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# Graph, Domain and Range of Absolute Value Functions
This is a step by step tutorial on how to graph functions with absolute value. Properties of the graph of these functions such as domain, range, x and y intercepts are also discussed. Free graph paper is available.
## Graph, Domain and Range of Absolute Value Functions; Examples with Detailed Solutions
Example 1: f is a function given by
f (x) = |x - 2|
1. Find the x and y intercepts of the graph of f.
2. Find the domain and range of f.
3. Sketch the graph of f.
Solution to Example 1
• a - The y intercept is given by
(0 , f(0)) = (0 ,|-2|) = (0 , 2)
• The x coordinate of the x intercepts is equal to the solution of the equation
|x - 2| = 0
which is
x = 2
• The x intercepts is at the point (2 , 0)
• b - The domain of f is the set of all real numbers
Since |x - 2| is either positive or zero for x = 2; the range of f is given by the interval [0 , +infinity).
• c - To sketch the graph of f(x) = |x - 2|, we first sketch the graph of y = x - 2 and then take the absolute value of y.
The graph of y = x - 2 is a line with x intercept (2 , 0) and y intercept (0 , -2). (see graph below)
• We next use the definition of the absolute value to graph f(x) = |x - 2| = | y |.
If y >= 0 then | y | = y , if y <0 then | y | = -y.
• For values of x for which y is positive, the graph of | y | is the same as that of y = x - 2. For values of x for which y is negative, the graph of | y | is a reflection on the x axis of the graph of y. The graph of y = x - 2 above has y negative on the interval (-infinity , 2) and it is this part of the graph that has to be reflected on the x axis. (see graph below).
• Check that the range is given by the interval [0 , +infinity), the domain is the set of all real numbers, the y intercept is at (0 , 2) and the x intercept at (2, 0).
Example 2: f is a function given by
f (x) = |(x - 2)2 - 4|
1. Find the x and y intercepts of the graph of f.
2. Find the domain and range of f.
3. Sketch the graph of f.
Solution to Example 2
• a - The y intercept is given by
(0 , f(0)) = (0 ,(-2)2 - 4) = (0 , 0)
• The x coordinates of the x intercepts are equal to the solutions of the equation
|(x - 2)2 - 4| = 0
which is solved
(x - 2)2 = 4
Which gives the solutions
x = 0 and x = 4
• The x intercepts is at the point (0 , 0) and (4 , 0)
• b - The domain of f is the set of all real numbers
Since |(x - 2)2 - 4| is either positive or zero for x = 4 and x = 0; the range of f is given by the interval [0 , +infinity).
• c - To sketch the graph of f(x) = |(x - 2)2 - 4|, we first sketch the graph of y = (x - 2)2 - 4 and then take the absolute value of y.
The graph of y = (x - 2)2 - 4 is a parabola with vertex at (2,-4), x intercepts (0 , 0) and (4 , 0) and a y intercept (0 , 0). (see graph below)
• The graph of f is given by reflecting on the x axis part of the graph of y = (x - 2)2 - 4 for which y is negative. (see graph below).
## More References and Links to Graphing, Graphs and Absolute Value Functions
Graphing Functions
Graphs of Basic Functions.
Absolute Value Functions.
Definition of the Absolute Value.
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##### Browse Questions
You can create printable tests and worksheets from these Grade 4 Number Properties questions! Select one or more questions using the checkboxes above each question. Then click the add selected questions to a test button before moving to another page.
The property that states any number multiplied by zero is zero is the Zero Property of Multiplication .
The Identity property of multiplication states when any number is multiplied by 1, the product is that number.
The property that states that the GROUPING of the factors does not change the product is the Associative Property of Multiplication.
The property that states that the ORDER in which two numbers are multiplied does not change the product is the Commutative Property .
Selena sorted coins into a stack of nickles and a stack of dimes, she uses distributive property to find the total. If each stack has 10 coins, which number expression will help Selena know the total value of her coins?
1. 10($0.10) +$0.05
2. 10 + ($0.10 +$0.05)
3. 10($0.10 +$0.05)
4. $0.10 +$0.05
Solve and tell what multiplication property is used.
2 x 7 x 5 = 5 x 7 x 2 = ?
Using the associative property, what is 6 + (4 + 2) equal to?
1. 2 + 6 + (4)
2. 4 + (6 + 2)
3. 12
Multiplying any number by 0 yields 0
1. Associative Property of Multiplication
2. Commutative Property of Multiplication
3. Identity Property of Multiplication
4. Zero Property of Multiplication
Factors can be multiplied in any order and the product remains the same.
2. Commutative Property of Multiplication
4. Associative Property of Multiplication
Solve using the DISTRIBUTIVE PROPERTY: $9 xx 67=$
When two numbers are multiplied together, the product is the same regardless of the order multiplicands.
3. commutative property of multiplication
4. identity property of multiplication
The property that states that when you add zero to any number, the sum is that number.
1. Associative property of multiplication
The Associative Property of Multiplication is also known as the Grouping
Property of Multiplication.
Which of the following statement describes the associative property of addition?
1. The sum of any number and zero is the original number,
2. The sum of two numbers is the same no matter which order they are in.
3. The sum of three or more numbers is always the same no matter how they are grouped.
4. The sum of two numbers times a third number is the same as the sum of each addend times the third number.
The property that states that when the order of two addends is changed, the sum is the same.
2. Commutative Property of Multiplication
3. Associative Property
4. Identity Property
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## Question:
S is the set of all 2 × 2 matrices of the form
$A=\begin{pmatrix} w & x \\ y & z \end{pmatrix}$, where wz − xy = 1.
Show that S is a group under matrix multiplication. Which element(s) have order 2? Prove that an element A has order 3 if w + z + 1 = 0.
## Step-by-step
The condition wz −xy = 1 is the same as |A| = 1; it follows that the set contains an identity element (with w = z = 1 and x = y = 0). Moreover, each matrix in S has an inverse and, since $|{A}^{−1}| |A| = | I | = 1$implies that $|{A}^{−1}| = 1$, the inverses also belong to the set.
If A and B belong to S then, since |AB| = |A| |B| = 1× 1 = 1, their product also belongs to S, i.e. the set is closed.
These observations, together with the associativity of matrix multiplication establish that the set S is, in fact, a group under this operation.
If A is to have order 2 then
$\begin{pmatrix} w & x \\ y & z \end{pmatrix}\begin{pmatrix} w & x \\ y & z \end{pmatrix}=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}.$
i.e. ${w}^{2} + xy = 1$, x(w + z) = 0, y(w + z) = 0, $xy+ {z}^{2} = 1$.
These imply that ${w}^{2} = {z}^{2}$ and that either z = −w or x = y = 0. If z = −w, then both
${w}^{2} + xy = 1$, from the above condition,
and $− {w}^{2} − xy = 1$, from wz − xy = 1.
This is not possible and so we must have x = y = 0, implying that w and z are either both +1 or both −1. The former gives the identity (of order 1), and so the matrix given by the latter, A = −I, is the only element in S of order 2.
If w + z + 1 = 0 (as well as xy = wz − 1), ${A}^{2}$ can be written as
${ A }^{ 2 }=\begin{pmatrix} { w }^{ 2 }+xy & x(w+z) \\ y(w+z) & xy+{ z }^{ 2 } \end{pmatrix}$
$=\begin{pmatrix} { w }^{ 2 }+wz-1 & -x \\ -y & wz-1+{z}^{2} \end{pmatrix}$
$=\begin{pmatrix} -w-1 & -x \\ -y & -z-1 \end{pmatrix}$.
Multiplying again by A gives
${ A }^{ 3 }=\begin{pmatrix} -w-1 & -x \\ -y & -z-1 \end{pmatrix}\begin{pmatrix} w & x \\ y & z \end{pmatrix}$
$=-\begin{pmatrix} w(w+1)+xy & (w+1)x+xz \\ wy+y(z+1) & xy+z(z+1) \end{pmatrix}$
$=-\begin{pmatrix} w(w+1)+wz-1 & x\times 0 \\ y\times 0 & wz-1+z(z+1) \end{pmatrix}$
$=-\begin{pmatrix} (w\times 0)-1 & 0 \\ 0 & (z\times 0)-1 \end{pmatrix}$
$=\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$
Thus A has order 3.
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# Trigonometric Equations Using Half Angle Formulas
## Simplifying all six trigonometric functions with half a given angle.
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Practice Trigonometric Equations Using Half Angle Formulas
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Trigonometric Equations Using Half Angle Formulas
As you've seen many times, the ability to find the values of trig functions for a variety of angles is a critical component to a course in Trigonometry. If you were given an angle as the argument of a trig function that was half of an angle you were familiar with, could you solve the trig function?
For example, if you were asked to find
$\sin 22.5^\circ$
would you be able to do it? Keep reading, and in this Concept you'll learn how to do this.
### Guidance
It is easy to remember the values of trigonometric functions for certain common values of $\theta$ . However, sometimes there will be fractional values of known trig functions, such as wanting to know the sine of half of the angle that you are familiar with. In situations like that, a half angle identity can prove valuable to help compute the value of the trig function.
In addition, half angle identities can be used to simplify problems to solve for certain angles that satisfy an expression. To do this, first remember the half angle identities for sine and cosine:
$\sin \frac{\alpha}{2} = \sqrt{\frac{1 - \cos \alpha}{2}}$ if $\frac{\alpha}{2}$ is located in either the first or second quadrant.
$\sin \frac{\alpha}{2} = - \sqrt{\frac{1 - \cos \alpha}{2}}$ if $\frac{\alpha}{2}$ is located in the third or fourth quadrant.
$\cos \frac{\alpha}{2} = \sqrt{\frac{1 + \cos \alpha}{2}}$ if $\frac{\alpha}{2}$ is located in either the first or fourth quadrant.
$\cos \frac{\alpha}{2} = - \sqrt{\frac{1 + \cos \alpha}{2}}$ if $\frac{\alpha}{2}$ is located in either the second or fourth quadrant.
When attempting to solve equations using a half angle identity, look for a place to substitute using one of the above identities. This can help simplify the equation to be solved.
#### Example A
Solve the trigonometric equation $\sin^2 \theta = 2 \sin^2 \frac{\theta}{2}$ over the interval $[0, 2\pi)$ .
Solution:
$\sin^2 \theta & = 2 \sin^2 \frac{\theta}{2} \\\sin^2 \theta & = 2 \left (\frac{1 - \cos \theta}{2} \right ) && \text{Half angle identity} \\1 - \cos^2 \theta & = 1 - \cos \theta && \text{Pythagorean identity} \\\cos \theta - \cos^2 \theta & = 0 \\\cos \theta (1 - \cos \theta) & = 0$
Then $\cos \theta = 0$ or $1 - \cos \theta = 0$ , which is $\cos \theta = 1$ .
$\theta = 0, \frac{\pi}{2}, \frac{3\pi}{2}, \text{or } 2\pi$ .
#### Example B
Solve $2 \cos^2 \frac{x}{2} = 1$ for $0 \le x < 2 \pi$
Solution:
To solve $2 \cos^2 \frac{x}{2} = 1$ , first we need to isolate cosine, then use the half angle formula.
$2 \cos^2 \frac{x}{2} & = 1 \\\cos^2 \frac{x}{2} & = \frac{1}{2} \\\frac{1 + \cos x}{2} & = \frac{1}{2} \\1 + \cos x & = 1 \\\cos x & = 0$
$\cos x = 0$ when $x = \frac{\pi}{2}, \frac{3 \pi}{2}$
#### Example C
Solve $\tan \frac{a}{2} = 4$ for $0^\circ \le a < 360^\circ$
Solution:
To solve $\tan \frac{a}{2} = 4$ , first isolate tangent, then use the half angle formula.
$\tan \frac{a}{2} & = 4 \\\sqrt{\frac{1 - \cos a}{1 + \cos a}} & = 4 \\\frac{1 - \cos a}{1 + \cos a} & = 16 \\16 + 16 \cos a & = 1 - \cos a \\17 \cos a & = - 15 \\\cos a & = - \frac{15}{17}$
Using your graphing calculator, $\cos a = - \frac{15}{17}$ when $a = 152^\circ, 208^\circ$
### Vocabulary
Half Angle Identity: A half angle identity relates the a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument.
### Guided Practice
1. Find the exact value of $\cos 112.5^\circ$
2. Find the exact value of $\sin 105^\circ$
3. Find the exact value of $\tan \frac{7 \pi}{8}$
Solutions:
1.
$\cos 112.5^\circ\\= \cos \frac{225^\circ}{2}\\= - \sqrt{\frac{1 + \cos 225^\circ}{2}} \\ = \sqrt{\frac{1 - \frac{\sqrt{2}}{2}}{2}}\\= - \sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}\\= - \sqrt{\frac{2 - \sqrt{2}}{4}}\\= - \frac{\sqrt{2 - \sqrt{2}}}{2}$
2.
$\sin 105^ \circ\\= \sin \frac{210^\circ}{2}\\= \sqrt{\frac{1 - \cos 210^\circ}{2}} \\= \sqrt{\frac{1 - \frac{\sqrt{3}}{2}}{2}}\\= \sqrt{\frac{\frac{2 - \sqrt{3}}{2}}{2}}\\= \sqrt{\frac{2 - \sqrt{3}}{4}}\\= \frac{\sqrt{2 - \sqrt{3}}}{2}$
3.
$\tan \frac{7 \pi}{8}\\= \tan \frac{1}{2} \cdot \frac{7 \pi}{4}\\= \frac{1 - \cos \frac{7 \pi}{4}}{\sin \frac{7 \pi}{4}} \\= \frac{1 - \frac{\sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}}\\= \frac{\frac{2 - \sqrt{2}}{2}}{- \frac{\sqrt{2}}{2}}\\= - \frac{2 - \sqrt{2}}{\sqrt{2}}\\= \frac{- 2 \sqrt{2} + 2}{2}\\= - \sqrt{2} +1$
### Concept Problem Solution
Knowing the half angle formulas, you can compute $\sin 22.5^\circ$ easily:
$\sin 22.5^\circ = \sin \left( \frac{45^\circ}{2} \right)\\=\sqrt{\frac{1-\cos 45^\circ}{2}}\\=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}}\\=\sqrt{\frac{\frac{2-\sqrt{2}}{2}}{2}}\\=\sqrt{\frac{2-\sqrt{2}}{4}}\\=\frac{\sqrt{2-\sqrt{2}}}{2}\\$
### Practice
Use half angle identities to find the exact value of each expression.
1. $\tan 15^\circ$
2. $\tan 22.5^\circ$
3. $\cot 75^\circ$
4. $\tan 67.5^\circ$
5. $\tan 157.5^\circ$
6. $\tan 112.5^\circ$
7. $\cos 105^\circ$
8. $\sin 112.5^\circ$
9. $\sec 15^\circ$
10. $\csc 22.5^\circ$
11. $\csc 75^\circ$
12. $\sec 67.5^\circ$
13. $\cot 157.5^\circ$
Use half angle identities to help solve each of the following equations on the interval $[0,2\pi)$ .
1. $3\cos^2(\frac{x}{2})=3$
2. $4\sin^2 x=8\sin^2(\frac{x}{2})$
### Vocabulary Language: English
Half Angle Identity
Half Angle Identity
A half angle identity relates a trigonometric function of one half of an argument to a set of trigonometric functions, each containing the original argument.
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Practice the questions of McGraw Hill Math Grade 8 Answer Key PDF Lesson 14.4 Slope to secure good marks & knowledge in the exams.
Exercises
FIND THE SLOPE
Question 1.
$$\frac{5.6}{10}$$,
Explanation:
Using two of the points on the line, we can find the slope of the line by
finding the rise and the run. The vertical change between two points is
called the rise and the horizontal change is called the run. The slope equals the rise divided by the run:
Slope =rise ÷ run = (y2 – y1) ÷ (x2 – x1),
We have given points as (-2, -5.1) and (8, 0.5),
So slope = (0.5 – (-5.1))/ (8 -(-2)) = 5.6/10.
Question 2.
–$$\frac{8.1}{14}$$,
Explanation:
Using two of the points on the line, we can find the slope of the line by
finding the rise and the run.
The vertical change between two points is called the rise and the horizontal change is called the run. The slope equals the rise divided by the run:
Slope =rise ÷ run = (y2 – y1) ÷ (x2 – x1),
We have given points as (-7, 7) and (7, -1.1),
So slope = (-1.1 – 7)/ (7 -(-7)) = -8.1/14.
Question 3.
y = 6x – 2
Slope is 6,
Explanation:
A slope of a line is the change in y coordinate with respect to
the change in x coordinate. The equation of the line is written in the
slope-intercept form, which is: y = mx + b,
where m represents the slope and b represents the y-intercept.
Therefore the given line y = 6x – 2 has slope 6.
Question 4.
y – x = 4
Slope is 1,
Explanation:
A slope of a line is the change in y coordinate with respect to
the change in x coordinate.
The equation of the line is written in the slope-intercept form,
which is: y = mx + b, where m represents the slope and b represents
the y-intercept. Therefore the given line y – x = 4, y = x + 4 has slope 1.
Question 5.
-3 = $$\frac{2}{3}$$x – y
Slope is $$\frac{2}{3}$$,
Explanation:
A slope of a line is the change in y coordinate with respect to
the change in x coordinate.
The equation of the line is written in the slope-intercept form,
which is: y = mx + b, where m represents the slope and
b represents the y-intercept.
Therefore the given line -3 = $$\frac{2}{3}$$x – y
y = $$\frac{2}{3}$$x + 3 has slope $$\frac{2}{3}$$.
Question 6.
x + y = -5
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# Difference between revisions of "2005 AMC 10B Problems/Problem 25"
## Problem
A subset $B$ of the set of integers from $1$ to $100$, inclusive, has the property that no two elements of $B$ sum to $125$. What is the maximum possible number of elements in $B$?
$\mathrm{(A)} 50 \qquad \mathrm{(B)} 51 \qquad \mathrm{(C)} 62 \qquad \mathrm{(D)} 65 \qquad \mathrm{(E)} 68$
## -Solutions-
### Solution 1
The question asks for the maximum possible number of elements. The integers from $1$ to $24$ can be included because you cannot make $125$ with integers from $1$ to $24$ without the other number being greater than $100$. The integers from $25$ to $100$ are left. They can be paired so the sum is $125$: $25+100$, $26+99$, $27+98$, $\ldots$, $62+63$. That is $38$ pairs, and at most one number from each pair can be included in the set. The total is $24 + 38 = \boxed{\mathrm{(C)}\ 62}$. Also, it is possible to see that since the numbers $1$ to $24$ are in the set there are only the numbers $25$ to $100$ to consider. As $62+63$ gives $125$, the numbers $25$ to $62$ can be put in subset $B$ without having two numbers add up to $125$. In this way, subset $B$ will have the numbers $1$ to $62$, and so the answer is $\boxed{\mathrm{(C)}\ 62}$.
#### Solution 1 Alternate Solution
Since there are 38 numbers that sum to $125$, there are $100-38=62$ numbers not summing to $125.$ ~mathboy282
### Solution 2 (If you have no time)
"Cut" $125$ into half. The maximum integer value in the smaller half is $62$. Thus the answer is $\boxed{\mathrm{(C)}\ 62}$.
### Solution 3
The maximum possible number of elements includes the smallest numbers. So, subset $B = \{1,2,3....n-1,n\}$ where n is the maximum number of elements in subset $B$. So, we have to find two consecutive numbers, $n$ and $n+1$, whose sum is $125$. Setting up our equation, we have $n+(n+1) = 2n+1 = 125$. When we solve for $n$, we get $n = 62$. Thus, the anser is $\boxed{\mathrm{(C)}\ 62}$.
~GentleTiger
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# 2Tvwo §5.4 Quadratic formulas 2!
§ 5.4 Quadratic formulas
1 / 21
Slide 1: Slide
WiskundeMiddelbare schoolhavo, vwoLeerjaar 2
This lesson contains 21 slides, with text slides.
Lesson duration is: 50 min
## Items in this lesson
§ 5.4 Quadratic formulas
#### Slide 1 -Slide
Removing brackets
§5.4 and §5.5 do NOT require you to remove brackets and write formulas shorter.
No, these paragraphs are about GRAPHS !
What have they got to do with §5.1 / §5.2 and §5.3 ?!
Well, the quadratic formulas from those paragraphs
can be used for making the graphs from §5.4 and §5.5.
#### Slide 2 -Slide
is a quadratic formula (it has a variable squared)
If we want to plot the graph of this formula, we should first make a table. Take the horizontal axis from -3 to 7.
y=2x28x
#### Slide 3 -Slide
Filling in a NEGATIVE number in a quadratic formula:
#### Slide 4 -Slide
Plotting the graph for a quadratic formula .....
requires a TABLE with x- and y-values, that we can FILL IN in a
COORDINATE SYSTEM.
Below you see the result. (By the way, we could do with less points.)
In case you think: how on earth do we get the y-outcomes for these
x values, then watch the next slide!
#### Slide 5 -Slide
Filling in numbers in a quadratic formula:
By the way:
+ we will discuss a few of these
calculations
+ normally we do these kinda
calculations in our heads and
+ write the results in the table!
#### Slide 6 -Slide
Plotting the graph for a quadratic formula
This table shows us we should take:
- the horizontal axis from -3 to 7 (with steps of 1)
- the vertical axis from - 10 to 45 (with steps of 5)
#### Slide 7 -Slide
Plotting the graph for a quadratic formula
Here
#### Slide 8 -Slide
Plotting the graph for a quadratic formula
Plot all points from the table
#### Slide 9 -Slide
Plotting the graph for a quadratic formula
Finally:
Connect them with a smooth line. line curve l We call this a PARABOLA.
Now a chess Joke!
#### Slide 11 -Slide
Upward or downward opening parabola
In the coordinate system on the
right, the green graphs are
upward opening parabolas
and the purple graphs are
downward opening parabolas.
#### Slide 12 -Slide
Upward or downward opening parabola
Graphs for a quadratic formula are
called parabolas
This parabola is an
upward opening parabola.
Notice that the number 2, before
the , is POSITIVE.
y=2x28x
x2
#### Slide 13 -Slide
Upward or downward opening parabola
You will find that you can tell from the formula if a parabola is upward or downward opening.
If the number in front of
is positive, it is upward opening
If the number in front of
is negative, it is downward opening.
x2
x2
#### Slide 14 -Slide
Vertex
The lowest point in an
upward opening parabola,
or the highest point in a
downward opening parabola
is called the vertex.
#### Slide 15 -Slide
Axis of symmetry
All parabolas have
reflection symmetry.
The vertical line through
the vertex is the axis of symmetry
All vertical lines have a formula
x = ....
One more chess joke follows.
#### Slide 20 -Slide
HOMEWORK
Make and correct §5.4
You may SKIP: 24 and 25
Then take pics and send in the lot in GC.
Make your teacher happy.
timer
25:00
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# Data And Statistics Standards Review Guide
Monday, May 26th: Data and Statistics Standard
Topics include: Histograms, Stem and Leaf Plots, Box and Whisker Plots, CEntral Tendencies, Mean vs. Median.
Math Notes:
Chapter 8 Problems 8-208-218-398-408-51, and 8-91
# Checkpoint 9A
Problem 9-39
Displays of Data: Histograms and Box Plots
Answers to problem 9-39: HistogramsA histogram is a method of showing data. It uses a bar to show the frequency (the number of times something occurs). The frequency measures something that changes numerically. (In a bar graph the frequency measures something that changes by category.) The intervals (called bins) for the data are shown on the horizontal axis and the frequency is represented by the height of a rectangle above the interval. The labels on the horizontal axis represent the lower end of each interval or bin. Example: Sam and her friends weighed themselves and here is their weight in pounds: 110, 120, 131, 112, 125, 135, 118, 127, 135, and 125. Make a histogram to display the information. Use intervals of 10 pounds. Solution: See histogram at right. Note that the person weighing 120 pounds is counted in the next higher bin. Box PlotsA box plot displays a summary of data using the median, quartiles, and extremes of the data. The box contains the “middle half” of the data. The right segment represents the top 25% of the data and the left segment represent the bottom 25% of the data. Example: Create a box plot for the set of data given in the previous example. Solution:Place the data in order to find the median (middle number) and the quartiles (middle numbers of the upper half and the lower half.) Based on the extremes, first quartile, third quartile, and median, the box plot is drawn. The interquartile range IQR = 131–118 = 13. Now we can go back to the original problem. The 0–1 bin contains the six students who do less than one hour of homework. The 1–2 bin contains the 10 students who do at least one hour but less than two hours. The 2–3 bin contains the seven students who do at least two hours but less than three hours. There are no students who do at least three hours and less than four. Two students did four hours and less than five. See the histogram above. The 0–5 bin contains two scores less than 5 points. The 5–10 bin contains the two scores of a least five but less than 10. The 10–15 bin contains the eight scores at least 10 but less than 15. The 15–20 bin contains the seven scores at least 15 but less than 20. See the histogram above. Place the ages in order: 46, 46, 47, 47, 48, 49, 50, 51, 52. The median is the middle age: 48. The first quartile is the median of the lower half of the ages. Since there are four lower-half ages, the median is the average of the middle two: . The third quartile is the median of the upper half ages. Again, there are four upper-half ages, so average the two middle ages: . The interquartile range is the difference between the third quartile and the first quartile: 50.5–46.5 = 4. See the box plot above. Place the scores in order: 70, 72, 75, 76, 80, 82, 85, 90, 93. The median is the middle score: 80. The lower quartile is the median of the lower half of the scores. Since there are four lower-half scores, the median is the average of the middle two: . The third quartile is the median of the upper half of the scores. Again, there are four upper-half scores, so average the two middle ages: . The interquartile range is the difference between the third quartile and the first quartile: 87.5–73.5 = 14. See the box plot above. Here are some more to try. For problems 1 through 6, create a histogram. For problems 7 through 12, create a box plot. State the quartiles and the interquartile range. 1. Number of heads showing in 20 tosses of three coins: 2, 2, 1, 3, 1, 0, 2, 1, 2, 1, 1, 2, 0, 1, 3, 2, 1, 3, 1, 2 2. Number of even numbers in 5 rolls of a dice done 14 times: 4, 2, 2, 3, 1, 2, 1, 1, 3, 3, 2, 2, 4, 5 3. Number of fish caught by 7 fishermen: 2, 3, 0, 3, 3, 1, 5 4. Number of girls in grades K-8 at local schools: 12, 13, 15, 10, 11, 12, 15, 11, 12 5. Number of birthdays in each March in various 2nd grade classes: 5, 1, 0, 0, 2, 4, 4, 1, 3, 1, 0, 4 6. Laps jogged by 15 students: 10, 15, 10, 13, 20, 14, 17, 10, 15, 20, 8, 7, 13, 15, 12 7. Number of days of rain: 6, 8, 10, 9, 7, 7, 11, 12, 6, 12, 14, 10 8. Number of times a frog croaked per minute: 38, 23, 40, 12, 35, 27, 51, 26, 24, 14, 38, 41, 23, 17 9. Speed in mph of 15 different cars: 30, 35, 40, 23, 33, 32, 28, 37, 30, 31, 29, 33, 39, 22, 30 10. Typing speed of 12 students in words per minute: 28, 30, 60, 26, 47, 53, 39, 42, 48, 27, 23, 86 11. Number of face cards pulled when 13 cards are drawn 15 times: 1, 4, 2, 1, 1, 0, 0, 2, 1, 3, 3, 0, 0, 2, 1 12. Height of 15 students in inches: 48, 55, 56, 65, 67, 60, 60, 57, 50, 59, 62, 65, 58, 70, 68
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# Basic Unit Conversions
The ability to convert between different units is an important aspect of math which is necessary for success in all of your science courses! For most science courses, you will be expected to know how to perform conversions between units with prefixes ranging from pico to mega. These larger and smaller units are defined in certain powers of 10, so making conversions is easy.
However, before we can convert between units, we need to know what each prefix represents. The following table provides the prefix, symbol, and meaning of the required units.
Prefix Symbol Multiplier Mega M 106 Kilo K 103 Centi C 10-2 Milli M 10-3 Micro μ 10-6 Nano N 10-9 Pico P 10-12
## CONVERTING FROM ONE UNIT TO ANOTHER
There are two common methods used for unit conversion.
### METHOD 1
• Subtract the power of the multiplier corresponding to the 2nd unit from that corresponding to the 1st unit.
• Move the decimal the number of places found in step one.
• If step one gives a positive result, the decimal moves to the right (this happens when we go from a larger prefix to a smaller one).
• If step one gives a negative result, the decimal moves to the left (this happens when we go from smaller prefix to larger one).
Example:
Convert 28 kg to mg.
Solution:
In order to convert 28 kg into mg, we would subtract the power of the multiplier for the 2nd unit (-3, since milligrams means we have 10-3 grams) from the first (3, since kilograms means we have 103 grams).
3 – (-3) = 6
Therefore, the decimal is moved 6 places to the right.
Answer: 28 000 000 mg, which can be rewritten as 2.8 x 107 mg in scientific notation.
Example:
Convert 127 nm to μ m.
Solution:
In order to convert 127 nm to μ m, we would subtract the power of the multiplier for the 2nd unit (-6, since micrometres means we have 10-6 metres) from the first (-9, since nanometers means we have 10-9 metres).
-9 – (-6)= -3
Therefore, the decimal is moved 3 places to the left.
Answer: 0.127 μ m, which can be rewritten as 1.27 x 10-1 μ m in scientific notation.
### METHOD 2
The second method is fairly similar to the first method. The number being converted is multiplied by a power of 10, which is determined by subtracting the power of the multiplier of the second unit from that of the first unit. The difference determines the power by which the value will be multiplied.
Example:
Convert 326 μ g to kg.
Solution:
Subtract the second multiplier’s power from that of the first (3, since kilograms means we have 103 grams, is subtracted from -6, since micrograms means we have 10-6 grams).
(-6) -3= -9
So, to perform the unit conversion, we multiply by 10-9.
Answer: 326 x 10-9kg, which can be written as 3.26 x 10-7 kg in scientific notation.
## CONVERTING MORE THAN ONE SET OF UNITS
More advanced conversions often require two different units to be converted. The best method is to convert one unit at a time.
Example:
Convert 18 km/hr into m/s.
Solution:
18 km/hr x 1000m/km x 1hr/60min x 1min/60sec = 5m/s
Answer: Therefore 18 km/hr = 5 m/s
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Try the fastest way to create flashcards
Question
# Solve by completing the square. Check your solutions. $x^{2}-x=2$
Solution
Verified
Step 1
1 of 3
Determine the solution to the equation.
\begin{align*} x^{2} - x + \dfrac{1}{4} & = 2 + \dfrac{1}{4} && {\text {find one-half of the coefficient of x then square it; add \dfrac{1}{4} to each side}} \\ \left (x - \dfrac{1}{2} \right)^{2} & = \dfrac{9}{4} && {\text {factor the perfect square trinomial}} \\\\ \sqrt{\left (x - \dfrac{1}{2} \right)^{2}} & = \sqrt{\dfrac{9}{4}} && {\text {square root property}} \\ x - \dfrac{1}{2} & = \pm \dfrac{3}{2} \\ x - \dfrac{1}{2} + \dfrac{1}{2} & = \dfrac{1}{2} \pm \dfrac{3}{2} && {\text {add \dfrac{1}{2} to both sides of the equation}} \\ x & = \dfrac{1}{2} \pm \dfrac{3}{2} \\\\ x & = \dfrac{1}{2} + \dfrac{3}{2} && {\text{write as two equations to simplify}} \\ x & = \dfrac{4}{2} \\ x & = 2 && {\text {solution to the equation}} \\\\ x & = \dfrac{1}{2} - \dfrac{3}{2}\\ x & = -\dfrac{2}{2} \\ x & = -1 && {\text {solution to the equation}} \end{align*}
## Recommended textbook solutions
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1st EditionISBN: 9781680330687Boswell, Larson
4,539 solutions
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# How do you find the sixth term of (x-y)^9?
May 25, 2017
$- 126 {x}^{4} {y}^{5}$
#### Explanation:
We have: ${\left(x - y\right)}^{9}$
Using the binomial theorem:
$= {x}^{9} + \left(\text{_1^9"") x^(8) (- y) + (""_2^9"") x^(7) (- y)^(2) + (""_3^9"") x^(6) (- y)^(3) + (""_4^9"") x^(5) (- y)^(4) + (""_5^9}\right) {x}^{4} {\left(- y\right)}^{5} + \ldots$
We only need to evaluate the sixth term of ${\left(x - y\right)}^{9}$:
= x^(9) + (""_1^9"") x^(8) (- y) + (""_2^9"") x^(7) (- y)^(2) + (""_3^9"") x^(6) (- y)^(3) + (""_4^9"") x^(5) (- y)^(4) + underline ((""_5^9"") x^(4) (- y)^(5)) + ...
Rightarrow "Sixth term" = (""_5^9"") x^(4) (- y)^(5)
$R i g h t a r r o w \text{Sixth term} = 126 \cdot {x}^{4} \cdot - {y}^{5}$
$R i g h t a r r o w \text{Sixth term} = - 126 {x}^{4} {y}^{5}$
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# 5 Essential Pythagorean Inequality Theorem Applications in Geometry
An Overview of the Pythagorean Inequality Theorem
The Pythagorean Inequality Theorem stands as a cornerstone in the study of geometry, taking the foundational concepts of the Pythagorean theorem further to classify triangles based on side lengths. It illuminates our understanding of triangles, enabling mathematicians and students alike to pinpoint whether a triangle is acute, obtuse, or right with simple calculations.
쉬운 목차
## Triangle Classification by Sides
Grasping the basics of triangle types is crucial before diving into the Pythagorean Inequality Theorem applications. Triangles are categorized as acute, obtuse, or right based on their internal angles. However, this theorem allows us to bypass angle measurements and directly use the side lengths to determine the triangle’s type.
### Pythagorean Theorem Fundamentals
The legendary Pythagorean Theorem asserts that for a right triangle, the square of the hypotenuse equals the sum of the squares of the other two sides. This formula (a^2 + b^2 = c^2) serves as the launching point for our theorem of interest.
#### Extending the Pythagorean Theorem
The Pythagorean Inequality Theorem stipulates that for any triangle, an acute one satisfies a^2 + b^2 > c^2, a right triangle meets a^2 + b^2 = c^2, and an obtuse triangle follows a^2 + b^2 < c^2. This provides a powerful tool for quick geometric analysis.
Discover more about the details of this theorem on Wikipedia.
Practical Applications and Examples
Consider a set of triangular sides measuring 3, 4, and 5 units. Applying the theorem, we find it’s a right triangle since 9 + 16 equals the square of 5. Another set of sides being 2, 2, and 3 units reveals an acute triangle since 4 + 4 is greater than 3 squared.
### Impacts on Structural Design
The implications of the Pythagorean Inequality Theorem applications extend to fields like architecting and computer graphics, ensuring precise geometric models and navigating systems.
Explore the interconnectedness of various geometric principles through green’s theorem explained: a comprehensive guide to vector calculus.
Enrichment Through Corollaries
Numerous corollaries that arise from the theorem deepen its significance, while related concepts like the Law of Cosines broaden our mathematical horizon.
Real-world Relevance
From astronomy to construction, the theorem greatly simplifies triangle classification and indirect distance measurement, proving invaluable in day-to-day problem-solving.
Educational Application Techniques
Instructors utilize the theorem in conjunction with engaging tools and real-world problems to bolster student understanding and appreciation of geometric concepts.
Conclusion: A Mathematical Staple
The universality of the Pythagorean Inequality Theorem applications is undeniable; its simplicity in classifying triangles makes it an essential component of geometry.
Future Research Trajectories
Mathematical advancements continually shine light on the theorem’s role in modern geometry, leading to fresh insights and broader applications.
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## Saturday, January 24, 2015
### Problems on normal shift and Refraction of Light
Normal shift happens when the object and the corresponding image are in two different mediums. Depending on the location of the object with respect to the medium there will be the normal shift. For example if the object is in the denser medium and observer is in the rarer medium, for the observer the object appears nearer to the surface. If the object is in the rarer medium and observer is in the denser medium, for the observer the object appears far from the original location.
Basing on the derivation is that we have made in the previous posts, we can use the equations that are derived for the normal shifts in the possible cases.
Problem and solution
A layer of oil 3 cm thick is floating on the layer of color water 5 cm thick. The refractive index of color water is 5 / 3. If the apparent depth of the two liquids together reuse 36 / 7 cm, what is the refractive index of oil ?
To solve this problem let us imagine a small colliery at the bottom of the vessel, below the two liquids. Let you yourself is the observer who is observing the oil from the surface. Therefore here in this case the object is in the denser medium and observer is in the rarer medium. As a result the object appears near to the surface. We can define refractive index in this case as the ratio of real depth to the apparent depth.
By applying this formula to both the liquids together we can get the refractive index of the given liquid as shown below.
Problem and solution
A fish is rising vertically to the surface of water in LA uniformly at a rate of 3 m/s. It observes a bird diving vertically towards the water at the rate of 9 m/s above it. If the refractive index of water is 4 /3, find the actual speed of the bird above the fish?
In this problem the fish is under the water and the bird in the air. Bird is the object and fishes the observer. As the object is the rarer medium and observer is the denser medium, for the observer objects appears away from its original position. Therefore the totally distance between the fish and the bird is sum of depth of the fish from the surface and the apparent height of the bird from the surface.
As the problem is dealing with the velocity, to get the velocity from the displacements we have to differentiate displacement once with respect to time.
We need to differentiate because rate of change of displacement is defined as velocity. In this problem the relative velocity of the bird with respect to fish is 9 m/s. We have to calculate the velocity of the bird with respect to air.
By simplifying the above equation we can get the velocity of the bird as shown.
Related Posts
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## Logarithm Base Changing Formula Proof
In a logarithmic expression, it is possible to change base using algebraic manipulation. For example, we can change
$log_416$ to $\dfrac{\log_216}{\log_24}$.
In this post, we are going to prove why it is possible to do such algebraic manipulation. The change of base above can be generalized as
$\log_ab = \dfrac{\log_cb}{\log_ca}$.
Theorem
$\log_ab = \dfrac{log_cb}{\log_ca}$.
Proof
If we let $\log_ab = x$, then by definition, $a^x = b$.
Now, take the logarithm to the base $c$ of both sides. That is
$log_c a^x = \log_cb$.
Simplifying the exponent, we have
$x \log_ca = \log_cb$.
Now, since $a \neq 1$, $\log_ca \neq 0$.
Therefore,
$x = \dfrac{\log_cb}{\log_ca}$
Thus,
$\log_ab = \dfrac{log_cb}{\log_ca}$
## Year 2015 in Review – Complete List of Posts
It’s the end of the year again, so let’s look at what we have learned so far in 2015. Below is the complete list of posts of mathematical proofs this year. Enjoy learning!
Year 2015 in Review – Complete List of Posts
You may also want to visit the Post List Page to explore the complete list of posts of Proofs from the Book.
## Proof that the nth root of a times nth root of b is equal to nth root of ab
Consider the following multiplications.
$\sqrt{4} \times \sqrt{16} = 2 \times 4 = 8$
$\sqrt{4 \times 16} = \sqrt{64} = 8$.
Also
$\sqrt[3]{27} \times \sqrt[3]{64} = 3 \times 4 = 12$
$\sqrt[3]{27} \times \sqrt[3]{64} = \sqrt[3]{1728} = 12$.
Try also several calculations like these and observe what happens.
From these calculations, we can observe that
$\sqrt{ab} = \sqrt{a} \times \sqrt{b}$
and
$\sqrt[3]{ab} = \sqrt[3]{a} \times \sqrt[3]{b}$
or, in general
$\sqrt[n]{ab} = \sqrt[n]{a} \times \sqrt[n]{b}$.
We now, prove our conjecture.
Theorem: $\sqrt[n]{a} \sqrt[n]{b} = \sqrt[n]{ab}$
Proof
If we multiply the expression on the left hand side by itself n times, we have
$(\sqrt[n]{a} \sqrt[n]{b})^n = (\sqrt[n]{a})^n (\sqrt[n]{b})^n = ab$.
On the other hand, a positive number ab has only a single positive root, $\sqrt[n]{ab}$.
Therefore,
$\sqrt[n]{a} \sqrt[n]{b} = \sqrt[n]{ab}$.
This completes our proof.
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# Math for 3rd Grade
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Teacher: Tom
Customers Who Have Viewed This Course: 2117
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#### 1 Course Introduction 00:43
Mr. P. introduces the course!
#### 2 Unit 1 Introduction 02:50
Mr. P. introduces the Unit 1 Review Sheet.
#### 3 Lesson 1 - Place Values to Millions 08:17
Lesson 1 Objective: we will identify the place values to the millions place (place value chart).
#### 4 Lesson 1 - Review Video 04:15
Mr. P. reviews the Lesson 1 problem set, quiz, and challenge.
#### 5 Lesson 2 - Value of a Digit 07:40
Lesson 2 - We will analyze how a digit’s placement in a number affects its value (place value chart).
#### 6 Lesson 2 - Review Video 04:58
Mr. P. reviews the Lesson 2 problem set, quiz, and challenge.
#### 7 Lesson 3 - Number Forms 08:32
Lesson 3 Objective: we will express whole numbers to the ten thousands place in standard, written, and expanded forms.
#### 8 Lesson 3 - Review Video 06:46
Mr. P. reviews the Lesson 3 problem set, quiz, and challenge.
#### 9 Lesson 4 - Bundle Units 06:37
Lesson 4 Objective: we will bundle units into tens and move them to the next greatest place value (place value chart).
#### 10 Lesson 4 - Review Video 08:30
Mr. P reviews the Lesson 4 problem set, quiz, and challenge.
#### 11 Lesson 5 - Analyze Digit Placement 10:19
Lesson 5 Objective: we will recognize that a digit represents ten times more than in the place value to its right (place value chart).
#### 12 Lesson 5 - Review Video 04:35
Mr. P. reviews the Lesson 5 problem set, quiz, and challenge.
#### 13 Lesson 6 - Compare Numbers 06:59
Lesson 6 Objective: we will compare whole numbers to the millions place (stacking).
#### 14 Lesson 6 - Review Video 07:50
Mr. P. reviews the Lesson 6 problem set, quiz, and challenge.
#### 15 Lesson 7 - Bookend Number Lines 10:13
Lesson 7 Objective: we will determine which two tens, hundreds, and thousands a given
whole number falls between (horizontal number line).
#### 16 Lesson 7 - Video Review 06:50
Mr. P. reviews the Lesson 7 problem set, quiz, and challenge.
#### 17 Lesson 8 - Round Numbers I 10:52
Lesson 8 Objective: we will round whole numbers to the thousands place (horizontal number line).
#### 18 Lesson 8 - Review Video 08:16
Mr. P. reviews the Lesson 8 problem set, quiz, and challenge.
#### 19 Lesson 9 - Round Numbers II 09:44
Lesson 9 Objective: we will round whole numbers to the thousands place (vertical number line).
#### 20 Lesson 9 - Review Video 07:01
Mr. P. reviews the Lesson 9 problem set, quiz, and challenge.
#### 21 Lesson 10 - Is it Reasonable? 10:15
Lesson 10 Objective: we will use our understanding of rounding to assess the reasonableness of given whole numbers.
#### 22 Lesson 10 - Review Video 09:39
Mr. P. reviews the Lesson 10 problem set, quiz, and challenge.
#### 23 Unit 2 Introduction 02:59
Mr. P. introduces Unit 2.
#### 24 Lesson 11 - Use Place Value to Add and Subtract 13:05
Lesson 11 Objective: we will find 1, 10 and 100 more and less than a given whole number.
#### 25 Lesson 11 - Review Video 06:45
Mr. P. reviews the Lesson 11 problem set, quiz, and challenge.
#### 26 Lesson 12 - Halves and Doubles 09:47
Lesson 12 Objective: we will find half of even whole numbers and double even whole numbers.
#### 26 Lesson 12 - Review Video 05:38
Mr. P. reviews the Lesson 12 problem set, quiz, and challenge.
#### 27 Lesson 13 - Add and Subtract Across Zero 07:18
Lesson 13 Objective: we will add and subtract single-digit whole numbers across zero (number line).
#### 28 Lesson 13 - Review Video 02:37
Mr. P. reviews the Lesson 13 problem set, quiz, and challenge.
#### 29 Lesson 14 - Dutch Algorithm I 08:21
Lesson 14 Objective: we will add and subtract whole numbers to the hundreds place (Dutch algorithm).
#### 30 Lesson 14 - Review Video 04:48
Mr. P. reviews the Lesson 14 problem set, quiz, and challenge.
#### 31 Lesson 15 - Dutch Algorithm II 09:41
Lesson 15 Objective: we will add and subtract whole numbers to the thousands place (Dutch algorithm).
#### 32 Lesson 15 - Review Video 07:57
Mr. P. reviews the Lesson 15 problem set, quiz, and challenge.
#### 33 Lesson 16 - Addition Word Problems 08:45
Lesson 16 Objective: we will solve one-step addition word problems.
#### 34 Lesson 16 - Review Video 08:24
Mr. P. reviews the Lesson 16 problem set, quiz, and challenge.
#### 35 Lesson 17 - Subtraction Word Problems 07:30
Lesson 17 Objective: we will solve one-step subtraction word problems (draw base-10 blocks).
#### 36 Lesson 17 - Review Video 08:06
Mr. P. reviews the Lesson 17 problem set, quiz, and challenge.
#### 37 Lesson 18 - 2-Step Word Problems 10:24
Lesson 18 Objective: we will solve two-step addition and subtraction word problems.
#### 38 Lesson 18 - Review Video 07:10
Mr. P. reviews the Lesson 18 problem set, quiz, and challenge.
#### 39 Unit 3 Introduction 02:44
Mr. P. introduces Unit 3.
#### 40 Lesson 19 - Multiples 08:58
Lesson 19 Objective: we will list a whole number’s multiples both vertically and horizontally.
#### 41 Lesson 19 - Review Video 07:10
Mr. P. reviews the Lesson 19 problem set, quiz, and challenge.
#### 42 Lesson 20 - Tile Area 06:15
Lesson 20 Objective: we will identify the correct equations for multiplicative comparisons when the greater value is unknown (bar model).
#### 43 Lesson 20 - Review Video 03:44
Mr. P. reviews the Lesson 20 problem set, quiz, and challenge.
#### 44 Lesson 21 - Multiplicative Comparisons I 06:27
Lesson 21 Objective: we will identify the correct equations for multiplicative comparisons when the greater value is unknown (bar model).
#### 45 Lesson 21 - Review Video 06:17
Mr. P. reviews the Lesson 21 problem set, quiz, and challenge.
#### 46 Lesson 22 - Multiplicative Comparisons II 09:19
Lesson 22 Objective: we will identify the correct scenarios for multiplicative comparisons when the greater value is unknown (bar model).
#### 47 Lesson 22 - Review Video 11:44
Mr. P. reviews the Lesson 22 problem set, quiz, and challenge.
#### 48 Lesson 23 - Fact Family 08:51
Lesson 23 Objective: we will understand division as an unknown factor problem (fact families).
#### 49 Lesson 23 - Review Video 03:02
Mr. P. reviews the Lesson 23 problem set, quiz, and challenge.
#### 50 Lesson 24 - Multiplicative Comparisons III 05:56
Lesson 24 Objective: we will identify the correct equations for multiplicative comparisons when the lesser value is unknown (bar model).
#### 51 Lesson 24 - Review Video 05:50
Mr. P. reviews the Lesson 24 problem set, quiz, and challenge.
#### 52 Lesson 25 - Multiplicative Comparisons IV 08:56
Lesson 25 Objective: we will identify the correct scenarios for multiplicative comparisons when the lesser value is unknown (bar model).
#### 53 Lesson 25 - Review Video 10:22
Mr. P. reviews the Lesson 25 problem set, quiz, and challenge.
#### 54 Lesson 26 - Compare Values I 07:22
Lesson 26 Objective: we will determine how many times more a greater value is than a lesser value (bar model).
#### 55 Lesson 26 - Review Video 07:44
Mr. P. reviews the Lesson 26 problem set, quiz, and challenge.
#### 56 Lesson 27 - Compare Values II 07:56
Lesson 27 Objective: we will determine how many times less a lesser value is than a greater value (bar model).
#### 57 Lesson 27 - Review Video 09:10
Mr. P. reviews the Lesson 27 problem set, quiz, and challenge.
#### 58 Lesson 28 - Factor Pairs 07:51
Lesson 28 Objective: we will find all factor pairs for a whole number in the range 1 – 100.
#### 59 Lesson 28 - Review Video 04:50
Mr. P. reviews the Lesson 28 problem set, quiz, and challenge.
#### 60 Lesson 29 - Factor Rainbows 09:07
Less 29 Objective: we will determine whether a whole number in the range 1 – 100 is a factor of another whole number (factor rainbow).
#### 61 Lesson 29 - Review Video 07:19
Mr. P. reviews the Lesson 29 problem set, quiz, and challenge.
#### 62 Lesson 30 - Factor Trees 10:08
Lesson 30 Objective: we will determine whether a whole number in the range 1 – 100 is prime or composite (factor tree).
#### 63 Lesson 30 - Review Video 08:50
Mr. P. reviews the Lesson 30 problem set, quiz, and challenge.
#### 64 Lesson 31 - Multiply by Multiples of 10 08:55
Lesson 31 Objective: we will multiply single-digit whole numbers by 10, 100, and 1,000 and recognize patterns.
#### 65 Lesson 31 - Review Video 04:51
Mr. P. reviews the Lesson 31 problem set, quiz, and challenge.
#### 66 Lesson 32 - Area Models I 07:54
Lesson 32 Objective: we will multiply 2-digit whole numbers by 1-digit whole numbers (area model).
#### 67 Lesson 32 - Review Video 06:56
Mr. P. reviews the Lesson 32 problem set, quiz, and challenge.
#### 68 Lesson 33 - Area Models II 10:30
Lesson 33 Objective: we will multiply 3-digit whole numbers by 1-digit whole numbers (area model).
#### 69 Lesson 33 - Review Video 09:08
Mr. P. reviews the Lesson 33 problem set, quiz, and challenge.
#### 70 Lesson 34 - Area Models III 14:00
Lesson 34 Objective: we will multiple 2-digit and 3-digit whole numbers by 2-digit whole numbers (area model).
#### 71 Lesson 34 - Review Video 08:22
Mr. P. reviews the Lesson 34 problem set, quiz, and challenge.
#### 72 Lesson 35 - Distributive Property of Multiplication 11:12
Lesson 35 Objective: we will apply the distributive property of multiplication.
#### 73 Lesson 35 - Review Video 06:41
Mr. P. reviews the Lesson 35 problem set, quiz, and challenge.
#### 74 Lesson 36 - Division Problems with Remainders 08:50
Lesson 36 Objective: we will solve division problems with remainders (array and area model).
#### 75 Lesson 36 - Review Video 08:36
Mr. P. reviews the Lesson 36 problem set, quiz, and challenge.
#### 76 Lesson 37 - Standard Division Algorithm I 10:33
Lesson 37 Objective: we will divide 2-digit whole numbers by 1-digit whole numbers (standard algorithm).
#### 77 Lesson 37 - Review Video 05:35
Mr. P. reviews the Lesson 37 problem set, quiz, and challenge.
#### 78 Lesson 38 - Standard Division Algorithm II 09:07
Lesson 38 Objective: we will divide 3-digit whole numbers by 1-digit whole numbers (standard algorithm).
#### 79 Lesson 38 - Review Video 06:45
Mr. P. reviews the Lesson 38 problem set, quiz, and challenge.
#### 80 Lesson 39 - Standard Division Algorithm III 11:34
Lesson 39 Objective: we will divide 4-digit whole numbers by 1-digit whole numbers (standard algorithm).
#### 81 Lesson 39 - Review Video 08:38
Mr. P. reviews the Lesson 39 problem set, quiz, and challenge.
#### 82 Lesson 40 - Division Word Problems 11:47
Lesson 40 Objective: we will solve division word problems.
#### 83 Lesson 40 - Review Video 07:12
Mr. P. reviews the Lesson 40 problem set, quiz, and challenge.
#### 84 Unit 4 Introduction 02:02
Mr. P. introduces Unit 4.
#### 85 Lesson 41 - Unit Fractions 06:17
Lesson 41 Objective: we will partition a variety of models in a variety of ways to show the unit fractions 1/2, 1/3, and 1/4.
#### 86 Lesson 41 - Review Video 04:52
Mr. P. reviews the Lesson 41 problem set, quiz, and challenge.
#### 87 Lesson 42 - Unit Fraction Sums 06:34
Lesson 42 Objective: we will decompose proper fractions into sums of unit fractions (number bond and bar model).
#### 88 Lesson 42 - Review Video 03:57
Mr. P. reviews the Lesson 42 problem set, quiz, and challenge.
#### 89 Lesson 43 - Unit Fractions Times Whole Numbers 05:24
Lesson 43: we will multiply unit fractions by whole numbers to build proper fractions.
#### 90 Lesson 43 - Review Video 04:25
Mr. P. reviews the Lesson 43 problem set, quiz, and challenge.
#### 91 Lesson 44 - Identify Equivalent Fractions 08:34
Lesson 44 Objective: we will identify equivalent fractions (bar model).
#### 92 Lesson 44 - Review Video 08:15
Mr. P. reviews the Lesson 44 problem set, quiz, and challenge.
#### 93 Lesson 45 - Multiply Equivalent Fractions 09:41
Lesson 45 Objective: we will identify equivalent fractions with multiplication (Magic-1).
#### 94 Lesson 45 - Review Video 06:32
Mr. P. reviews the Lesson 45 problem set, quiz, and challenge.
#### 95 Lesson 46 - Divide Equivalent Fractions 07:42
Lesson 46 Objective: we will identify equivalent fractions with division (Magic-1).
#### 96 Lesson 46 - Review Video 07:11
Mr. P. reviews the Lesson 46 problem set, quiz, and challenge.
#### 97 Lesson 47 - Whole Numbers as Fractions 06:21
Lesson 47 Objective: we will express whole numbers as fractions and connect them to equivalent fractions.
#### 98 Lesson 47 - Review Video 05:42
Mr. P. reviews the Lesson 47 problem set, quiz, and challenge.
#### 99 Lesson 48 - Compare Fractions I 05:52
Lesson 48 Objective: we will compare proper fractions with like denominators (bar model).
#### 100 Lesson 48 - Review Video 07:28
Mr. P. reviews the Lesson 48 problem set, quiz, and challenge.
#### 101 Lesson 49 - Compare Fractions II 08:02
Lesson 49 Objective: we will compare proper and improper fractions with like denominators.
#### 102 Lesson 49 - Review Video 06:46
Mr. P. reviews the Lesson 49 problem set, quiz, and challenge.
#### 103 Lesson 50 - Compare Fractions III 07:01
Lesson 50 Objective: we will compare two fractions with the same numerator but unlike denominators (bar model).
#### 104 Lesson 50 - Review Video 08:01
Mr. P. reviews the Lesson 50 problem set, quiz, and challenge.
#### 105 Lesson 51 - Add Fractions 06:01
Lesson 51 Objective: we will add fractions with like denominators.
#### 106 Lesson 51 - Review Video 03:34
Mr. P. reviews the Lesson 51 problem set, quiz, and challenge.
#### 107 Lesson 52 - Subtract Fractions 06:01
Lesson 52 Objective: we will subtract fractions with like denominators.
#### 108 Lesson 52 - Review Video 03:54
Mr. P. reviews the Lesson 52 problem set, quiz, and challenge.
#### 109 Lesson 53 - Improper Fractions to Mixed Numbers 06:59
Lesson 53 Objective: we will convert improper fractions into mixed numbers (number bond).
#### 110 Lesson 53 - Review Video 07:03
Mr. P. reviews the Lesson 53 problem set, quiz, and challenge.
#### 111 Lesson 54 - Fraction Word Problems 07:16
Lesson 54 Objective: we will solve fraction word problems (bar model and number line).
#### 112 Lesson 54 - Review Video 10:55
Mr. P. reviews the Lesson 54 problem set, quiz, and challenge.
#### 113 Unit 5 Introduction 02:02
Mr. P. introduces Unit 5.
#### 114 Lesson 55 - Tell Time 08:34
Lesson 55 Objective: we will tell and write time within one hour (clock model and number line).
#### 115 Lesson 55 - Review Video 07:53
Mr. P. reviews the Lesson 55 problem set, quiz, and challenge.
#### 116 Lesson 56 - Estimate Mass and Volume 06:38
Lesson 56 Objective: we will estimate liquid volumes and mass of objects using grams, kilograms, and liters.
#### 117 Lesson 56 - Review Video 05:40
Mr. P. reviews the Lesson 56 problem set, quiz, and challenge.
#### 118 Lesson 57 - Mass and Volume Word Problems 10:42
Lesson 57 Objective: we will solve word problems involving masses or volumes with the same units.
#### 119 Lesson 57 - Review Video 08:30
Mr. P. reviews the Lesson 57 problem set, quiz, and challenge.
#### 120 Lesson 58 - Scaled Picture Graphs 06:32
Lesson 58 Objective: we will interpret scaled picture graphs that represent data sets with several categories.
#### 121 Lesson 58 - Review Video 06:03
Mr. P. reviews the Lesson 58 problem set, quiz, and challenge.
#### 122 Lesson 59 - Scaled Picture Graph Word Problems 07:00
Lesson 59 Objective: we will solve word problems using information presented in scaled picture graphs.
#### 123 Lesson 59 - Review Video 07:39
Mr. P. reviews the Lesson 59 problem set, quiz, and challenge.
#### 124 Lesson 60 - Measurement 07:12
Lesson 60 Objective: we will measure everyday objects using rulers marked with halves and fourths of an inch.
#### 125 Lesson 60 - Review Video 05:04
Mr. P. reviews the Lesson 60 problem set, quiz, and challenge.
#### 126 Lesson 61 - Line Plots 07:51
Lesson 61 Objective: we will display ruler measurements in line plots.
#### 127 Lesson 61 - Review Video 06:19
Mr. P. reviews the Lesson 61 problem set, quiz, and challenge.
#### 128 Unit 6 Introduction 01:58
Mr. introduces Unit 6.
#### 129 Lesson 62 – Perimeter of Triangles 06:17
Lesson 62 Objective: we will calculate the perimeter of triangles.
#### 130 Lesson 62 - Review Video 04:31
Mr. P. reviews the Lesson 62 problem set, quiz, and challenge.
#### 132 Lesson 63 – Perimeter of Quadrilaterals 06:43
Lesson 63 Objective: we will calculate the perimeter of quadrilaterals.
#### 132 Lesson 63 - Review Video 05:12
Mr. P. reviews the Lesson 63 problem set, quiz, and challenge.
#### 133 Lesson 64 – Solve for an Unknown Side 08:49
Lesson 64 Objective: we will solve for an unknown side length when given the perimeter of triangles and quadrilaterals.
#### 134 Lesson 64 - Review Video 06:08
Mr. P. reviews the Lesson 64 problem set, quiz, and challenge.
#### 135 Lesson 65 – Rectangles with a Given Perimeter 08:09
Lesson 65 Objective: we will discover all the possible rectangles with a given perimeter.
#### 136 Lesson 65 - Review Video 11:26
Mr. P. reviews the Lesson 65 problem set, quiz, and challenge.
#### 137 Lesson 66 – Rectangles with a Given Area 07:27
Lesson 66 Objective: we will discover all the possible rectangles with a given area.
#### 138 Lesson 66 - Review Video 07:31
Mr. P. reviews the Lesson 66 problem set, quiz, and challenge.
#### 139 Lesson 67 - Express Area as Additive 07:07
Lesson 67 Objective: we will express area as additive (area model).
#### 140 Lesson 67 - Review Video 06:46
Lesson 67 Objective: we will express area as additive (area model).
#### 141 Lesson 68 - Identify Angles 09:05
Lesson 68 Objective: we will identify acute, right, and obtuse angles.
#### 142 Lesson 68 - Review Video 07:18
Mr. P. reviews the Lesson 68 problem set, quiz, and challenge.
#### 143 Lesson 69 - Types of Quadrilaterals 06:45
Lesson 69 Objective: we will recognize rhombuses, rectangles, and squares as examples of quadrilaterals.
#### 144 Lesson 69 - Review Video 08:30
Mr. P. reviews the Lesson 69 problem set, quiz, and challenge.
#### 145 Lesson 70 - Non-Traditional Quadrilaterals 07:39
Lesson 70 Objective: we will draw examples of quadrilaterals that do not belong to the following sub-categories: rhombus, rectangle, square.
#### 146 Lesson 70 - Review Video 07:02
Mr. P. reviews the Lesson 70 problem set, quiz, and challenge.
Course Description
• Informed by the Common Core State Standards for Mathematical Practice, this comprehensive 3rd grade math course features 70 lessons that emphasize critical analysis over rote computation. We’ll draw and make sense of models, analyze patterns, and explore a variety of non-standard algorithms to develop the number sense necessary for higher level math. This application-driven approach will arm your student with the skill-set necessary to independently tackle new concepts and confidently solve unfamiliar problems.
This course includes:
• 6 units
• 70 student packets
• 70 video-lessons
• 70 exercise review videos (Mr. P. reviews each problem step-by-step)
• approximately 18 hours of video-lessons
• a 2-question review at the start of each lesson
• “fast facts” fluency practice at the start of each lesson
• vocabulary words and definitions for each lesson
• a challenge question for each lesson
• a unit review sheet for each unit
Course Goals:
• This course explores a variety of models and non-standard algorithms designed to bolster conceptual understanding and strengthen number sense.
Target Audience:
• This video course is primarily intended for 7 – 10-year-olds from all education settings. It can function effectively both as a primary or supplemental curriculum.
Course Requirements:
• Students taking this course will need to have completed a 2nd grade math course or its equivalent.
Course Topics:
Unit 1 – Place Value and Rounding (10 lessons)
Lesson 1 – We will identify the place values to the millions place (place value chart).
Lesson 2 – We will analyze how a digit’s placement in a number affects its value (place value chart).
Lesson 3 – We will express whole numbers to the ten thousands place in standard, written, and expanded forms.
Lesson 4 – We will bundle units into tens and move them to the next greatest place value (place value chart).
Lesson 5 – We will recognize that a digit represents ten times more than in the place value to its right (place value chart).
Lesson 6 – We will compare whole numbers to the millions place (stacking).
Lesson 7 – We will determine which two tens, hundreds, and thousands a given
whole number falls between (horizontal number line).
Lesson 8 – We will round whole numbers to the thousands place (horizontal number line).
Lesson 9 – We will round whole numbers to the thousands place (vertical number line).
Lesson 10 – We will use our understanding of rounding to assess the reasonableness of given whole numbers.
Unit 2 – Addition and Subtraction (8 lessons)
Lesson 11 – We will find 1, 10 and 100 more and less than a given whole number.
Lesson 12 – We will find half of even whole numbers and double even whole numbers.
Lesson 13 – We will add and subtract single-digit whole numbers across zero (number line).
Lesson 14 – We will add and subtract whole numbers to the hundreds place (Dutch algorithm).
Lesson 15 – We will add and subtract whole numbers to the thousands place (Dutch algorithm).
Lesson 16 – We will solve one-step addition word problems.
Lesson 17 – We will solve one-step subtraction word problems (draw base-10 blocks).
Lesson 18 – We will solve two-step addition and subtraction word problems.
Unit 3 – Multiplication and Division (22 lessons)
Lesson 19 – We will list a whole number’s multiples both vertically and horizontally.
Lesson 20 – We will calculate the area of squares and rectangles (tile).
Lesson 21 – We will identify the correct equations for multiplicative comparisons when the greater value is unknown (bar model).
Lesson 22 – We will identify the correct scenarios for multiplicative comparisons when the greater value is unknown (bar model).
Lesson 23 – We will understand division as an unknown factor problem (fact family).
Lesson 24 – We will identify the correct equations for multiplicative comparisons when the lesser value is unknown (bar model).
Lesson 25 – We will identify the correct scenarios for multiplicative comparisons when the lesser value is unknown (bar model).
Lesson 26 – We will determine how many times more a greater value is than a lesser value (bar model).
Lesson 27 – We will determine how many times less a lesser value is than a greater value (bar model).
Lesson 28 – We will find all factor pairs for a whole number in the range 1 – 100.
Lesson 29 – We will determine whether a whole number in the range 1 – 100 is a factor of another whole number (factor rainbow).
Lesson 30 – We will determine whether a whole number in the range 1 – 100 is prime or composite (factor tree).
Lesson 31 – We will multiply single-digit whole numbers by 10, 100, and 1,000 and recognize patterns.
Lesson 32 – We will multiply 2-digit whole numbers by 1-digit whole numbers (area model).
Lesson 33 – We will multiply 3-digit whole numbers by 1-digit whole numbers (area model).
Lesson 34 – We will multiple 2-digit and 3-digit whole numbers by 2-digit whole numbers (area model).
Lesson 35 – We will apply the distributive property of multiplication.
Lesson 36 – We will solve division problems with remainders (array and area model).
Lesson 37 – We will divide 2-digit whole numbers by 1-digit whole numbers (standard algorithm).
Lesson 38 – We will divide 3-digit whole numbers by 1-digit whole numbers (standard algorithm).
Lesson 39 – We will divide 4-digit whole numbers by 1-digit whole numbers (standard algorithm).
Lesson 40 – We will solve division word problems.
Unit 4 – Fractions (14 lessons)
Lesson 41 – We will partition a variety of models in a variety of ways to show the unit fractions 1/2, 1/3, and 1/4.
Lesson 42 – We will decompose proper fractions into sums of unit fractions (number bond and bar model).
Lesson 43 – We will multiply unit fractions by whole numbers to build fractions.
Lesson 44 – We will identify equivalent fractions (bar model).
Lesson 45 – We will identify equivalent fractions with multiplication (Magic-1).
Lesson 46 – We will identify equivalent fractions with division (Magic-1).
Lesson 47 – We will express whole numbers as fractions and fractions as whole numbers.
Lesson 48 – We will compare proper fractions with like denominators (bar model).
Lesson 49 – We will compare proper and improper fractions with like denominators.
Lesson 50 – We will compare two fractions with like numerators but unlike denominators (bar model).
Lesson 51 – We will add fractions with like denominators.
Lesson 52 – We will subtract fractions with like denominators.
Lesson 53 – We will convert improper fractions into mixed numbers (number bond).
Lesson 54 – We will solve fraction word problems (bar model and number line).
Unit 5 – Time and Measurement (7 lessons)
Lesson 55 – We will calculate times within one hour (clock model and number line).
Lesson 56 – We will estimate liquid volumes and mass of objects using grams, kilograms, and liters.
Lesson 57 – We will solve a variety of word problems involving masses and volumes with like units.
Lesson 58 – We will interpret scaled picture graphs/pictographs.
Lesson 59 – We will solve word problems with information from scaled picture graphs/pictographs.
Lesson 60 – We will measure everyday objects using rulers marked with halves and fourths of an inch.
Lesson 61 – We will display ruler measurements in line plots.
Unit 6 – Geometry (9 lessons)
Lesson 62 – We will calculate the perimeter of triangles.
Lesson 63 – We will calculate the perimeter of quadrilaterals.
Lesson 64 – We will solve for an unknown side length when given the perimeter of triangles and quadrilaterals.
Lesson 65 – We will discover all the possible rectangles with a given perimeter.
Lesson 66 – We will discover all the possible rectangles with a given area.
Lesson 67 – We will express area as additive (area model).
Lesson 68 – We will identify acute, right, and obtuse angles.
Lesson 69 – We will recognize rhombuses, rectangles, and squares as examples of quadrilaterals.
Lesson 70 – We will draw examples of quadrilaterals that do not belong to the following sub-categories: rhombus, rectangle, square.
• Teacher: Tom
• Areas of expertise: 3rd - 5th grade math
• Education: Georgetown University
• Interests: Books, sports, travel, film
• Skills:
• Associations: Teach For America
• Issues I care about: Education, current events
Born and raised in Kansas City. After graduating from Georgetown University, I joined the Teach For America program in Los Angeles. For the last seven years I've taught 3rd - 5th grade math and ELA to students in Los Angeles and New York City. My application-driven lessons will arm your student with the skill-set necessary to independently tackle new concepts and confidently solve unfamiliar problems.
#### Unit 1 Review Sheet
A front and back one-sheet that reviews the content from Unit 1.
#### Unit 2 Review Sheet
A front and back one-sheet that reviews the content from Unit 2.
#### Unit 3 Review Sheet
A front and back one-sheet that reviews the content from Unit 3.
#### Unit 4 Review Sheet
A front and back one-sheet that reviews the content from Unit 4.
#### Unit 5 Review Sheet
A front and back one-sheet that reviews the content from Unit 5.
#### Unit 6 Review Sheet
A front and back one-sheet that reviews the content from Unit 6.
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ISEE Upper Level Math : How to find the surface area of a cube
Example Questions
← Previous 1
Example Question #1 : Solve For Surface Area
The length of the side of a cube is . Give its surface area in terms of .
Explanation:
Substitute in the formula for the surface area of a cube:
Example Question #1 : Solve Problems Involving Area, Volume And Surface Area Of Two And Three Dimensional Objects: Ccss.Math.Content.7.G.B.6
If a cube has one side measuring cm, what is the surface area of the cube?
Explanation:
To find the surface area of a cube, use the formula , where represents the length of the side. Since the side of the cube measures , we can substitute in for .
Example Question #1 : How To Find The Surface Area Of A Cube
Your friend gives you a puzzle cube for your birthday. If the length of one edge is 5 cm, what is the surface area of the cube?
Explanation:
Your friend gives you a puzzle cube for your birthday. If the length of one edge is 5 cm, what is the surface area of the cube?
To find the surface area of a cube, use the following formula:
This works, because we have 6 sides, each of which has an area of
Plug in our known to get our answer:
Example Question #2 : How To Find The Surface Area Of A Cube
A cube has a side length of , what is the surface area of the cube?
Explanation:
A cube has a side length of , what is the surface area of the cube?
Surface area of a cube can be found as follows:
Plug in our side length to find our answer:
Example Question #3 : How To Find The Surface Area Of A Cube
If one of the edges has a length of 6 inches, what is the surface area of the box?
Explanation:
If one of the edges has a length of 6 inches, what is the surface area of the box?
We can find the surface area of a square by squaring the length of the side and then multiplying it by 6.
Example Question #4 : How To Find The Surface Area Of A Cube
Find the surface area of a cube that has a width of 6cm.
Explanation:
To find the surface area of a cube, we will use the following formula:
where a is the length of any side of the cube.
Now, we know the width of the cube is 6cm. Because it is a cube, all sides are 6cm. That is why we can choose any side to substitute into the formula.
Now, knowing this, we can substitute into the formula. We get
Example Question #5 : How To Find The Surface Area Of A Cube
Find the surface area of a cube with a length of 7in.
Explanation:
To find the surface area of a cube, we will use the following formula:
where a is the length of any side of the cube. Note that all sides are equal on a cube. That is why we can use any length in the formula.
Now, we know the length of the cube is 7in.
Knowing this, we can substitute into the formula. We get
Example Question #6 : How To Find The Surface Area Of A Cube
Find the surface area of a cube with a width of 4cm.
Explanation:
To find the surface area of a cube, we will use the following formula.
where a is the length of any side of the cube.
Now, we know the width of the cube is 4in. Because it is a cube, all sides are equal (this is why we can use any length in the formula). So, we will use 4in in the formula. We get
Example Question #7 : How To Find The Surface Area Of A Cube
Find the surface area of a cube with a length of 12in.
Explanation:
To find the surface area of a cube, we will use the following formula:
where l is the length, and w is the width of the cube.
Now, we know the length of the cube is 12in. Because it is a cube, all lengths, widths, and heights are the same. Therefore, the width is also 12in.
Knowing this, we can substitute into the formula. We get
Example Question #8 : How To Find The Surface Area Of A Cube
While exploring an ancient ruin, you discover a small puzzle cube. You measure the side length to be . Find the cube's surface area.
Explanation:
While exploring an ancient ruin, you discover a small puzzle cube. You measure the side length to be . Find the cube's surface area.
To find the surface area, use the following formula:
← Previous 1
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# Math Snap
## PROBLEM
#### STEP 2
First, we need to find the remaining amount that Chloe has to pay. This can be calculated by subtracting the deposit from the total cost of the item. Let's denote the total cost as. $Remaining\, amount = - Deposit$
#### STEP 3
We don't know the total cost, but we can express the remaining amount in terms of the interest. The formula for simple interest is$Interest = Principal\, amount \\times Interest\, rate \\times Time$
#### STEP 4
In this case, the principal amount is the remaining amount, the interest rate is20.75%, and the time is3 years. So we can write$Interest = Remaining\, amount \\times20.75\% \\times3$
#### STEP 5
Rearranging the formula, we can express the remaining amount in terms of the interest$Remaining\, amount = Interest / (20.75\% \\times3)$
#### STEP 6
The problem asks for the amount payable after3 years, excluding the deposit. This is the same as the remaining amount plus the interest$Amount\, payable = Remaining\, amount + Interest$
#### STEP 7
Substituting the expression for the remaining amount from5 into the formula from6, we get$Amount\, payable = Interest / (20.75\% \\times3) + Interest$
#### STEP 8
This equation tells us that the amount payable depends on the interest. However, we don't know the interest. We can solve this problem by making an educated guess for the interest and adjusting it until the left and right sides of the equation match.
#### STEP 9
Let's start by guessing that the interest is R\$,000. Substituting this into the equation, we get$$Amount\, payable = R\,000 / (20.75\% \\times3) + R\,000$\$
#### STEP 10
Calculate the amount payable with this guess for the interest. $Amount\, payable = R\,000 / (0.2075 \\times3) + R\,000$
#### STEP 11
If the calculated amount payable is too high, try a lower guess for the interest. If it's too low, try a higher guess. Repeat this process until the calculated amount payable matches the actual amount payable.
##### SOLUTION
Once the correct interest has been found, the amount payable after years, excluding the deposit, is the calculated amount payable.
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# 6.1 – Ratios, Proportions, and the Geometric Mean
```6.1 – Ratios, Proportions, and the
Geometric Mean
If a and b are two numbers or quantities and b
does not equal zero, then the ratio of a to b is
a/b. The ratio of a to b can also be written as
a:b.
For example, the ratio of a side length of
Triangle ABC to a side length in Triangle DEF
can be written as 2/1 or 2:1.
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 1:
Simplify the ratio.
a. 64 m: 6 m
b. 5 ft / 20 in
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 1:
Simplify the ratio.
c. 24 yds to 3 yds
d. 150 cm: 6 m
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 2:
You are planning to paint a mural on a
rectangular wall. You know that the perimeter
of the wall is 484 feet and that the ratio of its
length to its width is 9:2. Find the area of the
wall.
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 3:
The measures of the angles in Triangle CDE
are in the extended ratio of 1:2:3. Find the
measures of the angles.
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 4:
The perimeter of a room is 48 feet and the ratio
of its length to its width is 7:5. Find the length
and width of the room.
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 5:
The measures of the angles in Triangle CDE
are in the extended ratio of 1:3:5. Find the
measures of the angles.
6.1 – Ratios, Proportions, and the
Geometric Mean
An equation that states that two ratios are
equal is called a proportion.
The numbers b and c are the means of the
proportion. The numbers a and d are the
extremes of the proportion.
6.1 – Ratios, Proportions, and the
Geometric Mean
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 6: Solve the proportion.
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 7: Solve the proportion.
As part of an environmental study, you need to
estimate the number of trees in a 150 acre
area. You count 270 trees in a 2 acre area and
you notice that the trees seem to be evenly
distributed. Estimate the total number of trees.
6.1 – Ratios, Proportions, and the
Geometric Mean
6.1 – Ratios, Proportions, and the
Geometric Mean
Example 8:
Find the geometric mean of 24 and 48.
Find the geometric mean of 12 and 27.
```
Temporal rates
16 Cards
Pi
12 Cards
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# 2.8 Graphical analysis of one-dimensional motion (Page 4/8)
Page 4 / 8
A graph of velocity vs. time of a ship coming into a harbor is shown below. (a) Describe the motion of the ship based on the graph. (b)What would a graph of the ship’s acceleration look like?
(a) The ship moves at constant velocity and then begins to decelerate at a constant rate. At some point, its deceleration rate decreases. It maintains this lower deceleration rate until it stops moving.
(b) A graph of acceleration vs. time would show zero acceleration in the first leg, large and constant negative acceleration in the second leg, and constant negative acceleration.
## Section summary
• Graphs of motion can be used to analyze motion.
• Graphical solutions yield identical solutions to mathematical methods for deriving motion equations.
• The slope of a graph of displacement $x$ vs. time $t$ is velocity $v$ .
• The slope of a graph of velocity $v$ vs. time $t$ graph is acceleration $a$ .
• Average velocity, instantaneous velocity, and acceleration can all be obtained by analyzing graphs.
## Conceptual questions
(a) Explain how you can use the graph of position versus time in [link] to describe the change in velocity over time. Identify (b) the time ( ${t}_{a}$ , ${t}_{b}$ , ${t}_{c}$ , ${t}_{d}$ , or ${t}_{e}$ ) at which the instantaneous velocity is greatest, (c) the time at which it is zero, and (d) the time at which it is negative.
(a) Sketch a graph of velocity versus time corresponding to the graph of displacement versus time given in [link] . (b) Identify the time or times ( ${t}_{a}$ , ${t}_{b}$ , ${t}_{c}$ , etc.) at which the instantaneous velocity is greatest. (c) At which times is it zero? (d) At which times is it negative?
(a) Explain how you can determine the acceleration over time from a velocity versus time graph such as the one in [link] . (b) Based on the graph, how does acceleration change over time?
(a) Sketch a graph of acceleration versus time corresponding to the graph of velocity versus time given in [link] . (b) Identify the time or times ( ${t}_{a}$ , ${t}_{b}$ , ${t}_{c}$ , etc.) at which the acceleration is greatest. (c) At which times is it zero? (d) At which times is it negative?
Consider the velocity vs. time graph of a person in an elevator shown in [link] . Suppose the elevator is initially at rest. It then accelerates for 3 seconds, maintains that velocity for 15 seconds, then decelerates for 5 seconds until it stops. The acceleration for the entire trip is not constant so we cannot use the equations of motion from Motion Equations for Constant Acceleration in One Dimension for the complete trip. (We could, however, use them in the three individual sections where acceleration is a constant.) Sketch graphs of (a) position vs. time and (b) acceleration vs. time for this trip.
A cylinder is given a push and then rolls up an inclined plane. If the origin is the starting point, sketch the position, velocity, and acceleration of the cylinder vs. time as it goes up and then down the plane.
## Problems&Exercises
Note: There is always uncertainty in numbers taken from graphs. If your answers differ from expected values, examine them to see if they are within data extraction uncertainties estimated by you.
(a) By taking the slope of the curve in [link] , verify that the velocity of the jet car is 115 m/s at $t=\text{20 s}$ . (b) By taking the slope of the curve at any point in [link] , verify that the jet car’s acceleration is $5\text{.}{\text{0 m/s}}^{2}$ .
(a) $\text{115 m/s}$
(b) $5\text{.}{\text{0 m/s}}^{2}$
Using approximate values, calculate the slope of the curve in [link] to verify that the velocity at $t=\text{10.0 s}$ is 0.208 m/s. Assume all values are known to 3 significant figures.
Using approximate values, calculate the slope of the curve in [link] to verify that the velocity at $t=\text{30.0 s}$ is 0.238 m/s. Assume all values are known to 3 significant figures.
$v=\frac{\left(\text{11.7}-6.95\right)×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{m}}{\left(40\text{.}\text{0 – 20}.0\right)\phantom{\rule{0.25em}{0ex}}\text{s}}=\text{238 m/s}$
By taking the slope of the curve in [link] , verify that the acceleration is $3\text{.}2 m{\text{/s}}^{2}$ at $t=\text{10 s}$ .
Construct the displacement graph for the subway shuttle train as shown in [link] (a). Your graph should show the position of the train, in kilometers, from t = 0 to 20 s. You will need to use the information on acceleration and velocity given in the examples for this figure.
(a) Take the slope of the curve in [link] to find the jogger’s velocity at $t=2\text{.}5 s$ . (b) Repeat at 7.5 s. These values must be consistent with the graph in [link] .
A graph of $v\left(t\right)$ is shown for a world-class track sprinter in a 100-m race. (See [link] ). (a) What is his average velocity for the first 4 s? (b) What is his instantaneous velocity at $t=5 s$ ? (c) What is his average acceleration between 0 and 4 s? (d) What is his time for the race?
(a) 6 m/s
(b) 12 m/s
(c) ${\text{3 m/s}}^{2}$
(d) 10 s
[link] shows the displacement graph for a particle for 5 s. Draw the corresponding velocity and acceleration graphs.
what are the calculations of Newton's third law of motiow
what is dark matter
(in some cosmological theories) non-luminous material which is postulated to exist in space and which could take either of two forms: weakly interacting particles ( cold dark matter ) or high-energy randomly moving particles created soon after the Big Bang ( hot dark matter ).
Usman
if the mass of a trolley is 0.1kg. calculate the weight of plasticine that is needed to compensate friction. (take g=10m/s and u=0.2)
what is a galaxy
what isflow rate of volume
flow rate is the volume of fluid which passes per unit time;
Rev
flow rate or discharge represnts the flow passing in unit volume per unit time
bhat
When two charges q1 and q2 are 6 and 5 coulomb what is ratio of force
When reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?
nehemiah
why is it proportional
i don't know
y
nehemiah
what are the relationship between distance and displacement
They are interchangeable.
Shii
Distance is scalar, displacement is vector because it must involve a direction as well as a magnitude. distance is the measurement of where you are and where you were displacement is a measurement of the change in position
Shii
Thanks a lot
Usman
I'm beginner in physics so I can't reason why v=u+at change to v2=u2+2as and vice versa
Usman
what is kinematics
praveen
kinematics is study of motion without considering the causes of the motion
Theo
The study of motion without considering the cause 0f it
Usman
why electrons close to the nucleus have less energy and why do electrons far from the nucleus have more energy
Theo
thank you frds
praveen
plz what is the third law of thermodynamics
third law of thermodynamics states that at 0k the particles will collalse its also known as death of universe it was framed at that time when it waa nt posible to reach 0k but it was proved wrong
bhat
I have not try that experiment but I think it will magnet....
Hey Rev. it will
Jeff
I do think so, it will
Chidera
yes it will
lasisi
If a magnet is in a pool of water, would it be able to have a magnetic field?.
yes Stella it would
Jeff
formula for electric current
Fokoua
what are you given?
Kudzy
what is current
Fokoua
I=q/t
saifullahi
Current is the flow of electric charge per unit time.
saifullahi
What are semi conductors
saifullahi
materials that allows charge to flow at varying conditions, temperature for instance.
Mokua
these are materials which have electrical conductivity greater than the insulators but less than metal, in these materials energy band Gap is very narrow as compared to insulators
Sunil
materials that allows charge to flow at varying conditions, temperature for instance.
Obasi
wao so awesome
Fokoua
At what point in the oscillation of beam will a body leave it?
Atambiri
what is gravitational force
what is meant by the term law
what is physics
it is branch of science that deal with interaction matter and energy is called physics . and physics is based in experiential observation and quentative measurement.
syed
to briefly understand the concept of physics start with history and a brief history of time by Stephen hawkings is what made me have interest in physics
ayesha
physics is a branch of science which deals with the study of matter, in relation to energy.
Frank
physics is a natural science that involve the study of matter and it's motion through space and time, along with related concept such as energy and force
Shodunke
Physics is the science of natural things. for instance, take classical laws which describe the principles of working of the macro realm and then take the quantum laws which describe the quantum realm. It relates everything in this universe –e. g when you see anything, actually photons penetrate.
Anshuman
why do isotopes of the same group undergo the same chemical reactions ?
Theo
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# Examples on Asymptotic Notation – Upper, Lower and Tight Bound
In this article, we will discuss some examples on asymptotic notation and mathematics behind it.
## Finding upper bound
Upper bound of any function is defined as follow:
Let f(n) and g(n) are two nonnegative functions indicating the running time of two algorithms. We say, g(n) is upper bound of f(n) if there exist some positive constants c and n0 such that 0 ≤ f(n) ≤ c.g(n) for all n ≥ n0. It is denoted as f(n) = Ο(g(n)).
### Examples on Upper Bound Asymptotic Notation
Example: Find upper bound of running time of constant function f(n) = 6993.
To find the upper bound of f(n), we have to find c and n0 such that 0 ≤ f (n) ≤ c.g(n) for all n ≥ n0
0 ≤ f (n) ≤ c × g (n)
0 ≤ 6993 ≤ c × g (n)
0 ≤ 6993 ≤ 6993 x 1
So, c = 6993 and g(n) = 1
Any value of c which is greater than 6993, satisfies the above inequalities, so all such values of c are possible.
0 ≤ 6993 ≤ 8000 x 1 → true
0 ≤ 6993 ≤ 10500 x 1 → true
Function f(n) is constant, so it does not depend on problem size n. So n0= 1
f(n) = O(g(n)) = O(1) for c = 6993, n0 = 1
f(n) = O(g(n)) = O(1) for c = 8000, n0 = 1 and so on.
Example: Find upper bound of running time of a linear function f(n) = 6n + 3.
To find upper bound of f(n), we have to find c and n0 such that 0 ≤ f (n) ≤ c × g (n) for all n ≥ n0
0 ≤ f (n) ≤ c × g (n)
0 ≤ 6n + 3 ≤ c × g (n)
0 ≤ 6n + 3 ≤ 6n + 3n, for all n ≥ 1 (There can be such infinite possibilities)
0 ≤ 6n + 3 ≤ 9n
So, c = 9 and g (n) = n, n0 = 1
Tabular Approach
0 ≤ 6n + 3 ≤ c × g (n)
0 ≤ 6n + 3 ≤ 7 n
Now, manually find out the proper n0, such that f (n) ≤ c.g (n)
From Table, for n ≥ 3, f (n) ≤ c × g (n) holds true. So, c = 7, g(n) = n and n0 = 3, There can be such multiple pair of (c, n0)
f(n) = O(g(n)) = O(n) for c = 9, n0 = 1
f(n) = O(g(n)) = O(n) for c = 7, n0 = 3
and so on.
Example: Find upper bound of running time of quadratic function f(n) = 3n2 + 2n + 4.
To find upper bound of f(n), we have to find c and n0 such that 0 ≤ f (n) ≤ c × g (n) for all n ≥ n0
0 ≤ f (n) ≤ c × g (n)
0 ≤ 3n2 + 2n + 4 ≤ c × g (n)
0 ≤ 3n2 + 2n + 4 ≤ 3n2 + 2n2 + 4n2
for all n ≥ 1:
0 ≤ 3n2 + 2n + 4 ≤ 9n2
0 ≤ 3n2 +2n + 4 ≤ 9n2
So, c = 9, g(n) = n2 and n0 = 1
Tabular approach:
0 ≤ 3n2 + 2n + 4 ≤ c.g (n)
0 ≤ 3n2 + 2n + 4 ≤ 4n2
Now, manually find out the proper n0, such that f(n) ≤ c.g(n)
From Table, for n ≥ 4, f(n) ≤ c × g (n) holds true. So, c = 4, g(n) = n2 and n0 = 4. There can be such multiple pair of (c, n0)
f(n) = O (g(n)) = O (n2) for c = 9, n0 = 1
f(n) = O (g(n)) = O (n2) for c = 4, n0 = 4
and so on.
Example: Find upper bound of running time of a cubic function f(n) = 2n3 + 4n + 5.
To find upper bound of f(n), we have to find c and n0 such that 0 ≤ f(n) ≤ c × g(n) for all n ≥ n0
0 ≤ f(n) ≤ c.g(n)
0 ≤ 2n3 + 4n + 5 ≤ c × g(n)
0 ≤ 2n3 + 4n + 5 ≤ 2n3+ 4n3 + 5n3, for all n ³ 1
0 ≤ 2n3 + 4n + 5 ≤ 11n2
So, c = 11, g(n) = n3 and n0 = 1
Tabular approach
0 ≤ 2n3 + 4n + 5 ≤ c × g(n)
0 ≤ 2n3 + 4n + 5 ≤ 3n3
Now, manually find out the proper n0, such that f(n) ≤ c × g(n)
From Table, for n ≥ 3, f(n) ≤ c × g(n) holds true. So, c = 3, g(n) = n3 and n0 = 3. There can be such multiple pair of (c, n0)
f(n) = O(g(n)) = O(n3) for c = 11, n0 = 1
f(n) = O(g(n)) = O(n3) for c =3, n0 = 3 and so on.
## Lower Bound
Lower bound of any function is defined as follow:
Let f(n) and g(n) are two nonnegative functions indicating the running time of two algorithms. We say the function g(n) is lower bound of function f(n) if there exist some positive constants c and n0 such that 0 ≤ c.g(n) ≤ f(n) for all n ≥ n0. It is denoted as f(n) = Ω (g(n)).
### Examples on Lower Bound Asymptotic Notation
Example: Find lower bound of running time of constant function f(n) = 23.
To find lower bound of f(n), we have to find c and n0 such that { 0 ≤ c × g(n) ≤ f(n) for all n ≥ n0 }
0 ≤ c × g(n) ≤ f(n)
0 ≤ c × g(n) ≤ 23
0 ≤ 23.1 ≤ 23 → true
0 ≤ 12.1 ≤ 23 → true
0 ≤ 5.1 ≤ 23 → true
Above all three inequalities are true and there exists such infinite inequalities
So c = 23, c = 12, c = 5 and g(n) = 1. Any value of c which is less than or equals to 23, satisfies the above inequality, so all such value of c are possible. Function f(n) is constant, so it does not depend on problem size n. Hence n0 = 1
f(n) = Ω (g(n)) = Ω (1) for c = 23, n0 = 1
f(n) = Ω (g(n)) = Ω (1) for c = 12, n0 = 1 and so on.
Example: Find lower bound of running time of a linear function f(n) = 6n + 3.
To find lower bound of f(n), we have to find c and n0 such that 0 ≤ c.g(n) ≤ f(n) for all n ≥ n0
0 ≤ c × g(n) ≤ f(n)
0 ≤ c × g(n) ≤ 6n + 3
0 ≤ 6n ≤ 6n + 3 → true, for all n ≥ n0
0 ≤ 5n ≤ 6n + 3 → true, for all n ≥ n0
Above both inequalities are true and there exists such infinite inequalities. So,
f(n) = Ω (g(n)) = Ω (n) for c = 6, n0 = 1
f(n) = Ω (g(n)) = Ω (n) for c = 5, n0 = 1
and so on.
Example: Find lower bound of running time of quadratic function f(n) = 3n2 + 2n + 4.
To find lower bound of f(n), we have to find c and n0 such that 0 ≤ c.g(n) ≤ f(n) for all n ³ n0
0 ≤ c × g(n) ≤ f(n)
0 ≤ c × g(n) ≤ 3n2 + 2n + 4
0 ≤ 3n2 ≤ 3n2 + 2n + 4, → true, for all n ≥ 1
0 ≤ n2 ≤ 3n2 + 2n + 4, → true, for all n ≥ 1
Above both inequalities are true and there exists such infinite inequalities.
So, f(n) = Ω (g(n)) = Ω (n2) for c = 3, n0 = 1
f(n) = Ω (g(n)) = Ω (n2) for c = 1, n0 = 1
and so on.
Example: Find lower bound of running time of quadratic function f(n) = 2n3 + 4n + 5.
To find lower bound of f(n), we have to find c and n0 such that 0 ≤ c.g(n) ≤ f(n) for all n ≥ n0
0 ≤ c × g (n) ≤ f(n)
0 ≤ c × g (n) ≤ 2n3 + 4n + 5
0 ≤ 2n3 ≤ 2n3 + 4n + 5 → true, for all n ≥ 1
0 ≤ n3 ≤ 2n3 + 4n + 5 → true, for all n ≥ 1
Above both inequalities are true and there exists such infinite inequalities.
So, f(n) = Ω (g(n)) = Ω (n3) for c = 2, n0 = 1
f(n) = Ω (g(n)) = Ω (n3) for c = 1, n0 = 1
and so on.
## Tight Bound
Tight bound of any function is defined as follow:
Let f(n) and g(n) are two nonnegative functions indicating running time of two algorithms. We say the function g(n) is tight bound of function f(n) if there exist some positive constants c1, c2, and n0 such that 0 ≤ c1× g(n) ≤ f(n) ≤ c2× g(n) for all n ≥ n0. It is denoted as f(n) = Θ (g(n)).
### Examples on Tight Bound Asymptotic Notation:
Example: Find tight bound of running time of constant function f(n) = 23.
To find tight bound of f(n), we have to find c1, c2 and n0 such that, 0 ≤ c1× g(n) ≤ f(n) ≤ c2 × g(n) for all n ≥ n0
0 ≤ c1× g(n) ≤ 23 ≤ c2 × g(n)
0 ≤ 22 ×1 ≤ 23 ≤ 24 × 1, → true for all n ≥ 1
0 ≤ 10 ×1 ≤ 23 ≤ 50 × 1, → true for all n ≥ 1
Above both inequalities are true and there exists such infinite inequalities.
So, (c1, c2) = (22, 24) and g(n) = 1, for all n ≥ 1
(c1, c2) = (10, 50) and g(n) = 1, for all n ≥ 1
f(n) = Θ (g (n)) = Θ (1) for c1 = 22, c2 = 24, n0 = 1
f(n) = Θ (g (n)) = Θ (1) for c1 = 10, c2 = 50, n0 = 1
and so on.
Example: Find tight bound of running time of a linear function f(n) = 6n + 3.
To find tight bound of f(n), we have to find c1, c2 and n0 such that, 0 ≤ c1× g(n) ≤ f(n) ≤ c2 × g(n) for all n ≥ n0
0 ≤ c1× g(n) ≤ 6n + 3 ≤ c2 × g(n)
0 ≤ 5n ≤ 6n + 3 ≤ 9n, for all n ≥ 1
Above inequality is true and there exists such infinite inequalities.
So, f(n) = Θ(g(n)) = Θ(n) for c1 = 5, c2 = 9, n0 = 1
Example: Find tight bound of running time of quadratic function f(n) = 3n2 + 2n + 4.
To find tight bound of f(n), we have to find c1, c2 and n0 such that, 0 ≤ c1 × g(n) ≤ f(n) ≤ c2 × g(n) for all n ≥ n0
0 ≤ c1 × g(n) ≤ 3n2 + 2n + 4 ≤ c2 × g(n)
0 ≤ 3n2 ≤ 3n2 + 2n + 4 ≤ 9n2, for all n ≥ 1
Above inequality is true and there exists such infinite inequalities. So,
f(n) = Θ(g(n)) = Θ(n2) for c1 = 3, c2 = 9, n0 = 1
Example: Find tight bound of running time of a cubic function f(n) = 2n3 + 4n + 5.
To find tight bound of f(n), we have to find c1, c2 and n0 such that, 0 ≤ c1 × g(n) ≤ f(n) ≤ c2 × g(n) for all n ≥ n0
0 ≤ c1 × g(n) ≤ 2n3 + 4n + 5 ≤ c2 × g(n)
0 ≤ 2n≤ 2n3 + 4n + 5 ≤ 11n3, for all n ≥ 1
Above inequality is true and there exists such infinite inequalities. So,
f(n) = Θ(g(n)) = Θ(n3) for c1 = 2, c2 = 11, n0 = 1
## General Problems
Example: Show that : (i) 3n + 2 = Θ(n) (ii) 6*2n + n2 = Θ(2n
(i) 3n + 2 = Θ(n)
To prove above statement, we have to find c1, c2 and n0 such that, 0 ≤ c1× g(n) ≤ f(n) ≤ c2 g(n) for all n ≥ n0
0 ≤ c1× g(n) ≤ 3n + 2 ≤ c2 × g(n)
0 ≤ 2n ≤ 3n + 2 ≤ 5n, for all n ≥ 1
So, f(n) = Θ(g(n)) = Θ(n) for c1 = 2, c2 = 5 n0 = 1
(ii) 6*2n + n2 = Θ(2n)
To prove above statement, we have to find c1, c2 and n0 such that, 0 ≤ c1× g(n) ≤ f(n) ≤ c2 g(n) for all n ≥ n0
0 ≤ c1× g(n) ≤ 6*2n + n2 ≤ c2 × g(n)
0 ≤ 6.2n ≤ 6*2n + n2 ≤ 7*2n, for all n ≥ 1
So, f(n) = Θ(g(n)) = Θ(2n) for c1 = 6, c2 = 7 n0 = 1
Example: Let f(n) and g(n) be asymptotically positive functions. Prove or disprove following. f(n) + g(n) = q(min(f(n), g(n))).
Function f(n) and g(n) are non-negative functions such that there exist f(n) ≥ 0 and g(n) ≥ 0, for all n ≥ n0.
For all such n ≥ n0, f(n) + g(n) ≥ f(n) ≥ 0
Similarly, f(n) + g(n) ≥ g(n) ≥ 0
Adding both inequalities, f(n) + g(n) ≥ min(f(n), g(n))
⇒ min(f(n), g(n)) ≤ c.( f(n) + g(n)) for all n ≥ n0 with c = 1.
From the definition of big oh, min(f(n), g(n)) = O(f(n) + g(n)).
However converse is not true. Let us take f(n) = n and g(n) = n2. i.e. min(f(n), g(n)) = n. It is easy to show that for any n0, c there is always exist n such that n < c(n + n2). So it is proved that min(f(n), g(n)) = Ω(f(n) + g(n)) is false.
Hence, the given statement is false.
Example: Prove that (n + a)b = Θ( nb ), b > 0
To prove that said statement, we show find positive constants c1, c2 and n0 such that 0 ≤ c1nb ≤ (n + a)b ≤ c2nb, for all n ≥ n0.
Here, a is constant, so for sufficient large value of n, n ≥ |a|, so
n + a ≤ n + | a |
This also implies, n + a ≤ n + |a| ≤ n + n
For further large value of n, n ≥ 2|a|, so |a| ≤ (n/2)
n + a ≥ n – |a| ≥ (n/2)
When n ≥ 2|a|, we have
0 = (1/2)n ≤ n + a ≤ 2n
Given that, b is positive constant, so raising the above equation to the power b does not change the relation,
0 = (n/2)b ≤ (n + a)b ≤ (2n)b
0 = (1/2)b (n)b ≤ (n + a)b ≤ (2)b (n)b
So the constants c1 = (1/2), c2 = 2 and n0 = 2|a| proves the given statement
Example: Is 2n+1 = Ο(2n) ? Explain.
To prove the given statement, we must find constants c and n0 such that 0 ≤ 2n+1 ≤ c.2n for all n ≥ n0.
2n+1 = 2 * 2n for all n. We can satisfy the given statement with c = 2 and n0 = 1.
Example: Find big theta and big omega notation of f(n) = 14 * 7 + 83
1. Big omega notation :
We have to find c and n0 such that 0 ≤ c × g(n) ≤ f(n) for all n ≥ n0
0 ≤ c × g(n) ≤ f(n)
0 ≤ c × g(n) ≤ 14 * 7 + 23
0 ≤ c × g(n) ≤ 181 for all n ³ 1
0 ≤ 181.1 ≤ 181
For c = 181, g(n) = 1 and n0 = 1
f(n) = Ω(g(n)) = Ω(1) for c = 181, n0 = 1
2. Big theta notation:
We have to find such c1, c2 and n0 such that,
0 ≤ c1 × g(n) ≤ f(n) ≤ c2 × g(n) for all n ≥ n0
0 ≤ c1 .g(n) ≤ f(n) ≤ c2 × g(n)
0 ≤ c1 × g(n) ≤ 181 ≤ c2 × g(n)
0 ≤ 180.1 ≤ 181 ≤ 182.1, for all n ³ 1
f(n) = Θ(g(n)) = Θ(1), for c1 = 180, c2 = 182, n0 = 1
Big oh notation is most commonly used in practice. In this course, we have analyzed many problems using big oh notation including various searching algoriths, sorting algoriths, reursive problems, etc.
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# HOW TO CONVERT PERCENTAGE INTO FRACTION
## About "How to convert percentage into fraction"
"How to convert percentage into fraction?" is the question having had by almost all the children who study math in primary level.
The process of converting percentage into fraction has been clearly explained in the picture given below.
## Steps involved
Step 1 :
Write the given percentage as fraction by taking 100 as denominator.
Step 2 :
If it is needed, the fraction can be simplified further. That's it.
In the process of converting percentage into fraction, we may have the following situations.
% (<100) -----------------> Proper fraction
% (>100) -----------------> Improper fraction / Mixed fraction
% (multiple of 100) ----> Integer
## Some more examples
To have better understanding on "How to convert percentage into fraction
Example 1 :
Convert the percentage given below into a proper fraction
24%
Solution :
24% = 24/100 = 6/25
Hence the proper fraction equal to the given percentage is 6/25
Example 2 :
Convert the percentage given below into a proper fraction
2.4%
Solution :
2.4% = 2.4/100 = 24/1000 = 3/125
Hence the proper fraction equal to the given percentage is 3/125
Example 3 :
Convert the percentage given below into a proper fraction
0.55%
Solution :
0.55% = 0.55/100 = 55/10000 = 11/2000
Hence the proper fraction equal to the given percentage is 11/2000
Example 4 :
Convert the percentage given below into a proper fraction
0.25%
Solution :
0.25% = 0.25/100 = 25/10000
Hence the proper fraction equal to the given percentage is 1/400
Example 5 :
Convert the percentage given below into an improper fraction
125%
Solution :
125% = 125/100 = 5/4
Hence the improper fraction equal to the given percentage is 5/4
Example 6 :
Convert the percentage given below into a mixed fraction
150%
Solution :
150% = 150/100 = 3/2 = 1 1/2
Hence the mixed fraction equal to the given percentage is 1 1/2
Example 7 :
Convert the percentage given below into a proper fraction
24.25%
Solution :
24.25% = 24.25/100 = 2425/1000 = 97/400
Hence the proper fraction equal to the given percentage is 97/400
Example 8 :
Convert the percentage given below into an integer
200%
Solution :
200% = 200/100 = 2
Hence the integer equal to the given percentage is 2
Example 9 :
Convert the percentage given below into a mixed fraction
250%
Solution :
250% = 250/100 = 5/2 = 2 1/2
Hence the mixed fraction equal to the given percentage is2 1/2
Example 10 :
Convert the percentage given below into a mixed fraction
125%
Solution :
125% = 125/100 = 5/4 = 1 1/4
Hence the mixed fraction equal to the given percentage is 5 1/4
After having gone through the stuff and examples, we hope that the students would have understood "How to convert percentage into fraction".
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# Solving $x^2-2x-3<0$
If i have to solve $$x^2-2x-3<0$$ I would do
$$x+1 < 0, \quad x-3<0$$
and end up getting $$x<-1$$ and $$x<3$$. Why is it wrong to use the same inequality sign? Shouldn’t both of the linear equations formed from factorising the quadratic use the same inequality sign as the quadratic?
• Did you try plugging some $x<-1$ into the original inequality? If you do that, I think you will agree that it doesn't work. Apr 12, 2019 at 10:51
– user65203
Apr 12, 2019 at 12:15
Hint: Sketch the graph of $$x²-2x -3$$. Notice that it is zero when $$x=-1$$ and $$x=3$$. Then consider the following areas: $$(-\infty,-1)$$, $$(-1,3)$$ and $$(3,\infty)$$.
Let's first assume
$$x^2-2x-3=0$$
$$(x+1)(x-3)=0$$
This is true when $$x=-1$$ or when $$x=3$$. That is, by equating each of the linear term with $$0$$ one by one.
Now let's come to
$$(x+1)(x-3)<0$$
So you are asking, to solve this why don't we do the same thing again, i.e. one by one making both the linear terms less than $$0$$.
But that's not the correct way to do it, because multiplying two negative numbers yields a positive number.
Suppose there are two variables $$A$$ and $$B$$ and it is given that,
$$AB<0$$
This is true when $$A<0$$ and $$B>0$$ because $$-ve\times +ve=-ve$$ or when $$A>0$$ and $$B<0$$ because $$+ve\times -ve=-ve$$.
Similarly
$$(x+1)(x-3)<0$$
Either when $$x+1<0$$ and $$x-3>0$$ or when $$x+1>0$$ and $$x-3<0$$. The former statement is absurd as there exist no real number which is less than $$-1$$ as well as greater than $$3$$. Hence the answer is $$x\in (-1,3)$$
• @UbaidHassan its not sensible to divide by $x$ until and unless you know whether $x$ is positive, negative or zero. And even if you somehow know what x is, i don't know how you will get -3<0?
– user585765
Apr 13, 2019 at 7:16
Option:
Completing the square:
$$(x-1)^2-4<0;$$
$$(x-1)^2 <4;$$
$$|x-1| <2;$$
$$-1 < x <3.$$
$$ab<0\iff a<0\land b<0$$ is wrong. You can easily find counterexamples.
The truth is
$$ab<0\iff (a<0\land b>0)\lor(a>0\land b<0).$$
A product is negative when the factors have opposite signs.
No, because of the rule of signs.
Anyway, there's a theorem on the sign of quadratic polynomials:
Let $$ax^2+bx+c\enspace (a\ne0$$) a quadrattic polynomial with real coefficients. Then $$ax^2+bx+c$$ has the sign of $$a$$, except between its (real) roots, if any.
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By Kathy Kuhl
Today we’ll finish looking at bean algebra, a simple way to make algebra concrete. We’ll start using negative numbers with our bean algebra. (This 4-part series began here.)
Often we rush our children into abstraction too soon when we teach math. But if our children practice at length with concrete objects, illustrate their discoveries, and write them down in algebra, it will make more sense. “Build it, draw it, write it,” and they are more likely to understand and to remember. (I discuss this in my math workshops; come hear me to learn more.)
When beginning algebra, we’re not going to include any fractions or decimals when we teach these new concepts, just integers. (Remember that integers are the whole numbers and their opposites.)
Quick review of earlier principles:
• Teach solving equations like puzzles. Search for the mystery number. The person inventing the puzzle (equation) begins by choosing a solution and putting that many beans in each container, snapping the lids on before showing the puzzle to others to solve. Let your kids build problems and see if they can stump you.
• White beans represent positive integers. Black beans represent negative integers. (If you use poker chips, use red for negative, since “in the red” means negative.)
• Both sides of the problem have the same value. We draw a line on a large piece of paper, put a ruler down the middle of the paper, or use two box lids, one for each side of the equation.
• Keep the two sides equal. When you change the value on one side, like adding 3 or dividing by 2 or tripling), you must do the same to the other side.
• Anti-matter matters! Remember the analogy (from Harold Jacobs’ Elementary Algebra) that we treat negative numbers like bits of anti-matter. When a bit of matter and antimatter come together, they destroy each other. (Kerboom!) It’s more fun than saying -1 + 1 = 0 or 7 + -7=0. (See this post if you don’t recall it.)
### To begin, we’ll construct a couple simple bean algebra problems…
with solutions that are negative. If you’re going to make up bean algebra problems for your children to solve, and if they’re going to make them up to challenge you and their siblings or classmates, you have to begin by choosing your solution.
For this first problem, I chose x = -10. I’ll leave the lids off the containers while building this so you can see that both sides are equal.
Showing x = -10
Remember to treat both sides the same. I decided to double both sides. So 2x = -20.
2x = -20 (We’ll put lids on before we give the equation to someone to solve.)
Next I’ll add -5 to each sides.
2x + -5 = -25 with lids off to show value of x
Now I snap on the lids, invite my child in to see the puzzle and see if he or she can solve it. First, the child would remove -5 from each side, leaving:
Then the child would divide the -20 beans into two groups and tell me I have -10 beans in each container. That’s very much like the equations in earlier posts with positive integers.
But students sometimes struggle with integer arithmetic when we add or subtract positive and negative numbers. So let’s build an equation with them. This time, I chose x = -5. Watch what happens when I add positive 1 to each side.
One positive bean cancels out one negative bean (annihilates, obliterates, destroys—great time to throw in vocabulary here) leaving:
How doe someone solve this? Suppose you put the lids on and your child walks in to solve the problem:
x + 1 = -4
The child tries to remove a white bean (positive number) from each side. But there is no white bean to remove from the right side!
However, this child remembers that adding one plus negative one is the same as adding nothing, so she adds a black and white bean to the right side:
Before those two bean annihilate each other, she can remove +1 (a white bean) from each side, leaving x = -5.
Finally, I’ll help you solve an equation without showing you the solution first.
We want to get those containers on one side by themselves so we can figure out what is in each of them. So we need to take away those 4 positive beans from each side.
But there are none to take on the right. So, we add 4 positive and 4 negative beans to the right. That won’t change the value of the right side, because they cancel each other out.
adding -4 and +4 to the right side won’t change its value
But before those beans explode, we remove 4 positive beans from each side, leaving
3x = -6
3x = -6
Now we divide the beans on the right into three equal groups, and we see there must be -2 in each container.
Try these and build your own. Let your child who hasn’t mastered simple linear equations build them.
Remember to draw them and write them as shown last time. Be sure to use small circles for positive numbers and small black dots for negative numbers, like this. The last problem would look like this. (I copied the drawing at 4 different stages, so you can see how it would change.)
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Pvillage.org
Informative blog about fresh lifehacks
# What is a betweenness in geometry?
## What is a betweenness in geometry?
Lesson Summary In this lesson, we focused on the idea of betweenness as a mathematical concept. We defined it as the quality of a point on a line being between two other points on the same line.
## How do you use betweenness of points?
The Betweenness Theorem for Points tells us that both definitions are equivalent, in the sense that for any three points A, B, C, the truth or falsity of A ∗ B ∗ C is the same no matter which definition we choose.
What is a real-life example of a point in geometry?
Point: Point refers to an exact location that is represented by a dot. Real-Life Examples: A location of a place in the Map. The tip of a needle.
What are real-life examples of points?
A few real-life examples of points are a pencil tip, the tip of a needle, or the location of a place on a map.
### What does distance mean in geometry?
Definition of a Distance The length along a line or line segment between two points on the line or line segment. If the point was “off of the line” you would dirve right by it.
### How do you calculate Betweenness?
To calculate betweenness centrality, you take every pair of the network and count how many times a node can interrupt the shortest paths (geodesic distance) between the two nodes of the pair. For standardization, I note that the denominator is (n-1)(n-2)/2.
What is the real life example of line?
What is a real world example of a line? Real-world examples of line segments are a pencil, a baseball bat, the cord to your cell phone charger, the edge of a table, etc. Think of a real-life quadrilateral, like a chessboard; it is made of four line segments.
What is the example of geometry?
For example: A triangle is a 3 sided shape, and the measure of its 3 interior angles is 180˚ A square, rectangle or quadrilateral are 4 sided shapes, and the measure of their interior angles is 360˚
## What is a real life example of a ray?
Lesson Summary An example of a ray is a sun ray in space; the sun is the endpoint, and the ray of light continues on indefinitely. In another example, a person hitting a tennis ball could cause it to travel in a ray if there were no resistance from the air; however, this can’t happen on earth due to friction.
## How to think of the betweenness of points?
You want to plant the tree at the right spot the first time to avoid awkward conversation and betweenness of points will help you do that. Think of your window as point A and your neighbor’s window as point C. We know a straight line can be drawn through any two points, so you treat the line of sight between two windows as line AC.
Which is the definition of betweenness of rays?
Def: Betweenness of Rays – A ray is between two others in the same halfrotation iff its coordinate is between their coordinates. (More briefly, OAOBOC iff a b>c.)
How is the betweenness of a graph defined?
The betweenness centrality of a graph is defined as where is the largest value of for any vertex in the given graph and is the maximum possible sum of differences in centrality for any graph of vertices which occur in star with the value times of the central vertex, that is, .
### What does the theorem of betweenness tell us?
The theorem of betweenness tells us that the length of AC is the sum of AB and BC. Now this is only true if B is between A and C, otherwise it would be false. By the same token, if you don’t have the length of AB but know BC and AC, you can then find the total length of AB.
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# Taken from Introduction to Probability
and Statistics
Thirteenth Edition
Chapter 12
Linear Regression and
Correlation
Introduction
• In Chapter 11, we used ANOVA to
investigate the effect of various factor-level
combinations (treatments) on a response x.
• Our objective was to see whether the
treatment means were different.
• In Chapters 12 and 13, we investigate a
response y which is affected by various
independent variables, xi.
• Our objective is to use the information
provided by the xi to predict the value of y.
Example
• Let y be a student’s college achievement,
measured by his/her GPA. This might be a
function of several variables:
– x1 = rank in high school class
– x2 = high school’s overall rating
– x3 = high school GPA
– x4 = SAT scores
• We want to predict y using knowledge of x1,
x2, x3 and x4.
Example
• Let y be the monthly sales revenue for a
company. This might be a function of
several variables:
– x2 = time of year
– x3 = state of economy
– x4 = size of inventory
• We want to predict y using knowledge of x1,
x2, x3 and x4.
Some Questions
• Which of the independent variables are
useful and which are not?
• How could we create a prediction
equation to allow us to predict y using
knowledge of x1, x2, x3 etc?
• How good is this prediction?
response y is a function of a single
independent variable, x.
A Simple Linear Model
• In Chapter 3, we used the equation of
a line to describe the relationship between
y and x for a sample of n pairs, (x, y).
• If we want to describe the relationship
between y and x for the whole
population, there are two models we can
choose
•Deterministic Model: y = a + bx
•Probabilistic Model:
–y = deterministic model + random error
–y = a + bx + e
A Simple Linear Model
• Since the bivariate measurements that
we observe do not generally fall exactly
on a straight line, we choose to use:
• Probabilistic Model:
– y = a + bx + e
– E(y) = a + bx
Points deviate from the
line of means by an amount
e where e has a normal
distribution with mean 0 and
variance s2.
The Random Error
• The line of means, E(y) = a + bx , describes
average value of y for any fixed value of x.
• The population of measurements is
generated as y deviates from
the population line
by e. We estimate a
and b using sample
information.
The Method of
Least Squares
• The equation of the best-fitting line
is calculated using a set of n pairs (xi, yi).
•We choose our
estimates a and b to
estimate a and b so
Bestthe
that fitting line :yˆ a + bx
vertical
distances
Choosea and of the
b to minimize
points from the 2line,
SSE ( y
are minimized. ˆ
y ) ( y a bx ) 2
Least Squares
Estimators
Calculatethe sumsof squares:
( x)2
( y ) 2
Sxx x 2
Syy y
2
n n
( x)( y )
Sxy xy
n
Bestfitting line : yˆ a + bx where
S xy
b and a y bx
S xx
Example
The table shows the math achievement test
scores for a random sample of n = 10 college
freshmen, along with their final calculus grades.
Student 1 2 3 4 5 6 7 8 9 10
Math test, x 39 43 21 64 57 47 28 75 34 52
Calculus grade, y 65 78 52 82 92 89 73 98 56 75
x 460 y 760
Use your
calculator to find x 23634 y 59816
2 2
## the sums and
sums of squares.
xy 36854
x 46 y 76
Example
(460) 2
Sxx 23634 2474
10
(760) 2
Syy 59816 2056
10
(460)(760)
Sxy 36854 1894
10
1894
b .76556 and a 76 .76556(46) 40.78
2474
Bestfitting line : yˆ 40.78 + .77 x
The Analysis of Variance
• The total variation in the experiment is
measured by the total sum of squares:
Total SS S yy ( y y )2
The Total SS is divided into two parts:
SSR (sum of squares for regression):
measures the variation explained by using
x in the model.
SSE (sum of squares for error):
measures the leftover variation not
explained by x.
The Analysis of Variance
We calculate
( S xy ) 2 18942
SSR
S xx 2474
1449.9741
SSE TotalSS - SSR
( S xy ) 2
S yy
S xx
2056 1449.9741
606.0259
The ANOVA Table
Total df = n -1 Mean Squares
Regression df = 1 MSR = SSR/(1)
Error df = n –1 – 1 = n - 2
MSE = SSE/(n-2)
Source df SS MS F
Regression 1 SSR SSR/(1) MSR/MS
E
Error n-2 SSE SSE/(n-2)
Total n -1 Total SS
The Calculus Problem
( S xy ) 2 18942
SSR 1449.9741
S xx 2474
( S xy ) 2
SSE Total SS - SSR S yy
S xx
2056 1449.9741 606.0259
Source df SS MS F
Regression 1 1449.9741 1449.9741 19.14
Error 8 606.0259 75.7532
Total 9 2056.0000
Testing the Usefulness
of the Model
• The first question to ask is whether the
independent variable x is of any use in
predicting y.
• If it is not, then the value of y does not
change, regardless of the value of x. This
implies that the slope of the line, b, is zero.
H 0 : b 0 versus H a : b 0
Testing the
Usefulness of the Model
• The test statistic is function of b, our best
estimate of b. Using MSE as the best
estimate of the random variation s2, we
obtain a t statistic.
b0
Test statistic: t which has a t distribution
MSE
S xx
MSE
with df n 2 or a confidenceinterval: b ta / 2
S xx
The Calculus Problem
• Is there a significant relationship between
a significant
scores at
linear relationship
the 5% level of significance?
between the calculus
H 0 : b 0 versusH a : b scores
0 for the
b0 .7656 population
0 of college
t 4.38
MSE/ S xx 75.7532 /freshmen.
2474
Reject H 0 when |t| > 2.306. Since t = 4.38 falls
into the rejection region, H 0 is rejected .
The F Test
• You can test the overall usefulness of
the model using an F test. If the model
is useful, MSR will be large compared to
the unexplained variation, MSE.
## To testH 0 : model is usefulin predicting y
MSR This test is
Test Statistic: F exactly
MSE equivalent to
Reject H 0 if F Fa with1 and n - 2 df . the t-test, with
t2 = F.
Minitab Output
Least squares
To test H 0 : b 0line
regression
Regression Analysis: y versus x
The regression equation is y = 40.8 + 0.766 x
Predictor Coef SE Coef T P
Constant 40.784 8.507 4.79 0.001
x 0.7656 0.1750 4.38 0.002
## S = 8.70363 R-Sq = 70.5% R-Sq(adj) = 66.8%
Analysis of Variance
Source DF SS MS F P
Regression 1 1450.0 1450.0 19.14 0.002
Residual Error 8 606.0 75.8
Total 9 2056.0
Regression
MSE coefficients, a and b t 2
F
Measuring the Strength
of the Relationship
• If the independent variable x is of useful in
predicting y, you will want to know how well
the model fits.
• The strength of the relationship between x
and y can be measured using:
S xy
Correlation coefficient : r
S xx S yy
2
S xy
SSR
Coefficient of determination : r 2
S xx S yy Total SS
Measuring the Strength
of the Relationship
• Since Total SS = SSR + SSE, r2 measures
the proportion of the total variation in the
responses that can be explained by using
the independent variable x in the model.
the percent reduction the total variation by
using the regression equation rather than
just using the sample mean y-bar to
estimate y.
SSR
For the calculus problem, r2 = .705 or r
2
## 70.5%. The model is working well! Total SS
Interpreting a
Significant Regression
• Even if you do not reject the null hypothesis
that the slope of the line equals 0, it does
not necessarily mean that y and x are
unrelated.
• Type II error—falsely declaring that the
slope is 0 and that x and y are unrelated.
• It may happen that y and x are perfectly
related in a nonlinear way.
Some Cautions
• You may have fit the wrong model.
• Extrapolation—predicting values of y
outside the range of the fitted data.
• Causality—Do not conclude that x causes
y. There may be an unknown variable at
work!
Checking the
Regression Assumptions
• Remember that the results of a regression
analysis are only valid when the necessary
assumptions have been satisfied.
1. The relationship between x and y is linear,
given by y = a + bx + e.
2. The random error terms e are independent
and, for any value of x, have a normal
distribution with mean 0 and variance s 2.
Diagnostic Tools
• We use the same diagnostic tools
used in Chapter 11 to check the
normality assumption and the
assumption of equal variances.
1. Normal probability plot of
residuals
2. Plot of residuals versus fit or
residuals versus variables
Residuals
• The residual error is the “leftover”
variation in each data point after the
variation explained by the regression
model has been removed.
Residual yi yˆi or yi a bxi
## • If all assumptions have been met,
these residuals should be normal, with
mean 0 and variance s2.
Normal Probability Plot
If the normality assumption is valid,
the plot should resemble a straight
line, sloping upward to the right.
Normal Probability Plot of the Residuals
(response is y)
99
95
## fail in the tails of the graph.
90
80
70
Percent
60
50
40
30
20
10
1
-20 -10 0 10 20
Residual
Residuals versus Fits
If the equal variance assumption is
valid, the plot should appear as a
random scatter around the zero
Residuals Versus the Fitted Values
(response is y)
center line. 15
10
## If not, you will see a pattern in the
5
residuals.
Residual
-5
-10
60 70 80 90 100
Fitted Value
Estimation and Prediction
• Once you have
determined that the regression line is
useful
used the diagnostic plots to check for
violation of the regression assumptions.
• YouEstimate
use the value
regression line to
of y for
a given value of x
Predict a particular value of y for a
given value of x.
Estimation and Prediction
Estimating a
particular value of y
when x = x0
Estimating the
average value of
y when x = x0
Estimation and Prediction
• The best estimate of either E(y) or y for
a given value x = x0 is
yˆ a + bx0
• Particular values of y are more difficult to
predict, requiring a wider range of values in the
prediction interval.
Estimation and
Prediction
To estimatethe averagevalueof y when x x0 :
1 ( x0 x ) 2
yˆ ta / 2 MSE +
n S xx
To predict a particularvalueof y when x x0 :
1 ( x0 x ) 2
yˆ ta / 2 MSE 1 + +
n S xx
The Calculus Problem
• Estimate the average calculus grade for
students whose achievement score is 50
with a 95% confidence interval.
Calculate yˆ 40.78424 + .76556(50) 79.06
1 (50 46) 2
yˆ 2.306 75.7532 +
10 2474
79.06 6.55 or 72.51to 85.61.
The Calculus Problem
• Estimate the calculus grade for a particular
student whose achievement score is 50 with
a 95% confidence interval.
## Calculate yˆ 40.78424 + .76556(50) 79.06
1 (50 46) 2
yˆ 2.306 75.75321 + +
10 2474
Notice how
79.06 21.11 or 57.95 to 100.17.much wider
this interval is!
Minitab Output
Confidence and prediction
intervals when x = 50
Predicted Values for New Observations
New Obs Fit SE Fit 95.0% CI 95.0% PI
1 79.06 2.84 (72.51, 85.61) (57.95,100.17)
## Values of Predictors for New Observations
New Obs x
1 50.0 Fitted Line Plot
y = 40.78 + 0.7656 x
## Green prediction 120
Regression
95% CI
95% PI
110
bands are always 100
S
R-Sq
8.70363
70.5%
## wider than red 90
80
y
confidence bands. 70
60
## Both intervals are 50
40
narrowest when x = 30
20 30 40 50 60 70 80
x
x-bar.
Correlation Analysis
• The strength of the relationship between x
and y is measured using the coefficient of
correlation:
S xy
Correlation coefficient : r
S xx S yy
• Recall from Chapter 3 that
(1) -1 r 1 (2) r and b have the same sign
(3) r 0 means no linear relationship
(4) r 1 or –1 means a strong (+) or (-)
relationship
Example
The table shows the heights and weights of
n = 10 randomly selected college football
players.
Player 1 2 3 4 5 6 7 8 9 10
Height, x 73 71 75 72 72 75 67 69 71 69
Weight, y 185 175 200 210 190 195 150 170 180 175
## Use your S xy 328 S xx 60.4 S yy 2610
calculator to find
328
the sums and r .8261
sums of squares. (60.4)(2610)
Football Players
Scatterplot of Weight vs Height
210
200
190
Weight
180
r = .8261
170
160
Strong positive
150
correlation
66 67 68 69 70 71 72 73 74 75
Height
As the player’s
height increases,
so does his
weight.
Some Correlation Patterns
• Use the Exploring Correlation applet to
r = 0; No r = .931; Strong
explore
correlationsome correlation patterns:
positive correlation
r = 1; Linear
relationship r = -.67; Weaker
negative correlation
Inference using r
• The population coefficient of correlation is
called r (“rho”). We can test for a significant
correlation between x and y using a t test:
## To testH0 : r 0 versusHa : r 0 This test is
exactly
n2 equivalent to
Test Statistic: t r the t-test for
1 r 2
the slope b0.
Reject H 0 if t ta / 2 or t ta / 2 with n - 2 df .
r .8261 Example
Is there a significant positive correlation
between weight and height in the
population of all college football players?
H0 : r 0 n2
Test Statistic: t r
Ha : r 0 1 r 2
8
Use the t-table with n-2 = 8 df .8261 4.15
to bound the p-value as p- 1 .8261 2
## value < .005. There is a
significant positive correlation.
Key Concepts
I. A Linear Probabilistic Model
1. When the data exhibit a linear relationship, the
appropriate model is y a + b x + e .
2. The random error e has a normal distribution
with mean 0 and variance s2.
II. Method of Least Squares
1. Estimates a and b, for a and b, are chosen to
minimize SSE, the sum of the squared
yˆ a + bx
.
2. The least squares estimates are b = Sxy/Sxx and
a y bx.
Key Concepts
III. Analysis of Variance
1. Total SS SSR + SSE, where Total SS Syy and
SSR (Sxy)2 / Sxx.
2. The best estimate of s 2 is MSE SSE / (n 2).
## IV. Testing, Estimation, and Prediction
1. A test for the significance of the linear regression—H0
: b 0 can be implemented using one of two test
statistics:
b MSR
t or F
MSE / S xx MSE
Key Concepts
2. The strength of the relationship between x and y can
be measured using
SSR
R
2
Total SS
which gets closer to 1 as the relationship gets
stronger.
3. Use residual plots to check for nonnormality,
inequality of variances, and an incorrectly fit model.
4. Confidence intervals can be constructed to estimate
the intercept a and slope b of the regression line and
to estimate the average value of y, E( y ), for a given
value of x.
5. Prediction intervals can be constructed to predict a
particular observation, y, for a given value of x. For a
given x, prediction intervals are always wider than
confidence intervals.
Key Concepts
V. Correlation Analysis
1. Use the correlation coefficient to measure the
relationship between x and y when both
variables are random:
S xy
r
S xx S yy
2. The sign of r indicates the direction of the
relationship; r near 0 indicates no linear
relationship, and r near 1 or 1 indicates a
strong linear relationship.
3. A test of the significance of the correlation
coefficient is identical to the test of the slope b.
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Nina May 21, 2022
# A student ran a distance of miles each day for 5 days. then the student ran a distance of miles each day for the next 5 days. what was the total distance in miles the student ran during these 10 days?
The time for answering the question is over
Hence, The total distance in miles the student ran during these 10 days is 38.75 miles.
<h3>What is distance?</h3>
Distance is the total length between two points.
To calculate the total distance in miles covered by the student, during these 10 days ran, we use the formula below
Formula:
• D = d₁T+d₂T........... Equation 1
Where:
• D = Total distance
• d₁ = distance covered each day in the first 5 days
• d₂ = distance covered each day in the next 5 days
From the question,
Given:
• d₁ = 3miles = 7/2 miles
• T = 5 Days
• d₂ = 4 miles = 17/4 miles
Substitute these values into equation 1
• D = [(7/2)×5]+[(17/4)×5]
• D = 17.5+21.25
• D = 38.75 miles
Hence, The total distance in miles the student ran during these 10 days is 38.75 miles.
Complete question: A student ran a distance of 3 miles each day for 5 days. then the student ran a distance of 4 miles each day for the next 5 days. what was the total distance in miles the student ran during these 10 days?
391
Cappelletti Serena
May 21, 2022
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# Lesson Plan: Migration (Math - Grade K)
Subject: Math
Lesson Objective: To compare two groups of objects
Common Core Standard: CCSS.MATH.CONTENT.K.CC.C.6- Identify whether the number of objects in one group is greater than, less than, or equal to the number of objects in another group, e.g., by using matching and counting strategies.
Materials:
Starter:
Say:
• How do you compare two groups of objects? (Allow the students to answer.)
Main:
Say:
• We compare two groups of objects by first figuring out how many objects are in each group. We can do this by pointing to each object in the group and counting it.
• Once you know how many objects are in each group, then you can compare them.
• The group with the number of objects in it that is bigger is the group with the greater number of objects.
• The group with the number of objects in it that is smaller is the group with less objects than the other group.
• For example, (draw two groups of objects on the board- one group with 4 and one group with 6) I have two groups of objects. Let’s count to see how many circles are in the first group. There are 4. Now, let’s count to see how many circles are in the second group. There are 6.
• Which number is bigger, 4 or 6? That’s right, 6 is bigger than 4. That means that there are a greater number of objects in the group with 6 circles.
• When comparing, you are usually asked which is greater than or less than. Greater than means the bigger number and less than means the smaller number.
• Today, you are going to be comparing groups of objects to figure out which group is greater than or less than the other group.
• I am going to give you a worksheet that has two groups of animals that migrate. You are going to compare those two groups.
• After you figure out how many animals are in each group, you will need to circle the group that answers the question. Some questions will ask you which group is greater than the other group and some questions will ask you which group is less than the other group.
• Does anyone have any questions?
Feedback:
Say:
• Who would like to share your answers? (Allow the students to share.)
Related lessons:
Migration - Writing (GR K)
Migration - Math (GR K)
Migration - Science (GR K)
Check out our complete Emergency Sub Plans Library!
Written by Kimberly Greacen, Education World® Contributing Writer
Kimberly is an educator with extensive experience in curriculum writing and developing instructional materials to align with Common Core State Standards and Bloom's Taxonomy.
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• Related Books
Composite Functions
• Two (or more) functions can interact with each other through good old arithmetic: addition, subtraction, multiplication, and division. These sorts of interactions are called arithmetic combinations. Here are the four types:
• Sum: (f+g)(x) = f(x) + g(x)
• Difference: (f−g)(x) = f(x) − g(x)
• Product: (fg)(x) = f(x) ·g(x)
• Quotient: (f/g)(x) = f(x)/g(x) [and it must be that g(x) ≠ 0]
• A much more interesting idea is to compose two functions. Instead of giving both functions the same input, we give the input to just one function. Then we take the first function's output, and plug that in to the second function. The second function is acting on the first function. [If this idea is confusing, make sure to watch the video where we see some analogies.]
• For function composition, we can use the notation of f °g. [Read as "f composed with g."] If f °g acts on x, we have (f°g ) (x). This means g acts on x first, then f acts on whatever results. [Notice how the functions act in order of closeness to the original input.]
• Another (much easier) way to see (f °g ) (x) is in the function notation format we're already used to:
⎛⎝ f °g ⎞⎠ (x) = f ⎛⎝ g(x) ⎞⎠ .
(Recommendation: If you see the ° notation [such as f °g], rewrite it in the "normal" format [such as f ( g(x) )]. This normally makes it easier to understand and solve problems.)
• Working with composite functions might seem intimidating at first, but it's really just about plugging in appropriately. Each function has its own "rule", so composing multiple functions just means using the rules in succession.
• This idea of plugging in is shown beautifully by the notation f ( g(x) ). The function g acts on x, then f acts on the resulting g(x). Since we almost certainly know what g(x) looks like from the problem, we just use that as input for f.
Composite Functions
Let f(x) = x2+1 and g(x) = 3x−2. What is (f+g)(x)? What is (f+g)(3)?
• The function (f+g)(x) is a new function created by an arithmetic combination of f(x) and g(x). It is defined as
(f+g)(x) = f(x) + g(x).
• To find (f+g)(x), just substitute into the above:
(f+g)(x) = [x2 +1] + [3x−2].
• Once you know (f+g)(x), just plug into it to find (f+g)(3). [Alternatively, you can find f(3) and g(3), then add them together. But in this case, we already had to figure out (f+g)(x), so we might as well use it.]
(f+g)(x) = x2 + 3x −1; (f+g)(3) = 17
Let f(x) = 2x2−2 and g(x) = 3x2−4. What is (f−g)(x)? What is (f−g)(−4)?
• The function (f−g)(x) is a new function created by an arithmetic combination of f(x) and g(x). It is defined as
(f−g)(x) = f(x) − g(x).
• To find (f−g)(x), just substitute into the above:
(f−g)(x) = [2x2 − 2] − [3x2−4].
• Once you know (f−g)(x), just plug into it to find (f−g)(−4). [Alternatively, you can find f(−4) and g(−4), then subtract one from the other. But in this case, we already had to figure out (f−g)(x), so we might as well use it.]
(f−g)(x) = −x2 + 2; (f−g)(−4) = −14
Let f(x) = 3x and g(x) = −2x2+11. What is (fg)(x)? What is (fg)(2)?
• The function (fg)(x) is a new function created by an arithmetic combination of f(x) and g(x). It is defined as
(fg)(x) = f(x) ·g(x).
• To find (fg)(x), just substitute into the above:
(fg)(x) = [3x] ·[−2x2+11].
• Once you know (fg)(x), just plug into it to find (fg)(2). [Alternatively, you can find f(2) and g(2), then multiply them together. But in this case, we already had to figure out (fg)(x), so we might as well use it.]
(fg)(x) = −6x3 + 33x; (fg)(2) = 18
Let f(x) = x2+1 and g(x) = 4x−8. What is ([f/g])(x)? What is ([f/g])(−5)? What is the domain of ([f/g]) (x)?
• The function ([f/g])(x) is a new function created by an arithmetic combination of f(x) and g(x). It is defined as
⎛⎝ f g ⎞⎠ (x) = f(x) g(x) .
• To find ([f/g])(x), just substitute into the above:
⎛⎝ f g ⎞⎠ (x) = [x2+1] [4x−8] .
• Once you know ([f/g])(x), just plug into it to find ([f/g])(−5). [Alternatively, you can find f(−5) and g(−5), then divide one by the other. But in this case, we already had to figure out (fg)(x), so we might as well use it.]
• To find the domain, look for any values that would "break" ([f/g])(x). In this case, the function will be broken only if the denominator equals 0. Thus, find the x-value such that 4x−8=0. This x-value is forbidden, but all others will work fine.
[f/g])(x) = [(x2+1)/(4x−8)]; ([f/g])(−5) = −[13/14]; Domain: x ≠ 2
Let f(x) = x2 and g(x) = x−3. Find f(g(x)) and g(f(x)).
• Plugging one function into another is called function composition. Using it as simple as substituting the plugged-in function for the variable that would normally be there.
• f(g(x)) = f(x−3) = (x−3)2
• g(f(x)) = g( x2 ) = (x2) − 3
f(g(x)) = x2 − 6x + 9; g(f(x)) = x2 − 3
Let f(x) = 4x+3 and g(x) = 1−x. Find (f°g)(x) and (g°f)(x).
• The ° notation denotes function composition. It says that the function on the left acts on the output of the function on the right.
• We can rewrite the ° notation in the function notation we're already used to:
⎛⎝ f°g ⎞⎠ (x) = f ⎛⎝ g(x) ⎞⎠ and ⎛⎝ g°f ⎞⎠ (x) = g ⎛⎝ f(x) ⎞⎠
From this, it's as simple as substituting the plugged-in function for the variable that would normally be there.
• (f°g)(x) = f(1−x) = 4(1−x)+3
• (g°f)(x) = g( 4x+3 ) = 1−(4x+3)
(f°g)(x) = −4x+7; (g°f)(x) = −4x−2
Let f(x) = |x+3| and g(x) = 2x−10. Find (f°g)(x) and (g°f)(x).
• The ° notation denotes function composition. It says that the function on the left acts on the output of the function on the right.
• We can rewrite the ° notation in the function notation we're already used to:
⎛⎝ f°g ⎞⎠ (x) = f ⎛⎝ g(x) ⎞⎠ and ⎛⎝ g°f ⎞⎠ (x) = g ⎛⎝ f(x) ⎞⎠
From this, it's as simple as substituting the plugged-in function for the variable that would normally be there.
• (f°g)(x) = f(2x−10) = |(2x−10) + 3|
• (g°f)(x) = g( |x+3| ) = 2·|x+3| −10
(f°g)(x) = |2x−7|; (g°f)(x) = 2·|x+3| −10
Let f(x) = [1/x] and g(x) = x−5. Find (f°g)(x) along with its domain.
• The ° notation denotes function composition. It says that the function on the left acts on the output of the function on the right. NOTE: This is even more important in this problem than in previous ones. Because one function acts on the output of the other, that output must exist. We have to be careful about the domains of functions at every step.
• We can rewrite the ° notation in the function notation we're already used to:
⎛⎝ f°g ⎞⎠ (x) = f ⎛⎝ g(x) ⎞⎠
From this, it's as simple as substituting the plugged-in function for the variable that would normally be there. NOTE: For a given x to be in the domain of (f°g)(x), it must be that x is in the domain of g and that g(x) is in the domain of f.
• (f°g)(x) = f(x−5) = [1/(x−5)]
• Because (f°g)(x) = f(x−5) = [1/(x−5)], it "breaks" when the denominator equals 0. Thus the domain of (f°g)(x) is x ≠ 5. [We don't have to worry about g(x) affecting the domain because the domain of g(x) is all numbers.]
(f°g)(x) = [1/(x−5)]; Domain: x ≠ 5
Let f(x) = x2−4 and g(x) = √{x−4}. Find (f°g)(x) along with its domain.
• The ° notation denotes function composition. It says that the function on the left acts on the output of the function on the right. NOTE: This is even more important in this problem than in previous ones. Because one function acts on the output of the other, that output must exist. We have to be careful about the domains of functions at every step.
• We can rewrite the ° notation in the function notation we're already used to:
⎛⎝ f°g ⎞⎠ (x) = f ⎛⎝ g(x) ⎞⎠
From this, it's as simple as substituting the plugged-in function for the variable that would normally be there. NOTE: For a given x to be in the domain of (f°g)(x), it must be that x is in the domain of g and that g(x) is in the domain of f.
• (f°g)(x) = f(√{x−4}) = (√{x−4})2−4
• Because g(x) = √{x−4}, it "breaks" when there is a negative inside the root. Thus the domain of g(x) is x ≥ 4. This domain is passed up to (f°g)(x) as well, since f must act upon g(x). So even though (f°g)(x) has no square root once simplified, it is still restricted by that domain.
(f°g)(x) = x−8; Domain: x ≥ 4
Let f(x) = 2x, g(x) = x2+4, and h(x) = 5x−20. What is (h °f °g °f) (x)?
• The ° notation denotes function composition. It says that the function to the left acts on the output of the function to the right.
• We can rewrite the ° notation in the function notation we're already used to:
⎛⎝ h °f °g °f ⎞⎠ (x) = h ⎛⎝ f ⎛⎝ g ⎛⎝ f(x) ⎞⎠ ⎞⎠ ⎞⎠
From this, it's just a matter of substituting the plugged-in function for the variable that would normally be there, then repeating the process until fully simplified.
• h( f (g (f(x) ) ) ) = h( f (g (2x) ) )
• h( f (g (2x) ) ) = h( f ([2x]2 + 4) ) ) = h( f (4x2 + 4) ) )
• h( f (4x2 + 4) ) ) = h( 2[4x2 +4] ) ) = h( 8x2 +8 )
• h( 8x2 +8 ) = 5 [ 8x2+8] − 20 = 40x2 +20
(h °f °g °f) (x) = 40x2 + 20
*These practice questions are only helpful when you work on them offline on a piece of paper and then use the solution steps function to check your answer.
Composite Functions
Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.
• Intro 0:00
• Introduction 0:04
• Arithmetic Combinations 0:40
• Basic Operations
• Definition of the Four Arithmetic Combinations
• Composite Functions 2:53
• The Function as a Machine 3:32
• Function Compositions as Multiple Machines 3:59
• Notation for Composite Functions 4:46
• Two Formats
• Another Visual Interpretation 7:17
• How to Use Composite Functions 8:21
• Example of on Function acting on Another
• Example 1 11:03
• Example 2 15:27
• Example 3 21:11
• Example 4 27:06
Transcription: Composite Functions
Hi--welcome back to Educator.com.0000
Today, we are going to talk about composite functions.0002
We are often going to have 2 or even more functions that interact with each other.0005
This lesson will explore the fundamental ways that functions can interact with each other.0009
First, we will look at how functions can interact through just arithmetic: addition, subtraction, multiplication, and division.0014
These sorts of interactions are called arithmetic combinations, because they are just using arithmetic.0019
Second, we will move on to a more complex idea, using one function's output as another function's input.0024
We call this idea composition of functions; if we want to talk about a specific example, we call that a composite function, when we put multiple functions together.0031
All right, let's go--let's say we have two functions, f and g, and f(x) = x2, and g(x) = x--nice, basic functions.0039
Now, it is easy to imagine creating a new function that just adds f and g together; we would call it f + g.0049
That is not very imaginative, but it makes sense.0055
It would give us the sum of the two functions: the new function, f + g (x), would be equal to x2 + x.0057
We are just adding the two functions together; so we know each function is x2 and x, so we just add them together.0066
That is a simple, basic idea; we are using a basic arithmetic operation, and we are just putting them together through that.0072
We have arithmetic; let's use it.0079
We could expand this idea to the other three basic operations.0081
We could do this with subtraction: f - g would become x2 - x; fg(x) would be x2 times x (fg being f times g,0084
just like when we say 3x, we mean 3 times x); and f/g would be x2/x--as simple as that.0093
Given two functions f and g, along with an x that is in the domain of both, we defined these four different arithmetic combinations.0101
If x means something for f(x), and x means something for g(x)--it doesn't fail, like if we had √x be one of them,0108
we couldn't plug in -3; but as long it is in the domain of both--it is a number that both of them can accept and work on--0115
all of these work really well; sum is f + g(x)--just break it down into adding the two together.0121
Difference is just subtracting one, just like normal subtraction.0127
Product--when we have fg, we read it as times: f times g; and quotient is f(x)/g(x).0132
And it also has to be that g(x) does not equal 0, because otherwise we could accidentally end up0141
blowing the world up when we divide by 0, since we are not allowed to divide by 0,0146
because it is nonsense and doesn't mean anything.0149
So, since you can't divide by 0, it is not going to be defined when g(x) is 0, since you would have to divide by 0.0151
But other than that, we are pretty good to go; if it means something, if it comes out as a normal output, it is defined;0157
that input is in the domain, and it is defined as an output; then we can just put the two together.0162
We just put what f(x) is together and put what g(x) is together with any arithmetic combination that we want to; great.0167
That is a nice, direct idea; an arithmetic combination makes sense.0174
We put in the same input to the two functions, and then we combine their outputs with some pre-decided arithmetic operation.0177
If it is sum, we do it through addition; if it is product we do it through multiplication; things like that.0182
But we can do something more interesting: we can compose one function with another.0188
Instead of giving both functions the same input, we give the input to one function.0193
Then we take the first function's output, and we plug that into the second function.0198
Input goes into one, and then an output comes out of that; and that immediately goes into the second function.0203
And then finally, we get an output of that; the second function is acting on the first function.0207
Many lessons back, we first introduced the idea of a function; and we talked about how we can view it as a machine.0213
It takes in inputs, and the function produces outputs: x goes into the machine, the function f;0219
and then it gets put out after having been acted on.0225
The function is some process; it does some transformation on x, so we get f(x), f having acted upon x.0228
All right, so that is the idea of it as a machine.0237
We can expand this idea into function composition.0240
Function composition is just linking multiple machines together in series; we just put multiple of them together.0243
The output of the first function goes directly into the second as its input.0248
Our first input goes into, say, g; and so, it is now g(x); and then we plug all of g(x) into f.0252
And so, we have all of g(x) now being acted upon by f; input into the first machine, and output comes out of that machine.0264
And then, we just jam that right into the second machine.0272
And if we wanted, we could string this up...3, 4, 5, 6, 7...we could string up as many of these machines in order as we wanted.0274
We could compose as many functions as we wanted to; but it is easy to start by thinking about it in terms of two functions being composed together.0280
We note the previous slide's composition, when it went into g first, and then went into f, as f composed with g.0287
This is just a little circle between them: f circle g...we read that as f composed with g.0295
If f composed with g acts on x, acts on some input x, we have f composed with g of x, just like we would normally.0301
We have created a new function out of putting the two together.0308
By linking those two machines together, it is effectively one larger machine that is doing a new way of working.0311
This means that in this machine, f composed with g, g will act on x first; and then f will act on whatever results.0318
So, we have f composed with g; and we can break it down into g goes first; then f goes on what results, the thing that comes out of that.0327
Now, notice: the functions act in order of closeness.0336
g goes on first; and then, f goes on second, because it is farther away.0340
We hit it with the things that are closest to the input that we are putting in.0348
So, g goes onto the x, and then f goes onto what results; and if we had even more stacked up,0352
whatever was even further to the left would act after that.0356
The functions act in order of closeness to the original input.0360
There is another, much easier, way to see f composed with g of x in the function notation format that we are already used to--0363
the thing that we have been using for quite a while now.0370
f composed with g of x is just f(g(x)); so f composed with g of x...remember how we broke it up:0372
this one went first; and then f acted second--well, that is what we have right here.0380
g is acting first, and then f is acting second--it makes a lot of sense.0387
I would personally recommend, any time you see this circle notation--this f composed with g stuff--rewrite it0392
in this normal format, the format that we are used to at this point, the f(g(x)).0399
This normally will make it easier to understand and solve problems; there are very few downsides to breaking it into this thing.0404
So, I would really recommend: any time you come up against a problem, and you are not quite sure what to do,0411
and it is this sort of thing: break it into f(g(x)), f acting on g(x)--0415
something acting on something else acting on the input that you are putting in.0420
This method, the second form of notation--this is really great as a way to look at things.0424
I really recommend that, when you see this, you break it into this thing right here; it will really help you understand what is going on.0429
Another way we can visually interpret it--this is a little hard to see what is going on here, but try to follow me0438
on what I am saying here--what we do is start with some x; we start with x, and then we apply g to that x.0444
g goes along and takes it to g(x); then, f comes along, and it hits this g(x), and it turns into f acting on g(x).0451
But we can also think of it as some new function that has been created, f composed with g.0460
We have created a new machine that can just go directly from our original x to the end result of f(g(x)).0466
It does both of these actions, both of these processes, in one thing; it is a machine that is built out of both of the machines inside of it.0476
We can look at it as stair-stepping across, or we can look at it as one new-built, giant leap,0483
where it does both of these actions in one jump.0489
All right, you can take steps across the pond, or you can take one giant leap.0492
But ultimately, they do the same thing; the leap has to be informed by the steps, though.0495
So, how do we actually use these composite functions?0502
We understand the idea behind them now; and it turns out that using them actually isn't that hard.0504
It is just important to understand the idea.0509
Each function has its own rule, like f(x) = x3 means to cube your input.0511
So, composing multiple functions just means using these rules in succession.0517
This idea is shown beautifully in that notation--that notation that I was talking about being the one I really recommend earlier, f(g(x)).0521
The function says that the function g acts on x; then, f acts on the resulting g(x).0528
So, g acts on x; that is what this says right here; and then f acts on the resulting g(x).0535
f acts on what we just had there; great.0543
Since we almost certainly know what g(x) is, and what f(x) is, from the problem, we just use that as an input for f.0546
We use the rules that we were given earlier, and we just apply them to these things.0552
Let's see an example: for example, if we had f(x) = x2 + 3 and g(x) = 2x - 2,0556
then f(g(x)) is equal to...well, what we see here--don't get tricked by the fact that we have x showing up multiple times.0564
Remember, it is just a placeholder: f(x) is just a way of saying f(whatever is in here), whatever f is acting on.0572
The thing that it is acting on will get squared; plus 3.0580
So, if it is acting on g(x), then what is g(x)? Well, g(x) is 2x - 2; so we are plugging in 2x - 2.0583
So then, we plug that in for f(x); f(x) becomes x2 + 3; so if what is inside of the box is 2x - 2,0592
it is going to be (2x - 2)2; so the box has the same process happen--it is just a new thing going on.0601
Instead of x going into it, it is just 2x - 2 going into it.0612
The same processes: it is taking the input, squaring it, and adding 3.0617
So, instead of taking an x, squaring it, and adding 3, we are taking in 2x - 2; we are squaring 2x - 2; and then we are adding 3.0621
So, if we wanted to, at this point we could expand (2x - 2)2 + 3; but this is really the key idea--0630
getting to this point of thinking of it as boxes; we are plugging in, based on boxes.0634
And we will see a bunch of examples using this idea later on.0639
But you want to think of it as we are just swapping out; we are using x as a placeholder.0642
It is not x that we are really attached to; f(x) is saying f of box, and then what happens to box;0647
f of placeholder, and then what happens to placeholder; f of input, and then what happens to input.0653
That is the way you want to think about it; and that makes it really easy to do composite functions.0659
All right, it is time for some examples.0663
So, f + g of 3; if we have f(x) = 2x + 3, and g(x) = x2 - 7, what would f + g of 3 be?0665
We do this: f(x) is 2x + 3; g(x) is x2 - 7; we have 2x + 3 + x2 - 7.0675
So, that becomes something; we could simplify it; but at this point, let's plug in x = 3.0686
We have that x = 3 is going to get plugged in, so we have 2(3) + 3 + 32 - 7.0692
6 + 3 + 9 - 7; 9 + 9 is 18; 18 - 7 is 11; so we have 11 as the answer here.0702
All right, the next one we will do with the color blue: g - f(1); what is g?0713
g is x2 - 7; what is f? it is -...and here is the key thing; it is not minus 2x; it is minus all of f;0718
not just the 2x, but minus (2x + 3); it is a whole quantity that we have to be subtracting.0728
Now, we will plug in; what happens when we plug in x = 1?0733
Well, we have 12 - 7 - (2(1) + 3); so that is 1 - 7 - 2 + 3, which is equal to -6 - 5, equals -11.0738
Great; the next one--I will do this one in green: fg(-2).0761
What is f? f is 2x + 3; and then, we are multiplying that by g; so it is that whole f, times the whole of g, x2 - 7.0766
So, (2x + 3)(x2 - 7); now let's plug in -2; when we plug in -2, we get 2(-2 + 3) (it has to be in that whole quantity),0776
times (-2)2, minus (oops, I accidentally made a plus sign) 7; great.0786
So, that is equal to 2 times -2, which is -4, plus 3; (-2)2 is 4 - 7; -4 + 3 gets us -1; 4 - 7 gets us -3; and we get positive 3.0795
Great; and finally, let's go back to red for our very last one, g/f(8).0813
So, what is g? g is x2 - 7, over all of f; so it is 2x + 3.0819
So, we plug in x = 8; 82 - 7, over 2(8) + 3; 82 is 64, minus 7, over 2 times 8 (is 16), plus 3.0827
64 - 7...we get 57, over 19; and it turns out that 57/19...19 times 2 will get up to 38; 19 times 3...we get up to 57.0844
So, 57/19...we get 3 once again, by chance; great.0854
Also, I want to point out: if we wanted to, we could have done this by figuring out what f was and what g was, separately.0859
So notice: f(3)...for example, we used the f + g(3) just to make a point here.0866
So, f(3) is equal to 2(3) + 3, which would be equal to 6 + 3, or 9.0876
g(3) is equal to 32 - 7, which equals 9 - 7, or 2; so f + g(3) is equal to, by the way we defined it, f(3) + g(3),0884
which is equal to 9, f(3), plus g(3) is 2, which equals 11; that is the exact same thing we got when we started by adding.0903
So, we can either put the functions together, and then plug in the variable;0914
or we can plug in the variable into each function and then add them together.0916
It depends on the way we want to approach it; sometimes it will be more useful to do it one way, sometimes more useful to do it the other way.0920
But it is important to notice that we can do it the other way.0924
The second example: we have the same functions that we just ended up using, f(x) = 2x + 3 and g(x) = x2 - 7.0927
What is f composed with g, first?--so we will start with the green for this one.0935
f composed with g of x; what was my recommendation? It was f(g(x)); that is the exact same thing.0939
It is another way to say this exact same thing, but it my opinion, makes it much easier to understand what it is going on.0948
f(g(x)): f(x) = 2x + 3, but we don't need that information yet; we need to plug in g(x) first.0957
f(x2 - 7); now, what is the x? Here is our x here, and x goes right here.0967
That means that f of box is equal to 2 times box plus 3.0979
The thing in the box now is x2 - 7; so that gives us f(x2 - 7) is 2 times the thing in the box, x2 - 7, plus 3.0983
And if we wanted to, we could expand that out.1000
We expand that out pretty easily; and we would get 2x2 - 14 + 3, which is 2x2 - 11; great.1003
The next one: blue for the next one: g composed with f(x) is much easier to write as g(f(x)).1015
What is f(x)? f(x) is 2x + 3, so that is what is going into g right now: 2x + 3.1024
And now, if we plug into g, g of box equals box squared minus 7.1032
So, g(2x + 3) is equal to (2x + 3)...that is the thing in the box...squared, minus 7.1040
We work that out; we get 4x2 (2x times 2x is 4x2) + 2x + 3, and (3 + 2)x is 6x, plus 6x is 12x,1048
plus 3 times 3 is 9, minus 7, which equals 4x2 + 12x + 2.1059
Great; OK, the next one--let's use red for this one: f composed with f--f composed with itself.1069
We can also write this as f(f(x)); what is f(x)? f(x) is 2x + 3, so f(2x + 3).1080
And now, the thing in our box is 2x + 3, so it is f of box equals 2 box + 3; so f(2x + 3) is 2(2x + 3) + 3.1089
Great; we just work this out: 2(2x + 3) is 4x + 6 + 3 = 4x + 9.1101
Here we are on the very last one; use green again--g composed with g(x); g(g(x))--it is much easier to see what is going on that way.1111
g(x) is x2; so now it is g acting on x2 - 7; remember, g of box is equal to box squared, minus 7.1121
So, if we are plugging in x2 - 7, it is going to be (x2 - 7)2 (remember, box squared minus 7), and then also - 7.1130
Great; so we square x2 - 7; x2 times x2 is x4;1139
x2 times -7 is 7x2, minus 7x2...another -7x2;1144
we have -7x2 + -7x2; that is minus 14x2;1149
and -7 times -7 is positive 49; and finally, minus 7; so it is x4 - 14x2 + 42; there we are.1154
There are a bunch of different function compositions, but it is not that hard, as long as we are plugging one thing into the other,1173
and remembering, in terms of the substitution: it is not about the letter x; it is about if we just had a box here.1177
If we just had a placeholder here--if we just had this thing to hold a space, then we saw what happened to that space when it was held open.1184
If we put in an input, what happens to the input?1190
We just happen to use x, because it is a convenient thing; we are used to using it as a placeholder.1192
But x isn't inherently important; it is just the idea of what happens to an input.1197
So, if we plug in something like 2x + 3, different things will happen than if we had just plugged in x.1202
Also, one other thing I want to point out: notice that, in general, f(g(x)) is not equal to (they are normally very different) g(f(x)).1207
For the most part, flipping the order that we do our function composition in gives us very, very different results.1226
Sometimes, it will end up being the same result; but mostly, if we take f(g) or g(f), it will be totally different.1234
So, mostly, f composed with g is a totally different function than g composed with f.1241
And this is just something to keep in mind--that composition order matters very much.1248
We can see it in what we got here: f(g(x)) got us 2x2 - 11, but g(f(x)) got us 4x2 + 12x + 2--totally different results.1253
So, the order that you put it in--the order that one function goes into another--the order of composition--has massive importance.1264
Great; the next example: let f(x) = x + 1, g(x) = 2x, h(x) = √(x + 1).1271
What is the domain of f, divided by g, composed with h, and then also of h composed with g composed with f?1278
Well, we will start with f(g) composed with h; now, the very first thing we want to do is figure out what f(g) is.1286
I am sorry, not f(g); what is f divided by g?1293
Well, f/g(x) is just equal to f(x)/g(x), except for when it is undefined, when g(x) is equal to 0.1296
So, what does this mean? Well, f(x) is x + 1, so (x + 1)/2x--there we are.1306
f/g(x) is equal to (x + 1)/2x; great, now we can just go back to what we are used to doing.1314
So, we use blue for this one; f/g(h(x))...it is much easier to see it written in that format.1322
f/g...what is h(x)?...f/g(√(x + 1)) = √(x + 1) + 1, over 2(√(x + 1)).1332
So, how do we figure out the domain? This right here is not our answer, but it is the function that comes out from that composition.1349
f divided by g composed with h comes out to be √(x + 1) + 1, divided by 2(√(x + 1)).1359
So, the domain is going to be everywhere where it doesn't break.1365
So, what things in here can break? Well, first, square root--any time we see square root, that breaks when a negative is inside.1373
If we have x + 1 going in in both cases, if x + 1 is less than 0 (that is to say, x + 1 is a negative value), then we have breaking.1391
It breaks--it is not defined, more formally--when x + 1 < 0, which is to say when x < -1.1400
So, that is one important point of information: x < -1 means failure.1409
Another failure point is if this bottom part is equal to 0; we have another break if 2√(x + 1) = 0.1414
And that is going to end up being x + 1 = 0, which means x = -1.1431
So, it fails if either of these conditions happens--if x is equal to -1, or x is less than -1, which is to say x ≤ -1.1436
So, its actual domain, the domain of this function, is x > -1.1450
It is all of the places where the function does not fail, where the function does not break.1459
The domain is everything that can go in; we know everything that breaks it, x < -1 and x = -1.1464
So, the domain is everything that does not break it, x > -1.1470
All right, what if we composed h with g with f?1475
All right, h composed with g composed with f might seem scary at first; but remember, we can break it into a much more pleasant, easy-to-work-with, h(g(f(x))).1479
So, first, what is f(x)? It is h of g of...what was f(x)?1490
f(x) is x + 1; so what is g(x)?...g of box is 2 times box, so g(x + 1)...it is still "h of," but g(x + 1) is going to be 2(x + 1).1495
So, let's simplify that inside just a little bit; it is h of 2...distribute; we get 2x + 2, so we have h(2x + 2), equals...1515
we plug that in here; remember, h of box is the square root of box + 1;1527
so h(2x + 2) is √(2x + 2 + 1), which is equal to √(2x + 3).1534
Great; so once again, this is not our answer; but it is going to help us figure out our answer.1549
The square root of (2x + 3) is what the function ends up being; that is what h composed with g composed with f of x is.1554
It is this thing right here; it is equal to the square root of (2x + 3).1565
So, when does √(2x + 3) break? Well, once again, it breaks--it fails--when there is a negative inside.1568
So, if 2x + 3 is negative, it breaks down; 2x + 3 < 0, so 2x < -3, which is when x is less than -3/2; we have failure.1578
It is going to be the reverse of that: everything that doesn't cause failure is the domain.1599
So, the domain is going to be everything that isn't x < -3/2, which is going to be everything greater than -3/2 or equal to it.1603
So, x is greater than or equal to -3/2, when it is big enough to not cause a negative to show up inside of that square root.1613
Great; the final example: The volume of a spherical balloon is given by volume = 4/3πr3.1625
The balloon starts being inflated at time t = 0, in seconds, and its radius, in centimeters, is given by r = 3√t.1633
OK, what does that mean? Let's try to figure it out really quickly.1641
We have a spherical balloon; well, a sphere is just a ball, so that is basically what we expect when we think of balloons.1644
This is making sense; and it is being inflated--it is being blown up; and at time t = 0 (that is just when we start), its radius is given by r = 3√t.1650
So, it starts at t = 0; and what is its radius at t = 0? r = 3√t, so 3√0, so its radius is 0.1661
So, it starts completely small; it is completely uninflated--it is just a dot at 0.1670
And then, from there, it inflates; it grows out from that point; it grows out from that moment in time.1675
Give the volume of the balloon as a function in time.1682
We blow into the balloon, and the radius expands, and the radius expands, and the radius expands.1685
And as the radius expands, there is now volume inside of the balloon.1689
What is the volume at 30 seconds?1693
All right, the first thing we need to do is give the volume of the balloon as a function of time.1696
Well, first, we might want to see these as functions, because right now, V = 4/3πr3, r = 3√t...they are not actually functions right now.1702
But we could easily turn them into functions: volume is really just a function of radius,1710
because the only thing that can vary in there is the radius.1715
Well, radius is a function based off of time, because the only thing that can vary in it is time, 3√t.1725
The volume of the balloon is a function of time; well, the volume of the balloon doesn't have time inside of it.1731
But we do know that volume has radius inside of it; and radius has time inside of it.1737
So, we can just put these together; we can compose them; and volume of radius of time will be equal to a function,1742
based off of time, that will give the volume of the balloon.1752
Let's see what that is: volume of 3√t...now we are just plugging in the radius at any given time.1755
And that is going to be the volume of 3√t; so if we plugged in our box for r, that gives us box cubed, times the other things.1763
So, it is going to be 4/3π times (3√t)3.1773
We simplify this out a bit; we get 4/3π times 33 times (√t)3.1780
Notice: 33 can cancel down to a squared and take this one out.1794
We have 4π times 32; well, 4π32...what is 32? 32 is 9.1801
4 times 9 is 36, so we have 36π.1810
What about √t3? Well, remember: √t2 (let's put it in a different color,1814
so we don't get it confused) would just be equal to t on its own; so √t3 is just one extra √t left over.1820
So, that gives us times t√t; and there we are.1830
This is the volume of this balloon, volume of radius of time; but it is also a way of seeing volume that is purely in terms of t.1834
t shows up; t shows up; but π is a constant; 36 is a constant; so what we have now is volume based purely off of time.1843
We have the first part of this question done.1852
The next part--volume at 30 seconds: well, we have two options for how to do this.1856
We could plug in, into the function that we just built, volume at 30 equals 36π times 30 times √30.1860
Or we could plug in volume of radius at time t, which would be volume of 3√30, which would be equal to 4/3π(3√30)3.1870
And it ends up being the case that these two things actually end up equaling the exact same thing.1891
Let's just fold them together: 36π times 30 times √30...that ends up being 1080π√30.1895
And if want to get this as an approximate value, something that we could actually know as a number,1907
as opposed to just having symbols that are precise and accurate, and exactly correct and right,1910
but hard to actually grasp as a single number and know what we are talking about,1915
we could get a pretty close thing, and we could round this to 8584 using a calculator.1919
What are the units that it comes in? It is centimeters cubed, because if radius is in centimeters,1926
and volume is centimeters cubed (and it makes sense, because we are talking about volume),1932
and length is centimeters, area is centimeters squared, and volume is centimeters cubed, at least if we are using centimeters.1937
If we are using meters, it is meters, meters squared, meters cubed.1944
If we are using inches or feet, it is square inches, square feet, and cubic inches and cubic feet.1947
All right, great; that completes it for composite functions.1952
I hope you have a much better understanding of what is going on.1954
Remember, when you see that circle, it means "composed with," but it is much easier1956
to break it into f of g of x, or g of f of x, depending on the order it goes in.1959
And remember, it is just going to be based off of the order that they are hitting the x in.1965
Whichever is closer goes first; so this becomes f of...g gets to act first, because it is closer to the x.1970
That is what that means; whichever is closer goes first, so it is whatever the order is with the circles.1979
But now, f(g(x)); f, circle, g(x) becomes f(g(x)); a, circle, b, circle, c(x) becomes a(b(c(x))).1984
Great; all right, I am glad to have taught you the composite functions.1995
I hope you can use it in a bunch of places; it will show up in a variety of things--it is really useful stuff here.1999
And we will see you at Educator.com later--goodbye!2003
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College Algebra (11th Edition)
$x=\dfrac{1+\sqrt{41}}{4}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $\log_2(2x-3)+\log_2(x+1)=1 ,$ use the properties of logarithms to simplify the logarithmic expressions on the left side. Then change to exponential form. Use the properties of equality and concepts on solving quadratic equations. Finally, do checking of the solutions with the original equation. $\bf{\text{Solution Details:}}$ Using the Product Rule of Logarithms, which is given by $\log_b (xy)=\log_bx+\log_by,$ the expression above is equivalent \begin{array}{l}\require{cancel} \log_2[(2x-3)(x+1)]=1 .\end{array} Since $\log_by=x$ is equivalent to $y=b^x$, the equation above, in exponential form, is equivalent to \begin{array}{l}\require{cancel} (2x-3)(x+1)=2^1 \\\\ (2x-3)(x+1)=2 .\end{array} Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 2x(x)+2x(1)-3(x)-3(1)=2 \\\\ 2x^2+2x-3x-3=2 \\\\ 2x^2+(2x-3x)+(-3-2)=0 \\\\ 2x^2-x-5=0 .\end{array} In the equation above, $a= 2 ,$ $b= -1 ,$ and $c= -5 .$ Using the Quadratic Formula which is given by $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a},$ then \begin{array}{l}\require{cancel} x=\dfrac{-(-1)\pm\sqrt{(-1)^2-4(2)(-5)}}{2(2)} \\\\ x=\dfrac{1\pm\sqrt{1+40}}{4} \\\\ x=\dfrac{1\pm\sqrt{41}}{4} .\end{array} Upon checking, only $x=\dfrac{1+\sqrt{41}}{4}$ satisfies the original equation.
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Algebra
# Prime numbers
Written by Prerit Jain
Updated on: 15 Aug 2023
## Introduction
Natural numbers known as prime numbers may only be divided by one (1) and by the number itself. In other terms, prime numbers are positive integers larger than one that only include the number itself and the number’s first digit as factors.
In this article we shall see how Euclid’s theorem and its proof, Wilson’s theorem, Euler’s theorem, and Goldbach’s conjecture and its status are related to prime numbers. We will see what composite numbers are and different topics and history of prime numbers.
## Theorems and Properties of Prime Numbers
● Euclid’s theorem and its proof
This theorem proves that there are infinite primes
Proof:
1.Assume there arefinite primes, withbeing the greatest.
2.Take into account the sum of them, plus one:.
3. cannot be divided by any of the p I by design.
4. Therefore, it is a prime itself or is divisible by a prime greater than proving that there are infinite primes.
● Wilson’s theorem and its proof
According to the Wilson Theorem,is considered a prime number if and only if the sum of all positive integers smaller thanis one less than a multiple of number. This applies to every natural number larger than 1 if .
Proof:
Given that and suppose is prime, and there are integers and in a way that
Reducing we get,
We rewrite the equation as
Suppose . We may take . If then factorization become trivial, suppose then and .
So,
Which means is 1.
● Euler’s theorem
The Euclid Euler Theorem states that an even perfect number may be expressed by the formula . where is a Mersenne prime number and is a prime number. It is the result of combining a Mersenne prime number with a power of two. A link between a Mersenne prime and an even perfect number is established by this theorem.
## Difference Between Prime Number and Composite Number
Numbers with more than two elements are known as composite numbers. Composite numbers are non-prime numbers that may be divided by more than two other numbers.
Examples of composite numbers:
12, factors are: 1,2,3,4,6.
14, factors are: 1,2,7,14
## Prime Numbers and Co-prime Numbers
The HCF (Highest Common Factor) of co-prime numbers, also known as substantially prime numbers, is 1. In other terms, two numbers are co-prime if they only share the number one as a common element.
A number that has just itself and the number one is said to be a prime number. Co-primes, on the other hand, are thought of in pairs, and two numbers are co-prime if they only have one common element.
Determine the factors of the number first, then the co-prime of that number. Next, pick any number and determine the variables that make up that number. The given number’s co-prime numbers are all those that have just the number one as a common element.
## What Does “Relatively Prime” Mean?
When two numbers have just one in common, or if there is no other value with the same value as one, you cannot divide them both and obtain zero as the remainder, two numbers are said to be relatively prime.
A and B are relatively prime numbers if the only component they have in common is 1. The pair (a, b) in this instance is considered to be relatively prime. These figures don’t necessarily have to be prime numbers. Additionally, two composite numbers, such as 9 and 10, can be substantially primes. Coprime or mutually prime numbers are other names for relatively prime numbers.
Finding the HCF of the numbers allows us to determine whether or not two numbers are relatively prime. The two integers are referred to as relatively primes if the HCF is 1. List all the contributing elements, then choose the one with the highest common factor between the two integers to determine their HCF.
## Prime Number Distribution and Density
The asymptotic distribution of prime numbers is described by the prime number theorem. It provides a broad overview of how primes are distributed among positive integers and notes that as positive integers get larger, the prevalence of primes declines. According to the informal theorem, there is a roughly equal chance that any randomly chosen positive integer between zero and a big number and the probability that a selected number is prime is .
The Gauss’s observation is:
This is also called the prime number theorem.
## The Sieve of Eratosthenes
The Sieve of Eratosthenes is a technique for identifying prime and composite numbers in a collection of numbers. Greek mathematician Eratosthenes developed this technique in the third century B.C.
Steps to find prime numbers from 1to 100:
Step 1: List all natural numbers from 1 to 100 in a row and a column, as illustrated in the image below.
Step 2: Put a cross over 1 in step 2 since it is neither a prime nor a composite.
Step 3: Next, cross all the multiples of 2, such as 4, 6, 8, 10, 12, and so on, and encircle the number 2, which is a prime number. due to the composite nature of all multiples of 2.
Step 4: Next, draw a circle over the number 3 and a cross over each of its multiples, including 6, 9, 15, 21, and so on. Since all of its multiples other than 3 are composite.
Step 5: Once more, circle the number 5 (which only has two components) and cross off all of its multiples.
Step 6: At this point, surround 7 and cut all of its multiples.
Step 7: Circle 11 and cross every multiple of 11
Step 8: Keep going until all of the numbers are crossed or surrounded.
## History of Prime Numbers
The study of prime numbers dates back thousands of years. Several findings on prime numbers were proven in Euclid’s “Elements,” which was published around 300 B.C. Euclid asserts that the number of prime integers is unlimited. The Fundamental Theorem of Arithmetic, which states that every integer can be expressed as a unique product of primes, is also shown by Euclid. In “Elements,” Euclid uses Mersenne primes to provide a solution to the conundrum of how to construct a perfect number, which is a positive integer equal to the sum of its positive divisors. A Mersenne prime is a prime that can be figured out using the formula .
No more study on prime numbers was done in the Dark Ages, when knowledge and science were banned. Mathematicians like Fermat, Euler, and Gauss started investigating the patterns seen in prime numbers in the 17th century. Math was transformed by the hypotheses and ideas advanced at the time, some of which remain unproven now.
## Prime Numbers Chart
The primes numbers charts for
1. between 1 to 100
2. between 1 to 200
3. between 1 to 1000,
Are as follows:
## How to Find Prime Numbers
● Prime factorization:
Step 1: Determine the provided number’s factors first.
Step 2: Determine how many components that number has.
Step 3: It is not a prime number if there are more than two elements.
● How to find if a large number is prime?
Step 1: Verify the number’s units’ placement. It is not a prime number if it ends in 0, 2, 4, 6, or 8.
The second step is to calculate the number’s digit sum. A number is not a prime number if the total is divisible by three.
Step 3: Once steps 1 and 2 have been proven to be false, locate the square root of the supplied integer.
Step 4: Subtract the provided integer from its square root by each prime number below it.
Step 5: A number is not a prime number if it can be divided by any prime number less than its square root; otherwise, it is a prime number.
## Is 1 a Prime Number?
In accordance with the definition of a prime number, a number must have precisely two components in order to qualify as a prime number. However, the sole component in number one is the number one itself. As a result, 1 is not regarded as a prime number.
Is 1 a composite number since its not a prime number?
The definition of composite numbers contains the solution to this question as well. A composite number is a natural number that has more than two positive elements, under the definition. However, 1 only has 1 component, which is 1 itself. Thus, one is not a composite number.
## Applications of Prime Numbers
● When it comes to cyber-age security, we frequently employ and rely on prime numbers.
● Utilizing the peculiar mathematical feature of primes for encryption and decryption
● To create error-correcting codes for use in communications, they are utilized. They make sure that automated message correction is delivered and received.
● The foundation for developing public-key cryptography techniques is primes.
● They’re employed in hash tables.
● They can also be employed to produce pseudorandom numbers.
● The design of rotor machines also makes use of primes. A number on one rotor is either coprime or prime to the number on another rotor. By doing so, the entire cycle may be generated before trying any different rotor locations.
● The RSA encryption system computes with primes.
## Famous Prime Numbers
● Mersenne primes
A prime number that may be expressed in the form is known as a Mersenne prime. For instance, the Mersenne prime number 31 can be represented as. The first five Mersenne primes are 3, 7, 31, 127, 8191.
● Sophie Germain primes
If both and are primes, then a prime p is said to be a Sophie Germain prime.
Some examples are: 2, 3, 5, 11, 23, 29, 41, 53, 83, 89, 113, 131, …
● Fermat primes
Fermat prime, a prime number of the form for , whereis a positive integer. For example, Fermat prime is which is prime.
## Some Facts about Prime Numbers
● The lowest prime number is 2.
● The only even prime number is 2, which is unique.
● The only consecutive prime numbers are 2 and 3.
● A whole number, excluding 0 and 1, is either a prime number or a composite number.
● Not every odd number is a prime number. Examples include 9, 15, etc.
● Beyond 5, no prime number ends in a 5.
● As the number increases, prime numbers become more uncommon.
## Conclusion
In our everyday lives as well as routine mathematical sums and difficulties, prime numbers are actually highly important. The world of factors and multiples has been given a magnificent gift by their inception. And it is due of their existence that we now know a great deal of unique information about other numbers, such as the rules for different numbers’ divisibility. Additionally, it has made it incredibly simple and pleasant to calculate complex sums. Most significantly, the prime numbers today serve as a cornerstone of the Number System, one of the branches and the core subject of mathematics, without which occasional pointless attempts at simple computations would have continued.
## Sample Examples
Example 1: Find the prime numbers in the following list: 5,28,4,9,39,44,121,47.
Solution 1:
The prime numbers are as follows:5,47
Since,28 has factors 1,2,14,7,4,28.
4 has 1,2,4.
9 has 1,3,9.
39 has 1,3,13,39.
44 has 1,2,22,4,11,44.
121 has 1,121,11
Example 2: What are the prime numbers between 10 and 20?
Solution 2:
The prime numbers are 11,13,17,19.
Example 3: What is the lowest prime number between 70 and 80?
Solution 3:
The lowest prime is 71.
Example 4: Find out if 101 is a prime number.
Solution 4:
Yes, since the only factors of the number are 1 and itself it is a prime number.
Example 5: Using the Fermat prime formula find the prime number whenis 13.
Solution 5:
The formula is since is 13, plugging in the value we get
## FAQs
1. Can a negative number be prime?
Ans: No, the prime numbers are always positive.
2. Is 2 the only even prime number? If yes, then why?
Ans: Yes, 2 is the only even prime number, because all other even numbers are a multiple of 2.
3. How long is the largest known prime number?
Ans: The largest known prime number to date has 24,862,048 digits.
4. What is the one number that is co-prime to all numbers?
Ans: 1 is the only number that is co-prime to all numbers.
5. What is a factor?
Ans: A factor is a number that completely divides another number with no remainder.
## References
Wells, D. G. (2005). Prime numbers. Hoboken: Wiley.
Ingham, A. E., & Ingham, A. E. (1990). The distribution of prime numbers (No. 30). Cambridge University Press.
Adleman, L. M. (1980, October). On distinguishing prime numbers from composite numbers. In 21st Annual Symposium on Foundations of Computer Science (sfcs 1980) (pp. 387-406). IEEE.
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# Difference between revisions of "2002 AIME II Problems/Problem 11"
## Problem
Two distinct, real, infinite geometric series each have a sum of $1$ and have the same second term. The third term of one of the series is $1/8$, and the second term of both series can be written in the form $\frac{\sqrt{m}-n}p$, where $m$, $n$, and $p$ are positive integers and $m$ is not divisible by the square of any prime. Find $100m+10n+p$.
## Solution 1
Let the second term of each series be $x$. Then, the common ratio is $\frac{1}{8x}$, and the first term is $8x^2$.
So, the sum is $\frac{8x^2}{1-\frac{1}{8x}}=1$. Thus, $64x^3-8x+1 = (4x-1)(16x^2+4x-1) = 0 \Rightarrow x = \frac{1}{4}, \frac{-1 \pm \sqrt{5}}{8}$.
The only solution in the appropriate form is $x = \frac{\sqrt{5}-1}{8}$. Therefore, $100m+10n+p = \boxed{518}$.
## Solution 2
Let the two sequences be $a, ar, ar^2 ... \text{ }an^2$ and $x, xy, xy^2 ... \text{ }xy^n$. We know for a fact that $ar = xy$. We also know that the sum of the first sequence = $\frac{a}{1-r} = 1$, and the sum of the second sequence = $\frac{x}{1-y} = 1$. Therefore we have $$a+r = 1$$$$x+y = 1$$$$ar=xy$$ We can then replace $r = \frac{xy}{a}$ and $y = \frac{ar}{x}$. We plug them into the two equations $a+r = 1$ and $x+y = 1$. We then get $$x^2 + ar = x$$$$a^2 + xy = a$$We subtract these equations, getting $$x^2 - a^2 + ar - xy = x-a$$Remember $$ar=xy$$, so $$(x-a)(x+a-1) = 0$$Then considering cases, we have either $x=a$ or $y=a$. This suggests that the second sequence is in the form $r, ra, ra^2...$, while the first sequence is in the form $a, ar, ar^2...$ Now we have that $ar^2 = \frac18$ and we also have that $a+r = 1$. We can solve for $r$ and the only appropriate value for $r$ is $\frac{1+\sqrt{5}}{4}$. All we want is the second term, which is $ar = \frac{ar^2}{r} = \frac{\frac18}{\frac{1+\sqrt{5}}{4}} = \frac{\sqrt{5} - 1}{8}$ solution by jj_ca888
## Solution 3
Let's ignore the "two distinct, real, infinite geometric series" part and focus on what it means to be a geometric series.
Let the first term of the series with the third term equal to $\frac18$ be $a,$ and the common ratio be $r.$ Then, we get that $\frac{a}{1-r} = 1 \implies a = 1-r,$ and $ar^2 = \frac18 \implies (1-r)(r^2) = \frac18.$
We see that this cubic is equivalent to $r^3 - r^2 + \frac18 = 0.$ Through experimenting, we find that one of the solutions is $r = \frac12.$ Using synthetic division leads to the quadratic $4x^2 - 2x - 1 = 0.$ This has roots $\dfrac{2 \pm \sqrt{4 - 4(4)(-1)}}{8},$ or, when reduced, $\dfrac{1 \pm \sqrt{5}}{4}.$
It becomes clear that the two geometric series have common ratio $\frac{1 + \sqrt{5}}{4}$ and $\frac{1 - \sqrt{5}}{4}.$ Let $\frac{1 + \sqrt{5}}{4}$ be the ratio that we are inspecting. We see that in this case, $a = \dfrac{3 - \sqrt{5}}{4}.$
Since the second term in the series is $ar,$ we compute this and have that $$ar = \left(\dfrac{3 - \sqrt{5}}{4} \right)\left(\dfrac{1+\sqrt{5}}{4}\right) = \dfrac{\sqrt{5} - 1}{8},$$for our answer of $100 \cdot 5 + 1 \cdot 10 + 8 = \boxed{518}.$
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The normal at point P on the ellipse ${{x}^{2}}+4{{y}^{2}}=16$ meets the x – axis at Q. If M is the midpoint of the line segment PQ, then the locus of M intersects the latus rectum of the given ellipse at the points:(a) $\left( \pm \dfrac{3\sqrt{5}}{2},\pm \dfrac{2}{7} \right)$ (b) $\left( \pm \dfrac{3\sqrt{5}}{2},\pm \dfrac{\sqrt{19}}{4} \right)$ (c) $\left( \pm 2\sqrt{3},\pm \dfrac{1}{7} \right)$ (d) $\left( \pm 2\sqrt{3},\pm \dfrac{4\sqrt{3}}{7} \right)$
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Hint: First of all, write the given equation of ellipse ${{x}^{2}}+4{{y}^{2}}=16$ in general form of the ellipse $\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$ by dividing 16 on both the sides we get $\dfrac{{{x}^{2}}}{16}+\dfrac{4{{y}^{2}}}{16}=1\Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1$. Now, the parametric coordinates of the ellipse at the point P $\left( a\cos \theta ,b\sec \theta \right)$ where “a is 4 and b is 2”. Then write the equation of normal passing through point P and to find the coordinates of point Q, put the y – coordinate of the equation of normal as 0 and find the value of x coordinate. Now, take the midpoint of P and Q to get the coordinates of M. Then find the locus of the midpoint which is the equation in x and y. After that, we are going to make the intersection of this locus with the latus rectum of the ellipse.
The equation of ellipse given in the above problem is equal to:
${{x}^{2}}+4{{y}^{2}}=16$
We know that the general form of equation of ellipse is equal to:
$\dfrac{{{x}^{2}}}{{{a}^{2}}}+\dfrac{{{y}^{2}}}{{{b}^{2}}}=1$
Now, we are going to write the given equation of ellipse in the above general form we get,
\begin{align} & \dfrac{{{x}^{2}}}{16}+\dfrac{4{{y}^{2}}}{16}=1 \\ & \Rightarrow \dfrac{{{x}^{2}}}{16}+\dfrac{{{y}^{2}}}{4}=1 \\ \end{align}
Now, we are going to write the parametric coordinates of P on the ellipse. We know that parametric coordinates on ellipse is equal to:
$\left( a\cos \theta ,b\sin \theta \right)$
In the above problem, “a is 4 and b is 2” so substituting these values of “a and b” in the above we get,
$\left( 4\cos \theta ,2\sin \theta \right)$
Now, we know that the parametric equation of normal on ellipse at point $\left( a\cos \theta ,b\sin \theta \right)$ is equal to:
$ax\sec \theta -by\text{cosec}\theta ={{a}^{2}}-{{b}^{2}}$
Now, substituting “a as 4 and b as 2” in the above equation we get,
\begin{align} & 4x\sec \theta -2y\text{cosec}\theta ={{\left( 4 \right)}^{2}}-{{\left( 2 \right)}^{2}} \\ & \Rightarrow 4x\sec \theta -2y\text{cosec}\theta =16-4 \\ & \Rightarrow 4x\sec \theta -2y\text{cosec}\theta =12 \\ \end{align}
Dividing 2 on both the sides we get,
\begin{align} & \dfrac{4x\sec \theta -2y\text{cosec}\theta }{2}=\dfrac{12}{2} \\ & \Rightarrow 2x\sec \theta -y\text{cosec}\theta =6.......Eq.(1) \\ \end{align}
Now, the normal from point P cuts the x axis at point Q which we are showing in the below figure:
In the above figure, we have shown the x and y axis by FG and EH respectively. Now, we are going to find the coordinates of point Q by putting y as 0 in eq. (1) we get,
\begin{align} & 2x\sec \theta -\left( 0 \right)\text{cosec}\theta =6 \\ & \Rightarrow 2x\sec \theta =6 \\ & \Rightarrow \dfrac{2x}{\cos \theta }=6 \\ & \Rightarrow x=3\cos \theta \\ \end{align}
Hence, we got the coordinates of point Q as $(3\cos \theta ,0)$. Now, we are going to find the midpoint of PQ by adding x coordinates of P and Q and then divide them by 2. Similarly, we will find the coordinates of y. The midpoint of PQ is denoted by M.
\begin{align} & M\left( \dfrac{4\cos \theta +3\cos \theta }{2},\dfrac{2\sin \theta }{2} \right) \\ & =M\left( \dfrac{7\cos \theta }{2},\sin \theta \right) \\ \end{align}
Now, writing the coordinates of Q and M in the above figure we get,
Now, we are going to find the locus of the midpoint M by equating x coordinate as “h” and y coordinate as “k”.
$h=\dfrac{7}{2}\cos \theta ,k=\sin \theta$
In the above, rewriting the first equation we get,
\begin{align} & \dfrac{2h}{7}=\cos \theta ......Eq.(2) \\ & k=\sin \theta ........Eq.(3) \\ \end{align}
We know that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ so using this relation in the above equations by squaring both the sides of eq. (2) and eq. (3) and then adding them we get,
$\dfrac{4{{h}^{2}}}{49}+{{k}^{2}}=1$
Now, locus is a relation between x and y so substituting h as x and k as y in the above equation we get,
$\dfrac{4{{x}^{2}}}{49}+{{y}^{2}}=1$…… Eq. (4)
In the above problem, we have to find the intersection of this locus with the latus rectum. We know that the equation of latus rectum is equal to $x=\pm ae$. In this equation, we need to find “e” of the given ellipse:
We know that:
$e=1-\dfrac{{{b}^{2}}}{{{a}^{2}}}$
Substituting “b” as 2 and “a” as 4 in the above equation we get,
\begin{align} & e=\sqrt{1-\dfrac{{{\left( 2 \right)}^{2}}}{{{\left( 4 \right)}^{2}}}} \\ & \Rightarrow e=\sqrt{1-\dfrac{4}{16}} \\ & \Rightarrow e=\sqrt{\dfrac{16-4}{16}} \\ & \Rightarrow e=\dfrac{\sqrt{12}}{4} \\ & \Rightarrow e=\dfrac{2\sqrt{3}}{4} \\ & \Rightarrow e=\dfrac{\sqrt{3}}{2} \\ \end{align}
Now, substituting the value of “a” as 4 and “e” from the above in $x=\pm ae$ we get,
\begin{align} & x=\pm 4\left( \dfrac{\sqrt{3}}{2} \right) \\ & \Rightarrow x=\pm 2\sqrt{3} \\ \end{align}
Substituting the above value of x in eq. (4) we get,
\begin{align} & \dfrac{4{{\left( 2\sqrt{3} \right)}^{2}}}{49}+{{y}^{2}}=1 \\ & \Rightarrow \dfrac{4\left( 12 \right)}{49}+{{y}^{2}}=1 \\ & \Rightarrow {{y}^{2}}=1-\dfrac{48}{49} \\ & \Rightarrow {{y}^{2}}=\dfrac{49-48}{49}=\dfrac{1}{49} \\ \end{align}
Taking square root on both the sides we get,
$y=\pm \dfrac{1}{7}$
Hence, we got the intersection of the locus of the midpoint and latus rectum as $\left( \pm 2\sqrt{3},\pm \dfrac{1}{7} \right)$.
Hence, the correct option is (c).
Note:
In the above problem, while taking the square root of ${{y}^{2}}=\dfrac{1}{49}$ on both the sides you might forget to write $\pm$ before $\dfrac{1}{7}$ but in this problem, all the options have $\pm$ sign so your answer won’t be wrong. But in this problem, you are having an advantage, it is not necessary that in every question you will get this advantage so make sure you won’t make this mistake.
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# Remainder Calculator
You can simply calculate the division of any number by an integer with our quotient and remainder calculator and obtain the result in the form of integers.
When we divide one number by another, the dividend is the number that is divided.
Smith, for example, has 9 Jellies for his 4 kids. He divides 9 jelly cubes by 4 kids, yielding 2 jelly cubes for each youngster and 1 jelly left over. In this case, the dividend is 9, the divisor is 4, the quotient is 2, and the remainder is 1.
9 = (4 x 2)+1
## Remainder Calculator Use
• This calculator is simple to use and run.
• Enter the dividend and divisor values in the appropriate fields and click the calculate button.
• The calculator will handle the rest.
• Both the remainder and the quotient will be shown as integers as a result.
• It should be noted that the dividend and divisor values must be integers.
## How to Calculate the Remainder
We will understand the process of finding the remainder by one of the example. By the end of this example you will be able to find the remainder manually without using any calculator.
Example : 247 by 3
Solution :
If you divide the integer 247÷3 on any simple calculator you will get the result as 82.3333333
So, how do we convert the result into an integer? The solution is easy; simply follow the instructions outlined below.
• Simply round the number whenever the result is in decimal form. In our case (247/3) the result is 82.3333333, hence we get 82 by rounding the value.
• Next, multiply the rounded number by the divisor, which is 82 x 3 = 246.
• Finally, subtract the acquired result from the original dividend i.e 247 – 246 = 1.
• The remainder value is 1
## Long Division with Remainder
Another way to find the remainder is to use the long division method. This method may seem complex at first glance, but we assure you that it is also the simplest. We guarantee that reading the instructions below will clear up any uncertainties you may have and that you will be able to perform the long division method independently.
We will understand this method by the same example that we discussed above (247/3).
Use the long division symbol or the long division bracket to set up the division problem. The dividend is the number you’re dividing, and the divisor is the number by which the dividend is divided. The dividend in this example is 243, while the divisor is 3. Inside the bracket, put the dividend (247) and outside the bracket, put the divisor (3).
Divide the dividend’s first two digits (24) by the divisor (3). So 24/3 equals 8.
So 24 divided by 3 equals 8 with a remainder of 0.
Now take down the remaining value of dividend (7) besides the remainder (0)
Divide the remaining value 7, by the same divisor 3, for a result of 2 with the remainder 1.
The graphic shows that the dividend is 247, the divisor is 3, the quotient is 82, and the remainder is 1.
For larger dividends, repeat the preceding stages until you have brought down every digit from the dividend and solved the problem.
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# Chapter 3 – Differentiation Rules
## Presentation on theme: "Chapter 3 – Differentiation Rules"— Presentation transcript:
Chapter 3 – Differentiation Rules
3.10 Linear Approximations and Differentials 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
If y = f(x), where f is a differentiable function, then the differential dx is an independent variable (can be given the value of any real number). The differential dy is defined in terms of dx by dy is the dependent variable that depends on values of x and dx. 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Let and be points on the graph of f and let dx = x. The corresponding change in y is Propagated error Measured Value Exact Value Measurement error 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
NOTE The approximation y dy becomes better as x becomes smaller. For more complicated functions, it may be impossible to compute y exactly. In such cases, the approximation by differentials is useful. The linear approximation can be written as 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 4 Find the differential of each function. 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 5 Find the differential dy. Evaluate dy for the given values of x and dx. 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 6 Compute y and dy for the given values of x and dx=x. Sketch a diagram showing the line segments with lengths dx, dy, and y. x = 1, x = 1 x = 0, x = 0.5 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 7 Use a linear approximation (or differentials) to estimate the given number. (2.001)5 e-0.015 tan44o 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 8 The radius of a circular disk is given as 24 cm with a maximum error in measurement of 0.2 cm. Use differentials to estimate the maximum error in the calculated area of the disk. What is the relative error? What is the percentage error? 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 9 Use differentials to find a formula for the approximate volume of a thin cylindrical shell with height h, inner radius r, and thickness r. What is the error involved in using the formula from part (a)? 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 10 One side of a right triangle is known to be 20 cm long and the opposite angle is measured as 30o, with a possible error of 1o. Use differentials to estimate the error in computing the length of the hypotenuse. What is the percentage error? 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 11 If current I passes through a resistor with resistance R, Ohm’s Law states that the voltage drop is V=RI. If V is constant and R is measured with a certain error, use differentials to show that the relative error in calculating I is approximately the same (in magnitude) as the relative error in R. 3.10 Linear Approximations and Differentials Dr. Erickson
3.10 Linear Approximations and Differentials
Example 12 When blood flows along a blood vessel, the flux F (volume of blood per unit time that flows past a given point) is proportional to the fourth power of the radius R of the blood vessel: (this is known as Poiseuille’s Law). A partially clogged artery can be expanded by an operation called angioplasty, in which a balloon-tipped catheter is inflated inside the artery in order to widen it and restore the normal blood flow. Show that the relative change in F is about four times the relative change in R. How will a 5% increase in the radius affect the flow of blood? 3.10 Linear Approximations and Differentials Dr. Erickson
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Master the 7 pillars of school success
# Trigonometric Ratios (The Basics)
Common Core Standard: G.SRT. 6 High School Geometry
Other Trigonometric Ratios
There are 3 co-functions:
Secant is the reciprocal of cosine.
Cosecant is the reciprocal of sine.
Cotangent is the reciprocal of tangent.
Cosecant = Hypotenuse/Opposite = 1/Sine
## Transcript: Trigonometric Ratios (The Basics)
Trigonometric ratios (The Basics) The trigonometric ratios are special measurements of a right triangle. In fact, trigon is Greek for triangle, and metric is Greek for measurement. Put these together and trigonometric means triangle measurement.
Triangles have six parts, three angles, and three sides.
Using trigonometric ratios enables you can to find any measure of a triangle if you know just two of the parts.
• The hypotenuse is always opposite the right angle.
• The adjacent side is the leg closest to the reference angle. ( Angle A )
• The opposite side is the leg opposite the reference angle ( Angle A )
### Trigonometric Ratios Chart
Hi welcome to MooMooMath. Today we are going to talk about some basic trig functions. There are three basic trig functions that we use in Geometry. They are sine, cosine, and tangent, and you will see those keys on the calculator. For these three functions you will use a right triangle. So let’s draw a right triangle, and we are going to lay out the basics of the three ratios. We will look from the perspective of the angle in the bottom left corner labeled A, which an acute angle. From angle A, we have to set up some ratios,based on the sides that we will label. The side opposite angle A is called the opposite side. The side opposite the right angle is the hypotenuse, and the third side is the adjacent side. So the three sides are the opposite, hypotenuse, and the adjacent leg. Using these three sides we will set up the three trigonometric ratios. The first ratio is the sin of the angle A, and the sin of angle A will equal the opposite side over the adjacent or O over A. The cosine of angle A is equal to the adjacent over the hypotenuse,so it will be abbreviated A over H. The third ratio is tangent of A, which is equal to opposite over adjacent or O over A. We have a cheat sheet. We have sine which is opposite over hypotenuse, cosine is adjacent over hypotenuse, and tangent is opposite over adjacent. You can remember the trig ratios using, SOH, CAH, TOA. S = Sine = Opposite over Hypotenuse,C= Cosine = Adjacent over hypotenuse, and T = Tangent = Opposite over Adjacent. Hope this video was helpful.
Math students use SOH CAH TOA in order to help them remember the trig functions.
If you have the following information: A right angle Length of opposite leg Length of the hypotenuse Use Sine Opposite Hypotenuse Remember using SOH A right angle Length of adjacent leg Length of the hypotenuse Use Cosine Adjacent Hypotenuse CAH A right angle Length of opposite leg Length of the adjacent Use Tangent Opposite Adjacent TOA
A
C
B
Related sites for Trig Ratios
Basic Trigonometry: Sin, Cos, Tan (mathbff) This video shows how to find sin, cos, and tan using SohCahToa
Trig Ratios Information,facts, and fun activities covering trig ratios
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# What is 51/499 as a decimal?
## Solution and how to convert 51 / 499 into a decimal
51 / 499 = 0.102
Converting 51/499 to 0.102 starts with defining whether or not the number should be represented by a fraction, decimal, or even a percentage. Fractions and decimals represent parts of a whole, sometimes representing numbers less than 1. Choosing which to use starts with the real life scenario. Fractions are clearer representation of objects (half of a cake, 1/3 of our time) while decimals represent comparison numbers a better (.333 batting average, pricing: \$1.50 USD). After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
## 51/499 is 51 divided by 499
The first step of teaching our students how to convert to and from decimals and fractions is understanding what the fraction is telling is. 51 is being divided into 499. Think of this as our directions and now we just need to be able to assemble the project! The numerator is the top number in a fraction. The denominator is the bottom number. This is our equation! Now we divide 51 (the numerator) into 499 (the denominator) to discover how many whole parts we have. Here's 51/499 as our equation:
### Numerator: 51
• Numerators are the top number of the fraction which represent the parts of the equation. Overall, 51 is a big number which means you'll have a significant number of parts to your equation. 51 is an odd number so it might be harder to convert without a calculator. Values closer to one-hundred make converting to fractions more complex. Let's take a look below the vinculum at 499.
### Denominator: 499
• Denominators are located at the bottom of the fraction, representing the total number of parts. 499 is a large number which means you should probably use a calculator. But 499 is an odd number. Having an odd denominator like 499 could sometimes be more difficult. Overall, two-digit denominators are no problem with long division. So without a calculator, let's convert 51/499 from a fraction to a decimal.
## Converting 51/499 to 0.102
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 499 \enclose{longdiv}{ 51 }$$
To solve, we will use left-to-right long division. This is the same method we all learned in school when dividing any number against itself and we will use the same process for number conversion as well.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 499 \enclose{longdiv}{ 51.0 }$$
We've hit our first challenge. 51 cannot be divided into 499! So we will have to extend our division problem. Add a decimal point to 51, your numerator, and add an additional zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 499 into 51 + 0 or 510.
### Step 3: Solve for how many whole groups you can divide 499 into 510
$$\require{enclose} 00.1 \\ 499 \enclose{longdiv}{ 51.0 }$$
How many whole groups of 499 can you pull from 510? 499 Multiply this number by 499, the denominator to get the first part of your answer!
### Step 4: Subtract the remainder
$$\require{enclose} 00.1 \\ 499 \enclose{longdiv}{ 51.0 } \\ \underline{ 499 \phantom{00} } \\ 11 \phantom{0}$$
If there is no remainder, you’re done! If you still have numbers left over, continue to the next step.
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
Sometimes you won't reach a remainder of zero. Rounding to the nearest digit is perfectly acceptable.
### Why should you convert between fractions, decimals, and percentages?
Converting between fractions and decimals is a necessity. Remember, fractions and decimals are both representations of whole numbers to determine more specific parts of a number. Same goes for percentages. So we sometimes overlook fractions and decimals because they seem tedious or something we only use in math class. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 51/499 into a decimal
Speed - Let's say you're playing baseball and a Major League scout picks up a radar gun to see how fast you throw. Your MPH will not be 90 and 51/499 MPH. The radar will read: 90.10 MPH. This simplifies the value.
### When to convert 0.102 to 51/499 as a fraction
Carpentry - To build a table, you must have the right measurements. When you stretch the tape measure across the piece of wood, you won't see 10.6 inches. You'll see a tick mark at 10 and 3/5 inches.
### Practice Decimal Conversion with your Classroom
• If 51/499 = 0.102 what would it be as a percentage?
• What is 1 + 51/499 in decimal form?
• What is 1 - 51/499 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.102 + 1/2?
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# 6.06 Purchases and change
Lesson
## Ideas
We need to remember how to calculate the total amount of money that we have. Let's try this problem.
### Examples
#### Example 1
Which of the following shows a total of \$22.50? A B Worked Solution Create a strategy Add the value of each note and coin using the vertical algorithm. Use this table of values to help you. Apply the idea For Option A, write the addition of values in a vertical algorithm and add the digits in each column. \begin{array}{c} & &2&0&.&0&0 \\ & &&2&.&0&0 \\ &+&&0&.&5&0 \\ \hline & &2&2 &. &5&0 \\ \hline \end{array} The total amount of notes and coins in Option A is \$22.50.
For Option B, write the addition of values in a vertical algorithm and add the digits in each column.
\begin{array}{c} & &2&0&.&0&0 \\ & &&5&.&0&0 \\ &&&1&.&0&0 \\ &+&&1&.&0&0 \\ \hline & &2&7 &. &0&0 \\ \hline \end{array}
The total amount of notes and coins in Option B is \$27.00. So the correct answer is option A. Idea summary The value of a coin is written on the coin. Silver coins are for cents, and gold coins are for dollars. All notes have dollar values. The value is written on the note. To add money we can write the values as decimals and add them using a vertical algorithm. ## Money and change Let's learn more about calculating change. Loading video... ### Examples #### Example 2 How much change would you receive from \$10 if you spent \\$7.50?
Worked Solution
Create a strategy
Use a number line and count back from 10 by 7.50.
Apply the idea
Plot 10 on the number line:
Count back by 7 to get to 3:
Count back by 0.5 to get to 2.5:
Idea summary
We can either count up or count back to find the difference between the amount of money we start with and the amount of money that we spent.
### Outcomes
#### MA2-5NA
uses mental and written strategies for addition and subtraction involving two-, three-, four and five-digit numbers
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Trying to discover out exactly how to transform 25/6 into a combined number or fraction? have I gained the answer because that you! In this guide, we"ll go you with the step-by-step process of convert an not correct fraction, in this situation 25/6, to a combined number. Check out on!
Want to quickly learn or show students exactly how to convert 25/6 to a blended number? play this an extremely quick and fun video clip now!
Before we begin, let"s revisit part basic portion terms so friend understand exactly what we"re handling here:
Numerator. This is the number above the fraction line. Because that 25/6, the molecule is 25.Denominator. This is the number below the fraction line. For 25/6, the denominator is 6.Improper fraction. This is a fraction where the numerator is higher than the denominator.Mixed number. This is a method of to express an improper portion by simple it to totality units and a smaller as whole fraction. It"s an essence (whole number) and also a proper fraction.
You are watching: 25/6 as a mixed number
Now let"s go through the measures needed to convert 25/6 come a combined number.
## Step 1: find the entirety number
We first want to find the entirety number, and also to execute this we divide the molecule by the denominator. Due to the fact that we are only interested in whole numbers, we ignore any type of numbers come the appropriate of the decimal point.
25/6= 4.1666666666667 = 4
Now the we have our totality number because that the combined fraction, we require to uncover our new numerator for the fraction part that the blended number.
## Step 2: get the brand-new numerator
To occupational this the end we"ll use the whole number us calculated in step one (4) and also multiply that by the original denominator (6). The an outcome of that multiplication is climate subtracted indigenous the initial numerator:
25 - (6 x 4) = 1
## Step 3: Our mixed fraction
We"ve currently simplified 25/6 come a mixed number. To see it, we just need to put the totality number along with our brand-new numerator and also original denominator:
4 1/6
## Step 4: simple our fraction
In this case, our portion (1/6) have the right to be streamlined down further. In order to carry out that, we have to calculate the GCF (greatest typical factor) that those two numbers. You can use our handy GCF calculator to work this out yourself if you desire to. We already did that, and also the GCF the 1 and 6 is 1.
We deserve to now divide both the brand-new numerator and the denominator through 1 to leveling this portion down to its lowest terms.
1/1 = 1
6/1 = 6
When we put that together, we have the right to see the our complete answer is:
4 1/6
Hopefully this tutorial has helped you come understand exactly how to convert any kind of improper portion you have actually into a combined fraction, finish with a entirety number and a suitable fraction. You"re cost-free to use our calculator below to work-related out more, however do try and learn how to carry out it yourself. It"s much more fun 보다 it seems, i promise!
If you uncovered this content advantageous in her research, please perform us a great favor and also use the tool listed below to make certain you correctly reference united state wherever you use it. Us really appreciate your support!
"What is 25/6 together a mixed number?". rewildtv.com. Accessed ~ above November 8, 2021. Https://rewildtv.com/calculator/improper-to-mixed/what-is-25-6-as-a-mixed-number/.
"What is 25/6 as a combined number?". rewildtv.com, https://rewildtv.com/calculator/improper-to-mixed/what-is-25-6-as-a-mixed-number/. Accessed 8 November, 2021.
See more: How Many Cups Are In 32 Fluid Ounces, 32 Ounces To Cups
What is 25/6 together a combined number?. rewildtv.com. Retrieved indigenous https://rewildtv.com/calculator/improper-to-mixed/what-is-25-6-as-a-mixed-number/.
## Improper portion to mixed Number
Enter one improper portion numerator and denominator
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# Statistics: Probability – Bayes Theorem
## Determining Probability Based on Prior Events
With my other probability articles I demonstrated how to determine the likelihood of independent and dependent events. In those examples, I used no prior knowledge of events, but in many scenarios when testing for probability on dependent events, one event has already occurred for which the probability has been determined, and you may use that knowledge to test the second, or subsequent events.
In this example I’ll demonstrate finding the probability of an event related to playing the card game 21. This game has one clear goal – to end with a total value of either 21 or value closer to it than the dealer, without going over 21. In this example game, I have one card and the dealer one card. My card is an Ace, meaning I currently have 11 (need 10 to make 21), and the dealer’s card which is face-up is a Jack (10, they need an Ace). Also, it was previously determined the probability me the player=11 after the dealer=11 is 1/32.
For this example game, I want to find the probability the dealer will “hit” 21.
For this to occur, the dealer will need to draw as their next card an Ace.
### Bayes’ Theorem
This formula determines the probability of future event(s) based on the occurrence of a previous event.
Using our example 21 game, here is the formula I’ll use.
Regarding the formula’s terminology, here is a brief overview:
• The probability of event2, given that event1 is…
• The probability of event1 given event2 multiplied probability of event2…
• Divided by…
• The probability of event1.
Now I’ll plug into the formula the example 21 game.
• The probability of dealer=11 (ace), given that player=11…
• The probability of player=11 given dealer=11 multiplied probability of dealer=11…
• Divided by…
• The probability of player=11.
P(D11|P11)=
Beginning with the numerator (top of the fraction), I will start completing the formula: P(P11|D11) * P(D11).
First, I’ll fill-in the first portion on the left using the probability previously determined from the beginning of this article = P(P11|D11), or the probability of player=11 given that (already occurred) dealer=11.
Next, I’ll determine the next (right) portion of the formula in the numerator: P(D11).
Therefore, I’ve determined the probability of the dealer=11 is 4/52 because 4 aces exist (matching criteria) out of 52 possible cards in the desk (All Possibilities).
Here is the original formula with the numerator updated:
Now, I will complete the formula for the denominator.
Since the probability in the numerator for P(D11) has already been determined (4/52) and results in the same probability as P(P11) – the player has the same probability of receiving 11 as the dealer, I’ll just copy that probability into the denominator.
Now I’m ready to complete the formula:
Therefore, using Bayes’ Theorem I’ve determined the likelihood of the dealer drawing 11 after I received an 11 is .7%.
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Michelle E. Chung
* EMAT6680 Assignment 3: Quadratic Equations
Let's think about quadratic equations. Consider graphs in the xc plane and xb plane.
1. When we vary 'b' :
Graph of Graph of Graph of The parabola opens down. The vertex of the parabola is (-0.5, 0.25). The axis of symmetry is x=-0.5. The parabola opens down. The vertex of the parabola is (0.5, 0.25). The axis of symmetry is x= 0.5. The parabola opens down. The vertex of the parabola is (-2.5, 6.25). The axis of symmetry is x=-2.5. The parabola opens up. The vertex of the parabola is (-50, -25). The axis of symmetry is x=-50. So, the graph of has following characteristics: The line of symmetry is The vertex of the parabola moves along Since this graph is passing the origin (0, 0), is always a tangent of the graph at (0, 0).
Back to the Top
2. When we vary 'a' :
Graph of Graph of Graph of The parabola opens down. The vertice of the parabolsaares(-0.5, 0.25) and (-0.1.5, 0.0625). The axixes of symmetry arex=-0.5 and x=-0.125. The parabolas opes up. The vertices of the parabolas are (0.5, -0.25) and (0.125, -0.0625). The axixes of symmetry are x=0.5 and x=0.125. When a=0, the graph of the equation would be the straight line, y=-x. So, the graph of has following characteristics: The line of symmetry of the graph is . When a=0, the parabola would be the line, y=-x. Since this graph is passing the origin (0, 0), y=-x would be a tangent of the graph at (0, 0).
Back to the Top
Consider the graphs of the cubic equation . Explore the pattern of roots in the xb, xc, or xd planes. (Okay, maybe look at the xa plane too!)
1. in the xb plane:
Graph of Graph of Graph of The graph of this equation is NOT a conic. It is just a ALGEBRAIC CURVE. As we see, y=-x is the slant asymptote of the equation. It is under aymptote for the top graph (we can say this by checking the values of a ordered pair on the graph - an absolute value of x-coordinate is always less than an absolute value of y-coordinate) and above asymptote for the bottom graph (similarly, we can say this because an absolute value of x-coordinate is always less than an absolute value of y-coordinate). The SKYBLUE is the graph of the derivative of the given equation. As we see, it is a hyperbola. When the constant term ('d') is -1, the graph of the equation is the reflection the RED GRAPH over the origin, (0,0). It also has y=-x as its slant asymptote. Again, the PURPLE is the graph of the derivative of the given equation. As we see, it is a hyperbola. As we see, the relative maxima and minima seem to be on the y=-2x. (maybe not...???) When the value of the constant term increases, the value of the relative maxima decreases. When the value of the constant term decreases, the value of the relative minima increases. When n=-0.1 When n=0.1 When n=0 When n=0, i.e. the value of constant term is 0, the graph is a hyperbola. The graph of this kind of equation is NOT a conic. It is just a ALGEBRAIC CURVE. It has three different shapes - when the constant term is positive, the graph is a ALGEBRAIC CURVE, not a conic, and most part of the graph is under y=-x - when the constant term is 0, the graph is a CONIC, actually a HYPERBOLA, and y=-x and y-axis are the aymptotes of the hyperbola - when the constant term is negative, the graph is a ALGEBRAIC CURVE, not a conic, and most part of the graph is above y=-x y=-x is the slant asymptote of the equation. When the value of the constant term increases, the value of the relative maxima decreases, and when the value of the constant term decreases, the value of the relative minima increases.
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# Mobius inversion formula
The Möbius Inversion Formula is a relation between pairs of arithmetic functions, each defined from the other by sums over divisors. Originally proposed by August Ferdinand Möbius in 1832, it has many uses in Number Theory and Combinatorics.
### The Formula
Let $g$ and $f$ be arithmetic functions and $\mu$ denote the Möbius Function. Then it follows that
$g(n)=\sum_{d|n}f(d)\leftrightarrow f(n)=\sum_{d|n}\mu(d)g\left(\frac{n}{d}\right)=\sum_{d|n}\mu\left(\frac{n}{d}\right)g(d).$
$\textit{Proof}$: Notice the double implication, so we have two directions to prove. We proceed with the proof of the backwards direction first. We have
$\sum_{d|n}\mu(d)g\left(\frac{n}{d}\right)=\sum_{d|n}\mu(d)\sum_{c|\frac{n}{d}}f(c)=\sum_{cd|n}\mu(d)f(c)=\sum_{c|n}f(c)\sum_{d|\frac{n}{c}}\mu(d).$
To finish, we will use the fact that
$\sum_{d|n}\mu(d)=1~\text{for}~n=1~,\sum_{d|n}\mu(d)=0~\text{for}~n>1.$
If we have $\frac{n}{c}=1\leftrightarrow n=c$ then we have
$\sum_{d|\frac{n}{c}}\mu(d)=\sum_{d|1}\mu(d)=1$
and that $\sum_{d|\frac{n}{c}}\mu(d)=0$ if otherwise. Hence by considering $n=c$ we get
$\sum_{c|n}f(c)\sum_{d|\frac{n}{c}}\mu(d)=\sum_{c|n}f(c)=f(n).$
The first direction is satisfied, and now we must prove the second. We see that
$\sum_{d|n}f(d)=\sum_{d|n}f\left(\frac{n}{d}\right)=\sum_{d|n}\sum_{c|\frac{n}{d}}\mu\left(\frac{n}{cd}\right)g(c)=\sum_{cd|n}\mu\left(\frac{n}{cd}\right)g(c)=\sum_{c|n}g(c)\sum_{d|\frac{n}{c}}\mu\left(\frac{n}{cd}\right)=g(n).$
Both directions have been proven, which completes our work $\square$
### Applications
One of the most common applications of the formula is by proving that
$n=\sum_{d|n}\varphi(d)$.
While there are some common combinatorial and group theoretic arguments one could use, a Möbius Inversion Formula solution also suffices. Clearly by choosing $g(n)=n$ and $f(n)=\varphi(n)$ the theorem is proven.
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# Testing Center
## Accuplacer Arithmetic Study Guide
I. Terms
Numerator: which tells how many parts you have (the number on top)
Denominator: which tells how many parts in the whole (the number on the bottom)
3
4
Example:
3
4
## 3 parts have a dot out of 4
Proper fraction: the top number is less than the bottom number.
1 7 9
Example:
, ,
3 10 19
Improper fraction: the top number is equal to or is larger than the bottom number.
3 9 8
Example:
, ,
2 4 8
Mixed Number: a whole number is written next to a proper fraction.
3 2
1
Example: 1 , 2 , 1 0
4 5
2
Common Denominator: is a number that can be divided evenly by all of the denominators
in the problem
Example:
3
9
12
4
12
## The common denominator for these fractions will be 12.
12 is also the least common denominator.
2
8
12
3
12
1
6
12
2
12
## Whole Number: is a number with no fraction, decimal or negative parts
Example: 1, 2, 3, 4, 945, 8224
Divisor (factor): is the number that you are dividing by
Dividend: is the number being divided
15
Example: 5 is the divisor, 15 is the dividend
5
15 5
(3 ways to write division)
5 15
Variable: a letter used for an unknown number, e.g. x or y
Equation: a mathematical way showing that two things are the same
Term: a number, variable or combination in an equation
Example:
Variable
Sum
5x 2 17
Equation:
Terms
II. Fractions
## A. Reducing Fractions to Lowest Terms
Step 1: Find a number that goes evenly into the numerator and the denominator of the
fraction.
Example:
## The number that will go in evenly is 8
48 8 6
64 8 8
Step 2: Check to see whether another number goes evenly into both the numerator and
denominator. Stop when there are no more numbers that can go into the fraction.
Example:
6 2 3
8 2 4
## B. Changing Mixed Numbers to Improper Fractions
Step 1: Multiply the denominator by the whole number.
3
Example: Change 2
to an improper fraction
4
2 4 8
Step 2: Add the result to the numerator
8 3 11
Step 3: Place the total over the denominator.
11
4
C. Adding and Subtracting Fractions with Different Denominators (Bottom Numbers)
Example 1:
3 2
4 3
3 3 9
4 3 12
2 4
8
3 4 12
9 8 17
5
1 *
12 12 12 12
## Step 1: Need to find the common
denominator for all fractions
subtract the fractions.
Example 2:
3 4 12
4 4 16
3 1 3
16 1 16
12 3
9
16 16 16
## *Remember to change improper fractions to a mixed number.
D. Multiplying Fractions
Step 1: Multiply the numerators across.
Step 2: Then multiply the denominators across.
Make sure the product is in lowest terms
3 3
4 16
3 5 15
4 6 24
15 3 5
24 3 8
2 2
2 1
3 5
2
2 8
3 3
2 7
1
5 5
8 7 56
3 5 15
## Step 2: Then multiply across
Step 3: Then change the improper fraction to a mixed number in lowest terms.
56
11
3
15
15
F. Dividing Fractions
The fraction that is right of the division sign will need to be turned upside down by
writing the numerator in the denominator and the denominator in the numerator. Then
follow the rules for multiplying fractions.
1 1
Example:
4 2
1 2 2
2 2 1
Simplify
4 1 4
4 2 2
Practice:
1) Change 4
3)
3
5
2
3
1
to an improper fraction. 2) Change
6
1
11
5
9
2
4)
5) 13
6)
2
1
3
2
3
2
1 5
7) 3
7 9
9)
42
to a mixed number.
16
7
10
8
3
2
7
3
7
8) 3 2
7
9
6
14
11
4
5
10) 3 5
5
6
III. Decimals
28.50
44.47
3075.60
3148.57
## Step 2: Then add or subtract
380.53
75.00
305.53
B. Multiplying Decimals
Example: Multiply 1.89 5.03
1.89
5.03
567
94500
9.5067
## Step 1: Multiply the decimals as you would do with whole numbers.
Step 2: Then count the number of spaces of each factor being multiplied.
Decimal places are the number of spaces to the right of the decimal
point. There are 2 in the top factor and 2 in the bottom factor, so the
decimal is placed 4 spaces from the right.
## C. Dividing a Decimal by a Whole Number
Place the decimal point directly above its position in the problem. Then divide the same
way as you divide whole numbers.
.037
Example: 73 2.701
219
511
511
0
D. Dividing Decimal by a Decimal Number
Example: 4.374 0.03
Dividend Divisor =
3 437.4
Step 1: Move the decimal point of the divisor as far right as you can go.
2 spaces in this example.
Step 2: Then move the decimal point in the dividend the same number of
places as the divisor
145.8
3 437.4
Step 3: Place the decimal point above its position in the problem. Then divide the
same way as divide whole numbers.
0.03 4.374
3
13
12
17
15
24
24
0
Practice:
11)
18.1
0.04
12)
0.97
5.6
16) 96 0.3992
18) 0.2601 9
17) 4 27.36
## 20) 2.03 4.466
IV. Percents
Percents are used to describe a part of something. Percents are used to figure out sales or the
amount of interest someone will pay on a loan. When converting a percent to its fraction form, it
will always have a denominator of 100.
A. Changing Decimal to Percents or Percents to Decimals
The important key is where to move the decimal point.
If changing from a decimal to percent, move the decimal point 2 places to the right
Example: 0.35 35%
0.8 80%
To change from percent to decimal, need to move the decimal point 2 places to the
left and drop the percent sign
Example: 30% 0.3
0.9% 0.009
B. Converting Fractions to Percent Form
Divide the bottom number of the fraction into the top number and move the point 2 places
to the right.
.75
3
Example:
0 . 7 5 75%
4 3.00
4
28
20
20
0
-or-
3
Example:
4
25
3 100%
4
1
1
75%
75%
1
## C. Converting Percents to Fraction Form
Write the percent as a fraction with 100 as the denominator. Then reduce the fraction to
the lowest terms.
Example: 85%
85 5 17
100 5 20
D. Finding the Percent of a Number
Example: What is 25% of \$6,500?
problem
multiply.
multiply.
1
6,500
4
6500
n
4
n \$1,625
n 25% \$6,500
n .25 6500
n \$1,625
## E. Finding What Percent One Number is of Another
There are key words to remember that will help you solve the problem it is asking you.
The word of in the sentence means to multiply
The word is means equal to.
The word what is the number you are trying to find which is represented by a letter.
Example: 9 is what percent of 45
9 a
9 45a
45 45
9
a
45
9 9
a
45 9
1
a
5
0.20 a
20% a
45
## Step 1: Divide both sides of the equation by same number to get
the variable alone
## Step 3: Change the fraction into a decimal
Step 4: Change the decimal into a percent
Therefore, 20% of 45 is 9.
## F. Finding a Number When a Percent of It is Given
Example: 20% of what number is 16?
20% x
16
20%a 16
20
a 16
100
1
16
a
5
1
1
16
5 a 5
5
1
a 1 6 5
a 80
Therefore, 20% of 80 is 16.
## Step 1: Change the percent to a fraction form
Step 2: Simplify the fraction
Step 3: Multiply both sides of the equation to remove the
fraction and get the variable alone
Practice:
Write the following in percent form.
6
21) 0.12
22)
23)
8
2
5
24)
0.233
25)
1.15
1)
25
6
2)
5
8
3)
4
15
4)
6)
25
56
7)
47
63
8)
11
21
9)
3
77
11)
0.724
12)
5.432
13)
134.64
14)
16)
95.6008
17)
6.84
18)
0.0289
19)
21) 12%
22) 75%
23) 40%
26) \$330
27) 0.5%
28) 70
1
6
9
26
10)
114
175
216.9527
15)
0.068
8.5
20)
2.2
24) 23.3%
5)
25) 115%
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# 1.2. division with remainder
In the previous section, we wrote a naive loop to determine if an integer is divisible by another without proving that such a loop would terminate. Instead of proving that the loop we wrote will eventually terminate, in this section we consider division with remainder and prove a stronger result: that division of an integer by a nonzero integer results in a unique quotient and remainder.
division with remainder. For an integer $a$ and a nonzero integer $b$, there exist unique integers $q$ and $r$such that $a = b \cdot q + r$ and $0 \le r < |b|$.
This is often called the division algorithm . In the output of the algorithm, the integer $q$ is called the quotient and the integer $r$ is called the remainder .
Case 1: when $a$ is nonnegative. Consider the set $$S = \{ a - b n \mid n \in \mathbb{Z} \}$$ consisting of all integers of the form $a - b n$ where $n$ is any integer. Now consider the subset $$S_+ = \{ s \in S \mid s \ge 0 \}$$ of elements in $S$ which are nonnegative.
Note that $a$ is an element of $S$ since $a = a - b(0)$. And since $a$ itself is nonnegative, it is in $S_+$, which means that $S_+$ is nonempty. By the well-ordering principle, there is a smallest element of $S_+$. By definition, this smallest element is of the form $r = a - b q$ for some integer $q$.
We know that $r$ is nonnegative, because it is an element in $S_+$. To complete the proof in this case, we need to show that $r < |b|$ or equivalently that $r - |b|$ is negative. This integer $r - |b|$ is an element of $S$ since it is equal to $a - b (q \pm 1)$, depending on the sign of $b$. However, it cannot be an element of $S_+$ since it is smaller than the smallest element $r$. This is only possible if $r - |b|$ is negative.
Case 2: $a$ is negative. In this case, we know that $-a$ is positive. Using the above, we can find unique integers $Q$ and $R$ such that $$-a = bQ + R \quad \text{and} \quad 0 \le R < |b|.$$
If $R = 0$, we set $r = 0$ and $q = -Q$, which gives $$a = b(-Q) - 0 = bq + r.$$
Otherwise, we set $r = -R + |b|$ and $q = -Q - |b| / b$, which gives $$a = b(-Q) - R = b(q + |b| / b) + (r - |b|) = bq + r.$$ Also, since $R > 0$, we know that $$0 < R < |b| \quad \longrightarrow \quad -|b| < -R < 0,$$ which implies that $$0 < -R + |b| < |b| \quad \longrightarrow \quad 0 < r < |b|.$$
## examples
1. What are the quotient and remainder when dividing $47$ by $10$?
The quotient is $4$ and the remainder is $7$ since $$47 = 10 (4) + 7 \quad \text{and} \quad 0 \le 7 < 10.$$
2. What are the quotient and remainder when dividing $47$ by $-10$?
The quotient is $-4$ and the remainder is $7$ since $$47 = -10 (-4) + 7 \quad \text{and} \quad 0 \le 7 < 10.$$
3. What are the quotient and remainder when dividing $-47$ by $10$?
The quotient is $-5$ and the remainder is $3$ since $$-47 = 10 (-5) + 3 \quad \text{and} \quad 0 \le 3 < 10.$$
4. What are the quotient and remainder when dividing $-47$ by $-10$?
The quotient is $5$ and the remainder is $3$ since $$-47 = -10 (5) + 3 \quad \text{and} \quad 0 \le 3 < 10.$$
5. What are the quotient and remainder when dividing by ?
## exercises
1. Write a naive div_rem function that takes two inputs (any integer, any nonzero integer) and returns a pair of integers consisting of the quotient and remainder resulting from dividing the first integer by the second. This function should only use addition, subtraction, multiplication, and comparison.
Some test values:
• div_rem(47, 10) = [4, 7]
• div_rem(47, -10) = [-4, 7]
• div_rem(-47, 10) = [-5, 3]
• div_rem(-47, -10) = [5, 3]
• div_rem(90, 10) = [9, 0]
• div_rem(90, -10) = [-9, 0]
• div_rem(-90, 10) = [-9, 0]
• div_rem(-90, -10) = [9, 0]
• div_rem(10, 47) = [0, 10]
• div_rem(10, -47) = [0, 10]
• div_rem(-10, 47) = [-1, 37]
• div_rem(-10, -47) = [1, 37]
• div_rem(0, 99) = [0, 0]
• div_rem(0, -99) = [0, 0]
solution
module Naive
def div_rem(a, b)
raise ZeroDivisionError if b == 0
if b < 0
quotient, remainder = div_rem(a, -b)
return [-quotient, remainder]
end
if a < 0
quotient, remainder = div_rem(-a, b)
return [-quotient, 0] if remainder == 0
return [-quotient - 1, -remainder + b.abs]
end
quotient = 0
until (remainder = a - quotient * b) < b.abs
quotient += 1
end
[quotient, remainder]
end
end
2. If $a$ is an integer and $b$ is a nonzero integer, prove that $b \mid a$ if and only if division of $a$ by $b$ yields a remainder of $0$.
solution
Suppose that $b \mid a$. Then $a = b n$ for some integer $n$, or equivalently that $a = b n + 0$. This satisfies the conditions for division with remainder, so we conclude that the remainder is equal to $0$, since division with remainder yields a unique remainder.
On the other hand, suppose that division with remainder gives a remainder of $0$, ie, that $a = b q + 0$. Then $a$ is a multiple of $b$, hence $b \mid a$.
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# FRACTION
, or Broken Number, in Arithmetic and Algebra, is a part, or some parts, of another number or quantity considered as a whole, but divided into a certain number of parts; as 3-4ths, which denotes 3 parts out of 4, of any quantity.
Fractions are usually divided into Vulgar, Decimal, Duodecimal, and Sexagesimal. For the last three sorts, see the respective words.
Vulgar Fractions, called also simple Fractions, are usually denoted by two numbers, the one set under the other, with a small line between them: thus 3/4 denotes the Fraction three-fourths, or 3 parts out of 4, of some| whole quantity considered as divided into 4 equal parts.
The lower number 4, is called the Denominator of the Fraction, shewing into how many parts the whole or integer is divided; and the upper number 3, is called the Numerator, and shews how many of those equal parts are contained in the Fraction. Hence it follows, that as the numerator is to the denominator, so is the Fraction itself, to the whole of which it is a Fraction; or as the denominator is to the numerator, so is the whole or integer, to the Fraction: thus, the integer being denoted by 1, as the Fraction.— And hence there may be innumerable Fractions all of the same value, as there may be innumerable quantities all in the same ratio, viz, of 4 to 3; such as 8 to 6, or 12 to 9, &c. So that if the two terms of any Fraction i. e. the numerator and denominator, be either both multiplied or both divided by any number, the resulting Fraction will still be of the same value: thus, 3/4 or 6/8 or 9/12 or 12/16 &c, are all of the same value with each other.
Fractional expressions are usually distinguished into Proper and Improper, Simple and Compound, and Mixt Numbers.
A Proper Fraction, is that whose numerator is less than the denominator; and consequently the Fraction is less than the whole or integer; as 3/4.
Improper Fraction, is when the numerator is either equal to, or greater than, the denominator; and consequently the Fraction either equal to, or greater than, the whole integer, as 4/4, which is equal to the whole; or 5/4, which is greater than the whole.
Simple Fractions, or Single Fractions, are such as consist of only one numerator, and one denominator; as 3/4, or 5/4, or 12/25.
Compound Fractions are Fractions of Fractions, and consist of several Fractions, connected together by the word of: as 2/3 of 3/4, or 1/2 of 2/3 of 3/4.
A Mixt Number consists of an integer and a Fraction joined together: as 1 3/4, or 12 2/3.
The arithmetic of Fractions consists in the Reduction, Addition, Subtraction, Multiplication, and Division of them.
Reduction of Fractions is of several sorts; as 1. To reduce a given whole number into a Fraction of any given denominator. Multiply the given integer by the proposed denominator, and the product will be the numerator. Thus, it is found that 3 = 6/2, and 5 = 20/4, or 7 = 35/5.
If no denominator be given, or it be only proposed to express the integer Fraction-wise, or like a Fraction; set 1 beneath it, for its denominator. So 3 = 3/1, and 5 = 5/1, and 7 = 7/1.
2. To reduce a given Fraction to another Fraction equal to it, that shall have a given denominator. Multiply the numerator by the proposed denominator, and divide the product by the former denominator, then the quotient set over the proposed denominator will form the Fraction required. Thus, if it be proposed to reduce 3/4 to an equal Fraction whose denominator shall be 8; then , and the numerator, so that 6/8 is the Fraction sought, being = 3/4, and having 8 for its denominator.
3. To Abbreviate, or reduce Fractions to lower terms. Divide their terms, i. e. numerator and denominator, by any number that will dividē them both without a remainder, so shall the quotients be the corresponding terms of a new Fraction, equal to the former, but in smaller numbers. In like manner abbreviate these new terms again, and so on till there be no number greater than 1 that will divide them without a remainder, and then the Fraction is said to be in its least terms. Thus, to abbreviate 15/60; first divide both terms by 5, and the Fraction becomes 3/12; next divide these by 3, and it becomes 1/4: so that 15/60 = <*>/12 = 1/4, which is in its least terms.
4. To reduce Fractions to other equivalent ones of the same denominator. Multiply each numerator, separately taken, by all the denominators except its own, and the products will be the new numerators; then multiply all the denominators continually together, for the common denominator, to these numerators. Thus, 2/3 and 4/5 reduce to 10/15 and 12/15; and 2/3, 3/4, and 4/5 reduce to 40/60, 45/60, and 48/60.
9 20 16) 180 (11 s 16 20 16 4 12 16) 48 (3 d 48
5. To find the value of a Fraction, in the known parts of its integer. Multiply always the numerator by the number of parts of the next inferior denomination, and divide the products by the denominator. So, to find the value of 9/16 of a pound sterling; multiply 9 by 20 for shillings, and dividing by 16, gives 11 for the shillings; then multiply the remainder 4 by 12 pence, and dividing by 16 gives 3 for pence: so that 11s. 3d. is the value of 9/16l. as required.
6. To reduce a mixt number to an equivalent improper Fraction. Multiply the integer by the denominator, and to the product add the numerator, for the new numerator, to be set over the same denominator as before. Thus 3 5/8 becomes 29/8.
7. To reduce an improper Fraction to its equivalent whole or mixt number. Divide the numerator by the denominator; so shall the quotient be the integral part, and the remainder set over the denominator will form the fractional part of the equivalent mixt number. Thus 29/8 reduces to 3 5/8, and 32/4 = 8.
8. To reduce a compound Fraction to a simple one. Multiply all the numerators together for the numerator, and all the denominators together for the denominator, of the simple Fraction sought. Thus, 1/2 of 3/4 = 3/8, and 2/3 of 4/5 of 7/9 = 56/135.
To reduce a Vulgar Fraction to a decimal. See Decimals. And for several other particulars concerning Reduction, as well as the other operations in Fractions; see my Arithmetic.
Addition of Fractions. First reduce the Fractions to their simplest form, and reduce them also to a common denominator, if their denominators are different; then add all the numerators together, and set the sum over the common denominator, for the sum of all the Fractions as required. Thus, ; And .
Subtraction of Fractions. Reduce the Fractions the same as for addition; then subtract the one numerator| from the other, and set the difference over the common denominator. So ; And .
To Multiply Fractions together. Reduce them all to the form of simple Fractions, if they are not so; then multiply all the numerators together for the numerator, and all the denominators together for the denominator of the product sought. Thus ; And .
To Divide Fractions. Divide the numerator by the numerator, and the denominator by the denominator, if they will exactly divide. Thus, .
But if they will not divide without a remainder, then multiply the dividend by the reciprocal of the divisor, that is, by the Fraction obtained by inverting or changing its terms. Thus, .
Algebraic Fractions, or Fractions in Species, are exactly similar to vulgar Fractions, in numbers, and all the operations are performed exactly in the same way; therefore the rules need not be repeated, and it may be sufficient here to set down a few examples to the foregoing rules. Thus,
1. The Fraction aab/bc abbreviates to aa/c.
2. , by dividing by 3a.
3. , by dividing by a - x. See Common Measure.
4. a/b and c/d become ad/bd and bc/bd, when reduced to a common denominator.
5. .
6. .
7. .
8. .
Continued Fraction, is used for a Fraction whose denominator is an integer with a Fraction, which latter Fraction has for its denominator an integer and a Fraction, and the same for this last Fraction again, and so on, to any extent, whether supposed to be infinitely continued, or broken off after any number of terms. Euler, Analys. Inf. vol. 1, p. 295. As , or , or . Or, using letters instead of numbers, . or .
When these series are not far extended, it is not difficult to collect them by common arithmetic.
Lord Brounker, it seems, was the first who considered Continued Fractions, or at least, who applied them to the quadrature of curves, in Wallis's Arith. Infin. prop. 191, vol. 1, p. 469 &c, where this author explains the manner of forming them, giving several numeral examples, in approximating ratios, as well as the geneneral series &c, as he denotes it. Huygens also used it for the like purpose, viz, to approximate the ratios of large numbers, in his Descrip. Autom. Planet. in Oper. Relig. p. 173 &c, edit. Amst. 1728. And a special treatise on Continued Fractions was given by Euler, in his Analys. Infin. vol. 1, pa. 295 &c.
This subject is perhaps capable of much improvement, though it has been rather neglected, as very little use has been made of it, except, by those authors, in approximating to the value of Fractions, and ratios, that are expressed in large numbers; besides a method of Goniometry by De Lagny, explained in the Introduction to my Logarithms, pa. 78; as also some use I have made of it in summing very slowly converging series, in my Tracts, p. 38 & seq.
As to the reducing of common Fractions, and ratios, that are expressed in large numbers, to Continued Fractions, it is no more than the common method of finding the greatest common measure of those two numbers, by dividing the greater by the less, and the last divisor always by the last remainder; for then the several quotients are the denominators of the Fractions, the numerators being always 1 or unity. Thus, to find approximating values of the Fraction 31415926535/10000000000, or to the ratio of 31415926535 to 10000000000, being the ratio of the circumference of a circle to its diameter, by means of a Continued Fraction; or, to change the said Common Fraction to a Continued Fraction: Dividing the greater term always by the less, the same as to find the greatest common measure of the said numbers or terms, the several quotients will be 3, 7, 15, 1, 292, 1, 1, &c, which, after the first, will be the denominators, to the common numerator 1; and therefore the said Fraction will be changed into this Continued Fraction, . Hence, stopping at any part of these single Fractions, one after another, will give several values of the proposed ratio, all successively nearer and nearer the truth, but alternately too great and too little. So, stopping| at 1/7, it is 3 1/7 = 22/7 = 3.142857 too great, or 22 to 7, the ratio of the circumference to the diameter as given by Archimedes. Again, stopping at 1/15, it is 3 1/(7 1/15) = 3 15/106 = 333/106 = 3.141509 &c, too little. But stopping at 1/1, it is (the ratio of Metius) = 3.1415929 &c, which is rather too great. And so on, always nearer and nearer, but alternately too great and too little.
And, in like manner is any algebraic Fraction thrown into a Continued Fraction. As the Fraction , which being in like manner divided, the quotients are a/a, b/b, g/c, d/d; which single Fractions being considered as denominators to other Fractions whose common numerator is 1, these will be the reciprocals of the former, and so will become a/a, b/b, c/g, d/d; and hence the proposed common Fraction is equal to this terminate Continued Fraction, .
On the other hand, any Continued Fraction being given, its equivalent common Fraction will be found, by beginning at the last denominator, or lowest end of the given Continued Fraction, and gradually collecting the Fractions backwards, till we arrive at the first, when the whole will thus be collected together into one common Fraction; as was done above in collecting the Fractions And in like manner the Continued Fraction collects into the Fraction .
When the given Continued Fraction is an infinite one, collect it successively, first one term, then two together, three together, &c, till the sum is sufficiently exact. Or, if these collected sums converge too slowly to the true value, having collected a few of the terms into successive sums, these being alternately too great and too little, the true value will be found as near as you please by the method of arithmetical means, explained in my Tracts, vol. 1, Tract 2, pa. 11.
Vanishing Fractions. Such Fractions as have both their numerator and denominator vanish, or equal to 0, at the same time, may be called Vanishing Fractions. We are not to conclude that such Fractions are equal to nothing, or have no value; for that they have a certain determinate value, has been shewn by the best ma- thematicians. The idea of such Fractions as these, first originated in a very severe contest among some French mathematicians, in which Varignon and Rolle were the two chief opposite combatants, concerning the then new or differential calculus, of which the latter gentleman was a strenuous opponent. Among other arguments against it, he proposed an example of drawing a tangent to certain curves at the point where the two parts cross each other; and as the fractional expression for the subtangent, by that method, had both its numerator and denominator equal to 0 at the point proposed, Rolle looked upon it as an absurd expression, and as an argument against the method of solution itself. The seeming mystery however was soon explained, and first of all by John Bernoulli. See an account of this affair in Montucla, Hist. Math. vol. 2, pa. 366.
Since that time, such kind of fractions have often been contemplated by mathematicians. As, by Maclaurin, in his Fluxions, vol. 2, pa. 698: Saunderson, in his Algebra, vol. 2, art. 469: De Moivre, in Miscel. Anal. pa. 165: Emerson, in his Algebra, pa. 212: and by many others. The same fractions have also proved a stumbling-block to more mathematicians than one, and the cause of more violent controversies: witness that between Powell and Waring, when they were competitors for the professorship at Cambridge. In the specimen of a work published on occasion of that competition, by Waring, was the fraction , which he said became 4 when p was = 1. This was struck at by Powell, as absurd, because when p = 1, then the fraction , which was one chief cause of his not succeeding to the professorship. Waring replied that (by common division) , when p is = 1. See the controversial pamphlets that passed between those two gentlemen at that time.
There are two modes of finding the value of such fractions, that have been given by the gentlemen above quoted. The one is by considering the terms of the fraction as two variable quantities, continually decreasing, till they both vanish together; or finding the ultimate value of the ratio denoted by the fraction. In this way of considering the matter, it appears that, as the terms of the fraction are supposed to decrease till they vanish, or become only equal to their fluxions or their increments, the value of the fraction at that state, will be equal to the fluxion or increment of the numerator divided by that of the denominator. Hence then, taking the example when x = 1; the fluxion of the numerator is , and of the denominator - x.; therefore , the value of the fraction when x = 1.——Or, thus, because x = 1, therefore ; then the fluxion of the numerator, - 4x3x., divided by| the fluxion of the denominator, or - x., gives 4x3 or 4, the same as before.
The other method is by reducing the given expression to another, or simple form, and then substituting the values of the letters. So in the above example , or , when x = 1; divide the numerator by the denominator, and it becomes , which when x = 1, becomes 4, for the given fraction, the same as before.—Again, to find the value of when x is = a, in which case both the numerator and denominator become = 0. Divide the numerator by the denominator, and the quotient is ; which when x = a, becomes , for the value of the fraction in that state of it.
· ·
Entry taken from A Mathematical and Philosophical Dictionary, by Charles Hutton, 1796.
This text has been generated using commercial OCR software, and there are still many problems; it is slowly getting better over time. Please don't reuse the content (e.g. do not post to wikipedia) without asking liam at holoweb dot net first (mention the colour of your socks in the mail), because I am still working on fixing errors. Thanks!
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# ML Aggarwal Sets Exe-5.2 Class 7 ICSE Maths Solutions
ML Aggarwal Sets Exe-5.2 Class 7 ICSE Maths Solutions. We Provide Step by Step Answer of Exe-5.2 Questions for Sets as council prescribe guideline for upcoming board exam. Visit official Website CISCE for detail information about ICSE Board Class-7.
## ML Aggarwal Sets Exe-5.2 Class 7 Maths Solutions
Board ICSE Publications Avichal Publishig Company (APC) Subject Maths Class 7th Chapter-5 Sets Writer ML Aggarwal Book Name Understanding Topics Solution of Exe-5.2 Questions Edition 2023-2024
### Sets Exe-5.2
ML Aggarwal Class 7 ICSE Maths Solutions
Page-112
#### Question- 1. Classify the following sets into empty set, finite set and infinite set. In case of (non-empty) finite sets, mention the cardinal number.
(i) {all colours of a rainbow}
(ii) {x | x is a prime number between 7 and 11}
(iii) {multiples of 5}
(iv) {all straight lines drawn in a plane}
(v) {x | x is a digit in the numeral 550131527}
(vi) {x | x is a letter in word SUFFICIENT}
(vii) {x | x = 4n, n ∈ I and x < 10}
(viii) {x | x ∈ N, x is a prime factor of 180}
(ix) {x : x is a vowel in the word WHY}
(x) {x : x = 5n, n ∈ W and x < 60}
(i) It is a finite set having 7 elements.
(ii) it is an empty set.
(iii) It is an infinite set having unlimited elements
(iv) It is an infinite set having unlimited number of elements.
(v) It is a finite set having 6 elements i.e 0, 1, 2, 3, 5, 7,
(vi) It is a finite set having 6 elements i.e S, U, F, I, E, N, T.
(vii) It is an infinite set having the set of integers i.e unlimited number of elements.
(viii) It is a finite set having 3 elements.
(ix) { x:x is a vowel in the word WHY}
It is an empty set as there is no vowel in the word why
(x) { x:x = 5n n ∈ W and x < 60 }
= { 5,10,15,20,25,30,35,40,45,50,55}
It is finite set and it has 12 elements.
### Sets Exe-5.2
ML Aggarwal Class 7 ICSE Maths Solutions
Page-113
#### Question 2. Which of the following describe the same sets:
(i) {vowels of English alphabet} and {e, a, u, i, o}
(ii) {a, b, d} and {d, a, b, b}
(iii) {letters of PUPPET} and {E, T, P, U}
(iv) {1, 2, 3} and {2, 3, 4}
(v) {1, 2, 3, 4, 5} and {x | x ∈ N, x ≤ 5}
(i) the given sets are the same sets.
(ii) These are the same sets.
(iii) the given sets are the same sets.
(iv) These are not the same sets.
(v) The given sets are the same sets.
#### Question 3. Find pairs/groups of equal sets from the following sets:
A = {0, 1, 2, 3}
B = {x : x2 < 10, x ∈ W}
C = {letters of word FOLLOW}
D = {days of a week}
E = {x | x ∈ W, x < 4}
F = {letters of word FLOW}
G = {Monday, Tuesday, ……… , Sunday}
H = {letters of word WOLF}
A = B = E because if we write B and E in tabular form, we get the
same elements.
C = F = H because the elements in a set can be rearranged as each set
can be written as { F,O,L,W} form.
D = G because if we write D in tabular we get the same elements.
#### Question- 4. Find pairs/groups of equivalent sets from the following sets.
A = {colours of a rainbow}
B = {letters of word GOOD}
C = {x : x is a digit in the numeral 371011489}
D = {letters of word TOM}
E = {x : x ∈ I, x2 < 10}
F = {months of a year}
G = {days of a week}
H = {x | x = 3n, n ∈ W and n < 12}
I = {all even numbers between 1 and 53}
J = {all letters of English alphabets}
A,C,E and G are equivalent sets as these all have same number of
elements i.e 7 elements.
B ⟺ as n (B) = 3 = n (D)
F ⟺ H as n (F) = 12 = n (H)
I ⟺ J as n (I) = 26 = n (j)
#### Question 5. In the following, find whether A ⊂ B or B ⊂ A or none of these:
(i) A = { 1, 2, 3}, B = {2, 3, 3, 3, 1, 3}
(ii) A = {2, 4, 6,….}, B = {all natural numbers}
(iii) A = {x | x ∈ I, x2 < 20}, B = {0, 1, 2, 3, 4}
(iv) A = {letters of KING}, B = {letters of QUEEN}
In the following find whether A ⊂ B or B ⊂ A or none of these
(i) A = { 1,2,3 } B = { 2,3,3,3,1,3} = {2,3,1}
A⊂B and B⊂A : i,e A = B
(ii) A = { 2,4,6, ….. } B = { all natural numbers }
= { 1,2,3,4,5,6,7……}
A⊂B but B⊄A
(iii) A = { x\x ∈ I , x2 < 20 } , B = { 0,1,2,3,4}
= { 0,1,4,9,16}
= { (0)2,(1)2,(2)2,(3)2,(4)2}
B = { 0,1,2,3,4}
B ⊂ A but A ⊄ B
(iv) A = { letters of king } = {K,I,N,G}
B = { letters of QUEEN } = { Q,U,E,N}
Here , A ⊄ B and B ⊄ C
Neither A ⊂ B nor B ⊂ A
#### Question 6. State whether each of the following statement is true or false for the sets A and B where
A = {letters of CLOUD} and B = {letters of KOLKATA}
(i) A ⊂ B
(ii) B ⊂ A
(iii) A ↔ B
A = {letters of CLOUD } = { C, L, O, U, D}
B = {letters of Kolkata } = { K, O, L, A, T }
(i) A ⊂ B : It is false because some elements of A are not the element of B .
(ii) B ⊂ A : It is false because some element of B is not a member
(iii) A ↔ B. It is true as n (A) = 5 n(B) .
#### Question -7. Write all the subsets of the following sets:
(i) Φ
(ii) {3, 5}
(iii) {2, 4, 6}
(i) Subset of Φ is Φ
(ii) Empty set is a subset of every set so, the subsets are Φ {3} , {5} {3 , 5}
(iii) Empty set is a subset of every set. So the subsets are Φ {2},{4},{6},{2,4} ,{4,6} ,(2,6) ,(2,4,6)
#### Question -8. If A = {x : x = 2n, n < 5}, then find A when
(i) ξ = N
(ii) ξ = W
(iii) ξ = I
(i) Natural numbers less than 5 are 1,2,3,4
Given x = 2n, putting n = 1,2,3,4 we get,
X = 2 × 1, 2×2, 2×3, 2× 4
= 2,4,6,8
The given set can be written as {2,4,6,8}
(Every set is a subset of universal set i.e. A ⊂ ξ )
(ii) whole numbers less than 5 are 0, 1, 2, 3, 4,
Given 2n i.e. 2 × 0, 2 ×1, 2 × 2, 2 × 3, 2 × 4,
i.e. 0,2,4,6,8
The given set i.e A can be written as (0,2,4,6,8)
(every set is a subset of universal set i.e A ⊂ ξ )
(iii) Integers less than 5 are …., -4,-3,-2,-1 ,0,1,2,3,4
Given 2 n i.e ……. 2 × -2, 2 × -1, 2× 0, 2 × 1, 2 × 2, 2 × 3, 2 ×4,
i.e …… -4 ,-2 0 2,4, 6,8
The given set i.e A can be written as { ……, -4, -2, 0 , 2, 4, 6, 8 }
(Every set is a subset of universal set i. e A ⊂ ξ )
— : End of ML Aggarwal Sets Exe-5.2 Class 7 ICSE Maths Solutions :–
Thanks
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# Equation of a Line Given Slope and a Point
### Linear Equations and Graphing
Sometimes we are asked to find the equation of a line given only the slope and a point on the line.
An airplane begins descending at the rate of 2000 feet per minute. After 1 min it is 28,000 feet above the ground. Assume the plane continues at the same rate of descent. What is the equation of the line that represents the plane’s decent.
#### Solution:
We will start by using the slope-intercept form of line equation: $\color{blue}{y=mx+b}$
Where $x$ represents the time and $y$ represents the distance from the ground.
We are told that the plane is descending at the rate of 2000 feet per minute. That would represent the slope of the equation.
$\color{blue}{m=\frac{-2000 ft.}{1 min.}}$ \begin{align*} y&=\color{blue}mx+b\\ y&=\color{blue}{-2000}x+b \end{align*}
Now we need to find a coordinate that will be on the line of the equation. Since the airplane was 28,000 feet above the ground after 1 minute, our coordinates are $(1,28000)$.
\begin{align*} \color{blue}y&=-2000\color{blue}x+b\\ \color{blue}{28,000}&={-2,000}*\color{blue}1+b \end{align*}
Now we can solve for b.
\begin{align*} 28,000&={-2,000}*1+b\\ 28,000\color{blue}{+2000}&={-2,000}\color{blue}{+2000}+b\\ \color{blue}{30,000}&=b \end{align*}
Substitute into the equation $y=mx+b$ to determine the equation of the line formed
$$\color{blue}{y=-2,000x+30,000}$$
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We've updated our
TEXT
Distance in the Coordinate Plane
Learning Outcomes
• Use the distance formula to find the distance between two points in the plane.
• Use the midpoint formula to find the midpoint between two points.
Derived from the Pythagorean Theorem, the distance formula is used to find the distance between two points in the plane. The Pythagorean Theorem, ${a}^{2}+{b}^{2}={c}^{2}$, is based on a right triangle where a and b are the lengths of the legs adjacent to the right angle, and c is the length of the hypotenuse. The relationship of sides $|{x}_{2}-{x}_{1}|$ and $|{y}_{2}-{y}_{1}|$ to side d is the same as that of sides a and b to side c. We use the absolute value symbol to indicate that the length is a positive number because the absolute value of any number is positive. (For example, $|-3|=3$. ) The symbols $|{x}_{2}-{x}_{1}|$ and $|{y}_{2}-{y}_{1}|$ indicate that the lengths of the sides of the triangle are positive. To find the length c, take the square root of both sides of the Pythagorean Theorem.
${c}^{2}={a}^{2}+{b}^{2}\rightarrow c=\sqrt{{a}^{2}+{b}^{2}}$
It follows that the distance formula is given as
${d}^{2}={\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}\to d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
We do not have to use the absolute value symbols in this definition because any number squared is positive.
A General Note: The Distance Formula
Given endpoints $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$, the distance between two points is given by
$d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$
Example: Finding the Distance between Two Points
Find the distance between the points $\left(-3,-1\right)$ and $\left(2,3\right)$.
Answer: Let us first look at the graph of the two points. Connect the points to form a right triangle. Then, calculate the length of d using the distance formula.
$\begin{array}{l}d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}\hfill \\ d=\sqrt{{\left(2-\left(-3\right)\right)}^{2}+{\left(3-\left(-1\right)\right)}^{2}}\hfill \\ =\sqrt{{\left(5\right)}^{2}+{\left(4\right)}^{2}}\hfill \\ =\sqrt{25+16}\hfill \\ =\sqrt{41}\hfill \end{array}$
Try It
Find the distance between two points: $\left(1,4\right)$ and $\left(11,9\right)$.
Answer: $\sqrt{125}=5\sqrt{5}$
Try it
In the graph below, you can move the points around the coordinate plane by clicking on them and dragging them. Try it out to see how the distance between them changes. Choose two points and answer the following questions:
1. Calculate the lengths of the sides of the triangle made by the two points you chose and the corner point connected to them by green dotted lines.
2. What parts of the distance formula are these lengths?
https://www.desmos.com/calculator/nrtjcgmy69
In the following video, we present more worked examples of how to use the distance formula to find the distance between two points in the coordinate plane. https://youtu.be/Vj7twkiUgf0
Example: Finding the Distance between Two Locations
Let’s return to the situation introduced at the beginning of this section. Tracie set out from Elmhurst, IL to go to Franklin Park. On the way, she made a few stops to do errands. Each stop is indicated by a red dot. Find the total distance that Tracie traveled. Compare this with the distance between her starting and final positions.
Answer: The first thing we should do is identify ordered pairs to describe each position. If we set the starting position at the origin, we can identify each of the other points by counting units east (right) and north (up) on the grid. For example, the first stop is 1 block east and 1 block north, so it is at $\left(1,1\right)$. The next stop is 5 blocks to the east so it is at $\left(5,1\right)$. After that, she traveled 3 blocks east and 2 blocks north to $\left(8,3\right)$. Lastly, she traveled 4 blocks north to $\left(8,7\right)$. We can label these points on the grid. Next, we can calculate the distance. Note that each grid unit represents 1,000 feet.
• From her starting location to her first stop at $\left(1,1\right)$, Tracie might have driven north 1,000 feet and then east 1,000 feet, or vice versa. Either way, she drove 2,000 feet to her first stop.
• Her second stop is at $\left(5,1\right)$. So from $\left(1,1\right)$ to $\left(5,1\right)$, Tracie drove east 4,000 feet.
• Her third stop is at $\left(8,3\right)$. There are a number of routes from $\left(5,1\right)$ to $\left(8,3\right)$. Whatever route Tracie decided to use, the distance is the same, as there are no angular streets between the two points. Let’s say she drove east 3,000 feet and then north 2,000 feet for a total of 5,000 feet.
• Tracie’s final stop is at $\left(8,7\right)$. This is a straight drive north from $\left(8,3\right)$ for a total of 4,000 feet.
Next, we will add the distances listed in the table.
From/To Number of Feet Driven
$\left(0,0\right)$ to $\left(1,1\right)$ 2,000
$\left(1,1\right)$ to $\left(5,1\right)$ 4,000
$\left(5,1\right)$ to $\left(8,3\right)$ 5,000
$\left(8,3\right)$ to $\left(8,7\right)$ 4,000
Total 15,000
The total distance Tracie drove is 15,000 feet or 2.84 miles. This is not, however, the actual distance between her starting and ending positions. To find this distance, we can use the distance formula between the points $\left(0,0\right)$ and $\left(8,7\right)$.
$\begin{array}{l}d=\sqrt{{\left(8 - 0\right)}^{2}+{\left(7 - 0\right)}^{2}}\hfill \\ =\sqrt{64+49}\hfill \\ =\sqrt{113}\hfill \\ =10.63\text{ units}\hfill \end{array}$
At 1,000 feet per grid unit, the distance between Elmhurst, IL to Franklin Park is 10,630.14 feet, or 2.01 miles. The distance formula results in a shorter calculation because it is based on the hypotenuse of a right triangle, a straight diagonal from the origin to the point $\left(8,7\right)$. Perhaps you have heard the saying "as the crow flies," which means the shortest distance between two points because a crow can fly in a straight line even though a person on the ground has to travel a longer distance on existing roadways.
Using the Midpoint Formula
When the endpoints of a line segment are known, we can find the point midway between them. This point is known as the midpoint and the formula is known as the midpoint formula. Given the endpoints of a line segment, $\left({x}_{1},{y}_{1}\right)$ and $\left({x}_{2},{y}_{2}\right)$, the midpoint formula states how to find the coordinates of the midpoint $M$.
$M=\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)$
A graphical view of a midpoint is shown below. Notice that the line segments on either side of the midpoint are congruent.
Example: Finding the Midpoint of the Line Segment
Find the midpoint of the line segment with the endpoints $\left(7,-2\right)$ and $\left(9,5\right)$.
Answer: Use the formula to find the midpoint of the line segment.
$\begin{array}{l}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\hfill&=\left(\frac{7+9}{2},\frac{-2+5}{2}\right)\hfill \\ \hfill&=\left(8,\frac{3}{2}\right)\hfill \end{array}$
Try It
Find the midpoint of the line segment with endpoints $\left(-2,-1\right)$ and $\left(-8,6\right)$.
Answer: $\left(-5,\frac{5}{2}\right)$
Example: Finding the Center of a Circle
The diameter of a circle has endpoints $\left(-1,-4\right)$ and $\left(5,-4\right)$. Find the center of the circle.
Answer: The center of a circle is the center or midpoint of its diameter. Thus, the midpoint formula will yield the center point.
$\begin{array}{l}\left(\frac{{x}_{1}+{x}_{2}}{2},\frac{{y}_{1}+{y}_{2}}{2}\right)\\ \left(\frac{-1+5}{2},\frac{-4 - 4}{2}\right)=\left(\frac{4}{2},-\frac{8}{2}\right)=\left(2,-4\right)\end{array}$
Try It
In the graph below, there is a circle with a diameter whose endpoints are (0,5) and (10,5).
1. Find the center of the circle.
2. Place a point at the center of the circle.
3. Now move the endpoints of the diameter by clicking on them and dragging them. Notice that the size and location of the circle changes.
4. Now graph the center point of a circle of any radius given that the equation for a circle of any radius with any center $(h,k)$ is $(x-h)^2+(y-k)^2 = r^2$.
https://www.desmos.com/calculator/j2blmpiid7
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NCERT Solutions For Class 6 Maths Chapter 9 Data Handling Exercise 9.1
Recording data, organisation of data, pictograph and its interpretation are the concepts discussed in NCERT Solutions For Class 6 Maths Chapter 9 Data Handling Exercise 9.1. Collection of numbers which gives useful information to the observer is called a data. The problems are solved in the best possible way to improve knowledge of concepts among students. For further reference, the students can download NCERT Solutions Class 6 Maths Chapter 9 Data Handling Exercise 9.1 and practice them to get a better hold on the problem solving methods.
NCERT Solutions for Class 6 Chapter 9: Data Handling Exercise 9.1 Download PDF
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Access NCERT Solutions for Class 6 Chapter 9: Data Handling Exercise 9.1
1. In a Mathematics test, the following marks were obtained by 40 students. Arrange these marks in a table using tally marks.
8 1 3 7 6 5 5 4 4 2 4 9 5 3 7 1 6 5 2 7 7 3 8 4 2 8 9 5 8 6 7 4 5 6 9 6 4 4 6 6
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(a) Find how many students obtained marks equal to or more than 7.
(b) How many students obtained marks below 4?
Solutions:
Marks Tally Marks Number of Students 1 || 2 2 ||| 3 3 ||| 3 4 || 7 5 | 6 6 || 7 7 5 8 Â |||| 4 9 Â ||| 3
(a) The students who got marks equal to or more than 7 are the students who got marks as either of 7, 8 and 9. Therefore number of these students
= 5 + 4 + 3
= 12
(b) The students who got marks below 4 are the students who got marks as either of 1, 2 and 3. Therefore number of these students are
= 2 + 3 + 3
= 8
2. Following is the choice of sweets of 30 students of Class VI.
(a) Arrange the names of sweets in a table using tally marks.
(b) Which sweet is preferred by most of the students?
Solutions:
By observing the choice of sweets of 30 students. We may construct the table as shown below:
Sweets Tally Marks Number of Students Ladoo | 11 Barfi ||| 3 Jalebi || 7 Rasgulla |||| 9 30
(b) The highest number of students preferred Ladoos. Hence, Ladoo is the most preferred sweet among students.
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3. Catherine threw a dice 40 times and noted the number appearing each time as shown below :
1 3 5 6 6 3 5 4 1 6 2 5 3 4 6 1 5 5 6 1 1 2 2 3 5 2 4 5 5 6 5 1 6 2 3 5 2 4 1 5
Make a table and enter the data using tally marks. Find the number that appeared.
(a) The minimum number of times
(b) The maximum number of times
(c) Find those numbers that appear an equal number of times.
Solutions:
Numbers Tally Marks Number of times 1 || 7 2 | 6 3 5 4 |||| 4 5 | 11 6 || 7
(a) The number that occurred for minimum number of times is 4
(b) The number that occurred for maximum number of times is 5
(c) 1 and 6 are the numbers that appear an equal number of times.
4. Following pictograph shows the number of tractors in five villages.
Observe the pictograph and answer the following questions.
(i) Which village has the minimum number of tractors?
(ii) Which village has the maximum number of tractors?
(iii) How many more tractors village C has as compared to village B.
(iv) What is the total number of tractors in all the five villages?
Solutions:
(i) Village D has the minimum number of tractors.
(ii) Village C has the maximum number of tractors.
(iii) Village B has 5 tractors
Village C has 8 tractors
= 8 – 5
= 3 tractors
Village C has 3 more tractors as compared to village B
(iv) Total number of tractors in all the villages = 6 + 5 + 8 + 3 + 6 = 28 tractors
5. The number of girl students in each class of a co-educational middle school is depicted by the pictograph:
Observe this pictograph and answer the following questions:
(a) Which class has the minimum number of girl students?
(b) Is the number of girls in Class VI less than the number of girls in Class V?
(c) How many girls are there in Class VII?
Solutions:
By observing the above table, there are 24, 18, 20, 14, 10, 16, 12 and 6 girls respectively from class I to VIII
(a) Class VIII has only 6 girls. Therefore, the minimum number of girl students are in Class VIII
(b) No. Class V has 10 girl students
Class VI has 16 girl students
Hence, the number of girls in Class VI are more than the number of girls in Class V
(c) The number of girls in Class VII are 12
6. The sale of electric bulbs on different days of a week is shown below:
Observe the pictograph and answer the following questions:
(a) How many bulbs were sold on Friday?
(b) On which day were the maximum number of bulbs sold?
(c) On which of the days same number of bulbs were sold?
(d) On which of the days minimum number of bulbs were sold?
(e) If one big carton can hold 9 bulbs. How many cartons were needed in the given week?
Solutions:
(a) Number of bulbs sold on Friday are 14 bulbs.
(b) On Sunday highest number of bulbs i.e 18 are sold. Thus maximum number of bulbs were sold on Sunday.
(c) On Wednesday and Saturday 8 bulbs are sold. Hence equal number of bulbs were sold on Wednesday and Saturday.
(d) Minimum number of bulbs were sold on Wednesday and Saturday i.e 8 bulbs.
(e) Total number of bulbs sold in a week = 12 + 16 + 8 + 10 + 14 + 8 + 18 = 86
7. In a village six fruit merchants sold the following number of fruit baskets in a particular season:
Observe this pictograph and answer the following questions:
(a) Which merchant sold the maximum number of baskets?
(b) How many fruit baskets were sold by Anwar?
(c) The merchants who have sold 600 or more number of baskets are planning to buy a godown for the next season. Can you name them?
Solutions:
From the pictograph, the number of fruit baskets sold by Rahim, Lakhanpal, Anwar, Martin, Ranjit Singh and Joseph are 400, 550, 700, 950, 800 and 450 respectively
(a) Martin sold the maximum number of fruit baskets i.e 950
(b) Anwar sold 700 fruit baskets
(c) Anwar, Martin and Ranjit Singh are the merchants who sold more than 600 fruit baskets. Hence, these are the merchants who are planning to buy a godown for the next season.
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# Function Notation and Linear Functions
## If y=mx+b and y=f(x) then f(x)=mx+b
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Growing Kids
Teacher Contributed
## Real World Applications – Algebra I
### Topic
How can we represent a kid’s growing height as a linear relationship?
### Student Exploration
Most doctors agree that the “normal” growth rate for children after the age of 2 is about \begin{align*}2 \ \frac{1}{2} \ inches\end{align*} or 6 centimeters per year until adolescence. Let’s represent this as a linear relationship using inches.
Let’s say a kid at 2 years old is 3 feet tall, or 36 inches. Using the information given, this kid will be 38.5 inches tall when 3 years old. Let’s write an equation representing this relationship using these two data points.
Given the information about the heights, we’d first have to calculate the slope (even though that was given to us). The slope would be,
\begin{align*}m &= \frac{(\text{the difference in height})}{(\text{the difference in age})}\\ m &= \frac{(38.5 - 36)}{(3 - 2)} = 2.5 \ inches \ per \ year\end{align*}
Now let’s use one of our data points and the slope to find the equation to represent this relationship. We’re going to use the slope-intercept form to substitute what we know so far.
\begin{align*}y &= mx + b\\ 36 &= (2.5)(2) + b \ \text{Now let’s solve for} \ “b.”\\ 36 &= 5 + b\\ 31 &= b\end{align*}
Our equation is: \begin{align*}y = 2.5x + 31\end{align*}
Now, this equation represents the linear relationship of a growing child after the age of 2. Looking at the equation, 31 is \begin{align*}b\end{align*}. This means that the \begin{align*}y-\end{align*}intercept is 31 inches. This can’t make sense, because then this would mean that a child was born at 31 inches! This also wouldn’t make sense when a kid hits puberty, because his/her growth spurt would be a lot faster!
Now let’s look at this linear relationship as a function. As you read from the concept, the \begin{align*}f(x)\end{align*} is the output. We can rewrite this relationship as \begin{align*}f(x) = 2.5x + 31\end{align*}. We can use this function to determine height at different ages.
If we were to find \begin{align*}f(5)\end{align*}, this means that we need to find the height of the child at 5 years old. Let’s figure it out:
\begin{align*}f(5) &= 2.5(5) + 31\\ f(5) &= 7.5 + 31\\ f(5) & = 38.5\end{align*}
This means that at 5 years old, the child will be 38.5 inches tall.
What’s \begin{align*}f(7)\end{align*} and what does it mean?
### Extension Investigation
Try asking a family member how tall you were at two different ages in your life, and practice finding the rate of change, or the slope between these two points. Would this equation make sense? Why or why not? Would this equation apply when you’re over 30 years old? Would you be getting taller at that age?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
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## Numbers and Sequences | Aster Classes
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, unit, exercise 2,
Question 1.
Prove that n2 – n divisible by 2 for every positive integer n.
We know that any positive integer is of the form 2q or 2q + 1 for some integer q.
Case 1: When n = 2 q
n2 – n = (2q)2 – 2q = 4q2 – 2q
= 2q (2q – 1)
In n2 – n = 2r
2r = 2q(2q – 1)
r = q(2q + 1)
n2 – n is divisible by 2
Case 2: When n = 2q + 1
n2 – n = (2q + 1)2 – (2q + 1)
= 4q2 + 1 + 4q – 2q – 1 = 4q2 + 2q
= 2q (2q + 1)
If n2 – n = 2r
r = q (2q + 1)
∴ n2 – n is divisible by 2 for every positive integer “n”
Question 2.
A milk man has 175 litres of cow’s milk and 105 litres of buffalow’s milk. He wishes to sell the milk by filling the two types of milk in cans of equal capacity. Calculate the following
(i) Capacity of a can
(ii) Number of cans of cow’s milk
(iii) Number of cans of buffalow’s milk.
175 litres of cow’s milk.
105 litres of goat’s milk.
H.C.F of 175 & 105 by using Euclid’s division algorithm.
175 = 105 × 1 + 70, the remainder 70 ≠ 0
Again using division algorithm,
105 = 70 × 1 + 35, the remainder 35 ≠ 0
Again using division algorithm.
70 = 35 × 2 + 0, the remainder is 0.
∴ 35 is the H.C.F of 175 & 105.
(i) ∴ The milk man’s milk can’s capacity is 35 litres.
(ii) No. of cow’s milk obtained = 17535 = 5 cans
(iii) No. of buffalow’s milk obtained = 10535= 3 cans
Question 3.
When the positive integers a, b and c are divided by 13 the respective remainders are 9,7 and 10. Find the remainder when a + 2b + 3c is divided by 13.
Given the positive integer are a, b and c
a = 13q + 9 (divided by 13 leaves remainder 9)
b = 13q + 7
c = 13q + 10
a + 2b + 3c = 13q + 9 + 2(13q + 7) + 3 (13q + 10)
= 13q + 9 + 26q + 14 + 39q + 30
= 78q + 53
When compare with a = 3q + r
= (13 × 6) q + 53
The remainder is 53
Question 4.
Show that 107 is of the form 4q +3 for any integer q.
Solution:
107 = 4 × 26 + 3. This is of the form a = bq + r.
Hence it is proved.
Question 5.
If (m + 1)th term of an A.P. is twice the (n + 1)th term, then prove that (3m + 1)th term is twice the (m + n + 1)th term.
tn = a + (n – 1)d
Given tm+1 = 2 tn+1
a + (m + 1 – 1)d = 2[a + (n + 1 – 1)d]
a + md = 2(a + nd) ⇒ a + md =2a + 2nd
md – 2nd = a
d(m – 2n) = a ….(1)
To Prove t(3m + 1) = 2(tm+n+1)
L.H.S. = t3m+1
= a + (3m + 1 – 1)d
= a + 3md
= d(m – 2n) + 3md (from 1)
= md – 2nd + 3md
= 4md – 2nd
= 2d (2m – n)
R.H.S. = 2(tm+n+1)
= 2 [a + (m + n + 1 – 1) d]
= 2 [a + (m + n)d]
= 2 [d (m – 2n) + md + nd)] (from 1)
= 2 [dm – 2nd + md + nd]
= 2 [2 md – nd]
= 2d (2m – n)
R.H.S = L.H.S
∴ t(3m+1) = 2 t(m+n+1)
Hence it is proved.
Question 6.
Find the 12th term from the last term of the A.P -2, -4, -6,… -100.
The given A.P is -2, -4, -6, …. 100
d = -4 – (-2) = -4 + 2 = – 2
Finding the 12 term from the last term
a = -100, d = 2 (taking from the last term)
n = 12
tn = a + (n – 1)d
t12 = – 100 + 11 (2)
= -100 + 22
= -78
∴ The 12th term of the A.P from the last term is – 78
Question 7.
Two A.P’s have the same common difference. The first term of one A.P is 2 and that of the other is 7. Show that the difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms.
Solution:
Let the two A.Ps be
AP1 = a1, a1 + d, a1 + 2d,…
AP2 = a2, a2 + d, a2 + 2d,…
In AP1 we have a1 = 2
In AP2 we have a2 = 7
t10 in AP1 = a1 + 9d = 2 + 9d ………….. (1)
t10 in AP2 = a2 + 9d = 7 + 9d …………… (2)
The difference between their 10th terms
= (1) – (2) = 2 + 9d – 7 – 9d
= -5 ………….. (I)
t21 m AP1 = a1 + 20d = 2 + 20d …………. (3)
t21 in AP2 = a2 + 20d = 7 + 20d ………… (4)
The difference between their 21 st terms is
(3) – (4)
= 2 + 20d – 7 – 20d
= -5 ……………. (II)
I = II
Hence it is Proved.
Question 8.
A man saved ₹16500 in ten years. In each year after the first he saved ₹100 more than he did in the preceding year. How much did he save in the first year?
Amount of saving in ten years = ₹ 16500
S10 = 16500, d= 100
Sn = n2 [2a + (n – 1)d]
S10 = 102 [2a + 9d]
16500 = 102 [2a + 900] = 5(2a + 900)
16500 = 10a + 4500 ⇒ 16500 – 4500 = 10a
12000 = 10a
a = 1200010 = 1200
Amount saved in the first year = ₹ 1200
Question 9.
Find the G.P. in which the 2nd term is 6–√and the 6th term is 9 6–√.
2nd term of the G.P = 6–√
t2 = 6–√
[tn = a rn-1]
a.r = 6–√ ….(1)
6th term of the G.P. = 9 6–√
a. r5 = 96–√ ……..(2)
Question 10.
The value of a motorcycle depreciates at a rate of 15% per year. What will be the value of the motorcycle 3 year hence, which is now purchased for ₹45,000?
Solution:
a = ₹45000
Depreciation = 15% for ₹45000
= 45000 × 15100
d = ₹6750 since it is depreciation
d = -6750
At the end of 1st year its value = ₹45000 – ₹6750
= ₹38250,
Again depreciation = 38250 × 15100 = 5737.50
At the end of 2nd year its value
= ₹38250 – ₹5737.50 = 32512.50
Again depreciation = 32512.50 × 15100 = 4876.88
At the end of the 3rd year its value
= 32512.50 – 4876.88 = 27635.63
∴ The value of the automobile at the 3rdyear
= ₹ 27636
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.10,
Question 1.
Euclid’s division lemma states that for positive integers a and b, there exist unique integers q and r such that a = bq + r, where r must satisfy ………………….
(1) 1 < r < b
(2) 0 < r < b
(3) 0 < r < 6
(4) 0 < r < b
Ans.
(3) 0 < r < b
Question 2.
Using Euclid’s division lemma, if the cube of any positive integer is divided by 9 then the possible remainders are ………………….
(1) 0, 1, 8
(2) 1, 4, 8
(3) 0, 1, 3
(4) 1, 3, 5
(1) 0, 1, 8
Hint: Let the +ve integer be 1, 2, 3, 4 …………
13 = 1 when it is divided by 9 the remainder is 1.
23 = 8 when it is divided by 9 the remainder is 8.
33 = 27 when it is divided by 9 the remainder is 0.
43 = 64 when it is divided by 9 the remainder is 1.
53 = 125 when it is divided by 9 the remainder is 8.
The remainder 0, 1, 8 is repeated.
Question 3.
If the H.C.F of 65 and 117 is expressible in the form of 65m – 117 , then the value of m is
(1) 4
(2) 2
(3) 1
(4) 3
(2) 2
Hint:
H.C.F. of 65 and 117
117 = 65 × 1 + 52
65 = 52 × 1 + 13
52 = 13 × 4 + 0
∴ 13 is the H.C.F. of 65 and 117.
65m – 117 = 65 × 2 – 117
130 – 117 = 13
∴ m = 2
Question 4.
The sum of the exponents of the prime factors in the prime factorization of 1729 is …………………….
(1) 1
(2) 2
(3) 3
(4) 4
(3) 3
Hint: 1729 = 7 × 13 × 19
Sum of the exponents = 1 + 1 + 1
= 3
Question 5.
The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is
(1) 2025
(2) 5220
(3) 5025
(4) 2520
(4) 2520
Hint:
L.C.M. = 23 × 32 × 5 × 7
= 8 × 9 × 5 × 7
= 2520
Question 6.
74k ≡ ______ (mod 100)
(1) 1
(2) 2
(3) 3
(4) 4
(1) 1
Hint:
74k ≡______ (mod 100)
y4k ≡ y4 × 1 = 1 (mod 100)
Question 7.
Given F1 = 1 , F2 = 3 and Fn = Fn-1 + Fn-2then F5 is ………….
(1) 3
(2) 5
(3) 8
(4) 11
(4) 11
Hint:
Fn = Fn-1 + Fn-2
F3 = F2 + F1 = 3 + 1 = 4
F4 = F3 + F2 = 4 + 3 = 7
F5 = F4 + F3 = 7 + 4 = 11
Question 8.
The first term of an arithmetic progression is unity and the common difference is 4. Which of the following will be a term of this A.P
(1) 4551
(2) 10091
(3) 7881
(4) 13531
(3) 7881
Hint:
t1 = 1
d = 4
tn = a + (n – 1)d
= 1 + 4n – 4
4n – 3 = 4551
4n = 4554
n = will be a fraction
It is not possible.
4n – 3 = 10091
4n = 10091 + 3 = 10094
n = a fraction
4n – 3 = 7881
4n = 7881 + 3 = 7884
n = 78844, n is a whole number.
4n – 3 = 13531
4n = 13531 – 3 = 13534
n is a fraction.
∴ 7881 will be 1971st term of A.P.
Question 9.
If 6 times of 6th term of an A.P is equal to 7 times the 7th term, then the 13th term of the A.P. is ………..
(1) 0
(2) 6
(3) 7
(4) 13
(1) 0
Hint:
6 t6 = 7 t7
6(a + 5d) = 7 (a + 6d) ⇒ 6a + 30d = 7a + 42d
30 d – 42 d = 7a – 6a ⇒ -12d = a
t13 = a + 12d (12d = -a)
= a – a = 0
Question 10.
An A.P consists of 31 terms. If its 16th term is m, then the sum of all the terms of this A.P. is
(1) 16 m
(2) 62 m
(3) 31 m
(4) 312 m
(3) 31 m
Hint:
t16 = m
S31 = 312 (2a + 30d)
= 312 (2(a + 15d))
(∵ t16 = a + 15d)
= 31(t16) = 31m
Question 11.
In an A.P., the first term is 1 and the common difference is 4. How many terms of the A.P must be taken for their sum to be equal to 120?
(1) 6
(2) 7
(3) 8
(4) 9
(3) 8
Here a = 1, d = 4, Sn = 120
Sn = n2[2a + (n – 1)d]
120 = n2 [2 + (n – 1)4] = n2 [2 + 4n – 4)]
= n2 [4n – 2)] = n2 × 2 (2n – 1)
120 = 2n2 – n
∴ 2n2 – n – 120 = 0 ⇒ 2n2 – 16n + 15n – 120 = 0
2n(n – 8) + 15 (n – 8) = 0 ⇒ (n – 8) (2n + 15) = 0
n = 8 or n = −152 (omitted)
∴ n = 8
Question 12.
A = 265 and B = 264 + 263 + 262 …. + 20 which of the following is true?
(1) B is 264 more than A
(2) A and B are equal
(3) B is larger than A by 1
(4) A is larger than B by 1
(4) A is larger than B by
A = 265
B = 264+63 + 262 + …….. + 20
= 2
= 1 + 22 + 22 + ……. + 264
a = 1, r = 2, n = 65 it is in G.P.
S65 = 1 (265 – 1) = 265 – 1
A = 265 is larger than B
Question 13.
The next term of the sequence 316,18,112,118 is ………..
(1) 124
(2) 127
(3) 23
(4) 181
(2) 127
Hint:
316,18,112,118
a = 316, r = 18 ÷ 316 = 18 × 163 = 23
The next term is = 118 × 23 = 127
Question 14.
If the sequence t1,t2,t3 … are in A.P. then the sequence t6,t12,t18 … is
(1) a Geometric Progression
(2) an Arithmetic Progression
(3) neither an Arithmetic Progression nor a Geometric Progression
(4) a constant sequence
(2) an Arithmetic Progression
Hint:
If t1, t2, t3, … is 1, 2, 3, …
If t6 = 6, t12 = 12, t18 = 18 then 6, 12, 18 … is an arithmetic progression
Question 15.
The value of (13 + 23 + 33 + ……. + 153) – (1 + 2 + 3 + …….. + 15) is …………….
(1) 14400
(2) 14200
(3) 14280
(4) 14520
(3) 14280
Hint:
1202 – 120 = 120(120 – 1)
120 × 119 = 14280
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.9,
Question 1.
Find the sum of the following series
(i) 1 + 2 + 3 + …….. + 60
(ii) 3 + 6 + 9 + …….. +96
(iii) 51 + 52 + 53 + …….. + 92
(iv) 1 + 4 + 9 + 16 + …….. + 225
(v) 62 + 72 + 82 + …….. + 212
(vi) 103 + 113 + 123 + …….. + 203
(vii) 1 + 3 + 5 + …… + 71
Solution:
(i) 1 + 2 + 3 + …….. + 60 = 60×612
[Using n(n+1)2 formula]
= 1830.
(ii) 3 + 6 + 9 + …….. + 96 = 3(1 + 2 + 3 + ……… + 32)
= 3×32×332
= 1584
(iii) 51 + 52 + 53 + …….. + 92 = (1 + 2 + 3 + ……. + 92) – (1 + 2 + 3 + …… + 50)
= 4278 – 1275
= 3003
(iv) 1 + 4 + 9 + 16 + …….. + 225 = 12 + 22 + 3+ 42 + ………… + 152
15×16×316
[using n(n+1)(2n+1)6] formula
= 1240
(v) 62 + 72 + 82 + …….. + 212 = 1 + 22 + 32 + 42 + ………… + 212 – (1 + 22 + ………… + 52)
= 21×22×436 – 5×6×116
= 3311 – 55
= 3256.
(vi) 103 = 113 + 123 + …….. + 203 = 13 + 23+ 33 + ………… + 203 – (13 + 23 + 33 + …………. + 93)
[Using (n(n+1)2)2 formula]
= 2102 – 452 = 44100 – 2025
= 42075
(vii) 1 + 3 + 5+ … + 71
Here a = 1; d = 3 – 1 = 2; l = 71
Question 2.
If 1 + 2 + 3 + …. + k = 325 , then find 13 + 23+ 33 + …………. + k3
1 + 2 + 3 + …. + k = 325
k(k+1)2 = 325 ……(1)
= 3252 (From 1)
= 105625.
Question 3.
If 13 + 23 + 33 + ………… + K3 = 44100 then find 1 + 2 + 3 + ……. + k
13 + 23 + 33 + ………….. + k3 = 44100
k(k+1)2 = 44100−−−−−√ = 210
1 + 2 + 3 + …… + k = k(k+1)2
= 210
Question 4.
How many terms of the series 13 + 23 + 33+ …………… should be taken to get the sum 14400?
13 + 23 + 33 + ……. + n3 = 14400
n(n+1)2 = 14400−−−−−√
n(n+1)2 = 120 ⇒ n2 + n = 240
n2 + n – 240 = 0
(n + 16) (n – 15) = 0
(n + 16) = 0 or (n – 15) = 0
n = -16 or n = 15 (Negative will be omitted)
∴ The number of terms taken is 15
Question 5.
The sum of the squares of the first n natural numbers is 285, while the sum of their cubes is 2025. Find the value of n.
12 + 22 + 32 + …. + n2 = 285
Question 6.
Rekha has 15 square colour papers of sizes 10 cm, 11 cm, 12 cm, …, 24 cm. How much area can be decorated with these colour papers?
Area of 15 square colour papers
= 102 + 112 + 122 + …. + 242
= (12 + 22 + 32 + …. + 242) – (12 + 22 + 92)
= 24×25×496−9×10×196
= 4 × 25 × 49 – 3 × 5 × 19
= 4900 – 285
= 4615
Area can be decorated is 4615 cm2
Question 7.
Find the sum of the series (23 – 1)+(43 – 33) + (63 – 153) + …….. to
(i) n terms
(ii) 8 terms
Sum of the series = (23 – 1) + (43 – 33) + (63 – 153) + …. n terms
= 23 + 43 + 63 + …. n terms – (13 + 33 + 53+ …. n terms) …….(1)
23 + 43 + 63 + …. n = ∑(23 + 43 + 63 + ….(2n)3]
∑ 23 (13 + 23 + 33 + …. n3)
= 8 (n(n+1)2)2
= 2[n (n + 1)]2
13 + 33 + 53 + ……….(2n – 1)3 [sum of first 2n cubes – sum of first n even cubes]
Substituting (2) and (3) in (1)
Sum of the series = 2n2 (n + 1)2 – n2 (2n + 1)2 + 2n2(n + 1)2
= 4n2 (n + 1)2 – n2 (2n + 1)2
= n2 [(4(n + 1)2 – (2n + 1)2]
= n2 [4n2 + 4 + 8n – 4n2 – 1 – 4n]
= n2 [4n + 3]
= 4n3 + 3n2
(ii) when n = 8 = 4(8)3 + 3(8)2
= 4(512) + 3(64)
= 2240.
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.8,
Question 1.
Which of the following sequences are in G.P?
(i) 3,9,27,81,…
(ii) 4,44,444,4444,…
(iii) 0.5,0.05,0.005,
(iv) 13,16,112, ………….
(v) 1, -5, 25,-125,…
(vi) 120, 60, 30, 18,…
(vii) 16, 4, 1, 14, ……….
Question 3.
In a G.P. 729, 243, 81,… find t7.
The G.P. is 729, 243, 81,….
Question 4.
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression
Solution:
G.P = x + 6, x + 12, x + 15
In G.P r = t2t1=t3t2
x+12x+6=x+15x+12
(x + 12)2 = (x + 6) (x + 5)
x2 + 24x + 144 = x2 + 6x + 15x + 90
24x – 21x = 90 – 144
3x = -54
x = −543 = -18
x = -18
Question 5.
Find the number of terms in the following G.P.
(i) 4,8,16,…,8192?
Here a = 4; r = 84 = 2
tn = 8192
a . rn-1 = 8192 ⇒ 4 × 2n-1 = 8192
2n-1 = 81924 = 2048
2n-1 = 211 ⇒ n – 1 = 11
n = 11 + 1 ⇒ n = 12
Number of terms = 12
(ii) 13, 19, 127, ……………, 12187
a = 13 ; r = 19 ÷ 13 = 19 × 31 = 13
n – 1 = 6 ⇒ n = 6 + 1 = 7
Number of terms = 7
Question 6.
In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.
Given, 9th term = 32805
a. rn-1 = 12187
t9 = 32805 [tn = arn-1]
a.r8 = 32805 …..(1)
6th term = 1215
a.r5 = 1215 …..(2)
Divide (1) by (2)
ar8ar5 = 328051215 ⇒ r3 = 6561243
= 218781 = 72927 = 2439 = 813
r3 = 27 ⇒ r3 = 33
r = 3
Substitute the value of r = 3 in (2)
a. 35 = 1215
a × 243 = 1215
a = 1215243 = 5
Here a = 5, r = 3, n = 12
t12 = 5 × 3(12-1)
= 5 × 311
∴ 12th term of a G.P. = 5 × 311
Question 7.
Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.
Solution:
t8 = 768 = ar7
r = 2
t10 = ar9 = ar7 × r × r
= 768 × 2 × 2 = 3072
Question 8.
If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.
a, b, c are in A.P.
t2 – t1 = t3 – t2
b – a = c – b
2b = a + c …..(1)
3a, 3b, 3c are in G.P.
From (1) and (2) we get
3a, 3b, 3c are in G.P.
Question 9.
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 572. Find the three terms.
Let the three terms of the G.P. be ar, a, ar
Product of three terms = 27
ar × a × ar = 27
a3 = 27 ⇒ a3 = 33
a = 3
Sum of the product of two terms taken at a time is 572
6r2 – 13r + 6 = 0
6r2 – 9r – 4r + 6 = 0
3r (2r – 3) -2(2r – 3) = 0
(2r – 3) (3r – 2) = 0
2r – 3 = 0 or 3r – 2 = 0
2r = 3 (or) 3r – 2 = 0
r = 32 (or) r = 23
∴ The three terms are 2, 3 and 92 or 92, 3 and 2
Question 10.
A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Starting salary (a) = ₹ 60000
Increased salary = 5% of starting salary
= 5100 × 60000
= ₹ 3000
Starting salary for the 2nd year = 60000 + 3000
= ₹ 63000
Year increase = 5% of 63000
= 5100 × 63000
= ₹ 3150
Starting salary for the 3rd year = 63000 + 3150
= ₹ 66150
60000, 63000, 66150,…. form a G.P.
a = 60000; r = 6300060000 = 6360 = 2120
tn = ann-1
t5 = (60000) (2120)4
= 60000 × 2120 × 2120 × 2120 × 2120
= 6×21×21×21×212×2×2×2
= 72930.38
5% increase = 5100 × 72930.38
= ₹ 3646.51
Salary after 5 years = ₹ 72930.38 + 3646.51
= ₹ 76576.90
= ₹ 76577
Question 11.
Sivamani is attending an interview for a job and the company gave two offers to him. Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.
Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
What is his salary in the 4th year with respect to the offers A and B?
Starting salary (a) = ₹ 20,000
Annual increase = 6% of 20000
= 5100 × 20000
= ₹ 1200
Salary for the 2nd year = ₹ 20000 + 1200
= ₹ 21200
Here a = 20,000; r = 2120020000 = 212200 = 106100 = 5350
n = 4 years
tn = arn-1
Salary at the end of 4th year = 23820
For B
Starting salary = ₹ 22000
(a) = 22000
Annual increase = 3% of 22000
= 3100 × 22000
= ₹ 660
Salary for the 2nd year = ₹ 22000 + ₹ 660
= ₹ 22,660
Here a = 22000; r = 2266022000
= 22662200 = 11331100 = 103100
Salary at the end of 4th year = 22000 × (103100)4-1
= 22000 × (103100)3
= 22000 × 103100 × 103100 × 103100
= 24039.99 = 24040
4th year Salary for A = ₹ 23820 and 4th year Salary for B = ₹ 24040
Question 12.
If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb-c × yc-a × za-b = 1
a, b, c are three consecutive terms of an A.P
∴ a = a, b = a + dand c = a + 2d respectively ….(1)
x, y, z are three consecutive terms of a G.P
∴ x = x, y = xr, z = xr2 respective ……(2)
L.H.S = xb-c × yc-a × za-b ( Substitute the values from 1 and 2 we get)
L.H.S = R.H.S
Hence it is proved
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.7,
Question 1.
Which of the following sequences are in G.P?
(i) 3,9,27,81,…
(ii) 4,44,444,4444,…
(iii) 0.5,0.05,0.005,
(iv) 13,16,112, ………….
(v) 1, -5, 25,-125,…
(vi) 120, 60, 30, 18,…
(vii) 16, 4, 1, 14, ……….
Question 2.
Write the first three terms of the G.P. whose first term and the common ratio are given below.
(i) a = 6, r = 3
a = 6, r = 3
ar = 6 × 3 = 18,
ar2 = 6 × 9 = 54
The three terms are 6, 18 and 54
(ii) a = 2–√, r = 2–√.
ar = 2–√ × 2–√ = 2,
ar2 = 2–√ × 2 = 2 2–√
The three terms are 2–√, 2 and 22–√
(iii) a = 1000, r = 25
ar = 1000 × 25 = 400,
ar2 = 1000 × 425 = 40 × 4 = 160
The three terms are 1000,400 and 160.
Question 3.
In a G.P. 729, 243, 81,… find t7.
The G.P. is 729, 243, 81,….
Question 4.
Find x so that x + 6, x + 12 and x + 15 are consecutive terms of a Geometric Progression
Solution:
G.P = x + 6, x + 12, x + 15
In G.P r = t2t1=t3t2
x+12x+6=x+15x+12
(x + 12)2 = (x + 6) (x + 5)
x2 + 24x + 144 = x2 + 6x + 15x + 90
24x – 21x = 90 – 144
3x = -54
x = −543 = -18
x = -18
Question 5.
Find the number of terms in the following G.P.
(i) 4,8,16,…,8192?
Here a = 4; r = 84 = 2
tn = 8192
a . rn-1 = 8192 ⇒ 4 × 2n-1 = 8192
2n-1 = 81924 = 2048
2n-1 = 211 ⇒ n – 1 = 11
n = 11 + 1 ⇒ n = 12
Number of terms = 12
(ii) 13, 19, 127, ……………, 12187
a = 13 ; r = 19 ÷ 13 = 19 × 31 = 1
tn = 12187
a. rn-1 = 12187
n – 1 = 6 ⇒ n = 6 + 1 = 7
Number of terms = 7
Question 6.
In a G.P. the 9th term is 32805 and 6th term is 1215. Find the 12th term.
Given, 9th term = 32805
a. rn-1 = 12187
t9 = 32805 [tn = arn-1]
a.r8 = 32805 …..(1)
6th term = 1215
a.r5 = 1215 …..(2)
Divide (1) by (2)
ar8ar5 = 328051215 ⇒ r3 = 6561243
= 218781 = 72927 = 2439 = 813
r3 = 27 ⇒ r3 = 33
r = 3
Substitute the value of r = 3 in (2)
a. 35 = 1215
a × 243 = 1215
a = 1215243 = 5
Here a = 5, r = 3, n = 12
t12 = 5 × 3(12-1)
= 5 × 311
∴ 12th term of a G.P. = 5 × 311
Question 7.
Find the 10th term of a G.P. whose 8th term is 768 and the common ratio is 2.
Solution:
t8 = 768 = ar7
r = 2
t10 = ar9 = ar7 × r × r
= 768 × 2 × 2 = 3072
Question 8.
If a, b, c are in A.P. then show that 3a, 3b, 3c are in G.P.
a, b, c are in A.P.
t2 – t1 = t3 – t2
b – a = c – b
2b = a + c …..(1)
3a, 3b, 3c are in G.P.
From (1) and (2) we get
3a, 3b, 3c are in G.P.
Question 9.
In a G.P. the product of three consecutive terms is 27 and the sum of the product of two terms taken at a time is 572. Find the three terms.
Let the three terms of the G.P. be ar, a, ar
Product of three terms = 27
ar × a × ar = 27
a3 = 27 ⇒ a3 = 33
a = 3
Sum of the product of two terms taken at a time is 572
6r2 – 13r + 6 = 0
6r2 – 9r – 4r + 6 = 0
3r (2r – 3) -2(2r – 3) = 0
(2r – 3) (3r – 2) = 0
2r – 3 = 0 or 3r – 2 = 0
2r = 3 (or) 3r – 2 = 0
r = 32 (or) r = 23
∴ The three terms are 2, 3 and 92 or 92, 3 and 2
Question 10.
A man joined a company as Assistant Manager. The company gave him a starting salary of ₹60,000 and agreed to increase his salary 5% annually. What will be his salary after 5 years?
Starting salary (a) = ₹ 60000
Increased salary = 5% of starting salary
= 5100 × 60000
= ₹ 3000
Starting salary for the 2nd year = 60000 + 3000
= ₹ 63000
Year increase = 5% of 63000
= 5100 × 63000
= ₹ 3150
Starting salary for the 3rd year = 63000 + 3150
= ₹ 66150
60000, 63000, 66150,…. form a G.P.
a = 60000; r = 6300060000 = 6360 = 2120
tn = ann-1
t5 = (60000) (2120)4
= 60000 × 2120 × 2120 × 2120 × 2120
= 6×21×21×21×212×2×2×2
= 72930.38
5% increase = 5100 × 72930.38
= ₹ 3646.51
Salary after 5 years = ₹ 72930.38 + 3646.51
= ₹ 76576.90
= ₹ 76577
Question 11.
Sivamani is attending an interview for a job and the company gave two offers to him.
Offer A: ₹20,000 to start with followed by a guaranteed annual increase of 6% for the first 5 years.
Offer B: ₹22,000 to start with followed by a guaranteed annual increase of 3% for the first 5 years.
What is his salary in the 4th year with respect to the offers A and B?
Starting salary (a) = ₹ 20,000
Annual increase = 6% of 20000
= 5100 × 20000
= ₹ 1200
Salary for the 2nd year = ₹ 20000 + 1200
= ₹ 21200
Here a = 20,000; r = 2120020000 = 212200 = 106100 = 5350
n = 4 years
tn = arn-1
Salary at the end of 4th year = 23820
For B
Starting salary = ₹ 22000
(a) = 22000
Annual increase = 3% of 22000
= 3100 × 22000
= ₹ 660
Salary for the 2nd year = ₹ 22000 + ₹ 660
= ₹ 22,660
Here a = 22000; r = 2266022000
= 22662200 = 11331100 = 103100
Salary at the end of 4th year = 22000 × (103100)4-1
= 22000 × (103100)3
= 22000 × 103100 × 103100 × 103100
= 24039.99 = 24040
4th year Salary for A = ₹ 23820 and 4th year Salary for B = ₹ 24040
Question 12.
If a, b, c are three consecutive terms of an A.P. and x, y, z are three consecutive terms of a G.P. then prove that xb-c × yc-a × za-b = 1
a, b, c are three consecutive terms of an A.P
∴ a = a, b = a + dand c = a + 2d respectively ….(1)
x, y, z are three consecutive terms of a G.P
∴ x = x, y = xr, z = xr2 respective ……(2)
L.H.S = xb-c × yc-a × za-b ( Substitute the values from 1 and 2 we get)
L.H.S = R.H.S
Hence it is proved
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.6,
Question 1.
Find the sum of the following
(i) 3, 7, 11,… up to 40 terms.
3,7,11,… up to 40 terms
First term (a) = 3
Common difference (d) = 7 – 3 = 4
Number of terms (n) = 40
Sn = n2 [2a + (n – 1) d]
S40 = 402 [6 + 39 × 4] = 20 [6 + 156]
= 20 × 162
S40 = 3240
(ii) 102,97, 92,… up to 27 terms.
Here a = 102, d = 97 – 102 = -5
n = 27
Sn = n2 [2a + (n – 1)d]
S27 = 272 [2(102) + 26(-5)]
= 272 [204 – 130]
= 272 × 74
= 27 × 37 = 999
S27 = 999
(iii) 6 + 13 + 20 + …. + 97
Here a = 6, d = 13 – 6 = 7, l = 97
= 97−67 + 1
= 917 + 1 =
13 + 1 = 14
Sn = n2 (a + l)
Sn = 142 (a + l)
Sn = 142 (6 + 97)
= 7 × 103
Sn = 721
Question 2.
How many consecutive odd integers beginning with 5 will sum to 480?
First term (a) = 5
Common difference (d) = 2
(consecutive odd integer)
Sn = 480
n2 [2a + (n-1) d] = 480
n2 [10 + (n-1)2] = 480
n + 24 = 0 or n – 20 = 0
n = -24 or n = 20
[number of terms cannot be negative]
∴ Number of consecutive odd integers is 20
Question 3.
Find the sum of first 28 terms of an A.P. whose nth term is 4n – 3.
Solution:
n = 28
tn = 4n – 3
t1 = 4 × 1 – 3 = 1
t2 = 4 × 2 – 3 = 5
t28 = 4 × 28 – 3
= 112 – 3 = 109
∴ a = 1, d = t2 – t1 = 5 – 1 = 4
l = 109.
Sn = n2 (2a+(n – 1)d)
S28 = 282 (2 × 1 + 27 × 4)
= 14(2 + 108)
= 14 × 110
= 1540
Question 4.
The sum of first n terms of a certain series is given as 2n2 – 3n . Show that the series is an A.P.
Let tn be nth term of an A.P.
tn = Sn – Sn-1
= 2n2 – 3n – [2(n – 1)2 – 3(n – 1)]
= 2n2 – 3n – [2(n2 – 2n + 1) – 3n + 3]
= 2n2 – 3n – [2n2 – 4n + 2 – 3n + 3]
= 2n2 – 3n – [2n2 – 7n + 5]
= 2n2 – 3n – 2n2 + 7n – 5
tn = 4n – 5
t1 = 4(1) – 5 = 4 – 5 = -1
t2 = 4(2) -5 = 8 – 5 = 3
t3 = 4(3) – 5 = 12 – 5 = 7
t4 = 4(4) – 5 = 16 – 5 = 11
The A.P. is -1, 3, 7, 11,….
The common difference is 4
∴ The series is an A.P.
Question 5.
The 104th term and 4th term of an A.P are 125 and 0. Find the sum of first 35 terms?
104th term of an A.P = 125
t104 = 125
[tn = a + (n – 1) d]
a + 103d = 125 …..(1)
4th term = 0
t4 = 0
a + 3d = 0 …..(2)
Sum of 35 terms = 612.5
Question 6.
Find the sum of ail odd positive integers less than 450.
Sum of odd positive integer less than 450
1 + 3 + 5 + …. 449
Here a = 1, d = 3 – 1 = 2,l = 449
Aliter: Sum of all the positive odd intergers
= n2
= 225 × 225
= 50625
Sum of the odd integers less than 450
= 50625
Question 7.
Find the sum of all natural numbers between 602 and 902 which are not divisible by 4?
First find the sum of all the natural’s number between 602 and 902
Here a = 603, d = 1, l = 901
Find the sum of all the numbers between 602 and 902 which are divisible by 4.
Here a = 604; l = 900; d = 4
Sum of the numbers which are not divisible
by 4 = Sn1 – Sn2
= 224848 – 56400
= 168448
Sum of the numbers = 168448
Question 8.
Raghu wish to buy a laptop. He can buy it by paying ₹40,000 cash or by giving it in 10 installments as ₹4800 in the first month, ₹4750 in the second month, ₹4700 in the third month and so on. If he pays the money in this fashion, find
(i) total amount paid in 10 installments.
(ii) how much extra amount that he has to pay than the cost?
Solution:
4800 + 4750 + 4700 + … 10 terms
Here a = 4800
(i) d = t2 – t1 = 4750 – 4800 = -50
n = 10
Sn = n2 (2a + (n – 1)d)
S10 = 102 (2 × 4800 + 9 × -50)
= 5 (9600 – 450)
= 5 × 9150 = 45750
Total amount paid in 10 installments = ₹ 45750.
(ii) The extra amount he pays in installments
= ₹ 45750 – ₹ 40,000
= ₹ 5750
Question 9.
A man repays a loan of ₹ 65,000 by paying ₹ 400 in the first month and then increasing the payment by ₹ 300 every month. How long will it take for him to clear the loan?
(i) Total loan amount = ₹ 65,000
Sn = 65,000
First month payment (a) = 400
Every month increasing ₹ 300
d = 300
Sn = n2 [2a + (n-1)d]
65000 = n2 [2(400) + (n – 1)300]
130000 = n [800 + 300n – 300]
= n [500 + 300n]
13000 = 500n + 300n2
Dividing by (100)
Number of installments will not be negative
∴ Time taken to pay the loan is 20 months.
Question 10.
A brick staircase has a total of 30 steps. The bottom step requires 100 bricks. Each successive step requires two bricks less than the previous step.
(i) How many bricks are required for the top most step?
(ii) How many bricks are required to build the stair case?
Solution:
100 + 98 + 96 + 94 + … 30 steps.
Here
a = 100
d = -2
n = 30
∴ Sn = n2 (2a + (n – 1)d)
S30 = 302 (2 × 100 + 29 × -2)
= 15(200 – 58)
= 15 × 142
= 2130
t30 = a + (n – 1)d
= 100 + 29 × -2
= 100 – 58
= 42
(i) No. of bricks required for the top step are 42.
(ii) No. of bricks required to build the stair case are 2130.
Question 11.
If S1, S2 , S3, ….Sm are the sums of n terms of m A.P.,s whose first terms are 1,2, 3…… m and whose common differences are 1,3,5,…. (2m – 1) respectively, then show that (S1 + S2 + S3 + ……. + Sm) = 12 mn(mn + 1)
First terms of an A.P are 1, 2, 3,…. m
The common difference are 1, 3, 5,….
(2m – 1)
By adding (1) (2) (3) we get
S1 + S2 + S3 + …… + Sm = n2 (n + 1) + n2 (3n + 1) + n2 (5n + 1) + ….. + n2 [n(2m – 1 + 1)]
= n2 [n + 1 + 3n + 1 + 5n + 1 ……. + n (2m – 1) + m)]
= n2 [n + 3n + 5n + ……. n(2m – 1) + m]
= n2 [n (1 + 3 + 5 + ……(2m – 1)) + m
= n2 [n(m2) (2m) + m]
= n2 [nm2 + m]
S1 + S2 + S3 + ……….. + Sm = mn2 [mn + 1]
Hint:
1 + 3 + 5 + ……. + 2m – 1
Sn = n2 (a + l)
= m2 (1 + 2m -1)
= m2 (2m)
Question 12.
Find the sum
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.5,
Question 1.
Check whether the following sequences are in A.P.?
(i) a – 3, a – 5, a – 7,…
a – 3, a – 5, a – 7…….
t2 – t1 = a – 5 – (a – 3)
= a – 5 – a + 3
= -2
t3 – t2 = a – 7 – (a – 5)
= a – 7 – a + 5
= -2
t2 – t1 = t3 – t2
(common difference is same)
The sequence is in A.P.
(ii) 12, 13, 14, 15, ……….
t2 – t1 = 13 – 12 = 2−36 = −16
t3 – t2 = 14 – 13 = 3−412 = −112
t2 – t1 ≠ t3 – t2
The sequence is not in A.P.
(iii) 9, 13, 17, 21, 25,…
t2 – t1 = 13 – 9 = 4
t3 – t2 = 17 – 13 = 4
t4 – t3 = 21 – 17 = 4
t5 – t4 = 25 – 21 = 4
Common difference are equal
∴ The sequence is in A.P.
(iv) −13, 0, 13, 23
t2 – t1 = 0 – (-13)
= 0 + 13 = 13
t3 – t2 = 13 – 0 = 13
t2 – t1 = t3 – t2
The sequence is in A.P.
(v) 1,-1, 1,-1, 1, -1, …
t2 – t1 = -1 – 1 = -2
t3 – t2 = 1 – (-1) = 1 + 1 = 2
t4 – t3 = -1-(1) = – 1 – 1 = – 2
t5 – t4 = 1 – (-1) = 1 + 1 = 2
Common difference are not equal
∴ The sequence is not an A.P.
Question 2.
First term a and common difference d are given below. Find the corresponding A.P. ?
(i) a = 5 ,d = 6
Here a = 5,d = 6
The general form of the A.P is a, a + d, a + 2d, a + 3d….
The A.P. 5, 11, 17, 23 ….
(ii) a = 7, d = -5
The general form of the A.P is a, a + d,
a + 2d, a + 3d… .
The A.P. 7, 2, -3, -8 ….
(iii) a = 34, d = 12
The general form of the A.P is a, a + d, a + 2d, a + 3d….
34,34 + 12,34 + 2(12), 34 + 3 (12)
The A.P. 34, 54, 74, …….
Question 3.
Find the first term and common difference of the Arithmetic Progressions whose nth terms are given below
(i) tn = -3 + 2n
(ii) tn = 4 – 7n
Solution:
(i) a = t1 = -3 + 2(1) = -3 + 2 = -1
d = t2 – t1
Here t2 = -3 + 2(2) = -3 + 4 = 1
∴ d = t2 – t1 = 1 – (-1) = 2
(ii) a = t1 = 4 – 7(1) = 4 – 7 = -3
d = t2 – t1
Here t2 = 4 – 7(2) = 4 – 14 – 10
∴ d = t2 – t1 = 10 – (-3) = -7
Question 4.
Find the 19th term of an A.P. -11, -15, -19,…
First term (a) = -11
Common difference (d) = -15 -(-11)
= -15 + 11 = -4
n = 19
tn = a + (n – 1) d
tn = -11 + 18(-4)
= -11 – 72
t19 = -83
19th term of an A.P. is – 83
Question 5.
Which term of an A.P. 16, 11, 6,1, ……….. is -54?
Solution:
A.P = 16, 11,6, 1, ………..
It is given that
tn = -54
a = 16, d = t2 – t1 = 11 – 16 = -5
∴ tn = a + (n – 1)d
-54 = 16 + (n – 1) (-5)
-54 = 16 – 5n + 5
21 – 5n = -54
-5n = -54 -21
-5n = -75
n = 755 =15
∴ 15th term is -54.
Question 6.
Find the middle term(s) of an A.P. 9, 15, 21, 27, …, 183.
First term (a) = 9
Last term (l) = 183
Common difference (d) = 15 – 9 = 6
= 183−96 + 1
= 1746 + 1
= 29 + 1
= 30
middle term = 15th term of
16th term
tn = a + (n – 1)d
t15 = 9 + 14(6)
= 9 + 84 = 93
t16 = 9 + 15(6)
= 9 + 90 = 99
The middle term is 93 or 99
Question 7.
If nine times ninth term is equal to the fifteen times fifteenth term, show that six times twenty fourth term is zero.
Solution:
Nine times ninth term = Fifteen times fifteenth term
9t9 = 15t15
9(a + 8d) = 5(a + 14d)
9a + 72d = 15a + 210
15a + 210d – 9a – 72d = 0
⇒ 6a + 138 d = 0
⇒ 6(a + 23 d) = 0
⇒ 6(a + (24 – 1)d) = 0
⇒ 6t24 = 0. Hence it is proved.
Question 8.
If 3 + k, 18 – k, 5k + 1 are in A.P. then find k?
3 + k, 18 – k, 5k + 1 are in AP
∴ t2 – t1 = t3 – t2 (common difference is same)
18 – k – (3 + k) = 5k + 1 – (18 – k)
18 – k – 3 – k = 5k + 1 – 18 + k
15 – 2k = 6k – 17
32 = 8k
k = 328 = 4
The value of k = 4
Question 9.
Find x, y and z, given that the numbers x, 10, y, 24, z are in A.P.
Solution:
A.P = x, 10, y, 24, z,…
d = t2 – t1 = 10 – x ………….. (1)
= t3 – t2 = y – 10 ………….. (2)
= t4 – t3 = 24 – y …………. (3)
= t5 – t4 = z – 24 ………….. (4)
(2) and (3)
⇒ y – 10 = 24 – y
2y = 24 + 10 = 34
y = 342 = 17
(1) and (2)
⇒ 10 – x = y – 10
10 – x = 17 – 10 = 7
-x = 7 – 10
-x = -3 ⇒ x = 3
From (3) and (4)
24 – y = z – 24
24 – 17 = z – 24
7 = z – 24
∴ z = 7 + 24 = 31
∴ Solutions x = 3
y = 17
z = 31
Question 10.
In a theatre, there are 20 seats in the front row and 30 rows were allotted. Each successive row contains two additional seats than its front row. How many seats are there in the last row?
Number of seats in the first row
(a) = 20
∴ t1 = 20
Number of seats in the second row
(t2) = 20 + 2
= 22
Number of seats in the third row
(t3) = 22 + 2
= 24
Here a = 20 ; d = 2
Number of rows
(n) = 30
tn = a + (n – 1)d
t30 = 20 + 29(2)
= 20 + 58
t30 = 78
Number of seats in the last row is 78
Question 11.
The sum of three consecutive terms that are in A.P. is 27 and their product is 288. Find the three terms.
Solution:
Let the three consecutive terms be a – d, a, a + d
Their sum = a – d + a + a + d = 27
3a = 27
a = 273 = 9
Their product = (a – d)(a)(a + d) = 288
= 9(a2 – d2) = 288
⇒ 9(9 – d2) = 288
⇒ 9(81 – d2) = 288
81 – d2 = 32
-d2 = 32 – 81
d2 = 49
⇒ d = ± 7
∴ The three terms are if a = 9, d = 7
a – d, a , a + d = 9 – 7, 9 + 7
A.P. = 2, 9, 16
if a = 9, d = -7
A.P. = 9 – (-7), 9, 9 + (-7)
= 16, 9, 2
Question 12.
The ratio of 6th and 8th term of an A.P is 7:9. Find the ratio of 9th term to 13th term.
Given : t6 : t8 = 7 : 9 (using tn = a + (n – 1)d
a + 5d : a + 7d = 7 : 9
9 (a + 5 d) = 7 (a + 7d)
9a + 45 d = 7a + 49d
9a – 7a = 49d – 45d
2a = 4d
a = 2d
To find t9 : t13
t9 : t13 = a + 8d : a + 12d
= 2d + 8d : 2d + 12d
= 10d : 14d
= 5 : 7
∴ t9 : t13 = 5 : 7
Question 13.
In a winter season, the temperature of Ooty from Monday to Friday to be in A.P. The sum of temperatures from Monday to Wednesday is 0° C and the sum of the temperatures from Wednesday to Friday is 18° C. Find the temperature on each of the five days.
Let the five days temperature be
(a – 2d), (a – d), a,(a + d) and (a + 2d)
Sum of first three days temperature = 0
a – 2d + a – d + a = 0
3a – 3d = 0
a – d = 0 …..(1)
Sum of the last three days temperature = 18°C
a + a + d + a + 2d = 18
3a + 3d = 18
(÷ by 3) ⇒ a + d = 6 ……(2)
Substitute to value of a = 3 in (2)
d = 3
The temperature in 5 days are
(3 – 6), (3 – 3), 3, (3 + 3) and (3 + 6)
-3°C, 0°C, 3°C, 6°C, 9°C
Question 14.
Priya earned ₹15,000 in the first month. Thereafter her salary increases by ₹1500 per year. Her expenses are ₹13,000 during the first year and the expenses increases by ₹900 per year. How long will it take her to save ₹20,000 per month.
Tabulate the given table
Monthly savings form an A.P.
2000, 2600, 3200 …..
a = 2000; d = 2600 – 2000 = 600
Given tn = 20,000
tn = a + (n – 1) d
20000 = 2000 + (n – 1) 600
20000 = 2000 + 600n – 600
= 1400 + 600n
20000 – 1400 = 600n
18600 = 600n
n = 18600600 = 31
He will take 31 years to save ₹ 20,000 per month.
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.4,
Question 1.
Find the next three terms of the following sequence.
(i) 8, 24, 72,…
(ii) 5, 1, -3, …
(iii) 14, 29, 316
(i) 216, 648, 1944 (This sequence is multiple of 3)
Next three terms are 216, 648, 1944
(ii) Next three terms are -7, -11, -15
(iii) Next three terms are 425,536 and 649
[using n(n+1)2]
Question 2.
Find the first four terms of the sequences whose nth terms are given by
(i) an = n3 – 2
(ii) an = (-1)n+1 n(n+1)
(iii) an = 2n2 – 6
Solution:
t= a= n-2
(i) a1 = 13 – 2 = 1 – 2 – 1
a2 = 23 – 2 = 8 – 2 = 6
a3 = 33 – 2 = 27 – 2 = 25
a4 = 43 – 2 = 64 – 2 = 62
∴ The first four terms are -1, 6, 25, 62, ……….
(ii) an = (-1)n+1 n(n + 1)
a1 = (-1)1+1 (1) (1 +1)
= (-1)2 (1) (2) = 2
a2 = (-1)2+1 (2) (2 + 1)
= (-1)3 (2) (3)= -6
a3 = (-1)3+1 (3) (3 + 1)
= (-1)4 (3) (4) = 12
a4 = (-1)4+1 (4) (4 + 1)
= (-1)5 (4) (5) = -20
∴ The first four terms are 2, -6, 12, -20,…
(iii) an = 2n2 – 6
a1 = 2(1)2 – 6 = 2 – 6 = -4
a2 = 2(2)2 – 6 = 8 – 6 = 2
a3 = 2(3)2 – 6 = 18 – 6 = 12
a4 = 2(4)2 – 6 = 32 – 6 = 26
∴ The first four terms are -4, 2, 12, 26, …
Question 3.
Find the nth term of the following sequences
(i) 2, 5, 10, 17, ……
(12 + 1);(22 + 1),(32 + 1),(42 + 1)….
nth term is n2 + 1
an = n2 + 1
(ii) 0,12,23 ……
(1−11), (2−12), (3−13) …..
nth term is n−1n
an = n−1n
(iii) 3,8,13,18,…….
[5(1) -2], [5(2) – 2], [5(3) – 2], [5(4) – 2] ….
The nth term is 5n – 2
an = 5n – 2
Question 4.
Find the indicated terms of the sequences whose nth terms are given by
(i) an = 5nn+2 ; a6 and a13
an = 5nn+2
a6 = 5(6)6+2 = 308 = 154
a13 = 5(13)13+2 = 5×1315 = 133
a6 = 154, a13 = 133
(ii) an = – (n2 – 4); a4 and a11
an = -(n2 – 4)
a4 = -(42 – 4)
= – (16 – 4)
= -12
a11 = -(112 – 4)
= – (121 – 4)
= – 117
a4 = -12 and a11 = -117
Question 5.
Find a8 and a15 whose nth term is an
Question 6.
If a1 = 1, a2 = 1 and an = 2an-1 + an-2, n > 3, n ∈ N, then find the first six terms of the sequence.
Solution:
a1 = 1, a2 = 1, an = 2an-1 + an-2
a3 = 2a(3-1) + a(3-2)
= 2a2 + a1
= 2 × 1 + 1 = 3
a4 = 2a(4-1) + a(4-2)
= 2a3 + a2
= 2 × 3 + 1 = 7
a5 = 2a(5-1) + a(5-2)
= 2a4 + a3
= 2 × 7 + 3 = 17
a6 = 2a(6-1) + a(6-2)
= 2a5 + a4
= 2 × 17 + 7
= 34 + 7
= 41
∴ The first six terms of the sequence are 1, 1, 3, 7, 17, 41 ………..
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.3,
Question 1.
Find the least positive value of x such that
(i) 71 = x (mod 8)
71 = 7 (mod 8)
∴ The value of x = 7
(ii) 78 + x = 3 (mod 5)
78 + x – 3 = 5n (n is any integer)
75 + x = 5n
(Let us take x = 5)
75 + 5 = 80 (80 is a multiple of 5)
∴ The least value of x is 5
(iii) 89 = (x + 3) (mod 4)
89 – (x + 3) = 4n
(n may be any integer)
89 – x – 3 = 4n
89 – x = 4n
86 – x is a multiple of 4
(84 is a multiple of 4)
86 – 2 = 4n
84 = 4n
The value of x is 2
(iv) 96 = x7 (mod 5)
96 – x7 = 5n (n may be any integer)
672 – x = 35n (multiple of 35 is 665)
672 – 7 = 665
∴ The value of x = 7
(v) 5x = 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = 6n+45
Substitute the value of n as 1, 6, 11, 16 …. as n values in x = 6n+45 which is divisible by 5.
2, 8, 14, 20,…………
The least positive value is 2.
Question 2.
If x is congruent to 13 modulo 17 then 7x -3 is congruent to which number modulo 17?
Solution:
x ≡ 13 (mod 17)
Let p be the required number …………. (1)
7x – 3 ≡ p (mod 17) ………….. (2)
From (1),
x – 13 = 17n for some integer M.
x – 13 is a multiple of 17.
x must be 30.
∴ 30 – 13 = 17
which is a multiple of 17.
From (2),
7 × 30 – 3 ≡ p (mod 17)
210 – 3 ≡ p (mod 17)
207 ≡ p (mod 17)
207 ≡ 3 (mod 17)
∴ P ≡ 3
Question 3.
Solve 5x ≡ 4 (mod 6)
5x – 4 = 6n (n may be any integer)
5x = 6n + 4
x = 6n+45
The value of n 1, 6, 11, 16 ……..
∴ The value of x is 2, 8, 14, 20 …………..
Question 4.
Solve 3x – 2 = 0 (mod 11)
Given 3x – 2 = 0(mod 11)
3x – 2 = 11n (n may be any integer)
3x = 2 + 11n
x = 11n+23
Substitute the value of n = 2, 5, 8, 11 ….
When n ≡ 2 ⇒ x = 22+23 = 243 = 8
When n = 5 ⇒ x = 55+23 = 573 = 19
When n = 8 ⇒ x = 88+23 = 903 = 30
When n = 11 ⇒ x = 121+23 = 1233 = 41
∴ The value of x is 8, 19, 30,41
Question 5.
What is the time 100 hours after 7 a.m.?
100 ≡ x (mod 12) Note: In a clock every 12 hours
100 ≡ 4 (mod 12) the numbers repeats.
The time repeat after 7 am is 7 + 4 = 11 o’ clock (or) 11 am.
Question 6.
What is time 15 hours before 11 p.m.?
Solution:
15 ≡ x (mod 12)
15 – x = 12n
15 – x is a multiple of 12 x must be 3.
∴ The time 15 hrs before 11 O’clock is 11 – 3 = 8 O’ clock i.e. 8 p.m
Question 7.
Today is Tuesday. My uncle will come after 45 days. In which day my uncle will be coming?
Number of days in a week = 7
45 ≡ x (mod 7)
45 ≡ 3 (mod 7)
The value of x must be 3.
Three days after tuesday is friday uncle will come on friday.
Question 8.
Prove that 2n + 6 × 9n is always divisible by 7 for any positive integer n.
Solution:
21 + 6 × 91 = 2 + 54 = 56 is divisible by 7
When n = k,
2k + 6 × 9k = 7 m [where m is a scalar]
⇒ 6 × 9k = 7 m – 2k …………. (1)
Let us prove for n = k + 1
Consider 2k+1 + 6 × 9k+1 = 2k+1 + 6 × 9k × 9
= 2k+1 + (7m – 2k)9 (using (1))
= 2k+1 + 63m – 9.2k = 63m + 2k.21 – 9.2k
= 63m – 2k (9 – 2) = 63m – 7.2k
= 7 (9m – 2k) which is divisible by 7
∴ 2n + 6 × 9n is divisible by 7 for any positive integer n
Question 9.
Find the remainder when 281 is divided by 17?
281 ≡ x(mod 17)
240 × 240 × 21 ≡ x(mod 17)
(24)10 × (24)10 × 21 ≡ x(mod 17)
(16)10 × (16)10 × 21 ≡ x(mod 17)
(162)5 × (162)5 × 21 ≡ x(mod 17)
= 1 × 1 × 2 (mod 17)
[(16)2 = 256 = 1 (mod 17)]
= 2 (mod 17)
281 = 2(mod 17)
∴ x = 2
The remainder is 2
Question 10.
The duration of flight travel from Chennai to London through British Airlines is approximately 11 hours. The airplane begins its journey on Sunday at 23:30 hours. If the time at Chennai is four and half hours ahead to that of London’s time, then find the time at London, when will the flight lands at London Airport?
Duration of the flight time = 11 hours
(Chennai to London)
Starting time on Sunday = 23 : 30 hour
Time difference is 4 12 horns ahead to london
The time to reach London airport = (10.30 – 4.30)
= 6 am
The first reach the london airport next day (monday) at 6 am
Tamilnadu, Samacheer Kalvi, 10th, Maths, Solutions, Chapter 2, Numbers and Sequences, Ex 2.2,
Question 1.
For what values of natural number n, 4th can end with the digit 6?
4n = (22)n = 22n
= 2n × 2n
2 is a factor of 4n
∴ 4n is always even.
Question 2.
If m, n are natural numbers, for what values of m, does 2n × 5n ends in 5?
Solution:
2n × 5m
2n is always even for all values of n.
5m is always odd and ends with 5 for all values of m.
But 2n × 5m is always even and ends in 0.
∴ 2n × 5m cannot end with the digit 5 for any values of m. No value of m will satisfy 2n × 5m ends in 5.
Question 3.
Find the H.C.F. of 252525 and 363636.
To find the HCF of 252525 and 363636 by using Euclid’s Division algorithm.
363636 = 252525 × 1 + 111111
The remainder 111111 ≠ 0
By division of Euclid’s algorithm
252525 = 111111 × 2 + 30303
The remainder 30303 ≠ 0
Again by division of Euclid’s algorithm
111111 = 30303 × 3 + 20202
The remainder 20202 ≠ 0
Again by division of Euclid’s algorithm.
30303 = 20202 + 10101
The remainder 10101 ≠ 0
Again by division of Euclid’s algorithm.
20202 = 10101 × 2 + 0
The remainder is 0
∴ The H.C.F. is 10101
Question 4.
If 13824 = 2a × 3b then find a and b?
Using factor tree method factorise 13824
13824 = 29 × 33
Given 13824 = 2a × 3b
Compare we get a = 9 and b = 3
Aliter:
13824 = 29 × 33
Compare with
13824 = 2a × 3b
The value of a = 9 b = 3
Question 5.
If p1x1 × p2x2 × p3x3 × p4x4 = 113400 where p1 p2, p3, p4 are primes in ascending order and x1, x2, x3, x4, are integers, find the value of p1,p2,p3,p4 and x1,x2,x3,x4.
Given 113400 = p1x1 × p2x2 × p3x3 × p4x4
Using tree method factorize 113400
113400 = 23 × 34 × 52 × 7
compare with
113400 = p1x1 × p2x2 × p3x3 × p4x4
P1 = 2, x1 = 3
P2 = 3, x2 = 4
P3 = 5, x3 = 2
P4 = 7, x4 = 1
Question 6.
Find the L.C.M. and H.C.F. of 408 and 170 by applying the fundamental theorem of Arithmetic.
Factorise 408 and 170 by factor tree method
Question 7.
Find the greatest number consisting of 6 digits which is exactly divisible by 24,15,36?
The greatest number of 6 digits is 999999.
The greatest number must be divisible by L.C.M. of 24, 15 and 36
24 = 23 × 3
15 = 3 × 5
36 = 22 × 32
L.C.M = 23 × 32 × 5
= 360
To find the greatest number 999999 must be subtracted by the remainder when 999999 is divided by 360
The greatest number in 6 digits = 999999 – 279
= 999720
Question 8.
What is the smallest number that when divided by three numbers such as 35, 56 and 91 leaves remainder 7 in each case?
Solution:
35 = 5 × 7
56 = 2 × 2 × 2 × 7
91 = 7 × 13
LCM of 35, 56, 91 = 5 × 7 × 2 × 2 × 2 × 13 = 3640
∴ Required number = 3647 which leaves remainder 7 in each case.
Question 9.
Find the least number that is divisible by the first ten natural numbers?
Find the L.C.M of first 10 natural numbers
The least number is 2520
Modular Arithmetic
Two integers “a” and “b” are congruence modulo n if they differ by an integer multiple of n. That b – a = kn for some integer k. This can be written as a = b (mod n).
Euclid’s Division Lemma and Modular Arithmetic
Let m and n be integers, where m is positive. By Euclid’s division Lemma we can write n = mq + r where 0 < r < m and q is an integer.
This n = r (mod m)
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gamomaniea1
2021-11-20
Solve the definite integral.
${\int }_{1}^{2}\left(\frac{3{y}^{2}+y-1}{{y}^{2}}\right)$
Steven Arredondo
Step 1
We have to solve the given definite integral:
${\int }_{1}^{2}\left(\frac{3{y}^{2}+y-1}{{y}^{2}}\right)dy$
Rewriting the integral and solving the given integral,
${\int }_{1}^{2}\left(\frac{3{y}^{2}+y-1}{{y}^{2}}\right)dy={\int }_{1}^{2}\left(\frac{3{y}^{2}}{{y}^{2}}+\frac{y}{{y}^{2}}-\frac{1}{{y}^{2}}\right)dy$
$={\int }_{1}^{2}\left(3+\frac{1}{y}-{y}^{-2}\right)dy$
$={\left[3y+\mathrm{ln}\left(y\right)-\frac{{y}^{-2+1}}{-2+1}\right]}_{1}^{2}$
$={\left[3y+\mathrm{ln}\left(y\right)+{y}^{-1}\right]}_{1}^{2}$
$={\left[3y+\mathrm{ln}\left(y\right)+\frac{1}{y}\right]}_{1}^{2}$
$=\left[3×2+\mathrm{ln}\left(2\right)+\frac{1}{2}-3×1-\mathrm{ln}\left(1\right)-1\right]$
$=\left[6+\mathrm{ln}\left(2\right)+\frac{1}{2}-3-0-1\right]$
$=\left[2+\frac{1}{2}+\mathrm{ln}\left(2\right)\right]$
$=\frac{5}{2}+\mathrm{ln}\left(2\right)$
Step 2
Hence, value of given definite integral is $\frac{5}{2}+\mathrm{ln}\left(2\right)$.
Alicia Washington
Step 1: Simplify $\frac{3{y}^{2}+y-1}{{y}^{2}}\to \frac{1}{y}+\frac{3{y}^{2}-1}{{y}^{2}}$.
${\int }_{1}^{2}\frac{1}{y}+\frac{3{y}^{2}-1}{{y}^{2}}dy$
Step 2: If f(x) is a continuous function from a to b, and if F(x) is its integral, then:
${\int }_{a}^{b}f\left(x\right)dx=F\left(x\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$
Step 3: In this case, $f\left(y\right)=\frac{1}{y}+\frac{3{y}^{2}-1}{{y}^{2}}$. Find its integral.
$\mathrm{ln}y+3y+\frac{1}{y}{\mid }_{1}^{2}$
Step 4: Since $F\left(y\right){\mid }_{a}^{b}=F\left(b\right)-F\left(a\right)$, expand the above into F(2)−F(1):
$\left(\mathrm{ln}2+3×2+\frac{1}{2}\right)-\left(\mathrm{ln}1+3×1+\frac{1}{1}\right)$
Step 5: Simplify.
$\frac{5}{2}+\mathrm{ln}2$
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# Factoring Trinomials bottoms-up method by umsymums38
VIEWS: 182 PAGES: 2
• pg 1
``` Factoring Trinomials:
bottoms-up method
Math 096 feb. 2009 Here are the steps to factor the trinomial ax2 + bx + c using the bottoms-up method. Step 1. Factor out the GCF. This should always be done first when factoring any polynomial. Step 2. Factor the trinomial x2 + bx + a · c Factoring this trinomial should be much easier since it has only x2 instead of ax2 . Step 3. Divide the constant term in each factor by a and reduce. See one of the examples if this doesn’t make sense. Step 4. Flip the bottoms up. That is, if any denominator remains make it a coefficient of x instead. Again, the examples will make this more clear. Step 5. Multiply the GCF by the result from step 4 and you are done. If you did not pull out a GCF then the result from step 4 is the answer.
Examples
1. Factor 6x2 − 17x + 10. Step 1. Factor out the GCF. Not necessary in this one. Step 2. Factor the trinomial x2 + bx + a · c x2 − 17x + 60 = (x − 12)(x − 5) Step 3. Divide the constant term in each factor by a and reduce. (x − 12)(x − 5) becomes (x −
12 )(x 6
− 5 ) = (x − 2)(x − 5 ) 6 6
Step 4. Flip the bottoms up. That is, if any denominator remains make it instead a coefficient of x.
5 (x − 2)(x − 6 ) becomes (x − 2)(6x − 5)
Step 5. Multiply the GCF by the result from step 4 and you are done. We did not have a GCF in step 1, so the answer is (x − 2)(6x − 5).
1
2. Factor 8x2 − 10x − 3. Step 1. Again, not necessary in this one. Step 2. x2 − 10x − 24 = (x − 12)(x + 2) Step 3. (x − 12)(x + 2) becomes (x −
12 )(x 8 1 + 2 ) = (x − 3 )(x + 4 ) 8 2
Step 4. (x − 3 )(x + 1 ) becomes (2x − 3)(4x + 1) 2 4 Step 5. We did not have a GCF in step 1, so the answer is (2x − 3)(4x + 1)
3. Factor 32x2 + 56x + 24. Step 1. The GCF is 8, so factor 8 out of every term to get 32x2 + 56x + 24 = 8(4x2 + 7x + 3) Now we ignore the GCF until the end and focus on factoring 4x2 + 7x + 3 Step 2. x2 + 7x + 12 = (x + 4)(x + 3)
3 Step 3. (x + 4)(x + 3) becomes (x + 4 )(x + 4 ) = (x + 1)(x + 3 ) 4 4
Step 4. (x + 1)(x + 3 ) becomes (x + 1)(4x + 3) 4 Step 5. The GCF is 8, so the answer is 8(x + 1)(4x + 3)
2
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Student Question
# int sqrt(x)/(x-4) dx Use substitution and partial fractions to find the indefinite integral
Indefinite integral are written in the form of int f(x) dx = F(x) +C
where: f(x) as the integrand
F(x) as the anti-derivative function of f(x)
C as the arbitrary constant known as constant of integration
To evaluate the integral problem: int sqrt(x)/(x-4)dx , we may apply u-substitution by letting:
u=sqrt(x) then u^2 =x and 2u du = dx
Plug-in the values, we get:
int sqrt(x)/(x-4)dx=int u/(u^2-4)* 2udu
= int (2u^2)/(u^2-4)du
To simplify, we may apply long division:(2u^2)/(u^2-4) =2 +8/(u^2-4)
To expand 8/(u^2-4) , we may apply partial fraction decomposition.
The pattern on setting up partial fractions will depend on the factors of the denominator. The factored form for the difference of perfect squares: (u^2-4)= (u-2)(u+2) .
For the linear factor (u-2) , we will have partial fraction: A/(u-2) .
For the linear factor (u+2) , we will have partial fraction: B/(u+2) .
The rational expression becomes:
8/(u^2-4) =A/(u-2) +B/(u+2)
Multiply both side by the LCD =(u-2)(u+2) .
(8/(u^2-4)) *(u-2)(u+2)=(A/(u-2) +B/(u+2)) *(u-2)(u+2)
8=A(u+2) +B(u-2)
We apply zero-factor property on (u-2)(u+2) to solve for values we can assign on u.
u-2=0 then u=2
u+2 =0 then u =-2
To solve for A , we plug-in u=2 :
8=A(2+2) +B(2-2)
8 =4A+0
8=4A
8/(4) = (4A)/4
A = 2
To solve for B , we plug-in u=-2 :
8=A(-2+2) +B(-2-2)
8 =0 -4B
8=-4B
8/(-4) = (-4B)/(-4)
B = -2
Plug-in A = 2 and B =-2 , we get the partial fraction decomposition:
8/(u^2-4)=2/(u-2) -2/(u+2)
Then the integral becomes:
int (2u^2)/(u^2-4)du= int [2+8/(u^2-4)]du
=int [2 +2/(u-2) -2/(u+2)]du
Apply the basic integration property: int (u+-v+-w) dx = int (u) dx +- int (v) dx+- int (w) dx .
int [2 +2/(u-2) -2/(u+2)]du =int 2du +int 2/(u-2) du int -2/(u+2)du
For the first integral, we may apply basic integration property: int a dx = ax+C.
int 2 du = 2u
For the second and third integral, we may apply integration formula for logarithm: int 1/u du = ln|u|+C .
int 2/(u-2) du =2ln|u-2|
int 2/(u+2) du =2ln|u+2|
Combining the results, we get:
int (2u^2)/(u^2-4)du = 2u +2ln|u-2| -2ln|u+2| +C
Apply logarithm property: n*ln|x| = ln|x^n| and ln|x| - ln|y| = ln|x/y|
int (2u^2)/(u^2-4)du = 2u + ln|(u-2)^2| - ln|(u+2)^2| +C
= 2u + ln|(u-2)^2/(u+2)^2| +C
Plug-in u =sqrt(x) on 2u + ln|(u-2)^2/(u+2)^2| +C , we get the indefinite integral as:
int sqrt(x)/(x-4)dx =2sqrt(x) +ln|(sqrt(x)-2)^2/(sqrt(x)+2)^2| +C
OR 2sqrt(x) +ln|(x-4sqrt(x)+4)/(x+4sqrt(x)+4)| +C
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# How many ways can 10 different colored beads be treated on a string?
Contents
Answer: This is called a cyclic permutation. The formula for this is simply (n-1)!/2, since all the beads are identical. Hence, the answer is 9!/2 = 362880/2 = 181440.
## How many necklaces can you make with 10 beads of colors?
This is easy: count all permutations of 10 beads, 10!, then divide by 20 because we counted each permutation 10 times due to rotation, and counted each of these twice because you can flip the necklace over. Thus the answer is 10!/20 = 181440.
## How many ways can 6 differently Coloured beads be threaded on a string?
Assuming that the beads are different, the first bead can be picked in 6 ways. Then the second bead can be picked in 5 ways. And the third bead can be picked in 4 ways, etc. Multiplying these together, we get 6*5*4*3*2*1 = 720 ways.
THIS IS EXCITING: Frequent question: What is 22 count cross stitch fabric?
## How many ways can 8 beads of different Colour be strung on a ring?
2520 Ways 8 beads of different colours be strung as a necklace if can be wear from both side.
## How many ways can 6 beads of different colours form a necklace?
When the necklace is unclasped and laid out with its ends separated, there are 6! = 720 distinct ways (permutations) to arrange the 6 different beads.
## How many ways can 10 different beads?
Answer: This is called a cyclic permutation. The formula for this is simply (n-1)!/2, since all the beads are identical. Hence, the answer is 9!/2 = 362880/2 = 181440.
## How many ways 8 different beads can be arranged to form a necklace?
The number of ways in which 8 different beads be strung on a necklace is. 2500. 2520.
## How many ways can 7 different colored beads be threaded on a string?
It would be 7! = 5040 diffrent necklaces.
## How many arrangements of beads are possible in a bracelet if there are 6 different designs of beads?
Since there are 6! linear arrangements of six distinct beads, the number of distinguishable circular arrangements is 6! 6=5!
## How many ways can 5 keys be put in a ring?
The number of ways of putting 5 different keys in a ring is very similar to that of arranging in a row except. ABCDE and EABCD are the same combinations when placed in a ring. Therefore, The answer is 24.
## How many ways 5 different beads can be arranged to form a necklace?
So, we have to divide 24 by 2. Therefore the total number of different ways of arranging 5 beads is 242=12 .
THIS IS EXCITING: Frequent question: How do you make fur yarn?
## How many necklaces can be formed with 6 white and 5 red beads if each necklace is unique how many can be formed?
5! but correct answer is 21.
## How many different way can 4 keys be arranged on a key ring that has no clasp?
Keep one key fixed in any position in the ring. So, the remaining 4 keys can be arranged in 4! = 24 ways.
|
# Associations in Categorical Data
ID: dusum-ruvuv
Illustrative Mathematics, CC BY 4.0
Subject: Algebra, Algebra 2
Standards: HSS-ID.B.5
17 questions
# Associations in Categorical Data
Classroom:
Due:
Student Name:
Date Submitted:
##### Cake or Pie
The table displays the dessert preference and dominant hand (left- or right-handed) for a sample of 300 people.
$\begin{array}{|c|c|c|c|} \hline \\[-1em] & \textbf{prefers cake} & \textbf{prefers pie} & \textbf{total} \\[-1em] \\ \hline \\[-1em] \textbf{left-handed} & 10 & 20 & 30 \\[-1em] \\ \hline \\[-1em] \textbf{right-handed} & 90 & 180 & 270 \\[-1em] \\ \hline \\[-1em] \textbf{total} & 100 & 200 & 300 \\[-1em] \\ \hline \end{array}$
For each of the calculations, describe the interpretation of the percentage in terms of the situation.
1) 10% from $\frac{10}{100} = 0.1$
2) 67% from $\frac{180}{270} \approx 0.67$
3) 30% from $\frac{90}{300} = 0.3$
##### Associations in Categorical Data
The two-way table displays data about 55 different locations. Scientists have a list of possible chemicals that may influence the health of the coral. They first look at how nitrate concentration might be related to coral health. The table displays the health of the coral (healthy or unhealthy) and nitrate concentration (low or high).
$\begin{array}{|l|c|c|c|} \hline \\[-1em] & \textbf{low nitrate concentration} & \textbf{high nitrate concentration} & \textbf{total}\\[-1em] \\ \hline \\[-1em] \textbf{healthy} & 20 & 5 & 25\\[-1em] \\ \hline \\[-1em] \textbf{unhealthy} & 8 & 22 & 30\\[-1em] \\ \hline \\[-1em] \textbf{total} & 28 & 27 & 55\\[-1em] \\ \hline \end{array}$
Complete the two-way relative frequency table for the data in the two-way table in which the relative frequencies are based on the total for each column. Find the values that belong in the labeled cells in the table, rounding to the nearest whole percent.
$\begin{array}{|c|c|c|} \hline \\[-1em] & \textbf{low nitrate concentration} & \textbf{high nitrate concentration} \\[-1em] \\ \hline \\[-1em] \textbf{healthy} & \text{A} & \text{B} \\[-1em] \\ \hline \\[-1em] \textbf{unhealthy} & \text{C} & \text{D} \\[-1em] \\ \hline \\[-1em] \textbf{total} & 100\% & 100\% \\[-1em] \\ \hline \end{array}$
4) Cell A
5) Cell B
6) Cell C
7) Cell D
8) When there is a low nitrate concentration, which had a higher relative frequency, healthy or unhealthy coral?
a)
Healthy coral
b)
Unhealthy coral
9) When there is a high nitrate concentration, is there a higher relative frequency of healthy or unhealthy coral?
a)
Healthy coral
b)
Unhealthy coral
10) Based on this data, is there a possible $\textbf{association}$ between coral health and the level of nitrate concentration?
True or false? Write below.
The scientists next look at how silicon dioxide concentration might be related to coral health. The relative frequencies based on the total for each column are shown in the table.
$\begin{array}{|l|c|c|} \hline \\[-1em] & \textbf{low silicon dioxide concentration} & \textbf{high silicon dioxide concentration} \\[-1em] \\ \hline \\[-1em] \textbf{healthy} & 44\% & 46\% \\[-1em] \\ \hline \\[-1em] \textbf{unhealthy} & 56\% & 54\% \\[-1em] \\ \hline \\[-1em] \textbf{total} & 100\% & 100\% \\[-1em] \\ \hline \end{array}$
12) Based on this data, is there a possible association between coral health and the level of silicon dioxide concentration?
True or false? Write below.
Jada surveyed 300 people from various age groups about their shoe preference. The two-way table summarizes the results of the survey.
$\begin{array}{|l|c|c|c|c|} \hline \\[-1em] & \textbf{prefers sneakers} & \textbf{prefers sneakers} & \textbf{prefers shoes that} & \textbf{total} \\ & \textbf{without laces} & \textbf{with laces} & \textbf{are not sneakers} & \\[-1em] \\ \hline \\[-1em] \textbf{4-10 years old} & 21 & 12 & 3 & 36 \\[-1em] \\ \hline \\[-1em] \textbf{11-17 years old} & 21 & 48 & 39 & 108\\[-1em] \\ \hline \\[-1em] \textbf{18-24 years old} & 15 & 54 & 87 & 156\\[-1em] \\ \hline \\[-1em] \textbf{total} & 57 & 114 & 129 & 300\\[-1em] \\ \hline \end{array}$
Jada concludes that there is a possible association between age and shoe preference.
True or false? Write below.
The two-way table summarizes data on writing utensil preference and the dominant hand for a sample of 100 people.
$\begin{array}{|l|c|c|c|} \hline \\[-1em] & \textbf{left-handed} & \textbf{right-handed} & \textbf{total}\\[-1em] \\ \hline \\[-1em] \textbf{prefers pen} & 7 & 82 & 89\\[-1em] \\ \hline \\[-1em] \textbf{prefers pencil} & 6 & 5 & 11\\[-1em] \\ \hline \\[-1em] \textbf{total} & 13 & 87 & 100\\[-1em] \\ \hline \end{array}$
16) Is there a possible association between dominant hand and writing utensil preference?
True or false? Write below.
|
# Derivative of arcsec (Inverse Secant) With Proof and Graphs
The derivative of the inverse secant function is equal to 1/(|x|√(x2-1)). We can prove this derivative using the Pythagorean theorem and algebra.
In this article, we will learn how to derive the inverse secant function. We will look at some basics, a graphical comparison of the function and its derivative, and some examples.
##### CALCULUS
Relevant for
Learning about the proof and graphs of the derivative of arcsec of x.
See proof
##### CALCULUS
Relevant for
Learning about the proof and graphs of the derivative of arcsec of x.
See proof
## Avoid confusion in using the denotations arcsec(x), sec-1(x), 1 / sec(x) , and secn(x)
It is important that we do not fall into the possible confusion that we may have when using different denotations $latex \text{arcsec}(x)$, $latex \sec^{-1}{(x)}$, $latex \frac{1 }{\sec{(x)}}$ and $latex \sec^{n}{(x)}$, as it can lead to derivation errors. Summarizing the definition of these symbols, we have
$latex \text{arcsec}(x) = \sec^{-1}{(x)}$
The symbols $latex \text{arcsec}$ and $latex \sec^{-1}$ are used interchangeably when calculating the inverse secant. $latex \text{arcsecant}$ is commonly used as the verbal symbol for the inverse secant function, while $latex \sec^{-1}$ is used as the mathematical symbol for the inverse secant function for a more formal setting.
However, in the case of the denotation $latex \sec^{-1}{(x)}$, we must consider that $latex -1$ is not an algebraic exponent of a secant. The $latex -1$ used for the inverse secant represents that the secant is inverse and not raised to $latex -1$.
Therefore,
$latex \sec^{-1}{(x)} \neq \frac{1}{\sec{(x)}}$
And givens such as $latex \sec^{2}{(x)}$ or $latex \sec^{n}{(x)}$, where n is any algebraic exponent of a non-inverse secant, MUST NOT use the inverse secant formula since in these givens, both the 2 and any exponent n are treated as algebraic exponents of a non-inverse secant.
## Proof of the Derivative of the Inverse Secant Function
In this proof, we will mainly use the concepts of a right triangle, the Pythagorean theorem, the trigonometric function of secant and tangent, and some basic algebra. Just like in the previous figure as a reference sample for a given right triangle, suppose we have that same triangle $latex \Delta ABC$, but this time, let’s change the variables for an easier illustration.
where for every one-unit of a side adjacent to angle y, there is a side $latex \sqrt{x^2-1}$ opposite to angle y and a hypothenuse x.
Using these components of a right-triangle, we can find the angle y by using Cho-Sha-Cao, particularly the secant function by using the hypothenuse x and its adjacent side.
$latex \sec{(\theta)} = \frac{hyp}{adj}$
$latex \sec{(y)} = \frac{x}{1}$
$latex \sec{(y)} = x$
Now, we can implicitly derive this equation by using the derivative of trigonometric function of secant for the left-hand side and power rule for the right-hand side. Doing so, we have
$latex \frac{d}{dx} (\sec{(y)}) = \frac{d}{dx} (x)$
$latex \frac{d}{dx} (\sec{(y)}) = 1$
$latex \frac{dy}{dx} (\sec{(y)}\tan{(y)}) = 1$
$latex \frac{dy}{dx} = \frac{1}{\sec{(y)}\tan{(y)}}$
Getting the tangent of angle y from our given right-triangle, we have
$latex \tan{(y)} = \frac{opp}{adj}$
$latex \tan{(y)} = \frac{\sqrt{x^2-1}}{1}$
$latex \tan{(y)} = \sqrt{x^2-1}$
We can then substitute $latex \sec{(y)}$ and $latex \tan{(y)}$ to the implicit differentiation of $latex \sec{(y)} = x$
$latex \frac{dy}{dx} = \frac{1}{\sec{(y)}\tan{(y)}}$
$latex \frac{dy}{dx} = \frac{1}{(x) \cdot \left(\sqrt{x^2-1}\right)}$
$latex \frac{dy}{dx} = \frac{1}{x\sqrt{x^2-1}}$
Now, since
$latex \sec{(y)} = x$
and
$latex hypothenuse = x$
We know that a negative hypotenuse cannot exist. Therefore, $latex \sec{(y)}$ in this case cannot be negative. That’s why the x multiplicand in the denominator of the derivative of the inverse secant must be considered an absolute value.
$latex \frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}}$
Therefore, algebraically solving for the angle y and getting its derivative, we have
$latex \sec{(y)} = x$
$latex y = \frac{x}{\sec}$
$latex y = \sec^{-1}{(x)}$
$latex \frac{dy}{dx} = \frac{d}{dx} \left( \sec^{-1}{(x)} \right)$
$latex \frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}}$
which is now the derivative formula for the inverse secant of x.
Now, for the derivative of an inverse secant of any function other than x, we may apply the derivative formula of inverse secant together with the chain rule formula. By doing so, we have
$latex \frac{dy}{dx} = \frac{d}{du} \sec^{-1}{(u)} \cdot \frac{d}{dx} (u)$
$latex \frac{dy}{dx} = \frac{1}{|u|\sqrt{u^2-1}} \cdot \frac{d}{dx} (u)$
where $latex u$ is any function other than x.
## Graph of Inverse Secant x VS. The Derivative of Inverse Secant x
The graph of the function
$latex f(x) = \sec^{-1}{(x)}$
is
And as we know by now, by deriving $latex f(x) = \sec^{-1}{(x)}$, we get
$latex f'(x) = \frac{1}{|x|\sqrt{x^2-1}}$
which has the following graph
Illustrating both graphs in one, we have
Using the graphs, it can be seen that the original function $latex f(x) = \sec^{-1}{(x)}$ has a domain of
$latex (-\infty,-1] \cup [1,\infty )$ or all real numbers except $latex -1 < x < 1$
and exists within the range of
$latex [0,\frac{\pi}{2}\big) \cup \big(\frac{\pi}{2},\pi]$ or $latex 0 \leq y \leq \pi$ except $latex \frac{\pi}{2}$
whereas the derivative $latex f'(x) = \frac{1}{|x|\sqrt{x^2-1}}$ has a domain of
$latex (-\infty,-1) \cup (1,\infty)$ or all real numbers except $latex -1 \leq x \leq 1$
and exists within the range of
$latex (0,\infty)$ or $latex y > 0$
## Examples
In the following examples, we will see how to derive compound inverse secant functions.
### EXAMPLE 1
What is the derivative of $latex f(x) = \sec^{-1}(2x)$?
Since we have a composite inverse secant function, we can use the chain rule to derive it.
Therefore, we consider $latex u=2x$ as the inner function, and we have $latex f(u)=\sec^{-1}(u)$ and applying the chain rule, we have:
$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$
$$\frac{dy}{dx}=\frac{1}{|u|\sqrt{u^2-1}} \times 2$$
Then, we substitute $latex u=2x$ back into the function and we have:
$$\frac{dy}{dx}=\frac{2}{|2x|\sqrt{(2x)^2-1}}$$
$$\frac{dy}{dx}=\frac{2}{|2x|\sqrt{4x^2-1}}$$
### EXAMPLE 2
Find the derivative of the function $latex F(x) = \sec^{-1}(x^2-5)$
To use the chain rule, we write the inverse secant function as $latex f (u) = \sec^{-1}(u)$, where $latex u = x^2-5$.
Therefore, we start by finding the derivative of the outer function $latex f(u)$:
$$\frac{d}{du} ( \sec^{-1}(u) ) = \frac{1}{|u|\sqrt{u^2-1}}$$
Now, we calculate the derivative of the inner function $latex g(x)=u=x^2-5$:
$$\frac{d}{dx}(g(x)) = \frac{d}{dx}(x^2-5)$$
$$\frac{d}{dx}(g(x)) = 2x$$
Then, we have to multiply the derivative of the outer function by the derivative of the inner function:
$$\frac{dy}{dx} = \frac{d}{du} (f(u)) \cdot \frac{d}{dx} (g(x))$$
$$\frac{dy}{dx} = \frac{1}{|u|\sqrt{u^2-1}} \cdot 2x$$
Finally, we substitute $latex u$ back in and simplify:
$$\frac{dy}{dx} = \frac{1}{|x^2-5|\sqrt{(x^2-5)^2-1}} \cdot 2x$$
$$\frac{dy}{dx} = \frac{2x}{|x^2-5|\sqrt{(x^2-5)^2-1}}$$
$$F'(x) = \frac{2x}{|x^2-5|\sqrt{x^4-10x^2+24}}$$
### EXAMPLE 3
Find the derivative of $latex f(x) = \sec^{-1}(\sqrt{x})$
In this case, the inner function is $latex u=\sqrt{x}$. Considering that we can write it as $latex u=x^{\frac{1}{2}}$, its derivative is:
$$\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}$$
If we apply the chain rule with $latex f(u)=\sec^{-1}(u)$, we have:
$$\frac{dy}{dx}=\frac{dy}{du} \frac{du}{dx}$$
$$\frac{dy}{dx}=\frac{1}{|u|\sqrt{u^2-1}} \times \frac{1}{2}x^{-\frac{1}{2}}$$
Substituting $latex u=\sqrt{x}$ and simplifying, we have:
$$\frac{dy}{dx}=\frac{1}{|\sqrt{x}|\sqrt{(\sqrt{x})^2-1}} \times \frac{1}{2}x^{-\frac{1}{2}}$$
$$\frac{dy}{dx}=\frac{1}{|\sqrt{x}|\sqrt{x-1}} \times \frac{1}{2}x^{-\frac{1}{2}}$$
$$\frac{dy}{dx}=\frac{1}{|\sqrt{x}|\sqrt{x-1}\sqrt{x}}$$
$$\frac{dy}{dx}=\frac{1}{|\sqrt{x}|\sqrt{x(x-1)}}$$
## Practice of derivatives of compound inverse secant inverse functions
Derivatives inverse secant quiz
You have completed the quiz!
|
# Active Calculus
## Section8.2Geometric Series
Many important sequences are generated by addition. In Preview Activity 8.2.1, we see an example of a sequence that is connected to a sum.
### Preview Activity8.2.1.
Warfarin is an anticoagulant that prevents blood clotting; often it is prescribed to stroke victims in order to help ensure blood flow. The level of warfarin has to reach a certain concentration in the blood in order to be effective.
Suppose warfarin is taken by a particular patient in a 5 mg dose each day. The drug is absorbed by the body and some is excreted from the system between doses. Assume that at the end of a 24 hour period, 8% of the drug remains in the body. Let $$Q(n)$$ be the amount (in mg) of warfarin in the body before the $$(n+1)$$st dose of the drug is administered.
1. Explain why $$Q(1) = 5 \times 0.08$$ mg.
2. Explain why $$Q(2) = (5+Q(1)) \times 0.08$$ mg. Then show that
\begin{equation*} Q(2) = (5 \times 0.08)\left(1+0.08\right) \text{mg}\text{.} \end{equation*}
3. Explain why $$Q(3) = (5+Q(2)) \times 0.08$$ mg. Then show that
\begin{equation*} Q(3) = (5 \times 0.08)\left(1+0.08+0.08^2\right) \text{mg}\text{.} \end{equation*}
4. Explain why $$Q(4) = (5+Q(3)) \times 0.08$$ mg. Then show that
\begin{equation*} Q(4) = (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3\right) \text{mg}\text{.} \end{equation*}
5. There is a pattern that you should see emerging. Use this pattern to find a formula for $$Q(n)\text{,}$$ where $$n$$ is an arbitrary positive integer.
6. Complete Table 8.2.1 with values of $$Q(n)$$ for the provided $$n$$-values (reporting $$Q(n)$$ to 10 decimal places). What appears to be happening to the sequence $$Q(n)$$ as $$n$$ increases?
### Subsection8.2.1Geometric Series
In Preview Activity 8.2.1 we encountered the sum
\begin{equation*} (5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \end{equation*}
for the long-term level of Warfarin in the patient’s system. This sum has the form
$$a+ar+ar^2+ \cdots + ar^{n-1}\tag{8.2.1}$$
where $$a=5 \times 0.08$$ and $$r=0.08\text{.}$$ Such a sum is called a finite geometric series with ratio $$r\text{.}$$
#### Activity8.2.2.
Let $$a$$ and $$r$$ be real numbers (with $$r \ne 1$$) and let
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \end{equation*}
In this activity we will find a shortcut formula for $$S_n$$ that does not involve a sum of $$n$$ terms.
1. Multiply $$S_n$$ by $$r\text{.}$$ What does the resulting sum look like?
2. Subtract $$rS_n$$ from $$S_n$$ and explain why
$$S_n - rS_n = a - ar^n\text{.}\tag{8.2.2}$$
3. Solve equation (8.2.2) for $$S_n$$ to find a simple formula for $$S_n$$ that does not involve adding $$n$$ terms.
Hint.
1. Distribute the factor of $$r\text{.}$$
2. Look for common terms in the two expressions being subtracted.
3. Observe that you can remove a factor of $$S_n$$ from $$S_n - rS_n\text{.}$$
The sum of the terms of a sequence is called a series. We summarize the result of Activity 8.2.2 in the following way.
A finite geometric series $$S_n$$ is a sum of the form
$$S_n = a + ar + ar^2 + \cdots + ar^{n-1}\text{,}\tag{8.2.3}$$
where $$a$$ and $$r$$ are real numbers such that $$r \ne 1\text{.}$$ The finite geometric series $$S_n$$ can be written more simply as
$$S_n = a+ar+ar^2+ \cdots + ar^{n-1} = \frac{a(1-r^n)}{1-r}\text{.}\tag{8.2.4}$$
We now apply Equation (8.2.4) to the example involving Warfarin from Preview Activity 8.2.1. Recall that
\begin{equation*} Q(n)=(5 \times 0.08)\left(1+0.08+0.08^2+0.08^3+ \cdots + 0.08^{n-1}\right) \text{mg}\text{,} \end{equation*}
so $$Q(n)$$ is a geometric series with $$a=5 \times 0.08 = 0.4$$ and $$r = 0.08\text{.}$$ Thus,
\begin{equation*} Q(n) = 0.4\left(\frac{1-0.08^n}{1-0.08}\right) = \frac{1}{2.3} \left(1-0.08^n\right)\text{.} \end{equation*}
Notice that as $$n$$ goes to infinity, the value of $$0.08^n$$ goes to 0. So,
\begin{equation*} \lim_{n \to \infty} Q(n) = \lim_{n \to \infty} \frac{1}{2.3} \left(1-0.08^n\right) = \frac{1}{2.3} \approx 0.435\text{.} \end{equation*}
Therefore, the long-term level of Warfarin in the blood under these conditions is $$\frac{1}{2.3}\text{,}$$ which is approximately 0.435 mg.
To determine the long-term effect of Warfarin, we considered a finite geometric series of $$n$$ terms, and then considered what happened as $$n$$ was allowed to grow without bound. In this sense, we were actually interested in an infinite geometric series (the result of letting $$n$$ go to infinity in the finite sum).
#### Definition8.2.2.
An infinite geometric series is an infinite sum of the form
$$a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{.}\tag{8.2.5}$$
The value of $$r$$ in the geometric series (8.2.5) is called the common ratio of the series because the ratio of the ($$n+1$$)st term, $$ar^n\text{,}$$ to the $$n$$th term, $$ar^{n-1}\text{,}$$ is always $$r\text{:}$$
\begin{equation*} \frac{ar^n}{ar^{n-1}} = r\text{.} \end{equation*}
Geometric series are common in mathematics and arise naturally in many different situations. As a familiar example, suppose we want to write the number with repeating decimal expansion
\begin{equation*} N=0.1212\overline{12} \end{equation*}
as a rational number. Observe that
\begin{align*} N \amp = 0.12 + 0.0012 + 0.000012 + \cdots\\ \amp = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \end{align*}
This is an infinite geometric series with $$a=\frac{12}{100}$$ and $$r = \frac{1}{100}\text{.}$$
By using the formula for the value of a finite geometric sum, we can also develop a formula for the value of an infinite geometric series. We explore this idea in the following activity.
#### Activity8.2.3.
Let $$r \ne 1$$ and $$a$$ be real numbers and let
\begin{equation*} S = a+ar+ar^2 + \cdots ar^{n-1} + \cdots \end{equation*}
be an infinite geometric series. For each positive integer $$n\text{,}$$ let
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1}\text{.} \end{equation*}
Recall that
\begin{equation*} S_n = a\frac{1-r^n}{1-r}\text{.} \end{equation*}
1. What should we allow $$n$$ to approach in order to have $$S_n$$ approach $$S\text{?}$$
2. What is the value of $$\lim_{n \to \infty} r^n$$ for $$|r| \gt 1\text{?}$$ for $$|r| \lt 1\text{?}$$ Explain.
3. If $$|r| \lt 1\text{,}$$ use the formula for $$S_n$$ and your observations in (a) and (b) to explain why $$S$$ is finite and find a resulting formula for $$S\text{.}$$
Hint.
1. Let $$n$$ increase without bound.
2. Think about what happens to powers of numbers that are less than or greater than 1.
3. Consider $$\frac{1-r^n}{1-r}$$ and how the numerator tends to 1 as $$n \to \infty$$ for certain values of $$r\text{.}$$
We can now find the value of the geometric series
\begin{equation*} N = \left(\frac{12}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right) + \left(\frac{12}{100}\right)\left(\frac{1}{100}\right)^2 + \cdots\text{.} \end{equation*}
Using $$a = \frac{12}{100}$$ and $$r = \frac{1}{100}\text{,}$$ we see that
\begin{equation*} N = \frac{12}{100} \left(\frac{1}{1-\frac{1}{100}}\right) = \frac{12}{100} \left(\frac{100}{99}\right) = \frac{4}{33}\text{.} \end{equation*}
The sum of a finite number of terms of an infinite geometric series is often called a partial sum of the series. Thus,
\begin{equation*} S_n = a+ar+ar^2 + \cdots + ar^{n-1} = \sum_{k=0}^{n-1} ar^k\text{.} \end{equation*}
is called the $$n$$th partial sum of the series $$\sum_{k=0}^{\infty} ar^k\text{.}$$ We summarize our recent work with geometric series as follows.
• An infinite geometric series is an infinite sum of the form
$$a + ar + ar^2 + \cdots = \sum_{n=0}^{\infty} ar^n\text{,}\tag{8.2.6}$$
where $$a$$ and $$r$$ are real numbers such that $$r \ne 0\text{.}$$
• The $$n$$th partial sum $$S_n$$ of an infinite geometric series is
\begin{equation*} S_n = a+ar+ar^2+ \cdots + ar^{n-1}\text{.} \end{equation*}
• If $$|r| \lt 1\text{,}$$ then using the fact that $$S_n = a\frac{1-r^n}{1-r}\text{,}$$ it follows that the sum $$S$$ of the infinite geometric series (8.2.6) is
\begin{equation*} S = \lim_{n \to \infty} S_n = \lim_{n \to \infty} a\frac{1-r^n}{1-r} = \frac{a}{1-r} \end{equation*}
#### Activity8.2.4.
The formulas we have derived for an infinite geometric series and its partial sum have assumed we begin indexing the sums at $$n=0\text{.}$$ If instead we have a sum that does not begin at $$n=0\text{,}$$ we can factor out common terms and use the established formulas. This process is illustrated in the examples in this activity.
1. Consider the sum
\begin{equation*} \sum_{k=1}^{\infty} (2)\left(\frac{1}{3}\right)^k = (2)\left(\frac{1}{3}\right) + (2)\left(\frac{1}{3}\right)^2 + (2)\left(\frac{1}{3}\right)^3 + \cdots\text{.} \end{equation*}
Remove the common factor of $$(2)\left(\frac{1}{3}\right)$$ from each term and hence find the sum of the series.
2. Next let $$a$$ and $$r$$ be real numbers with $$-1\lt r\lt 1\text{.}$$ Consider the sum
\begin{equation*} \sum_{k=3}^{\infty} ar^k = ar^3+ar^4+ar^5 + \cdots\text{.} \end{equation*}
Remove the common factor of $$ar^3$$ from each term and find the sum of the series.
3. Finally, we consider the most general case. Let $$a$$ and $$r$$ be real numbers with $$-1\lt r\lt 1\text{,}$$ let $$n$$ be a positive integer, and consider the sum
\begin{equation*} \sum_{k=n}^{\infty} ar^k = ar^n+ar^{n+1}+ar^{n+2} + \cdots\text{.} \end{equation*}
Remove the common factor of $$ar^n$$ from each term to find the sum of the series.
Hint.
1. Think about how $$r = \frac{1}{3}\text{.}$$
2. Note that $$ar^3+ar^4+ar^5 + \cdots = ar^3(1 + r + r^2 + \cdots)\text{.}$$
3. Compare your work in (b).
### Subsection8.2.2Summary
• An infinite geometric series is an infinite sum of the form
\begin{equation*} \sum_{k=0}^{\infty} ar^k \end{equation*}
where $$a$$ and $$r$$ are real numbers and $$r \neq 0\text{.}$$
• The $$n$$th partial sum of the geometric series $$\sum_{k=0}^{\infty} ar^k$$ is
\begin{equation*} S_n = \sum_{k=0}^{n-1} ar^k\text{.} \end{equation*}
A formula for the $$n$$th partial sum of a geometric series is
\begin{equation*} S_n = a \frac{1-r^n}{1-r}\text{.} \end{equation*}
If $$|r| \lt 1\text{,}$$ the infinite geometric series $$\sum_{k=0}^{\infty} ar^k$$ has the finite sum $$\frac{a}{1-r}\text{.}$$
### Exercises8.2.3Exercises
#### 1.Fourth term of a geometric sequence.
Find the $$4^{th}$$ term of the geometric sequence
$$-1 , -3.5 , -12.25 , ...$$
#### 2.A geometric series.
Find the sum of the series
$$\displaystyle 2 + \frac{2}{7} + \frac{2}{49} + ... + \frac{2}{7^{n-1}} + ...\text{.}$$
#### 3.A series that is not geometric.
Determine the sum of the following series.
\begin{equation*} \sum_{n=1}^\infty \left(\frac{3^n + 8^n}{12 ^n}\right) \end{equation*}
#### 4.Two sums of geometric sequences.
Find the sum of each of the geometric series given below. For the value of the sum, enter an expression that gives the exact value, rather than entering an approximation.
A. $$-15 + 5 - {5\over 3} + {5\over 9} - {5\over 27} + {5\over 81} - \cdots =$$
B. $$\sum\limits_{n=4}^{17} \left({1\over 2}\right)^n =$$
#### 5.
There is an old question that is often used to introduce the power of geometric growth. Here is one version. Suppose you are hired for a one month (30 days, working every day) job and are given two options to be paid.
Option 1.
You can be paid $500 per day or Option 2. You can be paid 1 cent the first day, 2 cents the second day, 4 cents the third day, 8 cents the fourth day, and so on, doubling the amount you are paid each day. 1. How much will you be paid for the job in total under Option 1? 2. Complete Table 8.2.3 to determine the pay you will receive under Option 2 for the first 10 days. 3. Find a formula for the amount paid on day $$n\text{,}$$ as well as for the total amount paid by day $$n\text{.}$$ Use this formula to determine which option (1 or 2) you should take. #### 6. Suppose you drop a golf ball onto a hard surface from a height $$h\text{.}$$ The collision with the ground causes the ball to lose energy and so it will not bounce back to its original height. The ball will then fall again to the ground, bounce back up, and continue. Assume that at each bounce the ball rises back to a height $$\frac{3}{4}$$ of the height from which it dropped. Let $$h_n$$ be the height of the ball on the $$n$$th bounce, with $$h_0 = h\text{.}$$ In this exercise we will determine the distance traveled by the ball and the time it takes to travel that distance. 1. Determine a formula for $$h_1$$ in terms of $$h\text{.}$$ 2. Determine a formula for $$h_2$$ in terms of $$h\text{.}$$ 3. Determine a formula for $$h_3$$ in terms of $$h\text{.}$$ 4. Determine a formula for $$h_n$$ in terms of $$h\text{.}$$ 5. Write an infinite series that represents the total distance traveled by the ball. Then determine the sum of this series. 6. Next, let’s determine the total amount of time the ball is in the air. 1. When the ball is dropped from a height $$H\text{,}$$ if we assume the only force acting on it is the acceleration due to gravity, then the height of the ball at time $$t$$ is given by \begin{equation*} H - \frac{1}{2}gt^2\text{.} \end{equation*} Use this formula to determine the time it takes for the ball to hit the ground after being dropped from height $$H\text{.}$$ 2. Use your work in the preceding item, along with that in (a)-(e) above to determine the total amount of time the ball is in the air. #### 7. Suppose you play a game with a friend that involves rolling a standard six-sided die. Before a player can participate in the game, he or she must roll a six with the die. Assume that you roll first and that you and your friend take alternate rolls. In this exercise we will determine the probability that you roll the first six. 1. Explain why the probability of rolling a six on any single roll (including your first turn) is $$\frac{1}{6}\text{.}$$ 2. If you don’t roll a six on your first turn, then in order for you to roll the first six on your second turn, both you and your friend had to fail to roll a six on your first turns, and then you had to succeed in rolling a six on your second turn. Explain why the probability of this event is \begin{equation*} \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)\text{.} \end{equation*} 3. Now suppose you fail to roll the first six on your second turn. Explain why the probability is \begin{equation*} \left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right) = \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) \end{equation*} that you to roll the first six on your third turn. 4. The probability of you rolling the first six is the probability that you roll the first six on your first turn plus the probability that you roll the first six on your second turn plus the probability that your roll the first six on your third turn, and so on. Explain why this probability is \begin{equation*} \frac{1}{6} + \left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right) + \left(\frac{5}{6}\right)^4\left(\frac{1}{6}\right) + \cdots\text{.} \end{equation*} Find the sum of this series and determine the probability that you roll the first six. #### 8. The goal of a federal government stimulus package is to positively affect the economy. Economists and politicians quote numbers like “$$k$$ million jobs and a net stimulus to the economy of $$n$$ billion of dollars.” Where do they get these numbers? Let’s consider one aspect of a stimulus package: tax cuts. Economists understand that tax cuts or rebates can result in long-term spending that is many times the amount of the rebate. For example, assume that for a typical person, 75% of her entire income is spent (that is, put back into the economy). Further, assume the government provides a tax cut or rebate that totals $$P$$ dollars for each person. 1. The tax cut of $$P$$ dollars is income for its recipient. How much of this tax cut will be spent? 2. In this simple model, we will say that the spent portion of the tax cut/rebate from part (a) then becomes income for another person who, in turn, spends 75% of this income. After this second round" of spent income, how many total dollars have been added to the economy as a result of the original tax cut/rebate? 3. This second round of spending becomes income for another group who spend 75% of this income, and so on. In economics this is called the multiplier effect. Explain why an original tax cut/rebate of $$P$$ dollars will result in multiplied spending of \begin{equation*} 0.75P(1+0.75+0.75^2+ \cdots )\text{.} \end{equation*} dollars. 4. Based on these assumptions, how much stimulus will a 200 billion dollar tax cut/rebate to consumers add to the economy, assuming consumer spending remains consistent forever. #### 9. Like stimulus packages, home mortgages and foreclosures also impact the economy. A problem for many borrowers is the adjustable rate mortgage, in which the interest rate can change (and usually increases) over the duration of the loan, causing the monthly payments to increase beyond the ability of the borrower to pay. Most financial analysts recommend fixed rate loans, ones for which the monthly payments remain constant throughout the term of the loan. In this exercise we will analyze fixed rate loans. When most people buy a large ticket item like car or a house, they have to take out a loan to make the purchase. The loan is paid back in monthly installments until the entire amount of the loan, plus interest, is paid. With a loan, we borrow money, say $$P$$ dollars (called the principal), and pay off the loan at an interest rate of $$r$$%. To pay back the loan we make regular monthly payments, some of which goes to pay off the principal and some of which is charged as interest. In most cases, the interest is computed based on the amount of principal that remains at the beginning of the month. We assume a fixed rate loan, that is one in which we make a constant monthly payment $$M$$ on our loan, beginning in the original month of the loan. Suppose you want to buy a house. You have a certain amount of money saved to make a down payment, and you will borrow the rest to pay for the house. Of course, for the privilege of loaning you the money, the bank will charge you interest on this loan, so the amount you pay back to the bank is more than the amount you borrow. In fact, the amount you ultimately pay depends on three things: the amount you borrow (called the principal), the interest rate, and the length of time you have to pay off the loan plus interest (called the duration of the loan). For this example, we assume that the interest rate is fixed at $$r$$%. To pay off the loan, each month you make a payment of the same amount (called installments). Suppose we borrow $$P$$ dollars (our principal) and pay off the loan at an interest rate of $$r$$% with regular monthly installment payments of $$M$$ dollars. So in month 1 of the loan, before we make any payments, our principal is $$P$$ dollars. Our goal in this exercise is to find a formula that relates these three parameters to the time duration of the loan. We are charged interest every month at an annual rate of $$r$$%, so each month we pay $$\frac{r}{12}$$% interest on the principal that remains. Given that the original principal is $$P$$ dollars, we will pay $$\left(\frac{0.0r}{12}\right)P$$ dollars in interest on our first payment. Since we paid $$M$$ dollars in total for our first payment, the remainder of the payment ($$M-\left(\frac{r}{12}\right)P$$) goes to pay down the principal. So the principal remaining after the first payment (let’s call it $$P_1$$) is the original principal minus what we paid on the principal, or \begin{equation*} P_1 = P - \left( M - \left(\frac{r}{12}\right)P\right) = \left(1 + \frac{r}{12}\right)P - M\text{.} \end{equation*} As long as $$P_1$$ is positive, we still have to keep making payments to pay off the loan. 1. Recall that the amount of interest we pay each time depends on the principal that remains. How much interest, in terms of $$P_1$$ and $$r\text{,}$$ do we pay in the second installment? 2. How much of our second monthly installment goes to pay off the principal? What is the principal $$P_2\text{,}$$ or the balance of the loan, that we still have to pay off after making the second installment of the loan? Write your response in the form $$P_2 = ( \ )P_1 - ( \ )M\text{,}$$ where you fill in the parentheses. 3. Show that $$P_2 = \left(1 + \frac{r}{12}\right)^2P - \left[1 + \left(1+\frac{r}{12}\right)\right] M\text{.}$$ 4. Let $$P_3$$ be the amount of principal that remains after the third installment. Show that \begin{equation*} P_3 = \left(1 + \frac{r}{12}\right)^3P - \left[1 + \left(1+\frac{r}{12}\right) + \left(1+\frac{r}{12}\right)^2 \right] M\text{.} \end{equation*} 5. If we continue in the manner described in the problems above, then the remaining principal of our loan after $$n$$ installments is $$P_n = \left(1 + \frac{r}{12}\right)^nP - \left[\displaystyle \sum_{k=0}^{n-1} \left(1+\frac{r}{12}\right)^k \right] M\text{.}\tag{8.2.7}$$ This is a rather complicated formula and one that is difficult to use. However, we can simplify the sum if we recognize part of it as a partial sum of a geometric series. Find a formula for the sum $$\displaystyle \sum_{k=0}^{n-1} \left(1+\frac{r}{12}\right)^k\text{.}\tag{8.2.8}$$ and then a general formula for $$P_n$$ that does not involve a sum. 6. It is usually more convenient to write our formula for $$P_n$$ in terms of years rather than months. Show that $$P(t)\text{,}$$ the principal remaining after $$t$$ years, can be written as $$P(t) = \left(P - \frac{12M}{r}\right)\left(1+\frac{r}{12}\right)^{12t} + \frac{12M}{r}\text{.}\tag{8.2.9}$$ 7. Now that we have analyzed the general loan situation, we apply formula (8.2.9) to an actual loan. Suppose we charge$1,000 on a credit card for holiday expenses. If our credit card charges 20% interest and we pay only the minimum payment of $25 each month, how long will it take us to pay off the$1,000 charge? How much in total will we have paid on this $1,000 charge? How much total interest will we pay on this loan? 8. Now we consider larger loans, e.g., automobile loans or mortgages, in which we borrow a specified amount of money over a specified period of time. In this situation, we need to determine the amount of the monthly payment we need to make to pay off the loan in the specified amount of time. In this situation, we need to find the monthly payment $$M$$ that will take our outstanding principal to $$0$$ in the specified amount of time. To do so, we want to know the value of $$M$$ that makes $$P(t) = 0$$ in formula (8.2.9). If we set $$P(t) = 0$$ and solve for $$M\text{,}$$ it follows that \begin{equation*} M = \frac{rP \left(1+\frac{r}{12}\right)^{12t}}{12\left(\left(1+\frac{r}{12}\right)^{12t} - 1 \right)}\text{.} \end{equation*} 1. Suppose we want to borrow$15,000 to buy a car. We take out a 5 year loan at 6.25%. What will our monthly payments be? How much in total will we have paid for this $15,000 car? How much total interest will we pay on this loan? 2. Suppose you charge your books for winter semester on your credit card. The total charge comes to$525. If your credit card has an interest rate of 18% and you pay $20 per month on the card, how long will it take before you pay off this debt? How much total interest will you pay? 3. Say you need to borrow$100,000 to buy a house. You have several options on the loan:
• 30 years at 6.5%
• 25 years at 7.5%
• 15 years at 8.25%.
1. What are the monthly payments for each loan?
2. Which mortgage is ultimately the best deal (assuming you can afford the monthly payments)? In other words, for which loan do you pay the least amount of total interest?
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# Workout the Mystery Number using Squaring
In this worksheet, students will identify a rule which is used to make a number and apply this rule to find a missing number.
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Key stage: KS 2
Curriculum topic: Verbal Reasoning
Curriculum subtopic: Number Logic
Difficulty level:
#### Worksheet Overview
Hi number detective! We’ve got a new challenge that we need your help with. Are you ready?
We need to work out how the numbers on the outside of the brackets create the numbers on the inside of the brackets and then apply the same rule to the third set of brackets.
10 (23) 3 5 (16) 6 2 ( __) 5
Be warned:
One of the numbers could be used more than once.
A third invisible number could be used.
One or both of the numbers could be squared (multiplied by themselves).
First, we need to find a way to make 23 using 10 and 3. Remember that we will probably have to consider one of the warnings above.
If we add 10 and 3 together we get 13. Then, if we add another lot of 10, we get to 23.
So the rule could be: add the two outside numbers and then add the first number again.
Let’s try this out for the second set of numbers: 5 + 5 + 6 = 16.
We’ve found the rule! Jackpot!
Let’s use this rule to work out the missing number in the final set: 2 + 2 + 5 = 9. Therefore, the missing number is 9.
Let’s try another:
5 (28) 3 3 (16) 7 4 ( __ ) 8
First, we need to find a way to make 28 using 5 and 3. Remember that we can use one of the warnings listed above.
Let’s try 5 + 3 = 8 and add the invisible number of 20 to get to 28.
So our rule could be: add the two numbers together and then add the invisible number 20.
Let’s see if this works for the second set of numbers: 3 + 7 = 10 but we need to use a different invisible number of 8 to get to our total.
Bad news, our rule didn’t work so we need to find another.
Let’s try (this means 5 x 5) which gives us 25 and then we need to add 3 to get to 28.
So our rule could be: square the first number and then add the second.
Let’s try this for the second set: is 9 then add 7 gives us 16.
Jackpot! We’ve found the correct rule.
Let’s use this rule to work out the missing number in the final set of numbers: 4² = 16 and then add 8 gives us 24. So the correct answer is 24.
Pssstt! Here's a handy hint to help you reach superstar status: Look out for the secret number and don't worry if the first rule you find doesn't work, just stay calm and try another one!
It’s now your turn to be a number detective and hunt down those missing numbers!
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# Row Echelon Form & Reduced Row Echelon Form
Watch the video for the definitions of echelon, row echelon and reduced row echelon:
Row Echelon and Reduced Row Echelon
## What is Echelon Form?
Echelon form means that the matrix is in one of two states:
• Row echelon form.
• Reduced row echelon form.
This means that the matrix meets the following three requirements:
1. The first number in the row (called a leading coefficient) is 1. Note: some authors don’t require that the leading coefficient is a 1; it could be any number. You may want to check with your instructor to see which version of this rule they are adhering to).
2. Every leading 1 is to the right of the one above it.
3. Any non-zero rows are always above rows with all zeros.
The following examples are of matrices in echelon form:
The following examples are not in echelon form:
Matrix A does not have all-zero rows below non-zero rows.
Matrix B has a 1 in the 2nd position on the third row. For row echelon form, it needs to be to the right of the leading coefficient above it. In other words, it should be in the fourth position in place of the 3.
Matrix C has a 2 as a leading coefficient instead of a 1.
Matrix D has a -1 as a leading coefficient instead of a 1.
Another way to think of a matrix in echelon form is that the matrix has undergone Gaussian elimination, which is a series of row operations.
## Uniqueness and Echelon Forms
The echelon form of a matrix isn’t unique, which means there are infinite answers possible when you perform row reduction. Reduced row echelon form is at the other end of the spectrum; it is unique, which means row-reduction on a matrix will produce the same answer no matter how you perform the same row operations.
## What is Row Echelon Form?
A matrix is in row echelon form if it meets the following requirements:
• The first non-zero number from the left (the “leading coefficient“) is always to the right of the first non-zero number in the row above.
• Rows consisting of all zeros are at the bottom of the matrix.
Technically, the leading coefficient can be any number. However, the majority of Linear Algebra textbooks do state that the leading coefficient must be the number 1. To add to the confusion, some definitions of row echelon form state that there must be zeros both above and below the leading coefficient. It’s therefore best to follow the definition given in the textbook you’re following (or the one given to you by your professor). If you’re unsure (i.e. it’s Sunday, your homework is due and you can’t get hold of your professor), it safest to use 1 as the leading coefficient in each row.
If the leading coefficient in each row is the only non-zero number in that column, the matrix is said to be in reduced row echelon form.
Row echelon forms are commonly encountered in linear algebra, when you’ll sometimes be asked to convert a matrix into this form. The row echelon form can help you to see what a matrix represents and is also an important step to solving systems of linear equations.
## Online Row Echelon Form Calculator
This online calculator will convert any matrix, and provides the row operations that get you from step to step. The following image (from the Old Dominion University Calculator shows how the matrix [01, 00, 59] is reduced to row echelon form with two simple row operations:
## What is Reduced Row Echelon Form?
Reduced row echelon form is a type of matrix used to solve systems of linear equations. Reduced row echelon form has four requirements:
• The first non-zero number in the first row (the leading entry) is the number 1.
• The second row also starts with the number 1, which is further to the right than the leading entry in the first row. For every subsequent row, the number 1 must be further to the right.
• The leading entry in each row must be the only non-zero number in its column.
• Any non-zero rows are placed at the bottom of the matrix.
## Transformation of a Matrix to Reduced Row Echelon Form
Any matrix can be transformed to reduced row echelon form, using a technique called Gaussian elimination. This is particularly useful for solving systems of linear equations. Most graphing calculators (like the TI-83) have a rref function which will transform a matrix into a reduced row echelon form. See: This article on the Colorado State University website for instructions on using the TI-89 and TI-83 to calculate reduced row echelon form.
This online calculator on the Old Dominion University website transforms a matrix that you input to reduced row echelon form.
Calculation by hand requires knowledge of elementary row operations, which are:
1. Interchange one row with another.
2. Multiply one row by a non-zero constant.
3. Replace one row with: one row, plus a constant, times another row.
In addition, it isn’t enough just to know the rules, you have to be able to look at the matrix and make a logical decision about which rule you’re going to use and when. You’re trying to get the matrix into reduced row echelon form, so you’ll also need to refer to the four requirements at the beginning of this article. If you have to convert a matrix to reduced row echelon form by hand, it’s a good idea to use one of the calculators above to check your work. In fact, of you use the ODU online calculator, it will even provide the row operations for you. The image below is the calculator’s conversion of the matrix [204,923]:
## What is Gaussian Elimination?
Gaussian elimination is a way to find a solution to a system of linear equations. The basic idea is that you perform a mathematical operation on a row and continue until only one variable is left. For example, some possible row operations are:
• Interchange any two rows
• Multiply one row by a non-zero constant (i.e. 1/3, -1, 5)
You can also perform more than one row operation at a time. For example, multiply one row by a constant and then add the result to the other row.
Following this, the goal is to end up with a matrix in reduced row echelon form where the leading coefficient, a 1, in each row is to the right of the leading coefficient in the row above it. In other words, you need to get a 1 in the upper left corner of the matrix. The next row should have a 0 in position 1 and a 1 in position 2. This gives you the solution to the system of linear equations.
## Gaussian Elimination Example
Solve the following system of linear equations using Gaussian elimination:
• x + 5y = 7
• -2x – 7y = -5
Step 1: Convert the equation into coefficient matrix form. In other words, just take the coefficient for the numbers and forget the variables for now:
Step 2: Turn the numbers in the bottom row into positive by adding 2 times the first row:
Step 3: Multiply the second row by 1/3. This gives you your second leading 1:
Step 4: Multiply row 2 by -5, and then add this to row 1:
That’s it!
In the first row, you have x = -8 and in the second row, y = 3. Note that x and y are in the same positions as when you converted the equation in step 1, so all you have to do is read the solution:
## What is the Rank of a Matrix?
The rank of a matrix is equal to the number of linearly independent rows. A linearly independent row is one that isn’t a combination of other rows.
The following matrix has two linearly independent rows (1 and 2). However, when the third row is thrown into the mix, you can see that the first row is now equal to the sum of the second and third rows. Therefore, the rank of this particular matrix is 2, as there are only two linearly independent rows.
The matrix rank will always be less than the number of non-zero rows or the number of columns in the matrix. If all of the rows in a matrix are linearly independent, the matrix is full row rank. For a square matrix, it is only full rank if its determinant is non-zero.
Figuring out the rank of a matrix by trying to determine by sight only how many rows or columns are linearly independent can be practically impossible. An easier (and perhaps obvious) way is to convert to row echelon form.
## How to Find the Matrix Rank
Finding the rank of a matrix is simple if you know how to find the row echelon matrix. To find the rank of any matrix:
1. Find the row echelon matrix.
2. Count the number of non-zero rows.
The above matrix has been converted to row echelon form with two non-zero rows. Therefore, the rank of the matrix is 2.
You can also find an excellent conversion tool on the Old Dominion University website.
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# Math-Linux.com
Knowledge base dedicated to Linux and applied mathematics.
Home > Mathematics > Interpolation > Lagrange interpolation polynomial
## Lagrange interpolation polynomial
In this section, we shall study the interpolation polynomial in the Lagrange form. Given a set of (n+1) data points and a function f, the aim is to determine a polynomial of degree n which interpolates f at the points in question.
### Interpolation
Given a set of data points , the points defined by are called points of interpolation. The points are the values of interpolation. To interpolate a function , we define the values of interpolation as follows:
### Lagrange interpolation polynomial
The purpose here is to determine the unique polynomial of degree , which verifies
The polynomial which meets this equality is Lagrange interpolation polynomial
where are polynomials of degree that form a basis of
### Properties of Lagrange interpolation polynomial and Lagrange basis
They are the polynomials which verify the following property:
They form a basis of the vector space of polynomials of degree at most equal to
By setting: , we obtain:
The set is linearly independent and consists of vectors. It is thus a basis of .
Finally, we can easily see that:
### Existence and uniqueness of Lagrange interpolation polynomials
Existence.
The proof is shown above. Actually, it corresponds to the construction of Lagrange interpolation polynomial with regard to the basis
of .
Uniqueness. Consider two elements and of which verify
Let . The polynomial has roots which are exactly the since
has then roots and . Therefore,
We have thus shown the existence and uniqueness of Lagrange interpolation polynomial.
### Error in Lagrange interpolation
Assume that and . Let be the closed set defined by (the smallest closed set containing and ).
Theorem.
Proof. There are two possible ways to prove this theorem: 1) the one encountered in The polynomial interpolation in the form of Newton and, 2) the following proof.
If , the problem is over:
Now, assume that and let us define the application as follows:
We also define the application :
is times differentiable and evaluates to zero at the points of the interval . By successively applying Rolle’s theorem, evaluates to zero at a point :
The derivative of can be easily calculated:
By setting , we have:
Therefore
We finally conclude that
Corollary.
Assume that and .
Alternatively stated:
### Example: computing Lagrange interpolation polynomials
Given a set of three data points , we shall determine the Lagrange interpolation polynomial of degree 2 which passes through these points.
First, we compute and :
Lagrange interpolation polynomial is:
### Scilab: computing Lagrange interpolation polynomial
The Scilab function lagrange.sci determines Lagrange interpolation polynomial. encompasses the points of interpolation and the values of interpolation. is the Lagrange interpolation polynomial.
lagrange.sci
function[P]=lagrange(X,Y)//X nodes,Y values;P is the numerical Lagrange polynomial interpolation
n=length(X);// n is the number of nodes. (n-1) is the degree
x=poly(0,"x");P=0;
for i=1:n, L=1;
for j=[1:i-1,i+1:n] L=L*(x-X(j))/(X(i)-X(j));end
P=P+L*Y(i);
end
endfunction
We thus have:
-->X=[0;2;4]; Y=[1;5;17]; P=lagrange(X,Y)
P = 1 + x^2
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We received a number of good solutions to this problem. Most of you noticed that the key to making rectangles that work for all bases is factorising quadratic equations. Luke from London Oratory School gave a good explanation of this:
The green section is a square, so its area is equal to $x^2$.
The red section consists of two rectangles with dimensions $a, x$ and $b, x$. Therefore the
red area is equal to $ax+bx$, which equals $x(a+b)$.
We can see that the area of the blue section will always have dimensions $a$ and $b$, so its area is equal to $ab$, if it ʻfillsʼ the gap created by the red area.
The total area is equal to the sum of these component areas.
Thus you can make a rectangle for all bases for equations of the form $x^2 + x(a+b) + ab$ where $a$ and $b$ are positive integers. The rectangle has dimensions $(x+a)$ by $(x+b)$.
So for the rectangle $x^2 + 7x + 12$, we can see that 1) $7=a+b$ and 2) $12=ab$.
$a=3 , b=4$ satisfies these equations.
So for any base we can make the rectangle with dimensions $(x+3)$ by $(x+4)$.
Vanessa and Annie sent us this solution:
Using 1 square and 12 sticks: You can make 7 different rectangles which work in all bases. These are:
$x(x+12)$, $(x+1)(x+11)$, $(x+2)(x+10)$, $(x+3)(x+9)$, $(x+4)(x+8)$, $(x+5)(x+7)$, $(x+6)(x+6)$.
You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $A+B=12$.
Using the same logic, if you have one square and 100 sticks, you can make 51 rectangles that work for all bases:
$x(x+100)$, $(x+1)(x+99)$, $(x+2)(x+98)$, ..., $(x+50)(x+50)$.
Using 1 square and 12 units: You can make 3 different rectangles which work in all bases:
$(x+1)(x+12)$, $(x+2)(x+6)$, $(x+3)(x+4)$.
You can make rectangles for all bases with dimensions $(x+A)(x+B)$ where $A$ and $B$ are positive integers and $AB=12$.
Using the same logic, if you have one square and 100 sticks, you can make 5 rectangles which work in all bases: $(x+1)(x+100)$, $(x+2)(x+50)$, $(x+4)(x+25)$, $(x+5)(x+20)$, $(x+10)(x+10)$.
Using one square, $p$ sticks, $q$ units, you can only make a rectangle which works for all bases if $p=A+B$ and $q=AB$ for $A$ and $B$ positive integers.
Extension: Using $n$ squares, we can make rectangles which work in all bases of dimensions $(ax+b)(cx+d)$ when $a$,$b$,$c$,$d$ are positive integers and $ac=n$. Then we will use $ad+bc$ sticks and $bd$ units. This will give rectangles which have the squares arranged in a rectangle. But we could arrange the squares in different ways, like L-shapes.
Well done!
|
# Math
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In algebra, adding and subtracting fractions is easy when you find the common denominator. This video shows you how to convert fractions for the common denominator. After you determine the common denominator, you can add and subtract fractions, including story problems, with ease.
### How to Find the Volume of a Solid with a Circular Cross-Section (Video)
Calculus allows you to calculate the volume of conical objects by dividing the object into an infinite number of circular cross-sections - geometrical shapes resembling pancakes or washers - and adding up the volume of all those cross-sections through integration. This video tutorial shows you how.
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Hong Kong
Stage 4 - Stage 5
# Tangents to Circles
Lesson
A tangent is a line that intersects the circumference of a circle in exactly one point, which we call the point of tangency.
A tangent is perpendicular to the radius from the point of tangency. Conversely, the perpendicular to a radius through the same endpoint is a tangent line.
## Multiple Tangents?
There can be more than one tangent on a circle. In fact there is basically an infinite number! The diagram below shows two tangents- $PM$PM and $PQ$PQ.
If two tangents are drawn from a common point, the tangents are equal.
Proof:
Let's start by drawing in radii from the points of tangency:
In $\triangle OMP$OMP and $\triangle OQP$OQP:
$OM=OQ$OM=OQ(radii in a circle are equal)
$OP$OP is common
$\angle OMP=\angle OQP$OMP=OQP$=$=$90^\circ$90° (tangents meet radii at right angles)
$\therefore$ $\triangle OMP$OMP$\cong$$\triangle OQP$OQP (RHS)
$\therefore$ $MP=QP$MP=QP (corresponding sides in congruent triangles are equal)
Remember, we aren't limited to the rules of circle geometry. We can use all our geometrical rules, including Pythagoras' theorem, congruency and similarity.
Let's look through some worked examples now to see this in action.
#### Worked Examples
##### Question 1
In this question we aim to prove that the tangent is perpendicular to the radius drawn from its point of contact.
In the diagram, $C$C is an arbitrary point on the line $AD$AD, and $B$B is the point at which the tangent meets the circle.
1. What can we say about the lines $OB$OB and $OC$OC?
$OB=OC$OB=OC
A
$OB>OC$OB>OC
B
|
# How to Multiply Fractions (+ 7 Engaging Activities)
Congratulations! You’ve succeeded in teaching multiplication (in six easy steps). But now it’s time to introduce your students to multiplying fractions.
Deep breaths.
Teachers and students alike might argue this concept is more daunting than leaping from subtraction and addition to multiplication.
Fortunately, strategies exist that should make learning how to multiply fractions much easier to grasp — and we’ve compiled them for you in a clear-cut guide.
But first, a quick refresher.
## What is multiplication?
Put simply, multiplication is adding the same number over and over.
Good news for your students: if they can add, they can multiply!
Instead of writing 1 + 1 + 1 + 1, there’s a much quicker way to write this addition problem: 1 × 4. Here are some examples:
In addition to multiplying whole numbers, you can also multiply by integers, decimals and, today, fractions.
## Defining three types of fractions
A fraction is generally composed of two parts:
• Numerator — the top number, which refers to how many parts (of a whole) you have.
• Denominator — the bottom number, which refers to the total number of parts making up the whole.
### 1. Proper fractions
A proper fraction has a numerator less than the denominator.
For example: ½, ⅔, ¾, ⅘, ⅚
### 2. Improper fractions
Though similar in structure, an improper fraction has a numerator greater than the denominator.
Note: When a numerator is equal to the denominator, it’s considered “improper” because you can change it into a whole number. The same rule applies to improper fractions such as ²⁶⁄₁₃ which, if reduced, become whole (i.e., two).
For example: ³⁄₂, ⁵⁄₃, ⁷⁄₆, ¹¹⁄₁₀, ⁸⁄₈
### 3. Mixed fractions
Unlike the first two, a mixed fraction is composed of a proper fraction and whole number.
For example: 3 ½, 7 ⅔, 2 ¾, 6 ⅘, 1 ⅚
Okay, so these weird, somehow-pizza-related numbers exist. But what are the rules for multiplying fractions?
## How to multiply fractions
Good things come in threes, including the three simple steps your students need to follow when multiplying fractions:
1. Multiply the numerators (top numbers)
2. Multiply the denominators (bottom numbers)
3. If needed, simplify or reduce the fraction
Let’s go through an example together!
### Area models for fraction multiplication
Perfect for the visual learners in your class, the area model effectively illustrates what one fraction times (or “of”) another looks like.
As you can see from the illustration below, creating an area model when multiplying fraction is easy:
1. Draw the fractions you’re multiplying in separate boxes, each using a different color
2. Combine the drawings into one box, using a new color for the parts that overlap
3. To write the product, ask yourself two questions:
• How many boxes have both colors? This will be your numerator
• How many boxes are there in total? This will be your denominator
### A catchy reminder
Oh! And if your students ever forget the steps, just remind them to sing this song:
Multiplying fractions? That’s no big problem.
Do top times top over bottom times bottom.
And before you say goodbye, don’t forget to simplify!
## Multiplying fractions with whole numbers
Multiplying whole numbers and fractions may stump your students. Why? Because it seems like there’s only one fraction instead of two.
But when multiplying fractions by whole numbers, students can rewrite them as fractions. Instead of the 4, for example, turn it into the fraction ⁴⁄₁.
Now it’s easier and clearer to multiply. For example, they can rewrite 2 × ⁵⁄₁₃:
Solve: 2 × ⁵⁄₁₃
Rewrite whole number as a fraction: ²⁄₁ × ⁵⁄₁₃
Multiply numerators: 2 × 5 = 10
Multiply denominators: 1 × 13 = 13
New fraction: ¹⁰⁄₁₃
Note: If students struggle with whole numbers, explain that they can think of a whole number as a top number, with the bottom number always being one.
## Multiplying improper fractions
You multiply improper fractions the same way you do proper ones. However, sometimes students can end up with improper fractions.
Take this problem as an example:
Solve: ⁵⁄₃ × ⁷⁄₆
Multiply numerators: 5 × 7 = 35
Multiply denominators: 3 × 6 = 18
New fraction: ³⁵⁄₁₈
If students are familiar with mixed fractions, they can change the improper fraction to a mixed one. In this case, that mixed number would be 1 ¹⁷⁄₁₈.
## Multiplying mixed fractions
Before multiplying fractions with mixed numbers, there are three steps students should know:
1. Convert any mixed fractions into improper fractions
2. Multiply the improper fractions
3. Convert the final product back into a mixed number
To complete the first step, teach your students how to make a mixed number “MAD”.
Remember: a mixed number is composed of a whole number and proper fraction. To complete step one and convert a mixed fraction into an improper one, you need to:
• Find the new numerator — Multiply the whole number with the denominator, then add the original numerator to it.
• Keep the same denominator — The denominator remains unchanged.
Multiply
Denominator
Step two, multiply the improper fractions as we illustrated before this section.
Step three, convert that improper fraction back into a mixed number. Here’s a little rhyme to help your students remember how to do this:
With an improper fraction,
division is the action!
Let’s use the example pictured earlier: ¹⁶⁄₅.
If division is the action, then you need to divide the numerator (16) by the denominator (5) and see if there’s a remainder.
Five goes into 16 three times evenly, for a total of 15. That means the whole number portion of the mixed number will be three.
But it didn’t divide perfectly, which you can see by the remainder of one. So, the fraction portion of the mixed number is the remainder over the denominator of the original fraction (⅕).
What that looks like as a mixed fraction is 3 ⅕.
For a more visual walkthrough of how to multiply mixed numbers, watch this four-minute video from Khan Academy:
And that’s what you need to know when multiplying fractions and mixed numbers.
## Two major errors students make with fraction multiplication
Though some students will quickly grasp your fraction multiplication lessons, others can struggle with these new concepts.
The earlier teachers catch these misconceptions, the sooner students can learn from and correct their errors.
According to the What Works Clearinghouse Institute of Education Sciences practice guide, “Developing Effective Fractions Instruction for Kindergarten Through 8th Grade,” these are some of the most common misconceptions in regards to learning how to multiply fractions.
### 1. Believing whole numbers have the same denominator as fraction in a problem
The guide’s panel of eight experts recognized this misconception can lead students to take a problem such as 4 – ⅜ and rewrite it as ⁴⁄₈ – ⅜, for an incorrect answer of ⅛.
When presented with a mixed number, students with such a misconception might add the whole number to the numerator, as in ³¹⁄₃ × ⁶⁄₇ = (³⁄₃ + ⅓) × ⁶⁄₇ = ⁴⁄₃ × ⁶⁄₇ = ²⁴⁄₂₁.
Helping students understand the relation between mixed numbers and improper fractions — and how to translate each into the other — is crucial for working with fractions.
Avoid the temptation to blast through foundational lessons.
Take the time your students need to help them understand the relationship between improper fractions and mixed numbers, and how to convert them from one to the other.
### 2. Leaving the denominator unchanged
Students can make the mistake of forgetting to multiply equal denominators. This is likely due to the fact you don’t have to touch equal denominators in fraction addition.
For example, they might see ⅔ × ⅓ and incorrectly answer ⅔ instead of ²⁄₉.
In the practice guide, expert panelists suggest “explaining the conceptual basis of fraction multiplication using unit fractions (e.g., ½ × ½ = half of a half = ¼).”
In particular, teachers can show that the problem ½ × ½ is actually asking what ½ of ½ is, which implies that the product must be smaller than either fraction being multiplied.
Verbalizing this misconception is helpful, but visualizing it is especially effective. Enter the fraction wall!
Fraction walls are a brilliant way to help students see what, in this case, an abstract one half of one half (i.e., one quarter) looks like.
Now you’re aware of many students’ pain points in terms of multiplying fractions, what’s next? Let’s explore ways to make your fraction lessons stick — and why worksheets may not be the best strategy.
### The death of worksheets?
Dr. Stephen Camarata, a specialist in child development and developmental disabilities, believes children’s natural curiosity is being derailed.
In an essay for Psychology Today, he expressed discontentment with the presumption that completing worksheets is directly associated with improvements in learning.
But the data on US achievement scores as compared to the rest of the world indicate otherwise. Ironically, as more and more worksheets are pushed in earlier and earlier grades and the more rote, boring homework is forced on developing minds, student outcomes in the US decline further.
A Pew Center for Research report indicated a decline in already dismal US achievement scores. Dr. Camarata highlighted:
Only 34% of fourth graders and 27% of eighth graders were rated as proficient in math in 2011 and this declined to 33% for 4th graders and 25% for 8th graders in 2015 (the most recent year these data are available).
There is no way to put a positive spin on these outcomes: Currently, more than two-thirds of fourth-graders and three in four eighth graders are not proficient in math. This ranks 38th in the world.
Does this mean schools should go worksheet-free? Not necessarily.
Correlation isn’t causation. In fact, many teachers and students have found success with worksheets.
However, educators need to realize education is rapidly changing, from worksheets to classroom technology.
So, here are a few creative ways to teach fraction multiplication — worksheet-free!
## 7 Engaging examples to teach multiplying fractions
### 1. Prodigy
Prodigy is a free, curriculum-aligned math game used by more than a million teachers and 50 million students around the world.
It offers content from every major math topic from 1st to 8th grade. With regard to multiplying fractions, Prodigy can help students learn how to:
• Multiply a fraction by a whole number
• Multiply two fractions
• Multiply a whole number by a missing fraction
• Multiply two fractions via word problems
• And more
A 2018 case study found that schools using Prodigy performed better on standardized tests and enjoyed an increase in test results.
Playing Prodigy will take your students on a fantasy-inspired learning journey that’s both fun and educational.
### 2. Flip-it fractions
For this activity, split your students into groups of four. Next, they’ll divide themselves in teams of two, one being Player A and the other Player B.
Give each group a deck of shuffled cards (aces = 1, jacks = 10, queens = 11, and kings = 12).
As in the picture above, each student will draw a numerator card (above the pencil) and a denominator card (below the pencil).
Both Player A’s will rewrite and multiply the fractions on paper, then simplify the product if possible. Once they’ve answered, Player B’s will do the exact same thing.
Depending on your schedule, you can assign the whole deck or give students a timer to complete as many as they can.
Have students hand in their answer sheets after the activity for you to mark, or go through 10+ questions together as a class.
Note: You can also call this game “Slam-it fractions,” replacing the cards and pencils with dominoes.
### 3. Fraction multiplication BINGO
Every student gets a fraction-filled bingo card along with small sheets of paper (or “bingo chips”) on which are corresponding fraction multiplication problems.
When you say “GO,” they can start solving each problem one by one, laying the chip on top of the correct fraction.
It’s up to you whether you want them to complete a line or the entire bingo card. You can also choose whether or not the bingo card answers are simplified.
For simplicity’s sake, you can give everyone the same bingo card with the same questions. That way, you can go through each problem afterwards and walk through how to solve them together.
Note: You can use this example and the ones that follow as formative assessments to gauge student learning and catch misconceptions early.
### 4. Word problems
Word problems are a wonderful way to make math lessons relevant to your students’ lives.
Multiplying fractions may seem foreign to them, but a simple story can change their entire perspective not just about fractions, but math as a whole.
Here’s a word problem example:
You have ½ a bag of chips in the cupboard, but ate ½ of it after dinner.
How much of the whole bag did you eat?
Granted, it’s a simple example. But a second ago, that fraction was just a number above and below a short line. Now, however, this “everyday” word problem has made multiplying fractions applicable to real life.
### 5. Fraction war
This two-person activity is adapted from the card game “War.”
Note: Many teachers are moving away from speed-based problem solving because it doesn’t necessarily demonstrate student understanding. You might also have kids in your class who get anxious under the pressure of doing math quickly and thus underperform. So, it’s best to gauge your class before playing this version.
Sitting side by side, each student will have half a deck of fraction cards (which you can download here).
With their pile of cards facedown, each student will draw a card at the same time.
The first student who multiplies the two fractions correctly adds those cards to their deck.
A student wins if they end up with the entire deck of fraction cards in their hand, or they have the majority of cards at the end of, for example, 20 minutes.
What’s more, the quick think-on-the-spot nature of this fraction activity can help improve your students’ mental math skills.
Alternatively, you can run a version of this game allowing everyone to play against you.
Who doesn’t love a chance to outsmart their teacher!?
Split your class into five groups.
You’ll go to the first group and have a one-question face-off, before moving to the second group and so on. Remember: the point of this version isn’t to answer before your students, but to help boost their mental math abilities.
This allows students to answer collectively, so as to avoid singling kids out who may be struggling with mental math.
### 6. Food fractions
Kids love food — it’s no secret! So why not incorporate it into your lesson plan?
A teacher in the Tweet above got her kids to practice multiplying fractions by converting delicious food recipes.
You can have each student choose their favorite food and multiply the ingredients to feed the entire class.
An incentive might help, too! For example, once everyone has converted their favorite food, choose a safe snack the class will love.
Pull up the original recipe. Now have your students work together to multiply ingredients and, if they do it properly, the whole class will get a homemade (or store-bought) baked good!
### 7) Multiplied fraction pennant
Cut out ribbon- or triangle-shaped sheets of paper for each student. At the top, write “I can multiply…”
Below that, you’ll include:
• An area model illustrating the fractions they’re multiplying
• The multiplication problem itself (with space to show their work)
• A space at the bottom that reads “My product reduces to…”
Once every student has completed and decorated their fraction multiplication pennant, glue or tape them to a string!
Not only will this active learning activity help enforce how to multiply fractions, but it’ll give your students a confidence boost. It’s their work up on the wall, a problem they solved, and everyone can see it.
It’ll look something like this:
View this post on Instagram
A post shared by Sara Greenfield (@math_charger) on
## Excited to teach multiplying fractions now?
We hope so!
Multiplying fractions can be a daunting task — to learn and teach.
Hopefully the thorough breakdown of different types of fractions, how to multiply them, and how to make teaching them fun helps enrich you and your students’ teaching and learning experience, respectively.
### Read next: How to Divide Fractions in 3 Easy Steps
#### Jordan Nisbet
Jordan crafts content for Prodigy — and wishes the game existed when he was in school. He's interested in education and passionate about helping build up the next generation!
## One thought on “How to Multiply Fractions (+ 7 Engaging Activities)”
1. Becky says:
Great article. Thanks for sharing. 🙂
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# How to Find a Formula of the Power of a Matrix
## Problem 8
Let $A= \begin{bmatrix} 1 & 2\\ 2& 1 \end{bmatrix}$.
Compute $A^n$ for any $n \in \N$.
Contents
## Plan.
We diagonalize the matrix $A$ and use this Problem.
### Steps.
1. Find eigenvalues and eigenvectors of the matrix $A$.
2. Diagonalize the matrix $A$.
3. Use the result of this Problem.
## Proof.
We first diagonalize the matrix $A$. We solve
\begin{align*}
\det(A-\lambda I) & =\begin{vmatrix}
1-\lambda & 2\\
2& 1-\lambda
\end{vmatrix} \\
&=(1-\lambda)^2-4=\lambda^2-2\lambda-3 =(\lambda+1)(\lambda-3)=0
\end{align*}
and obtain the eigenvalues $\lambda=-1, 3$.
To find an eigenvector $\mathbf{x}$ corresponding to $\lambda=-1$, we solve $(A+I)\mathbf{x}=\mathbf{0}$ or
$\begin{bmatrix} 2 & 2\\ 2& 2 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \end{bmatrix}= \begin{bmatrix} 0 \\ 0 \end{bmatrix}.$ We obtain an eigenvector $\mathbf{x}=\begin{bmatrix} 1 \\ -1 \end{bmatrix}$ corresponding to $\lambda=-1$.
Similarly, solving $(A-3I)\mathbf{y}=\mathbf{0}$, we obtain an eigenvector $\mathbf{y}=\begin{bmatrix} 1 \\ 1 \end{bmatrix}$
corresponding to $\lambda=3$.
Thus the invertible matrix $S=[\mathbf{x} \, \mathbf{y}]=\begin{bmatrix} 1 & 1 \\ -1& 1 \end{bmatrix}$ diagonalizes the matrix $A$, that is,
$S^{-1}AS =\begin{bmatrix} -1 & 0\\ 0& 3 \end{bmatrix} \text{ or equivalently } A=S\begin{bmatrix} -1 & 0\\ 0& 3 \end{bmatrix} S^{-1}.$ Then for each $n \in \N$, we have
\begin{align*}
A^n &= \left (S\begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix} S^{-1} \right)^n
=S \begin{bmatrix}
-1 & 0\\
0& 3
\end{bmatrix}^n S^{-1}
=S \begin{bmatrix}
(-1)^n & 0\\
0& 3^n
\end{bmatrix} S^{-1} \6pt] &= \begin{bmatrix} 1 & 1 \\ -1& 1 \end{bmatrix} \begin{bmatrix} (-1)^n & 0\\ 0& 3^n \end{bmatrix} \frac{1}{2} \begin{bmatrix} 1 & -1 \\ 1& 1 \end{bmatrix} =\frac{1}{2} \begin{bmatrix} (-1)^n+3^n & (-1)^{n+1}+3^n\\ (-1)^{n+1}+3^n& (-1)^n+3^n \end{bmatrix}. \end{align*} (See this problem for the details of these computations.) Therefore we obtained the formula \[A^n=\frac{1}{2} \begin{bmatrix} (-1)^n+3^n & (-1)^{n+1}+3^n\\ (-1)^{n+1}+3^n& (-1)^n+3^n \end{bmatrix}.
## Comment.
Another typical method to compute a power of a square matrix is mathematical induction. To use it, we need to first compute several small powers like $A^2$ and $A^3$ and guess the formula for $A^n$.
If you can guess the formula, then the mathematical induction part is not difficult. But for this specific problem, the formula is a bit complicated to guess as you can see from the solution above. Thus we used diagonalization trick.
Let $A=\begin{bmatrix} a & 0\\ 0& b \end{bmatrix}$. Show that (1) $A^n=\begin{bmatrix} a^n & 0\\ 0& b^n \end{bmatrix}$ for any...
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# Numbers in Motion: Before and After in the World of Math
Subject: Mathematics
## Objective
By the end of the lesson, students will show an understanding of numbers before and after a given number on the number line, focusing on moving up and down.
## Materials
• Large number line displayed in the classroom
• Number cards (0-20)
• Flashcards with visual representations of objects (e.g., fruits, stars, animals)
• Whiteboard and markers
## Introduction (15 minutes)
Begin the lesson by reviewing counting from 0 to 20 as a class.
Say: Good morning! I hope you're ready for an exciting math adventure. Today, we're going to explore numbers in motion. But before we begin, let's quickly count to 20 together. I'll start, and you continue. Ready?
*Count to 20 together*
Say: Fantastic! Give yourselves a round of applause. Now that our counting skills are warmed up let's jump into our lesson about numbers in motion on the number line.
Introduce the concept of a number line by showing a large number line displayed in the classroom.
Emphasize that it represents the order of numbers from 0 to 20.
Say: A number line is like a road that helps us understand the order of our numbers. And guess what? It goes from 0 to 20, just like our counting warm-up. See these numbers? Each one helps us know where we are and where we're going.
Do: Ask your students to take turns coming to the front and pointing to different numbers on the number line while the class counts together.
## Guided Practice (15 minutes)
Write a series of numbers on the whiteboard and ask students to come up individually to circle the number before or after the given number.
Demonstrate with a few examples:
1. Write the numbers 5, 6, 7. Ask students to circle the before and after numbers 5 and 7.
2. Write the numbers 8, 9, 10. Ask students to circle the before and after numbers 8 and 10.
Provide positive reinforcement and correct any mistakes as they arise.
## Activity 1: Before and After (20 minutes)
Distribute number cards (0-20) to each student. (If you do not have number card class sets, have your students make their own number cards using index cards. Place their custom number cards on a ring to use throughout the school year.)
Explain that you will call out a number.
Your students will raise the number before and after the called number using their number cards.
Demonstrate with a few examples:
1. If the number is 7, ask students to show cards 6 and 8.
2. If the number is 12, ask students to show cards 11 and 13.
Complete the activity as a class, in small groups or pairs, taking turns calling out numbers and identifying the ones before and after.
## Activity 2: Numbers in Motion Game (25 minutes)
Divide the class into two teams.
Place a set of flashcards with visual representations of objects (fruits, stars, animals) at one end of the classroom.
Each team member will take turns selecting a number card.
The student must then count the correct number of objects, walk the number card to the classroom number line, and hold the card to the corresponding number.
Demonstrate with a few examples:
1. If the number on the card is 13, ask students to hold the card under 13.
2. If the number on the card is 2, ask students to hold the card under 2.
Once confirmed, the student will return their number card to their team, where the next student will draw a new number card.
As your students' knowledge increases, increase the difficulty by asking them to identify the before and after numbers.
Demonstrate with a few examples:
1. If the number on the card is 18, ask students to identify 17 and 19.
2. If the number on the card is 1, ask students to identify 0 and 2.
Make the activity a relay race with a treasure box prize as your students improve their number skills.
Encourage teamwork and congratulate each team for their efforts.
## Closure (5 minutes)
Gather your students together and review the concept of numbers before and after on the number line.
Assign a small task for homework, such as drawing a simple number line at home and placing numbers before and after a given number.
## Assessment
• Observe students during the activities to assess their ability to identify numbers before and after on the number line.
• Review their participation in class discussions and activities.
• Check their homework assignments for understanding.
Written by Brooke Lektorich
Education World Contributor
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# Unitary Method
Go back to 'Ratio, Proportion, Percentages'
The unitary method, in essence, is all about finding the “per unit value”. The unitary method is a technique for solving a problem by first finding the value of a single unit, and then finding the necessary value by multiplying the single unit value. This thus becomes important in more advanced ideas like determining the amount of time, resources or capital required for certain types of problems, often which have their own constraints.
Let’s try and understand this using an example,
The cost of 5 apples is ₹ 30.
The cost of 1 apple is = 30 ÷ 5 = ₹ 6
The cost of 7 apples = 7 x 6 = ₹ 42
The unitary method forms an important part of more advanced mathematical concepts such as optimization and operations research where problem statements are converted into equations and then solved.
For example, you go to a supermarket and see two offers on the same product (say sugar) from 2 different brands:
• Brand A is offering 500g sugar for Rs. 80
• Brand B is offering 750g of sugar for Rs. 10
Which brand is the better deal? With Unitary method, your life would be made extremely simple!
## Speed distance and time
Suppose you travel in a car from Chennai to Bengaluru. You reach Bengaluru in 7 hours.
The distance between Chennai and Bengaluru is 350 Km.
The time taken to travel is 7 hrs.
The distance covered in 1 hour = 350 km/7 hr
This measure of distance per unit of time is called speed
Formula: Speed is calculated as distance/time taken.
In this case the speed of the car is 50 km/ hr.
(In this example, we assume that the car was travelling at the same speed through its journey. Similarly, in the next set of questions we will assume that the speed remains constant throughout the journey.)
## Tips and Tricks
• Not always do you have to use the unitary method to find the per-unit value.
If 32 items cost Rs 40. How much will 12 items cost?
Here finding the cost of each item as per the unitary method will be tedious (32 ÷ 40).
Instead use the trick from equivalent fractions to find an equivalent fraction to \begin{align} 32\over 40\end{align} with numerator 12:
This says you’ll get 12 items per Rs 15. That is, the cost of 12 items is Rs 15.
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Division of Rational Expressions
Invert, cancel, multiply, and reduce fractions with variables in the denominator
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Division of Rational Expressions
Suppose that the distance traveled by a hot air balloon in miles can be represented by 8x38x,\begin{align*}8x^3-8x,\end{align*} while the speed of the hot air balloon in miles per hour can be represented by x2+x\begin{align*}x^2+x\end{align*}. Would you be able to find an expression for the time it takes for the hot air balloon to cover a certain distance? Could you evaluate the expression that you found for x=2\begin{align*}x=2\end{align*}
Dividing Rational Expressions
Division of rational expressions works in the same manner as multiplication. A reminder of how to divide fractions is below.
For any rational expressions a0,b0,c0,d0\begin{align*}a \neq 0, b \neq 0, c \neq 0, d \neq 0\end{align*},
ab÷cdabdc=adbc\begin{align*}\frac{a}{b} \div \frac{c}{d} \rightarrow \frac{a}{b} \cdot \frac{d}{c}=\frac{ad}{bc}\end{align*}
Simplify the following division problems:
1. 9x242x2÷21x22x81\begin{align*}\frac{9x^2-4}{2x-2} \div \frac{21x^2-2x-8}{1}\end{align*}
9x242x2÷21x22x819x242x2121x22x8\begin{align*}\frac{9x^2-4}{2x-2} \div \frac{21x^2-2x-8}{1} \rightarrow \frac{9x^2-4}{2x-2} \cdot \frac{1}{21x^2-2x-8}\end{align*}
Repeat the process for multiplying rational expressions.
9x242x2121x22x89x242x2÷21x22x81(3x2)(3x2)2(x1)1(3x2)(7x+4)=3x214x26x8\begin{align*}\frac{9x^2-4}{2x-2} \cdot \frac{1}{21x^2-2x-8} & \rightarrow \frac{(3x-2)\cancel{(3x-2)}}{2(x-1)} \cdot \frac{1}{\cancel{(3x-2)}(7x+4)}\\ \frac{9x^2-4}{2x-2} \div \frac{21x^2-2x-8}{1} &= \frac{3x-2}{14x^2-6x-8}\end{align*}
1. x2+3x105x+15÷x2x2+2x3\begin{align*}\frac{x^2+3x-10}{5x+15} \div \frac{x-2}{x^2+2x-3}\end{align*}
x2+3x105x+15÷x2x2+2x3x2+3x105x+15x2+2x3x2\begin{align*}\frac{x^2+3x-10}{5x+15} \div \frac{x-2}{x^2+2x-3} \rightarrow \frac{x^2+3x-10}{5x+15} \cdot \frac{x^2+2x-3}{x-2}\end{align*}
Repeat the process for multiplying rational expressions.
x2+3x105x+15x2+2x3x2(x+5)(x2)5(x+3)x2(x+3)(x1)x2+3x105x+15÷x2x2+2x3(x+5)(x2)5(x+3)x2(x+3)(x1)=x+551(x1)=x+55x5=x+55x5\begin{align*}\frac{x^2+3x-10}{5x+15} \cdot \frac{x^2+2x-3}{x-2} & \rightarrow \frac{(x+5)(x-2)}{5(x+3)} \cdot \frac{x-2}{(x+3)(x-1)}\\ \frac{(x+5)\cancel{(x-2)}}{5\cancel{(x+3)}} \cdot \frac{\cancel{x-2}}{(\cancel{x+3})(x-1)}&=\frac{x+5}{5} \cdot \frac{1}{(x-1)}=\frac{x+5}{5x-5}\\ \frac{x^2+3x-10}{5x+15} \div \frac{x-2}{x^2+2x-3} &= \frac{x+5}{5x-5}\end{align*}
Consider the following real-world application:
Suppose Marciel is training for a running race. Marciel’s speed (in miles per hour) of his training run each morning is given by the function x39x\begin{align*}x^3-9x\end{align*}, where x\begin{align*}x\end{align*} is the number of bowls of cereal he had for breakfast (1x6).\begin{align*}(1 \le x \le 6).\end{align*} Marciel’s training distance (in miles), if he eats x\begin{align*}x\end{align*} bowls of cereal, is 3x29x\begin{align*}3x^2-9x\end{align*}. What is the function for Marciel’s time and how long does it take Marciel to do his training run if he eats five bowls of cereal on Tuesday morning?
timetimetimeIf xtime=distancespeed=3x29xx39x=3x(x3)x(x29)=3x(x3)x(x+3)(x3)=3x+3=5,then=35+3=38\begin{align*}\text{time} &= \frac{\text{distance}}{\text{speed}}\\ \text{time} &= \frac{3x^2-9x}{x^3-9x}=\frac{3x(x-3)}{x(x^2-9)}=\frac{3x\cancel{(x-3)}}{x(x+3)\cancel{(x-3)}}\\ \text{time} &= \frac{3}{x+3}\\ \text{If} \ x &= 5, \text{then}\\ \text{time} &= \frac{3}{5+3}=\frac{3}{8}\end{align*}
Marciel will run for 38\begin{align*}\frac{3}{8}\end{align*} of an hour.
Examples
Example 1
Earlier, you were told that the distance traveled by a hot air balloon in miles can be represented by 8x38x\begin{align*}8x^3-8x\end{align*}, while the speed of the hot air balloon in miles per hour can be represented by x2+x\begin{align*}x^2+x\end{align*}. What expression would represent the time it takes to cover a distance? What is the time it takes the balloon to travel a distance when x=2\begin{align*}x=2\end{align*}?
We can determine the expression for time by dividing distance by speed.
timetimetime=distancespeed=8x38xx2x=8x2(x1)x(x+1)=8x8\begin{align*}\text{time} &= \frac{\text{distance}}{\text{speed}}\\ \text{time} &= \frac{8x^3-8x}{x^2-x}=\frac{8x^2(x-1)}{x(x+1)}\\ \text{time} &= 8x-8 \end{align*}
The expression for the time it takes the balloon to cover a given distance is 8x8\begin{align*}8x-8\end{align*}.
Now, solve for x=2\begin{align*}x=2\end{align*}
timetimetime=8(2)8=168=8 hours\begin{align*}\text{time} &= 8(2)-8\\ \text{time} &= 16-8\\ \text{time} &= 8\ \mathrm{hours}\end{align*}
When x=2\begin{align*}x=2\end{align*}, the balloon takes 8 hours to travel the given distance.
Example 2
Simplify 15x230x+40÷3x62x28x\begin{align*} \frac{1}{5x^2-30x+40} \div \frac{3x-6}{2x^2-8x}\end{align*}.
15x230x+40÷3x62x28x=15x230x+402x28x3x6=15(x2)(x4)2x(x4)3(x2)=15(x2)(x4)2x(x4)3(x2)=2x5(x2)2\begin{align*} \frac{1}{5x^2-30x+40} \div \frac{3x-6}{2x^2-8x} &= \frac{1}{5x^2-30x+40} \cdot \frac{2x^2-8x}{3x-6}\\ &= \frac{1}{5(x-2)(x-4)} \cdot \frac{2x(x-4)}{3(x-2)}\\ &= \frac{1}{5(x-2) \cancel{(x-4)}} \cdot \frac{2x \cancel{(x-4)}}{3(x-2)}\\ &=\frac{2x}{5(x-2)^2} \end{align*}
Review
In 1–10, perform the indicated operation and reduce the answer to lowest terms.
1. \begin{align*}2xy \div \frac{2x^2}{y}\end{align*}
2. \begin{align*}\frac{x^2}{x-1} \div \frac{x}{x^2+x-2}\end{align*}
3. \begin{align*}\frac{a^2+2ab+b^2}{ab^2-a^2b} \div (a+b)\end{align*}
4. \begin{align*}\frac{3-x}{3x-5} \div \frac{x^2-9}{2x^2-8x-10}\end{align*}
5. \begin{align*}\frac{x^2-25}{x+3} \div (x-5)\end{align*}
6. \begin{align*}\frac{2x+1}{2x-1} \div \frac{4x^2-1}{1-2x}\end{align*}
7. \begin{align*}\frac{3x^2+5x-12}{x^2-9} \div \frac{3x-4}{3x+4}\end{align*}
8. \begin{align*}\frac{x^2+x-12}{x^2+4x+4} \div \frac{x-3}{x+2}\end{align*}
9. \begin{align*}\frac{x^4-16}{x^2-9} \div \frac{x^2+4}{x^2+6x+9}\end{align*}
10. \begin{align*}\frac{x^2+8x+16}{7x^2+9x+2} \div \frac{7x+2}{x^2+4x}\end{align*}
11. Maria’s recipe asks for \begin{align*}2 \frac{1}{2} \ \text{times}\end{align*} as much flour as sugar. How many cups of flour should she mix in if she uses \begin{align*}3 \frac{1}{3} \ \text{cups}\end{align*} of sugar?
12. George drives from San Diego to Los Angeles. On the return trip, he increases his driving speed by 15 miles per hour. In terms of his initial speed, by what factor is the driving time decreased on the return trip?
13. Ohm’s Law states that in an electrical circuit \begin{align*}I=\frac{V}{R_{tot}}.\end{align*} The total resistance for resistors placed in parallel is given by \begin{align*}\frac{1}{R_{tot}}=\frac{1}{R_1}+\frac{1}{R_2}.\end{align*} Write the formula for the electric current in terms of the component resistances: \begin{align*}R_1\end{align*} and \begin{align*}R_2\end{align*}.
To see the Review answers, open this PDF file and look for section 12.6.
Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Vocabulary Language: English Spanish
TermDefinition
rational function A ratio of two polynomials (a polynomial divided by another polynomial). The formal definition is: $f(x)=\frac{g(x)}{h(x)}, \text{where} \ h(x) \neq 0$.
reciprocal The reciprocal of a nonzero rational expression $\frac{a}{b}$ is $\frac{b}{a}$.
Least Common Denominator The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators.
Least Common Multiple The least common multiple of two numbers is the smallest number that is a multiple of both of the original numbers.
Rational Expression A rational expression is a fraction with polynomials in the numerator and the denominator.
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# Multiplying 2 digit by 2 digit numbers
Multiplying 2 digit by 2 digit numbers – Question 1
Multiply :
23 x 31 = ?
Explanation
Step I: Whenever we multiply a two digit number by another 2 digit number, we first multiply the first number by ones place of second number, and then by first number by Tens place of the second number.
First we need to multiply 23 with 1
Multiply 1 by 3 in ones place.
1 x 3 = 3
Write 3 in ones place.
Multiply 1 by 2 in tens place.
1 x 2 = 2
Write 2 in tens place.
Step II: Since the second number is a Tens value, we add a 0 or a cross to the answer obtained by multiplying first number by second digit
Multiply 23 with 3 and before multiplying write ‘x’ under 3
Multiply 3 by 3 in ones place.
3 x 3 = 9
Write 9 in tens place under 2
Multiply 3 by 2 in tens place.
3 x 2 = 6
Write 6 in hundreds place before 9
Step III: Now add both the parts which have been multiplied.
3 + 0 = 3
2 + 9 = 11
Since, the number is more than 9, we need to carry over 1 to the hundreds place.
Add hundreds place and the carry of 1
6 + 1 = 7
Hence, 23 x 31 = 713
Multiplying 2 digit by 2 digit numbers – Question 2
Multiply :
24 x 23 = ?
Explanation
Step I: Whenever we multiply a two digit number by another 2 digit number, we first multiply the first number by
ones place of second number, and then by first number by Tens place of the second number.
First we need to multiply 24 with 3
Multiply 3 by 4 in ones place.
3 x 4 = 12
Since, the number is more than 9, we need to carry over 1 to the tens place.
Write 2 in ones place.
Multiply 3 by 2 in tens place.
3 x 2 = 6
Add the carry of 1 to 6 i.e, 1 + 6 = 7
Write 7 in tens place.
Step II: Since the second number is a Tens value, we add a 0 or a cross to the answer obtained by multiplying first number by second digit
Multiply 24 with 2 and before multiplying write ‘x’ under 2
Multiply 2 by 4 in ones place.
2 x 4 = 8
Write 8 in tens place under 7
Multiply 2 by 2 in tens place.
2 x 2 = 4
Write 4 in hundreds place before 8
Step III: Now add both the parts which have been multiplied.
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# 4.4 Graphs of logarithmic functions (Page 5/8)
Page 5 / 8
Given a logarithmic function with the form $\text{\hspace{0.17em}}f\left(x\right)=a{\mathrm{log}}_{b}\left(x\right),$ $a>0,$ graph the translation.
1. Identify the vertical stretch or compressions:
• If $\text{\hspace{0.17em}}|a|>1,$ the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ is stretched by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units.
• If $\text{\hspace{0.17em}}|a|<1,$ the graph of $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ is compressed by a factor of $\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$ units.
2. Draw the vertical asymptote $\text{\hspace{0.17em}}x=0.$
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by multiplying the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ coordinates by $\text{\hspace{0.17em}}a.$
4. Label the three points.
5. The domain is $\text{\hspace{0.17em}}\left(0,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=0.$
## Graphing a stretch or compression of the parent function y = log b ( x )
Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=2{\mathrm{log}}_{4}\left(x\right)\text{\hspace{0.17em}}$ alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
Since the function is $\text{\hspace{0.17em}}f\left(x\right)=2{\mathrm{log}}_{4}\left(x\right),$ we will notice $\text{\hspace{0.17em}}a=2.$
This means we will stretch the function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{4}\left(x\right)\text{\hspace{0.17em}}$ by a factor of 2.
The vertical asymptote is $\text{\hspace{0.17em}}x=0.$
Consider the three key points from the parent function, $\text{\hspace{0.17em}}\left(\frac{1}{4},-1\right),$ $\left(1,0\right),\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(4,1\right).$
The new coordinates are found by multiplying the $\text{\hspace{0.17em}}y\text{\hspace{0.17em}}$ coordinates by 2.
Label the points $\text{\hspace{0.17em}}\left(\frac{1}{4},-2\right),$ $\left(1,0\right)\text{\hspace{0.17em}},$ and $\text{\hspace{0.17em}}\left(4,\text{2}\right).$
The domain is $\text{\hspace{0.17em}}\left(0,\text{\hspace{0.17em}}\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),\text{\hspace{0.17em}}$ and the vertical asymptote is $\text{\hspace{0.17em}}x=0.\text{\hspace{0.17em}}$ See [link] .
The domain is $\text{\hspace{0.17em}}\left(0,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=0.$
Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}\text{\hspace{0.17em}}{\mathrm{log}}_{4}\left(x\right)\text{\hspace{0.17em}}$ alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.
The domain is $\text{\hspace{0.17em}}\left(0,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=0.$
## Combining a shift and a stretch
Sketch a graph of $\text{\hspace{0.17em}}f\left(x\right)=5\mathrm{log}\left(x+2\right).\text{\hspace{0.17em}}$ State the domain, range, and asymptote.
Remember: what happens inside parentheses happens first. First, we move the graph left 2 units, then stretch the function vertically by a factor of 5, as in [link] . The vertical asymptote will be shifted to $\text{\hspace{0.17em}}x=-2.\text{\hspace{0.17em}}$ The x -intercept will be $\text{\hspace{0.17em}}\left(-1,0\right).\text{\hspace{0.17em}}$ The domain will be $\text{\hspace{0.17em}}\left(-2,\infty \right).\text{\hspace{0.17em}}$ Two points will help give the shape of the graph: $\text{\hspace{0.17em}}\left(-1,0\right)\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\left(8,5\right).\text{\hspace{0.17em}}$ We chose $\text{\hspace{0.17em}}x=8\text{\hspace{0.17em}}$ as the x -coordinate of one point to graph because when $\text{\hspace{0.17em}}x=8,\text{\hspace{0.17em}}$ $\text{\hspace{0.17em}}x+2=10,\text{\hspace{0.17em}}$ the base of the common logarithm.
The domain is $\text{\hspace{0.17em}}\left(-2,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=-2.$
Sketch a graph of the function $\text{\hspace{0.17em}}f\left(x\right)=3\mathrm{log}\left(x-2\right)+1.\text{\hspace{0.17em}}$ State the domain, range, and asymptote.
The domain is $\text{\hspace{0.17em}}\left(2,\infty \right),$ the range is $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and the vertical asymptote is $\text{\hspace{0.17em}}x=2.$
## Graphing reflections of f ( x ) = log b ( x )
When the parent function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ is multiplied by $\text{\hspace{0.17em}}-1,$ the result is a reflection about the x -axis. When the input is multiplied by $\text{\hspace{0.17em}}-1,$ the result is a reflection about the y -axis. To visualize reflections, we restrict $\text{\hspace{0.17em}}b>1,\text{\hspace{0.17em}}$ and observe the general graph of the parent function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ alongside the reflection about the x -axis, $\text{\hspace{0.17em}}g\left(x\right)={\mathrm{-log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ and the reflection about the y -axis, $\text{\hspace{0.17em}}h\left(x\right)={\mathrm{log}}_{b}\left(-x\right).$
## Reflections of the parent function y = log b ( x )
The function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{-log}}_{b}\left(x\right)$
• reflects the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ about the x -axis.
• has domain, $\text{\hspace{0.17em}}\left(0,\infty \right),$ range, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and vertical asymptote, $\text{\hspace{0.17em}}x=0,$ which are unchanged from the parent function.
The function $\text{\hspace{0.17em}}f\left(x\right)={\mathrm{log}}_{b}\left(-x\right)$
• reflects the parent function $\text{\hspace{0.17em}}y={\mathrm{log}}_{b}\left(x\right)\text{\hspace{0.17em}}$ about the y -axis.
• has domain $\text{\hspace{0.17em}}\left(-\infty ,0\right).$
• has range, $\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ and vertical asymptote, $\text{\hspace{0.17em}}x=0,$ which are unchanged from the parent function.
The center is at (3,4) a focus is at (3,-1), and the lenght of the major axis is 26
The center is at (3,4) a focus is at (3,-1) and the lenght of the major axis is 26 what will be the answer?
Rima
I done know
Joe
What kind of answer is that😑?
Rima
I had just woken up when i got this message
Joe
Rima
i have a question.
Abdul
how do you find the real and complex roots of a polynomial?
Abdul
@abdul with delta maybe which is b(square)-4ac=result then the 1st root -b-radical delta over 2a and the 2nd root -b+radical delta over 2a. I am not sure if this was your question but check it up
Nare
This is the actual question: Find all roots(real and complex) of the polynomial f(x)=6x^3 + x^2 - 4x + 1
Abdul
@Nare please let me know if you can solve it.
Abdul
I have a question
juweeriya
hello guys I'm new here? will you happy with me
mustapha
The average annual population increase of a pack of wolves is 25.
how do you find the period of a sine graph
Period =2π if there is a coefficient (b), just divide the coefficient by 2π to get the new period
Am
if not then how would I find it from a graph
Imani
by looking at the graph, find the distance between two consecutive maximum points (the highest points of the wave). so if the top of one wave is at point A (1,2) and the next top of the wave is at point B (6,2), then the period is 5, the difference of the x-coordinates.
Am
you could also do it with two consecutive minimum points or x-intercepts
Am
I will try that thank u
Imani
Case of Equilateral Hyperbola
ok
Zander
ok
Shella
f(x)=4x+2, find f(3)
Benetta
f(3)=4(3)+2 f(3)=14
lamoussa
14
Vedant
pre calc teacher: "Plug in Plug in...smell's good" f(x)=14
Devante
8x=40
Chris
Explain why log a x is not defined for a < 0
the sum of any two linear polynomial is what
Momo
how can are find the domain and range of a relations
the range is twice of the natural number which is the domain
Morolake
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
|
# Determine mean and standard deviation of first n terms of an A.P whose first term is 'a' and common difference d.
The arithmetic series of AP is $a_n= a+ (n-1)d$
$a_n =a+ (n-1) d$
Now, The arithematic series in two different ways .
Step 1:
$S_n =a_1 +(a_1+d) +(a_1+2d)+ .......(a_1+(n-2) d)+(a_1+(n-1) d)$----(1)
$S_n=(a_n-(n-1)d )+(a_n-(n-2)d)+.......(a_n-2d) +(a_n-d) +a_n$---(2)
Adding both sides of the equations
$2S_n=n (a_1+a_n)$
$S_n= \large\frac{n}{2} $$(a_1+a_n) \therefore a_n= a_1 +(n-1)d \therefore S_n= \large\frac{n}{2}$$ [a_1 +a_1 +(n-1)d]$
$\qquad=\large\frac{n}{2} $$[ 2a_1+(n-1)d] Sum of the series= \large\frac{n}{2}$$[2a_1+(n-1)d]$
Step 2:
$Mean =\large\frac{sum\;of\;the \;series}{n}$
$\qquad= \large\frac{\Large\frac{n}{2} [2a_1+(n-1) d]}{n}$
$\qquad= \large\frac{1}{2} $$[2a_1+(n-1)d] \qquad= a_1+\large\frac{(n-1)}{2}$$d$
$S.D= \sqrt {variance}$
Variance = $\large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i }{n}\bigg)^2$
As $a_n = a_1+(n-1)d$
Where $i=0$ to $n$
$\sum x_i ^2= a_1^2+(a_1+d)^2+(a_1+2d)^2+.....(a_1+(n-1)d)$
$=> na_1^2+d^2(1+4+9+16+......(n-1)^2) + 2a_1d( 1+2+3+4+........(n-1))$
$=> \sum x_i ^2 =na_1^2+d^2 \bigg[ \large\frac {n(n-1)(2n-1)}{6} \bigg]+$$2a_1d \bigg[\large\frac{n(n-1)}{2} \bigg] \large\frac{\sum x_i^2}{n}$$=a_1^2+\large\frac{d^2(n-1)(2n-1)}{6}$$+ a_1d (n-1)$
Variance = $\large\frac{\sum x_i^2}{n} - \bigg( \large\frac{\sum x_i }{n}\bigg)^2$
After solving the equation
$\qquad= \large\frac{d^2 (n^2-1)}{12}$
$\qquad= \sqrt { \large\frac{ d^2 (n^2-1)}{12}}$
$\qquad= d \sqrt {\large \frac{n^2-1}{12}}$
|
# How to Find LCM | Least Common Multiple, Definition, Meaning, Solved Problems
Contents
The full form of LCM is the Least Common Multiple. In mathematics, the least common multiple (LCM) is a method to find the lowest possible common number of two numbers that are divisible by both numbers. LCM can be calculated for two or more numbers. LCM is also commonly referred to as the Least Common Divisor (LCD).
There are three methods for determining the LCM of a given number: the listing method, prime factorization method, and division method. In this article, let us discuss everything about how to find the LCM of a given number with solved examples. Scroll down to find out more.
## What is the Least Common Multiple (LCM) in Maths?
The smallest positive number that is a multiple of two or more numbers.LCM Definition
For example, the LCM of 4 and 9 is 2 * 2 * 3 * 3 = 36.
Here, 4 is expressed as 2 * 2, and 9 is expressed as 3 * 3.
If we consider the multiples of 4 and 9, we get:
• Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 26,…
• Multiples of 9: 9, 18, 27, 36,…
The first common multiple for 4 and 9 is 36. Hence, the LCM of 4 and 9 is 36.
LCM is used for the addition and subtraction of two fractions. When the denominator value of the fractions is not the same, LCM is used to make the denominators equal. This makes the entire calculating process easy.
## Properties of LCM
There are certain properties of LCM that you should learn before knowing how to find LCM.
### 1. Associative Property of LCM
The associative property of LCM states that the LCM of A and B will be the same as the LCM of B and A.
LCM (A, B) = LCM (B, A)
For example, let’s consider A as 6 and B as 2. Here, 6 can be expressed as 2 * 3, and 2 can be expressed as 2.
So, 6 * 2 = 2 * 3 * 2.
Now, we will write the numbers in the exponent form and multiply the factors that have the highest power.
On writing the number in exponent form, we get:
6 = 2 * 3 = 21 * 31
2 = 21
So, the LCM of 6 and 2 is 21 * 31 = 6
And, LCM of 2 and 6 is also 6.
Hence, it is proved that LCM (6, 2) = LCM (2, 6) = 6.
### 2. Commutative Property of LCM
The commutative property is used while dealing with finding the LCM of 3 numbers. The commutative property of LCM states that,
LCM (A, B, C) = LCM (LCM (A, B), C) = LCM (A, LCM (B, C))
For example, let’s consider A as 3, B as 6, and C as 12. Here, 3 can be expressed as 3, 6 can be expressed as 2 * 3, and 12 can be expressed as 2 * 2 * 3.
Now, we will write the numbers in the exponent form and multiply the factors that have the highest power.
On writing the number in exponent form, we get:
3 = 31
6 = 2 * 3 = 21 * 31
12 = 2 * 2 * 3 = 22 * 31
So, the LCM of 3, 6, and 12 is 22 * 31 = 2 * 2 * 3 = 12
Now, the LCM of A and B, that is, LCM 3 and 6 is 31 * 21 = 6 and
LCM of (A, B) and C, that is, LCM of 6 and 12 is 22 * 31 = 2 * 2 * 3 = 12.
LCM (LCM (A, B), C) = LCM (LCM (3, 6), 12) = LCM (6, 12) = 12
LCM of B and C, that is, LCM of 6 and 12 is 22 * 31 = 2 * 2 * 3 = 12 and
LCM of A and LCM of (B, C), that is, LCM of 3 and 12 is 22 * 31 = 2 * 2 * 3 = 12.
LCM (A, LCM (B, C)) = LCM (3, LCM (6, 12)) = LCM (3, 12) = 12.
Hence, it is proved that LCM (3, 6, 12) = LCM (LCM (3, 6), 12) = LCM (3, LCM (6, 12)) = 12.
### 3. Distributive Property of LCM
The distributive property is also used while dealing with finding the LCM of 3 numbers. Distributive property of LCM states that,
LCM (dA, dB, dC) = d * LCM (A, B, C)
For example, let’s consider A as 5, B as 8, C as 13, and d to be any random variable.
Now, we will write the numbers in the exponent form and multiply the factors that have the highest power.
On writing the number in exponent form, we get:
5 = 51
8 = 2 * 2 * 2 = 23
13 = 131
LCM of 5, 8, and 13 is 51 * 23 * 131 = 5 * 2 * 2 * 2 * 13 = 520.
So, LCM (5d, 8d, 13d) = d * LCM (5, 8, 13) = 520
Hence, it is proved that LCM (5d, 8d, 13d) = d * LCM (5, 8, 13)
## How to Find the LCM?
There are three major methods for finding the LCM of two or more numbers. The methods are:
### 1. Division Method
To find the LCM using the division method, divide the given numbers by the smallest prime number, which is divisible by any of the given numbers. Then, the prime factors further obtained will be used to calculate the final LCM.
You can follow the following steps to find the LCM using the division method:
• Step 1: Write all the given numbers for which you have to find the LCM, separated by commas.
• Step 2: Now, find the smallest prime number which is divisible for any of the given two numbers.
• Step 3: If any number is not divisible, write that number in the next row just below it and proceed further.
• Step 4: Continue dividing the numbers obtained after each step by the prime numbers, until you get the result as 1 in the entire row.
• Step 5: Now, multiply all the prime numbers and the final result will be the LCM of the given numbers.
For example, you have to find the LCM of 12 and 5 using the division method.
So, LCM of 12 and 5 = 2 * 2* 3 * 5 = 50
### 2. Prime Factorization Method
To find the LCM of the given numbers using the prime factorization method, follow the steps given below:
• Step 1: Find the prime factors of the given numbers using the repeated division method explained above.
• Step 2: Write the prime factors in their exponent forms. Then multiply the prime factors having the highest power.
• Step 3: The final result after multiplication will be the LCM of the given numbers.
For example, you have to find the LCM of 18, 10, and 7 using the prime factorization method.
• Prime factorization of 18 can be expressed as 2 * 3 * 3 = 21 * 32
• Prime factorization of 10 can be expressed as 2 * 5 = 21 * 51
• Prime factorization of 7 can be expressed as 71
So, the LCM of 18, 10, and 7 = 21 * 32 * 51 * 71 = 2 * 3 * 3 * 5 * 7 = 630.
### 3. Listing Method
To find the LCM of the given numbers using the listing method, you can follow the following steps:
• Step 1: Write down the first few multiples of the given numbers separately.
• Step 2: Out of all the multiples of the numbers focus on the multiples which are common to all the given numbers.
• Step 3: Now, find out of all the common multiples, and take out the smallest common multiple. That will be the LCM of the given numbers
For example, you have to find the LCM of 8 and 5 using the listing method.
• Multiples of 8 are 8, 16, 24, 32, 40, 48, 64,…
• Multiples of 5 are 5, 10, 15, 20, 25, 30, 35, 40, 45,…
Here, it is clear that the least common multiple is 40.
So, the LCM of 8 and 5 is 40.
## Important LCM Formulas
There are two major LCM formulas, one for finding the LCM of integers and the other for finding the LCM of fractions.
Before moving forward and knowing the formulas, you should know about the HCF (Highest Common Factor).
HCF is the highest factor which is common among the factors of all the given numbers. It is also known as the greatest common divisor (GCD).
### 1. Formula for finding the LCM of the given integers
Let A and B be two given integers. So, the LCM of A and B can be calculated using the formula:
LCM (A, B) = (A * B) / HCF (A, B),
where HCF is the highest common factor or the greatest common divisor of A and B.
Another formula for finding the LCM of the given integers is:
A * B = LCM (A, B) * HCF (A, B), that is,
The product of the two given integers is equal to the product of their LCM and HCF.
### 2. Formula for finding the LCM of the given fractions
LCM = LCM of the numerator / HCF of the denominator
LCM List
## Solved Example Problems Based on LCM
Question 1: Find the LCM of 9 and 4 using the listing method.
Solution:
Multiples of 9 are 9, 18, 27, 36, 45, 54, 63, 72, 81, 90,…
Multiples of 4 are 4, 8, 12, 16, 20, 24, 28, 32, 36, 40,…
Here, it is clear that the least common multiple is 36.
So, the LCM of 9 and 4 is 36.
Question 2: What is the LCM of 16 and 21 using the prime factorization method?
Solution:
Prime factorization of 16 can be expressed as 2 * 2 * 2 * 2 = 24
Prime factorization of 21 can be expressed as 3 * 7 = 31 * 71
So, the LCM of 16 and 21 = 24 * 31 * 71 = 2 * 2 * 2 * 2 * 3 * 7 = 336.
Question 3: If the LCM and HCF of two numbers, 5 and B, are 45 and 1, respectively. Find B.
Solution:
As we know,
Product of two numbers = LCM * HCF
We are given that,
One of the numbers = 5, LCM = 45, and HCF = 1
So, 5 * B = 45 * 1
B = (45 * 1) / 5
B = 9
Hence, the other number is 9.
Question 4: Find the LCM of 24 and 45 using the division method.
Solution:
So, the LCM of 24 and 45 = 2 * 2 * 2 * 3 * 3 * 5 = 360.
Question 5: Find the LCM of 14, 22, and 18 using the prime factorization method.
Solution:
Prime factorization of 14 can be expressed as 2 * 7 = 21 * 71
I can express the prime factorization of 22 as 2 * 11 = 21 * 111
Prime factorization of 18 can be expressed as 2 * 3 * 3 = 21 * 32
So, the LCM of 14, 22, and 18 = 21 * 71 * 111 * 32 = 2 * 7 * 11 * 3 * 3 = 1386.
Question 6: If the HCF of two numbers, 42 and 9, is 3. Find the LCM.
Solution:
As we know,
LCM (A, B) = (A * B) / HCF (A, B)
We are given that,
Numbers = 42 and 9 and HCF = 3
So, LCM (42, 9) = (42 * 9) / HCF (42, 9)
= 378 / 3
= 126
Hence, the LCM of 42 and 9 is 126.
## FAQs on LCM
What do you mean by LCM?
LCM is the least common multiple. It is used to find the lowest possible common number that is divisible by all the numbers, for which you have to find the LCM.
How do LCM and HCF are connected?
LCM (A, B) = (A * B) / HCF (A, B),
where A and B are two integers.
This formula is used to find the LCM of the given integers.
Can LCM be calculated for only 2 numbers?
No, LCM can be calculated for more than two numbers as well. There should be at least 2 numbers for finding the LCM.
What are the methods to find the LCM?
There are three major methods for finding the LCM of two or more numbers. The methods are:
1. Division Method
2. Prime Factorization Method
3. Listing Method
What is the LCM of 2 and 8?
LCM of 2 and 8 is 8.
What are the properties of LCM?
There are three major properties of LCM. They are:
1. Associative property
2. Commutative property
3. Distributive property
What is the formula for finding the LCM of the fraction?
The formula for finding the LCM of a given fraction is:
LCM = LCM of the numerator / HCF of the denominator
We hope this article on LCM is helpful to you.
Practice Quiz
Questions: 1/2
|
# Building a concept: Angles
One concept that the Common Core requires of fourth-graders is that they develop a good understanding of what an angle is, what the different types of angles are, and the concept that a full circle is 360°. It is a relatively new concept for many students, so I work hard to try to build the foundation before I introduce some of the nitty-gritty math and protractor work.
Today we did a quick and easy activity that I thought you might be interested in. First of all, I punched out 6 inch circles from our Ellison die cutter. We certainly could have used a tracer and made our own circles, but I really wanted these to be precise so that we could get good and accurate folds.
I then set up our document camera to model and we got to work! Step one was to reinforce yesterday’s lesson which talked about circles being 360°. We also used the term rotation and practice rotating our bodies 90°, 180,° and so on. Today I really wanted to make the connection between the circle and measuring angles and the fraction unit we are finishing up.
The first step was to have the students label one side of their circle as 360°. I really wanted them to remember that the entire circle represents 360°. Earlier, I had shown them a nifty image that showed them what it would look like to have a 360° circle divided into 360 degrees…they were very impressed with all those tiny little angles!
We then flipped our circles over and began the process of dividing the circle into different sized pieces. Our first step was to fold it in half and talk about how taking 360 and dividing into two equal groups gave us two groups of 180°. We connected this to fractions by reviewing that two equal pieces are “halves”.
I had students label half of the circle 180 degrees and then had them shade it in. I then asked them to talk to a neighbor about what would happen if we folded it AGAIN. Students were pretty quick to realize that half of the 180 was the right angle we talked about yesterday. We folded the circle again, marked half of the blank section as 90 degrees and shaded it a new color. Students knew immediately that these were fourths, so we talked about how four, 90 degree angles would add up to 360 degrees, 2 of them would be
180 degrees, and so on.
At this point I reviewed the way we correctly mark angles (with an arc for most angles and a small square for right angles) and we then went on a “hunt” for right angles. For some reason, the concept of “bigger than” and “smaller than” a right angle concept is tricky for some students, so we really talked about “opening” and “closing” and I modeled with two rulers attached at the ends to show right angles, and then smaller and bigger angles. We’ll give them their “names” tomorrow! After students identified a bunch of right angles in the room, we talked about how important right angles are in building and in life to keep things straight and lined up!
We returned to our desks to divide our right angles in half to make 45’s and then again to make 22 1/2 degree angles! We had great talks about the “whole” circle, what it would be like to fold it into 1 degree angles, and how these sections relate to fractions!
I think students had a much better understanding of “estimating” with angles, and we’ll warm up tomorrow by doing some sketching to see what concepts stuck! Have a great day…and more angle activities will be coming up!
|
# ML Aggarwal Solutions Class 9 Mathematics Solutions for Trigonometric Ratios Exercise 17 in Chapter 17 - Trigonometric Ratios
Question 11 Trigonometric Ratios Exercise 17
Given A is an acute angle and 13 sin A 5, evaluate: (5 sin A – 2 cos A)/ tan A
The ratio of lengths of sides opposing the angle and the hypotenuse defines the sine of an angle. It is symbolised by the letter sinθ
The ratio of lengths of neighbouring sides to the angle and the hypotenuse defines the cosine of an angle. It is denoted by the symbol cosθ
The ratio of the length of the sides opposite the angle and the side adjacent to the angle defines the tangent of an angle. Tanθ is the symbol for it.
Let triangle ABC be a right-angled triangle at B and A is an acute angle
Given that 13 sin A = 5
Sin A = 5/13
AB/Ac = 5/13
Let AB = 5x
AC = 13 x
In right-angled triangle ABC,
Using Pythagoras theorem,
We get
\begin{array}{l} A C^{2}=A B^{2}+B C^{2} \\ B C^{2}=A C^{2}-B C^{2} \\ B C^{2}=(13 x)^{2}-(5 x)^{2} \\ B C^{2}=169 x^{2}-25 x^{2} \\ B C^{2}=144 x^{2} \\ B C=12 x \\ \sin A=5 / 13 \end{array}
Cos A = base/ hypotenuse
= BC/AC
= 12x/ 13x
= 12/13
Tan A = perpendicular/ base
= AB/BC
= 5x/ 12x
= 5/ 12
Now,
(5 sin A – 2 cos A)/ tan A = [(5) (5/13) – (2) (12/13)]/ (5/12)
= (1/13)/ (5/12)
= 12/65
Hence (5 sin A – 2 cos A)/ tan A = 12/65
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# ROUTH HURWITZ ANALYSIS
The Routh Hurwitz analysis tells you how many roots are located in the a) left-hand plane, b) right-hand plane, and c) on the jω-axis. The technique is illustrated here with an example. The Routh Hurwitz analysis involves creating an array of values. There are two special cases which may be involved, 1) a row of the array is all zeros (this happens in our example) and 2) a row begins with zero but is not all zeros (this does not occur in our example but is covered on the last page).
THE PROBLEM
Given the system equation
s 8 + 2s 7 + 8s 6 + 12s 5 + 20 s 4 + 16 s 3 + 16s 2 = 0
find the number of roots located in the left-hand plane, right-hand plane, and on the jω-axis.
FIND THE ZEROS
Since the equation can be factored by s2, there are two zeros in the polynomial, i.e. two roots at the origin.
s 2 s 6 + 2s 5 + 8s 4 + 12s 3 + 20s 2 + 16s + 16 = 0 STARTING THE ROUTH HURWITZ ARRAY
We can drop the s2 factor and consider only the remaining equation
(
)
s 6 + 2s 5 + 8s 4 + 12 s 3 + 20 s 2 + 16 s + 16 = 0
The order of this polynomial is n=6, so we will be construction an array having n+1 or 7 rows. Taking the coefficients, we construct the first two rows of the array in this order
1 8 20 16 2 12 16
http://www.teicontrols.com/notes
tomzap@eden.com
page 1 of 5
THE NEXT ROWS OF THE ROUTH HURWITZ ARRAY
Subsequently, each new row is derived from the two rows immediately above it, which I will call the working rows. The first column of the 2×2 matrix in our expressions is always the first column of the two working rows of the Routh Hurwitz array. The remaining column is the column of the working rows just to the right of the position of the unknown. The denominator of the expression is the first number of the lower of the two working rows. So now we have
1
8 20 16 0 b1 =
2 12 16 b1 b2 b3
1
8
− b2 = − b3 = 1 8 20 16 2 12 16 0 2 12 16
2 12 =2 2 1 20 2 16 = 12 2 1 16 2 2 0 = 16
SPECIAL CASE #1: WHEN A ROW IS ALL ZEROS
For the fourth row
1 8 20 16 2 12 16 0 2 12 16 0 c1 c2 c3
2 12 2 12 c1 = =0 2 2 16 − 2 16 c2 = =0 2 c3 = 0 −
and we have
1
8 20 16 0 0
2 12 16 2 12 16 0 0 0
We take elements of the last non-zero row and form an auxiliary equation:
2s (
n +1− K ) − 0
+ 12s ( n +1− K ) −2 + 16s ( n +1− K ) −4 = 0
where n is the order of the polynomial (6 in this case), and K is the row number of the row just above the 0s, from which these values are taken (3 in this case). Note the pattern of subtracting nothing from the first exponent, subtracting 2 from the second, and subtracting 4 from the third, etc. So the auxiliary equation becomes
2s 4 + 12s 2 + 16 = 0
http://www.teicontrols.com/notes
tomzap@eden.com
page 2 of 5
Now we take the derivative with respect to s
d ( 2s 4 + 12s 2 + 16) = 8s 3 + 24s + 0 ds
We replace row 4 with these new coefficients
1
8 20 16 0 0
2 12 16 2 12 16 8 24 0 THE FIFTH ROW
For the fifth row
1 2 2 8 d1
8 20 16 12 16 0 12 16 0 24 0 0 d 2 d3
− d1 = − d2 = d3 = 0
2 12 8 24 =6 8 2 16 8 8 0 = 16
So we have
1 2 2 8 6
8 20 16 12 16 12 16 24 0 16 0 0 0 0
http://www.teicontrols.com/notes
tomzap@eden.com
page 3 of 5
THE SIXTH ROW
For the sixth row
1 2 2 8 6 e1
8 20 16 12 16 12 16 24 0 16 0 e2 e3 0 0 0 0 e1 =
− e2 = 0 e3 = 0
8 24 6 16 = 2.667 6
So we have
1 2 2 8 6 2.667
8 20 16 12 16 12 16 24 0 16 0 0 0 0 0 0 0
THE SEVENTH ROW
For the seventh row
1 2 2 8 6 2.667 f1 1 2 2 8 6 2.667 16
8 20 16 12 16 0 12 16 0 24 0 0 16 0 0 0 0 0 f2 8 20 16 12 16 0 12 16 0 24 0 0 16 0 0 0 0 0 0 0 0
− f1 = f2 = 0
6 16 2.667 0 = 16 2.667
So we have
We now have the required n+1 rows.
http://www.teicontrols.com/notes
tomzap@eden.com
page 4 of 5
THE NUMBER OF ROOTS IN THE RHP
Count the number of sign changes in the first column (0 in this case). This is the number of roots in the right-hand plane.
THE NUMBER OF ROOTS ON THE jω-AXIS ω
Solving the auxiliary equation, we find 4 roots to that equation, so there are 4 roots on the jω-axis (so far).
THE NUMBER OF ROOTS IN THE LHP
Since n=6 and we have accounted for 4 of these roots, the remaining 2 roots are in the left-hand plane.
THE FINAL RESULT
Returning to the beginning of the problem, we had factored out an s2 from the original polynomial. These are zeros of the polynomial and are therefore on the jω-axis. And since the polynomial is in the denominator of the transfer function, these are additional roots on the jω-axis and are added to the other 4 roots we have found. So the final result is 6 roots on the jω-axis, 0 roots in the righthand plane and 2 roots in the left-hand plane.
SPECIAL CASE #2: WHAT IF A ROW BEGINS WITH 0?
This special case did not occur in our example, so it is covered here. If a row should begin with a zero and the row is not all zeros, we replace this leading zero with an ε. This ε is considered to be a small positive number and expressions containing ε that appear in column 1 of subsequent rows are determined to be positive or negative based on this characteristic.
http://www.teicontrols.com/notes
tomzap@eden.com
page 5 of 5
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# Lab46 Wiki
### Sidebar
information
support
math:exponents
~~TOC~~
Lab46 Tutorials
Exponents
Basic Rules & Properties
Rule 1
<latex>a^na^m = a^{n+m}</latex>
Rule 2
<latex>\frac{a^m}{a^n} = a^{m-n}</latex>
Rule 3
<latex>a^{-n} = \frac{1}{a^n} = (\frac{1}{a})^n</latex>
Rule 4
<latex>a^0 = 1, a \neq 0</latex>
Rule 5
<latex>(ab)^m = a^mb^m</latex>
Rule 6
<latex>(a^m)^n = a^{mn}</latex>
Derived
Take a perfect square number, let's use nine. Multiply the number by itself ( nine times is 81 ); this is written <latex>9*9=81</latex>. 9 times 9 can also be written as <latex>9^2=81</latex>, that is, 9 squared, or 9 raised to the second power. Since the substitution property allows 9 to be to be written as <latex>3*3=9</latex>, the previous could be written as <latex>(3*3)^2=81</latex>. That is, the product of 3 repeated 3 times, repeated another 9 (or <latex>3*3</latex>) times is 81. 3 times 3 can be written as <latex>3^2</latex>, therefore, <latex>(3^2)^2=81</latex>. Written out long-hand, the product of 3 squared raised to the second power looks like this: <latex>(3*3)*(3*3)=81 \to (3*3)^2 \to 3^{2*2} \to 3^4</latex>.
• <latex>9*9=81 \to 9^2=81 \to (3*3)^2=81 & \to (3^2)^2=81 \to 3^{2*2} = 81 \to 3^4 = 81</latex>
Rule 7
<latex>(\frac{a}{b})^m = \frac{a^m}{b^m}</latex>
Rule 8
<latex>|a^2| = |a|^2 = a^2</latex>
Rule 9
<latex>\sqrt{a} = a^\frac{1}{2}</latex>
Rule 10
<latex>\sqrt[n]{a} = a^\frac{1}{n}</latex>
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Posted by
Last updated on September 25th, 2020
The radical symbol (β) represents the square root of a number. You can multiply any two radicals that have the same indices (degrees of a root) together. If the radicals do not have the same indices, you can manipulate the equation until they do. Here is how to multiply radicals with or without coefficient.
## How to Multiply Radicals Without Coefficients
1. Radicals need to have the same index before you multiply them.
For Example: β(16) x β(4) = ?
2. Multiply the numbers under the radical signs.
For Example: β(16) x β(4) = β(64)
β(64) = 8. 64 is a perfect square because it is the product of 8 x 8. The square root of 64 is simply 8.
## How to Multiply Radicals with Coefficients
1. Multiply the coefficients.
For Example: Β 4β(3) x 3β(6) = 12β(Β ? )
• 4 x 3 = 12
2. Multiply the numbers inside the radicals.
For Example: 4β(3) x 3β(6) = 12β(3 x 6) = 12β(18)
3. Simplify the product.
For Example: 12β(18) = 12β(9 x 2) = 12β(3 x 3 x 2) = (12 x 3)β(2) = 36β(2)
## How to Multiply Radicals with Different Indices
1. Find the LCM (lowest common multiple) of the indices.Β To find the LCM of the indexes, find the smallest number that is evenly divisible by both indices. Find the LCM of the indices for the following equation:7β(5) xΒ 2β(2) =?
The indices are 5 and 2. 14 is the LCM of these two numbers because it is the smallest number that is evenly divisible by both 7 and 2. 14/7 = 2 and 14/2 = 7. To multiply the radicals, both of the indices will have to be 14.
2. Write each expression with the new LCM as the index.
Β 14β(5) xΒ 14β(2) =Β ?
3. Find the number that you would need to multiply each original index by finding the LCM.
– For the expression 7β(5), you’d need to multiply the index of 7 by 2 to get 14.
– For the expression 2β(2), you’d need to multiply the index of 2 by 7 to get 14.
4. Make this number the exponent of the number inside the radical.Β
2Β –>Β 14β(5) =Β 14β(5)2
7Β –>Β 14β(2) =Β 14β(2)7
5. Multiply the numbers inside the radicals by their exponents.Β Here’s how you do it:
14β(5)2Β =Β 14β(5 x 5) =Β 14β25
14β(2)7Β =Β 14β(2 x 2 x 2 x 2 x 2 x 2 x 2) = 14β128
6. Place these numbers under one radical.Β Here’s what the result would look like:Β 14β(128 x 25)
7. Multiply them.Β 14β(128 x 25) = 14β(3200).
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Sherpa
In this video, we take a look at linear equations and how to solve them using BIDMAS (Brackets, Indices, Division, Multiplication, Addition and Subtraction) and why it is so important to follow the golden rule of linear equations "Always do the SAME thing to both sides of the equation".
So we are looking at linear equations. We're going to look at what are they? So what do we mean by linear? We're going to look at bid Mass and how it affects the order in which we do things. And finally, we're going to see how the rule works with these things.
So we always have to do the same thing to both sides, and we're going to look at why we do that and what it means. So first of all, what are they? Well, the linear in linear equations, basically it refers to the highest power of x that is in our equation. So as an example, let's say we've got five x plus four equals 49. Well, in this case, we've got an x on its own, right?
There's no x squared or x cubed. We could think of this as x power of one. Anything from power of one is just itself. And so we could add a power of one there. It doesn't change anything, but it's just it allows us to think of this as in terms of powers, right?
So we would say this x has a power of one. And because it's got a power of one, because there are no x squares or x cubes or anything higher than that, we would say this is a linear equation. As an example, contrary to this, if we had say, I don't know, three x squared plus four x minus seven equals twelve. Well, in this case, because we've got an X squared here, we've got x to the power of two. The fact that we've got a power of two in our equation means that this equation is no longer linear, even though we've got x to the power of one in the equation.
That is trumped by the fact that we've got an X squared or an x to the power of two. And so we'd actually say this equation is quadratic.
You may have heard of the quadratic formula. That's why we call it a quadratic formula, is because it applies to quadratic equations. The equation part, the equation just means anything with an equal sign, right? So it's anything where we have to solve for a particular value for x and there are only certain values of x that will satisfy that. Right?
Bid Mass. The easiest way to understand how Bid Mass relates to solving linear equations is probably by thinking of a fairly simple example of a number machine, right? Let's go back to this simple example. We've got five x plus four equals 49.
And we know from Bid Mass that we do multiplication first and then we do addition afterwards, right? So we would do five times x and then we take that whole result and then we'd add it on to four. Well, if we think of this in terms of a number machine, we start off with x, we times x by five, and then we take that result and then we add fall onto it, and that then gives us the final result of 49. Well, that's how we can represent this equation in a kind of number machine. The tricky thing is, though, we want to find out what X is, right?
We want to go back down the opposite way. We want to kind of fight against the current, almost. We want to undo everything we've just done. And so we're kind of starting off with 49, we want to go back down the other way and we want to reverse this adding four. What's the opposite of adding four?
We know we have to minus four and then we take that result and we have to do the opposite of times by five. What's the opposite of timing by five? It's going to be dividing by five and that then undoes that times by five. And then once we do those two things, we should get back to where we started, right? We should get back to that value for X.
And whatever value we have here is going to be our value for X. We've kind of inadvertently solved the equation. We'll look at this in a couple of examples in the next slide. So that's been massive. That's how it relates to equations.
Finally, the rule. So the rule is we will do the same thing to both sides. And this becomes very obvious when we actually look at a very simple example, right? So we could do, say, I don't know, five x equals ten, right? Well, we know from our five times tables, right?
Five times what gives us ten. We know five times two gives us ten, right? We know X has to be two. What if we were to do the proper math way of solving this? Well, I want to get X on its own, right?
I want to end up with X equals something. And to do that, I need to therefore get rid of this five. And just like with our number machine, I always do the opposite. To do the reverse, I want to get rid of this five. And we can imagine that we're actually doing five times next year.
Well, the opposite of times by five is dividing by five, right? So I divide that side by five. And the whole reason I do that is to then cancel out that times by five. So if I cancel that out, I might just put it in red. Actually, if I cancelled out that times by five, the only thing I'm left with is going to be x, right?
I'm doing X times five divided by five. And obviously, if I make it five times bigger, then five times smaller, I get back to my original X. Well, if I don't do the same thing to the right hand side, I end up thinking that x equals ten, right? Which we know isn't true. We know the solution to this equation is x has to equal two so I have to do the same thing to both sides.
I have to divide this side by five as well. And if I do that, I do ten divided by five or ten divided by five is going to give me two. And again, we get our answer. So that's why we have that rule. We always do the same thing to both sides, right?
Let's look at a couple of examples. So, first of all, let's do question eight. So we want to solve five brackets. X minus six equals 65. Well, another way to think about this is we want to work our way in towards X, right?
Just like on the number machine, we work our way back towards X. We do the same thing here, right? So we kind of start off outside, further away from X, and then we move in and then we take care of the 6th and then we end up at X. Again, we're doing the opposite to bid mass, because with bid mass, we would take care of the minus six first and then times that whole thing by five. When we're doing the reverse, when we're working our way back towards X, we do the opposite of Bitmaster, we start outside the brackets.
We do that first and then we work inside the brackets at the very end. So the first thing we do is we take care of this five. Well, if I write this out slightly bigger, we're doing five times X minus six equals 65. And if I want to get rid of this five times, here what's the upset of X five. The opposite of times in by five is dividing by five.
And I do the same thing to both sides because it's a rule that we just covered. And the whole reason I do that is to cancel out that timing by five. So all I'm left with on the left hand side is going to be X minus six. On the right hand side, I've got to be 65 divided by five. Well, that's going to give me 13.
And the final thing we need to do is we need to get rid of this minus six, right? I want to end up at the end of the day, I want to do a load of maths and then I want to end up with X equals something, and that thing in that box is going to be my answer. So how do I get rid of that minus six? Well, what's the opposite of minus six? The opposite of minus six is plus six.
So I plastic both sides and again, the reason I do that is to cancel out that minus six. And so all I'm left with on the right hand side, on the left hand side is going to be X. Well, we also have to do plus six to the right hand side, we do 13 plus six and we get 19, and that is our answer. We kind of inadvertently arrived at our answer just by setting up these very simple rules. We get X equals 19 as our answer, right?
Final question, question 20. In this case, it's not quite as obvious, right? Which fully above question, we're knew that this left hand side was going to be the side that we wanted all the X's to be on, right? We said this is going to be on X side. We said we want to end up with X equals something.
Well, in the case of question 20, it's not as obvious, right? We've got X's on both sides and so do we want to get X equals something or do we want to get something equals X? Both are going to be valid, but we want to decide now what we're going to be aiming for. And I think because we've got more X's on the right hand side already, we've got five here and three here, we're going to say this is our X side. So if we want to get all the X's onto this side well, at the moment I've got a few exes over on the left hand side, so I want to get rid of this three X floating around here.
Well, to get rid of that three X, at the moment this three X is adding on to twelve, right? So what's the opposite of adding on three X? I just subtract three x So I subtract three X from both sides and I know therefore that's going to cancel out with that three X and I'm going to end up with just twelve on the left hand side and two X on the right hand side. And I saw that plus four lingering there. I can also try and get rid of this plus four, right?
Because I want to get at the end of the day, if this is my ex side, I want to end up with doing a load of maths and eventually ending up with something equals X. And so I need to get rid of this plus four. How do I get rid of that? Well, what's the opposite of plus four? It's going to be minus four, so I minus four from both sides.
Twelve minus four is going to give me eight. And again, the reason we do that is to cancel out that plus four. So we're left with eight equals two X. The final thing I need to do is get rid of two. Well, what's the two doing to X?
The two is timing the X. You can imagine there's a very small time sign in between the two and the X the opposite of times by T is going to be dividing by T, and so I can divide both sides by T and that's going to give me four on the left hand side. Again, the reason we do that is to cancel out that times in II on the right hand side and so we're left with four equals x. And that is my answer. That is the end of linear equations.
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# THE TRIGONOMETRIO SHOW
#### Storyboard Text
• Welcome viewers to our show! This is TRIGONOMETRIO again for another episode of making math easy. And for today's episode of our show, we will discuss solving different trigonometry problems. So without further ado, let's go!
• See, guys, it's that easy. So let's move on to our next problem.
• Good day, viewers! Hi, I'm Uno, and for our first problem, we have here an illustration, so take a closer look and let's find out how to solve this problem.
• FIND THE AREA OF THIS PIZZA.
• A = ?
• To find the area of a circle, you need to know the formula first. The formula for the area isA = πr² The given radius of the pizza is 5 inches. We'll substitute the 5 for the r, so our solution will look like this.A = π(5)²= 3.1416 x 25= 78.54 in²
• 5 inches
• Hola soy Dos! For our second problem, we are here at the park. Isn't it amazing? So, as you can see, I am beside a tree and we're going to find out how tall that tree is.
• Here we are trying to find the height of the tree. The distance between the tree and where I am standing is 6 ft., and the angle from where I'm standing to the tree is 70 degrees. We are going to use the formula of tanθ = Opposite/Adjacent.
• 70°
• 6ft.
• ?
• X
• In conclusion, the height of the tree is approximately 16.48 feet. I hope you understand that. Let's move on to the last problem.
• Solution:tan θ = opp/adjtan(70°)= x/6tan(70°)/1 = x/6 (cross multiply)x = 6 tan 70°x = 16.48 ft.
• 70°
• 6ft.
• ?
• X
• In this problem, we are going to find the length of the ladder, and that would be our missing side. To solve the problem, we are going to use the Pythagorean theorem. The formula for the Pythagorean theorem is a²+b²=c².
• Tres here, greetings! We are now on to our third and last problem. As you can see, we are here at my house. You can see the ladder leaning against the cabinet. And we are going to find out the length of this ladder.
• a = 6 feet
• b = 8 feet
• Our solution will be like this:a² + b² = c²(6)² + (8)² = c²36 + 64 = c²100 = c²√100 = √c210 = cTherefore, the length of the ladder is 10 feet.
• That's it for our final problem. It is easy, right? I hope you can now solve these kinds of problems on your own.
• a=6 feet
• b=8 feet
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Using corresponding angles and straight angles, find the measures of the angles formed by the intersection of parallel lines m and n cut by transversal l below. A transversal forms four pairs of corresponding angles. Corresponding angles are CONGRUENT (equal). Note: These shapes must either be similar or congruent… ∠2 ≅ ∠60° since they are corresponding angles, and m and n are parallel. Congruent angles can also be denoted without using specific angle measures by an equal number of arcs placed around the vertices of two angles, as shown below. If two corresponding angles of a transversal across parallel lines are right angles, what do you know about the figure? Corresponding angles are pairs of angles that lie on the same side of the transversal in matching corners. In other words, if a transversal intersects two parallel lines, the corresponding angles will be always equal. In quadrilateral ABCD above, ∠A≅∠C, ∠B≅∠D so, the quadrilateral is a parallelogram. If the corresponding angles of two lines cut by a transversal are congruent, then the lines are parallel. Corresponding Angles. This follows readily from the rigid-motion definition of congruence and from the statement that Corresponding Parts of Congruent Figures Are Congruent. Therefore PQR and MNO are congruent. The converse of the postulate is also true. Parallel lines m and n are cut by transversal l above, forming four pairs of congruent, corresponding angles: ∠1 ≅ ∠5, ∠2 ≅ ∠6, ∠3 ≅ 7, and ∠4 ≅ ∠8. Look at the pictures below to see what corresponding sides and angles look like. Parallel lines m and n are cut by transversal l above, forming four pairs of congruent, corresponding angles: ∠1 ≅ ∠5, ∠2 ≅ ∠6, ∠3 ≅ 7, and ∠4 ≅ ∠8. In certain situations, you can assume certain things about corresponding angles. Strategy: Proof by contradiction To prove this, we will introduce the technique of “proof by contradiction,” which will be very useful down the road. Whenever two lines intersect at a point the vertical angles formed are congruent. In the figure above, ∠DOF is bisected by OE so, ∠EOF≅∠EOD. Alternate exterior angles are CONGRUENT (equal). This statement is a biconditional, a statement that is true in either direction. Two polygons are congruent when their corresponding angles and corresponding sides are congruent. Angles that are both outside a set of lines and on opposite sides of the transversal. If two figures are similar, their corresponding angles are congruent (the same). ∠5 ≅ ∠120° since ∠1 and ∠5 are corresponding angles, and m and n are parallel. For angles, 'congruent' is similar to saying 'equals'. Two polygons are said to be similar when their corresponding angles are congruent. By the straight angle theorem , we can label every corresponding angle either α or β. Additionally, the three sides of PQR are equal to the three corresponding sides of MNO. The two lines above intersect at point O so, there are two pairs of vertical angles that are congruent. Congruent angles are angles that have the same measure. Imagine a transversal cutting across two lines. In the diagram below transversal l intersects lines m and n. ∠1 and ∠5 are a pair of corresponding angles. congruent, then corresponding pairs of sides and corresponding pairs of angles of the figures are congruent. This means that all congruent shapes are similar, but not all similar shapes are congruent. ∠8 ≅ ∠120° since ∠4 and ∠8 are corresponding angles, and m and n are parallel. You learn that corresponding angles are not congruent. For instance, take two figures that are similar, meaning they are the same shape but not necessarily the same size. Not only can congruent angles be appealing to the eye, they can also increase the structural integrity in construction. ∠7 and ∠5 form a straight angle, so∠7=60°. ∠3 and ∠4 form a straight angle, so∠4=120°. Two polygons are congruent when their corresponding angles and corresponding sides are congruent. In the figure above, PQR≅ MNO since ∠P≅∠M, ∠Q≅∠N, and ∠R≅∠O. The sides of the angles do not need to have the same length or open in the same direction to be congruent, they only need to have equal measures. Corresponding sides and angles are a pair of matching angles or sides that are in the same spot in two different shapes. The corresponding angles postulate states that if two parallel lines are cut by a transversal, the corresponding angles are congruent. The measure of angles A and B above are both 34° so angles A and B are congruent or ∠A≅∠B, where the symbol ≅ means congruent. But in geometry, the correct way to say it is "angles A and B are congruent". ∠1 and ∠2 form a straight angle, so∠1=120°. Can you possibly draw parallel lines with a transversal that creates a pair of corresponding angles, each measuring 181 °? The corresponding angles postulate states that if two parallel lines are cut by a transversal, the corresponding angles are congruent. By now, you must be well aware of a triangle till now that it is a 2-dimensional figure with three sides, three angles and three vertices. What are corresponding sides and angles? To determine the corresponding congruent parts of a triangle, we use the congruent markings of the triangle. Whenever an angle is bisected, two congruent angles are formed. Therefore △PQR and △MNO are congruent. Since Δ T U V and Δ C D E are congruent to each other, therefore, their corresponding sides and angles are exactly equal to each other. The sides of the angles do not need to have the same length or open in the same direction to be congruent, they only need to have equal measures. Congruent angles are angles that have the same measure. Additionally, the three sides of △PQR are equal to the three corresponding sides of △MNO.
.
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# 1.1: Compound Statements
[ "article:topic", "compound statements", "tautology", "authorname:thangarajahp", "license:ccbyncsa", "showtoc:yes" ]
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We can make a new statement from other statements; we call these compound propositions or compound statements
Example $$\PageIndex{1}$$:
1. It is not the case that all birds can fly. (This is the negation of the statement all birds can fly).
2. $$1+1=2$$ and "All birds can fly". (Here the connector "and" was used to create a new statement).
Note the following four basic ways to start with one or more propositions and use them to make a more elaborate compound statement. If $$p$$ and $$q$$ are statements
then here are four compound statements made from them:
1. $$\neg p$$, Not $$p$$ (i.e. the negation of $$p$$),
2. $$p \wedge q,\, p\, \textit{and}\, q$$,
3. $$p\vee q, \,p \,\textit{or} \,q$$ and
4. $$p \rightarrow q,\: \textit{If} \; p \, \textit{then}\, q.$$
Example $$\PageIndex{2}$$:
If $$p =$$ "You eat your supper tonight" and $$q =$$ "You get desert". Then
1. Not $$p$$ is "You don't eat your supper tonight".
2. $$p\, \textit{and}\, q$$ is "You eat your supper tonight and you get desert".
3. $$p \,\textit{or} \,q$$ is "You eat your supper tonight or you get desert".
4. $$\textit{If} \; p \, \textit{then}\, q$$ is "If you eat your supper tonight then you get dessert."
In English, we know these four propositions don't say the same thing. In logic, this is also the case, but we can make that clear by displaying the truth value possibilities. It is common to use a table to capture the possibilities for truth values of compound statements. We call such a table a truth table. Below are the possibilities: the first is the least profound. It says that a statement p is either true or false.
$$p$$
$$T$$
$$F$$
#### Negation
Truth tables are more useful in describing the possible truth values for various compound propositions. Consider the following truth table:
$$p$$ $$\neg p$$
$$T$$ $$F$$
$$F$$ $$T$$
The table above describes the truth value possibilities for the statements $$p$$ and $$\neg p$$, or "not p". As you can see, if $$p$$ is true then $$\neg p$$ is false and if $$p$$ false, the negation (i.e. not p) is true. $$\neg$$ is the mathematical notation used to mean "not."
Example $$\PageIndex{3}$$:
Consider the statement $$p$$: $$1 + 1 = 3$$.
Statement $$p$$ can either be true or false, not both.
$$\neg p$$ is "not $$p$$," or the negation of statement $$p$$.
$$\neg p$$ is $$1 + 1 \ne 3$$.
You can see that the negation of a proposition affects only the proposition itself, not any other assumptions.
#### Conjunction
Conjunction statements use two or more propositions. If two or more simple propositions are involved the truth table gets bigger. Below is the truth table for "and," otherwise known as a conjunction. When is an and statement true? As the truth table indicates, only when both of the component propositions are true is the compound conjunction statement true:
$$p$$ $$q$$ $$p \wedge q$$
$$T$$ $$T$$ $$T$$
$$T$$ $$F$$ $$F$$
$$F$$ $$T$$ $$F$$
$$F$$ $$F$$ $$F$$
Example $$\PageIndex{4}$$:
Consider statements $$p:= \,1 + 1 = 2$$ and $$q:=\,2 < 5$$.
Note that, $$p \wedge q$$ is true only if both $$p$$ and $$q$$ are both true.
Since statements $$p$$ and $$q$$ are both true, $$p \wedge q$$ is true.
#### Disjunction
Disjunction statements are compound statements made up of two or more statements and are true when one of the component propositions is true. They are called "Or Statements." In English, "or" is used in two ways:
1. If a person is looking for a house with 4 bedrooms or a short commute, a real estate agent might present houses with either 4 bedrooms or a short commute or both 4 bedrooms and a short commute. This is called an inclusive or.
2. If a person is asked whether they would like a Coke or a Pepsi, they are expected to choose between the two options. This is an exclusive or: "both" is not an acceptable case.
In logic, we use inclusive or statements
$$p$$ $$q$$ $$p \vee q$$
$$T$$ $$T$$ $$T$$
$$T$$ $$F$$ $$T$$
$$F$$ $$T$$ $$T$$
$$F$$ $$F$$ $$F$$
The $$p$$ or $$q$$ proposition is only false if both component propositions $$p$$ and $$q$$ are false.
Example $$\PageIndex{5}$$:
Consider the statement $$2 \leq -3$$
The statement reads "2 is less than or equal to -3", or "$$2 < -3 \vee 2 = -3$$" and can be broken into two component propositions:
1. Proposition $$p$$: $$2 < -3$$ (False)
2. Proposition $$q$$: $$2 = -3$$ (False)
Because propositions $$p$$ and $$q$$ are both false, the statement is false.
Example $$\PageIndex{6}$$:
Consider the statement $$2 \leq 5$$
The statement's two component propositions are:
1. Proposition $$p$$: $$2 < 5$$ (True)
2. Proposition $$q$$: $$2 = 5$$ (False)
Since proposition $$p$$ is true, the statement is true.
#### Conditional Statements
Consider the "if p then q" proposition. This is a conditional statement. Read the statements below. If these statements are made, in which instance is one lying (i.e. when is the overall statement false)?
Suppose, at suppertime, your mother makes the statement “If you eat your broccoli then you’ll get dessert.” Under what conditions could you say your mother is lying?
1. If you eat your broccoli but don't get dessert, she lied!
2. If you eat your broccoli and get dessert, she told the truth.
3. If you don’t eat your broccoli and you don’t get desert she told you the truth.
4. If you don’t eat your broccoli but you do get dessert we still think she told the truth. After all, she only outlined one condition that was supposed to get you desert, she didn’t say that was the only way you could earn dessert. Maybe you had cauliflower instead.
Note that the order in which the cases are presented in the truth table is irrelevant. The cases themselves are the important information, not their order relative to each other.
$$p$$ $$q$$ $$p \to q$$
$$T$$ $$F$$ $$F$$
$$T$$ $$T$$ $$T$$
$$F$$ $$F$$ $$T$$
$$F$$ $$T$$ $$T$$
It is important to notice that, if the first proposition is false, the conditional statement is true by default. A conditional statement is defined as being true unless a true hypothesis leads to a false conclusion.
Example $$\PageIndex{7}$$:
Consider the statement "If a closed figure has four sides, then it is a square." This is a false statement - why?
We can prove it using a counter-example: we draw a four-sided figure that is not a square. So there!
Example $$\PageIndex{8}$$:
Consider the statement "If $$2 = 3$$, then $$5 = 2$$"
Since $$2 \ne 3$$, it does not matter if $$5 = 2$$ is true or not, the conditional statement as a whole is true.
##### Converse of a conditional statement
Let P be a statement if p then q. Then the converse of P is if q then p.
Example $$\PageIndex{9}$$:
Consider the statement Q, "If a closed figure has four sides, then it is a square."
Then the converse of Q is "If it is a square then it is a closed figure with four sides".
##### Contrapositive of a Conditional Statement
Let P be a statement if p then q. Then the contrapositive of P is if $$\neg q$$ then $$\neg p.$$
Example $$\PageIndex{10}$$:
Consider the statement Q, "If a closed figure has four sides, then it is a square."
Then the converse of Q is "If it is not a square then it is not a closed figure with four sides".
#### Bi-Conditional Statements
Bi-conditional statements are conditional statements which depend on both component propositions. They read "p if and only if q" and are denoted $$p \leftrightarrow q$$ or "p iff q", which is logically equivalent to $$(p \to q) \wedge (q \to p)$$. These compound statements are true if both component propositions are true or both are false:
$$p$$ $$q$$ $$p \leftrightarrow q$$
$$T$$ $$T$$ $$T$$
$$T$$ $$F$$ $$F$$
$$F$$ $$T$$ $$F$$
$$F$$ $$F$$ $$T$$
Example $$\PageIndex{11}$$:
Consider the statement: "Two lines are perpendicular if and only if they intersect to form a right angle."
The component propositions are:
1. $$p$$: Two lines are perpendicular
2. $$q$$: [The lines] intersect to form a right angle
Logically, we can see that if two lines are perpendicular, then they must intersect to form a right angle. Also, we can see that if two lines form a right angle, then they are perpendicular.
If two lines are not perpendicular, then they cannot form a right angle. Conversely, if two lines do not form a right angle, they cannot be perpendicular. This is why, if both propositions in a biconditional statement are false, the statement itself is true!
#### Logically Equivalent Statements
Once we know the basic statement types and their truth tables, we can derive the truth tables of more elaborate compound statements. Below is the truth table for the proposition, not p or (p and q). First, we calculate the truth values for not p, then p and q and finally, we use these two columns of truth values to figure out the truth values for not p or (p and q).
$$p$$ $$q$$ $$\neg p$$ $$p \wedge q$$ $$\neg p \vee (p \wedge q)$$
$$T$$ $$T$$ $$F$$ $$T$$ $$T$$
$$T$$ $$F$$ $$F$$ $$F$$ $$F$$
$$F$$ $$T$$ $$T$$ $$F$$ $$T$$
$$F$$ $$F$$ $$T$$ $$F$$ $$T$$
So the proposition "not p or (p and q)" is only false if p is true and q is false. Does this seem familiar?
"If p then q" is only false if p is true and q is false as well.
$$p$$ $$q$$ $$p \to q$$
$$T$$ $$T$$ $$T$$
$$T$$ $$F$$ $$F$$
$$F$$ $$T$$ $$T$$
$$F$$ $$F$$ $$T$$
This has some significance in logic because if two propositions have the same truth table they are in a logical sense equal to each other – and we say that they are logically equivalent. So: $$\neg p \vee (p \wedge q) \equiv p \to q$$, or "Not p or (p and q) is equivalent to if p then q."
Example $$\PageIndex{12}$$:
Prove or disprove: for any mathematical statements $$p,q$$ and $$r,\, p\to(q \vee r)$$ is logically equivalent to $$\neg r \to ( p \to q).$$
$$p$$ $$q$$ $$r$$ $$q \vee r$$ $$p \to (q \vee r)$$ $$\neg r$$ $$p \to q$$ $$\neg r \to (p \to q)$$
T T T T T F T T
T T F T T T T T
T F T T T F F T
T F F F F T F F
F T T T T F T T
F T F T T T T T
F F T T T F T T
F F F F T T T T
Hence, $$p\to(q \vee r)$$ is logically equivalent to $$\neg r \to ( p \to q).$$
There are two cases in which compound statements can be made that result in either always true or always false. These are called tautologies and contradictions, respectively. Let's consider a tautology first, and then a contradiction:
Example $$\PageIndex{13}$$:
Consider the statement "$$(2 = 3) \vee (2 \ne 3)$$":
There are two component propositions:
1. $$p$$: $$2 = 3$$
2. $$\neg p$$: $$2 \ne 3$$
Clearly this statement is a tautology.
Let's make a truth table for general case $$p \vee (\neg p)$$:
$$p$$ $$\neg p$$ $$p \vee (\neg p)$$
$$T$$ $$F$$ $$T$$
$$F$$ $$T$$ $$T$$
As you can see, no matter what we do, this statement is always true. It is a tautology. Careful! This is not to say that this statement makes logical sense in English, but rather that, using logical mathematics, this statement is always true.
Example $$\PageIndex{14}$$:
Consider the statement "2 is even $$\wedge$$ 2 is odd"
There are two component propositions:
1. $$p$$: 2 is even
2. $$\neg p$$: 2 is odd
Clearly this statement is a contradiction.
Let's make a truth table for general case $$p \wedge (\neg p)$$:
$$p$$ $$\neg p$$ $$p \wedge (\neg p)$$
$$T$$ $$F$$ $$F$$
$$F$$ $$T$$ $$F$$
As you can see again, no matter what we do, this statement will always be false. It is a contradiction. These make more sense in English: 2 cannot be both even and odd, after all! Still, what matters is what we decide using logical mathematics.
Summary
Operation Notation Summary of truth values Negation $$\neg p$$ Opposite truth value of p Conjunction $$p \wedge q$$ True only when both p and q are true Disjunction $$p \vee q$$ False only when both p and q are false Conditional $$p \to q$$ False only when p is true and q is false Biconditional $$p\leftrightarrow q$$ True only when both p and q are true or both are false
New Notations & Definitions
• Negation: $$\neg$$ or "not"
• Conjunction: $$\wedge$$ or "and"
• Disjunction: $$\vee$$ or "or"
• Conditional: $$\to$$ or "implies" or "if/then"
• Bi-Conditional: $$\leftrightarrow$$ or "if and only if" or "iff"
• Counter-example: An example which disproves a mathematical proposition or statement.
• Logically Equivalent: $$\equiv$$ Two propositions which have the same truth table result.
• Tautology: A statement which is always true, and a truth table yields only true results.
• Contradiction: A statement which is always false, and a truth table yields only false results.
|
Synthetic Division
How to divide a polynomial by a binomial by doing the synthetic division: 2 examples and their solutions.
Example 1
Solution
The synthetic division is a way
to divide a polynomial by a binomial,
just like the long division.
The numerator is
x3 + 0x2 - 7x + 11.
(If there's a missing term, like the x2 term,
think its coefficient as 0.)
Then write the coefficients of the numerator terms
in descending order:
1 0 -7 11.
Draw an L shape form like this.
On the left side,
write the zero of the denominator (x - 2), 2.
Then, do the synthetic division.
Write 1
in ↓ direction.
↓: 1 = 1
Multiply 1 and the left 2.
Write 1⋅2 = 2
in ↗ direction.
↗: 1⋅2 = 2
in ↓ direction.
↓: 0 + 2 = 2
Multiply 2 and the left 2.
Write 2⋅2 = 4
in ↗ direction.
↗: 2⋅2 = 4
in ↓ direction.
↓: -7 + 4 = -3
Multiply -3 and the left 2.
Write -3⋅2 = -6
in ↗ direction.
↗: -3⋅2 = -6
in ↓ direction.
↓: 11 - 6 = 5
Draw another L shape form
that covers the right end number 5.
This 5 is the remainder.
This is the end of the calculation.
Let's write the answer from this.
The numbers below the L shape,
1 2 3,
are the coefficients of the quotient terms.
So the quotient is
x2 + 2x + 3.
The number in this L shape, 5,
is the remainder.
So write +, a fraction bar,
and 5 in the numerator.
The zero of the left 2 is
(x - 2).
So write (x - 2)
in the denominator.
From the synthetic division,
you can find that
(given) = x2 + 2x + 3 + 5/(x - 2).
So
x2 + 2x + 3 + 5/(x - 2)
Example 2
Solution
The numerator is
2x4 + x3 - 5x2 + 3x + 4.
Then write the coefficients of the numerator terms
in descending order:
2 1 -5 3 4.
Draw an L shape form like this.
On the left side,
write the zero of (x + 1), -1.
Then, do the synthetic division.
Write 2
in ↓ direction.
↓: 2 = 2
Multiply 2 and the left -1.
Write 2⋅(-1) = -2
in ↗ direction.
↗: 2⋅(-1) = -2
in ↓ direction.
↓: 1 - 2 = -1
Multiply -1 and the left -1.
Write (-1)⋅(-1) = 1
in ↗ direction.
↗: (-1)⋅(-1) = 1
in ↓ direction.
↓: -5 + 1 = -4
Multiply -4 and the left -1.
Write (-4)⋅(-1) = 4
in ↗ direction.
↗: (-4)⋅(-1) = 4
in ↓ direction.
↓: 3 + 4 = 7
Multiply 7 and the left -1.
Write 7⋅(-1) = 4
in ↗ direction.
↗: 7⋅(-1) = -7
in ↓ direction.
↓: 4 - 7 = -3
Draw another L shape form
that covers the right end number -3.
This -3 is the remainder.
This is the end of the calculation.
Let's write the answer from this.
The numbers below the L shape,
2 -1 -4 7,
are the coefficients of the quotient terms.
So the quotient is
2x3 - x2 - 4x + 7.
The number in this L shape, -3,
is the remainder.
So write -, a fraction bar,
and 3 in the numerator.
The zero of the left -1 is
(x + 1).
So write (x + 1)
in the denominator.
From the synthetic division,
you can find that
(given) = 2x3 - x2 - 4x + 7 - 3/(x + 1).
So
2x3 - x2 - 4x + 7 - 3/(x + 1)
|
# In an acute triangle, the square of the side opposite to an acute angle is equal to the sum of the squares of other two sides minus twice the product of one side and the projection of the other on first.
## Solution :
Given : In triangle ABC angle B is an acute angle and AD $$\perp$$ BC
To Prove : $${AC}^2$$ = $${AB}^2$$ + $${BC}^2$$ – 2BC.BD
Proof : Since $$\triangle$$ ADC is a right triangle, angled at D. Therefore, by Pythagoras theorem,
$${AC}^2$$ = $${AD}^2$$ + $${CD}^2$$
$$\implies$$ $${AC}^2$$ = $${AD}^2$$ + $${BC – BD}^2$$ (because DC = BC – BD)
$$\implies$$ $${AC}^2$$ = $${AD}^2$$ + $${BC}^2$$ + $${BD}^2$$ – 2BC.BD
Since, In triangle ADB, $${AB}^2$$ = $${AD}^2$$ + $${BD}^2$$
$$\implies$$ $${AC}^2$$ = $${AB}^2$$ + $${BC}^2$$ – 2BC.BD
|
# Search by Topic
#### Resources tagged with Addition & subtraction similar to Children's Mathematical Writing:
Filter by: Content type:
Stage:
Challenge level:
### Four Goodness Sake
##### Stage: 2 Challenge Level:
Use 4 four times with simple operations so that you get the answer 12. Can you make 15, 16 and 17 too?
### Sam's Quick Sum
##### Stage: 2 Challenge Level:
What is the sum of all the three digit whole numbers?
### The Deca Tree
##### Stage: 2 Challenge Level:
Find out what a Deca Tree is and then work out how many leaves there will be after the woodcutter has cut off a trunk, a branch, a twig and a leaf.
### Exploring Number Patterns You Make
##### Stage: 2 Challenge Level:
Explore Alex's number plumber. What questions would you like to ask? What do you think is happening to the numbers?
### The Clockmaker's Birthday Cake
##### Stage: 2 Challenge Level:
The clockmaker's wife cut up his birthday cake to look like a clock face. Can you work out who received each piece?
### The 24 Game
##### Stage: 2 Challenge Level:
There are over sixty different ways of making 24 by adding, subtracting, multiplying and dividing all four numbers 4, 6, 6 and 8 (using each number only once). How many can you find?
### Napier's Bones
##### Stage: 2 Challenge Level:
The Scot, John Napier, invented these strips about 400 years ago to help calculate multiplication and division. Can you work out how to use Napier's bones to find the answer to these multiplications?
### Abundant Numbers
##### Stage: 2 Challenge Level:
48 is called an abundant number because it is less than the sum of its factors (without itself). Can you find some more abundant numbers?
### Buckets of Thinking
##### Stage: 2 Challenge Level:
There are three buckets each of which holds a maximum of 5 litres. Use the clues to work out how much liquid there is in each bucket.
### Number Juggle
##### Stage: 2 Challenge Level:
Fill in the missing numbers so that adding each pair of corner numbers gives you the number between them (in the box).
### Magic Squares 4x4
##### Stage: 2 Challenge Level:
Fill in the numbers to make the sum of each row, column and diagonal equal to 34. For an extra challenge try the huge American Flag magic square.
### Calendar Calculations
##### Stage: 2 Challenge Level:
Try adding together the dates of all the days in one week. Now multiply the first date by 7 and add 21. Can you explain what happens?
### Amy's Dominoes
##### Stage: 2 Challenge Level:
Amy has a box containing domino pieces but she does not think it is a complete set. She has 24 dominoes in her box and there are 125 spots on them altogether. Which of her domino pieces are missing?
### A-magical Number Maze
##### Stage: 2 Challenge Level:
This magic square has operations written in it, to make it into a maze. Start wherever you like, go through every cell and go out a total of 15!
### Back to School
##### Stage: 2 Challenge Level:
Mr. Sunshine tells the children they will have 2 hours of homework. After several calculations, Harry says he hasn't got time to do this homework. Can you see where his reasoning is wrong?
### Oranges and Lemons
##### Stage: 2 Challenge Level:
On the table there is a pile of oranges and lemons that weighs exactly one kilogram. Using the information, can you work out how many lemons there are?
### Rocco's Race
##### Stage: 2 Short Challenge Level:
Rocco ran in a 200 m race for his class. Use the information to find out how many runners there were in the race and what Rocco's finishing position was.
### Clever Keys
##### Stage: 2 Short Challenge Level:
On a calculator, make 15 by using only the 2 key and any of the four operations keys. How many ways can you find to do it?
### Journeys in Numberland
##### Stage: 2 Challenge Level:
Tom and Ben visited Numberland. Use the maps to work out the number of points each of their routes scores.
### How Many Eggs?
##### Stage: 2 Challenge Level:
Peter, Melanie, Amil and Jack received a total of 38 chocolate eggs. Use the information to work out how many eggs each person had.
### One Wasn't Square
##### Stage: 2 Challenge Level:
Mrs Morgan, the class's teacher, pinned numbers onto the backs of three children. Use the information to find out what the three numbers were.
### Super Shapes
##### Stage: 2 Short Challenge Level:
The value of the circle changes in each of the following problems. Can you discover its value in each problem?
### Penta Post
##### Stage: 2 Challenge Level:
Here are the prices for 1st and 2nd class mail within the UK. You have an unlimited number of each of these stamps. Which stamps would you need to post a parcel weighing 825g?
### Highest and Lowest
##### Stage: 2 Challenge Level:
Put operations signs between the numbers 3 4 5 6 to make the highest possible number and lowest possible number.
### Doplication
##### Stage: 2 Challenge Level:
We can arrange dots in a similar way to the 5 on a dice and they usually sit quite well into a rectangular shape. How many altogether in this 3 by 5? What happens for other sizes?
### Arranging the Tables
##### Stage: 2 Challenge Level:
There are 44 people coming to a dinner party. There are 15 square tables that seat 4 people. Find a way to seat the 44 people using all 15 tables, with no empty places.
### Clock Face
##### Stage: 2 Challenge Level:
Where can you draw a line on a clock face so that the numbers on both sides have the same total?
### The Pied Piper of Hamelin
##### Stage: 2 Challenge Level:
This problem is based on the story of the Pied Piper of Hamelin. Investigate the different numbers of people and rats there could have been if you know how many legs there are altogether!
### Escape from the Castle
##### Stage: 2 Challenge Level:
Skippy and Anna are locked in a room in a large castle. The key to that room, and all the other rooms, is a number. The numbers are locked away in a problem. Can you help them to get out?
##### Stage: 2 Challenge Level:
Can you put plus signs in so this is true? 1 2 3 4 5 6 7 8 9 = 99 How many ways can you do it?
### Magic Triangle
##### Stage: 2 Challenge Level:
Place the digits 1 to 9 into the circles so that each side of the triangle adds to the same total.
### Fingers and Hands
##### Stage: 2 Challenge Level:
How would you count the number of fingers in these pictures?
### Pouring the Punch Drink
##### Stage: 2 Challenge Level:
There are 4 jugs which hold 9 litres, 7 litres, 4 litres and 2 litres. Find a way to pour 9 litres of drink from one jug to another until you are left with exactly 3 litres in three of the jugs.
### Sometimes We Lose Things
##### Stage: 2 Challenge Level:
Well now, what would happen if we lost all the nines in our number system? Have a go at writing the numbers out in this way and have a look at the multiplications table.
### Brothers and Sisters
##### Stage: 2 Challenge Level:
Cassandra, David and Lachlan are brothers and sisters. They range in age between 1 year and 14 years. Can you figure out their exact ages from the clues?
### Zios and Zepts
##### Stage: 2 Challenge Level:
On the planet Vuv there are two sorts of creatures. The Zios have 3 legs and the Zepts have 7 legs. The great planetary explorer Nico counted 52 legs. How many Zios and how many Zepts were there?
### Throw a 100
##### Stage: 2 Challenge Level:
Can you score 100 by throwing rings on this board? Is there more than way to do it?
### The Amazing Splitting Plant
##### Stage: 1 Challenge Level:
Can you work out how many flowers there will be on the Amazing Splitting Plant after it has been growing for six weeks?
### Make 100
##### Stage: 2 Challenge Level:
Find at least one way to put in some operation signs (+ - x ÷) to make these digits come to 100.
### It Was 2010!
##### Stage: 1 and 2 Challenge Level:
If the answer's 2010, what could the question be?
### Six Is the Sum
##### Stage: 2 Challenge Level:
What do the digits in the number fifteen add up to? How many other numbers have digits with the same total but no zeros?
### Big Dog, Little Dog
##### Stage: 1 Challenge Level:
Woof is a big dog. Yap is a little dog. Emma has 16 dog biscuits to give to the two dogs. She gave Woof 4 more biscuits than Yap. How many biscuits did each dog get?
### Hubble, Bubble
##### Stage: 2 Challenge Level:
Winifred Wytsh bought a box each of jelly babies, milk jelly bears, yellow jelly bees and jelly belly beans. In how many different ways could she make a jolly jelly feast with 32 legs?
##### Stage: 2 Challenge Level:
What happens when you add the digits of a number then multiply the result by 2 and you keep doing this? You could try for different numbers and different rules.
### Dividing a Cake
##### Stage: 2 Challenge Level:
Annie cut this numbered cake into 3 pieces with 3 cuts so that the numbers on each piece added to the same total. Where were the cuts and what fraction of the whole cake was each piece?
### On Target
##### Stage: 2 Challenge Level:
You have 5 darts and your target score is 44. How many different ways could you score 44?
### Machines
##### Stage: 2 Challenge Level:
What is happening at each box in these machines?
### Spiders and Flies
##### Stage: 1 Challenge Level:
There were 22 legs creeping across the web. How many flies? How many spiders?
### How Do You Do It?
##### Stage: 2 Challenge Level:
This group activity will encourage you to share calculation strategies and to think about which strategy might be the most efficient.
|
1. ## Finding the derivative
Determine the derivative of the function $\displaystyle f(x)=\frac{1}{\sqrt{x}}$ using first principles.
2. I don't know what you mean by first principles.
Think of f(x) as $\displaystyle x^{-\frac{1}{2}}$. Now use the power rule.
3. Its been awhile since I've done these.. but I think be first principles it means using the formular $\displaystyle f'(x)= lim (h\rightarrow0) \frac{f(x+h)-f(x)}{h}$
Therefore I have my first step as
$\displaystyle f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h}$
I then rationalized...
$\displaystyle f'(x)= lim (h\rightarrow0) \frac{1/{\sqrt{x+h}}-1/{\sqrt{x}}}{h} * \frac{1/{\sqrt{x+h}}+1/{\sqrt{x}}}{1/{\sqrt{x+h}}+1/{\sqrt{x}}}$
And I just can't seem to get a final derivative
4. Oh ok. I see what you meant now. You are using the formula limit definition of a derivative.
You are on the right track and so close!
Like you said, you multiply everything by the conjugate of the numerator. Just simplify your last step.
$\displaystyle \frac{\frac{1}{x+h}-\frac{1}{x}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}$
Now let's simplify the numerator again. Find a common denominator for the two terms and combine. The common denominator is x(x+h). So...
$\displaystyle \frac{\frac{x-(x+h)}{x(x+h)}}{h[\frac{1}{\sqrt{h+x}}+\frac{1}{\sqrt{x}}]}$.
Now when you simplify the numerator again, you'll notice you're left with a "-h" that can cancel with the "h" in the denominator leaving...
$\displaystyle \frac{-\frac{1}{x^2+xh}}{\frac{1}{\sqrt{h+x}}+\frac{1}{\s qrt{x}}}$
From here there is just a little more simplifying to go and you should get your final answer
5. Ok from the other thread you made I can tell you're still stuck. Try applying the limit now, meaning plug in h=0. From there you'll get a fraction divided by another fraction. Flip the bottom one and multiply. Remember $\displaystyle \frac{x^a}{x^b}=x^{a-b}$ no matter if a<b or b<a.
6. hmm thank you very much! and my mistake about the other post, I apologize. I think I may have come to final answer. I just wasn't sure before if I was allowed to sub in the h=0 yet.. but I think I ended up getting it. thanks again
7. No worries. You can show me your final answer and work for the rest of the problem and I'll check it. If you don't want to that's fine too.
8. $\displaystyle f'(x)=\frac{-\sqrt{x}}{2x^2}$
is what i ended up with
9. That's what I suspected. You need to simplify further.
$\displaystyle -\frac{x^{\frac{1}{2}}}{2x^2}=-\frac{1}{2}(x^{\frac{1}{2}}-x^2)=-\frac{1}{2}(x^{-\frac{3}{2}})$
If you do the normal power rule with $\displaystyle x^{-\frac{1}{2}}$, this is the answer you'll end up with.
|
# QUADRATIC SOLVER - NATURE OF THE ROOTS OF A QUADRATIC EQUATION, EXAMPLES, EXERCISE
if you have not already done so.
There, We presented the Derivation
Of Quadratic Formula in a lucid way.
An Example in applying the formula
is given. Another problem for practice
is also given.
That knowledge is a prerequisite here.
Here, we explain (derive) the Formulae
to decide the Nature of the roots of
We also present Solved Examples and
Exercise problems on application of
these Formulae.
## Nature of the roots of a Quadratic Equation :
By Quadratic Formula, the roots of ax2 + bx + c = 0
are α = {-b + √(b2 - 4ac)}⁄2a and β = {-b - √(b2 - 4ac)}⁄2a
Let (b2 - 4ac) be denoted by Δ (called Delta).
Then α = (-b + √Δ)⁄2a and β = (-b - √Δ)⁄2a
The nature of the roots (α and β) depends on Δ
Δ ( = b2 - 4ac) is called the DISCRIMINANT of ax2 + bx + c = 0.
Three cases arise depending on the value of Δ (= b2 - 4ac) is zero or positive or negative.
(i) If Δ ( = b2 - 4ac) = 0, then α = -b⁄2a and β = -b⁄2a i.e. the two roots are real and equal.
Thus ax2 + bx + c = 0 has real and equal roots, if Δ = 0
(ii) If Δ ( = b2 - 4ac) > 0, the roots are real and distinct.
(ii) (a) If Δ ( = b2 - 4ac) is a perfect square,
the roots are rational.
(ii) (b) otherwise, the roots are irrational.
(iii) If Δ ( = b2 - 4ac) < 0, √Δ is not real.It is called an imaginary number.
∴ α, β are imaginary when Δ is negative.
When Δ is negative, the roots are imaginary.
The roots of ax2 + bx + c = 0
(i) are real and equal if Δ ( = b2 - 4ac) = 0
(ii) are real and distinct if Δ ( = b2 - 4ac) > 0
(ii) (a) If Δ ( = b2 - 4ac) is a perfect square,
the roots are rational.
(ii) (b) otherwise, the roots are irrational.
(iii) are imaginary if Δ ( = b2 - 4ac) < 0
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### Example 1 : Quadratic Solver
Find the nature of the roots of the equations,
1. 5x2 - 2x - 7 = 0
2. 9x2 + 24x + 16 = 0
3. x2 + 6x - 5 = 0
4. x2 - x + 5 = 0
Solution of Example 1(i) of Quadratic Solver :
(i) The given equation is 5x2 - 2x - 7 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 5, b = -2 and c = -7
Discriminant = Δ = b2 - 4ac = (-2)2 - 4(5)(-7) = 4 + 140 = 144 = 122
Since the Discriminant is positive and a perfect square,
the roots of the given equation
are real, distinct and rational. Ans.
Solution of Example 1(ii) of Quadratic Solver :
(ii) The given equation is 9x2 + 24x + 16 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 9, b = 24 and c = 16
Discriminant = Δ = b2 - 4ac = (24)2 - 4(9)(16) = 576 - 576 = 0
Since the Discriminant is zero,
the roots of the given equation
are real and equal. Ans.
Solution of Example 1(iii) of Quadratic Solver :
(iii) The given equation is x2 + 6x - 5 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = 6 and c = -5
Discriminant = Δ = b2 - 4ac = (6)2 - 4(1)(-5) = 36 + 20 = 56
Since the Discriminant is positive and is not a perfect square,
the roots of the given equation
are real, distinct and irrational. Ans.
Solution of Example 1(iv) of Quadratic Solver :
(iv) The given equation is x2 - x + 5 = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = 1, b = -1 and c = 5
Discriminant = Δ = b2 - 4ac = (-1)2 - 4(1)(5) = 1 - 20 = -19
Since the Discriminant is negative,
the roots of the given equation
are imaginary. Ans. Great Deals on School & Homeschool Curriculum Books
### Example 2 : Quadratic Solver
For what values of k are the roots of kx2 + (k - 1)x + (k -1) = 0 equal ?For those values of k, find the equal roots.
Solution of Example 2 of Quadratic Solver :
The given equation is kx2 + (k - 1)x + (k -1) = 0
Comparing this equation with ax2 + bx + c = 0, we get
a = k, b = (k - 1) and c = (k -1)
Discriminant = Δ = b2 - 4ac = (k - 1)2 - 4k(k -1) = (k -1)(k - 1 - 4k)
= (k -1)(- 1 - 3k)
For roots to be equal, Discriminant = 0
⇒ (k -1)(- 1 - 3k) = 0 ⇒ (k -1) = 0 or (- 1 - 3k) = 0
k = 1 or -3k = 1 ⇒ k = 1 or k = -1⁄3 ⇒ k = 1, -1⁄3
roots = (-b ± 0)⁄2a = {-(k - 1)± 0}⁄2k = (1 - k)⁄2k
For k = 1, the equal roots are 0, 0.
For k = -1⁄3, the value of the equal roots is {1 - (-1⁄3)}⁄2(-1⁄3) = -2
Thus, for k = 1, the equal roots are 0, 0
and for k = -1⁄3, the equal roots are -2, -2. Ans.
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Solve the following problems of Quadratic Solver :
1. Find the nature of the roots of the equations,
1. x2 - 5x + 6 = 0
2. x2 - 16x + 64 = 0
3. x2 + 2x - 1 = 0
4. x2 + 4x + 5 = 0
2. If α, β are the roots of px2 + qx + r = 0, find the quadratic equation whose roots are(α + 1⁄β), (β + 1⁄α)
For Answers See at the bottom of the Page.
## Progressive Learning of Math : Quadratic Solver
Recently, I have found a series of math curricula
(Both Hard Copy and Digital Copy) developed by a Lady Teacher
who taught everyone from Pre-K students to doctoral students
and who is a Ph.D. in Mathematics Education.
This series is very different and advantageous
over many of the traditional books available.
These give students tools that other books do not.
Other books just give practice.
These teach students “tricks” and new ways to think.
These build a student’s new knowledge of concepts
from their existing knowledge.
These provide many pages of practice that gradually
increases in difficulty and provide constant review.
These also provide teachers and parents with lessons
on how to work with the child on the concepts.
The series is low to reasonably priced and include
Elementary Math curriculum
and
Algebra Curriculum.
|
# What is the base of 5?
Quinary (base-5 or pental) is a numeral system with five as the base. A possible origination of a quinary system is that there are five fingers on either hand. In the quinary place system, five numerals, from 0 to 4, are used to represent any real number. Each quinary digit has log25 (approx.
Also know, what is the base of a number?
In mathematics, a base or radix is the number of different digits or combination of digits and letters that a system of counting uses to represent numbers. For example, the most common base used today is the decimal system. Because “dec” means 10, it uses the 10 digits from 0 to 9.
What is the base of number system in computer?
We are used to using the base-10 number system, which is also called decimal. Other common number systems include base-16 (hexadecimal), base-8 (octal), and base-2 (binary).
## What is the base 16 number system?
The hexadecimal numeral system, also known as just hex, is a numeral system made up of 16 symbols (base 16). The standard numeral system is called decimal (base 10) and uses ten symbols: 0,1,2,3,4,5,6,7,8,9. Hexadecimal uses the decimal numbers and includes six extra symbols.
## What is the base of hexadecimal number system?
In mathematics and computing, hexadecimal (also base 16, or hex) is a positional numeral system with a radix, or base, of 16. It uses sixteen distinct symbols, most often the symbols 0–9 to represent values zero to nine, and A–F (or alternatively a–f) to represent values ten to fifteen.
## What do you mean by BCD?
Binary coded decimal (BCD) is a system of writing numerals that assigns a four-digit binary code to each digit 0 through 9 in a decimal (base-10) numeral. The four-bit BCD code for any particular single base-10 digit is its representation in binary notation, as follows: 0 = 0000. 1 = 0001. 2 = 0010.
## What is the 8421 code?
8421 code. 8421 code A weighted code in which each decimal digit 0 through 9 is represented by a four-bit codeword. The bit positions in each codeword are assigned weights, from left to right, of 8, 4, 2, and 1. See also binary-coded decimal, excess-3 code, biquinary code.
## What is the gray code?
The reflected binary code (RBC), also known just as reflected binary (RB) or Gray code after Frank Gray, is an ordering of the binary numeral system such that two successive values differ in only one bit (binary digit).
## What is a GREY code?
A Gray code is an encoding of numbers so that adjacent numbers have a single digit differing by 1. The term Gray code is often used to refer to a “reflected” code, or more specifically still, the binary reflected Gray code.
## What is meant by excess 3 code?
12. The Excess-3 code:- It is an important BCD code , is a 4 bit code and used with BCD numbers To convert any decimal numbers into its excess- 3 form ,add 3 to each decimal digit and then convert the sum to a BCD number As weights are not assigned, it is a kind of non weighted codes.
## What is the 2421 code?
2421 Code. This is a weighted code, its weight are 2, 4, 2 and 1. A decimal number is represented in 4-bit form and total weight of the four bits = 2 + 4 + 2 + 1 = 9. Hence the 2421 code represents decimal number 0 to 9.
## What is not a weighted code?
Weighted binary codes are those binary codes which obey the positional weight principle. Each position of the number represents a specific weight. Several systems of the codes are used to express the decimal digits 0 through 9. In these codes each decimal digit is represented by a group of four bits.
## What is 0 and 1 in computer language?
A binary code represents text, computer processor instructions, or any other data using a two-symbol system. The two-symbol system used is often the binary number system’s 0 and 1. The binary code assigns a pattern of binary digits, also known as bits, to each character, instruction, etc.
## What is a non weighted code?
Weighted Codes:-The weighted codes are those that obey the position weighting principle,which states that the position of each number represent a specific weight. In these codes each decimal digit is represented by a group of four bits. There are millions of weighted code The most common one is 8421/BCD Code.
## What is BCD code with example?
Short for Binary Coded Decimal, BCD is also known as packet decimal and is numbers 0 through 9 converted to four-digit binary. Using this conversion, the number 25, for example, would have a BCD number of 0010 0101 or 00100101. However, in binary, 25 is represented as 11001.
## What is the ascii?
ASCII (American Standard Code for Information Interchange) is the most common format for text files in computers and on the Internet. In an ASCII file, each alphabetic, numeric, or special character is represented with a 7-bit binary number (a string of seven 0s or 1s). 128 possible characters are defined.
## How many different characters can be represented in a 7 bit character set?
The ASCII table has 128 characters, with values from 0 through 127. Thus, 7 bits are sufficient to represent a character in ASCII; however, most computers typically reserve 1 byte, (8 bits), for an ASCII character.
## What is meant by a character set?
A defined list of characters recognized by the computer hardware and software. Each character is represented by a number. The ASCII character set, for example, uses the numbers 0 through 127 to represent all English characters as well as special control characters.
## What is the UTF 8 character set?
Summary. UTF-8 is a compromise character encoding that can be as compact as ASCII (if the file is just plain English text) but can also contain any unicode characters (with some increase in file size). UTF stands for Unicode Transformation Format. The ‘8’ means it uses 8-bit blocks to represent a character.
## What is a character set in HTML?
ASCII was the first character encoding standard (also called character set). ASCII defined 128 different alphanumeric characters that could be used on the internet: numbers (0-9), English letters (A-Z), and some special characters like ! UTF-8 (Unicode) covers almost all of the characters and symbols in the world.
## What is the character set of C?
C Character Set – C Programming Language. character:- It denotes any alphabet, digit or special symbol used to represent information. Use:- These characters can be combined to form variables. C uses constants, variables, operators, keywords and expressions as building blocks to form a basic C program.
## What is UTF 8 in HTML?
That meta tag basically specifies what character set is your website written with. Here is a definition of UTF-8: UTF-8 (U from Universal Character Set + Transformation Format—8-bit) is a character encoding capable of encoding all possible characters (called code points) in Unicode.
Categories FAQ
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# Steps to solve word problems in math
Steps to solve word problems in math can be found online or in math books. We will give you answers to homework.
## The Best Steps to solve word problems in math
Best of all, Steps to solve word problems in math is free to use, so there's no reason not to give it a try! There are different ways to solve word problems, but the best way to do it is by using an equation solver. This will help you to set up the problem in a way that is easy to understand and solve. Once you have the equation set up, you can plug in the values and solve for the answer.
Complex numbers are numbers that can be expressed in the form a + bi, where a and b are real numbers, and i is the imaginary unit. The real part of a complex number is the a value, and the imaginary part is the b value. To add or subtract complex numbers, you add or subtract the real parts and the imaginary parts separately. To multiply complex numbers, you need to use the distributive property and FOIL. To divide complex numbers, you need to multiply
To solve for the x intercept, you need to set y = 0 and solve for x. This can be done by using algebraic methods or by graphing the equation and finding the point where it crosses the x-axis.
There's nothing quite like a good math problem to get the brain going. And algebra problems are some of the best, especially when they're tricky. They can make you think in new and creative ways, and often have more than one correct answer. What's not to love?
Trigonometry is the branch of mathematics that deals with the relationships between the sides and angles of triangles. The trigonometry solver can be used to find the values of unknown sides and angles of a triangle.
There is no one definitive way to solve multi step equations, as there are a variety of methods that can be used depending on the equation and the desired solution. However, some general tips that may be helpful include: - manipulating the equation to isolate the variable of interest on one side - using inverse operations to solving the equation step by step - using algebraic methods such as factoring or completing the square If you are struggling to solve a particular equation, it may
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# 3.3.2 - The Standard Normal Distribution
3.3.2 - The Standard Normal Distribution
A special case of the normal distribution has mean $\mu = 0$ and a variance of $\sigma^2 = 1$. The 'standard normal' is an important distribution.
Standard Normal Distribution
A standard normal distribution has a mean of 0 and variance of 1. This is also known as a z distribution. You may see the notation $N(\mu, \sigma^2$) where N signifies that the distribution is normal, $\mu$ is the mean, and $\sigma^2$ is the variance. A Z distribution may be described as $N(0,1)$. Note that since the standard deviation is the square root of the variance then the standard deviation of the standard normal distribution is 1.
## Finding Probabilities of a Standard Normal Random Variable
As we mentioned previously, calculus is required to find the probabilities for a Normal random variable. Fortunately, we have tables and software to help us.
For any normal random variable, we can transform it to a standard normal random variable by finding the Z-score. Then we can find the probabilities using the standard normal tables.
Most statistics books provide tables to display the area under a standard normal curve. Look in the appendix of your textbook for the Standard Normal Table. We include a similar table, the Standard Normal Cumulative Probability Table so that you can print and refer to it easily when working on the homework.
Most standard normal tables provide the “less than probabilities”. For example, if $Z$ is a standard normal random variable, the tables provide $P(Z\le a)=P(Z<a)$, for a constant, $a$.
## Example 3-9: Probability 'less than'
Find the area under the standard normal curve to the left of 0.87.
There are two main ways statisticians find these numbers that require no calculus! Click on the tabs below to see how to answer using a table and using technology.
A typical four-decimal-place number in the body of the Standard Normal Cumulative Probability Table gives the area under the standard normal curve that lies to the left of a specified z-value. The probability to the left of z = 0.87 is 0.8078 and it can be found by reading the table:
1. Since z = 0.87 is positive, use the table for POSITIVE z-values.
2. Go down the left-hand column, label z to "0.8."
3. Then, go across that row until under the "0.07" in the top row.
You should find the value, 0.8078. Therefore,$P(Z< 0.87)=P(Z\le 0.87)=0.8078$
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
0.6 .7257 .7291 .7324 .7357 .7389 .7422 .7454 .7586 .7517 .7549
0.7 .7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
0.8 .7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
0.9 .8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
#### Using Minitab
To find the area to the left of z = 0.87 in Minitab...
1. From the Minitab menu select Calc> Probability Distributions> Normal.
2. Select Cumulative Probability.
3. In the Input constant box, enter 0.87. Click OK
You should see a value very close to 0.8078.
## Example 3-10: Probability 'greater than'
Find the area under the standard normal curve to the right of 0.87.
Based on the definition of the probability density function, we know the area under the whole curve is one. Since we are given the “less than” probabilities in the table, we can use complements to find the “greater than” probabilities. Therefore,
$P(Z>0.87)=1-P(Z\le 0.87)$.
Using the information from the last example, we have $P(Z>0.87)=1-P(Z\le 0.87)=1-0.8078=0.1922$
#### Using Minitab
Since we are given the “less than” probabilities when using the cumulative probability in Minitab, we can use complements to find the “greater than” probabilities. Therefore,
$P(Z>0.87)=1-P(Z\le 0.87)$.
Using the information from the last example, we have $P(Z>0.87)=1-P(Z\le 0.87)=1-0.8078=0.1922$
You can also use the probability distribution plots in Minitab to find the "greater than."
1. Select Graph> Probability Distribution Plot> View Probability and click OK.
2. In the pop-up window select the Normal distribution with a mean of 0.0 and a standard deviation of 1.0.
3. Select the Shaded Area tab at the top of the window.
4. Select X Value.
5. Enter 0.87 for X value.
6. Select Right Tail.
7. Click OK.
## Example 3-11: Probability 'between'
Find the area under the standard normal curve between 2 and 3.
To find the probability between these two values, subtract the probability of less than 2 from the probability of less than 3. In other words,
$P(2<Z<3)=P(Z<3)-P(Z<2)$
$P(Z<3)$ and $P(Z<2)$ can be found in the table by looking up 2.0 and 3.0.
For 3.0...
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
2.8 0.9974 0.9975 0.9976 0.9977 0.9977 0.9978 0.9979 0.9979 0.9980 0.9980
2.9 0.9981 0.9982 0.9982 0.9983 0.9984 0.9984 0.9985 0.9985 0.9986 0.9986
3.0 0.9987 0.9987 0.9987 0.9988 0.9988 0.9989 0.9989 0.9989 0.9990 0.9990
3.1 0.9990 0.9991 0.9991 0.9991 0.9992 0.9992 0.9992 0.9992 0.9993 0.9993
For 2.0...
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
1.8 0.9641 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706
1.9 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761 0.9767
2.0 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812 0.9817
2.1 0.9821 0.9826 0.9830 0.9834 0.9838 0.9842 0.9846 0.9850 0.9854 0.9857
$P(2 < Z < 3)= P(Z < 3) - P(Z \le 2)= 0.9987 - 0.9772= 0.0215$.
#### Using Minitab
To find the area between 2.0 and 3.0 we can use the calculation method in the previous examples to find the cumulative probabilities for 2.0 and 3.0 and then subtract.
$P(2 < Z < 3)= P(Z < 3) - P(Z \le 2)= 0.9987 - 0.9772= 0.0215$
You can also use the probability distribution plots in Minitab to find the "between."
1. Select Graph> Probability Distribution Plot> View Probability and click OK.
2. In the pop-up window select the Normal distribution with a mean of 0.0 and a standard deviation of 1.0.
3. Select the Shaded Area tab at the top of the window.
4. Select X Value.
5. Select Middle.
6. Enter 2.0 for X value 1 and 3.0 for X value 2.
7. Click OK.
## Percentiles of the Standard Normal Distribution
Recall from Lesson 1 that the $p(100\%)^{th}$ percentile is the value that is greater than $p(100\%)$ of the values in a data set. We can use the standard normal table and software to find percentiles for the standard normal distribution.
The intersection of the columns and rows in the table gives the probability. If we look for a particular probability in the table, we could then find its corresponding Z value.
## Example 3-12: Percentiles in the Standard Normal Distribution
Find the 10th percentile of the standard normal curve.
The question is asking for a value to the left of which has an area of 0.1 under the standard normal curve.
Since the entries in the Standard Normal Cumulative Probability Table represent the probabilities and they are four-decimal-place numbers, we shall write 0.1 as 0.1000 to remind ourselves that it corresponds to the inside entry of the table. We search the body of the tables and find that the closest value to 0.1000 is 0.1003. We look to the leftmost of the row and up to the top of the column to find the corresponding z-value.
The corresponding z-value is -1.28. Thus z = -1.28.
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
-1.3 0.0968 0.0951 0.934 0.0918 0.0901 0.0885 0.0869 0.0853 0.0838 0.0823
-1.2 0.1150 0.1131 0.1112 0.1093 0.1075 0.1056 0.1038 0.1020 0.1003 00985
-1.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.1251 0.1230 0.1210 0.1190 0.1170
-1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.1469 0.1446 0.1423 0.1401 0.1379
Therefore, the 10th percentile of the standard normal distribution is -1.28
#### Using Minitab
To find the 10th percentile of the standard normal distribution in Minitab...
1. Select Calc> Probability Distributions> Normal.
2. In the new window choose Inverse Cumulative Probability.
3. Enter 0.1 in the Input constant box.
4. Click OK.
You should see a value very close to -1.28.
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## Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous
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Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Total Gadha - Thursday, 27 Nove m be r 2008, 12:47 PM
Although the questions on progressions may not come directly in MBA exams, the theory behind progressions is used in every place where we need to sum up numbers. During your C AT 2010 preparations, the formula used in progressions should become second nature to you as they will save a lot of time. Also, the methods of summing up various types of progressions, arithmetic, geometric or otherwise, should be very clear to you so that you are able to instantly spot the type of series you are facing. Knowing the basic methods of progressions also helps you simplify a lot of complex series. So let's start with some basic progressions and their properties: Arithmetic Progression Numbers are said to be in Arithmetic Progression (A.P.) when the difference between any two consecutive numbers in the progression is constant i.e. in an Arithmetic Progression the numbers increase or decrease by a constant difference. Each of the following series forms an Arithmetical Progression: 2, 6, 10, 14... 10, 7, 4, 1, -2... a, a + d, a + 2d, a + 3d...
Example: 1. If the 7th term of an Arithmetical Progression is 23 and 12th term is 38 find the first term and the common difference. Answer: 7th term = 23 = a + 6d ---- (1) 12th term = 38 = a + 11d ---- (2) Solving (1) and (2) we get a = 5 and d = 3 2. How many numbers of the series -9, -6, -3 should we take so that their sum is equal to 66? Answer: n[-18 + (n - 1)3]/ 2 = 66 n2 - 7n - 44 = 0 --> n = 11 or -4. The series is -9, -6, -3, 0, 3, 6, 9, 12, 15, 18, 21.. We can see that the sum of first 7 terms is 0. The sum of next four terms after 7th terms gives us the sum. Otherwise, if we count 4 terms backward from -9 we'll get the sum as -66. 3. What is the value of k such that k + 1, 3k - 1, 4k + 1 are in AP? Answer: If the terms are in AP the difference between two consecutive terms will be the same. Hence, (3k - 1) - (k + 1) = (4k + 1) - (3k - 1) 2k - 2 = k + 2 --> k = 4.
TO INSERT ARITHMETIC MEANS BETWEEN TWO NUMBERS Let n arithmetic means m1 , m2 , m3 ... mn be inserted between two numbers a and b. Therefore, a, m1 , m2 , m3 , ... mn, b are in arithmetic progression. Let d be the common difference.
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Since b is the (n + 2)th term in the progression, b = a + (n + 1)d Whence d = (b - a)/(n + 1) Hence m1 = a + (b - a)/(n + 1), m2 = a + 2(b - a)/(n + 1).. and so on. Example: 4. If 10 arithmetic means are inserted between 4 and 37, find their sum. First Method: Let the means be m1 , m2 , m3 ... m1 0 . Therefore 4, m1 , m2 , m3 ... m1 0 , 37 are in AP and 37 is the 12th term in the arithmetic progression. Hence, 37 = 4 + 11d --> d = 3 Therefore means are 7, 10, 13 ... 34 and their sum is 205. Second Method: We know that in an APthe sum of first term + last term = sum of second term + second last term = the sum of third term + third last term = .. and so on. Therefore, 4 + 37 = m1 + m1 0 = m2 + m9 = m3 + m8 = m4 + m7 = m5 + m6 = 41. Hence m1 + m2 + m3 + m9 + m1 0 = (m1 + m1 0 ) + (m2 + m9 )...+ (m5 + m6 ) = 5 x 41 = 205.
Example: 5. The sum of three numbers in A.P. is 30, and the sum of their squares is 318. Find the numbers. Answer: Let the numbers be a - d, a, a + d Hence a - d + a + a + d = 30 or a = 10 The numbers are 10 - d, 10, 10 + d Therefore, (10 - d)2 + 102 + (10 + d)2 = 318 Or d = 3, therefore the numbers are 7, 10, and 13.
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I shall have to end here and leave the rest of it for my CBT Club students. I shall cover some problems based on this in the CBT Club this week.
You might also like: Absolute Value (Modulus) Simple and Compound Interest
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by pushpinde r rana - Tue sday, 9 De ce m be r 2008, 11:14 AM Hi TG ! This is m y first post in this forum although I am a re gular visitor of this awe som e we b site (for C AT aspirants) for quie t a som e tim e . W hat has provok e d m e to post is, for a couple of days I have be e n obse rving that the article s and the study m are rial you have share d is not ge tting loade d prope rly. Is this be cause of som e LAN spe e d or the article s have be e n m ove d out of he re ? R e gards, Pushpinde r.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by sum it jam wal - Tue sday, 17 March 2009, 10:41 PM hi Tg,
nice article .. . it would be nice if you could throw som e light on ..topics lik e ave rage s and m e an value s...m any proble m s in DI com e from sam e ..
R e gards, sum it
Find the sum... by Sam R ox - Tue sday, 21 April 2009, 05:16 PM Hi TG, I have be e n unable to ge t what is the sum of 1+4+6+5+11+6+... 200te rm s?? could you ple ase he lp m e out on this proble m ?? Thank s n re gards
Re: Find the sum... by Total Gadha - Tue sday, 28 April 2009, 01:03 AM
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Hi Sam ,
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Bre ak it into two se rie s- 1 + 6 + 11... and 4 + 5 + 6.. Total Gadha
pls help. by srinivasan ravi - Tue sday, 9 June 2009, 10:42 AM He re is an e x am ple of five conse cutive positive inte ge rs whose sum is 1000: 198 + 199 + 200 + 201 + 202 = 1000. Find the large st num be r of conse cutive positive inte ge rs whose sum is e x actly 1000. a)5 b)10 c)15 d)20 e )25 pls he lp m e solve this..thank s..
Re: pls help. by vik as sharm a - Tue sday, 9 June 2009, 04:44 PM hi ravi as pe r TG sir, a num be r can be writte n as sum 2 or m ore cose cutive nos is=no of odd idvisor-1; so he re odd divisor 4-1=3
## pls corre ct m e ?????
Re: pls help. by TG Te am - W e dne sday, 10 June 2009, 12:14 PM 1000 = 2 3 x 5 3 = 5 x 200 = 25 x 40 = 16 x 62.5 1000 = 198 + 199 + 200 + 201 + 202 ( 5 te rm s) 1000 = 55 + 56 + ... + 60 + 61 + 62 + 63 + 64 + 65 + ...+ 69 + 70 (16 te rm s) 1000 = 28 + 29 + 30 + ... + 38 + 39 + 40 + 41 + 42 + ... + 50 + 51 + 52 (25 te rm s)
25 is the large st num be r of conse cutive positive inte ge rs whose sum is e x actly 1000. Vik as I have shown above the thre e ways to write 1000 as sum of conse cutive positive inte ge rs.
Re: pls help. by srinivasan ravi - Saturday, 13 June 2009, 09:11 AM thank s k am al..but how did u find those 3 ways..is the re any m e thod??
Re: pls help. by srinivasan ravi - Saturday, 13 June 2009, 09:13 AM he lp m e solve this proble m How m any te rm s, at the m inim um of se que nce 1,1/2,1/3,1/4...., m ust be adde d toge the r for the ir sum to be not le ss than 3? (1) 16 (2) 17 (3) 18 (4) 20
Re: pls help. by gaurav jain - Friday, 3 July 2009, 08:56 AM hi ravi.
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u can solve the que stion with the he lp of the choice s give n. 1000=n/2 [2a +n-1] since d=1
whe n n=25 a is an inte ge r.. he nce thats the right answe r..
Hi ...Please help by harry se k hon - W e dne sday, 15 July 2009, 01:48 PM Hi TG, Thanx for the amazing stuff on AP,GP etc. Hi Everyone, Please help in solving the problem below:How many 3 digit numbers have the property that their digits taken from left to right form an AP or GP.? Options are: a)15 c)20 b)18 d)None of these
Re: Hi ...Please help by aashish biala - Thursday, 16 July 2009, 01:18 PM hi harsim ranje e t, i think the answe r to your que stion is 38 and he nce , the corre ct option would be none of the se .
Re: pls help. by W illiam W allace - Thursday, 16 July 2009, 08:50 PM Nice approach k am al
Re: pls help. by R onak k abani - Thursday, 6 August 2009, 12:37 PM Gaurav got ur approach - goodone k am al/W illiam can u plz e x plain how did u ge t those thre e ways of e x pre ssing the no 1000 is the re any ge ne ralise d way for finding a particular num be r as the sum of othe r num be rs Thank s in Advance R onak
## Doubt by Avishe k C hak raborty - Sunday, 9 August 2009, 01:23 AM TG Sir,
Firstly, I m ust say that the article is ve ry he lpful. I have one doubt.In e x am ple 10 we got p/q as 232/990 and the sum (p+q) is tak e n as 1222.My que stion is why are we not m inim izing this, I m e an shouldn't it be 116/445 and the sum 561? Plz corre ct m e if I am doing any m istak e . Thanx
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Sanjay J - Thursday, 13 August 2009, 04:35 PM Hi TG / Kam al,
In the following que stion five conse cutive positive inte ge rs whose sum is 1000: 198 + 199 + 200 + 201 + 202 = 1000. Find the large st num be r of conse cutive positive
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inte ge rs whose sum is e x actly 1000.
## Ple ase e x plain the approach to se le ct only 5, 25 and 16 as no of te rm s ...
Thank s
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by k um ar sourav - Friday, 14 August 2009, 10:22 AM Hi Sanjay, actually what k am al did is che ck out how m any tim e s a factor of 1000 can be sum m ate d to ge t 1000 as answe r.. so 1000=5*200 m e ans 200 whe n adde d 5 tim e s we ge t 10000.. and we k now 199+201=2(200).. lik e wise we will che ck which num be r hav m ax im um m ultiple s..he re one thing should be k e pt in m ind that lik e in 25*40...dat 25 num be r should be sm alle r tha 40...othe rwise it will e x te nd to ne gative digits thank s KS
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by abhishe k tripathi - Sunday, 16 August 2009, 09:01 PM hiiiiiii tg sir actulaay i face d proble m unde rtsanding 2 points can u e x aplain de m 1 one is u said tp/tq=(2p-1)/(2q-1).what doe s dis t signify is it da sum .and se cond one in da cat 03 que stion last ste p is 1+2/7(1+1/7+1/7square +.....)how did u arrive at da ans plz he lp m e out
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Nitin Kum ar - W e dne sday, 19 August 2009, 01:18 AM Hi TG Sir, Tell me how to solve below question: For how many integral values of P the three terms -1+P-X, P+X and -1+P+X can never be in HP?
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Ank ur gupta - Friday, 9 O ctobe r 2009, 12:27 AM sum m ation of re ciprocal of first N no's?
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by ank it dik shit - Friday, 23 O ctobe r 2009, 04:32 PM TG SIR ,, I fe e l m y sle f com fortable in the quantz n di se ction.. but u have provide d light to som e of m y we ak se ctions.. I can think Mathe m atics lik e a m anage r... C re dit goe s to you... I am pre paring for cat since aug '08 .. e nrolle d wid im s full tim e at cp and tim e basic te st se rie s... but what i ge t it he re in just 1000 buck s.. i have not found such .. e ve n afte r class room coaching... this is anothe r ge m of article s from TGs factory... I m ust say.. Kudos to TG Sir and his te am .. I wish all succe ss for u... Ple ase launch fe w m ore buk s on othe r topics lik e alge bra, logical re asoning, di, m ode rn m aths e tc.. be fore cat '09.. so that we can tak e be ne fit from it... O ne m ore sugge stion... for DI/LR .. ple ase add m ore the m e s on LR /DI in C At C BT C lub... i m e an lik e tournam e nts/gam e s/alligations .. if u will add.. case le ts/ blood re lation/ quantz base d re asoning e tc.. it will be m ore be ne fiacial to all gahdas in TG cat cbt club. R Egards, THE ANKIT DIKSHIT :p
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Ne ha Sharm a - Saturday, 24 O ctobe r 2009, 05:04 PM Hi TG,
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C an the m e thod of diffe re nce s be applie d to solve the se rie s whose te rm s follow this patte rn in re curring m anne r, e .g. 1,6,21,52,105...(200 te rm s) He re the patte rn is: 1 5 10 6 6 21 15 16 6 52 31 53 22 so the com m on diffe re nce is 6 afte r 3rd ste p. 105
## How to solve this using m e thod of diffe rnce s?
Re: Hi ...Please help by nave e n k iran - Sunday, 25 O ctobe r 2009, 01:39 AM The following are in AP 123,234,345..... 7 num be rs 135,246,357,.... 5 num be rs 147,258,369 3 num be rs 129 1 num be r The following are in GP 124,139,248 3 num be rs 111,222,333... 9 num be rs total = 28 he nce option "d". Ple ase corre ct m e if i'm wrong.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Nik hil Jangi - Sunday, 25 O ctobe r 2009, 07:02 PM Hi TG, Ple ase e x plain the fundas be hind harm onic progre ssions. Thank s & R e gards, Nik hil
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Abinash Sahoo - Monday, 26 O ctobe r 2009, 11:03 PM plz he lp i have a proble m he re .... in Q -11,GP 10+100+1000+........... he re r=10 n is nt de fine d. can we put S=ar*-1/r-1 ? i think it's unde fine d.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Abinash Sahoo - Monday, 26 O ctobe r 2009, 11:48 PM hi TG sir.. i have a doubt,plz e x plain.. in Q -15,arithm e tico-ge om e tric se rie s, how can the com m on ratio be 1/7. plz e x plain
Re: pls help. by nishchai ne vre k ar - Sunday, 8 Nove m be r 2009, 08:37 PM this form ula is from the com parison te st.. which says (sum m ation of 1/n whe re n range s from 1 to 2 k) > 1+k /2 he re R HS is = 3 right.. so k = 4 ... he nce sum m ation range s from 1 to 2 4.... he nce ans is 16
## a doubt....!! by raja k - Tue sday, 10 Nove m be r 2009, 12:54 PM
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TG sir,
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u r doping a fre at job and the se article s r re ally use full for be gine rs lik e m e ....!!
i have a doubt sir, if the re e x ists two diffe re nt AP se rie s with som e com m on value s.. the n how can we find the no of com m on te rm s in both the se rie s.. hope u would re ply soon..!!
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Subhash Me dhi - Monday, 26 April 2010, 02:49 AM De ar sir, C ould you k indly write an article on hype r-ge om e tric se rie s too? Ple ase do it for our sak e . R e gards, Subhash
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by shail m ishra - Thursday, 10 June 2010, 07:53 PM Hi Guys Ple ase he lp in solving this que stion. Find the sum of the following se rie s upto infinity. 1/1 + 1/3 + 1/6 + 1/10 + 1/15 + 1/21 ...... It se e m s to be a ve ry e asy que stion but i don't k now why is it not click ing. R e gards SM
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by TG Te am - Friday, 11 June 2010, 01:55 PM
Hi Shail See the nth term is: 2/n(n + 1) = 2[1/n - 1/n+1] So the required sum is S = 2[(1/1 - 1/2) + (1/2 - 1/3) + (1/3 - 1/4) + (1/4 - 1/5) + (1/5 - 1/6) + (1/6 - 1/7) + ....] = 2. Good question.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by siddharth jain - Thursday, 2 Se pte m be r 2010, 04:28 PM hiii... i m ne w to this site but can he lp u out.... 1x 1, 2x 3, 3x 7, 4x 13.... adding 1 to first digit and 2,4,6 to se cond digit of te rm s
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Pune e t Gulati - Friday, 3 Se pte m be r 2010, 04:41 PM not so im pre ssive as com pare d to othe r article s on total gadha
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by TG Te am - Tue sday, 14 Se pte m be r 2010, 04:17 PM
Hi Siddharth nth term of the second series i.e. 1, 3, 7, 13, ... = n2 - n + 1. So nth term of the given series is = n(n2 - n + 1) = n3 - n2 + n
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## And sum of which is given by = n(n + 1)(3n2 - n + 4)/12.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by pre e ti k um ari - Thursday, 5 May 2011, 05:19 PM hi could som e body ple ase he lp m e solve the following: conside r the se que nce whe re nth te rm , tn=n/(n+2), n=1, 2.... The value of t3*t4*t5....*t54 e quals a)2/495 b)2/477 c)12/55 d)1/485 e )1/2970
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by de stiny unrule d - Tue sday, 10 May 2011, 03:12 AM @ Pre e ti Kum ari I think it should be t3*t4*t5*.......*t53 not t54 t3*t4*t5*.......*t53 = (3/5)(4/6)(5/7)....(53/55) W e can se e that de nom inator of a te rm will cance l with num e rator of ne x t ne x t te rm . So, product = 3*4/(54*55) = 2/495
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by pre e ti k um ari - Tue sday, 10 May 2011, 09:03 AM thanx .. and ye s it is t53, not t54...m y m istak e !!
Re: pls help. by Jite ndra Soni - Tue sday, 6 Se pte m be r 2011, 05:29 PM se e if this he lps conside r the se rie s a-d, a-(d-1)... a-1, a, a+1, a+2, a+ 3...a + (d-1) , a+ d the n sum = a(2d+1) = 1000 = 2^3 X 5^3 ... now 2d+1 is odd in the LHS, he nce it can be one of the 3 factors 5, 25 or 125... this give s d = 2,12,62 & a = 200, 40 & 8 re spe ctive ly. Now the re is a condition that e ach te rm should be positive , he nce a-d > 0 is satisfie d in (a,d) = ( 200,2) & (40, 12). This ge ne rate s (198, 199, 200, 201, 202) (5 te rm s) and ( 28, 29.... 40....51, 52) (25 te rm s). The condition for +ve te rm s is not satisfie d in (a,d) = ( 8,62) so we can tak e a = 63 at le ast ( so that a-d =1). Now the num be r of te rm s will be 1000/63 which give s 15. sum thing :p he nce 16 te rm s in total he nce 63 ...70 ( 8 te rm s) and (55...62) (8 te rm s) - Je e tu
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by sonne l singh - W e dne sday, 21 Se pte m be r 2011, 03:27 AM can sm bdy plz answe r this que s.. Q ue s- Two distinct incre asing inte ge r arithe m atic progre ssions with sam e first te rm e qual to 1 are such that the product of the ir nth ne th is 2010. find the m ax im um possible value of n. 1) 2) 3) 4) 5) 2009 2 3 8 15
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by use r332 u - W e dne sday, 21 Se pte m be r 2011, 07:04 PM sir i am not able to solve this proble m ..so plz he lp m e 7 + 26 + 63 +124 + ......+999
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by sahil m adan - W e dne sday, 21 Se pte m be r 2011, 07:38 PM this se rie s can be writte n as 2^3-1 + 3^3-1 + 4^3-1 + .....+10^3-1
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So apply the form ula for sum m ation of cube s of n natural no's..I hope you can proce e d from he re ..
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by use r332 u - Thursday, 22 Se pte m be r 2011, 03:43 PM re ally thank s......
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Assassin C T - Saturday, 1 O ctobe r 2011, 03:50 PM I think answe r to this Q ue s- Two distinct incre asing inte ge r arithe m atic progre ssions with sam e first te rm e qual to 1 are such that the product of the ir nth ne th is 2010. find the m ax im um possible value of n. 1) 2) 3) 4) 5) 2009 2 3 8 15
is 3. {n(n-1)(d+D)/2} +2n =2010 is the e quation we ge t at last. For only n=2,3 d+D is inte ge r. Highe st value is 3 for n.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by sonne l singh - Sunday, 9 O ctobe r 2011, 02:33 AM but how did you ge t to the following e quation {n(n-1)(d+D)/2} +2n =2010
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by R ajase k aran R ajaram - Monday, 10 O ctobe r 2011, 10:50 PM
sonnel singh, I would go with option 4) 8. Consider two series with different difference between them i.e. d1 and d2. Series 1 --> 1+(n-1)*d1 = nth term of first series Series 2 --> 1+(n-1)*d2 = nth term of first series [1+(n-1)*d1]*[1+(n-1)*d2] = 1+ (n-1)*d1+(n-1)*d2+(n-1)*d1*d2 1+(n-1)[d1+d2+(n-1)*d1d2] = 2010 (n-1)[d1+d2+(n-1)*d1d2] = 2009 2009 = 41*7*7*1 n-1 can be 41 or 7 or 1 only. So n can be 42,8 or 2. 2 and 8 are in options. For maximum value i would go with 8.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by sonne l singh - Friday, 14 O ctobe r 2011, 03:37 PM thnk u so m uch R ajase k aran R ajaram
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Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by n k - Monday, 31 O ctobe r 2011, 01:04 PM Hi Eve ryone , C an you le t m e k now how to find com m on te rm s from two se rie s. For Ex : How m any inte ge rs do the following finite arithm e tic progre ssions have in com m on? 1,8, 15, 22, ..,2003 and 2, 13, 24, 35, ., 2004
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by TG Te am - Tue sday, 1 Nove m be r 2011, 03:43 PM
Hi n k First series has the numbers of the form: 7a + 1 and the second one of the form: 11b + 2 So first common term of the two series is the smallest number which satisfies both the above expressions and i.e. 57 and all other such numbers will be at a gap of 77. So we need to find the number of terms of an AP whose first term is 57, common difference is 77 and last term is 2003. Kamal Lohia
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by nisha rani - Tue sday, 1 Nove m be r 2011, 10:20 PM how do u find sum of following se rie s? 3/5+5/36+7/144+........+21/12100 he re tn=2n+1/(n(n+1))^2
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by King C asanov - W e dne sday, 2 Nove m be r 2011, 02:25 AM i gue ss u urse lf have give n the clue to solution... tn=2n+1/ (n(n+1))^2 now (1/n^2) - (1/(n+1)^2)== 2n+1/(n(n+1))^2
The KING
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by King C asanov - W e dne sday, 2 Nove m be r 2011, 11:16 PM TG Te am , Are the fundas (and topics) cove re d in C AT 2011 Q uant Le ssons com ple te as far as whole "im portant topics" of C AT are conce rne d?? And doe s a good unde rstanding of above m e ntione d le ssons he lp m e in ge tting into one of"II" ?(ye s I k now its a ve ry vague que stion, but ple ase do re ply ..asap) P.S. : ple ase te ll m e , whe the r m y above writte n que stions are gram m atically corre ct??
The KING
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by R ahul Kishore - Monday, 7 Nove m be r 2011, 12:01 PM Hi, How to find the n'th te rm of a random se rie s.
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Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Mohit Sharm a - Sunday, 1 April 2012, 11:34 AM Gr8 Article Sir, Your work is absolute ly fantastic. R e ally a he lpful site for MBA Aspirants. C an you ple ase solve this que stion for sir, Q ue s: If x ,y,z are in GP and a^x ,b^y and c^z are e qual, than a,b,c are in which progre ssion?
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by arsh arora - Sunday, 1 April 2012, 12:46 PM hi m ohit, try tak ing x y z as 1 2 and 4 and a,b and c as 16 4 and 2 u will se e thata b and c dnt follow any patte rn the re by no se rie s patte rn...
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Mohit Sharm a - Tue sday, 3 April 2012, 03:45 PM Thank s Arsh. for the he lp
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by arsh arora - Tue sday, 3 April 2012, 05:51 PM anytim e buddy!!!
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by TG Te am - W e dne sday, 4 April 2012, 12:07 PM
Hi Mohit Question you are asking is not correct. It should be going like this: If a, b, c are in GP and ax = by = c z, then x, y, z are in which progression? And the correct answer is : HP PS: This question was reported in previous year's CAT by many students. Kamal Lohia
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Mohit Sharm a - W e dne sday, 4 April 2012, 07:04 PM Thank s sir. I got this que stion from Am it Sharm a. I think wat u r saying m ust b corre ct.
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by tushar nijhawan - Sunday, 14 O ctobe r 2012, 01:38 AM
Re: Methods of Summation- A rithmetic Progression, Geometric Progression and Miscellaneous by Gaurav Jotriwal - Sunday, 14 O ctobe r 2012, 03:53 PM De ar TG sir, what will be the solution of: 10.11.12.13 + 11.12.13.14 + ....... + 96.97.98.99 R e gards Gaurav
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# Elementary Algebra Practice Test for the ACCUPLACER® test
0%
#### √ 4 × √ 5 =
Correct! Wrong!
4 5 = 20 = 2*2*5 = 2 5
#### 4(| − 3 − 2|) − 5 =
Correct! Wrong!
Start with the numbers inside the parentheses. 4(| − 3 − 2|) − 5 4(| − 5|) − 5 Remember that the two vertical lines around “- 3 - 2” mean “absolute value.” The absolute value of -5 is 5, so: 4⋅5 − 5 = 15
#### −2⁄3 + 1⁄6 ⋅ (−2) = ____
Correct! Wrong!
Following the “Order of Operations”, first multiply 16 by −2 to get −26 so it becomes: −23 + (−26) Reduce 26 to 1 and you have: −23 + (−13) = −1
#### Which of the following sequence of numbers lists the numbers from the least to the greatest?
Correct! Wrong!
There are several ways to approach a test item like this one. If the same numbers are given in each answer choice, simply convert them all to the same format. In this case, fractions and mixed numbers with the denominator of 6 would work well. Also note the simplification of the absolute value amount to −23, before changing to −46. (An absolute value is just the positive value of any number, so the negative of that would be negative.) 72 = 3 36 -56 = -56 −∣−23∣ = −46 12 = 48 = 36 Now, list them in the correct order and compare your listing with the answer choices to find the correct one. −56 < −46 < 36 < 3 36 For a test item in which the numbers in each of the answer choices are not the same, try to find errors that immediately “stick out” to you. For instance, in the choice −56 <12 < −∣−23∣ < 48 < 72, you’ll immediately notice that 12 is listed as less than 48 and you know those are equal, so that choice is wrong. There are pretty obvious inaccuracies in the other two incorrect choices for this problem, as well. So, you’d be able to narrow it down to only one possible choice, which you would need to check with the first process, above, to be sure.
#### 4(−7+5) − (−2)(6−11) = ____
Correct! Wrong!
4(−7+5) − (−2)(6−11) In a problem such as this one, it is necessary to follow the order of operations, PEMDAS, which stands for Parentheses, Exponents, Multiplication, Division, Addition, Subtraction. Perform these steps, from left to right, in this order. So, performing the operations inside parentheses first: 4(−7+5) − (−2)(6−11) 4⋅(−2) − (−2)⋅(−5) Then, since there are no exponents, do multiplication next. −8 − 10 −18 PEMDAS can be remembered with the pneumonic, “Please Excuse My Dear Aunt Sally.”
#### x(x2−3)−2x2+5 =
Correct! Wrong!
Distribute x by multiplying with the two terms inside the parentheses: x(x2−3)=x3−3x. Then, combine with the other terms: x3−3x−2x2+5. Finally, arrange terms in such a way that the exponents of the variable x are in descending order with the constant as the last term. Thus: x3−2x2−3x+5.
#### (5x+2y)2 =
Correct! Wrong!
Write the expression out as (5x+2y)(5x+2y). Then, multiply the expressions together by FOILing (multiple the First terms, the Outer terms, the Inner terms, and the Last terms). First terms: (5x)⋅(5x) = 25x2 Outer terms: (5x)⋅(2y)=10xy Inner terms: (2y)⋅(5x)=10xy Last terms: (2y)⋅(2y)=4y2 Combining all terms: 25x2+10xy+10xy+4y2 Add similar terms: 25x2+20xy+4y2
#### (x2+ 8x + 15)⁄(x + 3) =
Correct! Wrong!
Factor the numerator using the reverse FOIL method: x2+8x+15=(x+5)(x+3). The (x+3) term on both the numerator and the denominator cancels out and we are left with (x+5), the answer.
#### xy−2y2+5y=4
Correct! Wrong!
Rearrange the equation so only the term with the x (in this equation, the ‘xy’ term) is on the left side. Thus: xy= 2y2−5y+4 Then, divide both sides by y to get an expression for x. The equation becomes: x=2y−5+4y = 2y+4y
#### Evaluate the following expression for x = 6 and y = -1: 3y2−4x+2xy
Correct! Wrong!
Plug in the given values for x and y. Substituting: 3y2−4x+2xy 3(−1)2−4(6)+2(6)(−1) Perform the operations, from left to right: 3−24+(−12) −33
#### Solve the equation for x, given y = 3 9x−12+y2 = 5
Correct! Wrong!
Plug in the given value y. Then rearrange terms so that only x is on the left side. 9x−12+y2=5 9x−12+32=5 9x−12+9=5 9x=12−9+5=8 x= 89
#### What are the roots of the polynomial (x+1)(x2−x−6) = 0?
Correct! Wrong!
Find the roots of the polynomial by setting each expression equal to 0 and solving for x. For the expression (x+1) = 0: x+1 = 0 x=−1 Next, do the expression (x2−x−6)=0 x2−x−6 = 0 Factoring out: (x+2)(x−3) = 0 x = −2 and x = 3. Therefore, the roots are -1, -2 and 3
#### (-3)(2+6)⁄4 - 2 =
Correct! Wrong!
Start with the numbers inside the parentheses and then proceed with multiplication and division. (-3)(2+6)4-2 (-3)(8)4-2 = -244-2 = -6-2 = -8
#### Which of the following is equal to (x+3)(x2+3x−5)?
Correct! Wrong!
To solve (x+3)(x2+3x−5), multiply x by all of the terms in the second expression, and then multiply 3 by all of the terms in the second expression. Then, add to combine like terms, arranging the exponents in descending order. (x+3)(x2+3x−5)=(x3+3x2−5x)+(3x2+9x−15)=x3+6x2+4x−15
#### x = 3 and x = -3 are both solutions to which of the following equations?
Correct! Wrong!
Only the equation x2−9 = 0 holds true for the values of x=3 and x=-3. (3)2−9=9−9=0 and (−3)2−9=9−9=0
#### For x > 0, 3y⁄x - y⁄2x + 2y⁄4x =
Correct! Wrong!
Add the different terms (which are all in fraction form) by making sure they have the same denominator. Multiply the first term by 44 and the second term by 22 so all 3 terms have 4x as denominator. 3Yx - y2X + 2y4x 12y4x - 2y4x + 2y4x (12y-2y+2y)4x) = 12y4x Simplifying, 12y4x = 3yx
#### Which of the following correctly lists the numbers in order from greatest to least?ub>2
Correct! Wrong!
52 must be the greatest because it is the only given fraction that is greater than one. Next must be 34 because it is the only remaining positive fraction. Finally, −14 is closer to 0 than −12 is, so −14 is greater than −12.
#### Which of the following lists the fractions in order from greatest to least when x = -1?
Correct! Wrong!
Plug in x = -1 into all terms. This gives us the numbers: -3(2)(-1) = 32 1(2)(-1) = -12 2(-1) = -2 From the greatest to least, the order would be: 32, 23, -12, -2 or -32x > 23 > 12x > 2x
#### 3 - 7x⁄2 < 10
Correct! Wrong!
Our goal is to isolate the x. 3 - 7x2 < 10 3 - 3 - 7x2 < 10 - 3 (-27 ⋅ − 7x2 > 7 ⋅ (-27) x>−2. Flip the inequality sign because we divided by a negative number.
#### Solve the following system of equations for (x,y), 3x+4y = 25, x−2y = 5
Correct! Wrong!
Start with the second equation and isolate x: x−2y=5 x=5+2y Substitute this result to x in the first equation: 3(5+2y)+4y = 25 15+6y+4y = 25 10y = 25−15 10y = 10 y = 1010 = 1 Then, plug y = 1 into either equation (preferably the simpler one): x = 5+2(1) = 7
#### Which of the following is a factor of x2−x−12 ?
Correct! Wrong!
To factor the given polynomial: x2−x−12 Think of 2 numbers which will result to -12 when multiplied, and will result to -1 when added. (x+3)(x−4). Take note that: 3x-4 = -12, and 3+(-4) = -1. Either of (x+3) and (x−4) are factors, but only (x+3) is given in the choices; hence, it is the correct answer.
#### John is shopping for supplies for his office. Let x equal the number of pens he buys and y equal the number of pencils. If pens cost \$0.45 and pencils cost \$0.15, which of the following represents the total cost of his purchase?
Correct! Wrong!
If pens cost \$0.45 and pencils cost \$0.15, the total cost of his purchase is represented by the equation: TotalPurchase=\$0.45x+\$0.15y This is not given as an option. However, it should be noticed that both 0.45 and 0.15 are divisible by 3, meaning a 3 can be factored out of both terms: 3(\$0.15x+\$0.05y) which when expanded gives: \$0.45x+\$0.15y. Therefore, it is the correct answer.
#### Find the roots of the following polynomial: x2 + 2x − 15 = 0
Correct! Wrong!
The roots of a polynomial equation are the values that when substituted into the equation yield a true statement. ‘Finding roots’ is another way of describing, ‘solve for x.’ Note: The roots are not the same as the factors. The factors are the values that can be multiplied together to equal the original equation. You will find the factors first, then set each of them equal to 0 to find the roots. Finding the factors: x2+2x−15 = 0 Think of 2 numbers that result in -15 when multiplied, and result in 2 when added. That gives us the factors: (x−3)(x+5) = 0 Take note that: −3⋅5 = −15, and −3+5 = 2. Now, find the roots. Equate each factor to 0: x−3 = 0 x = 3 x+5 = 0 x = −5 The roots of the expression are 3 and -5.
#### 7 and -4 are the roots of which of the following polynomials?
Correct! Wrong!
To get the polynomial with 7 and -4 as roots, Use FOIL method and multiply the factors: (x−7)(x+4) = x2−3x−28 Hence, the polynomial x2−3x−28 has the roots 7 and -4.
#### Which of the following are factors of x2−16 = 0?
Correct! Wrong!
This is a problem on factoring a difference of squares because there is an x2 term, a subtraction sign, no x term, then a constant. Start with: (x+_)(x−_) The two numbers will be the same but opposite in sign, and will be the square root of 16. Hence: (x+4)(x−4) These are the factors of the given expression.
#### Stacy has a mixture of quarters and dimes in her pocket. The total value of the coins is \$2.25. If she has a total of 12 coins, how many of each coin does she have?
Correct! Wrong!
To solve this word problem, set up a system of equations with the given information. Let q = the number of quarters and d = the number of dimes. She has 12 coins total, so we know that: q+d = 12 (equation 1) q = 12−d We also know the total value of the coins is \$2.25, so we have: (equation 2) 0.25q+0.1d = 2.25 Plug in q from equation 1 into equation 2: 0.25(12−d)+0.1d = 2.25 3−0.25d+0.1d = 2.25 −0.15d = 2.25−3 d = −0.75−0.15 = 5 12−5 = 7 Therefore, there are 5 dimes and 7 quarters.
#### The sum of 2 numbers is 27. Three times the first number is equal to 3 less than the second number. What is the value of the larger number?
Correct! Wrong!
et up a system of equations with the given information. Let the 2 numbers be represented by x and y. Write “the sum of the two numbers is 27” as: x+y = 27 x = 27−y (equation 1) Write “3 times the first number is equal to 3 less than the second number” as: 3x = y−3 (equation 2) Solve equation 2 by using the value of x from equation 1: 3(27−y) = y−3 81−3y = y−3 −4y = −84 4y = 84 y = 21 x = 27−21 = 6
#### Write the following polynomial as a product of its factors: x2+3x−40
Correct! Wrong!
To factor the given polynomial x2+3x−40 Think of 2 numbers that result in -40 when multiplied, and result in 3 when added. (x−5)(x+8)
#### Steven rents a car for his vacation. The rental agency charges him \$78 per day for the car. Furthermore, if he drives more than 500 miles, there will be an additional charge of 20 cents per mile for each mile over 500. If Steven rents the car for 5 days and drives a total of 563 miles, how much does he owe the rental agency?
Correct! Wrong!
The total rental cost for the car is the rental for 5 days + the rental in excess of 500 miles Total rental cost = \$78(5)+\$0.20(563−500) = \$390+\$12.60 = \$402.60
#### Which of the following are possible values for x if 2x=(x2)
Correct! Wrong!
Try the given choices by plugging the values into the given equation. We can check each: For x = 0: 20 ≠ 02 For x = 1: 21 ≠ 12 For x = 2: 22 = 22 For x = 3: 23 ≠ 32 For x = 4: 24 = 42 For x = 5: 25 ≠ 52 So, “2 and 4” is the correct solution because all the other solutions contain at least one wrong answer.
#### √ 12 √ 3 is equal to all but which of the following?
Correct! Wrong!
12 3 can be simplified as 2*2*3 3 = 2 3 3 = 2(3) = 6 Each of the following are equal to 6: 36 = 6 9 4 = 3*2 = 6 12 144 = 12 * 12 = 6 Only 3 9 = 3*3 = 9, hence, not equal to 6 or 12 3
#### John’s business started with 25 people. Since then, they have brought on 10 additional employees every year. Which of the following equations gives the number of employees in John’s company (y) as a function of the number of years (x)?
Correct! Wrong!
The problem says that y is the number of employees in the company. It starts with 25 people. At the start: y = 25 For every year (x), 10 employees are added: y = 25+10x Write the expression in the normal format, that is, terms with variable first, then the constant: y = 10x+25
#### A rectangular pool is 50 feet wide. The pool is 50% longer than it is wide, and 1⁄ as deep as it is wide. What is the volume of the pool?
Correct! Wrong!
The formula for volume is Volume = length⋅width⋅depth. W = 50feet L = 50+50%(50) =50+25 = 75feet D = 15(50) = 10feet Volume = LWD = (50)(75)(10) = 37,500feet3
#### What is the product of (x+2)(x2−3x+5)?
Correct! Wrong!
Just like when multiplying two binomials, each term in the first set of parentheses must be multiplied by all of the terms in the second set. (x+2)(x2−3x+5)=x(x2−3x+5)+2(x2−3x+5)=(x3−3x2+5x)+(2x2−6x+10)=(x3−x2−x+10)
#### If the circumference of a circle is 10π, then what is its area?
Correct! Wrong!
The formula for the circumference of a circle is: C=2πr. It is given in this question that: C=10π Equating C=2πr=10π r=10π2π=5 The formula for the area of a circle is A=πr2. Plug in r=5: A=π(52)=25π.
#### A triangle has a height of 7 inches and an area of 35 inches2. What is the length of half of the base of the triangle?
Correct! Wrong!
The formula for the area of a triangle is Area = 12 ⋅ Base ⋅ Height. Plug in the given values: 35in2 = 12 ⋅ Base ⋅ 7in Base = (35)(2)7 = 10 The question asks for half of the length of the base, so the correct answer is 5inches.
#### Emily’s Shake Shack sells two types of milkshakes: chocolate and vanilla. If a chocolate shake costs \$2.75 and a vanilla shake costs \$2.50, which of the following equations would accurately give Emily’s revenue (R) based on the number of chocolate (C) and vanilla (V) shakes she sold?
Correct! Wrong!
Emily’s total revenue (R) equals total chocolate shakes sold (C) at \$2.75 plus total vanilla shakes sold (V) at \$2.50. Hence: R=\$2.75C+\$2.50V
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# Congruence and similarity | Lesson
## What do congruent and similar mean?
Congruent triangles have both the same shape and the same size. In the figure below, triangles $ABC$ and $DEF$ are congruent; they have the same angle measures and the same side lengths.
Similar triangles have the same shape, but not necessarily the same size. In the figure below, triangles $ABC$ and $XYZ$ are similar, but not congruent; they have the same angle measures, but not the same side lengths.
Note: If two objects are congruent, then they are also similar.
### What skills are tested?
• Determining whether two triangles are congruent
• Determining whether two triangles are similar
• Using similarity to find a missing side length
## What are the triangle congruence criteria?
Two triangles are congruent if they meet one of the following criteria.
: All three pairs of corresponding sides are equal.
: Two pairs of corresponding sides and the corresponding angles between them are equal.
: Two pairs of corresponding angles and the corresponding sides between them are equal.
: Two pairs of corresponding angles and one pair of corresponding sides (not between the angles) are equal.
: The pair of
and another pair of corresponding sides are equal in two right triangles.
## What are the triangle similarity criteria?
Two triangles are similar if they meet one of the following criteria.
: Two pairs of corresponding angles are equal.
: Three pairs of corresponding sides are proportional.
: Two pairs of corresponding sides are proportional and the corresponding angles between them are equal.
## Finding missing side lengths in similar triangles
The SSS similarity criterion allows us to calculate missing side lengths in similar triangles. For similar triangles $ABC$ and $XYZ$ shown below:
$\begin{array}{rl}XY& =k\left(AB\right)\\ \\ YZ& =k\left(BC\right)\\ \\ XZ& =k\left(AC\right)\\ \\ \frac{XY}{AB}& =\frac{YZ}{BC}=\frac{XZ}{AC}=k\end{array}$
To calculate a missing side length, we:
1. Write a proportional relationship using two pairs of corresponding sides.
2. Plug in known side lengths. We need to know $3$ of the $4$ side lengths to solve for the missing side length.
3. Solve for the missing side length.
TRY: PROPERTIES OF CONGRUENT TRIANGLES
Triangles $ABC$ and $DEF$ are shown above. If the two triangles are congruent, which of the following statements must be true?
TRY: CONGRUENCE CRITERIA
For triangles $LMN$ and $PQR$, $LM=PQ$, and $\mathrm{\angle }M$ has the same measure as $\mathrm{\angle }Q$. Which of the following statements, if true, is sufficient to show that the two triangles are congruent?
TRY: IDENTIFYING SIMILAR TRIANGLES
Triangle $ABC$ is shown above. Which of the following triangles are similar to Triangle $ABC$ ?
TRY: CALCULATING MISSING SIDE LENGTH
Triangles $RST$ and $XYZ$ are similar triangles. What is the length of $RT$ ?
## Things to remember
Congruent triangles have the same corresponding angle measures and side lengths. The triangle congruence criteria are:
• SSS (Side-Side-Side)
• SAS (Side-Angle-Side)
• ASA (Angle-Side-Angle)
• AAS (Angle-Angle-Side)
• HL (Hypotenuse-Leg, right triangle only)
Similar triangles have the same corresponding angle measures and proportional side lengths. The triangle similarity criteria are:
• AA (Angle-Angle)
• SSS (Side-Side-Side)
• SAS (Side-Angle-Side)
If triangles $ABC$ and $XYZ$ are similar, then their corresponding side lengths have the same ratio:
$\frac{XY}{AB}=\frac{YZ}{BC}=\frac{XZ}{AC}=k$
## Want to join the conversation?
• Thanks a lot really happed Geometry might not be straight forward, but once you focus you can really get it down.
• I was having trouble with this today TT but I am so glad I decided to do extra practice and get help, it really helps a lot! :)
• How does the last problem equal to 20? I get it up to singling the RT by itself but how is 30/12 x 8 equaling to 20? These answers should show the work vs. skipping the steps necessary to make everyone at any level understand.
• The first step is always to find the scale factor: the number you multiply the length of one side by to get the length of the corresponding side in the other triangle (assuming of course that the triangles are congruent).
In this case you have to find the scale factor from 12 to 30 (what you have to multiply 12 by to get to 30), so that you can multiply 8 by the same number to get to the length of RT. I like to figure out the equation by saying it in my head then writing it out:
'12 times the scale factor is 30'
12 * SF = 30
If you solve it algebraically (30/12) you get:
SF = 2.5
Now that we know the scale factor we can multiply 8 by it and get the length of RT:
8 * 2.5 = RT
RT = 20
If you want to know how this relates to the disjointed explanation above, 30/12 is like the ratio of the two known side lengths, and the other ratio would be RT/8. You can then equate these ratios and solve for the unknown side, RT.
30/12 = RT/8
2.5 = RT/8
RT = 20
OR
30/12 = RT/8
30 * 8 = 12 * RT
240 = 12 * RT
RT = 20
I hope that this isn't too late and that my explanation has helped rather than made things more confusing.
• Is the smaller triangle a 3-4-5 triangle?
• examples of congruent shapes
• According to you (SAS) shows similarity and congruence.
In this case, when we find two triangle that are SAS,
are they similar or congurent?
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How to Change Fractions to Decimals – dummies
Converting fractions to decimals isn’t difficult, but to do it, you need to know about decimal division. You also need to know how to deal with terminating and repeating decimals in your answer. Here are the steps to convert a fraction to a decimal:
\n
\n
1. Set up the fraction as a decimal division, dividing the numerator (top number) by the denominator (bottom number).
\n
2. \n
3. Attach enough trailing zeros to the numerator so that you can continue dividing until you find that the answer is either a terminating decimal or a repeating decimal.
\n
4. \n
\n
When the answer is a terminating decimal
\n
Sometimes, when you divide the numerator of a fraction by the denominator, the division eventually works out evenly. The result is a terminating decimal. The following examples show terminating decimals.
\n
Suppose you want to change the fraction 2/5 to a decimal. Here’s your first step:
\n \n
One glance at this problem, and it looks like you’re doomed from the start because 5 doesn’t go into 2. But watch what happens when you add a few trailing zeros. Notice that you can also place another decimal point in the answer just above the first decimal point. This step is important:
\n \n
Now you can divide because, although 5 doesn’t go into 2, 5 does go into 20 four times:
\n \n
You’re done! As it turns out, you only needed one trailing zero, so you can ignore the rest:
\n \n
Because the division worked out evenly, the answer is an example of a terminating decimal.
\n
As another example, suppose you want to find out how to represent 7/16 as a decimal. First, you attach three trailing zeros:
\n \n
In this case, three trailing zeros aren’t enough to get your answer, so you can attach a few more and continue:
\n \n
At last, the division works out evenly, so again the answer is a terminating decimal. Therefore, 7/16 = 0.4375.
\n
When the answer is a repeating decimal
\n
Sometimes when you try to convert a fraction to a decimal, the division never works out evenly. The result is a repeating decimal — that is, a decimal that cycles through the same number pattern forever.
\n
You may recognize these pesky little critters from your calculator, when an apparently simple division problem produces a long string of numbers.
\n
For example, to change 2/3 to a decimal, begin by dividing 2 by 3. Start out by adding three trailing zeros and see where it leads:
\n \n
At this point, you still haven’t found an exact answer. But you may notice that a repeating pattern has developed in the division. No matter how many trailing zeros you attach to the number 2, the same pattern will continue forever. This answer, 0.666. .., is an example of a repeating decimal. You can write 2/3 as
\n \n
The bar over the 6 means that in this decimal, the number 6 repeats forever. You can represent many simple fractions as repeating decimals. In fact, every fraction can be represented either as a repeating decimal or as a terminating decimal — that is, as an ordinary decimal that ends.
\n
Now suppose you want to find the decimal representation of 5/11. Here’s how this problem plays out:
\n \n
This time, the pattern repeats every other number — 4, then 5, then 4 again, and then 5 again, forever. Attaching more trailing zeros to the original decimal will only string out this pattern indefinitely. So you can write
\n \n
This time, the bar is over both the 4 and the 5, telling you that these two numbers alternate forever.
\n
Repeating decimals are an oddity, but they aren’t hard to work with. In fact, as soon as you can show that a decimal division is repeating, you’ve found your answer. Just remember to place the bar only over the numbers that keep on repeating.
“, ” description ” : ”
Converting fractions to decimals isn’t difficult, but to do it, you need to know about decimal division. You also need to know how to deal with terminating and repeating decimals in your answer. Here are the steps to convert a fraction to a decimal:
\n
\n
1. Set up the fraction as a decimal division, dividing the numerator (top number) by the denominator (bottom number).
\n
2. \n
3. Attach enough trailing zeros to the numerator so that you can continue dividing until you find that the answer is either a terminating decimal or a repeating decimal.
\n
4. \n
\n
When the answer is a terminating decimal
\n
Sometimes, when you divide the numerator of a fraction by the denominator, the division eventually works out evenly. The result is a terminating decimal. The following examples show terminating decimals.
\n
Suppose you want to change the fraction 2/5 to a decimal. Here’s your first step:
\n\n
One glance at this problem, and it looks like you’re doomed from the start because 5 doesn’t go into 2. But watch what happens when you add a few trailing zeros. Notice that you can also place another decimal point in the answer just above the first decimal point. This step is important:
\n\n
Now you can divide because, although 5 doesn’t go into 2, 5 does go into 20 four times:
\n\n
You’re done! As it turns out, you only needed one trailing zero, so you can ignore the rest:
\n\n
Because the division worked out evenly, the answer is an example of a terminating decimal.
\n
As another example, suppose you want to find out how to represent 7/16 as a decimal. First, you attach three trailing zeros:
\n\n
In this case, three trailing zeros aren’t enough to get your answer, so you can attach a few more and continue:
\n\n
At last, the division works out evenly, so again the answer is a terminating decimal. Therefore, 7/16 = 0.4375.
\n
When the answer is a repeating decimal
\n
Sometimes when you try to convert a fraction to a decimal, the division never works out evenly. The result is a repeating decimal — that is, a decimal that cycles through the same number pattern forever.
\n
You may recognize these pesky little critters from your calculator, when an apparently simple division problem produces a long string of numbers.
\n
For example, to change 2/3 to a decimal, begin by dividing 2 by 3. Start out by adding three trailing zeros and see where it leads:
\n\n
At this point, you still haven’t found an exact answer. But you may notice that a repeating pattern has developed in the division. No matter how many trailing zeros you attach to the number 2, the same pattern will continue forever. This answer, 0.666. .., is an example of a repeating decimal. You can write 2/3 as
\n\n
The bar over the 6 means that in this decimal, the number 6 repeats forever. You can represent many simple fractions as repeating decimals. In fact, every fraction can be represented either as a repeating decimal or as a terminating decimal — that is, as an ordinary decimal that ends.
\n
Now suppose you want to find the decimal representation of 5/11. Here’s how this problem plays out:
\n\n
This time, the pattern repeats every other number — 4, then 5, then 4 again, and then 5 again, forever. Attaching more trailing zeros to the original decimal will only string out this pattern indefinitely. So you can write
\n\n
This time, the bar is over both the 4 and the 5, telling you that these two numbers alternate forever.
\n
Repeating decimals are an oddity, but they aren’t hard to work with. In fact, as soon as you can show that a decimal division is repeating, you’ve found your answer. Just remember to place the bar only over the numbers that keep on repeating.
Read more : Hearts Game Rules
Mark Zegarelli is an instructor and math and test prep tutor in New Jersey. He is the author of Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies, and Calculus II For Dummies. In his spare time, he enjoys traveling and learning foreign languages.
“, ” authors ” : [ { “ authorId ” :9399, ” mention ” : ” Mark Zegarelli ”, ” type slug ” : ” mark-zegarelli ”, ” description ” : ”
Mark Zegarelli is an instructor and math and test prep tutor in New Jersey. He is the author of Basic Math & Pre-Algebra For Dummies, SAT Math For Dummies, ACT Math For Dummies, Logic For Dummies, and Calculus II For Dummies. In his spare time, he enjoys traveling and learning foreign languages.
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|
# Math Expressions Grade 3 Student Activity Book Unit 5 Lesson 5 Answer Key Comparison Problems with Misleading Language
This handy Math Expressions Grade 3 Student Activity Book Pdf Answer Key Unit 5 Lesson 5 Comparison Problems with Misleading Language provides detailed solutions for the textbook questions.
## Math Expressions Grade 3 Student Activity Book Unit 5 Lesson 5 Comparison Problems with Misleading Language Answer Key
What’s the Error?
Dear Math Students,
As part of my math homework, I solved this problem:
Carlos has 19 fish. He has 19 fewer fish than Daniel. How many fish does Daniel have?
Here is what I did: 19 – 14 = 5
Is my answer right? If not. please correct my work, and tell me what I did wrong.
Puzzled Penguin
Question 1.
Write an answer to Puzzled Penguin.
Number of fishes that Carlos contain = 19
But Daniel contain fewer than Daniel
Number of Fishes that Daniel contain = 19 -14 = 5
Solve Comparison Problems with Misleading Language
Solve each problem on a separate piece of paper.
Question 2.
Unknown Smaller Amount Daniel has 23 fish. He has 15 more fish than Carlos. How many fish does Carlos have?
Number of fishes that a Daniel have = 23
Number of fishes Daniel contain than Carlos = 15
Number of fishes that Carlos have = 23 -15 = 8
Therefore, Carlos have 8 fish.
Question 3.
Unknown Larger Amount Gina ran 12 laps. She ran 8 fewer laps than Bettina. How many laps did Bettina run?
Number of Laps that Gina ran =12
Number of Fewer laps Gina ran than Bettina = 8
Number of Laps that Bettina ran = 12 – 8 = 4
Bettina ran 4 laps.
Question 4.
Unknown Smaller Amount Bettina ran 20 laps. She ran 8 more laps than Gina. How many laps did Gina run?
Number of Laps that Bettina ran = 20
Number of laps Bettina ran more than Gina = 8
Number of Laps that Gina ran = 20 – 8 = 12
Thus, Gina ran 12 laps.
Question 5.
Unknown Larger Amount Sara read 18 books this month. She read 13 fewer books than Lupe. How many books did Lupe read this month?
Number of books read by Sara =18
Number of books Sara read than Lupe = 13 fewer
Number of Books Lupe read this month = 18 -13 = 5
Thus, Lupee read 5 books this month.
Solve Comparison Problems Without the Words More or Fewer
Question 6.
The coach brought 18 hockey sticks to practice. There were 23 players at practice. How many players didn’t get sticks?
Total Number of players at practice = 23
Number of hockey Sticks Coach brought to practice = 18
Number of players who didn’t get sticks to practice = 23 – 18 = 5
5 players didn’t have sticks.
Question 7.
At a meeting, 15 people had to stand because there were not enough chairs. There were 12 chairs. How many people came to the meeting?
Total number of people in a meeting = 15
Number of chairs = 12
Number of people came to meeting = 15 +12 =27
Hence, 27 people came to the meeting.
Question 8.
Jess had 16 apples. After he gave one to each of his cousins, he had 13 apples left. How many cousins does Jess have?
Number of apples jessie contain =16
Number of Apples left =13
Number of cousins jess contain = 16 -13 = 3
Therefore, Jess have 3 cousins.
Question 9.
At the park, 4 of the children could not swing because there were not enough swings. There were 20 children at the park. How many swings were on the swing set?
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# Find all $p,q$ such that $(5p+q) \mid 168$ and $(5p-q) \mid 168$
Let $p,q$ with $p < q$ be coprime positive integers. Find all $p,q$ such that $(5p+q) \mid 168$ and $(5p-q) \mid 168$.
I first factorized $168 = 2^3 \cdot 3 \cdot 7$ but didn't see how to use the system of divisibility to continue. How do we use the system of divisibility here in order to solve the question?
Hint: There are $16$ positive divisors of 168. There are $32$ positive or negative divisors of $168$.
You can construct the system \begin{align} 5p+q&=a\\ 5p-q&=b \end{align} where $a$ is a positive divisor of $168$ and $b$ is a positive or negative divisor. Solve the simultaneous system for $p$ and $q$, using the fractions which appear to figure out what $a$'s and $b$'s are possible.
• Note that $a$ and $b$ have to have the same parity as $a-b=2q$. That cuts the number of cases to check considerably. Also we need $a\equiv -b \pmod 5$ Once you satisfy both of those, there will be a solution. – Ross Millikan Jan 4 '17 at 20:31
The (positive) divisors of $168$ are
$$1,3,7,21,\\2,6,14,42,\\4,12,28,84,\\8,24,56,168$$
Let $a$ and $b$ be any of these or their negatives. If $a=5p+q$ and $b=5p-q$, then $a+b=10p$ and $a-b=2q$, so $a$ and $b$ must have the same parity and sum to a multiple of $10$. If $p$ and $q$ are coprime, then $\gcd(a+b,a-b)=\gcd(10p,2q)=2\gcd(5p,q)=2$, since $\gcd(p,q)=1$ and $5\not\mid q$ (because $5\not\mid168$).
This leaves a relatively short list of possibilies for $(a,b)$. Listing them with $a\gt b$, the possibilities are
$$(21,-1),\\(7,3),\\ (12,-2),(28,2),(8,2),(168,2)\\ (6,4),(56,-6),\\ (14,-4),(24,-14)\\ (42,8)$$
(where each row takes a number from the list at top of divisors and picks out all later numbers from that list that satisfy the conditions). These lead to $(p,q)$ pairs
$$(2,11),\\ (1,2),\\ (1,7),(3,13),(1,3),(17,83),\\ (1,1),(5,31),\\ (1,9),(1,19),\\ (5,17)$$
The OP's stipulation that $p\lt q$ eliminates $(1,1)$.
|
# What is kinetic energy?
When an object is in motion, it is capable of doing work. We say that the object has kinetic energy (KE) or energy of motion. Does it make sense to you that the object is capable of doing work? Take a look at the carts below.
The cart on the left is pushing the cart on the right and thus doing work on it or applying a force over a certain distance. The cart on the left has KE or energy of motion.
## kinetic energy formula
Suppose an object has mass m and is moving with a speed v, we define kinetic energy (KE) as
KE =
1 / 2
mv2
The unit for the mass or m is kg, the unit for the speed or v is m/s, and the unit for the kinetic energy or KE is joules.
## Derivation of kinetic energy formula
To derive the formula, we will need a few concepts.
• We will need the work formula.
Therefore, the 3 equations we need are
F = ma equation 1
v2 = v02 + 2ad equation 2
W = Fd equation 3
We will combine these 3 equations, do some math, and finally arrive at the formula for kinetic energy.
We will start with W = Fd since what we are looking is the work a moving object can do.
W = F × d
Using equation 3, substitute ma for F
W = ma × d equation 4
Now, use equation 2 to solve for a
Let v0 = 0. Why? Since the object is usually at rest before starting moving, the initial speed will be zero then.
Divide both sides by 2d
a =
v2 / 2d
Replace the value of a in equation 4
W = m
v2 / 2d
d
W = m
v2 / 2
Substitute KE for W since the work done by a moving object is called KE or kinetic energy.
KE = m
v2 / 2
KE =
mv2 / 2
## An important observation about the kinetic energy
Notice that the speed is squared. What happens if we double the speed?
Let v be the speed of a moving object. Let speed = 2v after the speed is doubled.
KE = m
(2v)2 / 2
KE = m
4v2 / 2
KE = 4
mv2 / 2
The 4 that you see means that the kinetic energy is quadrupled.
Whenever the speed is doubled, the kinetic energy is quadrupled.
This is an interesting finding. Here is what it means in terms of driving a vehicle. If you go from 40 miles per hour to 80 miles per hour, it will take four times as much work to stop the vehicle.
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
Recommended
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# Difference between revisions of "1983 AHSME Problems/Problem 18"
## Problem
Let $f$ be a polynomial function such that, for all real $x$, $f(x^2 + 1) = x^4 + 5x^2 + 3$. For all real $x, f(x^2-1)$ is
$\textbf{(A)}\ x^4+5x^2+1\qquad \textbf{(B)}\ x^4+x^2-3\qquad \textbf{(C)}\ x^4-5x^2+1\qquad \textbf{(D)}\ x^4+x^2+3\qquad \textbf{(E)}\ \text{none of these}$
## Solution
Let $y = x^2 + 1$. Then $x^2 = y - 1$, so we can write the given equation as \begin{align*}f(y) &= x^4 + 5x^2 + 3 \\ &= (x^2)^2 + 5x^2 + 3 \\ &= (y - 1)^2 + 5(y - 1) + 3 \\ &= y^2 - 2y + 1 + 5y - 5 + 3 \\ &= y^2 + 3y - 1.\end{align*} Then substituting $x^2 - 1$ for $y$, we get \begin{align*}f(x^2 - 1) &= (x^2 - 1)^2 + 3(x^2 - 1) - 1 \\ &= x^4 - 2x^2 + 1 + 3x^2 - 3 - 1 \\ &= x^4 + x^2 - 3.\end{align*} The answer is therefore $\boxed{\textbf{(B)}}$.
## Solution 2
Let $y=x^2.$ We have that
\begin{align*}f(x^2 + 1) &= x^4 + 5x^2 + 3 \\ &= y^2 + 5y + 3 \\ &= y^2 + 5y + 4 - 1 \\ &= (y+1)(y+4) -1 \\ &= (x^2 + 1)(x^2 + 4) - 1\end{align*}
Thus, we have $f(n) = n(n+3)-1.$
If we plug in $n = x^2-1$ we have
\begin{align*}f(x^2 - 1) &= (x^2 -1)(x^2 + 2)-1 \\ &= (x^4 + x^2 - 2) - 1 \\ &= \boxed{x^4 + x^2 -3} \end{align*}
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38500 in Words
# 38500 in Words
Edited By Team Careers360 | Updated on Feb 17, 2023 02:30 PM IST
## Introduction
38500 is written as “Thirty-Eight Thousand and Five Hundred. This can be better explained with an example. Assume the price of a table in a store is marked with MRP, 38500. It should be written as INR 38500. This will signify that the price of the table will be INR 38,500. Here, the number 38500 is the cardinal number of the table as it signifies that if we spend INR 38500 we can get the table in exchange for it.
This Story also Contains
## Some Fact About The Number 38500
• 38500 is an even number. By even number, we mean that it is divisible by 2.
• 38500 is a composite number. By composite number, we mean any number that has more than 2 factors.
## Place Value
The place value of any digit helps us to understand the numeric value of the digit on the basis of its position in the number. Now we have the number 38500. Let us know the place value of each of the digits in this number.
Digits Place Ten thousands Thousands Hundreds Tens Ones Digits 3 8 5 0 0
## Expansion of the number 38500
The number 38500 can be written as
\begin{aligned}
&3 \times \text { Ten thousand }+8 \times \text { Thousand }+5 \times \text { Hundred }+0 \times \text { Ten }+0 \times \text { One }
\\&=3 \times 10000+8 \times 1000+5 \times 100+0 \times 10+0 \times 1
\\&=38500
\\&=\text { Thirty Eight Thousand Five Hundred }
\end{aligned}
You can find out the detailed other number in words article list below:-
20000 in Words 350000 in Words 50 in Words 65000 in Words 18000 in Words 28000 in Words 150 in Words 80 in Words 49000 in Words 400 in Words 11800 in Words 13500 in Words 12 in Words 71000 in Words 55000 in Words 150000 in Words 24000 in Words 59000 in Words 900 in Words 27000 in Words 12000 in Words 14400 in Words 12400 in Words 300000 in Words 17 in Words 12800 in Words 75 in Words 33000 in Words 21000 in Words 1180 in Words 41000 in Words 36000 in Words 1999 in Words 28 in Words 57000 in Words 4500 in Words 85000 in Words 95000 in Words 1200 in Words 14500 in Words 83000 in Words 48000 in Words 49 in Words 500000 in Words 60 in Words 70 in Words 10500 in Words 1100 in Words 25 in Words 27000 in Words 31000 in Words 34000 in Words 400000 in Words 7500 in Words 800 in Words 15500 in Words 16500 in Words 19500 in Words 250000 in Words 43000 in Words 44 in Words 45 in Words 47000 in Words 63000 in Words 6500 in Words 9500 in Words 99 in Words 10000000 in Words 1400 in Words 140000 in Words 200 in Words 21 in Words 4 in Words 41 in Words 52000 in Words 110000 in Words 125000 in Words 130000 in Words 2 in Words 23 in Words 24 in Words 33 in Words 35 in Words 450000 in Words 48 in Words 51000 in Words 54000 in Words 58000 in Words 5900 in Words 7 in Words 700 in Words 95 in Words 96000 in Words 12600 in Words 1900 in Words 20500 in Words 2400 in Words 26 in Words 2800 in Words 31 in Words 42 in Words 47 in Words 56000 in Words 600000 in Words 66000 in Words 68000 in Words 84000 in Words 88000 in Words 98 in Words 99999 in Words 11200 in Words 1200000 in Words 21500 in Words 27 in Words 29 in Words 32 in Words 32500 in Words 3300 in Words 350 in Words 3540 in Words 3600 in Words 37500 in Words 43 in Words 46 in Words 61000 in Words 67000 in Words 76000 in Words 800000 in Words 9 in Words 99000 in Words 96 in Words 10 in Words 1050 in Words 1111 in Words 120 in Words 1250 in Words 13200 in Words 180000 in Words 2000000 in Words 2300 in Words 24500 in Words 25500 in Words 27500 in Words 29500 in Words 26500 in Words 31500 in Words 37 in Words 1 to 100 in Words 450 in Words 5000000 in Words 700000 in Words 72 in Words 77000 in Words 81000 in Words 85 in Words 86000 in Words 88 in Words 92000 in Words 93000 in Words 94 in Words 11100 in Words 11300 in Words 12200 in Words 123 in Words 13600 in Words 13900 in Words 15400 in Words 175000 in Words 1770 in Words 23600 in Words 29500 in Words 30500 in Words 33500 in Words 3700 in Words 3800 in Words 4200 in Words 42500 in Words 69000 in Words 7080 in Words 79000 in Words 82000 in Words 900000 in Words 97000 in Words 100000000 in Words 1000000000 in Words 3100 in Words 4700 in Words 4900 in Words 10700 in Words 1 to 20 in Words
1. What is a prime number?
Any number which is not divisible by any other number apart from 1 and itself, is known as a prime number.
2. Is 38500 a perfect square?
No, 38500 is not a perfect square
Any number that can be expressed as the multiplication of any other integer is known as a perfect square number
For example, 25 is a perfect square number, as it can be expressed by multiplying 5 twice
25 = 5\times5
3. Is 38500 a perfect cube?
No, 38500 is not a perfect cube.
If the number can be expressed by multiplying thrice by any other integer, it is known as a perfect cube number.
For example,
8 is a perfect cube number, as it can be expressed by multiplying 2 thrice.
8 = 2\times2\times2
4. What is an odd number?
Any number which is not divisible by 2 is known as an odd number.
5. What is a natural number?
The range of numbers starting from 1 to infinity is known as natural numbers.
6. What are whole numbers?
Numbers from the range of 0 to infinity are known as whole numbers.
Get answers from students and experts
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Evaluation of Square Roots
## Find the value of each square root.
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Evaluation of Square Roots
Have you ever loved a sport?
Miguel loves baseball. He is such a fan that he is volunteering all summer for the University. The University team, The “Wildcats” is an excellent team and Miguel is very excited to be helping out. He doesn’t even mind not being paid because he will get to see all of the games for free while he has the opportunity to learn more about baseball.
On the day of the first game, Miguel notices some big dark clouds as he rides his bike to the ball park. Sure enough as soon as the game is about to start, the rain begins. Like magic, a bunch of different people drag a huge tarp over the entire baseball infield. Miguel has never seen a tarp so big in his whole life.
He wonders how big the tarp actually is if it covers the entire infield. Miguel, being the fan that he is knows that the distance from one base to another, say 1st\begin{align*}1^{st}\end{align*} to 2nd\begin{align*}2^{nd}\end{align*} is 90 feet. If the infield is in the shape of a square, then how many square feet is the infield? How can he be sure that his answer is correct?
Miguel begins to figure this out in his head.
Can you figure this out? Squaring numbers and finding their square roots is just one way to solve this problem. This Concept will teach you all about square roots and squaring. Pay close attention and at the end of the Concept you will be able to figure out the size of the tarp.
### Guidance
Think about a square for a minute. We can look at a square in a couple of different ways. First, we can look at just the outline of the square.
When you look at this square, you can see only the outside, but we all know that the side of a square can be measured and for a square to be a square it has to have four congruent sides.
Do you remember what congruent means?
It means exactly the same. So if a square has congruent sides, then they are the same length.
Now let’s say the side of a square is 3 units long. That means that each side of the square is 3 units long. Look at this picture of a square.
We call a number like this one a square number because it makes up a square. 32\begin{align*}3^2\end{align*} is represented in this square.
How many units make up the entire square?
If we count, we can see that this square is made up of 9 units. It is the same answer as 32\begin{align*}3^2\end{align*}, because 32\begin{align*}3^2\end{align*} is equal to 9.
Do you see a connection?
Think back to exponents, when we square a number, we multiply the number by itself. All squares have congruent side lengths, so the side length of a square multiplied by itself will tell you the number of units in the square.
We square the side length to find the number of units in the square.
This Concept is all about square roots. A square root is the number that we multiply by itself, or square, to get a certain result. In fact, if you square a number, when you take the square root of that number you will be back to the original number again.
Let’s think about the square that we just looked at. The dimensions of the square is 3×3\begin{align*}3 \times 3\end{align*}. We square the three to find the units in the square. The square root of the 3×3\begin{align*}3 \times 3\end{align*} square is 3. This is the value that we would multiply by itself.
We can find the square root of a number. How do we do this?
Finding the square root is the inverse operation of squaring a number. Inverse operations are simply the opposite of each other. Subtraction and addition are inverse operations, because one “undoes” the other. Similarly, squaring and finding the square root are inverse operations. When we find the square root, we look for the number that, times itself, will produce a given number.
We also use a symbol to show that we are looking for the square root of a number. Here is the symbol for square root.
9\begin{align*}\sqrt{9}\end{align*}
If this were the problem, we would be looking for the square root of 9.
You could think of this visually as a square that has nine units in it. What would be the length of the side? It would be three.
You could also think of it using mental math to solve it. What number times itself is equal to nine. The answer is three.
When we find the square root of a number, we evaluate that square root.
25\begin{align*}\sqrt{25}\end{align*}
This problem is asking us for the square root of 25. What number times itself is equal to 25? If you don’t know right away, you can think about this with smaller numbers.
3×3=94×4=165×5=25\begin{align*}&3 \times 3 = 9\\ &4 \times 4 = 16\\ &5 \times 5 = 25\end{align*}
That’s it! The square root of 25 is 5.
We can also evaluate numbers where the square root is not a whole number.
7\begin{align*}\sqrt{7}\end{align*}
To find the square root of seven, we can think about which two squares it is closest to.
2×2=43×3=9\begin{align*}&2 \times 2 = 4\\ &3 \times 3 = 9\end{align*}
Seven is between four and nine, so we can say that the square root of seven is between 2 and 3.
Our answer would be that the 7\begin{align*}\sqrt{7}\end{align*} is between 2 and 3.
We can get a more exact number, but we aren’t going to worry about that for right now. Here is another one.
10\begin{align*}\sqrt{10}\end{align*}
The square root of ten is between which two numbers?
3×3=94×4=16\begin{align*}&3 \times 3 = 9\\ &4 \times 4 = 16\end{align*}
Our answer is that the 10\begin{align*}\sqrt{10}\end{align*} is between 3 and 4.
Now its time for you to try a few on your own. Evaluate each square root.
#### Example A
36\begin{align*}\sqrt{36}\end{align*}
Solution:6\begin{align*}6\end{align*}
#### Example B
49\begin{align*}\sqrt{49}\end{align*}
Solution:7\begin{align*}7\end{align*}
#### Example C
12\begin{align*}\sqrt{12}\end{align*}
Solution: Between 3 and 4.
Here is the original problem once again.
Miguel loves baseball. He is such a fan that he is volunteering all summer for the University. The University team, The “Wildcats” is an excellent team and Miguel is very excited to be helping out. He doesn’t even mind not being paid because he will get to see all of the games for free while he has the opportunity to learn more about baseball.
On the day of the first game, Miguel notices some big dark clouds as he rides his bike to the ball park. Sure enough as soon as the game is about to start, the rain begins. Like magic, a bunch of different people drag a huge tarp over the entire baseball infield. Miguel has never seen a tarp so big in his whole life.
He wonders how big the tarp actually is if it covers the entire infield. Miguel, being the fan that he is knows that the distance from one base to another, say 1st\begin{align*}1^{st}\end{align*} to 2nd\begin{align*}2^{nd}\end{align*} is 90 feet. If the infield is in the shape of a square, then how many square feet is the infield? How can he be sure that his answer is correct?
Miguel begins to figure this out in his head.
We can use what we know about squares to help us with this problem. We know that a square has four equal sides. This makes sense with baseball too. You want the distance from 1st\begin{align*}1^{st}\end{align*} to 2nd\begin{align*}2^{nd}\end{align*} base to be the same as from 3rd\begin{align*}3^{rd}\end{align*} to Home. Therefore, if you know the distance from one base to another is 90 feet, then you know each distance from base to base.
However, Miguel wants to figure out the size of the tarp. He can do this by squaring the distance from 1st\begin{align*}1^{st}\end{align*} to 2nd\begin{align*}2^{nd}\end{align*} base. This will give him the area of the square.
902=90×90=8100 squarefeet\begin{align*}90^2 = 90 \times 90 = 8100 \ square feet\end{align*}
This is the size of the tarp.
How can Miguel check the accuracy of his answer? He can do this by finding the square root of the area of the tarp. Remember that finding a square root is the inverse operation for squaring a number.
8100\begin{align*}\sqrt{8100}\end{align*}
To complete this, worry about the 81 and not the 8100. 81 is a perfect square. 9×9=81\begin{align*}9 \times 9 = 81\end{align*} so 90×90=8100\begin{align*}90 \times 90 = 8100\end{align*}
8100=90 ft\begin{align*}\sqrt{8100} = 90 \ ft\end{align*}
### Vocabulary
Here are the vocabulary words in this Concept.
Square
a four sided figure with congruent sides.
Congruent
exactly the same
Square Number
a number of units which makes a perfect square.
Square root
a number that when multiplied by itself equals the square of the number.
### Guided Practice
Here is one for you to try on your own.
64\begin{align*}\sqrt{64}\end{align*}
What is the square root of 64? What number times itself is 49? Let’s start where we left off with five.
6×6=367×7=498×8=64\begin{align*}&6 \times 6 = 36\\ &7 \times 7 = 49\\ &8 \times 8 = 64\end{align*}
That’s it! The square root of 64 is 8.
### Video Review
Here is a video for review.
### Practice
Directions: Evaluate each square root.
1. 16\begin{align*}\sqrt{16}\end{align*}
2. 25\begin{align*}\sqrt{25}\end{align*}
3. 1\begin{align*}\sqrt{1}\end{align*}
4. 49\begin{align*}\sqrt{49}\end{align*}
5. 144\begin{align*}\sqrt{144}\end{align*}
6. 81\begin{align*}\sqrt{81}\end{align*}
7. 169\begin{align*}\sqrt{169}\end{align*}
8. 121\begin{align*}\sqrt{121}\end{align*}
9. 100\begin{align*}\sqrt{100}\end{align*}
10. 256\begin{align*}\sqrt{256}\end{align*}
Directions: Name the two values each square root is in between.
11. 12\begin{align*}\sqrt{12}\end{align*}
12. 14\begin{align*}\sqrt{14}\end{align*}
13. 30\begin{align*}\sqrt{30}\end{align*}
14. 40\begin{align*}\sqrt{40}\end{align*}
15. 50\begin{align*}\sqrt{50}\end{align*}
16. \begin{align*}\sqrt{62}\end{align*}
17. \begin{align*}\sqrt{70}\end{align*}
18. \begin{align*}\sqrt{101}\end{align*}
19. \begin{align*}\sqrt{5}\end{align*}
20. \begin{align*}\sqrt{15}\end{align*}
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If students see that addition and altercation is similar because In multiplication you simply repeat the Dalton problem several times then they will have an easier time learning to multiply numbers. A way in which students can relate Dalton and multiplication Is by teaching them and having them work on grouping. By grouping the students will need to draw circles for the first number that Is being multiplied and starts Inside the circles for the second number that Is being multiplied.
For example In the student will need to draw 3 circles and then the student will need to draw 5 stars inside each circle. This way the student will be able to see that they are simply adding 5 three times. The commutative property states that the order in which you add or multiply two numbers does not affect the result. (ABA=baa) For example 3*5=5*3=15. A way that this property is connected to thinking strategies is by grouping. The teacher may have the students first draw 3 bubbles and 5 stars inside each bubble and then have them count the stars for the total of 15 stars.
Then the teacher can have the students draw 5 bubbles and put 3 stars inside each bubble ND once they have done this the teacher can once again make the students count the stars and they will realize that it once again equaled 15 stars, signifying that the two ways came out with the same answer, teaching them the commutative property. The associative law states that when you add or multiply numbers, the grouping of the numbers does not affect the result ((ABA)c=a(BC). For example (2*6)3=2(6*3)=36. The associative property can be worked out by drawing it out and grouping together.
For example for the (2*6)3=2(6*3) problem the students can draw 3 bubbles and raw 12 stars inside each bubble or draw out 2 bubbles and draw 18 stars inside each bubble, if the students count both of the different group of stars there will be 36 stars in each picture, therefore showing the students that the order In which the numbers are multiplied does not affect the outcome. The distributive law states that multiplying a number by a group of numbers added together Is the same as doing each multiplication separately. When the distributive property Is used you distribute a number to get the same answer. (b + c) = ABA + AC and (b + c)a = baa + ca) For example 2(3+4)= With the deliberate property the students can connect It to a thinking strategy Is by skip counting. For example In the problem 2(3+4) the students can either break the problem apart and do It separately or do It together, they can skip count by as 3 times and then by as 4 times and add the numbers or skip count by as 7 times, both will equal 14. One conceptual error that may be associated with addition and multiplication Is that students may rush themselves ND not look at the sign if it is addition or multiplication.
One way to help the worksheet using highlighters. Once the worksheet is handed out to the students the teacher can ask the students to take out their highlighters and when they are working out each problem they must first highlight the sign, whether it is addition or multiplication, this way they will take their time and look at the sign to correctly answer the problem. A second misconception associated with multiplication is that the students may not correctly work out the distributive law.
In a problem such as (2+4) they may forget that they must distribute the 3 to each number and instead do 3*2+4. A way to help the students not commit this error is to first hand them out a worksheet that they only need to write the next step they will take, such as 3(2+4)=3*2+3*4. A second way to help the students not commit this error is to have them draw an error from the number three to the number to and a second arrow from the number three to the number 4 for each problem, this way the students will remember that they must multiply the first number to each number inside the parenthesis first.
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