (vi) 98.53 ÷ 100
4. Find:
(i) 7.9 ÷ 1000 (ii) 26.3 ÷ 1000 (iii) 38.53 ÷ 1000
(iv) 128.9 ÷ 1000 (v) 0.5 ÷ 1000
5. Find:
(i) 7 ÷ 3.5 (ii) 36 ÷ 0.2 (iii) 3.25 ÷ 0.5
(iv) 30.94 ÷ 0.7 (v) 0.5 ÷ 0.25 (vi) 7.75 ÷ 0.25 (vii) 76.5 ÷ 0.15 (viii) 37.8 ÷ 1.4 (ix) 2.73 ÷ 1.3 (ix) 2.73 ÷ 1.3
6. A vehicle covers a distance of 43.2 km in 2.4 litres of petrol. How much distance will it cover in one litre of petrol?
### What have We Discussed?
1. We have learnt about fractions and decimals alongwith the operations of addition and subtraction on them, in the earlier class.
2. We now study the operations of multiplication and division on fractions as well as on decimals.
3. We have learnt how to multiply fractions. Two fractions are multiplied by multiplying their numerators and denominators seperately and writing the product as .
4. A fraction acts as an operator ‘of’. For example, of 2 is × 2 = 1.
5. (a) The product of two proper fractions is less than each of the fractions that are multiplied.
(b) The product of a proper and an improper fraction is less than the improper fraction and greater than the proper fraction.
(c) The product of two imporper fractions is greater than the two fractions.
6. A reciprocal of a fraction is obtained by inverting it upside down.
7. We have seen how to divide two fractions.
(a) While dividing a whole number by a fraction, we multiply the whole number with the reciprocal of that fraction.
For example,
(b) While dividing a fraction by a whole number we multiply the fraction by the reciprocal of the whole number.
For example,
(c) While dividing one fraction by another fraction, we multuiply the first fraction by the reciprocal of the other. So,
8. We also learnt how to multiply two decimal numbers. While multiplying two decimal numbers, first multiply them as whole numbers. Count the number of digits to the right of the decimal point in both the decimal numbers. Add the number of digits counted. Put the decimal point in the product by counting the digits from its rightmost place. The count should be the sum obtained earlier.
For example, 0.5 × 0.7 = 0.35
9. To multiply a decimal number by 10, 100 or 1000, we move the decimal point in the number to the right by as many places as there are zeros over 1.
Thus 0.53 × 10 = 5.3, 0.53 × 100 = 53, 0.53 × 1000 = 530
10. We have seen how to divide decimal numbers.
(a) To divide a decimal number by a whole number, we first divide them as whole numbers. Then place the decimal point in the quotient as in the decimal number.
For example, 8.4 ÷ 4 = 2.1
Note that here we consider only those divisions in which the remainder is zero.
(b) To divide a decimal number by 10, 100 or 1000, shift the digits in the decimal number to the left by as many places as there are zeros over 1, to get the quotient.
So, 23.9 ÷ 10 = 2.39, 23.9 ÷ 100 = 0 .239, 23.9 ÷ 1000 = 0.0239
(c) While dividing two decimal numbers, first shift the decimal point to the right by equal number of places in both, to convert the divisor to a whole number. Then divide. Thus, 2.4 ÷ 0.2 = 24 ÷ 2 = 12.
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Jun 18, 2023
Introduction
Calculating percentages is a fundamental skill that has a wide range of applications in daily life, from calculating discounts and marking grades to analyzing data and making financial decisions. While the concept of a percentage is simple, many people struggle with finding percentages of numbers due to a lack of understanding of the process involved. This article aims to guide readers through the steps involved in finding a percentage of a number, as well as offer tips and insights to make the process easier.
What is a Percentage?
A percentage is a fraction of a whole expressed as a fraction of 100. For example, 25% means 25 parts out of 100 or 0.25 as a decimal. Percentages are used in multiple fields, including science, business, and finance. In real life, percentages are used to calculate discounts, sales tax, tips, incentives, and more.
Step-by-Step Guide on Finding a Percentage of a Number
Finding the percentage of a number involves simple arithmetic operations. Here is a step-by-step guide:
1. Write the percentage as a fraction or decimal.
2. Multiply the number by the fraction or decimal.
3. The result is the percentage of the number.
For example, to find 30% of 500:
1. 30% = 0.3 (as a decimal).
2. 500 x 0.3 = 150.
3. Therefore, 30% of 500 is 150.
Different Methods for Calculating Percentages
There are various methods for calculating percentages, including:
1. Long division: This method involves dividing the number by 100 and then multiplying the quotient with the percentage.
2. Using a calculator: Most calculators have a percentage button that can be used to calculate the percentage of a number.
3. Estimation: This method involves rounding off the number to make the percentage easier to calculate mentally.
While each method has its benefits, long division is the most accurate method. Still, using a calculator may be more convenient for complex calculations. Estimation can be useful when making quick mental calculations, but may be less accurate.
Tips and Tricks for Quickly Finding Percentages Mentally
Here are some mental calculation techniques for quickly finding percentages:
1. Multiplying by 10: To find 10% of a number, move the decimal point one place to the left. To find 20%, double the 10% value. For example, 10% of 150 is 15, so 20% of 150 is 30.
2. Using multiples of 5: To find 5%, divide the number by 20. To find 25%, divide the number by 4. For example, 5% of 400 is 20 and 25% of 400 is 100.
3. Using proportionality: To find a percentage of a number that is not a multiple of 10, find a proportionate amount that is a multiple of 10 and adjust accordingly. For example, to find 15% of 200, find 10% (20) and half it to get 5%, then add 10% and 5% to arrive at 15% (30).
Common Mistakes and How to Avoid Them
Common mistakes include forgetting to convert the percentage to a decimal or fraction, using the wrong arithmetic operation, or confusing the percentage with the result. To avoid errors, double-check your work and use the correct order of operations.
Practice Problems and Quiz
Here are some practice problems to test your skills:
1. What is 40% of 200?
2. If 25% of a number is 250, what is the number?
3. What is 75% of 120?
Quiz:
1. What does the term percentage mean?
2. What is the formula for finding a percentage of a number?
3. What is the most accurate method for finding percentages?
4. How do you find 20% of 150?
5. What is a common error when finding percentages?
Conclusion
Calculating percentages is an essential skill that is used in many aspects of daily life. By following the simple steps outlined in this article, readers should now have a solid grasp of how to find percentages of numbers. With practice and familiarity, finding percentages will become an intuitive and easy process that can be done quickly and accurately.
By Riddle Reviewer
Hi, I'm Riddle Reviewer. I curate fascinating insights across fields in this blog, hoping to illuminate and inspire. Join me on this journey of discovery as we explore the wonders of the world together.
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## What is the Taylor series for sin?
sin�(x) = cos(x) sin��(x) = − sin(x) sin���(x) = − cos(x) sin(4)(x) = sin(x). Since sin(4)(x) = sin(x), this pattern will repeat. Next we need to evaluate the function and its derivatives at 0: sin(0) = 0 sin�(0) = 1 sin��(0) = 0 sin���(0) = −1 sin(4)(0) = 0.
How do you do error bounds?
To find the error bound, find the difference of the upper bound of the interval and the mean. If you do not know the sample mean, you can find the error bound by calculating half the difference of the upper and lower bounds.
### What is the expansion of the function Sinx?
The Maclaurin expansion of sinx is given by Sinx=x1!
How do you find the possible error?
How to Determine Greatest Possible Error
1. Step1: Identify the last nonzero digit to the right of the decimal point in the given measurement.
2. Step 2: Determine the precision.
3. Step 3: Divide the precision by 2 to determine the greatest possible error.
#### How do you find the error in calculus?
In these problems, we’ll typically take a derivative, and use the “dx” or “dy” part of the derivative as the error. Then, to get percent error, we’ll divide the error by the total amount and multiply by 100.
What is possible error?
The greatest possible error (GPE) is the largest amount a ballpark figure can miss the mark. It’s one half of the measuring unit you are using. For example: If measuring in centimeters, the GPE is 1/2 cm. …or in liters, the GPE is 1/2 L.
## What is the formula for maximum error?
The overall maximum error in any volume measured always comes from two measurements; Measurement 1 is the reading we take when we fill it to zero. Measurement 2 is the reading we take when we have let some out. Therefore the overall maximum error = 2 x 0.05 cm3 = 0.1 cm3.
How do you calculate approximation error?
Suppose a numerical value v is first approximated as x, and then is subsequently approximated by y. Then the approximate error, denoted Ea, in approximating v as y is defined as Ea = x − y. Similarly, the relative approximate error, denoted ϵa, is defined as ϵa = (x − y)/x = 1 − y/x.
### What are error bounds in calculus?
The expression means the maximum absolute value of the (n + 1) derivative on the interval between the value of x and c. The corollary says that this number is larger than the amount we need to add (or subtract) from our estimate to make it exact. This is the bound on the error.
How to find the error function of the Taylor series?
The error function is defined by e r f ( x) := 2 π ∫ 0 x e − t 2 d t. Find its Taylor expansion. f ( x) = ∑ n = 0 ∞ f ( n) ( a) n! ( x − a) n. However, the question doesn’t give a point a with which to center the Taylor series.
#### How does the Taylor series for sin x work?
These terms have the form of a power of x multiplied by a coefficient. When the terms in the series are added together, we can approximate a function at a specific value of x, provided the value lies within the interval of convergence for the function. We won’t show this, but the Taylor series for sin ( x) works for all values of x.
What is the error bound of the Taylor expansion?
This error bound (Rn (x)) is the maximum value of the (n+1)th term of the Taylor expansion, where M is an upper bound of the (n+1)th derivative for a < z < x.
## What is the Lagrange error bound of Taylor polynomial?
July Thomas contributed. The Lagrange error bound of a Taylor polynomial gives the worst case scenario for the difference between the estimated value of the function as provided by the Taylor polynomial and the actual value of the function.
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# Difference between revisions of "Spiral Explorations"
Alternative Fibonacci Spiral
Field: Geometry
Image Created By: [[Author:| ]]
Alternative Fibonacci Spiral
The picture on the right demonstrates a way of constructing the Fibonacci Spiral (explained below) using arcs. If you are not familiar with the fibonacci sequence, you can click here. Instead of using the squares to determine arcs, this uses circles to determine the arcs (their position and size) that make up the spiral. The traditional squares are overlaid to demonstrate the pattern: the ratio between circle sizes matches the ratio between square sizes.
# Basic Description
The basic definition of a spiral is a curve that winds in a gradually widening pattern. In this page, we are going more in depth on what a spiral is—what exactly causes a spiral to be defined as a spiral—and how to construct spirals. We are also exploring the relationships between different types of spirals. Each spiral starts with a circular pattern: one circle is tangent to a second slightly larger circle, then the second circle is tangent to a third circle, and so on. The spiral is made up of parts of each circle; each circle contains one arc that contributes to the spiral. One arc is the part of the circle from one tangent point to another tangent point. For example, one part of the spiral would be the arc from the point where the first circle is tangent with the second circle, to the point where the second circle is tangent with the third circle., and so on. Each arc is bigger than the previous arc so any given point is farther away from the center than the previous point. Together, the arcs form the spiral.
Shown above is a Fibonacci Spiral, in green, constructed from arcs on circles. The squares are included to show how parts of circles are used in spirals. The Fibonacci Spiral is created using arcs which are 1/4 of their respective circles. The radii of the circles fit the Fibonacci Sequence with the smallest circle having a radius of 1, the next largest being 2, then 3, then 5, and so on. Each circle has a radius along the length of one of the squares. The centers of the circles are rotated 90 degrees from each other at a distance that also fits the Fibonacci Sequence. That is, the center of the smallest circle is rotated 90 degrees and translated 1 unit away. Then, the next center is rotated 90 degrees and translated 2 units away. This pattern continues. This way, the circles are tangent to each other, and arcs become visible:
Shown above, on the left, the centers (shown in red) are rotated counterclockwise 90 degrees at a distance determined by the Fibonacci Sequence (1, 1, 2, 3, 5, 8, etc. units).
In the middle, circles were constructed around each of the points with radii starting at the second number in the Fibonacci Sequence (1, 3, 3, 5, 8, 13, etc. units).
On the right, the Fibonacci Spiral is constructed, shown in thick green. It is important to determine which direction the figure should spiral. The spiral shown above spirals counterclockwise away from the center. Each arc on the circles is 1/4 of the circle's circumference except the smallest arc, which is 1/2. The smallest arc is 1/2 of the circumference to clearly mark the beginning of the spiral. Any spiral where the radii are based on a predetermined set of numbers that start at one specific, unchangeable number and that get progressively bigger have beginnings. For example, the Fibonacci Spiral and the Archimedean Spiral start at a predetermined number—1 and $\sqrt {1}$ respectively—and get progressively bigger. However, spirals where the radii of the circles follow a pattern like the spiral shown below do not have a constant beginning. These spirals are constructed of arcs on circles where the ratio between the radii of the first (smallest) circle and the second circle is 1:2. As the spiral continues, the pattern continues; the radius of one circle is doubled to get the radius of the next circle. These spirals have no constant beginnings because the measurement of the smallest radius can simply be halved, resulting in a smaller arc.
The 1:2 spiral may look similar to the Fibonacci Spiral, but as both spirals get bigger, the differences between the radii increase, causing the spirals' shapes and sizes to differ. From the tenth to the seventeenth iteration, the Fibonacci Spiral follows the pattern 55; 89; 144; 233; 377; 610; 987; and 1,597. The spiral with the 1:2 ratio, with the first iteration of 1 unit, follows the pattern 512; 1,024; 2,048; 4,096; 8,192; 16,384; 32768; and 65,536. Obviously, the spiral with the 1:2 ratio is much larger than the Fibonacci Spiral. But what does this mean? How can it be measured?
It is possible to compare spirals based on the areas of their arc sectors and the lengths of the arcs that make up the spiral.
## Spiral Centers
Ratio spirals have no constant beginnings, but it is important to realize that they still have a center: a point that the spiral continuously travels away from. Similarly to the spirals with constant beginnings (the Fibonacci Spiral and the Archimedean Spiral), the center of this type of spiral is located at the first point on the spiral, the beginning, which is not constant for all spirals, but is present on all spirals. For example, think of a spiral with a ratio of the radii of the circles of 1:2. This first iteration of this spiral could have the smallest radius measuring 1 unit; however, it could also have a smallest radius of 1/2 units, or 1/4, or 1/8, etc. The location of the center is different in each of those cases, but the rule for finding the center of a spiral is as follows: It is the first point on the first arc of the spiral.
The Fibonacci Spiral is on the left, and a spiral with a 1:2 ratio is on the right. The orange segments connect the centers (in bright green) to random points on the spiral. The orange segments get progressively longer, demonstrating that spirals originate from this exact point. However, this one point is only one example of a center of a spiral. All spirals can have more that one center from which all points get father away. The easiest example of this is the Archimedean Spiral: all of the triangles meet at one vertex. Every point on the spiral gets progressively farther away from this point. Shown below, a 1:2 ratio spiral demonstrates the fact that all spirals have more than one center:
In the image, point G is a center. The requirement for all centers of spirals are as follows: the center must be closer to the first point on the first arc of the spiral, point D, than any other point on the spiral, for example, points E and I. To ensure that a center is closer to point D than any other point, one can draw a triangle connecting the three points. As long as the measure of angle GDE (or GDI) is greater than the measure of angle DEG (or DIG). The measure of segment GD must be greater than the measure of segment GE (or GI).
This is one type of center that all spirals have. The other type of center is the point around which the spiral revolves, which we will refer to as a rotational center. This point is the center of the circle used in creating the smallest arc on the spiral. The spiral as a whole appears to 'spiral' away from this one point. In the image above, point J is the rotational center.
The spiral in bold is the beginning of the well-known Archimedean Spiral. The pattern of the spiral is determined by corresponding points on right triangles that have a relationship to each other in the lengths of their hypotenuses, which is: $\sqrt {1}$, $\sqrt {2}$, $\sqrt {3}$, $\sqrt {4}$...etc. All of the right triangles meet at a point—a center of the spiral. The points on the spiral continuously get farther away from this point. The rotational center of the spiral is the uppermost small point. It is the center of the smallest circle.
# A More Mathematical Explanation
## Exploring 180° Rotation
Both the Fibonacci Spiral and the 1:2 ratio spiral have centers that are [...]
## Exploring 180° Rotation
Both the Fibonacci Spiral and the 1:2 ratio spiral have centers that are rotated 90° and then translated. We explored what would happen if we rotated the centers by 180 degrees (the centers would be on a straight line) but still translated using the previous pattern (the Fibonacci Sequence and the 1:2 pattern). The outcome is the following picture:
This spiral looks "tighter" than the Fibonacci Spiral; the spiral is more "controlled." The arcs that create the spiral now are 1/2 of the individual circle instead of 1/4 because the circles are tangent to each other in different places. When the centers are rotated 180°, they all lie upon a line. This line connects all of the centers and all of the tangent points of the circles. Because this line intersects each circle twice, there are only two possible places for two circles to be tangent to each other. In a spiral, three circles cannot be tangent in the same place, so the tangent points must alternate. When the line that connects the centers and tangent points is vertical, the alternation can be described like this: image the spiral on a coordinate grid. If a circle is tangent to "the bottom" (a point lower on the y-axis) of a smaller circle, then it must be tangent to a bigger circle at a point higher on the y-axis. Because of this alternation, the arcs make up 1/2 of their individual circles.
Next, instead of using the Fibonacci Sequence, we tried a 1:2 ratio. To summarize: the circle's centers are rotated 180 degrees from each other, or in other words, they are on a straight line. The radii of the circles are dependent on a 1:2 ratio. The circles are tangent to each other, so the distance between the centers is doubled each time another circle is created. Each bold arc is half of its individual circle instead of 1/4:
## Spiral Lengths and Areas
Spirals can be compared in terms of size and shape, but they can also be compared using "spiral area" and "spiral length."
### Spiral Length
Technically, it is impossible to find the length of a spiral because spirals go on forever, so the length would therefore be infinite. However, it is possible to find the length for iterations of the spiral. We came up with a formula to estimate the length of a spiral to a certain iteration. Each iteration would be added to find the desired spiral length after iteration n.
Note: These formulas serve as approximations because we are adding arcs at discrete (non-moving) intervals. Realistically, the radii increase gradually as the spiral progresses, not jumping from integer to integer, as is assumed here.
This formula works with all spirals that follow an iterative pattern; in other words, all spirals in which the increase from one circle to the next can be described using a ratio (so it does not work with the Fibonacci Sequence or the Archimedean Spiral, for example).
The formula for the spiral length estimate is as follows:
Spiral Length=x*[(absolute value of (180-a))/180]*π*[1+r+(r^2)+(r^3)...+(r^(n-1)]
x=radius of the first (smallest) circle
a=angle/degree in which the centers of the circles are rotated
r=ratio of increasing circles (written as an improper fraction, with the larger number in the numerator)
n=number of iterations
How we found this formula/why it works:
Basically, the spiral length is equal to the sum of the lengths of the arcs that make up that spiral. An arc is simply a fraction of the circumference of a circle. Therefore, the length of an arc is the circumference (2π*radius) multiplied by (number of degrees of the arc/360). If you look at the spirals on our page, you will notice that each arc takes up the following fraction of the entire circle: (the absolute value of (180-a)/360, with "a" equaling the angle in which the centers of the circles are rotated. Assuming the radius of the first circle is 1, the length of the first arc would be [(absolute value of (180-a))/360]*2π*1, which simplifies to [(absolute value of (180-a))/180]*π. The second arc would be the length of the first arc multiplied by the ratio in which the circles increase: [(absolute value of (180-a))/180]*π*r. The third arc would be the length of the second arc multiplied again by the ratio: [(absolute value of (180-a))/180]*π*r*r, or [(absolute value of (180-a))/180]*π*r^2. The pattern continues like this until the last arc: [(absolute value of (180-a))/180]*π*r^(n-1). Add all these together, and you get: {[(absolute value of (180-a))/180]*π} + {[(absolute value of (180-a))/180]*π*r} + {[(absolute value of (180-a))/180]*π*r^2}... + {[(absolute value of (180-a))/180]*π*r^(n-1)}. Because of the distributive property, you can simplify this to the following: [(absolute value of (180-a))/180]*π*[1+(r^2)+(r^2)+(r^(n-1))]. However, this equation only works if the radius of the first/smallest circle is equal to 1. If the radius of the first circle is any number other than one, this changes the length of the spiral. So we added "x" into the equation ("x" equals the radius of the first/smallest circle). If the radius of the first circle is any number other than one, you simply multiply the entire equation by that number. Therefore, the formula for spiral length is: x*[(absolute value of (180-a))/180]*π*[1+r+(r^2)+(r^3)...+(r^(n-1)]
Note: the formula explained above works with almost all spirals that follow a repetitive pattern (where the increase from one circle to the next can be explained using a ratio), but there is one exception. When the centers of the circles that form the spiral are rotated by 180 degrees, the formula is changed slightly. This is because [absolute value of (180-180)] is equal to 0, so the formula above would calculate the length of the spiral as zero, which is obviously not the length. For spirals in which the centers of the circles are rotated by 180, the formula has [absolute value of (180)] instead of [absolute value of (180-a)]. The entire formula would be [(absolute value of 180)/180]*π*[1+(r^2)+(r^2)+(r^(n-1))]. Since (absolute value 180)/180=1, it can be simplified to simply x*π*[1+(r^2)+(r^2)+(r^(n-1))]
### Spiral Area
Just like with spiral length, there is technically no spiral area. However, we wanted to come up with a way to estimate the area of the arc sectors that make up the spiral. The sum of the arc sectors that make up the spiral, as shown in the image below, is what approximate what we call "spiral area".
Note: Again, this is an approximation for similar reasons provided in "Spiral Length"
The formula for approximating spiral area is as follows:
Spiral Area=x*[(absolute value of (180-a))/360]*π*[1+(r^2)+(r^4)+ (r^6)...+(r^(n-1))^2]
(variables used in this formula are the same as those used in the formula for spiral length)
How we found this formula/why it works:
We came up with this formula in basically the same way we came up with the formula for spiral length: we found the area of each of the arc sectors, added them together, then simplified. Similar to how arc length is degrees multiplied by circumference, area of an arc sector is degrees multiplied by area of the circle, or degrees*π*radius^2. Assuming the radius of the first circle is 1, the area of the first arc sector would be [(absolute value of (180-a))/360)]*π*1^2. The radius of the second arc sector is the radius of the first multiplied by r (ratio in which the circles increase). So the area of the second arc sector would be [(absolute value of (180-a))/360)]*π*r^2. The radius of the third arc sector is the radius of the second multiplied again by r (radius of the third arc sector would be r^2). So the area of the third arc sector would be [(absolute value of (180-a))/360)]*π*((r^2)^2). Following this pattern, the area of the fourth would be [(absolute value of (180-a))/360)]*π*((r^3)^2) This continues until the last arc sector: [(absolute value of (180-a))/360)]*π*(n^(n-1))^2. If you add them together, then simplify in the same way that we simplified the spiral length formula, you get: [(absolute value of (180-a))/360)]*π*[1 + r +(r^2)^2 + (r^3)^2 + (n^(n-1))^2]. But if the radius of the first circle is any number other than 1, you have to multiply the entire equation by that number (shown as "x" in the formula), just like with the formula for spiral length. Therefore, the final formula that we came up with for spiral area is: x*[(absolute value of (180-a))/360]*π*[1+(r^2)+(r^4)+ (r^6)...+(r^(n-1))^2]
## Examples
We will approximate the spiral area and the spiral length of certain spirals.
### 1:2 Ratio Rotated 180°
This is a 1:2 ratio spiral with a 1st iteration circle with a radius of 1 cm.
Following the formula, the length of this spiral is
(1)*[180/180]*π*[1+(2)+(4)+(8)+(16)]
The absolute value of 180-180 is 0, but, following the rule for spirals with a 180° rotation, we will substitute 180° in. π(1+2+4+8+16) = π(31) ≈ 97.39 cm
The area of this spiral is
1*[(180/360]*π*[1+(2^2)+(2^4)+ (2^6)+(2^8)] (1/2)π*[1+4+16+64+256] ≈ 535.64 cm^2
### 1:2 Ratio Rotated 90°
Notice the effect on the length and area of the 1:2 ratio spiral when the centers are rotated 90° instead of 180°:
This is a 1:2 ratio spiral (where the smallest circle has a radius of 1) with the centers rotated 90°. Substituting the radii and the angle of rotation into the length equation results in the following equation: 1*[(absolute value of (180-90))/180]*π*[1+2+(2^2)+(2^3)+(2^4)] = (1/2)π*(31) ≈ 48.69 cm.
Area: 1*[(90/360]*π*[1+(2^2)+(2^4)+ (2^6)+(2^8)] = (1/4)π*(1+4+16+64+256) ≈ 267.82 cm^2
### 1:2 Ratio Rotated 45°
Now, instead of rotating the 1:2 ratio spiral by 90°, the spiral is rotated by 45°:
Substituting the radii and the angle of rotation results in this equation: 1*[(absolute value of (180-45))/180]*π*[1+2+(2^2)+(2^3)+(2^4)] = (135/180)π*(31) ≈ 73.04 cm
Area: 1*[(180-45)/360]*π*[1+(2^2)+(2^4)+ (2^6)+(2^8)] = (135/360)π*[341] ≈ 401.73 cm^2
### 3:4 Ratio Spiral Rotated 180°
This is a 3:4 ratio spiral with a 1st iteration circle with a radius of 1, and a rotation of 180°.
Following the formula, the length of this spiral is 1*[180/180]*π*[1+(4/3)+(4/3)^2+(4/3)^3+(4/3)^4] π*[1+(4/3)+(16/9)+(64/27)+(256/81)] ≈ 30.29 cm
Area: 1*[(180/360]*π*[1+((4/3)^2)+((4/3)^4)+ ((4/3)^6)+((4/3)^8)] = 32.27 cm^2
## Spiral Requirements
There are certain properties common to all spirals. They include:
1) All spirals are made of connected arcs belonging to tangent circles. If the circles are not tangent, then there is no place for the arcs to connect to each other. The result looks like a broken spiral, where the "pieces" are the unconnected arcs. The tangent points follow a pattern. For example, spirals where the centers of the circles are rotated 180° from each other have tangent points on alternating sides of the circle. This means that if the smallest circle is tangent to the second-smallest circle on "top" (if the circle was on a grid, the highest point on the y-axis), then the second circle will be tangent to the third circle on the "bottom" (the lowest point on the y-axis).
Spirals that are made of circles rotated 90° also follow a pattern. If the centers of the circles are rotated counterclockwise and the first circle is tangent to the second circle on "top," then the second circle will be tangent to the third circle on the left, the third will be tangent to the fourth on the bottom, and the fourth will be tangent to the fifth on the right. These patterns repeat as the spiral gets bigger.
2) Spirals can go on forever; they circle around a central point, and they get progressively farther away from a different central point. Each point of the spiral is farther away from this center of the spiral than the previous point. All spirals are never-ending because the patterns that the radii of the circles they are constructed of never end. The Fibonacci Sequence is infinite, for example. Also, spirals that are based off of ratios never end because those patterns can continue indefinitely, too. If the points do not get progressively farther away from a center, then the spiral appears to circle back, like in the pictures below:
This shape circles back to one point and it is obviously not a spiral.
This one seems like a spiral, since it is made of tangent circles, but it does not grow progressively bigger. The distance of L2 is shorter than L1, which means that the shape grew closer to the center than farther. This violates the personality of a spiral because it grew closer to the center at one point rather than growing farther. This may be why the shape seems more elliptical than round; it is lopsided and irregular. Also, it does not appear to have a rotational center because of this lopsidedness.
3) The proportion between every two circles that make up a spiral must not be 0 or 1. The ratio between one circle to the next circle determines how tight or wide the spiral is. If the ratio between the circles approaches 1, the spiral will be tighter (the arcs would be closer to the same size because the circles would be more similar in size ).
If the ratio between the circles approaches 0, then the spiral will be wider (one arc would be much bigger than the previous arc because the circles would be very different in size).
If the proportion is 0, then only one circle exists, so there is no tangent point and no place for the spiral to continue.
If the proportion is 1, then the circles are the same size. The circles must be tangent to each other to continue the spiral, but when a circle is tangent to a circle of the same size, then they are tangent at all points on the circles. The circles would be on top of each other. There would be no place for the spiral to continue here as well because there would be no visible pace for another arc.
4) Two subsequent circles cannot intersect If the subsequent circles intersect, then they are not tangent; circles that intersect cannot be tangent to each other.
# Why It's Interesting
These spirals are interesting because they occur in nature! Pages like Romanesco Broccoli and Fibonacci Numbers explore these phenomenon.
## Spirals in Plants
Spirals not only appear in nature but they play a big part in a plant's survival.
Phyllotaxis is the study of a plant's ordered position of leaves on its stem, and research shows that leaves are arranged in a spiral pattern that maximizes the space for each leaf, or how much light they receive.
The following picture is an example of a plant exhibiting a spiral pattern, the clearest representation is the angle on the lower left side:
Sunflowers themeselves are arranged in a spiral. The spiral arrangement gives them a biological advantage because it maximizes the quantity of seeds that can be stored in seed head.
Hence, we see that spirals help maximize a plant's chance of survival!
# References
Images were created in the Sketchpad by group members Natalia, Jane and Hannah
Further explorations include making actual formulas rather than approximations.
[[Description::The picture on the right demonstrates a way of constructing the Fibonacci Spiral (explained below) using arcs. If you are not familiar with the fibonacci sequence, you can click here. Instead of using the squares to determine arcs, this uses circles to determine the arcs (their position and size) that make up the spiral. The traditional squares are overlaid to demonstrate the pattern: the ratio between circle sizes matches the ratio between square sizes.|]]
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# Learning Target 8
Learning Target 8. Angle Formulas. Central Angle. Definition:. An angle whose vertex lies on the center of the circle. NOT A Central Angle (of a circle). Central Angle (of a circle). Central Angle (of a circle). Y. 110 . 110 . O. Z. Central Angle Theorem.
## Learning Target 8
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### Presentation Transcript
1. Learning Target 8 Angle Formulas Lesson 8-5: Angle Formulas
2. Central Angle Definition: An angle whose vertex lies on the center of the circle. NOT A Central Angle (of a circle) Central Angle (of a circle) Central Angle (of a circle) Lesson 8-5: Angle Formulas
3. Y 110 110 O Z Central Angle Theorem The measure of a center angle is equal to the measure of the intercepted arc. Center Angle Intercepted Arc Example: Give is the diameter, find the value of x and y and z in the figure. Lesson 8-5: Angle Formulas
4. Example: Find the measure of each arc. 4x + 3x + (3x +10) + 2x + (2x-14) = 360° 14x – 4 = 360° 14x = 364° x = 26° 4x = 4(26) = 104° 3x = 3(26) = 78° 3x +10 = 3(26) +10= 88° 2x = 2(26) = 52° 2x – 14 = 2(26) – 14 = 38° Lesson 8-5: Angle Formulas
5. Inscribed Angle Inscribed Angle: An angle whose vertex lies on a circle and whose sides are chords of the circle (or one side tangent to the circle). Examples: 3 1 2 4 Yes! No! No! Yes! Lesson 8-5: Angle Formulas
6. Intercepted Arc Intercepted Arc: An angle intercepts an arc if and only if each of the following conditions holds: 1. The endpoints of the arc lie on the angle. 2. All points of the arc, except the endpoints, are in the interior of the angle. 3. Each side of the angle contains an endpoint of the arc. Lesson 8-5: Angle Formulas
7. Inscribed Angle Theorem The measure of an inscribed angle is equal to ½ the measure of the intercepted arc. Y Inscribed Angle 110 55 Z Intercepted Arc An angle formed by a chord and a tangent can be considered an inscribed angle. Lesson 8-5: Angle Formulas
8. A F A ° ° 40 y D ° 50 B ° ° y 50 B ° x C ° C x E E Examples: Find the value of x and y in the fig. Lesson 8-5: Angle Formulas
9. An angle inscribed in a semicircle is a right angle. P 180 90 S R Lesson 8-5: Angle Formulas
10. A D 1 B C Interior Angle Theorem Definition: Angles that are formed by two intersecting chords. 2 E Interior Angle Theorem: The measure of the angle formed by the two intersecting chords is equal to ½ the sum of the measures of the intercepted arcs. Lesson 8-5: Angle Formulas
11. Example: Interior Angle Theorem 91 A C x° y° B D 85 Lesson 8-5: Angle Formulas
12. ° 1 y ° 2 y ° ° x 3 x ° y ° x Exterior Angles An angle formed by two secants, two tangents, or a secant and a tangent drawn from a point outside the circle. Two secants 2 tangents A secant and a tangent Lesson 8-5: Angle Formulas
13. Exterior Angle Theorem The measure of the angle formed is equal to ½ the difference of the intercepted arcs. Lesson 8-5: Angle Formulas
14. Example: Exterior Angle Theorem Lesson 8-5: Angle Formulas
15. D 6 C E Q 5 3 A F 2 1 4 G 30° 25° 100° Lesson 8-5: Angle Formulas
16. Inscribed Quadrilaterals If a quadrilateral is inscribed in a circle, then the opposite angles are supplementary. mDAB + mDCB = 180 mADC + mABC = 180 Lesson 8-5: Angle Formulas
More Related
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# Lecture 5
We’ve seen in the above, that $A\Rightarrow B$ and $B\Rightarrow A$ are different statements. It’s helpful to have a word relating them:
Definition: Consider a statement of the form $A\Rightarrow B$. Then the converse of that statement is the statement $B\Rightarrow A$.
The truth of an implication has very little to do with the truth of its converse, as we’ll see.
## Equivalence
Equivalence is the relationship between two statements of being both true, or both false.
Definition: Let $P$ and $Q$ be statements. We write $P\Leftrightarrow Q$, pronounced “$P$ is equivalent to $Q$” for the statement that $P$ is true if and only if $Q$ is true.
Sometimes people shorten “if and only if” to “iff”.
## Negation
Negation is a type of opposite:
Definition: Let $P$ be a statement. The negation of $P$, written $\neg P$ and often pronounced “not $P$”, is the statement “$P$ is false”.
We must be careful in thinking of negation as an opposite. For example, the negation of “Richard is happy” is “Richard is not happy”.
Most people would say that the “opposite” is “Richard is sad”. That’s not quite the same thing!
Similarly, the negation of “Alice is in front of Bob” is not “Alice is behind Bob”, but:
“Alice is not in front of Bob”.
Note that double negation doesn’t do anything: the statement $\neg(\neg P)$ is equivalent to $P$. Since statements are either true or false, if it’s not “not true”, it’s true.
The negation allows us to make sense of something very important about implication:
Definition: Consider a statement of the form $P\Rightarrow Q$. Its contrapositive is the statement $(\neg Q)\Rightarrow(\neg P)$.
The main useful fact about the contrapositive is that it’s equivalent to the original implication. This is something very familiar from everyday life. If my friend Mel says
“If I can come visit you this evening, then I’ll text you at lunchtime,”
then I might rephrase it to myself as:
“I don’t get a text at lunchtime, then Mel won’t visit this evening.”
But here’s a formal statement and proof, anyway.
#### Proposition
Let $P$ and $Q$ be statements. The statement $P\Rightarrow Q$ is equivalent to its contrapositive $(\neg Q)\Rightarrow(\neg P)$.
#### Proof
I’ll prove this using a truth table, showing what happens in all possibilities:
$P\;$ $\; Q\;$ $\; P\Rightarrow Q\;$ $\;\neg Q\;$ $\;\neg P\;$ $\;(\neg Q)\Rightarrow (\neg P)$
0 0 1 1 1 1
0 1 1 0 1 1
1 0 0 1 0 0
1 1 1 0 0 1
You see from this that $(\neg Q)\Rightarrow(\neg P)$ is true exactly when $P\Rightarrow Q$ is, and this proves that they’re equivalent.
## “And” and “Or”
There are other ways to combine statements, other than implication:
Definition: Let $P$ and $Q$ be statements. The statement $P\wedge Q$, pronounced “$P$ and $Q$”, is the statement that both $P$ and $Q$ are true.
The statement $P\vee Q$, pronounced “$P$ or $Q$”, is the statement that at least one of $P$ or $Q$ (and possibly both) is true.
We can make truth tables for both of them:
$P$ $Q$ $P\wedge Q$
0 0 0
0 1 0
1 0 0
1 1 1
$P$ $Q$ $P\vee Q$
0 0 0
0 1 1
1 0 1
1 1 1
While you’re all used to these words from everyday life, there can be vagueness about how “or” is used in English.
For example,
you can have your pie with chips or with mashed potatoes
is probably intended to mean “but not both”. In mathematical argument when we use “or” and mean “but not both”, we have to say so explicitly.
It is sometimes worth knowing that implication can be defined in terms of “or”:
#### Proposition
The statement $P\Rightarrow Q$ is equivalent to $(\neg P)\vee Q$.
#### Proof
The only way that the first statement can be false is if $P$ is true and $Q$ is false. But that’s also the only way that the second statement can be false, so they’re equivalent.
Note that that shows another style of proof of logical statements: by analysis rather than the “case bash” used in truth tables.
## Quantifiers
Lastly, we need to discuss how to make statements about general situations and particular examples. The phrases we use again and again are “for all” and “there exists”: these are called quantifiers.
The following symbols are in common use: \begin{aligned} {}\forall \quad&\text{for for all'';}\\ {}\exists \quad&\text{for there exists'';}\\ {}\quad\mathrm{s.t.}\quad\quad&\text{for such that''.}\end{aligned}
So, for example, $\forall n\in \mathbb{Z},\quad n^2-1 = (n+1)(n-1)$ is to be read as
“For all integers $n$, we have $n^2-1 = (n+1)(n-1)$.”
And $\exists x\in\mathbb{R}\quad\mathrm{s.t.}\quad x^2-3x-12=0$ is to be read as
“There exists a real number $x$ such that $x^2-3x-12=0$.”
It’s important that you get used to this notation. This is not because there is anything amazing about it, but because mathematics involves lots of general rules and particular examples: much of the mathematics you do for the next few years will require you to be able to deal with these things.
One thing you’ll have to get used to is situations with two or three quantifiers. These happen very frequently: “in general, there is always a particular example of such-and-such”, or “there is a particular amazing example which has the general property of such-and-such”.
For example, the statement $\forall n\in\mathbb{N},\quad\exists x\in\mathbb{R}\quad\mathrm{s.t.}\quad x^2 = n$ says that every natural number $n$ has a square root $x$.
The order of quantifiers is very important. If we swap over the two quantifiers in the last example, we get $\exists x\in\mathbb{R}\quad\mathrm{s.t.}\quad\forall n\in\mathbb{N},\quad x^2 = n.$ This says that there’s a particular number $x$ which has the property that $x$ is the square root of every natural number. And that’s nonsense.
Another thing that mathematicians have to do every day is understanding how negation interacts with quantifiers.
The negation of “all Teletubbies are red”is “not all Teletubbies are red”, which is equivalent to “there exists a Teletubby which is not red”.
Similarly, the negation of “there exists a dolphin who likes Beethoven”is “there does not exist a dolphin who likes Beethoven”,and that’s equivalent to “all dolphins do not like Beethoven”.
In symbols, \begin{aligned} {}\neg\left(\forall x\in X,\quad P(x)\right) \qquad&\text{is equivalent to}\qquad \exists x\in X\quad\mathrm{s.t.}\quad\neg P(x).\\ {}\neg\left(\exists x\in X\quad\mathrm{s.t.}\quad P(x)\right) \qquad&\text{is equivalent to}\qquad\forall x\in X,\quad \neg P(x).\end{aligned}
Perhaps you may want to remember that “negation swaps $\forall$ and $\exists$.” But being able to do it correctly by remembering what’s going on is much more important than remembering a slogan. After a while it should come to seem natural.
Suppose I am wondering whether all fish are slippery. If it’s true, I need to find some general reason why every single fish is slippery. If it’s false, I only need to find one single fish which isn’t slippery, and then I’ve proved it.
In general, if you have a general statement and you don’t know if whether it’s true or false, then it could either be:
• true, in which case you need to prove it in general (that’s a statement with a “$\forall$” in);
• false, in which case you need to find a counterexample (that’s a statement with a “$\exists$” in).
|
# Evaluate $\dfrac{\sin{3x}}{\sin{x}}$ $-$ $\dfrac{\cos{3x}}{\cos{x}}$ by Triple angle identities
The sine of triple angle and cosine of triple angle functions are involved in forming a triple angle trigonometric expression in this trigonometry problem. The involvement of triple angle trigonometric functions expresses that the triple angle trigonometric identities should be used to evaluate the given triple angle trigonometric expression.
Now, let’s learn how to find the value of the sine of three times $x$ divided by sine of angle $x$ minus cosine of three times $x$ divided by cosine of angle $x$ by using the triple angle trigonometric formulas.
### Expand the Triple angle Trigonometric functions
According to the triple angle identities, the sine of triple angle function $\sin{3x}$ can be expanded as per the sine triple angle identity and the cosine of triple angle function $\cos{3x}$ can also be expanded as per cosine triple angle identity.
$(1).\,\,$ $\sin{3x}$ $\,=\,$ $3\sin{x}$ $-$ $4\sin^3{x}$
$(2).\,\,$ $\cos{3x}$ $\,=\,$ $4\cos^3{x}$ $-$ $3\cos{x}$
Now, the triple angle trigonometric functions $\sin{3x}$ and $\cos{3x}$ can be replaced by their expansions in the trigonometric expression.
$\implies$ $\dfrac{\sin{3x}}{\sin{x}}$ $-$ $\dfrac{\cos{3x}}{\cos{x}}$ $\,=\,$ $\dfrac{3\sin{x}-4\sin^3{x}}{\sin{x}}$ $-$ $\dfrac{4\cos^3{x}-3\cos{x}}{\cos{x}}$
### Release expression from rational by simplification
The terms in the numerator of each term in the trigonometric expression can be divided by the corresponding denominator to start simplifying the trigonometric expression.
$=\,\,$ $\dfrac{3\sin{x}}{\sin{x}}$ $-$ $\dfrac{4\sin^3{x}}{\sin{x}}$ $-$ $\bigg(\dfrac{4\cos^3{x}}{\cos{x}}$ $-$ $\dfrac{3\cos{x}}{\cos{x}}\bigg)$
Now, continue the procedure of simplifying the trigonometric expression.
$=\,\,$ $\dfrac{3\sin{x}}{\sin{x}}$ $-$ $\dfrac{4\sin^3{x}}{\sin{x}}$ $-$ $\dfrac{4\cos^3{x}}{\cos{x}}$ $+$ $\dfrac{3\cos{x}}{\cos{x}}$
$=\,\,$ $\dfrac{3\cancel{\sin{x}}}{\cancel{\sin{x}}}$ $-$ $\dfrac{4\cancel{\sin^3{x}}}{\cancel{\sin{x}}}$ $-$ $\dfrac{4\cancel{\cos^3{x}}}{\cancel{\cos{x}}}$ $+$ $\dfrac{3\cancel{\cos{x}}}{\cancel{\cos{x}}}$
$=\,\,$ $3$ $-$ $4\sin^2{x}$ $-$ $4\cos^2{x}$ $+$ $3$
$=\,\,$ $3$ $+$ $3$ $-$ $4\sin^2{x}$ $-$ $4\cos^2{x}$
$=\,\,$ $6$ $-$ $4\sin^2{x}$ $-$ $4\cos^2{x}$
### Find the value of Trigonometric expression
The negative four is a common factor in both second and third terms. So, take out the negative four common from them.
$=\,\,$ $6$ $-$ $4 \times \big(\sin^2{x}+\cos^2{x}\big)$
The sum of squares of sine and cosine functions at an angle is one as per the Pythagorean identity of sine and cosine functions.
$=\,\,$ $6$ $-$ $4 \times (1)$
Now, find the value of the arithmetic expression by simplification.
$=\,\,$ $6$ $-$ $4 \times 1$
$=\,\,$ $6-4$
$=\,\,$ $2$
#### Another method
Evaluate $\dfrac{\sin{3x}}{\sin{x}}$ $-$ $\dfrac{\cos{3x}}{\cos{x}}$
Learn how to find the value of sine of three times angle $x$ divided by sine of $x$ minus cosine of three times $x$ divided by cosine of angle $x$ without using the triple angle trigonometric identities.
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Jun 26, 2023
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Difference between revisions of "2020 AMC 8 Problems/Problem 25"
Rectangles $R_1$ and $R_2,$ and squares $S_1,\,S_2,\,$ and $S_3,$ shown below, combine to form a rectangle that is 3322 units wide and 2020 units high. What is the side length of $S_2$ in units?
$[asy] draw((0,0)--(5,0)--(5,3)--(0,3)--(0,0)); draw((3,0)--(3,1)--(0,1)); draw((3,1)--(3,2)--(5,2)); draw((3,2)--(2,2)--(2,1)--(2,3)); label("R_1",(3/2,1/2)); label("S_3",(4,1)); label("S_2",(5/2,3/2)); label("S_1",(1,2)); label("R_2",(7/2,5/2)); [/asy]$
$\textbf{(A) }651 \qquad \textbf{(B) }655 \qquad \textbf{(C) }656 \qquad \textbf{(D) }662 \qquad \textbf{(E) }666$
Solution 1
For each square $S_{k}$, let the sidelength of this square be denoted by $s_{k}$.
As the diagram shows, $s_{1}+s_{2}+s_{3}=3322, s_{1}-s_{2}+s_{3}=2020.$ If we subtract the second equation from the first we will get $2s_{2}=1302$ or $s_{2}=\boxed{651\textbf{(A)}}$ ~icematrix, edits by starrynight7210
Solution 2
WLOG, assume that $S_1=S_3$ and $R_1=R_2$. Let the sum of the lengths of $S_1$ and $S_2$ be $x$ and let the length of $S_2$ be $y$. We have the system $$x+y =3322$$ $$x-y=2020$$
which we solve to find that $y=\textbf{(A) }651$.
-franzliszt
|
Stat Trek
Teach yourself statistics
Teach yourself statistics
# Multinomial Distribution Probability Calculator
The Multinomial Calculator makes it easy to compute multinomial probabilities. For help in using the calculator, read the Frequently-Asked Questions or review the Sample Problems.
To learn more, go to Stat Trek's tutorial on the multinomial distribution.
• First, specify the number of outcomes.
• Then, enter the probability and frequency for each outcome.
• Click Calculate to compute multinomial probability.
Instructions: To find the answer to a frequently-asked question, simply click on the question.
### What is a multinomial experiment?
A multinomial experiment is a statistical experiment that has the following characteristics:
• The experiment involves one or more trials.
• Each trial has a discrete number of possible outcomes.
• On any given trial, the probability that a particular outcome will occur is constant.
• All of the trials in the experiment are independent.
Tossing a pair of dice is a perfect example of a multinomial experiment. Suppose we toss a pair of dice three times. Each toss represents a trial, so this experiment would have 3 trials. Each toss also has a discrete number of possible outcomes - 2 through 12. The probability of any particular outcome is constant; for example, the probability of rolling a 12 on any particular toss is always 1/36. And finally, the outcome on any toss is not affected by previous or subsequent tosses; so the trials in the experiment are independent.
### What is a multinomial distribution?
A multinomial distribution is a probability distribution. It refers to the probabilities associated with each of the possible outcomes in a multinomial experiment.
For example, suppose we flip three coins and count the number of coins that land on heads. This multinomial experiment has four possible outcomes: 0 heads, 1 head, 2 heads, and 3 heads. Probabilities associated with each possible outcome are an example of a multinomial distribution, as shown below.
Outcome Probability
The table completely defines the probabilities associated with every possible outcome from this multinomial experiment. It is the multinomial distribution for this experiment.
### What is the number of outcomes?
The number of outcomes refers to the number of different results that could occur from a multinomial experiment. For example, suppose we roll a die. Each roll of the die can have six possible outcomes - 1, 2, 3, 4, 5, or 6. Similarly, the roll of two dice can have eleven possible outcomes - the numbers from 2 to 12.
### What is the probability of an outcome?
Each trial in a multinomial experiment can have a discrete number of outcomes. The likelihood that a particular outcome will occur in a single trial is the probability of the outcome.
For example, suppose we toss two dice. The probability of tossing a 2 is 1/36; the probability of tossing a 3 is 2/36, the probability of tossing 4 is 3/36, etc.
### What is the frequency of an outcome?
In a multinomial experiment, the frequency of an outcome refers to the number of times that an outcome occurs. For example, suppose we toss a single die. This experiment has 6 possible outcomes; the die could land on 1, 2, 3, 4, 5, or 6. Suppose that we roll the die four times and observe the following outcomes: we roll a 1, a 3, and a two 5's? The frequency for each outcome is shown in the table below.
Outcome Frequency
1 1
2 0
3 1
4 0
5 2
6 0
### What is the multinomial probability?
A multinomial probability refers to the probability of obtaining a specified frequency in a multinomial experiment. For example, suppose we toss a single die four times. We might ask: What is the probability that we roll a 1, a 3, and a two 5's?
The probability of getting this particular result would be very small: 0.00154. The easiest way to compute a multinomial probability is to use the Multinomial Calculator. To see how to compute multinomial probabilities by hand, go to Stat Trek's tutorial on the multinomial distribution.
### What is the relation between a multinomial and a binomial experiment?
A binomial experiment is actually a special case of a multinomial experiment. The binomial experiment is a multinomial experiment, in which each trial can have only two possible outcomes. The flip of a coin is a good example of a binomial experiment, since a coin flip can have only two possible outcomes - heads or tails. To learn more about binomial experiments, go to Stat Trek's tutorial on the binomial distribution.
## Sample Problem
1. Suppose you toss a pair of dice 10 times. What is the probability of getting the following outcome: two rolls of 7, two rolls of 6, and any other outcome on the remaining six rolls.
Hint: The probabilities associated with each roll of two dice are shown below.
Outcome Probability
2 1/36
3 2/36
4 3/36
5 4/36
6 5/36
7 6/36
8 5/36
9 4/36
10 3/36
11 2/36
12 1/36
Solution:
We know the following:
• The number of outcomes is 3. (Outcome 1 is a roll of 7, the Outcome 2 is a roll of 6, and Outcome 3 is any other roll.)
• Outcome 1: The probability is 6/36 (about 0.167), and the frequency is 2.
• Outcome 2: The probability is 5/36 (about 0.139), and the frequency is 2.
• Outcome 3: The probability is 25/36 (about .694), and the frequency is 6.
Therefore, we plug those numbers into the Multinomial Calculator and hit the Calculate button.
The calculator reports that the multinomial probability is about 0.076. Thus, in ten rolls of the dice, the probability of rolling 7 two times, 6 two times, and something else six times is about 0.076.
1. A bowl has 2 black marbles, 3 green marbles, and 5 white marbles. A marble is randomly selected and then put back in the bowl. Suppose this selection process is repeated five times. What is the probability that 3 white marbles, 1 green marble, and 1 black marble will be chosen?
Hint: On any given trial, the probability of choosing a black marble is 2/10; the probability of choosing a green marble is 3/10; and the probability of choosing a white marble is 5/10.
Solution:
We know the following:
• The number of outcomes is 3. (Outcome 1 is a black marble; Outcome 2, a green marble; and Outcome 3, a white marble.)
• Outcome 1: The probability is 0.2, and the frequency is 1.
• Outcome 2: The probability is 0.3, and the frequency is 1.
• Outcome 3: The probability is 0.5, and the frequency is 3.
Therefore, we plug those numbers into the Multinomial Calculator and hit the Calculate button.
The calculator reports that the multinomial probability is 0.15. Thus, the probability of selecting 1 black marble, 1 green marble, and 3 white marbles is 0.15.
## Problem?
Oops! Something went wrong.
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Vous êtes sur la page 1sur 11
6-1
Lesson 6
Angles of Elevation and Depression
Now we get a chance to apply all of our newly acquired skills in real life applications, otherwise known as word problems. The first section is in elevation and depression problems. I first encountered these in a Boy Scout Handbook many years ago. There was a picture of a tree, a boy, and several lines.
tree
5'
11'
30'
How tall is the tree?
Separating the picture into two triangles helps to clarify our ratios.
q
11
5
Using our trig. abilities:
in the "boy" triangle
tan q =
We find that q = 24.44°, so
5
11
q = 24.44°
= .4545
q
41
We could solve this with a X proportion (two ratios) as
X
5
=
and solve for X.
11
41
For the large triangle
The tree is 18.64'.
tan 24.44 = (41)(.4545) = X
18.64' = X
X
41
One of the key components in being a good problem solver is drawing a picture using all the data given. It turns a one dimensional group of words into a two dimensional picture. It is pretty obvious by the words themselves that an angle of elevation measures up, and an angle of depression down. Look at Figure 1.
Figure 1
elevation
depression
We assume that the line where the angle begins is perfectly flat, or horizontal.
6-2
Example 1
A campsite is 9.41 miles from a point directly below the mountain top. If the angle of
elevation is 12from the camp to the top of the mountain, how high is the mountain? top
mountain
9.41 mi.
12∞
campsite
You can now see a right triangle, with the side adjacent to the 12being 9.41 miles long. To find
the height of the mountain, which is the side opposite the 12angle, the tangent is the best choice. height
tan 12=
9.41 mi.
(9.41)(tan 12) = height
(9.41)(.2126) = height
2 miles = height
Example 2
Practice Problems
At a point 42.3 feet from the base of a building, the angle of elevation of the top is 75.
How tall is the building?
75∞
height
42.3'
tan 75=
building
(42.3)(tan 75) = height
(42.3)(3.7321) = height
157.87' = height of building
42.3'
1) How far from the door must a ramp begin in order to rise 3 feet with an 8∞ angle of elevation? 2) An A-frame cabin is 26.23 feet high at the center and the angle the floor makes with the base is 53∞15'. How
wide is the base?
Solutions
1)
3
8∞
3
tan 8=
X
X X tan 8= 3
X
=
3
X =
tan 8
3
.1405
X = 21.35 ft.
2)
26.23
53∞15'
tan 53.25=
X =
X
X
X =
26.23
X
26.23
tan 53.25
26.23
1.3392
5315' = 53.25
X = 19.59
2X = 39.18 ft
6-A
1)
Isaac's camp is 5,280 feet from a point directly beneath Mt. Monadnock. What is the hiking distance along the ridge if the angle of elevation is 25° 16'?
2)
How many feet higher is the top of the mountain than his campsite?
a
2
31
q
6
3
4
Express as a fraction.
3) csc q = 6) csc a = 4) sec q = 7) sec a = 5) cot q = 8) cot a = Express as a decimal. 9) sin q = 12) sin a = 10) cos q = 13) cos a = 11) tan q = 14) tan a =
15)
16)
Use your answers in #9-11 to find the measure of q.
Use your answers in #12-14 to find the measure of a.
Solve for the lengths of the sides and the measures of the angles.
17)
18)
a
59 29°
A
12
35.4°
a
B
C
F
G
a
19)
42.66°
20)
47 H
100
E
41°32'10"
a
D
6-B
1) The side of a lake has a uniform angle of elevation of 15∞30'. How far up the side of the lake does the water rise if, during the flood season, the height of the lake increases by 7.3 feet? 2) A building casts a shadow of 110 feet. If the angle of elevation from that point to the top of the building is 29∞3', find the height of the building. Express as a fraction. 3) csc q = 6) csc a = a 11 q 4) sec q = 7) sec a = 5) cot q = 8) cot a = 4.6 Express as a decimal. 9) sin q = 12) sin a = 10 10) cos q = 13) cos a = 11) tan q = 14) tan a = 15) Use your answers in #9-11 to find the measure of q. 16) Use your answers in #12-14 to find the measure of a.
Solve for the lengths of the sides and the measures of the angles.
17)
19)
M
18)
a
a
J
12
L
18°
29°
59
K
6
20)
q
67°
N
10.25
2
13
Q
q
a
P
6-C
1) From a point 120 feet from the base of a church, the angles of elevation of the top of the building and the top of a cross on the building are 38∞ and 43∞ respectively. Find the height of the cross. (The ground is flat.) 2) Find the height of the building as well as the height of the cross itself.
7.1
a
15
q
13.2
Express as a fraction.
3) csc q = 4) sec q = 5) cot q =
Express as a decimal.
9) sin q = 10) cos q = 11) tan q =
6) csc a = 7) sec a = 8) cot a = 12) sin a = 13) cos a = 14) tan a =
15)
Use your answers in #9-11 to find the measure of q. They may vary slightly.
16)
Use your answers in #12-14 to find the measure of a. They may vary slightly.
Solve for the lengths of the sides and the measures of the angles.
S
17)
18)
a
40°
T
R
36.2°
25
U
a
88
19)
150
20)
a
a
95
V
q
51.9°
7
W
X
6-D
1)
A campsite is 12.88 miles from a point directly below Mt. Adams. If the angle of elevation is 15.5from the camp to the top of the mountain, how high is the mountain?
2)
At a point 60.7 feet from the base of a building, the angle of elevation from that point to the top is 64.75. How tall is the building?
a
25
q
18.33
Express as a fraction. First find X.
3) csc q = 4) sec q = 5) cot q =
X Express as a decimal.
9) sin q = 10) cos q = 11) tan q =
6)
7)
8)
12)
13)
14)
csc a =
sec a =
cot a =
sin a =
cos a =
tan a =
15)
16)
Use your answers in #9-11 to find the measure of q.
Use your answers in #12-14 to find the measure of a.
Solve for the lengths of the sides and the measures of the angles.
2.24
10 .5
17)
18)
a
a
2
Z
Y
49.2°
q
A
19)
29.07°
20)
a
B
D
56
q
a
14
C
10
Test 6
Devan stands 926 meters from
a point directly below the peak of a mountain. The angle
of elevation between Devan and the top of the mountain
is 42°.
Use for questions 1 - 4:
1)
2)
Which equation can be used to find the height of the mountain (X)?
A)
C)
sin 42° =
cos 48° =
X
B)
926
926
926
X
D) tan 42° =
X
X
926
tan 42° =
What is the height of the mountain?
A) 833.8 m B) 1028.4 m C) 619.6 m D) 1383.9 m
3)
A tower 50 meters high is built on top of the mountain. What is the angle of elevation from Devan’s position to the top of the tower? (round decimal degrees to tenths)
A) 40°14’44” B) 43°42’ C) 57°15’ D) 46°20’08”
4)
If a bird flew from Devan’s position to the top of the mountain, how many meters would it travel?
A) 408.4 m B) 1246.1 m C) 1383.9 m D) 1280.8 m
Use for questions 5 - 8:
the base of a building to the top of the building the angle of elevation is 51°. From the same point to the top of a flag staff on the building the angle of elevation is 54°.
From a point 80 meters from
5)
6)
7)
What equation can be used to find the combined height (Y) of building and flagpole?
A) Y = 80 tan 51° C) Y = 80 tan 54°
B)
D)
Y = 80 sin 54°
Y =
tan 51°
80
What is the height of the building alone?
A)
C)
98.8 m
64.8 m
B) 110.1 m
D)
58.1 m
What is the height of the flagpole alone?
A) 15.1 m B) 45.3 m C) 4.2 m D) 11.3 m
8)
How long must a cable be in order to stretch from the observation point to the top of the building?
A) 102.9 m B) 127.1 m C) 136.1 m D) 50.3 m
Use for questions 9 & 10:
distance of 100 feet up a ramp to a bridge. The
angle of elevation of the ramp was 10°.
A car traveled a
9) How high was the bridge above road level? A) 17.4 ft. B) 98.5 ft. C) 10 ft. D) 100 ft. 10) What is the actual distance from the beginning of the ramp to the base of the bridge? A) 575 ft. B) 98.5 ft. C) 89.4 ft. D) 17.4 ft. 3 is the ratio for 11) 3 A) cos 45° B) cos 30° C) tan 60° D) tan 30° 12) Arcsin .8192 = A) 1.22 B) 35° C) 55° D) .9999 13) 46°21’02” = A) 46.21° B) 46.12° C) 46.35° D) 46.4° sin a 14) cos a is equal to A) tan a B) cot a C) sec a D) csc a 1 15) cos a is equal to A) csc a B) sec a C) sin a D) cos a
6-A
1)
D
25°16'
5,280'
Cos 25°16' = 5,280
D =
D
5,280
.9043
D = 5,838.77'
2)
D
M
25°16'
5,280'
a
2
31
q
6
3
4
Express as a fraction.
3)
4)
5)
csc q =
sec q =
cot q =
2
31
31
=
4
2
2
31
93
=
6
3
9
6
3
3
3
=
4 2
Express as a decimal.
4
9)
sin q =
2
31
6
3
10)
cos q =
2
31
4
11)
tan q =
6
3
=
.3592
= .9333
= .3849
Answers to 15 and 16 may vary slightly.
6)
7)
8)
12)
13)
14)
csc a =
sec a =
cot a =
sin a =
cos a =
tan a =
15)
The measure of q is 21.05°.
16)
The measure of q is 68.95°.
Solve for the lengths of the sides and the measures of the angles.
B
17)
Tan 54.6° =
12 18)
59 29°
54.6°
A 12 Tan 54.6° = B
12
D
16.89 = B
61°
35.4°
Sin 35.4° = 12
C
A
B
12
A =
Sin 35.4°
A = 20.72
F
G
F
19)
Tan 47.34° =
20)
42.66°
100
49°27'50"
100 100 Tan 47.34° = F
47 H
108.52 = F
E 47.34°
41°32'10"
Sin 42.66° = 100
E
100
E
=
41°32'10" = 41.54°
Sin 42.66°
E = 147.57
M
Tan 25°16' = 5,280
5,280 Tan 25°16' = M
2,492.09' = M
93
9
31
2
2
3
9
6
3
2
31
4
2
31
6
3
4
= .9333
=
=
.3592
2.598
Sin 61° =
D
59
(59) (Sin 61°) = D
51.6 = D
C
59
Cos 61° =
(59) (Cos 61°) = C
28.6 = C
G
41.54° =
47 Tan 41.54° = G
41.64 = G
47
Cos 41.54° = 47
H
H
=
47
Cos 41.54°
H
= 62.79
6-B
2)
29°3' = 29.05°
B
29°3'
110'
1)
15°30' = 15.5°
7.3'
X D
15°30'
7.3
sin 15.5 =
X
7.3
X =
sin 15.5
X = 27.32'
a
11
q
10
4.6
Tan 29.05° =
110 (Tan 29.05°) = B
B
110
61.1' = B
Express as a fraction.
3) csc q = 4) sec q = 5) cot q =
11
4.6
11
10
10
4.6
Express as a decimal.
9) sin q = 10) cos q = 11) tan q =
4.6
11
10
11
4.6
10
11
10
11
4.6
4.6
10
10
11
4.6
11
10
4.6
6)
7)
8)
csc a =
sec a =
cot a =
= .4182
= .9091
= .46
12)
13)
14)
sin a =
cos a =
tan a =
= .9091
= .4182
= 2.1739
15) The measure of q is 24.7°. 16) The measure of q is 65.3°.
Solve for the lengths of the sides and the measures of the angles.
M
59
K
Sin 29° =
17)
12
J 72°
61°
12 12 Tan 72° = 36.93
18°
L
12
K Sin 18° =
J
29°
59
J =
12 = 38.83
Sin 18°
6
Sin 23° = 10.25
20)
N
67°
N
10.25
N =
10.25 = 26.23
13
Sin 23°
23°
P Tan 67° =
P
10.25
50.24°
Q 2
39.76°
10.25 Tan 67° = P 24.15 = P
Tan 72° =
18)
M
59 Sin 29° = M
28.6 = M
L
Cos 29° =
59
59 Cos 29° = L
51.6 = L
2
13 = 7.2111
Tan q = 7.211
6
arctan q = 1.202
q = 50.24° a = 39.76°
(2 13 ) 2 + (6) 2 = Q 2 2 22 = Q 2
19)
6-C
1)
15)
16)
7.1
43°
120'
BC
a
15
q
13.2
The measure of q is 28.25°.
The measure of q is 61.75°.
BC
B
Tan 43° =
2)
Tan 38° =
120
120
BC = 120 Tan 43°
B
= 120 Tan 38°
B
BC = 112'
B
= 93.75' or 93" 9"
38°
120'
112' - 93.75' = 18.25' or 18' 3"
Express as a fraction.
3)
4)
5)
csc q =
sec q =
cot q =
15
7.1
15
13.2
13.2
7.1
Express as a decimal.
9)
10)
11)
sin q =
cos q =
tan q =
7.1
15
13.2
15
7.1
13.2
= .4733
= .88
= .5379
6)
7)
8)
12)
13)
14)
csc a =
sec a =
cot a =
sin a =
cos a =
tan a =
15
13.2
15
7.1
7.1
13.2
13.2
15
7.1
15
13.2
7.1
= .88
= .4733
=
1.8592
Solve for the lengths of the sides and the measures of the angles.
R
S Sin 40° =
17)
25
18)
40°
25 Sin 40° = R
T 53.8°
88
R = 16.07
R
S
36.2°
25
Cos 40° =
25
50°
U
25 Cos 40° = S
S = 19.15
W
Tan 38.1° =
19)
150 20)
150 Tan 38.1° = W
95 45.9°
38.1°
V
X
150
W = 117.62
Sin 51.9° = 150
44.1°
51.9°
V
150
7
W V=
Sin 51.9°
V = 190.61
U
88
Tan 53.8° =
88 Tan 53.8° = U
U = 120.24
88
Sin 36.2° =
T
88
T =
Sin 36.2°
T = 149.00
7 2 + X 2 = (
95
) 2
49
X
+ X 2
= 46
= 95
X
=
46
or 6.78
6.78
Tan q =
7
arctan .96 = 44.1
a = 90 - 44.1 = 45.9°
6-D
1)
15)
16)
Tan 15.5° =
M
15.5°
M = 3.57
12.88 mi.
M
12.88
(18.33) 2 + X 2 = 25 2
336 + X 2 = 625
X 2 = 625
X = 17
a
25
q
18.33
X = 17
The measure of q is 42.84°.
The measure of q is 47.16°.
Express as a fraction.
3) csc q = 4) sec q = 5) cot q =
25
17
25
18.33
18.33
17
Express as a decimal.
9) sin q = 10) cos q = 11) tan q =
17
25
18.33
25
17
18.33
2)
64.75°
60.7''
=
=
.68
.7332
= .9274
Tan 64.75° =
B
60.7
B B = 128.7'
6) csc a = 7) sec a = 8) cot a = 12) sin a = 13) cos a = 14) tan a =
25
18.33
25
17
17
18.33
18.33
25
17
25
18.33
17
=
=
.7332
.68
= 1.0782
Solve for the lengths of the sides and the measures of the angles.
2.24
10 .5
18)
17)
41.81°
2 2 + (2.24) 2 = Y 2
40.8°
3 = Y
2
2
Z
Y = 3
Cos q =
= .6667
49.2°
3
48.19°
arccos .6667 = 48.19°
a = 90 - 48.19 = 41.81°
C
Tan 29.07° =
56
19)
29.07°
20)
56 Tan 29.07° =C
a
D
B
31.13
= C
56
(56) 2 + (31.13) 2 = B 2
60.93°
q
64.07
= B
14
C
A
10
Tan 40.08° =
A
10.5
10.5 Tan 40.08° = A
9.06 = A
Sin 49.2° = 10.5
Z
Z
=
10.5
Sin 49.2°
Z
= 13.87
10
Tan q =
arctan .7143= 35.54°
a = 90 - 35.54 = 54.46°
14
10 2 + 14 2 = D 2
17.2 = D
|
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Section 10.4 : Convergence/Divergence of Series
4. Assume that the $$n$$th term in the sequence of partial sums for the series $$\displaystyle \sum\limits_{n = 0}^\infty {{a_n}}$$ is given below. Determine if the series $$\displaystyle \sum\limits_{n = 0}^\infty {{a_n}}$$ is convergent or divergent. If the series is convergent determine the value of the series.
${s_n} = \frac{{{n^2}}}{{5 + 2n}}$ Show Solution
There really isn’t all that much that we need to do here other than to recall,
$\sum\limits_{n = 0}^\infty {{a_n}} = \mathop {\lim }\limits_{n \to \infty } {s_n}$
So, to determine if the series converges or diverges, all we need to do is compute the limit of the sequence of the partial sums. The limit of the sequence of partial sums is,
$\mathop {\lim }\limits_{n \to \infty } {s_n} = \mathop {\lim }\limits_{n \to \infty } \frac{{{n^2}}}{{5 + 2n}} = \infty$
Now, we can see that this limit exists and is infinite. Therefore, we now know that the series, $$\sum\limits_{n = 0}^\infty {{a_n}}$$, diverges.
If you are unfamiliar with limits at infinity then you really need to go back to the Calculus I material and do some review of limits at infinity and L’Hospital’s Rule as we will be doing quite a bit of these kinds of limits off and on over the next few sections.
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## A Fraction Number Line Lesson
Fractions can really give students fits! How can we develop conceptual understanding with fractions?
I taught this lesson with a group of struggling 5th grade math learners during an after-school tutoring class. Each student stood at a whiteboard on the wall of my classroom. My goal was to help students conceptually understand why we must find a common denominator when adding fractions. To do this, I asked the students to construct their own number lines. All steps beyond #1 are directions I spoke to students as they wrote on their boards. These were the steps in my lesson:
1. I modeled how to make a fraction number line: arrows on both ends showing numbers go infinitely in both directions; small dashes pinpoint exact locations of numbers; begin with zero; space each mark evenly; starting at zero, build your number line and work across the number line rather than starting with a 0 and 1 and subdividing the line into equal size lengths. I wasn’t modeling fraction concepts per se; just creating the number line. (Previous number line work showed students struggling with this.)
2. Create a number line that shows thirds from 0 to 1.
3. Draw a parallel number line below the thirds number line. Put 0 and 1 directly under the 0 and 1 in the top number line. Now subdivide the new number line into sixths.
4. Draw arrows down to the second number line where fractions line up.
5. Solve 1/3 + 1/6 using one of the number lines.
6. Repeat the above steps with a number line showing fourths and then eighths. (See top photo.)
7. Add a third parallel number line showing sixths. (See bottom photo.)
8. Use the proper number line to solve 1/4 + 3/8.
Each time I asked the students to create something, I redirected with small hints and asked students to peer tutor until all students had a perfect model. We constantly discussed the models to help students make sense of their work. In step 7, we discussed how very few fractions aligned with the number line above (sixths and eighths.)
Students made excellent progress as shown in their work on the boards.
Inspiration for this lesson came from my esteemed colleague, Kristian Quiocho. Kristian is one of the most knowledgeable and passionate teachers I know.
## Teaching Evaluation Without the Teacher
Last year, it was my turn to be observed. After teaching for 27 years, I was ready to take a risk, try something new. So I went to my principal and asked, “Do I have to be in my classroom when you come do my observation?” As you can imagine, she didn’t exactly know how to respond. But since my boss was a good sport, and trusted me, she said, “Well, what do you have in mind?” After I explained, she decided to play along.
When the time came, my principal walked into my room and I walked out. It was a very good year, probably the best group of kids I’d ever had, a class of very mature fifth graders. One girl ran the class. She managed 32 students reviewing a language arts assignment. The students worked in groups, pairs, and held a whole class discussion. According to my principal, my observation went well. I had to take her word for it, since I wasn’t there.
So why would I do such a crazy thing?
About 12 years ago, I began my journey to make my students the most independent they possibly could be. I had read an archived newsletter at learningcentered.org that told the story of a sub who didn’t show up for school. The principal was walking the hallways, noticed the kids in one class working away diligently, but didn’t see any teacher. Upon inquiry of the students, the principal found out that the teacher hadn’t shown up, and neither had the substitute. So the kids just went about their business of learning. This struck me. Could my students do this? Would they?
So to me, the real assessment of my teaching is how my students work without me. And if that’s my goal, then why should I be in the room for my teacher observation?
Am I crazy? Did I get you thinking? Talk to me.
## Tim’s Projects
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Home > Math Shortcuts > Statistics Math Formulas
# Statistics Math Formulas
## Statistics Math Formulas :
Statistics Math Formulas :
SETS :
A = {2, 3, 4, 7, 8, 9, 12}
3 ∈ A
5 ∉ A
SUBSET:
B = {3, 8, 9} ⇒ B ⊆ A
C = {1, 5} ⇒ C ⊄ A
STATISTICS :
MEAN : The mean value is obtained the arithmetic mean or average of a set of numbers is expected value.
The mean value is calculated by adding up all the values, and then dividing that sum by the number of values .
Mean = Sum of all data values / Number of data values
Symbolically ,
Where (read as ” x bar”) is the mean of the set of x values, Σ x is the sum of all the x values, and n is the number of x values.
MEDIAN : The median is the middle value in a set of values. So to find the midian you need to order the numbers from largest to smallest and then you have to choose the value in the middle.
MODE : Mode is the value that the highest frequency in the data set.means values that occur most frequently and there can be more than one mode in a set.
numerical value that occurs most of the times.
F (Xmode) = max
INTERSECTION :
In intersection A ∩ B of two sets A and B is the set that contains all elements of B also belong to A (or similarly all elements of B that also belong to A) but no other elements. The symbol intersection is inverted U.
If Set A contain element A = {1,2,3} and set B contains B = {2,3,4} and the element in having common ares 2 and 3 and this intersection area formed a new set containing 2 and 3.
UNION :
The union of two sets A and B includes all elements which are members of either A or B. If sets A and B have any elements in common then this elements which are members of both sets are only include one in the union.
For example : If sets A contains the elements 1,2, and 3 and set B contains 2,3 and 4 the elements which are members of A or B are 1,2,3 and 4. This form a new set containing 1,2,3 and 4. when we write the union 2 and 3 are only listed once.
RELATIVE COMPLEMENT OF A IN B :
The relative complement of A in B denoted, B \ A, is the set of elements in B but not in A.
Symbolically :B \ A = {x | x ∈ B ∧ x ∉ A}
ABSOLUTE COMPLEMENT :
In a Set theory a complement of a set A refers to things not in A.
SYMMETRIC DIFFERENCE :
Operations on sets :
A ∪ A = A
A ∩ A = A
A ∪ B = B ∩ A
A ∩ B = B ∩ A
(A ∪ B) ∪ C = A ∪ (B ∪ C)
(A ∩ B) ∩ C = A ∩ (B ∩ C)
∪′ = ø
(A′)′ = A
A ∩ ø = ø
A ∩ U = A
A ∩ A′ = ø
(A ∪ B)′ = A′ ∩ B′
(A ∩ B)′ = A′ ∪ B′
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# How to Calculate Slope and Intercepts of a Line
The slope of a line measures how steep the line is. [1] You could also say it is the rise over the run; that is, how much the line rises vertically compared with how much it runs horizontally. Being able to find the slope of a line, or using the slope to find points on the line, is an important skill used in economics, [2] geoscience, [3] accounting/finance and other fields.
Method 1
Method 1 of 4:
### Using a Graph to Find the Slope
1. 1
Pick two points on the line. Draw dots on the graph to represent these points, and note their coordinates.
• Remember when graphing points to list the x-coordinate first, then the y-coordinate.
• For example, you might choose the points (-3, -2) and (5, 4).
2. 2
Determine the rise between the two points. To do this, you must compare the difference in y of the two points. Begin with the first point, the point that is the farthest left on the graph, and count up until you reach the y-coordinate of the second point.
• The rise can be positive or negative; that is, you can count up or down to find it.[4] If the line is moving up and to the right, the rise is positive. If the line is moving down and to the right, the rise is negative. [5]
• For example, if the y-coordinate of the first point is (-2), and the y-coordinate of the second point is (4), you will count up 6 points, so your rise is 6.
3. 3
Determine the run between the two points. To do this, you must compare the difference in x of the two points. Begin with the first point, the point that is farthest left on the graph, and count over until you reach the x-coordinate of the second point.
• To run is always positive; that is, you can only count from left to right, never right to left. [6]
• For example, if the x-coordinate of the first point is (-3), and the x-coordinate of the second point is (5), you will count over 8, so your run is 8.
4. 4
Make a ratio using the rise over the run to determine the slope. The slope is usually in fraction form, but it can also be a whole number.
• For example, if the rise is 6 and the run is 8, then your slope is ${\displaystyle {\frac {6}{8}}}$, which can be simplified to ${\displaystyle {\frac {3}{4}}}$.
Method 2
Method 2 of 4:
### Using Two Given Points to Find the Slope
1. 1
Set up the formula . In the formula, m = the slope, ${\displaystyle (x_{1},y_{1})}$ = the coordinates of the first point, ${\displaystyle (x_{2},y_{2})}$ = the coordinates of the second point.
• Remember that the slope is equal to ${\displaystyle {\frac {rise}{run}}}$. You are using this formula to find the change in y (rise) over the change in x (run). [7]
2. 2
Plug the x- and y-coordinates into the formula. Make sure you place the coordinates of the first point (${\displaystyle (x_{1},y_{1})}$) and the second point (${\displaystyle (x_{2},y_{2})}$) in the correct positions in the formula, or else you will not calculate the correct slope.
• For example, given the points (-3, -2) and (5, 4), your formula will look like this: ${\displaystyle m={\frac {4-(-2)}{5-(-3)}}}$.
3. 3
Complete the calculation and simplify, if possible. This will give you the slope as a fraction or whole number.
• For example, if your slope is ${\displaystyle m={\frac {4-(-2)}{5-(-3)}}}$you should calculate ${\displaystyle 4-(-2)=6}$ in the numerator (Remember when subtracting a negative number, you add.) and ${\displaystyle 5-(-3)=8}$ in the denominator. You can simplify ${\displaystyle {\frac {6}{8}}}$ to ${\displaystyle {\frac {3}{4}}}$, so ${\displaystyle m={\frac {3}{4}}}$.
Method 3
Method 3 of 4:
### Finding the y-intercept, Given the Slope and One Point
1. 1
Set up the formula . In the formula, y = the y-coordinate of any point on the line, m = slope, x = the x-coordinate of any point on the line, and b = the y-intercept.
• ${\displaystyle y=mx+b}$ is the equation of a line. [8]
• The y-intercept is the point at which the line crosses the y-axis.
EXPERT TIP
Math Instructor, City College of San Francisco
Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University.
Grace Imson, MA
Math Instructor, City College of San Francisco
Our Expert Agrees: If you have the slope and one point, plug them into the equation of the line. In y = mx + b, m is the slope, and the point coordinate will contain both x and y. Then, solve for b to find the y-intercept.
2. 2
Plug in the slope and the coordinates of one point in the line. Remember, the slope is equal to the rise over the run. If you need help finding the slope, see the instructions above.
• For example, if the slope is ${\displaystyle {\frac {3}{4}}}$, and on point on the line is (5,4), then the formula will look like this: ${\displaystyle 4={\frac {3}{4}}(5)+b}$.
3. 3
Complete the equation, solving for b. First multiply the slope and the x-coordinate. Subtract this number from both sides to solve for b.
• In the example problem the equation becomes ${\displaystyle 4=3{\frac {3}{4}}+b}$. Subtracting ${\displaystyle 3{\frac {3}{4}}}$ from both sides, you end up with ${\displaystyle {\frac {1}{4}}=b}$. So the y-intercept is ${\displaystyle {\frac {1}{4}}}$.
4. 4
Check your work. On a coordinate graph, plot your known point, then draw a line using the slope. To find the y-intercept, look for the point where the line crosses the y-axis.
• For example, if the slope is ${\displaystyle {\frac {3}{4}}}$, and one point is (5,4), draw a point at (5,4), then draw other points along the line by counting to the left 4 and down 3. When you draw a line through the points, you should see the line cross the y-axis just above the (0,0) coordinate.
Method 4
Method 4 of 4:
### Finding the x-intercept, Given the Slope and Y-intercept
1. 1
Set up the formula . In the formula, y = the y-coordinate of any point on the line, m = slope, x = the x-coordinate of any point on the line, and b = the y-intercept.
• ${\displaystyle y=mx+b}$ is the equation of a line. [9]
• The x-intercept is the point at which the line crosses the x-axis.
2. 2
Plug the slope and y-intercept into the formula. Remember, the slope is equal to the rise over the run. If you need help finding the slope, see the instructions above.
• For example, if the slope is ${\displaystyle {\frac {3}{4}}}$, and the y-intercept is ${\displaystyle {\frac {1}{4}}}$, the formula will look like this: ${\displaystyle y={\frac {3}{4}}x+{\frac {1}{4}}}$.
3. 3
Set y to 0. [10] You are looking for the x-intercept, the point at which the line crosses the x-axis. At this point, the y-coordinate will equal zero. So if we set y to 0, and solve for the corresponding x-coordinate, we will find the point (x, 0), which will be the x-intercept.
• In the example problem, the equation becomes ${\displaystyle 0={\frac {3}{4}}x+{\frac {1}{4}}}$.
4. 4
Complete the equation, solving for x. First subtract the y-intercept from both sides. Then divide both sides by the slope.
• In the example problem the equation becomes ${\displaystyle {\frac {-1}{4}}={\frac {3}{4}}x}$. Dividing both sides by ${\displaystyle {\frac {3}{4}}}$, you end up with ${\displaystyle {\frac {-4}{12}}=x}$. This simplifies to ${\displaystyle {\frac {-1}{3}}=x}$. So the point at which the line crosses the x-axis is ${\displaystyle ({\frac {-1}{3}},0)}$. So the x-intercept is ${\displaystyle {\frac {-1}{3}}}$.
5. 5
Check your work. On a coordinate graph, plot your y-intercept, then draw a line using the slope. To find the x-intercept, look for the point where the line crosses the x-axis.
• For example, if the slope is ${\displaystyle {\frac {3}{4}}}$, and the y-intercept is ${\displaystyle (0,{\frac {1}{4}})}$, draw a point at ${\displaystyle (0,{\frac {1}{4}})}$, then draw other points along the line by counting to the left 4 and down 3, and to the right 3 and up 4. When you draw a line through the points, you should see the line cross the x-axis just left of the (0,0) coordinate.
6. 6
Final Image:
## Community Q&A
Search
• Question
How do I write an equation for the line with a slope of 6 and y-intercept of -3?
Donagan
Use the formula y = mx + b, where m is the slope and b is the y-intercept.
• Question
How do I write the equation of a line, given the slope and y-intercept?
Donagan
Assuming you're talking about a linear (straight-line) equation, you would use the standard slope/y-intercept form, y = mx + b. Use the given slope for m, the coefficient of x. Use the given y-intercept for b, the constant in the equation. For example, if the slope is 3 and the y-intercept is -5, the equation would be y = mx + b = 3x + (-5) = 3x - 5.
• Question
What is intercept and how do I calculate this using a graph?
There are two types of intercept, x-intercept and y-intercept. When we say x-intercept, this means the line passes through the x-axis with coordinates (x,0). When we say y-intercept, the graph passes through the y-axis with coordinates (0,y). To find x or y intercepts, just observe where the line on the graph cuts the x or y axis respectively.
200 characters left
2. The workbook used to write this article was "y = ax + b.xlsx"
Co-authored by:
Math Instructor, City College of San Francisco
This article was co-authored by Grace Imson, MA. Grace Imson is a math teacher with over 40 years of teaching experience. Grace is currently a math instructor at the City College of San Francisco and was previously in the Math Department at Saint Louis University. She has taught math at the elementary, middle, high school, and college levels. She has an MA in Education, specializing in Administration and Supervision from Saint Louis University. This article has been viewed 355,099 times.
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Co-authors: 17
Updated: June 3, 2021
Views: 355,099
Article SummaryX
To find the slope of a line from a graph, first choose 2 points along the line and write down the X and Y coordinates for each. Next, find the rise by taking the difference between the 2 Y coordinates. If the line slopes up as it moves to the right, the rise will be positive. If it slopes down, the rise will be negative. Once you’ve found the rise, calculate the run by finding the difference between the 2 X coordinates, going from left to right. Finally, find the slope by dividing the rise by the run. Keep reading for more tips, including how to find the Y-intercept using the slope and 1 point!
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# Matrix and Math Magic Matrices
Topics: Matrix, Vector space, Multiplication Pages: 2 (320 words) Published: July 22, 2013
Math Magic – Matrices
Page 1
Working With Matrices:
A. Number Sense only deals with small matrices, usually 2 x 2 matrices. This page will look at 3 ways of manipulating matrices: Multiplying Matrices, Inverses, and Determinants. B. Multiplying Matrices 1. Unlike general multiplication, matrix multiplication is not commutative. Multiplying A x B and B x A will give different results. 2. The following will show how to multiply two 2x2 matrices:
a. In other words, you multiply the row of the matrix on the left with the column of the matrix on the right. Ex [1] , then d = _____?
a. Since we are looking for d, then we are only concerned with that spot. So d = 3(2) + 0(5) = 6. The answer is 6.
3. This method of multiplication can be expanded to matrices other that 2x2. However, due to space constraints, on a test, usually you will only see a 2 x2 matrix or smaller.
b. Some people have a hard time knowing which row and column to multiply. This is how I usually think of it. If I were to put the two matrices on top of each other, which row from the first matrix and which column from the second matrix would they intersect on. In this example, it is the second row of the first matrix, and the second column of the second matrix.
Math Magic – Matrices
Page 2
C. Inverses
1. In general, inverses are very difficult to compute. However, on 2x2 matrices, inverses are easy. 2. For a 2x2 matrix:
Ex [1]
, then c = ____.
D. Determinants
a. According to above, c = -4. The answer is -4.
a. In linear algebra, determinants have many uses. Like with inverses, finding the determinants can prove difficult. However, on a 2x2 matrix, determinants are easy. b. Determinants are usually abbreviated "det". So:
Ex [1]
a. The answer is 3(5)-(2)(-1) = 15+2 = 17.
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# Formula for Triangle Area
Earlier we find Area of Triangle by using the formula –
Area of Triangle = $$\sqrt{s(s-a)(s-b)(s-c)}$$
where s = $$a+b+c\over 2$$
and a, b, c are the sides of the triangle.
we have used this heron’s formula to find the area of a triangle when the lengths of its sides are given.
Here, we will learn what is the formula for triangle area in terms of coordinates of its vertices.
## Formula for Triangle Area
The area of triangle, the coordinates of whose vertices are A($$x_1,y_1$$), B($$x_2,y_2$$) and C($$x_3,y_3$$) is given by –
Area of Triangle ABC = $$1\over 2$$ |[$$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$]|
Example : Find the area of a triangle whose vertices are A(3, 2), B(11, 8) and C(8, 12).
Solution : Let A = ($$x_1, y_1$$) = (3, 2), B = ($$x_2, y_2$$) = (11, 8) and C = ($$x_3, y_3$$) = (8, 12) be the given points. Then,
Area of Triangle ABC = $$1\over 2$$ |[$$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$]|
$$\implies$$ A = |[$$3(8-12)+11(12-2)+8(2-8)$$]|
$$\implies$$ A = |(-12+110-48)| = 25 sq. units
Remarks :
(i) If the area of triangle joining three points is zero, then the points are collinear.
(ii) If altitude of any equilateral triangle is P, then its area = $$P^2\over {\sqrt{3}}$$. If ‘a’ be the side of equilateral triangle, then its area = ($$a^2\sqrt{3}\over 4$$)
Example : Prove that the area of triangle whose vertices are (t, t-2), (t+2, t+2) and (t+3, t) is independent of t.
Solution : Let A = ($$x_1, y_1$$) = (t, t-2), B = ($$x_2, y_2$$) = (t+2, t+2) and C = ($$x_3, y_3$$) = (t+3, t) be the vertices of given triangle. Then,
Area of Triangle ABC = $$1\over 2$$ |[$$x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)$$]|
$$\implies$$ A = |[$$t(t+2-t)+(t+2)(t-t+2)+(t+3)(t-2-t-2)$$]|
$$\implies$$ A = |(2t+2t+4-4t-12)| = |-4| = 4 sq. units
Clearly, area of triangle ABC is independent of t.
|
# Hypotenuse - RT
A triangle has a hypotenuse of 55 and an altitude to the hypotenuse of 33. What is the area of the triangle?
Result
S = (Correct answer is: N/A)
#### Solution:
$c=55 \ \\ h=33 \ \\ \ \\ h_{2}=c/2=55/2=\dfrac{ 55 }{ 2 }=27.5 \ \\ \ \\ h > c/2 \ \\ h > h_{2} \ \\ 33 > 27.2 \ \\ \ \\ S=c \cdot \ h / 2 \ \\ S=N/A$
Try calculation via our triangle calculator.
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Showing 1 comment:
Dr Math
Has no solution due to Thales theorem, that maximum height of a right triangle is half of the hypotenuse
Tips to related online calculators
#### You need to know the following knowledge to solve this word math problem:
We encourage you to watch this tutorial video on this math problem:
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# Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 2
Last Updated : 03 Mar, 2021
### Question 14. Find the mean of the following frequency distribution:
Solution:
Let’s consider the assumed mean (A) = 42
From the table it’s seen that,
A = 42 and h = 5
Mean = A + h x (Σfi ui/N)
= 42 + 5 x (-79/70)
= 42 – 79/14
= 42 – 5.643
= 36.357
### Question 15. For the following distribution, calculate mean using all suitable methods:
Solution:
By direct method
Mean = (sum/N) + A
= 848/64
= 13.25
By assuming mean method
Let the assumed mean (A) = 65
Mean = A + sum/N
= 6.5 + 6.75
= 13.25
### Question 16. The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost if living index.
Solution:
Let the assumed mean (A) = 1650
We have
A = 16, h = 100
Mean = A + h (sum/N)
= 1650 + (175/13)
= 21625/13
= 1663.46
### Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Solution:
(i) Direct method:
Mean = sum/N
= 3620/140
= 25.857
(ii) Assumed mean method:
Let the assumed mean = 25
Mean = A + (sum/N)
Mean = A + (sum/N)
= 25 + (120/140)
= 25 + 0.857
= 25.857
(iii) Step deviation method:
Let the assumed mean (A) = 25
Mean = A + h(sum/N)
= 25 + 10(12/140)
= 25 + 0.857
= 25.857
### Question 18. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f1 and f2.
Solution:
Given,
Sum of frequency = 50
5 + f1 + 10 + f2 + 7 + 8 = 50
f1 + f2 = 20
3f1 + 3f2 = 60 —(1) [Multiply both side by 3]
and mean = 62.8
Sum/N = 62.8
(30f1 + 70f2 + 2060)/50 = 62.8
30f1 + 70f2 = 3140 – 2060
30f1 + 70f2 = 1080
3f1 + 7f2 = 108 —(2) [divide it by 10]
Subtract equation (1) from equation (2)
3f1 + 7f2 – 3f1 – 3f2 = 108 – 60
4f2 = 48
f2 = 12
Put value of f2 in equation (1)
3f1 + 3(12) = 60
f1 = 24/3 = 8
f1 = 8, f2 = 12
### Question 19. The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.
Solution:
Given mean = 18,
Let the missing frequency be v
Mean = sum/N
18 = 752 + 20×44 + x
792 + 18x = 752 + 20x
2x = 40
x = 20
### Question 20. If the mean of the following distribution is 27. Find the value of p.
Solution:
Given mean = 27
Mean = sum/N
1245 + 15p43 + p = 27
1245 + 15p = 1161 + 27p
12p = 84
P =7
### Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?
Solution:
We may observe that class internals are not continuous
There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (xi) may be obtained by using the relation
xi = upperlimit + lowerclasslimit2
Class size (h) of this data = 3
Now taking 57 as assumed mean (a) we may calculated di, ui, fiui as follows
Now we have N
Sum = 25
Mean = A +h (sum/N)
= 57 + 3 (45/400)
= 57 + 3/16
= 57 + 0.1875
= 57.19
Clearly mean number of mangoes kept in packing box is 57.19
### Find the mean daily expenditure on food by a suitable method.
Solution:
We may calculate class mark (xi) for each interval by using the relation
xi = upperlimit + lowerclasslimit2
Class size = 50
Now, Taking 225 as assumed mean (xi) we may calculate di, ui, fiui as follows
Now we may observe that
N = 25
Sum = -7
225 + 50 (-7/25)
225 – 14 = 211
So, mean daily expenditure on food is Rs 211
### Find the mean concentration of SO2 in the air
Solution:
We may find class marks for each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Class size of this data = 0.04
Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows
From the table we may observe that
N = 30
Sum = -31
= 0.14 + (0.04)(-31/30)
= 0.099 ppm
So mean concentration of SO2 in the air is 0.099 ppm.
### Question 24. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Solution:
We may find class mark of each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Now, taking 16 as assumed mean (a) we may
Calculate di and fidi as follows
Now we may observe that
N = 40
Sum = -145
= 16 + (-145/40)
= 16 – 3.625
= 12.38
So mean number of days is 12.38 days, for which student was absent.
### Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.
Solution:
We may find class marks by using the relation
x = upperlimit + lowerclasslimit2x =
Class size (h) for this data = 10
Now taking 70 as assumed mean (a) wrong
Calculate di, ui, fiui as follows
Now we may observe that
N = 35
Sum = -2
= 70 + (-2/35)
= 70 – 4/7
= 70 – 0.57
= 69.43
So, mean literacy rate is 69.43%
### Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?
Solution:
We may observe that class internals are not continuous
There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (xi) may be obtained by using the relation
xi = upperlimit + lowerclasslimit2
Class size (h) of this data = 3
Now taking 57 as assumed mean (a) we may calculated di, ui, fiui as follows
Now we have N
Sum = 25
Mean = A +h (sum/N)
= 57 + 3 (45/400)
= 57 + 3/16
= 57 + 0.1875
= 57.19
Clearly mean number of mangoes kept in packing box is 57.19
### Find the mean daily expenditure on food by a suitable method.
Solution:
We may calculate class mark (xi) for each interval by using the relation
xi = upperlimit + lowerclasslimit2
Class size = 50
Now, Taking 225 as assumed mean (xi) we may calculate di, ui, fiui as follows
Now we may observe that
N = 25
Sum = -7
225 + 50 (-7/25)
225 – 14 = 211
So, mean daily expenditure on food is Rs 211
### Find the mean concentration of SO2 in the air
Solution:
We may find class marks for each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Class size of this data = 0.04
Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows
From the table we may observe that
N = 30
Sum = -31
= 0.14 + (0.04)(-31/30)
= 0.099 ppm
So mean concentration of SO2 in the air is 0.099 ppm.
### Question 24. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Solution:
We may find class mark of each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Now, taking 16 as assumed mean (a) we may
Calculate di and fidi as follows
Now we may observe that
N = 40
Sum = -145
= 16 + (-145/40)
= 16 – 3.625
= 12.38
So mean number of days is 12.38 days, for which student was absent.
### Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.
Solution:
We may find class marks by using the relation
x = upperlimit + lowerclasslimit2x =
Class size (h) for this data = 10
Now taking 70 as assumed mean (a) wrong
Calculate di, ui, fiui as follows
Now we may observe that
N = 35
Sum = -2
= 70 + (-2/35)
= 70 – 4/7
= 70 – 0.57
= 69.43
So, mean literacy rate is 69.43%
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# Difference between revisions of "2006 AMC 12B Problems/Problem 17"
## Problem
For a particular peculiar pair of dice, the probabilities of rolling $1$, $2$, $3$, $4$, $5$ and $6$ on each die are in the ratio $1:2:3:4:5:6$. What is the probability of rolling a total of $7$ on the two dice?
$\mathrm{(A)}\ \frac 4{63} \qquad \mathrm{(B)}\ \frac 18 \qquad \mathrm{(C)}\ \frac 8{63} \qquad \mathrm{(D)}\ \frac 16 \qquad \mathrm{(E)}\ \frac 27$
## Solution
The probability of getting an $x$ on one of these dice is $\frac{x}{21}$.
The probability of getting $1$ on the first and $6$ on the second die is $\frac 1{21}\cdot\frac 6{21}$. Similarly we can express the probabilities for the other five ways how we can get a total $7$. (Note that we only need the first three, the other three are symmetric.)
Summing these, the probability of getting a total $7$ is: $$2\cdot\left( \frac 1{21}\cdot\frac 6{21} + \frac 2{21}\cdot\frac 5{21} + \frac 3{21}\cdot\frac 4{21} \right) = \frac{56}{441} = \boxed{\frac{8}{63}}$$
See also 2016 AIME I Problems/Problem 2
|
10.09.2022
# Kylie has 80% of the total points possible in her science class. if 350 points are possible, how many points does kylie have?
0
30.05.2023, solved by verified expert
280
Step-by-step explanation: In order to figure out how many points Kyle has in her science class, we need to find 80% of all the possible points which is 350 points.
To find 80% of 350, first write 80% as a decimal by moving the decimal point two places to the left to get .80.
Next, the word "of" means multiply, so we multiply .80 by 350.
(.80) (350) = 280
Therefore, Kylie has 280 points in he science class.
### Faq
Mathematics
280
Step-by-step explanation: In order to figure out how many points Kyle has in her science class, we need to find 80% of all the possible points which is 350 points.
To find 80% of 350, first write 80% as a decimal by moving the decimal point two places to the left to get .80.
Next, the word "of" means multiply, so we multiply .80 by 350.
(.80) (350) = 280
Therefore, Kylie has 280 points in he science class.
Mathematics
Step-by-step explanation:
Score = Σ(tests) + Σ(exams) = (7 + 8 + 7 + 5) + (81 + 80 + x) = 35 + 161 + x = 196 + x
To receive an A, the student must score at least 0.9 × 350 = 315 points
To receive a B, the student must score at least 0.8 × 350 = 280 points.
1. Can the student receive an A?
2. Minimum score for an A
3. Minimum score for a B
Mathematics
a) The student cannot receive an A in the class.
b) The student must score 119 in the third exams to make an A. This is clearly not possible, since he cannot make 119 in a 100-points exam.
c) The student can make a B but he must score at least 84 in the third exam.
Step-by-step explanation:
To make an A, the student must score 315 (350 x 90%) in both home and the three exams.
The student who scored 35 (7 + 8 + 7 + 5 + 8) in the homework and 161 (81 + 80), getting a total of 196, is short by 119 (315 - 196) scores in making an A.
To make a B, the student must score 280 (350 x 80%) or higher but not reaching 315.
B ≥ 280 and < 315.
Since, the student had scored 196, he needs to score 84 and above to make a B in the last exam.
Mathematics
1. It is not possible, so the student cannot make an A, since he cannot make 119 scores in the third test.
2. Minimum score to make an A in the test is 315 - (35 + 161) = 119
3. The student can make a B, if he can score 84 and above in the third test.
Step-by-step explanation:
1. To get an A, a student must score 90% (0.9) and above
≥ 90% of 350 ≥ 315 scores,
Total Scores in the homework = 35 ((7 x 2) + (8 x 2) + 5) or 70% (35/50)
Total Scores in the two tests = 161 (81 + 80) or 80.5% (161/200)
Total scores by the student in homework and the two tests so far = 196 (35 + 161).
However, he needs to score 315 (350 x 90%) to make an A.
Therefore, he is short by 119 (350 - 196), he cannot make an A.
2. To get a B, the student must score ≥ 80% or < 90%,
ideally, the student must score 280 (80% of 350) or above.
He would need to score 84 (280 - 196) in the third test to make 280 scores or 80% average of the total.
Therefore, he can make a B if he can score 84 in the third test.
Mathematics
i) it is not possible for the student to receive an A in the class
ii) 119 points
iii) 84points
Step-by-step explanation:
Total exam scores = 350points
homework scores of 7, 8, 7, 5, and 8
Let the third exam score =y
Exam scores = 81, 80, x
i) To know if the student would get an A in class, we would find the third exam score
(Scores received by a student)/ (total scores) = least of the grade percentage to get an A
(7 + 8 + 7 +5 + 8 + 81 + 80 + x)/350 = 0.9
(196+x)/350 = 0.9
196+x = 350 × 0.9
196+x = 315
x = 315-196
x = 119
119 > 100
Since the maximum grade for each of the exam score is 100points, it is not possible for the student to receive an A in the class.
ii) Since the least of the grade percentage that would guarantee an A is 0.9, the minimum score on the third exam that will give an A = 119points
iii) (Scores received by a student)/ (total scores) = least of the grade percentage to get a B
(7 + 8 + 7 +5 + 8 + 81 + 80 + x)/350 = 0.8
(196+x)/350 = 0.8
196+x = 350 × 0.8 = 280
x = 280-196
x = 84
The minimum score on the third exam that will give a B = 84points
Mathematics
It is not possible for the student to receive an A grade in the class.
It is possible for the student to receive a B grade in the class.
Step-by-step explanation:
We are given that in the DBE 122 class, there are 350 possible points. These points come from 5 homework sets that are worth 10 points each and 3 exams that are worth 100 points each.
A student has received homework scores of 7, 8, 7, 5, and 8, and the first two exam scores are 81 and 80.
Firstly, we will calculate how many points have been scored by the student.
Number of possible points = 350
The points scored by the student in homework = 7 + 8 + 7 + 5 + 8 = 35 points.
The scores of the student on the two exams = 81 + 80 = 161 points
So, the total points scored by the students = 35 + 161 = 196 points.
As it is given in the question that if the grade percentage is 0.9 or higher then the student will get an A, i.e;
If the total possible points are 350 points;
This means that the student must have to score 315 points to get an A grade.
Till the second exam, the total points scored by the students are 196 points. If the student scored full 100 marks in the third exam, then the total points scored by the student will be = 196 + 100 = 296 points.
Since 296 < 315, this means that it is not possible for the student to receive an A in the class.
Also, it is given in the question that if the grade percentage is between 0.8 and 0.9 the student will get a B, i.e, the student must obtain a minimum of 80% to get B grade.
If the total possible points are 350 points;
This means that the student must have to score a minimum of 280 points to get a B grade.
Till the second exam, the total points scored by the students are 196 points. If the student scored full 100 marks in the third exam, then the total points scored by the student will be = 196 + 100 = 296 points.
Since 296 > 280, this means that it is possible for the student to receive a B grade in the class.
Mathematics
Answer: 440 grams for 1.54 is the better value
Explanation:
Take the price and divide by the number of grams
1.54 / 440 =0.0035 per gram
1.26 / 340 =0.003705882 per gram
0.0035 per gram < 0.003705882 per gram
Mathematics
The answer is in the image
Mathematics
The solution is in the following image
Mathematics
F=ma
where F=force
m=mass
a=acceleration
Here,
F=4300
a=3.3m/s2
m=F/a
=4300/3.3
=1303.03kg
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# Relation Between Continuity & Differentiability Notes | Study Mathematics (Maths) Class 12 - JEE
## JEE: Relation Between Continuity & Differentiability Notes | Study Mathematics (Maths) Class 12 - JEE
The document Relation Between Continuity & Differentiability Notes | Study Mathematics (Maths) Class 12 - JEE is a part of the JEE Course Mathematics (Maths) Class 12.
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E. Relation between Continuity & Differentiability
If a function f is derivable at x then f is continuous at x.
If f(x) is derivable for every point of its domain, then it is continuous in that domain .
The converse of the above result is not true :
"If f is continuous at x, then f may or maynot be derivable at x"
The functions f(x) = & g(0) = 0 are continuous at x = 0 but not derivable at x = 0.
Remark :
(a) Let f'+(a) = p & f'_(a) = q where p & q are finite then :
(i) p = q ⇒ f is derivable at x = a ⇒ f is continuous at x = a.
(ii) p ≠ q ⇒ f is not derivable at x = a but f is continuous at x = a
Differentiable ⇒ Continuous ; Non-differentiable Discontinuous
But Discontinuous ⇒ Non-differentiable .
(b) If a function f is not differentiable but is continuous at x = a it geometrically implies a sharp corner at x = a.
Ex.15 If f(x) = , then find the value of k so that f(x) becomes continuous at x = 0. Hence, find all the points where the functions is non-differentiable.
Sol. From the graph of f(x) it is clear that for the function to be continuous only possible value of k is 1.
Points of non-differentiability are x = 0, ±1.
Ex.16 If f(x) = where [.] denotes the greatest integer function.
Discuss the continuity and differentiability of f(x) in [0, 2).
Sol. Since 1 x - 1 < 1 then [x2 - 2x] = [(x - 1)2 - 1] = [(x - 1)2] - 1 = 0 - 1 = -1
Graph of f(x) :
It is clear from the graph that f(x) is discontinuous at x = 1 and not differentiable at x =1/2,and x= 1
Further details are as follows : ⇒
Hence, which shows f(x) is not differentiable at x = 1/2 (as RHD = 4 and LHD = –4) and x = 1 (as RHD = 0 and LHD = 8). Therefore, f(x) is differentiable, for x ∈ [0, 2) - {1/2, 1}
Ex.17 Suppose f (x) = . If f '' (1) exist then find the value of a2 + b2 + c2.
Sol. For continuity at x = 1 we leave f (1) = 1 and f (1+) = a + b + c
a + b + c = 1 ....(1)
for continuity of f ' (x) at x = 1 f ' (1) = 3; f ' (1+) = 2a + b
hence 2a + b = 3 ....(2)
f '' (1–) = 6; f '' (1+) = 2a for continuity of f '' (x) 2a = 6 ⇒ a = 3
from (2), b = – 3 ; c = 1. Hence a = 3, b = – 3 ; c = 1
Ex.18 Check the differentiability of the function f(x) = max {sin-1 |sin x|, cos-1 |sin x|}.
Sol. sin-1 |sin x| is periodic with period ⇒ sin-1 |sin x| =
Also cos-1 |sin x| = - sin-1 |sin x|
⇒ f(x) is not differentiable at
⇒ f(x) is not differentiable at
Ex.19 Find the interval of values of k for which the function f(x) = |x2 + (k - 1) |x| - k| is non differentiable at five points.
Sol.
f(x) = |x2 + (k – 1) |x| – k| = |(|x| – 1) (|x| + k)|
Also f(x) is an even function and f(x) is not differentiable at five points.
So |(x – 1) (x + k)| is non differentiable for two positive values of x.
⇒ Both the roots of (x – 1) (x + k) = 0 are positive.
⇒ k < 0 ⇒ k ∈ (–∝, 0).
Definition : A function f is differentiable at a if f'(a) exists. It is differentiable on an open interval (a,b) [or (a, ∝) or (–∝, a) or (– ∝, ∝)] if it is differentiable at every number in the interval.
Derivability Over An Interval : f(x) is said to be derivable over an interval if it is derivable at each & every point of the interval. f(x) is said to be derivable over the closed interval [a, b] if :
(i) for the points a and b, f'(a+) & f'(b -) exist &
(ii) for any point c such that a < c < b, f'(c+) & f'(c -) exist & are equal .
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# Data Science - Linear Functions
Mathematical functions are important to know as a data scientist, because we want to make predictions and interpret them.
## Linear Functions
In mathematics a function is used to relate one variable to another variable.
Suppose we consider the relationship between calorie burnage and average pulse. It is reasonable to assume that, in general, the calorie burnage will change as the average pulse changes - we say that the calorie burnage depends upon the average pulse.
Furthermore, it may be reasonable to assume that as the average pulse increases, so will the calorie burnage. Calorie burnage and average pulse are the two variables being considered.
Because the calorie burnage depends upon the average pulse, we say that calorie burnage is the dependent variable and the average pulse is the independent variable.
The relationship between a dependent and an independent variable can often be expressed mathematically using a formula (function).
A linear function has one independent variable (x) and one dependent variable (y), and has the following form:
y = f(x) = ax + b
This function is used to calculate a value for the dependent variable when we choose a value for the independent variable.
Explanation:
• f(x) = the output (the dependant variable)
• x = the input (the independant variable)
• a = slope = is the coefficient of the independent variable. It gives the rate of change of the dependent variable
• b = intercept = is the value of the dependent variable when x = 0. It is also the point where the diagonal line crosses the vertical axis.
## Linear Function With One Explanatory Variable
A function with one explanatory variable means that we use one variable for prediction.
Let us say we want to predict calorie burnage using average pulse. We have the following formula:
f(x) = 2x + 80
Here, the numbers and variables means:
• f(x) = The output. This number is where we get the predicted value of Calorie_Burnage
• x = The input, which is Average_Pulse
• 2 = Slope = Specifies how much Calorie_Burnage increases if Average_Pulse increases by one. It tells us how "steep" the diagonal line is
• 80 = Intercept = A fixed value. It is the value of the dependent variable when x = 0
## Plotting a Linear Function
The term linearity means a "straight line". So, if you show a linear function graphically, the line will always be a straight line. The line can slope upwards, downwards, and in some cases may be horizontal or vertical.
Here is a graphical representation of the mathematical function above:
### Graph Explanations:
• The horizontal axis is generally called the x-axis. Here, it represents Average_Pulse.
• The vertical axis is generally called the y-axis. Here, it represents Calorie_Burnage.
• Calorie_Burnage is a function of Average_Pulse, because Calorie_Burnage is assumed to be dependent on Average_Pulse.
• In other words, we use Average_Pulse to predict Calorie_Burnage.
• The blue (diagonal) line represents the structure of the mathematical function that predicts calorie burnage.
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Transcript
```Lesson 2 – 4: Solving Equations with Variables on Both Sides
2-4.1 - Solving equations with variables on both sides
To solve equation that have variables on both sides we must collect the variables terms so that they are on only
one side. We do this be adding or subtracting the variables to make one side zero. This is step 3 of the 5 for
solving equations
5 Steps to solve Equations:
Distribute to get rid of Parentheses
Combine like Terms
Get Variables to one side
Inverse of Multiplication and Division
Examples: Solve each equation
5 x 2 3x 4
7 k 4k 15
4k 4k
3k 15
k 5
3 x
3x
2x 2 4
2
2x 6
2
x3
2-4.2 - Simplifying both sides before solving
Examples: Solve each equation
2( y 6) 3 y
2 y 12 3 y
2 y
2y
12 y
3 5 x 2 x 2 2(1 x)
3 5 x 2 x 2 2 2 x
3 3x 4 2 x
3x
3x
3 4 5 x
4 4
7 5x
7
x
5
2-4.3 - Equations with infinitely many solutions or no solutions
Vocabulary:
Identity - an equation that is true for all values of the variable. It has infinite solutions
Contradiction - an equation that is not true for any value of the variable. It has no solutions
Sometimes equations are true no matter what value is substituted in for the variable. In this case, the equation is
called an identity and is said to have infinitely many solutions.
On the other hand, sometimes no matter what value is put into an equation for the variable nothing will work
and it has no solutions. This is said to be a contradiction.
Examples: Solve each equation
x 4 6 x 6 5x 2
5 x 4 4 5 x
5 x
5x
44
Infinite number of solutions
8 x 6 9 x 17 x
x 6 17 x
x
x
6 17
No solutions
```
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# Adding and Subtracting Fractions Help
based on 1 rating
By McGraw-Hill Professional
Updated on Sep 26, 2011
## Adding and Subtracting Fractions - Same and Different Denominators
### Same Denominators
When adding (or subtracting) fractions with the same denominators, add (or subtract) their numerators.
### Different Denominators
When the denominators are not the same, you have to rewrite the fractions so that they do have the same denominator. There are two common methods of doing this. The first is the easiest. The second takes more effort but can result in smaller quantities and less reducing. (When the denominators have no common divisors, these two methods are the same.)
### Finding A Common Denominator
The easiest way to get a common denominator is to multiply the first fraction by the second denominator over itself and the second fraction by the first denominator over itself.
In the first denominator is 2 and the second denominator is 7. Multiply by and multiply by .
#### Examples
Find practice problems and solutions at Adding and Subtracting Fractions Practice Problems - Set 1 and Set 2.
## Finding the Least Common Denominator (LCD)
Our goal is to add/subtract two fractions with the same denominator. In the previous examples and practice problems, we found a common denominator. Now we will find the least common denominator (LCD). For example in , we could compute
But we really only need to rewrite :
While 18 is a common denominator in the above example, 6 is the smallest common denominator. When denominators get more complicated, either by being large or having variables in them, you will find it easier to use the LCD to add or subtract fractions. The solution might require less reducing, too.
Find practice problems and solutions at Adding and Subtracting Fractions Practice Problems - Set 3.
## Prime Facorization
There are a couple of ways of finding the LCD. Take for example . We could list the multiples of 12 and 14—the first number that appears on each list will be the LCD:
12, 24, 36, 48, 60, 72, 84 and 14, 28, 42, 56, 70, 84 .
Because 84 is the first number on each list, 84 is the LCD for and . This method works fine as long as your lists are not too long. But what if your denominators are 6 and 291? The LCD for these denominators (which is 582) occurs 97th on the list of multiples of 6.
We can use the prime factors of the denominators to find the LCD more efficiently. The LCD will consist of every prime factor in each denominator (at its most frequent occurrence). To find the LCD for and factor 12 and 14 into their prime factorizations: 12 = 2 · 2 · 3 and 14 = 2 · 7. There are two 2s and one 3 in the prime factorization of 12, so the LCD will have two 2s and one 3. There is one 2 in the prime factorization of 14, but this 2 is covered by the 2s from 12. There is one 7 in the prime factorization of 14, so the LCD will also have a 7 as a factor. Once you have computed the LCD, divide the LCD by each denominator. Multiply each fraction by this number over itself.
#### Examples
Find practice problems and solutions at Adding and Subtracting Fractions Practice Problems - Set 4.
## Adding More than Two Fractions
Finding the LCD for three or more fractions is pretty much the same as finding the LCD for two fractions. Factor each denominator into its prime factorization and list the primes that appear in each. Divide the LCD by each denominator. Multiply each fraction by this number over itself.
#### Examples
Prime factorization of the denominators:
5 = 5 15 = 3 · 5 20 = 2 · 2 · 5
The LCD = 2 · 2 · 3 · 5 = 60
Prime factorization of the denominators:
10 = 2 · 5 12 = 2 · 2 · 3 18 = 2 · 3 · 3
LCD = 2 · 2 · 3 · 3 · 5 = 180
Find practice problems and solutions at
Practice problems for this concept can be found at: Algebra Fractions Practice Test.
150 Characters allowed
### Related Questions
#### Q:
See More Questions
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# What is 201/168 as a decimal?
## Solution and how to convert 201 / 168 into a decimal
201 / 168 = 1.196
Fraction conversions explained:
• 201 divided by 168
• Numerator: 201
• Denominator: 168
• Decimal: 1.196
• Percentage: 1.196%
201/168 or 1.196 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Both are used to handle numbers less than one or between whole numbers, known as integers. The difference between using a fraction or a decimal depends on the situation. Fractions can be used to represent parts of an object like 1/8 of a pizza while decimals represent a comparison of a whole number like \$0.25 USD. If we need to convert a fraction quickly, let's find out how and when we should.
201 / 168 as a percentage 201 / 168 as a fraction 201 / 168 as a decimal
1.196% - Convert percentages 201 / 168 201 / 168 = 1.196
## 201/168 is 201 divided by 168
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 201 is being divided into 168. Think of this as our directions and now we just need to be able to assemble the project! Fractions have two parts: Numerators and Denominators. This creates an equation. We must divide 201 into 168 to find out how many whole parts it will have plus representing the remainder in decimal form. This is our equation:
### Numerator: 201
• Numerators are the portion of total parts, showed at the top of the fraction. Overall, 201 is a big number which means you'll have a significant number of parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Large two-digit conversions are tough. Especially without a calculator. So how does our denominator stack up?
### Denominator: 168
• Denominators represent the total parts, located at the bottom of the fraction. 168 is a large number which means you should probably use a calculator. The good news is that having an even denominator makes it divisible by two. Even if the numerator can't be evenly divided, we can estimate a simplified fraction. Ultimately, don't be afraid of double-digit denominators. Now it's time to learn how to convert 201/168 to a decimal.
## How to convert 201/168 to 1.196
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 168 \enclose{longdiv}{ 201 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Solve for how many whole groups you can divide 168 into 201
$$\require{enclose} 00.1 \\ 168 \enclose{longdiv}{ 201.0 }$$
How many whole groups of 168 can you pull from 2010? 168 Multiply this number by 168, the denominator to get the first part of your answer!
### Step 3: Subtract the remainder
$$\require{enclose} 00.1 \\ 168 \enclose{longdiv}{ 201.0 } \\ \underline{ 168 \phantom{00} } \\ 1842 \phantom{0}$$
If there is no remainder, you’re done! If there is a remainder, extend 168 again and pull down the zero
### Step 4: Repeat step 3 until you have no remainder
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. And the same is true for percentages. It’s common for students to hate learning about decimals and fractions because it is tedious. But 201/168 and 1.196 bring clarity and value to numbers in every day life. Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 201/168 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 1.196 to 201/168 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 201/168 = 1.196 what would it be as a percentage?
• What is 1 + 201/168 in decimal form?
• What is 1 - 201/168 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 1.196 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 168 Denominator What is 201/158 as a decimal? What is 191/168 as a decimal? What is 201/159 as a decimal? What is 192/168 as a decimal? What is 201/160 as a decimal? What is 193/168 as a decimal? What is 201/161 as a decimal? What is 194/168 as a decimal? What is 201/162 as a decimal? What is 195/168 as a decimal? What is 201/163 as a decimal? What is 196/168 as a decimal? What is 201/164 as a decimal? What is 197/168 as a decimal? What is 201/165 as a decimal? What is 198/168 as a decimal? What is 201/166 as a decimal? What is 199/168 as a decimal? What is 201/167 as a decimal? What is 200/168 as a decimal? What is 201/168 as a decimal? What is 201/168 as a decimal? What is 201/169 as a decimal? What is 202/168 as a decimal? What is 201/170 as a decimal? What is 203/168 as a decimal? What is 201/171 as a decimal? What is 204/168 as a decimal? What is 201/172 as a decimal? What is 205/168 as a decimal? What is 201/173 as a decimal? What is 206/168 as a decimal? What is 201/174 as a decimal? What is 207/168 as a decimal? What is 201/175 as a decimal? What is 208/168 as a decimal? What is 201/176 as a decimal? What is 209/168 as a decimal? What is 201/177 as a decimal? What is 210/168 as a decimal? What is 201/178 as a decimal? What is 211/168 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 168 Denominator 202/168 as a percentage 201/169 as a percentage 203/168 as a percentage 201/170 as a percentage 204/168 as a percentage 201/171 as a percentage 205/168 as a percentage 201/172 as a percentage 206/168 as a percentage 201/173 as a percentage 207/168 as a percentage 201/174 as a percentage 208/168 as a percentage 201/175 as a percentage 209/168 as a percentage 201/176 as a percentage 210/168 as a percentage 201/177 as a percentage 211/168 as a percentage 201/178 as a percentage
|
# Lesson 2: Finding the Missing Lengths using the Pythagorean Theorem
By Donna Ventura
Students will use the Pythagorean Theorem to find the missing lengths of the sides of right triangles.
Use the Pythagorean Theorem to solve each of the problems.
Learning Objective: The lesson is aligned to the Common Core State Standards for Mathematics - 8.G.7 Geometry - Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
Materials Required: graph paper, calculator
## Lesson Procedure
Two-Dimensional Figures
1. The triangle shown here is a right triangle.
2. Side a equals 10 meters.
3. Side b equals 6 meters.
4. Use the Pythagorean Theorem a2 + b2 = c2, to find the length of side c of the right triangle. Calculate the length to the nearest tenth.
Answer: 11.7 meters
Three Dimensional Figures
1. The dimensions of a rectangular box are 4 cm x 5 cm x 3 cm. What is the length of the longest pole you could fit in the rectangular storage unit?
2. Let a = 4 cm, b = 5 cm, and c = 3 cm.
3. Use the formula = + + , to find the length of the longest pole (x). Calculate the length to the nearest tenth.
Answer: 7.1 centimeters
## Individual or Group Work:
Questions 1 - 4. Triangles drawn below are right triangles. Find the missing length for each right triangle. Calculate the length to the nearest tenth.
5. The dimensions of a rectangular storage unit are 4ft x 8ft x 2ft. What is the length of the longest pole you could fit in the rectangular storage unit? Calculate the length to the nearest tenth. 6. The dimensions of a rectangular juice box are 12 cm x 5 cm x 3 cm. What is the maximum length of a straw from the flex part of the straw to the end of the straw that can be used in the juice box? Calculate the length to the nearest tenth. Students should be able to apply the Pythagorean Theorem, a2 + b2 = c2 to determine unknown side lengths in right triangles in real-world problems with two-dimensional and three-dimensional figures.
## Answers
1. Rectangle 1: 8.6 feet
2. Rectangle 2: 8.2 feet
3. Rectangle 3: 9 meters
4. Rectangle 4: 16 meters
5. Rectangular Storage Unit: 9.2 feet
6. Rectangular Juice Box: 13.3 centimeters
|
# 2.2: Functions and Simple Graphs
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A large part of science may be described as the study of the dependence of one measured quantity on another measured quantity. The word function is used in this context in a special way. In previous examples, the word function may have been used as follows:
• The density of V. natriegens is a function of time.
• Light intensity is a function of depth below the surface in an ocean.
• Light intensity is a function of distance from the light source.
• The frequency of cricket chirps is a function of the ambient temperature.
• The percentage of females turtles from a clutch of eggs is a function of the incubation temperature.
## 2.2.1 Three definitions of “Function”.
Because of its prevalence and importance in science and mathematics, the word function has been defined several ways over the past three hundred years, and now is usually given a very precise formal meaning. More intuitive meanings are also helpful and we give three definitions of function, all of which will be useful to us.
The word variable means a symbol that represents any member of a given set, most often a set of numbers, and usually denotes a value of a measured quantity. Thus density of V. natriegens, time, percentage of females, incubation temperature, light intensity, depth and distance are all variables.
The terms dependent variable and independent variable are useful in the description of an experiment and the resulting functional relationship. The density of V. natriegens (dependent variable) is a function of time (independent variable). The percentage of females turtles from a clutch of eggs (dependent variable) is a function of the incubation temperature (independent variable).
Using the notion of variable, a function may be defined:
Definition 2.2.1 Function I Given two variables, x and y, a function is a rule that assigns to each value of x a unique value of y.
In this context, x is the independent variable, and y is the dependent variable. In some cases there is an equation that nicely describes the ‘rule’; in the percentage of females in a clutch of turtle eggs examples of the preceding section, there was not a simple equation that described the rule, but the rule met the definition of function, nevertheless.
The use of the words dependent and independent in describing variables may change with the context of the experiment and resulting function. For example, the data on the incubation of turtles implicitly assumed that the temperature was held constant during incubation. For turtles in the wild, however, temperature is not held constant and one might measure the temperature of a clutch of eggs as a function of time. Then, temperature becomes the dependent variable and time is the independent variable.
In Definition 2.2.1, the word ‘variable’ is a bit vague, and ‘a function is a rule’ leaves a question as to ‘What is a rule?’. A ‘set of objects’ or, equivalently, a ‘collection of objects’, is considered to be easier to understand than ‘variable’ and has broader concurrence as to its meaning. Your previous experience with the word function may have been that
Definition 2.2.2 Function II A function is a rule that assigns to each number in a set called the domain a unique number in a set called the range of the function.
Definition 2.2.2 is similar to Definition 2.2.1, except that ‘a number in a set called the domain’ has given meaning to independent variable and ‘a unique number in a set called the range’ has given meaning to dependent variable.
The word ‘rule’ is at the core of both definitions 2.2.1 Function I. and 2.2.2 Function II. and is still a bit vague. The definition of function currently considered to be the most concise is:
Definition 2.2.3 Function III A function is a collection of ordered number pairs no two of which have the same first number.
A little reflection will reveal that ‘a table of data’ is the motivation for Definition 2.2.3. A data point is actually a number pair. Consider the tables of data shown in Table 2.1 from V. natriegens growth and human population records. (16,0.036) is a data point. (64,0.169) is a data point. (1950,2.52) and ( 1980,4.45) are data points. These are basic bits of information for the functions. On the other hand, examine the data for cricket chirps in the same table, from Chapter 1. That also is a collection of ordered pairs, but the collection does not satisfy Definition 2.2.3. There are two ordered pairs in the table with the same first term – (66,102) and (66,103). Therefore the collection contains important information about the dependence of chirp frequency on temperature, although the collection does not constitute a function.
Table 2.1: Examples of tables of data
V. natriegens Growth pH 6.25
Time (min)Population Density
00.022
160.036
320.060
480.101
640.169
800.266
World Population
YearPopulation (billions)
19402.30
19502.52
19603.02
19703.70
19804.45
19905.30
20006.06
Cricket Chirps
Temperature ((^{circ} F))Chirps per Minute
67109
73136
78160
6187
66103
66102
67108
77154
74144
76150
In a function that is a collection of ordered number pairs, the first number of a number pair is always a value of the independent variable and a member of the domain and the second number is always a value of the dependent variable and a member of the range. Almost always in recording the results of an experiment, the numbers in the domain are listed in the column on the left and the numbers in the range are listed in the column on the right. Formally,
Definition 2.2.4 Domain and Range For Definition 2.2.3 of function, the domain is defined as the set of all numbers that occur as the first number in an ordered pair of the function and the range of the function is the set of all numbers that occur as a second number in an ordered pair of the function.
Example 2.2.1 Data for the percentage of U.S. population in 1955 that had antibodies to the polio virus as a function of age is shown in Table 2.2.1.1. The data show an interesting fact that a high percentage of the population in 1955 had been infected with polio. A much smaller percentage were crippled or killed by the disease.
Although Table 2.2.1.1 is a function, it is only an approximation to a perhaps real underlying function. The order pair, (17.5, 72), signals that 72 percent of the people of age 17.5 years had antibodies to the polio virus. More accurately, (17.5,72) signals that of a sample of people who had ages in the interval from age 15 to less than 20, the percentage who tested positive to antibodies to the polio virus was greater than or equal to 71.5 and less than 72.5.
Table 2.2.1.1 is a useful representation of an enormous table of data that lists for a certain instance of time during 1955, for each U.S. citizen, their age (measured perhaps in hours (minutes?, seconds?)), and whether they had HIV antibodies, ’yes’ or ’no,’ This table would not be a function, but for each age, the percent of people of that age who were HIV positive would be a number and those age-percent pairs would form a function. The domain of that function would be the finite set of ages in the U.S. population.
Age % 0.8 1.5 2.5 3.5 4.5 5.5 6.5 7.5 8.5 9.5 12.5 17.5 22.5 27.5 3 6 13 19 27 35 40 43 46 49 64 72 78 87
Remember that only a very few data ‘points’ were listed from the large number of possible points in each of the experiments we considered. There is a larger function in the background of each experiment.
Because many biological quantities change with time, the domain of a function of interest is often an interval of time. In some cases a biological reaction depends on temperature (percentage of females in a clutch of turtle eggs for example) so that the domain of a function may be an interval of temperatures. In cases of spatial distribution of a disease or light intensity below the surface of a lake, the domain may be an interval of distances.
It is implicit in the bacteria growth data that at any specific time, there is only one value of the bacterial density2 associated with that time. It may be incorrectly or inaccurately read, but a fundamental assumption is that there is only one correct density for that specific time. The condition that no two of the ordered number pairs have the same first term is a way of saying that each number in the domain has a unique number in the range associated with it.
All three of the definitions of function are helpful, as are brief verbal descriptions, and we will rely on all of them. Our basic definition, however, is the ordered pair definition, Definition 2.2.3.
## 2.2.2 Simple graphs.
Coordinate geometry associates ordered number pairs with points of the plane so that by Definition 2.2.3 a function is automatically identified with a point set in the plane called a simple graph:
Definition 2.2.5 Simple Graph
A simple graph is a point set, G, in the plane such that no vertical line contains two points of G.
The domain of G is the set of x-coordinates of points of G and the range of G is the set of all y-coordinates of points of G.
Note: For use in this book, every set contains at least one element.
The domain of a simple graph G is sometimes called the x-projection of G, meaning the vertical projection of G onto the x-axis and the range of G is sometimes called the y-projection of G meaning the horizontal projection of G onto the y-axis.
A review of the graphs of incubation temperature - percentage of females turtles in Figure 2.1.1 and Exercise Fig. 2.1.1 will show that in each graph at least one vertical line contains two points of the graph. Neither of these graphs is a simple graph, but the graphs convey useful information.
A circle is not a simple graph. As shown in Figure 2.2.1A there is a vertical line that contains two points of it. There are a lot of such vertical lines. The circle does contain a simple graph, and contains one that is ‘as large as possible’. The upper semicircle shown in Figure 2.2.1B is a simple graph. The points (-1,0) and (1,0) are filled to show that they belong to it. It is impossible to add any other points of the circle to this simple graph and still have a simple graph — thus it is ‘as large as possible’. An equation of the upper semicircle is
[y=sqrt{1-x^{2}}, quad-1 leq x leq 1]
The domain of this simple graph is [-1,1], and the range is [0,1]. Obviously the lower semicircle is a maximal simple graph also, and it has the equation
[y=-sqrt{1-x^{2}}, quad-1 leq x leq 1]
The domain is again [-1,1], and the range is [-1,0].
Figure (PageIndex{1}): A. A circle; a vertical line contains two point of the circle so that it is not a simple graph. B. A subset of the circle that is a simple graph. C. Another subset of the circle that is a simple graph. The simple graphs in (b) and (c) are maximal in the sense that any point from the circle added to the graphs would create a set that is not a simple graph — the vertical line containing that point would also contain a point from the original simple graph.
There is yet a third simple graph contained in the circle, shown in Figure 2.2.1C, and it is ‘as large as possible’. An equation for that simple graph is
[y=left{egin{aligned}
sqrt{1-x^{2}} & ext { if } &-1 leq x leq 0
-sqrt{1-x^{2}} & ext { if } & 0end{aligned} ight.]
The domain is [-1,1] and range is [-1,1] Because of the intuitive advantage of geometry, it is often useful to use simple graphs instead of equations or tables to describe functions, but again, we will use any of these as needed.
### Exercises for Section 2.2 Functions and Simple Graphs.
Exercise 2.2.1 Which of the tables shown in Table Ex. 2.2.1 reported as data describing the growth of V. natriegens are functions?
Table for Exercise 2.2.1 Hypothetical data for V. natriegens growing in Nutrient Broth.
TimeAbsTimeAbsTimeAbs
00.01800.01800.018
120.023120.023120.023
240.030240.030240.030
360.039360.039480.049
480.049480.049480.049
480.065600.065480.049
600.085720.065720.065
780.120870.065870.065
960.145960.080960.080
1100.1951100.0951100.095
1200.2401200.1201200.120
Exercise 2.2.2 For the following experiments, determine the independent variable and the dependent variable, and draw a simple graph or give a brief verbal description (your best guess) of the function relating the two.
1. A rabbit population size is a function of the number of coyotes in the region.
2. An agronomist, interested in the most economical rate of nitrogen application to corn, measures the corn yield in test plots using eight different levels of nitrogen application.
3. An enzyme, E, catalyzes a reaction converting a substrate, S, to a product P according to [mathrm{E}+mathrm{S} ightleftharpoons mathrm{ES} ightleftharpoons mathrm{E}+mathrm{P}] Assume enzyme concentration, [E], is fixed. A scientist measures the rate at which the product P accumulates at different concentrations, [S], of substrate.
4. A scientist titrates a 0.1 M solution of HCl into 5 ml of an unknown basic solution containing litmus (litmus causes the color of the solution to change as the pH changes).
Exercise 2.2.3 A table for bacterial density for growth of V. natriegens is repeated in Exercise Table 2.2.3. There are two functions that relate population density to time in this table, one that relates population density to time and another that relates population to time index.
1. Identify an ordered pair that belongs to both functions.
2. One of the functions is implicitly only a partial list of the order pairs that belong to it. You may be of the opinion that both functions have that property, but some people may think one is more obviously only a sample of the data. Which one?
3. What is the domain of the other function?
Table for Exercise 2.2.3 Data for V. natriegens growing in pH 6.25 nutrient broth.
pH 6.25
Time (min)Time Index (t)Population Density (B_t)
000.022
1610.036
3220.060
4830.101
6440.169
8050.266
Exercise 2.2.4 Refer to the graphs in Figure Ex. 2.2.4.
1. Which of the graphs are simple graphs?
2. For those that are not simple graphs,
1. Draw, using only the points of the graph, a simple graph that is ‘as large as possible’, meaning that no other points can be added and still have a simple graph.
2. Draw a second such simple graph.
3. Identify the domains and ranges of the two simple graphs you have drawn.
4. How many such simple graphs may be drawn?
Figure for Exercise 2.2.4 Graphs for Exercise 2.2.4. Some are simple graphs; some are not simple graphs.
Exercise 2.2.5 Make a table showing the ordered pairs of a simple graph contained in the graph in Figure Ex. 2.2.5 and that has domain
[{-1.5,-1.0,-0.5,0.0,0.5,1.0,1.5}]
How many such simple graphs are contained in the graph of Figure Ex. 2.2.5 and that have this domain?
Figure for Exercise 2.2.5 Graph for Exercise 2.2.5.
Exercise 2.2.6
1. Does every subset of the plane contain a simple graph?
2. Does every subset of the plane contain two simple graphs?
3. Is there a subset of the plane that contains two and only two simple graphs?
4. Is there a line in the plane that is not the graph of a function?
5. Is there a function whose graph is a circle?
6. Is there a simple graph in the plane whose domain is the interval [0,1] (including 0 and 1) and whose range is the interval [0,3]?
7. Is there a simple graph in the plane whose domain is [0,1] and whose range is the y-axis?
Exercise 2.2.7 A bit of a difficult exercise. For any location, (lambda) on Earth, let Annual Daytime at (lambda) , (AD(lambda)), be the sum of the lengths of time between sunrise and sunset at (lambda) for all of the days of the year. Find a reasonable formula for (AD(lambda)). You may guess or find data to suggest a reasonable formula, but we found proof of the validity of our formula a bit arduous. As often happens in mathematics, instead of solving the actual problem posed, we found it best to solve a ’nearby’ problem that was more tractable. The 365.24... days in a year is a distraction, the elliptical orbit of Earth is a downright hinderance, and the wobble of Earth on its axis can be overlooked. Specifically, we find it helpful to assume that there are precisely 366 days in the year (after all this was true about 7 or 8 million years ago), the Earth’s orbit about the sun is a circle, the Earth’s axis makes a constant angle with the plane of the orbit, and that the rays from the sun to Earth are parallel. We hope you enjoy the question.
## 2.2.3 Functions in other settings.
There are extensions of the function concept to settings where the ordered pairs are not ordered number pairs. A prime example of this is the genetic code shown in Figure 2.3. The relation is a true function (no two ordered pairs have the same first term), and during the translation of proteins, the ribosome and the transfer RNA’s use this function reliably.
Figure (PageIndex{2}): The genetic code (for human nuclear RNA). Sets of three nucleotides in RNA (codons) are translated into amino acids in the course of proteins synthesis. CAA codes for Gln (glutamine). ∗AUG codes for Met (methionine) and is also the START codon.
Explore 2.2.1 List three ordered pairs of the genetic code. What is the domain of the genetic code? What is the range of the genetic code?
The ordered pair concept is retained in the preceding example; the only change has been in the types of objects that are in the domain and range. When the objects get too far afield from simple numbers, the word transformation is sometimes used in place of function. The genetic code is a transformation of the codons into amino acids and start and stop signals.
Another commonly encountered extension of the kinds of objects in the domain of a function occurs when one physical or biological quantity is dependent on two others. For example, the widely known Charles’ Law in Chemistry can be stated as
[P=frac{n R T}{V}]
where (P) = pressure in atmospheres, n = number of molecular weights of the gas, (R) = 0.0820 Atmospheres/degree Kelvin-mol = 8.3 /degree Kelvin-mol (the gas constant), (T) = temperature in kelvins, and (V) = volume in liters. For a fixed sample of gas, the pressure is dependent on two quantities, temperature and volume. The domain is the set of all feasible temperature-volume pairs, the range is the set of all feasible pressures. The function in this case is said to be a function of two variables. The ordered pairs in the function are of the form
[((x, y), z), quad ext{or} quad( ext { (temperature, volume) }, ext { pressure })]
There may also be multivalued transformations. For example, doctors prescribe antibiotics. For each bacterial infection, there may be more than one antibiotic effective against that bacterium; there may be a list of such antibiotics. The domain would be a set of bacteria, and the range would be a set of lists of antibiotics.
### Exercises for Section 2.2.3 Functions in other settings.
Exercise 2.2.8 Describe the domain and range for each of the following transformations.
1. Bird identification guide book.
2. A judge sentences defendants to jail terms.
3. The time between sunrise and sunset.
4. Antibiotic side effects.
2 As measured, for example, by light absorbance in a spectrophotometer as discussed on page 4.
can be represented as the function. This mechansim can be extended to a wide variety of graphs types by slightly altering or enhancing the kind of function that represents the graph. Here are a few examples.
### Directed graph.
We could represent this as a Dgraph as follows:
### Disadvantages of using graphs as functions
• Cannot be extended to accomodate queries about the set of Vertices or the set of Edges.
• Depending upon the compiler that compiles the functions may not be very efficient. In fact the worst case time could be proportional to the number of vertices.
• The graph must be known statically at compile time.
## Graphs as arrays of adjacent vertexes.
In the rest of this note we will assume that Vertices are of type Int , and that the Vertices set is a finite range of the type Int . Thus a graph can be represented as follows:
## Graphs and Functions
This lesson is designed to introduce students to graphing functions. These activities can be done individually or in teams of as many as four students. Allow for 2-3 hours of class time for the entire lesson if all portions are done in class.
### Objectives
• have been introduced to plotting functions on the Cartesian coordinate plane
• have seen several categories of functions, including lines and parabolas
• Functions and Relationships
• The student demonstrates conceptual understanding of functions, patterns, or sequences including those represented in real-world situations.
• The student demonstrates algebraic thinking.
• Functions and Relationships
• The student demonstrates conceptual understanding of functions, patterns, or sequences including those represented in real-world situations.
• The student demonstrates algebraic thinking.
• Expressions and Equations
• Analyze and solve linear equations and pairs of simultaneous linear equations.
• Define, evaluate, and compare functions.
• Use functions to model relationships between quantities.
• Building Functions
• Build a function that models a relationship between two quantities
• Build new functions from existing functions
• Construct and compare linear, quadratic, and exponential models and solve problems
• Algebra
• Represent and analyze mathematical situations and structures using algebraic symbols
• Algebra
• Represent and analyze mathematical situations and structures using algebraic symbols
• Understand patterns, relations, and functions
• Use mathematical models to represent and understand quantitative relationships
• Algebra
• Competency Goal 4: The learner will use relations and functions to solve problems.
• Algebra
• Competency Goal 4: The learner will use relations and functions to solve problems.
• Number and Operations, Measurement, Geometry, Data Analysis and Probability, Algebra
• COMPETENCY GOAL 5: The learner will understand and use linear relations and functions.
• Algebra
• COMPETENCY GOAL 4: The learner will understand and use linear relations and functions.
• COMPETENCY GOAL 5: The learner will understand and use linear relations and functions.
• Algebra
• The student will demonstrate through the mathematical processes an understanding of numeric patterns, symbols as representations of unknown quantity, and situations showing increase over time.
• Algebra
• Standard 4-3: The student will demonstrate through the mathematical processes an understanding of numeric and nonnumeric patterns, the representation of simple mathematical relationships, and the application of procedures to find the value of an unknown.
• Standard 4-6: The student will demonstrate through the mathematical processes an understanding of the impact of data-collection methods, the appropriate graph for categorical or numerical data, and the analysis of possible outcomes for a simple event.
• Standard 4-4: The student will demonstrate through the mathematical processes an understanding of the relationship between two- and three-dimensional shapes, the use of transformations to determine congruency, and the representation of location and movement within the first quadrant of a coordinate system.
• Standard 4-4: The student will demonstrate through the mathematical processes an understanding of the relationship between two- and three-dimensional shapes, the use of transformations to determine congruency, and the representation of location and moveme
• Algebra
• The student will demonstrate through the mathematical processes an understanding of proportional relationships.
• Algebra
• The student will demonstrate through the mathematical processes an understanding of equations, inequalities, and linear functions.
• The student will demonstrate through the mathematical processes an understanding of the relationships between two variables within one population or sample.
• The student will demonstrate through the mathematical processes an understanding of the Pythagorean theorem the use of ordered pairs, equations, intercepts, and intersections to locate points and lines in a coordinate plane and the effect of a dilation in a coordinate plane.
• The student will demonstrate through the mathematical processes an understanding of the Pythagorean theorem the use of ordered pairs, equations, intercepts, and intersections to locate points and lines in a coordinate plane and the effect of a dilation
• Elementary Algebra
• Standard EA-1: The student will understand and utilize the mathematical processes of problem solving, reasoning and proof, communication, connections, and representation.
• Standard EA-3: The student will demonstrate through the mathematical processes an understanding of relationships and functions.
• Standard EA-4: The student will demonstrate through the mathematical processes an understanding of the procedures for writing and solving linear equations and inequalities.
• Standard EA-5: The student will demonstrate through the mathematical processes an understanding of the graphs and characteristics of linear equations and inequalities.
• Standard EA-6: The student will demonstrate through the mathematical processes an understanding of quadratic relationships and functions.
• Algebra
• The student will demonstrate through the mathematical processes an understanding of functions, systems of equations, and systems of linear inequalities.
• The student will demonstrate through the mathematical processes an understanding of quadratic equations and the complex number system.
• The student will demonstrate through the mathematical processes an understanding of algebraic expressions and nonlinear functions.
• Geometry
• 4.15.b The student will describe the path of shortest distance between two points on a flat surface.
• 4.16 The student will identify and draw representations of lines that illustrate intersection, parallelism, and perpendicularity.
• 4.15.b
• 4.16
• Probability and Statistics
• 7.17 The student, given a problem situation, will collect, analyze, display, and interpret data, using a variety of graphical methods, including frequency distributions line plots histograms stem-and-leaf plots box-and-whisker plots and scattergrams.
• 7.17 The student, given a problem situation, will collect, analyze, display, and interpret data, using a variety of graphical methods, including
• Patterns, Functions, and Algebra
• 8.14a The student will describe and represent relations and functions, using tables, graphs, and rules and
• 8.16 The student will graph a linear equation in two variables, in the coordinate plane, using a table of ordered pairs.
• 8.14 The student will
• 8.16 The student will graph a linear equation in two variables, in the coordinate plane, using a
• Algebra II
• AII.10 The student will investigate and describe through the use of graphs the relationships between the solution of an equation, zero of a function, x-intercept of a graph, and factors of a polynomial expression.
• AII.18 The student will identify conic sections (circle, ellipse, parabola, and hyperbola) from his/her equations. Given the equations in (h, k) form, the student will sketch graphs of conic sections, using transformations.
• AII.20 The student will identify, create, and solve practical problems involving inverse variation and a combination of direct and inverse variations.
• AII.10
• AII.18
• AII.20
### Textbooks Aligned:
• 7th
• [ Module 1 - Search and Rescue ] Section 4: Function Models
• Reason for Alignment: The Graphs and Functions lesson is a good follow up to the Introduction to Functions lesson, also aligned with this section of the text, by building on the graphing of functions. This one goes deeper into the vocabulary and algebra of functions. This lesson may take a while if completed together in class, but some students could move through it independently in a shorter time.
• [ Module 3 - The Mystery of Blacktail Canyon ] Section 2: Equations and Graphs
• Reason for Alignment: This is a detailed lesson on graphing functions. There are discussion suggestions, vocabulary and a Graph Sketcher Activity Worksheet already made up for practice. This lesson fits with the Graphit activity.
### Student Prerequisites
• Arithmetic: Student must be able to:
• perform integer and fractional arithmetic
• plot points on the Cartesian coordinate system
• read the coordinates of a point from a graph
• work with simple algebraic expressions
• perform basic mouse manipulations such as point, click and drag
• use a browser for experimenting with the activities
### Teacher Preparation
• Pencil and graph paper
• Copies of supplemental materials for the activities:
• Graph Sketcher Activity Worksheet
### Key Terms
constant functions Functions that stay the same no matter what the variable does are called constant functions constants In math, things that do not change are called constants. The things that do change are called variables. coordinate plane A plane with a point selected as an origin, some length selected as a unit of distance, and two perpendicular lines that intersect at the origin, with positive and negative direction selected on each line. Traditionally, the lines are called x (drawn from left to right, with positive direction to the right of the origin) and y (drawn from bottom to top, with positive direction upward of the origin). Coordinates of a point are determined by the distance of this point from the lines, and the signs of the coordinates are determined by whether the point is in the positive or in the negative direction from the origin coordinates A unique ordered pair of numbers that identifies a point on the coordinate plane. The first number in the ordered pair identifies the position with regard to the x-axis while the second number identifies the position on the y-axis function A function f of a variable x is a rule that assigns to each number x in the function's domain a single number f(x). The word "single" in this definition is very important graph A visual representation of data that displays the relationship among variables, usually cast along x and y axes. negative numbers Numbers less than zero. In graphing, numbers to the left of zero. Negative numbers are represented by placing a minus sign (-) in front of the number
### Lesson Outline
Remind students what has been learned in previous lessons that will be pertinent to this lesson and/or have them begin to think about the words and ideas of this lesson. You may ask the following questions:
• Can someone tell me what a function is?
• Will someone give me an example of a function?
• Will someone give me an example of something that is not a function?
Let the students know what it is they will be doing and learning today. Say something like this:
• Have the students try plotting points for several simple functions to ensure that they have some skill at plotting by hand. Even if graphing calculators are available, have the students plot points on graph paper - this is a skill that is important to practice by hand. Here are a few functions that might be assigned:
• Practice the students' function plotting skills by having them check their work from the previous activity by plotting the same functions using the Graph Sketcher Tool.
• Have the students investigate functions of the form y = _____ x + ____ using the Graph Sketcher Tool to determine what kinds of functions come from this form, and what changing each constant does to the function. Be sure to have them keep track of what they try and record their hypotheses and observations.
• Relate these graphs to the lesson on Linear Functions to demonstrate the rationale for the terms m = slope and b = intercept in the formula .
• You may wish to bring the class back together for a discussion of the findings. Once the students have been allowed to share what they found, summarize the results of the lesson.
### Alternate Outline
• Replace all Graph Sketcher activities with graphing calculator activities. Note: Depending on the graphing calculator, you might have to spend some additional time discussing setting the window ranges.
• Replace all Graph Sketcher activities with Simple Plot activities. Simple Plot is a point plotting activity, which requires that the students create tables of values for the functions before plotting.
• Limit investigations to functions with one operation as in the Function Machine lesson and/or to linear functions as in the Linear Functions lesson .
### Suggested Follow-Up
After these discussions and activities, students will have more experience with functions and graphing. The next lesson, Reading Graphs , shows the students that graphs can be used to convey lots of information about a given situation.
## 3.6 Graphs of Functions
In the last section we learned how to determine if a relation is a function. The relations we looked at were expressed as a set of ordered pairs, a mapping or an equation. We will now look at how to tell if a graph is that of a function.
The graph of a linear equation is a straight line where every point on the line is a solution of the equation and every solution of this equation is a point on this line.
A relation is a function if every element of the domain has exactly one value in the range. So the relation defined by the equation y = 2 x − 3 y = 2 x − 3 is a function.
If we look at the graph, each vertical dashed line only intersects the line at one point. This makes sense as in a function, for every x-value there is only one y-value.
If the vertical line hit the graph twice, the x-value would be mapped to two y-values, and so the graph would not represent a function.
This leads us to the vertical line test. A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point. If any vertical line intersects the graph in more than one point, the graph does not represent a function.
### Vertical Line Test
A set of points in a rectangular coordinate system is the graph of a function if every vertical line intersects the graph in at most one point.
If any vertical line intersects the graph in more than one point, the graph does not represent a function.
### Example 3.51
Determine whether each graph is the graph of a function.
#### Solution
ⓐ Since any vertical line intersects the graph in at most one point, the graph is the graph of a function.
ⓑ One of the vertical lines shown on the graph, intersects it in two points. This graph does not represent a function.
Determine whether each graph is the graph of a function.
Determine whether each graph is the graph of a function.
### Identify Graphs of Basic Functions
Compare the graph of y = 2 x − 3 y = 2 x − 3 previously shown in Figure 3.14 with the graph of f ( x ) = 2 x − 3 f ( x ) = 2 x − 3 shown in Figure 3.15. Nothing has changed but the notation.
### Graph of a Function
The graph of a function is the graph of all its ordered pairs, ( x , y ) ( x , y ) or using function notation, ( x , f ( x ) ) ( x , f ( x ) ) where y = f ( x ) . y = f ( x ) .
As we move forward in our study, it is helpful to be familiar with the graphs of several basic functions and be able to identify them.
We wrote linear equations in several forms, but it will be most helpful for us here to use the slope-intercept form of the linear equation. The slope-intercept form of a linear equation is y = m x + b . y = m x + b . In function notation, this linear function becomes f ( x ) = m x + b f ( x ) = m x + b where m is the slope of the line and b is the y-intercept.
The domain is the set of all real numbers, and the range is also the set of all real numbers.
### Linear Function
We will use the graphing techniques we used earlier, to graph the basic functions.
### Example 3.52
#### Solution
Notice that for any real number we put in the function, the function value will be b. This tells us the range has only one value, b.
### Identity Function
The next function we will look at is not a linear function. So the graph will not be a line. The only method we have to graph this function is point plotting. Because this is an unfamiliar function, we make sure to choose several positive and negative values as well as 0 for our x-values.
### Example 3.54
#### Solution
We choose x-values. We substitute them in and then create a chart as shown.
Looking at the result in Example 3.54, we can summarize the features of the square function. We call this graph a parabola. As we consider the domain, notice any real number can be used as an x-value. The domain is all real numbers.
The range is not all real numbers. Notice the graph consists of values of y never go below zero. This makes sense as the square of any number cannot be negative. So, the range of the square function is all non-negative real numbers.
### Square Function
The next function we will look at is also not a linear function so the graph will not be a line. Again we will use point plotting, and make sure to choose several positive and negative values as well as 0 for our x-values.
### Example 3.55
#### Solution
We choose x-values. We substitute them in and then create a chart.
Looking at the result in Example 3.55, we can summarize the features of the cube function. As we consider the domain, notice any real number can be used as an x-value. The domain is all real numbers.
The range is all real numbers. This makes sense as the cube of any non-zero number can be positive or negative. So, the range of the cube function is all real numbers.
### Cube Function
The next function we will look at does not square or cube the input values, but rather takes the square root of those values.
### Example 3.56
#### Solution
We choose x-values. Since we will be taking the square root, we choose numbers that are perfect squares, to make our work easier. We substitute them in and then create a chart.
### Example 3.57
#### Solution
We choose x-values. We substitute them in and then create a chart.
### Read Information from a Graph of a Function
In the sciences and business, data is often collected and then graphed. The graph is analyzed, information is obtained from the graph and then often predictions are made from the data.
We will start by reading the domain and range of a function from its graph.
Remember the domain is the set of all the x-values in the ordered pairs in the function. To find the domain we look at the graph and find all the values of x that have a corresponding value on the graph. Follow the value x up or down vertically. If you hit the graph of the function then x is in the domain.
Remember the range is the set of all the y-values in the ordered pairs in the function. To find the range we look at the graph and find all the values of y that have a corresponding value on the graph. Follow the value y left or right horizontally. If you hit the graph of the function then y is in the range.
### Example 3.58
Use the graph of the function to find its domain and range. Write the domain and range in interval notation.
#### Solution
To find the domain we look at the graph and find all the values of x that correspond to a point on the graph. The domain is highlighted in red on the graph. The domain is [ −3 , 3 ] . [ −3 , 3 ] .
To find the range we look at the graph and find all the values of y that correspond to a point on the graph. The range is highlighted in blue on the graph. The range is [ −1 , 3 ] . [ −1 , 3 ] .
Use the graph of the function to find its domain and range. Write the domain and range in interval notation.
Use the graph of the function to find its domain and range. Write the domain and range in interval notation.
We are now going to read information from the graph that you may see in future math classes.
First, a coördinate . A coördinate is a number. It labels a point on a line.
The coördinates 0, 1, 2, 3, etc. label those points. They are their "addresses."
A coördinate axis is a line with coördinates.
To label the points in a plane, we will need more than one coördinate axis, and we place them at right angles. Hence, they are called rectangular coördinate axes . And the coördinates on them are called rectangular coördinates . They are also called Cartesian coördinates , after the 17th century philosopher and mathematician René Descartes for he was one of the first to exploit the geometrical possibilities of coördinates.
Finally, the rectangular coördinates of a point are an ordered pair , ( x , y ). (2, 3) labels a point different from (3, 2). The x -coördinate is always entered first the y -coördinate, second.
Example 1. The graph of a function .
In which graph are the values of y a function of the values of x ?
In graph a). To each value of x there is one and only one value of y . Any straight line parallel to the y -axis will cut that graph only once.
The coördinate pairs of a function
Consider the function y = x 2 . The variable y now signifies the y -coördinate, and x , the x -coördinate. Therefore, every coördinate pair on the graph of that function is
In other words, the graph of a function
is that geometrical figure such that every coördinate pair on it is
then every coördinate pair on its graph has the form
Problem 1. The coördinate pairs on the graph of the following functions have what form?
a) y = 1 &minus x ( x , 1 &minus x ) b) y = ax + b ( x , ax + b ) c) y = 3 ( x , 3) d) y = f ( x ) ( x , f ( x ))
Problem 2. Let y = f ( x ). Write the second member of each coördinate pair.
a) (1, ?) (1, f (1)) b) (&minus1, ?) (&minus1, f (&minus1)) c) ( a , ?) ( a , f ( a )) d) ( t &minus 4, ?) ( t &minus 4, f ( t &minus 4))
Problem 3. Which of these points are on the graph of the function
a) (1, 4) No, because 4 is not equal to 2 · 1 2 .
b) (&minus1, 2) Yes. c) ( a , 2 a 2 ) Yes. d) ( b /2, b 2 /2) Yes.
a) The coördinate pair ( p , q ) is on the graph of the function
What is the relationship between the coördinates p , q ?
b) The coördinate pair ( r , s ) is on the graph of the function
What is the relationship between the coördinates r , s ?
Let y = f ( x ). Then the coördinates of every point on its graph are
( x , f ( x )). (Look at the first quadrant.)
f ( x ) -- or y -- is the height of the graph at x . It is the length of that vertical line.
Now, if f (&minus x ) = f ( x ) -- that is, if the height of the graph at &minus x is equal to the height of the graph at x -- then that point is shown at a ).
While if f (&minus x ) = &minus f ( x ) -- if the height of the graph at &minus x is the negative of the height of the graph at x -- then that point is shown at b ).
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## Modules
• Data
• Graph
• Data.Graph.Algorithm
• Data.Graph.Algorithm.DepthFirstSearch
• Data.Graph.Class.Bidirectional
• Data.Graph.Class.EdgeEnumerable
• Data.Graph.Class.VertexEnumerable
Graph each function in a viewing window [-2, 2] by [-1, 6].
(i) Which point is common to all four graphs ?
(ii) Analyze the functions for domain, range, continuity, increasing or decreasing behavior, symmetry, extrema, asymptotes and end behaviour.
(i) Every graph is passing through the point (0, 1).
(ii) Analyzing the function :
Domain is the defined value of x. For this function, the domain is all real numbers.
Every exponential functions are defined and continuous for all real numbers.
Increasing / decreasing :
Since the base is integer, the graph is increasing.
The graph is asymptotic to the x-axis as x approaches negative infinity
The graph increases without bound as x approaches positive infinity. So there is no extreama.
When x approaches x to ∞, f(x) = ∞
When x approaches x to - ∞, f(x) = 0
Graph each function in viewing windows [-2, 2] by [-1, 6]
(i) Which point is common to all four graphs ?
(ii) Analyze the functions for domain, range, continuity, increasing or decreasing behavior, symmetry, extrema, asymptotes and end behaviour.
(i) Every graph is passing through the point (0, 1).
(ii) Analyzing the function :
Domain is the defined value of x. For this function, the domain is all real numbers.
Every exponential functions are defined and continuous for all real numbers.
Increasing / decreasing :
Since the base is integer, the graph is increasing.
The graph is asymptotic to the x-axis as x approaches negative infinity
The graph increases without bound as x approaches positive infinity. So there is no extreama.
When x approaches x to -∞, f(x) = ∞
When x approaches x to ∞, f(x) = 0
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# 10th Class Mathematics Areas Related to Circles Area of Circle
Area of Circle
Category : 10th Class
AREA OF CIRCLE
FUNDAMENTALS
Perimeter and Area of a Circle
The distance covered by travelling once around a circle is called its perimeter, and in case of a circle, it is usually called its circumference. The circumference of a circle bears a constant ratio with its diameter. This constant ratio is denoted by the Greek letter $\pi$ (read as ‘pi’), m other words,
$\frac{circumference}{diameter}=\pi$
Or $circumference=\pi \times diameter$
$=\pi \times 2r$ (where r is the radius of the circle) $=2\pi r$
The great Indian mathematician Aryabhatta (A. D. 476 – 550) gave an approximate value of $\pi$ He stated that $\pi =\frac{62832}{20000}$, which is nearly equal to 3.1416. It is also interesting to note that using an identity of the great mathematical genius Srinivas Ramanujan (1887 – 1920) of India, mathematicians have been able to calculate the value of n correct to million places of decimals. As you known from Chapter I of Class IX, n is an irrational number and its decimal expansion is non – terminating and non – recurring (non – repeating). However, for practical purposes, we generally take the value of $\pi$ as $\frac{22}{7}$ or 3.14, approximately.
Areas of Sector and Segment of a Circle
The portion (or part) of the circular region enclosed by two radii and the corresponding arc is called sector of the circle and the portion (or part) of the circular regions enclosed between a chord and the corresponding arc is called a segment of the circle.
Thus, in figure shaded region OAPB is a sector of the circle with centre O. $\angle AOB$is called the angle of the sector. Note that in this figure, unshaded region OAQB is also a sector of the circle. For obvious reasons, OAPB is called the minor sector and OAQB is called the major sector. You can also see that angle of the major sector is ($360{}^\circ -\angle AOB$).
Now, look at figure in which AB is a chord of the circle with centre O. So, shaded region APB is a segment of the circle. You can also note that unshaded region AQB is another segment of the circle formed by the chord AB. For obvious reasons, APB is called the minor segment and AQB is called the major segment.
Remarks: When we write ‘segment’ and ‘sector’ we will mean the ‘minor segment’ and the ‘minor sector’ respectively, unless stated otherwise.
Calculating area of sector of a circle
Let OAPB be a sector of a circle with centre O and radius r (see figure). Let the degree measure of $\angle AOB$ be $\theta$.
You know that area of a circle (in fact of a circular regions or disc) is $\pi {{r}^{2}}$.
In a way, we can consider this circular region to be a sector forming an angle of $360{}^\circ$ (i.e., of degree measure 360) at the centre O. Now by applying the Unitary Method, we can arrive at the area of the sector OAPB as follows:
When degree measure of the angle at the centre is$360{}^\circ$, area of the sector $={{r}^{2}}$
So, when the degree measure of the angle at the centre is $1{}^\circ$, area of the sector $=\frac{\pi {{r}^{2}}}{360}$.
Therefore, when the degree measure of the angle at the centre is $\theta$, area of the sector $=\frac{\pi {{r}^{2}}}{360}\times \theta =\frac{\theta }{360}\times \pi {{r}^{2}}$
Thus, we obtain the following relation (or formula) for area of a sector of a circle:
Area of the sector of angle $\theta =\frac{\theta }{360}\times \pi {{r}^{2}}$,
Calculating length of arc of a circle
Let ‘r’ be the radius of the circle and $\theta$ the angle of the sector in degrees.
Now, a natural equation arises: Can we find the length of the arc APB corresponding to this sector OAPB? The answer is yes. Again, by applying the Unitary Method and taking the whole length of the circle (of angle $360{}^\circ$) as $2\pi r$, we can obtain the required length of the arc APB as $\frac{\theta }{360}\times 2\pi r$.
(ii) (ii)
Length of an arc of a sector of angle $\theta =\frac{\theta }{360}\times 2\pi r$
Calculating area of segment of a circle
Now let us take the case of the area of the segment APB of a circle with centre O and radius r (see above figure). You can see that:
Area of the segment APB = Area of the sector OAPB – Area of $\Delta \,OAB$
$=\frac{\theta }{360}\times \pi {{r}^{2}}-\text{area}\,\,\text{of}\,\,\Delta \,OAB$
Note: From figure (ii) to (iii) respectively, you can observe that:
Area of the major sector $OAQB=\pi {{r}^{2}}$ – Area of minor sector OAPB
And, Area of major segment $AQB=\pi {{r}^{2}}$– Area of the minor segment APB
Areas of Combinations of Plane Figure
So far, we have calculated the areas of different figures separately. Let us now try to calculate the areas of some combinations of plane figures. We come across these types of figures in our daily life and also in the form of various interesting designs. Flower beds, drain covers, window design, design on table covers, are some of such examples. We illustrate the process of calculating areas of these figures through some examples.
A quick look at relevant formula:
Circle and semi - Circle
Area of circle $=\pi {{r}^{2}}sq.$units.
1. Area of the semi-circle $=\frac{\pi {{r}^{2}}}{2}sq.$ units.
2. Circumference of the circle $=2\pi r$ units $=\pi d$units.
3. Circumference of the semicircle $=\left( \pi +2 \right)r$ units$=\frac{36r}{7}\text{units}$
(Where r is radius and d is diameter)
Circular Ring
Area of the ring $=\pi \left( {{R}^{2}}-{{r}^{2}} \right)=\pi \left( R+r \right)\left( R-r \right)$
(Where R and r are outer and inner radii of a ring and (R – r) is the width of the ring)
Sector and Segments
1. Arc length $\overset\frown{ACB}$
$1=\left( \frac{\theta }{360{}^\circ } \right)2\pi r$ units
1. Area of the sector OAPB
$A=\left( \frac{\theta }{360{}^\circ } \right)\pi {{r}^{1}}sq.$ units.
1. Perimeter of the sector $=[\left( 1+2r \right)$ units.
2. Area of the segment APB = (A – Area of the $\Delta \,AOB$) sq. units.
3. Perimeter of the segment ACB = (Arc length ACB + AB) units.
1. Distance covered by a wheel in one revolution = Circumference of the wheel.
2. Number of rotations made by a wheel in unit time $\text{=}\frac{\text{Distance}\,\,\text{covered}\,\,\text{by}\,\,\text{it}\,\,\text{in}\,\,\text{unit}\,\,\text{time}}{\text{Circumference}\,\,\text{of}\,\,\text{the}\,\,\text{wheel}}$
3. Angle made by minute hand in one minute $\text{=}\frac{360{}^\circ }{60}=6{}^\circ$.
4. Angle made by hour hand in one minute $\text{=}\frac{30{}^\circ }{60}={{\left( \frac{1}{2} \right)}^{0}}$.
Equilateral$\Delta$
1. Radius of in-circle (GE) of equilateral triangle $=\frac{1}{3}\left( \frac{\sqrt{3}a}{2} \right)=\frac{a}{2\sqrt{3}}$ units.
2. Circum - radius (AG) of equilateral triangle $=\frac{2}{3}\left( \frac{\sqrt{3}a}{2} \right)=\frac{a}{\sqrt{3}}$ units. (Where a is side of the triangle).
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