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# Two-point Form of a Line
We will discuss here about the method of finding the equation of a straight line in the two point form.
To find the equation of a straight line in the two point form,
Let AB be a line passing through two points A (x$$_{1}$$, y$$_{1}$$) and B (x$$_{2}$$, y$$_{2}$$).
Let the equation of the line be y = mx + c ................... (i), where m is the slope of the line and c is the y-intercept.
As (x$$_{1}$$, y$$_{1}$$) and (x$$_{2}$$, y$$_{2}$$) are points on the line AB, (x$$_{1}$$, y$$_{1}$$) and (x$$_{2}$$, y$$_{2}$$) satisfy (i).
Therefore, y$$_{1}$$ = mx$$_{1}$$ + c ................................ (ii)
and y$$_{2}$$ = mx$$_{2}$$ + c ................................ (iii)
Subtracting (iii) from (ii),
y$$_{1}$$ - y$$_{2}$$ = m(x$$_{1}$$ - x$$_{2}$$)
⟹ m = $$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$ ................................ (iv)
Substituting m = $$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$ in (ii),
y$$_{1}$$ = [$$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$]x$$_{1}$$ + c
⟹ c = y$$_{1}$$ - $$\frac{x_{1}(y_{1} - y_{2})}{ x_{1} - x_{2}}$$
c = $$\frac{ y_{1}(x_{1} - x_{2}) - x_{1}(y_{1} - y_{2})}{ x_{1} - x_{2}}$$
c = $$\frac{x_{1}y_{2} - x_{2}y_{1}}{ x_{1} - x_{2}}$$
Therefore, from (i),
y = [$$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$]x + $$\frac{x_{1}y_{2} - x_{2}y_{1}}{ x_{1} - x_{2}}$$
Subtracting y$$_{1}$$ from both sides of (v)
y - y$$_{1}$$ = [$$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$]x + $$\frac{x_{1}y_{2} - x_{2}y_{1}}{ x_{1} - x_{2}}$$
y - y$$_{1}$$ = [$$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$]x + $$\frac{x_{1}(y_{2} - y_{1})}{ x_{1} - x_{2}}$$
y - y$$_{1}$$ = $$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$(x + x$$_{1}$$)
The equation of the straight line passing through (x1, y1) and (x2, y2) is y - y$$_{1}$$ = $$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$(x + x$$_{1}$$)
Note: From (iv), the slope of the line joining the points (x$$_{1}$$, y$$_{1}$$) and (x$$_{2}$$, y$$_{2}$$) is $$\frac{y_{1} - y_{2}}{x_{1} - x_{2}}$$ i.e., $$\frac{Difference of y-coordinates}{difference of x-coordinates in the same order}$$
Solved example on two-point form of a line:
The equation of the line passing through the points (1, 1) and (-3, 2) is
y - 1 = $$\frac{1 - 2}{1 - (-3)}$$(x - 1)
⟹ y – 1 = -$$\frac{1}{4}$$(x – 1)
Also, y – 2 = $$\frac{2 - 1}{-3 - 1}$$(x + 3)
⟹ y – 2 = -$$\frac{1}{4}$$(x + 3)
However, the two equations are the same.
`
Equation of a Straight Line
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# Precalculus : Add, Subtract, Multiply, and Divide Functions
## Example Questions
### Example Question #1 : Algebra Of Functions
Fully expand the expression:
None of the other answers
Explanation:
The first step is to rewrite the expression:
Now that it is expanded, we can FOIL (First, Outer, Inner, Last) the expression:
First :
Outer:
Inner:
Last:
Now we can simply add up the values to get the expanded expression:
### Example Question #2 : Add, Subtract, Multiply, And Divide Functions
Evaluate
None of the other answers
Explanation:
When adding two expressions, you can only combine terms that have the same variable in them.
In this question, we get:
Now we can add each of the results to get the final answer:
### Example Question #3 : Add, Subtract, Multiply, And Divide Functions
Simplify the following expression:
.
Explanation:
First, we can start off by factoring out constants from the numerator and denominator.
The 9/3 simplifies to just a 3 in the numerator. Next, we factor the top numerator into , and simplify with the denominator.
We now have
### Example Question #4 : Add, Subtract, Multiply, And Divide Functions
Simplify the expression:
.
Explanation:
First, distribute the -5 to each term in the second expression:
Next, combine all like terms
to end up with
.
### Example Question #5 : Add, Subtract, Multiply, And Divide Functions
If and , what does equal?
Explanation:
We begin by factoring and we get .
Now, When we look at it will be .
We can take out from the numerator and cancel out the denominator, leaving us with .
### Example Question #6 : Add, Subtract, Multiply, And Divide Functions
If and , then what is equal to?
Explanation:
First, we must determine what is equal to. We do this by distributing the 3 to every term inside the parentheses,.
Next we simply subtract this from , going one term at a time:
Finally, combining our terms gives us .
### Example Question #7 : Add, Subtract, Multiply, And Divide Functions
Fully expand the expression:
Correct answer not listed
Explanation:
In order to fully expand the expression , let's first rewrite it as:
.
Then, using the FOIL(First, Outer, Inner, Last) Method of Multiplication, we expand the expression to:
First:
Outer:
Inner:
Last:
, which in turn
### Example Question #8 : Add, Subtract, Multiply, And Divide Functions
Simplify the following expression:
No correct answer listed
Explanation:
To simplify the above expression, we must combine all like terms:
Integers:
Putting all of the above terms together, we simplify to:
### Example Question #9 : Add, Subtract, Multiply, And Divide Functions
If and , what is ?
Explanation:
Given the information in the above problem, we know that:
Factoring the resulting fraction, we get:
### Example Question #10 : Add, Subtract, Multiply, And Divide Functions
Simplify the following:
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Patterns in Whole Numbers
# Patterns in Whole Numbers
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• Patterns in Whole Numbers
• Line
• Square
• Rectangle
• Triangle
• What’s Next?
In the previous segment, we learnt some tricks to add, subtract and multiply whole numbers. In this segment, we will look at another way to represent whole numbers.
## How can whole numbers be represented as patterns?
We have represented whole numbers on a number. These numbers can be represented using elementary shapes like line, square, rectangle and triangle.
Let us understand how this can be done using dots. That is we will arrange dots to form the patterns of elementary shapes.
## Dots to form a line
Dots can be arranged in a line to represent numbers.
For example, a single dot for 1, two dots for 2, three dots for 3 and so on.
Line to represent whole numbers
Dots to form a square
Dots arranged in a square can represent whole numbers.
The minimum number of dots that can represent a side of s square is 2. This means the smallest number represented as a square will be, 2 x 2, which is 4.
Dots to form a rectangle
A rectangle has unequal adjacent sides.
Thus, the smallest rectangle can be represented using 2 x 3 = 6 dots. This represents number 6.
Similarly, 2 x 5 = 10 represents the number 10 and so on.
2 x 3 = 6
2 x 4 = 8
2 x 5 =
3 x 4 =
Whole numbers as a rectangle
Dots to form a triangle
To use dots to form a triangle to represent a whole number, three rules have to be followed:
• The top row always has one dot.
• Moving from top to bottom, the number of dots in consecutive rows increases by one.
• Two sides of the triangle are equal.
Based on the rules the smallest number that can be represented as a triangle is 3. Similarly, the numbers that can be represented as triangles are 6, 10, 15 and so on
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# 1.4: Solving Right Triangles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Solve right triangles.
• Find the area of any triangle using trigonometry.
• Solve real-world problems that require you to solve a right triangle.
• Find angle measures using inverse trigonometric functions.
## Solving Right Triangles
You can use your knowledge of the Pythagorean Theorem and the six trigonometric functions to solve a right triangle. Because a right triangle is a triangle with a 90 degree angle, solving a right triangle requires that you find the measures of one or both of the other angles. How you solve will depend on how much information is given. The following examples show two situations: a triangle missing one side, and a triangle missing two sides.
Example 1: Solve the triangle shown below.
Solution:
We need to find the lengths of all sides and the measures of all angles. In this triangle, two of the three sides are given. We can find the length of the third side using the Pythagorean Theorem:
82+b264+b2b2b=102=100=36=±6b=6\begin{align*}8^2 + b^2 & = 10^2\\ 64 + b^2 & = 100\\ b^2 & = 36\\ b & = \pm 6 \Rightarrow b = 6\end{align*}
(You may have also recognized the “Pythagorean Triple,” 6, 8, 10, instead of carrying out the Pythagorean Theorem.)
You can also find the third side using a trigonometric ratio. Notice that the missing side, b\begin{align*}b\end{align*}, is adjacent to A\begin{align*}\angle{A}\end{align*}, and the hypotenuse is given. Therefore we can use the cosine function to find the length of b\begin{align*}b\end{align*}:
cos53.130.6b=adjacent sidehypotenuse=b10=b10=0.6(10)=6\begin{align*}\cos 53.13^\circ & = \frac{\text{adjacent side}}{\text{hypotenuse}} = \frac{b}{10}\\ 0.6 & = \frac{b}{10}\\ b & = 0.6(10) = 6\end{align*}
We could also use the tangent function, as the opposite side was given. It may seem confusing that you can find the missing side in more than one way. The point is, however, not to create confusion, but to show that you must look at what information is missing, and choose a strategy. Overall, when you need to identify one side of the triangle, you can either use the Pythagorean Theorem, or you can use a trig ratio.
To solve the above triangle, we also have to identify the measures of all three angles. Two angles are given: 90 degrees and 53.13 degrees. We can find the third angle using the Triangle Sum Theorem, 1809053.13=36.87\begin{align*}180 - 90 - 53.13 = 36.87^\circ\end{align*}.
Now let’s consider a triangle that has two missing sides.
Example 2: Solve the triangle shown below.
Solution:
In this triangle, we need to find the lengths of two sides. We can find the length of one side using a trig ratio. Then we can find the length of the third side by using a trig ratio with the same given information, not the side we solved for. This is because the side we found is an approximation and would not yield the most accurate answer for the other missing side. Only use the given information when solving right triangles.
We are given the measure of angle A\begin{align*}A\end{align*}, and the length of the side adjacent to angle A\begin{align*}A\end{align*}. If we want to find the length of the hypotenuse, c\begin{align*}c\end{align*}, we can use the cosine ratio:
cos40cos40ccos40c=adjacenthypotenuse=6c=6c=6=6cos407.83\begin{align*}\cos 40^\circ & = \frac{adjacent}{hypotenuse} = \frac{6}{c}\\ \cos 40^\circ & = \frac{6}{c}\\ c \cos 40^\circ & = 6\\ c & = \frac{6}{\cos 40^\circ} \approx 7.83\end{align*}
If we want to find the length of the other leg of the triangle, we can use the tangent ratio. This will give us the most accurate answer because we are not using approximations.
tan40a=oppositeadjacent=a6=6tan405.03\begin{align*}\tan 40^\circ & = \frac{opposite}{adjacent} = \frac{a}{6}\\ a & = 6 \tan 40^\circ \approx 5.03\end{align*}
Now we know the lengths of all three sides of this triangle. In the review questions, you will verify the values of c\begin{align*}c\end{align*} and a\begin{align*}a\end{align*} using the Pythagorean Theorem. Here, to finish solving the triangle, we only need to find the measure of B:1809040=50\begin{align*}\angle{B}: 180 - 90 - 40 = 50^\circ\end{align*}
Notice that in both examples, one of the two non-right angles was given. If neither of the two non-right angles is given, you will need a new strategy to find the angles.
## Inverse Trigonometric Functions
Consider the right triangle below.
From this triangle, we know how to determine all six trigonometric functions for both C\begin{align*}\angle{C}\end{align*} and T\begin{align*}\angle{T}\end{align*}. From any of these functions we can also find the value of the angle, using our graphing calculators. If you look back at #7 from 1.3, we saw that sin30=12\begin{align*}\sin 30^\circ = \frac{1}{2}\end{align*}. If you type 30 into your graphing calculator and then hit the \text{SIN}\begin{align*}\fbox{\text{SIN}}\end{align*} button, the calculator yields 0.5. (Make sure your calculator’s mode is set to degrees.)
Conversely, with the triangle above, we know the trig ratios, but not the angle. In this case the inverse of the trigonometric function must be used to determine the measure of the angle. These functions are located above the SIN, COS, and TAN buttons on the calculator. To access this function, press 2nd\begin{align*}2^{nd}\end{align*} and the appropriate button and the measure of the angle appears on the screen.
cosT=2425cos12425=T\begin{align*}\cos T = \frac{24}{25} \rightarrow \cos^{-1} \frac{24}{25} = T\end{align*} from the calculator we get
Example 3: Find the angle measure for the trig functions below.
a. sinx=0.687\begin{align*}\sin x = 0.687\end{align*}
b. tanx=43\begin{align*}\tan x = \frac{4}{3}\end{align*}
Solution: Plug into calculator.
a. sin10.687=43.4\begin{align*}\sin^{-1} 0.687 = 43.4^\circ\end{align*}
b. tan143=53.13\begin{align*}\tan^{-1} \frac{4}{3} = 53.13^\circ\end{align*}
Example 4: You live on a farm and your chore is to move hay from the loft of the barn down to the stalls for the horses. The hay is very heavy and to move it manually down a ladder would take too much time and effort. You decide to devise a make shift conveyor belt made of bed sheets that you will attach to the door of the loft and anchor securely in the ground. If the door of the loft is 25 feet above the ground and you have 30 feet of sheeting, at what angle do you need to anchor the sheets to the ground?
Solution:
From the picture, we need to use the inverse sine function.
sinθsinθsin1(sinθ)θ=25 feet30 feet=0.8333=sin10.8333=56.4\begin{align*}\sin \theta & = \frac{25 \ feet}{30 \ feet}\\ \sin \theta & = 0.8333\\ \sin^{-1} (\sin \theta) & = \sin ^{-1} 0.8333\\ \theta & = 56.4^\circ\end{align*}
The sheets should be anchored at an angle of 56.4\begin{align*}56.4^\circ\end{align*}.
## Finding the Area of a Triangle
In Geometry, you learned that the area of a triangle is A=12bh\begin{align*}A=\frac{1}{2}bh\end{align*}, where b\begin{align*}b\end{align*} is the base and h\begin{align*}h\end{align*} is the height, or altitude. Now that you know the trig ratios, this formula can be changed around, using sine.
Looking at the triangle above, you can use sine to determine h,sinC=ha\begin{align*}h, \sin C = \frac{h}{a}\end{align*}. So, solving this equation for h\begin{align*}h\end{align*}, we have asinC=h\begin{align*}a\sin C=h\end{align*}. Substituting this for h\begin{align*}h\end{align*}, we now have a new formula for area.
A=12absinC\begin{align*}A = \frac{1}{2} ab \sin C\end{align*}
What this means is you do not need the height to find the area anymore. All you now need is two sides and the angle between the two sides, called the included angle.
Example 5: Find the area of the triangle.
a.
Solution: Using the formula, A=12 absinC\begin{align*}A = \frac{1}{2} \ ab \sin C\end{align*}, we have
A=12813sin82=4130.990=51.494\begin{align*}A & = \frac{1}{2} \cdot 8 \cdot 13 \cdot \sin 82^\circ\\ & = 4 \cdot 13 \cdot 0.990\\ & = 51.494\end{align*}
Example 6: Find the area of the parallelogram.
Solution: Recall that a parallelogram can be split into two triangles. So the formula for a parallelogram, using the new formula, would be: A=212 absinC\begin{align*}A = 2 \cdot \frac{1}{2} \ ab \sin C\end{align*} or A=absinC\begin{align*}A = ab \sin C\end{align*}.
A=715sin65=95.162\begin{align*}A & = 7 \cdot 15 \cdot \sin 65^\circ\\ & = 95.162\end{align*}
## Angles of Elevation and Depression
You can use right triangles to find distances, if you know an angle of elevation or an angle of depression. The figure below shows each of these kinds of angles.
The angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object. For example, if you are standing on the ground looking up at the top of a mountain, you could measure the angle of elevation. The angle of depression is the angle between the horizontal line of sight and the line of sight down to an object. For example, if you were standing on top of a hill or a building, looking down at an object, you could measure the angle of depression. You can measure these angles using a clinometer or a theodolite. People tend to use clinometers or theodolites to measure the height of trees and other tall objects. Here we will solve several problems involving these angles and distances.
Example 7: You are standing 20 feet away from a tree, and you measure the angle of elevation to be 38\begin{align*}38^\circ\end{align*}. How tall is the tree?
Solution:
The solution depends on your height, as you measure the angle of elevation from your line of sight. Assume that you are 5 feet tall.
The figure shows us that once we find the value of T\begin{align*}T\end{align*}, we have to add 5 feet to this value to find the total height of the triangle. To find T\begin{align*}T\end{align*}, we should use the tangent value:
tan38tan38THeight of tree=oppositeadjacent=T20=T20=20tan3815.6320.63 ft\begin{align*}\tan 38^\circ & = \frac{opposite}{adjacent} = \frac{T}{20}\\ \tan 38^\circ & = \frac{T}{20}\\ T & = 20 \tan 38^\circ \approx 15.63\\ \text{Height of tree} & \approx 20.63 \ ft\end{align*}
The next example shows an angle of depression.
Example 8: You are standing on top of a building, looking at a park in the distance. The angle of depression is 53\begin{align*}53^\circ\end{align*}. If the building you are standing on is 100 feet tall, how far away is the park? Does your height matter?
Solution:
If we ignore the height of the person, we solve the following triangle:
Given the angle of depression is 53\begin{align*}53^\circ\end{align*}, A\begin{align*}\angle{A}\end{align*} in the figure above is 37\begin{align*}37^\circ\end{align*}. We can use the tangent function to find the distance from the building to the park:
tan37tan37d=oppositeadjacent=d100=d100=100tan3775.36 ft.\begin{align*}\tan 37^\circ & = \frac{opposite}{adjacent} = \frac{d}{100}\\ \tan 37^\circ & = \frac{d}{100}\\ d & = 100 \tan 37^\circ \approx 75.36\ ft.\end{align*}
If we take into account the height if the person, this will change the value of the adjacent side. For example, if the person is 5 feet tall, we have a different triangle:
tan37tan37d=oppositeadjacent=d105=d105=105tan3779.12 ft.\begin{align*}\tan 37^\circ & = \frac{opposite}{adjacent} = \frac{d}{105}\\ \tan 37^\circ & = \frac{d}{105}\\ d & = 105 \tan 37^\circ \approx 79.12\ ft.\end{align*}
If you are only looking to estimate a distance, then you can ignore the height of the person taking the measurements. However, the height of the person will matter more in situations where the distances or lengths involved are smaller. For example, the height of the person will influence the result more in the tree height problem than in the building problem, as the tree is closer in height to the person than the building is.
## Right Triangles and Bearings
We can also use right triangles to find distances using angles given as bearings. In navigation, a bearing is the direction from one object to another. In air navigation, bearings are given as angles rotated clockwise from the north. The graph below shows an angle of 70 degrees:
It is important to keep in mind that angles in navigation problems are measured this way, and not the same way angles are measured in trigonometry. Further, angles in navigation and surveying may also be given in terms of north, east, south, and west. For example, \begin{align*}N70^\circ E\end{align*} refers to an angle from the north, towards the east, while \begin{align*}N70^\circ W\end{align*} refers to an angle from the north, towards the west. \begin{align*}N70^\circ E\end{align*} is the same as the angle shown in the graph above. \begin{align*}N70^\circ W\end{align*} would result in an angle in the second quadrant.
Example 9: A ship travels on a \begin{align*}N50^\circ E\end{align*} course. The ship travels until it is due north of a port which is 10 nautical miles due east of the port from which the ship originated. How far did the ship travel?
Solution: The angle between \begin{align*}d\end{align*} and 10 nm is the complement of \begin{align*}50^\circ\end{align*}, which is \begin{align*}40^\circ\end{align*}. Therefore we can find \begin{align*}d\end{align*} using the cosine function:
\begin{align*}\cos 40^\circ & = \frac{adjacent}{hypotenuse} = \frac{10}{d}\\ \cos 40^\circ & = \frac{10}{d}\\ d \cos 40^\circ & = 10\\ d & = \frac{10}{\cos 40^\circ} \approx 13.05 \ nm\end{align*}
## Other Applications of Right Triangles
In general, you can use trigonometry to solve any problem that involves right triangles. The next few examples show different situations in which a right triangle can be used to find a length or a distance.
Example 10: The wheelchair ramp
In lesson 3 we introduced the following situation: You are building a ramp so that people in wheelchairs can access a building. If the ramp must have a height of 8 feet, and the angle of the ramp must be about \begin{align*}5^\circ\end{align*}, how long must the ramp be?
Given that we know the angle of the ramp and the length of the side opposite the angle, we can use the sine ratio to find the length of the ramp, which is the hypotenuse of the triangle:
\begin{align*}\sin 5^\circ & = \frac{8}{L}\\ L \sin 5^\circ & = 8\\ L & = \frac{8}{\sin 5^\circ} \approx 91.8 \ ft\end{align*}
This may seem like a long ramp, but in fact a \begin{align*}5^\circ\end{align*} ramp angle is what is required by the Americans with Disabilities Act (ADA). This explains why many ramps are comprised of several sections, or have turns. The additional distance is needed to make up for the small slope.
Right triangle trigonometry is also used for measuring distances that could not actually be measured. The next example shows a calculation of the distance between the moon and the sun. This calculation requires that we know the distance from the earth to the moon. In chapter 5 you will learn the Law of Sines, an equation that is necessary for the calculation of the distance from the earth to the moon. In the following example, we assume this distance, and use a right triangle to find the distance between the moon and the sun.
Example 11: The earth, moon, and sun create a right triangle during the first quarter moon. The distance from the earth to the moon is about 240,002.5 miles. What is the distance between the sun and the moon?
Solution:
Let \begin{align*}d =\end{align*} the distance between the sun and the moon. We can use the tangent function to find the value of \begin{align*}d\end{align*}:
\begin{align*}\tan 89.85^\circ & = \frac{d}{240,002.5}\\ d & = 240,002.5 \tan 89.85^\circ = 91,673,992.71 \ miles\end{align*}
Therefore the distance between the sun and the moon is much larger than the distance between the earth and the moon.
(Source: www.scribd.com, Trigonometry from the Earth to the Stars.)
## Points to Consider
• In what kinds of situations do right triangles naturally arise?
• Are there right triangles that cannot be solved?
• Trigonometry can solve problems at an astronomical scale as well as problems at a molecular or atomic scale. Why is this true?
## Review Questions
1. Solve the triangle.
2. Two friends are writing practice problems to study for a trigonometry test. Sam writes the following problem for his friend Anna to solve: In right triangle \begin{align*}ABC\end{align*}, the measure of angle \begin{align*}C\end{align*} is 90 degrees, and the length of side \begin{align*}c\end{align*} is 8 inches. Solve the triangle. Anna tells Sam that the triangle cannot be solved. Sam says that she is wrong. Who is right? Explain your thinking.
3. Use the Pythagorean Theorem to verify the sides of the triangle in example 2.
4. Estimate the measure of angle \begin{align*}B\end{align*} in the triangle below using the fact that \begin{align*}\sin B = \frac{3}{5}\end{align*} and \begin{align*}\sin 30^\circ = \frac{1}{2}\end{align*}. Use a calculator to find sine values. Estimate \begin{align*}B\end{align*} to the nearest degree.
5. Find the area of the triangle.
6. Find the area of the parallelogram below.
7. The angle of elevation from the ground to the top of a flagpole is measured to be \begin{align*}53^\circ\end{align*}. If the measurement was taken from 15 feet away, how tall is the flagpole?
8. From the top of a hill, the angle of depression to a house is measured to be \begin{align*}14^\circ\end{align*}. If the hill is 30 feet tall, how far away is the house?
9. An airplane departs city A and travels at a bearing of \begin{align*}100^\circ\end{align*}. City B is directly south of city A. When the plane is 200 miles east of city B, how far has the plan traveled? How far apart are City A and City B? What is the length of the slanted outer wall, \begin{align*}w\end{align*}? What is the length of the main floor, \begin{align*}f\end{align*}?
10. A surveyor is measuring the width of a pond. She chooses a landmark on the opposite side of the pond, and measures the angle to this landmark from a point 50 feet away from the original point. How wide is the pond?
11. Find the length of side \begin{align*}x\end{align*}:
12. A deck measuring 10 feet by 16 feet will require laying boards with one board running along the diagonal and the remaining boards running parallel to that board. The boards meeting the side of the house must be cut prior to being nailed down. At what angle should the boards be cut?
1. \begin{align*}\angle{A} & = 50^\circ\\ b & \approx 5.83\\ a & \approx 9.33\end{align*}
2. Anna is correct. There is not enough information to solve the triangle. That is, there are infinitely many right triangles with hypotenuse 8. For example:
3. \begin{align*}6^2 + 5.03^2 = 36 + 25.3009 = 61.3009 = 7.83^2\end{align*}.
4. \begin{align*}\angle{B \approx} 37^\circ\end{align*}
5. \begin{align*}A = \frac{1}{2} \cdot 10 \cdot 12 \cdot \sin 104^\circ = 58.218\end{align*}
6. \begin{align*}A = 4 \cdot 9 \cdot \sin 112^\circ = 33.379\end{align*}
9. The plane has traveled about 203 miles. The two cities are 35 miles apart.
12. \begin{align*}\tan \theta & = \frac{opposite}{adjacent}\\ \tan \theta & = 0.625\\ \theta & = 32^\circ\end{align*}
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# Class 8 NCERT Solutions – Chapter 16 Playing with Numbers – Exercise 16.1
Last Updated : 24 Nov, 2020
### Question 1.
3 A
+ 2 5
——–
B 2
——–
Solution:
A + 5 we get the unit digit as 2
if A = 7 , we get 7 + 5 = 12
The value of B = 1(carry) + 3 + 2 = 6
Hence the addition is
3 7
+ 2 5
——–
6 2
——–
Hence A = 7 and B = 6
### Question 2.
4 A
+ 9 8
————
C B 3
————
Solution:
A + 8 we get the unit digit as 3
if A = 5, we get A + 8 = 13
The values of B and C = 1(carry) + 4 + 9 = 14
1 is taken as carry
Hence the addition is
4 5
+ 9 8
————
1 4 3
————
Hence A = 5, B = 4 and C = 1
### Question 3.
1 A
X A
————-
9 A
————–
Solution:
A x A = A if A = 1 or 6
For A = 1
1 x 1 = 1 which is not equal to 9
For A = 6
6 x 6 = 36
1 x 6 = 6
Hence the multiplication is
1 6
X 6
————-
9 6
————–
Hence A = 6
### Question 4.
A B
+ 3 7
——–
6 A
——–
Solution:
A + 3 = 6
A = 6 – 3 = 3 = 2 + 1(carry)
A = 2
B + 7 we get the unit digit as 2
if B = 5 we get 5 + 7 = 12
Hence the addition is
2 5
+ 3 7
——–
6 2
———
Hence A = 2 and B = 5
### Question 5.
A B
X 3
————-
C A B
————-
Solution:
B x 3 = B
B = 0
A x 3 = CA
if A = 5 then 5 x 3 = 15
and C = 1
Hence the multiplication is
5 0
X 3
————-
1 5 0
————-
Hence A = 5 , B = 0 and C = 1
### Question 6.
A B
x 5
————-
C A B
————-
Solution:
B x 5 = B
B = 0 or 5
A x 5 = CA
if A = 5 then 5 x 5 = 25
and C = 2
Only possible when B = 0
Hence the multiplication is
5 0
X 5
————-
2 5 0
————-
Hence A = 5 , B = 0 and C = 2
### Question 7.
A B
× 6
————
B B B
————
Solution:
B x 6 = B
if B = 4
4 x 6 = 24
where 2 is carry
A x 6 = BB
if A = 7
7 x 6 = 42
Hence multiplication is
7 4
× 6
————
4 4 4
————-
Hence A = 7 and B = 4
### Question 8.
A 1
+ 1 B
———–
B 0
———–
Solution:
1 + B we get unit digit as 0
if B = 9
1 + 9 = 10
where 1 is the carry
1(carry) + A + 1 = B = 9
A + 2 = 9
A = 9 – 2 = 7
7 1
+ 1 9
———–
9 0
———–
Hence A = 7 and B = 9
### Question 9.
2 A B
+ A B 1
—————
B 1 8
—————
Solution:
B + 1 we get the unit digit as 8
if B = 7
7 + 1 = 8
A + 7 we get the tens digit as 1
if A = 4
4 + 7 = 11
where 1 is the carry
1(carry) + 2 + 4 = 7
2 4 7
+ 4 7 1
—————
7 1 8
—————-
Hence A = 4 and B = 7
### Question 10.
1 2 A
+ 6 A B
—————-
A 0 9
—————-
Solution:
A + B = 9
if A = 8 and B =1
8 + 1 = 9
2 + A = 0
if A = 8
2 + 8 = 10
where 1 is the carry
1(carry) + 1 + 6 = A = 8
Hence the addition is
1 2 8
+ 6 8 1
—————-
8 0 9
—————-
Hence A = 8 and B = 1
Previous
Next
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# Roots of a Quadratic Equation with Real Coefficients
## Roots of a Quadratic Equation with Real Coefficients
An equation of the form ax² + bx + c = 0 … (i)
Where a ≠ 0, a, b, c є R is called quadratic equation with real coefficients.
The quality D = b² – 4ac is known as the discriminant of the quadratic equation in (i) whose roots are given by $$\alpha =\frac{-b+\sqrt{{{b}^{2}}-4ac}}{2a}$$ and $$\beta =\frac{-b-\sqrt{{{b}^{2}}-4ac}}{2a}$$.
The nature of the roots is as given below:
1. The roots are real and distinct if D > 0.
2. The roots are real and equal if D = 0
3. The roots are complex with non-zero imaginary part if D < 0.
4. The roots are rational if a, b, c is rational, and D is a perfect square.
5. The roots are of the form p + √q (p, q є Q) if a, b, c is rational, and D is not a perfect square.
6. If a = 1, b, c є I and the roots are rational numbers, then these roots must be integers.
7. If a quadratic equation in x has more than two roots, then it is an identity in x that is a = b = c = 0.
Example: Solve for real x (x² – 1) (x + 2) + 1 = 0
Solution: we have,
x (x² – 1) (x + 2) + 1
⇒ x (x – 1) (x + 1) (x + 2) + 1 = 0
⇒ [ x(x+1)] [(x-1) (x+2)] + 1=0
⇒ (x² + x) (x² + x -2) +1 =0
⇒ y(y -2) + 1 = 0 where y = x² + x
⇒ y² -2y +1 = 0
⇒ (y – 1)² = 0
⇒ y= 1
⇒ x² + x = 1
⇒ x² + x – 1 = 0
⇒ $$x=\frac{-1\pm \sqrt{5}}{2}$$.
Hence, the roots of the given equations are $$\frac{1-+\sqrt{5}}{2}$$ and $$\frac{-1-\sqrt{5}}{2}$$.
|
# Real Mathematics – Geometry #4
Geometry has a sacred book: Elements. Author of Elements, Euclid, used only these tools when he discovered his geometry:
A compass and an unmarked ruler… He showed what can or can’t be done with them in geometry.
Dividing a Finite Straight Line into Two Equal Parts
We already know that one can draw a straight line segment between any two points. Let’s say we have a straight line between the points A and B. Euclid found an ingenious method to divide AB into two equal parts using his only two tools. Let’s assume that AB is 6 cm.
This way we will know that Euclid’s method works only if we find two parts that are 3 cm long.
According to Euclid one should take point A and point B as the centers of two equal circles with radius AB:
Euclid says that one should define the intersection points as C and D:
Then he suggests one to connect C to D:
At this point Euclid talks about two outcomes:
1. CD is perpendicular to AB.
2. CD divides AB into two equal parts.
As seen in the picture CD really divides AB into two equal parts. One can use this method and see that those two lines are perpendicular with a quadrant.
Dividing a Random Angle into Two Equal Parts
Obviously Euclid didn’t stop there and tried to figure out if it was possible to divide any given angle into two equal parts. Let’s consider straight lines AB and BC intersects and form a 90-degree angle:
Euclid says that one should take B (the intersection point) as center and draw a circle with random radius. This circle will intersect AB and BC at two points: D and E.
Now Euclid tells us to use our compass and draw two equal circles that have centers D and E:
As seen above, these circles intersect at two points. Let’s choose the point F and connect it with B:
Euclid claims that the straight line BF divides the angle into two equal parts:
We can see with our quadrant angle is divided into two equal parts of 45 degrees.
One wonders…
Is it possible to use Euclid’s methods and divide any given straight line into three equal parts?
Try to answer the same thing for an angle.
M. Serkan Kalaycıoğlu
|
# Does a rectangle add up to 360?
D
## Does a rectangle add up to 360?
Angles of Rectangle Properties There are four interior angles, each angle is a right angle. The sum of the interior angles of a rectangle is 360°.
Is a rectangle All 90 degrees?
A rectangle can be defined as a four-sided quadrilateral with all its four angles being 90°. In a rectangle, all the angles are equal and equal to 90 degrees. The diagonals of a rectangle are equal which is not equal in case of a parallelogram.
### Is every shape 360 degrees?
The sum of exterior angles in a polygon is always equal to 360 degrees. Therefore, for all equiangular polygons, the measure of one exterior angle is equal to 360 divided by the number of sides in the polygon.
Is each angle of a rectangle is a right angle?
(e) All the sides of a parallelogram are of equal length. (f) The opposite sides of a trapezium are parallel. a) True, each angle of rectangle is right angle .
## What is the sum of the angles for a rectangle?
As with any crossed quadrilateral, the sum of its interior angles is 720°, allowing for internal angles to appear on the outside and exceed 180°. A rectangle and a crossed rectangle are quadrilaterals with the following properties in common: Opposite sides are equal in length.
What is the sum of all angles of a rectangle?
360°
To find the sum of interior angles of a rectangle we can directly use the sum of angles formula. By putting the value of n = 4 we can easily get the answer. Sum of interior angles of a rectangle = ( n − 2) × 180° = (4 – 2) × 180° = 2 × 180° = 360°.
### Do rectangles have all right angles?
The angles of a rectangle are all congruent (the same size and measure.) Remember that a 90 degree angle is called a “right angle.” So, a rectangle has four right angles.
Are all 4 sided figures 360 degrees?
You would find that for every quadrilateral, the sum of the interior angles will always be 360°. Since the sum of the interior angles of any triangle is 180° and there are two triangles in a quadrilateral, the sum of the angles for each quadrilateral is 360°.
## Is angle of a rectangle is?
Opposite sides of a rectangle are the same length (congruent). The angles of a rectangle are all congruent (the same size and measure.) Remember that a 90 degree angle is called a “right angle.” So, a rectangle has four right angles. Opposite angles of a rectangle are congruent.
How many degrees does a rectangle have in it?
Since a rectangle has four angles of equal measure, the measure of each must be 360/4, or 90 degrees. Against this background, does a rectangle add up to 180? This rectangle has four 90 degree angles adding up to 360 degrees. Since the triangles are congruent each triangle has half as many degrees, namely 180.
### What are the internal angles of a rectangle?
A Rectangle is a four sided-polygon, having all the internal angles equal to 90 degrees. The two sides at each corner or vertex, meet at right angles.
How are the angles of a rectangle congruent?
The angles of a rectangle are all congruent (the same size and measure.) Remember that a 90 degree angle is called a “right angle.” So, a rectangle has four right angles. Opposite angles of a rectangle are congruent. Opposite sides of a rectangle are parallel.
## Which is the correct definition of a rectangle?
The distance between A and B or C and D is defined as the length (L), whereas the distance between B and C or A and D is defined as Width (W) of the given rectangle. A rectangle is a type of quadrilateral that has its parallel sides equal to each other and all the four vertices are equal to 90 degrees.
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# How do you graph y = sqrt (3x) + 4 by plotting point?
Oct 7, 2017
$y = \sqrt{3 x} + 4$
graph{sqrt(3x)+4 [-10.38, 9.62, -5, 5]}
$y = \sqrt{3} x + 4$
graph{sqrt3x+4 [-10.38, 9.62, -5, 5]}
#### Explanation:
1) If given equation is $y = \sqrt{3 x} + 4$
$x = 0 , y = 4$
$x = 1 , y = \sqrt{3} + 4$
$x = 2 , y = \sqrt{6} + 4$
$x = 3 , y = 7$
Four points are $\left(0 , 4\right) , \left(1 , 5.732\right) , \left(2 , 6.449\right) , \left(3 , 7\right)$
2) If given equation is $y = \sqrt{3} \cdot x + 4$
Giving values for x, find corresponding value of y from the above equation and plot in a graph.
$x = 0 , y = 4$
$x = 1 , y = \sqrt{3} + 4$
$x = 2 , y = 2 \sqrt{3} + 4$
$x = - 1 , y = - \sqrt{3} + 4$
Four points are $\left(0 , 4\right) , \left(1 , 5.732\right) , \left(2 , 7.464\right) , \left(- 1 , 2.268\right)$
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Select Page
Understanding 72 as a Fraction
When we talk about numbers, they can often be expressed in different forms. One of the common alternative forms for numbers is fractions. In this article, we will explore how to express the number 72 as a fraction, both in the simplest form and in decimal form.
Expressing 72 as a Fraction
To express 72 as a fraction, we can simply write it as:
72/1
This is because any whole number can be expressed as that number over 1 when written as a fraction. However, this form is not the simplest representation of 72 as a fraction.
Simplifying 72 As A Fraction
To simplify the fraction 72/1, we need to find the greatest common divisor (GCD) of the numerator (72) and the denominator (1), which is 1. Then we divide both the numerator and the denominator by 1:
Original Fraction Simplified Fraction
72/1 72/1
As we can see, the fraction 72/1 is already in its simplest form since the GCD of 72 and 1 is 1. Therefore, the simplified fraction of 72/1 remains 72/1.
Credit: www.amazon.com
Converting 72 to Decimal
Another way to represent the number 72 is in decimal form. To convert 72 to a decimal, we simply divide 72 by 1:
72 ÷ 1 = 72.000
So, the decimal representation of the whole number 72 is 72.000. It’s important to note that the decimal part is .000 because 72 is a whole number and does not have any fractional or decimal part.
In Conclusion
In conclusion, the number 72 can be expressed as the fraction 72/1, which is already in its simplest form. Additionally, when we convert 72 to a decimal, it becomes 72.000. Understanding how to express numbers in different forms allows us to work with them effectively in various mathematical and real-world contexts.
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# Length
```Length
and
UNITS OF
MEASUREMENT
LESSON TARGET:
Understand that there are different
measuring units and that those units
can be converted into other units.
1.
What is length?
2. When I want to know
how long something is,
what do I do?
Length can be measured in
different units.
millimeter
mm
centimeters
cm
meters
m
kilometer
km
When we look at a ruler which 2 units can
you spot on it ?
When we want to measure the length of a
wall, what unit do you think we could use?
What unit would we use if we wanted to
measure how far a car has driven?
measuring tools:
trundle
wheel
Ruler
measuring
tape
speedometer
CONVERSION:
E
N
O
E
K
A
T
E
E
G
N
W
A
N
H
E
C
H
D
W
N
S
A
I
T
N
I
T
I
O
N
I
N
U
S
U
R
T
R
E
N
E
V
E
H
N
M
T
O
E
O
C
R
N
U
A
S
O
A
T
N
ME
I
IT
10 mm = 1 cm
100 cm = 1 m
1000 m= 1 km
King
henry
died of a
Miserable
disease
Called
Measles
km
m
cm
mm
Km H D M D Cm Mm
Km H D M D Cm Mm
10 mm = 1 cm
100 cm = 1 m
1000 m= 1 km
Km H D M D Cm Mm
500. (m)
I measured 500m now I want to convert
it into km.
To do this we will use the following
steps.
Step 1: check
Step 2: point to where we want to go
There is always an invisible decimal point
at the and of numbers.
Step 3: Jump
Km H D M D Cm Mm
0,5 0 0. (m)
There is always an invisible decimal point
at the and of numbers.
After we completed these steps we can
move on to converting our number.
The spaces we counted between the
units will indicate how many spaces our
decimal point will move and the direction
that it will move.
Lets practice:
25cm to mm
50mm to cm
2000m to km
K H
D
M
D C
M
When we want to record measurements we write
down the number and the unit.
The length of this room is 6m and 35cm.
6m
Length (how long?)
35cm
width (how wide?)
How do we measure?
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
cm
```
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# How to find the area of the parallelogram?
Leonid Veselov
August 16, 2012
26998
Before we learn how to find the area of a parallelogram, we need to remember what a parallelogram is and what is called its height. A parallelogram is a quadrilateral whose opposite sides are in pairs parallel (lie on parallel straight lines). A perpendicular drawn from an arbitrary point on the opposite side to a line containing this side is called the height of the parallelogram.
The square, the rectangle and the rhombus are special cases of the parallelogram.
The area of the parallelogram is denoted by (S).
## Formulas for finding the area of a parallelogram
S = a * hwhere a is the base, h is the height that is drawn to the base.
S = a * b * sinα, where a and b are bases, and α is the angle between the bases of a and b.
S = p * rwhere p is a semi-perimeter, r is the radius of the circle, which is inscribed in the parallelogram.
The area of the parallelogram, which is formed by the vectors a and b, is equal to the modulus of the product of the specified vectors, namely:
S = | a x b |
Consider the example number 1: Given a parallelogram whose side is 7 cm and a height of 3 cm. How to find the area of the parallelogram, the formula for the solution we need.
Decision:
S = a * h
Thus, S = 7x3. S = 21. Answer: 21 cm2.
Consider the example number 2: Given the base of 6 and 7 cm, and also given the angle between the bases of 60 degrees. How to find the area of the parallelogram? The formula used to solve:
S = a * b * sinα
Thus, we first find the sine of the angle. Sine 60 = 0.5, respectively S = 6 * 7 * 0.5 = 21 Answer: 21 cm2.
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# Trig identity proofs solver
Here, we debate how Trig identity proofs solver can help students learn Algebra. Keep reading to learn more!
## The Best Trig identity proofs solver
Math can be a challenging subject for many students. But there is help available in the form of Trig identity proofs solver. There are a few steps to solving log equations. First, identify what the base of the log is. This can usually be done by looking at the coefficients in front of the log. Once the base is identified, use the following properties to solve the equation: - if the base is the same on both sides, then the arguments must be equal - if the base is different on both sides, then you can use algebra to solve for the argument - if there is only
Some math problems can be solved by scanning them and looking for patterns. This can be a quick and easy way to solve some problems without having to do any difficult calculations. However, this method will not work for all problems and some may require more work to solve.
There are a few different methods that can be used to solve quadratics by factoring. One method is to factor the quadratic equation into two linear equations. Another method is to use the quadratic formula. To factor a quadratic equation, one must first determine the greatest common factor of the terms. Then, the terms are divided by the greatest common factor and the equation is rewritten. The next step is to find two factors of the leading coefficient that add
Algebra is a fundamental tool in mathematics that allows us to solve equations and understand the relationships between variables. Basic algebra problems usually involve solving for a missing variable in an equation. For example, if we are given the equation "x + 5 = 10", we can solve for "x" by subtracting 5 from both sides of the equation to get "x = 5". Algebra can be used to solve for many different types of variables, making it a very powerful tool.
There are a few steps to solving linear equations: 1. First, you need to identify the equation's variables. 2. Next, you'll want to use algebraic methods to solve for the variable. 3. Once you've solved for the variable, you can plug the answer back into the equation to check your work.
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# GSEB Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.2
Gujarat Board Statistics Class 11 GSEB Solutions Chapter 4 Measures of Dispersion Ex 4.2 Textbook Exercise Questions and Answers.
## Gujarat Board Textbook Solutions Class 11 Statistics Chapter 4 Measures of Dispersion Ex 4.2
Question 1.
A shooter missed his target in the last 10 trials by the following distance (mm) during the practice session:
20, 32, 24, 41, 18, 27, 15, 36, 35, 25
Find the quartile deviation and coefficient of quartile deviation of such distance missed by the shooter.
Writing the measures of mistargets in ascending order :
15, 18, 20, 24, 25, 27, 32, 35, 36, 41
Here, n = 10
First Quartile :
Q1 = Value of $$\left(\frac{n+1}{4}\right)$$th observation
= Value of $$\left(\frac{10+1}{4}\right)$$ = 2.75 th observation
= Value of 2nd observation +0.75 (Value of 3rd observation – Value of 2nd observation)
= 18 + 0.75 (20 – 18)
= 18 + 0.75 (2)
= 18 + 1.50 = 19.50 mm
Third Quartile :
Q3 = Value of 3$$$\left(\frac{n+1}{4}\right)$$$ th observation
= Value of 3 (2.75) = 8.25th observation
= Value of 8th observation + 0.25 (Value of 9th observation – Value of 8th observation)
= 35 + 0.25 (36-35)
= 35 + 0.25 = 35.25 mm
Quartile deviation of measures of mistargets :
Qd = $$\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{2}$$
Putting Q3 = 35.25; Q1 = 19.50, we get
Qd = $$\frac{35.25-19.50}{2}=\frac{15.75}{2}$$ = 7.875 ≈ 7.88
Coefficient of quartile deviation = $$\frac{\mathrm{Q}_{3}-\mathrm{Q}_{1}}{\mathrm{Q}_{3}+\mathrm{Q}_{1}}$$
= $$\frac{35.25-19.50}{35.25+19.50}$$
= $$\frac{15.75}{54.75}$$
= 0.29
Question 2.
Find the quartile deviation and coefficient of quartile deviation of the marks from the following frequency distribution of marks of 43 students of a school;
First Quartile :
Q1 = Value of $$\left(\frac{n+1}{4}\right)$$th observation
= Value of $$\left(\frac{43+1}{4}\right)=\frac{44}{4}$$ = 11th observation
Referring to column cf,
Q1 = 20 marks
Third Quartile :
Q3 = Value of $$\left(\frac{n+1}{4}\right)$$th observation
= Value of 3(11)
= 33 rd observation Referring to column cf,
Q3 = 40 marks
Quartile deviation of marks:
Question 3.
The distribution of amount paid by 200 customers coming for snacks at a restaurant on a particular day is given below:
Find the quartile deviation and coefficient of quartile deviation of the amount paid by customers on the day.
Amount paid ₹ No. of customers f Cumulative frequency cf 0-50 25 25 50- 100 40 65 100-150 80 145 150-200 30 175 200 – 250 25 200 Total n = 200 –
First Quartile:
Q1 Class = Class that Includes $$\left(\frac{n}{4}\right)$$th observation
= Class that includes $$\left(\frac{200}{4}\right)$$ = 50th observation
Referring to column cf
Q1 class = 50 – 100
Now, Q1 = L + $$\frac{\left(\frac{n}{4}\right)-c f}{f}$$ × c
Putting L = 50: $$\left(\frac{n}{4}\right)$$ = 50: cf = 25: f = 40 and c = 50 In the formula,
Third Quartile:
Q3 Class = Class that Includes 3$$\left(\frac{n}{4}\right)$$th observation
= Class that Includes 3 (50)th = 150th observation
Referring to column cf. Q3 class = 150 – 200
Now. Q3 = L + $$\frac{3\left(\frac{n}{4}\right)-c f}{f}$$ × c
Putting L = 150: 3$$\left(\frac{n}{4}\right)$$ = 150: cf = 145; f = 30 and c = 50 in the formula.
|
# Pursuit Curve. Dog Chases Rabbit. Calculus 4.
(a) In Example 1.21, assume that $a$ is less than $b$ (so that $k$ is less than $1$) and find $y$ as a function of $x$. How far does the rabbit run before the dog catches him?
(b) Assume now that $a=b$, and find $y$ as a function of $x$. How close does the dog come to the rabbit?
Example 1.21
A rabbit begins at the origin and runs up the $y-axis$ with speed $a$ feet per second. At the same time, a dog runs at speed $b$ from the point $(c,0)$ in pursuit of the rabbit. What is the path of the dog?
Solution: At time $t$, measured from the instant both the rabbit and the dog start, the rabbit will be at the point $R=(0,at)$ and the dog at $D=(x,y)$. We wish to solve for $y$ as a function of $x$.
$$\frac{dy}{dx}=\frac{y-at}{x}$$
$$xy'-y=-at$$
$$xy''=-a\frac{dt}{dx}$$
Since the $s$ is a arc length along the path of the dog, it follows that $\frac{ds}{dt}=b$. Hence,
$$\frac{dt}{dx}=\frac{dt}{ds}\frac{ds}{dx}=\frac{-1}{b}\sqrt{1+(y')^2}$$
$$xy''=\frac{a}{b}\sqrt{1+(y')^2}$$
For convenience, we set $k=\frac{a}{b}$, $y'=p$, and $y''=\frac{dp}{dx}$
$$\frac{dp}{\sqrt{1+p^2}}=k\frac{dx}{x}$$
$$\ln\left({p+\sqrt{1+p^2}}\right)=\ln\left(\frac{x}{c}\right)^k$$
Now, solve for $p$:
$$\frac{dy}{dx}=p=\frac{1}{2}\Bigg(\left(\frac{x}{c}\right)^k-\left(\frac{c}{x}\right)^k\Bigg)$$
In order to continue the analysis, we need to know something about the relative sizes of $a$ and $b$. Suppose, for example, that $a \lt$ $b$ (so $k\lt$ $1$), meaning that the dog will certainly catch the rabbit. Then we can integrate the last equation to obtain:
$$y(x)=\frac{1}{2}\Bigg\{\frac{c}{k+1}\left(\frac{x}{c}\right)^{k+1}-\frac{c}{1-k}\left(\frac{c}{x}\right)^{k-1}\Bigg\}+D$$
Again, this is all I have to go on. I need to answer questions (a) and (b) stated at the top.
-
You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. There's an "edit" link under the question. – joriki Sep 20 '12 at 4:35
Thank you. I wasn't sure how to do that. The link you provided greatly helped me. I fixed everything the best I could. Can you or someone else help me now? Thanks again. – Pink Panda Sep 20 '12 at 15:53
I could really use a hint or something. – Pink Panda Sep 20 '12 at 18:53
I don't understand -- (a) and (b) refer to Example 1.18 but talk about variables $a$ and $b$ that don't occur in Example 1.18 -- but they do occur in Example 1.21. Perhaps you could explain more about the context? – joriki Sep 20 '12 at 21:59
By the way, you can get the appropriate size for a pair of parentheses (or other pairs of delimiters, like brackets, braces, absolute value bars) by preceding them with \left and \right, respectively. – joriki Sep 20 '12 at 22:08
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# How to Create a Tape Diagram
How to create a tape diagram? You can quickly solve word problems and better understand various math concepts with a tape diagram, and it also helps you simplify complex problems and understand mathematical relationships.
## 1. Introduction
Knowing how to draw and solve mathematic problems using a tape diagram is essential for every student. It helps students understand math concepts and depict relationships in a math problem. Students can use it to communicate their thoughts on any problem and sort out its contents and connections to find an easy and simple solution. These diagrams are mainly used to help students solve any word problem. Our guide will help you understand how to do a tape diagram.
## 2. How to Create a Tape Diagram
You can easily create a tape diagram by following a few basic steps. It visually illustrates complex problems in a taped segment divided into smaller rectangular sections. Students from all grades can use these diagrams to solve mathematic problems and equations. Knowing about tape diagrams is very helpful to students who move to new grades and deal with advanced math. Tape diagrams break down any complex word problem into simple parts. Follow these five steps to draw a tape diagram.
Step 1: Understand the topic
The first step to creating a tape diagram is determining why you need it or what problem you are trying to solve. Understanding the topic is about carefully reading the given problem statement or math equation. After that, you will understand the given problem and the steps to solve that problem. In math, you see various types of equations and problems, and that is why it is essential to analyze them before drawing a diagram to solve them.
Step 2: Divide the problem into sections
After reading the problem statement, the next step is to divide the problem into sections to find its easiest solution. When you divide the problem, you will understand which parts are essential for the diagram and which parts are already solved in the given data, so you don't have to solve it again.
Step 3: Draw the diagram
Start drawing the tape diagram after analyzing the problem and contemplating its possible solution. First, you have to draw a long rectangle tape like segment and then divide it into various sections in small boxes. The number of cells you draw depends on the given problem statement. You will need to create at least two sections, but there is no upper limit.
Step 4: Organize data
After your diagram is complete, it is time to add and organize data into the diagram. You can do this step from your understanding of the problem statement. Read the statement once again and add numbers, variables and other mathematical symbols to represent the problem visually. Once you organize data, the only remaining step will be finding the solution.
Step 5: Label the diagram
Solving the given problem will help you check whether your tape diagram is correct or not. To solve the problem, you have to label the diagram and solve the whole mathematical equation using the data provided by your tape diagram. You can use it for simple arithmetic operations or solve various complex problems.
## 3. How to Do a Tape Diagram with EdrawMax Online
The best choice is to use diagramming software such as EdrawMax Online to make your tape diagram. You can create tape diagrams from scratch, but it will take a lot of your time, and if you are drawing it for the first time, doing it on EdrawMax, where you can get any symbol or icon your want, will be a smart choice. With EdrawMax, you will get a template library and customization tools to edit your diagram, and you can easily export your diagram in any format you want.
Go to Diagram Type and select a template
After you sign in, navigate to General>Basic Diagram>Block Diagram. After that, you can click on it to get a blank canvas. You can start creating the diagram, and you can look for symbols in the symbol library to the left side of the canvas.
You can also look for templates in the EdrawMax Online template library. All you have to do is either go to templates or click the search bar. Type the diagram's name and get a comprehensive list of templates professionally made for you. You can easily make changes to these templates according to your requirements. Find more templates in Templates Community.
You can customize your tape diagram by using symbols from the library. You have to click and drag the symbols you need and add them to your diagram. You can also add text and change font styles and text colors. You can add shapes, and there is a toolbar at the top of the canvas with various customization options.
Save and Export in any format
You can save your drawing in all popular formats using EdrawMax Online. You have to go to files and click the save or export options. Please select the format you require, such as docs for a text file or jpeg to export it as an image.
Presentation
You can create presentations on EdrawMax Online for your project. You can add slides to explain your diagram, and you can style every slide differently to make it more creative.
## 4. Expert Tips for Creating a Tape Diagram
##### Tip 1: Draw after understanding the problem
When you draw a tape diagram, always remember you are doing it to quickly solve the given problem, not to create extra work for yourself. Understanding the problem statement comes first, and you can't draw a diagram first and then move towards the problem statement. Even if you are solving similar math equations, reading and analyzing the problem is always helpful before making a diagram for solving it.
##### Tip 2: Don’t make extra sections
The number of sections in the tape diagram affects the problem solution. You need the ideal number of cells to organize the data given in the problem statement before solving it. Having extra or fewer sections will get you to the wrong solution. Making tape diagrams is simple, and there is little chance of mistaking the number of cells.
##### Tip 3: Always review
Make sure you review your diagram before you turn in your assignment. When you review the diagram, you will have to check each step of your solution from start to finish in the given hierarchy. You can also ask your friends or instructors to review your diagram.
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# Direct and Inverse Proportions Class 8 Maths Notes Chapter 13
According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 11.
When two variables change in the same sense i.e., as one amount increases, the other amount also increases at the same rate it is called direct proportionality. When two variables such as x and y are given, y is directly proportional to x if there is a non-zero constant k. The constant ratio is called the constant of proportionality or proportionality constant.
## Inverse Proportions
If the value of variable x decreases or increases upon corresponding increase or decrease in the value of variable y, then we can say that variables x and
y are in inverse proportion.
For example : In the table below, we have variable y: Time taken (in minutes) reducing proportionally to the increase in value of variable x: Speed (in km/hour). Hence the two variables are in inverse proportion.
To know more about Direct and Inverse Proportion, visit here.
### Relation for Inverse Proportion
Considering two variables x and y,
xy=k or x=
$$\begin{array}{l}\frac{k}{y}\end{array}$$
establishes the relation for inverse proportionality between x and y, where k is a constant.
So if x and y are in inverse proportion, it can be said that
$$\begin{array}{l}\frac{x_{1}}{x_{2}}\end{array}$$
=
$$\begin{array}{l}\frac{y_{2}}{y_{1}}\end{array}$$
where y1 and y2 are corresponding values of variables x1 and x2
### Time and Work
It is important to establish the relationship between time taken and the work done in any given problem or situation. If time increases with increase in work, then the relation is directly proportional. In such a case we will use
$$\begin{array}{l}\frac{x_{1}}{y_{1}}\end{array}$$
=
$$\begin{array}{l}\frac{x_{2}}{y_{2}}\end{array}$$
to arrive at our solution.
However if they are inversely proportional we will use the relation
$$\begin{array}{l}\frac{x_{1}}{x_{2}}\end{array}$$
=
$$\begin{array}{l}\frac{y_{2}}{y_{1}}\end{array}$$
For example: In the table below, we have the number of students (x) that took a certain number of days (y) to complete a fixed amount of food supplies. Now we have to calculate the number of days it would take for an increased number of students to finish the identical amount of food.
Number of students 100 125 Number of days 20 y
We know that with greater number of people, the time taken to complete the food will be lesser, therefore we have an inverse proportionality relation between x and y here.
Hence by applying the formula, we have:
$$\begin{array}{l}\frac{100}{125}\end{array}$$
=
$$\begin{array}{l}\frac{y}{20}\end{array}$$
⇒ =
$$\begin{array}{l}\frac{20.100}{125}\end{array}$$
=16 days
## Introduction to Direct Proportions
### Direct Proportion
If the value of a variable x always increases or decreases with the respective increase or decrease in value of variable y, then it is said that the variables x and y are in direct proportion.
For example : In the table below, we have variable y – Cost (in Rs) always increasing when there is an increase in variable x – Weight of sugar (in kg). Likewise if the weight of sugar reduced, the cost would also reduce. Hence the two variables are in direct proportion.
To know more about Direct Proportion, visit here.
### Relation for Direct Proportion
Considering two variables x and y,
$$\begin{array}{l}\frac{x}{y}\end{array}$$
k or x=ky establishes the simple relation for direct proportion between x and y, where k is a constant.
So if x and y are in direct proportion, it can be said that
$$\begin{array}{l}\frac{x_{1}}{y_{1}}\end{array}$$
=
$$\begin{array}{l}\frac{x_{2}}{y_{2}}\end{array}$$
where y1 and y2 correspond to respective values of x1 and x2.
## Frequently asked Questions on CBSE Class 8 Maths Notes Chapter 13 Direct and Inverse Proportions
Q1
### What is a Direct Proportion?
Direct proportion or direct variation is the relation between two quantities where the ratio of the two is equal to a constant value.
Q2
### What is an Inverse Proportion?
Inverse proportion occurs when one value increases and the other decreases.
Q3
### What are the uses of Time and Work problems?
Time and work problems are important because there is a certain relationship between the number of persons doing the work, number of days or time taken by them to complete the work and the amount of work that is done.
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# Bellringer (copy at top of notes) #1.Simplify | -9 – (-5) | #2. Find the opposite and the reciprocal of 13/8. #3.Simplify 8 * 3 – 8 ÷ 4.
## Presentation on theme: "Bellringer (copy at top of notes) #1.Simplify | -9 – (-5) | #2. Find the opposite and the reciprocal of 13/8. #3.Simplify 8 * 3 – 8 ÷ 4."— Presentation transcript:
Bellringer (copy at top of notes) #1.Simplify | -9 – (-5) | #2. Find the opposite and the reciprocal of 13/8. #3.Simplify 8 * 3 – 8 ÷ 4
1-1 continued Properties of Real Numbers
1. Commutative Property of Addition a + b = b + a When adding two numbers, the order of the numbers does not matter. Examples of the Commutative Property of Addition 2 + 3 = 3 + 2(-5) + 4 = 4 + (-5)
2. Commutative Property of Multiplication a b = b a When multiplying two numbers, the order of the numbers does not matter. Examples of the Commutative Property of Multiplication 2 3 = 3 2(-3) 24 = 24 (-3)
3. Associative Property of Addition a + (b + c) = (a + b) + c When three numbers are added, it makes no difference which two numbers are added first. Examples of the Associative Property of Addition 2 + (3 + 5) = (2 + 3) + 5 (4 + 2) + 6 = 4 + (2 + 6)
4. Associative Property of Multiplication a(bc) = (ab)c When three numbers are multiplied, it makes no difference which two numbers are multiplied first. Examples of the Associative Property of Multiplication 2 (3 5) = (2 3) 5 (4 2) 6 = 4 (2 6)
5. Distributive Property a(b + c) = ab + ac Multiplication distributes over addition. Examples of the Distributive Property 2 (3 + 5) = (2 3) + (2 5) (4 + 2) 6 = (4 6) + (2 6)
6. Additive Identity Property The additive identity property states that if 0 is added to a number, the result is that number. Example: 3 + 0 = 0 + 3 = 3
7.Multiplicative Identity Property The multiplicative identity property states that if a number is multiplied by 1, the result is that number. Example: 5 1 = 1 5 = 5
8.Additive Inverse Property The additive inverse property states that opposites add to zero. 7 + (-7) = 0 and -4 + 4 = 0
9.Multiplicative Inverse Property The multiplicative inverse property states that reciprocals multiply to 1.
Ex.1 Identify which property that justifies each of the following. 4 (8 2) = (4 8) 2
Ex.2 Identify which property that justifies each of the following. 6 + 8 = 8 + 6
Ex.3 Identify which property that justifies each of the following. 12 + 0 = 12
Ex.4 Identify which property that justifies each of the following. 5(2 + 9) = (5 2) + (5 9)
Ex.5 Identify which property that justifies each of the following. 5 + (2 + 8) = (5 + 2) + 8
Ex.6 Identify which property that justifies each of the following.
Ex.7 Identify which property that justifies each of the following. 5 24 = 24 5
Ex.8 Identify which property that justifies each of the following. 18 + -18 = 0
Ex.9 Identify which property that justifies each of the following. -34 1 = -34
1-2 “Algebraic expressions” To evaluate an algebraic expression you plug in numbers for the variables and follow the order of operations Recall PEMDAS: 1. Parentheses 2. Exponents 3. Multiply/Divide 4. Add/Subtract
1. Evaluating algebraic expressions Ex.1 Evaluate 7x-3xy for x= -2 and y= 5 Ex. 2 Evaluate x+y÷x for x=4 and y=2
Your Turn! Try Ex. 3: Evaluate 3x-4y+x-y for x=4 and y= -2
Ex. 4 Evaluate (k-18) 2 – 4k for k=6 Ex. 5 Evaluate c 2 – d 2 for c= -3 and d= 5
Your Turn! Try Ex. 6 Evaluate c (3-d) – c 2 for c= -3 and d=5
2. Combining like terms A term is a number, variable or a number and a variable written together A coefficient is the number in a term –Ex. For 5y +10x 2, 5 and 10 are the coefficients Like terms have the same variables with the same exponents –Ex. 3t 2 and -4t 2
Ex.1 Simplify 4m 2 + 3m – 2m 2 Ex.2 –(r - t) + 3(r + 2t)
Your turn! Try Ex. 3 2h – 3k + 7(2h-3k)
Last Ex. Find the perimeter of this figure
Index Cards Unsimplified (green)Simplified (yellow) 3(2x+1) -812- 4x x 2 + x + x 2 - 0.5x.5x – x6x-5 5 – (4x-7)2x 2 + x
1-2 Homework Page 4 wb #1-11, #15-25
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# How do you write the equation of the line which has y-intercept (0, 5) and is perpendicular to the line with equation y = –3x + 1?
##### 1 Answer
May 29, 2016
$y = \frac{1}{3} x + 5$
#### Explanation:
The equation of a line in $\textcolor{b l u e}{\text{slope-intercept form}}$ is
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{y = m x + b} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
where m represents the gradient and b, the y-intercept.
The advantage of having the line in this form is that m and b can be extracted 'easily'
The equation : y = - 3x + 1 is in this form
hence m = - 3
If 2 lines are perpendicular then the product of their gradients .
${m}_{1} \text{ and " m_2", say,}$ is -1
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{m}_{1.} {m}_{2} = - 1} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
hence gradient of perpendicular line is $\frac{1}{3} \text{ since } \left(3 \times - \frac{1}{3} = - 1\right)$
We have y-intercept = b = 5 and m $= \frac{1}{3}$
$\Rightarrow y = \frac{1}{3} x + 5 \text{ is the equation}$
graph{(y+3x-1)(y-1/3x-5)=0 [-20, 20, -10, 10]}
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# 8.4 An Introduction to Functions: Linear Functions, Applications, and Models Part 1: Functions.
## Presentation on theme: "8.4 An Introduction to Functions: Linear Functions, Applications, and Models Part 1: Functions."— Presentation transcript:
8.4 An Introduction to Functions: Linear Functions, Applications, and Models Part 1: Functions
Relations Often we describe one quantity in terms of another, using ordered pairs. – When you fill your tank with gas, the total amount you pay is equal to the number of gallons multiplied by the price per gallon. If the value of y depends on the value of x, then y is the dependent variable and x is the independent variable. – The amount you pay depends on the number of gallons. When quantities are related in this way, we call it a relation. – A relation is a set of ordered pairs.
Functions A function is a relation in which for each value of the first component of the ordered pairs there is exactly one value of the second component. – For each value of x there is only one value of y. F = {(1, 2), (-2, 5), (3, -1)} G = {(-4, 1), (-2, 1), (-2, 0)} Which is a relation? Which is a function?
Function Notation When y is a function of x, we can use the notation f(x) which shows that y depends on x. – This is called function notation. – This does NOT mean multiply f and x! Note that f(x) is just another name for y. – y = 2x – 7 and f(x) = 2x – 7 are the same!
Using Function Notation Let f(x) = -x 2 + 5x – 3. Find the following: f(2) f(-1) f(2x)
Linear Functions A function that can be written in the form f(x) = mx + b for real numbers m and b is a linear function. – This is the same as y = mx + b!
Graphing Linear Functions Graph the linear function f(x) = -2x + 3.
Graph the linear function f(x) = 3.
Modeling with Linear Functions A companys cost of producing a product and the revenue from selling the product can be expressed as linear functions. The idea of break-even analysis then can be explained using the graphs of these functions. – When the cost equals the revenue, the company breaks even. – When cost is greater than revenue, the company loses money. – When the cost is less than revenue, the company makes money.
Analyzing Cost, Revenue, and Profit A company that produces DVDs of live concerts places an ad in a newspaper. The cost of the ad is \$100. Each DVD costs \$20 to produce and is sold for \$24. Express the cost C as a function of x, the number of DVDs produced. Express the revenue R as a function of x, the number of DVDs sold. When will the company break even (what value of x makes revenue equal cost)?
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# 1.2 Elasticity: Price elasticity of demand (PED)
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1 1.2 Elasticity: Price elasticity of demand (PED) Learning Outcomes Explain the concept of price elasticity of demand, understanding that it involves responsiveness of quantity demanded to a, along a given demand curve. Calculate PED using the following equation: PED = percentage change in quantity demanded / percentage. State that the PED value is treated as if it were positive although its mathematical value is usually negative. Explain using diagrams and PED values, the concepts of price elastic demand, price inelastic demand, unit elastic demand, and perfectly inelastic demand. Explain the determinants of PED, including the number and closeness of substitutes, the degree of necessity, time, and the proportion of income spent on the good. What is price elasticity of demand and how is it calculated? Price elasticity of demand (PED) measures the responsiveness of quantity demanded for a good to a change in its price. percentage change in quantity demanded It is calculated using the formula PED = percentage = % Qd % P is the Greek letter delta. It is used in mathematics and means change in. This formula is used when the changes are given in percentages. Calculation of PED using the formula PED = % Qd % P a step-by-step guide price elasticity of demand (PED) a measure of how quantity demanded responds to a in percentage terms quantity demanded the amount of a good consumers are willing and able to buy at a given price over a given period of time Price falls by 6% leading to an increase in quantity demanded of 9%. I will enter the percentage changes into the formula: PED = 9 6 simplify by dividing 9 by 6 (% P is negative because price has decreased) PED = 1.5 (PED is negative because a positive number divided by a negative number gives a negative number) When the changes are in raw values we use the formula: PED = Qd/Qd P/P ( Qd/Qd means Qd divided by Qd) ( P/P means P divided by P) Qd is the change in quantity demanded, Qd is the original quantity demanded, P is the and P is the original price. Calculation of PED using the formula PED = Qd/Qd Qd/Qd a step-bystep P/P guide Price increases from \$3 to \$9 causing quantity demanded to fall from 12,000 units to 4000 units. Qd = 8,000 units ( is negative because quantity demanded falls). Original Qd = 12, 000 units, P = \$3 to \$9 = \$6 (it is positive because price has increased) and Original price = \$3. Now the calculations are added into the formula: PED = 8000/12000 simplify by dividing 8000 by 12,000 and by dividing 6 by 3 6/3 PED = 0.67 Simplify by dividing 0.67 by 2 2 PED = 0.34 (PED is negative because a negative number divided by a positive number gives a negative number) 19
2 How to calculate an unknown value when all the other values are known a step-by-step guide Calculation of the change in quantity demanded caused by a when the original price, the, the original quantity demanded, and the value of PED are known. P = \$4, new price = \$6, original quantity demanded = 120 units, and PED = 0.5 The figures are entered into the formula PED = Qd/Qd P/P 0.5 = Qd/120 simplify by dividing 2 by 4 2/4 0.5 = Qd/ simplify by multiplying both sides of the equation by = Qd/120 simplify by multiplying both sides of the equation by = Qd New quantity demanded = original Qd Qd = 90 An increase in price from \$4 to \$6 causes quantity demanded to fall from 120 units to 90 units. inverse relationship a change in the value of one variable leads to an opposite change in direction in the value of the other variable. For example an increase in price leads to a fall in quantity demanded. law of demand states that there is a negative causal relationship between price and quantity demanded. As price rises quantity demanded falls. demand curve a graph that shows the relationship between price and quantity demanded absolute value the distance a number is from zero (e.g. the absolute value of 5 is 5 and the absolute value of -5 is 5) price elastic the percentage supplied > the percentage price inelastic the percentage supplied < the percentage price unit elastic the percentage change in quantity demanded / supplied = the percentage. PED/PES = 1 Why is the value of PED negative? A price increase is a positive change and it causes quantity demanded to fall which is a negative change and as price falls (negative change) quantity demanded rises (positive change). So price and quantity demanded have an inverse relationship reflecting the law of demand. In order to work out PED the percentage change in quantity demanded is divided by the percentage. When one value is positive the other value is negative. A negative number divided by a positive number gives a negative number (minus divided by plus is minus) and a positive divided by a negative gives a negative (plus divided by minus is minus). As the value of PED is always negative economists often ignore the minus sign. Explain using diagrams and PED values, the concepts of price elastic demand, price inelastic demand, unit elastic demand, and perfectly inelastic demand Figure 8.1a and 8.1b show demand curves for two different goods served during one week at a cafe. Price of burgers \$ Quantity of burgers Figure 8.1a Figure 8.1b D Price of chips \$ Portions of chips In Figure 8.1a an increase in price of 5% causes quantity demanded to fall by 10%: PED = 10/5 = ( ) 2. The % Qd > % P so the absolute value of PED is greater than 1 (ignoring the minus sign) and demand is price elastic. In Figure 8.1b a price rise of 20% causes quantity demanded to fall by 10%. PED = 10/20 = 0.5 (ignoring the minus sign). The % Qd < % P so the absolute value of PED is less than 1 and demand is price inelastic. When % Qd = % P the absolute value of PED is equal to 1 and demand is price unit elastic. D 20 Section 1: Microeconomics
3 1.2 Elasticity: Price elasticity of demand (PED) Price \$ Figure 8.2 Price \$ Figure Quantity demanded D In Figure 8.2 at a price of \$3 quantity demanded has no end. It is infinite. As price changes, quantity demanded falls from infinity to zero. To reflect this the demand curve is horizontal at price. As price changes, the change in quantity demanded is infinite. Any number divided by or into infinity equals infinity. A 5% increase in price for example causes quantity demanded to fall from infinity to zero. PED = % Qd % P PED = = ( means infinity). 5 Demand is perfectly elastic with respect to price. In Figure 8.3 a causes no change in quantity demanded. The change in quantity demanded = 0. Quantity demanded is 30 units at all prices. To reflect this the demand curve is vertical at 30 units. Any number divided by or into 0 equals 0. Price increases from \$2 to \$4. This is a 100% increase causing quantity demanded to change by 0. To reflect this the demand curve is vertical at 30 units. PED = % Qd % P = = 0. Demand is perfectly inelastic with respect to price. Model sentence: The value of PED is determined by the relative size of the percentage and quantity demanded: when the percentage change in quantity demanded is greater than the percentage, PED has an absolute value greater than 1 and demand is price elastic and when the percentage change in quantity demanded is less than the percentage, PED has an absolute value less than 1 and demand is price inelastic. Explain how PED is determined D Quantity demanded Quantity demanded of a good is more responsive to a change in its price when there are lots of close substitutes available for consumers to buy instead of the good. Demand is more price elastic. If a good is a necessity (e.g. oil) quantity demanded is less responsive to a change in its price and demand is more price inelastic. Demand for luxury goods is more price elastic than for necessities because consumption of luxury goods is not essential. perfectly elastic demand at a particular price quantity demanded is infinite but falls to nothing as price changes. The abosolute value of PED is equal to infinity. perfectly inelastic demand quantity demanded does not change as price changes. The absolute value of PED is equal to zero. substitutes in production two or more goods that can be produced by a firm necessity a consumer good or a producer good that is important to the producer or consumer and has few or no substitutes, therefore its PED tends to be inelastic. Also describes a good that is income inelastic. expenditure the price paid by buyers in exchange for goods and services. Total expenditure = price quantity purchased. consumption...use proportion...amount expenditure...spending/ money spent Glossary addictive something your body needs regularly and you cannot stop taking it If a good is addictive (e.g. tobacco) it is more difficult to reduce consumption following an increase in price. Therefore demand is more price inelastic. It is difficult for consumers to immediately change patterns of consumption. It takes time to find suitable substitutes and to ration use of a good following an increase in price. Over a short period of time demand is more price inelastic but becomes less price inelastic over time, as consumers are able to find substitutes and reduce consumption further. The price of some goods is very low. Expenditure on them makes up a very small proportion of a consumer s income so that even after a large increase in price there is little or no change in quantity demanded. If the price of a box of matches increases by 15% the change in quantity demanded would be much less than 15%, therefore demand is price inelastic. Model sentence: The more close substitutes there are available on the market the easier it is for consumers to switch expenditure and buy an alternative good, therefore demand for a good with many close substitutes will be more price elastic. 21
4 Test your understanding of this unit by answering the following questions Explain why the value of PED is negative. Calculate the PED when price falls by 25% causing quantity demanded to increase by 75%. Calculate PED when a price increase from \$4 to \$6 causes quantity demanded to fall from 1200 units to 1000 units. Quantity demanded = 150 units at \$16. Price increases to \$20 and PED for the good = Calculate the change in quantity demanded. Explain why as income rises PED for most goods becomes less elastic. Explain why demand for cigarettes is inelastic. Use a diagram to illustrate your answer. Learning Outcomes Calculate PED between two designated points on a demand curve using the PED equation. Explain why PED varies along a straight line demand curve and is not represented by the slope of the demand curve. Examine the role of PED for firms in making decisions regarding price changes and their effect on total revenue. Explain why PED for many primary goods is relatively low and the PED for manufactured goods is relatively high. Examine the significance of PED for government in relation to indirect taxes. designated...chosen/ selected diminishing...becoming smaller quantity demanded the amount of a good consumers are willing and able to buy at a given price over a given period of time price inelastic the percentage supplied < the percentage revenue the income a firm receives from consumers in exchange for goods (revenue = price quantity sold) price elastic the percentage supplied > the percentage Calculate and explain PED along a straight line demand curve (see Figure 9.1a/b) Price \$ PED > PED = 1 PED < Quantity Figure 9.1a Figure 9.1b As price increases from \$2 to \$4 quantity demanded falls from 40 to 30 units. (See pages for a detailed explanation of how to calculate PED.) PED = Qd/Qd P/P = 10/40 = 0.25 = ( ) PED < 1 so demand is price inelastic. Remember when stating 2/2 1 the value of PED that the minus sign is ignored it is treated as a positive. Revenue = price quantity. When price is \$2 revenue = \$2 40 = \$80. Total revenue (P Q) \$ Quantity (Q) When price is \$4 revenue = \$4 30 = \$120, an increase of \$40. As price rises from \$2 to \$4 revenue increases by \$40. As price falls from \$4 to \$2 revenue falls from \$120 to \$80, a fall of \$40. Therefore when demand is price inelastic an increase in price leads to an increase in revenue and a fall in price leads to a fall in revenue because the percentage is greater than the percentage change in quantity demanded. As price increases from \$6 to \$8 quantity demanded falls from 20 to 10 units. PED = 10/20 2/6 = / 20 = 0.5 = ( ) PED > 1 so demand is price elastic When price is \$6 revenue = \$6 20 = \$120. When price is \$8 revenue = \$8 10 = \$80, a fall of \$40. As price rises from \$6 to \$8 revenue falls from \$120 to \$80, a decrease of \$40. As price falls from \$8 to \$6 revenue increases from \$80 to \$120, a rise of \$40. Therefore, when demand is price elastic an increase in price leads to a fall in revenue and a fall in price leads to an increase in revenue, because the percentage change in quantity demanded is greater than the percentage. 22 Section 1: Microeconomics
5 1.2 Elasticity: Price elasticity of demand (PED) The slope remains constant moving along a straight line demand curve. The measure of the slope of the demand curve is in absolute terms. For example, a price increase of \$2 from \$2 to \$4 causes a fall in quantity demanded from 40 to 30 units and a price increase of \$2 from \$6 to \$8 causes a fall in quantity demanded from 20 to 10 units. In both cases price changes by \$2 causing quantity demanded to change by 10 units. PED, on the other hand, measures the relationship between a percentage and a percentage change in quantity demanded. It is measured in relative terms. A or quantity when price or quantity is low results in a relatively large percentage change. A or quantity when price or quantity is high results in a relatively small percentage change. For example, a price increase of \$2 from \$2 to \$4 is a 100% increase in price whereas a price increase of \$2 from \$6 to \$8 is only a 33.33% increase in price, even though the actual price in both cases changes by the same amount. Explain the relationship between PED and total revenue (see Figure 9.1a/b) At first, as price increases from zero the percentage > the percentage change in quantity demanded. Demand is price inelastic, therefore revenue increases as price rises. As price continues to rise it causes a movement up and along the demand curve. Price elasticity of demand becomes less price inelastic, because the difference between the rate at which price and quantity demanded changes begins to fall. While the percentage is greater than that of quantity demanded, revenue will continue to rise as price increases, but at a diminishing rate. This means that, as price increases, the addition to total revenue is positive, but the increase in total revenue is less than the increase gained by the previous increase in price. When the percentage = the percentage change in quantity demanded, total revenue is maximized. The value of PED = 1 and demand is unit price elastic. This occurs halfway along the straight line demand curve at a price of \$5 and a quantity of 25 units. As price increases above \$5 the percentage change in quantity demanded > the percentage, therefore revenue begins to fall. As price continues to rise, PED becomes more elastic. Total revenue falls at an increasing rate until price is \$10 and quantity demanded is 0 and there is no revenue. Revenue maximization If a firm s output and price is at a point on the demand curve where PED is inelastic the firm can increase revenue by increasing price and reducing output. If output and price is at a point where PED is elastic the firm can increase revenue by lowering price and increasing output. Therefore, in order to maximize revenue, a firm sets output or price where PED = 1 and demand is unit price elastic. Model sentence: When the absolute value of PED is less than 1 increasing price causes revenue to rise and when the absolute value of PED is more than 1 reducing price causes revenue to rise, therefore revenue is maximized when the absolute value of PED = 1. Examine the role of PED for firms in making decisions regarding price changes and their effect on total revenue A firm launching a new good wants to maximize revenue over the life of the good. The firm will try to get individual consumers to pay the most they are willing to pay. In other words the firm tries to steal as much consumer surplus as possible. Initially the firm charges a high price and a quantity of units are sold to less price-sensitive consumers who are willing and able to pay the high price. When demand of these consumers has been met and sales begin to fall the firm lowers price at a point on the demand curve where PED is inelastic, more consumers enter the market to buy the good, and revenue increases. Over time the firm continues to lower the price in order to increase quantity demanded and increase revenue. This pricing strategy is called skimming the market and is particularly used by firms producing new technological goods. In industries where there are firms producing branded goods a new firm might reduce the price of its new good in order to gain market share. As brand awareness rises and the good is established in the market, PED becomes more inelastic and the firm can raise the price and increase revenue. Why is PED for primary goods lower than the PED for manufactured goods? Primary goods such as wheat have fewer substitutes than manufactured goods. Primary goods are more likely to be necessities. Manufactured goods are often luxuries. In countries where incomes are relatively high consumers spend a lower proportion of income on primary goods and a higher proportion on manufactured Glossary slope the angle/gradient of the curve maximized made as great as possible demand curve a graph that shows the relationship between price and quantity demanded unit price elastic the percentage change in quantity demanded / supplied = the percentage. PED/PES = 1. output the quantity of goods produced by a firm, industry or economy consumer surplus the difference between the price a consumer is willing and able to pay and the price the consumer actually pays demand the amount of a good that consumers are willing and able to buy at each price market where buyers and sellers meet to exchange money for goods and services pricing strategy a plan made and used by a firm with the aim of increasing revenue and profits through the setting of price industry a group of firms that produce the same or similar goods or services branded goods goods that are identifiable as being the product of a particular firm usually through a distinctive label or logo market share the proportion of the market supply of a good or service that is controlled by a firm primary good a good that has not been processed and is in a raw state (e.g. fruit/wheat) substitutes in production two or more goods that can be produced by a firm manufactured goods goods produced from raw materials necessities consumer goods or producer goods that are important to the producer or consumer and few or no substitutes, therefore their PED tends to be inelastic. Also describes goods that are income inelastic. luxuries when income changes demand for a luxury good changes at a greater rate. Demand is relatively sensitive to changes in income. 23
6 unemployment occurs when there are people actively looking for work at the equilibrium wage rate but are not able to find work industry a group of firms that produce the same or similar goods or services tax revenue the income the government receives through the levying and collection of taxes duty... tax concept(s)...idea(s)/ theory/ies responsiveness...reaction/ sensitivity goods. Therefore quantity demanded of primary goods is less sensitive to a than quantity demanded of manufactured goods. Therefore the value of PED is likely to be lower and demand less price elastic for primary goods than for manufactured goods. Examine the significance of PED for government in relation to indirect taxes An indirect tax is a tax imposed on producers by the government. It is a tax placed on a good or service. Examples include duties on cigarettes, alcohol, fuel, and value added tax (VAT). When a government increases the duty on a good, the price of the good increases leading to a fall in quantity demanded. If demand is highly price elastic the rate of change in quantity demanded > the rate of change in price and a duty that raises price will cause a large decrease in sales increasing unemployment in the industry. The government places taxes on goods that are more price inelastic because the fall in quantity of goods bought is not as great, therefore there are not as many job losses and the tax revenue (tax per unit quantity sold) collected by the government from the sale of the goods will be greater. Test your understanding of this unit by answering the following questions Explain why revenue is maximized when the absolute value of PED = 1. Use a diagram to illustrate your answer. Explain why PED is lower for primary goods than for manufactured goods. Learning Outcomes Outline the concept of cross-price elasticity of demand, understanding that it involves responsiveness of demand for one good (and hence a shifting demand curve) to a change in the price of another good. Calculate XED using the following equation XED = percentage change in quantity demanded of good x percentage of good y Show that substitute goods have a positive value of XED and complementary goods have a negative value of XED. Explain that the (absolute) value of XED depends on the closeness of the relationship between two goods. Examine the implications of XED for businesses if prices of substitutes or complements change. cross-price elasticity of demand (XED) measures the responsiveness of demand for one good to a of another good in percentage terms complementary goods/ complements goods that are used together quantity demanded the amount of a good consumers are willing and able to buy at a given price over a given period of time demand curve a graph that shows the relationship between price and quantity demanded shifting...moving implications...effects/ outcomes relatively...comparatively 24 Section 1: Microeconomics Calculation of XED XED measures the responsiveness of demand for one good to a change in the price of another good. percentage change in quantity demanded of good x It is calculated using the formula XED = percentage of good y How does the price of one substitute affect the demand for the other? a step-by-step guide Goods X and Y are substitutes: one good can be used in place of the other. As the price of good Y increases it becomes relatively more expensive than good X. Some consumers buy good X in place of good Y. This leads to a fall in quantity demanded for good Y (a movement up and along the demand curve) and an increase in demand for good X (a shift up and to the right of the demand curve). As the price of good Y falls it becomes relatively cheaper than good X. Some consumers buy good Y in place of good X leading to an increase in quantity demanded for good Y (a movement down and along the demand curve) and a fall in demand for good (a shift down and to the left of the demand curve). Model sentence: In the case of two substitutes an increase in the price of one leads to a fall in quantity demanded of that good (a movement up and along its demand curve) and an increase in demand for the other good (a shift up and to the right of its demand curve).
7 Calculate XED and explain why it is always positive in the case of substitutes There are two barber shops in the high street, John s and Sam s. John increases the price of a haircut from 20 to 25. Some customers go to Sam s instead of John s. Quantity demanded of haircuts at John s falls and demand for haircuts at Sam s increases from 100 a week to 150. How to calculate XED from the information above a step-by-step guide Use the formula XED = QdX/QdX PY/PY QdX is the change in quantity demanded for service X, Qd is the original quantity demanded for service X, PY is the of service Y, and P is the original price of service Y. XED = 50/100 simplify by dividing 50 by 100 and 5 by 20 5/20 XED = 0.5 simplify by dividing 0.5 by XED = 2 There is a positive correlation between the price of one good and the demand for another when the goods are substitutes. An increase in price of one good causes an increase in demand for the other. A positive divided by a positive equals a positive. If John reduces the price some customers have their haircut at John s instead of Sam s leading to an increase in quantity demanded at John s and a fall in demand at Sam s. The fall in price of one service leads to a fall in demand for the other. A negative divided by a negative equals a positive. Therefore the value of XED for substitutes is always positive. The services are close substitutes: one can easily be consumed in place of the other. The demand for one service is very sensitive to a change in the price of the other. The greater the similarity between two goods the more responsive demand for one good is to a change in the price of the other and the higher the value of XED. 1.2 Elasticity: Cross-price elasticity of demand (XED) Glossary barber a shop where men s hair is cut positive correlation a relationship between two variables such that they move in the same direction rival business a competitor pricing strategy a plan made and used by a firm with the aim of increasing revenue and profits through the setting of price revenue the income a firm receives from consumers in exchange for goods (revenue = price quantity sold) profit the difference between total revenue and total cost correlation...link/ relationship consumed...used How does the XED of substitutes affect businesses? It is useful for a firm to know the effect on demand for their good when a rival business changes its price and the effect on demand for a rival s good when it changes price. This knowledge helps firms develop a pricing strategy that increases revenue and profit. How does the price of one complementary good affect the demand for the other? a step-by-step guide Goods X and Y are complements: they are used together. As the price of good Y falls quantity demanded increases (a movement down and along its demand curve). As more of good Y is sold demand for good X which is used with good Y increases (a shift up and to the right of the demand curve). As the price of good Y increases quantity demanded falls (a movement up and along its demand curve). As less of good Y is sold demand for good X decreases (a shift down and to the left of the demand curve). Model sentence: In the case of complementary goods an increase in the price of one good leads to a fall in quantity demanded of that good (a movement up and along its demand curve) and a fall in demand for the other good (a shift to the left of its demand curve). 25
8 inverse relationship a change in the value of one variable leads to an opposite change in direction in the value of the other variable unrelated goods goods that are not linked in their use Calculate XED and explain why it is always negative in the case of complements The price of using mobile phones falls and quantity demanded increases, leading to a movement down and along the demand curve. As more phones are sold the demand for apps (applications) increases causing the demand curve for apps to shift up and to the right. The price of mobile phones falls by 10% and this leads to a 25% increase in quantity demanded of apps. XED = % QdX % PY = 25% 10% = 2.5 Let s say the price of using mobile phones increases by 10% leading to a 25% fall in quantity demanded of apps. XED = 25% 10% = 2.5 The value of XED is always negative in the case of complements because of the inverse relationship between the price of one good and the demand of the other. A positive divided by a negative equals a negative. A negative divided by a positive equals a negative. Lots of people use apps on their mobile phones, therefore the number of phones sold has a large impact on the demand for apps. Demand for apps is highly responsive to changes in the price of mobile phones. The stronger the relationship between the two complementary goods the higher the negative value of XED. How does the XED of complements affect businesses? Trips to the cinema and confectionery and fizzy drinks are strong complements. The profit on the sale of confectionery and drinks at the cinema is very high. The owner knows how much on average each customer spends on these items. Reducing the price of cinema tickets to attract more customers will lead to an increase in demand for confectionery and drinks. It is possible that this pricing strategy might increase profits overall. Unrelated goods Goods are unrelated when an increase in the price of one good does not affect the demand for the other. concept(s)...idea(s)/ theory/ies shifting...moving Test your understanding of this unit by answering the following questions Explain why the value of XED is positive in the case of substitutes. Calculate the XED when price of one good falls from \$9 to \$6 causing quantity demanded for the other good to increase from 120,000 units to 200,000 units. Comment on the relationship between the two goods. As the price of petrol increases demand for more fuel-efficient cars will increase. Using the concept of cross-price elasticity of demand, comment on the validity of this comment. Learning Outcomes Outline the concept of income elasticity of demand, understanding that it involves responsiveness of demand (and hence a shifting demand curve) to a change in income. Calculate YED using the following equation percentage change in quantity demanded YED = percentage change in income Show that normal goods have a positive value of YED and inferior goods have a negative value of YED. Distinguish, with reference to YED, between necessity (income inelastic) goods and luxury (income elastic) goods. Examine the implications for producers and for the economy of a relatively low YED for primary goods, a relatively higher YED for manufactured goods, and an even higher YED for services. 26 Section 1: Microeconomics
9 How does a change in income affect demand in the case of normal goods? Income elasticity of demand measures the responsiveness of quantity demanded to a change in income. percentage change in quantity demanded It is calculated by using the formula YED = percentage change in income Price P e1 P e Figure 11.1 Calculate YED a step-by-step guide As income increases, demand for flights increases, leading to a shift up and to the right of the demand curve and excess demand (Q 2 Q e ) at the equilibrium price P e. Price increases to eliminate the excess. As price rises there is a movement up and along the supply curve and up and along the new demand curve D 1. Price rises until quantity demanded = quantity supplied at P e1. An increase in income causes price to rise from P e to P e1 and quantity demanded and supplied to increase from Q e to Q e1. A fall in income leads to a shift down and to the left of the demand curve and a fall in equilibrium price and quantity demanded and supplied. A consumer s income increases from \$30,000 to \$40,000 leading to an increase in quantity demanded from 120 to 160 units. The formula used to calculate YED when not given the changes in percentage terms is YED = Qd/Qd Y/Y Qd is the change in quantity demanded, Qd is the original quantity demanded, Y is the change in income, and Y is the original income. 40/120 YED = simplify by dividing 40 by 120 and by / YED = YED = 1 The proportional change in quantity demanded (0.33) = the proportional change in income (0.33) YED = 1 and demand is income unit elastic. There is a positive correlation between income and quantity demanded for normal goods (income and quantity demanded change in the same direction). Therefore the value of YED for a normal good is positive: a positive number divided by a positive number gives a positive number and a negative divided by a negative gives a positive. What is the difference between normal goods that are necessities and those that are luxuries? When the proportional change in quantity demanded < the proportional change in income YED is less than 1. This means that quantity demanded is relatively insensitive to changes in income. Demand is income inelastic and goods are described as necessities. For example, a fall of 3% in income leads to a 0.6% fall in the quantity of bread demanded. YED = % Qd % Y = = 0.2 D Q e Q e1 Q 2 Quantity D/S S D 1 YED is positive and less than 1, therefore bread is a necessity. 1.2 Elasticity: Income elasticity of demand (YED) income elasticity of demand a measure of how quantity demanded responds to a change in income in percentage terms quantity demanded the amount of a good consumers are willing and able to buy at a given price over a given period of time income the payment received by the factors of production (e.g. wages paid to labour, rent paid to the owners of land) demand the amount of a good that consumers are willing and able to buy at each price demand curve a graph that shows the relationship between price and quantity demanded excess demand occurs when quantity demanded is greater than quantity supplied equilibrium price the price at which the quantity consumers are willing and able to buy is equal to the quantity firms are willing and able to produce supply curve a graph that shows the relationship between price and quantity supplied quantity supplied the amount of a good that firms are willing and able to produce at a given price over a given period of time positive correlation a relationship between two variables such that they move in the same direction normal goods goods for which demand increases when income increases, and falls when income falls income inelastic demand for a good is income inelastic when the value of income elasticity of demand is positive and less than 1 necessities consumer goods or producer goods that are important to the producer or consumer and few or no substitutes, therefore their PED tends to be inelastic. Also describes goods that are income inelastic. responsiveness.reaction/ sensitivity eliminate...remove/get rid of 27
10 income elastic demand for a good is income elastic when the value of income elasticity of demand is greater than 1 luxuries when income changes demand for a luxury good changes at a greater rate. Demand is relatively sensitive to changes in income. inferior goods goods for which demand falls as income increases negative correlation a relationship between two variables such that they move in the opposite direction market where buyers and sellers meet to exchange money for goods and services manufactured goods goods produced from raw materials output the quantity of goods produced by a firm, industry or economy capital (goods) manufactured goods that are used in the production of other goods primary good a good that has not been processed and is in a raw state (e.g. fruit/wheat) direct taxes a tax that is paid directly by an individual or firm to the governmnent. For example income tax on wages and company profits. disposable income household income after direct taxation has been deducted industry a group of firms that produce the same or similar goods or services consumption...use When the proportional change in quantity demanded > the proportional change in income YED is greater than 1. This means that quantity demanded is highly responsive to changes in income. Demand is income elastic and goods are described as luxuries. For example, incomes increase by 4% leading to a 6% increase in the quantity of taxi journeys demanded. YED = % Qd % Y = 6 4 = 1.5 YED > 1, therefore a taxi journey is a luxury service. Model sentence: As quantity demanded becomes more responsive to changes in income, demand becomes more income elastic and the positive value of YED increases. Explain how a change in income affects demand in the case of inferior goods In the case of inferior goods, as income increases demand falls, leading to a shift down and to the left of the demand curve and a fall in quantity demanded. As income falls demand increases, leading to a shift up and to the right of the demand curve and an increase in quantity demanded. YED is negative for an inferior good because there is a negative correlation between income and quantity demanded (income and quantity demanded change in opposite directions). A positive divided by a negative gives a negative and a negative divided by a positive gives a negative. Calculation of YED An increase in national income of 3% leads to a 1% fall in the quantity of bus journeys demanded. YED = % Qd % Y = 1 3 = 0.33 YED is negative, therefore a bus journey is an inferior service. Explain the factors that affect YED Consumers react differently to changes in their income because the amount of benefit gained from the consumption of a good varies from person to person. Goods that are luxuries to some are necessities to others. If already on a very high income an increase in income might lead to no changes in the consumption of goods. In this case YED would be 0. (A number divided into zero = zero.) Chicken is a luxury good in China. As incomes increase, quantity demanded increases at a greater rate. As incomes continue to increase, the value of YED for chicken will begin to fall. In time chicken might become income inelastic. YED changes over the life of a good. Demand for mobile phones when first brought to market was income elastic. They were luxury goods. As price has fallen over time demand has become less income elastic. For many the good is no longer a luxury. Some models have become inferior goods. But in poorer countries demand will still be income elastic. YED depends upon how the good is described. For example, demand for bread in general is income inelastic and a necessity, but demand for specialist bread, such as organic raisin and walnut bread, is income elastic and a luxury. Explain how YED affects government and producers A business can predict what will happen to demand when income changes if it knows the YED. When incomes are expected to increase firms producing luxury manufactured goods, such as ipads, and services, such as foreign holidays, will have to make plans if the firm wants to increase output in order to meet the higher levels of demand. The firm will need to employ and train more workers and buy new capital. When incomes are expected to fall the same firms would have to consider reducing the size of the work force and the size of production. The firm might also consider producing an inferior good. Firms can reduce the risk of business failure by producing a range of goods with different YED values. 28 Section 1: Microeconomics
11 Why is demand for primary goods income inelastic? The demand for many primary goods is income inelastic. That is, the proportional change in income is greater than the proportional change in quantity demanded so the value of YED <1. As incomes rise the demand for many primary goods such as tea, coffee, and sugar increases, but by a proportionately smaller amount. If income rose by 30% the percentage change in quantity demanded of sugar or tea would be much smaller. 1.2 Elasticity: Price elasticity of supply (PES) implications...effects/ outcomes responsiveness...reaction/ sensitivity YED and tax Direct tax is a tax on income. It is called a direct tax because it goes directly from the payer of the tax to the government. Increasing tax reduces consumers disposable income and causes a fall in demand for necessities and luxury goods and an increase in demand for inferior goods. This will have an impact on employment in those industries producing goods that have high positive and high negative YEDs. Test your understanding of this unit by answering the following questions Organic bread has a YED of 4 and basic white bread has a YED of 0.1. Incomes are expected to increase by 5% next year. Calculate the percentage increase in quantity demanded for both types of bread. Discuss the implications of the expected increase in income for the bread-making industry. Learning Outcomes Explain the concept of price elasticity of supply, understanding that it involves responsiveness of quantity supplied to a along a given supply curve. Calculate PES using the following equation percentage change in quantity supplied PES = percentage Explain, using diagrams and PES values, the concepts of elastic supply, inelastic supply, unit elastic supply, perfectly elastic supply, and perfectly inelastic supply. Explain the determinants of PES, including time, mobility of factors of production, unused capacity, and ability to store stocks. Explain why the PES for primary goods is relatively low and the PES for manufactured goods is relatively high. Why is there a positive correlation between price and quantity supplied? As price increases, ceteris paribus, profit increases. Assuming that firms are profit maximizers they will allocate more resources to the production of the good that is now more profitable in order to increase output. As price rises output rises. When price falls profit falls and firms reduce output as it is now less profitable. Price elasticity of supply (PES) measures the responsiveness of quantity supplied to a and is percentage change in quantity supplied calculated using the formula PES = percentage Calculations of PES a step-by-step guide Price increases from \$8 to \$10 leading to an increase in quantity supplied from 8000 units to 9000 units. The formula used to calculate PES when the changes are in raw values is PES = Qs/Qs P/P Qs is the change in quantity supplied, Qs is the original quantity supplied, P is the change in price, and P is the original price. PES = 1000/8000 simplify by dividing 1000 by 8000 and 2 by 8 2/8 PES = simplify by dividing by PES = 0.5 The proportional change in quantity supplied (0.125) < the proportional (0.25) therefore PES < 1 and supply is price inelastic. perfectly elastic supply at a particular price quantity supplied is infinite but falls to nothing as price changes. The absolute value of PES is equal to infinity. perfectly inelastic supply quantity supplied does not change as price changes. PES equals zero. ceteris paribus Latin phrase meaning all other things being equal or all other things being held constant resources the inputs into the production process, the factors of production output the quantity of goods produced by a firm, industry or economy price elasticity of supply (PES) a measure of how quantity supplied responds to a change in price in percentage terms price inelastic the percentage supplied < the percentage 29
12 correlation...link/ relationship shift(s)...move(s) supply curve a graph that shows the relationship between price and quantity supplied demand curve a graph that shows the relationship between price and quantity demanded demand the amount of a good that consumers are willing and able to buy at each price perfectly inelastic supply quantity supplied does not change as price changes. PES equals zero. price unit elastic the percentage change in quantity demanded / supplied = the percentage. PED/PES = 1 perfectly elastic supply at a particular price quantity supplied is infinite but falls to nothing as price changes. The absolute value of PES is equal to infinity. factors of production the inputs into the production process (land, labour, capital and entrepreneurship) output the quantity of goods produced by a firm, industry or economy short run a period of time when at least one factor is variable and the others are fi xe d capital (goods) manufactured goods that are used in the production of other goods Glossary slope the angle/gradient of the curve When the changes are in percentage terms use the formula PES = % Qs % P Price falls by 2% leading to a fall in quantity supplied of 3%. PES = % Qs % P = 3 2 = 1.5 The percentage change in quantity supplied ( 3%) > the percentage ( 2%) therefore PES > 1 and supply is price elastic. There is a positive correlation between quantity supplied and price (they change in the same direction), therefore PES is positive (a positive divided by a positive gives a positive and a negative divided by a negative gives a positive). P S Figure 12.1 D S 2 D 2 S 3 Q Figure 12.1 shows a demand and supply diagram with 3 supply curves each representing different values of PES. The demand curve shifts up and to the right from D 1 to D 2. The rate of and quantity supplied caused by an increase in demand varies depending on the initial price and quantity supplied and the slope of the supply curve. Keeping the initial price and quantity supplied unchanged, as S 1 rotates clockwise the slope becomes less steep. The proportional caused by the same increase in demand falls and the proportional change in quantity supplied rises. Supply curve 1 Demand increases and the demand curve shifts up and to the right from D 1 to D 2. Price increases from \$5 to \$9 leading to no change in quantity supplied. Quantity supplied stays at 5 units. YED = Qs/Qs P/P = 0/5 4/5 = 0 = 0 (0 divided by any number = 0) 0.8 PES = 0, supply is perfectly inelastic with respect to price. Supply curve 2 The demand curve shifts up and to the right. Price increases from \$5 to \$7 leading to an increase in quantity supplied from 5 units to 7 units. PES = Qs/Qs P/P = 2/5 2/5 = 0.4 = 1 PES = 1, supply is price unit elastic. 0.4 Supply curve 3 Demand curve shifts up and to the right. Price does not change but quantity supplied increases from 5 units to 9 units. PES = Qs/Qs P/P = 4/5 = 0.8 = (any number divided by infinity = ) PES =. Supply is perfectly elastic with respect to price. Model sentence: The less responsive quantity supplied is to a the lower the value of PES and the more price inelastic the supply. 30 Section 1: Microeconomics Factors that determine how responsive quantity supplied is to a change in price Time In the very short run (the time period immediately after a price increase) the factors of production are fixed and a firm cannot increase output so supply is perfectly inelastic with respect to price. In the short run the quantity of labour is variable (changeable) but the quantities of capital and land are fixed (unchangeable). The firm can only increase output by adding more labour to existing capital so supply is price inelastic. Output can
13 be increased by a relatively small amount by only adding labour. In the long run the quantities of all factors are variable, therefore the firm can employ more labour and buy more capital in order increase output even more so supply becomes more price elastic. Over time quantity supplied becomes more responsive to a change in price. The value of PES increases, supply becomes more price elastic and the slope of the supply curve becomes less steep. For primary goods quantity supplied is less responsive to changes in price. For example, a farmer decides which crops to grow a long time before the goods come to market. It takes a long time to move resources away from the production of one crop to the production of another. It is not possible to change the quantity supplied in the short term therefore supply of agricultural goods is more price inelastic. Manufactured goods are likely to be more price elastic than agricultural goods because it is easier for firms producing manufactured goods to reallocate their factors to different production processes and thereby increase output. Capacity When a firm is operating at full capacity (all the firm s labour and capital is being used) it is difficult to increase output. Supply is more price inelastic. A firm is able to increase output if it has spare capacity. The greater the amount of capital and labour not being used by the firm the more responsive quantity supplied is to an increase in price so supply is more price elastic. Stocks Quantity supplied can be increased when the firm is able to hold lots of stock. These are goods held in storage. Goods can be released onto the market very quickly, therefore supply is more price elastic. Availability and mobility of resources In order to increase output a firm must get more resources. In a period of high economic activity unemployment is very low. Workers are in short supply. Raw materials might not be available. When resources needed for the production of a good are in short supply, quantity supplied of that good is less responsive to increases in its price and supply is more price inelastic. When there is a slowdown in economic activity demand for resources is lower. More are available for firms to use. Quantity supplied can be increased more easily so supply is more price elastic. There is a greater supply of unskilled labour than skilled. It is easier for firms that use unskilled labour to increase the size of the workforce in order to increase output, so supply is more price elastic. When specialized capital and skilled labour are needed, factors are less mobile. A firm will not be able to employ factors quickly in order to increase output so supply is more price inelastic. Model sentence: When the factors used in the production of a good are easily available, firms in an industry can quickly employ them in order to increase output in response to an increase in price and therefore supply is more price elastic. 1.2 Elasticity: Price elasticity of supply (PES) long run a conceptual moment in time when all factors are variable price elastic the percentage supplied > the percentage primary good a good that has not been processed and is in a raw state (e.g. fruit/wheat) resources the inputs into the production process, the factors of production price inelastic the percentage supplied < the percentage unemployment occurs when there are people actively looking for work at the equilibrium wage rate but are not able to find work short supply the amount demanded is greater than the amount supplied raw material the basic material from which a good is made demand the amount of a good that consumers are willing and able to buy at each price unskilled labour workers lacking in skills, training, and education skilled labour workers who are well trained, well educated, and who are experts in their field Glossary Test your understanding of this unit by answering the following questions capacity the amount of something a firm is able to Price of corn increases by 10% leading to an increase in quantity supplied of 1%. Comment on the make value of PES. Explain why PES becomes more elastic over time. Explain why PES is likely to be more inelastic during high levels of economic activity. spare...extra 31
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# SNAP Percentage Questions PDF [Most Important]
The percentage is an important topic in the Quant section of the SNAP Exam. You can also download this Free Percentage Questions for SNAP PDF with detailed answers by Cracku. These questions will help you practice and solve the Percentage questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practicing.
Question 1:Â Matthew scored 42 marks in biology, 51 marks in chemistry, 58 marks in mathematics, 35 marks in physics and 48 marks in English. The maximum marks a student can score in each subject are 60. How much overall percentage did Matthew get in this exam?
a)Â 76
b)Â 82
c)Â 68
d)Â 78
e)Â None of these
Solution:
Total marks obtained by Matthew
= 42 + 51 + 58 + 35 + 48 = 234
Maximum marks of the five subjects = 5 * 60 = 300
=> Required % = $\frac{234}{3} \times 100$ = 78Â %
Question 2:Â Aryan got 350 marks and Vidya scored 76 percent marks in the same test. If Vidya scored 296 marks more than Aryan, what were the maximum marks of the test ?
a)Â 650
b)Â 900
c)Â 850
d)Â 950
e)Â None of these
Solution:
Marks scored by Vidya = 350 + 296 = 646
Let the maximum marks be $x$
=> $\frac{76}{100} x = 646$
=> $x = \frac{646}{0.76} = 850$
Question 3:Â Last year there were 610 boys in a school. The number decreased by 20 percent this year. How many girls are there in the school if the number of girls is 175 percent of the total number of boys in the school this year ?
a)Â 854
b)Â 848
c)Â 798
d)Â 782
e)Â None of these
Solution:
Present population of boys = $\frac{80}{100} \times 610$ = 488
=> Number of girls = $\frac{175}{100} \times 488$
= $7 \times 122$ = 854
Instructions
<p “=””>Study the following information carefully to answer the questions that follow :Â <p “=””>There are two Trains, Train-A and Train-B. Both Trains have four different types of Coaches viz. General Coaches, Sleeper Coaches, First Class Coaches and AC Coaches. In Train A there are total 700 passengers. Train-B has thirty percent more passengers than Train A. Twenty percent of the passengers of Train-A are in General Coaches. One-fourth of the total number of passengers of Train-A are in AC coaches. Twenty three percent of the passengers of Train-A are in Sleeper Class Coaches. Remaining passengers of Train-A are in first class coaches. Total number of passengers in AC coaches in both the trains together is 480. Thirty percent of the number of passengers of Train-B is in Sleeper Class Coaches. Ten percent of the total passengers of Train-B are in first class coaches. Remaining passengers of Train-B are in general class coaches.
Question 4:Â Total number of passengers in General Class coaches in both the Trains together is approximately what percentage of total number of passengers in Train-B ?
a)Â 35
b)Â 42
c)Â 46
d)Â 38
e)Â 31
Solution:
Total passengers in train A = 700
=> Total passengers in train B = $\frac{130}{100} \times 700 = 910$
Number of passengers In train AÂ in the class :
General = $\frac{20}{100} \times 700 = 140$
AC = $\frac{1}{4} \times 700 = 175$
Sleeper = $\frac{23}{100} \times 700 = 161$
First = $700 – 140 – 175 – 161 = 224$
Number of passengers in train B in the class :
AC = $480 – 175 = 305$
Sleeper = $\frac{30}{100} \times 910 = 273$
First = $\frac{10}{100} \times 910 = 91$
General = $910 – 305 – 273 – 91 = 241$
Total number of passengers in General Class coaches in both the Trains together = 140 + 241 = 381
Total number of passengers in Train-BÂ = 910
=> Required % = $\frac{381}{910} \times 100$
= $41.86 \% \approx 42 \%$
Question 5:Â If tax on a commodity is reduced by 10%, total revenue remains unchanged. What is the percentage increase in its consumption?
a)Â $11\frac{1}{9}$
b)Â $20$%
c)Â $10$%
d)Â $9\frac{1}{11}$
e)Â None of these
Solution:
Revenue = consumption $\times$ tax amount
Let consumption = 10 and tax = 10
=> Revenue = $10 \times 10 = 100$
Now, after tax is reduced by 10 %, new tax = $10 – \frac{10}{100} \times 10$
= $10 – 1 = 9$
Total revenue remains unchanged
=> New consumption = $\frac{100}{9}$
$\therefore$ % increase in consumption = $\frac{\frac{100}{9} – 10}{10} \times 100$
= $\frac{10}{9} \times 10 = \frac{100}{9}$
= $11\frac{1}{9} \%$
Take SNAP mock tests here
Question 6:Â Sujata scored 2240 marks in an examination that is 128 marks more than the minimum passing percentage of 64%. What is the percentage of marks obtained by Meena if she scores 907 marks less than Sujata?
a)Â 35
b)Â 40
c)Â 45
d)Â 36
e)Â 48
Solution:
Let maximum marks in the examination = $100x$
Minimum passing marks = $\frac{64}{100} \times 100x = 64x$
Marks scored by Sujata = $2240$
Acc to ques,
=> $64x + 128 = 2240$
=> $64x = 2240 – 128 = 2112$
=> $x = \frac{2112}{64} = 33$
=> Maximum marks = $100 \times 33 = 3300$
Marks scored by Meena = $2240 – 907 = 1333$
$\therefore$ % marks obtained by Meena = $\frac{1333}{3300} \times 100$
= $40.39 \% \approx 40 \%$
Question 7: The difference between the 5/6 th of a number and 35 percent of the same is 1392 What will be 55% of that number ‘?
a)Â 2880
b)Â 1584
c)Â 1854
d)Â 1485
e)Â None of these
Solution:
Let the number be ‘n’.
According to question,
$\frac{5}{6}$n-35% of n = 1392.
$\frac{5}{6}n-\frac{35\times{n}}{100}=1392$.
$\frac{29\times{n}}{60}=1392$.
$n=2880$.
55 % of n = $\frac{55\times2880}{100}$.
=1584.
Hence, Option B is correct.
Question 8:Â Sohail scored 33 marks in English, 37 marks in Science, 28 marks in Mathematics, 26 marks in Hindi and 32 marks in Social studies. The maximum marks a student can score in each subject in 60. How much percentage did Sohail get in this exam ?
a)Â 52
b)Â 54
c)Â 48
d)Â 53
e)Â None of these
Solution:
So Sohail’s overall percentage marks-
=$\frac{33+37+28+26+32}{300}$ $\times$100
=$\frac{156}{300}$ $\times$100
=52
Question 9:Â In an examination the maximum aggregate marks that a student can get are 1020. In order to pass the exam a student is required to get 663 marks out of the aggregate marks. Shreya got 612 marks. By what percent did Shreva fail in the exam ?
a)Â 5
b)Â 8
c)Â 7
d)Â Cannot he determined
e)Â None of these
Solution:
Passing percentage=663/1020*100=65%
Shreya percentage=612/1020*100=60%
Therefore she failed by 5%
Question 10:Â Sushil got 103 marks in Hindi, 111 marks in Science, 98 marks in Sanskrit. 110 marks in Maths and 88 marks in English. If the maximum marks of each subject are equal and if Sushil scored 85 percent marks in all the subjects together, what is the maximum mark of each subject ?
a)Â 110
b)Â 120
c)Â 115
d)Â 100
e)Â None of these
Solution:
Total marks scored by sushil=103+111+98+110+88=510
Let total max. Marks be x.
Now, it is given that, 85% of x=total marks scored by sushil
I.e, 0.85*x=510
Therefore, x=510*100/85
x=600
This is for 5 subjects altogether.
Therefore, max. Marks for one subject = 600/5 =120 marks.
Option B is the right answer.
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# Common Core: 6th Grade Math : Recognize Opposite Signs of Numbers as Indicating Locations on Opposite Sides of 0 on the Number Line: CCSS.Math.Content.6.NS.C.6a
## Example Questions
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### Example Question #1 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #2 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #2 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #3 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #4 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #5 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #7 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #6 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #7 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
### Example Question #8 : Recognize Opposite Signs Of Numbers As Indicating Locations On Opposite Sides Of 0 On The Number Line: Ccss.Math.Content.6.Ns.C.6a
Select the number that is the opposite of
Possible Answers:
Correct answer:
Explanation:
If the original number is negative, the opposite number will be the positive value of that same number. Vise versa, if the original number is positive, the opposite number will be the negative value of the same number.
In this case, the opposite of is . As shown from the image, the numbers are on opposite sides of on the number line.
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Larry Riddle, Agnes Scott College
### Dihedral Group and Triangle Fractals
#### Dihedral Group
The six transformations of an equilateral triangle shown below form a finite group called the dihedral group of order 6. Transformations 2 and 3 are counterclockwise rotations by 120° and 240°, respectively. Transformations 4, 5, and 6 are reflections across the altitudes through the top, right, and left vertices, respectively.
The following table shows the result of combining one transformation with another. The one down the rows is done first, followed by the one across the columns. If we call the transformations Tn for n = 1, 2, 3, 4, 5, 6, then the table shows the result of the composition Tc•Tr, where r and c denote row and column.
$\begin{array}{*{20}{c}} {} & {\begin{array}{*{20}{c}} 1 & 2 & 3 & 4 & 5 & 6 \\ \end{array} } \\ {\begin{array}{*{20}{c}} 1 \\ 2 \\ 3 \\ 4 \\ 5 \\ 6 \\ \end{array} } & {\boxed{\begin{array}{*{20}{c}} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 6 & 4 & 5 \\ 3 & 1 & 2 & 5 & 6 & 4 \\ 4 & 5 & 6 & 1 & 2 & 3 \\ 5 & 6 & 4 & 3 & 1 & 2 \\ 6 & 4 & 5 & 2 & 3 & 1 \\ \end{array} }} \\ \end{array}$
So, for example, applying a reflection across the right corner of an equilateral triangle (#5) followed by a rotation by 120° (#2) is the same as doing a reflection of the triangle across the left corner (#6), as illustrated below. Hence T6 = T2•T5. Note that the composition order is not commutative, i.e. the order in which you do the operations is important.
In fact, you only need transformations 2 (rotation by 120°) and 4 (reflection across a vertical line) to generate the other three non-identity transformations:
to get transformation 3, repeat transformation 2 twice: T3 = T2•T2
to get transformation 5, do reflection 4 then rotation 2: T5 = T2•T4
to get transformation 6, do rotation 2 then reflection 4: T6 = T4•T2
This means that the functions in an iterated function system based on these types of triangle fractals can be generated by a combination of scaling by 1/2, rotating by 120°, and a vertical reflection. Well, except you must always remember to include the extra 180° rotation for the "middle" triangle.
#### IteratedFunctionSystem
As an example, consider the triangle fractal 253. We will take the initial set to be an equilateral triangle with vertices at the origin and (1,0).
The first function must scale by 1/2 and rotate by 120°. As the figure below shows, the triangle must also be translated 1/2 to the right.
\begin{aligned} {f_1}({\mathbf{x}}) &= \left[ {\begin{array}{*{20}{c}} {\cos {{(120)}^ \circ }} & { - \sin {{(120)}^ \circ }} \\ {\sin {{(120)}^ \circ }} & {\cos {{(120)}^ \circ }} \\ \end{array} } \right]\left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array} } \right]{\mathbf{x}} + \left[ {\begin{array}{*{20}{c}} {1/2} \\ {0} \\ \end{array} } \right] \\ &= \left[ {\begin{array}{*{20}{c}} { - 1/4} & {-\sqrt 3 /4} \\ { \sqrt 3 /4} & { - 1/4} \\ \end{array} } \right]{\mathbf{x}} + \left[ {\begin{array}{*{20}{c}} {1/2} \\ {0} \\ \end{array} } \right] \\ \end{aligned}
The second function must scale by 1/2, followed by a reflection across the vertical axis, and then a rotation by 120°. But because this is the middle triangle, the function must also do a rotation by 180°. As the figure below shows, this must then by followed by a translation by 1/2 to the right.
\begin{aligned} {f_2}({\mathbf{x}}) &= \left[ {\begin{array}{*{20}{c}} { - 1} & 0 \\ 0 & { - 1} \\ \end{array} } \right]\left[ {\begin{array}{*{20}{c}} {\cos {{(120)}^ \circ }} & { - \sin {{(120)}^ \circ }} \\ {\sin {{(120)}^ \circ }} & {\cos {{(120)}^ \circ }} \\ \end{array} } \right]\left[ {\begin{array}{*{20}{c}} { - 1} & 0 \\ 0 & 1 \\ \end{array} } \right]\left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array} } \right]{\mathbf{x}} + \left[ {\begin{array}{*{20}{c}} {1/2} \\ 0 \\ \end{array} } \right] \\ &= \left[ {\begin{array}{*{20}{c}} {-1/4} & {\sqrt 3 /4} \\ {\sqrt 3 /4} & { 1/4} \\ \end{array} } \right]{\mathbf{x}} + \left[ {\begin{array}{*{20}{c}} {1/2} \\ 0 \\ \end{array} } \right] \\ \end{aligned}
The third function must scale by 1/2, then do two rotations by 120° (for a total rotation of 240°). The top of the scaled triangle must then be translated to the midpoint of the right side of the original triangle.
\begin{aligned} {f_3}({\mathbf{x}}) &= \left[ {\begin{array}{*{20}{c}} {\cos {{(240)}^ \circ }} & { - \sin {{(240)}^ \circ }} \\ {\sin {{(240)}^ \circ }} & {\cos {{(240)}^ \circ }} \\ \end{array} } \right]\left[ {\begin{array}{*{20}{c}} {1/2} & 0 \\ 0 & {1/2} \\ \end{array} } \right]{\mathbf{x}} + \left[ {\begin{array}{*{20}{c}} {3/4} \\ {\sqrt 3 /4} \\ \end{array} } \right] \\ &= \left[ {\begin{array}{*{20}{c}} { - 1/4} & {\sqrt 3 /4} \\ { - \sqrt 3 /4} & { - 1/4} \\ \end{array} } \right]{\mathbf{x}} + \left[ {\begin{array}{*{20}{c}} {3/4} \\ {\sqrt 3 /4} \\ \end{array} } \right] \\ \end{aligned}
This IFS produces the following triangle fractal.
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# Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nthweek, her week, her weekly savings become Rs 20.75, find n. - Mathematics
Ramkali saved Rs 5 in the first week of a year and then increased her weekly saving by Rs 1.75. If in the nthweek, her week, her weekly savings become Rs 20.75, find n.
#### Solution
Given that,
a = 5
d = 1.75
an = 20.75
n = ?
an = a + (n − 1) d
20.75 = 5 + (n - 1) × 1.75
15.75 = (n - 1) × 1.75
(n - 1) = 15.75/1.75 = 1575/175
= 63/7 = 9
n - 1 = 9
n = 10
Hence, n is 10.
Concept: nth Term of an AP
Is there an error in this question or solution?
Chapter 5: Arithmetic Progressions - Exercise 5.2 [Page 107]
#### APPEARS IN
NCERT Class 10 Maths
Chapter 5 Arithmetic Progressions
Exercise 5.2 | Q 20 | Page 107
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# NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration
NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration, has been designed by the NCERT to test the knowledge of the student on the topic – Mensuration Introduction
### NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration
NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration
1. A square and a rectangular field with measurements as given in the figure have the same perimeter. Which field has a larger area?
Sol.: – We know that,
Area of Square = (Side)2
Area of Rectangular = Length × Breadth
As we don’t know Breadth of Rectangle, we need to find it first to calculate area.
We know that,
Perimeter of square = Perimeter of rectangular, (Mentioned in question)
Also,
Perimeter of square = Sum of all side = 4 × Side, And
Perimeter of rectangular = 2 × (Length + Breadth)
So,
4 × Side = 2 × (Length + Breadth)
4 × 60 = 2 × (80 + Breadth)
After solving this equation, we get
= 40m
Area of Square = (Side)2 = (60)2 = 60 × 60 = 3600 m2
And, Area of Rectangle = Length × Breadth = 40 × 80 = 3200 m2
Therefore, area of Square field is larger than Rectangular field.
2. Mrs. Kaushik has a square plot with the measurement as shown in the figure. She wants to construct a house in the middle of the plot. A garden is developed around the house. Find the total cost of developing a garden around the house at the rate of Rs 55 per m2.
Sol.: – Area of garden = Area of Total plot – Area of house
= 25 × 25 – 15 × 20
= 625 – 300
=325 m2
So, total cost of developing a garden around the house
= Total Area of garden × Cost of developing the garden per m2
= 325 × 55
= ₹ 17875
3. The shape of a garden is rectangular in the middle and semi circular at the ends as shown in the diagram. Find the area and the perimeter of this garden [Length of rectangle is 20 – (3.5 + 3.5) metres].
Sol.: – As we can see in the figure,
Perimeter of the garden = Perimeter of rectangle × perimeter of semi-circle
= (2 × Length of rectangle) + (2 × π × Radius of circle)
(Here, Length of rectangle = 20 – (3.5 + 3.5) = 13 m)
(Here, Radius of circle = 7/2 = 3.5 m)
So, Perimeter of the garden = (2 x 13) + (2 x (22/7) x 3.5)
= 26 + 22
= 48 m
And, Area of the garden = Area of rectangle + 2 × Area of semi-circle
= (Length × Breadth) + 2 × (πr²/2)
= (13 × 7) + 2 × (22×3.5²/2×7)
= 91 + 38.5
=129.5 m2
4. A flooring tile has the shape of a parallelogram whose base is 24 cm and the corresponding height is 10 cm. How many such tiles are required to cover a floor of area 1080 m2? (If required you can split the tiles in whatever way you want to fill up the corners).
Sol.: – We know that,
Area of parallelogram = Base × Height
= 24 × 10
= 240 cm2
Area of floor = 1080 m2
= 1080 × 10000 m(As 1m2= 10000 cm2 )
= 10800000 m2
So, Number of tiles required to cover the floor = (Area of floor/Area of parallelogram)
= 10800000/240
= 45000 tiles
5. An ant is moving around a few food pieces of different shapes scattered on the floor. For which food-piece would the ant have to take a longer round? Remember, circumference of a circle can be obtained by using the expression c = 2πr, where r is the radius of the circle.
Sol.: – For checking on which food-piece the ant would take longer round; we need to find Perimeter of all food-piece.
a). Perimeter of 1st food-piece = circumference of semi-circle + D
= πr + D
= [(22 × 1.4)/7] + 2.8
= 7.2 cm
b). Perimeter of 2nd food-piece = circumference of semi-circle + 3 sides of rectangle
= [(22 × 1.4)/7] + 1.5 +1.5 +2.8
= 10.2 cm
c). Perimeter of 3rd food-piece = circumference of semi-circle + 2 sides of triangle
= [(22 × 1.4)/7] + 2 + 2
= 8.4 cm
Therefore, Perimeter of food piece (b) is largest. So, ‘Ant’ will take longest round around food piece (b).
The next Exercise for NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.2 – Mensuration can be accessed by clicking here
Download NCERT Solutions for Class 8 Maths Chapter 11 Exercise 11.1 – Mensuration
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# How do you combine (x-10)/(x-8)-(x+10)/(8-x)?
Feb 4, 2017
See the entire solution process below:
#### Explanation:
First, we need to put the fractions over common denominators. We can do this by multiplying the first fraction by the appropriate form of $1$:
$\left(\frac{- 1}{-} 1 \times \frac{x - 10}{x - 8}\right) - \frac{x + 10}{8 - x} \to$
$\left(\frac{- 1 \times \left(x - 10\right)}{- 1 \times \left(x - 8\right)}\right) - \frac{x + 10}{8 - x} \to$
$\frac{- x + 10}{- x + 8} - \frac{x + 10}{8 - x} \to$
$\frac{- x + 10}{8 - x} - \frac{x + 10}{8 - x}$
Now that there is a common denominator for each fraction we can subtract the numerators:
$\frac{- x + 10 - x - 10}{8 - x}$
$\frac{- x - x + 10 - 10}{8 - x}$
$\frac{- 2 x + 0}{8 - x}$
$\frac{- 2 x}{8 - x}$
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# Define Ellipses
Define Ellipses In mathematics , geometry is define as the warehouse of several figures that are expressed through different types of expression. In geometry there is one figure that is define as the Ellipse. According to the definition of ellipse , it is a set of points that are on a plane and sum of the distance from one point to two other fixed points is equal to a constant value. If there is a point p and two fixed points k1 , k2 then according to the ellipse equation it will express as p k1 + p k2 = c here c is describe as a constant value. Here these two fixed points are define as the focal points of the ellipse. Means point k1 and k2 are focal points of ellipse. As we define the ellipse there are basically two axis that are known as major axis and minor axis and these axis are end on the endpoints that are situated on the ellipse and these endpoints on the ellipse are define as the vertices.
## Know More About :- Find Surface Area Of A Cone
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These two major and minor axis are perpendicular to each other means these axis are created 90' angle between them. The mid point on the major axis is called as center of the ellipse. Minor axis also have the end points that are also situated on the ellipse and these end points are called as the co vertices. We can also describe the endpoints as the meeting point or intersection point of the ellipse and major axis and as well as co vertices are define as the intersection point of minor axis and ellipse. We can express the ellipse in for of expression that are standard form of the ellipse as follows: As we know an ellipse is a close two dimensional curve and there are two axis that are different in lengths. x 2 / a 2 + y 2 / b 2 = 1. In the above expression a and b are two axis of the ellipse that are of different lengths in which a is greater in length than b so a is define as the major or semi major axis and b is called as minor or semi minor axis. We can also express ellipse equation in form of major axis and its eccentricity e when the center of the ellipse is same as origin of x y plane means center of ellipse is the origin of x y plane then expression for ellipse is expressed as e=(1b2/a2)1/2. when in the x y plane , x axis shows as a major axis of the ellipse then expression for ellipse is expressed as: f = 1 b / a, where f is called as flattening.
## Learn More :- How to Find the Surface Area of a Cube
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There are some other ways to define the ellipse expression as : (x h) 2 / a 2 + (y k) 2 / b 2 = 1. In the above expression a and b should always be positive and h , k are real numbers. In the ellipse expression we know that there are two focal points of ellipse that are situated on the axis and one focal point is described as the origin. Some of people define central of ellipse as origin so there are some distance between these oints that are calculated as : c = (a 2 b 2) 1 / 2 where c define the distance.
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6.4 Average Project Shum Science for Kids
Contents
# Mode
The mode is the score that occurs most frequently. For example, the mode in the set {6, 3, 2, 3, 7, 8, 2, 3, 6, 2} is 2, because it occurs 3 times, more time than the other numbers.
In the above histogram example, the mode is the most recurring score.
Here, the most recurring score is 6 secs, which occurs 8 times.
# Median
The median is the middle score of a sorted set. For example, the median of 1, 3, 3, 3, 4, 7, 5, 9, 5, 7, 1, 8, 3, 2, 5, 5, 6. First, the set should be arranged from smallest to largest, then the middle selected, i.e. {1, 1, 2, 3, 3, 3, 3, 4, 5, 5, 5, 5, 6, 7, 7, 8, 9}. Of the 17 numbers, the middle is number 8, or the number 4.
Where there are odd number of scores, the median is put down to 1 possibility. However, where there are an even number of scores, the median can have 2 possibilities. The median is the mean of these 2 numbers (the mean is explained below).
# Mean
The mean is the mathematical average, calculated by summing up the scores, and dividing by the number of scores. For example, for the set {5, 2, 6, 4, 8, 9, 12, 2}, the mean is calculated as $\dfrac{5+2+6+4+8+9+12+2}{8}=\dfrac{48}{8}=6$.
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# Factors and Multiples
Factors and Multiples – Definition
Factors – The factors of a numbers are those numbers by which that number is completely divisible, with a remainder of 0.
Multiple – A multiple is the product of two or more factors.
### Examples of Factors and Multiples
Example 1
Determine whether 4 is the factor of 32 ?
Explanation
The factors of a numbers are those numbers by which that number is completely divisible, with a remainder of 0
When we divide 32 by 4
i.e, 32 ÷ 4
We get,
Quotient = 8
Remainder = 0
Here we will see that the remainder is 0 because 32 is exactly divisible by 4
Hence, we can say that 4 is the factor of 32 .
Example 2
Determine whether 32 is the multiple of 6 ?
Explanation
If 32 is the multiple of 6 , then the remainder is 0
when 32 is divided by 6 remainder should be equal to 0.
When we divide 32 by 6
i.e, 32 ÷ 6
Quotient = 5
Remainder = 2
Here we will see that the remainder is 2
Hence, we can say that 32 is not the multiple of 6
Example 3
Determine whether 6 is the factor of 40 ?
Explanation
The factors of a numbers are those numbers by which that number is completely divisible, with a remainder of 0
When we divide 40 by 6
i.e, 40 ÷ 6
We get,
Quotient = 6
Remainder = 4
Here we will see that the remainder is 4 because 40 is not exactly divisible by 6
Hence, we can say that 6 is not the factor of 40 .
Example 4
Determine whether 36 is the multiple of 6 ?
Explanation
If 36 is the multiple of 6 , then the remainder is 0
when 36 is divided by 6 remainder should be equal to 0.
When we divide 36 by 6
i.e, 36 ÷ 6
Quotient = 6
Remainder = 0
Here we will see that the remainder is 0
Hence, we can say that 36 is the multiple of 6
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A vector algebra is an algebra where the terms are denoted by vectors and operations are performed corresponding to algebraic expressions. The "Distributive Law" is the BEST one of all, but needs careful attention. However, if you convert the subtraction to an addition, you can use the commutative law - both with normal subtraction and with vector subtraction. According to Newton's law of motion, the net force acting on an object is calculated by the vector sum of individual forces acting on it. Thus, A – B = A + (-B) Multiplication of a Vector. This law is known as the associative law of vector addition. The elements are often numbers but could be any mathematical object provided that it can be added and multiplied with acceptable properties, for example, we could have a vector whose elements are complex numbers.. Vector addition and subtraction is simple in that we just add or subtract corresponding terms. Commutative Property: a + b = b + a. In practice, to do this, one may need to make a scale diagram of the vectors on a paper. These quantities are called vector quantities. The first is a vector sum, which must be handled carefully. The unit vectors i and j are directed along the x and y axes as shown in Fig. What is Associative Property? Subtraction of Vectors. We'll learn how to solve this equation in the next section. So, the 3× can be "distributed" across the 2+4, into 3×2 and 3×4. Vector addition is commutative, i. e. . They include addition, subtraction, and three types of multiplication. A vector is a set of elements which are operated on as a single object. Characteristics of Vector Math Addition. You can regard vector subtraction as composition of negation and addition. 5. Recall That Vector Addition Is Associative: (u+v)+w=u+(v+w), For All U, V, W ER". Commutative Law- the order of addition does not matter, i.e, a + b = b + a; Associative law- the sum of three vectors has nothing to do with which pair of the vectors are added at the beginning. Matrix subtraction is not associative (neither is subtraction of real numbers) Scalar Multiplication. This is what it lets us do: 3 lots of (2+4) is the same as 3 lots of 2 plus 3 lots of 4. Thus vector addition is associative. (Vector addition is also associative.) A.13 shows A to be the vector sum of Ax and Ay.That is, AA A=+xy.The vectors Ax and Ay lie along the x and y axes; therefore, we say that the vector A has been resolved into its x and y components. As shown, the resultant vector points from the tip The above diagrams show that vector addition is associative, that is: The same way defined is the sum of four vectors. Properties.Several properties of vector addition are easily verified. This can be illustrated in the following diagram. The matrix can be any order; ... X is a column vector containing the variables, and B is the right hand side. (This definition becomes obvious when is an integer.) Vector addition is commutative, just like addition of real numbers. Vector addition involves only the vector quantities and not the scalar quantities. For example, X & Y = X + (&Y), and you can rewrite the last equation When adding vectors, all of the vectors must have ... subtraction is to find the vector that, added to the second vector gives you the first vector ! We can multiply a force by a scalar thus increasing or decreasing its strength. Subtraction of a vector B from a vector A is defined as the addition of vector -B (negative of vector B) to vector A. 1. If you start from point P you end up at the same spot no matter which displacement (a or b) you take first. • Vector addition is commutative: a + b = b + a. Vector quantities are added to determine the resultant direction and magnitude of a quantity. *Response times vary by subject and question complexity. Is (u - V) - W=u-(v - W), For All U, V, WER”? Vector addition is associative in nature. ! (a + b) + c = a + (b + c) Vector Subtraction Mathematically, Vector quantities also satisfy two distinct operations, vector addition and multiplication of a vector by a scalar. Associative law is obeyed in vector addition while not in vector subtraction. Scalar-vector multiplication. The sum of two vectors is a third vector, represented as the diagonal of the parallelogram constructed with the two original vectors as sides. Vector subtraction is similar. If $a$ and $b$ are numbers, then subtraction is neither commutative nor associative. ( – ) = + (– ) where (–) is the negative of vector . Vector subtraction is similar to vector addition. 1. This … Adding Vectors, Rules final ! Two vectors of different magnitudes cannot give zero resultant vector. Using the technique of Fig. Distributive Law. This is the triangle law of vector addition . For matrix multiplication, the number of columns in the first matrix must be equal to the number of rows in the second matrix. We define subtraction as addition with the opposite of a vector: $$\vc{b}-\vc{a} = \vc{b} + (-\vc{a}).$$ This is equivalent to turning vector $\vc{a}$ around in the applying the above rules for addition. Resolution of vectors. Vector Subtraction. In mathematics, particularly in linear algebra, matrix multiplication is a binary operation that produces a matrix from two matrices. Associative law is obeyed by - (A) Addition of vectors. i.e. If two vectors and are to be added together, then 2. Each form has advantages, so this book uses both. acceleration vector of the mass. A) Let W, X, Y, And Z Be Vectors In R”. The resultant vector, i.e. This fact is known as the ASSOCIATIVE LAW OF VECTOR ADDITION. Vectors are entities which has magnitude as well as direction. Vector addition is associative:- While adding three or more vectors together, the mutual grouping of vector does not affect the result. Vector Addition is Commutative. (Here too the size of $$0$$ is the size of $$a$$.) This is called the Associative Property of Addition ! You can move around the points, and then use the sliders to create the difference. Subtracting a vector from itself yields the zero vector. Notes: When two vectors having the same magnitude are acting on a body in opposite directions, then their resultant vector is zero. The process of splitting the single vector into many components is called the resolution of vectors. the vector , is the vector that goes from the tail of the first vector to the nose of the last vector. Vector addition (and subtraction) can be performed mathematically, instead of graphically, by simply adding (subtracting) the coordinates of the vectors, as we will see in the following practice problem. (If The Answer Is No, Justify Your Answer By Giving A Counterexample.) associative law. By a Real Number. This video shows how to graphically prove that vector addition is associative with addition of three vectors. Such as with the graphical method described here. Justify Your Answer. VECTOR AND MATRIX ALGEBRA 431 2 Xs is more closely compatible with matrix multiplication notation, discussed later. Let these two vectors represent two adjacent sides of a parallelogram. Vector subtraction does not follow commutative and associative law. And we write it like this: Following is an example that demonstrates vector subtraction by taking the difference between two points – the mouse location and the center of the window. COMMUTATIVE LAW OF VECTOR ADDITION: Consider two vectors and . However, in the case of multiplication, vectors have two terminologies, such as dot product and cross product. Is Vector Subtraction Associative, I.e. As an example, The result of vector subtraction is called the difference of the two vectors. Associative law states that result of, numbers arranged in any manner or group, will remain same. Consider two vectors and . Health Care: Nurses At Center Hospital there is some concern about the high turnover of nurses. We also find that vector addition is associative, that is (u + v) + w = u + (v + w ). Vector Addition is Associative. Another operation is scalar multiplication or scalar-vector multiplication, in which a vector is multiplied by a scalar (i.e., number), which is done by multiplying every element of the vector by the scalar. The resulting matrix, known as the matrix product, has the number of rows of the first and the number of columns of the second matrix. The head-to-tail rule yields vector c for both a + b and b + a. The vector $$\vec a + \vec b$$ is then the vector joining the tip of to $$\vec a$$ the end-point of $$\vec b$$ . A scalar is a number, not a matrix. We can add two forces together and the sum of the forces must satisfy the rule for vector addition. We construct a parallelogram. We construct a parallelogram : OACB as shown in the diagram. Let these two vectors represent two adjacent sides of a parallelogram. If is a scalar then the expression denotes a vector whose direction is the same as , and whose magnitude is times that of . We will find that vector addition is commutative, that is a + b = b + a . Vector operations, Extension of the laws of elementary algebra to vectors. The second is a simple algebraic addition of numbers that is handled with the normal rules of arithmetic. Median response time is 34 minutes and may be longer for new subjects. It can also be shown that the associative law holds: i.e., (1264) ... Vector subtraction. 8:24 6 Feb 2 Clearly, &O = OX + O = X &(&X) = XX + (&X) = O. ... subtraction, multiplication on vectors. This property states that when three or more numbers are added (or multiplied), the sum (or the product) is the same regardless of the grouping of the addends (or the multiplicands).. Grouping means the use of parentheses or brackets to group numbers. Vector addition is commutative:- It means that the order of vectors to be added together does not affect the result of addition. The applet below shows the subtraction of two vectors. $$\vec a\,{\rm{and}}\,\vec b$$ can equivalently be added using the parallelogram law; we make the two vectors co-initial and complete the parallelogram with these two vectors as its sides: Addition and Subtraction of Vectors 5 Fig. Note that we can repeat this procedure to add any number of vectors. Worked Example 1 ... Add/subtract vectors i, j, k separately. Multiplication of a vector by a positive scalar changes the length of the vector but not its direction. For any vectors a, b, and c of the same size we have the following. ... Vector subtraction is defined as the addition of one vector to the negative of another. Associative property involves 3 or more numbers. Well, the simple, but maybe not so helpful answer is: for the same reason they don’t apply to scalar subtraction. A.13. For question 2, push "Combine Initial" to … Vector addition is commutative and associative: + = + , ( + )+ = +( + ); and scalar multiplication is distributive: k( + ) = k +k . VECTOR ADDITION. The vector $-\vc{a}$ is the vector with the same magnitude as $\vc{a}$ but that is pointed in the opposite direction. When a vector A is multiplied by a real number n, then its magnitude becomes n times but direction and unit remains unchanged. Question 2. … the resultant vector notation, discussed later c = a + b and b is the same as and... Performed corresponding to algebraic expressions, vectors have two terminologies, such as dot product and cross product Xs more. 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# Total positive solution of a linear equation?
I have a linear equation in four variable
$a + b + c + d = 10$
given that $a, b, c, d > 0$ and $a,b,c,d <= 6$ and $a,b,c, d$ are integer
what are total possible solution ?
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@Clayton - yes! – AppDeveloper Jan 30 '13 at 23:38
Let's first enumerate the solutions with $a \le b \le c \le d$.
You must have $d \ge 10/4 = 2.5$, so $d = 3$, $4$, $5$ or $6$.
With $d=6$, $a+b+c=4$, and it's easy to see that $a=1$, $b=1$, $c=2$.
With $d=5$, $a+b+c=5$, and you could have $a=1$, $b=1$, $c=3$ or $a=1$, $b=2$, $c=2$.
With $d=4$, $a+b+c=6$, and you could have $a=1$, $b=1$, $c=4$, or $a=1$, $b=2$, $c=3$ or $a=2$, $b=2$,$c=2$.
With $d=3$, $a+b+c=7$, and you could have $a=1$, $b=3$,$c=3$ or $a=2$,$b=2$,$c=3$.
Thus the solutions with $a \le b \le c \le d$ are $[a,b,c,d] = [1,1,2,6]$, $[1,1,3,5]$, $[1,2,2,5]$, $[1,1,4,4]$, $[1,2,3,4]$, $[2,2,2,4]$, $[1,3,3,3]$, $[2,2,3,3]$. Take all permutations of those and you get all the positive integer solutions.
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This is too painful. 7 is close enough to 10 that you should consider counting the complement. – Calvin Lin Jan 30 '13 at 23:59
Hint: Use the Principle of Inclusion and Exclusion. In a general solution of positive integers $w + x + y + z = 10$, notice that at most one of the them can be at least 7.
Additional Hint: If $a\geq 7$, there is only 1 solution.
Hence, by PIE, there are ${9 \choose 3} - 1 - 1 - 1 - 1 = 80$ solutions.
For Poseidonium. Assuming we want to solve $a+b+c+d = 20, a, b, c, d \leq 6$.
Let sets $A, B, C, D$ be the ways such that $a \geq 7$, etc. By PIE, we have
$$|A \cup B \cup C \cup D| = \sum |A| - \sum |A\cap B| + \sum |A \cap B \cap C| - |A \cap B \cap C \cap D|$$
This is equal to $4 \times { 13 \choose 3} - 6 \times {7 \choose 3} + 6 \times {1 \choose 3} - 0 = 934$. Hence, the number of ways is ${19 \choose 3} - |A \cup B \cup C \cup D| = 35$.
[I would have approached this by solving $w + x + y + z = 4$, using the substitution $w =6-a$ etc, subject to $0 \leq w \leq 5$, which is clearly a non-restriction since the sum is only 4, thus there are ${4+3 \choose 3} = 35$ solutions.]
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a lil more help? – AppDeveloper Jan 30 '13 at 23:46
@AppDeveloper Do you know what PIE is? Have you applied it before? – Calvin Lin Jan 30 '13 at 23:55
@Poseidonium Thanks! Strangely I'm not used to the positive integer version, and much more familiar with the $\geq 0$ complicated version lol. – Calvin Lin Jan 31 '13 at 0:03
@Poseidonium Do you mean replace 10 by 30? If so, I'd do it directly by saying there are no solutions. – Calvin Lin Jan 31 '13 at 1:03
@Poseidonium I've shown how to 'properly' apply PIE using $a+b+c+d = 20$ as an example. – Calvin Lin Jan 31 '13 at 1:11
I would do by a combinatorial approach.
Let's say you have 10 books a bookshelf and you have 3 stands to partition the books. (a/b/c/d books and each / being the stand)
You don't put stands next to each other. (a,b,c,d > 0)
You don't want a partition bigger than 6.. (a,b,c,d <= 6)
We further break it down into simpler parts and slowly work to our solution.
Part 1
If we are not restricted to a maximum of 6 books, we can put the 3 dividers between any 2 books and that gives us 9 options, ie. ${}^9{C_3}=84$
Unfortunately, this is not our answer as there will be some cases when the partitioning has more than 6 books.
Part 2 (which are the cases which are not acceptable?)
If there is more than 6, it has to be the partitions of 1/1/1/7, 1/1/7/1, 1/7/1/1 or 7/1/1/1.
So, there are 4 unacceptable cases
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# 2020 USAMO Problems/Problem 1
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 1
Let $ABC$ be a fixed acute triangle inscribed in a circle $\omega$ with center $O$. A variable point $X$ is chosen on minor arc $AB$ of $\omega$, and segments $CX$ and $AB$ meet at $D$. Denote by $O_1$ and $O_2$ the circumcenters of triangles $ADX$ and $BDX$, respectively. Determine all points $X$ for which the area of triangle $OO_1O_2$ is minimized.
## Solution
Let $E$ be midpoint $AD.$ Let $F$ be midpoint $BD \implies$ $$EF = ED + FD = \frac {AD}{2} + \frac {BD}{2} = \frac {AB}{2}.$$ $E$ and $F$ are the bases of perpendiculars dropped from $O_1$ and $O_2,$ respectively.
Therefore $O_1O_2 \ge EF = \frac {AB}{2}.$
$$CX \perp O_1O_2, AX \perp O_1O \implies \angle O O_1O_2 = \angle AXC$$ $\angle AXC = \angle ABC (AXBC$ is cyclic) $\implies \angle O O_1O_2 = \angle ABC.$
Similarly $\angle BAC = \angle O O_2 O_1 \implies \triangle ABC \sim \triangle O_2 O_1O.$
The area of $\triangle OO_1O_2$ is minimized if $CX \perp AB$ because $$\frac {[OO_1O_2]} {[ABC]} = \left(\frac {O_1 O_2} {AB}\right)^2 \ge \left(\frac {EF} {AB}\right)^2 = \frac {1}{4}.$$ vladimir.shelomovskii@gmail.com, vvsss
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KIRCHHOFF'S CURRENT LAW
In this post, I will try to teach you perfectly about the Kirchhoff’s Current Law. The things that we are going to discuss here are given below.
• INTRODUCTION
• STATEMENT
• EXAMPLE
• PRACTICE PROBLEMS
INTRODUCTION
We will now consider the law named for Gustav Robert Kirchhoff (two h’s and two f’s), a German university professor who was born about the time when Ohm was doing his experimental work. This axiomatic law is known as Kirchhoff’s Current Law (abbreviated KCL).
STATEMENT
“The algebraic sum of all the currents entering any node is equal to zero”
This law represents a mathematical statement of the fact that charge cannot accumulate at a node. A node is not a circuit element, and it certainly cannot store, destroy, or generate charge. Hence, the currents must sum to zero. A hydraulic analogy is sometimes useful here:
for example, consider three water pipes joined in the shape of a Y. We define three “currents” as flowing into each of the three pipes. If we insist that water is always flowing, then obviously we cannot have three positive water currents, or the pipes would burst. This is a result of our defining currents independent of the direction that water is actually flowing.
Therefore, the value of either one or two of the currents as defined must be negative.
A compact expression for Kirchhoff’s current law is
which is just a shorthand statement for
i1 + i2 + i3 +……..+ iN = 0
EXAMPLE
Consider the node shown in the figure below. The algebraic sum of the four currents entering the node must be zero:
iA + iB + (-iC) + (-iD) = 0
It is evident that the law could be equally well applied to the algebraic sum of the currents leaving the node:
( -iA )+ ( -iB )+ ( iC ) + ( iD ) = 0
We might also wish to equate the sum of the currents having reference arrows directed into the node to the sum of those directed out of the node:
iA + iB = iC + iD
PRACTICE PROBLEMS
PRACTICE PROBLEMS NO: 1
Q: For the circuit that is given below, compute the current through resistor R if it is known that the voltage source supplies a current of 3A.
Identify the goal of the problem
The current through resistor R3, labeled as i on the circuit diagram.
Collect the known information
This current flows from the top node of R3 which is connected to three other branches. The current flowing into the node from each branch will add to form the current i.
Devise a plan
If we label the current through R1, we may write a KCL equation at the top node of resistors R2 and R3
Construct an appropriate set of equation
Summing the currents flowing into the node:
iR1 – 2 – i + 5 = 0
The currents flowing into this node are shown in the expanded diagram below.
Determine if additional information is required
We see that we have one equation but two unknowns, which means we need to obtain an additional equation. At this point, the fact that we know the 10V source is supplying 3A comes in handy: KCL shows us that this is also the current iR1 .
Now after substituting, we find that
i = 3 – 2 + 5 = 6 A
It is always worth the effort to recheck our work. Also, we can attempt to evaluate whether at least the magnitude of the solution is reasonable. In this case, we have two sources —– one supplies 5 A, and the other supplies 3 A. There are no other sources, independent or dependent. Thus, we would not expect to find any current in the circuit in excess of 8 A.
PRACTICE PROBLEMS NO: 2
Q) Count the number of branches and nodes in the circuit given below. If ix = 3A and the 18 V source delivers 8 A current, What is the value of RA.
Branches = 5 (As there are 5 elements in the circuit)
Nodes = 3
we have given,
Ivs1 = 8A
Ivs = 3A
To find RA = ?
by using KCL, writing KCL equation for the above circuit:
8 – IRA – 3 + 13 = 0
IRA = 18A
Now by using ohm’s law,
V=IR
R = I/V
R= 18/18
R = 1 Ω
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# Examples of Regular Expression
### Example 1:
Solution:
In a regular expression, the first symbol should be 1, and the last symbol should be 0. The r.e. is as follows:
R = 1 (0+1)* 0
### Example 2:
Write the regular expression for the language starting and ending with a and having any having any combination of b’s in between.
Solution:
The regular expression will be:
R = a b* b
### Example 3:
Write the regular expression for the language starting with a but not having consecutive b’s.
Solution: The regular expression has to be built for the language:
1. L = {a, aba, aab, aba, aaa, abab, …..}
The regular expression for the above language is:
R = {a + ab}*
### Example 4:
Write the regular expression for the language accepting all the string in which any number of a’s is followed by any number of b’s is followed by any number of c’s.
Solution: As we know, any number of a’s means a* any number of b’s means b*, any number of c’s means c*. Since as given in problem statement, b’s appear after a’s and c’s appear after b’s. So the regular expression could be:
1. R = a* b* c*
### Example 5:
Write the regular expression for the language over ∑ = {0} having even length of the string.
Solution:
The regular expression has to be built for the language:
L = {ε, 000000000000, ……}
The regular expression for the above language is:
1. R = (00)*
### Example 6:
Write the regular expression for the language having a string which should have atleast one 0 and alteast one 1.
Solution:
The regular expression will be:
R = [(0 + 1)* 0 (0 + 1)* 1 (0 + 1)*] + [(0 + 1)* 1 (0 + 1)* 0 (0 + 1)*]
### Example 7:
Describe the language denoted by following regular expression
r.e. = (b* (aaa)* b*)*
Solution:
The language can be predicted from the regular expression by finding the meaning of it. We will first split the regular expression as:
r.e. = (any combination of b’s) (aaa)* (any combination of b’s)
L = {The language consists of the string in which a’s appear triples, there is no restriction on the number of b’s}
### Example 8:
Write the regular expression for the language L over ∑ = {0, 1} such that all the string do not contain the substring 01.
Solution:
The Language is as follows:
1. L = {ε, 01001110100, …..}
The regular expression for the above language is as follows:
1. R = (10*)
### Example 9:
Write the regular expression for the language containing the string over {0, 1} in which there are at least two occurrences of 1’s between any two occurrences of 1’s between any two occurrences of 0’s.
Solution: At least two 1’s between two occurrences of 0’s can be denoted by (0111*0)*.
Similarly, if there is no occurrence of 0’s, then any number of 1’s are also allowed. Hence the r.e. for required language is:
R = (1 + (0111*0))*
### Example 10:
Write the regular expression for the language containing the string in which every 0 is immediately followed by 11.
Solution:
The regular expectation will be:
R = (011 + 1)*
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Height of a Binary Tree
ANKIT MITTAL
Last Updated: May 13, 2022
Introduction
A binary tree is an important data structure with applications in computer science, from operating systems to building servers. Using binary trees can reduce the complexity of various problems to a large extent.
The depth of a tree node is equal to the number of edges from the node to the tree's root node. A root node has a depth of 0.
The height of a tree node is equal to the number of edges on the longest path from the node to a leaf. A leaf node has a height of 0.
In this blog, we will be covering a very common question on Binary trees: How can we calculate the height of a binary tree?
Let’s understand it using an example.
Consider the binary tree given below:
In the tree given above,
Height - the height of a root node is 5 since the longest path from the root node to any of the leaf nodes is 5.
Depth - the depth of the root node will be 0 since we are at the root node.
The longest path is coloured, i.e. 1->2->4->6->7.
The path for getting the height of the binary tree will be traversed as shown below :
If we carefully try to observe the depth and height of 2, then as per the definition
Height - the height will be 4 (7->6->4->2)
Depth - if we compute the depth then it will be 2 (1->2).
Now, let’s understand the approach behind this.
Height of a Binary tree
First, let us understand the concept of the height of a binary tree. The height of a binary tree is the maximum distance from the root node to any leaf node. First, let's discuss the recursive approach to calculate the height of a binary tree.
1. We will start from the root node and initially, the height will be 0.
2. We will recursively calculate the height of the left subtree.
3. Similarly, we will calculate the height of the right subtree.
4. Now I will calculate the height of the current node by adding 1 to the max height between the right and left subtree.
Algorithm
We can solve this problem using a recursive algorithm which will be used to traverse the binary tree. This traversal is similar to depth-first search in the graph algorithm. Let us understand this algorithm with the help of the code given below:
Output :
Time Complexity - O(N), as we are traversing every node once, the time complexity to compute the height of a binary tree is of the order N, where N is the number of nodes present in the tree.
Space Complexity -The space complexity is O(H), where H is the height of the binary tree considering the stack space used during recursion while checking for symmetric subtrees. In the worst case(skewed binary tree), H can be equal to N, where N is the number of nodes in the tree.
This problem given above can be modified as - Given the node’s address, we have to find the height of that particular node in the tree. We can simply find the given node and start traversing to the left and right. The function to compute the following is :
Modified Height of a binary tree
Output :
Time Complexity - O(N), as we are traversing every node once, the time complexity to compute the height of a binary tree is N, where N is the number of nodes present in the tree.
Space Complexity - The space complexity is O(H), where H is the height of the binary tree considering the stack space used during recursion while checking for symmetric subtrees. In the worst case(skewed binary tree), H can be equal to N, where N is the number of nodes in the tree.
Depth of a node in a binary tree
Now let us calculate the depth of a node in a binary tree -
We can simply calculate the depth of the tree by subtracting the height of the given node by the height of the root node. i.e -
Height(root) - Height(given node)
We can also calculate the depth of the binary tree by applying the recursive algorithm. We initiate the depth to be 0 initially when we are present at the root node. On every recursive call we add one to the depth and return when the given element for which depth has to be calculated is equal to the current element.
Output :
Time Complexity - O(N), as we are traversing every node once, the time complexity to compute the height of a binary tree is N, where N is the number of nodes present in the tree. In the worst case our node will be leaf node and we will end up traversing all the nodes.
Space Complexity - The space complexity is O(H), where H is the height of the binary tree considering the stack space used during recursion while checking for symmetric subtrees. In the worst case(skewed binary tree), H can be equal to N, where N is the number of nodes in the tree.
FAQs
Q1) What is a binary search tree?
A1) A binary Search Tree or BST is a tree used to sort, retrieve, and search data. It is also a non-linear data structure in which the nodes are arranged in a particular order.
• The left subtree of a node has nodes that are only lesser than that node’s key.
• The right subtree of a node has nodes that are only greater than that node’s key.
• Each node has distinct keys, and duplicates are not allowed in the Binary Search Tree.
• The left and right subtree must also be a binary search tree.
Q2) Is DFS & preorder traversal the same?
Preorder traversal is another variant of DFS. Depth-first-search (DFS) traverses the search tree as much as possible before proceeding to the next sibling, which is similar to pre-order traversal.
Key Takeaways
In this article, we discussed the solution to achieve the height/depth of a binary tree. This is an easy recursive problem on binary trees and can be solved using a simple application of recursion. This concept is a very important topic and has application in other questions for binary tree data structure. You can also refer to our course on master tree data structure to understand Trees.
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# From a solid cylinder whose height is 2.4 cm and diameter 1.4 cm,
Question:
From a solid cylinder whose height is $2.4 \mathrm{~cm}$ and diameter $1.4 \mathrm{~cm}$, a conical cavity of the same height and same diameter is hollowed out. Find the total surface area of the remaining solid to the nearest $\mathrm{cm}^{2}$.
Solution:
For cylinder part :
Height = 2.4 cm and diameter = 1.4 cm
$\Rightarrow$ Radius $(r)=0.7 \mathrm{~cm}$
$\therefore$ Total surface area of the cylindrical part
$=2 \times \frac{\mathbf{2 2}}{\mathbf{7}} \times \frac{\mathbf{7}}{\mathbf{1 0}}[2.4+0.7] \mathrm{cm}^{2}$
$=\frac{\mathbf{4 4}}{\mathbf{1 0}} \times 3.1 \mathrm{~cm}^{2}=\frac{\mathbf{4 4} \times \mathbf{3 1}}{\mathbf{1 0 0}}=\frac{\mathbf{1 3 6 4}}{\mathbf{1 0 0}} \mathbf{c m}^{2}$
For conical part :
Base radius (r) = 0.7 cm and height (h) = 2.4 cm
$\therefore$ Slant height $(\ell)=\sqrt{\mathbf{r}^{2}+\mathbf{h}^{2}}=\sqrt{(\mathbf{0 . 7})^{2}+(\mathbf{2 . 4})^{2}}$
$=\sqrt{\mathbf{0 . 4 9}+\mathbf{5 . 7 6}}=\sqrt{\mathbf{6 . 2 5}}=2.5 \mathrm{~cm}$
$\therefore$ Curved surface area of the conical part
=\pi \mathrm{r} \ell=\frac{\boldsymbol{2 2}}{\boldsymbol{7}} \times 0.7 \times 2.5 \mathrm{~cm}^{2}=\frac{\mathbf{5 5 0}}{\mathbf{1 0 0}} \mathrm{cm}^{2}
Base area of the conical part
$\pi r^{2}=\frac{22}{7} \times\left(\frac{7}{10}\right)^{2} \mathrm{~cm}^{2}=\frac{22 \times 7}{100} \mathrm{~cm}^{2}=\frac{154}{100} \mathrm{~cm}^{2}$
Total surface area of the remaining solid
$=[$ (Total surface area of cylinderical part) $+$ (Curved surface area of conical part) $-($ Base a of the conical part $)]=\left[\frac{1364}{100}+\frac{550}{100}-\frac{154}{100}\right] \mathrm{cm}^{2}$
$=\frac{\mathbf{1 7 6 0}}{\mathbf{1 0 0}} \mathrm{cm}^{2}=17.6 \mathrm{~cm}^{2}$
Hence, total surface area to the nearest $\mathrm{cm}^{2}$ is $18 \mathrm{~cm}^{2}$.
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## What is the relationship between the factors and the roots?
The factor theorem states that if x = c is a root, (x – c) is a factor. For example, look at the following roots: If x = –1/2, (x – (–1/2)) is your factor, which you write as (x + 1/2). If x = –3 is a root, (x – (–3)) is a factor, which you write as (x + 3).
## How are a quadratic function roots related to its factors?
It’s factors are (x – 7)(x + 4). The roots are the x-values that make our expression equal 0. In order for x2 – 3x – 28 to equal 0, either of our factors need to equal 0, since 0 times anything is 0.
READ: What do you wear under leathers on a hot day?
What is the relationship between the roots of a quadratic equation?
The sum of the roots of a quadratic equation is equal to the negation of the coefficient of the second term, divided by the leading coefficient. The product of the roots of a quadratic equation is equal to the constant term (the third term), divided by the leading coefficient.
What is the relation between roots and coefficients of a quadratic equation?
Solution: Let α and β be the roots of the given equation. Sometimes the relation between roots of a quadratic equation is given and we are asked to find the condition i.e., relation between the coefficients a, b and c of quadratic equation. This is easily done using the formula α + β = -ba and αβ = ca.
### How are the factors of a quadratic related to the zeros or roots of a quadratic?
The zero product principle states that if two factors multiply to zero, at least one of the factors must equal zero. The values that make a polynomial equal zero are known as the roots. To find the zeros/roots of a quadratic: factor the equation, set each of the factors to 0, and solve for x.
READ: What contributions did Isaac Asimov make?
### What is the difference between roots and factors of a quadratic equation?
If we set this equal to a number c, we have a quadratic equation. you can say (x – a) and (x – b) are factors. But the numbers a and b are the roots. The root a is a number that you can substitute in for x and get a true statement.
Where are the roots in a quadratic equation?
The roots of a function are the x-intercepts. By definition, the y-coordinate of points lying on the x-axis is zero. Therefore, to find the roots of a quadratic function, we set f (x) = 0, and solve the equation, ax2 + bx + c = 0.
What is the relationship between zeros and roots?
A zero ( or root ) of a function is an input, whose corresponding output is zero. Let P be a polynomial, and let c be an input to P .
## What the difference is between roots solutions and zeros?
A root of an equation is a value at which the equation is satisfied. Roots the equation f(x)= x3+ x2– 3x – ex=0 are the x values of the points A, B, C and D. At these points, the value of the function becomes zero; therefore, the roots are called zeroes.
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1. ## Total Momentum?
The question:
In reconstructing an automobile accident, investigators study the total momentum, both before and after the accident, of the vehicles involved. The total momentum of two vehicles moving in the same direction is found by multiplying the weight of each vehicle by its speed and then adding the results. For example, if one vehicle weighs 3000 pounds and is traveling at 35 miles per hour, and another weighs 2500 pounds and is traveling at 45 miles per hour in the same direction, then the total momentum is 3000 35 + 2500 45 = 217,500
In this exercise we study a collision in which a vehicle weighing 3720 pounds ran into the rear of a vehicle weighing 2480 pounds.
(a) After the collision, the larger vehicle was traveling at 30 miles per hour, and the smaller vehicle was traveling at 50 miles per hour. Find the total momentum of the vehicles after the collision.
235600
(b) The smaller vehicle was traveling at 20 miles per hour before the collision, but the speed V, in miles per hour, of the larger vehicle before the collision is unknown. Find a formula expressing the total momentum of the vehicles (B) before the collision as a function of V.
$\displaystyle B=$
(c) The principle of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision. Using this principle with parts (a) and (b), determine at what speed the larger vehicle was traveling before the collision. (Enter your answer to the nearest whole number.)
___mph
2. (b)
After Collision: $\displaystyle 3720*30 + 2480*50 = 235,600$
Before Collision: $\displaystyle 3720*V + 2480*20= 3720v+49600$
Formula for finding momentum: $\displaystyle B(V)=3720V+49600$
(c)
Conservation of momentum:
$\displaystyle 235,600=3720V+49600$
$\displaystyle 186,000=3720V$
$\displaystyle V=50$mph
3. Originally Posted by MathBane
The question:
In reconstructing an automobile accident, investigators study the total momentum, both before and after the accident, of the vehicles involved. The total momentum of two vehicles moving in the same direction is found by multiplying the weight of each vehicle by its speed and then adding the results. For example, if one vehicle weighs 3000 pounds and is traveling at 35 miles per hour, and another weighs 2500 pounds and is traveling at 45 miles per hour in the same direction, then the total momentum is 3000 35 + 2500 45 = 217,500
In this exercise we study a collision in which a vehicle weighing 3720 pounds ran into the rear of a vehicle weighing 2480 pounds.
(a) After the collision, the larger vehicle was traveling at 30 miles per hour, and the smaller vehicle was traveling at 50 miles per hour. Find the total momentum of the vehicles after the collision.
235600
(b) The smaller vehicle was traveling at 20 miles per hour before the collision, but the speed V, in miles per hour, of the larger vehicle before the collision is unknown. Find a formula expressing the total momentum of the vehicles (B) before the collision as a function of V.
$\displaystyle B=$
(c) The principle of conservation of momentum states that the total momentum before the collision equals the total momentum after the collision. Using this principle with parts (a) and (b), determine at what speed the larger vehicle was traveling before the collision. (Enter your answer to the nearest whole number.)
___mph
For (a) 30(3720) + 50 (2480) = ...
(b) not sure
(c) 2480 (20) +3720 V = 235600
v= ...
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# How Many Lines of Symmetry Does a Hexagon Have?
A hexagon is a six-sided polygon with straight sides and angles. It is one of the most common shapes in nature, and it appears in various forms, from snowflakes to honeycombs. Hexagons can be regular or irregular, convex or concave, but they all share a unique property: symmetry.
Symmetry is an essential concept in mathematics and geometry, referring to a characteristic of a shape or object being the same on both sides of an axis or plane. In other words, if you divide a shape into two identical parts by a line or plane, then it has symmetry. The number of lines or planes that divide a shape into symmetrical parts is called its lines of symmetry.
So, how many lines of symmetry does a hexagon have? The answer depends on the type of hexagon and its symmetry properties. Let’s explore the different cases in detail.
Regular Hexagon
A regular hexagon is a hexagon with six equal sides and angles. It has six lines of symmetry, meaning that you can divide it into six symmetrical parts by six different lines passing through its center and vertices. Each line of symmetry bisects two opposite sides and angles of the hexagon and passes through the center of the hexagon.
To visualize the lines of symmetry, imagine drawing a regular hexagon on a piece of paper and connecting its opposite vertices, forming three pairs of parallel lines. Then, draw three more lines connecting the midpoints of each pair of opposite sides, forming an equilateral triangle in the center of the hexagon. Finally, draw three more lines connecting each vertex to the midpoint of the opposite side, forming six smaller equilateral triangles. Each of these lines is a line of symmetry of the hexagon.
Irregular Hexagon
An irregular hexagon is a hexagon with sides and angles of different lengths and measures. It may or may not have lines of symmetry, depending on its shape and properties. In general, an irregular hexagon has at least one line of symmetry if it is equilateral or if it has opposite sides parallel and equal in length.
For example, a parallelogram is an irregular hexagon with two pairs of opposite sides parallel and equal in length. It has two lines of symmetry, passing through the midpoints of its opposite sides and bisecting its angles. Similarly, a trapezoid is an irregular hexagon with two parallel sides and two non-parallel sides of different lengths. It has one line of symmetry passing through its parallel sides’ midpoint and dividing it into two symmetrical parts.
If an irregular hexagon has no lines of symmetry, it is called asymmetric or chiral. In other words, you cannot divide it into two identical parts by any line or plane passing through its center or vertices. Examples of asymmetric hexagons include hexagons with sides and angles randomly arranged or hexagons with one or more sides missing.
Conclusion
In summary, the number of lines of symmetry of a hexagon depends on its type and properties. A regular hexagon has six lines of symmetry, while an irregular hexagon may have one, two, or none, depending on its shape and symmetry properties. Knowing the lines of symmetry of a hexagon can help us understand its geometric properties, such as its area, perimeter, and angles, and find its mirror images or rotations. Symmetry is an essential concept in geometry and math, and hexagons are a fascinating example of symmetrical shapes in nature and art.
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NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6
# NCERT Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.6
Exercise 2.6
Ex 2.6 Class 7 Maths Question 1.
Find:
(i) 0.2 × 6
(ii) 8 × 4.6
(iii) 2.71 × 5
(iv) 20.1 × 4
(v) 0.05 × 7
(vi) 211.02 × 4
(vii) 2 × 0.86
Solution:
(i) 0.2 × 6
∵ 2 × 6 = 12 and we have 1 digit right to the decimal point in 0.2.
Thus 0.2 × 6 = 1.2
(ii) 8 × 4.6
∵ 8 × 46 = 368 and there is one digit right to the decimal point in 4.6.
Thus 8 × 4.6 = 36.8
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(iii) 2.71 × 5
∵ 271 × 5 = 1355 and there are two digits right to the decimal point in 2.71.
Thus 2.71 × 5 = 13.55
(iv) 20.1 × 4
∵ 201 × 4 = 804 and there is one digit right to the decimal point in 20.1.
∵ 20.1 × 4 = 80.4
(v) 0.05 × 7
∵ 5 × 7 = 35 and there are 2 digits right to the decimal point in 0.05.
Thus 0.05 × 7 = 0.35
(vi) 211.02 × 4
∵ 21102 × 4 = 84408 and there are 2 digits right to the decimal point in 211.02.
Thus 211.02 × 4 = 844.08
(vii) 2 × 0.86
∵ 2 × 86 = 172 and there are 2 digits right to the decimal point in 0.86.
Thus 2 × 0.86 = 1.72
Ex 2.6 Class 7 Maths Question 2.
Find the area of rectangle whose length is 5.7 cm and breadth is 3 cm.
Solution:
Length = 5.7 cm
Area of rectangle = length × breadth
= 5.7 × 3 = 17.1 cm2
Hence, the required area =17.1 cm2
Ex 2.6 Class 7 Maths Question 3.
Find:
(i) 1.3 × 10
(ii) 36.8 × 10
(iii) 153.7 × 10
(iv) 168.07 × 10
(v) 31.1 × 100
(vi) 156.1 × 100
(vii) 3.62 × 100
(viii) 43.07 × 100
(ix) 0.5 × 10
(x) 0.08 × 10
(xi) 0.9 × 100
(xii) 0.03 × 1000
Solution:
Ex 2.6 Class 7 Maths Question 4.
A two-wheeler covers a distance of 55.3 km in one litre of petrol. How much distance will it cover is 10 litres of petrol?
Solution:
Distance covered in 1 litre = 55.3 km
Distance covered in 10 litres = 55.3 × 10 km
Hence, the required distance = 553 km
Ex 2.6 Class 7 Maths Question 5.
Find:
(i) 2.5 ×0.3
(ii) 0.1 × 51.7
(iii) 0.2 × 316.8
(iv) 1.3 × 3.1
(v) 0.5 × 0.05
(vi) 11.2 × 0.15
(vii) 1.07 × 0.02
(viii) 10.05 × 1.05
(ix) 100.01 × 1.1
(x) 100.01 × 1.1
Solution:
(i) 0 2.5 × 0.3
∵ 25 × 3 = 75 and there are 2 digits (1 + 1) right to the decimal points in 2.5 and 0.3.
Thus 2.5 × 0.3 = 0.75
(ii) 0.1 × 51.7
∵ 1 × 517 = 517 and there are two digits (1 + 1) right to the decimal places in 0.1 and 51.7.
(iii) 0.2 × 316.8
∵ 2 × 3168 = 6336 and there are 2 digits (1 + 1) right to the decimal points in 0.2 and 316.8.
Thus 0.2 × 316.8 = 63.36.
(iv) 1.3 × 3.1
∵ 13 × 31 = 403 and there are 2 digits (1 + 1) right to the decimal points in 1.3 and 3.1.
Thus 1.3 × 3.1 – 4.03
(v) 0.5 × 0.05
∵ 5 × 5 = 25 and there are 3 digits (1 + 2) right to the decimal points in 0.5 and 0.05.
Thus 0.5 × 0.05 = 0.025
(vi) 11.2 × 0.15
∵ 112 × 15 = 1680 and there are 3 digits (1 + 2) right to the decimal points in 11.2 and 0.15.
Thus 11.2 × 0.15 = 1.680
(vii) 1.07 × 0.02
∵ 107 × 2 = 214 and there are 4-digits (2 + 2) right to the decimal places is 1.07 × 0.02.
Thus 1.07 × 0.02 = 0.0214
(viii) 10.05 × 1.05
∵ 1005 × 105 = 105525 and there are 4 digits (2 + 2) right to the decimal places in 10.05 × 1.05.
Thus 10.05 × 1.05 = 10.5525
(ix) 101.01 × 0.01
∵ 10101 × 1 = 10101 and there are 4 digits (2 + 2) right to the decimal places in 101.01 and 0.01.
Thus 101.01 × 0.01 = 1.0101
(x) 100.01 × 1.1
∵ 10001 × 11 = 110011 and there are 3 digits (2 + 1) right to the decimal points in 100.01 and 1.1.
Thus 100.01 × 1.1 = 110.011.
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# Module 4: MATH AND DATA ANALYSIS
## Learning Objectives
After reading this lesson, you should be able to:
1. solve an ordinary differential equation using MATLAB.
## What is a differential equation?
An equation that consists of derivatives is called a differential equation. Differential equations have applications in all areas of science and engineering. A mathematical formulation of most physical and engineering problems leads to differential equations. So, it is important for engineers and scientists to know how to set up differential equations and solve them. Differential equations are of two types:
1. ordinary differential equations (ODE) and
2. partial differential equations (PDE).
An ordinary differential equation is that in which all the derivatives are with respect to a single independent variable. Examples of ordinary differential equations include:
1. $$\displaystyle\frac{d^2y}{dx^2} + 2\frac{dy}{dx} + y = 0$$, $$\displaystyle\frac{dy}{dx}(0) = 2,\ \ y(0) = 4$$,
2. $$\displaystyle\frac{d^3y}{dx^3} + 3\frac{d^2y}{dx^2} + 5\frac{dy}{dx} + y = \sin x,$$ $$\displaystyle\frac{d^2y}{dx^2}(0)=12$$ $$\displaystyle\frac{dy}{dx}(0)=2$$, $$y(0) = 4$$
3. $$\displaystyle\frac{d^2y}{dx^2}-5y=6(100-x)$$, $$\displaystyle y(0)=5$$, $$\displaystyle y(100)=27$$
Ordinary differential equations are classified in terms of order and degree. The order of an ordinary differential equation is the same as the highest derivative and the degree of an ordinary differential equation is the power of the highest derivative. Thus, the differential equation,
$\displaystyle x^3\frac{d^3y}{dx^3}+x^2\frac{d^2y}{dx^2}+x\frac{dy}{dx}+xy=e^x$
is of order 3 and degree 1, whereas the differential equation
$\displaystyle \left( \frac{dy}{dx}+1 \right)^2+x^2\frac{dy}{dx}=\sin x$
is of order 1 and degree 2.
## How do I set up and solve a differential equation?
An engineer’s approach to differential equations is different from that of a mathematician. While the latter is interested in the mathematical solution, an engineer should be able to interpret and implement the result physically. So, an engineer’s approach can be divided into three phases:
1. formulation of a differential equation from a given physical situation,
2. solving the differential equation using given conditions, and
3. interpreting the results physically for implementation.
To solve a differential equation, we need
1. the differential equation, deq,
2. the initial/boundary condition(s), IV, and
3. the independent variable, x.
In MATLAB, the function to solve an ordinary differential equation exactly is dsolve(). A usage example is dsolve(deq,IV,x). The number of initial or boundary conditions needed is the same as the order of the differential equation.
### Example 1
Analytically solve the following differential equation using MATLAB.
$\displaystyle2\frac{dy}{dx}=-3y+5e^x, \ \ y(0)=5$
Solution
Note in Example 1 that the $$\displaystyle\frac{dy}{dx}$$ part of the equation is entered as diff(y,x). This is possible because we have defined the symbolic function y(x). Therefore, both y and x are defined as symbolic variables in MATLAB. Also, note the use of the double equals (==) to separate the left- and right-hand sides of the differential equation. Recall that we used this same syntax when entering a left- and right-hand side of an equation when solving for its roots in Lesson 4.3. This must be done as MATLAB always associates a single equals sign (=) with assigning a value to a variable, which we have already done at the beginning of that line (deq =). In other words, when entering an equation with a left- and right-hand side, use double equals.
The simplify() function, in Example 1, is used to make the output aesthetically pleasing. It will try to algebraically simplify a mathematical expression that you give it as an input. Look at the code without simplify() to see the difference!
## How do I solve a higher order ODE?
The dsolve() function is used to solve ordinary differential equations of all orders. However, you must make sure to place all the initial and boundary conditions correctly into the function. Look at Example 2 that shows how to solve a typical second order ordinary differential equation.
### Example 2
Using the dsolve() function, solve the following ordinary differential equation.
$\displaystyle3\frac{d^y}{dx^2}+5\frac{dy}{dx}+7y=11e^{-13x}, \ \ y(0)=19, \ \ \frac{dy}{dx}=17$
Solution
When inputting the equation, look at how the $$\displaystyle\frac{d^2y}{dx^2}$$ part of the equation is entered as diff(y,x,2). Other than that, the only changes made to the dsolve() function when compared to Example 1 are the number of initial values used to solve the differential equation. Since this ODE is second order, we require two initial values.
The solution to this differential equation is rather long, and one can see that using MATLAB to solve it saved some time and effort.
## What are the limitations of using the dsolve() function?
As a programmer, one of the main pitfalls that you may experience is that the dsolve() function may not output a solution to a given differential equation. The dsolve() function uses symbolic manipulations to solve differential equations, and although very advanced, the algorithm cannot solve all ordinary differential equations (most do not have explicit solutions). You may receive this warning message in the Command Window:
Warning: Unable to find explicit solution.
If this is the case, solve the differential equation using a numerical method. Two (of the many) MATLAB functions that numerically solve a differential equation are ode45() and ode23(). For more information on using these functions, use the MATLAB documentation.
## Lesson Summary of New Syntax and Programming Tools
Task Syntax Example Usage Solve a differential equation analytically dsolve() dsolve(deq,IV,x) Numerically solve a differential equation ode45() ode45(deq,tspan,IV) Numerically solve a differential equation ode23() ode23(deq,tspan,IV)
## Multiple Choice Quiz
(1). The MATLAB function for solving an initial value ordinary differential equation is
(a) ode()
(b) diff()
(c) dsolve()
(d) diffsolve()
(2). The most appropriate choice for defining the following differential equation
$7\frac{dy}{dt} + 3y = 4, \ \ y(0) = 2$
for use with the dsolve() function is
(a) deq = 7diff(y,x) + 3*y == 4
(b) deq = 7Dy + 3y == 4
(c) deq = 7* diff(y,x) + 3*y == 4
(d) deq = 7diff(y,x) + 3*y = 4
(3). Complete the code to solve the initial value ordinary differential equation.
(a) dsolve(eqn1,inCond1,'x')
(b) dsolve(eqn1,inCond1,x)
(c) dsolve(eqn1,inCond1,'y')
(d) dsolve(inCond1,eqn1,'x')
(4). To solve
$\displaystyle\frac{d^2y}{dx^2}=5x(30-x), \ \ y(0)=5, \ \ \frac{dy}{dx}(30)=7$ the most appropriate MATLAB line of code to add to the following program is
(a) dsolve(diff(y,x,2) == 5*x*(30-x), IV, x)
(b) dsolve(diff(y,x,2) == 5*x*(30-x), IV, 'y')
(c) dsolve(D2y == 5*x*(30-x), IV, 'x')
(d) dsolve(D2y == 5*x*(30-x), IV)
(5). A MATLAB function that finds the numerical solution to an ordinary differential equation is
(a) ode45()
(b) oda555()
(c) dsolve()
(d) trapz()
## Problem Set
(1). Solve the following initial value differential equation in MATLAB. $\displaystyle3\frac{dy}{dx} + 6y = e^{- x}, \ \ y(0) = 6$.
Find $$y(10)$$.
(2). Solve the following initial value differential equation in MATLAB.
$\displaystyle7\frac{d^{2}y}{dx^{2}} + 11\frac{dy}{dx} + 13y = \sin(x), \ \ y(0) = 6, \ \ y'(0) = 17$
Find $$y(10)$$ and $$y'(10)$$. Plot $$y$$ as a function $$x$$from $$x =0$$ to $$x=20$$.
(3). Solve the following boundary value differential equation in MATLAB.
$\frac{d^{2}y}{dx^{2}} - 3 \times 10^{- 6}y = 7.5 \times 10^{- 7}(100 - x), \ \ y(0) = 0, \ \ y(100) = 0$
Find $$y(15)$$, and the maximum value of $$y$$.
(4). A ball bearing at $$1200K$$ is allowed to cool down in air at an ambient temperature of $$300K$$. Assuming heat is lost only due to radiation, the differential equation for the temperature of the ball is given by
$\displaystyle\frac{d\theta}{dt} = - 2.2067 \times 10^{- 12}\left( \theta^{4} - 81 \times 10^{8} \right), \ \ \theta(0) = 1200K$
where $$\theta$$ is in $$K$$ and $$t$$ in seconds. Using MATLAB, find the temperature, the rate of change of temperature, and the rate at which heat is lost at $$t = 480$$ seconds. The rate at which the heat is lost (in Watts) is given by
$\text{Rate at which heat is lost} = 2.42 \times \text{1}\text{0}^{- 10}\ (\theta^{4} - 81 \times 10^{8})$
(5). Pollution in lakes can be a serious issue as they are used for recreational use. One is generally interested in knowing that if the concentration of a particular pollutant is above acceptable levels. The differential equation governing the concentration of pollution in a lake as a function of time is given by
$25 \times 10^{6}\frac{dC}{dt} + 1.5 \times 10^{6}C = 0$
If the initial concentration of the pollutant is $$10^{7}\text{parts}/m^{3}$$, and the acceptable level is $$5 \times 10^{6}\text{parts}/m^{3}$$, how long will it take for the pollution level to decrease to an acceptable level? Plot the concentration of the pollutant as a function of time from the initial time to the time it takes the pollution level to decrease to the acceptable level.
(6). The speed (rad/s) of a motor without damping for a voltage input of 20 V is given by
$\displaystyle20 = (\text{0.02})\ \frac{dw}{dt} + (0.06)w$
If the initial speed is zero $$\displaystyle(w(0)=0)$$ , what is the speed of the motor at $$t = 0.8s$$? What is the angular acceleration of the motor at $$t = 0.8s$$.
(7). For a solid steel shaft to be shrunk-fit into a hollow hub, the solid shaft needs to be contracted. Initially at a room temperature of $$27\ ^{o}C$$, the solid shaft is placed in a refrigerated chamber that is maintained at $$- 33\ ^{o}C$$. The differential equation governing the change of temperature of the solid shaft $$\theta$$ is given by
$\displaystyle\frac{d\theta}{dt} = - 5.33 \times 10^{- 6}\begin{pmatrix} - 3.69 \times 10^{- 6}\theta^{4} + 2.33 \times 10^{- 5}\theta^{3} + 1.35 \times 10^{- 3}\theta^{2} \\ + 5.42 \times 10^{- 2}\theta + 5.588 \\ \end{pmatrix}(\theta + 33)$
Using MATLAB, find the temperature and the rate of change of temperature after the steel shaft has been in the chamber for 12 hours.
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# Is an acute angle 0 to 90 degrees?
In geometry and mathematics, acute angles are angles whose measurements fall between 0 and 90 degrees or has a radian of fewer than 90 degrees. When the term is given to a triangle as in an acute triangle, it means that all angles in the triangle are less than 90 degrees.
## Is an acute angle 0 to 90 degrees?
In geometry and mathematics, acute angles are angles whose measurements fall between 0 and 90 degrees or has a radian of fewer than 90 degrees. When the term is given to a triangle as in an acute triangle, it means that all angles in the triangle are less than 90 degrees.
### Is used to draw angles from 0 to 90 degrees?
Use an adjustable triangle to measure angles between 0 and 90 degrees by placing it on the horizontal axis and adjusting the hinged edge till it lines up with the angled line you want to measure. If you need to copy an angle precisely, couple a compass with a ruler to measure the angle and draw an identical one.
#### What type of angle is 0 degrees?
An angle that measures as zero degrees (0°) is called a zero angle or zero radian. The condition of zero angle, the two lines or two rays are in the same direction with a common point called vertex.
Is 0 degrees an acute angle?
Is 0 degrees an acute angle? Yes, 0 degrees is an acute angle. This is the perfect example of acute angles as the range of acute angles starts from 0 and ends with values which are less than 90. The measure 0 degrees is also called zero angle.
Why acute angle is less than 90?
An angle is formed when two rays meet at a vertex. When this angle measures less than 90°, it is termed an acute angle. In the figure shown below, the angle formed between ‘Ray 1’ and ‘Ray 2’ is an acute angle. Also, when a right angle is divided into two it forms two acute angles.
## What is 90 degrees in a circle?
The Full Circle A Full Circle is 360° Half a circle is 180° (called a Straight Angle) Quarter of a circle is 90° (called a Right Angle)
### Which tool can be used to draw a 90 angle?
A protractor is a measuring instrument that can be used to draw a 90-degree angle and can be used for measuring other angles as well.
#### What does angle 0 look like?
An angle that does not form a vertex or measure 0 degrees is called a zero angle. It is also called zero radians. A zero angle is formed when both the rays or arms of the angle are pointing towards the same direction and vertex as just a point without any space.
How do you make a zero angle?
When two lines are in same direction with a vertex, then there is no angle between them. Therefore, the angle between them is a zero angle.
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# Measuring 3D Shapes: Volume and Surface Area Formulas
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## 12 Questions
### What is the formula for calculating the surface area of a cube?
$6s^2$
### How is the surface area of a cylinder calculated?
$2 ext{π}r(r + l) + 2 ext{π}r^2$
### What is the formula for finding the surface area of a sphere?
$4 ext{π}r^2$
Prism
### If a cylinder has a radius of 5 units and a height of 10 units, what is its surface area?
$200 ext{π}$ square units
### How is the volume of a cylinder calculated?
$Volume = \pi r^2 \times h$
### What is the formula to calculate the volume of a cube?
$Volume = s^3$
### What formula is used to find the volume of a cone?
$Volume = rac{1}{3} \pi r^2 \times h$
### How do you calculate the surface area of a three-dimensional shape?
By adding the area of all faces and bases
### What is the formula for finding the volume of a sphere?
$Volume = 4rac{1}{3} \pi r^3$
Cylinder
## Measuring Shapes in Three Dimensions: Volume and Surface Area
In the world of geometry, the study of three-dimensional shapes, known as mensuration, deals with the quantification of their volume and surface area. These two properties are fundamental to our understanding of the physical properties and behavior of objects around us.
### Volume of 3D Shapes
Volume refers to the amount of space a three-dimensional shape occupies. It's a measure of the object's size in three dimensions. Some common three-dimensional shapes with known volume formulas include:
Cubes and Cube-like Shapes
The volume of a cube can be calculated using the formula:
[ Volume = s^3 ]
where (s) is the side length.
Prisms
For a prism, the volume can be calculated using the formula:
[ Volume = A \times h ]
where (A) is the area of the base and (h) is the height.
Cylinders
The volume of a cylinder is given by the formula:
[ Volume = \pi r^2 \times h ]
where (r) is the radius and (h) is the height.
Cones and Cone-like Shapes
The volume of a cone is calculated using the formula:
[ Volume = \frac{1}{3} \pi r^2 \times h ]
where (r) is the radius of the base and (h) is the height.
Spheres
The volume of a sphere can be calculated using the formula:
[ Volume = \frac{4}{3} \pi r^3 ]
### Surface Area of 3D Shapes
Surface area refers to the total exposed surface of a three-dimensional shape. It can influence the object's properties such as heat transfer, friction, and mass. Here are some common geometric shapes and their surface area formulas:
Cubes and Cube-like Shapes
The surface area of a cube is calculated using the formula:
[ Surface\ Area = 6 \times s^2 ]
where (s) is the side length.
Prisms
For a prism, the surface area can be calculated using the formula:
[ Surface\ Area = 2A + (A_b + A_t) ]
where (A) is the area of the base, (A_b) is the area of the base's bottom face, and (A_t) is the area of the base's top face.
Cylinders
The surface area of a cylinder can be calculated using the formula:
[ Surface\ Area = 2 \pi r(r + l) + 2 \pi r^2 ]
where (r) is the radius and (l) is the length of the cylinder.
Cones and Cone-like Shapes
The surface area of a cone can be calculated using the formula:
[ Surface\ Area = \pi r(r + l) + \pi r^2 ]
where (r) is the radius and (l) is the slant height.
Spheres
The surface area of a sphere can be calculated using the formula:
[ Surface\ Area = 4 \pi r^2 ]
These formulas are essential tools for designers, architects, engineers, and anyone seeking to understand and manipulate the properties of geometric objects in three dimensions. By using these formulas, you can accurately measure and compare the volume and surface area of various shapes in the world around you.
Explore the formulas for calculating the volume and surface area of common three-dimensional shapes like cubes, prisms, cylinders, cones, and spheres. Learn how to quantify these properties fundamental to geometry and physical sciences.
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# Percentiles Help (page 2)
(not rated)
By — McGraw-Hill Professional
Updated on Aug 26, 2011
## Percential Points, Ranks, and Inversion
### Percential Points
We get around the foregoing conundrum by defining a scheme for calculating the positions of the percentile points in a set of ranked data elements. A set of ranked data elements is a set arranged in a table from ''worst to best,'' such as that in Table 4-1. Once we have defined the percentile positioning scheme, we accept it as a convention, ending all confusion forever and ever.
So – suppose we are given the task of finding the position of the pth percentile in a set of n ranked data elements. First, multiply p by n, and then divide the product by 100. This gives us a number i called the index:
i = pn/100
Here are the rules:
• If i is not a whole number, then the location of the pth percentile point is i + 1.
• If i is a whole number, then the location of the pth percentile point is i + 0.5.
### Percential Ranks
If we want to find the percentile rank p for a given element or position s in a ranked data set, we use a different definition. We divide the number of elements less than s (call this number t) by the total number of elements n, and multiply this quantity by 100, getting a tentative percentile p*:
p* = 100 t/n
Then we round p* to the nearest whole number between, and including, 1 and 99 to get the percentile rank p for that element or position in the set.
Percentile ranks defined in this way are intervals whose centers are at the percentile boundaries as defined above. The 1st and 99th percentile ranks are often a little bit oversized according to this scheme, especially if the population is large. This is because the 1st and 99th percentile ranks encompass outliers, which are elements at the very extremes of a set or distribution.
### Percentile Inversion
Once in a while you'll hear people use the term ''percentile'' in an inverted, or upside-down, sense. They'll talk about the ''first percentile'' when they really mean the 99th, the ''second percentile'' when they really mean the 98th, and so on. Beware of this! If you get done with a test and think you have done well, and then you're told that you're in the ''4th percentile,'' don't panic. Ask the teacher or test administrator, ''What does that mean, exactly? The top 4%? The top 3%? The top 3.5%? Or what?'' Don't be surprised if the teacher or test administrator is not certain.
## Percentile Practice Problems
#### Practice 1
Where is the 56th percentile point in the data set shown by Table 4-1?
#### Solution 1
There are 1000 students (data elements), so n = 1000. We want to find the 56th percentile point, so p = 56. First, calculate the index:
i = (56 × 1000)/100
= 56,000/100
= 560
This is a whole number, so we must add 0.5 to it, getting i + 0.5 = 560.5. This means the 56th percentile is the boundary between the ''560th worst'' and ''561st worst'' test papers. To find out what score this represents, we must check the cumulative absolute frequencies in Table 4-1. The cumulative frequency corresponding to a score of 25 is 531 (that's less than 560.5); the cumulative frequency corresponding to a score of 26 is 565 (that's more than 560.5). The 56th percentile point thus lies between scores of 25 and 26.
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# Fractions
* If a cake is cut into five pieces and three pieces are taken out of five pieces, then these three pieces are known as .
* Thus the relation to represent some part to the whole part is known as fraction.
* The upper part of the fraction is called Numerator and the lower part is called Denominator.
* For example in the fraction , 4 is the numerator and 7 is the denominator.
* Fraction is always represented in its lowest terms.
* To reduce a fraction to its lowest term, both numerator and denominator are to be divided by the highest common factor.
* For example 3 is the highest common factor of the numerator and denominator of . When both 6 and 9 are divided by 3, the fraction becomes .
Comparison of Fractions:
When comparing two or more fractions, the following points are to be remembered.
* When the denominators of the fractions are same, the fraction having largest numerator is the largest of all the fractions.
* When the numerators of the fractions are same, the fraction having smallest denominator is the largest of all the fractions.
* When neither numerators nor denominators of the fractions are same then they are converted into equivalent fractions of the same denominator by taking the LCM of the denominators of the given fractions and then compared.
* If the difference of the numerators and denominators of each of the given fractions be same then the fraction of the greatest numerator is the greatest and the fraction of the least numerators is the smallest.
* For example,
the difference between the numerator and the denominator of all the fractions is same which is 3 in numerator, 11 is the biggest and 4 is the smallest
Therefore, is the biggest and is the smallest fraction.
Cross multiplication Method:
* Two fractions can be compared with this method.
* The numerator of the first fraction is multiplied by the denominator of the second fraction and the numerator of the second fraction is multiplied by the denominator of the first fraction.
* For example,
are to be compared then
5 × 7 = 35 and 4 × 9 = 36
since 36 is greater than 35 and in 36 the numerator 0f is included.
Posted Date : 21-11-2020
గమనిక : ప్రతిభ.ఈనాడు.నెట్లో కనిపించే వ్యాపార ప్రకటనలు వివిధ దేశాల్లోని వ్యాపారులు, సంస్థల నుంచి వస్తాయి. మరి కొన్ని ప్రకటనలు పాఠకుల అభిరుచి మేరకు కృత్రిమ మేధస్సు సాంకేతికత సాయంతో ప్రదర్శితమవుతుంటాయి. ఆ ప్రకటనల్లోని ఉత్పత్తులను లేదా సేవలను పాఠకులు స్వయంగా విచారించుకొని, జాగ్రత్తగా పరిశీలించి కొనుక్కోవాలి లేదా వినియోగించుకోవాలి. వాటి నాణ్యత లేదా లోపాలతో ఈనాడు యాజమాన్యానికి ఎలాంటి సంబంధం లేదు. ఈ విషయంలో ఉత్తర ప్రత్యుత్తరాలకు, ఈ-మెయిల్స్ కి, ఇంకా ఇతర రూపాల్లో సమాచార మార్పిడికి తావు లేదు. ఫిర్యాదులు స్వీకరించడం కుదరదు. పాఠకులు గమనించి, సహకరించాలని మనవి.
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# If $\theta$ lies in the third quadrant, then the expression $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}(\dfrac{\pi }{4} - \dfrac{\theta }{2})$ equals 2.A. TrueB. False
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Hint: Here we will solve the given equation using the trigonometric identities and check whether the value of expression equals 2 or not.
As we know that $\sin 2\theta = 2\sin \theta \cos \theta \to (1)$
$\Rightarrow \sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } + 4{\cos ^2}(\dfrac{\pi }{4} - \dfrac{\theta }{2}) \to (2)$
Putting the value of $\sin 2\theta$ as in equation (1) in equation (2) then we have
$\Rightarrow \sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } = \sqrt {4{{\sin }^2}\theta ({{\sin }^2}\theta + {{\cos }^2}\theta )} = \sqrt {4{{\sin }^2}\theta } = 2\left| {\sin \theta } \right|.$
Now as per question, $\theta$ lies in the third quadrant, so $\sin \theta < 0$ and $\left| {\sin \theta } \right| = - \sin \theta$
Hence $\sqrt {4{{\sin }^4}\theta + {{\sin }^2}2\theta } = - 2\sin \theta$
And $4{\cos ^2}(\dfrac{\pi }{4} - \dfrac{\theta }{2}) = 2[1 + \cos \left( {\dfrac{\pi }{2} - \theta } \right)] = 2 + 2\sin \theta {\text{ [}}\because {\text{cos2}}\theta {\text{ = 2co}}{{\text{s}}^2}\theta - 1] \to (3)$
$\Rightarrow - 2\sin \theta + 2 + 2\sin \theta = 2$
Note: Equation (1) and (3) is trigonometry identity, while solving any question always try to expand the expression by putting the value of trigonometry identities. Remember $\left| x \right| = x$ if $x \geqslant 0$ and $\left| x \right| = - x$ if $x \leqslant 0$. In the third quadrant only $\tan \theta$ and $\cot \theta$ is positive.
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# M21-04
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16 Sep 2014, 01:10
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The table below shows the number of members of five pool clubs in 2001 and 2002 respectively:
Which of the clubs experienced the greatest percentage increase in membership from 2001 to 2002?
A. club A
B. club B
C. club C
D. club D
E. club E
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Joined: 02 Sep 2009
Posts: 58427
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16 Sep 2014, 01:10
Official Solution:
The table below shows the number of members of five pool clubs in 2001 and 2002 respectively:
Which of the clubs experienced the greatest percentage increase in membership from 2001 to 2002?
A. club A
B. club B
C. club C
D. club D
E. club E
The absolute increase in membership was 4 for club A, 2 for club B, -5 for club C, 2 for club D, and 3 for club E. To compare the percentage increases we have to compare fractions $$\frac{4}{54}$$, $$\frac{2}{62}$$, $$-\frac{5}{60}$$, $$\frac{2}{48}$$, and $$\frac{3}{31}$$. $$\frac{3}{31}$$ is the largest ($$\frac{3}{31}$$ is approximately $$\frac{1}{10} = \frac{4}{40} \gt \frac{4}{54} \gt \frac{2}{48} = \frac{1}{24}$$).
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08 Nov 2017, 03:01
Take the ratio and compare. E wins.
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### Show Tags
02 Feb 2018, 21:09
Instead of calculating all the ratios, you can eliminate some quickly.
Club C decreased in size so can be instantly eliminated.
Club E started with the smallest number (31) and increased by 3. Use that as a comparison for other clubs.
- Club B started with more members (62) and increased by fewer (2), so it won't beat E.
- Likewise Club D started with more members (48) and increased by fewer (2), so it won't beat E either.
- Club A started with more members (54) but increased by more (4) so it's not immediately obvious.
You then just have to calculate whether 4/54 is larger than 3/31. You will probably have time to do this long-hand, or you can estimate if you're confident.
Re: M21-04 [#permalink] 02 Feb 2018, 21:09
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# M21-04
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You are viewing an older version of this Concept. Go to the latest version.
Consistent and Inconsistent Linear Systems
Systems with parallel or intersecting lines
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Consistent and Inconsistent Linear Systems
What if you were given a system of equations like and ? How could you rewrite these equations to determine the number of solutions the system has? After completing this Concept, you'll be able to identify whether a system of equations like this one is an inconsistent one, a consistent one, or a dependent one.
Guidance
As we saw in Section 7.1, a system of linear equations is a set of linear equations which must be solved together. The lines in the system can be graphed together on the same coordinate graph and the solution to the system is the point at which the two lines intersect.
Or at least that’s what usually happens. But what if the lines turn out to be parallel when we graph them?
If the lines are parallel, they won’t ever intersect. That means that the system of equations they represent has no solution. A system with no solutions is called an inconsistent system.
And what if the lines turn out to be identical?
If the two lines are the same, then every point on one line is also on the other line, so every point on the line is a solution to the system. The system has an infinite number of solutions, and the two equations are really just different forms of the same equation. Such a system is called a dependent system.
But usually, two lines cross at exactly one point and the system has exactly one solution:
A system with exactly one solution is called a consistent system.
To identify a system as consistent, inconsistent, or dependent, we can graph the two lines on the same graph and see if they intersect, are parallel, or are the same line. But sometimes it is hard to tell whether two lines are parallel just by looking at a roughly sketched graph.
Another option is to write each line in slope-intercept form and compare the slopes and intercepts of the two lines. To do this we must remember that:
• Lines with different slopes always intersect.
• Lines with the same slope but different intercepts are parallel.
• Lines with the same slope and the same intercepts are identical.
Example A
Determine whether the following system has exactly one solution, no solutions, or an infinite number of solutions.
Solution
We must rewrite the equations so they are in slope-intercept form
The slopes of the two equations are different; therefore the lines must cross at a single point and the system has exactly one solution. This is a consistent system.
Example B
Determine whether the following system has exactly one solution, no solutions, or an infinite number of solutions.
Solution
We must rewrite the equations so they are in slope-intercept form
The slopes of the two equations are the same but the intercepts are different; therefore the lines are parallel and the system has no solutions. This is an inconsistent system.
Example C
Determine whether the following system has exactly one solution, no solutions, or an infinite number of solutions.
Solution
We must rewrite the equations so they are in slope-intercept form
The lines are identical; therefore the system has an infinite number of solutions. It is a dependent system.
Watch this video for help with the Examples above.
Vocabulary
• A system with no solutions is called an inconsistent system. For linear equations, this occurs with parallel lines.
• A system where the two equations overlap at one, multiple, or infinitely many points is called a consistent system.
• Coincident lines are lines with the same slope and intercept. The lines completely overlap.
• When solving a system of coincident lines, the resulting equation will be without variables and the statement will be true. You can conclude the system has an infinite number of solutions. This is called a consistent-dependent system.
Guided Practice
Determine whether the following system of linear equations has zero, one, or infinitely many solutions:
What kind of system is this?
Solution:
It is easier to compare equations when they are in the same form. We will rewrite the first equation in slope-intercept form.
Since the two equations have the same slope, but different -intercepts, they are different but parallel lines. Parallel lines never intersect, so they have no solutions.
Since the lines are parallel, it is an inconsistent system.
Practice
Express each equation in slope-intercept form. Without graphing, state whether the system of equations is consistent, inconsistent or dependent.
Vocabulary Language: English
substitute
substitute
In algebra, to substitute means to replace a variable or term with a specific value.
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# RD Sharma Solutions for Class 11 Chapter 10 - Sine and Cosine Formulae and their Applications Exercise 10.1
Exercise 10.1 of Chapter 10, we shall discuss problems based on the law of sines or sine rule. Highly experienced subject experts at BYJU’S having vast knowledge of concepts develop the solutions, which match the understanding ability of the students. Students can use RD Sharma Class 11 Solutions for Maths as the best study material which consists of solutions explained in the most simple language which any student can understand. The solutions of this exercise are made available in the RD Sharma Class 11 Maths pdf, which can be downloaded easily from the below given links.
## Download the Pdf of RD Sharma Solutions for Class 11 Maths Exercise 10.1 Chapter 10 – Sine and Cosine Formulae and their Applications
### Access answers to RD Sharma Solutions for Class 11 Maths Exercise 10.1 Chapter 10 – Sine and Cosine Formulae and their Applications
1. If in a ∆ABC, ∠A = 45o, ∠B = 60o, and ∠C = 75o; find the ratio of its sides.
Solution:
Given: In ABC, ∠A = 45o, ∠B = 60o, and ∠C = 75o
By using the sine rule, we get
a: b: c = 2: √6: (1+√3)
Hence the ratio of the sides of the given triangle is a: b: c = 2: √6: (1+√3)
2. If in any ∆ABC, ∠C = 105o, ∠B = 45o, a = 2, then find b.
Solution:
Given: In ∆ABC, ∠C = 105o, ∠B = 45o, a = 2
We know in a triangle,
∠A + ∠B + ∠C = 180°
∠A = 180° – ∠B – ∠C
Substituting the given values, we get
∠A = 180° – 45° – 105°
∠A = 30°
By using the sine rule, we get
3. In ∆ABC, if a = 18, b = 24 and c = 30 and ∠C = 90o, find sin A, sin B and sin C.
Solution:
Given: In ∆ABC, a = 18, b = 24 and c = 30 and ∠C = 90o
By using the sine rule, we get
In any triangle ABC, prove the following:
Solution:
By using the sine rule we know,
= RHS
Hence proved.
5. (a – b) cos C/2 = C sin (A – B)/2
Solution:
By using the sine rule we know,
= RHS
Hence proved.
Solution:
By using the sine rule we know,
Solution:
By using the sine rule we know,
cos (A + B)/2 = cos (A/2 + B/2) = cos A/2 cos B/2 + sin A/2 sin B/2
cos (A – B)/2 = cos (A/2 – B/2) = cos A/2 cos B/2 – sin A/2 sin B/2
Substituting the above equations in equation (vi) we get,
Solution:
By using the sine rule we know,
11. b sin B – c sin C = a sin (B – C)
Solution:
By using the sine rule we know,
So, c = k sin C
Similarly, a = k sin A
And b = k sin B
We know,
Now let us consider LHS:
b sin B – c sin C
Substituting the values of b and c in the above equation, we get
k sin B sin B – k sin C sin C = k (sin2 B – sin2 C) ……….(i)
We know,
Sin2 B – sin2 C = sin (B + C) sin (B – C),
Substituting the above values in equation (i), we get
k (sin2 B – sin2 C) = k (sin (B + C) sin (B – C)) [since, π = A + B + C ⇒ B + C = π –A]
The above equation becomes,
= k (sin (π –A) sin (B – C)) [since, sin (π – θ) = sin θ]
= k (sin (A) sin (B – C))
From sine rule, a = k sin A, so the above equation becomes,
= a sin (B – C)
= RHS
Hence proved.
12. a2 sin (B – C) = (b2 – c2) sin A
Solution:
By using the sine rule we know,
So, c = k sin C
Similarly, a = k sin A
And b = k sin B
We know,
Now let us consider RHS:
(b2 – c2) sin A …
Substituting the values of b and c in the above equation, we get
(b2 – c2) sin A = [(k sin B)2 – ( k sin C)2] sin A
= k2 (sin2 B – sin2 C) sin A………. (i)
We know,
Sin2 B – sin2 C = sin (B + C) sin (B – C),
Substituting the above values in equation (i), we get
= k2 (sin (B + C) sin (B – C)) sin A [since, π = A + B + C ⇒ B + C = π –A]
= k2 (sin (π –A) sin (B – C)) sin A
= k2 (sin (A) sin (B – C)) sin A [since, sin (π – θ) = sin θ]
Rearranging the above equation we get
= (k sin (A))( sin (B – C)) (k sin A)
From sine rule, a = k sin A, so the above equation becomes,
= a2 sin (B – C)
= RHS
Hence proved.
= RHS
Hence proved.
14. a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = 0
Solution:
By using the sine rule we know,
a = k sin A, b = k sin B, c = k sin C
Let us consider LHS:
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B)
Substituting the values of a, b, c from sine rule in above equation, we get
a (sin B – sin C) + b (sin C – sin A) + c (sin A – sin B) = k sin A (sin B – sin C) + k sin B (sin C – sin A) + k sin C (sin A – sin B)
= k sin A sin B – k sin A sin C + k sin B sin C – k sin B sin A + k sin C sin A – k sin C sin B
Upon simplification, we get
= 0
= RHS
Hence proved.
Upon simplification we get,
= k2 [sin A sin (B – C) + sin B sin (C – A) + sin C sin (A – B)]
We know, sin (A – B) = sin A cos B – cos A sin B
Sin (B – C) = sin B cos C – cos B sin C
Sin (C – A) = sin C cos A – cos C sin A
So the above equation becomes,
= k2 [sin A (sin B cos C – cos B sin C) + sin B (sin C cos A – cos C sin A) + sin C (sin A cos B – cos A sin B)]
= k2 [sin A sin B cos C – sin A cos B sin C + sin B sin C cos A – sin B cos C sin A + sin C sin A cos B – sin C cos A sin B)]
Upon simplification we get,
= 0
= RHS
Hence proved.
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# Linear Functions and Relations
## Presentation on theme: "Linear Functions and Relations"— Presentation transcript:
Linear Functions and Relations
Chapter 2 Linear Functions and Relations
In Chapter 2, You Will… Move from simplifying variable expressions and solving one-step equations and inequalities to working with two variable equations and inequalities. Learn how to represent function relationships by writing and graphing linear equations and inequalities. By graphing data and trend lines, you will understand how the slope of a line can be interpreted in real-world situations.
2-1 Relations and Functions
What You’ll Learn … To graph relations. To identify functions.
Example 1 Graphing a Relation
A relation is a set of pairs of input and output numbers. Example 1 Graphing a Relation [(-2,4), (3,-2), (-1,0), (1,5)] [(0,4),(-2,3),(-1,3),(-2,2),(1,-3)]
Finding Domain and Range
(2,4),(3,4.5),(4,7.5),(5,7),(6,5),(6,7.5) D= _____________ R= _____________ The domain of a relation is the set of all inputs, or x-coordinates of the ordered pairs. The range of a relation is the set of all outputs, or y-coordinates of the ordered pairs. D= _____________ R= _____________
Using a Mapping (-2,-1) (-1,-1) (-2,1) (6,3) -2 -1
Another way to show a relation is to use a mapping diagram, which links elements of the domain with corresponding elements of the range. -1 1 6 3
Example 3 Making a Mapping Diagram
(0,2) (1,3) (2,4) (2,8) (-1,5) (0,8) (-1,3) (-2,3)
Are the x's different? A function is a relation that assigns exactly one value in the range to each value in the domain X Y 1 -3 6 -2 9 -1 3 X Y -4 -1 3
Example 4 Identifying Functions
-2 5 -1 3 4 -1 2 3 -1 3 5
One way you can tell whether a relation is a function is to analyze the graph of the relation using the vertical line test. If any vertical line passes through more than one point of the graph, the relation is NOT a function.
Which are Functions?
Function Notation Another way to write a function
y = 3x + 4 is f(x)= 3x + 4. You read f(x) as “f of x” or “f is a function of x”. The notations g(x) and h(x) also indicate functions of x.
Evaluating Functions Function Rule A function rule is an equation that describes a function. You can think of a function rule as an input-output machine Input Output
Evaluating a Function Function Rule y = 3x + 4 3x + 4 X Y 1 2 3 x y
Input Output X Y 1 2 3 3x + 4 x y
Evaluating a Function Rule
f(n)= -3n – 10 Find f(6). g(x) = -2x² + 7 Find g(6).
Example 6 Real World Connection
The area of a square tile is a function of the length of a side of the square. Write a function rule for the area of a square. Evaluate the function for a square tile with side length 3.5 in.
7-6 Function Operations 2.01 Use the composition of functions to model and solve problems; justify results. What you’ll learn … To add, subtract, multiply and divide functions To find the composite of two numbers
Function Operations Addition (f+g)(x) = f(x)+g(x)
Multiplication (fg)(x) = f(x) g(x) Subtraction (f-g)(x) = f(x) – g(x) Division (x)= , g(x)≠0 f g f(x) g(x)
Example 1 Adding and Subtracting Functions
Let f(x) = 3x +8 and g(x) = 2x-12. Find f+g and f - g and their domain. Let f(x) = 5x2 - 4x and g(x) = 5x+1. Find f+g and f - g and their domain.
Example 2 Multiplying and Dividing Functions
Let f(x) = x2 - 1 and g(x) = x+1. Find fg and and their domain. f g Let f(x) = 6x2 +7x - 5 and g(x) = 2x-1. Find fg and and their domain. f g
Composition of Functions
The composition of function g with function f is written as g°f and is defined as (g°f)(x)= g(f(x)), where the domain of g°f consists of the values a in the domain of f such that f(a) is in the domain of g. (g°f)(x) = g( f(x) ) Evaluate the inner function f(x) first. 2. Then use your answer as the input of the outer function g(x).
Example 3 Composition of Functions
Let f(x) = x-2 and g(x) = x2. Find (g°f)(-5). Let f(x) = x-2 and g(x) = x2. Find (f°g)(x) and evaluate (f°g)(-5).
Example 4a Real World Connection
Suppose you are shopping in the store in the photo. You have a coupon worth \$5 off any item. Use functions to model discounting an item by 20% and to model applying the coupon. Use a composition of your two functions to model how much you would pay for an item if the clerk applies the discount first and then the coupon. Use a composition of your two functions to model how much you would pay for an item if the clerk applies the coupon first and then the discount. How much more is any item if the clerk applies the coupon first?
Example 4b Real World Connection
A store is offering a 10% discount on all items. In addition, employees get a 25% discount. Write a composite function to model taking the 10% discount first. Write a composite function to model taking the 25% discount first. Suppose you are an employee. Which discount would you prefer to take first?
7-7 Inverse Relations and Functions
2.01 Use the composition of functions to model and solve problems; justify results. What you’ll learn … To find the inverse of a relation or function.
The Inverse of a Function
If a relation maps element a of its domain to element b of its range., the inverse relation “undoes” the relation and maps b back to a. Relation r Inverse of r 1 2 1.2 1.4 1.6 1.9 1.2 1.4 1.6 1.9 1 2
Example 1 Finding the Inverse of a Relation
Find the inverse of relation s. Graph s and its inverse. x -1 1 y 2 3 4
Example 2 Interchanging x and y
Find the inverse of y = x2 + 3. Does y = x2 + 3 define a function? Is its inverse a function? Explain. Find the inverse of y = 3x - 10. Is its inverse a function? Explain.
Example 3 Graphing a Relation and Its Inverse
Graph y= x2 + 3 and its inverse, y = +√x -3 . Graph y= 3x-10 and its inverse.
The inverse of a function is denoted by f-1
The inverse of a function is denoted by f-1. Read f-1 as “the inverse of f” or as “f inverse”. The notation f(x) is used for functions, but f-1(x) may be a relation that is not a function.
Example 4a Finding an Inverse Function
Consider the function f(x) = √x+1. Find the domain and range of f. Find f-1. Find the domain and range of f-1. Is f-1 a function? Explain.
Example 4b Finding an Inverse Function
Consider the function f(x) = 10 – 3x. Find the domain and range of f. Find f-1. Find the domain and range of f-1. Find f-1(f(3)). Find f-1(f(2)).
Composite of Inverse Functions
If f and f-1 are inverse functions then, (f-1°f)(x) and (f°f-1)(x) = x.
Example 6 Composite of Inverse Functions
For f(x) = 5x + 11, find (f-1°f)(777). For f(x) = 5x + 11, find (f°f-1)(-5802).
2-2 Linear Equations What you’ll learn … To graph linear equations.
To write equations of lines.
Graphing Linear Equations
A function whose graph is a line is a linear function. You can represent a linear function with a linear equation, such as y=3x+2. A solution is any ordered pair (x,y) that makes the equation true. Because the y depends on the value of x, y is called the dependent variable and the x is called the independent variable.
Example 1 Graphing a Linear Equation
Graph the equations using a table. y=-3x y=½x+3 x y x y
The y intercept of a line is the point in which the line crosses the y-axis.
The x intercept of a line is the point in which the line crosses the x-axis. The standard form of a linear equation is Ax +By = C, where A,B and C are real numbers and A and B are not both zero.
Example 2 Real World Connection
The equation 3x +2y =120 models the number of passengers who can sit in a train car, where x is the number of adults and y is the number of children. Graph the equation. Describe the domain and range. Explain what the x and y intercepts represent.
Slope The slope of a non-vertical line is the ratio of the vertical change to a corresponding horizontal change. You can calculate the slope by subtracting the corresponding coordinates of two points on the line.
Slope Formula = Vertical change (rise) Horizontal change (run) y2 – y1
x2 – x1 =
Example 3 Finding Slope Find the slope of the line through the points
(3,2) and (-9,6). Find the slope of the line through the points (5,2) and (-6,2).
Point-Slope Form When you know the slope and a point on a line, you can use the point-slope form to write an equation of the line. y – y1 = m (x - x1)
Example 4 Writing an Equation Given the Slope and a Point
Write in standard form an equation of the line with slope -½ through the point (8,-1). Write in standard form an equation of the line with slope 2 through the point (4, -2).
Example 5 Writing an Equation Given Two Points
Write in point slope form an equation of the line through (1,5) and (4,-1). Write in point slope form an equation of the line through (-2,-1) and (-10,17).
Slope Intercept Form Another form of the equation of a line is slope intercept form, which you can use to find the slope by examining the equation. y= mx +b Slope y intercept
Example 6 Finding Slope Using Slope-Intercept Form
Find the slope of 4x + 3y = 7. Find the slope of ½x + ¾y = 1
Summary: Equations of Lines
Point Slope Form y-y1 = m(x-x1) y- 2= -3(x+4) Standard Form Ax + By = C 3x + y = -10 Slope Intercept Form y = mx + b y = -3x - 10
Special Slopes Vertical Line Horizontal Line Zero Slope
Undefined Slope
Special Slopes Perpendicular Lines Parallel Lines Have same slopes
Have reciprocal slopes
Example 7 Writing an Equation of a Perpendicular or Parallel Line
Write an equation perpendicular to y=5x-3 And through the point (-1,3). Write an equation parallel to y=2/3x+5/8 and through the point (2,1).
2-4 Using Linear Models What you’ll learn …
2.04 Create and use best-fit mathematical models of linear, exponential, and quadratic functions to solve problems involving sets of data. a) Interpret the constants, coefficients, and bases in the context of the data. b) Check the model for goodness-of-fit and use the model, where appropriate, to draw conclusions or make predictions. What you’ll learn … To write linear equations that model real-world data. To make predictions from linear models.
Example 1a Real World Connection
Jacksonville, Florida has an elevation of 12 ft above sea level. A hot air balloon taking off from Jacksonville rises 50 ft/min. Write an equation to model the balloon’s elevation as a function of time. Graph the equation.
Example 1b Real World Connection
Suppose a balloon begins descending at a rate of 20 ft/min from an elevation of 1350 ft. Write an equation to model the balloon’s elevation as a function of time. What is true about the slope of this line? Graph the equation. Interpret the h-intercept.
Example 2a Real World Connection
A candle is 6 inches tall after burning for 1hour. After 3 hours, it is 5½ inches tall. Write a linear equation to model the height y of the candle after burning x hours. What does the slope represent ? The y intercept? Graph the equation.
Example 2b Real World Connection
Another candle is 7 inches tall after burning for 1hour. After 2 hours, it is 5 inches tall. Write a linear equation to model the height y of the candle after burning x hours. How tall will the candle be after burning 11 hours? What was the original height of the candle? When will the candle burn out?
Scatter Plots A scatter plot is a graph that relates two different sets of data by plotting the data as ordered pairs. You can use a scatter plot to determine a relationship between the data sets. A trend line is a line that approximates the relationship between the data sets of a scatter plot. You can use a trend line to make predictions.
Correlation Strong, Positive Correlation Weak, Positive Correlation
No Correlation Weak, Negative Correlation Strong, Negative Correlation
Calculator Steps Enter data into lists. Turn on Stat Plot. Zoom 9
Stat Edit Turn on Stat Plot. 2nd y = Zoom 9 Turn on Diagnostic 2nd Zero Find the trend line. Stat Calc 4
Trend Line Make a scatter plot of the data. Draw a trend line.
Fat Calories 6 267 7 260 10 220 19 388 20 430 27 550 36 633 Make a scatter plot of the data. Draw a trend line. Estimate the number of calories in a fast-food item that has 14g of fat.
Trend Line Make a scatter plot of the data on the calculator.
Draw a trend line. Predict the wingspan of a hawk that is 28 inches long. Hawk Length Wingspan Cooper’s 21 36 Crane 41 Gray 18 38 Harris’s 24 46 Roadside 16 31 Broad-winged 19 39 Short-tailed 17 35 Swanson’s’
2-5 Absolute Value Functions and Graphs
2.08 Use equations and inequalities with absolute value to model and solve problems; justify results. a) Solve using tables, graphs, and algebraic properties. What you’ll learn … To graph absolute value functions.
Graphing Absolute Value Functions
A function of the form f(x) = mx+b +c, where m≠0, is an absolute value function. The vertex of a function is a point where the function reaches a maximum or minimum.
Example 1 Graphing an Absolute Value Function
Graph y= 3x+12 Graph y= - x
Example 2 Using a Graphing Calculator
Graph y= 3 - ½x Vertex _________ Graph y= - 3x+4 +6 Vertex __________
Example 4 Real World Connection
Suppose you pass the Betsy Ross House halfway along your trip to school each morning. You walk at a rate of one city block per minute. Sketch a graph of your trip to school based on your distance and time from the Betsy Ross House. Blocks from Ross House Minutes before arrival Minutes after departure
Example 4 Real World Connection
Suppose you ride your bicycle to school at a rate of three city blocks per minute. How would the graph of your trip to school change? Sketch a new graph. Blocks from Ross House Minutes before arrival Minutes after departure
2-6 Vertical and Horizontal Translations
2.08 Use equations and inequalities with absolute value to model and solve problems; justify results. a) Solve using tables, graphs, and algebraic properties. What you’ll learn … To analyze vertical translations. To analyze horizontal translations.
Translating Graphs Vertically
A translation is an operation that shifts a graph horizontally, vertically or both. It results in a graph of the same slope and size, in a different position. Vertical Translation Horizontal Translation
Example 1 Comparing Graphs
Compare the graphs y = x and y = x -3 y = x and y = x +5
Rules of Translations Given f(x), then –f(x) is a reflection about the x axis. Given f(x), then f(x) + k moves up k units. Given f(x), then f(x) - k moves down k units. Given f(x), then f(x+k) moves left k units. Given f(x), then f(x-k) moves right k units.
A family of functions is a group of functions with common characteristics. A parent function is the simplest function with these characteristics. A parent function and one or more translations make up a family of functions.
Example 2 Graphing a Vertical Translation
For each function, identify the parent function and the value of k. Then graph the function by translating the parent function. y = x – 1 y = 3x + 5 y = x – 3 y = - x +2 y = x y = - x
Example 3a Writing Equations for Vertical Translations
Write an equation for each translation. y = 2x, 4 units down. y = 3x , 2 units down.
Example 3b Writing Equations for Vertical Translations
Write an equation for each translation given y = x .
Example 4 Graphing a Horizontal Translation
For each function, identify the parent function and the value of k. Then graph the function by translating the parent function. y = x + 3 y = - x - 2 y = x - 1 y = - x + ¾
Example 5 Writing Equations for Horizontal Translations
Write an equation for each translation given y = x .
Example 6 Real World Connection
Describe a possible translation of Figures A and B in the Nigerian textile design below. A B
2-7 Two- Variable Inequalities
2.08 Use equations and inequalities with absolute value to model and solve problems; justify results. a) Solve using tables, graphs, and algebraic properties. What you’ll learn … To graph linear inequalities. To graph absolute value inequlaities.
Graphing Linear Inequalities
A linear inequality is an inequality in two variables whose graph is a region of the coordinate plane that is bounded by a line. To graph a linear inequality, first graph the boundary line. Then decide which side of the line contains solutions to the inequality and whether the boundary line is included.
Example 1a Graphing an Inequality
y > 2x + 3 m= ____ b = ____
Example 1b Graphing an Inequality
3x - 5y ≥ 10 m= ____ b = ____
Example 2 Real World Connection
At least 35 performers of the Big Tent Circus are in the grand finale. Some pile into cars, while others balance on bicycles. Seven performers are in each car, and five performers are on each bicycle. Draw a graph showing all the possible combination of cars and bicycles that could be use in the finale.
Example 3 Graphing Absolute Value Inequalities
y ≤ x – -y + 3 > x + 1
Example 4 Writing Inequalities
Write an inequality for each graph. The boundary line is given.
In Chapter 2, You Should Have
Moved from simplifying variable expressions and solving one-step equations and inequalities to working with two variable equations and inequalities. Learned how to represent function relationships by writing and graphing linear equations and inequalities. By graphing data and trend lines, you should understand how the slope of a line can be interpreted in real-world situations.
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## SectionAlgebraic Fractions
In the previous section, we saw a useful application of factoring polynomials, namely solving polynomial equations. In this section, we will explore another important application of factoring: algebraic fractions. Algebraic fractions are a special type of function, and factoring will allow us to simplify them as well as perform the four basic operations on them.
###### In this section, you will...
• define and simplify algebraic fractions
• identify the domain of algebraic fractions
• add, subtract, multiply, and divide algebraic fractions
### SubsectionIdentifying Restrictions and Simplifying
Algebraic fractions, or rational functions, have the form
\begin{equation*} r(x)=\frac{p(x)}{q(x)}\text{,} \end{equation*}
where $p(x)$ and $q(x)$ are polynomials and $q(x)\neq 0\text{.}$ The domain of an algebraic fraction will be identified with the domain or the rational function it defines and consists of all real numbers $x$ except those where the denominator $q(x)=0\text{.}$ Restrictions on the domain are real numbers for which the expression is undefined. For example, consider the function
\begin{equation*} f(x)=\frac{(x-1)(x-3)}{(x-2)(x-3)}\text{.} \end{equation*}
This is an algebraic fraction since (once expanded) both the numerator and denominator are polynomials. To find the domain, we need to consider what values of $x$ will make the denominator equal to zero; in other words, the domain of this function is everything except the value(s) that solve $0=(x-2)(x-3)\text{.}$ Applying the zero-factor property, we have
\begin{equation*} \begin{aligned} x-2\amp =0 \: \: \: \: \: \amp \text{ or }\: \: \: \: \: x-3\amp =0\\ x\amp =2\amp x\amp =3 \end{aligned} \end{equation*}
Therefore, the original function is defined for any real number except $2$ and $3\text{.}$ We can express its domain using notation.
\begin{equation*} x\neq 2,3 \: \: \: \: \: \text{ or }\: \: \: \: \: (-\infty, 2)\cup(2,3)\cup(3,\infty) \end{equation*}
The restrictions to the domain of an algebraic fraction are determined by the denominator. Once the restrictions are determined we can cancel factors and obtain an equivalent function.
It is important to note that $1$ is not a restriction to the domain because the expression is defined as $0$ when the numerator is $0\text{.}$ In fact, $x=1$ is a root. This function is graphed below.
Notice that there is a vertical asymptote at the restriction $x=2$ and the graph is left undefined at the restriction $x=3$ as indicated by the open dot, or hole, in the graph. Graphing algebraic fractions in general is beyond the scope of this course. However, it is useful at this point to know that the restrictions are an important part of the graph of algebraic fractions.
###### Example176
State the restrictions and simplify the algebraic fraction $g(x)=\frac{24x^7}{6x^5}\text{.}$
Solution
Since $g(x)$ is an algebraic fraction, it is undefined, or restricted, where its denominator is zero. Since the only solution of $6x^5=0$ is $x=0\text{,}$ the function is undefined where $x$ is $0\text{.}$
\begin{equation*} g(0)=\frac{24(0)^7}{6(0)^5}=\frac{0}{0} \: \: \: \alert{\text{Undefined}} \end{equation*}
Therefore, the domain consists of all real numbers $x\text{,}$ where $x\neq 0\text{.}$ With this understanding, we can simplify by reducing the rational expression to lowest terms. Cancel common factors.
\begin{equation*} g(x)=\frac{24x^7}{6x^5}=\frac{4x^7}{x^5}=4x^2 \end{equation*}
Our answer is $g(x)=4x^2\text{,}$ where $x\neq 0\text{.}$
###### Example177
State the restrictions and simplify the algebraic fraction $g(x)=\frac{2x^2+5x-3}{4x^2-1}\text{.}$
Solution
Before we can identify the restrictions and simplify the function, we need to factor both the numerator and the denominator.
\begin{equation*} \begin{aligned} g(x)\amp =\frac{2x^2+5x-3}{4x^2-1}\\ \amp =\frac{(2x-1)(x+3)}{(2x+1)(2x-1)} \end{aligned} \end{equation*}
Since $g(x)$ is an algebraic fraction, it is undefined, or restricted, where its denominator is zero. Applying the zero-factor principle, we have
\begin{equation*} \begin{aligned} 2x+1\amp =0 \: \: \: \: \: \amp \text{ or }\: \: \: \: \: 2x-1\amp =0\\ 2x\amp =-1\amp 2x\amp =1\\ x\amp =-\frac{1}{2}\amp x\amp =\frac{1}{2} \end{aligned} \end{equation*}
Therefore, the domain consists of all real numbers $x\text{,}$ where $x\neq -\frac{1}{2},\frac{1}{2}\text{.}$ With this understanding, we can simplify by reducing the rational expression to lowest terms. Cancel common factors.
\begin{equation*} g(x)=\frac{(2x-1)(x+3)}{(2x+1)(2x-1)}=\frac{x+3}{2x+1} \end{equation*}
Our answer is $g(x)=\frac{x+3}{2x+1}\text{,}$ where $x\neq -\frac{1}{2},\frac{1}{2}\text{.}$
Occasionally we will have to factor out a negative one to simplify an algebraic fraction as much as possible.
###### Example178
State the restrictions and simplify the algebraic fraction $g(x)=\frac{25-x^2}{x^2-10x+25}\text{.}$
Solution
First, we factor the numerator and denominator.
\begin{equation*} \begin{aligned} g(x)\amp =\frac{25-x^2}{x^2-10x+25}\\ \amp =\frac{(5-x)(5+x)}{(x-5)(x-5)}\\ \amp = \frac{\alert{-1\cdot (x-5)}(5+x)}{(x-5)(x-5)}\\ \amp=\frac{-1(5+x)}{(x-5)}\\ \amp = -\frac{x+5}{x-5}\\ \end{aligned} \end{equation*}
Our answer is $g(x)=-\frac{x+5}{x-5}\text{,}$ where $x\neq 5\text{.}$
It is important to remember that we can only cancel factors of a product. A common mistake is to cancel terms.
###### Example179
State the restrictions and simplify $f(x)=\frac{x-2x^2}{4x^4-x^2}\text{.}$
Solution
Before we can identify the restrictions and simplify the function, we need to factor both the numerator and the denominator.
\begin{equation*} \begin{aligned} f(x)\amp= \frac{x-2x^2}{4x^4-x^2}\\ \amp= \frac{x(1-2x)}{x^2(2x+1)(2x-1)} \\ \amp= \frac{-x(2x-1)}{x^2(2x+1)(2x-1)} \end{aligned} \end{equation*}
Since $f(x)$ is an algebraic fraction, it is undefined, or restricted, where its denominator is zero. Applying the zero-factor principle, we find that the restrictions are $x=0,\frac{1}{2},-\frac{1}{2}\text{.}$
Therefore, the domain consists of all real numbers $x\text{,}$ where $x\neq 0,\frac{1}{2},-\frac{1}{2}\text{.}$ With this understanding, we can simplify by reducing the rational expression to lowest terms. Cancel common factors.
\begin{equation*} f(x)=\frac{-x(2x-1)}{x^2(2x+1)(2x-1)}=\frac{-1}{x(2x+1)} \end{equation*}
Our answer is $f(x)=-\frac{1}{x(2x+1)}\text{,}$ where $x\neq 0,\frac{1}{2},-\frac{1}{2}\text{.}$
In some examples, we will make a broad assumption that the denominator is nonzero. When we make that assumption, we do not need to determine the restrictions.
###### Example180
Given $f(x)=x^2-2x+5\text{,}$ simplify $\frac{f(x)-f(3)}{ax-3a}\text{.}$ Assume all denominators are nonzero.
Solution
First, we calculate $f(3)\text{.}$
\begin{equation*} \begin{aligned} f(3)\amp = (3)^2-2(3)+5\\ \amp = 9-6+5\\ \amp =3+5\\ \amp =8. \end{aligned} \end{equation*}
Now we substitute into the quotient.
\begin{equation*} \begin{aligned} \frac{f(x)-f(3)}{ax-3a}\amp = \frac{x^2-2x+5-8}{ax-3a}\\ \amp = \frac{x^2-2x-3}{ax-3a}\\ \amp =\frac{(x+1)(x-3)}{a(x-3)}\\ \amp =\frac{x+1}{a} \end{aligned} \end{equation*}
Our answer is $\frac{f(x)-f(3)}{ax-3a}=\frac{x+1}{a}\text{.}$
### SubsectionMultiplying and Dividing Algebraic Fractions
Recall from Numbers and Operations that to multiply two fractions, we can multiply the numerators together, multiply the denominators together, and then reduce. Multiplying rational expressions is performed in a similar manner. In general, given polynomials $P, Q, R,$ and $S\text{,}$ where $Q\neq 0$ and $S\neq 0\text{,}$ we have
\begin{equation*} \frac{P}{Q}\cdot\frac{R}{S}=\frac{PR}{QS}\text{.} \end{equation*}
The restrictions to the domain of a product consist of the restrictions of each function.
###### Example181
Given $f(x)=\frac{9x^2-25}{x-5}$ and $g(x)=\frac{x^2-2x-15}{3x+5}\text{,}$ find $f(x)\cdot g(x)$ and determine the restrictions on the domain.
Solution
In this case, the domain of $f$ consists of all real numbers except $5\text{,}$ and the domain of $g$ consists of all real numbers except $-\frac{5}{3}\text{.}$ Therefore, the domain of the product consists of all real numbers except $5$ and $-\frac{5}{3}\text{.}$ Multiply the functions and then simplify the result.
\begin{equation*} \begin{aligned} f(x)\cdot g(x) \amp =\frac{9x^2-25}{x-5}\cdot \frac{x^2-2x-15}{3x+5}\\ \amp =\frac{(3x+5)(3x-5)}{x-5}\cdot\frac{(x-5)(x+3)}{3x+5}\\ \amp =\frac{(3x+5)(3x-5)(x-5)(x+3)}{(x-5)(3x+5)}\\ \amp=\frac{(3x-5)(x+3)}{1}\\ \amp =(3x-5)(x+3) \end{aligned} \end{equation*}
Our answer is $f(x)\cdot g(x)=(3x-5)(x+3)$ where $x\neq 5,-\frac{5}{3}\text{.}$
To divide two fractions, we multiply by the reciprocal of the divisor. Dividing rational expressions is performed in a similar manner. In general, given polynomials $P, Q, R\text{,}$ and $S\text{,}$ where $Q\neq 0\text{,}$ $R\neq 0\text{,}$ and $S\neq 0\text{,}$ we have
\begin{equation*} \frac{P}{Q}\div\frac{R}{S}=\frac{\frac{P}{Q}}{\frac{R}{S}}=\frac{P}{Q}\cdot \frac{S}{R}=\frac{PS}{QR}\text{.} \end{equation*}
The restrictions to the domain of a quotient will consist of the restrictions of each function as well as the restrictions on the reciprocal of the divisor.
###### Example182
Given $f(x)=\frac{2x^2+13x-7}{x^2-4x-21}$ and $g(x)=\frac{2x^2+5x-3}{49-x^2}\text{,}$ find $\frac{f(x)}{g(x)}$ and determine the restrictions on the domain.
Solution
We'll start by setting up the quotient and factoring each polynomial.
\begin{equation*} \begin{aligned} f(x)\div g(x) \amp =\frac{2x^2+13x-7}{x^2-4x-21} \div \frac{2x^2+5x-3}{49-x^2}\\ \amp= \frac{(2x-1)(x+7)}{(x+3)(x-7)} \div \frac{(2x-1)(x+3)}{(7+x)(7-x)} \end{aligned} \end{equation*}
To find the restrictions, we need to find the restrictions of $f(x)\text{,}$ $g(x)\text{,}$ as well as the reciprocal of $g(x)\text{.}$ The restrictions of $f(x)$ are $-3$ and $7\text{,}$ and the restrictions of $g(x)$ are $7$ and $-7\text{,}$ and the reciprocal of $g(x)$ has restrictions of $-3$ and $\frac{1}{2}\text{.}$ Therefore, the domain of this quotient consists of all real numbers except $-3, \frac{1}{2}, 7$ and $-7\text{.}$
Next, to simplify the quotient, we perform the division and reduce.
\begin{equation*} \begin{aligned} f(x)\div g(x) \amp =\frac{2x^2+13x-7}{x^2-4x-21} \div \frac{2x^2+5x-3}{49-x^2}\\ \amp= \frac{(2x-1)(x+7)}{(x+3)(x-7)} \div \frac{(2x-1)(x+3)}{(7+x)(7-x)} \\ \amp= \frac{(2x-1)(x+7)}{(x+3)(x-7)} \cdot \frac{(7+x)(7-x)}{(2x-1)(x+3)} \\ \amp= \frac{(2x-1)(x+7)(7+x)(7-x)}{(x+3)(x-7)(2x-1)(x+3)} \\ \amp= \frac{(x+7)(7+x)(-1)}{(x+3)(x+3)} \end{aligned} \end{equation*}
Our answer is $\left(\frac{f}{g}\right)(x)=-\frac{(x+7)^2}{(x+3)^2}$ where $x\neq -3,\frac{1}{2},7,-7\text{.}$
Given $f(x)=\frac{2x+5}{3x^2+14x-5}$ and $g(x)=\frac{6x^2+13x-5}{x+5}\text{,}$ find $\left(\frac{f}{g}\right)(x)$ and determine the restrictions on the domain.
Solution
Our answer is $\left(\frac{f}{g}\right)(x)=\frac{1}{(3x-1)^2}$ where $x\neq -5,-\frac{5}{2},\frac{1}{3}\text{.}$
### SubsectionAdding and Subtracting Algebraic Fractions
Recall from Numbers and Operations that to multiply two fractions, we need a common denominator and then we can simply add or subtract the numerators and write the result over the common denominator. When working with rational expressions, the common denominator will be a polynomial. In general, given polynomials $P, Q\text{,}$ and $R\text{,}$ where $Q\neq 0\text{,}$ we have the following:
\begin{equation*} \frac{P}{Q}\pm\frac{R}{Q}=\frac{P\pm R}{Q}\text{.} \end{equation*}
The set of restrictions to the domain of a sum or difference of rational expressions consists of the restrictions to the domains of each expression.
###### Example184
Perform the subtraction $\frac{4x}{x^2-64}-\frac{3x+8}{x^2-64}$ and keep track of restrictions on the domain.
Solution
Since the denominators are the same, we can subtract the numerators and write the result over the common denominator. Take care to distribute the negative $1\text{.}$
\begin{equation*} \begin{aligned} \frac{4x}{x^2-64}-\frac{3x+8}{x^2-64}\amp = \frac{4x-(3x+8)}{x^2-64}\\ \amp = \frac{4x-3x-8}{x^2-64}\\ \amp = \frac{x-8}{(x+8)(x-8)}\\ \amp = \frac{1}{x+8} \end{aligned} \end{equation*}
Our solution is $\frac{1}{x+8}$ for all real numbers $x\neq \pm 8\text{.}$
To add or subtract rational expressions with unlike denominators, we first find equivalent expressions with common denominators. We do this just as we have with fractions. If the denominators of fractions are relatively prime, then the least common denominator (LCD) is their product. For example,
\begin{equation*} \frac{1}{x}+\frac{1}{y} \implies \text{LCD}=x\cdot y=xy. \end{equation*}
Multiply each fraction by the appropriate form of $1$ to obtain equivalent fractions with a common denominator.
\begin{equation*} \begin{aligned} \frac{1}{x}+\frac{1}{y}\amp =\frac{1}{x}\cdot\alert{\frac{y}{y}}+\frac{1}{y}\cdot\alert{\frac{x}{x}}\\ \amp = \frac{y}{xy}+\frac{x}{xy}\\ \amp = \frac{y+x}{xy} \end{aligned} \end{equation*}
In general, given polynomials $P, Q, R\text{,}$ and $S\text{,}$ where $Q\neq 0$ and $S\neq 0\text{,}$ we have the following:
\begin{equation*} \frac{P}{Q}\pm\frac{R}{S}=\frac{PS\pm QR}{QS} \end{equation*}
###### Example185
Given $f(x)=\frac{5x}{3x+1}$ and $g(x)=\frac{2}{x+1}\text{,}$ find $f(x)+g(x)$ and determine the restrictions.
Solution
Here the LCD is the product of the denominators $(3x+1)(x+1)\text{.}$ We multiply by the appropriate factors to obtain rational expressions with a common denominator before adding.
\begin{equation*} \begin{aligned} (f+g)(x)\amp = f(x)+g(x)\\ \amp =\frac{5x}{3x+1}+\frac{2}{x+1}\\ \amp =\frac{5x}{3x+1}\cdot\alert{\frac{(x+1)}{(x+1)}}+\frac{2}{x+1}\cdot\alert{\frac{(3x+1)}{(3x+1)}}\\ \amp = \frac{5x(x+1)}{(3x+1)(x+1)}+\frac{2(3x+1)}{(x+1)(3x+1)}\\ \amp=\frac{5x(x+1)+2(3x+1)}{(3x+1)(x+1)}\\ \amp = \frac{5x^2+5x+6x+2}{(3x+1)(x+1)}\\ \amp =\frac{5x^2+11x+2}{(3x+1)(x+1)}\\ \amp =\frac{(5x+1)(x+2)}{(3x+1)(x+1)}\\ \end{aligned} \end{equation*}
The domain of $f$ consists all real numbers except $-\frac{1}{3}\text{,}$ and the domain of $g$ consists of all real numbers except $-1\text{.}$ Therefore, the domain of $f + g$ consists of all real numbers except $-1$ and $-\frac{1}{3}\text{.}$
Our solution is $\frac{(5x+1)(x+2)}{(3x+1)(x+1)}$ for all real numbers $x\neq -1,-\frac{1}{3}\text{.}$
It is not always the case that the LCD is the product of the given denominators. Typically, the denominators are not relatively prime; thus determining the LCD requires some thought. Begin by factoring all denominators. The LCD is the product of all factors with the highest power.
###### Example186
Given $f(x)=\frac{3x}{3x-1}$ and $g(x)=\frac{4-14x}{3x^2-4x+1}\text{,}$ find $f-g$ and determine the restrictions.
Solution
To determine the LCD, factor the denominator of $g\text{.}$
\begin{equation*} \begin{aligned} (f-g)(x)\amp = f(x)-g(x)\\ \amp =\frac{3x}{3x-1} -\frac{4-14x}{3x^2-4x+1}\\ \amp=\frac{3x}{3x-1} -\frac{4-14x}{(3x-1)(x-1)}\\ \end{aligned} \end{equation*}
In this case the LCD$=(3x-1)(x-1)\text{.}$ Multiply $f$ by $1$ in the form of $\frac{(x-1)}{(x-1)}$ to obtain equivalent algebraic fractions with a common denominator and then subtract.
\begin{equation*} \begin{aligned} (f-g)(x)\amp = f(x)-g(x)\\ \amp =\frac{3x}{3x-1}\cdot\alert{\frac{(x-1)}{(x-1)}} -\frac{4-14x}{(3x-1)(x-1)}\\ \amp=\frac{3x(x-1)-4+14x}{(3x-1)(x-1)}\\ \amp =\frac{(3x-1)(x+4)}{(3x-1)(x-1)}\\\\ \amp = \frac{(x+4)}{(x-1)} \end{aligned} \end{equation*}
The domain of $f$ consists all real numbers except $\frac{1}{3}\text{,}$ and the domain of $g$ consists of all real numbers except $1$ and $\frac{1}{3}\text{.}$ Therefore, the domain of $f - g$ consists of all real numbers except $1$ and $\frac{1}{3}\text{.}$
Our solution is $\frac{x+4}{x-1}$ for all real numbers $x\neq 1, \frac{1}{3}\text{.}$
Simplify $\frac{-2x}{x+6}-\frac{3x}{6-x}-\frac{18(x-2)}{x^2-36}$ and determine the restrictions.
Solution
Our solution is $\frac{x+6}{x-6}$ for all real numbers $x\neq \pm 6\text{.}$
Simplify $\frac{x+1}{(x-1)^2}-\frac{2}{x^2-1}-\frac{4}{(x+1)(x-1)^2}$ and determine the restrictions.
Solution
Our solution is $\frac{1}{x-1}$ for all real numbers $x\neq \pm 1\text{.}$
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# 15: More Transformations © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules.
## Presentation on theme: "15: More Transformations © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules."— Presentation transcript:
15: More Transformations © Christine Crisp “Teach A Level Maths” Vol. 2: A2 Core Modules
Trig Transformations Combined Translations x axis translation of –a and y axis translation of b x axis translation of +a and y axis translation of b Note Opposite sign Note Opposite sign
Trig Transformations Stretches The function is obtained from by a stretch of scale factor ( s.f. ) k, parallel to the y -axis. The function is obtained from by a stretch of scale factor ( s.f. ), parallel to the x -axis.
Trig Transformations Reflections Reflections in the axes Reflecting in the x -axis changes the sign of y Reflecting in the y -axis changes the sign of x
Trig Transformations e.g. 1 Sketch the graph of the function is a stretch of s.f. 2, parallel to the y -axis. Solution: We can use the fact that is a stretch of. xysin2
Trig Transformations e.g. 1 Sketch the graph of the function Solution: We can use the fact that is a stretch of. xysin2 is a stretch of s.f. 2, parallel to the y -axis. The scale factor of the stretch gives the amplitude of the function.
Trig Transformations e.g. 2 Sketch the graph of the function Solution: is a stretch of s.f., parallel to the x -axis. So,
Trig Transformations e.g. 2 Sketch the graph of the function Solution: is a stretch of s.f., parallel to the x -axis. So, The period of is or radians.
Trig TransformationsExercises 1. Give the equation of the function that is shown on the sketch below. Ans:
Trig Transformations Solution: A stretch of s.f. 2 parallel to the x -axis. Sketch both functions on the same axes for the interval 2. Describe in words the transformation Exercises
Trig Transformations Solution: 3.Sketch the graph of for showing the scales clearly. What is the period of the function? The period is Exercises
Trig Transformations Reflection in the x -axis Every y -value changes sign when we reflect in the x -axis e.g. So, x x In general, a reflection in the x -axis is given by
Trig Transformations then (iii) a reflection in the x -axis (i) a stretch of s.f. 2 parallel to the x -axis then (ii) a translation of e.g.3 Find the equation of the graph which is obtained from by the following transformations, sketching the graph at each stage. ( Start with ).
Trig Transformations Solution: (i) a stretch of s.f. 2 parallel to the x -axis stretch
Trig Transformations Brackets aren’t essential here but they make it clearer. (ii) a translation of : translate
Trig Transformations (ii) a translation of : translatereflect x x (iii) a reflection in the x -axis
Trig Transformations Exercises 1. Describe the transformations that map the graphs of the 1 st of each function given below onto the 2nd. Sketch the graphs at each stage. (a) to ( Draw for ) (b) y = cosx to y = 2cos(x – 30)
Trig Transformations Solutions: (a) to Translation Stretch s.f. parallel to the x -axis
Trig Transformations Vertical stretch factor 2 (b) to Solutions: Translation parallel to the x -axis
Trig Transformations (i) (ii) The diagram shows part of the curve with equation. Copy the diagram twice and on each diagram sketch one of the following: x y
Trig Transformations Solution: (ii) x y xy (i)
Trig Transformations In an earlier section, we met stretches. is a stretch of scale factor ( s.f. ) k, parallel to the y -axis e.g. is a stretch of s.f. 2, parallel to the y -axis Reminder: ( multiplied by k )
Trig Transformations is a stretch of scale factor ( s.f. ), parallel to the x -axis. e.g. is a stretch of s.f. parallel to the x -axis. ( x multiplied by k )
Trig Transformations e.g. 1 Sketch the graph of the function is a stretch of s.f. 2, parallel to the y -axis. Solution: We can use the fact that is a stretch of. xysin2
Trig Transformations e.g. 1 Sketch the graph of the function Solution: We can use the fact that is a stretch of. xysin2 is a stretch of s.f. 2, parallel to the y -axis. The scale factor of the stretch gives the amplitude of the function.
Trig Transformations e.g. 2 Sketch the graph of the function Solution: is a stretch of s.f., parallel to the x -axis. So,
Trig Transformations e.g. 2 Sketch the graph of the function Solution: is a stretch of s.f., parallel to the x -axis. So, The period of is or radians.
Trig TransformationsExercises 1. Give the equation of the function that is shown on the sketch below. Ans:
Trig Transformations Solution: A stretch of s.f. 2 parallel to the x -axis. Sketch both functions on the same axes for the interval 2. Describe in words the transformation Exercises
Trig Transformations Solution: 3.Sketch the graph of for showing the scales clearly. What is the period of the function? The period is Exercises
Trig Transformations
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# Evaluate the definite integral of f=1/(x^2+2x+5) from 0 to 1 .
giorgiana1976 | Student
To evaluate the definite integral, we'll re-write the denominator.
x^2+2x+5 = x^2+2x+4+1
We'll group the terms: x^2+2x+1 = (x+1)^2
The denominator will become:
(x+1)^2 + 4
We'll write the integral:
Int f(x)dx = Int dx/(x^2+2x+5)
Int dx/(x^2+2x+5) = Int dx/[(x+1)^2 + 4]
We'll substitute x+1 by t.
x+1 = t
(x+1)' = 1*dx
t' = dt
So, dx = dt
We'll re-write the integral in t:
Int dx/[(x+1)^2 + 4] = Int dt/(t^2+4)
We'll factorize by 4, at denominator:
Int dt/4(t^2/4 + 1) = (1/4)*Int dt/[(t/2)^2 + 1]
(1/4)*Int dt/[(t/2)^2 + 1] = 2/4*arctg (t/2) = (1/2)*arctg (t/2)
But t = x+1
For x = 0 => t = 1
For x = 1 => t = 2
Now, we'll apply Leibniz-Newton formula:
Intdx/(x^2+2x+5), from 0 to 1 = F(2) - F(1)
F(2) = (1/2)*arctg (2/2) = (1/2)*(pi/4)
F(1) = (1/2)*arctg (1/2)
F(2) - F(1) = (1/2)*[pi/4 - arctg (1/2)]
Intdx/(x^2+2x+5), from 0 to 1 = (1/2)*[pi/4 - arctg (1/2)]
neela | Student
To evaluate Integral f(x) dx x = 0 to1.
f(x) = 1/(x^2+2x+5)
Solution:
f(x) = 1/(x+1)^2 +4
Put x+1 = t. Then dx = dt, when x= 0 to 1, t= 1 to 2.
{Integral f(x) dx , x=0 to 1} = {integral dt/(t+2^2) , t = 1 to 2.
= (1/2){arc tan t/2 , t= 1, 2}, as Integral dx/(x^2+a^2) = (1/a) arctan (x/a).
= {(1/2) arc tan (2/2) - (1/2) arc tan (1/2)}
= (1/2{ arctan1 - arctan(1/2)}
= (1/2){pi/4 - aerctan(1/2)}
= (1/2){0.785398163 - 0.463647609}
= 0.160875277 radians.
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How to find the total number of pages which a book has when the clues given indicate a range?
This problem doesn't seem very complicated but I got stuck at trying to understand what is the meaning of the last clue involving an integer and a range. Can somebody help me?
The problem is as follows:
Marina is reading a novel. The first day she read a third of the book, the second day she read the fourth parts of what it was left, the third day she read half of what it was left to read, the fourth day she read the fifth parts of what it was still left to read, the fifth day she decided to end the novel and found that it was left less than $$70$$ pages. If she always read an integer number of pages and never read less than $$14$$ pages. How many pages did the novel had?
The alternatives given in my book were as follows:
$$\begin{array}{ll} 1.&\textrm{360 pages}\\ 2.&\textrm{240 pages}\\ 3.&\textrm{180 pages}\\ 4.&\textrm{300 pages}\\ 5.&\textrm{210 pages}\\ \end{array}$$
I'm lost with this problem. What would be the correct way to go?. So far what I attempted to do was the following:
I thought that the number of pages that the book has to be $$x$$.
Since it said that the first day she read a third of the book I defined it as:
$$\frac{1}{3}x$$
On the second day it is said that she read the fourth parts of what it was left so that would account for:
$$\frac{1}{4}\left(x-\frac{1}{3}x\right)=\frac{1}{4}\left(\frac{2x}{3}\right)=\frac{x}{6}$$
The third day:
$$\frac{1}{2}\left(x-\frac{x}{6}\right)=\frac{1}{2}\left(\frac{5x}{6}\right)=\frac{5x}{12}$$
The fourth day:
$$\frac{1}{5}\left(x-\frac{5x}{12}\right)=\frac{1}{5}\left(\frac{7x}{12}\right)=\frac{7x}{60}$$
The fifth day:
She decides to end reading the novel but, what it was left was less than $$70$$ pages.
So this would translate as:
$$x-\frac{7x}{60}<70$$
This would become into:
$$\frac{53x}{60}<70$$
Therefore:
$$53x<4200$$
$$x<\frac{4200}{53}$$
However this fraction is not an integer.
There is also another piece of information which mentioned that she always read no less than $$14$$ pages.
If during the first day she read a third of the novel then this would be:
$$\frac{1}{3}x>14$$
So $$x>42$$
But, on the fourth day she read:
$$\frac{7x}{60}>14$$
Therefore:
$$x>120$$
How come x can be greatest than $$42$$ and at the same time $$120$$?. Am I understanding this correctly?.
If I were to select the greatest value and put it in the range which I found earlier:
$$120
and round to the nearest integer:
$$120
Which doesn't make sense.
If it were $$\frac{5x}{12}>14$$
$$x>33$$ (rounded to the nearest integer)
Which would be:
$$33
But again this doesn't produce an reasonable answer within the specified range in the answers. Did I overlooked something or perhaps didn't understood something right?. Can somebody help me with this inequation problem?.
Compute the fractions of the book read per day:
• On day 1, $$\frac13$$ of the novel was read, leaving $$\frac23$$.
• On day 2, $$\frac23×\frac14=\frac16$$ was read, leaving $$\frac23×\frac34=\frac12$$.
• On day 3, $$\frac12×\frac12=\frac14$$ was read, the same fraction being left.
• On day 4, $$\frac14×\frac15=\frac1{20}$$ was read, leaving $$\frac14×\frac45=\frac15$$ that was finished off on day 5.
Letting $$x$$ be the number of pages in the book, because at most 70 pages were left on day 5 we have $$\frac15x<70$$ or $$x<350$$. Because at least 14 pages were read per day, including day 4, we have $$\frac1{20}x>14$$ or $$x>280$$. Only option 4 satisfies both inequalities, so the novel had 300 pages.
• I would really like to understand what you meant. But the thing is you ommited some steps and I don't get very clearly where do those fractions come?. I don't get the idea in the second step for day 2. I understand the part of $\frac{1}{4}\times\frac{2}{3}$ but why should I multiply $\frac{2}{3} \times \frac{3}{4}$? In other words why should be multiplied what it is left instead of summing them?. – Chris Steinbeck Bell Mar 2 at 21:33
Your method is good, but you made a mistake on the third day. Indeed, she read a third of what was left. So she read :
$$\frac{1}{2}\left( x - \frac{1}{3}x-\frac{1}{6}x \right) = \frac{1}{2}\frac{1}{2}x = \frac{1}{4}x$$
So on the fourth day, she read $$\frac{1}{5}\left(1 - \frac{1}{2} - \frac{1}{4}\right)x = \frac{1}{5}\frac{1}{4}x = \frac{1}{20}x$$
So on the fifth days, there's : $$x\left(1 - \frac{1}{3} - \frac{1}{6} - \frac{1}{4} - \frac{1}{20}\right) = \frac{1}{5}{x}$$ Pages left.
To sum up :
First day : $$\frac{1}{3}x$$
Second day : $$\frac{1}{6}x$$
Third day : $$\frac{1}{4}x$$
Fourth day : $$\frac{1}{20}x$$
You want $$\frac{x}{20} \ge 14$$ so $$x\ge 280$$. Furthermore, you need $$\frac{1}{5}x< 70$$ so $$x<350$$.
The only possibility now is answer 4 : 300 pages.
• Thanks for the confidence boost. I really needed it. I think the source of my confusion was that I did not consider the passage mentions each day individually and not the number of pages that had been read until that day. Hence for each day I need to account what it was read on the prior day. Because of this for the third day $\frac{1}{2}\left(x-\left(\frac{x}{3}+\frac{x}{6}\right)\right)$ as $\frac{1x}{3}$ accounts for day 1 and $\frac{1x}{6}$ accounts for day two, so they must be summed up and so on until get to the last day. Did I understood this part correctly? – Chris Steinbeck Bell Mar 2 at 21:12
• Now I get to some confusion, why do you use $\leq$ and $\geq$ with $<$ and $>$ almost interchangeably? By the time I got to $\frac{x}{5}$ it is obviously stated in the problem that is less than $70$. But the passage also mentions that everyday she read no less than $14$ pages wouldn't this meant that I can use the other found quantities as well?. Let's say $\frac{1}{4}x\geq 14$ hence $x\geq 56$ isn't it?. But also meant that on the first day $\frac{1}{3}x\geq 14$ so $x \geq 42$. I'm kind of confused, having found these inequalities don't produce contradictory results?. – Chris Steinbeck Bell Mar 2 at 21:24
• Or just because they're inequalities it is possible to have different answers. It turns that I may need to look for the one which produces a highest number so I can reduce the boundary and find what number of pages they're asking. This problem in particular offered alternatives, but could this have been solved without any? by only concluding something from what it is given? – Chris Steinbeck Bell Mar 2 at 21:25
• Yes you understood correctly ! I use both < and \le because I don't really care about an exact boundary, I just need a good enough one to answer the question, but there is no logic in the use of one or the other. – aleph0 Mar 3 at 20:44
• Thanks for that but I have a question. Does it mean that given the conditions at itself it cannot be found the number of pages without checking the alternatives given?. – Chris Steinbeck Bell Mar 5 at 22:29
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# Ordering Rational Positive Numbers Worksheet
A Rational Amounts Worksheet might help your kids become a little more knowledgeable about the ideas behind this rate of integers. Within this worksheet, college students are able to solve 12 diverse troubles associated with reasonable expression. They will likely learn to grow a couple of amounts, class them in couples, and find out their items. They are going to also process simplifying reasonable expression. After they have learned these ideas, this worksheet will certainly be a beneficial device for furthering their scientific studies. Ordering Rational Positive Numbers Worksheet.
## Logical Figures can be a ratio of integers
There are 2 types of phone numbers: rational and irrational. Logical numbers are considered total numbers, whilst irrational figures usually do not repeat, and possess an limitless amount of digits. Irrational amounts are low-zero, no-terminating decimals, and rectangular beginnings which are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To outline a realistic variety, you need to understand just what a rational amount is. An integer is actually a total variety, and a logical variety is actually a rate of two integers. The proportion of two integers is definitely the quantity on the top separated from the variety on the bottom. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be produced in to a small percentage
A rational variety includes a numerator and denominator that are not zero. This means that they may be expressed as a small percentage. Along with their integer numerators and denominators, logical figures can also have a adverse benefit. The negative benefit ought to be placed to the left of and its total importance is its length from absolutely no. To make simpler this illustration, we shall state that .0333333 is really a fraction that could be created like a 1/3.
Along with negative integers, a realistic number can even be produced in to a small percentage. For example, /18,572 is a logical number, although -1/ is just not. Any small percentage composed of integers is realistic, as long as the denominator does not include a and will be created for an integer. Likewise, a decimal that ends in a position can be another logical amount.
## They make perception
In spite of their brand, realistic amounts don’t make a lot perception. In math, these are one entities using a exclusive duration on the number series. Which means that if we count something, we can easily order the size and style by its rate to its initial volume. This keeps real even if there are limitless realistic figures between two specific phone numbers. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To obtain the period of a pearl, for example, we could add up its thickness. Just one pearl weighs in at 10 kilograms, which is actually a logical variety. Additionally, a pound’s weight equates to twenty kilos. Therefore, we must be able to separate a pound by 15, without be concerned about the length of one particular pearl.
## They may be expressed like a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal variety can be written being a a number of of two integers, so four times several is the same as seven. A comparable dilemma involves the recurring small fraction 2/1, and each side must be split by 99 to get the right response. But how can you create the conversion? Here are some cases.
A logical number can also be designed in great shape, including fractions and a decimal. One way to symbolize a realistic quantity inside a decimal is to split it into its fractional equal. There are actually three ways to split a logical number, and each one of these methods results in its decimal counterpart. One of those approaches is to split it into its fractional equivalent, and that’s what’s known as the terminating decimal.
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### Playing with Numbers-Solutions Ex. 3.5
CBSE Class –VI Mathematics
NCERT Solutions
Chapter 3 Playing With Numbers (Ex. 3.5)
Question 1. Which of the following statements are true?
(a) If a number is divisible by 3, it must be divisible by 9.
(b) If a number is divisible by 9, it must be divisible by 3.
(c) If a number is divisible by 18, it must be divisible by both 3 and 6.
(d) If a number is divisible by 9 and 10 both, then it must be divisible by 90.
(e) If two numbers are co-prime, at least one of them must be prime.
(f) All numbers which are divisible by 4 must also be divisible by 8.
(g) All numbers which are divisible by 8 must also be divisible by 4.
(h) If a number is exactly divides two numbers separately, it must exactly divide their sum.
(i) If a number is exactly divides the sum of two numbers, it must exactly divide the two numbers separately.
Answer: Statements (b), (c), (d), (g) and (h) are true.
Question 2. Here are two different factor trees for 60. Write the missing numbers.
(a)
(b)
Answer: Sol.
(a)
(b)
Question 3. Which factors are not included in the prime factorization of a composite number?
Answer:
1 is not included in the prime factorization of a composite number.
Question 4. Write the greatest 4-digit number and express it in terms of its prime factors.
Answer: The greatest 4-digit number is 9999.
The prime factors of 9999 are 3 x 3 x 11 x 101.
Question 5. Write the smallest 5-digit number and express it in terms of its prime factors.
Answer: The smallest 5-digit number is 10000.
The prime factors of 10000 are 2 x 2 x 2 x 2 x 5 x 5 x 5 x 5.
Question 6. Find all the prime factors of 1729 and arrange them in ascending order. Now state the relation, if any, between, two consecutive prime numbers.
Answer:
Prime factors of 1729 are 7 x 13 x 19.
The difference of two consecutive prime factors is 6.
Question 7. The product of three consecutive numbers is always divisible by 6. Verify this statement with the help of some examples.
Answer: Among the three consecutive numbers, there must be one even number and one multiple of 3. Thus, the product must be multiple of 6.
Example:
(i) 2 x 3 x 4 = 24
(ii) 4 x 5 x 6 = 120
Question 8. The sum of two consecutive odd numbers is always divisible by 4. Verify this statement with the help of some examples.
Answer: 3 + 5 = 8 and 8 is divisible by 4.
5 + 7 = 12 and 12 is divisible by 4.
7 + 9 = 16 and 16 is divisible by 4.
9 + 11 = 20 and 20 is divisible by 4.
Question 9. In which of the following expressions, prime factorization has been done:
(a) 24 = 2 x 3 x 4
(b) 56 = 7 x 2 x 2 x 2
(c) 70 = 2 x 5 x 7
(d) 54 = 2 x 3 x 9
Answer: In expressions (b) and (c), prime factorization has been done.
Question 10. Determine if 25110 is divisible by 45.
[Hint: 5 and 9 are co-prime numbers. Test the divisibility of the number by 5 and 9.]
Answer: The prime factorization of 45 = 5 x 9
25110 is divisible by 5 as ‘0’ is at its unit place.
25110 is divisible by 9 as sum of digits is divisible by 9.
Therefore, the number must be divisible by 5 x 9 = 45
Question 11. 18 is divisible by both 2 and 3. It is also divisible by 2 x 3 = 6. Similarly, a number is divisible by 4 and 6. Can we say that the number must be divisible by 4 x 6 = 24? If not, give an example to justify your answer.
Answer: No. Number 12 is divisible by both 6 and 4 but 12 is not divisible by 24.
Question 12. I am the smallest number, having four different prime factors. Can you find me?
Answer: 2 x 3 x 5 x 7 = 210
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# Statistics – Regression Intercept Confidence Interval
Regression Intercept Confidence Interval, is a way to determine closeness of two factors and is used to check the reliability of estimation.
## Formula
R=β0±t(1−α2,n−k−1)×SEβ0R=β0±t(1−α2,n−k−1)×SEβ0
Where −
· β0β0 = Regression intercept.
· kk = Number of Predictors.
· nn = sample size.
· SEβ0SEβ0 = Standard Error.
· αα = Percentage of Confidence Interval.
· tt = t-value.
### Example
Problem Statement:
Compute the Regression Intercept Confidence Interval of following data. Total number of predictors (k) are 1, regression intercept β0β0 as 5, sample size (n) as 10 and standard error SEβ0SEβ0 as 0.15.
Solution:
Let us consider the case of 99% Confidence Interval.
Step 1: Compute t-value where α=0.99α=0.99.
=t(1−α2,n−k−1)=t(1−0.992,10−1−1)=t(0.005,8)=3.3554=t(1−α2,n−k−1)=t(1−0.992,10−1−1)=t(0.005,8)=3.3554
Step 2: ≥≥Regression intercept:
=β0+t(1−α2,n−k−1)×SEβ0=5−(3.3554×0.15)=5−0.50331=4.49669=β0+t(1−α2,n−k−1)×SEβ0=5−(3.3554×0.15)=5−0.50331=4.49669
Step 3: ≤≤Regression intercept:
=β0−t(1−α2,n−k−1)×SEβ0=5+(3.3554×0.15)=5+0.50331=5.50331=β0−t(1−α2,n−k−1)×SEβ0=5+(3.3554×0.15)=5+0.50331=5.50331
As a result, Regression Intercept Confidence Interval is 4.496694.49669 or 5.503315.50331for 99% Confidence Interval.
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# Go Math Grade 3 Answer Key Chapter 4 Multiplication Facts and Strategies Assessment Test
This chapter can improve student’s math skills, by referring to the Go Math Grade 3 Answer Key Chapter 4 Multiplication Facts and Strategies Assessment Test, and with the help of this Go Math Grade 3 Assessment Test Answer Key, students can score good marks in the examination.
Go Math Grade 3 Answer Key Chapter 4 contains all the topics of chapter 4 which helps to test the student’s knowledge. Through this assessment test, students can check their knowledge. This assessment test is also helpful for the teachers to know how much a student understood the topics.
Chapter 4: Multiplication Facts and Strategies Assessment Test
### Test – Page 1 – Page No. 41
Question 1.
Alberto packed 8 apples in each of 4 boxes. How many apples did Alberto pack?
Draw circles to model the problem. Then solve.
_____ apples
Answer: 8 × 4 = 32
Explanation:
Given that each box contains 8 apples
Total no of apples in 4 boxes = 8 × 4 = 32 apples.
Question 2.
For numbers 2a–2d, select True or False for each multiplication sentence.
a. 3 × 8 = 24
i. True
ii. False
Explanation: multiplication of 3 × 8 is 24.
Question 2.
b. 5 × 8 = 48
i. True
ii. False
Explanation: Multiplication of 5 × 8 is 40 not 48.
Question 2.
c. 7 × 8 = 56
i. True
ii. False
Explanation: Multiplication of 7 × 8 = 56.
Question 2.
d. 9 × 8 = 81
i. True
ii. False
Explanation: Multiplication of 9 × 8 is 72 not 81.
Question 3.
Peggy is putting flowers in vases. She puts either 2 or 3 flowers in each vase. If Peggy has a total of 12 flowers, how many different ways can she place them all in the vases?
Write multiplication sentences to show your work.
______ different ways
Explanation: 2 × 6 = 12; She can 2 flowers in 6 vases each or 4 × 3 = 12; she can put 3 flowers in 4 vases each or (2 × 3) + (3 × 2) = 12, she can put 2 flowers in 3 vases each and 3 flowers in 2 vases.
Question 4.
Dean plants 7 corn plants in each of 5 rows. How many corn plants does Dean plant?
______ corn plants
Answer: 7 × 5 = 35 plants
Explanation:
Given that Dean has planted 7 corn plants in each row.
The total no of rows = 5, therefore total no of plants was planted = 7 × 5 = 35.
### Test – Page 2 – Page No. 42
Question 5.
Circle groups to show 4 × (2 × 2).
Question 6.
Rebecca keeps all of her pairs of gloves in a drawer. Select the number of gloves that Rebecca could have in the drawer. Mark all that apply.
Options:
a. 5
b. 4
c. 6
d. 11
e. 12
Answer: Options b, c, and e
Explanation: Given that Rebecca kept all of her pairs of gloves in a drawer. Since it was mentioned pairs and pairs can be only in even number. Possible answers could be 4, 6, and 12.
Question 7.
Hal completed the table to describe the product of a mystery one-digit factor and each number.
Part A
Give all of the possible numbers that could be Hal’s mystery one-digit factor.
Answer: 1, 3, 5, 7, 9.
Question 7.
Part B
Explain how you know that you have selected all of the correct possibilities.
Answer: The products alternate between even and odd, the mystery factor must be an odd number. We have selected all of the odd one-digit numbers.
### Test – Page 3 – Page No. 43
Question 8.
Yuri used toothpicks to make 6 separate octagons. An octagon has 8 sides. How many toothpicks did Yuri use?
______ toothpicks
Answer: 6 * 8 = 48 toothpicks.
Explanation: One octagon has 8 sides, to make one octagon Yuri need 8 toothpicks
Therefore, to make 6 octagons she needs 6*8 = 48 toothpicks.
Question 9.
Maria practiced soccer 5 days last week. She practiced 2 hours each day. How many hours did Maria practice soccer last week?
______ hours
Answer: 5 × 2 = 10 hrs
Explanation:
Given that Maria practiced for 5 days, each day she practiced for 2 hrs.
Total no of hrs she practiced in 5 days = 5 × 2 = 10 hrs.
Question 10.
Break apart the array to show 5 × 7 = (5 × 2) + (5 × 5).
Question 11.
Circle the symbol that makes the multiplication sentence true.
9 × 5 (9 × 4) × 1
Answer: 9 × 5 > (9 × 4) × 1
Explanation: 9 × 5 > (9 × 4) × 1, Since 45 > (36) +1 => 45 > 37.
9 × 5 (9 × 4) × 1
### Test – Page 4 – Page No. 44
Question 12.
Lori has 18 new stamps to add to her collection. She displays the stamps on pages of an album in groups of either 3, 6, or 9 stamps. How many different ways can she display the 18 new stamps?
______ different ways
Explanation:
Set 1: 3 of 6, placing 3 stamps on 6 pages each
Set 2: 6 of 3, placing 6 stamps on 3 pages each
Set 3: 2 of 9, placing 2 stamps in 9 pages each
Set 4: 1 of 3, 1 of 6, and 1 of 9, placing 1 stamp in 3 pages and 1 in 6 pages, and 1 in 9 pages each. Which in total 18 stamps
Set 5: 2 of 6 and 2 of 3, placing 2 stamps in 6 pages and 2 stamps in 3 pages each, in total 18 stamps
Set 6: 1 of 9 and 3 of 3, placing 1 stamp in 9 pages and 3 stamps in 3 pages each in total 18 stamps.
Set 7: 1 of 6 and 4 of 3, placing 1 stamp in 6 and 4 stamps in 3 pages each, in total 18 stamps.
Question 13.
A shop owner sells 3-wheel baby strollers. She checks the air in the tires on 4 different strollers. How many tires does she check in all?
Use the array to explain how you know your answer is correct.
Answer: 4 * 3 = 12 tires
Explanation:
Here in the array, we can group 3 rows of 4,
3 × 4 = 12. Then count by threes to check.
Question 14.
Max arranges all of his toy cars in 9 equal rows, with 9 cars in each row. How many toy cars does Max have?
______ cars
Answer: 9 * 9 = 81 cars
Explanation:
Given that Max arranged all of his toy cars in 9 equal rows
Each row has 9 cars, therefore total no of cars = 9 * 9 = 81.
Question 15.
Deanna, Amy, and Pam pick the same number of peaches at an orchard. They each set their peaches in 4 equal piles with 6 peaches in each pile.
Write a multiplication sentence that shows how many peaches they picked.
Answer: 3 × (4 × 6) = 72
Explanation: Given that 3 persons (Deanna, Amy, and Pam) pick the same number of peaches at an orchard
Each set 4 equal piles with 6 peaches in each pile, total no of peaches per person = 4 × 6
No of peaches from all three = 3 × ( 4 × 6 ) = 72.
### Test – Page 5 – Page No. 45
Question 16.
Kate is baking 5 apple pies for the bake sale. She uses 3 red apples and 2 green apples in each pie. How many apples does Kate use? Show your work.
Explanation:
No of apple pies baked by kate = 5
In each pie she used 3 red apples and 2 green apples
So total no of apples used in a pie = 3+2 = 5
Therefore total no of apples used to make 5 apple pies = 5 ×(3+2) = 25.
Question 17.
For numbers 17a–17d, select True or False for each equation.
a. 2 × 7 = 16
i. True
ii. False
Explanation: Multiplication of 2 × 7 = 14 not 16.
Question 17.
b. 4 × 7 = 21
i. True
ii. False
Explanation: Multiplication of 4 × 7 = 28 not 21.
Question 17.
c. 6 × 7 = 42
i. True
ii. False
Explanation: Multiplication of 6 × 7 is 42.
Question 17.
d. 7 × 7 = 49
i. True
ii. False
Explanation: Multiplication of 7 × 7 is 49.
Question 18.
Circle the number that makes the multiplication sentence true.
10 × = 70
______
Explanation:
10 × = 70.
Question 19.
For numbers 19a–19d, select Yes or No to indicate whether the sum or product is equal to 9 × 4.
a. (5 × 4) + (4 × 4)
i. yes
ii. no
Explanation: 9 × 4 is 36 and (5 × 4) + (4 × 4) = (20) + (16) = 36.
Question 19.
b. 5 + (4 × 5)
i. yes
ii. no
Explanation: 9 × 4 is 36 and 5 + (4 × 5) = 25.
Question 19.
c. (3 × 3) + (2 × 2)
i. yes
ii. no
Explanation: 9 × 4 is 36 and (3 × 3) + (2 × 2) = 9 + 4 = 13.
Question 19.
d. 4 × (5 + 4)
i. yes
ii. no
Explanation: 9 × 4 is 36 and 4 × (5 + 4) = 4 × 9 = 36.
### Test – Page 6 – Page No. 46
Question 20.
A rollercoaster car can fit 6 people. How many people can fit in a rollercoaster that is 9 cars long?
______ people
Answer: 9 × 6 = 54 people.
Explanation:
Give that a rollercoaster car can fit 6 people
Number of people can fit if it having 9 cars = 9 × 6 = 54.
Question 21.
Write a multiplication sentence using the following numbers and symbols.
Answer: 2 × (4 × 7) = 56.
Question 22.
Debbie started a table showing a multiplication pattern.
Part A
Complete the table. Describe a pattern you see in the products.
Answer: In the given pattern it is adding 6 to the value in the next cell.
Explanation:Â
Question 22.
Part B
If you multiplied 6 × 73, would the product be an even number or an odd number? Use the table to explain your reasoning.
Answer: 6 × 73 = 438.
Explanation: When any number is multiplied by 6 which is an even number the product is even, so the product of 6 × 73 = 438 is even. For example, if we take the table started by Debbie, we can see that both even and odd numbers are multiplied by 6 which even and the final result of all the products are even.
Question 23.
Use the number line to show the product of 3 × 8.
3 × 8 = ______
Answer: 3 × 8 = 24.
Explanation:
From the number line 3 × 8 which is 8 + 8 + 8 = 24.
Conclusion:
This assessment test helps students to check their math skills. Go Math Grade 3 Chapter 10 questions are explained in detail that students can understand easily.
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# How to multiply using the Urdhva Tiryagbyham technique?
Urdhva Tiryagbyham is a shortcut method/ Sutra 3 for the multiplication of all types of numbers in Vedic Mathematics. The English translation for the same is Vertically and Cross-wise. For numbers up to 3 digits, we can easily use the normal process of this method. Yet for numbers having more than 3 digits, the Vinculum process is used to convert the negative numbers into normal numbers.
Let us start by learning one method at a time.
## Shortcut method for multiplication of 2 numbers.
We will follow this procedure from the left to the right.
Let us take two numbers- 22 and 24
The steps:
1. First, we will multiply the first digits of both numbers. Therefore we will multiply 2*2 which will give us 4. Write it separately.
2. Then cross-multiply both digits of both numbers. That is we will multiply 2*4 which will give us 8 and 2*2 which will give us 4.
4. The last step is to multiply the last digits of both the numbers i.e 2*4 which gives us 8.
5. For our final answer, we will place the digits according to their place value i.e 8 in the unit’s place. Also as we have a double-digit number in our tens place i.e 12 we will carry forward the 1 and add it to the number which is placed in the hundreds place (1 will be carry forwarded and added to 4 which we had written separately in Step 1.-1+4=5(in the hundreds place)
6. Therefore, our final answer is 528.
Let us try with another example:
Let us take two numbers- 43 and 54
The steps:
1. First, we will multiply the first digits of both numbers. Therefore we will multiply 4*5 which will give us 20. Write it separately.
2. Then cross-multiply both digits of both numbers. That is, we will multiply 4*4 which will give us 16, and 3*5 which will give us 15.
4. The last step is to multiply the last digits of both the numbers i.e 3*4 which is equal to 12.
5. For our final answer, we will place the digits according to their place value i.e 2(from 3*4=12) in the unit’s place. For the ten’s place, we already have the number 31 but 1 from the above 12 will be carry forwarded and added to 31 and therefore we have 32.
Now from 32, we will leave 2 in the tens place and 3 will be carry forwarded and added to the 20 that we wrote separately in the 1st step giving us 23.
• Therefore, finally appending all the digits our final answer is 2322.
## Shortcut method for multiplication of 3-digit numbers.
We will follow this procedure from the left to the right
• Let us take 2 numbers – 157 and 225
• Here the cross multiplications are a little tricky therefore make sure to concentrate.
• First, we will simply multiply the 1st digits of both the numbers i.e 1*2 which gives us 2.
• Then we will cross multiply the 1st two digits of both numbers and add their answers. Therefore, (1*2) +(2*5) = 12 is our answers. Make sure to write these numbers separately so that you don’t mix them up while finding out the final answer.
• Then, we will cross multiply the first and last digits of both numbers and vertically multiply the center digits of the same and add all three of them together. Therefore, <(1*5)+(5*2)+(7*2)>=29 is our answer for this equation.
• Again, we will cross multiply the 2nd and 3rd digits of the numbers. (5*5)+(7*2)>= 39
• For the final equation, we will have to multiply the last digits of the numbers vertically. Therefore, our last equation summarizes to (7*5)= 35.
• We have answers from all the equations. Lets’ write them down – 2,12,29,39,35.
• To solve this we will use the carry forward method.
• Doing this, our final answer will come out to be as 35325.
## Urdhva Tiryagbyham – One more Example
We will follow this procedure from the left to the right
• Let us take 2 numbers – 777 and 483
• Here the cross multiplications are a little tricky therefore make sure to concentrate.
• First, we will simply multiply the 1st digits of both the numbers i.e (7*4) which gives us 28.
• Then we will cross multiply the 1st two digits of both numbers and add their answers. Therefore, (7*8) +(7*4) = 84 is our answer. Make sure to write these numbers separately so that you don’t mix them up while finding out the final answer.
• Then, we will cross multiply the first and last digits of both numbers and vertically multiply the center digits of the same and add all three of them together. Therefore, (7*3)+(7*8)+(7*4)=105 is our answer for this equation.
• Again, we will cross multiply the 2nd and 3rd digits of the numbers. (7*3)+(7*8)= 77
• For the final equation, we will have to multiply the last digits of the numbers vertically. Therefore, our last equation summarizes to (7*3)= 21.
• We have answers from all the equations. Lets’ write them down – 28,84,105,77,21
• To solve this we will use the carry forward method.
• Doing this, our final answer will come out to be as 375291.
• Shortcut method for multiplication of 4-digit numbers.
We will follow this procedure from the left to the right
• Let us take two numbers – 2345 and 3456
• We will start with vertical multiplication of 1st two digits of the numbers(2*3)
• Then we will crosswise multiply and add 1st 2 digits of both numbers <(3*4)+(3*3)>
• Crosswise multiply and add 1st 3 digits of both numbers <(2*5)+(3*4)+(4*3)>
• Crosswise multiply and add all the digits <(2*6)+(3*5)+(4*4)+5*3)>
• Crosswise multiply and add the last 3 digits <(3*6)+(4*5)+(5*4)>
• Crosswise multiply and add the last 2 digits<(4*6)+(5*5)>
• Vertical multiplication of last digits (5*6).
The final answer for the above equations using the carry forward method will be 8104320
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# Partial Fractions
A LevelAQAEdexcelOCR
## Partial Fractions
Algebraic fractions that have a denominator with more than one linear factor can be split, or decomposed, into partial fractions. This can be particularly useful when integrating complex expressions or with binomial expansions.
Questions involving partial fractions will only ever involve a numerator that is a constant or a linear term and the denominator will only ever include up to three terms with the most complex being squared linear terms.
A LevelAQAEdexcelOCR
## Types of Partial Fraction
There are 3 types of partial fraction that you will see in A level Maths:
1. Considering a fraction where the denominator has two linear terms, we can split up the fraction into two separate fractions,
$\dfrac{5}{\textcolor{limegreen}{(x+1)} \textcolor{blue}{(x+2)}}$ can be split up into $\dfrac{A}{\textcolor{limegreen}{(x+1)}}+\dfrac{B}{\textcolor{blue}{(x+2)}}$
2. Considering a fraction where the denominator has three linear terms, we can split up the fraction into three separate fractions,
$\dfrac{3x+4}{\textcolor{limegreen}{(x+1)}\textcolor{blue}{(x+2)}\textcolor{orange}{(x-1)}}$ can be split up into $\dfrac{A}{\textcolor{limegreen}{(x+1)}}+\dfrac{B}{\textcolor{blue}{(x+2)}} +\dfrac{C}{\textcolor{orange}{(x-1)}}$
3. Considering a fraction where the denominator has one linear term and a repeated term, we can split up the fraction into three separate fractions,
$\dfrac{2x+1}{{\textcolor{limegreen}{(x+1)}}^2 \textcolor{blue}{(x+2)}}$ can be split up into $\dfrac{A}{{\textcolor{limegreen}{(x+1)}}^2}+\dfrac{B}{\textcolor{limegreen}{(x+1)}} +\dfrac{C}{\textcolor{blue}{(x+2)}}$
A LevelAQAEdexcelOCR
## Expressing Fractions as Partial Fractions
To find the coefficients in the numerators in the partial fractions, $A$, $B$ and $C$, there are 2 different methods you can use: Substitution or Equating Coefficients.
Example: Express $\dfrac{5x+1}{(x-1)(x+1)(x+2)}$ as partial fractions.
Step 1: Write the fraction as partial fractions with unknown constants, and put it over a common denominator:
\begin{aligned} \dfrac{5x+1}{(x-1)(x+1)(x+2)} &\equiv \dfrac{A}{(x-1)} + \dfrac{B}{(x+1)} + \dfrac{C}{(x+2)} \\[1.2em] &\equiv \dfrac{A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)}{(x-1)(x+1)(x+2)} \end{aligned}
Step 2: Cancel down the denominators:
$5x+1 \equiv A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)$
Step 3: Use the Substitution or Equating Coefficients to find the values of the unknown constants, $A$, $B$ and $C$:
Substitution:
Substitute values of $x$ inside the brackets to make them $0$, so that you are left with only one of $A$, $B$ and $C$:
Substitute in $x = 1$:
\begin{aligned} 5(1) + 1 &= A(1+1)(1+2) \\ 6 &= 6A \\ \textcolor{red}{A} &\textcolor{red}{=} \textcolor{red}{1} \end{aligned}
Substitute in $x = -1$:
\begin{aligned} 5(-1) + 1 &= B(-1-1)(-1+2) \\ -4 &= -2B \\ \textcolor{blue}{B} &\textcolor{blue}{=} \textcolor{blue}{2} \end{aligned}
Substitute in $x = -2$:
\begin{aligned} 5(-2) + 1 &= C(-2-1)(-2+1) \\ -9 &= 3C \\ \textcolor{limegreen}{C} &\textcolor{limegreen}{=} \textcolor{limegreen}{-3} \end{aligned}
Equating Coefficients:
Equate coefficients and then solve them simultaneously to find the values of the unknown constants, $A$, $B$ and $C$:
$5x+1 \equiv A(x+1)(x+2) + B(x-1)(x+2) + C(x-1)(x+1)$
$\equiv A(x^2 + 3x + 2) + B(x^2 + x - 2) + C(x^2 - 1)$
$\equiv (A + B + C)x^2 + (3A + B)x + (2A - 2B - C)$
Equating $x^2$ coefficients gives:
$0 = A + B + C$
Equating $x$ coefficients gives:
$5 = 3A + B$
Equating constant terms gives:
$1 = 2A - 2B - C$
Then, solve these simultaneously, which gives $\textcolor{red}{A = 1}$, $\textcolor{blue}{B = 2}$ and $\textcolor{limegreen}{C = -3}$
Step 4: Replace the constant terms, $A$, $B$ and $C$ in the original identity:
$\dfrac{5x+1}{(x-1)(x+1)(x+2)} \equiv \dfrac{\textcolor{red}{1}}{(x-1)} + \dfrac{\textcolor{blue}{2}}{(x+1)} \textcolor{limegreen}{-} \dfrac{\textcolor{limegreen}{3}}{(x+2)}$
Note: The unknown constants may not always be this nice – you may get questions with constant terms that are fractions.
A LevelAQAEdexcelOCR
## Note:
• You may need to combine the methods of Substitution and Equating Coefficients for some examples, such as fractions with repeated terms.
• You may be given a fraction that is not split up into brackets in the denominator, so you will need to factorise it. Also, look out for the difference of two squares.
A LevelAQAEdexcelOCR
## Example: Repeated Terms
Express $\dfrac{2x-3}{x(x-2)^2}$ as partial fractions.
[3 marks]
Write the fraction as partial fractions with unknown constants, and put it over a common denominator:
\begin{aligned} \dfrac{2x-3}{x(x-2)^2} &\equiv \dfrac{A}{(x-2)^2} + \dfrac{B}{x-2} + \dfrac{C}{x} \\[1.1em] &\equiv \dfrac{Ax + Bx(x-2) + C(x-2)^2}{x(x-2)^2} \end{aligned}
Then, cancel the denominators:
$2x-3 \equiv Ax + Bx(x-2) + C(x-2)^2$
Substitute in $x=0$:
\begin{aligned} -3 &= C(-2)^2 \\ C &= - \dfrac{3}{4} \end{aligned}
Substitute in $x=2$:
\begin{aligned} 2(2) - 3 &= 2A \\ A &= \dfrac{1}{2} \end{aligned}
Now, equate coefficients of $x^2$:
\begin{aligned} 0 &= B + C \\ B &= \dfrac{3}{4} \end{aligned}
Finally, replace the values of $A$, $B$ and $C$ into the original identity:
$\dfrac{2x-3}{x(x-2)^2} \equiv \dfrac{1}{2(x-2)^2} + \dfrac{3}{4(x-2)} - \dfrac{3}{4x}$
A LevelAQAEdexcelOCR
## Partial Fractions Example Questions
Question 1: Express $\dfrac{8}{9x^2 - 16}$ as partial fractions.
[3 marks]
A Level AQAEdexcelOCR
$\dfrac{8}{9x^2 - 16} = \dfrac{8}{(3x+4)(3x-4)}$
\begin{aligned} \dfrac{8}{(3x+4)(3x-4)} &\equiv \dfrac{A}{(3x+4)} + \dfrac{B}{(3x-4)} \\[1.1em] &\equiv \dfrac{A(3x-4) + B(3x+4)}{(3x+4)(3x-4)} \end{aligned}
$8 \equiv A(3x-4) + B(3x+4)$
Substitute in $x=\dfrac{4}{3}$:
\begin{aligned} 8 &= 8B \\ B &= 1 \end{aligned}
Substitute in $x=- \dfrac{4}{3}$:
\begin{aligned} 8 &= -8A \\ A &= -1 \end{aligned}
Then, replace the values of $A$ and $B$ into the identity:
$\dfrac{8}{9x^2 - 16} \equiv - \dfrac{1}{(3x+4)} + \dfrac{1}{(3x-4)}$
Gold Standard Education
Question 2: Express $\dfrac{x}{(x-1)(x-2)(x-3)}$ as partial fractions.
[3 marks]
A Level AQAEdexcelOCR
$\dfrac{x}{(x-1)(x-2)(x-3)} \equiv \dfrac{A}{(x-1)} + \dfrac{B}{(x-2)} + \dfrac{C}{(x-3)}$
$\equiv \dfrac{A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)}{(x-1)(x-2)(x-3)}$
$x = A(x-2)(x-3) + B(x-1)(x-3) + C(x-1)(x-2)$
Substitute in $x = 1$:
\begin{aligned} 1 &= A(1-2)(1-3) \\ A &= \dfrac{1}{2} \end{aligned}
Substitute in $x = 2$:
\begin{aligned} 2 &= B(2-1)(2-3) \\ B &= -2 \end{aligned}
Substitute in $x = 3$:
\begin{aligned} 3 &= C(3-1)(3-2) \\ C &= \dfrac{3}{2} \end{aligned}
Then, replace the values of $A$, $B$ and $C$ into the identity:
$\dfrac{x}{(x-1)(x-2)(x-3)} \equiv \dfrac{1}{2(x-1)} - \dfrac{2}{(x-2)} + \dfrac{3}{2(x-3)}$
Gold Standard Education
Question 3: Express $\dfrac{10}{x^2(x+1)}$ as partial fractions.
[3 marks]
A Level AQAEdexcelOCR
\begin{aligned} \dfrac{10}{x^2(x+1)} &\equiv \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{(x+1)} \\[1.1em] &\equiv \dfrac{A(x+1) + Bx(x+1) + Cx^2}{x^2(x+1)} \end{aligned}
$10 \equiv A(x+1) + Bx(x+1) + Cx^2$
Substitute in $x = 0$:
\begin{aligned} 10 &= A(0+1) \\ A &= 10 \end{aligned}
Substitute in $x = -1$:
\begin{aligned} 10 &= C(-1)^2 \\ C &= 10 \end{aligned}
Equate coefficients of $x^2$:
\begin{aligned} 0 &= B + C \\ B &= -10 \end{aligned}
Then, replace the values of $A$, $B$ and $C$ into the identity:
$\dfrac{10}{x^2(x+1)} \equiv \dfrac{10}{x^2} - \dfrac{10}{x} + \dfrac{10}{(x+1)}$
|
# How do you graph h(x)=ln(x+1)?
Jan 2, 2017
$h \left(x\right)$ is the standard function $\ln \left(x\right)$ shifted (transformed) one unit left (negative) on the $x$-axis
#### Explanation:
$h \left(x\right) = \ln \left(x + 1\right)$
$\ln \left(x\right)$ is defined for $x > 0 \to h \left(x\right)$ is defined for $x + 1 > 0$
$\therefore h \left(x\right)$ is defined for $x > - 1$
$\ln \left(1\right) = 0 \to h \left(x\right) = 0$ for $x + 1 = 1$
$\therefore h \left(x\right) = 0$ for $x = 0$
$h \left(x\right)$ is the standard function $\ln \left(x\right)$ shifted (transformed) one unit left (negative) on the $x$-axis
The graph of $h \left(x\right)$ is shown below:
graph{ln(x+1) [-10, 10, -5, 5]}
|
Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Document related concepts
Transcript
```Order of Operations rules for arithmetic and algebra
that describe what sequence to
involving more than one operation
The Rules
Step 1: Do operations inside grouping symbols such
as parentheses (), brackets [], and braces {},
and operations separated by fraction bars.
Parentheses within parentheses are called
nested parentheses (( )).
Step 2: Evaluate Powers (exponents) or roots.
Step 3: Perform multiplication or division in order
by reading the problem from left to right.
Step 4: Perform addition or subtraction in order by
reading the problem from left to right.
Order of Operations - WHY?
Imagine if two different people wanted to evaluate the
same expression two different ways...
#1 does each step left to right:
21 6 3 5
21 6 3 5
27 3 5
27 3 5
9 5
45
#2 uses the order of operations
The rules for order of
operations exist so
that everyone can
perform the same
consistent operations
and achieve the same
results. Method 2 is
the correct method.
21 6 3 5
21 6 3 5
21 2 5
21 2 5
21 10
31
Order of Operations - WHY?
• Can you imagine what it would be like if
calculations were performed differently by
various financial institutions?
• What if doctors prescribed different doses
of medicine using the same formulas but
achieving different results?
Order of Operations: Example 1
Evaluate without grouping symbols
54 6 18 2
54 6 18 2
Divide.
9 18 2
This expression has no parentheses and no
exponents.
• First solve any multiplication or
division parts left to right.
• Then solve any addition or subtraction
parts left to right.
Multiply.
9 36
45
The order of operations must be followed
each time you rewrite the expression.
Order of Operations: Example 2
Expressions with powers
25 6
• Firs,t solve exponents
(powers).
25 6
• Second, solve multiplication
or division parts left to right.
2
2
Exponents (powers)
2 25 6
Multiply.
• Then, solve any addition or
subtraction parts left to right.
50 6
Subtract.
44
The order of operations must be followed
each time you rewrite the expression.
Order of Operations: Example 3
Evaluate with grouping symbols
3 4 8 2
• First, solve parts inside grouping
symbols according to the order
of operations.
2
3 42 8 2
3 42 6
3 16 6
48 6
8
Grouping
symbols
• Solve any exponent (Powers).
Subtract.
Exponents (powers)
Multiply.
Divide.
• Then, solve multiplication or
division parts left to right.
• Then solve any addition or
subtraction parts left to right.
The order of operations must be followed
each time you rewrite the expression.
Order of Operations: Example 4
Expressions with fraction bars
Work above the
fraction bar.
3 4
2 (18 4)
2
Work below the
fraction bar.
Exponents (powers)
3 4 2
2 (18 4)
Grouping symbols
Multiply.
316
2 (14)
48
16
Simplify
: Divide.
4816 3
Order of Operations: Example 5
Evaluate variable expressions
( x y 5) n
Evaluate when x=2, y=3, and n=4:
1) Substitute in the values for the variables
3
Inside
grouping
symbols:
6
Exponents (powers)
(2 33 5) 42 6
(2 27 5) 42 6
Subtract.
Continue
with the
rest:
2
(29 5) 42 6
Exponents (powers)
24 4 2 6
Subtract.
24 16 6
|
# The Graph y = mx + c has Slope m
Video Solutions to help Grade 6 students learn the slope of a line joining any two distinct points of the graph of y = mx + c has slope, m
## New York State Common Core Math Module 4, Grade 8, Lesson 17
### Lesson 17 Student Outcomes
• Students show that the slope of a line joining any two distinct points of the graph of y = mx + c has slope, m.
• Students transform the standard form of an equation into y = - a/b x + c/b.
### Lesson 17 Summary
The line joining two distinct points of the graph of the linear equation y = mx + b has slope m. The m of y = mx + b is the number that describes the slope. For example, in the equation y = -2x + 4 the slope of the graph of the line is -2.
### Lesson 17 Classwork
Exercises
1. Find at least three solutions to the equation y = 2x, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.
2. Find at least three solutions to the equation y = 3x - 1, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.
3. Find at least three solutions to the equation y = 3x + 1, and graph the solutions as points on the coordinate plane. Connect the points to make a line. Find the slope of the line.
4. The graph of the equation y = 7x - 3 has what slope?
5. The graph of the equation y = -3/4 x - 3 has what slope?
6. You have \$20 in savings at the bank. Each week, you add \$2 to your savings. Let y represent the total amount of money you have saved at the end of weeks. Write an equation to represent this situation and identify the slope of the equation. What does that number represent?
7. A friend is training for a marathon. She can run 4 miles in 28 minutes. Assume she runs at a constant rate. Write an equation to represent the total distance, y, your friend can run in minutes. Identify the slope of the equation. What does that number represent?
8. Four boxes of pencils cost \$5. Write an equation that represents the total cost, y, for x boxes of pencils. What is the slope of the equation? What does that number represent?
9. Solve the following equation for y: 9x - 3y = 15, then identify the slope of the line.
10. Solve the following equation for y: 5x + 9y = 6, then identify the slope of the line.
11. Solve the following equation for y: ax + by = c, then identify the slope of the line.
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Review question
# Can we solve these equations for $a, b, x$ and $y$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource
Ref: R6871
## Solution
Find all values of $a$, $b$, $x$ and $y$ that satisfy \begin{align*} a+b &= 1 \\ ax+by &= \frac{1}{3} \\ ax^2+by^2 &= \frac{1}{5} \\ ax^3+by^3 &= \frac{1}{7}. \end{align*}
[Hint: you may wish to start by multiplying the second equation by $x+y$.]
Let’s do what the hint suggests, and multiply the second equation by $x+y$, which yields $(ax+by)(x+y) = \frac{1}{3}(x+y),$ that is, $ax^2+by^2+(a+b)xy = \frac{1}{3}(x+y).$
Substituting in the information from the first and third equations yields $\frac{1}{5}+xy=\frac{1}{3}(x+y),$ or, rearranging, $x+y-3xy-\frac{3}{5}=0.$
Now let’s try the same trick with the third equation; we have $(ax^2+by^2)(x+y) = \frac{1}{5}(x+y),$ that is, $ax^3+by^3+xy(ax+by) = \frac{1}{5}(x+y).$
Substituting in the information from the second and fourth equations, we find that $\frac{1}{7}+\frac{xy}{3}=\frac{1}{5}(x+y),$ or, rearranging, $x+y-\frac{5}{3}xy-\frac{5}{7}=0.$
Now let $xy = z$. This gives us
$$$x+y-3z-\frac{3}{5}=0, \label{eq:1}$$$ and $$$x+y-\frac{5}{3}z-\frac{5}{7}=0. \label{eq:2}$$$
so subtracting these gives us $-\dfrac{4z}{3}+ \dfrac{4}{35}=0,$ which means that $z = \dfrac{3}{35}$.
So $y = \dfrac{3}{35x}$, and thus from (1), $x+\dfrac{3}{35x} - 3\dfrac{3}{35}-\dfrac{3}{5}=0.$ This is the quadratic equation $35x^2-30x+3=0.$ Since $x$ and $y$ are interchangeable, we take the solution to be \begin{align*} x &= \frac{30+4\sqrt{30}}{70}=\frac{3}{7}+\frac{2\sqrt{30}}{35},\\ y &= \frac{30-4\sqrt{30}}{70}=\frac{3}{7}-\frac{2\sqrt{30}}{35}. \end{align*}
We now wish to find the values of $a$ and $b$ for these values of $x$ and $y$. The second simultaneous equation with the above values of $x$ and $y$ looks like $\frac{3a}{7}+\frac{2\sqrt{30}a}{35}+\frac{3b}{7}-\frac{2\sqrt{30}b}{35}=\frac{1}{3},$ and substituting in $b=1-a$ from the first of the simultaneous equations yields $\frac{4\sqrt{30}a}{35}+\frac{3}{7}-\frac{2\sqrt{30}}{35}=\frac{1}{3}.$ Rearranging, we find that $a=\frac{1}{2}-\frac{5}{6\sqrt{30}}=\frac{1}{2}-\frac{\sqrt{30}}{36},$ and therefore $b=1-a=1-\frac{1}{2}+\frac{\sqrt{30}}{36}=\frac{1}{2}+\frac{\sqrt{30}}{36}.$ Notice that if we’d chosen $x$ and $y$ to be the other way around then $a$ and $b$ would also have been swapped.
How do we know this solution works without substituting our values in to all 4 equations to check?
How would our answer change if we were given the additional constraint that $ax^4+by^4 = \dfrac{1}{9}$?
This GeoGebra file helps us to see what is happening.
We are plotting here the three curves $ax+(1-a)y = \dfrac{1}{3}, ax^2+(1-a)y^2 = \dfrac{1}{5}$ and $ax^3+(1-a)y^3 = \dfrac{1}{7}$.
As we vary $a$, can we get the three curves to all pass through the same point?
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# Difference between revisions of "2015 AMC 8 Problems/Problem 7"
Each of two boxes contains three chips numbered $1$, $2$, $3$. A chip is drawn randomly from each box and the numbers on the two chips are multiplied. What is the probability that their product is even?
$\textbf{(A) }\frac{1}{9}\qquad\textbf{(B) }\frac{2}{9}\qquad\textbf{(C) }\frac{4}{9}\qquad\textbf{(D) }\frac{1}{2}\qquad \textbf{(E) }\frac{5}{9}$
### Solution
We can instead calculate the probability that their product is odd, and subtract this from $1$. In order to get an odd product, we have to draw an odd number from each box. We have a $\frac{2}{3}$ probability of drawing an odd number from one box, so there is a $\left ( \frac{2}{3} \right )^2=\frac{4}{9}$ probability of having an odd product. Thus, there is a $1-\frac{4}{9}=\boxed{\textbf{(E)}~\frac{5}{9}}$ probability of having an even product.
### Solution 2
You can also make this problem into a spinner problem. You have the first spinner with $3$ equally divided
sections, $1, 2$ and $3.$ You make a second spinner that is identical to the first, with $3$ equal sections of
$1$,$2$, and $3$. If the first spinner lands on $1$, to be even, it must land on two. You write down the first
combination of numbers $(1,2)$. Next, if the spinner lands on $2$, it can land on any number on the second
spinner. We now have the combinations of $(1,2) ,(2,1), (2,2), (2,3)$. Finally, if the first spinner ends on $3$, we
have $(3,2).$ Since there are $3*3=9$ possible combinations, and we have $5$ evens, the final answer is
$\boxed{\textbf{(E) }\frac{5}{9}}$.
|
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# Find the slope of the line passing through the points A(2,3) and B(4,7).
Last updated date: 22nd Jul 2024
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Hint: The slope is also known as gradient which describes the direction and the steepness of the line. The slope of a line in the plane containing the X and Y axis is generally represented by the letter m, and is defined as the change in the Y coordinate divided by the corresponding change in the x coordinate, between two distinct points on the line. Mathematically, it is expressed as following,
\begin{align} &\text{we have two given points }({{x}_{1}},{{y}_{1}})\text{ and }({{x}_{2}},{{y}_{2}})\text{ then slope ''m'' is given by,} \\ &\text{m= }\tan \theta =\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\ \end{align}
Here, $\theta$ is the angle made by the line joining the two given points with the X axis in an anticlockwise direction.
In the given question we have ${{x}_{1}}=2,{{y}_{1}}=3\text{ and }{{x}_{2}}=4,{{y}_{2}}=7.$
\begin{align} & \text{we have two given points }({{x}_{1}},{{y}_{1}})\text{ and }({{x}_{2}},{{y}_{2}})\text{ then slope ''m'' is given by,} \\ & \text{m= }\tan \theta =\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}} \\ \end{align}
\begin{align} & \Rightarrow m=\dfrac{7-3}{4-2} \\ & \Rightarrow m=\dfrac{4}{2}=2 \\ \end{align}
We have two given points $({{x}_{1}},{{y}_{1}})\text{ and }({{x}_{2}},{{y}_{2}})$ then the slope “m” is given by,
$m=\tan \theta =\dfrac{{{y}_{2}}-{{y}_{1}}}{{{x}_{2}}-{{x}_{1}}}$
The concept of slope is very important in mathematics as well as in engineering as it applies directly to grades or gradients in geography and civil engineering. Through trigonometry, the slope of the line is related to its angle of incline $\theta$ by the tangent function which I have mentioned in the solution.
|
Question
# Solve the following equation:${x^2} + xy + xz = 18, \\ {y^2} + yz + yz + 12 = 0, \\ {z^2} + zx + zy = 30. \\ \\ {\mathbf{A}}.\,x = \pm 2,y = \mp 2,z = \pm 4 \\ {\mathbf{B}}.\,x = \pm 3,y = \mp 2,z = \pm 5 \\ {\mathbf{C}}.\,\,x = \pm 4,y = \pm 5,z = \pm 5 \\ {\mathbf{D}}.\,{\text{None of the above}} \\$
Hint: In order to solve this question, we have to solve the three equations given in the question. We can take something common then solve it by eliminating or dividing.
The given equations are,
${x^2} + xy + xz = 18$,
Taking $x$ as common from LHS we get,
$x(x + y + z) = 18\,\,\,\,\,\,...({\text{i}}) \\ \\ {y^2} + yz + yz + 12 = 0 \\$
Taking $y$ as common from LHS we get,
$y(x + y + z) = - 12\,\,\,\,\,\,\,\,\,...({\text{ii}})$
${z^2} + zx + zy = 30$
Taking $z$ as common from LHS we get,
$z(x + y + z) = 30\,\,\,\,\,\,\,\,\,...({\text{iii}})$
Dividing (i) by (ii) we get,
$\dfrac{x}{y} = \dfrac{{ - 18}}{{12}} = \dfrac{{ - 3}}{2} \\ y = - \dfrac{2}{3}x\;\;{\text{ }}\;.......\left( a \right) \\$
Dividing (i) by (iii) we get,
$\dfrac{x}{z} = \dfrac{{18}}{{30}} = \dfrac{3}{5}$
Then,
$z = \dfrac{5}{3}x\,\,\,\,\,\,......(b)$
Substituting $(a)$ and $(b)$ in (i) we get,
$x\left( {x - \dfrac{2}{3}x + \dfrac{5}{3}x} \right) = 18$
On solving above we get,
$2{x^2} = 18 \\ {x^2} = 9 \\$
Therefore either $x = 3{\text{ }}$ or $x = - 3$
That is $x = \pm 3$
On putting the value of $x$ in $(a)$ and $(b)$ we get,
$y = - \dfrac{2}{3}( \pm 3) = \mp 2 \\ \\ z = \dfrac{5}{3}( \pm 3) = \pm 5 \\$
Therefore,
$x = \pm 3,y = \mp 2,z = \pm 5$
Hence the correct option is B.
Note: In this question we have taken $x,y,z$ common from the equation ${\text{(i),(ii),(iii)}}$ then we got various equations which can be used to find the values of $x,y,z$ 5as done above . Therefore we can find the values of various variables . We can multiply, divide, add or subtract between these equations to get the desired results.
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# AP Statistics Curriculum 2007 Limits Poisson2Bin
(Difference between revisions)
Revision as of 01:33, 3 February 2008 (view source)IvoDinov (Talk | contribs)← Older edit Revision as of 01:34, 3 February 2008 (view source)IvoDinov (Talk | contribs) (→Examples)Newer edit → Line 9: Line 9: ===Examples=== ===Examples=== - The [[About_pages_for_SOCR_Distributions | Binomial distribution]] can be approximated well by Poisson when $n$ is large and $p$ is small with $np < 7$. This is true because + The [[About_pages_for_SOCR_Distributions | Binomial distribution]] can be approximated well by Poisson when $n$ is large and $p$ is small with $np < 10$, as stated above. This is true because $\lim_{n \rightarrow \infty} [itex] \lim_{n \rightarrow \infty} {n \choose x} p^x(1-p)^{n-x}=\frac{\lambda^x e^{-\lambda}}{x!}$, where {n \choose x} p^x(1-p)^{n-x}=\frac{\lambda^x e^{-\lambda}}{x!} [/itex], where Line 19: Line 19:
+ ===References=== ===References===
## General Advance-Placement (AP) Statistics Curriculum - Poisson as Approximation to Binomial Distribution
### Poisson as Approximation to Binomial Distribution
np < 10
$n \geq 20$ and $p \leq 0.05$.
### Examples
The Binomial distribution can be approximated well by Poisson when n is large and p is small with np < 10, as stated above. This is true because $\lim_{n \rightarrow \infty} {n \choose x} p^x(1-p)^{n-x}=\frac{\lambda^x e^{-\lambda}}{x!}$, where λ = np. Here is an example. Suppose $2\%$ of a certain population have Type AB blood. Suppose 60 people from this population are randomly selected. The number of people X among the 60 that have Type AB blood follows the Binomial distribution with n = 60,p = 0.02. The figure below represents the distribution of X. This figure also shows P(X = 0).
• Note: This distribution can be approximated well with Poisson with λ = np = 60(0.02) = 1.2. The figure below is approximately the same as the figure above (the width of the bars is not important here. The height of each bar represents the probability for each value of X which is about the same for both distributions).
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# How do you solve 10b^2 = 27b - 18 by factoring?
Aug 21, 2015
The solutions are:
color(blue)(b=6/5
color(blue)(b=3/2
#### Explanation:
$10 {b}^{2} - 27 b + 18 = 0$
We can Split the Middle Term of this expression to factorise it and thereby find solutions.
In this technique we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 10 \cdot 18 = 180$
and
${N}_{1} + {N}_{2} = b = - 27$
After trying out a few numbers we get ${N}_{1} = - 12$ and ${N}_{2} = - 15$
$\left(- 12\right) \cdot \left(- 15\right) = 180$, and $\left(- 12\right) + \left(- 15\right) = - 27$
$10 {b}^{2} - 27 b + 18 = 10 {b}^{2} - 15 b - 12 b + 18$
$10 {b}^{2} - 15 b - 12 b + 18 = 0$
$5 b \left(2 b - 3\right) - 6 \left(2 b - 3\right) = 0$
$\left(5 b - 6\right) \left(2 b - 3\right) = 0$ is the factorised form of the expression.
Now we equate the factors to zero.
5b-6=0, color(blue)(b=6/5
2b-3=0, color(blue)(b=3/2
|
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# Horizontal line test
The horizontal line test is a method used in mathematics to determine if a function is one-to-one by checking if any horizontal line intersects the graph more than once. If a horizontal line crosses the graph at more than one point, the function is not bijective, meaning it fails the test. This test is crucial for understanding the invertibility of functions, enhancing your grasp of advanced algebra and calculus.
#### Create learning materials about Horizontal line test with our free learning app!
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## Horizontal Line Test – Definition
The Horizontal Line Test is a method used in mathematics to determine if a function is one-to-one. This test is particularly useful in understanding whether each input value of a function maps uniquely to one output value.
### What is the Horizontal Line Test?
The Horizontal Line Test helps you identify if a function is injective, meaning that every element of the function's codomain is mapped by at most one element of its domain. This is important for understanding if the function has an inverse that is also a function.When applying the Horizontal Line Test, you draw horizontal lines across the graph of the function. If any horizontal line crosses the graph more than once, the function is not one-to-one. If each horizontal line touches the graph at most once, the function is one-to-one and passes the test.
Example: Consider the function $$f(x) = x^3$$. When you graph this function and apply the Horizontal Line Test, you will notice that no horizontal line intersects the graph more than once. Hence, $$f(x) = x^3$$ is a one-to-one function.
### Horizontal Line Test Explained Step-by-Step
Follow these steps to perform the Horizontal Line Test:
• Step 1: Draw the graph of the function you want to test.
• Step 2: Place a ruler horizontally across the graph.
• Step 3: Slide the ruler from the top of the graph to the bottom.
• Step 4: Observe the points where the ruler intersects the graph.
• Step 5: If any horizontal line intersects the graph more than once, the function is not one-to-one. If each horizontal line intersects the graph at most once, the function is one-to-one.
You can use graphing calculators or software to make this process easier.
More on Injective Functions: An injective function, or one-to-one function, ensures distinct outputs for distinct inputs. Mathematically, a function $$f: A \rightarrow B$$ is injective if, for every $$a_1, a_2 \in A$$, whenever $$f(a_1) = f(a_2)$$, it follows that $$a_1 = a_2$$. This property is critical for functions to be invertible.Conversely, functions that are not injective map at least two different elements of their domain to the same element in the codomain. For example, the function $$f(x) = x^2$$ is not injective over the set of all real numbers because both $$x = -1$$ and $$x = 1$$ map to $$y = 1$$.Injective functions are also significant in higher mathematics disciplines such as algebra and calculus, where they play a crucial role in function composition and transformation properties.
## Horizontal Line Test for Inverse Functions
The Horizontal Line Test is a technique used to determine if a function has an inverse that is also a function. This is achieved by verifying whether the function is one-to-one.
### Applying the Horizontal Line Test to Determine Inverses
To find out if a function has an inverse that is also a function, you need to confirm if every output has a unique input. The Horizontal Line Test simplifies this process:
• Step 1: Draw the graph of the function.
• Step 2: Use a horizontal line to test.
• Step 3: Observe if any horizontal line intersects the graph more than once. If it does, the function does not have an inverse that is also a function.
When you follow these steps, you can determine if the function is one-to-one. If each horizontal line intersects the graph at most once, then the function is one-to-one.
Example: Consider the function $$f(x) = x^2$$ on the interval $$x \ge 0$$. Graphing this and applying the Horizontal Line Test, you will find that no horizontal line intersects the graph more than once. Hence, $$f(x) = x^2$$ restricted to $$x \ge 0$$ is one-to-one.
You can also perform the Horizontal Line Test with graphing software for more complex functions.
More on Inverses: For a function to have an inverse that is also a function, it must be one-to-one and onto. This property is essential in many areas of mathematics including calculus and linear algebra. A function $$f: A \rightarrow B$$ having an inverse means there exists a function $$f^{-1}: B \rightarrow A$$ such that $$f(f^{-1}(y)) = y$$ for every $$y \in B$$ and $$f^{-1}(f(x)) = x$$ for every $$x \in A$$.
### Examples of Horizontal Line Test for Inverse Functions
Let's consider a variety of examples to apply the Horizontal Line Test and understand its utility in determining one-to-one functions.First, take the function $$f(x) = \sin(x)$$ over the interval $$[0, \pi]$$. When applying the test, each horizontal line intersects the graph at most once within this interval. Thus, $$\sin(x)$$ is one-to-one over $$[0, \pi]$$ and has an inverse within this interval: $f^{-1}(x) = \arcsin(x), \; -1 \le x \le 1$Next, consider the function $$g(x) = e^x$$. Plotting the graph and applying the Horizontal Line Test, you will observe that each horizontal line intersects the graph exactly once, proving that $$e^x$$ is one-to-one and has an inverse: $g^{-1}(x) = \ln(x), \; x > 0$
Example: For the function $$h(x) = x^3 -3x + 2$$, the graph shows that horizontal lines may intersect more than once, particularly around the turning points. Thus, this function is not one-to-one, and hence $$h(x)$$ doesn't have an inverse that is also a function.
Using the Horizontal Line Test can help students in identifying non-obvious properties of functions. In situations where the function is complex or has multiple intervals, breaking it down into simpler intervals where the function behaves more predictably can aid understanding. This deep dive into the more intricate examples can further solidify the concepts and show the underlying consistency of mathematical principles.Remember that these principles also apply to more complex and higher-degree polynomials. Performing the test step-by-step and ensuring observations at each stage ensure accurate analysis.
## Horizontal Line Test Example
The Horizontal Line Test provides a simple way to identify if a function is one-to-one. This section will present examples to help you understand how to apply this test.
### Simple Horizontal Line Test Example
Let's start with a basic example. Consider the function $$f(x) = x^3$$. The graph of this function is a curve that continues indefinitely in both the positive and negative directions.To apply the Horizontal Line Test:
• Draw horizontal lines across the graph.
• Observe the points of intersection.
For $$f(x) = x^3$$:
• Every horizontal line intersects the curve only once.
This means that each output has a unique input value, indicating that $$f(x) = x^3$$ is a one-to-one function.
Example Table:
Function One-to-One $$f(x) = x^3$$ Yes
For functions where the graph is not obvious, you can use graphing calculators or software to help illustrate.
### Horizontal Line Test Example in Functions
Consider the function $$g(x) = x^2$$. The graph of this function is a parabola that opens upwards. Applying the Horizontal Line Test:
• Draw horizontal lines across the parabola.
• Observe the points of intersection.
For $$g(x) = x^2$$:
• Some horizontal lines intersect the graph at two points.
This indicates that $$g(x) = x^2$$ is not a one-to-one function because multiple inputs map to the same output. Restricting the domain of $$g(x)$$ to non-negative values (i.e., $$x \geq 0$$) can make this function one-to-one. Then, applying the Horizontal Line Test:
• Every horizontal line intersects the graph at most once.
Example Table:
Function Initial Domain One-to-One Restricted Domain One-to-One $$g(x) = x^2$$ All real numbers No $$x \geq 0$$ Yes
Dive deeper into injective functions: An injective function, often termed as a one-to-one function, ensures each output is produced by exactly one input. Mathematically, a function $$f: A \rightarrow B$$ is injective if, for all $$a_1, a_2 \in A$$, whenever $$f(a_1) = f(a_2)$$, it follows that $$a_1 = a_2$$. This property is fundamental when it comes to inverting functions.Consider another example function $$h(x) = e^x$$:
• For every input $$x$$, there is a unique output $$e^x$$.
• Every horizontal line intersects the graph only once.
This confirms that $$h(x)$$ is one-to-one and thus has an inverse function $$h^{-1}(x) = \ln(x)$$.Injective functions play a crucial role in many areas of mathematics, including algebra and calculus, particularly when dealing with function composition and transformation properties.
## Horizontal Line Test One to One Functions
The Horizontal Line Test is an essential tool in mathematics to determine if a function is one-to-one. This helps to identify functions that have unique output values for each unique input value.
### Using Horizontal Line Test to Identify One to One Functions
To apply the Horizontal Line Test and find out if a function is one-to-one, follow these steps:
• Step 1: Draw the graph of the function in question.
• Step 2: Draw horizontal lines across the graph at various points.
• Step 3: Observe whether any horizontal line intersects the graph more than once.
If a horizontal line crosses the graph more than once, the function is not one-to-one. If each horizontal line intersects the graph at most once, the function is one-to-one.
Consider the function $$f(x) = x^2$$. When you draw its graph, you'll notice that horizontal lines can intersect the parabola at two points. Hence, without any domain restrictions, $$f(x) = x^2$$ is not one-to-one.However, if we restrict the domain to non-negative values of $$x$$, i.e., $$x \geq 0$$, then each horizontal line will intersect at most once. This means $$f(x) = x^2$$ is one-to-one on the interval $$x \geq 0$$.
Use graphing software or calculators for complex functions to simplify the process of the Horizontal Line Test.
Horizontal Line Test: A method used to determine if a function is injective (one-to-one). This test involves drawing horizontal lines across the graph of a function to see if each line intersects the graph at most once.
### Example of Horizontal Line Test in One to One Functions
Let's apply the Horizontal Line Test to a few functions.Case 1: Consider the function $$f(x) = x^3$$. Drawing the graph and applying the Horizontal Line Test, we observe that each horizontal line intersects the graph exactly once.
• This implies $$f(x) = x^3$$ is one-to-one.
Function One-to-One $$f(x) = x^3$$ Yes
Consider the function $$g(x) = e^x$$. This exponential function is always increasing, and no horizontal line will intersect the graph more than once. Therefore, $$e^x$$ is a one-to-one function, and it also has an inverse function $$g^{-1}(x) = \ln(x)$$. This property is crucial in understanding the transformation and composition of exponential and logarithmic functions. Functions like these are used extensively in higher-level mathematics and have applications in sciences and engineering.
The Horizontal Line Test can also help to determine if polynomial functions of higher degrees are one-to-one or not.
## Horizontal line test - Key takeaways
• Horizontal Line Test: A method used to determine if a function is injective (one-to-one) by drawing horizontal lines across the graph to see if each line intersects the graph at most once.
• Injective Function: A function where every element of the codomain is mapped by at most one element of the domain, ensuring distinct outputs for distinct inputs.
• Horizontal Line Test for Inverse Functions: This test is used to determine if a function has an inverse that is also a function by confirming the function is one-to-one.
• Steps to Perform Horizontal Line Test: Draw the function's graph, place a ruler horizontally, slide the ruler from top to bottom, observe intersections, and check if any line intersects the graph more than once.
• Examples: The function f(x) = x^3 passes the Horizontal Line Test and is one-to-one, while g(x) = x^2 is not one-to-one over all reals but becomes one-to-one if restricted to non-negative values.
#### Flashcards in Horizontal line test 12
###### Learn with 12 Horizontal line test flashcards in the free StudySmarter app
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What is the horizontal line test?
The horizontal line test determines if a function is injective (one-to-one). If any horizontal line intersects the graph of the function at more than one point, the function fails the test and is not injective. If it intersects at most once, the function passes and is injective. This test is useful for determining if a function has an inverse.
Why is the horizontal line test important in mathematics?
The horizontal line test is important in mathematics because it determines whether a function is injective (one-to-one). If no horizontal line intersects the graph of the function more than once, the function has an inverse, ensuring each output is unique to one input.
How does the horizontal line test determine if a function is one-to-one?
The horizontal line test determines if a function is one-to-one by checking whether any horizontal line drawn through the graph of the function intersects it more than once. If no horizontal line intersects the graph more than once, the function is one-to-one.
Can the horizontal line test be used on all types of functions?
Yes, the horizontal line test can be used on all types of functions to determine if they are injective (one-to-one). However, it is primarily applied to real-valued functions to check if each horizontal line intersects the graph at most once, indicating the function is injective.
Can a function pass the vertical line test but fail the horizontal line test?
Yes, a function can pass the vertical line test but fail the horizontal line test. Passing the vertical line test means the function is indeed a function, while failing the horizontal line test indicates the function is not injective, or one-to-one.
## Test your knowledge with multiple choice flashcards
What is the purpose of the Horizontal Line Test?
What is the Horizontal Line Test used for?
Why is the function $$g(x) = x^2$$ not considered one-to-one on all real numbers?
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NCERT Solutions: Arithmetic Progressions (Exercise 5.1)
NCERT Solutions for Class 6 Maths - Arithmetic Progressions (Exercise 5.1)
Q1. In which of the following situations, does the list of numbers involved make an arithmetic progression and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
We can write the given condition as:
• Taxi fare for 1 km = 15
• Taxi fare for first 2 kms = 15+8 = 23
• Taxi fare for first 3 kms = 23+8 = 31
• Taxi fare for first 4 kms = 31+8 = 39 and so on……
Thus, 15, 23, 31, 39 … forms an A.P. because every next term is 8 more than the preceding term.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Let the volume of air in a cylinder, initially, be litres.
• In each stroke, the vacuum pump removes 1/4th of air remaining in the cylinder at a time.
• Or we can say, after every stroke, 1-1/4 = 3/4th part of air will remain.
• Therefore, volumes will be V, 3V/4 , (3V/4)2, (3V/4)3…and so on
Clearly, we can see here, the adjacent terms of this series do not have a common difference between them. Therefore, this series is not an A.P.
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
We can write the given condition as:
• Cost of digging a well for first metre = Rs.150
• Cost of digging a well for first 2 metres = Rs.150+50 = Rs.200
• Cost of digging a well for first 3 metres = Rs.200+50 = Rs.250
• Cost of digging a well for first 4 metres =Rs.250+50 = Rs.300 and so on.
Clearly, 150, 200, 250, 300 … forms an A.P. with a common difference of 50 between each term.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
We know that if Rs. P is deposited at r% compound interest per annum for n years, the amount of money will be:
• P(1+r/100)n
Therefore, after each year, the amount of money will be;
10000(1+8/100), 10000(1+8/100)2, 10000(1+8/100)3……
Clearly, the terms of this series do not have a common difference between them. Therefore, this series is not an A.P.
Q2. Write first four terms of the A.P. when the first term a and the common difference are given as follows:
(i) a = 10, d = 10
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5 …
a1 = a = 10
a2 = a1+d = 10+10 = 20
a3 = a2+d = 20+10 = 30
a4 = a3+d = 30+10 = 40
a5 = a4+d = 40+10 = 50
And so on…
Therefore, the A.P. series will be 10, 20, 30, 40, 50 …
And First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = – 2, d = 0
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5
a1 = a = -2
a2 = a1+d = – 2+0 = – 2
a3 = a2+d = – 2+0 = – 2
a4 = a3+d = – 2+0 = – 2
Therefore, the A.P. series will be – 2, – 2, – 2, – 2 …
And, First four terms of this A.P. will be – 2, – 2, – 2 and – 2.
(iii) a = 4, d = – 3
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5
a1 = a = 4
a2 = a1+d = 4-3 = 1
a3 = a2+d = 1-3 = – 2
a4 = a3+d = -2-3 = – 5
Therefore, the A.P. series will be 4, 1, – 2 – 5 …
And, first four terms of this A.P. will be 4, 1, – 2 and – 5.
(iv) a = – 1, d = 1/2
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5
a2 = a1+d = -1+1/2 = -1/2
a3 = a2+d = -1/2+1/2 = 0
a4 = a3+d = 0+1/2 = 1/2
Thus, the A.P. series will be-1, -1/2, 0, 1/2
And First four terms of this A.P. will be -1, -1/2, 0 and 1/2.
(v) a = – 1.25, d = – 0.25
Let us consider, the Arithmetic Progression series be a1, a2, a3, a4, a5
a1 = a = – 1.25
a2 = a1 + d = – 1.25-0.25 = – 1.50
a3 = a2 + d = – 1.50-0.25 = – 1.75
a4 = a3 + d = – 1.75-0.25 = – 2.00
Therefore, the A.P series will be 1.25, – 1.50, – 1.75, – 2.00 ……..
And first four terms of this A.P. will be – 1.25, – 1.50, – 1.75 and – 2.00.
Q3. For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …
Ans.
(i) Given series, 3, 1, – 1, – 3 …
First term, a = 3
Common difference, d = Second term – First term
⇒ 1 – 3 = -2
d = -2
(ii) Given series, – 5, – 1, 3, 7 …
First term, a = -5
Common difference, d = Second term – First term
⇒ ( – 1)-( – 5) = – 1+5 = 4 =d
(iii) Given series, 1/3, 5/3, 9/3, 13/3 ….
First term, a = 1/3
Common difference, d = Second term – First term
⇒ 5/3 – 1/3 = 4/3 = d
(iv) Given series, 0.6, 1.7, 2.8, 3.9 …
First term, a = 0.6
Common difference, d = Second term – First term
⇒ 1.7 – 0.6
1.1 = d
Q4. Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a
(xi) a, a2, a3, a4
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 12, 32, 52, 72
(xv) 12, 52, 72, 73
Ans.
(i) Given, 2, 4, 8, 16 …
Here, the common difference is:
a2a1 = 4 – 2 = 2
a3a2 = 8 – 4 = 4
a4a3 = 16 – 8 = 8
Since, an+1aor the common difference is not the same every time.
Therefore, the given series are not forming an A.P.
(ii) Given, 2, 5/2, 3, 7/2 ….
Here,
a2a1 = 5/2-2 = 1/2
a3a2 = 3-5/2 = 1/2
a4a3 = 7/2-3 = 1/2
Since, an+1an or the common difference is same every time.
Therefore, d = 1/2 and the given series are in A.P.
The next three terms are;
a5 = 7/2+1/2 = 4
a6 = 4 +1/2 = 9/2
a7 = 9/2 +1/2 = 5
(iii) Given, -1.2, – 3.2, -5.2, -7.2 …
Here,
a2a1 = (-3.2)-(-1.2) = -2
a3a2 = (-5.2)-(-3.2) = -2
a4a3 = (-7.2)-(-5.2) = -2
Since, an+1an or common difference is same every time.
Therefore, d = -2 and the given series are in A.P.
Hence, next three terms are;
a5 = – 7.2-2 = -9.2
a6 = – 9.2-2 = – 11.2
a7 = – 11.2-2 = – 13.2
(iv) Given, -10, – 6, – 2, 2 …
Here, the terms and their difference are;
a2a1 = (-6)-(-10) = 4
a3a2 = (-2)-(-6) = 4
a4a3 = (2 -(-2) = 4
Since, an+1an or the common difference is same every time.
Therefore, d = 4 and the given numbers are in A.P.
Hence, next three terms are;
a5 = 2+4 = 6
a6 = 6+4 = 10
a7 = 10+4 = 14
(v) Given, 3, 3+√2, 3+2√2, 3+3√2
Here,
a2a1 = 3+√2-3 = √2
a3a2 = (3+2√2)-(3+√2) = √2
a4a3 = (3+3√2) – (3+2√2) = √2
Since, an+1an or the common difference is same every time.
Therefore, d = √2 and the given series forms a A.P.
Hence, next three terms are;
a5 = (3+√2) +√2 = 3+4√2
a6 = (3+4√2)+√2 = 3+5√2
a7 = (3+5√2)+√2 = 3+6√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
Here,
a2a1 = 0.22-0.2 = 0.02
a3a2 = 0.222-0.22 = 0.002
a4a3 = 0.2222-0.222 = 0.0002
Since, an+1an or the common difference is not same every time.
Therefore, and the given series doesn’t forms a A.P.
(vii) 0, -4, -8, -12 …
Here,
a2a1 = (-4)-0 = -4
a3a2 = (-8)-(-4) = -4
a4a3 = (-12)-(-8) = -4
Since, an+1an or the common difference is same every time.
Therefore, d = -4 and the given series forms a A.P.
Hence, next three terms are;
a5 = -12-4 = -16
a6 = -16-4 = -20
a7 = -20-4 = -24
(viii) -1/2, -1/2, -1/2, -1/2 ….
Here,
a2a1 = (-1/2) – (-1/2) = 0
a3a2 = (-1/2) – (-1/2) = 0
a4a3 = (-1/2) – (-1/2) = 0
Since, an+1an or the common difference is same every time.
Therefore, d = 0 and the given series forms a A.P.
Hence, next three terms are;
a5 = (-1/2)-0 = -1/2
a6 = (-1/2)-0 = -1/2
a7 = (-1/2)-0 = -1/2
(ix) 1, 3, 9, 27 …
Here,
a2a1 = 3-1 = 2
a3a2 = 9-3 = 6
a4a3 = 27-9 = 18
Since, an+1an or the common difference is not same every time.
Therefore, and the given series doesn’t form a A.P.
(x) a, 2a, 3a, 4a
Here,
a2a1 = 2a= a
a3a2 = 3a-2a = a
a4a3 = 4a-3a = a
Since, an+1an or the common difference is same every time.
Therefore, d = a and the given series forms a A.P.
Hence, next three terms are;
a5 = 4a+a = 5a
a6 = 5a+a = 6a
a7 = 6a+a = 7a
(xi) a, a2, a3, a4 ...
Here,
a2a1 = a2a = a(a-1)
a3a2 = a a= a2(a-1)
a4a3 = a4a= a3(a-1)
Since, an+1an or the common difference is not same every time.
Therefore, the given series doesn’t forms a A.P.
(xii) √2, √8, √18, √32 …
Here,
a2a1 = √8-√2 = 2√2-√2 = √2
a3a2 = √18-√8 = 3√2-2√2 = √2
a4a3 = 4√2-3√2 = √2
Since, an+1an or the common difference is same every time.
Therefore, d = √2 and the given series forms a A.P.
Hence, next three terms are;
a5 = √32+√2 = 4√2+√2 = 5√2 = √50
a6 = 5√2+√2 = 6√2 = √72
a7 = 6√2+√2 = 7√2 = √98
(xiii) √3, √6, √9, √12 …
Here,
a2a1 = √6-√3 = √3×√2-√3 = √3(√2-1)
a3a2 = √9-√6 = 3-√6 = √3(√3-√2)
a4a3 = √12 – √9 = 2√3 – √3×√3 = √3(2-√3)
Since, an+1an or the common difference is not same every time.
Therefore, the given series doesn’t form a A.P.
(xiv) 12, 32, 52, 72
Or, 1, 9, 25, 49 …..
Here,
a2a1 = 9−1 = 8
a3a= 25−9 = 16
a4a3 = 49−25 = 24
Since, an+1an or the common difference is not same every time.
Therefore, the given series doesn’t form an A.P.
(xv) 12, 52, 72, 73 …
Or 1, 25, 49, 73 …
Here,
a2a1 = 25−1 = 24
a3a= 49−25 = 24
a4a3 = 73−49 = 24
Since, an+1an or the common difference is same every time.
Therefore, d = 24 and the given series forms a A.P.
Hence, next three terms are:
a5 = 73+24 = 97
a6 = 97+24 = 121
a= 121+24 = 145
The document NCERT Solutions for Class 6 Maths - Arithmetic Progressions (Exercise 5.1) is a part of the Class 10 Course Mathematics (Maths) Class 10.
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# Subsequences
## 1. Introduction
This page is about subsequences of a sequence. A subsequence of a sequence $( a n )$ is an infinite collection of numbers from $( a n )$ in the same order that they appear in that sequence.
The main theorem on subsequences is that every subsequence of a convergent sequence $( a n )$ converges to the same limit as $( a n )$. This (together with the Theorem on Uniqueness of Limits) is the main tool in showing a sequence does not converge.
It is harder to use subsequences to prove a sequence does converge and much easier to make mistakes in this direction. Just knowing a sequence $( a n )$ has a convergent subsequence says nothing about the convergence of $( a n )$. We discuss these topics with an example in another web page.
## 2. The Theorem on Subsequences
If $( a n )$ is a sequence, such as $1 1 , 1 2 , 1 3 , 1 4 , …$ we may obtain a subsequence by selecting some of the terms in the sequence. The only rules are that the order of the terms in the original sequence must be preserved and we must select an infinite number of terms. For example, from the above sequence we may select $b 1 = a 1 = 1$, $b 2 = a 2 = 1 2$, $b 3 = a 4 = 1 4$, $b 4 = a 8 = 1 8$, and so on. This idea of subsequence is rather simple and very natural. Unfortunately there are some difficulties with notation that tends to obscure the main idea and can make the manipulation of subsequences a little more tricky.
Definition of subsequence.
Suppose $( a n )$ is a sequence and $f : ℕ → ℕ$ is an increasing function (i.e., ) then $( a n )$ and $f$ define a new sequence
$b n = a f ( n )$
with elements
$b 1 = a f ( 1 ) , b 2 = a f ( 2 ) , b 3 = a f ( 3 ) , b 4 = a f ( 4 ) , …$
The sequence $( b n )$ is called a subsequence of $( a n )$.
Example.
For the sequence $( a n )$ given by $a n = 1 n$ we obtain the subsequence $b n = 1 2 n$ by taking as our increasing function $f ( n ) = 2 n$, since $b n = a f ( n ) = a 2 n = 1 2 n$.
The ideas of selecting infinitely many terms from a sequence and using an increasing function to enumerate a selection of terms are equivalent. If we have an infinite set of terms we can enumerate them with a function $f$. Conversely, an increasing function $f$ defines a selection of terms with indices in the image of $f$. The following lemma is useful in switching between these two views.
Lemma.
Let $f : ℕ → ℕ$ be increasing. Then $f ( n ) ⩾ n$ for all $n ∈ ℕ$.
Proof.
By induction on $n$. Suppose for the sake of this proof that $ℕ = { 0 , 1 , 2 , 3 , … }$ (a similar argument starting with one works if you prefer $0$ not to be a natural number) and let the induction hypothesis be $f ( n ) ⩾ n$. Then $f ( 0 ) ∈ ℕ$ and so $f ( 0 ) ⩾ 0$, so the base case is established. Now inductively assume $f ( n ) ⩾ n$ and observe that $f ( n +1 ) > f ( n )$ as $f$ is increasing. So $f ( n +1 ) ⩾ f ( n ) + 1 ⩾ n + 1$ by our hypothesis, and the induction step is proved.
Theorem on Subsequences.
Suppose $a n → l ∈ ℝ$ as $n → ∞$ and $( b n )$ is a subsequence of $( a n )$ defined by $b n = a f ( n )$. Then $b n → l$ as $n → ∞$.
Proof.
Subproof.
Let $ε > 0$ be arbitrary.
Let $N ∈ ℕ$ satisfy .
Subproof.
Let $n ∈ ℕ$ be arbitrary.
Subproof.
Suppose $n ⩾ N$.
Then $f ( n ) ⩾ n ⩾ N$ by the lemma.
So $| b n - l | = | a f ( n ) - l | < ε$.
So $n ⩾ N ⇒ | b n - l | < ε$.
So .
So .
So .
Part of the power of this theorem is that the subsequence $( b n )$ converges to the same limit. This will be combined with the Theorem on Uniqueness of Limits to allow us to prove non-convergence results.
Example.
We use these ideas to re-prove the assertion that the sequence $a n = ( -1 ) n$ doesn't converge. Define subsequences $b n = a 2 n + 1$ and $c n = a 2 n$, and observe that $b n = -1$ for all $n$ and $c n = 1$ for all $n$, so $b n → -1$ and $c n → 1$ as $n → ∞$, as these are constant sequences. Now suppose (to get a contradiction) that $( a n )$ converges to some limit $l$. Then by the Theorem on Subsequences both $b n → l$ and $c n → l$ as these are subsequences of $( a n )$. By the Theorem on Uniqueness of Limits, $-1 = l$ and $1 = l$. But this implies $1 = -1$ which is false.
Example.
Consider the sequence defined by $a n = cos ( n )$. This is proved to be non-convergent as follows.
First a subsequence $( b n )$ is selected consisting of all $a n$ for which . It may not be immediately obvious that there are infinitely many such $n$ but this is easily proved as for each $k ∈ ℕ$ the integer part of $2 k π$ is such an $n$. Note for such $n$ that $cos ( n ) > cos ( 1 ) > 0$.
Next a subsequence $( c n )$ is selected consisting of all $a n$ for which . Again, note that for each $k ∈ ℕ$ the integer part of $( 2 k + 1 ) π$ is such an $n$. Note also that for such $n$ that $cos ( n ) < - cos ( 1 ) < 0$.
Finally, suppose $a n → l ∈ ℝ$. Then by the Theorem on Subsequences both $b n → l$ and $c n → l$. But by the Theorem Giving Bounds on Limits we have $l ⩾ cos ( 1 )$ as $l$ is the limit of $( b n )$. Also $l ⩽ - cos ( 1 )$ as $l$ is the limit of $( c n )$. But this is impossible as $- cos ( 1 ) < 0 < cos ( 1 )$.
Be careful to note the $⩾$ and $⩽$ here! In particular this argument wouldn't have worked if we have simply shown that $c n < 0 < b n$ for all $n$ for then it might have been possible that $l = 0$.
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# How do you factor x^3 + 6x^2 - 9x - 54 by grouping?
Jul 23, 2016
(x+6)(x-3)(x+3)
#### Explanation:
Group the terms into 'pairs' as follows.
$\left[{x}^{3} + 6 {x}^{2}\right] + \left[- 9 x - 54\right]$
Now, factorise each 'pair'
$\textcolor{red}{{x}^{2}} \left(x + 6\right) \textcolor{red}{- 9} \left(x + 6\right)$
Take out the common factor of (x +6)
$\left(x + 6\right) \textcolor{red}{\left({x}^{2} - 9\right)}$
$\textcolor{red}{{x}^{2} - 9} \text{ is a "color(blue)"difference of squares}$ and in general factorises.
$\textcolor{red}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{a}^{2} - {b}^{2} = \left(a - b\right) \left(a + b\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
${x}^{2} = {\left(x\right)}^{2} \text{ and " 9=(3)^2rArra=x" and } b = 3$
$\Rightarrow \textcolor{red}{{x}^{2} - 9} = \left(x - 3\right) \left(x + 3\right)$
Pulling this altogether we get.
${x}^{3} + 6 {x}^{2} - 9 x - 54 = \left(x + 6\right) \left(x - 3\right) \left(x + 3\right)$
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Triangle Congruence and Relationship of Angles Quiz
DexterousCouplet
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6 Questions
If two triangles have two sides and an included angle equal to a given angle, when can we conclude that the triangles are congruent?
When the two sides included in the angle are equal in length.
What can we conclude about the angles in two triangles if the sides opposite the included angle are equal?
The angles opposite the included angle are also equal.
In which scenario can we conclude that two triangles are not congruent?
When only one pair of sides and one angle are equal in both triangles.
What should be verified to ensure the congruence of two triangles beyond checking sides and angles corresponding to a given angle?
The measurement of at least one additional angle and side must be equal.
How do we verify if two triangles are congruent using different angles and sides?
By examining corresponding pairs of sides and angles in each triangle.
What is an essential step suggested in the text for understanding and applying triangular relationships effectively?
Practicing by verifying relationships in various triangles with different measures.
Study Notes
• The text is about understanding relationships between angles in a triangle.
• Two triangles, each with two sides and an included angle equal to a given angle, are being compared.
• The text explains that if the two sides included in the angle are equal in both triangles, then the two triangles are congruent.
• The text also explains that if the sides opposite the included angle are equal in both triangles, then the two angles opposite the included angle are also equal.
• The text uses the example of a 60 degree angle to illustrate this concept.
• The sides included in the 60 degree angle are 55 and 65, and the opposite sides are 80 and 80.
• The text shows that since both pairs of sides are equal in length, the triangles are congruent.
• The text also mentions the need to check other angles in the triangles to make sure they are equal as well.
• The text mentions that this concept can be extended to the verification of other triangular relationships using different angles and sides.
• The text encourages the reader to practice and try to understand the concept by verifying the relationships in other triangles.
Test your understanding of triangle congruence and the relationship between angles in triangles with this quiz. Learn how to determine if triangles are congruent by comparing side lengths and included angles. Explore how to verify if angles opposite equal sides are also equal in triangles.
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Equivalent Expressions Calculator
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Equivalent expression is said to be equivalent if two algebraic expressions are equalized by substituting the values of the variables. We use the symbol “ = “ or “ $\equiv$ “ to denote the equivalent expression.
Equivalent Expressions Calculator is used to find the equivalent expressions when two algebraic expressions are equivalent.
An expression is given below as a default input for this calculator. By applying distributive property and expanding the expression, get the simplified expression on clicking "Submit".
## Step by Step Calculation
Step 1 :
Observe the given expressions and expand the expression using distributive property.
Step 2 :
Simplify the expression to get the answer.
## Example Problems
1. ### Find the equivalent expression for this: 3y(x + 9)
Step 1 :
The given algebra expression is 3y(x + 9)
Now we need to expand by applying distributive property.
Step 2 :
So, 3y(x + 9) = 3y • x + 3y • 9
= 3xy + 27y
Therefore, 3y(x + 9) $\equiv$ 3xy + 27y.
3xy + 27y.
2. ### Find the equivalent expression for this: (x + 4)(x + 3)
Step 1 :
The given algebra expression is (x + 4)(x + 3)
Now we need to expand by applying distributive property.
Step 2 :
So, (x + 4)(x + 3) = x • (x + 3) + 4 • (x + 3)
= x • x + x • 3 + 4 • x + 4 • 3
= x2 + 3x + 4x + 12
= x2 + 7x + 12
Therefore, (x + 4)(x + 3) $\equiv$ x2 + 7x + 12.
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# Power of Power Index Law: The Ultimate Simplification Hack
As an experienced mathematics tutor, I’ve seen countless students struggle with simplifying complex expressions involving indices. However, the Power of Power Index Law is a powerful tool that can make the process much more manageable. In this article, we’ll explore the intricacies of this essential law, investigate its applications, and discover how it can become your ultimate simplification hack.
## Understanding the Fundamentals of the Power of Power Index Law
Before we dive into the various applications of the Power of Power Index Law, let’s first grasp its definition and how it works.
### Defining the Power of Power Index Law
The Power of Power Index Law states that when a power is raised to another power, you can simplify the expression by multiplying the indices. Mathematically, it can be expressed as:
$\displaystyle (a^m)^n = a^{m \times n}$
Where $a$ is the base, and $m$ and $n$ are the indices.
### The Importance of the Power of Power Index Law
The Power of Power Index Law is a crucial tool for simplifying complex expressions involving indices. By applying this law, you can reduce the number of steps required to evaluate an expression, making the process more efficient and less prone to errors.
## Exploring the Applications of the Power of Power Index Law
Now that we have a solid understanding of the Power of Power Index Law, let’s delve into some of its practical applications.
### Simplifying Expressions with Multiple Powers
One of the most common applications of the Power of Power Index Law is simplifying expressions with multiple powers. For example, consider the following expression:
$\displaystyle (x^2)^3$
By applying the Power of Power Index Law, we can simplify this expression as follows:
$\displaystyle (x^2)^3 = x^{2 \times 3} = x^6$
### Evaluating Expressions with Negative Indices
Another useful application of the Power of Power Index Law is evaluating expressions with negative indices. For instance, consider the following expression:
$\displaystyle \left(\frac{1}{x^2}\right)^3$
We can rewrite this expression using the negative index rule:
$\displaystyle \left(\frac{1}{x^2}\right)^3 = (x^{-2})^3$
Now, by applying the Power of Power Index Law, we get:
$\displaystyle (x^{-2})^3 = x^{-2 \times 3} = x^{-6} = \frac{1}{x^6}$
### Simplifying Expressions with Fractional Indices
The Power of Power Index Law can also be used to simplify expressions with fractional indices. For example, consider the following expression:
$\displaystyle \left(x^\frac{1}{2}\right)^\frac{1}{3}$
By applying the Power of Power Index Law, we can simplify this expression as follows:
$\displaystyle \left(x^\frac{1}{2}\right)^\frac{1}{3} = x^{\frac{1}{2} \times \frac{1}{3}} = x^\frac{1}{6}$
## Mastering the Power of Power Index Law: Tips and Tricks
Now that we’ve seen some of the applications of the Power of Power Index Law, let’s explore some tips and tricks to help you apply it more effectively.
### Breaking Down Complex Expressions
When faced with a complex expression involving multiple powers, break it down into smaller, more manageable parts. Apply the Power of Power Index Law to each part separately, and then combine the results to obtain the simplified expression.
### Using Parentheses for Clarity
When working with expressions involving multiple powers and indices, it’s essential to use parentheses to clearly indicate the order of operations. This will help you avoid confusion and ensure that you apply the Power of Power Index Law correctly.
### Practice, Practice, Practice
As with any mathematical concept, practice is key to mastering the Power of Power Index Law. Work on a variety of problems involving different types of expressions, and focus on applying the law accurately and efficiently.
## Common Mistakes to Avoid
While the Power of Power Index Law is a powerful tool, students often make some common mistakes when applying it. Here are a few to watch out for:
### Forgetting to Multiply the Indices
One of the most common mistakes is forgetting to multiply the indices when applying the Power of Power Index Law. Remember, the law states that you multiply the indices, not add them.
### Misapplying the Law to Expressions with Different Bases
Another frequent mistake is attempting to apply the Power of Power Index Law to expressions with different bases. The law only holds when the base is consistent for both powers.
### Confusing the Order of Operations
Lastly, it’s crucial to keep the order of operations in mind when applying the Power of Power Index Law. Remember to simplify any expressions within parentheses first, and then apply the law to the resulting expression.
## Real-World Applications of the Power of Power Index Law
The Power of Power Index Law is not just a theoretical concept; it has numerous real-world applications across various fields.
### Scientific Notation
In scientific notation, very large or very small numbers are expressed as a product of a number between 1 and 10 and a power of 10. The Power of Power Index Law is often used to simplify calculations involving scientific notation.
### Computer Science
In computer science, the Power of Power Index Law is frequently used in algorithms that involve exponentiation, such as cryptography and prime number generation.
### Physics and Engineering
In physics and engineering, the Power of Power Index Law is applied to solve problems involving exponential growth and decay, such as population growth models and radioactive decay.
## Conclusion
In conclusion, the Power of Power Index Law is an indispensable tool that every student should have in their mathematical arsenal. By understanding its functionality and applying it correctly, you can simplify even the most intricate expressions involving indices with ease.
Remember to practice regularly, break down complex expressions into smaller components, and avoid common mistakes like forgetting to multiply the indices or misapplying the law to expressions with different bases.
Discover more enlightening videos by visiting our YouTube channel!
## Power of Power Index Law: The Ultimate Evaluation Hack
As an experienced mathematics tutor, I’ve seen countless students struggle with simplifying complex expressions involving indices. However, one powerful tool that can make the process…
## Power of Power Index Law: The Ultimate Equations Hack
As an experienced mathematics tutor, I’ve witnessed numerous students grapple with simplifying intricate expressions involving indices. However, the Power of Power Index Law is a…
## Induction Made Simple: The Ultimate Guide
“Induction Made Simple: The Ultimate Guide” is your gateway to mastering the art of mathematical induction, demystifying a powerful tool in mathematics. This ultimate guide…
## Mastering Integration by Parts: The Ultimate Guide
Welcome to the ultimate guide on mastering integration by parts. If you’re a student of calculus, you’ve likely encountered integration problems that seem insurmountable. That’s…
## How to Tackle Negative Bases in Indices Brackets
As an experienced mathematics tutor, I’ve encountered many students who struggle with understanding and simplifying negative bases in indices brackets. This concept can be challenging…
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# How do you solve (3^{4} \div 9) + 32- ( 5\times 10) + 6?
Mar 29, 2018
See below.
#### Explanation:
To simplify this expression, we need to use a process called the Order of Operations.
This is more commonly known as $P E M D A S$, an acronym to describe the necessary steps (in order) of solving such an expression:
$P$ - Parenthesis
$E$ - Exponents
$M D$ - Multiply/Divide (Left to right)
$A S$ - Add/Subtract (Left to right)
Knowing this, we can being to simplify:
$\left({3}^{4} \div i \mathrm{de} 9\right) + 32 - \left(5 \times 10\right) + 6$
$= \left(81 \div i \mathrm{de} 9\right) + 32 - \left(5 \times 10\right) + 6$
$= \left(9\right) + 32 - \left(50\right) + 6$
$= 41 - \left(50\right) + 6$
$= - 9 + 6$
$= - 3$
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# 11.1 Transformations (I)
11.1 Transformations (I)
11.1.1 Transformation
A transformation is a one-to-one correspondence or mapping between points of an object and its image on a plane.
11.1.2 Translation
1. A translation is a transformation which moves all the points on a plane through the same distance in the same direction.
2. Under a translation, the shape, size and orientation of object and its image are the same.
3. A translation in a Cartesian plane can be represented in the form $\left(\begin{array}{l}a\\ b\end{array}\right),$ whereby, a represents the movement to the right or left which is parallel to the x-axis and b represents the movement upwards or downwards which is parallel to the y-axis.
Example 1
:
Write the coordinates of the image of A (–2, 4) under a translation $\left(\begin{array}{l}\text{}4\\ -3\end{array}\right)$ and B (1, –2) under a translation $\left(\begin{array}{l}-5\\ \text{}3\end{array}\right)$ .
Solution
:
A’ = [–2 + 4, 4 + (–3)] = (2, 1)
B’ = [1 + (–5), –2 + 3] = (–4, 1)
Example 2
:
Point moved to point K’ (3, 8) under a translation $\left(\begin{array}{l}-4\\ \text{}3\end{array}\right).$
What are the coordinates of point K?
Solution
:
$K\left(x,\text{}y\right)\to \left(\begin{array}{l}-4\\ \text{}3\end{array}\right)\to K\text{'}\left(3,\text{}8\right)$
The coordinates of K = [3 – (– 4), 8 – 3]
= (7, 5)
Therefore the coordinates of K are (7, 5).
11.1.3 Reflection
1. A reflection is a transformation which reflects all points of a plane in a line called the axis of reflection.
2. In a reflection, there is no change in shape and size but the orientation is changed. Any points on the axis of reflection do not change their positions.
Example 3:
11.1.4 Rotation
1. A rotation is a transformation which rotates all points on a plane about a fixed point known as the centre of rotation through a given angle in a clockwise or anticlockwise direction.
2. In a rotation, the shape, size and orientation remain unchanged.
3. The centre of rotation is the only point that does not change its position.
Example 4
:
Point A (3, –2) is rotated through 90o clockwise to A’ and 180o anticlockwise to A1 respectively about origin.
State the coordinates of the image of point A.
Solution
:
Image A’ = (–2, 3)
Image A= (–3, 2)
11.1.5 Isometry
1. An isometry is a transformation that preserves the shape and size of an object.
2.Translation, reflection and rotation and a combination of it are isometries.
11.1.6 Congruence
1. Congruent figures have the same size and shape regardless of their orientation.
2. The object and the image obtained under an isometry are congruent.
### 1 thought on “11.1 Transformations (I)”
1. can you double check the example 2 ?? I think there was some mistake in the equation
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# Difference between revisions of "2016 AMC 10B Problems/Problem 23"
## Problem
In regular hexagon $ABCDEF$, points $W$, $X$, $Y$, and $Z$ are chosen on sides $\overline{BC}$, $\overline{CD}$, $\overline{EF}$, and $\overline{FA}$ respectively, so lines $AB$, $ZW$, $YX$, and $ED$ are parallel and equally spaced. What is the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$?
$\textbf{(A)}\ \frac{1}{3}\qquad\textbf{(B)}\ \frac{10}{27}\qquad\textbf{(C)}\ \frac{11}{27}\qquad\textbf{(D)}\ \frac{4}{9}\qquad\textbf{(E)}\ \frac{13}{27}$
## Solution 1
We draw a diagram to make our work easier: $[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); label("A",A,SW); label("B",B,SE); label("C",C,ESE); label("D",D,NE); label("E",E,NW); label("F",F,WSW); label("W",W,ENE); label("X",X,ESE); label("Y",Y,WSW); label("Z",Z,WNW); [/asy]$
Assume that $AB$ is of length $1$. Therefore, the area of $ABCDEF$ is $\frac{3\sqrt 3}2$. To find the area of $WCXYFZ$, we draw $\overline{CF}$, and find the area of the trapezoids $WCFZ$ and $CXYF$.
$[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("A",A,SW); label("B",B,SE); label("C",C,ESE); label("D",D,NE); label("E",E,NW); label("F",F,WSW); label("W",W,ENE); label("X",X,ESE); label("Y",Y,WSW); label("Z",Z,WNW); [/asy]$
From this, we know that $CF=2$. We also know that the combined heights of the trapezoids is $\frac{\sqrt 3}3$, since $\overline{ZW}$ and $\overline{YX}$ are equally spaced, and the height of each of the trapezoids is $\frac{\sqrt 3}6$. From this, we know $\overline{ZW}$ and $\overline{YX}$ are each $\frac 13$ of the way from $\overline{CF}$ to $\overline{DE}$ and $\overline{AB}$, respectively. We know that these are both equal to $\frac 53$.
We find the area of each of the trapezoids, which both happen to be $\frac{11}6 \cdot \frac{\sqrt 3}6=\frac{11\sqrt 3}{36}$, and the combined area is $\frac{11\sqrt 3}{18}$.
We find that $\dfrac{\frac{11\sqrt 3}{18}}{\frac{3\sqrt 3}2}$ is equal to $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.
## Solution 2
$[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(F--C--B--E--D--A); label("A",A,SW); label("B",B,SE); label("C",C,ESE); label("D",D,NE); label("E",E,NW); label("F",F,WSW); label("W",W,ENE); label("X",X,ESE); label("Y",Y,WSW); label("Z",Z,WNW); [/asy]$
First, like in the first solution, split the large hexagon into 6 equilateral triangles. Each equilateral triangle can be split into three rows of smaller equilateral triangles. The first row will have one triangle, the second three, the third five. Once you have drawn these lines, it's just a matter of counting triangles. There are $22$ small triangles in hexagon $ZWCXYF$, and $9 \cdot 6 = 54$ small triangles in the whole hexagon.
Thus, the answer is $\frac{22}{54}=\boxed{\textbf{(C)}\ \frac{11}{27}}$.
## Solution 3 (Similar Triangles)
$[asy] pair A,B,C,D,E,F,W,X,Y,Z; A=(0,0); B=(1,0); C=(3/2,sqrt(3)/2); D=(1,sqrt(3)); E=(0,sqrt(3)); F=(-1/2,sqrt(3)/2); W=(4/3,2sqrt(3)/3); X=(4/3,sqrt(3)/3); Y=(-1/3,sqrt(3)/3); Z=(-1/3,2sqrt(3)/3); pair G = (0.5, sqrt(3)*3/2); draw(A--B--C--D--E--F--cycle); draw(W--Z); draw(X--Y); draw(E--G--D); draw(F--C); label("A",A,SW); label("B",B,SE); label("C",C,ESE); label("D",D,NE); label("E",E,NW); label("F",F,WSW); label("W",W,ENE); label("X",X,ESE); label("Y",Y,WSW); label("Z",Z,WNW); label("G",G,N); [/asy]$ Extend $\overline{EF}$ and $\overline{CD}$ to meet at point $G$, as shown in the diagram. Then $\triangle GZW \sim \triangle GFC$. Then $[GZW] = \left(\dfrac53\right)^2[GED]$ and $[GFC] = 2^2[GED]$. Subtracting $[GED]$, we find that $[EDWZ] = \dfrac{16}{9}[GED]$ and $[EDCF] = 3[GED]$. Subtracting again, we find that $$[ZWCF] = [EDCF] - [EDWZ] = \dfrac{11}{9}[GED].$$Finally, $$\dfrac{[WCXYFZ]}{[ABCDEF]} = \dfrac{[ZWCF]}{[EDCF]} = \dfrac{\dfrac{11}{9}[GED]}{3[GED]} = \textbf{(C) } \dfrac{11}{27}.$$
## Solution 4 (Extending Lines)
Refer to the diagram from Solution 1.
Let us start by connecting points $F$ to $C$ to create a new line segment $FC$. We drop a perpendicular line segment from $E$ to side $FC$ at point $P$. Since $\angle FED = 120$, and $\angle PED = 90$, we know that $\angle FEP = 30.$
Therefore, $\triangle EFP$ is a 30-60-90 triangle. Assume that $EP = \sqrt{3}$. Therefore, we know that $FP = 1$, $EF = 2$.
Let us draw $PQ$ such that $PQ$ is perpendicular to $ZW$. Since the distance between the parallel lines are equal, we know that $PQ$ is half of the distance between the parallel lines. Hence, $PQ$ will be one-third the length of $EP$.
From this, we also know that $EQ = EP - PQ = EP - \frac{1}{3} EP = \frac{2}{3} EP$.
Hence, we know that $EQ = \frac{2}{3} EP = \frac{2\sqrt{3}}{3}.$ Since $\angle EQZ = 90, FEP = 30$ we know that $\triangle ZEQ$ is also a 30-60-90 triangle. From this, we know that $ZQ = \frac{2}{3}$.
Let us drop another perpendicular line segment from $D$ to point $T$ such that $DT$ is perpendicular to $ZW$.
Since $ED = QT$ and since $ED$ is the side length of the regular hexagon, we know that $ED = EF = QT = 2.$
By symmetry, we also know that $WT = \frac{2}{3}$.
Therefore, we can find the length of $ZW = ZQ + QT + QW = \frac{2}{3} + 2 + \frac{2}{3}$. Hence, we know that $ZW = \frac{10}{3}.$
Now, we can find the area of trapezoid $EDZW = \frac{ED + ZW}{2} \cdot EQ = \frac{2 + \frac{10}{3}}{2} \cdot \frac{2\sqrt{3}}{3} = \frac{16\sqrt{3}}{9}.$
By symmetry, we know that the area of trapezoid $YZAB = \frac{16\sqrt{3}}{9}.$
Using the area of a hexagon formula, we get that the area of regular hexagon $ABCDEF = \frac{3\sqrt{3}}{2} \cdot 2^2 = 6\sqrt{3}.$
Hence, the area of hexagon $WCXYFZ = 6\sqrt{3} - 2\left(\frac{16\sqrt{3}}{9}\right) = \frac{22\sqrt{3}}{9}.$
Hence, the ratio of the area of hexagon $WCXYFZ$ to the area of hexagon $ABCDEF$ is $\dfrac{\frac{22\sqrt{3}}{9}}{6\sqrt{3}} = \boxed{\frac{11}{27} \implies C}.$
~yk2007 :D
2016 AMC 10B (Problems • Answer Key • Resources) Preceded byProblem 22 Followed byProblem 24 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AMC 10 Problems and Solutions
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# Negative Fractions
Question: Where should I put the negative sign when I am writing a fraction like negative two thirds?
Answer: As long as you write only one negative sign, it does not matter where you put it.
Two ideas are useful to keep in mind during the explanation that follows:
– Subtraction is the same thing as the addition of a negative.
– The negative of a number can be created by multiplying the number by negative one.
These principles apply to fractions as well, so:
$-\dfrac{3}{5}\\*~\\*~\\*=(-1)(\dfrac{3}{5})\\*~\\*~\\*=(\dfrac{-1}{~1})(\dfrac{3}{5})\\*~\\*~\\*=\dfrac{-3}{~5}$
Placing the negative sign before the entire fraction (subtracting the fraction) is equivalent to adding the same fraction, but with a negative numerator. But this is not the only option…
Recall how equivalent fractions are created: multiply the original fraction by a fraction that equals one, where numerator and denominator have the same value. If we multiply the above result by 1 in the form of a negative one divided by itself to create another equivalent fraction:
$\dfrac{-3}{~5}\\*~\\*~\\*=(\dfrac{-1}{-1})(\dfrac{-3}{~5})\\*~\\*~\\*=\dfrac{~3}{-5}$
we end up with the negative sign in the denominator. Summarizing these equivalent fraction results:
$-\dfrac{3}{5}~=~\dfrac{-3}{~5}~=~\dfrac{~3}{-5}$
As long as there is only one negative sign, either in front of the fraction, or in the numerator, or in the denominator, the fraction represents a negative quantity. As the examples above illustrate, you are welcome to move the negative sign around from where it is to either of the other two positions… whichever is most convenient for you.
If there are two fractions being subtracted, and you move the subtraction sign into the numerator or denominator of the fraction being subtracted, put a plus sign where the subtraction sign was (subtracting is the same as adding a negative):
$\dfrac{3}{5}~-~\dfrac{x}{5}$
$=~\dfrac{3}{5}~+~\dfrac{-x}{5}$
$=~\dfrac{3-x}{5}$
However, note that the negative sign must be applied to the entire numerator, or the entire denominator. So, in cases with more than one term in the numerator or denominator, the negative sign will have to be distributed if it is moved to the numerator or denominator:
$-\dfrac{3-x}{x-5}~=~\dfrac{-(3-x)}{x-5}~=~\dfrac{-3+x}{x-5}~=~\dfrac{x-3}{x-5}$
or
$-\dfrac{3-x}{x-5}~=~\dfrac{3-x}{-(x-5)}~=~\dfrac{3-x}{-x+5}~=~\dfrac{3-x}{5-x}$
If the last step in the above two examples surprised or puzzled you, my post on Negative Differences may help clear up the confusion.
If you have other questions (like the one at the top of this post), please ask them in a comment!
Published
Categorized as Concepts
## By Whit Ford
Math tutor since 1992. Former math teacher, product manager, software developer, research analyst, etc.
1. RJKS says:
where did the #1 come from in your problems?
1. If I am interpreting your question correctly, you are asking where the (-1) came from in the second line of the first example above. There are a couple of principles involved:
a) 1 is the multiplicative identity, which means that any quantity can be multiplied by 1 without changing it. Therefore, a factor of (1) can be introduced into ANY product without changing the result, or any number can be represented as the product of 1 and itself. For example, 3 = (1)(3) or (2)(3) = (2)(1)(3).
b) The additive “opposite” of a number is (-1) times the number. Additive opposites always add to zero. So, (3) + (-3) = 0 … the first number (3), multiplied by negative one (-1)(3), produces its additive opposite (-3). Added together they result in a sum of zero. Or, (-5) + (5) = 0 … the first number (-5), multiplied by negative one (-1)(-5), produces its additive opposite (5). Added together they result in a sum of zero.
So, the first example in the post above starts with a negative fraction, which using the ideas from the explanation above, can be described as (-1)(3/5) (the additive opposite of positive 3/5). A negative sign in front of the fraction can be interpreted as “multiply the positive fraction by (-1) to make it a negative number” if you wish, since (-1)(3/5) = – (3)/(5) = (-3)/(5) = (3)/(-5).
1. RJKS says:
yes, it very much does! ( thank you very much( It also surprised me how quickly you answered that))
2. I am in 7th grade and I want to understand why if you have 5-(-9) that it equals 14.
1. Subtraction is a pain, because it is not commutative or associative like addition:
3 – 2 does NOT = 2 – 3
4 – 3 – 2 does NOT = 4 – (3 – 2)
To get around this problem, and make subtraction problems easier to re-arrange without changing their result, we can rewrite subtraction as the addition of a negative:
3 – 2
= 3 + (-2)
= (-2) + 3 using the commutative property of addition
Negative 2 is the “additive opposite” of 2. When “additive opposites” are added together, the result will always be zero. In other words, if I add a number (any number) to its negative, the result MUST always be zero. Thinking in terms of a number line, a number and its negative will always be on opposite sides of zero, and at equal distances from zero. “Negating” a number will flip it from one side of zero to the other on the number line. If I negate a number twice, it flips to the other size of zero twice, which puts it right back where it started.
So, to answer your question using a number line, 9 is located nine units to the RIGHT of zero. Negating it, by putting a “-” in front of it, flips it to the other side of zero so that (-9) is nine units to the LEFT of zero. Negating that quantity, -(-9), flips it back so that it is now nine units to the RIGHT of zero, or equal to the original quantity: +9.
5 – (-9)
= 5 + (- (-9)) After changing subtraction into the addition of a negative
= 5 + 9 After using the reasoning above to change a double negative into a positive
= 14
I hope that helps!
1. So, if it is subtraction and a negitive has a () around it, it flips to addition and it turns positive?
2. The parentheses are only there to help clarify things. Two negative signs in a row cancel each other out, with or without parentheses, as long as no other variable or number is between the two negatives:
$-~-3~=~3\\* -(-3)~=~3\\* (-(-3))~=~3$
$5-3~=~2$
$5-~-3~=~8\\* 5-(-3)~=~8$
2. 5 – (-9)
= 5 + (- (-9)) After changing subtraction into the addition of a negative
= 5 + 9 After using the reasoning above to change a double negative into a positive
= 14
3. Kavitha says:
Fractions are of the form p/q where p and are natural numbers and q is not equal to zero. If so, why negative numbers is used in fractions
4. Brian Riggs says:
Thanks! Negative sign on a fraction help. Very short and simple. 50 years old, studying for college Accurate test. God Bless!
5. Mia :) says:
if there’s only one negative fraction, will my answer be positive or negative?
1. I’m not sure I understand your question. Do you mean something like -3/4? Or are you asking about a situation that has multiple terms, one of which is a negative fraction?
6. The equation is y = 7 – 3/5x, find slope. Obviously slope is -3/5 but in the question on the paper, the (-) is in front of 3/5 making the negative to be able to be either on the 3 or the 5. I think this would affect the slope majorly but I am not sure, I was wondering if it would affect it and how would one know which number the negative sign is on whether the 3 or the 5.
1. Think of the slope as Rise / Run.
If you start from the origin (0, 0), Rise by -3, then Run by +5, you will end up at (5, -3) and the slope from the origin to that point is -3/5.
Now start from (5, -3), and this time put the negative sign by the 5, so Rise by 3, then Run by -5… you are back at the origin. This tells you that the two slopes are equivalent, since they both keep you on the same line.
So a a slope of (-3)/(5) = (3)/(-5) = – (3/5). No matter which version you use, if you move from one point on the line using that slope you will end up at another point that DOES lie on the line… although perhaps in the other direction from the starting point.
7. Lori says:
Will the answer always be negative when there is a negative fraction with a negative exponent?
8. Charley says:
Which fraction is equivalent to negative 3 over 7 ?
1. (-3)/7 = 3/(-7) = (-)(3/7) All three are equivalent
9. Debra Tate says:
How do we solve:
4 +3squared ÷(-1)(4)(2) -1
1. Debra,
What you typed can be simplified, but not solved, as it does not include a variable or an equal sign with something on the other side.
I am unsure whether you intended the expression to be a single fraction, as in
$\dfrac{4+3^2}{(-1)(4)(2)-1}$
Or interpreted as your calculator would
$4 + \dfrac{3^2}{-1}\cdot (4)(2) - 1$
They will have different results
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Question Video: Differentiating Logarithmic Functions Using the Chain Rule | Nagwa Question Video: Differentiating Logarithmic Functions Using the Chain Rule | Nagwa
# Question Video: Differentiating Logarithmic Functions Using the Chain Rule Mathematics • Third Year of Secondary School
## Join Nagwa Classes
Find ππ¦/ππ₯, given that π¦ = ln(π₯Β² + 7).
01:53
### Video Transcript
Find ππ¦ by ππ₯, given that π¦ equals the natural logarithm of π₯ squared plus seven.
π¦ is the composition of functions. Itβs the composition of the natural logarithm function and the function which takes π₯ to π₯ squared plus seven. And so this is a natural candidate for the chain rule. Let π§ equal π₯ squared plus seven. Then, π¦ equals the natural logarithm of π§.
Now, the chain rule tells us that ππ¦ by ππ₯ is ππ¦ by ππ§ times ππ§ by ππ₯. Letβs apply this rule to our problem. We need to find ππ¦ by ππ§, which is the derivative of the natural logarithm of π§ with respect to π§. And the derivative of the natural logarithm function is the reciprocal function. So ππ¦ by ππ§ is one over π§.
Now, how about ππ§ by ππ₯? Well, π§ is π₯ squared plus seven. And differentiating this with respect to π₯, we get two π₯. We can write this as a fraction as two π₯ over π§. But weβre not quite done yet because we have ππ¦ by ππ₯ in terms of both π₯ and π§ and really would like it in terms of π₯ alone if thatβs possible. And it is possible. We can substitute π₯ squared plus seven for π§. And doing this, we get our final answer two π₯ over π₯ squared plus seven.
This is a special case of the more general rule that the derivative of the logarithm of a function π is the derivative of that function π prime divided by the function π. This is a very useful rule and it pops up now and again. And it is itself a special case of the more general chain rule.
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# Area - Mathematics Form 1 Notes
## Introduction
### Units of Areas
• The area of a plane shape is the amount of the surface enclosed within its boundaries. It is normally measured in square units. For example, a square of sides 5 cm has an area of
5 x 5 = 25 cm
• A square of sides 1 m has an area of 1 m, while a square of side 1 km has an area of 1 km
### Conversion of Units of Area
1 m² =1 mx 1 m
= 100 cm x 1 00 cm
= 10 000 cm²
1 km² = 1 km x 1 km
= 1 000m x 1 000 m
=1 000 000 m²
1 are = 10 m x 10 m
=100m²
1 hectare (ha) = 100 Ares
=10000 m²
## Area of a Regular Plane Figures
### Areas of a Rectangle
Area, A = 5 x 3 cm
=15
m2
Hence, the area of the rectangle, A =L X W square units, where l is the length and b breadth.
### Area of a Triangle
Area of a triangle
A =1/2bh square units
### Area of a Parallelogram
Area = 1/2bh +1/2bh
=bh square units
Note:
• This formulae is also used for a rhombus
### Area of a Trapezium
The figure below shows a trapezium in which the parallel sides are a units and b units, long. The perpendicular distance between the two parallel sides is h units.
Area of a triangle ABD =1/2 ah square units
Area of triangle DBC = ½ bh square units
Therefore area of trapezium ABCD =1/2 ah +1/2 bh
= 1/2h (a + b) square units.
Thus, the area of a trapezium is given by a half the sum of the length of parallel sides multiplied by the perpendicular distance between them.
That is, area of trapezium =1/2 (a+b)h
### Area of a Circle
The area A of a circle of radius r is given by: A = πr2
#### The Area of a Sector
• A sector is a region bounded by two radii and an arc.
Suppose we want to find the area of the shaded part in the figure below
The area of the whole circle is πr²
The whole circle subtends 360ͦ at the centre.
Therefore, 360corresponds to πr²
1 ͦ corresponds to 1/360 ͦ x πr²
60 ͦ corresponds to 60 ͦ /360
ͦ x πr²
Hence, the area of a sector subtending an angle θ at the centre of the circle is given by
A = θ x πr2
360
0
Example
Find the area of the sector of a circle of radius 3 cm if the angle subtended at the centre is 140 ͦ (take π=22/7)
Solution
Area A of a sector is given by
A = θ/360 x πr2
Here, r = 3 cm and θ =1400
Therefore, A= 140/3600 x 22/7 x 3 x 3
= 11 cm²
Example
The area of a sector of a circle is 38.5 cm². Find the radius of the circle if the angle subtended at the centre is 900 (Take π = 22/7)
Solution
From the formula a = θ/360 x πr², we get 90/360 x 22/7 x r² = 38.5
Therefore, r² =
38.5 x 360 x 7
90 x 22
Thus, r = 7
Example
The area of a circle radius 63 cm is 41 58 cm². Calculate the angle subtended at the centre of the circle. (Take π=22/7)
Solution
Using a =θ/360 x πr²,
θ=
4158 x 7 x 360
22 x 63 x 63
1200
## Surface Area of Solids
• Consider a cuboid ABCDEFGH shown in the figure below. If the cuboid is cut through a plane parallel to the ends, the cut surface has the same shape and size as the end faces. PQRS is a plane. The plane is called the cross-section of the cuboid
• A solid with uniform cross-section is called a prism. The following are some of the prisms. The following are some of the prisms.
• The surface area of a prism is given by the sum of the area of the surfaces.
• The figure below shows a cuboid of length l, breath b and height h. its area is given by;
A=2lb+2bh+2hl
=2(lb + bh + hl)
For a cube of side 2cm;
A =2(3x2²)
=24 cm²
Example
Find the surface area of a triangular prism shown below.
Area of the triangular surfaces = ½ x 5 x 12 x 2cm²
=60 cm²
Area of the rectangular surfaces=20 x 13 +5 x 20 +12 x 20
=260 + 100 + 240 = 600cm²
Therefore, the total surface area= (60+600) cm²
=660 cm²
### Cylinder
• A prism with a circular cross-section is called a cylinder, see the figure below.
• If you roll a piece of paper around the curved surface of a cylinder and open it out, you will get a rectangle whose breath is the circumference and length is the height of the cylinder. The ends are two circles. The surface area S of a cylinder with base and height h is therefore given by;
S=2πrh + 2πr²
Example
Find the surface area of a cylinder whose radius is 7.7 cm and height 12 cm.
Solution
S =2π(7.7) x 12 + 2π(7.7)² cm²
=2 π(7.7) x 12 + (7.7)² cm²
=2 x 7.7π (12 + 7.7) cm²
=2 x 7.7 x π(1 9.7) cm²
=15.4π (19.7) cm²
=953.48 cm²
## Area of Irregular Shapes
The area of irregular shape cannot be found accurately, but it can be estimated. As follows;
1. Draw a grid of unit squares on the figure or copy the figure on such a grid, see the figure below
2. Count all the unit squares fully enclosed within the figure.
3. Count all partially enclosed unit squares and divide the total by two, i.e.., treat each one of them as half of a unit square.
4. The sum of the numbers in (ii) and (ii) gives an estimate of the areas of the figure.
From the figure, the number of full squares is 9
Number of partial squares= 18
Total number of squares = 9 + 18/2
=18
Approximate area = 18 sq. units.
## Past KCSE Questions on the Topic
1. Calculate the area of the shaded region below, given that AC is an arc of a circle centre B. AB=BC=14cm CD=8cm and angle ABD = 750 (4 mks)
2. The scale of a map is 1:50000. A lake on the map is 6.16cm2. find the actual area of the lake in hactares. (3mks)
3. The figure below is a rhombus ABCD of sides 4cm. BD is an arc of circle centre C. Given that ∠ABC = 1380. Find the area of shaded region. (3mks)
4. The figure below sows the shape of Kamau’s farm with dimensions shown in meters
Find the area of Kamau’s farm in hectares (3mks)
5. In the figure below AB and AC are tangents to the circle centre O at B and C respectively, the angle AOC = 600
Calculate
1. The length of AC
6. The figure below shows the floor of a hall. A part of this floor is in the shape of a rectangle of length 20m and width 16m and the rest is a segment of a circle of radius 12m. Use the figure to find
1. The size of angle COD (2mks)
2. The area of figure DABCO (4mks)
3. Area of sector ODC (2mks)
4. Area of the floor of the house. (2mks)
7. The circle below whose area is 1 8.05cm2 circumscribes a triangle ABC where AB = 6.3cm, BC = 5.7cm and AC = 4.8cm. Find the area of the shaded part
8. In the figure below, PQRS is a rectangle in which PS=1 0k cm and PQ = 6k cm. M and N are midpoints of QR and RS respectively
1. Find the are of the shaded region (4 marks)
2. Given that the area of the triangle MNR = 30 cm2. find the dimensions of the rectangle (2 marks)
3. Calculate the sizes of angles and giving your answer to 2 decimal places (4 marks)
9. The figure below shows two circles each of radius 10.5 cm with centres A and B. the circles touch each other at T.
Given that angle XAD =angle YBC = 1600 and lines XY, ATB and DC are parallel, calculate the area of:
1. The minor sector AXTD (2 marks)
2. Figure AXYBCD (6marks)
3. The shaded region (2 marks)
10. The floor of a room is in the shape of a rectangle 1 0.5 m long by 6 m wide. Square tiles of length 30 cm are to be fitted onto the floor.
1. Calculate the number of tiles needed for the floor.
2. A dealer wishes to buy enough tiles for fifteen such rooms. The tiles are packed in cartons each containing 20 tiles. The cost of each carton is Kshs. 800. Calculate
1. the total cost of the tiles.
2. If in addition, the dealer spends Kshs. 2,000 and Kshs. 600 on transport and subsistence respectively, at what price should he sell each carton in order to make a profit of 12.5% (Give your answer to the nearest Kshs.)
11. The figure below is a circle of radius 5cm. Points A, B and C are the vertices of the triangle ABC in which ∠ABC = 60o and ∠ACB=50o which is in the circle. Calculate the area of ∠ABC
12. Mr.Wanyama has a plot that is in a triangular form. The plot measures 1 70m, 1 90m and 21 0m, but the altitudes of the plot as well as the angles are not known. Find the area of the plot in hectares
13. Three sirens wail at intervals of thirty minutes, fifty minutes and thirty five minutes. If they wail together at 7.18a.m on Monday, what time and day will they next wail together?
14. A farmer decides to put two-thirds of his farm under crops. Of this, he put a quarter under maize and four-fifths of the remainder under beans. The rest is planted with carrots. If 0.9acres are under carrots, find the total area of the farm
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AC and DC are used widely in physics. AC stands for alternating current while DC stands for direct current. AC and DC are the types of electric current flow.
In this article, we will learn about the definition of AC and DC, the conversion of AC and DC & DC to AC along a lot of examples. We shall also discuss the applications of AC and DC also the difference between AC and DC.
## What are AC and DC?
AC stands for alternating current and is defined as the electric current which periodically reverses direction and changes its magnitude continuously with time. It can also be defined as the electric charge changing the direction periodically. SI unit is volt-amperes.
DC stands for direct currentand is defined as an electric current that has the same direction and magnitude all the time. It is also defined as the electric charge that flows in one direction. SI unit is volt-amperes.
## How to convert AC to DC?
To convert alternating current in the direct current we use formula,
DC = 0.636 x AC
Here, AC = Alternating current, DC = Direct current, and 0.636 = constant. To get the accurate results, you can use AC to DC Calculator.
Let us take some examples in order to understand this conversion.
Example 1
Convert alternating current to direct current if AC = 275 volts?
Solution
Step 1:write down the formula.
DC = 0.636 x AC
Step 2:Put the value of AC in the formula.
DC = 0.636 x AC
DC = 0.636 x 275
DC = 174.9 volts
Example 2
Convert alternating current to direct current if AC = 2670 volts?
Solution
Step 1:write down the formula.
DC = 0.636 x AC
Step 2:Put the value of AC in the formula.
DC = 0.636 x AC
DC = 0.636 x 2670
DC = 1698.12 volts
Example 3
Convert alternating current to direct current if AC = 0.2670 volts?
Solution
Step 1:write down the formula.
DC = 0.636 x AC
Step 2:Put the value of AC in the formula.
DC = 0.636 x AC
DC = 0.636 x 0.2670
DC = 0.169812 volts
## How to convert DC to AC?
To convert direct current in alternating current we use formula,
AC = DC / 0.636
Here, AC = Alternating current, DC = Direct current, and 0.636 = constant.
Let us take some examples in order to understand this conversion.
Example 1
Convert direct current in the alternating current if DC = 75 volts?
Solution
Step 1:write down the formula.
AC = DC / 0.636
Step 2:Put the value of DC in the formula.
AC = DC / 0.636
AC = 75 / 0.636
AC = 117.92 volts
Example 2
Convert direct current in the alternating current if DC = 6270 volts?
Solution
Step 1:write down the formula.
AC = DC / 0.636
Step 2:Put the value of DC in the formula.
AC = DC / 0.636
AC = 6270 / 0.636
AC = 9858.49 volts
Example 3
Convert direct current in the alternating current if DC = 0.2670 volts?
Solution
Step 1:write down the formula.
AC = DC / 0.636
Step 2:Put the value of AC in the formula.
AC = DC / 0.636
AC = 0.2670 / 0.636
DC = 0.4198 volts
## Difference between AC and DC
• In Ac,currentmodifications its pathduring flow whilein DC,current does now no longermodifications its pathduring flow and staysconstant.
• The AC has a frequency that suggests how generally the path of current flow changes during flow while the frequency of the direct current is 0because it doesnot change the path of flow.
• The electricityelement of AC is zeroto at least onewhile DC is Constant Zero.
• The AC is generated via (by) the alternator whilst DC is generated via Photovoltaic cells, generators, and batteries.
• The AC load may be capacitive, inductive, or resistive however the load on DC is usually resistive.
• The DC graph has a constant line displayingvalue and the path is constantwhilst the AC may be a sinusoidal wave, rectangular wave, or triangular wave.
• The AC transformed into DC with the use of a tool named rectifier whilst the DC transformed into AC named inverter.
• AC is broadlyutilized inthe businessdevice and purchaser electronics like AC, Freezer, Cooler, washing machine, lights, fans While DC is utilized indigitaldevices and small gadgets like clocks, laptops, mobile phones, Sensors.
• Ac may be transmitted over a lengthy distance with a few losses whilst DC may be transmitted very long distance with very low loss the use of HVDC.
## Applications of AC and DC
AC and DC have some applications.
### Applications of AC
• AC is used for long-distance transmission for Offices and Homes.
• Energy Loss in AC is much less so extensivelyutilized in transmission.
• The AC may betransformedright into anexcessive voltage to low voltage and occasional to excessive voltage successfullythe use of the transformer.
• AC energy is utilized inlargepackages and home equipment like Freezers, AC. Dishwashers, washing machines, Fans, Bulbs.
### Applications of DC
• DC is extensivelyutilized in small digitaldevices and gadgets.
• DC isn’tproper for long-distance transmission howevergarage of DCis simple in the shape of a Battery.
• DC energy is utilized in Cell phones, laptops, radio, and differentdigital gadgets.
• DC Current isutilized in flashlights.
• DC are utilized in EV and hybrid motors and automobile.
## Summary
AC and DC are widely used in the flow of current. AC and DC are used in a large number of appliances used in our daily life. AC stands for alternating current while DC stands for direct current. In AC electric current reverses the direction periodically while in DC electric current the electric current remains the same.
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# Solve the following equation $\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
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Hint: Rearrange the given equation such that the unknown terms arranged on one side and all other terms are arranged on other side and then compare both sides and with further simplification we can determine the value of unknown terms.
Given: $\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
To find: the value of ‘x’
Step 1: Firstly we will determine the number of unknown terms in the given equation.
$\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
So here we have only one unknown term so we require only one equation to determine the value of ‘x’
Step 2: multiply both sides of the equation with 2 such that all the denominator part get reduced and we get a simple linear equation
$\dfrac{{x - 1}}{2} - x + 13 = 5 - x$
$2 \times (\dfrac{{x - 1}}{2} - x + 13) = (5 - x) \times 2$
Opening the bracket and we get
$(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2$
Step 3: rearranging the terms such that the unknown term arranged on one side and all other terms are arranged on other side, that is
$(x - 1) - 2 \times x + 2 \times 13 = 5 \times 2 - x \times 2$
$x - 2x - 1 + 26 = 10 - 2x$
$x - 2x + 2x = 10 + 1 - 26$
On further simplification, we get
$x - 2x + 2x = 10 + 1 - 26$
$x = - 15$
Hence, on solving the given equation we determined the value of ‘x’ and it is equal to $x = - 15$
Note: We have different solution methods for different types of equation
We can use the substitution method
We can use the elimination method without multiplication.
We can use the elimination method with multiplication.
There might be the possibility of infinite solution or no solution.
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LCM that 4 and 6 is the the smallest number amongst all usual multiples of 4 and also 6. The first few multiples that 4 and also 6 space (4, 8, 12, 16, 20, 24, . . . ) and (6, 12, 18, 24, . . . ) respectively. There room 3 generally used approaches to find LCM that 4 and also 6 - by listing multiples, by element factorization, and also by division method.
You are watching: Find the least common multiple of 4 and 6
1 LCM that 4 and also 6 2 List the Methods 3 Solved Examples 4 FAQs
Answer: LCM the 4 and 6 is 12.
Explanation:
The LCM of 2 non-zero integers, x(4) and y(6), is the smallest optimistic integer m(12) that is divisible through both x(4) and y(6) without any type of remainder.
The approaches to discover the LCM that 4 and also 6 are defined below.
By Listing MultiplesBy element Factorization MethodBy department Method
### LCM that 4 and also 6 through Listing Multiples
To calculation the LCM the 4 and also 6 by listing out the common multiples, we deserve to follow the given listed below steps:
Step 1: list a couple of multiples of 4 (4, 8, 12, 16, 20, 24, . . . ) and 6 (6, 12, 18, 24, . . . . )Step 2: The common multiples native the multiples the 4 and 6 are 12, 24, . . .Step 3: The smallest common multiple the 4 and 6 is 12.
∴ The least common multiple the 4 and 6 = 12.
### LCM the 4 and 6 by prime Factorization
Prime administer of 4 and also 6 is (2 × 2) = 22 and (2 × 3) = 21 × 31 respectively. LCM of 4 and 6 have the right to be acquired by multiply prime determinants raised to their respective greatest power, i.e. 22 × 31 = 12.Hence, the LCM that 4 and 6 by prime factorization is 12.
### LCM the 4 and also 6 by department Method
To calculate the LCM of 4 and also 6 by the division method, we will divide the numbers(4, 6) by their prime components (preferably common). The product of these divisors provides the LCM that 4 and 6.
Step 3: proceed the procedures until only 1s space left in the last row.
The LCM of 4 and 6 is the product of all prime numbers on the left, i.e. LCM(4, 6) by division method = 2 × 2 × 3 = 12.
## FAQs top top LCM the 4 and also 6
### What is the LCM of 4 and also 6?
The LCM the 4 and also 6 is 12. To find the least usual multiple (LCM) that 4 and 6, we require to discover the multiples of 4 and 6 (multiples the 4 = 4, 8, 12, 16; multiples that 6 = 6, 12, 18, 24) and also choose the smallest multiple the is specifically divisible through 4 and also 6, i.e., 12.
### If the LCM of 6 and also 4 is 12, uncover its GCF.
LCM(6, 4) × GCF(6, 4) = 6 × 4Since the LCM the 6 and also 4 = 12⇒ 12 × GCF(6, 4) = 24Therefore, the GCF = 24/12 = 2.
### What is the Relation between GCF and also LCM that 4, 6?
The adhering to equation can be used to express the relation between GCF and LCM of 4 and also 6, i.e. GCF × LCM = 4 × 6.
See more: What Is A Female Eagle Called ? What Is A Mother Eagle Called
### What is the least Perfect Square Divisible through 4 and 6?
The least number divisible by 4 and 6 = LCM(4, 6)LCM of 4 and also 6 = 2 × 2 × 3 ⇒ the very least perfect square divisible by every 4 and 6 = LCM(4, 6) × 3 = 36 Therefore, 36 is the compelled number.
### What space the techniques to uncover LCM of 4 and also 6?
The generally used techniques to find the LCM that 4 and also 6 are:
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# Question #0ed08
May 5, 2016
$\left(x , y\right) = \left(- 4 , 1\right)$
#### Explanation:
Performing the multiplication, first, we have
${A}^{2} = {\left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right)}^{2} = \left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right) \left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right) = \left(\begin{matrix}11 & 8 \\ 4 & 3\end{matrix}\right)$
$x A = x \left(\begin{matrix}3 & 2 \\ 1 & 1\end{matrix}\right) = \left(\begin{matrix}3 x & 2 x \\ x & x\end{matrix}\right)$
$y I = y \left(\begin{matrix}1 & 0 \\ 0 & 1\end{matrix}\right) = \left(\begin{matrix}y & 0 \\ 0 & y\end{matrix}\right)$
Then, summing, we have
${A}^{2} + x A + y I = \left(\begin{matrix}11 & 8 \\ 4 & 3\end{matrix}\right) + \left(\begin{matrix}3 x & 2 x \\ x & x\end{matrix}\right) + \left(\begin{matrix}y & 0 \\ 0 & y\end{matrix}\right)$
$= \left(\begin{matrix}3 x + y + 11 & 2 x + 8 \\ x + 4 & x + y + 3\end{matrix}\right) = \left(\begin{matrix}0 & 0 \\ 0 & 0\end{matrix}\right)$
Equating entries with corresponding indices, we get the following system of equations from the bottom row:
$\left\{\begin{matrix}x + 4 = 0 \\ x + y + 3 = 0\end{matrix}\right.$
From the first equation, we have $x = - 4$
Substituting this into the second equation, we get
$- 4 + y + 3 = 0$
$\implies y = 1$
If we substitute these values in for $x$ and $y$ we find that they fulfill the given equation, and thus we have $\left(x , y\right) = \left(- 4 , 1\right)$
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# S.2 Confidence Intervals
S.2 Confidence Intervals
Let's review the basic concept of a confidence interval.
Suppose we want to estimate an actual population mean $$\mu$$. As you know, we can only obtain $$\bar{x}$$, the mean of a sample randomly selected from the population of interest. We can use $$\bar{x}$$ to find a range of values:
$\text{Lower value} < \text{population mean}\;\; \mu < \text{Upper value}$
that we can be really confident contains the population mean $$\mu$$. The range of values is called a "confidence interval."
## Example S.2.1
### Should using a hand-held cell phone while driving be illegal?
There is little doubt that you have seen numerous confidence intervals for population proportions reported in newspapers over the years.
For example, a newspaper report (ABC News poll, May 16-20, 2001) was concerned about whether or not U.S. adults thought using a hand-held cell phone while driving should be illegal. Of the 1,027 U.S. adults randomly selected for participation in the poll, 69% believed it should be illegal. The reporter claimed that the poll's "margin of error" was 3%. Therefore, the confidence interval for the (unknown) population proportion p is 69% ± 3%. That is, we can be really confident that between 66% and 72% of all U.S. adults think using a hand-held cell phone while driving a car should be illegal.
## General Form of (Most) Confidence Intervals
The previous example illustrates the general form of most confidence intervals, namely:
$\text{Sample estimate} \pm \text{margin of error}$
The lower limit is obtained by:
$\text{the lower limit L of the interval} = \text{estimate} - \text{margin of error}$
The upper limit is obtained by:
$\text{the upper limit U of the interval} = \text{estimate} + \text{margin of error}$
Once we've obtained the interval, we can claim that we are really confident that the value of the population parameter is somewhere between the value of L and the value of U.
So far, we've been very general in our discussion of the calculation and interpretation of confidence intervals. To be more specific about their use, let's consider a specific interval, namely the "t-interval for a population mean µ."
### (1-α)100% t-interval for the population mean $$\mu$$
If we are interested in estimating a population mean $$\mu$$, it is very likely that we would use the t-interval for a population mean $$\mu$$.
t-Interval for a Population Mean
The formula for the confidence interval in words is:
$\text{Sample mean} \pm (\text{t-multiplier} \times \text{standard error})$
and you might recall that the formula for the confidence interval in notation is:
$\bar{x}\pm t_{\alpha/2, n-1}\left(\dfrac{s}{\sqrt{n}}\right)$
#### Note that:
• the "t-multiplier," which we denote as $$t_{\alpha/2, n-1}$$, depends on the sample size through n - 1 (called the "degrees of freedom") and the confidence level $$(1-\alpha)\times100%$$ through $$\frac{\alpha}{2}$$.
• the "standard error," which is $$\frac{s}{\sqrt{n}}$$, quantifies how much the sample means $$\bar{x}$$ vary from sample to sample. That is, the standard error is just another name for the estimated standard deviation of all the possible sample means.
• the quantity to the right of the ± sign, i.e., "t-multiplier × standard error," is just a more specific form of the margin of error. That is, the margin of error in estimating a population mean µ is calculated by multiplying the t-multiplier by the standard error of the sample mean.
• the formula is only appropriate if a certain assumption is met, namely that the data are normally distributed.
Clearly, the sample mean $$\bar{x}$$, the sample standard deviation s, and the sample size n are all readily obtained from the sample data. Now, we need to review how to obtain the value of the t-multiplier, and we'll be all set.
#### How is the t-multiplier determined?
As the following graph illustrates, we put the confidence level $1-\alpha$ in the center of the t-distribution. Then, since the entire probability represented by the curve must equal 1, a probability of α must be shared equally among the two "tails" of the distribution. That is, the probability of the left tail is $\frac{\alpha}{2}$ and the probability of the right tail is $\frac{\alpha}{2}$. If we add up the probabilities of the various parts $(\frac{\alpha}{2} + 1-\alpha + \frac{\alpha}{2})$, we get 1. The t-multiplier, denoted $$t_{\alpha/2}$$, is the t-value such that the probability "to the right of it" is $\frac{\alpha}{2}$:
It should be no surprise that we want to be as confident as possible when we estimate a population parameter. This is why confidence levels are typically very high. The most common confidence levels are 90%, 95%, and 99%. The following table contains a summary of the values of $$\frac{\alpha}{2}$$ corresponding to these common confidence levels. (Note that the"confidence coefficient" is merely the confidence level reported as a proportion rather than as a percentage.)
Confidence levels
Confidence Coefficient $(1-\alpha)$ Confidence Level $(1-\alpha) \times 100$ $(1-\dfrac{\alpha}{2})$ $\dfrac{\alpha}{2}$
0.90 90% 0.95 0.05
0.95 95% 0.975 0.025
0.99 99% 0.995 0.005
## Minitab® – Using Software
The good news is that statistical software, such as Minitab, will calculate most confidence intervals for us.
Let's take an example of researchers who are interested in the average heart rate of male college students. Assume a random sample of 130 male college students were taken for the study.
The following is the Minitab Output of a one-sample t-interval output using this data.
### One-Sample T: Heart Rate
#### Descriptive Statistics
N Mean StDev SE Mean 95% CI for $\mu$
130 73.762 7.062 0.619 (72.536, 74.987)
$\mu$: mean of HR
In this example, the researchers were interested in estimating $$\mu$$, the heart rate. The output indicates that the mean for the sample of n = 130 male students equals 73.762. The sample standard deviation (StDev) is 7.062 and the estimated standard error of the mean (SE Mean) is 0.619. The 95% confidence interval for the population mean $\mu$ is (72.536, 74.987). We can be 95% confident that the mean heart rate of all male college students is between 72.536 and 74.987 beats per minute.
## Factors Affecting the Width of the t-interval for the Mean $\mu$
Think about the width of the interval in the previous example. In general, do you think we desire narrow confidence intervals or wide confidence intervals? If you are not sure, consider the following two intervals:
• We are 95% confident that the average GPA of all college students is between 1.0 and 4.0.
• We are 95% confident that the average GPA of all college students is between 2.7 and 2.9.
Which of these two intervals is more informative? Of course, the narrower one gives us a better idea of the magnitude of the true unknown average GPA. In general, the narrower the confidence interval, the more information we have about the value of the population parameter. Therefore, we want all of our confidence intervals to be as narrow as possible. So, let's investigate what factors affect the width of the t-interval for the mean $$\mu$$.
Of course, to find the width of the confidence interval, we just take the difference in the two limits:
Width = Upper Limit - Lower Limit
What factors affect the width of the confidence interval? We can examine this question by using the formula for the confidence interval and seeing what would happen should one of the elements of the formula be allowed to vary.
$\bar{x}\pm t_{\alpha/2, n-1}\left(\dfrac{s}{\sqrt{n}}\right)$
What is the width of the t-interval for the mean? If you subtract the lower limit from the upper limit, you get:
$\text{Width }=2 \times t_{\alpha/2, n-1}\left(\dfrac{s}{\sqrt{n}}\right)$
Now, let's investigate the factors that affect the length of this interval. Convince yourself that each of the following statements is accurate:
• As the sample mean increases, the length stays the same. That is, the sample mean plays no role in the width of the interval.
• As the sample standard deviation s decreases, the width of the interval decreases. Since s is an estimate of how much the data vary naturally, we have little control over s other than making sure that we make our measurements as carefully as possible.
• As we decrease the confidence level, the t-multiplier decreases, and hence the width of the interval decreases. In practice, we wouldn't want to set the confidence level below 90%.
• As we increase the sample size, the width of the interval decreases. This is the factor that we have the most flexibility in changing, the only limitation being our time and financial constraints.
### In Closing
In our review of confidence intervals, we have focused on just one confidence interval. The important thing to recognize is that the topics discussed here — the general form of intervals, determination of t-multipliers, and factors affecting the width of an interval — generally extend to all of the confidence intervals we will encounter in this course.
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Solving inverse functions
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Solve inverse functions
Are you struggling with Solving inverse functions? In this post, we will show you how to do it step-by-step. In mathematics, "solving for x" refers to the process of finding the value of an unknown variable in an equation. In most equations, the variable is represented by the letter "x." Fractions can be used to solve for x in a number of ways. For example, if the equation is 2x + 1 = 7, one can isolated the x term by subtracting 1 from each side and then dividing each side by 2. This would leave x with a value of 3. In some cases, more than one step may be necessary to solve for x. For example, if the equation is 4x/3 + 5 = 11, one would first need to multiply both sides of the equation by 3 in order to cancel out the 4x/3 term. This would give 12x + 15 = 33. From there, one could subtract 15 from each side to find that x = 18/12, or 1.5. As these examples demonstrate, solving for x with fractions is a matter of careful algebraic manipulation. With a little practice, anyone can master this essential math skill.
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# OBJECTIVE AFTER STUDYING THIS SECTION, YOU WILL BE ABLE TO FIND THE SURFACE AREAS OF CIRCULAR SOLIDS 12.3 Surface Areas of Circular Solids.
## Presentation on theme: "OBJECTIVE AFTER STUDYING THIS SECTION, YOU WILL BE ABLE TO FIND THE SURFACE AREAS OF CIRCULAR SOLIDS 12.3 Surface Areas of Circular Solids."— Presentation transcript:
OBJECTIVE AFTER STUDYING THIS SECTION, YOU WILL BE ABLE TO FIND THE SURFACE AREAS OF CIRCULAR SOLIDS 12.3 Surface Areas of Circular Solids
Cylinders A cylinder resembles a prism in having two congruent parallel bases. The bases are circles. If we look at the net of a cylinder, we can see two circles and a rectangle. The circumference of the circle is the length of the rectangle and the height is the width.
Theorem The lateral area of a cylinder is equal to the product of the height and the circumference of the base where C is the circumference of the base, h is the height of the cylinder, and r is the radius of the base.
Definition The total area of a cylinder is the sum of the cylinder’s lateral area and the areas of the two bases.
Cone A cone resembles a pyramid but its base is a circle. The slant height and the lateral edge are the same in a cone. Slant height (italicized l) height radius
Theorem The lateral area of a cone is equal to one-half the product of the slant height and the circumference of the base where C is the circumference of the base, l is the slant height, and r is the radius of the base.
Definition The total area of a cone is the sum of the lateral area and the area of the base.
Sphere A sphere is a special figure with a special surface-area formula. (A sphere has no lateral edges and no lateral area).
Postulate where r is the sphere’s radius
Example 1 Find the total area of the figure 5 6
Example 2 Find the total area of the figure 5 6
Example 3 Find the total area of the figure 5
Summary Explain in your own words how to find the surface area of a cylinder? Homework: worksheet
Download ppt "OBJECTIVE AFTER STUDYING THIS SECTION, YOU WILL BE ABLE TO FIND THE SURFACE AREAS OF CIRCULAR SOLIDS 12.3 Surface Areas of Circular Solids."
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## Presentation on theme: ""— Presentation transcript:
Intervals of Continuity
Even though not all functions are continuous on all real numbers (i.e., everywhere), we can still talk about the continuity of a function in terms of intervals. As you may recall, there are three kinds of intervals: open intervals, closed intervals, half-open (or half-closed) intervals. Definition of f(x) is continuous on an open interval: We say a function f(x) is continuous on an open interval of (a, b) if there are no points of discontinuity on the interval (a, b). That is, f(x) is continuous at any x where a < x < b. We say a function f(x) is continuous on a closed interval of [a, b] if i. f(x) is continuous on the open interval (a, b), ii. f(a) and f(b) are defined, and iii. limxa f(x) = f(a) and limxb f(x) = f(b). We say a function f(x) is continuous on a half-open interval of [a, b) if i. f(x) is continuous on the open interval (a, b), and ii. f(a) is defined and limxa f(x) = f(a). We say a function f(x) is continuous on a half-open interval of (a, b] if i. f(x) is continuous on the open interval (a, b), and ii. ______________________________. b a a b b a b a b a + b a b a +
Continuous on its Domain
Recall that functions such as f(x) = x1/2 and f(x) = log x are not continuous on all real numbers, nevertheless, they are continuous at every number in their domains. For example, the domain of f(x) = x1/2 is [0, ) and its interval of continuity is also [0, ). Similarly, the domain of f(x) = log x is (0, ) and its interval of continuity is also (0, ). Another example is f(x) = 1/x, where its graph has two pieces and there is a vertical asymptote at x = 0. However, since its domain is all real numbers except 0, i.e., (, 0) (0, ), and we can see that the left piece of f(x) = 1/x is continuous on (, 0) where as the right piece is continuous on (0, ), we still say that the function is continuous on its domain. So what is an example of a function which is not continuous on its domain? There are many functions which are not continuous on their domains. For example, the integer function, f(x) = [x]. The domain is all real numbers (plug any real number x into the integer function, it will yield back the greatest integer ≤ x). However, the function is not continuous on its domain since it is discontinuous at every integer. Another example is the sign function, f(x) = sgn x = where given any real number x, this function will yield back 1 if x is positive, 1 if x is negative, and 0 is x is zero. As you can see, its domain is all real numbers, but it’s not continuous at x = 0.
How to Give the Intervals of Continuity of a Function
If a function is continuous on all real numbers, e.g., f(x) = x2, then we say its interval of continuity (IOC) is (, ). If a function is continuous on its domain, then its interval(s) of continuity is same as its domain. For example, i) f(x) = x1/2 IOC = [0, ); ii) f(x) = 1/x IOC = (, 0) (0, ). If a function is not continuous on its domain, e.g., f(x) = sgn x, we must say the IOC is (, 0) (0, ) despite that the domain is (, ). This implies the interval notation for the domain is not necessarily same as the intervals of continuity. Example For the function with graph below, give the domain and intervals of continuity using the least number of intervals as possible. Domain: ____________________________ IOC: _______________________________ 3 y = f(x) 2 1 –6 –5 –4 –3 –2 –1 1 2 3 4 5 6 7 8 9 10 11 –1 –2 –3 Domain: (-oo, -1) U (-1,5) U (5, oo) IOC: (-oo, -4] U (-4,-1) U (-1, 1)U (1, 3) U (3, 5) U (5, 8) U [8.oo)
Properties of Functions which are Continuous on their Domains
Theorem: The following types of functions are continuous at every number in their domains: i) polynomial functions ii) rational functions iii) root functions iv) trigonometric functions v) exponential functions vi) logarithmic functions vii) absolute-value functions viii) inverse trigonometric functions Properties: If f(x) and g(x) are two functions continuous at every number in their respective domains, and let D be the domain of f(x) + g(x) and E be the domain of f(g(x)), then: i)f(x) + g(x), f(x) – g(x) and f(x)∙g(x) are continuous on D, ii)f(x)/g(x) is continuous on D – {x | g(x) = 0}, and iii)f(g(x)) is continuous on E. Continuous on all real numbers Continuous only on their domains Polynomial Rational Odd-indexed root Even-indexed root sin, cos tan, cot, sec, csc tan–1, cot–1 sin–1, cos–1, sec–1, csc–1 Exponential Logarithmic Absolute-value Example Let f(x) = sin x and g(x) = x2 – 1 . Find the intervals of continuity of the following functions: 1. f(x) + g(x) 2. f(x) – g(x) 3. f(x)∙g(x) 4. f(g(x)) 5. g(f(x)) 6. f(x)/g(x) 7. g(x)/f(x)
How Do We Find the Limit of a Function Without the Graph?
There are several ways we can find the limit of a function f as x approaches a without knowing its graph. One obvious way is to use numbers close to a. See the following examples: 1. limx1 (x2 + 1) = ___ 2. limx2 1/x2 = ___ x 1.9 1.99 1.999 2 2.001 2.01 2.1 1/x2 x 0.9 0.99 0.999 1 1.001 1.01 1.1 x2 + 1 3. limx3 (x – 3)/|x – 3| = ___ 4. limx2 1/(x2 – 4) = ___ x 2.9 2.99 2.999 3 3.001 3.01 3.1 x 1.9 1.99 1.999 2 2.001 2.01 2.1 5. limx–1 5/(x + 1) = ___ 6. limx0 6 sin x/x = ___ (where x in radians) x –1.1 –1.01 –1.001 –1 –.999 –.99 –.9 x –.1 –.01 –.001 0.001 0.01 0.1 7. limx2– 7/(x – 2) = ___ 8. limx2+ 8/(x + 2) = ___ 9. limx∞ 3x/(2x – 9) = ___ x 1.9 1.99 1.999 2 x 2 2.001 2.01 2.1 x 100 1,000 10,000
Implication of Continuity on Limits
Recall that if f(x) is continuous at x = a, then the limit of f(x) as x approach a must exist and it must equal to f(a). Therefore, when we need to evaluate limxa f(x) and if we know f(x) is continuous at x = a, all we need to do is to evaluate f(a), i.e., whatever f(a) is, is the limit! Direct Substitution Property: If f(x) is continuous at x = a, then limxa f(x) = f(a). Rule of Thumb of Evaluating Limit When you evaluate the limit of a function f(x) as x approaches a, just plug in a into the f(x) first. That is, when limxa f(x) is asked, just do f(a). With this property, it allows us to evaluate the limit by using the so-called “plug-in” method. That is, as long as we know f(x) is continuous at a, even if we don’t know the graph of f(x), we can just evaluate f(a) and that will be limit! We don’t need to use any numbers close to a, hence, it is a much better and faster way of finding the limit than the “tabular” method used on page 5. Examples: 1. limx1 (x2 + 1) = 2 2. limx2 1/x2 = 1/22 = ¼ 3. limx3+ (x2 – 2)/(x + 1) = 7/4 4. limx– cos x = The rule of thumb above really is another way (actually, my way) to state the Direct Substitution Property. The difference is: here we don’t even need to care whether f(x) is continuous at a or not. By Direction Substitution Property By My Rule of Thumb Example 1: x2 + 1 is a quadratic function. Quadratic functions are a type of polynomial functions and polynomial functions are continuous everywhere, therefore it’s okay to plug the 1 into x, and obtain = 2 as the limit. Example 1: = 2 Example 2: 1/x2 is a rational function with a vertical asymptote at x = 0. So 1/x2 is discontinuous at x = 0 but it’s continuous elsewhere, therefore it’s okay to plug the 2 into x, and obtain 1/22 = ¼ as the limit. Example 2: 1/22 = 4
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# How do you find the derivative of y = e^cosh(2x)?
Jun 7, 2016
$\frac{d}{\mathrm{dx}} \left({e}^{\cosh \left(2 x\right)}\right) = {e}^{\cosh \left(2 x\right)} \sinh \left(2 x\right) 2$
#### Explanation:
$\frac{d}{\mathrm{dx}} \left({e}^{\cosh \left(2 x\right)}\right)$
Applying the chain rule, $\frac{\mathrm{df} \left(u\right)}{\mathrm{dx}} = \frac{\mathrm{df}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$
$L e t , \cosh \left(2 x\right) = u$
$= \frac{d}{\mathrm{du}} \left({e}^{u}\right) \frac{d}{\mathrm{dx}} \left(\cosh \left(2 x\right)\right)$
We know,
$\frac{d}{\mathrm{du}} \left({e}^{u}\right) = {e}^{u}$
$\frac{d}{\mathrm{dx}} \left(\cosh \left(2 x\right)\right) = \sinh \left(2 x\right) 2$
So,
$\frac{d}{\mathrm{dx}} \left(\cosh \left(2 x\right)\right) = \sinh \left(2 x\right) 2$
substituted back,$u = \cosh \left(2 x\right)$
we get,
${e}^{\cosh \left(2 x\right)} \sinh \left(2 x\right) 2$
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# Question Video: Finding the Probability of Getting an Odd Number When Rolling a Die Mathematics
For a single roll of a fair six-sided die, find the probability of getting an odd number.
02:20
### Video Transcript
For a single roll of a fair six-sided die, find the probability of getting an odd number.
We’ll draw this normal fair six-sided die. What would be the possible outcomes if we roll this die? We can roll a one, we can roll a two, we can roll a three or a four, or we can roll a five. Our last option would be a six. We’re trying to find the probability that on this row, a single row, we would get an odd number.
Finding the probability of rolling an odd number would be equal to finding all of the odd outcomes over the total outcomes. One is an odd number, three is an odd number, and so is five. So we can say for our odd outcomes, there would be three different odd outcomes that would be our numerator. And on a six-sided die, there would be six total outcomes. Our denominator for total outcomes would be six.
But we want to write our probability in simplest terms. And right now, we’ve written three over six. We recognize that three and six share a factor. We can divide the numerator and the denominator by three. Three divided by three equals one; six divided by three equals two. Our probability after we’ve simplified comes out to one-half. The probability of rolling an odd number for a single row of a fair sided die is one-half.
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# Synthetic Division of Polynomials
The volume of a rectangular prism is 2x3+5x2−x−6. Determine if 2x+3 is the length of one of the prism's sides.
# Synthetic Division
Synthetic division is an alternative to long division. It can also be used to divide a polynomial by a possible factor, x−k. However, synthetic division cannot be used to divide larger polynomials, like quadratics, into another polynomial.
Let's use synthetic division to divide 2x4−5x3−14x2+47x−30 by x−2.
Using synthetic division, the setup is as follows:
To “read” the answer, use the numbers as follows:
Therefore, 2 is a solution, because the remainder is zero. The factored polynomial is 2x3−x2−16x+15. Notice that when we synthetically divide by k, the “leftover” polynomial is one degree less than the original. We could also write
(x−2)(2x3−x2−16x+15)=2x4−5x3−14x2+47x−30.
Now, let's determine if 4 is a solution to f(x)=5x3+6x2−24x−16.
Using synthetic division, we have:
The remainder is 304, so 4 is not a solution. Notice if we substitute in x=4, also written f(4), we would have f(4)=5(4)3+6(4)2−24(4)−16=304. This leads us to the Remainder Theorem.
Remainder Theorem: If f(k)=r, then r is also the remainder when dividing by (x−k).
This means that if you substitute in x=k or divide by k, what comes out of f(x) is the same. r is the remainder, but it is also the corresponding y−value. Therefore, the point (k,r) would be on the graph of f(x).
Finally, let's determine if (2x−5) is a factor of 4x4−9x2−100.
If you use synthetic division, the factor is not in the form (x−k). We need to solve the possible factor for zero to see what the possible solution would be. Therefore, we need to put $$\ \frac{5}{2}$$ up in the left-hand corner box. Also, not every term is represented in this polynomial. When this happens, you must put in zero placeholders. In this problem, we need zeros for the x3−term and the x−term.
This means that $$\ \frac{5}{2}$$ is a zero and its corresponding binomial, (2x−5), is a factor.
Example 1
Earlier, you were asked to determine if 2x+3 is the length of one of the prism's sides.
Solution
If 2x+3 divides evenly into 2x3+5x2−x−6 then it is the length of one of the prism's sides.
If we want to use synthetic division, notice that the factor is not in the form (x−k). Therefore, we need to solve the possible factor for zero to see what the possible solution would be. If 2x+3=0 then x=$$\ -\frac{3}{2}$$. Therefore, we need to put $$\ -\frac{3}{2}$$ up in the left-hand corner box.
When we perform the synthetic division, we get a remainder of 0. This means that (2x+3) is a factor of the volume. Therefore, it is also the length of one of the sides of the rectangular prism.
Example 2
Divide x3+9x2+12x−27 by (x+3). Write the resulting polynomial with the remainder (if there is one).
Solution
Using synthetic division, divide by -3.
The answer is $$\ x^{2}+6 x-6-\frac{9}{x+3}$$.
Example 3
Divide 2x4−11x3+12x2+9x−2 by (2x+1). Write the resulting polynomial with the remainder (if there is one).
Solution
Using synthetic division, divide by $$\ -\frac{1}{2}$$.
The answer is $$\ 2 x^{3}-12 x^{2}+18 x-\frac{2}{2 x+1}$$
Example 4
Is 6 a solution for f(x)=x3−8x2+72? If so, find the real-number zeros (solutions) of the resulting polynomial.
Solution
Put a zero placeholder for the x−term. Divide by 6.
The resulting polynomial is x2−2x−12. While this quadratic does not factor, we can use the Quadratic Formula to find the other roots.
$$\ x=\frac{2 \pm \sqrt{2^{2}-4(1)(-12)}}{2}=\frac{2 \pm \sqrt{4+48}}{2}=\frac{2 \pm 2 \sqrt{13}}{2}=1 \pm \sqrt{13}$$
The solutions to this polynomial are $$\ 6,1+\sqrt{13} \approx 4.61 \text { and } 1-\sqrt{13} \approx-2.61$$.
# Review
Use synthetic division to divide the following polynomials. Write out the remaining polynomial.
1. (x3+6x2+7x+10)÷(x+2)
2. (4x3−15x2−120x−128)÷(x−8)
3. (4x2−5)÷(2x+1)
4. (2x4−15x3−30x2−20x+42)÷(x+9)
5. (x3−3x2−11x+5)÷(x−5)
6. (3x5+4x3−x−2)÷(x−1)
7. Which of the division problems above generate no remainder? What does that mean?
8. What is the difference between a zero and a factor?
9. Find f(−2) if f(x)=2x4−5x3−10x2+21x−4.
10. Now, divide 2x4−5x3−10x2+21x−4 by (x+2) synthetically. What do you notice?
Find all real zeros of the following polynomials, given one zero.
1. 12x3+76x2+107x−20; −4
2. x3−5x2−2x+10; −2
3. 6x3−17x2+11x−2; 2
Find all real zeros of the following polynomials, given two zeros.
1. x4+7x3+6x2−32x−32; −4, −1
2. 6x4+19x3+11x2−6x; 0, −2
# Answers for Review Problems
To see the Review answers, open this PDF file and look for section 6.10.
# Vocabulary
Term Definition
Oblique Asymptote An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Oblique Asymptotes An oblique asymptote is a diagonal line marking a specific range of values toward which the graph of a function may approach, but generally never reach. An oblique asymptote exists when the numerator of the function is exactly one degree greater than the denominator. An oblique asymptote may be found through long division.
Remainder Theorem The remainder theorem states that if f(k)=r, then r is the remainder when dividing f(x) by (x−k).
Synthetic Division Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.
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# Checking Solutions to Inequalities
## Substitute values for variables to check inequality solutions
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Solving Basic Inequalities
The average weight gain of an infant, after 6 months of age, is one pound a month, until the age of 2. If the average 6-month-old weighs 16 pounds, up to what age would an infant weigh 25 pounds or less?
### Basic Inequalities
Solving a linear inequality is very similar to solving a linear equality, or equation. There are a few very important differences. We no longer use an equal sign. There are four different inequality signs, shown below.
\begin{align*} < \end{align*} Less than
\begin{align*} > \end{align*} Greater than
\begin{align*} \le \end{align*} Less than or equal to
\begin{align*} \ge \end{align*} Greater than or equal to
Notice that the line underneath the \begin{align*}\le\end{align*} and \begin{align*}\ge\end{align*} signs indicates “equal to.” The inequality \begin{align*}x>-1\end{align*} would be read, “\begin{align*}x\end{align*} is greater than -1.” We can also graph these solutions on a number line. To graph an inequality on a number line, shading is used. This is because an inequality is a range of solutions, not just one specific number. To graph \begin{align*}x>-1\end{align*}, it would look like this:
Notice that the circle at -1 is open. This is to indicate that -1 is not included in the solution. A < sign would also have an open circle. If the inequality was a \begin{align*}\ge\end{align*} or \begin{align*}\le\end{align*} sign, then the circle would be closed, or filled in. Shading to the right of the circle shows that any number greater than -1 will be a solution to this inequality.
Let's determine whether \begin{align*}x=-8\end{align*} is a solution to \begin{align*}\frac{1}{2}x+6>3\end{align*}.
Plug in -8 for \begin{align*}x\end{align*} and test this solution.
\begin{align*}\frac{1}{2}(-8)+6 &> 3\\ -4+6 &> 3\\ 2 &> 3\end{align*}
Of course, 2 cannot be greater than 3. Therefore, this is not a valid solution.
Now, let's solve the following basic inequalities.
1. Solve and graph the solution to \begin{align*}2x-5 \le 17\end{align*}.
For the most part, solving an inequality is the same as solving an equation. The major difference will be addressed in problem #2 below. This inequality can be solved just like an equation.
\begin{align*}& 2x-\bcancel{5} \le 17\\ & \underline{\quad \ + \bcancel{5} \ +5 \; \;}\\ & \quad \ \frac{\bcancel{2}x}{\bcancel{2}} \le \frac{22}{2}\\ & \qquad x \le 2\end{align*}
Test a solution, \begin{align*}x = 0: 2(0)-5 \le 17 \rightarrow -5 \le 17 \checkmark\end{align*}
Plotting the solution, we get:
Always test a solution that is in the solution range. It will help you determine if you solved the problem correctly.
1. Solve and graph \begin{align*}-6x+7 \le - 29\end{align*}.
When solving inequalities, be careful with negatives. Let’s solve this problem the way we normally would an equation.
\begin{align*}& -6x+\bcancel{7} \le -29\\ & \underline{\qquad \ \ -\bcancel{7} \quad \ -7 \; \; \; \;}\\ & \qquad \frac{-\bcancel{6}x}{-\bcancel{6}x} \le \frac{-36}{-6}\\ & \qquad \quad \ x \le 6\end{align*}
Let’s check a solution. If \begin{align*}x\end{align*} is less than or equal to 6, let’s test 1.
\begin{align*}-6(1)+7 & \le -29\\ -6+7 & \le -29\\ 1 & \bcancel{\le} -29\end{align*}
This is not a true inequality. To make this true, we must flip the inequality. Therefore, whenever we multiply or divide by a negative number, we must flip the inequality sign. The answer to this inequality is actually \begin{align*}x\ge 6\end{align*}. Now, let’s test a number in this range.
\begin{align*}-6(10)+7 & \le - 29\\ -60+7 & \le -29\\ -60 & \le -29\end{align*}
This is true. The graph of the solution is:
### Examples
#### Example 1
Earlier, you were asked to find the last age (the maximum age) that an infant would weigh 25 pounds or less.
First, write an inequality. Let m represent the age of the infant, in months. Remember, when you get the final answer, you must add 6 for the initial weight of the infant at 6 months.
\begin{align*}16 + m \le 25 \\ m \le 9 \end{align*}
Adding 6, we have \begin{align*}m \le 15\end{align*}. So, up to 15 months, the average baby should weigh 25 pounds or less.
#### Example 2
Is \begin{align*}x = -5\end{align*} a solution to \begin{align*}-3x+7>12\end{align*}?
Plug -5 into the inequality.
\begin{align*}-3(-5)+7 &>12\\ 15+7&>12\end{align*}
This is true because 22 is larger than 12. -5 is a solution.
#### Example 3
Solve and graph the solution to \begin{align*}\frac{3}{8}x+5<26\end{align*}.
No negatives with the \begin{align*}x-\end{align*}term, so we can solve this inequality like an equation.
\begin{align*}& \frac{3}{8}x+\bcancel{5}<26\\ & \underline{\quad \ -\bcancel{5} \ \ -5 \; \; \; \; \; \;}\\ & \ \xcancel{\frac{8}{3} \cdot \frac{3}{8}}x<21 \cdot \frac{8}{3}\\ & \qquad \ x<56\end{align*}
Test a solution, \begin{align*}x = 16: \frac{3}{8}(16)+5 < 26 \checkmark\end{align*}
\begin{align*}6+5 < 26\end{align*}
The graph looks like:
#### Example 4
Solve and graph the solution to \begin{align*}11<4-x\end{align*}.
In this inequality, we have a negative \begin{align*}x-\end{align*}term. Therefore, we will need to flip the inequality.
\begin{align*}& \ \ 11<\bcancel{4}-x\\ & \underline{-4 \ \ - \bcancel{4} \; \; \; \; \; \; \;}\\ & \frac{7}{-1} < \frac{\bcancel{-}x}{\bcancel{-1}}\\ & -7>x\end{align*}
Test a solution, \begin{align*}x = -10: 11 < 4-(-10) \checkmark\end{align*}
\begin{align*}11 < 14\end{align*}
Notice that we flipped the inequality sign when we divided by -1. Also, this equation can also be written \begin{align*}x< -7\end{align*}.
Here is the graph:
### Review
Solve and graph each inequality.
1. \begin{align*}x+5>-6\end{align*}
2. \begin{align*}2x \ge 14\end{align*}
3. \begin{align*}4<-x\end{align*}
4. \begin{align*}3x-4 \le 8\end{align*}
5. \begin{align*}21-8x<45\end{align*}
6. \begin{align*}9>x-2\end{align*}
7. \begin{align*}\frac{1}{2}x+5 \ge 12\end{align*}
8. \begin{align*}54 \le -9x\end{align*}
9. \begin{align*}-7<8+\frac{5}{6}x\end{align*}
10. \begin{align*}10-\frac{3}{4}x<-8\end{align*}
11. \begin{align*}4x+15 \ge 47\end{align*}
12. \begin{align*}0.6x-2.4<4.8\end{align*}
13. \begin{align*}1.5>-2.7-0.3x\end{align*}
14. \begin{align*}-11<12x+121\end{align*}
15. \begin{align*}\frac{1}{2}-\frac{3}{4}x \le -\frac{5}{8}\end{align*}
For questions 16 and 17, write the inequality statement given by the graph below.
To see the Review answers, open this PDF file and look for section 1.10.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
inequality An inequality is a mathematical statement that relates expressions that are not necessarily equal by using an inequality symbol. The inequality symbols are $<$, $>$, $\le$, $\ge$ and $\ne$.
solution A solution to an equation or inequality should result in a true statement when substituted for the variable in the equation or inequality.
Variable A variable is a symbol used to represent an unknown or changing quantity. The most common variables are a, b, x, y, m, and n.
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# Test yourself now
High marks in science are the key to your success and future plans. Test yourself and learn more on Siyavula Practice.
## 1.9 Chapter summary (EMAR)
Presentation: 2DR6
• $$\mathbb{N}$$: natural numbers are $$\left\{1; 2; 3; \ldots\right\}$$
• $$\mathbb{N}_0$$: whole numbers are $$\left\{0; 1; 2; 3; \ldots\right\}$$
• $$\mathbb{Z}$$: integers are $$\left\{\ldots; -3; -2; -1; 0; 1; 2; 3; \ldots\right\}$$
• A rational number is any number that can be written as $$\frac{a}{b}$$ where $$a$$ and $$b$$ are integers and $$b\ne 0$$.
• The following are rational numbers:
• Fractions with both numerator and denominator as integers
• Integers
• Decimal numbers that terminate
• Decimal numbers that recur (repeat)
• Irrational numbers are numbers that cannot be written as a fraction with the numerator and denominator as integers.
• If the $$n^{\text{th}}$$ root of a number cannot be simplified to a rational number, it is called a surd.
• If $$a$$ and $$b$$ are positive whole numbers, and $$a<b$$, then $$\sqrt[n]{a}<\sqrt[n]{b}$$.
• A binomial is an expression with two terms.
• The product of two identical binomials is known as the square of the binomial.
• We get the difference of two squares when we multiply $$\left(ax+b\right)\left(ax-b\right)$$
• Factorising is the opposite process of expanding the brackets.
• The product of a binomial and a trinomial is:
$\left(A+B\right)\left(C+D+E\right)=A\left(C+D+E\right)+B\left(C+D+E\right)$
• Taking out a common factor is the basic factorisation method.
• We often need to use grouping to factorise polynomials.
• To factorise a quadratic we find the two binomials that were multiplied together to give the quadratic.
• The sum of two cubes can be factorised as: ${x}^{3}+{y}^{3}=\left(x+y\right)\left({x}^{2}-xy+{y}^{2}\right)$
• The difference of two cubes can be factorised as: ${x}^{3}-{y}^{3}=\left(x-y\right)\left({x}^{2}+xy+{y}^{2}\right)$
• We can simplify fractions by incorporating the methods we have learnt to factorise expressions.
• Only factors can be cancelled out in fractions, never terms.
• To add or subtract fractions, the denominators of all the fractions must be the same.
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# How do you add -9/8+7/4?
Sep 20, 2016
$- \frac{9}{8} + \frac{7}{4} = \frac{5}{8}$
#### Explanation:
WE have denominators $4$ and $8$ here and their GCD is $8$, hence we can add them by converting them to common GCD.
$- \frac{9}{8} + \frac{7}{4}$
= $- \frac{9}{8} + \frac{7 \times 2}{4 \times 2}$
= $- \frac{9}{8} + \frac{14}{8}$
= $\frac{- 9 + 14}{8}$
= $\frac{5}{8}$
Sep 21, 2016
$\frac{5}{8}$
#### Explanation:
Fraction$\to \left(\text{count")/("size indicator")->("numerator")/("denominator}\right)$
$\textcolor{w h i t e}{.}$
Size indicator is how many of what you are counting to make a whole 1 of something.
$\textcolor{w h i t e}{.}$
,~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
$\textcolor{b l u e}{\text{Point 1}}$
We need to add/subtract counts but we can only do this 'directly' if the 'size indicators' (denominators) are the same.
$\textcolor{b l u e}{\text{Point 2}}$
Multiply by 1 and you do not change the intrinsic value. However, 1 comes in many forms. So we can change the way something looks but not change its intrinsic value.
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Given:$\text{ } - \frac{9}{8} + \frac{7}{4}$
Change the order:
$\frac{7}{4} - \frac{9}{8}$
This is the same in value as:
$\left(\frac{7}{4} \textcolor{m a \ge n t a}{\times 1}\right) - \frac{9}{8}$
But write 1 as $1 = \frac{2}{2}$ giving:
$\left(\frac{7}{4} \textcolor{m a \ge n t a}{\times \frac{2}{2}}\right) - \frac{9}{8} \text{ "->" } \frac{7 \times 2}{4 \times 2} - \frac{9}{8}$
$\frac{14}{8} - \frac{9}{8} = \frac{5}{8}$
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# How do you use substitution to integrate (r^2) √(r^(3) +3)dr?
##### 1 Answer
Mar 25, 2018
$\int {r}^{2} / \sqrt{{r}^{3} + 3} \mathrm{dr} = \frac{2}{3} \sqrt{{r}^{3} + 3} + C$
#### Explanation:
So, we want to determine
$\int {r}^{2} / \sqrt{{r}^{3} + 3} \mathrm{dr}$
The substitution should be picked in such a way that the differential of what we've picked will show up in the integral.
The best choice is $u = {r}^{3} + 3$, as
$\mathrm{du} = 3 {r}^{2} \mathrm{dr}$
$3 {r}^{2} \mathrm{dr}$ does not show up in this exact form in the integral; however, $\frac{1}{3} \mathrm{du} = {r}^{2} \mathrm{dr}$ shows up in the numerator of the integral. Thus, we can substitute and now have
$\frac{1}{3} \int \frac{\mathrm{du}}{\sqrt{u}} = \frac{1}{3} \int {u}^{- \frac{1}{2}} \mathrm{du}$ (We've factored $\frac{1}{3}$ outside of the integral)
Integrate:
$\left(\frac{1}{3}\right) \left(2\right) \sqrt{u} + C$
Rewriting in terms of $x$ yields
$\int {r}^{2} / \sqrt{{r}^{3} + 3} \mathrm{dr} = \frac{2}{3} \sqrt{{r}^{3} + 3} + C$
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## Riddler Express
While watching batter spread out on my waffle iron, and thinking back to a recent conversation I had with Friend-of-The-Riddler™ Benjamin Dickman, I noticed the following sequence of numbers:
$$1, 4, 4, 0, 4, 8, 0, 0, 4, 4, 8, …$$
Before you ask — yes, you can find this sequence on the On-Line Encyclopedia of Integer Sequences. However, for the full Riddler experience, I urge you to not look it up. See if you can find the next few terms in the sequence, as well as the pattern.
Now, for the actual riddle: Once you’ve determined the pattern, can you figure out the average value of the entire sequence?
## Solution
Let the sequence be denoted by $s(n)$. You can find more about the sequence here 😊. Then $s(n)$ represents the number of ways an integer $n$ can be expressed as a sum of two squares (positive, negative, or zero). That is, $s(n)$ denotes the number of solutions in integers $(x,y$ to the equation $x^2 +y^2 = n$. For example, $s(5)=8$ since the solutions to $x^2 + y^2 = 5$ are $(1,2)$, $(2,1)$, $(-1,2)$, $(2,-1)$, $(1,-2)$, $(-2,1)$, $(-1,-2)$, and $(-2,-1)$. Because $s(n) = 0$ whenever $n$ has the form $4k - 1$, $s(n)$ is a very erratic function.Thankfully, the problem is about the average value of $s(n)$ as $n \rightarrow \infty$. If we define $t(n)$ to be the number of solutions in integers to $x^2 + y^2 \leq n$, then the average of $s(k)$ for $0 \leq k \leq n$ is
$$\frac{s(0) + s(1) + \dots + s(n)}{n+1} = \frac{t(n)}{n+1}$$
The code (using brute force) for calculating $t(n)$ and $t(n)/(n+1)$ is given below:
from math import sqrt, ceil
def t(n):
t, l = 0, ceil(sqrt(n))+1
for i in range(0, n+1):
for j in range(-l, l):
for k in range(-l, l):
if j**2 + k**2 == i:
t += 1
return t, t/(n+1)
print(list(map(t, [1,2,3,4,5,10,20,50,100])))
Here is a table of $t(n)$ and $t(n)/(n+1)$ for a few values of $n$:
$n$12345102050100
$t(n)$59913213769161317
$\frac{t(n)}{n+1}$2.532.252.63.53.363.293.163.15
$\textbf{Theorem:}$ $\lim_{n \rightarrow \infty} \frac{t(n)}{n+1} = \pi$.
$\textbf{Proof}.$ The proof from the reference below is based on a geometric interpretation of $t(n)$, and is due to Carl Friedrich Gauss $(1777–1855)$: $t(n)$ is the number of points in or on a circle $x^2 + y^2 = n$ with integer coordinates. For example, $t(10) = 37$ since the circle centered at the origin of radius $\sqrt{10}$ contains $37$ lattice points as illustrated in the figure below:
If we draw a square with area $1$ centered at each of the $37$ points, then the total area (in grey) of the squares is also $t(n)$. Thus we would expect the area of the squares to be approximately the area of the circle, or in general, $t(n)$ to be approximately $\pi(\sqrt{n})^2 = \pi n$. If we expand the circle of radius $\sqrt{n}$ by half the length of the diagonal $\sqrt{2}/2$ of a square of area $1$, then the expanded circle contains all the squares. If we contract the circle by the same amount, then the contracted circle would be contained in the union of all the squares as seen in the figure below:
Thus,
$$\pi(\sqrt{n} - \sqrt{2}/2)^2 \leq t(n) \leq \pi(\sqrt{n} + \sqrt{2}/2)^2$$
Dividing each term by $n+1$ and applying the squeeze theorem for limits yields the desired result.
### References
Charming Proofs: A Journey into Elegant Mathematics
## Riddler Classic
The finals of the sport climbing competition has eight climbers, each of whom compete in three different events: speed climbing, bouldering and lead climbing. Based on their time and performance, each of the eight climbers is given a ranking (first through eighth, with no ties allowed) in each event, as well as a corresponding score ($1$ through $8$, respectively).
The three scores each climber earns are then multiplied together to give a final score. For example, a climber who placed second in speed climbing, fifth in bouldering and sixth in lead climbing would receive a score of $2 × 5 × 6$, or $60$ points. The gold medalist is whoever achieves the lowest final score among the eight finalists.
What is the highest (i.e., worst) score one could achieve in this event and still have a chance of winning (or at least tying for first place overall)?
## Computational Solution
From the $\textbf{Monte-Carlo simulation}$ below, we see that the worst score one could achieve and still have a chance of winning is $\textbf{48}$.
from random import shuffle
from operator import mul
from functools import reduce
def max_min_score(np, ne, runs = 1000000):
max_min_score = 0
for _ in range(runs):
ranks = [list(range(1,np+1)) for i in range(ne)]
for i in range(ne):
shuffle(ranks[i])
scores = [reduce(mul, [ranks[j][i] for j in range(ne)]) for i in range(np)]
min_score = min(scores)
if min_score > max_min_score:
max_min_score = min_score
return max_min_score
print(max_min_score(8,3))
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# Steps to solve Quadratic or polynomial inequalities
## Steps to solve Quadratic or polynomial inequalities
ax2+bx+c > 0
or
ax2+bx+c < 0
or
ax2+bx+c $\geq$ 0
or
ax2+bx+c $\leq$ 0
1) Obtain the Quadratic equation inequation
2) Pull all the terms having on one side and Simplify the equation in the form given above
3) Simplify the equation in the form given above
4) Find the roots of the quadratic equation using any of the method and write in this form
(x-a)(x-b)
5) Plot these roots on the number line .This divide the number into three segment
5) Start from LHS side of the number and Substitute some value in the equation in all the three segments to find out which segments satisfy the equation
6) Write down the solution set in interval form
Example
1) Simpify the inequality which means factorizing the equation in case of quadratic equalities
example
x2-5x+6 > 0
Which can be simpified as
(x-2)(x-3) > 0
2) Now plot those points on Number line clearly
3) Now start from left of most left point on the Number line and look out the if inequalities looks good or not. Check for greater ,less than and equalities at all the end points
So in above case of
x2-5x+6 > 0
We have two ends points 2 ,3
Case 1
So for x < 2 ,Let take x=1,then (1-2)(1-3) > 0
2 > 0
So it is good
So This inequalties is good for x < 2
Case 2
Now for x =2,it makes it zero,so not true. Now takes the case of x > 2 but less 3.Lets takes 2.5
(2.5-2)(2.5-3) > 0
-.25 > 0
Which is not true so this solution is not good
Case 3
Now lets take the right most part i.e x > 3
Lets take x=4
(4-2)(4-3) > 0
2> 0
So it is good.
Now the solution can either be represented on number line or we can say like this
$(-\infty,2)\cup (3,\infty)$
## Steps to solve the pair of Linear inequation or quadratic equation
1) Solving them is same as steps given above to solve the linear inequation and quadratic equation
2) Solve both the inequation in the pair
3) Look for the intersection of the solution and give the solution in interval set
## Some Important points to note
1) We cannot have zero in denominator
2) We should be checking for equalities at all the end points
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