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# The Volume of Geometrical Figures
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Last updated date: 15th Jul 2024
Total views: 236.1k
Views today: 5.36k
## What is Volume?
The volume of an object or a closed surface is a mathematical quantity that shows how often three-dimensional space it occupies. The volume is measured in cubic units such as m3, cm3, and so on.
Volume is sometimes spoken to as capacity. The volume of a cylindrical jar, for example, is used to calculate how much water it can contain.
### Volume of Cuboid
A cuboid is a solid geometrical object with 6 faces.
Volume of the cuboid(V) = length × breadth × height
Volume of the cuboid (V)= l × b × h
### Volume of Cube
A cube is a solid geometrical object with 6 faces.
All the sides of the cube are equal in length.
Volume of the cube(V) = length × breadth × height
Volume of the cube(V) = s × s × s (where s is the side of the cube)
Volume of the cube(V) = s3
Volume of the cube (V)= s3
### Volume by Counting Unit Cubes
We know that Volume is defined as a space occupied by a three-dimensional figure.
Let’s understand the concept of counting unit cubes with the help of the following examples:
Example: Find the volume of the given figure. Take the volume of each small cube as 1cm3.
Solution
Step 1: We have to number the cubes.
A total of 6 cubes are present in the given figure.
So, cubes are numbered from 1 to 6.
Step 2: Calculate the number of layers in the given figure.
A total of 2 layers are present in the given figure.
Step3: Calculate the volume of each layer
Volume of layer= (Number of cubes in the layer × volume of small cube)
Volume of layer 1 = 3 × 1cm3 = 3 cm3
Volume of layer 2 = 3 × 1cm3 = 3 cm3
Step4 : Calculate the Total volume
Total volume= Volume of layer 1 + Volume of layer 2
So, Total Volume of figure = 3 cm3 + 3 cm3
Total Volume of figure = 6 cm3.
### Solved Questions
1. Find the volume of the given figure. Take the volume of each small cube as 1cm3.
Ans: Layer 1 contains 6 cubes,
So, Volume of layer 1 = 6 × 1cm3 = 6 cm3
Layer 2 contains 6 cubes,
So, Volume of layer 2 = 6 × 1cm3 = 6 cm3
Layer 3 contains 12 cubes, (6 in front + 6 in back)
So, Volume of layer 3 = 12 × 1cm3 = 12 cm3
So, Total Volume of figure = 6 cm3+6 cm3+12 cm3
Total Volume of figure = 24 cm3.
2. Find the volume of the cube having side 3 cm.
Ans: Volume of cube(V) = s3
Volume of cube(V) = (3)3
Volume of cube(V) = 3 cm × 3 cm × 3 cm
Volume of cube(V) = 27 cm3
3.Find the volume of the cuboid having l = 6 cm, b = 4 cm and h = 5 cm.
Ans: Volume of cuboid(V) = length × breadth × height
Volume of cuboid(V) = 6 cm × 4 cm × 5 cm
Volume of cuboid(V) = 120 cm3.
4. What is the volume of the pictures given below. Take the volume of each small cube as 1cm3.
Ans: For figure A:
There is a total of 5 cubes
So, the Volume of figure A = 5 × 1cm3 = 5 cm3
For figure B:
There are a total of 12 cubes
(6 cubes in layer 1+ 6 cubes in layer 2)
So, Volume of figure B = 12 × 1cm3 = 12 cm3
### Fun Facts:
• When you link two points with only a line segment, you get a one-dimensional object that can only be measured in length.
• Two-dimensional figures are flat and have two dimensions i.e length and width. The area of a two-dimensional figure can be calculated by using length and width.
• The objects we come across on a daily basis are solid, three-dimensional objects with the following dimensions: length, width, and depth. The volume for three-dimensional objects is used to estimate their size.
### Summary
In this article we have discussed the concept of volume. First, we discussed volume definition, the volume of cube and cuboid concept and formulas, fun facts, and finally solved the problems. We have learned how to find the volume of figures by counting the unit cubes.
### Learning by Doing
1. Jerry was flying a kite that was yellow in shade. Suddenly, a powerful wind blew, and his kite became tangled in a huge tree. Jerry-built a ladder made of cubical boxes. Now let us count the volume of the ladder to help Jerry in getting his yellow kite. Take the volume of each small cube as 1cm3.
2. Tim wants to pack his old books in the cubical box. Help Tim in finding the volume of the cubical box on the side 5 cm.
3. Miss Mary went shopping and bought too many clothes. Now, Miss Mary wants to place all her clothes in her cuboidal cupboard. Let's help Miss Mary in finding the volume of the cuboidal cupboard whose l = 4 cm, b = 3 cm, and h = 6 cm.
## FAQs on The Volume of Geometrical Figures
1. Define Area?
The region covered by two-dimensional shapes is referred to as the area. The area of various shapes is determined by their dimensions(length and width). It is calculated in square meters(m2, cm2, etc).
2. Find the volume of the cube having a side of 6 cm?
Volume of cube(V) = s3
Volume of cube(V) = (6)3
Volume of cube(V) = 6 cm × 6 cm × 6 cm
Volume of cube(V) = 216 cm3
3. Find the volume of the cuboid having l = 5 cm, b = 4 cm and h = 10 cm?
Volume of cuboid(V) = length × breadth × height
Volume of cuboid(V) = 5 cm × 4 cm × 10 cm
Volume of cuboid(V) = 200 cm3.
4. Count the number of cubes in the given shape?
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# 3 DIFFERENTIATION RULES DIFFERENTIATION RULES 3 4 The
• Slides: 57
3 DIFFERENTIATION RULES
DIFFERENTIATION RULES 3. 4 The Chain Rule In this section, we will learn about: Differentiating composite functions using the Chain Rule.
CHAIN RULE Suppose you are asked to differentiate the function § The differentiation formulas you learned in the previous sections of this chapter do not enable you to calculate F’(x).
CHAIN RULE Observe that F is a composite function. In fact, if we let and let u = g(x) = x 2 + 1, then we can write y = F(x) = f (g(x)). That is, F = f ◦ g.
CHAIN RULE We know how to differentiate both f and g. So, it would be useful to have a rule that shows us how to find the derivative of F = f ◦ g in terms of the derivatives of f and g.
CHAIN RULE It turns out that the derivative of the composite function f ◦ g is the product of the derivatives of f and g. This fact is one of the most important of the differentiation rules. It is called the Chain Rule.
CHAIN RULE It seems plausible if we interpret derivatives as rates of change. Regard: § du/dx as the rate of change of u with respect to x § dy/du as the rate of change of y with respect to u § dy/dx as the rate of change of y with respect to x
CHAIN RULE If u changes twice as fast as x and y changes three times as fast as u, it seems reasonable that y changes six times as fast as x. So, we expect that:
THE CHAIN RULE If g is differentiable at x and f is differentiable at g(x), the composite function F = f ◦ g defined by F(x) = f(g(x)) is differentiable at x and F’ is given by the product: F’(x) = f’(g(x)) • g’(x) § In Leibniz notation, if y = f(u) and u = g(x) are both differentiable functions, then:
COMMENTS ON THE PROOF Let ∆u be the change in corresponding to a change of ∆x in x, that is, ∆u = g(x + ∆x) - g(x) Then, the corresponding change in y is: ∆y = f(u + ∆u) - f(u)
COMMENTS ON THE PROOF It is tempting to write: Equation 1
COMMENTS ON THE PROOF The only flaw in this reasoning is that, in Equation 1, it might happen that ∆u = 0 (even when ∆x ≠ 0) and, of course, we can’t divide by 0.
COMMENTS ON THE PROOF Nonetheless, this reasoning does at least suggest that the Chain Rule is true. § A full proof of the Chain Rule is given at the end of the section.
CHAIN RULE Equations 2 and 3 The Chain Rule can be written either in the prime notation (f ◦ g)’(x) = f’(g(x)) • g’(x) or, if y = f(u) and u = g(x), in Leibniz notation:
CHAIN RULE Equation 3 is easy to remember because, if dy/du and du/dx were quotients, then we could cancel du. However, remember: § du has not been defined § du/dx should not be thought of as an actual quotient
CHAIN RULE E. g. 1—Solution 1 Find F’(x) if § One way of solving this is by using Equation 2. § At the beginning of this section, we expressed F as F(x) = (f ◦ g))(x) = f(g(x)) where and g(x) = x 2 + 1.
CHAIN RULE § Since we have E. g. 1—Solution 1
CHAIN RULE E. g. 1—Solution 2 We can also solve by using Equation 3. § If we let u = x 2 + 1 and then:
CHAIN RULE When using Equation 3, we should bear in mind that: § dy/dx refers to the derivative of y when y is considered as a function of x (called the derivative of y with respect to x) § dy/du refers to the derivative of y when considered as a function of u (the derivative of y with respect to u)
CHAIN RULE For instance, in Example 1, y can be considered as a function of x and also as a function of u § Note that: .
NOTE In using the Chain Rule, we work from the outside to the inside. § Equation 2 states that we differentiate the outer function f [at the inner function g(x)] and then we multiply by the derivative of the inner function.
CHAIN RULE Differentiate: a. y = sin(x 2) b. y = sin 2 x Example 2
CHAIN RULE Example 2 a If y = sin(x 2), the outer function is the sine function and the inner function is the squaring function. So, the Chain Rule gives:
CHAIN RULE Example 2 b Note that sin 2 x = (sin x)2. Here, the outer function is the squaring function and the inner function is the sine function. Therefore,
CHAIN RULE Example 2 b The answer can be left as 2 sin x cos x or written as sin 2 x (by a trigonometric identity known as the double-angle formula).
COMBINING THE CHAIN RULE In Example 2 a, we combined the Chain Rule with the rule for differentiating the sine function.
COMBINING THE CHAIN RULE In general, if y = sin u, where u is a differentiable function of x, then, by the Chain Rule, Thus,
COMBINING THE CHAIN RULE In a similar fashion, all the formulas for differentiating trigonometric functions can be combined with the Chain Rule.
COMBINING CHAIN RULE WITH POWER RULE Let’s make explicit the special case of the Chain Rule where the outer function is a power function. § If y = [g(x)]n, then we can write y = f(u) = un where u = g(x). § By using the Chain Rule and then the Power Rule, we get:
POWER RULE WITH CHAIN RULE Rule 4 If n is any real number and u = g(x) is differentiable, then Alternatively,
POWER RULE WITH CHAIN RULE Notice that the derivative in Example 1 could be calculated by taking n = ½ in Rule 4.
POWER RULE WITH CHAIN RULE Example 3 Differentiate y = (x 3 – 1)100 § Taking u = g(x) = x 3 – 1 and n = 100 in the rule, we have:
POWER RULE WITH CHAIN RULE Example 4 Find f’ (x) if § First, rewrite f as f(x) = (x 2 + x + 1)-1/3 § Thus,
POWER RULE WITH CHAIN RULE Example 5 Find the derivative of § Combining the Power Rule, Chain Rule, and Quotient Rule, we get:
CHAIN RULE Example 6 Differentiate: y = (2 x + 1)5 (x 3 – x + 1)4 § In this example, we must use the Product Rule before using the Chain Rule.
CHAIN RULE Thus, Example 6
CHAIN RULE Example 6 Noticing that each term has the common factor 2(2 x + 1)4(x 3 – x + 1)3, we could factor it out and write the answer as:
CHAIN RULE Example 7 Differentiate y = esin x § Here, the inner function is g(x) = sin x and the outer function is the exponential function f(x) = ex. § So, by the Chain Rule:
CHAIN RULE We can use the Chain Rule to differentiate an exponential function with any base a > 0. § Recall from Section 1. 6 that a = eln a. § So, ax = (eln a)x = e(ln a)x.
CHAIN RULE Thus, the Chain Rule gives because ln a is a constant.
CHAIN RULE Formula 5 Therefore, we have the formula:
CHAIN RULE Formula 6 In particular, if a = 2, we get:
CHAIN RULE In Section 3. 1, we gave the estimate § This is consistent with the exact Formula 6 because ln 2 ≈ 0. 693147
CHAIN RULE The reason for the name ‘Chain Rule’ becomes clear when we make a longer chain by adding another link.
CHAIN RULE Suppose that y = f(u), u = g(x), and x = h(t), where f, g, and h are differentiable functions, then, to compute the derivative of y with respect to t, we use the Chain Rule twice:
CHAIN RULE Example 8 If § Notice that we used the Chain Rule twice.
CHAIN RULE Example 9 Differentiate y = esec 3θ § The outer function is the exponential function, the middle function is the secant function and the inner function is the tripling function. § Thus, we have:
HOW TO PROVE THE CHAIN RULE Recall that if y = f(x) and x changes from a to a + ∆x, we defined the increment of y as: ∆y = f(a + ∆x) – f(a)
HOW TO PROVE THE CHAIN RULE According to the definition of a derivative, we have:
HOW TO PROVE THE CHAIN RULE So, if we denote by ε the difference between the difference quotient and the derivative, we obtain:
HOW TO PROVE THE CHAIN RULE However, § If we define ε to be 0 when ∆x = 0, then ε becomes a continuous function of ∆x.
HOW TO PROVE THE CHAIN RULE Equation 7 Thus, for a differentiable function f, we can write: § ε is a continuous function of ∆x. § This property of differentiable functions is what enables us to prove the Chain Rule.
PROOF OF THE CHAIN RULE Equation 8 Suppose u = g(x) is differentiable at a and y = f(u) at b = g(a). If ∆x is an increment in x and ∆u and ∆y are the corresponding increments in u and y, then we can use Equation 7 to write ∆u = g’(a) ∆x + ε 1 ∆x = [g’(a) + ε 1] ∆x where ε 1 → 0 as ∆x → 0
PROOF OF THE CHAIN RULE Equation 9 Similarly, ∆y = f’(b) ∆u + ε 2 ∆u = [f’(b) + ε 2] ∆u where ε 2 → 0 as ∆u → 0.
PROOF OF THE CHAIN RULE If we now substitute the expression for ∆u from Equation 8 into Equation 9, we get: So,
PROOF OF THE CHAIN RULE As ∆x→ 0, Equation 8 shows that ∆u→ 0. So, both ε 1 → 0 and ε 2 → 0 as ∆x→ 0.
PROOF OF THE CHAIN RULE Therefore, This proves the Chain Rule.
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# What Is 8 Thousands Divided By 10?
## How do you divide a number by 10?
Here’s the rule for dividing by 10: move the decimal point one place to the left.
Place value is the value of a digit based on its location in the number.
Beginning with a decimal point and moving left, we have the ones, tens, hundreds, thousands, ten thousands, hundred thousands, and millions.
We could go on forever!.
## How is 9 Ten Thousands written?
Answer: The standard form of nine ten thousand is 90,000.
## How do you divide a number by 100?
To divide a whole or a decimal number by 100, move the decimal point two places to the left. Note that although a whole number does not have a decimal point, we can always add it at the end of the number. (For example, 35 and 35. are the same numbers.)
## What does 10 times the value mean?
So in the number 44,000 the digit in the ten thousands place is 10 times the value of the digit in the thousands place because you have moved one place to the left.
## Why do you add a zero when multiplying by 10?
In the second multiplication, one has to add a zero in the ones place. This is because we’re actually multiplying by a multiple of ten (such as 70 or 40).
## How many hundreds are in a thousand?
NumberNameHow Many1,000one thousandten hundreds10,000ten thousandten thousands100,000one hundred thousandone hundred thousands1,000,000one millionone thousand thousands1 more row
## What is 4 thousands divided by 10?
Divide Multiple Copies Of One Unit By 10 Change each thousand for 10 smaller units. 4 thousands can be changed to be 40 hundreds because 4 thousands and 40 hundreds are equal. 4 thousands ÷ 10 is 4 hundreds because 4 thousands unbundled becomes 40 hundreds. 40 hundreds divided by 10 is 4 hundreds.
## What is 7 thousands 4 tens divided by 10?
Step-by-step explanation: 7000 + 40 (7 thousands + 4 Tens) Divide that by 10.
## How do you divide a whole number by a power of 10?
To divide by a power of 10, simply move the decimal to the left the same number of places as the exponent or as the number of zeros. Example: (Note: The decimal of a whole number is always to the right of the one’s place.) Another Way to Indicate Division by a power of 10 is to multiply by 10 to a negative exponent.
## How many zeros does 10 thousand have?
410,000/Number of zeros
## In which number does the 5 represent a value 10 times?
35,187The 5 value is at thousand. A value 10 times the value represented by the 5 in 35,187 is given by any number placed in ten thousand place.
## How is 26 thousands 13 Hundreds written?
13 hundred means 1300. So, 26 thousands 13 hundreds means 26000+1300=27300.
## What does 15 TENS mean?
If you have 15 tens this means that you are adding 10, 15 times or multiplying 10 by 15, which gives you 150. If you have 7 ones, you are adding 1, 7 times or multiplying 1 by 7. Once you add 150 and 7 you gain 157.
## How do you explain dividing by 100?
To divide by 100, move each digit two place value columns to the right. If the number ends in two ‘0’ digits in the tens and units columns, dividing by 100 has the same effect as removing these digits. To divide by 100, move all digits two place value columns to the right.
## What is a ten number?
What is number ten? In mathematics, the number 10 represents a quantity or value of 10. The whole number between 9 and 11 is 10. The number name of 10 is ten. Little Sera is showing 10 fingers.
## How many hundreds are there in 5000?
50 hundreds is another way to say 50 x 100, which is 5000.
## What is 7 ten times the value of?
According to question, the value of 7 ten times the value of 7 in the number 1273. So, 7 should be in the hundreds place. for example , it must be look like 1723.
## How many thousands are in a million?
1000 thousandsA million is 1000 thousands, a billion is 1000 millions, and a trillion is 1000 billions.
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# How do you solve x+y=8, y-x=8 by graphing?
May 4, 2018
Answer is #(0,8), intersection of graphs
#### Explanation:
First, we make $y$ the subject of the equation:
$x + y = 8$
$y = - x + 8$-----> by subtracting $x$ both sides
$y - x = 8$
$y = x + 8$----> by adding $x$ both sides
Now we have:
$y = - x + 8$ (red line graph)
$y = x + 8$ (blue line graph)
Create a graph to locate the intersection of the equations. The intersection of the system of equations is the solution.
by plotting the graph, see the attached, we get intersection at $\left(0 , 8\right)$, hence the solution is $\left(0 , 8\right)$
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# How to subtract rational numbers with different denominators ?
In this chapter we will learn to subtract rational numbers with different denominators.
In the end of chapter, we have also solved some problems for better conceptual understanding.
## Subtracting Rational numbers with different denominators
To subtract the numbers, we have to first make all the denominators same.
This can be done by using LCM concept.
Follow the below steps to subtract the rational numbers;
(a) Find LCM of denominators
(b) Multiply each rational number to make denominator equal to LCM
(c) Now simply subtract the numerator and retain the same denominator.
I hope you understood the above three steps. Let us see some solved examples for better clarity.
Example 01
Subtract \mathtt{\frac{15}{7} -\frac{2}{3}}
Solution
Observe that both rational numbers have different denominators.
To subtract, follow the below steps;
(a) Find LCM of denominator
LCM (7, 3) = 21
(b) Multiply each rational number to make denominator 21.
(i) Fraction 15 / 7
Multiply numerator & denominator by 3.
\mathtt{\Longrightarrow \ \frac{15}{7}}\\\ \\ \mathtt{\Longrightarrow \ \frac{15\times 3}{7\times 3}}\\\ \\ \mathtt{\Longrightarrow \ \frac{45}{21}}
(ii) Fraction 2/3
Multiply numerator and denominator by 7
\mathtt{\Longrightarrow \ \frac{2}{3}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2\times 7}{3\times 7}}\\\ \\ \mathtt{\Longrightarrow \ \frac{14}{21}}
We have got the fraction with same denominator. Now simply subtract the numerator and get the solution.
(c) Subtract the numerator
\mathtt{\Longrightarrow \frac{45}{21} -\frac{14}{21}}\\\ \\ \mathtt{\Longrightarrow \ \frac{45-14}{21}}\\\ \\ \mathtt{\Longrightarrow \ \frac{31}{21}}
Hence, 31/21 is the solution of given subtraction.
Example 02
Subtract \mathtt{\frac{9}{20} -\frac{1}{4}}
Solution
Follow the below steps;
(a) Find LCM of denominators.
LCM (20, 4) = 20
(b) Multiply each rational number to make denominator 20.
(i) Fraction 9 / 20
The denominator is already 20. No need to do anything.
(ii) Fraction 1 / 4
Multiply numerator and denominator by 5.
\mathtt{\Longrightarrow \ \frac{1}{4}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1\times 5}{4\times 5}}\\\ \\ \mathtt{\Longrightarrow \ \frac{5}{20}}
We have got fractions with same denominator. Now simply subtract the numerator to get the solution.
(c) Subtract the numerator
\mathtt{\Longrightarrow \frac{9}{20} -\frac{5}{20}}\\\ \\ \mathtt{\Longrightarrow \ \frac{9-5}{20}}\\\ \\ \mathtt{\Longrightarrow \ \frac{4}{20}}
Here we got 4 / 20 as solution.
The fraction can be further reduced by dividing numerator and denominator by 4.
\mathtt{\Longrightarrow \ \frac{4}{20}}\\\ \\ \mathtt{\Longrightarrow \ \frac{4\div 4}{20\div 4}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{5}}
Hence, 1/5 is the solution.
Example 03
Subtract \mathtt{\frac{1}{15} -\frac{2}{9}}
Solution
Follow the below steps;
(a) Find LCM of denominators.
LCM (15, 9 ) = 45
(b) Multiply each rational number to make denominator equal to 45.
(i) Fraction 1/ 15
Multiply numerator and denominator by 3
\mathtt{\Longrightarrow \ \frac{1}{15}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1\times 3}{15\times 3}}\\\ \\ \mathtt{\Longrightarrow \ \frac{3}{45}}
(ii) Fraction 2/9
Multiply numerator and denominator by 5.
\mathtt{\Longrightarrow \ \frac{2}{9}}\\\ \\ \mathtt{\Longrightarrow \ \frac{2\times 5}{9\times 5}}\\\ \\ \mathtt{\Longrightarrow \ \frac{10}{45}}
We have got the fractions with same denominator. Now simply subtract the numerator to get the solution.
(c) Subtract the numerator
\mathtt{\Longrightarrow \frac{3}{45} -\frac{10}{45}}\\\ \\ \mathtt{\Longrightarrow \ \frac{3-10}{45}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-7}{45}}
Hence, -7/45 is the solution of given expression.
Example 04
Subtract \mathtt{\frac{10}{23} -\frac{-6}{5}}
Solution
Do the following steps;
(a) Find LCM of denominators
LCM (23, 5) = 115
(b) Multiply the rational numbers to make denominator equals 115
(i) Fraction 10 / 23
Multiply numerator and denominator by 5.
\mathtt{\Longrightarrow \ \frac{10}{23}}\\\ \\ \mathtt{\Longrightarrow \ \frac{10\times 5}{23\times 5}}\\\ \\ \mathtt{\Longrightarrow \ \frac{50}{115}}
(ii) Fraction -6 / 5
Multiply numerator and denominator by 23
\mathtt{\Longrightarrow \ \frac{-6}{5}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-6\times 23}{5\times 23}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-138}{115}}
We have got both fractions with same denominator. Now simply subtract the numerator.
(c) Subtracting the numerators
\mathtt{\Longrightarrow \frac{50}{115} -\frac{-138}{115}}\\\ \\ \mathtt{\Longrightarrow \ \frac{50-( -138)}{115}}\\\ \\ \mathtt{\Longrightarrow \ \frac{50\ +\ 138}{115}}\\\ \\ \mathtt{\Longrightarrow \ \frac{188}{115}}
Hence 188/115 is the solution.
Example 05
Subtract \mathtt{\frac{-1}{6} -\frac{-1}{3}}
Solution
(a) Find LCM of denominators.
LCM (6, 3) = 6
(b) Multiply each rational number to make denominator equals to 6.
(i) Rational number -1/6
The denominator is already 6, so no need to do anything.
(ii) Rational number -1/3
Multiply numerator and denominator by 2.
\mathtt{\Longrightarrow \ \frac{-1}{3}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-1\times 2}{3\times 2}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-2}{6}}
(c) Now subtract the numerators.
\mathtt{\Longrightarrow \frac{-1}{6} -\frac{-2}{6}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-1-( -2)}{6}}\\\ \\ \mathtt{\Longrightarrow \ \frac{-1\ +\ 2}{6}}\\\ \\ \mathtt{\Longrightarrow \ \frac{1}{6}}
Hence, 1/6 is the solution of given subtraction.
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# How to Calculate the Probability of Combinations
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Lesson Transcript
Chad has taught Math for the last 9 years in Middle School. He has a M.S. in Instructional Technology and Elementary Education.
To calculate the probability of a combination, you will need to consider the number of favorable outcomes over the number of total outcomes. Combinations are used to calculate events where order does not matter. In this lesson, we will explore the connection between these two essential topics.
## Combinations
Note: The formulas in this lesson assume that we have no replacement, which means items cannot be repeated.
Combinations are a way to calculate the total outcomes of an event where order of the outcomes does not matter. To calculate combinations, we will use the formula nCr = n! / r! * (n - r)!, where n represents the total number of items, and r represents the number of items being chosen at a time.
To calculate a combination, you will need to calculate a factorial. A factorial is the product of all the positive integers equal to and less than your number. A factorial is written as the number followed by an exclamation point. For example, to write the factorial of 4, you would write 4!. To calculate the factorial of 4, you would multiply all of the positive integers equal to and less than 4. So, 4! = 4 * 3 * 2 * 1. By multiplying these numbers together, we can find that 4! = 24.
Let's look at another example of how we would write and solve the factorial of 9. The factorial of 9 would be written as 9!. To calculate 9!, we would multiply 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, and that equals 362,880.
## Combinations Formula
Looking at the equation to calculate combinations, you can see that factorials are used throughout the formula. Remember, the formula to calculate combinations is nCr = n! / r! * (n - r)!, where n represents the number of items, and r represents the number of items being chosen at a time. Let's look at an example of how to calculate a combination.
There are ten new movies out to rent this week on DVD. John wants to select three movies to watch this weekend. How many combinations of movies can he select?
In this problem, John is choosing three movies from the ten new releases. 10 would represent the n variable, and 3 would represent the r variable. So, our equation would look like 10C3 = 10! / 3! * (10 - 3)!.
The first step that needs to be done is to subtract 10 minus 3 on the bottom of this equation. 10 - 3 = 7, so our equation looks like 10! / 3! * 7!.
Next, we need to expand each of our factorials. 10! would equal 10 * 9 * 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1 on the top, and 3! * 7! would be 3 * 2 * 1 * 7 * 6 * 5 * 4 * 3 * 2 * 1. The easiest way to work this problem is to cancel out like terms. We can see that there is a 7, 6, 5, 4, 3, 2 and 1 on both the top and bottom of our equation. These terms can be cancelled out. We now see that our equation has 10 * 9 * 8 left on top and 3 * 2 * 1 left on bottom. From here, we can just multiply. 10 * 9 * 8 = 720, and 3 * 2 * 1 = 6. So, our equation is now 720 / 6.
To finish this problem, we will divide 720 by 6, and we get 120. John now knows that he could select 120 different combinations of new-release movies this week.
## Probability
To calculate the probability of an event occurring, we will use the formula: number of favorable outcomes / the number of total outcomes.
Let's look at an example of how to calculate the probability of an event occurring. At the checkout in the DVD store, John also purchased a bag of gumballs. In the bag of gumballs, there were five red, three green, four white and eight yellow gumballs. What is the probability that John drawing at random will select a yellow gumball?
John knows that if he adds all the gumballs together, there are 20 gumballs in the bag. So, the number of total outcomes is 20. John also knows that there are eight yellow gumballs, which would represent the number of favorable outcomes. So, the probability of selecting a yellow gumball at random from the bag is 8 out of 20.
All fractions, however, must be simplified. So, both 8 and 20 will divide by 4. So, 8/20 would reduce to 2/5. John knows that probability of him selecting a yellow gumball from the bag at random is 2/5.
## Probability of Combinations
To calculate the number of total outcomes and favorable outcomes, you might have to calculate a combination. Remember, a combination is a way to calculate events where order does not matter.
Let's look at an example. To enjoy his movies, John decides to order a pizza. Looking at the menu, John sees the Pizza King offers eight different topping (four meat and four vegetables). The toppings are: pepperoni, ham, bacon, sausage, peppers, mushrooms, onions and olives. John has a coupon for a 3-topping pizza. Choosing ingredients at random, what is the probability of John selecting a pizza with meat only?
John is looking for the probability of selecting a meat-only pizza. In order to do so, he will need to calculate the total number of favorable outcomes over the total outcomes possible. Let's first calculate the total number of outcomes. To calculate the total outcomes, we will use the formula for combinations because the order of the pizza toppings does not matter. The formula for combinations is nCr = n! / r! * (n - r)!, where n represents the number of items, and r represents the number of items being chosen at a time.
John is selecting three toppings from the eight offered by Pizza King. 8 would represent our n term, and 3 would represent our r term. So, our equation will look like 8C3 = 8! / 3! * (8 - 3)!.
To solve this equation, we need to subtract 8 - 3 = 5. So, our equation now looks like 8! / 3! * 5!. Next, we need to expand each of these factorials. 8! would equal 8 * 7 * 6 * 5 * 4 * 3 * 2 * 1, divided by 3! * 5!, which would equal 3 * 2 * 1 * 5 * 4 * 3 * 2 * 1.
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# What is 42 percent of 88 + Solution with Free Steps
42% of 88 is 36.96, so this equals 36.96. Simply multiply 88 by 0.42 to obtain this result.
In practise, a 42% of 88 estimation can be really useful. Consider purchasing a laptop battery from Amazon as an example. You wish to purchase a laptop battery from Amazon, but they cost \$65 each. An Askari bank is making a deal public. If you use their Mastercard to purchase laptop accessories from Amazon, you will receive a discount. We know that 36.96 is 42 percent of 88. It is possible to determine that the laptop battery is now available for 51.04 dollars, down from 88 dollars. Therefore, you could save 36.96 dollars by utilizing the online service of Askari Bank.
The primary purpose of this question is to calculate 42 percent of 88
What Is 42 percent of 88?
42% of 88 is 36.96, which can also be calculated by multiplying 42 by 88 and then dividing the result by 100.
Multiply 42 by 88 and then divide by 100 to calculate the value of 42% of 88. The mathematical instructions below will aid you in figuring out what 42 percent of 88 is.
## How To Calculate 42 percent of 88?
To determine what 42% of 88 is, use the following simple math steps:
### Step 1
42% of 88 can be expressed as follows:
42 percent of 88 = 42% x 88
### Step 2
Substituting 1/100 for the percent sign:
42 percent of 88 = ( 42 x 1/100 ) x 88
### Step 3
The result of reversing the order of (42 x 1/100) x 88 is:
42 percent of 88 = ( 42 x 88 ) / 100
### Step 4
Now, the answer to the preceding equation is:
42 percent of 88 = ( 3696 ) / 100
### Step 5
Dividing 3696 by 100:
42 percent of 88 = 36.96
However, 42% of 88 is equivalent to 36.96.
The following pie chart indicates that 42 percent of 88 equals 36.96.
Figure 1: Pie Chart of 42 percent of 88
The blue portion of the pie chart demonstrates that 42 percent of 88 is equivalent to the value of 36.96 while the pink area of the graph indicates 58% of 88 or an absolute value of 51.04
In mathematics, a percentage is a number or proportion that may be expressed as a fraction of 100. A percentage is calculated by dividing a number by the entire number and multiplying the result by 100. The percentage symbol is used to represent it.
All the Mathematical drawings/images are created using GeoGebra.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.4: Special Right Triangles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Learn and use the 45-45-90 triangle ratio.
• Learn and use the 30-60-90 triangle ratio.
## Review Queue
Find the value of the missing variables. Simplify all radicals.
1. Is 9, 12, and 15 a right triangle?
2. Is 3, 33\begin{align*}3 \sqrt{3}\end{align*}, and 6 a right triangle?
Know What? A baseball diamond is a square with sides that are 90 feet long. Each base is a corner of the square. What is the length between 1st\begin{align*}1^{st}\end{align*} and 3rd\begin{align*}3^{rd}\end{align*} base and between 2nd\begin{align*}2^{nd}\end{align*} base and home plate? (the red dotted lines in the diagram).
## Isosceles Right Triangles
There are two special right triangles. The first is an isosceles right triangle.
Isosceles Right Triangle: A right triangle with congruent legs and acute angles. This triangle is also called a 45-45-90 triangle (after the angle measures).
ABC\begin{align*}\triangle ABC\end{align*} is a right triangle with:
mAAB¯¯¯¯¯¯¯¯mB=90AC¯¯¯¯¯¯¯¯=mC=45\begin{align*}m \angle A &= 90^\circ\\ \overline {AB} & \cong \overline{AC}\\ m \angle B &= m \angle C = 45^\circ\end{align*}
Investigation 8-2: Properties of an Isosceles Right Triangle
Tools Needed: Pencil, paper, compass, ruler, protractor
1. Draw an isosceles right triangle with 2 inch legs and the 90\begin{align*}90^\circ\end{align*} angle between them.
2. Find the measure of the hypotenuse, using the Pythagorean Theorem. Simplify the radical.
22+228c=c2=c2=8=42=22\begin{align*}2^2 + 2^2 &= c^2\\ 8 &= c^2\\ c &= \sqrt{8} = \sqrt{4 \cdot 2} = 2 \sqrt{2}\end{align*}
What do you notice about the length of the legs and hypotenuse?
3. Now, let’s say the legs are of length x\begin{align*}x\end{align*} and the hypotenuse is h\begin{align*}h\end{align*}. Use the Pythagorean Theorem to find the hypotenuse. How is it similar to your answer in #2?
x2+x22x2x2=h2=h2=h\begin{align*}x^2 + x^2 &= h^2\\ 2x^2 &= h^2\\ x \sqrt{2} &= h\end{align*}
45-45-90 Theorem: If a right triangle is isosceles, then its sides are x:x:x2\begin{align*}x:x:x \sqrt{2}\end{align*}.
For any isosceles right triangle, the legs are x\begin{align*}x\end{align*} and the hypotenuse is always x2\begin{align*}x \sqrt{2}\end{align*}. Because the three angles are always 45,45,\begin{align*}45^\circ, 45^\circ,\end{align*} and 90\begin{align*}90^\circ\end{align*}, all isosceles right triangles are similar.
Example 1: Find the length of the missing sides.
a)
b)
Solution: Use the x:x:x2\begin{align*}x:x:x \sqrt{2}\end{align*} ratio.
a) TV=6\begin{align*}TV = 6\end{align*} because it is equal to ST\begin{align*}ST\end{align*}. So, SV=62=62\begin{align*}SV = 6 \cdot \sqrt{2} = 6 \sqrt{2}\end{align*}.
b) AB=92\begin{align*}AB = 9 \sqrt{2}\end{align*} because it is equal to AC\begin{align*}AC\end{align*}. So, BC=922=92=18\begin{align*}BC = 9 \sqrt{2} \cdot \sqrt{2} = 9 \cdot 2 = 18\end{align*}.
Example 2: Find the length of x\begin{align*}x\end{align*}.
a)
b)
Solution: Use the x:x:x2\begin{align*}x:x:x \sqrt{2}\end{align*} ratio.
a) 122\begin{align*}12 \sqrt{2}\end{align*} is the diagonal of the square. Remember that the diagonal of a square bisects each angle, so it splits the square into two 45-45-90 triangles. 122\begin{align*}12 \sqrt{2}\end{align*} would be the hypotenuse, or equal to x2\begin{align*}x \sqrt{2}\end{align*}.
12212=x2=x\begin{align*}12 \sqrt{2} &= x \sqrt{2}\\ 12 &= x\end{align*}
b) Here, we are given the hypotenuse. Solve for x\begin{align*}x\end{align*} in the ratio.
x2x=16=16222=1622=82\begin{align*}x \sqrt{2} &= 16\\ x &= \frac{16}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{16 \sqrt{2}}{2} = 8 \sqrt{2}\end{align*}
In part b, we rationalized the denominator which we learned in the first section.
## 30-60-90 Triangles
The second special right triangle is called a 30-60-90 triangle, after the three angles. To draw a 30-60-90 triangle, start with an equilateral triangle.
Investigation 8-3: Properties of a 30-60-90 Triangle
Tools Needed: Pencil, paper, ruler, compass
1. Construct an equilateral triangle with 2 inch sides.
2. Draw or construct the altitude from the top vertex to form two congruent triangles.
3. Find the measure of the two angles at the top vertex and the length of the shorter leg.
The top angles are each 30\begin{align*}30^\circ\end{align*} and the shorter leg is 1 in because the altitude of an equilateral triangle is also the angle and perpendicular bisector.
4. Find the length of the longer leg, using the Pythagorean Theorem. Simplify the radical.
12+b21+b2b2b=22=4=3=3\begin{align*}1^2 + b^2 &= 2^2\\ 1+b^2 &= 4\\ b^2 &= 3\\ b &= \sqrt{3}\end{align*}
5. Now, let’s say the shorter leg is length x\begin{align*}x\end{align*} and the hypotenuse is 2x\begin{align*}2x\end{align*}. Use the Pythagorean Theorem to find the longer leg. How is this similar to your answer in #4?
x2+b2x2+b2b2b=(2x)2=4x2=3x2=x3\begin{align*}x^2 + b^2 &= (2x)^2\\ x^2+b^2 &= 4x^2\\ b^2 &= 3x^2\\ b &= x \sqrt{3}\end{align*}
30-60-90 Theorem: If a triangle has angle measures 30,60\begin{align*}30^\circ, 60^\circ\end{align*} and 90\begin{align*}90^\circ\end{align*}, then the sides are x:x3:2x\begin{align*}x:x \sqrt{3}:2x\end{align*}.
The shortest leg is always x\begin{align*}x\end{align*}, the longest leg is always x3\begin{align*}x \sqrt{3}\end{align*}, and the hypotenuse is always 2x\begin{align*}2x\end{align*}. If you ever forget these theorems, you can still use the Pythagorean Theorem.
Example 3: Find the length of the missing sides.
a)
b)
Solution: In part a, we are given the shortest leg and in part b, we are given the hypotenuse.
a) If x=5\begin{align*}x=5\end{align*}, then the longer leg, b=53\begin{align*}b=5 \sqrt{3}\end{align*}, and the hypotenuse, c=2(5)=10\begin{align*}c=2(5)=10\end{align*}.
b) Now, 2x=20\begin{align*}2x=20\end{align*}, so the shorter leg, f=202=10\begin{align*}f = \frac{20}{2} = 10\end{align*}, and the longer leg, g=103\begin{align*}g=10 \sqrt{3}\end{align*}.
Example 4: Find the value of x\begin{align*}x\end{align*} and y\begin{align*}y\end{align*}.
a)
b)
Solution: In part a, we are given the longer leg and in part b, we are given the hypotenuse.
a) \begin{align*}x \sqrt{3} = 12\!\\ x = \frac{12}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{12 \sqrt{3}}{3} = 4 \sqrt{3}\!\\ \text{The hypotenuse is}\\ y = 2(4 \sqrt{3}) = 8 \sqrt{3}\end{align*}
b) \begin{align*}2x=16\!\\ x = 8\!\\ \text{The longer leg is}\!\\ y = 8 \cdot \sqrt 3 = 8 \sqrt{3}\end{align*}
Example 5: A rectangle has sides 4 and \begin{align*}4 \sqrt{3}\end{align*}. What is the length of the diagonal?
Solution: If you are not given a picture, draw one.
The two lengths are \begin{align*}x, x \sqrt{3}\end{align*}, so the diagonal would be \begin{align*}2x\end{align*}, or \begin{align*}2(4) = 8\end{align*}.
If you did not recognize this is a 30-60-90 triangle, you can use the Pythagorean Theorem too.
\begin{align*}4^2 + \left( 4 \sqrt{3} \right )^2 &= d^2\\ 16 + 48 &= d^2\\ d &= \sqrt{64} = 8\end{align*}
Example 6: A square has a diagonal with length 10, what are the sides?
Solution: Draw a picture.
We know half of a square is a 45-45-90 triangle, so \begin{align*}10=s \sqrt{2}\end{align*}.
\begin{align*}s \sqrt{2} &= 10\\ s &= \frac{10}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}}= \frac{10 \sqrt{2}}{2}=5 \sqrt{2}\end{align*}
Know What? Revisited The distance between \begin{align*}1^{st}\end{align*} and \begin{align*}3^{rd}\end{align*} base is one of the diagonals of the square. So, it would be the same as the hypotenuse of a 45-45-90 triangle. Using our ratios, the distance is \begin{align*}90 \sqrt{2} \approx 127.3 \ ft\end{align*}. The distance between \begin{align*}2^{nd}\end{align*} base and home plate is the same length.
## Review Questions
• Questions 1-4 are similar to Example 1-4.
• Questions 5-8 are similar to Examples 5 and 6.
• Questions 9-23 are similar to Examples 1-4.
• Questions 24 and 25 are a challenge.
1. In an isosceles right triangle, if a leg is 4, then the hypotenuse is __________.
2. In a 30-60-90 triangle, if the shorter leg is 5, then the longer leg is __________ and the hypotenuse is ___________.
3. In an isosceles right triangle, if a leg is \begin{align*}x\end{align*}, then the hypotenuse is __________.
4. In a 30-60-90 triangle, if the shorter leg is \begin{align*}x\end{align*}, then the longer leg is __________ and the hypotenuse is ___________.
5. A square has sides of length 15. What is the length of the diagonal?
6. A square’s diagonal is 22. What is the length of each side?
7. A rectangle has sides of length 6 and \begin{align*}6 \sqrt{3}\end{align*}. What is the length of the diagonal?
8. Two (opposite) sides of a rectangle are 10 and the diagonal is 20. What is the length of the other two sides?
For questions 9-23, find the lengths of the missing sides. Simplify all radicals.
Challenge For 24 and 25, find the value of \begin{align*}y\end{align*}. You may need to draw in additional lines. Round all answers to the nearest hundredth.
1. \begin{align*}4^2+4^2 = x^2\!\\ {\;} \quad \ \ 32 = x^2\!\\ {\;} \qquad \ x = 4 \sqrt{2}\end{align*}
2. \begin{align*}3^2+y^2 = 6^2\!\\ {\;} \qquad y^2 = 27\!\\ {\;} \qquad \ y = 3 \sqrt{3}\end{align*}
3. \begin{align*}x^2 + x^2 = \left ( 10 \sqrt{2} \right )^2\!\\ {\;} \quad \ 2x^2 = 200\!\\ {\;} \qquad x^2 = 100\!\\ {\;} \qquad \ x = 10\end{align*}
4. Yes, \begin{align*}9^2 + 12^2 = 15^2 \rightarrow 81+144 = 225\end{align*}
5. Yes, \begin{align*}3^2 + \left( 3 \sqrt{3} \right )^2 = 6^2 \rightarrow 9+27 = 36\end{align*}
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GSEB Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.3
Gujarat Board GSEB Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.3 Textbook Questions and Answers.
Gujarat Board Textbook Solutions Class 7 Maths Chapter 4 Simple Equations Ex 4.3
Question 1.
Solve the following equations.
(a) 2y + $$\frac { 5 }{ 2 }$$ = $$\frac { 37 }{ 2 }$$
(b) 5t + 28 = 10
(c) $$\frac { a }{ 5 }$$ + 3 = 2
(d) $$\frac { q }{ 4 }$$ + 7 = 5
(e) $$\frac { 5 }{ 2 }$$x = – 10
(f) $$\frac { 5 }{ 2 }$$x = $$\frac { 25 }{ 4 }$$
(g) 7m + $$\frac { 19 }{ 2 }$$ = 13
(h) 6z + 10 = – 2
(i) $$\frac { 3l }{ 2 }$$ = $$\frac { 2 }{ 3 }$$
Solution:
(a) We have
2y + $$\frac { 5 }{ 2 }$$ = $$\frac { 37 }{ 2 }$$
2y = $$\frac { 37 }{ 2 }$$ – $$\frac { 5 }{ 2 }$$ = $$\frac { 37-5 }{ 2 }$$
Transposing $$\frac { 5 }{ 2 }$$ to R.H.S.
= $$\frac { 32 }{ 2 }$$ = 16
or y = $$\frac { 16 }{ 2 }$$ = 8
Dividing both sides by 2,
Thus, y = 8 is the required solution.
(b) We have 5t + 28 = 10
or 5t = 10 – 28
Transposing 28 to R.H.S.
or 5t = – 18 or t = – $$\frac { 18 }{ 5 }$$
Dividing both sides by 5,
Thus, t = – $$\frac { 18 }{ 5 }$$ is the required solution.
(c) We have $$\frac { a }{ 5 }$$ + 3 = 2 a
or $$\frac { a }{ 5 }$$ = 2 – 3
Transposing 3 to R.H.S.
or $$\frac { a }{ 5 }$$ = (- 1) x 5
Multiplying both sides by 5, we have:
or $$\frac { a }{ 5 }$$ x 5 = (- 1) x 5
or a = – 5
Thus, a = – 5 is the required solution.
(d) $$\frac { q }{ 4 }$$ + 7 = 5
or $$\frac { q }{ 4 }$$ = 5 – 7
Transposing 7 to R.H.S.
or $$\frac { q }{ 4 }$$ = – 2
Multiplying both sides by 4, we have:
or $$\frac { q }{ 4 }$$ x 4 = – 2 x 4
or q = – 8
Thus, q = – 8 is the required solution.
(e) $$\frac { 5 }{ 2 }$$x = – 10
Multiplying both sides by 2, we have
or $$\frac { 5x }{ 2 }$$ x 2 = – 10 x 2 = – 20
or 5x = – 20
or $$\frac { 5x }{ 5 }$$ = $$\frac { – 20 }{ 5 }$$
Dividing both sides by 5,
i.e. x = – 4
∴ x = – 4 is the required solution.
(f) We have $$\frac { 5 }{ 2 }$$x = $$\frac { 25 }{ 4 }$$
Multiplying both sides by 2, we have
$$\frac { 5 }{ 2 }$$x × 2 = $$\frac { 25 }{ 4 }$$ x 2
$$\frac { 5 }{ 2 }$$x × $$\frac { 25 }{ 2 }$$
Dividing both sides by 5, we have
$$\frac { 5x }{ 5 }$$ = $$\frac { 25 }{ 2 }$$ x $$\frac { 1 }{ 5 }$$
or x = $$\frac { 5 }{ 2 }$$
Thus, x= $$\frac { 5 }{ 2 }$$ is the required solution.
(g) We have 7m + $$\frac { 19 }{ 2 }$$ = 13
or 7m = 13 – $$\frac { 19 }{ 2 }$$
Transposing $$\frac { 19 }{ 2 }$$ from L.H.S. to R.H.S.
or 7m = $$\frac { 26-19 }{ 2 }$$
or 7m = $$\frac { 7 }{ 2 }$$
Dividing both sides by 7, we have
$$\frac { 7m}{ 7 }$$ = $$\frac { 7 }{ 2 }$$ x $$\frac { 1 }{ 7 }$$
or m = $$\frac { 1 }{ 2 }$$
Thus, m = $$\frac { 1 }{ 2 }$$ is the required solution.
(h) We have 6z + 10 = – 2
or 6z = – 2 – 10
Transposing 10 to R.H.S.
or 6z = -12
Dividing both sides by 6, we have
$$\frac { 6z }{ 6 }$$ = $$\frac { -12 }{ 6 }$$
or z = – 2
Thus, z = – 2 is the required solution.
(i) We have $$\frac { 3l }{ 2 }$$ = $$\frac { 2 }{ 3 }$$
Multiplying both sides by 2, we have
$$\frac { 3l }{ 2 }$$ x 2 = $$\frac { 2 }{ 3 }$$ x 2
Dividing both sides by 3, we have
$$\frac { 3l }{ 3 }$$ = $$\frac { 4 }{ 3 }$$ x $$\frac { 1 }{ 3 }$$
or l = $$\frac { 4 }{ 9 }$$
Thus, l = $$\frac { 4 }{ 9 }$$ is the required solution.
(j) We have $$\frac { 2b }{ 3 }$$ – 5 = 3
or $$\frac { 2b }{ 3 }$$
Transposing – 5 to R.H.S.
Multiplying both sides by $$\frac { 3 }{ 2 }$$, we have
∴ $$\frac { 2b }{ 3 }$$ x $$\frac { 3 }{ 2 }$$ = $$\frac { 8 }{ 2 }$$ x 3 = 12
or b = 12
Thus, b = 12 is the required solution.
Question 2.
Solve the following equations.
(a) 2(x + 4) = 12
(b) 3(n – 5) = 21
(c) 3(n – 5) = – 21
(d) – 4(2 + x) = 8
(e) 4(2 – x) = 8
Solution:
(a) We have 2(x + 4) = 12
or x + 4 = $$\frac { 12 }{ 2 }$$ = 6
[Dividing both sides by 2]
or x = 6 – 4 = 2
[Transposing 4 to R.H.S.]
∴ x = 2 is the required solution.
(b) We have 3(n – 5) = 21
or n – 5 = $$\frac { 21 }{ 3 }$$ = 7
[Dividing both sides by 3]
or n = 7 + 5 = 12
[Transposing – 5 to R.H.S.]
Thus, n = 12 is the required solution.
(c) We have 3(n – 5) = – 21
or n – 5 = $$\frac { – 21 }{ 3 }$$ = – 7
[Dividing both sides by 3]
or n = – 7 + 5 = – 2
[Transposing – 5 to R.H.S.]
Thus, n = – 2 is the required solution.
(d) We have – 4(2 + x) = 8
or $$\frac { -4(2+x) }{ -4 }$$ = $$\frac { 8 }{ -4 }$$
[Dividing both sides by – 4]
or 2 + x = – 2
x = – 2 – 2
[Transposing 2 to R.H.S.]
x = – 4
Thus, x = – 4 is the required solution.
(e) We have 4(2 – x) = 8
or 2 – x = $$\frac { 8 }{ 4 }$$ =2
[Dividing both sides by 4]
or – x = 2 – 2
[Transposing 2 to R.H.S.]
or – x = 0
or x – 0
Thus, x = 0 is the required solution.
Question 3.
Solve the following equations:
(a) 4 = 5(p – 2)
(b) – 4 = 5(p – 2)
(c) 16 = 4 + 3(t + 2)
(d) 4 + 5(p – 1) = 34
(e) 0 = 16 + 4(m – 6)
Solution:
(a) 4 = 5(p – 2)
Interchanging the sides, we have
5(p – 2) = 4
Dividing both sides by 5, we have
$$\frac { 5(p-2) }{ 5 }$$ = $$\frac { 4 }{ 5 }$$
or p – 2 = $$\frac { 4 }{ 5 }$$ or p = $$\frac { 4 }{ 5 }$$ + 2
[Transposing (- 2) from L.H.S. to R.H.S.]
or p = $$\frac { 4+10 }{ 5 }$$ = $$\frac { 14 }{ 5 }$$
∴ p = $$\frac { 14 }{ 5 }$$ is the required solution.
(b) – 4 = 5(p – 2)
Interchanging the sides, we have
5(p – 2) = – 4
Dividing both sides by 5, we have
$$\frac { 5(p-2) }{ 5 }$$ = – $$\frac { 4 }{ 5 }$$
or p – 2 = – $$\frac { 4 }{ 5 }$$ or p = – $$\frac { 4 }{ 5 }$$ + 2
[Transposing (-2) from L.H.S. to R.H.S.]
= $$\frac { -4+10 }{ 5 }$$ = $$\frac { 6 }{ 5 }$$
Thus, p = $$\frac { 6 }{ 5 }$$ is the required solution.
(c) 16 = 4 + 3(t + 2)
Interchanging the sides, we have
4 + 3(t + 2) = 16
or 3(t + 2)= 16 – 4 = 12
[Transposing 4 to R.H.S.]
or $$\frac { 3(t+2) }{ 3 }$$ = $$\frac { 12 }{ 3 }$$
[Dividing both sides by 3]
or t + 2 =4
or t = 4 – 2
[Transposing 2 to R.H.S.]
or t = 2
Thus, t = 2 is the required solution.
(d) We have 4 + 5(p – 1) = 34
or 5(p – 1) = 34 – 4 = 30
[Transposing 4 to R.H.S.]
or $$\frac { 5(p-1) }{ 5 }$$ = $$\frac { 30 }{ 5 }$$
[Dividing both sides by 5]
or p – 1 = 6
or p = 6 + 1 = 7
[Transposing 1 to R.H.S.]
Thus, p = 7 is the required solution.
(e) We have 0 = 16 + 4(m – 6)
Interchanging the sides, we have
16 + 4(m – 6) = 0
or 4(m – 6) = – 16
[Transposing 16 from L.H.S. to R.H.S.]
Dividing both sides by 4, we have
$$\frac { 4(m-6) }{ 4 }$$ = $$\frac { – 16 }{ 4 }$$
or m – 6 = – 4
or m = – 4 + 6
[Transposing – 6 from L.H.S. to R.H.S.] or m = 2
Thus, m = 2 is the required solution.
Question 4.
(a) Construct 3 equations starting with x = 2.
(b) Construct 3 equations starting with x = – 2.
Solution:
(a) Starting with x = 2
I. x = 2
Multiplying both sides by 5, we have
5 × x = 5 × 2
or 5x = 10
Subtracting 3 from both sides, we have
5x – 3 = 10 – 3
or 5x – 3 = 7
II. x = 2
Multiplying both sides by 7, we have
7 × x = 7 × 2
or 7x = 14
Adding 5 to both sides, we have
7x + 5 = 14 + 5
or 7x + 5 = 19
III. x = 2
Dividing both sides by 3, we have
$$\frac { x }{ 3 }$$ = $$\frac { 2 }{ 3 }$$
Subtracting 4 from both sides, we have
$$\frac { x }{ 3 }$$ – 4 = $$\frac { 2 }{ 3 }$$ – 4
or $$\frac { x }{ 3 }$$ – 4 = $$\frac { 2-12 }{ 3 }$$ = $$\frac { -10 }{ 3 }$$
or $$\frac { x }{ 3 }$$ – 4 = $$\frac { -10 }{ 3 }$$
(b) Starting with x = – 2
I.x = – 2
Adding 8 to both sides, we have
x + 8 = – 2 + 8 or x + 8 = 6
II. x = – 2
Subtracting 10 from both sides, we have x – 10 = – 2 – 10 or x – 10 = – 12
III. x = – 2
Multiplying both sides by 8, we have
8 × x = (- 2) x 8
or 8x = – 16
Subtracting 2 from both sides, we have
8x – 2 = – 16 – 2
or 8x – 2 = – 18
|
# Solving Quadratic Equations by Factoring06:09 minutes
Video Transcript
## TranscriptSolving Quadratic Equations by Factoring
Every day in San Francisco, on Pier 39, there is a street performer named FOIL. People pass by him, but no one seems to notice the inconspicuous man. FOIL has a secret. At night, he morphs into a superhero, armed with a sparkling wit and powerful tools: factors, sums, the Zero Factor Property and most importantly, his powerful calculator wrist. What’s that, you ask? Hold on to that thought. The mayor has just made an announcement: his two daughters have been kidnapped by a pair of bad guys and they are holding the girls by the dock but can he save the girls and foil the crime just in time?
FOIL will need his calculator, his super smarts and how to solve quadratic equations by factoring. The girls are locked up in a container secured by a very complicated code. It won’t be easy to crack this code. To figure it out, FOIL must find the solutions for this quadratic equation.
### Zero Factor Property
Let’s start by having a look at the Zero Factor Property. In this example, either one or both of the factors have to be zero to get a true statement. It's important that you always set the expression equal to zero, otherwise this rule won’t work. Set each set of parentheses equal to zero. If you use opposite operations, you should get two answers for x, -6 and 1. When you plug either of these two values into the original equation, you should get 0. Remember to always check your solutions.
### Reverse FOIL Method
Now that we understand the main concept of the Zero Factor Property, let’s look at the quadratic equation FOIL has to solve. To factor this quadratic equation we can use the reverse foil method. We have to find the values for the variables 'm' and 'n'. In our example, m(n) = -6 and mx + nx = 1x. Therefore, 'm' plus 'n' have to be equal to 1. So now we have to find factors of -6 that sum to 1. As you can see, the product of -2 and 3 is equal to -6 and the sum of -2 and 3 is equal to 1. Perfect, we found the values for 'm' and 'n', so we can subtitute the variable 'm' and 'n' in the reverse FOIL method with these values.
So we factored our quadratic equation into two binomials (x-2) and (x+3). As always, it’s a good idea to check your work by FOILing. Good job, the factorization is proved. But, FOIL's work is not done. Although he's figured out the factors, he doesn't have a solution. To calculate the solution and free the girls, he must use the Zero Factor Property. He sets each binomial factor to zero and solves for 'x' in each case. By using opposite operations, he gets two answers for 'x', -3 and 2.
Let’s try some more problems and investigate some strategies to make these calculations easier to work with. For this first problem, modify the equation to write it in standard form to set the equation equal to zero. Now use the reverse of FOIL to factor, and then apply the Zero Factor Property to find the solutions for the variable. For this equation there is just one solution, x equals 2. I know you're anxious to find out what’s happening on the dock, but be patient.
Let’s work out one last example. This problem looks different because it has no constant. To solve, you can factor out the greatest common factor, which is 'x'. The first solution for 'x' is zero, and the second solution is 8. Don't forget to check your answer by substituting the values for 'x' into the original equation to make sure you get a true statment because even super heros can make mistakes!
Now, back to the kidnapping. FOIL enters the code and saves the girls. Now his big moment has come. Time to get famous. But no one is exactly sure what he looks like maybe he should think about a new material for his costume.
## Videos in this Topic
Quadratic Equations / Functions (11 Videos)
## Solving Quadratic Equations by Factoring Exercise
### Would you like to practice what you’ve just learned? Practice problems for this video Solving Quadratic Equations by Factoring help you practice and recap your knowledge.
• #### State the solutions of the equation $(x+6)(x-1)=0$.
##### Hints
Sure $4\times 0=0$ as well as $0\times 5=0$.
Here you see an example for the solution of an equation.
You can check your solutions by putting them into the starting equation.
##### Solution
The zero factor property says that if a product $a\times b$ is equal to zero, then one of its factors, either $a$ or $b$, is equal to zero.
We can use this fact to solve factorized quadratic equations. For example,
$(x+6)(x-1)=0$.
By the zero factor property, we know that either $x+6=0$ or $x-1=0$. Using opposite operations we get
$\begin{array}{rcl} x+6 & = & ~0\\ \color{#669900}{-6} & &\color{#669900}{-6}\\ x & = & ~-6 \end{array}$
as well as
$\begin{array}{rcl} x-1 & = & ~0\\ \color{#669900}{+1} & &\color{#669900}{+1}\\ x & = & ~1 \end{array}$
So, we can conclude that the equation above has two solutions: $x=-6$ or $x=1$.
• #### Solve the quadratic equation.
##### Hints
To factor a quadratic term $ax^2+bx+c$, find the factors of $a\times c$ which sum to $b$.
Use the zero factors property: if a product is zero, then one of its factors must be zero.
To solve linear equations use opposite operations.
##### Solution
If we could factor the quadratic term $x^2+x-6=(x+m)(x+n)$, then we would be able to solve this equation using the zero factor property.
With the reverse FOIL method we can determine $m$ and $n$ as follows:
1. $(x+m)(x+n)=x^2+nx+mx+mn=x^2+(m+n)x+mn$.
2. So we have to find the $m$ and $n$ which statisfy $m\times n=-6$ and $m+n=1$.
3. Let's check:
• $-1\times 6=-6$ but $-1+6=5$
• $1\times (-6)=-6$ but $1-6=-5$
• $-2\times 3=-6$ and $-2+3=1$ $~~~~~$✓
Now we know that $x^2+x-6=(x-2)(x+3)$ and thus the equation above is equivalent to
$(x-2)(x+3)=0$.
Thus we have to solve $x+6=0$ and $x-1=0$. Using opposite operations we get
$\begin{array}{rcl} x-2 & = & ~0\\ \color{#669900}{+2} & &\color{#669900}{+2}\\ x & = & ~2 \end{array}$
as well as
$\begin{array}{rcl} x+3 & = & ~0\\ \color{#669900}{-3} & &\color{#669900}{-3}\\ x & = & ~-3 \end{array}$
Thus the solutions are $x=2$ or $x=-3$.
• #### Determine the solutions of the factorized equations.
##### Hints
Use the zero factors property: if a product is zero, then one of its factors must be zero.
Each factor is linear. To solve a linear equation use opposite operations.
You can solve an equation like $x^2-8x=0$ as follows:
1. Factor out an $x$ to get $x(x-8)=0$.
2. Thus either $x=0$ or $x-8=0$, which is equivalent to $x=8$.
##### Solution
$(x-1)(x-2)=0$: either $x-1=0$ or $x-2=0$. This leads to the following solutions
$\begin{array}{rcl} x-1 & = & ~0\\ \color{#669900}{+1} & &\color{#669900}{+1}\\ x & = & ~1 \end{array}$
or
$\begin{array}{rcl} x-2 & = & ~0\\ \color{#669900}{+2} & &\color{#669900}{+2}\\ x & = & ~2 \end{array}$
$~$
$(2x+4)(x-5)=0$: either $2x+4=0$ or $x-5=0$. So we solve these equations to get
$\begin{array}{rcl} 2x+4 & = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ 2x & = & ~-4 \\ \color{#669900}{\div 2} & &\color{#669900}{\div 2}\\ x&=&~-2 \end{array}$
or
$\begin{array}{rcl} x-5 & = & ~0\\ \color{#669900}{+5} & &\color{#669900}{+5}\\ x & = & ~5 \end{array}$
$~$
$(x+7)(-x+3)=0$: either $x+7=0$ or $-x+3=0$. Again we use opposite operations to solve those equations
$\begin{array}{rcl} x+7 & = & ~0\\ \color{#669900}{-7} & &\color{#669900}{-7}\\ x & = & ~-7 \end{array}$
or
$\begin{array}{rcl} -x+3 & = & ~0\\ \color{#669900}{+x} & &\color{#669900}{+x}\\ 3 & = & ~x \end{array}$
$~$
Let's have a look at a last example $x^2+4x=0$ which isn't factored yet. You can factor out $x$ to get $x(x+4)=0$. So either $x=0$ or $x+4=0$. We still have to solve the second equation:
$\begin{array}{rcl} x+4 & = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ x & = & ~-4 \end{array}$
• #### Multiply the two binomials using the FOIL method.
##### Hints
FOIL is a mnemonic device for multiplying binomials:
• First
• Outer
• Inner
• Last
Here you see an example using FOIL:
$(x-5)(x+2)=$
• F: $x\times x=x^2$
• O: $x\times 2=2x$
• I: $-5\times x=-5x$
• L: $-5\times 2=-10$
Add and combine the like terms. For the example above we get
$x^2+2x-5x-10=x^2-3x-10$.
##### Solution
FOIL is a mnemonic device for multiplying two binomials:
First, Outer, Inner, Last.
$~$
$(2x-1)(x+1)=$
• F: $2x\times x=2x^2$
• O: $2x\times 1=2x$
• I: $-1\times x=-x$
• L: $-1\times 1=-1$
Now we add the terms and combine the like terms to get $2x^2+2x-x-1=2x^2+x-1$.
$(x-5)(2x+2)=$
• F: $x\times 2x=2x^2$
• O: $x\times 2=2x$
• I: $-5\times 2x=-10x$
• L: $-5\times 2=-10$
We still have to add the terms to $2x^2+2x-10x-10$ and combine the like terms $2x^2-8x-10$.
$(x+3)(x-3)=$
• F: $x\times x=x^2$
• O: $x\times (-3)=-3x$
• I: $3\times x=3x$
• L: $3\times (-3)=-9$
Next add the terms $x^2-3x+3x-9$ and combine the like terms $x^2-9$.
$(x-3)(x-3)=$
• F: $x\times x=x^2$
• O: $x\times (-3)=-3x$
• I: $-3\times x=-3x$
• L: $(-3)\times (-3)=9$
Almost done: we just have to add the terms to get $x^2-3x-3x+9=x^2-6x+9$.
• #### Check the factorization.
##### Hints
FOIL is a mnemonic device for multiplying binomials:
• First
• Outer
• Inner
• Last
Here you see an example for using FOIL.
$2x$ and $x$ are like terms and can be combined to get $2x+x=3x$.
Two terms are correct.
##### Solution
FOIL is a mnemonic device for multiplying binomials. The letters stand for
• First
• Outer
• Inner
• Last
Let's have a look at the example beside.
• F: $x\times x=x^2$
• O: $x\times 3=3x$
• I: $-2\times x=-2x$
• L: $-2\times 3=-6$
$x^2+3x-2x-6$
and combing the like terms gives
$x^2+x-6$.
So we have together
$x^2+x-6=(x-2)(x+3)$.
• #### Determine the code.
##### Hints
To factor a quadratic term $x^2+bx+c$, find the factors of $c$ which sum to $b$.
For the equation above, check all factors of $m\times n=-28$ and pick the pair $m$ and $n$ satisfying $m+n=-3$.
If you know the factorization, for example $(x-2)(x+15)$, just solve the two equations:
• $x-2=0$
• $x+15=0$
Check your solution: put both solutions for $x$ into the equation above to see if they satisfy it.
##### Solution
To solve this quadratic equation we factorize the quadratic term $x^2-3x-28$.
For this we check the factors $m\times n=-28$ and choose those which sum to $m+n=-3$:
• $m=-1$ and $n=28$ but $m+n=27$
• $m=1$ and $n=-28$ but $m+n=-27$
• $m=-2$ and $n=14$ but $m+n=12$
• $m=2$ and $n=-14$ but $m+n=-12$
• $m=-4$ and $n=7$ but $m+n=3$
• $m=4$ and $n=-7$ and $m+n=-3$ $~~~~~$✓
Thus $x^2-3x-28=(x+4)(x-7)$ and so the equation above is equivalent to
$(x+4)(x-7)=0$.
We conclude that either $x+4=0$ or $x-7=0$. Using opposite operations we get
$\begin{array}{rcl} x+4 & = & ~0\\ \color{#669900}{-4} & &\color{#669900}{-4}\\ x & = & ~-4 \end{array}$
as well as
$\begin{array}{rcl} x-7 & = & ~0\\ \color{#669900}{+7} & &\color{#669900}{+7}\\ x & = & ~7 \end{array}$
Now we know Grandma Millers Code: $x=-4$ and $x=7$.
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# LCM HCF Divisibility Questions Trick
LCM HCF Divisibility Questions Trick: Here you can check shortcut trick to solve questions based on LCM & HCF division. In various competitive exams, questions like "find the least number which when devided by x, leaves remainder y". The solution to these problems is quite easy if you know the proper basic method. It takes a little time to find the solution to these problems. Here you can check solution to these type of problems with examples.
The solution is explained here with example.
## LCM HCM Based Questions Method of Solution
Example: 1.
Find the least number which should be added to 2497 so that the sum is exactly divisible by 5, 6, 4 and 3:
Solution:
First we have to find the LCM of 5, 6, 4 and 3.
5 = 5
6 = 2, 3
4 = 2, 2
3 = 3
So L.C.M. of 5, 6, 4 and 3 = 60.
When we divide 2497 by 60, we find that the remainder is 37.
So the Number that to be added = 60 - 37 = 23.
Example: 2.
Find the least number which when increased by 5 is divisible by 24, 32, 36 and 54 each.
Solution:
First we have to find the LCM of 24, 32, 36, 54
24 = 2, 2, 2, 3
32 = 2, 2, 2, 2, 2
36 = 2, 2, 3, 3
54 = 2, 3, 3, 3
So LCM = 864
So Required number = (L.C.M. of 24, 32, 36, 54) - 5 = 864 - 5 = 859.
Example: 3.
Find the least multiple of 7, which leaves a remainder of 4, when divided by 6, 9, 15 and 18.
Solution:
First Find LCM.
6 = 2, 3
9 = 3, 3
15 = 3, 5
18 = 2, 3, 3
LCM of 6, 9, 15 and 18 is 90.
Let required number be 90x + 4, which is a multiple of 7.
The Least value of k for which (90x + 4) is divisible by 7 is k = 4.
So the Required number is = (90*4) + 4 = 364.
Example: 4.
Find the least multiple of 23 which when devided by 18, 21 and 24 leaves a remainder 7, 10, 13 respectively.
Solution:
First we have to find the LCM of 18, 21 & 24.
18 = 2 × 3 × 3
21 = 3 × 7
24 = 2 × 2 × 2 × 3
So LCM of 18, 21 & 24 = 504
Now check the given numbers remainder respectively.
18-7=11
21-10=11
24-13=11
We find it that there is 11 difference in each number.
Now the formula is
Required Number = LCM*X-11
Required Number = 504*X-11
by Hit & Trial method:
when putting X=6
We get
Required number = 3013 Ans.
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Reciprocal Relations of Trigonometric Ratios
Reciprocal relations of trigonometric ratios are explained here to represent the relationship between the three pairs of trigonometric ratios as well as their reciprocals.
There are three reciprocal trigonometric functions, making a total of six including cosine, sine, and tangent. The reciprocal cosine function is secant: sec(theta)=1/cos(theta). The reciprocal sine function is cosecant, csc(theta)=1/sin(theta). The reciprocal tangent function is cotangent, expressed two ways: cot(theta)=1/tan(theta) or cot(theta)=cos(theta)/sin(theta).
Let OMP be a right-angled triangle at M and ∠MOP = θ.
According to the definition of trigonometric ratios we have,
• sin θ = perpendicular/hypotenuse = MP/PO ………….. (i)
and csc θ = hypotenuse/perpendicular = PO/MP ………….. (ii)
From (i) sin θ = 1/(PO/MP)
⇒ sin θ = 1/csc θ ………………… (A)
Again, from (ii) csc θ = 1/(MP/PO)
⇒ csc θ = 1/sin θ ………………… (B)
From (A) and (B) we conclude that
sin θ and csc θ are reciprocal of each other.
• cos θ = adjacent/hypotenuse = OM/OP ………….. (iii)
and sec θ = hypotenuse/adjacent = OP/OM ………….. (iv)
From (iii) cos θ = 1/(OP/OM)
⇒ cos θ = 1/sec θ ………………… (C)
Again, from (iv) sec θ = 1/(OM/OP)
⇒ sec θ = 1/cos θ ………………… (D)
From (C) and (D) we conclude that
cos θ and sec θ are reciprocal of each other.
• tan θ = perpendicular/adjacent = MP/OM ………….. (v)
and cot θ = adjacent/perpendicular = OM/MP ………….. (vi)
From (v) tan θ = 1/(OM/MP)
⇒ tan θ = 1/cot θ ………………… (E)
Again, from (vi) cot θ = 1/(MP/OM)
⇒ cot θ = 1/tan θ ………………… (F)
From (E) and (F) we conclude that
tan θ and cot θ are reciprocal of each other.
To find values of trig functions we can use these reciprocal relationships to solve different types of problems.
Note: From the above discussion about the reciprocal trigonometric functions we get;
1. sin θ ∙ csc θ = 1
2. cos θ ∙ sec θ = 1
3. tan θ ∙ cot θ = 1
Information Source:
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# A Triangle Problem from Caucasus
### Solution 1
Let $b+c-a=2x,$ $c+a-b=2y$ and $a+b-c=2z.$ The problem constraint is equivalent to
$\displaystyle \frac{yz}{(x+y)(x+z)}\gt 0$ and $\displaystyle \frac{x(x+y+z)}{(x+y)(x+z)}\gt 0.$
We need to prove that
$\displaystyle \frac{zx}{(y+z)(y+x)}\gt 0$ and $\displaystyle \frac{y(x+y+z)}{(y+z)(y+x)}\gt 0$
and
$\displaystyle \frac{xy}{(z+x)(z+y)}\gt 0$ and $\displaystyle \frac{z(x+y+z)}{(z+x)(z+y)}\gt 0.$
Now, by weakening the given inequality,
\displaystyle\begin{align}\left(\frac{yz}{(x+y)(x+z)}\right)\left(\displaystyle \frac{x(x+y+z)}{(x+y)(x+z)}\right)\gt 0\;&\Rightarrow\\ \frac{xyz(x+y+z)}{(x+y)^2(x+z)^2}\gt 0\;&\Rightarrow\\ xyz(x+y+z)\gt 0. \end{align}
Let $\displaystyle \frac{zx}{(y+z)(y+x)}=A$ and $\displaystyle \frac{y(x+y+z)}{(y+z)(y+x)}=B.$ Then $\displaystyle AB=\frac{xyz(x+y+z)}{(y+z)^2(y+x)^2}\gt 0,$ whereas $\displaystyle A+B=\frac{(y+z)(y+x)}{(y+z)(y+x)}=1,$ which implies that $A,B\gt 0.$ Similarly, $\displaystyle \frac{xy}{(z+x)(z+y)}\gt 0$ and $\displaystyle \frac{z(x+y+z)}{(z+x)(z+y)}\gt 0.$
### Solution 2
WLOG, we may assume that $a,b,c$ are positive. We have $\displaystyle \left|\frac{b^2+c^2-a^2}{2bc}\right|\lt 1,$ so that there is an angle $\alpha\in (0^{\circ},180^{\circ})$ such that $\displaystyle \cos\alpha=\frac{b^2+c^2-a^2}{2bc},$ implying
$a^2=b^2+c^2-2bc\cos\alpha.$
If we form a triangle with sides $b, c$ and angle $\alpha$ in-between then, according to the Law of Cosines, the side opposite $\alpha$ will be exactly $a.$ This means that $a,b,c$ form a triangle and the Law of Cosines applied twice implies the required two inequalities.
### Acknowledgment
Leo Giugiuc has kindly messaged me this problem from the 2017 Caucasus Mathematics Olympiad (#6), and a fancy solution of his (Solution 1).
Computer assisted Solution 3 is by N. N. Taleb.
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# Multiplication
Four bags with three marbles per bag gives twelve marbles (4 × 3 = 12).
Multiplication can also be thought of as scalin'. Here we see 2 bein' multiplied by 3 usin' scalin', givin' 6 as a holy result.
Animation for the bleedin' multiplication 2 × 3 = 6.
4 × 5 = 20. The large rectangle is made up of 20 squares, each 1 unit by 1 unit.
Area of a holy cloth 4.5m × 2.5m = 11.25m2; 41/2 × 21/2 = 111/4
Multiplication (often denoted by the cross symbol ×, by the oul' mid-line dot operator , by juxtaposition, or, on computers, by an asterisk *) is one of the feckin' four elementary mathematical operations of arithmetic, with the other ones bein' addition, subtraction, and division. The result of a multiplication operation is called a product.
The multiplication of whole numbers may be thought of as repeated addition; that is, the oul' multiplication of two numbers is equivalent to addin' as many copies of one of them, the feckin' multiplicand, as the quantity of the feckin' other one, the bleedin' multiplier. G'wan now. Both numbers can be referred to as factors.
${\displaystyle a\times b=\underbrace {b+\cdots +b} _{a{\text{ times}}}}$
For example, 4 multiplied by 3, often written as ${\displaystyle 3\times 4}$ and spoken as "3 times 4", can be calculated by addin' 3 copies of 4 together:
${\displaystyle 3\times 4=4+4+4=12}$
Here, 3 (the multiplier) and 4 (the multiplicand) are the oul' factors, and 12 is the bleedin' product.
One of the main properties of multiplication is the feckin' commutative property, which states in this case that addin' 3 copies of 4 gives the oul' same result as addin' 4 copies of 3:
${\displaystyle 4\times 3=3+3+3+3=12}$
Thus the oul' designation of multiplier and multiplicand does not affect the feckin' result of the multiplication.[1]
Systematic generalizations of this basic definition define the oul' multiplication of integers (includin' negative numbers), rational numbers (fractions), and real numbers.
Multiplication can also be visualized as countin' objects arranged in a holy rectangle (for whole numbers) or as findin' the bleedin' area of a feckin' rectangle whose sides have some given lengths, for the craic. The area of a feckin' rectangle does not depend on which side is measured first—a consequence of the feckin' commutative property.
The product of two measurements is a bleedin' new type of measurement. Would ye swally this in a minute now?For example, multiplyin' the oul' lengths of the bleedin' two sides of a rectangle gives its area. Such a feckin' product is the subject of dimensional analysis.
The inverse operation of multiplication is division. For example, since 4 multiplied by 3 equals 12, 12 divided by 3 equals 4. Indeed, multiplication by 3, followed by division by 3, yields the oul' original number, grand so. The division of a number other than 0 by itself equals 1.
Multiplication is also defined for other types of numbers, such as complex numbers, and for more abstract constructs, like matrices, bejaysus. For some of these more abstract constructs, the oul' order in which the bleedin' operands are multiplied together matters. Sufferin' Jaysus. A listin' of the oul' many different kinds of products used in mathematics is given in Product (mathematics).[verification needed]
## Notation and terminology
× ⋅
Multiplication signs
In UnicodeU+00D7 × MULTIPLICATION SIGN (×)
U+22C5 DOT OPERATOR (⋅)
Different from
Different fromU+00B7 · MIDDLE DOT
U+002E . FULL STOP
In arithmetic, multiplication is often written usin' the multiplication sign (either × or${\displaystyle \times }$) between the oul' terms (that is, in infix notation).[2] For example,
${\displaystyle 2\times 3=6}$ ("two times three equals six")
${\displaystyle 3\times 4=12}$
${\displaystyle 2\times 3\times 5=6\times 5=30}$
${\displaystyle 2\times 2\times 2\times 2\times 2=32}$
There are other mathematical notations for multiplication:
• To reduce confusion between the oul' multiplication sign × and the feckin' common variable x, multiplication is also denoted by dot signs,[3] usually a middle-position dot (rarely period):
5 ⋅ 2 or 5 , the shitehawk. 3
The middle dot notation, encoded in Unicode as U+22C5 DOT OPERATOR, is now standard in the oul' United States and other countries where the oul' period is used as a holy decimal point. Be the hokey here's a quare wan. When the oul' dot operator character is not accessible, the feckin' interpunct (·) is used. Would ye swally this in a minute now?In other countries that use a holy comma as a decimal mark, either the bleedin' period or an oul' middle dot is used for multiplication.[citation needed]
Historically, in the bleedin' United Kingdom and Ireland, the feckin' middle dot was sometimes used for the feckin' decimal to prevent it from disappearin' in the oul' ruled line, and the feckin' period/full stop was used for multiplication. However, since the oul' Ministry of Technology ruled to use the feckin' period as the oul' decimal point in 1968,[4] and the oul' SI standard has since been widely adopted, this usage is now found only in the feckin' more traditional journals such as The Lancet.[5]
• In algebra, multiplication involvin' variables is often written as a bleedin' juxtaposition (e.g., xy for x times y or 5x for five times x), also called implied multiplication.[6] The notation can also be used for quantities that are surrounded by parentheses (e.g., 5(2) or (5)(2) for five times two). This implicit usage of multiplication can cause ambiguity when the concatenated variables happen to match the bleedin' name of another variable, when a bleedin' variable name in front of a parenthesis can be confused with a feckin' function name, or in the correct determination of the oul' order of operations.[citation needed]
• In vector multiplication, there is a bleedin' distinction between the oul' cross and the dot symbols. The cross symbol generally denotes the feckin' takin' a cross product of two vectors, yieldin' a vector as its result, while the bleedin' dot denotes takin' the dot product of two vectors, resultin' in an oul' scalar.[citation needed]
In computer programmin', the asterisk (as in 5*2) is still the feckin' most common notation. Be the holy feck, this is a quare wan. This is due to the oul' fact that most computers historically were limited to small character sets (such as ASCII and EBCDIC) that lacked a holy multiplication sign (such as ⋅ or ×), while the bleedin' asterisk appeared on every keyboard. Sufferin' Jaysus. This usage originated in the oul' FORTRAN programmin' language.[citation needed]
The numbers to be multiplied are generally called the "factors". The number to be multiplied is the "multiplicand", and the feckin' number by which it is multiplied is the "multiplier". Here's a quare one. Usually, the oul' multiplier is placed first and the oul' multiplicand is placed second;[1] however sometimes the bleedin' first factor is the multiplicand and the bleedin' second the multiplier.[7] Also, as the feckin' result of multiplication does not depend on the order of the factors, the oul' distinction between "multiplicand" and "multiplier" is useful only at a holy very elementary level and in some multiplication algorithms, such as the bleedin' long multiplication. Jasus. Therefore, in some sources, the bleedin' term "multiplicand" is regarded as an oul' synonym for "factor".[8] In algebra, a number that is the bleedin' multiplier of a bleedin' variable or expression (e.g., the 3 in 3xy2) is called a holy coefficient.
The result of a multiplication is called a feckin' product, you know yourself like. When one factor is an integer, the oul' product is a bleedin' multiple of the oul' other or of the bleedin' product of the bleedin' others. Thus 2 × π is an oul' multiple of π, as is 5133 × 486 × π. A product of integers is a feckin' multiple of each factor; for example, 15 is the oul' product of 3 and 5 and is both a multiple of 3 and a holy multiple of 5.[citation needed]
## Computation
The Educated Monkey – a feckin' tin toy dated 1918, used as a feckin' multiplication "calculator". Bejaysus. For example: set the monkey's feet to 4 and 9, and get the product – 36 – in its hands.
Many common methods for multiplyin' numbers usin' pencil and paper require a multiplication table of memorized or consulted products of small numbers (typically any two numbers from 0 to 9), the hoor. However, one method, the oul' peasant multiplication algorithm, does not. The example below illustrates "long multiplication" (the "standard algorithm", "grade-school multiplication"):
23958233
× 5830
———————————————
00000000 ( = 23,958,233 × 0)
71874699 ( = 23,958,233 × 30)
191665864 ( = 23,958,233 × 800)
+ 119791165 ( = 23,958,233 × 5,000)
———————————————
139676498390 ( = 139,676,498,390 )
In some countries such as Germany, the feckin' above multiplication is depicted similarly but with the bleedin' original product kept horizontal and computation startin' with the oul' first digit of the bleedin' multiplier:[9]
23958233 · 5830
———————————————
119791165
191665864
71874699
00000000
———————————————
139676498390
Multiplyin' numbers to more than a couple of decimal places by hand is tedious and error-prone. Common logarithms were invented to simplify such calculations, since addin' logarithms is equivalent to multiplyin'. C'mere til I tell ya. The shlide rule allowed numbers to be quickly multiplied to about three places of accuracy, you know yourself like. Beginnin' in the early 20th century, mechanical calculators, such as the Marchant, automated multiplication of up to 10-digit numbers, like. Modern electronic computers and calculators have greatly reduced the oul' need for multiplication by hand.
### Historical algorithms
Methods of multiplication were documented in the feckin' writings of ancient Egyptian, Greek, Indian,[citation needed] and Chinese civilizations.
The Ishango bone, dated to about 18,000 to 20,000 BC, may hint at an oul' knowledge of multiplication in the feckin' Upper Paleolithic era in Central Africa, but this is speculative.[10][verification needed]
#### Egyptians
The Egyptian method of multiplication of integers and fractions, which is documented in the bleedin' Rhind Mathematical Papyrus, was by successive additions and doublin'. Me head is hurtin' with all this raidin'. For instance, to find the feckin' product of 13 and 21 one had to double 21 three times, obtainin' 2 × 21 = 42, 4 × 21 = 2 × 42 = 84, 8 × 21 = 2 × 84 = 168, the hoor. The full product could then be found by addin' the oul' appropriate terms found in the bleedin' doublin' sequence:[11]
13 × 21 = (1 + 4 + 8) × 21 = (1 × 21) + (4 × 21) + (8 × 21) = 21 + 84 + 168 = 273.
#### Babylonians
The Babylonians used a sexagesimal positional number system, analogous to the modern-day decimal system, enda story. Thus, Babylonian multiplication was very similar to modern decimal multiplication, you know yerself. Because of the bleedin' relative difficulty of rememberin' 60 × 60 different products, Babylonian mathematicians employed multiplication tables. These tables consisted of a bleedin' list of the first twenty multiples of a feckin' certain principal number n: n, 2n, ..., 20n; followed by the multiples of 10n: 30n 40n, and 50n. Jesus, Mary and Joseph. Then to compute any sexagesimal product, say 53n, one only needed to add 50n and 3n computed from the feckin' table.[citation needed]
#### Chinese
38 × 76 = 2888
In the mathematical text Zhoubi Suanjin', dated prior to 300 BC, and the feckin' Nine Chapters on the feckin' Mathematical Art, multiplication calculations were written out in words, although the oul' early Chinese mathematicians employed Rod calculus involvin' place value addition, subtraction, multiplication, and division. The Chinese were already usin' a decimal multiplication table by the oul' end of the oul' Warrin' States period.[12]
### Modern methods
Product of 45 and 256. Note the bleedin' order of the bleedin' numerals in 45 is reversed down the left column. Sufferin' Jaysus. The carry step of the oul' multiplication can be performed at the final stage of the feckin' calculation (in bold), returnin' the feckin' final product of 45 × 256 = 11520. Sufferin' Jaysus. This is a variant of Lattice multiplication.
The modern method of multiplication based on the bleedin' Hindu–Arabic numeral system was first described by Brahmagupta. Whisht now and listen to this wan. Brahmagupta gave rules for addition, subtraction, multiplication, and division, that's fierce now what? Henry Burchard Fine, then a feckin' professor of mathematics at Princeton University, wrote the followin':
The Indians are the inventors not only of the oul' positional decimal system itself, but of most of the oul' processes involved in elementary reckonin' with the bleedin' system. Jesus, Mary and Joseph. Addition and subtraction they performed quite as they are performed nowadays; multiplication they effected in many ways, ours among them, but division they did cumbrously.[13]
These place value decimal arithmetic algorithms were introduced to Arab countries by Al Khwarizmi in the oul' early 9th century and popularized in the oul' Western world by Fibonacci in the 13th century.[14]
#### Grid method
Grid method multiplication, or the box method, is used in primary schools in England and Wales and in some areas[which?] of the feckin' United States to help teach an understandin' of how multiple digit multiplication works. An example of multiplyin' 34 by 13 would be to lay the feckin' numbers out in a feckin' grid as follows:
× 30 4
10 300 40
3 90 12
and then add the bleedin' entries.
### Computer algorithms
The classical method of multiplyin' two n-digit numbers requires n2 digit multiplications, so it is. Multiplication algorithms have been designed that reduce the bleedin' computation time considerably when multiplyin' large numbers. Holy blatherin' Joseph, listen to this. Methods based on the feckin' discrete Fourier transform reduce the bleedin' computational complexity to O(n log n log log n). Here's a quare one for ye. In 2016, the factor log log n was replaced by a function that increases much shlower, though still not constant.[15] In March 2019, David Harvey and Joris van der Hoeven submitted a feckin' paper presentin' an integer multiplication algorithm with an oul' complexity of ${\displaystyle O(n\log n).}$[16] The algorithm, also based on the fast Fourier transform, is conjectured to be asymptotically optimal.[17] The algorithm is not practically useful, as it only becomes faster for multiplyin' extremely large numbers (havin' more than 2172912 bits).[18]
## Products of measurements
One can only meaningfully add or subtract quantities of the bleedin' same type, but quantities of different types can be multiplied or divided without problems, what? For example, four bags with three marbles each can be thought of as:[1]
[4 bags] × [3 marbles per bag] = 12 marbles.
When two measurements are multiplied together, the product is of a feckin' type dependin' on the bleedin' types of measurements. C'mere til I tell ya now. The general theory is given by dimensional analysis. In fairness now. This analysis is routinely applied in physics, but it also has applications in finance and other applied fields.
A common example in physics is the oul' fact that multiplyin' speed by time gives distance. For example:
50 kilometers per hour × 3 hours = 150 kilometers.
In this case, the hour units cancel out, leavin' the bleedin' product with only kilometer units.
Other examples of multiplication involvin' units include:
2.5 meters × 4.5 meters = 11.25 square meters
11 meters/seconds × 9 seconds = 99 meters
4.5 residents per house × 20 houses = 90 residents
## Product of a feckin' sequence
### Capital pi notation
The product of a sequence of factors can be written with the feckin' product symbol, which derives from the bleedin' capital letter ${\displaystyle \textstyle \prod }$ (pi) in the Greek alphabet (much like the same way the capital letter ${\displaystyle \textstyle \sum }$ (sigma) is used in the feckin' context of summation).[19][20] Unicode position U+220F contains a holy glyph for denotin' such a product, distinct from U+03A0 Π , the letter. The meanin' of this notation is given by:
${\displaystyle \prod _{i=1}^{4}i=1\cdot 2\cdot 3\cdot 4,}$
that is
${\displaystyle \prod _{i=1}^{4}i=24.}$
The subscript gives the feckin' symbol for a holy bound variable (i in this case), called the "index of multiplication", together with its lower bound (1), whereas the bleedin' superscript (here 4) gives its upper bound. Whisht now and eist liom. The lower and upper bound are expressions denotin' integers. The factors of the feckin' product are obtained by takin' the feckin' expression followin' the bleedin' product operator, with successive integer values substituted for the oul' index of multiplication, startin' from the feckin' lower bound and incremented by 1 up to (and includin') the oul' upper bound. For example:
${\displaystyle \prod _{i=1}^{6}i=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot 6=720.}$
More generally, the feckin' notation is defined as
${\displaystyle \prod _{i=m}^{n}x_{i}=x_{m}\cdot x_{m+1}\cdot x_{m+2}\cdot \,\,\cdots \,\,\cdot x_{n-1}\cdot x_{n},}$
where m and n are integers or expressions that evaluate to integers, bejaysus. In the case where m = n, the bleedin' value of the feckin' product is the same as that of the feckin' single factor xm; if m > n, the feckin' product is an empty product whose value is 1—regardless of the bleedin' expression for the feckin' factors.
#### Properties of capital pi notation
By definition,
${\displaystyle \prod _{i=1}^{n}x_{i}=x_{1}\cdot x_{2}\cdot \ldots \cdot x_{n}.}$
If all factors are identical, a product of n factors is equivalent to exponentiation:
${\displaystyle \prod _{i=1}^{n}x=x\cdot x\cdot \ldots \cdot x=x^{n}.}$
Associativity and commutativity of multiplication imply
${\displaystyle \prod _{i=1}^{n}{x_{i}y_{i}}=\left(\prod _{i=1}^{n}x_{i}\right)\left(\prod _{i=1}^{n}y_{i}\right)}$ and
${\displaystyle \left(\prod _{i=1}^{n}x_{i}\right)^{a}=\prod _{i=1}^{n}x_{i}^{a}}$
if a is a bleedin' nonnegative integer, or if all ${\displaystyle x_{i}}$ are positive real numbers, and
${\displaystyle \prod _{i=1}^{n}x^{a_{i}}=x^{\sum _{i=1}^{n}a_{i}}}$
if all ${\displaystyle a_{i}}$ are nonnegative integers, or if x is a positive real number.
### Infinite products
One may also consider products of infinitely many terms; these are called infinite products. Arra' would ye listen to this shite? Notationally, this consists in replacin' n above by the Infinity symbol ∞. The product of such an infinite sequence is defined as the limit of the product of the first n terms, as n grows without bound, would ye swally that? That is,
${\displaystyle \prod _{i=m}^{\infty }x_{i}=\lim _{n\to \infty }\prod _{i=m}^{n}x_{i}.}$
One can similarly replace m with negative infinity, and define:
${\displaystyle \prod _{i=-\infty }^{\infty }x_{i}=\left(\lim _{m\to -\infty }\prod _{i=m}^{0}x_{i}\right)\cdot \left(\lim _{n\to \infty }\prod _{i=1}^{n}x_{i}\right),}$
provided both limits exist.[citation needed]
## Properties
Multiplication of numbers 0–10. Holy blatherin' Joseph, listen to this. Line labels = multiplicand. X-axis = multiplier. Y-axis = product.
Extension of this pattern into other quadrants gives the reason why a bleedin' negative number times an oul' negative number yields an oul' positive number.
Note also how multiplication by zero causes a holy reduction in dimensionality, as does multiplication by a bleedin' singular matrix where the feckin' determinant is 0, would ye swally that? In this process, information is lost and cannot be regained.
For real and complex numbers, which includes, for example, natural numbers, integers, and fractions, multiplication has certain properties:
Commutative property
The order in which two numbers are multiplied does not matter:
${\displaystyle x\cdot y=y\cdot x.}$[21][22][23]
Associative property
Expressions solely involvin' multiplication or addition are invariant with respect to the bleedin' order of operations:
${\displaystyle (x\cdot y)\cdot z=x\cdot (y\cdot z)}$[21][23]
Distributive property
Holds with respect to multiplication over addition. This identity is of prime importance in simplifyin' algebraic expressions:
${\displaystyle x\cdot (y+z)=x\cdot y+x\cdot z}$[21][23]
Identity element
The multiplicative identity is 1; anythin' multiplied by 1 is itself. Be the holy feck, this is a quare wan. This feature of 1 is known as the feckin' identity property:
${\displaystyle x\cdot 1=x}$[21][23]
Property of 0
Any number multiplied by 0 is 0. This is known as the zero property of multiplication:
${\displaystyle x\cdot 0=0}$[21]
Negation
−1 times any number is equal to the feckin' additive inverse of that number.
${\displaystyle (-1)\cdot x=(-x)}$ where ${\displaystyle (-x)+x=0}$
–1 times –1 is 1.
${\displaystyle (-1)\cdot (-1)=1}$
Inverse element
Every number x, except 0, has an oul' multiplicative inverse, ${\displaystyle {\frac {1}{x}}}$, such that ${\displaystyle x\cdot \left({\frac {1}{x}}\right)=1}$.[24]
Order preservation
Multiplication by a feckin' positive number preserves the oul' order:
For a > 0, if b > c then ab > ac.
Multiplication by a holy negative number reverses the oul' order:
For a < 0, if b > c then ab < ac.
The complex numbers do not have an orderin' that is compatible with both addition and multiplication.[25][26]
Other mathematical systems that include an oul' multiplication operation may not have all these properties, like. For example, multiplication is not, in general, commutative for matrices and quaternions.[21]
## Axioms
In the feckin' book Arithmetices principia, nova methodo exposita, Giuseppe Peano proposed axioms for arithmetic based on his axioms for natural numbers.[27] Peano arithmetic has two axioms for multiplication:
${\displaystyle x\times 0=0}$
${\displaystyle x\times S(y)=(x\times y)+x}$
Here S(y) represents the oul' successor of y; i.e., the oul' natural number that follows y. The various properties like associativity can be proved from these and the feckin' other axioms of Peano arithmetic, includin' induction. For instance, S(0), denoted by 1, is a bleedin' multiplicative identity because
${\displaystyle x\times 1=x\times S(0)=(x\times 0)+x=0+x=x.}$
The axioms for integers typically define them as equivalence classes of ordered pairs of natural numbers, the cute hoor. The model is based on treatin' (x,y) as equivalent to xy when x and y are treated as integers. Jesus, Mary and holy Saint Joseph. Thus both (0,1) and (1,2) are equivalent to −1, the hoor. The multiplication axiom for integers defined this way is
${\displaystyle (x_{p},\,x_{m})\times (y_{p},\,y_{m})=(x_{p}\times y_{p}+x_{m}\times y_{m},\;x_{p}\times y_{m}+x_{m}\times y_{p}).}$
The rule that −1 × −1 = 1 can then be deduced from
${\displaystyle (0,1)\times (0,1)=(0\times 0+1\times 1,\,0\times 1+1\times 0)=(1,0).}$
Multiplication is extended in a similar way to rational numbers and then to real numbers.[citation needed]
## Multiplication with set theory
The product of non-negative integers can be defined with set theory usin' cardinal numbers or the oul' Peano axioms. Here's a quare one for ye. See below how to extend this to multiplyin' arbitrary integers, and then arbitrary rational numbers. Me head is hurtin' with all this raidin'. The product of real numbers is defined in terms of products of rational numbers; see construction of the oul' real numbers.[citation needed]
## Multiplication in group theory
There are many sets that, under the operation of multiplication, satisfy the axioms that define group structure. These axioms are closure, associativity, and the oul' inclusion of an identity element and inverses.
A simple example is the set of non-zero rational numbers. Here we have identity 1, as opposed to groups under addition where the bleedin' identity is typically 0. Note that with the bleedin' rationals, we must exclude zero because, under multiplication, it does not have an inverse: there is no rational number that can be multiplied by zero to result in 1. Listen up now to this fierce wan. In this example, we have an abelian group, but that is not always the feckin' case.
To see this, consider the set of invertible square matrices of an oul' given dimension over a feckin' given field, to be sure. Here, it is straightforward to verify closure, associativity, and inclusion of identity (the identity matrix) and inverses. However, matrix multiplication is not commutative, which shows that this group is non-abelian.
Another fact worth noticin' is that the integers under multiplication do not form an oul' group—even if we exclude zero, for the craic. This is easily seen by the oul' nonexistence of an inverse for all elements other than 1 and −1.
Multiplication in group theory is typically notated either by an oul' dot or by juxtaposition (the omission of an operation symbol between elements). Sufferin' Jaysus listen to this. So multiplyin' element a by element b could be notated as a ${\displaystyle \cdot }$ b or ab. When referrin' to an oul' group via the oul' indication of the bleedin' set and operation, the oul' dot is used. For example, our first example could be indicated by ${\displaystyle \left(\mathbb {Q} /\{0\},\,\cdot \right)}$.[citation needed]
## Multiplication of different kinds of numbers
Numbers can count (3 apples), order (the 3rd apple), or measure (3.5 feet high); as the feckin' history of mathematics has progressed from countin' on our fingers to modellin' quantum mechanics, multiplication has been generalized to more complicated and abstract types of numbers, and to things that are not numbers (such as matrices) or do not look much like numbers (such as quaternions).
Integers
${\displaystyle N\times M}$ is the oul' sum of N copies of M when N and M are positive whole numbers. G'wan now and listen to this wan. This gives the bleedin' number of things in an array N wide and M high. Generalization to negative numbers can be done by
${\displaystyle N\times (-M)=(-N)\times M=-(N\times M)}$ and
${\displaystyle (-N)\times (-M)=N\times M}$
The same sign rules apply to rational and real numbers.[citation needed]
Rational numbers
Generalization to fractions ${\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}}$ is by multiplyin' the bleedin' numerators and denominators respectively: ${\displaystyle {\frac {A}{B}}\times {\frac {C}{D}}={\frac {(A\times C)}{(B\times D)}}}$. Would ye swally this in a minute now?This gives the bleedin' area of an oul' rectangle ${\displaystyle {\frac {A}{B}}}$ high and ${\displaystyle {\frac {C}{D}}}$ wide, and is the oul' same as the bleedin' number of things in an array when the bleedin' rational numbers happen to be whole numbers.[21][22]
Real numbers
Real numbers and their products can be defined in terms of sequences of rational numbers.
Complex numbers
Considerin' complex numbers ${\displaystyle z_{1}}$ and ${\displaystyle z_{2}}$ as ordered pairs of real numbers ${\displaystyle (a_{1},b_{1})}$ and ${\displaystyle (a_{2},b_{2})}$, the feckin' product ${\displaystyle z_{1}\times z_{2}}$ is ${\displaystyle (a_{1}\times a_{2}-b_{1}\times b_{2},a_{1}\times b_{2}+a_{2}\times b_{1})}$. Here's a quare one. This is the feckin' same as for reals ${\displaystyle a_{1}\times a_{2}}$ when the bleedin' imaginary parts ${\displaystyle b_{1}}$ and ${\displaystyle b_{2}}$ are zero.[citation needed]
Equivalently, denotin' ${\displaystyle {\sqrt {-1}}}$ as ${\displaystyle i}$, we have ${\displaystyle z_{1}\times z_{2}=(a_{1}+b_{1}i)(a_{2}+b_{2}i)=(a_{1}\times a_{2})+(a_{1}\times b_{2}i)+(b_{1}\times a_{2}i)+(b_{1}\times b_{2}i^{2})=(a_{1}a_{2}-b_{1}b_{2})+(a_{1}b_{2}+b_{1}a_{2})i.}$[21][22]
Alternatively, in trigonometric form, if ${\displaystyle z_{1}=r_{1}(\cos \phi _{1}+i\sin \phi _{1}),z_{2}=r_{2}(\cos \phi _{2}+i\sin \phi _{2})}$, then${\textstyle z_{1}z_{2}=r_{1}r_{2}(\cos(\phi _{1}+\phi _{2})+i\sin(\phi _{1}+\phi _{2})).}$[21]
Further generalizations
See Multiplication in group theory, above, and Multiplicative group, which for example includes matrix multiplication. Jaysis. A very general, and abstract, concept of multiplication is as the oul' "multiplicatively denoted" (second) binary operation in a feckin' rin'. An example of a holy rin' that is not any of the oul' above number systems is a bleedin' polynomial rin' (you can add and multiply polynomials, but polynomials are not numbers in any usual sense.)
Division
Often division, ${\displaystyle {\frac {x}{y}}}$, is the feckin' same as multiplication by an inverse, ${\displaystyle x\left({\frac {1}{y}}\right)}$. Multiplication for some types of "numbers" may have correspondin' division, without inverses; in an integral domain x may have no inverse "${\displaystyle {\frac {1}{x}}}$" but ${\displaystyle {\frac {x}{y}}}$ may be defined. Story? In a division rin' there are inverses, but ${\displaystyle {\frac {x}{y}}}$ may be ambiguous in non-commutative rings since ${\displaystyle x\left({\frac {1}{y}}\right)}$ need not be the same as ${\displaystyle \left({\frac {1}{y}}\right)x}$.[citation needed]
## Exponentiation
When multiplication is repeated, the oul' resultin' operation is known as exponentiation, you know yourself like. For instance, the feckin' product of three factors of two (2×2×2) is "two raised to the third power", and is denoted by 23, a two with a feckin' superscript three. Be the holy feck, this is a quare wan. In this example, the number two is the feckin' base, and three is the exponent.[28] In general, the feckin' exponent (or superscript) indicates how many times the oul' base appears in the oul' expression, so that the oul' expression
${\displaystyle a^{n}=\underbrace {a\times a\times \cdots \times a} _{n}}$
indicates that n copies of the base a are to be multiplied together. This notation can be used whenever multiplication is known to be power associative.[29]
## Notes
1. ^ a b c Devlin, Keith (January 2011). Be the holy feck, this is a quare wan. "What Exactly is Multiplication?". Mathematical Association of America. Archived from the feckin' original on May 27, 2017, would ye swally that? Retrieved May 14, 2017, to be sure. With multiplication you have an oul' multiplicand (written second) multiplied by a feckin' multiplier (written first)
2. ^ Khan Academy (2015-08-14), Intro to multiplication | Multiplication and division | Arithmetic | Khan Academy, archived from the original on 2017-03-24, retrieved 2017-03-07
3. ^ Khan Academy (2012-09-06), Why aren't we usin' the bleedin' multiplication sign? | Introduction to algebra | Algebra I | Khan Academy, archived from the feckin' original on 2017-03-27, retrieved 2017-03-07
4. ^ "Victory on Points". Nature, Lord bless us and save us. 218 (5137): 111. C'mere til I tell yiz. 1968, would ye believe it? Bibcode:1968Natur.218S.111.. Here's a quare one for ye. doi:10.1038/218111c0.
5. ^ "The Lancet – Formattin' guidelines for electronic submission of manuscripts" (PDF). Retrieved 2017-04-25.
6. ^ Announcin' the TI Programmable 88! (PDF). Chrisht Almighty. Texas Instruments. Chrisht Almighty. 1982, like. Archived (PDF) from the feckin' original on 2017-08-03. Retrieved 2017-08-03.
7. ^ Crewton Ramone, the cute hoor. "Multiplicand and Multiplier", what? Crewton Ramone's House of Math, that's fierce now what? Archived from the original on 26 October 2015, bejaysus. Retrieved 10 November 2015..
8. ^ Chester Litvin (2012). C'mere til I tell ya. Advance Brain Stimulation by Psychoconduction, would ye believe it? pp. 2–3, 5–6. G'wan now and listen to this wan. ISBN 978-1-4669-0152-0 – via Google Book Search.
9. ^ "Multiplication". www.mathematische-basteleien.de, be the hokey! Retrieved 2022-03-15.
10. ^ Pletser, Vladimir (2012-04-04), bejaysus. "Does the feckin' Ishango Bone Indicate Knowledge of the Base 12? An Interpretation of an oul' Prehistoric Discovery, the feckin' First Mathematical Tool of Humankind". arXiv:1204.1019 [math.HO].
11. ^ "Peasant Multiplication", game ball! www.cut-the-knot.org. Jesus, Mary and Joseph. Retrieved 2021-12-29.
12. ^ Qiu, Jane (7 January 2014). "Ancient times table hidden in Chinese bamboo strips". Nature. C'mere til I tell ya. doi:10.1038/nature.2014.14482. S2CID 130132289. C'mere til I tell ya now. Archived from the feckin' original on 22 January 2014. Retrieved 22 January 2014.
13. ^ Fine, Henry B. (1907), enda story. The Number System of Algebra – Treated Theoretically and Historically (PDF) (2nd ed.). Arra' would ye listen to this shite? p. 90.
14. ^ Bernhard, Adrienne, Lord bless us and save us. "How modern mathematics emerged from an oul' lost Islamic library", that's fierce now what? www.bbc.com. Retrieved 2022-04-22.
15. ^ Harvey, David; van der Hoeven, Joris; Lecerf, Grégoire (2016), you know yerself. "Even faster integer multiplication". Holy blatherin' Joseph, listen to this. Journal of Complexity. Jesus, Mary and holy Saint Joseph. 36: 1–30, the shitehawk. arXiv:1407.3360. doi:10.1016/j.jco.2016.03.001, to be sure. ISSN 0885-064X. S2CID 205861906.
16. ^ David Harvey, Joris Van Der Hoeven (2019). Integer multiplication in time O(n log n) Archived 2019-04-08 at the oul' Wayback Machine
17. ^ Hartnett, Kevin (11 April 2019). Bejaysus this is a quare tale altogether. "Mathematicians Discover the bleedin' Perfect Way to Multiply". Quanta Magazine. Arra' would ye listen to this shite? Retrieved 2020-01-25.
18. ^ Klarreich, Erica. Be the hokey here's a quare wan. "Multiplication Hits the oul' Speed Limit". C'mere til I tell ya. cacm.acm.org. Arra' would ye listen to this shite? Archived from the original on 2020-10-31. Right so. Retrieved 2020-01-25.
19. ^ Weisstein, Eric W. Arra' would ye listen to this shite? "Product", what? mathworld.wolfram.com. Retrieved 2020-08-16.
20. ^ "Summation and Product Notation". math.illinoisstate.edu, what? Retrieved 2020-08-16.
21. "Multiplication - Encyclopedia of Mathematics". Me head is hurtin' with all this raidin'. encyclopediaofmath.org. Retrieved 2021-12-29.
22. ^ a b c "multiplication". Sure this is it. planetmath.org. Retrieved 2021-12-29.
23. ^ a b c d Biggs, Norman L. Story? (2002). Discrete Mathematics. Bejaysus. Oxford University Press. p. 25. Stop the lights! ISBN 978-0-19-871369-2.
24. ^ Weisstein, Eric W. Holy blatherin' Joseph, listen to this. "Multiplicative Inverse". mathworld.wolfram.com. Retrieved 2022-04-19.
25. ^ Angell, David, bedad. "ORDERING COMPLEX NUMBERS... Here's another quare one. NOT*" (PDF). web.maths.unsw.edu.au, the cute hoor. Retrieved 29 December 2021.{{cite web}}: CS1 maint: url-status (link)
26. ^ "Total orderin' on complex numbers". Jesus Mother of Chrisht almighty. Mathematics Stack Exchange. Arra' would ye listen to this. Retrieved 2021-12-29.
27. ^ "Peano arithmetic". PlanetMath, begorrah. Archived from the original on 2007-08-19. Retrieved 2007-06-03.
28. ^ Weisstein, Eric W. "Exponentiation". mathworld.wolfram.com. Here's a quare one. Retrieved 2021-12-29.
29. ^ "general associativity". planetmath.org. Be the hokey here's a quare wan. Retrieved 2021-12-29.
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# What is the maximum number of rectangular blocks measuring 3 inches by 2 inches by 1 inch that can be packed into a cube-shaped box whose interior measures 6inches on an edge?A. 24B. 8C. 30D. 36
Last updated date: 14th Jun 2024
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Hint: In order to solve this problem we will find the volume of both the blocks and assume that n number of boxes gets fitted into the big box then we will equate the volume of n number of boxes with the bigger volume to get the value of n. Doing this will solve your problem and will give you the right answer.
Complete step-by-step solution:
It is given that,
Dimensions of the Rectangular block are (l, b, h) = (3, 2, 1).
And that is the length of an edge of the cube-shaped box = 6.
To find the maximum number of such rectangular blocks that can be packed into the above cube-shaped box;
Let ′n′ be the maximum number of rectangular blocks.
The volume of ′n′ rectangular blocks
$\Rightarrow n ×l × b × h$.
$\Rightarrow n ×3 × 2 × 1$
$\Rightarrow 6 × n$
Volume of the above cube of side (s=6) = ${s^3}$ = ${6^3}$
Equating the volume of the above, we get
$\Rightarrow 6 × n =${6^3} \Rightarrow n = {6^2} \Rightarrow n = 36\$
Therefore, the maximum number of such rectangular blocks that can be packed into the above cube-shaped box is ′36′.
Note: When you get to solve such problems you need to know that a cube is a three-dimensional solid object bounded by six square faces, facets, or sides, with three meetings at each vertex whereas cuboid is a 3D shape. Cuboids have six faces, which form a convex polyhedron. Broadly, the faces of the cuboid can be any quadrilateral. Whenever you need to calculate the number of small things fitted in a big thing then you have to assume the variables and get the total volume and get the value of the variable to get the number. Doing this will give you the right answers.
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## FANDOM
1,183 Pages
A special right triangle is a right triangle with some regular feature that makes calculations on the triangle easier, or for which simple formulas exist. For example, a right triangle may have angles that form a simple ratio, such as 45-45-90. This is called an "angle based" right triangle. A "side based" right triangle is one in which the lengths of the sides form a whole number ratio, such as 3-4-5. Knowing the ratios of the angles or sides of these special right triangles allows one to quickly calculate various lengths in geometric problems without resorting to more advanced methods.
## Angle-based
"Angle-based" special right triangles are specified by the integer ratio of the angles of which the triangle is composed. The integer ratio of the angles of these triangles are such that the larger (right) angle equals the sum of the smaller angles: $m:n:(m+n)\,$. The side lengths are generally deduced from the basis of the unit circle or other geometric methods. This form is most interesting in that it may be used to rapidly reproduce the values of trigonometric functions for the angles 30°, 45°, & 60°.
### 45-45-90 triangle
Constructing the diagonal of a square results in a triangle whose three angles are in the ratio $1:1:2\,$. With the three angles adding up to 180° (π) the angles respectively measure 45° $(\frac{\pi}{4}),$ 45° $(\frac{\pi}{4}),$ and 90° $(\frac{\pi}{2}).$ The sides are in the ratio
$1:1:\sqrt{2}.\,$
A simple proof. Say you have such a triangle with legs a and b and hypotenuse c. Suppose that a = 1. Since two angles measure 45°, this is an isosceles triangle and we have b = 1. The fact that $c=\sqrt{2}$ follows immediately from the Pythagorean theorem.
### 30-60-90 triangle
This is a triangle whose three angles are in the ratio $1:2:3\,$, and respectively measure 30°, 60°, and 90°. Since this triangle is half of an equilateral triangle, some refer to this as the hemieq triangle. The designation 30-60-90 is not only cumbersome, it references the degree, an arbitrary division of angular measure. The sides are in the ratio $1-\sqrt3-2$.
The proof of this fact is clear using trigonometry. Although the geometric proof is less apparent, it is equally trivial:
Draw an equilateral triangle ABC with side length 2 and with point D as the midpoint of segment BC. Draw an altitude line from A to D. Then ABD is a 30-60-90 (Hemieq) triangle with hypotenuse of length 2, and base BD of length 1.
The fact that the remaining leg AD has length $\sqrt{3}$ follows immediately from the Pythagorean theorem.
## Side-based
All of the special side based right triangles possess angles which are not necessarily rational numbers, but whose sides are always of integer length and form a Pythagorean triple. They are most useful in that they may be easily remembered and any multiple of the sides produces the same relationship.
### Common Pythagorean triples
There are several Pythagorean triples which are very well known, including:
$3:4:5\,$
$5:12:13\,$
$6:8:10\,$ (a multiple of the 3:4:5 triple)
$8:15:17\,$
$7:24:25\,$
The smallest of these (and its multiples, 6:8:10, 9:12:15,...) is the only right triangle with edges in arithmetic progression. Triangles based on Pythagorean triplets are Heronian and therefore have integer area.
### Fibonacci triangles
Starting with 5, every other Fibonacci number {0,1,1,2,3,5,8,13,21,34,55,89,144, 233,377, 710,...} is the length of the hypotenuse of a right triangle with integral sides, or in other words, the largest number in a Pythagorean triple. The length of the longer leg of this triangle is equal to the sum of the three sides of the preceding triangle in this series of triangles, and the shorter leg is equal to the difference between the preceding bypassed Fibonacci number and the shorter leg of the preceding triangle.
The first triangle in this series has sides of length 5, 4, and 3. Skipping 8, the next triangle has sides of length 13, 12 (5 + 4 + 3), and 5 (8 − 3). Skipping 21, the next triangle has sides of length 34, 30 (13 + 12 + 5), and 16 (21 − 5). This series continues indefinitely and approaches a limiting triangle with edge ratios:
$\sqrt{5}:2:1$.
This right triangle is sometimes referred to as a dom, a name suggested by Andrew Clarke to stress that this is the triangle obtained from dissecting a domino along a diagonal. The dom forms the basis of the aperiodic pinwheel tiling proposed by John Conway and Charles Radin.
### Almost-isosceles Pythagorean triples
Isosceles right-angled triangles can not have sides with integer values. However, infinitely many almost-isosceles right triangles do exist. These are right-angled triangles with integral sides for which the lengths of the non-hypotenuse edges differ by one.[1] Such almost-isosceles right-angled triangles can be obtained recursively using Pell's equation:
a0 = 1, b0 = 2
an = 2bn-1 + an-1
bn = 2an + bn-1
an is length of hypotenuse, n=1, 2, 3,... . The smallest Pythagorean triples resulting are:
$3:4:5\,$
$20:21:29\,$
$119:120:169\,$
$696:697:985\,$
## Calculating common trig functions
Special triangles are used to aid in calculating common trig functions, as below:
30 $\frac{\pi}{6}$ $\frac{1}{2}$ $\frac{\sqrt3}{2}$ $\frac{\sqrt3}{3}$
45 $\frac{\pi}{4}$ $\frac{\sqrt2}{2}$ $\frac{\sqrt2}{2}$ 1
60 $\frac{\pi}{3}$ $\frac{\sqrt3}{2}$ $\frac{1}{2}$ $\sqrt3$
90 $\frac{\pi}{2}$ 1 0 -
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The trick to solving Cyclic Quadrilateral MCQs Quiz is to understand its properties and then practice applying them to Cyclic Quadrilateral objective questions. Here is a collection of questions for the practice of candidates preparing the Mensuration Topic for competitive exams. Mensuration is a commonly featured Topic in most competitive exams and hence candidates should be well prepared for it. Therefore, we also have a few tips and tricks for candidates to save time and increase the accuracy of Cyclic Quadrilateral question answers.
The length of two parallel sides of a trapezium are 53 cm and 68 cm respectively, and the distance between the parallel sides is 16 cm. Find the area of the trapezium.
1. 968 cm2
2. 972 cm2
3. 988 cm2
4. 1024 cm2
Option 1 : 968 cm2
Detailed Solution
Area of the Trapezium = 1/2 × (Sum of the parallel sides) × (Distance between parallel sides)
⇒ 1/2 × (53 + 68) × 16
⇒ 1/2 × 121 × 16
∴ Area of the Trapezium = 968 cm2
PQRS is a cyclic quadrilateral in which PQ = x cm, QR = 16.8 cm, RS = 14 cm, PS = 25.2 cm, and PR bisects QS. What is the value of x?
1. 24
2. 21
3. 28
4. 18
Option 2 : 21
Detailed Solution
From the following figure
As we know,
ΔPOQ∼ΔSOR
PQ/SR = OQ/OR
OQ/OR = x/14
Given, OQ = OS
OS/OR = x/14 --- (1)
As we know,
ΔPOS∼ΔQOR
OS/OR = PS/QR
OS/OR = 25.2/16.8 --- (2)
From equation (1) and equation (2)
x/14 = 25.2/16.8
x = (25.2 × 14)/16.8
∴ x = 21 cm
ABCD is a cyclic quadrilateral in which AB = 16.5 cm, BC = x cm, CD = 11 cm, AD = 19.8 cm, BD is bisected by AC at O. What is the value of x?
1. 13.8 cm
2. 12.4 cm
3. 12.8 cm
4. 13.2 cm
Option 4 : 13.2 cm
Detailed Solution
From the following figure
∠ACD = ∠ABD (∵ Angle made by a chord in same segment are equal)
∠BDC = ∠BAC (∵ Angle made by a chord in same segment are equal)
So, ΔAOB ∼ ΔDOC by AA.
AB/DC = OB/OC
⇒ OB/OC = 16.5/11
Given, OB = OD
OD/OC = 16.5/11 ...1)
As we know,
⇒ OD/OC = 19.8/x ...2)
From equation (1) and equation (2)
19.8/x = 16.5/11
⇒ x = (11 × 19.8)/16.5
⇒ x = 13.2 cm
The diagonals of a rectangle are inclined to one side of the rectangle at 25°. The acute angle formed between the diagonals is:
1. 25°
2. 50°
3. 55°
4. 40°
Option 2 : 50°
Detailed Solution
Figure:
Calculation:
As the diagonals of a rectangle intersect each other,
⇒ AO = OB
⇒ ∠OBA = ∠OAB = 25° [∵ Angle opposite to equal side are equal]
By angle sum property in ΔAOB,
⇒ ∠AOB + ∠OAB + ∠OBA = 180°
⇒ ∠AOB + 25° + 25° = 180°
⇒ ∠AOB = 130°
By linear pair property,
⇒ ∠DOA + ∠AOB = 180°
⇒ ∠DOA + 130° = 180°
⇒ ∠DOA = 50°
∴ Both diagonals make 50° angle with each other.
BART is a cyclic quadrilateral. Diagonals BR and AT meet at P. If ∠BPA = 110° and ∠RAT = 30°, then ∠BTA measures
1. 80°
2. 70°
3. 30°
4. 55°
Option 1 : 80°
Detailed Solution
∠BPT = 180° – 110° = 70°
⇒ ∠TAR = ∠TBR = 30° ( Angles made at the circumference by same arc are equal)
In ΔBPT,
⇒ ∠BTA = 180° – (70° + 30°)
= 180° – 100°
= 80°
In the following figure, if angles ∠ABC = 95∘ ∠FED = 115∘ (not to scale). Then the angle ∠APC is equal to?:
1. 120∘
2. 150∘
3. 155
4. 135∘
Option 2 : 150∘
Detailed Solution
Given:
∠ABC = 95° and ∠FED = 115°
Formula:
The sum of the opposite angle of cycle quadrilateral is 180°.
Calculation:
Join A to F
∠ABC + ∠AFC = 180
⇒ 95 + ∠AFC = 180
⇒ ∠AFC = 180 - 95
⇒ ∠AFC = 85
⇒ ∠FAD + 115 = 180
⇒ ∠FAD = 180 - 115
In ΔAFP
⇒ ∠AFP + ∠FAP + ∠FPA = 180
⇒ 85 + 65 + ∠FPA = 180
⇒ ∠FPA = 180 - 150
⇒ ∠FPA = 30
∠APC + ∠FPA = 180 (liner pair)
⇒ ∠APC + 30 = 180
⇒ ∠APC = 180 - 30
∴ ∠APC = 150
A cyclic quadrilateral ABCD is such that AB = BC, AD = DC and AC, BD intersect at O and BD is angle bisector of ∠B. If ∠CAD = 46°, then the measure of ∠AOB is equal to:
1. 84°
2. 86°
3. 80°
4. 90°
Option 4 : 90°
Detailed Solution
Given -
A cyclic quadrilateral ABCD is such that AB = BC, AD = DC and AC and BD intersect at O and ∠CAD = 46°
Concept used -
∠A + ∠C = ∠B + ∠D = 180°
Solution -
∠ADC = 180° - 92° = 88°
∠ABC = 180° - 88° = 92°
AB = BC
⇒ ∠BAC = ∠BCA = 88°/2 = 44°
BD is angle bisector of ∠B.
⇒ ∠ABO = 46°
⇒ ∠AOB = 180° - (46° + 44°) = 90°
∴ ∠AOB is 90°.
ABCD is a cyclic quadrilateral in which sides AD and BC are produced to meet at P, and sides DC and AB meet at Q when produced. If ∠A = 60° and ∠ABC = 72°, then ∠DPC –∠BQC = ?
1. 30°
2. 36°
3. 24°
4. 40°
Option 2 : 36°
Detailed Solution
∠A = 60° and ∠ABC = 72°
In ΔAPB
∠PAB + ∠ABP + ∠APB = 180°
∠APB = 180° – 60° – 72° = 48°
∠APB = ∠DPC = 48°
As we know, In cyclic quadrilateral, sum of opposite angles are 180°.
∠ADC = 180° – 72° = 108°
∠DAQ + ∠ADQ + ∠AQD = 180°
∠AQD = 180° – 60° – 108° = 12°
Difference between ∠DPC and ∠BQC = 48° – 12° = 36°
PQRS is a cyclic quadrilateral diagonals PR and QS intersect each other at O. If ∠POS = 112° and ∠PRS = 40°. Then, what is the measure of ∠QPR?
1. 75°
2. 72°
3. 64°
4. 68°
Option 2 : 72°
Detailed Solution
GIVEN:
PQRS is a cyclic quadrilateral diagonals PR and QS intersect each other at O.
∠POS = 112° and ∠PRS = 40°
CONCEPT USED:
Inscribed angles subtended by the same arc are equal.
An exterior angle of a triangle is equal to the sum of the two opposite interior angles.
⇒ a° + c° = d°
CALCULATION:
We know that –
Inscribed angles subtended by the same arc are equal.
∠PQS = ∠PRS = 40°
Now, In ΔPOQ,
∠QPO + ∠PQO + ∠POQ = 180°
⇒ ∠QPO + ∠PQO + (180° – ∠POS) = 180°
⇒ ∠QPO + 40° + (180° – 112°) = 180°
⇒ ∠QPO = 112° – 40°
⇒ ∠QPO = 72°
⇒ ∠QPO = ∠QPR = 72°
∴ The measure of ∠QPR is 72°.
In the figure given below, ABCD is a cyclic quadrilateral, O is the center of the circle and PT is a tangent at point C. If AB = BC, ∠PCD = 55° and ∠ADO = 25°, then what is the measure of ∠BAD?
1. 85°
2. 80°
3. 75°
4. 90°
Option 1 : 85°
Detailed Solution
∠PCD = ∠CAD = 55° (From alternate segment theorem)
∠OCD = ∠ODC = 90° – 55° = 35° (OC = OD)
∠ADC = 25° + 35° = 60°
⇒ ∠ABC = 180° – 60° = 120°
∠BAC = ∠BCA (AB = BC)
In ∆ABC:
∠BAC + ∠BCA + ∠ABC = 180°
2∠BAC + 120° = 180°
∠BAC = (180° – 120°)/2 = 30°
Now,
∠BAD = 55° + 30° = 85°
In the given figure, ABCD is a cyclic quadrilateral where diagonals intersect at P such that ∠DBC is 50° and ∠BAC is 40°. Find the value of ∠BCD
1. 40°
2. 50°
3. 45°
4. 90°
Option 4 : 90°
Detailed Solution
Given:
∠DBC = 50°
∠BAC = 40°
Concept used:
Angles in the same segment will be equal.
ABCD is a cyclic quadrilateral, so the sum of a pair of two opposite angles will be 180°.
∠A + ∠C = 180°
∠B + ∠D = 180°
Calculations:
⇒ ∠DAC = ∠DBC (Angles in the same segment will be equal.)
⇒ ∠DAC = 50°
In ΔBCD,
⇒ ∠BCD + ∠DBC + ∠BDC = 180°
⇒ ∠BCD + 50° + 40° = 180°
⇒ ∠BCD = 90°
∴ The value of ∠BCD is 90°.
Which of the following can not be a measure of an angle of a cyclic quadrilateral?
1. 90°
2. 150°
3. 170°
4. 180°
Option 4 : 180°
Detailed Solution
Concept Used:
The angles of a cyclic quadrilateral are Supplementary.
Calculation:
The sum of measures of opposite angles = 180°
⇒ The sum of two angles is 180°
⇒ Every angle of a cyclic quadrilateral need to be less than 180°
∴ One angle of a cyclic quadrilateral cannot be 180°.
Sides AB and DC of a cyclic quadrilateral ABCD are produced to meet at E, and sides AD and BC are produced to meet at F. ∠ADC = 75°, and ∠BEC = 52°, then the difference between ∠BAD and ∠AFB is:
1. 21°
2. 31°
3. 22°
4. 23°
Option 2 : 31°
Detailed Solution
∠ADC = 75° and ∠BEC = 52°
As we know, In a cyclic quadrilateral, the sum of opposite angles are 180°.
⇒ ∠ABC = 180° – 75° = 105°
⇒ ∠ABC + ∠CBE = 180° Δ [straight line]
⇒ ∠CBE = 180° – 105° = 75°
In ΔBEC
∠CBE + ∠BEC + ∠ECB = 180°
⇒ ∠ECB = 180° °– 75° – 52° °= 53°
⇒ ∠ECB + ∠BCD = 180° [straight line]
⇒ ∠BCD = 180° – 53° = 127°
⇒ ∠BAD + ∠BCD = 180°
⇒ ∠BAD = 180° – 127° = 53°
In ΔAFB
∠BAF + ∠ABF + ∠AFB = 180°
⇒ ∠AFB = 180° – 53° – 105° = 22°
∴ Difference between ∠BAD and ∠AFB = 53° – 22° = 31°
PQRS is a cyclic quadrilateral. If ∠P is thrice ∠R and ∠S is 5 times ∠Q, what is the sum of ∠Q and ∠R?
1. 75°
2. 65°
3. 72°
4. 70°
Option 1 : 75°
Detailed Solution
As we know
⇒ ∠P + ∠R = 180° [∵ ∠P : ∠R = 3 : 1]
⇒ 4 ∠R = 180°
⇒ ∠R = 45°
Similarly,
⇒ ∠S + ∠Q = 180° [∵ ∠S : ∠Q = 5 : 1]
⇒ 6∠Q = 180°
⇒ ∠Q = 30°
⇒ ∠R + ∠Q = 45° + 30° = 75°
ABCD is a cyclic trapezium with AB || DC and AB is the diameter of the circle. If ∠CAB = 30° then ∠ADC is –
1. 40°
2. 70°
3. 120°
4. 150°
Option 3 : 120°
Detailed Solution
Given:
∠CAB = 30°
AB || DC and AB is the diameter of the circle
Concept used:
The angle in a semi-circle is 90°.
Opposite angles of a cyclic quadrilateral are supplementary.
Calculation:
ΔABC is right angled triangle (Angle in a semi-circle)
⇒ ∠ABC = 90° - ∠CAB = 90° - 30° = 60°
Now,
∵ ABCD is a cyclic quadrilateral,
Opposite angles of cyclic quadrilateral are supplementary.
⇒ ∠ABC + ∠ADC = 180°
⇒ 60° + ∠ADC = 180°
A circle touches all the sides of a quadrilateral PQRS, whose sides, PQ = 2 cm, QR = 3 cm and RS = 4 cm. What is the length of PS?
1. 3 cm
2. 2 cm
3. 1 cm
4. 4 cm
Option 1 : 3 cm
Detailed Solution
In the figure,
⇒ PW = PZ (∵ Tangents drawn from an external point to a circle are of same length)
Suppose, PW = PZ = x cm
Similarly,
⇒ QW = QX
⇒ RX = RY
⇒ SY = SZ
∵ PW = x cm,
⇒ QW = PQ – PW = (2 – x) cm
Now, QW = QX = (2 – x) cm
∵ QX = (2 – x) cm,
⇒ RX = QR – QX = 3 – (2 – x) = (1 + x) cm
Now, RX = RY = (1 + x) cm
∵ RY = (1 + x) cm,
⇒ SY = RS – RY = 4 – (1 + x) = (3 – x) cm
Now, SY = SZ = (3 – x) cm
∴ Length of PS = PZ + SZ = x + (3 – x) = 3 cm
PQRS is a cyclic quadrilateral in which PQ = 14.4 cm, QR = 12.8 cm and SR = 9.6 cm. If PR bisects QS, what is the length of PS?
1. 16.4 cm
2. 13.6 cm
3. 19.2 cm
4. 15.8 cm
Option 3 : 19.2 cm
Detailed Solution
From the following figure
As we know,
ΔPOQ ∼ ΔSOR
⇒ PQ/SR = OQ/OR
⇒ OQ/OR = 14.4/9.6
⇒ OQ/OR = 3/2
Given, OQ = OS
⇒ OS/OR = 3/2 ----(1)
As we know,
ΔPQO ∼ ΔSRO
⇒ OS/OR = PS/QR
⇒ OS/OR = PS/12.8 ----(2)
From equation (1) and equation (2)
⇒ PS/12.8 = 3/2
⇒ PS = (12.8 × 3)/2
⇒ PS = 19.2 cm
Theorem: (remember this result):
If either of the diagonals of a cyclic quadrilateral bisects the other diagonal, then the opposite side of the quadrilateral are in the same ratio, see the figure,
PS/QR = PQ/SR
⇒ PS/12.8 = 14.4/9.6
⇒ PS = 19.2 cm
ABCD is a cycle quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 158°. Then ∠BAC is equal to:
1. 40°
2. 38°
3. 50°
4. 68°
Option 4 : 68°
Detailed Solution
We know that in cyclic quadrilateral the sum of opposite angle = 180°
⇒ ∠ABC + ∠ADC = 180°
⇒ ∠ABC + 158° = 180°
⇒ ∠ABC = 180° − 158° = 22°
In ΔABC
We know that angle made by diameter is 90°
∴ ∠ACB = 90°
Now, ∠BAC + ∠ABC + ∠ACB = 180°
⇒ ∠BAC + 22° + 90° = 180°
∴ ∠BAC = 180° − 112° = 68°
In the given figure, AB is the diameter of the given circle. If ∠DAC ∶ ∠DCA = 3 ∶ 1 and ∠CAB = 30°, then find the value of ∠DAC.
1. 15°
2. 30°
3. 45°
4. 60°
Option 3 : 45°
Detailed Solution
Given:
∠DAC ∶ ∠DCA = 3 ∶ 1
∠CAB = 30°
Concept used:
The sum of opposite angles of a cyclic quadrilateral = 180°
Sum of all the angles of a triangle = 180°
If a triangle is inscribed inside a semicircle such that diameter is one of its sides, then the given triangle is a right-angled triangle
Calculations:
Let the ∠DAC and ∠DCA be 3x and x respectively
ΔABC is a right-angled triangle
⇒ ∠ACB = 90°
Sum of all the angles of the ΔABC = ∠ACB + ∠CBA + ∠CAB
⇒ 90° + ∠CBA + 30° = 180°
⇒ ∠CBA = 60°
⇒ ∠ADC + 60° = 180°
Sum of all the angles of the ΔADC = ∠ADC + ∠DCA + ∠DAC
⇒ 120° + x + 3x = 180°
⇒ 4x = 60°
⇒ x = 15°
∠DAC = 3x
⇒ 3 × 15°
⇒ 45°
∴ The measure of ∠DAC is 45°
1. πr2
2. 2r
3. (r + 2)
4. πr2/2
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# Author Archives: duran216
## SLOPE and Y-INTERCEPT
### PROPORTIONAL KAHOOT:
https://play.kahoot.it/#/k/f5f48c8d-2ea4-451c-ad57-8eb0fa1c2728
### y = mx + b
m = Slope (how steep the line is)
b = the Y Intercept (where the line crosses the Y axis)\
### How do you find “m” and “b“?
• b is easy: just see where the line crosses the Y axis.
• m (the Slope) needs some calculation. Remember we like to write slope like a fraction.
## Example 1)
The fastest and easiest thing to find first when looking at a graph is the y-intercept (b).
Here we see that the line crosses the y-axis at positive 1.
So, b = 1 .
Now to find the slope (m) we will use rise over run:
• Rise is how far up
• Run is how far along
In this example the rise is 2 and the run is 1.
So, m = 2/1 .
Now that I know m = 2/1 and b = 1 I can plug them into the equation for slope intercept form y = mx + b.
y = (2/1) x + 1
^^^^^ This is the equation of the line.
For more information on y = mx + b click here.
# Practice Problems
Get your Slope Intercept on!
# SLOPE
Standard
Slope is about how steep a line is.
When finding the slope from a line on a graph we use the method of rise over run .
• Rise is how far up
• Run is how far along
You can also think of it as the change in y over the change in x.
EXAMPLE #1:
The slope here is 4/6 which can simplify to 2/3.
EXAMPLE #2:
In this example the slope is 3/5.
For more information on slope click here.
# Practice Problems
Get your SLOPE on!
TEST YOUR SELF ON LINES HERE!!!!
# Y – INTERCEPT
Standard
## Y Intercept
Where a straight line crosses the Y axis of a graph.
### Example:
In the above diagram the line crosses the Y axis at 1.
So the Y intercept is equal to 1.
For more information on intercepts click here.
## Practice Problems
Get your Y-intercept on!
TEST YOUR SELF ON LINES HERE!!!!
# Standard Form of a Linear Equation:
The “Standard Form” for writing down a Linear Equation is
Ax + By = C
A shouldn’t be negative, A and B shouldn’t both be zero, and A, B and C should be integers.
### y = 3x + 2
Bring 3x to the left:
-3x + y = 2
Multiply all by -1:
3x – y = -2
Note: A=3, B=-1, C=-2
This form:
Ax + By + C = 0
is sometimes called “Standard Form”, but is more properly called the “General Form”.
ALGEBRA REVIEW KAHOOT:
https://play.kahoot.it/#/?quizId=8af8cbf5-78da-4760-8b40-ccc70cd54a6e
# Slope Kahoot:
https://play.kahoot.it/#/k/31ab2ba7-803c-45a4-b504-ccc96ce3ea17
# Y-Intercept Kahoot:
https://play.kahoot.it/#/k/e4aa37b7-7aac-4f4c-9d8c-4fe790b1c78d
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# What is a Rectangular Fraction
Let us know about What is a Rectangular Fraction. The rectangular fraction model is one of the more practical ways of representing fractions. We start with a rectangle that represents the whole sum, and divide it into equal parts . Each part is a unit fraction.
Also, what does the power of 10 to 4 mean?
Example: 10 4 = 10 × 10 × 10 × 10 = 10,000 .
Here, what is 4 to the power of 4?
Answer: The value of 4 to 4 power i.e. 4 4 is 256 . Let us calculate the value from 4 to the 4th power, that is, 4 4 . Thus, 4 4 can be written as 4 × 4 × 4 × 4 = 256.
Also to know what is 3 to the power of 4? Answer: The value of 4 to 3rd power i.e. 4 3 is 64 .
What is the 10th to 15th power called?
positive forces
## o What is 3 of the power?
The power of 3 is 3 the same as 0 . extended to . x. In y , 3 is the base (x) and 0 is the exponent (y). Therefore, you can solve problem 3. can also be written as 0 . You will also get the answer of 0 to the power of 3 (0 to 3rd power) if you type 3 then x y then 0 and then = on your scientific calculator.
### How do you type to the power of 4?
How to Type the Fourth Power Symbol
1. Type number 4. ,
2. Highlight number 4.
3. Format the font and change it to “Superscript”. On the “Format” menu, choose “Font” and change the style to “Superscript” by choosing an option from the menu or clicking the Superscript check box.
#### What does 3 to the third power mean?
When a number is to the ‘third power’, it means that you are multiplying the number by itself three times.
#### What is the highest power of 100 out of 10!?
We know that to make a 2 we need a 5 and a 10. is 24. If you saw the highest power of 100 in 10! is equal to 5. to the highest power of .
#### What does the power of 10 to 8 mean?
108 = 100,000,000 . (ten crore) 10-7=0.0000001. (one millionth)
#### What is the largest number?
Despite there being more than the number of atoms in the universe, trying to prove that your integer is bigger than someone else’s integer has been going on for centuries. NS The largest regular number referenced is a googolplex (10 googol ), which is 10. Works as 10 ^ 100 .
#### What is 4 to the power of negative 3 as a fraction?
Minus 3 = 4. To the power of 3 – 4 = / 1 is 81 .
To stick to 4 to the power of negative 3 as an example, put 3 for the base and enter -4 as the index, aka the exponent, or power.
#### How do you write 3 to the power of 5?
Answer: On raising the value of 5 to the power of 3, 5 . happens 3 = 125 .
#### What is 0 to the exponent of 3?
Answer 3: Any number to the power of zero always gives one . x a * x – a = x a * 1/x a : This means that any number x 0 = 1.
#### What is the symbol of pi?
pi, in mathematics, is the ratio of the circumference of a circle to its diameter. The symbol to represent the ratio was devised in 1706 by the British mathematician William Jones and later popularized by the Swiss mathematician Leonhard Euler.
#### Is the cube 3 or 4?
learn cube numbers
#### What does symbol 4 mean?
“From almost prehistoric times, the number four was employed to signify what was concrete , that which could be touched and felt. Its association with the cross (the four dots) made it a classic symbol of wholeness and universality. , a symbol that attracted everyone towards him.
#### How do you symbolize the third power?
1) Press the “Alt” key on your keyboard, and don’t let go. 2) Hold down “Alt” on your keyboard, type the number “252” , which is the number of the letter or symbol “³” in the ASCII table.
#### What does the third power mean in mathematics?
In mathematics, an expression to the third power means raising a number or expression to an exponent of 3 or 3 .
#### What number is 120% of 20?
Answer: 120% of 20 is 24 .
#### What is the highest power of 100 in a 15 factorial?
So 15 out of 100 at the highest power! is 24 and the correct option is B.
#### What does the 10th to 8th power look like?
An exponent is the number of times to use a number in a multiplication. Therefore, the 10th to the 8th power is 100,000,000. This is solved by the equation 10 x 10 x 10 x 10 x 10 x 10 x 10 x 10 .
#### What is the factorial of 10?
What is the factorial of 10? The value of the factorial of 10 is 3628800 , i.e. 10! = 10 × 9 × 8 × 7 × 6 × 5 × 4 × 3 × 2 × 1 = 3628800.
#### What does 10 mean?
In angel numerology, ten represents the magic number , and in many cultures and religions, it is considered very happy, and the number ten is destined to make a person happy. Tens signify the end of the old and the beginning of a new, happy future, harmony in all its aspects.
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## Can probabilities be displayed in a pie graph?
Probability in Pie This means that if we randomly select an observation from the data used to create the pie chart, the probability of it taking on a specific value is exactly equal to the size of that category’s slice in the pie chart.
## What do pie graphs represent?
Pie charts can be used to show percentages of a whole, and represents percentages at a set point in time. Unlike bar graphs and line graphs, pie charts do not show changes over time. The following pages describe the different parts of a pie chart.
## What is a pie chart used for?
A pie chart shows how some total amount is divided among distinct categories as a circle (the namesake pie) divided into radial slices. Each category is associated with a single slice whose size corresponds with the category’s proportion of the total.
## How do you explain a pie chart in a presentation?
How do you describe a pie chart? A pie chart divides data into separate sections to show which individual parts make up the whole. To describe the chart, compare each “slice” of the chart to the others to determine what share of the total each category has.
## What are the 3 ways in presenting data?
In this article, the techniques of data and information presentation in textual, tabular, and graphical forms are introduced.
## Why are pie charts good for percentages?
Pie charts work best for values around 25%, 50% or 75%. It’s easier for readers to spot these percentages in a pie chart than in a stacked bar or column chart. Pie charts are not the best choice if you want readers to compare the size of shares. This is especially true if the differences between the shares are small.
## How do I manually create a pie chart?
The total value of the pie chart is always 100%. Each portion in the circle shows a fraction or percentage of the total. Pie chart is a circular graph which is used to represent data….Construction of Pie Chart.
Activity No. of Hours Measure of central angle
Study 4 (4/24 × 360)° = 60°
T. V. 1 (1/24 × 360)° = 15°
Others 3 (3/24 × 360)° = 45°
## How do you solve a pie chart problem?
You can solve Pie chart problems using the properties of a circle and the basic concept of percentages. A whole circle contains 360 degrees. In a pie chart, these 360 degrees corresponds to the total of the values represented in the chart.
## How do you put percentages into a pie chart?
To display percentage values as labels on a pie chart
2. On the design surface, right-click on the pie and select Show Data Labels.
3. On the design surface, right-click on the labels and select Series Label Properties.
4. Type #PERCENT for the Label data option.
## How do you show percentages on a pie chart in R?
Pie chart in R with percentage Note that the round function allows you to modify the number of decimals. An alternative to display percentages on the pie chart is to use the PieChart function of the lessR package, that shows the percentages in the middle of the slices.
## Does a pie chart have to equal 100?
Pie charts are designed to show parts of a whole, so any sum below or above 100% doesn’t represent the entire picture. Finally, when it comes to legends, pie charts don’t generally need one.
## What is a limitation of using a pie charts?
Disadvantages of a Pie Chart If too many pieces of data are used, pie chart becomes less effective. They themselves may become crowded and hard to read if there are too many pieces of data, and even if you add data labels and numbers may not help here.
## How many slices should a pie chart have?
How to make pie charts look better
• Don’t use more than five sections. Too many skinny slices are hard to read.
• Place the largest slices from “12” at the top (like on a clock) and work your way around the circle. Like this:
• Avoid comparing one pie chart to another.
• Don’t use 3-D pie charts.
## When should you avoid a pie chart Doughnut chart?
You should avoid using pie charts when: Your data values are not distinctly separated; data analysis using a pie chart becomes restricted when dealing with data points of similar sizes. You need to compare data for more than one metric. You need to showcase specific data values and facilitate a part-to-part comparison.
## How do you read a pie graph?
To interpret a pie chart, compare groups.
1. When you interpret one pie chart, look for differences in the size of the slices.
2. When you compare multiple pie charts, look for differences in the size of slices for the same categories in all the pie charts.
## Which are the attributes of a pie chart?
A pie chart (or a circle chart) is a circular statistical graphic, which is divided into slices to illustrate numerical proportion. In a pie chart, the arc length of each slice (and consequently its central angle and area), is proportional to the quantity it represents.
## What makes a pie chart unique?
Pie Chart vs Bar Chart Bar charts are easier to read when you’re comparing categories or looking at change over time. The only thing bar charts lack is the whole-part relationship that makes pie charts unique. Pie charts imply that if one wedge gets bigger, the other has to be smaller.
## Which of the following is a major advantage of pie chart?
Advantages of pie chart: summarize a large data set into visual form. pie charts permit a visual check of the reasonableness or accuracy of the calculation. pie charts are visually simpler than other types of graphs. size of the circle can be made proportional to the quantity it represents.
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# HLT 362V Week 5 Assignment Workbook Exercise 14 and 19, 23 and 24, 29 and 35
## HLT 362V Week 5 Assignment Workbook Exercise 14 and 19, 23 and 24, 29 and 35
### HLT 362V Week 5 Assignment Workbook Exercise 14 and 19, 23 and 24, 29 and 35
Click here to ORDER an A++ paper from our Verified MASTERS and DOCTORATE WRITERS:HLT 362V Week 5 Assignment Workbook Exercise 14 and 19, 23 and 24, 29 and 35
Exercise # 14: Understanding Simple Linear Regression
According to the study narrative and Figure 1 in the Flannigan et al. (2014) study, does the APLS UK formulae under- or overestimate the weight of children younger than 1 year of age? Provide a rationale for your answer.
Using the values a = 3.161 and b = 0.502 with the novel formula in Figure 1, what is the predicted weight in kilograms (kg) for a child at 9 months of age? Show your calculations.
Using the values a = 3.161 and b = 0.502 with the novel formula in Figure 1, what is the predicted weight in kilograms for a child at 2 months of age? Show your calculations.
In Figure 2, the formula for calculating y (weight in kg) is Weight in kg = (0.176 × age in months) + 7.241. Identify the y intercept and the slope in this formula.
Using the values a = 7.241 and b = 0.176 with the novel formula in Figure 2, what is the predicted weight in kilograms for a child 3 years of age? Show your calculations.
Using the values a = 7.241 and b = 0.176 with the novel formula in Figure 2, what is the predicted weight in kilograms for a child 5 years of age? Show your calculations.
In Figure 3, some of the actual mean weights represented by the blue line with squares are above the dotted straight line for the novel formula, but others are below the straight line. Is this an expected finding? Provide a rationale for your answer.
In Figure 3, the novel formula is (Weight in kilograms = (0.331 × Age in months) – 6.868. What is the predicted weight in kilograms for a child 10 years old? Show your calculations.
Was the sample size of this study adequate for conducting simple linear regression? Provide a rationale for your answer.
Describe one potential clinical advantage and one potential clinical problem with using the three novel formulas presented in Figures 1, 2, and 3 in a PICU setting.
Exercise # 19: Understanding Pearson Chi-Square
According to the relevant study results section of the Darling-Fisher et al. (2014) study, what categories are reported to be statistically significant?
What level of measurement is appropriate for calculating the χ2 statistic? Give two exam¬ples from Table 2 of demographic variables measured at the level appropriate for χ2.
What is the χ2 for U.S. practice region? Is the χ2 value statistically significant? Provide a rationale for your answer. X2= 29.68; p= <.00
What is the df for provider type? Provide a rationale for why the df for provider type pre¬sented in Table 2 is correct.
Is there a statistically significant difference for practice setting between the Rapid Assessment for Adolescent Preventive Services (RAAPS) users and nonusers? Provide a rationale for your answer.
State the null hypothesis for provider age in years for RAAPS users and RAAPS nonusers.
Should the null hypothesis for provider age in years developed for Question 6 be accepted or rejected? Provide a rationale for your answer.
Describe at least one clinical advantage and one clinical challenge of using RAAPS as described by Darling-Fisher et al. (2014).
How many null hypotheses are rejected in the Darling-Fisher et al. (2014) study for the results presented in Table 2? Provide a rationale for your answer.
A statistically significant difference is present between RAAPS users and RAAPS nonusers for U.S. practice region, χ2 = 29.68. Does the χ2 result provide the location of the difference? Provide a rationale for your answer.
Exercise # 23
What is the r value for the relationship between Hamstring strength index 60°/s and the Shuttle run test? Is this r value significant? Provide a rationale for your answer.
Consider r = 1.00 and r = -1.00. Which r value is stronger? Provide a rationale for your answer.
Describe the direction of the relationship between the Hamstring strength index 60°/s and the Shuttle run test.
Without using numbers, describe the relationship between the Hamstring strength index 120°/s and the Triple hop index.
Which variable has the weakest relationship with the Quadriceps strength index 120°/s? Provide a rationale for your answer.
Which of the following sets of variables has the strongest relationship?
a. Hamstring strength index 120°/s and the Hop index
b. Quadriceps strength index 60°/s and the Carioca test
c. Quadriceps strength index 120°/s and the Side step test
d. Quadriceps strength index 60°/s and the Triple hop index
In Table 5, two r values are reported as r = -0.498 and r = -0.528.
Describe each r value in words, indicating which would be more statistically significant, and provide a rationale for your answer.
The researchers stated that the study showed a positive, significant correlation between Quadriceps strength indices and pre- and postoperative functional stability. Considering the data presented in the Table 5, do you agree with their statement? Provide a rationale for your answer.
The researchers stated that no significant relationship could be described between Hamstring strength indices 60°/s and functional stability. Given the data in Table 5, explain why not.
Consider the relationship reported for the Quadriceps strength index 120°/s and the Hop index (r = 0.744**, p = 0.000 ). What do these r and p values indicate related to statistical significance and clinical importance?
Exercise # 24
What is the r value listed for the relationship between variables 4 and 9?
Describe the correlation r = -0.32** using words. Is this a statistically significant correlation? Provide a rationale for your answer.
Calculate the percentage of variance explained for r = 0.53. Is this correlation clinically important? Provide a rationale for your answer.
According to Table 2, r = 0.15 is listed as the correlation between which two items? Describe this relationship. What is the effect size for this relationship, and what size sample would be needed to detect this relationship in future studies?
Calculate the percentage of variance explained for r = 0.15. Describe the clinical importance of this relationship.
Which two variables in Table 2, have the weakest correlation, or r value? Which relationship is the closest to this r value? Provide a rationale for your answer.
Is the correlation between LOT-R Total scores and Avoidance-Distraction coping style statistically significant? Is this relationship relevant to practice? Provide rationales for your answers.
Is the correlation between variables 9 and 4 significant? Is this correlation relevant to practice? Provide a rationale for your answer.
Consider two values, r = 0.08 and r = -0.58. Describe them in relationship to each other. Describe the clinical importance of both r values.
Examine the Pearson r values for LOT-R Total, which measured Optimism with the Task and Emotion Coping Styles. What do these results indicate? How might you use this information in your practice?
BONUS QUESTION
One of the study goals was to examine the relationship between optimism and psychopathology. Using the data in Table 2, formulate an opinion regarding the overall correlation between optimism and psychopathology. Provide a rationale for your answer.
Exercise # 29: Calculating Simple Linear Regression
If you have access to SPSS, compute the Shapiro-Wilk test of normality for the variable age (as demonstrated in Exercise 26). If you do not have access to SPSS, plot the frequency distributions by hand. What do the results indicate?
State the null hypothesis where age at enrollment is used to predict the time for comple¬tion of an RN to BSN program.
What is b as computed by hand (or using SPSS)?
What is a as computed by hand (or using SPSS)?
Write the new regression equation.
How would you characterize the magnitude of the obtained R2 value? Provide a rationale for your answer.
How much variance in months to RN to BSN program completion is explained by knowing the student’s enrollment age?
What was the correlation between the actual y values and the predicted y values using the new regression equation in the example?
Write your interpretation of the results as you would in an APA-formatted journal.
Given the results of your analyses, would you use the calculated regression equation to predict future students’ program completion time by using enrollment age as x? Provide a rationale for your answer.
Exercise # 35: Calculating Pearson Chi-Square
Do the example data in Table 35-2 meet the assumptions for the Pearson χ2 test? Provide a rationale for your answer.
2. Compute the χ2 test. What is the χ2 value? 11.931
Is the χ2 significant at α =0.05? Specify how you arrived at your answer.
If using SPSS, what is the exact likelihood of obtaining the χ2 value at least as extreme as or as close to the one that was actually observed, assuming that the null hypothesis is true?
Using the numbers in the contingency table, calculate the percentage of antibiotic users who tested positive for candiduria.
Using the numbers in the contingency table, calculate the percentage of non-antibiotic users who tested negative for candiduria.
Using the numbers in the contingency table, calculate the percentage of veterans with candiduria who had a history of antibiotic use.
Using the numbers in the contingency table, calculate the percentage of veterans with candiduria who had no history of antibiotic use.
Write your interpretation of the results as you would in an APA-formatted journal.
Was the sample size adequate to detect differences between the two groups in this example? Provide a rationale for your answer.
• Click here to ORDER an A++ paper from our Verified MASTERS and DOCTORATE WRITERS:HLT 362V Week 5 Assignment Workbook Exercise 14 and 19, 23 and 24, 29 and 35
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# How do you solve (r+13)/12=1?
Jul 23, 2017
$r = - 1$
#### Explanation:
$\text{multiply both sides of the equation by 12 the}$
$\text{denominator of the fraction}$
${\cancel{12}}^{1} \times \frac{r + 13}{\cancel{12}} ^ 1 = 12 \times 1$
$\Rightarrow r + 13 = 12$
$\text{subtract 13 from both sides}$
$r \cancel{+ 13} \cancel{- 13} = 12 - 13$
$\Rightarrow r = - 1$
$\textcolor{b l u e}{\text{As a check}}$
substitute this value into the left side of the equation and if equal to the right side then it is the solution.
$\frac{- 1 + 13}{12} = \frac{12}{12} = 1 = \text{ right side}$
$\Rightarrow r = - 1 \text{ is the solution}$
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# Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line Exercise 14(E)
The last exercise has problems covering all the concepts discussed in this chapter. In order to get a strong overall understanding of this chapter, Selina Solutions for Class 10 Maths is the right tool. Students can use this for any quick references or doubt clearance, as per their convenience. The answers of this exercise can be accessed in the Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line Exercise 14(E) PDF, in the link given below.
## Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line Exercise 14(E) Download PDF
### Access other exercises of Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line
Exercise 14(A) Solutions
Exercise 14(B) Solutions
Exercise 14(C) Solutions
Exercise 14(D) Solutions
### Access Selina Solutions Concise Maths Class 10 Chapter 14 Equation of a Line Exercise 14(E)
#### Exercise 14(E) Page No: 202
1. Point P divides the line segment joining the points A (8, 0) and B (16, -8) in the ratio 3: 5. Find its co-ordinates of point P.
Also, find the equation of the line through P and parallel to 3x + 5y = 7.
Solution:
Given points, A (8, 0) and B (16, -8)
By section formula, the co-ordinates of the point P which divides AB in the ratio 3: 5 is given by
= (11, -3) = (x1, y1)
Given line equation is,
3x + 5y = 7
5y = -3x + 7
y = (-3/5) x + 7/5
So, the slope of this line = -3/5
The line parallel to the line 3x + 5y = 7 will have the same slope,
Hence, the slope of the required line = Slope of the given line = -3/5
Thus,
The equation of the required line is
y – y1 = m(x – x1)
y + 3 = (-3/5)(x – 11)
5y + 15 = -3x + 33
3x + 5y = 18
2. The line segment joining the points A(3, -4) and B (-2, 1) is divided in the ratio 1: 3 at point P in it. Find the co-ordinates of P. Also, find the equation of the line through P and perpendicular to the line 5x – 3y + 4 = 0.
Solution:
Given points, A (3, -4) and B (-2, 1)
By section formula, the co-ordinates of the point P which divides AB in the ratio 1: 3 is given by
= (7/4, -11/4) = (x1, y1)
Given line equation is,
5x – 3y + 4 = 0
3y = 5x + 4
y = (5/3) x + 4/3
So, the slope of this line = 5/3
The line perpendicular to the given line will have slope
Slope of the required line = -1/(5/3) = -3/5
Hence,
The equation of the required line is given by
y – y1 = m(x – x1)
20y + 55 = -12x + 21
12x + 20y + 34 = 0
6x + 10y + 17 = 0
3. A line 5x + 3y + 15 = 0 meets y-axis at point P. Find the co-ordinates of point P. Find the equation of a line through P and perpendicular to x – 3y + 4 = 0.
Solution:
As the point P lies on y-axis,
Putting x = 0 in the equation 5x + 3y + 15 = 0, we get
5(0) + 3y + 15 = 0
y = -5
Hence, the co-ordinates of the point P are (0, -5).
Given line equation,
x – 3y + 4 = 0
3y = x + 4
y = (1/3)x + 4/3
Slope of this line = 1/3
From the question, the required line equation is perpendicular to the given equation: x – 3y + 4 = 0.
So, the product of their slopes is -1.
Slope of the required line = -1/(1/3) = -3
And,
(x1, y1) = (0, -5)
Therefore,
The required line equation is
y – y1 = m(x – x1)
y + 5 = -3(x – 0)
3x + y + 5 = 0
4. Find the value of k for which the lines kx – 5y + 4 = 0 and 5x – 2y + 5 = 0 are perpendicular to each other.
Solution:
Given,
kx – 5y + 4 = 0
5y = kx + 4
⇒ y = (k/5)x + 4/5
So, the slope of this line = m1 = k/5
And, for 5x – 2y + 5 = 0
⇒ 2y = 5x + 5
y = (5/2) x + 5/2
Slope of this line = m2 = 5/2
As, the lines are perpendicular to each other
m1 x m2 = -1
(k/5) x (5/2) = -1
k = -2
5. A straight line passes through the points P (-1, 4) and Q (5, -2). It intersects the co-ordinate axes at points A and B. M is the mid-point of the segment AB. Find:
(i) the equation of the line.
(ii) the co-ordinates of A and B.
(iii) the co-ordinates of M.
Solution:
(i) Given points, P (-1, 4) and Q (5, -2)
Slope of PQ = (-2 – 4)/ (5 + 1) = -6/6 = -1
Equation of the line PQ is given by,
y – y1 = m(x – x1)
y – 4 = -1(x + 1)
y – 4 = -x – 1
x + y = 3
(ii) For point A (on x-axis), y = 0.
So, putting y = 0 in the equation of PQ, we have
x = 3
Hence, the co-ordinates of point A are (3, 0).
For point B (on y-axis), x = 0.
So, putting x = 0 in the equation of PQ, we have
y = 3
Hence, the co-ordinates of point B are (0, 3).
(iii) M is the mid-point of AB.
Thus, the co-ordinates of point M are
(3+0/2, 0+3/2) = (3/2, 3/2)
6. (1, 5) and (-3, -1) are the co-ordinates of vertices A and C respectively of rhombus ABCD. Find the equations of the diagonals AC and BD.
Solution:
Given, A = (1, 5) and C = (-3, -1) of rhombus ABCD.
We know that in a rhombus, diagonals bisect each other at right angle.
Let’s take O to be the point of intersection of the diagonals AC and BD.
Then, the co-ordinates of O are
Slope of AC = (-1 – 5)/ (-3 – 1) = -6/-4 = 3/2
Then, the equation of the line AC is
y – y1 = m (x – x1)
y – 5 = (3/2) (x – 1)
2y – 10 = 3x – 3
3x – 2y + 7 = 0
Now, the line BD is perpendicular to AC
Slope of BD = -1/(slope of AC) = -2/3
And, (x1, y1) = (-1, 2)
Hence, equation of the line BD is
y – y1 = m (x – x1)
y – 2 = (-2/3) (x + 1)
3y – 6 = -2x – 2
2x + 3y = 4
7. Show that A (3, 2), B (6, -2) and C (2, -5) can be the vertices of a square.
(i) Find the co-ordinates of its fourth vertex D, if ABCD is a square.
(ii) Without using the co-ordinates of vertex D, find the equation of side AD of the square and also the equation of diagonal BD.
Solution:
Given, A (3, 2), B (6, -2) and C (2, -5)
Now, by distance formula
AB = √[(6 – 3)2 + (-2 – 2)2] = √(9 + 16) = 5
BC = √[(2 – 6)2 + (-5 + 2)2] = √(16 + 9) = 5
Thus, AC = BC
Then,
Slope of AB = (-2 – 2)/ (6 – 3) = -4/3
Slope of BC = (-5 + 2)/ (2 – 6) = -3/-4 = ¾
Slope of AB x Slope of BC = -4/3 x ¾ = -1
Hence, AB ⊥ BC
Therefore, A, B, C can be the vertices of a square.
(i) Slope of AB = (-2 – 2)/ (6 – 3) = -4/3 = slope of CD
So, the equation of CD is
y – y1 = m (x – x1)
y + 5 = -4/3(x – 2)
3y + 15 = -4x + 8
3y = -4x – 7
4x + 3y + 7 = 0 … (1)
Now, slope of BC = (-5 + 2)/ (2 – 6) = -3/-4 = ¾ = Slope of AD
So, the equation of the line AD is
y – y1 = m (x – x1)
y – 2 = (3/4) (x – 3)
4y – 8 = 3x – 9
3x – 4y = 1 …. (2)
Now, D is the point of intersection of CD and AD.
Solving (1) and (2),
4 x (1) + 3 x (2) ⇒
16x + 12y + 9x – 12y = -28 + 3
25x = -25
x = -1
Putting value of x in (1), we get
4(-1) + 3y + 7 = 0
3y = -3
y = -1
Therefore, the co-ordinates of point D are (-1, -1).
(ii) From the equation (2)
The equation of the line AD is,
3x – 4y = 1
Slope of BD = (-1 + 2) / (-1 – 6) = (1 / -7) = (-1 / 7)
The equation of the diagonal BD is
y – y1 = m (x – x1)
⇒y + 1 = -1 / 7 (x + 1)
⇒7y + 7 = – x – 1
⇒x + 7y + 8 = 0
8. A line through origin meets the line x = 3y + 2 at right angles at point X. Find the co-ordinates of X.
Solution:
The given line equation is
x = 3y + 2 … (1)
3y = x – 2
y = 1/3 x – 2/3
So, slope of this line is 1/3.
And, the required line intersects the given line at right angle.
Thus, slope of the required line = -1/(1/3) = -3
And. the required line passes through (0, 0) = (x1, y1)
So, the equation of the required line is:
y – y1 = m(x – x1)
y – 0 = -3(x – 0)
3x + y = 0 … (2)
Next,
Point X is the intersection of the lines (1) and (2).
Using (1) in (2), we get,
3(3y + 2) + y = 0
9y + 6 + y = 0
10y = -6
y = -6/10 = -3/5
And, finally
x = 3(-3/5) + 2 = -9/5 + 2 = 1/5
Thus, the co-ordinates of the point X are (1/5, -3/5).
9. A straight line passes through the point (3, 2) and the portion of this line, intercepted between the positive axes, is bisected at this point. Find the equation of the line.
Solution:
Let the line intersect the x-axis at point A (x, 0) and y-axis at point B (0, y).
Since, P is the mid-point of AB, we have:
(x/2, y/2) = (3, 2)
x = 6, y = 4
Thus, A = (6, 0) and B = (0, 4)
Slope of line AB = (4 – 0)/ (0 – 6) = 4/-6 = -2/3
And, let (x1, y1) = (6, 0)
So, the required equation of the line AB is given by
y – y1 = m(x – x1)
y – 0 = (-2/3) (x – 6)
3y = -2x + 12
2x + 3y = 12
10. Find the equation of the line passing through the point of intersection of 7x + 6y = 71 and 5x – 8y = -23; and perpendicular to the line 4x – 2y = 1.
Solution:
Given line equations are,
7x + 6y = 71 ⇒ 28x + 24 = 284 … (1)
5x – 8y = -23 ⇒ 15x – 24y = -69 … (2)
On adding (1) and (2), we have
43x = 215
x = 5
From (2), we get
8y = 5x + 23 = 25 + 23 = 48
⇒ y = 6
Hence, the required line passes through the point (5, 6).
Given, 4x – 2y = 1
2y = 4x – 1
y = 2x – (1/2)
So, the slope of this line = 2
And, the slope of the required line = -1/2 [As the lines are perpendicular to each other]
Thus, the required equation of the line is
y – y1 = m(x – x1)
y – 6 = (-1/2) (x – 5)
2y – 12 = -x + 5
x + 2y = 17
11. Find the equation of the line which is perpendicular to the line x/a – y/b = 1 at the point where this line meets y-axis.
Solution:
The given line equation is,
x/a – y/b = 1
y/b = x/a – 1
y = (b/a)x – b
The slope of this line = b/a
So, the slope of the required line = -1/(b/a) = – a/b
Let the line intersect at point P (0, y) on the y-axis.
So, putting x = 0 in the equation x/a – y/b = 1, we get
0 – y/b = 1
y = -b
Hence, P = (0, -b) = (x1, y1)
Therefore,
The equation of the required line is
y – y1 = m (x – x1)
y + b = (-a/b) (x – 0)
by + b2 = -ax
ax + by + b2 = 0
12. O (0, 0), A (3, 5) and B (-5, -3) are the vertices of triangle OAB. Find:
(i) the equation of median of triangle OAB through vertex O.
(ii) the equation of altitude of triangle OAB through vertex B.
Solution:
(i) Let’s consider the median through O meets AB at D. So, D will be the mid-point of AB.
Co-ordinates of point D are:
The slope of OD = (1 – 0)/ (-1 – 0) = -1
And, (x1, y1) = (0, 0)
Hence, the equation of the median OD is
y – y1 = m(x – x1)
y – 0 = -1(x – 0)
x + y = 0
(ii) The altitude through vertex B is perpendicular to OA.
We have, slope of OA = (5 – 0)/ (3 – 0) = 5/3
Then, the slope of the required altitude = -1/(5/3) = -3/5
Hence, the equation of the required altitude through B is
y – y1 = m(x – x1)
y + 3 = (-3/5) (x + 5)
5y + 15 = -3x – 15
3x + 5y + 30 = 0
13. Determine whether the line through points (-2, 3) and (4, 1) is perpendicular to the line 3x = y + 1.
Does the line 3x = y + 1 bisect the line segment joining the two given points?
Solution:
Let A = (-2, 3) and B = (4, 1)
Slope of AB = m1 = (1 – 3)/(4 + 2) = -2/6 = -1/3
So, the equation of line AB is
y – y1 = m1(x – x1)
y – 3 = (-1/3) (x + 2)
3y – 9 = -x – 2
x + 3y = 7 … (1)
Slope of the given line 3x = y + 1 is 3 = m2.
It’s seen that, m1 x m2 = -1
Thus, the line through points A and B is perpendicular to the given line.
Given line is 3x = y +1 … (2)
The co-ordinates of the mid-point of AB are
Now, Let’s check if point P satisfies the line equation (2)
3(1) = 2 + 1
3 = 3
Hence, the line 3x = y + 1 bisects the line segment joining the points A and B.
14. Given a straight line x cos 30o + y sin 30o = 2. Determine the equation of the other line which is parallel to it and passes through (4, 3).
Solution:
Given line equation, x cos 30o + y sin 30o = 2
So, the slope of this line = -√3
Slope of a line which is parallel to this given line = -√3
Let (4, 3) = (x1, y1)
Therefore, the equation of the required line is given by
y – y1 = m1 (x – x1)
y – 3 = -√3 (x – 4)
√3x + y = 4√3 + 3
15. Find the value of k such that the line (k – 2)x + (k + 3)y – 5 = 0 is:
(i) perpendicular to the line 2x – y + 7 = 0
(ii) parallel to it.
Solution:
Given line equation,
(k – 2)x + (k + 3)y – 5 = 0 …. (1)
(k + 3)y = -(k – 2)x + 5
y =
Slope of this line = m1 = (2 – k)/ (k + 3)
(i) Given, 2x – y + 7 = 0
y = 2x + 7 = 0
Slope of this line = m2 = 2
Given that, line (1) is perpendicular to 2x – y + 7 = 0
m1 x m2 = -1
(2 – k)/ (k + 3) x 2 = -1
4 – 2k = -k – 3
k = 7
(ii) Line (1) is parallel to 2x – y + 7 = 0
So, m1 = m2
(2 – k)/ (k + 3) = 2
2 – k = 2k + 6
3k = -4
k = -4/3
16. The vertices of a triangle ABC are A (0, 5), B (-1, -2) and C (11, 7). Write down the equation of BC. Find:
(i) the equation of line through A and perpendicular to BC.
(ii) the co-ordinates of the point, where the perpendicular through A, as obtained in (i), meets BC.
Solution:
Slope of BC = (7 + 2)/(11 +1) = 9/12 = 3/4
Then the equation of the line BC is
y – y1 = m(x – x1)
y + 2 = ¾ (x + 1) where x1 = -1 and y1 = -2
4y + 8 = 3x + 3
3x – 4y = 5 …. (1)
(i) Slope of line perpendicular to BC will be = -1/(3/4) = -4/3
So, the required equation of the line through A (0, 5) and perpendicular to BC is given by
y – y1 = m1 (x – x1)
y – 5 = (-4/3) (x – 0)
3y – 15 = -4x
4x + 3y = 15 …. (2)
(ii) Hence, the required point will be the point of intersection of lines (1) and (2).
Solving (1) & (2),
(1) ⇒ 9x – 12y = 15
(2) ⇒ 16x + 12y = 60
Now, adding the above two equations, we get
25x = 75
x = 3
Substituting the value of x in equation (1) we get,
4y = 3x – 5 = 9 – 5 = 4
y = 1
Thus, the co-ordinates of the required point is (3, 1).
17. From the given figure, find:
(i) the co-ordinates of A, B and C.
(ii) the equation of the line through A and parallel to BC.
Solution:
(i) A = (2, 3), B = (-1, 2), C = (3, 0)
(ii) Slope of BC = (0 – 2)/ (3 + 1) = -2/4 = -1/2
Slope of the line which is parallel to BC = Slope of BC = -1/2
(x1, y1) = (2, 3)
Hence, the required equation of the line through A and parallel to BC is given by
y – y1 = m1 (x – x1)
y – 3 = (-1/2) (x – 2)
2y – 6 = -x + 2
x + 2y = 8
18. P (3, 4), Q (7, -2) and R (-2, -1) are the vertices of triangle PQR. Write down the equation of the median of the triangle through R.
Solution:
We know that, the median, RX through R will bisect the line PQ.
The co-ordinates of point X are
Slope of RX = (1 + 1)/ (5 + 2) = 2/7 = m1
(x1, y1) = (-2, -1)
Then, the required equation of the median RX is given by
y – y1 = m1(x – x1)
y + 1 = (2/7) (x + 2)
7y + 7 = 2x + 4
7y = 2x – 3
19. A (8, -6), B (-4, 2) and C (0, -10) are vertices of a triangle ABC. If P is the mid-point of AB and Q is the mid-point of AC, use co-ordinate geometry to show that PQ is parallel to BC. Give a special name of quadrilateral PBCQ.
Solution:
P is the mid-point of AB. Hence, the co-ordinates of point P are
Q is the mid-point of AC. So, the co-ordinate of point Q are
Slope of PQ = (-8 + 2)/ (4 – 2) = -6/2 = -3
Slope of BC = (-10 – 2)/ (0 + 4) = -12/4 = -3
As, the slope of PQ = Slope of BC,
Therefore, PQ || BC
Also,
Slope of PB = (-2 – 2)/ (2 + 4) = -2/3
Slope of QC = (-8 + 10)/ (4 – 0) = 1/2
So, PB is not parallel to QC as their slopes are not equal
Thus, PBCQ is a trapezium.
20. A line AB meets the x-axis at point A and y-axis at point B. The point P (-4, -2) divides the line segment AB internally such that AP: PB = 1: 2. Find:
(i) the co-ordinates of A and B.
(ii) the equation of line through P and perpendicular to AB.
Solution:
(i) Let’s assume the co-ordinates of point A, lying on x-axis be (x, 0) and the co-ordinates of point B (lying on y-axis) be (0, y).
Given,
P = (-4, -2) and AP: PB = 1:2
By section formula, we get
-4 = 2x/3 and -2 = y/3
x = -6 and y = -6
Hence, the co-ordinates of A and B are (-6, 0) and (0, -6).
(ii) Slope of AB = (-6 – 0)/ (0 + 6) = -6/6 = -1
Slope of the required line perpendicular to AB = -1/-1 = 1
Here, (x1, y1) = (-4, -2)
Therefore, the required equation of the line passing through P and perpendicular to AB is given by
y – y1 = m(x – x1)
y + 2 = 1(x + 4)
y + 2 = x + 4
y = x + 2
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# Lcm Of 12 And 15
The smallest numbers of common multiples for two numbers is called their least common multiple.LCM is almost GCF evaluation but for greatest number here we are finding lowest common multiple. There are two methods for finding LCM which are multiplication method and division method.LCM for given numbers is used in fraction additions and simplifications.The product of greatest common diviosrs (GCD) and Least common multiple (LCM) of two numbers is equal to the product of given numbers. the least common multiple (LCM) when two integers a and b is a smallest positive integer which is multiple of both of a and b. which can be divided by a and b without a remainder. If a or b will be 0, then there will be no such positive integer, then LCM for (ab) is zero.
## Methods For Finding Least Common Muliples:
Multiplication Method :
Step 1: Finding the multiples of first number
Step 2: Finding the multiples of second number
Step 3: Note the common multiples for both numbers given.
Step 4: LCM for numbers given are least common multiples
Division method:
Step 1: Dividing given numbers by common multiple
Step 2: Dividing till get zero or no divisor
Step 3: multiplying all factor then we get LCM
## Example Problems for LCM:
Example 1 :
Find the l.c.m. of , 12 and 15.
Solution :
Multiples of 12 : 12, 24 , 36, 48 , 60, 72 , 84
Multiples of 15: 15,30,45,60,75,90
Common multiples factor: 60
Lowest common multiples: 60
Example 2:
Find the lowest common multiple of 12 and 15.
Solution:
3¦ 12,15 (common divisor 3)
4¦ 4,5 ( no common divisor 4)
5¦ 1,5(no common divisor 5)
1,1
l.c.m. = 3×4×5 = 60
Example 3:
Find lowest common multiple of 12
Solution:
2 ¦12 (common divisor 2)
2 ¦6 (common divisor 2)
3 ¦3 (common divisor 3)
1 (No common divisor)
l.c.m. = 2 × 2 × 3 = 12.
Example 3:
Find lowest common multiple of 15
Solution:
3 ¦15 (common divisor 3)
5 ¦5 ( no common divisor 5)
1
l.c.m. = 3× 5 = 15.
Example 4:
Solution:
For adding two unlike fractions we take LCM for 12, 15 are 60
`5/12 +2/15` = `(5*5)/(12*5) +(2*4)/(15*4)` (multiply denominator with number which get LCM)
=`25/60+8/60`
=`33/60`
Example 5:
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# A boat takes 3 hours to travel 30 km down a river, then 5 hours to return. How fast is the river flowing?
• The average time of the trip, which cancels out the current, is 4 hours, so the boat's motor is pushing it at 7.5 km/h. You can figure out the rest.
• Here's what we know:
distance = rate X time. Let s = the speed of the river, r = the speed of the boat relative to the river. Then
30 Km = (r + s) X 3 [The speed down-river is the boat's speed plus the river's speed]
and
30 Km = (r - s) X 5 [The speed up-river is the boat's speed minus the river's speed]
Since 30 Km = 30 Km, (r + s) X 3 = (r - s) X 5
Solving for s:
3r + 3s = 5r - 5s
8s = 2r
s = r/4
Putting this back in the first equation,
30 Km = (r + r/4) X 3.
Solving for r:
15r/4 = 30 Km
So r = 8 Km/hr (speed of the boat) and
s = 2 Km/hr (speed of the river)
Source(s): high school algebra
• 3x30 = 90km.
90/5=18km speed while returning.
30-18=12
River helps boat while going and slows it down while returning. Therefore, 12/2=6km change is the on boat's speed. Therefore, river's flowing speed is 6km.
• Fast enough to slow it by 2 hrs
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Introduction
Calculus is the study of the following concepts.
1. Limits
2. Continuity
3. Derivatives
4. Integrals
If you understand these four core concepts then you understand calculus. The remaining parts of calculus are just theorems and applications of these four concepts. In this article we are going to focus on these core concepts. The goal of the article is to give a basic description of these core concepts and how they fit together.
Numbers
There are different types of numbers. A number system is a set of a certain kind of number. Examples of number systems are natural numbers, integers, rational numbers, real numbers and complex numbers.
In introductory calculus we work with real numbers. Real numbers is just a fancy name for the number system that you are used to working with. It is possible to have a calculus that works with complex numbers. Therefore we usually specify that we are working with real numbers.
Functions
You are probably already familiar with the concept of a function. A function is a rule that associates an element of a set with elements of another set. Usually the sets are the sets of real numbers but they don't have to be. A real function is a function that associates a real number with a real number. A complex function is a function that associates a complex number with a complex number. When it comes to introductory calculus we are only concerned with real functions. Therefore from this point the terms function and real function will be used interchangeably.
A function is usually denoted by f(x). Here f(x) is the value that the function associates with the number x. A function does not have to be defined for all values of x. For example the function 1/x is not defined when x is 0.
A graph of a function is the set of all points (x, f(x)). You have probably already seen a graph of a function.
Absolute value
A number can be positive or negative. The absolute value of a number |x| is equal to the numerical value of x disregarding the sign of the number. Perhaps an example will shed more light. The absolute value of 42 or |42| is equal to 42. The absolute value of -13 or |-13| is equal to 13.
The absolute value of a number is the distance between the number and 0. The absolute value of a number can be thought of as the magnitude of the number. The absolute value is used when formally defining the concept of a limit.
Limit
A limit is one of the core concepts of analysis. To understand and in fact to be able to define the concepts of continuity, derivatives and integrals you have to understand limits. Before we can define limits we first need to introduce two new symbols. Traditionally these are the lower case greek letters delta (δ) and epsilon (ε).
Intuitively a limit of a function f(x) at a point x0 is the value that f(x) approaches if x gets arbitrarily close to x0 without x actually landing on x0. Here is the formal definition of a limit.
The function f(x) approaches the value L at x0 if for every ε > 0 there exists a number δ > 0 such that |f(x0 + δ) - L| < ε.
The above definition can be a bit tricky to wrap your head around. I suggest finding a math book that goes over the concept of limits and that has examples that you can do yourself. Remember that to be able to understand the rest of calculus you have to understand limits.
Limits have a special notation. The limit of the function f(x) as x approaches c is written in the following way.
Continuity
Continuity is a core concept in calculus. Intuitively it is easy to see if a function is continuous or not. A function is continuous if it is possible to draw the graph of a function without lifting the pencil. To be able to mathematically define the concept of continuity you have to understand the concept of a limit.
We can formally define the concept of continuity in the following way.
A function f(x) is continuous at a point c if the limit of f(x) as x approaches c exists and that the limit is equal to f(x). Or using mathematical notation the function f(x) is continuous at the point c if the following holds.
Slope of a line
Before we look at derivatives it is good to review how the slope of a line is calculated. Imagine that we have a line between the points (x1, y1) and (x2,y2). Let k be the slope of the line between the two points. Then the slope of the line is calculated in the following manner.
Next assume that we have a function f(x). Let x1 and x2 be two points on the x axis. Then (x1, f(x1)) and (x2, f(x2)) are two points that lie on the graph of f(x). It is now possible to calculate the slope of the line between the points f(x1)) and (x2, f(x2)). The slope of the line is calculated in the following manner.
Let x0 be a point on the x axis. Let h be a number that is not equal to 0. Then x0 and (x0 + h) are two different numbers. We can now again calculate the slope of the line between the points (x0, f(x0)) and (x0 + h, f(x0 + h)).
This last equation is important because it shows up in the definition of the derivative.
Derivaties
The derivative is one of the core concepts of calculus. Consider the following limit.
Simplifying the denominator we get the following limit.
If the limit exists then the function f(x) is said to be differentiable at the point x0 and the derivative of the function f(x) at the point x0 is equal to the value of the limit.
Since the derivative is one of the core concepts in it has it's own notation. The derivative of the function f(x) at the point x can be written as f'(x), Df(x) or df/dx.
It is possible to show that a function that is differentiable at a point is continuous at that same point. Therefore a function that is differentiable will also be continuous.
The derivative f'(x) of the function f(x) is also a function. Sometimes it is possible to take the derivative of f'(x). This is called the second derivative of f(x) and is denoted by f''(x). The nth derivative of f(x) is denoted by f(n)(x).
The derivative has a geometric interpretation. The derivative of a function f(x) at a point is the slope of the tangent line to the function f(x) at that point.In fact the tangent line of the function f(x) at the point x is defined as the line that goes through the point (x, f(x)) and whose slope is equal to f'(x).
Primitive functions
A function F(x) is said to be a primitive function of f(x) if the derivative of F(x) is equal to f(x). If F(x) is a primitive function to f(x) then F(x) + C will also be a primitive function to f(x). Primitive functions are important due to the fundamental theorem of analysis which we will come to later.
Step functions
Before we can introduce the concept of an integral we need to introduce step functions. A step function is a function where the x-axis can be divided into a finite number of pieces. On each piece the step function is constant. In other words a step function is piecewise constant.
Intervals
Let a and b be two numbers such that a < b. Then the interval between a and b is defined as follows.
[a,b] = x such that a ≤ x ≤ b.
Integrals
Another core concept in calculus are integrals. First we define the integral of a step function. To calculate the integral of a step function on an interval on the x axis we first divide the interval into subintervals in such a way that the step function is constant on each subinterval. Assume there are n subintervals and that the subintervals have the following x coordinates.
[x0,x1],[x1,x2], ..., [xn-1,xn].
Then the integral of the step function is defined in the following manner. The integral is equal to the sum of the following term for each subinterval. Take the width of the subinterval times the value of the function in the sub interval.
Therefore the integral can be viewed as the area under the curve of the function. However negative parts of the function subtract area from the curve. If the step function is never negative then the integral is in fact equal to the area under the curve.
Let the integral of a step function f(x) be denoted by I(f).
We now turn to the problem of defining an integral for a function that is not a step function. Assume that we have a function f(x) that is continuous on an interval [a,b]. Then it is possible to find a step function L(x) such that L(x) ≤ f(x) for every x in [a,b]. L(x) is a lower bound for f(x). It is also possible to find a step function U(x) such that f(x) ≤ U(x) for every x in [a,b].
Let f(x) be a function that is defined on the interval [a,b]. If for any ε there exists step functions L(x) and U(x) such that
|I(U) - I(L)| < ε
then the function f(x) is said to be integratable on the interval [a,b]. Furthermore the integral of f(x) is equal to the value that lies between I(L) and I(U) when ε approaches 0.
More informally you can think of the function f(x) being approximated with a step function. The approximation gets better and better as ε gets smaller. The integral of the function f(x) is then equal to the integral of the step function that is the approximation of f(x). And we already have a way of defining the integral of a step function so we are done.
Since integrals are so important in calculus they have their own notation. The integral of the function f(x) from a to b is denoted as follows.
Principal theorem of analysis
Let S(x) be the integral of the function f(t) on the interval [a,x]. The S(x) is differentiable and S'(x) = f(x). This is known as the principal theorem of analysis. Another way of stating this is that S(x) is a primitive function to f(x).
Differential equations
A differential equation is an equation where the solution is a real function. The following is an example of a differential equation.
y'(x) + k * y(x) + 3 = 0
Here y(x) is the unknown function. Once you find a solution to a differential equation it is easy to check that it is correct. Just take the derivative of the function and plug it into the equation. If the left hand side of the equation equals the right hand side than you have found a solution.
Factorial
The factorial of a number n is defined is defined as follows.
n! = n*(n-1)*(n-2)*...*3*2*1
The factorial is a useful shorthand mathematical notation. The factorial is used when defining Maclaurin polynomials and Taylor polynomials.
Maclaurin and Taylor approximations
Assume that you have a continuous function f(x). It is possible to approximate f(x) using a polynomial.
p(x) = p0 + p1x + p2x2 + ... + pnxn
The issue now is to find the values of the coefficients in the polynomial. If we want to approximate a function around the origin then we can use the Maclaurin polynomial. The Maclaurin polynomial of order n of the function f(x) is defined as follows.
If we want to approximate the function around the point a then we can use a Taylor polynomial instead. The Taylor polynomial of order n is defined as follows.
Conclusion
We have now gone through and defined limits, continuity, derivatives and integrals which are the core concepts of calculus. Furthermore we have looked at differential equations, Maclaurin polynomials and Taylor polynomials. This is what is usually covered in your first calculus course.
If you are interested in learning calculus I highly recommend getting a introductory book on calculus. The purpose of this article is to give an overview of how the core concepts of calculus fit together. Most of the definitions have been simplified. Another reason to get a book is that typesetting of mathematics in books is much better than what can be achieved on the web. Buy or borrow a calculus book. You won't regret it.
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We've updated our
TEXT
Arithmetic Sequences and Series
Arithmetic Sequences
An arithmetic sequence is a sequence of numbers in which the difference between the consecutive terms is constant.
Learning Objectives
Calculate the nth term of an arithmetic sequence and describe the properties of arithmetic sequences
Key Takeaways
Key Points
• The behavior of the arithmetic sequence depends on the common difference $d$.
• Arithmetic sequences can be finite or infinite.
Key Terms
• arithmetic sequence: An ordered list of numbers wherein the difference between the consecutive terms is constant.
• infinite: Boundless, endless, without end or limits; innumerable.
An arithmetic progression, or arithmetic sequence, is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence $5, 7, 9, 11, 13, \cdots$ is an arithmetic sequence with common difference of $2$.
• $a_1$: The first term of the sequence
• $d$: The common difference of successive terms
• $a_n$: The $n$th term of the sequence
The behavior of the arithmetic sequence depends on the common difference $d$. If the common difference, $d$, is:
• Positive, the sequence will progress towards infinity ($+\infty$)
• Negative, the sequence will regress towards negative infinity ($-\infty$)
Note that the first term in the sequence can be thought of as $a_1+0\cdot d,$ the second term can be thought of as $a_1+1\cdot d,$ the third term can be thought of as $a_1+2\cdot d,$and so the following equation gives $a_n$: [latex-display]a_n= a_1+(n−1) \cdot d[/latex-display] Of course, one can always write out each term until getting the term sought—but if the 50th term is needed, doing so can be cumbersome.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 9/10 - 3/10 = 3/5 = 0.6
Spelled result in words is three fifths.
### How do you solve fractions step by step?
1. Subtract: 9/10 - 3/10 = 9 - 3/10 = 6/10 = 2 · 3/2 · 5 = 3/5
For adding, subtracting, and comparing fractions, it is suitable to adjust both fractions to a common (equal, identical) denominator. The common denominator you can calculate as the least common multiple of both denominators - LCM(10, 10) = 10. In practice, it is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 10 × 10 = 100. In the next intermediate step, , cancel by a common factor of 2 gives 3/5.
In words - nine tenths minus three tenths = three fifths.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Cupcakes
In a bowl was some cupcakes. Janka ate one third and Danka ate one quarter of cupcakes. a) How many of cookies ate together? b) How many cookies remain in a bowl? Write the results as a decimal number and in notepad also as a fraction.
• Michael
Michael had a bar of chocolate. He ate 1/2 of it and gave away 1/3. What fraction had he left?
Pupils doing research which a winter sport do their classmates most popular. They found that 2/5 of classmates would most like to play hockey, skate prefer 2/9 pupils, 3/10 students prefer skiing and 1/15 classmates don't like any winter sport. What propo
• Mountain
Mountain has an elevation of 7450 meters and in the morning is the middle portion thereof in the clouds. How many meters of height is in the sky if below the clouds are 2,000 meters, and above clouds are two-fifths of the mountain's elevation?
• Mr. Peter
Mr. Peter mowed 2/7 of his lawn. His son mowed 1/4 of it. Who mowed the most? How much of the lawn still need to be mowed?
• Bitoo and Reena
Bitoo ate 3/5 part of an apple and the remaining part was eaten by his sister Reena. How much part of an apple did Renna eat? Who had the larger share? By how much?
• From a
From a 1 meter ribbon, Ericka cut 2/4 meter for her hat and another 1/4 meter for her bag. How long was the remaining piece?
• Ali bought
Ali bought 5/6 litre of milk. He drank 1/2 litre and his brother drank 1/6 litre. How much litre of milk left?
• Math test
Brayden was solving some math problems for the math team. He answered 2 math problems. Matthew answered 3, John answered 1 reasoning. Matthew 1/2 times as many. Brayden said that 2/6. Is he correct? Why or why not? Be sure to explain your answer.
• Emily
Emily had 20 minutes to do a three-problem quiz. She spent 11 3/4 minutes on question A and 5 1/2 minutes on question B. How much time did she have left for question C?
• Metal rod
You have a metal rod that’s 51/64 inches long. The rod needs to be trimmed. You cut 1/64 inches from one end and 1/32 inches from the other end. Next, you cut the rod into 6 equal pieces. What will be the final length of each piece?
• Two ribbons
The total length of the two ribbons is 13 meters. If one ribbon is 7 and 5/8 meters long, what is the length of the other ribbon?
• Interior designer
To make draperies an interior designer needs 11 1/4 yards of material for the den and 8 1/2 yards for the living room. If material comes only in 20 yard bolts, how much will be left over after completing both sets of draperies?
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• This is an assessment test.
• To draw maximum benefit, study the concepts for the topic concerned.
• Kindly take the tests in this series with a pre-defined schedule.
## Geometry and Mensuration: Level 2 Test 7
Congratulations - you have completed Geometry and Mensuration: Level 2 Test 7. You scored %%SCORE%% out of %%TOTAL%%. You correct answer percentage: %%PERCENTAGE%% . Your performance has been rated as %%RATING%%
Question 1
In the following figure, find ∠ADC.
A 55o B 27.5o C 60o D 30o
Question 1 Explanation:
∠ADC = 70 + (180 – 70)/2 = 55+70
Thus ∠ADC = (180-125)/2 = 55/2 = 27.5
Correct option is (b)
Question 2
An equilateral triangle and a regular hexagon have equal perimeters. If the area of the triangle is 2 cm2, then the area of the hexagon is
A 2 cm2 B 3 cm2 C 4 cm2 D 6 cm2
Question 2 Explanation:
Let the individual sides of the hexagon be 1 cm
Thus the perimeter =6 and the side of the equilateral triangle is 2cm.
Thus the area of the equilateral triangle = √3/4 x 4 = √3cm
Thus the area of the hexagon= {(6 x √3)/4} x 12 = (3/2)√3
Thus the ratio of the triangle to hexagon is 2:3.
Thus the area of the hexagon = 2 x 3/2 = 3 cm.
The correct option is (b)
Question 3
In a right angled ABC, ABC= 90o; BN is perpendicular to AC, AB= 6cm, AC= 10 cm. Then AN: NC is
A 3: 4 B 9: 16 C 3: 16 D 1: 4
Question 3 Explanation:
Before starting calculation, one can eliminate two choices = a
and since 1 cannot be broken down in the required ratios.
By Pythagoras theorem the third side = 8 cm
The length of BN = 48/10 =4.8 cm
Thus the length of CN according to Pythagoras theorem is √40.96 =6.4 cm
The ratio = (10-6.4): 6.4 = 9:16
Correct option is (b)
Question 4
In a triangle PQR, R = 90o and Q is mid-point of RP. The value of PS2 – QS2 is equal to
A PQ2 B 2PQ2 C 3PR2 D 4RQ2
Question 4 Explanation:
By Pythagoras Theorem
PS2 = PR2 + RS2
QS2 = QR2 + RS2
PS2 – QS2
= PR2 + RS2 – QR2 –RS2
= PR2 + QR2
= (PR– QR) (PR + QR)
=(2QR – QR) (2QR + QR)
= QR × 3QR = 3QR2
Question 5
If the interior angle of a regular polygon is double the measure of exterior angle, then the number of sides of the polygon is
A 6 B 8 C 10 D 12
Question 5 Explanation:
$\displaystyle \begin{array}{l}\begin{array}{*{35}{l}} Let\text{ }the\text{ }number\text{ }of\text{ }sides\text{ }of\text{ }the\text{ }polygon\text{ }=\text{ }n \\ We\text{ }know\text{ }that\text{ }the\text{ }interior\text{ }angle\text{ }= \\ \end{array}\\=\left( \frac{2n-4}{n} \right)\times {{90}^{o}}\\And\,\,the\,\,exterior\,\,angle\,\,=\frac{{{360}^{o}}}{n}\\\therefore \,\,from\,\,question\,\,\\\frac{2n-4}{n}\times {{90}^{o}}=\frac{2\times 360}{n}\\\Rightarrow 2n-4=8\\\Rightarrow 2n=12\\\Rightarrow n=6=Number\,\,\,of\,\,\,sides\end{array}$
Once you are finished, click the button below. Any items you have not completed will be marked incorrect.
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# 2.1 Writing Equations.
## Presentation on theme: "2.1 Writing Equations."— Presentation transcript:
2.1 Writing Equations
You evaluated and simplified algebraic expressions.
Translate sentences into equations. Translate equations into sentences. Then/Now
formula Vocabulary
A. Translate this sentence into an equation.
Translate Sentences into Equations A. Translate this sentence into an equation. A number b divided by three is equal to six less than c. b divided by three is equal to six less than c. = c – 6 Answer: Example 1
B. Translate this sentence into an equation.
Translate Sentences into Equations B. Translate this sentence into an equation. Fifteen more than z times six is y times two minus eleven. Fifteen more than z times six is y times two minus eleven. z × = y × – Answer: The equation is z = 2y – 11. Example 1
A. Translate this sentence into an equation
A. Translate this sentence into an equation. A number c multiplied by six is equal to two more than d. A. 6c = d + 2 B. 2c = d + 6 C. c = d + 2 D. c = 6(d + 2) Example 1
B. Translate this sentence into an equation
B. Translate this sentence into an equation. Three less than a number a divided by four is seven more than 3 times b. A. B. C. D. Example 1
Use the Four-Step Problem-Solving Plan
JELLY BEANS A jelly bean manufacturer produces 1,250,000 jelly beans per hour. How many hours does it take them to produce 10,000,000 jelly beans? Understand You know that 1,250,000 jelly beans are produced each hour. You want to know how many hours it will take to produce 10,000,000 jelly beans. Plan Write an equation to represent the situation. Let h represent the number of hours needed to produce the jelly beans. Example 2
Find h mentally by asking, “What number times 125 equals 1000?” h = 8
Use the Four-Step Problem-Solving Plan 1,250, times hours equals 10,000,000. 1,2500, × h = ,000,000 Solve 1,250,000h = 10,000,000 Find h mentally by asking, “What number times 125 equals 1000?” h = 8 Answer: It will take 8 hours to produce 10,000,000 jellybeans. Check If 1,250,000 jelly beans are produced in one hour, then 1,250,000 x 8 or 10,000,000 jelly beans are produced in 8 hours. The answer makes sense. Example 2
A person at the KeyTronic World Invitational Type-Off typed 148 words per minute. How many minutes would it take to type 3552 words? A. 148 minutes B. 30 minutes C minutes D. 24 minutes Example 2
Words Perimeter equals four times the length of a side.
Write a Formula GEOMETRY Translate the sentence into a formula. The perimeter of a square equals four times the length of a side. Words Perimeter equals four times the length of a side. Variables Let P = perimeter and s = length of a side. Perimeter equals four times the length of a side. Formula P = s Answer: The formula is P = 4s. Example 3
Translate the sentence into a formula
Translate the sentence into a formula. The area of a circle equals the product of and the square of the radius r. A. A = + r2 B. A = r2 C. A = 2r D. A = 2r + Example 3
Group Work on Board Skills practice #1-10
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Line Equation-Slope-Distance-Midpoint-Y intercept Calculator
Enter 2 points below or 1 point and the slope of the line equation and press the appropriate button
Point 1: (x1 = , y1 = ) Slope:
Point 2: (x2 = , y2 = ) b:
Given the two points you entered of (1, 4) and (5, 6), we need to calculate 8 items:
Item 1: Calculate the slope and point-slope form:
Slope (m) = y2 - y1 x2 - x1
Slope (m) = 6 - 4 5 - 1
Slope (m) = 2 4
Since the slope is not fully reduced, we reduce numerator and denominator by the (GCF) of 2
Slope = (2/2)/(4/2)
Slope = 1/2
Calculate the point-slope form using the formula below:
y - y1 = m(x - x1)
y - 4 = 1/2(x - 1)
Item 2: Calculate the line equation that both points lie on.
The standard equation of a line is y = mx + b where m is our slope, x and y are points on the line, and b is a constant.
Rearranging that equation to solve for b, we get b = y - mx. Using the first point that you entered = (1, 4) and the slope (m) = 1/2 that we calculated, let's plug in those values and evaluate:
b = 4 - (1/2 * 1)
b = 4 - (1/2)
b = 8 2
-
1 2
b = 7 2
This fraction is not reduced. Using our GCF Calculator, we see that the top and bottom of the fraction can be reduced by 7
Our reduced fraction is:
1 0.29
Now that we have calculated (m) and (b), we have the items we need for our standard line equation:
y = 1/2x + 1/0.29
Item 3: Calculate the distance between the 2 points you entered.
Distance = Square Root((x2 - x1)2 + (y2 - y1)2)
Distance = Square Root((5 - 1)2 + (6 - 4)2)
Distance = Square Root((42 + 22))
Distance = √(16 + 4)
Distance = √20
Distance = 4.4721
Item 4: Calculate the Midpoint between the 2 points you entered. Midpoint is denoted as follows:
Midpoint =
x2 + x1 2
,
y2 + y1 2
Midpoint =
1 + 5 2
,
4 + 6 2
Midpoint =
6 2
,
10 2
Midpoint = (3, 5)
Item 5: Form a right triangle and calculate the 2 remaining angles using our 2 points:
Using our 2 points, we form a right triangle by plotting a 3rd point (5,4)
Our first triangle side = 5 - 1 = 4
Our second triangle side = 6 - 4 = 2
Using the slope we calculated, Tan(Angle1) = 0.5
Angle1 = Atan(0.5)
Angle1 = 26.5651°
Since we have a right triangle, we only have 90 degrees left, so Angle2 = 90 - 26.5651° = 63.4349
Item 6: Calculate the y intercept of our line equation
The y intercept is found by setting x = 0 in the line equation y = 1/2x + 1/0.29
y = 1/2(0) + 1/0.29
y = 1/0.29
Item 7: Determine the parametric equations for the line determined by (1, 4) and (5, 6)
Parametric equations are written in the form (x,y) = (x0,y0) + t(b,-a)
Plugging in our numbers, we get
(x,y) = (1,4) + t(5 - 1,6 - 4)
(x,y) = (1,4) + t(4,2)
x = 1 + 4t
y = 4 + 2t
Calculate Symmetric Equations:
x - x0 z
y - y0 b
Plugging in our numbers, we get:
x - 1 4
y - 4 2
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# 7.3: Double-Angle, Half-Angle, and Reduction Formulas
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Skills to Develop
• Use double-angle formulas to find exact values.
• Use double-angle formulas to verify identities.
• Use reduction formulas to simplify an expression.
• Use half-angle formulas to find exact values.
Bicycle ramps made for competition (see Figure $$\PageIndex{1}$$) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be $$\theta$$ such that $$\tan \theta=\dfrac{5}{3}$$. The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.
Figure $$\PageIndex{1}$$: Bicycle ramps for advanced riders have a steeper incline than those designed for novices.
#### Using Double-Angle Formulas to Find Exact Values
In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where $$\alpha=\beta$$. Deriving the double-angle formula for sine begins with the sum formula,
$\sin(\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta. \nonumber$
If we let $$\alpha=\beta=\theta$$, then we have
\begin{align*} \sin(\theta+\theta)&= \sin \theta \cos \theta+\cos \theta \sin \theta\\ \sin(2\theta)&= 2\sin \theta \cos \theta. \end{align*}
There are three options for the double-angle formula for cosine. First, starting from the sum formula, $$\cos(\alpha+\beta)=\cos \alpha \cos \beta−\sin \alpha \sin \beta$$,and letting $$\alpha=\beta=\theta$$, we have
\begin{align*} \cos(\theta+\theta)&= \cos \theta \cos \theta-\sin \theta \sin \theta\\ \cos(2\theta)&= {\cos}^2 \theta - {\sin}^2 \theta. \end{align*}
Using one of the Pythagorean Identities, we can expand this double-angle formula for cosine and get two more variations. The first variation is:
\begin{align*} \cos(2\theta)&= {\cos}^2 \theta-{\sin}^2 \theta\\ &= (1-{\sin}^2 \theta)-{\sin}^2 \theta \\ &= 1 - 2\sin^2 \theta. \end{align*}
The second variation is:
\begin{align*} \cos(2\theta)&= {\cos}^2 \theta-{\sin}^2 \theta\\ &= {\cos}^2 \theta-(1-{\cos}^2 \theta)\\ &= 2 {\cos}^2 \theta-1 \end{align*}
To derive the double-angle formula for tangent, replacing $$\alpha=\beta=\theta$$ in the sum formula gives
\begin{align*} \tan(\alpha+\beta)&= \dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\\ \tan(\theta+\theta)&= \dfrac{\tan \theta+\tan \theta}{1-\tan \theta \tan \theta}\\ \tan(2\theta)&= \dfrac{2\tan \theta}{1-{\tan}^2 \theta}. \end{align*}
DOUBLE-ANGLE FORMULAS
The double-angle formulas are summarized as follows:
\begin{align*} \sin(2\theta)&= 2 \sin \theta \cos \theta\\ \cos(2\theta)&= {\cos}^2 \theta-{\sin}^2 \theta =1-2 {\sin}^2 \theta = 2{\cos}^2 \theta-1\\ \tan(2\theta)&= \dfrac{2 \tan \theta}{1-{\tan}^2\theta} \end{align*}
How to: Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value
1. Draw a triangle to reflect the given information.
2. Determine the correct double-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.
Example $$\PageIndex{1}$$: Using a Double-Angle Formula to Find the Exact Value Involving Tangent
Given that $$\tan \theta=−\dfrac{3}{4}$$ and $$\theta$$ is in quadrant II, find the following:
1. $$\sin(2\theta)$$
2. $$\cos(2\theta)$$
3. $$\tan(2\theta)$$
Solution
If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given $$\tan \theta=−\dfrac{3}{4}$$,such that $$\theta$$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because $$\theta$$ is in the second quadrant, the adjacent side is on the x-axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:
\begin{align*} {(-4)}^2+{(3)}^2&= c^2\\ 16+9&= c^2\\ 25&= c^2\\ c&= 5 \end{align*}
Now we can draw a triangle similar to the one shown in Figure $$\PageIndex{2}$$.
Figure $$\PageIndex{2}$$
1. Let’s begin by writing the double-angle formula for sine.
$$\sin(2\theta)=2 \sin \theta \cos \theta$$
We see that we to need to find $$\sin \theta$$ and $$\cos \theta$$. Based on Figure $$\PageIndex{2}$$, we see that the hypotenuse equals $$5$$, so $$\sin θ=35$$, $$\sin θ=35$$, and $$\cos θ=−45$$. Substitute these values into the equation, and simplify.
Thus,
\begin{align*} \sin(2\theta)&= 2\left(\dfrac{3}{5}\right)\left(-\dfrac{4}{5}\right)\\ &= -\dfrac{24}{25} \end{align*}
2. Write the double-angle formula for cosine.
$$\cos(2\theta)={\cos}^2 \theta−{\sin}^2 \theta$$
Again, substitute the values of the sine and cosine into the equation, and simplify.
\begin{align*} \cos(2\theta)&= {\left(-\dfrac{4}{5}\right)}^2-{\left(\dfrac{3}{5}\right)}^2\\ &= \dfrac{16}{25}-\dfrac{9}{25}\\ &= \dfrac{7}{25} \end{align*}
3. Write the double-angle formula for tangent.
$$\tan(2\theta)=\dfrac{2 \tan \theta}{1−{\tan}^2\theta}$$
In this formula, we need the tangent, which we were given as $$\tan \theta=−\dfrac{3}{4}$$. Substitute this value into the equation, and simplify.
\begin{align*} \tan(2\theta)&= \dfrac{2\left(-\dfrac{3}{4}\right )}{1-{\left(-\dfrac{3}{4}\right)}^2}\\ &= \dfrac{-\dfrac{3}{2}}{1-\dfrac{9}{16}}\\ &= -\dfrac{3}{2}\left(\dfrac{16}{7}\right)\\ &= -\dfrac{24}{7} \end{align*}
Exercise $$\PageIndex{1}$$
Given $$\sin \alpha=\dfrac{5}{8}$$,with $$\theta$$ in quadrant I, find $$\cos(2\alpha)$$.
$$\cos(2\alpha)=\dfrac{7}{32}$$
Example $$\PageIndex{2}$$: Using the Double-Angle Formula for Cosine without Exact Values
Use the double-angle formula for cosine to write $$\cos(6x)$$ in terms of $$cos(3x)$$.
Solution
\begin{align*} \cos(6x)&= \cos(3x+3x)\\ &= \cos 3x \cos 3x-\sin 3x \sin 3x\\ &= {\cos}^2 3x-{\sin}^2 3x \end{align*}
Analysis
This example illustrates that we can use the double-angle formula without having exact values. It emphasizes that the pattern is what we need to remember and that identities are true for all values in the domain of the trigonometric function.
### Using Double-Angle Formulas to Verify Identities
Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.
Example $$\PageIndex{3}$$: Using the Double-Angle Formulas to Verify an Identity
Verify the following identity using double-angle formulas: $$1+\sin(2\theta)={(\sin\theta+\cos\theta)}^2 \nonumber$$
Solution
We will work on the right side of the equal sign and rewrite the expression until it matches the left side.
\begin{align*} {(\sin \theta+\cos \theta)}^2&= {\sin}^2 \theta+2 \sin \theta \cos \theta+{\cos}^2 \theta\\ &= ({\sin}^2 \theta+{\cos}^2 \theta)+2 \sin \theta \cos \theta\\ &= 1+2 \sin \theta \cos \theta\\ &= 1+\sin(2\theta) \end{align*}
Analysis
This process is not complicated, as long as we recall the perfect square formula from algebra:
${(a\pm b)}^2=a^2\pm 2ab+b^2 \nonumber$
where $$a=\sin \theta$$ and $$b=\cos \theta$$. Part of being successful in mathematics is the ability to recognize patterns. While the terms or symbols may change, the algebra remains consistent.
Exercise $$\PageIndex{2}$$
Verify the identity: $${\cos}^4 \theta−{\sin}^4 \theta=\cos(2\theta)$$.
$${\cos}^4 \theta−{\sin}^4 \theta=({\cos}^2 \theta+{\sin}^2 \theta)({\cos}^2 \theta−{\sin}^2 \theta)=\cos(2\theta)$$
Example $$\PageIndex{4}$$: Verifying a Double-Angle Identity for Tangent
Verify the identity: $$\tan(2 \theta)=2\cot \theta−\tan \theta$$
Solution
In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.
\begin{align*} \tan(2\theta)&= \dfrac{2 \tan \theta}{1-{\tan}^2 \theta} \qquad \text{Double-angle formula}\\ &= \dfrac{2 \tan \theta\left (\dfrac{1}{\tan \theta}\right)}{(1-{\tan}^2 \theta)\left (\dfrac{1}{\tan \theta}\right )} \qquad \text{Multiply by a term that results in desired numerator}\\ &= \dfrac{2}{\dfrac{1}{\tan \theta}-\dfrac{ {\tan}^2 \theta}{\tan \theta}}\\ &= \dfrac{2}{\cot \theta-\tan \theta} \qquad \text {Use reciprocal identity for } \dfrac{1}{\tan \theta} \end{align*}
Analysis
Here is a case where the more complicated side of the initial equation appeared on the right, but we chose to work the left side. However, if we had chosen the left side to rewrite, we would have been working backwards to arrive at the equivalency. For example, suppose that we wanted to show
\begin{align*} \dfrac{2\tan \theta}{1-{\tan}^2 \theta}&= \dfrac{2}{\cot \theta-\tan \theta} \\ \text{Lets work on the right side}\\ \dfrac{2}{\cot \theta-\tan \theta}&= \frac{2}{\frac{1}{\tan \theta }-\tan \theta }\left ( \frac{\tan \theta }{\tan \theta } \right )\\ &= \dfrac{2 \tan \theta}{\dfrac{1}{\tan \theta}(\tan \theta)-\tan \theta(\tan \theta)}\\ &= \dfrac{2 \tan \theta}{1-{\tan}^2 \theta} \end{align*}
When using the identities to simplify a trigonometric expression or solve a trigonometric equation, there are usually several paths to a desired result. There is no set rule as to what side should be manipulated. However, we should begin with the guidelines set forth earlier.
Exercise $$\PageIndex{3}$$
Verify the identity: $$\cos(2\theta)\cos \theta={\cos}^3 \theta−\cos \theta {\sin}^2 \theta$$.
$$\cos(2 \theta)\cos \theta=({\cos}^2 \theta−{\sin}^2 \theta) \cos \theta={\cos}^3 \theta−\cos \theta {\sin}^2 \theta$$
#### Use Reduction Formulas to Simplify an Expression
The double-angle formulas can be used to derive the reduction formulas, which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.
We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with $$\cos(2\theta)=1−2 {\sin}^2 \theta$$. Solve for $${\sin}^2 \theta$$:
\begin{align*} \cos(2\theta)&= 1-2 {\sin}^2 \theta\\[5pt] 2 {\sin}^2 \theta&= 1-\cos(2\theta)\\[5pt] {\sin}^2 \theta&= \dfrac{1-\cos(2\theta)}{2} \end{align*}
Next, we use the formula $$\cos(2\theta)=2 {\cos}^2 \theta−1$$. Solve for $${\cos}^2 \theta$$:
\begin{align*} \cos(2\theta)&= 2 {\cos}^2 \theta-1\\[5pt] 1+\cos(2\theta)&= 2 {\cos}^2 \theta\\[5pt] \dfrac{1+\cos(2\theta)}{2}&= {\cos}^2 \theta \end{align*}
The last reduction formula is derived by writing tangent in terms of sine and cosine:
\begin{align*} \tan^2 \theta &= \frac{\sin^2 \theta}{\cos^2 \theta} \\[5pt] &= \dfrac{\dfrac{1-\cos(2\theta)}{2}}{\dfrac{1+\cos(2\theta)}{2}} \qquad \tag{Substitute the reduction formulas} \\[5pt] &= \left(\dfrac{1-\cos(2 \theta)}{2}\right)\left(\dfrac{2}{1+\cos(2 \theta)}\right) \\[5pt] &= \dfrac{1-\cos(2 \theta)}{1+\cos(2 \theta)} \end{align*}
REDUCTION FORMULAS
The reduction formulas are summarized as follows:
${\sin}^2 \theta=\dfrac{1−\cos(2 \theta)}{2}$
${\cos}^2 \theta=\dfrac{1+\cos(2 \theta)}{2}$
${\tan}^2 \theta=\dfrac{1−\cos(2 \theta)}{1+\cos(2 \theta)}$
Example $$\PageIndex{5}$$: Writing an Equivalent Expression Not Containing Powers Greater Than 1
Write an equivalent expression for $${\cos}^4 x$$ that does not involve any powers of sine or cosine greater than $$1$$.
Solution
We will apply the reduction formula for cosine twice.
\begin{align*} {\cos}^4 x&= {({\cos}^2 x)}^2\\[5pt] &= {\left(\dfrac{1+\cos(2x)}{2}\right)}^2\qquad \text{Substitute reduction formula}\\[5pt] &= \dfrac{1}{4}(1+2\cos(2x)+{\cos}^2 (2x))\\[5pt] &= \dfrac{1}{4}+\dfrac{1}{2} \cos(2x)+\dfrac{1}{4}\left (\dfrac{1+{\cos}^2(2x)}{2}\right )\qquad \text{ Substitute reduction formula for } {\cos}^2 x\\[5pt] &= \dfrac{1}{4}+\dfrac{1}{2} \cos(2x)+\dfrac{1}{8}+\dfrac{1}{8} \cos(4x)\\[5pt] &= \dfrac{3}{8}+\dfrac{1}{2} \cos(2x)+\dfrac{1}{8} \cos(4x) \end{align*}
Analysis
The solution is found by using the reduction formula twice, as noted, and the perfect square formula from algebra.
Example $$\PageIndex{6}$$: Using the Power-Reducing Formulas to Prove an Identity
Use the power-reducing formulas to prove $${\sin}^3(2x)=\left[ \dfrac{1}{2} \sin(2x) \right] [ 1−\cos(4x)$$
Solution
We will work on simplifying the left side of the equation:
\begin{align*} {\sin}^3(2x)&= [\sin(2x)][{\sin}^2(2x)]\\[5pt] &= \sin(2x)\left [\dfrac{1-\cos(4x)}{2}\right ]\qquad \text{Substitute the power-reduction formula.}\\[5pt] &= \sin(2x)\left(\dfrac{1}{2}\right)[1-\cos(4x)]\\[5pt] &= \dfrac{1}{2}[\sin(2x)][1-\cos(4x)] \end{align*}
Analysis
Note that in this example, we substituted $$\dfrac{1−\cos(4x)}{2}$$ for $${\sin}^2(2x)$$. The formula states $${\sin}^2 \theta=\dfrac{1−\cos(2\theta)}{2}$$
We let $$\theta=2x$$, so $$2\theta=4x$$.
Exercise $$\PageIndex{4}$$
Use the power-reducing formulas to prove that $$10{\cos}^4 x=\dfrac{15}{4}+5\cos(2x)+\dfrac{5}{4}\cos(4x)$$.
\begin{align*} 10{\cos}^4 x&= 10{({\cos}^2x)}^2\\[5pt] &= 10{\left[\dfrac{ 1+\cos(2x)}{2} \right]}^2\qquad \text{Substitute reduction formula for } {\cos}^2x\\[5pt] &= \dfrac{10}{4}[1+2\cos(2x)+{\cos}^2(2x)]\\[5pt] &= \dfrac{10}{4}+\dfrac{10}{2}\cos(2x)+\dfrac{10}{4}\left(\dfrac{1+{\cos}^2(2x)}{2}\right)\qquad \text{ Substitute reduction formula for } {\cos}^2 x\\[5pt] &= \dfrac{10}{4}+\dfrac{10}{2} \cos(2x)+\dfrac{10}{8}+\dfrac{10}{8}\cos(4x)\\[5pt] &= \dfrac{30}{8}+5\cos(2x)+\dfrac{10}{8}\cos(4x)\\[5pt] &= \dfrac{15}{4}+5\cos(2x)+\dfrac{5}{4}\cos(4x) \end{align*}
#### Using Half-Angle Formulas to Find Exact Values
The next set of identities is the set of half-angle formulas, which can be derived from the reduction formulas and we can use when we have an angle that is half the size of a special angle. If we replace $$\theta$$ with $$\dfrac{\alpha}{2}$$,the half-angle formula for sine is found by simplifying the equation and solving for $$\sin\left(\dfrac{\alpha}{2}\right)$$. Note that the half-angle formulas are preceded by a $$\pm$$ sign. This does not mean that both the positive and negative expressions are valid. Rather, it depends on the quadrant in which $$\dfrac{\alpha}{2}$$ terminates.
The half-angle formula for sine is derived as follows:
\begin{align*} {\sin}^2 \theta&= \dfrac{1-\cos(2\theta)}{2}\\ {\sin}^2\left(\dfrac{\alpha}{2}\right)&= \dfrac{1-\left(\cos 2\cdot \dfrac{\alpha}{2}\right)}{2}\\ &= \dfrac{1-\cos \alpha}{2}\\ \sin \left(\dfrac{\alpha}{2}\right)&= \pm \sqrt{\dfrac{1-\cos \alpha}{2}} \end{align*}
To derive the half-angle formula for cosine, we have
\begin{align*} {\cos}^2 \theta&= \dfrac{1+\cos(2\theta)}{2}\\ {\cos}^2\left(\dfrac{\alpha}{2}\right)&= \dfrac{1+\cos\left(2\cdot \dfrac{\alpha}{2}\right)}{2}\\ &= \dfrac{1+\cos \alpha}{2}\\ \cos\left(\dfrac{\pi}{2}\right)&= \pm \sqrt{\dfrac{1+\cos \alpha}{2}} \end{align*}
For the tangent identity, we have
\begin{align*} {\tan}^2 \theta&= \dfrac{1-\cos(2\theta)}{1+\cos(2\theta)}\\ {\tan}^2\left(\dfrac{\alpha}{2}\right)&= \dfrac{1-\cos\left(2\cdot \dfrac{\alpha}{2}\right)}{1+\cos\left(2\cdot \dfrac{\alpha}{2}\right)}\\ \tan\left(\dfrac{\alpha}{2}\right)&= \pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}} \end{align*}
HALF-ANGLE FORMULAS
The half-angle formulas are as follows:
\begin{align} \sin\left(\dfrac{\alpha}{2}\right)&=\pm \sqrt{\dfrac{1-\cos \alpha}{2}} \label{halfsine} \\[5pt] \cos \left(\dfrac{\alpha}{2} \right) &=\pm \sqrt{\dfrac{1+\cos \alpha}{2}} \\[5pt] \tan\left(\dfrac{\alpha}{2}\right) &=\pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}} =\dfrac{\sin \alpha}{1+\cos \alpha} =\dfrac{1-\cos \alpha}{\sin \alpha}\end{align}
Example $$\PageIndex{7}$$
Using a Half-Angle Formula to Find the Exact Value of a Sine Function. Find $$\sin(15°)$$ using a half-angle formula.
Solution
Since $$15°=\dfrac{30°}{2}$$,we use the half-angle formula for sine (Equation \ref{halfsine}):
\begin{align*} \sin \dfrac{30^{\circ}}{2}&= \sqrt{\dfrac{1-\cos 30^{\circ}}{2}}\\ &= \sqrt{\dfrac{1-\dfrac{\sqrt{3}}{2}}{2}}\\ &= \sqrt{\dfrac{\dfrac{2-\sqrt{3}}{2}}{2}}\\ &= \sqrt{\dfrac{2-\sqrt{3}}{4}}\\ &= \dfrac{\sqrt{2-\sqrt{3}}}{2} \end{align*}
Remember that we can check the answer with a graphing calculator.
Analysis
Notice that we used only the positive root because $$\sin(15°)$$ is positive.
Howto: Given the tangent of an angle and the quadrant in which the angle lies, find the exact values of trigonometric functions of half of the angle.
1. Draw a triangle to represent the given information.
2. Determine the correct half-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.
Example $$\PageIndex{8}$$: Finding Exact Values Using Half-Angle Identities
Given that $$\tan \alpha=\dfrac{8}{15}$$ and $$α$$ lies in quadrant III, find the exact value of the following:
1. $$\sin\left(\dfrac{\alpha}{2}\right)$$
2. $$\cos\left(\dfrac{\alpha}{2}\right)$$
3. $$\tan\left(\dfrac{\alpha}{2}\right)$$
Solution
Using the given information, we can draw the triangle shown in Figure $$\PageIndex{3}$$. Using the Pythagorean Theorem, we find the hypotenuse to be 17. Therefore, we can calculate $$\sin \alpha=−\dfrac{8}{17}$$ and $$\cos \alpha=−\dfrac{15}{17}$$.
Figure $$\PageIndex{3}$$
1. Before we start, we must remember that if $$α$$ is in quadrant III, then $$180°<\alpha<270°$$,so $$\dfrac{180°}{2}<\dfrac{\alpha}{2}<\dfrac{270°}{2}$$. This means that the terminal side of $$\dfrac{\alpha}{2}$$ is in quadrant II, since $$90°<\dfrac{\alpha}{2}<135°$$. To find $$\sin \dfrac{\alpha}{2}$$,we begin by writing the half-angle formula for sine. Then we substitute the value of the cosine we found from the triangle in Figure $$\PageIndex{3}$$ and simplify. \begin{align*} \sin \dfrac{\alpha}{2}&= \pm \sqrt{\dfrac{1-\cos \alpha}{2}}\\ &= \pm \sqrt{\dfrac{1-(-\dfrac{15}{17})}{2}}\\ &= \pm \sqrt{\dfrac{\dfrac{32}{17}}{2}}\\ &= \pm \sqrt{\dfrac{32}{17}\cdot \dfrac{1}{2}}\\ &= \pm \sqrt{\dfrac{16}{17}}\\ &= \pm \dfrac{4}{\sqrt{17}}\\ &= \dfrac{4\sqrt{17}}{17} \end{align*} We choose the positive value of $$\sin \dfrac{\alpha}{2}$$ because the angle terminates in quadrant II and sine is positive in quadrant II.
2. To find $$\cos \dfrac{\alpha}{2}$$,we will write the half-angle formula for cosine, substitute the value of the cosine we found from the triangle in Figure $$\PageIndex{3}$$, and simplify. \begin{align*} \cos \dfrac{\alpha}{2}&= \pm \sqrt{\dfrac{1+\cos \alpha}{2}}\\ &= \pm \sqrt{\dfrac{1+\left(-\dfrac{15}{17}\right)}{2}}\\ &= \pm \sqrt{\dfrac{\dfrac{2}{17}}{2}}\\ &= \pm \sqrt{\dfrac{2}{17}\cdot \dfrac{1}{2}}\\ &= \pm \sqrt{\dfrac{1}{17}}\\ &= -\dfrac{\sqrt{17}}{17} \end{align*} We choose the negative value of $$\cos \dfrac{\alpha}{2}$$ because the angle is in quadrant II because cosine is negative in quadrant II.
3. To find $$\tan \dfrac{\alpha}{2}$$,we write the half-angle formula for tangent. Again, we substitute the value of the cosine we found from the triangle in Figure $$\PageIndex{3}$$ and simplify. \begin{align*} \tan \dfrac{\alpha}{2}&= \pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}\\ &= \pm \sqrt{\dfrac{1-\left(-\dfrac{15}{17}\right)}{1+\left(-\dfrac{15}{17}\right)}}\\ &= \pm \sqrt{\dfrac{\dfrac{32}{17}}{\dfrac{2}{17}}}\\ &= \pm \sqrt{\dfrac{32}{2}}\\ &= -\sqrt{16}\\ &= -4 \end{align*} We choose the negative value of $$\tan \dfrac{\alpha}{2}$$ because $$\dfrac{\alpha}{2}$$ lies in quadrant II, and tangent is negative in quadrant II.
Exercise $$\PageIndex{5}$$
Given that $$\sin \alpha=−\dfrac{4}{5}$$ and $$\alpha$$ lies in quadrant IV, find the exact value of $$\cos \left(\dfrac{\alpha}{2}\right)$$.
$$-\dfrac{2}{\sqrt{5}}$$
Example $$\PageIndex{9}$$: Finding the Measurement of a Half Angle
Now, we will return to the problem posed at the beginning of the section. A bicycle ramp is constructed for high-level competition with an angle of $$θ$$ formed by the ramp and the ground. Another ramp is to be constructed half as steep for novice competition. If $$tan θ=53$$ for higher-level competition, what is the measurement of the angle for novice competition?
Solution
Since the angle for novice competition measures half the steepness of the angle for the high level competition, and $$\tan \theta=\dfrac{5}{3}$$ for high competition, we can find $$\cos \theta$$ from the right triangle and the Pythagorean theorem so that we can use the half-angle identities. See Figure $$\PageIndex{4}$$.
\begin{align*} 3^2+5^2&=34\\ c&=\sqrt{34} \end{align*}
Figure $$\PageIndex{4}$$
We see that $$\cos \theta=\dfrac{3}{\sqrt{34}}=\dfrac{3\sqrt{34}}{34}$$. We can use the half-angle formula for tangent: $$\tan \dfrac{\theta}{2}=\sqrt{\dfrac{1−\cos \theta}{1+\cos \theta}}$$. Since $$\tan \theta$$ is in the first quadrant, so is $$\tan \dfrac{\theta}{2}$$.
\begin{align*} \tan \dfrac{\theta}{2}&= \sqrt{\dfrac{1-\dfrac{3\sqrt{34}}{34}}{1+\dfrac{3\sqrt{34}}{34}}}\\ &= \sqrt{\dfrac{\dfrac{34-3\sqrt{34}}{34}}{\dfrac{34+3\sqrt{34}}{34}}}\\ &= \sqrt{\dfrac{34-3\sqrt{34}}{34+3\sqrt{34}}}\\ &\approx 0.57 \end{align*}
We can take the inverse tangent to find the angle: $${\tan}^{−1}(0.57)≈29.7°$$. So the angle of the ramp for novice competition is $$≈29.7°$$.
Media
Access these online resources for additional instruction and practice with double-angle, half-angle, and reduction formulas.
## Key Equations
Double-angle formulas $$\sin(2\theta)=2\sin \theta \cos \theta$$ $$\cos(2\theta)={\cos}^2 \theta−{\sin}^2 \theta$$ $$=1−2{\sin}^2 \theta$$ $$=2{\cos}^2 \theta−1$$ $$\tan(2\theta)=\dfrac{2\tan \theta}{1−{\tan}^2 \theta}$$ Reduction formulas $${\sin}^2 \theta=\dfrac{1−\cos(2\theta)}{2}$$ $${\cos}^2 \theta=\dfrac{1+\cos(2\theta)}{2}$$ $${\tan}^2 \theta=\dfrac{1−\cos(2\theta)}{1+\cos(2\theta)}$$ Half-angle formulas $$\sin \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1−\cos \alpha}{2}}$$ $$\cos \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1+\cos \alpha}{2}}$$ $$\tan \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1−\cos \alpha}{1+\cos \alpha}}$$ $$=\dfrac{\sin \alpha}{1+\cos \alpha}$$ $$=\dfrac{1−\cos \alpha}{\sin \alpha}$$
## Key Concepts
• Double-angle identities are derived from the sum formulas of the fundamental trigonometric functions: sine, cosine, and tangent. See Example $$\PageIndex{1}$$, Example $$\PageIndex{2}$$, Example $$\PageIndex{3}$$, and Example $$\PageIndex{4}$$.
• Reduction formulas are especially useful in calculus, as they allow us to reduce the power of the trigonometric term. See Example $$\PageIndex{5}$$ and Example $$\PageIndex{6}$$.
• Half-angle formulas allow us to find the value of trigonometric functions involving half-angles, whether the original angle is known or not. See Example $$\PageIndex{7}$$, Example $$\PageIndex{8}$$, and Example $$\PageIndex{9}$$.
#### Contributors
This page titled 7.3: Double-Angle, Half-Angle, and Reduction Formulas is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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# Velocity and Speed: Tutorials with Examples
Examples with explanations on the concepts of average speed and average velocity of moving object. More problems and their solutions can be found in this website.
The average speed is a scalar quantity (magnitude) that describes the rate of change (with the time) of the distance of a moving object.
average speed = distance time
The average velocity is a vector quantity (magnitude and direction) that describes the rate of change (with the time) of the position of a moving object.
average velocity = change in position time = displacement time
Example 1: An object moves from A to D along the red path as shown below in 41 minutes and 40 seconds.
a) Find the average speed of the object in m/s
b) Find the average velocity of the object in m/s
Solution:
a) Using the given scale (1km per division); the total distance d is given by
d = AB + BC + CD = 2 + 5 + 2 = 9 km
average speed = distance time = 9 km 41 mn + 40 s = 9000 m (41*60 + 40) s = 9000 m 2500 s = 3.6 m/s
b) The final and initial and positions of the moving object are used to find the displacement. The distance from A (initial position) to D (final position) is equal to AD = 5 km. The displacement is the vector AD whose magnitude if 5 km and its direction is to the east.
average velocity = displacement time = 5 km 41 mn + 40 s = 5000 m 2500 s = 2.5 m/s
The average velocity is a vector whose magnitude is 2.5 m/s and its direction is to the east.
Example 2: An object moves, along a line, from point A to B to C and then back to B again as shown in the figure below in half an hour.
a) Find the average speed of the moving object in km/h.
b) Find the magnitude of the average velocity of the object in km/h.
Solution:
a) The total distance d covered by the object is
d = AB + BC + CB = 5 km + 4 km + 4 km = 13 km
average speed = distance time = 13 km 0.5 hour = 26 km/h
b) The magnitude of the displacement is equal to the distance from A (initial position) to B (final position) which is equal to 5 km.
average velocity = displacement time = 5 km 0.5 hour = 10 km/h
Example 3: An fast object moves from point A to B to C to D and then back to A along the rectangle shown in the figure below in 5 seconds.
a) Find the average speed of the moving object in m/s.
b) Find the velocity of the object in m/s.
Solution:
a) The total distance d is equal to the perimeter of the rectangle. Using the given scale,
d = 2 AB + 2 BC = 10 + 6 = 16 km
average speed = distance time = 16 km 5 seconds = 16000 m 5 seconds = 3200 m/s
b) Since the moving object starts at point A and finish at A, there is no change in the position of the object and therefore the displacement is equal to zero.
average velocity = displacement time = 0 5 second = 0
Example 4: A person walks, for two hours, from point A to B to C along a circular field as shown in the figure below.
a) Find the average speed of the person in km/h.
b) Find the velocity of the person.
Solution:
a) The total distance d is equal to half the circumference of the circle and given by
d = (1/2)(2 * Pi * 3) = 3 Pi
average speed = distance time = 3 Pi km 2 hours = 1.5 Pi km/h = 4.7 km/h
b) The magnitude of the displacement D is equal to the diameter AC of the circle and is given by
D = 2 * 3 = 6 Km with direction to the East
average velocity = displacement time = 6 km 2 hours = 3 km/h
Example 5: A person walks for one hour and 12 minutes, from point A to point B, along a circular field as shown in the figure.
a) Find the average speed of the person in km/h.
b) Find the magnitude of the displacement of the person in km/h.
Solution:
a) The total distance d is equal to the quarter the circumference of the circle and given by
d = (1/4)(2 * Pi * 3) = 1.5 Pi
average speed = distance time = 1.5 Pi km 1 hour + 12 minutes = 1.5 Pi km 1 hour + 12/60 hour = 1.5 Pi km 1.2 hour
= 1.25 Pi km/h = 3.9 km/h
b) The magnitude of the displacement D is equal to the hypotenuse AB of the right angle ABO as shown below
Use Pythagora's theorem to find AB as follows
AB
2 = 32 + 32 = 18
D = AB = 3√2 km
average velocity = displacement time = 3√2 km 1 hour + 12 minutes = 3√2 km 1.2 hours
= 2.5√2 km/h = 3.5 km/h
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## What do you mean by sets?
A set is a group or collection of objects or numbers, considered as an entity unto itself. Sets are usually symbolized by uppercase, italicized, boldface letters such as A, B, S, or Z. Each object or number in a set is called a member or element of the set.
## What is the meaning of sets in math grade 7?
Sets. A set is a collection of unique objects i.e. no two objects can be the same. Objects that belong in a set are called members or elements. Elements of set can be anything you desire – numbers, animals, sport teams. Representing Sets.
## What are sets of numbers?
The sets of numbers are defined as follows: Real numbers. any number that is rational or irrational. Rational numbers. any number that can be written as the ratio of two integers and that is terminating or repeating in decimal form.
You might be interested: Quick Answer: How To Teach Math To Toddlers?
## What is set Class 11?
A set is a well-defined collection of objects, whose elements are fixed and cannot vary. It means set doesn’t change from person to person. Like for example, the set of natural numbers up to 7 will remain the same as {1,2,3,4,5,6,7}.
## What is sets and its types?
Set is defined as a well-defined collection of objects. These objects are referred to as elements of the set. Different types of sets are classified according to the number of elements they have. Basically, sets are the collection of distinct elements of the same type.
## What are the types of sets?
Types of a Set
• Finite Set. A set which contains a definite number of elements is called a finite set.
• Infinite Set. A set which contains infinite number of elements is called an infinite set.
• Subset.
• Proper Subset.
• Universal Set.
• Empty Set or Null Set.
• Singleton Set or Unit Set.
• Equal Set.
## How do you introduce a set?
An introduction of sets and its definition in mathematics. The concept of sets is used for the foundation of various topics in mathematics. To learn sets we often talk about the collection of objects, such as a set of vowels, set of negative numbers, a group of friends, a list of fruits, a bunch of keys, etc.
## What is the symbol for empty set?
Empty Set: The empty set (or null set) is a set that has no members. Notation: The symbol ∅ is used to represent the empty set, { }.
## What are the uses of sets in our daily life?
The purpose of sets is to house a collection of related objects. They are important everywhere in mathematics because every field of mathematics uses or refers to sets in some way. They are important for building more complex mathematical structure.
## How do you classify numbers?
The classifications of numbers are: real number, imaginary numbers, irrational number, integers, whole numbers, and natural numbers. Real numbers are numbers that land somewhere on a number line. Imaginary numbers are numbers that involve the number i, which represents sqrt{ -1}.
## What is the symbol for all real numbers?
R = real numbers, Z = integers, N=natural numbers, Q = rational numbers, P = irrational numbers.
## What sets of numbers does 10 belong to?
Question 1091250: What set of numbers does – 10 belong? It is an integer. It is also a real number.
## Is class 11 maths tough?
Yes CBSE class 11 and 12 maths is tough. you have to solve all questions with full efforts and have to study hard. If you are search of these books then you can go at online site sastabooks.
## What is the power of a set?
In mathematics, the power set (or powerset) of a set S is the set of all subsets of S, including the empty set and S itself. The notation 2S is used because given any set with exactly two elements, the powerset of S can be identified with the set of all functions from S into that set.
## What is the first chapter of maths class 11?
NCERT Solutions for Class 11 Maths Chapter 1 Sets
Section Name Topic Name
1.2 Sets and their Representations
1.3 The Empty Set
1.4 Finite and Infinite Sets
1.5 Subsets
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# Zero divided by zero must be equal to zero
What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0$
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $a=cb$, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$\therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$. Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
• Ever look at $\frac{\sin x}{x}$ as $x\to0$? That would be a limit that looks like $\frac{0}{0}$ yet equals one. Commented Dec 1, 2015 at 15:36
• @JBKing: Limits are not the same as evaluations. The former fails to address this question. Commented Dec 1, 2015 at 18:42
• Lennart I understand not knowing basic TeX but you could have tried to copypaste the syntax in my edit to the first part before in posting the second... Commented Dec 1, 2015 at 21:04
• 'Suppose "fish" is not equal to 42. That means "fish" is equal to some x for which x =/= 42. But there is no such x, therefore "fish" is equal to 42.' Commented Dec 2, 2015 at 4:48
• 0/0 + 1/1 = (0*1)/(0*1) + (1*0)/1*0) Why do you think you can multiply both numerator and denominator of a ratio by 0 and the ratio's value will preserve? Commented Dec 2, 2015 at 10:24
The error is in the very first implication. If $0/0$ is not equal to zero, there is no reason why $0/0$ must equal some $x$. There is no reason to believe that we can do this division and get a number.
Therefore you have a proof that $0/0$ cannot equal any nonzero $x$. Combine this with a proof that $0/0$ cannot equal zero, and you have proved that $0/0$ is not a number.
• This. Merely saying something doesn't make it true or exist (or denial make it false either), and writing something down does't make it well-formed. Commented Dec 2, 2015 at 23:54
• There is no proof above that 0/0 cannot equal to 0. Or more precisely, there is a proof but it is wrong, which has been indicated. Commented Dec 7, 2015 at 22:27
• Waiting for a general theorem of non-numbers Commented Aug 8, 2023 at 10:25
Your proof assumes that $0/0$ is a number since your argument involves arithmetic operations.
Now, if we assume the usual rules of arithmetic for the rationals, why can't we do this: $$\frac00+\frac11=\frac{0\times1+1\times0}{1\times0}=\frac00.$$ Then $$1=\frac11=0.$$
• Thanks, this is exactly what I was looking for. Then one could construct an elementary proof of the proposition: There is no real number x such that 0/0 = x. Commented Dec 1, 2015 at 15:44
• Best elementary proof by far. Commented Dec 1, 2015 at 16:07
• This proof presupposes that $\frac{1}{1}=\frac{1}{1}\frac{0}{0}$, which is needed to add the fractions. But this isn't the case, unless $\frac{0}{0}=1$ (and OP proved that it isn't). i.e., this argument does not prove that $\frac{0}{0}\neq 0$. Commented Dec 1, 2015 at 16:39
• @vadim123: that depends on how you define what $\mathbb Q$ is, and how you define the addition of rationals. I personally define $$\frac ab+\frac cd$$ to be $$\frac{ad+bc}{bd}.$$ Commented Dec 1, 2015 at 19:01
• This is why these sorts of insanities show up when trying to reconcile theories of physics. 0/0 is like a black hole - it is simultaneously enormous and yet occupies no space; attempt to use it like a number and it devours everything, which is exactly what numbers aren't supposed to do
– J...
Commented Dec 4, 2015 at 11:47
In order to prove anything whatsoever about $\frac{0}{0}$, we need a definition of $\frac{a}{b}$ where $a$ and $b$ are, say, integers. The correct definition is $\frac{a}{b} = ab^{-1}$, where the definition of $b^{-1}$ is the number (which is unique, when it exists, according to some basic properties of numbers) such that $bb^{-1} = 1$. But then one can see (again, assuming some basic properties of numbers) that $0^{-1}$ does not exist, so $\frac{0}{0}$ isn't even defined.
• The mere existence of $0^{-1}$ is contradictory by ring theory arguments (what you refer to as "some basic properties of numbers", correct me if I'm wrong), but that's apparently not what the OP is talking about. Commented Dec 2, 2015 at 20:51
• @AdarHefer, more precisely, the existence of $0^{-1}$ implies that $1=0$, and hence that the ring is terminal. i.e. $0_R$ is invertible iff $R$ is the terminal ring. Commented Dec 3, 2015 at 13:24
• I would say that any argument involving addition and multiplication invokes "ring theory." As others pointed out, the issue is that $\frac{0}{0}$ is not even a number, and I was trying to make the point that it's not a number by definition, rather than for some other reason. A lot of mathematics can be demystified by understanding that everything goes back to definitions. This is a difficult point to get across e.g. to calculus students, because, well, calculus has at its core profound definitions. Commented Dec 3, 2015 at 15:50
You can define $\frac{0}{0}$ to be anything you want; it can be $0$, or $1$, or $\pi$. But here's the catch: we want the choice of value for $\frac{0}{0}$ to be compatible with the usual laws of arithmetic.
From this standpoint, what you've really proved is that if $\frac{0}{0}$ is anything except $0$, then the law $2\frac{0}{0} = \frac{2 \cdot 0}{0}$ cannot hold. Therefore, the law: $a\frac{b}{c} = \frac{ab}{c}$ cannot hold, either. This suggests that simply defining $\frac{0}{0}=0$ might actually be a good idea. Unfortunately, this breaks another law of arithmetic, namely $\frac{a}{a} = 1.$
Since we cannot have the best of both worlds, perhaps it is best to simply leave $\frac{0}{0}$ undefined.
Actually, I think the best solution is to define division not of real numbers, but of affine subsets of $\mathbb{R}$. These are: the singleton subsets of $\mathbb{R}$, the empty subset, and $\mathbb{R}$ itself. So by passing to the affine subsets, we've effectively adjoined two new "points", namely $\emptyset$ and $\mathbb{R}$.
Now define that given affine subsets $Y$ and $X$ of $\mathbb{R}$, we have:
$$\frac{Y}{X} = \{r \in \mathbb{R} \mid Y \supseteq rX\}$$
You can check that the result of dividing one affine subset by another will itself always be affine.
Under these conventions, we have:
$$\frac{0}{0} = \mathbb{R}, \qquad \frac{1}{0} = \emptyset$$
This justifies the intuition that trying to divide $0$ by $0$ is somehow different from trying to divide a non-zero number by $0$.
Full disclosure: although passing to the affine subsets works algebraically, I'm not entirely sure how to put a topology or uniform structure or metric on the collection of affine subsets of $\mathbb{R}$. Until we can figure out how to do this, my proposed solution probably isn't that useful.
• Your proposed solution is completely unworkable! For instance, $3/2$ is no longer equal to $1.5$; according to your definition, it equals $\{1.5\}$. And $(1/2)/2 = \{\{0.25\}\}$. Have you really thought this through? Commented Dec 2, 2015 at 19:35
• @TonyK, I've edited. Commented Dec 3, 2015 at 1:11
• The best part of this answer is the start: you can (or “one could”) define 0/0 how you like, but the problem is how to do it usefully – and of course we want a definition which plenty of people will accept. Commented Dec 4, 2015 at 0:04
In my experience, when people have a problem with $\frac 0 0$ it generally means they have not developed a clear idea of the way mathematics forms models.
### The character of mathematical systems
At first, every expression a pupil meets can be evaluated to a number they understand, but fairly soon $\frac 0 0$ and $\sqrt -1$ turn up, yet they cling to the feeling that “if you can write it, it must mean something”. The point has to be made that we define what expressions mean in such a way as to help us solve problems, and sometimes these definitions just can’t be made to cover all cases without implying weird behaviour, and so we agree to consider some cases undefined. This makes maths sound a rather arbitrary process, but the constraint of solving problems actually makes it about the least contingent human activity there is. Of course, having made some definitions, we experiment with their implications, which gives rise to new problems, so pure mathematics feeds off itself as well as off applications. We discover that we can create things in an inevitable way.
### Algorithmic v. solution seeking operations
We can distinguish algorithmic operations such as addition, multiplication and raising to powers, from solution-seeking operations such as subtraction, division and the taking of roots. Operations in the first class are defined by some terminating algorithm (such as repeated increment, addition or multiplication), often in the form of a recursive definition, which we show terminates. Operations in the second class are defined as solutions to equations, i.e. they are required to yield numbers which, when combined using previously defined operations, yield a certain result.
More generally, we may have conditions which are not written as equations, and we may be operating on other entities than ‘numbers’ (whatever they may be!1).
### New numbers to make insoluble equations soluble
Sometimes we find that our condition can be satisfied by members of the set defined so far. For example, given numbers $a$ and $b$, we want to solve $a + n =b$ for $n$, and call that $b - a$; then, since $1 + 2 = 3$, we say $3 - 1 = 2$. At other times, we find that there is no such solution, e.g. there is no natural (counting) number $n$ such that $3 + n = 2$ — this is when we ask ourselves if we can define $2 - 3$. Thus we move from natural to integers.
Given a type of equation we are trying to solve, we hope we can define the new ‘numbers’ in a useful way:
• They should include the old numbers, or something very like them.
• We want to be able to apply the existing operations to them.
• We want them to obey they same rules that the old numbers did.
You can define $\frac 0 0$ to be anything you want; it can be 0, or 1, or π. But here's the catch: we want the choice of value for $\frac 0 0$ to be compatible with the usual laws of arithmetic.
Actually, though you can define it as you want, it may make you rather unpopular if you insist on something really quirky, so perhaps it is better to say “one could” etc.
### Two ways of formalising a new number system
Typically we find that obeying the existing rules means that several expressions, such as $2 - 3$ and $3 - 4$, should have the same value. One way we deal with that is to define equivalence classes of pairs of numbers (if the operation is binary) that should yield the same value; we then show that we can define the various operations on the pairs so that they respect the classes and work as in the original situation.
Another way to deal with it is to introduce one or two new symbols and a syntax for combining them: we might then define integers as formal expressions of the form $+n$ or $-n$, and show that these formal expressions behave as desired. Similarly, introducing $i$ lets us define complex numbers.
These are two important ways to build a formal ‘working model’ of a number system, but there are many others, such as Dedekind cuts, von Neumann’s construction of $\mathbb N$ as ‘free sets’ and Conway’s construction of Surreal numbers.
### The error in the question
The problem with $\frac 0 0$ in this question arises partly from failing to recognise that the formal expressions we introduce have only those properties we (consistently) define them to have and anything that follows from those. A bigger problem, though, is not realising that the formal expressions do not automatically mean anything, until we have shown that they reproduce the old numbers as a subset. Once we realise that they do not automatically mean anything, it is easier to accept that our definition may exclude expressions like $\frac 0 0$ (or call them ‘undefined’) in order to ensure desirable behaviour of the new number system.
Another error in the question as first formulated was to assume that the new expressions behave like the numbers they were constructed from, rather than considering that as something to be investigated.
### The solution set approach
Goblin’s answer, (after making the essential point that we decide how to define expressions) went on to suggest an alternative definition of division in terms of the solution set of $a \times n = b$. This suggests you consider $\frac 0 0$ as $\mathbb R$ (why not $\mathbb Q$ or $\mathbb C$ … or even $\mathbb Z$ or $\mathbb N$ ?), but it does not help you deal with $n^2 = -1$, which needs the approach described above.
### Notes
1 As Dedekind asked, Was sind und was sollen die Zahlen? “What are numbers and what should they be?” or just possibly “… what is the point of them”!
• Good answer. $\;$ Commented Dec 4, 2015 at 14:55
• I'm not good at math at all but I could understand this answer. +1 :) Commented Dec 5, 2015 at 22:54
• Regarding the last paragraph before the Notes: I like to think of $0_\mathbb{R}$ as different from $0_\mathbb{Q}$, which is, in turn, different from $0_\mathbb{C}$. So $\frac{0_\mathbb{R}}{0_\mathbb{R}} = \mathbb{R}$, but $\frac{0_\mathbb{Q}}{0_\mathbb{Q}} = \mathbb{Q}.$ Commented May 5, 2016 at 10:40
How do you define a fraction?
The definition I know is that for any number $a \neq 0$ we say the symbol $\frac{1}{a}$ satisfies $a \cdot \frac{1}{a} = 1$. One can show that this is well defined for any $a \neq 0$ and unique. Further, the symbol $\frac{b}{a}$ for any numbers $b,a$ with $a\neq 0$ is defined as $\frac{b}{a}:=b \cdot \frac{1}{a}$.
If $\frac{1}{0}$ would exists, then clearly by definition we would have $$\frac{0}{0}=0 \cdot \frac{1}{0} = 1.$$ However, the symbol $\frac{1}{a}$ is not definied for $a=0$, because there exists no number $b$ with $0 \cdot b = 1$, so there is no value to which we can assign the symbol $\frac{1}{0}$.
Example of the above definition:
We can compute $\frac{1}{2}$, because we know
$$2 \cdot \frac{1}{2} = \frac{1}{2} +\frac{1}{2} =1$$ and since
$$2 \cdot 0.5 = 0.5 + 0.5 = 1$$ it follows that $$\frac{1}{2} = 0.5$$ holds.
• I feel like this answer would tie together better if your example said "2x1/2 =1" and "2x0.5 = 1". I mean it's obvious they're equal, but I did a double take for a sec, purely based on the different format.
– Jeff
Commented Dec 2, 2015 at 3:21
Commented Dec 2, 2015 at 9:44
If $\frac00 = 0$ then $0\times0=0$ would work as a proof. The problem is that also $3\times0=0$ and $7\times0$ and so on would all work.
So $\frac00=0,\space\frac00=1,\space\frac00=3,\space\frac00=$ just about anything.
There are too many possible results, we want such operations to return defined results (one and only one result), otherwise the operation itself is not actually defined. In fact, dividing any number by $0$ is 'undefined', 'illegal'.
When dividing a non-zero integer by zero, we have the same problem. only, in that case, we simply have no-results.
ie: $\frac60 =$ number which multiplied by $0$ equals $6$.
We can say $\frac60=\infty$ only to cut short on a theorem, stating that the limit of $\frac mn$ with m integer and $n$ going to $0$ is infinite. It is obviously a notation-artifact. Otherwise we should say that $\infty\times0$ equals $6$, and $7$, and $8$, and $9$...
• I would say that this presupposes that 0/0 = 1 (or the ring definition of division). If 0/0 = 0, then, for example, 3*0 = 0 <=> (3*0)/0 = 0/0 <=> 3*(0/0) = 0/0 <=> 3*0 = 0 <=> 0 = 0. It leads to nowhere, but not to 0/0 = 3. Commented Dec 1, 2015 at 18:15
The first fault is in the very statement. You assume that there is a $0/0$ in the first place, you have to have that before you even assume that it's not equal to zero.
Then even if there were a $0/0$ you're assuming that there is a $2{0\over0}$ and then that it is equal to ${2\cdot 0\over0}={0\over0}$.
Finally you assume that if $2x=x$ you must have that $x=0$, even if you've got this far there's no reason to think that $x$ is a number and must obey the rules for numbers.
In the first case you use the identity ${a\over b}+{c\over d} = {ad+bc\over bd}$ which is only shown to be true if $bd\ne0$.
In the second case you use the identity $a{b\over c} = {ab\over c}$ which is only shown to be true if $c\ne 0$.
That is you use twice identities where the prerequisite is not fulfilled.
The multiplicative semigroup of a field is not a group, but only an inverse semigroup. In the context of semigroups, we call $y$ an inverse element of $x$, if $x=xyx$ and $y=yxy$. An inverse semigroup is a semigroup where each element has an unique inverse element, such that an inverse operation $^{-1}$ can be defined via $x=xyx \land y=yxy \Leftrightarrow y=x^{-1}$. The unique multiplicative inverse element of $0$ in this context is $0$, so $\frac{0}{0}=0\cdot 0^{-1}=0\cdot 0=0$. But note that we also have $\frac{1}{0}=0$.
Note however that the rules for dealing with fractions get more complicated, because the sum of two symbolic fractions is not necessarily a pure symbolic fraction: $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}+0_d\frac{a}{b}+0_b\frac{c}{d}$. Here $0_x:=1-\frac{x}{x}$ is idempotent, because $\frac{x}{x}$ is idempotent, i.e. $0_x^2=1-2\frac{x}{x}+\frac{x^2}{x^2}=1-\frac{x}{x}=0_x$. The details of this standard representation are are worked out in a paper by Bergstra et al.
Inverse semigroups are the appropriate generalization of groups if we look at partial bijections instead of total bijections. In the context of rings, a ring whose multiplicative semigroup is an inverse semigroup is called a strongly regular ring. The category of strongly regular rings is related to the category of skew fields in a similar way as the integers are related to the prime numbers (and the same applies to the relation of the category of regular commutative rings to the category of fields). The name meadow has also been proposed for such rings. I have collected some scattered informations on those subjects in a blog post.
If we add the universal formula $a = 0 \lor (ab = ac \Rightarrow b = c)$ characterizing integral domains to the equations characterizing regular commutative rings, then we end up with a finite set of universal formulas characterizing the category of fields, as I learned from a paper by Hiroakira Ono. From this we can learn that the class of fields is closed under isomorphism, substructures and ultraproducts, see theorem 2.20 in chapter V "Connections with Model Theory" of Burris and Sankappanavar's A Course in Universal Algebra.
Now goblin decided to give an answer with a slightly different spin, but goblin has played with meadows himself in the past. He probably noticed that inversion defined like this is not a continuous function, and so kept on searching. I decided to mention this, not in order to avoid partial functions, but in order to point out that partial functions (and especially partial bijections) do have nice properties on their own, which are different from the properties of total functions.
It is important to keep in mind that this definition of $\frac{x}{0}$ leads to discontinuous operations, so it cannot be used to evaluate limits. And for meromorphic functions, using this definition would even destroy part of the inherent structure, but declaring division by zero as undefined will do so too.
Actually there are solid arguments why $0/0$ can be defined as $0$.
Here is a graphic of the function $f(x,y)=x/y$:
The function is odd against both $x$ and $y$ variables.
Along the axis $x=0$ it is constant zero.
Along the axis $y=0$ it is unsigned infinity, but its Cauchy principal value is constant zero.
Along the diagonal $y=x$ it is constant 1.
Along the diagonal $y=-x$ it is constant -1.
As all odd functions its behavior around zero does not have any preference to positive or negative values. As all odd analytic functions, this function in zero(if defined) can either be zero or unsigned infinity. But unsigned infinity is not a part of affinely extended real line or complex plane, which is usually used in analysis (it is part of projectively extended real line/complex plane). As such, the only variant that remains is $0$.
Note also that we can define a Cauchy principal value of a two-variable function in a point $(x_0,y_0)$ as a limit of the following integral:
$$\lim_{r\to 0}\frac 1{2\pi}\int_0^{2\pi}f(x_0+ r \sin (t),y_0+r \cos(t)) dt$$
This integrates the values of a function over a circle around a point. For analytic functions as long as the radius of the circle r comes to 0, the value of the integral comes to the value of the function in $(x_0,y_0)$. For our function, in $x_0=0, y_0=0$ the value of the integral is constant zero (because the function is odd) and does not depend on r. Consequently, the limit is also zero.
We can extend our analysis to complex numbers, but we again will get the same result, $0$.
• I had the impression that there is a qualitative difference between the statements "the principal Cauchy value of $p$ is $q$" and "$p = q$"; specifically, that the first statement is weaker and does not imply the second statement. Have I misunderstood something? Commented Jan 6, 2016 at 19:57
Using the definition of a Ring, it is pretty simple to show that $$\forall a, 0\times a=a\times 0=0$$ Thus, If we were to define division (only in the case where $c=ab=ba$): $$\forall a, a=\frac{0}{0}$$ Which can be true if and only if your ring is the Zero ring $$\{0\}$$ Therefore, when $\frac{0}{0}$ exists, it is actually $0$ (but there aren't any other numbers in the ring)
• The second statement does not follow from axioms of ring. Division is not defined on rings. Commented Dec 18, 2015 at 11:49
• You are right, I corrected my answer Commented Dec 18, 2015 at 14:03
• You can define division this way, but it is not the only possible way, actually. One can define 0/0=0 and have everything fine in a ring (not only empty ring). Commented Dec 19, 2015 at 11:24
The answer to this might be better understood in layman's terms. Going back to grade-school level word puzzles, $\frac{x}{0}$ means:
If you have $x$ items, and you distribute them equally among $0$ people, how many items does each have?
The answer is not a number—it is "I can't answer that" because it rests on an invalid assumption, exposed by noticing that each doesn't refer to any people. The correct response is "each who?" or "there are no people!"
Just because a math problem is expressed using normal numbers and operators, taking the form of other valid math problems to which an answer is possible, doesn't mean that it must have an answer nor that the answer is a number. Division by 0 really means "don't divide."
What's the result of a division where you don't divide? Whatever the answer is, it's $NaN$—not a number.
Using a logic analogy (since math is ultimately just a specialized branch of logic), if someone asks you a question that normally has a true/false answer, such as "Have you stopped beating your wife?" but the premise is wrong because you have never done so, then neither true nor false is a correct answer. It is either the case that the answer is outside the domain of Boolean, or there is no answer at all. One cannot do further Boolean or true-false logic on such a non-Boolean. Attempting to do so is a category mistake.
Similarly, doing any number math with the result of the requested operation $\frac{x}{0}$ is also a category mistake. Proceeding as though it must equal some number $n$ is to act like "What does the color blue taste like?" is a sensible question. But blue doesn't taste like anything, because it isn't a flavor.
• If you don't have a temperature, what is your temperature?
• If you punch no one in the nose, how mad will they all be?
• If you forgot to buy a coffee cup, does your empty cup have 0 ml caffeinated or 0 ml decaffeinated coffee in it?
• Of all the times you've successfully committed suicide, which one hurt the most?
• Were all the parents you never had blue-eyed or green-eyed?
• How many ccs does a little white lie occupy?
• That's silly. What people? You've completely missed the point. To the downvoters: comments please? This is an awesome answer! Commented Dec 6, 2015 at 20:01
• It's silly to make such a comment. Future down voters may see my comment before downvoting so it had some utility. I understand stackexchange. Commented Dec 6, 2015 at 20:34
• I think you're missing the point. It's silly to talk about nonexistent humans. You're having trouble breaking out of your familiar, but invalid, frame of reference. That is the exact point my answer makes. You glibly gave an answer to a nonsense question as if you were showing me up! That's why I referred to logic, not math, and tried to take it out of the realm of numbers via a word problem, to prompt seeing the answer through clearer thinking. Commented Dec 6, 2015 at 20:56
• No assuming allowed. Math. Fake proof? Strikethrough. Real proof, with logic. Commented Dec 9, 2015 at 1:55
• He doesn't want a fake proof. He wants to debunk a fake proof. Commented Dec 9, 2015 at 8:08
The problem is that you assume $\frac00$ is = some $x$ belong to the real number group.$\frac00$ is just not defined for you to assume it to be $=x$.
Basically its one of the seven indeterminate forms of mathematics.Look here for the complete list-https://en.wikipedia.org/wiki/Indeterminate_form#List_of_indeterminate_forms
• No, $\frac{0}{0}$ isn't defined in the way that it does not equal any $x$. It just isn't defined. Commented Dec 7, 2015 at 13:39
Try to cancel 0/0 with different numbers each time you will get different solutions for 0/0 . Suppose 1×0/1×0 you get solution 1. Suppose 2×0/0×1 you get solution 2 . Suppose n×0/0×1 you get solution n. ∵0/0 has infinitely many solutions 0/0 is not defined.
0/0 is indeterminate. Which means it has infinite number of values.
(0)(1) = 0 => 0/0 = 1
(0)(2) = 0 => 0/0 = 2
(0)(4) = 0 => 0/0 = 3 .....
From above it can be shown that 0/0 has infinite number of values which make it meaningless. Hence the term.
Well, then $n \cdot 0=0\; \forall \mathbb{R} \implies \frac{0}{0}=n=\frac{1}{n}$ so, $\frac{0}{0}$ is pretty much any real number. This is definitely not the case. This is why it is undefined.
• I don't accept this argument. If 0/0 = 0, then n*0 = 0 <=> (n*0)/0 = 0/0 <=> n*(0/0) = 0/0 <=> n*0 = 0. It leads nowhere. Commented Dec 1, 2015 at 17:54
• I would say this definitely is the case, which why it fails to be (uniquely) well-defined. Commented Dec 1, 2015 at 18:41
• That is another way to look at it. No matter how you look at it, it doesn't make sense. @Lennart Commented Dec 2, 2015 at 1:29
• And where did I accept that $\frac{0}{0}=0$??? It's your argument that doesn't make sense to me. @Lennart Commented Dec 2, 2015 at 1:35
• I don't know why you guys keep downvoting this answer. Maybe this is what they call 'reputation bias'. Commented Dec 3, 2015 at 3:35
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# How do you write 3 tens?
## How do you write 3 tens?
We write numbers using only ten symbols (called Digits)….Example: “35” means 3 Tens and 5 Ones, which is also 3 × 10 + 5 × 1.
Tens Ones
3 5
## What is the standard form of 5 tens 5 tens?
5 tens is equal to 50. So, then what is 50 + 50? Just like 5 dollars + 5 dollars = 10 dollars (and you might want to trade it for a \$10 bill now), 5 tens + 5 tens = 10 tens (and they have \$100 bills, too).
How is 5 tens written?
We know that 1 ten means 10, so 5 tens means 5 * 10 = 50.
How many 1’s are there from 1 to 100?
From \[91\] to \[100\], 1 the digit 1 appears two times for 91 and 100. Hence, the digit 1 appears in numbers from \[1\] to \[100\] for 21 times.
### What is 20 tens the same as?
Hence, 200 is equal to 20 tens.
### How many tens are there in a 100?
10 tens
There are 10 tens in 100.
What is the value of 5 Tens?
Place value is the value of a digit according to its position in the number such as ones, tens, hundreds, and so on. For example, 5 in 3458 represents 5 tens, or 50.
How much is 5 tens more than 50 ones?
5 tens more than 50 tens is 150.
## How to write 1 ten and 2 ones?
It says we have 1 Ten and 2 Ones, which makes 12. This can also be written as 1 × 10 + 2 × 1. What if we have 1 Ten, but no Ones? We show “no Ones” by putting a zero there: We have to put a Zero in the Ones place or “10” looks like “1”. A Hundred Or More When we have more than 99 items, we start another column – the “hundreds” column.
## Which is correct three million or two hundred fifty one thousand?
Going back to our example, 3 becomes “three million,” and 251 becomes “two hundred fifty-one thousand.” All together, we get “three million, two hundred fifty-one thousand, four hundred sixty-nine.” If the number isn’t whole, like 0.42, the process is just a little bit different.
How to write TENS, units and tens in math?
There are 6 tens, so we write a 6 in the tens column . Finally, we count how many units we have. There are 3 units remaining, so we write a 3 in the units column . The digits in the place value columns then form our total number. We have 4, 6 and then 3. Therefore, we have the number 463 (four hundred and sixty-three).
How many sets of hundreds are in the tens place?
It tells you there are 3 sets of hundreds in the hundreds place. The 6 is in the tens place. It tells you there are 6 sets of The 5 is in the ones place.
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# In the given figure on a square handkerchief, nine circular designs each of radius $7cm$ are made. Find the area of the remaining of the handkerchief.
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Hint: Find the relation between the radius of the circle and the side of the square and then calculate the required areas.
Let the radius of the circle is $r$ and the side of the square handkerchief is $S$. The radius is given in the question. So, we have:
$\Rightarrow r = 7cm$
As we can see from the figure, each side of the square is completely covered by three circles.
Thus the side of the square will be the sum of the lengths of the diameter of these three circles. But the circles are of equal diameters, then we have:
$\Rightarrow s = 2r + 2r + 2r, \\ \Rightarrow s = 2(7) + 2(7) + 2(7) \\ \Rightarrow s = 14 + 14 + 14, \\ \Rightarrow s = 42 \\$
Thus the side of the square is $42 cm$ . And we know the formula for the area of square which is ${s^2}$. So we have:
$\Rightarrow {A_{square}} = {s^2}, \\ \Rightarrow {A_{square}} = {\left( {42} \right)^2}, \\ \Rightarrow {A_{square}} = 1764c{m^2} \\$
Area of circle is $\pi {r^2}$. And there are $9$ circles in the square. So, the total area of all the circles is:
$\Rightarrow {A_{circles}} = 9\pi {r^2}, \\ \Rightarrow {A_{circles}} = 9 \times \dfrac{{22}}{7} \times {\left( 7 \right)^2}, \\ \Rightarrow {A_{circles}} = 1386c{m^2}. \\$
Therefore, the area of the remaining part of the handkerchief is:
$\Rightarrow {A_{remaining}} = {A_{square}} - {A_{circles}}, \\ \Rightarrow {A_{remaining}} = 1764 - 1386, \\ \Rightarrow {A_{remaining}} = 378c{m^2}. \\$
Thus, the area of the remaining portion of the handkerchief is $378c{m^2}$.
Note: In such cases, when one of the standard geometrical figures is inscribed in another standard figure, finding the relation between the sides of both the figures is the key point to solve the question.
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# SAT / ACT Prep Online Guides and Tips
Trigonometry and radians are new additions to the SAT Math section! Do you love SOHCAHTOA and ${π}$ angle measurements? Do you hate trigonometry and radians and don't know what SOHCAHTOA or ${π}/{2}$ means? No matter how you feel about SAT trigonometry, there is no need to stress. In this guide, I'll let you know everything you need to know about trigonometry and radians for the SAT Math test and guide you through some practice problems.
## Trigonometric Formulas: Sine, Cosine, Tangent
Although trigonometry makes up less than 5% of all math questions, you still want to get those questions right, and you won't be able to answer any trigonometry questions correctly without knowing the following formulas:
Find the sine of an angle given the measures of the sides of the triangle.
$$\sin(x)={\(Measure\: of\: the\: opposite\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}$$
In the figure above, the sine of the labeled angle would be ${a}/{h}$
Find the cosine of an angle given the measures of the sides of the triangle.
$$\cos(x)={\(Measure\: of\: the\: adjacent\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}$$
In the figure above, the cosine of the labeled angle would be ${b}/{h}$.
Find the tangent of an angle given the measures of the sides of the triangle.
$$\tan(x)={\(Measure\: of\: the\: opposite\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: adjacent\: side\: to\: the\: angle)}$$
In the figure above, the tangent of the labeled angle would be ${a}/{b}$.
A helpful memory trick is an acronym: SOHCAHTOA.
Sine equals Opposite over Hypotenuse
You should also know the complementary angle relationship for sine and cosine, which is $\sin(x°)=\cos(90°−x°)$.
## How to Apply Trigonometry Skills on SAT Math
There are two main trigonometry questions types you'll see on the test. I'll teach you how to address each.
Question type 1 will ask you to find the sine, cosine, or tangent and using the measures of the sides of the triangle. In order to answer these questions, you will need to use a diagram (that means drawing one if it's not given to you). Let's walk through this example:
Triangle ABC is a right triangle where angle B measures 90°; the hypotenuse is 5 and side AB is 4. What is cosine A?
First, set up this triangle using the given information:
Then, identify the information you need. In this case, the question asked for the cosine A. We know, based on the previous formulas that $\cos(A)={\(Measure\: of\: the\: adjacent\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}$. Identify the pieces you need: the angle, the adjacent side to the angle, and the hypotenuse:
We have all the information we need, so we just need to put it into the formula:
$\cos(A)={\(Measure\: of\: the\: adjacent\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}={4}/{5}$.
${4}/{5}$ is the answer.
A slightly harder version of this question might ask you for sine A instead of cosine A. If you look back at the diagram, you'll notice we don't know what the measure of the opposite side to angle A is (which is what we need to find sine A).
In that case, we need to use the Pythagorean theorem (or our knowledge of 3-4-5 right triangles) to find the measure of the opposite side to angle A (BC).
$$BC=√{(5^2)-(4^2)}=√{(25)-(16)}=√{9}=3$$
Know that we know that side BC is 3, we just need to put it into the formula:
$$\sin(A)={\(Measure\: of\: the\: opposite\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}={3}/{5}$$
Question type 2 will ask you to find the sine, cosine, or tangent of an angle using a different given sine, cosine, or tangent of an angle. Similarly to question type one, to answer these questions, you'll need to use a diagram (that means drawing one if it's not given to you). Check out this example:
In a right ABC triangle, where B is the right angle, $\cos(A)={4}/{5}$. What is the sin(C)?
You want to attack these problems by drawing a diagram, but first you need to figure out what should go where. Use the cosine formula to figure out how to draw the diagram.
$$\cos(A)={\(Measure\: of\: the\: adjacent\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}={4}/{5}$$
Meaure of adjacent side (AB) = 4
Measure of the hypotenuse (AC) = 5
You may notice it's the same triangle from the previous example. In this case, we want to find cosine C. We know, based on the previous formulas that $\sin(C)={Measure\: of\: the\: opposite\: side\: to\: the\: angle}/{Measure\: of\: the\: hypotenuse}$. Identify the pieces you need: the angle, the adjacent side to the angle, and the hypotenuse.
$$\sin(C)={\(Measure\: of\: the\: opposite\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}={4}/{5}$$
${4}/{5}$ is the answer.
A slightly harder version of this question might ask you for tangent C instead of sine C. If you look back at the diagram, you'll notice we don't know what the measure of the adjacent side to angle C is (which is what we need to find tan A).
In that case, we need to use the Pythagorean theorem (or our knowledge of 3-4-5 right triangles) to find the measure of the adjacent side to angle C (BC).
$$BC=√{(5^2)-(4^2)}=√{(25)-(16)}=√{9}=3$$
Know that we know that side BC is 3, we just need to put it into the formula:
$$\tan(C)={\(Measure\: of\: the\: opposite\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: adjacent\: side\: to\: the\: angle)}={4}/{3}$$
Now that we know how to apply the necessary formulas to tackle trig questions, let's try to apply them to some real SAT practice problems.
## SAT Trigonometry Practice Problems
### Example #1
Answer Explanation: Triangle ABC is a right triangle with its right angle at B. Therefore, AC is the hypotenuse of right triangle ABC, and AB and BC are the legs of right triangle ABC. According to the Pythagorean theorem,
$$AB=√(202)−(162)=√(400)−(256)=√144=12$$
Since triangle DEF is similar to triangle ABC, with vertex F corresponding to vertex C, the measure of angle F equals the measure of angle C. Therefore, $\sinF=\sinC$. From the side lengths of triangle ABC, $\sin C={\(Measure\: of\: the\: opposite\: side\: to\: the\: angle)}/{\(Measure\: of\: the\: hypotenuse)}={\AB}/{\AC}={12}/{20}={3}/{5}$. Therefore, $\sinF={3}/{5}$.
The final answer is ${3}/{5}$ or .6.
### Example #2
Answer Explanation: There are two ways to solve this. The quicker way is if you know the complementary angle relationship for sine and cosine, which is $\sin(x°)=\cos(90°−x°)$. Therefore, $\cos(90°−x°)={4}/{5}$ or 0.8.
However, you can also solve this problem by constructing a diagram using the given information. It's a right triangle (which it has to be to use sine/cosine), and the sine of angle x is ${4}/{5}$ if $\sine={\(opposite\: side)}/{\hypotenuse}$ then the opposite side is 4 long, and the hypotenuse is 5 long:
Since two of the angles of the triangle are of measure x° and 90°, the third angle must have the measure $180°−90°−x°=90°−x°$. From the figure, $\cos(90°−x°)$, which is equal the ${adjacent\: side}/{the\: hypotenuse}$, is also ${4}/{5}$ or 0.8.
### Example #3
Answer Explanation: Similarly to the other trigonometry problem, there are two ways to solve this problem.
The quicker way is to realize that x and y are complementary angles (add up to 90°). Then, using the complementary angle relationship for sine and cosine, which is $\sin(x°)=\cos(90°−x°)$, you realize that $\cos(y°)=0.6$.
However, you can also solve this problem by constructing a diagram using the given information. It's a right triangle (which it has to be to use sine/cosine), and the sine of angle x is 0.6. Therefore, the ratio of the side opposite the x° angle to the hypotenuse is .6.
The side opposite the x° angle is the side adjacent to the y° angle. $\cos(y°)={\(the\: side\: adjacent\: to\: the\: y°\: angle)}/{\(the\: hypotenuse)}={6}/{10}$, is equal to .6.
Radians will only account for a small portion (around 5%) of SAT math questions, but you still want to get those questions right! Radians are one of the trickier concepts. What do you need to know about radian measure?
The bare bones definition: Radian is a measure of an angle (just as degree is a measure of angle).
The in-depth/conceptual version: Radian is a measure of an angle that is based on the length of the arc that the angle intercepts on the unit circle. That sounds like gibberish I know. Let me break it down. A unit circle is a circle with a radius of 1 unit. See picture:
Gustavb/Wikimedia
The circumference (or length around) this unit circle is ${2π}$, since ${C=2πr}$, and r=1.
If the measure of an angle were 360°, the radian measure would be ${2π}$ since the length of the arc that the 360° angle intercepts on the unit circle would be the whole circumference of the circle (which we already established was ${2π}$). Here are some good basic radian measures to have memorized:
Degrees Radians (exact) 30° ${π}/{6}$ 45° ${π}/{4}$ 60° ${π}/{3}$ 90° ${π}/{2}$
### How to Convert Between Angle Measure in Degrees and Radians
To go from degrees to radians, you need to multiply by ${π}$, divide by 180°. Here is how to convert 90° to radians:
$${90°π}/{180°}$$
$$={π}/{2}$$
To go from radians to degrees, you need to multiply by 180°, divide by ${π}$. Here is how to convert ${π}/{4}$ to degrees:
$${({π}/{4})(180°)}/{π}$$
$$={({180°π}/{4})/{π}$$
$$=45°$$
### How to Evaluate Trigonometric Functions at Benchmark Angle Measures
The benchmark angle measures (as defined by the College Board) are 0, ${π}/{6}$, ${π}/{4}$, ${π}/{3}$, ${π}/{2}$ radians which are equal to the angle measures 0°, 30°, 45°, 60°, and 90°, respectively.
You need to be able to use these with the trigonometric functions described in the above trigonometry section (sine, cosine, and tangent). You will not be asked for values of trigonometric functions that require a calculator.
Remember, the complementary angle relationship for sine and cosine, which is $\sin(x°)=\cos(90°−x°)$ will be $\sin(x)=\cos({π}/{2}−x)$ when converted into radians.
### Example #1
Answer Explanation: The correct answer is 6. By the distance formula, the length of radius OA is $√{((√3)^2)+(1^2)}=√{3+1}=√{4}=2$. Thus, $\sin(∠AOB)={1}/{2}$.
Therefore ∠AOB is 30°, which is equal to $30({π}/{180})={π}/{6}$ radians. Hence, the value of a is 6.
### Example #2
Answer Explanation: A complete rotation around a point is 360° or ${2π}$ radians. Since the central angle AOB has measure ${5π}/{4}$ radians, it represents $/{2π}={5}/{8}$ of a complete rotation around point O. Therefore, the sector formed by central angle AOB has area equal to ${5}/{8}$ the area of the entire circle. The answer is ${5}/{8}$ or in decimal form .625.
### Example #3
Which of the following is equivalent to $\cos({3π}/{10})$?
A) $\-cos ({π}/{5})$
B) $\sin ({7π}/{10})$
C) $\-sin ({π}/{5})$
D) $\sin ({π}/{5})$
Answer Explanation: To answer this question correctly, you need to both understand trigonometry and radians. Sine and cosine are related by the equation $\sin(x)=\cos({π}/{2}-x)$.
In order to find out what the equivalent to $\cos({3π}/{10})$ is, you need to change ${3π}/{10}$ into the form ${π}/{2}-x$. To do that, you need to set up an equation:
$${3π}/{10}= {π}/{2}-x$$
Then, solve for x.
$${3π}/{10}-{π}/{2}=-x$$
$${3π}/{10}-{5π}/{10}=-x$$
$$-{2π}/{10}=-x$$
$${2π}/{10}=x$$
$${π}/{5}=x$$
Therefore, $\cos({3π}/{10})=\cos({π}/{2}-{π}/{5})=\sin({π}/{5})$. D is the correct answer.
## Test Yourself on SAT Trigonometry Questions!
### Practice #1
In triangle DCE, the measure of angle C is 90°, $\DC=5$ and $\CE=12$. What is the value of $\sin(D)$?
### Practice #2
In a right triangle, $\cos({π}/{2}-x)={6}/{8}$. What is $\sin(x)$?
### Practice #3
In circle O, central angle AOB has a measure of ${3π}/{4}$ radians. The area of the sector formed by central angle AOB is what fraction of the area of the circle?
Answers: #1: ${12}/{13}$, #2: ${6}/{8}$, 3) ${3}/{8}$
## What's Next?
Now that you know how to handle trigonometry and radians, make sure you're prepared for all of the other math topics you'll see on the SAT. All of our math guides will take you through strategies and practice problems for all the topics covered on the math section, from integers to ratios, circles to polygons (and more!).
Feeling anxious about test day? Make sure you know exactly what to do and bring to ease your mind and settle your nerves before it's time to take your SAT.
Running out of time on the SAT math section? Look no further than our guide to help you beat the clock and maximize your SAT math score.
Angling to get a perfect score? Check out our guide to getting a perfect 800, written by a perfect-scorer.
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# What algorithm is used to calculate GCD of two integers?
Updated: 4/28/2022
Wiki User
11y ago
There are two main methods:
1. Choose one of the numbers to be the dividend of a division and the other to be the divisor.
2. Perform the division
3. Ignore the quotient and keep the remainder
4. If the remainder is zero, the last divisor is the GCD
5. Replace the dividend by the divisor
6. Replace the divisor by the last remainder
7. Repeat from step 2.
It doesn't matter which number is the dividend and which is the divisor of the first division, but if the larger is chosen as the divisor, the first run through the steps above will swap the two over so that the larger becomes the dividend and the smaller the divisor - it is better to choose the larger as the dividend in the first place.
• Prime factorisation
Express the numbers in their prime factorisations in power format.
Multiply the common primes to their lowest power together to get the GCD.
The first is limited to two numbers, but the latter can be used to find the gcd of any number of numbers.
Examples:
GCD of 500 and 240:
• Euclid's method:
500 ÷ 240 = 2 r 20
240 ÷ 20 = 6 r 0
gcd = 20
• Prime factorisation:
500 = 22 x 53
240 = 24 x 3 x 5
gcd = 22 x 5 = 20
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11y ago
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Related questions
### What is Euclid's Algorithm?
Euclid's algorithm is a popular algorithm to compute the GCD of two numbers. Algorithm: Gcd(a,b) = Gcd(b, a mod b), where a>=b and Gcd(a,0) = a Say we want to find the GCD of 72 and 105. 105 mod 72 = 33, so GCD(72,105) = GCD(33,72) 72 mod 33 = 6, so GCD(33,72) = GCD(6,33) 33 mod 6 = 3 so GCD(6,33) = GCD(3,6) 6 mod 3 = 0 so GCD(3,6) = GCD(0,3) = 3. So the GCD of 72 and 105 is 3.
### What is the least common multiple of decimals and fractions in C?
To calculate the least common multiple (lcm) of decimals (integers) and fractions you first need to calculate the greatest common divisor (gcd) of two integers: int gcd (int a, int b) { int c; while (a != 0) { c = a; a = b % a; b = c; } return b; } With this function in place, we can calculate the lcm of two integers: int lcm (int a, int b) { return a / gcd (a, b) * b; } And with this function in place we can calculate the lcm of two fractions (a/b and c/d): int lcm_fraction (int a, int b, int c, int d) { return lcm (a, c) / gcd (b, d); }
### How do you write a algorithm that gives the GCD of two given numbers?
algorithm GCD (a, b) is:while (a b) doif a > b then a := a - b else b := b - aend whilereturn a
### What is GCD in C Plus Plus?
A C++ implementation of the Binary GCD (Stern's) algorithm is shown in the Related Link below.
### What is the greatest common factor of 275 and 375?
Euclid's Algorithm (http://www.cs.berkeley.edu/~vazirani/s99cs170/notes/lec3.pdf). the mod function (or %, as used here) is equal to the remainder of x/y. In this first case, 375 mod 275 = the remainder of 375/275, 375/275 is 1 r100 thus 375%275=100. gcd(375,275) => gcd(275,375%275) = gcd(275,100) =>gcd(100,275%100) = gcd(100,75) => gcd(75,100%75) = gcd(75,25) => gcd(25,75%25) = gcd(25,0) ===> gcd is 25.
### Find the greatest common divisor in of 11 plus 7i and 18-i?
Use the division algorithm. If b = pa + r, then gcd(b,a) = gcd(a,r). Then you can apply the division algorithm again with a = qr + r' and gcd(a,r) = gcd(r, r'). Note that each time the square norm of the remainder gets smaller and smaller, so eventually this process will terminate and you can get the answer. Here, it should be 1.
### A program to find GCD andLCM of two numbers?
// recursive algorithm to return gcd using Euclid's Algorithm int gcd (int a, int b) { if (a<0) a= -a; if (b<0) b= -b; if (a<b) { int tmp; tmp= a; a= b; b= tmp; } if (b == 0) return a; return gcd (b, a%b); } // LCM using gcd int LCM (int a, int b) { int t; t = a*b; if (t<0) t=-t; return t / gcd (a, b); }
### Write a program to find gcd using recursive method in java?
for two positive integers: public static int gcd(int i1, int i2) { // using Euclid's algorithm int a=i1, b=i2, temp; while (b!=0) { temp=b; b=a%temp; a=temp; } return a; }
### Examples of Euclid's algorithm for finding GCD of large numbers?
Euclid's algorithm is a time-tested method for finding the greatest common divisor (GCD) of two numbers. It's based on the principle that the greatest common divisor of two numbers also divides their difference. This algorithm is efficient and works well for large numbers, making it a practical choice in numerous applications. The algorithm operates in a recursive or iterative manner, continually reducing the problem size until it reaches a base case. Here’s how Euclid's algorithm works: print (gcd (a, b) ) # Output: 3ere >a>b , subtract b from a. Replace a with (a−b). Repeat this process until a and b become equal, at which point, a (or b) is the GCD of the original numbers. A more efficient version of Euclid’s algorithm, known as the Division-based Euclidean Algorithm, operates as follows: Given two numbers a and b, where >a> b, find the remainder of a divided by b, denoted as r. Replace a with b and b with r. Repeat this process until b becomes zero. The non-zero remainder, a, is the GCD of the original numbers. In this example, even though a and b are large numbers, the algorithm quickly computes the GCD. The division-based version of Euclid’s algorithm is more efficient than the subtraction-based version, especially for large numbers, as it reduces the problem size more rapidly. Euclid's algorithm is a fundamental algorithm in number theory, with applications in various fields including cryptography, computer science, and engineering. Its efficiency and simplicity make it a powerful tool for computing the GCD, even for large numbers.
### What is the gcd of 5 over 8?
That only applies to integers. The GCF of 5 and 8 is 1.
### How do you write a program to read two integers and print the greater common divisor?
#include<stdio.h> int gcd (int a, int b) { if (a==0) return b; if (b==0) return a; return a<b ? gcd (a, b%a) : gcd (b, a%b); } int main (void) { int a, b; printf ("Enter two integers: ") scanf ("%d\n", &a); scanf ("%d\n", &b); printf ("The GCD of %d and %d is %d\n", a, b, gcd (a, b)); return 0; }
### How can you find the LCM in c without using any loop and condition?
The LCM can be calculated without using any loop or condition as follows: int lcm (int a, int b) { return a / gcd (a, b) * b; } The problem is that the typical implementation for the GCD function uses Euclid's algorithm, which requires a conditional loop: int gcd (int a, int b) { while (b!=0) b ^= a ^= b ^= a %= b; return a; } So the question is really how do we calculate the GCD without a conditional loop, not the LCM. The answer is that we cannot. There are certainly alternatives to Euclid's algorithm, but they all involve conditional loops. Although recursion isn't technically a loop, it still requires a conditional expression to terminate the recursion.
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# What is the Prime Factorization Of 46?
• Equcation for number 46 factorization is: 2 * 23
• It is determined that the prime factors of number 46 are: 2, 23
## Is 46 A Prime Number?
• No the number 46 is not a prime number.
• Forty-six is a composite number. Because 46 has more divisors than 1 and itself.
## How To Calculate Prime Number Factorization
• How do you calculate natural number factors? To get the number that you are factoring just multiply whatever number in the set of whole numbers with another in the same set. For example 7 has two factors 1 and 7. Number 6 has four factors 1, 2, 3 and 6 itself.
• It is simple to factor numbers in a natural numbers set. Because all numbers have a minimum of two factors(one and itself). For finding other factors you will start to divide the number starting from 2 and keep on going with dividers increasing until reaching the number that was divided by 2 in the beginning. All numbers without remainders are factors including the divider itself.
• Let's create an example for factorization with the number nine. It's not dividable by 2 evenly that's why we skip it(Remembe 4,5 so you know when to stop later). Nine can be divided by 3, now add 3 to your factors. Work your way up until you arrive to 5 (9 divided by 2, rounded up). In the end you have 1, 3 and 9 as a list of factors.
## Mathematical Information About Numbers 4 6
• About Number 4. Four is linear. It is the first composite number and thus the first non-prime number after one. The peculiarity of the four is that both 2 + 2 = 4 and 2 * 2 = 4 and thus 2^2 = 4. Four points make the plane of a square, an area with four sides. It is the simplest figure that can be deformed while keeping it's side lengths, such as the rectangle to parallelogram. Space let's us arrange equidistantly a maximum of four points. These then form a tetrahedron (tetrahedron), a body with four identical triangular faces. Another feature of the four is the impossibility of an algebraic equation of higher degree than four square roots using simple arithmetic and basic operations dissolve.
• About Number 6. Six is the smallest composite number with two distinct prime factors, and the third triangular number. It is the smallest perfect number: 6 = 1 + 2 + 3 and the faculty of 3 is 6 = 3! = 1 * 2 * 3, which is remarkable, because there is no other three numbers whose product is equal to their sum. Similarly 6 = sqrt(1 ^ 3 + 2 + 3 ^ 3 ^ 3). The equation x ^ 3 + Y ^ 3 ^ 3 + z = 6xyz is the only solution (without permutations) x = 1, y = 2 and z = 3. Finally 1/1 = 1/2 + 1/3 + 1/6. The cube (from the Greek) or hexahedron (from Latin) cube is one of the five Platonic solids and has six equal areas. A tetrahedron has six edges and six vertices an octahedron. With regular hexagons can fill a plane without gaps. Number six is a two-dimensional kiss number.
## What is a prime number?
Prime numbers or primes are natural numbers greater than 1 that are only divisible by 1 and with itself. The number of primes is infinite. Natural numbers bigger than 1 that are not prime numbers are called composite numbers.
## What is Prime Number Factorization?
• In mathematics, factorization (also factorisation in some forms of British English) or factoring is the decomposition of an object (for example, a number, a polynomial, or a matrix) into a product of other objects, or factors, which when multiplied together give the original. For example, the number 15 factors into primes as 3 x 5, and the polynomial x2 - 4 factors as (x - 2)(x + 2). In all cases, a product of simpler objects is obtained. The aim of factoring is usually to reduce something to basic building blocks, such as numbers to prime numbers, or polynomials to irreducible polynomials.
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# A bag contains 4 red balls and 4 black balls. Another bag contains 2 red and 6 black. One of the bags is selected at random and a ball is drawn from the bag which is found to be red. Find the probability that the ball is drawn from the first bag.
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Toolbox:
• Given $E_1, E_2, E_3.....E_n$ are mutually exclusive and exhaustive events, we can find the conditional probability P(E_i|A) for any event A associated w/ $E_i$ using the Bayes theorem as follows: $$\;P(E_i/A)$$=$$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$$
Step 1:
Let $E_1$ be the event that the first bag is selected, and $E_2$ the event that the second bag is selected.
Since $E_1$ and $E_2$ are mutually exclusive and exhaustive,
$P(E_1) = P(E_2) = \large\frac{1}{2}$
Let A be the event of drawing a red ball.
Given that Bag1 has 4 red and 4 black balls, n(Bag1) = 8, and n(R1) = 4.
Similarly for Bag2, n(Bag2) = 8 and n(R2) = 2.
$\Rightarrow$ P (drawing a red from Bag 1) = $\large \frac{4}{8} = \frac{1}{2}$ =$P (A/E_1)$
$\Rightarrow$ P (drawing a red from Bag 2) = $\large \frac{2}{8} = \frac{1}{4}$ = $P(A/E_2)$
Step 2:
To find the probability of that the ball is drawn from the first bag, let's use Bayes theorem:
$$\;P(E_i/A)$$=$$\large \frac{P(E_i)P(A/E_i)}{\sum_{i=1}^{n}\;P(E_i)P(A/E_i)}$$
Therefore the probability that the ball is drawn from the first bag = $P(E_1/A) = \large\frac{P(E_1)(P(A|E_1)}{P(E_1)P(A|E_1) + P(E_2)+P(A|E_2)}$
$P(E_1)(P(A/E_1) = \large\frac{1}{2}$$\times$$ \large \frac{1}{2} = \frac{1}{4}$
$P(E_1)P(A/E_1) + P(E_2)+P(A/E_2) = \large\frac{1}{2}$$\times$$\large\frac{1}{2} + \frac{1}{2}$$\times \large\frac{1}{4} \qquad\qquad\qquad\qquad\qquad\qquad\;\;= \large\frac{2+1}{8} \qquad\qquad\qquad\qquad\qquad\qquad\;\;=\large \frac{3}{8} Therefore P(E_1/A) = \large \frac{\Large \frac{1}{4}}{\Large \frac{3}{8}}$$ =\large \frac{2}{3}$
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### 4. Curves and Coordinates
#### a. Coordinate Equations
Equations C-16 and C-17 are general equations for computing coordinates using direction and distance from a known point, Figure C-26.
Equation C-16 Equation C-17
Figure C-26. Coordinate Computation
Direction (Dir) may be either a bearing or azimuth.
Curve point coordinates can be computed using these equations from a base point. Since the radial chord method uses the BC as one end of all the chords, it can also be used as the base point for coordinate computations.
#### b. Computation Process
Assuming we start with the tangents and PI, then fit a curve, the general process is as follows:
Figure C-27
The original tangent lines have directions; PC has coordinates.
Figure C-28
A curve is fit to the tangents.
End points are at distance T from the PI along the tangents.
Figure C-29
Compute coordinates of BC using back-direction of the tangent BC-PI and T.
Equation C-18 Equation C-19
Figure C-30
Compute coordinates of EC using direction of the tangent PI-EC and T.
These will be use for a later math check.
Equation C-20 Equation C-21
Figure C-31
Use a curve point's deflection angle to compute the direction if its radial chord from the BC.
Equation C-22
δ is positive for right deflections, negative for left.
Using the direction and chord length, compute the point's coordinates.
Equation C-23 Equation C-24
#### c. Example
Continuing with the previous example problem.
Summary of given and computed curve data:
Δ = 55°00'00" R = 500.00 ft D = 11°27'33.0" L = 479.965 ft T = 260.284 ft LC = 461.749 ft E = 63.691 ft M = 56.494 ft
Point Station PI 25+00.00 BC 22+39.716 EC 27+19.681 Bk = 27+60.284 Ah
Additional information: Azimuth of the initial tangent is 75°40'10"; coordinates of the PI are 1000.00 N, 5000.00' E.
Compute coordinates of the BC:
Compute the coordinates of the EC:
Set up Equatons C-21 through C-24 for this curve.
This is the Radial Chord table computed previously:
Curve Point Arc dist, li, (ft) Defl angle,δi Radial chord, c EC 27+19.681 Bk 479.965 27°30'00.0" 461.748 27+00 460.284 26°22'20.4" 444.203 26+00 360.284 20°38'33.9" 352.540 25+00 260.284 14°54'47.4" 257.355 24+00 160.284 9°11'01.0" 159.599 23+00 60.284 3°27'14.5" 60.248 BC 22+39.716 0.000 0°00'00.0" 0.000
Add three more columns for direction and coordinates:
Curve Point Azimuth, Azi North, Ni East, Ei EC 27+19.681 Bk 27+00 26+00 25+00 24+00 23+00 BC 22+39.716
Complete the table using the three equations for this curve
At 22+39.716, we're still at the BC so the coordinates don't change.
At 23+00:
At 24+00:
and so on for the rest of the curve points.
The completed curve table is:
Curve Point Azimuth, Azi North, Ni East, Ei EC 27+19.681 Bk 103°10'10.0 830.375 5197.419 27+00 102°02'30.4" 842.904 5182.244 26+00 96°18'43.9" 896.816 5098.218 25+00 90°34'57.4" 932.959 5005.157 24+00 84°51'11.0" 949.894 4906.770 23+00 79°07'24.5" 946.944 4806.981 BC 22+39.716 75°40'10" 935.576 4747.815
Math check: the coordinates computed for the EC in the table should be the same as the EC coordinates computed from the PI. Within rounding error, that's the case here.
#### d. Summary
The radial chord method lends itself nicely to computing curve point coordinates. The computations are not complex, although they are admittedly tedious.
Once coordinates are computed, field stakeout is much more flexible using Coordinate Geometry (COGO).
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# Chi-Square Calculator
You can use this chi-square calculator as part of a statistical analysis test to determine if there is a significant difference between observed and expected frequencies.
To use the calculator, simply input the true and expected values (on separate lines) and click on the "Calculate" button to generate the results.
## What is a Chi-square Test?
A chi-square test is a popular statistical analysis tool that is employed to identify the extent to which an observed frequency differs from the expected frequency.
Let's look at an example.
Let's say you are a college professor. The 100 students you teach complete a test that is graded on a scale ranging from 2 (lowest possible grade) through to 5 (highest possible grade). In advance of the test, you expect 25% of the students to achieve a 5, 45% to achieve a 4, 20% to achieve a 3, and 10% to get a 2.
After the test, you grade the papers. You can then use the chi-square test to determine the extent to which your predicted grades differed from the actual grades.
## How to Calculate a Chi-square
The chi-square value is determined using the formula below:
X2 = (observed value - expected value)2 / expected value
Returning to our example, before the test, you had anticipated that 25% of the students in the class would achieve a score of 5. As such, you expected 25 of the 100 students would achieve a grade 5. However, in reality, 30 students achieved a score of 5. As such, the chi-square calculation is as follows:
X2 = (30 - 25)2 / 25 = (5)2 / 25 = 25 / 25 = 1
## An In-depth Example of the Chi-square Calculator
Let's take a more in-depth look at the paper grading example.
The grade distribution for the 100 students you tested were as follows: 30 received a 5, 25 received a 4, 40 received a 3, and 5 received a 2.
• a.) We can now determine how many students were expected to receive each grade per the forecast distribution.
• Grade 2: 0.10 * 100 = 10
• Grade 3: 0.20 * 100 = 20
• Grade 4: 0.45 * 100 = 45
• Grade 5: 0.25 * 100 = 25
• b.) We can use this information to determine the chi-square value for each grade.
• Grade 2: X2 = (5 - 10)2 / 10 = 2.5
• Grade 3: X2 = (40 - 20)2 / 20 = 20
• Grade 4: X2 = (25 - 45)2 / 45 = 8.89
• Grade 5: X2 = (30 - 25)2 / 25 = 1
• c.) Finally, we can sum the chi-square values: X2 = 2.5 + 20 + 8.89 + 1 = 32.39
You may also be interested in our P-Value Calculator or T-Value Calculator
Chi-Square Calculator
P-Value from Chi-Square Calculator
Use this tool to calculate the p-value for a given chi-square value and degrees of freedom.
You can use this chi-square calculator as part of a statistical analysis test to determine if there is a significant difference between observed and expected frequencies.
To use the calculator, simply input the true and expected values (on separate lines) and click on the "Calculate" button to generate the results.
## What is a Chi-square Test?
A chi-square test is a popular statistical analysis tool that is employed to identify the extent to which an observed frequency differs from the expected frequency.
Let's look at an example.
Let's say you are a college professor. The 100 students you teach complete a test that is graded on a scale ranging from 2 (lowest possible grade) through to 5 (highest possible grade). In advance of the test, you expect 25% of the students to achieve a 5, 45% to achieve a 4, 20% to achieve a 3, and 10% to get a 2.
After the test, you grade the papers. You can then use the chi-square test to determine the extent to which your predicted grades differed from the actual grades.
## How to Calculate a Chi-square
The chi-square value is determined using the formula below:
X2 = (observed value - expected value)2 / expected value
Returning to our example, before the test, you had anticipated that 25% of the students in the class would achieve a score of 5. As such, you expected 25 of the 100 students would achieve a grade 5. However, in reality, 30 students achieved a score of 5. As such, the chi-square calculation is as follows:
X2 = (30 - 25)2 / 25 = (5)2 / 25 = 25 / 25 = 1
## An In-depth Example of the Chi-square Calculator
Let's take a more in-depth look at the paper grading example.
The grade distribution for the 100 students you tested were as follows: 30 received a 5, 25 received a 4, 40 received a 3, and 5 received a 2.
• a.) We can now determine how many students were expected to receive each grade per the forecast distribution.
• Grade 2: 0.10 * 100 = 10
• Grade 3: 0.20 * 100 = 20
• Grade 4: 0.45 * 100 = 45
• Grade 5: 0.25 * 100 = 25
• b.) We can use this information to determine the chi-square value for each grade.
• Grade 2: X2 = (5 - 10)2 / 10 = 2.5
• Grade 3: X2 = (40 - 20)2 / 20 = 20
• Grade 4: X2 = (25 - 45)2 / 45 = 8.89
• Grade 5: X2 = (30 - 25)2 / 25 = 1
• c.) Finally, we can sum the chi-square values: X2 = 2.5 + 20 + 8.89 + 1 = 32.39
You may also be interested in our P-Value Calculator or T-Value Calculator
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# How to establish traversability
Which of these diagrams can be drawn without taking your pencil off the paper and without going over the same line twice?
(Note: This is the famous Konigsberg bridges question which Euler posed in 1735, and is the very first problem solved about networks.)
Diagram 1
## Key terms relating to traversability
We can model the problems above as graphs by introducing a vertex (where two lines meet) and by regarding the lines as edges.
Diagram 2
• If we can find a walk, essentially a sequence of edges, that goes over all the edges without repeating an edge, then the graph is traversable.
• Whether or not a graph is traversable depends on the number of odd vertices it has.
• The number of edges that meet at a vertex is called the degree of the vertex.
• An odd vertex is a vertex with an odd degree. An even vertex is a vertex with an even degree.
Diagram 3
A graph will have an even number of odd vertices, because the sum of the degrees of all the vertices is twice the number of the edges.
A connected graph with even vertices only is traversable, and we can start and finish at the same vertex. It is called Eulerian. Further, we can choose any vertex to be the start/finish.
A connected graph with exactly two odd vertices is traversable as long as we start at one of the odd vertices and finish at the other. It is called semi-Eulerian.
A connected graph with more than two odd vertices is not traversable.
## An example of traversability: Route inspection
### Problem 1
Hone is a postie who has to cycle along all the roads shown in this network. The lengths of the roads in metres are shown on this network. Can he do this without cycling along any road twice?
Diagram 4
All the vertices are even, so Hone is able to transverse the network. He can start and return to any point. One possible route is AEBCDFEDBGCFA. The length of this postal route is 6000 m or 6 km (the sum of the lengths of the edges).
### Problem 2
New Zealand Post add one more road to Hone’s route that goes directly between A and D. This road passes through a tunnel under the road EF so there is no vertex where these edges (AD and EF) cross on the graph.
Diagram 5
(There are two odd vertices, A and D, so Hone could traverse the network by starting at A and finishing at D or vice versa. The distance travelled is now 7200 m.)
b) Hone would ideally like to start and finish at the same place so that he can drive to the starting point, cycle round the route, and get back to his car. Find a way Hone can do this that minimises the distance he cycles.
If he wants to do this, and start at A, he will have to traverse the graph from A to D and then find the shortest route back to A. We can see from the graph that the shortest way back from D to A is DEA, which adds 1000m to his route.
One possible way of doing this is AEBGCFEDBCDFADEA, and its length is 8200 m. There are many possible routes.
### Problem 3
New Zealand Post adds another road to Hone’s round that goes directly from G to F.
Diagram 6
Hone is unable to transverse the network as there are four odd vertices, A, D, F, and G.
b) Explain how Hone could complete his round, starting and finishing at the same point.
To complete his round, and start and finish at the same point, Hone will have to repeat the edges that connect the four odd vertices: A, D, F, and G.
He can do this by repeating one of the following pairs of paths:
• G–F and A–D: The shortest route to do this is 1600m (GF and DEA)
• G–D and A–F: The shortest route to do this is 1500m (GBD and AF)
• G–A and D–F: The shortest route to do this is 1700m (GFA and DF)
The middle pairing gives the shortest extra distance.
One possible shortest route is AEBGCFEDBCDFGBDAFA, and the length is now 9300 m.
Last updated May 1, 2012
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# Sums of Arithmetic Series
## Sum of numbers whose consecutive terms form an arithmetic sequence.
Estimated17 minsto complete
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Practice Sums of Arithmetic Series
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Arithmetic Series
While it is possible to add arithmetic series one term at a time, it is not feasible or efficient when there are a large number of terms. What is a clever way to add up all the whole numbers between 1 and 100?
#### Watch This
http://www.youtube.com/watch?v=Dj1JZIdIwwo James Sousa: Arithmetic Series
#### Guidance
The key to adding up a finite arithmetic series is to pair up the first term with the last term, the second term with the second to last term and so on. The sum of each pair will be equal. Consider a generic series:
When you pair the first and the last terms and note that the sum is:
When you pair up the second and the second to last terms you get the same sum:
The next logical question to ask is: how many pairs are there? If there are terms total then there are exactly pairs. If happens to be even then every term will have a partner and will be a whole number. If happens to be odd then every term but the middle one will have a partner and will include a pair that represents the middle term with no partner. Here is the general formula for arithmetic series:
where is the common difference for the terms in the series.
Example A
Add up the numbers between one and ten (inclusive) in two ways.
Solution: One way to add up lists of numbers is to pair them up for easier mental arithmetic.
Another way is to note that . There are 5 pairs of 11 which total 55.
Example B
Evaluate the following sum.
Solution: The first term is -2, the last term is 23 and there are 6 terms making 3 pairs. A common mistake is to forget to count the 0 index.
Example C
Try to evaluate the sum of the following geometric series using the same technique as you would for an arithmetic series.
Solution:
The real sum is:
When you try to use the technique used for arithmetic sequences you get:
It is important to know that geometric series have their own method for summing. The method learned in this concept only works for arithmetic series.
Concept Problem Revisited
Gauss was a mathematician who lived hundreds of years ago and there is an anecdote told about him when he was a young boy in school. When misbehaving, his teacher asked him to add up all the numbers between 1 and 100 and he stated 5050 within a few seconds.
You should notice that and that there are exactly 50 pairs that sum to be 101. .
#### Vocabulary
An arithmetic series is a sum of numbers whose consecutive terms form an arithmetic sequence.
#### Guided Practice
1. Sum the first 15 terms of the following arithmetic sequence.
2. Sum the first 100 terms of the following arithmetic sequence.
3. Evaluate the following sum.
1. The initial term is -1 and the common difference is .
2. The initial term is -7 and the common difference is 3.
3. The initial term is -312 and the common difference is 2.
#### Practice
1. Sum the first 24 terms of the sequence
2. Sum the first 102 terms of the sequence
3. Sum the first 85 terms of the sequence
4. Sum the first 97 terms of the sequence
5. Sum the first 56 terms of the sequence
6. Sum the first 91 terms of the sequence
Evaluate the following sums.
7.
8.
9.
10.
11.
12.
13.
14.
15.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 12.4.
### Vocabulary Language: English
arithmetic series
arithmetic series
An arithmetic series is the sum of an arithmetic sequence, a sequence with a common difference between each two consecutive terms.
common difference
common difference
Every arithmetic sequence has a common or constant difference between consecutive terms. For example: In the sequence 5, 8, 11, 14..., the common difference is "3".
series
series
A series is the sum of the terms of a sequence.
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# Chapter 7 - Integration Techniques - 7.2 Integration by Parts - 7.2 Exercises: 19
$= x{\tan ^{ - 1}}x - \frac{1}{2}\ln \,\left( {1 + {x^2}} \right) + C$
#### Work Step by Step
$\begin{gathered} \int_{}^{} {{{\tan }^{ - 1}}xdx} \hfill \\ \hfill \\ set\,\,\,the\,\,substitution \hfill \\ \hfill \\ v = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \to \,\,\,\,\,dv = dx \hfill \\ u = {\tan ^{ - 1}}x\,\, \to \,\,\,\,\,du = \frac{1}{{1 + {x^2}}} \hfill \\ \hfill \\ use\,\,uv - \int_{}^{} {vdu} \hfill \\ \hfill \\ {\text{replacing the values }}{\text{in the equation}} \hfill \\ \hfill \\ = x{\tan ^{ - 1}}x - \int_{}^{} {\frac{x}{{1 + {x^2}}}} dx \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ x{\tan ^{ - 1}}x - \frac{1}{2}\int_{}^{} {\frac{{2x}}{{1 + {x^2}}}dx} \hfill \\ \hfill \\ integrate\,\,{\text{using }}\int {\frac{{du}}{u} = \ln u + C} \hfill \\ \hfill \\ = x{\tan ^{ - 1}}x - \frac{1}{2}\ln \,\left( {1 + {x^2}} \right) + C \hfill \\ \hfill \\ \end{gathered}$
After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.
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Browse Questions
# Solution of the differential equation $\tan y\sec^2x dx+\tan x\sec^2y dy=0$ is $(A)\;\tan x+\tan y=c \quad (B)\;\tan x-\tan y=c \quad(C)\;\frac{\tan x}{\tan y}=c \quad (D\;\tan x.tan y=c$
Toolbox:
• A linear differential equation of the form $\large\frac{dy}{dx}$$=f(x) can be solved by seperating the variables and then integrating it. • \int \large\frac{dx}{x}$$=\log |x|+c$
$\tan y \sec ^2 xdx+\tan x\sec ^2 y dy=0$
This can be written as: $\tan y \sec ^2 x dx=-\tan x \sec ^2 y dy$
Seperate the variables
$\large\frac{\sec ^2 y dy}{\tan y}=-\large\frac{\sec ^2 x dx}{\tan x}$
Put $\tan y=t$ and $\tan x=u$ on differentiating with respect to x we get,
$\sec ^2 ydy=dt$ and $\sec^2 xdx=du$
Substituting this we get,
$\large\frac{dt}{t}=-\frac{du}{u}$
Integrate on both sides,
$\int \large\frac{dt}{t}=-\int \frac{du}{u}$
$=\log |t|=-\log |u|+\log |c|$
Substitute for t and u,
$\log |\tan y|=-\log |tan x|+\log c$
$\log (\tan y).(\tan x)=\log c$
=>$\tan x.\tan y=c$
Hence $D$ is the correct answer
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# Why Probability? - PowerPoint PPT Presentation
Why Probability?
1 / 16
Why Probability?
## Why Probability?
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. Why Probability? • Probability theory describes the likelihood of observing various outcomes for a given population • Statistics uses rules of probability as a tool for making inferences about or describing a population using data from a sample
2. Some Concepts • Experiment: the process by which an observation is obtained • e.g. the roll of a dice • Event: the outcome of an experiment. • e.g. observe a 1; observe an odd number (1,3,5) • Simple Event: an event that cannot be decomposed • Sample Space (S): set of all events
3. Definition of Probability • The probability of an event A is a measure of our belief that the experiment will result in event A. • If we repeat an experiment N times, and event A occurs n times, then P(A) = n / N • Computing probabilities in this way is infeasible in practice, but implications are useful
4. Probability Rules For an event A: • P(A) is between 0 and 1, inclusive • If A contains t simple events, then P(A) = P(E1) + P(E2) + … + P(Et) For a sample space S with s simple events: P(S) = P(E1) + P(E2) + … + P(Es) = 1
5. Event Composition • The Intersection of events A and B is the event that both A and B occur • denoted AB or A∩B • The Union of events A and B is the event that A or B or both occur • denoted AUB
6. Event Composition • A and B are mutually exclusive if there are no simple events in A∩B. If A,B are mutually exclusive then: (1) P(A∩B) = 0 (2) P(AUB) = P(A) + P(B) • The complement of an event A consists of all simple events that are not in A • denoted
7. Conditional Probability • In some cases events are related, so that if we know event A has occurred then we learn more about an event B • Example: Roll a die A: observe an even number (2,4,6) B: observe a number less than 4 (1,2,3) if we know nothing else then P(B) = 3/6 = 1/2 But if we know A has occurred then P(B | A) = 1/3
8. Conditional Probability • More generally, we can express the conditional probability of B given that A has occurred as: • We can rewrite this formula to get the Multiplicative Rule of Probability:
9. Independence • Events are not always be related. Events A and B are independent if and only if: • If A and B are independent, then from Multiplicative Rule of Probability:
10. Rules of Probability Given 2 events A and B: • Additive Probability: • If A and B are mutually exclusive then • P(AB)=0 • P(A+B) = P ( A )+ P ( B ) • Total Probability: for mutually exclusive B1, B2, …
11. Bayes Rule • Take into account prior information when computing probabilities • Let S1, S2, S3,…Sk represent k mutually exclusive, only possible states of nature with prior probabilities P(S1), P(S2),…P(Sk). If an event A occurs, the posterior probability of Si given A is the conditional probability
12. Random Variables • X is a random variable if value that it assumes depends on the random outcome of an experiment • A random variable may be • Discrete: countable number of values • Continuous: infinite number of values
13. Discrete Probability Distribution • The probability distribution for a discrete random variable is a formula, table or graph that provides p(x), the probability associated with observing x • Rules for probability distribution: • 0 <= p(x) <= 1 • ∑ p(x) = 1
14. Expected Value • Expected value (or population mean) of a random variable x with the probability distribution p(x) is Intuition:expected value is weighted average of x
15. Variance of a Random Variable • The variance of random variable x with probability distribution p(x) and expected value E(x)= is given as • The Standard Deviation of random variable x is equal to the square root of its variance.
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##### Which is greater in each of the following:(i) (ii) (iii) (iv) (v)
In this question we have to compare two given fraction i.e. to find whether the given two fractions are equal or one is greater or less than the other.
Concept: It is easier to compare fractions with like denominators. So, we convert all the fractions into like denominators and the fraction which has greater denominator will be greater than the other fraction.
The parts of the given question are solved below:
(i) We have,
Writing the fractions in the form of like denominators we get,
Now,
And,
Since,
15 is greater than 4
Hence,
is greater than
(ii) We have,
Writing the fractions in the form of like denominators we get,
And,
Since,
-5 is greater than -8
Hence,
is greater than
(iii) We have,
Now, Writing the fractions in the form of like denominators we get,
And,
Since,
-8 is greater than -9
Hence,
is greater than
(iv) Here,
We have,
Now, in the question the fractions given already have like denominators.
But, one of the given fractions is positive and the other fraction is negative.
Since, positive is always greater than Negative
Hence,
is greater than
(v) Here,
We have,
First convert the mixed fraction into improper fraction. An improper fraction is the one whose numerator is bigger than the denominator. On doing so, we get,
And,
Since,
-115 is greater than -133
Hence,
is greater than
11
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# Perimeter Lesson
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Category: Education
## Presentation Description
No description available.
## Presentation Transcript
### Perimeter:
Perimeter By Elizabeth Hill Third Grade
### PowerPoint Presentation:
What are we learning? M3M3 Students will understand and measure the perimeter of geometric figures and Determine the perimeter of a geometric figure by measuring and summing the lengths of the sides The State of Georgia Says…
### PowerPoint Presentation:
In Kid Terms I can find the perimeter of different shapes and objects.
### PowerPoint Presentation:
What is perimeter? Perimeter is the measurement of the distance all the way around an object.
### PowerPoint Presentation:
What is the perimeter? Let’s see how many squares it takes to get all the way around. It is 16 all the way around! How far is it all the way around?
### PowerPoint Presentation:
What is the perimeter? Let’s see how many squares it takes to get all the way around. It is 24 all the way around! How far is it all the way around?
### PowerPoint Presentation:
What is the perimeter? Work with your partner to find the perimeter of the 3 shapes? 4 4 5 5 18 3 3 3 3 12 3 3 6 1 3 2 18
Day 2
### PowerPoint Presentation:
What are we learning? M3M3 Students will understand and measure the perimeter of geometric figures and Determine the perimeter of a geometric figure by measuring and summing the lengths of the sides The State of Georgia Says…
### PowerPoint Presentation:
In Kid Terms I can find the perimeter of different shapes and objects.
### PowerPoint Presentation:
Perimeter Song Perimeter measures around the shape. Around the shape, Around the shape, Perimeter measures around the shape. Just add up all the sides.
### Finding Perimeter:
Finding Perimeter P = 14 5 3 6 What is perimeter? How do you find the perimeter of a shape? Remember the P in perimeter reminds you to “plus”! P = S + S + S P = 3 + 5 + 6 Find the perimeter of this shape.
### Finding Perimeter:
Finding Perimeter P = 22 6 8 8 What is perimeter? How do you find the perimeter of a shape? Remember the P in perimeter reminds you to “plus”! P = S + S + S P = 8 + 8 + 6 Find the perimeter of this shape.
### Finding Perimeter:
Finding Perimeter P = 28 5 3 3 What is perimeter? How do you find the perimeter of a shape? Remember the P in perimeter reminds you to “plus”! P = S+S+S+S+S+S+S+S P = 5+5+3+3+3+3+3+3 Find the perimeter of this shape. 5 3 3 3 3
### Finding Perimeter:
Finding Perimeter P = 28 9 3 1 What is perimeter? How do you find the perimeter of a shape? Remember the P in perimeter reminds you to “plus”! P = S+S+S+S+S+S+S+S P = 3+3+1+1+9+9+2 Find the perimeter of this shape. 9 3 1 2
Day 3
### PowerPoint Presentation:
What are we learning? M3M3 Students will understand and measure the perimeter of geometric figures and Determine the perimeter of a geometric figure by measuring and summing the lengths of the sides The State of Georgia Says…
### PowerPoint Presentation:
In Kid Terms I can find the perimeter of different shapes and objects.
### PowerPoint Presentation:
Perimeter Song Perimeter measures around the shape. Around the shape, Around the shape, Perimeter measures around the shape. Just add up all the sides.
### Finding Perimeter:
Finding Perimeter P = 18 6 3 6 What is perimeter? How do you find the perimeter of a shape? Remember the P in perimeter reminds you to “plus”! P = S + S + S P = 3 + 3 + 6 + 6 Find the perimeter of this shape. 3
### Finding Perimeter:
Finding Perimeter P = 20 8 2 8 What is perimeter? How do you find the perimeter of a shape? Remember the P in perimeter reminds you to “plus”! P = S + S + S P = 2+2+8+8 Find the perimeter of this shape. 2
### PowerPoint Presentation:
What do you know about the sides of rectangles Rectangles have 2 long sides that are the same length and 2 short sides that are the same length.
### Finding Perimeter:
Finding Perimeter P = 14 9 What is perimeter? How do you find the perimeter of a shape? Remember the P in perimeter reminds you to “plus”! P = S + S + S + S P = 9 + 5 + 9 + 5 Find the perimeter of this shape. 5 If this side is 9, then we know the opposite side is _____ If that side is 5, then we know the opposite side is _____ If you only add the 9 and the 5 together then you are only finding half of the perimeter.
### PowerPoint Presentation:
Tricky, Tricky! Find the perimeter of this shape. 4
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# Geometric proof – bisector of triangle
Let triangle ABC (denoted △ABC\triangle ABC) be an isosceles triangle in the plane where |AB|=|AC||AB|=|AC|. We try to prove that the perpendicular bisector of the segment BCBC coincides with the bisector of ∠A\angle A (i.e. they are the same line when extend) and that it cuts the line BCBC in half. The diagram shows the configuration of △ABC\triangle ABC .
My proof is as follows.
By definition of the angle bisector of ∠A\angle A, we have ∠BAO=∠CAO=12∠A.\angle BAO = \angle CAO =\frac{1}{2}\angle A. Let the bisector of ∠A\angle A intersect the line segment BCBC at the point O.O.
As △ABC\triangle ABC is isosceles, we can say that ∠OBA=∠ACO=α (say) ,\angle OBA =\angle ACO = \alpha \text{ (say) }, (assume without proof) and so by side angle side (SAS) (assume without proof) we have shown that △OBA≅△OCA.\triangle OBA \cong \triangle OCA.
This means that |BO|=|OC||BO|=|OC| and that ∠COB=∠BOA=\angle COB = \angle BOA= since both ∠COB=∠BOA\angle COB =\angle BOA and the fact that ∠COB+∠BOA=π,\angle COB + \angle BOA=\pi, ( since ∠COB\angle COB is a straight angle) we conclude that ∠AOB=∠COB=π/2.\angle AOB =\angle COB =\pi /2 .
So we have shown that the bisector of ∠A\angle A is actually the same line as the perpendicular bisector of the segment ACAC if both lines are extended infinitely. ◻\ \ \ \ \square
Is this proof valid, or are there any mistakes or parts i’m missing?
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It seems that you made an error – the bisector is of A
– Moti
2 days ago
Apart of this the proof seems OK
– Moti
2 days ago
Yeah it’s because when I wrote the proof myself the vertices were ordered differently but I just pulled this picture of an isosceles triangle off the internet.
– Ben
yesterday
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# NCERT Solutions for Class 10 Maths Chapter 11- Constructions
NCERT Solutions for Class 10 Maths Chapter 11 Constructions are provided in a detailed manner, where one can find a step-by-step solution to all the questions for fast revisions. Solutions for the 11th chapter of NCERT class 10 maths are prepared by subject experts under the guidelines of NCERT to assist students in their board exam preparations. Get free NCERT Solutions for Class 10 Maths, Chapter 11 – Constructions at BYJU’S to accelerate the exam preparation. All the questions of NCERT exercises are solved using diagrams step-by-step procedure for construction. Solutions of NCERT help students boost their concepts and clear doubts.
### Class 10 Maths Chapter 11 Exercise 11.1 Page: 220
In each of the following, give the justification of the construction also:
1. Draw a line segment of length 7.6 cm and divide it in the ratio 5 : 8. Measure the two parts. Construction Procedure: A line segment with a measure of 7.6 cm length is divided in the ratio of 5:8 as follows.
1. Draw line segment AB with the length measure of 7.6 cm
2. Draw a ray AX that makes an acute angle with line segment AB.
3. Locate the points i.e.,13 (= 5 + 8) points, such as A1, A2, A3, A4 …….. A13, on the ray AX such that it becomes AA1 = A1A2 = A2A3 and so on.
4. Join the line segment and the ray, BA13.
5. Through the point A5, draw a line parallel to BA13 which makes an angle equal to ∠AA13B
6. The point A5 which intersects the line AB at point C.
7. C is the point divides line segment AB of 7.6 cm in the required ratio of 5:8.
8. Now, measure the lengths of the line AC and CB. It comes out to the measure of 2.9 cm and 4.7 cm respectively.
Justification:
The construction of the given problem can be justified by proving that AC/CB = 5/ 8 By construction,
we have A5C || A13B.
From Basic proportionality theorem for the triangle AA13B,
we get AC/CB =AA5/A5A13….. (1)
From the figure constructed, it is observed that AA5 and A5A13 contain 5 and 8 equal divisions of line segments respectively.
Therefore, it becomes AA5/A5A13=5/8… (2)
Compare the equations (1) and (2),
we obtain AC/CB = 5/ 8
Hence, Justified.
2. Construct a triangle of sides 4 cm, 5 cm and 6 cm and then a triangle similar to it whose sides are 2/3 of
the corresponding sides of the first triangle.
Construction Procedure:
1. Draw a line segment AB which measures 4 cm, i.e., AB = 4 cm.
2. Take the point A as centre, and draw an arc of radius 5 cm.
3. Similarly, take the point B as its centre, and draw an arc of radius 6 cm .
4. The arcs drawn will intersect each other at point C.
5. Now, we obtained AC = 5 cm and BC = 6 cm and therefore ΔABC is the required triangle.
6. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.
7. Locate 3 points such as A1, A2, A3 (as 3 is greater between 2 and 3) on line AX such that it becomes AA1= A1A2 = A2A3.
8. Join the point BA3 and draw a line through A2 which is parallel to the line BA3 that intersect AB at point B’.
9. Through the point B’, draw a line parallel to the line BC that intersect the line AC at C’.
10. Therefore, ΔAB’C’ is the required triangle.
Justification:
The construction of the given problem can be justified by proving that
AB’ =(2/3)AB
B’C’ = (2/3)BC
AC’ = (2/3)AC
From the construction, we get B’C’ || BC
∴ ∠ AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC,
∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAAB’ and ΔAAB, ∠A2AB’=∠A3AB (Common)
From the corresponding angles, we get,
∠AA2B’=∠AA3B
Therefore, from the AA similarity criterion, we obtain
Δ AA2B’ and AA3B
So, AB’/AB = AA2/AA3
Therefore, AB’/AB = 2/3 ……. (2)
From the equations (1) and (2),
we get AB’/AB= B’C’/BC = AC’/ AC = 2/3
This can be written as AB’ = (2/3)AB
B’C’ = (2/3) BC
AC’= (2/3) AC
Hence, justified.
3. Construct a triangle with sides 5 cm, 6 cm and 7 cm and then another triangle whose sides are 7/5 of the corresponding sides of the first triangle
Construction Procedure:
1. Draw a line segment AB =5 cm.
2. Take A and B as centre, and draw the arcs of radius 6 cm and 5 cm respectively.
3. These arcs will intersect each other at point C and therefore ΔABC is the required triangle with the length of sides as 5 cm, 6 cm, and 7 cm respectively.
4. Draw a ray AX which makes an acute angle with the line segment AB on the opposite side of vertex C.
5. Locate the 7 points such as A1, A2, A3, A4 A5, A6, A7 (as 7 is greater between 5and 7), on line AX such that it becomes AA1 = A1A2 = A2A3 = A3A4 = A4A5 = A5A6 = A6A7.
6. Join the points BA5 and draw a line from A7 to BA5 which is parallel to the line BA5 where it intersect the extended line segment AB at point B’.
7. Now, draw a line from B’ the extended line segment AC at C’ which is parallel to the line BC and it intersects to make a triangle.
8. Therefore, ΔAB’C’ is the required triangle.
Justification:
The construction of the given problem can be justified by proving that
AB’ = (7/5) AB
B’C’ = (7/5) BC
AC’= (7/5) AC
From the construction, we get B’C’ || BC
∴ ∠ AB’C’ = ∠ABC (Corresponding angles)
In ΔAB’C’ and ΔABC, ∠ABC = ∠AB’C (Proved above)
∠BAC = ∠B’AC’ (Common)
∴ ΔAB’C’ ∼ ΔABC (From AA similarity criterion)
Therefore, AB’/AB = B’C’/BC= AC’/AC …. (1)
In ΔAA7B’ and ΔAA5B, ∠A7AB’=∠A5AB (Common)
From the corresponding angles, we get,
∠AA7B’=∠AA5B
Therefore, from the AA similarity criterion, we obtain
Δ AA2B’ and AA3B
So, AB’/AB = AA5/AA7
Therefore, AB /AB’ = 5/7 ……. (2)
From the equations (1) and (2), we get
AB’/AB= B’C’/BC = AC’/ AC = 7/5
This can be written as
AB’ = (7/5) AB
B’C’ = (7/5) BC
AC’= (7/5) AC
Hence, justified.
4. Construct an isosceles triangle whose base is 8 cm and altitude 4 cm and then another triangle whose sides are 1$\frac{1}{2}$ times the corresponding sides of the isosceles triangle
Construction Procedure:
1. Draw a line segment BC with the measure of 8 cm.
2. Now draw the perpendicular bisector of the line segment BC and intersect at the point D
3. Take the point D as centre and draw an arc with the radius of 4 cm which intersect the perpendicular bisector at the point A
4. Now join the lines AB and AC and the triangle is the required triangle.
5. Draw a ray BX which makes an acute angle with the line BC on the side opposite to the vertex A.
6. Locate the 3 points B1,B2 and B3 on the ray BX such that BB1 = B1B2 = B2B3
7. Join the points B2C and draw a line from B3 which is parallel to the line B2C where it intersect the extended line segment BC at point C’.
8. Now, draw a line from C’ the extended line segment AC at A’ which is parallel to the line AC and it intersects to make a triangle.
9. Therefore, ΔA’BC’ is the required triangle.
Justification:
The construction of the given problem can be justified by proving that
A’B = (3/2) AB
BC’ = (3/2) BC
A’C’= (3/2) AC
From the construction, we get A’C’ || AC
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
In ΔA’BC’ and ΔABC, ∠B = ∠B (common)
∠A’BC’ = ∠ACB
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Therefore, A’B/AB = BC’/BC= A’C’/AC
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
A’B/AB = BC’/BC= A’C’/AC = 3/2
Hence, justified.
5. Draw a triangle ABC with side BC = 6 cm, AB = 5 cm and ∠ABC = 60°. Then construct a triangle whose sides are 3/4 of the corresponding sides of the triangle ABC.
Construction Procedure:
1. Draw a ΔABC with base side BC = 6 cm, and AB = 5 cm and ∠ABC = 60°.
2. Draw a ray BX which makes an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 3 and 4), such as B1, B2, B3, B4, on line segment BX.
4. Join the points B4C and also draw a line through B3, parallel to B4C intersecting the line segment BC at C’.
5. Draw a line through C’ parallel to the line AC which intersects the line AB at A’.
6. Therefore, ΔA’BC’ is the required triangle
Justification:
The construction of the given problem can be justified by proving that Since the scale factor is ¾, we need to prove
A’B = (3/4) AB
BC’ = (3/4) BC
A’C’= (3/4) AC
From the construction, we get
A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore, A’B/AB = BC’/BC= A’C’/AC
So, it becomes A’B/AB = BC’/BC= A’C’/AC = 3/4
Hence, justified.
6. Draw a triangle ABC with side BC = 7 cm, ∠ B = 45°, ∠ A = 105°. Then, construct a triangle whose sides are 4/3 times the corresponding sides of ∆ ABC.
To find ∠C:
Given: ∠B = 45°, ∠A = 105°
We know that, Sum of all interior angles in a triangle is 180°.
∠A + ∠B + ∠C = 180°
105° + 45° + ∠C = 180°
∠C = 180° − 150°
∠C = 30°
So, from the property of triangle, we get ∠C = 30°
Construction Procedure:
The required triangle can be drawn as follows.
1. Draw a ΔABC with side measures of base BC = 7 cm, ∠B = 45°, and ∠C = 30°.
2. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.
3. Locate 4 points (as 4 is greater in 4 and 3), such as B1, B2, B3, B4, on the ray BX.
4. Join the points B3C.
5. Draw a line through B4 parallel to B3C which intersects the extended line BC at C’.
6. Through C’, draw a line parallel to the line AC that intersects the extended line segment at C’.
7. Therefore, ΔA’BC’ is the required triangle.
Justification:
The construction of the given problem can be justified by proving that Since the scale factor is 4/3, we need to prove
A’B = (4/3) AB
BC’ = (4/3) BC
A’C’= (4/3) AC
From the construction, we get
A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio, it becomes
Therefore, A’B/AB = BC’/BC= A’C’/AC
So, it becomes A’B/AB = BC’/BC= A’C’/AC = 4/3
Hence, justified.
7. Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle.
Given: The sides other than hypotenuse are of lengths 4 cm and 3 cm. It defines that the sides are perpendicular to each other Construction Procedure: The required triangle can be drawn as follows.
1. Draw a line segment BC =3 cm.
2. Now measure and draw ∠B = 90°
3. Take B as centre and draw an arc with a radius of 4 cm and intersects the ray at the point B.
4. Now, join the lines AC and the triangle ABC is the required triangle.
5. Draw a ray BX makes an acute angle with BC on the opposite side of vertex A.
6. Locate 5 such as B1, B2, B3, B4, on the ray BX such that such that BB1 = B1B2 = B2B3=B3B4 = B4B5
7. Join the points B3C.
8. Draw a line through B5 parallel to B3C which intersects the extended line BC at C’.
9. Through C’, draw a line parallel to the line AC that intersects the extended line AB at A’.
10. Therefore, ΔA’BC’ is the required triangle.
Justification:
The construction of the given problem can be justified by proving that Since the scale factor is 5/3,
we need to prove
A’B = (5/3) AB
BC’ = (5/3) BC
A’C’= (5/3) AC
From the construction, we get A’C’ || AC
In ΔA’BC’ and ΔABC,
∴ ∠ A’C’B = ∠ACB (Corresponding angles)
∠B = ∠B (common)
∴ ΔA’BC’ ∼ ΔABC (From AA similarity criterion)
Since the corresponding sides of the similar triangle are in the same ratio,
So, it becomes A’B/AB = BC’/BC= A’C’/AC = 5/3
Hence, justified.
### Class 10 Maths Chapter 11 Exercise 11.2 Page: 221
In each of the following, give the justification of the construction also:
1. Draw a circle of radius 6 cm. From a point 10 cm away from its centre, construct the pair of tangents to the circle and measure their lengths.
Construction Procedure: The construction to draw a pair of tangents to the given circle is as follows.
1. Draw a circle with radius = 6 cm with centre O.
2. Locate a point P, which is 10 cm away from O.
3. Join the points O and P through line
4. Draw the perpendicular bisector of the line OP.
5. Let M be the mid-point of the line PO.
6. Take M as centre and measure the length of MO
7. The length MO is taken as radius and draw a circle.
8. The circle drawn with the radius of MO, intersect the previous circle at point Q and R.
9. Join PQ and PR.
10. Therefore, PQ and PR are the required tangents.
Justification:
The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 6cm with centre O.
To prove this, join OQ and OR represented in dotted lines.
From the construction, ∠PQO is an angle in the semi-circle.
We know that angle in a semi-circle is a right angle, so it becomes,
∴ ∠PQO = 90° Such that ⇒ OQ ⊥ PQ
Since OQ is the radius of the circle with radius 6 cm, PQ must be a tangent of the circle.
Similarly, we can prove that PR is a tangent of the circle.
Hence, justified.
2. Construct a tangent to a circle of radius 4 cm from a point on the concentric circle of radius 6 cm and measure its length. Also, verify the measurement by actual calculation.
Construction Procedure: For the given circle, the tangent can be drawn as follows.
1. Draw a circle of 4 cm radius with centre “ O”.
2. Again, take O as centre draw a circle of radius 6 cm.
3. Locate a point P on this circle
4. Join the points O and P through lines such that it becomes OP.
5. Draw the perpendicular bisector to the line OP
6. Let M be the mid-point of PO.
7. Draw a circle with M as its centre and MO as its radius
8. The circle drawn with the radius OM, intersect the given circle at the points Q and R.
9. Join PQ and PR.
10. PQ and PR are the required tangents.
From the construction, it is observed that PQ and PR are of length 4.47 cm each.
It can be calculated manually as follows
In ∆PQO, Since PQ is a tangent, ∠PQO = 90°. PO = 6cm and QO = 4 cm
Applying Pythagoras theorem in ∆PQO,
we obtain PQ2 + QO2 = PO2
PQ2 + (4)2= (6)2
PQ2 + 16 = 36
PQ2= 36 − 16
PQ2= 20
PQ= 2√5 PQ
= 4.47 cm
Therefore, the tangent length PQ = 4.47
Justification:
The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 4 cm with centre O.
To prove this, join OQ and OR represented in dotted lines.
From the construction, ∠PQO is an angle in the semi-circle.
We know that angle in a semi-circle is a right angle, so it becomes,
∴ ∠PQO = 90° Such that
⇒ OQ ⊥ PQ Since OQ is the radius of the circle with radius 4 cm, PQ must be a tangent of the circle.
Similarly, we can prove that PR is a tangent of the circle.
Hence, justified.
3. Draw a circle of radius 3 cm. Take two points P and Q on one of its extended diameter each at a distance of 7 cm from its centre. Draw tangents to the circle from these two points P and Q
Construction Procedure:
The tangent for the given circle can be constructed as follows.
1. Draw a circle with a radius of 3cm with centre “O”.
2. Draw a diameter of a circle and it extends 7 cm from the centre and mark it as P and Q.
3. Draw the perpendicular bisector of the line PO and mark the midpoint as M.
4. Draw a circle with M as centre and MO as radius
5. Now join the points PA and PB in which the circle with radius MO intersects the circle of circle 3cm.
6. Now PA and PB are the required tangents.
7. Similarly, from the point Q, we can draw the tangents.
8. From that, QC and QD are the required tangents.
Justification:
The construction of the given problem can be justified by proving that PQ and PR are the tangents to the circle of radius 3 cm with centre O.
To prove this, join OA and OB.
From the construction, ∠PAO is an angle in the semi-circle.
We know that angle in a semi-circle is a right angle, so it becomes,
∴ ∠PAO = 90° Such that ⇒ OA ⊥ PA
Since OA is the radius of the circle with radius 3 cm, PA must be a tangent of the circle.
Similarly, we can prove that PB, QC and QD are the tangents of the circle.
Hence, justified
4. Draw a pair of tangents to a circle of radius 5 cm which is inclined to each other at an angle of 60°
Construction Procedure:
The tangents can be constructed in the following manner:
1. Draw a circle of radius 5 cm and with centre as O.
2. Take a point Q on the circumference of the circle and join OQ.
3. Draw a perpendicular to QP at point Q.
4. Draw a radius OR, making an angle of 120° (180° − 60°) with OQ.
5. Draw a perpendicular to RP at point R.
6. Now both the perpendiculars intersect at point P.
7. Therefore, PQ and PR are the required tangents at an angle of 60°.
Justification:
The construction can be justified by proving that ∠QPR = 60°
By our construction ∠OQP = 90°
∠ORP = 90° And ∠QOR = 120°
We know that the sum of all interior angles of a quadrilateral= 360°
∠OQP + ∠QOR + ∠ORP + ∠QPR=90° + 120° + 90° + ∠QPR= 360°
Therefore, ∠QPR = 60°
Hence Justified
5. Draw a line segment AB of length 8 cm. Taking A as centre, draw a circle of radius 4 cm and taking B as centre, draw another circle of radius 3 cm. Construct tangents to each circle from the centre of the other circle.
Construction Procedure: The tangent for the given circle can be constructed as follows.
1. Draw a line segment AB = 8 cm.
2. Take A as centre and draw a circle of radius 4 cm
3. Take B as centre, draw a circle of radius 3 cm
4. Draw the perpendicular bisector of the line AB and the midpoint is taken as M.
5. Now, take M as centre draw a circle with the radius of MA or MB which the intersects the circle at the points P, Q, R and S.
6. Now join AR, AS, BP and BQ
7. Therefore, the required tangents are AR, AS, BP and BQ
Justification:
The construction can be justified by proving that AS and AR are the tangents of the circle (whose centre is B with radius is 3 cm) and BP and BQ are the tangents of the circle (whose centre is A and radius is 4 cm).
From the construction, to prove this, join AP, AQ, BS, and BR.
∠ASB is an angle in the semi-circle.
We know that an angle in a semi-circle is a right angle.
∴ ∠ASB = 90° ⇒ BS ⊥ AS
Since BS is the radius of the circle, AS must be a tangent of the circle.
Similarly, AR, BP, and BQ are the required tangents of the given circle.
6. Let ABC be a right triangle in which AB = 6 cm, BC = 8 cm and ∠ B = 90°. BD is the perpendicular from B on AC. The circle through B, C, D is drawn. Construct the tangents from A to this circle.
Construction Procedure: The tangent for the given circle can be constructed as follows
1. Draw the line segment with base BC =8cm
2. Measure the angle 90° at the point B, such that ∠ B = 90°.
3. Take B as centre and draw an arc with a measure of 6cm
4. Let the point be A where the arc intersects the ray
5. Join the line AC.
6. Therefore, ABC be the required triangle.
7. Now, draw the perpendicular bisector to the line BC and the midpoint is marked as E.
8. Take E as centre and BE or EC measure as radius draw a circle.
9. Join A to the midpoint E of the circle
10. Now, again draw the perpendicular bisector to the line AE and the midpoint is taken as M
11. Take M as Centre and AM or ME measure as radius, draw a circle.
12. This circle intersects the previous circle at the points B and Q
13. Join the points A and Q
14. Therefore, AB and AQ are the required tangents
Justification:
The construction can be justified by proving that AG and AB are the tangents to the circle. From the construction, join EQ.
∠AQE is an angle in the semi-circle.
We know that an angle in a semi-circle is a right angle.
∴ ∠AQE = 90° ⇒ EQ ⊥ AQ
Since EQ is the radius of the circle, AQ has to be a tangent of the circle.
Similarly, ∠B = 90° ⇒ AB ⊥ BE Since BE is the radius of the circle, AB has to be a tangent of the circle. Hence, justified.
7. Draw a circle with the help of a bangle. Take a point outside the circle. Construct the pair of tangents from this point to the circle.
Construction Procedure: The required tangents can be constructed on the given circle as follows.
1. Draw a circle with the help of a bangle.
2. Draw two non-parallel chords such as AB and CD
3. Draw the perpendicular bisector of AB and CD
4. Take the centre as O where the perpendicular bisector intersects.
5. To draw the tangents, take a point P outside the circle.
6. Join the points O and P.
7. Now draw the perpendicular bisector of the line PO and midpoint is taken as M
8. Take M as centre and MO as radius draw a circle.
9. Let the circle intersects intersect the circle at the points Q and R
10. Now join PQ and PR
11. Therefore, PQ and PR are the required tangents.
Justification:
The construction can be justified by proving that PQ and PR are the tangents to the circle.
Since, O is the centre of a circle,
we know that the perpendicular bisector of the chords passes through the centre.
Now, join the points OQ and OR.
We know that perpendicular bisector of a chord passes through the centre.
It is clear that the intersection point of these perpendicular bisectors is the centre of the circle.
Since, ∠PQO is an angle in the semi-circle.
We know that an angle in a semi-circle is a right angle.
∴ ∠PQO = 90° ⇒ OQ ⊥ PQ
Since OQ is the radius of the circle, PQ has to be a tangent of the circle.
Similarly, ∴ ∠PRO = 90° ⇒ OR ⊥ PO Since OR is the radius of the circle,
PR has to be a tangent of the circle
Therefore, PQ and PR are the required tangents of a circle.
Also Access NCERT Exemplar for Class 10 Maths Chapter 11 CBSE Notes for Class 10 Maths Chapter 11
## NCERT Solutions for Class 10 Maths Chapter 11 Constructions
Topics present in NCERT Solutions for Class 10 Maths Chapter 11 includes division of a line segment, constructions of tangents to a circle, line segment bisector and many more. Students in class 9, study some basics of constructions like drawing the perpendicular bisector of a line segment, bisecting an angle, triangle construction etc. Using Class 9 concepts, students in Class 10 will learn about some more constructions along with the reasoning behind that work.
NCERT Class 10, Chapter 11-Constructions is a part of Geometry. Over the past few years, geometry consists a total weightage of 15 marks in the final exams. Construction is a scoring chapter of geometry section. In the previous year exam, one question of 4 marks being asked from this chapter.
List of Exercises in class 10 Maths Chapter 11
Exercise 11.1 Solutions (7 Questions)
Exercise 11.2 Solutions (7 Questions)
The NCERT solutions for Class 10 for the 11th chapter of Maths is all about construction of line segments, division of a Line Segment and Construction of a Circle, Constructions of Tangents to a circle using analytical approach. Also students have to provide justification of each answer.
The topics covered in Maths Chapter 11 Constructions are:
Exercise Topic 11.1 Introduction 11.2 Division of a Line Segment 11.3 Construction of Tangents to a Circle 11.4 Summary
Some of the ideas applied in this chapter:
1. The locus of a point that moves in an identical distance from 2 points, is normal to the line joining both the points.
2. Perpendicular or Normal means right angles whereas, bisector cuts a line segment in two half.
3. The design of different shapes utilizing a pair of compasses and straightedge or ruler.
Key Features of NCERT Solutions for Class 10 Maths Chapter 11 Constructions
• NCERT solutions can also prove to be of valuable help to students in their assignments and preparation of boards and competitive exams.
• Each question is explained using diagrams which makes learning more interactive.
• Easy and understandable language used in NCERT solutions.
• Provide detailed solution using an analytical approach.
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# Hobbit Math: Elementary Problem Solving 5th Grade
[Photo by OliBac. Visit OliBac’s photostream for more.]
The elementary grades 1-4 laid the foundations, the basics of arithmetic: addition, subtraction, multiplication, division, and fractions. In grade 5, students are expected to master most aspects of fraction math and begin working with the rest of the Math Monsters: decimals, ratios, and percents (all of which are specialized fractions).
Word problems grow ever more complex as well, and learning to explain (justify) multi-step solutions becomes a first step toward writing proofs.
This installment of my elementary problem solving series is based on the Singapore Primary Mathematics, Level 5A. For your reading pleasure, I have translated the problems into the world of J.R.R. Tolkien’s classic, The Hobbit.
[Note: No decimals or percents here. Those are in 5B, which will need an article of its own. But first I need to pick a book. I’m thinking maybe Naya Nuki…]
### Printable Worksheet
In case you’d like to try your hand at the problems before reading my solutions, I’ve put together a printable worksheet:
### An Unexpected Party
Bilbo had 3 times as many apple tarts as mince pies in his larder. If he had 24 more apple tarts than mince pies, how many of the pastries (both tarts and pies) did he have altogether?
Long before ratios are specifically introduced, students get plenty of practice with the informal ratios twice as many and three times as many. In each case, we draw the smaller group as one unit, and the larger group as whatever number of units make the “times as many.”
We know there are 24 more tarts than pies, so:
2 units = 24
1 unit = 24 $\div$ 2 = 12
And finally, we count how many units we have in all:
4 units = 4 $\times$ 12 = 48
There are 48 pastries altogether.
### Roast Mutton
The three trolls had 123 pieces of gold. Tom had 15 pieces of gold more than Bert. Bert had 3 pieces fewer than William. How many pieces of gold did William have?
This is a comparison problem, where they tell us how much more or less one thing is than another. To model it, we will draw a bar for each thing we’re comparing. The left edges of the bars line up, making it easy to compare which is larger or smaller.
In this case, we will need three bars — one for each troll’s amount of gold. We’ll label each treasure with the troll’s name. (My kids would usually just put the troll’s initial.) Notice that Bert has the smallest number of pieces. Tom has Bert’s amount plus 15 more, and William has Bert’s amount plus 3 more.
We could make an algebra equation:
$3x + 18 = 123$
But we won’t. Remember, this is a 5th grade problem! Instead, we’ll use the “stealth algebra” of our diagram to help us think through the numbers.
All the gold together is 123 pieces. We can imagine taking away the extra bits, reducing each of the other trolls’ loot to match Bert’s stash:
3 units = 123 – 18 = 105
1 unit = 105 $\div$ 3 = 35
And now, because I forgot to put a question mark in my drawing, I have to go back and read the problem again. Oh, yes: we need to find William’s amount.
1 unit + 3 = 35 + 3 = 38
### Over Hill and Under Hill
The Great Goblin had twice as many goblin soldiers as his cousin, the Gross Goblin. How many soldiers must the Great Goblin send to his cousin so that they will each have 1200 goblin soldiers?
What a cool problem! It almost seems like they haven’t given enough information, doesn’t it?
Students who are not used to bar diagrams often get confused by transfer problems, which have a beginning situation and then something is moved from one person to another to set up the end of the story. Desperate children will grab any number and guess at the answer: “Send half of them. 1200 $\div$ 2…”
Let’s see what the bar diagrams tell us. First, the starting ratio:
And this is what we want to end up with:
But how can we get there? A useful problem-solving tool is to work backwards. In this case, we’ll start with the end of the story, since that’s where they gave us a number to work with. We know how many goblins will be in each army, which means we can find the total number of soldiers:
All goblin soldiers = 1200 $\times$ 2 = 2400
Aha! Moving some of the soldiers around wouldn’t change the total number, so there must have been 2400 soldiers in the first diagram, too:
3 units = 2400
1 unit = 2400 $\div$ 3 = 800
But how many soldiers need to move? The diagram makes the answer clear: We need to move half a unit:
1/2 unit = 800 $\div$ 2 = 400
And to double-check:
1 unit + 400 = 800 + 400 = 1200
The Great Goblin sent 400 soldiers to his cousin.
### Alternate Solution
There are usually many ways to approach any word problem. Kitten chose a method similar to, but shorter than, the way I did it. She started with the simple 2:1 ratio, and then figured out that the Great Goblin would have to send 1/4 of his soldiers:
Since the “extra” unit had to be divided in half, she divided all the other units in half, too. Finally, she realized that the three smaller units remaining in the Great Goblin’s army must be 1200 soldiers:
3 units = 1200
1 unit = 1200 $\div$ 3 = 400
The Great Goblin sent 400 soldiers to his cousin.
### Riddles in the Dark
Gollum caught 10 small fish. He divided the fish to make 4 equal meals. How many fish did he eat at each meal?
The 5th grade unit on fractions begins by making explicit something that in earlier grades has only been implied — the connection between fractions and division. I tell my students, “The division symbol looks like a little fraction, with dots for the numbers. Let that remind you: Every fraction is a division problem, top divided by bottom. And any division problem can be written as a fraction.”
Gollum had 10 fish, split into 4 equal groups. Each group is 1/4 of the whole 10:
1 unit = 10 $\div$ 4 = 10/4
And putting that into simplest form:
10/4 = 5/2 = 2 $\frac{1}{2}$
Gollum ate $2 \frac{1}{2}$ fish at each meal.
### Queer Lodgings
Beorn baked a large loaf of wholegrain bread. He ate 1/3 of the loaf himself (with plenty of honey!), and he sliced 1/2 of the same loaf to feed the dwarves and Bilbo. What fraction of the loaf was left?
Bar diagrams help students see the need for a common denominator. Unless all the pieces are the same size, it doesn’t make much sense to say, “We have one piece left.”
Beorn cut 1/3 of the loaf from one side, and 1/2 of the loaf from the other:
What size is the white piece in the middle? It is clearly half of the middle third. Let’s get a common denominator by cutting all of the thirds in half.
Aha! Half of a third is the same as a sixth.
$1 - \frac{1}{3} - \frac{1}{2} = 1 - \frac{2}{6} - \frac{3}{6} = \frac {1}{6}$
### Alternate Solution
Again, Kitten doesn’t think the way I do. She finds adding fractions much easier than subtracting them, so she reasoned, “First I’ll find out how much they ate. Then I can see what is left.”
$\frac{1}{3} + \frac{1}{2} = \frac{2}{6} + \frac{3}{6} = \frac {5}{6}$
$1 - \frac{5}{6} = \frac {1}{6}$
### Barrels Out of Bond
The Elvenking had a barrel of fine wine. His butler poured 3/4 gallon of it into a small keg. He drank 1/2 of the keg and gave the other half to his friend, the chief of the guards. How much wine did the Elvenking’s butler drink?
I teach my students to connect the idea of multiplication with the preposition “of” (see If It Ain’t Repeated Addition, What is It? and A Mathematical Trauma). In this problem, we need to find 1/2 of 3/4 of a gallon, which means:
$\frac{1}{2} \times \frac{3}{4} = ?$
This could be done on a single bar, but I find it easier to draw two:
Notice that the line that splits the keg in half also bisects the unit above it (the second fourth of the gallon). Let’s make an equivalent fraction by splitting all of the units in half:
$\frac{1}{2} \times \frac{3}{4} = \frac{1}{2} \times \frac{6}{8} = \frac{3}{8}$
The butler drank 3/8 of a gallon of wine.
### Inside Information
2/3 of the items in the dragon Smaug’s treasure were made of gold. 1/4 of the remaining part was precious gems. What fraction of Smaug’s treasure was precious gems?
As students gain skill in working with fractions, the problems grow correspondingly more complex. Now we introduce the fraction of the remaining part problem. In this type of problem, what was a part of the original whole becomes a new “whole thing” to be cut up into parts of its own.
Here, Smaug’s treasure is divided into thirds, and one of these thirds is our remaining part. We show that it is being treated separately by drawing a new bar below the first, connected with lines to its original position.
We are interested in 1/4 of the remaining part, and we want to know what fraction of the original bar (the whole treasure) it would be:
Well, if our remaining third is cut into fourths, we could make a common denominator by cutting all of the thirds the same way. What size pieces would we have then?
$\frac {1}{4} \ of \ \frac {1}{3} = \frac {1}{4} \times \frac {1}{3} = \frac {1}{4} \times \frac {4}{12} = \frac {1}{12}$
1/12 of Smaug’s treasure was precious gems.
### The Battle of Five Armies, Part 1
When the Elvenking heard the dragon had been killed, he set out to claim a share of the treasure. 2/5 of his army were archers. 1/2 of the remainder fought with spears, and the rest carried swords. If 300 soldiers carried swords, how many elves marched out with the Elvenking?
Here is another remaining part puzzle. In this case, our original bar (representing the Elvenking’s army) is divided into fifths, two of which are archers. The other three fifths become our remainder bar, of which 1/2 fight with spears:
The last chunk is the swordsmen (swords-elves?), and there are 300 of them. That means we can find the size of our remaining part:
1 unit = 300
2 units = 300 $\times$ 2 = 600
If the remaining part of the army is 600, then on the original bar:
3 units = 600
1 unit = 600 $\div$ 3 = 200
5 units = 200 $\times$ 5 = 1000
There were 1,000 elves marching with the Elvenking.
[Note: For more practice with the remaining part, try your hand at Solving Complex Story Problems and Solving Complex Story Problems II.]
### The Battle of Five Armies, Part 2
The bowman Bard gathered a small army of 600 survivors from the town of Esgaroth, which the dragon had destroyed. The ratio of archers to swordsmen was 2:3. How many archers followed Bard to the Lonely Mountain?
Now our students have matured, graduating from the simple three times as many situations to full-fledged ratios. Without bar diagrams (or similar pictorial methods), middle school students find ratios an abstract and difficult subject, but a diagram makes it easy to see relationships.
The ratio tells how many parts (units) to draw for each group. 2:3 means 2 units of archers and 3 units of swordsmen:
5 units = 600
1 unit = 600 $\div$ 5 = 120
2 units = 120 $\times$ 2 = 240
240 archers followed Bard to the Lonely Mountain.
### The Return Journey
The dwarves rewarded Bilbo with two chests of gold, silver, and small gems — 6000 pieces of treasure altogether. There were twice as many pieces of gold as there were gems. There were 600 more pieces of silver than gems. How much of each type of treasure did Bilbo receive?
At the end of our textbook, we meet another comparison problem, like the earlier one with the trolls. Of course, we hope our students have learned something in the interim, so this problem will take a few extra steps — for example, we are asked to find the amount for all three types of treasure, not just for one of them.
We are comparing the number of pieces of gold, gems, and silver, so we will need three bars. The fewest are the gems. There are twice as many (notice the ratio?) pieces of gold. And the pieces of silver match the gems plus 600 more:
As we did earlier in the troll problem, we first remove the “extra” pieces. This will let us work with the unknown units by themselves:
4 units = 6000 $-$ 600 = 5400
1 unit = 5400 $\div$ 4 = ?
Excuse me while I do the long division
1 unit = 1350
2 units = 1350 $\times$ 2 = 2700
1 unit + 600 = 1950
And on complex problems, it’s always a good idea to double-check:
1350 + 2700 + 1950 = 6000
(Yes, it really is important to check! I made a mental math error as I was typing this post, which I caught only because I took the time to add up my answers and see if they made sense.)
Bilbo received 2,700 pieces of gold; 1,350 gems; and 1,950 pieces of silver.
### Practice and Learn
To get more practice creating bar diagrams, your students may enjoy these online tutorials:
And for some fun practice with fractions, geometry, decimals, percentages, averages, and more:
The Hobbit (1977)
### Update: My New Book
You can help prevent math anxiety by giving your children the mental tools they need to conquer the toughest story problems.
Read expanded explanations of the Hobbit word problems—and many more!—in Word Problems from Literature: An Introduction to Bar Model Diagrams. Now available at all your favorite online bookstores.
And there’s a paperback Student Workbook, too.
## 10 thoughts on “Hobbit Math: Elementary Problem Solving 5th Grade”
1. Fantastic! I love the work you put into not only this unit, but also the post itself.
2. Great idea even though I’m not a fan of The Hobbit! My 10 year old future mathematician loves to read ; don’t think she’s read The Hobbit, but I’ll show her your worksheets and see what she thinks!
3. Great post. I love it. Thanks
4. This is a wonderful blog. I look forward to staying updated during the upcoming school year.
5. Thanks for the encouragement, everyone! I had fun putting this together.
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# Manhattan GMAT Challenge Problem of the Week – 2 Nov 2010
by Manhattan Prep, Nov 2, 2010
Here is a new Challenge Problem! If you want to win prizes, try entering our Challenge Problem Showdown. The more people enter our challenge, the better the prizes.
## Question
If the diagonal of rectangle Z is d, and the perimeter of rectangle Z is p, what is the area of rectangle Z, in terms of d and p?
(A) [pmath](d^2-p)/3[/pmath]
(B) [pmath](2d^2-p)/2[/pmath]
(C) [pmath](p-d^2)/2[/pmath]
(D) [pmath](12d^2-p^2)/8[/pmath]
(E) [pmath](p^2-4d^2)/8[/pmath]
Lets take an algebraic approach. The first step is to realize that we should create variables for the simplest things about this rectangle: its length and its width. The reason is that we can express all of these secondary features (diagonal, perimeter, and area) in terms of length and width. Then we can look for a relationship between these expressions.
We can call the length of rectangle a and its width b. Now we can write equations relating a and b to the diagonal, perimeter, and area, respectively:
Diagonal: [pmath]a^2 + b^2 = d^2[/pmath], by the Pythagorean Theorem
Perimeter: [pmath]2(a + b) = p[/pmath]
Area: Area = ab = ? (in terms of p and d)
So we are looking to manipulate the first two equations to isolate ab on one side. The expression involving p and d on the other side will be our answer.
To avoid square roots, lets square the second equation (for p) and see what we get.
[pmath]2(a + b) = p[/pmath]
[pmath]4(a + b)^2 = p^2[/pmath]
[pmath]4(a^2 +2ab + b^2) = p^2[/pmath]
[pmath]4a^2 +8ab + 4b^2 = p^2[/pmath]
Comparing this equation to the first equation (for d), we hopefully notice that the [pmath]a^2[/pmath] and [pmath]b^2[/pmath] terms can be made to line up and cancel. We multiply the first equation by 4, to begin with.
[pmath]a^2 + b^2 = d^2[/pmath]
[pmath]4a^2 + 4b^2 = 4d^2[/pmath]
Now we can line up and subtract:
[pmath]4a^2 +8ab + 4b^2 = p^2[/pmath]
-[pmath][4a^2 + 0 + 4b^2 = 4d^2][/pmath]
yields
[pmath]8ab=p^2-4d^2[/pmath]
Now just divide by 8 to isolate ab:
[pmath]ab=(p^2-4d^2)/8[/pmath]
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Priyanka Madiraju — Published On February 9, 2021
This article was published as a part of the Data Science Blogathon.
## Introduction
In the previous post, we have defined Probability Distributions and briefly discussed different Discrete Probability distributions. In this post, we will continue learning about probability distributions through Continuous Probability Distributions.
## Definition
If you recall from our previous discussion, continuous random variables can take an infinite number of values over a given interval. For example, in the interval [2, 3] there are infinite values between 2 and 3. Continuous distributions are defined by the Probability Density Functions(PDF) instead of Probability Mass Functions. The probability that a continuous random variable is equal to an exact value is always equal to zero. Continuous probabilities are defined over an interval. For instance, P(X = 3) = 0 but P(2.99 < X < 3.01) can be calculated by integrating the PDF over the interval [2.99, 3.01]
## List of Continuous Probability Distributions
We discuss the most commonly used continuous probability distributions below:
### 1. Continuous Uniform Distribution
Uniform distribution has both continuous and discrete forms. Here, we discuss the continuous one. This distribution plots the random variables whose values have equal probabilities of occurring. The most common example is flipping a fair die. Here, all 6 outcomes are equally likely to happen. Hence, the probability is constant.
Consider the example where a = 10 and b = 20, the distribution looks like this:
The PDF is given by,
where a is the minimum value and b is the maximum value.
### 2. Normal Distribution
This is the most commonly discussed distribution and most often found in the real world. Many continuous distributions often reach normal distribution given a large enough sample. This has two parameters namely mean and standard deviation.
This distribution has many interesting properties. The mean has the highest probability and all other values are distributed equally on either side of the mean in a symmetric fashion. The standard normal distribution is a special case where the mean is 0 and the standard deviation of 1.
It also follows the empirical formula that 68% of the values are 1 standard deviation away, 95% percent of them are 2 standard deviations away, and 99.7% are 3 standard deviations away from the mean. This property is greatly useful when designing hypothesis tests(https://www.statisticshowto.com/probability-and-statistics/hypothesis-testing/).
The PDF is given by,
where μ is the mean of the random variable X and σ is the standard deviation.
### 3. Log-normal Distribution
This distribution is used to plot the random variables whose logarithm values follow a normal distribution. Consider the random variables X and Y. Y = ln(X) is the variable that is represented in this distribution, where ln denotes the natural logarithm of values of X.
The PDF is given by,
where μ is the mean of Y and σ is the standard deviation of Y.
### 4. Student’s T Distribution
The student’s t distribution is similar to the normal distribution. The difference is that the tails of the distribution are thicker. This is used when the sample size is small and the population variance is not known. This distribution is defined by the degrees of freedom(p) which is calculated as the sample size minus 1(n – 1).
As the sample size increases, degrees of freedom increases the t-distribution approaches the normal distribution and the tails become narrower and the curve gets closer to the mean. This distribution is used to test estimates of the population mean when the sample size is less than 30 and population variance is unknown. The sample variance/standard deviation is used to calculate the t-value.
The PDF is given by,
where p is the degrees of freedom and Γ is the gamma function. Check this link for a brief description of the gamma function.
The t-statistic used in hypothesis testing is calculated as follows,
where x̄ is the sample mean, μ the population mean and s is the sample variance.
### 5. Chi-square Distribution
This distribution is equal to the sum of squares of p normal random variables. p is the number of degrees of freedom. Like the t-distribution, as the degrees of freedom increase, the distribution gradually approaches the normal distribution. Below is a chi-square distribution with three degrees of freedom.
The PDF is given by,
where p is the degrees of freedom and Γ is the gamma function.
The chi-square value is calculated as follows:
where o is the observed value and E represents the expected value. This is used in hypothesis testing to draw inferences about the population variance of normal distributions.
### 6. Exponential Distribution
Recall the discrete probability distribution we have discussed in the Discrete Probability post. In the Poisson distribution, we took the example of calls received by the customer care center. In that example, we considered the average number of calls per hour. Now, in this distribution, the time between successive calls is explained.
The exponential distribution can be seen as an inverse of the Poisson distribution. The events in consideration are independent of each other.
The PDF is given by,
where λ is the rate parameter. λ = 1/(average time between events).
To conclude, we have very briefly discussed different continuous probability distributions in this post. Feel free to add any comments or suggestions below.
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Question Video: Finding the Diagonal of a Square given Its Area | Nagwa Question Video: Finding the Diagonal of a Square given Its Area | Nagwa
# Question Video: Finding the Diagonal of a Square given Its Area Mathematics
The area of the square π΄π΅πΆπ· is 9/2 cmΒ². Find the length of π΅π·.
04:20
### Video Transcript
The area of the square π΄π΅πΆπ· is nine over two centimeters squared. Find the length of line π΅π·.
Almost immediately, we probably think the area of a square formula area equals side squared, which means the area of our square, nine over two, is equal to the side squared. If we take the square root of both sides, the square root of a side squared equals π , a side, and the side of this square measures the square root of nine over two centimeters. We can write that as the square root of nine over the square root of two. Weβre only interested in the positive square root since weβre dealing with distance. So we can say the square root of nine equals three.
And weβve now found the side length of every side of this square. But thatβs in fact not what this question is looking for. Weβre interested in the length of the diagonal π΅π·. Because we know that this is a square, we could say that angle π΅πΆπ· measures 90 degrees, which means line π΅π· is the hypotenuse of a right triangle. And we can use the Pythagorean theorem to find the length of π΅π·. Since π΅π· is the hypotenuse, thatβs the variable πΆ and the Pythagorean theorem, π΅π· squared will be equal to the sum of the other two sides squared, which is three over the square root of two squared plus three over the square root of two squared.
We know that three over the square root of two squared is nine over two because we were given that to start with. And so we have nine over two plus nine over two. π΅π· squared equals nine over two plus nine over two. Nine over two plus nine over two is 18 over two, which reduces to nine. Weβll take the square root of both sides. The square root of π΅π· squared is just line π΅π·. And the square root of nine is plus or minus three. But since weβre dealing with distance, we only want the positive square root, which is three. And that means the diagonal π΅πΆ measures three centimeters.
There is one other way we could work this out without using the Pythagorean theorem. To do this, we need to remember a few things. Number one, a square is a type of rhombus. Number two, we find the area of a rhombus by multiplying π times π and dividing by two where π and π are the diagonals of a rhombus. Number three, the diagonals of a square are equal to one another. Therefore, another way to find the area of a square is to say the diagonal squared divided by two.
And if you know all of that, you can say the area of this square is equal to π΅π· squared over two. And then you can plug in nine over two for the area because that was given to us. If nine over two is equal to π΅ squared over two, then nine equals π΅π· squared. And we take the square root of both sides of this equation to show that π΅π· equals three. This second method requires far less calculation but depends on you remembering these facts about a square and a rhombus. But both methods confirm that the length of π΅π·, the diagonal of this square, is three centimeters.
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# T4T Is it Odd or Even?
Click to access fully formatted lesson and materials:
Lesson excerpt:
Understand the meaning of odd and even numbers
Theoretical Foundation: As students in second grade begin to look at the properties and attributes of numbers, they begin to have an understanding of odd and even. It is not enough for students to just be able to identify an odd or an even number, students should build a conceptual understanding of why a number is classified as odd or even.
Estimated Time: 40 minutes
Materials: 2 dice per student (or pair of students), 1-6 or 0-9 counters, recording and sorting pages, Even Steven and Odd Todd, by Kathryn Cristaldi
Description:
1. Introduce the vocabulary “odd” and “even”.
2. Read and discuss the book Even Steven and Odd Todd. As you read the book, draw or model the numbers of items that are modeled in the story. As you do this, be sure to put them into 2 equal groups and a left over (if odd).
3. After reading the story, have students identify different times in the story in which an item was odd or was even. Be sure to have students explain their reasoning.
4. Give each student (or pair of students) two dice and at least 12 (18 counters if using the 0-9 dice) counters.
5. Have students roll the dice and count out the number of counters that are shown on the dice.
6. Once students have the correct number of counters counted out, they are to record the number of counters on their recording sheet.
7. Next, have the students separate the counters into two equal groups (on the recording sheet) until all counters are in the circles or until there is only one left over.
8. Have the students decide whether or not the number on the dice is odd or even and record their findings on their recording sheet.
9. Continue this for 12 turns.
10. Create a class chart, with numbers 1-12, have students to help fill in the odd/even column.
Differentiation Suggestions:
1. Use 6 dice so the numbers can go up to 36. This will allow students to begin to see the patterns in the digits that make up odd/even numbers.
2. Allow students to work with a partner. Partners can take turns rolling dice, counting out counters and recording.
3. Break this lesson into two smaller lessons.
Probing Questions:
1. Do you notice any patterns with the numbers that are odd or even?
2. What happens every time you add one to the number?
3. What happens when you take one (or two) away from the number that you rolled? Is it still even/odd? Does that happen every time?
Assessment:
1. Does the student see any patterns to the way odd and even numbers are classified?
2. Do the students understand why numbers are classified as odd or even?
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# Addition of like algebraic terms
A mathematical operation of adding two or more like algebraic terms to obtain sum of them is called the addition of like algebraic terms.
## Introduction
In mathematics, it is essential for every learner to know the mathematical operation of performing summation with two or more like algebraic terms in some cases. Like algebraic terms have a literal factor commonly and it allows us to combine them as an algebraic term and it is also in the same form.
### Example
$2xy$ and $3xy$ are two like algebraic terms. The addition of these two like algebraic terms can be done in two simple steps.
01
#### Display a Plus between terms
A plus sign should be displayed between every two like terms.
$2xy+3xy$
02
#### Obtain summation of them
Like algebraic terms have the same literal factor. So, take the literal factor common from them. In this example, $xy$ is the common literal factor of them.
$2xy+3xy = (2+3)xy$
Now, add the numbers and then multiply the sum of them by their common literal factor.
$\implies 2xy+3xy = 5xy$
### More Examples
Observe the mathematical procedure in the following examples to learn how to add two or more like algebraic terms.
$(1)\,\,\,\,\,\,$ $3x^2+4x^2+5x^2$ $=$ $(3+4+5)x^2$ $=$ $12x^2$
$(2)\,\,\,\,\,\,$ $7abc+3abc$ $=$ $(7+3)abc$ $=$ $10abc$
$(3)\,\,\,\,\,\,$ $c^2d^2+20c^2d^2+11c^2d^2+5c^2d^2$ $=$ $(1+20+11+5)c^2d^2$ $=$ $37c^2d^2$
$(4)\,\,\,\,\,\,$ $7e^3d^4f^5+2e^3d^4f^5$ $=$ $(7+2)e^3d^4f^5$ $=$ $9e^3d^4f^5$
$(5)\,\,\,\,\,\,$ $2g+3g+7g+4g+5g$ $=$ $(2+3+7+4+5)g$ $=$ $21g$
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# Varsity Math, Week 52
## ________________
With the new academic year underway, it’s time for the grueling tryouts for this year’s first string on the Varsity Math team. Here’s a sample of what the competitors are up against.
## ________________
### Mean Triangle
You are given a right triangle with the property that the length of the median to the hypotenuse is the geometric mean (the square root of the product) of the lengths of the legs. The hypotenuse is ten units in length.
What is the length of the shorter leg?
### As Easy as 4132
Let’s call a sequence of four real numbers a “sumo sequence” if, whenever you add any one of them to the product of the other three, the sum is exactly two.
How many sumo sequences are there?
## Solutions to week 51
Rhombarium. There are many ways to attack this problem, but here’s one particularly simple one. Start by noticing that the vertices of the rhombic dodecahedron highlighted in the diagram at right lie at the vertices of a perfect cube. Moreover, if you imagine removing that cube from the rhombic dodecahedron, six identical square pyramids remain. Since each original rhombic face is planar, if you rotate any one of the square pyramids 180° around an edge of the (removed) cube, one triangular face of that pyramid exactly coincides with a triangular face of the adjacent pyramid, and the square faces end up at right angles to each other. Continuing with other square pyramids similarly, you can see that the six square pyramids precisely assemble into a cube identical to the removed one. Hence, the volume of the rhombarium is just twice the volume of that inscribed cube, whose edge is the short diagonal of the rhombic faces. With an edge length of 5√3 and a long diagonal √2 the length of the short diagonal, the Pythagorean theorem shows that the sides of the right triangle formed by two half-diagonals and one edge of a rhombus are 5, 5√2, and 5√3 cm, respectively. Hence, the short diagonal is 10cm, the removed cube has a volume of one liter, and the total volume of the rhombarium is two liters.
Well Trained. First, imagine starting at the main depot with 32 watt-hours (Wh) of charge before charging — enough so that the train will make it all the way around the track no matter what happens. Then the graph of the charge level in the train’s batteries as a function of distance traveled around the track, including the charging, is shown below. Note that the lowest battery charge occurs just before charging at the first station eight meters beyond the main depot. If the train started there with zero charge in its battery (which can be visualized just by mentally moving the x-axis up so it just touches the graph at this point) the train will have non-zero charge in its battery all the way around, and will come back exactly to this point with zero charge ready to charge up again at this station, since it both acquires and expends 32 Wh in making it around the track once. There is only one lowest point on the graph, so the unique place the train can start on an empty battery is 8 meters past the main depot.
Coach Newton elaborates:
This argument actually shows something much more general. In any closed circuit, if the resources consumed matches the resources acquired in the execution of a single circuit, then there will be someplace in the circuit that the process can start with zero resources and yet complete the entire circuit without running out of resources. This could be a vehicle with fuel, or a carnivalgoer with tickets, or any setting in which some resource is both acquired and consumed in various places on a closed path.
## Recent Weeks
Links to all of the puzzles and solutions are on the Complete Varsity Math page.
Come back next week for answers and more puzzles.
[asciimathsf]
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# Factors of 40: Prime Factorization, Methods, Tree, and Examples
Factors of 40 are a list of numbers that give a whole number quotient and zero remainders when divided. Or when two numbers are multiplied to produce the number 40, those two numbers will be called the factors of 40.
Figure 1 – All possible Factors of 40
A factor can never be in decimal or fraction form. As the number 40 is an even composite number it will have more than 2 factors. The number 40 has 16 total factors. 8 are positive factors and the rest 8 are negative factors.
This article will let you understand the concept of factors, techniques to calculate the factors, prime factorization, factor tree, factor pair, and its examples.
## What Are the Factors of 40?
The factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40. This means when these numbers divide 40 they produce a whole number quotient and zero remainders.
Note that these factors can also be called the divider since they divide the number 40, which is the dividend, to produce an answer. The list of factors for the number 40 are:
Factors of 40: 1, 2, 4, 5, 8, 10, 20, 40Â
## How To Calculate the Factors of 40?
You can calculate the factors of 40 with two methods:
1. Division Method
2. Multiplication Method
For the division method you follow these steps:
$\dfrac{40}{1}=40$
$\dfrac{40}{2}=20$
$\dfrac{40}{4}=10$
$\dfrac{40}{5}=8$
$\dfrac{40}{8}=5$
$\dfrac{40}{10}=4$
$\dfrac{40}{20}=2$
$\dfrac{40}{40}=1$
To calculate the factors using the division method you will take the smallest integer i.e 1. Then you will divide the number 40 by 1. Since 1 is a factor of 40 it will you a whole number in the quotient and zero remainders.Â
To find the list of all of the factors of 40 you will keep on repeating this process for all consecutive integers starting from 1 to 40. As any factor of a number can never be greater than the number itself.Â
So from the above steps, we can list the factors of 40 as:
Positive factors of 40= 1, 2, 4, 5, 8, 10, 20, 40Â
Since every number has both positive and negative factors so we can also list the negative factors of 40 as:
Negative factors of 40 = -1, -2, -4, -5, -8, -10,- 20, -40
If you want to find the factors using an alternate method then you will opt for the multiplication method. To find the factors of 40 through this method you will follow these steps:
1 x 40 = 40Â
In this method, we will multiply any 2 numbers and if the product of these numbers yields 40 then we will consider those numbers as the factors of 40. We will repeat this process until we have multiplied all of the numbers from 1 to 40.
## Factors of 40 by Prime Factorization
When the prime factors of any number are multiplied together to give that number then it is called Prime Factorization. As we have already discussed that 40 is a composite number so we can easily find its prime factorization.
To find the prime factorization of 40 we will follow these steps:
$\dfrac{40}{2}=20$
$\dfrac{20}{2}=10$
$\dfrac{10}{2}=5$
$\dfrac{5}{5}=1$
For prime factorization, you will use the smallest prime number that divides the number 40. In this case, it is 2 so we will divide 40 by 2 and the answer will be further divided by 2 until we get a non-decimal number.Â
Once we get a decimal number we will shift to the next prime number that divides the existing number. We will keep on repeating this process until we get 1 in the answer. We can list all the prime factors like this:
2 x 2 x 2 x 5 = 40Â Â
Figure 2 – Prime Factorization of 40
## Factor Tree of 40
To demonstrate the prime factors we use a factor tree. At each step, we divide a composite number into its factors and keep on repeating the process until we don’t find a prime number or 1.
The factor tree of the number 40 is as given below:
Figure 3 – Factor Tree of 40
## Factors of 40 in Pairs
To find the factor pair of the number 40 we multiply any 2 numbers with each other. If the answer of those two numbers is 40 then both the multiplicand and multiplier will be known as the factor pairs of the number 40.
We can list the factor pairs by finding them in this way:
1 x 40 = 40Â
2 x 20 = 40Â
4 x 10 = 40Â
5 x 8 = 40Â
There is no need to repeat the factors again and again. So the factor pairs of 40 can be written as:
Factor pairs:Â (1,40), (2,20), (4,10), and (5,8)
Since 40 has both positive and negative pairs so we can also calculate all of the negative pairs:
-1 x -40 = 40Â
-2 x -20 = 40Â
 -4 x -10 = 40Â
-5 x -8 = 40
So we can write the negative pair factors as:
Negative Pairs: (-1,-40), (-2,-20), (-4,-10), and (-5,-8).
## Factors of 40 Solved Examples
### Example 1
Audrey is a music teacher and has been assigned the Christmas choir.Â
40 children want to participate in this activity. So Audrey must put all of the students in small equal groups in such a way that no one is left behind. Each group must have more than 5 students but less than 10. Can you help Audrey?
### Solution
As we know that the Factors of 40 are 1, 2, 4, 5, 8, 10, 20, and 40.
According to this list, we know that the only factor greater than 5 and less than 10 is 8.
So Audrey will make each choir group consisting of 8 children so no one is left out.
### Example 2
Write both positive and negative factors of the number 40.
### Solution
We can find the factors of 40 by either division or multiplication method. The factors list of the number 40 is as given below:
Factor List of 40 = 1, 2, 4, 5, 8, 10, 20, 40
Similarly, we can find the negative factors of 40 too:
Negative factors of 40 = -1, -2, -4, -5, -8, -10,- 20, -40Â
### Example 3
Calculate the Prime Factorization of the number 40.
### Solution
The Prime factorization of 40 can be calculated in this way:
$\dfrac{40}{2}=20$
$\dfrac{20}{2}=10$
$\dfrac{10}{2}=5$
$\dfrac{5}{5}=1$
Hence, we can write it as:
2 x 2 x 2 x 5 = 40Â
Images/mathematical drawings are created with GeoGebra.
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Community Primary School
Together we Thrive
# Maths
My Maths
I have set another challenge for you on My Maths! x
Activity 1
Telling the time to the hour.
When we tell the time, there are two hands on the clock - the hour hand, which is short, and the minute hand which is long.
Look at this clock and have a think about the questions below.
How many hands can you see on the clock?
Do you know how to use them to find out what time it is?
Which hand is for the hours and which hand is for the minutes?
What time does the clock say in this picture?
If the party starts at 5 o'clock, what will the clock show?
Use this picture to help you if you find this tricky.
Now have another go. You can ask an adult at home to help you.
The first clock shows 7 o'clock.
The second clock shows 2 o'clock.
Did you get them right?
Now let's practice!
Activity 2
Yesterday we had a go at telling the time to the hour.
Today we are going to learn to tell the time to the half hour.
Remember that a clock has two hands - the hour hand and the minute hand. Do you remember how to tell which one is which?
What time does the Teacher say that assembly will start?
Can you see the minute hand on the classroom clock in the picture? Where is it pointing?
This shows us it is half past. When the minute hand is on 6 we know it is half past.
What is the time on the clock in the picture? The hour hand will have gone a little bit past the hour we are on.
Have a go at these, with help from an adult if you need it.
Now practice telling the time to half an hour!
Activity 3
Today we are going to think about how long it might take to do a simple activity, and to understand how long seconds are compared to minutes.
You will need a stopwatch to time the activities.
Here is the list of your activities and how long you have to complete each one. You need to look at the activity and how long you have to do it. Then guess how many times you can do it, before completing it.
You can print this off and fill it in if you wish.
You will need an adult to time you.
There is a space at the bottom for you to make up your own activity.
Activity 4
Yesterday we timed ourselves doing activities for a number of seconds or minutes.
Today we are going to use this understanding to help us estimate some different activities.
Remember that some activities might only take seconds, some might take minutes, but some activities might take hours.
I estimate:
What about these activities. Do you think they will take seconds, minutes or hours?
Now have a go at today's quick task:
Activity 5
We have spent time this week learning about seconds, minutes and hours. Today we are going to Compare time.
Look at what the teacher is saying here.
How long did it take the class to tidy the room yesterday?
Was that a long time, or a short time?
Do you think that they should be able to do it more quickly?
What time would be faster than 20 minutes to tidy the classroom?
If George tidies his desk in 30 seconds, is this faster or slower than 1 minute?
Now try Activity 5 - Comparing Time
Top
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# Writing the Equation of Hyperbolas
Remember the two patterns for hyperbolas:
We can write the equation of a hyperbola by following these steps:
1. Identify the center point (h, k)
2. Identify a and c
3. Use the formula c2 = a2 + b2 to find b (or b2)
4. Plug h, k, a, and b into the correct pattern.
5. Simplify
Sometimes you will be given a graph and other times you might just be told some information.
Let's try a few.
1. Find the equation of a hyperbola whose vertices are at (-1, -1) and (-1, 7) and whose foci are at (-1, 8) and (-1, -2).
To start, let's graph the information we have:
We can tell that it is a vertical hyperbola. Let's find our center point next and mark it. If we want, we can also draw in a rough hyperbola just to make it easier to visualize:
The center point is (-1, 3). To find a, we'll count from the center to either vertex. a = 4. To find c, we'll count from the center to either focus. c = 5
We'll use the formula c2 = a2 + b2 to find b. To do that, we'll sub in a = 4 and c = 5 then solve for b.
c2 = a2 + b2
52 = 42 + b2
25 = 16 + b2
9 = b2 We need to take the square root.
b = 3
We have all our information: h = -1, k = 3, a = 4, b = 3. Since it's a vertical hyperbola, we'll choose that formula and substitute in our information.
And simplify:
2. Find the equation of this hyperbola:
We can tell that it is a horizontal hyperbola. Let's find our center point next and mark it.k
The center point is (1, 2). To find a, we'll count from the center to either vertex. a = 2. To find c, we'll count from the center to either focus. c = 6
We'll use the formula c2 = a2 + b2 to find b. To do that, we'll sub in a = 2 and c = 6 then solve for b.
c2 = a2 + b2
62 = 22 + b2
36 = 4 + b2
32 = b2
To find b, we would need to take the square root, but it won't come out evenly. That's okay, though, because the pattern needs b2, so we can just substitute in 32 for b2.
We have all our information: h = 1, k = 2, a = 2, b2= 32 . Since it's a horizontal hyperbola, we'll choose that formula and substitute in our information.
And simplify:
Practice: Find the equation of each parabola:
1) Vertices: (2, 1) and (2, -5) Foci: (2, 3) and (2, -7)
2) Vertices: (0, 1) and (6, 1) Foci: (-1, 1) and (7, 1)
3) Vertices: (1, 0) and (3, 0) Foci: (-1, 0) and (5, 0)
Related Links: Math Fractions Factors
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# RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions 2022 | Download Free PDF
RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions: In this exercise, the students will study the problems based on the area of a polygon. Our Mathematics expert tutors have created the problems from the RS Aggarwal textbook to ensure that the students are thorough with their basic concepts & assist clear their queries. RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions are the best study material that is prepared in a simple manner by using shortcut methods to solve problems.
The students can refer to the textbook & download the PDF easily from the link given below. These solutions can build a good foundation for students & they can get well-prepared for answering more tough questions correctly in the Maths final exam. Practicing different types of questions related to the area of a polygon enables the students to boosts their reasoning as well as logical skills. The students will find a summary for a quick revision & formulae based on the area of a polygon.
## Download RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions
RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions
## Important Definition for RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions
• Area of Polygon
The area of any given polygon whether it a square, triangle, quadrilateral, rectangle, parallelogram or rhombus, hexagon, or pentagon is defined as the region occupied by it in a two-dimensional plane. The areas or formulas for areas of different types of polygon depend on their shapes. For instance, to find the area of the triangle, we have to know the length of its base & height.
• Find Area of Polygon with n-sides
We split the figure into triangles, squares, trapezium, etc to find the area of a polygon that is not regular or its formula is not defined. The purpose is to visualize the given geometry as a combination of geometries for which we know how to calculate the area. We then calculate the area for each of the parts & then add them up to get the area of the polygon.
• Area of Polygon Formulas
Here are some formulas for different types of polygons:
Triangle = ½ × base × height
Square = side2
Rectangle = length × width
Pentagon = 5/2 × side length × distance from the center of sides to the center of the pentagon
Rhombus = ½ × product of diagonals
Hexagon = (3√3)/2 × distance from the center of sides to the center of the hexagon
## Benefits of RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions
• RS Aggarwal Class 8 Maths Chapter 18 Ex 18.3 Solutions are easily accessible in PDF format & from any device.
• These exercise solutions are considered as the best study refresher that provides a quick revision of the topics of an area of a polygon along with important definitions & formulas.
• The solutions enable the students to get the clarification of every topic covered in this exercise & can able to solve any question in the Maths final exams.
Know more at the official website.
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If you share 45 between 5, you will get 9.
# Multiplication and Division 1
This Math quiz is called 'Multiplication and Division 1' and it has been written by teachers to help you if you are studying the subject at elementary school. Playing educational quizzes is an enjoyable way to learn if you are in the 3rd, 4th or 5th grade - aged 8 to 11.
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Multiplication and division are easier if you know your times tables. You can practice them by playing our Elementary school Times Tables Quizzes. However you still need other strategies to multiply numbers and to divide them. Some problems can be made simpler, for example to multiply 7 by 9 just multiply it by 10 and take 7 away. That makes it much easier to work out.
Take this quiz to find out how good your strategies are when multiplying or dividing numbers.
1.
What happens when a number is multiplied by 1?
The number is doubled
The number does not change
The number is halved
The digit moves one place to the right
A number does not change when it is divided by 1 either
2.
What number do I get if I share 45 between 5?
4
5
9
10
Share is another way of saying divide
3.
30 x 4 = ?
12
60
120
150
If you know 3 x 4 is 12 then just multiply by 10
4.
500 ÷ 10 = ?
5
50
250
5,000
To divide a number by 10 move the decimal point one place to the left
5.
What number do you get if you double 35?
60
70
80
90
35 x 2 = 70
6.
How many 6s in 18?
2
3
5
6
18 ÷ 6 = 3
7.
Which number sentence is incorrect?
12 x 4 = 48
4 x 12 = 48
48 ÷ 12 = 4
12 ÷ 4 = 48
12 ÷ 4 = 3 not 48!
8.
What number do you get if you double 18?
36
38
40
42
Remember that double is the same as multiply by 2
9.
How do we calculate double a number?
Multiply by 10
Divide by 2
Multiply by 2
Divide by 3
Double means the same as x 2
10.
A tower is 4 bricks tall; how many bricks in a tower 3 times taller?
1
3
12
15
4 x 3 = 12
Author: Amanda Swift
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# roots of complex polynomial - tricks
What tricks are there for calculating the roots of complex polynomials like
$$p(t) = (t+1)^6 - (t-1)^6$$
$t = 1$ is not a root. Therefore we can divide by $(t-1)^6$. We then get
$$\left( \frac{t+1}{t-1} \right)^6 = 1$$
Let $\omega = \frac{t+1}{t-1}$ then we get $\omega^6=1$ which brings us to
$$\omega_k = e^{i \cdot k \cdot \frac{2 \pi}{6}}$$
So now we need to get the values from t for $k = 0,...5$.
How to get the values of t from the following identity then?
\begin{align} \frac{t+1}{t-1} &= e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\ (t+1) &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - e^{i \cdot 2 \cdot \frac{2 \pi}{6}} \\ 1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t\cdot e^{i \cdot 2 \cdot \frac{2 \pi}{6}} - t \\ 1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}} &= t \cdot (e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1) \\ \end{align}
And now?
$$t = \frac{1+e^{i \cdot 2 \cdot \frac{2 \pi}{6}}}{e^{i \cdot 2 \cdot \frac{2 \pi}{6}}-1}$$
So I've got six roots for $k = 0,...5$ as follows
$$t = \frac{1+e^{i \cdot k \cdot \frac{2 \pi}{6}}}{e^{i \cdot k \cdot \frac{2 \pi}{6}}-1}$$
Is this right? But how can it be that the bottom equals $0$ for $k=0$?
I don't exactly know how to simplify this:
$$\frac{ \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} } + 1 }{ 1 - \frac{1}{ e^{i \cdot k \cdot \frac{2 \pi}{6}} }}$$
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Yes, it is right. And divide top and bottom by $e^{\pi i k/6}$. On top you get $2\cos(k\pi/6)$. On the bottom you get $2i\sin(k\pi/6)$. The answers simplify to $-i\cot(k\pi/6)$. – André Nicolas Feb 2 '12 at 23:57
Notice that $t=1$ is not a root. Divide by $(t-1)^6$.
If $\omega$ is a root of $z^6 - 1$, then a root of the original equation is given by $\frac{t+1}{t-1} = \omega$.
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So I've got the roots of $z^6 -1$ as $\omega_k = e^{i\cdot k\cdot\frac{2\pi}{6}}$. How to conclude now $t$ for example from $\frac{t+1}{t-1}=\omega_1$? – meinzlein Feb 2 '12 at 23:24
Try multiplying by $t-1$... – Aryabhata Feb 2 '12 at 23:32
I tried and edited the original post accordingly.. – meinzlein Feb 2 '12 at 23:42
Looks right.... – Aryabhata Feb 2 '12 at 23:54
Well a collegue of mine tried the way mentioned by André Nicolas and got only five roots: $$t_{1}=0$$ $$t_{2,3} = \pm \sqrt{3} i,$$ $$t_{4,5} = \pm \sqrt{\frac{1}{3}} i$$But I got 6 roots (see original post).. Where's the problem? – meinzlein Feb 2 '12 at 23:58
Note that $$(t+1)^6 - (t-1)^6=((t+1)^3-(t-1)^3)((t+1)^3+(t-1)^3)$$ (difference of squares).
When you simplify the first term in the product on the right, there is no $t^3$ term and no $t$ term! The second term in the product simplifies to $2t^3+6t$.
Remark: The solution by Arhabhata is the right one, it works if we replace $6$ by $n$. And when we set $\frac{t-1}{t+1}=e^{2\pi i k/n}$, where $k=1,2,\dots,n-1$, and solve for $t$, we get $-i$ times cotangents.
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# Solving Problems By Elimination
Felix may notice that now both equations have a constant of 25, but subtracting one from another is not an efficient way of solving this problem.
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The elimination method of solving systems of equations is also called the addition method.
To solve a system of equations by elimination we transform the system such that one variable "cancels out".
So let’s now use the multiplication property of equality first.
You can multiply both sides of one of the equations by a number that will result in the coefficient of one of the variables being the opposite of the same variable in the other equation. Notice that the first equation contains the term 4y, and the second equation contains the term y.
Example 3: $$\begin 2x - 5y &= 11 \\ 3x 2y &= 7 \end$$ Solution: In this example, we will multiply the first row by -3 and the second row by 2; then we will add down as before.
$$\begin &2x - 5y = 11 \color\ &\underline \end\ \begin &\underline} \text\ &19y = -19 \end$$ Now we can find: back into first equation: $$\begin 2x - 5\color &= 11 \ 2x - 5\cdot\color &= 11\ 2x 5 &= 11\ \color &\color \color \end$$ The solution is $(x, y) = (3, -1)$.
Recall that a false statement means that there is no solution.
If both variables are eliminated and you are left with a true statement, this indicates that there are an infinite number of ordered pairs that satisfy both of the equations. A theater sold 800 tickets for Friday night’s performance. Combining equations is a powerful tool for solving a system of equations.
Substituting the value of y = 3 in equation (i), we get 2x 3y = 11 or, 2x 3 × 3 = 11or, 2x 9 = 11 or, 2x 9 – 9 = 11 – 9or, 2x = 11 – 9or, 2x = 2 or, x = 2/2 or, x = 1Therefore, x = 1 and y = 3 is the solution of the system of the given equations. Solve 2a – 3/b = 12 and 5a – 7/b = 1 Solution: The given equations are: 2a – 3/b = 12 ……………
(iv) Multiply equation (iii) by 5 and (iv) by 2, we get 10a – 15c = 60 …………… (vi) Subtracting (v) and (vi), we get or, c = 58 /-29 or, c = -2 But 1/b = c Therefore, 1/b = -2 or b = -1/2 Subtracting the value of c in equation (v), we get 10a – 15 × (-2) = 60 or, 10a 30 = 60 or, 10a 30 - 30= 60 - 30 or, 10a = 60 – 30 or, a = 30/10 or, a = 3 Therefore, a = 3 and b = 1/2 is the solution of the given system of equations. x/2 2/3 y = -1 and x – 1/3 y = 3 Solution: The given equations are: x/2 2/3 y = -1 …………… (ii) Multiply equation (i) by 6 and (ii) by 3, we get; 3x 4y = -6 …………… (iv) Solving (iii) and (iv), we get; or, y = -15/5 or, y = -3 Subtracting the value of y in (ii), we get; x - 1/3̶ × -3̶ = 3 or, x 1 = 3 or, x = 3 – 1 or, x = 2 Therefore, x = 2 and y = -3 is the solution of the equation.
## Comments Solving Problems By Elimination
• ###### The Elimination Method
The elimination method for solving systems of linear equations uses the addition. Example. Problem. Use elimination to solve the system. x – y = −6. x + y = 8.…
• ###### Simultaneous Equatuions by Elimination, Maths First, Institute.
This method for solving a pair of simultaneous linear equations reduces one equation to one that. This method is known as the Gaussian elimination method.…
• ###### Solving by Elimination 1
Let's just do one and you'll see how it works See how these guys are the same, but with a different sign?…
• ###### The elimination method for solving linear systems Algebra 1.
Another way of solving a linear system is to use the elimination method. In the elimination method you either add or subtract the equations to get an equation in.…
• ###### Solving systems of equations by elimination video Khan.
An old video where Sal introduces the elimination method for systems of linear equations. Let's explore a few more methods for solving systems of equations. found skills to tackle a word problem, our newly found skills in elimination.…
• ###### Elimination method - free math help -
Elimination method for solving systems of linear equations with examples, solutions and exercises.…
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Work with numbers 11-19 to gain foundations for place value.
• CCSS.Math.Content.K.NBT.A.1 Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (such as 18 = 10 + 8); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones.
________________________________________________________________________
If you are not a math person, haven’t studied math in many years, or have any amount of “math fear” the words BASE TEN may be one of those things that make you sweat and tremble. In general, I would venture to guess that many of us have heard about Base Ten, but have little to no idea what it really means.
Base Ten is the number system that we commonly use that describes the place of each number (ones, tens, hundreds, thousands, etc.).
Take a look at a number like 4,352
The 2 is in the one’s place, the 5 is in the ten’s place, the 3 is in the hundred’s place and the 4 is in the thousand’s place. Each of those number is 10 times the value to the right of it (thus the idea of Base Ten- each place increases by a multiple of 10).
One of the common ways that teachers are currently teaching Base Ten is by introducing Base Ten Blocks like those below.
For the most part, I think these manipulatives are too sophisticated for pre-k children but they will be introduced to these in kindergarten and will probably use them quite extensively.
If I remember correctly, ones are called “bits”, tens are called “rods”, hundreds are called “flats” and thousands are called “blocks”. Children begin to create a “rod” by putting 10 bits together, a “flat” by putting 10 rods together and so on. There are all sorts of interesting and innovative ways teachers are incorporating these into their math teaching.
How can we support the early concepts associated with Base Ten for younger children? The best way we prepare children to understand place value is to reinforce counting, cardinality, ordinality, and one-to-one correspondence. There are better manipulatives for younger children (Unifix cubes, and Cuisenaire Rods, for instance) that can reinforce these concepts through exploration and play.
2 Replies to “Common Core – Numbers & Operations in Base Ten”
1. Base ten blocks are an excellent classroom and home resource for learners.
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# How many ways can you make 50p?
## How many ways can you make 50p?
> Total 12 different ways.
### How many different ways are there of paying exactly 20p using only 1p and 2p coins?
There are 41 ways to make 20p – too many to write here!
How many ways can you make 10p?
eleven ways
Problem 2. Show that you can make up 10 pence in eleven ways using 10p, 5p, 2p and 1p coins.
What coins make 50p?
A combination of different coins is used to make 50p to pay at a car park. Bill pays at a machine to park the car. One at a time, he puts in 5p, 5p, 10p, 10p, then 20p to make 50p.
## Can you make 10p with 3 coins?
There are two 5p coins left, which can make 10p. The three 1p coins make 3p in total but there is not a 3 pence coin. We can use two of the 1p coins to make a 2p coin and then leave the final third 1 pence coin as a 1p.
### How do you get 10 coins?
A dollar is worth 100 cents and each dollar bill is the same value as 100 penny coins. Each dime is worth 10 cents so ten dimes make one dollar because 10 x 10 = 100 cents. Each nickel is worth 5 cents so twenty nickels make one dollar because 20 x 5 = 100 cents.
How many pennies are in 10 pounds?
10 pounds (1500 copper pennies) 1909-1982.
What shape is a 20p coin?
equilateral curve heptagon
To help identification and avoid confusion with similar sized coins the 20p is seven sided and, like the 50p, an equilateral curve heptagon. The shape, with its constant rolling diameter, means that it is readily acceptable in vending machines. The 20p coin is legal tender for amounts up to £10.
## How much money is a pound of pennies?
For the purists, there are about 145 copper cents / pennies in a pound.
### Can you add 5p coins to a 10p coin?
We have: two 10p coins, a 5p coin and two 1p coins. We start by adding the coins with the largest value. We will add the two 10p coins. Two 10p coins are worth 20p. There is a 20 pence coin and so we will write this down. The 5p coin cannot be added to two 1p coins to make a new coin.
How much money can you make from 2p coins?
The two 2p coins plus the 1p coin can be replaced with a 5p coin. The total of the money is 55 pence. Instead of using the original combination of coins, we can also make 55p from a single 50p coin and a single 5p coin. In this example we have three 10p coins, two 5p coins and three 2p coins.
Which is more a pound or a 50p coin?
£1 is equivalent to a 50p coin, two 20p coins plus a 10p coin. £1 is equivalent to five 20p coins. £1 is equivalent to ten 10p coins. Any combination of coins that equals 100 pence is equivalent to one pound.
## How to make an amount out of coins?
Start by adding the largest value coins. Here we have six 5p coins and three 1p coins. We will try to add the coins to make any of the coins listed at the top. We can add four 5p coins to make 20p. The remaining two 5p coins make 10p. The three 1p coins make 3p but there is no 3p coin. We can add two of the 1p coins to make 2p.
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# How to Prove & Derive Trigonometric Identities
Coming up next: Solving Oblique Triangles Using the Law of Cosines
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• 0:01 Trigonometric Identities
• 0:36 The Tangent
• 2:25 The Double-Angle Identities
• 3:55 The Half-Angle Identities
• 6:36 Lesson Summary
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
After watching this video lesson, you will learn how some of our trigonometric identities are derived. You will also see how some identities naturally lead to the others.
## Trigonometric Identities
In trigonometry, we have a bunch of trigonometry identities, or true statements about trig functions. Think of these as definitions if you will. They tell you how to describe certain trig functions in other terms.
We use our trigonometry identities to help us simplify more complicated trig problems and prove other trig statements. What's really neat about some of our identities is that we can easily prove them from the other identities. So, if you ever forget some of them, you could derive them yourself if you remember the proofs that you're about to see. Are you ready to begin? Get your thinking cap on!
## The Tangent
The first one we are going to see is the tangent function. Remember that you've already learned that the tangent function also happens to be the sine function divided by the cosine function. How did they come up with this? We can easily derive this using our definitions for each of those functions. Make sure your thinking cap is still on, as this requires a bit of thinking.
First, our definitions - we know that our sine function is defined as opposite over hypotenuse, our cosine function is adjacent over hypotenuse, and our tangent function is opposite over adjacent. Recall that these definitions are based on the right triangle where the hypotenuse is the hypotenuse side, the adjacent is the side closest to the angle, and the opposite is the side opposite to the angle.
We are going to use these definitions to show how we can go from sine over cosine to the tangent function. We begin with our sine/cosine. We then insert our definitions. We get (opposite/hypotenuse) / (adjacent/hypotenuse).
Using our knowledge of dividing fractions, we turn this into a multiplication problem by flipping the bottom fraction. We get (opposite/hypotenuse) * (hypotenuse/adjacent). Now, we can go ahead and cancel or simplify what we can. We see a hypotenuse in the numerator and denominator. We can go ahead and cancel these.
What are we left with? We are left with opposite/adjacent. Which function does this define? Why, isn't it the tangent function? And there we have it; we have derived the tangent function from sine/cosine. The whole process looks like this:
Pretty cool, huh?
## The Double-Angle Identities
Now, let's look at something slightly more complicated, but not more difficult. We're going to derive our double-angle identities from our sum and difference identities. Recall that our double-angle identities are these:
And our sum identities are these:
The difference identities are the same as the sum identities except that the signs are opposite. What this means is that where you see a plus sign now, you will see a minus sign, and where you see a minus sign now, you will see a plus sign. If you don't see a sign in front of something, it stays the same.
For this part of the lesson, we are only going to look at the sum identities that you see. We can derive our double-angle identities from our sum identities by simply setting the angles alpha and beta equal to each other. If alpha and beta were both x, then alpha plus beta will become 2x. If we plug in x for both alpha and beta, we will get these for our sum identities:
All we did was plug in x for both alpha and beta and then simplified our expressions. We applied our algebra skills to combine like terms. Do you recognize the formulas that we ended up with? Why, aren't they our double-angle identities?
Yes, they are indeed! If you ever forget your double-angle identities, but you remember your sum identities, then you can easily find the double-angle identities by simply setting both angles in the sum identities to the same value.
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# How do you factor y= 2x^2 - 9x – 18 ?
Dec 30, 2015
$y = \left(2 x + 3\right) \left(x - 6\right)$
#### Explanation:
If the factored quadratic is expressed as $\left(a x + b\right) \left(c x + d\right)$ then the general form of a quadratic is $y = a c {x}^{2} + \left(b c + a d\right) x + b d$
We therefore need to look for factors of $2$ and $\left(- 18\right)$ that will combine to give $\left(- 9\right)$
Possible factors of $2$ are only $2$ and $1$
Possible factors of $\left(- 18\right)$ are $- 9$ and $2$ or $- 6$ and $3$
$2 \cdot \left(- 6\right) + 1 \cdot 3 = - 9$ so these are the correct factors.
Hence $y = \left(2 x + 3\right) \left(x - 6\right)$
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## Intermediate Algebra (12th Edition)
$\dfrac{2-9\sqrt{2}}{3}$
$\bf{\text{Solution Outline:}}$ To simplify the given expression, $\dfrac{12-9\sqrt{72}}{18} ,$ simplify the radicand that contains a factor that is a perfect power of the index Then, find the $GCF$ of all the terms and express all terms as factors using the $GCF.$ Finally, cancel the $GCF$ in all the terms. $\bf{\text{Solution Details:}}$ Writing the radicand as an expression containing a factor that is a perfect power of the index and extracting the root of that factor result to \begin{array}{l}\require{cancel} \dfrac{12-9\sqrt{36\cdot2}}{18} \\\\= \dfrac{12-9\sqrt{(6)^2\cdot2}}{18} \\\\= \dfrac{12-9(6)\sqrt{2}}{18} \\\\= \dfrac{12-54\sqrt{2}}{18} .\end{array} The $GCF$ of the coefficients of the terms, $\{ 12,-54,18 \},$ is $6$ since it is the highest number that can divide all the given coefficients. Writing the given expression as factors using the $GCF$ results to \begin{array}{l}\require{cancel} \dfrac{6\cdot2+6\cdot(-9)\sqrt{2}}{6\cdot3} .\end{array} Cancelling the $GCF$ in every term results to \begin{array}{l}\require{cancel} \dfrac{\cancel6\cdot2+\cancel6\cdot(-9)\sqrt{2}}{\cancel6\cdot3} \\\\= \dfrac{2-9\sqrt{2}}{3} .\end{array}
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How to Find a Vector’s Components - dummies
# How to Find a Vector’s Components
You can convert from the magnitude/angle way of specifying a vector to the coordinate way of expression. Doing so is essential for the kinds of operations you can expect to execute on vectors, such as when adding vectors.
For example, you have one vector at 15 degrees and one at 19 degrees, and you want to add them together. How the heck do you do that? If you were to convert them into their coordinates, (a, b) and (c, d), the answer would be trivial because you only have to add the x and y coordinates to get the answer: (a + c, b + d).
To see how to convert between the two ways of looking at vectors, take a look at vector v in the figure. The vector can be described as having a magnitude v at an angle of theta.
To convert this vector into the coordinate way of looking at vectors, you have to use the trigonometry shown in the figure. The x coordinate equals v cos theta, and the y coordinate equals v sin theta:
vx = v cos theta
vy = v sin theta
Keep these relationships in mind because you’ll come across them often in physics questions.
## Sample question
1. Suppose that you’ve walked away from the origin so you’re now 5.0 kilometers from the origin, at an angle of 45 degrees. Resolve that into vector coordinates.
The correct answer is (3.5, 3.5) km.
1. Apply the equation vx = v cos theta to find the x coordinate. That’s 5.0 cos 45 degrees, or 3.5.
2. Apply the equation vy = v sin theta to find the y coordinate. That’s 5.0 sin 45 degrees, or 3.5.
## Practice questions
1. Resolve a vector 3.0 meters long at 15 degrees into its components.
2. Resolve a vector 9.0 meters long at 35 degrees into its components.
3. Resolve a vector 6.0 meters long at 125 degrees into its components.
4. Resolve a vector 4.0 meters long at 255 degrees into its components.
Following are answers to the practice questions:
1. (2.9, 0.8) m
1. Apply the equation vx = v cos theta to find the x coordinate: 3.0 x cos 15 degrees, or 2.9.
2. Apply the equation vy = v sin heta to find the y coordinate: 3.0 x sin 15 degrees, or 0.8.
2. (7.4, 5.2) m
1. Apply the equation vx = v cos theta to find the x coordinate: 9.0 x cos 35 degrees, or 7.4.
2. Apply the equation vy = v sin theta to find the y coordinate: 9.0 x sin 35 degrees, or 5.2.
3. (–3.4, 4.9) m
1. Apply the equation vx = v cos theta to find the x coordinate: 6.0x cos 125 degrees, or –3.4.
2. Apply the equation vy = v sin theta to find the y coordinate: 6.0 x sin 125 degrees, or 4.9.
4. (–1.0, –3.9) m
1. Apply the equation vx = v cos theta to find the x coordinate: 4.0 x cos 255 degrees, or –1.0.
2. Apply the equation vy = v sin theta to find the y coordinate: 4.0 x sin 255 degrees, or –3.9.
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Review question
# Can we sketch the graph of $y=x(x+1)(x-2)^4$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R8387
## Solution
The curve $C$ has equation $y=x(x+1)(x-2)^4.$ Show that the gradient of $C$ is $(x-2)^3(6x^2+x-2)$ and find the coordinates of all the stationary points.
We can write $y=(x^2+x)(x-2)^4$ before differentiating using the product and chain rules. This yields \begin{align*} \frac{dy}{dx}&=(2x+1)(x-2)^4+4(x^2+x)(x-2)^3 \\ &=(x-2)^3\left[(2x+1)(x-2)+4x(x+1)\right] \\ &=(x-2)^3(6x^2+x-2). \end{align*}
Alternatively, we could use the Product Rule for three functions, that is, if $y = uvw$, then $y' = u'vw + uv'w + uvw'$. \begin{align*} \frac{dy}{dx}&=(x+1)(x-2)^4+x(x-2)^4+4x(x+1)(x-2)^3 \\ &=(x-2)^3\left[(x+1)(x-2)+x(x-2)+4x(x+1)\right] \\ &=(x-2)^3\left[x^2-x-2+x^2-2x+4x^2+4x\right] \\ &=(x-2)^3(6x^2+x-2). \end{align*}
To find the $x$ coordinates of the stationary points we solve $\frac{dy}{dx}=(x-2)^3(6x^2+x-2)=0 = (x-2)^3 (3x+2)(2x-1),$ which has solutions given by $x=2 \text{ or } x=-\frac{2}{3} \text{ or } x=\frac{1}{2}.$ Substituting these values back in to $y$ we find the stationary points of $C$ are at $(2,0), \left(\frac{1}{2}, \frac{243}{64}\approx 3.8\right), \left(-\frac{2}{3}, -\frac{8192}{729}\approx-11.2\right).$
Determine the nature of each stationary point and sketch $C$.
As this function is a degree six polynomial (sextic) with positive coefficient of $x^6$ we know that for $x$ large and negative it is decreasing, and for $x$ large and positive it is increasing. Between these values the curve undulates so the first and last stationary points must be minima and the middle one a maximum.
Alternatively we could calculate the gradient at the integer values $x= -1, \: 0, \: 1, \:$ and $\: 3$ to verify the nature of the points.
We also know where all the $x$ intercepts are, and have calculated the stationary points: joining the dots gives us a sketch of our curve.
To determine the nature of the stationary point, we might be tempted to differentiate again and find the sign of this at the points where $\dfrac{dy}{dx} = 0$: $\frac{d^2y}{dx^2}=3(x-2)^2(6x^2+x-2)+(x-2)^3(12x+1).$
If we choose to do this we should notice that if $\dfrac{dy}{dx} = 0$, then the first term here is bound to be zero, which reduces the workload, but is still more effort than necessary.
We might also think that at $x=2$ there is a point of inflection because $\frac{d^2y}{dx^2}=0$ here. We can see from the expression for $y$ that the curve has a quadruple root at $x=2$, so we expect the curve to touch the $x$ axis here. (Note that a triple root would indicate a point of inflection.)
In separate diagrams draw sketches of the curves whose equations are:
1. $y^2=x(x+1)(x-2)^4$;
2. $y=x^2(x^2+1)(x^2-2)^4$.
In each case, you should pay particular attention to the points where the curve meets the $x$ axis.
To sketch the graph of (i), first notice that we must have $x(x+1)(x-2)^4 \geq 0$ in order for $y$ to take real values. This means that (i) is not defined for the region $-1 < x <0$.
The $x$-intercepts are the same, and the general shape will be similar to our first graph.
The difference is that every original $y$-value needs to be square-rooted for this graph, which will therefore be symmetrical about the $x$ axis. This leads us to
For sketch (ii), first notice that the roots for $y = 0$ have changed, that is, the $x$-intercepts will be different.
Also, notice that $y\geq 0$ for all $x$, and that we will have symmetry about the $y$ axis, since swapping $-x$ for $x$ gives the same $y$-value. .
We see that the roots are at $x=0$ and $x=\pm\sqrt{2}$, and now we have all we need to sketch the graph.
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# AP Calculus AB : Analysis of curves, including the notions of monotonicity and concavity
## Example Questions
### Example Question #1 : Analysis Of Curves, Including The Notions Of Monotonicity And Concavity
Find the coordinates of all local extrema for , and specify whether each is a local maximum or local minimum.
No local extrema
is a local maximum.
is a local minimum.
is a local maximum.
is a local minimum.
is a local maximum.
is a local minimum.
is a local maximum.
is a local minimum.
Explanation:
To find the coordinates of the local extrema of a function, we need to find the critical points of its first derivative.
Since is a polynomial, we can find its derivative term by term. The first 3 terms can be differentiated using the power rule, , and the constant multiple rule, .
The last term is a constant, and its derivative is zero.
Applying these rules, we find the first derivative:
Now we need to find the critical points. To do this, we set the first derivative equal to zero and solve. Factoring is the best method in this problem.
Now that we have the critical points, we need to determine for each one, whether it is a maximum, minimum, or neither. We use the first derivative line test to determine this.
For , we will test the interval before it, , and the interval after it, , and find whether they are increasing or decreasing.
For the interval , we will test and find whether is positive or negative.
Since is positive, the original function is increasing before the critical point, .
Now we will test the interval after .
For the interval, , we will test and find whether is positive or negative.
Since is negative, the original function is decreasing in the interval following .
Since the function is increasing before , and decreasing afterward, we can conclude that a maximum occurs at .
Now we find the value of this maximum, .
Thus is a local maximum.
Now we will determine whether a maximum or minimum occurs at .
We know that is decreasing before , but we still need to determine what happens afterward.
For the interval , we will test , and find whether is positive or negative.
Since is positive, the original function is increasing following .
Since is decreasing before, and increasing after , we can conclude that a minimum occurs at .
Now we need to find the value of this minimum, .
Thus is a local minimum.
So our answer is:
is a local maximum.
is a local minimum.
### Example Question #2 : Analysis Of Curves, Including The Notions Of Monotonicity And Concavity
Find the intervals of concavity for the function
Concave down:
Concave up:
Concave down:
Concave up:
Concave down:
Concave up:
Concave down:
Concave up:
Concave down:
Concave up:
Concave down:
Concave up:
Explanation:
Concavity refers to the "curving" of the function. While the first derivative describes when the function is increasing or decreasing (instantaneous rate of change of ), the second derivative describes concavity, the instantaneous rate of change of .
The first derivative is like velocity, (moving forward or backward), while the second derivative is like acceleration (speeding up or slowing down).
Since we need to find the intervals of concavity, we will find the second derivative and work with it.
First we must find the first derivative using the power rule and constant multiple rule for each term of .
This gives:
Now to find the second derivative, we take the derivative of .
The same derivative rules apply:
Now that we have the second derivative, , we must find its critical points. We do this by setting , and solving.
The best method for this case is factoring.
The greatest common factor is ,
Inside the parentheses is a quadratic expression that can be factored like so:
Setting each factor equal to zero we find the following:
Now we know the critical points for the second derivative. To find the intervals of concavity, we test a point in each interval around these critical points and find whether the second derivative is positive or negative in that interval.
For the interval , we can test .
Using the factored form of the second derivative is easier than using the polynomial form, since the arithmetic involves fewer large numbers.
Since is negative, the original function is Concave Down in the interval .
Now for the interval , we can test .
We will combine the fractions inside the parentheses by getting the common denominator.
Multiplying the fractions gives:
Since is positive, the original function is Concave Up on the interval .
For the interval , we can test .
Since is negative, the original function is Concave Down on the interval .
Finally, for the interval , we can test .
Since is positive, the original function is Concave Up on the interval .
Summarizing the results, the intervals of concavity are:
Concave Down:
Concave Up:
### Example Question #5 : Finding Regions Of Concavity And Convexity
At the point where , is increasing or decreasing, and is it concave up or down?
Decreasing, concave up
Increasing, concave down
Increasing, concave up
There is no concavity at that point.
Decreasing, concave down
Increasing, concave up
Explanation:
To find if the equation is increasing or decreasing, we need to look at the first derivative. If our result is positive at , then the function is increasing. If it is negative, then the function is decreasing.
To find the first derivative for this problem, we can use the power rule. The power rule states that we lower the exponent of each of the variables by one and multiply by that original exponent.
Remember that anything to the zero power is one.
Plug in our given value.
Is it positive? Yes. Then it is increasing.
To find the concavity, we need to look at the second derivative. If it is positive, then the function is concave up. If it is negative, then the function is concave down.
Repeat the process we used for the first derivative, but use as our expression.
For this problem, we're going to say that since, as stated before, anything to the zero power is one.
Notice that as anything times zero is zero.
As you can see, there is no place for a variable here. It doesn't matter what point we look at, the answer will always be positive. Therefore this graph is always concave up.
This means that at our given point, the graph is increasing and concave up.
### Example Question #1 : Analysis Of Curves, Including The Notions Of Monotonicity And Concavity
Consider the function:
On what intervals is increasing? Consider all real numbers.
Explanation:
To answer this question, one first needs to find and then find the critical points of the function (i.e. where . Finally, one would need to determine the sign of for the intervals between the critical points.
For the given function:
.
Therefore, when and . So, the intervals to consider are:
To determine the sign of , pick any number for the given interval and evaluate at that number.
Therefore, is increasing on the intervals and since is greater than zero on these intervals.
### Example Question #1 : Analysis Of Curves, Including The Notions Of Monotonicity And Concavity
A function, , is concave up on the intervals and with and .
Which of the following must be true?
Two or more of the other answers.
Explanation:
On the domain , we know that the derivative begins positive, and because the concavity is positive, we know the derivative is increasing. Thus, the derivative stays positive for this entire interval, and the function increases from 2 to 4. Thus, must be greater than .
In the case of the interval , we know that the derivative is increasing, but it starts out negative. Thus, perhaps the derivative only increased from -1 to -0.5 in this interval, and the function would have decreased the entire time. In this case, would be less than , so we can't really say anything about these values.
For the remaining two, there's not any clear way to relate the functions at and . While we know needs to be bigger than 1, we don't know by how much. Similarly, while we know needs to be bigger than -1, we don't know by how much. Thus, it's completely possible that and .
As for and , we know even less. In between the two intervals, our function could have shot up a million, or shot down by the same amount. Thus, there's no safe comparison we can make between these two values.
### Example Question #541 : Derivatives
Find the intervals on which is increasing.
Explanation:
To find the intervals where the function is increasing, we need to find the points at which its slope changes from positive to negative and vice versa. The first derivative, which is the slope at any point, will help us.
First, we find the derivative of , using the power rule for each term. Recall that the power rule says
Also, the constant multiple rule will apply to the coefficients of each term. The constant multiple rule simply says that any constant factor of a term will "carry" to the derivative of that term. For example:
Lastly, the derivative of a constant is zero. This will result in the last term, , dropping off as we take the derivative.
Applying these rules, we find the derivative
Where the derivative is positive (blue line in graph), the tangent line to the original function is angled up. Where the derivative is negative (red line in graph), the slope of the tangent line is angled down.
The points where the tangent line's slope transitions from negative to positive or vice versa, are called the critical points. At these points, the tangent line becomes a horizontal line with a slope of zero (green line in graph). In other words, the "critical points" occur when the derivative is zero. These points will be the endpoints of our intervals of increasing and decreasing. To find the critical points, we will set the derivative equal to zero and solve for x. In this problem factoring is the best method:
Now that we have found the critical points, we need to know whether the original function is increasing or decreasing in the intervals between them. We will do so by testing a point in each interval and determining whether the derivative is positive or negative at that point. This is called the first derivative line test.
For the interval , we will test .
(Note: I will use the factored form of the derivative, but we could also use the polynomial version. Both will give the same result)
Since the derivative is negative at this point, we can conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .
For the interval , we will test .
Since the derivative is positive at this point, we can again conclude that the derivative is positive for the whole interval. Thus, the original function is increasing on .
For the interval , we will test .
Since the derivative is negative at this point, we can again conclude that the derivative is negative for the whole interval. Thus, the original function is decreasing on .
Lastly, for the interval , we will test .
Since the derivative is positive at this point, we know that the derivative is positive for the whole interval. Thus the original function is increasing on .
From these 4 results, we now know the answer. The function is increasing on the intervals .
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Bridges in Mathematics Grade 3 Student Book Unit 3 Module 4 Answer Key
The solutions to Bridges in Mathematics Grade 3 Student Book Answer Key Unit 3 Module 4 can help students to clear their doubts quickly.
Bridges in Mathematics Grade 3 Student Book Answer Key Unit 3 Module 4
Bridges in Mathematics Grade 3 Student Book Unit 3 Module 4 Session 1 Answer Key
Round, Estimate & Find the Sum
Before you start adding numbers, it is a good idea to estimate what their sum will be. That way, you can tell if your final answer is reasonable. Round each pair of numbers to the nearest ten and then add the rounded numbers to estimate the sum. Then use the standard algorithm to find the exact sum.
Question 1.
The sum will be about _____
Exact sum
(Use the algorithm)
First Round both the given numbers ,Then estimate the given by adding them and get the result as 610. Also, find out the Exact sum of the actual given two numbers using the algorithm as 605.
Question 2.
The sum will be about _____
Exact sum
(Use the algorithm)
First Round both the given numbers ,Then estimate the given by adding them and get the result as 1020. Also, find out the Exact sum of the actual given two numbers using the algorithm as 1021.
Question 3.
The sum will be about _____
Exact sum
(Use the algorithm)
First Round both the given numbers ,Then estimate the given by adding them and get the result as 1000. Also, find out the Exact sum of the actual given two numbers using the algorithm as 1001.
Question 4.
The sum will be about _____
Exact sum
(Use the algorithm)
First Round both the given numbers ,Then estimate the given by adding them and get the result as 980. Also, find out the Exact sum of the actual given two numbers using the algorithm as 982.
Bridges in Mathematics Grade 3 Student Book Unit 3 Module 4 Session 2 Answer Key
Use the standard algorithm to solve each problem. Then solve it a different way. Label your method. Circle the strategy that seemed quicker and easier.
ex
Question 1.
51 + 29 =
Standard Algorithm
Different Strategy
Strategy: ___________
First add both the given numbers 51 , 29 and get the result as 90. Then, write the result with a different strategy.
Question 2.
Standard Algorithm
Different Strategy
Strategy: ___________
First add both the given numbers 198, 56 and get the result as 254. Then, write the result with a different strategy.
Question 3.
348 + 578 =
Standard Algorithm
Different Strategy
Strategy: ___________
First add both the given numbers 348, 578 and get the result as 926. Then, write the result with a different strategy.
Question 4.
Standard Algorithm
Different Strategy
Strategy: ___________
First add both the given numbers 34 , 56 and 72 and get the result as 119. Then, write the result with a different strategy.
Use the standard algorithm to solve each problem. Then solve it a different way. Label your method. Circle the strategy that seemed quicker and easier.
a.
Standard Algorithm
Different Strategy
Strategy: ______
First add both the given numbers 63 ,36 and get the result as 99. Then, write the result with a different strategy.
b.
149 + 253
Standard Algorithm
Different Strategy
Strategy: ______
First add both the given numbers 149 , 253 and get the result as 402. Then, write the result with a different strategy.
c.
53 + 28 + 72 =
Standard Algorithm
Different Strategy
Strategy: ______
First add both the given numbers 53,28 ,72 and get the result as 153. Then, write the result with a different strategy.
d.
Standard Algorithm
Different Strategy
Strategy: ______
First add both the given numbers 379 , 272 and get the result as 651. Then, write the result with a different strategy.
e.
Standard Algorithm
Different Strategy
Strategy: ______
First add both the given numbers 512 and 365 and get the result as 877. Then, write the result with a different strategy.
Bridges in Mathematics Grade 3 Student Book Unit 3 Module 4 Session 3 Answer Key
Running Robots
Question 1.
Fill in the bubble to show the best estimate for each problem:
a.
30
40
50
60
First round the both numbers , then subtract the 30 from 80 and get the result 50 which is as shown below.
b.
30
40
50
60
First round the both numbers , then subtract the 170 from 200 and get the result 30 which is as shown below.
Question 2.
Solve each subtraction problem below. You may use the regrouping strategy (standard algorithm) or any other strategy for addition that is efficient for you. Be sure to show your work.
a. Nina and Ricardo designed robots. Nina’s robot can run for 235 minutes before the batteries need to be recharged. Ricardo’s robot can run for 187 minutes before the batteries need recharging. How much longer can Nina’s robot run than Ricardo’s?
Number of minutes that Ricardo’s robot can run = 187
Number of minutes that Nina’s robot can run = 235
Number of minutes that Nina’s robot can run than Ricardo’s = ?
= 235 – 187
= 48 minutes.
b. Kiran and Brenda also designed robots. Kiran’s robot can walk 307 meters before the batteries need to be recharged. Brenda’s robot can walk 268 meters before the batteries need to be recharged. How much farther can Kiran’s robot walk than Brenda’s?
Number of minutes taken for design the Robot for Kiran = 307
Number of minutes taken for design the Robot for Brenda’s = 268
Number of minutes taken for Kiran than Brenda’s = ?
= 307 – 268
= 39 minutes.
Bridges in Mathematics Grade 3 Student Book Unit 3 Module 4 Session 4 Answer Key
Subtraction Strategies
Use the standard algorithm to solve each problem. Then solve it a different way. Label your method. Circle the strategy that seemed quicker and easier.
Question 1.
75 – 24 =
Standard algorithm
Different Strategy
Strategy: _________
First subtract the given numbers 24 from 75 and get the result as 51.Then , write the given number with a different strategy as shown below.
Question 2.
Standard algorithm
Different Strategy
Strategy: _________
First subtract the given numbers 129 from 243 and get the result as 114. Then , write the given number with a different strategy as shown below.
Question 3.
512 – 339 =
Standard algorithm
Different Strategy
Strategy: _________
First subtract the given numbers 339 from 512 and get the result as 173. Then , write the given number with a different strategy as shown below.
Question 4.
Standard algorithm
Different Strategy
Strategy: _________
First subtract the given numbers 326 from 649 and get the result as 323. Then , write the given number with a different strategy as shown below.
More Subtraction Strategies
Use the standard algorithm to solve each problem. Then solve it a different way. Label your method. Circle the strategy that seemed quicker and easier.
Question 1.
Standard algorithm
Different Strategy
Strategy: ________
First subtract the given numbers 45 from 91 and get the result as 46. Then , write the given number with a different strategy as shown below.
Question 2.
253 – 149
Standard algorithm
Different Strategy
Strategy: ________
First subtract the given numbers 149 from 253 and get the result as 104. Then , write the given number with a different strategy as shown below.
Question 3.
265 – 174
Standard algorithm
Different Strategy
Strategy: ________
First subtract the given numbers 174 from 265 and get the result as 91. Then , write the given number with a different strategy as shown below.
Question 4.
Standard algorithm
Different Strategy
Strategy: ________
First subtract the given numbers 251 from 374 and get the result as 123. Then , write the given number with a different strategy as shown below.
Question 5.
Standard algorithm
Different Strategy
Strategy: ________
Akira and Taro are participating in a bike-a-thon to raise money for people who lost their homes in an earthquake. For every mile they ride, they will earn $1 to help rebuild and refurnish homes. Show your work with numbers, sketches, or words as you help Akira and Taro figure out how far they have ridden and how much money they have earned. Then, write an equation for each problem. Question 1. In one week, Akira rode her bike 87 miles. Taro rode his bike 51 miles. How many more miles did Akira ride? Equation: ___________________ Answer: Number of miles rode by Akira = 87 miles Number of miles rode by Taro = 51 miles Number of more miles ride by Akira = x = ? Equation : x = 87 – 51 = 36 miles Question 2. After two weeks, Akira and Taro biked a total 276 miles. After 3 weeks, they had biked a total of 413 miles. How many miles did they ride in the third week? (Hint: How much farther had they biked at the end of the third week than at the end of the second week?) Equation: _______________________ Answer: Number of miles biked by Akira and taro after 2 weeks = 276 Number of miles biked by Akira and taro after 3 weeks = 413 Number of miles biked by both at the end of 3 week than end of second week = x = ? Equation : x = 413 – 276 x = 137 Question 3. Akira and Taro want to raise$537. After four weeks, they have raised $498. How much more money do they need to earn? Equation: _____________________________ Answer: Amount of money wants to raise by both Akira and Taro =$537
Amount of money raised after 4 weeks = $498 Amount of money both needed to earn = x = ? Equation : x = 537 – 498 =$ 39
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# How do you find a standard form equation for the line with (5,4) perpendicular to the line 3x+2y=7?
Sep 15, 2017
$2 x - 3 y = - 2$
#### Explanation:
$3 x + 2 y = 7$ has a slope of $- \frac{3}{2} \textcolor{w h i t e}{\text{xxxx}}$ see Note 1
All lines perpendicular to $3 x + 2 y = 7$ have a slope of $\frac{2}{3} \textcolor{w h i t e}{\text{xxxx}}$see Note 2
If such a perpendicular line goes through $\left(5 , 4\right)$
then we can write it equation in slope point form as:
$y - 4 = \frac{2}{3} \left(x - 5\right) \textcolor{w h i t e}{\text{xxxxxxxxxxxxx}}$see Note 3
This can be converted into standard form as:
2x-3y=-2color(white)("xxxxxxxxxxxxxxxx"see Note 4
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note 1
A relation in the form $A x + B y = C$ has a slope of $- \frac{A}{B}$;
in this case $A = 3$ and $B = 2$
If you are not familiar with this rule, you can convert given relation $3 x + 2 y = 7$ into slope-intercept form:
$2 y = - 3 x + 7$
$y = \left(- \frac{3}{2}\right) x + \frac{7}{2}$ with slope $\left(- \frac{3}{2}\right)$ and y-intercept $\frac{7}{2}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note 2
If a line has a slope of $m$ then all line perpendicular to it have a slope of $\left(- \frac{1}{m}\right)$
In this case $- \frac{1}{- \frac{3}{2}} = \frac{2}{3}$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note 3
A line with slope $m$ through a point $\left({x}_{0} , {y}_{0}\right)$
has a slope-point form:
$\textcolor{w h i t e}{\text{XXX}} y - {y}_{0} = m \left(x - {x}_{0}\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Note 4
"standard form" for a linear equation is
$\textcolor{w h i t e}{\text{XXX}} A x + B y = C$ with $A , B , C \in \mathbb{Z} , A \ge 0$
converting $y - 4 = \frac{2}{3} \left(x - 5\right)$ into this form:
$\textcolor{w h i t e}{\text{XXX}} 3 \left(y - 4\right) = 2 \left(x - 5\right)$
$\textcolor{w h i t e}{\text{XXX}} 3 y - 12 = 2 x - 10$
$\textcolor{w h i t e}{\text{XXX}} - 2 x + 3 y = 2$
$\textcolor{w h i t e}{\text{XXX}} 2 x - 3 y = - 2$
=============================================
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# Area and Perimeter Related to Arcs of a Circle
Yes! The pizza in your hand is a perfect example of a circle, didn’t bother to ever notice? Well in this section we shall be using it as our benchmark example to derive some interesting concepts of arcs of a circle. We will see in this section how to calculate area and perimeter of circle and arcs of a circle.
## Circle
Going by the basic definition, it is a closed plane geometric shape. Now getting into the technical jargon, a circle is a locus of a point moving in a plane in such a way that its distance from a fixed point is always constant. In layman terms, a round shape is often referred to as a circle. Point seeking attention here is that the fixed distance is the radius of the circleMost famous examples of a circle in a line are pizza, chapatti, wheel etc.
Basics of a circle
## Circumference/Perimeter of a Circle
A perimeter of any geometrical figure is the length of the outer boundary of the shape. Similarly, in case of a circle, it’s perimeter is termed as circumference. On practical grounds, let’s take an example of a wheel, the distance covered by the wheel in one complete revolution will be the circumference of the wheel.
The formula for the circumference of a circle is C= 2πR
where C= circumference,
R = Radius of the circle,
π = It is constant pronounced as “pi” with a value of 22/7 or 3.1416…
## Arcs of a Circle
Now that we are done with the circumference of the circle, what is an arc? Arc is a part of the circumference of a circle. If the length is zero, it will be merely a point on the boundary of the circle. And if it is of length, it will be the circumference of the circle i.e. an arc of length 2πR.
Length of the arc of a circle = (θ/360o ) x 2πR
θ=Angle subtended by an arc at the centre of the circle, measured in degrees. If you are working with angles measured in radians instead of in degrees, then go an extra mile converting it in degrees with the conversion factor: (180o/π) x θ
An arc of length L with a circle of radius R
### Derivation of Length of an Arc of a Circle
• Step 1: Draw a circle with centre O and assume radius. Let it be R.
• Step 2: Now, point to be noted here is that the circumference of circle i.e. arc of length 2πR subtends an angle of 360o at centre.
• Step 3: Going by the unitary method an arc of length 2πR subtends an angle of 360o at the centre, Therefore; an arc subtending angle θ at the centre will be of length: (θ/360o ) x 2πR
### Area of Arcs of a Circle or Sectors of a Circle
When we talk about area enclosed by arcs of a circle it is actually the space enclosed between the ends of the arc and the centre of the circle specifically area enclosed by an arc is the area of the sector of the circle. A sector of a circle is like a slice of a pizza where whole pizza is the complete circle.
Slice of a pizza same as a sector of a circle where pizza is the complete circle in the present case
### Derivation for the Area of a Circle
• Step 1: Consider a pizza of equal-sized slices.
• Step 2: Arrange the slices such that they form a rectangle in the following manner:
Visualising area of a circle using Area of Rectangle
• Step 3: Now, as we can see from the figure, the breadth of the rectangle is R, that is the radius of a circle and length is πR, which is half of the circumference of the circle – reason being that we have arranged slices in an inverted manner, alternatively half the number of slices will contribute to length on each side. Hence, the area of the circle comes out to be πR2.
### Derivation for Area of an Arc
Following the unitary method the area of the arc subtending an angle of 360o at the centre, the angle subtended by a complete circle is πR2 then the arc suspending angle of θ will be:
Area enclosed by an arc of a circle or Area of a sector = (θ/360o ) x πR2
We have seen in this section how we are supposed to calculate area and perimeter of circle and arc. As we know mathematics is not a spectator sport so we also got through its application in some practical examples of area and perimeter related to circle and arc. Now I am sure you will be able to “calculate” the biggest slice of pizza for yourself.
## Solved Example for You
Q: Find length of an arc of a circle of radius 14cm subtending an angle of 30o
Solution: Length of an arc of circle is (θ/360o )x 2πR
θ= 30o; R=14 cm
Length of arc= (30o/360o )x 2(22/7)x14
= 88/12 cm =7.33 cm
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NCERT Solutions for Class 8 Maths Chapter 15 Exercise 15.3 – Introduction to Graphs
NCERT Solutions for Class 8 Maths Chapter 15 Exercise 15.3 – Introduction to Graphs, has been designed by the NCERT to test the knowledge of the student on the topic – Some Applications of Introduction to Graphs
NCERT Solutions for Class 8 Maths Chapter 15 Exercise 15.3 – Introduction to Graphs
NCERT Solutions for Class 8 Maths Chapter 15 Exercise 15.3 – Introduction to Graphs
Draw the graphs for the following tables of values, with suitable scales on the axes.
a) Cost of apples
Sol.: – By taking, ‘Number of Apple’ on X-Axis and ‘Cost (in ₹)’ on Y-Axis.
1 unit of x-axis = 1 Apple
1 unit of y-axis = Rs 5
b) Distance travelled by a car
Sol.: – By taking, Time (in Hours) on x-axis and Distances (in km) on y-axis.
1 unit of x-axis = 0.5 Hours
1 unit of y-axis = 40 km
i) How much distance did the car cover during the period 7.30 a.m. to 8 a.m?
Sol.: – Distance cover during the period 7.30 a.m. to 8 a.m
= Distance at 8 a.m. – Distance at 7.30 a.m.
= 120 km – 100 km
= 20 km
ii) What was the time when the car had covered a distance of 100 km since it’s start?
Sol.: – At 7.30 a.m. the car had covered a distance of 100 km since it’s start.
c) Interest on deposits for a year.
Sol.: – By taking, Deposit (in Rs) on x-axis and Simple interest (in Rs) on y-axis.
2 unit of x-axis = Rs 1000
1 unit of y-axis = Rs 50
i) Does the graph pass through the origin?
Sol.: – Yes, the graph passes through the origin.
ii) Use the graph to find the interest on Rs 2500 for a year.
Sol.: – The interest on Rs 2500 for a year will be Rs 200.
iii) To get an interest of Rs 280 per year, how much money should be deposited?
Sol.: – To get an interest of Rs 280 per year, we should deposit a sum of Rs 3500.
2. Draw a graph for the following.
i)
Is it a linear graph?
Sol.: – By taking, Side of square (in cm) on x-axis and Perimeter (in cm2) on y-axis.
1 unit of x-axis = 1 cm
1 unit of y-axis = 5 cm2
It is a linear graph.
ii)
Is it a linear graph?
Sol.: – By taking, Side of square (in cm) on x-axis and Area (in cm2) on y-axis.
1 unit of x-axis = 1 cm
1 unit of y-axis = 5 cm2
It is not a non-linear graph.
The next Exercise for NCERT Solutions for Class 8 Maths Chapter 16 Exercise 16.1 – Playing with Numbers can be accessed by clicking here
Download NCERT Solutions for Class 8 Maths Chapter 15 Exercise 15.3 – Introduction to Graphs
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# How do you find the acceleration of a block on a pulley?
## How do you find the tension between two blocks on a pulley?
Calculate the tension in the rope using the following equation: T = M x A. Four example, if you are trying to find T in a basic pulley system with an attached mass of 9g accelerating upwards at 2m/s² then T = 9g x 2m/s² = 18gm/s² or 18N (newtons).
## How do you calculate effort force of a pulley?
We can calculate the effort force of the pulley by dividing the load by the number of ropes.
## What is an example of a pulley?
Examples of pulleys include: Elevators use multiple pulleys in order to function. A cargo lift system that allows for items to be hoisted to higher floors is a pulley system. Wells use the pulley system to hoist the bucket out of the well.
## How do you calculate torque in a pulley system?
To calculate load torque, multiply the force (F) by the distance away from the rotational axis, which is the radius of the pulley (r). If the mass of the load (blue box) is 20 Newtons, and the radius of the pulley is 5 cm away, then the required torque for the application is 20 N x 0.05 m = 1 Nm.
## How does the mass of a pulley affect acceleration?
The larger the mass of the pulley the less the acceleration of the object. If you know the the mass and moment of inertia of the pulley then you can calculate the acceleration. Note that for the most common pulley shapes (e.g. disc, hoop and disc, mostly hoop), the acceleration will be independent of the radius.
## How do you solve a tension problem?
1. Step 1: Identify the direction of the tension force.
2. Step 2: Identify any other forces on the same axis as the tension force.
3. Step 3: Identify the acceleration along the tension force axis.
4. Step 4: Using Newton’s second law.
5. Step 5: Check units to be sure they are in NewtonsN
## What is the formula of tension in the string?
Solution: We know that the force of tension is calculated using the formula T = mg + ma.
## What is the tension between 2 blocks?
It is a force defined for strings, rope, or springs; tread like objects which experience tension on stretching. The tension between two blocks can be found by knowing the net forces acting on the two blocks attached to the string, we can calculate the tension exerted on the string due to the two blocks.
## Is tension the same on both sides of a pulley?
The tension of an “ideal cord” that runs through an “ideal pulley” is the same on both sides of the pulley (and at all points along the cord).
## Is tension constant in a pulley?
It changes direction continuously, in infinitesimal small increments. At any point, though, the tension vectors are essentially equal and opposite, so the tension is considered to be constant as the rope wraps around the pulley.
## What is the formula for effort force?
Answer: Mechanical Advantage (MA) = Resistance force (FR)/ Effort force (FE) = 50/15 MA = 3.33 Page 2 In this equation, the distance between the load and fulcrum is called the Resistance-Arm (Lr), while the distance from the fulcrum to the effort is called the Effort-Arm (Le).
## How do you calculate block and tackle?
If there are n of these parts of the rope supporting the load FB, then a force balance on the moving block shows that the tension in each of the parts of the rope must be FB/n. This means the input force on the rope is FA=FB/n. Thus, the block and tackle reduces the input force by the factor n.
## Does distance between pulleys matter?
The maximum recommended separation between pulley’s center is 15 to 20 multiple of the smaller pulley pitch diameter. The larger the distance between perspective pulley’s could resulting in significant differences in belt tension relative to any stretch of the belt.
## What are the 3 types of pulleys?
There are three main types of pulleys: fixed, movable, and compound. A fixed pulley’s wheel and axle stay in one place. A good example of a fixed pulley is a flag pole: When you pull down on the rope, the direction of force is redirected by the pulley, and you raise the flag.
## What are the 4 types of pulleys?
Movable pulley. Compound Pulley. Cone Pulley. Block and Tackle pulley.
## What is the simplest pulley?
A rope, wheel, and axel are the components of a simple pulley. The wheel is attached to the axel, which is the center of the machine and which allows the wheel to spin freely. The rope is then run around the outside of the wheel so that it touches approximately 50% of the wheel’s circumference.
## Does a bigger pulley increase speed?
The bigger your driven pulley (vs the driver), the slower it will turn but the more torque it will produce. The smaller your driven pulley (vs the driver), the faster it will turn but the less torque it will produce.
## Which pulley is fastest?
Pulleys of Different or Same Size If a smaller pulley turns a larger one, the larger one will turn slower, but with more power available at the shaft. If a bigger pulley turns a smaller one, the smaller one will turn much faster than the bigger one but with less power available at the shaft.
## What is pulley ratio?
The ratio of any two drive accessories can be determined by dividing the diameter of the smaller pulley into the larger one. For example, if a 6-inch water-pump pulley is used in conjunction with a 7-inch crank pulley, the pulley ratio is 1.16:1.
## How do you calculate motor rpm and pulley size?
In the example below, the pump RPM is 1070, for full output, while the motor is 1750 RPM. If the drive pulley on the engine is 4 inches in diameter, we need to calculate 4/. 315 = 12.70. This means that the pump pulley must be 12.70 inches, in diameter, to run the pump at 1070 rpm.
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# lecture 3 Slides.pptx
28. May 2023
1 von 13
### lecture 3 Slides.pptx
• 1. Lecture 3 Topics: Outlier in Box Plot Mean, Median and Spread data SADAF SALEEM Department of Computer Sciences GIFT University, Pakistan Course Title: Probability Theory Course Code: MATH-313 Program: BS MATH Semester: Spring 2023
• 2. Box-and-Whisker Plot Outlier: • Sometime in box and whisker plot we mark another point called outlier. • In a box plot, an asterisk (*) identifies an outlier. • An outlier is a value that is inconsistent with the rest of the data. It is defined as a value that is more than 1.5 times the interquartile range smaller than Q1 or larger than Q3.
• 3. Box-and-Whisker Plot Example 1: 10.2, 14.1, 14.4. 14.4, 14.4, 14.5, 14.5, 14.6, 14.7, 14.7, 14.7, 14.9, 15.1, 15.9, 16.4 Mark any outlier if exist. Solution: Step 1: arrange data Step 2: Find Q2, Q1, Q3 • Q2 = 14.6 • Q1 = 14.4 • Q3 = 14.9 Step 3: Find IQR (inter quartile range) IQR = 14.9 – 14.4 = 0.5 Step 4: Calculated Lower limit and Upper limit • Q1 – 1.5 ×IQR = 14.4 – 0.75 = 13.65 • Q3 + 1.5×IQR = 14.9 + 0.75 = 15.65. Step 5: Mark outliers Then the outliers are at: 10.2, 15.9, and 16.4
• 4. Box-and-Whisker Plot Example 1: 21, 23, 24, 25, 29, 33, 49 Mark any outlier if exist. Solution: Step 1: arrange data Step 2: Find Q2, Q1, Q3 • Q2 = 25 • Q1 = 23; • Q3 = 33 Step 3: Find IQR (inter quartile range) IQR = 33 – 23 = 10 Step 4: Calculated Lower limit and Upper limit • Q1 – 1.5 ×IQR = 23 – 1.5×10 = 23 – 15 = 8 • Q3 + 1.5×IQR = 33 + 1.5×10 = 33 + 15 = 48 Step 5: Mark outliers
• 5. Box-and-Whisker Plot Example 3: Let the data range be 199, 201, 236, 269,271,278,283,291, 301, 303, and 341. Mark any outlier if exist. Solution: • Q2 = 278 • Q1 = 236 • Q3 = 301 Step 3: Find IQR (inter quartile range) IQR = 301-236=65 Step 4: Calculated Lower limit and Upper limit • Q1 – 1.5 ×IQR = 138.5 • Q3 + 1.5×IQR =398.5 No outlier exist
• 6. Box-and-Whisker Plot Example 3: The box plot below shows the amount spent for books and supplies per year by students at four-year public colleges. a. Estimate the median amount spent. b. Estimate the first and third quartiles for the amount spent. c. Estimate the interquartile range for the amount spent. d. Beyond what point is a value considered an outlier? e. Identify any outliers and estimate their value.
• 9. Sample Mean • Average value of the sample • Influenced by extreme values • Represented by 𝑥 • Formula: • Example 1: • 1.7, 2.2, 3.9, 3.11, 14.7 • 𝒙 = 1.7+2.2+3.9+3.11+14.7 5 • 𝒙 = 5.12
• 10. Sample Mean: Class Practice
• 11. Sample Median • Central tendency of the Sample • Don't influenced by extreme values • Represented by 𝑥 • Formula: • Arrange observation in increasing order • Example 1: 1.7, 2.2, 3.9, 3.11, 14.7 • 1.7, 2.2, 3.11, 3.90, 14.7 • 𝒙 = 𝟑. 𝟏𝟏
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# 2001 USAMO Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Each of eight boxes contains six balls. Each ball has been colored with one of $n$ colors, such that no two balls in the same box are the same color, and no two colors occur together in more than one box. Determine, with justification, the smallest integer $n$ for which this is possible.
### Solution 1
We claim that $n=23$ is the minimum. Consider the following construction (replacing colors with numbers) which fulfills this:
$$\begin{tabular}{|r|r|r|r|r|r|} \hline 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1 & 7 & 8 & 9 & 10 & 11 \\ \hline 1 & 12 & 13 & 14 & 15 & 16 \\ \hline 2 & 7 & 12 & 17 & 18 & 19 \\ \hline 3 & 8 & 13 & 17 & 20 & 21 \\ \hline 4 & 9 & 14 & 17 & 22 & 23 \\ \hline 5 & 10 & 15 & 18 & 20 & 22 \\ \hline 6 & 11 & 16 & 19 & 21 & 23 \\ \hline \end{tabular}$$
Suppose a configuration exists with $n \le 22$.
Suppose a ball appears $5$ or more times. Then the remaining balls of the $5$ boxes must be distinct, so that there are at least $n \ge 5 \cdot 5 + 1 = 26$ balls, contradiction. If a ball appears $4$ or more times, the remaining balls of the $4$ boxes must be distinct, leading to $5 \cdot 4 + 1 = 21$ balls. The fifth box can contain at most four balls from the previous boxes, and then the remaining two balls must be distinct, leading to $n \ge 2 + 21 = 23$, contradiction.
However, by the Pigeonhole Principle, at least one ball must appear $3$ times. Without loss of generality suppose that $1$ appears three times, and let the boxes that contain these have balls with colors $\{1,2,3,4,5,6\},\{1,7,8,9,10,11\},\{1,12,13,14,15,16\}$. Each of the remaining five boxes can have at most $3$ balls from each of these boxes. Thus, each of the remaining five boxes must have $3$ additional balls $> 16$. Thus, it is necessary that we use $\le 22 - 16 = 6$ balls to fill a $3 \times 5$ grid by the same rules.
Again, no balls may appear $\ge 4$ times, but by Pigeonhole, one ball must appear $3$ times. Without loss of generality, let this ball have color $17$; then the three boxes containing $17$ must have at least $2 \cdot 3 + 1 = 7$ balls, contradiction.
Therefore, $n = 23$ is the minimum.
### Solution 2
Similar to the above solution, no color can appear 4 times or more.
We can use PIE. Let $S_1,S_2,\cdots ,S_8$ be the colors in each of the $8$ boxes. Then $n=|\cup_{i=1}^8S_i|$. By PIE we know that$$|\cup_{i=1}^8S_i|=\sum_{i=1}^8 |S_i|-\sum_{1\leq iNote however that no color can appear 4 times or more, so all the items after $\sum_{1\leq i must be equal to 0, so$\[|\cup_{i=1}^8S_i|=\sum_{i=1}^8 |S_i|-\sum_{1\leq iNow note that for each $i$, $|S_i|=6,$ so $\sum_{i=1}^8 |S_i|=48.$ Also no pair of colors appears twice, so for any $i,j, |S_i\cap S_j\leq 1|,$ and so $\sum_{1\leq i Thus, $n\geq 48-28+\sum_{1\leq i Given that there are $n$ colors, to minimize the number of colors in 3 boxes means each color is in either 2 or 3 boxes, and since there are 48 total spots, this means there are $48-2n$ colors in 3 boxes and $3n-48$ colors in 2. Thus $n\geq 20+48-2n,$ so $3n\geq 68$, and $n\geq23$ and we are done. ### Solution 3 Let $m_{i,j}$ be the number of balls which are the same color as the $j^{\text{th}}$ ball in box $i$ (including that ball). For a fixed box $i$, $1\leq i\leq 8$, consider the sums $\[S_i = \sum_{j=1}^6 m_{i,j}\quad\text{and}\quad s_i = \sum_{j=1}^6 \frac{1}{m_{i,j}}.$$ For each fixed $i$, since no pair of colors is repeated, each of the remaining seven boxes can contribute at most one ball to $S_i$. Thus, $S_i\leq 13$. It follows by the convexity of $f(x) = 1/x$ that $s_i$ is minimized when one of the $m_{i,j}$ is equal to 3 and the other five equal to 2. Hence $s_i\geq 17/6$. Note that $$n = \sum_{i=1}^8 \sum_{j=1}^6 \frac{1}{m_{i,j}}\geq 8\cdot\frac{17}{6} = \frac{68}{3}.$$ Hence there must be 23 colors. The construction for $n = 23$ is above.
2001 USAMO (Problems • Resources) Preceded byFirst question Followed byProblem 2 1 • 2 • 3 • 4 • 5 • 6 All USAMO Problems and Solutions
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Problem Solving Lesson #3 MMW
3. Problem Solving 3.1. Inductive and Deductive Reasoning 3.2. Problem Solving with Patterns 3.3. Problem-Solving Strategies
Use different types of reasoning to justify statements and arguments made about mathematics and mathematical concepts (K) Write clear and logical proofs (K) Solve problems involving patterns and recreational problems following Polya’s four steps (S) Organize one’s methods and approaches for proving and solving problems (V)
Try this out! Pick a number. Multiply the number by 9. Add 15 to the product. Divide the sum by 3. Subtract 5.
Use different numbers. What have you noticed?
Inductive reasoning
It is the process of reaching a general conclusion by examining specific examples.
When you examine a list of numbers and predict the next number in the list according to some pattern you have observed, you are using inductive reasoning.
The conclusion formed by using inductive reasoning is called a conjecture.
A conjecture is an idea that may or may not be correct.
Examples: 1.
a. b.
Use inductive reasoning to predict the next number in each of the following lists. 5, 10, 15, 20, ? 1, 4, 9, 16, 25, ?
The Pendulum Scientist often use inductive reasoning. Galileo Galilei (1564-1642) used inductive reasoning to discover that the time required for a pendulum to complete one swing.
A length of 10 inches has been designated as 1 unit.
Length of pendulum, in units
Period of pendulum, in heartbeats
1
1
4
2
9
3
16
4
25
5
Length of pendulum, in units
Period of pendulum, in heartbeats
1
1
4
2
9
3
16
4
25
5
If a pendulum has a length of 49 units, what is its period?
If the length of the pendulum is quadrupled, what happens to its period?
The Tsunami
A tsunami is a sea wave produced by an underwater earthquake. The height of a tsunami as it approaches land depends on its velocity. Velocity of tsunami, in feet per second
Height of tsunami, in feet
6
4
9
9
12
16
15
25
18
36
21
49
Velocity of tsunami, in feet per second
Height of tsunami, in feet
6
4
9
9
12
16
15
25
18
36
21
49
What happens to the height of a tsunami when its velocity is doubled?
What should be the height of tsunami if its velocity is 30 ft per second?
Deductive Reasoning
It is a process of reaching a conclusion by applying general assumptions, procedures, or principles.
From general to specific
Examples: Use deductive reasoning to make a conjecture. Pick a number. Multiply the number by 10, add 8 to the product, divide the sum by 2, and subtract 4.
10n + 8 - 4 2
Solve a Logic Puzzle
Each of the four friends Donna, Sarah, Nikki, and Cris, has a different pet (fish, cat, dog, and snake). From the following clues, determine the pet of each individual. Sarah is older than her friend who owns the cat and younger than her friend who owns the dog. Nikki and her friend who owns the snake are both of the same age and are the youngest members of their group. Donna is older than her friend who owns the fish.
Problem Solving with Patterns Fibonacci Sequence Jacques Binet in 1543 was able to find a formula for the nth Fibonacci number.
Fn =
1 1+ 5 𝑛 [( ) 5 2
Find Fib (50).
1− 5 𝑛 ( ) ] 2
The Pascal’s Triangle
Blaise Pascal (1623-1662) introduced the Pascal’s Triangle
(𝑥 + 𝑦)3 = 𝑥 3 + 3𝑥 2 𝑦 + 3𝑥𝑦 2 + 𝑦 3
(𝑥 + 𝑦)6 = 𝑥 6 + 6𝑥 5 𝑦 + 15𝑥 4 𝑦 2 + 20𝑥 3 𝑦 3 + 15𝑥 2 𝑦 4 + 6𝑥𝑦 5 + 𝑦 6
Find (𝑥 + 𝑦)7
Solve this problem.
An agency charged Php 15,000 for a 3-day and 2-night tour in Macau and Php 20,000 for the same tour with a side trip to Hong Kong. Ten persons joined the trip, which enable them to collect Php 170,000. How many tourists made a side trip to Hong Kong?
POLYA’S PROBLEM SOLVING STRATEGY
George Polya (1887-1985) ◦ George Polya was a Hungarian who immigrated to the United States in 1940. His major contribution is for his work in problem solving.
Polya’s Four-Step Problem-Solving Strategy Understand the problem. Devise a plan. Carry-out a plan. Review the solution.
Polya’s First Principle: Understand the Problem Do you understand all the words used in stating the problem? What are you asked to find or show? Can you restate the problem in your own words? Can you think of a picture or a diagram that might help you understand the problem? Is there enough information to enable you to find a solution?
Polya’s Second Principle: Devise a plan Guess and check Look for a pattern Make and orderly Draw a picture list Solve a simpler Eliminate problem possibilities Use a model Use symmetry Work backward Consider special Use a formula cases Be ingenious Use direct reasoning Solve an equation
Polya’s third Principle: Carry out the plan Work carefully. Keep an accurate and neat record of all your attempts. Realize that some of your initial plans will not work and that you may have to devise another plan or modify your existing plan.
Polya’s Fourth Principle: Look back Ensure that the solution is consistent with the facts of the problem. Interpret the solution in the context of the problem. Ask yourself whether there are generalizations of the solution that could apply to other problems.
Examples: 1. If two ladders are placed end to end, their combined height is 31.5 feet. One ladder is 6.5 feet shorter than the other ladder. What are the heights of the two ladders?
Examples: The number of ducks and pigs in a field totals 35. The total number of the legs among them is 98. Assuming each duck as exactly two legs and each pig has exactly four legs, determine how many ducks and how many pigs are in the field.
Seatwork: By pairs (3 points each)
A
hat and a jacket together cost \$100. The jacket costs \$90 more than the hat. What are the cost of the hat and the cost of the jacket?
The four children in the Rivera family are Reynaldo, Ramiro, Shakira, and Sasha. The ages of the two teenagers are 13 and 15. The ages of the younger children are 5 and 7. From the following clues, determine the age of each of the children.
Reynaldo is older than Ramiro. Sasha is younger than Shakira. Sasha is 2 years older than Ramiro. Shakira is older than Reynaldo.
On
three examinations, Dana received scores of 82, 91, and 76. What score does Dana need on the fourth examination to raise her average to 85?
An airline flies from Manila to Davao with a stopover in Cebu. It charges a fare of Php 1,800 Manila to Cebu and a fare Php 3,500 from Manila to Davao. In one flight, a total of 180 passengers boarded the plane in Manila and the collected fare totalled Php 494,000. How many got off the plane in Cebu?
HOMEWORK: 1. Determine whether the argument is an example of deductive reasoning or inductive reasoning. a.
b.
c.
Samantha got an A on each of her four math tests, so she will get an A on her next math test. All amoeba multiply by dividing. I have named the amoeba shown in my microscope, Amelia. Therefore, Amelia multiplies by dividing. We had rain each day for the last five days, so it will rain today.
2. Erick is 25 years older than Edwin. In 15 years, Erick will be as twice as old as Edwin will be. What will their ages be then? 3. The sum of two numbers is 89 and their difference is 41. Find the number.
4. A tank can be filled by pipe A in 5 hours and by pipe B in 8 hours. It can be emptied by pipe C in 6 hours. How long will it take the two pipes to fill the tank if the third pipe is drawing water at the same time? 5. Engr. Cruz has invested Php 500,000. Part of it was invested at 4% and the rest at 5%. The total annual income from the two investments is Php 95,000. How much is invested at each of these rates?
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Hong Kong
Stage 2
# Adding and subtracting Fractions
Lesson
Use the interactive below to represent two fractions with different denominators. Try setting one to be $\frac{1}{4}$14 and one to be $\frac{1}{5}$15. How can we add these together?
You might have noticed that adding or subtracting fractions uses our skills of finding Lowest Common Multiples, and Equivalent Fractions. You can apply those skills to add and subtract fractions without using the model.
#### Example
Evaluate: $\frac{3}{4}+\frac{2}{5}$34+25
Think: Find the LCM between $4$4 and $5$5 and find equivalent fractions. Then we will be able to add them, just like in the applet above.
Do: The LCM between $4$4 and $5$5 is $20$20
$\frac{3}{4}+\frac{2}{5}$34+25 $=$= $\frac{15}{20}+\frac{8}{20}$1520+820 $=$= $\frac{23}{20}$2320
#### Worked Examples
##### QUESTION 1
Evaluate $\frac{5}{9}-\frac{4}{9}$5949.
Write your answer in its simplest form.
##### QUESTION 2
Evaluate $\frac{6}{35}+\frac{3}{7}-\frac{8}{7}$635+3787.
Write your answer in its simplest form.
##### QUESTION 3
Evaluate $\frac{9}{10}-\frac{3}{7}$91037.
Write your answer in its simplest form.
##### QUESTION 4
Evaluate $\frac{2}{3}+\frac{4}{5}-\frac{1}{2}$23+4512.
Write your answer in the simplest form possible.
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# A Module is Irreducible if and only if It is a Cyclic Module With Any Nonzero Element as Generator
## Problem 434
Let $R$ be a ring with $1$.
A nonzero $R$-module $M$ is called irreducible if $0$ and $M$ are the only submodules of $M$.
(It is also called a simple module.)
(a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.
(b) Determine all the irreducible $\Z$-modules.
## Proof.
### (a) Prove that a nonzero $R$-module $M$ is irreducible if and only if $M$ is a cyclic module with any nonzero element as its generator.
$(\implies)$ Suppose that $M$ is an irreducible module.
Let $a\in M$ be any nonzero element and consider the submodule $(a)$ generated by the element $a$.
Since $a$ is a nonzero element, the submodule $(a)$ is non-zero. Since $M$ is irreducible, this yields that
$M=(a).$ Hence $M$ is a cyclic module generated by $a$. Since $a$ is any nonzero element, we conclude that the module $M$ is a cyclic module with any nonzero element as its generator.
$(\impliedby)$ Suppose that $M$ is a cyclic module with any nonzero element as its generator.
Let $N$ be a nonzero submodule of $M$. Since $N$ is non-zero, we can pick a nonzero element $a\in N$. By assumption, the non-zero element $a$ generates the module $M$.
Thus we have
$(a) \subset N \subset M=(a).$ It follows that $N=M$, and hence $M$ is irreducible.
### (b) Determine all the irreducible $\Z$-modules.
By the result of part (a), any irreducible $\Z$-module is generated by any nonzero element.
We first claim that $M$ cannot contain an element of infinite order. Suppose on the contrary $a\in M$ has infinite order.
Then since $M$ is irreducible, we have
$M=(a)\cong \Z.$ Since $\Z$-module $\Z$ has, for example, a proper submodule $2\Z$, it is not irreducible. Thus, the module $M$ is not irreducible, a contradiction.
It follows that any irreducible $\Z$-module is a finite cyclic group.
(Recall that any $\Z$-module is an abelian group.)
We claim that its order must be a prime number.
Suppose that $M=\Zmod{n}$, where $n=ml$ with $m,l > 1$.
Then
$(\,\bar{l}\,)=\{l+n\Z, 2l+n\Z, \dots, (m-1)l+n\Z\}$ is a proper submodule of $M$, and it is a contradiction.
Thus, $n$ must be prime.
We conclude that any irreducible $\Z$-module is a cyclic group of prime order.
## Related Question.
Here is another problem about irreducible modules.
Problem. Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module.
Prove that the $R$-module $M$ is irreducible if and only if $M$ is isomorphic to $R/I$, where $I$ is a maximal ideal of $R$, as an $R$-module.
For a proof of this problem, see the post “A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$.“.
### More from my site
• A Module $M$ is Irreducible if and only if $M$ is isomorphic to $R/I$ for a Maximal Ideal $I$. Let $R$ be a commutative ring with $1$ and let $M$ be an $R$-module. Prove that the $R$-module $M$ is irreducible if and only if $M$ is isomorphic to $R/I$, where $I$ is a maximal ideal of $R$, as an $R$-module. Definition (Irreducible module). An […]
• Fundamental Theorem of Finitely Generated Abelian Groups and its application In this post, we study the Fundamental Theorem of Finitely Generated Abelian Groups, and as an application we solve the following problem. Problem. Let $G$ be a finite abelian group of order $n$. If $n$ is the product of distinct prime numbers, then prove that $G$ is isomorphic […]
• Can $\Z$-Module Structure of Abelian Group Extend to $\Q$-Module Structure? If $M$ is a finite abelian group, then $M$ is naturally a $\Z$-module. Can this action be extended to make $M$ into a $\Q$-module? Proof. In general, we cannot extend a $\Z$-module into a $\Q$-module. We give a counterexample. Let $M=\Zmod{2}$ be the order […]
• Torsion Submodule, Integral Domain, and Zero Divisors Let $R$ be a ring with $1$. An element of the $R$-module $M$ is called a torsion element if $rm=0$ for some nonzero element $r\in R$. The set of torsion elements is denoted $\Tor(M)=\{m \in M \mid rm=0 \text{ for some nonzero} r\in R\}.$ (a) Prove that if $R$ is an […]
• Submodule Consists of Elements Annihilated by Some Power of an Ideal Let $R$ be a ring with $1$ and let $M$ be an $R$-module. Let $I$ be an ideal of $R$. Let $M'$ be the subset of elements $a$ of $M$ that are annihilated by some power $I^k$ of the ideal $I$, where the power $k$ may depend on $a$. Prove that $M'$ is a submodule of […]
• Annihilator of a Submodule is a 2-Sided Ideal of a Ring Let $R$ be a ring with $1$ and let $M$ be a left $R$-module. Let $S$ be a subset of $M$. The annihilator of $S$ in $R$ is the subset of the ring $R$ defined to be $\Ann_R(S)=\{ r\in R\mid rx=0 \text{ for all } x\in S\}.$ (If $rx=0, r\in R, x\in S$, then we say $r$ annihilates […]
• Ascending Chain of Submodules and Union of its Submodules Let $R$ be a ring with $1$. Let $M$ be an $R$-module. Consider an ascending chain $N_1 \subset N_2 \subset \cdots$ of submodules of $M$. Prove that the union $\cup_{i=1}^{\infty} N_i$ is a submodule of $M$. Proof. To simplify the notation, let us […]
• Short Exact Sequence and Finitely Generated Modules Let $R$ be a ring with $1$. Let $0\to M\xrightarrow{f} M' \xrightarrow{g} M^{\prime\prime} \to 0 \tag{*}$ be an exact sequence of left $R$-modules. Prove that if $M$ and $M^{\prime\prime}$ are finitely generated, then $M'$ is also finitely generated. […]
### 1 Response
1. 06/10/2017
[…] the post “A module is irreducible if and only if it is a cyclic module with any nonzero element as generator” for a […]
##### Finitely Generated Torsion Module Over an Integral Domain Has a Nonzero Annihilator
(a) Let $R$ be an integral domain and let $M$ be a finitely generated torsion $R$-module. Prove that the module...
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## What is the formula for sum of n terms?
An example of AP is natural numbers, where the common difference is 1. Therefore, to find the sum of natural numbers, we need to know the formula to find it.Sum of N Terms of AP And Arithmetic Progression.
Sum of n terms in APn/2[2a + (n – 1)d]
Sum of natural numbersn(n+1)/2
## How do you find the sum?
To do this, add the two numbers, and divide by 2. Multiply the average by the number of terms in the series. This will give you the sum of the arithmetic sequence. So, the sum of the sequence 10, 15, 20, 25, 30 is 100.
Sum
## What is the sum of a number?
The result of adding two or more numbers. Example: 9 is the sum of 2, 4 and 3. (because 2 + 4 + 3 = 9).
## What is the sum of a series?
The n-th partial sum of a series is the sum of the first n terms. The sequence of partial sums of a series sometimes tends to a real limit. If this happens, we say that this limit is the sum of the series. A series can have a sum only if the individual terms tend to zero.
## What is the sum of 5?
NumberRepeating Cycle of Sum of Digits of Multiples
5{5,1,6,2,7,3,8,4,9}
6{6,3,9,6,3,9,6,3,9}
7{7,5,3,1,8,6,4,2,9}
8{8,7,6,5,4,3,2,1,9}
## What is the sum of numbers from 1 to 100?
The sum of the numbers 1-100 would be equal to the number of pairs (50) multiplied by the sum of each pair (101), or 50 x 101 = 5,050.
## How do you estimate the sum of two numbers?
Estimating a Sum. A quick way to estimate the sum of two numbers is to round each number and then add the rounded numbers.
## What does sum mean?
noun. the aggregate of two or more numbers, magnitudes, quantities, or particulars as determined by or as if by the mathematical process of addition: The sum of 6 and 8 is 14. a particular aggregate or total, especially with reference to money: The expenses came to an enormous sum.
## What does [] mean in math?
if using this in sets, () stands for open interval and [] means closed interval and {} is used to denote specific elements. Closed interval means the extreme numbers of the set are included in it and open means they arent. For example, (9,10] is the set of all numbers between 9 and 10 excluding 9 and including 10.
## What does || mean in math?
The symbol || indicates that two lines are parallel . “||”symbol is used for parallel lines in mathematics means two lines they never meet.
## How do you find the sum of the first 20 terms?
Exercise. Calculate the sum of the first 20 terms of the arithmetic sequence whose formula for the nth term is: un=1+(n−1)×4. Answers Without Working.Formula 2. Given an arithmetic sequence, we can calculate the sum of its first n terms, Sn, using the formula: Sn=n2(2.
### Releated
#### Write an equation for the polynomial graphed below
What is the formula for a polynomial function? A polynomial is a function of the form f(x) = anxn + an−1xn−1 + + a2x2 + a1x + a0 . The degree of a polynomial is the highest power of x in its expression. What are examples of polynomial functions? Basic knowledge of polynomial functions Polynomial […]
#### Equation of vertical line
How do you write an equation for a vertical and horizontal line? Horizontal lines go left and right and are in the form of y = b where b represents the y intercept. Vertical lines go up and down and are in the form of x = a where a represents the shared x coordinate […]
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# Question 13 & 14 Exercise 4.5
Solutions of Question 13 & 14 of Exercise 4.5 of Unit 04: Sequence and Series. This is unit of A Textbook of Mathematics for Grade XI is published by Khyber Pakhtunkhwa Textbook Board (KPTB or KPTBB) Peshawar, Pakistan.
If $y=\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}+\ldots$ where $0<x<3$, then show that $x=\dfrac{3 y}{1+y}$.
Adding 1 to both sides of the given series, we get $$1+y=1+\dfrac{x}{3}+\dfrac{x^2}{3^2}+\dfrac{x^3}{3^3}$$ Now the series on R.H.S of the above is geometric series with $a_1=1$,
$r=\dfrac{x}{3}$, and $|r|=\dfrac{x}{3}<1$ because $0<x<3$.
Thus infinite sum exists and is given by $S_{\infty}=\dfrac{a_1}{1-r}$, putting $a_1, \quad r$, we get $$S_{\infty}=\dfrac{1}{1-\dfrac{x}{3}}=\dfrac{3}{3-x}$$ putting in (i), we have \begin{align}1+y&=\dfrac{3}{3-x} \\ \Rightarrow \quad 3-x&=\dfrac{3}{1+y}\\ \Rightarrow \quad x&=3-\dfrac{3}{1+y} \\ \Rightarrow \quad x&=\dfrac{3+3 y-3}{1+y} \\ \Rightarrow \quad x&=\dfrac{3 y}{1+y}\end{align} which is required result.
A ball rebound to half the height from which it is dropped. If it is dropped from $10 \mathrm{ft}$, how far does it travel from the moment it dropped until the moment of its eighth bounce?
\begin{align}S&=10+[10(\dfrac{1}{2})+10(\dfrac{1}{2})]+ \\ & {[10(\dfrac{1}{2})^2+10(\dfrac{1}{2})^2]+\ldots} \\ & +[10(\dfrac{1}{2})^7+10(\dfrac{1}{2})^7] \\ S&=10+2[10(\dfrac{1}{2})+10(\dfrac{1}{2})^2+10(\dfrac{1}{2})^7]...(i)\end{align} The sequence is bracket is geometric sequence, with $a_1=10(\dfrac{1}{2}), r=\dfrac{1}{2}, n=7$.
Then $S_7=\dfrac{10(\dfrac{1}{2})[1-(\dfrac{1}{2})^7]}{1-\dfrac{1}{2}}$ \begin{align}\Rightarrow S_7&=10[1-\dfrac{1}{2^7}] \\ \Rightarrow S_7&=10,0.9921875\\ &=9.921875\end{align} Putting (ii) in (i), we get
\begin{align}S&=10+2(9.921875) \\ & =10+19.84375\\ S&=29.84375\quad \text{ feet approximately}\\ S&=29\dfrac{27}{32}\text{ft}\end{align}
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# How To Find Slope Intercept Form With 2 Points
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
How To Find Slope Intercept Form With 2 Points – Among the many forms used to depict a linear equation, one that is commonly used is the slope intercept form. You may use the formula for the slope-intercept to find a line equation assuming that you have the slope of the straight line and the y-intercept, which is the coordinate of the point’s y-axis where the y-axis crosses the line. Read more about this particular linear equation form below.
## What Is The Slope Intercept Form?
There are three main forms of linear equations: the standard one, the slope-intercept one, and the point-slope. Although they may not yield the same results when utilized in conjunction, you can obtain the information line produced more quickly using the slope intercept form. As the name implies, this form makes use of the sloped line and it is the “steepness” of the line indicates its value.
This formula can be used to determine the slope of straight lines, y-intercept, or x-intercept, where you can utilize a variety available formulas. The equation for a line using this formula is y = mx + b. The straight line’s slope is represented by “m”, while its y-intercept is represented with “b”. Every point on the straight line is represented by an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” need to remain variables.
## An Example of Applied Slope Intercept Form in Problems
For the everyday world in the real world, the slope intercept form is commonly used to illustrate how an item or problem changes in an elapsed time. The value of the vertical axis is a representation of how the equation handles the intensity of changes over the value given via the horizontal axis (typically the time).
A simple example of this formula’s utilization is to discover how much population growth occurs in a certain area as the years pass by. If the area’s population grows annually by a fixed amount, the point values of the horizontal axis will grow by a single point each year and the point values of the vertical axis will rise to reflect the increasing population by the fixed amount.
You may also notice the beginning point of a problem. The starting value occurs at the y-value of the y-intercept. The Y-intercept represents the point at which x equals zero. If we take the example of the problem mentioned above the beginning value will be at the point when the population reading begins or when the time tracking begins along with the changes that follow.
The y-intercept, then, is the place when the population is beginning to be recorded in the research. Let’s suppose that the researcher starts with the calculation or measure in the year 1995. The year 1995 would represent”the “base” year, and the x=0 points would be in 1995. So, it is possible to say that the population of 1995 will be the “y-intercept.
Linear equations that employ straight-line formulas are nearly always solved in this manner. The starting value is represented by the yintercept and the rate of change is represented through the slope. The most significant issue with the slope-intercept form typically lies in the horizontal variable interpretation in particular when the variable is attributed to one particular year (or any other kind or unit). The first step to solve them is to make sure you understand the variables’ meanings in detail.
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# AP Statistics Curriculum 2007 Distrib Dists
## General Advance-Placement (AP) Statistics Curriculum - Geometric, HyperGeometric, Negative Binomial Random Variables and Experiments
### Geometric
• Definition: The Geometric Distribution is the probability distribution of the number X of Bernoulli trials needed to get one success, supported on the set {1, 2, 3, ...}. The name geometric is a direct derivative from the mathematical notion of geometric series.
• Mass Function: If the probability of successes on each trial is P(success)=p, then the probability that x trials are needed to get one success is $P(X = x) = (1 - p)^{x-1} \times p$, for x = 1, 2, 3, 4,....
• Expectation: The Expected Value of a geometrically distributed random variable X is ${1\over p}.$
• Variance: The Variance is ${1-p\over p^2}.$
### HyperGeometric
The hypergeometric distribution is a discrete probability distribution that describes the number of successes in a sequence of n draws from a finite population without replacement. An experimental design for using Hypergeometric distribution is illustrated in this table:
Type Drawn Not-Drawn Total Defective k m-k m Non-Defective n-k N+k-n-m N-m Total n N-n N
• Explanation: Suppose there is a shipment of N objects in which m are defective. The Hypergeometric Distribution describes the probability that in a sample of n distinctive objects drawn from the shipment exactly k objects are defective.
• Mass function: The random variable X follows the Hypergeometric Distribution with parameters N, m and n, then the probability of getting exactly k successes is given by
$P(X=k) = {{{m \choose k} {{N-m} \choose {n-k}}}\over {N \choose n}}.$
This formula for the Hypergeometric Mass Function may be interpreted as follows: There are ${{N}\choose{n}}$ possible samples (without replacement). There are ${{m}\choose{k}}$ ways to obtain k defective objects and there are ${{N-m}\choose{n-k}}$ ways to fill out the rest of the sample with non-defective objects.
The mean and variance of the hypergeometric distribution have the following closed forms:
Mean: $n \times m\over N$
Variance: ${ {nm\over N} ( 1-{m\over N} ) (N-n)\over N-1}$
#### Examples
• SOCR Activity: The SOCR Ball and Urn Experiment provides a hands-on demonstration of the utilization of Hypergeometric distribution in practice. This activity consists of selecting n balls at random from an urn with N balls, R of which are red and the other N - R green. The number of red balls Y in the sample is recorded on each update. The distribution and moments of Y are shown in blue in the distribution graph and are recorded in the distribution table. On each update, the empirical density and moments of Y are shown in red in the distribution graph and are recorded in the distribution table. Either of two sampling models can be selected with the list box: with replacement and without replacement. The parameters N, R, and n can be varied with scroll bars.
• A lake contains 1,000 fish; 100 are randomly caught and tagged. Suppose that later we catch 20 fish. Use SOCR Hypergeometric Distribution to:
• Compute the probability mass function of the number of tagged fish in the sample of 20.
• Compute the expected value and the variance of the number of tagged fish in this sample.
• Compute the probability that this random sample contains more than 3 tagged fish.
• Hypergeometric distribution may also be used to estimate the population size: Suppose we are interested in determining the population size. Let N = number of fish in a particular isolated region. Suppose we catch, tag and release back M=200 fish. Several days later, when the fish are randomly mixed with the untagged fish, we take a sample of n=100 and observe m=5 tagged fish. Suppose p=200/N is the population proportion of tagged fish. Notice that when sampling fish we sample without replacement. Thus, hypergeometric is the exact model for this process. Assuming the sample-size (n) is < 5% of the population size(N), we can use binomial approximation to hypergeometric. Thus if the sample of n=100 fish had 5 tagged, the sample-proportion (estimate of the population proportion) will be $\hat{p}={5\over 100}=0.05$. Thus, we can estimate that $0.05=\hat{p}={200\over N}$, and $N\approx 4,000$, as shown on the figure below.
### Negative Binomial
The family of Negative Binomial Distributions is a two-parameter family; p and r with 0 < p < 1 and r > 0. There are two (identical) combinatorial interpretations of Negative Binomial processes (X or Y).
#### X=Trial index (n) of the rth success, or Total # of experiments (n) to get r successes
• Probability Mass Function: $P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} \!$, for n = r,r+1,r+2,.... (n=trial number of the rth success)
• Mean: $E(X)= {r \over p}$
• Variance: $Var(X)= {r(1-p) \over p^2}$
#### Y = Number of failures (k) to get r successes
• Probability Mass Function: $P(Y=k) = {k+r-1 \choose k}\cdot p^r \cdot (1-p)^k \!$, for k = 0,1,2,.... (k=number of failures before the rth successes)
• $Y \sim NegBin(r, p)$, the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial.
• Mean: $E(Y)= {r(1-p) \over p}$.
• Variance: $Var(Y)= {r(1-p) \over p^2}$.
• Note that X = Y + r, and E(X) = E(Y) + r, whereas VAR(X)=VAR(Y).
#### Application
Suppose Jane is promoting and fund-raising for a presidential candidate. She wants to visit all 50 states and she's pledged to get all electoral votes of 6 states before she and the candidate she represents are satisfied. In every state, there is a 30% chance that Jane will be able to secure all electoral votes and 70% chance that she'll fail.
• What's the probability mass function of the number of failures (k=n-r) to get r=6 successes?
In other words, What's the probability mass function that the last 6th state she succeeds to secure all electoral votes happens to be the at the nth state she campaigns in?
NegBin(r, p) distribution describes the probability of k failures and r successes in n=k+r Bernoulli(p) trials with success on the last trial. Looking to secure the electoral votes for 6 states means Jane needs to get 6 successes before she (and her candidate) is happy. The number of trials (i.e., states visited) needed is n=k+6. The random variable we are interested in is X={number of states visited to achieve 6 successes (secure all electoral votes within these states)}. So, n = k+6, and $X\sim NegBin(r=6, p=0.3)$. Thus, for $n \geq 6$, the mass function (giving the probabilities that Jane will visit n states before her ultimate success is:
$P(X=n) = {n-1 \choose r-1}\cdot p^r \cdot (1-p)^{n-r} = {n - 1 \choose r-1} \cdot 0.3^6 \cdot 0.7^{n-r}$
• What's the probability that Jane finishes her campaign in the 10th state?
Let $X\sim NegBin(r=6, p=0.3)$, then $P(X=10) = {10-1 \choose 6-1}\cdot 0.3^6 \cdot 0.7^{10-6} = 0.022054.$
• What's the probability that Jane finishes campaigning on or before reaching the 8th state?
$P(X\leq 8) = 0.011292$
• Suppose the success of getting all electoral votes within a state is reduced to only 10%, then X~NegBin(r=6, p=0.1). Notice that the shape and domain the Negative-Binomial distribution significantly chance now (see image below)!
What's the probability that Jane covers all 50 states but fails to get all electoral votes in any 6 states (as she had hoped for)?
$P(X\geq 50) = 0.632391$
### Negative Multinomial Distribution (NMD)
The Negative Multinomial Distribution is a generalization of the two-parameter Negative Binomial distribution (NB(r,p)) to $m\ge 1$ outcomes. Suppose we have an experiment that generates $m\ge 1$ possible outcomes, $\{X_0,\cdots,X_m\}$, each occurring with probability $\{p_0,\cdots,p_m\}$, respectively, where with 0 < pi < 1 and $\sum_{i=0}^m{p_i}=1$. That is, $p_0 = 1-\sum_{i=1}^m{p_i}$. If the experiment proceeds to generate independent outcomes until $\{X_0, X_1, \cdots, X_m\}$ occur exactly $\{k_0, k_1, \cdots, k_m\}$ times, then the distribution of the m-tuple $\{X_1, \cdots, X_m\}$ is Negative Multinomial with parameter vector $(k_0,\{p_1,\cdots,p_m\})$. Notice that the degree-of-freedom here is actually m, not (m+1). That is why we only have a probability parameter vector of size m, not (m+1), as all probabilities add up to 1 (so this introduces one relation). Contrast this with the combinatorial interpretation of Negative Binomial (special case with m=1):
X˜NegativeBinomial(NumberOfSuccesses = r,ProbOfSuccess = p),
X=Total # of experiments (n) to get r successes (and therefore n-r failures);
X˜NegativeMultinomial(k0,{p0,p1}),
X=Total # of experiments (n) to get k0 (dafault variable) and nk0 outcomes of 1 other possible outcome (X1).
#### Negative Multinomial Summary
• Probability Mass Function: $P(k_0, \cdots, k_m) = \left (\sum_{i=0}^m{k_i}-1\right)!\frac{p_0^{k_0}}{(k_0-1)!} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, or equivalently:
$P(k_0, \cdots, k_m) = \Gamma\left(\sum_{i=1}^m{k_i}\right)\frac{p_0^{k_0}}{\Gamma(k_0)} \prod_{i=1}^m{\frac{p_i^{k_i}}{k_i!}}$, where Γ(x) is the Gamma function.
• Mean (vector): $\mu=E(X_1,\cdots,X_m)= (\mu_1=E(X_1), \cdots, \mu_m=E(X_m)) = \left ( \frac{k_0p_1}{p_0}, \cdots, \frac{k_0p_m}{p_0} \right)$.
• Variance-Covariance (matrix): Cov(Xi,Xj) = {cov[i,j]}, where
$cov[i,j] = \begin{cases} \frac{k_0 * p_i * p_j}{p_0 * p_0},& i\not= j,\\ \frac{k_0* p_i * (p_i + p_0)}{p_0 * p_0},& i=j.\end{cases}$.
#### Cancer Example
The Probability Theory Chapter of the EBook shows the following example using 400 Melanoma (skin cancer) Patients where the Type and Site of the cancer are recorded for each subject, as in the Table below.
Type Site Totals Head and Neck Trunk Extremities Hutchinson's melanomic freckle 22 2 10 34 Superficial 16 54 115 185 Nodular 19 33 73 125 Indeterminant 11 17 28 56 Column Totals 68 106 226 400
The sites (locations) of the cancer may be independent, but there may be positive dependencies of the type of cancer for a given location (site). For example, localized exposure to radiation implies that elevated level of one type of cancer (at a given location) may indicate higher level of another cancer type at the same location. We want to use the Negative Multinomial distribution to model the sites cancer rates and try to measure some of the cancer type dependencies within each location.
Let's denote by yi,j the cancer rates for each site ($0\leq i \leq 2$) and each type of cancer ($0\leq j \leq 3$). For each (fixed) site ($0\leq i \leq 2$), the cancer rates are independent Negative Multinomial distributed random variables. That is, for each column index (site) the column-vector X
X = {X1,X2,X3NMD(k0,{p1,p2,p3}).
Different columns (sites) are considered to be different instances of the random multinomially distributed X vector. Then we have the following estimates:
$\hat{\mu}_{i,j} = \frac{x_{i,.}\times x_{.,j}}{x_{.,.}}$
$x_{i,.} = \sum_{j=0}^{3}{x_{i,j}}$
$x_{.,j} = \sum_{i=0}^{2}{x_{i,j}}$
$x_{.,.} = \sum_{i=0}^{2}\sum_{j=0}^{3}{{x_{i,j}}}$
Example: $\hat{\mu}_{1,1} = \frac{x_{1,.}\times x_{.,1}}{x_{.,.}}=\frac{34\times 68}{400}=5.78$
• Variance-Covariance: For a single column vector, X = {X1,X2,X3NMD(k0,{p1,p2,p3}), covariance between any pair of Negative Multinomial counts (Xi and Xj) is:
$cov[X_i,X_j] = \begin{cases}k_0 * p_i * p_j / (p_0 * p_0),& i\not= j,\\ k_0* p_i * (p_i + p_0) / (p_0 * p_0),& i=j.\end{cases}$.
Example: For the first site (Head and Neck, i=0), suppose that $X=\left \{X_1=5, X_2=1, X_3=5\right \}$ and X˜NMD(k0 = 10,{p1 = 0.2,p2 = 0.1,p3 = 0.2}). Then:
$p_0 = 1 - \sum_{i=1}^3{p_i}=0.5$
NMD(X | k0,{p1,p2,p3}) = 0.00465585119998784
$cov[X_1,X_3] = \frac{10 * 0.2 * 0.2}{0.5^2}=1.6$
$\mu_2=\frac{k_0 p_2}{p_0} = \frac{10\times 0.1}{0.5}=2.0$
$\mu_3=\frac{k_0 p_3}{p_0} = \frac{10\times 0.2}{0.5}=4.0$
$corr[X_2,X_3] = \left (\frac{\mu_2 \times \mu_3}{(k_0+\mu_2)(k_0+\mu_3)} \right )^{\frac{1}{2}}$ and therefore, $corr[X_2,X_3] = \left (\frac{2 \times 4}{(10+2)(10+4)} \right )^{\frac{1}{2}} = 0.21821789023599242.$
You can also use the interactive SOCR negative multinomial distribution calculator to compute these quantities, as shown on the figure below.
• There is no MLE estimate for the NMD k0 parameter (see this reference). However, there are approximate protocols for estimating the k0 parameter, see the example below.
• Correlation: correlation between any pair of Negative Multinomial counts (Xi and Xj) is:
$Corr[X_i,X_j] = \begin{cases} \left (\frac{\mu_i \times \mu_j}{(k_0+\mu_i)(k_0+\mu_j)} \right )^{\frac{1}{2}} = \left (\frac{p_i p_j}{(p_0+p_i)(p_0+p_j)} \right )^{\frac{1}{2}}, & i\not= j, \\ 1, & i=j.\end{cases}$.
• The marginal distribution of each of the Xi variables is negative binomial, as the Xi count (considered as success) is measured against all the other outcomes (failure). But jointly, the distribution of $X=\{X_1,\cdots,X_m\}$ is negative multinomial, i.e., $X \sim NMD(k_0,\{p_1,\cdots,p_m\})$ .
Notice that the pair-wise NMD correlations are always positive, where as the correlations between multinomail counts are always negative. Also note that as the parameter k0 increases, the paired correlations go to zero! Thus, for large k0, the Negative Multinomial counts Xi behave as independent Poisson random variables with respect to their means $\left ( \mu_i= k_0\frac{p_i}{p_0}\right )$.
#### Parameter estimation
• Estimation of the mean (expected) frequency counts (μj) of each outcome (Xj):
The MLE estimates of the NMD mean parameters μj are easy to compute.
If we have a single observation vector $\{x_1, \cdots,x_m\}$, then $\hat{\mu}_i=x_i.$
If we have several observation vectors, like in this case we have the cancer type frequencies for 3 different sites, then the MLE estimates of the mean counts are $\hat{\mu}_j=\frac{x_{j,.}}{I}$, where $0\leq j \leq J$ is the cancer-type index and the summation is over the number of observed (sampled) vectors (I).
Example, for the cancer data above, we have the following MLE estimates for the expectations for the frequency counts:
Hutchinson's melanomic freckle type of cancer (X0) is $\hat{\mu}_0 = 34/3=11.33$.
Superficial type of cancer (X1) is $\hat{\mu}_1 = 185/3=61.67$.
Nodular type of cancer (X2) is $\hat{\mu}_2 = 125/3=41.67$.
Indeterminant type of cancer (X3) is $\hat{\mu}_3 = 56/3=18.67$.
• Estimation of the k0 (gamma) parameter:
There is no MLE for the k0 parameter; however, there is a protocol for estimating k0 using the chi-squared goodness of fit statistic. In the usual chi-squared statistic:
$\Chi^2 = \sum_i{\frac{(x_i-\mu_i)^2}{\mu_i}}$
we can replace the expected-means (μi) by sample-means ()$\hat{\mu_i}$ and replace denominators by the corresponding negative multinomial variances. Then we get the following test statistic for negative multinomial distributed data:
$\Chi^2(k_0) = \sum_{i}{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$.
Now we can derive a simple method for estimating the k0 parameter by varying the values of k0 in the expression Χ2(k0) and matching the values of this statistic with the corresponding asymptotic chi-squared distribution. The following protocol summarizes these steps using the cancer data above:
df = (# rows – 1)(# columns – 1) = (3-1)*(4-1) = 6
• Mean Counts Estimates: The mean counts estimates (μj) for the 4 different cancer types are:
$\hat{\mu}_{.,1} = \frac{x_{.,1}}{x_{.,.}}=\frac{185}{400}=0.4625$
$\hat{\mu}_{.,2} = \frac{x_{.,2}}{x_{.,.}}=\frac{185}{400}=0.4625$
$\hat{\mu}_{.,3} = \frac{x_{.,3}}{x_{.,.}}=\frac{185}{400}=0.4625$
• Thus, we can solve the equation above Χ2(k0) = 5.261948 for the single variable of interest -- the unknown parameter k0. This solution is an asymptotic chi-squared distribution driven estimate of the parameter k0.
$\Chi^2(k_0) = \sum_{i=1}^3{\frac{(x_i-\hat{\mu_i})^2}{\hat{\mu_i} \left (1+ \frac{\hat{\mu_i}}{k_0} \right )}}$.
Χ2(k0) = ....
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# 2015 AMC 8 Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem
Onkon wants to cover his rooms floor with his Favourite red carpet. How many square yards of red carpet are required to cover a rectangular floor that is $12$ feet long and $9$ feet wide? (There are 3 feet in a yard.)
$\textbf{(A) }12\qquad\textbf{(B) }36\qquad\textbf{(C) }108\qquad\textbf{(D) }324\qquad \textbf{(E) }972$
## Solutions
### Solution 1
First, we multiply $12\cdot9$. To get that you need $108$ square feet of carpet you need to cover. Since there are $9$ square feet in a square yard, you divide $108$ by $9$ to get $12$ square yards, so our answer is $\bold{\boxed{\textbf{(A)}~12}}$.
### Solution 2
Since there are $3$ feet in a yard, we divide $9$ by $3$ to get $3$, and $12$ by $3$ to get $4$. To find the area of the carpet, we then multiply these two values together to get $\boxed{\textbf{(A)}~12}$.
~savannahsolver
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circles-lesson-6-spheres
Interactive video lesson plan for: Circles Lesson 6 "Spheres"
Activity overview:
Geometry Teachers Never Spend Time Trying to Find Materials for Your Lessons Again!
http://geometrycoach.com/Geometry-Lesson-Plans/?pa=MOOMOOMATH
Note Taking Guide:
http://www.moomoomath.com/High-school-geometry-help.html#.UqfaFdJDuSo
Common Core Standard G.GMD.3
Spheres
Note taking guide
http://www.moomoomath.com/High-school-geometry-help.html#.UqfaFdJDuSo
This is a long video so use the video guide to jump to the problems you want help with.
Video Guide
0:08 Definition of a sphere
1:16 Radius of a sphere explained.
2:56 Definition of the "Great Circle"
3:27 Definition of hemisphere and a brief explanation of hemisphere
4:55 Surface area of a sphere formula
5:00 Volume of a sphere formula
5:32 Directions for finding the surface area of a sphere.
6:54 Sample Problem: . Find the surface area of a sphere with a radius of 9 in. Give both the exact answer and the rounded decimal answer.
7:52 Solution: Find the volume of a sphere with a diameter of 10cm.
10:21 Solution: Find the radius of a sphere with .volume of 72 meters cubed
13:00 The circumference of a great circle is 25 inches. Find the surface area and volume of the sphere.
17:57 Solution: A spherical balloon has a surface area of 16 in2. Find the volume of the sphere.
20:23 A spherical balloon has an initial radius of 5 in. When more air is added, the radius becomes 10 in. Explain how the surface area and volume change as the radius changes.
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Tagged under: geometry lessons,moomoomath,geometry ,spheres,volume sphere,volume formula sphere,radius sphere,math ,surface area sphere,Common Core Standard G.GMD.3
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# Scenario: Franca solving a story problem with the students.
## Context – public primary school, national curriculum
The Italian National Curriculum emphasizes the development of students’ number sense through realistic mathematical problems where the estimate of numbers can be performed. The mental calculation has taken great importance and the students should manage the control of their reasoning. Moreover the National Curriculum underlines that students should be able to solve mathematical problems with different strategies.
Franca is teaching volume measures to her pupils. She has already explained the litre and its submultiples as a measure of capacity. In this lesson she wants to show her pupils how to use mental calculation in a story problem context to give them a chance to manage the change of unit in meaningful way.
## Scenario
Franca tells her students the following story problem:
In a kindergarten a teacher wants to give every child 200ml of chamomile. She has some thermos. The capacity of each thermos is 500ml. How many thermos it is necessary to fill to give the chamomile to all the 18 children?
She writes the data on the blackboard in this way:
500ml thermos
18 b/i
200ml every child
and asks to all the pupils how they would start to solve the problem. Manuele says that they have to find how much chamomile it needs in all and that the operation is 200 times 18 (200 x 18). Franca writes on the blackboard 18 x 200ml and asks what is the result of this multiplication. Manuele answers that the amount of chamomile is found. Then Franca tells Manuele to calculate 200 x 18. Manuele says 18x2 is 36 therefore you obtain 3600. Franca writes to the blackboard 3600ml and asks to all the pupils how many lits are 3600ml. Manule says it is 3,6 lits. Franca asks Mattia the reason why it is so. The student says that 3600 are millilitres. So the teacher underlines that you need to divide by 1000 to convert mls to lits. Then the teacher asks how many thermos are needed if every thermos is filled with ½ lit of chamomile. A student answers that 6 thermos are needed, but Franca forces the pupils to tell their reasoning. So another pupil says that it needs to divide 3600 by 500. Franca acknowledges that it is correct but she would like the pupils found a strategy to do the operation in mental way. Then Franca asks to the pupils:
Franca: If a thermos is filled with ½ a lit how many thermos are necessary for one litre?
Student: Two
Franca: Then what can I do?
The students try to find the answer telling numbers randomly until a student says that you need six thermos. Franca asks for the reason behind this answer. The student answers that if a thermos contains ½ a litre then 6 thermos contain 3 lits, and that he did a multiplication by two to find this result. Franca wants to know why he did a multiplication. Another student explains that the classmate made a multiplication because if a thermos is filled with ½ lit, for one lit you need two thermos. Franca approves this answer. Then she asks to do 3,6 x 2 in their mind. She pushes the students to make (3 + 0.6) x 2 = 3 x 2 + 0.6 x 2= 6 + 1.2 = 7.2 and when they find the result, Franca writes 6, 1.2 and 7.2 on the blackboard. She points out that 7.2 are thermos and then she asks to her students how is it possible to fill 7.2 thermos. She writes on the blackboard 7 and 0.2 saying that the question is what “0.2 thermos” means. Then a student answers that 0.2 is half a thermos. Franca asks:
Franca: How much is one half?
Some students: 0.5!
Franca: So half thermos is 0.5 not 0.2.
She writes 0.5 on the blackboard. While she’s saying that 0.2 is less than one half and a student says that it is a quarter. So she asks what a quarter is in decimal numbers. The students try some numbers randomly until one says 0.25 and Franca writes it on the blackboard. Then she explains that 0.25 is ¼ of a thermos because 25 times 4 is 100. Then she encourages the students to compare 0.2 with 0.25 to find which is bigger. The students easily find that 0.2 is less than 0.25 and then Franca wants to know from them how to calculate 0.2 of 500ml. Manuele says 500 divided by 0.2. Franca writes it on the blackboard but she says that the result of the division is more than 500 and that it is impossible to have more than 500ml of chamomile in a thermos. Then she asks which is the fraction of 0.2 knowing that 0.25 is ¼. She writes on the blackboard 0.25= ¼ and 0.2 = 1/? She reminds her students that two is contained five times in ten. Then some students say 1/5. She writes it on the blackboard. Then she explains that is easy to calculate 1/5 of 500. The students find that is 100. Franca asks which is the operation and the students answer that 500:5. Then Franca writes on the blackboard 7 thermos and 100ml. Again Franca asks what it is needed to fill 8 thermos. At the end some students answer that 400ml are needed.
# Knowledge Quartet Coding Commentary
## Franca is bringing the pupils step by step to the problem solution. In so doing she makes clear which are her purposes related to the way to solve the problem and to use the mental calculation.
First of all we can recognize the flexibility she asks from her students in changing the unit of measurement to perform a mental calculation (in the solution process Franca converts 3600ml in 3.6lit and 500ml in ½ lit to reduce the size of the numbers). This flexibility in making conversion of units opens the way to make easier the calculations and to link the decimal representation of rational numbers with the representation by fractions.
The number of thermos is found by a sort of change of unit of measurement. When Franca asks how many thermos are needed for one litre and then for three lits, if one thermos is half a litre, this reasoning can be understood like a change of kind of units from lits to thermos. This strategy gives the opportunity to extend the meaning of unit of measurement and to perform a calculation in a simpler way than to divide 3600 by 500. Moreover it makes clearer why the result of the operation is the number of thermos. The division between 3600 and 500 is among numbers with different kind of unit, lits and lits/thermos, so the result is the number of thermos but it must be made explicit to the students, as Franca does, because it is not so obvious for them.
Another step of the solution that emphasizes the purpose of Franca can be found in the transformation of 7.2 thermos in millilitres. The research on mathematics education underlines the troubles that student find in giving sense to the numbers in the school problems (Schoenfeld, 1988). Franca makes a realistic consideration separating 7 from 0.2. Seven is a whole number that can properly represent the number of thermos but 0.2 is a part of a thermos. How much is it this part? Again Franca tries to force her pupils to make explicit their reasoning and moreover she encourages them to find a strategy to make calculation by mind. Then 0.2 is transformed in 1/5 by progressive adaptations. First the students find the fraction corresponding to 0.5, then the fraction corresponding to 0.25 and finally the students find the fraction corresponding to 0.2. Franca wants the students have a number sense, knowing that 0.5 is a half then that 0.25 is half of a half ( a quarter) they can grasp the solution by continuous adjustments.
Franca’s awareness of the purposes in this lesson is about the kinds of mathematical learning the students should achieve. The good subject matter knowledge that the teacher shows in this episode allows to connect different parts of the contents of mathematics teaching (story problems, units of measurement, fractions) and this connection is made following some close purposes. The story problem is not a tool to perform calculation in an algorithmic way, but a tool to develop some competences as the number sense and the more useful representation of a number in a context where mental calculation are a target of mathematical instruction (Arcavi, 1994; Sowder, 1992). Franca considers different strategies to confront numbers and measures. The focus is on the strategies to adopt and not on the procedures to perform. Moreover Franca wants her students to be aware of the meaning of each number they find in the process of solving the problem, because this is one good way to find a number sense and a more economic strategy for solving the problem.
## References
Arcavi, A. (1994) Symbol sense: Informal sense-making in formal mathematics. For the Learning of Mathematics, 14(3), 24-35
Schoenfeld, A.H. (1988) When good teaching leads to bad results: The disaster of “well taught” mathematics courses. Educational Psychologist 23(2), 145-166.
Sowder, J. (1992). Estimation and number sense. In D.A. Grouws (Ed) Handbook for research on mathematical teaching and learning . New York: McMillan.
AP: Scenario 2
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Home MATHEMATICS TOPIC 3: STATISTICS ~ MATHEMATICS FORM 3
# TOPIC 3: STATISTICS ~ MATHEMATICS FORM 3
2171
18
Mean
Calculating the Mean from a Set of Data, Frequency Distribution Tables and Histogram
Calculate the mean from a set of data, frequency distribution tables and histogram
Measures of central tendency:
After collecting the data, organizing it and illustrating it by means of diagrams, there is a need to calculate certain statistical measures to describe the data more precisely. There are various types of measures of central tendency – the arithmetic mean (or simply the mean), the median, and the mode. Once the measures of central tendency are found, it is easier to compare two or more sets of data.
The arithmetic mean
When people are asked to find the measure of central tendency of some numbers, they usually find the total of the numbers, and then divide this total by however many numbers there are. This type of measure of central tendency is the arithmetic mean. If the n values are x1+x2+x3 ………+xn then the arithmetic mean is = x1 + x2+ ….. + xn/n
Example 1
The masses of some parcels are 5kg, 8kg, 20kg and 15kg. Find the mean mass of the parcels.
Solution
Total mass = (5 + 8 + 20 + 15) kg = 48kg
The number of parcels = 4
The mean mass = 48kg ÷ 4 = 12kg
The arithmetic mean used as measure of central tendency can be misleading as can be seen in the following example.
Example 2
John
and Mussa played for the local cricket team. In the last six batting
innings, they scored the following number of runs.John: 64, 0, 1, 2, 4,
1;Mussa: 15, 20, 13, 11 , 10, 3.Find the mean score of each player.
Which player would you rather have in your team? Give a reason.
Solution
John’s mean = (64 + 0 + 1 + 2 + 4 + 1) ÷ 6 = 12
Mussa’s mean = (15 + 20 + 13 + 11 + 10 + 3) ÷ 6 = 12
Each
player has the same mean score. However, observing the individual
scores suggests that they are different types of player. If you are
looking for a steady reliable player, you would probably choose Mussa.
Often it is possible to use the mean of one set of numbers to find the mean of another set of related numbers.
Suppose a number a is added to or subtracted from all the data. Then a is added to or subtracted from the mean.
Suppose
the n values are 𝑥!+𝑥! + 𝑥! ………+𝑥!. Multiply each by a, and
we obtain 𝑎𝑥!+𝑎𝑥! + 𝑎𝑥! ………+𝑎𝑥!. So we see that the mean
has been multiplied by a.
Interpreting the Mean Obtained from a Set Data, Frequency Distribution Tables and Histogram
Interpret the mean obtained from a set data, frequency distribution tables and histogram
Measures of central tendency from frequency tables
If the data has already been put into a frequency table, the calculation of the measures of central tendency is slightly easier.
Exercise 1
Juma rolled a six- sided die 50 times. The scores he obtained are summarized in the following table.Calculate the mean score
Score (x) 1 2 3 4 5 6 Frequency (f) 8 10 7 5 12 8
Solution
10 scores of 2 give a total 10 x 2 = 20
8 scores of 1 gives a total 8 x 1 = 8
And so on, giving a total score of
8 x 1 +10 x 2+7 x 3 + 5 x 4 + 12 x 5 + 8 x 6 = 177
The total frequency = 8 + 10 + 7 + 5 + 12 + 8 = 50
The mean score = 177 ÷ 50 = 3.34
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# 7.12: Binomial Theorem and Expansions
Difficulty Level: At Grade Created by: CK-12
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Practice Binomial Theorem and Expansions
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Multiplying binomials is not a terribly difficult exercise, but it can certainly be time consuming with higher exponents, for example:
Calculate: (x3)2\begin{align*}(x - 3)^2\end{align*} is pretty easy:
(x3)(x3)\begin{align*}(x - 3) \cdot (x - 3)\end{align*}
(x26x+9)\begin{align*}(x^2 -6x +9)\end{align*}
BUT...
Calculate: (x3)5\begin{align*}(x - 3)^5\end{align*}
This is: (x3)(x3)(x3)(x3)(x3)\begin{align*}(x - 3) \cdot (x - 3) \cdot (x - 3) \cdot (x - 3) \cdot (x - 3)\end{align*}
First we "foil" the first two terms to get (x26x+9)\begin{align*}(x^2 -6x +9)\end{align*}
2nd, we multiply (x26x+9)\begin{align*}(x^2 -6x +9)\end{align*} by (x3)\begin{align*}(x - 3)\end{align*}, yielding: (x39x2+27x27)\begin{align*}(x^3 -9x^2 +27x -27)\end{align*}
3rd, we multiply (x39x2+27x27)\begin{align*}(x^3 -9x^2 +27x -27)\end{align*} by (x3)\begin{align*}(x - 3)\end{align*}
... and so on.
Definitely do-able, but a nightmare of a job, particularly by hand.
Isn't there an easier way?
Embedded Video:
### Guidance
There is a particular pattern in combinations that is seen in the expansion of polynomials of the form (x + y)n.
This pattern is most commonly displayed in a triangle:
This triangle is referred to as Pascal’s triangle, named after mathematician Blaise Pascal, although other mathematicians before him worked with these numbers. The numbers in the triangle can be used to generate more rows: notice that if you add two consecutive numbers, you get the number between and below them in the next row.
We can generalize this pattern as follows: (nr1)+(nr)=(n+1r)\begin{align*}\binom{n} {r - 1} + \binom{n} {r} = \binom{n + 1} {r}\end{align*}.
Binomial Expansion
To expand a binomial is to multiply all of the factors. The resulting polynomial is in standard form. For example:
(x + y)2 = (x + y) (x + y) = x2 + xy + xy + y2 = x2 + 2xy + y2
If we expand (x + y)3 , we get:
(x + y)3 = (x + y) (x + y) (x + y)
= (x + y) (x2 + 2xy + y2)
= x3 + 2x2y + xy2 + x2y + 2xy2 + y3
= x3 + 3x2y + 3xy2 + y3
Notice that the coefficients of each polynomial correspond to a row of Pascal’s triangle.
Also notice that the exponents of x descend, and the exponents of y ascend with each term. These are key aspects of the Binomial Theorem.
The Binomial Theorem
The binomial theorem can be stated using a summation:
(x+y)n=r=0n((nr)xnryr)\begin{align*}(x + y)^n = \sum_{r = 0}^n \left (\binom{n} {r} x^{n - r} y^r \right)\end{align*}
This is a very succinct way of summarizing the pattern in a binomial expansion. Let’s return to (x + y)3 to see how the theorem works.
(x+y)3\begin{align*}(x + y)^3\end{align*} =(30)x30y0+(31)x31y1+(32)x32y2+(33)x33y3\begin{align*}= \binom{3} {0}x^{3 - 0} y^0 + \binom{3} {1}x^{3 -1}y^1 + \binom{3} {2}x^{3 - 2}y^2 + \binom{3} {3}x^{3 -3}y^3\end{align*}
=1x31+3x2y+3xy2+1x0y3\begin{align*}= 1x^3 \cdot 1 + 3x^2y + 3xy^2 + 1 \cdot x^0 \cdot y^3\end{align*}
=x3+3x2y+3xy2+y3\begin{align*}= x^3 + 3x^2y + 3xy^2 + y^3\end{align*}
Again, the exponents on x descend from 3 to 0. The exponents on y ascend from 0 to 3. The coefficients on the terms correspond to row 3 of Pascal’s triangle. These coefficients are, not surprisingly, referred to as binomial coefficients!
Given this theorem, we can expand any binomial without having to multiply all of the factors.
Finding a specific term in a Binomial Expansion
Finding a term in an expansion can be used to answer a particular kind of probability question.
Consider an experiment, in which there are two possible outcomes, such as flipping a coin. If we flip a coin over and over again, this is referred to as a Bernoulli trial. In each flip (“experiment”), the probability of getting a head is 0.5, and the probability of getting a tail is 0.5. (Note: this is true for flipping a coin, but not for other situations. That is, it’s not always “50-50 chance!) Now say we flip a coin 25 times. What is the probability of getting exactly 10 heads?
The answer to this question is a term of a binomial expansion. That is, the probability of getting 10 heads from 25 coin tosses is: (2510)(0.5)10(0.5)150.0974\begin{align*} \binom{25}{10} (0.5)^{10}(0.5)^{15}\approx 0.0974\end{align*}, or about a 10% chance.
#### Example A
Use the binomial theorem to expand each polynomial:
a.(2x + a)4 b. (x - 3)5
Solution
a.(2x + a)4
= 1(2x)4 (a)0 + 4(2x)3 (a)1 + 6(2x)2(a)2 + 1(2x)0 (a)4
= 16x4 + 32x3a + 24x2a2 + 8x'a3 + a4
Note that it is easier to simply use the numbers from the appropriate row of the triangle than to write out all of the coefficients as combinations. However, if n is large, it may be easier to use the combinations.
b. (x - 3)5
= 1(x)5 (-3)0 + 5(x)4 (-3)1 + 10(x)3(-3)2 + 10(x)2(-3)3 + 5(x)1 (-3)4 + 1(x)0 (-3)5
= x5 - 1x4a + 90x3 - 270x2 + 405x - 243
Notice that in this expansion, the terms alternate signs. This is the case because the second term in the binomial is -3. When expanding this kind of polynomial, be careful with your negatives!
We can also use the Binomial Theorem to identify a particular term or coefficient.
#### Example B
Identify the 3rd term of the expansion of (2x + 3)6.
Solution
The 3rd term is \begin{align*}\mathit \ \binom{6}{2} (2x)^{4}3^{2}=15\cdot 16x^{4} \cdot 9 = 2160x^{4}\end{align*}
Keep in mind that row 6 of Pascal’s triangle starts with \begin{align*}\mathit \ \binom{6}{0}\end{align*} , so the coefficient of the third term in the expansion is \begin{align*}\mathit \ \binom{6}{2}\end{align*}. Also keep in mind that the first term includes (2x)6 and 30 , so the third term includes (2x)4 and 32.
#### Example C
What is the coefficient of \begin{align*}x^2\end{align*} in the expansion of \begin{align*}(3x +3)^3\end{align*}
Solution
Use the binomial theorem for the second term:
\begin{align*}_3C_2(3x)^2(3)^{3-2}\end{align*}
\begin{align*}3(3x)^2(3)^1 = 81\end{align*}
### Vocabulary
The Binomial Theorem is an efficient formula for calculating the expansion of binomials.
Binomial Expansion is the process of raising a binomial such as (x + 2) to a power. The process can be time consuming, particularly with higher exponents.
Pascal's Triangle is a pyramid of sorts constructed with the coefficients of binomials as they are expanded. It is a convenient reference.
### Guided Practice
1) Use Pascal's triangle to find the coefficient of \begin{align*}x^2y\end{align*} in \begin{align*}(x - y)^3\end{align*}
2) Expand: \begin{align*}(x + 3)^3\end{align*}
3) Expand: \begin{align*}(3x - 3)^6\end{align*}
4) Find the coefficient of\begin{align*}x^3\end{align*} in \begin{align*}(x - y)^3\end{align*}
Solutions
1) First, create Pascal's triangle. Remember to add 1's on the ends and the sum of the two numbers above to get the new numbers.
\begin{align*}1\end{align*}
\begin{align*}1 - 1\end{align*}
\begin{align*}1 - 2 - 1\end{align*}
\begin{align*}1 - 3 - 3 - 1\end{align*}
We only had to go up to the third row, we are looking for the 2nd term, so we count over \begin{align*} 1+ 1\end{align*} from the left to get our coefficient of 3. Don't forget to start with the 1st term.
We use the formula for the second term:
\begin{align*}_3C_2 x^2(-y)^{3-2}\end{align*}
\begin{align*}3x^2(-y)^1\end{align*}
So we get an answer of: -3
2) Start by expanding out each term separately:
Use the formula for term 0:
\begin{align*}_3C_3x^3(3)^{3-3} = x^3(3)^0 = x ^3\end{align*}
...and term 1:
\begin{align*}_3C_2x^2(3)^{3-2} = 3x^2(3)^1 = 9x ^2\end{align*}
...and term 2:
\begin{align*}_3C_1x^1(3)^{3-1} = 3x^1(3)^2 = 27x \end{align*}
...and term 3:
\begin{align*}_3C_0x^0(3)^{3-0} = x^0(3)^3 = 27\end{align*}
So we end up with:
\begin{align*}x^3 + 9x^2 + 27x + 27\end{align*}
3) We will expand out each term separately:
We use the formula for term 0:
\begin{align*}_6C_6 (3x)^6(-3)^{6 - 6} = (3x)^6(-3)^0 = 729x^6\end{align*}
...and term 1:
\begin{align*}_6C_5 (3x)^5(-3)^{6 - 5} = 6(3x)^5(-3)^1 = -4374x^5\end{align*}
...and term 2:
\begin{align*}_6C_4 (3x)^4(-3)^{6 - 4} = 15(3x)^4(-3)^2 = 10,935x^4\end{align*}
...and term 3:
\begin{align*}_6C_3 (3x)^3(-3)^{6 - 3} = 20(3x)^3(-3)^3 = -14,580x^3\end{align*}
...and term 4:
\begin{align*}_6C_2 (3x)^2(-3)^{6 - 2} = 15(3x)^2(-3)^4 = 10,935x^2\end{align*}
...and term 5:
\begin{align*}6C_1 (3x)^1(-3)^{6 - 1} = 6(3x)^1(-3)^5 = -4374x\end{align*}
...and term 6:
\begin{align*}_6C_0 (3x)^0(-3)^{6 - 0} = (3x)^0(-3)^6 = 729\end{align*}
The expanded polynomial is:
\begin{align*}729x^6 - 4374x^5 + 10,935x^4 - 14,580x^3 + 10,935x^2 - 4374x + 729\end{align*}
4) We use the formula for the third term:
\begin{align*}_3C_3 x^3(-y)^{3 - 3}\end{align*}
\begin{align*}x^3(-y)^0\end{align*}
### Practice
Questions
1. Expand: \begin{align*}(x+3a)^{4}\end{align*}
2. Expand: \begin{align*}(y + \frac{1}{2})^5\end{align*}
3. Expand: \begin{align*}(2x-a)^{6}\end{align*}
4. Expand: \begin{align*}(x + y)^6\end{align*}
5. Expand: \begin{align*}(3x + 1)^5\end{align*}
6. Expand: \begin{align*}(x + y)^5\end{align*}
7. Expand: \begin{align*} (2x + 2)^4\end{align*}
Find the Term
1. Find the 3rd term in the expansion \begin{align*}(3x+2a)^{9}\end{align*}.
2. Find the 7th term in the expansion of \begin{align*}(4x - \frac{1}{2}a)^{10}\end{align*}.
Use Pascal's Triangle to find the coefficient:
1. What is the coefficient of \begin{align*}x^3y^2\end{align*} in the expansion of \begin{align*}(x + y)^5\end{align*}?
2. What is the coefficient of \begin{align*}x^4\end{align*} in the expansion\begin{align*}(3x + 1)^4\end{align*}?
3. What is the coefficient of \begin{align*}x\end{align*} in the expansion:\begin{align*}(2x + 2)^5\end{align*}?
What is the coefficient of the expansions?
1. What is the coefficient of \begin{align*}x^2\end{align*} in the expansion of \begin{align*}(x + 1) ^6\end{align*}?
2. What is the coefficient of \begin{align*}x^5\end{align*} in the expansion of \begin{align*}(2x + 1) ^5\end{align*}?
3. What is the coefficient of \begin{align*}x\end{align*} in the expansion of \begin{align*}(3x + 2) ^3\end{align*}?
4. What is the coefficient of \begin{align*}x^6\end{align*} in the expansion of \begin{align*}(x + y) ^6\end{align*}?
5. What is the coefficient of \begin{align*}x\end{align*} in the expansion of \begin{align*}(2x + 1) ^6\end{align*}?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Bernoulli trial
In probability, a Bernoulli trial is an experiment with exactly two possible outcomes
Binomial Expansion
A binomial expansion is the process of raising a binomial such as $(x + 2)$ to a power. The process can be time consuming when completed manually, particularly with higher exponents.
Binomial Theorem
The binomial theorem is an efficient formula for calculating the expansion of binomials. It states that $(x + y)^n = \sum_{r = 0}^n \left (\binom{n} {r} x^{n - r} y^r \right)$.
combination
Combinations are distinct arrangements of a specified number of objects without regard to order of selection from a specified set.
Pascal's Triangle
Pascal's triangle is a triangular array of numbers constructed with the coefficients of binomials as they are expanded. The ends of each row of Pascal's triangle are ones, and every other number is the sum of the two nearest numbers in the row above.
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1. ## sovle equation
If is the focal length of a lens and is the distance of an object from the lens and is the distance of the image from the lens then
(a) Solve this equation for the distance of the image in terms of the other two quantities. Simplify your answer and put the denominator of the simplified fraction into the answer box (notice the numerator) (b) If the focal length is and the distance of the object from the lens is what is the distance of the image from the lens? (c) If the focal length is and the distance of the object from the lens is very large, for example what is the distance of the image from the lens approximately? (this explains the name focal length)
2. Hello, lilikoipssn!
If $f$ is the focal length of a lens
and $u$ is the distance of an object from the lens
and $v$ is the distance of the image from the lens,
then: . $\frac{1}{u} + \frac{1}{v}\;=\;\frac{1}{f}$
(a) Solve this equation for $v.$
Muliply by $uvf\!:\quad fv + fu \:=\:uv\quad\Rightarrow\quad uv-fv \:=\:fu$
Factor: . $(u-f)v \:=\:fu \quad\Rightarrow\quad\boxed{ v \:=\:\frac{fu}{u-f}}$
(b) If $f = 8$ cm and $u = 38$ cm, find $v.$
We have: . $v \;=\;\frac{(8)(38)}{38-8} \;=\;\frac{304}{30} \;=\;\boxed{\frac{152}{15}\text{ cm}}$
(c) If $f = 8$ cm, and $u$ is very large, for example, 1,000,000 cm
find $v$ approximately. (This explains the name focal length)
We have: . $v \;=\;\frac{(8)(1,\!000,\!000)}{1,\!000,\!000 -8} \;\approx\;\frac{8,\!000,\!000}{1,\!000,\!000} \;=\;8$
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# What is Ln in Math | Log vs Ln
We always heard about the logarithms log and ln. So it is important to know about their meaning and differences. Note that log represents the common logarithm and ln represents the natural logarithm. In this post, we will learn about them.
## What is log?
The notation log is referred to the short form of the logarithm. log denotes the common logarithm and it is with base 10.
For example, $\log 2$ is the natural logarithm of $2$. That is, $\log 2=\log_{10} 2$.
As we know that the logarithm of $x$ with base $x$ is equal to $1$, that is, $\log_a a=1$, we must have that $\log_{10} 10=1$. Thus, the common logarithm of $10$ is equal to $1$.
## What is ln?
The notation ln is used to denote the natural logarithm and it is with base $e$. Here, $e$ is the irrational number defined by the following convergent series:
$e=\sum_{n=0}^\infty \dfrac{1}{n!}$ $=1+\dfrac{1}{1!}+\dfrac{1}{2!}+\dfrac{1}{3!}+\cdots$
The value of $e$ is approximately equal to $2.71828$.
$e \approx 2.71828$
## Question Answer on log vs ln
Question 1: Find the value of e to the power lnx, that is, evaluate $e^{\ln x}$.
Let us assume that $y=e^{\ln x}$. We need to find the value of $y$.
Takning $\ln$ both sides of $y=e^{\ln x}$, we have that
$\ln y = \ln (e^{\ln x})$
$\Rightarrow \ln y =\ln x \ln e$
$\Rightarrow \ln y =\ln x$ as we know that $\ln e=1$.
$\Rightarrow y = x$.
Thus, we have shown that the value of e to the power ln x is equal to x.
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Note: We know that the equation of the line is $\displaystyle \frac{x}{a} + \frac{y}{b} = 1$ where $\displaystyle a$ is the x-intercept and $\displaystyle b$ is the y-intercept.
Question 1: Find the equation to the straight line:
(i) cutting off intercepts $\displaystyle 3 \text{ and } 2$ from the axes.
(ii) cutting off intercepts $\displaystyle - 5 \text{ and } 6$ from the axes.
i) Here $\displaystyle a = 3 \hspace{0.3cm} b = 2$
Therefore the equation of the line:
$\displaystyle \frac{x}{3} + \frac{y}{2} = 1$
$\displaystyle \Rightarrow 2x + 3y - 6 = 0$
ii) Here $\displaystyle a = -5 \hspace{0.3cm} b = 6$
Therefore the equation of the line:
$\displaystyle \frac{x}{-5} + \frac{y}{6} = 1$
$\displaystyle \Rightarrow 6x - 5y + 30 = 0$
$\displaystyle \\$
Question 2: Find the equation of the straight line which passes through $\displaystyle (1,-2)$ and cuts off equal intercepts on the axes.
Here $\displaystyle a = b$
Therefore the equation of the line:
$\displaystyle \frac{x}{a} + \frac{y}{a} = 1$
$\displaystyle \Rightarrow x+y = a$
Since the line passes through $\displaystyle ( 1, -2)$ we get
$\displaystyle 1+(-2) = a \hspace{0.5cm} \Rightarrow a = -1$
Hence the equation of the line is $\displaystyle x+y +1 = 0$
$\displaystyle \\$
Question 3: Find the equation to the straight line which, passes through the point $\displaystyle (5,6)$ and has intercepts on the axes (i) equal in magnitude and both positive. (ii) equal in magnitude but opposite in sign.
i) Here $\displaystyle a = b$
Therefore the equation of the line:
$\displaystyle \frac{x}{a} + \frac{y}{a} = 1$
$\displaystyle \Rightarrow x+y = a$
Since the line passes through $\displaystyle ( 5,6)$ we get
$\displaystyle 5+6 = a \hspace{0.5cm} \Rightarrow a = -11$
Hence the equation of the line is $\displaystyle x+y = 11$
ii) Here $\displaystyle b=-a$
Therefore the equation of the line:
$\displaystyle \frac{x}{a} + \frac{y}{-a} = 1$
$\displaystyle \Rightarrow x-y = a$
Since the line passes through $\displaystyle ( 5,6)$ we get
$\displaystyle 5-6 = a \hspace{0.5cm} \Rightarrow a = -1$
Hence the equation of the line is $\displaystyle x+y = -1$
$\displaystyle \\$
Question 4: For what values of $\displaystyle a \text{ and } b$ the -intercepts cut off on the coordinate axes by the line $\displaystyle ax+by+8=0$ are equal in length but opposite in signs to those cut off by the line $\displaystyle 2x - 3y + 6 = 0$ on the axes.
Given $\displaystyle 2x - 3y + 6 = 0$
$\displaystyle \Rightarrow \frac{2x}{-6} - \frac{3y}{-6} + \frac{6}{-6} = 0$
$\displaystyle \Rightarrow \frac{x}{-3} + \frac{y}{2} = 1$
$\displaystyle \text{ Therefore the x-intercept } = a = -3$ $\displaystyle \text{ and y-intercept } = b = 2$
We also have $\displaystyle ax+by+8=0$
$\displaystyle \Rightarrow \frac{ax}{-8} + \frac{by}{-8} + \frac{8}{-8} = 0$
$\displaystyle \Rightarrow \frac{x}{(-8/a)} + \frac{y}{(-8/b)} = 1$
$\displaystyle \text{ Therefore the x-intercept } = \frac{-8}{a}$ $\displaystyle \text{ and y-intercept } = \frac{-8}{b}$
$\displaystyle \therefore \frac{-8}{a} = -(-3) \Rightarrow a = \frac{-8}{3}$
$\displaystyle \text{ and } \frac{-8}{b} = -(2) \Rightarrow b = 4$
$\displaystyle \\$
Question 5: Find the equation to the straight line which cuts off equal positive intercepts on the axes and their product is $\displaystyle 25$.
Here $\displaystyle a = b \hspace{0.5cm} ab = 25$
Solving: $\displaystyle ab = 25 \hspace{0.5cm} \Rightarrow a^2 = 25 \hspace{0.5cm} \Rightarrow a = \pm 5$
Since the intercepts are positive, we get $\displaystyle a = 5$
Therefore the equation of the line:
$\displaystyle \frac{x}{5} + \frac{y}{5} = 1$
$\displaystyle \Rightarrow x+y = 5$
$\displaystyle \\$
Question 6: Find the equation of the line which passes through the point $\displaystyle (- 4, 3)$ and the portion of the line intercepted between the axes is divided internally in the ratio $\displaystyle 5 : 3$ by this point.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given $\displaystyle (-4,3)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ in the ratio of $\displaystyle 5:3$
$\displaystyle \therefore -4 = \frac{3 \times a + 5 \times 0}{5+3}$ $\displaystyle \Rightarrow a = \frac{-32}{3}$
$\displaystyle \text{ Similarly } 3 = \frac{3 \times 0 + 5 \times b}{5+3}$ $\displaystyle \Rightarrow b = \frac{24}{5}$
Since the equation of line passing through $\displaystyle ( -4, 3)$, therefore
$\displaystyle \frac{x}{\frac{-32}{3}} + \frac{y}{\frac{24}{5}} = 1$
$\displaystyle \Rightarrow \frac{-3x}{32} + \frac{5y}{24} = 1$
$\displaystyle \Rightarrow \frac{-3x}{4} + \frac{5y}{3} = 8$
$\displaystyle \Rightarrow 9x-20y+96=0$
$\displaystyle \\$
Question 7: A straight line passes through the point $\displaystyle (\alpha , \beta )$ and this point bisects the portion of the line intercepted between the axes. Show that the equation of the straight ling is $\displaystyle \frac{x}{2\alpha} + \frac{y}{2\beta} = 1$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given $\displaystyle (\alpha, \beta)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ in the ratio of $\displaystyle 1:1$
$\displaystyle \therefore \alpha = \frac{a+0}{2} \Rightarrow a = 2 \alpha$
$\displaystyle \text{ Similarly } \beta = \frac{0+b}{2} \Rightarrow b = 2 \beta$
Hence the equation of line is:
$\displaystyle \frac{x}{2\alpha} + \frac{y}{2\beta} = 1$
Hence proved.
$\displaystyle \\$
Question 8: Find the equation of the line which passes through the point $\displaystyle (3,4)$ and is such that the portion of it intercepted between the axes is divided by the point in the ratio $\displaystyle 2 : 3$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given $\displaystyle P(3,4)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ in the ratio of $\displaystyle 2:3$
i.e. $\displaystyle AP:BP = 2:3$
$\displaystyle \therefore 3 = \frac{2 \times 0 + 3 \times a}{2+3} \Rightarrow a = 5$
$\displaystyle \text{ Similarly } 4 = \frac{2 \times b + 3 \times 0}{2+3} \Rightarrow b = 10$
Since the equation of line passing through $\displaystyle P( 3,4)$, therefore
$\displaystyle \frac{x}{5} + \frac{y}{10} = 1$
$\displaystyle \Rightarrow 2x+y = 10$
$\displaystyle \\$
Question 9: Point $\displaystyle R (h, k)$ divides a line segment between the axes in the ratio $\displaystyle 1 : 2$. Find the equation of the line.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given $\displaystyle P(h, k)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ in the ratio of $\displaystyle 1:2$
i.e. $\displaystyle AP:BP = 1:2$
$\displaystyle \therefore h = \frac{1 \times 0 + 2 \times a}{1+2}$ $\displaystyle \Rightarrow a = \frac{3h}{k}$
$\displaystyle \text{ Similarly } k = \frac{1 \times b + 2 \times 0}{1+2}$ $\displaystyle \Rightarrow b = 3k$
Since the equation of line passing through $\displaystyle P( h,k)$, therefore
$\displaystyle \frac{x}{(3h/2)} + \frac{y}{3k} = 1$
$\displaystyle \frac{2x}{3h} + \frac{y}{3k} = 1$
$\displaystyle \Rightarrow 2kx+hy - 3hk=0$
$\displaystyle \\$
Question 10: Find the equation of the straight line which passes through the point $\displaystyle (- 3, 8)$ and cuts off positive intercepts on the coordinate axes whose sum is $\displaystyle 7$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given: $\displaystyle a+b = 7 \hspace{0.5cm} b = 7 - a$.
The line also passes through $\displaystyle ( -3, 8 )$. Therefore
$\displaystyle \frac{-3}{a} + \frac{8}{7-a} = 1$
$\displaystyle \Rightarrow -3( 7-a) + 8a =a ( 7-a)$
$\displaystyle \Rightarrow -21 + 3a + 8a = 7a - a^2$
$\displaystyle \Rightarrow a^2 + 4a - 21 = 0$
$\displaystyle \Rightarrow ( a-3)(a+7) = 0$
$\displaystyle \Rightarrow a = 3 \ or \ a = -7$
Since the intercepts are positive we get $\displaystyle a = 3$
$\displaystyle \therefore b = 7 - 3 = 4$
Hence the equation of the line is:
$\displaystyle \frac{x}{3} + \frac{y}{4} = 1$
$\displaystyle \Rightarrow 4x + 3y = 12$
$\displaystyle \\$
Question 11: Find the equation to the straight line which passes through the point $\displaystyle (- 4, 3)$ and is such that the portion of it between the axes is divided by the point in the ratio $\displaystyle 5 : 3$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given $\displaystyle (-4,3)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ in the ratio of $\displaystyle 5:3$
$\displaystyle \therefore -4 = \frac{3 \times a + 5 \times 0}{5+3}$ $\displaystyle \Rightarrow a = \frac{-32}{3}$
$\displaystyle \text{ Similarly } 3 = \frac{3 \times 0 + 5 \times b}{5+3}$ $\displaystyle \Rightarrow b = \frac{24}{5}$
Since the equation of line passing through $\displaystyle ( -4, 3)$, therefore
$\displaystyle \frac{x}{\frac{-32}{3}} + \frac{y}{\frac{24}{5}} = 1$
$\displaystyle \Rightarrow \frac{-3x}{32} + \frac{5y}{24} = 1$
$\displaystyle \Rightarrow \frac{-3x}{4} + \frac{5y}{3} = 8$
$\displaystyle \Rightarrow 9x-20y+96=0$
$\displaystyle \\$
Question 12: Find the equation of a line which passes through the point $\displaystyle (22, - 6)$ and is such that the intercept on x-axis exceeds the intercept on y-axis by $\displaystyle 5$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given: $\displaystyle a=b+5 \hspace{0.5cm} b = a - 5$
The line also passes through $\displaystyle (22, -6 )$. Therefore
$\displaystyle \frac{22}{a} + \frac{-6}{a-5} = 1$
$\displaystyle \Rightarrow 22(a-5)-6a=a(a-5)$
$\displaystyle \Rightarrow 22a - 110 - 6a = a^2 - 5a$
$\displaystyle \Rightarrow a^2 - 21 a + 110=0$
$\displaystyle \Rightarrow (a-11)(a-10) = 0$
$\displaystyle \Rightarrow a = 11 \ \text{ or } \ \ \ \ a = 10$
$\displaystyle \Rightarrow b = 11-5 = 6 \ \ \ \text{ or } \ \ \ \ b = 10-5 = 5$
Hence the equation of the lines are:
$\displaystyle \frac{x}{11} + \frac{y}{6} = 1 \text{ or } \frac{x}{10} + \frac{y}{5} = 1$
$\displaystyle \Rightarrow 6x+11y = 66 \text{ or } \Rightarrow x+2y = 10$
$\displaystyle \\$
Question 13: Find the equation of the line, which passes through $\displaystyle P (1, - 7)$ and meets the axes at $\displaystyle A \text{ and } B$ respectively so that $\displaystyle 4 AP - 3 BP =0$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given $\displaystyle 4 AP - 3 BP =0 \hspace{0.5cm} \Rightarrow AP : BP = 3: 4$
Given $\displaystyle P(1,-7)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ in the ratio of $\displaystyle 3:4$
$\displaystyle \therefore 1 = \frac{3 \times 0 + 4 \times a}{3+4}$ $\displaystyle \Rightarrow a = \frac{7}{4}$
$\displaystyle \text{ Similarly } -7 = \frac{3 \times b + 4 \times 0}{3+4}$ $\displaystyle \Rightarrow b = \frac{-49}{3}$
Since the equation of line passing through $\displaystyle P( 3,4)$, therefore
$\displaystyle \frac{x}{(7/4)} + \frac{y}{(-49/3)} = 1$
$\displaystyle \Rightarrow \frac{4x}{7} - \frac{3y}{49} = 1$
$\displaystyle \Rightarrow 28x-3y=49$
$\displaystyle \\$
Question 14: Find the equation of the line passing through the point $\displaystyle (2,2)$ and cutting off intercepts on the axes whose sum is $\displaystyle 9$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given: $\displaystyle a+b=9 \hspace{0.5cm} b = 9-a$
The line also passes through $\displaystyle (2,2 )$. Therefore
$\displaystyle \frac{2}{a} + \frac{2}{9-a} = 1$
$\displaystyle \Rightarrow 18-2a+2a=9a-a^2$
$\displaystyle \Rightarrow a^2-9a+18=0$
$\displaystyle \Rightarrow (a-3)(a-6)=0$
$\displaystyle \Rightarrow a = 3 \ \text{ or } \ \ \ \ a = 6$
$\displaystyle \Rightarrow b = 9-3 = 6 \ \ \ \text{ or } \ \ \ \ b = 9-6 = 3$
Hence the equation of the lines are:
$\displaystyle \frac{x}{3} + \frac{y}{6} = 1 \text{ or } \frac{x}{6} + \frac{y}{3} = 1$
$\displaystyle \Rightarrow 2x+y = 6 \text{ or } \Rightarrow x+2y = 6$
$\displaystyle \\$
Question 15: Find the equation of the straight line which passes through the point $\displaystyle P(2, 6)$ and cuts the coordinate axes at the point $\displaystyle A \text{ and } B$ respectively so that $\displaystyle \frac{AP}{BP} = \frac{2}{3}$ .
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given $\displaystyle AP : BP = 2:3$
Given $\displaystyle P(2, 6)$ divides the $\displaystyle A(a,0) \text{ and } B( 0, b)$ in the ratio of $\displaystyle 2:3$
$\displaystyle \therefore 2 = \frac{2 \times 0 + 3 \times a}{2+3}$ $\displaystyle \Rightarrow a = \frac{10}{3}$
$\displaystyle \text{ Similarly } 6 = \frac{2 \times b + 3 \times 0}{2+3}$ $\displaystyle \Rightarrow b = 15$
Since the equation of line passing through $\displaystyle P( 2,6)$, therefore
$\displaystyle \frac{x}{(10/3)} + \frac{y}{15} = 1$
$\displaystyle \Rightarrow \frac{3x}{10} + \frac{y}{15} = 1$
$\displaystyle \Rightarrow 9x+2y=30$
$\displaystyle \\$
Question 16: Find the equations of the straight lines each of which passes through the point $\displaystyle (3,2)$ and cuts off intercepts $\displaystyle a \text{ and } b$ respectively on x and y-axes such that $\displaystyle a -b =2$.
Let the intercepts be $\displaystyle A(a,0) \text{ and } B(0,b)$ .
Given: $\displaystyle a-b=2 \hspace{0.5cm} a= b+2$
The line also passes through $\displaystyle (3,2 )$. Therefore
$\displaystyle \frac{3}{b+2} + \frac{2}{b} = 1$
$\displaystyle \Rightarrow 3b+ 2b + 4 = b^2 + 2b$
$\displaystyle \Rightarrow b^2 - 3b - 4 = 0$
$\displaystyle \Rightarrow (b+1)(b-4) =0$
$\displaystyle \Rightarrow b = -1 \ or \ \ \ \ b = 4$
$\displaystyle \Rightarrow a = -1+2=1 \ \ \ or \ \ \ \ a = 4+2 = 6$
Hence the equation of the lines are:
$\displaystyle \frac{x}{1} + \frac{y}{-1} = 1 \text{ or } \frac{x}{6} + \frac{y}{4} = 1$
$\displaystyle \Rightarrow x-y = 1 \text{ or } \Rightarrow 2x+3y=12$
$\displaystyle \\$
Question 17: Find the equations of the straight lines which pass through the origin and trisect the portion of the straight line $\displaystyle 2x + 3y = 6$ which is intercepted between the axes.
Given line $\displaystyle 2x + 3y = 6$
$\displaystyle \text{ x-intercept } = A ( 3,0)$
$\displaystyle \text{ y-intercept } = B ( 0,2)$
$\displaystyle P$ divides $\displaystyle AB$ in the ratio of $\displaystyle 2:1$
$\displaystyle \text{ Coordinates of } P = \Big( \frac{2 \times 3 + 1 \times 0}{2+1} , \frac{2 \times 0 + 1 \times 2}{2+1} \Big) = \Big( 2 , \frac{2}{3} \Big)$
$\displaystyle Q$ divides $\displaystyle AB$ in the ratio of $\displaystyle 1:2$
$\displaystyle \text{ Coordinates of } Q = \Big( \frac{1 \times 3 + 2 \times 0}{1+2} , \frac{1 \times 0 + 1 \times 2}{1+2} \Big) = \Big( 1 , \frac{4}{3} \Big)$
$\displaystyle \text{ Slope of } OP = \frac{\frac{2}{3}-0}{2-0} = \frac{1}{3}$
Therefore equation of $\displaystyle OP$:
$\displaystyle y - 0 = \frac{1}{3} ( x - 0) \hspace{0.5cm} \Rightarrow x-3y=0$
$\displaystyle \text{ Slope of } OQ = \frac{\frac{4}{3}-0}{1-0} = \frac{1}{3}$
$\displaystyle \text{ Similarly, equation of } OQ$:
$\displaystyle y - 0 = \frac{4}{3} ( x - 0) \hspace{0.5cm} \Rightarrow 4x-3y=0$
$\displaystyle \\$
Question 18: Find the equation of the straight line passing through the point $\displaystyle (2, 1)$ and bisecting the portion of the straight line $\displaystyle 3x -5y =15$ lying between the axes.
Given $\displaystyle 3x -5y =15$
Therefore intercepts of $\displaystyle x \text{ and } y$ axis are $\displaystyle A ( 5,0) \text{ and } B ( 0, -3)$
$\displaystyle \text{ Mid point M of } AB = \Big( \frac{5+0}{2} , \frac{0-3}{2} \Big) = \Big( \frac{5}{2} , \frac{-3}{2} \Big)$
$\displaystyle \text{ Slope of line passing through M and ( 2, 1)} = \frac{1 - ( \frac{-3}{2}) }{2-(\frac{5}{2})} = \frac{\frac{5}{2}}{\frac{-1}{2}} = -5$
Therefore equation of $\displaystyle OP$:
$\displaystyle y - 0 = -5 ( x - 2) \hspace{0.5cm} \Rightarrow y-1=-5x+10 \hspace{0.5cm} \Rightarrow 5x+y = 11$
$\displaystyle \\$
Question 19: Find the equation of the straight tine passing through the origin and bisecting the portion of the line $\displaystyle ax +by + c = 0$ intercepted between the coordinate axes.
Given $\displaystyle ax +by + c = 0$
When $\displaystyle x = 0, y = \frac{-c}{b}$ $\displaystyle \Rightarrow$ $\displaystyle \text{ y-intercept } B( 0, \frac{-c}{b} )$
When $\displaystyle y = 0 , x = \frac{-c}{a}$ $\displaystyle \Rightarrow$ $\displaystyle \text{ y-intercept } A( \frac{-c}{a} , 0)$
$\displaystyle \text{ Therefore midpoint of } AB = ( \frac{ \frac{-c}{a}+0 }{2} , \frac{ 0+ \frac{-c}{b} }{ 2 } ) = ( \frac{-c}{2a} , \frac{-c}{2b} )$
$\displaystyle \text{ Slope of } AB = \frac{ \frac{-c}{2b}-0 }{ \frac{-c}{2a}-0 } = \frac{a}{b}$
$\displaystyle \text{ Similarly, equation of } OQ$:
$\displaystyle y - 0 = \frac{a}{b} ( x - 0) \hspace{0.5cm} \Rightarrow ax-by = 0=0$
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Rational root theorem
(Redirected from Rational root test)
In algebra, the rational root theorem (or rational root test) states a constraint on rational solutions (or roots) of a polynomial equation
$a_nx^n+a_{n-1}x^{n-1}+\cdots+a_0 = 0\,\!$
with integer coefficients.
If a0 and an are nonzero, then each rational solution x, when written as a fraction x = p/q in lowest terms (i.e., the greatest common divisor of p and q is 1), satisfies
The rational root theorem is a special case (for a single linear factor) of Gauss's lemma on the factorization of polynomials. The integral root theorem is a special case of the rational root theorem if the leading coefficient an = 1.
Proofs
A proof
Let P(x) = anxn + an−1xn−1 + ... + a1x + a0 for some a0, ..., anZ, and suppose P(p/q) = 0 for some coprime p, qZ:
$P\left(\tfrac{p}{q}\right) = a_n\left(\tfrac{p}{q}\right)^n + a_{n-1}\left(\tfrac{p}{q}\right)^{n-1} + \cdots + a_1\left(\tfrac{p}{q}\right) + a_0 = 0.$
If we multiply both sides by qn, shift the constant term to the right hand side, and factor out p on the left hand side, we get
$\qquad p(a_np^{n-1} + a_{n-1}qp^{n-2} + \cdots + a_1q^{n-1}) = -a_0q^n.$
We see that p times the integer quantity in parentheses equals −a0qn, so p divides a0qn. But p is coprime to q and therefore to qn, so by (the generalized form of) Euclid's lemma it must divide the remaining factor a0 of the product.
If we instead shift the leading term to the right hand side and factor out q on the left hand side, we get
$\qquad q(a_{n-1}p^{n-1} + a_{n-2}qp^{n-2} + \cdots + a_0q^{n-1}) = -a_np^n.$
And for similar reasons, we can conclude that q divides an.[1]
Proof using Gauss's lemma
Should there be a nontrivial factor dividing all the coefficients of the polynomial, then one can divide by the greatest common divisor of the coefficients so as to obtain a primitive polynomial in the sense of Gauss's lemma; this does not alter the set of rational roots and only strengthens the divisibility conditions. That lemma says that if the polynomial factors in ℚ[X], then it also factors in ℤ[X] as a product of primitive polynomials. Now any rational root p/q corresponds to a factor of degree 1 in ℚ[X] of the polynomial, and its primitive representative is then qx − p, assuming that p and q are coprime. But any multiple in ℤ[X] of qx − p has leading term divisible by q and constant term divisible by p, which proves the statement. This argument shows that more generally, any irreducible factor of P can be supposed to have integer coefficients, and leading and constant coefficients dividing the corresponding coefficients of P.
Example
For example, every rational solution of the equation
$3x^3 - 5x^2 + 5x - 2 = 0\,\!$
must be among the numbers symbolically indicated by
± $\tfrac{1,2}{1,3}\,,$
which gives the list of 8 possible answers:
$1, -1, 2, -2, \frac{1}{3}, -\frac{1}{3}, \frac{2}{3}, -\frac{2}{3}\,.$
These root candidates can be tested using the Horner's method (for instance). In this particular case there is exactly one rational root. If a root candidate does not satisfy the equation, it can be used to shorten the list of remaining candidates.[2] For example, x = 1 does not satisfy the equation as the left hand side equals 1. This means that substituting x = 1 + t yields a polynomial in t with constant term 1, while the coefficient of t3 remains the same as the coefficient of x3. Applying the rational root theorem thus yields the following possible roots for t:
$t=\pm\tfrac{1}{1,3}$
Therefore,
$x = 1+t = 2, 0, \frac{4}{3}, \frac{2}{3}$
Root candidates that do not occur on both lists are ruled out. The list of rational root candidates has thus shrunk to just x = 2 and x = 2/3.
If a root r1 is found, Horner's method will also yield a polynomial of degree n − 1 whose roots, together with r1, are exactly the roots of the original polynomial. It may also be the case that none of the candidates is a solution; in this case the equation has no rational solution. If the equation lacks a constant term a0, then 0 is one of the rational roots of the equation.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
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# Percents as Decimals
## Rewrite percents as decimals.
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Practice Percents as Decimals
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Percents as Decimals
Have you ever purchased organic vegetables? Do you know what a CSA is?
Casey has moved from milk to organic vegetables. She has heard of a CSA - community supported agriculture and is wondering how many people actually belong to a CSA. She surveyed the 90 students in her class and found that 30% of the students belong to a CSA. That means that 70% don't belong to one.
If Casey wanted to write these percents as decimals how could she do it? Do you know?
This Concept is all about converting percents to decimals. By the end of the Concept you will know how Casey can accomplish this goal.
### Guidance
Now that you understand how fractions, decimals and percents are related, we can look at the relationship between them in more detail. Let’s look at the relationship between decimals and percents first.
How are decimals and percents related?
Just as percents are out of 100, decimals can also be written out of 100. When we have a decimal with two decimal places, it is also representing a quantity out of 100.
.34 = 34%
These two quantities are equivalent. The decimal .34 means 34 hundredths or 34 out of 100. 34% means 34 out of 100. Because decimals and percents are both parts of a whole, we can write percents as decimals.
How do we write percents as decimals?
We write a percent as a decimal by thinking “out of 100 means two decimal places.” You can drop the percent sign and move the decimal point two places to the left. Then the percent will be written as a decimal.
\begin{align*}45\%\end{align*}
The % sign is just like two decimal places since both mean out of one hundred or hundredths. We drop the % sign and insert the decimal two decimal places to the left.
\begin{align*}45\% = .45\end{align*}
Our answer is \begin{align*}.45\end{align*}.
Write 5% as a decimal
First, we drop the % sign and move the decimal point in two places. OOPS! This one doesn’t have two places. That’s okay, we can add a zero in for the missing place.
\begin{align*}5\%\end{align*} becomes \begin{align*}.05\end{align*}
Our answer is \begin{align*}.05\end{align*}.
Practice a few of these on your own. Write each percent as a decimal.
#### Example A
\begin{align*}17\%\end{align*}
Solution:\begin{align*}.17\end{align*}
#### Example B
\begin{align*}25\%\end{align*}
Solution:\begin{align*}.25\end{align*}
#### Example C
\begin{align*}75\%\end{align*}
Solution:\begin{align*}.75\end{align*}
Now back to Casey and the organic vegetables. Here is the original problem once again.
Casey has moved from milk to organic vegetables. She has heard of a CSA - community supported agriculture and is wondering how many people actually belong to a CSA. She surveyed the 90 students in her class and found that 30% of the students belong to a CSA. That means that 70% don't belong to one.
If Casey wanted to write these percents as decimals how could she do it? Do you know?
Now you should know how to write each percent as a decimal. Let's start with the 30%.
To convert 30% to a decimal we drop the % sign and move the decimal point two places to the left.
\begin{align*}30\% = .30\end{align*}
Next, we can convert 70%.
We drop the % sign and move the decimal two places to the left.
\begin{align*}70\% = .70\end{align*}
### Vocabulary Review
Here are the vocabulary words in this Concept.
Percent
means out of 100, it is a quantity written with a % sign, and is a part of a whole (100)
Fraction
a part of a whole, related to decimals and percents.
Decimal
a part of a whole shown by a decimal point, hundredths means two decimal places.
### Guided Practice
Here is one for you to try on your own.
Write 15% as a decimal.
First, we drop the % sign and move the decimal point in two places to the left, to represent the hundredths place in the number.
15% becomes .15
### Video Review
Here are videos for review.
http://www.mathplayground.com/howto_perfracdec.html – Converting fractions and decimals to percents
### Practice
Directions: Write each percent as a decimal.
1. 54%
2. 11%
3. 6%
4. 12%
5. 89%
6. 83%
7. 19%
8. 4%
9. 9%
10. 32%
11. 65%
12. 88%
13. 78%
14. 67.5%
15. 18.2%
### Vocabulary Language: English
Decimal
Decimal
In common use, a decimal refers to part of a whole number. The numbers to the left of a decimal point represent whole numbers, and each number to the right of a decimal point represents a fractional part of a power of one-tenth. For instance: The decimal value 1.24 indicates 1 whole unit, 2 tenths, and 4 hundredths (commonly described as 24 hundredths).
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# GRE Quantitative Comparison Tip #3 – Logic over Algebra
In previous posts (Tip #1 – Dealing with Variables, Tip #2: Striving for Equality), I have discussed two approaches when tackling Quantitative Comparison (QC) questions involving variables. Those approaches are:
1) Apply algebraic techniques
2) Plug in numbers
In those posts, I noted that the algebraic approach is typically the faster and more reliable approach.
In today’s post, we’ll examine a third strategy that can sometimes be the fastest and easiest approach. We’ll call this the “logical approach.”
To set things up, please consider the following QC question:
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
The algebraic approach to this question looks something like this:
First multiply both sides by 35 (the least common multiple of 5 and 7) to get:
Then add 5x to both sides to get:
Then add 42 to both sides to get:
And, finally, divide both sides by 19 to get:
So, now we’re comparing x and 3, and the given information tells us that x is greater than 3.
This means the correct answer must be A.
Now let’s take the original question and use logic to solve it (in about 5 seconds).
Column A: If x > 3, then 2x > 6, which means that 2x-6 must be positive.
Column B: If x > 3, then 3-x must be negative.
So, the two columns can be rewritten as:
From here, we can see that Column A is always positive and Column B is always negative. As such, Column A will always be greater than Column B. So, the correct answer is A.
Let’s try another one. See if you can solve it in your head.
A. The quantity in Column A is greater
B. The quantity in Column B is greater
C. The two quantities are equal
D. The relationship cannot be determined from the information given
For this question, I’ll leave the algebraic approach to you.
Let’s apply some logic.
First, we’re told that . In order to apply some logic, let’s refer to the denominator as “something.” In other words, 18y divided by “something” equals 3. Well, we know that 18y divided by 6y equals 3, so that “something” must equal 6y.
In other words, it must be the case that 7y-x = 6y
Now consider the fact that 7y-x = 6y. If we now refer to “x” as “something,” we can see that 7y minus “something” equals 6y. Since we know that 7y-y=6y, we can see that “something” must equal y.
In other words x = y.
Now that we have concluded that x=y, we’ll return to the original question:
If x=y, we can see that the answer here must be C.
So, although the algebraic approach is typically the superior approach for quantitative comparison questions involving variables, be sure to take a moment to see whether the problem can be solved by applying a little logic.
Heres’ the whole series of QC tips:
Tip #1: Dealing with Variables
Tip #2: Striving for Equality
Tip #3: Logic over Algebra
Tip #4: Comparing in Parts
Tip #5: Estimation with a Twist
## Author
• Brent Hanneson is a master tutor with over 20 years of teaching experience. He developed all the math content for Magoosh Test Prep. Brent plays ice hockey in his free time.
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# Functions in Set Theory
I am a PhD student of mathematics. I have complete MS in math from the University of Pakistan and have been writing online since 2020.
## Functions in Set Theory
Introduction:
The concept of a function is one of the most key ones in mathematics. The word "map," "mapping," "transformation," and many more all refer to the same thing; which one to use in a given context is typically chosen by tradition and the user's mathematical background. Gottfried Wilhelm Leibniz used the word "function" for the first time in a text he sent in 1673 to refer to a number associated to a curve's points, such as a coordinate or slope.:
Before defining the term function of a set, it is necessary to understand the terminology of image.
Image of a Function:
The collection of all possible output values is known as a image of a function in mathematics.
Definition of Function:
Assume that we assign a distinct element from a set X to each element of a set Y. This collection of mappings is known as a function from X into Y. Set X is referred to as the function's domain, while Set Y is referred to as the target set or codomain.
Mathematical Representation of Function:
In most cases, symbols are used to represent functions. Let f is a function from set X to set Y write,
f: X →Y
Example of Function:
Consider the function f (x) = x^2, i.e., f assigns to each real number its square. Then the image of 2 is 4, and so we may write f (2) = 4.
Functions as Relations:
A relation from X to Y (i.e., a subset of X × Y) is defined by the function f: X→ Y such that each x ∈ X is a member of a unique ordered pair (x, y) in f.
Although we do not distinguish between a function and its graph, we will still refer to f as a collection of ordered pairs as "graph of f". Additionally, as the graph of f is a relation, its picture can be drawn, just like relations in general, and this picture-based representation is sometimes referred to as the graph of f.
Additionally, the geometric requirement that each vertical line cross the graph at exactly one point is identical to the defining condition of a function, which states that each a X belongs to a certain pair (x, y) in f.
Composition of Functions:
Suppose that a functions f: X → Y and g: K → L; that is, where the codomain of f is the domain of g. Then we may define a new function from X to L, called the composition of f and g and written gof, as follows: (gof) (x) ≡ g (f (x))
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## Types of Functions:
There are three types of functions.
• One-to-One Function.
• Onto Function.
• Invertible Function.
## One-to-One Function (Injective)
Definition:
A function f: X → Y is said to be one-to-one (1-1) if different elements in the domain X have distinct images. The same thing may also be expressed as that f is one-to-one if f (x) = f (x`) implies x = x`.
Example:
The function f(x) = x + 2 is a one-to-one function because it produces different output for a different input of x. Put x = 1
F(x) = 1+2
F(x) = 3
Put x = 7
F(x) = 7+2
F(x) = 9 and so on….
## Onto Function (SURJECTIVE)
Definition:
A function f: X → Y is said to be onto function if each element of Y is the image of some element of X. In other words, f: X → Y is onto if the image of f is the entire codomain, i.e., if f (X) = Y. In such a case we say that f is a function from X onto Y or that f maps X onto Y.
## Invertible Function
Definition:
A function f: X → Y is invertible if it’s inverse relation f ^-1 is a function from X to Y. In other words, if a function f satisfies the conditions of into as well as onto then the function f is said to be invertible function.
Theorem:
A function f: X → Y is invertible if and only if f is both one-to-one and onto.
## Geometrical Characterization of One-to-One and Onto Functions:
Now think about functions with the notation f: R→ R because such graph of function can be plotted in the Cartesian plane R2 and because functions can be recognized by their graphs. if the concepts of being one-to-one and onto have any geometrical significance.
Accordingly,
(1) If each horizontal line crosses the graph of f at most once, then f:R →R is one-to-one.
(2) Each horizontal line must intersect the graph of f at one or more points in order for it to be an onto function.
Specifically, each horizontal line will cross the graph if f is both one-to-one and onto, or invertible at exactly one point.
This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.
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