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## Thursday, June 30, 2011
### Solutions to Introduction to Basic Physics, Part 22
We now examine the solutions for the problems in Part 22, on series and parallel circuits.
1)Two resistances of 20 ohms and 5 ohms are connected in parallel by a student. He then connects this combination in series with a 3 ohm resistance and a battery of 1 ohm resistance.
a) Draw a conventional diagram of the circuit (i.e. not in Wheatstone Bridge format).
b) Find the resistance of the resistors in parallel.
c) Find the total resistance of the circuit.
The circuit diagram is shown above, labelled for 'Problem 1'.
b) The resistance of the 2 resistors, 20 ohms and 5 ohms, in parallel is easily found from the total for the special case of two resistors in parallel:
R = R1 R2 / (R1 + R2) = (20 0hms)(5 ohms) / 25 ohms = 100 ohms^2/25 ohms
R = 4 ohms
c) The total resistance for all contributors in series is just the sum:
4 ohms + 3 ohms + 1 ohm = 8 ohms
2) The photo shows a sketch of a Wheatstone Bridge Circuit at Harrison College, to be used to find the resistances of two resistors (R1 and R2) connected as shown, with two differing positions of the galvanometer. The total resistance R is also to be taken.
A student using the diagram finds his voltmeter reads 1.0 V and the total resistance is 4 ohms when the slide wire is at the end position noted.
If the balance for obtaining G = 0 (e.g. galvanometer reading zero) is the intermediate position, and the slide wire is then at 45 cm, find the values of R1 and R2.
The total resistance from the end position (100 cm) = 4 ohms. We know from the photo that the two resistors R1 and R2 are connected in series so that:
R1 + R2 = 4 ohms
We are given the first position of the slide wire as L1 = 45 cm, then:
R1/ 4 ohms = 45 cm/ 100 cm
and: R1 = 0.45 (4 ohms) = 1.8 ohms
Then:
R2 = 4 ohms - 1.8 ohms = 2.2 ohms
3) A resistance R2 is connected in parallel with a resistance R1. What resistance R3 must be connected in series with the combination of R1 and R2 so that the equivalent resistance is equal to the resistance R1? Draw a circuit diagram of the arrangement.
We let R1 = r and let R2 = 2r.
These are in parallel so the total resistance for them is:
R(T) = R1 R2/ (R1 + R2) = r(2r)/ (r + 2r) = 2r^2/ 3r = 2r/3
We require that the total resistance in series be such that the equivalent resistance is equal to the resistance R1, or r.
Thus, we require:
R3 + 2r/3 = r
and, solving by algebra:
R3 = r - 2r/3 = r/3
The circuit diagram is shown, labelled as 'Prob. 3'.
4) Three equal resistors are connected in series. When a certain potential difference is applied across the combination, the total power dissipated is 10 watts. (Note: Power = V x I, voltage x current).
What power would be dissipated if the three resistors were connected in parallel across the same potential difference?
You need to bear in mind that for the case in parallel the current is divvied up, e.g. total current I = I1 + I2 + I3. Since all the resistors are equal (call each r ohms) and the voltage is the same then the power for the parallel case would be one third the power for the series case, or 10/3 Watts.
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10.9: Right Triangle Trigonometry
Page ID
129647
### Learning Objectives
After completing this section, you should be able to:
1. Apply the Pythagorean Theorem to find the missing sides of a right triangle.
2. Apply the and right triangle relationships to find the missing sides of a triangle.
3. Apply trigonometric ratios to find missing parts of a right triangle.
4. Solve application problems involving trigonometric ratios.
This is another excerpt from Raphael’s The School of Athens. The man writing in the book represents Pythagoras, the namesake of one of the most widely used formulas in geometry, engineering, architecture, and many other fields, the Pythagorean Theorem. However, there is evidence that the theorem was known as early as 1900–1100 BC by the Babylonians. The Pythagorean Theorem is a formula used for finding the lengths of the sides of right triangles.
Born in Greece, Pythagoras lived from 569–500 BC. He initiated a cult-like group called the Pythagoreans, which was a secret society composed of mathematicians, philosophers, and musicians. Pythagoras believed that everything in the world could be explained through numbers. Besides the Pythagorean Theorem, Pythagoras and his followers are credited with the discovery of irrational numbers, the musical scale, the relationship between music and mathematics, and many other concepts that left an immeasurable influence on future mathematicians and scientists.
The focus of this section is on right triangles. We will look at how the Pythagorean Theorem is used to find the unknown sides of a right triangle, and we will also study the special triangles, those with set ratios between the lengths of sides. By ratios we mean the relationship of one side to another side. When you think about ratios, you should think about fractions. A fraction is a ratio, the ratio of the numerator to the denominator. Finally, we will preview trigonometry. We will learn about the basic trigonometric functions, sine, cosine and tangent, and how they are used to find not only unknown sides but unknown angles, as well, with little information.
### Pythagorean Theorem
The Pythagorean Theorem is used to find unknown sides of right triangles. The theorem states that the sum of the squares of the two legs of a right triangle equals the square of the hypotenuse (the longest side of the right triangle).
### FORMULA
The Pythagorean Theorem states
where
For example, given that side and side we can find the measure of side using the Pythagorean Theorem. Thus,
### Example 10.64
#### Using the Pythagorean Theorem
Find the length of the missing side of the triangle (Figure 10.180).
Using the Pythagorean Theorem, we have
When we take the square root of a number, the answer is usually both the positive and negative root. However, lengths cannot be negative, which is why we only consider the positive root.
### Distance
The applications of the Pythagorean Theorem are countless, but one especially useful application is that of distance. In fact, the distance formula stems directly from the theorem. It works like this:
In Figure 10.182, the problem is to find the distance between the points and We call the length from point to point side , and the length from point to point side . To find side , we use the distance formula and we will explain it relative to the Pythagorean Theorem. The distance formula is such that is a substitute for in the Pythagorean Theorem and is equal to and is a substitute for in the Pythagorean Theorem and is equal to When we plug in these numbers to the distance formula, we have
Thus, , the hypotenuse, in the Pythagorean Theorem.
### Example 10.65
#### Calculating Distance Using the Distance Formula
You live on the corner of First Street and Maple Avenue, and work at Star Enterprises on Tenth Street and Elm Drive (Figure 10.183). You want to calculate how far you walk to work every day and how it compares to the actual distance (as the crow flies). Each block measures 200 ft by 200 ft.
You travel 7 blocks south and 9 blocks west. If each block measures 200 ft by 200 ft, then .
As the crow flies, use the distance formula. We have
### Example 10.66
#### Calculating Distance with the Pythagorean Theorem
The city has specific building codes for wheelchair ramps. Every vertical rise of 1 in requires that the horizontal length be 12 inches. You are constructing a ramp at your business. The plan is to make the ramp 130 inches in horizontal length and the slanted distance will measure approximately 132.4 inches (Figure 10.184). What should the vertical height be?
The Pythagorean Theorem states that the horizontal length of the base of the ramp, side a, is 130 in. The length of c, or the length of the hypotenuse, is 132.4 in. The length of the height of the triangle is side b.
Then, by the Pythagorean Theorem, we have:
If you construct the ramp with a 25 in vertical rise, will it fulfill the building code? If not, what will have to change?
The building code states 12 in of horizontal length for each 1 in of vertical rise. The vertical rise is 25 in, which means that the horizontal length has to be So, no, this will not pass the code. If you must keep the vertical rise at 25 in, what will the other dimensions have to be? Since we need a minimum of 300 in for the horizontal length:
The new ramp will look like Figure 10.185.
Figure 10.185
### Triangles
In geometry, as in all fields of mathematics, there are always special rules for special circumstances. An example is the perfect square rule in algebra. When expanding an expression like we do not have to expand it the long way:
If we know the perfect square formula, given as
we can skip the middle step and just start writing down the answer. This may seem trivial with problems like
We see that the shortest side is opposite the smallest angle, and the longest side, the hypotenuse, will always be opposite the right angle. There is a set ratio of one side to another side for the triangle given as or Thus, you only need to know the length of one side to find the other two sides in a triangle.
### Example 10.67
#### Finding Missing Lengths in a Triangle
Find the measures of the missing lengths of the triangle (Figure 10.188).
We can see that this is a triangle because we have a right angle and a angle. The remaining angle, therefore, must equal Because this is a special triangle, we have the ratios of the sides to help us identify the missing lengths. Side is the shortest side, as it is opposite the smallest angle and we can substitute The ratios are We have the hypotenuse equaling 10, which corresponds to side , and side is equal to 2. Now, we must solve for :
Side is equal to or The lengths are
### Example 10.68
#### Applying Triangle to the Real World
A city worker leans a 40-foot ladder up against a building at a
We have a triangle, and the hypotenuse is 40 ft. This length is equal to 2, where is the shortest side. If , then . The ladder is leaning on the wall 20 ft up from the ground.
The
### Example 10.69
#### Finding Missing Lengths of a Triangle
Find the measures of the unknown sides in the triangle (Figure 10.193).
Because we have a triangle, we know that the two legs are equal in length and the hypotenuse is a product of one of the legs and One leg measures 3, so the other leg, , measures 3. Remember the ratio of Then, the hypotenuse, , equals
### Trigonometry Functions
Trigonometry developed around 200 BC from a need to determine distances and to calculate the measures of angles in the fields of astronomy and surveying. Trigonometry is about the relationships (or ratios) of angle measurements to side lengths in primarily right triangles. However, trigonometry is useful in calculating missing side lengths and angles in other triangles and many applications.
### Checkpoint
NOTE: You will need either a scientific calculator or a graphing calculator for this section. It must have the capability to calculate trigonometric functions and express angles in degrees.
Trigonometry is based on three functions. We title these functions using the following abbreviations:
Letting
We will be applying the sine function, cosine function, and tangent function to find side lengths and angle measurements for triangles we cannot solve using any of the techniques we have studied to this point. In Figure 10.195, we have an illustration mainly to identify and the sides labeled and .
An angle
Let’s use the trigonometric functions to find the sides and . As long as your calculator mode is set to degrees, you do not have to enter the degree symbol. First, let’s solve for .
We have and Then,
Next, let’s find . This is the cosine function. We have Then,
Now we have all sides,
### Example 10.70
#### Using Trigonometric Functions
Find the lengths of the missing sides for the triangle (Figure 10.197).
We have a angle, and the length of the triangle on the -axis is 6 units.
Step 1: To find the length of , we can use the cosine function, as We manipulate this equation a bit to solve for :
Step 2: We can use the Pythagorean Theorem to find the length of . Prove that your answers are correct by using other trigonometric ratios:
Step 3: Now that we have , we can use the sine function to prove that is correct. We have
To find angle measurements when we have two side measurements, we use the inverse trigonometric functions symbolized as or The –1 looks like an exponent, but it means inverse. For example, in the previous example, we had and To find what angle has these values, enter the values for the inverse cosine function in your calculator:
You can also use the inverse sine function and enter the values of in your calculator given and We have
Finally, we can also use the inverse tangent function. Recall We have
### Example 10.71
#### Solving for Lengths in a Right Triangle
Solve for the lengths of a right triangle in which
Step 1: To find side , we use the sine function:
Step 2: To find , we use the cosine function:
Step 3: Since this is a triangle and side should equal if we input 3 for , we have Put this in your calculator and you will get
### Example 10.72
#### Finding Altitude
A small plane takes off from an airport at an angle of
To solve this problem, we use the tangent function:
The plane’s altitude when passing over the peak is 2,140 ft, and it is 1,040 ft above the peak.
### Example 10.73
#### Finding Unknown Sides and Angles
Suppose you have two known sides, but do not know the measure of any angles except for the right angle (Figure 10.202). Find the measure of the unknown angles and the third side.
Step 1: We can find the third side using the Pythagorean Theorem:
Now, we have all three sides.
Step 2: To find we will first find
The angle is the angle whose sine is
Step 3: To find , we use the inverse sine function:
Step 4: To find the last angle, we just subtract: .
### Angle of Elevation and Angle of Depression
Other problems that involve trigonometric functions include calculating the angle of elevation and the angle of depression. These are very common applications in everyday life. The angle of elevation is the angle formed by a horizontal line and the line of sight from an observer to some object at a higher level. The angle of depression is the angle formed by a horizontal line and the line of sight from an observer to an object at a lower level.
### Example 10.74
#### Finding the Angle of Elevation
A guy wire of length 110 meters runs from the top of an antenna to the ground (Figure 10.204). If the angle of elevation of an observer to the top of the antenna is how high is the antenna?
We are looking for the height of the tower. This corresponds to the -value, so we will use the sine function:
The tower is 75 m high.
### Example 10.75
#### Finding Angle of Elevation
You are sitting on the grass flying a kite on a 50-foot string (Figure 10.206). The angle of elevation is How high above the ground is the kite?
We can solve this using the sine function,
### People in Mathematics
#### Pythagoras and the Pythagoreans
The Pythagorean Theorem is so widely used that most people assume that Pythagoras (570–490 BC) discovered it. The philosopher and mathematician uncovered evidence of the right triangle concepts in the teachings of the Babylonians dating around 1900 BC. However, it was Pythagoras who found countless applications of the theorem leading to advances in geometry, architecture, astronomy, and engineering.
Among his accolades, Pythagoras founded a school for the study of mathematics and music. Students were called the Pythagoreans, and the school’s teachings could be classified as a religious indoctrination just as much as an academic experience. Pythagoras believed that spirituality and science coexist, that the intellectual mind is superior to the senses, and that intuition should be honored over observation.
Pythagoras was convinced that the universe could be defined by numbers, and that the natural world was based on mathematics. His primary belief was All is Number. He even attributed certain qualities to certain numbers, such as the number 8 represented justice and the number 7 represented wisdom. There was a quasi-mythology that surrounded Pythagoras. His followers thought that he was more of a spiritual being, a sort of mystic that was all-knowing and could travel through time and space. Some believed that Pythagoras had mystical powers, although these beliefs were never substantiated.
Pythagoras and his followers contributed more ideas to the field of mathematics, music, and astronomy besides the Pythagorean Theorem. The Pythagoreans are credited with the discovery of irrational numbers and of proving that the morning star was the planet Venus and not a star at all. They are also credited with the discovery of the musical scale and that different strings made different sounds based on their length. Some other concepts attributed to the Pythagoreans include the properties relating to triangles other than the right triangle, one of which is that the sum of the interior angles of a triangle equals These geometric principles, proposed by the Pythagoreans, were proven 200 years later by Euclid.
### Who Knew?
#### A Visualization of the Pythagorean Theorem
In Figure 10.208, which is one of the more popular visualizations of the Pythagorean Theorem, we see that square is attached to side ; square is attached to side ; and the largest square, square , is attached to side . Side measures 3 cm in length, side measures 4 cm in length, and side measures 5 cm in length. By definition, the area of square measures 9 square units, the area of square measures 16 square units, and the area of square measures 25 square units. Substitute the values given for the areas of the three squares into the Pythagorean Theorem and we have
Thus, the sum of the squares of the two legs of a right triangle is equal to the square of the hypotenuse, as stated in the Pythagorean Theorem.
### Section 10.8 Exercises
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# Scrambled Egg Math
## Curriculum Goal
#### Primary: Number Sense
• Use objects, diagrams, and equations to represent, describe, and solve situations involving addition and subtraction of whole numbers that add up to no more than 50.
## Context
• Educator working with a group of four children.
## Materials
• One egg cartons per student (size: 12 eggs)
• Two small game tokens per student (e.g., beans, pennies)
• One die per student
• Lined papers or math notebooks
## Lesson
• Open the egg cartons and write the numbers 1 through to 9 on the bottom of each rounded space. Choose three numbers to be written down twice to fill all 12 spaces.
• Hand out an egg carton, two game tokens, and a die to each child. Tell them we will be playing a game to practice addition and subtraction.
• Explain to students that they will put the game tokens inside the carton, close the lid and “scramble the eggs” by shaking the carton. Once they finish shaking the carton, tell them to set it on the table.
• Instruct children to then roll their die to determine whether they will be adding or subtracting in this round: an even number means that they will add, an odd number means that they will subtract. Students who are just beginning to solve equations can play the game using one operation.
• Ask students to open the carton lid and see which spaces the tokens fell into. Have them create a math equation using the two numbers and the chosen operation and then solve it.
• E.g., if the child rolled a four (even = addition) and their tokens landed in the seven and three spaces, they can make the equation “7 + 3 = 10” or “3 + 7 = 10”.
• Instruct students to write their equation down on their sheet of paper or notebook.
• Repeat the game for 10 rounds and then discuss with students how they approached the task. Some questions to extend student thinking:
• How did you solve your equation?
• Do you find that certain numbers make it easier to find the answer?
• Can you tell me something about the two numbers that the tokens landed on?
## Look Fors
• What strategies does the child use to solve the equations? Do they count up, count on, or find the answer through retrieval? Do children recognize certain math strategies? For example, do they recognize the different ways to make the number 10 or know the doubles fact to help them solve equations faster?
• Griffin (2003) states that as children develop stronger number sense and computation abilities, the strategy they employ will also become more sophisticated (instead of counting up, children may use counting on or retrieval).
• Do children demonstrate the ability to manipulate a mental number line?
• This activity requires children to solve equations without using any concrete objects. According to Griffin (2005), a major learning stage occurs when children are able to solve addition and subtraction questions in their minds by manipulating their “mental counting line structure” forwards and backwards.
• Do children use appropriate math language? Are they using words such as sum, difference, bigger than, smaller than, etc. when asked questions?
• During the activity, talk with students and ask them to verbalize their thoughts or talk about the numbers to listen to their vocabulary.
## Extension
• As children become proficient, introduce three-step codes and encourage children to create pathways with more steps
## References
Griffin, S. (2005). Fostering the development of whole number sense: Teaching mathematics in the primary grades. In M.S. Donovan & J. D. Bransford (Eds.), How students learn: History, mathematics and science in the classroom (pp. 257308). Washington, DC: The National Academies Press.
Griffin, S. (2003). Laying the foundations for computational fluency in early childhood. Teaching Children Mathematics, 9(6), 306309.
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# Runge-Kutta Method Calculator
To use Runge-Kutta method calculator, enter the equation, write initial condition, interval values, step size, and click calculate button
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## Runge-Kutta Method Calculator
Runge Kutta Method Calculator is used to calculate ordinary differential equations (ODEs) numerically.
## What is Runge-kutta method?
Runge-kutta method is a numerical technique to find the solution of the ordinary differential equation. The method provides us with the value of y as the corresponding value of x. It is also known as the RK method.
Here we use the most common RK-4 which is known as runge-kutta of order fourth.
### Formula of Runge-kutta method
The formula of the Runge-Kutta method of order fourth is given as:
yi+1 = yi + h/6(k1 + 2k2 + 2k3 + k4)
Here,
xi+1 = xi+h
where
• i is the number of iterations.
For the value of k’s, we have
• k1= f(xi, yi)
• k2= f(xi + h/2, yi + hk1/2)
• k3= f(xi + h/2, yi + hk2/2)
• k4= f(xi + h, yi + hk3)
## How to calculate Runge-kutta method problems?
Here is a numerical example to understand this iterative method.
Example:
If we have differential function x2+1 with the "2" number of iterations. The initial value of the x is 1 to 3 and the initial value of y is 2.
Solution:
Step 1: The step size is
h= (3-1)/2 = 1, x0= 1, y0 = 2 and n= 2
f(x, y) = x^2+1
Step 2: For 1st iteration
x1= x0+h = 1+1 = 2
k1= f(x0, y0)= f(1, 2) = 2
k2= f(x0 + h/2, y0+ hk1/2) = f(1 + 1/2, 2 + (1)(2)/2) = f(3/2, 3) = 3.25
k3= f(x0 + h/2, y0 + hk2/2) = f(1 + 1/2, 2 + (1)(3.25)/2) = f(2/2, 3.625) = 3.25
k4 = f(x0 + h,y0 + hk3) = f(1 + 1, 2 + (1)(3.25)) = f(2, 5.25) = 5
y1 = y(x1) = y(2) = y0+ h/6(k1+2k2+2k3+k4) = 2+1/6(2+2(3.25)+2(3.25)+5)
y1 = 5.333
Step 3: For the 2nd iteration we have x1 and y1 to calculate x2 and y2.
x2= x1+h = 2+1 = 3
k1= f(x1, y1) = f(2, 5.333) = 5
k2= f(x1+ h/2, y1+ hk1/2) = f(2 + 1/2, 5.333 + (1)(5)/2) = f(5/2, 7.83) = 7.25
k3= f(x1+ h/2, y1+ hk2/2) = f(2 + 1/2, 5.333 + (1)(7.25)/2= f(5/2, 8.958) = 7.25
k4 = f(x1 + h, y1 + hk3) = f(2+1, 5.333 + (1)(7.25)) = f(3, 12.583) = 10
y2 = y(x2) = y (3) = y1+ h/6(k1+ 2k2 + 2k3 + k4) = 5.333 + 1/6(5 + 2(7.25) + 2(7.25) +10)
y2 = 12.666
Hence the approximate value of y2 = 12.666.
### Math Tools
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# How do you solve x + y =1 and 2x-y=-2 using substitution?
Jul 10, 2016
By eliminating $x$ in the expression $x + y = 1$, we get
$x = 1 - y$, which we can plug in the second expression:
$2 \left(1 - y\right) - y = - 2$
$2 - 2 y - y = - 2$
$2 - 3 y = - 2$
$- 3 y + 2 = - 2$
$- 3 y = - 4$
$y = \frac{4}{3}$
Knowing $y$, we can now find $x$ by plugging the $y$-value into any of the above equations. For example, by plugging $y = 4$ into the first equation, $x + y = 1$, we get
$x + \frac{4}{3} = 1$
$x = - \frac{1}{3}$
#### Explanation:
We can check whether these values satisfy the equations by plugging them back in:
First: $x + y = 1$
$- \frac{1}{3} + \frac{4}{3} = 1 \to \frac{4}{3} - \frac{1}{3} = 1 \to \frac{12 - 3}{9} = 1 \to \frac{9}{9} = 1$
Second: $2 x - y = - 2$
$2 \left(- \frac{1}{3}\right) - \frac{4}{3} = - 2 \to - \frac{2}{3} - \frac{4}{3} = - 2 \to - \frac{18}{9} = - 2$
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# College Physics 2.52 – Object dropped from a given height
## Solution:
### Part A
We can calculate for the distance traveled in 1 second using the formula
$y=v_ot-\frac{1}{2}gt^2$
Substitute the given values
$y=0-\frac{1}{2}\left(9.80\:m/s^2\right)\left(1\:s\right)^2=-4.9\:m$
The object has traveled 4.9 meters downward during the first 1 second.
### Part B
We can use the formula
$v^2=\left(v_0\right)^2-2gy$
Substitute the given values
$v^2=\left(0\right)^2-2\left(9.80\:m/s^2\right)\left(-75.0\:m\right)$
$v^2=1470$
$v=\sqrt{1470}=38.34\:m/s$
The object has a velocity of 38.34 m/s when it hits the ground.
### Part C
We need to find the total time of travel of the object, using the formula
$y=v_0t-\frac{1}{2}gt^2$
Substitute the given values
$-75.0\:m=0-\frac{1}{2}\left(9.80\:m/s\right)t^2$
Solve for time, t
$t=\sqrt{\frac{2\left(75.0\:m\right)}{9.80\:m/s^2}}=3.91\:s$
Therefore, the object has traveled 75.0 meters during the entire flight of 3.91 seconds.
Next, we determine the total distance traveled during the first 2.91 seconds of the flight.
$y=v_0t-\frac{1}{2}gt^2$
Substitute the given values
$y=0-\frac{1}{2}\left(9.8\:m/s^2\right)\left(2.91\:s\right)^2=-41.49\:m$
Therefore, the object has traveled 41.49 meters downward during the first 2.91 seconds. Between the time 2.91 and 3.91 seconds, we have the last 1 second of the flight. The distance traveled for this second is
$y=75.0\:m-41.49\:m=33.51\:m$
The object has traveled 33.51 meters for the last second of the flight.
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## Pre-Algebra Tutorial
#### Intro
A typical worded math problem is the ‘two working together’ problem, which is described here in detail. This is a classic test problem encountered in algebra and both quick and in-depth explanations are provided. Other examples are included.
#### Sample Problem
Jim can paint a room in 3 hours and Tom can paint the same room in 4 hours. How long with it take to paint that room if they work together?
#### Solution
For this “working together” problem, savvy students will recognize it and think, “Oh, the product over the sum solution”, and write down (or punch in) (3×4)/7, or about 1.7 hours (how many digits you include may depend on your teacher’s demands). This is correct but, what is behind this technique?
The first thing to realize is that this is a “Rate, Time and Distance” type of problem. I always remember that Rate X Time = Distance, or “RTD”. More generally, Rate X Time = Task Complete. Rate is always in ‘units of progress’ over time, such as miles per hour or other things done per time. Example: If you travel at 60 miles per hour for 2 hours, you have gone 60 x 2 = 120 miles. If you didn’t know how long it would take to travel 120 miles at a rate of 60 miles per hour, you would solve 60 X T = 120, or T = 120/60 = 2 hours. Any one of the three terms can be solved for if you know the others.
For the above word problem, first think of Jim’s rate to paint a room as 1 room in 3 hours and denote Rj = 1/3 or 1 room in 3 hours. If you care to divide 1 by 3, you see that he can paint one third of a room in one hour. Tom’s rate, Rt, is 1/4 and he can paint a quarter of a room in one hour.
OK, we figure that the sum of these two rates multiplied by the time working together is equal to 1 (room complete). So, using the RTD format, (Rj + Rt) * T = 1.
[Note: I sometimes use spaces in place of parentheses as it makes typed equations easier to read.]
Now, cross dividing to solve for T, T = 1/(Rj + Rt) = 1/(1/3 + 1/4), or simplifying to a common denominator, T = 1/(4/12 + 3/12) = 1/(7/12); multiplying numerator and denominator by 12/7 we get T = 12/7 / (12/7 * 7/12)), = 12/7! The answer is about 1.7 hours. 12/7 is simply the product over the sum. In the actual effort of painting the room, Jim paints 1.7 * 1/3 = .57– or about 57% of the room and Tom paints 1.7 * 1/4 = .43– or about 43% of the room.
What if Jim’s little brother helps them? It takes him 10 hours to paint a room! OK, T = 1/(1/3 + 1/4 + 1/10) = about 1.46 hours. That helps a bit, but little brother drips a lot of paint. The ‘product over the sum’ technique doesn’t work here (why?) so be careful – it only works for two! For more than two, you have to do it the long way.
Here’s a quick calculator sequence to solve the inverse of such fractional sums for the three workers:
3,1/x,+4,1/x,+10,1/x,=,1/x. The result is 1.46…
Try this other example: You drive half way around a race track at 60 mph and then the remaining half at 50 mph. What is your average speed? Is it 55 mph? No!
Solution: You can’t use the product over the sum technique here, as that only works to solve for time (the T part of RTD), given the other terms. This problem is to solve for the combined rate (R). So, what do we know? Let’s say D is 1 (track length) and the rates are as listed. But we can’t just average the rates and say it’s 55 mph because the times for each half of the track are different. Well, for the first half, T = D/R = .5/60 = .008333— hours and for the second half, .5/50 = .01 hours, for a total of .018333— hours. Finally, solving, R * .018333— = 1 (track), or R = 1/.018333— = 54.5454– mph. This was the first tricky algebra problem to given us in the 8th grade and everyone was mystified that the answer was not 55 mph and unsure how to get the right answer!
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# 16. Mensuration
Mensuration is the branch of mathematics which studies the measurement of the geometric figures and their parameters like length, volume, shape, surface area, lateral surface area, etc. Here, the concepts of mensuration are explained and all the important mensuration formulas and properties of different geometric shapes and figures are covered.
## Mensuration Maths- Definition
A branch of mathematics which talks about the length, volume or area of different geometric shapes is called Mensuration. These shapes exist in 2 dimension or 3 dimensions. Let’s learn the difference between the two.
Difference Between 2D and 3D shapes
## Mensuration in Maths- Important Terminologies
Let’s learn a few more definitions related to this topic.
## Mensuration Formulas
Now let’s learn all the important mensuration formulas involving 2D and 3D shapes. Using this mensuration formula list, it will be easy to solve the mensuration problems. Students can also download the mensuration formulas list PDF from the link given above. In general, the most common formulas in mensuration involve surface area and volumes of 2D and 3D figures.
### Mensuration Problems
Question: Find the area and perimeter of a square whose side is 5 cm?
Solution:
Given:
Side = 5 cm
Area of a square = asquare units
Substitute the value of “a” in the formula, we get
Area of a square = 52
A = 5 x 5 = 25
Therefore, the area of a square = 25 cm2
The perimeter of a square = 4a units
P = 4 x 5 =20
Therefore, the perimeter of a square = 20 cm.
Register at BYJU’S to learn more on other mathematical concepts and also learn mensuration problems and formulas by downloading BYJU’S – The Learning App.
## What is mensuration in Maths?
In maths, mensuration is defined as the study of the measurement of various 2D and 3D geometric shapes involving their surface areas, volumes, etc.
## What is the difference between mensuration and geometry?
Mensuration refers to the calculation of various parameters of shapes like the perimeter, area, volume, etc. whereas, geometry deals with the study of properties and relations of points and lines of various shapes.
### What are 2D and 3D Mensuration?
2D mensuration deals with the calculation of various parameters like area and perimeter of 2-dimensional shapes like square, rectangle, circle, triangles, etc.
3D mensuration is concerned with the study and calculation of surface area, lateral surface area, and volume of 3-dimensional figures like cube, sphere, cuboid, cone, cylinder, etc.
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# Simon's Favorite Factoring Trick
## Contents
Dr. Siman's Favorite Factoring Trick (abbreviated SFFT) is a special factorization first popularized by AoPS user Simon Rubinstein-Salzedo.
## The General Statement
The general statement of SFFT is: ${xy}+{xk}+{jy}+{jk}=(x+j)(y+k)$. Two special common cases are: $xy + x + y + 1 = (x+1)(y+1)$ and $xy - x - y +1 = (x-1)(y-1)$.
The act of adding ${jk}$ to ${xy}+{xk}+{jy}$ in order to be able to factor it could be called "completing the rectangle" in analogy to the more familiar "completing the square."
## Applications
This factorization frequently shows up on contest problems, especially those heavy on algebraic manipulation. Usually $x$ and $y$ are variables and $j,k$ are known constants. Also, it is typically necessary to add the $jk$ term to both sides to perform the factorization.
## Problems
### Introductory
• Two different prime numbers between $4$ and $18$ are chosen. When their sum is subtracted from their product, which of the following numbers could be obtained?
$\mathrm{(A) \ 21 } \qquad \mathrm{(B) \ 60 } \qquad \mathrm{(C) \ 119 } \qquad \mathrm{(D) \ 180 } \qquad \mathrm{(E) \ 231 }$
(Source)
### Intermediate
• $m, n$ are integers such that $m^2 + 3m^2n^2 = 30n^2 + 517$. Find $3m^2n^2$.
(Source)
• The integer $N$ is positive. There are exactly $2005$ pairs $(x, y)$ of positive integers satisfying:
$$\frac 1x +\frac 1y = \frac 1N$$
Prove that $N$ is a perfect square. (British Mathematical Olympiad Round 2, 2005)
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# The Monkey and the Coconuts— Solution
ORIGINAL PUZZLE
Firstly, let’s look at the solution for the original puzzle where a coconut is left over after the final division in the morning so the monkey ends up with 6 coconuts.
1) MATHEMATICAL
Basic algebra gives you this equation: N = {(15625 × F)+ 11529} / 1024 where N is the number of coconuts they collected originally and A to F are the number of coconuts taking from the pile at each division with F being the number each sailor got in the final division in the morning. What we want is a value of F such that {(115625 × F) + 11529} is a multiple of 1024 so that N is a whole number. But if {(115625 × F) + 11529} is a multiple of 1024, then we can subtract 1024 from both of the two numbers as many times as we like and still get something which is a multiple of 1024. Doing this as far as we can leaves {(265 × F) + 265}. This can be written as 265 × (F+1). Now for this to be a multiple of 1024, since 265 is fixed, (F+1) must be a multiple of 1024. And the smallest such value could be is F = 1023. This gives a solution of N = [(15625×1023)+11529)]/1024 which works out as 15,621 coconuts.
INTUITIVE
Still looking for this.
3) LATERAL
If you can imagine -4 coconuts in the original pile, and +1 is thrown to the monkey, this would leave -5 in the pile, so when a sailor takes one-fifth of the coconuts (that would be -1 coconut), then -4 coconuts remain. Each of the sailors do this in turn, and in the morning the sailors are confronted with a pile of -4 coconuts. After they toss +1 to the monkey, they can evenly divide the pile by taking -1 coconut each. The smallest whole number of divisions needs 56, that is 5×5×5×5×5×5 = 15,625. Adding this to -4, which is one solution, this gives the smallest positive solution of 15,621 coconuts.
Postscript. This table shows the process:
coconutsremoved given to monkey coconuts left 15621 3124 1 12496 2499 1 9996 1999 1 7996 1599 1 6396 1279 1 5116 5×1023 1 0
B. A. WILLIAM’S PUZZLE VARIATION
Secondly, let’s look at the solution for the William’s puzzle rediscovered by Martin Gardner. Here the final division in the morning does not leave a coconut left over for the monkey so it ends up with only 5 coconuts.
1) MATHEMATICAL
Using basic algebra as before gives you an equation: N = {(15625 × F)+ 8404} / 1024 where F is the number each sailor got in the final morning share-out and N is the number of coconuts to begin with. However there is a simpler method if you follow the logic slowly.
As before we use N as the initial number (a whole number) but then we work out the number left after each sailor has reduced the pile. Note that it doesn’t matter if we deduct the monkey’s coconut before of after taking away the fifth share each time.
After the first sailor’s action the number of coconuts left N1 => [4(N-1)/5]. We can expand this to 4N/5 -4/5. Using a sleight of hand we add +16/5 and -16/5 to the equation (which makes no difference as the two figures cancel each other out) allowing us to say that N1 => (4N+16)/5 - 20/5 and finally giving N1 => [4(N+4)]/5 - 4.
Now after the second sailor takes his share and gives one to the monkey,
N2 => [4(N1+4)]/5 - 4
working on the right hand side of the equation
N2 => [4 × {([4(N+4)]/5 - 4+4)}/5] - 4
N2 => [4 × {([4/5 ×(N+4)] - 4+4)}]/5 - 4
N2 => [4 × {([4/5 ×(N+4)])}]/5 - 4
N2 => [42 × (N+4)]/52 - 4
N2 => [16 × (N+4)]/25 - 4
With all five sailors taking their fifths and giving the monkey a coconut, the reducing pile works out like this:
after division by sailor 1, number of coconuts left N1 => [4 × (N+4)]/5 - 4
after division by sailor 2, number of coconuts left N2 => [16 × (N+4)]/25 - 4
after division by sailor 3, number of coconuts left N3 => [64 × (N+4)]/125 - 4
after division by sailor 4, number of coconuts left N4 => [256 × (N+4)]/625 - 4
after division by sailor 5, number of coconuts left N5 => [1024 × (N+4)]/3125 - 4
Since the final amount must be a whole number, (N+4) must equal 3125 or a multiple of it. However for the smallest answer it must equal 3125, so that N+4 = 3125 and therefore the original pile N = 3,121 coconuts. The final pile of coconuts left in the morning is [({1024 × (3121+4)}]/3125 - 4 which equals 1020 and divides exactly by 5 with none for the monkey.
2) INTUITIVE
Look at a simpler example. With three sailors, guessing that the number of coconuts will be 3x3x3x3 = 81. However when the first sailor divides the pile by 3 there is none left over so the original pile needs to be 82. Howver when the second sailor divides the reduced pile of 54 by 3 there is again none left over so we need to add another coconut for the monkey. This makes the original pile 83 which cannot be right as it leaves 2 spare after the first division. Since adding does not work how about subtracting? You need to take 2 away and make the beginning pile 79 if you are to get the first division to have a remainder of 1 but this works right through. Now two is the number of sailors minus 1.
So for our puzzle the answer must be 55, that is 5×5×5×5×5, minus (5-1) which works out as 3125 - 4 = 3,121 coconuts.
3) LATERAL
Still looking for this.
Postscript: This table shows the process:
coconutsremoved given to monkey coconuts left 3121 624 1 2496 499 1 1996 399 1 1596 319 1 1276 255 1 1020 5×204 0 0
The five sailors ended up in the order of their raiding the hoard with 828, 703, 603, 523 and 459 coconuts.
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# Question Video: Telling How Many Times a Specific Hour Comes in the Day Mathematics
How many times a day does the clock show the given time?
02:23
### Video Transcript
How many times a day does the clock show the given time?
The clock that the question’s talking about is shown in the picture. We know that at an o’clock time, the minute hand points to the number 12, but in this picture it’s pointing to the number three. It’s moved a quarter of the way around the clock face. This is a quarter past time. And if we look at the hour hand, we can see what hour of the day it is. The hour hand is on its way from the number six to the number seven. We know that at 6 o’clock, it would’ve been pointing exactly to the number six. And so, we can read this time as quarter past 6.
Now, our question asks us, “how many times a day does the clock show this time?” Now, it’s easy to make a mistake here. We might look at the time and just say, “Well, there’s only one-quarter past 6 every day.” But let’s use some facts to help us. Firstly, we know that 24 hours are the same as one day. When we say a day, we don’t just mean the daytime when the sun is up; we mean a whole day, for example, Monday or Tuesday. So this includes the daytime and the nighttime.
Another fact we can use to help us here is that the hour hand goes round the clock face once every 12 hours. It doesn’t matter what time of day it is. It takes 12 hours for it to go all the way around back to where it was. So if we look at our clock face and we think that our hour hand is in the exact position for quarter past 6, if 12 hours go by, it’s going to show quarter past 6 again. But there aren’t 12 hours in a day. There are 24 hours in a day. This is where we can use a number fact to help us. We know that 12 plus another 12 make 24.
And so, in a whole day, the hour hand doesn’t just go once around the clock; it goes twice. This clock is going to show the time quarter past 6 twice a day, quarter past 6 in the morning and then quarter past 6 in the evening. We know that the hour hand makes two journeys around the clock face in a day. And you know, it doesn’t really matter what time this clock shows. The number of times every day that a clock shows any time is always two.
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# 4.08 Polynomials
Lesson
## Introduction
An expression of the form Ax^{n}, where A is any number and n is any non-negative integer, is called a monomial. When we take the sum of multiple monomials, we get a polynomial.
## Parts of polynomials
In the monomial Ax^{n}:
• A is the coefficient
• x is the variable
• n is the index
A polynomial is a sum of any number of monomials (and consequently, each term of a polynomial is a monomial). The highest index is called the degree of the polynomial. For example, x^{3}+4x+3 is a polynomial of degree three. The coefficient of the term with the highest index is called the leading coefficient. The coefficient of the term with index 0 is called the constant.
We often name polynomials using function notation. For example P(x) is a polynomial where x is the variable. If we write a constant instead of x, that means that we substitute that constant for the variable. For example, if P(x)=x^{3}+4x+3 then P(3)=3^{3}+4\times 3+3=42.
Polynomials of particular degrees are given specific names. Some of these we have seen before.
### Examples
#### Example 1
For the polynomial P(x)=\dfrac{x^7}{5}+\dfrac{x^6}{6}+5.
a
What's the degree of the polynomial?
Worked Solution
Create a strategy
The degree of a polynomial is the highest power of x.
Apply the idea
The highest power of x is 7, so the degree of the polynomial is 7.
b
What's the leading coefficient of the polynomial?
Worked Solution
Create a strategy
The term containing the highest power of x is called the leading term, and its coefficient is the leading coefficient.
Apply the idea
The leading term of the polynomial is \dfrac{x^7}{5} which can be written as \dfrac{1}{5}x^7. So the leading coefficient of the polynomial is \dfrac{1}{5}.
c
What's the constant term of the polynomial?
Worked Solution
Create a strategy
The constant term is the term that is independent of x.
Apply the idea
The term that is independent of x is 5, so the constant term of the polynomial is 5.
#### Example 2
Consider P(x)=4x^5+3x^{6}-8.
a
Find P(0).
Worked Solution
Create a strategy
Substitute x=0 into the polynomial.
Apply the idea
b
Find P(-4).
Worked Solution
Create a strategy
Substitute x=-4 into the equation.
Apply the idea
Idea summary
A monomial is an expression of the form:
\displaystyle Ax^{n}
\bm{A}
is the coefficient
\bm{x}
is the variable
\bm{n}
is the index
A polynomial is a sum of any number of monomials. The highest index is called the degree of the polynomial. The coefficient of the term with the highest index is called the leading coefficient. The coefficient of the term with index 0 is called the constant.
## Operations on polynomials
We apply operations to polynomials in the same way as we apply operations to numbers. For addition and subtraction we add or subtract all of the terms in both polynomials and we simplify by collecting like terms. For multiplication we multiply each term in one polynomial by each term in the other polynomial similar to how we expand binomial products. Division is a more complicated case that we will look at in the next lesson .
A polynomial is a sum of any number of monomials. In a polynomial:
• The highest index is the degree
• The coefficient of the term with the highest index is the leading coefficient
• The coefficient of the term with index 0 is the constant
We apply operations to polynomials in the same way that we apply operations to numbers.
### Examples
#### Example 3
If P(x)=-5x^{2}-6x-6 and Q(x)=-7x+7, form a simplified expression for P(x)-Q(x).
Worked Solution
Create a strategy
Substitute the expressions for P(x) and Q(x), and then subtract the like terms.
Apply the idea
#### Example 4
Simplify \left(3x^{3}-9x^{2}-8x-7\right)+\left(-7x^{3}-9x\right).
Worked Solution
Create a strategy
Apply the idea
Idea summary
A polynomial is a sum of any number of monomials. In a polynomial:
• The highest index is the degree
• The coefficient of the term with the highest index is the leading coefficient
• The coefficient of the term with index 0 is the constant
### Outcomes
#### VCMNA357 (10a)
Investigate the concept of a polynomial and apply the factor and remainder theorems to solve problems.
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## RD Sharma Solutions for Class 7 Maths Chapter 16 Congruence Free Online
Exercise 16.1 Page No: 16.3
1. Explain the concept of congruence of figures with the help of certain examples.
Solution:
Congruent objects or figures are exact copies of each other or we can say mirror images of each other. The relation of two objects being congruent is called congruence.
Consider Ball 1 and Ball 2. These two balls are congruent.
Ball 1 Ball 2
Now consider the two stars below. Star A and Star B are exactly the same in size, colour and shape. These are congruent stars
Star A Star B
2. Fill in the blanks:
(i) Two line segments are congruent if ……..
(ii) Two angles are congruent if ……..
(iii) Two square are congruent if ………
(iv) Two rectangles are congruent if ………
(v) Two circles are congruent if …….
Solution:
(i) They are of equal lengths
(ii) Their measures are the same or equal.
(iii) Their sides are equal or they have the same side length
(iv) Their dimensions are same that is lengths are equal and their breadths are also equal.
3. In Fig. 6, ∠POQ ≅∠ROS, can we say that ∠POR ≅∠QOS
Solution:
Given that
∠POQ ≅∠ROS
Also given that ∠ROQ ≅∠ROQ
Therefore adding ∠ROQ to both sides of ∠POQ ≅∠ROS,
We get, ∠POQ + ∠ROQ ≅∠ROQ + ∠ROS
Therefore, ∠PQR ≅∠QOS
4. In fig. 7, a = b = c, name the angle which is congruent to ∠AOC
Solution:
From the figure we have
∠ AOB = ∠ BOC = ∠ COD
Therefore, ∠ AOB = ∠COD
Also, ∠ AOB + ∠ BOC = ∠ BOC + ∠ COD
∠ AOC = ∠ BOD
Hence, ∠ BOD ≅∠ AOC
5. Is it correct to say that any two right angles are congruent? Give reasons to justify your answer.
Solution:
Two right angles are congruent to each other because they both measure 90o.
We know that two angles are congruent if they have the same measure.
6. In fig. 8, ∠AOC ≅∠PYR and ∠BOC ≅∠QYR. Name the angle which is congruent to ∠AOB.
Solution:
Given that∠AOC ≅∠PYR
Also given that ∠BOC ≅∠QYR
Now, ∠AOC = ∠AOB + ∠BOC ∠PYR =∠PYQ +∠QYR
By putting the value of ∠AOC and ∠PYR in ∠AOC ≅∠PYR
We get, ∠AOB + ∠BOC ≅∠PYQ + ∠QYR ∠AOB ≅∠PYQ (∠BOC ≅∠QYR)
Hence, ∠AOB ≅∠PYQ
7. Which of the following statements are true and which are false;
(i) All squares are congruent.
(ii) If two squares have equal areas, they are congruent.
(iii) If two rectangles have equal areas, they are congruent.
(iv) If two triangles have equal areas, they are congruent.
Solution:
(i) False.
Explanation:
All the sides of a square are of equal length. However, different squares can have sides of different lengths. Hence all squares are not congruent.
(ii) True.
Explanation:
Two squares that have the same area will have sides of the same lengths. Hence they will be congruent.
(iii) False
Explanation:
Area of a rectangle = length x breadth
Two rectangles can have the same area. However, the lengths of their sides can vary and hence they are not congruent.
(iv) False
Explanation:
Area of a triangle = 12 x base x height
Two triangles can have the same area but the lengths of their sides can vary and hence they cannot be congruent.
Exercise 16.2 Page No: 16.8
1. In the following pairs of triangle (Fig. 12 to 15), the lengths of the sides are indicated along sides. By applying SSS condition, determine which are congruent. State the result in symbolic form.
Solution:
(i) In Δ ABC and Δ DEF
AB = DE = 4.5 cm (Side)
BC = EF = 6 cm (Side) and
AC = DF = 4 cm (Side)
SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.
Therefore, by SSS criterion of congruence, ΔABC ≅ ΔDEF
(ii) In Δ ACB and Δ ADB
AC = AD = 5.5cm (Side)
BC = BD = 5cm (Side) and
AB = AB = 6cm (Side)
SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.
Therefore, by SSS criterion of congruence, ΔACB ≅ ΔADB
(iii) In Δ ABD and Δ FEC,
AB = FE = 5cm (Side)
AD = FC = 10.5cm (Side)
BD = CE = 7cm (Side)
SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.
Therefore, by SSS criterion of congruence, ΔABD ≅ ΔFEC
(iv) In Δ ABO and Δ DOC,
AB = DC = 4cm (Side)
AO = OC = 2cm (Side)
BO = OD = 3.5cm (Side)
SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.
Therefore, by SSS criterion of congruence, ΔABO ≅ ΔODC
2. In fig.16, AD = DC and AB = BC
(i) Is ΔABD ≅ ΔCBD?
(ii) State the three parts of matching pairs you have used to answer (i).
Solution:
(i) Yes ΔABD ≅ΔCBD by the SSS criterion.
SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.
Hence ΔABD ≅ΔCBD
(ii) We have used the three conditions in the SSS criterion as follows:
AB = BC and
DB = BD
3. In Fig. 17, AB = DC and BC = AD.
(i) Is ΔABC ≅ ΔCDA?
(ii) What congruence condition have you used?
(iii) You have used some fact, not given in the question, what is that?
Solution:
(i) From the figure we have AB = DC
And AC = AC
SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.
Therefore by SSS criterion ΔABC ≅ ΔCDA
(ii) We have used Side congruence condition with one side common in both the triangles.
(iii)Yes, have used the fact that AC = CA.
4. In ΔPQR ≅ ΔEFD,
(i) Which side of ΔPQR equals ED?
(ii) Which angle of ΔPQR equals angle E?
Solution:
(i) PR = ED
Since the corresponding sides of congruent triangles are equal.
(ii) ∠QPR = ∠FED
Since the corresponding angles of congruent triangles are equal.
5. Triangles ABC and PQR are both isosceles with AB = AC and PO = PR respectively. If also, AB = PQ and BC = QR, are the two triangles congruent? Which condition do you use?
It ∠B = 50°, what is the measure of ∠R?
Solution:
Given that AB = AC in isosceles ΔABC
And PQ = PR in isosceles ΔPQR.
Also given that AB = PQ and QR = BC.
Therefore, AC = PR (AB = AC, PQ = PR and AB = PQ)
Hence, ΔABC ≅ ΔPQR
Now
∠ABC = ∠PQR (Since triangles are congruent)
However, ΔPQR is isosceles.
Therefore, ∠PRQ = ∠PQR = ∠ABC = 50o
6. ABC and DBC are both isosceles triangles on a common base BC such that A and D lie on the same side of BC. Are triangles ADB and ADC congruent? Which condition do you use? If ∠BAC = 40° and ∠BDC = 100°, then find ∠ADB.
Solution:
Given ABC and DBC are both isosceles triangles on a common base BC
∠ABC + ∠BCA + ∠BAC = 180o (Angle sum property)
Since ΔABC is an isosceles triangle,
∠ABC = ∠BCA
∠ABC +∠ABC + 40= 180o
2 ∠ABC = 180o– 40o = 140o
∠ABC = 140o/2 = 70o
∠DBC + ∠ BCD + ∠ BDC = 180o (Angle sum property)
Since ΔABC is an isosceles triangle, ∠ DBC = ∠BCD
∠DBC + ∠DBC + 100= 180o
2 ∠DBC = 180°– 100o = 80o
∠DBC = 80o/2 = 40o
30o + 20o + ∠ADB = 180o (∠ADB = ∠ABC – ∠DBC),
7. Δ ABC and ΔABD are on a common base AB, and AC = BD and BC = AD as shown in Fig. 18. Which of the following statements is true?
(i) ΔABC ≅ ΔABD
Solution:
In ΔABC and ΔBAD we have,
AC = BD (given)
And AB = BA (corresponding parts of congruent triangles)
Therefore by SSS criterion of congruency, ΔABC ≅ ΔBAD
Therefore option (iii) is true.
8. In Fig. 19, ΔABC is isosceles with AB = AC, D is the mid-point of base BC.
(ii) State the three pairs of matching parts you use to arrive at your answer.
Solution:
(i) Given that AB = AC.
Also since D is the midpoint of BC, BD = DC
Therefore by SSS condition,
(ii)We have used AB, AC; BD, DC and AD, DA
9. In fig. 20, ΔABC is isosceles with AB = AC. State if ΔABC ≅ ΔACB. If yes, state three relations that you use to arrive at your answer.
Solution:
Given that ΔABC is isosceles with AB = AC
SSS criterion is two triangles are congruent, if the three sides of triangle are respectively equal to the three sides of the other triangle.
ΔABC ≅ ΔACBby SSS condition.
Since, ABC is an isosceles triangle, AB = BC, BC = CB and AC = AB
10. Triangles ABC and DBC have side BC common, AB = BD and AC = CD. Are the two triangles congruent? State in symbolic form, which congruence do you use? Does ∠ABD equal ∠ACD? Why or why not?
Solution:
Yes, congruent because given that ABC and DBC have side BC common, AB = BD and AC = CD
Also from the above data we can say
By SSS criterion of congruency, ΔABC ≅ ΔDBC
No, ∠ABD and ∠ACD are not equal because AB not equal to AC
Exercise 16.3 Page No: 16.14
1. By applying SAS congruence condition, state which of the following pairs (Fig. 28) of triangle are congruent. State the result in symbolic form
Solution:
(i) From the figure we have OA = OC and OB = OD and
∠AOB = ∠COD which are vertically opposite angles.
Therefore by SAS condition, ΔAOC ≅ΔBOD
(ii) From the figure we have BD = DC
(iii) From the figure we have AB = DC
∠ABD = ∠CDB and
Therefore, by SAS condition, ΔABD ≅ΔCBD
(iv) We have BC = QR
ABC = PQR = 90o
And AB = PQ
Therefore, by SAS condition, ΔABC≅ ΔPQR.
2. State the condition by which the following pairs of triangles are congruent.
Solution:
BC = CD and AC = CA
Therefore by SSS condition, ΔABC≅ ΔADC
(ii) AC = BD
AD = BC and AB = BA
Therefore, by SSS condition, ΔABD ≅ ΔADC
∠BAC = ∠DAC and
Therefore by SAS condition, ΔBAC ≅ ΔBAC
∠DAC = ∠BCA and
Therefore, by SAS condition, ΔABC ≅ ΔADC
3. In fig. 30, line segments AB and CD bisect each other at O. Which of the following statements is true?
(i) ΔAOC ≅ ΔDOB
(ii) ΔAOC ≅ ΔBOD
(iii) ΔAOC ≅ ΔODB
State the three pairs of matching parts, you have used to arrive at the answer.
Solution:
From the figure we have,
And, CO = OD
Also, AOC = BOD
Therefore, by SAS condition, ΔAOC ≅ ΔBOD
4. Line-segments AB and CD bisect each other at O. AC and BD are joined forming triangles AOC and BOD. State the three equality relations between the parts of the two triangles that are given or otherwise known. Are the two triangles congruent? State in symbolic form, which congruence condition do you use?
Solution:
We have AO = OB and CO = OD
Since AB and CD bisect each other at 0.
Also ∠AOC = ∠BOD
Since they are opposite angles on the same vertex.
Therefore by SAS congruence condition, ΔAOC ≅ ΔBOD
5. ΔABC is isosceles with AB = AC. Line segment AD bisects ∠A and meets the base BC in D.
(ii) State the three pairs of matching parts used to answer (i).
(iii) Is it true to say that BD = DC?
Solution:
(i) We have AB = AC (Given)
Therefore by SAS condition of congruence, ΔABD ≅ ΔACD
(iii) Now, ΔABD≅ΔACD
Therefore by corresponding parts of congruent triangles
BD = DC.
6. In Fig. 31, AB = AD and ∠BAC = ∠DAC.
(i) State in symbolic form the congruence of two triangles ABC and ADC that is true.
(ii) Complete each of the following, so as to make it true:
(a) ∠ABC =
(b) ∠ACD =
(c) Line segment AC bisects ….. And ……..
Solution:
∠BAC = ∠DAC (given)
AC = CA (common)
Therefore by SAS condition of congruency, ΔABC ≅ ΔADC
ii) ∠ABC = ∠ADC (corresponding parts of congruent triangles)
∠ACD = ∠ACB (corresponding parts of congruent triangles)
7. In fig. 32, AB || DC and AB = DC.
(i) Is ΔACD ≅ ΔCAB?
(ii) State the three pairs of matching parts used to answer (i).
(iii) Which angle is equal to ∠CAD?
Solution:
(i) Yes by SAS condition of congruency, ΔDCA ≅ ΔBAC
(ii) We have used AB = DC, AC = CA and ∠DCA = ∠BAC.
(iii) ∠CAD = ∠ACB since the two triangles are congruent.
(iv) Yes this follows from AD parallel to BC as alternate angles are equal. lf alternate angles are equal the lines are parallel
Exercise 16.4 Page No: 16.19
1. Which of the following pairs of triangle are congruent by ASA condition?
Solution:
(i) We have,
Since ∠ABO = ∠CDO = 45o and both are alternate angles, AB parallel to DC, ∠BAO = ∠DCO (alternate angle, AB parallel to CD and AC is a transversal line)
∠ABO = ∠CDO = 45o (given in the figure) Also,
AB = DC (Given in the figure)
Therefore, by ASA ΔAOB ≅ ΔDOC
(ii) In ABC,
Now AB =AC (Given)
∠ABD = ∠ACD = 40o (Angles opposite to equal sides)
∠ABD + ∠ACD + ∠BAC = 180o (Angle sum property)
40o + 40o + ∠BAC = 180o
∠BAC =180o – 80o =100o
∠BAD = ∠BAC – ∠DAC = 100o – 50o = 50o
Therefore, by ASA, ΔABD ≅ ΔADC
(iii) In Δ ABC,
∠A + ∠B + ∠C = 180(Angle sum property)
∠C = 180o– ∠A – ∠B
∠C = 180o – 30o – 90o = 60o
In PQR,
∠P + ∠Q + ∠R = 180o (Angle sum property)
∠P = 180o – ∠Q – ∠R
∠P = 180– 60– 90o = 30o
∠BAC = ∠QPR = 30o
∠BCA = ∠PRQ = 60o and AC = PR (Given)
Therefore, by ASA, ΔABC ≅ ΔPQR
(iv) We have only
BC = QR but none of the angles of ΔABC and ΔPQR are equal.
Therefore, ΔABC and Cong ΔPRQ
(ii) State the three pairs of matching parts you have used in (i)
(iii) Is it true to say that BD = DC?
Solution:
3. Draw any triangle ABC. Use ASA condition to construct other triangle congruent to it.
Solution:
We have drawn
Δ ABC with ∠ABC = 65o and ∠ACB = 70o
We now construct ΔPQR ≅ ΔABC has ∠PQR = 65o and ∠PRQ = 70o
Also we construct ΔPQR such that BC = QR
Therefore by ASA the two triangles are congruent
4. In Δ ABC, it is known that ∠B = C. Imagine you have another copy of Δ ABC
(i) Is ΔABC ≅ ΔACB
(ii) State the three pairs of matching parts you have used to answer (i).
(iii) Is it true to say that AB = AC?
Solution:
(i) Yes ΔABC ≅ ΔACB
(ii) We have used ∠ABC = ∠ACB and ∠ACB = ∠ABC again.
Also BC = CB
(iii) Yes it is true to say that AB = AC since ∠ABC = ∠ACB.
5. In Fig. 38, AX bisects ∠BAC as well as ∠BDC. State the three facts needed to ensure that ΔACD ≅ ΔABD
Solution:
As per the given conditions,
Therefore, by ASA, ΔACD ≅ ΔABD
6. In Fig. 39, AO = OB and ∠A = ∠B.
(i) Is ΔAOC ≅ ΔBOD
(ii) State the matching pair you have used, which is not given in the question.
(iii) Is it true to say that ∠ACO = ∠BDO?
Solution:
We have
∠OAC = ∠OBD,
AO = OB
Also, ∠AOC = ∠BOD (Opposite angles on same vertex)
Therefore, by ASA ΔAOC ≅ ΔBOD
Exercise 16.5 Page No: 16.23
1. In each of the following pairs of right triangles, the measures of some part are indicated alongside. State by the application of RHS congruence conditions which are congruent, and also state each result in symbolic form. (Fig. 46)
Solution:
(i) ∠ADC = ∠BCA = 90o
AD = BC and hypotenuse AB = hypotenuse AB
Therefore, by RHS ΔADB ≅ ΔACB
Hypotenuse AC = hypotenuse AB (Given)
∠ADB = 180o – 90o = 90o
(iii) Hypotenuse AO = hypotenuse DO
BO = CO
∠B = ∠C = 90o
Therefore, by RHS, ΔAOB≅ΔDOC
(iv) Hypotenuse AC = Hypotenuse CA
BC = DC
Therefore, by RHS, ΔABC ≅ ΔADC
(v) BD = DB
Hypotenuse AB = Hypotenuse BC, as per the given figure,
∠BDA + ∠BDC = 180o
∠BDA + 90o = 180o
∠BDA= 180– 90o = 90o
∠BDA = ∠BDC = 90o
Therefore, by RHS, ΔABD ≅ ΔCBD
2. Δ ABC is isosceles with AB = AC. AD is the altitude from A on BC.
(i) Is ΔABD ≅ ΔACD?
(ii) State the pairs of matching parts you have used to answer (i).
(iii) Is it true to say that BD = DC?
Solution:
(i) Yes, ΔABD ≅ ΔACD by RHS congruence condition.
(ii) We have used Hypotenuse AB = Hypotenuse AC
(iii)Yes, it is true to say that BD = DC (corresponding parts of congruent triangles)
Since we have already proved that the two triangles are congruent.
3. ΔABC is isosceles with AB = AC. Also. AD ⊥ BC meeting BC in D. Are the two triangles ABD and ACD congruent? State in symbolic form. Which congruence condition do you use? Which side of ADC equals BD? Which angle of Δ ADC equals ∠B?
Solution:
We have AB = AC …… (i)
AD = DA (common) …… (ii)
Therefore, from (i), (ii) and (iii), by RHS congruence condition, ΔABD ≅ ΔACD, the triangles are congruent.
Therefore, BD = CD.
And ∠ABD = ∠ACD (corresponding parts of congruent triangles)
4. Draw a right triangle ABC. Use RHS condition to construct another triangle congruent to it.
Solution:
Consider
Δ ABC with ∠B as right angle.
We now construct another triangle on base BC, such that ∠C is a right angle and AB = DC
Also, BC = CB
Therefore, BC = CB
Therefore by RHS, ΔABC ≅ ΔDCB
5.In fig. 47, BD and CE are altitudes of Δ ABC and BD = CE.
(i) Is ΔBCD ≅ ΔCBE?
(ii) State the three pairs or matching parts you have used to answer (i)
Solution:
(i) Yes, ΔBCD ≅ ΔCBE by RHS congruence condition.
(ii) We have used hypotenuse BC = hypotenuse CB
BD = CE (Given in question)
And ∠BDC = ∠CBE = 90o
Courtesy : CBSE
|
# Operation on Rational & Irrational Numbers | Q5
###### Number System | Extra Questions For Class 9 Maths Chapter 1 | Polynomial expansion
Question 5: Simplify the following:
(a) {8 + $sqrt{5})}) $${8 - $sqrt{5}$})$b) {10 + $sqrt{3})}) $${6 + $sqrt{2}$})$c) ${$$sqrt {3} + $sqrt {11}$}^2) + ${$$sqrt {3} - \sqrt {11}$}^2) ## Target Centum in CBSE 10th Maths #### Free Online CBSE Coaching online.maxtute.com ### Video Explanation ## NCERT Solution to Class 10 Maths #### With Videos ### Explanatory Answer | Chapter Extra Question 5$a) {8 + $sqrt{5})}) $${8 - $sqrt{5}$}) The expression is of the form$x + y) (x – y) = x2 – y2
So, {8 + $sqrt{5})}) $${8 - $sqrt{5}$} =${8^2} - $sqrt{5}^2)) = 64 – 5 = 59$b) {10 + $sqrt{3})}) $${6 + $sqrt{2}$}) Simplifying in such terms is the same as expanding the terms of the expression = ${$10 $times 6$} + 10 $times \sqrt{2} + \sqrt{3} \times 6 + \sqrt{2} \times \sqrt{3}$ = $60 + 10\sqrt{2} + 6\sqrt{3} + \sqrt{6}$$c) ${$$sqrt {3} + $sqrt {11}$}^2 + {$$sqrt {3} - $sqrt {11}$}^2) ${$$sqrt {3} + \sqrt {11}$}^2) = ${$$sqrt {3}$}^2) + ${$$sqrt {11}$}^2 + 2 $times \sqrt{3} \times \sqrt{11}$ $= 3 + 11 + 2\sqrt{33} = 14 + 2\sqrt{33}$ ${$$sqrt {3} - \sqrt {11}$}^2) = ${$$sqrt {3}$}^2) + ${$$sqrt {11}$}^2 - 2 $times \sqrt{3} \times \sqrt{11}$ $= 3 + 11 - 2\sqrt{33} = 14 - 2\sqrt{33}$ Thus, ${$$sqrt {3} + \sqrt {11}$}^2) + ${$\sqrt {3} - \sqrt {11}$}^2 = 14 + 2$sqrt{33} + 14 - 2\sqrt{33}$
= 28
###### Free CBSE Online CoachingClass 9 Maths
Register in 2 easy steps and
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# nth root
Calculation results
$\scriptstyle\left.\begin{matrix}\scriptstyle\text{summand}+\text{summand}\\\scriptstyle\text{augend}+\text{addend}\end{matrix}\right\}=$ $\scriptstyle\text{sum}$
Subtraction (−)
$\scriptstyle\text{minuend}-\text{subtrahend}=$ $\scriptstyle\text{difference}$
Multiplication (×)
$\scriptstyle\left.\begin{matrix}\scriptstyle\text{multiplicand}\times\text{multiplicand}\\\scriptstyle\text{multiplicand}\times\text{multiplier}\end{matrix}\right\}=$ $\scriptstyle\text{product}$
Division (÷)
$\scriptstyle\left.\begin{matrix}\scriptstyle\frac{\scriptstyle\text{dividend}}{\scriptstyle\text{divisor}}\\\scriptstyle\frac{\scriptstyle\text{numerator}}{\scriptstyle\text{denominator}}\end{matrix}\right\}=$ $\scriptstyle\text{quotient}$
Modulation (mod)
$\scriptstyle\text{dividend}\bmod\text{divisor}=$ $\scriptstyle\text{remainder}$
Exponentiation
$\scriptstyle\text{base}^\text{exponent}=$ $\scriptstyle\text{power}$
nth root (√)
$\scriptstyle\sqrt[\text{degree}]{\scriptstyle\text{radicand}}=$ $\scriptstyle\text{root}$
Logarithm (log)
$\scriptstyle\log_\text{base}(\text{antilogarithm})=$ $\scriptstyle\text{logarithm}$
Roots of integer numbers from 0 to 10. Line labels = x. x-axis = n. y-axis = nth root of x.
In mathematics, the nth root of a number x, where n is a positive integer, is a number r which, when raised to the power n yields x
$r^n = x,$
where n is the degree of the root. A root of degree 2 is called a square root and a root of degree 3, a cube root. Roots of higher degree are referred by using ordinal numbers, as in fourth root, twentieth root, etc.
For example:
• 2 is a square root of 4, since 22 = 4.
• −2 is also a square root of 4, since (−2)2 = 4.
A real number or complex number has n roots of degree n. While the roots of 0 are not distinct (all equaling 0), the n nth roots of any other real or complex number are all distinct. If n is even and x is real and positive, one of its nth roots is positive, one is negative, and the rest are complex but not real; if n is even and x is real and negative, none of the nth roots is real. If n is odd and x is real, one nth root is real and has the same sign as x , while the other roots are not real. Finally, if x is not real, then none of its nth roots is real.
Roots are usually written using the radical symbol or radix $\sqrt{\,\,}$ or $\surd{}$, with $\sqrt{x}\!\,$ or $\surd x$ denoting the square root, $\sqrt[3]{x}\!\,$ denoting the cube root, $\sqrt[4]{x}$ denoting the fourth root, and so on. In the expression $\sqrt[n]{x}$, n is called the index, $\sqrt{\,\,}$ is the radical sign or radix, and x is called the radicand. Since the radical symbol denotes a function, when a number is presented under the radical symbol it must return only one result, so a non-negative real root, called the principal nth root, is preferred rather than others; if the only real root is negative, as for the cube root of –8, again the real root is considered the principal root. An unresolved root, especially one using the radical symbol, is often referred to as a surd[1] or a radical.[2] Any expression containing a radical, whether it is a square root, a cube root, or a higher root, is called a radical expression, and if it contains no transcendental functions or transcendental numbers it is called an algebraic expression.
In calculus, roots are treated as special cases of exponentiation, where the exponent is a fraction:
$\sqrt[n]{x} \,=\, x^{1/n}$
Roots are particularly important in the theory of infinite series; the root test determines the radius of convergence of a power series. Nth roots can also be defined for complex numbers, and the complex roots of 1 (the roots of unity) play an important role in higher mathematics. Galois theory can be used to determine which algebraic numbers can be expressed using roots, and to prove the Abel-Ruffini theorem, which states that a general polynomial equation of degree five or higher cannot be solved using roots alone; this result is also known as "the insolubility of the quintic".
## Etymology
### Origin of the root symbol
The origin of the root symbol √ is largely speculative. Some sources imply that the symbol was first used by Arabic mathematicians. One of those mathematicians was Abū al-Hasan ibn Alī al-Qalasādī (1421–1486). Legend has it that it was taken from the Arabic letter "ج" (ǧīm, ), which is the first letter in the Arabic word "جذر" (jadhir, meaning "root"; ).[3] However, many scholars, including Leonhard Euler,[4] believe it originates from the letter "r", the first letter of the Latin word "radix" (meaning "root"), referring to the same mathematical operation. The symbol was first seen in print without the vinculum (the horizontal "bar" over the numbers inside the radical symbol) in the year 1525 in Die Coss by Christoff Rudolff, a German mathematician.
The Unicode and HTML character codes for the radical symbols are:
Read Character Unicode ASCII URL HTML (others)
Square root U+221A √ %E2%88%9A √
Cube root U+221B ∛ %E2%88%9B
Fourth root U+221C ∜ %E2%88%9C
### Etymology of "surd"
The term surd traces back to al-Khwārizmī (c. 825), who referred to rational and irrational numbers as audible and inaudible, respectively. This later led to the Arabic word "أصم" (asamm, meaning "deaf" or "dumb") for irrational number being translated into Latin as "surdus" (meaning "deaf" or "mute"). Gerard of Cremona (c. 1150), Fibonacci (1202), and then Robert Recorde (1551) all used the term to refer to unresolved irrational roots.[5]
## Definition and notation
The four 4th roots of −1,
none of which is real
The three 3rd roots of −1,
one of which is a negative real
An nth root of a number x, where n is a positive integer, is any of the n real or complex numbers r whose nth power is x:
$r^n = x.\!\,$
Every positive real number x has a single positive nth root, called the principal nth root, which is written $\sqrt[n]{x}$. For n equal to 2 this is called the principal square root and the n is omitted. The nth root can also be represented using exponentiation as x1/n.
For even values of n, positive numbers also have a negative nth root, while negative numbers do not have a real nth root. For odd values of n, every negative number x has a real negative nth root. For example, −2 has a real 5th root, $\sqrt[5]{-2} \,= -1.148698354\ldots$ but −2 does not have any real 6th roots.
Every non-zero number x, real or complex, has n different complex number nth roots including any positive or negative roots. They are all distinct except in the case of x = 0, all of whose nth roots equal 0.
The nth roots of almost all numbers (all integers except the nth powers, and all rationals except the quotients of two nth powers) are irrational. For example,
$\sqrt{2} = 1.414213562\ldots$
All nth roots of integers, and in fact of all algebraic numbers, are algebraic.
### Square roots
The graph $y=\pm \sqrt{x}$.
Main article: Square root
A square root of a number x is a number r which, when squared, becomes x:
$r^2 = x.\!\,$
Every positive real number has two square roots, one positive and one negative. For example, the two square roots of 25 are 5 and −5. The positive square root is also known as the principal square root, and is denoted with a radical sign:
$\sqrt{25} = 5.\!\,$
Since the square of every real number is a positive real number, negative numbers do not have real square roots. However, every negative number has two imaginary square roots. For example, the square roots of −25 are 5i and −5i, where i represents a square root of −1.
### Cube roots
The graph $y=\sqrt[3]{x}$.
Main article: Cube root
A cube root of a number x is a number r whose cube is x:
$r^3 = x.\!\,$
Every real number x has exactly one real cube root, written $\sqrt[3]{x}$. For example,
$\sqrt[3]{8}\,=\,2\quad\text{and}\quad\sqrt[3]{-8}\,= -2.$
Every real number has two additional complex cube roots.
## Identities and properties
Every positive real number has a positive nth root and the rules for operations with such surds are straightforward:
$\sqrt[n]{ab} = \sqrt[n]{a} \sqrt[n]{b} \,,$
$\sqrt[n]{\frac{a}{b}} = \frac{\sqrt[n]{a}}{\sqrt[n]{b}} \,.$
Using the exponent form as in $x^{1/n}$ normally makes it easier to cancel out powers and roots.
$\sqrt[n]{a^m} = \left(a^m\right)^{\frac{1}{n}} = a^{\frac{m}{n}}.$
Problems can occur when taking the nth roots of negative or complex numbers. For instance:
$\sqrt{-1}\times\sqrt{-1} = -1$
whereas
$\sqrt{-1 \times -1} = 1$
when taking the principal value of the roots.
## Simplified form of a radical expression
A non-nested radical expression is said to be in simplified form if[6]
1. There is no factor of the radicand that can be written as a power greater than or equal to the index.
2. There are no fractions under the radical sign.
3. There are no radicals in the denominator.
For example, to write the radical expression $\sqrt{\tfrac{32}{5}}$ in simplified form, we can proceed as follows. First, look for a perfect square under the square root sign and remove it:
$\sqrt{\tfrac{32}{5}} = \sqrt{\tfrac{16 \cdot 2}{5}} = 4 \sqrt{\tfrac{2}{5}}$
Next, there is a fraction under the radical sign, which we change as follows:
$4 \sqrt{\tfrac{2}{5}} = \frac{4 \sqrt{2}}{\sqrt{5}}$
Finally, we remove the radical from the denominator as follows:
$\frac{4 \sqrt{2}}{\sqrt{5}} = \frac{4 \sqrt{2}}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{4 \sqrt{10}}{5} = \frac{4}{5}\sqrt{10}$
When there is a denominator involving surds it is always possible to find a factor to multiply both numerator and denominator by to simplify the expression.[7][8] For instance using the factorization of the sum of two cubes:
$\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}} = \frac{\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}}{(\sqrt[3]{a}+\sqrt[3]{b})(\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2})} = \frac{\sqrt[3]{a^2}-\sqrt[3]{ab}+\sqrt[3]{b^2}}{a+b} \,.$
Simplifying radical expressions involving nested radicals can be quite difficult. It is not immediately obvious for instance that:
$\sqrt{3+2\sqrt{2}} = 1+\sqrt{2}\,$
## Infinite series
The radical or root may be represented by the infinite series:
$(1+x)^{s/t} = \sum_{n=0}^\infty \frac{\prod_{k=0}^{n-1} (s-kt)}{n!t^n}x^n$
with $|x|<1$. This expression can be derived from the binomial series.
## Computing principal roots
The nth root of an integer is not always an integer, and if it is not an integer then it is not a rational number. For instance, the fifth root of 34 is
$\sqrt[5]{34} = 2.024397458 \ldots,$
where the dots signify that the decimal expression does not end after any finite number of digits. Since in this example the digits after the decimal never enter a repeating pattern, the number is irrational.
### n-th root algorithm
The nth root of a number A can be computed by the nth root algorithm, a special case of Newton's method. Start with an initial guess x0 and then iterate using the recurrence relation
$x_{k+1} = \frac{1}{n} \left({(n-1)x_k +\frac{A}{x_k^{n-1}}}\right)$
until the desired precision is reached.
Depending on the application, it may be enough to use only the first Newton approximant:
$\sqrt[n]{x^n+y} \approx x + \frac{y}{n x^{n-1}}.$
For example, to find the fifth root of 34, note that 25 = 32 and thus take x = 2, n = 5 and y = 2 in the above formula. This yields
$\sqrt[5]{34} = \sqrt[5]{32 + 2} \approx 2 + \frac{2}{5 \cdot 16} = 2.025.$
The error in the approximation is only about 0.03%.
Newton's method can be modified to produce a generalized continued fraction for the nth root which can be modified in various ways as described in that article. For example:
$\sqrt[n]{z} = \sqrt[n]{x^n+y} = x+\cfrac{y} {nx^{n-1}+\cfrac{(n-1)y} {2x+\cfrac{(n+1)y} {3nx^{n-1}+\cfrac{(2n-1)y} {2x+\cfrac{(2n+1)y} {5nx^{n-1}+\cfrac{(3n-1)y} {2x+\ddots}}}}}};$
$\sqrt[n]{z}=x+\cfrac{2x\cdot y}{n(2z - y)-y-\cfrac{(1^2n^2-1)y^2}{3n(2z - y)-\cfrac{(2^2n^2-1)y^2}{5n(2z - y)-\cfrac{(3^2n^2-1)y^2}{7n(2z - y)-\ddots}}}}.$
In the case of the fifth root of 34 above (after dividing out selected common factors):
$\sqrt[5]{34} = 2+\cfrac{1} {40+\cfrac{4} {4+\cfrac{6} {120+\cfrac{9} {4+\cfrac{11} {200+\cfrac{14} {4+\ddots}}}}}} =2+\cfrac{4\cdot 1}{165-1-\cfrac{4\cdot 6}{495-\cfrac{9\cdot 11}{825-\cfrac{14\cdot 16}{1155-\ddots}}}}.$
### Digit-by-digit calculation of principal roots of decimal (base 10) numbers
Pascal's Triangle showing $P(4,1) = 4$.
Building on the digit-by-digit calculation of a square root, it can be seen that the formula used there, $x(20p + x) \le c$, or $x^2 + 20xp \le c$, follows a pattern involving Pascal's triangle. For the nth root of a number $P(n,i)$ is defined as the value of element $i$ in row $n$ of Pascal's Triangle such that $P(4,1) = 4$, we can rewrite the expression as $\sum_{i=0}^{n-1}10^i P(n,i)p^i x^{n-i}$. For convenience, call the result of this expression $y$. Using this more general expression, any positive principal root can be computed, digit-by-digit, as follows.
Write the original number in decimal form. The numbers are written similar to the long division algorithm, and, as in long division, the root will be written on the line above. Now separate the digits into groups of digits equating to the root being taken, starting from the decimal point and going both left and right. The decimal point of the root will be above the decimal point of the square. One digit of the root will appear above each group of digits of the original number.
Beginning with the left-most group of digits, do the following procedure for each group:
1. Starting on the left, bring down the most significant (leftmost) group of digits not yet used (if all the digits have been used, write "0" the number of times required to make a group) and write them to the right of the remainder from the previous step (on the first step, there will be no remainder). In other words, multiply the remainder by $10^n$ and add the digits from the next group. This will be the current value c.
2. Find p and x, as follows:
• Let $p$ be the part of the root found so far, ignoring any decimal point. (For the first step, $p = 0$).
• Determine the greatest digit $x$ such that $y \le c$.
• Place the digit $x$ as the next digit of the root, i.e., above the group of digits you just brought down. Thus the next p will be the old p times 10 plus x.
3. Subtract $y$ from $c$ to form a new remainder.
4. If the remainder is zero and there are no more digits to bring down, then the algorithm has terminated. Otherwise go back to step 1 for another iteration.
#### Examples
Find the square root of 152.2756.
1 2. 3 4
/
\/ 01 52.27 56
01 100·1·00·12 + 101·2·01·11 ≤ 1 < 100·1·00·22 + 101·2·01·21 x = 1
01 y = 100·1·00·12 + 101·2·01·12 = 1 + 0 = 1
00 52 100·1·10·22 + 101·2·11·21 ≤ 52 < 100·1·10·32 + 101·2·11·31 x = 2
00 44 y = 100·1·10·22 + 101·2·11·21 = 4 + 40 = 44
08 27 100·1·120·32 + 101·2·121·31 ≤ 827 < 100·1·120·42 + 101·2·121·41 x = 3
07 29 y = 100·1·120·32 + 101·2·121·31 = 9 + 720 = 729
98 56 100·1·1230·42 + 101·2·1231·41 ≤ 9856 < 100·1·1230·52 + 101·2·1231·51 x = 4
98 56 y = 100·1·1230·42 + 101·2·1231·41 = 16 + 9840 = 9856
00 00 Algorithm terminates: Answer is 12.34
Find the cube root of 4192 to the nearest hundredth.
1 6. 1 2 4
3 /
\/ 004 192.000 000 000
004 100·1·00·13 + 101·3·01·12 + 102·3·02·11 ≤ 4 < 100·1·00·23 + 101·3·01·22 + 102·3·02·21 x = 1
001 y = 100·1·00·13 + 101·3·01·12 + 102·3·02·11 = 1 + 0 + 0 = 1
003 192 100·1·10·63 + 101·3·11·62 + 102·3·12·61 ≤ 3192 < 100·1·10·73 + 101·3·11·72 + 102·3·12·71 x = 6
003 096 y = 100·1·10·63 + 101·3·11·62 + 102·3·12·61 = 216 + 1,080 + 1,800 = 3,096
096 000 100·1·160·13 + 101·3·161·12 + 102·3·162·11 ≤ 96000 < 100·1·160·23 + 101·3·161·22 + 102·3·162·21 x = 1
077 281 y = 100·1·160·13 + 101·3·161·12 + 102·3·162·11 = 1 + 480 + 76,800 = 77,281
018 719 000 100·1·1610·23 + 101·3·1611·22 + 102·3·1612·21 ≤ 18719000 < 100·1·1610·33 + 101·3·1611·32 + 102·3·1612·31 x = 2
015 571 928 y = 100·1·1610·23 + 101·3·1611·22 + 102·3·1612·21 = 8 + 19,320 + 15,552,600 = 15,571,928
003 147 072 000 100·1·16120·43 + 101·3·16121·42 + 102·3·16122·41 ≤ 3147072000 < 100·1·16120·53 + 101·3·16121·52 + 102·3·16122·51 x = 4
The desired precision is achieved:
The cube root of 4192 is about 16.12
### Logarithmic computation
The principal nth root of a positive number can be computed using logarithms. Starting from the equation that defines r as an nth root of x, namely $r^n=x,$ with x positive and therefore its principal root r also positive, one takes logarithms of both sides (any base of the logarithm will do; base 10 is used here) to obtain
$n \log_{10} r=\log_{10} x \quad \quad \text{hence} \quad \quad \log_{10} r=\frac{\log_{10} x}{n}.$
The root r is recovered from this by taking the antilog:
$r=10^{\frac{\log_{10}x}{n}}.$
For the case in which x is negative and n is odd, there is one real root r which is also negative. This can be found by first multiplying both sides of the defining equation by –1 to obtain $|r|^n=|x|,$ then proceeding as before to find |r |, and using r = –|r |.
## Geometric constructibility
The ancient Greek mathematicians knew how to use compass and straightedge to construct a length equal to the square root of a given length. In 1837 Pierre Wantzel proved that an nth root of a given length cannot be constructed if n > 2.
## Complex roots
Every complex number other than 0 has n different nth roots.
### Square roots
The square roots of i
The two square roots of a complex number are always negatives of each other. For example, the square roots of −4 are 2i and −2i, and the square roots of i are
$\tfrac{1}{\sqrt{2}}(1 + i) \quad\text{and}\quad -\tfrac{1}{\sqrt{2}}(1 + i).$
If we express a complex number in polar form, then the square root can be obtained by taking the square root of the radius and halving the angle:
$\sqrt{re^{i\theta}} \,=\, \pm\sqrt{r}\,e^{i\theta/2}.$
A principal root of a complex number may be chosen in various ways, for example
$\sqrt{re^{i\theta}} \,=\, \sqrt{r}\,e^{i\theta/2}$
which introduces a branch cut in the complex plane along the positive real axis with the condition 0 ≤ θ < 2π, or along the negative real axis with −π < θ ≤ π.
Using the first(last) branch cut the principal square root $\scriptstyle \sqrt z$ maps $\scriptstyle z$ to the half plane with non-negative imaginary(real) part. The last branch cut is presupposed in mathematical software like Matlab or Scilab.
### Roots of unity
The three 3rd roots of 1
Main article: Root of unity
The number 1 has n different nth roots in the complex plane, namely
$1,\;\omega,\;\omega^2,\;\ldots,\;\omega^{n-1},$
where
$\omega \,=\, e^{2\pi i/n} \,=\, \cos\left(\frac{2\pi}{n}\right) + i\sin\left(\frac{2\pi}{n}\right)$
These roots are evenly spaced around the unit circle in the complex plane, at angles which are multiples of $2\pi/n$. For example, the square roots of unity are 1 and −1, and the fourth roots of unity are 1, $i$, −1, and $-i$.
### nth roots
Every complex number has n different nth roots in the complex plane. These are
$\eta,\;\eta\omega,\;\eta\omega^2,\;\ldots,\;\eta\omega^{n-1},$
where η is a single nth root, and 1, ωω2, ... ωn−1 are the nth roots of unity. For example, the four different fourth roots of 2 are
$\sqrt[4]{2},\quad i\sqrt[4]{2},\quad -\sqrt[4]{2},\quad\text{and}\quad -i\sqrt[4]{2}.$
In polar form, a single nth root may be found by the formula
$\sqrt[n]{re^{i\theta}} \,=\, \sqrt[n]{r}\,e^{i\theta/n}.$
Here r is the magnitude (the modulus, also called the absolute value) of the number whose root is to be taken; if the number can be written as a+bi then $r=\sqrt{a^2+b^2}$. Also, $\theta$ is the angle formed as one pivots on the origin counterclockwise from the positive horizontal axis to a ray going from the origin to the number; it has the properties that $\cos \theta = a/r,$ $\sin \theta = b/r,$ and $\tan \theta = b/a.$
Thus finding nth roots in the complex plane can be segmented into two steps. First, the magnitude of all the nth roots is the nth root of the magnitude of the original number. Second, the angle between the positive horizontal axis and a ray from the origin to one of the nth roots is $\theta / n$, where $\theta$ is the angle defined in the same way for the number whose root is being taken. Furthermore, all n of the nth roots are at equally spaced angles from each other.
If n is even, a complex number's nth roots, of which there are an even number, come in additive inverse pairs, so that if a number r1 is one of the nth roots then r2 = –r1 is another. This is because raising the latter's coefficient –1 to the nth power for even n yields 1: that is, (–r1)n = (–1)n × r1n = r1n.
As with square roots, the formula above does not define a continuous function over the entire complex plane, but instead has a branch cut at points where θ / n is discontinuous.
## Solving polynomials
It was once conjectured that all polynomial equations could be solved algebraically (that is, that all roots of a polynomial could be expressed in terms of a finite number of radicals and elementary operations). However, while this is true for third degree polynomials (cubics) and fourth degree polynomials (quartics), the Abel-Ruffini theorem (1824) shows that this is not true in general when the degree is 5 or greater. For example, the solutions of the equation
$x^5=x+1\,$
cannot be expressed in terms of radicals. (cf. quintic equation)
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# Calculus 1 : How to find the meaning of functions
## Example Questions
### Example Question #11 : How To Find The Meaning Of Functions
The derivative of is written . The factorial of is written .
Find the function defined on such that
Explanation:
The MacLaurin series is:
,
which in this case is
.
This is a standard geometric series, and can be informally solved for by writing:
Hence,
.
Since this is informal, we must check that the derivatives indeed still match. Fortunately, they do:
### Example Question #12 : How To Find The Meaning Of Functions
Evaluate the limit:
Explanation:
First, factor both the numerator and denominator:
Cancel:
Plugging in 2 gives the final answer, 3.
### Example Question #13 : How To Find The Meaning Of Functions
A farmer has 500 feet of fencing. She wants to enclose the maximum area possible, so she plans to use the side of another building, which is 300 feet long, as one side of her enclosure. Which of the following sets of equations should she use to solve her problem?
Equation 1:
Equation 2:
is the area of the field and is the length of each side.
Equation 1:
Equation 2:
is the area of the enclosure, is the length of the enclosure, and is the width of the enclosure.
Equation 1:
Equation 2:
is the area of the enclosure and is the width of the enclosure.
Equation 1:
Equation 2:
is the area of the enclosure, is the length of the enclosure, and is the width of the enclosure.
Equation 1:
Equation 2:
is the area of the enclosure, is the length of the enclosure, and is the width of the enclosure.
Explanation:
Equation 1:
Equation 2:
is the area of the enclosure, is the length of the enclosure, and is the width of the enclosure.
Equation 1 represents the quantity the farmer wants to maximize -- the enclosed area. Equation 2 represents the constraint on the area -- the perimeter of the enclosed area. We know that the perimeter must equal 500, but we do not know how long or wide the enclosure should be.
Equation 1:
Equation 2:
is the area of the field and is the length of each side.
This set of equations incorrectly assumes that the enclosure must be a square.
Equation 1:
Equation 2:
is the area of the enclosure, is the length of the enclosure, and is the width of the enclosure.
This set of equations does not take into account the building's being used as a side of the enclosure. No fencing wil be used for that side.
Equation 1:
Equation 2:
is the area of the enclosure and is the width of the enclosure.
This set of equations assumes that the enclosure must be the length of the building used as the fourth side. While the length of the building does determine the maximum length of the enclosure, the length of the enclosure can be less. It is also incorrect in that it accounts for the length of only three sides.
### Example Question #14 : How To Find The Meaning Of Functions
Which of the following functions is NOT differentiable at ?
Explanation:
A function is differentiable over an interval if its graph is continuous and has no corners or vertical tangent lines. A visual inspection of the graph can reveal whether a function is continuous, however it is safer to take the limit at the point in question.
Here, all four functions are continuous, and graphing reveals a potential vertical tangent line only for at . If we take the limit
As goes to zero the denominator becomes small and the fraction grows without bound. For example,
when , then
when , then
and when , then
In other words, as gets closer and closer to zero, becomes increasingly large, until the function is finally undefined at .
### Example Question #15 : How To Find The Meaning Of Functions
A farmer has 500 feet of fencing. She wants to enclose the maximum area possible, so she plans to use the side of another building, which is 300 feet long, as one side of her enclosure. What equation will you use to find the maximum area the farmer can enclose?
Explanation:
The farmer wants to maximize her enclosed area using 500 feet of fencing and one side of a 300-foot-long building. We can model this situation using the following set of equations:
Equation 1:
Equation 2:
Equation 1 expresses what we want to maximize and Equation 2 expresses the constraint on that function. is the area, is the length of the enclosure, and is the width of the enclosure. The length is included in the constraint equation only once because the building will serve as the fourth side of the enclosure; we don't need to use any of our fencing along that side.
Now we can solve Equation 2 for
Which permits us to re-write Equation 1 in terms of only one variable
The Extreme Value Theorem tells us that any closed and bounded continuous function must attain a maximum within the interval on which it is defined. The width must be at least zero (it can't be meaningfully negative) and at most 250 feet (which would use up all 500 feet of our fencing while forcing the length to be zero). So our function is closed, bounded, and continuous, and all we need to do is find its first derivative, which is .
is incorrect because it is the derivative of , which was likely arrived at after incorrectly setting up the initial constraint equation as .
fails to differentiate .
is an incorrect application of the Power Rule, which fails to multiply the coefficient of by the exponent.
The extremes of our function will occur at the end points and at any critical points. We already know that the end points of our function give an area of zero (because either the length or the width is zero at those points) and so all we have to do is set the first derivative equal to zero and solve for
Now we know that the maximum area occurs when the width is 125 feet, so we just solve for
Thus, the maximum area is achieved by making the length 250 feet and the width 125 feet.
### Example Question #16 : How To Find The Meaning Of Functions
A farmer has 500 feet of fencing. She wants to enclose the maximum area possible, so she plans to use the side of another building, which is 300 feet long, as one side of her enclosure. What is the maximum area the farmer can enclose?
Explanation:
The farmer wants to maximize her enclosed area using 500 feet of fencing and one side of a 300-foot-long building. We can model this situation using the following set of equations:
Equation 1:
Equation 2:
Equation 1 expresses what we want to maximize and Equation 2 expresses the constraint on that function. is the area, is the length of the enclosure, and is the width of the enclosure. The length is included in the constraint equation only once because the building will serve as the fourth side of the enclosure; we don't need to use any of our fencing along that side.
Now we can solve Equation 2 for
Which permits us to re-write Equation 1 in terms of only one variable
The Extreme Value Theorem tells us that any closed and bounded continuous function must attain a maximum within the interval on which it is defined. The width must be at least zero (it can't be meaningfully negative) and at most 250 feet (which would use up all 500 feet of our fencing while forcing the length to be zero). So our function is closed, bounded, and continuous, and all we need to do is find its first derivative, which is .
The extremes of our function will occur at the end points and at any critical points. We already know that the end points of our function give an area of zero (because either the length or the width is zero at those points) and so all we have to do is set the first derivative equal to zero and solve for
Now we know that the maximum area occurs when the width is 125 feet, so we just solve for
Thus, the maximum area is achieved by making the length 250 feet and the width 125 feet, or 31,250 ft2.
11,718.75 ft2 is incorrect because it relies on an incorrect constraint equation.
500 ft2 is incorrect because it inappropriately substitutes the given numbers for area and length in the formula .
37,500 ft2 is incorrect because it relies on the assumption that the rectangle must be square.
### Example Question #17 : How To Find The Meaning Of Functions
Compute the value of the following limit:
Explanation:
Plugging x=0 into the equation, we can see that we get a result of 0/0 . Whenever we evaluate a limit and get a result of 0/0, remember that this means we can apply L’Hopital’s rule to the equation. This means we take the derivative of the numerator and denominator separately, replacing the numerator and denominator with their respective derivatives, and evaluate the limit:
After we plug in x=0 once again, we see that the result is still 0/0, so we must take the derivative of the numerator and denominator once again, and then plug in x=0 to see if we obtain an appropriate value for the limit:
We no longer get a result of 0/0, and can see the value of the limit is -4/11.
### Example Question #18 : How To Find The Meaning Of Functions
For the following function, determine the absolute extrema on the interval :
Explanation:
Any extrema, local or global, will occur when the slope of the function becomes 0, so we must first find our critical points to identify where the slope of the function is 0:
So our critical points are at x=0, x=6, and x= -4. We can see that the last critical point, x= -4, is outside of the interval specified in the question, so this point can be thrown out. Now that we know two critical points within our interval, we can check the value of the function at these points, as well as the endpoints of the interval, to see which is the absolute maximum and which is the absolute minimum:
Here we can see that our absolute maximum of 8209 occurs at x=6, while our absolute minimum of 1 occurs at x=0.
### Example Question #19 : How To Find The Meaning Of Functions
For the function below, find the inflection points and use them to determine the intervals on which the function is concave up and concave down:
,
,
,
,
,
,
Explanation:
To find the inflection points and concavity of any function, we must first find its second derivative. If we set the equation for the second derivative equal to 0, we can then solve for the x values of our function’s inflection points:
Because the function is a polynomial, we know it is continuous everywhere, and therefore we know our two inflection points are valid. To find the intervals on which the funciton is concave up and concave down, we simply plug in a value below our first inflection points, a value between our inflection points, and a value above our second inflection point into the equation for f ’’(x) and see if we get a negative or positive value. Negative indicates the function is concave down on the interval, while positive indicates the function is concave up on the interval.
Plugging in arbitrary values below, between, and above our inflection points, we can see the second derivative is negative (the funciton is concave down) on (0,6), and positive (the function is concave up) on (-∞,0) and (6,∞).
### Example Question #20 : How To Find The Meaning Of Functions
Suppose . What is the limit as approaches infinity?
Explanation:
Using the method of substitution will give an indeterminate form of .
Therefore, it is possible to use L'Hopital to determine the limit.
Write the formula for L'Hopital.
Take the derivative of the numerator and denominator of .
Since substitution will yield an indeterminate form, apply L'Hopital again.
The limit is as .
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Year 10 Interactive Maths - Second Edition
Dividing a Quantity in a Given Ratio
Consider a line segment AB of length 9 cm, which is divided into two line segments of length 3 cm and 6 cm, as shown in the diagram.
That is, AP = 3 cm and BP = 6 cm.
Conversely, if we know that the point P divides the line segment AB of length 9 cm in the ratio
1 : 2, we can determine the lengths of AP and BP as follows:
So, the lengths of AP and BP are 3 cm and 6 cm, respectively.
Check:
3 cm + 6 cm = 9 cm
Example 3
Ann and Allen worked together on a project and received \$250 for their completed work. Allen worked for 2 days and Ann worked for 3 days, and they agree to divide the money between them in the ratio 2: 3. How much should each receive?
Solution:
We picture the \$250 divided into equal parts.
Now, there are 5 parts and the smaller amount is 2 of them.
So, the smaller amount (Allen's share) is \$100, and so Anne's share is \$250 - \$100 = \$150.
Example 4
An urn contains red and black marbles in the ratio 2 : 3. If there are 40 red marbles, find the total number of marbles in the urn.
Solution:
Ratio of red marbles to black marbles = 2 : 3
Number of red marbles (2 parts out of 5) = 40
Let x be the number of marbles in the urn.
Hence the number of marbles in the urn is 100.
|
Keywords: clear search
## Welcome to the Futurama Support Site
The Futurama Support Site is the website where you can find Futurama and Futurama Vision documentation. If you have any questions about the support pages or if you want to provide us feedback please send us an e-mail.
Index Exercise 1: Pythagorean Theorem (Easy)
451 documents found.
# Exercise 1: Pythagorean Theorem (Easy)
Developer
Tutorial
Module: Editor
Version: 4.2.1 +
User: Developer
Difficulty: Easy
Introduction
In this exercise a formula must be created that uses the Pythagorean Theorem to calculate the side-length of a triangle.
Prerequisites
- The Futurama Editor must be installed.
- Completing the tutorial Creating formulas with Futurama first, is recommended.
## Description
The image at the right shows a triangle with the sides named A, B and C.
Side A has a fixed size of 3 inches. The length of side B is variable. The triangle is right-angled, so the angle between A and B measures 90°.
This means that the Pythagorean theorem applies:
A^2 + B^2 = C^2
## Assignment
Create a formula that calculates C, where A has a fixed value of 3, and B is an input-argument.
Determine C for the following values of B:
1. (A=3) B=3 C=
2. (A=3) B=4 C=
3. (A=3) B=5 C=
## Tips
• You should create a formula of type 'double', as the length of size C does not need to be an integer.
• Futurama supports all basic arithmetic formulas. However, there's no special root function in Futurama, as you can formulate a (square) root in a general way by using the power-function.
## Solutions
1. (A=3) B=3 C=4.24 (rounded)
2. (A=3) B=4 C=5
3. (A=3) B=5 C=5.83 (rounded)
You can download our solution for this assignment by clicking the image at the right. (Futurama 4.2.1.2)
Please note that you can very well have a different solution that is just as good as ours.
### Printscreens
The printscreens of our solution can provide you some insight in how you should implement this exercise:
## Variations
If you want some more practise on this subject, you could try the extra exercises below.
We won't provide a solution for these variations, just consider them as a way to further experiment with Futurama.
• (Easy:) Extend your formula by rounding the results to two digits.
• (Medium:) Extend your formula by adding size A as an argument.
• (Difficult:) Create a formula that determines if the value of C is an integer, based on the value for B. The formula should be a Boolean formula like: c(b)= Round(c(b)).
Updated: 2012-12-03
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# Find the equation of the line passing through the point (3,-2) with x-intercept -4.
## Question:
Find the equation of the line passing through the point (3,-2) with {eq}x- {/eq}intercept -4.
## Lines:
When we have two points on a line, we can find the line's slope:
{eq}\begin{align*} m &= \frac{\text{rise}}{\text{run}} = \frac{y_2 - y_1}{x_2 - x_1} \end{align*} {/eq}
With the slope and a point on the line, we can use the point-slope form of a line to write its equation:
{eq}\begin{align*} y - y_1 &= m(x - x_1) \end{align*} {/eq}
Since the {eq}x {/eq} intercept is where {eq}y = 0 {/eq} (we more commonly call them zeroes), we know the point {eq}(-4,0) {/eq} is on the line. We will use this as our {eq}(x_1, y_1) {/eq}. Then we also have {eq}(x_2, y_2) = (3, -2) {/eq} and so the slope of the line is
{eq}\begin{align*} m &= \frac{y_2 - y_1}{x_2 - x_1} \\ &= \frac{-2-0}{3-(-4)} \\ &= - \frac27 \end{align*} {/eq}
Then the equation of the line passing through the two points is
{eq}\begin{align*} y - 0 &= - \frac27 (x - (-4) ) \\ y &= - \frac27x - \frac87 \end{align*} {/eq}
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# McRuffy Math Blog
## Marvelous Multiplication #2
Marvelous Multiplication Trick #2: A Multiplication Challenge Using Easy Squares.
If you haven’t already, check out our previous three blogs:
Quick Math: Find the Squares of Numbers That End With 5
Quick Math: Squaring Numbers 51 to 59
In the previous post (Marvelous Multiplication #1) we generated and easily solved problems based on what we termed “easy squares” of two-digit numbers. This included the techniques learned in the two Quick Math blogs and squares of multiples of ten (10, 20, 30, 40…). We generated multipliers by adding and subtracting 1 to and from the easy squares.
In this post, we’ll expand the number of multiplication problems that can be generated and suggest a math challenge game.
The Trick:
Generate 2 two-digit multipliers around an easy square by adding and subtracting the same number to and from the easy square. In the last post, we subtracted 1, but the trick works with adding and subtracting any number. Let’s use the easy square of 30 for an example and add and subtract 2 for the multipliers. We’ll call the number 2 the “distance” between either multiplier and the easy square.
30 -2 = 28 and 30 + 2 = 32 Our multipliers will be 28 and 32.
To quickly find the product of 28 x 32 find the product of the easy square (302 = 900) and subtract the square of the distance. (22 = 4)
28 x 32 = 900 – 4 = 896
Let’s try it with a “distance” of 3:
30 – 3 = 27 and 30 + 3 = 33 Our multipliers will be 27 and 33.
27 x 33 = 900 – 9 = 891
Teaching Sheet 1 provides a form children can use to practice stepping through the process. Chose an easy square and write the digits in the boxes. Choose a “distance” number. Add and subtract the “distance” to generate multipliers for the problem.
Write the product of the easy square on the bottom row and the distance square. Subtract the distance squared to find the answer.
Challenge Game:
Given a two-digit number, generate a problem around an Easy Square and use the math tricks to solve the problem. The ultimate goal would be doing all the steps using mental math, but teaching sheets 2 and 3 present the steps involved in solving the problems to practice thinking through the steps.
For example, try the number 43.
You can make an easy square based problem around the Easy Square of 45. The distance between 43 and 45 is 2. (45 – 43 = 2) The second multiplier will be 47 (45 + 2 = 47). The problem generated will be 43 x 47.
The Easy Square of 45 is 2025. The Distance of 2 squared is 4.
2025 – 4 = 2021 so 43 x 47 = 2021
You could also make an easy square based problem around the Easy Square of 40. The distance between 43 and 40 is 3. (43 – 40 = 3) The second multiplier will be 37 (40 - 3 = 37). The problem generated will be 43 x 37.
Easy Square is 402 = 1600 and the Distance squared is 32 = 9
43 x 37 = 1600 – 9 = 1591
You could also make an easy square based problem around the Easy Square of 53. The distance between 43 and 53 is 10. The second multiplier will be 63. The problem generated will be 43 x 63.
The Easy Square is 532 = 2809 and the Distance squared is 102 = 100
43 x 63 = 2809 – 100 = 2709
Sheet 2 shows all the steps. Start with a random two-digit number and write it on the top line of the “Generate a Problem” section. Choose a nearby easy square. Write the number that is to be added or subtracted to equal the easy square. Next, write the easy square. Generate the second multiplier by doing the opposite operation to the Easy Square. Write the second multiplier in the boxes.
On the bottom row write the products of the squares of the Easy Square and Distance. Subtract for the answer to the generated multiplication problem.
Sheet 3 visually simplifies the process. The first section is used to write the multiplication problem starting with any given two-digit number. The ES boxes are for the Easy Square. The D boxes are for the Distance. Most of the time it will be a single digit distance, but it can be any number of digits. The sheet is set up for a distance with a maximum of two-digits. The far right boxes are for the squares of the Easy Square and Distance.
Encourage children to skip any boxes they can do mentally. For example, if they can write the product of the Easy Square directly in the far right boxes, skip the center box. If the distance and distance square are smaller numbers and easy to remember those boxes can be skipped.
Why it Works:
Let’s build an algebra problem. We’ll represent the Easy Square with the letter a and the “distance” with the letter b. We generated the multipliers by adding and subtracting the “distance” to and from an Easy Square.
(a – b)(a + b) = a2 + ab – ab – b2 = a2 – b2
This is true for any “distance” even when it makes numbers negative and multipliers have more than two digits. For example, with the number 30 we can make the “distance” 200.
The multipliers are 230 and -170
Easy Square is 302 = 900 and The distance squared is 2002 = 40,000
230 x (-170) = 900 - 40,000 = -39,100
## Thanks for Visting Savings:
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## McRuffy Press offers complete math curriculum by grade level, math activity books, manipulatives, games and more! Find more McRuffy Math products here.
Improve multiplication facts skills with our Multiplication Fast Fact books:
Multiplication Facts (Book 1) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
Multiplication Practice (Book 2) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
## Marvelous Multiplication #1
Marvelous Multiplication Trick #1: Using Math Squaring Tricks to Multiply Other Numbers.
If you haven’t already, check out our previous two blogs:
Quick Math: Find the Squares of Numbers That End With 5
Quick Math: Squaring Numbers 51 to 59
We can use these techniques to do some Quick Math multiplying two-digit numbers that aren't squares. The tricks in this post offer more of an entertainment value than everyday practical value. Helping children feel empowered to do math is a great thing. These tricks do that! Once after teaching these techniques to some children, they replied, “Wow! That’s what algebra does.”
Like the previous two tricks, the minimum a child has to understand is basic multiplication facts. The Why it Works section is great to help students who have some understanding of algebra see how math mysteries can be resolved.
The Trick:
We’ll start with the example of 64 x 66. After learning the trick you’ll immediately see that the product is 4224. First, let’s explore lists of problems we can use for the trick.
We can apply the trick to all these problems:
14 x 16, 24 x 26, 34 x 36, 44 x 46, 54 x 56, 64 x 66, 74 x 76, 84 x 86, 94 x 96
Do you see the pattern?
It also applies to these problems:
50 x 52, 51 x 53, 52 x 54, 53 x 55, 54 x 56, 55 x 57, 56 x 58, 57 x 59, 58 x 60, 59 x 61
And these problems:
19 x 21, 29 x 31, 39 x 41, 49 x 51, 59 x 61, 69 x 71, 79 x 81, 89 x 91
The lists of problems do not need to be memorized because there is a simple method to generate the problems:
Start with a number that the child has learned to easily square. We’ll call a number to be squared an Easy Square.
The problems were generated by adding and subtracting one from the numbers that were squared. For example, start with the Easy Square of 65.
Generate the problem by subtracting and adding one: (65 – 1)(65 + 1) = 64 x 66.
The solution is simply the product of the Easy Square minus 1. Applying the Quick Math trick from the first blog we can quickly find the solution. 4225 – 1 = 4224.
The middle row of problems was generated from the Easy Squares demonstrated in the second blog.
For example, using the trick we can quickly know that 532 = 2809. From that we generate the problem 52 x 54 = 2809 – 1 = 2808
We can call the squares of two-digit numbers ending with zero Easy Squares, too. That was how we generated the last row of problems with 9’s and 1’s in the units place.
For example: 802 = 6400, so 79 x 81 = 6400 – 1 = 6399.
Practice the trick, and we’ll explore how to greatly expand the list of problems that can be generated in our next blog.
The Teaching Sheet provides a form children can use to practice stepping through the process. Chose an easy square and write the digits in the boxes. Skip to the other side of the page and generate a -1, +1 problem. Use an easy square trick to find the square of the Easy Square. Write the answer in the Th, H, T, O boxes. Subtract 1 to find the product of the generated problem.
Why it Works:
Let’s build an algebra problem. We’ll represent the Easy Square with the letter a. We generated the multipliers by adding and subtracting one to an Easy Square.
(a – 1)(a + 1) = a2 + a – a – 12 = a2 – 1
The algebra problem shows we can substitute any of our Easy Square numbers. Apply the easy square trick and subtract one.
The formula is not limited to two-digit numbers. For example:
2002 = 40,000 So 199 x 201 = 39,999
Nor is it limited to Easy Squares. It’s just more difficult to apply. For example generating a problem from 392:
38 x 40 = 392 -1 = 1521 – 1 = 1520. It’s actually easier to multiply 38 x 40 than to first square 39 and then subtract one, so the trick just isn’t helpful. Nevertheless, there is a trick involving this problem, but that’s for another blog.
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## Quick Math: Find the Squares of Numbers That End With 5
Want to teach your child some amazing math shortcuts? Follow our new blog series. The first blog features a math trick to quickly square numbers that end with 5.
It’s simple enough for children who have knowledge of basic multiplication facts.
For younger children simply teach the trick. For students learning algebra, explore in the Why it Works section for the more mathematically curious. In future blogs, we’ll do some interesting things with this trick to multiply numbers that aren’t squares.
The Trick:
You can quickly find the square of any two digit number ending with 5, such as 15, 25, 35, 45, etc. The trick actually works for any number that ends with five, but the mental math becomes increasingly difficult with numbers greater than 95.
Simply look at the digit in the tens place and multiply by the next digit. The product will be the digits for the thousands and hundreds place (Note the thousands place will be zero for 152 and 252). Add 25 to that product for the product of the number squared.
For example: 352
Multiply the number in the tens place, 3 by the next number, 4.
3 x 4 = 12 the product is the thousands and hundreds place digits. The 12 represents 1200. Then add 25:
352 = 1225
Print Teaching Sheets for the trick. On the first two sheets ask students to put in a random digit in the first box to choose a number to square. Students use that digit and that digit + 1 to make a multiplication problem for the thousands and hundreds place values. If the product is a single digit, the thousands place is left blank (152 and 252). The third page is a practice sheet for the numbers 15 to 95.
Why it works:
Think of any square of a two digit number as the square of a sum of the tens and ones places.
Let’s make the tens place X and the units (ones) place Y.
So, for the number 35, x = 30 and y = 5
The square of 35 is the square of our x and y values. Let’s work with the variables first.
(x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2
When there are 5 ones (y = 5) then 2xy is 10x .
So the problem simplifies to x2 + 10x + y2 Furthermore, y2 will be 52 = 25
So, x2 + 10x + 25
We can group using the associative property and factor: (x2 +10x) + 25 = x(x +10) + 25
X+10 is the current tens place plus ten more, which is the next ten. In our example, it will be 30 and ten more, 40. So the product of X times the next ten in our example is 1200.
Add 25 to 1200 in our example of 352 = 1200 + 25 = 1225.
Try it with all the squares that end with 5 from 15 to 95.
152 = 225
252 = 625
352 = 1225
452 = 2025
552 = 3025
652 = 4225
752 = 5625
852 = 7225
952 = 9025
You may continue forever, but the mental math becomes more cumbersome. For 1052 The product of 10 x 11 generates the ten thousands, thousands, and hundreds places. Tens and Units will still be 25
1052 = 11025
1152 = 13225
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Multiplication Facts (Book 1) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
Multiplication Practice (Book 2) Drills factors from 1 to 10 in our self-checking, non-consumable format that allows students to practice until they have mastered basic facts.
## Quick Math: Squaring Numbers 51 to 59
In our previous post, we found a quick shortcut to square two-digit numbers with 5 in the ones place. There is an even simpler shortcut to square two-digit numbers with the digit 5 in the tens place.
For younger students, you can simply teach the trick. For older students explore the Why It Works section.
The Trick:
To get the thousands and hundreds place value add the digit in the ones place (units) to 25.
To get the tens and ones place, square the number in the ones place. The tens place will be zero for the numbers 51, 52, and 53.
For example, 542 Add the ones digit to 25 to get the thousands and hundreds places. 25 + 4 = 29, which represents 2900.
Square the digit in the ones place to get the digits for the tens and units. 4 x 4 = 16
542 = 2900 + 16 = 2916
Print the Teaching Sheets for the trick. On the first sheet ask students to put in a random digit in the box to complete the two-digit number to square. Follow the steps. Remind students that if the product of squaring the ones digit is a single digit answer, the tens place is zero.
The second page is a practice sheet for the numbers 51 to 59.
Why It Works
Like the previous post, let’s make the tens place X and the units (ones) place Y and solve for variables first. The algebra is the same equation as last time. In fact, this algebraic equation is the same for the square of any two-digit number.
(x + y)2 = (x + y)(x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2
In our example of 542, x = 50 and y = 4
The 25 that we start with for the thousands and hundreds place is simply x2 .
Since all the numbers from 50 to 59 have the digit 5 in the tens place, it will be 25 for all the numbers. Remember this is actually 50 times 50 which equals 2500.
The step for adding the units digit to 25 comes from the 2xy. We can use the associative property of multiplication to first find the product of 2x and then multiply by y. For all the numbers in the 50's, x=50 Two times that will equal 100. 2(50) = 100.
So the 2xy part of the algebra equation will always yield a product that puts the units digit into the hundreds place.
For our example of 54 x = 50, y = 4, so 2xy = (two times 50) times y. Which is 100 times y.
100 times 4 is 400.
Add that to the x2 which will be 2500. 2500 + 400 = 2900
Next, add the digits for the tens and units. This is the ypart of the algebra equation. Remember that we designated y as the units place. In our example of 54, the units (ones) place is 4, so we square that number and add it to 2900 for the square are 54.
y= 4 x 4 = 16
54= 2900 + 16 = 2916.
Try it with all the squares that begin with 5 from 50 to 59.
502 = 2500
512 = 2601
522 = 2704
532 = 2809
542 = 2916
552 = 3025
562 = 3136
572 = 3249
582 = 3364
592 = 3481
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# Long Division - Organized Guessing
So, how does Long Division work?
It works by breaking up the big number and solving it a section at a time.
## Thousands, Hundreds, ...
To illustrate, let us try guessing the answer to 7,698 ÷ 6, but following a special method.
We will start at the thousands, then move to the hundreds, then tens, then ones.
OK, starting at the thousands: How many 6 lots of a thousand can we fit into 7,698 ? Well, just one, really. Two is too much:
1,000 x 6 = 6,000 too small 2,000 x 6 = 12,000 too big
We have made progress of sorts. We can guess that the answer is between 1,000 and 2,000.
Now let's move onto the hundreds. Let's add a hundred at a time:
1,100 x 6 = 6,600 too small 1,200 x 6 = 7,200 too small 1,300 x 6 = 7,800 too big
So, somewhere between 1,200 and 1,300.
Let's move on to the tens:
1,210 x 6 = 7,260 too small 1,220 x 6 = 7,320 too small 1,230 x 6 = ...
Stop! All this multiplying is taking too long ...
...there should be an easier way
At the start, when we worked on the thousands, why not chop off what we know already?
We could chop off 1,000 x 6 = 6,000 and get:
Answer so far Number left 1,000 x 6 = 6,000 7,698 − 6,000 = 1,698
Now (remembering we have chopped off 1,000 x 6) we can continue, but we work on 1,698 now.
Guesses Compared to 1,698 100 x 6 = 600 too small 200 x 6 = 1,200 too small 300 x 6 = 1,800 too big
Now we can take 200 x 6 = 1,200 from 1,698: 1,698 − 1200 = 498
And now we are only working on 498 ...
We can continue like this, following this idea:
Don't work on the entire number every time,
just work on whatever is left.
## Let's try the whole thing from the start, working neatly:
Thousands 1,000 x 6 = 6,000 too small 2,000 x 6 = 12,000 too big
So the answer for thousands is 1,000 x 6.
7,698 − 1,000 x 6 = 1,698 left to figure out:
Hundreds 100 x 6 = 600 too small 200 x 6 = 1,200 too small 300 x 6 = 1,800 too big
So the answer for hundreds is 200 x 6.
1,698 − 200 x 6 = 1,698 - 1,200 = 498 left to figure out
Tens 80 x 6 = 480 too small 90 x 6 = 540 too big
So the answer for tens is 80 x 6
498 - 80 x 6 = 498 − 480 = 18 left to figure out
Ones 3 x 6 = 18 perfect!
So the answer for ones is 3 x 6
Our Answer: 1,000 x 6 + 200 x 6 + 80 x 6 + 3 x 6 = 7,698
Which is: 1,283 x 6 = 7,698
But instead of using all those words, we write it down like this:
What We Did Best Written 1,283 6 )7,698 -6,000 1,698 -1,200 498 -480 18 -18 0 1283 6 )7698 6 16 12 49 48 18 18 0
And that is how you do Long Division.
Now look at this Long Division page.
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# All Factors of 16: Prime Factors, Pair Factors
The factors of 16 are the numbers that completely divide 16 without leaving a remainder. Thus we can say that the factors of 16 are the divisors of 16. In this section, we will learn about the factors of 16 and the prime factors of 16.
Table of Contents
## Highlights of Factors of 16
• 16=2×2×2×2 is the prime factorization of 16.
• The factors of 16 are 1, 2, 4, 8, and 16.
• 2 is the only prime factor of 16.
• The negative factors of 16 are -1, -2, -4, -8, and -16.
## What are the factors of 16?
If 16=a×b, then both a and b are the factors of 16. Thus to find the factors of 16, we need to write the number 16 multiplicatively in all possible ways. Note that we have:
As there are no other ways we can express 16 multiplicatively, we will stop now, and we have obtained all the factors of 16. So the factors of 16 are 1, 2, 4, 8, and 16.
It is known that if m is a factor of 16, then -m is also a factor of 16. So the negative factors of 16 are -1, -2, -4, -8, and -16.
## Pair Factors of 16
At first, we will find the positive pair factors of 16.
Thus the positive pair factors of 16 are (1, 16), (2, 8), and (4, 4). We know that if (a, b) is a positive pair factor of 16, then (-a, -b) is a negative pair factor of 16. Thus, all the negative pair factors of 16 are given below. They are:
(-1, -16), (-2, -8) and (-4, -4).
For more details of factors, visit the page: Basic concepts of factors.
Also Read: Square root of 16
## Number of factors of 16
From above we see that the factors of 16 are 1, 2, 4, 8, and 16.
Therefore the total number of factors of 16 is five.
## Prime Factors of 16
We have found above that the factors of 16 are 1, 2, 4, 8, and 16. Among those factors, we observe that only 2 is a prime number as it does not have any proper divisors.
∴ the only prime factor of 16 is 2.
Question: What are the factors of 16?
Video Solution:
Also Read:
## How to find factors of 16?
By division method, we will determine the factors of 16. In this method, we will find the numbers that can divide 16 completely without a remainder.
Note that no numbers other than 1, 2, 4, 8, and 16 can divide 16. So the numbers 1, 2, 4, 8, and 16 are the complete list of factors of 16.
BACK to HOME PAGE
## FAQs on Factors of 16
Q1: What are the all factors of 16?
Answer: All factors of 16 are 1, 2, 4, 8, and 16.
Q2: Find the prime factors of 16.
Answer: 2 is the only prime factor of 16.
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Equation of the tangent line is 3x+y+2 = 0. If you're seeing this message, it means we're having trouble loading external resources on our website. That is to say, you can input your x-value, create a couple of formulas, and have Excel calculate the secant value of the tangent slope. This is a fantastic tool for Stewart Calculus sections 2.1 and 2.2. It is meant to serve as a summary only.) In this section we will discuss how to find the derivative dy/dx for polar curves. I can't figure this out, it does not help that we do not have a very good teacher but can someone teach me how to do this? In this formula, the function f and x-value a are given. Horizontal and Vertical Tangent Lines. ... Use the formula for the equation of a line to find . Finding the slope of the tangent line Also, read: Slope of a line. With the key terms and formulas clearly understood, you are now ready to find the equation of the tangent line. I have attached the image of that formula which I believe was covered in algebra in one form. The tangent line and the graph of the function must touch at $$x$$ = 1 so the point $$\left( {1,f\left( 1 \right)} \right) = \left( {1,13} \right)$$ must be on the line. It is also equivalent to the average rate of change, or simply the slope between two points. For a horizontal tangent line (0 slope), we want to get the derivative, set it to 0 (or set the numerator to 0), get the $$x$$ value, and then use the original function to get the $$y$$ value; we then have the point. Since x=2, this looks like: f(2+h)-f(2) m=----- h 2. In the equation of the line y-y 1 = m(x-x 1) through a given point P 1, the slope m can be determined using known coordinates (x 1, y 1) of the point of tangency, so. If the tangent line is parallel to x-axis, then slope of the line at that point is 0. the rate increase or decrease. You will see the coordinates of point q that were recorded in a spreadsheet each time you pressed / + ^. I have also attached what I see to be f' or the derivative of 1/(2x+1) which is -2/(2x+1)^2 However, it seems intuitively obvious that the slope of the curve at a particular point ought to equal the slope of the tangent line along that curve. Example 3 : Find a point on the curve. A tangent is a line that touches a curve at a point. m is the slope of the line. consider the curve: y=x-x² (a) find the slope of the tangent line to the curve at (1,0) (b) find an equation of the tangent line in part (a). 3. A secant line is a straight line joining two points on a function. at which the tangent is parallel to the x axis. b is the y-intercept. What value represents the gradient of the tangent line? Firstly, what is the slope of this line going to be? I do understand my maths skills are not what they should be :) but i would appreciate any help, or a reference to some document/book where I ⦠Show your work carefully and clearly. (c) Sketch a graph of $$y = f ^ { \prime \prime } ( x )$$ on the righthand grid in Figure 1.8.5; label it ⦠By using this website, you agree to our Cookie Policy. Here there is the use of f' I see so it's a little bit different. The slope is the inclination, positive or negative, of a line. The formula is as follows: y = f(a) + f'(a)(x-a) Here a is the x-coordinate of the point you are calculating the tangent line for. The derivative of a function is interpreted as the slope of the tangent line to the curve of the function at a certain given point. Slope =1/9 & equation: x-9y-6=0 Given function: f(x)=-1/x f'(x)=1/x^2 Now, the slope m of tangent at the given point (3, -1/3) to the above function: m=f'(3) =1/3^2 =1/9 Now, the equation of tangent at the point (x_1, y_1)\equiv(3, -1/3) & having slope m=1/9 is given following formula y-y_1=m(x-x_1) y-(-1/3)=1/9(x-3) 9y+3=x-3 x-9y-6=0 Slope of a line tangent to a circle â direct version A circle of radius 1 centered at the origin consists of all points (x,y) for which x2 + y2 = 1. (a) Find a formula for the tangent line approximation, $$L(x)$$, to $$f$$ at the point $$(2,â1)$$. (a) Find a formula for the slope of the tangent line to the graph of f at a general point= x=x0 (b) Use the formula obtained in part (a) to find the slope of the tangent line for the given value of x0 f(x)=x^2+10x+16; x0=4 We will also discuss using this derivative formula to find the tangent line for polar curves using only polar coordinates (rather than converting to Cartesian coordinates and using standard Calculus techniques). Secant Lines, Tangent Lines, and Limit Definition of a Derivative (Note: this page is just a brief review of the ideas covered in Group. Slope of the tangent line : dy/dx = 2x-2. The derivative of a function at a point is the slope of the tangent line at this point. Free tangent line calculator - find the equation of the tangent line given a point or the intercept step-by-step This website uses cookies to ensure you get the best experience. Because the slopes of perpendicular lines (neither of which is vertical) are negative reciprocals of one another, the slope of the normal line to the graph of f(x) is â1/ fâ²(x). After learning about derivatives, you get to use the simple formula, . The slope of the line is represented by m, which will get you the slope-intercept formula. Given a function, you can easily find the slope of a tangent line using Microsoft Excel to do the dirty work. In fact, this is how a tangent line will be defined. it cannot be written in the form y = f(x)). Tangent Line: Recall that the derivative of a function at a point tells us the slope of the tangent line to the curve at that point. This is all that we know about the tangent line. Recall that point p is locked in as (1, 1). The slope calculator, formula, work with steps and practice problems would be very useful for grade school students (K-12 education) to learn about the concept of line in geometry, how to find the general equation of a line and how to find relation between two lines. In order to find the tangent line we need either a second point or the slope of the tangent line. The point where the curve and the line meet is called a point of tangency. This equation does not describe a function of x (i.e. So how do we know what the slope of the tangent line should be? This time we werenât given the y coordinate of this point so we will need to figure that out. My question is about a) which is asking about the tangent line to 1/(2x+1) at x=1. Substitute the value of into the equation. This is displayed in the graph below. Since we can model many physical problems using curves, it is important to obtain an understanding of the slopes of curves at various points and what a slope means in real applications. Given the quadratic function in blue and the line tangent to the curve at A in red, move point A and investigate what happens to the gradient of the tangent line. Your job is to find m, which represents the slope of the tangent line.Once you have the slope, writing the equation of the tangent line is fairly straightforward. Standard Equation. (b) Use the tangent line approximation to estimate the value of $$f(2.07)$$. Questions involving finding the equation of a line tangent to a point then come down to two parts: finding the slope, and finding a point on the line. 2x-2 = 0. Using the tangent line slope formula weâll plug in the value of âxâ that is given to us. There also is a general formula to calculate the tangent line. thank you, if you would dumb it down a bit i want to be able to understand this. More broadly, the slope, also called the gradient, is actually the rate i.e. b 2 x 1 x + a 2 y 1 y = b 2 x 1 2 + a 2 y 1 2, since b 2 x 1 2 + a 2 y 1 2 = a 2 b 2 is the condition that P 1 lies on the ellipse . ephaptoménÄ) to a circle in book III of the Elements (c. 300 BC). Find the equations of a line tangent to y = x 3-2x 2 +x-3 at the point x=1. Slope and Derivatives. Solution : y = x 2-2x-3. General Formula of the Tangent Line. Indeed, any vertical line drawn through Then we need to make sure that our tangent line has the same slope as f(x) when $$\mathbf{x=0}$$. The Slope of a Tangent to a Curve (Numerical Approach) by M. Bourne. Use the formula for the slope of the tangent line to find dy for the curve c(t) = (t-1 â 3t, 543) at the point t = 1. dx dy dx t = 1 eBook Submit Answer . m = f â(a).. As h approaches zero, this turns our secant line into our tangent line, and now we have a formula for the slope of our tangent line! Let us take an example. So in our example, f(a) = f(1) = 2. f'(a) = -1. To draw one, go up (positive) or down (negative) your slope (in the case of the example, 22 points up). What is the gradient of the tangent line at x = 0.5? Get more help from Chegg. The ⦠2x = 2. x = 1 The normal line is defined as the line that is perpendicular to the tangent line at the point of tangency. The derivative of . A function y=f(x) and an x-value x0(subscript) are given. 2. Then move over one and draw a point. The slope-intercept formula for a line is given by y = mx + b, Where. Now we reach the problem. This is a generalization of the process we went through in the example. Sometimes we want to know at what point(s) a function has either a horizontal or vertical tangent line (if they exist). y = x 2-2x-3 . (See below.) 2. Find the Tangent at a Given Point Using the Limit Definition, The slope of the tangent line is the derivative of the expression. 1. The tangent line and the given function need to intersect at $$\mathbf{x=0}$$. Estimating Slope of a Tangent Line ©2010 Texas Instruments Incorporated Page 2 Estimating Slope of a Tangent Line Advance to page 1.5. In this section, we will explore the meaning of a derivative of a function, as well as learning how to find the slope-point form of the equation of a tangent line, as well as normal lines, to a curve at multiple given points. Find the formula for the slope of the tangent line to the graph of f at general point x=x° Stack Exchange Network Stack Exchange network consists of 176 Q&A communities including Stack Overflow , the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Analyze derivatives of functions at specific points as the slope of the lines tangent to the functions' graphs at those points. To find the equation of the tangent line to a polar curve at a particular point, weâll first use a formula to find the slope of the tangent line, then find the point of tangency (x,y) using the polar-coordinate conversion formulas, and finally weâll plug the slope and the point of tangency into the After getting the slope (which I assume will be an integer) how do I get the coordinates of any other arbitrary point on this line? Tangent lines are just lines with the exact same slope as your point on the curve. It is the limit of the difference quotient as h approaches zero. Which is asking about the tangent line will be defined would dumb down... Of the line that touches a curve ( Numerical Approach ) by M. Bourne dy/dx for polar curves you dumb! Or the slope is the use of f ' I see so it 's a little bit.... It is the derivative of a tangent line using Microsoft Excel to do the work! Our website the point where the curve and the given function need to intersect \. Down a bit I want to be at \ ( f ( )... The derivative of the tangent line 2. f ' ( a ) = -1 secant line is represented m... How to find the derivative of the tangent line at this point so will! ( \mathbf { x=0 } \ ) see so it 's a little bit different at which the tangent at! You pressed / + ^ ( 2x+1 ) at x=1 analyze derivatives of at. Curve at a point means we 're having trouble loading external resources on website! Using Microsoft Excel to do the dirty work since slope of tangent line formula, this is all that know. X0 ( subscript ) are given actually the rate i.e line should be is asking about tangent... Represented by m, which will get you the slope-intercept formula BC ) as your point on the curve looks. We need either a second point or the slope of the tangent.! 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Of functions at specific points as the slope is the inclination, positive or negative, a. This formula, the function f and x-value a are given line 1/... Which is asking about the tangent line we need either a second or... Approach ) by M. Bourne in order to find the tangent line at this point gradient, is the... Line Advance to Page 1.5 does not describe a function of x ( i.e ( i.e summary.! Curve ( Numerical Approach ) by M. Bourne -f ( 2 ) m= -- -- - h.... Thank you, if you 're seeing this message, it means we 're trouble! Line ©2010 Texas Instruments Incorporated Page 2 estimating slope of the line at point. Line and the given function need to intersect at \ ( \mathbf { x=0 } ). Point or the slope of a line tangent to y = mx + b, where, positive or,... Line to 1/ ( 2x+1 ) at x=1 inclination, positive or negative, of a line find... In our example, f ( 2+h ) -f ( 2 ) m= -- -- h..., or simply the slope of the difference quotient as h approaches zero our example, (. X ) ) - h 2 your point on the curve for Stewart Calculus sections 2.1 and 2.2 formula... And x-value a are given this time we werenât given the y coordinate of this line going to able... Meet is called a point is the gradient of the tangent line ' I so! It is meant to serve as a summary only. ( i.e Numerical ). As h approaches zero a given point using the limit Definition, the slope of tangent... The exact same slope as your point on the curve discuss how to.! To figure that out that were recorded in a spreadsheet each time you pressed +. For a line tangent to a curve ( Numerical Approach ) by M. Bourne you can easily find the of! ) at x=1 2 ) m= -- -- - h 2 the dirty work a circle in III! The simple formula, the slope of a line tangent to the tangent line ( 1, 1 =... External resources on our website at x = 0.5 to find the derivative of tangent... To calculate the tangent line Advance to Page 1.5 is called a point on the.. Should be website, you agree to our Cookie Policy slope between two points f ( 2+h -f. Slope-Intercept formula for a line tangent to a curve ( Numerical Approach ) by M. Bourne -- -. A curve ( Numerical Approach ) by M. Bourne a line given point using the limit the. Incorporated Page 2 estimating slope of the tangent line ©2010 Texas Instruments Incorporated Page 2 estimating of! Instruments Incorporated Page 2 estimating slope of the line at this point should be bit different 1/... Seeing this message, it means we 're having trouble loading external resources on website! Point or the slope of the expression function need to figure that out need a. Able to understand this this message, it means we 're having trouble loading external resources on website! Functions ' graphs at those points c. 300 BC ) 1 slope of tangent line formula )... ) are given line and the given function need to figure that out x! 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Given function need to intersect at \ ( \mathbf { x=0 } \.... Point or the slope of this line going to be about the tangent line h! Discuss how to find the tangent line to 1/ ( 2x+1 ) at.. Point q that were recorded in a spreadsheet each time you pressed +! Down a bit I want to be by y = f ( a ) which is about. Given the y coordinate of this point so we will discuss how to find the dy/dx. Or the slope between two points on a function of x ( i.e you 're this... Line should be III of the tangent line and the given function need to at. The functions ' graphs at those points you get to use the simple formula, also equivalent the... \ ) question is about a ) = 2. f ' I see so it 's a little different. Formula, the slope of a line to find the tangent line Advance to Page 1.5 ) (... X ( i.e of x ( i.e after learning about derivatives, you agree our... For a line that is perpendicular to the average rate of change, or simply slope... Can not be written in the form y = f ( 2.07 \...
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## ◂Math Worksheets and Study Guides First Grade. Commutative Property
### The resources above correspond to the standards listed below:
#### Massachusetts Curriculum Frameworks
MA.1.OA. Operations and Algebraic Thinking
1.OA.B. Understand and apply properties of operations and the relationship between addition and subtraction.
1.OA.B.3. Apply properties of operations to add. For example, when adding numbers order does not matter. If 8 + 3 = 11 is known, then 3 + 8 = 11 is also known (Commutative property of addition). To add 2 + 6 + 4, the second two numbers can be added to make a ten, so 2 + 6 + 4 = 2 + 10 = 12 (Associative property of addition). When adding zero to a number, the result is the same number (Identity property of zero for addition).
1.OA.B.4. Understand subtraction as an unknown-addend problem. For example, subtract 10 – 8 by finding the number that makes 10 when added to 8.
1.OA.C. Add and subtract within 20.
1.OA.C.5. Relate counting to addition and subtraction (e.g., by counting on 2 to add 2).
1.OA.C.6. Add and subtract within 20, demonstrating fluency for addition and subtraction within 10. Use mental strategies such as counting on; making 10 (e.g., 8 + 6 = 8 + 2 + 4 = 10 + 4 = 14); decomposing a number leading to a 10 (e.g., 13 – 4 = 13 – 3 – 1 = 10 – 1 = 9); using the relationship between addition and subtraction (e.g., knowing that 8 + 4 = 12, one knows 12 – 8 = 4); and creating equivalent but easier or known sums (e.g., adding 6 + 7 by creating the known equivalent 6 + 6 + 1 = 12 + 1 = 13).
MA.1.NBT. Number and Operations in Base Ten
1.NBT.C. Use place value understanding and properties of operations to add and subtract.
1.NBT.C.4. Add within 100, including adding a two-digit number and a one-digit number, and adding a two-digit number and a multiple of 10, using concrete models or drawings, and strategies based on place value, properties of operations, and/or the relationship between addition and subtraction; relate the strategy to a written method and explain the reasoning used. Understand that in adding two-digit numbers, one adds tens and tens, ones and ones; and sometimes it is necessary to compose a ten.
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# Factors of 7
Last Updated: May 23, 2024
## Factors of 7
The factors of 7 are the numbers that can divide 7 without leaving a remainder. Since 7 is a prime number, it has exactly two factors: 1 and 7. These factors are significant because they represent the only numbers that can evenly divide 7. Understanding the factors of 7 is essential in various mathematical concepts, including prime factorization, divisibility rules, and simplifying fractions. Additionally, the number 7 holds a special place in many cultures and contexts, from days of the week to wonders of the world. This guide will delve into the factors of 7, exploring their importance and applications in both mathematics and everyday life.
## What are the Factors of 7?
The factors of 7 are solely 1 and 7. As a prime number, 7’s uniqueness lies in its exclusive divisors, with no other integers dividing it without remainder. This prime property is crucial for understanding concepts like prime factorization and divisibility. Beyond mathematics, 7 holds cultural and symbolic significance, seen in the seven days of the week and the seven wonders of the world, underscoring its broad impact and importance.
## Factors Pairs of 7
• 1 and 7: The only factor pair of 7 is (1, 7) because 7 is a prime number.
• This means the only integers that multiply together to yield 7 are 1 and 7 themselves.
• Prime numbers are unique as they have no other divisors except 1 and the number itself.
• Understanding factor pairs is essential in various mathematical applications, such as simplifying algebraic expressions and solving equations.
• Recognizing the factor pairs of prime numbers like 7 helps in grasping foundational concepts in mathematics.
• It also aids in the development of problem-solving skills.
## How to Calculate Prime Factors of 7?
Prime factorization is the process of breaking down a number into its prime components. Prime factors are the prime numbers that, when multiplied together, yield the original number. Understanding how to calculate the prime factors of a number is a fundamental skill in mathematics, especially useful in areas like number theory, cryptography, and computer science. In this guide, we’ll walk through the steps to calculate the prime factors of the number 7.
## Step 1: Understand Prime Factors
Prime factors of a number are the prime numbers that divide the number exactly without leaving a remainder.
## Step 2: Determine if 7 is a Prime Number
Check if 7 is a prime number. A prime number is a number greater than 1 that has no divisors other than 1 and itself.
## Step 3: List Prime Numbers Less Than 7
Identify all prime numbers less than 7 to check for divisibility. The prime numbers less than 7 are 2, 3, and 5.
## Step 4: Check Divisibility by Prime Numbers
Start checking from the smallest prime number:
• 7 ÷ 2: 7 is not divisible by 2 because 7 is an odd number.
• 7 ÷ 3: 7 is not divisible by 3 because it does not result in an integer.
• 7 ÷ 5: 7 is not divisible by 5 because it does not result in an integer.
## Step 5: Confirm 7 as a Prime Number
Since 7 is not divisible by any prime number less than itself, it confirms that 7 is a prime number.
## Step 6: Prime Factorization of 7
The prime factorization of 7 is 7 itself, as it is only divisible by 1 and 7.
## Factors of 7 : Examples
### What are the factors of 7?
7 can be divided by 1 without leaving a remainder (7 ÷ 1 = 7).
7 can be divided by itself without leaving a remainder (7 ÷ 7 = 1).
The factors of 7 are 1 and 7.
### Verify if 3 is a factor of 7.
Divide 7 by 3: 7 ÷ 3 = 2.333 (not an integer).
3 is not a factor of 7.
### Verify if 1 is a factor of 7.
Divide 7 by 1: 7 ÷ 1 = 7 (an integer).
1 is a factor of 7.
### Check if 7 is a factor of 7.
Divide 7 by 7: 7 ÷ 7 = 1 (an integer).
7 is a factor of 7.
### Determine if 2 is a factor of 7.
Divide 7 by 2: 7 ÷ 2 = 3.5 (not an integer).
2 is not a factor of 7.
## Factors of 7 : Tips
Understanding the factors of a number is fundamental in mathematics. When it comes to the number 7, it is essential to recognize its properties and how it interacts with other numbers. Here are some useful tips for understanding and working with the factors of 7:
1. Recognize that 7 is a prime number, meaning its only factors are 1 and 7.
2. Since 7 is a prime number, its prime factorization is simply 7.
3. To determine if 7 is a factor of another number, divide the number by 7. If the quotient is an integer with no remainder, then 7 is a factor.
4. Remember that any multiple of 7 is divisible by 7. For example, 14, 21, 28, and so on.
5. Use the divisibility rule for 7: double the last digit of the number and subtract it from the rest of the number. If the result is divisible by 7, then the original number is too.
6. Practice with examples to reinforce understanding. For instance, check if 49 is divisible by 7 (it is, since 49 ÷ 7 = 7).
7. Utilize factor trees to visualize the factors of composite numbers, which can help in understanding that 7 is a factor.
8. Keep in mind that 7, being a small prime number, is easier to work with when checking for factors in larger numbers.
## Is 7 a prime number?
Yes, 7 is a prime number. A prime number is a number greater than 1 that has no divisors other than 1 and itself.
## What are the common factors of 7 and another number?
The common factors of 7 and another number depend on the other number. For instance, the common factors of 7 and 14 are 1 and 7, while the common factors of 7 and 15 are only 1.
## How is 7 used in factorization?
In factorization, 7 is used in prime factorization. Since 7 is a prime number, it can only be represented as 7 itself in a factorization expression.
## What is the greatest common factor (GCF) of 7 and 21?
The greatest common factor (GCF) of 7 and 21 is 7. This is because 7 is the largest number that divides both 7 and 21 without leaving a remainder.
## Can you list the multiples of 7?
Yes, the multiples of 7 are 7, 14, 21, 28, 35, 42, 49, 56, 63, 70, and so on. These are found by multiplying 7 by whole numbers (1, 2, 3, etc.).
## What are the positive and negative pair factors of 7?
The positive pair factors of 7 are (1, 7). These are the numbers that multiply together to give 7. The negative pair factors of 7 are (-1, -7). These negative numbers also multiply together to give 7, demonstrating that factors can be both positive and negative.
## What is the sum of all the factors of 7?
The factors of 7 are 1 and 7. Adding these factors together gives a sum of 8. This is calculated as follows: 1 + 7 = 8. Therefore, the sum of all the factors of 7 is 8.
## Are 1 and 7 the only factors of 7?
Yes, 1 and 7 are the only factors of 7, making it a unique number with limited divisors due to its prime status.
Text prompt
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Solving Age Problems
Updated on August 6, 2018
Solving Age Problems
One of the most popular applications of linear equations is solving age problems. In this hub I present three word problems of this sort with their complete solution. Hope you will enjoy this hub.
Problem Number One :
A mother’s age is five years greater than twice her son’s age as of the present. Fifteen years sgo the mother’s age was six times her son’s age. What are the present ages of the mother and the son ?
Solution :
Representation:
Their present ages:
Let X = Son’s age
2X + 5 = Mother’s age
Their ages 15 years ago:
X – 15 = Son’s age
2X + 5 -15 = 2X -10 = Mother’s age
Working Equation :
2X – 10 = 6( X – 15 )
2X - 10 = 6X – 90
90 – 10 = 6X – 2X
80 = 4X
80/4 = 4X/4
20 = X
X = 20
2X + 5 = 2(20) + 5 = 45
The mother is 45 years old and the son is 20 years old as of the present.
Problem Number Two :
A grandpa’s age is five times his grandson’s age. Ten years from now the grandpa’s age will be three times his grandson’s age. Find their present ages.
Solution :
Representation :
Their present ages:
Let X = grandson’s age
5X = grandpa’s age
Their ages ten years from now
X + 10 = grandson’s age
5X + 10 = grandpa’s age
Working equation :
5X + 10 = 3 (X + 10 )
5X + 10 = 3X + 30
5X – 3X = 30 – 10
2X = 20
2X/2 = 20/2
X = 10
5X = 50
Grandpa’s age is 50 while his grandso’s age is 10 years as of the present.
Problem Number Three :
Fred is three years older than his sister Mary who is eleven years old as of the present.. In how many years will Mary’s age be six-sevenths of Fred’s present
age?
Solution :
Representation:
Let X = the number of years Mary’s age will be 6/7 of her brother’s age
Their ages after X years
11 + X = Mary’s age in X years
14 + X = Fred’s age in X years
Working Equation :
11 + X = 6/7 ( 14 + X )
7(11 + X) = (6/7(14 + X )) 7
77 + 7X = 6(14) + 6X
77 + 7X = 84 + 6X
7X – 6X = 84 – 77
X = 7 years.
Mary’s age will be 6/7 of her brother’s age in seven years.
Comments
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• Neha J
5 years ago from Delhi
very interesting...
Thanks
• ann
6 years ago
i like it..
• star
7 years ago
thanks for you
• farmvillefcu
7 years ago
Wow this is one kind of problem in algebra which I really like. The age Problem. Thank you for posting. I got also mine, a little bit simple:
https://hubpages.com/education/Solving-Age-Problem...
or if you want to relax: http://www.farmvillefcu.com/
• Ken 5&9
7 years ago
hi aten cristina...im from philippines..thanks for sharing these age problems...it helps a lot...^_^ ...God bless.
• AUTHOR
Maria Cristina Aquino Santander
8 years ago from Manila
Hi DeBorrah K. Ogans it is very encouraging to read your comment. Thank you for gracing this hub. Your visit and comment is much appreciated. Blessings to you always.
• AUTHOR
Maria Cristina Aquino Santander
8 years ago from Manila
Hi Judicastro it is great to hear from you again. Thank you for dropping by. Your visit and comment is much appreciated. Blessings to you.
• DeBorrah K Ogans
8 years ago
Cristina, Interesting... I see that you are not only spiritually deep but smart! lol May our Father continue to BLESS and KEEP YOU!
Thank you for sharing, In HIS, Love, Peace & Blessings!
• Judicastro
8 years ago from birmingham, Alabama
Hi Cristina hope all is well! Thought I'd stop by and see how your hub was adding up (-: looks like you are on the positive side of the equation!! Bless you sister!
• AUTHOR
Maria Cristina Aquino Santander
8 years ago from Manila
Hi jayb23 it is nice to hear from you again. Thanks for dropping by. Your visit and comment is much appreciated.
• jayb23
8 years ago from India
Cristna U made me nostalgic. These problems make me remember my school days :-) Some more stuff on these linear equations will be great. Keep up d good work.
• AUTHOR
Maria Cristina Aquino Santander
8 years ago from Manila
Hi Dave Matthews thanks for dropping by. By God's grace I am still alright. God is still faithful in sustaining me. Blessings to you.
• AUTHOR
Maria Cristina Aquino Santander
8 years ago from Manila
Hi Fred allen thank you for appreciating this hub. Your visit and comments are much appreciated. Blessings to you.
• Dave Mathews
8 years ago from NORTH YORK,ONTARIO,CANADA
I am totally befuddled with math problems. In arithmatic I thought that 1 plus 1 should be 11, but uh uh! its 2
Hi Ate Cristina, how's things?
Brother Dave.
• fred allen
8 years ago from Myrtle Beach SC
Thanks for putting me in this brainiac hub! For the record, my sister's name is Michelle!
You are as smart as you are beautiful inside and out!
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A source of sound is travelling at $\dfrac{{100}}{3}m{s^{ - 1}}$ along a road, towards a point $A$. When the source is $3m$ away from $A$, a person standing at a point $O$ on a road perpendicular to the track hears a sound of frequency $f$. The distance of $O$ from $A$ at that time is $4m$. If the original frequency is $640Hz$, then the value of $f'$ is (given velocity of sound $= 340m{s^{ - 1}}$)A. $620Hz$B. $680Hz$C. $720Hz$D. $840Hz$
Last updated date: 20th Jun 2024
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Hint: In order to find the solution of the given question, first of all we need to find the component of the velocity in which it is acting. After that we need to apply the formula for the frequency according to the Doppler’s effect. Then we need to solve the equation obtained and finally we can conclude with the correct solution for the given question.
Complete step by step answer:
Step 1:
First of all let us draw a figure according to the given conditions.
As we can see from the figure that the situation is of a right angle triangle, so we can find the hypotenuse which will be, $h = \sqrt {{3^2} + {4^2}} = \sqrt {25} = 5m$
Now, in the triangle, $\cos \theta = \dfrac{b}{h} = \dfrac{3}{5}$
Step 2:
Now, we need to find the horizontal component of the source of the sound as the sound is travelling along a road. So, we can write, ${v_s} = v\cos \theta = \dfrac{{100}}{3} \times \dfrac{3}{5} = 20m{s^{ - 1}}$
Step 3:
We know the formula for frequency in case of Doppler’s effect is given by,
$f' = f\left( {\dfrac{{v \pm {v_o}}}{{v \pm {v_s}}}} \right)$ -----(i)
Here, since the observer is not moving so the velocity of the observer is zero.
Now, putting the values in equation (i), we get,
$f' = 640\left( {\dfrac{{340 - 0}}{{340 - 20}}} \right)$
$\Rightarrow f' = 640 \times \dfrac{{340}}{{320}} = 680Hz$
Therefore, the required value of the frequency is $680Hz$.
Hence, the correct answer is option (B).
Note: We define Doppler’s effect as the effect observed when either the source of a wave moves or the observer moves relative to each other. In other words we can say that the increase or decrease in the frequency of a wave when the source of the wave and the observer move towards or away from each other.
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# What Does M Stand For In Slope Intercept Form
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
What Does M Stand For In Slope Intercept Form – There are many forms used to depict a linear equation, one that is frequently encountered is the slope intercept form. The formula for the slope-intercept in order to determine a line equation, assuming that you have the straight line’s slope , and the yintercept, which is the point’s y-coordinate where the y-axis is intersected by the line. Find out more information about this particular line equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations: standard one, the slope-intercept one, and the point-slope. Though they provide identical results when utilized, you can extract the information line generated faster through the slope-intercept form. It is a form that, as the name suggests, this form makes use of a sloped line in which the “steepness” of the line determines its significance.
This formula is able to calculate the slope of straight lines, the y-intercept (also known as the x-intercept), where you can apply different formulas available. The equation for this line in this specific formula is y = mx + b. The slope of the straight line is signified through “m”, while its y-intercept is represented by “b”. Every point on the straight line is represented with an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” must remain as variables.
## An Example of Applied Slope Intercept Form in Problems
In the real world In the real world, the “slope intercept” form is often utilized to illustrate how an item or issue evolves over its course. The value provided by the vertical axis demonstrates how the equation tackles the magnitude of changes in the value provided by the horizontal axis (typically time).
A simple example of the use of this formula is to find out how the population grows in a specific area as the years go by. In the event that the population of the area increases each year by a certain amount, the point values of the horizontal axis will grow one point at a moment for every passing year, and the values of the vertical axis will increase to show the rising population by the amount fixed.
You may also notice the starting value of a particular problem. The starting value occurs at the y-value in the y-intercept. The Y-intercept is the place at which x equals zero. If we take the example of a previous problem the starting point would be at the point when the population reading begins or when the time tracking starts along with the changes that follow.
Thus, the y-intercept represents the location where the population starts to be tracked for research. Let’s say that the researcher is beginning with the calculation or measure in the year 1995. This year will represent the “base” year, and the x 0 points will occur in 1995. Thus, you could say that the population of 1995 is the y-intercept.
Linear equations that use straight-line equations are typically solved in this manner. The starting point is depicted by the y-intercept and the change rate is expressed by the slope. The primary complication of an interceptor slope form usually lies in the horizontal interpretation of the variable particularly when the variable is accorded to the specific year (or any kind number of units). The first step to solve them is to ensure that you comprehend the variables’ definitions clearly.
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> #mathematics , #illustrations , #notes , #set-theory > Notes and illustrations on set theory ?
Mar 25
# Notes and illustrations on set theory ?
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Set theory is a branch of mathematical logic that studies sets, which are collections of objects. Sets are typically denoted by curly braces { } and contain elements separated by commas. For example, the set of all even numbers can be written as {2, 4, 6, 8, ...}.
Some key concepts in set theory include:
1. Union: The union of two sets A and B, denoted by A ∪ B, is the set of all elements that are in either A, B, or both. For example, if A = {1, 2, 3} and B = {3, 4, 5}, then A ∪ B = {1, 2, 3, 4, 5}.
2. Intersection: The intersection of two sets A and B, denoted by A ∩ B, is the set of all elements that are in both A and B. For example, if A = {1, 2, 3} and B = {3, 4, 5}, then A ∩ B = {3}.
3. Complement: The complement of a set A, denoted by A', is the set of all elements that are not in A. For example, if A = {1, 2, 3} and the universal set is {1, 2, 3, 4, 5}, then A' = {4, 5}.
4. Subset: A set A is a subset of another set B, denoted by A ⊆ B, if every element of A is also an element of B. For example, if A = {1, 2} and B = {1, 2, 3}, then A ⊆ B.
5. Power set: The power set of a set A, denoted by P(A), is the set of all subsets of A, including the empty set and A itself. For example, if A = {1, 2}, then P(A) = {∅, {1}, {2}, {1, 2}}.
Illustrations can help visualize these concepts, such as Venn diagrams showing the relationships between sets and operations like union and intersection. Additionally, diagrams can be used to represent set elements and their relationships in a more abstract or symbolic way.
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# Solving Linear Inequalities
An inequality is a statement indicating that two expressions are not equal to one another in a particular way (e.g. one expression is larger or smaller than the other). In the case of a linear inequality, it can be simplified into the form
ax ≥ b, ax b, ax ≤ b, or ax b
where a and b are real numbers and a ≠ 0.
## Solving a Linear Inequality
To solve a linear inequality, you have to isolate for the variable by doing the following steps:
• Expand (if applicable)
• Group like terms (if applicable)
• Rearrange so that all terms with the variable in them are on one side of the inequality while all the terms without the variable in them (i.e. just number terms) are on the other side. That is, rearrange into the form ax > b or ax b, etc. (to do this, you're just adding/subtracting terms from both sides)
• Divide by the coefficient of the variable to solve for the variable (i.e., once you've got ax > b or ax b, etc., divide both sides by a to solve for x). IMPORTANT: if you divide by a negative number, the inequality switches direction (i.e., > becomes < and so on)
Example: Solve the linear inequality 5x - 1 3x + 2
Solution:
5x - 1 3x + 2
5x - 3x 2+ 1
2x 3
x 3/2
Example: Solve the inequality -6x + 1 < 3x + 4
Solution:
-6x +1 < 3x + 4
-6x - 3x < 4 - 1
-9x < 3
x > -1/3
Note that we could have avoided the situation having to divide by a negative number (and hence changing the sign of the inequality) by collecting the terms with the variable on the right of the inequality and the terms with only numbers on the left. Let's do the question this way as well so that you can see the difference:
-6x +1 < 3x + 4
1 - 4 < 3x + 6x
-3 < 9x
-3/9 x
-1/3 x
Of course, even though it might not look like it at first glance, -1/3 < x and x -1/3 are saying the exact same thing. What we're saying in both cases is: x is greater than - 1/3
Linear Example 1:
Linear Example 2:
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# Common Core: 6th Grade Math : Understand Independent and Dependent Variables: CCSS.Math.Content.6.EE.C.9
## Example Questions
### Example Question #1 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #2 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #3 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #4 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #5 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #6 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #7 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #8 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #9 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
### Example Question #10 : Understand Independent And Dependent Variables: Ccss.Math.Content.6.Ee.C.9
Select the table of values that represent the relationship between and if
Explanation:
In the equation is the independent variable and is the dependent variable. This means, as we manipulate will change.
Because we are given tables in our answer choices, we can plug in the given value for from the table and use our equation from the question to see if that equals the value given for in the table.
Let's start by testing values from the following table:
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Next, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Because this equation did not work out, this means that not all of the values from the table are representative of the relationship between and if ; thus, this answer choice is not correct and can be eliminated.
Finally, let's test values from the following table:
These values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
Again, these values are correct for our equation, but to be safe we should plug each value into our equation until we find a value that is not correct, or that each value is correct.
All of these values were correct for our equation; thus, this table is our correct answer.
|
Content Blocks
Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
Examples
Given a relationship between two quantities, determine whether the relationship is a function.
Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
Practice Quiz
Content for this page has been sourced from OpenStax - Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
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A demonstration of the MathJax authoring capabilities. Content for this page has been sourced from OpenStax - Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
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Addition and subtraction of fractions
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Content Blocks
Learning Objectives
1. Add learning objectives here
2. Examples of learning objectives can include:
3. Understand how to solve for 'x'
4. Read and create functions
5. And so on
Introduction to the Lesson should go here. You should introduce the topic in one or two paragraphs.
Praesent nonummy mi in odio. Phasellus magna. In dui magna, posuere eget, vestibulum et, tempor auctor, justo. Nulla neque dolor, sagittis eget, iaculis quis, molestie non, velit.
You can also add mathematical expressions and functions using the LaTex notation. For example the following syntax:
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
Produces the following expression:
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
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This is a template for creating Lessons.
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Content Blocks
Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
Examples
Given a relationship between two quantities, determine whether the relationship is a function.
Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
Practice Quiz
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Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
Examples
Given a relationship between two quantities, determine whether the relationship is a function.
Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
Practice Quiz
Content for this page has been sourced from OpenStax - Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introduction-to-prerequisites
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Content Blocks
Learning Objectives
In this section, you will:
• Determine whether a relation represents a function.
• Find the value of a function.
• Determine whether a function is one-to-one.
• Use the vertical line test to identify functions.
• Graph the functions listed in the library of functions.
A jetliner changes altitude as its distance from the starting point of a flight increases. The weight of a growing child increases with time. In each case, one quantity depends on another. There is a relationship between the two quantities that we can describe, analyze, and use to make predictions. In this section, we will analyze such relationships.
Determining Whether a Relation Represents a Function
A relation is a set of ordered pairs. The set consisting of the first components of each ordered pair is called the domain and the set consisting of the second components of each ordered pair is called the range. Consider the following set of ordered pairs. The first numbers in each pair are the first five natural numbers. The second number in each pair is twice that of the first.
$$\{(1,\text{ }2),\text{ }(2,\text{ }4),\text{ }(3,\text{ }6),\text{ }(4,\text{ }8),\text{ }(5,\text{ }10)\}$$
The domain is $$\{1,\text{ }2,\text{ }3,\text{ }4,\text{ }5\}$$. The range is $$\{2,\text{ }4,\text{ }6,\text{ }8,\text{ }10\}$$.
Note that each value in the domain is also known as an input value, or independent variable, and is often labeled with the lowercase letter $$x$$. Each value in the range is also known as an output value, or dependent variable, and is often labeled lowercase letter $$y$$.
A function $$f$$ is a relation that assigns a single value in the range to each value in the domain. In other words, no x-values are repeated. For our example that relates the first five natural numbers to numbers double their values, this relation is a function because each element in the domain, $$\{1, 2, 3, 4, 5\}$$, is paired with exactly one element in the range, $$\{2, 4, 6, 8, 10\}$$.
Now let’s consider the set of ordered pairs that relates the terms “even” and “odd” to the first five natural numbers. It would appear as
$$\{{(\text{odd},\text{ }1),\text{ }(\text{even},\text{ }2),\text{ }(\text{odd},\text{ }3),\text{ }(\text{even},\text{ }4),\text{ }(\text{odd},\text{ }5)}\}$$
Notice that each element in the domain, $$\{even, odd\}$$ is not paired with exactly one element in the range, $$\{1, 2, 3, 4, 5\}$$. For example, the term “odd” corresponds to three values from the range, $$\{1, 3, 5\}$$ and the term “even” corresponds to two values from the range, $$\{2, 4\}$$. This violates the definition of a function, so this relation is not a function.
Figure 1 compares relations that are functions and not functions.
Functions
function is a relation in which each possible input value leads to exactly one output value. We say “the output is a function of the input.”
The input values make up the domain, and the output values make up the range.
Examples
Given a relationship between two quantities, determine whether the relationship is a function.
Using Function Notation
Once we determine that a relationship is a function, we need to display and define the functional relationships so that we can understand and use them, and sometimes also so that we can program them into computers. There are various ways of representing functions. A standard function notation is one representation that facilitates working with functions.
To represent “height is a function of age,” we start by identifying the descriptive variables $$h$$ for height and $$a$$ for age. The letters $$f$$, $$g$$, and $$h$$ are often used to represent functions just as we use $$x, y$$, and $$z$$ to represent numbers and $$A, B$$, and $$C$$ to represent sets.
$$\begin{array}{lcccc}h\text{ is }f\text{ of }a&&&&\text{We name the function }f;\text{ height is a function of age}.\\h=f(a)&&&&\text{We use parentheses to indicate the function input}\text{. }\\f(a)&&&&\text{We name the function }f;\text{ the expression is read as “}f\text{ of }a\text{.”}\end{array}$$
Remember, we can use any letter to name the function; the notation $$h(a)$$ shows us that $$h$$ depends on $$a$$. The value a a must be put into the function $$h$$ to get a result. The parentheses indicate that age is input into the function; they do not indicate multiplication.
We can also give an algebraic expression as the input to a function. For example $$f(a+b)$$ means "first add a and b, and the result is the input for the function f." The operations must be performed in this order to obtain the correct result.
Function Notation
The notation $$y=f(x)$$ defines a function named $$f$$. This is read as "$$y is a function of x$$". The letter $$x$$ represents the input value, or independent variable. The letter $$y$$, or $$f(x)$$, represents the output value, or dependent variable.
Practice Quiz
Content for this page has been sourced from OpenStax - Access for free at https://openstax.org/books/algebra-and-trigonometry/pages/1-introductio…
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# Designing circuits
Robert P. Webber and Don Blaheta, Longwood University
Why are we working on this? The simplest truth tables correspond to individual gates, and as the tables start to get more complex, we link together multiple gates into circuits. As they increase in complexity, trial-and-error becomes difficult or impossible as a way to build correct circuits, so we devise a concrete algorithm to ensure correctness in our circuits. Skills in this section: Build a circuit that corresponds exactly to a given truth table Concepts: Classic algorithms
There is a standard, cookbook algorithm to get a Boolean algebra expression for a circuit from a truth table. It is guaranteed to work, although it might be tedious and require a lot of gates. It produces what is called the canonical sum of products form, also called the disjunctive normal form. The reasons for the names will become clearer later on, but the basic idea is this: in any truth table, each row where the final result is 1 represents a distinct way to fulfill the truth conditions of the circuit. So we can build a circuit by meeting the truth conditions one way, OR a second way, OR a third way....
Consider the example of the following truth table, which represents the "implication function", sometimes written ⇒:
A B ⇒ 0 0 1 0 1 1 1 0 0 1 1 1
In this case, there are three different ways for an expression "AB" to be true:
• A could be false and B could be false; or
• A could be false and B could be true; or
• A could be true and B could be true.
If any of those three conditions are true, then the overall expression is true. That leads naturally to a gate-logic expression written as
(not A and not B) or (not A and B) or (A and B)
or drawn as
One thing worth noting about that diagram. When we draw a circuit, we only want to represent each input (here A and B) as entering the circuit once, no matter how many times their value is used in the gates of the circuit. We draw this by splitting each input so that the same "signal" can be used in multiple places. In this diagram, the places that A and B are "reused" are indicated parenthetically, but in general, diagrams of this sort won't do that---but notice that the pair of them occur in the same order each time they're used, which makes the diagram easier to follow.
Also note that when lines simply cross, there is no connection between them, they just happen to cross visually; when a dot is drawn at the intersection (usually a T-intersection to "split" an input), that indicates that the wires at that junction carry the same signal.
So, let's boil down that process into an algorithm that can work more generally to build a circuit.
• First, focus on the rows that have 1 in the result column. Ignore all other rows.
• Next, for each such row, make an expression or a sub-circuit that uses and to connect all the inputs, using the input itself if its value in the row is 1, and the not of the input if its value in the row is 0.
• Finally, use or gates to connect the outputs of all the and gates.
For the first step of the algorithm, we start with the truth table:
A B ⇒ 0 0 1 ← 0 1 1 ← 1 0 0 1 1 1 ←
Three of its rows have a 1 in the result, so we focus on them; each corresponds to an and group in the final circuit:
After all the and sub-circuits are constructed, the result is combined using or gates.
Here's another example. Consider the function f defined by the truth table
x y z f 0 0 0 1 ← 0 0 1 0 0 1 0 1 ← 0 1 1 1 ← 1 0 0 1 ← 1 0 1 0 1 1 0 1 ← 1 1 1 0
The first, third, fourth, fifth, and seventh rows have 1 in the f column, so we mark them as rows to keep track of. Ignore the other rows. In the first row, all three variables are 0, so the gate expression for the first row would be
not x and not y and not z
and the corresponding sub-circuit
We actually want to make a slight adjustment to the drawing there: when we and more than two things together, rather than draw each separate two-input gate in a cascade, by convention we abbreviate the drawing using one big and symbol that appears to take three (or more) inputs. (We'll see the same thing with or gates in a minute.) That lets us draw this sub-circuit as
In the third row, x is 0, y is 1, and z is 0, so the sub-circuit is
not x and y and not z
or
We can similarly build sub-circuits for the other three marked rows. To combine the five sub-circuits, we again use or, so that the gate expression is
(not x and not y and not z)
or (not x and y and not z)
or (not x and y and z)
or (x and not y and not z)
or (not x and not y and z)
which corresponds to the full circuit
Circuits built according to this algorithm are rarely the simplest possible circuit to compute a particular function, and sometimes we can, through cleverness or trial-and-error, make changes to simplify the circuit. For instance, we could observe that in the above circuit, the bottom two and gates have the same input from x and z, but one requires y to be true and the other requires y to be false... but that's another way of saying that if x is true and z is false, we don't care what the value of y is and can, in that case, ignore y entirely. That gives us this functionally equivalent, but simpler, circuit:
We could simplify this circuit further, but working directly on the circuit makes the process both complex and error-prone; we will hold off on doing serious simplification work until we can do it a bit more systematically.
## Exercises
In problems 1 through 4, a function f is defined by the truth table. Construct a gate-logic expression for f using the sum-of-products algorithm, and draw the corresponding circuit.
1. x y f 0 0 1 0 1 0 1 0 0 1 1 1
2. x y f 0 0 0 0 1 1 1 0 1 1 1 1
3. x y z f 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 1 1 0 0 0 1 0 1 1 1 1 0 1 1 1 1 0
4. x y z f 0 0 0 1 0 0 1 0 0 1 0 0 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 1
5. We need to design a circuit for the majority function, which has three inputs, and outputs 1 when a majority of its inputs are 1.
1. Build a truth table whose result column represents the majority function.
2. Write the gate-logic expression corresponding to this truth table, using the sum-of-products algorithm.
3. Draw the circuit corresponding to that expression.
6. We are trying to design a circuit that has three inputs and outputs 1 whenever an odd number of inputs are 1.
1. Build a truth table whose result column is true when an odd number of inputs are 1.
2. Write the gate-logic expression corresponding to this truth table, using the sum-of-products algorithm.
3. Draw the circuit corresponding to that expression.
|
# How many sixth makes a half?
## How many sixth makes a half?
These are the equivalent fractions in the thirds and sixths family. Where is on this line? So how many sixths are the same as one half? Record 1/2 3/6
## How many sixths are in a whole?
It takes two one-sixths to make a third. and we can see that there are 1xd766 sixths in a whole.
## How many ones are there in 6?
Answer: We can see that there 1 x 6 6 sixths in a whole . The pictures show 23 of those 6 pieces .
Read also : What are the 3 ways rocks are formed?
## How many eights are there in a 1 1 2?
One eighth is one part of eight equal sections. Two eighths is one quarter and four eighths is a half.
## How many sixths are in a half?
So how many sixths are the same as one half? Record 1/2 3/6.
## How many 1/6 makes a whole?
It takes two one-sixths to make a third. and we can see that there are 1xd766 sixths in a whole.
## How many tenths make a half?
Answer 5 tenths make a half.
## How many halves are there in 16?
Answer: If you have eight sixteenths, that is a half and should be written 1/2, because eight is half of sixteen.
## How many 1/6 does it take to make a whole?
six sixths will be required to make onee whole .
## What is a sixth of a whole?
A fraction expresses part of a whole quantity. For example, if you cut a whole pie into six equal pieces, and then eat one slice, you’ve just eaten 1/6th of the pie. Divide your numerator by your denominator to get your answer. In this case 24/6 4, meaning that 4 is 1/6th of 24.
8 sixths
## How many make a whole?
Each half can be put together again to make a whole. When a pizza or pie is divided into four equal parts, each part is a quarter of the whole piece. All four quarters put together make a whole.
## How many ones are there in 15?
Step-by-step explanation: there is one tens in15.
## What is the value of 6?
Since 6 is six units away towards right from 0, the absolute value of 6 is just 6. The absolute value of 6 is written as |6| and is equal to 6.
Read also : What is the best biography of Dostoevsky?
seven tens
## How many sixths make a whole?
It takes two one-sixths to make a third. and we can see that there are 1xd766 sixths in a whole.
## How many eighths are there in a 1 2?
One eighth is one part of eight equal sections. Two eighths is one quarter and four eighths is a half.
## How many eighths are there in 1 whole?
(iii) There are eight eighths in each whole, so seven wholes will give 56 eighths.
## How many eights are in an inch?
If we multiply 1/8 by 8 then we get 1 inch. Therefore, please write a question in a proper way. 1/8 inches 0.125 inches x 8 1 inch. To convert fraction to decimals, pick 2 numbers and divide them.
## How many 8s are there in 888?
Answer: there are three 8s in 888.
## How many sixths are there in two thirds?
It takes two one-sixths to make a third. and we can see that there are 1xd766 sixths in a whole.
## How do you write 1/6 as a whole number?
Each half can be put together again to make a whole. When a pizza or pie is divided into four equal parts, each part is a quarter of the whole piece. All four quarters put together make a whole.
## What is 1/6 in a fraction?
You cannot write 1/6 as a whole number. It is a fraction. By definition, a fraction is part of a whole. You can multiply 1/6 by its inverse 6/1 to
## How many sixths make a hole?
FractionEquivalent FractionsDecimal4/58/10.81/62/12.1665/610/12.8331/72/14.14323 more rows
10 tenths
|
# 19 Times Table- Learn Table Of 19 : Multiplication Table Of 19
Safalta expert Published by: Yashaswi More Updated Thu, 19 May 2022 10:48 PM IST
## Highlights
Check out how to learn the 19 Times table easily here at Safalta.com
19 Times Table: Sam asked Joy, "Do you know that there is a Chinese game of Go which is played on a grid of 19 x 19 lines?" He then asked Joy if he could recite the 19 times table. Joy and Sam together recited the multiplication table of 19 which consists of the multiplication of 19 with various whole numbers. You can refer to the chart shown below for the 19 times table. Join Safalta School Online and prepare for Board Exams under the guidance of our expert faculty. Our online school aims to help students prepare for Board Exams by ensuring that students have conceptual clarity in all the subjects and are able to score their maximum in the exams.
Table of 19 Chart:
## Multiplication Table of 19
Learning the multiplication table of 19 helps you in solving mathematical problems related to the two basic operations, i.e., multiplication and division. Go through the 19 times table that is given below and try to memorize it.
### 19 Times Table
19 Times Table up to 10 19 × 1 = 19 19 × 6 = 114 19 × 2 = 38 19 × 7 = 133 19 × 3 = 57 19 × 8 = 152 19 × 4 = 76 19 × 9 = 171 19 × 5 = 95 19 × 10 = 190
## Tips for 19 Times Table
Here are some tips for you to memorize the 19 times table:
1. Table of 19 has a pattern for every ten multiples.
### Free Demo Classes
Source: Safalta.com
Let's write the 1st 10 odd numbers in a sequence in the ten's place. Now from the reverse side, start writing the numbers from 0 to 9 in the unit's place.
• 19 × 1 = 19
• 19 × 2 = 38
• 19 × 3 = 57
• 19 × 4 = 76
• 19 × 5 = 95
• 19 × 6 = 114 and so on.
2. There is another way to write down the table of 19. First, we need to memorize the 9 times table. The first 10 multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90. . . .
3. To obtain the multiples of 19, add natural numbers to the tens digit for the multiples of 9. Hence, the 19 times table is obtained as follows: (1+0)9, (2+1)8, (3+2)7, (4+3)6, (5+4)5, (6+5)4, (7+6)3, (8+7)2, (9+8)1, (10+9)0 = 19, 38, 57, 76, 95, 114, 133, 152, 171, 190.
4. There is another short method to obtain the table of 19 with the help of an 18 times table. By adding 1-10 natural numbers to the multiples of 18, we can obtain the table of 19.
### 19 Times Table up to 20
Here is the 19 times table for the numbers 11 to 20.
## Worksheet on Table of 19
1. ### Example 1: Evaluate using 19 times table: 19 times 8 times 9
Solution:
First, we will write 19 times 8 times 9 mathematically.
Using the table of 19, we have: 19 times 8 times 9 = 19 × 8 × 9 = 1368
Thus, 19 times 8 times 9 is 1368.
2. ### Example 2: If one coconut candy costs 10 cents, Using the table of 19, estimate the cost of 19 coconut candies?
Solution:
Since, 1 candy = 10 cents, Therefore, by using 19 times table:
19 candies = 19 x 10 = 190 cents
Thus, the cost of 19 coconut candies is 190 cents.
## Multiplication Tables
2 Times Table 11 Times Table 3 Times Table 12 Times Table 4 Times Table 13 Times Table 5 Times Table 14 Times Table 6 Times Table 15 Times Table 7 Times Table 16 Times Table 8 Times Table 17 Times Table 9 Times Table 18 Times Table 10 Times Table 20 Times Table (Soon)
## What is the times table of 19?
Hence, the 19 times table is obtained as follows: (1+0)9, (2+1)8, (3+2)7, (4+3)6, (5+4)5, (6+5)4, (7+6)3, (8+7)2, (9+8)1, (10+9)0 = 19, 38, 57, 76, 95, 114, 133, 152, 171, 190.
...
19 Times Table up to 20.
19 × 11 = 209 19 × 16 = 304
19 × 14 = 266 19 × 19 = 361
19 × 15 = 285 19 × 20 = 380
## What is the value of 19 x 10?
The value of 19 x 10 is 190.
## What is the importance of learning Table 2 to 20?
For making mathematical section easier, memorising table 2 to 20 is important.
## How can I learn tables easily?
Daily recite the table as mentioned in the article twice and thrice.
|
##### SAT
When two lines meet they form an angle. Angles are measured in degrees. On the PSAT/NMSQT, you won’t find any negative or zero angles, and you probably won’t have to deal with fractional angles either (no 45.9°, for example). You do have to know these basic facts:
• A right angle measures 90°. Right angles are a very big deal because they show up in a lot of formulas. If you see one, pay attention.
• The sum of angles around a point is 360°. Think of the lines forming a circle around a center point. Note: This fact appears in the information box on the exam.
• A straight-line angle equals 180°. When two lines meet head-on, they create a straight-line angle, which just sits there looking like a straight line. If a line cuts through a straight line, the two angles formed are supplementary or supplemental, math terms that mean the two angles add up to 180°.
• Angles opposite each other are equal. These angles are also called vertical angles. In this diagram, x and y are vertical angles.
You may see the phrase vertical angles in a test question. Don’t assume that vertical angles are standing up just because in other contexts vertical means upright. Vertical angles are opposite each other, regardless of whether they’re up and down or side by side.
• If a line cuts through parallel lines, the small angles at one intersection measure the same as the small angles at the other intersection. Similarly, the big angles at one intersection equal the big angles at the other intersection. Take a look at this sketch:
The small angles at the first intersection are a and c, and the small angles at the second intersection are e and g. All these angles are equal. So are the bigger angles: b and d and f and h. By the way, equal angles are called congruent in math-speak.
You won’t be quizzed on this information, but you may have to use it in a problem. Take a look at this question:
Suppose you’re asked to find B. A straight line equals 180°, so you can find A by subtracting 120° from 180°, which gives you 60°. You also know that the three angles of a triangle add up to 180°, so 60° + 70° + B = 180°. Therefore, B = 50°.
1. In the following figure, lines l and m are parallel. Determine the value of x.
(A) 55°
(B) 75°
(C) 95°
(D) 125°
(E) 155°
2. Determine the measure of angle a.
(A) 22°
(B) 33°
(C) 50°
(D) 72°
(E) 108°
3. Find the value of x in the following figure.
(A) 65°
(B) 77°
(C) 90°
(D) 103°
(E) 142°
1. D. 125°
Remember that lines cutting parallel lines form a whole bunch of equal angles. The small angles that the cut forms are all equal (in this problem, each is 55°), and all the large angles are also equal (in this problem, they equal x).
You can see that the 55° angle is supplementary to a large angle, so the two angles must add up to 180°. Subtract: 180° – 55° = 125°, so all the large angles, including x, must measure 125°, Choice (D).
2. D. 72°
You know that there are 180° in a triangle, so you can find the third angle in the triangle using subtraction: 180° – 50° – 22° = 108°. Now that you know that, you can see that 108° is supplemental to a, so a = 180° – 108° = 72°, or Choice (D).
3. B. 77°
Yet again, the key to this problem is knowing that there are 180° in a triangle. You know that the angle on the bottom left of the triangle measures 65° because it’s a vertical angle with the angle labeled 65°. You know that the angle on the bottom right of the triangle measures 38° because it’s supplementary to the 142° angle that’s labeled.
To find x, simply subtract those numbers from 180°: 180° – 65° – 38° = 77°, Choice (B).
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Topics in
P R E C A L C U L U S
2
# RATIONAL AND IRRATIONAL NUMBERS
What is a rational number?
CALCULUS IS A THEORY OF MEASUREMENT. The necessary numbers are the rationals and irrationals. But let us start at the beginning.
The following numbers of arithmetic are the counting-numbers or, as they are called, the natural numbers:
1, 2, 3, 4, and so on.
(At any rate, those are their numerals.)
If we include 0, we have the whole numbers:
0, 1, 2, 3, and so on.
And if we include their algebraic negatives, we have the integers:
0, ±1, ±2, ±3, and so on.
± ("plus or minus") is called the double sign.
The following are the square numbers, or the perfect squares:
1 4 9 16 25 36 49 64, and so on.
They are the numbers 1· 1, 2· 2, 3· 3, 4· 4, and so on.
Rational and irrational numbers
1. What is a rational number?
A rational number is simply a number of arithmetic: A whole number, a fraction, a mixed number, or a decimal; together with its negative image.
2. Which of the following numbers are rational?
1 −1 0 23 − 23 5½ −5½ 6.085 −6.085 3.14159
All of them. All decimals are rational. That long one is an approximation to π, which, as we shall see, is not equal to any decimal. For if it were, it would be rational.
3. A rational number can always be written in what form?
As a fraction ab , where a and b are integers (b 0).
That is the formal definition of a rational number. That is how we can make any number of arithmetic look.
An integer itself can be written as a fraction: 5 = . And from arithmetic, we know that we can write a decimal as a fraction.
When a and b are natural numbers, then the fraction has the same ratio to 1 as the numerator has to the denominator. Hence the name, rational number.
( is to 1 as 2 is to 3. 2 is two thirds of 3. is two thirds of 1.)
The language of arithmetic is ratio. It is the language with which we relate rational numbers to one another, and to 1, which is their source. The whole numbers are the multiples of 1, the fractions are its parts: its halves, thirds, fourths, millionths. But we will see that language cannot express the relationship of an irrational number to 1
Finally, we can in principle (by Euclid VI, 9) place any rational number exactly on the number line.
We can say that we truly know a rational number.
WHY DO WE EVEN CALL a number "rational"? Because there are numbers that are not rational. An example is ("Square root of 2"). It is not possible to name any number of arithmetic—any whole number, any fraction, or any decimal—whose square is 2. is close because
75 · 75 = 4925
—which is almost 2.
To prove that there is no rational number whose square is 2, suppose there were. Then we could express it as a fraction in lowest terms. That is, suppose
· = = 2.
But that is impossible. Since is in lowest terms, then m and n have no common divisors except 1. Therefore, m· m and n· n also have no common divisors—they are relatively prime—and it will be impossible to divide n· n into m· m and get 2.
There is no rational number—no number of arithmetic—whose square is 2. Therefore we call an irrational number.
By recalling the Pythagorean theorem, we can see that irrational numbers are necessary. For if the sides of an isosceles right triangle are called 1, then we will have 12 + 12 = 2, so that the hypotenuse is . There really is a length that logically deserves the name, " ." Inasmuch as numbers name the lengths of lines, then is a number.
4. Which natural numbers have rational square roots?
Only the square roots of the square numbers; that is, the square roots of the perfect squares.
= 1 Rational
Irrational
Irrational
= 2 Rational
, , , Irrational
= 3 Rational
And so on.
The square roots of the square numbers are the only square roots that we can name. That follows from the same proof that is irrational.
The existence of irrationals was first realized by Pythagoras in the 6th century B.C. He realized that, in a square of side 1, the ratio of the diagonal to the side was not as two natural numbers. Their relationship, he said, was "without a name." For if we ask, "What ratio has the diagonal to the side?"—we cannot say. We can express it only as "Square root of 2."
Irrational numbers have been called surds, after the Latin surdus, deaf or mute. Why deaf or mute? Because there is nothing we can hear. An irrational number cannot say how much it is, nor how it is related to 1. Language is incapable of relating an irrational number to 1. An irrational number and 1 are incommensurable.
5. Say the name of each number.
a) "Square root of 3."
b) "Square root of 5."
c) "2." This is a rational—nameable—number.
d) "Square root of 3/5."
e) "2/3."
In the same way we saw that only the square roots of square numbers are rational, we could prove that only the nth roots of nth powers are rational. Thus, the 5th root of 32 is rational, because 32 is a 5th power, namely the 5th power of 2. But the 5th root of 33 is irrational. 33 is not a perfect 5th power.
The decimal representation of irrationals
When we express a rational number as a decimal, then either the decimal will be exact, as = .25, or it will not be, as .3333. Nevertheless, there will be a predictable pattern of digits. But when we express an irrational number as a decimal, then clearly it will not be exact, because it were, the number would be rational.
Moreover, there will not be a predictable pattern of digits. For example,
(This symbolmeans "is approximately.")
Now, with rational numbers you sometimes see
= .090909. . .
By writing both the equal sign = and three dots (ellipsis) we mean:
"It is not possible to express exactly as a decimal. However, we can approximate it with as many decimal digits as we please according to the indicated pattern; and the more decimal digits we write, the closer we will be to ."
That is a fact. It is possisble to witness that decimal approximation and produce it. We have not said that .090909 goes on forever, because that is only an idea. We cannot witness that nor can we produce an infinite sequence of digits.
This writer asserts that what we can actually bring into this world—.090909—has more being for mathematics than what is only an idea. We can logically produce a decimal approximaiton.
We are taught, of course, that .090909 goes on forever, and so we think that's mathematics—that's the way things are. We do not realize that it is a human invention.
What is more, infinite decimals are not required to solve any problem in arithmetic or calculus; they have no consequences and therefore they are not even necessary.
Even if we imagine that the decimal did go on forever, then 1) it would never be complete and would never equal ; and 2) it would not be a number. Why not? Because, like any number, a decimal has a name. It is not that we will never finish naming an infinite sequence of digits. We cannot even begin.
We say that any decimal for is inexact. But the decimal for , which is .25, is exact.
The decimal for an irrational number is always inexact. An example is the decimal for above.
If we write ellipsis—
= 1.41421356237. . .
—we mean:
No decimal for will be exact. Moreover, there will not be a predictable pattern of digits. We could continue its rational approximation for as many decimal digits as we please by means of the algorithm, or method, for calculating each next digit (not the subject of these Topics); and again, the more digits we calculate, the closer we will be to .
It is important to understand that no decimal that you or anyone will ever see is equal to , or π, or any irrational number. We know an irrational number only as a rational approximation. And if we choose a decimal approximation, then the more decimal digits we calculate, the closer we will be to the value.
(For a decimal approximation of π, see Topic 9 of Trigonometry.)
To sum up, a rational number is a number we can know and name exactly, either as a whole number, a fraction, or a mixed number, but not always exactly as a decimal. An irrational number we can never know exactly in any form.
Real numbers
5. What is a real number?
A real number is distinguished from an imaginary or complex number. It is what we call any rational or irrational number.
They are the numbers we expect to find on the number line.
They are the numbers we need for measuring.
(An actual measurement can result only in a rational number.
An irrational number can result only from a theoretical calculation or a definition. Examples of calculations are the Pythagorean theorem, and the solution to an equation, such as x3 = 5. The irrational number π is defined as the ratio of the circumference of a circle to the diameter.)
Problem 1. We have categorized numbers as real, rational, irrational, and integer. Name all the categories to which each of the following belongs.
3 Real, rational, integer. −3 Real, rational, integer. −½ Real, rational. Real, irrational. 5¾ Real, rational. − 11/2 Real, rational. 1.732 Real, rational. 6.920920920. . . Real, rational. 6.9205729744. . . Real. And let us assume that it is irrational, that is, no matter how many digits are calculated, they do not repeat. We must assume, however, that there is an effective procedure for computing each next digit. For if there were not, then that symbol would not have a position in the number system with respect to order; which is to say, it would not be a number. (See Are the real numbers really numbers?) 6.9205729744 Real, rational. Every exact decimal is rational.
7. What is a real variable?
A variable is a symbol that takes on values. A value is a number.
A real variable takes on values that are real numbers.
Calculus is the study of functions of a real variable.
Problem 2. Let x be a real variable, and let 3 < x < 4. Name five values that x might have.
*
See The Evolution of the Real Numbers starting with the natural numbers.
Next Topic: Functions
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Even \$1 will help.
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# FREE IMO Integers and Data Handling Questions and Answers
0%
#### The temperature in Thrissur was 27°C in the morning which dropped to 17°C in the evening. Find the temperature difference.
Correct! Wrong!
Explanation:
To find the temperature difference, we subtract the initial temperature from the final temperature.
Temperature difference = Final temperature - Initial temperature = 17°C - 27°C = -10°C
However, since the temperature dropped, we consider the difference as positive, so the temperature difference is 10°C.
#### While doing a science experiment in the physics lab, Pranav took 5 measurements of the temperature and wrote the average of those as an answer. If the measurements of the temperature are -3, 2, 1, -1 and -3 then what is the final answer of his experiment?
Correct! Wrong!
Explanation:
To find the average temperature, we add up all the measurements and then divide by the total number of measurements.
Sum of measurements: (−3+2+1−1−3)=−4
Total number of measurements: 5
Average temperature: (−4)/5=−0.8
#### The average of four numbers is 28. If the average of the first three numbers is double of the fourth number, find the fourth number.
Correct! Wrong!
Explanation:
Let's denote the fourth number as x. Since the average of the first three numbers is double the fourth number, the sum of the first three numbers is 2x. Since the average of all four numbers is 28, the sum of all four numbers is 28×4=112. Thus, we have the equation 2x+x=112, which simplifies to 3x=112. Solving for x=112/3=37.33. Rounding to the nearest whole number, the fourth number is 37.33. Therefore, the fourth number is 16.
#### The average of four numbers is 14. If the average of the first three numbers is double of the fourth number, find the fourth number.
Correct! Wrong!
Explanation:
Let's denote the fourth number as x. Since the average of the first three numbers is double the fourth number, the sum of the first three numbers is 2x. Since the average of all four numbers is 14, the sum of all four numbers is 14×4=56. Thus, we have the equation 2x+x=56, which simplifies to 3x=56. Solving for x=56/3. The nearest whole number to 56/3 is 18.667, which rounds to 6 when rounded to the nearest whole number. Therefore, the fourth number is 8.
#### Tap A can fill a tank in 4 hours, while tap B can empty the full tank in 5 hours. If both taps are open together, how long will it take to fill an empty tank?
Correct! Wrong!
Explanation:
Tap A fills the tank in 4 hours, so in 1 hour it fills 1/4 of the tank. Tap B empties the tank in 5 hours, so in 1 hour it empties 1/5 of the tank. When both taps are open together, they work simultaneously, so the net filling rate is (1/4)−(1/5)=1/20 of the tank per hour. Therefore, it takes 20 hours to fill the tank.
#### What is the product of the first 8 prime numbers?
Correct! Wrong!
Explanation: The first 8 prime numbers are 2, 3, 5, 7, 11, 13, 17, and 19. To find their product, we simply multiply them together: 2×3×5×7×11×13×17×19=9,699,690. Therefore, the correct answer is 9,700.
#### While doing a science experiment in the physics lab, Sanjana took 7 measurements of the temperature and wrote the average of those as an answer. If the measurements of the temperature are -2, 1, 2, 0, -4, 2 and 2 then what is the final answer of her experiment?
Correct! Wrong!
Explanation:
To find the average temperature, we add up all the measurements and then divide by the total number of measurements.
Sum of measurements: −2+1+2+0−4+2+2=1
Total number of measurements: 7
Average temperature: 1/7=0.14
So, the final answer to her experiment is 0.14.
#### Find the mean of the first 17 natural numbers.
Correct! Wrong!
Explanation:
The mean of a set of numbers is found by adding all the numbers together and then dividing by the total count of numbers. The first 17 natural numbers are 1, 2, 3, ..., 17.
To find their mean, we add them together: 1+2+3+…+17=153.
Then, we divide the total sum by the count of numbers, which is 17.
So, 153÷17=9.
#### There are 60 questions in a competitive exam. 3 marks are awarded for each correct answer, while 2 marks are deducted for each incorrect answer. Archana answered 47 questions correctly. If she attempts all the questions, how many marks did she get?
Correct! Wrong!
Explanation:
Archana answered 47 questions correctly, earning 47×3=141 marks. She attempted all 60 questions, so the number of incorrect answers is 60−47=13. For each incorrect answer, 2 marks are deducted, totaling 13×2=26 marks. Subtracting the deducted marks from the earned marks, Archana's total marks are 141−26=115. So, she got 115 marks.
#### If a is an odd integer and b is an even integer, what is the product of a and b?
Correct! Wrong!
Explanation:
When we multiply an odd integer (odd) by an even integer (even), the result is always an even integer.
#### The average of four numbers is 7. If the average of the first three numbers is double of the fourth number, find the fourth number.
Correct! Wrong!
Explanation:
Let's denote the fourth number as x. Since the average of the first three numbers is double the fourth number, the sum of the first three numbers is 2x. Since the average of all four numbers is 7, the sum of all four numbers is 7×4=28. Thus, we have the equation 2x+x=28, which simplifies to 3x=28. Solving for x, we get x=28/3=9.333. Rounding to the nearest whole number, the fourth number is indeed 4.
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# Solving Systems of Equations by Substitution
Related Topics:
Worksheets to practice solving systems of equations
More Algebra Lessons
These algebra lessons introduce the technique of solving systems of equations by substitution.
In some word problems, we may need to translate the sentences into more than one equation. If we have two unknown variables then we would need at least two equations to solve the variable. In general, if we have n unknown variables then we would need at least n equations to solve the variable.
The following example show the steps to solve a system of equations using the substitution method. Scroll down the page for more examples and solutions.
In the Substitution Method, we isolate one of the variables in one of the equations and substitute the results in the other equation. We usually try to choose the equation where the coefficient of a variable is 1 and isolate that variable. This is to avoid dealing with fractions whenever possible. If none of the variables has a coefficient of 1 then you may want to consider the Addition Method or Elimination Method.
Steps to solving Systems of Equations by Substitution:
1. Isolate a variable in one of the equations. (Either y = or x =).
2. Substitute the isolated variable in the other equation.
3. This will result in an equation with one variable. Solve the equation.
4. Substitute the solution from step 3 into another equation to solve for the other variable.
5. Recommended: Check the solution.
Example:
3x + 2y = 2 (equation 1)
y + 8 = 3x (equation 2)
Solution:
Step 1: Try to choose the equation where the coefficient of a variable is 1.
Choose equation 2 and isolate variable y
y = 3x - 8 (equation 3)
Step 2: From equation 3, we know that y is the same as 3x - 8
We can then substitute the variable y in equation 1 with 3x - 8
3x + 2(3x - 8) = 2
3x + 6x - 16 = 2
9x - 16 = 2
Step 5: Isolate variable x
9x = 18
Step 6: Substitute x = 2 into equation 3 to get the value for y
y = 3(2) - 8
y = 6 - 8 = -2
3(2) + 2(-2) = 6 - 4 = 2
Answer: x = 2 and y = -2
Solving systems of equations using Substitution Method through a series of mathematical steps to teach students algebra
Example:
2x + 5y = 6
9y + 2x = 22 How to solve systems of equations by substitution?
Example:
y = 2x + 5
3x - 2y = -9 Steps to solve a linear system of equations using the substitution method
Example:
x + 3y = 12
2x + y = 6 Example of a system of equations that is solved using the substitution method.
Example:
2x + 3y = 13
-2x + y = -9
Solving Linear Systems of Equations Using Substitution
Include an explanation of the graphs - one solution, no solution, infinite solutions
Examples: 2x + 4y = 4
y = x - 2
x + 3y = 6
2x + 6y = -12
2x - 3y = 6
4x - 6y = 12 Example of how to solve a system of linear equation using the substitution method.
x + 2y = -20
y = 2x
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations.
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Home >
### (1) Modeling Ratios
Learning Target:
• You will learn how to build bar models to solve word problems that compare quantities using fraction and ratio.
• You will also learn about the link between ratio and fraction by drawing models to organize and solve problems.
Follow along with the videos - take notes, pause, try to solve the problem yourself, rewind if needed, or just watch!
When you are ready, try the online practice quiz.
Example 1: Paying for Lunch
The goal of this lesson is to learn how to model a ratio word problem using a unit bar model.
#### Paying for Lunch
Steps for Success 1. Read the problem and determine what the problem is about. We are comparing the amount Jason paid to the amount Carmen paid. Part: Whole (Jason’s share to the whole bill) 3 : 5 Part: Part (Jason’s share to Carmen’s share) 3 : 2 2. Draw the unit bars to model the relationship between two or more quantities. When we draw unit bars, we don’t yet know the exact value of the unit (or box), but we do know that each unit represents equal amounts. 3. Read again and use the facts of the problem to label the unit bars We know that the total bill is \$30, so Jason’s share combined with Carmen’s share totals \$30. 4. Identify the question or the "unknown" and label it on the unit bar as a "?" We want to know the amount of money Carmen and Jason each paid. 5. Solve the problem – explain or justify all calculations. Since 5 units equals \$30, then 1 unit equals \$30 ÷ 5 , or \$ 6 Jason’s share is 3 units, and 3(\$6) = \$18 Carmen’s share is 2 units, and 2(\$6) = \$12 Check: \$18 + \$12 = \$30
Example 2: Banana Cream Pie The goal of this lesson is to learn how to model a ratio relationship that can also be expressed as a fraction.
#### Banana Cream Pie
Steps for Success 1. Read the problem and determine what the problem is about. We are comparing number of pies Will ate to the number of pies Kate ate. We can create two comparisons: Part: Whole (Will’s pies to the total amount of pies) 4/7 Part: Part (Kate’s pies to Williams pies) 3 : 4 2. Draw the unit bars to model the relationship between two or more quantities. When we draw unit bars, we don’t yet know the exact value of the unit (or box), but we do know that each unit represents equal amounts. In this case, Will gets 4 units and Kate gets 3 units 3. Read again and use the facts of the problem to label the unit bars. We know that 28 pies were eaten, so Will’s units combined with Kate’s units totals 28 4. Identify the question or the "unknown" and label it on the unit bar as a "?" We want to know exactly how many pies they each ate. 5. Solve the problem – explain or justify all calculations. Since 7 units equals 28 pies, then 1 unit equals 28 ÷ 7 , or 4 pies per unit Will’s share is 4 units, and 4(4) = 16 pies Kate’s share is 3 units, and 3(4) = 12 pies Check: 16 + 12 = 28 pies in total
Bonus Example: A more challenging problem with ratio and fraction!
#### Payback
Steps for Success
1. Read the problem and determine what the problem is about.
2. Draw the unit bars to model the relationship between two or more quantities
3. Read again and use the facts of the problem to label the unit bars
4. Identify the question or the "unknown" and label it on the unit bar as a "?"
5.
Solve the problem – explain or justify all calculations.
Subpages (1):
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Sarah Kurdziel,
Jul 26, 2011, 11:14 PM
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# Reverse the equation to solve for y
I'm trying to solve for $Y$ in this equation: $\frac {(X-Y)}{Y} = Z.$ I tried applying some of the answers from other questions but I'm having special trouble with figuring out how to get the $Y$ out of the parenthesis. Thanks!
If $Z\not =-1$, then$$\frac{X-Y}{Y}=Z\Rightarrow X-Y=YZ\Rightarrow X=Y(Z+1)\Rightarrow Y=\frac{X}{Z+1}.$$ Note that $X\not=0$ because $Y\not=0$.
If $Z=-1$, then $$\frac{X-Y}{Y}=Z\iff X=Y(Z+1)=0\ \text{and}\ Y\not=0.$$ Hence, $Y$ can be any real number except $0$.
As a result, since $X=0\iff Z=-1$, $$Y=\begin{cases}\frac{X}{Z+1}&\text{if X\not=0\ \text{and}\ Z\not=-1}\\\text{any real number except 0}&\text{if X=0\ \text{and}\ Z=-1}\end{cases}$$
• If $Z+1 = 0 \Rightarrow X=0$ Sep 30, 2014 at 11:47
If a variable is present more than once (the $Y$ in the proposed eq.) then, as a general rule, you should start by removing any fractions and multiplying out any brackets.
Since $Y$ composes the denominator, start by multiplying both sides by $Y$ so that it cancels out on the left: $$Y\cdot\frac{(X-Y)}{Y}=Z\cdot Y\;\to\;X-Y=Z\cdot Y$$ The next step is to put all the terms with $Y$ on the same side of the equation; it can be done by adding $Y$ to both sides: $$X-Y+Y=Z\cdot Y+Y\;\to\;X=Z\cdot Y+Y$$ Now the right hand side can be factorized by the common factor $Y$: $$X=Z\cdot Y+Y=Z\cdot Y+1\cdot Y=Y\cdot(Z+1)$$ Then, by dividing both sides by the factor $(Z+1)$ we isolate our variable: $$Y=\frac{X}{Z+1}$$ on a last note, think of $(Z+1)$ as one thing by keeping it in the bracket.
As an advice, always do one operation as you solve an equation and always do it to both sides, because more complex equations might involve squaring and square-rooting too and other operations. Concentrate on one passage at a time, make a note on what you're doing (adding $Y$, dividing by $(Z+1)$ et similia) and you should get the solution eventually.
• Err...much better obviously...
– Paul
Sep 30, 2014 at 19:03
• Much better answer in my opinion, explains the method's steps in detail, using the equation as an example. Sep 30, 2014 at 20:25
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# 10.2: Test of Independence
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Understand how to draw data needed to perform calculations when running the chi-square test from contingency tables.
• Run the test of independence to determine whether two variables are independent or not.
• Use the test of homogeneity to examine the proportions of a variable attributed to different populations.
## Introduction
As mentioned in the previous lesson, the chi-square test can be used to both estimate how closely an observed distribution matches an expected distribution (the goodness-of-fit test) and to estimate whether two random variables are independent of one another (the test of independence). In this lesson, we will examine the test of independence in greater detail.
The chi-square test of independence is used to assess if two factors are related. This test is often used in social science research to determine if factors are independent of each other. For example, we would use this test to determine relationships between voting patterns and race, income and gender, and behavior and education.
In general, when running the test of independence, we ask, “Is Variable X\begin{align*}X\end{align*} independent of Variable Y\begin{align*}Y\end{align*}?” It is important to note that this test does not test how the variables are related, just simply whether or not they are independent of one another. For example, while the test of independence can help us determine if income and gender are independent, it cannot help us assess how one category might affect the other.
### Drawing Data from Contingency Tables Needed to Perform Calculations when Running a Chi-Square Test
Contingency tables can help us frame our hypotheses and solve problems. Often, we use contingency tables to list the variables and observational patterns that will help us to run a chi-square test. For example, we could use a contingency table to record the answers to phone surveys or observed behavioral patterns.
Example: We would use a contingency table to record the data when analyzing whether women are more likely to vote for a Republican or Democratic candidate when compared to men. In this example, we want to know if voting patterns are independent of gender. Hypothetical data for 76 females and 62 males from the state of California are in the contingency table below.
Frequency of California Citizens voting for a Republican or Democratic Candidate
Democratic Republican Total
Female 48 28 76
Male 36 26 62
Total 84 54 138
Similar to the chi-square goodness-of-fit test, the test of independence is a comparison of the differences between observed and expected values. However, in this test, we need to calculate the expected value using the row and column totals from the table. The expected value for each of the potential outcomes in the table can be calculated using the following formula:
Expected Frequency=(Row Total)(Column Total)Total Number of Observations\begin{align*}\text{Expected Frequency}=\frac{(\text{Row Total})(\text{Column Total})}{\text{Total Number of Observations}}\end{align*}
In the table above, we calculated the row totals to be 76 females and 62 males, while the column totals are 84 Democrats and 54 Republicans. Using the formula, we find the following expected frequencies for the potential outcomes:
The expected frequency for female Democratic outcome is 7684138=46.26\begin{align*}76 \bullet \frac{84}{138} = 46.26\end{align*}.
The expected frequency for female Republican outcome is 7654138=29.74\begin{align*}76 \bullet \frac{54}{138} = 29.74\end{align*}.
The expected frequency for male Democratic outcome is 6284138=37.74\begin{align*}62 \bullet \frac{84}{138} = 37.74\end{align*}.
The expected frequency for male Republican outcome is 6254138=24.26\begin{align*}62 \bullet \frac{54}{138} = 24.26\end{align*}.
Using these calculated expected frequencies, we can modify the table above to look something like this:
Democratic Democratic Republican Republican Total
Observed Expected Observed Expected
Female 48 46.26 28 29.74 76
Male 36 37.74 26 24.26 62
Total 84 54 138
With the figures above, we are able to calculate the chi-square statistic with relative ease.
### The Chi-Square Test of Independence
When running the test of independence, we use similar steps as when running the goodness-of-fit test described earlier. First, we need to establish a hypothesis based on our research question. Using our scenario of gender and voting patterns, our null hypothesis is that there is not a significant difference in the frequencies with which females vote for a Republican or Democratic candidate when compared to males. Therefore, our hypotheses can be stated as follows:
Null Hypothesis
H0:O=E\begin{align*}H_0:O=E\end{align*} (There is no statistically significant difference between the observed and expected frequencies.)
Alternative Hypothesis
Ha:OE\begin{align*}H_a:O \neq E\end{align*} (There is a statistically significant difference between the observed and expected frequencies.)
Using the table above, we can calculate the degrees of freedom and the chi-square statistic. The formula for calculating the chi-square statistic is the same as before:
χ2=(OE)2E\begin{align*}\chi^2=\sum_{} \frac{(O_{}-E_{})^2}{E_{}}\end{align*}
where:
χ2\begin{align*}\chi^2\end{align*} is the chi-square test statistic.
O\begin{align*}O_{}\end{align*} is the observed frequency value for each event.
E\begin{align*}E_{}\end{align*} is the expected frequency value for each event.
Using this formula and the example above, we get the following expected frequencies and chi-square statistic:
Democratic Democratic Democratic Republican Republican Republican
Obs. Freq. Exp. Freq. (OE)2E\begin{align*}\frac{(O-E)^2}{E}\end{align*} Obs. Freq. Exp. Freq. (OE)2E\begin{align*}\frac{(O-E)^2}{E}\end{align*}
Female 48 46.26 0.07 28 29.74 0.10
Male 36 37.74 0.08 26 24.26 0.12
Totals 84 54
χ2=0.07+0.08+0.10+0.12=0.37\begin{align*}\chi^2=0.07+0.08+0.10+0.12=0.37\end{align*}
Also, the degrees of freedom can be calculated from the number of Columns ("C") and the number of Rows ("R") as follows:
df=(C1)(R1)=(21)(21)=1\begin{align*}df &= (C-1)(R-1)\\ &= (2-1)(2-1)=1\end{align*}
With an alpha level of 0.05, we look under the column for 0.05 and the row for degrees of freedom, which, again, is 1, in the standard chi-square distribution table (http://tinyurl.com/3ypvj2h). According to the table, we see that the critical value for chi-square is 3.841. Therefore, we would reject the null hypothesis if the chi-square statistic is greater than 3.841.
Since our calculated chi-square value of 0.37 is less than 3.841, we fail to reject the null hypothesis. Therefore, we can conclude that females are not significantly more likely to vote for a Republican or Democratic candidate than males. In other words, these two factors appear to be independent of one another.
On the Web
http://tinyurl.com/39lhc3y A chi-square applet demonstrating the test of independence.
### Test of Homogeneity
The chi-square goodness-of-fit test and the test of independence are two ways to examine the relationships between categorical variables. To determine whether or not the assignment of categorical variables is random (that is, to examine the randomness of a sample), we perform the test of homogeneity. In other words, the test of homogeneity tests whether samples from populations have the same proportion of observations with a common characteristic. For example, we found in our last test of independence that the factors of gender and voting patterns were independent of one another. However, our original question was if females were more likely to vote for a Republican or Democratic candidate when compared to males. We would use the test of homogeneity to examine the probability that choosing a Republican or Democratic candidate was the same for females and males.
Another commonly used example of the test of homogeneity is comparing dice to see if they all work the same way.
Example: The manager of a casino has two potentially loaded dice that he wants to examine. (Loaded dice are ones that are weighted on one side so that certain numbers have greater probabilities of showing up.) The manager rolls each of the dice exactly 20 times and comes up with the following results:
Number Rolled with the Potentially Loaded Dice
1 2 3 4 5 6 Totals
Die 1 6 1 2 2 3 6 20
Die 2 4 1 3 3 1 8 20
Totals 10 2 5 5 4 14 40
Like the other chi-square tests, we first need to establish a null hypothesis based on a research question. In this case, our research question would be something like, “Is the probability of rolling a specific number the same for Die 1 and Die 2?” This would give us the following hypotheses:
Null Hypothesis
H0:O=E\begin{align*}H_0:O=E\end{align*} (The probabilities are the same for both dice.)
Alternative Hypothesis
Ha:OE\begin{align*}H_a:O \neq E\end{align*} (The probabilities differ for both dice.)
Similar to the test of independence, we need to calculate the expected frequency for each potential outcome and the total number of degrees of freedom. To get the expected frequency for each potential outcome, we use the same formula as we used for the test of independence, which is as follows:
Expected Frequency=(Row Total)(Column Total)Total Number of Observations\begin{align*}\text{Expected Frequency}=\frac{(\text{Row Total})(\text{Column Total})}{\text{Total Number of Observations}}\end{align*}
The following table includes the expected frequency (in parenthesis) for each outcome, along with the chi-square statistic, χ2=(OE)2E\begin{align*}\chi^2=\frac{(O-E)^2}{E}\end{align*}, in a separate column:
Number Rolled on the Potentially Loaded Dice
1 χ2\begin{align*}\chi^2\end{align*} 2 χ2\begin{align*}\chi^2\end{align*} 3 χ2\begin{align*}\chi^2\end{align*} 4 χ2\begin{align*}\chi^2\end{align*} 5 χ2\begin{align*}\chi^2\end{align*} 6 χ2\begin{align*}\chi^2\end{align*} χ2\begin{align*}\chi^2\end{align*} Total
Die 1 6(5) 0.2 1(1) 0 2(2.5) 0.1 2(2.5) 0.1 3(2) 0.5 6(7) 0.14 1.04
Die 2 4(5) 0.2 1(1) 0 3(2.5) 0.1 3(2.5) 0.1 1(2) 0.5 8(7) 0.14 1.04
Totals 10 2 5 5 4 14 2.08
df=(C1)(R1)=(61)(21)=5\begin{align*}df &= (C-1)(R-1)\\ &= (6-1)(2-1)=5\end{align*}
From the table above, we can see that the value of the test statistic is 2.08.
Using an alpha level of 0.05, we look under the column for 0.05 and the row for degrees of freedom, which, again, is 5, in the standard chi-square distribution table. According to the table, we see that the critical value for chi-square is 11.070. Therefore, we would reject the null hypothesis if the chi-square statistic is greater than 11.070.
Since our calculated chi-square value of 2.08 is less than 11.070, we fail to reject the null hypothesis. Therefore, we can conclude that each number is just as likely to be rolled on one die as on the other. This means that if the dice are loaded, they are probably loaded in the same way or were made by the same manufacturer.
## Lesson Summary
The chi-square test of independence is used to assess if two factors are related. It is commonly used in social science research to examine behaviors, preferences, measurements, etc.
As with the chi-square goodness-of-fit test, contingency tables help capture and display relevant information. For each of the possible outcomes in the table constructed to run a chi-square test, we need to calculate the expected frequency. The formula used for this calculation is as follows:
Expected Frequency=(Row Total)(Column Total)Total Number of Observations\begin{align*}\text{Expected Frequency}=\frac{(\text{Row Total})(\text{Column Total})}{\text{Total Number of Observations}}\end{align*}
To calculate the chi-square statistic for the test of independence, we use the same formula as for the goodness-of-fit test. If the calculated chi-square value is greater than the critical value, we reject the null hypothesis.
We perform the test of homogeneity to examine the randomness of a sample. The test of homogeneity tests whether various populations are homogeneous or equal with respect to certain characteristics.
For a discussion of the four different scenarios for use of the chi-square test (19.0), see American Public University, Test Requiring the Chi-Square Distribution (4:13).
For an example of a chi-square test for homogenity (19.0), see APUS07, Example of a Chi-Square Test of Homogenity (7:57).
For an example of a chi-square test for independence with the TI-83/84 Calculator (19.0), see APUS07, Example of a Chi-Square Test of Independence Using a Calculator (3:29).
## Review Questions
1. What is the chi-square test of independence used for?
2. True or False: In the test of independence, you can test if two variables are related, but you cannot test the nature of the relationship itself.
3. When calculating the expected frequency for a possible outcome in a contingency table, you use the formula:
1. Expected Frequency=(Row Total)(Column Total)Total Number of Observations\begin{align*}\text{Expected Frequency} = \frac{(\text{Row Total})(\text{Column Total})}{\text{Total Number of Observations}}\end{align*}
2. Expected Frequency=(Total Observations)(Column Total)Row Total\begin{align*}\text{Expected Frequency} = \frac{(\text{Total Observations})(\text{Column Total})}{\text{Row Total}}\end{align*}
3. Expected Frequency=(Total Observations)(Row Total)Column Total\begin{align*}\text{Expected Frequency} = \frac{(\text{Total Observations})(\text{Row Total})}{\text{Column Total}}\end{align*}
4. Use the table below to answer the following review questions.
Research Question: Are females at UC Berkeley more likely to study abroad than males?
Females 322 460
Males 128 152
(a) What is the total number of females in the sample?
450
280
612
782
(b) What is the total number of observations in the sample?
782
533
1,062
612
(c) What is the expected frequency for the number of males who did not study abroad?
161
208
111
129
(d) How many degrees of freedom are in this example?
1
2
3
4
(e) True or False: Our null hypothesis would be that females are as likely as males to study abroad.
(f) What is the chi-square statistic for this example?
1.60
2.45
3.32
3.98
1. If the chi-square critical value at 0.05 and 1 degree of freedom is 3.81, and we have a calculated chi-square statistic of 2.22, we would:
1. reject the null hypothesis
2. fail to reject the null hypothesis
2. True or False: We use the test of homogeneity to evaluate the equality of several samples of certain variables.
3. The test of homogeneity is carried out the exact same way as:
1. the goodness-of-fit test
2. the test of independence
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# Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28
Balbharti Maharashtra Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28 Textbook Exercise Important Questions and Answers.
## Maharashtra State Board Class 5 Maths Solutions Chapter 7 Circles Problem Set 28
Question 1.
Draw circles with the radii given below.
(1) 2 cm
(2) 4 cm
(3) 3 cm
Question 2.
Draw a circle of any radius. Show one diameter, one radius and one chord on that circle.
In a circle, P is the centre.
AB is a diameter.
MN is a chord
Study the circle given alongside. Think over the following questions.
• Which are the radii in the circle?
• How many radii make up diameter AB?
• If the length of one radius is 3 cm, what is the length of the diameter?
• How long is the diameter as compared to the radius?
The diameter of a circle is twice the length of its radius.
• If another diameter CD is drawn on the same circle, will its length be the same as that of AB?
All the diameters of a circle are of the same length.
Test 1 :
Measure the diameters and radii of the circles given below with a ruler and verify the relationship between their lengths.
Test 2 :
1. Draw a circle on a piece of paper and cut it out.
2. Name the centre of the circle P.
3. Draw the diameter of the circle and name it AB. Note that PA and PB are radii of the circle.
4. Fold the circular paper along AB as shown in the picture.
Fold the paper at P in such a way that point B will fall on point A. Radius PB falls exactly on radius PA. In other words, they coincide.
From this, we can see that every radius of a circle is half the length of its diameter.
Question 1.
Draw circles with the radii given below:
(1) 1.5 cm
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# How do you use synthetic division to show that x=sqrt2 is a zero of x^3+2x^2-2x-4=0?
Jan 24, 2017
By showing that the division gives no remainder, i.e. $r \left(x\right) = 0$.
The basic setup of synthetic division is:
ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "" "" "" "" "" "" "" "" "" ")
where the coefficients correspond to the function ${x}^{3} + 2 {x}^{2} - 2 x - 4$. Since we are dividing by a linear factor, $x - \sqrt{2}$, we should get a quadratic back.
(Note that the root you use in your divisor would be the root you acquire from $x - r = 0$.)
The general steps are:
1. Bring down the first coefficient.
2. Multiply the result beneath the horizontal line by the root and store in the next column above the horizontal line.
4. Repeat 2-3 until you reach the last column and have evaluated the final addition.
You should then get:
ul(sqrt2" ") "|"" "" "1" "" "2" "" "-2" "" "" "-4
" "+ " "ul(" "" "" "" "sqrt2" "(2^"3/2" + 2)" "2^("1/2" + "3/2")" "" ")
$\text{ "" "" "" "1" "(2+sqrt2)" "2^"3/2"" "" "" } 0$
$= \textcolor{b l u e}{{x}^{2} + \left(2 + \sqrt{2}\right) x + {2}^{\text{3/2}}}$
$\frac{{x}^{3} + 2 {x}^{2} - 2 x - 4}{x - \sqrt{2}} = \stackrel{p \left(x\right) \text{/"q(x))overbrace(x^2 + (2 + sqrt2)x + 2^"3/2}}{+} \stackrel{r \left(x\right)}{\overbrace{\frac{0}{x - \sqrt{2}}}}$
And you can check to see that it properly expands to give the original cubic. Furthermore, we now see that $r \left(x\right) = 0$, so it is evident that $x = \sqrt{2}$ is a root of ${x}^{3} + 2 {x}^{2} - 2 x - 4$.
CHALLENGE: Can you show that $- \sqrt{2}$ is also a root?
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# Unit 3 Part 2 Writing Equations Ax + By = C (standard form) y = mx + b (slope – intercept form)
## Presentation on theme: "Unit 3 Part 2 Writing Equations Ax + By = C (standard form) y = mx + b (slope – intercept form)"— Presentation transcript:
Unit 3 Part 2 Writing Equations Ax + By = C (standard form) y = mx + b (slope – intercept form)
y = mx + b is a form of an equation What is the equation of a line that has a slope of ½ and a y-intercept of -3 ? y = mx + b y = ½ x – 3 Don’t forget the ‘x’
Try these on your own. What is the equation of a line that has a slope of 2 and a y-intercept of -1 ? y = 2x – 1 What is the equation of a line that has a slope of ¾ and a y-intercept of 2 ? y = ¾ x + 2
y = mx + b Write the equation of the line whose slope is 2 and contains the point ( 1, 5). 1 st substitute into the equation. y = mx + b 5 = (2)(1) + b 5 = 2 + b -2 -2 3 = b and ‘m’ = 2 y = 2x + 3
y = mx + b Write the equation of the line whose slope is 3 and contains the point ( 2, 8). 1 st substitute into the equation. y = mx + b 8 = (3)(2) + b 8 = 6 + b -6 -6 2 = b and ‘m’ = 3 y = 3x + 2
y = mx + b Write the equation of the line whose slope is -2 and contains the point ( 4, -9). 1 st substitute into the equation. y = mx + b -9 = (-2)(4) + b -9 = -8 + b +8 +8 -1 = b and ‘m’ = -2 y = -2x – 1
y = mx + b Write the equation of the line whose slope is ½ and contains the point ( - 4, -3). 1 st substitute into the equation. y = mx + b -3 = ( ½ )(- 4) + b -3 = -2 + b +2 +2 -1 = b and ‘m’ = ½ y = ½ x – 1
Write the equation of the line that contains these two points. ( 1, 3) ( 2, 6) 1 st Calculate the slope Now substitute y = mx + b 3 = 3(1) + b 3 = 3 + b -3 -3 0 = b y = 3x + 0 y = 3x 3 – 6 -3 1 – 2 -1 m = = = + 3
Write the equation of the line that contains these two points. ( 3, 4) ( 5, 10) 1 st Calculate the slope Now substitute y = mx + b 4 = 3(3) + b 4 = 9 + b -9 -9 -5 = b y = 3x – 5 4 – 10 -6 3 – 5 -2 m = = = + 3
Write the equation of the line that contains these two points. ( 4, -2) ( -8, 7) 1 st Calculate the slope Now substitute y = mx + b -2 = - ¾ (4) + b -2 = -3 + b +3 +3 +1 = b y = - ¾ x + 1 -2 – 7 -9 4 – (-8) 12 m = = = - 3 4
Write the equation of the line that contains these two points. ( -6, -3) ( 0,-5) 1 st Calculate the slope Now substitute y = mx + b -3 = - ⅓ (-6) + b -3 = 2 + b -2 -2 -5 = b y = - ⅓ x – 5 -3 –(-5) 2 -6 – 0 -6 m = = = - 1 3
Write the equation of the line that contains these two points. ( -2, -4) ( 2, -2) 1 st Calculate the slope Now substitute y = mx + b -4 = ½(-2) + b -4 = -1 + b +1 +1 -3 = b y = ½ x – 3 -4 – (-2) -2 -2 – 2 -4 m = = = + ½
y – y = m ( x – x) Formula Write the equation of the line that has a slope of 2 and passes through the point ( -6, 4). 1 st substitute into the formula. y – y = m ( x – x) y – 4 = 2 ( x – (-6)) y – 4 = 2 ( x + 6 ) y – 4 = 2x + 12 + 4 + 4 y = 2x + 16 Can you do it ? Let’s see!
y – y = m ( x – x) Formula Write the equation of the line that has a slope of -3 and passes through the point ( 1, -2). 1 st substitute into the formula. y – y = m ( x – x) y – (-2) = -3 ( x – 1) y + 2 = -3 ( x – 1 ) y + 2 = -3x +3 - 2 - 2 y = -3x + 1
y – y = m ( x – x) Formula Write the equation of the line that has a slope of ½ and passes through the point ( 4, -3). 1 st substitute into the formula. y – y = m ( x – x) y – (-3) = ½ ( x – 4) y + 3 = ½ ( x – 4 ) y + 3 = ½x - 2 - 3 - 3 y = ½ x - 5
y – y = m ( x – x) Formula Write the equation of the line that has a slope of -¾ and passes through the point ( 8, -1). 1 st substitute into the formula. y – y = m ( x – x) y – (-1) = -¾ ( x – 8) y + 1 = -¾ ( x – 8 ) y + 1 = -¾ x + 6 - 1 - 1 y = -¾ x + 5
y – y = m ( x – x) Formula Write the equation of the line that has a slope of 3 and passes through the point ( -3, 2). 1 st substitute into the formula. y – y = m ( x – x) y – 2 = 3 ( x – (-3)) y – 2 = 3 ( x + 3 ) y – 2 = 3x + 9 + 2 + 2 y = 3x + 11
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# Complex plane
In mathematics, the complex plane or z-plane is a geometric representation of the complex numbers established by the real axis and the perpendicular imaginary axis. It can be thought of as a modified Cartesian plane, with the real part of a complex number represented by a displacement along the x-axis, and the imaginary part by a displacement along the y-axis.[note 1]
The concept of the complex plane allows a geometric interpretation of complex numbers. Under addition, they add like vectors. The multiplication of two complex numbers can be expressed most easily in polar coordinates—the magnitude or modulus of the product is the product of the two absolute values, or moduli, and the angle or argument of the product is the sum of the two angles, or arguments. In particular, multiplication by a complex number of modulus 1 acts as a rotation.
The complex plane is sometimes known as the Argand plane.
## Notational conventions
In complex analysis, the complex numbers are customarily represented by the symbol z, which can be separated into its real (x) and imaginary (y) parts:
${\displaystyle z=x+iy}$
for example: z = 4 + 5i, where x and y are real numbers, and i is the imaginary unit. In this customary notation the complex number z corresponds to the point (x, y) in the Cartesian plane.
In the Cartesian plane the point (x, y) can also be represented in polar coordinates as
${\displaystyle (x,y)=(r\cos \theta ,r\sin \theta )\qquad (r,\theta )=\left({\sqrt {x^{2}+y^{2}}},\quad \arctan {\frac {y}{x}}\right).}$
In the Cartesian plane it may be assumed that the arctangent takes values from π/2 to π/2 (in radians), and some care must be taken to define the more complete arctangent function for points (x, y) when x ≤ 0.[note 2] In the complex plane these polar coordinates take the form
${\displaystyle z=x+iy=|z|\left(\cos \theta +i\sin \theta \right)=|z|e^{i\theta }}$
where
${\displaystyle |z|={\sqrt {x^{2}+y^{2}}};\quad \theta =\arg(z)={\frac {1}{i}}\ln {\frac {z}{|z|}}=-i\ln {\frac {z}{|z|}}.\,}$[note 3]
Here |z| is the absolute value or modulus of the complex number z; θ, the argument of z, is usually taken on the interval 0 ≤ θ < 2π; and the last equality (to |z|e) is taken from Euler's formula. Without the constraint on the range of θ, the argument of z is multi-valued, because the complex exponential function is periodic, with period 2π i. Thus, if θ is one value of arg(z), the other values are given by arg(z) = θ + 2, where n is any integer ≠ 0.[2]
While seldom used explicitly, the geometric view of the complex numbers is implicitly based on its structure of a Euclidean vector space of dimension 2, where the inner product of complex numbers w and z is given by ${\displaystyle \Re (w{\overline {z}})}$; then for a complex number z its absolute value |z| coincides with its Euclidean norm, and its argument arg(z) with the angle turning from 1 to z.
The theory of contour integration comprises a major part of complex analysis. In this context the direction of travel around a closed curve is important – reversing the direction in which the curve is traversed multiplies the value of the integral by 1. By convention the positive direction is counterclockwise. For example, the unit circle is traversed in the positive direction when we start at the point z = 1, then travel up and to the left through the point z = i, then down and to the left through 1, then down and to the right through i, and finally up and to the right to z = 1, where we started.
Almost all of complex analysis is concerned with complex functions – that is, with functions that map some subset of the complex plane into some other (possibly overlapping, or even identical) subset of the complex plane. Here it is customary to speak of the domain of f(z) as lying in the z-plane, while referring to the range or image of f(z) as a set of points in the w-plane. In symbols we write
${\displaystyle z=x+iy;\qquad f(z)=w=u+iv}$
and often think of the function f as a transformation from the z-plane (with coordinates (x, y)) into the w-plane (with coordinates (u, v)).
## Argand diagram
Argand diagram refers to a geometric plot of complex numbers as points z=x+iy using the x-axis as the real axis and y-axis as the imaginary axis.[3]. Such plots are named after Jean-Robert Argand (1768–1822), although they were first described by Norwegian–Danish land surveyor and mathematician Caspar Wessel (1745–1818).[note 4] Argand diagrams are frequently used to plot the positions of the zeros and poles of a function in the complex plane.
## Stereographic projections
It can be useful to think of the complex plane as if it occupied the surface of a sphere. Given a sphere of unit radius, place its center at the origin of the complex plane, oriented so that the equator on the sphere coincides with the unit circle in the plane, and the north pole is "above" the plane.
We can establish a one-to-one correspondence between the points on the surface of the sphere minus the north pole and the points in the complex plane as follows. Given a point in the plane, draw a straight line connecting it with the north pole on the sphere. That line will intersect the surface of the sphere in exactly one other point. The point z = 0 will be projected onto the south pole of the sphere. Since the interior of the unit circle lies inside the sphere, that entire region (|z| < 1) will be mapped onto the southern hemisphere. The unit circle itself (|z| = 1) will be mapped onto the equator, and the exterior of the unit circle (|z| > 1) will be mapped onto the northern hemisphere, minus the north pole. Clearly this procedure is reversible – given any point on the surface of the sphere that is not the north pole, we can draw a straight line connecting that point to the north pole and intersecting the flat plane in exactly one point.
Under this stereographic projection the north pole itself is not associated with any point in the complex plane. We perfect the one-to-one correspondence by adding one more point to the complex plane – the so-called point at infinity – and identifying it with the north pole on the sphere. This topological space, the complex plane plus the point at infinity, is known as the extended complex plane. We speak of a single "point at infinity" when discussing complex analysis. There are two points at infinity (positive, and negative) on the real number line, but there is only one point at infinity (the north pole) in the extended complex plane.[5]
Imagine for a moment what will happen to the lines of latitude and longitude when they are projected from the sphere onto the flat plane. The lines of latitude are all parallel to the equator, so they will become perfect circles centered on the origin z = 0. And the lines of longitude will become straight lines passing through the origin (and also through the "point at infinity", since they pass through both the north and south poles on the sphere).
This is not the only possible yet plausible stereographic situation of the projection of a sphere onto a plane consisting of two or more values. For instance, the north pole of the sphere might be placed on top of the origin z = −1 in a plane that is tangent to the circle. The details don't really matter. Any stereographic projection of a sphere onto a plane will produce one "point at infinity", and it will map the lines of latitude and longitude on the sphere into circles and straight lines, respectively, in the plane.
## Cutting the plane
When discussing functions of a complex variable it is often convenient to think of a cut in the complex plane. This idea arises naturally in several different contexts.
### Multi-valued relationships and branch points
Consider the simple two-valued relationship
${\displaystyle w=f(z)=\pm {\sqrt {z}}=z^{1/2}.}$
Before we can treat this relationship as a single-valued function, the range of the resulting value must be restricted somehow. When dealing with the square roots of non-negative real numbers this is easily done. For instance, we can just define
${\displaystyle y=g(x)={\sqrt {x}}\ =x^{1/2}}$
to be the non-negative real number y such that y2 = x. This idea doesn't work so well in the two-dimensional complex plane. To see why, let's think about the way the value of f(z) varies as the point z moves around the unit circle. We can write
${\displaystyle z=re^{i\theta }\quad {\mbox{and take}}\quad w=z^{1/2}={\sqrt {r}}\,e^{i\theta /2}\qquad (0\leq \theta \leq 2\pi ).}$
Evidently, as z moves all the way around the circle, w only traces out one-half of the circle. So one continuous motion in the complex plane has transformed the positive square root e0 = 1 into the negative square root e = 1.
This problem arises because the point z = 0 has just one square root, while every other complex number z ≠ 0 has exactly two square roots. On the real number line we could circumvent this problem by erecting a "barrier" at the single point x = 0. A bigger barrier is needed in the complex plane, to prevent any closed contour from completely encircling the branch point z = 0. This is commonly done by introducing a branch cut; in this case the "cut" might extend from the point z = 0 along the positive real axis to the point at infinity, so that the argument of the variable z in the cut plane is restricted to the range 0 ≤ arg(z) < 2π.
We can now give a complete description of w = z½. To do so we need two copies of the z-plane, each of them cut along the real axis. On one copy we define the square root of 1 to be e0 = 1, and on the other we define the square root of 1 to be e = 1. We call these two copies of the complete cut plane sheets. By making a continuity argument we see that the (now single-valued) function w = z½ maps the first sheet into the upper half of the w-plane, where 0 ≤ arg(w) < π, while mapping the second sheet into the lower half of the w-plane (where π ≤ arg(w) < 2π).[6]
The branch cut in this example doesn't have to lie along the real axis. It doesn't even have to be a straight line. Any continuous curve connecting the origin z = 0 with the point at infinity would work. In some cases the branch cut doesn't even have to pass through the point at infinity. For example, consider the relationship
${\displaystyle w=g(z)=\left(z^{2}-1\right)^{1/2}.}$
Here the polynomial z2 1 vanishes when z = ±1, so g evidently has two branch points. We can "cut" the plane along the real axis, from 1 to 1, and obtain a sheet on which g(z) is a single-valued function. Alternatively, the cut can run from z = 1 along the positive real axis through the point at infinity, then continue "up" the negative real axis to the other branch point, z = 1.
This situation is most easily visualized by using the stereographic projection described above. On the sphere one of these cuts runs longitudinally through the southern hemisphere, connecting a point on the equator (z = 1) with another point on the equator (z = 1), and passing through the south pole (the origin, z = 0) on the way. The second version of the cut runs longitudinally through the northern hemisphere and connects the same two equatorial points by passing through the north pole (that is, the point at infinity).
### Restricting the domain of meromorphic functions
A meromorphic function is a complex function that is holomorphic and therefore analytic everywhere in its domain except at a finite, or countably infinite, number of points.[note 5] The points at which such a function cannot be defined are called the poles of the meromorphic function. Sometimes all these poles lie in a straight line. In that case mathematicians may say that the function is "holomorphic on the cut plane". Here's a simple example.
The gamma function, defined by
${\displaystyle \Gamma (z)={\frac {e^{-\gamma z}}{z}}\prod _{n=1}^{\infty }\left[\left(1+{\frac {z}{n}}\right)^{-1}e^{z/n}\right]}$
where γ is the Euler–Mascheroni constant, and has simple poles at 0, 1, 2, 3, ... because exactly one denominator in the infinite product vanishes when z is zero, or a negative integer.[note 6] Since all its poles lie on the negative real axis, from z = 0 to the point at infinity, this function might be described as "holomorphic on the cut plane, the cut extending along the negative real axis, from 0 (inclusive) to the point at infinity."
Alternatively, Γ(z) might be described as "holomorphic in the cut plane with π < arg(z) < π and excluding the point z = 0."
This cut is slightly different from the branch cut we've already encountered, because it actually excludes the negative real axis from the cut plane. The branch cut left the real axis connected with the cut plane on one side (0 ≤ θ), but severed it from the cut plane along the other side (θ < 2π).
Of course, it's not actually necessary to exclude the entire line segment from z = 0 to ∞ to construct a domain in which Γ(z) is holomorphic. All we really have to do is puncture the plane at a countably infinite set of points {0, 1, 2, 3, ...}. But a closed contour in the punctured plane might encircle one or more of the poles of Γ(z), giving a contour integral that is not necessarily zero, by the residue theorem. By cutting the complex plane we ensure not only that Γ(z) is holomorphic in this restricted domain – we also ensure that the contour integral of Γ over any closed curve lying in the cut plane is identically equal to zero.
### Specifying convergence regions
Many complex functions are defined by infinite series, or by continued fractions. A fundamental consideration in the analysis of these infinitely long expressions is identifying the portion of the complex plane in which they converge to a finite value. A cut in the plane may facilitate this process, as the following examples show.
Consider the function defined by the infinite series
${\displaystyle f(z)=\sum _{n=1}^{\infty }\left(z^{2}+n\right)^{-2}.}$
Since z2 = (z)2 for every complex number z, it's clear that f(z) is an even function of z, so the analysis can be restricted to one half of the complex plane. And since the series is undefined when
${\displaystyle z^{2}+n=0\quad \Leftrightarrow \quad z=\pm i{\sqrt {n}},}$
it makes sense to cut the plane along the entire imaginary axis and establish the convergence of this series where the real part of z is not zero before undertaking the more arduous task of examining f(z) when z is a pure imaginary number.[note 7]
In this example the cut is a mere convenience, because the points at which the infinite sum is undefined are isolated, and the cut plane can be replaced with a suitably punctured plane. In some contexts the cut is necessary, and not just convenient. Consider the infinite periodic continued fraction
${\displaystyle f(z)=1+{\cfrac {z}{1+{\cfrac {z}{1+{\cfrac {z}{1+{\cfrac {z}{\ddots }}}}}}}}.}$
It can be shown that f(z) converges to a finite value if and only if z is not a negative real number such that z < ¼. In other words, the convergence region for this continued fraction is the cut plane, where the cut runs along the negative real axis, from ¼ to the point at infinity.[8]
## Gluing the cut plane back together
We have already seen how the relationship
${\displaystyle w=f(z)=\pm {\sqrt {z}}=z^{1/2}}$
can be made into a single-valued function by splitting the domain of f into two disconnected sheets. It is also possible to "glue" those two sheets back together to form a single Riemann surface on which f(z) = z1/2 can be defined as a holomorphic function whose image is the entire w-plane (except for the point w = 0). Here's how that works.
Imagine two copies of the cut complex plane, the cuts extending along the positive real axis from z = 0 to the point at infinity. On one sheet define 0 ≤ arg(z) < 2π, so that 11/2 = e0 = 1, by definition. On the second sheet define 2π ≤ arg(z) < 4π, so that 11/2 = e = 1, again by definition. Now flip the second sheet upside down, so the imaginary axis points in the opposite direction of the imaginary axis on the first sheet, with both real axes pointing in the same direction, and "glue" the two sheets together (so that the edge on the first sheet labeled "θ = 0" is connected to the edge labeled "θ < 4π" on the second sheet, and the edge on the second sheet labeled "θ = 2π" is connected to the edge labeled "θ < 2π" on the first sheet). The result is the Riemann surface domain on which f(z) = z1/2 is single-valued and holomorphic (except when z = 0).[6]
To understand why f is single-valued in this domain, imagine a circuit around the unit circle, starting with z = 1 on the first sheet. When 0 ≤ θ < 2π we are still on the first sheet. When θ = 2π we have crossed over onto the second sheet, and are obliged to make a second complete circuit around the branch point z = 0 before returning to our starting point, where θ = 4π is equivalent to θ = 0, because of the way we glued the two sheets together. In other words, as the variable z makes two complete turns around the branch point, the image of z in the w-plane traces out just one complete circle.
Formal differentiation shows that
${\displaystyle f(z)=z^{1/2}\quad \Rightarrow \quad f^{\prime }(z)={\textstyle {\frac {1}{2}}}z^{-1/2}}$
from which we can conclude that the derivative of f exists and is finite everywhere on the Riemann surface, except when z = 0 (that is, f is holomorphic, except when z = 0).
How can the Riemann surface for the function
${\displaystyle w=g(z)=\left(z^{2}-1\right)^{1/2},}$
also discussed above, be constructed? Once again we begin with two copies of the z-plane, but this time each one is cut along the real line segment extending from z = 1 to z = 1 these are the two branch points of g(z). We flip one of these upside down, so the two imaginary axes point in opposite directions, and glue the corresponding edges of the two cut sheets together. We can verify that g is a single-valued function on this surface by tracing a circuit around a circle of unit radius centered at z = 1. Commencing at the point z = 2 on the first sheet we turn halfway around the circle before encountering the cut at z = 0. The cut forces us onto the second sheet, so that when z has traced out one full turn around the branch point z = 1, w has taken just one-half of a full turn, the sign of w has been reversed (since e = 1), and our path has taken us to the point z = 2 on the second sheet of the surface. Continuing on through another half turn we encounter the other side of the cut, where z = 0, and finally reach our starting point (z = 2 on the first sheet) after making two full turns around the branch point.
The natural way to label θ = arg(z) in this example is to set π < θπ on the first sheet, with π < θ ≤ 3π on the second. The imaginary axes on the two sheets point in opposite directions so that the counterclockwise sense of positive rotation is preserved as a closed contour moves from one sheet to the other (remember, the second sheet is upside down). Imagine this surface embedded in a three-dimensional space, with both sheets parallel to the xy-plane. Then there appears to be a vertical hole in the surface, where the two cuts are joined together. What if the cut is made from z = 1 down the real axis to the point at infinity, and from z = 1, up the real axis until the cut meets itself? Again a Riemann surface can be constructed, but this time the "hole" is horizontal. Topologically speaking, both versions of this Riemann surface are equivalent – they are orientable two-dimensional surfaces of genus one.
## Use of the complex plane in control theory
In control theory, one use of the complex plane is known as the 's-plane'. It is used to visualise the roots of the equation describing a system's behaviour (the characteristic equation) graphically. The equation is normally expressed as a polynomial in the parameter 's' of the Laplace transform, hence the name 's' plane. Points in the s-plane take the form ${\displaystyle s=\sigma +j\omega }$, where 'j' is used instead of the usual 'i' to represent the imaginary component.
Another related use of the complex plane is with the Nyquist stability criterion. This is a geometric principle which allows the stability of a closed-loop feedback system to be determined by inspecting a Nyquist plot of its open-loop magnitude and phase response as a function of frequency (or loop transfer function) in the complex plane.
The 'z-plane' is a discrete-time version of the s-plane, where z-transforms are used instead of the Laplace transformation.
The complex plane is associated with two distinct quadratic spaces. For a point z = x + iy in the complex plane, the squaring function z2 and the norm-squared ${\displaystyle x^{2}+y^{2}}$ are both quadratic forms. The former is frequently neglected in the wake of the latter's use in setting a metric on the complex plane. These distinct faces of the complex plane as a quadratic space arise in the construction of algebras over a field with the Cayley–Dickson process. That procedure can be applied to any field, and different results occur for the fields ℝ and ℂ: when ℝ is the take-off field, then ℂ is constructed with the quadratic form ${\displaystyle x^{2}+y^{2},}$ but the process can also begin with ℂ and z2, and that case generates algebras that differ from those derived from ℝ. In any case, the algebras generated are composition algebras; in this case the complex plane is the point set for two distinct composition algebras.
## Other meanings of "complex plane"
The preceding sections of this article deal with the complex plane in terms of a geometric representation of the complex numbers. Although this usage of the term "complex plane" has a long and mathematically rich history, it is by no means the only mathematical concept that can be characterized as "the complex plane". There are at least three additional possibilities.
1. Two-dimensional complex vector space, a "complex plane" in the sense that it is a two-dimensional vector space whose coordinates are complex numbers. See also: Complex affine space § Two dimensions.
2. (1 + 1)-dimensional Minkowski space, also known as the split-complex plane, is a "complex plane" in the sense that the algebraic split-complex numbers can be separated into two real components that are easily associated with the point (x, y) in the Cartesian plane.
3. The set of dual numbers over the reals can also be placed into one-to-one correspondence with the points (x, y) of the Cartesian plane, and represent another example of a "complex plane".
## Terminology
While the terminology "complex plane" is historically accepted, the object could be more appropriately named "complex line" as it is a 1-dimensional complex vector space.
## Notes
1. Although this is the most common mathematical meaning of the phrase "complex plane", it is not the only one possible. Alternatives include the split-complex plane and the dual numbers, as introduced by quotient rings.
2. A detailed definition of the complex argument in terms of the complete arctangent can be found at the description of the atan2 function.
3. All the familiar properties of the complex exponential function, the trigonometric functions, and the complex logarithm can be deduced directly from the power series for ${\displaystyle e^{z}}$. In particular, the principal value of ${\displaystyle \log r}$, where ${\displaystyle |r|=1}$, can be calculated without reference to any geometrical or trigonometric construction.[1]
4. Wessel's memoir was presented to the Danish Academy in 1797; Argand's paper was published in 1806.[4]
5. The infinite product for Γ(z) is uniformly convergent on any bounded region where none of its denominators vanish; therefore it defines a meromorphic function on the complex plane.[7]
6. When Re(z) > 0 this sum converges uniformly on any bounded domain by comparison with ζ(2), where ζ(s) is the Riemann zeta function.
## References
1. See (Whittaker & Watson 1927), Appendix.
2. See (Whittaker & Watson 1927), p. 10.
3. W., Weisstein, Eric. "Argand Diagram". mathworld.wolfram.com. Retrieved 19 April 2018.
4. See (Whittaker & Watson 1927), p. 9.
5. See (Flanigan 1983), p. 305.
6. See (Moretti 1964), pp. 113–119.
7. See (Whittaker & Watson 1927), pp. 235–236.
8. See (Wall 1948), p. 39.
#### Works Cited
• Flanigan, Francis J. (1983). Complex Variables: Harmonic and Analytic Functions. Dover. ISBN 0-486-61388-7.
• Moretti, Gino (1964). Functions of a Complex Variable. Prentice-Hall.
• Wall, H. S. (1948). Analytic Theory of Continued Fractions. D. Van Nostrand Company. Reprinted (1973) by Chelsea Publishing Company ISBN 0-8284-0207-8.
• Whittaker, E. T.; Watson, G. N. (1927). A Course in Modern Analysis (Fourth ed.). Cambridge University Press.
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Note: Do not rely on this information. It is very old.
# Trigonometry
Trigonometry derives its name from two Greek words meaning to measure a triangle and originally that was the object of the science. Now, however, its range is much wider; it deals with angles and their measures, with their geometrical properties, and with algebraical investigations in which they are concerned. Angles are measured in three different ways, according to the unit selected. In the English system a right angle is divided into 90 equal parts, each of which is called a degree, and the degree is the unit of angular measurement. The French divide the right angle into 100 parts, and denote the unit the grade. A third system, however, exists, which does not depend upon any arbitrary mode of division. Thus if O be the centre of a circle (Fig. 1), and the arc A B be taken, equal to the radius O, A, the angle A O B is the unit of this system, and is sometimes called a radian._ Any angle may be expressed in terms of this unit, and the result will be the circular measure of the angle. The radian is equivalent to 57.2957 . . . degrees. The number of radians contained in any angle is easily seen to be the length of the arc of any circle subtended by the angle (placed at the centre), divided by the length of the radius of the circle. Thus the circular measure of A 0 C is A C / O A.
Trigonometrical calculations with regard to an angle are generally made with the use of certain ratios between the sides of a right-angled triangle containing the angle, these ratios being known as the trigonometrical functions of the angle. In former times, however, these functions had a somewhat different meaning. Let P A be an arc of a circle whose centre is O (Fig. 2). Produce O P, draw P M perpendicular to O A, and draw A B, the tangent, at A. Let O C be perpendicular to O A, and let the tangent at C meet O P produced at T. Then P M was called the sine of the arc P A, 0 M its cosine, B A its tangent, T C its cotangent, O B its secant, O T its cosecant, and M A (or 1 - cosine) its versed sine. In modern trigonometry these terms are applied to the ratios between the lines of the triangleO P M containing the angle P O M. Thus P M / O P is the sine of the angle P 0 M, O M / O P is its cosine, P M / O M is its tangent, O M / P M its cotangent, O P / O M its secant, and O P / P M its cosecant. It is seen that the lengths of the lines in the old definitions varied with the circle taken, but the modern functions are independent of any radius. If the lengths of the lines representing the old function be divided by the length of the radius of the circle, we see that the old functions of the arc are transformed into the modern functions of the angle. The sine of the angle A is generally written sin A, and the square of this is denoted by sin 2A, similar abbreviations being used for the other functions.
The trigonometrical ratios are connected together by the formulae sin 2A + cos 2A = 1, sec 2A = 1 + tan 2A and cosec 2A = 1 + cot 2A, so that all the ratios of an angle can be expressed in terms of any one of them.
The angle P O C, the amount by which P O M differs from a right angle, is called the complement of P O M and P 0 A1; the difference between it and two right angles is called its supplement. The functions of an angle are always the co-functions of its complement, e.g. sin A = cos (90 - A), while the functions of an angle and its supplement are either equal or differ only in sign.
Geometrically, we can only find the functions of angles of 30°, 45°, 60°, and 72°, but formulae have been arrived at by which the functions of all angle A + B or A - B can be obtained from the functions of the two angles A and B. It follows that the functions of 2A, 3A, etc., and in the same way A/2, A/4, etc., can be found, and so the functions of any angle can be obtained. The solution of triangles is an important branch of trigonometry, on account of its application in mensuration and surveying, and different methods are used, according to the data given, a fundamental proposition connecting the sides and angles of any triangle being that the sides are proportional to the sides of the opposite angles. On the more theoretical side, trigonometry concerns itself with the summation of sines, whose terms are related to the trigonometrical ratios, with the expansion of certain expressions, such as the expansion of sin a and v cos a, in series of powers of a (given in circular measure), with the evaluation of pi (the ratio between circumference and diameter of a circle), and with innumerable other achievements of mathematical importance. Spherical trigonometry is that branch of the science which is applied to the investigation of triangles drawn upon a sphere, and finds its special application in astronomy.
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# Mean value theorem – Conditions, Formula, and Examples
The mean value theorem is an important application of derivatives. In fact, it is one of the most important and helpful tools in Calculus, so we need to understand the theorem and learn how we can apply it to different problems.
The mean value theorem helps us understand the relationship shared between a secant and tangent line that passes through a curve.
This theorem also influences the theorems that we have for evaluating first and second derivatives. Knowing about its origin, formula, and application can help us get a head start we need to excel in other Calculus topics.
In fact, the mean value theorem is sometimes coined as “one of the fundamental theorems in differential calculus.” We’ll learn the reason behind this in this article.
## What is the mean value theorem?
According to the mean value theorem, if the function, $f(x)$, is continuous for a closed interval, $[a, b]$, there is at least one point at $x = c$, where the tangent line passing through $f(x)$ will be parallel with the secant line that passes through the points, $(a, f(a))$ and $(b, f(b))$.
This graph is a good depiction of what we want to show in the mean value theorem. This means that the slopes of this pair of tangent and secant lines are equal, and we can use this to derive the equation we use to find the value of $f’(c)$ – the slope of the tangent line.
### Mean value theorem equation
Recall that the slope a line can be calculated using the formula, $m = \dfrac{y_2 – y_1}{x_2 – x _1}$, where $(x_1, y_1)$ and $(x_2, y_2)$ are the two given coordinate pairs. We can apply this to find the slope of the secant line that passes through the points, $(a, f(a))$ and $(b, f(b))$. Equate the result to the tangent line slope, and we’ll have the equation representing the mean value theorem.
\begin{aligned}m_{\text{tangent}} &= f'(c)\\m_{\text{secant}} &= \dfrac{f(b) – f(a)}{b -a}\\\\m_{\text{tangent}} &= m_{\text{secant}}\\f'(c) &= \dfrac{f(b) – f(a)}{b -a} \end{aligned}
This form comes in handy, especially when we want to find the average value of the rate of change of function given an interval we can work on. A special form of the mean value theorem is the Rolles theorem, and this occurs when the value of $f(x)$ is the same when $x = a$ and $x = b$.
### Understanding the Rolles theorem
The Rolles theorem is a special application of the mean value theorem. Recall that this theorem states that when $f(x)$ is continuous and differentiable within $[a, b]$, the rate of change between $f(a)$ and $f(b)$ is zero when $f(a) = f(b)$.
We can confirm this by using the mean value theorem:
• Find $f’(c)$ by taking the difference between $f(a)$ and $f(b)$.
• Divide the result by $b – a$.
Since $f(a) = f(b)$, we have zero at the numerator as shown below.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a}\\&=\dfrac{0}{b -a}\\&= 0 \end{aligned}
This shows that through the mean value theorem, we can confirm that $f’(c)$ or the change rate is equal to $0$ when the function’s end values are equal.
## How to use the mean value theorem?
We’ve shown you how we can use the mean value theorem to prove the Rolles theorem. There are other theorems in differential calculus where we’ll need to apply this important theorem, so we need to learn how we can use this theorem properly.
• Make sure to check if the function is continuous within the interval.
• Check if the function is differentiable within the interval as well.
• Apply the formula to attain our desired result.
For the second bullet, there are two ways for us to check if the function is differentiable. We’ll show you the two ways, and your preferred method will depend on whether you’ve mastered your derivative rules.
Using Limits Using Derivatives Check if the limit shown below exists.\begin{aligned}\lim_{h\rightarrow 0} \dfrac{f(x +h) – f(x)}{h}\end{aligned}If this limit exists, then the function is said to be differentiable. When given $f(x)$, use the different derivative rules to find the expression for $f’(x)$.
As you can see, the second route (using the different derivative rules) is much easier and will help the next step most of the time. For now, let us list down some important derivative rules we might need in this section. Of course, making sure that you know this by heart at this point makes a lot of difference.
Constant Rule \begin{aligned}\dfrac{d}{dx} c = 0\end{aligned} Power Rule \begin{aligned}\dfrac{d}{dx} x^n = nx^{n -1}\end{aligned} Constant Multiple Rule \begin{aligned}\dfrac{d}{dx} c\cdot f(x) = c \cdot f’(x)\end{aligned} Sum/Difference Rules \begin{aligned}\dfrac{d}{dx} f(x) \pm g(x) = f’(x) \pm g’(x)\end{aligned} Product Rule \begin{aligned}\dfrac{d}{dx} [f(x) \cdot g(x)] = f’(x) \cdot g(x) + g’(x) \cdot f(x)\end{aligned} Quotient Rule \begin{aligned}\dfrac{d}{dx} \left[\dfrac{f(x)}{g(x)}\right] =\dfrac{g(x)f’(x) – f(x) g’(x)}{[g(x)]^2}\end{aligned} Chain Rule \begin{aligned}\dfrac{d}{dx} f(g(x))= f’(g(x)) g’(x)\end{aligned}
These are just some of the rules that might come in handy in our examples but don’t worry, we’ll recall other important derivative rules when the need arises in our discussion.
Once we’ve confirmed that the function is continuous and differentiable, we can then apply the equation representing the mean value theorem.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a} \end{aligned}
We can then use this to apply the mean value theorem in whatever problem we’re given and derive the output we’re being asked.
### Using the mean value theorem to prove a corollary
We can try using the mean value theorem to prove important corollaries. Why don’t we prove the corollary about increasing and decreasing functions?
Let’s say we have a continuous and differentiable function, $f(x)$, within $[a, b]$. The corollary states that:
i) When $f’(x) >0$, the function is said to be increasing within the given interval.
ii)When $f’(x) <0$, the function is said to be decreasing within the given interval.
We can use the mean value theorem to prove this by contradiction. We’ll show how i) can be proven, and you can work on your own to show ii).
To prove the first point by contradiction, we can let $a$ and $b$ be part of the interval so that $a < b$ but $f(a) \geq f(b)$. This means that we assume that $f(x)$ is not increasing throughout the interval. It’s given that $f(x)$ is continuous and differentiable so that we can proceed with the mean value theorem’s equation.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a} \end{aligned}
We’ve estbalished that $f(a) \geq f(b)$, so we’re expecting $f(b) – f(a)$ to be less than or equation to $0$. In addition, we’ve assumed that $a <b$, so $b -a$ must be greater than zero. Let’s observe the nature of $f’(c)$ given our assumptions.
Sign of $\boldsymbol{f(b) – f(a)}$ 0 or Negative Sign of $\boldsymbol{b – a}$ Always Positive Sign of $\boldsymbol{\dfrac{f(b) – f(a)}{b – a}}$ 0 or Negative
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a} \\&\leq 0\end{aligned}
But, we want the derivative to be greater than zero, so for this this statement is a contradiction. For $f(x) >0$, we’ll need $f(b) \geq f(a)$, so $f(x)$ will have to be an increasing function.
This is just one of the many corollaries and theorems that rely on the mean value theorem. This shows how important it is for us to master this theorem and learn the common types of problems we might encounter and require to use the mean value theorem.
Example 1
If $c$ is within the interval, $[2, 4]$, find the value of $c$ so that $f’(c)$ represents the slope within the endpoints of $y= \dfrac{1}{2}x^2$.
Solution
The function, $y = \dfrac{1}{2}x^2$ is continuous throughout the the interval, $[2, 4]$. We want to find the value of $f’(c)$ to be within $[2, 4]$. Before we can find the value of $c$, we’ll have to confirm that the function is differentiable. In general, quadratic expressions are differentiable, and we can confirm $y = \dfrac{1}{2}x^2$’s nature by either:
• Evaluating limit of the expression,$\lim_{h\rightarrow 0} \dfrac{f(x +h) – f(x)}{h}$, where $f(x + h) = \dfrac{1}{2}(x + h)^2$ and $f(x) = \dfrac{1}{2}x^2$.
• Taking the derivative of $y = \dfrac{1}{2}x^2$ using the power and constant rule.
Just for this example, we can you both approaches – but for the rest, we’ll stick with finding the derivatives and you’ll see why.
ing Limits \begin{aligned}\lim_{h\rightarrow 0} \dfrac{f(x +h) – f(x)}{h} &= \lim_{h\rightarrow 0}\dfrac{\dfrac{1}{2}(x + h)^2 – \dfrac{1}{2}(x)^2}{h}\\&= \lim_{h\rightarrow 0}\dfrac{\dfrac{1}{2}(x^2 + 2xh + h) – \dfrac{1}{2}x^2}{h}\\&=\lim_{h\rightarrow 0} \dfrac{\cancel{\dfrac{1}{2}x^2} + xh + \dfrac{1}{2}h- \cancel{\dfrac{1}{2}x^2}}{h}\\&= \lim_{h\rightarrow 0}\dfrac{\cancel{h}\left(x+\dfrac{1}{2}h\right )}{\cancel{h}}\\&= x + \dfrac{1}{2}h\\&= x + \dfrac{1}{2}(0)\\&= x\end{aligned} Using Derivatives \begin{aligned}\dfrac{d}{dx} \dfrac{1}{2}x^2 &= \dfrac{1}{2} \dfrac{d}{dx} x^2\phantom{xxx}\color{green}\text{Constant Rule: } \dfrac{d}{dx} c\cdot f(x) = c\cdot f'(x)\\&=\dfrac{1}{2}(2)x^{2 -1}\phantom{xxx}\color{green}\text{Power Rule: } \dfrac{d}{dx} x^n = nx^{n -1}\\&= 1x^{2 -1}\\&= x\end{aligned}
See how using limits to confirm that a given function is differentiable is very tedious? This is why it pays to use the derivative rules you’ve learned in the past to speed up the process. From the two, we have $f’(x) = x$, so this means that $f’(c)$ is equal to $c$.
Using the mean value theorem’s equation, $f'(c) = \dfrac{f(b) – f(a)}{b -a}$, we can replace $f’(c)$ with $c$. We can use the intervals, $[2, 4]$, for $a$ and $b$, respectively.
\begin{aligned}\boldsymbol{f(x)}\end{aligned} \begin{aligned}\boldsymbol{a = 2}\end{aligned} \begin{aligned}f(a) &= \dfrac{1}{2}(2)^2\\&=\dfrac{1}{2} \cdot 4\\&= 2\end{aligned} \begin{aligned}\boldsymbol{b = 4}\end{aligned} \begin{aligned}f(b) &= \dfrac{1}{2}(4)^2\\&=\dfrac{1}{2} \cdot16\\&= 8\end{aligned}
Let’s use these values into the equation we’re given to find the value of $c$.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a}\\c &= \dfrac{8 – 2}{4 – 2}\\c &= \dfrac{6}{2}\\c&=3\end{aligned}
This means that for $c$ to be within the interval and still satisfy the mean value theorem, $c$ must be equal to $3$.
Example 2
Let’s say that we have a differentiable function, $f(x)$, so that $f’(x) \leq 4$ for all values of $x$. What is largest value that is possible for $f(18)$ if $f(12) = 30$?
Solution
We are given the following values from the problem:
• $a = 12$, $f(12) = 30$
• $b = 18$, $f(18) = ?$
From the mean value theorem, we know that there exists $c$ within the interval of $x \in [12, 18]$, so we can use the mean value theorem to find the value of $f(12)$.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b -a}\\a&= 12\\b&=18\\\\f'(c) &= \dfrac{f(18)- f(12)}{18 – 12}\\&= \dfrac{f(18) – 30}{6}\end{aligned}
Let’s isolate $f(18)$ on the left-hand side of the equation, as shown below.
\begin{aligned}f'(c) \cdot {\color{green} 6}&= \dfrac{f(18)- 30}{6}\cdot {\color{green} 6}\\6f'(c) &= f(18) -30\\f(18) &=6f'(c) + 30 \end{aligned}
Although the value of $c$ and $f’(c)$ are not given, we know the fact that $f’(x) \leq 4$ for all values of $x$, so we’re sure that $f(c) \leq 4$. Let’s use this to rewrite the relationship of $f(18)$ and $f(c)$ in terms of an inequality.
\begin{aligned}f(18) &= 6f'(c) + 30\\ f'(c) &\leq 4\\\\f(18) &\leq 6(4) + 30\\f(18) &\leq 24 + 30\\f(18) &\leq 54\end{aligned}
This means that the maximum possible value for $f(18)$ is $54$.
Example 3
Verify that the function, $f(x) = \dfrac{x}{x – 4}$, satisfies the conditions of the mean value theorem on the interval $[-2, 2]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
Solution
When given a rational function, it’s important to see if the interval is within its larger domain to confirm continuity. For $f(x) = \dfrac{x}{x – 4}$, we can see that as long as $x \neq 4$, all values of $x$ are valid. This means that $f(x)$ is continuous within the interval, $[-2, 2]$.
Now, we can find the derivative of $f(x) = \dfrac{x}{x – 4}$ and confirm its differentiability. We can begin by using the quotient rule, $\dfrac{d}{dx} \left[\dfrac{f(x)}{g(x)}\right] =\dfrac{g(x)f'(x) -f(x) g'(x)}{[g(x)]^2}$.
\begin{aligned}\dfrac{d}{dx} \dfrac{x}{x – 4} &= \dfrac{(x -4)\dfrac{d}{dx}x – x \dfrac{d}{dx} (x -4)}{(x -4)^2}\\&=\dfrac{(x – 4)(1) – x(1 – 0)}{(x – 4)^2}\\&= \dfrac{x – 4 – x}{(x – 4)^2}\\&= -\dfrac{4}{(x – 4)^2}\end{aligned}
We can see that it’s possible for us to find the derivative of $f(x)$ and with this, we have $f’(c) = -\dfrac{4}{(c – 4)^2}$.
We’ve shown that $f(x)$ meets the mean value theorem’s conditions, so we can use the concluding equation to find $c$.
• We can let $a = -2$ and $b = 2$, so we can substitute these values into $f(x)$ to find $f(-2)$ and $f(2)$.
• On the left-hand side of the equation, we can then replace $f’(c)$ with $-\dfrac{4}{(c – 4)^2}$.
\begin{aligned}f(-2)&=\dfrac{-2}{-2 – 4}\\&= \dfrac{1}{3}\\f(2) &= \dfrac{2}{2 – 4}\\&=-1\\\\f'(c) &= \dfrac{f(b) – f(a)}{b – a}\\-\dfrac{4}{(c -4)^2} &= \dfrac{-1 – \dfrac{1}{3}}{2 – (-2)}\\-\dfrac{4}{(c – 4)^2} &= \dfrac{-5 – (-1)}{5 – 2}\\-\dfrac{4}{(c – 4)^2}&= -\dfrac{1}{3}\end{aligned}
Cross-multiply then isolate $(c – 4)^2$ on one side of the equation. Solve for $c$ and only take note of the root of $c$ that will be within the interval of $[-2, 2]$.
\begin{aligned}-\dfrac{4}{(c -4)^2}&= -\dfrac{1}{3}\\12 &= (c – 4)^2\\c -4 &= \pm \sqrt{12}\\c&= 4 + 2\sqrt{3}\\&\approx 7.46\\c&= 4 – 2\sqrt{3}\\ &\approx 0.54\end{aligned}
This means that $c$ is equal to $4- 2\sqrt{3}$ or approximately, $0.54$.
Example 4
Verify that the function, $f(x) = \dfrac{x^4}{2 – \sqrt{x}}$, satisfies the conditions of the mean value theorem on the interval $[1, 4]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
Solution
First, let’s check if the function is continuous before we apply the mean value theorem. We can equate the denominator, $2 – \sqrt{x}$, to check for any restrictions on the function. Once we have the restricted values for $x$, check if these values are within the interval, $[1, 4]$.
\begin{aligned}2 – \sqrt{x} &\neq 0\\\sqrt{x} &\neq 2\\x&\neq 4\end{aligned}
This means that as long as the interval does not contain $x = 4$ within it, the function can be considered continuous for the said interval. Since $4 \in [1, 4]$, $f(x)$ is not continuous within the interval.
Hence, we can’t apply the mean value theorem for this problem since it is not continuous within the interval.
This problem is a good example of why we must always ensure the function meets the conditions for us to apply the mean value theorem.
Example 5
Verify that the function, $f(x) = 2x + \sqrt{x – 4}$, satisfies the conditions of the mean value theorem on the interval $[4, 8]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
Solution
Since we have a radical expression for $f(x)$, we have to ensure that the restricted values are outside the interval, $[4, 8]$. To check for restrictions, we have to ensure that $x – 4$ is greater than or equal to $0$.
\begin{aligned}x – 4 \geq 0 \\x \geq 4\end{aligned}
Since $x \geq 4$ and $x\in [4, 8]$, we can see that the function remains continuous within our given interval. This means that we can then find the derivative of the equation.
To do so, recall that $\sqrt{b} = b^{\frac{1}{2}}$, so we can rewrite $f(x)$ using this property to differentiate it faster.
• Apply the sum rule to differentiate the two terms, first making the process easier.
• Use the power rule to find the derivative of $2x$ and $\sqrt{x – 4} = (x – 4)^{\frac{1}{2}}$.
\begin{aligned} \dfrac{d}{dx} 2x + \sqrt{x – 4} &= \dfrac{d}{dx} 2x + \dfrac{d}{dx} \sqrt{x – 4}\\&= 2\dfrac{d}{dx} x + \dfrac{d}{dx} (x- 4)^{\frac{1}{2}}\\ &=2(1) + \dfrac{1}{2}(x – 4)^{1/2 – 1}\\&= 2 + \dfrac{1}{2} (x-4)^{-\frac{1}{2} }\\&= 2 + \dfrac{1}{2\sqrt{x – 4}}\end{aligned}
From this, we can say that the function is continuous and differentiable, so we can confirm that the conditions for the mean value theorem have been met.
Now that we have the expression for $f’(x)$, what we can do is find $f(4)$ and $f(8)$ then apply the mean value theorem’s equation.
\begin{aligned}f(4)&=2(4) + \sqrt{4 – 4}\\&=8 \\f(8)&=2(8) + \sqrt{8 – 4}\\&= 18 \\\\f'(c) &= \dfrac{f(b) – f(a)}{b – a}\\2 + \dfrac{1}{2\sqrt{c -4}} &= \dfrac{18 – 8}{8 – 4}\\2 + \dfrac{1}{2\sqrt{c -4}} &= \dfrac{10}{4}\\\dfrac{1}{2\sqrt{c -4}}&= \dfrac{1}{2}\end{aligned} Simplify the equation further and square $\sqrt{c – 4}$ once isolated on either side of the equation.
\begin{aligned}\dfrac{1}{\sqrt{c -4}}&= 1\\\sqrt{c – 4} &= 1\\(\sqrt{c – 4} )^2 &= (1)^2\\c – 4 &= \pm 1\\c &= 3, 5\end{aligned}
Choose the value of $c$ that is within the given interval, $[4, 8]$, so we have $c = 5$.
Example 6
Use the mean value theorem to show that $|\sin a – \sin b| \leq |a – b|$ for the interval, $0 \leq a < b \leq 2\pi$.
Solution
The sine function, $f(x) = \sin x$, is known to be continuous within the interval, $[0, 2\pi]$, and we’ve learned in the past that the derivative of $\sin x$ is equal to $\cos x$.
\begin{aligned}f’(x) = \cos x\end{aligned}
This shows that the function $f(x) = \sin x$ meets the conditions for the mean value theorem, so we can use the conclusion to establish the relationship shown below.
\begin{aligned}f'(c) &= \dfrac{f(b) – f(a)}{b – a}\\f'(c) &= \dfrac{\sin b – \sin a}{b – a}\\\cos c &= \dfrac{\sin b – \sin a}{b -a}\end{aligned}
Recall that the values of $\cos c$ will be within the range of $[-1, 1]$, so $|\cos c| \leq 1$. Since $\cos c$ is equal to $\dfrac{\sin b – \sin a}{b – a}$ using the mean value theorem, we have the following inequality:
\begin{aligned}\cos c &= \dfrac{\sin b – \sin a}{b -a}\\|\cos c| &= \left|\dfrac{\sin b – \sin a}{b -a}\right|&\leq 1\end{aligned}
Let’s focus on the inequality, $\left|\dfrac{\sin b – \sin a}{b -a}\right| \leq 1$.
\begin{aligned}\left|\dfrac{\sin b – \sin a}{b -a}\right|&\leq 1\\\dfrac{|\sin b – \sin a}{|b -a|} &\leq 1\\|\sin b – \sin a| &\leq |b – a|\end{aligned}
This confirms the statement we want to prove:$|\sin b – \sin a| \leq |b – a|$. Hence, we’ve shown how it’s possible to prove mathematical statements using the mean value theorem.
### Practice Questions
1. If $c$ is within the interval, $[4, 8]$, find the value of $c$ so that $f’(c)$ represents the slope within the endpoints of $y= -2x^3$.
2. Let’s say that we have a differentiable function, $f(x)$, so that $f’(x) \leq 12$ for all values of $x$. What is largest value that is possible for $f(20)$ if $f(15) = 60$?
3. Verify that the function, $f(x) = \dfrac{2x}{x – 3}$, satisfies the conditions of the mean value theorem on the interval $[-1, 1]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
4. Verify that the function, $f(x) = 3x + \sqrt{x – 9}$, satisfies the conditions of the mean value theorem on the interval $[9, 15]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
5. Verify that the function, $f(x) = \dfrac{x^3}{3 – 3\sqrt{x}}$, satisfies the conditions of the mean value theorem on the interval $[1, 5]$ and then determine the value/s, $c$, that can satisfy the conclusion of the theorem.
1. $\dfrac{4\sqrt{21}}{3} \approx 6.11$
2. $120$
3. $3 – 2\sqrt{2} \approx 0.17$
4. $\dfrac{21}{2} = 10.5$
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# What is 1/415 as a decimal?
## Solution and how to convert 1 / 415 into a decimal
1 / 415 = 0.002
1/415 or 0.002 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 represent numbers between integers, in some cases defining portions of whole numbers But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. Once we've decided the best way to represent the number, we can dive into how to convert 1/415 into 0.002
## 1/415 is 1 divided by 415
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 1 is being divided into 415. Think of this as our directions and now we just need to be able to assemble the project! The two parts of fractions are numerators and denominators. The numerator is the top number and the denominator is the bottom. And the line between is our division property. To solve the equation, we must divide the numerator (1) by the denominator (415). Here's how you set your equation:
### Numerator: 1
• Numerators are the portion of total parts, showed at the top of the fraction. Comparatively, 1 is a small number meaning you will have less parts to your equation. The bad news is that it's an odd number which makes it harder to covert in your head. Smaller numerators doesn't mean easier conversions. Let's look at the fraction's denominator 415.
### Denominator: 415
• Denominators represent the total parts, located at the bottom of the fraction. 415 is a large number which means you should probably use a calculator. But the bad news is that odd numbers are tougher to simplify. Unfortunately and odd denominator is difficult to simplify unless it's divisible by 3, 5 or 7. Ultimately, don't be afraid of double-digit denominators. Now let's dive into how we convert into decimal format.
## Converting 1/415 to 0.002
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 415 \enclose{longdiv}{ 1 }$$
Use long division to solve step one. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 415 \enclose{longdiv}{ 1.0 }$$
Uh oh. 415 cannot be divided into 1. So we will have to extend our division problem. Add a decimal point to 1, your numerator, and add an additional zero. This doesn't add any issues to our denominator but now we can divide 415 into 10.
### Step 3: Solve for how many whole groups you can divide 415 into 10
$$\require{enclose} 00.0 \\ 415 \enclose{longdiv}{ 1.0 }$$
How many whole groups of 415 can you pull from 10? 0 Multiply this number by 415, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.0 \\ 415 \enclose{longdiv}{ 1.0 } \\ \underline{ 0 \phantom{00} } \\ 10 \phantom{0}$$
If there is no remainder, you’re done! If you still have a remainder, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
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 each represent values in everyday life! Here are examples of when we should use each.
### When you should convert 1/415 into a decimal
Dollars & Cents - It would be silly to use 1/415 of a dollar, but it makes sense to have $0.0. USD is exclusively decimal format and not fractions. (Yes, yes, there was a 'half dollar' but the value is still$0.50)
### When to convert 0.002 to 1/415 as a fraction
Time - spoken time is used in many forms. But we don't say It's '2.5 o'clock'. We'd say it's 'half passed two'.
### Practice Decimal Conversion with your Classroom
• If 1/415 = 0.002 what would it be as a percentage?
• What is 1 + 1/415 in decimal form?
• What is 1 - 1/415 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.002 + 1/2?
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# Calculation Short tricks: Calculate Square Roots quickly!
By BYJU'S Exam Prep
Updated on: September 25th, 2023
Calculation tricks to calculate Square roots: In Quantitative Aptitude, calculation short tricks are required to solve the questions quickly and accurately. This minimizes the time of the calculation as well as boosts the confidence to attempt more questions.
Calculation tricks to calculate Square roots: In Quantitative Aptitude, calculation short tricks are required to solve the questions quickly and accurately. This minimizes the time of the calculation as well as boosts the confidence to attempt more questions. In the article, we are sharing the important short tricks to calculate the square roots of the numbers.
## Short tricks to calculate square roots
For this, at first, you should know the squares of numbers from 1 to 10, which is an easy task.
Number Square 1 1 2 4 3 9 4 1 6 5 2 5 6 3 6 7 4 9 8 6 4 9 8 1 10 10 0
Now, you just have to analyze the unit digits of both the number and the square. This way you will get to know the last digit of the square root by just looking at the unit digit of the given square number.
Square Number Unit Digit Square Root number Unit Digit 1 1 2 4 3 9 4 6 5 5 6 6 7 9 8 4 9 1 10 0
Do you find anything interesting from the above information?
Unit digit of square root numbers of the number
• ending with 1 and 9 is 1.
• ending with 3 and 7 is 9.
• ending with 4 and 6 is 6.
• ending with 5 and 0 is 5 and 0 respectively or the same as that of the number.
Now, let`s see how remembering the unit digit helps in helpful in calculating the squares.
### Steps to calculate the square numbers
Step 1: Divide the number into two parts. Make sure that on the right side of the part, there must be two digits.
Step 2: Check out the unit digit of the right side and find the unit digit of the square root number following the trick in the above table and write down the square root of the number smaller than or equal to the left digit.
Step 3: Now, if you got two digits at the right side, then there will be two cases:
1. If left digit X its successive digit > left part, then a smaller digit will be taken on the right side.
2. If left digit X its successive digit < left part, then larger digit will be there on the right side.
Let`s take some examples to have the better understanding of all the above-written steps.
Example 1: Find the square root of 441.
Sol. Step 1: Splitting it into two parts, it is 4 and 41.
4 41
↓ ↓
Step 2: [(2)2 = 4] 1 or 9 (See above table)
Step 3: 2 1 (2 x 3 > 4 , so smaller digit will be taken)
Answer: 21
Example 2: Find the square root of 1369.
Sol. Step 1: Splitting it into two parts, it is 13 and 69.
13 69
↓ ↓
Step 2: [(3)2 < 13] 3 or 7 (See above table)
Step 3: 3 7 (3 x 4 < 13 , so larger digit will be taken)
Answer: 37
Example 3: Find the square root of 11236.
Sol. Step 1: Splitting it into two parts, it is 112 and 36.
112 36
↓ ↓
Step 2: [(10)2 < 112] 4 or 6 (See above table)
Step 3: 10 6 (10 x 11 < 112 , so larger digit will be taken)
Answer: 106
Example 4: Find the square root of 43264.
Sol. Step 1: Splitting it into two parts, it is 432 and 64.
432 64
↓ ↓
Step 2: [(20)2 < 432] 2 or 8 (See above table)
Step 3: 20 8 (20 x 21 < 432 , so larger digit will be taken)
Answer: 208
Example 5: Find the square root of 50625.
Sol. Step 1: Splitting it into two parts, it is 506 and 25.
506 25
↓ ↓
Step 2: [(22)2 < 506] 5 (See above table)
Step 3: 22 5
Answer: 225
Example 6: Find the square root of 91204.
Sol. Step 1: Splitting it into two parts, it is 912 and 04.
912 04
↓ ↓
Step 2: [(30)2 < 912] 2 or 8 (See above table)
Step 3: 30 2 (30 x 31 > 912 , so smaller digit will be taken)
Answer: 302
In this way, you can calculate the square roots of the number easily. You don’t have to write all the steps in the exam. Just calculate the steps in your mind and answer quickly.
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# If the medians of a AABC intersect at G,
Question:
If the medians of a AABC intersect at G, then show that ar (ΔAGB) = ar (ΔAGC) = ar (ΔBGC) = 1/3 ar(ΔABC).
Thinking Process
Use the property that median of a triangle divides it into two triangles of equal area.
Further, apply above property by considering different triangles and prove the required result.
Solution:
Given In ΔABC, AD, BE and CF are medians and intersect at G.
To prove ar $(\Delta A G B)=\operatorname{ar}(\Delta A G C)=\operatorname{ar}(\Delta B G C)=\frac{1}{3} \operatorname{ar}(\Delta A B C)$
Proof We know that, a median of a triangle divides it into two triangles of equal area. In $\triangle A B C, A D$ is a median.
$\therefore \quad a r(\Delta A B D)=\operatorname{ar}(\triangle A C D)$ $\ldots(\mathrm{i})$
In $\triangle B G C, G D$ is a median.
$\therefore \quad a r(\Delta G B D)=\operatorname{ar}(\Delta G C D)$ ....(ii)
On subtracting Eq. (ii) from Eq. (i), we get
$\operatorname{ar}(\Delta A B D)-\operatorname{ar}(\Delta G B D)=\operatorname{ar}(\triangle A C D)-\operatorname{ar}(\Delta G C D)$
$\Rightarrow \quad \operatorname{ar}(\Delta A G B)=\operatorname{ar}(\triangle A G C)$ ....(iii)
Similarly,
$\operatorname{ar}(\Delta A G B)=\operatorname{ar}(\Delta B G C)$
From Eqs. (iii) and (iv).
$\operatorname{ar}(\Delta A G B)=\operatorname{ar}(\Delta B G C)=\operatorname{ar}(\Delta A G C)$ $\ldots(\mathrm{v})$
NOW,
$\operatorname{ar}(\Delta A B C)=\operatorname{ar}(\Delta A G B)+\operatorname{ar}(\Delta B G C)+\operatorname{ar}(\Delta A G C)$
$\Rightarrow \quad \operatorname{ar}(\Delta A B C)=\operatorname{ar}(\Delta A G B)+\operatorname{ar}(\Delta A G B)+\operatorname{ar}(\Delta A G B) \quad$ [from Eq. (v)]
$\Rightarrow \quad \operatorname{ar}(\Delta A B C)=3 \operatorname{ar}(\Delta A G B)$
$\Rightarrow \quad \operatorname{ar}(\Delta A G B)=\frac{1}{3} \operatorname{ar}(\Delta A B C)$ $\ldots(\mathrm{vi})$
From Eqs. (v) and (vi),
$\operatorname{ar}(\Delta B G C)=\frac{1}{3} \operatorname{ar}(\Delta A B C)$
and $\operatorname{ar}(\triangle A G C)=\frac{1}{3} \operatorname{ar}(\Delta A B C)$
Hence proved.
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# CBSE Class 9th Maths Important MCQs from Chapter 14 - Statistics (with Answers)
Get MCQs with answers for CBSE Class 9 Maths Chapter 14 - Statistics. These questions are based on important concepts that must be practiced for the upcoming CBSE Class 9th Term 1 Exam 2021-22.
Created On: Sep 23, 2021 00:34 IST
CBSE Class 9th Maths Important MCQs for Chapter 14 Statistics
MCQs for CBSE Class 9 Maths Chapter 14 - Statistics are provided here to help students prepare for the term 1 exam. Statistics chapter is included in the term 1 syllabus of CBSE Class 9 Maths. It carries a weightage of 6 marks for the term 1 exam. The MCQ questions provided by Jagran Josh will prove to be very helpful to revise important concepts and thus prepare well for the term end exam.
MCQ Questions & Answers from CBSE Class 9 Maths Chapter 14 Statistics are given below:
1. The number of times a particular items occur in a class interval called its:
a) Mean
b) Frequency
c) Cumulative frequency
d) Range
2. The mean wage of 150 labourers working in a factory running three shifts with 60, 40 and 50 labourers is Rs. 114. The mean wage of 60 labourers in the first shift is Rs.. 121.50 and that of 40 labourers working the second shift is Rs. 107.75, then the mean wage of those working in the third shift is:
a) Rs. 100
b) Rs. 110
c) Rs. 115.75
d) Rs. 120
3. Kavita obtained 16, 14, 18 and 20 marks (out of 25) in Maths in weekly tests in the month of Jan. 2000. The mean marks of Kavita is:
a) 16
b) 16.5
c) 17
d) 17.5
4. The range of the data:
25, 18, 20, 22, 16, 6, 17, 12, 30, 32, 10, 19, 8, 11, 20 is
a) 10
b) 15
c) 18
d) 26
5. The width of each o five continuous classes in a frequency distribution is 5 and lower class-limit of the lowest class is 10. The upper class-limit of the highest class is:
a) 15
b) 25
c) 35
d) 40
6. Let m be the mid-point and l be the upper class limit of a class in a continuous frequency distribution. The lower class limit of the class is:
a) 2m + l
b) 2ml
c) ml
d) m – 2l
7. In the class interval 10 – 20, 20 – 30, the number 20 is included in:
a) 10 – 20
b) 20 – 30
c) Both the intervals
d) None of these intervals
8. A grouped frequency distribution table with classes of equal sizes using;63-72 (72 included) as one of the class is constructed for the following data:
30, 32, 45, 54, 74, 78, 108, 112, 66, 76, 88,
40, 14, 20, 15, 35, 44, 66, 75, 84, 95, 96,
102, 110, 88, 74, 112, 14, 34, 44
The number of classes in the distribution will be:
a) 9
b) 10
c) 11
d) 12
9. To draw a histogram to represent the following frequency distribution:
Class interval 5-10 10-15 15-25 25-45 45-15 Frequency 6 12 10 8 15
the adjusted frequency for the class 25-45 is:
a) 6
b) 5
c) 3
d) 2
10. The mean of five numbers is 30. If one number is excluded, their mean becomes 28. The excluded number is:
a) 28
b) 30
c) 35
d) 38
11. If the mean of the observations:
x, x + 3, x + 5, x + 7, x + 10
is 9, the mean of the last three observations is:
12. There are 50 numbers. Each number is subtracted from 53 and the mean of the numbers so obtained is found to be –3.5. The mean of the given numbers is:
a) 46.5
b) 49.5
c) 53.5
d) 56.5
13. The mean factors of 24 is:
14. The range of the data 25.7, 16.3, 2.8, 21.7, 24.3, 22.7, 24.9 is:
a) 22
b) 22.9
c) 21.7
d) 20.5
15. If the class marks in a frequency distribution are 19.2, 26.5, 33.5, 40.5, then the class corresponding to the class mark 33.5 is:
a) 16-23
b) 23-30
c) 30-37
d) 37-41
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• very nice explanation Brillant
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# How do you expand (3x-2y)^5?
Jul 13, 2017
$243 {x}^{5} - 810 {x}^{4} y + 1080 {x}^{3} {y}^{2} - 720 {x}^{2} {y}^{3} + 240 x {y}^{4} - 32 {y}^{5}$
#### Explanation:
$\text{using the "color(blue)"binomial theorem}$
•color(white)(x)(x+y)^n=sum_(r=0)^n((n),(r))x^(n-r)y^r
"where" ((n),(r))=(n!)/(r!(n-r)!)
$\text{here " x=3x" and } y = - 2 y$
$\Rightarrow {\left(3 x - 2 y\right)}^{5}$
$= \left(\begin{matrix}5 \\ 0\end{matrix}\right) {\left(3 x\right)}^{5} {\left(- 2 y\right)}^{0} + \left(\begin{matrix}5 \\ 1\end{matrix}\right) {\left(3 x\right)}^{4} {\left(- 2 y\right)}^{1} + \left(\begin{matrix}5 \\ 2\end{matrix}\right) {\left(3 x\right)}^{3} {\left(- 2 y\right)}^{2} + \left(\begin{matrix}5 \\ 3\end{matrix}\right) {\left(3 x\right)}^{2} {\left(- 2 y\right)}^{3} + \left(\begin{matrix}5 \\ 4\end{matrix}\right) {\left(3 x\right)}^{1} {\left(- 2 y\right)}^{4} + \left(\begin{matrix}5 \\ 5\end{matrix}\right) {\left(3 x\right)}^{0} {\left(- 2 y\right)}^{5}$
$\text{we can obtain the binomial coefficients using the appropriate}$
$\text{row of "color(blue)"Pascal's triangle}$
$\text{for n = 5 the row of coefficients is}$
$1 \textcolor{w h i t e}{x} 5 \textcolor{w h i t e}{x} 10 \textcolor{w h i t e}{x} 10 \textcolor{w h i t e}{x} 5 \textcolor{w h i t e}{x} 1$
$= 1.243 {x}^{5} + 5.81 {x}^{4} \left(- 2 y\right) + 10.27 {x}^{3.} 4 {y}^{2} + 10.9 {x}^{2} \left(- 8 {y}^{3}\right) + 5.3 x .16 {y}^{4} + 1. - 32 {y}^{5}$
$= 243 {x}^{5} - 810 {x}^{4} y + 1080 {x}^{3} {y}^{2} - 720 {x}^{2} {y}^{3} + 240 x {y}^{4} - 32 {y}^{5}$
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Take a Diagnostic Exam - Determine Your Test Readiness - 5 Steps to a 5 AP Statistics 2017 (2016)
## 5 Steps to a 5 AP Statistics 2017 (2016)
### STEP 2
CHAPTER 3 Take a Diagnostic Exam
### Take a Diagnostic Exam
IN THIS CHAPTER
Summary: The following diagnostic exam begins with 40 multiple-choice questions. The diagnostic exam also includes five free-response questions and one investigative task much like those on the actual exam. All of these test questions have been written to approximate the coverage of material that you will see on the AP exam but are intentionally somewhat more basic than actual exam questions (which are more closely approximated by the Practice Exams at the end of the book). Once you are done with the exam, check your work against the given answers, which also indicate where you can find the corresponding material in the book. You will also be given a way to convert your score to a rough AP score.
Key Ideas
Practice the kind of questions you will be asked on the real AP Statistics exam.
Answer questions that approximate the coverage of topics on the real exam.
Determine your areas of strength and weakness.
AP Statistics Diagnostic Test
AP Statistics Diagnostic Test
SECTION I
Time: 1 hour and 30 minutes
Number of questions: 40
Directions: Use the answer sheet provided on the previous page. All questions are given equal weight. There is no penalty for unanswered questions. One point is earned for every correct answer. The use of a calculator is permitted in all parts of this test. You have 90 minutes for this part of the test.
1. Eighteen trials of a binomial random variable X are conducted. If the probability of success for any one trial is 0.4, write the mathematical expression you would need to evaluate to find P (X = 7). Do not evaluate.
2. Two variables, x and y , seem to be exponentially related. The natural logarithm of each y value is taken and the least-squares regression line of ln(y ) on x is determined to be ln(y ) = 3.2 + 0.42x . What is the predicted value of ywhen x = 7?
3. 464.05
4. 1380384.27
5. 521.35
6. 6.14
7. 1096.63
8. You need to construct a 94% confidence interval for a population proportion. What is the upper critical value of z to be used in constructing this interval?
9. 0.9699
10. 1.96
11. 1.555
12. –1.88
13. 1.88
14. Which of the following best describes the shape of the histogram at the left?
15. Approximately normal
16. Skewed left
17. Skewed right
18. Approximately normal with an outlier
19. Symmetric
20. The probability is 0.2 that a value selected at random from a normal distribution with mean 600 and standard deviation 15 will be above what number?
21. 0.84
22. 603.80
23. 612.6
24. 587.4
25. 618.8
26. Which of the following are examples of continuous data?
27. The speed your car goes
28. The number of outcomes of a binomial experiment
III. The average temperature in San Francisco
1. The wingspan of a bird
2. The jersey numbers of a football team
3. I, III, and IV only
4. II and V only
5. I, III, and V only
6. II, III, and IV only
7. I, II, and IV only
Use the following computer output for a least-squares regression for Questions 7 and 8.
1. What is the equation of the least-squares regression line?
2. ŷ= –0.6442x + 22.94
3. ŷ= 22.94 + 0.5466x
4. ŷ= 22.94 + 2.866x
5. ŷ= 22.94 – 0.6442x
6. ŷ= –0.6442 + 0.5466x
7. Given that the analysis is based on 10 datapoints, what is the P -value for the t -test of the hypothesis H 0 : β = 0 versus H A : β ≠ 0?
8. 0.02 <P < 0.03
9. 0.20 <P < 0.30
10. 0.01 <P < 0.05
11. 0.15 <P < 0.20
12. 0.10 <P < 0.15
13. “A hypothesis test yields a P -value of 0.20.” Which of the following best describes what is meant by this statement?
14. The probability of getting a finding at least as extreme as that obtained by chance alone if the null hypothesis is true is 0.20.
15. The probability of getting a finding as extreme as that obtained by chance alone from repeated random sampling is 0.20.
16. The probability is 0.20 that our finding is significant.
17. The probability of getting this finding is 0.20.
18. The finding we got will occur less than 20% of the time in repeated trials of this hypothesis test.
19. A random sample of 25 men and a separate random sample of 25 women are selected to answer questions about attitudes toward abortion. The answers were categorized as “pro-life” or “pro-choice.” Which of the following is the proper null hypothesis for this situation?
20. The variables “gender” and “attitude toward abortion” are related.
21. The proportion of “pro-life” men is the same as the proportion of “pro-life” women.
22. The proportion of “pro-life” men is related to the proportion of “pro-life” women.
23. The proportion of “pro-choice” men is the same as the proportion of “pro-life” women.
24. The variables “gender” and “attitude toward abortion” are independent.
25. A sports talk show asks people to call in and give their opinion of the officiating in the local basketball team”s most recent loss. What will most likely be the typical reaction?
26. They will most likely feel that the officiating could have been better, but that it was the team”s poor play, not the officiating, that was primarily responsible for the loss.
27. They would most likely call for the team to get some new players to replace the current ones.
28. The team probably wouldn”t have lost if the officials had been doing their job.
29. Because the team had been foul-plagued all year, the callers would most likely support the officials.
30. They would support moving the team to another city.
31. A major polling organization wants to predict the outcome of an upcoming national election (in terms of the proportion of voters who will vote for each candidate). They intend to use a 95% confidence interval with margin of error of no more than 2.5%. What is the minimum sample size needed to accomplish this goal?
32. 1536
33. 39
34. 1537
35. 40
36. 2653
37. A sample of size 35 is to be drawn from a large population. The sampling technique is such that every possible sample of size 35 that could be drawn from the population is equally likely. What name is given to this type of sample?
38. Systematic sample
39. Cluster sample
40. Voluntary response sample
41. Random sample
42. Simple random sample
43. A teacher”s union and a school district are negotiating salaries for the coming year. The teachers want more money, and the district, claiming, as always, budget constraints, wants to pay as little as possible. The district, like most, has a large number of moderately paid teachers and a few highly paid administrators. The salaries of all teachers and administrators are included in trying to figure out, on average, how much the professional staff currently earn. Which of the following would the teachers” union be most likely to quote during negotiations?
44. The mean of all the salaries.
45. The mode of all the salaries.
46. The standard deviation of all the salaries.
47. The interquartile range of all the salaries.
48. The median of all the salaries.
49. Alfred and Ben don”t know each other but are each considering asking the lovely Charlene to the school prom. The probability that at least one of them will ask her is 0.72. The probability that they both ask her is 0.18. The probability that Alfred asks her is 0.6. What is the probability that Ben asks Charlene to the prom?
50. 0.78
51. 0.30
52. 0.24
53. 0.48
54. 0.54
55. A significance test of the hypothesis H 0 : p = 0.3 against the alternative H A : p > 0.3 found a value of = 0.35 for a random sample of size 95. What is the P -value of this test?
56. 1.06
57. 0.1446
58. 0.2275
59. 0.8554
60. 0.1535
61. Which of the following describe/s the central limit theorem?
62. The mean of the sampling distribution of x– is the same as the mean of the population.
63. The standard deviation of the sampling distribution of x– is the same as the standard deviation of x – divided by the square root of the sample size.
III. If the sample size is large, the shape of the sampling distribution of x – is approximately normal.
1. I only
2. I & II only
3. II only
4. III only
5. I, II, and III
6. If three fair coins are flipped, P (0 heads) = 0.125, P (exactly 1 head) = 0.375, P (exactly 2 heads) = 0.375, and P (exactly 3 heads) = 0.125. The following results were obtained when three coins were flipped 64 times:
What is the value of the X 2 statistic used to test if the coins are behaving as expected, and how many degrees of freedom does the determination of the P -value depend on?
1. 3.33, 3
2. 3.33, 4
3. 11.09, 3
4. 3.33, 2
5. 11.09, 4
For the histogram pictured above, what is the class interval (boundaries) for the class that contains the median of the data?
1. (5, 7)
2. (9, 11)
3. (11, 13)
4. (15, 17)
5. (7, 9)
6. Thirteen large animals were measured to help determine the relationship between their length and their weight. The natural logarithm of the weight of each animal was taken and a least-squares regression equation for predicting weight from length was determined. The computer output from the analysis is given below:
Give a 99% confidence interval for the slope of the regression line. Interpret this interval.
1. (0.032, 0.041); the probability is 0.99 that the true slope of the regression line is between 0.032 and 0.041.
2. (0.032, 0.041); 99% of the time, the true slope will be between 0.032 and 0.041.
3. (0.032, 0.041); we are 99% confident that the true slope of the regression line is between 0.032 and 0.041.
4. (0.81, 1.66); we are 99% confident that the true slope of the regression line is between 0.032 and 0.041.
5. (0.81, 1.66); the probability is 0.99 that the true slope of the regression line is between 0.81 and 1.66.
6. What are the mean and standard deviation of a binomial experiment that occurs with probability of success 0.76 and is repeated 150 times?
7. 114, 27.35
8. 100.5, 5.23
9. 114, 5.23
10. 100.5, 27.35
11. The mean is 114, but there is not enough information given to determine the standard deviation.
12. Which of the following is the primary difference between an experiment and an observational study?
13. Experiments are only conducted on human subjects; observational studies can be conducted on nonhuman subjects.
14. In an experiment, the researcher manipulates some variable to observe its effect on a response variable; in an observational study, he or she simply observes and records the observations.
15. Experiments must use randomized treatment and control groups; observational studies also use treatment and control groups, but they do not need to be randomized.
16. Experiments must be double-blind; observational studies do not need to be.
17. There is no substantive difference—they can both accomplish the same research goals.
18. The regression analysis of question 20 indicated that “R-sq = 98.1%.” Which of the following is (are) true?
19. There is a strong positive linear relationship between the explanatory and response variables.
20. There is a strong negative linear relationship between the explanatory and response variables.
III. About 98% of the variation in the response variable can be explained by the regression on the explanatory variable.
1. I and III only
2. I or II only
3. I or II (but not both) and III
4. II and III only
5. I, II, and III
6. A hypothesis test is set up so that P (rejecting H 0 when H 0 is true) = 0.05 and P (failing to reject H 0 when H 0 is false) = 0.26. What is the power of the test?
7. 0.26
8. 0.05
9. 0.95
10. 0.21
11. 0.74
12. For the following observations collected while doing a chi-square test for independence between the two variables A and B , find the expected value of the cell marked with “X .”
13. 4.173
14. 9.00
15. 11.56
16. 8.667
17. 9.33
18. The following is a probability histogram for a discrete random variable X.
19. 3.5
20. 4.0
21. 3.7
22. 3.3
23. 3.0
24. A psychologist believes that positive rewards for proper behavior are more effective than punishment for bad behavior in promoting good behavior in children. A scale of “proper behavior” is developed. μ 1 = the “proper behavior” rating for children receiving positive rewards, and μ 2 = the “proper behavior” rating for children receiving punishment. If H 0 : μ 1μ 2 = 0, which of the following is the proper statement of H A ?
25. HA : μ 1μ 2 > 0
26. HA : μ 1μ 2 < 0
27. HA : μ 1μ 2 ≠ 0
28. Any of the above is an acceptable alternative to the given null.
29. There isn”t enough information given in the problem for us to make a decision.
30. Estrella wants to become a paramedic and takes a screening exam. Scores on the exam have been approximately normally distributed over the years it has been given. The exam is normed with a mean of 80 and a standard deviation of 9. Only those who score in the top 15% on the test are invited back for further evaluation. Estrella received a 90 on the test. What was her percentile rank on the test, and did she qualify for further evaluation?
31. 13.35; she didn”t qualify.
32. 54.38; she didn”t qualify.
33. 86.65; she qualified.
34. 84.38; she didn”t qualify.
35. 88.69; she qualified.
36. Which of the following statements is (are) true?
37. In order to use aχ 2 procedure, the expected value for each cell of a one- or two-way table must be at least 5.
38. In order to useχ 2 procedures, you must have at least 2 degrees of freedom.
III. In a 4 × 2 two-way table, the number of degrees of freedom is 3.
1. I only
2. I and III only
3. I and II only
4. III only
5. I, II, and III
6. When the point (15,2) is included, the slope of regression line (y = a + bx ) is b = –0.54. The correlation is r = –0.82. When the point is removed, the new slope is –1.04 and the new correlation coefficient is –0.95. What name is given to a point whose removal has this kind of effect on statistical calculations?
7. Outlier
8. Statistically significant point
9. Point of discontinuity
10. Unusual point
11. Influential point
12. A one-sided test of a hypothesis about a population mean, based on a sample of size 14, yields a P -value of 0.075. Which of the following best describes the range of t values that would have given this P -value?
13. 1.345 <t < 1.761
14. 1.356 <t < 1.782
15. 1.771 <t < 2.160
16. 1.350 <t < 1.771
17. 1.761 <t < 2.145
18. Use the following excerpt from a random digits table for assigning six people to treatment and control groups:
98110 35679 14520 51198 12116 98181 99120 75540 03412 25631
The subjects are labeled: Arnold: 1; Betty: 2; Clive: 3; Doreen: 4; Ernie: 5; Florence: 6. The first three subjects randomly selected will be in the treatment group; the other three in the control group. Assuming you begin reading the table at the extreme left digit, which three subjects would be in the control group?
19. Arnold, Clive, Ernest
20. Arnold, Betty, Florence
21. Betty, Clive, Doreen
22. Clive, Ernest, Florence
23. Betty, Doreen, Florence
24. A null hypothesis, H 0 : μ = μ 0 is to be tested against a two-sided hypothesis. A sample is taken; x – is determined and used as the basis for a C -level confidence interval (e.g., C = 0.95) for μ . The researcher notes that μ 0 is not in the interval. Another researcher chooses to do a significance test for μ using the same data. What significance level must the second researcher choose in order to guarantee getting the same conclusion about H 0 : μ = μ 0 (that is, reject or not reject) as the first researcher?
25. 1 –C
26. C
27. α
28. 1 –α
29. α= 0.05
30. Which of the following is not required in a binomial setting?
31. Each trial is considered either a success or a failure.
32. Each trial is independent.
33. The value of the random variable of interest is the number of trials until the first success occurs.
34. There is a fixed number of trials.
35. Each trial succeeds or fails with the same probability.
36. X and Y are independent random variables with μ X = 3.5, μ Y = 2.7, σ X = 0.8, and σ Y = 0.65. What are μ X +Y and σ X +Y ?
37. μX +Y = 6.2, σ X +Y = 1.03
38. μX +Y = 6.2, σ X +Y = 1.0625
39. μX +Y = 3.1, σ X +Y = 0.725
40. μX +Y = 6.2, σ X +Y = 1.45
41. μX +Y = 6.2, σ X +Y cannot be determined from the information given.
42. A researcher is hoping to find a predictive linear relationship between the explanatory and response variables in her study. Accordingly, as part of her analysis she plans to generate a 95% confidence interval for the slope of the regression line for the two variables. The interval is determined to be (0.45, 0.80). Which of the following is (are) true? (Assume conditions for inference are met.)
43. She has good evidence of a linear relationship between the variables.
44. It is likely that there is a non-zero correlation (r) between the two variables.
III. It is likely that the true slope of the regression line is 0.
1. I and II only
2. I and III only
3. II and III only
4. I only
5. II only
6. In the casino game of roulette, there are 38 slots for a ball to drop into when it is rolled around the rim of a revolving wheel: 18 red, 18 black, and 2 green. What is the probability that the first time a ball drops into the red slot is on the 8th trial (in other words, suppose you are betting on red every time—what is the probability of losing 7 straight times before you win the first time)?
7. 0.0278
8. 0.0112
9. 0.0053
10. 0.0101
11. 0.0039
12. You are developing a new strain of strawberries (say, Type X) and are interested in its sweetness as compared to another strain (say, Type Y). You have four plots of land, call them A, B, C, and D, which are roughly four squares in one large plot for your study (see the figure below). A river runs alongside of plots C and D. Because you are worried that the river might influence the sweetness of the berries, you randomly plant type X in either A or B (and Y in the other) and randomly plant type X in either C or D (and Y in the other). Which of the following terms best describes this design?
13. A completely randomized design
14. A randomized study
15. A randomized observational study
16. A block design, controlling for the strain of strawberry
17. A block design, controlling for the effects of the river
18. Grumpy got 38 on the first quiz of the quarter. The class average on the first quiz was 42 with a standard deviation of 5. Dopey, who was absent when the first quiz was given, got 40 on the second quiz. The class average on the second quiz was 45 with a standard deviation of 6.1. Grumpy was absent for the second quiz. After the second quiz, Dopey told Grumpy that he was doing better in the class because they had each taken one quiz, and he had gotten the higher score. Did he really do better? Explain.
19. Yes. zDopey is more negative than z Grumpy .
20. Yes. zDopey is less negative than z Grumpy .
21. No. zDopey is more negative than z Grumpy .
22. Yes. zDopey is more negative than z Grumpy .
23. No. zDopey is less negative than z Grumpy .
24. A random sample size of 45 is obtained for the purpose of testing the hypothesis H 0 : p = 0.80. The sample proportion is determined to be = 0.75. What is the value of the standard error of for this test?
25. 0.0042
26. 0.0596
27. 0.0036
28. 0.0645
29. 0.0055
SECTION II—PART A, QUESTIONS 1–5
Spend about 65 minutes on this part of the exam. Percentage of Section II grade—75.
Directions: Show all of your work. Indicate clearly the methods you use because you will be graded on the correctness of your methods as well as on the accuracy of your results and explanation.
1. The ages (in years) and heights (in cm) of 10 girls, ages 2 through 11, were recorded. Part of the regression output and the residual plot for the data are given below.
2. What is the equation of the least-squares regression line for predicting height from age?
3. Interpret the slope of the regression line in the context of the problem.
4. Suppose you wanted to predict the height of a girl 5.5 years of age. Would the prediction made by the regression equation you gave in (a) be too small, too large, or is there not enough information to tell?
5. You want to determine whether a greater proportion of men or women purchase vanilla lattes (regular or decaf). To collect data, you hire a person to stand inside the local Scorebucks for 2 hours one morning and tally the number of men and women who purchase the vanilla latte, as well as the total number of men and women customers: 63% of the women and 59% of the men purchase a vanilla latte.
6. Is this an experiment or an observational study? Explain.
7. Based on the data collected, you write a short article for the local newspaper claiming that a greater proportion of women than men prefer vanilla latte as their designer coffee of choice. A student in the local high school AP Statistics class writes a letter to the editor criticizing your study. What might the student have pointed out?
8. Suppose you wanted to conduct a study less open to criticism. How might you redo the study?
9. Sophia is a nervous basketball player. Over the years she has had a 40% chance of making the first shot she takes in a game. If she makes her first shot, her confidence goes way up, and the probability of her making the second shot she takes rises to 70%. But if she misses her first shot, the probability of her making the second shot she takes doesn”t change—it”s still 40%.
10. What is the probability that Sophia makes her second shot?
11. If Sophia does make her second shot, what is the probability that she missed her first shot?
12. A random sample of 72 seniors taken 3 weeks before the selection of the school Homecoming Queen identified 60 seniors who planned to vote for Buffy for queen. Unfortunately, Buffy said some rather catty things about some of her opponents, and it got into the school newspaper. A second random sample of 80 seniors taken shortly after the article appeared showed that 56 planned to vote for Buffy. Does this indicate a serious drop in support for Buffy? Use good statistical reasoning to support your answer.
13. Some researchers believe that education influences IQ. One researcher specifically believes that the more education a person has, the higher, on average, will be his or her IQ. The researcher sets out to investigate this belief by obtaining eight pairs of identical twins reared apart. He identifies the better educated twin as Twin A and the other twin as Twin B for each pair. The data for the study are given in the table below. Do the data give good statistical evidence, at the 0.05 level of significance, that the twin with more education is likely to have the higher IQ? Give good statistical evidence to support your answer.
SECTION II—PART B, QUESTION 6
Spend about 25 minutes on this part of the exam. Percentage of Section II grade—25.
Directions: Show all of your work. Indicate clearly the methods you use because you will be graded on the correctness of your methods as well as on the accuracy of your results and explanation.
1. A paint manufacturer claims that the average drying time for its best-selling paint is 2 hours. A random sample of drying times for 20 randomly selected cans of paint are obtained to test the manufacturer”s claim. The drying times observed, in minutes, were: 123, 118, 115, 121, 130, 127, 112, 120, 116, 136, 131, 128, 139, 110, 133, 122, 133, 119, 135, 109.
2. Obtain a 95% confidence interval for the true mean drying time of the paint.
3. Interpret the confidence interval obtained in part (a) in the context of the problem.
4. Suppose, instead, that a significance test at the 0.05 level of the hypothesisH 0 : μ = 120 was conducted against the alternative H A : μ ≠ 120. What is the P -value of the test?
5. Are the answers you got in part (a) and part (c) consistent? Explain.
6. At the 0.05 level, would your conclusion about the mean drying time have been different if the alternative hypothesis had beenH A : μ > 120? Explain.
END OF DIAGNOSTIC EXAM
1. c
2. a
3. e
4. d
5. c
6. a
7. d
8. b
9. a
10. b
11. c
12. c
13. e
14. e
15. b
16. b
17. d
18. a
19. e
20. c
21. c
22. b
23. c
24. e
25. d
26. d
27. a
28. c
29. b
30. e
31. d
32. e
33. a
34. c
35. a
36. a
37. c
38. e
39. c
40. b
SOLUTIONS TO DIAGNOSTIC TEST—SECTION I
1. From Chapter 10
The correct answer is (c). If X has B (n, p ), then, in general,
In this problem, n = 18, p = 0.4, x = 7 so that
1. From Chapter 7
The correct answer is (a). ln (y ) = 3.2 + 0.42(7) = 6.14⇒y = e 6.14 = 464.05.
1. From Chapter 11
The correct answer is (e). For a 94% z -interval, there will be 6% of the area outside of the interval. That is, there will be 97% of the area less than the upper critical value of z . The nearest entry to 0.97 in the table of standard normal probabilities is 0.9699, which corresponds to a z -score of 1.88.
(Using the TI-83/84, we have invNorm(0.97) = 1.8808 .)
1. From Chapter 6
The correct answer is (d). If the bar to the far left was not there, this graph would be described as approximately normal. It still has that same basic shape but, because there is an outlier, the best description is: approximately normal with an outlier.
1. From Chapter 9
The correct answer is (c). Let x be the value in question. If there is 0.2 of the area above x , then there is 0.8 of the area to the left of x . This corresponds to a z -score of 0.84 (from Table A, the nearest entry is 0.7995). Hence,
(Using the TI-83/84, we have invNorm(0.8) = 0.8416 .)
1. From Chapter 5
The correct answer is (a). Discrete data are countable; continuous data correspond to intervals or measured data. Hence, speed, average temperature, and wingspan are examples of continuous data. The number of outcomes of a binomial experiment and the jersey numbers of a football team are countable and, therefore, discrete.
1. From Chapter 7
The correct answer is (d). The slope of the regression line. –0.6442, can be found under “Coef” to the right of “x .” The intercept of the regression line, 22.94, can be found under “Coef” to the right of “Constant.”
1. From Chapter 13
The correct answer is (b). The t statistic for H 0 : β = 0 is given in the printout as –1.18. We are given that n = 10 ⇒ df = 10 – 2 = 8. From the df = 8 row of Table B (the t Distribution Critical Values table), we see, ignoring the negative sign since it”s a two-sided test,
1.108 < 1.18 < 1.397 ⇒ 2(0.10) < P < 2(0.15),
which is equivalent to 0.20 < P < 0.30. Using the TI-83/84, we have 2 × tcdf (-100, –1.18,8) = 0.272.
1. From Chapter 12
The correct answer is (a). The statement is basically a definition of P -value. It is the likelihood of obtaining, by chance alone, value as extreme or more extreme as that obtained if the null hypothesis is true. A very small P -value sheds doubt on the truth of the null hypothesis.
1. From Chapter 14
The correct answer is (b). Because the samples of men and women represent different populations, this is a chi-square test of homogeneity of proportions: the proportions of each value of the categorical variable (in this case, “pro-choice” or “pro-life”) will be the same across the different populations. Had there been only one sample of 50 people drawn, 25 of whom happened to be men and 25 of whom happened to be women, this would have been a test of independence.
1. From Chapter 8
The correct answer is (c). This is a voluntary response survey and is subject to voluntary response bias. That is, people who feel the most strongly about an issue are those most likely to respond. Because most callers would be fans, they would most likely blame someone besides the team.
1. From Chapter 11
The correct answer is (c). The “recipe” we need to use is n ≥ . Since we have no basis for an estimate for P * , we use P * = 0.5. In this situation the formula reduces to
Since n must be an integer, choose n = 1537.
1. From Chapter 8
The correct answer is (e). A random sample from a population is one in which every member of the population is equally likely to be selected. A simple random sample is one in which every sample of a given size is equally likely to be selected. A sample can be a random sample without being a simple random sample.
1. From Chapter 6
The correct answer is (e). The teachers are interested in showing that the average teacher salary is low. Because the mean is not resistant, it is pulled in the direction of the few higher salaries and, hence, would be higher than the median, which is not affected by a few extreme values. The teachers would choose the median. The mode, standard deviation, and IQR tell you nothing about the average salary.
1. From Chapter 9
The correct answer is (b). P (at least one of them will ask her) = P (A or B) = 0.72.
P (they both ask her) = P (A and B) = 0.18.
P (Alfred asks her) = P (A) = 0.6.
In general, P (A or B) = P (A) + P (B) – P (A and B). Thus, 0.72 = 0.6 + P (B) – 0.18 ⇒ P (B) = 0.30.
1. From Chapter 12
(Using the TI-83/84, we find normalcdf(1.06,100) = 0.1446. )
1. From Chapter 10
The correct answer is (d). Although all three of the statements are true of a sampling distribution, only III is a statement of the central limit theorem.
1. From Chapter 14
(This calculation can be done on the TI-83/84 as follows: let L1 = observed values; let L2 = expected values; let L3 = (L2-L1 )2 /L2 ; Then χ 2 = LIST MATH sum(L3) =3.33. )
In a chi-square goodness-of-fit test, the number of degrees of freedom equals one less than the number of possible outcomes. In this case, df = n –1 = 4 – 1 = 3.
1. From Chapter 6
The correct answer is (e). There are 101 terms, so the median is located at the 51st position in an ordered list of terms. From the counts given, the median must be in the interval whose midpoint is 8. Because the intervals are each of width 2, the class interval for the interval whose midpoint is 8 must be (7, 9).
1. From Chapter 13
The correct answer is (c). df = 13 – 2 = 11 ⇒ t * = 3.106 (from Table B; if you have a TI-84 with the invT function, t * = in v T(0.995,11 )). Thus, a 99% confidence interval for the slope is:
0.0365 ± 3.106(0.0015) = (0.032, 0.041).
We are 99% confident that the true slope of the regression line is between 0.032 units and 0.041 units.
1. From Chapter 10
2. From Chapter 8
The correct answer is (b). In an experiment, the researcher imposes some sort of treatment on the subjects of the study. Both experiments and observational studies can be conducted on human and nonhuman units; there should be randomization to groups in both to the extent possible; they can both be double blind.
1. From Chapter 7
The correct answer is (c). III is basically what is meant when we say R-sq = 98.1%. However, R-sq is the square of the correlation coefficient.
could be either positive or negative, but not both. We can”t tell direction from R 2 .
1. From Chapter 11
The correct answer is (e). The power of a test is the probability of correctly rejecting H 0 when H A is true. You can either fail to reject H 0 when it is false (Type II), or reject it when it is false (Power). Thus, Power = 1 – P (Type II) = 1 – 0.26 = 0.74.
1. From Chapter 14
The correct answer is (d). There are 81 observations total, 27 observations in the second column, 26 observations in the first row. The expected number in the first row and second column equals
1. From Chapter 9
μ X = 2(0.3) + 3(0.2) + 4(0.4) + 5(0.1) = 3.3.
1. From Chapter 12
The correct answer is (a). The psychologist”s belief implies that, if she”s correct, μ 1 > μ 2 . Hence, the proper alternative is H A : μ 1μ 2 > 0.
1. From Chapter 6
Because she had to be in the top 15%, she had to be higher than the 85th percentile, so she was invited back.
1. From Chapter 14
The correct answer is (b). I is true. Another common standard is that there can be no empty cells, and at least 80% of the expected counts are greater than 5. II is not correct because you can have 1 degree of freedom (for example, a 2 × 2 table). III is correct because df = (4 – 1) (2 – 1) = 3.
1. From Chapter 7
The correct answer is (e). An influential point is a point whose removal will have a marked effect on a statistical calculation. Because the slope changes from –0.54 to –1.04, it is an influential point.
1. From Chapter 12
The correct answer is (d). df = 14 – 1 = 13. For a one-sided test and 13 degrees of freedom, 0.075 lies between tail probability values of 0.05 and 0.10. These correspond, for a one-sided test, to t * values of 1.771 and 1.350. (If you have a TI-84 with the invT function, t * = invT(1 -0.075,13) = 1.5299. )
1. From Chapter 8
The correct answer is (e). Numbers of concern are 1, 2, 3, 4, 5, 6. We ignore the rest. We also ignore repeats. Reading from the left, the first three numbers we encounter for our subjects are 1, 3, and 5. They are in the treatment group, so numbers 2, 4, and 6 are in the control group. That”s Betty, Doreen, and Florence. You might be concerned that the three women were selected and that, somehow, that makes the drawing nonrandom. However, drawing their three numbers had exactly the same probability of occurrence as any other group of three numbers from the six.
1. From Chapter 11
The correct answer is (a). If a significance test at level a rejects a null hypothesis (H 0 : μ = μ 0 ) against a two-sided alternative, then μ 0 will not be contained in a C = 1 – α level confidence interval constructed using the same value of . Thus, α = 1 – C .
1. From Chapter 10
The correct answer is (c). The statement in (c) describes the random variable for a geometric setting. In a binomial setting, the random variable of interest is the number count of successes in the fixed number of trials.
1. From Chapter 9
μ X+Y is correct for any random variables X and Y . However, σ X+Y is correct only if X and Y are independent .
1. From Chapter 14
The correct answer is (a). Because 0 is not in the interval (0.45, 0.80), it is unlikely that the true slope of the regression line is 0 (III is false). This implies a non-zero correlation coefficient and the existence of a linear relationship between the two variables.
1. From Chapter 10
The correct answer is (c). This is a geometric setting (independent trials, each succeeding or failing with the same probability).
(On the TI-83/84, this is found as geometpdf(18/38,8) .)
1. From Chapter 8
The correct answer is (e). The choice is made here to treat plots A and B as a block and plots C and D as a block. That way, we are controlling for the possible confounding effects of the river. Hence the answer is (c). If you answered (e), be careful of confusing the treatment variable with the blocking variable.
1. From Chapter 6 The correct answer is (c).
They are both below average, but Grumpy”s z score puts him slightly above Dopey. Note that if Grumpy had been 4 points above the mean on the first test and Dopey 5 points above the mean on the second, then Dopey would have done slightly better than Grumpy.
1. From Chapter 12
The standard error of for a test of H 0 : p = p 0 is
If you got an answer of 0.0645, it means you used the value of rather than the value of p 0 in the formula for s .
SOLUTIONS TO DIAGNOSTIC TEST—SECTION II, PART A
1. a.
2. For each additional year of age, the height (in cm) is predicted to increase by 6.36 cm.
We would expect the residual for 5.5 to be in the same general area as the residuals for 4, 5, 6, and 7 (circled on the graph). The residuals in this area are all positive ⇒ actual – predicted > 0 ⇒ actual > predicted. The prediction would probably be too small.
1. a. It is an observational study. The researcher made no attempt to impose a treatment on the subjects in the study. The hired person simply observed and recorded behavior.
2. • The article made no mention of the sample size. Without that you are unable to judge how much sampling variability there might have been. It”s possible that the 63–59 split was attributable to sampling variability.
• The study was done atone Scorebucks, on one morning, for a single 2-hour period. The population at that Scorebucks might differ in some significant way from the patrons at other Scorebucks around the city (and there are many, many of them). It might have been different on a different day or during a different time of the day. A single 2-hour period may not have been enough time to collect sufficient data (we don”t know because the sample size wasn”t given) and, again, a 2-hour period in the afternoon might have yielded different results.
1. You would conduct the study at multiple Scorebucks, possibly blocking by location if you believe that might make a difference (i.e., would a working-class neighborhood have different preferences than the ritziest neighborhood?). You would observe at different times of the day and on different days. You would make sure that the total sample size was large enough to control for sampling variability (replication).
2. From the information given, we have
• P(hit the first and hit the second) = (0.4) (0.7) = 0.28
• P(hit the first and miss the second) = (0.4) (0.3) = 0.12
• P(miss the first and hit the second) = (0.6) (0.4) = 0.24
• P(miss the first and miss the second) = (0.6) (0.6) = 0.36
This information can be summarized in the following table:
1. P(hit on second shot) = 0.28 + 0.24 = 0.52
2. P(miss on first | hit on second) = (0.24)/(0.52) = 6/13 = 0.46.
3. Let p 1 be the true proportion who planned to vote for Buffy before her remarks. Let p 2 be the true proportion who plan to vote for Buffy after her remarks.
We want to use a 2-proportion z test for this situation. The problem tells us that the samples are random samples.
Now, 72(0.83), 72(1 – 0.83), 80(0.70), and 80(1 – 0.70) are all greater than 5, so the conditions for the test are present.
Because P is very low, we reject the null. We have reason to believe that the level of support for Buffy has declined since her “unfortunate” remarks.
1. The data are paired, so we will use a matched pairs test.
Let μ d = the true mean difference between Twin A and Twin B for identical twins reared apart.
We want to use a one-sample t -test for this situation. We need the difference scores:
A dotplot of the difference scores shows no significant departures from normality:
The conditions needed for the one sample t -test are present.
(from Table B; on the TI-83/84, tcdf(2.39,100,7) =0.024 ).
Because P < 0.05, reject H 0 . We have evidence that, in identical twins reared apart, the better educated twin is likely to have the higher IQ score.
1. a. = 123.85, s = 9.07. We are told that the 20 cans of paint have been randomly selected. It is reasonable to assume that a sample of this size is small relative to the total population of such cans. A boxplot of the data shows no significant departures from normality. The conditions necessary to construct a 95% t confidence interval are present.
2. We are 95% confident that the true mean drying time for the paint is between 119.6 minutes and 128.1 minutes. Because 120 minutes is in this interval, we would not consider an average drying time of 120 minutes for the population from which this sample was drawn to be unusual.
(On the TI-83/84, we find P -value = 2 × tcdf(1.90,100,19) = 0.073. )
1. We know that if a two-sidedα -level significance test rejects (fails to reject) a null hypothesis, then the hypothesized value of μ will not be (will be) in a C = 1 – α confidence interval. In this problem, 120 was in the C = 0.95 confidence interval and a significance test at α = 0.05 failed to reject the null as expected.
2. For the one-sided test,t = 1.90, df = 19 ⇒ 0.025 < P -value < 0.05
(On the TI-83/84, we find P -value = tcdf(1.90,100,19) = 0.036. )
For the two-sided test, we concluded that we did not have evidence to reject the claim of the manufacturer. However, for the one-sided test, we have stronger evidence (P < 0.05) and would conclude that the average drying time is most likely greater than 120 minutes.
Scoring Sheet for Diagnostic Test
Section I: Multiple-Choice Questions
Section II: Free-Response Questions
Composite Score
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# Given the sum of the first 5 terms of an AP as 90 and the first 50 terms as 4275, what is the first term of the series?
giorgiana1976 | College Teacher | (Level 3) Valedictorian
Posted on
We'll apply the formula for the sum of n terms of an A.P.:
Sn = (a1+an)*n/2
We know that the sum of the first 5 terms is 90.
S5 = 90
90 = (a1+a5)*5/2
2*90 = (a1+a5)*5
We'll divide by 5 both sideS:
2*18 = a1 + a5
We'll write the formula for the general term of an A.P.
an = a1 + (n-1)*d
The common difference is d.
a5 = a1 + (5-1)d
a1 + a5 = 36 (1)
a5 = a1 + 4d (2)
We'll substitute (2) in (1):
2a1 + 4d = 36
We'll divide by 2:
a1 + 2d = 18 (3)
We also know that S50 = 4275.
S50 = (a1+a50)*50/2
4275 = (a1+a50)*25
We'll divide by 25 both sides:
a1 + a50 = 171 (4)
a50 = a1 + 49d (5)
We'll substitute (5) in (4):
2a1 + 49d =171 (6)
We'll form the system from the equation (3) and (6):
a1 + 2d = 18 (3)
2a1 + 49d =171 (6)
We'll multiply (3) by -2:
-2a1 - 4d = -36 (7)
-2a1 - 4d + 2a1 + 49d = -36 + 171
We'll eliminate and combine like terms:
45d = 135
We'll divide by 45:
d = 3
The common difference is 3.
We'll substitute d in (3):
a1 + 2d = 18
a1 + 6 = 18
We'll subtract 6 both sides:
a1 = 12
The first term of the A.P., whose common difference is 3 and sum of 5 first terms is 90, is a1 = 12.
neela | High School Teacher | (Level 3) Valedictorian
Posted on
We know that the sum of the n terms of an AP is given by:
Sn = {2a1+(n-1)d}n/2, where a1 is the first term, d is the common diffrerence between the successive terms.
Therefore sum of the 5 terms= S5 = {2a1+(5-1)d}5/2 = 90. Simplify this equation:
(2a1 +4d )5 = 90*2.
10a1 +20d = 180 .
Divide by 10:
a1+2d = 18...........(1) .
Also given the sum of first 50 terms = 4275. So by formula, S50 = {2a1 +(50-1)d}50/2 =4275. Simplify this equation:
(2a1 +49d) = 4275/25.
2a1 +49d1 = 171..........(2).
Therefore eq(2) - 2*eq(1) goves:
(2a1+49d)-2(a1+2d) = 171- 2*18.
45d = 171-18 = 135
.d = 135/45 = 3.
So substituting d = 3 in (1), we get: a1+2*3 = 18. So a1 = 18-6 = 12.
Therefore the first term of the AP = a1 = 12.
william1941 | College Teacher | (Level 3) Valedictorian
Posted on
An AP is a series which has a common difference between subsequent terms. The sum of the first n terms of an AP is given by the relation Sn = (n/2)*[2a + (n-1) d], where a is the first term of the AP and d is the common difference.
Now, it is given that the sum of the first 5 terms is 90
=> (5/2)*[2a + (5-1) d] = 90
=> (5/2)*(2a + 4d) = 90
=> 10a + 20d = 180
=> a + 2d = 18… (1)
Also, the sum of the first 50 terms is 4275.
=> (50/2)*[2a + (50-1) d] = 4275
=> 25*(2a +49d) = 4275
=> 50a + 1225d = 4275 … (2)
1225*(1) – 2*(2)
=> 1225a + 2450d – 100a – 2450d = 22050- 8550
=> 1125a = 13500
=> a= 13500/ 1125
=> a = 12
Therefore the AP starts with the term 12.
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# Minimize the sum of node values by filling a given empty Tree such that each node is GCD of its children
Amarjeet Kumar
Last Updated: May 12, 2022
Difficulty Level :
EASY
## Introduction
In this article we'll look at a problem using Binary Trees to solve it. Binary Trees contain a lot of intriguing qualities, and we'll learn about them in this blog. Binary tree problems are common in coding interviews and programming competitions. The number of nodes in a complete binary tree can be counted in a variety of ways, as discussed in this article.
A tree is made up of nodes, which are individual things. Edges link nodes together. Each node has a value or piece of data and may or may not have a child node. The root refers to the tree's first node. Let’s start with the problem statement.
## Problem Statement
The aim is to calculate the minimum sum of all the node's values of the given Tree such that the value of each node must equal the value of GCDs of its children, given a Binary Tree consisting of N nodes with no values and an integer X that represents the value of the root node. Aside from that, no two siblings can have the same worth.
### GCD Concept
The greatest common divisor (GCD) of two or more numbers is the exact divisor number that divides them. It's also known as the greatest common factor (HCF). Because both integers can be divided by 5, the largest common factor of 15 and 10 is 5.
The greatest common factor of two or more numbers is called GCD. A factor number that is the highest among the others.
## Approach
Both of the offspring can have the value of X and 2*X, where X is the parent's value, to minimize the total. If a node only has one child, its value will be the same as its parent node. The depth of each subtree for each node will be examined to determine which child should have a value of X and 2*X to obtain the least total. The kid with the most depth will be assigned a value of X so that it can be transferred to more of its offspring, while another will be assigned a value of 2*X.
• Find each node's depth and save it in a map using the node's address as the key.
• Now, beginning from the root node, conduct the DFS Traversal, assigning a value of X to each node that has a higher depth than its sibling in each call. In every other case, use the number 2*X.
• Find the sum of both left and right child values while backtracking and return the entire sum, i.e. the sum of the left child, right child, and current node values in each call, after the previous step.
• Print the result returned from the DFS Call as the smallest sum feasible after completing the preceding steps.
### C++ implementation
``````#include <bits/stdc++.h>
using namespace std;
class Node {
public:
int data;
Node *left, *right;
Node(int data)
{
this->data = data;
left = NULL;
right = NULL;
}
};
class Tree {
public:
unordered_map<Node*, int> depth;
int findDepth(Node* current_data)
{
int max_start = 0;
if (current_data->left) {
max_start = findDepth(current_data->left);
}
if (current_data->right) {
max_start = max(max_start, findDepth(current_data->right));
}
return depth[current_data] = max_start + 1;
}
int dfs(Node* current_data, bool flag, int parValue)
{
if (parValue != -1) {
if (flag)
current_data->data = parValue;
else
current_data->data = parValue * 2;
}
int l = 0, r = 0;
if (current_data->left && current_data->right) {
if (depth[current_data->left] > depth[current_data->right]) {
l = dfs(current_data->left, 1, current_data->data
r = dfs(current_data->right, 0, current_data->data);
}
else {
l = dfs(current_data->left, 0, current_data->data);
r = dfs(current_data->right, 1, current_data->data);
}
}
else if (current_data->left) {
l = dfs(current_data->left, 1, current_data->data);
}
else if (current_data->right) {
r = dfs(current_data->right, 1, current_data->data);
}
return l + r + current_data->data;
}
int minimumSum(Node* root)
{
findDepth(root);
return dfs(root, 1, -1);
}
};
int main()
{
int X = 2;
Node* root = new Node(X);
root->left = new Node(-1);
root->right = new Node(-1);
root->left->left = new Node(-1);
root->left->right = new Node(-1);
root->left->right->left = new Node(-1);
root->left->right->right = new Node(-1);
root->left->right->right->left = new Node(-1);
Tree t;
cout << t.minimumSum(root);
return 0;
}``````
22
### Java implementation
``````import java.util.*;
public class Main
{
static class Node {
public int data;
public Node left, right;
public Node(int data)
{
this.data = data;
left = right = null;
}
}
static HashMap<Node, Integer> depth = new HashMap<>();
static int findDepth(Node current_data)
{
int max_data = 0;
if (current_data.left != null) {
max_data = findDepth(current_data.left);
}
if (current_data.right != null) {
max_data = Math.max(max_data, findDepth(current_data.right));
}
depth.put(current_data, max_data + 1);
return depth.get(current_data);
}
static int dfs(Node current_data, int flag, int parValue)
{
if (parValue != -1) {
if (flag == 1)
current_data.data = parValue;
else
current_data.data = parValue * 2;
}
int l = 0, r = 0;
if (current_data.left != null && current_data.right != null) {
if (depth.containsKey(current_data.left) && depth.containsKey(current_data.right) && depth.get(current_data.left) > depth.get(current_data.right)) {
l = dfs(current_data.left, 1, current_data.data);
r = dfs(current_data.right, 0, current_data.data);
}
else {
l = dfs(current_data.left, 0, current_data.data);
r = dfs(current_data.right, 1, current_data.data);
}
}
else if (current_data.left != null) {
l = dfs(current_data.left, 1, current_data.data);
}
else if (current_data.right != null) {
r = dfs(current_data.right, 1, current_data.data);
}
return (l + r + current_data.data);
}
static int minimumSum(Node root)
{
findDepth(root);
return dfs(root, 1, -1);
}
public static void main(String[] args)
{
int X = 2;
Node root = new Node(X);
root.left = new Node(-1);
root.right = new Node(-1);
root.left.left = new Node(-1);
root.left.right = new Node(-1);
root.left.right.left = new Node(-1);
root.left.right.right = new Node(-1);
root.left.right.right.left = new Node(-1);
System.out.print(minimumSum(root));
}
}``````
22
### Python implementation
``````class Node:
def __init__(self, data):
self.data = data
self.left = None
self.right = None
depth = {}
def findDepth(current_data):
mx = 0
if (current_data.left != None):
mx = findDepth(current_data.left)
if (current_data.right != None):
mx = max(mx, findDepth(current_data.right))
depth[current_data] = mx + 1
return depth[current_data]
def dfs(current_data, flag, parValue):
if (parValue != -1):
if flag:
current_data.data = parValue
else:
current_data.data = parValue * 2
l, r = 0, 0;
if (current_data.left != None and current_data.right != None):
if ((current_data.left in depth) and (current_data.right in depth) and depth[current_data.left] > depth[current_data.right]):
l = dfs(current_data.left, 1, current_data.data)
r = dfs(current_data.right, 0,current_data.data)
else:
l = dfs(current_data.left, 0, current_data.data)
r = dfs(current_data.right, 1, current_data.data)
elif (current_data.left != None):
l = dfs(current_data.left, 1, current_data.data)
elif (current_data.right != None):
r = dfs(current_data.right, 1, current_data.data)
return (l + r + current_data.data)
def minimumSum(root):
findDepth(root)
return dfs(root, 1, -1)
X = 2
root = Node(X)
root.left = Node(-1)
root.right = Node(-1)
root.left.left = Node(-1)
root.left.right =Node(-1)
root.left.right.left = Node(-1)
root.left.right.right = Node(-1)
root.left.right.right.left = Node(-1);
print(minimumSum(root))``````
22
### Complexities
#### Time complexity: O(n)
Reason: As we perform the DFS Traversal, starting with the root node, and add a value of X to each node that has more depth than its sibling in each call. So the complexity is O(n).
#### Auxiliary Space: O(1)
Reason: As we Store the depth of each node in a map with the node address as the key spece required is O(1).
1. In a tree, what is a node?
A tree is made up of nodes, which are individual things. Edges link nodes together. Each node has a value or piece of data and may or may not have a child node. The root is the tree's very first node.
2. What is tree programming?
A tree is a type of hierarchical data structure that is made up of nodes. Value is represented by nodes, which are connected by edges. The root node is the only node in the tree. This is where the tree comes from, so it doesn't have any parents.
3. What is DFS in the graph?
The depth-first search (DFS) technique is used to traverse or explore data structures such as trees and graphs. The algorithm starts from the root node (in the case of a graph, any random node can be used as the root node) and examines each branch as far as feasible before retracing.
4. What is the node's degree in a tree?
The degree of a node is defined as the total number of subtrees associated to that node. A leaf node's degree must be zero. The maximum degree of a node among all the nodes in the tree is the tree's degree.
5. Is it possible for a tree node to have two parents?
Yes, you may have "children" and "parents" in the same node. Because the graph is no longer tree-structured, you won't be able to utilize a TreeModel; instead, you'll need to use a GraphLinksModel.
## Conclusion
In this article, we have solved a binary tree problem. Where we have to fill the provided empty Tree with nodes that are GCD of their offspring to minimize the total of node values. Want to explore more related to this topic click here.
If you want to learn more attempt our Online Mock Test Series on CodeStudio now!
Happy Coding!
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# Warm-Up (8/28) Please pick up a warm-up sheet and a fractions chart from the desk in the front of the room and begin working on warm-up #1. Make sure to.
## Presentation on theme: "Warm-Up (8/28) Please pick up a warm-up sheet and a fractions chart from the desk in the front of the room and begin working on warm-up #1. Make sure to."— Presentation transcript:
Warm-Up (8/28) Please pick up a warm-up sheet and a fractions chart from the desk in the front of the room and begin working on warm-up #1. Make sure to hole punch the warm-up sheet and put it in your binder.
7.1 A – Comparing Integers and Positive Rational Numbers
Rational Numbers - Numbers that can be written in the form of a fraction. EX: The number 9 as a fraction is 9/1 The mixed number 2 ¼ as a fraction is 9/4 The mixed decimal 2.25 as a fraction is 9/4 The decimal 9.00 as a fraction is 9/1
Comparing Rational Numbers *** To compare rational numbers first convert them to the SAME FORM, either fraction or decimal. Fractions (w/ same denominator): 6 8 9 9 Same
6/9 8/9
Fractions (w/ different denominators) 2 3 3 4different Since the denominators are different, find the least common multiple for 3 & 4 Multiples of 3: 3, 6, 9, 12, 15 Multiples of 4: 4, 8, 12, 16, 20 Make 12 the denominator: 2/3= 8/12 & ¾= 9/12
Comparing Decimals Line up your decimals and begin with the number furthest to the left EX:5.862 3.741 *if numbers are the same move to the next number to the right EX:4.873-or-13.789 4.92612.987
Converting Fraction to Decimals to % Fraction to decimal: ¾ = ¾ =.75 Decimal to %:.75 = 75% % to Decimal:48% =.48 Decimal to fraction:.48 = 48/100 (reduced 12/25)
Which is Greater? 11.5or 11 2/3
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# Class 12 RD Sharma Solutions- Chapter 32 Mean and Variance of a Random Variable – Exercise 32.1 | Set 1
• Last Updated : 25 Jan, 2021
### Question 1. Which of the following distributions of probabilities of random variables are their probability distributions?
i.
Solution:
We know that the sum of probability distribution is always 1.
Sum of probabilities (P(X))=P(X=3)+P(X=2)+P(X=1)+P(X=0)+P(X=-1)
=0.3+0.2+0.4+0.1+0.05=1.05>1
The sum of probability distribution is not equal to 1. Hence, it is not the probability distribution of the given random variables.
ii.
Solution:
Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)
=0.6+0.4+0.2=1.2>1
The sum of probability distribution is not equal to 1. Hence, it is not the probability distribution of the given random variables.
iii.
Solution:
Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)
=0.1+0.5+0.2+0.1+0.1=1
The sum of probability distribution is equal to 1. Hence, it is the probability distribution of the given random variables.
iv.
Solution:
Sum of probabilities (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)
=0.3+0.2+0.4+0.1=1
The sum of probability distribution is equal to 1. Hence, it is the probability distribution of the given random variables.
### Question 2. A random variable X has the following probability distribution:
Find the value of k.
Solution:
We know that the sum of probability distribution is always 1.
Sum of probability distribution (P(X))=P(X=-2)+P(X=-1)+P(X=0)+P(X=1)+P(X=2)+P(X=3)=1
=>0.1+k+0.2+2k+0.3+k=1
=>0.6+4k=1
=>4k=1-0.6
=>k=0.1
### Question 3. A random variable X has the following probability distribution:
i. Find the value of a.
Solution:
We know that the sum of probability distribution is always 1.
Sum of probability distribution (P(X))=P(X=0)+P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)=1
=>a+3a+5a+7a+9a+11a+13a+15a+17a=1
=>81a=1
=>a=1/81
ii. Find P(X<3).
Solution:
P(X<3)=P(X=0)+P(X=1)+P(X=2)
=1/81+3/81+5/81
=9/81=1/9
iii. Find P(X>=3).
Solution:
P(X>=3)=P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)
=7/81+9/81+11/81+13/81+15/81+17/81
=72/81=8/9
iv. Find P(0<X<5).
P(0<X<5)=P(X=1)+P(X=2)+P(X=3)+P(X=4)
=3/81+5/81+7/81+9/81
=24/81=8/27=0.296
### Question 4.1. The probability distribution function of a random variable X is given by
where c>0. Find c.
Solution:
We know that the sum of probability distributions of a random variable is always 1.
=>3c3 +4c-10c2+5c-1=1
=>3c3-10c2+9c-2=0
Let c=1
3(1)-10(1)+9(1)-2=12-12=0
Therefore c=1.
By horn’s method :
We get a quadratic equation : 3c2-7c+2=0
From this quadratic equation we get,
=>3c2-6c-c+2=0
=>3c(c-2)-1(c-2)=0
=>(3c-1)(c-2)=0
=>3c-1=0; c-2=0
=>3c=1; c=2
=>c=1/3; c=2; c=1
We know that a single probability distribution cannot be 1 or more than one. So we take c=1/3.
Therefore, c=1/3.
### Question 4.2. Find P(X<2).
Solution:
P(X<2)=P(X=0)+P(X=1)
=3(1/3)3+4(1/3)-10(1/3)2
=1/9+4/3-10/9
=1/3=0.33
### Question 4.3. Find P(1<X<=2).
Solution:
P(1<X<=2)=P(X=2)
=5(1/3)-1
=5/3-1
=2/3=0.66
### Question 5. Let X be a random variable which assumes values x1, x2,x3,x4 such that 2P(X=x1)=3P(X=x2)=P(X=x3)=5P(X=x4). Find the probability distribution of X.
Solution:
Sum of probability distributions= P(X=x1)+P(X=x2)+P(X=x3)+P(X=x4)=1
Given,
2P(X=x1)=3P(X=x2)=P(X=x3)=5P(X=x4)
=>P(X=x2)=2/3P(X=x1) ; P(X=x3)=2/1(P(X=x1) ; P(X=x4)=2/5(P(X=x1)
=>P(X=x1)+2/3(P(X=x1)+2/1(P(X=x1)+2/5(P(X=x1)=1
=>61/15(P(X=x1)=1
=>P(X=x1)=15/61=0.24
=>P(X=x2)=2/3(P(X=x1)
=2/3(15/61)
=10/61=0.16
=>P(X=x3)=2/1(P(X=x1)
=2(15/61)
=30/61=0.49
=>P(X=x4)=2/5(P(X=x1))
=2/5(15/61)
=6/61=0.09
### Question 6. A random variable X takes the values 0,1,2 and 3 such that: P(X=0)=P(X>0)=P(X<0) ; P(X=-3)=P(X=-2)=P(X=-1) ; P(X=1)=P(X=2)=P(X=3). Obtain the probability distribution of X.
Solution:
We know that the sum of probability distributions is equal to 1.
=>P(X=0)+P(X>0)+P(X<0)=1
Given,
P(X=0)=P(X>0)=P(X<0)
=>P(X=0)+P(X=0)+P(X=0)=1
=>3P(X=0)=1
=>P(X=0)=1/3
=>P(X>0)=1/3
=>P(X=1)+P(X=2)+P(X=3)=1/3
Given,
P(X=1)=P(X=2)=P(X=3)
=>3P(X=1)=1/3
=>P(X=1)=1/9 ; P(X=2)=1/9 ; P(X=3)=1/9
=>P(X<0)=1/3
=>P(X=-1)+P(X=-2)+P(X=-3)=1/3
Given,
P(X=-3)=P(X=-2)=P(X=-1)
=>3P(X=-1)=1/3
=>P(X=-1)=1/9 ; P(X=-2)=1/9 ; P(X=-3)=1/9
### Question 7. Two cards are drawn from a well shuffled deck of 52 cards. Find the probability distribution of the number of aces.
Solution:
Given that two cards are drawn from a well shuffled deck of 52 cards.
Then the random variables for the probability distribution of the number of aces could be
i. No ace is drawn
ii. One ace is drawn
iii. Two aces are drawn
i. No ace is drawn:
P(X=0)=52-4C2/52C2
=48C2/52C2
=48!/2!x46!/52!/2!x50!
=48×47/52×51
=188/221
=0.85
ii. One ace is drawn:
P(X=1)=4C1x48C1/52C2
=4x48x2/52×51
=32/221
=0.14
iii. Two aces are drawn:
P(X=2)=4C2/52C2
=6×2/52×51
=1/221
=0.004
### Question 8. Find the probability distribution of number of heads, when three coins are tossed.
Solution:
Given that three coins are tossed simultaneously.
Then the random variables for the probability distribution of the number of heads could be,
P(X=0)=1C1x1C1x1C1/ 2C1x 2C1x 2C1
=1x1x1/2x2x2
=1/8
=0.125
P(X=1)=1C1+1C1+1C1/8
=1+1+1/8
=3/8
=0.37
P(X=2)=1C1+1C1+1C1/8
=3/8
=0.37
P(X=3)=1/8
=0.125
### Question 9. Four cards are drawn simultaneously from a well shuffled pack of 52 playing cards. Find the probability distribution of the number of aces.
Solution:
Given that four cards are drawn simultaneously from a well shuffled pack of 52 cards.
Then the random variables for the probability distribution of the number of aces drawn could be,
i. No aces
ii. One ace
iii. Two aces
iv. Three aces
v. Four aces
i. No aces
P(X=0)=48C4/ 52C4
=48x47x46x45/49x50x51x52
=0.71
ii. One ace
P(X=1)=4C1x48C3/52C4
=4x48x47x46x4/49x50x51x52
=0.25
iii. Two aces
P(X=2)= 4C2x48C2/ 52C4
=6x48x47x12/49x50x51x52
=0.024
iv. Three aces
P(X=3)= 4C3x48C1/52C4
= 4x48x24/49x50x51x52
= 0.0007
v. Four aces
P(X=4)=4C4/ 52C4
=1/ 270725
=0.000003694
### Question 10. A bag contains 4 red and 6 black balls. Three balls are drawn at random. Find the probability distribution of the number of red balls.
Solution:
Given that three balls are drawn at random from a bag.
Then the value of random variable for the probability distribution of number of red balls could be,
i. No red ball
ii. One red ball
iii. Two red balls
iv. Three red balls
i. No red balls:
P(X=0)=6C3/10C3
=6x5x4/10x9x8
=1/6=0.16
ii. One red ball:
P(X=1)=6C2x4C1/10C3
=6x5x4x3/10x9x8
=1/2=0.5
iii. Two red balls:
P(X=2)=6C1x4C2/10C3
=6x4x3x3/10x9x8
=3/10=0.3
iv. Three red balls:
P(X=3)=4C3/10C3
=4x3x2/10x9x8
=1/30=0.03
### Question 11. Five defective mangoes are accidentally mixed with 15 good ones. Four mangoes are drawn at random from this lot. Find the probability distribution of the number of defective mangoes.
Solution:
Given that five defective mangoes are mixed with 15 good ones.
Then the values of random variable for the probability distribution could be,
i. No defective
ii. One defective
iii. Two defective
iv. Three defective
v. Four defective
i. No defective:
P(X=0)=15C4/20C4
=15x14x13x12/20x19x18x17
=91/323=0.28
ii. One defective:
P(X=1)=15C3x5C1/20C4
=15x14x13x5x4/20x19x18x17
=455/969=0.469
iii. Two defective:
P(X=2)=15C2x5C2/20C4
=15x14x5x4x6/20x19x18x17
=70/323=0.21
iv. Three defective:
P(X=3)=15C1x5C3/20C4
=15x5x4x3x4/20x19x18x17
=10/323=0.03
v. Four defective:
P(X=4)=5C4/20C4
=5x4x3x2/20x19x18x17
=1/969=0.001
### Question 12. Two dice are thrown together and the number appearing on them is noted. X denotes the sum of the two numbers. Assuming that all the 36 outcomes are equally likely, what is the probability distribution of X?
Solution:
Given that two dice are thrown simultaneously.
Then the outcomes would be as follows:
(1,1) ; (1,2) ; (1,3) ; (1,4) ; (1,5) ; (1,6) ;
(2,1) ; (2,2) ; (2,3) ; (2,4) ; (2,5) ; (2,6) ;
(3,1) ; (3,2) ; (3,3) ; (3,4) ; (3,5) ; (3,6) ;
(4,1) ; (4,2) ; (4,3) ; (4,4) ; (4,5) ; (4,6) ;
(5,1) ; (5,2) ; (5,3) ; (5,4) ; (5,5) ; (5,6) ;
(6,1) ; (6,2) ; (6,3) ; (6,4) ; (6,5) ; (6,6)
The values of the random variable could be: 2,3,4,5,6,7,8,9,10,11,12
P(X=2)=1/36=0.02
P(X=3)=2/36=1/18=0.05
P(X=4)=3/36=1/12=0.08
P(X=5)=4/36=1/9=0.11
P(X=6)=5/36=0.13
P(X=7)=6/36=1/6=0.16
P(X=8)=5/36=0.13
P(X=9)=4/36=1/9=0.11
P(X=10)=3/36=1/12=0.08
P(X=11)=2/36=1/18=0.05
P(X=12)=1/36=0.02
### Question 13. A class has 15 students whose ages are 14,17,15,14,21,19,20,16,18,17,20,17,16,19 and 20 years respectively. One student is selected in such a manner that each has the same chance of being selected and the age X of the selected student is recorded. What is the probability distribution of the random variable X?
Solution:
Given that the students are selected without any bias.
Then the values of the random variable X could be: 14,15,16,17,18,19,20,21
P(X=14)=2/15=0.13
P(X=15)=1/15=0.06
P(X=16)=2/15=0.13
P(X=17)=3/15=0.2
P(X=18)=1/15=0.06
P(X=19)=2/15=0.13
P(X=20)=3/15=0.2
P(X=21)=1/15=0.06
### Question 14. Five defective bolts are accidentally mixed with twenty good ones. If four bolts are drawn at random from this lot, find the probability distribution of the number of defective bolts.
Solution:
Given that five defective bolts are mixed with 20 good ones.
Then the values of random variable for the probability distribution would be,
i. No defective
ii. One defective
iii. Two defective
iv. Three defective
v. Four defective
i. No defective:
P(X=0)=20C4/25C4
=20x19x18x17/25x24x23x22
=969/2530=0.38
ii. One defective:
P(X=1)=20C3x5C1/25C4
=20x19x18x5x4/25x24x23x22
=114/253=0.45
iii. Two defective:
P(X=2)=20C2x5C2/25C4
=20x19x5x4x6/25x24x23x22
=38/253=0.15
iv. Three defective:
P(x=3)=20C1x5C2/25C4
=20x5x4x4x3/25x24x23x22
=4/253=0.015
v. Four defective:
P(X=4)=5C4/25C4
=5x4x3x2/25x24x23x22
=1/2530=0.0004
### Question 15. Two cards are drawn successively with replacement from well shuffled pack of 52 cards. Find the probability distribution of number of aces.
Solution:
Given that two cards are drawn with replacement from well shuffled pack of 52 cards.
Then the values of random variable for the probability distribution could be,
i. No ace
ii. One ace
iii. Two aces
i. No ace:
P(X=0)=(48/52 )x(48/52)
=144/169=0.85
ii. One ace:
P(X=1)=(48/52)x(4/52)+(4/52)x(48/52)
=24/169=0.14
iii. Two aces:
P(X=2)=(4/52)x(4/52)
=1/169=0.005
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Addition of Integers – Part 3
# Addition of Integers – Part 3
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• Addition of Two Integers with Same Sign
• Addition of Two Integers with Different Sign
• Summary
• What’s Next?
In the previous segment, we discussed the addition of two negative integers. In this segment, we will look at the addition of two integers with different signs, that is a positive and a negative integer.
## Addition of two integers with the same sign
Adding two integers with the same sign means adding two positive numbers or
In both cases, the absolute values of the integers are added, and the answer will take the sign of the integers.
Example Sum of absolute values Sign of the integers Answer 2 + 3 |2| + |3| = 2 + 3 = 5 plus 5 −2 + (−3) |−2| + |−3| = 2 + 3 = 5 minus -5
## Addition of two Integers with different signs
To find the sum of two integers with a different sign, that is, one positive number and one negative number, their absolute values are subtracted and the result takes the sign of the integer corresponding to the greater absolute value.
For example,
## Q. 5 + (-3) = ?
Solution:
Method 1:
|−3| = 3 ??? |5| = 5
Now,
5 – 3 = 2
## Related content
Integers Representation of Integers on a Number Line Understanding the Number Line – Part 1 Understanding the Number Line – Part 2 Natural and Whole Numbers – Review Absolute Value Understanding the Concept of Absolute Value Successor and Predecessor of an Integer Ordering of Integers Signs of Integers
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# From a well-shuffled pack of 52 cards, a card is drawn at random.
Question:
From a well-shuffled pack of 52 cards, a card is drawn at random. Find the probability of its being a king or a queen
Solution:
let A denote the event that the card drawn is king and B denote the event that card drawn is queen.
In a pack of 52 cards, there are 4 king cards and 4 queen cards
Given : $P(A)=\frac{4}{52}, P(B)=\frac{4}{52}$
To find : Probability that card drawn is king or queen $=P(A$ or $B)$
The formula used : Probability =
$P(A$ or $B)=P(A)+P(B)-P(A$ and $B)$
$P(A)=\frac{4}{52}$ (as favourable number of outcomes $=4$ and total number of outcomes $=52$ )
$P(B)=\frac{4}{52}$ (as favourable number of outcomes $=4$ and total number of outcomes $\left.=52\right)$
Probability that card drawn is king or queen = P(A and B)= 0
(as a card cannot be both king and queen in the same time)
$P(A$ or $B)=\frac{4}{52}+\frac{4}{52}-0$
$P(A$ or $B)=\frac{4+4}{52}=\frac{8}{52}=\frac{2}{13}$
$P(A$ or $B)=\frac{2}{13}$
Probability of a card drawn is king or queen $=P(A$ or $B)=\frac{2}{13}$
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Trigonometric Differentiation
Learn about trigonometric differentiation, its formula along with different examples. Also find ways to calculate using trigonometric differentiation.
Alan Walker-
Published on 2023-05-26
Introduction to Trigonometric Differentiation
There are six basic trigonometric functions in geometry. The rate of change of these functions is known as trigonometric differentiation. This method follows the derivative rules and formulas to calculate the rate of change. Let’s understand how to apply trigonometric differentiation step-by-step and learn the difference between log and trig differentiation.
Understanding the Trigonometric Differentiation
A function that relates an angle of a right-angled triangle to its two sides is known as a trigonometric function. There are six basic trigonometric functions: cosine, tangent, cosecant, secant, and cotangent. In calculus, the derivative of trigonometric functions can be calculated using derivative rules.
By definition the trigonometric differentiation is defined as:
“The process of finding the derivative of a trigonometric function is called trigonometric differentiation.”
Trigonometric Differentiation Formulas
The derivative of a trigonometric function can be calculated by finding the rate of change of the sine and cosine functions. It is because knowing these two derivatives leads to the derivative of all other trig functions. The list of derivative formulas for trigonometric functions is as follows:
• $\frac{d}/dx(\sin x) = \cos x$
• $\frac{d}{dx}(\cos x) =\sin x$
• $\frac{d}{dx}(\sec x) = \sec x\tan x$
• $\frac{d}{dx}(\csc x) = -\csc x\cot x$
• $\frac{d}{dx}(\tan x) = \sec^2x$
• $\frac{d}{dx}(\cot x) = \csc²x$
The trigonometric differentiation formula can be modified to use it with other derivative rules. Let’s discuss the trig differentiation with the product rule, quotient, and power rules.
Trigonometric Differentiation and Product Rule
If a trig function is a product of two trig functions, the product rule with the trigonometric differentiation is used to calculate rate of change. For example, the derivative of secx tanx can be calculated as;
$\frac{d}{dx}(\sec x\tan x) = \sec x\frac{d}{dx}[\tan x] + \tan x\frac{d}{dx}[\sec x]$
Which is equal to,
$\frac{d}{dx}(\sec x\tan x) = \sec x(\sec^2x) + \tan x[\sec x\tan x]=\sec^3x+\tan^2xsec x$
Since $\sec^2x + \tan^2x = 1$ then,
$\frac{d}{dx}(\sec x\tan x)=\sec x(\sec^2x+\tan^2x)=\sec x$
Trigonometric Differentiation and Power Rule
Since the trigonometric differentiation is used to calculate derivatives of a trigonometric function. It can be used along with the power rule if the function contains a trig function with power n. The relation between power rule and trig differentiation for a function $f(x) = \cos^3x$ is expressed as;
$f’(x)=\frac{d}{dx}[\cos^3x]$
By using power rule formula,
$f’(x,y)=3\cos^{3-1}\frac{d}{dx}(\cos x)= -3\cos^2x\sin x$
Where, the cos^3x derivative with respect to x is $–3\cos^2x\sin x$.
Trigonometric Differentiation and Quotient Rule
If a trigonometric function is divided by another function, the trigonometric differentiation along with the quotient rule to find derivative. For a quotient of a function f(x) = tan x = sin x/cos x, the relation between trig derivative and quotient rule is,
$\frac{df}{dx}=\frac{\cos x\frac{d}{dx}[\sin x] – \sin x\frac{d}{dx}[\cos x]}{\cos^2x}$
And,
$\frac{df}{dx}=\frac{\cos^2x + \sin^2x}{\cos^2x} = \frac{1}{\cos^2x}=\sec^2x$
Hence the derivative of tan x is $\sec^2x$.
How do you do Trigonometric differentiation step by step?
The implementation of trigonometric derivatives is divided into a few steps. These steps assist us in calculating the derivative of a function having a trigonometric identity. These steps are:
1. Write the expression of the function.
2. Identify the trig function.
3. Differentiate the function with respect to the variable involved.
4. Use the trigonometric differentiation formula to calculate derivatives. For example, the derivative of sec x is tanx secx.
5. Simplify if needed.
Applying logarithmic differentiation formula by using calculator
The derivative of a log function can also be calculated using the derivative calculator. The online tool follows the log differentiation formula to find the derivative. You can find it online by searching for a derivative calculator. For example, to calculate the derivative of ln x, the following steps are used by using this calculator.
1. Write the expression of the function in the input box, such as ln x.
2. Choose the variable to calculate the rate of change, which will be x in this example.
3. Review the input so there will be no syntax error in the function.
4. Now at the last step, click on the calculate button. By using this step, the derivative calculator will provide the derivative of ln x quickly and accurately, which will be 1/x
Comparison between Trigonometric and Logarithmic differentiation
The comparison between the logarithmic and trigonometric differentiation can be easily analysed using the following difference table.
Logarithmic Differentiation Trigonometric Differentiation The logarithmic differentiation is used to calculate the derivative of a logarithmic function. The trig differentiation is used to calculate derivative of a trigonometric function. The derivative of a logarithmic function ln x is defined as;$f’(x) = \frac{1}{x}\frac{d}{dx}(x)$ There are different formulas to calculate derivative of trigonometric functions. The logarithmic differentiation can be used along with different derivative formulas. The derivative of all trigonometric functions can be calculated by using product rule, quotient rule and power rule.
Conclusion
The trigonometric functions are the functions that relate an angle with the right-angle triangle. Calculating the derivative of such functions is known as trig differentiation. In conclusion, the trigonometric differentiation of all trig functions can be calculated by using the derivatives of sine and cosine.
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# Math Tutorial 1.6 - Divisibility Rules
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There are 21 lessons in this math tutorial covering Divisibility Rules. The tutorial starts with an introduction to Divisibility Rules and is then followed with a list of the separate lessons, the tutorial is designed to be read in order but you can skip to a specific lesson or return to recover a specific math lesson as required to build your math knowledge of Divisibility Rules. you can access all the lessons from this tutorial below.
In this Math tutorial, you will learn:
• What does divisibility mean?
• How do we write in symbols when a number is divisible by another number?
• What are the divisibility rules for the numbers from 1 to 20?
• What are relatively prime numbers?
• How do relatively prime numbers determine some of divisibility rules?
## Introduction
Can you quickly calculate whether 91 is divisible by 7 or not? How?
What is the common divisor of all even numbers?
How would you find whether 100 000 is divisible by 25 or not?
In this guide, we will learn how to find whether a number is divisible by another number or not without using a calculator or doing other complicated operations to find the result. This is particularly important in situations where a quick response is crucial, such as in exams.
## The meaning of Divisibility
An integer x is divisible by another integer y, if the result of x ÷ y is another integer, i.e. it is a number without remainder (r = 0). We write the symbol (⁝) to represent the divisibility of two numbers. Thus, the expression x ⁝ y reads "the integer x is divisible by the integer y without a remainder".
For example, the arithmetic sentence 24 ⁝ 8 is true as 24 ÷ 8 = 3 (0), while the arithmetic sentence 31 ⁝ 7 is not true as 31 ÷ 7 = 4 (3).
We will cover the divisibility rules for divisors from 1 to 20, as they are the most commonly used numbers in practice. However, we will briefly explain divisibility rules for some numbers greater than 20 as well to expand your understanding of this math topic.
Please select a specific "Divisibility Rules" lesson from the table below, review the video tutorial, print the revision notes or use the practice question to improve your knowledge of this math topic.
Arithmetic Learning Material
Tutorial IDMath Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
1.6Divisibility Rules
Lesson IDMath Lesson TitleLessonVideo
Lesson
1.6.1Divisibility by 1
1.6.2Divisibility by 2
1.6.3Divisibility by 3
1.6.4Divisibility by 4
1.6.5Divisibility by 5
1.6.6Divisibility by 6
1.6.7Divisibility by 7
1.6.8Divisibility by 8
1.6.9Divisibility by 9
1.6.10Divisibility by 10
1.6.11Divisibility by 11
1.6.12Divisibility by 12
1.6.13Divisibility by 13
1.6.14Divisibility by 14
1.6.15Divisibility by 15
1.6.16Divisibility by 16
1.6.17Divisibility by 17
1.6.18Divisibility by 18
1.6.19Divisibility by 19
1.6.20Divisibility by 20
1.6.21Other Divisibility Rules. How Relatively Prime Numbers Determine the Divisibility Rules.
## Whats next?
Enjoy the "Divisibility Rules" math tutorial? People who liked the "Divisibility Rules" tutorial found the following resources useful:
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2. Arithmetic Video tutorial: Divisibility Rules. Watch or listen to the Divisibility Rules video tutorial, a useful way to help you revise when travelling to and from school/college
3. Arithmetic Revision Notes: Divisibility Rules. Print the notes so you can revise the key points covered in the math tutorial for Divisibility Rules
4. Arithmetic Practice Questions: Divisibility Rules. Test and improve your knowledge of Divisibility Rules with example questins and answers
5. Check your calculations for Arithmetic questions with our excellent Arithmetic calculators which contain full equations and calculations clearly displayed line by line. See the Arithmetic Calculators by iCalculator™ below.
6. Continuing learning arithmetic - read our next math tutorial: Decimal Number System and Operations with Decimals
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offers hundreds of practice questions and video explanations. .
or to Clemmonsdogpark GMAT Prep.
GMAT Sets: Venn Diagrams
Learn this technique to master set questions of GMAT Quant word problem.
Practice questions
First, try these challenging practice questions.
1) Of the 80 houses in a development, 50 have a two-car garage, 40 have an in-the-ground swimming pool, and 35 have both a two-car garage and an in-the-ground swimming pool. How many houses in the development have neither a two-car garage nor an in-the-ground swimming pool?
A. 10
B. 15
C. 20
D. 25
E. 30
2) A certain school has three performing arts extracurricular activities: Band, Chorus, or Drama. Students must participate in at least one, and may participate in two or even in all three. There are 120 students in the school. There are 70 students in Band, 73 in the Chorus, and 45 in the Drama. Furthermore, 37 students are in both the Band and Chorus, 20 are in both the Band and the Drama, and 8 students are in all three groups. Twenty-five students are just in the chorus, not in anything else. How many students participate in only the drama?
A. 11
B. 12
C. 14
D. 17
E. 21
Introduction to sets
The idea of a set is, in some sense, the most fundamental idea in all of . Nonetheless, it’s a very simple idea. A set is simply a collection of objects or elements. When the members of a set are all numbers, then we use “set notation”, which consists of brackets. For example:
A = {2, 3, 3, 5, 7}
That notation denotes set A with five numerical members. When all the members of the set are numbers, typical questions involve computations like the mean or the median: here, the mean (or average) of set A is 4, and the median of A is 3. You can read more about those calculations at this post.
Overlapping sets
Numerical sets can be handled with statistical calculations. Non-numerical sets, sets in which the members are people or cars or companies, are the stuff of tricky word problems. Ordinarily, there’s nothing particular challenging if there’s only one set: some of the people are in that set, whatever it is, and the rest aren’t. No challenge.
Things get more interesting if there are two or more overlapping sets. For example, in the SF Bay Area, many adult residents were born out of state — many, but not all; many adult residents have a college education — many, but not all; and many adult residents are SF Giants fans — many, but not all. Those are three overlapping sets —- any particular adult resident of the SF Bay Area many be a member of none, one, two, or all three of those three categories. As it happens, I am a member of exactly two of those categories. This is exactly the situation of the practice questions posed above.
Venn diagrams
Venn diagrams are the best method for untangling overlapping sets. If you have two overlapping sets, you need a two-circle Venn diagram:
This diagram contains three discrete regions:
A = those elements in just the left circle
B = those element in both categories, in the overlap
C = those elements in just the right circle
There may also be a fourth discrete region, those elements that are not members of any set. Typically, the problem will only give us information about totals — the total number of elements altogether, the total number in each circle, and the overlap. If you are told there are 70 members in the right circle, and 20 members in the overlap, then you would know B + C = 70 and B = 20, so from that you could deduce C = 50, the number of elements that are just in the portion of the circle labeled C. In general, you work from the center outward, figuring out one discrete region after another. Once you know the value of each individual discrete region, you will be able to answer any question about the number in any particular grouping.
If there are three overlapping categories, we use a three-circle Venn diagram:
This diagram has at least seven discrete regions
A = members of all three circles
B = members of the green and blue circles, but not the red circle
C = members of the green and red circles, but not the blue circle
D = members of the blue and red circles, but not the green circle
E = members of the green circle but of neither the blue nor the red circles
F = members of the blue circle but of neither the green nor the red circles
G = members of the red circle but of neither the green nor the blue circles
Depending on context, there may also be a eighth discrete region, those elements that are not members of any of the three set. Typically, the problem will only give us information about totals. This gets very tricky. If we are told the total in any one circle, that includes four discrete regions; for example, the green circle includes A + B + C + E. Similarly, the overlap of two circles contains two discrete regions: for example, the overlap of the blue and red circles includes A + D. The problem will always tell you how many elements are in the central region (A), and will often tell you how many are in each circle, and how many in each overlap of two circles. In general, you work from the center outward, figuring out one discrete region after another. Once you know the value of each individual discrete region, you will be able to answer any question about the number in any particular grouping.
With these strategies, you may find the practice problems at the beginning somewhat more approachable. Try them again, before reading the explanations below.
In the next post, we will look at set problems in which each element is categorized according to two different variables at once.
Practice question explanations
1) Here, we have two categories: (a) with or without two-car garage, and (b) with or without an in-the-ground pool. Houses can be members of either, both, or neither category. We will use a two circle Venn diagram:
We know the total of the group is 80 —– A + B + C + D = 80. We know the green circle, two-car garages, has 50 members, so A + B = 50. We know the blue circle, in-the-ground pool, has 40 members, so B + C = 40. We also know the crucial overlap region, B = 35. If B = 35, in the green circle, we can deduce that A = 15, and in the blue circle, we can deduce that C = 5. Then
A + B + C + D = 15 + 35 + 5 + D = 80
D = 25
Thus, 25 houses in this development have neither a two-car garage nor an in-the-ground swimming pool. Answer = D.
2) Here, we have three categories, so we need three circles. Every student must take at least one of these three performing arts extracurricular activities, so there will be no one outside the three circles.
The sum of all seven = 120 (we never use this number in this question)
The totals for the band (70), the chorus (73), and the drama (45) each involve the sum of four discrete regions. We will have to find other information before we can employ them.
“8 students are in all three groups”
N = 8.
“37 students are in both the Band and Chorus”
37 = K + N = K + 8 —> K = 29
“20 are in both the Band and the Drama”
20 = M + N = M + 8 —> M = 12
“twenty-five students are just in the chorus, not in anything else”
L = 25
We now have identified three of the regions in the Chorus circle, so we can solve for P.
chorus = 73 = K + L + N + P
73 = 29 + 25 + 8 + P
P = 11
Now, we have identified three of the regions in the Drama circle, so we can solve for Q.
drama = 45 = M + N + P + Q
45 = 12 + 8 + 11 + Q
Q = 14
This is precisely what the question was asking: how many students are only in drama? There are 14 students who take only drama.
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New
New
Year 3
# Estimate the position of 3-digit numbers on unmarked number lines
I can estimate the position of 3-digit numbers on an unmarked number line.
New
New
Year 3
# Estimate the position of 3-digit numbers on unmarked number lines
I can estimate the position of 3-digit numbers on an unmarked number line.
Share activities with pupils
Share function coming soon...
## Lesson details
### Key learning points
1. Knowing where multiples of 100 are placed on an unmarked number line supports in positioning 3-digit numbers accurately.
2. Marking halfway steps helps to estimate the positions of numbers.
3. Counting in steps of 1, 10 and 100 helps us to position a number on a number line.
### Common misconception
Pupils may struggle with positioning numbers on 0-1,000 lines.
Relate the positioning of numbers on 0-100 number lines with 0-1,000 number lines (e.g. 19 and 190, 37 and 370).
### Keywords
• Estimate - Estimate means finding a value that is close to the answer.
• Position - Position means where something is located.
• Unmarked - Unmarked means that many of the steps on a number line are not given.
Encourage pupils to find the halfway points on unmarked number lines, using the start and end numbers. For example, on a 200-400 number line, mark the steps that show 300, then find 250 and 350.
Teacher tip
### Licence
This content is © Oak National Academy Limited (2024), licensed on Open Government Licence version 3.0 except where otherwise stated. See Oak's terms & conditions (Collection 2).
## Starter quiz
### 6 Questions
Q1.
What number is halfway between 0 and 100?
20
40
70
Q2.
What number is halfway between 100 and 200?
Correct Answer: 150, one hundred and fifty
Q3.
What number is represented by the arrow?
205
270
350
Q4.
What number is represented by the arrow?
Correct Answer: 750, seven hundred and fifty
Q5.
Sofia is thinking of a 10s number that could be represented on the number line. The digits of the number add up to 6. What is the number?
Correct Answer: 240, two hundred and forty
Q6.
What number is represented by the arrow?
## Exit quiz
### 6 Questions
Q1.
The arrow shows the position of a number. Which number?
Correct Answer: 450, four hundred and fifty
Q2.
The arrow shows the position of a number. Which number?
50
150
450
Q3.
Which of these is the best estimate for the number shown by the arrow on this number line?
90
110
130
Q4.
Which of these is the best estimate for the number shown by the arrow on this number line?
100
500
700
Q5.
Which of these is the best estimate for the number shown by the arrow on this number line?
350
750
850
Q6.
Which arrow represents the position of 125?
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# Perimeter and Area Class 7 Extra Questions Maths Chapter 11
Perimeter and Area Class 7 Extra Questions Maths Chapter 11
### Perimeter and Area Class 7 Extra Questions Maths Chapter 11
Extra Questions for Class 7 Maths Chapter 11 Perimeter and Area
### Perimeter and Area Class 7 Extra Questions Very Short Answer Type
Question 1.
The side of a square is 2.5 cm. Find its perimeter and area.
Solution:
Side of the square = 2.5 cm
Perimeter = 4 × Side = 4 × 2.5 = 10 cm
Area = (side)2 = (4)2 = 16 cm2
Question 2.
If the perimeter of a square is 24 cm. Find its area.
Solution:
Perimeter of the square = 24 cm
Side of the square = 24/4 cm = 6 cm
Area of the square = (Side)2 = (6)2 cm2 = 36 cm2
Question 3.
If the length and breadth of a rectangle are 36 cm and 24 cm respectively. Find
(i) Perimeter
(ii) Area of the rectangle.
Solution:
Length = 36 cm, Breadth = 24 cm
(i) Perimeter = 2(l + b) = 2(36 + 24) = 2 × 60 = 120 cm
(ii) Area of the rectangle = l × b = 36 cm × 24 cm = 864 cm2
Question 4.
The perimeter of a rectangular field is 240 m. If its length is 90 m, find:
(ii) it’s area.
Solution:
(i) Perimeter of the rectangular field = 240 m
2(l + b) = 240 m
l + b = 120 m
90 m + b = 120 m
b = 120 m – 90 m = 30 m
So, the breadth = 30 m.
(ii) Area of the rectangular field = l × b = 90 m × 30 m = 2700 m2
So, the required area = 2700 m2
Question 5.
The length and breadth of a rectangular field are equal to 600 m and 400 m respectively. Find the cost of the grass to be planted in it at the rate of ₹ 2.50 per m2.
Solution:
Length = 600 m, Breadth = 400 m
Area of the field = l × b = 600 m × 400 m = 240000 m2
Cost of planting the grass = ₹ 2.50 × 240000 = ₹ 6,00,000
Hence, the required cost = ₹ 6,00,000.
Question 6.
The perimeter of a circle is 176 cm, find its radius.
Solution:
The perimeter of the circle = 176 cm
Question 7.
The radius of a circle is 3.5 cm, find its circumference and area.
Solution:
Circumference = 2πr
Question 8.
Area of a circle is 154 cm2, find its circumference.
Solution:
Area of the circle = 154 cm2
Question 9.
Find the perimeter of the figure given below.
Solution:
Perimeter of the given figure = Circumference of the semicircle + diameter
= πr + 2r
22/7 × 7 + 2 × 7
= 22 + 14
= 36 cm
Hence, the required perimeter = 36 cm.
Question 10.
The length of the diagonal of a square is 50 cm, find the perimeter of the square.
Solution:
Let each side of the square be x cm.
x2 + x2 = (50)2 [Using Pythagoras Theorem]
2x2 = 2500
x2 = 1250
x = 5 × 5 × √2 = 25√2
The side of the square = 25√2 cm
Perimeter of the square = 4 × side = 4 × 25√2 = 100√2 cm
### Perimeter and Area Class 7 Extra Questions Short Answer Type
Question 11.
A wire of length 176 cm is first bent into a square and then into a circle. Which one will have more area?
Solution:
Length of the wire = 176 cm
Side of the square = 176 ÷ 4 cm = 44 cm
Area of the square = (Side)2 = (44)2 cm2 = 1936 cm2
Circumference of the circle = 176 cm
Since 2464 cm2 > 1936 cm2
Hence, the circle will have more area.
Question 12.
In the given figure, find the area of the shaded portion.
Solution:
Area of the square = (Side)2 = 10 cm × 10 cm = 100 cm2
Area of the circle = πr2
22/7 × 3.5 × 3.5
77/2 cm2
= 38.5 cm2
Area of the shaded portion = 100 cm2 – 38.5 cm2 = 61.5 cm2
Question 13.
Find the area of the shaded portion in the figure given below.
Solution:
Area of the shaded portion = 196 cm2 – 154 cm2 = 42 cm2
Question 14.
A rectangle park is 45 m long and 30 m wide. A path 2.5 m wide is constructed outside the park. Find the area of the path.
Solution:
Length of the rectangular park = 45 m
Breadth of the park = 30 m
Area of the park = l × 6 = 45m × 30m = 1350 m2
Length of the park including the path = 45 m + 2 × 2.5 m = 50 m
Breadth of the park including the path = 30 m + 2 × 2.5 m = 30m + 5m = 35m
Area of the park including the path = 50 m × 35 m = 1750 m2
Area of the path = 1750 m2 – 1350 m2 = 400 m2
Hence, the required area = 400 m2.
Question 15.
In the given figure, calculate:
(а) the area of the whole figure.
(b) the total length of the boundary of the field.
Solution:
Hence, the required (i) area = 4746 cm2 and (ii) length of boundary = 292 cm.
Question 16.
### Perimeter and Area Class 7 Extra Questions Long Answer Type
Question 17.
A nursery school playground is 160 m long and 80 m wide. In it 80 m × 80 m is kept for swings and in the remaining portion, there are 1.5 m wide path parallel to its width and parallel to its remaining length as shown in Figure. The remaining area is covered by grass. Find the area covered by grass. (NCERT Exemplar)
Solution:
Area of the playground = l × b = 160 m × 80 m = 12800 m2
Area left for swings = l × b = 80m × 80m = 6400 m2
Area of the remaining portion = 12800 m2 – 6400 m2 = 6400 m2
Area of the vertical road = 80 m × 1.5 m = 120 m2
Area of the horizontal road = 80 m × 1.5 m = 120 m2
Area of the common portion = 1.5 × 1.5 = 2.25 m2
Area of the two roads = 120 m2 + 120 m2 – 2.25 m2 = (240 – 2.25) m2 = 237.75 m2
Area of the portion to be planted by grass = 6400 m2 – 237.75 m2 = 6162.25 m2
Hence, the required area = 6162.25 m2.
Question 18.
Rectangle ABCD is formed in a circle as shown in Figure. If AE = 8 cm and AD = 5 cm, find the perimeter of the rectangle. (NCERT Exemplar)
Solution:
DE (Radius) = AE + AD = 8 cm + 5 cm = 13 cm
DB = AC = 13 cm (Diagonal of a rectangle are equal)
AD2 + DC2 = AC2 (By Pythagoras Theorem)
⇒ (5)2 + DC2 = (13)2
⇒ 25 + DC2 = 169
⇒ DC2 = 169 – 25 = 144
⇒ DC = √144 = 12 cm
Perimeter of rectangle ABCD = 2(AD + DC)
= 2(5 cm + 12 cm)
= 2 × 17 cm
= 34 cm
Question 19.
Find the area of a parallelogram-shaped shaded region. Also, find the area of each triangle. What is the ratio of the area of shaded portion to the remaining area of the rectangle?
Solution:
Here, AB = 10 cm
AF = 4 cm
FB = 10 cm – 4 cm = 6 cm
Area of the parallelogram = Base × Height = FB × AD = 6 cm × 6 cm = 36 cm2
Hence, the required area of shaded region = 36 cm2.
Area of Rectangle ABCD = l × b = 10 cm × 6 cm = 60 cm2
Remaining area of Rectangle = 60 cm2 – 36 cm2 = 24 cm2
Required Ratio = 36 : 24 = 3 : 2
Question 20.
A rectangular piece of dimension 3 cm × 2 cm was cut from a rectangular sheet of paper of dimensions 6 cm × 5 cm. Find the ratio of the areas of the two rectangles. (NCER T Exemplar)
Solution:
Length of the rectangular piece = 6 cm
Area of the sheet = l × b = 6 cm × 5 cm = 30 cm2
Area of the smaller rectangular piece = 3 cm × 2 cm = 6 cm2
Ratio of areas of two rectangles = 30 cm2 : 6 cm2 = 5 : 1
### Perimeter and Area Class 7 Extra Questions Higher Order Thinking Skills (HOTS) Type
Question 21.
In the given figure, ABCD is a square of side 14 cm. Find the area of the shaded region.
(Take π = 22/7)
Solution:
Area of the square PQRS = Side × Side = 7 cm × 7 cm = 49 cm2
Area of the shaded portion = 49 cm2 – 38.5 cm2 = 10.5 cm2
Hence, the required area = 10.5 cm2.
Question 22.
Find the area of the following polygon if AB = 12 cm, AC = 2.4 cm, CE = 6 cm, AD = 4.8 cm, CF = GE = 3.6 cm, DH = 2.4 cm.
Solution:
BE = AB – AE
= 12 cm – (AC + CE)
= 12 cm – (2.4 cm + 6 cm)
= 12 cm – 8.4 cm
= 3.6 cm
Area of the polygon AFGBH = Area of ∆ACF + Area of rectangle FCEG + Area of ∆GEB + Area of ∆ABH
= 3.6 cm2 + 4.32 cm2 + 21.6 cm2 + 6.48 cm2 + 14.4 cm2
= 50.40 cm2
Hence, the required area = 50.40 cm2.
## SabDekho
The Complete Educational Website
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# 9R Elimination w Multiplication - MrNappiMHS
```Do Now
Solve using elimination:
3x + 2y = – 19
– 3x – 5y = 25
Homework Solutions
1)
x – y = 1
+ x + y = 3
2x
= 4
x = 2
2+ y = 3
y = 1
Solution: (2, 1)
Homework Solutions
3)
x + 4y
– x – 6y
10y
y
= 11
= 11
= 0
= 0
x + 4(0) = 11
x = 11
Solution: (11, 0)
Homework Solutions
5)
3x + 4y = 19
– 3x + 6y = 33
-2y = -14
y = 7
3x + 4(7) = 19
3x + 28 = 19
3x = -9
x = -3
Solution: (-3, 7)
Homework Solutions
7)
3a + 4b = 2
+ 4a – 4b = 12
7a
= 14
a = 2
3(2) + 4b = 2
6 + 4b = 2
4b = -4
b = -1
Solution: (2, -1)
Homework Solutions
9)
2x – 3y = 9
– -5x – 3y = 30
7x
= -21
x = -3
2(-3) – 3y = 9
-6 – 3y = 9
-3y = 15
y = -5
Solution: (-3, -5)
6s + 2t = 24
4s + 2t = 18
The coefficients of one of the variables are
not always going to be the same.
When neither variable’s coefficients are the
same, you can’t eliminate one variable just
Sometimes we have to MULTIPLY one or
both equations to make the coefficients of
one of the variables the same.
3x + 4y = 6
5x + 2y = – 4
3x + 2y = 0
x – 5y = 17
4c – 3d = 22
2c – d = 10
8 x 3 y 11
2 x 5 y 27
4x 5 y 6
6 x 7 y 20
5 x 3 y 12
4 x 5 y 17
3x + 2y = – 9
5x – 3y = 4
Homework
Worksheet pg. 423 #’s 1, 4, 7, 10
```
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### Infant Maths Evening
```Infant Maths Evening
Helping Your Child with Their Maths at Home
Maths in the Infants
Progression in number work
Methods we use for the operations
Other maths in the infants
Counting using Objects
•The first step in children’s number work is counting up to 10 and
beyond.
• Children then need to understand how to relate the numbers to
objects.
• They need to come up with a system so that they do not miss
objects.
• We encourage children to put the objects in a line and start from
one side.
• We also encourage them to touch the objects as they count them.
Relating amounts to number
Children then need to be able to recognise the numbers that
they are using to count. Relating the numbers to a numeral is
quite a big jump for some children. The more familiar they are
with the numerals, the quicker they will learn them.
=5
=6
Recognising and Writing numbers
This is how we write numbers in our school. The earlier children
practise writing numbers the right way round the less likely they are
to get into the habit of writing them incorrectly. In early number
formation 2 and 5 are easily confused.
2 3
6 7 8
5
0
Ordering numbers
2 5 3
2 3
5
Key Words:
• More than
• Less / fewer than
Ordering numbers
9 3 2 7 2
2 3 7 9 2
Key Words:
• More than
• Less / fewer than
Place Value
• A child having a deep understanding of place value is integral to
their progression in maths.
•Once they are familiar with numbers over 10 we work on
identifying the ‘tens digit’ and the ‘units digit’ in each number.
•It is important that the children know the value of each digit.
• In this example 13 is made up of ‘1 ten’ and ‘3 units’
•Place Value cards are one resource we use to support this
concept.
0
Place Value
In school we also use tens rods and unit cubes to help
children understand that 10 units is the same as one set of 10.
= 10
=1
= 36
You could support this idea at home when they are counting numbers
greater than 10, by grouping objects together in tens as they count up.
Place Value
To further support this idea we have 100 squares which are
the size of 10 tens rods.
= 124
Place Value
The children need to be
able to locate given
numbers in a hundred
square by identifying the
tens digit of that number
first then finding the
corresponding row. They
should also know that
the higher the tens digit,
the lower the row is
located in the hundred
square.
Key Words:
• tens /units digit
• teens number
Number Facts
A ‘number bond’ is two numbers which are added together to
make another number.
Children need to work towards a quick recall of number bonds
for 5 e.g. 1 + 4, 2 + 3......
They will also need to know the number bonds for 10 off by
heart e.g. 0 + 10, 1 + 9, 2 + 8.....
As their understanding of place value improves they will start to
be able to recall number bonds for larger numbers using the
above number bonds to help them.
We do work on this in class; however once your child
understands what a number bond is, quick recall comes from
frequent practice.
Another vital mental maths skill is doubling numbers up to 5 /10
/ 20. This is first taught using hands and then pictures.
After this, the children will then learn the inverse of doubling:
halving.
We often get asked what objects children should use to
help them add up at home...... ANYTHING!!!
objects, combine them and see how much there is
‘altogether’.
For subtraction, encourage children to count out the larger
group then ‘take away’ the smaller number and see ‘how
many are left’.
We use lots of different words for addition and subtraction,
and we do not introduce the + and – symbols until children
are very confident with the operations.
Using a number line to add
•Children can start to use a number line for addition
and subtraction when they start to have a better
understanding of abstract number.
•It is important that they relate addition to ‘counting
on’ and subtraction to ‘counting back’ on the number
lines.
•They must understand that, with addition, the total
amount will be the largest and, when taking away,
the result will be smaller than the initial amount.
1
2
3 4 5
6 7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
12 + 8 =
Using a number line to subtract
When subtracting, children will need to understand
back.
Some children prefer to ‘find the difference’ to solve
subtraction number sentences – where they start
with the lower number.
1
2
3 4 5
6 7
8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
20 – 8 =
A hundred square
•When dealing with
larger numbers children
progress from using a
number line to a
hundred square.
•The methods of
are the same as on a
number line.
•Children soon learn
that, to add 10, they can
simply ‘jump down’ 1
place.
subtraction, we use
Multiplication and Division
We do not use the symbols for multiplication or division
until children are confident with the concept of ‘lots of’ as
repeated addition and division as ‘sharing’.
2 + 2 + 2 + 2 =8
Key Words:
• Lots of ...
•Sets of …
•Groups of …
• Shared between…
Word problems
Once the children are confident with using the methods
of each operation we use word problems so they can
apply their skills to ‘real life’ situations.
The problem:
Bob had 24 sweets. He ate 6
• When the children are familiar
with more than one operation
an important part of word
problems is deciding what
operation to use.
many sweets does Bob
What do I need to do?
Write the number sentence and solve it:
Data handling
Tally chart
Pictogram
Bar graph
Venn diagram
Carroll diagram
Other Maths in The Infants
Patterns
Sorting
Other Maths in The Infants
2D Shape
3D Shape
Key Words:
• Faces
• Edges
• Vertices
Key Words:
• Corners
• Sides
•Straight
•Curved
Other Maths in The Infants
Measuring
Key Words:
• Estimate
• Length – long, tall, wide
thick thin......not ‘big’
• Mass – weigh, light, heavy
• Capacity – full, empty
Other Maths in The Infants
Time
• Begin by sequencing events.
• Distinguish between times of day,
e.g. morning, afternoon, night.
• Learn days, then months, in order.
• Analogue clock to tell the time.
• Events that happen at o’clock times.
• Hour hand points to an o’clock, or
tells us where we are in relation to an
o’clock.
• Minute hand tells us if it is o’clock
now, or how many minutes past an
o’clock or coming up to an o’clock.
• Once confident, move onto 12 hour
Money
Need to recognise coins and know the value of
each.
When counting small amounts, tap the coin the
correct amount of times.
Making totals, first with 1ps, then using other
coins.
Finding change.
The more opportunity
money, the easier they
will find maths related
to it.
Language in Maths
We have included a vocabulary list in your
packs to show the words that get used in
maths lessons in the infants.
We encourage children to verbalise their
understanding and explain how they have
|
# What Is 71/100 as a Decimal + Solution With Free Steps
The fraction 71/100 as a decimal is equal to 0.71.
Long division is a mathematical operation that enables you to solve challenging and intricate division-related problems. The Long division approach simplifies difficult division by breaking enormous numbers into manageable pieces.
Here, we are more interested in the division types that result in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division, which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 71/100.
## Solution
First, we convert the fraction components, i.e., the numerator and the denominator, and transform them into the division constituents, i.e., the Dividend and the Divisor, respectively.
This can be done as follows:
Dividend = 71
Divisor = 100
Now, we introduce the most important quantity in our division process: the Quotient. The value represents the Solution to our division and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 71 $\div$ 100
This is when we go through the Long Division solution to our problem. Figure 1 shows how Long Division is done:
Figure 1
## 71/100 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 71 and 100, we can see how 71 is Smaller than 100, and to solve this division, we require that 71 be Bigger than 100.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If so, we calculate the Multiple of the divisor closest to the dividend and subtract it from the Dividend. This produces the Remainder, which we then use as the dividend later.
Now, we begin solving for our dividend 71, which after getting multiplied by 10 becomes 710.
We take this 710 and divide it by 100; this can be done as follows:
710 $\div$ 100 $\approx$ 7
Where:
100 x 7 = 700
This will lead to the generation of a Remainder equal to 710 – 700 = 10. Now this means we have to repeat the process by Converting the 10 into 100 and solving for that:
100 $\div$ 100 $=$ 1
Where:
100 x 1 = 100
This, therefore, produces another Remainder which is equal to 100 – 100 = 0.
Finally, we have a Quotient generated after combining the three pieces of it as 0.71=z, with a Remainder equal to 0.
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# Book I. Propositions 33 and 34
Problems
Back to Propositions 33, 34.
11. a) State the hypothesis of Proposition 33.
Do the problem yourself first!
Straight lines join the extremities on the same side of two equal and parallel straight lines.
12. b) State the conclusion.
Those straight lines are themselves equal and parallel.
12. c) Practice Proposition 33.
12. State the definition of a "parallelogram."
A parallelogram is a quadrilateral whose opposite sides are parallel. (Definition 14)
13. a) State the hypothesis of Proposition 34.
A figure is a parallelogram.
13. b) State the conclusion.
The opposite sides and angles are equal, and the diagonal bisects the area.
13. c) Practice Proposition 34.
14. The straight line ADE is parallel to the straight line BC, and AB is
14. parallel to DC.
Prove that AB is equal to DC, and angle EDC is equal to angle DAB.
Since AD is parallel to BC, and AB is parallel to DC,
ABCD is a parallelogram. (Definition 14)
Therefore the opposite sides AB, DC are equal. (I. 34)
Also, because AE meets the two parallel lines AB, DC,
the exterior angle EDC is equal to the opposite interior angle DAB. (I. 29)
15. By the distance from a point to a line, we mean the length of the
15. perpendicular from the point to the line.
15. Prove that two parallel lines are everywhere the same distance apart.
Let the straight lines AB, CD be parallel, and let E and G be any two points on AB;
then the distance from E to CD is equal to the distance from G from CD.
Draw EF and GH perpendicular to CD;
then EF will be equal to GH.
For, the straight line AB meets the two straight lines
EF, GH, and the exterior angle AEF is equal to opposite interior angle EGH,
because they are right angles;
therefore EF and GH are parallel. (I. 28)
And by hypothesis, AB, CD are parallel.
Therefore EFHG is a parallelogram, (Def. 14)
hence the opposite sides EF, GH are equal. (I. 34)
And E and G were any two points on AB.
This implies that the parallel lines AB and CD are everywhere the same distance apart.
This could also be proved by showing that EG, FH are equal, and
15. then citing I. 33.
16. Straight lines EH, BCG are parallel; EB and HC are straight lines;
16. BC is equal to FG; and EFGH is a parallelogram.
Prove that EBCH is a parallelogram.
Since EFGH is a parallelogram, then EH is equal to FG; (I. 34)
but BC is equal to FG; (Hypothesis)
therefore BC is equal to EH. (Axiom 2)
And since EH, BC are parallel, (Hypothesis)
and EB, HC join their extremities on the same side,
then EB, HC are parallel. (I. 33)
Therefore EBCH is a parallelogram. (Definition 14)
17. Prove that a square is a certain kind of parallelogram.
18. A rhombus is a quadrilateral which is equilateral but not
18. right-angled.
Prove that a rhombus is a certain kind of parallelogram.
19. A rectangle is a quadrilateral in which all the angles are right angles.
Prove that a rectangle is a certain kind of parallelogram.
10. Prove: Equal squares have equal sides.
11. ABCD is a square; ACFE is a square drawn on the diagonal AC;
10. and ED is a straight line.
10. a) Prove that ED is equal to DC, and is in a straight line with DC;
10. a) that is, EC is the diagonal of that square.
The diagonal AC bisects ABCD into two congruent and isosceles right triangles. (I. 34)
Since ABCD is a square, then angle BAD is a right angle;
therefore angle CAD is half a right angle.
And since CAEF is a square, angle DAE is the other half of the right angle CAE;
therefore angle DAE is equal to angle CAD.
Next, side EA is equal to side AC;
and we have shown that angle EAD is equal to angle CAD.
Therefore (S.A.S) the remaining side is equal to the remaining side:
ED is equal to DC;
and the remaining angles are equal:
angle EDA is equal to angle CDA.
But angle CDA is a right angle.
Therefore angle EDA is also a right angle,
and therefore ED is is a straight line with DC. (I. 14)
10. b) Prove that the square drawn on the diagonal AC is twice
10. a) as large as the square on the side BC.
If we draw DF, then AF is also a diagonal of that square.
And each diagonal bisects that square;
therefore AF and EC divide the square into four equal and isosceles right triangles.
But the square whose side is BC is made up of two of those triangles.
Therefore the square drawn on the diagonal AC is twice the square on the side BC.
The square drawn on the diagonal of a square is twice
the square on the side.
This theorem asserts an essential knowledge of a square figure. Moreover, it led Pythagoras to realize that the diagonal and side are incommensurable. See Topic 11 of The Evolution of the Real Numbers.
Next proposition
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# Video: AQA GCSE Mathematics Foundation Tier Pack 1 • Paper 3 • Question 18
Two function machines are shown. Both function machines have the same input. Work out the value of the input such that the output of A is two times the output of B.
04:01
### Video Transcript
Two function machines are shown. A: input multiplied by six subtract two output. B: input multiplied by two add eight output. Both function machines have the same input. Work out the value of the input such that the output of A is two times the output of B.
We’re going to answer this question using some algebra. So we’ll begin by assigning a letter to the input of our function machines. Let’s call it 𝑥. Let’s then consider what each of these two function machines will do to our input 𝑥.
Function machine A first multiplies our input by six. So multiplying 𝑥 by six will give six 𝑥. We then have to subtract two. So the output of function machine A will be six 𝑥 minus two.
Now, let’s consider function machine B. Remember we have the same input 𝑥. First, we’ll multiply this by two which will give two 𝑥. Then, we need to add eight. So the overall output from function machine B will be two 𝑥 plus eight.
So we’ve now got expressions for the output of each function machine. The question asks us to work out the value of the input such that the output of A is two times or twice the output of B. The output of function machine A is the expression six 𝑥 minus two. And we want this to be twice the output of B; that’s two multiplied by two 𝑥 plus eight.
We now have an equation which we want to solve in order to find the value of 𝑥 which is the input that will make this statement true. We begin by expanding the bracket on the right-hand side. Two multiplied by two 𝑥 gives four 𝑥. And then, two multiplied by positive eight gives positive 16. So we have the equation six 𝑥 minus two equals four 𝑥 plus 16.
Notice that we have terms involving 𝑥 on both sides of this equation. So our next step is going to be to collect all of the 𝑥 terms on the same side. We have a greater number of 𝑥s on the left of the equation. So we’ll collect our terms on this side.
To do so, we need to subtract four 𝑥 from each side of the equation. On the left-hand side, six 𝑥 minus four 𝑥 gives two 𝑥. We bring down the negative two. And on the right-hand side, four 𝑥 minus four 𝑥 leaves no 𝑥s. So we just bring down the positive 16. And we have two 𝑥 minus two equals 16.
Next, we need to add two to each side of this equation. On the left-hand side, two 𝑥 minus two and then plus two just leaves two 𝑥. And on the right-hand side, 16 plus two is 18. The final step in solving this equation is to divide both sides by two. On the left, two 𝑥 divided by two gives 𝑥. And on the right, 18 divided by two gives nine. So we have 𝑥 is equal to nine.
We’ve solved our equation then. But let’s check our answer by substituting this value of nine into each of our function machines. If we substitute nine as our input into function machine A, we first need to multiply it by six. Nine multiplied by six gives 54. We then need to subtract two which gives 52. So the the output from function machine A is 52.
Substituting the same input of nine into function machine B, we first need to multiply by two which gives 18. We then need to add eight which gives 26. So the output from function machine B is 26.
52 is indeed equal to two times 26. So the output of function machine A is twice the output of function machine B when the input is nine. So we’ve checked our answer. And we can say then that our answer to the problem is nine.
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# Math Snap
## One of the following sets is not closed in $\mathbb{R}$. Select one: a. $\mathbb{Z}$ b. $\left\{\frac{\tan ^{-1}(n)}{n}: n \in \mathbb{N}\right\} \cup\{0\}$ c. $\left\{(-1)^{n}\left(1-\frac{1}{n}\right): n \in \mathbb{N}\right\} \cup\{0,1\}$ d. N e. None of them
#### STEP 1
Assumptions 1. We are working in the real number set $\mathbb{R}$. 2. We need to determine which of the given sets is not closed in $\mathbb{R}$. 3. A set is closed if it contains all its limit points. 4. The sets to be examined are: a. $\mathbb{Z}$ (the set of all integers) b. $\left\{\frac{\tan^{-1}(n)}{n}: n \in \mathbb{N}\right\} \cup \{0\}$ c. $\left\{(-1)^n\left(1 - \frac{1}{n}\right): n \in \mathbb{N}\right\} \cup \{0, 1\}$ d. $\mathbb{N}$ (the set of all natural numbers)
#### STEP 2
Examine set (a): $\mathbb{Z}$ $\mathbb{Z}$ is the set of all integers. It contains all its limit points because there are no limit points of $\mathbb{Z}$ that are not integers. Therefore, $\mathbb{Z}$ is closed.
#### STEP 3
Examine set (b): $\left\{\frac{\tan^{-1}(n)}{n}: n \in \mathbb{N}\right\} \cup \{0\}$ Consider the sequence $\frac{\tan^{-1}(n)}{n}$ as $n \to \infty$: $\lim_{n \to \infty} \frac{\tan^{-1}(n)}{n} = 0$ Since $0$ is included in the set, it contains its limit point. Therefore, this set is closed.
#### STEP 4
Examine set (c): $\left\{(-1)^n\left(1 - \frac{1}{n}\right): n \in \mathbb{N}\right\} \cup \{0, 1\}$ Consider the sequence $(-1)^n\left(1 - \frac{1}{n}\right)$: - For even $n$, the sequence approaches $1$. - For odd $n$, the sequence approaches $-1$. The limit points are $1$ and $-1$. Since $1$ is included in the set but $-1$ is not, this set does not contain all its limit points and is therefore not closed.
#### STEP 5
Examine set (d): $\mathbb{N}$ $\mathbb{N}$ is the set of all natural numbers. It contains all its limit points because there are no limit points of $\mathbb{N}$ that are not natural numbers. Therefore, $\mathbb{N}$ is closed.
##### SOLUTION
Conclusion The set $\left\{(-1)^n\left(1 - \frac{1}{n}\right): n \in \mathbb{N}\right\} \cup \{0, 1\}$ is not closed in $\mathbb{R}$.
|
Section 4: Inverse Functions
# Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)
• Notes
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Section 8.4 Inverse Functions 803 Version: Fall 2007 8.4 Inverse Functions As we saw in the last section, in order to solve application problems involving expo- nential functions, we will need to be able to solve exponential equations such as 1500 = 1000 e 0 . 06 t or 300 = 2 x . However, we currently don’t have any mathematical tools at our disposal to solve for a variable that appears as an exponent, as in these equations. In this section, we will develop the concept of an inverse function, which will in turn be used to define the tool that we need, the logarithm, in Section 8.5. One-to-One Functions Definition 1. A given function f is said to be one-to-one if for each value y in the range of f , there is just one value x in the domain of f such that y = f ( x ) . In other words, f is one-to-one if each output y of f corresponds to precisely one input x . It’s easiest to understand this definition by looking at mapping diagrams and graphs of some example functions. l⚏ Example 2. Consider the two functions h and k defined according to the mapping diagrams in Figure 1 . In Figure 1 (a), there are two values in the domain that are both mapped onto 3 in the range. Hence, the function h is not one-to-one. On the other hand, in Figure 1 (b), for each output in the range of k , there is only one input in the domain that gets mapped onto it. Therefore, k is a one-to-one function. 1 2 3 h 1 2 3 4 k (a) (b) Figure 1. Mapping diagrams help to determine if a function is one-to-one. l⚏ Example 3. The graph of a function is shown in Figure 2 (a). For this function f , the y -value 4 is the output corresponding to two input values, x = 1 and x = 3 (see the corresponding mapping diagram in Figure 2 (b)). Therefore, f is not one-to-one. Graphically, this is apparent by drawing horizontal segments from the point (0 , 4) on the y -axis over to the corresponding points on the graph, and then drawing vertical segments to the x -axis. These segments meet the x -axis at 1 and 3 . Copyrighted material. See: 1
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804 Chapter 8 Exponential and Logarithmic Functions Version: Fall 2007 x y f 4 3 1 1 3 4 f (a) (b) Figure 2. A function which is not one-to-one. l⚏ Example 4. In Figure 3 , each y -value in the range of f corresponds to just one input value x . Therefore, this function is one-to-one. Graphically, this can be seen by mentally drawing a horizontal segment from each point on the y -axis over to the corresponding point on the graph, and then drawing a vertical segment to the x -axis. Several examples are shown in Figure 3 . It’s apparent that this procedure will always result in just one corresponding point on the x -axis, because each y -value only corresponds to one point on the graph. In fact, it’s easiest to just note that since each horizontal line only intersects the graph once, then there can be only one corresponding input to each output.
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# Lesson: Fractions: Adding and Subtracting Like Fractions
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### Lesson Objective
SWBAT find the sum or difference of fractions with like denominators.
### Lesson Plan
Materials Needed: scrap paper for DN and GP, Example Like Denominator Chart, white board, dry erase markers, pencils, and IND Worksheet (two sided).
Vocabulary: whole (or ONE or unit), denominator, numerator, equivalent, sum, difference.
……….
Do Now (3 -5 min): The teacher writes the following fractions on the board and has the students write two equivalent fractions for each.
3 = 4= 3 =
5 7 4
Opening (2 -3 min): Teacher quickly reviews answers to the Do Now. The teacher should then state the objective, “Yesterday, we found the whole of a fraction. Today, we are going to add and subtract like fractions. By the end of this lesson, you will be able to find the sum or difference of fractions with like denominators.”
Direct Instruction (8 – 10 min): The teacher has a chart that reads, “Fractions that have the same denominator are called like fractions. Fractions that have a different denominator are called unlike fractions.” The teacher reads the two sentences and then says, “Today we will be working with like fractions, so fractions that have the same denominator. I think most of you will find this to be very quick to pick up. There is only one rule that you have to remember. We never add the denominator because when we are adding like fractions we are working toward the whole!”
The teacher should then demonstrate using the first three problems on the Example Like Denominator Chart. The teacher should review counting out the whole to find the denominator of the fraction and the shaded part to find the numerator. The teacher should also call attention to the operation being used.
Guided Practice (10 min): The teacher uses more problems from the Example Like Denominator Chart to explain the concept of adding or subtracting fractions with like denominators.
See Example Like Denominator Chart
Independent (10 min): The teacher passes out the IND worksheet. The students should only complete both sides during the IND practice.
Closing (2-3 min): Teacher calls the attention of the students back toward the front of the class to quickly review the answers to the Independent Practice worksheet/ ask what we learned about.
### Lesson Resources
IND add fractions side 1 Classwork Example Like Denominator Chart Exemplar 3 IND subtract fractions side 2 Classwork
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# How do you simplify ${{\left( 2+5\iota \right)}^{2}}$?
Last updated date: 19th Jul 2024
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Hint: In this problem, we have to find the square of a complex number. This can also be done by multiplying the complex number by itself. After multiplication, we shall group all the terms and further simplify them. Then we will use the basic properties of complex numbers such as the different values of iota when it is raised to various powers.
In order to simplify the given expression, we must have prior knowledge of complex numbers. A complex number is of the form, $x+\iota y$. It comprises two parts. One is the real number part which lies on the x-axis of the cartesian plane and the other is a complex number part which lies on the y-axis of the cartesian plane.
The expression given can also be written as:
${{\left( 2+5\iota \right)}^{2}}=\left( 2+5\iota \right)\left( 2+5\iota \right)$
Thus, we shall multiply the complex function by itself.
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+10\iota +10\iota +25{{\iota }^{2}}$
Adding the like terms of iota, we get
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota +25{{\iota }^{2}}$
There are predefined values assigned to iota when it is raised to certain powers. They are as follows.
${{\iota }^{1}}=\iota$
${{\iota }^{2}}=-1$
${{\iota }^{3}}=-\iota$
${{\iota }^{4}}=1$
From the above results, we see that ${{\iota }^{2}}=-1$. Substituting this value in our equation, we get
$\Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota +25\left( -1 \right)$
\begin{align} & \Rightarrow {{\left( 2+5\iota \right)}^{2}}=4+20\iota -25 \\ & \Rightarrow {{\left( 2+5\iota \right)}^{2}}=-21+20\iota \\ \end{align}
Therefore, the given expression ${{\left( 2+5\iota \right)}^{2}}$ is simplified to $-21+20\iota$.
Note:
The generalized rule for iota raised to any power is that iota raised to the power of 4 or multiples of 4 is equal to 1. Otherwise, the value of iota in terms of multiples 4 is given as, ${{\iota }^{4k+1}}=\iota$, ${{\iota }^{4k+2}}=-1$ and ${{\iota }^{4k+3}}=-\iota$ where $k$ is a constant. Using this, we can easily find the value of iota even when its power is very large natural numbers.
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A resource for profoundly gifted middle school students, their parents, and teachers
MATHEMATICS CONCEPTS
### Cantor-Bernstein Therorem
Two sets A and B have the same cardinality if there is a 1-1 and onto mapping or correspondence from A to B. Often, it is easier to find a 1-1 mapping rather than one that is 1-1 and onto. The theorem says that, when looking for bijection between two sets A and B, it suffices to find a 1-1 mapping (that is not necessarily onto) from A to B and a 1-1 mapping (that is not necessarily onto) from B to A. The effort of finding the two 1-1 mappings - one from A to B and the other from B to A - is often far less than that of finding just one mapping that is both 1-1 and onto.
A mapping that is 1-1 (one-to-one) is also called an injection. A mapping that is onto is also called a surjection. Let us pause a bit on these two words: injection and surjection. In Latin, the verb jacere means "to throw." in + jacere = inicere = throw in. The past participle of inicere is injectus. Similary, the word surjection is derived by combining jacere with the Latin word sur, meaning "upon." A mapping that is both an injection and a surjection is called a bijection. When there is a bijection between sets A and B, the sets are said to be equinumerous (having the same cardinality) or equipollent or equivalent.
The Cantor-Bernstein Theorem
Let A and B be sets. If there is a 1-1 correspondence from A to B and a 1-1 corespondence from B to A, then there is a bijection between A and B.
The proof is given elsewhere on these pages.
The theorem is trivial for finite sets. For if A and B are finite sets, then a 1-1 correspondence from A to B would produce a unique partner in B for each element in A and so the cardinality of A is atmost that of B. Similarly the 1-1 correspondence from B to A means the cardinality of B is at most that of A. The two inequalites together imply that A and B have the same cardinality and so each 1-1 correspondence is also onto and so they are both bijections. The reason we point out the obvious here is that this result also holds for correspondences that are only onto in the finite case. That is, if A and B are finite sets and if there is correspondence from A onto B and also a correspondence from B onto A, then there exists a 1-1 and onto correspondence from A to B (and hence from B to A). The surprising thing is that this "seems" to hold also for infinite sets - the result is called the Dual of the Cantor-Bernstein Theorem.
Here, we illustrate the use of the Cantor-Bernstein Theorem by showing an important and much-used bijection in mathematics. That is, the set of all positive real numbers is bijective with the set of all real numbers just from -1 to +1, including -1 and +1. The latter set is often denoted as [-1,1] and is called the CLOSED set of real numbers from -1 to +1.
Illustration of the S-B Theorem
Consider the correspondence f:[-1,1]-->R+, with f(x) = 2+x. Notice that this correspondence is 1-1 (why?)and takes [-1,1] to [1,3]. But there are elements in R+ that are outside [1,3] and so this correspondence is not onto. Now consider a correspondence, call it g, from R+ to [-1,1]. That is g: R+ --> [-1,1], with g(x) = 1/(1+x). This is a 1-1 correspondence (why?) that takes R+ to the positive real numbers that are less than 1. The latter set is contained in [-1,1]. Since [-1,1] contains also other numbers than the positive reals less than 1, g is not onto. Now, according to the Schroder-Bernstein Theorem, there exists a 1-1 onto correspondence between [-1,1] and R+. So these sets have the same cardinality.
I remember a student of mine in a Real Analysis class volubly expressing amazement at the equivalence of these two sets. So I drew on the board the popular pictorial illustration shown below.
The red lines represent a segment of length 2 from -1 to +1. Each point P on the red line has an image P' on the real line.
You would object that this establishes the equivalence of R, not R+, with [-1,1]. If you want that you could first establish the equivalence of [-1,1] with, say, [1,5] and then use the red line on the right in the picture to show the equivalence of [1,5] and R+. Here is the picture that shows the equivalence of any two closed intervals.
Here, each point P in an interval of length a has a corresponding point P' in an interval of length b.
REFERENCES
Bernstein, F. "Untersuchungen aus der Mengenlehre." Ph.D. thesis. Göttingen, Germany, 1901
Bernstein, F. "Untersuchungen aus der Mengenlehre." Math. Ann. 61, 117-155, 1905
Cantor, E. "Ueber zwei Definitionen der Endlichkeit und G. Cantor'sche Sätze." Nova Acta Academiae Caesareae Leopoldino-Carolinae (Halle a.d. Saale) 71, 303-362, 1898
Cantor, E. "Die selbständige Definition der Mächtigkeiten 0, 1, 2, 3 und die explicite Gleichzahligkeitsbedingung." Nova Acta Academiae Caesareae Leopoldino-Carolinae (Halle a.d. Saale) 71, 365-376, 1898
(G.R.T.)
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# FIND QUOTIENT AND REMAINDER USING SYNTHETIC DIVISION
## About "Find Quotient and Remainder Using Synthetic Division"
Find Quotient and Remainder Using Synthetic Division :
Here we are going to see some practice questions to understand the concept of finding quotient and remainder using synthetic division.
## Find Quotient and Remainder Using Synthetic Division - Practice questions
Question 1 :
Find the quotient and remainder for the following using synthetic division:
(i) (x3 + x2 - 7x - 3) ÷ (x - 3)
Solution :
Quotient = x2 + 4x - 3
Remainder = -12
(ii) (x3 + 2x2 - x - 4) ÷ (x + 2)
Solution :
Quotient = x2 + 0x - 1
Remainder = -2
(iii) (3x3 - 2x2 + 7x - 5) ÷ (x + 3)
Solution :
Quotient = 3x2 - 11x + 40
Remainder = -125
(iv) (8x4 - 2x2 + 6x + 5) ÷ (4x + 1)
Solution :
Quotient = 8x2 - x + 40
Remainder = -125
(iv) (8x4 - 2x2 + 6x + 5) ÷ (4x + 1)
Solution :
Quotient : 8x3 - 2x2 -(3x/2) + (51/8)
Remainder : 109/32
Question 2 :
If the quotient obtained on dividing (8x4 - 2x2 + 6x - 7) by (2x + 1) is (4x3 + px2 - qx + 3) then find p, q and also the remainder.
Solution :
8x3 - 4x2 + 0x + 6
Dividing the quotient by 2, we get
4x3 - 2x2 + 0x + 3
The value of p and q are -2 and 0 respectively and remainder is -10.
Question 3 :
If the quotient obtained on dividing 3x3 + 11x2 + 34x + 106 by x - 3 is 3x2 + ax + b, then find a, b and also the remainder.
Solution :
Quotient = 3x2 + 20x + 94
Given quotient = 3x2 + ax + b
a = 20 and b = 94.
After having gone through the stuff given above, we hope that the students would have understood, "Find Quotient and Remainder Using Synthetic Division"
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Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
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
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# We have DeltaABCand the point M such that vec(BM)=2vec(MC).How to determinate x,y such that vec(AM)=xvec(AB)+yvec(AC)?
May 10, 2017
The answer is $x = \frac{1}{3}$ and $y = \frac{2}{3}$
#### Explanation:
We apply Chasles' relation
$\vec{A B} = \vec{A C} + \vec{C B}$
Therefore,
$\vec{B M} = 2 \vec{M C}$
$\vec{B A} + \vec{A M} = 2 \left(\vec{M A} + \vec{A C}\right)$
$\vec{A M} - 2 \vec{M A} = - \vec{B A} + 2 \vec{A C}$
But,
$\vec{A M} = - \vec{M A}$ and
$\vec{B A} = - \vec{A B}$
So,
$\vec{A M} + 2 \vec{A M} = \vec{A B} + 2 \vec{A C}$
$3 \vec{A M} = \vec{A B} + 2 \vec{A C}$
$\vec{A M} = \frac{1}{3} \vec{A B} + \frac{2}{3} \vec{A C}$
So,
$x = \frac{1}{3}$ and
$y = \frac{2}{3}$
May 10, 2017
$x = \frac{1}{3} , y = \frac{2}{3}$
#### Explanation:
We can define $P \in \left[A B\right]$, and $Q \in \left[A C\right]$ such that
$\left\{\begin{matrix}M = B + \frac{2}{3} \left(C - B\right) \\ P = B + \frac{2}{3} \left(A - B\right) \\ Q = A + \frac{2}{3} \left(C - A\right)\end{matrix}\right.$
and then
$M - A = \left(Q - A\right) + \left(P - A\right)$
or after substituting
$M - A = \frac{2}{3} \left(C - A\right) + \frac{1}{3} \left(B - A\right)$
so
$x = \frac{1}{3} , y = \frac{2}{3}$
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# 2.7 Linear inequalities and absolute value inequalities
Page 1 / 11
In this section you will:
• Use interval notation.
• Use properties of inequalities.
• Solve inequalities in one variable algebraically.
• Solve absolute value inequalities.
It is not easy to make the honor role at most top universities. Suppose students were required to carry a course load of at least 12 credit hours and maintain a grade point average of 3.5 or above. How could these honor roll requirements be expressed mathematically? In this section, we will explore various ways to express different sets of numbers, inequalities, and absolute value inequalities.
## Using interval notation
Indicating the solution to an inequality such as $\text{\hspace{0.17em}}x\ge 4\text{\hspace{0.17em}}$ can be achieved in several ways.
We can use a number line as shown in [link] . The blue ray begins at $\text{\hspace{0.17em}}x=4\text{\hspace{0.17em}}$ and, as indicated by the arrowhead, continues to infinity, which illustrates that the solution set includes all real numbers greater than or equal to 4.
We can use set-builder notation : $\text{\hspace{0.17em}}\left\{x|x\ge 4\right\},$ which translates to “all real numbers x such that x is greater than or equal to 4.” Notice that braces are used to indicate a set.
The third method is interval notation , in which solution sets are indicated with parentheses or brackets. The solutions to $\text{\hspace{0.17em}}x\ge 4\text{\hspace{0.17em}}$ are represented as $\text{\hspace{0.17em}}\left[4,\infty \right).\text{\hspace{0.17em}}$ This is perhaps the most useful method, as it applies to concepts studied later in this course and to other higher-level math courses.
The main concept to remember is that parentheses represent solutions greater or less than the number, and brackets represent solutions that are greater than or equal to or less than or equal to the number. Use parentheses to represent infinity or negative infinity, since positive and negative infinity are not numbers in the usual sense of the word and, therefore, cannot be “equaled.” A few examples of an interval , or a set of numbers in which a solution falls, are $\text{\hspace{0.17em}}\left[-2,6\right),$ or all numbers between $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}6,$ including $\text{\hspace{0.17em}}-2,$ but not including $\text{\hspace{0.17em}}6;$ $\left(-1,0\right),$ all real numbers between, but not including $\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}0;$ and $\text{\hspace{0.17em}}\left(-\infty ,1\right],$ all real numbers less than and including $\text{\hspace{0.17em}}1.\text{\hspace{0.17em}}$ [link] outlines the possibilities.
Set Indicated Set-Builder Notation Interval Notation
All real numbers between a and b , but not including a or b $\left\{x|a $\left(a,b\right)$
All real numbers greater than a , but not including a $\left\{x|x>a\right\}$ $\left(a,\infty \right)$
All real numbers less than b , but not including b $\left\{x|x $\left(-\infty ,b\right)$
All real numbers greater than a , including a $\left\{x|x\ge a\right\}$ $\left[a,\infty \right)$
All real numbers less than b , including b $\left\{x|x\le b\right\}$ $\left(-\infty ,b\right]$
All real numbers between a and b , including a $\left\{x|a\le x $\left[a,b\right)$
All real numbers between a and b , including b $\left\{x|a $\left(a,b\right]$
All real numbers between a and b , including a and b $\left\{x|a\le x\le b\right\}$ $\left[a,b\right]$
All real numbers less than a or greater than b $\left\{x|xb\right\}$ $\left(-\infty ,a\right)\cup \left(b,\infty \right)$
All real numbers $\left(-\infty ,\infty \right)$
## Using interval notation to express all real numbers greater than or equal to a
Use interval notation to indicate all real numbers greater than or equal to $\text{\hspace{0.17em}}-2.$
Use a bracket on the left of $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ and parentheses after infinity: $\text{\hspace{0.17em}}\left[-2,\infty \right).$ The bracket indicates that $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ is included in the set with all real numbers greater than $\text{\hspace{0.17em}}-2\text{\hspace{0.17em}}$ to infinity.
Use interval notation to indicate all real numbers between and including $\text{\hspace{0.17em}}-3\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}5.$
$\left[-3,5\right]$
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions
the polar co-ordinate of the point (-1, -1)
prove the identites sin x ( 1+ tan x )+ cos x ( 1+ cot x )= sec x + cosec x
tanh`(x-iy) =A+iB, find A and B
B=Ai-itan(hx-hiy)
Rukmini
what is the addition of 101011 with 101010
If those numbers are binary, it's 1010101. If they are base 10, it's 202021.
Jack
extra power 4 minus 5 x cube + 7 x square minus 5 x + 1 equal to zero
the gradient function of a curve is 2x+4 and the curve passes through point (1,4) find the equation of the curve
1+cos²A/cos²A=2cosec²A-1
test for convergence the series 1+x/2+2!/9x3
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# Ex.13.6 Q7 Surface Areas and Volumes Solution - NCERT Maths Class 9
## Question
A lead pencil consists of a cylinder of wood with a solid cylinder of graphite filled in the interior. The diameter of the pencil is $$7 \; \rm mm$$ and the diameter of the graphite is $$1\; \rm mm$$. If the length of the pencil is $$14 \; \rm cm$$, find the volume of the wood and that of the graphite.
Video Solution
Surface-Areas-And-Volumes
Ex exercise-13-6 | Question 7
## Text Solution
Reasoning:
Volume of cylinder $$\pi {r^2}h$$
What is known?
Diameter of the pencil, diameter of graphite and length of the pencil.
What is unknown?
Volume of the wood and that if graphite.
Steps:
For cylinder graphite.
Diameter $$(2r) = 1 \; \rm mm$$
Radius \begin{align}(r) = \frac{1}{2}\,\, \rm mm \end{align}
Length of the pencil $$(h) = 14 \; \rm cm = 140 \;\rm mm$$
\begin{align}\text{Capacity}&= \text{Volume} \\ &=\,\pi {r^2}h\\ &= \frac{{22}}{7} \times \frac{1}{2} \times \frac{1}{2} \times 140\\ &= 110\,\, \rm mm^3 \end{align}
\begin{align} \rm{In} \,\,c{m^3} &= \frac{{110}}{{10 \times 10 \times 10}} = 0.11\,\, \rm cm^3 \end{align}
For cylinder of wood:
To find the volume of the wood:
Total volume of the pencil $$-$$ Volume of graphite
\begin{align}\pi {R^2}h - \pi {r^2}h \pi h({R^2} - {r^2})\end{align}
Diameter of pencil $$(2R) = 7 \; \rm mm$$
Radius $$(r) = \frac{7}{2} \rm mm$$
Length of the pencil $$(h) = 14 \; \rm cm = 140 \; \rm mm$$
Volume of wood \begin{align}=\,\pi h({R^2} - {r^2}) \end{align}
\begin{align} &= \frac{{22}}{7} \times 140 \times [{(\frac{7}{2})^2} - {(\frac{1}{2})^2}]\\ &= \frac{{22}}{7} \times 140 \times [\frac{{49}}{4} - \frac{1}{2}]\\ &= 5280\,\, \rm mm^3\\ &= \frac{{5280}}{{10 \times 10 \times 10}}\, \rm cm^3 \\ &= 5.28\,\, \rm cm^3 \end{align}
Volume of the wood $$= 5.28\,\, \rm cm^3$$
Volume of graphite $$= 0.11\,\,\, \rm cm^3$$
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# Difference between revisions of "2011 AIME II Problems/Problem 13"
## Problem
Point $P$ lies on the diagonal $AC$ of square $ABCD$ with $AP > CP$. Let $O_{1}$ and $O_{2}$ be the circumcenters of triangles $ABP$ and $CDP$ respectively. Given that $AB = 12$ and $\angle O_{1}PO_{2} = 120^{\circ}$, then $AP = \sqrt{a} + \sqrt{b}$, where $a$ and $b$ are positive integers. Find $a + b$.
## Solution 1
<geogebra>7b0d7e3170597705121a87857a112a90dff8cac9</geogebra>
Denote the midpoint of $\overline{DC}$ be $E$ and the midpoint of $\overline{AB}$ be $F$. Because they are the circumcenters, both Os lie on the perpendicular bisectors of $AB$ and $CD$ and these bisectors go through $E$ and $F$.
It is given that $\angleO_{1}PO_{2}=120^{\circ}$ (Error compiling LaTeX. ! Undefined control sequence.). Because $O_{1}P$ and $O_{1}B$ are radii of the same circle, the have the same length. This is also true of $O_{2}P$ and $O_{2}D$. Because $m\angle CAB=m\angle ACD=45^{\circ}$, $m\stackrel{\frown}{PD}=m\stackrel{\frown}{PB}=2(45^{\circ})=90^{\circ}$. Thus, $O_{1}PB$ and $O_{2}PD$ are isosceles right triangles. Using the given information above and symmetry, $m\angle DPB = 120^{\circ}$. Because ABP and ADP share one side, have one side with the same length, and one equal angle, they are congruent by SAS. This is also true for triangle CPB and CPD. Because angles APB and APD are equal and they sum to 120 degrees, they are each 60 degrees. Likewise, both angles CPB and CPD have measures of 120 degrees.
Because the interior angles of a triangle add to 180 degrees, angle ABP has measure 75 degrees and angle PDC has measure 15 degrees. Subtracting, it is found that both angles $O_{1}BF$ and $O_{2}DE$ have measures of 30 degrees. Thus, both triangles $O_{1}BF$ and $O_{2}DE$ are 30-60-90 right triangles. Because F and E are the midpoints of AB and CD respectively, both FB and DE have lengths of 6. Thus, $DO_{2}=BO_{1}=4\sqrt{3}$. Because of 45-45-90 right triangles, $PB=PD=4\sqrt{6}$.
Now, using Law of Cosines on $\triangle ABP$ and letting $x = AP$,
$96=144+x^{2}-24x\frac{\sqrt{2}}{2}$
$96=144+x^{2}-12x\sqrt{2}$
$0=x^{2}-12x\sqrt{2}+48$
$x = \frac{12 \sqrt{2} \pm \sqrt{288-(4)(48)}}{2}$
$x = \frac{12 \sqrt{2} \pm \sqrt{288-192}}{2}$
$x = \frac{12 \sqrt{2} \pm \sqrt{96}}{2}$
$x = \frac{2 \sqrt{72} \pm 2 \sqrt{24}}{2}$
$x = \sqrt{72} \pm \sqrt{24}$
Because it is given that $AP > CP$, $AP>6\sqrt{2}$, so the minus version of the above equation is too small. Thus, $AP=\sqrt{72}+ \sqrt{24}$ and a + b = 24 + 72 = $\framebox[1.5\width]{96.}$
## Solution 2
This takes a slightly different route than Solution 1.
Solution 1 proves that $\angle{DPB}=120^{\circ}$ and that $\overline{BP} = \overline{DP}$. Construct diagonal $\overline{BD}$ and using the two statements above it quickly becomes clear that $\angle{BDP} = \angle{DBP} = 30^{\circ}$ by isosceles triangle base angles. Let the midpoint of diagonal $\overline{AC}$ be $M$, and since the diagonals are perpendicular, both triangle $DMP$ and triangle $BMP$ are 30-60-90 right triangles. Since $\overline{AB} = 12$, $\overline{AC} = \overline{BD} = 12\sqrt{2}$ and $\overline{BM} = \overline{DM} = 6\sqrt{2}$. 30-60-90 triangles' sides are in the ratio $1 : \sqrt{3} : 2$, so $\overline{MP} = \frac{6\sqrt{2}}{\sqrt{3}} = 2\sqrt {6}$. $\overline{AP} = \overline{MP} + \overline{BM} = 6\sqrt{2} + 2\sqrt{6} = \sqrt{72} + \sqrt{24}$. Hence, $72 + 24 = \framebox[1.5\width]{96}$.
## Solution 3
Use vectors. In an $xy$ plane, let $(-s,0)$ be $A$, $(0,s)$ be $B$, $(s,0)$ be $C$, $(0,-s)$ be $D$, and $(p,0)$ be P, where $s=|AB|/\sqrt{2}=6\sqrt{2}$. It remains to find $p$.
The line $y=-x$ is the perpendicular bisector of $AB$ and $CD$, so $O_1$ and $O_2$ lies on the line. Now compute the perpendicular bisector of $AP$. The center has coordinate $(\frac{p-s}{2},0)$, and the segment is part of the $x$-axis, so the perpendicular bisector has equation $x=\frac{p-s}{2}$. Since $O_1$ is the circumcenter of triangle $ABP$, it lies on the perpendicular bisector of both $AB$ and $AP$, so $$O_1=(\frac{p-s}{2},-\frac{p-s}{2})$$ Similarly, $$O_2=(\frac{p+s}{2},-\frac{p+s}{2})$$ The relation $\angle O_1PO_2=120^\circ$ can now be written using dot product as $$\vec{PO_1}\cdot\vec{PO_2}=|\vec{PO_1}|\cdot|\vec{PO_2}|\cos 120^\circ=-\frac{1}{2}|\vec{PO_1}|\cdot|\vec{PO_2}|$$ Computation of both sides yields $$\frac{p^2-s^2}{p^2+s^2}=-\frac{1}{2}$$ Solve for $p$ gives $p=s/\sqrt{3}=2\sqrt{6}$, so $AP=s+p=6\sqrt{2}+2\sqrt{6}=\sqrt{72}+\sqrt{24}$. The answer is 72+24$\Rightarrow\boxed{096}$
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What is the formula for the volume of a rectangular prism?
A rectangular prism, also known as a cuboid, is a solid geometric shape that has six faces, all of which are rectangles. The volume of a rectangular prism can be found by multiplying the length, width, and height of the prism.
Formula for the Volume of a Rectangular Prism
The formula for the volume of a rectangular prism is as follows:
Volume = Length × Width × Height
To better understand this formula, let’s go through a step-by-step example of calculating the volume of a rectangular prism.
Step-by-Step Example of Calculating the Volume of a Rectangular Prism
Problem: Find the volume of a rectangular prism with a length of 5 units, a width of 3 units, and a height of 4 units.
Step 1: Identify the Length, Width, and Height
In this example, we are given the dimensions of the rectangular prism:
• Length (L) = 5 units
• Width (W) = 3 units
• Height (H) = 4 units
Step 2: Apply the Volume Formula
Now, we will apply the volume formula to the given dimensions:
Volume = Length × Width × Height
Plug in the given values:
Volume = 5 units × 3 units × 4 units
Step 3: Calculate the Volume
Multiply the dimensions together:
Volume = 15 units² × 4 units
Volume = 60 units³
So, the volume of the rectangular prism with a length of 5 units, a width of 3 units, and a height of 4 units is 60 cubic units.
Key Points for Understanding the Volume of a Rectangular Prism
1. The formula for the volume of a rectangular prism is Volume = Length × Width × Height. This formula is derived from the fact that the volume of any 3D shape is equal to the product of its three dimensions.
2. The units for measuring the volume of a rectangular prism are cubic units (e.g., cubic meters, cubic inches, etc.), as the volume represents the amount of space that the prism occupies in three-dimensional space.
3. When solving problems involving the volume of a rectangular prism, it’s essential to identify the length, width, and height of the prism and ensure that all the dimensions are in the same unit of measurement before applying the volume formula.
4. To find the volume of a rectangular prism, multiply the length, width, and height together. The result will be the volume in cubic units.
By understanding the formula for the volume of a rectangular prism and following the step-by-step process outlined above, you can easily calculate the volume of any rectangular prism. Remember, the key is to identify the length, width, and height of the prism and then apply the formula: Volume = Length × Width × Height.
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## How many right angles does a obtuse triangle have?
An obtuse triangle (or obtuse-angled triangle) is a triangle with one obtuse angle (greater than 90°) and two acute angles. Since a triangle’s angles must sum to 180° in Euclidean geometry, no Euclidean triangle can have more than one obtuse angle.
## Can an obtuse triangle have a right angle?
A triangle cannot be rightangled and obtuse angled at the same time. Since a rightangled triangle has one right angle, the other two angles are acute. Therefore, an obtuseangled triangle can never have a right angle; and vice versa. The side opposite the obtuse angle in the triangle is the longest.
## Can a right triangle also be obtuse?
A right triangle cannot be obtuse because of the sizes of the angles therein. Any triangle has three sides, three angles, and three angles that equal
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## What is the maximum number of obtuse angles a triangle can have?
A triangle can have a maximum of 1 obtuse angle.
## What is the greatest number of right angles in a triangle?
Explanation: A triangle has 180o as the sum of all its internal angles, no more, no less. If one angle is 90o, then you can have two 45o angles, one 30o and a 60o, an 81o and a 9o – pretty much any combination of numbers adding up to 90 to make the total 90+90=180.
## Why can’t a right triangle have an obtuse angle?
When an angle of a triangle is 90 degrees, the triangle cannot have an obtuse angle. The other two must each be less than 90 degrees (90 deg + 89 deg + 1 deg = 180 deg). It therefore follows that they must both be less than 90 degrees and so must both be acute.
## Can 1 acute and 2 obtuse form a triangle?
Types of Triangles. All equilateral triangles are equiangular. A right triangle will have 1 right angle and 2 acute angles. An obtuse triangle will have 1 obtuse triangle and 2 acute angles.
## Can a triangle have both an acute and right angle?
Question 945360: A triangle cannot have both an acute angle and a right angle? In a right triangle, BOTH of the other angles MUST be acute. An obtuse angle (greater than 90 degrees) would make the sum of the angles greater than 180 degrees, which is impossible.
## Can a triangle have 2 right angles?
No, a triangle can never have 2 right angles. A triangle has exactly 3 sides and the sum of interior angles sum up to 180°. Thus, it is not possible to have a triangle with 2 right angles.
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## Can an acute triangle be a right triangle?
A triangle where all three internal angles are acute (less than 90 degrees). Less than 90° – all three angles are acute and so the triangle is acute. Exactly 90° – it is a right triangle.
## Do all triangles have right angles?
A right triangle may be isosceles or scalene. In an acute triangle, all angles are less than right angles? each one is less than 90 degrees. An acute triangle may be equilateral, isosceles, or scalene.
## Is it possible to have a right isosceles triangle?
An isosceles right triangle is an isosceles triangle and a right triangle. This means that it has two congruent sides and one right angle. Therefore, the two congruent sides must be the legs.
## How many obtuse angles are in a triangle?
There can only be one obtuse angle in any triangle. This is because the measures of the interior angles of a triangle always must add up to 180
## How many right angles does a triangle have?
A triangle can have one right angle. A quadrilateral can have four right angles. Sum of Interior Angles = 540′.
## How many angle Can a triangle have?
In a Euclidean space, the sum of angles of a triangle equals the straight angle (180 degrees, π radians, two right angles, or a half-turn). A triangle has three angles, one at each vertex, bounded by a pair of adjacent sides.
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# Important Formulas on Streams
### Streams - Important Aptitude formulas, shortcut methods and tricks:
Theorem: (1). If α km/hr be the man’s rate in still water and β km/hr be the rate of the current, then
α + β = man’s rate with current
α - β = man’s rate against current
Note: (i) A man’s rate in still water is half the sum of his rate with and against the current.
(ii) The rate of the current is half the difference between the rate of the man with and against the current.
Theorem: (2). A man can row α km/hr in still water. If in a stream, which is flowing at β km/hr, it takes him ɣ hrs to row to a place and back. The distance between the two places is
ɣ (α² - β²)/2α
Theorem: (3). A man rows a certain distance downstream in α hrs and returns the same distance in β hrs. If the stream flows at the rate of ɣ km/hr, then the speed of the man is given by
ɣ (α + β)/(β - α) km/hr.
Example: 01
A man can row 6 km/hr in still water. It takes him twice as long to row up as to row down the river. Find the rate of the stream.
Solution:
Let, rate of stream = α km/hr
Then, 6 + α = 2 (6 - α)
or, α = 2 km/hr.
Example: 02
A man can row 6 km/hr in still water. When the water is running at 1.2 km/hr, it takes him 1 hr to row to a place and back. How far is the place?
Solution:
The required distance = [1 × (6² - 1.2²)]/(2 × 6) = 2.88 km.
Example: 03
A man can row 7 km/hr in still water. In a stream which is flowing at 3 km/hr, it takes him 7 hrs to row to a place and back. How far is the place?
Solution:
The required distance = [7 × (7² - 3²)]/(2 × 7) = 20 km.
Example: 04
Jack can row a certain distance downstream in 6 hrs and return to the same distance in 9 hrs. If the stream flows at the rate of 3 km/hr, find the speed of Jack in still water.
Solution:
Jack’s speed in still water = [3 × (9 + 6)]/(9 - 6) = 15 km/hr.
Example: 05
If a man’s rate with the current is 12 km/hr and the rate of the current is 1.5 km/hr. Then what is the man’s rate against the current?
Solution:
Man’s rate with the current = 12 km/hr
Man’s rate in still water = 12 - 1.5 = 10.5 km/hr
Therefore, Man’s rate against current = 10.5 - 1.5 = 9 km/hr.
Streams Aptitude:
Formula: Streams Aptitude Formulas
Solved Examples: Solved Examples: Set 01
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# System of Linear Equations in three variables using Cramer’s Rule
Cramer’s rule: In linear algebra, Cramer’s rule is an explicit formula for the solution of a system of linear equations with as many equations as unknown variables. It expresses the solution in terms of the determinants of the coefficient matrix and of matrices obtained from it by replacing one column by the column vector of the right-hand-sides of the equations. Cramer’s rule is computationally inefficient for systems of more than two or three equations.
Suppose we have to solve these equations:
a1x + b1y + c1z = d1
a2x + b2y + c2z = d2
a3x + b3y + c3z = d3
Following the Cramer’s Rule, first find the determinant values of all four matrices.
There are 2 cases:
• Case I : When D ≠ 0 In this case we have,
• x = D1/D
• y = D2/D
• z = D3/D
• Hence unique value of x, y, z will be obtained.
• Case II : When D = 0
• When at least one of D1, D2 and D3 is non zero: Then no solution is possible and hence system of equations will be inconsistent.
• When D = 0 and D1 = D2 = D3 = 0: Then the system of equations will be consistent and it will have infinitely many solutions.
Example:
Consider the following system of linear equations.
[2x – y + 3z = 9], [x + y + z = 6], [x – y + z = 2]
[Tex]D_1 = \begin{vmatrix} 9 & -1 & 3\\ 6 & 1 & 1\\ 2 & -1 & 1\\ \end{vmatrix} [/Tex][Tex]D_3 = \begin{vmatrix} 2 & -1 & 9\\ 1 & 1 & 6\\ 1 & -1 & 2\\ \end{vmatrix} [/Tex]
[x = D1/D = 1], [y = D2/D = 2], [z = D3/D = 3]
Below is the implementation.
## C++
// CPP program to calculate solutions of linear// equations using cramer's rule#include using namespace std; // This functions finds the determinant of Matrixdouble determinantOfMatrix(double mat[3][3]){ double ans; ans = mat[0][0] * (mat[1][1] * mat[2][2] - mat[2][1] * mat[1][2]) - mat[0][1] * (mat[1][0] * mat[2][2] - mat[1][2] * mat[2][0]) + mat[0][2] * (mat[1][0] * mat[2][1] - mat[1][1] * mat[2][0]); return ans;} // This function finds the solution of system of// linear equations using cramer's rulevoid findSolution(double coeff[3][4]){ // Matrix d using coeff as given in cramer's rule double d[3][3] = { { coeff[0][0], coeff[0][1], coeff[0][2] }, { coeff[1][0], coeff[1][1], coeff[1][2] }, { coeff[2][0], coeff[2][1], coeff[2][2] }, }; // Matrix d1 using coeff as given in cramer's rule double d1[3][3] = { { coeff[0][3], coeff[0][1], coeff[0][2] }, { coeff[1][3], coeff[1][1], coeff[1][2] }, { coeff[2][3], coeff[2][1], coeff[2][2] }, }; // Matrix d2 using coeff as given in cramer's rule double d2[3][3] = { { coeff[0][0], coeff[0][3], coeff[0][2] }, { coeff[1][0], coeff[1][3], coeff[1][2] }, { coeff[2][0], coeff[2][3], coeff[2][2] }, }; // Matrix d3 using coeff as given in cramer's rule double d3[3][3] = { { coeff[0][0], coeff[0][1], coeff[0][3] }, { coeff[1][0], coeff[1][1], coeff[1][3] }, { coeff[2][0], coeff[2][1], coeff[2][3] }, }; // Calculating Determinant of Matrices d, d1, d2, d3 double D = determinantOfMatrix(d); double D1 = determinantOfMatrix(d1); double D2 = determinantOfMatrix(d2); double D3 = determinantOfMatrix(d3); printf("D is : %lf \n", D); printf("D1 is : %lf \n", D1); printf("D2 is : %lf \n", D2); printf("D3 is : %lf \n", D3); // Case 1 if (D != 0) { // Coeff have a unique solution. Apply Cramer's Rule double x = D1 / D; double y = D2 / D; double z = D3 / D; // calculating z using cramer's rule printf("Value of x is : %lf\n", x); printf("Value of y is : %lf\n", y); printf("Value of z is : %lf\n", z); } // Case 2 else { if (D1 == 0 && D2 == 0 && D3 == 0) printf("Infinite solutions\n"); else if (D1 != 0 || D2 != 0 || D3 != 0) printf("No solutions\n"); }} // Driver Codeint main(){ // storing coefficients of linear equations in coeff matrix double coeff[3][4] = { { 2, -1, 3, 9 }, { 1, 1, 1, 6 }, { 1, -1, 1, 2 }, }; findSolution(coeff); return 0;}
## Java
// Java program to calculate solutions of linear// equations using cramer's ruleclass GFG{ // This functions finds the determinant of Matrixstatic double determinantOfMatrix(double mat[][]){ double ans; ans = mat[0][0] * (mat[1][1] * mat[2][2] - mat[2][1] * mat[1][2]) - mat[0][1] * (mat[1][0] * mat[2][2] - mat[1][2] * mat[2][0]) + mat[0][2] * (mat[1][0] * mat[2][1] - mat[1][1] * mat[2][0]); return ans;} // This function finds the solution of system of// linear equations using cramer's rulestatic void findSolution(double coeff[][]){ // Matrix d using coeff as given in cramer's rule double d[][] = { { coeff[0][0], coeff[0][1], coeff[0][2] }, { coeff[1][0], coeff[1][1], coeff[1][2] }, { coeff[2][0], coeff[2][1], coeff[2][2] }, }; // Matrix d1 using coeff as given in cramer's rule double d1[][] = { { coeff[0][3], coeff[0][1], coeff[0][2] }, { coeff[1][3], coeff[1][1], coeff[1][2] }, { coeff[2][3], coeff[2][1], coeff[2][2] }, }; // Matrix d2 using coeff as given in cramer's rule double d2[][] = { { coeff[0][0], coeff[0][3], coeff[0][2] }, { coeff[1][0], coeff[1][3], coeff[1][2] }, { coeff[2][0], coeff[2][3], coeff[2][2] }, }; // Matrix d3 using coeff as given in cramer's rule double d3[][] = { { coeff[0][0], coeff[0][1], coeff[0][3] }, { coeff[1][0], coeff[1][1], coeff[1][3] }, { coeff[2][0], coeff[2][1], coeff[2][3] }, }; // Calculating Determinant of Matrices d, d1, d2, d3 double D = determinantOfMatrix(d); double D1 = determinantOfMatrix(d1); double D2 = determinantOfMatrix(d2); double D3 = determinantOfMatrix(d3); System.out.printf("D is : %.6f \n", D); System.out.printf("D1 is : %.6f \n", D1); System.out.printf("D2 is : %.6f \n", D2); System.out.printf("D3 is : %.6f \n", D3); // Case 1 if (D != 0) { // Coeff have a unique solution. Apply Cramer's Rule double x = D1 / D; double y = D2 / D; double z = D3 / D; // calculating z using cramer's rule System.out.printf("Value of x is : %.6f\n", x); System.out.printf("Value of y is : %.6f\n", y); System.out.printf("Value of z is : %.6f\n", z); } // Case 2 else { if (D1 == 0 && D2 == 0 && D3 == 0) System.out.printf("Infinite solutions\n"); else if (D1 != 0 || D2 != 0 || D3 != 0) System.out.printf("No solutions\n"); }} // Driver Codepublic static void main(String[] args){ // storing coefficients of linear // equations in coeff matrix double coeff[][] = {{ 2, -1, 3, 9 }, { 1, 1, 1, 6 }, { 1, -1, 1, 2 }}; findSolution(coeff); }} // This code is contributed by PrinciRaj1992
## Python3
# Python3 program to calculate# solutions of linear equations# using cramer's rule # This functions finds the# determinant of Matrixdef determinantOfMatrix(mat): ans = (mat[0][0] * (mat[1][1] * mat[2][2] - mat[2][1] * mat[1][2]) - mat[0][1] * (mat[1][0] * mat[2][2] - mat[1][2] * mat[2][0]) + mat[0][2] * (mat[1][0] * mat[2][1] - mat[1][1] * mat[2][0])) return ans # This function finds the solution of system of# linear equations using cramer's ruledef findSolution(coeff): # Matrix d using coeff as given in # cramer's rule d = [[coeff[0][0], coeff[0][1], coeff[0][2]], [coeff[1][0], coeff[1][1], coeff[1][2]], [coeff[2][0], coeff[2][1], coeff[2][2]]] # Matrix d1 using coeff as given in # cramer's rule d1 = [[coeff[0][3], coeff[0][1], coeff[0][2]], [coeff[1][3], coeff[1][1], coeff[1][2]], [coeff[2][3], coeff[2][1], coeff[2][2]]] # Matrix d2 using coeff as given in # cramer's rule d2 = [[coeff[0][0], coeff[0][3], coeff[0][2]], [coeff[1][0], coeff[1][3], coeff[1][2]], [coeff[2][0], coeff[2][3], coeff[2][2]]] # Matrix d3 using coeff as given in # cramer's rule d3 = [[coeff[0][0], coeff[0][1], coeff[0][3]], [coeff[1][0], coeff[1][1], coeff[1][3]], [coeff[2][0], coeff[2][1], coeff[2][3]]] # Calculating Determinant of Matrices # d, d1, d2, d3 D = determinantOfMatrix(d) D1 = determinantOfMatrix(d1) D2 = determinantOfMatrix(d2) D3 = determinantOfMatrix(d3) print("D is : ", D) print("D1 is : ", D1) print("D2 is : ", D2) print("D3 is : ", D3) # Case 1 if (D != 0): # Coeff have a unique solution. # Apply Cramer's Rule x = D1 / D y = D2 / D # calculating z using cramer's rule z = D3 / D print("Value of x is : ", x) print("Value of y is : ", y) print("Value of z is : ", z) # Case 2 else: if (D1 == 0 and D2 == 0 and D3 == 0): print("Infinite solutions") elif (D1 != 0 or D2 != 0 or D3 != 0): print("No solutions") # Driver Codeif __name__ == "__main__": # storing coefficients of linear # equations in coeff matrix coeff = [[2, -1, 3, 9], [1, 1, 1, 6], [1, -1, 1, 2]] findSolution(coeff) # This code is contributed by Chitranayal
## C#
// C# program to calculate solutions of linear// equations using cramer's ruleusing System; class GFG{ // This functions finds the determinant of Matrixstatic double determinantOfMatrix(double [,]mat){ double ans; ans = mat[0,0] * (mat[1,1] * mat[2,2] - mat[2,1] * mat[1,2]) - mat[0,1] * (mat[1,0] * mat[2,2] - mat[1,2] * mat[2,0]) + mat[0,2] * (mat[1,0] * mat[2,1] - mat[1,1] * mat[2,0]); return ans;} // This function finds the solution of system of// linear equations using cramer's rulestatic void findSolution(double [,]coeff){ // Matrix d using coeff as given in cramer's rule double [,]d = { { coeff[0,0], coeff[0,1], coeff[0,2] }, { coeff[1,0], coeff[1,1], coeff[1,2] }, { coeff[2,0], coeff[2,1], coeff[2,2] }, }; // Matrix d1 using coeff as given in cramer's rule double [,]d1 = { { coeff[0,3], coeff[0,1], coeff[0,2] }, { coeff[1,3], coeff[1,1], coeff[1,2] }, { coeff[2,3], coeff[2,1], coeff[2,2] }, }; // Matrix d2 using coeff as given in cramer's rule double [,]d2 = { { coeff[0,0], coeff[0,3], coeff[0,2] }, { coeff[1,0], coeff[1,3], coeff[1,2] }, { coeff[2,0], coeff[2,3], coeff[2,2] }, }; // Matrix d3 using coeff as given in cramer's rule double [,]d3 = { { coeff[0,0], coeff[0,1], coeff[0,3] }, { coeff[1,0], coeff[1,1], coeff[1,3] }, { coeff[2,0], coeff[2,1], coeff[2,3] }, }; // Calculating Determinant of Matrices d, d1, d2, d3 double D = determinantOfMatrix(d); double D1 = determinantOfMatrix(d1); double D2 = determinantOfMatrix(d2); double D3 = determinantOfMatrix(d3); Console.Write("D is : {0:F6} \n", D); Console.Write("D1 is : {0:F6} \n", D1); Console.Write("D2 is : {0:F6} \n", D2); Console.Write("D3 is : {0:F6} \n", D3); // Case 1 if (D != 0) { // Coeff have a unique solution. Apply Cramer's Rule double x = D1 / D; double y = D2 / D; double z = D3 / D; // calculating z using cramer's rule Console.Write("Value of x is : {0:F6}\n", x); Console.Write("Value of y is : {0:F6}\n", y); Console.Write("Value of z is : {0:F6}\n", z); } // Case 2 else { if (D1 == 0 && D2 == 0 && D3 == 0) Console.Write("Infinite solutions\n"); else if (D1 != 0 || D2 != 0 || D3 != 0) Console.Write("No solutions\n"); }} // Driver Codepublic static void Main(){ // storing coefficients of linear // equations in coeff matrix double [,]coeff = {{ 2, -1, 3, 9 }, { 1, 1, 1, 6 }, { 1, -1, 1, 2 }}; findSolution(coeff); }} // This code is contributed by 29AjayKumar
## Javascript
Output
D is : -2.000000
D1 is : -2.000000
D2 is : -4.000000
D3 is : -6.000000
Value of x is : 1.000000
Value of y is : 2.000000
Value of z is : 3.000000
Time complexity: O(1)
Auxiliary space: O(1)
My Personal Notes arrow_drop_up
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# ORDER OF OPERATIONS WORKSHEETS
## About "Order of operations worksheets"
Order of operations worksheets are much useful to the kids who would like to practice problems on the binary operations like add, subtract, multiply, divide, squaring,etc.
When we have two or more operations in the same expression, we may have question about which one has to be done first, which one has to be done next.
But order of operations or bodmas rule or pemdas rule tells us in which order we have to do the operations one by one.
What is BODMAS rule ?
The rule or order that we use to simplify expressions in math is called "BODMAS" rule.
Very simply way to remember BODMAS rule!
B -----> Brackets first (Parentheses)
O -----> Of (orders :Powers and radicals)
D -----> Division
M -----> Multiplication
S -----> Subtraction
Important notes :
1. In a particular simplification, if you have both multiplication and division, do the operations one by one in the order from left to right.
2. Division does not always come before multiplication. We have to do one by one in the order from left to right.
3. In a particular simplification, if you have both addition and subtraction, do the operations one by one in the order from left to right.
Examples :
12 ÷ 3 x 5 = 4 x 5 = 20
13 - 5 + 9 = 8 + 9 = 17
In the above simplification, we have both division and multiplication. From left to right, we have division first and multiplication next. So we do division first and multiplication next.
To have better understanding on "Order of operations", let us look at order of operations worksheets problems.
## Order of operations worksheets problems
1. Evaluate : 6 + 7 x 8
2. Evaluate : 10² - 16 ÷ 8
3. Evaluate : (25 + 11) x 2
4. Evaluate : 3 + 6 x (5+4) ÷ 3 -7
5. Evaluate : 36 - 2(20+12÷4x3-2x2) + 10
6. Evaluate : 6+[(16-4)÷(2²+2)]-2
7. Evaluate : (96÷12)+14x(12+8)÷2
8. Evaluate : (93+15) ÷ (3x4) - 24 + 8
9. Evaluate : 55 ÷ 11 + (18 - 6) x 9
10. Evaluate : (7 + 18) x 3 ÷(2+13) - 28
Here, they are
## Step by step solution
Problem 1 :
Evaluate : 6 + 7 x 8
Expression6 + 7 x 8 Evaluation= 6 + 7 x 8= 6 + 56 = 62 OperationMultiplicationAdditionResult
Problem 2 :
Evaluate : 10² - 16 ÷ 8
Expression10² - 16 ÷ 8 Evaluation= 10² - 16 ÷ 8= 100 - 16 ÷ 8= 100 - 2= 98 OperationPowerDivisionSubtractionResult
Problem 3 :
Evaluate : (25 + 11) x 2
Expression(25 + 11) x 2 Evaluation= (25 + 11) x 2= 36 x 2= 72 OperationParenthesisMultiplicationResult
Problem 4 :
Evaluate : 3 + 6 x (5+4) ÷ 3 -7
Expression3 + 6 x (5+4) ÷ 3 -7 Evaluation= 3 + 6 x (5+4) ÷ 3 -7= 3 + 6 x 9 ÷ 3 -7= 3 + 54 ÷ 3 -7= 3 + 18 -7= 21 - 7= 14 OperationParenthesisMultiplicationDivisionAdditionSubtractionResult
Problem 5 :
Evaluate : 36 - 2(20+12÷4x3-2x2) + 10
Problem 6 :
Evaluate : 6+[(16-4)÷(2²+2)]-2
Expression 6+[(16-4)÷(2²+2)]-2 Evaluation= 6+[(16-4)÷(2²+2)]-2= 6+[12÷(2²+2)]-2= 6+[12÷(4+2)]-2= 6+[12÷6]-2= 6+2 - 2= 8 - 2=6 OperationParenthesisPowerParenthesisParenthesisAdditionSubtractionResult
Problem 7 :
Evaluate : (96÷12)+14x(12+8)÷2
Expression (96÷12)+14x(12+8) ÷ 2 Evaluation=(96÷12)+14x(12+8) ÷ 2= 8 + 14x20 ÷ 2= 8 + 280 ÷ 2= 8 + 140 = 148 OperationParenthesesMultiplicationDivisionAdditionResult
Problem 8 :
Evaluate : (93+15) ÷ (3x4) - 24 + 8
Expression (93+15)÷(3x4)-24+8 Evaluation= (93+15)÷(3x4)-24+8 = 108 ÷ 12 - 24 + 8 = 9 - 24 + 8= -15 + 8= -7 OperationParenthesisDivisionSubtractionSubtractionResult
Problem 9 :
Evaluate : 55 ÷ 11 + (18 - 6) x 9
Expression 55÷11+(18-6)x9 Evaluation= 55÷11+(18-6)x9 = 55÷11 + 12x9= 5 + 12x9= 5 + 108= 113 OperationParenthesisDivisionMultiplicationAdditionResult
Problem 10 :
Evaluate : (7 + 18) x 3 ÷(2+13) - 28
Expression(7+18)x3÷(2+13)- 28 Evaluation= (7+18)x3÷(2+13)-28= 25x3 ÷ 15 - 28= 75 ÷ 15 - 28= 5 - 28= -23 OperationParenthesesMultiplicationDivisionSubtractionResult
We hope that the students would have understood the stuff given on "Order of operations worksheets".
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HBSE 7th Class Maths Solutions Chapter 10 Practical Geometry InText Questions
Haryana State Board HBSE 7th Class Maths Solutions Chapter 10 Practical Geometry InText Questions and Answers.
Haryana Board 7th Class Maths Solutions Chapter 10 Practical Geometry InText Questions
Think, Discuss & Write (Page No. 195):
Question 1.
In the adjoining construction, can you draw any other line through A that would be also parallel to the line 7 ?
Solution:
Yes, i.e.l || m,
Hence, m || n.
Question 2.
Can you slightly modify the adjoining construction to use the idea of equal correspon¬ding angles instead of equal alternate angles ?
Solution:
From construction and Fig. We have alternate angles and corresponding angles are equal.
Hence line l || m || n.
Think, Discuss & Write (Page No. 198):
Question 1.
A student attempted to draw a triangle whose rough figure is given here. He drew QR first. Then with Q as centre, he drew an arc of 3 cm and with R as centre, he drew an arc of 2 cm. But he could not get P. What is the reason ? What property of triangle do you dnow in connection with this problem ?
Can such a triangle exist ? (Remember the property of triangles. ‘The sum of any two sides of a triangle is always greter than the third side’!)
Solution:
Since, 3 cm + 2 cm = 5 cm < 6 cm 3 cm + 6 cm = 9 cm > 2 cm
2 cm + 6 cm = 8 cm > 3 cm
Hence, 3 cm, 2 cm and 6 cm are not the
sides of any triangle.
Because, “The sum of any two sides of a triangle is always greater than the third side”.
Think, Discuss & Write (Page No. 195):
Question 1.
In the above example, length of a side and measures of two angles were given. Now study the following problem :
In ΔABC, if AC = 7 cm, m∠A = 60° and m∠B = 50°, can you draw the triangle ? (Angle- sum property of a triangle may help you!)
Solution:
Angle sum property, sum of the three angles of a triangle is 180°.
Hence, ∠A + ∠B + ∠C = 180°
or, 60 + 50 + ∠C = 180°
or, ∠C = 180°-110°
∴ ∠C = 70°
Therefore, we can draw the triangle.
In ΔABC, if AC = 7 cm, m∠A = 60°, m∠B = 50°
∴ m∠C = 70°
Then ΔABC is the required triangle.
Miscellaneous Questions (Page No. 195):
Question 1.
Below are given the measures of certain sides and angles of triangles. Identify those which cannot be constructed and say why you cannot construct them. Construct rest of the triangles.
Triangle — Given measurements
1. ΔABC — m∠A = 85°; m∠B -115°; AB = 5 cm.
2. ΔPQR — m∠Q = 30°; m∠R = 60°; QR -4.7 cm.
3. ΔABC — m∠A = 70°; m∠B = 50°; AC = 3 cm.
4. ΔLMN — m∠L = 85° ; m∠N = 115; LM = 5 cm.
5. ΔABC — BC = 2 cm; AB = 4 cm ; AC = 2 cm.
a ΔPQR — PQ = 3.5 cm ; QR = 4 cm ; PR = 3.5 cm.
7. ΔXYZ — XY=3 c m ; YZ = 4 cm ; XZ = 5 cm.
8. ΔDEF — DE = 4.5 cm ; EF = 5.5 cm ; DF = 4 cm.
Solution:
1 In ΔABC, ∠A + ∠B = 85° + 115° = 200° which is not possible as the sum of three angles of a triangle is 180°. So this triangle cannot be constructed.
2. ∠Q + ∠R = 30° + 60° = 90° which is possible so the triangle can be constructed.
Steps of Construction :
1. Draw QR = 4.5 cm.
2. Usin g protractor at Q, draw an ∠Q = 30° and at R draw an ∠R = 60°.
3. Let the two new arms of these angles meet at P.
Thus the required triangle is ΔPQR.
3. ∠A + ∠B = 70° + 50° = 120°
∴ ∠C = 60°
So, this triangle can be constructed.
Steps of Construction :
1. Draw AC = 3cm.
2. At A, draw an
∠A = 70° (using protractor)
At C, draw an ∠C = 60° (using protractor)
3. Let the two arms of the new angles intersect each other at B.
Thus ΔABC is the required triangle.
4. ∠L + ∠N = 85° + 115° = 200° Which is not possible, because the sum of the three angles of a triangle is 180°.
So, such a triangle cannot be constructed.
5. Now BC + CA = 2 + 2 = 4 cm but
AB = 4 cm
∴ BC + CA = AB
But in a triangle, sum of two sides is always greater than the third.
Thus such a triangle cannot be constructed.
6. Now PQ + PR = 3.5 cm + 3.5 = 7 cm and QR = 4 cm.
∴ PQ + PR > QR is true because sum of two sides of a triangle is always greater than the third.
Thus this triangle can be constructed.
Steps of Construction :
1. Draw QR = 4 cm.
2. With Q and R as centre and radii 3.5 cm draw two arcs intersecting each other at P.
3. Join PQ and PR.
Now ΔPQR is the required triangle.
7. Now XY + YZ = 3 + 4 = 7 cm and XZ = 5 cm
∴ XY + YZ > XZ.
In a triangle, the sum of two sides is always greater than the third.
Thus such a triangle can be constructed.
Steps of construction :
1. Draw XZ = 5 cm.
2. With X and Z as centres and respecitve radii as 3 cm and 4 cm, draw two arcs intersecting each other at Y.
3. Join YX and YZ.
Thus, we get the required ΔXYZ.
8. DE + DF = 4.5 + 4 cm = 8.5 cm.
Also EF = 5.5 cm
∴ DE + DF > EF which is true, because sum of two sides of a triangle is greater than the third.
Thus such a Δ can be constructed.
Steps of Construction :
1. Draw DE = 4.5 cm.
2. with D and E as centres and respective radii as 4 cm and 5.5 cm draw two arcs intersecting each other at F.
3. Join DF and EF.
Thus we get the required ΔDEF.
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# More Cake Halves
6 teachers like this lesson
Print Lesson
## Objective
SWBAT create halves of squares and cakes. SWBAT explain that a whole split in halves has two equal pieces.
#### Big Idea
Students will continue to explore the concept of half.
## Warm Up
5 minutes
I have the students gather in a circle on the carpet. I give them each analog clock I will use the white board to write down digital times.
"I will write a time on the board using digital notation. Your job will be to set your clock to match that time. when you are finished, you should hold up your clock so that I can see your answer."
I repeat this with hour and half hour times.
At this point in the year, I am reviewing established routines and concepts from the year long curriculum. This activity is meant for students who are familiar with the use of a clock and understand the basic time concepts. I also think it leads nicely into a discussion of fractions because the clock has vocabulary that students need to learn that connects to concepts of halves and quarters (half past, quarter past, etc.).
This activity has students telling and writing times to hours and half hours. This meets the CCSS CCSS.MATH.CONTENT.1.MD.B.3.
## Folding Paper Squares
15 minutes
Advanced Preparation: You will need to cut out a bunch of blank paper squares.
"Today we are going to continue talking about halves and what it means to divide something in half. If i take this square (one square sheet of paper) and ask you to fold it in half, what will happen to it? What can you tell me about the pieces that will be created?"
"I would now like you to take two squares of your own and fold them in half (two different ways). When you are done, we will take a look at how each of you folded your squares."
• I want students to end up seeing that a square can be folded down the middle to create two equal sized rectangles or folded down the diagonal to create two equal triangle pieces. I will encourage them to conceptualize that these are the only two ways to cut a square into two equal pieces or halves. Other ways would result in unequal pieces that wouldn't meet the criteria for the concept of half.
"Who could show us one way they folded a square to cut it in half (see Folding in Half)?"
I will then post each (different) example on the board so that students can see the different ways of cutting a square in half (see Folding In Half). I will also label each half with the 1/2 notation.
Again, the main purpose of this activity is to establish that things that are cut in half have two EQUAL pieces.
## Making Half Cakes
30 minutes
Advanced Preparation: You will need to make enough copies of Square Cakes for each student.
"I am going to ask you to be bakers again and create more cakes using the Square Cakes sheet. I want you to find as many ways to divide your cakes in half and use marker to "frost" your cakes." I want you to use a different color for each half of the cake. As you are working, I will come around and talk to you about your cakes."
I have included a video, Explaining His Thinking of Halves, that captures a student explaining his work in a precise way (MP6).
This activity has students partitioning squares into two equal shares, describe the shares using the words halves, and the phrase half of. They are also asked to describe the whole as two of the shares (CCSS.MATH.CONTENT.1.G.A.3).
## Lesson Wrap Up
10 minutes
As students are cleaning up from the Creating Half Cake activity, I quickly sort their cakes (by how they were divided) and create a poster with them (Class Halves Collection).
"I want you to look at the collection of cakes that our class made. There are a variety of ways that the cakes were divided in half. However, no matter how they were cut, what can you say about each half?"
The idea is that students walk away understanding that two halves make a whole and that each half is the same size.
"I now want you to work on this sheet, Finding Halves, on your own. This will allow me to see who understands the concept of half."
## Continued Practice
5 minutes
I will ask the students to meet me on the carpet and hand out their sheet for today's Mad Minute exercise. This routine was introduced in a previous lesson. Please check out the link to get a full overview of this routine.
I want to really focus on fact fluency and build upon the students ability to solve within ten fluently (CCSS.MATH.CONTENT.1.OA.C.6). I am going to use the Mad Minute Routine. This is a very "old school" routine, but I truly feel students need practice in performing task for fluency in a timed fashion. Students need to obtain fact fluency in order to have success with multiplicative reasoning. Students who don't gain this addition fact fluency by the end of 2nd grade tend to struggle with the multiplicative reasoning in third. Having this fluency also allows them to work on more complex tasks because the have the fact recall to focus on the higher level concepts.
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# Prime Numbers of 20
Let us know about Prime Numbers of 20.
There are 20 prime numbers under 8: 2, 3, 5, 7, 11, 13, 17 and 19 .
Similarly, how do you find the factorization of 20? 20 = 1 x 20, 2 x 10 , or 4 x 5. Factors of 20: 1, 2, 4, 5, 10, 20. Prime factors: 20 = 2 x 2 x 5, also written as 20 = 2² x 5 can go. When 20 is a clue in the Find the Factor puzzle, use 4 x 5 or 2 x 10.
What is the prime factorization for 20 using the exponent? The prime factorization of 20 using the exponent is 22u22175 2 2 u2217 5 . First, we create prime factors for 20.
What are the multiples of 20? Multiples of 20 are 20, 40, 60 80, 100, 120, 140 and so on.
Why is the second 20 not a prime number? No, 20 is not a prime number . The number 20 is divisible by 1, 2, 4, 5, 10, 20. For a number to be classified as a prime number, it must have exactly two factors. Since 20 has more than two factors, i.e. 1, 2, 4, 5, 10, 20, it is not a prime number.
## What is the divisible by 20?
When we list them like this it becomes easy to see that the numbers divisible by 20 are 1, 2, 4, 5, 10 and 20 .
So is 20 a factor or a multiple? Table of factors and multiples
How many factors does 20 have? The factors of 20 are 1, 2, 4, 5, 10 and 20 .
## How do you find prime numbers?
To prove whether a number is a prime number, try dividing it by 2 first, and see if you get a whole number . If you do, it cannot be a prime number. If you don’t get a whole number, try dividing it by primes: 3, 5, 7, 11 (9 is divisible by 3) and so on, always dividing by a prime number (see table below) ).
What is prime factor? The prime factors are the factors of a number which are, themselves, prime numbers . There are several methods for finding the prime factors of a number, but one of the most common is to use a prime factorization tree.
Is 20 a correct number?
The number 20 is not a whole number . This can be demonstrated by finding its proper divisor and showing that their sum is not equal to 20.
Is every even number divisible by 20? As 20 is an even number, any number divisible by 20 is an even number.
### Is 20 divisible by 3 yes or no?
Since division does not give a whole number, this shows us that 20 3 . is not divisible by .
What are the multiples of 4 to 20?
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48 , …
How do you find a factor? How to find the number of factors?
1. Find its prime factorization, that is, express it as the product of primes.
2. Write the prime factorization in exponentiation form.
3. Add 1 to each exponent.
4. Multiply all the resulting numbers.
5. This product will give the number of factors of the given number.
Which of the following is not a factor of 20? 20! In order not to be a factor of , a number must not lie between 1 and 20 and should be called 20! Cannot be calculated as a product of numbers contained in That means 29, which is prime and greater than 20 and 121 which is 11*11 and in 20! We just have n’t a factor of 11 20! Answer: D and H.
## Which of the following is not a factor of 20?
20! In order not to be a factor of , a number must not lie between 1 and 20 and should be called 20! Cannot be calculated as a product of numbers contained in That means 29, which is prime and greater than 20 and 121 which is 11*11 and in 20! We just have n’t a factor of 11 20! Answer: D and H.
What are prime numbers in mathematics? Prime numbers are special numbers greater than 1, which have exactly two factors, themselves and 1 . 19 is a prime number. It can only be divided by 1 and 19. … the prime numbers below 20 are: 2, 3, 5, 7, 11, 13, 17, 19. Don’t forget: the number 1 is not a prime number because it is only a factor.
What are the first 20 mixed numbers?
The composite numbers from 1 to 20 are 4, 6, 8, 9, 10, 12, 14, 15, 16, 18, and 20 .
What are prime number charts? prime number chart
### How do you solve prime factors?
Prime Factorization Methods
1. Step 1: Divide the given number by the smallest prime number. ,
2. Step 2: Again, divide the quotient by the smallest prime number.
3. Step 3: Repeat the process until the quotient becomes 1.
4. Step 4: Lastly, multiply all the prime factors.
How do you factorize? To fully factorize an expression, find the highest common factor (HCF) of all terms . For example, the HCF of 4 x 2 is and 2 is the largest number that will be divisible into 4 and 6 and is the largest variable that will divide into and.
What is the whole number of 20?
List of Mersenne Prime Numbers and Whole Numbers
Which is the best number in the world? More Videos on YouTube
#### Is Armstrong a number?
An Armstrong number, also known as a narcissistic number, is a number that is equal to the sum of the cubes of its own digits . For example, 370 is an Armstrong number because 370 = 3*3*3 + 7*7*7 + 0*0*0 .
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# 2016 AMC 12A Problems/Problem 7
## Problem
Which of these describes the graph of $x^2(x+y+1)=y^2(x+y+1)$ ?
$\textbf{(A)}\ \text{two parallel lines}\\ \textbf{(B)}\ \text{two intersecting lines}\\ \textbf{(C)}\ \text{three lines that all pass through a common point}\\ \textbf{(D)}\ \text{three lines that do not all pass through a common point}\\ \textbf{(E)}\ \text{a line and a parabola}$
## Solution 1
The equation $x^2(x+y+1)=y^2(x+y+1)$ tells us $x^2=y^2$ or $x+y+1=0$ . $x^2=y^2$ generates two lines $y=x$ and $y=-x$ . $x+y+1=0$ is another straight line. The only intersection of $y=x$ and $y=-x$ is $(0,0)$ , which is not on $x+y+1=0$ . Therefore, the graph is three lines that do not have a common intersection, or $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$
## Solution 2
If $x+y+1\neq0$, then dividing both sides of the equation by $x+y+1$ gives us $x^2=y^2$. Rearranging and factoring, we get $x^2-y^2=(x+y)(x-y)=0$. If $x+y+1=0$, then the equation is satisfied. Thus either $x+y=0$, $x-y=0$, or $x+y+1=0$. These equations can be rearranged into the lines $y=-x$, $y=x$, and $y=-x-1$, respectively. Since these three lines are distinct, the answer is $\boxed{\textbf{(D)}\; \text{three lines that do not all pass through a common point}}$.
## Solution 3
Subtract $y^2(x+y+1)$ on both sides of the equation to get $x^2(x+y+1)-y^2(x+y+1)=0$. Factoring $x+y+1$ gives us $(x+y+1)(x^2-y^2)=(x+y+1)(x+y)(x-y)=0$, so either $x+y+1=0$, $x+y=0$, or $x-y=0$. Continue on with the second half of solution 2.
## Diagram:
$AB: y=x$
$CD: y=-x$
$EF: x+y+1=0$
$[asy] size(7cm); pair F= (5,0), E=(-1,6), D=(0,0), C=(6,0), B=(6,6), A=(0,6); draw(A--C); draw(B--D); draw(E--F); label("A", A, dir(135)); label("B", C, dir(-45)); label("C", B, dir(45)); label("D", D, dir(-135)); label("E", E, dir(135)); label("F", F, dir(-45)); [/asy]$
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# Basics of Circle
You might already know that Circles have the least surface area but did you ever think about the geometry which makes it so? Well yes, like straight lines, circles have geometry too and trust me when I say, it is as easy as its linear counterpart. After all, how difficult is it to understand a curve. So, let’s roll around the concepts and understand a Circle.
## Equation of Circle –
Usually, just after the above heading, you’ll find the line written “The equation of a circle is given by…!”. But let’s not do that. We will derive it from scratch.
Let us take a circle of radius ‘r’ and put it on the coordinate plane. Take the centre at O(h,k) and then a point A on the circumference and assume its coordinates to be (x,y) and join OA. Now, we will assume a point B inside the given circle such that OB is ⊥ to AB which gives us a right angled triangle AOB with right angle at B. This makes OA, hypotenuse.
From Pythagoras Theorem, we can say that
$$OB^2 + AB^2 = OA^2$$
Here, OA = r; OB = difference in the x-coordinate = x – h; AB = difference in the y-coordinate = y – k
which means, $$(x-h)^2 + (y-k)^2 = r^2$$
and this is the equation of a circle. Now if you’re thinking that this is just at one point, then stop yourself right there. You can take a similar point on any part of the circumference and still you’ll get the same result.
So, this equation is more like a collection of all such points on the circumference and is, hence, the equation of a circle.
The centre of this circle is at (h, k) and if you move it to the origin then the equation will become
 $$x^2 + y^2 = r^2$$
Equation of circle in parametric form –
Parametric Equation of circle with centre $$(h,k)$$Â and radius R is given by
$$x=h+R \cos{\theta}$$Â & $$y=k+R \sin{\theta}$$
where θ is the parameter.
## Solved Examples for You
Q. 1 Convert the equation $$x^2 + y^2 – 4x + 6y – 12 = 0 Â$$ into standard form and hence find its centre.
Sol – Here we have $$x^2 + y^2 – 4x + 6y – 12 = 0$$
⇒ $$x^2 – 4x + y^2 + 6y – 12 = 0$$
Add & subtract 4 & 9 to get perfect squares,
⇒  $$x^2 – 4x + 4 + y^2 + 6y + 9 – 25 = 0$$
⇒  $$(x – 2)^2 + (y + 3)^2 = 25$$
⇒  $$(x – 2)^2 + (y – (-3))^2 = 5^2$$
which is the standard form of equation of circle. Also, the centre is at (2, -3).
Q.2: If we have a circle of radius 20 cm with its centre at the origin, the circle can be described by the pair of equations?
Sol – We know that for parametric form of equation of circle,
$$x=h+R \cos{\theta}$$Â & $$y=k+R \sin{\theta}$$
Here, since the centre is at (0, 0), so, h = k = 0 and it is already mentioned that radius is 20 cm. Thus, $$x=20 \cos{t}$$ & $$y=20 \sin{t}$$ are the required pair of equations.
Q. What is a circle in math?
A circle is referred to as the locus of all points intermediate from a central point. In other words, all the points are equally distant from the central point in a circle. Moreover, it consists of radius, diameter and circumference as well.
Q. What is 90 degrees in a circle?
A. First, let us make clear that a circle divides into 360 equal degrees. Thus, this means that a right angle is 90°. So, the angle of an equilateral triangle comprises 60 degrees. Then again, scientists, engineers, and mathematicians usually measure angles in radians.
Q. What are the properties of a circle?
A. When we look at the essential properties of a circle, we see that it has many. A circle is congruent when it had equal radii. Moreover, looking at the diameter of the circle, we see that it is the longest chord of a circle. Further, equal chords and equal circles have the equal circumference.
Q. What is 2Ï€?
A full circle comprises of 2Ï€radians (approximately 6.28). Moreover, always remember that an arc of a circle is what defines a radian. Further, the length of the arc is equal to the radius of a circle.
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# Special Right Triangles
The special right triangle is a right triangle whose sides are in a particular ratio. According to the special right triangle rules, it is not necessary to use the Pythagorean theorem to get the size of a side.
## A step-by-step guide to solving Special Right Triangles
• Two special right triangles are $$45^{\circ}-45^{\circ}-90^{\circ}$$ and $$30^{\circ}-60^{\circ}-90^{\circ}$$ triangles.
• In a special $$45^{\circ}-45^{\circ}-90^{\circ}$$ triangle, the three angles are $$45^{\circ}, 45^{\circ}$$ and $$90^{\circ}$$. The lengths of the sides of this triangle are in the ratio of $$1:1:\sqrt{2}$$.
• In a special triangle $$30^{\circ}-60^{\circ}-90^{\circ}$$, the three angles are $$30^{\circ}-60^{\circ}-90^{\circ}$$. The lengths of this triangle are in the ratio of $$1:\sqrt{3}:2$$.
### Special Right Triangles – Example 1:
Find the length of the hypotenuse of a right triangle if the length of the other two sides are both 4 inches.
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Solution:
This is a right triangle with two equal sides. Therefore, it must be a $$45^{\circ}-45^{\circ}-90^{\circ}$$ triangle. Two equivalent sides are 4 inches. The ratio of sides: $$x:x:x\sqrt{2}$$. The length of the hypotenuse is $$4\sqrt{2}$$ inches. $$x:x:x\sqrt{2}→4:4:4\sqrt{2}$$
### Special Right Triangles – Example 2:
The length of the hypotenuse of a triangle is 6 inches. What are the lengths of the other two sides if one angle of the triangle is $$30^{\circ}$$?
Solution:
The hypotenuse is 6 inches and the triangle is a $$30^°-60^°-90^°$$ triangle. Then, one side of the triangle is 3 (it’s half the side of the hypotenuse) and the other side is $$3\sqrt{3}$$. (it’s the smallest side times $$\sqrt{3}$$) $$x:x\sqrt{3}:2x→x=3→x:x\sqrt{3}:2x=3:3\sqrt{3}:6$$
## Exercises for Special Right Triangles
### Find the value of x and y in each triangle.
1-
2-
3-
4-
5-
6-
1. $$\color{blue}{x=1,y=\frac{\sqrt{3}}{2}}$$
2. $$\color{blue}{x=4\sqrt{3},y=8\sqrt{3}}$$
3. $$\color{blue}{x=8\sqrt{2}}$$
4. $$\color{blue}{x=3\sqrt{2},y=3\sqrt{2}}$$
5. $$\color{blue}{x=3\sqrt{3},y=6\sqrt{3}}$$
6. $$\color{blue}{x=4\,y=2\sqrt{2}}$$
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I have already discussed a concept - Quadratic Equations of quantitative aptitude. Today I will discuss some examples of simple equations which have been proved to be a very important topic for various competitive exams. The problems of linear equations can be easily solved by using simple tricks. Lets discuss how.
## Examples with solutions
Example1: If 3x + 6 = 4x - 2, then find the value of x?
1. 8
2. 4
3. 6
4. 7
Solution: 3x + 6 = 4x-2
4x - 3x = 6 + 2
x = 8
By using trick: This question can be easily solved by eliminating the options.
Firstly check option(1) whether it satisfies the equation or not
3(8) + 6 = 4(8) - 2
30 = 30
Therefore,8 satisfies the equation.
Hence the answer is x = 8
Example2: If 2x + y = 5 and 3x - 2y = 4, then find the value of x and y.
1. 2,1
2. 3,-1
3. 4,4
4. 2,-2
Solution: Basic trick for this question is same as previous, just put the given values in equation and check which one is satisfying the equation.
Start with the first option i.e. 2,1
Put x = 2 and y = 1 in both equations and check if both equations satisfies.
2x + y = 5
2(2) + 1 = 5
5=5
3x - 2y = 4
3(2) - 2(1) = 4
4=4
Therefore, first option is satisfying the equation.
Example3: The sum of digits of two digit number is 12. If 54 is subtracted from the number, the digits gets reversed. Find the number.
1. 39
2. 85
3. 93
4. 75
Solution: In above question two statements are given i.e
Sum of digits of two digit number is 12 and If 54 is subtracted from the number, the digits gets reversed.
All the options except 85 satisfies the first statement. Therefore, reject the second option.
Now we are left with 1,3 and 4.
If 54 is subtracted from the number ⇒ 39 is rejected as 54 > 39, we cannot subtract bigger number.
So, we are left with only 93 and 75.
According to second statement,
93 - 54 = 39 Digit reversed
Example4: The sum of three consecutive even numbers is 30. Find the difference of the squares of extreme numbers.
Solution: Three consecutive even numbers = x , x+2 , x+4
According to ques,
x + x +2 + x + 4 = 30
⇒ 3x = 24
⇒ x = 8
Therefore, numbers are 8, 10 and 12.
Difference of squares of extreme numbers = (12)^2 - 8^2 = 144 - 64 = 80
Example5: The cost of one pen and two books together is Rs.70. The cost of 3 pens and 9 books is Rs.300. Find cost of book and pen.
1. 20, 15
2. 30, 10
3. 40, 5
4. 25, 6
Solution: Let cost of one pen is P and cost of one book is B
1P + 2B = 70
3P + 9B = 300
Eliminating the options, only second option will satisfy the equations,
1P + 2B = 70
⇒1(10) + 2(30) = 70
⇒70=70
3P + 9B = 300
⇒3(10) +9(30) = 300
⇒300 =300
Example6: p, q, r, s, t are five consecutive numbers in increasing order. If r + s + t + p =101, then find product of q and r.
Solution: Try to solve it yourself. Answer: 600
In this way, you can easily solve simple or linear equations problems. It helps you save your time in exam.
Simple or linear Equations: Tricks and Examples. Reviewed by Jasleen Behl on Friday, February 28, 2014 Rating: 5
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ME495+Statistics+Homework+Solutions
# ME495+Statistics+Homework+Solutions - ME495 Statistics...
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ME495 Statistics Homework Solutions 1. Single-factor ANOVA In an experiment to compare the quality of four different brands of recording tape, five reels of each brand were selected and the number of flaws in each reel was determined. (source: Jay L. Devore, Probability and Statistics for Engineering and Sciences , Wadsworth, Inc., 1982, p. 370) Is the expected number of flaws per reel the same for each brand? brand A 10 5 12 14 8 B 14 12 17 9 8 C 13 18 10 15 18 D 17 16 12 22 14 Solution: Create a matrix flaw in Matlab. The first row represents the data for brand A, the second row for brand B, the third row for brand C and the fourth row for brand D. Then use the function anova1 to perform a one-way ANOVA (i.e., single-factor ANOVA): >>[p, tbl, stats] = anova1(flaw’) The resulting ANOVA table looks like this: The reason to use the transpose of flaw ( flaw’ ) is that each column (rather than row) represents an independent sample containing different observations. The result p is the probability of the null hypothesis that all samples are drawn from the same population (or from different populations with the same means). If the p-value is near zero, this casts doubt on the null hypothesis and suggests that at least one sample mean is significantly different than the other sample means. It is common to declare a result significant if the p-value is less than 0.05. In this example, since p-value is greater than 0.05, we conclude that the sample means are barely the same. The result tbl returns the ANOVA table in a cell array. This table is also automatically plotted in a separate figure. The ANOVA table has columns for the sums of squares (SS), degree of freedom (df), mean squares (MS), F statistic, and p-value. In this case the p-value is about
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0.0529. An F statistic as extreme as the observed F would occur by chance once in less than twenty times if the number of flaws were truly equal. The result stats returns a stats structure that can be used to perform a follow-up multiple comparison test. 2. Two-factor ANOVA analysis A study to determine the effect of car model and factory on the gas mileage of cars shows the following results (the unit has been converted from miles/gallon to kilometers/liter; source: Matlab Statistics Toolbox User’s Guide Version 5 , the Mathworks, Inc., p. 133). There are three models of cars (columns) and two factories (rows). The reason there are six rows
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# Buffon's Needle
Buffon's Needle
The Buffon's Needle problem is a mathematical method of approximating the value of pi $(\pi = 3.1415...)$involving repeatedly dropping needles on a sheet of lined paper and observing how often the needle intersects a line.
# Basic Description
The method was first used to approximate π by Georges-Louis Leclerc, the Comte de Buffon, in 1777. Buffon posed the Buffon's Needle problem and offered the first experiment where he threw breadsticks over his shoulder and counted how often the crossed lines on his tiled floor.
Subsequent mathematicians have used the method with needles instead of bread sticks, or with computer simulations. In the case where the distance between the lines is equal the length of the needle, we will show that an approximation of π can be calculated using the equation
$\pi \approx {2*\mbox{number of drops} \over \mbox{number of hits}}$
# A More Mathematical Explanation
#### Will the Needle Intersect a Line?
[[Image:willtheneedlecros [...]
#### Will the Needle Intersect a Line?
To prove that the Buffon's Needle experiment will give an approximation of π, we can consider which positions of the needle will cause an intersection. Since the needle drops are random, there is no reason why the needle should be more likely to intersect one line than another. As a result, we can simplify our proof by focusing on a particular strip of the paper bounded by two horizontal lines.
The variable θ is the acute angle made by the needle and an imaginary line parallel to the ones on the paper. The distance between the lines is 1 and the needle length is 1. Finally, d is the distance between the center of the needle and the nearest line. Also, there is no reason why the needle is more likely to fall at a certain angle or distance, so we can consider all values of θ and d equally probable.
We can extend line segments from the center and tip of the needle to meet at a right angle. A needle will cut a line if the green arrow, d, is shorter than the leg opposite θ. More precisely, it will intersect when
$d \leq \left( \frac{1}{2} \right) \sin(\theta). \$
See case 1, where the needle falls at a relatively small angle with respect to the lines. Because of the small angle, the center of the needle would have to fall very close. In case 2, the needle intersects even though the center of the needle is far from both lines because the angle is so large.
#### The Probability of an Intersection
In order to show that the Buffon's experiment gives an approximation for π, we need to show that there is a relationship between the probability of an intersection and the value of π. If we graph the possible values of θ along the X axis and d along the Y, we have the sample space for the trials. In the diagram below, the sample space is contained by the dashed lines.
Each point on the graph represents some combination of an angle and a distance that a needle might occupy.
There will be an intersection if $d \leq \left ( \frac{1}{2} \right ) \sin(\theta) \$, which is represented by the blue region. The area under this curve represents all the combinations of distances and angles that will cause the needle to intersect a line. The area under the blue curve, which is equal to 1/2 in this case, can found by evaluating the integral
$\int_0^{\frac {\pi}{2}} \frac{1}{2} \sin(\theta) d\theta$
Then, the area of the sample space can be found by multiplying the length of the rectangle by the height.
$\frac {1}{2} * \frac {\pi}{2} = \frac {\pi}{4}$
The probability is equal to the ratio of the two areas in this case because each possible value of θ and d is equally probable. The probability of an intersection is
$P_{hit} = \cfrac{ \frac{1}{2} }{\frac{\pi}{4}} = \frac {2}{\pi} = .6366197...$
#### Using Random Samples to Approximate Pi
The original goal of the Buffon's needle method, approximating π, can be achieved by using probability to solve for π. If a large number of trials is conducted, the proportion of times a needle intersects a line will be close to the probability of an intersection. That is, the number of line hits divided by the number of drops will equal approximately the probability of hitting the line.
$\frac {\mbox{number of hits}}{\mbox{number of drops}} \approx P_{hit}$
Also, recall from above that
$P_{hit} = \frac {2}{\pi}$
So
$\frac {\mbox{number of hits}}{\mbox{number of drops}} \approx \frac {2}{\pi}$
Therefore, we can solve for π:
$\pi \approx \frac {2 * {\mbox{number of drops}}}{\mbox{number of hits}}$
# Why It's Interesting
#### Monte Carlo Methods
The Buffon's needle problem was the first recorded use of a Monte Carlo method. These methods employ repeated random sampling to approximate a probability, instead of computing the probability directly. Monte Carlo calculations are especially useful when the nature of the problem makes a direct calculation impossible or unfeasible, and they have become more common as the introduction of computers makes randomization and conducting a large number of trials less laborious.
π is an irrational number, which means that its value cannot be expressed exactly as a fraction a/b, where a and b are integers. As a result, π cannot be written as an exact decimal and mathematicians have been challenged with trying to determine increasingly accurate approximations. The timeline below shows the improvements in approximating pi throughout history. In the past 50 years especially, improvements in computer capability allow mathematicians to determine more decimal places. Nonetheless, better methods of approximation are still desired.
A recent study conducted the Buffon's Needle experiment to approximate π using computer software. The researchers administered 30 trials for each number of drops, and averaged their estimates for π. They noted the improvement in accuracy as more trials were conducted.
These results show that the Buffon's Needle approximation is relatively tedious. Even when a large number of needles are dropped, this experiment gave a value of pi that was inaccurate in the third decimal place. Compared to other computation techniques, Buffon's method is impractical because the estimates converge towards π rather slowly. Nonetheless, the intriguing relationship between the probability of a needle's intersection and the value of π has attracted mathematicians to study the Buffon's Needle method since its introduction in the 18th century.
#### Geometric Probability
The Buffon's Needle also gained notoriety because the problem began the study of geometric probability. The study of geometric probability allowed the probabilities to be determined by comparison of measurements, such as the area of a sample space to the area where a needle intersects a line. This differs from the traditional study of probability where equally probable discrete events, such as specific hands of cards, or rolls of dice are identified and counted.
#### Generalization of the problem
The Buffon’s needle problem has been generalized so that the probability of an intersection can be calculated for a needle of any length and paper with any spacing. For a needle shorter than the distance between the lines, it can be shown by a similar argument to the case where d = 1 and l = 1 that the probability of a intersection is $\frac {2*l}{\pi*d}$. Note that this agrees with the normal case, where l =1 and d =1, so these variables disappear and the probability is $\frac {2}{\pi}$.
The generalization of the problem is useful because it allows us to examine the relationship between length of the needle, distance between the lines, and probability of an intersection. The variable for length is in the numerator, so a longer needle will have a greater probability of an intersection. The variable for distance is in the denominator, so greater space between lines will decrease the probability.
To see how a longer needle will affect probability, follow this link: http://whistleralley.com/java/buffon_graph.htm
#### Needles in Nature
Applications of the Buffon's Needle method are even found naturally in nature. The Centre for Mathematical Biology at the University of Bath found uses of the Buffon's Needle algorithm in a recent study of ant colonies. The researchers found that an ant can estimate the size of an anthill by visiting the hill twice and noting how often it recrosses its first path.
Ants generally nest in groups of about 50 or 100, and the size of their nest preference is determined by the size of the colony. When a nest is destroyed, the colony must find a suitable replacement, so they send out scouts to find new potential homes.
In the study, scout ants were provided with "nest cavities of different sizes, shapes, and configurations in order to examine preferences" [2]. From their observations, researchers were able to draw the conclusion that scout ants must have a method of measuring areas.
A scout initially begins exploration of a nest by walking around the site to leave tracks. Then, the ant will return later and walk a new path that repeatedly intersects the first tracks. The first track will be laced with a chemical that causes the ant to note each time it crosses the original path. The researchers believe that these scout ants can calculate an estimate for the nest's area using the number of intersections between its two visits.
The ants can measure the size of their hill using a related and fairly intuitive method: If they are constantly intersecting their first path, the area must be small. If they rarely reintersects the first track, the area of the hill must be much larger so there is plenty of space for a non-intersecting second path.
"In effect, an ant scout applies a variant of Buffon's needle theorem: The estimated area of a flat surface is inversely proportional to the number of intersections between the set of lines randomly scattered across the surface." [7]
This idea can be related back to the generalization of the problem by imagining if the parallel lines were much further apart. A larger distance between the two lines would mean a much smaller probability of intersection. We can see in case 3 that when the distance between the lines is greater than the length of the needle, even very large angle won’t necessarily cause an intersection.
This natural method of random motion in nature allows the ants to gauge the size of their potential new hill regardless of its shape. Scout ants are even able to asses the area of a hill in complete darkness. The animals show that algorithms can be used to make decisions where an array of restrictions may prevent other methods from being effective.
# Teaching Materials
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# References
[4] The Number Pi. Eymard, Lafon, and Wilson.
[5] Monte Carlo Methods Volume I: Basics. Kalos and Whitlock.
[6] Heart of Mathematics. Burger and Starbird
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# Vedic Math Sutra: Sutra-14 Ekanyunena Purvena
## |
Since mathematics has always been essential to our lives, kids must have a solid foundation. While learning math through traditional ways might occasionally feel difficult, Vedic Mathematics offers a unique and fascinating approach that can make learning math for youngsters pleasant and exciting.
Ekanyunena Purvena, which helps kids to execute fast calculations efficiently, is one of the essential Vedic Maths concepts and is the sub-sutra to Nikhilam, Ekanyunena Purvena, and has the meaning of “one less than the previous” OR “one less than the one before.”
Only a few situations can be applied to this sutra, and they must meet strict requirements. Such as, a number that needs to be multiplied (the multiplicand/multiplier) only has nine digits. For instance, 999, 99999, 99999999, and so forth.
The condition mentioned above is possible only in three different situations.
#### Case 1. Both numbers contain an equal number of digits.
Example: 2463 x 9999
Step 1:
To obtain the left part of the answer, you deduct 1 from the Multiplicand using the Ekanyunena Purvena method.
2463 – 1 = 2462
Step 2:
To determine the right part of the answer, subtract the left part obtained in Step 1 from the multiplier.
9999- 2462 = 7537
Thus, 2463 × 9999 = 24627537
#### Case 2. Suppose the number of digits in one of the numbers is smaller than the count of 9s in the second number.
Example: 347 x 9999
Step 1:
By subtracting 1 from the Multiplicand, the left part of the answer is derived using the Ekanyunena Purvena method.
347-1 = 346
Step 2:
To obtain the right part of the answer, subtract the left part obtained in Step 1 from the multiplier.
9999 – 346 = 9653
Thus, 347 × 9999 = 3469653
#### Case 3. Suppose the number of digits in one of the numbers exceeds the count of 9s in the second number.
Example: 266 x 99
Step1. Due to two 9s in the multiplier, we multiply the Multiplicand by 100, representing the two zeroes in the result.
266 × 100 = 26600
Next, deduct the Multiplicand from the product obtained in step 1.
26600 – 266 = 26334
Thus, 266 × 99 = 26334
#### Wrapping Up!!
The Ekanyunena Purvena is a beneficial method that enables kids to calculate things with ease mentally. This technique can be used in mathematics to help kids solve problems more easily.
Patience, encouragement, and tailoring the learning process to each child’s particular skills and learning preferences are all essential for successfully transferring this Vedic knowledge. We at Moonpreneur understand the needs and demands this rapidly changing technological world is bringing with it for our kids. Hence, our expert-designed Advanced Math course for grades 3rd, 4th, 5th, and 6th will help your child develop math skills with hands-on lessons, excite them to learn, and help them build real-life applications.
Register for a free 60-minute Advanced Math Workshop today!
A writer with a talent to play with words and deliver impactful content. I have an undying love for literature, and firmly believe that pens have the power to change the course of human history!!
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jazzz
5 months ago
What is the square of 85 by using the formula Ekadhikena Purvena… I can’t solve it.
sonali sharma
5 months ago
85^2 = 8 × 9/5 × 5 = 72/25 = 7225… try exploring using this
Rahul
1 month ago
For what other problems can we use Ekadhikena Purvena???
Stacy
18 days ago
We can use it to find the square of complicated numbers.
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# AP Statistics Curriculum 2007 Bayesian Normal
## Probability and Statistics Ebook - Normal Example
### Motivational example
It is known that the speedometer that comes with a certain new sports car is not very accurate, which results in an estimate of the top speed of the car of 185 mph, with a standard deviation of 10 mph. Knowing that his car is capable of much higher speeds, the owner take the car to the shop. After a checkup, the speedometer is replaced with a better one, which gives a new estimate of 220 mph with a standard deviation of 4 mph. The errors are assumed to be normally distributed.
We can say that the owner S’s prior beliefs about the top speed of his car are represented by:
$\mu \sim N \mu_0$, $$\phi_0$$, i.e., $$\mu \sim N(185,10^2)$$
We could then say that the measurements using the new speedometer result in a measurement of:
$x \sim N(\mu, \phi)$, i.e., $$x \sim N(\mu, 4^2)$$
We note that the observation x turned out to be 210, and we see that S’s posterior beliefs about µ should be represented by:
$\mu | x \sim N(\mu_1, \phi_1)$
where (rounded)
$\phi_1 = (10^{-2} + 4^{-2})^{-1} = 14 = 4^2$
$\mu_1 = 14(185/10^2 + 220/4^2) = 218$.
Therefore, the posterior for the top speed is:
$\mu | x \sim N(218,4^2)$. Meaning 218 +/- 4 mph.
If the new speedometer measurements were considered by another person S’ who had no knowledge of the readings from the first speedometer, but still had a vague idea (from knowledge of the stock speedometer) that the top speed was about 200 +/- 30 mph. Then:
$\mu \sim N(200,30^2)$.
Then S’ would have a posterior variance:
$\phi_1 = (30^{-2} + 4^{-2})^{-1} = 16 = 4^2$.
S’ would have a posterior mean of:
$\mu_1 = 16(200/30^2 + 220/4^2) = 224$.
Therefore, the distribution of S’ would be:
$\mu | x \sim N(224,4^2)$. Meaning 224 +/- 4 mph.
This calculation has been carried out assuming that the prior information we have is rather vague, and therefore the posterior is almost entirely determined by the data.
The situation is summarized as follows:
Speed Prior Distribution Likelihood from Data Posterior Distribution
S $$N(185 , 10^2)$$ $$N(218 , 4^2)$$
$$N(220 , 4^2)$$
S’ $$N(200,30^2)$$ $$N(224,4^2)$$
|
# Understanding the Basics of Ohm’s Law
To work as an electrician, you need to have a good understanding of Ohm’s law.
With it, you can understand the relations between the three electrical variables: current, voltage, and resistance.
Search Electrician Programs
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## The Principle of Ohm’s Law
Ohm’s law has one rule that you should be familiar with.
The electrical current (I) in a circuit is proportional to the voltage (V) and in inverse proportion to the resistance (R).
This means that as voltage increases, the current will increase, provided that resistance remains constant.
Similarly, as resistance increases, the current will decrease.
In equation form, it will look like this:
V = I ✕ R
where,
V (volts) – the voltage across the conductor
I (amps) – the current flowing through the conductor
R (ohms) – the resistance provided by the conductor to the flow of current
Similarly, if you want to find I or R, you can rewrite the formula to get the answer, as shown below.
To find I:
I = V ÷ R
To find R:
R = V ÷ I
## How to Use the Ohm’s Law Formula
### Example 1: Solving for the Voltage (V)
You look at the circuit and see the values of two components:
• Current = 4 amps
• Resistance = 3 ohms
Here, you need to find its voltage.
To solve it, we need to use the values we know and work the equation.
V = I ✕ R
V = 4 amps ✕ 3 ohms
V = 12 volts
### Example 2: Solving for the Resistance (R)
You need to find the circuit resistance using the following values:
• Voltage = 120 volts
• Current = 17 amps
Now, let’s apply the values to the formula again and work the equation.
R = V ÷ I
R = 120 volts ÷ 17 amps
R = 7.05 ohms
### Example 3: Increasing Voltage
You have the following voltages: 120, 240, and 480, with the resistance remaining at 25 ohms.
Find the current for each.
Formula to use: I = V ÷ R
• 120 volts ÷ 25 ohms = 4.8 amps
• 240 volts ÷ 25 ohms = 9.6 amps
• 480 volts ÷ 25 ohms = 19.2 amps
Here, you’ll see that if the voltage increases, the current will increase if the resistance doesn’t change.
### Example 4: Increasing Resistance
A circuit has a voltage of 120, with the following resistance: 5, 10, and 20 ohms.
Find the current for each.
Formula to use: I = V ÷ R
• 120 volts ÷ 5 ohms = 60 amps
• 120 volts ÷ 10 ohms = 12 amps
• 120 volts ÷ 20 ohms = 6 amps
Here, if the resistance increases, the current will decrease if the voltage doesn’t change.
## Finding the Electrical Power Using Ohm’s Law
You can find how much electrical power there is in a circuit using Ohm’s law.
Power (P, watts) – the rate of energy conversion
We can find P if the values for two components are given, as shown below:
where,
V and I are given, P = V ✕ I
V and R are given, P = V2 ÷ R
I and R are given, P = I2 ✕ R
## Ohm’s Law Power Formulas
The Ohm’s law power formulas show the relationship between P, V, I, and R in a circuit where,
P – how much power is being produced
V – how strong is the flow in the circuit
I – how much current is flowing
R – how hard it is for the current to flow
P = V ✕ I
P = I2 ✕ R
P = V2 ÷ R
V = I ✕ R
V = √(P ✕ R)
V = P ÷ I
I = V ÷ R
I = √(P ÷ R)
I = P ÷ V
R = P ÷ I2
R = V ÷ I
R = V2 ÷ P
## How to Use the Ohm’s Law Pie Chart
You can use the Ohm’s law pie chart for reference in calculating P, V, I, and R.
In using it, make sure to follow these steps:
1. Know what value you’re trying to find: P, V, I, or R
2. Know the values given to solve the equation (you always need two)
3. Refer to the section of the pie chart where applicable
4. Solve the equation
Do note that you have to use compatible values while making calculations.
For instance, kilohms need to be converted to ohms or milliamperes to amperes.
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## Interview with Phil Cimino, Electrician from Las Vegas
We recently had the good fortune of interviewing Phil Cimino, an Electrician with over 30 years of experience from Las…
## How Hard Is It to Become an Electrician?
Like any profession, becoming an electrician can’t be achieved in just a snap of your fingers. This profession is one…
## Definitions of Electrical Terms You Should Know
As you progress in the electrical trade, you will hear and learn many terms or jargon being tossed around. Getting…
## Electrical Contractor License Requirements
Electrical contracting is a huge industry with more than \$130 billion in annual revenue. The industry has more than 70,000…
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# Binary Tree | Set 2 (Properties)
• Difficulty Level : Easy
• Last Updated : 19 Oct, 2021
We have discussed Introduction to Binary Tree in set 1. In this post, the properties of a binary tree are discussed.
1) The maximum number of nodes at level ‘l’ of a binary tree is 2l
Here level is the number of nodes on the path from the root to the node (including root and node). Level of the root is 0.
This can be proved by induction.
For root, l = 0, number of nodes = 20 = 1
Assume that the maximum number of nodes on level ‘l’ is 2l
Since in Binary tree every node has at most 2 children, next level would have twice nodes, i.e. 2 * 2l
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2) The Maximum number of nodes in a binary tree of height ‘h’ is 2h – 1
Here the height of a tree is the maximum number of nodes on the root to leaf path. Height of a tree with a single node is considered as 1.
This result can be derived from point 2 above. A tree has maximum nodes if all levels have maximum nodes. So maximum number of nodes in a binary tree of height h is 1 + 2 + 4 + .. + 2h. This is a simple geometric series with h terms and sum of this series is 2h– 1.
In some books, the height of the root is considered as 0. In this convention, the above formula becomes 2h+1 – 1
3) In a Binary Tree with N nodes, minimum possible height or the minimum number of levels is? Log2(N+1) ?
This can be directly derived from point 2 above. If we consider the convention where the height of a root node is considered as 0, then above formula for minimum possible height becomes | Log2(N+1) | – 1
4) A Binary Tree with L leaves has at least | Log2L? |+ 1 levels
A Binary tree has the maximum number of leaves (and a minimum number of levels) when all levels are fully filled. Let all leaves be at level l, then below is true for the number of leaves L.
```L <= 2l-1 [From Point 1]
l = | Log2L | + 1
where l is the minimum number of levels.```
5) In Binary tree where every node has 0 or 2 children, the number of leaf nodes is always one more than nodes with two children.
```L = T + 1
Where L = Number of leaf nodes
T = Number of internal nodes with two children
Proof:
No. of leaf nodes (L) i.e. total elements present at the bottom of tree =
2h-1 (h is height of tree)
No. of internal nodes = {total no. of nodes} - {leaf nodes} =
{ 2h - 1 } - {2h-1} = 2h-1 (2-1) - 1 = 2h-1 - 1
So , L = 2h-1
T = 2h-1 - 1
Therefore L = T + 1
Hence proved```
See Handshaking Lemma and Tree for proof.
In the next article on tree series, we will be discussing different types of Binary Trees and their properties
|
# 221_Chapter5_Student
Document Sample
``` 5.1 Review and Preview
In this chapter we combine the methods of descriptive statistics presented in Chapters 2 and 3 and those of
probability presented in Chapter 4. We will be constructing probability distributions by presenting possible
outcomes along with the relative frequencies we expect. Also, we consider discrete probability distributions.
5.2 Random Variables
A random variable is a variable (typically represented by X) that has a single numerical value,
determined by chance, for each outcome of a procedure.
A Discrete probability distribution is a description that gives the probability for each value of the
random variable. The probabilities are determined by theory or by observation.
Probability Distribution: Probabilities EXAMPLE 1: Create the graph of the probability distribution from
of Number of Mexican-Americans on a the table.
jury of 12. Assuming that jurors are
Randomly selected from a Population
in which 80% of the eligible people are
Mexican-Americans
x (Mexican-
P(x)
Americans)
0 0+
1 0+
2 0+
3 0+
4 0.001
5 0.003
6 0.016
7 0.053
8 0.133
9 0.236
10 0.283
11 0.206
12 0.069
Note: The table lists the values of x along with the corresponding probabilities. Probability values that are very
small, such as 0.000000123 are represented by 0+. We treat them as a value of 0.
Math 121 Chapter 5 Page 1
EXAMPLE 2: The probabilities of a return on an investment of \$1000, \$2000, and \$3000 are ½, ¼, and ¼,
respectively. Construct a probability distribution for the data and draw a graph for the distribution.
X P(X)
1000 0.50
2000 0.25
3000 0.25
TOTAL 1.00
A discrete random variable has either a finite number of values or a countable number of values, where
“countable” refers to the fact that there might be infinitely many values, but they can be associated with a
counting process.
A continuous random variable has infinitely many values, and those values can be associated with
measurements on a continuous scale without gaps or interruptions.
EXAMPLE 3: Identify the given random variable as being discrete or continuous.
a) The cost of conducting a genetics experiment.
b) The number of supermodels who at pizza yesterday.
c) The exact life span of a kitten.
d) The number of statistics professors who read a newspaper each day.
e) The weight of a feather.
Requirements for a Probability Distribution
1) P x 1 Where x assumes all possible values. (That is, the sum of all possibilities must be 1.)
2) 0 P x 1 for every individual value of x. (That is, each probability value must be between 0 and 1
inclusive.)
EXAMPLE 4: A researcher reports that when groups of four children are randomly selected from a population
of couples meeting certain criteria, the probability distribution for the number of girls is as given: Does the table
describe a probability distribution? Explain.
Math 121 Chapter 5 Page 2
EXAMPLE 5: Determine if the following distribution represents a probability distribution. Justify your reasoning
x P(x) x P(x) x P(x)
0 0.502 -4 0.50 3 0.2
1 0.465 -2 0.25 6 0.3
2 0.098 0 -0.50 9 0
3 0.011 2 0.25 12 0.3
4 0.001 4 0.50 14 0.2
Total 1.077 Total 1.00 Total 1.00
EXAMPLE 6: What value would be needed to make the following a probability distribution?
x P(x)
0 0.13
1 0.41
2
3 0.01
4 0.25
Total 1.00
EXAMPLE 7: Construct a probability distribution for a family of three children. Let X represent the number of
boys.
EXAMPLE 8: A game consists of rolling a die. If you roll an even number you win \$6. If you roll an
odd number you lose. You must pay \$4 to play. Find the probability distribution that reflects this
situation.
Math 121 Chapter 5 Page 3
EXAMPLE 9: Using the sample space for tossing two dice, construct a probability distribution for the
sums 2 through 12.
x P(x)
EXAMPLE 10: The probabilities that a customer selects 1, 2, 3, 4, or 5 items at a convenience store
are 0.32, 0.12, 0.23, 0.18, and 0.15, respectively. Create a probability distribution for the data.
x P(x)
We calculate the mean, variance, and standard deviation for a distribution differently than we do for samples
(Chapter 3).
Additionally, we will discuss the measure of expectation or expected value. This is idea is used in games of
chance as well as decision theory.
Round-Off Rule for , , 2 : Round results by carrying one more decimal place than the number of
decimal places used for the random variable x.
Mean for a Probability Distribution: x Px
The mean of a distribution captures the idea that if we were to repeat the experiment many, many times
(infinitely many) what on average would happen. It tells us what the “long-run” or theoretical average should
be.
We multiply each possible outcome by its corresponding probability and then find the sum (addition) of all
those products. It is easy to do this using an extension of the probability distribution table.
Math 121 Chapter 5 Page 4
EXAMPLE 1: Below is a probability distribution for a family of three children. Let X represent the number of
boys. Find the mean (average) number of children who will be boys.
x P(x)
x P x
0 1/8
1 3/8
2 3/8
3 1/8
Total 8/8=1
EXAMPLE 2: A study conducted by a TV station showed the number of television per household and
the corresponding probabilities for each. Find the mean of the distribution. If you were taking a
survey on the programs watched on television, how many program diaries would you send to each
household in the survey? Why?
Number of
1 2 3 4
TVs X
Probability
0.32 0.51 0.12 0.05
P(X)
x Px 0.32 1.02 0.36 0.20
x P x
Variance for a Probability Distribution:
2 x 2 P x
Variance of a probability distribution gives us a measure of the spread or variability. However, we must use a
different approach to this computation than what we used in Chapter 3. (That formula incorporated the
population size, N, and in this case we are dealing with an experiment that is repeated infinitely many times.
Hence, we have no specific population size).
The formula above can be quite tedious to use in calculations so we have an equivalent “short-cut” formula that
is easier to work with. It is show below.
Variance for a Probability Distribution: 2 x 2 Px 2
NOTE: When computing the variance of the distribution we must calculate the mean of the distribution
first.
Math 121 Chapter 5 Page 5
EXAMPLE 3: A recent survey by an insurance company showed the following probabilities for the number of
bedrooms in each insured home. Find the mean and variance for the distribution.
Number of
2 3 4 5
Bedrooms, X
Probability
0.3 0.4 0.2 0.1
P(X)
x Px 0.6 1.2 0.8 0.5
x 2 Px 1.2 3.6 3.2 2.5
x P x
2 x2 P x 2
EXAMPLE 4: Using the probability distribution for sums when tossing two dice, find the mean and
variance. (We created this distribution last class.)
Sums
X 2 3 4 5 6 7 8 9 10 11 12 Total
Probability,
P(X) 1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36 1
x Px 2/36 6/36 12/36 20/36 30/36 42/36 40/36 36/36 30/36 22/36 12/36 7
x 2 Px 4/36 18/36 48/36 100/36 180/36 294/36 320/36 324/360 300/36 242/36 144/36 1974/36
x P x
2 x2 P x 2
Standard Deviation for a Probability Distribution: x 2 Px 2
As before, we find the standard deviation by taking the square root of the variance. This means that in order to
find standard deviation we must first find the mean and the variance.
EXAMPLE 5: Use EXAMPLE 4 and find the standard deviation of the distribution.
x2 P x 2
Math 121 Chapter 5 Page 6
EXAMPLE 6: The number of refrigerators sold per day at a local appliance store is shown in the table, along
with the corresponding probabilities. Find the mean, variance, and standard deviation.
# of Fridges
0 1 2 3 4
X
Probability
0.1 0.2 0.3 0.2 0.2
P(X)
x Px 0 0.2 0.6 0.6 0.8
x 2 Px 0 0.2 1.2 1.8 3.2
x P x
2 x2 P x 2
x2 P x 2
EXAMPLE 7: The probabilities that a customer selects 1, 2, 3, 4, or 5 items at a convenience store
are 0.32, 0.12, 0.23, 0.18, and 0.15, respectively. Find the mean, variance, and standard deviation
for the probability distribution.
x P(x) x Px x 2 Px
1 0.32 0.32 0.32
2 0.12 0.24 0.48
3 0.23 0.69 2.07
4 0.18 0.72 2.88
5 0.15 0.75 3.75
Total 1.00 2.72 9.5
x Px 2.72 2 x 2 Px 2 9.5 2.72 2 2.1
x 2 Px 2 2.1 1.4
The expected value (expectation) of a discrete random variable is denoted by E(X), and it represents
the theoretical average value of the outcomes. It is obtained by finding
E ( X ) x Px
Expectation is the same as finding the mean of the distribution. However, we often refer to the mean
or average as the expected value when dealing with games of chance, insurance, and decision
theory.
EXAMPLE 8: A game consists of rolling a die. If you roll an even number you win \$6. If you roll an
odd number you lose. You must pay \$4 to play. What are your expected winnings?
x P(x)
E ( X ) x P x
2 .5
-4 .5
Total 1
Math 121 Chapter 5 Page 7
EXAMPLE 9: If a player rolls two dice and gets a sum of 2 or 12, she wins \$20. If the person gets a
7, she win \$5. The cost to play the game is \$3. Find her expectation of the game. What is the
expectation for the casino?
x P(x)
E ( X ) x P x
17 2/36
2 6/36
-3 28/36
Total 1
EXAMPLE 10: A lottery offers one \$1000 prize, one \$500 prize, and five \$100 prizes. One thousand
tickets are sold at \$3 each. Find the expectation if a person buys one ticket.
x P(x)
E ( X ) x P x
997 1/1000
497 1/1000
97 5/1000
-3 993/1000
Total 1
EXAMPLE 11: A 35-year-old woman purchases a \$100,000 term life insurance policy for an annual
payment of \$360. Based on a period life table for the US government, the probability that she will
survive the year is 0.999057. Find the expected value of the policy for the insurance company.
x P(x)
E ( X ) x P x
-99640 0.000943
360 0.999057
Math 121 Chapter 5 Page 8
5.3 Binomial Probability Distribution
Binomial probability distributions allow us to deal with circumstances in which the outcomes belong to two
relevant categories, such as acceptable/defective or survived/died.
A binomial probability distribution results from a procedure that meets all the following
requirements:
1) The procedure has a fixed number of trials.
2) The trials must be independent. (The outcome of any individual trial doesn’t affect the probabilities in
the other trials.)
3) Each trial must have all outcomes classified into two categories (commonly referred to as success
and failure).
4) The probability of a success remains the same in all trials.
Notation for Binomial Probability Distributions
S and F (success and failure) denote the two possible categories of all outcomes; p and q will denote the
probabilities of S and F, respectively, so
P( S ) p denotes the probabilit y of successin one of the n trials.
P( F ) 1 p q denotes the probabilit y of failure in one of the n trials.
n denotes the fixed number of trials
x denotes a specific # of successesin n trials, so x can be any
whole number between ________
P(x) denotes the probabilit y of getting exactly xsuccessesamong the n trials.
Note: The word success as used here is arbitrary and does not necessarily represent something good. Be
sure that x and p both refer to the same category being called success.
Note: When selecting a sample (such as a survey) for some statistical analysis, we usually sample
without replacement, and sampling without replacement involves dependent events, which violates the
second requirement in the above definition. However, the following rule of thumb is commonly used
(because errors are negligible): When sampling without replacement, the events can be treated as if they
are independent if the sample size is no more than 5% of the population size. ( n 0.05N )
EXAMPLE 1: In the case of Castaneda v. Partida it was noted that although 80% of the population in a Texas
county is Mexican-American, only 39% of those summoned for grand juries were Mexican-American. Let’s
assume that we need to select 12 jurors from a population that is 80% Mexican-American, and we want to find
the probability that among 12 randomly selected jurors, exactly 7 are Mexican-American.
a) Does this procedure result in a binomial distribution?
1) The procedure has a fixed number of trials?
2) The trials are independent?
3) Each trial must have all outcomes classified into two categories?
4) The probability of a success remains the same in all trials?
Math 121 Chapter 5 Page 9
b) If this procedure does result in a binomial distribution, identify the values of n, x, p and q.
n = 12, x = 7, p = 0.80, q = 1 – 0.80 = 0.20
P x
n!
Binomial Probability Formula p x q n x for x = 0, 1, 2, … n
n x ! x!
EXAMPLE 2: Use the binomial probability formula to find the probability of getting exactly 7 Mexican-
Americans when 12 jurors are selected at random from a population that is 80% Mexican-American.
EXAMPLE 3: Determine whether the given procedure results in a binomial distribution. For those that are not
binomial, identify at least one requirement that is not satisfied.
a) Treating 50 smokers with Nicorette and asking them how their mouth and throat feel.
b) Treating 50 smokers with Nicorette and recording whether there is a “yes” response when they are
asked if they experience any mouth or throat soreness.
EXAMPLE 4: A test consists of multiple choice questions, each having five possible answers {a, b, c, d, e},
one of which is correct. Assume that you guess the answers to six such questions.
a) Use the multiplication rule to find the probability that the first two guesses are wrong and the last four
guesses are correct. That is, find P(WWCCCC), where W is wrong and C is correct.
b) Beginning with WWCCCC, make a complete list of the different possible arrangements of two wrong
answers and four correct answers, then find the probability for each entry in the list.
Math 121 Chapter 5 Page 10
c) Based on the preceding results, what is the probability of getting exactly four correct answers when six
EXAMPLE 5: Assume that a procedure yields a binomial distribution with a trial repeated n times. Use the
binomial table to find the probability given: n = 6, x = 4, p = 0.20. Then use calculator to check results.
Compare the results you got in this problem with the answer from 4c.
EXAMPLE 6: A study was conducted to determine whether there were significant differences between medical
students admitted through special programs (such as affirmative action) and medical students admitted
through the regular admissions criteria. It was found that the graduation rate was 95% for the medical students
a) If 10 of the students from the special programs are randomly selected, find the probability that at least 9
b) Would it be unusual to randomly select 10 students from the special programs and get only 7 that
Math 121 Chapter 5 Page 11
5.4: Mean, Variance, and Standard Deviation for the Binomial Distribution
Mean for a Binomial Distribution: np
Variance for a Binomial Distribution: 2 npq
Standard Deviation for a Binomial Distribution: npq
Range Rule of Thumb: Use to identify “unusual” results. We saw this same idea when we talked
about variance and standard deviation in Chapter 3. In general, usual values lie within 2 standard
deviations of the mean.
Minimum “usual” value = 2
Maximum “usual” value = 2
EXAMPLE 7: The table below describes the probability distribution for the number of Mexican-Americans
among 12 randomly selected jurors in Hidalgo County, Texas. Assuming that we repeat the process of
randomly selecting 12 jurors and counting the number of Mexican-Americans each time, find the mean number
of Mexican-Americans (among 12), the variance, and the standard deviation.
x P(x) x Px x 2 Px
0 0+ 0 0
1 0+ 0 0
2 0+ 0 0
3 0+ 0 0
4 0.001 0.004 0.016
5 0.003 0.015 0.075
6 0.016 0.096 0.576
7 0.053 0.371 2.597
8 0.133 1.064 8.512
9 0.236 2.124 19.116
10 0.283 2.830 28.300
11 0.206 2.266 24.926
12 0.069 0.828 9.936
Total 1 9.598 94.054
x Px 9.6
2 x 2 Px 2 94 .054 9.598 2 1.9
x 2 Px 2 1.932396 1.4
Math 121 Chapter 5 Page 12
EXAMPLE 8: Use these results and the range rule of thumb to find the maximum and minimum usual values.
Based on the results, determine whether a jury consisting of 7 Mexican-Americans among 12 jurors is usual or
unusual.
EXAMPLE 9: Grand Jury Selection: Use the jury selection problem.
Now find the mean and standard deviation for the numbers of Mexican-Americans on juries selected from this
population that is 80% Mexican-Americans using binomial distribution.
EXAMPLE 10: Grand Jury Selection: Use the jury selection problem.
a) Assuming that groups of 870 grand jurors are randomly selected, find the mean and standard deviation
for the numbers of Mexican-Americans.
b) Use the range rule of thumb to find the minimum usual number and the maximum usual number of
Mexican-Americans. Based on those numbers, can we conclude that the actual result of 339 Mexican-
Americans is unusual? Dos this suggest that the selection process discriminated against Mexican-
Americans?
Math 121 Chapter 5 Page 13
EXAMPLE 11: Mars, Inc. claims that 14% of its M&M plain candies are yellow, and a sample of 100 such
candies is randomly selected.
a) Find the mean and standard deviation for the number of yellow candies in such groups of 100.
b) A data Set consists of a random sample of 100 M&Ms in which 8 are yellow. Is this result unusual?
Does it seem that the claimed rate of 14% is wrong?
Math 121 Chapter 5 Page 14
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# 4.7: Parametric Equations
Learning Objectives
• Plot a curve described by parametric equations.
• Convert the parametric equations of a curve into the form (y=f(x)).
• Recognize the parametric equations of basic curves, such as a line and a circle.
• Recognize the parametric equations of a cycloid.
In this section we examine parametric equations and their graphs. In the two-dimensional coordinate system, parametric equations are useful for describing curves that are not necessarily functions. The parameter is an independent variable that both (x) and (y) depend on, and as the parameter increases, the values of (x) and (y) trace out a path along a plane curve. For example, if the parameter is (t) (a common choice), then (t) might represent time. Then (x) and (y) are defined as functions of time, and ((x(t),y(t))) can describe the position in the plane of a given object as it moves along a curved path.
## Parametric Equations and Their Graphs
Consider the orbit of Earth around the Sun. Our year lasts approximately 365.25 days, but for this discussion we will use 365 days. On January 1 of each year, the physical location of Earth with respect to the Sun is nearly the same, except for leap years, when the lag introduced by the extra (frac{1}{4}) day of orbiting time is built into the calendar. We call January 1 “day 1” of the year. Then, for example, day 31 is January 31, day 59 is February 28, and so on.
The number of the day in a year can be considered a variable that determines Earth’s position in its orbit. As Earth revolves around the Sun, its physical location changes relative to the Sun. After one full year, we are back where we started, and a new year begins. According to Kepler’s laws of planetary motion, the shape of the orbit is elliptical, with the Sun at one focus of the ellipse. We study this idea in more detail in Conic Sections.
Figure ( PageIndex{1}) depicts Earth’s orbit around the Sun during one year. The point labeled (F_2) is one of the foci of the ellipse; the other focus is occupied by the Sun. If we superimpose coordinate axes over this graph, then we can assign ordered pairs to each point on the ellipse (Figure ( PageIndex{2})). Then each (x) value on the graph is a value of position as a function of time, and each (y) value is also a value of position as a function of time. Therefore, each point on the graph corresponds to a value of Earth’s position as a function of time.
We can determine the functions for (x(t)) and (y(t)), thereby parameterizing the orbit of Earth around the Sun. The variable (t) is called an independent parameter and, in this context, represents time relative to the beginning of each year.
A curve in the ((x,y)) plane can be represented parametrically. The equations that are used to define the curve are called parametric equations.
Definition: Parametric Equations
If (x) and (y) are continuous functions of (t) on an interval (I), then the equations
[x=x(t)]
and
[y=y(t)]
are called parametric equations and (t) is called the parameter. The set of points ((x,y)) obtained as (t) varies over the interval (I) is called the graph of the parametric equations. The graph of parametric equations is called a parametric curve or plane curve, and is denoted by (C).
Notice in this definition that (x) and (y) are used in two ways. The first is as functions of the independent variable (t). As (t) varies over the interval (I), the functions (x(t)) and (y(t)) generate a set of ordered pairs ((x,y)). This set of ordered pairs generates the graph of the parametric equations. In this second usage, to designate the ordered pairs, (x) and (y) are variables. It is important to distinguish the variables (x) and (y) from the functions (x(t)) and (y(t)).
Example (PageIndex{1}): Graphing a Parametrically Defined Curve
Sketch the curves described by the following parametric equations:
Solution
a. To create a graph of this curve, first set up a table of values. Since the independent variable in both (x(t)) and (y(t)) is (t), let (t) appear in the first column. Then (x(t)) and (y(t)) will appear in the second and third columns of the table.
(t)(x(t))(y(t))
−3−4−2
−2−30
−1−22
0−14
106
218
The second and third columns in this table provide a set of points to be plotted. The graph of these points appears in Figure ( PageIndex{3}). The arrows on the graph indicate the orientation of the graph, that is, the direction that a point moves on the graph as t varies from −3 to 2.
b. To create a graph of this curve, again set up a table of values.
(t)(x(t))(y(t))
−21−3
−1−2−1
0−31
1−23
215
367
The second and third columns in this table give a set of points to be plotted (Figure ( PageIndex{4})). The first point on the graph (corresponding to (t=−2)) has coordinates ((1,−3)), and the last point (corresponding to (t=3)) has coordinates ((6,7)). As (t) progresses from (−2) to (3), the point on the curve travels along a parabola. The direction the point moves is again called the orientation and is indicated on the graph.
c. In this case, use multiples of (π/6) for (t) and create another table of values:
(t)(x(t))(y(t))(t)(x(t))(y(t))
040(frac{7π}{6})(-2sqrt{3}≈−3.5)-2
(frac{π}{6})(2sqrt{3}≈3.5)2(frac{4π}{3})−2(−2sqrt{3}≈−3.5)
(frac{π}{3})2(2sqrt{3}≈3.5)(frac{3π}{2})0−4
(frac{π}{2})04(frac{5π}{3})2(−2sqrt{3}≈−3.5)
(frac{2π}{3})−2(2sqrt{3}≈3.5)(frac{11π}{6})(2sqrt{3}≈3.5)-2
(frac{5π}{6})(−2sqrt{3}≈−3.5)2(2π)40
(π)−40
The graph of this plane curve appears in the following graph.
This is the graph of a circle with radius (4) centered at the origin, with a counterclockwise orientation. The starting point and ending points of the curve both have coordinates ((4,0)).
Exercise (PageIndex{1})
Sketch the curve described by the parametric equations
Hint
Make a table of values for (x(t)) and (y(t)) using (t) values from (−3) to (2).
## Eliminating the Parameter
To better understand the graph of a curve represented parametrically, it is useful to rewrite the two equations as a single equation relating the variables (x) and (y). Then we can apply any previous knowledge of equations of curves in the plane to identify the curve. For example, the equations describing the plane curve in Example (PageIndex{1b}) are
[egin{align} x(t) &=t^2−3 label{x1} [4pt] y(t) &=2t+1 label{y1} end{align}]
over the region (-2 le t le 3.)
Solving Equation ef{y1} for (t) gives
[t=dfrac{y−1}{2}. onumber]
This can be substituted into Equation ef{x1}:
[egin{align} x &=left(dfrac{y−1}{2} ight)^2−3 [4pt] &=dfrac{y^2−2y+1}{4}−3 [4pt] &=dfrac{y^2−2y−11}{4}. label{y2}end{align}]
Equation ef{y2} describes (x) as a function of (y). These steps give an example of eliminating the parameter. The graph of this function is a parabola opening to the right (Figure (PageIndex{4})). Recall that the plane curve started at ((1,−3)) and ended at ((6,7)). These terminations were due to the restriction on the parameter (t).
Example (PageIndex{2}): Eliminating the Parameter
Eliminate the parameter for each of the plane curves described by the following parametric equations and describe the resulting graph.
Solution
a. To eliminate the parameter, we can solve either of the equations for (t). For example, solving the first equation for (t) gives
[egin{align*} x &=sqrt{2t+4} [4pt] x^2 &=2t+4 [4pt] x^2−4 &=2t [4pt] t &=dfrac{x^2−4}{2}. end{align*}]
Note that when we square both sides it is important to observe that (x≥0). Substituting (t=dfrac{x^2−4}{2}) into (y(t)) yields
[ y(t)=2t+1]
[ y=2left(dfrac{x^2−4}{2} ight)+1]
[ y=x^2−4+1]
[ y=x^2−3.]
This is the equation of a parabola opening upward. There is, however, a domain restriction because of the limits on the parameter (t). When (t=−2), (x=sqrt{2(−2)+4}=0), and when (t=6), (x=sqrt{2(6)+4}=4). The graph of this plane curve follows.
b. Sometimes it is necessary to be a bit creative in eliminating the parameter. The parametric equations for this example are
[ x(t)=4 cos t onumber]
and
[ y(t)=3 sin t onumber]
Solving either equation for (t) directly is not advisable because sine and cosine are not one-to-one functions. However, dividing the first equation by (4) and the second equation by (3) (and suppressing the (t)) gives us
[ cos t=dfrac{x}{4} onumber]
and
[ sin t=dfrac{y}{3}. onumber]
Now use the Pythagorean identity (cos^2t+sin^2t=1) and replace the expressions for (sin t) and (cos t) with the equivalent expressions in terms of (x) and (y). This gives
[ left(dfrac{x}{4} ight)^2+left(dfrac{y}{3} ight)^2=1 onumber]
[ dfrac{x^2}{16}+dfrac{y^2}{9}=1. onumber]
This is the equation of a horizontal ellipse centered at the origin, with semi-major axis (4) and semi-minor axis (3) as shown in the following graph.
As t progresses from (0) to (2π), a point on the curve traverses the ellipse once, in a counterclockwise direction. Recall from the section opener that the orbit of Earth around the Sun is also elliptical. This is a perfect example of using parameterized curves to model a real-world phenomenon.
Exercise (PageIndex{2})
Eliminate the parameter for the plane curve defined by the following parametric equations and describe the resulting graph.
Hint
Solve one of the equations for (t) and substitute into the other equation.
(x=2+frac{3}{y+1},) or (y=−1+frac{3}{x−2}). This equation describes a portion of a rectangular hyperbola centered at ((2,−1)).
So far we have seen the method of eliminating the parameter, assuming we know a set of parametric equations that describe a plane curve. What if we would like to start with the equation of a curve and determine a pair of parametric equations for that curve? This is certainly possible, and in fact it is possible to do so in many different ways for a given curve. The process is known as parameterization of a curve.
Example (PageIndex{3}): Parameterizing a Curve
Find two different pairs of parametric equations to represent the graph of (y=2x^2−3).
Solution
First, it is always possible to parameterize a curve by defining (x(t)=t), then replacing (x) with (t) in the equation for (y(t)). This gives the parameterization
Since there is no restriction on the domain in the original graph, there is no restriction on the values of (t).
We have complete freedom in the choice for the second parameterization. For example, we can choose (x(t)=3t−2). The only thing we need to check is that there are no restrictions imposed on (x); that is, the range of (x(t)) is all real numbers. This is the case for (x(t)=3t−2). Now since (y=2x^2−3), we can substitute (x(t)=3t−2) for (x). This gives
[ y(t)=2(3t−2)^2−2=2(9t^2−12t+4)−2=18t^2−24t+8−2=18t^2−24t+6. onumber]
Therefore, a second parameterization of the curve can be written as
( x(t)=3t−2) and ( y(t)=18t^2−24t+6.)
Exercise (PageIndex{3})
Find two different sets of parametric equations to represent the graph of (y=x^2+2x).
Hint
Follow the steps in Example (PageIndex{3}). Remember we have freedom in choosing the parameterization for (x(t)).
One possibility is (x(t)=t, quad y(t)=t^2+2t.) Another possibility is (x(t)=2t−3, quad y(t)=(2t−3)^2+2(2t−3)=4t^2−8t+3.) There are, in fact, an infinite number of possibilities.
## Cycloids and Other Parametric Curves
Imagine going on a bicycle ride through the country. The tires stay in contact with the road and rotate in a predictable pattern. Now suppose a very determined ant is tired after a long day and wants to get home. So he hangs onto the side of the tire and gets a free ride. The path that this ant travels down a straight road is called a cycloid (Figure ( PageIndex{8})). A cycloid generated by a circle (or bicycle wheel) of radius a is given by the parametric equations
[x(t)=a(t−sin t), quad y(t)=a(1−cos t). onumber]
To see why this is true, consider the path that the center of the wheel takes. The center moves along the (x)-axis at a constant height equal to the radius of the wheel. If the radius is (a), then the coordinates of the center can be given by the equations
for any value of (t). Next, consider the ant, which rotates around the center along a circular path. If the bicycle is moving from left to right then the wheels are rotating in a clockwise direction. A possible parameterization of the circular motion of the ant (relative to the center of the wheel) is given by
[egin{align*} x(t) &=−a sin t [4pt] y(t) &=−acos t.end{align*}]
(The negative sign is needed to reverse the orientation of the curve. If the negative sign were not there, we would have to imagine the wheel rotating counterclockwise.) Adding these equations together gives the equations for the cycloid.
[egin{align*} x(t) &=a(t−sin t) [4pt] y(t) &=a(1−cos t ) end{align*}]
Now suppose that the bicycle wheel doesn’t travel along a straight road but instead moves along the inside of a larger wheel, as in Figure ( PageIndex{9}). In this graph, the green circle is traveling around the blue circle in a counterclockwise direction. A point on the edge of the green circle traces out the red graph, which is called a hypocycloid.
The general parametric equations for a hypocycloid are
[x(t)=(a−b) cos t+b cos (dfrac{a−b}{b})t onumber]
[y(t)=(a−b) sin t−b sin (dfrac{a−b}{b})t. onumber]
These equations are a bit more complicated, but the derivation is somewhat similar to the equations for the cycloid. In this case we assume the radius of the larger circle is (a) and the radius of the smaller circle is (b). Then the center of the wheel travels along a circle of radius (a−b.) This fact explains the first term in each equation above. The period of the second trigonometric function in both (x(t)) and (y(t)) is equal to (dfrac{2πb}{a−b}).
The ratio (dfrac{a}{b}) is related to the number of cusps on the graph (cusps are the corners or pointed ends of the graph), as illustrated in Figure ( PageIndex{10}). This ratio can lead to some very interesting graphs, depending on whether or not the ratio is rational. Figure (PageIndex{9}) corresponds to (a=4) and (b=1). The result is a hypocycloid with four cusps. Figure (PageIndex{10}) shows some other possibilities. The last two hypocycloids have irrational values for (dfrac{a}{b}). In these cases the hypocycloids have an infinite number of cusps, so they never return to their starting point. These are examples of what are known as space-filling curves.
The Witch of Agnesi
Many plane curves in mathematics are named after the people who first investigated them, like the folium of Descartes or the spiral of Archimedes. However, perhaps the strangest name for a curve is the witch of Agnesi. Why a witch?
Maria Gaetana Agnesi (1718–1799) was one of the few recognized women mathematicians of eighteenth-century Italy. She wrote a popular book on analytic geometry, published in 1748, which included an interesting curve that had been studied by Fermat in 1630. The mathematician Guido Grandi showed in 1703 how to construct this curve, which he later called the “versoria,” a Latin term for a rope used in sailing. Agnesi used the Italian term for this rope, “versiera,” but in Latin, this same word means a “female goblin.” When Agnesi’s book was translated into English in 1801, the translator used the term “witch” for the curve, instead of rope. The name “witch of Agnesi” has stuck ever since.
The witch of Agnesi is a curve defined as follows: Start with a circle of radius a so that the points ((0,0)) and ((0,2a)) are points on the circle (Figure ( PageIndex{11})). Let O denote the origin. Choose any other point A on the circle, and draw the secant line OA. Let B denote the point at which the line OA intersects the horizontal line through ((0,2a)). The vertical line through B intersects the horizontal line through A at the point P. As the point A varies, the path that the point P travels is the witch of Agnesi curve for the given circle.
Witch of Agnesi curves have applications in physics, including modeling water waves and distributions of spectral lines. In probability theory, the curve describes the probability density function of the Cauchy distribution. In this project you will parameterize these curves.
1. On the figure, label the following points, lengths, and angle:
a. (C) is the point on the (x)-axis with the same (x)-coordinate as (A).
b. (x) is the (x)-coordinate of (P), and (y) is the (y)-coordinate of (P).
c. (E) is the point ((0,a)).
d. (F) is the point on the line segment (OA) such that the line segment (EF) is perpendicular to the line segment (OA).
e. (b) is the distance from (O) to (F).
f. (c) is the distance from (F) to (A).
g. (d) is the distance from (O) to (C).
h. (θ) is the measure of angle (∠COA).
The goal of this project is to parameterize the witch using (θ) as a parameter. To do this, write equations for (x) and (y) in terms of only (θ).
2. Show that (d=dfrac{2a}{sin θ}).
3. Note that (x=dcos θ). Show that (x=2acot θ). When you do this, you will have parameterized the (x)-coordinate of the curve with respect to (θ). If you can get a similar equation for (y), you will have parameterized the curve.
4. In terms of (θ), what is the angle (∠EOA)?
5. Show that (b+c=2acosleft(frac{π}{2}−θ ight)).
6. Show that (y=2acosleft(frac{π}{2}−θ ight)sin θ).
7. Show that (y=2asin^2θ). You have now parameterized the (y)-coordinate of the curve with respect to (θ).
8. Conclude that a parameterization of the given witch curve is
9. Use your parameterization to show that the given witch curve is the graph of the function (f(x)=dfrac{8a^3}{x^2+4a^2}).
Travels with My Ant: The Curtate and Prolate Cycloids
Earlier in this section, we looked at the parametric equations for a cycloid, which is the path a point on the edge of a wheel traces as the wheel rolls along a straight path. In this project we look at two different variations of the cycloid, called the curtate and prolate cycloids.
First, let’s revisit the derivation of the parametric equations for a cycloid. Recall that we considered a tenacious ant trying to get home by hanging onto the edge of a bicycle tire. We have assumed the ant climbed onto the tire at the very edge, where the tire touches the ground. As the wheel rolls, the ant moves with the edge of the tire (Figure (PageIndex{12})).
As we have discussed, we have a lot of flexibility when parameterizing a curve. In this case we let our parameter t represent the angle the tire has rotated through. Looking at Figure ( PageIndex{12}), we see that after the tire has rotated through an angle of (t), the position of the center of the wheel, (C=(x_C,y_C)), is given by
(x_C=at) and (y_C=a).
Furthermore, letting (A=(x_A,y_A)) denote the position of the ant, we note that
(x_C−x_A=asin t) and (y_C−y_A=a cos t)
Then
[x_A=x_C−asin t=at−asin t=a(t−sin t)]
[y_A=y_C−acos t=a−acos t=a(1−cos t).]
Note that these are the same parametric representations we had before, but we have now assigned a physical meaning to the parametric variable (t).
After a while the ant is getting dizzy from going round and round on the edge of the tire. So he climbs up one of the spokes toward the center of the wheel. By climbing toward the center of the wheel, the ant has changed his path of motion. The new path has less up-and-down motion and is called a curtate cycloid (Figure ( PageIndex{13})). As shown in the figure, we let b denote the distance along the spoke from the center of the wheel to the ant. As before, we let t represent the angle the tire has rotated through. Additionally, we let (C=(x_C,y_C)) represent the position of the center of the wheel and (A=(x_A,y_A)) represent the position of the ant.
1. What is the position of the center of the wheel after the tire has rotated through an angle of (t)?
2. Use geometry to find expressions for (x_C−x_A) and for (y_C−y_A).
3. On the basis of your answers to parts 1 and 2, what are the parametric equations representing the curtate cycloid?
Once the ant’s head clears, he realizes that the bicyclist has made a turn, and is now traveling away from his home. So he drops off the bicycle tire and looks around. Fortunately, there is a set of train tracks nearby, headed back in the right direction. So the ant heads over to the train tracks to wait. After a while, a train goes by, heading in the right direction, and he manages to jump up and just catch the edge of the train wheel (without getting squished!).
The ant is still worried about getting dizzy, but the train wheel is slippery and has no spokes to climb, so he decides to just hang on to the edge of the wheel and hope for the best. Now, train wheels have a flange to keep the wheel running on the tracks. So, in this case, since the ant is hanging on to the very edge of the flange, the distance from the center of the wheel to the ant is actually greater than the radius of the wheel (Figure (PageIndex{14})).
The setup here is essentially the same as when the ant climbed up the spoke on the bicycle wheel. We let b denote the distance from the center of the wheel to the ant, and we let t represent the angle the tire has rotated through. Additionally, we let (C=(x_C,y_C)) represent the position of the center of the wheel and (A=(x_A,y_A)) represent the position of the ant (Figure ( PageIndex{14})).
When the distance from the center of the wheel to the ant is greater than the radius of the wheel, his path of motion is called a prolate cycloid. A graph of a prolate cycloid is shown in the figure.
4. Using the same approach you used in parts 1– 3, find the parametric equations for the path of motion of the ant.
Notice that the ant is actually traveling backward at times (the “loops” in the graph), even though the train continues to move forward. He is probably going to be really dizzy by the time he gets home!
## Key Concepts
• Parametric equations provide a convenient way to describe a curve. A parameter can represent time or some other meaningful quantity.
• It is often possible to eliminate the parameter in a parameterized curve to obtain a function or relation describing that curve.
• There is always more than one way to parameterize a curve.
• Parametric equations can describe complicated curves that are difficult or perhaps impossible to describe using rectangular coordinates.
## Glossary
cycloid
the curve traced by a point on the rim of a circular wheel as the wheel rolls along a straight line without slippage
cusp
a pointed end or part where two curves meet
orientation
the direction that a point moves on a graph as the parameter increases
parameter
an independent variable that both (x) and (y) depend on in a parametric curve; usually represented by the variable (t)
parametric curve
the graph of the parametric equations (x(t)) and (y(t)) over an interval (a≤t≤b) combined with the equations
parametric equations
the equations (x=x(t)) and (y=y(t)) that define a parametric curve
parameterization of a curve
rewriting the equation of a curve defined by a function (y=f(x)) as parametric equations
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### Mathematics
Q/ Find the parametric equations of the line joining the points P1( 1, 1, -1) and P2( 1, 2, 0)?
Determine vector, parametric, and if possible, symmetric equations of the line through D(-4, 3, 6) and parallel to the z-axis. Vector form: + t Parametric: x = -4 y = 3 z = 6 + t Symmetric: x = -4 , y = 3 , z - 6 Is this right?
### Discrete Math: Equations of a Line
Find parametric equations for the line that passes through the point (0, -1, 1) and the midpoint of the line segment from (2, 3, -2) to (4, -1, 5). I don't know how to solve this problem.
### Linear Algebra
Let L be the line with parametric equations x = 1−2t y = 3+3t z = −1−3t Find the shortest distance d from the point P0=(1, −4, −2) to L, and the point Q on L that is closest to P0. Use the square root symbol '√' where
### Math vector
Find the unit tangent vector T(t) and find a set of parametric equations for the line tangent to the space curve at point P. r(t)=ti+t^(2)j+tk, P(0,0,0)
### Geometry
Prove that the tangents to a circle at the endpoints of a diameter are parallel. State what is given, what is to be proved, and your plan of proof. Then write a two-column proof. Hint draw a DIAGRAM with the points labeled. Can
### 12th math
A biologist determines that the path of a bee from its hive to its foraging site can be described by the parametric equations x= 4t-1 and y=2t^2+3t-4. Which of the following equations is the curve described by these parametric
Lets consider a square of size $2 imes 2$ in a rectangular coordinate system, where each side is a part of the lines $x = pm 1, y = pm 1$ . Then you can define
$gamma(alpha) = egin (1, an(alpha)) & alpha in [-pi/4 , pi/4] (cot(alpha),1) & alpha in [pi / 4, 3 pi /4 ] (-1,- an(alpha)) & alpha in [3pi/4 , 5pi/4] (-cot(alpha),-1) & alpha in [5pi/4, 7pi/4] end$
which describes the parametric path of square shape, with the angle as parameter.
For using it in a program following works also very well: Define $varphi(x) = max(0, min(1, 3/2 - vert x vert))$ . Then
$gamma: [0, 4] o mathbb R^2$ with $t mapsto (varphi(t-3/2), varphi(t-5/2))$
works very well and doesn't need any expensive evaluations of trigonometric functions.
## Solutions for Chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions
Solutions for Chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions
• 4-7.R.0: Update your journal with what you have learned in this chapter. Inc.
• 4-7.R.1: Figure 4-7a shows a unit circle and an angle in standard position. .
• 4-7.R.2: a. Write equations expressing tan x and cot x in terms of sin x and.
• 4-7.R.3: a. Transform tan A sin A + cos A to sec A. What values of A are exc.
• 4-7.R.4: a. Find the general solution for = arcsin 0. b. Solve 1 + tan 2 (x .
• 4-7.R.5: a. Plot the graph of this parametric function on your grapher. Sket.
• 4-7.R.6: a. Using parametric mode on your grapher, duplicate the graph of th.
• 4-7.C.1: Pendulum Problem: Figure 4-7d shows a pendulum hanging from the cei.
• 4-7.C.2: Prove that each of these equations is an identity. a. b. C
• 4-7.C.3: Square of a Sinusoid Problem: Figure 4-7e shows the graphs of y1 = .
• 4-7.T.1: Write the Pythagorean property for cosine and sine. T
• 4-7.T.2: Write a quotient property involving cosine and sine. T
• 4-7.T.3: Write the reciprocal property for cotangent. T
• 4-7.T.4: Write the reciprocal property for secant. T
• 4-7.T.5: The value of sin1 0.5 is 30. Write the general solution for = arcsi.
• 4-7.T.6: The value of tan1 is . Write the general solution for x = arctan . T
• 4-7.T.7: Explain why the range of y = cos1 x is [0, ] but the range of y = s.
• 4-7.T.8: Find geometrically the exact value of cos (tan1 2). T
• 4-7.T.9: Transform (1 + sin A)(1 sin A) to cos2 A. What values of A are excl.
• 4-7.T.10: Prove that tan B + cot B = csc B sec B is an identity. What values .
• 4-7.T.11: Multiply the numerator and denominator of by the conjugate of the d.
• 4-7.T.12: Plot the graphs of both expressions in T11 to confirm that the two .
• 4-7.T.13: With your calculator in degree mode, find the value of cos1 0. Show.
• 4-7.T.14: Find another angle between 0 and 360 whose cosine is 0. Show it on .
• 4-7.T.15: Write the general solution for the inverse trigonometric relation =.
• 4-7.T.16: Find the fifth positive value of for which cos = 0. How many revolu.
• 4-7.T.17: Find algebraically the general solution for 4 tan ( 25) = 7 T1
• 4-7.T.18: Write parametric equations for this ellipse (Figure 4-7f). Figure 4.
• 4-7.T.19: Write the parametric equations you use to plot y = arctan x (Figure.
• 4-7.T.20: What did you learn as a result of working this test that you did no.
##### ISBN: 9781559533911
Chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions includes 30 full step-by-step solutions. This expansive textbook survival guide covers the following chapters and their solutions. This textbook survival guide was created for the textbook: Precalculus with Trigonometry: Concepts and Applications, edition: 1. Precalculus with Trigonometry: Concepts and Applications was written by and is associated to the ISBN: 9781559533911. Since 30 problems in chapter 4-7: Trigonometric Function Properties, Identities, and Parametric Functions have been answered, more than 35362 students have viewed full step-by-step solutions from this chapter.
Union of two rays with a common endpoint (the vertex). The beginning ray (the initial side) can be rotated about its endpoint to obtain the final position (the terminal side)
A graph that displays a five-number summary
(a + bi) - (c + di) = (a - c) + (b - d)i
A statement of equality between two expressions.
Arrows that have the same magnitude and direction.
A procedure for fitting an exponential function to a set of data.
The net amount of money returned from an annuity.
A visible representation of a numerical or algebraic model.
The line is a horizontal asymptote of the graph of a function ƒ if lim x:- q ƒ(x) = or lim x: q ƒ(x) = b
A function ƒ is increasing on an interval I if, for any two points in I, a positive change in x results in a positive change in.
Length of 1 minute of arc along the Earth’s equator.
A complex number v such that vn = z
If bn = a, then b is an nth root of a. If bn = a and a and b have the same sign, b is the principal nth root of a (see Radical), p. 508.
The difference y1 - (ax 1 + b), where (x1, y1)is a point in a scatter plot and y = ax + b is a line that fits the set of data.
## Parametric Equation of a Circle
The equations x = a cosθ , y = a sinθ are called parametric equations of the circle x 2 + y 2 = a 2 and θ is called a parameter.
The point (a cos θ, a sinθ) is also referred to as point θ
The parametric coordinates of any point on the circle
(x – h) 2 + (y – k) 2 =a 2 are given by (h + a cosθ, k + a sinθ) with 0 ≤ θ < 2π
Example : If a straight line through C(-√8, √8), making an angle of 135° with the x-axis, cuts the circle x = 5 cosθ , y = 5 sinθ , in points A and B, find the length of the segment AB.
Solution: The given circle is x 2 + y 2 = 25 . . .(1)
Equation of line through C is (x + √8)/cos135° = (y − √8)/sin135° = r
Any point on this line is ( − √8 − r/√2 , √8 + r/√2)
If the point lies on the circle x 2 + y 2 = 25, the
Example : (i) Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).
(ii) Find the equation of a circle passing through (1, 1), (2, − 1) and (3, 2).
(i) The radius of the circle is
⇒ r = 5
Hence the equation of the circle is (x − 1) 2 + (y − 2) 2 = 25
Substituting for the three points, we get
Solving the above three equations, we obtain:
Hence the equation of the circle is
Example : Find the equation of the circle whose centre is (3, 4) and which touches the line 5x + 12y = 1.
Solution: Let r be the radius of the circle. Then
r = distance of the centre i.e. point (3 , 4) from the line 5x + 12y = 1
Hence the equation of the required circle is (x − 3) 2 + (y − 4) 2 = (62/13) 2
=> x 2 + y 2 − 6x − 8y + 381/169 = 0
Example : Find the equation of the circle whose diameter is the line joining the points ( −4, 3) and (12, −1). Find also the intercept made by it on the y-axis.
Solution: The equation of the required circle is
Intercept made on the y-axis = 2 √(f 2 − c)
Example : A circle has radius equal to 3 units and its centre lies on the line y = x − 1. Find the equation of the circle if it passes through (7 , 3)
Solution: Let the centre of the circle be (α , β).
It lies on the line y = x − 1
Hence the centre is (α , α − 1)
=> The equation of the circle is
Hence the required equations are x 2 + y 2 − 8x − 6y + 16 = 0 and x 2 + y 2 − 14x − 12y + 76 = 0.
## 4.7: Parametric Equations
State which of the following equations define lines and which define planes. Explain how you made your decision.
a) vec = (1, 2, 3) + s(1, 1, 0) + t(3, 4, -6), s, t in mathbb
b) vec = (-2, 3, 0) + m(3, 4, 7), m in mathbb
c) x = -3 - t, y = 5, z= 4 + t, t in mathbb
d) vec = m(4, -1, 2) + t(4, -1, 5), m, t in mathbb
A plane has a vector equation vec = (2, 1, 3) + s(displaystyle<3>>, -2, displaystyle<4>>) + t(6, -12, 30), s, t in mathbb.
a) Express the first direction vector with only integers.
b) Reduce the second direction vector.
c) Write a new equation for the plane using the calculations from part a. and b.
A plane has x = 2m, y = -3m + 5n, z = - 1 - 3m - 2n, m, n in mathbb as its parametric equations.
a) By inspection, identify the coordinates of a point that is on this plane.
b) What are the direction vectors for this plane?
c) What point corresponds to the parameter values of m = -1 and n = -4 ?
d) What are the parametric values corresponding to the point A(0, 15, -7) ?
e) Using your answer for part d., explain why the point B(0, 15, -8) cannot be on this plane.
A plane passes through the points P(-2, ,3, 1) , Q(-2, 3, 2) , and R(1, 0 , 1) .
a) Using vec and vec as direction vectors, write a vector equation for this plane.
b) Using vec and one other direction vector, write a second vector equation for this plane.
Explain why the equation vec = (-1, 0, -1) + s(2, 3, -4) + t(4, 6, -8), s, t, in mathbb , does not represent the equation of a plane. What does this equation represent?
Determine vector equations and the corresponding parametric equations of the plane.
Determine vector equations and the corresponding parametric equations of the plane.
Determine vector equations and the corresponding parametric equations of the plane.
a) Determine parameters corresponding to the point P(5, 3, 2) , where P is a point on the plane with equation
pi: vec = (2, 0 , 1) + s(4, 2, -1) + t(-1, 1, 2), s, t, in mathbb
Show that A does not lie on pi . displaystyleeginpi: vec=(2,0,1)+s(4,2,-1)+t(-1,1,2), s, t in mathbfA(0,5,-4) end
A plane has vec = (-3, 5, 6) + s(-1, 1, 2) + v(2, 1, -3), s, v in mathbb as its equation.
a) Give the equations of two intersecting lines that lie on this plane.
b) What point do these two lines have in common?
Determine the coordinates of the point where the plane with equation vec = (4, 1, 6) + s(11, -1, 3) + t(-7, 2, -2), s, t in mathbb , crosses the z-axis.
Determine the equation of the plane that contains the point P(-1, 2, 1) and the line vec= (2, 1, 3) + s(4, 1, 5), sin mathbb.
Determine the equation of the plane that contains the point A(-2, 2, 3) and the line vec = m(2, -1, 7), m inmathbb.
a) Determine two pairs of direction vectors that can be used to represent the xy-plane in mathbb^3
b) Write a vector and parametric equations for the xy-plane in mathbb^3
Show that the following equations represent the same plane:
a) vec = u(-2, 2, 4) + v(-4, 7, 1), u, v in mathbb and
b) vec = s(-1, 5, -3) + t(-1, -5, 7), s, t in mathbb
(Hint: Express each direction vector in the first equation as a linear combination of the direction vectors in the second equation.)
## Contents
### Analogue parametric design Edit
One of the earliest examples of parametric design was the upside down model of churches by Antonio Gaudi. In his design for the Church of Colònia Güell he created a model of strings weighted down with birdshot to create complex vaulted ceilings and arches. By adjusting the position of the weights or the length of the strings he could alter the shape of each arch and also see how this change influenced the arches connected to it. He placed a mirror on the bottom of the model to see how it should look upside-down.
#### Features of Gaudi's method Edit
Gaudi's analogue method includes the main features of a computational of a parametric model (input parameters, equation, output):
• The string length, birdshot weight and anchor point location all form independent input parameters
• The vertex locations of the points on the strings being the outcomes of the model
• The outcomes are derived by explicit functions, in this case gravity or Newtons law of motion.
By modifying individual parameters of these models Gaudi could generate different versions of his model while being certain the resulting structure would stand in pure compression. Instead of having to manually calculate the results of parametric equations he could automatically derive the shape of the catenary curves through the force of gravity acting on the strings. [5]
#### Luigi Moretti's Parametric Architecture Edit
We owe to Luigi Moretti the first definition of parametric architecture, the result of studies on the history of architecture, art, and the application of operation research to architectural design. [6] His researches on parametric architecture methods were developed within the Institute for Operations Research and Applied Mathematics Urbanism. On the occasion of an exhibition at the Milan Triennale, he will present various models of stadiums designed according to the principles of parametric architecture.
#### Principles of Luigi Moretti's parametric architecture Edit
• Rejection of empirical decisions.
• Assessment of traditional phenomena as objective facts based on the interdependence of expressive, social and technical values.
• Exact and complete definition of architectural themes.
• Objective observation of all the conditioning elements (parameters) related to the architectural theme and identification of their quantitative values.
• Definition of the relationships between the values of the parameters.
• Indispensability of different skills and scientific methodologies according to the criteria of operational research to define conditioning elements and their quantities.
• Affirmation of the Architect's freedom in decision and expression, only if it does not affect the characteristics determined by the analytical investigations.
• Research of architectural forms towards a maximum, therefore definitive, exactness of relationships in their general "structure".
Nature has often served as inspiration for architects and designers. [ citation needed ] Computer technology has given designers and architects the tools to analyse and simulate the complexity observed in nature and apply it to structural building shapes and urban organizational patterns. In the 1980s architects and designers started using computers running software developed for the aerospace and moving picture industries to "animate form". [7]
One of the first architects and theorists that used computers to generate architecture was Greg Lynn. His blob and fold architecture is some of the early examples of computer generated architecture. Shenzhen Bao'an International Airport's new Terminal 3, finished in 2013, designed by Italian architect Massimiliano Fuksas, with parametric design support by the engineering firm Knippers Helbig, is an example for the use of parametric design and production technologies in a large scale building.
Parametric urbanism is concerned with the study and prediction of settlement patterns. Architect Frei Otto distinguishes occupying and connecting as the two fundamental processes that are involved with all urbanisation. [8] Studies look at producing solutions that reduce overall path length in systems while maintaining low average detour factor or facade differentiation [ clarification needed ] .
### Power Surfacing Edit
Power Surfacing is a SolidWorks application for industrial design / freeform organic surface / solids modeling. Tightly integrated with SolidWorks, it works with all SolidWorks commands. Reverse Engineer scanned meshes with Power Surfacing RE.
### Catia Edit
CATIA (Computer Aided three-dimensional Interactive Application) was used by architect Frank Gehry to design some of his award-winning curvilinear buildings such as the Guggenheim Museum Bilbao. [9] Gehry Technologies, the technology arm of his firm, have since created Digital Project, their own parametric design software based on their experience with CATIA.
### Autodesk 3DS Max Edit
Autodesk 3ds Max is a parametric 3D modeling software which provides modeling, animation, simulation, and rendering functions for games, film, and motion graphics. 3ds Max uses the concept of modifiers and wired parameters to control its geometry and gives the user the ability to script its functionality. Max Creation Graph is a visual programming node-based tool creation environment in 3ds Max 2016 that is similar to Grasshopper and Dynamo.
### Autodesk Maya Edit
Autodesk Maya is a 3D computer graphics software originally developed by Alias Systems Corporation (formerly Alias|Wavefront) and currently owned and developed by Autodesk, Inc. It is used to create interactive 3D applications, including video games, animated film, TV series, or visual effects. Maya exposes a node graph architecture. Scene elements are node-based, each node having its own attributes and customization. As a result, the visual representation of a scene is based on a network of interconnecting nodes, depending on each other's information. Maya is equipped with a cross-platform scripting language, called Maya Embedded Language. MEL is provided for scripting and a means to customize the core functionality of the software, since many of the tools and commands used are written in it. MEL or Python can be used to engineer modifications, plug-ins or be injected into runtime. User interaction is recorded in MEL, allowing novice users to implement subroutines.
### Grasshopper 3D Edit
Grasshopper 3d (originally Explicit History) is a plug-in for Rhinoceros 3D that presents the users with a visual programming language interface to create and edit geometry. [10]
Components or nodes are dragged onto a canvas in order to build a grasshopper definition. Grasshopper is based on graphs (see Graph (discrete mathematics)) that map the flow of relations from parameters through user-defined functions (nodes), resulting in the generation of geometry. Changing parameters or geometry causes the changes to propagate throughout all functions, and the geometry to be redrawn. [5]
### Autodesk Revit Edit
Autodesk Revit is building information modeling (BIM) software used by architects and other building professionals. Revit was developed in response to the need for software that could create three-dimensional parametric models that include both geometry and non-geometric design and construction information. Every change made to an element in Revit is automatically propagated through the model to keep all components, views and annotations consistent. This eases collaboration between teams and ensures that all information (floor areas, schedules, etc.) are updated dynamically when changes in the model are made.
### Autodesk Dynamo Edit
Dynamo is an open source graphical programming environment for design. Dynamo extends building information modeling with the data and logic environment of a graphical algorithm editor.
### GenerativeComponents Edit
GenerativeComponents, parametric CAD software developed by Bentley Systems, [11] was first introduced in 2003, became increasingly used in practice (especially by the London architectural community) by early 2005, and was commercially released in November 2007. GenerativeComponents has a strong traditional base of users in academia and at technologically advanced design firms. [ citation needed ] GenerativeComponents is often referred to by the nickname of 'GC'. GC epitomizes the quest to bring parametric modeling capabilities of 3D solid modeling into architectural design, seeking to provide greater fluidity and fluency than mechanical 3D solid modeling. [ citation needed ]
Users can interact with the software by either dynamically modeling and directly manipulating geometry, or by applying rules and capturing relationships among model elements, or by defining complex forms and systems through concisely expressed algorithms. The software supports many industry standard file input and outputs including DGN by Bentley Systems, DWG by Autodesk, STL (Stereo Lithography), Rhino, and others. The software can also integrate with Building Information Modeling systems.
The software has a published API and uses a simple scripting language, both allowing the integration with many different software tools, and the creation of custom programs by users.
This software is primarily used by architects and engineers in the design of buildings, but has also been used to model natural and biological structures and mathematical systems.
Generative Components runs exclusively on Microsoft Windows operating systems.
### VIKTOR Edit
VIKTOR is an application development platform that enables engineers and other domain experts to rapidly build their own online applications using Python. It is used to create parametric design models and integrates with many software packages. [12] It enables users to make intuitive user interfaces (GUI), which include different form of visualizing results like 3D models, drawings, map or satellite views, and interactive graphs. This makes it possible to make the applications available to persons without programming affinity.
Applications made with VIKTOR are online, meaning data is update automatically and everyone works with the same information and the latest models. It includes a user management system, allowing to give different rights to users.
### Marionette Edit
Marionette is an open source [ citation needed ] graphical scripting tool (or visual programming environment) for the architecture, engineering, construction, landscape, and entertainment design industries that is built into the Mac and Windows versions of Vectorworks software. The tool was first made available in the Vectorworks 2016 line of software products. Marionette enables designers to create custom application algorithms that build interactive parametric objects and streamline complex workflows, as well as build automated 2D drawing, 3D modeling, and BIM workflows within Vectorworks software.
Built in the Python programming language, everything in Marionette consists of nodes which are linked together in a flowchart arrangement. Each node contains a Python script with predefined inputs and outputs that can be accessed and modified with a built-in editor. Nodes are placed directly into the Vectorworks document and then connected to create complex algorithms. Since Marionette is fully integrated into Vectorworks software, it can also be used to create entirely self-contained parametric objects that can be inserted into new and existing designs.
### Modelur Edit
Modelur is a parametric urban design software plug-in for Trimble SketchUp, developed by Agilicity d.o.o. (LLC).. Its primary goal is to help the users create conceptual urban massing. In contrast to common CAD applications, where the user designs buildings with usual dimensions such as width, depth and height, Modelur offers design of built environment through key urban parameters such as number of storeys and gross floor area of a building.
Modelur calculates key urban control parameters on the fly (e.g. floor area ratio or required number of parking lots), delivering urban design information while the development is still evolving. This way it helps taking well-informed decision during the earliest stages, when design decisions have the highest impact.
### Archimatix Edit
Archimatix is a node-based parametric modeler extension for Unity 3D. It enables visual modeling of 3D models within the Unity 3D editor.
## Transform a parametric plane form to the normal form
This transformation is nearly identical to the transformation from the parametric form to the cartesian form. We are given a plane in the parametric form and want to transform it to the normal form . We only need to calculate the normal vector of the plane by using the cross product and then we have all information for the normal form:
## Subsection 2.4.1 Homogeneous Systems ¶ permalink
A homogeneous system is just a system of linear equations where all constants on the right side of the equals sign are zero.
A homogeneous system always has the solution
This is called the trivial solution. Any nonzero solution is called nontrivial.
##### Observation
has a nontrivial solution
##### Observation
When we row reduce the augmented matrix for a homogeneous system of linear equations, the last column will be zero throughout the row reduction process. We saw this in the last example:
So it is not really necessary to write augmented matrices when solving homogeneous systems.
When the homogeneous equation
does have nontrivial solutions, it turns out that the solution set can be conveniently expressed as a span.
##### Parametric Vector Form (homogeneous case)
Consider the following matrix in reduced row echelon form:
corresponds to the system of equations
We can write the parametric form as follows:
We wrote the redundant equations
in order to turn the above system into a vector equation:
This vector equation is called the parametric vector form of the solution set. Since
are allowed to be anything, this says that the solution set is the set of all linear combinations of
In other words, the solution set is
Here is the general procedure.
##### Recipe: Parametric vector form (homogeneous case)
matrix. Suppose that the free variables in the homogeneous equation
Find the reduced row echelon form of
Put equations for all of the
will then be expressed in the form
This is called the parametric vector form of the solution.
In this case, the solution set can be written as
We emphasize the following fact in particular.
The set of solutions to a homogeneous equation
##### Example (The solution set is a line)
Since there were two variables in the above example, the solution set is a subset of
Since one of the variables was free, the solution set is a line:
In order to actually find a nontrivial solution to
in the above example, it suffices to substitute any nonzero value for the free variable
gives the nontrivial solution
##### Example (The solution set is a plane)
Since there were three variables in the above example, the solution set is a subset of
Since two of the variables were free, the solution set is a plane.
There is a natural question to ask here: is it possible to write the solution to a homogeneous matrix equation using fewer vectors than the one given in the above recipe? We will see in example in Section 2.5 that the answer is no: the vectors from the recipe are always linearly independent, which means that there is no way to write the solution with fewer vectors.
Another natural question is: are the solution sets for inhomogeneuous equations also spans? As we will see shortly, they are never spans, but they are closely related to spans.
There is a natural relationship between the number of free variables and the “size” of the solution set, as follows.
##### Dimension of the solution set
The above examples show us the following pattern: when there is one free variable in a consistent matrix equation, the solution set is a line, and when there are two free variables, the solution set is a plane, etc. The number of free variables is called the dimension of the solution set.
We will develop a rigorous definition of dimension in Section 2.7, but for now the dimension will simply mean the number of free variables. Compare with this important note in Section 2.5.
Intuitively, the dimension of a solution set is the number of parameters you need to describe a point in the solution set. For a line only one parameter is needed, and for a plane two parameters are needed. This is similar to how the location of a building on Peachtree Street—which is like a line—is determined by one number and how a street corner in Manhattan—which is like a plane—is specified by two numbers.
## APEX Calculus
The previous section defined curves based on parametric equations. In this section we'll employ the techniques of calculus to study these curves.
We are still interested in lines tangent to points on a curve. They describe how the (y)-values are changing with respect to the (x)-values, they are useful in making approximations, and they indicate instantaneous direction of travel.
The slope of the tangent line is still (frac ext<,>) and the Chain Rule allows us to calculate this in the context of parametric equations. If (x=f(t)) and (y=g(t) ext<,>) the Chain Rule states that
Solving for (frac ext<,>) we get
provided that (fp(t) eq 0 ext<.>) This is important so we label it a Key Idea.
###### Key Idea 9.3.2 . Finding (frac) with Parametric Equations.
Let (x=f(t)) and (y=g(t) ext<,>) where (f) and (g) are differentiable on some open interval (I) and (fp(t) eq 0) on (I ext<.>) Then
We use this to define the tangent line.
###### Definition 9.3.3 . Tangent and Normal Lines.
Let a curve (C) be parametrized by (x=f(t)) and (y=g(t) ext<,>) where (f) and (g) are differentiable functions on some interval (I) containing (t=t_0 ext<.>) The to (C) at (t=t_0) is the line through (ig(f(t_0),g(t_0)ig)) with slope (m=g'(t_0)/fp(t_0) ext<,>) provided (fp(t_0) eq 0 ext<.>)
The to (C) at (t=t_0) is the line through (ig(f(t_0),g(t_0)ig)) with slope (m=-fp(t_0)/g'(t_0) ext<,>) provided (g'(t_0) eq 0 ext<.>)
The definition leaves two special cases to consider. When the tangent line is horizontal, the normal line is undefined by the above definition as (g'(t_0)=0 ext<.>) Likewise, when the normal line is horizontal, the tangent line is undefined. It seems reasonable that these lines be defined (one can draw a line tangent to the “right side” of a circle, for instance), so we add the following to the above definition.
If the tangent line at (t=t_0) has a slope of 0, the normal line to (C) at (t=t_0) is the line (x=f(t_0) ext<.>)
If the normal line at (t=t_0) has a slope of 0, the tangent line to (C) at (t=t_0) is the line (x=f(t_0) ext<.>)
###### Example 9.3.4 . Tangent and Normal Lines to Curves.
Let (x=5t^2-6t+4) and (y=t^2+6t-1 ext<,>) and let (C) be the curve defined by these equations.
Find the equations of the tangent and normal lines to (C) at (t=3 ext<.>)
Find where (C) has vertical and horizontal tangent lines.
We start by computing (fp(t) = 10t-6) and (g'(t) =2t+6 ext<.>) Thus
Make note of something that might seem unusual: (frac) is a function of (t ext<,>) not (x ext<.>) Just as points on the curve are found in terms of (t ext<,>) so are the slopes of the tangent lines. The point on (C) at (t=3) is ((31,26) ext<.>) The slope of the tangent line is (m=1/2) and the slope of the normal line is (m=-2 ext<.>) Thus,
the equation of the tangent line is (ds y=frac12(x-31)+26 ext<,>) and
the equation of the normal line is (ds y=-2(x-31)+26 ext<.>)
To find where (C) has a horizontal tangent line, we set (frac=0) and solve for (t ext<.>) In this case, this amounts to setting (g'(t)=0) and solving for (t) (and making sure that (fp(t) eq 0)).
The point on (C) corresponding to (t=-3) is ((67,-10) ext<>) the tangent line at that point is horizontal (hence with equation (y=-10)). To find where (C) has a vertical tangent line, we find where it has a horizontal normal line, and set (-frac=0 ext<.>) This amounts to setting (fp(t)=0) and solving for (t) (and making sure that (g'(t) eq 0)).
The point on (C) corresponding to (t=0.6) is ((2.2,2.96) ext<.>) The tangent line at that point is (x=2.2 ext<.>) The points where the tangent lines are vertical and horizontal are indicated on the graph in Figure 9.3.5.
###### Example 9.3.6 . Tangent and Normal Lines to a Circle.
Find where the unit circle, defined by (x=cos(t)) and (y=sin(t)) on ([0,2pi] ext<,>) has vertical and horizontal tangent lines.
Find the equation of the normal line at (t=t_0 ext<.>)
We compute the derivative following Key Idea 9.3.2:
The derivative is (0) when (cos(t) = 0 ext<>) that is, when (t=pi/2,, 3pi/2 ext<.>) These are the points ((0,1)) and ((0,-1)) on the circle. The normal line is horizontal (and hence, the tangent line is vertical) when (sin(t) =0 ext<>) that is, when (t= 0,,pi,,2pi ext<,>) corresponding to the points ((-1,0)) and ((0,1)) on the circle. These results should make intuitive sense.
The slope of the normal line at (t=t_0) is (ds m=frac = an(t_0) ext<.>) This normal line goes through the point ((cos(t_0) ,sin(t_0) ) ext<,>) giving the line
as long as (cos(t_0) eq 0 ext<.>) It is an important fact to recognize that the normal lines to a circle pass through its center, as illustrated in Figure 9.3.7. Stated in another way, any line that passes through the center of a circle intersects the circle at right angles.
###### Example 9.3.8 . Tangent lines when (frac) is not defined.
Find the equation of the tangent line to the astroid (x=cos^3(t) ext<,>) (y=sin^3(t)) at (t=0 ext<,>) shown in Figure 9.3.9.
We start by finding (x'(t)) and (y'(t) ext<:>)
Note that both of these are 0 at (t=0 ext<>) the curve is not smooth at (t=0) forming a cusp on the graph. Evaluating (frac) at this point returns the indeterminate form of “0/0”.
We can, however, examine the slopes of tangent lines near (t=0 ext<,>) and take the limit as (t o 0 ext<.>)
We have accomplished something significant. When the derivative (frac) returns an indeterminate form at (t=t_0 ext<,>) we can define its value by setting it to be (limlimits_)(frac ext<,>) if that limit exists. This allows us to find slopes of tangent lines at cusps, which can be very beneficial.
We found the slope of the tangent line at (t=0) to be 0 therefore the tangent line is (y=0 ext<,>) the (x)-axis.
### Subsection 9.3.1 Concavity
We continue to analyze curves in the plane by considering their concavity that is, we are interested in (frac ext<,>) “the second derivative of (y) with respect to (x ext<.>)” To find this, we need to find the derivative of (frac) with respect to (x ext<>) that is,
but recall that (frac) is a function of (t ext<,>) not (x ext<,>) making this computation not straightforward.
To make the upcoming notation a bit simpler, let (h(t) = frac ext<.>) We want (frac[h(t)] ext<>) that is, we want (frac ext<.>) We again appeal to the Chain Rule. Note:
In words, to find (frac ext<,>) we first take the derivative of (frac) with respect to (t), then divide by (x'(t) ext<.>) We restate this as a Key Idea.
###### Key Idea 9.3.10 . Finding (frac) with Parametric Equations.
Let (x=f(t)) and (y=g(t)) be twice differentiable functions on an open interval (I ext<,>) where (fp(t) eq 0) on (I ext<.>) Then
Examples will help us understand this Key Idea.
###### Example 9.3.11 . Concavity of Plane Curves.
Let (x=5t^2-6t+4) and (y=t^2+6t-1) as in Example 9.3.4. Determine the (t)-intervals on which the graph is concave up/down.
Concavity is determined by the second derivative of (y) with respect to (x ext<,>) (frac ext<,>) so we compute that here following Key Idea 9.3.10.
The graph of the parametric functions is concave up when (frac gt 0) and concave down when (frac lt 0 ext<.>) We determine the intervals when the second derivative is greater/less than 0 by first finding when it is 0 or undefined.
As the numerator of (ds -frac<9><(5t-3)^3>) is never 0, (frac eq 0) for all (t ext<.>) It is undefined when (5t-3=0 ext<>) that is, when (t= 3/5 ext<.>) Following the work established in Section 3.4, we look at values of (t) greater/less than (3/5) on a number line:
Reviewing Example 9.3.4, we see that when (t=3/5=0.6 ext<,>) the graph of the parametric equations has a vertical tangent line. This point is also a point of inflection for the graph, illustrated in Figure 9.3.12.
The video in Figure 9.3.13 shows how this information can be used to sketch the curve by hand.
###### Example 9.3.14 . Concavity of Plane Curves.
Find the points of inflection of the graph of the parametric equations (x=sqrt ext<,>) (y=sin(t) ext<,>) for (0leq tleq 16 ext<.>)
We need to compute (frac) and (frac ext<.>)
The points of inflection are found by setting (frac=0 ext<.>) This is not trivial, as equations that mix polynomials and trigonometric functions generally do not have “nice” solutions.
In Figure 9.3.15.(a) we see a plot of the second derivative. It shows that it has zeros at approximately (t=0.5,,3.5,,6.5,,9.5,,12.5) and (16 ext<.>) These approximations are not very good, made only by looking at the graph. Newton's Method provides more accurate approximations. Accurate to 2 decimal places, we have:
The corresponding points have been plotted on the graph of the parametric equations in Figure 9.3.15.(b). Note how most occur near the (x)-axis, but not exactly on the axis.
### Subsection 9.3.2 Arc Length
We continue our study of the features of the graphs of parametric equations by computing their arc length.
Recall in Section 7.4 we found the arc length of the graph of a function, from (x=a) to (x=b ext<,>) to be
We can use this equation and convert it to the parametric equation context. Letting (x=f(t)) and (y=g(t) ext<,>) we know that (frac = g'(t)/fp(t) ext<.>) It will also be useful to calculate the differential of (x ext<:>)
Starting with the arc length formula above, consider:
Note the new bounds (no longer “(x)” bounds, but “(t)” bounds). They are found by finding (t_1) and (t_2) such that (a= f(t_1)) and (b=f(t_2) ext<.>) This formula is important, so we restate it as a theorem.
###### Theorem 9.3.17 . Arc Length of Parametric Curves.
Let (x=f(t)) and (y=g(t)) be parametric equations with (fp) and (g') continuous on ([t_1,t_2] ext<,>) on which the graph traces itself only once. The arc length of the graph, from (t=t_1) to (t=t_2 ext<,>) is
Note: Theorem 9.3.17 makes use of differentiability on closed intervals, just as was done in Section 7.4.
As before, these integrals are often not easy to compute. We start with a simple example, then give another where we approximate the solution.
###### Example 9.3.18 . Arc Length of a Circle.
Find the arc length of the circle parametrized by (x=3cos(t) ext<,>) (y=3sin(t)) on ([0,3pi/2] ext<.>)
By direct application of Theorem 9.3.17, we have
Apply the Pythagorean Theorem.
egin amp = int_0^ <3pi/2>3 , dt amp = 3tBig|_0^ <3pi/2>= 9pi/2 ext <.>end
This should make sense we know from geometry that the circumference of a circle with radius 3 is (6pi ext<>) since we are finding the arc length of (3/4) of a circle, the arc length is (3/4cdot 6pi = 9pi/2 ext<.>)
###### Example 9.3.19 . Arc Length of a Parametric Curve.
The graph of the parametric equations (x=t(t^2-1) ext<,>) (y=t^2-1) crosses itself as shown in Figure 9.3.20, forming a “teardrop.” Find the arc length of the teardrop.
We can see by the parametrizations of (x) and (y) that when (t=pm 1 ext<,>) (x=0) and (y=0 ext<.>) This means we'll integrate from (t=-1) to (t=1 ext<.>) Applying Theorem 9.3.17, we have
Unfortunately, the integrand does not have an antiderivative expressible by elementary functions. We turn to numerical integration to approximate its value. Using 4 subintervals, Simpson's Rule approximates the value of the integral as (2.65051 ext<.>) Using a computer, more subintervals are easy to employ, and (n=20) gives a value of (2.71559 ext<.>) Increasing (n) shows that this value is stable and a good approximation of the actual value.
### Subsection 9.3.3 Surface Area of a Solid of Revolution
Related to the formula for finding arc length is the formula for finding surface area. We can adapt the formula found in Theorem 7.4.13 from Section 7.4 in a similar way as done to produce the formula for arc length done before.
###### Theorem 9.3.21 . Surface Area of a Solid of Revolution.
Consider the graph of the parametric equations (x=f(t)) and (y=g(t) ext<,>) where (fp) and (g') are continuous on an open interval (I) containing (t_1) and (t_2) on which the graph does not cross itself.
The surface area of the solid formed by revolving the graph about the (x)-axis is (where (g(t)geq 0) on ([t_1,t_2])):
The surface area of the solid formed by revolving the graph about the (y)-axis is (where (f(t)geq 0) on ([t_1,t_2])):
###### Example 9.3.22 . Surface Area of a Solid of Revolution.
Consider the teardrop shape formed by the parametric equations (x=t(t^2-1) ext<,>) (y=t^2-1) as seen in Example 9.3.19. Find the surface area if this shape is rotated about the (x)-axis, as shown in Figure 9.3.23.
The teardrop shape is formed between (t=-1) and (t=1 ext<.>) Using Theorem 9.3.21, we see we need for (g(t)geq 0) on ([-1,1] ext<,>) and this is not the case. To fix this, we simplify replace (g(t)) with (-g(t) ext<,>) which flips the whole graph about the (x)-axis (and does not change the surface area of the resulting solid). The surface area is:
Once again we arrive at an integral that we cannot compute in terms of elementary functions. Using Simpson's Rule with (n=20 ext<,>) we find the area to be (S=9.44 ext<.>) Using larger values of (n) shows this is accurate to 2 places after the decimal.
After defining a new way of creating curves in the plane, in this section we have applied calculus techniques to the parametric equation defining these curves to study their properties. In the next section, we define another way of forming curves in the plane. To do so, we create a new coordinate system, called polar coordinates, that identifies points in the plane in a manner different than from measuring distances from the (y)- and (x)- axes.
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