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## Precalculus (6th Edition) Blitzer
Published by Pearson
# Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 782: 55
#### Answer
$-60$.
#### Work Step by Step
Put $u=-2i+3j,v=6i-j$, $w=-3i$ in the expression ${{\left\| u+v \right\|}^{2}}-{{\left\| u-v \right\|}^{2}}$. Then, \begin{align} & {{\left\| u+v \right\|}^{2}}-{{\left\| u-v \right\|}^{2}}={{\left\| -2i+3j+6i-j \right\|}^{2}}-\left\| -2i+3j-\left( 6i-j \right) \right\| \\ & ={{\left\| 4i+2j \right\|}^{2}}-{{\left\| -8i+4j \right\|}^{2}} \end{align} Therefore \begin{align} & {{\left\| u+v \right\|}^{2}}-{{\left\| u-v \right\|}^{2}}={{4}^{2}}+{{2}^{2}}-\left( {{\left( -8 \right)}^{2}}+{{4}^{2}} \right) \\ & =16+4-64-16 \\ & =20-80 \\ & =-60 \end{align}
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# How do I solve systems of equations and substitution word problems?
justaguide | College Teacher | (Level 2) Distinguished Educator
Posted on
Systems of equations can be solved to determine the value of the variables used in the equations. There are broadly two ways of solving a system of linear equations, by elimination and by substitution.
Let us take a simple example to illustrate how the two methods can be used for solving the same system of equations.
The system of equations to be solved is:
x - y = 7 ...(1)
x + y = 9 ...(2)
Adding (1) and (2) gives x - y + x + y = 7 + 9
=> 2x = 16
As you can see, the variable y has been eliminated and this allows the value of x to be determined.
=> x = 8
Similarly, (1) - (2) gives x - y - x - y = 7 - 9
=> -2y = -2
=> y = 1
The solution of the system of equations is (8, 1).
To solve, the system using substitution, write one of the equations such that it is possible to express one of the variables in terms of the other.
x - y = 7
=> x = y + 7
Now substitute this for x in (2)
x + y = 9
=> y + 7 + y = 9
=> 2y = 2
=> y = 1
Similarly from (1) we can get y = x - 7
Substituting this for x in (2)
x + x - 7 = 9
=> 2x = 16
=> x = 8
This gives the same solution as the earlier method.
malkaam | Student, Undergraduate | (Level 1) Valedictorian
Posted on
There are majorly two ways of solving the systems of equations, preferably linear equations. They are substitution method or elimination method. Each of these has a distinct format of solution.
In order to solve substitution word problems, we use the following method, and to explain the method, let’s take up an example.
For example there are two equations:
y=x+1 and y=-2x-4
Now,
In order to find the value of x and y, we first substitute the value of any one variable from one equation by isolating the variable (say y in equation (i), and use it in the other to find out the value of one variable by putting it in equation (ii), then use that value derived (say, value of x) and put it in equation (i) to find out the value of the second variable( say, y).
Let,
y=x+1 ----(i)
y=-2x-4----(ii)
As given in the above explanation, we will now see if the variable y is already isolated or not in equation (i). Since it is already isolated, we take that value of y and insert it in place of y in the (ii) equation:
x+1 = 2x-4
1=2x-4-x
1+4=2x-x
2=x
or x=2
Now we put the value of x in equation (i) to derive the value of y:
y=x+1
y= 2+1
y=3
Therefore, the values of x and y have been derived by using the substitution method, which is merely based on substituting the value of one variable from one equation and using it in the other to derive the value of the other.
Educator Approved
rachellopez | Student, Grade 12 | (Level 1) Valedictorian
Posted on
A system of equations can be solved in a couple different ways, but a common way is by substituting. When you are given a system of equations, you have two equations with the same variables that you have to solve for. An example would be
2x-3y=-2
4x+y=24
How you would solve this using substitution is to solve for one variable and put that answer into the other equation to get the second variable. For this example, I'll solve for y first with the second equation.
y=-4x+24
Put this into the y value for the other equation to solve for x.
2x-3(-4x+24)=-2
2x-(-12x+72)=-2
14x-72=-2
14x=70
x=5
Now you can put in this numerical value of x to get the numerical value of y (put it into any equation you want) and you get y=4.
For substitution into word problems, you are usually given an equation, or you'll be asked to create one, and you can simply put in the values it gives you. If the word problem says solve x+y when x=5 and y=4, you just put those numbvers into the probelm.
Sources: |
Patterns In Shapes Maths Grade 2 | Orchids International School
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Patterns
# Shapes and Patterns for Class 2 Maths
When a group of shapes or a single shape is repeated, it is called a pattern in the given shapes.
In this learning concept, the students will learn about shapes and patterns for class 2. They will learn about the mathematical patterns in nature.
In this learning concept, the students will learn:
• Growing pattern
• Repeating pattern
Each concept is explained to class 2 maths students using illustrations, examples, and mind maps. Students can assess their learning by solving the two printable worksheets given at the pages’ end.
Download the shapes and patterns worksheet for class 2 and check the solutions for the concept of shapes and patterns for class 2 maths provided in PDF format.
A pattern is something that is arranged in a specific manner and repeated over and over again.
## Patterns Around Us
In our surroundings, we can find so many different patterns. Some of them are man-made, and some of them are present in nature.
Following are some examples of patterns which are commonly found in nature.
Following are examples of man-made patterns.
We can classify patters as:
1. Growing Patterns
2. Repeating Patterns
## Growing Pattern
If the shapes in the pattern keep on growing in number, then it is called the growing pattern.
### Example:
In this pattern, the number of triangles is increasing by one.
## Growing Pattern in Nature
The peacock’s tail is an example of a growing pattern in nature.
Here the design of the feather is shown on right. The design of the feather is repeated many times in the tail of the peacock.
## Repeating Pattern
If the shapes in the pattern are repeating, then it is called a repeating pattern.
### Example:
In this pattern, after a heart shape, there is a hexagon.
## Mathematical Patterns in Nature
The Honey hive of bees is an example of a repeating pattern in nature.
Here, hexagonal shapes are repeating themselves to form a hive.
• - |
# How do you find the quotient of (x^2+x-20)div(x-4)?
Jun 12, 2017
$x + 5$
#### Explanation:
$\textcolor{w h i t e}{w w w w w} x + 5$
x-4 )bar(x^2+x-20)
$\textcolor{w h i t e}{w w w} \underline{\textcolor{red}{-} {x}^{2} \textcolor{red}{+} 4 x} \text{ } \leftarrow$ subtract
$\textcolor{w h i t e}{w w w w w w w w} 5 x - 20$
$\textcolor{w h i t e}{w w n \ldots . . w w} \textcolor{red}{-} \underline{5 x \textcolor{red}{+} 20}$
$\textcolor{w h i t e}{w w w w w w w w w w w w w w} 0 \text{ }$ remainder
$\left({x}^{2} + x - 20\right) \div \left(x - 4\right) = x + 5$
Jun 12, 2017
$x + 5$
#### Explanation:
$\text{factorise the numerator and simplify}$
$\Rightarrow \frac{{x}^{2} + x - 20}{x - 4}$
$= \frac{\left(x + 5\right) \cancel{\left(x - 4\right)}}{\cancel{\left(x - 4\right)}}$
$= x + 5 \leftarrow \textcolor{red}{\text{ quotient}}$
Jun 12, 2017
Factorise and cancel like factors.
$x + 5$
#### Explanation:
In a division such as $48 \div 6$, we could find the quotient by finding factors:
$\frac{48}{6} = \frac{8 \times 6}{6} \text{ } \leftarrow$ like factors can cancel $\text{ } \frac{6}{6} = 1$
$\frac{8 \times \cancel{6}}{\cancel{6}} = 8$
In the same way we can find the quotient by factorising the numerator:
$\frac{{x}^{2} + x - 20}{\left(x - 4\right)} = \frac{\left(x + 5\right) \left(x - 4\right)}{\left(x - 4\right)}$
Cancel the like factors:
$\frac{\left(x + 5\right) \cancel{\left(x - 4\right)}}{\cancel{\left(x - 4\right)}}$
$= x + 5$ |
# How do you write an equation of a line that contains the given point (-5,5) and is perpendicular to the given line y=-5x+9?
##### 1 Answer
Jan 14, 2017
$\left(y - 5\right) = \frac{1}{5} \left(x + 5\right)$
or
$y = \frac{1}{5} x + 6$
#### Explanation:
To find the equation of the line perpendicular to the given line and going through the given point we will use the point-slope formula. We have been given a point so what we are missing is the slope.
The given line is in slope-intercept form.
The slope-intercept form of a linear equation is:
$y = \textcolor{red}{m} x + \textcolor{b l u e}{b}$
Where $\textcolor{red}{m}$ is the slope and color(blue)(b is the y-intercept value.
Therefore we know the slope of the given line, $\textcolor{red}{m}$ is $\textcolor{red}{- 5}$ - it is the coefficient of the $x$ term.
A perpendicular line will have a slope which is the negative inverse of this line, or color(red)(m = - 1/-5 = 1/5.
We can now use the point-slope formula to write the equation for the perpendicular line.
The point-slope formula states: $\left(y - \textcolor{red}{{y}_{1}}\right) = \textcolor{b l u e}{m} \left(x - \textcolor{red}{{x}_{1}}\right)$
Where $\textcolor{b l u e}{m}$ is the slope and $\textcolor{red}{\left(\left({x}_{1} , {y}_{1}\right)\right)}$ is a point the line passes through.
Substituting the point we were given and the slope we calculated gives:
$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{\frac{1}{5}} \left(x - \textcolor{red}{- 5}\right)$
$\left(y - \textcolor{red}{5}\right) = \textcolor{b l u e}{\frac{1}{5}} \left(x + \textcolor{red}{5}\right)$
or, we can solve for $y$ to convert to the more familiar slope-intercept form:
$y - \textcolor{red}{5} = \textcolor{b l u e}{\frac{1}{5}} x + \left(\textcolor{b l u e}{\frac{1}{5}} \times \textcolor{red}{5}\right)$
$y - \textcolor{red}{5} = \textcolor{b l u e}{\frac{1}{5}} x + 1$
$y - \textcolor{red}{5} + 5 = \textcolor{b l u e}{\frac{1}{5}} x + 1 + 5$
$y - 0 = \textcolor{b l u e}{\frac{1}{5}} x + 6$
$y = \textcolor{b l u e}{\frac{1}{5}} x + 6$ |
# Lesson 9: Side Length Quotients in Similar Triangles
Let’s find missing side lengths in triangles.
## 9.1: Two-three-four and Four-five-six
Triangle $A$ has side lengths 2, 3, and 4. Triangle $B$ has side lengths 4, 5, and 6. Is Triangle $A$ similar to Triangle $B$?
## 9.2: Quotients of Sides Within Similar Triangles
Your teacher will assign you one of the three columns in the second table.
Triangle $ABC$ is similar to triangles $DEF$, $GHI$, and $JKL$. The scale factors for the dilations that show triangle $ABC$ is similar to each triangle are in the table.
triangle scale factor length of short side length of medium side length of long side
row 1 $ABC$ 1 4 5 7
row 2 $DEF$ 2
row 3 $GHI$ 3
row 4 $JKL$ $\frac{1}{2}$
triangle (long side) $\div$ (short side) (long side) $\div$ (medium side) (medium side) $\div$ (short side)
row 1 $ABC$ $\frac{7}{4}$ or $1.75$
row 2 $DEF$
row 3 $GHI$
row 4 $JKL$
1. Find the side lengths of triangles $DEF$, $GHI$, and $JKL$. Record them in the first table.
2. For all four triangles, find the quotient of the triangle side lengths assigned to you and record them in the second table. What do you notice about the quotients?
## 9.3: Using Side Quotients to Find Side Lengths of Similar Triangles
Triangles $ABC$, $EFD$, and $GHI$ are all similar. The side lengths of the triangles all have the same units. Find the unknown side lengths.
## Summary
If two polygons are similar, then the side lengths in one polygon are multiplied by the same scale factor to give the corresponding side lengths in the other polygon. For these triangles the scale factor is 2:
Here is a table that shows relationships between the short and medium length sides of the small and large triangle.
small triangle large triangle
row 1 medium side 4 8
row 2 short side 3 6
row 3 (medium side) $\div$ (short side) $\frac{4}{3}$ $\frac{8}{6} = \frac{4}{3}$
The lengths of the medium side and the short side are in a ratio of $4:3$. This means that the medium side in each triangle is $\frac43$ as long as the short side.
This is true for all similar polygons; the ratio between two sides in one polygon is the same as the ratio of the corresponding sides in a similar polygon.
We can use these facts to calculate missing lengths in similar polygons. For example, triangles $A’B’C’$ and $ABC$ shown here are similar. Let's find the length of segment $B’C’$.
In triangle $ABC$, side $BC$ is twice as long as side $AB$, so this must be true for any triangle that is similar to triangle $ABC$. Since $A'B'$ is 1.2 units long and $2\boldcdot 1.2 = 2.4$, the length of side $B’C’$ is 2.4 units. |
# Vedic Maths Tutorial (interactive
)
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Vedic Maths Tutorial
Vedic Maths is based on sixteen sutras or principles. These principles are general in nature and can be applied in many ways. In practice many applications of the sutras may be learned and combined to solve actual problems. These tutorials will give examples of simple applications of the sutras, to give a feel for how the Vedic Maths system works. These tutorials do not attempt to teach the systematic use of the sutras. For more advanced applications and a more complete coverage of the basic uses of the sutras, we recommend you study one of the texts available. N.B. The following tutorials are based on examples and exercises given in the book 'Fun with figures' by Kenneth Williams, which is a fun introduction some of the applications of the sutras for children. If you are having problems using the tutorials then you could always read the instructions. Tutorial 1 Tutorial 2 Tutorial 3 Tutorial 4 Tutorial 5 Tutorial 6 Tutorial 7
Tutorial 1
Use the formula ALL FROM 9 AND THE LAST FROM 10 to perform instant subtractions.
q
For example 1000 - 357 = 643 We simply take each figure in 357 from 9 and the last figure from 10.
So the answer is 1000 - 357 = 643 And thats all there is to it! This always works for subtractions from numbers consisting of a 1 followed by noughts: 100; 1000; 10,000 etc.
q
Similarly 10,000 - 1049 = 8951
q
For 1000 - 83, in which we have more zeros than figures in the numbers being subtracted, we simply suppose 83 is 083. So 1000 - 83 becomes 1000 - 083 = 917
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Vedic Maths Tutorial (interactive)
Try some yourself: 1) 1000 - 777 2) 1000 - 283 3) 1000 - 505 4) 10,000 - 2345 5) 10000 - 9876 6) 10,000 - 1101 7) 100 - 57 8) 1000 - 57 9) 10,000 - 321 10) 10,000 - 38 Total Correct
= = = = = = = = = = =
Reset Test
Tutorial 2
Using VERTICALLY AND CROSSWISE you do not need to the multiplication tables beyond 5 X 5.
q
Suppose you need 8 x 7 8 is 2 below 10 and 7 is 3 below 10. Think of it like this:
The answer is 56. The diagram below shows how you get it.
You subtract crosswise 8-3 or 7 - 2 to get 5, the first figure of the answer. And you multiply vertically: 2 x 3 to get 6, the last figure of the answer. That's all you do:
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Vedic Maths Tutorial (interactive)
See how far the numbers are below 10, subtract one number's deficiency from the other number, and multiply the deficiencies together.
q
7 x 6 = 42
Here there is a carry: the 1 in the 12 goes over to make 3 into 4. Multply These: 1) 8 8x 2) 9 7x 3) 8 9x 4) 7 7x 5) 9 9x 6) 6 6x
Total Correct =
Reset Test
Here's how to use VERTICALLY AND CROSSWISE for multiplying numbers close to 100.
q
Suppose you want to multiply 88 by 98. Not easy,you might think. But with VERTICALLY AND CROSSWISE you can give the answer immediately, using the same method as above. Both 88 and 98 are close to 100. 88 is 12 below 100 and 98 is 2 below 100. You can imagine the sum set out like this:
As before the 86 comes from subtracting crosswise: 88 - 2 = 86 (or 98 - 12 = 86: you can subtract either way, you will always get the same answer). And the 24 in the answer is just 12 x 2: you multiply vertically. So 88 x 98 = 8624 This is so easy it is just mental arithmetic. Try some:
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Vedic Maths Tutorial (interactive)
1) 87 98 x
2) 88 97 x
3) 77 98 x
4) 93 96 x
5) 94 92 x
6) 64 99
7) 98 97 x
Total Correct =
Reset Test
Multiplying numbers just over 100.
q
103 x 104 = 10712 The answer is in two parts: 107 and 12, 107 is just 103 + 4 (or 104 + 3), and 12 is just 3 x 4.
q
Similarly 107 x 106 = 11342 107 + 6 = 113 and 7 x 6 = 42
Again, just for mental arithmetic Try a few:
1) 102 x 107 = 1) 106 x 103 = 1) 104 x 104 = 4) 109 x 108 = 5) 101 x123 = 6) 103 x102 = Total Correct = Return to Index
Reset Test
Tutorial 3
The easy way to add and subtract fractions. Use VERTICALLY AND CROSSWISE to write the answer straight down!
q
Multiply crosswise and add to get the top of the answer: 2 x 5 = 10 and 1 x 3 = 3. Then 10 + 3 = 13. The bottom of the fraction is just 3 x 5 = 15.
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Vedic Maths Tutorial (interactive)
You multiply the bottom number together. So:
q
Subtracting is just as easy: multiply crosswise as before, but the subtract:
q
Try a few:
Reset Test
Tutorial 4
A quick way to square numbers that end in 5 using the formula BY ONE MORE THAN THE ONE BEFORE.
q
752 = 5625 752 means 75 x 75. The answer is in two parts: 56 and 25. The last part is always 25. The first part is the first number, 7, multiplied by the number "one more", which is 8: so 7 x 8 = 56
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Vedic Maths Tutorial (interactive)
q
Similarly 852 = 7225 because 8 x 9 = 72.
Try these:
1) 452 = 2) 652 = 3) 952 = 4) 352 = 5) 152 = Total Correct =
Reset Test
Method for multiplying numbers where the first figures are the same and the last figures add up to 10.
q
32 x 38 = 1216 Both numbers here start with 3 and the last figures (2 and 8) add up to 10. So we just multiply 3 by 4 (the next number up) to get 12 for the first part of the answer. And we multiply the last figures: 2 x 8 = 16 to get the last part of the answer. Diagrammatically:
q
And 81 x 89 = 7209 We put 09 since we need two figures as in all the other examples.
Practise some:
1) 43 x 47 = 2) 24 x 26 = 3) 62 x 68 =
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Vedic Maths Tutorial (interactive)
4) 17 x 13 = 5) 59 x 51 = 6) 77 x 73 = Total Correct = Return to Index
Reset Test
Tutorial 5
An elegant way of multiplying numbers using a simple pattern.
q
21 x 23 = 483 This is normally called long multiplication but actually the answer can be written straight down using the VERTICALLY AND CROSSWISE formula. We first put, or imagine, 23 below 21:
There are 3 steps: a) Multiply vertically on the left: 2 x 2 = 4. This gives the first figure of the answer. b) Multiply crosswise and add: 2 x 3 + 1 x 2 = 8 This gives the middle figure. c) Multiply vertically on the right: 1 x 3 = 3 This gives the last figure of the answer. And thats all there is to it.
q
Similarly 61 x 31 = 1891
q
6 x 3 = 18; 6 x 1 + 1 x 3 = 9; 1 x 1 = 1
Try these, just write down the answer:
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Vedic Maths Tutorial (interactive)
1) 14 21 x
2) 22 31 x
3) 21 31 x
4) 21 22 x
5) 32 21 x
Total Correct =
Reset Test
Multiply any 2-figure numbers together by mere mental arithmetic! If you want 21 stamps at 26 pence each you can easily find the total price in your head. There were no carries in the method given above. However, there only involve one small extra step.
q
21 x 26 = 546
The method is the same as above except that we get a 2-figure number, 14, in the middle step, so the 1 is carried over to the left (4 becomes 5). So 21 stamps cost £5.46. Practise a few: 1) 21 47 x 2) 23 43 x 3) 32 53 x 4) 42 32 x 5) 71 72 x
Total Correct =
q
Reset Test
33 x 44 = 1452 There may be more than one carry in a sum:
Vertically on the left we get 12. Crosswise gives us 24, so we carry 2 to the left and mentally get 144. Then vertically on the right we get 12 and the 1 here is carried over to the 144 to make 1452.
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Vedic Maths Tutorial (interactive)
6) 32 56 x
7) 32 54 x
8) 31 72 x
9) 44 53 x
10) 54 64 x
Total Correct =
Reset Test
Any two numbers, no matter how big, can be multiplied in one line by this method. Return to Index
Tutorial 6
Multiplying a number by 11. To multiply any 2-figure number by 11 we just put the total of the two figures between the 2 figures.
q
26 x 11 = 286 Notice that the outer figures in 286 are the 26 being multiplied. And the middle figure is just 2 and 6 added up.
q
So 72 x 11 = 792
Multiply by 11:
1) 43 = 2) 81 = 3) 15 = 4) 44 = 5) 11 = Total Correct =
q
Reset Test
77 x 11 = 847 This involves a carry figure because 7 + 7 = 14 we get 77 x 11 = 7147 = 847.
Multiply by 11:
1) 88 =
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Vedic Maths Tutorial (interactive)
2) 84 = 3) 48 = 4) 73 = 5) 56 = Total Correct =
q
Reset Test
234 x 11 = 2574 We put the 2 and the 4 at the ends. We add the first pair 2 + 3 = 5. and we add the last pair: 3 + 4 = 7.
Multiply by 11:
1) 151 = 2) 527 = 3) 333 = 4) 714 = 5) 909 = Total Correct = Return to Index
Reset Test
Tutorial 7
Method for diving by 9.
q
23 / 9 = 2 remainder 5 The first figure of 23 is 2, and this is the answer. The remainder is just 2 and 3 added up!
q
43 / 9 = 4 remainder 7 The first figure 4 is the answer and 4 + 3 = 7 is the remainder - could it be easier?
Divide by 9:
1) 61 =
remainder
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Vedic Maths Tutorial (interactive)
2) 33 = 3) 44 = 4) 53 = 5) 80 = Total Correct =
q
remainder remainder remainder remainder
Reset Test
134 / 9 = 14 remainder 8 The answer consists of 1,4 and 8. 1 is just the first figure of 134. 4 is the total of the first two figures 1+ 3 = 4, and 8 is the total of all three figures 1+ 3 + 4 = 8.
Divide by 9:
6) 232 = 7) 151 = 8) 303 = 9) 212 = 10) 2121 = Total Correct =
q
remainder remainder remainder remainder remainder
Reset Test
842 / 9 = 812 remainder 14 = 92 remainder 14 Actually a remainder of 9 or more is not usually permitted because we are trying to find how many 9's there are in 842. Since the remainder, 14 has one more 9 with 5 left over the final answer will be 93 remainder 5
Divide these by 9:
1) 771 = 2) 942 = 3) 565 = 4) 555 = 5) 777 =
remainder remainder remainder remainder remainder
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Vedic Maths Tutorial (interactive)
6) 2382 = 7) 7070 = Total Correct = Return to Index
remainder remainder
Reset Test
Instructions |
The following steps will help us to solve quadratic equations by factoring:
Step I: Clear all the fractions and brackets, if necessary.
Step II: Transpose all the terms to the left hand side to get an equation in the form ax$$^{2}$$ + bx + c = 0.
Step III: Factorize the expression on the left hand side.
Step IV: Put each factor equal to zero and solve.
1. Solve the quadratic equation 6m$$^{2}$$ – 7m + 2 = 0 by factorization method.
Solution:
⟹ 6m$$^{2}$$ – 4m – 3m + 2 = 0
⟹ 2m(3m – 2) – 1(3m – 2) = 0
⟹ (3m – 2) (2m – 1) = 0
⟹ 3m – 2 = 0 or 2m – 1 = 0
⟹ 3m = 2 or 2m = 1
⟹ m = $$\frac{2}{3}$$ or m = $$\frac{1}{2}$$
Therefore, m = $$\frac{2}{3}$$, $$\frac{1}{2}$$
2. Solve for x:
x$$^{2}$$ + (4 – 3y)x – 12y = 0
Solution:
Here, x$$^{2}$$ + 4x – 3xy – 12y = 0
⟹ x(x + 4) - 3y(x + 4) = 0
or, (x + 4) (x – 3y) = 0
⟹ x + 4 = 0 or x – 3y = 0
⟹ x = -4 or x = 3y
Therefore, x = -4 or x = 3y
3. Find the integral values of x (i.e., x ∈ Z) which satisfy 3x$$^{2}$$ - 2x - 8 = 0.
Solution:
Here the equation is 3x$$^{2}$$ – 2x – 8 = 0
⟹ 3x$$^{2}$$ – 6x + 4x – 8 = 0
⟹ 3x(x – 2) + 4(x – 2) = 0
⟹ (x – 2) (3x + 4) = 0
⟹ x – 2 = 0 or 3x + 4 = 0
⟹ x = 2 or x = -$$\frac{4}{3}$$
Therefore, x = 2, -$$\frac{4}{3}$$
But x is an integer (according to the question).
So, x ≠ -$$\frac{4}{3}$$
Therefore, x = 2 is the only integral value of x.
4. Solve: 2(x$$^{2}$$ + 1) = 5x
Solution:
Here the equation is 2x^2 + 2 = 5x
⟹ 2x$$^{2}$$ - 5x + 2 = 0
⟹ 2x$$^{2}$$ - 4x - x + 2 = 0
⟹ 2x(x - 2) - 1(x - 2) = 0
⟹ (x – 2)(2x - 1) = 0
⟹ x - 2 = 0 or 2x - 1 = 0 (by zero product rule)
⟹ x = 2 or x = $$\frac{1}{2}$$
Therefore, the solutions are x = 2, 1/2.
5. Find the solution set of the equation 3x$$^{2}$$ – 8x – 3 = 0; when
(i) x ∈ Z (integers)
(ii) x ∈ Q (rational numbers)
Solution:
Here the equation is 3x$$^{2}$$ – 8x – 3 = 0
⟹ 3x$$^{2}$$ – 9x + x – 3 = 0
⟹ 3x(x – 3) + 1(x – 3) = 0
⟹ (x – 3) (3x + 1) = 0
⟹ x = 3 or x = -$$\frac{1}{3}$$
(i) When x ∈ Z, the solution set = {3}
(ii) When x ∈ Q, the solution set = {3, -$$\frac{1}{3}$$}
6. Solve: (2x - 3)$$^{2}$$ = 25
Solution:
Here the equation is (2x – 3)$$^{2}$$ = 25
⟹ 4x$$^{2}$$ – 12x + 9 – 25 = 0
⟹ 4x$$^{2}$$ – 12x - 16 = 0
⟹ x$$^{2}$$ – 3x - 4 = 0 (dividing each term by 4)
⟹ (x – 4) (x + 1) = 0
⟹ x = 4 or x = -1
`
Formation of Quadratic Equation in One Variable
Word Problems on Quadratic Equations by Factoring
Worksheet on Formation of Quadratic Equation in One Variable
Worksheet on Nature of the Roots of a Quadratic Equation
Worksheet on Word Problems on Quadratic Equations by Factoring |
# Area of RT
In the right triangle has orthogonal projections of legs to the hypotenuse lengths 15 cm and 9 cm. Determine the area of this triangle.
S = 139.4 cm2
### Step-by-step explanation:
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Comparison of Unit Rates
## Use <, > or = to compare unit rates.
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Practice Comparison of Unit Rates
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Comparison of Unit Rates
Credit: Amy
Source: https://www.flickr.com/photos/stella12/17265462765/in/
Connie is decorating cookies for the school bake sale. She has worked for 3 hours and completed 150 cookies. Her friend, Daniel, is also participating in the sale. He decorated 98 cookies in 2 hours. Which cookie decorator was faster, Connie or Daniel?
In this concept, you will learn to compare unit rates.
### Guidance
unit rate is a comparison of two measurements, one of which has a value of 1. Given the number of units, the unit rate can be used to calculate a total rate. Unit rates can be used for comparison purposes.
Here is an example.
Felicia needs to buy sugar. She could buy a 16-ounce box of sugar for $1.12, or she could buy a 24-ounce box of sugar for$1.44. Which is the better buy? How much cheaper, in cents per ounce, is that buy?
First, find the unit price for the 16-oz box by writing a fraction.
161.12ouncesdollars\begin{align*}\frac{16}{1.12}\frac{ounces}{dollars}\end{align*}
Next, recognize the answer must be per ounce and reduce the fraction to its lowest terms by reducing ounces to 1.
16÷161.12÷16=1.07\begin{align*}\frac{16\div 16}{1.12\div 16}=\frac{1}{.07}\end{align*} ouncedollars\begin{align*}\frac{ounce}{dollars}\end{align*}
Then, find the unit price per ounce for the 24-ounce box. Remember to keep the units consistent. In this case, ounces in the numerator and dollars in the denominator.
241.44ouncesdollars\begin{align*}\frac{24}{1.44}\frac{ounces}{dollars}\end{align*}
Next, reduce this fraction to its lowest terms.
24÷241.44÷24=1.06ouncedollars\begin{align*}\frac{24\div 24}{1.44\div 24}=\frac{1}{.06}\frac{ounce}{dollars}\end{align*}
Then, compare the two unit rates.
1 ounce for 0.07 >\begin{align*}>\end{align*} 1 ounce for0.06
The 16-oz box is more expensive than the 24-oz box.
Next, subtract one unit price from the other to find the difference in price.
.07 - .06 = .01
The answer is that the best buy is the 24-oz box because it is 0.01 per ounce cheaper than the 16-oz box. ### Guided Practice Compare: 3:1 and 6 to 2. First, write 3:1 as a fraction. 31\begin{align*}\frac{3}{1}\end{align*} Next, reduce to lowest terms. 31=3\begin{align*}\frac{3}{1}=3\end{align*} Then, write 6 to 2 as a fraction and reduce to lowest terms. 62=3\begin{align*}\frac{6}{2}=3\end{align*} Next, compare the values. 3 = 3 The answer is that the rates are equal. ### Examples #### Example 1 Which rate is higher? 4 to 5 or 1620\begin{align*}\frac{16}{20}\end{align*} First, write the fractions,and reduce. 45\begin{align*}\frac{4}{5}\end{align*} 1620=16÷420÷4=45\begin{align*}\frac{16}{20}=\frac{16\div 4}{20\div 4}=\frac{4}{5}\end{align*} Next, compare. 45=45\begin{align*}\frac{4}{5}=\frac{4}{5}\end{align*} The answer is that the rates are equal. #### Example 2 Matt can melt 18 m&ms in his mouth without chewing in 3 minutes. It takes Kelly 5 minutes to melt 15. Who melts the m&ms faster, Matt or Kelly? First, write the fractions. Matt 183m&msminutes\begin{align*}\frac{18}{3}\frac{m\&ms}{minutes}\end{align*} Kelly 155m&msminutes\begin{align*}\frac{15}{5}\frac{m\&ms}{minutes}\end{align*} Next, calculate each of the unit rates recognizing that the word "faster" applies to minutes and reducing that time to 1. Matt 183=61\begin{align*}\frac{18}{3}=\frac{6}{1}\end{align*} Matt can melt 6 m&ms in 1 minute. Kelly 155=31\begin{align*}\frac{15}{5}=\frac{3}{1}\end{align*} Kelly can melt 3 m&ms in 1 minute. Then, compare the unit rates. 6 >\begin{align*}>\end{align*} The answer is that Matt melts m&ms faster. Example 3 Frank read 8 books in three weeks. It took Bonnie 4 weeks to read 10 books. Who is the faster reader? First, write the fractions being sure to keep the units consistent. Frank 83booksweeks\begin{align*}\frac{8}{3}\frac{books}{weeks}\end{align*} Bonnie 104booksweeks\begin{align*}\frac{10}{4}\frac{books}{weeks}\end{align*} Next, reduce to unit rates by reducing the times to 1.Faster can refer to time, not books. Frank 83=2.67\begin{align*}\frac{8}{3}=2.67\end{align*} Frank read 2.67, or 223\begin{align*}\frac{2}{3}\end{align*}books per week. Bonnie 104=2.5\begin{align*}\frac{10}{4}=2.5\end{align*} Bonnie read 2.5, or 212\begin{align*}\frac{1}{2}\end{align*}books per week. Then, compare. 2.67 >\begin{align*}>\end{align*} 2.5 The answer is that Frank is the faster reader. ### Follow Up License: CC BY-NC 3.0 Remember Connie and Daniel and the cookie decorating? Connie completed 150 cookies in three hours, and it took Daniel 2 hours to decorate 98 cookies. Who is the faster cookie decorator? First, write the fractions with consistent units. Connie 1503cookieshours\begin{align*}\frac{150}{3}\frac{cookies}{hours}\end{align*} Daniel 982cookieshours\begin{align*}\frac{98}{2}\frac{cookies}{hours}\end{align*} Next, reduce to lowest terms. 1503=50\begin{align*}\frac{150}{3}=50\end{align*} Connie decorated 50 cookies per hour. 982=49\begin{align*}\frac{98}{2}=49\end{align*} Daniel decorated 49 cookies per hour. Then, compare. 50>\begin{align*}>\end{align*}49 The answer is that Connie decorated faster. ### Video Review ### Explore More Determine whether each is equivalent or not. 1. 60 units in 3 minutes and 80 units in 4 minutes 2.16 for 8 pounds and $22 for 10 pounds 3. 50 kilometers in 2 hours and 75 kilometers in 3 hours Solve each problem. 4. Max paid$45 for 15 gallons of gasoline. What was the cost per gallon of gasoline?
5. Mr. Brown paid $8.28 for 12 cans of green beans. Express this cost as a unit price. 6. A train travels 480 kilometers in 3 hours. Express this speed as a unit rate. 7. Mrs. Jenkins paid$50 for 40 square feet of carpeting. What was the cost per square foot for the carpeting?
8. A copy machine can produce 310 copies in 5 minute. How many copies can the machine produce per minute?
Nadia needs to buy some cheddar cheese. An 8-ounce package of cheddar cheese costs $2.40. A 12-ounce package of cheddar cheese costs$3.36.
9. Find the unit price for the 8-ounce package.
10. Find the unit price for the 12-ounce package.
11. Which is the better buy? How many cents per ounce cheaper is the better buy?
Joe drove 141 miles in 3 hours. His cousin Amy drove 102 miles in 2 hours. Assume both cousins were driving at constant speeds.
12. How fast was Joe driving, in miles per hour?
13. How fast was Amy driving, in miles per hour?
14. Who was driving at a faster rate of speed?
15. How much faster was the faster person traveling?
### Vocabulary Language: English
Per
Per
Per means "for each". It is a word that indicates a rate is being used. |
Lesson video
In progress...
Hi I'm Miss Kidd-Rossiter, and I'm going to be taking your lesson today on dividing into a ratio.
It's going to build on the work that we've already done laying the groundwork for dividing into ratio.
Before we get started, please make sure you're in a nice quiet area and that you've got no distractions.
You're also going to need something to write with and something to write on.
If you can a ruler would be really helpful for today's lesson, 'cause you're going to be drawing lots of nice diagrams. If you need to pause the video here to get anything, then please do.
If not, let's get going.
So for today's Try this activity, you've got to share 100 pounds between three charities so that each charity gets an exact whole amount of pounds.
When you've done that, you need to express the sharing as a ratio and then figure out what fraction of the £100 is donated to each charity.
So pause the video here and have a go at this activity.
Excellent work well done.
There's absolutely loads of combinations here of money that you could have given away.
So I can't possibly go through all of them, but let's go through one example together.
I'm going to give Charity A, the first charity £30.
I'm going to give Charity B £50, and I'm going to give Charity C £20.
Let's just double check that that adds up to my £100.
So 30 add 50 is 80 add more 20 is a hundred pounds.
So that's correct.
So I'm just going to write it a bit bigger, 30 pounds to 50 pounds to 20 pounds.
How else could I write this ratio? Could I simplify it? Tell me now.
Excellent, I could couldn't I? It would be 3:5:2.
So that's my ratio in its simplest form.
What fraction then of the 100 pounds does Charity A get? So Charity A will get 30 pounds out of the 100 pounds, which we can simplify to three tenths.
What about Charity B then? So Charity B will get 50 pounds out of the 100 pounds, which simplifies to five tenths, doesn't it? Or we can simplify it further to a half, so Charity B is getting half the money and then Charity C is getting 20 pounds out of the 100 pounds, which we know simplifies to two tenths or one fifth.
So Charity C is getting one fifth of the money.
Can we see how the two tents, the five tenths and the three tenths might relate to this ratio here? Pause the video now and think about that.
Excellent.
I can see that Charity A is three parts of the ratio out of a total of 10, B is five parts out of a total of 10 and C is two parts out of a total of 10.
Excellent work, well done.
Let's have a look at the connect part of today's lesson then.
So we can represent dividing 60 pounds between charities in the ratio, one to three to six using a bar model.
So this is building on the work that we've already done on using bar models.
You can see there that I've drawn my one part of the ratio here at the top.
I've drawn my three parts here and I've drawn my six parts here.
As we've mentioned previously, it's really important that all the pieces of this bar model are the same exact size.
So they're the same width and they're the same height because each individual part of my ratio is worth the same amount.
We are then asked, what is the donation to each charity and what fraction of the total is donated to each charity.
So let's have a go at this together.
So we've got 60 pounds and we're sharing it in the ratio 1:3:6.
So how many parts of my ratio do I have in total? Tell me now.
Excellent.
We have 10 don't we? So to find what one part is, we'll do 60 pounds divided by our 10 parts.
And that tells us that one part of the ratio must be worth six pounds.
So I can write that in here, each part has got to be worth six pounds.
So if each part is worth six pounds, how much is each charity getting? Pause the video now and work that out.
Excellent! So let's look at the first charity then.
So we'll call that Charity A, the one that's getting one part of the ratio.
So how much does Charity A get? Tell me now.
Excellent.
Six pounds.
How much does Charity B get, tell me now.
Excellent.
18 pounds because we do six pounds, times three don't we? 'Cause we've got three parts.
And then Charity C, how much does Charity C get? Tell me now.
Excellent.
36 pounds because we've got the six pounds and we've got six parts, haven't we? Brilliant.
So what fraction of the total is donated to each charity then? So Charity A is getting six pounds out of 60, which we know is one tenth.
So that's how much Charity A is getting.
I need remember to put in my units, don't I? Charity B is getting 18 pounds out to 60 pounds, which we know simplifies to three tenths and Charity C is getting 36 pounds out of 60 pounds, which we know simplifies to six tenths.
And again, we could simplify that further to three fifths.
What you're going to do now is pause the video and have a go at creating your own bar models to represent dividing 60 pounds between three charities in the ratios that are on the screen.
Then you've also got the question, what fraction of the largest share is the smallest share.
So pause the video here and have a go at this activity.
Excellent work, well done.
We're going to go through these quite quickly because I don't want to just be doing the same thing over and over again.
So the first one then, we've got our 60 pounds and we're dividing it between six parts in total so that gives us the one part is 10 pounds.
So that means that Charity A is getting 10 pounds, Charity B is getting 20 pounds and Charity C is getting 30 pounds.
So let's do this last one together.
What fraction of the largest share is the smallest share? So what is our smallest share first of all? Tell me now.
Excellent, it's 10 pounds and what is our largest share? Tell me now.
Excellent.
It's 30 pounds isn't it? So 10 pounds out of 30 pounds, we simplify to one third.
Excellent.
Second one then.
So this time we're dividing our 60 pounds into 12 equal parts.
So that means that each one part is worth five pounds and that means then that Charity A will get 15 pounds, Charity B will get 20 pounds and Charity C will get 25 pounds.
So fraction then, the smallest share is 15 pounds out of the largest share, which is 25 pounds.
And we can simplify that down to three fifths.
Excellent.
The third one then, we're going to look at together.
So we're sharing in the ratio N:N:N.
Now we don't know what this number is.
This N could represent any number, it could represent one.
So we could draw each part the same size there.
Remember you're doing it with a ruler and it will be much neater than mine.
So we're saying that this part here is N.
Could be that N is representing two, in which case it would be like this.
Could be that N is representing four.
N could be any number here but what's important is that each bar that you draw is the same size.
So Charity A, Charity B and Charity C are getting an equal share of the 60 pounds.
And again, yours would be much neater than mine it would be drawn with a ruler and would be exactly the same size.
So if we've got that, these are 60 pounds, but we've got our 60 pounds divided by three because each share is equal so we're dividing it into the three charities equally, which tells us that each share will be 20 pounds.
So each charity, Charity A, and Charity B and Charity C will each receive 20 pounds.
So what fraction of the largest share is the smallest share where they're all the same.
So 20 divided by 20 gives us one.
You're now going to apply what you've learned to the independent tasks so pause the video here, navigate to the independent task and give it your best go.
When you're ready, resume the video.
Good luck.
Excellent work on that independent task, I hope you gave it a really good go.
I'm going to put the answers on the screen now and some of them we'll talk through.
So pause the video at any point to mark your work.
Here's the first set of questions.
So please pause the video here and check what you've done.
If you wrote anything different for what you notice, that's absolutely fine.
You could have worded it completely differently to me and I'm sure it's still a great explanation.
Match the descriptions of the sharing then.
So the ones that are coloured the same are matched together.
So that's this one here and this one here and then this one, this one and this one.
We're moving on to the explore activity now, Antoni and Binh are each sharing the same amount of money between two charities.
Antoni is sharing his money in the ratio 1:3, Binh is sharing her money in the ratio 1:5.
Antoni then says, "Together we're sharing in the ratio 2:8." Do you agree with Antony? And if you do or don't, can you explain your reasoning using a bar model? So pause the video here and have a go at this task.
Antoni said he'll share his money in the ratio 1:3, so we could represent that using this bar model, couldn't we? And Binh said she's sharing her money in the ratio 1:5.
So we could represent using this bar model.
Do we agree that together, we're sharing in the ratio 2:8? Pause the video now and tell me what you think.
Excellent.
Even though we have two red parts and eight blue parts there, we can't say that it's the ratio 2:8 because we can quite clearly see that each of Binh's parts is much smaller than each of Antoni's parts.
And we know that in a ratio, all the parts have got to be the same size.
That's it for today's lesson so thank you so much for all your hard work on dividing into a ratio.
I hope you've enjoyed it as much as I have.
Please don't forget to go and take the end of lesson quiz so that you can show me what you've learned and I hope to see you again soon for some more maths.
Bye. |
# Difference between revisions of "2012 AIME I Problems/Problem 12"
## Problem 12
Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$
## Solution
### Solution 1
Without loss of generality, set $CB = 1$. Then, by the Angle Bisector Theorem on triangle $DCB$, we have $CD = \frac{8}{15}$. We apply the Law of Cosines to triangle $DCB$ to get $1 + \frac{64}{225} - \frac{8}{15} = BD^{2}$, which we can simplify to get $BD = \frac{13}{15}$.
Now, we have $\cos \angle B = \frac{1 + \frac{169}{225} - \frac{64}{225}}{\frac{26}{15}}$ by another application of the Law of Cosines to triangle $DCB$, so $\cos \angle B = \frac{11}{13}$. In addition, $\sin \angle B = \sqrt{1 - \frac{121}{169}} = \frac{4\sqrt{3}}{13}$, so $\tan \angle B = \frac{4\sqrt{3}}{11}$.
Our final answer is $4+3+11 = \boxed{018}$.
### Solution 2
(This solution does not use the Angle Bisector Theorem or the Law of Cosines, but it uses the Law of Sines and more trig)
Find values for all angles in terms of $\angle B$. $\angle CEB = 150-B$, $\angle CED = 30+B$, $\angle CDE = 120-B$, $\angle CDA = 60+B$, and $\angle A = 90-B$.
Use the law of sines on $\triangle CED$ and $\triangle CEB$:
In $\triangle CED$, $\frac{8}{\sin 30} = \frac{CE}{\sin (120-B)}$. This simplifies to $16 = \frac{CE}{\sin (120-B)}$.
In $\triangle CEB$, $\frac{15}{\sin 30} = \frac{CE}{\sin B}$. This simplifies to $30 = \frac{CE}{\sin B}$.
Solve for $CE$ and equate them so that you get $16\sin (120-B) = 30\sin B$.
From this, $\frac{8}{15} = \frac{\sin B}{\sin (120-B)}$.
Use a trig identity on the denominator on the right to obtain: $\frac{8}{15} = \frac{\sin B}{\sin 120 \cos B - \cos 120 \sin B}$
This simplifies to $\frac{8}{15} = \frac{\sin B}{\frac{\sqrt{3}\cos B}{2} + \frac{\sin B}{2}} = \frac{\sin B}{\frac{\sqrt{3} \cos B + \sin B}{2}} = \frac{2\sin B}{\sqrt{3}\cos B + \sin B}$
This gives $8\sqrt{3}\cos B+8\sin B=30\sin B$ Dividing by $\cos B$, we have ${8\sqrt{3}}= 22\tan B$
$\tan{B} = \frac{8\sqrt{3}}{22} = \frac{4\sqrt{3}}{11}$. Our final answer is $4 + 3 + 11 = \boxed{018}$.
### Solution 3
Let $X$ be the foot of the perpendicular from $D$ to $\overline{BC}$, and let $Y$ be the foot of the perpendicular from $E$ to $\overline{BC}$.
Now let $EY=x$. Clearly, triangles $EYB$ and $DXB$ are similar with $\frac{BE}{BD}=\frac{15}{15+8}=\frac{15}{23}=\frac{EY}{DX}$, so $DX=\frac{23}{15}x$.
Since triangles $CDX$ and $CEY$ are 30-60-90 right triangles, we can easily find other lengths in terms of $x$. For example, we see that $CY=x\sqrt{3}$ and $CX=\frac{\frac{23}{15}x}{\sqrt{3}}=\frac{23\sqrt{3}}{45}x$. Therefore $XY=CY-CX=x\sqrt{3}-\frac{23\sqrt{3}}{45}x=\frac{22\sqrt{3}}{45}x$.
Again using the fact that triangles $EYB$ and $DXB$ are similar, we see that $\frac{BX}{BY}=\frac{XY+BY}{BY}=\frac{XY}{BY}+1=\frac{23}{15}$, so $BY=\frac{15}{8}XY=\frac{15}{8}*\frac{22\sqrt{3}}{45}=\frac{11\sqrt{3}}{2}$.
Thus $\tan \angle B = \frac{x}{\frac{11\sqrt{3}}{12}x}=\frac{4\sqrt{3}}{11}$, and our answer is $4+3+11=\boxed{018}$.
### Solution 4
(Another solution without trigonometry)
Extend $CD$ to point $F$ such that $\overline{AF} \parallel \overline{CB}$. It is then clear that $\triangle AFD$ is similar to $\triangle BCD$.
Let $AC=p$, $BC=q$. Then $\tan \angle B = p/q$.
With the Angle Bisector Theorem, we get that $CD=\frac{8}{15}q$. From 30-60-90 $\triangle CAF$, we get that $AF=\frac{1}{\sqrt{3}}p$ and $FD=FC-CD=\frac{2}{\sqrt{3}}p-\frac{8}{15}q$.
From $\triangle AFD \sim \triangle BCD$, we have that $\frac{FD}{CD}=\frac{FA}{CB}=\frac{\frac{2}{\sqrt{3}}p-\frac{8}{15}q}{\frac{8}{15}q}=\frac{\frac{1}{\sqrt{3}}p}{q}$. Simplifying yields $\left(\frac{p}{q}\right)\left(\frac{2\sqrt{3}}{3}*\frac{15}{8}-\frac{\sqrt{3}}{3}\right)=1$, and $\tan \angle B=\frac{p}{q}=\frac{4\sqrt{3}}{11}$, so our answer is $4+3+11=\boxed{018}$. |
# I.S.I. 10+2 Subjectives Solution (2 problems)
P164. Show that the area of the bounded region enclosed between the curves $(y^3=x^2)$ and $(y=2-x^2)$, is $(2\frac{2}{15})$.
Solution:
Note that $(y=x^{\frac{2}{3}})$ is an even function (green line).
P165. Find the area of the region in the xy plane, bounded by the graphs of $(y=x^2)$, x+y = 2 and $(y=-\sqrt {x})$
Solution:
The parabola and straight line intersects at (1,1) (we find that by solving the $(y=x^2)$ and x+y=2)
Thus the area is found by adding area under parabola (from 0 to 1) and area under straight line (from 1 to 2).
$(\int^1_0 x^2,dx=\left[\frac{x^3}{3}\right]^1_0=\frac{1}{3})$ (area under parabola)
area under straight line above 'x' axis is the triangle with height 1 unit and base 1 unit (from x=1 to x=2, area under x+y=2)
that area = $(\frac{1}{2}\times 1\times 1=\frac{1}{2})$
Thus total area above x axis (of the required region) is $(\frac{1}{2}+\frac{1}{3}=\frac{5}{6})$
Now we come to the region below 'x' axis.
x+y = 2 and $(y=-\sqrt{x})$ intersect at (4, -2) (found by solving the two equations). We calculate the area under the curve $(y=-\sqrt{x})$ from x=0 to x=4 and subtract from it the area of the triangle with base from x=2 to x=4 and height =2 (hence the area of the triangle to be subtracted is 2 sq unit).
Area under the square root curve is
$(|\int^4_0 -\sqrt{x},dx|=\int^4_0 \sqrt{x},dx= \left[\frac {x^{\frac{1}{2}+1}} {\frac{1}{2}+1}\right]^4_0)$.
=$(\frac {2}{3}\times 8=\frac {16}{3})$
Delete 2 square unit from this and add the area computed before (above 'x' axis).
Area = $(\frac {16}{3} - 2 + \frac{5}{6} = \frac{25}{6})$ (ANS)
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### Cheenta. Passion for Mathematics
Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject. |
# Difference between revisions of "2019 AMC 8 Problems/Problem 24"
## Problem 24
In triangle $ABC$, point $D$ divides side $\overline{AC}$ so that $AD:DC=1:2$. Let $E$ be the midpoint of $\overline{BD}$ and let $F$ be the point of intersection of line $BC$ and line $AE$. Given that the area of $\triangle ABC$ is $360$, what is the area of $\triangle EBF$?
$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("A",A,N); dot(B); label("B", B,SW);dot(C); label("C",C,SE); dot(DD); label("D",DD,NE); dot(EE); label("E",EE,NW); dot(FF); label("F",FF,S); [/asy]$
$\textbf{(A) }24\qquad\textbf{(B) }30\qquad\textbf{(C) }32\qquad\textbf{(D) }36\qquad\textbf{(E) }40$
## Solution 1
Draw $X$ on $\overline{AF}$ such that $XD$ is parallel to $BC$. That makes triangles $BEF$ and $EXD$ congruent since $BE = ED$. $FC=3XD$ so $BC=4BF$. Since $AF=3EF$ ($XE=EF$ and $AX=\frac13 AF$, so $XE=EF=\frac13 AF$), the altitude of triangle $BEF$ is equal to $\frac{1}{3}$ of the altitude of $ABC$. The area of $ABC$ is $360$, so the area of $BEF=\frac{1}{3} \cdot \frac{1}{4} \cdot 360=\boxed{\textbf{(B) }30}$
## Solution 2 (Mass Points)
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(7cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.28, xmax = 6.28, ymin = -5.49, ymax = 5.73; /* image dimensions */ pen wrwrwr = rgb(0.3803921568627451,0.3803921568627451,0.3803921568627451); /* draw figures */ draw((0.28,2.39)--(-2.8,-1.17), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(3.78,-1.05), linewidth(2) + wrwrwr); draw((3.78,-1.05)--(0.28,2.39), linewidth(2) + wrwrwr); draw((-2.8,-1.17)--(1.2887445398528459,1.3985482236874887), linewidth(2) + wrwrwr); draw((0.28,2.39)--(-0.7199623188673492,-1.1320661821070033), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.1,2.93), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((4.48,-1.28), 0.46818799642878506), linewidth(2) + wrwrwr); draw(circle((1.98,1.56), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-3.36,-1.62), 0.46818799642878517), linewidth(2) + wrwrwr); draw(circle((0.16,0.14), 0.46818799642878495), linewidth(2) + wrwrwr); draw(circle((-0.74,-1.81), 0.46818799642878495), linewidth(2) + wrwrwr); /* dots and labels */ dot((0.28,2.39),dotstyle); label("A", (0.36,2.59), NE * labelscalefactor); dot((-2.8,-1.17),dotstyle); label("B", (-2.72,-0.97), NE * labelscalefactor); dot((3.78,-1.05),dotstyle); label("C", (3.86,-0.85), NE * labelscalefactor); dot((1.2887445398528459,1.3985482236874887),dotstyle); label("D", (1.36,1.59), NE * labelscalefactor); dot((-0.7199623188673492,-1.1320661821070033),dotstyle); label("F", (-0.64,-0.93), NE * labelscalefactor); dot((-0.2815567696989588,0.41208536204620183),linewidth(4pt) + dotstyle); label("E", (-0.2,0.57), NE * labelscalefactor); label("2", (-0.18,2.81), NE * labelscalefactor,wrwrwr); label("1", (4.4,-1.39), NE * labelscalefactor,wrwrwr); label("3", (1.9,1.45), NE * labelscalefactor,wrwrwr); label("3", (-3.44,-1.73), NE * labelscalefactor,wrwrwr); label("6", (0.08,0.03), NE * labelscalefactor,wrwrwr); label("4", (-0.82,-1.93), NE * labelscalefactor,wrwrwr); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
First, when we see the problem, we see ratios, and we see that this triangle basically has no special properties (right, has medians, etc.) and this screams mass points at us.
First, we assign a mass of $2$ to point $A$. We figure out that $C$ has a mass of $1$ since $2\times1 = 1\times2$. Then, by adding $1+2 = 3$, we get that point $D$ has a mass of $3$. By equality, point $B$ has a mass of $3$ also.
Now, we add $3+3 = 6$ for point $E$ and $3+1 = 4$ for point $F$.
Now, $BF$ is a common base for triangles $ABF$ and $EBF$, so we figure out that the ratios of the areas is the ratios of the heights which is $\frac{AE}{EF} = 2:1$. So, $EBF$'s area is one third the area of $ABF$, and we know the area of $ABF$ is $\frac{1}{4}$ the area of $ABC$ since they have the same heights but different bases.
So we get the area of $EBF$ as $\frac{1}{3}\times\frac{1}{4}\times360 = \boxed{\textbf{(B) }30}$ -Brudder
Note: We can also find the ratios of the areas using the reciprocal of the product of the mass points of $EBF$ over the product of the mass points of $ABC$ which is $\frac{2\times3\times1}{3\times6\times4}\times360$ which also yields $\boxed{\textbf{(B) }30}$ -Brudder
## Solution 3
$\frac{BF}{FC}$ is equal to $\frac{\textrm{The area of triangle ABE}}{\textrm{The area of triangle ACE}}$. The area of triangle $ABE$ is equal to $60$ because it is equal to on half of the area of triangle $ABD$, which is equal to one third of the area of triangle $ABC$, which is $360$. The area of triangle $ACE$ is the sum of the areas of triangles $AED$ and $CED$, which is respectively $60$ and $120$. So, $\frac{BF}{FC}$ is equal to $\frac{60}{180}$=$\frac{1}{3}$, so the area of triangle $ABF$ is $90$. That minus the area of triangle $ABE$ is $\boxed{\textbf{(B) }30}$. ~~SmileKat32
## Solution 4 (Similar Triangles)
Extend $\overline{BD}$ to $G$ such that $\overline{AG} \parallel \overline{BC}$ as shown: $[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label("A", A, N); label("B", B, WSW); label("C", C, ESE); label("D", D, dir(0)*1.5); label("E", E, SE); label("F", F, S); label("G", G, ENE); [/asy]$ Then $\triangle ADG \sim \triangle CDB$ and $\triangle AEG \sim \triangle FEB$. Since $CD = 2AD$, triangle $CDB$ has four times the area of triangle $ADG$. Since $[CDB] = 240$, we get $[ADG] = 60$.
Since $[AED]$ is also $60$, we have $ED = DG$ because triangles $AED$ and $ADG$ have the same height and same areas and so their bases must be the congruent. Thus triangle $AEG$ has twice the side lengths and therefore four times the area of triangle $BEF$, giving $[BEF] = (60+60)/4 = \boxed{\textbf{(B) }30}$.
$[asy] size(8cm); pair A, B, C, D, E, F, G; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); G = (4.5, 3); draw(A--B--C--A--G--B); draw(A--F); label("A", A, N); label("B", B, WSW); label("C", C, ESE); label("D", D, dir(0)*1.5); label("E", E, SE); label("F", F, S); label("G", G, ENE); label("60", (A+E+D)/3); label("60", (A+E+B)/3); label("60", (A+G+D)/3); label("30", (B+E+F)/3); [/asy]$ (Credit to MP8148 for the idea)
## Solution 5 (Area Ratios)
$[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("A", A, N); label("B", B, WSW); label("C", C, ESE); label("D", D, dir(0)*1.5); label("E", E, SSE); label("F", F, S); label("60", (A+E+D)/3); label("60", (A+E+B)/3); label("120", (D+E+C)/3); label("x", (B+E+F)/3); label("120-x", (F+E+C)/3); [/asy]$ As before we figure out the areas labeled in the diagram. Then we note that $$\dfrac{EF}{AE} = \dfrac{x}{60} = \dfrac{120-x}{180}.$$Solving gives $x = \boxed{\textbf{(B) }30}$. (Credit to scrabbler94 for the idea)
## Solution 6 (Coordinate Bashing)
Let $ADB$ be a right triangle, and $BD=CD$
Let $A=(-2\sqrt{30}, 0)$
$B=(0, 4\sqrt{30})$
$C=(4\sqrt{30}, 0)$
$D=(0, 0)$
$E=(0, 2\sqrt{30})$
$F=(\sqrt{30}, 3\sqrt{30})$
The line $\overleftrightarrow{AE}$ can be described with the equation $y=x-2\sqrt{30}$
The line $\overleftrightarrow{BC}$ can be described with $x+y=4\sqrt{30}$
Solving, we get $x=3\sqrt{30}$ and $y=\sqrt{30}$
Now we can find $EF=BF=2\sqrt{15}$
$[\bigtriangleup EBF]=\frac{(2\sqrt{15})^2}{2}=\boxed{\textbf{(B) }30}\blacksquare$
-Trex4days
## Solution 7
$[asy] /* Geogebra to Asymptote conversion, documentation at artofproblemsolving.com/Wiki go to User:Azjps/geogebra */ import graph; size(15cm); real labelscalefactor = 0.5; /* changes label-to-point distance */ pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ pen dotstyle = black; /* point style */ real xmin = -6.61, xmax = 16.13, ymin = -6.4, ymax = 6.42; /* image dimensions */ /* draw figures */ draw(circle((0,0), 5), linewidth(2)); draw((-4,-3)--(4,3), linewidth(2)); draw((-4,-3)--(0,5), linewidth(2)); draw((0,5)--(4,3), linewidth(2)); draw((12,-1)--(-4,-3), linewidth(2)); draw((0,5)--(0,-5), linewidth(2)); draw((-4,-3)--(0,-5), linewidth(2)); draw((4,3)--(0,2.48), linewidth(2)); draw((4,3)--(12,-1), linewidth(2)); draw((-4,-3)--(4,3), linewidth(2)); /* dots and labels */ dot((0,0),dotstyle); label("E", (0.27,-0.24), NE * labelscalefactor); dot((-5,0),dotstyle); dot((-4,-3),dotstyle); label("B", (-4.45,-3.38), NE * labelscalefactor); dot((4,3),dotstyle); label("D", (4.15,3.2), NE * labelscalefactor); dot((0,5),dotstyle); label("A", (-0.09,5.26), NE * labelscalefactor); dot((12,-1),dotstyle); label("C", (12.23,-1.24), NE * labelscalefactor); dot((0,-5),dotstyle); label("G", (0.19,-4.82), NE * labelscalefactor); dot((0,2.48),dotstyle); label("I", (-0.33,2.2), NE * labelscalefactor); dot((0,0),dotstyle); label("E", (0.27,-0.24), NE * labelscalefactor); dot((0,-2.5),dotstyle); label("F", (0.23,-2.2), NE * labelscalefactor); clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); /* end of picture */ [/asy]$
Let $A[\Delta XYZ]$ = $\text{Area of Triangle XYZ}$
$A[\Delta ABD]: A[\Delta DBC] :: 1:2 :: 120:240$
$A[\Delta ABE] = A[\Delta AED] = 60$ (the median divides the area of the triangle into two equal parts)
Construction: Draw a circumcircle around $\Delta ABD$ with $BD$ as is diameter. Extend $AF$ to $G$ such that it meets the circle at $G$. Draw line $BG$.
$A[\Delta ABD] = A[\Delta ABG] = 120$ (Since $\square ABGD$ is cyclic)
But $A[\Delta ABE]$ is common in both with an area of 60. So, $A[\Delta AED] = A[\Delta BEG]$.
\therefore $A[\Delta AED] \cong A[\Delta BEG]$ (SAS Congruency Theorem).
In $\Delta AED$, let $DI$ be the median of $\Delta AED$.
Which means $A[\Delta AID] = 30 = A[\Delta EID]$
Rotate $\Delta DEA$ to meet $D$ at $B$ and $A$ at $G$. $DE$ will fit exactly in $BE$ (both are radii of the circle). From the above solutions, $\frac{AE}{EF} = 2:1$.
$AE$ is a radius and $EF$ is half of it implies $EF$ = $\frac{radius}{2}$.
Which means $A[\Delta BEF] \cong A[\Delta DEI]$
Thus $A[\Delta BEF] = \boxed{\textbf{(B) }30}$
~phoenixfire & flamewavelight
## Solution 8
$[asy] import geometry; unitsize(2cm); pair A,B,C,DD,EE,FF, M; B = (0,0); C = (3,0); M = (1.45,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("A",A,N); dot(B); label("B", B,SW);dot(C); label("C",C,SE); dot(DD); label("D",DD,NE); dot(EE); label("E",EE,NW); dot(FF); label("F",FF,S); draw(EE--M,StickIntervalMarker(1,1)); label("M",M,S); draw(A--DD,invisible,StickIntervalMarker(1,1)); dot((DD+C)/2); draw(DD--C,invisible,StickIntervalMarker(2,1)); [/asy]$ Using the ratio of $\overline{AD}$ and $\overline{CD}$, we find the area of $\triangle ADB$ is $120$ and the area of $\triangle BDC$ is $240$. Also using the fact that $E$ is the midpoint of $\overline{BD}$, we know $\triangle ADE = \triangle ABE = 60$. Let $M$ be a point such $\overline{EM}$ is parellel to $\overline{CD}$. We immediatley know that $\triangle BEM \sim BDC$ by $2$. Using that we can conclude $EM$ has ratio $1$. Using $\triangle EFM \sim \triangle AFC$, we get $EF:AE = 1:2$. Therefore using the fact that $\triangle EBF$ is in $\triangle ABF$, the area has ratio $\triangle BEF : \triangle ABE=1:2$ and we know $\triangle ABE$ has area $60$ so $\triangle BEF$ is $\boxed{\textbf{(B) }30}$. - fath2012
## Solution 9 (Menelaus's Theorem)
$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(B--DD);dot(A); label("A",A,N); dot(B); label("B", B,SW);dot(C); label("C",C,SE); dot(DD); label("D",DD,NE); dot(EE); label("E",EE,NW); dot(FF); label("F",FF,S); [/asy]$ By Menelaus's Theorem on triangle $BCD$, we have $$\dfrac{BF}{FC} \cdot \dfrac{CA}{DA} \cdot \dfrac{DE}{BE} = 3\dfrac{BF}{FC} = 1 \implies \dfrac{BF}{FC} = \dfrac13 \implies \dfrac{BF}{BC} = \dfrac14.$$ Therefore, $$[EBF] = \dfrac{BE}{BD}\cdot\dfrac{BF}{BC}\cdot [BCD] = \dfrac12 \cdot \dfrac 14 \cdot \left( \dfrac23 \cdot [ABC]\right) = \boxed{\textbf{(B) }30}.$$
## Solution 10 (Graph Paper)
$[asy] unitsize(2cm); pair A,B,C,D,E,F,a,b,c,d,e,f; A = (2,3); B = (0,2); C = (2,0); D = (2/3)*A+(1/3)*C; E = (B+D)/2; F = intersectionpoint(B--C,A--A+2*(E-A)); a = (0,0); b = (1,0); c = (2,1); d = (1,3); e = (0,3); f = (0,1); draw(a--C,dashed); draw(f--c,dashed); draw(e--A,dashed); draw(a--e,dashed); draw(b--d,dashed); draw(A--B--C--cycle); draw(A--F); draw(B--D); dot(A); label("A",A,NE); dot(B); label("B",B,dir(180)); dot(C); label("C",C,SE); dot(D); label("D",D,dir(0)); dot(E); label("E",E,SE); dot(F); label("F",F,SW); [/asy]$ Note: If graph paper is unavailable, this solution can still be used by constructing a small grid on a sheet of blank paper.
As triangle $ABC$ is loosely defined, we can arrange its points such that the diagram fits nicely on a coordinate plane. By doing so, we can construct it on graph paper and be able to visually determine the relative sizes of the triangles.
As point $D$ splits line segment $\overline{AC}$ in a $1:2$ ratio, we draw $\overline{AC}$ as a vertical line segment $3$ units long. Point $D$ is thus $1$ unit below point $A$ and $2$ units above point $C$. By definition, Point $E$ splits line segment $\overline{BD}$ in a $1:1$ ratio, so we draw $\overline{BD}$ $2$ units long directly left of $D$ and draw $E$ directly between $B$ and $D$, $1$ unit away from both.
We then draw line segments $\overline{AB}$ and $\overline{BC}$. We can easily tell that triangle $ABC$ occupies $3$ square units of space. Constructing line $AE$ and drawing $F$ at the intersection of $AE$ and $BC$, we can easily see that triangle $EBF$ forms a right triangle occupying $\frac{1}{4}$ of a square unit of space.
The ratio of the areas of triangle $EBF$ and triangle $ABC$ is thus $\frac{1}{4}\div3=\frac{1}{12}$, and since the area of triangle $ABC$ is $360$, this means that the area of triangle $EBF$ is $\frac{1}{12}\times360=\boxed{\textbf{(B) }30}$. ~emerald_block
Additional note: There are many subtle variations of this triangle; this method is one of the more compact ones. ~i_equal_tan_90
## Solution 11
$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF,G; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); G = (1.5,0); draw(A--B--C--cycle); draw(A--FF); draw(B--DD); draw(G--DD); label("A",A,N); label("B", B,SW); label("C",C,SE); label("D",DD,NE); label("E",EE,NW); label("F",FF,S); label("G",G,S); [/asy]$ We know that $AD = \dfrac{1}{3} AC$, so $[ABD] = \dfrac{1}{3} [ABC] = 120$. Using the same method, since $BE = \dfrac{1}{2} BD$, $[ABE] = \dfrac{1}{2} [ABD] = 60$. Next, we draw $G$ on $\overline{BC}$ such that $\overline{DG}$ is parallel to $\overline{AF}$ and create segment $DG$. We then observe that $\triangle AFC \sim \triangle DGC$, and since $AD:DC = 1:2$, $FG:GC$ is also equal to $1:2$. Similarly (no pun intended), $\triangle DBG \sim \triangle EBF$, and since $BE:ED = 1:1$, $BF:FG$ is also equal to $1:1$. Combining the information in these two ratios, we find that $BF:FG:GC = 1:1:2$, or equivalently, $BF = \dfrac{1}{4} BC$. Thus, $[BFA] = \dfrac{1}{4} [BCA] = 90$. We already know that $[ABE] = 60$, so the area of $\triangle EBF$ is $[BFA] - [ABE] = \boxed{\textbf{(B) }30}$. ~i_equal_tan_90
## Solution 12 (Fastest Solution if you have no time)
The picture is misleading. Assume that the triangle ABC is right.
Then find two factors of $720$ that are the closest together so that the picture becomes easier in your mind. Quickly searching for squares near $720$ to use difference of squares, we find $24$ and $30$ as our numbers. Then the coordinates of D are $(10,16)$(note, A=0,0). E is then $(5,8)$. Then the equation of the line AE is $-16x/5+24=y$. Plugging in $y=0$, we have $x=\dfrac{15}{2}$. Now notice that we have both the height and the base of EBF.
Solving for the area, we have $(8)(15/2)(1/2)=30$.
## Solution 13
$AD : DC = 1:2$, so $ADB$ has area $120$ and $CDB$ has area $240$. $BE = ED$ so the area of $ABE$ is equal to the area of $ADE = 60$. Draw $\overline{DG}$ parallel to $\overline{AF}$.
Set area of BEF = $x$. BEF is similar to BDG in ratio of 1:2
so area of BDG = $4x$, area of EFDG=$3x$, and area of CDG$=240-4x$.
CDG is similar to CAF in ratio of 2:3 so area CDG = $4/9$ area CAF, and area AFDG=$5/4$ area CDG.
Thus $60+3x=5/4(240-4x)$ and $x=30$. ~EFrame
## Solution 14 - Geometry & Algebra
$[asy] unitsize(2cm); pair A,B,C,DD,EE,FF; B = (0,0); C = (3,0); A = (1.2,1.7); DD = (2/3)*A+(1/3)*C; EE = (B+DD)/2; FF = intersectionpoint(B--C,A--A+2*(EE-A)); draw(A--B--C--cycle); draw(A--FF); draw(DD--FF,blue); draw(B--DD);dot(A); label("A",A,N); dot(B); label("B", B,SW);dot(C); label("C",C,SE); dot(DD); label("D",DD,NE); dot(EE); label("E",EE,NW); dot(FF); label("F",FF,S); [/asy]$
We draw line $FD$ so that we can define a variable $x$ for the area of $\triangle BEF = \triangle DEF$. Knowing that $\triangle ABE$ and $\triangle ADE$ share both their height and base, we get that $ABE = ADE = 60$.
Since we have a rule where 2 triangles, ($\triangle A$ which has base $a$ and vertex $c$), and ($\triangle B$ which has Base $b$ and vertex $c$)who share the same vertex (which is vertex $c$ in this case), and share a common height, their relationship is : Area of $A : B = a : b$ (the length of the two bases), we can list the equation where $\frac{ \triangle ABF}{\triangle ACF} = \frac{\triangle DBF}{\triangle DCF}$. Substituting $x$ into the equation we get:
$$\frac{x+60}{300-x} = \frac{2x}{240-2x}$$. $$(2x)(300-x) = (60+x)(240-x).$$ $$600-2x^2 = 14400 - 120x + 240x - 2x^2.$$ $$480x = 14400.$$ and we now have that $\triangle BEF=30.$ ~$\bold{\color{blue}{onionheadjr}}$
## Video Solutions
https://youtu.be/Ns34Jiq9ofc —DSA_Catachu
https://www.youtube.com/watch?v=nm-Vj_fsXt4 - Happytwin (Another video solution)
## Solution 16 (Straightforward Solution)
$[asy] size(8cm); pair A, B, C, D, E, F; B = (0,0); A = (2, 3); C = (5, 0); D = (3, 2); E = (1.5, 1); F = (1.25, 0); draw(A--B--C--A--D--B); draw(A--F); draw(E--C); label("A", A, N); label("B", B, WSW); label("C", C, ESE); label("D", D, dir(0)*1.5); label("E", E, SSE); label("F", F, S); label("60", (A+E+D)/3); label("60", (A+E+B)/3); label("120", (D+E+C)/3); label("x", (B+E+F)/3); label("120-x", (F+E+C)/3); [/asy]$ Since $AD:DC=1:2$ thus $\triangle ABD=\frac{1}{3} \cdot 360 = 120.$
Similarly, $\triangle DBC = \frac{2}{3} \cdot 360 = 240.$
Now, since $E$ is a midpoint of $BD$, $\triangle ABE = \triangle AED = 120 \div 2 = 60.$
We can use the fact that $E$ is a midpoint of $BD$ even further. Connect lines $E$ and $C$ so that $\triangle BEC$ and $\triangle DEC$ share 2 sides.
We know that $\triangle BEC=\triangle DEC=240 \div 2 = 120$ since $E$ is a midpoint of $BD.$
Let's label $\triangle BEF$ $x$. We know that $\triangle EFC$ is $120-x$ since $\triangle BEC = 120.$
Note that with this information now, we can deduct more things that are needed to finish the solution.
Note that $\frac{EF}{AE} = \frac{120-x}{180} = \frac{x}{60}.$ because of triangles $EBF, ABE, AEC,$ and $EFC.$
We want to find $x.$
This is a simple equation, and solving we get $x=\boxed{\textbf{(B)}30}.$
~mathboy282, an expanded solution of Solution 5, credit to scrabbler94 for the idea.
## Note
This question is extremely similar to 1971 AHSME Problem 26. |
Starting with maths: Patterns and formulas
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# Starting with maths: Patterns and formulas
Although there may be many occasions when you are given a formula to use, sometimes you may need to devise your own formulas, for example if you use a spreadsheet on a computer at home or at work. This section looks at the process of devising a formula in more detail.
Part of a spreadsheet that has been constructed to record monthly income and expenditure is shown below. It is similar to a balance sheet that you might draw up by hand and includes the monthly income and outgoings, the totals and the overall balance. It is used to keep track of expenditure, particularly to try and ensure that the balance remains positive. However the spreadsheet has been stored on a computer rather than on paper and is updated regularly. One of the advantages of using a computer spreadsheet is that you can insert formulas into the spreadsheet to carry out the necessary calculations automatically, without using your calculator.
A spreadsheet is made up of rows and columns of cells. The columns are identified by letters and the rows by numbers. This enables you to identify each cell in the spreadsheet. For example the number 110.56 is in column B, row 11, so this cell is B11.
Figure 11
Cells can be found from their reference, by looking down the column and across the row. So cell A3 can be found by looking down column A and across row 3. This cell contains the word ‘Salary’. Notice that cells can contain either text or numbers.
• a.What is contained in the following cells?
• i.A5
• ii.A12
• iii.B15
• iv.B1
• b.What is the reference for the cells that contain the following?
• i.The number 792.59
• ii.The word ‘Food’
### Discussion
• a.
• i.The cell that is in both column A and row 5 contains the words ‘Total Income’.
• ii.Looking down column A and across row 12, cell A12 contains the word ‘Other’.
• iii.The cell that is in column B, row 15 contains the number 83.36.
• iv.The cell that is in column B, row 1 contains the heading ‘Amount /£’. This indicates that the entries in this column are amounts of money and that they are measured in £. This heading could also have been written as ‘Amount in £’ or ‘Amount (£)’.
• b.
• i.792.59 is in column B and row 13, so its reference is B13.
• ii.‘Food’ is in column A and row 9, so its reference is A9.
Now have another look at the spreadsheet and see if you can work out what information it shows. For example, if you look at row 3, this shows that the monthly salary is £850.28. Although cell B3 only contains the number 850.28, you know that this is measured in pounds from the heading in cell B1. Overall, the spreadsheet shows the different items that make up the monthly income and where money has been spent over the month as well. If you were keeping these records by hand, you would then need to calculate the total income, the total outgoings and the balance.
For example, to find the total income for the month, you would need to add the salary (£850.28) to the other income (£25.67). This gives the total income of £875.95. In other words, to calculate the value in cell B5, you need to add the values in cells B3 and B4 together. This can be written as the following formula:
value in B5 = value in B3 + value in B4
## Activity 11: Finding the balance
Formulas have also been used to calculate the total monthly outgoings and the balance. If you were working these calculations out by hand, what would you do? Check by comparing your answers with the values in cell B13 and B15. Try to write down the formulas for these cells in the form ‘value in B13 = …’.
### Discussion
To calculate the total monthly outgoings, you need to add up the individual outgoings on ‘Rent’, ‘Food’, ‘Transport’, ‘Regular Bills’ and ‘Other’. So the formula will be:
value in B13 = value in B8 + value in B9 + value in B10 + value in B11 + value in B12
To find the balance, you need to take the outgoings away from the total income. So the formula will be:
value in B15 = value in B5 − value in B13
To put these formulas in the spreadsheet, you can type the formula directly into the relevant cell, starting with an equals sign to show that you are entering a formula rather than a word.
Figure 12
Notice that in cell B13, a shorthand form for the sum has been entered. You could have typed in = B8+B9+B10+B11+B12, but it is also acceptable to use the shorthand form, = SUM(B8:B12), which is shown here. This instruction tells the computer to add together the values in all the cells from B8 to B12.
One of the advantages of using a spreadsheet is that if you change some of the numbers, all the calculations that use that particular number will be automatically updated to reflect the change. For example, in this budget the amounts for the salary, rent and the regular bills are likely to remain the same from one month to the next and may only be updated once or twice a year. However, the amounts for food, transport, other bills and other income probably will change from month to month. These values can be changed easily on the spreadsheet and the revised balance produced immediately.
The formulas used in a spreadsheet are shown in the diagram below. Note that * is the notation for multiplication and / is the notation for division.
Figure 13
• (a) The values in columns C and D are displayed to two decimal places as they represent an amount of money. What values will be displayed in cell C3 and cell D3?
• (b) What do you think is being calculated in the cells in column C? Can you suggest a suitable heading for this column, to be entered into C1?
• (c) Cell D5 should contain the total of the values in D2, D3 and D4. Write down a formula that could be entered in cell D5. What does this value represent?
### Discussion
• (a) The formula in C3 is:
• value in C3 = value in B3 × 0.175.
• So the value in C3 = 16.49 × 0.175 = 2.885 75.
• Although the full value is kept in the cell, it will only be displayed to 2 d.p. as 2.89.
• The value in D3 is obtained by adding together the values in B3 and C3.
• So, value in D3 = value in B3 + value in C3 = 16.49 + 2.885 75 which is 19.375 75. This will be displayed to 2 d.p. as 19.38.
• (b) The numbers in column C are 0.175 or 17.5% of the values in column B. You may know that 17.5 % is the main UK VAT rate, so it looks as if column C represents the amount of VAT paid on the different items described in column A. A suitable heading might be VAT/£ or VAT in £ or VAT (£).
• (c) The total can be found by adding together the values in cells D2, D3 and D4. So the formula ‘= D2+D3+D4’ could be entered into cell D5. Alternatively, you could use ‘=SUM(D2:D4)’. This represents the total cost (including VAT) of the radio, kettle and fan.
With these changes, the spreadsheet will look like the following.
Figure 14 |
# How do use the discriminant to find all values of b for which the equation 2x^2-bx-9=0 has one real root?
##### 1 Answer
Aug 30, 2017
See a solution process below:
#### Explanation:
The quadratic formula states:
For $a {x}^{2} + b x + c = 0$, the values of $x$ which are the solutions to the equation are given by:
$x = \frac{- b \pm \sqrt{{b}^{2} - 4 a c}}{2 a}$
The discriminate is the portion of the quadratic equation within the radical: ${\textcolor{b l u e}{b}}^{2} - 4 \textcolor{red}{a} \textcolor{g r e e n}{c}$
If the discriminate is:
- Positive, you will get two real solutions
- Zero you get just ONE solution
- Negative you get complex solutions
To find the value of $b$ where there will be just ONE soluition, we set the discriminate equal to $0$, substitute for $a$ and $c$ and solve for $b$:
Substitute:
$\textcolor{red}{2}$ for $\textcolor{red}{a}$
$\textcolor{b l u e}{- b}$ for $\textcolor{b l u e}{b}$
$\textcolor{g r e e n}{- 9}$ for $\textcolor{g r e e n}{c}$
${\textcolor{b l u e}{- b}}^{2} - \left(4 \cdot \textcolor{red}{2} \cdot \textcolor{g r e e n}{- 9}\right) = 0$
${\textcolor{b l u e}{- b}}^{2} - \left(- 72\right) = 0$
${\textcolor{b l u e}{- b}}^{2} + 72 = 0$
${\textcolor{b l u e}{- b}}^{2} + 72 - \textcolor{red}{72} = 0 - \textcolor{red}{72}$
${\textcolor{b l u e}{- b}}^{2} + 0 = - 72$
${\textcolor{b l u e}{- b}}^{2} = - 72$
$\textcolor{red}{- 1} \cdot {\textcolor{b l u e}{- b}}^{2} = \textcolor{red}{- 1} \cdot - 72$
${b}^{2} = 72$
$\sqrt{{b}^{2}} = \pm \sqrt{72}$
$b = \pm \sqrt{36 \cdot 2}$
$b = \pm \sqrt{36} \sqrt{2}$
$b = \pm 6 \sqrt{2}$ |
# The Meaning of the Equals Sign: Level 3
## Indicator of progress
Students display an understanding of the equals sign in writing number sentences.
Many students do not understand the mathematical meaning of the equals sign: that the expressions on either side have the same value. Instead they believe that an equals sign indicates where to write an answer. This has implications for their work in the Number dimension and also for their success with algebra. From the first years of school, teachers are encouraged to stress the equality meaning of the equals sign.
See more about the equals sign.
### Illustration 1: Students' interpretations of equals sign
Students were asked to find the missing number in the open number sentence 7 + 6 = ? + 5.
The first two students, Luke and Cameron, do not understand the meaning of the equals sign in the number sentence, believing that it is requiring them to write an answer to the expression on the left side.
By contrast, Fiona and Chris recognise that the two sides of the number sentence must be equal.
### Illustration 2: Common incorrect uses of equals sign
The equals sign is often used with meanings other than equality, so the correct meaning needs emphasising.
For example, it often indicates just an association as in "MATHS = FUN" or "CONFIDENCE = SUCCESS".
Students often use it to indicate they have worked out an answer e.g. when calculating 3 × (14 + 36), students frequently write 14 + 36 = 50 × 3 = 150. This is a false mathematical statement, because it implies that 14 + 36 = 150.
As a diagnostic item, teachers can ask questions such as this:
Ed had to find how many points his AFL footy team would have if they had scored 8 goals and 5 behinds. Ed wrote:
8 × 6 = 48 + 5 = 53
Did Ed make a mistake? What advice would you give Ed?
Correct working: (8 × 6) + 5 = 53
### Illustration 3: Calculators often use = as 'work it out'
Many calculators use the button labelled = for 'work it out now'. For example, if I enter 13 × 7 × 11 and press = , I will get the answer 1001.
Some more advanced calculators label the 'work it out' button as EXE (for execute), because they use = in its equality meaning.
Spreadsheets use = to indicate that a formula is being written.
Uses of = in the everyday world and in technology can reinforce students’ understanding of mathematical ideas, or limit conceptions. Pointing these conflicting uses out to students enables them to address this issue.
## Teaching strategies
Misuse of the equals sign is not difficult to change, but it is preferable to make sure students have good habits right from the start. Reinforcing the correct interpretation of the equals sign should occur regularly, with students being reminded in any contexts where it arises.
At every level, the usual reading of = by teachers and by students should be 'is equal to' rather than 'makes' or 'gives'.
Activity 1: Equations from a mat uses simple concrete materials in an open and creative way to prompt students to make equations of many different types. This activity can be undertaken from the earliest years of school, adjusting the numbers and relationships that are observed.
Activity 2: Matching card game with equivalent statements is a card matching activity, which reinforces the idea that there are many different ways of writing a mathematical sentence.
Activity 3: What goes in the box? shows expressions on both sides of a mathematical statement, emphasising the equality that it represents.
### Activity 1: Equations from a mat
Make 'mats' out of Cuisenaire rods or other materials and write the resulting equalities in many ways. Cuisenaire rods are excellent because they are finely manufactured, but Unifix blocks or centicubes are good replacements. Make a rod of given length by connecting the right number of blocks or centicubes together.
For example, the 12-mat below could lead to many equations. It is called a 12-mat because it shows the length of 12 in many different ways. Students should aim for a variety of equations. They can check each other's work. Equations should be written with 'answers' on the right or left, and also without single number ('answers') on either side. Discussion and correction (if necessary) is essential.
Encourage two variations:
• Having the 'answer' (12) on either the right or the left (e.g. 12 = 3 × 4 as well as 3 × 4 = 12)
• Having an expression on both sides (e.g. 12 – 7 = 1 + 4)
12 = 3 × 4, 3 × 4 = 12 (row 1)
3 + 3 + 3 + 3 = 12, 4 × 3 = 12 (row 2)
7 + 1 + 4 = 12, 12 – 7 = 1 + 4 (row 3)
4 + 4 + 4 = 10 + 2 (row 1 and row 4)
3 + 3 + 3 + 3 = 7 + 1 + 4 (row 2 and row 3)
This Cuisenaire mat activity can be carried out at many levels, using many different materials. Even a simple mat showing that a 2 rod and a 3 rod are the same length as a 5 rod leads to 8 variations of correct equations.
2 + 3 = 5,
3 + 2 = 5,
5 = 2 + 3,
5 = 3 + 2,
5 – 2 = 3,
3 = 5 – 2,
2 = 5 – 3,
5 – 3 = 2.
Even when students only know the + and = signs, they can write both 2 + 3 = 5 and 5 = 2 + 3, and in each case read the sign as 'is equal to' rather than 'gives' or 'makes'.
### Activity 2: Matching card game with equivalent statements
Students should recognise equality statements with 'hidden' numbers that are equivalent.
For example, 3 × 4 = ? ; 4 × ? = 12; ? = 3 × 4; ? × 3 = 12 are all equivalent mathematical statements.
Laminated cards with equality statements can be used for a game that students play in pairs or small groups, matching up and collecting sets of equality statements. A set of cards (PDF - 30Kb) can be downloaded, in which there are sets of 4 equivalent statements. Alternatively you can design your own relating to current Number topics.
How to play:
Remove one card from the deck, and deal out the remainder to the players. Players take it in turns to remove one card (without looking) from those held by the person on their right. They then try to make pairs of equivalent statements from the cards they hold. Once a pair is made it is set aside. A player's objective is to pair up all the cards in his/her hand.
### Activity 3: What goes in the box?
The main purpose of this activity is to help students to recognise the meaning of the equals sign and that the two sides of the number sentence must be equal numbers.
The first part of the activity is an awareness-raising activity. Students then complete a set of number sentences to reinforce the learning. Click here for resource sheet: What goes in the box? (Word - 35Kb)
Part A. Discussion with teacher modelling the number sentence with concrete materials
Present this scenario:
Kim was asked to find the missing number in the number sentence 5 + 7 = ? + 9
Kim wrote 5 + 7 = 12 + 9
• Why do you think Kim wrote this?
• Work out the value of the left side.
• Work out the value of the right side.
• Is Kim’s complete number sentence true?
• What number would you put in the box to make the two sides equal?
Remember that both sides must equal 12 because 5 + 7 = 12.
Teachers can model the number sentences in this discussion, eg by making a rod from 5 yellow and 7 red Unifix and comparing that with a rod made of 9 blue Unifix. What do we need to make the second rod the same length as the first?
Part B. Resource Sheet Instructions
• Find the missing numbers in these number sentences. The first one is done for you to show you how to fill in the table.
• Remember that the left side and the right side of your number sentence must be equal to each other.
• Check by working out the left side and working out the right side. Are they equal?
My number sentence
Left side
Right side
Equal?
5 + ? = 3 + 4
5 + 2 = 3 + 4
5 + 2 = 7
3 + 4 = 7
Yes
9 + ? = 6 + 8
9 + ? = 12 + 8
9 + ? = 12 + 6
10 + ? = 7 + 11
? + 6 = 9 + 9
? + 11 = 13 + 6
7 + 21 = ? + 11
10 +15 = ? + 8
23 + 7 = 18 + ?
17 + 15 = 8 + ?
## References
Kieran, C. (1992). The Learning and Teaching of School Algebra, In D.A. Grouws (Ed.), Handbook of research on mathematics teaching and learning (pp. 390-419). New York: Macmillan.
MacGregor, M., & Stacey, K. (1999). A flying start to algebra. Teaching Children Mathematics. 6(2), 78-85.
Stacey, K and MacGregor, M (1997) Building Foundations for Algebra, Mathematics Teaching in the Middle School. 2 (4, February). |
# How to resolve forces in mechanics?
Jul 19, 2022
When two objects interact, they exert forces upon each other. These forces can be described as vectors, which have both magnitude and direction. The magnitude of a force is the amount of force that is exerted, while the direction is the direction in which the force is exerted. In order to resolve a force, you need to determine both the magnitude and direction of the force.
To do this, you can use the law of physics known as the law of superposition. This law states that the net force on an object is the sum of all the individual forces that are acting on the object. This means that you can add up all the individual forces to find the net force.
The direction of the force is determined by the direction of the vector. To find the direction, you can use a vector diagram. A vector diagram is a diagram that shows the magnitude and direction of a vector. To draw a vector diagram, you need to draw a line from the origin to the point where the vector ends. The length of the line is the magnitude of the vector, and the direction of the line is the direction of the vector.
Once you have drawn the vector diagram, you can determine the direction of the force by looking at the angle that the vector makes with the horizontal axis. If the vector is pointing to the right, then the force is to the right. If the vector is pointing to the left, then the force is to the left. If the vector is pointing up, then the force is up. And if the vector is pointing down, then the force is down.
Now that you know the magnitude and direction of the force, you can resolve the force into its component parts. To do this, you need to find the x- and y-components of the force. The x-component is the force that is exerted in the horizontal direction, while the y-component is the force that is exerted in the vertical direction.
To find the x-component of the force, you need to multiply the magnitude of the force by the cosine of the angle that the force makes with the horizontal axis. To find the y-component of the force, you need to multiply the magnitude of the force by the sine of the angle that the force makes with the horizontal axis.
Once you have found the x- and y-components of the force, you can add them together to find the net force. The net force is the vector sum of all the individual forces that are acting on the object.
## Other related questions:
### Q: How do you do resolving forces?
A: There are a few steps to resolving forces:
1. Draw a free body diagram of the object you are trying to find the forces on.
2. Label the forces on the diagram.
3. Find the sum of the forces in the x-direction.
4. Find the sum of the forces in the y-direction.
5. Set the sum of the forces in the x-direction equal to zero and solve for the unknown force.
6. Set the sum of the forces in the y-direction equal to zero and solve for the unknown force.
### Q: What is resolution of forces in mechanics?
A: In mechanics, resolution of forces is the process of decomposing a force into its component forces. This can be done by considering the force as a vector and using vector addition to determine its components.
### Q: How do you resolve a level physics forces?
A: There is no one definitive answer to this question. Some methods that could be used to resolve forces include using Newton’s laws of motion, using vector addition or subtraction, using the law of universal gravitation, or using conservation of energy.
### Q: How do you resolve forces into perpendicular components?
A: There are a few ways to do this, but one approach is to use the Pythagorean theorem. In this case, you would take the magnitude of the force (F) and divide it into two components, one along the x-axis and one along the y-axis. The x-component would be:
Fx = F * cos(theta)
And the y-component would be:
Fy = F * sin(theta)
Where theta is the angle between the force and the x-axis. |
Home | | Maths 8th Std | Medians of a Triangle
Medians of a Triangle
A median of a triangle is a line segment from a vertex to the midpoint of the side opposite that vertex.
Medians of a Triangle
A median of a triangle is a line segment from a vertex to the midpoint of the side opposite that vertex.
In the figure is a median of ∆ABC.
Are there any more medians for ∆ABC ? Yes, since there are three vertices in a triangle, one can identify three medians in a triangle.
Example 5.16
In the figure, ABC is a triangle and AM is one of its medians. If BM = 3.5 cm, find the length of the side BC.
Solution:
AM is median M is the midpoint of BC.
Given that, BM = 3.5 cm, hence BC = twice the length BM = 2 × 3.5 cm = 7 cm.
Activity
1. Consider a paper cut-out of a triangle. (Let us have an acute-angled triangle, to start with). Name it, say ABC.
2. Fold the paper along the line that passes through the point A and meets the line BC such that point B falls on C. Make a crease and unfold the sheet.
3. Mark the mid point M of BC.
4. You can now draw the median AM, if you want to see it clearly. (Or you can leave it as a fold).
5. In the same way, fold and draw the other two medians.
6. Do the medians pass through the same point?
Now you can repeat this activity for an obtuse - angled triangle and a right triangle.
What is the conclusion? We see that,
The three medians of any triangle are concurrent.
1. Centroid
The point of concurrence of the three medians in a triangle is called its Centroid, denoted by the letter G.Interestingly, it happens to be the centre of mass of the triangle. One can easily verify this fact. Take a sticut out of triangle of paper. It can be balanced horizontally at this point on a finger tip or a pencil tip.
Should you fold all the three medians to find the centroid? Now you can explore among yourself the following questions:
(i) How can you find the centroid of a triangle?
(ii) Is the centroid equidistant from the vertices?
(iii) Is the centroid of a triangle always in its interior?
(iv) Is there anything special about the medians of an (a) Isosceles triangle? (b) Equilateral triangle?
Properties of the centroid of a triangle
The location of the centroid of a triangle exhibits some nice properties.
• It is always located inside the triangle.
• We have already seen that it serves as the Centre of gravity for any triangular lamina.
• Observe the figure given. The lines drawn from each vertex to G form the three triangles ∆ABG, ∆BCG, and ∆CAG.
Surprisingly, the areas of these triangles are equal.
That is, the medians of a triangle divide it into three smaller triangles of equal area!
The centroid of a triangle splits each of the medians in two segments, he one closer to the vertex being twice as long as the other one.
This means the centroid divides each median in a ratio of 2:1. (For example, GD is ⅓ of PD).
(Try to verify this by paper folding).
Example 5.17
In the figure G is the centroid of the triangle XYZ.
(i) If GL = 2.5 cm, find the length XL.
(ii) If YM = 9.3 cm, find the length GM.
Solution:
(i) Since G is the centroid, XG : GL = 2 : 1 which gives XG : 2.5 = 2 : 1.
Therefore, we get 1 × (XG) = 2 × (2.5)
XG = 5 cm.
Hence, length XL = XG + GL = 5 + 2.5 = 7.5 cm.
(ii) If YG is of 2 parts then GM will be 1 part. (Why?)
This means YM has 3 parts.
3 parts is 9.3 cm long. So GM (made of 1 part) must be 9.3 ÷ 3 = 3.1 cm.
Example 5.18
ABC is a triangle and G is its centroid. If AD=12 cm, BC=8 cm and BE=9 cm, find the perimeter of ∆BDG .
Solution:
ABC is a triangle and G is its centroid. If,
The perimeter of ∆BDG = BD+GD+BG = 4+4+6 = 14 cm
Solution 2:
ABC is a triangle and G is its centroid. If,
GD = 1/3 of AD = 1/3(12) = 4 cm and BE = 9 cm
BG = 2/3 of BE = 2/3 (9) = 6cm .
Also D is a midpoint of BC
BD = 1/2 of BC = 1/2 (8) = 4cm.
The perimeter of ∆BDG = BD+GD+BG = 4+4+6 = 14 cm
Tags : Geometry | Chapter 5 | 8th Maths , 8th Maths : Chapter 5 : Geometry
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
8th Maths : Chapter 5 : Geometry : Medians of a Triangle | Geometry | Chapter 5 | 8th Maths |
# Multiplication
Multiplying a 2 digit number by a 2 digit number can be done as mental maths. Mnemonics are one way to approach it.
For example, 37 x 84 = 3108 ; and that involves:
1. Knowing that the right side 7 times the right side 4 helps to get the rightmost digit of the answer: 7 x 4 = 28 ; and so the 8 is the right side digit of the answer:
...8
The left side digit of that 7 x 4 calculation needs to be remembered temporarily: the 'carrier'.
Then the 3 of 37 x 84 and the 4 of 37 x 84 need a calculation; and
the 7 of 37 x 84 and the 8 of 37 x 84 need a calculation; and again,
the right side of each answer is what matters (and will be added to the carrier that was 2).
So, 3 x 4 = 12 but we just want to note the 2; and
7 x 8 = 56 but we just want the 6; then
that 2 plus that 6 equal 8; and add to that the carrier (from the early on 7 x 4 = 28) of 2 to get 10.
Write the 0 as the second digit of the answer; and note the 1 (right side of 10) as a carrier.
Now, we look at the two recent calculations but take an interest in their left side result rather than their right digit result:
The 3 x 4 = 12 has the 1 noted; and
the 7 x 8 = 56 has the 5 noted; and then they are summed and the carrier of 1 is also added:
1 + 5 + 1 = 7 ; but we see above that the actual answer is 3108; and
the missing part is that the 3 and 8 of 37 x 84 need calculating:
3 x 8 = 24 (and the 4 needs adding to that total we were just looking at): 7 + 4 = 11.
Write the right side 1 as the third digit of the answer; and use the left side 1 as a carrier.
Add that 1 to the 2 of '3 x 8 = 24' to get the 3 that is the fourth digit of the answer:
We now have 3108 .
The effort to learn the results of some common calculations would speed up the process of doing these maths steps.
The initial question is: 37 x 84 = ?
What if the left side digits (3 and 8) are expressed as consonants; and the right side digits (7 and 4) are expressed as vowels, according to the following table:
Digit Syllable Vowel 0 L o 1 J a 2 B u 3 N oo 4 D oh 5 F ay 6 S ee 7 G ie 8 H e 9 W i
So 37 x 84 becomes NieHoh; and then memorised results tables can help with some of the maths we were looking at earlier.
The 7 x 4 = 28 early on in those steps is basic maths that people learn at school; and the 3 x 8 near the end of the steps to work towards the left side digit of the answer 3108 is also quite basic.
The part in the middle is trickier: taking the right side or taking the left side of different calculations.
The NieHoh word could be mentally rearranged into NohHie so that Noh is like '3 x 4' and Hie is like '7 x 8' which occur in the steps we looked at earlier.
If you have a table where Noh is written next to NohB, and you memorised that table then, any multiplication challenge involving 3 x 4 can remind you of the 'B' which is the 'right side' 2 sub result needed in the calculation.
Also in the table, next to Noh you would find aNoh where the 'a' is the 'left side' 1 carrier that you would want in this type of calculation.
Furthermore, NH (the consonants of the word) could be wrote learned to give the answer 24 (the 3 x 8 calculation near the end). In this way, there is less thinking to do while the calculation is going on in your head.
Left_Carrier X_Consonant Y_Vowel Right_Carrier x y Result o L o L 0 0 00 o L a L 0 1 00 o L u L 0 2 00 o L oo L 0 3 00 o L oh L 0 4 00 o L ay L 0 5 00 o L ee L 0 6 00 o L ie L 0 7 00 o L e L 0 8 00 o L i L 0 9 00 o J o L 1 0 00 o J a J 1 1 01 o J u B 1 2 02 o J oo N 1 3 03 o J oh D 1 4 04 o J ay F 1 5 05 o J ee S 1 6 06 o J ie G 1 7 07 o J e H 1 8 08 o J i W 1 9 09 o B o L 2 0 00 o B a B 2 1 02 o B u D 2 2 04 o B oo S 2 3 06 o B oh H 2 4 08 a B ay L 2 5 10 a B ee B 2 6 12 a B ie D 2 7 14 a B e S 2 8 16 a B i H 2 9 18 o N o L 3 0 00 o N a N 3 1 03 o N u S 3 2 06 o N oo W 3 3 09 a N oh B 3 4 12 a N ay F 3 5 15 a N ee H 3 6 18 u N ie J 3 7 21 u N e D 3 8 24 u N i G 3 9 27 o D o L 4 0 00 o D a D 4 1 04 o D u H 4 2 08 a D oo B 4 3 12 a D oh S 4 4 16 u D ay L 4 5 20 u D ee D 4 6 24 u D ie H 4 7 28 oo D e B 4 8 32 oo D i S 4 9 36 o F o L 5 0 00 o F a F 5 1 05 a F u L 5 2 10 a F oo F 5 3 15 u F oh L 5 4 20 u F ay F 5 5 25 oo F ee L 5 6 30 oo F ie F 5 7 35 oh F e L 5 8 40 oh F i F 5 9 45 o S o L 6 0 00 o S a S 6 1 06 a S u B 6 2 12 a S oo H 6 3 18 u S oh D 6 4 24 oo S ay L 6 5 30 oo S ee S 6 6 36 oh S ie B 6 7 42 oh S e H 6 8 48 ay S i D 6 9 54 o G o L 7 0 00 o G a G 7 1 07 a G u D 7 2 14 u G oo J 7 3 21 u G oh H 7 4 28 oo G ay F 7 5 35 oh G ee B 7 6 42 oh G ie W 7 7 49 ay G e S 7 8 56 ee G i N 7 9 63 o H o L 8 0 00 o H a H 8 1 08 a H u S 8 2 16 u H oo D 8 3 24 oo H oh B 8 4 32 oh H ay L 8 5 40 oh H ee H 8 6 48 ay H ie S 8 7 56 ee H e D 8 8 64 ie H i B 8 9 72 o W o L 9 0 00 o W a W 9 1 09 a W u H 9 2 18 u W oo G 9 3 27 oo W oh S 9 4 36 oh W ay F 9 5 45 ay W ee D 9 6 54 ee W ie N 9 7 63 ie W e B 9 8 72 e W i J 9 9 81 |
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# Foundation of mathematical work
4∗𝑥
Let’s consider a function 𝑝(𝑥) = 𝑥 . It is clear that the function is not defined at 𝑥 = 0 (there is not y-value on its graph), in
other words, a value (output) of the function does not exist at 𝑥 = 0 . Now, take a limit of 𝑝(𝑥) as 𝑥 goes to zero:
4∗𝑥
lim 𝑝(𝑥) = lim = lim 4 = 4
𝑥→0 𝑥→0 𝑥 𝑥→0
The algebraic trick (operation) is used to find a limit of 𝑝(𝑥) as 𝑥 goes to zero. This limit is equal to four; however, it does not
mean that an output of the function 𝑝(𝑥) exists at 𝑥 = 0. Therefore, this limit cannot be used to define a value (output) of the
function 𝑝(𝑥) at 𝑥 = 0, because 𝑝(𝑥) does not have any output at 𝑥 = 0.
A slope of some original function is derived by the definition of the derivative (limit) which has the following form if 𝑔(𝑥) is the
original function and ℎ is a distance between two x-values of 𝑔(𝑥) :
## 𝑑 𝑔(𝑥 + ℎ) − 𝑔(𝑥) 𝑔(𝑥 + ℎ) − 𝑔(𝑥)
𝑔(𝑥) = 𝑔′ (𝑥) = lim = lim
𝑑𝑥 ℎ→0 ((𝑥 + ℎ) − 𝑥) ℎ→0 ℎ
An algebraic form of a slope of the original function is a quotient in which the numerator is a difference between two y-values
of the original function and the denominator is a difference between two x-values of the original function. This quotient can be
expressed as a function of ℎ for every x-value. Let’s assume 𝑔(𝑥) = 𝑥 2 , then its derivative is:
′ (𝑥)
(𝑥 + ℎ)2 − 𝑥 2 (𝑥 2 + 2 ∗ 𝑥 ∗ ℎ + ℎ2 ) − 𝑥 2
𝑔 = lim = lim
ℎ→0 ℎ ℎ→0 ℎ
Hence, the quotient in this limit is an algebraic form defining a value of the slope of the original function at any value of 𝑥.
Therefore, this quotient can be taken as a function of ℎ at, say, 𝑥 = 1 :
(12 + 2 ∗ 1 ∗ ℎ + ℎ) − 12 (1 + 2 ∗ ℎ + ℎ2 ) − 1 2 ∗ ℎ + ℎ2
= =
ℎ ℎ ℎ
The value (output) of this function of ℎ at ℎ = 0 is undefined; therefore, the value (output) does not exist:
2 ∗ 0 + 02 0
= = 𝑢𝑛𝑑𝑒𝑓𝑖𝑛𝑒𝑑 (𝑛𝑜𝑡 𝑦 − 𝑣𝑎𝑙𝑢𝑒 𝑜𝑛 𝑡ℎ𝑒 𝑔𝑟𝑎𝑝ℎ)
0 0
This quotient (function of ℎ ) defines a value of the slope of the original function 𝑔(𝑥) at 𝑥 = 1. As seen, the slope of 𝑔(𝑥) at
𝑥 = 1 is not defined if ℎ = 0, in other words, the slope does not exist at 𝑥 = 1 if ℎ = 0 ! However, the limit of the function of ℎ
as ℎ goes to zero can be found by using the algebraic trick (operation):
2 ∗ ℎ + ℎ2 ℎ ∗ (2 + ℎ) 1 ∗ (2 + ℎ)
lim = lim = lim = lim (2 + ℎ) = 2
ℎ→0 ℎ ℎ→0 ℎ ℎ→0 1 ℎ→0
Note: ℎ is replaced with zero, but not as small as needed (infinitesimal) number
But this limit does not define a value (output) of the function of ℎ at ℎ = 0; therefore, this limit does not define the slope of the
original function 𝑔(𝑥) at 𝑥 = 1, but define the slope of its “tangent” line. I think that some people mix the formulas for limits
with the calculations defining values of limits. The formulas contain an element (ℎ → 0 or other) showing that some variable (ℎ
or other) goes to zero, some number or infinity, but such a variable is replaced with zero, some number or infinity in the
calculations to define values of limits.
Conclusion:
Whether the limit in the definition of the derivative can define a value of any slope of any original function (except a value of its
“tangent” line slope) if the quotient, expressing the slope of the original function in this limit, does not have any value at ℎ = 0?
Now, let’s go even deeper. As far as I know, the general rule in Calculus is that derivatives (two-sided limits) cannot be taken
at endpoints if such endpoints are last possible points of a function or closed intervals (in case of discontinuity); however, a
basement of such impossibility looks unsatisfying for the following reason. A derivative of some original function expresses a
value of such an original function slope at some point, but this value of the original function slope does not depend on ℎ as it
was proved on the previous page. As you can see, the original function derivatives are defined at ℎ = 0 by using the algebraic
trick (truncation of ℎ in the denominator and numerator); therefore, the derivative is absolutely independent of ℎ.
As a result of ℎ = 0, all values of derivatives of original functions are outside of the tangents induced by the quotients in the
definition of the derivative (limit) for all functions! Hence, it seems to be somehow misleading to use a term “tangent” for “the
slope of the function tangent line” because the values of the slopes of the function tangent lines are outside of the quotients
which express tangents in the definition of the derivative (limit) for all functions. Seemingly, “tangent” was recklessly left from
the initial definition of the derivative when, if I do not mistake, ℎ went to zero but was not equal to zero, i.e. ℎ expressed
infinitesimal values. In case of ℎ as an infinitesimal, it was understandable to apply “tangent” for “the slope of the function
tangent line”.
In terms of the tangents in the definitions of the derivative (limits), it is clear that these tangents have the geometrical and
algebraic basements. The algebraic basement consists of a quotient, and the geometrical one consists of an appropriate angle
of a triangle. Concerning the tangents in the definitions of the derivative (limits), these tangents have the algebraic basement
as the quotients in these limits. Therefore, if the derivative is defined by the use of the algebraic trick as ℎ = 0, then the
derivative kind of loses its geometrical basement because the quotient does not have any value at ℎ = 0, but saves its algebraic
basement. In this way, the current derivatives are clearly algebraic constructions which depend only on the structure of the
original function formulas used for the quotients in the definitions of the derivative (limits). Therefore, the derivatives (two-
sided limits) can be taken at endpoints of the original functions and also at endpoints of closed intervals!
Summary:
As it is proved above, it appears that the derivatives, i.e. the limits in the definitions of the derivative, express non-existing
rates because existing rates are always defined by quotients! For example, it is evident that one cannot get any density (physical
parameter) by dividing zero kilograms of some matter by zero cubic meters of volume which is true for any other analogous
example. However, one can algebraically calculate some value for density by using the algebraic tricks even if one uses zero
kilogram for mass and zero cubic meters for volume (just need formula(s) of original function(s)). Some value for density in this
case is just an algebraic result which does not express any real density value but may express a slope of “tangent” line of some
original function. This example nicely illustrates that values of slope of “tangent” lines of original functions are not existing (real)
rates. However, slopes of “tangent” lines of original functions can be good approximations for existing (real) rates.
To sum up, the derivatives, i.e. slopes of original functions “tangent” lines, should be separated from existing rates (real physical
parameters)! The next question which the author of this mathematical work tries to answer is:
“Whether the derivatives as non-existing rates can lead to meaningful errors in calculations with physical parameters (mass,
density, speed, acceleration etc.) in integrals?”
Defining errors in Integrals
As it is shown in the foundation, the derivatives are proved to express non-existing rates, including physical parameters defined
by quotients, but are good approximations of existing rates, including appropriate physical parameters. So if 𝑔(𝑥) = 𝑥 2 is the
original function, then its derivative (non-existing rate) at 𝑥 = 1 is 2, whereas the existing rate should be expressed as 2 + ℎ
where ℎ is some value (number) but not zero. Hence, the derivative of the original function 𝑔(𝑥) = 𝑥 2 expressing existing rates
must be 2 ∗ 𝑥 + ℎ.
In according to Fundamental Theorem of Calculus1:
𝐹(𝑥) = ∫ 𝑓(𝑥) 𝑑𝑥
## where 𝐹(𝑥) is some original function and 𝑓(𝑥) is its derivative
If 𝐹(𝑥) is an original function and 𝑓(𝑥) is its derivative, then the definition of the derivative (limit)2 is:
## 𝐹(𝑥 + ℎ) − 𝐹(𝑥) 𝐹(𝑥 + ℎ) − 𝐹(𝑥)
𝐹 ′ (𝑥) = 𝑓(𝑥) = lim = lim
ℎ→0 ℎ ℎ→0 (𝑥 + ℎ) − 𝑥
As you understand, ℎ in this definition is replaced with zero at the end of the limit calculations (not an infinitesimal value as it
was a long time ago as the author was told). The derivative produced by this definition expresses non-existing rates as it is
proved in the foundation of this work. Please remember these points across all this work.
Now, let’s consider the definite integral from a to b as the limit of Riemann’s sum3 that may be expressed by the next formula:
𝑏 𝑛 𝑛
𝑏−𝑎
𝐹(𝑏) − 𝐹(𝑎) = ∫ 𝑓(𝑥) 𝑑𝑥 = lim ∑ 𝑓(𝑥𝑖 ) ∗ ∆𝑥 = lim ∑ 𝑓(𝑥𝑖 ) ∗
𝑛→∞ 𝑛→∞ 𝑛
𝑎 𝑖=1 𝑖=1
𝑏−𝑎
where ∆𝑥 = 𝑛
is equal to zero, so dividing some number by infinity results in zero
## And, it is time to look at Riemann’s sum as a sequence of its terms:
𝑛
𝑏−𝑎 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎
lim ∑ 𝑓(𝑥𝑖 ) ∗ = 𝑓(𝑥1 ) ∗ + 𝑓(𝑥2 ) ∗ + ⋯ + 𝑓(𝑥𝑛 ) ∗
𝑛→∞ 𝑛 𝑛 𝑛 𝑛
𝑖=1
𝑏−𝑎
You can see that every term of Riemann’s sum is multiplied by 𝑛 that is equal to zero if 𝑛 is infinity; therefore, 𝑓(𝑥𝑖 ) in every
term is multiplied by zero. Hence, every term is equal to zero. Consequently, all Riemann’s sum terms add up to zero! Also, ∆𝑥
in the limit of Riemann’s sum is a differential 𝑑𝑥 in the definite integral as it can be seen:
𝑏−𝑎
= ∆𝑥 = 𝑑𝑥 = ℎ = 0
𝑛
The last equality shows that the differentials in the definite integral are equal to zero. Therefore, I conclude 𝑑𝑥 and ∆𝑥 as well
as ∆𝑦, ∆𝑧 in the limit of Riemann’s sum play a role of an algebraic trick because multiplying by zero results in zero or
undetermined form that does not make any physical sense for calculating physical parameters in my opinion. For example,
calculating mass by using a density of a material and an integral defining a volume does not make sense in my opinion because
every term of Riemann’s sum in the limit expresses zero volume if ∆𝑉 is zero:
𝑛
## lim ∑ 𝑓(𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 ) ∗ ∆𝑉 = 𝑓(𝑥1 , 𝑦1 , 𝑧1 ) ∗ 0 + 𝑓(𝑥2 , 𝑦2 , 𝑧2 ) ∗ 0 + ⋯ + 𝑓(𝑥𝑛 , 𝑦𝑛 , 𝑧𝑛 ) ∗ 0 = 0
𝑛→∞
𝑖=1
1
https://en.wikipedia.org/wiki/Fundamental_theorem_of_calculus
2
https://en.wikipedia.org/wiki/Derivative
3
https://en.wikipedia.org/wiki/Riemann_sum
If a density and an original formula 𝑓(𝑥𝑖 , 𝑦𝑖 , 𝑧𝑖 ) are given, one could calculate a mass of the figure by adding the terms of
Riemann’s sum. However, both ∆𝑥, ∆𝑦, ∆𝑧 and 𝑑𝑥 are equal to zero that makes every term zero! Here is a diagram of some 3D
figure:
Series 1
20
10
0
0
5
10
15
20
25
30
35
40
45
50
55
Series 1
Now, turn to 𝑓(𝑥𝑖 ) as an element of Riemann’s sum term. In the limit (definite integral) of Riemann’s sum, 𝑓(𝑥𝑖 ) is a function
which is a derivative of some original function 𝐹(𝑥). The original function 𝐹(𝑥) can be a function expressing distance, speed,
acceleration, density or other physical parameters. Assume that 𝐹(𝑡) expresses distance with respect to time, then 𝑓(𝑡𝑖 )
expresses its derivative, i.e. speed with respect to time. It means that 𝑓(𝑡𝑖 ) expresses some value of speed for every t-value if
such a value can be determined.
Based on the foundation of this work, 𝑓(𝑡𝑖 ) expresses non-existing rates, but speed is a rate of change of the distance function
𝐹(𝑡) and a parameter used in physics. Therefore, 𝑓(𝑡𝑖 ) expresses non-existing speed for every t-value of the original function
𝐹(𝑡); however, 𝑓(𝑡𝑖 ) gives good approximations of real speed for every t-value of 𝐹(𝑡).
Next, figure out which errors can be caused by non-existing values (approximations) of speed used in the definite integrals,
that is, in the limits of Riemann’s sum. At first, consider the definition of the derivative (limit) applied to the original function
𝐹(𝑡) with ℎ and ∆𝑡 each of which, as it is proved above, is equal to zero. Assume that 𝐹(𝑡) = 𝑡 2 , then its derivative 𝑓(𝑡) is:
(𝑡 + ℎ)2 − 𝑡 2
𝐹 ′ (𝑡) = 𝑓(𝑡) = lim = lim (2 ∗ 𝑡 + ℎ) = 2 ∗ 𝑡
ℎ→0 ℎ ℎ→0
Now, replace 𝑓(𝑥𝑖 ) with 2 ∗ 𝑡 in the limit (definite integral) of Riemann’s sum:
𝑏 𝑛
2 2
𝑏−𝑎 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎
𝐹(𝑏) − 𝐹(𝑎) = 𝑏 − 𝑎 = ∫ 2 ∗ 𝑡 𝑑𝑡 = lim ∑ 2 ∗ 𝑡𝑖 ∗ = 2 ∗ 𝑡1 ∗ + 2 ∗ 𝑡2 ∗ + ⋯ + 2 ∗ 𝑡𝑛 ∗
𝑛→∞ 𝑛 𝑛 𝑛 𝑛
𝑎 𝑖=1
where a is an initial time and b is a finish time (first and last endpoints) within one chosen interval
𝑏−𝑎
If 𝑑𝑡 = ∆𝑡 = 𝑛
= 0 , then every term in the limit of Riemann’s sum is equal to zero; hence, the whole series is equal to zero:
𝑛
𝑏−𝑎 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎
lim ∑ 2 ∗ 𝑡𝑖 ∗ = 2 ∗ 𝑡1 ∗ + 2 ∗ 𝑡2 ∗ + ⋯ + 2 ∗ 𝑡𝑛 ∗ = 2 ∗ 𝑡1 ∗ 0 + 2 ∗ 𝑡2 ∗ 0 + ⋯ + 2 ∗ 𝑡𝑛 ∗ 0 = 0
𝑛→∞ 𝑛 𝑛 𝑛 𝑛
𝑖=1
Moreover, every 2 ∗ 𝑡𝑖 expresses a non-existing value of the speed as the derivative of 𝐹(𝑥) because existing values of the speed
are expressed by 2 ∗ 𝑡 + ℎ as it was proved above. As a result, there is a paradoxical situation that every term of the series is
zero and a non-existing value of the speed is used in every term. However, if 2 ∗ 𝑡𝑖 is an approximation of an existing value of
the speed at 𝑡𝑖 , what is an error if one calculates a passed distance during the interval from a to b by using the definite integral?
To be mentioned, a signed area under the derivative function curve increases if 2 ∗ 𝑡𝑖 + ℎ is used instead of 2 ∗ 𝑡𝑖 . Next, assign
Series 1 as a signed area under the derivative function curve 2 ∗ 𝑡𝑖 , and Series 2 as a signed area of ((2 ∗ 𝑡𝑖 + ℎ) − 2 ∗ 𝑡𝑖 ). In
other words, Series 2 is the limit of Riemann’s sum (or simply infinite series) in which 𝑓(𝑡𝑖 ) is replaced with just ℎ:
𝑏 𝑛
𝑏−𝑎 𝑏−𝑎 𝑏−𝑎 𝑏−𝑎
∫ ℎ 𝑑𝑡 = lim ∑ ℎ ∗ =ℎ∗ +ℎ∗ + ⋯+ ℎ ∗
𝑛→∞ 𝑛 𝑛 𝑛 𝑛
𝑎 𝑖=1
Notice that all signed area is positive if the derivative function curve is above the t – axis during the chosen interval.
𝑏−𝑎
So if ignore that every term is multiplied by zero remembering 𝑛 = 0 , that is, allege that all these terms can be added to
some number except zero, and assume that ℎ is as small as needed, then another problem appears. This problem is that there
are infinitely many terms during any chosen interval but adding up infinitely many as-small-as-needed (except zero) terms is
always resulting into infinity! Therefore, Series 2 diverges, that is, its limit of the partial sums is always infinity. In this case, it
does not matter how small (except zero) ℎ is.
140
120
100
80
60
40
20
0
5 10 15 20 25 30 35 40 45 50 55 60
Series 1 Series 2
However, it is seen that the brown area is not infinite on the diagram above. Under such circumstances, it is impossible to define
an error by using Series 2. Notice that the limit of nth-terms of both Series 1 and Series 2 is not zero if n goes to infinity;
therefore, such series diverge4!
If allege that the terms of the limit (definite integral) of Riemann’s sum are not zeros and can be added to some number, then
another paradoxical point emerges. This point is that Series 1 diverges in this case, i.e. its limit of the partial sums is infinity,
because there are infinitely many terms which are not zeros. But the definite integrals give signed areas within some closed
intervals (see the diagram of Series 1 and Series 2 above). As seen, these areas are not infinite. All of this does not make any
physical sense in my opinion.
Let’s algebraically consider a difference between multiplying an as-small-as-needed number by infinity and adding infinitely
many as-small-as-needed equal positive numbers, for example:
## 0.0000000001 ∗ ∞ 𝑎𝑛𝑑 0.0000000001 + 0.0000000001 + 0.0000000001 + ⋯ + 0.0000000001
4
https://en.wikipedia.org/wiki/Divergent_series
If I understand right, these two operations must result in infinity, and it does not matter which small numbers (except zero) are
chosen! However, adding infinitely many terms of Riemann’s sum (definite integral) results in different amounts of signed area,
volume, mass during chosen intervals. Notice that it makes physical sense only if ∆𝑥 ≠ 0 in my opinion.
In other words, multiplying any number (except zero) by infinity is always giving infinity. Therefore, multiplying any amount
(except zero amount) of matter by infinity is always resulting into infinite amount of matter if I do not mistake! The same is
about adding any amount (except zero) of matter infinitely many times. As an example of the limit (definite integral) of
Riemann’s sum in which all terms are equal to each other is area under a straight line parallel to the horizontal axis (see diagram
below).
## Series (area under straight line)
1 2 3 4 5 6 7 8 9 10
Final Conclusions:
If summarize all the points, ℎ in the definition of the derivative (limit) must be some infinitesimal number so that the derivatives
can express existing (real) physical parameters in my opinion.
To get the definite integrals making a physical sense, ∆𝑥, ∆𝑦, ∆𝑧 must be some infinitesimal value (except zero); however, it
leads to removing the infinite number of Riemann’s sum terms in its limit (definite integral).
At present, the calculations related with derivatives and integrals are solved only mathematically by using the algebraic tricks
and allegations in my opinion. At still, some physical sense is lost in these operations. I think the main problem that does not
allow to make physical sense in these operations is the application of infinity. In other words, infinity does not work in these
operations so that they can make sense. |
# 4 8 Triangles 4 8 Isoscelesand Equilateral Triangles
• Slides: 25
4 -8 Triangles 4 -8 Isoscelesand Equilateral Triangles Warm Up Lesson Presentation Lesson Quiz Holt Geometry
4 -8 Isosceles and Equilateral Triangles Warm Up 1. Find each angle measure. 60°; 60° True or False. If false explain. 2. Every equilateral triangle is isosceles. True 3. Every isosceles triangle is equilateral. False; an isosceles triangle can have only two congruent sides. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Objectives Prove theorems about isosceles and equilateral triangles. Apply properties of isosceles and equilateral triangles. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Vocabulary legs of an isosceles triangle vertex angle base angles Holt Geometry
4 -8 Isosceles and Equilateral Triangles Recall that an isosceles triangle has at least two congruent sides. The congruent sides are called the legs. The vertex angle is the angle formed by the legs. The side opposite the vertex angle is called the base, and the base angles are the two angles that have the base as a side. 3 is the vertex angle. 1 and 2 are the base angles. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Holt Geometry
4 -8 Isosceles and Equilateral Triangles Reading Math The Isosceles Triangle Theorem is sometimes stated as “Base angles of an isosceles triangle are congruent. ” Holt Geometry
4 -8 Isosceles and Equilateral Triangles Example 1: Astronomy Application The length of YX is 20 feet. Explain why the length of YZ is the same. The m YZX = 180 – 140, so m YZX = 40°. Since YZX X, ∆XYZ is isosceles by the Converse of the Isosceles Triangle Theorem. Thus YZ = YX = 20 ft. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Check It Out! Example 1 If the distance from Earth to a star in September is 4. 2 1013 km, what is the distance from Earth to the star in March? Explain. 4. 2 1013; since there are 6 months between September and March, the angle measures will be approximately the same between Earth and the star. By the Converse of the Isosceles Triangle Theorem, the triangles created are isosceles, and the distance is the same. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Example 2 A: Finding the Measure of an Angle Find m F = m D = x° Isosc. ∆ Thm. m F + m D + m A = 180 ∆ Sum Thm. Substitute the x + 22 = 180 given values. Simplify and subtract 2 x = 158 22 from both sides. x = 79 Divide both sides by 2. Thus m F = 79° Holt Geometry
4 -8 Isosceles and Equilateral Triangles Example 2 B: Finding the Measure of an Angle Find m G. m J = m G Isosc. ∆ Thm. (x + 44) = 3 x 44 = 2 x Substitute the given values. Simplify x from both sides. Divide both sides by 2. Thus m G = 22° + 44° = 66°. x = 22 Holt Geometry
4 -8 Isosceles and Equilateral Triangles Check It Out! Example 2 A Find m H = m G = x° Isosc. ∆ Thm. m H + m G + m F = 180 ∆ Sum Thm. Substitute the x + 48 = 180 given values. Simplify and subtract 2 x = 132 48 from both sides. x = 66 Divide both sides by 2. Thus m H = 66° Holt Geometry
4 -8 Isosceles and Equilateral Triangles Check It Out! Example 2 B Find m N. m P = m N Isosc. ∆ Thm. (8 y – 16) = 6 y 2 y = 16 y = 8 Substitute the given values. Subtract 6 y and add 16 to both sides. Divide both sides by 2. Thus m N = 6(8) = 48°. Holt Geometry
4 -8 Isosceles and Equilateral Triangles The following corollary and its converse show the connection between equilateral triangles and equiangular triangles. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Holt Geometry
4 -8 Isosceles and Equilateral Triangles Example 3 A: Using Properties of Equilateral Triangles Find the value of x. ∆LKM is equilateral. Equilateral ∆ equiangular ∆ (2 x + 32) = 60 2 x = 28 x = 14 Holt Geometry The measure of each of an equiangular ∆ is 60°. Subtract 32 both sides. Divide both sides by 2.
4 -8 Isosceles and Equilateral Triangles Example 3 B: Using Properties of Equilateral Triangles Find the value of y. ∆NPO is equiangular. Equiangular ∆ equilateral ∆ 5 y – 6 = 4 y + 12 y = 18 Holt Geometry Definition of equilateral ∆. Subtract 4 y and add 6 to both sides.
4 -8 Isosceles and Equilateral Triangles Check It Out! Example 3 Find the value of JL. ∆JKL is equiangular. Equiangular ∆ equilateral ∆ 4 t – 8 = 2 t + 1 2 t = 9 t = 4. 5 Definition of equilateral ∆. Subtract 4 y and add 6 to both sides. Divide both sides by 2. Thus JL = 2(4. 5) + 1 = 10. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Remember! A coordinate proof may be easier if you place one side of the triangle along the x -axis and locate a vertex at the origin or on the y-axis. Holt Geometry
4 -8 Isosceles and Equilateral Triangles Example 4: Using Coordinate Proof Prove that the segment joining the midpoints of two sides of an isosceles triangle is half the base. Given: In isosceles ∆ABC, X is the mdpt. of AB, and Y is the mdpt. of AC. Prove: XY = Holt Geometry 1 AC. 2
4 -8 Isosceles and Equilateral Triangles Example 4 Continued Proof: Draw a diagram and place the coordinates as shown. By the Midpoint Formula, the coordinates of X are (a, b), and Y are (3 a, b). By the Distance Formula, XY = √ 4 a 2 = 2 a, and AC = 4 a. 1 Therefore XY = AC. 2 Holt Geometry
4 -8 Isosceles and Equilateral Triangles Check It Out! Example 4 What if. . . ? The coordinates of isosceles ∆ABC are A(0, 2 b), B(-2 a, 0), and C(2 a, 0). X is the midpoint of AB, and Y is the midpoint of AC. Prove ∆XYZ is isosceles. y Proof: A(0, 2 b) Draw a diagram and place the coordinates as shown. X Y Z B(– 2 a, 0) Holt Geometry x C(2 a, 0)
4 -8 Isosceles and Equilateral Triangles Check It Out! Example 4 Continued By the Midpoint Formula, the coordinates. of X are (–a, b), the coordinates. of Y are (a, b), and the coordinates of Z are (0, 0). By the Distance y Formula, XZ = YZ = √a 2+b 2. A(0, 2 b) So XZ YZ and ∆XYZ is isosceles. X Y Z B(– 2 a, 0) Holt Geometry x C(2 a, 0)
4 -8 Isosceles and Equilateral Triangles Lesson Quiz: Part I Find each angle measure. 1. m R 28° 2. m P 124° Find each value. 3. x 5. x Holt Geometry 20 4. y 26° 6
4 -8 Isosceles and Equilateral Triangles Lesson Quiz: Part II 6. The vertex angle of an isosceles triangle measures (a + 15)°, and one of the base angles measures 7 a°. Find a and each angle measure. a = 11; 26°; 77° Holt Geometry |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Families of Lines
## Lines sharing a point or slope
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Families of Lines
Suppose you were tasked with finding a line parallel or perpendicular to y=4x3\begin{align*}y=-4x-3\end{align*} and passing through the point (2, 1). How could you find the equation of this line? After completing this Concept, you'll be able to write equations of perpendicular and parallel lines.
### Watch This
CK-12 Foundation: 0505S Equations of Parallel and Perpendicular Lines (H264)
### Guidance
We can use the properties of parallel and perpendicular lines to write an equation of a line parallel or perpendicular to a given line. You might be given a line and a point, and asked to find the line that goes through the given point and is parallel or perpendicular to the given line. Here’s how to do this:
1. Find the slope of the given line from its equation. (You might need to re-write the equation in a form such as the slope-intercept form.)
2. Find the slope of the parallel or perpendicular line—which is either the same as the slope you found in step 1 (if it’s parallel), or the negative reciprocal of the slope you found in step 1 (if it’s perpendicular).
3. Use the slope you found in step 2, along with the point you were given, to write an equation of the new line in slope-intercept form or point-slope form.
#### Example A
Find an equation of the line perpendicular to the line y=3x+5\begin{align*}y=-3x+5\end{align*} that passes through the point (2, 6).
Solution
The slope of the given line is -3, so the perpendicular line will have a slope of 13\begin{align*}\frac{1}{3}\end{align*}.
Now to find the equation of a line with slope 13\begin{align*}\frac{1}{3}\end{align*} that passes through (2, 6):
Start with the slope-intercept form: y=mx+b\begin{align*}y=mx+b\end{align*}.
Plug in the slope: y=13x+b\begin{align*}y=\frac{1}{3}x+b\end{align*}.
Plug in the point (2, 6) to find b\begin{align*}b\end{align*}: 6=13(2)+bb=623b=163513\begin{align*}6=\frac{1}{3}(2)+b \Rightarrow b=6-\frac{2}{3} \Rightarrow b=\frac{16}{3} \to 5 \frac{1}{3}\end{align*}.
The equation of the line is y=13x+513\begin{align*}y=\frac{1}{3}x + 5\frac{1}{3}\end{align*}.
#### Example B
Find the equation of the line parallel to 6x5y=12\begin{align*}6x-5y=12\end{align*} that passes through the point (-5, -3).
Solution
Rewrite the equation in slope-intercept form: 6x5y=125y=6x12y=65x125\begin{align*}6x-5y=12 \Rightarrow 5y=6x-12 \Rightarrow y=\frac{6}{5}x-\frac{12}{5}\end{align*}.
The slope of the given line is 65\begin{align*}\frac{6}{5}\end{align*}, so we are looking for a line with slope 65\begin{align*}\frac{6}{5}\end{align*} that passes through the point (-5, -3).
Start with the slope-intercept form: y=mx+b\begin{align*}y=mx+b\end{align*}.
Plug in the slope: y=65x+b\begin{align*}y=\frac{6}{5}x+b\end{align*}.
Plug in the point (-5, -3): n3=65(5)+b3=6+bb=3\begin{align*}n-3=\frac{6}{5}(-5)+b \Rightarrow -3=-6+b \Rightarrow b=3\end{align*}
The equation of the line is y=65x+3\begin{align*}y=\frac{6}{5}x+3\end{align*}.
Investigate Families of Lines
A family of lines is a set of lines that have something in common with each other. Straight lines can belong to two types of families: one where the slope is the same and one where the y\begin{align*}y-\end{align*}intercept is the same.
Family 1: Keep the slope unchanged and vary the y\begin{align*}y-\end{align*}intercept.
The figure below shows the family of lines with equations of the form y=2x+b\begin{align*}y=-2x+b\end{align*}:
All the lines have a slope of –2, but the value of b\begin{align*}b\end{align*} is different for each line.
Notice that in such a family all the lines are parallel. All the lines look the same, except that they are shifted up and down the y\begin{align*}y-\end{align*}axis. As b\begin{align*}b\end{align*} gets larger the line rises on the y\begin{align*}y-\end{align*}axis, and as b\begin{align*}b\end{align*} gets smaller the line goes lower on the y\begin{align*}y-\end{align*}axis. This behavior is often called a vertical shift.
Family 2: Keep the y\begin{align*}y-\end{align*}intercept unchanged and vary the slope.
The figure below shows the family of lines with equations of the form y=mx+2\begin{align*}y=mx+2\end{align*}:
All the lines have a y\begin{align*}y-\end{align*}intercept of two, but the slope is different for each line. The steeper lines have higher values of m\begin{align*}m\end{align*}.
#### Example C
Write the equation of the family of lines satisfying the given condition.
a) parallel to the x\begin{align*}x-\end{align*}axis
b) through the point (0, -1)
c) perpendicular to 2x+7y9=0\begin{align*}2x+7y-9=0\end{align*}
d) parallel to x+4y12=0\begin{align*}x+4y-12=0\end{align*}
Solution
a) All lines parallel to the x\begin{align*}x-\end{align*}axis have a slope of zero; the y\begin{align*}y-\end{align*}intercept can be anything. So the family of lines is y=0x+b\begin{align*}y=0x+b\end{align*} or just y=b\begin{align*}y=b\end{align*}.
b) All lines passing through the point (0, -1) have the same y\begin{align*}y-\end{align*}intercept, b=1\begin{align*}b = -1\end{align*}. The family of lines is: y=mx1\begin{align*}y=mx-1\end{align*}.
c) First we need to find the slope of the given line. Rewriting 2x+7y9=0\begin{align*}2x+7y-9=0\end{align*} in slope-intercept form, we get y=27x+97\begin{align*}y=-\frac{2}{7}x+\frac{9}{7}\end{align*}. The slope of the line is 27\begin{align*}-\frac{2}{7}\end{align*}, so we’re looking for the family of lines with slope 72\begin{align*}\frac{7}{2}\end{align*}.
The family of lines is y=72x+b\begin{align*}y=\frac{7}{2}x+b\end{align*}.
d) Rewrite x+4y12=0\begin{align*}x+4y-12=0\end{align*} in slope-intercept form: y=14x+3\begin{align*}y=-\frac{1}{4}x+3\end{align*}. The slope is 14\begin{align*}-\frac{1}{4}\end{align*}, so that’s also the slope of the family of lines we are looking for.
The family of lines is y=14x+b\begin{align*}y=-\frac{1}{4}x+b\end{align*}.
Watch this video for help with the Examples above.
CK-12 Foundation: Equations of Parallel and Perpendicular Lines
### Guided Practice
Find the equation of the line perpendicular to x5y=15\begin{align*}x-5y=15\end{align*} that passes through the point (-2, 5).
Solution
Re-write the equation in slope-intercept form: \begin{align*}x-5y=15 \Rightarrow -5y=-x+15 \Rightarrow y=\frac{1}{5}x-3\end{align*}.
The slope of the given line is \begin{align*}\frac{1}{5}\end{align*}, so we’re looking for a line with slope -5.
Start with the slope-intercept form: \begin{align*}y=mx+b\end{align*}.
Plug in the slope: \begin{align*}y=-5x+b\end{align*}.
Plug in the point (-2, 5): \begin{align*}5=-5(-2)+b \Rightarrow b=5-10 \Rightarrow b=-5\end{align*}
The equation of the line is \begin{align*}y=-5x-5\end{align*}.
### Explore More
1. Find the equation of the line parallel to \begin{align*}5x-2y=2\end{align*} that passes through point (3, -2).
2. Find the equation of the line perpendicular to \begin{align*}y=-\frac{2}{5}x-3\end{align*} that passes through point (2, 8).
3. Find the equation of the line parallel to \begin{align*}7y+2x-10=0\end{align*} that passes through the point (2, 2).
4. Find the equation of the line perpendicular to \begin{align*}y+5=3(x-2)\end{align*} that passes through the point (6, 2).
5. Line \begin{align*}S\end{align*} passes through the points (2, 3) and (4, 7). Line \begin{align*}T\end{align*} passes through the point (2, 5). If Lines \begin{align*}S\end{align*} and \begin{align*}T\end{align*} are parallel, name one more point on line \begin{align*}T\end{align*}. (Hint: you don’t need to find the slope of either line.)
6. Lines \begin{align*}P\end{align*} and \begin{align*}Q\end{align*} both pass through (-1, 5). Line \begin{align*}P\end{align*} also passes through (-3, -1). If \begin{align*}P\end{align*} and \begin{align*}Q\end{align*} are perpendicular, name one more point on line \begin{align*}Q\end{align*}. (This time you will have to find the slopes of both lines.)
7. Write the equation of the family of lines satisfying the given condition.
1. All lines that pass through point (0, 4).
2. All lines that are perpendicular to \begin{align*}4x+3y-1=0\end{align*}.
3. All lines that are parallel to \begin{align*}y-3=4x+2\end{align*}.
4. All lines that pass through the point (0, -1).
8. Name two lines that pass through the point (3, -1) and are perpendicular to each other.
9. Name two lines that are each perpendicular to \begin{align*}y=-4x-2\end{align*}. What is the relationship of those two lines to each other?
10. Name two perpendicular lines that both pass through the point (3, -2). Then name a line parallel to one of them that passes through the point (-2, 5).
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 5.5. |
# The Eyeball Theorem
### What is this about?
17 October 2014, Created with GeoGebra
### The Eyeball Theorem
Let there be two circles $C(A, R_{A})$ and $C(B, R_{B}),$ one with center $A$ and radius $R_{A},$ the other with center $B$ and radius $R_{B}.$ Assume the tangents from $A$ to $C(B, R_{B})$ intersect $C(A, R_{A})$ at points $P$ and $Q,$ whereas the tangents from $B$ to $C(A, R_{A})$ intersect $C(B, R_{B})$ at points $R$ and $S.$
Then $PQ = RS.$
### Proof 1
Angles $AMB$ and $ANB$ are right. Therefore, the quadrilateral $AMNB$ is cyclic, which implies
(1)
$\angle MAN = \angle MBN,$
since the two angles subtend the same arc. Now,
$\angle PMR = \angle MAP/2 = \angle MAN/2\\ \angle PNR = \angle RBN/2 = \angle MBN/2$
since the two angles are formed by a tangent and a chord. Along with (1) this gives
(2)
$\angle PMR = \angle PNR,$
which means that the quadrilateral $MPRN$ is also cyclic. From here,
(3)
$\angle NMR = \angle NPR,$
while, for a similar reason in the circle circumscribing the quadrilateral $AMNB,$
(4)
$\angle NAB = \angle NMB = \angle NMR,$
so that from (3) and (4) we conclude that
$\angle NPR = \angle NAB.$
In other words, $PR$ is parallel to $AB.$ Since $PQ \perp AB$ and $RS \perp AB$, and because of the symmetry in $AB,$ the quadrilateral $PQSR$ is a rectangle, so that $PQ = RS,$ indeed.
### Proof 2
(NRich Program, University of Cambridge)
Let $C$ be the midpoint of $PQ.$ $C$ lies on $AB,$ and therefore $\Delta ACP$ is right. As such, it is similar to the right triangle $ANB,$ since the two share an angle. From the similarity of the triangles we derive the proportion $CP/AP = BN/AB,$ which means that
(5)
$CP = R_{A}R_{B}/AB,$ or $PQ = 2\cdot R_{A}R_{B}/AB.$
The latter expression is symmetric in $A$ and $B.$ Therefore also $RS = 2\cdot R_{A}R_{B}/AB.$ Q.E.D.
### Proof 2'
We could have used (5) a little differently. Assume the circle $C(A, R_{A})$ is fixed as is the point $B.$ What happens when $R_{B}$ changes? Obviously (or, e.g., because all circles are similar), $DR,$ where $D$ is the center of $RS,$ changes linearly in proportion to $R_{B}.$ From (5), the same is true of $CP.$ The rates are bound to be the same, since when $R_{A} = R_{B},$ we trivially have $CP = DR.$ Therefore, the latter identity holds for any $R_{B}.$
### Proof 3
(E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Problem 185)
Triangles $AMU$ and $BNU$ are similar, from which
(6)
$AM/AU = BN/BU$
On the other hand, $AM = AP$ and $BN = BR,$ such that (6) implies
(6')
$AP/AU = BR/BU$
Because of the symmetry of the configuration, $UV$ is orthogonal to $AB$ and is, therefore, parallel to $PQ$ and $RS.$ From here triangles $AUV$ and $APQ$ are similar as are triangles $BUV$ and $BRS.$ In combination with (6'), we obtain
$PQ/UV = AP/AU = BR/BU = RS/UV,$
which simply means that $PQ = RS.$
### Proof 3'
Floor van Lamoen has suggested a different ending following (6'). The latter says that $PR$ is parallel to $AB$ as $P$ and $R$ divide the sides $AU$ and $BU$ of triangle $ABU$ in the same ratio. The same holds for $QS:$ $QS\parallel AB.$ It follows that $QSRP$ is a rectangle which proves the theorem.
Floor also came up with an entirely novel proof, which, although, lengthy brings to light additional features of the configuration, see Proof 4.
### Proof 4
(By Floor van Lamoen, March 2006, private communication)
Extend $PR$ to $TU$ with $T$ on circle $A$ and $U$ on circle $B.$ Let $TM$ and $UN$ meet in $V.$
Note that $\angle MAN = \angle MBN$ and thus
$\angle VTU = \angle MTP = \frac{1}{2}\angle MAN = \frac{1}{2}\angle MBN = \angle RUN = \angle VUT,$
so triangle $TUV$ is isosceles with $VT = VU.$
We also see that
$\angle MBN = \angle VTU + \angle TUV$
so that $\angle MBN$ and $\angle NVM$ add to $180^{\circ}$, implying that $V$ lies on the circle with diameter $AB.$
Since inscribed angles $TVA$ and $KVA$ subtend equal arcs they are equal: $\angle TVA = \angle KVA$ and, similarly, $\angle UVB = \angle LVB$. By reflection through $AV$ and $BV$ respectively, we see that in fact $VT = VK = VL = VU.$
Let $W$ be the reflection of $V$ through $AB.$ $\Delta WMN$ is isosceles just as $\Delta KLV.$ Hence
$\angle WVN = \angle WMN = \angle WNM = \angle WVM,$
so $VW$ bisects angle $TVU,$ and $VW$ is perpendicular to $PR,$ and thus $PR$ is parallel to $AB.$ By symmetry $PQSR$ is a rectangle.
### Proof 5
Talk about the length of a proof. Here is quite a short one also by Floor van Lamoen (private communication, March 28, 2006):
With a reference to the following diagram
Note that $\angle RNA = \angle RTN$ by inscribed angles.
By isosceles $\Delta NBR,$ $\angle BNR = \angle BRN.$
So
\begin{align} \angle RTN + \angle TRN &= \angle RTN + \angle BRN\\ &= \angle RNA + \angle BNR\\ &= 90^{\circ}. \end{align}
This shows that $TR$ is a diameter of circle $(B).$ Hence $\angle RST$ is right. By symmetry $PQRS$ is a rectangle.
Remark: This proof makes an unwarranted assumption. Can you locate it?
### Proof 6
Dao Thanh Oai has noticed that the Eyeball theorem is a consequence of a Japanese sangaku, concerning a convex quadrilateral $ABCD.$ If the quadrilateral is cyclic, the incenters of triangles $BCD,$ $ACD,$ $ABD,$ and $ABC$ form a rectangle. As applied to the Eyeball theorem, the quadrilateral in question is $MKLN$ whereas points $P,$ $Q,$ $R,$ and $S$ are the incenters of the corresponding triangles. I'll show that this is true of $P$ and $\Delta MKN.$ Three other points are treated similarly.
Indeed, in circle $(AB),$ $\angle MAN=\angle MKN.$ In circle $(A),$ $\angle MKP=\frac{1}{2}\angle MAP=\frac{1}{2}\angle MAN.$ It follows that $\angle MKP=\angle PKN$ as both equal half of $\angle MKN.$ This means that $KP$ is the angle bisector of $\angle MKN.$
Points $M$ and $K$ are symmetric in $AB,$ implying that arcs $MA$ and $AK$ of circle $(AB)$ are equal, so that too $\angle MNA=\angle ANK,$ which shows that NP is the bisector of $\angle MNK,$ making $P$ the incenter of $\Delta MKN.$
### Remark
There is another interesting fact with ophthalmological connotations that I dubbed the Squinting Eyes Theorem. There are actually several more, see the list at the bottom of the page.
### References
1. E. J. Barbeau, M. S. Klamkin, W. O. J. Moser, Five Hundred Mathematical Challenges, MAA, 1995
2. J. Konhauser, D. Velleman, S. Wagon, Which Way Did the Bicycle Go?, MAA, 1996, #33
3. D. Wells, Curious and Interesting Geometry, Penguin Books, 1991
### Problems with Ophthalmological Connotations
• Eye-to-Eye Theorem I
• Eye-to-Eye Theorem II
• The Squinting Eyes Theorem
• Eyeballing a ball
• Praying Eyes Theorem
• Focus on the Eyeball Theorem
• Eyeball Theorem Rectified
• Bespectacled Eyeballs Extension
• Shedding Light on the Ball for Eyeballing
• Eyeballs Projected
• Archimedean Siblings out of Wedlock, i.e., Arbelos
• Rectified, Halved, Sheared, Eyeballs Still Surprise |
# SSAT Elementary Level Math : How to find the perimeter of a rectangle
## Example Questions
### Example Question #1382 : Common Core Math: Grade 4
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #81 : Apply Area And Perimeter Formulas For Rectangles: Ccss.Math.Content.4.Md.A.3
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #1384 : Common Core Math: Grade 4
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #31 : How To Find Perimeter
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #81 : How To Find The Perimeter Of A Rectangle
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #81 : Solving For Length
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #171 : Solve Problems Involving Measurement And Conversion Of Measurements
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #1387 : Common Core Math: Grade 4
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #81 : Apply Area And Perimeter Formulas For Rectangles: Ccss.Math.Content.4.Md.A.3
What is the length of a room with a perimeter of and a width of
Explanation:
We have the perimeter and the width, so we can plug those values into our equation and solve for our unknown.
Subtract from both sides
Divide by both sides
### Example Question #152 : Quadrilaterals
What is the length of a yard with a perimeter of and a width of |
## Create your own teaching resources
Teacherbot 02 Sep, 15:53:03
Lesson Plan: Estimating and Counting Objects up to 100
Main Focus: The main focus of this lesson is to help students develop the skills of estimating and counting a number of objects up to 100.
Prior Knowledge: Students should have prior knowledge of reciting and reading numbers up to 100 and counting objects, matching one-to-one.
Key Vocabulary: estimate, count, more, less
Curriculum Objectives: - N2.1A: Count in steps of 2 from 0, in steps of 5 from 0, and in steps of 10 from 0. - N2.1C: Recognize patterns in digits when counting in 2s from 0 (and so identify odd and even numbers), 5s from 0, and 10s from 0, from any number forwards and backwards.
Materials Needed: - Counting objects (e.g., counters, cubes, buttons, etc.) - Whiteboard or chart paper - Markers - Worksheets or activity sheets
Procedure:
1. Introduction (5 minutes): - Begin the lesson by reviewing the concept of counting and reciting numbers up to 100. Ask students to recite numbers from 1 to 100 together as a class. - Introduce the concept of estimating and explain that estimating means making a guess or approximation of a number.
2. Estimating Activity (10 minutes): - Show students a jar filled with a certain number of objects (e.g., buttons, counters, etc.). - Ask students to estimate how many objects are in the jar. Write their estimates on the whiteboard or chart paper. - Count the objects in the jar together as a class and compare the actual count with the estimates. Discuss which estimates were close and which were far off.
3. Counting in Steps Activity (15 minutes): - Divide the class into small groups and provide each group with a set of counting objects. - Instruct the groups to count the objects in steps of 2, 5, and 10. Encourage them to use the objects to physically count and practice counting aloud. - After each round of counting, have the groups share their results with the class. Write the numbers on the whiteboard or chart paper. - Discuss the patterns that emerge when counting in steps of 2, 5, and 10. Help students identify odd and even numbers.
4. Independent Practice (15 minutes): - Distribute worksheets or activity sheets to each student. - Instruct students to estimate and count the number of objects in each question. - Circulate the classroom to provide assistance and monitor student progress.
5. Conclusion (5 minutes): - Gather the class together and review the concepts of estimating and counting in steps of 2, 5, and 10. - Ask students to share any patterns or observations they made during the lesson. - End the lesson by emphasizing the importance of estimating and counting accurately in real-life situations.
Note: This lesson plan can be modified and adapted to suit the specific needs and abilities of the second-grade students. |
# 9.3: Multiplication of Monomials by Polynomials
Difficulty Level: Basic Created by: CK-12
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Did you know that the formula for the volume of a pyramid is \begin{align*}V= \frac{1}{3}Bh\end{align*}, where \begin{align*}B\end{align*} is the area of the base of the pyramid and \begin{align*}h\end{align*} is the pyramid's height? What if the area of the base of a pyramid were \begin{align*}x^2 + 6x + 8\end{align*} and the height were \begin{align*}9x\end{align*}? What would the volume of the pyramid be? In this Concept, you'll learn how to multiply a polynomial by a monomial so that you can answer questions such as this.
### Guidance
When multiplying polynomials together, we must remember the exponent rules we learned in previous Concepts, such as the Product Rule. This rule says that if we multiply expressions that have the same base, we just add the exponents and keep the base unchanged. If the expressions we are multiplying have coefficients and more than one variable, we multiply the coefficients just as we would any numbers. We also apply the product rule on each variable separately.
#### Example A
Multiply \begin{align*}(2x^2 y^3) \times (3x^2 y)\end{align*}.
Solution:
\begin{align*}(2x^2 y^3) \times (3x^2 y)=(2\cdot3) \times (x^2 \cdot x^2) \times (y^3 \cdot y)=6x^4 y^4\end{align*}
Multiplying a Polynomial by a Monomial
This is the simplest of polynomial multiplications. Problems are like the one above.
#### Example B
Multiply the following monomials.
(a) \begin{align*}(2x^2)(5x^3)\end{align*}
(c) \begin{align*}(3xy^5)(-6x^4y^2)\end{align*}
(d) \begin{align*}(-12a^2b^3c^4)(-3a^2b^2)\end{align*}
Solutions:
(a) \begin{align*}(2x^2)(5x^3)=(2 \cdot 5)\cdot (x^2 \cdot x^3) = 10x^{2+3}=10x^5\end{align*}
(c) \begin{align*}(3xy^5)(-6x^4y^2)=-18x^{1+4}y^{5+2}=-18x^5y^7\end{align*}
(d) \begin{align*}(-12a^2b^3c^4)(-3a^2b^2) = 36a^{2+2}b^{3+2}c^4=36a^4b^5c^4\end{align*}
To multiply monomials, we use the Distributive Property.
Distributive Property: For any expressions \begin{align*}a, \ b\end{align*}, and \begin{align*}c\end{align*}, \begin{align*}a(b+c)=ab+ac\end{align*}.
This property can be used for numbers as well as variables. This property is best illustrated by an area problem. We can find the area of the big rectangle in two ways.
One way is to use the formula for the area of a rectangle.
\begin{align*}Area \ of \ the \ big \ rectangle & = Length \times Width\\ Length & = a, \ Width = b + c\\ Area & = a \times (b + c)\end{align*}
The area of the big rectangle can also be found by adding the areas of the two smaller rectangles.
\begin{align*}Area \ of \ red \ rectangle & = ab\\ Area \ of \ blue \ rectangle & = ac\\ Area \ of \ big \ rectangle & = ab + ac\end{align*}
This means that \begin{align*}a(b+c)=ab+ac\end{align*}.
In general, if we have a number or variable in front of a parenthesis, this means that each term in the parenthesis is multiplied by the expression in front of the parenthesis.
\begin{align*}a(b+c+d+e+f+\ldots)=ab+ac+ad+ae+af+ \ldots\end{align*} The “...” means “and so on.”
#### Example C
Multiply \begin{align*}2x^3 y(-3x^4 y^2+2x^3 y-10x^2+7x+9)\end{align*}.
Solution:
\begin{align*}& 2x^3 y(-3x^4 y^2+2x^3 y-10x^2+7x+9)\\ & = (2x^3 y)(-3x^4 y^2 )+(2x^3 y)(2x^3 y)+(2x^3 y)(-10x^2 )+(2x^3 y)(7x)+(2x^3 y)(9)\\ & = -6x^7 y^3+4x^6 y^2-20x^5 y+14x^4 y+18x^3 y\end{align*}
### Guided Practice
Multiply \begin{align*}-2a^2b^4(3ab^2+7a^3b-9a+3)\end{align*}.
Solution:
Multiply the monomial by each term inside the parenthesis:
\begin{align*}& -2a^2b^4(3ab^2+7a^3b-9a+3)\\ & = (-2a^2b^4)(3ab^2)+(-2a^2b^4)(7a^3b)+(-2a^2b^4)(-9a)+(-2a^2b^4)(3)\\ & = -6a^3b^6-14a^5b^5+18a^5b^4-6a^2b^4\end{align*}
### Practice
Multiply the following monomials.
1. \begin{align*}(2x)(-7x)\end{align*}
2. \begin{align*}4(-6a)\end{align*}
3. \begin{align*}(-5a^2b)(-12a^3b^3)\end{align*}
4. \begin{align*}(-5x)(5y)\end{align*}
5. \begin{align*}y(xy^4)\end{align*}
6. \begin{align*}(3xy^2z^2)(15x^2yz^3)\end{align*}
Multiply and simplify.
1. \begin{align*}x^8 (xy^3+3x)\end{align*}
2. \begin{align*}2x(4x-5)\end{align*}
3. \begin{align*}6ab(-10a^2 b^3+c^5)\end{align*}
4. \begin{align*}9x^3(3x^2-2x+7)\end{align*}
5. \begin{align*}-3a^2b(9a^2-4b^2)\end{align*}
Mixed Review
1. Give an example of a fourth degree trinomial in the variable \begin{align*}n\end{align*}.
2. Find the next four terms of the sequence \begin{align*}1,\frac{3}{2},\frac{9}{4},\frac{28}{8}, \ldots\end{align*}
3. Reece reads three books per week.
1. Make a table of values for weeks zero through six.
2. Fit a model to this data.
3. When will Reece have read 63 books?
1. Write 0.062% as a decimal.
2. Evaluate \begin{align*}ab\left ( a+\frac{b}{4} \right )\end{align*} when \begin{align*}a=4\end{align*} and \begin{align*}b=-3\end{align*}.
3. Solve for \begin{align*}s\end{align*}: \begin{align*}3s(3+6s)+6(5+3s)=21s\end{align*}.
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# 3.6: Basic Counting Rules
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Understand the definition of random sampling.
• Calculate ordered arrangements using factorials.
• Calculate combinations and permutations.
• Calculate probabilities with factorials.
Inferential Statistics is a method of statistics that consists of drawing conclusions about a population based on information obtained from a subset or sample of the population. The main reason a sample of the population is only taken rather than the entire population (a census) is because it is less costly and it can be done more quickly than a census. In addition, because of the inability to actually reach everyone in a census, a sample can actually be more accurate than a census.
Once a statistician decides that a sampling is appropriate, the next step is to decide how to select the sample. That is, what procedure should we use to select the sample from the population? The most important characteristic of any sample is that it must be a very good representation of the population. It would not make sense to use the average height of basketball players to make an inference about the average height of the entire US population. It would not be also reasonable to estimate the average income of the entire state of California by sampling the average income of the wealthy residents of Beverly Hills. Therefore, the goal of sampling is to obtain a representative sample. For now, we will only study one powerful way of taking a sample from a population. It is called random sampling.
## Random Sampling
A random sampling is a procedure in which each sample of a given size is equally likely to be the one selected. Any sample that is obtained by random sampling is called a random sample.
In other words, if \begin{align*}n\end{align*} elements are selected from a population in such a way that every set of \begin{align*}n\end{align*} elements in the population has an equal probability of being selected, then the \begin{align*}n\end{align*} elements form a random sample.
Example:
Suppose you randomly select \begin{align*}4\;\mathrm{cards}\end{align*} from an ordinary deck of \begin{align*}52\;\mathrm{cards}\end{align*} and all the cards selected are kings. Would you conclude that the deck is still an ordinary deck or do you conclude that the deck is not an ordinary one and probably contains more than \begin{align*}4\end{align*} kings?
Solution:
The answer depends on how the cards were drawn. It is possible that the \begin{align*}4\end{align*} kings were intentionally put on top of the deck and hence drawing \begin{align*}4\end{align*} kings is not unusual, it is actually certain. However, if the deck was shuffled well, getting \begin{align*}4\end{align*} kings is highly improbable. The point of this example is that if you want to select a random sample of \begin{align*}4\;\mathrm{cards}\end{align*} to draw an inference about a population, the \begin{align*}52\;\mathrm{cards}\end{align*} deck, it is important that you know how the sample was selected from the deck.
Example:
Suppose a lottery consists of \begin{align*}100\end{align*} tickets and one winning ticket is to be chosen. What would be a fair method of selecting a winning ticket?
Solution:
First we must require that each ticket has an equal chance of winning. That is, each ticket must have a probability of \begin{align*}1/100\end{align*} of being selected. One fair way of doing that is mixing all the tickets in a container and blindly picking one ticket. This is an example of random sampling.
However, this method would not be too practical if we were dealing with a very large population, say a million tickets, and we were asked to select \begin{align*}5\end{align*} winning tickets. There are several standard procedures for obtaining random samples using a computer or a calculator.
Sometimes experiments have so many simple events that it is impractical to list them. However, in some experiments we can develop a counting rule, with the use of tree diagrams that can aid us to do that. The following examples show how that is done.
Example:
Suppose there are six balls in a box. They are identical except in color. Two balls are red, three are blue, and one is yellow. We will draw one ball, record its color, and set it aside. Then we will draw another one, record its color. With the aid of a tree diagram, calculate the probability of each outcome of the experiment.
Solution:
We first draw a tree diagram to aid us see all the possible outcomes of this experiment.
The tree diagram shows us the two stages of drawing two balls without replacing them back into the box. In the first stage, we pick a ball blindly. Since there are \begin{align*}2\end{align*} red, \begin{align*}3\end{align*} blue, and \begin{align*}1\end{align*} yellow, then the probability of getting a red is \begin{align*}2/6.\end{align*} The probability of getting a blue is \begin{align*}3/6\end{align*} and the probability of getting a yellow is \begin{align*}1/6.\end{align*}
Remember that the probability associated with the second ball depends on the color of the first ball. Therefore, the two stages are not independent. To calculate the probabilities of getting the second ball, we look back at the tree diagram and observe the followings.
There are eight possible outcomes for the experiment:
RR: red on the \begin{align*}1^{st}\end{align*} and red on the \begin{align*}2^{nd}\end{align*}
RB: red on the \begin{align*}1^{st}\end{align*} and blue on the \begin{align*}2^{st}\end{align*}
And so on. Here are the rest,
\begin{align*}RY, BR, BB, BY, YR, YB.\end{align*}
Next, we want to calculate the probabilities of each outcome.
\begin{align*}P(R\ 1^{st}\ \text{and}\ R\ 2^{st} ) & = P(RR) = 2/6 \cdot 1/5 = 2/30\\ P(R\ 1^{st}\ \text{and}\ B\ 2^{st} ) & = P(RB) = 2/6 \cdot 3/5 = 6/30\\ P(RY) & = 2/6 \cdot 1/5 = 2/30\\ P(BR) & = 3/6 \cdot 2/5 = 6/30\\ P(YB) & = 3/6 \cdot 2/5 = 6/30\\ P(YB) & = 3/6 \cdot 1/5 = 3/30\\ P(YB) & = 1/6 \cdot 2/5 = 2/30\\ P(YB) & = 1/6 \cdot 3/5 = 3/30\end{align*}
Notice that all the probabilities must add up to \begin{align*}1\end{align*}, as they should.
The method used to solve the example above can be generalized to any number of stages. This method is called the Multiplicative Rule of Counting.
## The Multiplicative Rule of Counting
(I) If there are \begin{align*}n\end{align*} possible outcomes for event \begin{align*}A\end{align*} and \begin{align*}m\end{align*} possible outcomes for event \begin{align*}B\end{align*}, then there are a total of \begin{align*}nm\end{align*} possible outcomes for the series of events \begin{align*}A\end{align*} followed by \begin{align*}B\end{align*}.
Another way of stating it:
(II) You have \begin{align*}k\end{align*} sets of elements, \begin{align*}n_1\end{align*} in the first set, \begin{align*}n_2\end{align*} in the second set,..., and \begin{align*}n_k\end{align*} in the \begin{align*}k\end{align*}th set. Suppose you want to take one sample from each of the \begin{align*}k\end{align*} sets. The number of different samples that can be formed is the product
\begin{align*}n_1 n_2 n_3 \ldots n_k\end{align*}
Example:
A restaurant offers a special dinner menu every day. There are three entrées to choose from, five appetizers, and four desserts. A costumer can only select one item from each category. How many different meals can be ordered from the special dinner menu?
Solution:
Let’s summarize what we have.
Entrees:3
Appetizer: 5
Dessert: 4
We use the multiplicative rule above to calculate the number of different dinner meals that can be selected. We simply multiply all the number of choices per item together:
\begin{align*}(3)(5)(4) = 60\end{align*}
There are \begin{align*}60\end{align*} different dinners that can be ordered by the customers.
Example:
Here is a classic example. In how many different ways can you seat \begin{align*}8\end{align*} people at a dinner table?
Solution:
For the first seat, there are eight choices. For the second, there are seven remaining choices, since one person has already been seated. For the third seat, there are \begin{align*}6\end{align*} choices, since two people are already seated. By the time we get to the last seat, there is only one seat left. Therefore, using the multiplicative rule above, we get
\begin{align*}(8)(7)(6)(5)(4)(3)(2)(1) = 40,320\end{align*}
The multiplication pattern above appears so often in statistics that it has its own name and its own symbol. So we say “eight factorial,” and we write \begin{align*}8!\end{align*}.
## Factorial Notation
\begin{align*}n! = n(n - 1)(n - 2) \ldots 1\end{align*}
Example:
Suppose there are \begin{align*}30\end{align*} candidates that are competing for three executive positions. How many different ways can you fill the three positions?
Solution:
This is a more difficult problem than the examples above and we will use the second version of the Multiplicative Rule of Counting. We need to analyze it in the following way:
The executive positions can be denoted by \begin{align*}k = 3\end{align*} sets of elements that correspond to
\begin{align*}n_1 =\end{align*} The number of candidates that are available to fill the first position
\begin{align*}n_2 =\end{align*} The number of candidates remaining to fill the second position
\begin{align*}n_3 =\end{align*} The number of candidates remaining to fill the third position
Hence,
\begin{align*} n_1 = 30\\ n_2 = 29\\ n_3 = 28\end{align*}
The number of different ways to fill the three positions is
\begin{align*}n_1 n_2 n_3 = (30)(29)(28) = 24,360\ \text{possible positions.}\end{align*}
The arrangement of elements in distinct order, as the example above shows, is called the permutation. Thus, from the example above there are \begin{align*}24,360\end{align*} possible permutations of three positions drawn from a set of \begin{align*}30\end{align*} elements.
## Counting Rule for Permutations
The number of ways to arrange in order \begin{align*}n\end{align*} different objects within \begin{align*}r\end{align*} positions is
\begin{align*}P^n_r = \frac{n!} {(n- r)!}\end{align*}
Example:
Let’s go back to the previous example but this time we want to compute the number of ordered seating arrangements we have for \begin{align*}8\end{align*} people for only \begin{align*}5\end{align*} seats.
Solution:
In this case, we are considering a total of \begin{align*}n = 8\end{align*} people and we wish to arrange \begin{align*}r = 5\end{align*} of these people to be seated. Substituting into the permutation equation,
\begin{align*}P^n_r & = \frac{n!} {(n - r)!} = \frac{8!} {(8 -5)!} \\ & = \frac{8!} {3!} \\ & = \frac{40,320} {6} \\ & =6720\end{align*}
Another way of solving this problem is to use the Multiplicative Rule of Counting,
Since there are only \begin{align*}5\end{align*} seats available for \begin{align*}8\end{align*} people, then for the first seat, there are eight people. For the second seat, there are seven remaining people, since one person has already been seated. For the third seat, there are \begin{align*}6\end{align*} people, since two people are already seated. For the fifth seat, there are \begin{align*}4\end{align*} people. After that we run out of seats. Thus
\begin{align*}(8)(7)(6)(5)(4) = 6720.\end{align*}
Of course, the permutation rule is more powerful since it has the advantage of using the factorial. Most scientific calculators can do factorials permutations, so make sure to know how to do them on your calculator.
Example:
The board of directors at The Orion Foundation has \begin{align*}13\end{align*} members. Three officers will be elected from the \begin{align*}13\end{align*} members to hold the positions of a provost, a general director and a treasure. How many different slates of three candidates are there, if each candidate must specify which office he or she wishes to run for?
Solution:
Each slate is a list of one person for each of three positions, the provost, the general director and the treasure. If, for example, Mr. Smith, Mr. Hale, and Ms. Osborn wish to be on a slate together, there are several different slates possible, depending on which one will run for provost, general director and treasurer. So we are not just asking for the number of different groups of three names on a slate but we are also asking for a specific order, since it makes a difference which name is listed in which position.
So,
\begin{align*}n = 13\\ r = 3 \end{align*}
Using the permutation formula,
\begin{align*}P^n_r = \frac{n!} {(n - r)!} = \frac{13!} {(13 - 3)!} = 1716\end{align*}
There are \begin{align*}1716\end{align*} different slates of officers.
Notice that in our previous examples, the order of people or objects was taken into account. What if the order is not important? For example, in the previous example for electing three officers, what if we wish to choose \begin{align*}3\end{align*} members of the \begin{align*}13-\end{align*}member board to attend a convention. Here, we are more interested in the group of three but we are not interested in their order. In other words, we are only concerned with different combinations of \begin{align*}13\end{align*} people taken \begin{align*}3\end{align*} at a time. The permutation rule will not work here since order is not important. We have a new formula that will compute different combinations.
## Counting Rule for Combinations
The number of combinations of \begin{align*}n\end{align*} objects taken \begin{align*}r\end{align*} at a time is
\begin{align*}C^n_r = \frac{n!} {r!(n - r)!}\end{align*}
It is important to notice the difference between permutations and combinations. When we consider grouping and order, we use permutations. But when we consider grouping with no particular order, we use combinations.
Example:
Back to our example above. How many different groups of three are there, taken out of \begin{align*}13\end{align*} people?
Solution:
As explained in the previous paragraph, we are interested in combinations rather than permutations of \begin{align*}13\end{align*} people taken \begin{align*}3\end{align*} at a time. We use the combination formula
\begin{align*}C^n_r & = \frac{n!} {r!(n - r)!} \\ C^{13}_3 & = \frac{13!} {3!(13 - 3)!} = \frac{13!} {3!10!} = 286\end{align*}
There are \begin{align*}286\end{align*} different groups of \begin{align*}3\end{align*} to go to the convention.
In the above computation you can see that the difference between the formulas for \begin{align*}nCr\end{align*} and \begin{align*}nPr\end{align*} is in the factor \begin{align*}r!\end{align*} in the denominator of the fraction. Since \begin{align*}r!\end{align*} is the number of different orders of \begin{align*}r\end{align*} things, and combinations ignore order, then we divide by the number of different orders.
Example:
You are taking a philosophy course that requires you to read \begin{align*}5\end{align*} books out of a list of \begin{align*}10\end{align*} books. You are free to select any five books and read them in whichever order that pleases you. How many different combinations of \begin{align*}5\end{align*} books are available from a list of \begin{align*}10\end{align*}?
Solution:
Since considerations of order in which the books are selected are not important, we compute the number of combinations of \begin{align*}10\end{align*} books taken \begin{align*}5\end{align*} at a time. We use the combination formula
\begin{align*}C^n_r & = \frac{n!} {r!(n - r)!}\\ C^{10}_5 & = \frac{10!} {5!(10 - 5)!} = 256\end{align*}
There are \begin{align*}252\end{align*} different groups of \begin{align*}5\end{align*} books that can be selected from a list of \begin{align*}10\end{align*} books.
## Technology Note
The TI-83/84 calculators and the EXCEL spreadsheet have commands for factorials, permutations, and combinations.
Using the TI-83/84 Calculators
Press [MATH] and then choose PRB (Probability). You will see the following choices, among others: \begin{align*}nPr, nCr,\end{align*} and ! The screens show the menu and the proper uses of these commands.
Using EXCEL
In Excel the above commands are entered as follows:
• \begin{align*}=\end{align*} PERMUT (10,2)
• \begin{align*}=\end{align*} COMBIN (10,2)
• \begin{align*}=\end{align*} FACT (10)
## Lesson Summary
1. Inferential Statistics is a method of statistics that consists of drawing conclusions about a population based on information obtained from a subset or sample of the population.
2. A random sampling is a procedure in which each sample of a given size is equally likely to be the one selected.
3. The Multiplicative Rule of Counting states: if there are \begin{align*}n\end{align*} possible outcomes for event \begin{align*}A\end{align*} and \begin{align*}m\end{align*} possible outcomes for event \begin{align*}B\end{align*}, then there are a total of \begin{align*}nm\end{align*} possible outcomes for the series of events \begin{align*}A\end{align*} followed by \begin{align*}B\end{align*}.
4. The factorial, ' ! ', means \begin{align*}n! = n(n - 1)(n - 2) \ldots 1.\end{align*}
5. The number of permutations (ordered arrangements) of \begin{align*}n\end{align*} different objects within \begin{align*}r\end{align*} positions is \begin{align*}P^n_r = \frac{n!} {(n r)!} \end{align*}
6. The number of combinations (unordered arrangements) of \begin{align*}n\end{align*} objects taken \begin{align*}r\end{align*} at a time is \begin{align*}C^n_r = \frac{n!} {r!(n - r)!}\end{align*}
## Review Questions
1. Determine the number of simple events when you toss a coin the following number of times: (Hint: as the numbers get higher, you will need to develop a systematic method of counting all the outcomes)
1. Twice
2. Three times
3. Five times
4. Look for a pattern in the results of a) through c) and try to figure out the number of outcomes for tossing a coin \begin{align*}n\end{align*} times.
2. Flying into Los Angeles from Washington DC, you can choose one of three airlines and can choose either first class or economy. How many travel options do you have?
3. How many different \begin{align*}5-\end{align*}card hands can be chosen from a \begin{align*}52-\end{align*}card deck?
4. Suppose an automobile license plate is designed to show a letter of the English alphabet, followed by a five-digit number. How many different license plates can be issued?
1. \begin{align*}4\end{align*}
2. \begin{align*}8\end{align*}
3. \begin{align*}32\end{align*}
4. \begin{align*}2^n\end{align*}
1. \begin{align*}6\end{align*}
2. \begin{align*}2,598,960\end{align*}
3. \begin{align*}2,600,000\end{align*}
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# Find the area of the region enclosed by the parabola
Question:
Find the area of the region enclosed by the parabola $x^{2}=y$, the line $y=x+2$ and $x$-axis
Solution:
The area of the region enclosed by the parabola, $x^{2}=y$, the line, $y=x+2$, and $x$-axis is represented by the shaded region $\mathrm{OACO}$ as
The point of intersection of the parabola, x2 = y, and the line, y = x + 2, is A (–1, 1) and C(2, 4).
Area of OACO $=\int_{-1}^{2}(x+2) d x-\int_{-1}^{2} x^{2} d x$
$\Rightarrow$ Area of $\mathrm{OACO}=\left[\frac{x^{2}}{2}+2 x\right]_{-1}^{2}-\frac{1}{3}\left[x^{3}\right]_{-1}^{2}$
$\Rightarrow$ Area of $\mathrm{OACO}=\left[\left\{\frac{(2)^{2}}{2}+2(2)\right\}-\left\{\frac{(-1)^{2}}{2}+2(-1)\right\}\right]-\frac{1}{3}\left[(2)^{3}-(-1)^{3}\right]$
$\Rightarrow$ Area of $\mathrm{OACO}=\left[2+4-\left(\frac{1}{2}-2\right)\right]-\frac{1}{3}(8+1)$
$\Rightarrow$ Area of $\mathrm{OACO}=6+\frac{3}{2}-3$
$\Rightarrow$ Area of $\mathrm{OACO}=3+\frac{3}{2}=\frac{9}{2}$ square units |
Step by step NCERT Solutions for Class 10 Maths Chapter 6 Triangles. This solutions covers all questions of Chapter 6 Triangles Class 10 as per CBSE Board guidelines from the latest NCERT book for class 10 maths. Following topics and sub-topics in Chapter 6 Triangles are covered.
6.1 Introduction
6.2 Similar Figures,
6.3 Similarity of Triangles,
6.4 Criteria for Similarity of Triangles,
6.5 Areas of Similar Triangles,
6.6 Pythagoras Theorem and
6.7 Summary.
## Triangles NCERT Solutions – Class 10 Maths
### NCERT Solutions for Class 10 Maths Chapter 6
Exercise 6.1 : Solutions of Questions on Page Number : 122
Q1 : Fill in the blanks using correct word given in the brackets:-
(i) All circles are __________. (congruent, similar)
(ii) All squares are __________. (similar, congruent)
(iii) All __________ triangles are similar. (isosceles, equilateral)
(iv) Two polygons of the same number of sides are similar, if (a) their corresponding angles are __________ and (b) their corresponding sides are __________. (equal, proportional)
(i) Similar
(ii) Similar
(iii) Equilateral
(iv) (a) Equal
(b) Proportional
Q2 : Give two different examples of pair of
(i) Similar figures
(ii)Non-similar figures
(i) Two equilateral triangles with sides 1 cm and 2 cm
Two squares with sides 1 cm and 2 cm
(ii) Trapezium and square
Triangle and parallelogram
Q3 : State whether the following quadrilaterals are similar or not:
Quadrilateral PQRS and ABCD are not similar as their corresponding sides are proportional, i.e. 1:2, but their corresponding angles are not equal.
Exercise 6.2 : Solutions of Questions on Page Number : 128
Q1 :In figure.6.17. (i) and (ii), DE || BC. Find EC in (i) and AD in (ii).
(i)
(ii)
(i)
Let EC = x cm
It is given that DE || BC.
By using basic proportionality theorem, we obtain
(ii)
It is given that DE || BC.
By using basic proportionality theorem, we obtain
Q2 : E and F are points on the sides PQ and PR respectively of a ΔPQR. For each of the following cases, state whether EF || QR.
(i) PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm and FR = 2.4 cm
(ii) PE = 4 cm, QE = 4.5 cm, PF = 8 cm and RF = 9 cm
(iii)PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm and PF = 0.63 cm
(i)
Given that, PE = 3.9 cm, EQ = 3 cm, PF = 3.6 cm, FR = 2.4 cm
(ii)
PE = 4 cm, QE = 4.5 cm, PF = 8 cm, RF = 9 cm
(iii)
PQ = 1.28 cm, PR = 2.56 cm, PE = 0.18 cm, PF = 0.36 cm
Q3 : In the following figure, if LM || CB and LN || CD, prove that
In the given figure, LM || CB
By using basic proportionality theorem, we obtain
Q4 : In the following figure, DE || AC and DF || AE. Prove that
In ΔABC, DE || AC
Q5 : In the following figure, DE || OQ and DF || OR, show that EF || QR.
In Δ POQ, DE || OQ
Q6 : In the following figure, A, B and C are points on OP, OQ and OR respectively such that AB || PQ and AC || PR. Show that BC || QR.
In Δ POQ, AB || PQ
Q7 : Using Basic proportionality theorem, prove that a line drawn through the mid-points of one side of a triangle parallel to another side bisects the third side. (Recall that you have proved it in Class IX).
Consider the given figure in which PQ is a line segment drawn through the mid-point P of line AB, such thatBy using basic proportionality theorem,we obtain
Or, Q is the mid-point of AC.
Q8 : Using Converse of basic proportionality theorem, prove that the line joining the mid-points of any two sides of a triangle is parallel to the third side. (Recall that you have done it in Class IX).
Consider the given figure in which PQ is a line segment joining the mid-points P and Q of line AB and AC respectively.
i.e., AP = PB and AQ = QC
It can be observed that
Hence, by using basic proportionality theorem, we obtain
Q9 : ABCD is a trapezium in which AB || DC and its diagonals intersect each other at the point O. Show that
Draw a line EF through point O, such that
By using basic proportionality theorem, we obtain
In ΔABD,
So, by using basic proportionality theorem, we obtain
From equations (1) and (2), we obtain
Q10 : The diagonals of a quadrilateral ABCD intersect each other at the point O such that Show that ABCD is a trapezium.
Let us consider the following figure for the given question.
Draw a line OE || AB
In ΔABD, OE || AB
By using basic proportionality theorem, we obtain
However, it is given that
⇒ EO || DC [By the converse of basic proportionality theorem]
⇒ AB || OE || DC
⇒ AB || CD
∴ ABCD is a trapezium.
Exercise 6.3 : Solutions of Questions on Page Number : 138
Q1 : State which pairs of triangles in the following figure are similar? Write the similarity criterion used by you for answering the question and also write the pairs of similar triangles in the symbolic form:
(i)
(ii)
(iii)
(iv)
(v)
(vi)
(i) ∠A = ∠P = 60°
∠B = ∠Q = 80°
∠C = ∠R = 40°
Therefore, ΔABC ∼ ΔPQR [By AAA similarity criterion]
(ii)
∠F=∠R (Each30°)
∴ΔDEF˜ΔPQR[By AAA similarity criterion]
Q2 : In the following figure, ΔODC ˜ ΔOBA, ∠ BOC = 125° and ∠ CDO = 70°. Find ∠ DOC, ∠ DCO and ∠ OAB
DOB is a straight line.
∴ ∠ DOC + ∠ COB = 180°
⇒ ∠ DOC = 180° – 125°
= 55°
In ΔDOC,
∠ DCO + ∠ CDO + ∠ DOC = 180°
(Sum of the measures of the angles of a triangle is 180º.)
⇒ ∠ DCO + 70º + 55º = 180°
⇒ ∠ DCO = 55°
It is given that ΔODC ∠¼ ΔOBA.
∴ ∠ OAB = ∠ OCD [Corresponding angles are equal in similar triangles.]
⇒ ∠ OAB = 55°
Q3 : Diagonals AC and BD of a trapezium ABCD with AB || DC intersect each other at the point O. Using a similarity criterion for two triangles, show that
In ΔDOC and ΔBOA,
∠CDO = ∠ABO [Alternate interior angles as AB || CD]
∠DCO = ∠BAO [Alternate interior angles as AB || CD]
∠DOC = ∠BOA [Vertically opposite angles]
∴ ΔDOC ˜ ΔBOA [AAA similarity criterion]
Q4 : In the following figure, Show that
In ΔPQR, ∠PQR = ∠PRQ
∴ PQ = PR (i)
Given,
Q5 : S and T are point on sides PR and QR of ΔPQR such that ∠ P = ∠ RTS. Show that ΔRPQ ˜ ΔRTS.
In ΔRPQ and ΔRST,
∠ RTS = ∠ QPS (Given)
∠ R = ∠ R (Common angle)
∴ ΔRPQ ˜ ΔRTS (By AA similarity criterion)
Q6 : In the following figure, if ΔABE ≅ ΔACD, show that ΔADE ˜ ΔABC.
It is given that ΔABE ≅ ΔACD.
∴ AB = AC [By CPCT] (1)
And, AD = AE [By CPCT] (2)
[Dividing equation (2) by (1)]
∠A = ∠A [Common angle]
∴ ΔADE ˜ΔABC [By SAS similarity criterion]
Q7 : In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
(i) ΔAEP˜ ΔCDP
(ii) ΔABD ˜ ΔCBE
(iv) ΔPDC ˜ ΔBEC
(i)
In ΔAEP and ΔCDP,
∠ AEP = ∠ CDP (Each 90°)
∠ APE = ∠ CPD (Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ˜ ΔCDP
(ii)
In ΔABD and ΔCBE,
∠ ADB = ∠ CEB (Each 90°)
∠ ABD = ∠ CBE (Common)
Hence, by using AA similarity criterion,
ΔABD ˜ ΔCBE
(iii)
∠ AEP = ∠ ADB (Each 90°)
∠ PAE = ∠ DAB (Common)
Hence, by using AA similarity criterion,
(iv)
In ΔPDC and ΔBEC,
∠ PDC = ∠ BEC (Each 90°)
∠ PCD = ∠ BCE (Common angle)
Hence, by using AA similarity criterion,
ΔPDC ˜ ΔBEC
Q8 : E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. Show that ΔABE ˜ ΔCFB
In ΔABE and ΔCFB,
∠ A = ∠ C (Opposite angles of a parallelogram)
∠ AEB = ∠ CBF (Alternate interior angles as AE || BC)
∴ ΔABE ˜ ΔCFB (By AA similarity criterion)
Q9 : In the following figure, ABC and AMP are two right triangles, right angled at B and M respectively, prove that:
(i) ΔABC ˜ ΔAMP
(ii)
In ΔABC and ΔAMP,
∠ABC = ∠AMP (Each 90°)
∠A = ∠A (Common)
∴ ΔABC ˜ ΔAMP (By AA similarity criterion)
Q10 : CD and GH are respectively the bisectors of ∠ACB and ∠EGF such that D and H lie on sides AB and FE of ΔABC and ΔEFG respectively. If ΔABC ˜ ΔFEG, Show that:
(i)
(ii) ΔDCB ˜ ΔHGE
(iii) ΔDCA ˜ ΔHGF
It is given that ΔABC ˜ ΔFEG.
∴ ∠A = ∠F, ∠B = ∠E, and ∠ACB = ∠FGE
∠ACB = ∠FGE
∴ ∠ACD = ∠FGH (Angle bisector)
And, ∠DCB = ∠HGE (Angle bisector)
In ΔACD and ΔFGH,
∠A = ∠F (Proved above)
∠ACD = ∠FGH (Proved above)
∴ ΔACD ˜ ΔFGH (By AA similarity criterion)
In ΔDCB and ΔHGE,
∠DCB = ∠HGE (Proved above)
∠B = ∠E (Proved above)
∴ ΔDCB ˜ ΔHGE (By AA similarity criterion)
In ΔDCA and ΔHGF,
∠ACD = ∠FGH (Proved above)
∠A = ∠F (Proved above)
∴ ΔDCA ˜ ΔHGF (By AA similarity criterion)
Q11 : In the following figure, E is a point on side CB produced of an isosceles triangle ABC with AB = AC. If AD ⊥ BC and EF ⊥ AC, prove that ΔABD ˜ ΔECF
It is given that ABC is an isosceles triangle.
∴ AB = AC
⇒ ∠ ABD = ∠ ECF
In ΔABD and ΔECF,
∠ ADB = ∠ EFC (Each 90°)
∠ BAD = ∠ CEF (Proved above)
∴ ΔABD ˜ ΔECF (By using AA similarity criterion)
Q12 : Sides AB and BC and median AD of a triangle ABC are respectively proportional to sides PQ and QR and median PM of ΔPQR (see the given figure). Show that ΔABC ˜ ΔPQR.
Median divides the opposite side.
Given that,
In ΔABD and ΔPQM,
(Proved above)
∴ ΔABD ˜ ΔPQM (By SSS similarity criterion)
⇒ ∠ABD = ∠PQM (Corresponding angles of similar triangles)
In ΔABC and ΔPQR,
∠ABD = ∠PQM (Proved above)
∴ ΔABC ˜ ΔPQR (By SAS similarity criterion)
Q13 : D is a point on the side BC of a triangle ABC such that ∠ADC = ∠BAC. Show that
∠ACD = ∠BCA (Common angle)
∴ ΔADC ˜ ΔBAC (By AA similarity criterion)
We know that corresponding sides of similar triangles are in proportion.
Q14 : Sides AB and AC and median AD of a triangle ABC are respectively proportional to sides PQ and PR and median PM of another triangle PQR. Show that
Given that,
Let us extend AD and PM up to point E and L respectively, such that AD = DE and PM = ML. Then, join B to E, C to E, Q to L, and R to L.
We know that medians divide opposite sides.
Therefore, BD = DC and QM = MR
Also, AD = DE (By construction)
And, PM = ML (By construction)
In quadrilateral ABEC, diagonals AE and BC bisect each other at point D.
Therefore, quadrilateral ABEC is a parallelogram.
∴ AC = BE and AB = EC (Opposite sides of a parallelogram are equal)
Similarly, we can prove that quadrilateral PQLR is a parallelogram and PR = QL, PQ = LR
It was given that
∴ ΔABE ˜ ΔPQL (By SSS similarity criterion)
We know that corresponding angles of similar triangles are equal.
∴ ∠BAE = ∠QPL … (1)
Similarly, it can be proved that ΔAEC ˜ ΔPLR and
∠CAE = ∠RPL … (2)
Adding equation (1) and (2), we obtain
∠BAE + ∠CAE = ∠QPL + ∠RPL
⇒ ∠CAB = ∠RPQ … (3)
In ΔABC and ΔPQR,
(Given)
∠CAB = ∠RPQ [Using equation (3)]
∴ ΔABC ˜ ΔPQR (By SAS similarity criterion)
Q15 : A vertical pole of a length 6 m casts a shadow 4m long on the ground and at the same time a tower casts a shadow 28 m long. Find the height of the tower.
Let AB and CD be a tower and a pole respectively.
Let the shadow of BE and DF be the shadow of AB and CD respectively.
At the same time, the light rays from the sun will fall on the tower and the pole at the same angle.
Therefore, ∠DCF = ∠BAE
And, ∠DFC = ∠BEA
∠CDF = ∠ABE (Tower and pole are vertical to the ground)
∴ ΔABE ˜ ΔCDF (AAA similarity criterion)
Therefore, the height of the tower will be 42 metres.
Q16 : If AD and PM are medians of triangles ABC and PQR, respectively where
It is given that ΔABC ∼ ΔPQR
We know that the corresponding sides of similar triangles are in proportion.
… (1)
Also, ∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)
Since AD and PM are medians, they will divide their opposite sides.
… (3)
From equations (1) and (3), we obtain
… (4)
In ΔABD and ΔPQM,
∠B = ∠Q [Using equation (2)]
[Using equation (4)]
∴ ΔABD ˜ ΔPQM (By SAS similarity criterion)
Exercise 6.4 : Solutions of Questions on Page Number : 143
Q1 : Let and their areas be, respectively, 64 cm2 and 121 cm2. If EF = 15.4 cm, find BC.
Q2 : Diagonals of a trapezium ABCD with AB || DC intersect each other at the point O. If AB = 2CD, find the ratio of the areas of triangles AOB and COD.
Since AB || CD,
∴ ∠OAB = ∠OCD and ∠OBA = ∠ODC (Alternate interior angles)
In ΔAOB and ΔCOD,
∠AOB = ∠COD (Vertically opposite angles)
∠OAB = ∠OCD (Alternate interior angles)
∠OBA = ∠ODC (Alternate interior angles)
∴ ΔAOB ˜ ΔCOD (By AAA similarity criterion)
Q3 : In the following figure, ABC and DBC are two triangles on the same base BC. If AD intersects BC at O, show that
Let us draw two perpendiculars AP and DM on line BC.
We know that area of a triangle =
In ΔAPO and ΔDMO,
∠APO = ∠DMO (Each = 90°)
∠AOP = ∠DOM (Vertically opposite angles)
∴ ΔAPO ˜ ΔDMO (By AA similarity criterion)
Q4 : If the areas of two similar triangles are equal, prove that they are congruent.
Let us assume two similar triangles as ΔABC ˜ ΔPQR.
Q5 : D, E and F are respectively the mid-points of sides AB, BC and CA of ΔABC. Find the ratio of the area of ΔDEF and ΔABC.
D and E are the mid-points of ΔABC.
Q6 : Prove that the ratio of the areas of two similar triangles is equal to the square
of the ratio of their corresponding medians.
Let us assume two similar triangles as ΔABC ˜ ΔPQR. Let AD and PS be the medians of these triangles.
∴ΔABC ˜ ΔPQR
…(1)
∠A = ∠P, ∠B = ∠Q, ∠C = ∠R … (2)
Since AD and PS are medians,
∴ BD = DC =
And, QS = SR =
Equation (1) becomes
… (3)
In ΔABD and ΔPQS,
∠B = ∠Q [Using equation (2)]
And, [Using equation (3)]
∴ ΔABD ˜ ΔPQS (SAS similarity criterion)
Therefore, it can be said that
… (4)
From equations (1) and (4), we may find that
And hence,
Q7 : Prove that the area of an equilateral triangle described on one side of a square is equal to half the area of the equilateral triangle described on one of its diagonals.
Let ABCD be a square of side a.
Therefore, its diagonal
Two desired equilateral triangles are formed as ΔABE and ΔDBF.
Side of an equilateral triangle, ΔABE, described on one of its sides = a
Side of an equilateral triangle, ΔDBF, described on one of its diagonals
We know that equilateral triangles have all its angles as 60 º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
Q8 : ABC and BDE are two equilateral triangles such that D is the mid-point of BC. Ratio of the area of triangles ABC and BDE is
(A) 2 : 1
(B) 1 : 2
(C) 4 : 1
(D) 1 : 4
We know that equilateral triangles have all its angles as 60 º and all its sides of the same length. Therefore, all equilateral triangles are similar to each other. Hence, the ratio between the areas of these triangles will be equal to the square of the ratio between the sides of these triangles.
Let side of ΔABC = x
Therefore, side of
Hence, the correct answer is (C).
Q9 : Sides of two similar triangles are in the ratio 4 : 9. Areas of these triangles are in the ratio
(A) 2 : 3
(B) 4 : 9
(C) 81 : 16
(D) 16 : 81
If two triangles are similar to each other, then the ratio of the areas of these triangles will be equal to the square of the ratio of the corresponding sides of these triangles.
It is given that the sides are in the ratio 4:9.
Therefore, ratio between areas of these triangles =
Hence, the correct answer is (D).
Exercise 6.5 : Solutions of Questions on Page Number : 150
Q1 : Sides of triangles are given below. Determine which of them are right triangles? In case of a right triangle, write the length of its hypotenuse.
(i) 7 cm, 24 cm, 25 cm
(ii) 3 cm, 8 cm, 6 cm
(iii) 50 cm, 80 cm, 100 cm
(iv) 13 cm, 12 cm, 5 cm
(i) It is given that the sides of the triangle are 7 cm, 24 cm, and 25 cm.
Squaring the lengths of these sides, we will obtain 49, 576, and 625.
49 + 576 = 625
Or,
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest side of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 25 cm.
(ii) It is given that the sides of the triangle are 3 cm, 8 cm, and 6 cm.
Squaring the lengths of these sides, we will obtain 9, 64, and 36.
However, 9 + 36 ≠ 64
Or, 32 + 62 ≠ 82
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.
(iii)Given that sides are 50 cm, 80 cm, and 100 cm.
Squaring the lengths of these sides, we will obtain 2500, 6400, and 10000.
However, 2500 + 6400 ≠ 10000
Or, 502 + 802 ≠ 1002
Clearly, the sum of the squares of the lengths of two sides is not equal to the square of the length of the third side.
Therefore, the given triangle is not satisfying Pythagoras theorem.
Hence, it is not a right triangle.
(iv)Given that sides are 13 cm, 12 cm, and 5 cm.
Squaring the lengths of these sides, we will obtain 169, 144, and 25.
Clearly, 144 +25 = 169
Or,
The sides of the given triangle are satisfying Pythagoras theorem.
Therefore, it is a right triangle.
We know that the longest side of a right triangle is the hypotenuse.
Therefore, the length of the hypotenuse of this triangle is 13 cm.
Q2 : PQR is a triangle right angled at P and M is a point on QR such that PM ⊥ QR. Show that PM2 = QM x MR.
Q4 : ABC is an isosceles triangle with AC = BC. If AB2 = 2 AC2, prove that ABC is a right triangle.
Given that,
Q5 : ABC is an equilateral triangle of side 2a. Find each of its altitudes.
Let AD be the altitude in the given equilateral triangle, ΔABC.
We know that altitude bisects the opposite side.
∴ BD = DC = a
In an equilateral triangle, all the altitudes are equal in length.
Therefore, the length of each altitude will be.
Q6 : Prove that the sum of the squares of the sides of rhombus is equal to the sum of the squares of its diagonals.
In ΔAOB, ΔBOC, ΔCOD, ΔAOD,
Applying Pythagoras theorem, we obtain
Q7 : In the following figure, O is a point in the interior of a triangle ABC, OD ⊥ BC, OE ⊥ AC and OF ⊥ AB. Show that
(i) OA2 + OB2 + OC2 – OD2 – OE2 – OF2 = AF2 + BD2 + CE2
(ii) AF2 + BD2 + CE2 = AE2 + CD2 + BF2
Join OA, OB, and OC.
(i) Applying Pythagoras theorem in ΔAOF, we obtain
Similarly, in ΔBOD,
Similarly, in ΔCOE,
(ii) From the above result,
Q8 : A ladder 10 m long reaches a window 8 m above the ground. Find the distance of the foot of the ladder from base of the wall.
Let OA be the wall and AB be the ladder.
Therefore, by Pythagoras theorem,
Therefore, the distance of the foot of the ladder from the base of the wall is
6m.
Q9 : A guy wire attached to a vertical pole of height 18 m is 24 m long and has a stake attached to the other end. How far from the base of the pole should the stake be driven so that the wire will be taut?
Let OB be the pole and AB be the wire.
By Pythagoras theorem,
Therefore, the distance from the base is m.
Q10 : An aeroplane leaves an airport and flies due north at a speed of 1,000 km per hour. At the same time, another aeroplane leaves the same airport and flies due west at a speed of 1,200 km per hour. How far apart will be the two planes after hours?
Distance travelled by the plane flying towards north in
Similarly, distance travelled by the plane flying towards west in
Let these distances be represented by OA and OB respectively.
Applying Pythagoras theorem,
Distance between these planes after, AB =
Therefore, the distance between these planes will be km after.
Q11 : Two poles of heights 6 m and 11 m stand on a plane ground. If the distance between the feet of the poles is 12 m, find the distance between their tops.
Let CD and AB be the poles of height 11 m and 6 m.
Therefore, CP = 11 – 6 = 5 m
From the figure, it can be observed that AP = 12m
Applying Pythagoras theorem for ΔAPC, we obtain
Therefore, the distance between their tops is 13 m.
Q12 : D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2
Applying Pythagoras theorem in ΔACE, we obtain
Q13 : The perpendicular from A on side BC of a ΔABC intersect BC at D such that DB = 3 CD. Prove that 2 AB2 = 2 AC2 + BC2
Applying Pythagoras theorem for ΔACD, we obtain
Applying Pythagoras theorem in ΔABD, we obtain
Q14 : In an equilateral triangle ABC, D is a point on side BC such that BD = BC. Prove that 9 AD2 = 7 AB2.
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC ==
And, AE =
Given that, BD = BC
∴ BD =
DE = BE – BD =
Applying Pythagoras theorem in ΔADE, we obtain
⇒ 9 AD2 = 7 AB2
Q15 : In an equilateral triangle, prove that three times the square of one side is equal to four times the square of one of its altitudes.
Let the side of the equilateral triangle be a, and AE be the altitude of ΔABC.
∴ BE = EC = =
Applying Pythagoras theorem in ΔABE, we obtain
AB2 = AE2 + BE2
4AE2 = 3a2
⇒ 4 × (Square of altitude) = 3 × (Square of one side)
Q16 : Tick the correct answer and justify: In ΔABC, AB = cm, AC = 12 cm and BC = 6 cm.
The angle B is:
(A) 120° (B) 60°
(C) 90° (D) 45°
Given that, AB =cm, AC = 12 cm, and BC = 6 cm
It can be observed that
AB2 = 108
AC2 = 144
And, BC2 = 36
AB2 +BC2 = AC2
The given triangle, ΔABC, is satisfying Pythagoras theorem.
Therefore, the triangle is a right triangle, right-angled at B.
∴ ∠B = 90°
Hence, the correct answer is (C).
Exercise 6.6 : Solutions of Questions on Page Number : 152
Q1 : In the given figure, PS is the bisector of ∠QPR of ΔPQR. Prove that .
Let us draw a line segment RT parallel to SP which intersects extended line segment QP at point T.
Given that, PS is the angle bisector of ∠QPR.
∠QPS = ∠SPR … (1)
By construction,
∠SPR = ∠PRT (As PS || TR) … (2)
∠QPS = ∠QTR (As PS || TR) … (3)
Using these equations, we obtain
∠PRT = ∠QTR
∴ PT = PR
By construction,
PS || TR
By using basic proportionality theorem for ΔQTR,
Q2 : In the given figure, D is a point on hypotenuse AC of ΔABC, DM ⊥ BC and DN ⊥ AB, Prove that:
(i) DM2 = DN.MC
(ii) DN2 = DM.AN
(i)Let us join DB.
We have, DN || CB, DM || AB, and ∠B = 90°
∴ DMBN is a rectangle.
∴ DN = MB and DM = NB
The condition to be proved is the case when D is the foot of the perpendicular drawn from B to AC.
∴ ∠CDB = 90°
⇒ ∠2 + ∠3 = 90° … (1)
In ΔCDM,
∠1 + ∠2 + ∠DMC = 180°
⇒ ∠1 + ∠2 = 90° … (2)
In ΔDMB,
∠3 + ∠DMB + ∠4 = 180°
⇒ ∠3 + ∠4 = 90° … (3)
From equation (1) and (2), we obtain
∠1 = ∠3
From equation (1) and (3), we obtain
∠2 = ∠4
In ΔDCM and ΔBDM,
∠1 = ∠3 (Proved above)
∠2 = ∠4 (Proved above)
∴ ΔDCM ˜ ΔBDM (AA similarity criterion)
⇒ DM2 = DN × MC
(ii) In right triangle DBN,
∠5 + ∠7 = 90° … (4)
In right triangle DAN,
∠6 + ∠8 = 90° … (5)
D is the foot of the perpendicular drawn from B to AC.
⇒ ∠5 + ∠6 = 90° … (6)
From equation (4) and (6), we obtain
∠6 = ∠7
From equation (5) and (6), we obtain
∠8 = ∠5
In ΔDNA and ΔBND,
∠6 = ∠7 (Proved above)
∠8 = ∠5 (Proved above)
∴ ΔDNA ˜ ΔBND (AA similarity criterion)
⇒ DN2 = AN × NB
⇒ DN2 = AN × DM (As NB = DM)
Q3 : In the given figure, ABC is a triangle in which ∠ ABC> 90° and AD ⊥ CB produced. Prove that AC2 = AB2 + BC2 + 2BC.BD.
Applying Pythagoras theorem in ΔADB, we obtain
AB2 = AD2 + DB2 … (1)
Applying Pythagoras theorem in ΔACD, we obtain
AC2 = AD2 + (DB + BC)2
AC2 =AD2 + DB2 + BC2 + 2DB x BC
AC2= AB2 + BC2 + 2DB x BC [Using equation (1)]
Q4 : In the given figure, ABC is a triangle in which ∠ ABC < 90° and AD ⊥ BC. Prove that AC2 = AB2 + BC2 – 2BC.BD.
Applying Pythagoras theorem in ΔADB, we obtain
⇒ AD2 = AB2 – DB2 … (1)
Applying Pythagoras theorem in ΔADC, we obtain
AB2 – BD2 + DC2 = AC2 [Using equation (1)]
AB2 – BD2 + (BC – BD)2 = AC2
AC2 = AB2 – BD2 + BC2 + BD2 -2BC x BD
= AB2 + BC2 – 2BC x BD
Q5 : In the given figure, AD is a median of a triangle ABC and AM ⊥ BC. Prove that:
(i)
(ii)
(iii)
(i) Applying Pythagoras theorem in ΔAMD, we obtain
AM2 + MD2 = AD2 … (1)
Applying Pythagoras theorem in ΔAMC, we obtain
AM2 + MC2 = AC2
AM2 + (MD + DC)2 = AC2
(AM2 + MD2) + DC2 + 2MD.DC = AC2
AD2 + DC2 + 2MD.DC = AC2 [Using equation (1)]
Using the result,, we obtain
(ii) Applying Pythagoras theorem in ΔABM, we obtain
AB2 = AM2 + MB2
= (AD2 – DM2) + MB2
= (AD2 – DM2) + (BD – MD)2
= AD2 – DM2 + BD2 + MD2 – 2BD × MD
= AD2 + BD2 – 2BD × MD
(iii)Applying Pythagoras theorem in ΔABM, we obtain
AM2 + MB2 = AB2 … (1)
Applying Pythagoras theorem in ΔAMC, we obtain
AM2 + MC2 = AC2 … (2)
Adding equations (1) and (2), we obtain
2AM2 + MB2 + MC2 = AB2 + AC2
2AM2 + (BD – DM)2 + (MD + DC)2 = AB2 + AC2
2AM2+BD2 + DM2 – 2BD.DM + MD2 + DC2 + 2MD.DC = AB2 + AC2
2AM2 + 2MD2 + BD2 + DC2 + 2MD ( – BD + DC) = AB2 + AC2
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What Is The Cube Root Of 64?
• Step by step simplification process to get cube roots radical form and derivative:
• First we will find all factors under the cube root: 64 has the cube factor of 64.
• Let's check this with ∛64*1=∛64. As you can see the radicals are not in their simplest form.
• Now extract and take out the cube root ∛64 * ∛1. Cube of ∛64=4 which results into 4∛1
• All radicals are now simplified. The radicand no longer has any cube factors.
Determine The Cubed Root Of 64?
• The cubed root of sixty-four ∛64 = 4
How To Calculate Cube Roots
• The process of cubing is similar to squaring, only that the number is multiplied three times instead of two. The exponent used for cubes is 3, which is also denoted by the superscript³. Examples are 4³ = 4*4*4 = 64 or 8³ = 8*8*8 = 512.
• The cubic function is a one-to-one function. Why is this so? This is because cubing a negative number results in an answer different to that of cubing it's positive counterpart. This is because when three negative numbers are multiplied together, two of the negatives are cancelled but one remains, so the result is also negative. 7³ = 7*7*7 = 343 and (-7)³ = (-7)*(-7)*(-7) = -343. In the same way as a perfect square, a perfect cube or cube number is an integer that results from cubing another integer. 343 and -343 are examples of perfect cubes.
Mathematical Information About Numbers 6 4
• About Number 6. Six is the smallest composite number with two distinct prime factors, and the third triangular number. It is the smallest perfect number: 6 = 1 + 2 + 3 and the faculty of 3 is 6 = 3! = 1 * 2 * 3, which is remarkable, because there is no other three numbers whose product is equal to their sum. Similarly 6 = sqrt(1 ^ 3 + 2 + 3 ^ 3 ^ 3). The equation x ^ 3 + Y ^ 3 ^ 3 + z = 6xyz is the only solution (without permutations) x = 1, y = 2 and z = 3. Finally 1/1 = 1/2 + 1/3 + 1/6. The cube (from the Greek) or hexahedron (from Latin) cube is one of the five Platonic solids and has six equal areas. A tetrahedron has six edges and six vertices an octahedron. With regular hexagons can fill a plane without gaps. Number six is a two-dimensional kiss number.
• About Number 4. Four is linear. It is the first composite number and thus the first non-prime number after one. The peculiarity of the four is that both 2 + 2 = 4 and 2 * 2 = 4 and thus 2^2 = 4. Four points make the plane of a square, an area with four sides. It is the simplest figure that can be deformed while keeping it's side lengths, such as the rectangle to parallelogram. Space let's us arrange equidistantly a maximum of four points. These then form a tetrahedron (tetrahedron), a body with four identical triangular faces. Another feature of the four is the impossibility of an algebraic equation of higher degree than four square roots using simple arithmetic and basic operations dissolve.
What is a cube root?
In arithmetic and algebra, the cube of a number n is its third power: the result of the number multiplied by itself twice: n³ = n * n * n. It is also the number multiplied by its square: n³ = n * n².
This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also n raised to the one-third power.
Both cube and cube root are odd functions: (-n)³ = -(n³). The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 2³ = 8 or (x + 1)³.
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You might think about work as, reading books and doing homework. You may have also heard the sentence “hard work leads to success”. But, in physics work has its own meaning and it is different from the previously described example.
## What is Work?
In physics, work is done when a force is applied to an object and it travels some distance. For example, you kicked a ball with a certain amount of force and ball traveled few meters, then you had done a certain amount of work on the ball.
## How is Work measured?
Work is the scalar quantity, means it doesn’t need direction to define it. Work is measured in joule or newton-meter. This is the same unit in which energy is measured. The work done on an object is found by multiplying the force exerted on an object with the distance that it traveled.
Work = Force*distance
But, there are also some complications in calculating the work which may sometimes confuse you. Here are the few things to remember when calculating the work.
1. If you are applying a force on an object but it is not moving, then work will be zero. Because object didn’t travel any distance. For example, you are pushing a wall with all of your force, but the wall is not moving. So, your work done will be zero.
2. If you are applying a force on an object and it is moving in a path that brings it back from where it started. Then, work done on that object is also zero. Because the net distance traveled by the object is zero.
For example, a race car is starting from a starting point and traveling in a circular path and it comes back again at the same starting point. Then, work done by the car is zero because it didn’t travel any net distance.
Example: A man is pushing a box with a force 50 newtons. The box has traveled the distance of 100 meters. Calculate the work done by the man.
Solution:
Applied Force = 50 N
Distance covered = 100 meters
Work = Force*distance = 50*100 = 5000 joules or N-m
## Facts
• If a force is applied in opposite direction to the covered distance, then the work is said to be negative work.
• An object that falls from a height also performs work due to gravitational force and height.
• In space when an object is accelerated to a certain speed then the force is removed from it. But, the object doesn’t lose its speed because there is no any air friction in space. So, work done by objects flying in space is also zero. |
# Math 5 - Act. 21: The Amazing Inch and Measuring Up!
Group Size:
Small Groups
Summary:
In this activity, students will gain a knowledge and understanding of both the metric and customary systems of measuring length with rulers, meter / yardsticks, and tape measures.
Materials:
For each student:
• (2) 3" x 12" pieces of oak tag
• ruler with standard and metric measurements
• “Measuring in Feet and Inches” chart
• Enlarged Inch labeled
• Enlarged Inch
For teacher:
• plastic overhead ruler (to nearest 1/8 inch)
Literature:
How Tall How Short How Faraway by David A. Adler
Measuring by Sheila Cato
Attachments
Background For Teachers:
Use accurate terminology to demonstrate and explain Metric and Customary Units of Length Measurement. Encourage students to also use correct terminology.
Vocabulary
Customary: A system of measurement used in the United States. The system includes units for measuring length, capacity, and weight.
Units of Length Measurement in Customary System: one foot (ft or ') = 12 inches (in or ") one yard (yd) = 36 inches = 3 feet 1 mile (mi) = 5,280 feet = 1,760 yards
Intended Learning Outcomes:
1. Become mathematical problem solvers.
2. Make mathematical connections.
Instructional Procedures:
Invitation to Learn
Have students work in partners/cooperative groups to list as many occupations as possible where tools are used for measuring. Students should also list what tools are used with each job. Discuss why it is important to learn how to use these tools in real life situations.
Instructional Procedures
1. Each student receives one of the 3" x 12" pieces of oak tag.
2. Students are then instructed that this paper represents one magic “amazing inch.” As the teacher models, students follow the teacher’s example:
1. Teacher will fold the paper in half once. Then draw a line on the fold, and label it 1/2. The left side edge of the paper will be labeled 0/2, while the right side edge will be labeled 2/2.
2. While students continue to follow the teacher’s example, the teacher will fold the paper in half again. Students will be asked how many equal parts there are. A line a littler shorter will be drawn on the additional folds and labeled 1/4, 2/4, and 3/4. The edges will be appropriately labeled 0/4 and 4/4.
3. One more fold and with shorter lines drawn on the folds which will be labeled 1/8, 2/8, 3/8, 4/8, 5/8, 6/8, and 7/8. The edges will be labeled 0/8 and 8/8.
3. During the folding process, the teacher will review how each time the “magic inch” is folded each of the new sections are equal in size.
4. Teacher places a transparent plastic ruler on the overhead and points to the different lines on the ruler asking individual students (or popcorn style where students quietly call out answers), what part of the inch each line represents.
5. Students will apply the concept of the magic inch to an enlarged inch ruler on a sheet of paper by labeling it as was done with the “magic inch” and with the overhead.
6. Students will practice estimating and measuring by using paper rulers which are ruled to 1 inch, 1/2 inch, 1/4 inch, and 1/8 inch. (It would be a good idea to have students practice first with the 1 inch ruler. Then, after some practice use the 1/2 inch ruler. Continue to practice, then progress to the 1/4 inch ruler and finally after more practice the 1/8 inch ruler). Answers are recorded on a copy of the “Measuring in Feet and Inches” Worksheet.
7. Students will then partner and, using a ruler, review with each other what the different lines on the ruler represent.
8. Working in partners or groups, the class will then estimate the measurement of several items in the classroom and record their answers on the “Measuring in Feet and Inches” worksheet. After estimating, students will measure and record the exact measurement of each item.
9. Students label and glue a copy of an enlarged inch in their journals.
10. Students will explain in their journals how 2/4 and 4/8 are equal to 1/2, and 2/8 is equal to 1/4, etc. (tie in with fractions).
Curriculum Integration
Journal writing: Discuss how measurement is used in occupations. Use rulers as an additional way to teach fractions. Students, working with a partner, measure themselves and each other, then make a half-size selfportrait on butcher paper. They may draw clothes, hair, etc., to make the half-size me look like themselves. Construction paper and yarn may also be used for clothes and hair. (This activity may be done by customary, to the nearest inch, or metric measuring, to the nearest centimeter.)
Extensions:
“Inches” or “Metric Measurement” Games
Social Studies—Students interview parents or other relative to discover how measurement is used in their jobs. Graph the class results.
Math—The measuring activities may also be done in the metric system.
Homework & Family Connections
Students take a 3" x 12" piece of tag board home and teach parents or siblings how to fold and label their own “magic inch.”
Students estimate and then measure items in their homes and record answers on “Measuring in Feet and Inches” worksheet.
Assessment Plan:
Students will estimate the measurement of several items, then measure them and record answers on a copy of the “Measuring in Feet and Inches” worksheet.
Students write the correct measurements on an enlarged inch (blackline included).
Author:
Utah LessonPlans
Created Date :
Sep 03 2003 16:37 PM
55725 |
# Question #1 Explanation | Finding p-value of z-score with population distribution
Question #1: What percentage of the population has an IQ score less than 105?
## Explanation
"p-values" are a core concept of statistics. In simple terms, they representing the "probability" of a given event occurring. They enable us to determine the "statistical significance" of the given event, which will be used in conjunction with confidence intervals and hypothesis tests.
p-value stands for probability-value, and is a number between 0.000 and 1.000 representing the probability of an event occurring in a population.
When able to do so, the best way to find a p-value is with a z-score (also referred to as "z-test statistic").
To find our p-value for the below scenario...
Question #1: What percentage of the population has an IQ score less than 105?
...we're first going to calculate the z-score for the IQ score of 105, and then we'll find the corresponding p-value in the z-table!
### Calculating the z-score
To calculate z-score, we'll use the below formula:
What do each of these variables mean?
x is the data point that you want to obtain the z-score of.
μ is the mean value of the given population.
σ is the standard deviation of the given population’s values.
#### Plugging in the mean (μ) and standard deviation (σ)
In Z-scores, we stated the following about the distribution of IQ scores:
"Well, it is known that all IQ scores follow on a normal distribution like this...
...with a mean of 100...
...and a standard deviation of 15."
Therefore, we can plug in 100 for μ...
...and 15 for σ!
#### Plugging in the data point (x)
Above we stated that we got an IQ score of 105.
"You've always been curious how smart you are compared to the average person, so you decide to take an IQ test. After taking the test, you obtain a score of 105."
Therefore, we'll plug in 105 for x.
#### Solving for z-score
When we solve this out, we get 0.33!
### How to associate a p-value to your z-score
Cool! You've got a z-score of 0.33! But in reference to our prompt...
Question #1: What percentage of the population has an IQ score less than 105?
...how do we utilize that to figure out what percentage of the population has an IQ score below 105?
In other words, what percentage of the area under the normal distribution curve is to the left of our z-score of 0.33?
To find this, we'll use z-table! (Don't sleep on this table. It's super clutch.)
All we need to do is find "0.3" in the left-hand column of the right table (representing 0.33)...
...and then "0.03" in the top row (representing 0.33)...
...to locate our p-value of 0.6293!
Notice how at the top of the z-table, the area to the left of the "z" on the x-axis is filled in?
That's because...
The p-values in the z-table are telling you the area under the normal distribution curve to the left of your z-score!
So in the case of our 105 IQ score on the distribution of all IQ scores...
...our p-value is 0.6293, which means that 62.93% of the area under the curve occurs to the left of our 105 IQ score!
In other words, to answer our original question...
Question #1: What percentage of the population has an IQ score less than 105?
...62.93% of the population has an IQ score less than 105! |
# Unit 2 Project - Math II
Sarah Lamm - Math II - Period 3
# Activity 1: Graphing
In this activity I used a table that showed the sound beats per minute from songs in a mash-up. I graphed the beats per minute as a function of time, and then I found the piecewise function that corresponds to the graph.
### "5 Minute Mash-Up as a Piecewise Function Graph"
The second part of this activity was finding the piecewise function of the graph. What is a piecewise function? A piecewise function is used to graph a relationship of functions. Below is the piecewise function for the graph above.
# Activity 2: Modeling
In this activity I found the average BPM of the 4 songs but I also added a crescendo for the last 2 minutes of the song. Once I gathered my information I created a graph. The average BPM is 121 for all 4 songs
### "5 Minute Mash-Up with Crescendo"
Below if the function for the graph. The y = 122 shows the constant BPM of the song for the first 3 minutes. The y = 121.5 + 5^(x-3) shows the exponential growth of the BPM after the 3rd minute. The number 121.5 is the average of all the songs' BPMs and is where the exponential growth starts at. 5 is part of the parent function which shows the rate of exponential growth and 3 is the horizontal translation.
What is the BPM at....
3 minutes? 123
4 minutes? 126
5 minutes? 146
# Activity 3: Designing
In this activity I have compiled 4 songs and collected their BPM. Now I will create a graph and piecewise function to show my mash-up
### "My 5 Minute Mash-Up Graph"
Below is the piecewise function for the graph of my mash up. Notice that the greater than or equal to signs, as well as the less than or equal to sign correspond to the closed circles on the graph.
Here is the graph of my mashup with a crescendo. This technique averages the BPM of all the songs for the first 3 minutes and then plugs the information into a formula for the last 2 minutes. The formula for this mash-up's crescendo was 132 + 5^ x - 3. |
Jump to a New ChapterIntroduction to the SAT IIContent and Format of the SAT II Math IICStrategies for SAT II Math IICMath IIC FundamentalsAlgebraPlane GeometrySolid GeometryCoordinate GeometryTrigonometryFunctionsStatisticsMiscellaneous MathPractice Tests Are Your Best Friends
7.1 Prisms 7.2 Solids That Aren’t Prisms 7.3 Relating Length, Surface Area, and Volume 7.4 Inscribed Solids
7.5 Solids Produced by Rotating Polygons 7.6 Key Formulas 7.7 Review Questions 7.8 Explanations
Inscribed Solids
An inscribed solid is a solid fit inside another solid, with the edges of the two solids touching. The figures below are, from left to right, a cylinder inscribed in a sphere, a sphere inscribed in a cube, and a rectangular solid inscribed in a sphere.
Math IIC questions that involve inscribed solids don’t require any additional techniques than those you’ve already learned. What these questions do require is an ability to visualize inscribed solids and the awareness of how certain line segments relate to both solids in a given figure.
Most often, an inscribed solid question will present a figure of an inscribed solid and give you information about one of the solids. For example, you may be given the radius of a cylinder and then be asked to find the volume of the other solid using the figure as your guide. Use what you know about the radius of the cylinder to find the dimensions of the other solid. Take a look at this example:
In the figure below, a cube is inscribed in a cylinder. If the length of the diagonal of the cube is 4 and the height of the cylinder is 5, what is the volume of the cylinder?
The formula for the volume of a cylinder is πr2(h). It is given in the question that h = 5, but there is no value given for r. So in order to solve for the volume of the cylinder, we need to first find the value of r.
The key step in this problem is to recognize that the diagonal of a face of the cube is also the diameter, or twice the radius, of the cylinder. To see this, draw a diagonal, d, in either the top or bottom face of the cube.
In order to find this diagonal, which is the hypotenuse in a 45-45-90 triangle, we need the length of an edge of the cube, or s. We can find s from the diagonal of the cube (not to be mistaken with the diagonal of a face of the cube), since the formula for the diagonal of a cube is s with s being the length of an edge of the cube. The question gives us the diagonal of the cube as 4 so it follows that s = 4. This means that the diagonal along a single face of the cube is 4 (using the special properties of a 45-45-90 triangle). Therefore, the radius of the cylinder is 4/ 2 = 2 Plug that back into the formula for the volume of the cylinder, and you get π (2)2 5 = 40π.
The Rules of Inscribed Solids
Math IIC questions involving inscribed solids are much easier to solve when you know how the lines of different solids relate to one another. For instance, the previous example showed that when a cube is inscribed in a cylinder, the diagonal of a face of the cube is equal to the diameter of the cylinder. The better you know the rules of inscribed solids, the better you’ll do on these questions. Here are the rules of inscribed solids that most commonly appear on the Math IIC.
Cylinder Inscribed in a Sphere
The diameter of the sphere is equal to the diagonal of the cylinder’s height and diameter.
Sphere Inscribed in a Cube
The diameter of the sphere is equal to the length of cube’s edge.
Sphere Inscribed in a Cylinder
The cylinder and the sphere have the same radius.
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# Algebra 1 : How to solve one-step equations
## Example Questions
### Example Question #81 : Linear Equations
Solve for :
Possible Answers:
Correct answer:
Explanation:
To solve for , you must get it alone on one side of the equation. To do so, subtract from both sides.
### Example Question #82 : Linear Equations
Ashley and friends ate dinner at a restaurant together, and they split the bill evenly. If each person paid , how much was the total bill?
Possible Answers:
Correct answer:
Explanation:
Let be the total bill.
Since there is a total of people, we can write the following equation:
To solve for , multiply both sides by .
The total bill cost .
### Example Question #83 : Linear Equations
Aaron was baking cupcakes. The recipe calls for cups of flour. If he has already put in cups of flour, how many more cups of flour does he need?
Possible Answers:
Correct answer:
Explanation:
Let be the number of cups of flour he needs to still add.
Since he has already added cups, we can write the following equation:
To solve for , subtract both sides by :
### Example Question #84 : Linear Equations
Last Monday, Adam had . Over the course of the week, he earned some money babysitting. He now has . How much money did he earn babysitting?
Possible Answers:
Correct answer:
Explanation:
Let be the amount he earned babysitting.
Since he started out with , we can write the following equation:
To solve for , subtract from both sides.
Adam earned babysitting.
### Example Question #85 : Linear Equations
Anna ran more miles than Nelson ran. If Anna ran miles, how many miles did Nelson run?
Possible Answers:
The miles that Nelson ran cannot be determined.
Correct answer:
Explanation:
Let be the number of miles that Nelson ran.
Since we know that Anna ran more miles than Nelson did, we can write the following expression to tell us how many miles Anna ran in terms of :
Since we know that Anna ran a total of miles, we can then write the following equation:
Nelson ran miles.
### Example Question #86 : Linear Equations
Steve has remaining after paying for a salad. How much money did Steve have before buying the salad?
Possible Answers:
Correct answer:
Explanation:
Let be the amount of money Steve had before he paid for the salad.
Since he spent , we can then write the following equation:
To solve for , add to both sides.
Steve had before paying for the salad.
### Example Question #87 : Linear Equations
If a box of chocolates costs , how many boxes of chocolate can you buy with ?
Possible Answers:
Correct answer:
Explanation:
Let be the number of boxes of chocolate that you can buy.
Since we know that each box costs , we can write the following equation:
To solve for , divide both sides by .
You can buy boxes of chocolate with .
### Example Question #88 : Linear Equations
Solve for :
Possible Answers:
Correct answer:
Explanation:
To solve for , multiply both sides by .
### Example Question #89 : Linear Equations
Solve for :
Possible Answers:
Correct answer:
Explanation:
To solve for , add to both sides of the equation.
### Example Question #90 : Linear Equations
Solve for :
Possible Answers:
Correct answer:
Explanation:
To solve for , divide both sides by . |
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Amala, Bina, and Gouri invest money in the ratio $3:4:5$ in fixed deposits having respective annual interest rates in the ratio $6:5:4$. what is their total interest income (in Rs) after a year, if Bina’s interest income exceeds Amala’s by Rs $250$?
1. $6350$
2. $7250$
3. $7000$
4. $6000$
Let money invested by Amala, Bina, and Gouri be $3x, 4x,$ and $5x$. And the annual interest rates be $6y,5y,$ and $4y$ respectively.
We know that,
• Interest income $\propto$ Amount invested
• Interest income $\propto$ Interest rate
Therefore, interest income must be in the ratio of the product of their amount invested and interest rate.
• Amala’s interest income $= 3x \times 6y = 18xy$
• Bina’s interest income $= 4x \times 5y = 20xy$
• Gouri’s interest income $= 5x \times 4y = 20xy$
Bina’s interest income exceeds Amala’s by $\text{Rs}. 250$
$20xy – 18xy = 250$
$\Rightarrow 2xy = 250$
$\Rightarrow xy = 125$
$\therefore$ Total interest income after a year $= 18xy + 20xy + 20xy$
$\quad = 58xy = 58 \times 125 = \text{Rs}. 7250.$
$\textbf{Short Method:}$
$\begin{array}{lccc} & \text{Amala} & \text{Bina} & \text{Gouri} \\ \text{Invest} & 3 & 4 & 5 \\ \text{Interest rate} & 6 & 5 & 4 \\ \text{Interest income} & {\color{Red} {18}} & {\color{Blue} {20}} & 20 \end{array}$
According to the question,
• $2 \longrightarrow 250$
• $1 \longrightarrow 125$
Therefore, total interest income (in Rs) after a year $= (18 + 20 + 20) \times 125 = 58 \times 125 = \text{Rs}. 7250.$
Correct Answer $: \text{B}$
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# Vertical Angles: Theorem, Proof, Vertically Opposite Angles
Studying vertical angles is an essential topic for anyone who wishes to master math or any related subject that employs it. It's tough work, but we'll make sure you get a handle on these theories so you can make the grade!
Don’t feel discouraged if you don’t recollect or don’t have a good grasp on these theories, as this blog will help you study all the essentials. Additionally, we will help you understand the secret to learning faster and increasing your grades in arithmetic and other prevailing subjects today.
## The Theorem
The vertical angle theorem stipulates that whenever two straight lines intersect, they create opposite angles, known as vertical angles.
These opposite angles share a vertex. Furthermore, the most important point to remember is that they also measure the same! This refers that irrespective of where these straight lines cross, the angles opposite each other will consistently share the exact value. These angles are referred as congruent angles.
Vertically opposite angles are congruent, so if you have a value for one angle, then it is feasible to find the others employing proportions.
### Proving the Theorem
Proving this theorem is relatively straightforward. Primarily, let's draw a line and name it line l. After that, we will draw another line that intersects line l at some point. We will call this second line m.
After drawing these two lines, we will label the angles created by the intersecting lines l and m. To prevent confusion, we labeled pairs of vertically opposite angles. Therefore, we label angle A, angle B, angle C, and angle D as follows:
We understand that angles A and B are vertically contrary reason being that they share the equivalent vertex but don’t share a side. If you recall that vertically opposite angles are also congruent, meaning that angle A is identical angle B.
If we look at angles B and C, you will notice that they are not joined at their vertex but close to each other. They share a side and a vertex, signifying they are supplementary angles, so the sum of both angles will be 180 degrees. This instance repeats itself with angles A and C so that we can summarize this in the following way:
∠B+∠C=180 and ∠A+∠C=180
Since both sums up to equal the same, we can add these operations as follows:
∠A+∠C=∠B+∠C
By eliminating C on both sides of the equation, we will be left with:
∠A=∠B
So, we can conclude that vertically opposite angles are congruent, as they have the same measure.
## Vertically Opposite Angles
Now that we have studied about the theorem and how to prove it, let's talk explicitly about vertically opposite angles.
### Definition
As we stated, vertically opposite angles are two angles created by the intersection of two straight lines. These angles opposite each other satisfy the vertical angle theorem.
However, vertically opposite angles are never next to each other. Adjacent angles are two angles that have a common side and a common vertex. Vertically opposite angles at no time share a side. When angles share a side, these adjacent angles could be complementary or supplementary.
In the case of complementary angles, the sum of two adjacent angles will equal 90°. Supplementary angles are adjacent angles whose addition will equal 180°, which we just used to prove the vertical angle theorem.
These concepts are appropriate within the vertical angle theorem and vertically opposite angles because supplementary and complementary angles do not satisfy the characteristics of vertically opposite angles.
There are several characteristics of vertically opposite angles. Regardless, odds are that you will only need these two to secure your examination.
1. Vertically opposite angles are always congruent. Hence, if angles A and B are vertically opposite, they will measure the same.
2. Vertically opposite angles are never adjacent. They can share, at most, a vertex.
### Where Can You Locate Opposite Angles in Real-Life Situations?
You may think where you can find these concepts in the real world, and you'd be surprised to observe that vertically opposite angles are quite common! You can locate them in various daily objects and circumstances.
For instance, vertically opposite angles are created when two straight lines overlap each other. Right in your room, the door installed to the door frame creates vertically opposite angles with the wall.
Open a pair of scissors to make two intersecting lines and adjust the size of the angles. Track intersections are also a great example of vertically opposite angles.
Finally, vertically opposite angles are also discovered in nature. If you watch a tree, the vertically opposite angles are created by the trunk and the branches.
Be sure to observe your environment, as you will detect an example next to you.
## Puttingit All Together
So, to summarize what we have considered so far, vertically opposite angles are made from two overlapping lines. The two angles that are not adjacent have the same measure.
The vertical angle theorem states that when two intersecting straight lines, the angles created are vertically opposite and congruent. This theorem can be tried out by depicting a straight line and another line overlapping it and using the concepts of congruent angles to complete measures.
Congruent angles means two angles that measure the same.
When two angles share a side and a vertex, they can’t be vertically opposite. However, they are complementary if the addition of these angles equals 90°. If the sum of both angles equals 180°, they are deemed supplementary.
The total of adjacent angles is consistently 180°. Thus, if angles B and C are adjacent angles, they will always add up to 180°.
Vertically opposite angles are quite common! You can locate them in several daily objects and situations, such as doors, windows, paintings, and trees. |
## Elementary Linear Algebra 7th Edition
The solution of this system is $$x=\cos\theta,\quad y=\sin\theta.$$
Here we treat the sines and cosines as ordinary coefficients. So to solve this problem we will follow the usual steps for solving a system of two equations in two variables: Step 1: Express $y$ from the second equation: $$(\cos\theta)y=(\sin\theta)x$$ so we have $$y=\frac{\sin\theta}{\cos\theta}x = (\tan\theta) x.$$ Step 2: Put this into the first equation and solve for $x$: $$(\cos\theta)x+x\sin\theta\tan\theta = 1 \Rightarrow x\left(\cos\theta+\sin\theta\tan\theta\right)=1$$ Now remember that $\tan\theta = \sin\theta/\cos\theta$ so for the expression in the bracket on the left we have $$\cos\theta + \sin\theta\frac{\sin\theta}{\cos\theta} = \cos\theta+ \frac{\sin^2\theta}{\cos\theta} = \frac{\cos^2\theta+\sin^2\theta}{\cos\theta}.$$ Now remember that for any $\theta$ we have that $\sin^2\theta+ \cos^2\theta = 1$ so that the expression in the bracket is finally equal to $$\frac{1}{\cos\theta}.$$ returning to the equation we now have $$x\frac{1}{\cos\theta} = 1$$ and this gives $$x=\cos\theta$$ Step 3: Now use this equality to calculate $y$: $$y=\frac{\sin\theta}{\cos\theta}x = \frac{\sin\theta}{\cos\theta}\times\cos\theta = \sin\theta.$$ |
# Difference between revisions of "2011 AIME I Problems/Problem 4"
## Problem 4
In triangle $ABC$, $AB=125$, $AC=117$ and $BC=120$. The angle bisector of angle $A$ intersects $\overline{BC}$ at point $L$, and the angle bisector of angle $B$ intersects $\overline{AC}$ at point $K$. Let $M$ and $N$ be the feet of the perpendiculars from $C$ to $\overline{BK}$ and $\overline{AL}$, respectively. Find $MN$.
## Solution 1
Extend ${CM}$ and ${CN}$ such that they intersects lines ${AB}$ at points $P$ and $Q$, respectively. Since ${BM}$ is the angle bisector of angle B,and ${CM}$ is perpendicular to ${BM}$ ,so , $BP=BC=120$, M is the midpoint of ${CP}$ .For the same reason,$AQ=AC=117$,N is the midpoint of ${CQ}$. Hence$MN=\frac{PQ}{2}$.But $PQ=BP+AQ-AB=120+117-125=112$,so$MN=\boxed{56}$.
## Solution 2
[This solution appears to have said A where it should have said C and vice versa.] Let $I$ be the incenter of $ABC$. Now, since $AM \perp LC$ and $AN \perp KB$, we have $AMIN$ is a cyclic quadrilateral. Consequently, $\frac{MN}{\sin \angle MIN} = 2R = AI$. Since $\angle MIN = 90 - \frac{\angle BAC}{2} = \cos \angle IAK$, we have that $MN = AI \cdot \cos \angle IAK$. Letting $X$ be the point of contact of the incircle of $ABC$ with side $AC$, we have $AX=MN$ thus $MN=\frac{117+120-125}{2}=\boxed{56}$
## Solution 3 (Bash)
Project $I$ onto $AC$ and $BC$ as $D$ and $E$. $ID$ and $IE$ are both in-radii of $\triangle ABC$ so we get right triangles with legs $r$ (the in-radius length) and $s - c = 56$. Since $IC$ is the hypotenuse for the 4 triangles ($\triangle INC, \triangle IMC, \triangle IDC,$ and $\triangle IEC$), $C, D, M, I, N, E$ are con-cyclic on a circle we shall denote as $\omega$ which is also the circumcircle of $\triangle CMN$ and $\triangle CDE$. To find $MN$, we can use the Law of Cosines on $\angle MON \implies MN^2 = 2R^2(1 - \cos{2\angle MCN})$ where $O$ is the center of $\omega$. Now, the circumradius $R$ can be found with Pythagorean Theorem with $\triangle CDI$ or $\triangle CEI$: $r^2 + 56^2 = (2R)^2$. To find $r$, we can use the formula $rs = [ABC]$ and by Heron's, $[ABC] = \sqrt{181 \cdot 61 \cdot 56 \cdot 64} \implies r = \sqrt{\frac{61 \cdot 56 \cdot 64}{181}} \implies 2R^2 = \frac{393120}{181}$. To find $\angle MCN$, we can find $\angle MIN$ since $\angle MCN = 180 - \angle MIN$. $\angle MIN = \angle MIC + \angle NIC = 180 - \angle BIC + 180 - \angle AIC = 180 - (180 - \frac{\angle A + \angle C}{2}) + 180 - (180 - \frac{\angle B + \angle C}{2}) = \frac{\angle A + \angle B + \angle C}{2} + \frac{\angle C}{2}$. Thus, $\angle MCN = 180 - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2}$ and since $\angle A + \angle B + \angle C = 180$, we have $\angle A + \angle B + \angle C - \frac{\angle A + \angle B + \angle C}{2} - \frac{\angle C}{2} = \frac{\angle A + \angle B}{2}$. Plugging this into our Law of Cosines formula gives $MN^2 = 2R^2(1 - \cos{\angle A + \angle B}) = 2R^2(1 + \cos{\angle C})$. To find $\cos{\angle C}$, we use LoC on $\triangle ABC \implies cos{\angle C} = \frac{120^2 + 117^2 - 125^2}{2 \cdot 117 \cdot 120} = \frac{41 \cdot 19}{117 \cdot 15}$. Our formula now becomes $MN^2 = \frac{393120}{181} + \frac{2534}{15 \cdot 117}$. After simplifying, we get $MN^2 = 3136 \implies MN = \boxed{056}$. $\square$
--lucasxia01
## Solution 4
Because $\angle CMI = \angle CNI = 90$, $CMIN$ is cyclic.
Ptolemy on CMIN:
$CN*MI+CM*IN=CI*MN$
$CI^2(\cos \angle ICN \sin \angle ICM + \cos \angle ICM \sin \angle ICN) = CI * MN$
$MN = CI \sin \angle MCN$ by angle addition formula.
$\angle MCN = 180 - \angle MIN = 90 - \angle BCI$.
Let $H$ be where the incircle touches $BC$, then $CI \cos \angle BCI = CH = \frac{a+b-c}{2}$. $a=120, b=117, c=125$, for a final answer of $\boxed{056}$. |
# Place Value - Ones, Tens, Hundreds and Thousands
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Place Value - Ones, Tens, Hundreds and Thousands
CCSS.MATH.CONTENT.3.NBT.A
## Basics on the topicPlace Value - Ones, Tens, Hundreds and Thousands
### In This Video
Skylar and Henry are invited to a block party - but to find the location they need to use place value. Using a thousands place value chart, and place value blocks - thousands, they’re able to find the house and go to the block party. However, it’s a little different than they’re used to!
### Place Value to the Thousands Example
To find place value to the thousands, we can use a place value chart to thousands to help us. Let’s take a look at the number two thousand and ninety-seven to practice place value to thousands.
Starting at the thousand place, use two thousands place value blocks, and write two thousands underneath as this represents the thousands place value.
Now move to the hundreds place. There is the number zero here, so the place value of the hundreds is zero hundreds. Write this underneath the hundreds place.
Now move to the tens place. There is a number nine here, so use nine tens blocks. The place value of the tens is nine tens. Write this underneath the hundreds place.
Now move to the ones place. There is a number seven here, so use seven ones blocks. The place value of the ones is seven ones. Write this underneath the ones place.
### Place Value to Thousands - Summary
You can use a place value chart - thousands, and place value blocks to thousands to help you identify the value of each digit in numbers up to the thousands place. Remember, each place value shows the value of the digit in that place!
You can find a place value to thousands worksheet below for extra practice, where you can use place value charts to thousands to help solve the problems!
### TranscriptPlace Value - Ones, Tens, Hundreds and Thousands
Skylar and Henry are invited to a party at two thousand ninety-seven Block Avenue. However, houses on Block Avenue use base ten blocks to represent their numbers. To find the location, Skylar and Henry will use place value - ones, tens, hundreds, and thousands. Place value is the value of a digit based on its position in a number. We can use a place value chart and base ten blocks to help identify the value of a digit in a number with four-digits. Let's take a look at the number one thousand, two hundred thirty-five. In the thousands place, use a thousands block because one is in the thousands place. The value of one in the thousands place is one thousand. In the hundreds place, use two hundreds blocks because two is in the hundreds place. The value of two in the hundreds place is two hundreds. In the tens place, use three tens blocks because three is in the tens place. The value of three in the tens place is three tens. Finally, in the ones place, use five ones blocks because five is in the ones place. The value of five in the ones place is five ones. Now that we have looked at place value up to the thousands place, let's help Skylar and Henry find the location of the party! On Block Avenue, they need to find two thousand ninety seven. Use a place value chart to set up the number. In the thousands place, use two thousands blocks because two is in the thousands place. What is the value of the thousands place? The value is two thousands. In the hundreds place, use zero hundreds blocks because zero is in the hundreds place. We use zero to hold the place value when there is no value for it. What is the value of the hundreds place? The value is zero hundreds. In the tens place, use nine tens blocks because nine is in the tens place. What is the value of the tens place? The value is nine tens. Finally, in the ones place, use seven ones blocks because seven is in the ones place. What is the value of the ones place? The value is seven ones. Skylar and Henry will look for the location with base ten blocks like this above the door! While they make their way, let's review! Remember, base ten blocks can help identify the place value of numbers. The thousands place shows how many thousands there are. The hundreds place shows how many hundreds there are. The tens place shows how many tens there are. The ones place shows how many ones there are. "Henry, I don't recall block parties being like this. Do you?" "No Skylar, I really don't." |
Congruent Shapes
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## What Are Congruent Shapes?
Congruent shapes are shapes that have the same shape and same size.
Each pair above is made of congruent shapes
Congruent shapes can have different color, or rotation. They can even be flipped
As long as they have the same shape and size, they're still called congruent.
If you put the blue circle above on top of the green one, the blue circle completely covers the other circle.
The blue and green circles have:
1️⃣ the same shape ✅
2️⃣ the same size ✅
These shapes are congruent.
### Another Pair of Circles
Are these circles congruent?
Let's check:
1️⃣ Are they the same shape?
Yes, they're both circles. ✅
2️⃣ Are they of the same size?
One circle is small. The second circle is bigger.
No, they're not the same size. ❌
These shapes are not congruent.
### Are These Shapes Congruent?
Are they congruent?
1️⃣ Are they the same shape?
They both have the same number of sides.
But the first shape has 4 equal sides, while the sides of the second shape are not all equal.
So no, they're not the same shape. ❌
These shapes are not congruent.
Tip: We don't look at the color of the shapes when checking for congruence.
### Another Example
Are these pentagons congruent?
Let's check.
1️⃣ Are they the same shape?
They both have 5 sides, so yes.
2️⃣ Are they the same size?
Probably, but we're not sure.
The purple shape doesn't look exactly like the green one.
What if we rotate the green shape?
Do their sizes look the same now?
Yes, they have the same size and shape.
This means that these shapes are congruent. ✅
Tip: Congruent shapes match even if you flip, slide, or turn them.
Excellent job learning about congruent shapes.
Now, complete the practice to master recognizing congruent shapes on your own.
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AP Calculus: 10-Step Guide to Curve Sketching
What can calculus tell us about curve sketching? It turns out, quite a lot! In this article, you’ll see a list of the 10 key characteristics that describe a graph. While you may not be tested on your artistic ability to sketch a curve on the AP Calculus exams, you will be expected to determine these specific features of graphs.
Guide to Curve Sketching
The ten steps of curve sketching each require a specific tool. But some of the steps are closely related. In the list below, you’ll see some steps grouped if they are based on similar methods.
Algebra and pre-calculus
1. Domain and Range
2. y-Intercept
3. x-Intercept(s)
4. Symmetry
5. Limits
6. Vertical Asymptote(s)
7. Horizontal and/or Oblique Asymptote(s)
8. First Derivative
9. Increase/Decrease
10. Relative Extrema
11. Second Derivative
12. Concavity
13. Inflection Points
Some books outline these steps differently, sometimes combining items together. So it’s not uncommon to see “The Eight Steps for Curve Sketching,” etc.
Let’s briefly review what each term means. More details can be found at AP Calculus Exam Review: Analysis of Graphs, for example.
Step 1. Determine the Domain and Range
The domain of a function f(x) is the set of all input values (x-values) for the function.
The range of a function f(x) is the set of all output values (y-values) for the function.
Methods for finding the domain and range vary from problem to problem. Here is a good review.
Step 2. Find the y-Intercept
The y-intercept of a function f(x) is the point where the graph crosses the y-axis.
This is easy to find. Simply plug in 0. The y-intercept is: (0, f(0)).
Step 3. Find the x-Intercept(s)
An x-intercept of a function f(x) is any point where the graph crosses the x-axis.
To find the x-intercepts, solve f(x) = 0.
Step 4. Look for Symmetry
A graph can display various kinds of symmetry. Three main symmetries are especially important: even, odd, and periodic symmetry.
• Even symmetry. A function is even if its graph is symmetric by reflection over the y-axis.
• Odd symmetry. A function is odd if its graph is symmetric by 180 degree rotation around the origin.
• Periodicity. A function is periodic if an only if its values repeat regularly. That is, if there is a value p > 0 such that f(x + p) = f(x) for all x in its domain.
The algebraic test for even/odd is to plug in (-x) into the function.
• If f(-x) = f(x), then f is even.
• If f(-x) = –f(x), then f is odd.
On the AP Calculus exams, periodicity occurs only in trigonometric functions.
Step 5. Find any Vertical Asymptote(s)
A vertical asymptote for a function is a vertical line x = k showing where the function becomes unbounded.
For details, check out How do you find the Vertical Asymptotes of a Function?.
Step 6. Find Horizontal and/or Oblique Asymptote(s)
A horizontal asymptote for a function is a horizontal line that the graph of the function approaches as x approaches ∞ or -∞.
An oblique asymptote for a function is a slanted line that the function approaches as x approaches ∞ or -∞.
Both horizontal and oblique asymptotes measure the end behavior of a function. For details, see How do you find the Horizontal Asymptotes of a Function? and How do you find the Oblique Asymptotes of a Function?.
Step 7. Determine the Intervals of Increase and Decrease
A function is increasing on an interval if the graph rises as you trace it from left to right.
A function is decreasing on an interval if the graph falls as you trace it from left to right.
The first derivative measures increase/decrease in the following way:
• If f '(x) > 0 on an interval, then f is increasing on that interval.
• If f '(x) < 0 on an interval, then f is decreasing on that interval.
Step 8. Locate the Relative Extrema
The term relative extrema refers to both relative minimum and relative maximum points on a graph.
A graph has a relative maximum at x = c if f(c) > f(x) for all x in a small enough neighborhood of c.
A graph has a relative minimum at x = c if f(c) < f(x) for all x in a small enough neighborhood of c.
The relative maxima (plural of maximum) and minima (plural of minimum) are the “peaks and valleys” of the graph. There can be many relative maxima and minima in any given graph.
Relative extrema occur at points where f '(x) = 0 or f '(x) does not exist. Use the First Derivative Test to classify them.
This graph increases, reaching a relative maximum, then decreases into the relative minimum, and finally increases afterwards.
Step 9. Determine the Intervals of Concavity
Concavity is a measure of how curved the graph of the function is at various points. For example, a linear function has zero concavity at all points, because a line simply does not curve.
A graph is concave up on an interval if the tangent line falls below the curve at each point in the interval. In other words, the graph curves “upward,” away from its tangent lines.
A graph is concave down on an interval if the tangent line falls above the curve at each point in the interval. In other words, the graph curves “downward,” away from its tangent lines.
Here’s one way to remember the definitions: “Concave up looks like a cup, and concave down looks like a frown.”
The second derivative measures concavity:
• If f ''(x) > 0 on an interval, then f is concave up on that interval.
• If f ''(x) < 0 on an interval, then f is concave down on that interval.
Step 10. Locate the Inflection Points
Any point at which concavity changes (from up to down or down to up) is called a point of inflection.
Any point where f ''(x) = 0 or f ''(x) does not exist is a possible point of inflection. Look for changes in concavity to determine if these are actual points of inflection.
This graph shows a change in concavity, from concave down to concave up. The inflection point is where the transition occurs.
Final Thoughts
This short article only outlines the steps for accurate curve sketching. Now it’s up to you to familiarize yourself with the various methods and tools that will help you to analyze the graph of any function.
Author
• Shaun earned his Ph. D. in mathematics from The Ohio State University in 2008 (Go Bucks!!). He received his BA in Mathematics with a minor in computer science from Oberlin College in 2002. In addition, Shaun earned a B. Mus. from the Oberlin Conservatory in the same year, with a major in music composition. Shaun still loves music -- almost as much as math! -- and he (thinks he) can play piano, guitar, and bass. Shaun has taught and tutored students in mathematics for about a decade, and hopes his experience can help you to succeed!
By the way, Magoosh can help you study for both the SAT and ACT exams. Click here to learn more!
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Precalculus by Richard Wright
Are you not my student and
has this helped you?
This book is available
to download as an epub.
You are my hiding place; you will protect me from trouble and surround me with songs of deliverance. Psalms 32:7 NIV
# 1-06 Graphs of Parent Functions
Summary: In this section, you will:
• Identify the graphs of parent functions.
• Graph piecewise functions.
SDA NAD Content Standards (2018): PC.4.1
Imagine trying to fix a car with only one wrench. You might have problems because the nuts and bolts are different sizes. To do your best work on a car you need a lot of different tools in your toolbox. In a similar way, to do precalculus and higher mathematics, you need a toolbox full of functions. These are called tool-kit, or parent, functions.
## Parent Functions
Constant function f(x) = c
• Domain is all real numbers.
• Range is the set {c} that contains this single element.
• Neither increasing or decreasing.
• Symmetric over the y-axis.
Linear function f(x) = x
• Domain is all real numbers.
• Range is all real numbers.
• Increases from (−∞, ∞).
• Symmetric about the origin.
Absolute value function f(x) = |x|
• Domain is all real numbers.
• Range is [0, ∞).
• Decreasing on (−∞, 0) and increasing on (0, ∞).
• Symmetric over the y-axis
Quadratic function f(x) = x2
• Domain is all real numbers.
• Range is only nonnegative real numbers, [0, ∞).
• Decreasing over (−∞, 0) and increasing on (0, ∞).
• Symmetric over the y-axis.
Cubic function f(x) = x3
• Domain is all real numbers.
• Range is all real numbers.
• Increasing on (−∞, ∞).
• Symmetric about the origin.
Reciprocal function $$f(x) = \frac{1}{x}$$
• Domain is all real numbers except 0, {x|x ≠ 0}.
• Range is all real numbers except 0, {y|y ≠ 0}.
• Decreasing on (−∞, 0) and (0, ∞).
• Symmetric about the origin and over the lines y = x and y = −x.
Reciprocal squared function $$f(x) = \frac{1}{x^2}$$
• Domain is all real numbers except 0, {x|x ≠ 0}.
• Range is only positive real numbers, (0, ∞).
• Increasing on (−∞, 0) and decreasing on (0, ∞).
• Symmetric over the y-axis.
Square root function $$f(x) = \sqrt{x}$$
• Domain is 0 or greater, [0, ∞).
• Range is 0 or greater, [0, ∞).
• Increasing on (0, ∞).
• No symmetry.
Cube root function $$f(x) = \sqrt[3]{x}$$
• Domain is all real numbers.
• Range is all real numbers.
• Increasing over (−∞, ∞).
• Symmetric about the origin.
###### Graph on a Graphing Calculator
All graphing calculators are different, but have similar commands. The following instructions are for the TI-84.
TI-83/84
1. Press the Y= button
2. Enter the equation
3. Press the GRAPH button
1. If the axis are not centered, press ZOOM, then choose zStandard.
2. If the graph is not visible, press ZOOM, then choose ZoomFit
3. The window range can also be set by pressing WINDOW
4. To copy the graph to your paper, press 2ND TABLE and plot the points on your paper
NumWorks
1. Press the home button and select Grapher
2. Select the Expressions tab at the top
3. Add or edit the function
4. Select the Graph tab at the top
5. The zoom options are at the top
1. Auto: should show most of the graph
2. Axes: lets you enter the values to set the visible window
3. Navigate: lets you use the arrows to move the graph around
4. Zoom with the + and keys.
6. To copy the graph to your paper, select the Table tab at the top and plot the points on your paper
#### Example 1: Identify the Parent Function
Identify the parent function of f(x) = 3x3 + 1. Then graph it on a graphing calculator.
###### Solution
Because the function f(x) = 3x3 + 1 has x3, its parent function is cubic.
#### Example 2: Identify the Parent Function
Identify the parent function of $$f(x) = -\sqrt{x - 4}$$. Then graph it on a graphing calculator.
###### Solution
Because the function $$f(x) = -\sqrt{x - 4}$$ has $$\sqrt{x}$$, its parent function is square root.
##### Try It 1
Identify the parent function of $$f(x) = -\frac{4}{x^2}$$. Then graph it on a graphing calculator.
###### Answer
Reciprocal squared;
## Graph Piecewise Functions
Piecewise functions were discussed and evaluated in lesson 01-04. Remember that they are made up of several different equations each with its own domain interval. They were evaluated by first deciding which domain the value of x was in and then evaluating that equation. Graphing piecewise functions is similar. Start by marking off each section of the domain on the x-axis. Then graph each equation only in its domain interval.
###### Graph a Piecewise Function
1. Mark the boundaries on the x-axis of the intervals for each piece of the domain.
2. For each piece of the domain, graph on the corresponding equation. Do not graph two functions over one interval because it would violate the criteria of a function.
#### Example 3: Graph a Piecewise Function
Sketch a graph of the function.
f(x) = \left\{\begin{align} x^2 &, \text{ if } x≤1 \\ 3 &, \text{ if } 1 < x ≤ 2 \\ x &, \text{ if } x > 2 \end{align}\right.
###### Solution
Before graphing the equations, start by marking the domains on the x-axis. Do something like draw light vertical dotted lines at x = 1 and x = 2.
Each of the component functions is from the library of parent functions, so their shapes are known. However, only draw the portion of the graph in each domain. At the edges of the domain, draw a filled dot when the endpoint is included because of a less-than-or-equal-to or greater-than-or-equal-to sign; draw a open dot when the point is not included due to a less-than or greater-than sign. Figure 16 shows the individual pieces of the graph, then figure 17 shows the complete graph.
Analysis
Note that the graph does pass the vertical line test even at x = 1 and x = 2 because the points (1, 3) and (2, 2) are not part of the graph of the function, though (1, 1) and (2, 3) are.
##### Try It 2
Sketch a graph of f(x) = \left\{\begin{align} x + 2 &, \text{ if } x < -1 \\ |x| &, \text{ if } x ≥ -1 \end{align}\right.
##### Lesson Summary
###### Graph on a Graphing Calculator
All graphing calculators are different, but have similar commands. The following instructions are for the TI-84.
TI-83/84
1. Press the Y= button
2. Enter the equation
3. Press the GRAPH button
1. If the axis are not centered, press ZOOM, then choose zStandard.
2. If the graph is not visible, press ZOOM, then choose ZoomFit
3. The window range can also be set by pressing WINDOW
4. To copy the graph to your paper, press 2ND TABLE and plot the points on your paper
NumWorks
1. Press the home button and select Grapher
2. Select the Expressions tab at the top
3. Add or edit the function
4. Select the Graph tab at the top
5. The zoom options are at the top
1. Auto: should show most of the graph
2. Axes: lets you enter the values to set the visible window
3. Navigate: lets you use the arrows to move the graph around
4. Zoom with the + and keys.
6. To copy the graph to your paper, select the Table tab at the top and plot the points on your paper
###### Graph a Piecewise Function
1. Mark the boundaries on the x-axis of the intervals for each piece of the domain.
2. For each piece of the domain, graph on the corresponding equation. Do not graph two functions over one interval because it would violate the criteria of a function.
Helpful videos about this lesson.
## Practice Exercises
Identify the parent function and then use a graphing utility to graph the function. Be sure to choose an appropriate viewing window.
1. $$f(x) = \frac{2}{3}x - \frac{1}{3}$$
2. g(x) = −x2 − 4
3. $$h(x) = 2\sqrt{x}$$
4. $$j(x) = \frac{1}{x+1}$$
5. k(x) = −|x| + 4
6. Sketch a graph of the piecewise function.
7. f(x) = \left\{\begin{align} -3x - 2 &, \text{ if } x < 1 \\ \frac{1}{2}x - \frac{3}{2} &, \text{ if } x ≥ 1 \end{align}\right.
8. f(x) = \left\{\begin{align} \frac{1}{x} &, \text{ if } x < -1 \\ \sqrt{x+1}-1 &, \text{ if } x ≥ -1 \end{align}\right.
9. f(x) = \left\{\begin{align} -x^3 &, \text{ if } x < 0 \\ x^2 &, \text{ if } x ≥ 0 \end{align}\right.
10. f(x) = \left\{\begin{align} |x + 4| + 1 &, \text{ if } x ≤ 0 \\ x &, \text{ if } x > 0 \end{align}\right.
11. Identify the parent function.
12. Problem Solving
13. A secretary is paid \$14 per hour for regular time and time-and-a-half for overtime. The weekly wage function is
W(h) = \left\{\begin{align} 14h &, \text{ if } 0 < h ≤ 40 \\ 21(h - 40) + 560 &, \text { if } h > 40\end{align}\right.
where h is the number of hours worked in a week.
1. Find W(30), W(40), W(50), W(60).
2. The company decreased the regular work week to 35 hours. What is the new weekly wage function?
14. Mixed Review
15. (1-05) Use the graph of the function to estimate the intervals on which the function is increasing or decreasing.
16. (1-05) Find zeros of f(x) = x2 − 4.
17. (1-05) Find the average rate of change from [x, x + h] for f(x) = 2x2.
18. (1-04) Evaluate the function g(x) = 2x + 3 at the indicated values g(−1), g(2), g(a), g(a + h)
19. (1-04) Find the domain of the function using interval notation: $$h(x) = 3\sqrt{x - 2}$$
20. (1-02) Find the (a) radius and (b) equation of the circle with center (2, 3) and point on the circle (4, 5). Then (c) graph the circle.
### Answers
1. Linear;
2. Quadratic;
3. Square root;
4. Reciprocal;
5. Absolute value;
6. Cubic
7. Reciprocal squared
8. Absolute Value
9. Square Root
10. 420, 560, 770, 980; W(h) = \left\{\begin{align} 14h &, \text{ if } 0 < h ≤ 35 \\ 21(h - 35) + 490 &, \text { if } h > 35\end{align}\right.
11. Increasing: [1, ∞), never decreases
12. −2, 2
13. 4x + 2h
14. 1, 7, 2a + 3, 2a + 2h + 3
15. [2, ∞)
16. $$2\sqrt{2}$$; (x − 2)2 + (y − 3)2 = 8; |
# 3.2 Domain and range (Page 2/11)
Page 2 / 11
Before we begin, let us review the conventions of interval notation:
• The smallest number from the interval is written first.
• The largest number in the interval is written second, following a comma.
• Parentheses, ( or ), are used to signify that an endpoint value is not included, called exclusive.
• Brackets, [ or ], are used to indicate that an endpoint value is included, called inclusive.
See [link] for a summary of interval notation.
## Finding the domain of a function as a set of ordered pairs
Find the domain of the following function: .
First identify the input values. The input value is the first coordinate in an ordered pair . There are no restrictions, as the ordered pairs are simply listed. The domain is the set of the first coordinates of the ordered pairs.
$\left\{2,3,4,5,6\right\}$
Find the domain of the function:
$\left\{\left(-5,4\right),\left(0,0\right),\left(5,-4\right),\left(10,-8\right),\left(15,-12\right)\right\}$
$\left\{-5,\text{\hspace{0.17em}}0,\text{\hspace{0.17em}}5,\text{\hspace{0.17em}}10,\text{\hspace{0.17em}}15\right\}$
Given a function written in equation form, find the domain.
1. Identify the input values.
2. Identify any restrictions on the input and exclude those values from the domain.
3. Write the domain in interval form, if possible.
## Finding the domain of a function
Find the domain of the function $\text{\hspace{0.17em}}f\left(x\right)={x}^{2}-1.$
The input value, shown by the variable $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ in the equation, is squared and then the result is lowered by one. Any real number may be squared and then be lowered by one, so there are no restrictions on the domain of this function. The domain is the set of real numbers.
In interval form, the domain of $\text{\hspace{0.17em}}f\text{\hspace{0.17em}}$ is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right).$
Find the domain of the function: $\text{\hspace{0.17em}}f\left(x\right)=5-x+{x}^{3}.$
$\left(-\infty ,\infty \right)$
Given a function written in an equation form that includes a fraction, find the domain.
1. Identify the input values.
2. Identify any restrictions on the input. If there is a denominator in the function’s formula, set the denominator equal to zero and solve for $\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$ . If the function’s formula contains an even root, set the radicand greater than or equal to 0, and then solve.
3. Write the domain in interval form, making sure to exclude any restricted values from the domain.
## Finding the domain of a function involving a denominator
Find the domain of the function $\text{\hspace{0.17em}}f\left(x\right)=\frac{x+1}{2-x}.$
When there is a denominator, we want to include only values of the input that do not force the denominator to be zero. So, we will set the denominator equal to 0 and solve for $\text{\hspace{0.17em}}x.$
$\begin{array}{ccc}\hfill 2-x& =& 0\hfill \\ \hfill -x& =& -2\hfill \\ \hfill x& =& 2\hfill \end{array}$
Now, we will exclude 2 from the domain. The answers are all real numbers where $\text{\hspace{0.17em}}x<2\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}x>2\text{\hspace{0.17em}}$ as shown in [link] . We can use a symbol known as the union, $\text{\hspace{0.17em}}\cup ,$ to combine the two sets. In interval notation, we write the solution: $\left(\mathrm{-\infty },2\right)\cup \left(2,\infty \right).$
Find the domain of the function: $\text{\hspace{0.17em}}f\left(x\right)=\frac{1+4x}{2x-1}.$
$\left(-\infty ,\frac{1}{2}\right)\cup \left(\frac{1}{2},\infty \right)$
Given a function written in equation form including an even root, find the domain.
1. Identify the input values.
2. Since there is an even root, exclude any real numbers that result in a negative number in the radicand. Set the radicand greater than or equal to zero and solve for $\text{\hspace{0.17em}}x.$
3. The solution(s) are the domain of the function. If possible, write the answer in interval form.
## Finding the domain of a function with an even root
Find the domain of the function $\text{\hspace{0.17em}}f\left(x\right)=\sqrt{7-x}.$
When there is an even root in the formula, we exclude any real numbers that result in a negative number in the radicand.
Set the radicand greater than or equal to zero and solve for $\text{\hspace{0.17em}}x.$
$\begin{array}{ccc}\hfill 7-x& \ge & 0\hfill \\ \hfill -x& \ge & -7\hfill \\ \hfill x& \le & 7\hfill \end{array}$
Now, we will exclude any number greater than 7 from the domain. The answers are all real numbers less than or equal to $\text{\hspace{0.17em}}7,\text{\hspace{0.17em}}$ or $\text{\hspace{0.17em}}\left(-\infty ,7\right].$
find to nearest one decimal place of centimeter the length of an arc of circle of radius length 12.5cm and subtending of centeral angle 1.6rad
factoring polynomial
find general solution of the Tanx=-1/root3,secx=2/root3
find general solution of the following equation
Nani
the value of 2 sin square 60 Cos 60
0.75
Lynne
0.75
Inkoom
when can I use sin, cos tan in a giving question
depending on the question
Nicholas
I am a carpenter and I have to cut and assemble a conventional roof line for a new home. The dimensions are: width 30'6" length 40'6". I want a 6 and 12 pitch. The roof is a full hip construction. Give me the L,W and height of rafters for the hip, hip jacks also the length of common jacks.
John
I want to learn the calculations
where can I get indices
I need matrices
Nasasira
hi
Raihany
Hi
Solomon
need help
Raihany
maybe provide us videos
Nasasira
Raihany
Hello
Cromwell
a
Amie
What do you mean by a
Cromwell
nothing. I accidentally press it
Amie
you guys know any app with matrices?
Khay
Ok
Cromwell
Solve the x? x=18+(24-3)=72
x-39=72 x=111
Suraj
Solve the formula for the indicated variable P=b+4a+2c, for b
Need help with this question please
b=-4ac-2c+P
Denisse
b=p-4a-2c
Suddhen
b= p - 4a - 2c
Snr
p=2(2a+C)+b
Suraj
b=p-2(2a+c)
Tapiwa
P=4a+b+2C
COLEMAN
b=P-4a-2c
COLEMAN
like Deadra, show me the step by step order of operation to alive for b
John
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
yah
immy |
# Problem Set #11 solutions
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1 Find three nonisomorphic graphs with the same degree sequence (1, 1, 1,,, 3). There are several such graphs: three are shown below Show that two projections of the Petersen graph are isomorphic. The isomorphism of these two different presentations can be seen fairly easily: pick any vertex from the first graph to associate with the center vertex of the second graph. Then, if we associate its neighbors in any order with the vertices at 1 o clock, 4 o clock, and 8 o clock on the second representation, the association of the remaning 6 vertices should be reasonably straightforward. Note: an explicit association is necessary. The fact that both graphs are conencted with every vertex of degree 3 is not sufficient there are in fact 19 different connected 10-vertex graphs in which every vertex has degree 3. These graphs can all be seen at html Suppose G is a graph with p vertices and q edges whose smallest degree is δ and whose largest degree is δ +. Show that δ q p δ+. For any vertex v in the graph, we know that δ deg v δ +. If we sum this inequality over all vertices v, we get δ deg v Since the argument of the sums in the lower and upper bounds are constant, these summations simply serve to multiply their arguments by V (G) = G = p; likewise, the middle term in the enequality we know to be G = q. The above inequality can thus be simplified to pδ q pδ + And thus δ q p δ At the beginning of a math department meeting, there are 10 persons and a total of 6 handshakes. We shall represent this situation by way of a graph in which people are represented by vertices and handshakes between people are represented by edges between those vertices. Thus, this situation is described by a graph G with 10 vertices and 6 edges. δ + Page 1 of 5 April 10, 008
2 (a) Explain why there must be a person who shakes hands at least six times. In the graph being used to represent this problem, there are 10 vertices and 6 edges. By the result determined in exercise 5.1.1, we know that in G, δ + G = 6 = 5.. Since G 10 δ+ must be an integer, we thus know that δ + is at least 6, so there is some vertex in G with degree 6. Translating the edges incident to a vertex into handshakes in which the associated person has taken part, we see that the person associated with this vertex has shaken at least 6 hands. (b) Explain why there are not exactly three people who shake an odd number of hands. We know that deg v = G = 5. Let us partition V (G) into two sets: put vertices of odd degree into A, and even degree into B. Then 5 = deg v deg v = v A deg v + v B Suppose there are exactly three people who shake an odd number of hands, so A = 3. Then v A deg v is a sum of three odd nubmers, so v A deg v is odd. In order for the sum of v A deg v and v B deg v to be 5 (an even number), v B deg v must also be odd. However, by the definition of B, v B deg v is a sum of even numbers, which must be even, leading to a contradtction During a summer vacation, nine students promise to keep in touch by writing letters, so each student promises to write letters to three of the others. (a) Is it possible to arrange for each student to write three letters and also to receive three letters? There are several ways to do this; we can represent this as a graph with students represented by vertices and letter-writing represented as edges (actually, since correspondence is one-way, a better tool for this purpose is a directed graph, in which edges have orientation, but we can make do with undirected edges). We want a graph in which each vertex has degree 6 (to represent sending 3 letters and receiving 3). The easiest way to do this is by numbering our vertices v 1,...,v 9 and connecting each v i to v i+1, v i+, and v i+3, wrapping around on the addition so that v 8, for instance, is conencted to v 9, v 1, and v. Then, we have outbound connections from each vertex to its three successors, and inbound from its three predecessors in our arbitrary ordering. (b) Is it possible to arrange for each student to write letters to three other students and also to receive letters from the same three students? Let us represent this as a graph G in which the vertices are students and edges represent two-way communication; that is, writing to another and being written to by them. In such a scheme, we want each vertex to have degree 3, since each student is in communication with 3 other students. Thus, G = v G deg v = 9 3 = 7. We thus have the surprising and contrary result that G must have 7 edges. Since no graph has a nonintegral number of edges, no such graph G can possibly be constructed and the situation described is impossible. Page of 5 April 10, 008
3 Show that in any group of 10 people there are either four mutual friends or three mutual strangers. Representing people as vertices of a graph, and acquaintance of two people as an edge between their associated nodes, this question is equivalent to the claim that for any 10-vertex graph G, there are either 4 mutually adjacent points (a K 4 subgraph of G), or 3 mutually non-adjacent points (a K 3 subgraph of G c ). We will start with a simple (and useful!) lemma: Lemma 1. If G 6, then either K 3 is a subgraph of G or K 3 is a subgraph of G c. Proof. Pick an arbitrary vertex v in G. There are two cases to be dealt with: Case I: deg v 3. Thus, v has at least three neighbors, which we shall call u 1, u, and u 3. Since v u 1 and v u, a K 3 would be formed among v, u 1, and u if u 1 u. Similar arguments will guarantee the existence of a K 3 if u u 3 or u 1 u 3. Thus, to prevent the appearance of a K 3 in v, we would require that u 1 u, u u 3, and u 1 u 3. But then, these three vertices will be mutually non-adjacent, and thus form a K 3 in G c. Case II: deg v. Then, deg v (5 ) = 3 in G c, so the argument presented in Case I, applied to vertex v in G c instead of in G, proves that there is a K 3 in either G c or (G c ) c, which is just G. Now we can prove our main result: Theorem 1. If G 10, then either K 4 is a subgraph of G or K 3 is a subgraph of G c. Proof. Pick an arbitrary vertex v in G. There are two cases to be dealt with: Case I: deg v 6. Thus, v has at least six neighbors. By the lemma above, among these six vertices there must be either three mutually non-adjacent vertices or three mutually adjacent vertices. If there are three mutually non-adjacent vertices, one of the structures we sought is present (a K 3 in G c ); if, on the other hand, there are three mutually adjacent vertices among these six vertices, let us call them u 1, u, and u 3. By mutual adjacency, u 1 u, u 1 u 3, and u u 3. Since all three are neighbors of v, it is also the case that v u 1, v u, and v u 3. Thus, since all 6 possible edges among v, u 1, u, and u 3 exist, these vertices form a K 4 subgraph of G. Case II: deg v 5. Then there are at least 9 5 = 4 vertices to which v is not adjacent: call them u 1, u, u 3, and u 4. If any u i u j, then v, u i, and u j would be mutually nonadjacent (forming a K 3 subgraph of G c ), if, on the other hand every u i u j, then u 1, u, u 3, and u 4 would be mutually adjacent, forming a K 4 subgraph of G. This result can be slightly improved to hold whenever G 9 by combining the above argument with certain impossibility results on the parity of vertex degrees (see (b), 5.1.0(b)). It is not true when G = 8, and specific 8-vertex graphs can be produced containing neither a K 4 nor three mutually non-adjacent vertices. Page 3 of 5 April 10, 008
4 5... For any vertex v of a simple graph G with δ k, show that G has a path of length k with initial vertex v. Let us give v the alternative name u 0, and let us choose a neighbor of u 0 arbitrarily and call it u 1. Then, since deg u 1 k, u 1 is incident to at least k edges, at least k 1 of which do not go to u 0. Pick an arbitrary such edge, and label its other endpoint u, so that we now have a chain of distinct adjacent vertices u 0, u 1, u. Since deg u k, u is incident to at least k edges which do not go to u 0 or u 1 ; we choose one and label the vertex at the other end u 3. We continue in like manner: for each vertex u i, we know it is incident on k edges, and there are i endpoints we wish to avoid (u 0, u 1, u,..., u i 1 ). Thus, of these k edges, k i of them are acceptable, so we pick one of those arbitrarily and label its endpoint u i+1. This process will yield a sequence of adjacent distinct vertices u 0,u 1,...,u k ; it cannot be extended any further by these means since u k s k neighbors might very well be exactly those vertices already in the path. However, we have successfully produced a length-k path by these means. This argument may also be phrased by induction on k: the inductive hypothesis guarantees the existence of u 0,u 1,...,u k 1, and all that remains is to show that a single-node extension to some point u k is possible If a graph G has exactly two components that are both complete graphs with k and p k vertices respectively (1 k p 1) then: A complete graph on k nodes has ( ) k = k k edges; a complete graph on p k nodes has ( ) p k = k pk+p p+k edges. Together the two components thus have ( ) ( k + p k ) = k pk+p p edges. (a) For which values of k is the number of edges a minimum? We can treat this as a good old calculus or algebra minimization problem. We want to minimize k pk+p p where p is a constant and k a variable in [1,p 1]. Quick inspection of the quadratic rveals that it s a concave-upwards parabola with axis k = 1p. Thus, the maximizing value of this continuous function is p ; since as the points closest to k is actually an integer, we have k = p and k = p minimal as possible. The graph so produced has ( ) ( p + p ) edges. (b) For which values of k is the number of edges a maximum? As above, this is the maximum vsalue on an interval for an upwards-concave parabola. Since such a function s values increase with distance from the vertex of the parabola, one of the endpoints of our domain will maximize the function k pk+p p. In fact, both of them do: both k = 1 and k = p 1 yield a graph with ( ) p 1 edges Show that any two longest paths in a connected graph have at least one vertex in common. Suppose we have two vertex-disjoint paths of length n. We shall show that a longer path can be constructed, which is equivalent to stating that two longest paths cannot be vertex-disjoint. Page 4 of 5 April 10, 008
5 Let our two paths of length n be denoted u 0 u 1 u u n and v 0 v 1 v v n, and we operate under the assumption that all of these vertices are distinct. Now, let us consider a path from u 0 to v 0 given by u 0 = w 0 w 1 w w k = v 0. The vertices in this path are obviously not distinct from the u i and v i ; in particular, we know at least one u i occurs and at least one v i occurs. Thus, considering the subsequence of the path {w i } consisting only of vertices from the two other paths, we have a sequence consisting of at least one u i and at least one v i. Thus, somewhere in this sequence is a pair of consecutive terms u i, v j. Extrapolating from this to the path of which it is a subsequence, we know that there is some subpath u i = w t w t+1 w t+l = v j in which none of the vertices except the first and last are in either of the paths {u i } and {v j }. Now we have two possibilities: Case I: i j. We may construct the following path by splicing together disjoint pieces of paths we already know about: v 0 v 1 v v j = w t+l w t+l 1 w t = u i u i+1 u n This path is built up of fragments of length j, l and n i in order, so the path has length j + l + (n i) = n + l + (j i) > n. Case II: i > j. We may construct the following path by splicing together disjoint pieces of paths we already know about: u 0 u 1 u u i = w t w t+1 w t+l = v j v j+1 v n This path is built up of fragments of length i, l and n j in order, so the path has length i + l + (n j) = n + l + (i j) > n. Thus, in either case we can construct a path longer than n, so disjoint paths {u k } and {v k } cannot be the longest paths in the graph. Thus, if we have two longest paths in the graph, they are not vertex-disjoint Show that if G is disconnected, G c is connected. Since G is disconnected, we can find vertices u and v in G such that no path exists between u and v. In particular, u and v are non-adjacent. Let us consider an arbitrary third vertex w in G. If w were adjacent to both u and v, u and v would have a path between them. Thus, w is non-adjacent to at least one of u and v in G. Since non-adjacency becomes adjacency in the complement, we know that in G c, u is adjacent to v and w is adjacent to at least one of u or v. Thus, w is connected to both u and v in G c. Since the above argument depended on no special properties of w, it is true for every vertex of G distinct from u and v. Thus, every vertex in G c is connected to u and v; by transitivity of connectivity, this means every vertex in G c is connected to every other vertex, so G c is connected. Page 5 of 5 April 10, 008
### A Sublinear Bipartiteness Tester for Bounded Degree Graphs
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# Absolute value inequality - Examples, Exercises and Solutions
## Inequality
When you come across signs like $<$ or ,$>$ you will know it is an inequality.
The result of the inequality will be a certain range of values that you will have to find.
An important rule to keep in mind: when you double or divide both sides of the operation, the sign of the inequality is reversed!
## Absolute Value Inequality
We can solve absolute value inequalities in $2$ ways:
In the geometric method and in the algebraic method.
## Practice Absolute value inequality
### Exercise #1
Solve the following inequality:
5x+8<9
### Step-by-Step Solution
This is an inequality problem. The inequality is actually an exercise we solve in a completely normal way, except in the case that we multiply or divide by negative.
Let's start by moving the sections:
5X+8<9
5X<9-8
5X<1
We divide by 5:
X<1/5
And this is the solution!
x<\frac{1}{5}
### Exercise #2
Solve the inequality:
5-3x>-10
### Step-by-Step Solution
Inequality equations will be solved like a regular equation, except for one rule:
If we multiply the entire equation by a negative, we will reverse the inequality sign.
We start by moving the sections, so that one side has the variables and the other does not:
-3x>-10-5
-3x>-15
Divide by 3
-x>-5
Divide by negative 1 (to get rid of the negative) and remember to reverse the sign of the equation.
x<5
5 > x
### Exercise #3
Which diagram represents the solution to the inequality below? 5-8x<7x+3
### Step-by-Step Solution
First, we will move the elements:
5-8x>7x+3
5-3>7x+8x
2>13x
We divide the answer by 13, and we get:
x > \frac{2}{13}
### Exercise #4
What is the solution to the following inequality?
$10x-4≤-3x-8$
### Step-by-Step Solution
In the exercise, we have an inequality equation.
We treat the inequality as an equation with the sign -=,
And we only refer to it if we need to multiply or divide by 0.
$10x-4 ≤ -3x-8$
We start by organizing the sections:
$10x+3x-4 ≤ -8$
$13x-4 ≤ -8$
$13x ≤ -4$
Divide by 13 to isolate the X
$x≤-\frac{4}{13}$
Answer A is with different data and therefore was rejected.
Answer C shows a case where X is greater than$-\frac{4}{13}$, although we know it is small, so it is rejected.
Answer D shows a case (according to the white circle) where X is not equal to$-\frac{4}{13}$, and only smaller than it. We know it must be large and equal, so this answer is rejected.
Therefore, answer B is the correct one!
### Exercise #5
Solve the inequality:
8x+a < 3x-4
### Step-by-Step Solution
Solving an inequality equation is just like a normal equation. We start by trying to isolate the variable (X).
It is important to note that in this equation there are two variables (X and a), so we may not reach a final result.
8x+a<3x-4
We move the sections
8x-3x<-4-a
We reduce the terms
5x<-4-a
We divide by 5
x< -a/5 -4/5
And this is the solution!
x < -\frac{1}{5}a-\frac{4}{5}
### Exercise #1
Given:
\left|x+2\right|<3
Which of the following statements is necessarily true?
-5 < x < 1
### Exercise #2
Given:
\left|x-4\right|<8
Which of the following statements is necessarily true?
-4 < x < 12
### Exercise #3
Given:
\left|x+4\right|>13
Which of the following statements is necessarily true?
x>9 o x<-17
### Exercise #4
What is the solution to the inequality shown in the diagram?
### Video Solution
$3 ≤ x$
### Exercise #5
Which inequality is represented by the numerical axis below?
-7 < x ≤ 2
### Exercise #1
Given:
\left|x-5\right|>11
Which of the following statements is necessarily true?
x>16 o x<-6
### Exercise #2
Given:
\left|x-5\right|>-11
Which of the following statements is necessarily true?
No solution
### Exercise #3
When are the following inequalities satisfied?
3x+4<9
3 < x+5
### Video Solution
-2 < x < 1\frac{2}{3}
### Exercise #4
Find when the inequality is satisfied:
-3x+15<3x<4x+8
2.5 < x
### Exercise #5
Which diagram corresponds to the inequality below?
$40x+57≤5x-13≤25x+7$
What is its solution? |
19
Q:
# 14 persons are seated around a circular table. Find the probability that 3 particular persons always seated together.
A) 11/379 B) 21/628 C) 24/625 D) 26/247
Explanation:
Total no of ways = (14 – 1)! = 13!
Number of favorable ways = (12 – 1)! = 11!
So, required probability = $11!×3!13!$ = $39916800×66227020800$ = $24625$
Q:
When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less than 12?
A) 35/36 B) 17/36 C) 15/36 D) 1/36
Explanation:
When two dice are thrown simultaneously, the probability is n(S) = 6x6 = 36
Required, the sum of the two numbers that turn up is less than 12
That can be done as n(E)
= { (1,1), (1,2), (1,3), (1,4), (1,5), (1,6)
(2,1), (2,2), (2,3), (2,4), (2,5), (2,6)
(3,1), (3,2), (3,3), (3,4), (3,5), (3,6)
(4,1), (4,2), (4,3), (4,4), (4,5), (4,6)
(5,1), (5,2), (5,3), (5,4), (5,5), (5,6)
(6,1), (6,2), (6,3), (6,4), (6,5) }
= 35
Hence, required probability = n(E)/n(S) = 35/36.
3 632
Q:
In a purse there are 30 coins, twenty one-rupee and remaining 50-paise coins. Eleven coins are picked simultaneously at random and are placed in a box. If a coin is now picked from the box, find the probability of it being a rupee coin?
A) 4/7 B) 2/3 C) 1/2 D) 5/6
Explanation:
Total coins 30
In that,
1 rupee coins 20
50 paise coins 10
Probability of total 1 rupee coins = 20C11
Probability that 11 coins are picked = 30C11
Required probability of a coin now picked from the box is 1 rupee = 20C11/30C11 = 2/3.
7 1035
Q:
In a box, there are 9 blue, 6 white and some black stones. A stone is randomly selected and the probability that the stone is black is ¼. Find the total number of stones in the box?
A) 15 B) 18 C) 20 D) 24
Explanation:
We know that, Total probability = 1
Given probability of black stones = 1/4
=> Probability of blue and white stones = 1 - 1/4 = 3/4
But, given blue + white stones = 9 + 6 = 15
Hence,
3/4 ----- 15
1 ----- ?
=> 15 x 4/3 = 20.
Hence, total number of stones in the box = 20.
10 1077
Q:
What is the probability of an impossible event?
A) 0 B) -1 C) 0.1 D) 1
Explanation:
The probability of an impossible event is 0.
The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.
The probability of a certain event is 1.
9 1491
Q:
In a box, there are four marbles of white color and five marbles of black color. Two marbles are chosen randomly. What is the probability that both are of the same color?
A) 2/9 B) 5/9 C) 4/9 D) 0
Explanation:
Number of white marbles = 4
Number of Black marbles = 5
Total number of marbles = 9
Number of ways, two marbles picked randomly = 9C2
Now, the required probability of picked marbles are to be of same color = 4C2/9C2 + 5C2/9C2
= 1/6 + 5/18
= 4/9.
9 1820
Q:
A bag contains 3 red balls, 5 yellow balls and 7 pink balls. If one ball is drawn at random from the bag, what is the probability that it is either pink or red?
A) 2/3 B) 1/8 C) 3/8 D) 3/4
Explanation:
Given number of balls = 3 + 5 + 7 = 15
One ball is drawn randomly = 15C1
probability that it is either pink or red =
14 1673
Q:
Two letters are randomly chosen from the word TIME. Find the probability that the letters are T and M?
A) 1/4 B) 1/6 C) 1/8 D) 4
Explanation:
Required probability is given by P(E) =
19 2395
Q:
Two dice are rolled simultaneously. Find the probability of getting the sum of numbers on the on the two faces divisible by 3 or 4?
A) 3/7 B) 7/11 C) 5/9 D) 6/13
Explanation:
Here n(S) = 6 x 6 = 36
E={(1,2),(1,5),(2,1),(2,4),(3,3),(3,6),(4,2),(4,5),(5,1),(5,4),(6,3) ,(6,6),(1,3),(2,2),(2,6),(3,1),(3,5), (4,4),(5,3),(6,2)}
=> n(E)=20
Required Probability n(P) = n(E)/n(S) = 20/36 = 5/9. |
Overview of Fractions
by Ron Kurtus (updated 18 January 2022)
A fraction is a part of the whole. It is a form of number that indicates the quotient or division of two whole numbers, except 0.
The top number of a fraction is called the numerator and the bottom number is the denominator. An improper fraction is one where the numerator is larger than the denominator. A mixed number is a whole number plus a proper fraction. An improper fraction should be reduced to a mixed number.
Questions you may have include:
• What is a description of a fraction?
• What are examples of mixed numbers?
• How do you reduce an improper fraction?
This lesson will answer those questions.
Fraction definition
A fraction indicates a part of the whole or the quotient of two whole numbers, except 0.
Note: If the numerator is 0, the number is not a fraction. For example, 0/5 = 0. The denominator cannot be 0, since dividing by 0 is forbidden.
Fraction is part of the whole
If you would take a pie and divide it into five equal pieces, then one piece would be one-fifth (1/5) of the whole pie. In this way, a fraction indicates it is part of the whole. Two pieces of this pie would be 2/5 of the whole pie.
Since a fraction is part of the whole, a proper positive fraction is always less than 1.
Incomplete division
Typically, you divide a number by a smaller number, such as 6 ÷ 2 = 3. If you try to divide a number by a larger number, the result is called a fraction. It is incomplete division. For example, 2 ÷ 3 = 2/3, which is a fraction.
Note that the the ÷ sign almost looks like a fraction, with the dots representing numbers. Also, the / sign is often used as division instead of the ÷ sign.
Numerator and denominator
The top number in a fraction is called the numerator. The bottom number is called the denominator. I don't know of any easy way to remember that, but since the words are used so much when dealing with fractions, it is worthwhile to learn them.
Can't use 0 in a fraction
If the numerator in a quotient is 0, then the number cannot be a fraction, since 0 divided by any number is 0. For example 0/5 = 0.
On the other hand, you cannot have a denominator equal to 0. You cannot have 3/0, because that is 3 ÷ 0, and you cannot divide by 0. That is a basic rule in arithmetic.
Mixed numbers
A mixed number is a whole number plus a fraction. 5 2/3 is a mixed number. It is called "five and two-thirds" and is the same as the sum of 5 + 2/3.
Other examples of mixed numbers include 1 1/2, 3200 7/10 and 15 23/24.
Reducing an improper fraction
An improper fraction is a fraction where the numerator is larger than the denominator. For example, 7/5 is an improve fraction, since 7 is greater than 5.
An improper fraction can be reduced to a mixed number, consisting of a whole number and a fraction. This is done by dividing numerator by the denominator. The remainder of the division is the numerator of the new fraction.
Consider the improper fraction 37/5. If you divide 37 by 5, you will get an answer of 7 and a remainder of 2. Thus your mixed number is 7 2/5.
Examples
Since most improper fractions do not entail large numbers, you can usually perform the division in your head, as opposed to using long division.
Consider 5/4. 5 divided by 4 equals 1 with a remainder of 1.
5/4 = 1 1/4
32/3 is 32 divided by 3 equals 10 with a remainder of 2.
32/3 = 10 2/3
126 divided by 25 equals 5 with a remainder of 1.
126/25 = 5 1/25
No remainder
If there is no remainder, the improper fraction simply becomes a whole number:
27/9 = 3
64/4 = 16
Summary
A fraction indicates a part of the whole or the quotient of two whole numbers except 0. The top number of a fraction is called the numerator and the bottom number is the denominator. An improper fraction is one where the numerator is larger than the denominator. A mixed number is a whole number plus a proper fraction. An improper fraction should be reduced to a mixed number.
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Find the roots for y = (2x-3)(x+5) (The x values)
Question
Find the roots for y = (2x-3)(x+5) (The x values)
in progress 0
3 years 2021-08-20T00:55:17+00:00 1 Answers 0 views 0
Answers ( )
1. Answer:
x =
Step-by-step explanation:
We know that we have to find the roots of the equation – which is another way of saying the x-intercepts, or the points that are on the x-axis that the graph passes through. All points on the x-axis have a y-value of 0. So, in order to find the roots for this equation, we need to find which values of x make the y equal 0.
The equation has already been factored out. So, we just need to find which values of x for each of the expressions in the parentheses will make the result 0. To do this, set the expressions in the parentheses to 0 and isolate x.
1) First, let’s look at (2x-3). Write the equation 2x-3 = 0. Then, isolate x:
Therefore, is one of the roots.
2) Next, let’s look at (x+5). Write the equation x+5 = 0. Then, isolate x:
Therefore, -5 is a root as well.
The roots of the equation would be and -5. |
# How do you find the first and second derivative of ln(x^2-4)?
Oct 2, 2016
$f ' \left(x\right) = \frac{2 x}{{x}^{2} - 4}$
$f ' ' \left(x\right) = \frac{- 2 \left({x}^{2} + 4\right)}{{x}^{2} - 4} ^ 2$
#### Explanation:
let $f \left(x\right) = \ln \left({x}^{2} - 4\right)$
$\textcolor{\mathmr{and} a n \ge}{\text{Reminder }} \textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\ln x\right) = \frac{1}{x}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
and more generally. $\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\frac{d}{\mathrm{dx}} \left(\ln \left(g \left(x\right)\right)\right) = \frac{g ' \left(x\right)}{g} \left(x\right)} \textcolor{w h i t e}{\frac{a}{a}} |}}}$
$\Rightarrow f ' \left(x\right) = \frac{2 x}{{x}^{2} - 4}$
To find the second derivative, use the $\textcolor{b l u e}{\text{quotient rule}}$
Given $f \left(x\right) = g \frac{x}{h \left(x\right)} \text{ then}$
$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{f ' \left(x\right) = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{a}{a}} |}}} \ldots \ldots \left(A\right)$
here $g \left(x\right) = 2 x \Rightarrow g ' \left(x\right) = 2$
and $h \left(x\right) = {x}^{2} - 4 \Rightarrow h ' \left(x\right) = 2 x$
substitute these values into (A)
$f ' ' \left(x\right) = \frac{\left({x}^{2} - 4\right) .2 - 2 x .2 x}{{x}^{2} - 4} ^ 2$
$= \frac{2 {x}^{2} - 8 - 4 {x}^{2}}{{x}^{2} - 4} ^ 2 = \frac{- 2 {x}^{2} - 8}{{x}^{2} - 4} ^ 2 = \frac{- 2 \left({x}^{2} + 4\right)}{{x}^{2} - 4} ^ 2$ |
# What is a local maximum?
## What is a local maximum?
The local maximum is a point within an interval at which the function has a maximum value. The absolute maxima is also called the global maxima and is the point across the entire domain of the given function, which gives the maximum value of the function.
## What is the difference between local minimum and minimum?
Local and Absolute Extrema An absolute minimum, also called a global minimum, occurs when a point is lower than any other point on the function. A local minimum, also called a relative minimum, occurs when a point is lower than the points surrounding it.
Can a local min be higher than a local Max?
This example shows that a local minimum may have a value that is greater than a local maximum. In the example, the local maximum value of f is 0, while the local minimum value of f is 4. This is perfectly fine.
How do you find the local maximum?
To find the local maximum, we must find where the derivative of the function is equal to 0. Given that the derivative of the function yields using the power rule . We see the derivative is never zero. However, we are given a closed interval, and so we must proceed to check the endpoints.
### What is a local minimum and maximum on a graph?
Looking at a graph, the local maxima and minima are the points where the graph flattens out and changes from increasing to decreasing, or vice versa. When the graph is flat, that means the slope is zero. We can find out when the slope is zero using the derivative.
### What is a local minimum in math?
minimum, in mathematics, point at which the value of a function is less than or equal to the value at any nearby point (local minimum) or at any point (absolute minimum); see extremum.
What is the difference between maximum and local maximum?
Maximum is the greatest element in a set or a range of a function. Global maximum is the greatest value among the overall elements of a set or values of a function. Local maximum is the greatest element in a subset or a given range of a function.
What is difference between absolute maximum and local maximum?
An absolute maximum occurs at the x value where the function is the biggest, while a local maximum occurs at an x value if the function is bigger there than points around it (i.e. an open interval around it).
#### Can local maximum be less than local minimum?
Yes, for instance the secant function x↦1cosx has local minima of value 1 at even multiples of π, and local maxima of value −1 at odd multiples of π.
#### How do you find the local minimum?
To find the local minimum of any graph, you must first take the derivative of the graph equation, set it equal to zero and solve for . To take the derivative of this equation, we must use the power rule, . We also must remember that the derivative of a constant is 0.
How to calculate local maximum and minimum?
it is less than 0, so −3/5 is a local maximum At x = +1/3: y” = 30 (+1/3) + 4 = +14 it is greater than 0, so +1/3 is a local minimum (Now you can look at the graph.) Words A high point is called a maximum (plural maxima ). A low point is called a minimum (plural minima ). The general word for maximum or minimum is extremum (plural extrema ).
How do you find local Min and Max?
– f ( x) = a ( x − h) 2 + k {\\displaystyle f (x)=a (x-h)^ {2}+k} – If your function is already given to you in this form, you just need to recognize the variables a {\\displaystyle a}, h {\\displaystyle h} and k {\\displaystyle k}. – To review how to complete the square, see Complete the Square.
## How to find the local Max and Min?
– ‘center’ — Indicate only the center element of a flat region as the local maximum. – ‘first’ — Indicate only the first element of a flat region as the local maximum. – ‘last’ — Indicate only the last element of a flat region as the local maximum. – ‘all’ — Indicate all the elements of a flat region as the local maxima.
## What is local Max and Min?
The general word for maximum or minimum is extremum (plural extrema ). We say local maximum (or minimum) when there may be higher (or lower) points elsewhere but not nearby. Is it a maximum or minimum?
Begin typing your search term above and press enter to search. Press ESC to cancel. |
## Geometry: Common Core (15th Edition)
The third side would be a leg measuring $84$.
We can find the third side by using the Pythagorean theorem, which states that $a^2 + b^2 = c^2$, where $a$ and $b$ are the legs of the right triangle and $c$ is the hypotenuse. Let's plug in what we know into the Pythagorean theorem: $13^2 + 85^2 = c^2$ Evaluate the exponents: $169 + 7225 = c^2$ Add to simplify: $7394 = c^2$ If we take the square root of this number, the answer would not be rational, so we cannot be looking for the value of $c$. Let's see if $85$ may be the hypotenuse, so we may be looking for one of the legs of the right triangle. Let's set up the equation such that $85$ becomes the hypotenuse: $13^2 + b^2 = 85^2$ Evaluate the exponents: $169 + b^2 = 7225$ Subtract $169$ from each side of the equation to move constants to one side of the equation: $b^2 = 7056$ Take the positive square root to solve for $b$: $b = 84$ The third side would be a leg measuring $84$. |
NSW Syllabuses
# Mathematics K–10 - Stage 1 - Number and Algebra Addition and Subtraction
### Outcomes
#### A student:
• MA1-1WM
describes mathematical situations and methods using everyday and some mathematical language, actions, materials, diagrams and symbols
• MA1-2WM
uses objects, diagrams and technology to explore mathematical problems
• MA1-3WM
supports conclusions by explaining or demonstrating how answers were obtained
• MA1-5NA
uses a range of strategies and informal recording methods for addition and subtraction involving one- and two-digit numbers
### Content
• Students:
• Represent and solve simple addition and subtraction problems using a range of strategies, including counting onpartitioning and rearranging parts (ACMNA015)
• use the terms 'add', 'plus', 'equals', 'is equal to', 'take away', 'minus' and the 'difference between'
• use concrete materials to model addition and subtraction problems involving one- and two-digit numbers
• use concrete materials and a number line to model and determine the difference between two numbers, eg
• recognise and use the symbols for plus (+), minus (–) and equals (=)
• record number sentences in a variety of ways using drawings, words, numerals and mathematical symbols
• recognise, recall and record combinations of two numbers that add to 10
• create, record and recognise combinations of two numbers that add to numbers up to and including 9
• model and record patterns for individual numbers by making all possible whole-number combinations, eg
$$\begin{array}{c}5 + 0 = 5\\4 + 1 = 5\\3 + 2 = 5\\2 + 3 = 5\\1 + 4 = 5\\0 + 5 = 5\end{array}$$ (Communicating, Problem Solving)
• describe combinations for numbers using words such as 'more', 'less' and 'double', eg describe 5 as 'one more than four', 'three combined with two', 'double two and one more' and 'one less than six' (Communicating, Problem Solving)
• create, record and recognise combinations of two numbers that add to numbers from 11 up to and including 20
• use combinations for numbers up to 10 to assist with combinations for numbers beyond 10 (Problem Solving)
• investigate and generalise the effect of adding zero to a number, eg 'Adding zero to a number does not change the number'
• use concrete materials to model the commutative property for addition and apply it to aid the recall of addition facts, eg 4 + 5 = 5 + 4
• relate addition and subtraction facts for numbers to at least 20, eg 5 + 3 = 8, so 8 – 3 = 5 and 8 – 5 = 3
• use and record a range of mental strategies to solve addition and subtraction problems involving one- and two-digit numbers, including:
• counting on from the larger number to find the total of two numbers
• counting back from a number to find the number remaining
• counting on or back to find the difference between two numbers
• using doubles and near doubles, eg 5 + 7: double 5 and add 2
• combining numbers that add to 10, eg 4 + 7 + 8 + 6 + 3: first combine 4 and 6, and 7 and 3, then add 8
• bridging to 10, eg 17 + 5: 17 and 3 is 20, then add 2 more
• using place value to partition numbers, eg 25 + 8: 25 is 20 + 5, so 25 + 8 is 20 + 5 + 8, which is 20 + 13
• choose and apply efficient strategies for addition and subtraction (Problem Solving)
• use the equals sign to record equivalent number sentences involving addition, and to mean 'is the same as', rather than as an indication to perform an operation, eg 5 + 2 = 3 + 4
• check given number sentences to determine if they are true or false and explain why,
eg 'Is 7 + 5 = 8 + 4 true? Why or why not?' (Communicating, Reasoning)
### Language
Students should be able to communicate using the following language: counting on, counting back, combine, plus, add, take away, minus, the difference between, total, more than, less than, double, equals, is equal to, is the same as, number sentence, strategy.
The word 'difference' has a specific meaning in this context, referring to the numeric value of the group. In everyday language, it can refer to any attribute. Students need to understand that the requirement to carry out subtraction can be indicated by a variety of language structures. The language used in the 'comparison' type of subtraction is quite different from that used in the 'take away' type.
Students need to understand the different uses for the = sign, eg 4 + 1 = 5, where the = sign indicates that the right side of the number sentence contains 'the answer' and should be read to mean 'equals', compared to a statement of equality such as 4 + 1 = 3 + 2, where the = sign should be read to mean 'is the same as'.
### National Numeracy Learning Progression links to this Mathematics outcome
When working towards the outcome MA1‑5NA the sub-elements (and levels) of Quantifying numbers (QuN7-QuN8), Additive strategies (AdS2-AdS6) and Number patterns and algebraic thinking (NPA4) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning.
The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression.
### Outcomes
#### A student:
• MA1-1WM
describes mathematical situations and methods using everyday and some mathematical language, actions, materials, diagrams and symbols
• MA1-2WM
uses objects, diagrams and technology to explore mathematical problems
• MA1-3WM
supports conclusions by explaining or demonstrating how answers were obtained
• MA1-5NA
uses a range of strategies and informal recording methods for addition and subtraction involving one- and two-digit numbers
### Content
• Students:
• Explore the connection between addition and subtraction (ACMNA029)
• use concrete materials to model how addition and subtraction are inverse operations
• use related addition and subtraction number facts to at least 20,
eg 15 + 3 = 18, so 18 – 3 = 15 and 18 – 15 = 3
• Solve simple addition and subtraction problems using a range of efficient mental and written strategies (ACMNA030)
• use and record a range of mental strategies to solve addition and subtraction problems involving two-digit numbers, including:
• the jump strategy on an empty number line
• the split strategy, eg record how the answer to 37 + 45 was obtained using the split strategy
$$\begin{array}{r}30 + 40 = 70\\7 + 5 = 12\\ \text{so }70 + 12 = 82\end{array}$$
• an inverse strategy to change a subtraction into an addition, eg 54 – 38: start at 38, adding 2 makes 40, then adding 10 makes 50, then adding 4 makes 54, and so the answer is 2 + 10 + 4 = 16
• select and use a variety of strategies to solve addition and subtraction problems involving one- and two-digit numbers
• perform simple calculations with money, eg buying items from a class shop and giving change (Problem Solving)
• check solutions using a different strategy (Problem Solving)
• recognise which strategies are more efficient and explain why (Communicating, Reasoning)
• explain or demonstrate how an answer was obtained for addition and subtraction problems,
eg show how the answer to 15 + 8 was obtained using a jump strategy on an empty number line
(Communicating, Reasoning)
### Background Information
It is appropriate for students in Stage 1 to use concrete materials to model and solve problems, for exploration and for concept building. Concrete materials may also help in explanations of how solutions were obtained.
Addition and subtraction should move from counting and combining perceptual objects, to using numbers as replacements for completed counts with mental strategies, to recordings that support mental strategies (such as jump, split, partitioning and compensation).
Subtraction typically covers two different situations: 'taking away' from a group, and 'comparison' (ie determining how many more or less when comparing two groups). In performing a subtraction, students could use 'counting on or back' from one number to find the difference. The 'counting on or back' type of subtraction is more difficult for students to grasp than the 'taking away' type. Nevertheless, it is important to encourage students to use 'counting on or back' as a method of solving comparison problems once they are confident with the 'taking away' type.
In Stage 1, students develop a range of strategies to aid quick recall of number facts and to solve addition and subtraction problems. They should be encouraged to explain their strategies and to invent ways of recording their actions. It is also important to discuss the merits of various strategies in terms of practicality and efficiency.
Jump strategy on a number line – an addition or subtraction strategy in which the student places the first number on an empty number line and then counts forward or backwards, first by tens and then by ones, to perform a calculation. (The number of jumps will reduce with increased understanding.)
Jump strategy method: eg 46 + 33
Jump strategy method: eg 79 – 33
Split strategy – an addition or subtraction strategy in which the student separates the tens from the units and adds or subtracts each separately before combining to obtain the final answer.
Split strategy method: eg 46 + 33
$$\begin{array}{ll} 46 + 33 &= 40 + 6 + 30 + 3\\ &= 40 + 30 + 6 + 3 \\ &= 70 + 9 \\ &= 79 \end{array}$$
Inverse strategy – a subtraction strategy in which the student adds forward from the smaller number to obtain the larger number, and so obtains the answer to the subtraction calculation.
Inverse strategy method: eg 65 – 37
start at 37
then add 20 to make 60
then add 5 to make 65
and so the answer is 3 + 20 + 5 = 28
An inverse operation is an operation that reverses the effect of the original operation. Addition and subtraction are inverse operations; multiplication and division are inverse operations.
### Language
Students should be able to communicate using the following language: plus, add, take away, minus, the difference between, equals, is equal to, empty number line, strategy.
Some students may need assistance when two tenses are used within the one problem, eg 'I had six beans and took away four. So, how many do I have now?'
The word 'left' can be confusing for students, eg 'There were five children in the room. Three went to lunch. How many are left?' Is the question asking how many children are remaining in the room, or how many children went to lunch?
### National Numeracy Learning Progression links to this Mathematics outcome
When working towards the outcome MA1‑5NA the sub-elements (and levels) of Quantifying numbers (QuN7-QuN8), Additive strategies (AdS6-AdS7), Understanding money (UnM3) and Number patterns and algebraic thinking (NPA3) describe observable behaviours that can aid teachers in making evidence-based decisions about student development and future learning.
The progression sub-elements and indicators can be viewed by accessing the National Numeracy Learning Progression. |
# Converse of The Pythagorean Theorem
Written by
Malcolm McKinsey
Fact-checked by
Paul Mazzola
## What is the converse of the Pythagorean theorem?
The Pythagorean theorem allows you to find the length of any side of a right triangle if you know the lengths of the other two sides. It can be viewed in another way, as the Converse Of The Pythagorean theorem, to determine if a given triangle is a right triangle just by knowing the lengths of its three sides.
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## Pythagorean theorem
To understand the converse of the Pythagorean Theorem, you need to know and recall the Pythagorean Theorem itself:
This formula works for any right triangle ABC where a and b are legs and c is the hypotenuse. The theorem works for all right triangles, so if you know any two lengths (say, a and c), you can find the unknown length (in our example, b). That is a useful application of the Pythagorean theorem.
## The law of cosines
Before we leap ahead, let's make sure we see the special application of the Law of Cosines in the Pythagorean Theorem.
First, here is the Law of Cosines for △ABC where aa and bb are legs and cc is the hypotenuse, with ∠C the right angle opposite the hypotenuse:
The cosine of 90° is 0, which leaves you with 2ab × 0, so the entire expression, 2ab cos(∠C) = 0 and can be removed, leaving just the Pythagorean Theorem.
## Converse of the Pythagorean theorem
Suppose, though, we start at the "other end." We have three sides ab, and c, but are not certain △ABC is a right triangle. In that case, we can apply the converse of the Pythagorean theorem, which states:
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## Applying the converse of the Pythagorean theorem
Suppose you are given the lengths of three sides of a triangle and asked to determine if it is a right triangle. A common reason for this might be in architecture or in engineering, like getting the correct length of guy-wires bracing an important bridge. Right angles in engineering are very strong.
You know the three sides (legs a and b, hypotenuse c) are as follows:
If ${a}^{2}+{b}^{2}={c}^{2}$, then the triangle has to be a right triangle and the guy-wires have to be the perfect and safe length to hold up the bridge.
So you put the three lengths into the Pythagorean Theorem formula:
What a relief! The guy-wires are all the correct length to keep the bridge at a safe, sturdy right angle.
## Lesson summary
Today you reviewed what the Pythagorean theorem is and why it is useful, how to write and use the formula for the Pythagorean theorem (${a}^{2}+{b}^{2}={c}^{2}$), and learned how the Pythagorean theorem is one application of the law of cosines.
You also learned what the converse of the Pythagorean theorem is; namely, any triangle in which the square of the longest side of a triangle is equal to the sum of the squares of the other two sides must be a right triangle.
Related articles |
# Algebra Slope Intercept Problems
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
Algebra Slope Intercept Problems – There are many forms employed to represent a linear equation among the ones most frequently found is the slope intercept form. You can use the formula for the slope-intercept in order to determine a line equation, assuming you have the straight line’s slope as well as the y-intercept. It is the point’s y-coordinate where the y-axis meets the line. Learn more about this specific linear equation form below.
## What Is The Slope Intercept Form?
There are three fundamental forms of linear equations, namely the standard slope, slope-intercept and point-slope. While they all provide the same results , when used however, you can get the information line produced more efficiently through the slope-intercept form. The name suggests that this form employs a sloped line in which it is the “steepness” of the line determines its significance.
This formula can be used to determine the slope of a straight line, the y-intercept (also known as the x-intercept), which can be calculated using a variety of formulas that are available. The equation for a line using this formula is y = mx + b. The slope of the straight line is indicated through “m”, while its intersection with the y is symbolized with “b”. Each point of the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula the “x” and the “y” have to remain as variables.
## An Example of Applied Slope Intercept Form in Problems
In the real world in the real world, the slope intercept form is used frequently to represent how an item or problem changes in it’s course. The value that is provided by the vertical axis represents how the equation tackles the extent of changes over the amount of time indicated via the horizontal axis (typically the time).
One simple way to illustrate the application of this formula is to discover how much population growth occurs in a certain area in the course of time. In the event that the population in the area grows each year by a specific fixed amount, the values of the horizontal axis increases by a single point each year and the point worth of the vertical scale will rise to show the rising population by the set amount.
Also, you can note the beginning value of a question. The beginning value is located at the y value in the yintercept. The Y-intercept represents the point at which x equals zero. In the case of a previous problem the beginning point could be the time when the reading of population starts or when the time tracking begins along with the associated changes.
This is the location when the population is beginning to be documented to the researchers. Let’s assume that the researcher is beginning with the calculation or measurement in the year 1995. Then the year 1995 will be”the “base” year, and the x=0 points will occur in 1995. Thus, you could say that the population of 1995 is the y-intercept.
Linear equations that use straight-line equations are typically solved this way. The initial value is expressed by the y-intercept and the rate of change is represented as the slope. The main issue with the slope intercept form is usually in the horizontal variable interpretation especially if the variable is associated with one particular year (or any kind of unit). The most important thing to do is to ensure that you understand the definitions of variables clearly. |
+0
0
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+473
Find the sum of all integral values of $c$ with $c\le 25$ and $c\ge1$ for which the equation $y=x^2-17x-c$ has two integral roots.
Aug 25, 2023
#1
+177
+1
This quadratic is in the form of $$y = x^2 - 17x - c$$. We can find the solutions of a quadratic equation by utilizing the quadratic formula.
$$x_{1, 2} = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 * -c}}{2*1} \\ x_{1, 2} = \frac{17 \pm \sqrt{289 + 4c}}{2}$$
All solutions are of this particular form. In order for this family of quadratics to have any chance of having integer roots, the radicand $$289 + 4c$$ must be a perfect square. We can combine this with the constraint information $$1 < c < 25$$ to narrow the options significantly.
$$1 < c < 25 \\ 4 < 4c < 100 \\ 293 < 289 + 4c < 389$$
In other words, any candidate perfect square must lie between 293 and 389. 172 = 289 and the lower bound is 293. This means that the radicand must have a perfect square greater than 17. 202 = 400 and the upper bound is 389. This means that the radicand must have a perfect square less than 20. This restricts the options considerably to either 289 + 4c = 182 or 289 + 4c = 192. We can find the corresponding value for c and determine whether or not this yields integer solutions.
$$289 + 4c = 18^2 \\ 289 + 4c = 324 \\ 4c = 35 \\ c = \frac{35}{4} \not\in \mathbb{Z}$$ $$289 + 4c = 19^2 \\ 289 + 4c = 361 \\ 4c = 72 \\ c = 18 \in \mathbb{Z}$$
We have now determined that, when c = 18, the radicand it a perfect square! Now, we should check that it generates integer solutions.
$$\text{Let } c = 18: \\ \Delta = 289 + 4c = 19^2 \\ x_{1, 2} = \frac{17 \pm \sqrt{19^2}}{2} \\ x_1 = \frac{17 + 19}{2} \text{ or } x_2 = \frac{17 - 19}{2}$$
At this point, it is clear that the solutions are integers because the numerator is even.
When 1 < c < 25, c = 18 is the only value that generated integer solutions of the given family of quadratics, so the sum of all possible integer values for c is 18.
Aug 25, 2023
#1
+177
+1
This quadratic is in the form of $$y = x^2 - 17x - c$$. We can find the solutions of a quadratic equation by utilizing the quadratic formula.
$$x_{1, 2} = \frac{-(-17) \pm \sqrt{(-17)^2 - 4 * -c}}{2*1} \\ x_{1, 2} = \frac{17 \pm \sqrt{289 + 4c}}{2}$$
All solutions are of this particular form. In order for this family of quadratics to have any chance of having integer roots, the radicand $$289 + 4c$$ must be a perfect square. We can combine this with the constraint information $$1 < c < 25$$ to narrow the options significantly.
$$1 < c < 25 \\ 4 < 4c < 100 \\ 293 < 289 + 4c < 389$$
In other words, any candidate perfect square must lie between 293 and 389. 172 = 289 and the lower bound is 293. This means that the radicand must have a perfect square greater than 17. 202 = 400 and the upper bound is 389. This means that the radicand must have a perfect square less than 20. This restricts the options considerably to either 289 + 4c = 182 or 289 + 4c = 192. We can find the corresponding value for c and determine whether or not this yields integer solutions.
$$289 + 4c = 18^2 \\ 289 + 4c = 324 \\ 4c = 35 \\ c = \frac{35}{4} \not\in \mathbb{Z}$$ $$289 + 4c = 19^2 \\ 289 + 4c = 361 \\ 4c = 72 \\ c = 18 \in \mathbb{Z}$$
We have now determined that, when c = 18, the radicand it a perfect square! Now, we should check that it generates integer solutions.
$$\text{Let } c = 18: \\ \Delta = 289 + 4c = 19^2 \\ x_{1, 2} = \frac{17 \pm \sqrt{19^2}}{2} \\ x_1 = \frac{17 + 19}{2} \text{ or } x_2 = \frac{17 - 19}{2}$$
At this point, it is clear that the solutions are integers because the numerator is even.
When 1 < c < 25, c = 18 is the only value that generated integer solutions of the given family of quadratics, so the sum of all possible integer values for c is 18.
The3Mathketeers Aug 25, 2023 |
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# Definition of the Inverse of Trigonometric Ratios
## Solving for an angle given two sides of a right triangle.
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Definition of the Inverse of Trigonometric Ratios
You are in band practice one day when something catches your eye. One of your fellow students has a musical instrument that you haven't seen played before - a triangle. This is an instrument that is a metal triangle that the musician plays by striking it. Walking over to your classmate, you ask him about the instrument. During a conversation about the triangle, you wonder if it would be possible to make different types of triangle instruments to make different sounds.
You ask your science teacher about this, and she is excited that you have taken an interest in the topic. Together you decide to devise some new instruments based on different length sides of triangles, and then try them out to see how they sound. To begin, your teacher asks you to create a list of a few different sides of triangle lengths, and then list the interior angles of the triangles you come up with.
You immediately get to work and generate a list. The first triangle has sides of 12 cm, 35 cm, and 37 cm, and is a right triangle. However, you realize that you aren't sure of the interior angles of the triangles. You are about to get out a paper and pencil to start plotting the triangles when you start to wonder if there might be an easier way to use math in finding the measurement of the angles instead of plotting them and measuring by hand.
As it turns out, there is a way to do this. Read this Concept on inverse trig functions, and then you'll know how to find the measures of the interior angles, as well as find the interior angles of the first triangle you devised for your list.
### Guidance
In this Concept, we'll discuss and apply examples of the inverse functions for trig ratios.
Recall from Chapter 1, the ratios of the six trig functions and their inverses, with regard to the unit circle.
These ratios can be used to find any \begin{align*}\theta\end{align*} in standard position or in a triangle.
In a sense, this is a way of "undoing" a trig function. Before, to find a trig function, you would use the ratio of two sides. Now, by using the inverse trig ratio, you can find angles when you need them.
Let's investigate this by doing a few examples.
#### Example A
Find the measure of the angles below.
a.
b.
Solution: For part a, you need to use the sine function and part b utilizes the tangent function. Because both problems require you to solve for an angle, the inverse of each must be used.
a. \begin{align*}\sin x=\frac{7}{25} \rightarrow \sin^{-1} \frac{7}{25} = x \rightarrow x = 16.26^\circ\end{align*}
b. \begin{align*}\tan x=\frac{40}{9} \rightarrow \tan^{-1} \frac{40}{9} = x \rightarrow x = 77.32^\circ\end{align*}
The trigonometric value \begin{align*}\tan \theta = \frac{40}{9}\end{align*} of the angle is known, but not the angle. In this case the inverse of the trigonometric function must be used to determine the measure of the angle. The inverse of the tangent function is read “tangent inverse” and is also called the arctangent relation. The inverse of the cosine function is read “cosine inverse” and is also called the arccosine relation. The inverse of the sine function is read “sine inverse” and is also called the arcsine relation.
#### Example B
Find the angle, \begin{align*}\theta\end{align*} , in standard position.
Solution: The \begin{align*}\tan \theta = \frac{y}{x}\end{align*} or, in this case, \begin{align*}\tan \theta = \frac{8}{-11}\end{align*} . Using the inverse tangent, you get \begin{align*}\tan^{-1} -\frac{8}{11} = -36.03^\circ\end{align*} . This means that the reference angle is \begin{align*}36.03^\circ\end{align*} . This value of \begin{align*}36.03^\circ\end{align*} is the angle you also see if you move counterclockwise from the -x axis. To find the corresponding angle in the second quadrant (which is the same as though you started at the +x axis and moved counterclockwise), subtract \begin{align*}36.03^\circ\end{align*} from \begin{align*}180^\circ\end{align*} , yielding \begin{align*}143.97^\circ\end{align*} .
Recall that inverse trigonometric functions are also used to find the angle of depression or elevation.
#### Example C
A new outdoor skating rink has just been installed outside a local community center. A light is mounted on a pole 25 feet above the ground. The light must be placed at an angle so that it will illuminate the end of the skating rink. If the end of the rink is 60 feet from the pole, at what angle of depression should the light be installed?
Solution: In this diagram, the angle of depression, which is located outside of the triangle, is not known. Recall, the angle of depression equals the angle of elevation. For the angle of elevation, the pole where the light is located is the opposite and is 25 feet high. The length of the rink is the adjacent side and is 60 feet in length. To calculate the measure of the angle of elevation the trigonometric ratio for tangent can be applied.
The angle of depression at which the light must be placed to light the rink is \begin{align*}22.6^\circ\end{align*}
### Vocabulary
Trigonometric Inverse: The trigonometric inverse is a function that undoes a trig function to give the original argument of the function. It can also be used to find an angle from the ratio of two sides.
### Guided Practice
1. Find the value of the missing angle.
2. Find the value of the missing angle.
3. What is the value of the angle with its terminal side passing through (-14, -23)?
Solutions:
1. \begin{align*}\cos \theta = \frac{12}{17} \rightarrow \cos^{-1} \frac{12}{17} = 45.1^\circ\end{align*}
2. \begin{align*}\sin \theta = \frac{25}{36} \rightarrow \sin^{-1} \frac{31}{36} = 59.44^\circ\end{align*}
3. This problem uses tangent inverse. \begin{align*}\tan x = \frac{-23}{-14} \rightarrow x = \tan^{-1} \frac{23}{14} = 58.67^\circ\end{align*} (value graphing calculator will produce). However, this is the reference angle. Our angle is in the third quadrant because both the \begin{align*}x\end{align*} and \begin{align*}y\end{align*} values are negative. The angle is \begin{align*}180^\circ + 58.67^\circ = 238.67^\circ\end{align*} .
### Concept Problem Solution
Since you now know about inverse trigonometric ratios, you know that you can apply the inverse of a trig function to help solve this problem. For example, it is straightforward to apply the tangent function:
You can find the other angle the in a similar manner, this time using the sine function:
### Practice
Find the measure of angle A in each triangle below.
Use inverse tangent to find the value of the angle with its terminal side passing through each of the given points.
1. (-3,4)
2. (12,13)
3. (-4, -7)
4. (5, -4)
5. (-6, 10)
6. (-3, -10)
7. (6, 8)
Use inverse trigonometry to solve each problem.
1. A 30 foot building casts a 60 foot shadow. What is the angle that the sun hits the building?
2. Over 3 miles (horizontal), a road rises 100 feet (vertical). What is the angle of elevation?
3. An 80 foot building casts a 20 foot shadow. What is the angle that the sun hits the building?
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rahul-had-some-bananas-and-he-divided-them-into-two-lots
Q:) Rahul had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of ₹ 2 for 3 bananas and the second lot at the rate of ₹ 1 per banana and got a total of ₹400. If he had sold the first lot at the rate of ₹ 1 per banana and the second lot at the rate of ₹ 4 for 5 bananas, his total collection would have been ₹460. Find the total number of bananas he had.
Solution: Let Rahul have bananas in first lots = x
And Let number of bananas in second lots = y
First condition
Amount of money for lot A =
Amount of money for lot B = 1.y
Multiply by 3 in equation
⇒ 2x + 3y = 1200 —— (i)
Secon condition
Amount of money for lot A = 1.x
Amount of money for lot B =
Multiply by 5 in equation
5x + 4y = 2300 —–(ii)
Multiply by 5 in (i) and 2 in (ii) and substracting (i) to (ii)
5(2x + 3y) – 2(5x + 4y) = 1200 – 2300
⇒ 10x + 15 y – 10x – 20y = -1100
⇒ – 5y = -1100
⇒ y = 220
Putting the value of x in (i)
2x + 3×220 = 1200
⇒ 2x = 1200 – 660
⇒ 2x = 540
⇒ x = 540/2 = 270
Number of bananas in lot A = 270
Number of bananas in lot B = 220
Ruhi invested a certain amount of money in two schemes |
# How do you write a linear equation that goes through (1,-6) and is parallel to the line x + 2y = 6?
Mar 5, 2018
$x + 2 y + 11 = 0$
is the required equation of the line.
#### Explanation:
$x + 2 y = 6 \text{ " " } \left(1\right)$
The slope of the line $= - \frac{1}{2}$
The slope of a line parallel to line (1) will be same
$\text{slope} = m = - \frac{1}{2}$
Now, the equation of the line through $\left(1 , - 6\right)$ with a slope $m = - \frac{1}{2}$ by using the point-slope form is
$y - {y}_{1} = m \left(x - {x}_{1}\right)$
$y - \left(- 6\right) = - \frac{1}{2} \left(x - 1\right)$
$y + 6 = - \frac{1}{2} \left(x - 1\right)$
$2 y + 12 = - x + 1$
or
$x + 2 y + 11 = 0$ |
# how to find the missing side of a right triangle
Posted by on Jan 11, 2021 in Uncategorized | 0 comments
Find the measure of each side indicated. Share practice link. Example 1. Here you can enter two known sides or angles and calculate unknown side ,angle or area. ASA. Set up an equation using a sohcahtoa ratio. The legs are labeled 6 and 9. 65 plus 90 is 155. Effortless Math services are waiting for you. We are going to focus on two specific cases. Finding an Angle in a Right Angled Triangle Angle from Any Two Sides. How Do You Find the Missing Side of a Right Angled Triangle? In the video below, you’ll learn how to deal with harder problems, including how to solve for … The cosine rule can be used to find a missing side when all sides and an angle are involved in the question. What are the rules of Pythagoras? The tangent ratio is not only used to identify a ratio between two sides of a right triangle, but it can also be used to find a missing side length. Sides "a" and "b" are the perpendicular sides and side "c" is the hypothenuse. Step-by-step explanations are provided for each calculation. Finding Sides of a Triangle. Separate worksheets or one double-sided. The Law of Cosines is useful for many types of applied problems. Example. The theorem states that the hypotenuse of a right triangle can be easily calculated from the lengths of the sides. Articles. The factors are the lengths of the sides and one of the two angles, other than the right angle. In this tutorial I show you how to find a length of a side of a non right-angled triangle by using the Cosine Rule. Then Find the length of the side marked x in the following triangle: Show Solution. Such a triangle can be solved by using Angles of a Triangle to find the other angle, and The Law of Sines to find each of the other two sides. *Response times vary by subject and question complexity. Example of finding Side Length. Sal is given a right triangle with an acute angle of 65° and a leg of 5 units, and he uses trigonometry to find the two missing sides. So, if you know the lengths of two sides, all you have to do is square the two lengths, add the result, … Pythagoras’ Theorem (The Pythagorean Theorem) The hypotenuse is the longest side of a right triangle, and is located opposite the right angle. Jayden's WorkSheridan's Worka2 + b2 = c2a2 + b2 = c272 + 132 = c272 + b2 = … 0. When we know 2 sides of the right triangle, use the Pythagorean theorem. Since we know the hypotenuse and want to find the side opposite of the 53° angle, we are dealing with sine. Use the Pythagorean theorem to find the missing side of a triangle. It's the third one. Adjacent, Opposite and Hypotenuse, in a right triangle is shown below. kingarabsavage33 is waiting for your help. See answer MetallicRose is waiting for your help. The formula is as follows: c 2 = a 2 + b 2. c = â(a 2 + b 2) What Is the Name of a Triangle With Two Equal Sides? In a right triangle, the side that is opposite of the 90° angle is the longest side of the triangle, and is called the hypotenuse. So, if you know the lengths of two sides, all you have to do is square the two lengths, add the result, then take the square root of the sum to get the length of the hypotenuse. Use the Pythagorean theorem to find the missing side of a triangle. Example 2. Enter the length of any two sides and leave the side to be calculated blank. These are the four steps to follow: Step 1 Find the names of the two sides we are using, one we are trying to find and one we already know, out of Opposite, Adjacent and Hypotenuse. That is exactly what we are going to learn . In a right-angled triangle, the sides of the triangle are named as the opposite, adjacent, and hypotenuse sides. So how many more centimeters, what length must this missing side be to get us a total distance around the outside of 24 centimeters. Trigonometry Calculator - Right Triangles. You can use this equation to figure out the length of one side if you have the lengths of the other two. Use the law of sines to find the missing measurements of the triangles in these examples. The formula to follow is a^2+b^2=c^2 so side a to the second power (squared) plus side b to the second power (squared) cosec θ = Hypotenuse side / Opposite side. A right triangle is shown. To play this quiz, please finish editing it. Add your answer and earn points. Print; Share; Edit; Delete; Host a game. Recall the three main trigonometric functions. A triangle with two equal sides and one side that is longer or shorter than the others is called an isosceles triangle. Sides of the Right-Angled Triangle: There are three sides to a triangle. After you are comfortable writing sine, cosine, tangent ratios you will often use sohcahtoa to find the sides of a right triangle. This was created as a revision aid for year 11s. Finding Missing Sides and Angles in Right Triangles DRAFT. Practice. To solve a triangle means to find the length of all the sides and the measure of all the angles. Sides of a right triangle; Height of a right triangle; Bisector of a right triangle; Median of a right triangle; Height, Bisector and Median of an equilateral triangle ; All geometry formulas for any triangles; Parallelogram. This calculator is for a right triangle only! Complete the quick and easy checkout process. You can do this if you are given the opposite angle and the two other sides. Calculating the length of another side of a triangle If you know the length of the hypotenuse and one of the other sides, you can use Pythagorasâ theorem to find the length of the third side. Password will be generated automatically and sent to your email. He provides an individualized custom learning plan and the personalized attention that makes a difference in how students view math. cos θ = Adjacent side / Hypotenuse side. â I want to calculate: Input two elements of a right triangle use letter r to input square root. If you can’t immediately see which Pythagorean triple family a triangle belongs to, don’t worry: you can always use the following step-by-step method to pick the family and find the missing side. Explain your reasoning. To enter a value, click inside one of the text boxes. So … New questions in Mathematics. You need only two given values in the case of: one side and one angle; two sides; area and one side; Remember that if you know two angles, it's not enough to find the sides of the triangle. Click on the "Calculate" button to solve for all unknown variables. Angle C and angle 3 cannot be entered. Finding the missing length of a triangle using pythagorean theorem . by Reza about 9 months ago in After registration you can change your password if you want. Homework. Find a Right Triangle’s Missing Side Using Triangle Families. Draw the triangle on your paper labeling the two sides adjacent to the right angle, or legs, “ a ” and “ b .” Label the hypotenuse, or third side “ c .” Set Up Your Equation Set up your equation so that Add your answer and earn points. sin $$(θ=\frac{opposite}{hypotenuse}$$. Sides "a" and "b" are the perpendicular sides and side "c" is the hypothenuse. Mckenzie and Cara both tried to find the missing side of the right triangle. In case you need them, here are the Trig Triangle Formula Tables, the Triangle Angle Calculator is also available for angle only calculations. They help us to create proportions for finding missing side lengths! Read more. What is the angle between the ladder and the wall? The ladder leans against a wall as shown. Find AC in the following triangle. Pythagorean Theorem. See other my FREE resources in this set. Finding an angle of a non right angled triangle . We can find an unknown angle in a right-angled triangle, as long as we know the lengths of two of its sides. Finding the missing side of triangle (no rating) 0 customer reviews. Step #3: Enter the two known lengths of the right triangle. See and . s i n ( 53) = o p p o s i t e h y p o t e n u s e s i n ( 53) = x 12. For example, an area of a right triangle is equal to 28 in² and b = 9 in. Finding the length of a side of a non right angled triangle. Finish Editing. The hypotenuse is the longest side of a right triangle. We even offer a free study guide that easy explains the steps easily on one page! Triangle Sides Calculator: This calculator calculates for the length of one side of a right triangle given the length of the other two sides. Learn how to use trig functions to find an unknown side length in a right triangle. 78% average accuracy. This quiz is incomplete! Over 10,000 reviews with an average rating of 4.5 out of 5, Schools, tutoring centers, instructors, and parents can purchase Effortless Math eBooks individually or in bulk with a credit card or PayPal. Law of Sines $$\frac{sin(A)}{a} = \frac{sin(B)}{b} = \frac{sin(C)}{c}$$ Examples: Law of sines. This calculator is designed to give the two unknown factors in a right triangle, assuming two factors are known. Immediately receive the download link and get the eBook in PDF format. Step #4: Tap the "Calculate Unknown" button. sine $$45^\circ=\frac{AC}{8}â8 Ãsin 45^\circ=AC$$,now use a calculator to find sin $$45^\circ$$. Select which side of the right triangle you wish to solve for (Hypotenuse c, Leg a, or Leg b). You may adjust the accuracy of your results. Mckenzie and Cara both tried to find the missing side of the right triangle. He works with students individually and in group settings, he tutors both live and online Math courses and the Math portion of standardized tests. Two of its sides to enter a value, click inside one of the more famous mathematical formulas is (. Known variables ( sides a, or 45, how to find the missing side of a right triangle, 45, 90, an! Browser, redraw the triangle using Tangent resources on our website simple matter if two sides are known input root! Which the sides satisfy this condition is a pretty simple matter if two sides are known we 're having loading. Determining which side you know and which side of right triangle applied problems you are for... Just solve the equation: step 3 and a test-prep expert who has been tutoring students 2008. Email address will not be entered missing sides and the measure of the other two triangle, use sohcahtoa decide! 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Pq=6.5Cm, QR=9.7cm and PR = c cm value you know one side ’ s measure which the satisfy! Response times vary by subject and question complexity theorem to find an unknown side length of right... ( 53 ) x = 11.98 their dreams when all sides and an in! In how students view math missing length of a triangle there are three to... # 4: Tap the calculate unknown side length in a right triangle we can simplify left-hand! Side how do you find the missing side lengths are we must use the cosine rule can an! That easy explains the steps easily on one page triangle trig calculator fill in 2 fields in the but! Their standardized test scores -- and attend the colleges of their dreams just solve the:! Created: Oct 27, 2017 | Updated: Sep 16, 2019 values and press calculate following! On calculate the eBook you wish to purchase by searching for the test or title this...: enter the length of a right triangle you must use one of the sides leave! 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# Lessons from a Triangle¶
OK, let’s look at what we have learned from writing a drawTriangle and drawOctagon function. The first thing you had to figure out was that you needed to modify the parameter to the range function based on the number of sides in the polygon. The next thing, and this may have been the most challenging part for you was to figure out how much to turn each time. The following table summarizes what you probably learned very nicely. If you didn’t, look at the table and then go back to the previous page and see if you can finish drawTriangle and drawOctagon.
Shape
Sides
range()
Angle
Triangle
3
3
360/3 = 120
Square
4
4
360/4 = 90
Octagon
8
8
360/8 = 45
Looking at the table above you can really see that there is a pattern. If you know the number of sides you want, the rest can be figured out from there. This leads us to the next problem solving stage of this exercise, generalization. Why write a separate function for every kind of polygon when you can just write a single function that can be used to draw many different polygons?
Our new function drawPolygon will have three parameters, a turtle and the length of the side just like we have in the previous functions, but now we will add an additional parameter: numSides.
Here’s the starting point for the drawPolygon function, see if you can fill in the details on your own.
# Finally a Circle¶
Now that we have our drawPolygon function working we are almost done with the drawCircle function. In fact if you tried to make a 100 sided polygon, you probably noticed that it looked suspiciously close to a circle. The only thing we’ll want to do differently is to remember that we don’t usually specify a circle according to the number of sides it has, but rather we specify a circle by its radius.
So, here is our next problem solving challenge. Given a radius, how can we transform the radius into the parameters we need for the drawPolygon function? Let’s figure this out in two steps. First what should we use as the number of sides? Second, what should we use as the side length? To simplify the problem we can make a simple assumption about the number of sides. Let’s just assume that 360 sides is going to be good enough for any circle. Plus, it just makes sense that a circle has 360 degrees, one side for each degree should give us a nice round circle.
Now that we have the number of sides figured out, we can use a bit of math to figure out the length of each of those sides. Since we are given the radius we know how large the circumference of the circle is. Right? You may remember from math class that the circumference of a circle is 2 * pi * radius. If we know the circumference and the number of sides (360), then the length of each side should be an equal proportion of the circumference. circumference / 360.
Here’s the final solution:
One annoying limitation of this drawCircle function is that it draws the circle to the right of the turtle’s current position and heading. As a short aside, how could you change one line of the code to draw the circle to the left of the turtle’s current position and heading? What we would really like is to have the circle centered on the turtle’s current position when drawCircle is called. Modify the program above to make it so.
Finally, write another function drawFilledCircle that takes a turtle, a radius, and a color to fill in the circle.
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# 7 Best Tips for Teaching the Order of Operations
Teaching the order of operations is a fundamental skill that students must learn in mathematics. Without an understanding of the order of operations, students will struggle with more advanced concepts in mathematics when they move through the grades.
#### Teaching the Order of Operations Rule (PEMDAS or GEMDAS)
The order of operations, also known as the PEMDAS rule, is a set of rules that determine the order in which mathematical operations should be performed. The PEDMAS rule stands for Parentheses, Exponents, Division, Multiplication, Addition, and Subtraction.
#### Using Real-Life Examples for Teaching the Order of Operations
When teaching the order of operations, it is important to start with the basics. Begin by explaining what the PEDMAS rule stands for and why it is important. Use real-life examples to demonstrate how the rule is used in everyday situations. One of the best real life situations for order of operations is budgeting.
I like to explain to students when planning a budget, you need to be able to perform mathematical operations in the correct order. For example, if you have a total income of \$1,000 and want to save 20% of it, you would first multiply \$1,000 by 0.20 to find the amount you need to save: \$1,000 x 0.20 = \$200. You would then subtract \$200 from \$1,000 to find the amount you have left for expenses: \$1,000 – \$200 = \$800.
#### Examples of Mathematical Expressions
Once students have a basic understanding of the PEDMAS rule, introduce them to examples of mathematical expressions that require the rule to be applied. For example, consider the expression 3 + 5 x 2. Starting with simple problems will make it easier for them to go back if they make an error, and then progress to more difficult expressions.
#### Using a Variety of Examples and Practice Problems
It is important to use a variety of examples when teaching the order of operations. Use simple examples at first and gradually increase the complexity of the expressions. Encourage students to use mental math whenever possible, as this will help them to develop their problem-solving skills. In addition to using examples, it is also important to provide students with practice problems.
#### Games and Activities
One helpful way to teach the order of operations is to use games or activities. For example, create a game in which students have to solve a series of expressions using the PEDMAS rule. Alternatively, use a group activity in which students work together to solve a complex expression, with each student responsible for one or two operations. Click here to see all of the no prep resources I have for the order of operations.
#### Emphasizing the Importance of Following the Rule
When teaching the order of operations, it is important to emphasize the importance of following the rule. Remind students that the order of operations is not optional and that failing to follow the rule can lead to incorrect answers.
Encourage students to take their time when solving expressions and to double-check their work to make sure they followed the steps properly. This will relate back to real-life examples because if they calculate something wrong, especially with finances it could be a mistake that causes them to be short on money.
#### Assessing Students’ Understanding
Finally, it is important to assess students’ understanding of the PEDMAS rule. Use quizzes, tests, or other assessment methods to evaluate their knowledge and provide feedback on their progress. Encourage students to ask questions if they are unsure of anything, and be prepared to provide additional support or guidance as needed. This can be a difficult skill for them because it’s different than what they’re most likely used to, so it will require some reteaching with some students.
Teaching the order of operations is a critical skill that students must learn in mathematics. By using a variety of examples, practice problems, and games, students can develop a strong understanding of the PEDMAS rule and its importance in mathematics. By emphasizing the importance of following the rule and providing feedback on students’ work, teachers can help their students to develop strong problem-solving skills and achieve success in mathematics. |
# GRAPHS OF EXPONENTIAL FUNCTIONS
GRAPHS OF EXPONENTIAL FUNCTIONS
By Nancy Marcus
In this section we will illustrate, interpret, and discuss the graphs of exponential functions. We will also illustrate how you can use graphs to HELP you solve exponential problems.
Stretch and Shrink: The following examples discuss the difference between the graph of f(x) and f(Cx).
Example 15: Graph the function and the function on the same rectangular coordinate system. and answer the following questions about each graph:
1.In what quadrants in the graph of the function located? In what quadrants is the graph of the function . located?
2.What is the x-intercept and the y-intercept on the graph of the function ? What is the x-intercept and the y-intercept on the graph of the function ?
3.Find the point (2, f(2)) on the graph of and find (2, g(2)) on the graph of . What do these two points have in common?
4.Describe the relationship between the two graphs.
5.Write g(x) in terms of f(x).
6.How would you moved the graph of so that it would be superimposed on the graph of ? When you moved the graph, where would the point (0, 1) on be after the move?
1.You can see that the both graphs are located in quadrants I and II.
2.You can see that neither of the graphs crosses the x-axis; therefore, neither of the graphs has an x-intercept. Notice that both graphs cross the y-axis at 1 because .
3.The point is located on the graph of . The point is located on the graph of . This is a significant stretch.
4.Even though the graph of g(x) looks difference from the graph of f(x), both graphs have the same shape. The graph of g(x) is located above the graph of f(x) for all positive values of x, and the graph of g(x) is located below the graph of f(x) for all negative values of x. Both graphs cross the y-axis at the same point.
The graph of is a result of stretching and shrinking the graph of . For example, for every positive value of x the value of g(x) is larger than the value of f(x). For every negative value of x, just the opposite is true. When the value of x is 0, both functions values are the same.
5. =
6.The point (0, 1) on the graph of does not move. The point (5, 148.4) would be stretched to (5, 3,269). The point (-3, 0.05) would be shrunk to (-3, 0.0000003).
Example 16: Graph the function and the function on the same rectangular coordinate system. and answer the following questions about each graph:
1.In what quadrants in the graph of the function located? In what quadrants is the graph of the function located?
2.What is the x-intercept and the y-intercept on the graph of the function ? What is the x-intercept and the y-intercept on the graph of the function ?
3.Find the point (2, f(2)) on the graph of and find (2, g(2)) on the graph of . What do these two points have in common?
4.Describe the relationship between the two graphs.
5.Write g(x) in terms of f(x).
6.Describe how you would move the graph of moved so that it would be superimposed on the graph of . Where would the point (0, 1) on the graph of wind up on after the move?
1.Both graphs are located in quadrants I and II. This means that the function values for both functions will always be positive.
Neither of the graphs cross the x-axis. This means that there is no value of x that will cause the function value to be zero.
2.Both graphs cross the y-axis at y = 1.
3.The point is located on the graph of . The point is located on the graph of .
4.Both graphs have the same shape. The graph of g(x) is located below the graph of f(x) in the first quadrant (for all positive values of x), and the graph of g(x) is located above the graph of f(x) in the second quadrant (for all negative values of x).
5. =
6.=.The point (0, 1) would stay in the same place. The point (1, 2.718282) on the graph of f(x) would be moved to (1, 1.284025). The point ( -1, 0.367879) would be moved to (-1, 0.778801).
If you would like to review another example, click on Example.
[Exponential Rules] [Logarithms]
[Algebra] [Trigonometry ] [Complex Variables] |
# Chapter 03.01: Prerequisites to Numerical Methods for Solving Nonlinear Equations
## Learning Objectives
After successful completion of this lesson, you should be able to:
1) find the solutions of quadratic equations,
2) derive the formula for the solution of quadratic equations,
3) solve simple physical problems involving quadratic equations.
## What are quadratic equations, and how do we solve them?
A quadratic equation has the form
$ax^{2} + bx + c = 0,\ \text{where } a \neq 0$
The solution to the above quadratic equation is given by
$x = \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a}$
So the equation has two roots, and depending on the value of the discriminant, $$b^{2} - 4ac$$, the equation may have real, complex, or repeated roots.
$\text{If } b^{2} - 4ac < 0,\ \text{the roots are complex.}$
$\text{If } b^{2} - 4ac > 0,\ \text{the roots are real.}$
$\text{If } b^{2} - 4ac = 0,\ \text{the roots are real and repeated.}$
### Example 1
Derive the solution to $$ax^{2} + bx + c = 0$$.
Solution
$ax^{2} + bx + c = 0\;\;\;\;\;\;\;\;\;\;\;\; (E1.1)$
Dividing both sides by $$a$$, $$\left( a \neq 0 \right)$$, we get
$x^{2} + \frac{b}{a}x + \frac{c}{a} = 0$
Note if $$a = 0$$, the solution to
$ax^{2} + bx + c = 0$
is
$x = - \frac{c}{b}$
Rewrite Equation (E1.1) for $$a\neq0$$
$x^{2} + \frac{b}{a}x + \frac{c}{a} = 0$
as
$\left( x + \frac{b}{2a} \right)^{2} - \frac{b^{2}}{4a^{2}} + \frac{c}{a} = 0$
$\begin{split} \left( x + \frac{b}{2a} \right)^{2}\ &= \frac{b^{2}}{4a^{2}} - \frac{c}{a}\\ &= \frac{b^{2} - 4ac}{4a^{2}} \end{split}$
$\begin{split} x + \frac{b}{2a} &= \pm \sqrt{\frac{b^{2} - 4ac}{4a^{2}}}\\ &= \pm \frac{\sqrt{b^{2} - 4ac}}{2a}\end{split}$
$\begin{split} x &= - \frac{b}{2a} \pm \frac{\sqrt{b^{2} - 4ac}}{2a}\\ &= \frac{- b \pm \sqrt{b^{2} - 4ac}}{2a} \end{split}\;\;\;\;\;\;\;\;\;\;\;\; (E1.2)$
### Example 2
A ball is thrown down at 50 mph from the top of a building. The building is 420 feet tall. Derive the equation that would let you find the time the ball takes to reach the ground.
Solution
The distance $$s$$ covered by the ball is given by
$s = ut + \frac{1}{2}gt^{2}\;\;\;\;\;\;\;\;\;\;\;\; (E2.1)$
where
$u = \text{initial velocity (ft/s)}$
$g = \text{acceleration due to gravity (ft}/\text{s}^2)$
$t = \text{time (s)}$
Given
$\begin{split} u &= {50}\frac{\text{miles}}{\text{hour}} \times \frac{1\text{ hour}}{3600\text{ s}} \times \frac{5280\text{ ft}}{1\text{ mile}}\\ &= { 73.33\ }\frac{\text{ft}}{\text{s}}\end{split}$
$g = \text{32.2}\frac{\text{ft}}{\text{s}^{2}}$
$s = 420\ \text{ft}$
we have from Equation (2.1)
$420 = 73.33t + \frac{1}{2}\left( 32.2 \right)\ t^{2}$
$16.1t^{2} + 73.33t - 420 = 0$
The above equation is a quadratic equation, the solution of which would give the time it would take the ball to reach the ground. The solution of the quadratic equation is
$\begin{split} t\ &= \frac{- 73.33 \pm \sqrt{73.33^{2} - 4 \times 16.1 \times ( - 420)}}{2(16.1)} \\ &= 3.315, - 7.870\end{split}$
Since $$t > 0,$$ the valid value of time $$t$$ is $$3.315\ \text{s.}$$
## Multiple Choice Test
(1). The value of $$x$$ that satisfies $$f\left( x \right) = 0$$ is called most suitably the
(A) root of an equation $$f\left( x \right) = 0$$
(B) root of a function $$f\left( x \right)$$
(C) zero of an equation $$f\left( x \right) = 0$$
(D) none of the above
(2). A quadratic equation has _______ root(s).
(A) one
(B) two
(C) three
(D) four
(3). For a certain cubic equation, at least one of the roots is known to be a complex root. How many complex roots does the cubic equation have?
(A) one
(B) two
(C) three
(D) cannot be determined
(4). An equation such as $$\tan x = x$$ has _______ root(s).
(A) zero
(B) one
(C) two
(D) infinite
(5). A polynomial of order n has __________ zeros.
(A) $$n - 1$$
(B) $$n$$
(C) $$n + 1$$
(D) $$n + 2$$
(6). The velocity of a body is given by $$v(t) = 5e^{- t} + 4$$, where t is in seconds and $$v$$ is in m/s. The velocity of the body is $$6$$ m/s at t = ____________ seconds.
(A) $$0.1823$$
(B) $$0.3979$$
(C) $$0.9163$$
(D) $$1.609$$
For complete solution, go to
http://nm.mathforcollege.com/mcquizzes/03nle/quiz_03nle_background_solution.pdf
## Problem Set
(1). Find the roots of the equation $$2x^{2} + 5x =-3$$.
Answer: $$-1.5,\ -1$$
(2). Find the roots of the equation $$2x^{2} + 4x + 4 = 0$$.
Answer: $$-1+i,\ -1-i$$
(3). Using any method, estimate the smallest non-negative root of the equation $$\tan(x) = x$$. Plot the function, $$\tan(x)$$ and $$x$$ to help you to find an approximate answer.
Answer: $$x=0$$
(4). You are working for ‘DOWN THE TOILET COMPANY’ that makes floats for ABC commodes. The ball has a specific gravity of $$0.6$$ and has a radius of $$5.5$$ cm. You are asked to find the depth to which the ball will get submerged when floating in water.
The equation that gives the depth $$x$$ (unit of $$x$$ is m) to which the ball is submerged under water is given by
$x^{3} - 0.165x^{2} + 3.993 \times 10^{- 4} = 0$
Note that the cubic equation will have three roots. Can some of these roots be complex. If so how many complex roots can it have? If more than one root of the above equation is real, how do you choose the acceptable root? Or are all the real roots physically acceptable?
Answer: The roots are $$0.062378,\ 0.14636,\ -0.043737$$. Only $$x=0.062378$$ is acceptable as the root needs to lie between $$0$$ and the diameter, $$D=0.11m$$.
(5). A ball is thrown down at $$50$$ mph from the top of a building. The building is $$420$$ feet tall. The time in seconds the ball would take to reach the ground is given by
${16.1}t^{2} + {73.33}t - {420} = 0{,\ t} \geq 0$
a) How many roots does the quadratic equation have?
b) Which of the two roots is a valid answer?
c) How much time would the ball take to reach the ground if it was just let go rather than thrown?
d) If the quadratic equation gave you complex roots, what would be your conclusion?
Answer: $$a)\ 2\ \ b)\ 3.3149\ s\ \ c)\ 5.1075\ s\ \ d)\ \text{Modeling of eqn must be wrong.}$$
(6). A straight steel ruler that is $$12^{\prime\prime}$$ long at the initial temperature of $$80^{\circ}$$F contracts in length when dipped in a cold liquid. It is allowed to reach steady state. If the reduction in the length at steady state is found to be $$0.001^{\prime\prime}$$, find the nonlinear equation (you do not have to solve it) that will allow you to find the temperature of the liquid. The reduction in the length, $$\Delta L$$ is given by
$\Delta L = L\int_{T_{\text{initial}}}^{T_{\text{liquid}}}{\alpha dT}$
where
$L = \text{initial length of the ruler,}$
$\alpha = - 1.2 \times 10^{- 11}T^{2} + 6.2 \times 10^{- 9}T + 6 \times 10^{- 6}$
$(\text{units of}\ \alpha\ \text{are}\ in/in/^{\circ}F,\ \text{and units of temperature are}\ ^{\circ}F)$
$T_{\text{liquid}}= \text{temperature of the liquid,}$
$T_{\text{initial}} = \text{initial temperature of the steel ruler.}$
Answer: $$-4.8 \times 10^{-11} (T_{liquid})^{3}+3.72\times10^{-8}(T_{liquid})^{2}+7.2\times10^{-5}(T_{liquid})-4.9735\times10^{-3}=0$$ |
12:28 pm Uncategorized
# The Complement of a Set: Understanding the Basics
When it comes to set theory, one fundamental concept that often arises is the complement of a set. The complement of a set refers to the elements that are not included in the set. In other words, it is everything outside of the set. Understanding the complement of a set is crucial in various fields, including mathematics, computer science, and statistics. In this article, we will delve into the basics of the complement of a set, explore its properties, and provide real-world examples to illustrate its significance.
## What is a Set?
Before we dive into the complement of a set, let’s first establish what a set is. In mathematics, a set is a collection of distinct objects, which are referred to as elements. These elements can be anything, such as numbers, letters, or even other sets. Sets are denoted by curly braces, and the elements are listed inside the braces, separated by commas. For example, consider the set A:
A = {1, 2, 3, 4}
In this case, the set A contains the elements 1, 2, 3, and 4.
## The Complement of a Set
Now that we have a clear understanding of what a set is, let’s explore the complement of a set. The complement of a set A, denoted as A’, is the set of all elements that are not in A. In other words, it consists of everything outside of the set A. To represent the complement of a set, we often use the universal set, which is the set of all possible elements in a given context.
For example, let’s consider the universal set U as the set of all integers:
U = {..., -3, -2, -1, 0, 1, 2, 3, ...}
Now, suppose we have a set A defined as:
A = {1, 2, 3, 4}
The complement of set A, denoted as A’, would be:
A' = {x | x ∉ A}
In this case, the complement of set A would include all the integers that are not present in set A.
## Properties of the Complement of a Set
The complement of a set possesses several important properties that are worth exploring. Understanding these properties can help us gain a deeper insight into the concept and its applications. Let’s take a closer look at some of these properties:
### 1. Identity Property
The identity property states that the complement of the complement of a set is the set itself. In other words, taking the complement of a set twice results in the original set. Mathematically, it can be represented as:
(A')' = A
This property is analogous to the concept of double negation in logic.
### 2. Union Property
The union property states that the complement of the union of two sets is equal to the intersection of their complements. Mathematically, it can be represented as:
(A ∪ B)' = A' ∩ B'
This property allows us to simplify complex set operations by working with the complements instead.
### 3. Intersection Property
The intersection property states that the complement of the intersection of two sets is equal to the union of their complements. Mathematically, it can be represented as:
(A ∩ B)' = A' ∪ B'
Similar to the union property, this property allows us to simplify set operations by working with the complements.
## Real-World Examples
Now that we have explored the basics and properties of the complement of a set, let’s examine some real-world examples to better understand its practical applications.
### Example 1: Student Clubs
Consider a university with various student clubs. Let’s say there are three clubs: Club A, Club B, and Club C. The universal set U in this context would be all the students in the university. Now, suppose we have the following sets:
A = {students in Club A}
B = {students in Club B}
C = {students in Club C}
The complement of set A, denoted as A’, would be all the students who are not part of Club A. Similarly, the complements of sets B and C, denoted as B’ and C’ respectively, would represent the students who are not part of Club B and Club C. These complements can be useful in analyzing the distribution of students across different clubs and identifying students who are not involved in any club.
### Example 2: Online Shopping
Consider an online shopping platform that offers a wide range of products. Let’s say there are three categories of products: Electronics, Clothing, and Home Decor. The universal set U in this context would be all the products available on the platform. Now, suppose we have the following sets:
E = {Electronics}
C = {Clothing}
H = {Home Decor}
The complement of set E, denoted as E’, would be all the products that are not classified as Electronics. Similarly, the complements of sets C and H, denoted as C’ and H’ respectively, would represent the products that are not classified as Clothing and Home Decor. These complements can be useful in analyzing the distribution of products across different categories and identifying products that do not fall into any specific category.
## Summary
The complement of a set is a fundamental concept in set theory that refers to the elements that are not included in the set. It is denoted as the set of all elements outside of the given set. Understanding the complement of a set is crucial in various fields, including mathematics, computer science, and statistics. Some key takeaways from this article include:
• A set is a collection of distinct objects, denoted by curly braces.
• The complement of a set, denoted as A’, is the set of all elements that are not in A.
• The complement of a set possesses properties such as the identity property, union property, and intersection property.
• Real-world examples, such as student clubs and online shopping, can help illustrate the practical applications of the complement of a set.
By understanding the complement of a
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# Prove that the points A(1, 7), B (4, 2),
Question:
Prove that the points A(1, 7), B (4, 2), C(−1, −1) D (−4, 4) are the vertices of a square.
Solution:
The distance $d$ between two points $\left(x_{1}, y_{\mathrm{I}}\right)$ and $\left(x_{2}, y_{2}\right)$ is given by the formula
$d=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
In a square all the sides are equal in length. Also, the diagonals are equal in length in a square.
Here the four points are A(17), B(4, 2), C(1, −1) and D(4, 4).
First let us check if all the four sides are equal.
$A B=\sqrt{(1-4)^{2}+(7-2)^{2}}$
$=\sqrt{(-3)^{2}+(5)^{2}}$
$=\sqrt{9+25}$
$A B=\sqrt{34}$
$B C=\sqrt{(4+1)^{2}+(2+1)^{2}}$
$=\sqrt{(5)^{2}+(3)^{2}}$
$=\sqrt{25+9}$
$B C=\sqrt{34}$
$C D=\sqrt{(-1+4)^{2}+(-1-4)^{2}}$
$=\sqrt{(3)^{2}+(-5)^{2}}$
$=\sqrt{9+25}$
$C D=\sqrt{34}$
$A D=\sqrt{(1+4)^{2}+(7-4)^{2}}$
$=\sqrt{(5)^{2}+(3)^{2}}$
$=\sqrt{25+9}$
$A D=\sqrt{34}$
Since all the sides of the quadrilateral are the same it is a rhombus.
For the rhombus to be a square the diagonals also have to be equal to each other.
$A C=\sqrt{(1+1)^{2}+(7+1)^{2}}$
$=\sqrt{(2)^{2}+(8)^{2}}$
$=\sqrt{4+64}$
$A C=\sqrt{68}$
$B D=\sqrt{(4+4)^{2}+(2-4)^{2}}$
$=\sqrt{(8)^{2}+(-2)^{2}}$
$=\sqrt{64+4}$
$B D=\sqrt{68}$
Since the diagonals of the rhombus are also equal to each other the rhombus is a square.
Hence the quadrilateral formed by the given points is a. |
# Buoyant Force Formula with Solved Examples
In this tutorial, we will explore the buoyant force formula through a problem-solving approach.
You may have seen a ship or a boat floating on water. But what makes some objects float or sink in fluids?
Let’s try an experiment.
Attach a dense object to a spring scale and measure its weight. Then, immerse it in water and measure its weight again. You will notice that the scale reading in water is lower than that in air.
Why is that?
We can infer that there is an upward force acting on the object in the fluid (for example, water) that counteracts the downward gravitational force (weight) resulting in a lower weight, called apparent weight, in the fluid.
This is a common phenomenon that happens in all fluids when an object with a different density is submerged in it.
The force exerted by a fluid on an object that is completely or partially submerged in it is called buoyant force and its magnitude is equal to the weight of the fluid displaced by the object.
This upward force, which is called the buoyant force, is not a new type of force.
It arises from the pressure difference between the upper and lower surfaces of a submerged object in a fluid.
## Origin of the buoyant force formula:
Consider a body submerged (say, a cube) in a fluid. The fluid exerts forces on all sides of the cube.
The forces in the horizontal direction are equal and opposite, so they cancel out. (This is because they are at the same depth from the fluid surface, so the pressure is the same.)
The forces in the vertical direction are different because the top and bottom sides are at different depths of the fluid.
We know that the pressure increases with depth, so the force on the bottom side is larger than the force on the top side. This results in a net upward force, which is called the buoyant force.
## But how is the buoyancy force formula derived?
To find the buoyant force, let’s simplify the problem by considering a cube of height $h$ and base area $A$ that is completely submerged in a fluid with density $\rho_f$.
As shown in the figure, the fluid exerts vertical forces on the cube due to the water pressure at different depths. The force $F_1$ acts downward on the top surface of the cube, while the force $F_2$ acts upward on the bottom surface of the cube.
These are the only vertical forces that the fluid applies to the cube.
The horizontal forces cancel out since they are equal and opposite.
By adding the vertical forces, we get a net upward force, which is the buoyant force. We can derive the formula for the buoyant force as follows: \begin{align*} F_{net} &= F_2-F_1\\ \\&=P_2A-P_1A\\ \\&=(\rho_f gh_2)A-(\rho_g g h_1)A\\ \\&=\rho_f g(h_2-h_1)A\\ \\&=\rho_f V_{dis}g \end{align*} In the above derivation, we used the following facts: the pressure at a depth $h$ below the fluid surface is $P=P_0+\rho_f gh$ where $P_0$ is the surface pressure.
Therefore, the buoyant force formula is the product of the fluid density $\rho_f$, the displaced fluid volume $V_{dis}$, and the gravitational constant $g$. $F_b=\rho_f V_{dis} g$
We can also use the definition of density as the ratio of mass to volume, $\rho=\frac mV$, to find another useful equation for buoyant force as $F_b=m_f g$ where, in this version, $m_f$ is the mass of the fluid displaced by the object submerged in it.
Example (1): A solid cylinder with base area $A=20\,{\rm cm^2}$ is placed completely underwater as shown in the figure. Calculate the difference between the forces exerted by the water on the upper and lower bases of the cylinder. ($g=10\,{\rm m/s^2}$ and $\rho_{water}=1\,{\rm g/cm^3}$).
Solution: The water exerts two forces $F_1=P_1 A$ and $F_2=P_2 A$ on the upper and lower bases of the cylinder, respectively. The pressure at depth $h$ of a fluid with density $\rho$ is given by $P=\rho gh$.
The upper base is at depth $h_1=0.1\,{\rm m}$, so the pressure at that depth is $P_1=1000\times 10\times 0.1=1000\,{\rm Pa}$.
The lower base is at depth $h_2=0.5\,{\rm m}$, so the pressure at that depth is $P_2=1000\times 10\times 0.5=5000\,{\rm Pa}$.
The difference in forces is the upward buoyant force exerted on the cylinder by the water. $\Delta F=F_2-F_1=(P_2-P_1)A$, so we have \begin{align*}\Delta F&=F_2-F_1=(P_2-P_1)A\\&=(5000-1000)20\times(0.01)^2 \\&=8\quad {\rm N}\end{align*}
We can also find it using the buoyant force equation as $F_b=\rho_{water}V_{body}g$, where $V_{body}$ is the volume of the cylinder submerged in the water (here, the whole cylinder is submerged). \begin{align*}F_b&=\rho_{water}V_{body}g\\ &=(1000)(40\times 20)\times 10^{-5}\times 10\\&=8\quad {\rm N}\end{align*} The volume of the cylinder is the product of the base area ($A=20\,{\rm cm^2}$) and the height ($h=40\,{\rm cm}$).
The buoyant force formula was derived for a simple shape, a cube, but it can be shown that it is valid for any shape.
## Key points about the buoyant force formula:
• The mass in the formula is the mass of the fluid that is displaced by the object, not the mass of the object itself.
• The formula works for any shape, because any shape can be broken down into many tiny cubes, and the formula applies to each cube.
• The buoyant force only depends on the density of the fluid $\rho_f$, not the density of the object. It does not matter what material the object is made of, whether it is wood or iron. The buoyant force is the same as long as the object displaces the same amount of fluid.
The example below shows what is meant by displaced volume in the buoyancy force equation.
Example (2): A rectangular container of ethanol with density $\rho_{eth}=800\,{\rm kg/m^3}$ floats in the liquid of density $\rho_{f}=1200\,{\rm kg/m^3}$. Height and base of the container are $H=6\,{\rm cm}$ and $A=20\,{\rm cm^2}$, respectively. $h=4\,{\rm cm}$ of its height is in the fluid. Find the buoyancy force exerted on the container.
Solution: The buoyant force is the weight of the fluid that is displaced by an object submerged in it. We can write it as $F_b=m_f g$ or $F_b=\rho_f V_{dis}g$, where $m_f$ is the mass of the displaced fluid, and $V_{dis}$ is the volume of the object under the fluid surface.
In this problem, the volume of the object under the fluid surface is $V_{dis}=A\times h$, where $A$ is the base area and $h$ is the height of the object. Substituting this into the buoyant force formula, we get \begin{align*}F_b&=\rho_{f}V_{dis}g\\&=\rho_f \times (Ah)g\\&=(1200)(20\times 4)\times 10^{-5}\times 10\\&=9.6\quad {\rm N}\end{align*}
This is the buoyant force that acts upward on the object.
The buoyant force is based on Archimedes' principle, which states that:
Any object that is wholly or partially submerged in a fluid experiences a buoyant force equal to the weight of the fluid displaced by the object.
Example(3): A piece of wood with density $\rho_{wood}=0.6\,{\rm g/cm^3}$ floats on the surface of the water. What is the ratio of the portion of the wood submerged in the water to that of outside the water? ($\rho_{water}=1\,{\rm g/cm^3}$)
Solution: The wood floats on water because its density is lower than that of water. The weight of the wood is balanced by the upward buoyant force exerted by the water. Using the buoyant force formula and the weight formula, we get: \begin{align*} \text{buoyant force}&=\text{weight}\\ \\ \rho_{fluid}V_{dis}g&=\rho_{body}V_{body}g\\ \\ \Rightarrow \frac{V_{dis}}{V_{body}}&=\frac{\rho_{body}}{\rho_{fluid}}\\ \\ &=\frac{0.6}{1}\\ \\ \Rightarrow V_{fluid}&=0.6\, V_{body}\end{align*} Here, $V_{dis}$ is the volume of the water displaced by the wood, and $V_{body}$ is the total volume of the wood.
To find the volume of the wood that is above the water, we can use the following equation: \begin{align*} V_{wood}&=V_{wood-in}+V_{wood-out}\\ \\V&=0.6V+V_{wood-out} \\ \\\Rightarrow& V_{wood-out}=0.4V\end{align*} Here, $V_{out}$ is the volume of the wood that is outside the water.
Therefore, the volume of the wood that is inside the water is $V_{wood-in}=0.6V$ and the volume of the wood that is outside the water is $V_{wood-out}=0.4V$. Their ratio is $\frac 32$.
The buoyancy formula can be used to find the density of a fluid if we know the density of an object that floats in it. For example, consider the following problem:
Example (4): A piece of wood floats in water so that two-thirds of its volume $V$ is beneath the water level. When the same piece of wood is placed in oil, the submerged volume is $0.8V$. Find the density of (a) the wood and (b) the oil.
Solution: The object floats on the fluids (water and oil) so the downward gravitation force and upward buoyant force must balance each other.
Recall that the buoyancy force formula is the fluid density times the volume of the part of the object beneath the fluid level times the gravitational constant, i.e., $F_b=\rho_f V_{dis}g$.
In the case of water, the volume of the fluid displaced by the wood is, i.e., $V_{dis}=2/3 V$, since two-thirds of the wood's volume is submerged. Therefore, equating the weight and the buoyant force, we get: \begin{align*} F_b&=mg\\ \rho_{water}V_{dis}g&=\rho_{wood}Vg\\ (1000)(2/3 V)&=\rho_{wood}\times V\\ \Rightarrow \rho_{wood}&=\frac{2000}{3}\quad{\rm \frac{kg}{m^3}}\end{align*}
In the case of oil, the volume of the fluid displaced by the wood is $V_{dis}=0.8 V$, since $80\%$ of the wood's volume is submerged. Using the same logic, we get: \begin{align*} F_b&=mg\\ \rho_{oil}V_{dis}g&=\rho_{wood}Vg\\ \\\rho_{oil}(0.8 V)&=\underbrace{\rho_{wood}}_{2000/3}\times V\\ \\ \Rightarrow \rho_{oil}&=834 \quad{\rm \frac{kg}{m^3}}\end{align*} Therefore, the density of the wood is $\frac{2000}{3}\,\rm kg/m^3$ and the density of the oil is $834\,\rm kg/m^3$.
Another application of the buoyant force equation is to find the apparent weight of objects in fluids. Consider the following example:
Example (5): An iron object has a density of $7.8\,{\rm g/cm^3}$ and appears 200 N lighter in water than in air.
(a) What is the volume of the object?
(b) How much does it weigh in the air?
Solution: Since the body has become lighter in water, there must be an upward force acting on the object, which cancels some downward weight force. In fluids, this force is called floating or buoyancy force.
Note: 200 N is the buoyant force that acts on the iron object when it is submerged in water. It is the force that makes the object appear lighter in water than in air. You can find this force by using Archimedes’ principle.
(a) According to Archimedes’ principle, the buoyant force is equal to the weight of the fluid displaced by the object. Therefore, we can use the following equation to find the volume of the object: \begin{align*} F_b &= \rho_{water} \times V_{object} \times g \\ \\ 200&=1000\times V_{object}\times 10 \\ \\ \Rightarrow V_{object}&=\frac{2}{100}\quad {\rm m^3}\end{align*} Note that $V_{object}$ is volume of the water displaced by the iron or explicitly, part of the iron underwater (All of it).
(b) The weight of the object in air is given by the product of its mass and gravity. We can find its mass by multiplying its density and volume: $W=\rho Vg=(7800)\left(\frac{2}{100}\right)(10)=1560\,{\rm N}$ where, here, $V$ is the actual volume of the object.
Example (6): An object with a mass of 150 kg is thrown into the water and displaces a volume of 100 liters of water. Will the object sink or float?
Solution: To see what happens after throwing, we need to calculate the forces acting on the object in the water: the upward buoyant force and the downward weight.
We know that the buoyant force is equal to the weight of the fluid displaced by the object. Here, the volume displaced by the object is $100$ liters, which we need to convert to SI units of volume, $m^3$, as $1\,{\rm L}=10^{-3}\,m^3$.
Using the buoyant force formula, we get \begin{align*} F_b&=\rho_{water}\times V_{dis}\times g\\&=(1000)(100\times 10^{-3})(10)\\&=1000\quad {\rm N}\end{align*} The weight of the object, directed downward, is given by $W=mg=1500\,{\rm N}$. Applying Newton’s second law, we find the net force on the object: $F_{net}=1000\uparrow-1500\downarrow=-500\downarrow\,{\rm N}$ Since weight is greater than the upward buoyant force, the object will sink.
Example (7): A raft has a volume of 6 cubic meters and two-thirds of its volume is submerged in seawater with a density of $1025\,{\rm kg/m^3}$. What is the buoyant force on the raft?
Solution: We can use the buoyant force equation $F_B=\rho V_{dis}g$, where $V_{dis}$ is the amount of the volume of the object under the fluid's level and $\rho$ is the density of the surrounding fluid. \begin{align*} F_B&=\rho V_{dis} g\\ &=(1025)\left(\frac {2}{3} 6\right)(9.8)\\&=40180\quad {\rm N}\end{align*}
## Summary:
This article explained how to use the buoyant force equation, which states that the buoyant force is equal to the weight of the fluid displaced by the object.
The article also provided examples of how to apply the equation to different situations, such as finding the volume, weight, or net force of an object in a fluid.
The article used a problem-solving approach to illustrate the steps and calculations involved in each example.
Author: Ali Nemati
Page Created: 1/31/2021
Last Update: Feb 2, 2024 |
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# ISI MStat PSB 2006 Problem 6 | Counting Principle & Expectations
This is a very beautiful sample problem from ISI MStat PSB 2006 Problem 6 based on counting principle . Let's give it a try !!
## Problem- ISI MStat PSB 2006 Problem 6
Let $$Y_{1}, Y_{2}, Y_{3}$$ be i.i.d. continuous random variables. For i=1,2, define $$U_{i}$$ as
$$U_{i}= \begin{cases} 1 , & Y_{i+1}>Y_{i} \\ 0 , & \text{otherwise} \end{cases}$$
Find the mean and variance of $$U_{1}+U_{2}$$ .
### Prerequisites
Basic Counting Principle
Probability
Continuous random variable
## Solution :
$$E(U_1+U_2)=E(U_1)+E(U_2)$$
Now $$E(U_1)=1 \times P(Y_2>Y_1) = \frac{1}{2}$$, as there are only two cases either $$Y_2>Y_1$$ or $$Y_2<Y_1$$.
Similarly , $$E(U_2)= \frac{1}{2}$$
So, $$E(U_1+U_2)= 1$$
$$Var(U_1+U_2)=Var(U_1)+Var(U_2)+2Cov(U_1,U_2)$$ .
$$Var(U_1)=E({U_1}^2)-{E(U_1)}^2=1^2 \times P(Y_2>Y_1) - {\frac{1}{2}}^2 = \frac{1}{2}-\frac{1}{4}=\frac{1}{4}$$
Similarly ,$$Var(U_2)=\frac{1}{4}$$
$$Cov(U_1,U_2)=E(U_1 U_2)-E(U_1)E(U_2)=1 \times P(Y_3>Y_2>Y_1) - {\frac{1}{2}}^2 = \frac{1}{3!}-{\frac{1}{2}}^2$$ ( as there are 3! possible arrangements of $$Y_i's$$ keeping inequalities fixed .
Therefore , $$Var(U_1+U_2) = 2 \times \frac{1}{4} + 2 \times (\frac{1}{6}- \frac{1}{4}) = \frac{1}{3}$$
## Food For Thought
Find the same under the condition that $$Y_i's$$ are iid poission random variables . |
How to Find Tangent Lines
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A tangent line to a curve touches the curve at only one point, and its slope is equal to the slope of the curve at that point. You can estimate the tangent line using a kind of guess-and-check method, but the most straightforward way to find it is through calculus. The derivative of a function gives you its slope at any point, so by taking the derivative of the function that describes your curve, you can find the slope of the tangent line then solve for the other constant to get your answer.
Things You'll Need
• Pencil
• Paper
• Calculator
• Write down the function for the curve whose tangent line you need to find. Determine at which point you want to take the tangent line (e.g., x = 1).
• Take the derivative of the function using the derivative rules. There are too many to summarize here; you can find a list of the rules of derivation under the Resources section, however, in case you need a refresher:
Example: If the function is f(x) = 6x^3 + 10x^2 - 2x + 12, the derivative would be as follows:
f'(x) = 18x^2 + 20x - 2
Note that we represent the derivative of the original function by adding the ' mark, so that f'(x) is the derivative of f(x).
• Plug the x-value for which you need the tangent line into f'(x) and calculate what f'(x) will be at that point.
Example: If f'(x) is 18x^2 + 20x - 2 and you need the derivative at the point where x = 0, then you would plug 0 into this equation in place of x to obtain the following:
f'(0) = 18 (0)^2 + 20(0) - 2
so f'(0) = -2.
• Write out an equation of the form y = mx + b. This will be your tangent line. m is the slope of your tangent line and it's equal to your result from step 3. You don't know b yet, however, and will need to solve for it. Continuing the example, your initial equation based on step 3 would be y = -2x + b.
• Plug the x-value you used to find the slope of the tangent line back into your original equation, f(x). This way, you can determine the y-value of your original equation at this point, then use it to solve for b in your tangent line equation.
Example: If x is 0, and f(x) = 6x^3 + 10x^2 - 2x + 12, then f(0) = 6(0)^3 + 10(0)^2 - 2(0) + 12. All terms in this equation go to 0 except for the last one, so f(0) = 12.
• Substitute the result from step 5 for y in your tangent line equation, then substitute the x-value you used in step 5 for x in your tangent line equation and solve for b.
Example: You know from a prior step that y = -2x + b. If y = 12 when x = 0, then 12 = -2(0) + b. The only possible value for b that will give a valid result is 12, therefore b = 12.
• Write out your tangent line equation, using the m and b values you have found.
Example: You know m = -2 and b = 12, so y = -2x + 12.
References
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## Maria M
### Teaching Style
I love tutoring and consider myself to be effective at it. I approach it with enthusiasm. I apply theory to practical applications in my career and life. My approach is one of ease. I consider mathematics to be easy, as long as you accept what I call "the rules of the game"; i.e., there are certain principles that have to be accepted, and not questioned, then everything else falls in place. I believe that all people have the capability of learning, and I love the opportunity to provide a positive experience to students.
### Experience Summary
I have been tutoring for more than 5 years. My main focus is mathematics, but I also tutor students in Spanish I through IV. I have tutored Spanish, Algebra I, Algebra II, Geometry, Trigonometry, and Pre-Calculus. I have recently helped several students prepare for the SAT math test. I enjoy making a difference in the student’s life.
### Credentials
Type Subject Issued-By Level Year
Other Spanish Native speaker Fluent Current
Certification Project Management Studies Project Management Institute PMP Certification 2006
Degree Civil Engineering California State University at Long Beach MSCE 1987
Other Mathematics El Camino College AA 1981
## Justin J
### Teaching Style
I am a very enthusiastic tutor and, as I stated in the previous section, believe teaching in such a way that the student gains a true mastery of the given subject. In a sense, I believe in each student understanding a particular concept in their own unique way that is consistent with their unique thinking processes. I strive to do this by relating a given concept analogously to something the student already fundamentally understands. I also create new problems to solve that probe the students’ progress and comprehension. I try to make these problems as realistic as possible to make them interesting to the student.
### Experience Summary
Being a Ph.D. candidate and appointed research assistant, I very much realize the importance of true comprehension of a subject. Not only do I hold two B.S. degrees in Applied Mathematics and Chemical Engineering but I also carry credits to the equivalent of a Non-Thesis Masters in Applied Mathematics and am one course shy of a minor in Chemistry. I have tutored groups of students for Sylvan Learning Centers, conducted one-on-one sessions with Educational Enrichment and independently, and prepared and delivered lectures for advanced undergraduate chemical engineering courses. I am capable and experienced in tutoring all levels of mathematics, chemistry, chemical engineering, physics and materials science.
### Credentials
Type Subject Issued-By Level Year
Degree Chemical Engineering University of Florida Ph.D.--in progress 2010
Degree Chemical Engineering NC State University B.S. with Honors 2005
Degree Applied Mathematics NC State University B.S. with Honors 2005
## Janet J
### Teaching Style
I enjoy teaching; it is my best talent. I understand that math is not a "favorite subject" to a lot of students. Working one on one in a tutoring situation gets past the negative block students have in a math classroom. Several of my private students have worked with me for several years. Their parents said that they felt more confident in the class room and in testing situations because of their work with me. In addition to tutoring in traditional school math subjects, I have also worked with several students on PSAT and SAT preparation. All of these students achieved a higher score on these tests after working with me one on one.
### Experience Summary
I taught 34 years with the Guilford County School System, in Greensboro, NC. I retired in June 2007, and was asked to fill 2 interim positions in the Spring and Fall semesters of 2008. I taught all courses from general math through Honors Pre-calculus. I have been privately tutoring at home for the last 7 or 8 years, to supplement my income.
### Credentials
Type Subject Issued-By Level Year
Degree Secondary Mathematics University of NC-Greensboro BS 1972
## Robert R
### Teaching Style
I’ve always been interested in the application of math and science to the solution of real world problems. This led me to a very satisfying career in engineering. Therefore my approach to teaching is very application oriented. I like to relate the subject to problems that the students will encounter in real life situations. I've generally only worked with older students; high school or college age or older mature adults who have returned to school to get advance training or learn a new trade.
### Experience Summary
I’ve always been interested in math and science; especially in their application to solving real world problems. This led me to a very satisfying career in engineering. I have a BS in electrical engineering from General Motors Institute (now Kettering University) and an MS in electrical engineering from Marquette University. I am a registered professional engineer in Illinois. I have over 30 years of experience in the application, development, and sales/field support of electrical/electronic controls for industrial, aerospace, and automotive applications. I’m currently doing consulting work at Hamilton-Sundstrand, Delta Power Company, and MTE Hydraulics in Rockford. I also have college teaching and industrial training experience. I have taught several courses at Rock Valley College in Electronic Technology, mathematics, and in the Continuing Education area. I’ve done industrial technical training for Sundstrand, Barber Colman, and others. I’ve also taught math courses at Rasmussen College and Ellis College (online course). I’ve also been certified as an adjunct instructor for Embry-Riddle Aeronautical University for math and physics courses. I've tutored my own sons in home study programs. I'm currently tutoring a home schooled student in math using Saxon Math. I hope to do more teaching/tutoring in the future as I transition into retirement.
### Credentials
Type Subject Issued-By Level Year
Degree Electrical Engineering Marquette University MS 1971
Degree Electrical Engineering GMI (Kettereing University) BS 1971
## Boris B
### Teaching Style
I believe that excellence in teaching comes from the teacher's adaptation to student's specific needs. In particular, for some students, visualization may be the essential component for them to understand a certain concept or an idea in mathematics, and by furnishing examples and various proofs with pictures enables the student to learn the concept, whereas for other students it may be the algebraic equation that allows them to see a certain idea. In the first couple of sessions, I probe for the specific needs of the student and then am able to connect with that student so that he/she feels comfortable with the subject. I am a patient teacher and believe that all students are able to grasp the subject. I teach in a disciplined manner, so that the topic presented is coherent and follows a logical flow. I make sure that the theoretical concepts are internalized in a concrete example for the student. Above all, I carry a positive disposition wherever I go and encourage students to enjoy math.
### Experience Summary
I am a graduate of GaTech, with a Math and Psychology BA degrees. Aside from my two majors I have minors in Philosophy and Cognitive Science. During my years of high school and college, I have tutored students in Mathematics -- be it in calculus or statistics, or the math portion of the SAT. At the moment I work part-time at the Korean after school program, called Daekyo America Inc., as a math instructor for both high school and middle school. I have participated in various Mathematical Competitions, and have won numerous awards, including the Grand Prize Winner in USAMTS (United States Mathematical Talent Search).
### Credentials
Type Subject Issued-By Level Year
Degree Applied Mathematics GaTech Bachelors 2004
Degree Psychology GaTech BA 2004
## Timothy T
### Teaching Style
When I tutor a student, I seek first to understand the student and how he/she thinks. I find it is very important to have good rapport and communication with the student so I understand how he/she views the subject and the difficulties of it. Next I try to make "conceptual bridges" from what they know to what they are having difficulty understanding. This process usually teaches me about seeing the subject from a new point of view. I try to achieve a fine balance between guiding and directing the student’s thoughts on the topic with following the student in their own line of thinking of the subject. The student needs to learn to have confidence in his own thoughts on the subject and in his own ability to master it.
### Experience Summary
Over the past four years, I have tutored high school and middle school students in math, algebra, calculus, chemistry, SAT Math, and general study skills. My preference is to tutor math, algebra, calculus, physics, physical science, chemistry, and programming. Math is the subject for which I have the greatest passion. I also participate in the homeschooling of four of my children (13, 11, 8, 6). I have mentored my 13 yr old son in Algebra I & II, Chemistry, Elementary Math, and Middle-school Physical Science, and taught elementary math to my 11, 8, and 6 year olds. Additionally, I read and review history lessons to my kids. I completed my MS in Electrical Engineering in 2006 from The University of Texas at Arlington and my BS in Electrical Engineering and a BA in Philosophy from Rice University. I have recent experience as a student having completed Cellular Biology II at St. Petersburg College in Fall 2011.
### Credentials
Type Subject Issued-By Level Year
Degree Electrical Engineering Univ. of Texas - Arlington Masters 2006
Certification Design for Six Sigma Honeywell International DFSS - Green Belt 2003
Degree Electrical Engineering Rice University BSEE 1989
Degree Philosophy Rice University BA 1989
## Vinod V
### Teaching Style
The cornerstone of my teaching philosophy and personal teaching goals is to help students develop their own thinking skills. I believe all students should leave the school armed with the ability to think for them selves, to think critically and to think creatively. Understanding how people learn is one of the significant aspects of teaching. This is linked to their “knowledge” background and maturity. The key to teaching is to relate to the audience by starting from what they know and building upon it. As a teacher I am totally involved with the class, dedicated to my students and 100% prepared to devote time and energy for their intellectual growth. Love for teaching evokes passion and dedication within me. I believe that the enthusiasm of a motivated teacher rubs off on his/her students, who derive the inspiration and encouragement which actuates their desire to learn. A good teacher should have sound fundamentals and command over the concepts. Fundamentals are the foundation intrinsic for mastering the subject; only teachers who are strong in fundamentals will be able to pass it on to their students. I believe that my strong command over the fundamentals will rub off on my students. I believe that the role of a teacher is that of a leader where you have to show the path, motivate, encourage, and lead by example. In short, my success lies in seeing my students succeed.
### Experience Summary
My enthusiasm and love for education can be gauged from the fact that I pursued three Masters degrees in three distinct but related fields. One cannot pursue engineering as a profession without having an affinity for Math and Analysis. Math was a passion for me from my young days and still very much remains so. I have a thorough knowledge and understanding of math. Right from my school days I was involved and loved to teach math. I invariably obtained A+ scores in whatever math test I took in my lifetime. For instance my GRE math score was above 95% of test takers' scores. I have taught Middle school, High school and under-graduate students in Algebra, Geometry, Trigonometry, Quadratic Equations, Applied Probability and Calculus.
### Credentials
Type Subject Issued-By Level Year
Degree City Planning Kansas State University MRCP 2002
Degree Engineering Anna University ME 2000
Degree Civil Engineering Institution of Engineers BE 1994
## Farrah F
### Teaching Style
I enjoy teaching and working with students. I now understand that all children learn on their own level and at their own pace. I am very easy going, but I insist that students do their homework and stay after school for extra help if they need it. I grade work off of understanding. It is not always finding the correct answer, but the process to get to the correct answer. I believe that a good teacher makes a good student and I strive to be that good teacher and student.
### Experience Summary
I enjoyed tutoring while I was in college working in the math lab for three years. However, now that I have been actually teaching my very own students I have a better understanding of the basics of mathematics and I am a much better student myself.
### Credentials
Type Subject Issued-By Level Year
Certification Mathematics 6-12 Florida Department of Education Leon County 2005
Degree Mathematics Florida A&M University BS 2004
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Ft. Lauderdale
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#### Please solve RD Sharma class 12 chapter Algebra of matrices exercise 4.3 question 31 maths textbook solution
Answer: Hence, proved $A^{3}-4 A^{2}+A=0$
Hint: Matrix multiplication is only possible, when the number of columns of first matrix is equal to the number of rows of second matrix.
Given: $A=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$
Prove: $A^{3}-4 A^{2}+A=0$
Consider,$A^{2}=A A$
$A^{2}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}4+3 & 6+6 \\ 2+2 & 3+4\end{array}\right]=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right] \\\\ A^{3}=A^{2} A\\\\ A^{2} A=\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]\\\\ A^{3}=\left[\begin{array}{cc}14+12 & 21+24 \\ 8+7 & 12+14\end{array}\right]=\left[\begin{array}{ll}26 & 45 \\ 15 & 26\end{array}\right]$
Now putting value of $A^{3}, A^{2}$ and $A$ in the equation $A^{3}-4 A^{2}+A$we get,
$=\left[\begin{array}{cc}26 & 45 \\ 15 & 26\end{array}\right]-4\left[\begin{array}{cc}7 & 12 \\ 4 & 7\end{array}\right]+\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]$
$=\left[\begin{array}{ll} 26 & 45 \\ 15 & 26 \end{array}\right]-\left[\begin{array}{ll} 28 & 48 \\ 16 & 28 \end{array}\right]+\left[\begin{array}{ll} 2 & 3 \\ 1 & 2 \end{array}\right] \\ =\left[\begin{array}{ll} 0 & 0 \\ 0 & 0 \end{array}\right] \\\\ =0 \\ \text { So, } A^{3}-4 A^{2}+A=0$
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14
Dec 21
## Sum of random variables and convolution
### Link between double and iterated integrals
Why do we need this link? For simplicity consider the rectangle $A=\left\{ a\leq x\leq b,c\leq y\leq d\right\} .$ The integrals
$I_{1}=\underset{A}{\int \int }f(x,y)dydx$
and
$I_{2}=\int_{a}^{b}\left( \int_{c}^{d}f(x,y)dy\right) dx$
both are taken over the rectangle $A$ but they are not the same. $I_{1}$ is a double (two-dimensional) integral, meaning that its definition uses elementary areas, while $I_{2}$ is an iterated integral, where each of the one-dimensional integrals uses elementary segments. To make sense of this, you need to consult an advanced text in calculus. The difference notwithstanding, in good cases their values are the same. Putting aside the question of what is a "good case", we concentrate on geometry: how a double integral can be expressed as an iterated integral.
It is enough to understand the idea in case of an oval $A$ on the plane. Let $y=l\left( x\right)$ be the function that describes the lower boundary of the oval and let $y=u\left( x\right)$ be the function that describes the upper part. Further, let the vertical lines $x=m$ and $x=M$ be the minimum and maximum values of $x$ in the oval (see Chart 1).
Chart 1. The boundary of the oval above the green line is described by u(x) and below - by l(x)
We can paint the oval with strokes along red lines from $y=l\left( x\right)$ to $y=u\left(x\right) .$ If we do this for all $x\in \left[ m,M\right] ,$ we'll have painted the whole oval. This corresponds to the representation of $A$ as the union of segments $\left\{ y:l\left( x\right) \leq y\leq u\left( x\right) \right\}$ with $x\in \left[ m,M\right] :$
$A=\bigcup\limits_{m\leq x\leq M}\left\{ y:l\left( x\right) \leq y\leq u\left( x\right) \right\}$
and to the equality of integrals
(double integral)$\underset{A}{\int \int }f(s,t)dsdt=\int_{m}^{M}\left(\int_{l\left( x\right) }^{u(x)}f(x,y)dy\right) dx$ (iterated integral)
### Density of a sum of two variables
Assumption 1 Suppose the random vector $\left( X,Y\right)$ has a density $f_{X,Y}$ and define $Z=X+Y$ (unlike the convolution theorem below, here $X,Y$ don't have to be independent).
From the definitions of the distribution function $F_{Z}\left( z\right)=P\left( Z\leq z\right)$ and probability
$P\left( A\right) =\underset{A}{\int \int }f_{X,Y}(x,y)dxdy$
we have
$F_{Z}\left( z\right) =P\left( Z\leq z\right) =P\left( X+Y\leq z\right) =\underset{x+y\leq z}{\int \int }f_{X,Y}(x,y)dxdy.$
The integral on the right is a double integral. The painting analogy (see Chart 2)
Chart 2. Integration for sum of two variables
suggests that
$\left\{ (x,y)\in R^{2}:x+y\leq z\right\} =\bigcup\limits_{-\infty
Hence,
$\int_{-\infty }^{z}f_{Z}\left( z\right) dz=F_{Z}\left( z\right)=\int_{R}\left( \int_{-\infty }^{z-x}f_{X,Y}(x,y)dy\right) dx.$
Differentiating both sides with respect to $z$ we get
$f_{Z}\left( z\right) =\int_{R}f_{X,Y}(x,z-x)dx.$
If we start with the inner integral that is with respect to $x$ and the outer integral $-$ with respect to $y,$ then similarly
$f_{Z}\left( z\right) =\int_{R}f_{X,Y}(z-y,y)dy.$
Exercise. Suppose the random vector $\left( X,Y\right)$ has a density $f_{X,Y}$ and define $Z=X-Y.$ Find $f_{Z}.$ Hint: review my post on Leibniz integral rule.
### Convolution theorem
In addition to Assumption 1, let $X,Y$ be independent. Then $f_{X,Y}(x,y)=f_{X}(x)f_{Y}\left( y\right)$ and the above formula gives
$f_{Z}\left( z\right) =\int_{R}f_{X}(x)f_{Y}\left( z-x\right) dx.$
This is denoted as $\left( f_{X}\ast f_{Y}\right) \left( z\right)$ and called a convolution.
The following may help to understand this formula. The function $g(x)=f_{Y}\left( -x\right)$ is a density (it is non-negative and integrates to 1). Its graph is a mirror image of that of $f_{Y}$ with respect to the vertical axis. The function $h_{z}(x)=f_{Y}\left( z-x\right)$ is a shift of $g$ by $z$ along the horizontal axis. For fixed $z,$ it is also a density. Thus in the definition of convolution we integrate the product of two densities $f_{X}(x)f_{Y}\left( z-x\right) .$ Further, to understand the asymptotic behavior of $\left( f_{X}\ast f_{Y}\right) \left( z\right)$ when $\left\vert z\right\vert \rightarrow \infty$ imagine two bell-shaped densities $f_{X}(x)$ and $f_{Y}\left( z-x\right) .$ When $z$ goes to, say, infinity, the humps of those densities are spread apart more and more. The hump of one of them gets multiplied by small values of the other. That's why $\left(f_{X}\ast f_{Y}\right) \left( z\right)$ goes to zero, in a certain sense.
The convolution of two densities is always a density because it is non-negative and integrates to one:
$\int_{R}f_{Z}\left( z\right) dz=\int_{R}\left( \int_{R}f_{X}(x)f_{Y}\left(z-x\right) dx\right) dz=\int_{R}f_{X}(x)\left( \int_{R}f_{Y}\left(z-x\right) dz\right) dx$
Replacing $z-x=y$ in the inner integral we see that this is
$\int_{R}f_{X}(x)dx\int_{R}f_{Y}\left( y\right) dy=1.$
1
May 18
## Density of a sum of independent variables is given by convolution
This topic is pretty complex because it involves properties of integrals that economists usually don't study. I provide this result to be able to solve one of UoL problems.
## General relationship between densities
Let $X,Y$ be two independent variables with densities $f_X,f_Y$. Denote $f_{X,Y}$ the joint density of the pair $(X,Y)$.
By independence we have
(1) $f_{X,Y}(x,y)=f_X(x)f_Y(y).$
Let $Z=X+Y$ be the sum and let $f_Z,\ F_Z$ be its density and distribution function, respectively. Then
(2) $f_Z(z)=\frac{d}{dz}F_Z(z)$.
These are the only simple facts in this derivation. By definition,
(3) $F_Z(z)=P(Z\le z)=P(X+Y\le z)$.
For the last probability in (3) we have a double integral
$P(X+Y\le z)=\int\int_{x+y\le z}f_{X,Y}(x,y)dxdy$.
Using (1), we replace the joint probability by the product of individual probabilities and the double integral by the repeated one:
(4) $P(X+Y\le z)=\int\int_{x+y\le z}f_X(x)f_Y(y)dxdy=\int_R\int_{-\infty}^{z-x}f_X(x)f_Y(y)dxdy$
$=\int_Rf_X(x)\left(\int_{-\infty}^{z-x}f_Y(y)dy\right)dx$.
The geometry is explained in Figure 1. The area $x+y\le z$ is limited by the line $y=z-x$. In the repeated integral, we integrate first over red lines from $-\infty$ to $z-x$ and then in the outer integral over all $x\in R$.
Figure 1. Area of integration
(3) and (4) imply
$F_Z(z)=\int_Rf_X(x)\left(\int_{-\infty}^{z-x}f_Y(y)dy\right)dx$.
Finally, using (2) we differentiate both sides to get
(5) $f_Z(z)=\int_Rf_X(x)f_Y(z-x)dx$.
This is the result. The integral on the right is called a convolution of functions $f_X,f_Y$.
Remark. Existence of density (2) follows from existence of $f_X,f_Y$, although we don't prove this fact.
Exercise. Convolution is usually denoted by $(f*g)(z)=\int_Rf(x)g(z-x)dx$. Prove that
1. $(f*g)(z)=(g*f)(z)$.
2. $\int_R|(f*g)(z)|dz\le \int_R|f(x)|dx\int_R|g(x)|dx$.
3. If $X$ is uniformly distributed on some segment, then $(f_X*f_X)(z)$ is zero for large $z$. |
Lecture 10 - Consequences of flt
# Summary
We started today by finishing off the proof of Fermat's Little Theorem, then proceeded to use this theorem to give various properties which make life modulo p nice. In particular, we saw a test which can be used to prove a number is composite without actually factoring the number! We also saw that there are certain composite numbers which behave much like prime numbers do.
# The Proof of flt
To prove flt, last class period we picked up a number a satisfying $(a,p) = 1$, then explained why the sets
(1)
\begin{align} \begin{split} \{1,2,\cdots,p-1\}\\ \{a,2a,\cdots,(p-1)a\} \end{split} \end{align}
are the same modulo p. This came in two parts: first showing the second set consisted of distinct elements mod p, and then showing that the elements of the second set were all relatively prime to p. You can go back to the notes from last class to see how we proved these results.
With this equality of sets (modulo p) established, we now notice that the products of all elements in both sets must be the same
(2)
\begin{align} 1\cdot 2 \cdot \cdots \cdot (p-1) \equiv a \cdot 2a \cdot \cdots \cdot (p-1)a \mod{p}. \end{align}
Now the left-hand side is just $(p-1)! \equiv -1 \mod{p}$ (using Wilson's Theorem), whereas the right hand side is just [{$a^{p-1}(p-1)! \equiv -a^{p-1} \mod{p}$]]. Hence we get
(3)
\begin{align} -1 \equiv -a^{p-1} \mod{p} \end{align}
or, equivalently, $a^{p-1} \equiv 1 \mod{p}$. $\square$
# Some Useful Corollaries
There are several nice things that can be taken away from flt. The first is an extension of flt to all bases, not just those relatively prime to p.
Corollary: If p is prime, then for any integer a we have $a^p \equiv a \mod{p}$.
Proof: Let a be an integer. If $(a,p) = 1$ then flt says that $a^{p-1} \equiv 1 \mod{p}$. Multiplying this equation on both sides by a, we have $a^p \equiv a \mod{p}$.
If, on the other hand, we have $(a,p) \neq 1$, then we must have $p \mid a$ (note: we need p to be prime in order to make this conclusion. Why?). Therefore we get $a \equiv 0 \mod{p}$. But since a and 0 are congruent mod p, we must then have $a^p \equiv 0^p \equiv 0 \mod{p}$ as well. Hence in this case, we again have $a^p \equiv a \mod{p}$.
$\square$
Another useful application let's us compute large exponentials very easily
Corollary: If p is prime and $p \nmid a$, and if $b \equiv c \mod{p-1}$, then $a^b \equiv a^c \mod{p}$.
Proof: To see that this is true, notice that $b \equiv c \mod{p-1}$ means there is some $k \in \mathbb{Z}$ so that $b = c + (p-1)k$. Hence we have
(4)
\begin{align} a^b \equiv a^{c+(p-1)k} \equiv a^c \left(a^{p-1})^{k} \equiv a^c \cdot 1^k \equiv a^c \mod{p}. \end{align}
$\square$
#### Example: Reducing Exponents
Suppose someone asked you to compute $1376427^{1376427} \mod{11}$. We start by reducing the base modulo 11, which is easy to do because we have the "alternating sum" trick:
(5)
\begin{align} 1376427 \equiv 7 - 2 + 4 - 6 + 7 - 3 + 1 \equiv 8 \mod{11}. \end{align}
Hence
(6)
\begin{align} 1376427^{1376427} \equiv 8^{1376427} \mod{11}. \end{align}
This is certainly easier to compute, but the exponent we're dealing with is ridiculously huge. To improve this situation, the previous corollary let's us instead replace the exponent with any number congruence to 1376427 mod 10 (note: we need that $(8,11) = 1$ if we want to use this exponent-reduction trick). We'll choose 7, since that's easiest.
(7)
\begin{align} 8^{1376427} \equiv 8^7 \mod{11}. \end{align}
Now to compute this last number, I just calculate successive squares of 8. In fact, to make life easier I'll compute successive squares of -3, since $-3 \equiv 8 \mod{11}$:
(8)
\begin{split} (-3)^2 &\equiv 9 \equiv -2 \mod{11}\\ (-3)^4 &\equiv (-2)^2 \equiv 4 \mod{11}. \end{split}
Hence we have
(9)
\begin{align} 8^7 \equiv (-3)^7 \equiv (-3)^4(-3)^2(-3) \equiv 4 \cdot -2 \cdot (-3) \equiv 2 \mod{11} \end{align}
$\square$
There is also a corollary to this result that can help us in proving that a number is composite. Amazingly, we can do this without factoring the number at all! This is incredibly sweet.
Corollary: If there exists an integer a so that $a^n \not\equiv a \mod{n}$, then n is not prime.
Proof:
This is just the contrapositive of our first corollary. $\square$
#### Example: Proving composite-ness without factoring
We'll compute $2^{90} \mod{91}$. To do this, we'll use successive squaring:
(10)
\begin{align} \begin{split} 2^2 &\equiv 4 \mod{91}\\ 2^4 &\equiv 4^2 \equiv 16 \mod{91}\\ 2^8 &\equiv 16^2 \equiv 256 \mod{91} \equiv 74 \mod{91}\\ 2^{16} &\equiv 74^2 \equiv (-17)^2 \equiv 289 \equiv 16 \mod{91}\\ 2^{32} &\equiv 16^2 \equiv 74 \mod{91}\\ 2^{64} &\equiv 74^2 \equiv 16 \mod{91}. \end{split} \end{align}
Hence we have
(11)
\begin{align} 2^{90} \equiv 2^{64+16+8+2} \equiv 16 \cdot 16 \cdot 74 \cdot 4 \equiv 74 \cdot 74 \cdot 4 \equiv 16 \cdot 4 \equiv 64 \mod{91}. \end{align}
After all this math, it's now easy to see that $2^{91} \equiv 64 \cdot 2 \not\equiv 2 \mod{91}$. With $a = 2$ and $n = 91$ in the previous corollary, this tells us that 91 isn't prime. $\square$
# Fake Primes
You should be careful not to think that this last corollary is a perfect test for primality, as there are numbers composite numbers n which satisfy
(12)
\begin{align} a^n \equiv a \mod{n} \quad \quad \mbox{ for all integers }a. \end{align}
Since this congruence is a property that primes have, we think of composites which share this property as "behaving like prime numbers." In that sense, these composite numbers are fake prime numbers; even though they walk like primes and talk like primes, they aren't actually prime.
There are degrees to which a composite number can fake being a prime. For instance, we have the following
Definition: A composite number is called a psuedoprime if $2^n \equiv 2 \mod{n}$.
These kinds of fake primes satisfy an flt-like congruence when the base is 2, so they are "somewhat fake" primes.
#### Example: 645 is psuedoprime
We'll show that $645 = 3\cdot 5 \cdot 43$ is a pseudoprime. To do this, we need to show that $2^{645} \equiv 2 \mod{645}$. Now we could do this using successive squaring, but since we know the factorization of 645 we could instead verify that the following congruences all hold
(13)
\begin{align} \begin{split} 2^{645} &\equiv 2 \mod{3}\\ 2^{645} &\equiv 2 \mod{5}\\ 2^{645} &\equiv 2 \mod{43}. \end{split} \end{align}
Why is this the same thing as checking $2^{645} \equiv 2 \mod{645}$, you ask? Our good old friend the CRT.
Ok, let's check these congruences. First, we know that $2^{2} \equiv 1 \mod{3}$ using flt, and so we have
(14)
\begin{align} 2^{645} \equiv 2^{2\cdot 322 + 1} \equiv \left(2^2\right)^{322}2 \equiv 1^{322}2 \equiv 2 \mod{3}. \end{align}
Likewise we know that $2^{4} \equiv 1 \mod{5}$ by flt, and so
(15)
\begin{align} 2^{645} \equiv 2^{4\cdot 161 + 1} \equiv \left(2^4\right)^{161}2 \equiv 2 \mod{5}. \end{align}
Finally, since $2^{42} \equiv 1 \mod{43}$ by flt, we have
(16)
\begin{align} 2^{645} \equiv 2^{42\cdot 15+ 15} \equiv \left(2^{42}\right)^{15}2^{15} \equiv 2^{15} \mod{43}. \end{align}
Using successive squaring, we have
(17)
\begin{split} 2^2 &\equiv 4 \mod{43}\\ 2^4 &\equiv (4)^2 \equiv 16 \mod{43}\\ 2^8 &\equiv (16)^2 \equiv 256 \equiv 41 \equiv -2 \mod{43}. \end{split}
Hence we get
(18)
\begin{align} 2^{15} \equiv 2^{8+4+2+1} \equiv 2^8 2^42^22^1 \equiv -2 \cdot 16\cdot 4 \cdot 2 \equiv (-16)(16) \equiv 2 \mod{43}. \end{align}
Now that we know that $2^{645} \equiv 2$ modulo 3,5 and 43, the CRT says that we must have $2^{645} \equiv 2 \mod{3\cdot 5\cdot 43}$.
$\square$
There are other composite numbers which are a kind of "strong" psuedoprimes, in the sense that they satisfy the conditions of psuedoprime numbers but with 2 replaced by an arbitrary integer a.
Definition: A Carmichael number is a composite number n such that every integer a satisfies $a^{n} \equiv a \mod{n}$.
#### Example: 561 is Carmichael
In class we showed that $561 = 3\cdot 11 \cdot 17$ is Carmichael. To do this, we showed that for an arbitrary integer a with $(a,561) = 1$, the congruences
(19)
\begin{align} \begin{split} a^{561} &\equiv a \mod{3}\\ a^{561} &\equiv a \mod{11}\\ a^{561} &\equiv a \mod{17} \end{split} \end{align}
are satisfied. To do this, notice first that since $(a,561) = 1$ we must also have $(a,3) = (a,11) = (a,17) = 1$. This has to be the case since any common divisor of a with any of 3,7 of 17 would mean that a has a common divisor with 561, something we already know doesn't exist. Because a is relatively prime to all these prime numbers, then, we can apply flt in each case, as follows:
(20)
\begin{align} \begin{split} a^{561} &\equiv a^{2\cdot 280+1} \equiv a \mod{3}\\ a^{561} &\equiv a^{10\cdot 56+1} \equiv a \mod{11}\\ a^{561} &\equiv a^{16\cdot 35+1} \equiv a \mod{17}. \end{split} \end{align}
$\square$
Notice that this same technique can be used to show that a number $n = p_1\cdots p_k$ such that [{$p_i-1 \mid n-1$]] for every i is a Carmichael number.
# A Preview of Next Class
We finished the class today by asking if we could find a generalization of flt that applies to composites. More specifically, for prime numbers flt says
For every integer a with $(a,p) = 1$, the congruence $a^{p-1} \equiv 1 \mod{p}$.
We want to know whether there is a "magic exponent" e so that
For every integer a with $(a,n) = 1$, the congruence $a^{e} \equiv 1 \mod{n}$. |
During AMC testing, the AoPS Wiki is in read-only mode. No edits can be made.
# Difference between revisions of "2019 AMC 10A Problems/Problem 19"
## Problem
What is the least possible value of $$(x+1)(x+2)(x+3)(x+4)+2019$$where $x$ is a real number?
$\textbf{(A) } 2017 \qquad\textbf{(B) } 2018 \qquad\textbf{(C) } 2019 \qquad\textbf{(D) } 2020 \qquad\textbf{(E) } 2021$
## Solution 1
Grouping the first and last terms and two middle terms gives $(x^2+5x+4)(x^2+5x+6)+2019$, which can be simplified to $(x^2+5x+5)^2-1+2019$. Noting that squares are nonnegative, and verifying that $x^2+5x+5=0$ for some real $x$, the answer is $\boxed{\textbf{(B) } 2018}$.
## Solution 2
Let $a=x+\tfrac{5}{2}$. Then the expression $(x+1)(x+2)(x+3)(x+4)$ becomes $\left(a-\tfrac{3}{2}\right)\left(a-\tfrac{1}{2}\right)\left(a+\tfrac{1}{2}\right)\left(a+\tfrac{3}{2}\right)$.
We can now use the difference of two squares to get $\left(a^2-\tfrac{9}{4}\right)\left(a^2-\tfrac{1}{4}\right)$, and expand this to get $a^4-\tfrac{5}{2}a^2+\tfrac{9}{16}$.
Refactor this by completing the square to get $\left(a^2-\tfrac{5}{4}\right)^2-1$, which has a minimum value of $-1$. The answer is thus $2019-1=\boxed{\textbf{(B) }2018}$.
## Solution 3 (calculus)
Similar to Solution 1, grouping the first and last terms and the middle terms, we get $(x^2+5x+4)(x^2+5x+6)+2019$.
Letting $y=x^2+5x$, we get the expression $(y+4)(y+6)+2019$. Now, we can find the critical points of $(y+4)(y+6)$ to minimize the function:
$\frac{d}{dx}(y^2+10y+24)=0$
$2y+10=0$
$2y(y+5)=0$
$y=-5,0$
To minimize the result, we use $y=-5$. Hence, the minimum is $(-5+4)(-5+6)=-1$, so $-1+2019 = \boxed{\textbf{(B) }2018}$.
Note: We could also have used the result that minimum/maximum point of a parabola $y = ax^2 + bx + c$ occurs at $x=-\frac{b}{2a}$.
## Solution 4
The expression is negative when an odd number of the factors are negative. This happens when $-2 < x < -1$ or $-4 < x < -3$. Plugging in $x = -\frac32$ or $x = -\frac72$ yields $-\frac{15}{16}$, which is very close to $-1$. Thus the answer is $-1 + 2019 = \boxed{\textbf{(B) }2018}$.
## Solution 5 (using the answer choices)
Using the answer choices, we see that choices $C$, $D$, and $E$ are impossible, since $(x+1)(x+2)(x+3)(x+4)$ can actually be negative (as seen when e.g. $x = -\frac{3}{2}$). Plug in $x = -\frac{3}{2}$ to see that it becomes $2019 - \frac{15}{16}$, so round this to $\boxed{\textbf{(B) }2018}$.
We can also see that the limit of the function is at least -1 since at the minimum, two of the numbers are less than 1, but two are between 1 and 2.
## Video Solution
For those who want a video solution: https://www.youtube.com/watch?v=Mfa7j2BoNjI |
# Sean Penn Under The Microscope (03/26/2020)
How will Sean Penn get by on 03/26/2020 and the days ahead? Let’s use astrology to complete a simple analysis. Note this is for entertainment purposes only – don’t get too worked up about the result. I will first find the destiny number for Sean Penn, and then something similar to the life path number, which we will calculate for today (03/26/2020). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology enthusiasts.
PATH NUMBER FOR 03/26/2020: We will analyze the month (03), the day (26) and the year (2020), turn each of these 3 numbers into 1 number, and add them together. Here’s how it works. First, for the month, we take the current month of 03 and add the digits together: 0 + 3 = 3 (super simple). Then do the day: from 26 we do 2 + 6 = 8. Now finally, the year of 2020: 2 + 0 + 2 + 0 = 4. Now we have our three numbers, which we can add together: 3 + 8 + 4 = 15. This still isn’t a single-digit number, so we will add its digits together again: 1 + 5 = 6. Now we have a single-digit number: 6 is the path number for 03/26/2020.
DESTINY NUMBER FOR Sean Penn: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Sean Penn we have the letters S (1), e (5), a (1), n (5), P (7), e (5), n (5) and n (5). Adding all of that up (yes, this can get tedious) gives 34. This still isn’t a single-digit number, so we will add its digits together again: 3 + 4 = 7. Now we have a single-digit number: 7 is the destiny number for Sean Penn.
CONCLUSION: The difference between the path number for today (6) and destiny number for Sean Penn (7) is 1. That is less than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A GOOD RESULT. But don’t go jumping for joy yet! As mentioned earlier, this is not at all guaranteed. If you want to see something that people really do vouch for, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
Introduction
Quartiles
Percentiles
Quantiles
7. Quantiles
Standard Deviation and Variance
## Instruction
Percentiles and quartiles are specific types of quantiles. More generally, quantiles are values that partition (or split) a set of values into approximately equal-sized subsets. Common quantiles have special names:
• The 2-quantile is called the median.
• The 4-quantiles are called quartiles.
• The 10-quantiles are called deciles.
• The 100-quantiles are called percentiles.
• The 1,000-quantiles are called permilles.
The formula for p-th q-quantile of an n-element set is analogous to the one presented for percentiles:
\lceil\frac{p}{q}n\rceil
Let's calculate the deciles of the following set:
1, 4, 6, 8, 10, 12, 14, 16, 18, 20
There are exactly 10 elements and 10 deciles, so each decile corresponds to a single element.
Decile d Position Decile value
1^\text{st} \lceil\frac{1}{10}10\rceil=\lceil1\rceil=1 1
2^\text{nd} \lceil\frac{2}{10}10\rceil=\lceil2\rceil=2 4
3^{rd} \lceil\frac{3}{10}10\rceil=\lceil3\rceil=3 6
4^\text{th} \lceil\frac{4}{10}10\rceil=\lceil4\rceil=4 8
5^\text{th} \lceil\frac{5}{10}10\rceil=\lceil5\rceil=5 10
6^\text{th} \lceil\frac{6}{10}10\rceil=\lceil6\rceil=6 12
7^\text{th} \lceil\frac{7}{10}10\rceil=\lceil7\rceil=7 14
8^\text{th} \lceil\frac{8}{10}10\rceil=\lceil8\rceil=8 16
9^\text{th} \lceil\frac{9}{10}10\rceil=\lceil9\rceil=9 18
10^\text{th} \lceil\frac{10}{10}10\rceil=\lceil10\rceil=10 20 |
Categories
## Triangle Problem | PRMO-2018 | Problem No-24
Try this beautiful Trigonometry Problem based on Triangle from PRMO -2018, Problem 24.
## Triangle Problem – PRMO 2018- Problem 24
If $\mathrm{N}$ is the number of triangles of different shapes (i.e. not similar) whose angles are all integers (in degrees), what is $\mathrm{N} / 100$ ?
,
• $15$
• $22$
• $27$
• $32$
• $37$
Trigonometry
Triangle
Integer
## Suggested Book | Source | Answer
Pre College Mathematics
#### Source of the problem
Prmo-2018, Problem-24
#### Check the answer here, but try the problem first
$27$
## Try with Hints
#### First Hint
Given that $\mathrm{N}$ is the number of triangles of different shapes. Therefore the different shapes of triangle the angles will be change . at first we have to find out the posssible orders of the angles that the shape of the triangle will be different…
Now can you finish the problem?
#### Second Hint
case 1 : when $x \geq 1$ & $y \geq 3 \geq 1$
$$x+y+z=180$$
$={ }^{179} \mathrm{C}_{2}=15931$
Case 2 : When two angles are same
$$2 x+y=180$$
1,1,178
2,2,176
$\vdots$
89,89,2
#### Solution
But we have one case $60^{\circ}, 60^{\circ}, 60^{\circ}$
$$\text { Total }=89-1=88$$
Such type of triangle $=3(88)$
When 3 angles are same $=1(60,60,60)$
So all distinct angles’s triangles
$$\begin{array}{l} =15931-(3 \times 88)-1 \ \neq 3 ! \ =2611 \end{array}$$
Now, distinct triangle $=2611+88+1$
$=2700 \ N=2700 \ \frac{N}{100}=27 \$
Categories
## Value of Sum | PRMO – 2018 | Question 16
Try this beautiful Problem based on Value of Sum from PRMO 2018, Question 16.
## Value of Sum – PRMO 2018, Question 16
What is the value of $\sum_{1 \leq i<j \leq 10 \atop i+j=\text { odd }}(i-j)-\sum_{1 \leq i<j \leq 10 \atop i+j=\text { even }}(i-j) ?$
• $50$
• $53$
• $55$
• $59$
• $65$
### Key Concepts
Odd-Even
Sum
integer
Answer:$55$
PRMO-2018, Problem 16
Pre College Mathematics
## Try with Hints
We have to find out the sum . Now substitite $i=1,2,3…9$ and observe the all odd-even cases……
Can you now finish the problem ……….
$i=1 \Rightarrow$$1+(2+4+6+8+10-3-5-7-9) =1-4+10=7 i=2 \Rightarrow$$0 \times 2+(3+5+7+9-4-6-8-10)$
$=-4$
$i=3 \Rightarrow$$1 \times 3+(4+6+8+10-5-7-9) =3-3+10=10 i=4 \Rightarrow$$ 0 \times 4+(5+7+9-6-8-10)=-3$
$i=5 \Rightarrow $$1 \times 5+(6+8+10-7-9)=5-2+10 =13 i=6 \Rightarrow$$ 0 \times 6+(7+9-8-10)=-2$
$i=7 \Rightarrow $$1 \times 7+(8+10-9)=7-1+10=16 i=8 \Rightarrow$$ 0 \times 8+(9-10)=-1$
$i=9 \Rightarrow$$1 \times 9+(10)=19 Can you finish the problem…….. Therefore S =(7+10+13+16+19)-(4-3-2-1) =55 ## Subscribe to Cheenta at Youtube Categories ## Chessboard Problem | PRMO-2018 | Problem No-26 Try this beautiful Chessboard Problem based on Chessboard from PRMO – 2018. ## Chessboard Problem – PRMO 2018- Problem 26 What is the number of ways in which one can choose 60 units square from a 11 \times 11 chessboard such that no two chosen square have a side in common? , • $56$ • $58$ • $60$ • $62$ • $64$ ### Key Concepts Game problem Chess board combination ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-26 #### Check the answer here, but try the problem first $62$ ## Try with Hints #### First Hint Total no. of squares =121 Out of these, 61 squares can be placed diagonally. From these any 60 can be selected in { }^{61} C_{60} ways =61 Now can you finish the problem? #### Second Hint From the remaining 60 squares 60 can be chosen in any one way Total equal to { }^{61} \mathrm{C}{60}+{ }^{60} \mathrm{C}{60}=61+1=62 ## Subscribe to Cheenta at Youtube Categories ## Measure of Angle | PRMO-2018 | Problem No-29 Try this beautiful Trigonometry Problem based on Measure of Angle from PRMO -2018. ## Measure of Angle – PRMO 2018- Problem 29 Let D be an interior point of the side B C of a triangle ABC. Let l_{1} and l_{2} be the incentres of triangles A B D and A C D respectively. Let A l_{1} and A l_{2} meet B C in E and F respectively. If \angle B l_{1} E=60^{\circ}, what is the measure of \angle C l_{2} F in degrees? , • $25$ • $20$ • $35$ • $30$ • $45$ ### Key Concepts Trigonometry Triangle Angle ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-29 #### Check the answer here, but try the problem first $30$ ## Try with Hints #### First Hint According to the questations at first we draw the picture . We have to find out the value of \angle C l_{2} F. Now at first find out $\angle AED$ and $\angle AFD$ which are the exterioe angles of $\triangle BEL_1$ and $\triangle CL_2F$. Now sum of the angles is 180^{\circ} Now can you finish the problem? #### Second Hint \angle E A D+\angle F A D=\angle E A F=\frac{A}{2} \angle A E D=60^{\circ}+\frac{B}{2} \angle A F D=\theta+\frac{C}{2} Therefore \quad \ln \Delta A E F: \frac{A}{2}+60^{\circ}+\frac{B}{2}+\theta+\frac{C}{2}=180^{\circ} 90^{\circ}+60^{\circ}+\theta=180^{\circ} (as sum of the angles of a Triangle is 180^{\circ} Therefore \quad \theta=30^{\circ} ## Subscribe to Cheenta at Youtube Categories ## Good numbers Problem | PRMO-2018 | Question 22 Try this beautiful good numbers problem from Number theory from PRMO 2018, Question 22. ## Good numbers Problem – PRMO 2018, Question 22 A positive integer k is said to be good if there exists a partition of {1,2,3, \ldots, 20} into disjoint proper subsets such that the sum of the numbers in each subset of the partition is k. How many good numbers are there? • 4 • 6 • 8 • 10 • 2 ### Key Concepts Number theorm good numbers subset ## Check the Answer Answer:6 PRMO-2018, Problem 22 Pre College Mathematics ## Try with Hints What is good numbers ? A good number is a number in which every digit is larger than the sum of digits of its right (all less significant bits than it). For example, 732 is a good number, 7>3+2 and 3>2 . Given that k is said to be good if there exists a partition of {1,2,3, \ldots, 20} into disjoint proper subsets such that the sum of the numbers in each subset of the partition is k. Now at first we have to find out sum of these integers {1,2,3, \ldots, 20}. Later create some partitions such that two partitions be disjoint set and sum of the numbers of these partitions be good numbers Can you now finish the problem ………. Sum of numbers equals to \frac{20 \times 21}{2}=210 \& 210=2 \times 3 \times 5 \times 7 So \mathrm{K} can be 21,30,35,47,70,105 Can you finish the problem…….. Case 1 : \mathrm{A}=\{1,2,3,4,5,16,17,18,19,20\}, \mathrm{B}=\{6,7,8,9,10,11,12,13,14,15\} Case 2 : A=\{20,19,18,13\}, B=\{17,16,15,12,10\}, C=\{1,2,3,4,5,6,7,8,9,11,14\} Case 3 : \mathrm{A}=\{20,10,12\}, \mathrm{B}=\{18,11,13\}, \mathrm{C}=\{16,15,9,2\}, \mathrm{D}=\{19,8,7,5,3\}, \mathrm{E}=\{1,4,6,14,17\} Case 4 : A=\{20,10\}, B=\{19,11\}, C=\{18,12\}, D=\{17,13\}, E=\{16,14\}, F=\{1,15,5\}, G=\{2,3,4,6,7,8\} Case 5 : A=\{20,15\}, B=\{19,16\}, C=\{18,17\}, D=\{14,13,8\}, E=\{12,11,10,2\}, F=\{1,3,4,5,6,7,9\} Case 6 : A=\{1,20\}, B=\{2,19\}, C=\{3,18\} \ldots \ldots \ldots \ldots, J=\{10,11\} Therefore Good numbers equal to 6 ## Subscribe to Cheenta at Youtube Categories ## Polynomial Problem | PRMO-2018 | Question 30 Try this beautiful Polynomial Problem from Number theorm from PRMO 2018, Question 30. ## Polynomial Problem – PRMO 2018, Question 30 Let P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n} be a polynomial in which a_{i} is non-negative integer for each \mathrm{i} \in{0,1,2,3, \ldots, \mathrm{n}} . If \mathrm{P}(1)=4 and \mathrm{P}(5)=136, what is the value of \mathrm{P}(3) ? • 30 • 34 • 36 • 39 • 42 ### Key Concepts Number theorm Polynomial integer ## Check the Answer Answer:34 PRMO-2018, Problem 30 Pre College Mathematics ## Try with Hints Given that P(x)=a_{0}+a_{1} x+a_{2} x^{2}+\ldots .+a_{n} x^{n} where \mathrm{P}(1)=4 and \mathrm{P}(5)=136. Now we have to find out P(3). Therefore if we put x=1 and x=5 then we will get two relations . Using these relations we can find out a_0 , a_1, a_2 . Can you now finish the problem ………. a_{0}+a_{1}+a_{2}+\ldots \ldots+a_{n}=4 \Rightarrow a_{i} \leq 4 a_{0}+5 a_{1}+5^{2} a_{2}+\ldots+a 5^{n} a_{n}=136 \Rightarrow a_{0}=1+5 \lambda \Rightarrow a_{0}=1 Can you finish the problem…….. Hence 5 a_{1}+5^{2} a_{2}+\ldots \ldots+5^{n} a_{n}=135 a_{1}+5 a_{2}+\ldots 5^{n-1} a_{n-1}=27 \Rightarrow a_{1}=5 \lambda+2 \Rightarrow a_{1}=2 \Rightarrow 5 a_{2}+\ldots .5^{n-1} a_{n-1}=25 a_{2}+5 a_{3}+\ldots .5^{n-2} a_{n-2}=5 \Rightarrow a_{2}=5 \lambda \Rightarrow a_{2}=0 a_{3}+5 a_{4}+\ldots \ldots \ldots+5^{n-3} a_{n-3}=1 a_{3}=1 \Rightarrow a_{4}+5 a_{5}+\ldots .+5^{n-4} a_{n-3}=0 a_{4}=a_{5}=\ldots . a_{n}=0 Hence P(n)=x^{3}+2 x+1 P(3)=34 ## Subscribe to Cheenta at Youtube Categories ## Digits Problem | PRMO – 2018 | Question 19 Try this beautiful Digits Problem from Number theorm from PRMO 2018, Question 19. ## Digits Problem – PRMO 2018, Question 19 Let N=6+66+666+\ldots \ldots+666 \ldots .66, where there are hundred 6 ‘s in the last term in the sum. How many times does the digit 7 occur in the number N ? • 30 • 33 • 36 • 39 • 42 ### Key Concepts Number theorm Digits Problem integer ## Check the Answer Answer:33 PRMO-2018, Problem 19 Pre College Mathematics ## Try with Hints Given that \mathrm{N}=6+66+666+……. \underbrace{6666 …..66}_{100 \text { times }} If you notice then we can see there are so many large terms. but we have to find out the sum of the digits. but since the number of digits are large so we can not calculate so eassily . we have to find out a symmetry or arrange the number so that we can use any formula taht we can calculate so eassily. if we multiply $\frac{6}{9}$ then it becomes =\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}] Can you now finish the problem ………. \mathrm{N}=\frac{6}{9}[9+99+\ldots \ldots \ldots \ldots+\underbrace{999 \ldots \ldots \ldots .99}_{100 \text { times }}] =\frac{6}{9}\left[(10-1)+\left(10^{2}-1\right)+…….+\left(10^{100}-1\right)\right] =\frac{6}{9}\left[\left(10+10^{2}+…..+10^{100}\right)-100\right] Can you finish the problem…….. =\frac{6}{9}\left[\left(10^{2}+10^{3}+\ldots \ldots \ldots+10^{100}\right)-90\right] =\frac{6}{9}\left(10^{2} \frac{\left(10^{99}-1\right)}{9}\right)-60 =\frac{200}{27}\left(10^{99}-1\right)-60 =\frac{200}{27}\underbrace{(999….99)}_{99 \text{times}}-60 =\frac{1}{3}\underbrace{(222…..200)}_{99 \mathrm{times}}-60 =\underbrace{740740 \ldots \ldots .7400-60}_{740 \text { comes } 33 \text { times }} =\underbrace{740740 \ldots \ldots .740}_{32 \text { times }}+340 \Rightarrow 7 comes 33 times ## Subscribe to Cheenta at Youtube Categories ## Chocolates Problem | PRMO – 2018 | Problem No. – 28 Try this beautiful Problem on Combinatorics from integer based on chocolates from PRMO -2018 ## Chocolates Problem – PRMO 2018- Problem 28 Let N be the number of ways of distributing 8 chocolates of different brands among 3 children such that each child gets at least one chocolate, and no two children get the same number of chocolates. Find the sum of the digits of \mathrm{N}. , • $28$ • $90$ • $24$ • $16$ • $27$ ### Key Concepts Combination Combinatorics Probability ## Suggested Book | Source | Answer #### Suggested Reading Pre College Mathematics #### Source of the problem Prmo-2018, Problem-28 #### Check the answer here, but try the problem first $24$ ## Try with Hints #### First Hint we have to distribute $8$ chocolates among $3$ childrens and the condition is Eight chocolets will be different brands that each child gets at least one chocolate, and no two children get the same number of chocolates. Therefore thr chocolates distributions will be two cases as shown below….. Now can you finish the problem? #### Second Hint case 1:(5,2,1) Out of $8$ chocolates one of the boys can get $5$ chocolates .So $5$ chocolates can be choosen from $8$ chocolates in $8 \choose 5$ ways. Therefore remaining chocolates are $3$ . Out of $3$ chocolates another one of the boys can get $2$ chocolates .So $2$ chocolates can be choosen from $3$ chocolates in $3 \choose 2$ ways. Therefore remaining chocolates are $1$ . Out of $1$ chocolates another one of the boys can get $1$ chocolates .So $1$ chocolates can be choosen from $1$ chocolates in $1 \choose 1$ ways. Therefore number of ways for first case will be $8 \choose 5$ $\times$ $3 \choose 2$ $\times$ $1 \choose 1$$\times$ 3!=\frac{8}{2!.5!.1!}$$\times 3$
Case 2:$(4,3,1)$
Out of $8$ chocolates one of the boys can get $4$ chocolates .So $4$ chocolates can be choosen from $8$ chocolates in $8 \choose 4$ ways.
Therefore remaining chocolates are $4$ . Out of $4$ chocolates another one of the boys can get $3$ chocolates .So $3$ chocolates can be choosen from $4$ chocolates in $4 \choose 3$ ways.
Therefore remaining chocolates are $1$ . Out of $1$ chocolates another one of the boys can get $1$ chocolates .So $1$ chocolates can be choosen from $1$ chocolates in $1 \choose 1$ ways.
Categories
## Trigonometry | PRMO-2018 | Problem No-14
Try this beautiful Problem on Trigonometry from PRMO -2018
## Trigonometry Problem – PRMO 2018- Problem 14
If $x=\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \ldots . \cos 89^{\circ}$ and $y=\cos 2^{\circ} \cos 6^{\circ} \cos 10^{\circ} \ldots \ldots \cos 86^{\circ},$ then what is the integer nearest to $\frac{2}{7} \log _{2}(\mathrm{y} / \mathrm{x}) ?$
,
• $28$
• $19$
• $24$
• $16$
• $27$
Trigonometry
Logarithm
## Suggested Book | Source | Answer
Pre College Mathematics
#### Source of the problem
Prmo-2018, Problem-14
#### Check the answer here, but try the problem first
$19$
## Try with Hints
#### First Hint
Given that $x=\cos 1^{\circ} \cos 2^{\circ} \cos 3^{\circ} \cdots \cos 89^{\circ}$
$\Rightarrow x=\cos 1^{\circ}\cos 89^{\circ}\cos 2^{\circ}\cos 88^{\circ}……..\cos 45^{\circ}$
$\Rightarrow x=\cos 1^{\circ}\sin 1^{\circ}\cos 2^{\circ}\cos 2^{\circ}……..\cos 45^{\circ}$ ( as cos(90-x)=sin x)
$\Rightarrow x=\frac{1}{2^{44}} (2 \cos 1^{\circ} \sin 1^{\circ}) (2\cos 2^{\circ}\sin 2^{\circ})……..\cos 45^{\circ}$
$\Rightarrow x =\frac{ \cos 45^{\circ} \cdot \sin 2^{\circ} \cdot \sin 4^{\circ} \cdots \sin 88}{ 2^{44}}$ (as sin 2x= 2 sin x cos x)
Now can you finish the problem?
#### Second Hint
Therefore we can say that
$x=\frac{1}{2^{1 / 2}} \times \frac{\sin 4^{\circ} \cdot \sin 8^{\circ} \cdots \sin 88^{\circ}}{2^{44} \times 2^{22}}$
$=\frac{\sin \left(90-86^{\circ}\right) \cdot \sin \left(90-84^{\circ}\right) \cdots \sin (90-2)}{2^{66} \cdot 2^{1 / 2}}$
$=\frac{\cos 2^{\circ} \cdot \cos 6^{\circ} \cdot \cos 86^{\circ}}{2^{133 / 2}}$
$x=\frac{y}{2^{133 / 2}}$
Can you finish the problem…?
#### Third Hint
$\frac{y}{x}=2^{133 / 2}$
$\frac{2}{7} \log \left(\frac{y}{x}\right)=\frac{2}{7} \times \log _{2}(2)^{133 / 2}$
$=\frac{2}{7} \times \frac{133}{2}$
$=19$
Categories
## External Tangent | AMC 10A, 2018 | Problem 15
Try this beautiful Problem on Geometry based on External Tangent from AMC 10 A, 2018. You may use sequential hints to solve the problem.
## External Tangent – AMC-10A, 2018- Problem 15
Two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$, as shown in the diagram. The distance $A B$ can be written in the form $\frac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. What is $m+n ?$
,
• $21$
• $29$
• $58$
• $69$
• $93$
Geometry
Triangle
Pythagoras
## Suggested Book | Source | Answer
Pre College Mathematics
#### Source of the problem
AMC-10A, 2018 Problem-15
#### Check the answer here, but try the problem first
$69$
## Try with Hints
#### First Hint
Given that two circles of radius 5 are externally tangent to each other and are internally tangent to a circle of radius 13 at points $A$ and $B$. we have to find out the length $AB$.
Now join $A$ & $B$ and the points $Y$ & $Z$. If we can show that $\triangle XYZ \sim \triangle XAB$ then we can find out the length of $AB$.
Now can you finish the problem?
#### Second Hint
now the length of $YZ=5+5=10$ (as the length of the radius of smaller circle is $5$) and $XY=XA-AY=13-5=8$. Now $YZ|| AB$.therefore we can say that $\triangle XYZ \sim \triangle XAB$. therefore we can write $\frac{X Y}{X A}=\frac{Y Z}{A B}$
Now Can you finish the Problem?
#### Third Hint
From the relation we can say that $\frac{X Y}{X A}=\frac{Y Z}{A B}$
$\Rightarrow \frac{8}{13}=\frac{10}{AB}$
$\Rightarrow AB=\frac{13\times 10}{8}$
$\Rightarrow AB=\frac{65}{4}$ which is equal to $\frac{m}{n}$
Therefore $m+n=65+4=69$ |
# A point moves in a manner that the sum of the squares of its distance from thr origin and the point (2,-3) is always 19.show that the locus of moving point is a circle.find the equation to the circle?
Jul 31, 2018
${x}^{2} + {y}^{2} - 2 x + 3 y - 3 = 0$.
#### Explanation:
Let the variable point be $P \left(x , y\right)$ and the given fixed points be
$O \left(0 , 0\right) \mathmr{and} A \left(2 , - 3\right)$.
It is given that, $P {O}^{2} + P {A}^{2} = 19$.
Using the distance formula, we have,
$\therefore \left\{{\left(x - 0\right)}^{2} + {\left(y - 0\right)}^{2}\right\} + \left\{{\left(x - 2\right)}^{2} + {\left(y + 3\right)}^{2}\right\} = 19$.
$\therefore 2 {x}^{2} + 2 {y}^{2} - 4 x + 6 y + 4 + 9 = 19 , \mathmr{and} ,$
${x}^{2} + {y}^{2} - 2 x + 3 y - 3 = 0$.
Rewriting after completing squares,
${\left(x - 1\right)}^{2} + {\left(y + \frac{3}{2}\right)}^{2} = 3 + 1 + \frac{9}{4} = \frac{25}{4} = {\left(\frac{5}{2}\right)}^{2}$.
This exhibits that the locus of the point is a circle having
centre at $\left(1 , - \frac{3}{2}\right)$ and radius $\frac{5}{2}$.
It may be interesting to note that the centre of the circle is the
mid-point of $O A$. |
# Thread: Rectangle Inscribed in Semicircle
1. ## Rectangle Inscribed in Semicircle
A rectangle is Inscribed in a semicircle of radius 2.
Let P = (x, y) be the point in quadrant 1 that is a vertex of the rectangle and is on the circle.
(a) Express the area A of the rectangle as a function of x.
(b) Express the perimeter p of the rectangle as a function of x.
2. Originally Posted by symmetry
A rectangle is Inscribed in a semicircle of radius 2.
Let P = (x, y) be the point in quadrant 1 that is a vertex of the rectangle and is on the circle.
(a) Express the area A of the rectangle as a function of x.
(b) Express the perimeter p of the rectangle as a function of x.
Hello,
to (a)
The area of the rectangle is calculated in general by:
$\displaystyle A=length \times width$
According to your problem the area is:
$\displaystyle A=2x \cdot y$. Now use Pythagoran theorem:
$\displaystyle x^2+y^2=r^2 \Longleftrightarrow y=\sqrt{r^2-x^2}$. Thus:
$\displaystyle A=2x \cdot \sqrt{r^2-x^2}$. That means with r = 2:
$\displaystyle A=2x \cdot \sqrt{4-x^2}$
to (b):
The perimeter of a rectangle is calculated in general:
$\displaystyle p=2 \cdot {length} + 2 \cdot {width}$. Plug in the values you know:
$\displaystyle p=2 \cdot 2x + 2 \cdot \sqrt{4-x^2}= 4x+ 2 \cdot \sqrt{4-x^2}$
3. ## ok
Another great reply. Tell me, how do you make those images?
,
,
,
,
,
# area of a rectangle inscribed in a semicircle
Click on a term to search for related topics. |
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### Quadrilaterals Class 9th Mathematics Part Ii MHB Solution
##### Quadrilaterals Class 9th Mathematics Part Ii MHB Solution
###### Practice Set 5.1
Question 1.
Diagonals of a parallelogram WXYZ intersect each other at point O. If ∠ XYZ = 135° then what is the measure of ∠XWZ and ∠YZW? If l(OY)= 5 cm then l(WY)=?
Given ZX and WY are the diagonals of the parallelogram
∠ XYZ = 135° ⇒ ∠ XWZ = 135° as the opposite angels of a parallelogram are congruent.
∠YZW + ∠ XWZ = 180° as the adjacent angels of the parallelogram are supplementary.
⇒ ∠YZW = 180° - 135° = 45°
Length of OY = 5 cm then length of WY = WO + OY = 5+5 = 10 cm
(diagonals of the parallelogram bisect each other. So, O is midpoint of WY)
Question 2.
In a parallelogram ABCD, If ∠A = (3x +12)°, ∠B = (2x -32) ° then find the value of x and then find the measures of ∠C and ∠D.
∠A = (3x +12)°
∠B = (2x -32) °
∠A + ∠B = 180° (supplementary angles of the ∥gram)
(3x +12) + (2x -32) = 180
5x – 20 = 180°
5x = 200°
∴ x= 40°
∠A = (3 × 40) +12 = 120 + 12
= 132
⇒ ∠C = 132° (opposite ∠s are congruent)
Similarly, ∠B = 2× 40 – 32
= 80 - 32°
= 48°
⇒ ∠D = 48°(opposite ∠s are congruent)
Question 3.
Perimeter of a parallelogram is 150 cm. One of its sides is greater than the other side by 25 cm. Find the lengths of all sides.
perimeter of parallelogram = 150cm
Let the one side of parallelogram be x cm then
Acc. To the given condition
Other side is (x+25) cm
Perimeter of parallelogram = 2(a+b)
150 = 2( x+x+25)
150 = 2(2x+25)
One side is 25cm and the other side is 50cm.
Question 4.
If the ratio of measures of two adjacent angles of a parallelogram is 1 : 2, find the measures of all angles of the parallelogram.
Given that the ratio of measures of two adjacent angles of a parallelogram= 1 : 2
If one ∠ is x other would be 180 – x as the adjacent ∠s of a parallelogram are supplementary.
Other ∠ is 120° .
The measure of all the angles are 60 °, 120 °, 60 ° and 120 ° where 60 ° and 120 ° are adjacent ∠s and 60 ° and 60 ° are congruent opposite angles.
Question 5.
Diagonals of a parallelogram intersect each other at point O. If AO = 5, BO = 12 and AB = 13 then show that ABCD is a rhombus.
The figure is given below:
Given AO =5, BO = 12 and AB = 13
In Δ AOB, AO2 + BO2 = AB2
∵ 52 + 122 = 132
252 + 1442 = 1692
so by the Pythagoras theorem
Δ AOB is right angled at ∠ AOB.
But ∠ AOB + ∠ AOD forms a linear pair so the given parallelogram is rhombus whose diagonal bisects each other at 90°.
Question 6.
In the figure 5.12, PQRS and ABCR are two parallelograms. If ∠P = 110° then find the measures of all angles of ABCR.
given PQRS and ABCR are two ∥gram.
∠P = 110° ⇒ ∠R = 110°
(opposite ∠s of parallelogram are congruent)
Now if , ∠R = 110° ⇒ ∠ B = 110°
∠B + ∠A = 180°
(adjacent ∠s of a parallelogram are supplementary)
⇒ ∠A = 70° ⇒ ∠C = 70°
(opposite ∠s of parallelogram are congruent)
Question 7.
In figure 5.13 ABCD is a parallelogram. Point E is on the ray AB such that BE = AB then prove that line ED bisects seg BC at point F.
Given, ABCD is a parallelogram
And BE = AB
But AB = DC (opposite sides of the parallelogram are equal and parallel)
⇒ DC = BE
In Δ BEF and ∠DCF
∠DFC = ∠BFE (vertically opposite angles)
∠DFC = ∠ BFE (alternate ∠s on the transversal BC with AB and DC as ∥ )
And BE = AB (given)
Δ BEF ≅ ∠DCF (by AAS criterion)
⇒ BF =FC (corresponding parts of the congruent triangles)
⇒ F is mid-point of the line BC. Hence proved.
###### Practice Set 5.2
Question 1.
In figure 5.22, ABCD is a parallelogram, P and Q are midpoints of side AB and DC respectively, then prove APCQ is a parallelogram.
Given AB ∥ to DC and AB = DC as ABCD is ∥gram.
⇒ AP ∥CQ (parts of ∥ sides are ∥) & 1/2 AB = 1/2 DC
⇒ AP = QC (P and Q are midpoint of AB and DC respectively)
⇒ AP = PB and DQ = QC
Hence APCQ is a parallelogram as the pair of opposite sides is = and ∥.
Question 2.
Using opposite angles test for parallelogram, prove that every rectangle is a parallelogram.
Opposite angle property of parallelogram says that the opposite angles of a parallelogram are congruent.
Given a rectangle which had at least one angle as 90°.
If ∠ A is 90° and AD = BC (opposite sides of rectangle are ∥ and =)
AB is transversal
⇒ ∠ A + ∠B = 180 (angles on the same side of transversal is 180°)
But ∠B + ∠C is 180 (AD ∥ BC, opposite sides of rectangle)
⇒ ∠ A = ∠C = 90°
Since opposite ∠s are equal this rectangle is a parallelogram too.
Question 3.
In figure 5.23, G is the point of concurrence of medians of ΔDEF. Take point H on ray DG such that D-G-H and DG = GH, then prove that GEHF is a parallelogram.
Given G is the point of concurrence of medians of Δ DEF so the medians are divided in the ratio of 2:1 at the point of concurrence. Let O be the point of intersection of GH AND EF.
The figure is shown below:
⇒ DG = 2 GO
But DG = GH
⇒ 2 GO = GH
Also DO is the median for side EF.
⇒ EO = OF
Since the two diagonals bisects each other
⇒ GEHF is a ∥gram.
Question 4.
Prove that quadrilateral formed by the intersection of angle bisectors of all angles of a parallelogram is a rectangle. (Figure 5.24)
Given ABCD is a parallelogram
AR bisects ∠BAD, DP bisects ∠ADC , CP bisects ∠BCD and BR bisects ∠CBA
1/2 ∠ABC = ∠RBA
∠BAR + ∠RBA = 1/2 × 180° = 90°
⇒ Δ ARB is right angled at ∠ R since its acute interior angles are complementary.
Similarly Δ DPC is right angled at ∠ P and
Also in Δ COB , ∠BOC = 90° ⇒ ∠POR = 90° (vertically opposite angles)
Similarly in ΔADS , ∠ASD = 90° = ∠PSR (vertically opposite angles)
Since vertically opposite angles are equal and measures 90° the quadrilateral is a rectangle.
Question 5.
In figure 5.25, if points P, Q, R, S are on the sides of parallelogram such that AP = BQ = CR = DS then prove that PQRS is a parallelogram.
Given ABCD is a parallelogram so
and DC = AB and DC ∥ AB
also AP = BQ = CR = DS
⇒ AS = CQ and PB = DR
in ΔAPS and Δ CRQ
∠ A = ∠C (opposite ∠s of a parallelogram are congruent)
AS = CQ
AP = CR
ΔAPS ≅ Δ CRQ( SAS congruence rule)
⇒ PS = RQ (c.p.c.t.)
Similarly PQ= SR
Since both the pair of opposite sides are equal
PQRS is ∥gram.
###### Practice Set 5.3
Question 1.
Diagonals of a rectangle ABCD intersect at point O. If AC = 8 cm then find BO and if ∠CAD = 35° then find ∠ACB.
The diagonals of a rectangle are congruent to each other and bisects each other at the point of intersection so since AC = 8 cm
⇒ BD = 8 cm and
O is point of intersection so DO = OB = AO = OC = 4 cm
⇒ ∠ACB = 35 °
(since AB ∥ DC and AC is transversal ∴ ∠CAD and ∠ACB are pair of alternate interior angle.)
Question 2.
In a rhombus PQRS if PQ = 7.5 then find QR. If ∠QPS = 75° then find the measure of ∠PQR and ∠SRQ.
⇒ all the sides are congruent /equal
⇒ PQ = QR = 7.5
Also ∠QPS = 75° (given)
⇒∠QPS = 75° (opposite angles are congruent)
But ∠QPS + ∠PQR = 180° (adjacent angles are supplementary)
⇒ ∠PQR = 105°
∴ ∠SRQ = 105° (opposite angles)
Question 3.
Diagonals of a square IJKL intersects at point M, Find the measures of ∠IMJ, ∠JIK and ∠LJK.
The given quadrilateral is a square
⇒ all the angles are 90°
∴ ∠JIK = 90°
Since the diagonals are to each other ∠IMJ = 90°
Since the diagonals os a square are bisectors of the angles also
∠LJK = ∠IJL = 1/2 × 90° = 45°
Question 4.
Diagonals of a rhombus are 20 cm and 21 cm respectively, then find the side of rhombus and its perimeter.
Let the diagonal AC = 20cm and BD = 21
AB2 = BO2 + AO2
AB2 = (10.5)2 + (10)2
(the diagonals of a rhombus bisect each other at 90°)
AB2 = 110.25 + 100
AB = √210.25 = 14.5cm (side of the rhombus)
Perimeter = 4a = 14. 5 × 4 = 58cm
Question 5.
State with reasons whether the following statements are ‘true’ or ‘false’.
(i) Every parallelogram is a rhombus.
(ii) Every rhombus is a rectangle.
(iii) Every rectangle is a parallelogram.
(iv) Every square is a rectangle.
(v) Every square is a rhombus.
(vi)Every parallelogram is a rectangle.
(i) False.
Explanation: Every Parallelogram cannot be the rhombus as the diagonals of a rhombus bisects each other at 90° but this is not the same with every parallelogram. Hence the statement if false.
(ii) False.
Explanation: In a rhombus all the sides are congruent but in a rectangle opposite sides are equal and parallel. Hence the given statement is false.
(iii) True.
Explanation: The statement is true as in a rectangle opposite angles and adjacent angles all are 90°. And for any quadrilateral to be parallelogram the opposites angles should be congruent.
(iv) True.
Explanation: Every square is a rectangle as all the angles of the square at 90° , diagonal bisects each other and are congruent , pair of opposite sides are equal and parallel . Hence every square is a rectangle is true statement.
(v) True.
Explanation: The statement is true as all the test of properties of a rhombus are meet by square that is diagonals are perpendicular bisects each other , opposite sides are parallel to each other and the diagonals bisects the angles.
(vi) False.
Explanation:
Every parallelogram is a rectangle is not true as rectangle has each angle of 90° measure but same is not the case with every parallelogram.
###### Practice Set 5.4
Question 1.
In IJKL, side IJ || side KL ∠I = 108° ∠K = 53° then find the measures of ∠J and ∠L.
IJ ∥ KL and IL is transversal
∠I + ∠L = 180° (adjacent angles on the same side of the transversal)
⇒ ∠L = 180° – 108° = 72°
Now again IJ ∥ KL and JK is transversal
∠J + ∠K= 180 ° (adjacent angles on the same side of the transversal)
⇒ ∠K = 180° – 53° = 127°
Question 2.
In ABCD, side BC || side AD, side AB ≅ sided DC If ∠A = 72° then find the measures of ∠B, and ∠D.
⇒ the quadrilateral is a parallelogram (pair of opposite sides are equal and parallel)
∠ A = 72°
⇒ ∠C = 72° (opposite angles of parallelogram are congruent)
∠B = 180° – 72° = 108° (adjacent angles of a parallelogram are supplementary)
∠D = 108° (opposite angles of parallelogram are congruent)
Question 3.
In ABCD, side BC < side AD (Figure 5.32) side BC || side AD and if side BE ≅ side CD then prove that ∠ABC ≅ ∠DCB.
The figure of the question is given below:
Construction: we will draw a segment ∥ to BA meeting BC in E through point D.
And AB ∥ ED (construction)
⇒ AB = DE (distance between parallel lines is always same)
Hence ABDE is parallelogram
⇒ ∠ABE ≅ ∠DEC (corresponding angles on the same side of transversal)
And segBA ≅ seg DE (opposite sides of a ∥gram)
But given BA ≅ CD
So seg DE ≅ seg CD
⇒∠CED ≅ ∠DCE ( ∵ Δ CED is isosceles with CE = CD)
(Angle opposite to opposite sides are equal)
⇒ ∠ABC ≅ ∠DCB
###### Practice Set 5.5
Question 1.
In figure 5.38, points X, Y, Z are the midpoints of side AB, side BC and side AC of ΔABC respectively. AB = 5 cm, AC = 9 cm and BC = 11 cm. Find the length of XY, YZ, XZ.
Given X , Y and Z is the mid-point of AB, BC and AC.
Length of AB = 5 cm
So length of ZY = 1/2 × AB =1/2 × 5=2.5 cm (line joining mid-point of two sides of a triangle is parallel of the third side and is half of it)
Similarly, XZ = 1/2 × BC = 1/2 × 11= 5.5cm
Similarly, XY = 1/2 × AC = 1/2 × 9 = 4.5cm
Question 2.
In figure 5.39, PQRS and MNRL are rectangles. If point M is the midpoint of side PR then prove that,
i. SL = LR. Ii. LN = 1/2SQ.
The two rectangle PQRS and MNRL
In Δ PSR,
∠ PSR = ∠ MLR = 90°
∴ ML ∥ SP when SL is the transversal
M is the midpoint of PR (given)
By mid-point theorem a parallel line drawn from a mid-point of a side of a Δ meets at the Mid-point of the opposite side.
Hence L is the mid-point of SR
⇒ SL= LR
Similarly if we construct a line from L which is parallel to SR
This gives N is the midpoint of QR
Hence LN∥ SQ and L and N are mis points of SR and QR respectively
And LN = 1/2 SQ (mid-point theorem)
Question 3.
In figure 5.40, ΔABC is an equilateral triangle. Points F,D and E are midpoints of side AB, side BC, side AC respectively. Show that ΔEFD is an equilateral triangle.
Given F, D and E are mid-point of AB, BC and AC of the equilateral ΔABC ∴ AB =BC = AC
So by mid-point theorem
Line joining mid-points of two sides of a triangle is 1/2 of the parallel third side.
∴ FE = 1/2 BC =
Similarly, DE = 1/2 AB
And FD = 1/2 AC
But AB =BC = AC
⇒ 1/2 AB = 1/2 BC = 1/2 AC
⇒ DE = FD = FE
Since all the sides are equal ΔDEF is a equilateral triangle.
Question 4.
In figure 5.41, seg PD is a median of ΔPQR, Point T is the midpoint of seg PD. Produced QT intersects PR at M. Show that
[Hint : draw DN || QM.]
PD is median so QD = DR (median divides the side opposite to vertex into equal halves)
T is mid-point of PD
⇒ PT = TD
In ΔPDN
T is mid-point and is ∥ to TM (by construction)
⇒TM is mid-point of PN
PM =MN……………….1
Similarly in ΔQMR
QM ∥ DN (construction)
D is mid –point of QR
⇒ MN = NR…………………..2
From 1 and 2
PM = MN = NR
Or PM = 1/3 PR
hence proved
###### Problem Set 5
Question 1.
Choose the correct alternative answer and fill in the blanks.
If all pairs of adjacent sides of a quadrilateral are congruent then it is called ....
A. rectangle
B. parallelogram
C. trapezium
D. rhombus
As per the properties of a rhombus:- A rhombus is a parallelogram in which adjacent sides are equal(congruent).
Question 2.
Choose the correct alternative answer and fill in the blanks.
If the diagonal of a square is 12√2 cm then the perimeter of square is ......
A. 24 cm
B. 24√2 cm
C. 48 cm
D. 48√2 cm
Here d= 12√2 = √2 s where s is side of square
Given diagonal = 20 cm
⇒ s =
Therefore, perimeter of the square is 4s = 4 x 12
= 48cm. (C)
Question 3.
Choose the correct alternative answer and fill in the blanks.
If opposite angles of a rhombus are (2x)° and (3x - 40)° then value of x is .......
A. 100°
B. 80°
C. 160°
D. 40°
As rhombus is a parallelogram with opposite angles equal
⇒ 2x = 3x -40
x= 40°
Question 4.
Adjacent sides of a rectangle are 7 cm and 24 cm. Find the length of its diagonal.
Adjacents sides are 7cm and 24 cm
In a rectangle angle between the adjacent sides is 90°
⇒ the diagonal is hypotenuse of right Δ
By pythagorus theorem
Hypotenuse2 = side2 + side2
Hypotenuse2 = 49 + 576 =
length of the diagonal = 25cm
Question 5.
If diagonal of a square is 13 cm then find its side.
given Diagonal of the Square = 13cm
The angle between each side of the square is 90°
Using Pythagoras theorem
Hypotenuse2 = side2 + side2
Side = cm
Question 6.
Ratio of two adjacent sides of a parallelogram is 3 : 4, and its perimeter is 112 cm. Find the length of its each side.
In a parallelogram opposite sides are equal
Let the sides of parallelogram be x and y
2x+ 2y =112 and given
⇒
⇒7y = 224
y= 32
x= 24
four sides of the parallelogram are 24cm , 32 cm, 24cm, 32cm.
Question 7.
Diagonals PR and QS of a rhombus PQRS are 20 cm and 48 cm respectively. Find the length of side PQ.
According to the properties of Rhombus diagonals of the rhombus bisects each other at 90°
In the rhombus PQRS
SO = OQ = 10 cm
PO= OR = 12cm
So in ΔPOQ
∠ POQ = 90°
⇒ PQ is hypotenuse
By Pythagoras theorem,
102 + 242 =PQ2
100 + 576 = PPQ2
676 = PQ2
26cm = PQ Ans
Question 8.
Diagonals of a rectangle PQRS are intersecting in point M. If ∠QMR = 50° then find the measure of ∠MPS.
The figure is given below:
Given PQRS is a rectangle
⇒ PS ∥ QR (opposite sides are equal and parallel)
QS and PR are transversal
So ∠ QMR = ∠ MPS (vertically opposite angles)
Given ∠ QMR = 50°
∴ ∠ MPS = 50°
Question 9.
In the adjacent Figure 5.42, if seg AB || seg PQ, seg AB ≅ seg PQ, seg AC || seg PR, seg AC seg PR then prove that, seg BC || seg QR and seg BC seg QR.
Given
AB ∥ PQ
AB ≅ PQ ( or AB = PQ )
⇒ ABPQ is a parallelogram (pair of opposite sides is equal and parallel)
⇒ AP ∥ BQ and AP ≅ BQ……………..1
Similarly given,
AC ∥ PR and AC ≅ PR
⇒ACPR is a parallelogram (pair of opposite sides is equal and parallel)
⇒ AP ∥ CR and AP ≅ CR ……………………2
From 1 and 2 we get
BQ ∥ CR and BQ ≅ CR
Hence BCRQ is a parallelogram with a pair of opposite sides equal and parallel.
Hence proved.
Question 10.
In the Figure 5.43, ABCD is a trapezium. AB || DC. Points P and Q are midpoints of seg AD and seg BC respectively.
Then prove that, PQ || AB and
Given AB ∥ DC
P and Q are mid points of AD and BC respectively.
Construction :- Join AC
The figure is given below:
P is mid point of AD and PQ is ∥ DC the part of PQ which is PO is also ∥ DC
By mid=point theorem
A line from the mid-point of a side of Δ parallel to third side, meets the other side in the mid-point
⇒ O is mid-point of AC
⇒ PO = 1/2 DC…………..1
Similarly in Δ ACB
Q id mid-point of BC and O is mid –point of AC
⇒ OQ∥ AB and OQ = 1/2 AB………………2
PO + OQ = 1/2 (DC+ AB)
PQ = 1/2 (AB +DC)
And PQ ∥ AB
Hence proved.
Question 11.
In the adjacent figure 5.44, ABCD is a trapezium. AB || DC. Points M and N are midpoints of diagonal AC and DB respectively then prove that MN || AB.
Given AB ∥ DC
M is mid-point of AC and N is mid-point of DB
Given ABCD is a trapezium with AB ∥ DC
P and Q are the mid-points of the diagonals AC and BD respectively
The figure is given below:
To Prove:- MN ∥ AB or DC and
In ΔAB
AB || CD and AC cuts them at A and C, then
∠1 = ∠2 (alternate angles)
Again, from ΔAMR and ΔDMC,
∠1 = ∠2 (alternate angles)
AM = CM (since M is the mid=point of AC)
∠3 = ∠4 (vertically opposite angles)
From ASA congruent rule,
ΔAMR ≅ ΔDMC
Then from CPCT,
AR = CD and MR = DM
Again in ΔDRB, M and N are the mid points of the sides DR and DB,
then PQ || RB
⇒ PQ || AB
⇒ PQ || AB and CD ( ∵ AB ∥ DC)
Hence proved.
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Important-formula
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Back to the class
Section 7.1 #6: Set up the definite integral that gives the area in the region bounded by $y_1=(x-1)^3$ and $y_2=x-1$:
Solution: First draw the region:
We will use a $\mathrm{d}x$ integral. It will require two such integrals. For the left region, the function on top is $y_1$ and the function on the bottom is $y_2$ and in the right region, the function on top is $y_2$ and the function on the bottom is $y_1$. Therefore we may compute the area with $$\mathrm{Area}=\mathrm{LeftArea}+\mathrm{RightArea}=\displaystyle\int_0^1 (x-1)^3 - (x-1) \mathrm{d}x + \displaystyle\int_1^2 (x-1) - (x-1)^3 \mathrm{d}x.$$ Section 7.1 #16: Find the area of the region bounded by $y=x^2$ and $y=6-x$ by integrating...
a.) with respect to $x$, and
b.) with respect to $y$.
c.) Compare your results. Which was simpler?
Solution: First draw the region:
For a.), the top function is always the linear function and the bottom function is the quadratic. Therefore the area is given by $$\begin{array}{ll} \mathrm{Area}&=\displaystyle\int_{-3}^2 (6-x) - x^2 \mathrm{d}x \\ &= 6x - \dfrac{x^2}{2} - \dfrac{x^3}{3} \Bigg|_{-3}^2 \\ &= \left( 12 - \dfrac{4}{2} - \dfrac{8}{3} \right) - \left( -18 - \dfrac{9}{2} + \dfrac{27}{3} \right) \\ &= \dfrac{125}{6}. \end{array}$$ For b.), the right and left functions change at $(2,4)$. First rewrite $y=x^2$ as $x=\pm \sqrt{y}$ and rewrite $y=6-x$ as $x=6-y$. Between $y=0$ and $y=4$, the function $x=\sqrt{y}$ is on the right and the function $x=-\sqrt{y}$ is on the left. Between $y=4$ and $y=9$, the function $x=6-y$ is on the right and the function $x=-\sqrt{y}$ is on the left. Therefore, $$\begin{array}{ll} \mathrm{Area} &= \displaystyle\int_0^4 \sqrt{y} - (-\sqrt{y}) \mathrm{d}y + \displaystyle\int_4^9 (6-y) - (-\sqrt{y}) \mathrm{d}y \\ &=2\displaystyle\int_0^4 y^{\frac{1}{2}} \mathrm{d}y + \displaystyle\int_4^9 6-y+y^{\frac{1}{2}} \mathrm{d}y \\ &= 2 \left[ \dfrac{y^{\frac{3}{2}}}{\frac{3}{2}} \right|_0^4 + \left[ 6y - \dfrac{y^2}{2} + \dfrac{y^{\frac{3}{2}}}{\frac{3}{2}} \right|_4^9 \\ &=\dfrac{4}{3} \left[ 4^{\frac{3}{2}} - 0 \right] + \left[ \left(54 - \dfrac{81}{2} + \dfrac{2}{3}9^{\frac{3}{2}} \right) - \left( 24 - \dfrac{16}{2} + \dfrac{2}{3} 4^{\frac{3}{2}} \right)\right] \\ &= \dfrac{32}{3} + \left[ \dfrac{63}{2} - \dfrac{64}{3} \right] \\ &= \dfrac{125}{6}. \end{array}$$ For c.), we got the same result both ways and the $\mathrm{d}x$ integral was much simpler!
Section 7.1 #17: Sketch the region bounded by the graphs of the equations and find the area of region: $y=x^2-1$, $y=-x+2$, $x=0$, and $x=1$.
Solution: First find intersection points of the first two curves by setting them equal and solving:
$$x^2-1 = -x+2,$$ which is algebraically equivalent to $$x^2+x-3=0.$$ Solving this equation (using the quadratic formula) yields the roots $$x= \dfrac{-1 \pm \sqrt{1^2-4(1)(3)}}{2} = -\dfrac{1}{2} \pm \dfrac{\sqrt{13}}{2}.$$ Now sketch the region:
It is easier to do this as a $\mathrm{d}x$ integral: compute $$\begin{array}{ll} \mathrm{Area} &= \displaystyle\int_0^1 (-x+2) - (x^2-1) \mathrm{d}x \\ &= \displaystyle\int_0^1 -x^2-x+3 \mathrm{d}x \\ &= \left. -\dfrac{x^3}{3} - \dfrac{x^2}{2} + 3x \right|_0^1 \\ &= \left( - \dfrac{1}{3} - \dfrac{1}{2} + 3 \right) - 0 \\ &= \dfrac{13}{6}. \end{array}$$ Section 7.1 #21: Sketch the region bounded by the graphs of the equations and find the area of region: $y=x$, $y=2-x$, $y=0$.
Solution: First find the intersection points between $y=x$ and $y=2-x$ by setting them equal: $$x=2-x,$$ hence $$x=1.$$ Therefore $(1,1)$ is the intersection point of the two curves. Now sketch the region:
It is easier to do this integral as a $\mathrm{d}y$ (notice that if we did a $\mathrm{d}x$, then it would require two integrals). The "right" function is $x=2-y$ and the "left" function is $x=y$. Therefore we compute $$\begin{array}{ll} \mathrm{Area} &= \displaystyle\int_0^1 (2-y) - y \mathrm{d}y \\ &=\displaystyle\int_0^1 2-2y \mathrm{d}y \\ &=2y - y^2 \Bigg|_0^1 \\ &= (2-1) - 0 \\ &= 1. \end{array}$$
Section 7.2 #22: Sketch the region bounded by the graphs of the equations and find the area of region: $y=\dfrac{4}{x^3}$, $y=0$, $x=1$, and $x=4$.
Solution: Sketch the region:
This is easier to do as a $\mathrm{d}x$ integral, so compute $$\begin{array}{ll} \displaystyle\int_1^4 \dfrac{4}{x^3} \mathrm{d}x &= 4 \displaystyle\int_1^4 x^{-3} \mathrm{d}x \\ &= -\dfrac{4}{2} x^{-2} \Bigg|_1^4 \\ &= -\dfrac{4}{2} \dfrac{1}{4^2} + \dfrac{4}{2} \dfrac{1}{1} \\ &= 2 - \dfrac{1}{8} \\ &= \dfrac{15}{8}. \end{array}$$ |
# Maths Year 2 Spring Addition and Subtraction
Each unit has everything you need to teach a set of related skills and concepts.
First time using Hamilton Maths?
The PowerPoint incorporates step-by-step teaching, key questions, an in-depth mastery investigation, problem-solving and reasoning questions - in short, everything you need to get started.
All the other resources are there to support as-and-when required. Explore at your leisure - and remember that we are always here to answer your questions.
## Unit 1 Use facts, patterns, PV to add/subtract (suggested as 3 days)
### Objectives
Add/subtract numbers using facts, patterns and place value
Unit 1: ID# 2417
National Curriculum
PV (vi)
Hamilton Objectives
6. Use place value and number facts to solve problems, e.g. 60 – = 20.
7. Know securely number pairs for all the numbers up to and including 20.
8. Know different unit patterns when adding or subtracting in numbers up to 100.
9. Add two or three single-digit numbers, using number facts and counting up.
13. Show that addition of 2 numbers can be done in any order (commutative) and subtraction cannot.
### Planning and Activities
Day 1 Teaching
Use cards on the Interactive Whiteboard to form: 6 + 5 + 3 + 5 + 4 = [ ]. We can change the order to make it easier. Children discuss with a partner. Take feedback, drawing out pairs that make 10. Move cards round to show:
5 + 5 + 6 + 4 + 3 = [ ], then rewrite as:
10 + 10 + 3 = [ ], then with the answer:
10 + 10 + 3 = 23. It was easy. We just used pairs to 10 and place value! Repeat with other similar additions.
Group Activities
-- Pick five or more 1–10 number cards to add; identify pairs to 10, doubles, and number facts to help add them.
-- Roll three or more dice and identify pairs to 10, doubles, and number facts to help add.
Day 2 Teaching
Write: 7 + 7, 10 + 6, 9 + 3, 6 + 4, 15 + 5, 4 + 7, 18 + 5, 20 + 7. Discuss how some can be solved using place value, putting 10s and 1s together, like 10 + 6. Some use number facts. Some use counting on. Sort into two groups: 1) counting on; 2) place value/ number facts. See important note on teaching/ activities document.
Group Activities
Use the ‘Cross Additions’ in-depth problem-solving investigation below as today’s group activity.
Or, use this activity:
-- Sort and solve additions according to how they might best be solved (counting on, place value or number facts).
Day 3 Teaching
Write: 26 – 6, 30 – 7, 13 – 4, 20 – 8, 9 – 2, 14 – 4, 23 – 7. Discuss and sort into groups: those best worked out by counting back, and those most efficiently solved using place value/number facts.
Group Activities
-- Sort and solve subtractions according to how they might best be solved (counting back, place value or number facts).
### You Will Need
• Number cards or text boxes on the Interactive Whiteboard
• 1–10 number cards (enough for each pair)
• 1–10 or 0-9 dice
• Coloured whiteboard pens
• Place Value cards
• ‘Using number facts and place value for addition’ sheet (see resources)
• Two hoops
• 0–100 beaded lines (see resources)
• Number cards (see resources)
• ‘Using number facts and place value for subtraction’ sheet (see resources)
### Mental/Oral Maths Starters
Day 1
Pairs to 10 (pre-requisite skills)
Suggested for Day 2
Pairs to 20 (simmering skills)
Suggested for Day 3
Saying 1 or 2 more than a 2-digit number (simmering skills)
### Worksheets
Day 1
Look for known number facts to help efficiently add a series of single-digit numbers.
Day 2
Code a series of additions according to how they might best be solved (counting on, place value or number facts).
Day 3
Code a series of subtractions according to how they might best be solved (counting back, place value or number facts).
### Mastery: Reasoning and Problem-Solving
• Write an addition of five different numbers, all less than 10, with a total of 28.
• Which one number less than 10 fits the gaps?
6 + ☐ + 7 + ☐ + ☐ = 25
• Solve each of these additions using a different method. Say how you did each one.
30 + 9 =
17 + 5 =
4 + 7 + 6 =
• Solve each of these subtractions using a different method. Say how you did each one.
25 – 5 =
14 – 6 =
58 – 4 =
Using number facts including pairs to 10 children find totals of five numbers less than 10 to give different totals.
### Extra Support
Spot the 10
Adding three single-digit numbers, spotting a pair with a total of 10
## Unit 2 Use number line & 100 grid to +/- (suggested as 4 days)
### Objectives
Use number line & 100 grid to add and subtract
Unit 2: ID# 2423
National Curriculum
Hamilton Objectives
10. Add a 2-digit number and tens; add two 2-digit numbers that total less than 100 by counting on in tens and ones.
11. Count back in ones or tens or use number facts to take away, e.g. 27 – 3, or 54 – 20.
### Planning and Activities
Day 1 Teaching
Use a 1–100 landmarked line to model adding two 10s to 45, and then another 4 to complete 45 + 24. Model adding other pairs of numbers: 65 + 24, 52 + 30, 52 + 33, 72 + 33 and 46 + 32, using the landmarked line.
Group Activities
Use the ‘Lines of Numbers’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
-- Add 2-digit numbers where the ones bridge through a 10.
Day 2 Teaching
Show a 1–100 grid. Spider and Fly will help us subtract! Write 54 – 23. Where do we need to put Spider? 54. Subtract 20, jumping back 10 to 44 and another 10 to 34. Then count back 3 with Fly to end up on 31. Repeat with 86 – 32 and 77 – 25.
Group Activities
-- Subtract 2-digit numbers using a 1–100 grid and Spider and Fly.
-- Find mystery subtraction numbers following the movements of Spider and Fly.
Day 3 Teaching
Write 65 – 24 and discuss how to solve this. We count back 20 using Spider and another 4 using Fly. Model solving this using a landmarked line, drawing a hop back to subtract 20, then a second hop to subtract 4. Repeat for other subtractions.
Group Activities
-- Subtract 2-digit numbers using landmarked and beaded lines.
-- Predict which subtractions will have an answer less than a given number.
Day 4 Teaching
Show 1–100 square on the Interactive Whiteboard. Put Spider on 47 and write 47 – 32 =. How can Spider and Fly help us? Children discuss with a partner. Remind children how Spider and Fly like to move. Spider jumps back 3 tens to 17. Fly slides back the rest. So 47 – 32 = 15! Repeat with other subtractions.
Group Activities
-- Subtract 2-digit numbers using a 1–100 grid and Spider and Fly.
-- Subtract 2-digit numbers using landmarked or beaded lines.
### You Will Need
• ‘0 to 100 landmarked lines’ (see resources)
• ‘0 to 100 beaded lines’ (see resources)
• ‘Adding 2-digit numbers’ (see resources)
• Counters (or Plastic fly and spider)
• 1–100 square on the Interactive Whiteboard
• Additional activity sheets (see resources)
• Cubes
• Calculators
• Dice labelled 10, 20, 30, 40, 20, 30 and 1–6 dice
### Mental/Oral Maths Starters
Day 1
Count on and back in 10s (pre-requisite skills)
Day 2
Add and subtract multiples of 10 to/ from 2-digit numbers (pre-requisite skills)
Suggested for Day 3
Bridging 10 (simmering skills)
Suggested for Day 4
Counting on in 10s (simmering skills)
### Worksheets
Day 1
Day 2
Use a 1–100 grid to subtract 2-digit numbers.
Day 3
Use landmarked lines to subtract 2-digit numbers.
Day 4
Use a 1–100 grid to subtract 2-digit numbers.
### Mastery: Reasoning and Problem-Solving
Write the missing number in each bar diagram:
Diagram 1
? 54 23
Diagram 2
? 32 36
Diagram 3
? 85 14
Write the missing number in each bar diagram:
Diagram 1
78 ? 23
Diagram 2
46 32 ?
Diagram 3
95 ? 14
In-depth Investigation: Lines of Numbers
Children create sequences based on Fibonacci and explore patterns of odd and even numbers, making connections with general statements about addition.
### Extra Support
Secret Spider
Adding and subtracting 10 using the 1-100 grid
## Unit 3 Find money totals: solve word problems (suggested as 3 days)
### Objectives
Find money totals: solve word problems
Unit 3: ID# 2437
National Curriculum
AS (i) (ii) (iii) (v)
Meas (iii) (iv) (v)
Hamilton Objectives
27. Recognise/use symbols for pounds (£) and pence (p); combine amounts, find different combinations of coins that give the same amount.
28. Solve simple problems in a practical context; add and subtract pence and pounds, including finding and giving change.
15. Solve problems involving addition and subtraction of numbers, quantities and measures, using recall of number facts and appropriate models and images.
### Planning and Activities
Day 1 Teaching
Use BBC website to play a game of ‘Igloo shopping’. Children have a pot of money and choose coins to pay for different objects. Check on screen. Move to the hard level to select and pay amounts like £4.23.
Group Activities
Use the ‘Coin Trios’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
-- Play an online shopping game to make given amounts, offering alternative combinations.
-- Use coins to make given amounts, offering alternative combinations.
Day 2 Teaching
Set up a shop with priced items. Children select two items and work out the total using a familiar strategy based on adding 10ps and then the 1ps to the larger starting number. Repeat. Make errors in your addition as the teacher and have children correct you, modelling the method.
Group Activities
-- Shop for two items with 2-digit prices, adding to find the total.
-- Play Cuddly Toy Sale game online, adding two 2-digit amounts.
Day 3 Teaching
Display a Word problem. Discuss the strategies that might be needed to solve it. Take suggestions. Write a number sentence to represent the problem. Children display a subtraction or addition symbol in response to your reading more word problems to show which operation they would use to solve them.
Group Activities
-- Sort and solve addition and subtraction word problems.
-- Create addition and subtraction word problems.
### You Will Need
• Pots of coins, cubes & drawing materials
• Mini whiteboards and pens
• Priced items ‘to sell in your shop’
• Items to sell & blank tags
• Paper with + and – signs
• 1–100 square on the Interactive Whiteboard
• ‘Word problems’ sheet (see resources)
• Counters (or a plastic spider and fly)
• 1–100 number grid (see resources)
• 0–100 landmarked line (see resources)
### Mental/Oral Maths Starters
Day 1
Coin recognition (pre-requisite skills)
Day 2
Ten more/ten less (pre-requisite skills)
Suggested for Day 3
Saying 1 or 2 less than a 2-digit number (simmering skills)
### Worksheets
Day 1
Suggest ways to make given amounts, using at least 3 coins.
Day 2
Muffin shop: combine the costs of muffins and icing to find the total cost of a cake.
Day 3
Identify the operation for a series of word problems, then solve them.
### Mastery: Reasoning and Problem-Solving
• Amit makes £1.15 in three different ways using different coins. He never uses any ‘copper’ coins.
Suggest these three ways. One way uses 5 coins. Can you find it?
• Choose two of these cards.
Find and write their total.
How many different totals can you make?
34p 53p
44p 25p
In-depth Investigation: Coin Trios
Children make amounts of money using three coins, speculating whether larger amounts of money can be made in more ways or not.
### Extra Support
Coin Counting On
Finding the total of pairs of coins (up to 20p)
## Unit 4 Add and double by partitioning (suggested as 3 days)
### Objectives
Unit 4: ID# 2447
National Curriculum
PV (vi)
Hamilton Objectives
6. Use place value and number facts to solve problems, e.g. 60 – = 20.
10. Add a 2-digit number and tens; add two 2-digit numbers that total less than 100 by counting on in tens and ones.
15. Solve problems involving addition and subtraction of numbers, using recall of number facts and appropriate models and images.
### Planning and Activities
Day 1 Teaching
Launch the ITP Place value; use 2 pairs of cards to show double 23. Click to partition, then model doubling 20 and doubling 3.
Record: 23 + 23
= 20 + 20 + 3 + 3
= 40 + 6
= 46. Repeat for double 34, 42, 46. Model halving 46 in a similar way. Halve 24, 66 and 34.
Group Activities
Use the ‘Jack’s amazing beanstalk’ in-depth problem-solving investigation below as today’s group activity.
Or, use these activities:
-- Play Hit the Button online: doubling and halving game, applying partitioning.
-- Play halving and doubling ‘pairs’ game.
Day 2 Teaching
Model partitioning to add 34 + 23. Use the same method as for doubling and record in same way:
34 + 23
= 30 + 20 + 4 + 3
= 50 + 7
= 57. Repeat for 45 + 24, 36 + 22 and 46 + 25.
Group Activities
-- Play an adding two 2-digit numbers card game: collaborative calculations.
-- Add two 2-digit prices to find the total cost.
-- Play an adding two 2-digit numbers card game: first to answer challenge.
Day 3 Teaching
In pairs, children make 65 and 24 using place value cards. Mimic on ITP Place Value. Combine the 10s and 1s to find a total.
Record: 65 + 24
= 60 + 20 + 5 + 4
= 80 + 9
= 89. OR we can count up in 10s and 1s (so add 20 then 4). Show an empty/ landmarked number line jotting to work this out. Which do the children prefer? Why? Share other sums discussing strategies.
Group Activities
-- Play an adding two 2-digit numbers card game: first to answer challenge.
-- Suggest pairs of numbers that make a given total.
### You Will Need
• ITP: Place Value
• Place value cards
• Internet access
• Blue and green cards
• 10s Number cards
• 1s Number cards
• Whiteboards and pens
### Mental/Oral Maths Starters
Day 1
Double multiples of 5 to 50 (pre-requisite skills)
Day 2
Add multiples of 10 (pre-requisite skills)
Suggested for Day 3
Number facts (simmering skills)
### Worksheets
Day 1
Calculate and match pairs of halves and doubles.
Day 2
Add 2-digit numbers using partitioning and record jottings.
Day 3
Choose and ue different strategies to add pairs of 2-digit numbers (partitioning or counting on in 10s then 1s).
### Mastery: Reasoning and Problem-Solving
• Double 13 = Δ
Double Δ is ✦
Double ✦ is ☐
Find the value of Δ, ✦ and ☐
• Fill in the missing numbers:
65 + 24
add the 10s: 60 + ☐ = ☐
add the 1s: ☐ + 4 = ☐
so 65 + 24 = ☐
46 + 35
☐ + 30 = ☐
6 + ☐ = ☐
so 46 + 35 = ☐
• Explain why it is probably easier to do these two additions in different ways:
a) 65 + 21
b) 56 + 35
Now find both totals
In-depth Investigation: Jack's Amazing Beanstalk
Children double numbers and add to a running total, and then look for patterns to predict how much a beanstalk will grow each day. |
##### Geometry: 1001 Practice Problems For Dummies (+ Free Online Practice)
The first four Pythagorean triple triangles are the favorites of geometry problem-makers. These triples — especially the first and second in the list that follows — pop up all over the place in geometry books. (Note: The first two numbers in each of the triple triangles are the lengths of the legs, and the third, largest number is the length of the hypotenuse).
Here are the first four Pythagorean triple triangles:
• The 3-4-5 triangle
• The 5-12-13 triangle
• The 7-24-25 triangle
• The 8-15-17 triangle
You’d do well to memorize these Fab Four so you can quickly recognize them on tests.
## Forming irreducible Pythagorean triple triangles
As an alternative to counting sheep some night, you may want to see how many other Pythagorean triple triangles you can come up with.
The first three on the above list follow a pattern. Consider the 5-12-13 triangle, for example. The square of the smaller, odd leg
is the sum of the longer leg and the hypotenuse (12 + 13 = 25). And the longer leg and the hypotenuse are always consecutive numbers. This pattern makes it easy to generate as many more triangles as you want. Here’s what you do:
1. Take any odd number and square it.
2. Find the two consecutive numbers that add up to this value.
40 + 41 = 81
You can often just come up with the two numbers off the top of your head, but if you don’t see them right away, just subtract 1 from the result in Step 1 and then divide that answer by 2:
That result and the next larger number are your two numbers.
3. Write the number you squared and the two numbers from Step 2 in consecutive order to name your triple.
You now have another Pythagorean triple triangle: 9-40-41.
Here are the next few Pythagorean triple triangles that follow this pattern:
This list is endless — capable of dealing with the worst possible case of insomnia. And note that each triangle on this list is irreducible; that is, it’s not a multiple of some smaller Pythagorean triple triangle (in contrast to the 6-8-10 triangle, for example, which is not irreducible because it’s the 3-4-5 triangle doubled).
When you make a new Pythagorean triple triangle (like the 6-8-10) by blowing up a smaller one (the 3-4-5), you get triangles with the exact same shape. But every irreducible Pythagorean triple triangle has a shape different from all the other irreducible triangles.
## Forming further Pythagorean triple triangles
The 8-15-17 triangle is the first Pythagorean triple triangle that doesn’t follow the pattern mentioned previously. Here’s how you generate triples that follow the 8-15-17 pattern:
1. Take any multiple of 4.
Say you choose 12.
2. Square half of it.
3. Take the number from Step 1 and the two odd numbers on either side of the result in Step 2 to get a Pythagorean triple triangle.
12-35-37
The next few triples in this infinite set are
By the way, you can use this process for the other even numbers (the non-multiples of 4) such as 10, 14, 18, and so on. But you get a triangle such as the 10-24-26 triangle, which is the 5-12-13 Pythagorean triple triangle blown up to twice its size, rather than an irreducible, uniquely-shaped triangle. |
# 2-dimensional Shapes and 3-Dimensional Shapes
Last Updated on July 15, 2020 by Alabi M. S.
MATHEMATICS
THEME: Mensuration and Geometry
TOPIC: Shapes
PERFORMANCE OBJECTIVES
By the end of the lesson, the pupils should have attained the following objectives (cognitive, affective and psychomotor):
1. name two and three dimensional shapes;
1. they should be able to draw line of symmetry;
1. identify and appreciate 3 – dimensional shapes;
1. solve quantitative aptitude problems relating to polygon.
ENTRY BEHAVIOR
The pupils are required to already have taught and learned types of triangles and quadrilateral.
INSTRUCTIONAL MATERIALS
The teacher will teach the lesson with the aid of charts of two and three dimensional figures.
METHOD OF TEACHING
MAIN REFERENCE MATERIALS
1. Prime Mathematics book 6 (PMB 296);
1. New Method Mathematics book 6 (NMM 213).
CONTENT OF THE LESSON/LESSON NOTE
LESSON 1
PERIOD:
DATE:
TIME:
• INTRODUCTION 2 – DIMENSIONAL SHAPES NOT EXCEEDING OCTAGON
2 – Dimensional shapes have length and width (breadth). Polygons are 2-dimensional shapes. Polygons have a straight sides. They are made of straight lines, and the shape is “closed” (all the lines connect up).
The name of each shape tells you the number of sides it has.
• TYPES OF POLYGONS
A regular polygon has all angles equal and all sides equal, otherwise it is irregular.
LESSON EVALUATION
LESSON 2
PERIOD:
DATE:
TIME:
• CONTENT – TRIANGLE AND QUADRILATERAL
Triangle
Triangle is a three sided figure (shape). It is classified by number of sides and size of angles or both.
Quadrilateral is four – sided figure (shape). There are special types of quadrilateral.
LESSON EVALUATION
1. ___________ is a three sided figure. a. Circle b. Pentagon c. Rectangle d. Triangle
2. ___________ is a four sided figure. a. Circle b. Pentagon c. Rectangle d. Triangle
3. ___________ has an equal sides and angles. a. Square b. Rectangle c. Kite d. triangle
4. Draw the lines of symmetry on triangles and rectangles.
5. Mention the number of lines of symmetry of triangles and rectangles.
LESSON 3
PERIOD:
DATE:
TIME:
• CONTENT – Pentagon means 5 sides, hexagon means 6, heptagon means 7 and octagon means 8. The shape below is pentagon. It has five sides.
Pentagon Hexagon
Heptagon Octagon
LESSON EVALUATION
1. Draw the lines of symmetry each of the above figures.
2. Mention the number of lines of symmetry of each figures above.
3. Complete the table below:
LESSON 4
PERIOD:
DATE:
TIME:
• CONTENT – 3 – DIMENSIONAL SHAPES
3 – Dimensional shapes have height, width and depth.
Examples are cube, cuboid, cylinder, sphere, cone, triangular prism, square – base pyramid, and regular tetrahedron.
Further exercise will be taken from the listed reference materials.
PRESENTATION
To deliver the lesson, the teacher adopts the following steps:
• To introduce the lesson, the teacher revises the previous lesson- time. Based on this, he/she asks the pupils some questions;
• Guides the pupils to recall the names and features of given shapes;
• Pupils Activities – Identify a given plane shapes in terms of its name and essential features e.g. square, rectangle, triangle, rhombus, parallelograms etc.
• Guides pupils to identify lines of symmetry, number of sides, number of angles, etc.
• Pupils Activities – Using paper cut outs to determine the number of lines of symmetry, number of angles and number of sides.
• Guides pupils relate 3 – Dimensional to the world as around them;
• Pupils Activities – Identify and appreciate 3 – dimensional shapes e.g. Cube, cone, triangular, prison i.e. Closed or solid object. Examples of cube – sugar, maggi, cone – ice cream, cuboid – boxes, etc.
• Evaluation
CONCLUSION
• To conclude the lesson for the week, the teacher revises the entire lesson and links it to the following week’s lesson (scale drawing).
Smart Teachers Plan Lesson Notes - ClassRoomNotes support teachers with hands-on lesson plans/notes, printable and thoughtful teaching resources. @ClassRoomNotes - We always love to hear from you always. Stay connected with your classroom.
error: |
FutureStarr
A 1 4 Squared As a Fraction
# 1 4 Squared As a Fraction
via GIPHY
An alternative method for finding a common denominator is to determine the least common multiple (LCM) for the denominators, then add or subtract the numerators as one would an integer. Using the least common multiple can be more efficient and is more likely to result in a fraction in simplified form. In the example above, the denominators were 4, 6, and 2. The least common multiple is the first shared multiple of these three numbers. (
### Use
Unlike adding and subtracting integers such as 2 and 8, fractions require a common denominator to undergo these operations. One method for finding a common denominator involves multiplying the numerators and denominators of all of the fractions involved by the product of the denominators of each fraction. Multiplying all of the denominators ensures that the new denominator is certain to be a multiple of each individual denominator. The numerators also need to be multiplied by the appropriate factors to preserve the value of the fraction as a whole. This is arguably the simplest way to ensure that the fractions have a common denominator. However, in most cases, the solutions to these equations will not appear in simplified form (the provided calculator computes the simplification automatically). Below is an example using this method.
Use this fractions calculator to easily perform calculations with fractions. Add, subtract, multiply, and divide fractions, as well as raise a fraction to power (fraction or not). Supports evaluation of mixed fractions (e.g. "2 1/3") and negative fractions (e.g. "-2/3"). Use "pi" or "π" for the number Pi. Powerful advanced mode for evaluating whole expressions with fractions. (Source: www.gigacalculator.com)
### Work
First, simply input the values a,b,c,d for the fractions $$\frac{a}{b}$$ and $$\frac{c}{d}$$, then the mathematical operation you wish to perform (+, -, x, /). The calculator will instantly and accurately perform the operation and give the answer in the simplest form. You can also use the calculator to check your work that you’ve done manually.
In mathematics, some problems mix fractions with whole numbers. Learn how to multiply fractions with whole numbers, explore the rule for multiplying these two different types of numbers, and gain understanding by working on an example. (Source: study.com)
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# What is a complement in math sets?
## What is a complement in math sets?
What Is the Complement of a Set? The complement of set A is defined as a set that contains the elements present in the universal set but not in set A. For example, Set U = {2,4,6,8,10,12} and set A = {4,6,8}, then the complement of set A, A′ = {2,10,12}.
### What is the complement of a ∩ B?
A Intersection B Complement is equal to the union of the complements of the sets A and B. Mathematically, it is written as (A ∩ B)’ = A’ U B’. It is one of the important De-Morgan’s Law of sets.
#### What is the complement of set Q?
The complement of set A, denoted by A’ , is the set of all elements in the universal set that are not in A. The number of elements of A and the number of elements of A ‘ make up the total number of elements in U . Q ‘ = {–3, –2, –1, 2, 3}.
What is the complement of A and B?
Definition. If A and B are sets, then the relative complement of A in B, also termed the set difference of B and A, is the set of elements in B but not in A.
What is a complement of a number?
A. The number derived by subtracting a number from a base number. For example, the tens complement of 8 is 2. In set theory, complement refers to all the objects in one set that are not in another set.
## How do you find the complement?
To find the complement of an angle, subtract that angle’s measurement from 90 degrees. The result will be the complement. The measure of the complementary angle is 50 degrees.
### How do you find AUB )’?
The number of elements in A union B can be calculated by counting the elements in A and B and taking the elements that are common only once. The formula for the number of elements in A union B is n(A U B) = n(A) + n(B) – n(A ∩ B).
#### What is the complement of 0?
Ones’ complement
Bits Unsigned value Ones’ complement value
0000 0010 2 2
0000 0001 1 1
0000 0000 0 0
1111 1111 255 −0
How do you find AB in a set?
An Example To see how the difference of two sets forms a new set, let’s consider the sets A = {1, 2, 3, 4, 5} and B = {3, 4, 5, 6, 7, 8}. To find the difference A – B of these two sets, we begin by writing all of the elements of A, and then take away every element of A that is also an element of B.
How do you find r complement?
To find r’s complement, just add 1 to the calculated (r-1)’s complement. Step 1: Identify the base (or) radix. Here r = 8. Step 2: Since 7 is the largest digit in the number system, subtract each digit of given number from 7 i.e. if it’s a three digit number, subtract the number from 777.
## What is the complement of a in math?
The complement of A, denoted by A’, consists of all students in The Kewl school that are not in Mrs. Glosser’s class. Recall that a Universal Set is the set of all elements under consideration, denoted by capital , and that all other sets are subsets of the Universal Set.
### How do you find the complement of a universal set?
To make it more clear consider a universal set U U of all natural numbers less than or equal to 20. Let the set A A which is a subset of U U be defined as the set which consists of all the prime numbers. Now the complement of this set A consists of all those elements which is present in the universal set but not in A A.
#### What is the complement of 5 and 6 in set theory?
All elements (from a Universal set) NOT in our set. Then the complement of {5,6} is {1,2,3,4}. Symbol is a little dash in the top-right corner. Or a little “C” in the top-right corner. Together the set and its complement make the Universal set.
What is the double complement of a set?
The complement of the set A′, where A′ itself is the complement of A, the double complement of A is thus A itself. In earlier example, U = {1, 2, 3, 4, 5} and A = {4, 5} then A’ = {1, 2, 3 }. complement of A’ = {4, 5} which is equals to set A. Law of Empty set and Universal Set
Begin typing your search term above and press enter to search. Press ESC to cancel. |
# Sum of Perfect Squares Formula
Before knowing what is the sum of the perfect squares formula, first, let us recall what are perfect squares. A perfect square is a number that can be written as the square of a number. 1, 4, 9, 16, 25, 36, etc are some perfect squares as they can be expressed as 12, 22, 32, 42, 52, 62, etc respectively. The sum of perfect squares formula is used to find the sum of two or more perfect squares without adding them manually.
## What Is the Sum of Perfect Squares Formula?
We have two types of formulas for finding the sum of perfect squares. One is the formula to find the sum of two perfect squares, the other formula is to find the sum of the first "n" perfect squares.
• ### The formula for finding the sum of the squares for first "n" natural numbers is:12 + 22 + 32 + ... + n2 = [ n (n + 1) (2n + 6) ] / 6
Let us see the applications of the sum of perfect squares formulas in the following section.
## Solved Examples Using Sum of Perfect Squares Formula
### Example 1: Find the sum of squares of 101 and 99.
Solution:
To find: The sum of squares of 101 and 99. i.e., 1012 + 992.
Using the sum of perfect squares formula:
a2 + b2 = (a + b)2 - 2ab
Substitute a = 101 and b = 99 in the above formula:
1012 + 992 = (101 + 99)2 - 2(101)(99)
= (200)2 - 2 (9999)
= 40000 - 19998
= 20,002
### Example 2: Find the sum of squares of the first 25 natural numbers.
Solution:
To find: The sum of the first 25 natural numbers.
Substitute n = 25 in the sum of the perfect squares formula of the first n natural numbers:
12 + 22 + 32 + ... + n2 = [ n (n + 1) (2n + 6) ] / 6
12 + 22 + 32 + ... + 252 = [ 25 (25 + 1) (2(25) + 1) ] / 6
= ( 25 × 26 × 51 ) / 6
= 5,525
Answer: The sum of the first 25 natural numbers = 5,525. |
# How do you find the axis of symmetry, graph and find the maximum or minimum value of the function y= 3(x+2)^2-1?
Jul 17, 2018
Minimum (-2, -1); " axis of symmetry: " x = -2
#### Explanation:
Given: $y = 3 {\left(x + 2\right)}^{2} - 1$
The given function is a quadratic equation that is the graph of a parabola.
When the equation is in vertex form: $y = a {\left(x - h\right)}^{2} + k$,
the $\text{vertex": (h, k); " axis of symmetry: } x = h$
The parabola has a minimum when $a > 0$ and a maximum when $a < 0$
the $\text{vertex": (-2, -1); " axis of symmetry: } x = - 2$
the vertex is a minimum since $a = 3 > 0$
To graph just find additional points using point-plotting. Since $x$ is the independent variable, you can select any value of $x$ and calculate the corresponding value of $y$:
$\underline{\text{ "x" "|" "y" }}$
$- 4 \text{ "|" "11" }$
$- 3 \text{ "|" "2" }$
$- 1 \text{ "|" "2" }$
$\text{ "0" "|" "11" }$ This is the $y$-intercept
Graph of $3 {\left(x + 2\right)}^{2} - 1$:
graph{3(x+2)^2 - 1 [-5, 3, -5, 12]} |
# Angles in a Pentagon
Angles in a Pentagon
• A pentagon is a 5-sided shape.
• Angles in pentagons always add up to 540°.
• To find a missing angle in a pentagon, add up the 4 known angles and subtract this from 540°.
• We first add the 4 known angles: 120° + 100 + 100 + 110° = 430°.
• We then subtract this from 540°: 540 – 430 = 110 and so the missing angle is 110°.
Angles in a pentagon always add up to 540°.
• Angles in a pentagon always add to 540°.
• A regular pentagon is a pentagon that has 5 angles that are all the same size.
• To find the size of each angle in a regular pentagon, we simply divide 540° equally into 5 parts.
• 540° ÷ 5 = 108° and so each angle in a regular pentagon is 108°.
# Angles in Pentagons
## What do Angles in a Pentagon Add Up To?
The sum of the 5 interior angles in any pentagon is always equal to 540°.
We can see that the angles in the pentagon below all add up to 540°.
## How to Find a Missing Angle in a Pentagon
To find a missing angle in a pentagon, add up all of the other known angles then subtract this sum from 540°.
For example, here is a pentagon with known angles of 110°, 130°, 80° and 160°.
The first step is to add the known angles together.
110 + 130 + 80 + 160 = 480°.
The second step is to subtract this total from 540°.
540° – 480° = 60° and so, the missing angle is 60°.
Here is an example of finding more than one missing angle in a pentagon.
In this example, the angles in the pentagon have lines on them. This tells us that some angles are the same size.
The two angles that both have one line on them are equal to each other. Therefore angle a = 120°.
We also know that angles b and c are equal to each other because they both have 2 lines on them.
To work out the size of angles b and c, we will first find the sum of the three angles above.
100° + 120 ° + 120° = 340°.
We can subtract this from 540° to get 200°.
This means that angles b and c must add together to make 200°. Since they are both the same size, we will just divide 200° by 2.
Angle b = 100° and angle c = 100°.
## Angles in a Regular Pentagon
Each interior angle in a regular pentagon is equal to 108°. This is because the sum of all 5 interior angles in any pentagon is 540°. In a regular pentagon, all five angles are of equal size and so, we divide 540° by 5 to get 108°.
Below is a regular pentagon.
If a shape is regular, this means that all of its sides are the same length and all of its angles are the same size.
Angles in a pentagon add up to 540°.
Since all 5 angles are the same size, the total of 540° is shared evenly into 5 equal parts.
540° ÷ 5 = 108° and so each angle is 108°.
We can check the result by adding our angles.
108° + 108° + 108° + 108° + 108° = 540°.
## Formula to Find the Angles in a Pentagon
The formula for the sum of interior angles in a polygon is (n-2) × 180°, where n is the number of sides. A pentagon has 5 sides and so, n = 5. The formula (n-2) × 180° becomes 3 × 180° = 540°. Therefore the sum of angles in a pentagon is 540°.
The formula works because it tells us how many triangles can be formed inside each shape.
The number of triangles that can be drawn is equal to 2 less than the number of sides, or n – 2.
Each triangle contributes 180° to the sum of the interior angles and so, we have (n-2) × 180°.
## Why do Angles in a Pentagon Add to 540?
Angles in a pentagon add to 540° because three triangles can be made inside any pentagon by drawing lines from one corner to each of the other corners. Each triangle contains 180° and 3 × 180° = 540°.
Below is a pentagon divided into 3 triangles.
These triangles have been formed by taking one corner of the shape and drawing straight lines to each of the other corners.
Each triangle contains 180°.
180° + 180° +180° = 540° and so, the sum of the three triangles is 540°.
The angles in all three triangles form the interior angles of the pentagon and so the sum of angles in a pentagon equals 540°.
## Exterior Angles of a Pentagon
Exterior angles of all pentagons add up to 360°. In a regular pentagon, each exterior angle is 72°. This is because each angle is the same size and 360° ÷ 5 = 72°.
Exterior angles of all polygons always add up to 360°.
We can see the 5 exterior angles of a regular pentagon marked below.
Exterior angles add up to 360° because each polygon can be shrunk down until it forms a point. All that is left are the exterior angles. Angles around a point add to 360° and so, the exterior angles add to 360°.
This is shown below.
Now try our lesson on How to Find the Area of a Parallelogram where we learn how to find the area of a parallelogram.
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# Exponential Terms Raised to an Exponent
## Multiply to raise exponents to other exponents
Estimated7 minsto complete
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Progress
Practice Exponential Terms Raised to an Exponent
Progress
Estimated7 minsto complete
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Power Rule for Exponents
Can you simplify an expression where an exponent has an exponent? For example, how would you simplify ?
### Watch This
James Sousa: Properties of Exponents
### Guidance
In the expression , the is called the base and the is called the exponent. Exponents are often referred to as powers. When an exponent is a positive whole number, it tells you how many times to multiply the base by itself. For example:
• .
There are many rules that have to do with exponents (often called the Laws of Exponents) that are helpful to know so that you can work with expressions and equations that involve exponents more easily. Here you will learn a rule that has to do with raising a power to another power.
RULE: To raise a power to a new power, multiply the exponents.
Evaluate .
Solution: .
Simplify .
Solution: .
Evaluate .
Solution: .
Simplify .
Solution: .
#### Concept Problem Revisited
. Notice that the power rule applies even when a number has been raised to more than one power. The overall exponent is 24 which is .
### Vocabulary
Base
In an algebraic expression, the base is the variable, number, product or quotient, to which the exponent refers. Some examples are: In the expression , ‘2’ is the base. In the expression , ‘’ is the base.
Exponent
In an algebraic expression, the exponent is the number to the upper right of the base that tells how many times to multiply the base times itself. Some examples are:
In the expression , ‘5’ is the exponent. It means to multiply 2 times itself 5 times as shown here: .
In the expression , ‘4’ is the exponent. It means to multiply times itself 4 times as shown here: .
Laws of Exponents
The laws of exponents are the algebra rules and formulas that tell us the operation to perform on the exponents when dealing with exponential expressions.
### Guided Practice
You know you can rewrite as and then calculate in order to find that . This concept can also be reversed. To write 32 as a power of 2, . There are 5 twos; therefore, . Use this idea to complete the following problems.
1. Write 81 as a power of 3.
2. Write as a power of 3.
3. Write as a power of 2.
1.
There are 4 threes. Therefore
2.
There are 2 threes. Therefore .
Apply the law of exponents for power to a power-multiply the exponents.
Therefore
3.
There are 2 twos. Therefore
Apply the law of exponents for power to a power-multiply the exponents.
Apply the law of exponents for power to a power-multiply the exponents.
Therefore
### Practice
Simplify each of the following expressions.
1. True or false:
2. True or false:
3. Write 64 as a power of 4.
4. Write as a power of 2.
5. Write as a power of 3.
6. Write as a power of 3.
7. Write as a power of 5.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 6.3. |
13112an2.5-6
# 13112an2.5-6 - c Kendra Kilmer Section 2.2 The Limit of a...
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Unformatted text preview: c Kendra Kilmer January 11, 2012 Section 2.2 - The Limit of a Function Definitions: Left-Hand Limit: We write lim f (x) = K if f (x) is close to K whenever x is close to, but to the left of c. x→c− Right-Hand Limit: We write lim f (x) = L if f (x) is close to L whenever x is close to, but to the right of c. x→c+ (Two-Sided) Limit: We write lim f (x) = M if the functional value f (x) is close to M whenever x is close, but not x→c equal, to c (on either side of c). Note: For a (two-sided) limit to exist, the limit from the left and the limit from the right must be equal. That is Example 1: Use the graph below to find the following limits: 6 5 4 3 f(x) 2 1 −6 −5 −4 −3 −2 −1 a) b) 123 −1 −2 −3 −4 −5 −6 lim f (x) f) lim f (x) lim f (x) g) lim f (x) x→−3− x→−3+ x→1 x→4− c) lim f (x) h) lim f (x) d) lim f (x) i) lim f (x) x→−3 x→1− x→4+ x→4 e) lim f (x) x→1+ 5 4 56 c Kendra Kilmer January 11, 2012 Example 2: Sketch the graph of an example of a function f that satisfies all of the given conditions. • lim f (x) = 1 x→0 • lim f (x) = −2 x→3− • lim f (x) = 2 x→3+ • f (0) = −1, f (3) = 1 tan 3x . Confirm your result graphically. x→0 tan 5x Example 3: Use a table of values to estimate lim x2 + x √ Example 4: Use a table of values to estimate lim . Confirm your result graphically. x→0 x3 + x2 Section 2.2 Highly Suggested Homework Problems: 1, 3, 5, 7, 9, 11, 13, 15, 17, 19 6 ...
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# What Is 8/25 as a Decimal + Solution With Free Steps
The fraction 8/25 as a decimal is equal to 0.32.
In the form of p/q, a fraction is an expression that can be used to indicate the relationship between two integers.
A Fraction is a name for the mathematical representation of something divided into two or more portions or parts. For example, the Denominator and the Numerator are the two parts of a fraction. Usually, solving fractions using multiples other than their fractional representations is difficult. But turning them into division is a simple solution.
Instead of using the Multiples method in this case, we solve these fractions using the Long Division method. Finally, we receive the outcome using this method in decimal values.
So here, we are using the Long division method to find the decimal equivalent of the fraction 8/25 in this problem.
## Solution
The Dividend, which is being divided, and the Divisor, which is the number performing the dividing, are the first two parts of our fraction that we shall dissect. The procedure is as follows:
Dividend = 8
Divisor = 25
Quotient and Remainder are two more division-specific terminologies that might be used. The result of dividing is the solution or Quotient.
It can be stated as follows:
Quotient = Dividend $\div$ Divisor = 8 $\div$ 25
The Remainder, on the other hand, stands for a term that is left over after partial division. And the remainder is also used as a dividend for upcoming iterations in the division.
Let’s get into the Long Division solution of our fraction 8/25 as we are using this technique to solve this division:
Figure 1
## 8/25 Long Division Method
The first step in applying the long division method to solve a fraction is to represent the fraction as a division:
8 $\div$ 25
First, determining if the Dividend is greater than the Divisor is the first step in long division. We have to use a decimal point if the divisor is greater. We must add a zero to the dividend’s right to achieve this. If the dividend is greater, we can omit the decimal point.
In the above scenario, 8 is smaller than 25, which means the Divisor is smaller than the Dividend. Hence we require a Decimal Point to proceed.
So next, add a 0 to the dividend and a decimal to the quotient as follows:
80 $\div$ 25 $\approx$ 3
Where:
25 x 3 = 75
To determine the Remainder, subtract the two values given below:
80 – 75 =5
As a remainder 5 is obtained, the procedure is repeated by adding zero to dividends right and making it 50:
50 $\div$ 25 = 2
Where:
25 x 2 = 50
Reminder:
25 – 25 = 0
As a result of this division, we have 0 remainders. Therefore, it signifies that the fraction has been completely solved and that no more operations are required. As a result, we have a Quotient of 0.32 with no remainder.
Images/mathematical drawings are created with GeoGebra. |
Class 7 - Maths - Perimeter and Area
Exercise 11.1
Question 1:
The length and breadth of a rectangular piece of land are 500 m and 300 m respectively.
Find:
(i) Its area. (ii) The cost of the land, if 1 m2 of the land costs Rs 10,000.
Given: Length of a rectangular piece of land = 500 m
and breadth of a rectangular piece of land = 300 m
(i) Area of a rectangular piece of land = Length * Breadth
= 500 * 300
= 1,50,000 m2
(ii) Since, the cost of 1 m2 land = Rs 10,000
Therefore, the cost of 1,50,000 m2 land = 10,000 * 1,50,000 = Rs 1,50,00,00,000
Question 2:
Find the area of a square park whose perimeter is 320 m.
Given: Perimeter of square park = 320 m
=> 4 * side = 320
=> side = 320/4 = 80 m
Now, Area of square park = side * side = 80 * 80 = 6400 m2
Thus, the area of the square park is 6400 m2.
Question 3:
Find the breadth of a rectangular plot of land, if its area is 440 m2 and the length is 22 m. Also find its perimeter.
Area of rectangular park = 440 m2
=> length * breadth = 440 m2
=> 22 * breadth = 440
=> breadth = 440/22 = 20 m
Now, Perimeter of rectangular park = 2(length + breadth)
= 2(22 + 20)
= 2 * 42 = 84 m
Thus, the perimeter of rectangular park is 84 m.
Question 4:
The perimeter of a rectangular sheet is 100 cm. If the length is 35 cm, find its breadth. Also find the area.
Perimeter of the rectangular sheet = 100 cm
=> 2(length + breadth) = 100 cm
=> 2(35 + breadth) = 100
=> 35 + breadth = 100/2
=> 35 + breadth = 50
=> breadth = 50 - 35
Now, Area of rectangular sheet = length * breadth
= 35 * 15 = 525 cm2
Thus, breadth and area of rectangular sheet are 15 cm and 525 cm2 respectively.
Question 5:
The area of a square park is the same as of a rectangular park. If the side of the square park is 60 m and the length of the rectangular park is 90 cm,
find the breadth of the rectangular park.
Given: The side of the square park = 60 m
The length of the rectangular park = 90 m
According to the question,
Area of square park = Area of rectangular park
=> side * side = length * breadth
=> 60 * 60 = 90 * breadth
=> 3600 = 90 * breadth
=> breadth = 3600/90 = 40 m
Thus, the breadth of the rectangular park is 40 m.
Question 6:
A wire is in the shape of a rectangle. Its length is 40 cm and breadth is 22 cm. If the same wire is rebent in the shape of a square,
what will be the measure of each side. Also find which shape encloses more area?
According to the question,
Perimeter of square = Perimeter of rectangle
=> 4 * side = 2(length + breadth)
=> 4 * side = 2(40 + 22)
=> 4 * side = 2 * 62
=> 4 * side = 124
=> side = 124/4
=> side = 31 cm
Thus, the side of the square is 31 cm.
Now, Area of rectangle = length * breadth = 40 * 22 = 880 cm2
and area of square = side * side = 31 * 31 = 961 cm2
Therefore, on comparing, the area of square is greater than that of rectangle.
Question 7:
The perimeter of a rectangle is 130 cm. If the breadth of the rectangle is 30 cm, find its length. Also, find the area of the rectangle.
Perimeter of rectangle = 130 cm
=> 2(length + breadth) = 130 cm
=> 2(length + 30) = 130
=> length + 30 =130/2
=> length + 30 = 65
=> length = 65 – 30 = 35 cm
Now area of rectangle = length * breadth = 35 * 30 = 1050 cm2
Thus, the area of rectangle is 1050 cm2.
Question 8:
A door of length 2 m and breadth 1 m is fitted in a wall.The length of the wall is 4.5 m and the breadth is 3.6 m.
Find the cost of white washing the wall, if the rate of white washing the wall is Rs 20 per m2.
Area of rectangular door = length * breadth = 2 m * 1 m = 2 m2
Area of wall including door = length * breadth = 4.5 m * 3.6 m = 16.2 m2
Now, Area of wall excluding door = Area of wall including door – Area of door
= 16.2 – 2 = 14.2 m2
Since the rate of white washing of 1 m2 the wall = Rs 20
Therefore, the rate of white washing of 14.2 m2 the wall = 20 * 14.2 = Rs 284
Thus, the cost of white washing the wall excluding the door is Rs 284.
Exercise 11.2
Question 1:
Find the area of each of the following parallelograms:
We know that the area of parallelogram = base * height
(a) Here base = 7 cm and height = 4 cm
So, area of parallelogram = 7 * 4 = 28 cm2
(b) Here base = 5 cm and height = 3 cm
So, area of parallelogram = 5 * 3 = 15 cm2
(c) Here base = 2.5 cm and height = 3.5 cm
So, area of parallelogram = 2.5 * 3.5 = 8.75 cm2
(d) Here base = 5 cm and height = 4.8 cm
So, area of parallelogram = 5 * 4.8 = 24 cm2
(e) Here base = 2 cm and height = 4.4 cm
So, area of parallelogram = 2 * 4.4 = 8.8 cm2
Question 2:
Find the area of each of the following triangles:
We know that the area of triangle = (1/2) * base * height
(a) Here, base = 4 cm and height = 3 cm
So, area of triangle = (1/2) * 4 * 3 = 12/2 = 6 cm2
(b) Here, base = 5 cm and height = 3.2 cm
Area of triangle = (1/2) * 5 * 3.2 = 16/2 = 8 cm2
(c) Here, base = 3 cm and height = 4 cm
Area of triangle = (1/2) * 3 * 4 = 12/2 = 6 cm2
(d) Here, base = 3 cm and height = 2 cm
Area of triangle = (1/2) * 3 * 2 = 6/2 = 3 cm2
Question 3:
Find the missing values:
We know that the area of parallelogram = base * height
(a) Here, base = 20 cm and area = 246 cm2
Area of parallelogram = base * height
=> 246 = 20 * height
=> height = 246/20 = 12.3 cm
(b) Here, height = 15 cm and area = 154.5 cm2
Area of parallelogram = base * height
=> 154.5 = base * 15
=> base = 154.5/15 = 10.3 cm
(c) Here, height = 8.4 cm and area = 48.72 cm2
Area of parallelogram = base * height
=> 48.72 = base * 8.4
=> base = 48.72/8.4 = 5.8 cm
(d) Here, base = 15.6 cm and area = 16.38 cm2
Area of parallelogram = base * height
=> 16.38 = 15.6 * height
=> height = 16.38/15.6 = 1.05 cm
Thus, the missing values are:
Question 4:
Find the missing values:
We know that the area of triangle = (1/2) * base * height
In first row, base = 15 cm and area = 87 cm2
=> 87 = (1/2) * 15 * height
=> height = (87 * 2)/15
=> height = 174/15 = 11.6 cm
In second row, height = 31.4 mm and area = 1256 mm2
=> 1256 = (1/2) * base * 31.4
=> base = (1256 * 2)/31.4
=> base = 1512/31.4
=> base = (1512 * 10)/314
=> base = 15120/314 = 80 mm
In third row, base = 22 cm and area = 170.5 cm2
=> 170.5 = (1/2) * 22 * height
=> height = (170.5 * 2)/22
=> height = 341/22 = 15.5 cm
Thus, the missing values are:
Question 5:
PQRS is a parallelogram (given figure). QM is the height from Q to SR and QN is the height from Q to PS.
If SR = 12 cm and QM = 7.6 cm. Find:
(a) the area of the parallelogram PRS
(b) QN, if PS = 8 cm
Given: SR = 12 cm, QM= 7.6 cm, PS = 8 cm.
(a) Area of parallelogram = base * height
= 12 * 7.6 = 91.2 cm2
(b) Area of parallelogram = base * height
=> 91.2 = 8 * QN
=> QN = 91.2/8 = 11.4 cm
Question 6:
DL and BM are the heights on sides AB and AD respectively of parallelogram ABCD (given figure).
If the area of the parallelogram is 1470 cm2, AB = 35 cm and AD = 49 cm, find the length of BM and DL.
Given: Area of parallelogram = 1470 cm2
Base (AB) = 35 cm and base (AD) = 49 cm
Since Area of parallelogram = base * height
=> 1470 = 35 * DL
=> DL = 1470/35
=> DL = 42 cm
Again, Area of parallelogram = base * height
=> 1470 = 49 * BM
=> BM = 1470/49
=> BM = 30 cm
Thus, the lengths of DL and BM are 42 cm and 30 cm respectively.
Question 7:
Δ ABC is right angled at A (Given figure). AD is perpendicular to BC. If AB = 5 cm, BC = 13 cm and AC = 12 cm, find the area of Δ ABC.
Also, find the length of AD.
In right angles triangle BAC, AB = 5 cm and AC = 12 cm
Area of triangle = (1/2) * base * height
= (1/2) * AB * AC
= (1/2) * 5 * 12
= 60/2
= 30 cm2
Now, in Δ ABC,
Area of triangle ABC = (1/2) * BC * AD
=> 30 = (1/2) * 13 * AD
=> 13 * AD = 30 * 2
=> 13 * AD = 60
Question 8:
Δ ABC is isosceles with AB = AC = 7.5 cm and BC = 9 cm (given figure). The height AD from A to BC, is 6 cm.
Find the area of Δ ABC. What will be the height from C to AB i.e. CE?
In Δ ABC, AD = 6 cm and BC = 9 cm
Area of triangle = (1/2) * base * height
= (1/2) * BC * AD
= (1/2) * 9 * 6
= 54/2
= 27 cm2
Again, Area of triangle = (1/2) * base * height
=> 27 = (1/2) * AB * CE
=> 27 = (1/2) * 7.5 * CE
=> CE = (27 * 2)/7.5
=> CE = 54/7.5
=> CE = (54 * 10)/75
=> CE = 540/75
=> CE = 7.2 cm
Thus, the height from C to AB i.e., CE is 7.2 cm.
Exercise 11.3
Question 1:
Find the circumference of the circles with the following radius: (Take π = 22/7)
(a) 14 cm (b) 28 mm (c) 21 cm
(a) Circumference of the circle = 2πr = 2 * 22/7 * 14 = 2 * 22 * 2 = 88 cm
(b) Circumference of the circle = 2πr = 2 * 22/7 * 28 = 2 * 22 * 4 = 176 cm
(c) Circumference of the circle = 2πr = 2 * 22/7 * 21 = 2 * 22 * 3 = 132 cm
Question 2:
Find the area of the following circles, given that: (Take π = 22/7)
(a) radius = 14 mm (b) diameter = 49 m (c) radius 5 cm
(a) Area of circle = πr2 = π * r * r = 22/7 * 14 * 14 = 22 * 14 * 7 = 616 mm2
(b) Diameter = 49 m
So, radius = 49/2 = 24.5 m
Area of circle = πr2 = π * r * r = 22/7 * 24.5 * 24.5 = 22 * 24.5 * 3.5 = 616 = 1886.5 m2
(c) Area of circle = πr2 = π * r * r = 22/7 * 5 * 5 = (22 * 25)/7 = 550/7mm2
Question 3:
If the circumference of a circular sheet is 154 m, find its radius. Also find the area of the sheet. (Take π = 22/7)
Circumference of the circular sheet = 154 m
=> 2πr = 154
=> r = 154/2π
=> r = 77/π
=> r = 77/(22/7)
=> r = (77 * 7)/22
=> r = (7 * 7)/2
=> r = 49/2
=> r = 24.5 m
Now Area of circular sheet = πr2
= (22/7) * 24.5 * 24.5
= 22 * 3.5 * 24.5
= 1886.5 m2
Thus, the radius and area of circular sheet are 24.5 m and 1886.5 m2 respectively.
Question 4:
A gardener wants to fence a circular garden of diameter 21 m. Find the length of the rope he needs to purchase, if he makes 2 rounds of fence.
Also, find the costs of the rope, if it cost Rs 4 per meter. (Take π = 22/7)
Diameter of the circular garden = 21 m
So, radius of the circular garden = 21/2 m
Now Circumference of circular garden = 2πr
= 2 * 22/7 * 21/2
= 22 * 3
= 66 m
The gardener makes 2 rounds of fence so the total length of the rope of fencing = 2 * 2πr
= 2 * 66 = 132 m
Since, the cost of 1 meter rope = Rs 4
Therefore, cost of 132 meter rope = Rs 4 * 132 = Rs 528
Question 5:
From a circular sheet of radius 4 cm, a circle of radius 3 cm is removed. Find the area of the remaining sheet. (Take π = 3.14)
Radius of circular sheet (R) = 4 cm and
radius of removed circle (r) = 3 cm
Area of remaining sheet = Area of circular sheet – Area of removed circle
= πR2 - πr2
= π(R2 - r2)
= π(42 - 32)
= π(16 - 9)
= 7π
= 7 * 3.14
= 21.98 cm2
Thus, the area of remaining sheet is 21.98 cm2.
Question 6:
Saima wants to put a lace on the edge of a circular table cover of diameter 1.5 m. Find the length of the lace required and also
find its cost if one meter of the lace costs Rs 15. (Take π = 3.14)
Diameter of the circular table cover = 1.5 m
So, radius of the circular table cover = 1.5/2 m
Circumference of circular table cover = 2πr
= 2 * 3.14 * 1.5/2
= 1.5 * 3.14
= 4.71 m
Therefore the length of required lace is 4.71 m.
Now the cost of 1 m lace = Rs 15
Then the cost of 4.71 m lace = 15 * 4.71 = Rs 70.65
Hence, the cost of 4.71 m lace is Rs 70.65.
Question 7:
Find the perimeter of the adjoining figure, which is a semicircle including its diameter.
From the figure, diameter D = 10 cm
So, radius r = 10/2 = 5 cm
According to question,
Perimeter of figure = Circumference of semi-circle + diameter
= πr + D
= 22/7 * 5 + 10
= 110/7 + 10
= (110 + 70)/7
= 180/7
= 25.71 cm
Thus, the perimeter of the given figure is 25.71 cm.
Question 8:
Find the cost of polishing a circular table-top of diameter 1.6 m, if the rate of polishing is Rs 15/m2. (Take π = 3.14)
Diameter of the circular table top = 1.6 m
So, radius of the circular table top = 1.6/2 = 0.8 m
Area of circular table top =πr2
= 3.14 * 0.8 * 0.8
= 2.0096 m2
Now cost of 1 m2 polishing = Rs 15
Then cost of 2.0096 m2 polishing = 15 * 2.0096 = Rs 30.14 (approx.)
Thus, the cost of polishing a circular table top is Rs 30.14 (approx.)
Question 9:
Shazli took a wire of length 44 cm and bent it into the shape of a circle. Find the radius of that circle. Also find its area.
If the same wire is bent into the shape of a square, what will be the length of each of its sides?
Which figure encloses more area, the circle or the square? (Take π = 22/7)
Total length of the wire = 44 cm
=> the circumference of the circle = 44
=> 2πr = 44
=> πr = 44/2
=> πr = 22
=> r = 22/π
=> r = 22/(22/7)
=> r = (22 * 7)/22
=> r = 7 cm
Now Area of the circle = πr2
= 22/7 * 7 * 7
= 22 * 7
= 154 cm2
Now the wire is converted into square.
Then perimeter of square = 44 cm
=> 4 * side = 44
=> side = 44/4
=> side = 11 cm
Now area of square = side * side = 11 * 11 = 121 cm2
Therefore, on comparing, the area of circle is greater than that of square, so the circle
enclosed more area.
Question 10:
From a circular card sheet of radius 14 cm, two circles of radius 3.5 cm and a rectangle of length 3 cm and breadth 1 cm are removed (as shown in the adjoining figure).
Find the area of the remaining sheet. (Take π = 22/7)
Radius of circular sheet (R) = 14 cm and Radius of smaller circle (r) = 3.5 cm
Length of rectangle (l) = 3 cm and breadth of rectangle (b) = 1 cm
According to question,
Area of remaining sheet = Area of circular sheet– (Area of two smaller circle + Area
of rectangle)
= πR2 – [2(πr2) + (l * b)]
= 22/7 * 14 *1 4 – [(2 * 22/7 * 3.5 * 3.5) + (3 *1)]
= 22 * 7 *1 4 – [(2 * 22 * 0.5 * 3.5) + 3]
= 22 x 14 x 2 – [44 * 0.5 * 3.5 + 3]
= 616 – 80
= 536 cm2
Therefore the area of remaining sheet is 536 cm2.
Question 11:
A circle of radius 2 cm is cut out from a square piece of an aluminum sheet of side 6 cm. What is the area of the left over aluminum sheet? (Take π = 3.14)
Radius of circle = 2 cm and side of aluminum square sheet = 6 cm
According to question,
Area of aluminum sheet left = Total area of aluminum sheet – Area of circle
= side * side – πr2
= 6 * 6 – 22/7 * 2 * 2
= 36 – 88/7
= 36 – 12.56
= 23.44 cm2
Therefore, the area of aluminum sheet left is 23.44 cm2.
Question 12:
The circumference of a circle is 31.4 cm. Find the radius and the area of the circle. (Take π = 3.14)
The circumference of the circle = 31.4 cm
=> 2πr = 31.4
=> 2 * 3.14 * r = 31.4
=> 62.8 * r = 31.4/(2 * 3.14)
=> r = 10/2
=> r = 5 cm
Then area of the circle = πr2
= 3.14 x 5 x 5
= 78.5 cm2
Therefore, the radius and the area of the circle are 5 cm and 78.5 cm2 respectively.
Question 13:
A circular flower bed is surrounded by a path 4 m wide. The diameter of the flower bed is 66 m. What is the area of this path? (Take π = 3.14)
Diameter of the circular flower bed = 66 m
So, radius of circular flower bed = 66/2 = 33 m
Radius of circular flower bed with 4 m wide path (R) = 33 + 4 = 37 m
According to the question,
Area of path = Area of bigger circle – Area of smaller circle
= πR2 – πr2
= π(R2 – r2)
= π(372 – 332)
= π(37 + 33)*(37 - 33) [(a2 – b2) = (a - b)*(a + b)]
= 3.14 * 70 * 4
= 879.20 m2
Therefore, the area of the path is 879.20 m2
Question 14:
A circular flower garden has an area of 314 m2. A sprinkler at the centre of the garden can cover an area that has a radius of 12 m.
Will the sprinkler water the entire garden? (Take π = 3.14)
Circular area by the sprinkler = πr2
= 3.14 * 12 * 12
= 3.14 * 144
= 452.16 m2
Area of the circular flower garden = 314 m2
Since Area of circular flower garden is smaller than area by sprinkler.
Therefore, the sprinkler will water the entire garden.
Question 15:
Find the circumference of the inner and the outer circles, shown in the adjoining figure. (Take π = 3.14)
Radius of outer circle (r) = 19 m
So, circumference of outer circle = 2πr
= 2 * 3.14 * 19
= 119.32 m
Now radius of inner circle (r’) = 19 – 10 = 9 m
So, circumference of inner circle = 2πr’
= 2 * 3.14 * 9
= 56.52 m
Therefore, the circumferences of inner and outer circles are 56.52 m and 119.32 m
respectively.
Question 16:
How many times a wheel of radius 28 cm must rotate to go 352 m? (Take π = 22/7)
Let wheel must be rotate n times of its circumference.
Radius of wheel = 28 cm and Total distance = 352 m = 35200 cm
So, distance covered by wheel = n * circumference of wheel
=> 35200 = n * 2πr
=> 35200 = n * 2 * 22/7 * 28
=> 35200 = n * 2 * 22 * 4
=> 176n = 35200
=> n = 35200/176
=> n = 200 revolutions
Thus, wheel must rotate 200 times to go 352 m.
Question 17:
The minute hand of a circular clock is 15 cm long. How far does the tip of the minute hand move in 1 hour? (Take π = 22/7)
In 1 hour, minute hand completes one round means makes a circle.
Radius of the circle (r) = 15 cm
Circumference of circular clock = 2πr
= 2 * 3.14 * 15
= 94.2 cm
Therefore, the tip of the minute hand moves 94.2 cm in 1 hour.
Exercise 11.4
Question 1:
A garden is 90 m long and 75 m broad. A path 5 m wide is to be built outside and around it. Find the area of the path.
Also find the area of the garden in hectares.
Length of rectangular garden = 90 m and breadth of rectangular garden = 75 m
Outer length of rectangular garden with path = 90 + 5 + 5 = 100 m
Outer breadth of rectangular garden with path = 75 + 5 + 5 = 85 m
Outer area of rectangular garden with path = length * breadth = 100 * 85 = 8,500 m2
Inner area of garden without path = length * breadth = 90 * 75 = 6,750 m2
Now, Area of path = Area of garden with path – Area of garden without path
= 8,500 – 6,750
= 1,750 m2
Since, 1 m2 = 1/10000 hectares
Therefore, 6,750 m2 = 6750/10000 = 0.675 hectares
Hence, the area of the garden is 0.675 hectares.
Question 2:
A 3 m wide path runs outside and around a rectangular park of length 125 m and breadth 65 m. Find the area of the path.
Length of rectangular park = 125 m
Breadth of rectangular park = 65 m and
Width of the path = 3 m
Length of rectangular park with path = 125 + 3 + 3 = 131 m
Breadth of rectangular park with path = 65 + 3 + 3 = 71 m
Area of path = Area of park with path – Area of park without path
= (AB * AD) – (EF * EH)
= (131 * 71) – (125 * 65)
= 9301 – 8125
= 1,176 m2
Thus, area of path around the park is 1,176 m2.
Question 3:
A picture is painted on a cardboard 8 cm long and 5 cm wide such that there is a margin of 1.5 cm along each of its sides.
Find the total area of the margin.
Length of painted cardboard = 8 cm and breadth of painted card = 5 cm
Since, there is a margin of 1.5 cm long from each of its side.
Therefore reduced length = 8 – (1.5 + 1.5) = 8 – 3 = 5 cm
And reduced breadth = 5 – (1.5 + 1.5) = 5 – 3 = 2 cm
Area of margin = Area of cardboard (ABCD) – Area of cardboard (EFGH)
= (AB * AD) – (EF * EH)
= (8 * 5) – (5 * 2)
= 40 – 10
= 30 cm2
Thus, the total area of margin is 30 cm2.
Question 4:
A verandah of width 2.25 m is constructed all along outside a room which is 5.5 m long and 4 m wide. Find:
(i) the area of the verandah.
(ii) the cost of cementing the floor of the verandah at the rate of Rs 200 per m2.
(i) The length of room = 5.5 m and width of the room = 4 m
The length of room with verandah = 5.5 + 2.25 + 2.25 = 10 m
The width of room with verandah = 4 + 2.25 + 2.25 = 8.5 m
Area of verandah = Area of room with verandah – Area of room without verandah
= Area of ABCD – Area of EFGH
= (AB * AD) – (EF * EH)
= (10 * 8.5) – (5.5 * 4)
= 85 – 22
= 63 m2
(ii) The cost of cementing 1 m2 the floor of verandah = Rs 200
The cost of cementing 63 m2 the floor of verandah = 200 * 63 = Rs 12,600
Question 5:
A path 1 m wide is built along the border and inside a square garden of side 30 m. Find:
(i) the area of the path.
(ii) the cost of planting grass in the remaining portion of the garden at the rate of Rs 40 per m2.
(i) Side of the square garden = 30 m and
Width of the path along the border = 1 m
Side of square garden without path = 30 – (1 + 1) = 30 – 2 = 28 m
Now Area of path = Area of ABCD – Area of EFGH
= (AB * AD) – (EF * EH)
= (30 * 30) – (28 * 28)
= 900 – 784
= 116 m2
(ii) Area of remaining portion = 28 * 28 = 784 m2
The cost of planting grass in 1 m2 of the garden = Rs 40
The cost of planting grass in 784 m2 of the garden = Rs 40 * 784 = Rs 31,360
Question 6:
Two cross roads, each of width 10 m, cut at right angles through the centre of a rectangular park of length 700 m and breadth 300 m and parallel to its sides.
Find the area of the roads. Also find the area of the park excluding cross roads. Give the answer in hectares.
Here, PQ = 10 m and PS = 300 m, EH = 10 m and EF = 700 m
And KL = 10 m and KN = 10 m
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[Since KLMN is taken twice, which is to be subtracted]
= PS * PQ + EF * EH – KL * KN
= (300 * 10) + (700 * 10) – (10 * 10)
= 3000 + 7000 – 100
= 9,900 m2
Area of road in hectares, 1 m2 = 1/10000 hectares
So, 9,900 m2 = 9900/10000 = 0.99 hectares
Now, Area of park excluding cross roads = Area of park – Area of road
= (AB * AD) – 9,900
= (700 * 300) – 9,900
= 2,10,000 – 9,900
= 2,00,100 m2
= 200100/10000 hectares
= 20.01 hectares
Question 7:
Through a rectangular field of length 90 m and breadth 60 m, two roads are constructed which are parallel to the sides and cut each other at
right angles through the centre of the fields. If the width of each road is 3 m, find:
(i) the area covered by the roads.
(ii) the cost of constructing the roads at the rate of Rs 110 per m2.
(i) Here, PQ = 3 m and PS = 60 m, EH = 3 m and EF = 90 m and KL = 3 m and KN = 3 m
Area of roads = Area of PQRS + Area of EFGH – Area of KLMN
[Since KLMN is taken twice, which is to be subtracted]
= PS * PQ + EF * EH – KL * KN
= (60 * 3) + (90 * 3) – (3 * 3)
= 180 + 270 – 9
= 441 m2
(ii) The cost of 1 m2 constructing the roads = Rs 110
The cost of 441 m2 constructing the roads = Rs 110 * 441 = Rs 48,510
Therefore, the cost of constructing the roads = Rs 48,510
Question 8:
Pragya wrapped a cord around a circular pipe of radius 4 cm (adjoining figure) and cut off the length required of the cord.
Then she wrapped it around a square box of side 4 cm (also shown). Did she have any cord left? (Take π = 3.14)
Radius of pipe = 4 cm
Wrapping cord around circular pipe = 2πr
= 2 * 3.14 * 4 = 25.12 cm
Again, wrapping cord around a square = 4 * side
= 4 * 4 = 16 cm
Remaining cord = Cord wrapped on pipe – Cord wrapped on square
= 25.12 – 16
= 9.12 cm
Thus, she has left 9.12 cm cord.
Question 9:
The adjoining figure represents a rectangular lawn with a circular flower bed in the middle. Find:
(i) the area of the whole land. (ii) the area of the flower bed.
(iii) the area of the lawn excluding the area of the flower bed.
(iv) the circumference of the flower bed.
Length of rectangular lawn = 10 m,
breadth of the rectangular lawn = 5 m
And radius of the circular flower bed = 2 m
(i) Area of the whole land = length * breadth
= 10 * 5 = 50 m2
(ii) Area of flower bed = πr2
= 3.14 * 2 * 2 = 12.56 m2
(iii) Area of lawn excluding the area of the flower bed = area of lawn – area of flower bed
= 50 – 12.56
= 37.44 m2
(iv) The circumference of the flower bed = 2πr
= 2 * 3.14 * 2 = 12.56 m
Question 10:
In the following figures, find the area of the shaded portions:
(i) Here, AB = 18 cm, BC = 10 cm, AF = 6 cm, AE = 10 cm and BE = 8 cm
Area of shaded portion = Area of rectangle ABCD – (Area of ∆ FAE + area of ∆ EBC)
= (AB * BC) – (1/2 * AE * AF + 1/2 * BE * BC)
= (18 * 10) – (1/2 * 10 * 6 + 1/2 * 8 * 10)
= 180 – (30 + 40)
= 180 – 70
= 110 cm2
(ii) Here, SR = SU + UR = 10 + 10 = 20 cm, QR = 20 cm
PQ = SR = 20 cm, PT = PS – TS = 20 – 10 cm
TS = 10 cm, SU = 10 cm, QR = 20 cm and UR = 10 cm
Area of shaded region = Area of square PQRS – Area of ∆ QPT – Area of ∆ TSU – Area of ∆ UQR
= (SR * QR) – 1/2 * PQ * PT – 1/2 * ST * SU – 1/2 * QU * QR
= 20 * 20 – 1/2 * 20 * 10 – 1/2 * 10 * 10 – 1/2 * 20 * 10
= 400 – 100 – 50 - 100
= 400 – 250 = = 150 cm2
Question 11:
Find the area of the equilateral ABCD. Here, AC = 22 cm, BM = 3 cm, DN = 3 cm and BM ^ AC, DN ^ AC.
From the figure, AC = 22 cm, BM = 3 cm, DN = 3 cm
Area of quadrilateral ABCDF = Area of ∆ ABC + Area of ∆ ADC
= 1/2 * AC * BM + 1/2 * AC * DN
= 1/2 * 22 * 3 + 1/2 * 22 * 3
= 3 * 11 + 3 * 11
= 33 + 33
= 66 cm2
Thus, the area of quadrilateral ABCD is cm2. |
# How do you find an angle between two points on the edge of a circle?
I have two points on the circumference of circle and I also know the center of the circle. I want to calculate the angle between those two points which are on the circumference of circle.
Is this formula is suitable to this situation?
$$\tan(\theta) = \frac{y_2-y_1}{x_2-x_1}$$
where $(x_1,y_1)$ are the one of the circumference point $(x_2,y_2)$ is the other point on the circumference.
-
Well, you should shift your circle first so that it's centered at the origin, before you can use the formula for the tangent of the difference of two angles... – J. M. Aug 23 '12 at 11:44
The two points along with the center of the circle form an isosceles triangle. Two sides are radii of the circle. The base of the triangle is a line segment connecting the two points. Bisect the base with a line segment from the base to the center of the circle. Now we have two triangles, each with hypotenuse being a radius of the circle and the base being half the distance between the two points. The angle is half the angle that we want. sine of the half angle is the opposite side of the triangle (half the distance between the two points) over the hypotenuse (radius of the circle). The solution is then
angle = 2 x arcsin (0.5 x |P1 - P2| / radius)
-
You have an isosceles triangle.
You can use cosine formula for calculation the angle.
$$c^2 = a^2 + b^2 -2ab \cos(\alpha)$$
$a$ and $b$ are sides next to the angle $\alpha$, which are the radius of the center $r$. $c$ is the distance between the two points $P_1$ and $P_2$. So we get:
$$\left|P_1 - P_2\right|^2 = 2r^2-2r^2 \cos(\alpha)$$
$$\frac{2r^2-\left|P_1 - P_2\right|^2}{2r^2} = \cos(\alpha)$$
$$\alpha = \cos^{-1}\left(\frac{2r^2-\left|P_1 - P_2\right|^2}{2r^2}\right)$$
-
I'm having a hard time understanding what is meant by |P1 - P2|^2. Could you elaborate? – WebWanderer Feb 27 '15 at 17:18
@WebWanderer $|P1 - P2|$ is the distance between the 2 points $P1$ and $P2$. You calculate the distance by subtracting the points and calculate the length of the resulting vector. E.g. if the two points are $\left(\frac{1}{2}\right)$ and $\left(\frac{3}{2}\right)$, the distance is $\left|\left(\frac{1}{2}\right) - \left(\frac{3}{2}\right) \right| = \left|\left(\frac{-2}{0}\right) \right| = \sqrt((-2)^2 + 0^2) = \sqrt{4} = 2$. – user38034 Feb 27 '15 at 17:30
Ah, ok. Thanks! – WebWanderer Feb 27 '15 at 17:31 |
# Slope
## Understand slope as the steepness of a line.
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Slope (3.2)
Suppose you have a toy airplane, and upon takeoff, it rises 5 feet for every 6 feet that it travels along the horizontal. What would be the slope of its ascent? Would it be a positive value or a negative value? In this Concept, you'll learn how to determine the slope of a line by analyzing vertical change and horizontal change so that you can handle problems such as this one.
### Guidance
The pitch of a roof, the slant of a ladder against a wall, the incline of a road, and even your treadmill incline are all examples of slope.
The slope of a line measures its steepness (either negative or positive).
For example, if you have ever driven through a mountain range, you may have seen a sign stating, “10% incline.” The percent tells you how steep the incline is. You have probably seen this on a treadmill too. The incline on a treadmill measures how steep you are walking uphill. Below is a more formal definition of slope.
The slope of a line is the vertical change divided by the horizontal change.
In the figure below, a car is beginning to climb up a hill. The height of the hill is 3 meters and the length of the hill is 4 meters. Using the definition above, the slope of this hill can be written as \begin{align*}\frac{3 \ meters}{4 \ meters}=\frac{3}{4}\end{align*}. Because \begin{align*}\frac{3}{4}=75\%\end{align*}, we can say this hill has a 75% positive slope.
Similarly, if the car begins to descend down a hill, you can still determine the slope.
\begin{align*}Slope=\frac{vertical \ change}{horizontal \ change}=\frac{-3}{4}\end{align*}
The slope in this instance is negative because the car is traveling downhill.
Another way to think of slope is: \begin{align*}slope=\frac{rise}{run}\end{align*}.
When graphing an equation, slope is a very powerful tool. It provides the directions on how to get from one ordered pair to another. To determine slope, it is helpful to draw a slope-triangle.
Using the following graph, choose two ordered pairs that have integer values such as (–3, 0) and (0, –2). Now draw in the slope triangle by connecting these two points as shown.
The vertical leg of the triangle represents the rise of the line and the horizontal leg of the triangle represents the run of the line. A third way to represent slope is:
\begin{align*}slope=\frac{rise}{run}\end{align*}
Starting at the left-most coordinate, count the number of vertical units and horizontal units it took to get to the right-most coordinate.
\begin{align*}slope=\frac{rise}{run}=\frac{-2}{+3}=-\frac{2}{3}\end{align*}
#### Example A
Find the slope of the line graphed below.
Solution: Begin by finding two pairs of ordered pairs with integer values: (1, 1) and (0, –2).
Draw in the slope triangle.
Count the number of vertical units to get from the left ordered pair to the right.
Count the number of horizontal units to get from the left ordered pair to the right.
\begin{align*}Slope=\frac{rise}{run}=\frac{+3}{+1}=\frac{3}{1}\end{align*}
A more algebraic way to determine a slope is by using a formula. The formula for slope is:
The slope between any two points \begin{align*}(x_1,y_1 )\end{align*} and \begin{align*}(x_2,y_2)\end{align*} is: \begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}\end{align*}.
\begin{align*}(x_1,y_1)\end{align*} represents one of the two ordered pairs and \begin{align*}(x_2,y_2)\end{align*} represents the other. The following example helps show this formula.
#### Example B
Using the slope formula, determine the slope of the equation graphed in Example A.
Solution: Use the integer ordered pairs used to form the slope triangle: (1, 1) and (0, –2). Since (1, 1) is written first, it can be called \begin{align*}(x_1,y_1)\end{align*}. That means \begin{align*}(0,-2)=(x_2,y_2)\end{align*}
Use the formula: \begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{-2-1}{0-1}=\frac{-3}{-1}=\frac{3}{1}\end{align*}
As you can see, the slope is the same regardless of the method you use. If the ordered pairs are fractional or spaced very far apart, it is easier to use the formula than to draw a slope triangle.
Types of Slopes
Slopes come in four different types: negative, zero, positive, and undefined. The first graph of this Concept had a negative slope. The second graph had a positive slope. Slopes with zero slopes are lines without any steepness, and undefined slopes cannot be computed.
Any line with a slope of zero will be a horizontal line with equation \begin{align*}y = some \ number\end{align*}.
Any line with an undefined slope will be a vertical line with equation \begin{align*}x = some \ number\end{align*}.
We will use the next two graphs to illustrate the previous definitions.
#### Example C
To determine the slope of line \begin{align*}A\end{align*}, you need to find two ordered pairs with integer values.
(–4, 3) and (1, 3). Choose one ordered pair to represent \begin{align*}(x_1,y_1)\end{align*} and the other to represent \begin{align*}(x_2,y_2)\end{align*}.
Now apply the formula: \begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{3-3}{1-(-4)}=\frac{0}{1+4}=0\end{align*}.
To determine the slope of line \begin{align*}B\end{align*}, you need to find two ordered pairs on this line with integer values and apply the formula.
(5, 1) and (5, –6)
\begin{align*}slope=\frac{y_2-y_1}{x_2-x_1}=\frac{-6-1}{5-5}=\frac{-7}{0}=Undefined\end{align*}
It is impossible to divide by zero, so the slope of line \begin{align*}B\end{align*} cannot be determined and is called undefined.
### Guided Practice
Find the slope of each line in the graph below:
Solution:
For each line, identify two coordinate pairs on the line and use them to calculate the slope.
For the green line, one choice is \begin{align*} (0, 2)\end{align*} and \begin{align*} (5, 0)\end{align*}. This results in a slope of:
\begin{align*} \text{slope}=\frac{0-2}{5-0}=-\frac{2}{5}\end{align*}
For the blue line, one choice is \begin{align*} (6, 1)\end{align*} and \begin{align*} (7, 1)\end{align*}. This results in a slope of:
\begin{align*} \text{slope}=\frac{1-1}{7-6}=\frac{0}{1}=0\end{align*}
The slopes can be seen in this graph:
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Slope and Rate of Change (13:42)
1. Define slope.
2. Describe the two methods used to find slope. Which one do you prefer and why?
3. What is the slope of all vertical lines? Why is this true?
4. What is the slope of all horizontal lines? Why is this true?
Using the graphed coordinates, find the slope of each line.
In 8 – 20, find the slope between the two given points.
1. (–5, 7) and (0, 0)
2. (–3, –5) and (3, 11)
3. (3, –5) and (–2, 9)
4. (–5, 7) and (–5, 11)
5. (9, 9) and (–9, –9)
6. (3, 5) and (–2, 7)
7. \begin{align*}\left (\frac{1}{2},\frac{3}{4}\right )\end{align*} and (–2, 6)
8. (–2, 3) and (4, 8)
9. (–17, 11) and (4, 11)
10. (31, 2) and (31, –19)
11. (0, –3) and (3, –1)
12. (2, 7) and (7, 2)
13. (0, 0) and \begin{align*}\left (\frac{2}{3},\frac{1}{4}\right )\end{align*}
14. Determine the slope of \begin{align*}y=16\end{align*}.
15. Determine the slope of \begin{align*}x=-99\end{align*}.
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# Difference between revisions of "2021 AIME I Problems/Problem 8"
## Problem
Find the number of integers $c$ such that the equation $$\left||20|x|-x^2|-c\right|=21$$has $12$ distinct real solutions.
## Solution 1
Let $y = |x|.$ Then the equation becomes $\left|\left|20y-y^2\right|-c\right| = 21$, or $\left|y^2-20y\right| = c \pm 21$. Note that since $y = |x|$, $y$ is nonnegative, so we only care about nonnegative solutions in $y$. Notice that each positive solution in $y$ gives two solutions in $x$ ($x = \pm y$), whereas if $y = 0$ is a solution, this only gives one solution in $x$, $x = 0$. Since the total number of solutions in $x$ is even, $y = 0$ must not be a solution. Hence, we require that $\left|y^2-20y\right| = c \pm 21$ has exactly $6$ positive solutions and is not solved by $y = 0.$
If $c < 21$, then $c - 21$ is negative, and therefore cannot be the absolute value of $y^2 - 20y$. This means the equation's only solutions are in $\left|y^2-20y\right| = c + 21$. There is no way for this equation to have $6$ solutions, since the quadratic $y^2-20y$ can only take on each of the two values $\pm(c + 21)$ at most twice, yielding at most $4$ solutions. Hence, $c \ge 21$. $c$ also can't equal $21$, since this would mean $y = 0$ would solve the equation. Hence, $c > 21.$
At this point, the equation $y^2-20y = c \pm 21$ will always have exactly $2$ positive solutions, since $y^2-20y$ takes on each positive value exactly once when $y$ is restricted to positive values (graph it to see this), and $c \pm 21$ are both positive. Therefore, we just need $y^2-20y = -(c \pm 21)$ to have the remaining $4$ solutions exactly. This means the horizontal lines at $-(c \pm 21)$ each intersect the parabola $y^2 - 20y$ in two places. This occurs when the two lines are above the parabola's vertex $(10,-100)$. Hence we have: $$-(c + 21) > -100$$ $$c + 21 < 100$$ $$c < 79$$
Hence, the integers $c$ satisfying the conditions are those satisfying $21 < c < 79.$ There are $\boxed{057}$ such integers. Note: Be careful of counting at the end, you may mess up and get 59.
## Solution 2 (also graphing)
Graph $y=|20|x|-x^2|$ (If you are having trouble, look at the description in the next two lines and/or the diagram in solution 3). Notice that we want this to be equal to $c-21$ and $c+21$.
We see that from left to right, the graph first dips from very positive to $0$ at $x=-20$, then rebounds up to $100$ at $x=-10$, then falls back down to $0$ at $x=0$.
The positive $x$ are symmetric, so the graph re-ascends to $100$ at $x=10$, falls back to $0$ at $x=10$, and rises to arbitrarily large values afterwards.
Now we analyze the $y$ (varied by $c$) values. At $y=k<0$, we will have no solutions, as the line $y=k$ will have no intersections with our graph.
At $y=0$, we will have exactly $3$ solutions for the three zeroes.
At $y=n$ for any $n$ strictly between $0$ and $100$, we will have exactly $6$ solutions.
At $y=100$, we will have $4$ solutions, because local maxima are reached at $x= \pm 10$.
At $y=m>100$, we will have exactly $2$ solutions.
To get $12$ distinct solutions for $y=|20|x|-x^2|=c \pm 21$, both $c +21$ and $c-21$ must produce $6$ solutions.
Thus $0 and $c+21<100$, so $c \in \{ 22, 23, \dots , 77, 78 \}$ is required.
It is easy to verify that all of these choices of $c$ produce $12$ distinct solutions (none overlap), so our answer is $\boxed{057}$.
## Solution 3 (Piecewise Functions: Analyses and Graphs)
We take cases for the outermost absolute value, then rearrange: $$\left|20|x|-x^2\right|-c=\pm21.$$ Let $f(x)=\left|20|x|-x^2\right|.$ We will rewrite $f(x)$ as a piecewise function without using any absolute value: $$f(x) = \begin{cases} \left|-20x-x^2\right| & \text{if} \ x \le 0 \begin{cases} 20x+x^2 & \text{if} \ x\le-20 \\ -20x-x^2 & \text{if} \ -20 0 \begin{cases} 20x-x^2 & \text{if} \ 020 \end{cases} \end{cases}.$$ We graph $f(x)$ as shown below, with some key points labeled. The fact that $f(x)$ is an even function ($f(x)=f(-x)$ holds for all real numbers $x,$ from which the graph of $f(x)$ is symmetric about the $y$-axis) should facilitate the process of graphing.
Graph in Desmos: https://www.desmos.com/calculator/fwvhtltxjr
Since $f(x)-c=\pm21$ has $12$ distinct real solutions, it is clear that each case has $6$ distinct real solutions geometrically. We need to shift the graph of $f(x)$ down by $c$ units:
1. For $f(x)-c=21$ to have $6$ distinct real solutions, we get $0
2. For $f(x)-c=-21$ to have $6$ distinct real solutions, we get $21
Taking the intersection of these two cases gives $21 from which there are $79-21-1=\boxed{057}$ such integers $c.$
~MRENTHUSIASM
## Solution 4
Removing the absolute value bars from the equation successively, we get $$\left||20|x|-x^2|-c\right|=21$$ $$|20|x|-x^2|= c \pm21$$ $$20|x|-x^2 = \pm c \pm 21$$ $$x^2 \pm 20x \pm c \pm21 = 0$$
The discriminant of this equation is $$\sqrt{400-4(\pm c \pm 21)}$$
Equating the discriminant to $0$, we see that there will be two distinct solutions to each of the possible quadratics above only in the interval $-79 < c < 79$. However, the number of zeros the equation $ax^2+b|x|+k$ has is determined by where $ax^2+bx+k$ and $ax^2-bx+k$ intersect, namely at $(0,k)$. When $k<0$, $a>0$, $ax^2+b|x|+k$ will have only $2$ solutions, and when $k>0$, $a>0$, then there will be $4$ real solutions, if they exist at all. In order to have $12$ solutions here, we thus need to ensure $-c+21<0$, so that exactly $2$ out of the $4$ possible equations of the form $ax^2+b|x|+k$ given above have y-intercepts below $0$ and only $2$ real solutions, while the remaining $2$ equations have $4$ solutions. This occurs when $c>21$, so our final bounds are $21, giving us $\boxed{057}$ valid values of $c$.
## Remark
The graph of $F(x)=\left||20|x|-x^2|-c\right|$ is shown here in Desmos: https://www.desmos.com/calculator/i6l98lxwpp
Move the slider around for $21 to observe how the graph of $F(x)$ intersects the line $y=21$ for $12$ times.
~MRENTHUSIASM |
# 4-4 How Do I Express And Use Numbers In Scientific Notation? Warm Up
## Presentation on theme: "4-4 How Do I Express And Use Numbers In Scientific Notation? Warm Up"— Presentation transcript:
4-4 How Do I Express And Use Numbers In Scientific Notation? Warm Up
Course 3 Warm Up Lesson Presentation
4-4 Scientific Notation Warm Up 1. 10 , 10 , 10 , 10 10 , 10 , 10 , 10
Course 3 4-4 Scientific Notation Warm Up Order each set of numbers from least to greatest. , 10 , 10 , 10 4 –2 –1 10 , 10 , 10 , 10 4 –1 –2 2. 8 , 8 , 8 , 8 2 –2 3 8 , 8 , 8 , 8 3 2 –2 3. 2 , 2 , 2 , 2 3 –6 –4 1 2 , 2 , 2 , 2 3 1 –4 –6
Course 3 4-4 Scientific Notation Learn to express large and small numbers in scientific notation and to compare two numbers written in scientific notation.
Course 3 4-4 Scientific Notation Vocabulary scientific notation
Course 3 4-4 Scientific Notation An ordinary quarter contains about 97,700,000,000,000,000,000,000 atoms. The average size of an atom is about centimeters across. The length of these numbers in standard notation makes them awkward to work with. Scientific notation is a shorthand way of writing such numbers.
Course 3 4-4 Scientific Notation To express any number in scientific notation, write it as the product of a power of ten and a number greater than or equal to 1 but less than 10. In scientific notation the number of atoms in a quarter is 9.77 1022, and the size of each atom is 3.0 10–8 centimeters across.
Example 1: Translating Scientific Notation to Standard Notation
Course 3 4-4 Scientific Notation Example 1: Translating Scientific Notation to Standard Notation Write the number in standard notation. 1.35 105 1.35 10 5 10 = 100,000 5 1.35 100,000 Think: Move the decimal right 5 places. 135,000 A positive exponent means move the decimal to the right, and a negative exponent means move the decimal to the left. Helpful Hint
4-4 Scientific Notation 2.7 10–3 1 2.7 10–3 10 = 1000 1 2.7 1000
Course 3 4-4 Scientific Notation Example 1B: Translating Scientific Notation to Standard Notation Continued Write the number in standard notation. 2.7 10–3 2.7 10–3 10 = –3 1 1000 2.7 1 1000 2.7 1000 Divide by the reciprocal. 0.0027 Think: Move the decimal left 3 places.
4-4 Scientific Notation 2.01 104 2.01 104 10 = 10,000
Course 3 4-4 Scientific Notation Example 1C: Translating Scientific Notation to Standard Notation Continued Write the number in standard notation. 2.01 104 2.01 104 10 = 10,000 4 2.01 10,000 20,100 Think: Move the decimal right 4 places.
4-4 Scientific Notation Write the number in standard notation.
Course 3 4-4 Scientific Notation Check It Out: Example 1A Write the number in standard notation. 2.87 109 2.87 10 9 10 = 1,000,000,000 9 2.87 1,000,000,000 2,870,000,000 Think: Move the decimal right 9 places.
4-4 Scientific Notation Write the number in standard notation.
Course 3 4-4 Scientific Notation Check It Out: Example 1B Write the number in standard notation. 1.9 10–5 1.9 10 –5 –5 1 10 = 100,000 1.9 1 100,000 1.9 100,000 Divide by the reciprocal. Think: Move the decimal left 5 places.
4-4 Scientific Notation Write the number in standard notation.
Course 3 4-4 Scientific Notation Check It Out: Example 1C Write the number in standard notation. 5.09 108 5.09 108 10 = 100,000,000 8 5.09 100,000,000 509,000,000 Think: Move the decimal right 8 places.
Course 3 4-4 Scientific Notation A number less than 1 will have a negative exponent when written in scientific notation. Helpful Hint
Course 3 4-4 Scientific Notation Additional Example 2: Translating Standard Notation to Scientific Notation Write in scientific notation. Think: The decimal needs to move 3 places to get a number between 1 and 10. 7.09 7.09 10 Set up scientific notation. Think: The decimal needs to move left to change 7.09 to , so the exponent will be negative. So written in scientific notation is 7.09 10–3. Check 10–3 = 7.09 =
Course 3 4-4 Scientific Notation Check It Out: Example 2 Write in scientific notation. Think: The decimal needs to move 4 places to get a number between 1 and 10. 8.11 8.11 10 Set up scientific notation. Think: The decimal needs to move left to change 8.11 to , so the exponent will be negative. So written in scientific notation is 8.11 10–4. Check 10 = 8.11 = –4
4-4 Scientific Notation Example 3: Application
Course 3 4-4 Scientific Notation Example 3: Application A pencil is 18.7 cm long. If you were to lay 10,000 pencils end-to-end, how many millimeters long would they be? Write the answer in scientific notation. 1 centimeter = 10 millimeters 18.7 centimeters = 187 millimeters Multiply by 10. 187 mm 10,000 Find the total length. 1,870,000 mm Multiply.
4-4 Scientific Notation 1.87 10 Set up scientific notation.
Course 3 4-4 Scientific Notation Example 3 Continued 1.87 10 Set up scientific notation. Think: The decimal needs to move right to change 1.87 to 1,870,000, so the exponent will be positive. Think: The decimal needs to move 6 places. In scientific notation the 10,000 pencils would be 1.87 106 mm long, laid end-to-end.
4-4 Scientific Notation Check It Out: Example 3
Course 3 4-4 Scientific Notation Check It Out: Example 3 An oil rig can hoist 2,400,000 pounds with its main derrick. It distributes the weight evenly between 8 wire cables. What is the weight that each wire cable can hold? Write the answer in scientific notation. Find the weight each cable is expected to hold by dividing the total weight by the number of cables. 2,400,000 pounds ÷ 8 cables = 300,000 pounds per cable Each cable can hold 300,000 pounds. Now write 300,000 pounds in scientific notation.
Check It Out: Example 3 Continued
Course 3 4-4 Scientific Notation Check It Out: Example 3 Continued 3.0 10 Set up scientific notation. Think: The decimal needs to move right to change 3.0 to 300,000, so the exponent will be positive. Think: The decimal needs to move 5 places. In scientific notation, each cable can hold 3.0 105 pounds.
Additional Example 4: Life Science Application
Course 3 4-4 Scientific Notation Additional Example 4: Life Science Application A certain cell has a diameter of approximately 4.11 x 10-5 meters. A second cell has a diameter of 1.5 x 10-5 meters. Which cell has a greater diameter? 4.11 x x 10-5 10-5 = 10-5 Compare powers of 10. Compare the values between 1 and 10. 4.11 > 1.5 4.11 x 10-5 > 1.5 x 10-5 The first cell has a greater diameter.
4-4 Scientific Notation Check It Out: Example 4
Course 3 4-4 Scientific Notation Check It Out: Example 4 A certain cell has a diameter of approximately 5 x 10-3 meters. A second cell has a diameter of 5.11 x 10-3 meters. Which cell has a greater diameter? 5 x x 10-3 10-3 = 10-3 Compare powers of 10. Compare the values between 1 and 10. 5 < 5.11 5 x 10-3 < 5.11 x 10-3 The second cell has a greater diameter.
4-4 Scientific Notation Lesson Quiz: Part I
Course 3 4-4 Scientific Notation Lesson Quiz: Part I Write each number in standard notation. 104 17,200 10–3 0.0069 Write each number in scientific notation. 5.3 10–3 5.7 107 4. 57,000,000 5. Order the numbers from least to greatest. T 2 10–4, 9 10–5, 7 10–5 7 10–5, 9 10–5, 2 10–4
4-4 Scientific Notation Lesson Quiz: Part II
Course 3 4-4 Scientific Notation Lesson Quiz: Part II 6. A human body contains about 5.6 microliters of blood. Write this number in standard notation. 5,600,000
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# Right Angle Triangles and SOH CAH TOA
This topic is about right angle triangles and the memory aid SOH CAH TOA. This topic is typically found in (Canadian) high school mathematics.
Right Angle Triangles
A right angle triangle has three sides and three angles with one of the three angles being a 90 degree (right) angle. The three angles inside a (right) triangle add up to 180 degrees. Also, the sides of the right angled triangle obey the formula (Pythagorean Theorem) where , and are side lengths of the triangle. (Recall that is the hypotenuse and the longest side of a right angle triangle.
Here is an example of a right angled triangle:
Source: http://kineticmaths.com/images/7/76/Findangle.jpg
Finding Angles and Side Lengths
When dealing with triangles, we sometimes do not know all the angles and all the side lengths of a triangle. If we are given just enough information (i.e. 2 angles and 1 side length), we can use the following formulas to determine the unknown angle(s) and the unknown side length(s).
The sine of an angle (theta) is the ratio of the side length opposite to the angle over the length of the hypotenuse. The equation is as follows.
The cosine of an angle (theta) is the ratio of the side length which is adjacent to the angle over the length of the hypotenuse. Note that the adjacent side to the angle is NOT the hypotenuse.
The tangent of an angle (theta) is the ratio of the side length which is opposite to the angle over the side length which is adjacent to the angle .
SOH CAH TOA Memory Aid
The three formulas can be hard to memorize and follow. A very neat memory aid is SOH CAH TOA. In SOH, the letter S refers to the sine function, the letter O refers to the side length opposite to the angle and H refers to the hypotenuse side length. The same logic applies to CAH and TOA.
Here is a picture which provides a good summary.
Source: http://www.mathwarehouse.com/trigonometry/images/sohcohtoa/sohcahtoa-all.png
Examples
These examples will show how SOH CAH TOA can help in finding unknown side lengths and angles of right angled triangles. The pictures are my own and from my phone.
Example One
In the given picture, we have a right angle triangle with known side lengths and just the known right angle which is angle C.
Referring to the angle we can determine the ratios using sine, cosine and tangent.
We have the ratios but not the angle . To find the angle we need to use the respective inverse trigonometric function. For example, the inverse sine function would be used to find the angle given a (valid) numeric value .
Note that the inverse function of the original function is the argument inside the function. In this case, the inverse trigonometric function of the trigonometric function is the angle . Since the angle is found, we can now find the angle B. Angle B would be .
Example Two
In this second example, we have only one known side length and know two of three angles in this right triangle.
The side is known at 2 units with angle L at and angle being a right angle. The angle can be determined as .
One way of finding the side length is to use the tangent ratio. In this case, is adjacent to the angle and the side is the opposite side at length 2. Solving for gives us:
Another way of solving for is using the tangent of . The adjacent side of angle is with length 2 and the opposite side length is .
Assuming the tangent formula is used correctly when solving for the unknown side length , the choice of (acute) angle does not matter too much.
If we had wanted to find the length of the hypotenuse instead of , we would use the cosine of . |
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