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## IB CRASH COURSE FOR MAY SESSION 2024
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# Mastering the Greater Than and Less Than Signs: Tips and Tricks
Have you ever stared at the greater than and less than signs, feeling lost in a sea of symbols? Don’t worry, you’re not alone! These mathematical operators are essential for comparing numbers and values, but they can be confusing to use. Knowing how to master these signs is crucial for acing math tests or creating charts and graphs. In this blog post, we’ll explore the ins and outs of using the greater than and less than signs with ease. We’ll provide tips, tricks, and insights into how to remember which symbol means what. Get ready to level up your math game with our comprehensive guide!
## What do the greater than and less than signs mean?
The greater than and less than signs are mathematical symbols used for comparing numbers, values, or variables. The greater than sign (>), often called an open triangle, means that the value on the left side is larger or more significant than the value on the right.
On the other hand, the less than sign (<), also known as a closed triangle, shows that the number on its left is smaller in magnitude compared to its counterpart on its right. These operators can be combined with equal signs (=) to indicate “greater than or equal to” (≥) and “less than or equal to” (≤).
In essence, these simple yet powerful symbols allow us to express relationships between two quantities effectively. For instance, you can use them when comparing prices of items at a store or finding out which athlete ran faster during a race.
While they may seem basic concepts in math, mastering their usage could save you from making critical errors while solving complex problems. So always remember: > means bigger and < stands for smaller!
Also Read : Literary Techniques: Mastering the Artistry of Language in Literature
## When to use the greater than and less than signs?
The greater than and less than signs are used to represent mathematical comparisons between two quantities. These symbols indicate which quantity is larger or smaller in value, respectively. Knowing when to use these signs can be essential for solving equations, interpreting data, and making decisions based on numerical information.
One common application of the greater than and less than signs is in comparing measurements or values. For example, if you want to compare the heights of two people, you would use the greater than or less than sign depending on which person is taller.
Another situation where you might use these symbols is in analyzing trends over time. By comparing data from different periods using the greater than or less than sign, you can identify whether a trend is increasing or decreasing.
In addition to quantitative analysis, these signs are also useful for logical reasoning. In logic puzzles and arguments that involve comparisons between ideas or statements, the greater than and less than signs can help clarify relationships between them.
Understanding when to use the greater-than and less-than symbols requires an ability to interpret numerical information accurately as well as an understanding of how they fit into broader patterns of logical reasoning.
## Tips for using the greater than and less than signs
Using the greater than and less than signs may seem simple, but they can be tricky to use correctly. Here are some tips that will help you master these symbols:
• Always remember the direction: The greater than symbol (>) always points towards the larger value while the less than symbol (<) points towards the smaller value.
• Use them in math equations: The symbols are commonly used in math equations to show which number is larger or smaller.
• Compare values in data analysis: In data analysis, we often compare two sets of numbers to determine which one is bigger or smaller using these signs.
• Know when to use an equals sign: When comparing values that could be equal, it’s important to include an equals sign with either > or < so that it becomes >= or <=
• Avoid confusion with open and closed circles: When graphing on a number line, open circles mean “less than” or “greater than,” while closed circles represent “less than or equal to” and “greater than or equal to.”
By keeping these tips in mind, you’ll have no problem using greater-than and less-than signs accurately!
## Tricks for using the greater than and less than signs
Tricks for using the greater than and less than signs can come in handy when you need to compare numbers or values. One trick is to remember that the alligator always wants to eat the bigger number. The open mouth of the alligator faces towards the larger value, indicating that it wants to consume it.
Another helpful trick is to use simple math equations alongside the signs. For example, if you want to compare 7 and 12, write out “7 + __ < 12.” Then solve for what goes in the blank space: “5.” So your sentence would read “7 + 5 < 12,” which is true.
It’s also important to keep in mind that these signs can be used for more than just numerical comparisons. They can also be used with words or phrases, such as “apples > oranges” or “summer < winter.”
A useful tip is to double-check your work before submitting anything. It’s easy for mistakes like accidentally flipping around the signs or forgetting a negative symbol to go unnoticed if you don’t take a second look at your work.
Incorporating these tricks into your use of greater than and less than signs will help ensure accuracy and make comparisons easier overall.
## How to remember the difference between the two signs?
Are you tired of confusing the greater than and less than signs? Don’t worry, you’re not alone. Many people struggle with remembering which sign is which. But fear not, there are a few helpful tricks that can aid in differentiating between the two symbols.
One common method is to think of the symbols as hungry alligators. The alligator always wants to eat more, so its mouth faces towards the larger number. Therefore, if we have 5 < 10, we see that the alligator’s mouth is facing towards 10 because it’s bigger.
Another trick involves using an acronym: LAME (Less Than => More). This means that when we see “less than”, it indicates that something is smaller and therefore pointing towards a lesser value or quantity.
Alternatively, some find it easier to visualize these signs on a number line where going from left to right represents increasing values. In this case, anything above another point will be labeled as greater while any points below are labeled as less than.
Ultimately finding what works for you best will help make mastering these mathematical operators even easier!
Also Read : Mastering 4 Quarts to Cups Conversion: Your Ultimate Guide
## Conclusion
Mastering the greater than and less than signs is a crucial skill that every student must learn. Whether you’re solving math problems or comparing numbers in everyday life situations, these symbols are essential tools that can help simplify your work.
Remember to always use the correct symbol when expressing comparisons between two values, keeping in mind that the wider opening points towards the larger value. Keep practicing and applying these tips and tricks until they become second nature to you.
With enough practice and patience, anyone can master this fundamental concept of mathematics. So go ahead and start using these symbols with confidence!
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### Section 4.14 : Business Applications
5. The production costs, in dollars, per week of producing $$x$$ widgets is given by,
$C\left( x \right) = 4000 - 32x + 0.08{x^2} + 0.00006{x^3}$
and the demand function for the widgets is given by,
$p\left( x \right) = 250 + 0.02x - 0.001{x^2}$
What is the marginal cost, marginal revenue and marginal profit when $$x = 200$$ and $$x = 400$$? What do these numbers tell you about the cost, revenue and profit?
Show All Steps Hide All Steps
Start Solution
First, we need to get the revenue and profit functions. From the notes for this section we know that these functions are,
\begin{align*}{\mbox{Revenue : }} & R\left( x \right) = x\,p\left( x \right) = 250x + 0.02{x^2} - 0.001{x^3}\\ {\mbox{Profit : }} & P\left( x \right) = R\left( x \right) - C\left( x \right) = - 4000 + 282x - 0.06{x^2} - 0.00106{x^3}\end{align*} Show Step 2
From the notes in this section we know that the marginal cost, marginal revenue and marginal profit functions are simply the derivative of the cost, revenue and profit functions so let’s start with those.
\begin{align*}C'\left( x \right) & = - 32 + 0.16x + 0.00018{x^2}\\ R'\left( x \right) & = 250 + 0.04x - 0.003{x^2}\\ P'\left( x \right) & = 282 - 0.12x - 0.00318{x^2}\end{align*} Show Step 3
The marginal cost, marginal revenue and marginal profit for each value of $$x$$ is then,
\require{bbox} \bbox[2pt,border:1px solid black]{\begin{align*}C'\left( {200} \right) & = 7.2 & \hspace{0.75in}R'\left( {200} \right) & = 138 & \hspace{0.5in}P'\left( {200} \right) & = 130.8\\ C'\left( {400} \right) & = 60.8 & \hspace{0.5in}R'\left( {400} \right) & = - 214 & \hspace{0.5in}P'\left( {400} \right) & = - 274.8\end{align*}} Show Step 4
From these computations we can see that producing the 201st widget will cost approximately $7.2 and will add approximately$138 in revenue and $130.8 in profit. Likewise, producing the 401st widget will cost approximately$60.8 and will see a decrease of approximately $214 in revenue and a decrease of$274.8 in profit.
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Mathematics
# Evaluate $i)\displaystyle \int \dfrac{x^2 + 1}{x^4 + 1}$ dx$ii)\displaystyle \int \dfrac{dx}{x^2 + 1}$
##### SOLUTION
$i)\displaystyle \int{\cfrac{{x}^{2} + 1}{{x}^{4} + 1} dx}$
$= \displaystyle\int{\cfrac{1 + \cfrac{1}{{x}^{2}}}{{x}^{2} + \cfrac{1}{{x}^{2}}} dx}$
$= \displaystyle \int{\cfrac{1 + \cfrac{1}{{x}^{2}}}{{\left( x - \cfrac{1}{x} \right)}^{2} + 2} dx}$
Let
$x - \cfrac{1}{x} = t \Rightarrow \left( 1 + \cfrac{1}{{x}^{2}} \right) dx = dt$
Therefore, the integral become-
$\int{\cfrac{dt}{{t}^{2} + {\left( \sqrt{2} \right)}^{2}}}$
$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{t}{\sqrt{2}} \right)} + C$
Substituting the value of $t$, we get
$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{\left( 1 + \cfrac{1}{{x}^{2}} \right)}{\sqrt{2}} \right)} + C$
$= \cfrac{1}{\sqrt{2}} \tan^{-1}{\left( \cfrac{{x}^{2} + 1}{\sqrt{2} {x}^{2}} \right)} + C$
$ii)\displaystyle \int{\cfrac{1}{{x}^{2} + 1} dx}$
$\displaystyle \int{\cfrac{1}{{x}^{2} + {\left( 1 \right)}^{2}} dx}$
$= \tan^{-1}{x} + C$
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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86
#### Realted Questions
Q1 Single Correct Medium
$\displaystyle\int^{\pi/2}_0\dfrac{\sin^nx}{(\sin^nx+\cos^nx)}dx=?$
• A. $\dfrac{\pi}{2}$
• B. $1$
• C. $0$
• D. $\dfrac{\pi}{4}$
1 Verified Answer | Published on 17th 09, 2020
Q2 Single Correct Medium
$\displaystyle \int\dfrac{(\cos x)^{n-1}}{(\sin x)^{n+1}}dx=$
• A. $\displaystyle \frac{-\cot^{n}x}{n+1}+c$
• B. $\displaystyle \frac{\cot^{n}x}{n}+c$
• C. $\displaystyle \frac{\cot^{n}x}{n+1}+c$
• D. $\displaystyle \frac{-\cot^{n}x}{n}+c$
1 Verified Answer | Published on 17th 09, 2020
Q3 Single Correct Hard
Statement-l $\displaystyle \int_{0}^{\pi/2}\frac{dx}{1+\tan^{5}x}=\frac{\pi}{4}$
Statement 2:$\displaystyle \int_{0}^{a}f(x)dx=\int_{0}^{a}f(a+x)dx= \int_{0}^{\pi/2}\displaystyle \frac{dx}{1+\tan^{3}x}=\int_{0}^{\pi/2}\displaystyle \frac{d_{X}}{1+\cot^{3}x}=\frac{\pi}{4}$
• A. Statement 1 is True, Statement 2 is True; Statement 2 is a correct exlanation for Statement 1
• B. Statement 1 is True, Statement 2 is True; Statement 2 Not a correct exlanation for Statement 1
• C. Statement 1 is False, Statement 2 is True
• D. Statement 1 is True, Statement 2 is False
1 Verified Answer | Published on 17th 09, 2020
Q4 Single Correct Medium
$\displaystyle \int \dfrac{dt}{t + \sqrt{a^2 - t^2}}$ equal to
• A. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log (t + \sqrt{a^2 - t^2}) + k$
• B. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log \sqrt{a + \sqrt{a^2 - t^2}} + k$
• C. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log (a + \sqrt{a^2 - t^2}) + k$
• D. $\dfrac{1}{2} \sin^{-1} \left(\dfrac{t}{a} \right) + \log \sqrt{t + \sqrt{a^2 - t^2}} + k$
Consider two differentiable functions $f(x), g(x)$ satisfying $\displaystyle 6\int f(x)g(x)dx=x^{6}+3x^{4}+3x^{2}+c$ & $\displaystyle 2 \int \frac {g(x)dx}{f(x)}=x^{2}+c$. where $\displaystyle f(x)>0 \forall x \in R$
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# 7.5 Matrices and matrix operations (Page 4/10)
Page 4 / 10
We proceed the same way to obtain the second row of $\text{\hspace{0.17em}}AB.\text{\hspace{0.17em}}$ In other words, row 2 of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ times column 1 of $\text{\hspace{0.17em}}B;\text{\hspace{0.17em}}$ row 2 of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ times column 2 of $\text{\hspace{0.17em}}B;\text{\hspace{0.17em}}$ row 2 of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ times column 3 of $\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$ When complete, the product matrix will be
$AB=\left[\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{11}+{a}_{12}\cdot {b}_{21}+{a}_{13}\cdot {b}_{31}\\ \end{array}\\ {a}_{21}\cdot {b}_{11}+{a}_{22}\cdot {b}_{21}+{a}_{23}\cdot {b}_{31}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{12}+{a}_{12}\cdot {b}_{22}+{a}_{13}\cdot {b}_{32}\\ \end{array}\\ {a}_{21}\cdot {b}_{12}+{a}_{22}\cdot {b}_{22}+{a}_{23}\cdot {b}_{32}\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}\begin{array}{l}{a}_{11}\cdot {b}_{13}+{a}_{12}\cdot {b}_{23}+{a}_{13}\cdot {b}_{33}\\ \end{array}\\ {a}_{21}\cdot {b}_{13}+{a}_{22}\cdot {b}_{23}+{a}_{23}\cdot {b}_{33}\end{array}\right]$
## Properties of matrix multiplication
For the matrices $\text{\hspace{0.17em}}A,B,\text{}$ and $\text{\hspace{0.17em}}C\text{\hspace{0.17em}}$ the following properties hold.
• Matrix multiplication is associative: $\text{\hspace{0.17em}}\left(AB\right)C=A\left(BC\right).$
• Matrix multiplication is distributive: $\begin{array}{l}\begin{array}{l}\\ \text{\hspace{0.17em}}C\left(A+B\right)=CA+CB,\end{array}\hfill \\ \text{\hspace{0.17em}}\left(A+B\right)C=AC+BC.\hfill \end{array}$
Note that matrix multiplication is not commutative.
## Multiplying two matrices
Multiply matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and matrix $\text{\hspace{0.17em}}B.$
First, we check the dimensions of the matrices. Matrix $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ has dimensions $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$ and matrix $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ has dimensions $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$ The inner dimensions are the same so we can perform the multiplication. The product will have the dimensions $\text{\hspace{0.17em}}2\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2.$
We perform the operations outlined previously.
## Multiplying two matrices
Given $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}B:$
1. Find $\text{\hspace{0.17em}}AB.$
2. Find $\text{\hspace{0.17em}}BA.$
1. As the dimensions of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}2\text{}×\text{}3\text{\hspace{0.17em}}$ and the dimensions of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}3\text{}×\text{}2,\text{}$ these matrices can be multiplied together because the number of columns in $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ matches the number of rows in $\text{\hspace{0.17em}}B.\text{\hspace{0.17em}}$ The resulting product will be a $\text{\hspace{0.17em}}2\text{}×\text{}2\text{\hspace{0.17em}}$ matrix, the number of rows in $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ by the number of columns in $\text{\hspace{0.17em}}B.$
2. The dimensions of $\text{\hspace{0.17em}}B\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}3×2\text{\hspace{0.17em}}$ and the dimensions of $\text{\hspace{0.17em}}A\text{\hspace{0.17em}}$ are $\text{\hspace{0.17em}}2×3.\text{\hspace{0.17em}}$ The inner dimensions match so the product is defined and will be a $\text{\hspace{0.17em}}3×3\text{\hspace{0.17em}}$ matrix.
Is it possible for AB to be defined but not BA ?
Yes, consider a matrix A with dimension $\text{\hspace{0.17em}}3\text{\hspace{0.17em}}×\text{\hspace{0.17em}}4\text{\hspace{0.17em}}$ and matrix B with dimension $\text{\hspace{0.17em}}4\text{\hspace{0.17em}}×\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$ For the product AB the inner dimensions are 4 and the product is defined, but for the product BA the inner dimensions are 2 and 3 so the product is undefined.
## Using matrices in real-world problems
Let’s return to the problem presented at the opening of this section. We have [link] , representing the equipment needs of two soccer teams.
Wildcats Mud Cats
Goals 6 10
Balls 30 24
Jerseys 14 20
We are also given the prices of the equipment, as shown in [link] .
Goal $300 Ball$10 Jersey $30 We will convert the data to matrices. Thus, the equipment need matrix is written as $E=\left[\begin{array}{c}6\\ 30\\ 14\end{array}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\begin{array}{c}10\\ 24\\ 20\end{array}\right]$ The cost matrix is written as $C=\left[\begin{array}{ccc}300& 10& 30\end{array}\right]$ We perform matrix multiplication to obtain costs for the equipment. The total cost for equipment for the Wildcats is$2,520, and the total cost for equipment for the Mud Cats is $3,840. Given a matrix operation, evaluate using a calculator. 1. Save each matrix as a matrix variable $\text{\hspace{0.17em}}\left[A\right],\left[B\right],\left[C\right],...$ 2. Enter the operation into the calculator, calling up each matrix variable as needed. 3. If the operation is defined, the calculator will present the solution matrix; if the operation is undefined, it will display an error message. #### Questions & Answers Need help solving this problem (2/7)^-2 Simone Reply what is the coefficient of -4× Mehri Reply -1 Shedrak the operation * is x * y =x + y/ 1+(x × y) show if the operation is commutative if x × y is not equal to -1 Alfred Reply An investment account was opened with an initial deposit of$9,600 and earns 7.4% interest, compounded continuously. How much will the account be worth after 15 years?
lim x to infinity e^1-e^-1/log(1+x)
given eccentricity and a point find the equiation
12, 17, 22.... 25th term
12, 17, 22.... 25th term
Akash
College algebra is really hard?
Absolutely, for me. My problems with math started in First grade...involving a nun Sister Anastasia, bad vision, talking & getting expelled from Catholic school. When it comes to math I just can't focus and all I can hear is our family silverware banging and clanging on the pink Formica table.
Carole
I'm 13 and I understand it great
AJ
I am 1 year old but I can do it! 1+1=2 proof very hard for me though.
Atone
hi
Not really they are just easy concepts which can be understood if you have great basics. I am 14 I understood them easily.
Vedant
find the 15th term of the geometric sequince whose first is 18 and last term of 387
I know this work
salma
The given of f(x=x-2. then what is the value of this f(3) 5f(x+1)
hmm well what is the answer
Abhi
If f(x) = x-2 then, f(3) when 5f(x+1) 5((3-2)+1) 5(1+1) 5(2) 10
Augustine
how do they get the third part x = (32)5/4
make 5/4 into a mixed number, make that a decimal, and then multiply 32 by the decimal 5/4 turns out to be
AJ
how
Sheref
can someone help me with some logarithmic and exponential equations.
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
I'm not sure why it wrote it the other way
Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
oops. ignore that.
so you not have an equal sign anywhere in the original equation?
hmm
Abhi
is it a question of log
Abhi
🤔.
Abhi
I rally confuse this number And equations too I need exactly help
salma
But this is not salma it's Faiza live in lousvile Ky I garbage this so I am going collage with JCTC that the of the collage thank you my friends
salma
Commplementary angles
hello
Sherica
im all ears I need to learn
Sherica
right! what he said ⤴⤴⤴
Tamia
hii
Uday
hi
salma
hi
Ayuba
Hello
opoku
hi
Ali
greetings from Iran
Ali
salut. from Algeria
Bach
hi
Nharnhar
what is a good calculator for all algebra; would a Casio fx 260 work with all algebra equations? please name the cheapest, thanks.
a perfect square v²+2v+_
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# Algebra : Download Mathematics Study Notes Free PDF For REET Exam
Mathematics is an equally important section for CTET, MPTET, KVS & DSSSB Exams and has even more abundant importance in some other exams conducted by central or state govt. Generally, there are questions asked related to basic concepts, Facts and Formulae of the Algebra.
To let you make the most of Mathematics section, we are providing important facts related to the Algebra. At least 2-3 questions are asked from this topic in most of the teaching exams. We wish you all the best of luck to come over the fear of the Mathematics section.
How to Overcome Exam Fever, Especially When You Fear Maths
## Algebra
Algebra
It is branch of mathematics that substitutes letters for numbers. Algebra gives different methods of solve equations.
Variables
A variable is represented by either a sign or a letter. Its value may not be the same in every equation.
Example: 2x – 10 = 0; here, will get x = 5
3x – 3 = 0; here, he will get x = 10.
Here, x can have different values in different equations.
Therefore, x is variable
Constants
A constant always has fixed values. Every real number is a constant.
Example: 2, 5, 7 etc.
Expressions
An expression is a combination of constant and/or variables connected to each other by mathematical operators (addition, subtraction, multiplication and division).
Example: 3x + 5, 2y² – 4x 5, etc.
Terms
The parts of an expression are separated from one another by plus or minus sign and are called terms of that expression.
Example: in 3x + 5, 3x and 5 are both expressions, as they are separated by a plus sign.
In 2y² – 4x + 5, 2y², 4x and 5 are separated by plus and minus signs, Therefore, they all are terms.
Mathematics Study Notes For All Teaching Exams
Like Terms
Two or more terms are said to be alike if their algebraic factors are the same.
Example: 3x²y, 7x²y and 10x²y are like terms.
In the expression 2xy + 3x – 4y – 7xy, 2xy and 7xy are like terms.
Unlike Terms
Two or more terms are said to be unalike if their algebraic factors are different
Example: 3x²y, 7xy and 10xy² are unlike terms.
Factors
When numbers and variables multiply to form a product, each quantity is called a factor of the product.
Example: Factors of 3xy are 3, x and y.
Coefficients
The numerical part of term is called its coefficient.
Example: In 6x³, 6 is the coefficient of x³.
Get free Study material for REET Exam
Polynomials
The algebraic expression having one or more terms, each of which consists of a constant multiplied by one or more variables raised to a negative integral power, is called a polynomial.
A polynomial can have any finite number of terms.
Monomials, binomials and trinomials are the types of polynomials.
Monomials
The algebraic expression having a single term is called a monomial.
Example: 4x is a monomial expression.
Binomials
The algebraic expression having two terms is called a binomial.
Example: 4x³ + 7x is binomial expression
Trinomials
The algebraic expression having three terms is called a trinomial.
Example: 4x³ + 2y² – x is a trinomial expression.
Addition and subtraction of Algebraic Expressions
The addition or subtraction of algebraic expressions can be simplified by combining the like terms. In this method, coefficient or combined according to their signs, keeping the same algebraic factors.
Example:
Practice REET Subject wise Quiz
Multiplication of Algebraic Expressions
By the distributive law, the product of algebraic expressions is calculated.
Example:
1. Find 2x²y × (7x²y – 2xy²).
We have
2x²y × (7x²y – 2xy²)
= 2x²y × 7x²y – 2x²y × 2xy²
= 14x⁴y² – 4x³y³
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Question Video: Finding the Ratio between Two Terms in a Binomial Expansion | Nagwa Question Video: Finding the Ratio between Two Terms in a Binomial Expansion | Nagwa
# Question Video: Finding the Ratio between Two Terms in a Binomial Expansion Mathematics • Third Year of Secondary School
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Consider the expansion of (8π₯ + 2π¦)Β²Β³. Find the ratio between the eighth and the seventh terms.
02:17
### Video Transcript
Consider the expansion of eight π₯ plus two π¦ to the power of 23. Find the ratio between the eighth and the seventh terms.
In order to actually find the answer to this and actually find the ratio between the eighth and the seventh terms, what weβre actually gonna do is actually find the ratio between consecutive terms in a binomial expansion. Now in order to find this ratio, what we actually have is a formula that we can use. And the formula tells us that if we have a binomial expansion in the form π plus π to the power of π, then the ratio of two consecutive terms is equal to π minus π plus one over π multiplied by π over π.
Okay, great, so we now have a formula for actually finding the ratio between two consecutive terms. So what we need to do now is actually use this to find the ratio between our eighth and seventh terms. So what we have is the ratio between the eighth and seventh terms is equal to and now here is the point but we have to actually work out so what are our π and what are our π.
Well, our π is equal to seven and our π is equal to 23 because this is the exponent that the actual parenthesis raise to. And then, π is equal to eight π₯ because thatβs our first term and π is equal to two π¦ because this is our second time. Okay, great, so weβve now got all our unknowns, we can actually put them into our formula and find the ratio of our terms. So we get that the ratio between the eighth and seventh terms is equal to 23 β because thatβs our π β minus seven β our π β plus one all divided by seven because again thatβs our π. And then, this is multiplied by π over π β so two π¦ over eight π₯.
Okay, great, so now, letβs start to simplify. So this gives us 17 over seven multiplied by two π¦ over eight π₯. So now as weβre multiplying fractions, we just multiply the numerators and denominators, which gives us 34π¦ over 56π₯. So then, finally, we divide the numerator and denominator by two because thatβs a factor of both 34 and 56.
So therefore, we can say that the ratio between the eighth and the seventh terms is equal to 17π¦ over 28π₯.
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## Pages
Showing posts with label inequality. Show all posts
Showing posts with label inequality. Show all posts
## Saturday, November 10, 2012
### Introduction to Inequalities and Interval Notation
All of the steps that we have learned for solving linear equations are the same for solving linear inequalities except one. We may add or subtract any real number to both sides of an inequality and we may multiply or divide both sides by any positive real number. The only new rule comes from multiplying or dividing by a negative number.
So whenever we divide or multiply by a negative number we must reverse the inequality. It is easy to forget to do this so take special care to watch out for negative coefficients.
Notice that we obtain infinitely many solutions for these linear inequalities. Because of this we have to present our solution set in some way other than a big list. The two most common ways to express solutions to an inequality are by graphing them on a number line and interval notation.
Note: We use the following symbol to denote infinity:
Tip: Always use round parentheses and open dots for inequalities without the equal and always use square brackets and closed dots for inequalities with the equal.
J. Redden on G+
## Saturday, November 3, 2012
### Linear Inequalities (Two Variables)
When graphing an equation like y = 3x − 6 we know that it will be a line. The graph of a linear inequality such as y >= 3x − 6, on the other hand, gives us a region of ordered pair solutions.
Not only do the points on the line satisfy this linear inequality - so does any point in the region that we have shaded. This line is the boundary that separates the plane into two halves - one containing all the solutions and one that does not. Therefore, from the above graph, both (0, 0) and (−2, 4) should solve the inequality.
Use a test point not on the boundary to determine which side of the line to shade when graphing solutions to a linear inequality. Usually the origin is the easiest point to test as long as it is not a point on the boundary.
Graph the solution set.
If the test point yields a true statement shade the region that contains it. If the test point yields a false statement shade the opposite side.
When graphing strict inequalities, inequalities without the equal, the points on the line will not satisfy the inequality; hence, we will use a dotted line to indicate this. Otherwise, the steps are the same.
Graph the solution set.
Given the graph determine the missing inequality.
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Motion Along a Straight Line
Instantaneous Velocity and Speed
Learning Objectives
By the end of this section, you will be able to:
• Explain the difference between average velocity and instantaneous velocity.
• Describe the difference between velocity and speed.
• Calculate the instantaneous velocity given the mathematical equation for the velocity.
• Calculate the speed given the instantaneous velocity.
We have now seen how to calculate the average velocity between two positions. However, since objects in the real world move continuously through space and time, we would like to find the velocity of an object at any single point. We can find the velocity of the object anywhere along its path by using some fundamental principles of calculus. This section gives us better insight into the physics of motion and will be useful in later chapters.
Instantaneous Velocity
The quantity that tells us how fast an object is moving anywhere along its path is the instantaneous velocity, usually called simply velocity. It is the average velocity between two points on the path in the limit that the time (and therefore the displacement) between the two points approaches zero. To illustrate this idea mathematically, we need to express position x as a continuous function of t denoted by x(t). The expression for the average velocity between two points using this notation is $\stackrel{\text{–}}{v}=\frac{x\left({t}_{2}\right)-x\left({t}_{1}\right)}{{t}_{2}-{t}_{1}}$. To find the instantaneous velocity at any position, we let ${t}_{1}=t$ and ${t}_{2}=t+\text{Δ}t$. After inserting these expressions into the equation for the average velocity and taking the limit as $\text{Δ}t\to 0$, we find the expression for the instantaneous velocity:
$v\left(t\right)=\underset{\text{Δ}t\to 0}{\text{lim}}\frac{x\left(t+\text{Δ}t\right)-x\left(t\right)}{\text{Δ}t}=\frac{dx\left(t\right)}{dt}.$
Instantaneous Velocity
The instantaneous velocity of an object is the limit of the average velocity as the elapsed time approaches zero, or the derivative of x with respect to t:
$v\left(t\right)=\frac{d}{dt}x\left(t\right).$
Like average velocity, instantaneous velocity is a vector with dimension of length per time. The instantaneous velocity at a specific time point ${t}_{0}$ is the rate of change of the position function, which is the slope of the position function $x\left(t\right)$ at ${t}_{0}$. (Figure) shows how the average velocity $\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}$ between two times approaches the instantaneous velocity at ${t}_{0}.$ The instantaneous velocity is shown at time ${t}_{0}$, which happens to be at the maximum of the position function. The slope of the position graph is zero at this point, and thus the instantaneous velocity is zero. At other times, ${t}_{1},{t}_{2}$, and so on, the instantaneous velocity is not zero because the slope of the position graph would be positive or negative. If the position function had a minimum, the slope of the position graph would also be zero, giving an instantaneous velocity of zero there as well. Thus, the zeros of the velocity function give the minimum and maximum of the position function.
In a graph of position versus time, the instantaneous velocity is the slope of the tangent line at a given point. The average velocities $\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{{x}_{\text{f}}-{x}_{\text{i}}}{{t}_{\text{f}}-{t}_{\text{i}}}$ between times $\text{Δ}t={t}_{6}-{t}_{1},\text{Δ}t={t}_{5}-{t}_{2},\text{and}\phantom{\rule{0.2em}{0ex}}\text{Δ}t={t}_{4}-{t}_{3}$ are shown. When $\text{Δ}t\to 0$, the average velocity approaches the instantaneous velocity at $t={t}_{0}$.
Finding Velocity from a Position-Versus-Time Graph
Given the position-versus-time graph of (Figure), find the velocity-versus-time graph.
The object starts out in the positive direction, stops for a short time, and then reverses direction, heading back toward the origin. Notice that the object comes to rest instantaneously, which would require an infinite force. Thus, the graph is an approximation of motion in the real world. (The concept of force is discussed in Newton’s Laws of Motion.)
Strategy
The graph contains three straight lines during three time intervals. We find the velocity during each time interval by taking the slope of the line using the grid.
Solution
Time interval 0 s to 0.5 s: $\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.5\phantom{\rule{0.2em}{0ex}}\text{m}-0.0\phantom{\rule{0.2em}{0ex}}\text{m}}{0.5\phantom{\rule{0.2em}{0ex}}\text{s}-0.0\phantom{\rule{0.2em}{0ex}}\text{s}}=1.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$
Time interval 0.5 s to 1.0 s: $\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\phantom{\rule{0.2em}{0ex}}\text{m}-0.0\phantom{\rule{0.2em}{0ex}}\text{m}}{1.0\phantom{\rule{0.2em}{0ex}}\text{s}-0.5\phantom{\rule{0.2em}{0ex}}\text{s}}=0.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$
Time interval 1.0 s to 2.0 s: $\stackrel{\text{–}}{v}=\frac{\text{Δ}x}{\text{Δ}t}=\frac{0.0\phantom{\rule{0.2em}{0ex}}\text{m}-0.5\phantom{\rule{0.2em}{0ex}}\text{m}}{2.0\phantom{\rule{0.2em}{0ex}}\text{s}-1.0\phantom{\rule{0.2em}{0ex}}\text{s}}=-0.5\phantom{\rule{0.2em}{0ex}}\text{m/s}$
The graph of these values of velocity versus time is shown in (Figure).
The velocity is positive for the first part of the trip, zero when the object is stopped, and negative when the object reverses direction.
Significance
During the time interval between 0 s and 0.5 s, the object’s position is moving away from the origin and the position-versus-time curve has a positive slope. At any point along the curve during this time interval, we can find the instantaneous velocity by taking its slope, which is +1 m/s, as shown in (Figure). In the subsequent time interval, between 0.5 s and 1.0 s, the position doesn’t change and we see the slope is zero. From 1.0 s to 2.0 s, the object is moving back toward the origin and the slope is −0.5 m/s. The object has reversed direction and has a negative velocity.
Speed
In everyday language, most people use the terms speed and velocity interchangeably. In physics, however, they do not have the same meaning and are distinct concepts. One major difference is that speed has no direction; that is, speed is a scalar.
We can calculate the average speed by finding the total distance traveled divided by the elapsed time:
$\text{Average speed}=\stackrel{\text{–}}{s}=\frac{\text{Total distance}}{\text{Elapsed time}}.$
Average speed is not necessarily the same as the magnitude of the average velocity, which is found by dividing the magnitude of the total displacement by the elapsed time. For example, if a trip starts and ends at the same location, the total displacement is zero, and therefore the average velocity is zero. The average speed, however, is not zero, because the total distance traveled is greater than zero. If we take a road trip of 300 km and need to be at our destination at a certain time, then we would be interested in our average speed.
However, we can calculate the instantaneous speed from the magnitude of the instantaneous velocity:
$\text{Instantaneous speed}=|v\left(t\right)|.$
If a particle is moving along the x-axis at +7.0 m/s and another particle is moving along the same axis at −7.0 m/s, they have different velocities, but both have the same speed of 7.0 m/s. Some typical speeds are shown in the following table.
Speeds of Various Objects*Escape velocity is the velocity at which an object must be launched so that it overcomes Earth’s gravity and is not pulled back toward Earth.
Speed m/s mi/h
Continental drift ${10}^{-7}$ $2\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-7}$
Brisk walk 1.7 3.9
Cyclist 4.4 10
Sprint runner 12.2 27
Rural speed limit 24.6 56
Official land speed record 341.1 763
Speed of sound at sea level 343 768
Space shuttle on reentry 7800 17,500
Escape velocity of Earth* 11,200 25,000
Orbital speed of Earth around the Sun 29,783 66,623
Speed of light in a vacuum 299,792,458 670,616,629
Calculating Instantaneous Velocity
When calculating instantaneous velocity, we need to specify the explicit form of the position function x(t). For the moment, let’s use polynomials $x\left(t\right)=A{t}^{n}$, because they are easily differentiated using the power rule of calculus:
$\frac{dx\left(t\right)}{dt}=nA{t}^{n-1}.$
The following example illustrates the use of (Figure).
Instantaneous Velocity Versus Average Velocity
The position of a particle is given by $x\left(t\right)=3.0t+0.5{t}^{3}\phantom{\rule{0.2em}{0ex}}\text{m}$.
1. Using (Figure) and (Figure), find the instantaneous velocity at $t=2.0$ s.
2. Calculate the average velocity between 1.0 s and 3.0 s.
Strategy(Figure) gives the instantaneous velocity of the particle as the derivative of the position function. Looking at the form of the position function given, we see that it is a polynomial in t. Therefore, we can use (Figure), the power rule from calculus, to find the solution. We use (Figure) to calculate the average velocity of the particle.
Solution
1. $v\left(t\right)=\frac{dx\left(t\right)}{dt}=3.0+1.5{t}^{2}\phantom{\rule{0.2em}{0ex}}\text{m/s}$.
Substituting t = 2.0 s into this equation gives $v\left(2.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left[3.0+1.5{\left(2.0\right)}^{2}\right]\phantom{\rule{0.2em}{0ex}}\text{m/s}=9.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$.
2. To determine the average velocity of the particle between 1.0 s and 3.0 s, we calculate the values of x(1.0 s) and x(3.0 s):
$x\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left[\left(3.0\right)\left(1.0\right)+0.5{\left(1.0\right)}^{3}\right]\phantom{\rule{0.2em}{0ex}}\text{m}=3.5\phantom{\rule{0.2em}{0ex}}\text{m}$
$x\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=\left[\left(3.0\right)\left(3.0\right)+0.5{\left(3.0\right)}^{3}\right]\phantom{\rule{0.2em}{0ex}}\text{m}=22.5\phantom{\rule{0.2em}{0ex}}\text{m.}$
Then the average velocity is
$\stackrel{\text{–}}{v}=\frac{x\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)-x\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}{t\left(3.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)-t\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)}=\frac{22.5-3.5\phantom{\rule{0.2em}{0ex}}\text{m}}{3.0-1.0\phantom{\rule{0.2em}{0ex}}\text{s}}=9.5\phantom{\rule{0.2em}{0ex}}\text{m/s}\text{.}$
Significance
In the limit that the time interval used to calculate $\stackrel{\text{−}}{v}$ goes to zero, the value obtained for $\stackrel{\text{−}}{v}$ converges to the value of v.
Instantaneous Velocity Versus Speed
Consider the motion of a particle in which the position is $x\left(t\right)=3.0t-3{t}^{2}\phantom{\rule{0.2em}{0ex}}\text{m}$.
1. What is the instantaneous velocity at t = 0.25 s, t = 0.50 s, and t = 1.0 s?
2. What is the speed of the particle at these times?
Strategy
The instantaneous velocity is the derivative of the position function and the speed is the magnitude of the instantaneous velocity. We use (Figure) and (Figure) to solve for instantaneous velocity.
Solution
1. $v\left(t\right)=\frac{dx\left(t\right)}{dt}=3.0-6.0t\phantom{\rule{0.2em}{0ex}}\text{m/s}$
2. $v\left(0.25\phantom{\rule{0.2em}{0ex}}\text{s}\right)=1.50\phantom{\rule{0.2em}{0ex}}\text{m/s,}\phantom{\rule{0.5em}{0ex}}v\left(0.5\phantom{\rule{0.2em}{0ex}}\text{s}\right)=0\phantom{\rule{0.2em}{0ex}}\text{m/s,}\phantom{\rule{0.5em}{0ex}}v\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)=-3.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$
3. $\text{Speed}=|v\left(t\right)|=1.50\phantom{\rule{0.2em}{0ex}}\text{m/s},0.0\phantom{\rule{0.2em}{0ex}}\text{m/s,}\phantom{\rule{0.2em}{0ex}}\text{and}\phantom{\rule{0.2em}{0ex}}3.0\phantom{\rule{0.2em}{0ex}}\text{m/s}$
Significance
The velocity of the particle gives us direction information, indicating the particle is moving to the left (west) or right (east). The speed gives the magnitude of the velocity. By graphing the position, velocity, and speed as functions of time, we can understand these concepts visually (Figure). In (a), the graph shows the particle moving in the positive direction until t = 0.5 s, when it reverses direction. The reversal of direction can also be seen in (b) at 0.5 s where the velocity is zero and then turns negative. At 1.0 s it is back at the origin where it started. The particle’s velocity at 1.0 s in (b) is negative, because it is traveling in the negative direction. But in (c), however, its speed is positive and remains positive throughout the travel time. We can also interpret velocity as the slope of the position-versus-time graph. The slope of x(t) is decreasing toward zero, becoming zero at 0.5 s and increasingly negative thereafter. This analysis of comparing the graphs of position, velocity, and speed helps catch errors in calculations. The graphs must be consistent with each other and help interpret the calculations.
(a) Position: x(t) versus time. (b) Velocity: v(t) versus time. The slope of the position graph is the velocity. A rough comparison of the slopes of the tangent lines in (a) at 0.25 s, 0.5 s, and 1.0 s with the values for velocity at the corresponding times indicates they are the same values. (c) Speed: $|v\left(t\right)|$ versus time. Speed is always a positive number.
Check Your Understanding The position of an object as a function of time is $x\left(t\right)=-3{t}^{2}\phantom{\rule{0.2em}{0ex}}\text{m}$. (a) What is the velocity of the object as a function of time? (b) Is the velocity ever positive? (c) What are the velocity and speed at t = 1.0 s?
(a) Taking the derivative of x(t) gives v(t) = −6t m/s. (b) No, because time can never be negative. (c) The velocity is v(1.0 s) = −6 m/s and the speed is $|v\left(1.0\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=6\phantom{\rule{0.2em}{0ex}}\text{m/s}$.
Summary
• Instantaneous velocity is a continuous function of time and gives the velocity at any point in time during a particle’s motion. We can calculate the instantaneous velocity at a specific time by taking the derivative of the position function, which gives us the functional form of instantaneous velocity v(t).
• Instantaneous velocity is a vector and can be negative.
• Instantaneous speed is found by taking the absolute value of instantaneous velocity, and it is always positive.
• Average speed is total distance traveled divided by elapsed time.
• The slope of a position-versus-time graph at a specific time gives instantaneous velocity at that time.
Conceptual Questions
There is a distinction between average speed and the magnitude of average velocity. Give an example that illustrates the difference between these two quantities.
Average speed is the total distance traveled divided by the elapsed time. If you go for a walk, leaving and returning to your home, your average speed is a positive number. Since Average velocity = Displacement/Elapsed time, your average velocity is zero.
Does the speedometer of a car measure speed or velocity?
If you divide the total distance traveled on a car trip (as determined by the odometer) by the elapsed time of the trip, are you calculating average speed or magnitude of average velocity? Under what circumstances are these two quantities the same?
Average speed. They are the same if the car doesn’t reverse direction.
How are instantaneous velocity and instantaneous speed related to one another? How do they differ?
Problems
A woodchuck runs 20 m to the right in 5 s, then turns and runs 10 m to the left in 3 s. (a) What is the average velocity of the woodchuck? (b) What is its average speed?
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Sketch the velocity-versus-time graph from the following position-versus-time graph.
Given the following velocity-versus-time graph, sketch the position-versus-time graph.
An object has a position function x(t) = 5t m. (a) What is the velocity as a function of time? (b) Graph the position function and the velocity function.
A particle moves along the x-axis according to $x\left(t\right)=10t-2{t}^{2}\phantom{\rule{0.2em}{0ex}}\text{m}$. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b) What is the instantaneous speed at these times? (c) What is the average velocity between t = 2 s and t = 3 s?
a. $v\left(t\right)=\left(10-4t\right)\text{m/s}$; v(2 s) = 2 m/s, v(3 s) = −2 m/s; b. $|v\left(2\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=2\phantom{\rule{0.2em}{0ex}}\text{m/s},|v\left(3\phantom{\rule{0.2em}{0ex}}\text{s}\right)|=2\phantom{\rule{0.2em}{0ex}}\text{m/s}$; (c) $\stackrel{\text{–}}{v}=0\phantom{\rule{0.2em}{0ex}}\text{m/s}$
Unreasonable results. A particle moves along the x-axis according to $x\left(t\right)=3{t}^{3}+5t\text{}$. At what time is the velocity of the particle equal to zero? Is this reasonable?
Glossary
instantaneous velocity
the velocity at a specific instant or time point
instantaneous speed
the absolute value of the instantaneous velocity
average speed
the total distance traveled divided by elapsed time
License
University Physics Volume 1 Copyright © 2016 by cnxuniphysics is licensed under a Creative Commons Attribution 4.0 International License, except where otherwise noted.
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Saturday, November 14, 2009
M5M1 b - Developing Area Formulas (4)
M5M1. Students will extend their understanding of area of geometric plane figures.
b. Derive the formula for the area of a parallelogram.
As discussed in the previous post, through activities like finding the area of L-shaped region, students can develop the understanding that "when we are given an unfamiliar shape, we may still be able to calculate its area by somehow making a familiar shape (or a collection of familiar shapes)." Moreover, students can develop the following strategies to make a familiar shapes:
* divide the given shape up into several familiar shapes
* cut and re-arrange to make a familiar shape
* make-it-bigger
Now they are ready to tackle this standard.
In many textbooks, students are asked to find the area of parallelograms like the one shown below using what they already know:
Some students will count the number of unit squares, making appropriate adjustments when only a part of a unit square is inside the parallelogram. Other students will try to change the parallelogram to a rectangle, a familiar shape they already know how to calculate the area of. The typical way that this is accomplished by cutting a triangular segment from one end of the parallelogram and moving it to the other side, as shown below:
Since this rectangle is 6 cm wide and 4 cm long, area can be calculated by 6 x 4, or 24 cm2.
In most textbooks, this method is then generalized to derive the formula for calculating the area of parallelograms: Area of Parallelograms = Base x Height. So, is this the end? Have we successfully addressed this particular standard? I argue that the formula at this stage is an overgeneralization. Students at this point may have difficulty calculating the area of parallelograms like the following:
Some students will try to create a new rectangle like before and notice that "the height (in red) stops here!"
Others might try to turn the figure and make a rectangle like this:
Unfortunately, they can't determine the length and the width of this new rectangle other than actually measuring them, which isn't possible if the figure isn't drawn to scale. Even if the figure is drawn to scale, actually measuring the length and the width will introduce measurement errors. So, what can students do? Actually, there are a lot of things they can do using the understanding they developed through the L-shape lesson. Here are some possibilities:
Note that (a), (c) and (d) use the "cut and re-arrange" strategy, (b) uses the "divide up" strategy, and (e) uses the "make-it-bigger" strategy. In (b), (c) and (d), the "familiar" shape students created are parallelograms that can be changed to rectangles by cutting and re-arranging right triangles.
Some of you may be wondering about (e) since students have not learned how to calculate the area of triangles. In this case, instead of calculating the area of each triangle, this student actually pushed together the two triangles that were used to make a bigger rectangle. The two triangles will make a rectangle whose dimensions are 5 cm by 6 cm.
Actually, some students may use this make-it-bigger strategy with the first parallelograms. If they did, then, this "slanted" parallelograms do not pose any challenge to them since they can use exactly the same strategy to this one as well. This strategy could have been used to derive the formula for calculating the area of parallelograms, too. Look at the figure below:
The area of the original parallelogram (un-shaded part in the figure on the left) can be calculated by subtracting the area of shaded rectangle (in the middle figure) from the large rectangle. However, this difference is really the area of the yellow rectangle in the figure on the right. That means that the area of the parallelogram is the same as the area of rectangle you can build on the base whose length is the distance between the base and its opposite side, or more accurately, the distance between the parallel lines containing the base and its opposite side. If we consider this distance between the base and its opposite side as height, we still have the same formula, Area of Parallelogram = Base x Height.
The important idea here, though, is what constitute as the height. The height of a parallelogram is the distance between the base and its opposite side, and the distance between two parallel lines is the length of a perpendicular segment connecting them. It is not the length of the adjacent side to the base. In case of a rectangle, which is a special type of parallelograms, the adjacent side may be used as the height because it is perpendicular to the base. However, that is not generally the case in parallelograms. Thus, understanding what the height of a parallelogram is may be the most important aspect of deriving the formula. Unfortunately, students don't understand this idea because they aren't asked to grapple with parallelograms like the second one we saw above, or derive the formula through the make-it-bigger strategy. I hope you will seriously consider giving your students this challenge as they try to derive the formula for calculating the area of parallelograms.
Saturday, November 7, 2009
M5M1 - Developing Area Formulas (3)
M5M1. Students will extend their understanding of area of geometric plane figures.
As we discussed in the previous post, the GPS expects students to determine the area of rectangles and squares by counting or calculation. Then, in Grade 5, students are expected to derive and use formulas to determine the area of parallelograms, triangles, and circles. Interestingly, there is nothing about area mentioned in Grade 4. It is listed as one of the "Concepts/Skills to Maintain," but there is no specific standard about the area measurement in Grade 4. Many people might wonder about the feasibility of fifth graders actually deriving the area formulas of parallelograms and triangles on their own. Do they have enough background knowledge? What background knowledge do they need to increase the likelihood of their deriving those formulas?
In a previous post on the idea of teaching through problem solving (April, 2009), how children can learn through problem solving new mathematical ideas. Those mathematical ideas are the ones that will serve as the bridge between M3M4 (area of rectangles and squares) and M5M1 (area of parallelograms, triangles, and circles). As we will see shortly, those specific understandings will be used over and over to derive the formulas. So, in Grade 3, finding the area of L-shapes may be simply a complex application of what they learned, but, in Grade 5, the focus should be on ways of thinking involved in calculating the area. If those understandings are made explicit, students are much more likely to be successful in deriving the area formulas. So, I encourage you to read that post again (or for the first time, if you have not read it before).
By the way, element (f) of this standard says, "Find the area of a polygon (regular and irregular) by dividing it into squares, rectangles, and/or triangles and find the sum of the areas of those shapes." Actually, this element is simply one of the strategies developed in the L-shape lesson, that is, sub-dividing the given unfamiliar shape into a collection of familiar shapes. The only difference is what shapes are available to students as familiar shapes. When students work on the L-shape problem, they only knew how to calculate the area of rectangles and squares. However, after students have learned the formulas for the area of parallelograms and triangles, students can also use those figures. So, in case you are wondering if you can afford to spend an extra time to discuss something that is not explicitly mentioned in the GPS, the L-shape lesson does address the GPS directly, too.
Tuesday, November 3, 2009
M3M4 - Developing Area Formulas (2)
M3M4. Students will understand and measure the area of simple geometric figures (squares and rectangles).
a. Understand the meaning of the square unit and measurement in area.
b. Model (by tiling) the area of a simple geometric figure using square units (square inch, square foot, etc.).
c. Determine the area of squares and rectangles by counting, addition, and multiplication with models.
Once students understand that area is the amount of space inside any geometric figures, we are ready to start thinking about ways to measure the area of various shapes. The next step is to pick a unit and actually "cover" shapes to see how many units will be needed. So, what should we use as a unit? Although we will eventually use squares as units, we may want to think about using anything that can cover the plane without a hole or an overlap. Also, using a familiar objects might be helpful to focus students' attention on the process of area measurement. One such familiar object might be index cards. Students can measure the area of the surface of desks or any other large rectangular regions.
If students have many index cards available to them, they will cover the rectangular region in many different ways. Here are three possibilities.
In this particular example, no matter how you cover the rectangle, it takes 24 small rectangles. So, we can say that the area of the rectangle is 24 units.
After measuring the area by actually covering rectangles with units, many students will realize that some ways of covering the given shape is easier to count than others. For example, the arrangements like the one on the left requires us to actually count all of the units to determine how many units were used. On the other hand, since the other two arrangements will result in equal groups (either rows or columns), we can use multiplication to find the area (either 4x6 or 8x3).
At this point, you might want to give students only 3 or 4 unit pieces to see if they can think about ways of calculating the area. A common error at this stage is to do the following:
and .
So, the area is 4x3=12 units. It is important for students to understand here why they cannot rotate the unit as they measure how many units will fit in each dimension of the rectangle. What we are trying to do when we measure the second dimension is how many rows (in this example) of 4 units there are. If we turn the unit as shown on the right, we are no longer counting the number of rows of 4 units.
You may want to ask students what we can do to avoid this type of confusion. Some students will realize that if we use a square as a unit, then it doesn't matter whether we rotate it since squares have 4 equal sides. You can then introduce that the standard units of area measurement are squares with unit length on each side, e.g., 1 cm, 1 in, 1 ft, etc.. Each unit square is said to have the area of 1 cm2, 1 in2, 1 ft2, etc., respectively. Actually, I am not sure exactly how the GPS wants these standards units of area to be handled. Unlike the units for volume, these area units are not mentioned in the GPS. However, it seems strange not to talk about the units when we are talking about the area of rectangles.
By using unit squares, we can also make it easier to determine the number of units that fit along each dimension of a rectangle by simply measuring their lengths. So, if a rectangle is 5 inches wide and 8 inches long, that means we can fit 5 1-inch squares along one row and there will be 8 rows. Therefore, we can multiply 5 and 8 to get 40 cm2. It is important that students understand that when 2 lengths are multiplied together, the product mysteriously becomes the area measurement. The two lengths we are measuring are simply telling us how many unit squares will fit along each side of the given rectangle.
Also note that students are not introduced to letters as variables until Grade 5, the formula should be written (if it is to be written at all) as, Area of Rectangle = Length x Width. Again, it is important to emphasize that this formula is to calculate the area of rectangles. Some students (and adults, unfortunately) will say that area is "length x width," but it is only a formula for a specific shape. Area is the amount of space inside a shape, no matter what the shape is.
Creative Commons
Elaboration of Georgia Performance Standards by Tad Watanabe is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License.
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# if
Question:
If $A=\left[\begin{array}{rr}3 & 1 \\ -1 & 2\end{array}\right]$, show that $A^{2}-5 A+7 l=0$ use this to find $A^{4}$.
Solution:
Given : $A=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$
Now,
$A^{2}=A A$
$\Rightarrow A^{2}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{cc}9-1 & 3+2 \\ -3-2 & -1+4\end{array}\right]$
$\Rightarrow A^{2}=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]$
$A^{2}-5 A+7 I$
$\Rightarrow A^{2}-5 A+7 I=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-5\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]+7\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow A^{2}-5 A+7 I=\left[\begin{array}{cc}8 & 5 \\ -5 & 3\end{array}\right]-\left[\begin{array}{cc}15 & 5 \\ -5 & 10\end{array}\right]+\left[\begin{array}{ll}7 & 0 \\ 0 & 7\end{array}\right]$
$\Rightarrow A^{2}-5 A+7 I=\left[\begin{array}{cc}8-15+7 & 5-5+0 \\ -5+5+0 & 3-10+7\end{array}\right]$
$\Rightarrow A^{2}-5 A+7 I=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$
Hence proved.
Given : $A^{2}-5 A+7 I=0$
$\Rightarrow A^{2}=5 A-7 I$
$\Rightarrow A^{3}=A(5 A-7 I) \quad$ (Multilpying by $A$ on both sides)
$\Rightarrow A^{3}=5 A^{2}-7 A I$
$\Rightarrow A^{3}=5(5 A-7 I)-7 A \quad$ [From eq. (1)]
$\Rightarrow A^{3}=25 A-35 I-7 A$
$\Rightarrow A^{3}=18 A-35 I$
$\Rightarrow A^{4}=(18 A-35 I) A$ (Multilpying by $A$ on both sides)
$\Rightarrow A^{4}=18 A^{2}-35 A$
$\Rightarrow A^{4}=18(5 A-7 I)-35 A$ [From eq. (1)]
$\Rightarrow A^{4}=90 A-126 I-35 A$
$\Rightarrow A^{4}=55 A-126 I$
$\Rightarrow A^{4}=55\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]-126\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
$\Rightarrow A^{4}=\left[\begin{array}{cc}165 & 55 \\ -55 & 110\end{array}\right]-\left[\begin{array}{cc}126 & 0 \\ 0 & 126\end{array}\right]$
$\Rightarrow A^{4}=\left[\begin{array}{cc}165-126 & 55-0 \\ -55-0 & 110-126\end{array}\right]$
$\Rightarrow A^{4}=\left[\begin{array}{cc}39 & 55 \\ -55 & -16\end{array}\right]$
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# Rational And Irrational Numbers Worksheet With Answer Key
A Rational Figures Worksheet may help your son or daughter be a little more familiar with the methods associated with this ratio of integers. With this worksheet, college students are able to resolve 12 distinct troubles associated with logical expression. They will figure out how to grow a couple of numbers, class them in sets, and find out their goods. They will also training simplifying rational expressions. When they have enhanced these principles, this worksheet might be a important instrument for furthering their scientific studies. Rational And Irrational Numbers Worksheet With Answer Key.
## Reasonable Numbers can be a rate of integers
There are two kinds of numbers: irrational and rational. Realistic amounts are understood to be complete numbers, while irrational figures usually do not replicate, and also have an endless variety of digits. Irrational numbers are non-absolutely nothing, no-terminating decimals, and sq . roots which are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To define a rational variety, you need to realize exactly what a logical amount is. An integer can be a complete quantity, along with a reasonable number is really a proportion of two integers. The proportion of two integers is the amount at the top separated from the amount on the bottom. For example, if two integers are two and five, this would be an integer. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be created in a small percentage
A reasonable number carries a numerator and denominator that are not zero. Consequently they may be indicated like a fraction. Along with their integer numerators and denominators, rational phone numbers can also have a unfavorable benefit. The unfavorable benefit must be positioned to the left of and its complete importance is its extended distance from absolutely no. To easily simplify this instance, we shall claim that .0333333 can be a fraction which can be published as a 1/3.
In addition to adverse integers, a reasonable variety can also be made into a small percentage. For instance, /18,572 can be a rational amount, whilst -1/ is not really. Any portion consisting of integers is realistic, as long as the denominator fails to have a and may be composed as an integer. Likewise, a decimal that ends in a point is yet another realistic number.
## They can make sensation
Regardless of their name, reasonable numbers don’t make very much sensation. In mathematics, these are solitary entities using a special length about the number collection. This means that once we count up some thing, we can order the shape by its ratio to the initial volume. This keeps accurate even though you will find infinite reasonable amounts between two certain phone numbers. In other words, numbers should make sense only if they are ordered. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To get the length of a pearl, as an example, we might add up its width. A single pearl weighs 10 kilos, and that is a rational number. Moreover, a pound’s excess weight equals twenty kgs. As a result, we should be able to break down a pound by 10, without concern yourself with the size of one particular pearl.
## They could be expressed being a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal variety could be created as a a number of of two integers, so 4 times 5 is equivalent to 8-10. The same problem requires the recurring small fraction 2/1, and either side must be separated by 99 to obtain the appropriate solution. But how would you make your conversion? Here are some good examples.
A reasonable quantity can also be printed in great shape, including fractions plus a decimal. One method to symbolize a reasonable amount within a decimal is usually to divide it into its fractional equal. There are 3 ways to break down a logical number, and each of these approaches yields its decimal counterpart. One of those approaches is always to separate it into its fractional counterpart, and that’s what’s known as a terminating decimal.
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Similarity, Ratios, and Proportions
SIMILARITY, RATIOS, and PROPORTIONS
by Dr. Carol JVF Burns (website creator)
Follow along with the highlighted text while you listen!
• PRACTICE (online exercises and printable worksheets)
Suppose you have a triangle that you'd like to enlarge.
That is, you want to keep the shape exactly the same, but you want it to be bigger.
Perhaps you want an enlarged copy where the length of each side is three times the size of the original,
as shown below:
Or, perhaps you have a quadrilateral that you'd like to reduce.
That is, you want to keep the shape exactly the same, but you want it to be smaller.
Perhaps you want a reduced copy where the length of each side is half the size of the original, as shown below:
In both cases, you are keeping the angles exactly the same,
and you are multiplying all the sides by an appropriate scaling factor.
When you made it three times as big, the scaling factor was $\,3\,$.
When you made it half as big, the scaling factor was $\,0.5\,$.
This idea of keeping the shape the same, but changing the size is made precise by the concept of similarity.
Very roughly, two geometric figures are said to be similar when they have the same shape, but not necessarily the same size.
To make the concept of similarity precise, we first need to review ratios and proportions.
DEFINITION ratio
Let $\,a\,$ and $\,b\,$ be real numbers, with $\,b\neq 0\,$.
The ‘ratio of $\,a\,$ to $\,b\,$’ is the quotient $\displaystyle\,\frac{a}{b}\,$.
A ratio automatically brings a scaling factor into the picture, as follows.
Start by giving the ‘ratio of $\,a\,$ to $\,b\,$’ a simpler name, $\,s\,$.
(We're calling it $\,s\,$ because it will be our scaling factor.)
$\displaystyle s := \frac{a}{b}$
(The notation $\,:=\,$ is frequently used in mathematics to mean equals, by definition.)
Solving for $\,a\,$ gives $\,a=sb\,$.
So, $\,b\,$ has been scaled by $\,s\,$ to give $\,a\,$.
If $\,s=3\,$, then $\,a=3b\,$, so that $\,a\,$ is three times bigger than $\,b\,$.
If $\,s=0.5\,$, then $\,a=0.5b\,$, so that $\,a\,$ is half as big as $\,b\,$.
When we scale geometric figures, equal ratios automatically enter the picture.
Consider the picture below, where a quadrilateral has been scaled by a factor of $\,s\,$ (here, $\,s\,$ is greater than one) to get a new quadrilateral:
Notice that $\,A=sa\,$ and $\,B=sb\,$ and $\,C=sc\,$ and $\,D=sd\,$.
Solving for $\,s\,$ in each of these four equations gives an equality of four ratios: $$\cssId{s38}{s = \frac{A}{a}} \cssId{s39}{= \frac{B}{b}} \cssId{s40}{= \frac{C}{c}} \cssId{s41}{= \frac{D}{d}}$$ Thus, equality of ratios arises very naturally in any scaling situation.
An equality of ratios is called a proportion:
DEFINITION proportion
A proportion is an equality of ratios.
Therefore, a proportion takes the form: $$\cssId{s48}{\frac{A}{a} = \frac{B}{b}}$$
Using just a little bit of algebra, many useful equalities arise from proportions:
THEOREM equivalent proportions
Let $\,a\,$, $\,b\,$, $\,A\,$, and $\,B\,$ be nonzero real numbers.
Then, the following are equivalent:
• $\displaystyle\frac{A}{a} = \frac{B}{b}$
• $\displaystyle\frac{a}{A} = \frac{b}{B}$
• $\displaystyle\frac{a}{b} = \frac{A}{B}$
• $\displaystyle\frac{b}{a} = \frac{B}{A}$
Think about these proportions in the context of a scaled quadrilateral: Some of the proportions, like $$\cssId{s60}{\frac{A}{a} = \frac{B}{b}}\ \cssId{s61}{\text{ and }}\ \cssId{s62}{\frac{a}{A} = \frac{b}{B}}$$ compare ratios between the two quadrilaterals. Some of the proportions, like $$\cssId{s65}{\frac{a}{b} = \frac{A}{B}}\ \cssId{s66}{\text{ and }}\ \cssId{s67}{\frac{b}{a} = \frac{B}{A}}$$ compare ratios within the two quadrilaterals. In both cases, attaching words to the proportions may be useful. For example, you can read $\ \displaystyle\frac{A}{a} = \frac{B}{b}\$ as ‘$\,A\,$ is to $\,a\,$ as $\,B\,$ is to $\,b\,$’.
Cross-multiplying is a useful technique whenever you find yourself working with proportions:
DEFINITION cross-multiplying
The process of going from the proportion $\displaystyle\,\frac{A}{a} = \frac{B}{b}\,$ to the equivalent statement $\,Ab = aB\,$ is called cross-multiplying.
Cross-multiplying is nothing other than an application of the Multiplication Property of Equality,
where both sides of the equation are multiplied by the product of the original denominators: $$\begin{gather} \cssId{s83}{\frac{A}{a} = \frac{B}{b}}\cr\cr \cssId{s84}{\frac{A}{a}\cdot (ab) = \frac{B}{b}\cdot (ab)}\cr\cr \cssId{s85}{Ab = Ba} \end{gather}$$
DEFINITION the phrase: ‘corresponding sides are proportional’
Suppose a triangle with sides $\,a\,$, $\,b\,$, and $\,c\,$ is scaled to get a new triangle
with corresponding sides $\,A\,$, $\,B\,$, and $\,C\,$.
In this situation, the phrase corresponding sides are proportional is used to describe the equality of ratios: $$\cssId{s90}{\frac{A}{a} = \frac{B}{b} = \frac{C}{c}}$$ This same terminology is used when scaling any polygon.
Finally, we are ready for the precise definition of similar polygons:
DEFINITION similar polygons
Two polygons are similar if and only if there exists a correspondence between their vertices such that corresponding sides are proportional and corresponding angles are equal.
In general, to prove that two polygons are similar, you must check two different things:
• that corresponding angles are equal
• that corresponding sides are proportional
It is an extremely useful result that for triangles (and triangles alone!) it suffices to check only that angles are equal;
Of course, if two angles are equal, then the third angles must be equal, and so we have the (unproved) theorem:
AA SIMILARITY THEOREM
Two triangles are similar if and only if
two angles of one triangle are equal to two angles of the other triangle.
This is NOT true for polygons other than triangles.
For example, look at the pictures below.
All the angles are the same, but the polygons are not similar.
Master the ideas from this section
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Lesson Video: Operations on Events: Difference | Nagwa Lesson Video: Operations on Events: Difference | Nagwa
Lesson Video: Operations on Events: Difference Mathematics • Third Year of Preparatory School
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In this video, we will learn how to find the probability of the difference of two events.
17:39
Video Transcript
In this video, we will learn how to find the probability of the difference of two events. This is written as the probability of π΄ minus π΅, where π΄ and π΅ are two events such that π΄ minus π΅ is all outcomes that are in event π΄ but not in event π΅. We will begin this video by introducing some key notation and formulae.
In the Venn diagram shown representing the events π΄ and π΅, the probability of π΄ is represented by the shaded section. The intersection of events π΄ and π΅ is represented on a Venn diagram by the overlap between the circles. These are the outcomes that occur in event π΄ and event π΅. Combining these two definitions leads us to the difference formula, which states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅. This can be represented on a Venn diagram as shown.
We could use the same method to prove that the probability of π΅ minus π΄ is equal to the probability of π΅ minus the probability of π΄ intersection π΅. On a Venn diagram, this can be shown by shading the region that is in circle π΅ but not circle π΄. We will now look at a couple of examples where we need to use the difference formula.
Suppose π΄ and π΅ are two events. Given that the probability of π΄ is 0.3 and the probability of π΄ intersection π΅ is 0.03, determine the probability of π΄ minus π΅.
Letβs begin by recalling the notation shown in this question. Firstly, we have the intersection of events π΄ and π΅. This is all the outcomes that occur in event π΄ and in event π΅. This can be represented on a Venn diagram as shown. The probability of π΄ minus π΅ is known as the difference and could be represented on a Venn diagram as shown. It is all of the outcomes in event π΄ that are not in event π΅.
We recall that the difference formula states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅. Using the values given in the question, the probability of π΄ minus π΅ is equal to 0.3 minus 0.03. This is equal to 0.27.
In our next example, we will look at a problem in context.
A ball is drawn at random from a bag containing 12 balls each with a unique number from one to 12. Suppose π΄ is the event of drawing an odd number and π΅ is the event of drawing a prime number. Find the probability of π΄ minus π΅.
We are asked in this question to find the probability of π΄ minus π΅. And we can do this using the difference formula. This states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅. We are told in the question that there are 12 balls, each with a unique number from one to 12 as shown. We are told that π΄ is the event of drawing an odd number. There are six of these in the bag, the numbers one, three, five, seven, nine, and 11. This means that the probability of event π΄ selecting an odd-numbered ball is six out of 12 or six twelfths. Dividing the numerator and denominator by six, we see that this simplifies to one-half.
We are also told that π΅ is the event of drawing a prime number. We know that a prime number has exactly two factors, the number one and itself. The prime numbers between one and 12 are two, three, five, seven, and 11. As there are five of these, the probability of event π΅ is five out of 12 or five twelfths. At this stage, we have the probability of π΄, but we donβt have the probability of π΄ intersection π΅. The intersection of two events is all outcomes that occur in both of the events. In this case, the numbers three, five, seven, and 11 are both odd numbers and prime numbers. The probability of π΄ intersection π΅ is therefore equal to four twelfths, which in turn simplifies to one-third.
We can now calculate the probability of π΄ minus π΅ by subtracting one-third from one-half. Using equivalent fractions, this is the same as three-sixths minus two-sixths. The probability of π΄ minus π΅ is therefore equal to one-sixth. In context, the probability of drawing an odd number that is not a prime number is one-sixth.
We could also represent this by listing all the outcomes on a Venn diagram. We have already mentioned that the numbers three, five, seven, and 11 are both odd and prime. There are two other odd numbers between one and 12, the numbers one and nine. And there is one other prime number, the number two. The numbers four, six, eight, 10, and 12 are neither odd nor prime. As a result, we write these outside of the two circles representing event π΄ and event π΅.
Two of the numbers are in the section represented by the probability of π΄ minus π΅. These are the numbers one and nine, as they are odd but not prime. This confirms that the probability of drawing one of these balls is two out of 12, which simplifies to one-sixth.
Before looking at our next example, we will recall one of our other probability formulae. The probability of the complement of event π΄, denoted π of π΄ prime or π of π΄ bar, is the probability of event π΄ not occurring. This satisfies the formula the probability of π΄ prime is equal to one minus the probability of π΄. We will now look at an example where we need to use this.
Suppose π΄ and π΅ are two events. Given that π΄ intersection π΅ is the empty set, the probability of π΄ prime is 0.66, and the probability of π΅ prime is 0.79, find the probability of π΅ minus π΄.
Before trying to answer this question, letβs recall some of the notation. We know that π΄ prime and π΅ prime are the complement of events π΄ and π΅, respectively. And we also know that the probability of the complement of event π΄ is equal to one minus the probability of event π΄. Using the information given, we can therefore calculate the probability of event π΄ along with the probability of event π΅.
Firstly, we have 0.66 is equal to one minus the probability of π΄. Rearranging this equation, we have the probability of π΄ is equal to one minus 0.66. This is equal to 0.34. In the same way, 0.79 is equal to one minus the probability of event π΅. The probability of π΅ is therefore equal to one minus 0.79, which is equal to 0.21.
We are also told that π΄ intersection π΅ is equal to the empty set. This means that there are no elements in event π΄ and event π΅. And we can therefore say that the two events are mutually exclusive. And the probability of π΄ intersection π΅ is therefore equal to zero. When representing this on a Venn diagram, there is no overlap as shown. We can fill in the fact that the probability of event π΄ is 0.34 and the probability of event π΅ is 0.21. We can complete the Venn diagram by filling in the probability that neither event π΄ nor event π΅ occur. This is equal to 0.45.
We are asked to find the probability of π΅ minus π΄. And using the difference formula, we know this is equal to the probability of π΅ minus the probability of π΄ intersection π΅. Substituting in the values we know, this is equal to 0.21 minus zero, which is just equal to 0.21. This leads us onto an important rule. If two events π΄ and π΅ are mutually exclusive, then the probability of π΅ minus π΄ is simply equal to the probability of π΅. Likewise, the probability of π΄ minus π΅ is equal to the probability of π΄.
Before looking at one final example, letβs recall the addition rule of probability. The addition rule of probability states that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ intersection π΅. This can be represented using Venn diagrams as shown. We will now look at one final example.
Suppose π΄ and π΅ are events in a sample space which consists of equally likely outcomes. Given that π΄ contains six outcomes, the probability of π΄ union π΅ is three-quarters, the probability of π΅ is one-half, and the total number of outcomes is 20, find the probability that only one of the events π΄ or π΅ occurs.
Letβs begin by looking at how we can represent the probability that only one of the events π΄ and π΅ occurs on a Venn diagram. The probability that only event π΄ occurs is shaded in pink. We know that this can be written using the difference formula. It is the probability of π΄ minus π΅. And this is equal to the probability of π΄ minus the probability of π΄ intersection π΅. The probability that only event π΅ occurs is shaded in blue. And this is denoted by the probability of π΅ minus π΄. This is equal to the probability of π΅ minus the probability of π΄ intersection π΅. In order to answer this question, we will need to find the sum of these two values.
Letβs now consider the information we are given in this question. We are told that there are 20 outcomes in total and that event π΄ contains six of these. The probability of event π΄ is therefore equal to six out of 20. Dividing the numerator and denominator by two, this simplifies to three-tenths. We are also told that the probability of π΄ union π΅ is three-quarters and the probability of π΅ is one-half. We now have both the probabilities of event π΄ and π΅ but not the probability of π΄ intersection π΅.
We can calculate this by using the addition rule of probability, which states that the probability of π΄ union π΅ is equal to the probability of π΄ plus the probability of π΅ minus the probability of π΄ intersection π΅. This can be rearranged as shown. Substituting in our values, we have the probability of π΄ intersection π΅ is equal to three-tenths plus a half minus three-quarters. Our three fractions have a common denominator of 20. So we can rewrite this as six over 20 plus 10 over 20 minus 15 over 20. This is equal to one over 20.
The probability of π΄ intersection π΅ is one twentieth. We can now use this value together with the probabilities of π΄ and π΅ to calculate the probability of π΄ minus π΅ and the probability of π΅ minus π΄. The probability of π΄ minus π΅ is equal to three-tenths minus one twentieth. This is equal to five twentieths. The probability of π΅ minus π΄ is equal to one-half minus one twentieth. And this is equal to nine twentieths. The probability that only one of the events π΄ or π΅ occurs is therefore equal to five twentieths plus nine twentieths. Adding our numerators gives us fourteen twentieths, which in turn simplifies to seven-tenths or 0.7. The probability that only one of the events π΄ or π΅ occurs is seven-tenths.
An alternative way to calculate this would be to list the number of outcomes on our Venn diagram. Since the probability of π΄ intersection π΅ is one twentieth and there are a total of 20 outcomes, there is one outcome in the intersection of π΄ and π΅. We were told that π΄ contains six outcomes. Therefore, five of these will occur in just event π΄. Since the probability of event π΅ is one-half, there are 10 outcomes in event π΅. And nine of these must occur in just event π΅. As nine plus one plus five equals 15, there must be five outcomes that are not in event π΄ nor event π΅. This confirms that 14 of the outcomes are in only one of event π΄ or π΅. And 14 out of 20 simplifies to seven-tenths.
We will now summarize the key points from this video. In this video, we used the difference rule of probability together with other probability formulae to solve a variety of problems. The difference rule of probability states that the probability of π΄ minus π΅ is equal to the probability of π΄ minus the probability of π΄ intersection π΅. For any two events π΄ and π΅, it is also true that the probability of π΅ minus π΄ is equal to the probability of π΅ minus the probability of π΄ intersection π΅.
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# What is 24/30 as a percentage?
The given example here is of converting 24/30 as a percentage. Converting a fraction to a percentage is a very crucial process that you need to learn and here is a complete guide for you to convert the given fraction to a percentage. The steps are explained for you so you can learn the two different methods in which a fraction can be converted to a percentage.
What is 24/30 as a percentage? The answer is 24/30 as a percentage = 80%
## Basic concepts
1) A fraction is made of two different parts. They are the numerator and the denominator.
2) A numerator of a fraction is defined as the part which lies above the line of fraction while a denominator on the other hand is present below the line of division.
3) In the given fraction 24/30, the numerator is 24 and the denominator is 30.
4) When we are initiating the process of division, the numerator becomes the dividend and the denominator is converted into the divisor.
5) What is a percentage – a percentage is described as expressing any digit or value in terms of 100. We can also define it as expressing the given value in a fraction of 100. So if we say 20% it means 20/100 and if we say 78% it means 78/100.
## Calculations to show 24/30 as a percentage
Two different methods are involved in the conversion of the fraction to a percentage.
### Method #1
Step 1
You need to convert the denominator of the given fraction so that it turns to 100. To do so, you have to divide 100 by the denominator 30.
#### 100÷30=3.333334
$\frac{100}{30}=3.333334$
Step 2
Now, you need to balance both the denominator and the numerator by the obtained result.
Hence,
#### (24×3.3334)/(30×3.3334) = 80/100
$\frac{(24\times3.3334)}{(30\times3.3334)}=\frac{80}{100}$
Thus the required result is 80%.
#### 24/30 as a percentage = 80%
Also read: What is 34/40 as a percentage?
### Method #2
Step1
Here in the second method, you have to divide the numerator by the denominator. The numerator 24 becomes the dividend and 30 is the divisor.
So,
#### 24÷30= 0.8
Step 2
Now you need to multiply 100 with the obtained decimal value and you shall get the required value in percentage.
#### 0.8×100=80%
Thus there are two different methods of conversion of the given fraction to a percentage. In both cases, the answer was 80% and thus both the methods are valid in the conversion of the fraction to a percentage. You must practice more and then you will be able to solve the sums without any trouble.
#### 24/30 as a percentage = 80%
Also read: What is 45/50 as a percentage?
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# Weighted mean
The weighted mean, also called weighted average, is used a lot by teachers.You will learn how to calculate the weighted average using a real life example.
The power of weighted average is noticeable with a final exam. Since the final exam usually carries more weight, this encourages students to work hard to improve their grade on the final exam.
Even if students got some bad grades in the beginning, they may still pass the class with a good grade if they ace the final exam.
This happened to me a lot and teachers love to see growth on a final exam and reward students as a result.
## A real-life example showing how to find the weighted mean
Here is how a teacher may decide to grade his class. 10% for homework assignments, 15% for quizzes, 25% for a midterm, and 50% for the final exam
Notice that 10% + 15% + 25% + 50% = 100%
Suppose the maximum score is 100 and a student gets the following scores throughtout the year:
Homework assignments: 40, 50, 60, 50, 40, 30
Quizzes: 50, 60, 20, 50, 40, 80, 30, 70
Midterm: 65
Final exam: 95
What is the weighted mean?
First, find the average score for homework assignments
40 + 50 + 60 + 50 + 40 + 30 / 6
270 / 6
= 45
Then, find the average score for the quizzes
50 + 60 + 20 + 50 + 40 + 80 + 30 + 70 / 8
400 / 8
= 50
Now you can calculate the weighted mean
weighted average = 10% × 45 + 15% × 50 + 25% × 65 + 50% × 95
weighted average = 0.10 × 45 + 0.15 × 50 + 0.25 × 65 + 0.50 × 95
weighted average = 4.5 + 7.5 + 16.25 + 47.5
weighted average = 75.75
As you can see, despite having lots of bad grades on homework assignments and quizzes and failing the midterm, the student still passed.
In some cases, the weight may be given not as a percent, but a number may be assigned to each section of the examination.
Homework assignments: 1
Quizzes: 1
Midterm: 2
Final exam: 3
For the same example above, weighted average =
1 × 45 + 1 × 50 + 2 × 65 + 3 × 95 / 7
weighted average =
45 + 50 + 130 + 285 / 7
weighted average =
510 / 7
= 72.85
## Weighted mean formulas
When the weight is expressed as a percent, the formula can be expressed as shown below:
weighted average = percent1 × x1 + percent2 × x2 + ... + percentn × xn
When the weight is expressed as a number, the formula can be expressed as shown below:
Weighted average =
weight1 × x1 + weight2 × x2 + ... + weightn × xn / weight1 + weight2 + ... + weightn
Let weighted average = X and let weight = w
X =
w1 × x1 + w2 × x2 + ... + wn × xn / w1 + w2 + ... + wn
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
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#### Need Help?
Get in touch with us
Sep 17, 2022
## Key Concepts
• Define the vertical motion model.
• Assess the fit of a function by analyzing residuals.
• Explain to fit a quadratic function to data.
## Vertex form of the quadratic function
• The function f(x) = a(x−h)2+k , a≠0 is called the vertex form of a quadratic function
• The vertex of the graph g is (h, k).
• The graph of f(x) = a(x−h)2+k is a translation of the function f(x) = ax2 that is translated h units horizontally and k units vertically.
## The standard form of the quadratic function
• The standard form of a quadratic function is ax2+bx+c = 0 , a≠0
• The axis of symmetry of a standard form of quadratic function f(x) = ax2+bx+c is the line x = −b/2a.
• The y-intercept of f(x) is c.
• The x-coordinate of the graph of f(x) = ax2+bx+c is –b/2a.
• The vertex of f(x) = ax2+bx+c is (–b/2a, f(–b/2a)).
### Vertical motion model
When George throws the ball, it moves in a parabolic path.
So, we can relate such real-life situations with quadratic functions.
If the ball was hit with an initial velocity v0, and h0 be the initial height where the ball was hit. The height h (in feet) of the ball after some time (t seconds) can be calculated by the quadratic function:
h = -16t2+v0t+h0.
This is called the vertical motion model.
### Assess the fit of the function by analyzing residuals
A shopkeeper increases the cost of each item according to a function −8×2+95x+745−8×2+95x+745. Find how well the function fits the actual revenue.
Step 1: Find the predicted value for each price increase using the function.
For x=0, −8(0)2+95(0)+745 = 745
For x=1, −8(1)2+95(1)+745 = 832
For x=2, −8(2)2+95(2)+745 = 903
For x=3, −8(3)2+95(3)+745 = 958
For x=4, −8(4)2+95(4)+745 = 997
Subtract the predicted value from the actual revenues to find the residues.
Residual = observed – predicted
Step 2: Make a scatterplot of the data and graph the function on the same coordinate grid.
Step 3: Make a residual plot to show the fit of the function of the data.
Step 4: Assess the fit of the function using the residual plot.
The residual plot shows both positive and negative residuals, which indicates a generally good model.
### Fit a quadratic function to data
We know that to find the equation of the straight line that best fits a set of data, we use linear regression.
Quadratic regression is a method used to find the equation of the parabola (quadratic function) that best fits data.
Step 1: Using the graphing calculator, enter the values of x and y
Step 2: Use the quadratic regression feature.
R-squared is the coefficient of the determination.
The closes R2 is to 1, the better the equation matches the given data points.
Step 3: Graph the data and quadratic regression.
## Exercise
1. A rectangular wall has a length seven times the breadth. It also has a 4-ft wide brick border around it. Write a quadratic function to determine the area of the wall.
2. The data are modelled by f(x) = -2x2+16.3x+40.7. What does the graph of the residuals tell you about the fit of the model?
3. Write a function h to model the vertical motion, given h(t) = -16t2+v0t+h0. Find the maximum height if the initial vertical velocity is 32 ft/s, and the initial height is 75 ft.
4. Using a graphic calculator to find a quadratic regression for the data set.
### What have we learned
• We can relate to real life situations using quadratic functions
### Concept Map
We can relate to real−life situations using quadratic functions.
• To find the height of an object, we can use the vertical motion model.
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#### Special Right Triangles: Types, Formulas, with Solved Examples.
Learn all about special right triangles- their types, formulas, and examples explained in detail for a better understanding. What are the shortcut ratios for the side lengths of special right triangles 30 60 90 and 45 45 90? How are these ratios related to the Pythagorean theorem? Right Angle Triangles A triangle with a ninety-degree […]
#### Ways to Simplify Algebraic Expressions
Simplify algebraic expressions in Mathematics is a collection of various numeric expressions that multiple philosophers and historians have brought down. Talking of algebra, this branch of mathematics deals with the oldest concepts of mathematical sciences, geometry, and number theory. It is one of the earliest branches in the history of mathematics. The study of mathematical […]
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UW AMATH 352 - Lecture 3: Matrices and Linear Equations
Unformatted text preview:
Lecture 3: Matrices and Linear EquationsAMath 352Fri., Apr. 21 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear EquationsSolve for x and y:x + 2y = 33x − y = 2.Use eq. 1 to eliminate x in eq. 2. Subtract 3 times eq. 1 from eq. 2:3x − y − 3(x + 2y) = 2 − 3 · 3 ⇒ −7y = −7 ⇒ y = 1.Now substitute y = 1 into eq. 1 and solve for x:x + 2 · 1 = 3 ⇒ x = 1.Now check:1 + 2 · 1?= 3 yes3 · 1 − 1?= 2 yes2 / 17Linear Equations, Cont.This technique can be extended to any number of linear equationsin the same number of unknowns. Solve for x, y, and z:x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.Use eq. 1 to eliminate x in eqs. 2 and 3. Subtract 2 times the first eq.from the second; add the first eq. to the third:2x + 3y + 4z − 2(x + 2y + 3z) = 5 − 2 · 4 ⇒ −y − 2z = −3.−x − 6y + z + (x + 2y + 3z) = 2 + 4 ⇒ −4y + 4z = 6.3 / 17Linear Equations, Cont.This technique can be extended to any number of linear equationsin the same number of unknowns. Solve for x, y, and z:x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.Use eq. 1 to eliminate x in eqs. 2 and 3. Subtract 2 times the first eq.from the second; add the first eq. to the third:2x + 3y + 4z − 2(x + 2y + 3z) = 5 − 2 · 4 ⇒ −y − 2z = −3.−x − 6y + z + (x + 2y + 3z) = 2 + 4 ⇒ −4y + 4z = 6.3 / 17Linear Equations, Cont.x + 2y + 3z = 4−y − 2z = −3−4y + 4z = 6.Subtract 4 times the second eq. from the third to find:−4y + 4z − 4(−y − 2z) = 6 − 4 · (−3) ⇒ 12z = 18 ⇒ z =32.Substitute this into the second eq. to find:−y − 2 ·32= −3 ⇒ y = 0.Substitute for y and z in the first eq. to find:x + 2 · 0 + 3 ·32= 4 ⇒ x = −12.4 / 17Linear Equations, Cont.x + 2y + 3z = 4−y − 2z = −3−4y + 4z = 6.Subtract 4 times the second eq. from the third to find:−4y + 4z − 4(−y − 2z) = 6 − 4 · (−3) ⇒ 12z = 18 ⇒ z =32.Substitute this into the second eq. to find:−y − 2 ·32= −3 ⇒ y = 0.Substitute for y and z in the first eq. to find:x + 2 · 0 + 3 ·32= 4 ⇒ x = −12.4 / 17Linear Equations, Cont.x + 2y + 3z = 4−y − 2z = −3−4y + 4z = 6.Subtract 4 times the second eq. from the third to find:−4y + 4z − 4(−y − 2z) = 6 − 4 · (−3) ⇒ 12z = 18 ⇒ z =32.Substitute this into the second eq. to find:−y − 2 ·32= −3 ⇒ y = 0.Substitute for y and z in the first eq. to find:x + 2 · 0 + 3 ·32= 4 ⇒ x = −12.4 / 17Linear Equations, Cont.x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.Check: x = −12, y = 0, z =32.−12+92?= 4 yes2 · (−12) + 4 ·32?= 5 yes12+32?= 2 yes.5 / 17Matrix NotationWe don’t have to write x, y, and z so many times!x + 2y + 3z = 42x + 3y + 4z = 5−x − 6y + z = 2.can be written as1 2 32 3 4−1 −6 1xyz=452.The 3 by 3 array of numbers on the left is a matrix, and theproduct of this matrix with the vector [x, y, z]Tof unknowns isjust the left-hand side of the above equations:1 2 32 3 4−1 −6 1xyz=x + 2y + 3z2x + 3y + 4z−x − 6y + z.6 / 17Solving Equations Using Matrix NotationAppend the right-hand side vector to the matrix:1 2 3 | 42 3 4 | 5−1 −6 1 | 2.Eliminate x from the second and third eqs. by subtracting 2 timesthe first eq. from the second and adding the first eq. to the third:1 2 3 | 40 −1 −2 | −30 −4 4 | 6.Eliminate y from the third eq. by subtracting 4 times the secondeq.:1 2 3 | 40 −1 −2 | −30 0 12 | 18.7 / 17Solving Equations Using Matrix NotationAppend the right-hand side vector to the matrix:1 2 3 | 42 3 4 | 5−1 −6 1 | 2.Eliminate x from the second and third eqs. by subtracting 2 timesthe first eq. from the second and adding the first eq. to the third:1 2 3 | 40 −1 −2 | −30 −4 4 | 6.Eliminate y from the third eq. by subtracting 4 times the secondeq.:1 2 3 | 40 −1 −2 | −30 0 12 | 18.7 / 17Solving Equations Using Matrix NotationAppend the right-hand side vector to the matrix:1 2 3 | 42 3 4 | 5−1 −6 1 | 2.Eliminate x from the second and third eqs. by subtracting 2 timesthe first eq. from the second and adding the first eq. to the third:1 2 3 | 40 −1 −2 | −30 −4 4 | 6.Eliminate y from the third eq. by subtracting 4 times the secondeq.:1 2 3 | 40 −1 −2 | …
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# UW AMATH 352 - Lecture 3: Matrices and Linear Equations
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VISUALISING SOLID SHAPES EXERCISE 15.1 1. Identify the nets which can be used to make cubes (cut out copies of the nets and try it): (i) (ii) (iii) (iv) (v) (vi) 2. Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them. Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box. Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7. 3. Can this be a net for a die? Explain your answer. MATHEMATICS 4. Here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.) Play this game You and your friend sit back-to-back. One of you reads out a net to make a 3-D shape, while the other attempts to copy it and sketch or build the described 3-D object. 15.4 DRAWING SOLIDS ON A FLAT SURFACE Your drawing surface is paper, which is flat. When you draw a solid shape, the images are somewhat distorted to make them appear three-dimensional. It is a visual illusion.You will find here two techniques to help you. 15.4.1 Oblique Sketches Here is a picture of a cube (Fig 15.11). It gives a clear idea of how the cube looks like,
# Visualising SolidShapes
### 15.1 Introduction: Plane Figures and Solid Shapes
In this chapter, you will classify figures you have seen in terms of what is known as dimension.
In our day to day life, we see several objects like books, balls, ice-cream cones etc., around us which have different shapes. One thing common about most of these objects is that they all have some length, breadth and height or depth.
That is, they all occupy space and have three dimensions.
Hence, they are called three dimensional shapes.
Do you remember some of the three dimensional shapes (i.e., solid shapes) we have seen in earlier classes?
Match the shape with the name:
Try to identify some objects shaped like each of these.
By a similar argument, we can say figures drawn on paper which have only length and breadth are called two dimensional (i.e., plane) figures. We have also seen some two dimensional figures in the earlier classes.
Match the 2 dimensional figures with the names (Fig 15.2):
Note: We can write 2-D in short for 2-dimension and 3-D in short for
3-dimension.
### 15.2 Faces, Edges and Vertices
Do you remember the Faces, Vertices and Edges of solid shapes, which you studied earlier? Here you see them for a cube:
The 8 corners of the cube are its vertices. The 12 line segments that form the skeleton of the cube are its edges. The 6 flat square surfaces that are the skin of the cube are its faces.
### Do This
Complete the following table:
Can you see that, the two dimensional figures can be identified as the faces of thethree dimensional shapes? For example a cylinder has two faces which are circles, and a pyramid, shaped like this has triangles as its faces.
We will now try to see how some of these 3-D shapes can be visualised on a 2-D surface, that is, on paper.
In order to do this, we would like to get familiar with three dimensional objects closely. Let us try forming these objects by making what are called nets.
### 15.3 Nets for Building 3-D Shapes
Take a cardboard box. Cut the edges to lay the box flat. You have now a net for that box. A net is a sort of skeleton-outline in 2-D [Fig154 (i)], which, when folded [Fig154 (ii)], results in a 3-D shape [Fig154 (iii)].
Here you got a net by suitably separating the edges. Is the reverse process possible?
Here is a net pattern for a box (Fig 15.5). Copy an enlarged version of the net and try to make the box by suitably folding and gluing together. (You may use suitable units). The box is a solid. It is a 3-D object with the shape of a cuboid
Similarly, you can get a net for a cone by cutting a slit along its slant surface (Fig 15.6).
You have different nets for different shapes. Copy enlarged versions of the nets given (Fig 15.7)and try to make the 3-D shapes indicated.(You may
also like to prepare skeleton models using strips of cardboard fastened with paper clips).
We could also try to make a net for making a pyramid like the Great Pyramid in Giza (Egypt) (Fig 15.8). That pyramid has a square base and triangles on the four sides.
See if you can make it with the given net (Fig 15.9).
### Try These
Here you find four nets (Fig 15.10). There are twocorrectnets among them to make a tetrahedron. See if you can work out which nets will make a tetrahedron.
Fig 15.10
#### Exercise 15.1
1. Identify the nets which can be used to make cubes (cut out copies of the nets and try it):
2. Dice are cubes with dots on each face. Opposite faces of a die always have a total of seven dots on them.
Here are two nets to make dice (cubes); the numbers inserted in each square indicate the number of dots in that box.
Insert suitable numbers in the blanks, remembering that the number on the opposite faces should total to 7.
3. Can this be a net for a die?
4. here is an incomplete net for making a cube. Complete it in at least two different ways. Remember that a cube has six faces. How many are there in the net here? (Give two separate diagrams. If you like, you may use a squared sheet for easy manipulation.)
5. Match the nets with appropriate solids:
Play this game
You and your friend sit back-to-back. One of you reads out a net to make a 3-D shape, while the other attempts to copy it and sketch or build the described 3-D object.
### 15.4 Drawing Solids on a Flat Surface
Your drawing surface is paper, which is flat. When you draw a solid shape, the images are somewhat distorted to make them appear three-dimensional. It is a visual illusion. You will find here two techniques to help you.
Fig 15.11
#### 15.4.1 Oblique Sketches
Here is a picture of a cube (Fig 15.11). It gives a clear idea of how the cube looks like, when seen from the front. You do not see certain faces. In the drawn picture, the lengths are not equal, as they should be in a cube. Still, you are able to recognise it as a cube. Such a sketch of a solid is called an oblique sketch.
How can you draw such sketches? Let us attempt to learn the technique.
You need a squared (lines or dots) paper. Initially practising to draw on these sheets will later make it easy to sketch them on a plain sheet (without the aid of squared lines or dots!) Let us attempt to draw an oblique sketch of a 3 × 3 × 3 (each edge is 3 units) cube (Fig 15.12).
In the oblique sketch above, did you note the following?
(i) The sizes of the front faces and its opposite are same; and
(ii) The edges, which are all equal in a cube, appear so in the sketch, though the actual measures of edges are not taken so.
You could now try to make an oblique sketch of a cuboid (remember the faces in this case are rectangles)
Note: You can draw sketches in which measurements also agree with those of a given solid. To do this we need what is known as an isometric sheet. Let us try to make a cuboid with dimensions 4 cm length, 3 cm breadth and 3 cm height on given isometric sheet.
#### 15.4.2 Isometric Sketches
Have you seen an isometric dot sheet? (A sample is given at the end of the book). Such a sheet divides the paper into small equilateral triangles made up of dots or lines. To draw sketches in which measurements also agree with those of the solid, we can use isometric dot sheets. [Given on inside of the back cover (3rd cover page).]
Let us attempt to draw an isometric sketch of a cuboid of dimensions 4 × 3 × 3 (which means the edges forming length, breadth and height are 4, 3, 3 units respectively) (Fig 15.13).
Note that the measurements are of exact size in an isometric sketch; this is not so in the case of an oblique sketch.
Example1Hereis an oblique sketch of a cuboid [Fig 15.14(i)]. Draw an isometric sketch that matches this drawing.
Solution Here is the solution [Fig 15.14(ii)]. Note how the measurements are taken care of.
Fig. 15.14(i)
Fig. 15.14(ii)
How many units have you taken along (i) ‘length’? (ii) ‘breadth’? (iii) ‘height’? Do they match with the units mentioned in the oblique sketch?
### Exercise 15.2
1. Use isometric dot paper and make an isometric sketch for each one of the given shapes:
2. The dimensions of a cuboid are 5 cm, 3 cm and 2 cm. Draw three different isometric sketches of this cuboid.
3. Three cubes each with 2 cm edge are placed side by side to form a cuboid. Sketch an oblique or isometric sketch of this cuboid.
4. Make an oblique sketch for each one of the given isometric shapes:
5. Give (i) an oblique sketch and (ii) an isometric sketch for each of the following:
(a) A cuboid of dimensions 5 cm, 3 cm and 2 cm. (Is your sketch unique?)
(b) A cube with an edge 4 cm long.
An isometric sheet is attached at the end of the book. You could try to make on it some cubes or cuboids of dimensions specified by your friend.
### Do This
Sometimes when you look at combined shapes, some of them may be hidden from your view.
Here are some activities you could try in your free time to help you visualise some solid objects and how they look. Take some cubes and arrange them as shown in
Fig 15.16.
Now ask your friend to guess how many cubes there are when observed from the view shown by the arrow mark.
### Try These
Try to guess the number of cubes in the following arrangements (Fig 15.17).
Such visualisation is very helpful. Suppose you form a cuboid by joining such cubes. You will be able to guess what the length, breadth and height of the cuboid would be.
Example 2 If two cubes of dimensions 2 cm by 2cm by 2cm are placed side by side, what would the dimensions of the resulting cuboid be?
Fig 15.18
Solution As you can see (Fig 15.18) when kept side by side, the length is the only measurement which increases, it becomes 2 + 2 = 4 cm.
The breadth = 2 cm and the height = 2 cm.
### Try These
1.Two dice are placed side by side as shown: Can you say what the total would be on the face opposite to
Fig 15.19
(a) 5 + 6 (b) 4 + 3
(Remember that in a die sum of numbers on opposite faces is 7)
2.Three cubes each with 2 cm edge are placed side by side to form a cuboid. Try to make an oblique sketch and say what could be its length, breadth and height.
### 15.5 Viewing Different Sections of a Solid
Now let us see how an object which is in 3-D can be viewed in different ways.
15.5.1 One Way to view an Object is by Cutting or Slicing
Slicing game
Here is a loaf of bread (Fig 15.20). It is like a cuboid with a square face. You ‘slice’ it with a knife.
When you give a ‘vertical’ cut, you get several pieces, as shown in the Figure 15.20. Each face of the piece is a square! We call this face a cross-section of the whole bread. The cross section is nearly a square in this case.
Beware! If your cut is not ‘vertical’ you may get a different cross section! Think about it. The boundary of the cross-section you obtain is a plane curve. Do you notice it?
Fig 15.20
A kitchen play
Have you noticed cross-sections of some vegetables when they are cut for the purposes of cooking in the kitchen? Observe the various slices and get aware of the shapes that result as cross-sections.
Play this
Make clay (or plasticine) models of the following solids and make vertical or horizontal cuts. Draw rough sketches of the cross-sections you obtain. Name them wherever you can.
### Exercise 15.3
1. What cross-sections do you get when you give a
(i) vertical cut (ii) horizontal cutto the following solids?
(a) A brick (b) A round apple (c) A die
(d) A circular pipe (e) An ice cream cone
#### 15.5.2 Another Way is by Shadow Play
Shadows are a good way to illustrate how three-dimensional objects can be viewed in two dimensions. Have you seen a shadow play? It is a form of entertainment using solid articulated figures in front of an illuminated back-drop to create the illusion of moving images. It makes some indirect use of ideas in Mathematics.
Fig. 15.22
You will need a source of light and a few solid shapes for this activity. (If you have an overhead projector, place the solid under the lamp and do these investigations.)
Fig. 15.23
Keep a torchlight, right in front of a Cone. What type of shadow does it cast on the screen? (Fig 15.23)
The solid is three-dimensional; what is the dimension of the shadow?
If, instead of a cone, you place a cube in the above game, what type of shadow will you get?
Experiment with different positions of the source of light and with different positions of the solid object. Study their effects on the shapes and sizes of the shadows you get.
Here is another funny experiment that you might have tried already: Place a circular plate in the open when the Sun at the noon time is just right above it as shown in Fig 15.24 (i). What is the shadow that you obtain?
Will it be same during
Study the shadows in relation to the position of the Sun and the time of observation.
#### Exercise 15.4
1. A bulb is kept burning just right above the following solids. Name the shape of the shadows obtained in each case. Attempt to give a rough sketch of the shadow. (You may try to experiment first and then answer these questions).
2. Here are the shadows of some 3-D objects, when seen under the lamp of an overhead projector. Identify the solid(s) that match each shadow. (There may be multiple answers for these!)
3. Examine if the following are true statements:
(i) The cube can cast a shadow in the shape of a rectangle.
(ii) The cube can cast a shadow in the shape of a hexagon.
#### 15.5.3 A Third Way is by Looking at it from Certain Angles to Get Different Views
One can look at an object standing in front of it or by the side of it or from above. Each time one will get a different view (Fig 15.25).
Here is an example of how one gets different views of a given building. (Fig 15.26)
You could do this for figures made by joining cubes.
Try putting cubes together and then making such sketches from different sides.
### Try These
1. For each solid, the three views (1), (2), (3) are given. Identify for each solid the corresponding top, front and side views.
2.Draw a view of each solid as seen from the direction indicated by the arrow.
### What have We Discussed?
1. The circle, the square, the rectangle, the quadrilateral and the triangle are examples of plane figures; the cube, the cuboid, the sphere, the cylinder, the cone and the pyramid are examples of solid shapes.
2. Plane figures are of two-dimensions (2-D) and the solid shapes are of three-dimensions (3-D).
3. The corners of a solid shape are called its vertices; the line segments of its skeleton are its edges; and its flat surfaces are its faces.
4. A net is a skeleton-outline of a solid that can be folded to make it. The same solid can have several types of nets.
5. Solid shapes can be drawn on a flat surface (like paper) realistically. We call this 2-D representation of a 3-D solid.
6. Two types of sketches of a solid are possible:
(a) An oblique sketch does not have proportional lengths. Still it conveys all important aspects of the appearance of the solid.
(b) An isometric sketch is drawn on an isometric dot paper, a sample of which is given at the end of this book. In an isometric sketch of the solid the measurements kept proportional.
7. Visualising solid shapes is a very useful skill. You should be able to see ‘hidden’ parts of the solid shape.
8. Different sections of a solid can be viewed in many ways:
(a) One way is to view by cutting or slicing the shape, which would result in the cross-section of the solid.
(b) Another way is by observing a 2-D shadow of a 3-D shape.
(c) A third way is to look at the shape from different angles; the front-view, the side-view and the top-view can provide a lot of information about the shape observed.
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## I need to factor theses polynomials 9x^2-16 anda^3-64
I need to factor these two polynimials :9x^2-16 anda^3-64
The polynomials, 9x2 - 16 and a3 - 64 are the differences of two perfect squares and two perfect cubes, respectively.
Let's begin by recognizing the factoring pattern for the difference of two perfect squares:
a2 - b2 = (a + b)(a - b)
Let's apply this factoring pattern to our polynomial, 9x2 - 64, substuting 9x2 for a2 and 16 for b2 which gives us 3x for a and 4 for b.
Plugging in these values for a and b, our factored expression is (3x + 4)(3x - 4).
Now let's consider the factoring pattern for the difference of two perfect cubes:
a3 - b3 = (a - b)(a2 + ab + b2)
Let's use this factoring pattern to factor our polynomial, a3 - 64, substituting a for a and 4 for b.
Plugging in thesen values for a and b, our factored expression is (a - 4)(a2 + 4a + 16).
Most generally the difference of two n-th powers factors as,
an - bn = (a - b)(an-1 + an-2b+ an-3b2... + a2bn-3 + abn-2 + bn-1)
If n is composite, then the second factor can be factored further, which is a straight forward exercise using the distributive property.
9x2 - 16 = (3x)2 - 42= (3x - 4)(3x + 4)
a3 - 64 = a3 - 43 = (a - 4)(a2 + 4a + 42) = (a - 4)(a2 + 4a + 16)
Hello Tanesha,
First we'll factorize 9x2 - 16. If you look at the problem it is based on factoring a difference of two squares
[(a2 - b2) = (a + b) (a - b)]. Note that the sum of two squares DOES NOT factor.
When you have the difference of two bases being squared, it factors as the product of the sum and the difference of the bases that are being squared.
9x2 - 16
= (3x)2 - (4)2(form of a difference of two squares)
= (3x + 4)(3x - 4) (factor as the product of the sum and the differnce of the bases)(answer)
To check your answer multiply (3x + 4)(3x -4) using FOIL method and you'll get back 9x2 - 16.
Now let's do a3 - 64. As you can see that it is factoring a difference of two cubes
(a3 - b3) = (a - b)(a2 + ab + b2).
When you have the difference of two cubes, you get a product of a binomial and a trinomial. The binomial is the differnce of that bases that are cubed. Trinomial is the first base square, the second term is the opposite of the product of the two bases, and the third term is the second base squared.
a3 - 64
= (a)3 - (4)3 (form of a difference of two cubes)
= (a - 4) [(a)2 + (a)(4) + (4)2] (binomial is difference of bases)
(trinomial is first base squared, plus product of bases, plus second base squared)
= (a - 4)(a2 + 4a + 16) (answer)
To check your answer multiply binomial and trinomial and you'll get back a3 - 64.
Good luck factorizing!
General Information:
For x± # x ± # you have a ( x ± # )( x ± # ) The two numbers at the end have to add together to get to # x and have to multiply together to get to the ending #
When you have a number in front of the x2 that number must be split up so that when you multiply ( x ± # )( x ± # ) your "x"s will get you back to that number. So if you have a problem that starts with 6x2 you can have either ( 1x ± # )( 6x ± # ) OR ( 2x ± # )( 3x ± # ) because 1*6=6 and 2*3=6.
9x- 16
The first thing that we have to worry about it the 9. You can have either (9x ± #)(1x ± #) OR (3x ± #)(3x ± #). What I would usually do from here is figure out what multiplies together to get sixteen: 1*16, 2*8, 4*4, 8*2, and 16*1 (order matters!). Then (becasue we have to take into account the numbers in front of x), I would plug them into each of the equations until I found one where the singular "x"s canceled out. That would give us our 9x2 +0x -16
Because I've been at this a while, I know that 16 is a perfect square (4*4=16), So, I would choose the second of the two potential answer bases (3x ± #)(3x ± #) to use. Because 4 is the same either way, and I know that they need to cancel out in the middle, one would be positive and the other would be negative to look like this: (3x + 4)(3x - 4) then, like any good problem, you should multiply it back out and check that yes (3x + 4)(3x - 4) = 9x2-16
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# Section11.5The Arc Length Parameter and Curvature¶ permalink
In normal conversation we describe position in terms of both time and distance. For instance, imagine driving to visit a friend. If she calls and asks where you are, you might answer “I am 20 minutes from your house,” or you might say “I am 10 miles from your house.” Both answers provide your friend with a general idea of where you are.
Currently, our vector–valued functions have defined points with a parameter $t\text{,}$ which we often take to represent time. Consider Figure 11.5.1(a), where $\vrt = \la t^2-t,t^2+t\ra$ is graphed and the points corresponding to $t=0,\ 1$ and $2$ are shown. Note how the arc length between $t=0$ and $t=1$ is smaller than the arc length between $t=1$ and $t=2\text{;}$ if the parameter $t$ is time and $\vec r$ is position, we can say that the particle traveled faster on $[1,2]$ than on $[0,1]\text{.}$
Now consider Figure 11.5.1(b), where the same graph is parametrized by a different variable $s\text{.}$ Points corresponding to $s=0$ through $s=6$ are plotted. The arc length of the graph between each adjacent pair of points is 1. We can view this parameter $s$ as distance; that is, the arc length of the graph from $s=0$ to $s=3$ is 3, the arc length from $s=2$ to $s=6$ is 4, etc. If one wants to find the point 2.5 units from an initial location (i.e., $s=0$), one would compute $\vec r(2.5)\text{.}$ This parameter $s$ is very useful, and is called the arc length parameter.
How do we find the arc length parameter?
Start with any parametrization of $\vec r\text{.}$ We can compute the arc length of the graph of $\vec r$ on the interval $[0,t]$ with \begin{equation*} \text{ arc length } = \int_0^t\norm{\vec r\,'(u)}\ du. \end{equation*}
We can turn this into a function: as $t$ varies, we find the arc length $s$ from $0$ to $t\text{.}$ This function is $$s(t) = \int_0^t \norm{\vec r\,'(u)}\ du. \label{eq_vvfarc}\tag{11.5.1}$$
This establishes a relationship between $s$ and $t\text{.}$ Knowing this relationship explicitly, we can rewrite $\vec r(t)$ as a function of $s\text{:}$ $\vec r(s)\text{.}$ We demonstrate this in an example.
##### Example11.5.4Finding the arc length parameter
Let $\vec r(t) = \la 3t-1,4t+2\ra\text{.}$ Parametrize $\vec r$ with the arc length parameter $s\text{.}$
Solution
Things worked out very nicely in Example 11.5.4; we were able to establish directly that $s=5t\text{.}$ Usually, the arc length parameter is much more difficult to describe in terms of $t\text{,}$ a result of integrating a square–root. There are a number of things that we can learn about the arc length parameter from Equation (11.5.1), though, that are incredibly useful.
First, take the derivative of $s$ with respect to $t\text{.}$ The Fundamental Theorem of Calculus (see Theorem 5.4.6) states that $$\frac{ds}{dt}=s\,'(t) = \norm{\vrp(t)}. \label{eq_vvfarc3}\tag{11.5.2}$$
Letting $t$ represent time and $\vec r(t)$ represent position, we see that the rate of change of $s$ with respect to $t$ is speed; that is, the rate of change of “distance traveled” is speed, which should match our intuition.
The Chain Rule states that \begin{align*} \frac{d\vec r}{dt} \amp = \frac{d\vec r}{ds}\cdot\frac{ds}{dt}\\ \vrp(t) \amp = \vrp(s)\cdot \norm{\vrp(t)}. \end{align*}
Solving for $\vrp(s)\text{,}$ we have $$\vrp(s) = \frac{\vrp(t)}{\norm{\vrp(t)}} = \vec T(t), \label{eq_vvfarc2}\tag{11.5.3}$$ where $\vec T(t)$ is the unit tangent vector. Equation (11.5.3) is often misinterpreted, as one is tempted to think it states $\vrp(t) = \vec T(t)\text{,}$ but there is a big difference between $\vrp(s)$ and $\vrp(t)\text{.}$ The key to take from it is that $\vrp(s)$ is a unit vector. In fact, the following theorem states that this characterizes the arc length parameter.
# Subsection11.5.1Curvature
Consider points $A$ and $B$ on the curve graphed in Figure 11.5.7(a). One can readily argue that the curve curves more sharply at $A$ than at $B\text{.}$ It is useful to use a number to describe how sharply the curve bends; that number is the curvature of the curve.
We derive this number in the following way. Consider Figure 11.5.7(b), where unit tangent vectors are graphed around points $A$ and $B\text{.}$ Notice how the direction of the unit tangent vector changes quite a bit near $A\text{,}$ whereas it does not change as much around $B\text{.}$ This leads to an important concept: measuring the rate of change of the unit tangent vector with respect to arc length gives us a measurement of curvature.
##### Definition11.5.10Curvature
Let $\vec r(s)$ be a vector–valued function where $s$ is the arc length parameter. The curvature $\kappa$ of the graph of $\vec r(s)$ is \begin{equation*} \kappa = \snorm{\frac{d\vec T}{ds}} = \snorm{\vec T\,'(s)}. \end{equation*}
If $\vec r(s)$ is parametrized by the arc length parameter, then \begin{equation*} \vec T(s) = \frac{\vrp(s)}{\norm{\vrp(s)}} \text{ and } \vec N(s) = \frac{\vec T\,'(s)}{\norm{\vec T\,'(s)}}. \end{equation*}
Having defined $\norm{\vec T\,'(s)} =\kappa\text{,}$ we can rewrite the second equation as $$\vec T\,'(s) = \kappa\vec N(s). \label{eq_curvature}\tag{11.5.4}$$
We already knew that $\vec T\,'(s)$ is in the same direction as $\vec N(s)\text{;}$ that is, we can think of $\vec T(s)$ as being “pulled” in the direction of $\vec N(s)\text{.}$ How “hard” is it being pulled? By a factor of $\kappa\text{.}$ When the curvature is large, $\vec T(s)$ is being “pulled hard” and the direction of $\vec T(s)$ changes rapidly. When $\kappa$ is small, $T(s)$ is not being pulled hard and hence its direction is not changing rapidly.
We use Definition 11.5.10 to find the curvature of the line in Example 11.5.4.
##### Example11.5.11Finding the curvature of a line
Use Definition 11.5.10 to find the curvature of $\vrt = \la 3t-1,4t+2\ra\text{.}$
Solution
While the definition of curvature is a beautiful mathematical concept, it is nearly impossible to use most of the time; writing $\vec r$ in terms of the arc length parameter is generally very hard. Fortunately, there are other methods of calculating this value that are much easier. There is a tradeoff: the definition is “easy” to understand though hard to compute, whereas these other formulas are easy to compute though it may be hard to understand why they work.
We practice using these formulas.
##### Example11.5.13Finding the curvature of a circle
Find the curvature of a circle with radius $r\text{,}$ defined by $\vec c(t) = \la r\cos(t) ,r\sin(t) \ra\text{.}$
Solution
Example 11.5.13 gives a great result. Before this example, if we were told “The curve has a curvature of 5 at point $A\text{,}$” we would have no idea what this really meant. Is 5 “big” — does is correspond to a really sharp turn, or a not-so-sharp turn? Now we can think of 5 in terms of a circle with radius 1/5. Knowing the units (inches vs. miles, for instance) allows us to determine how sharply the curve is curving.
Let a point $P$ on a smooth curve $C$ be given, and let $\kappa$ be the curvature of the curve at $P\text{.}$ A circle that:
• passes through $P\text{,}$
• lies on the concave side of $C\text{,}$
• has a common tangent line as $C$ at $P$ and
• has radius $r=1/\kappa$ (hence has curvature $\kappa$)
is the osculating circle, or circle of curvature, to $C$ at $P\text{,}$ and $r$ is the radius of curvature. Figure 11.5.14 shows the graph of the curve seen earlier in Figure 11.5.7 and its osculating circles at $A$ and $B\text{.}$ A sharp turn corresponds to a circle with a small radius; a gradual turn corresponds to a circle with a large radius. Being able to think of curvature in terms of the radius of a circle is very useful.
(The word “osculating” comes from a Latin word related to kissing; an osculating circle “kisses” the graph at a particular point. Many beautiful ideas in mathematics have come from studying the osculating circles to a curve.)
##### Example11.5.15Finding curvature
Find the curvature of the parabola defined by $y=x^2$ at the vertex and at $x=1\text{.}$
Solution
##### Example11.5.17Finding curvature
Find where the curvature of $\vec r(t) = \la t, t^2, 2t^3\ra$ is maximized.
Solution
# Subsection11.5.2Curvature and Motion
Let $\vec r(t)$ be a position function of an object, with velocity $\vvt = \vrp(t)$ and acceleration $\vec a(t)=\vrp'(t)\text{.}$ In Section 11.4 we established that acceleration is in the plane formed by $\vec T(t)$ and $\vec N(t)\text{,}$ and that we can find scalars $a_\text{T}$ and $a_\text{N}$ such that \begin{equation*} \vat = a_\text{T} \vec T(t) + a_\text{N} \vec N(t). \end{equation*}
Theorem 11.4.13 gives formulas for $a_\text{T}$ and $a_\text{N}\text{:}$ \begin{equation*} a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) \text{ and } a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}}. \end{equation*}
We understood that the amount of acceleration in the direction of $\vec T$ relates only to how the speed of the object is changing, and that the amount of acceleration in the direction of $\vec N$ relates to how the direction of travel of the object is changing. (That is, if the object travels at constant speed, $a_\text{T} =0\text{;}$ if the object travels in a constant direction, $a_\text{N} =0\text{.}$)
In Equation (11.5.2) at the beginning of this section, we found $s\,'(t) = \norm{\vvt}\text{.}$ We can combine this fact with the above formula for $a_\text{T}$ to write \begin{equation*} a_\text{T} = \frac{d}{dt}\Big(\norm{\vvt}\Big) = \frac{d}{dt}\big( s\,'(t)\big) = s\,''(t). \end{equation*}
Since $s\,'(t)$ is speed, $s\,''(t)$ is the rate at which speed is changing with respect to time. We see once more that the component of acceleration in the direction of travel relates only to speed, not to a change in direction.
Now compare the formula for $a_\text{N}$ above to the formula for curvature in Theorem 11.5.12: \begin{equation*} a_\text{N} = \frac{\norm{\vvt\times \vat}}{\norm{\vvt}} \text{ and } \kappa = \frac{\norm{\vrp(t)\times\vrp'(t)}}{\norm{\vrp(t)}^3}=\frac{\norm{\vvt\times \vat}}{\norm{\vvt}^3} . \end{equation*}
Thus \begin{align*} a_\text{N} \amp = \kappa \norm{\vvt}^2\\ \amp = \kappa\Big(s\,'(t)\Big)^2 \end{align*}
This last equation shows that the component of acceleration that changes the object's direction is dependent on two things: the curvature of the path and the speed of the object.
Imagine driving a car in a clockwise circle. You will naturally feel a force pushing you towards the door (more accurately, the door is pushing you as the car is turning and you want to travel in a straight line). If you keep the radius of the circle constant but speed up (i.e., increasing $s\,'(t)$), the door pushes harder against you ($a_\text{N}$ has increased). If you keep your speed constant but tighten the turn (i.e., increase $\kappa$), once again the door will push harder against you.
Putting our new formulas for $a_\text{T}$ and $a_\text{N}$ together, we have \begin{equation*} \vat = s\,''(t)\vec T(t) + \kappa\norm{\vvt}^2\vec N(t). \end{equation*}
This is not a particularly practical way of finding $a_\text{T}$ and $a_\text{N}\text{,}$ but it reveals some great concepts about how acceleration interacts with speed and the shape of a curve.
The minimum radius of the curve in a highway cloverleaf is determined by the operating speed, as given in the table in Figure 11.5.22. For each curve and speed, compute $a_\text{N}\text{.}$
Solution
We end this chapter with a reflection on what we've covered. We started with vector–valued functions, which may have seemed at the time to be just another way of writing parametric equations. However, we have seen that the vector perspective has given us great insight into the behavior of functions and the study of motion. Vector–valued position functions convey displacement, distance traveled, speed, velocity, acceleration and curvature information, each of which has great importance in science and engineering.
# Subsection11.5.3Exercises
Terms and Concepts
In the following exercises, a position function \vrt is given, where $t=0$ corresponds to the initial position. Find the arc length parameter $s\text{,}$ and rewrite \vrt in terms of $s\text{;}$ that is, find $\vec r(s)\text{.}$
In the following exercises, a curve $C$ is described along with 2 points on $C\text{.}$
1. Using a sketch, determine at which of these points the curvature is greater.
2. Find the curvature $\kappa$ of $C\text{,}$ and evaluate $\kappa$ at each of the 2 given points.
In the following exercises, find the value of $x$ or $t$ where curvature is maximized.
In the following exercises, find the radius of curvature at the indicated value.
In the following exercises, find the equation of the osculating circle to the curve at the indicated $t$-value.
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Introduction to Tensors
Atta ur Rehman Shah PhD Student Advanced Composite Materials Lab. CHANGWON NATIONAL UNIVERSITY
First Impression
Here is the first definition of tensor found on page 11 of Synges Tensor Calculus.
This type of definition doesnt offer any understanding for a student who is initially trying to find out what a tensor is. The student will close the book and prefer to play a game on his mobile. Hence I will try to start from the simple concepts of scalars and vectors and guide towards Tensors.
ADVANCED COMPOSITE MATERIALS LAB
Pre-requisites
What are scalars and vectors? Scalars are just numbers (magnitude) e.g. Temperature, mass, density etc.
Vectors are numbers with one direction. e.g. Velocity, Force, Displacement
ADVANCED COMPOSITE MATERIALS LAB
Pre-requisites
A vector can be multiplied with a scalar to obtain another vector with same direction but different magnitude. A vector can be multiplied to another vector to obtain a scalar. (Dot product) A vector can be multiplied to another vector to obtain a vector perpendicular to the plane first two vectors. (Cross product) What if we want to change both magnitude and direction of a given vector? (Not perpendicular as in cross product)
We need another entity
ADVANCED COMPOSITE MATERIALS LAB
Introducing Tensors
Lets pause to introduce some terminology. We will rename the familiar quantities in the following way: Scalar: Tensor of rank 0. (magnitude only 1 component) Vector: Tensor of rank 1. (magnitude and one direction 3 components)
This terminology is suggestive. Why stop at rank 1? Why not go onto rank 2, rank 3, and so on. Dyad: Tensor of rank 2. (magnitude and two directions 32 = 9 components) Triad: Tensor of rank 3. (magnitude and three directions 33 = 27 components) Tensor of rank n. (magnitude and n-directions 3n components)
ADVANCED COMPOSITE MATERIALS LAB
Introducing Tensors
We will now merely state that:
if we form the inner product of a vector and a tensor of rank 2 (a dyad), the result will be another vector with both a new magnitude and a new direction. A tensor of rank 2 is defined as a system that has a magnitude and two directions associated with it. It has 9 components. For now, we will use an example from classical electrodynamics to illustrate the point just made.
ADVANCED COMPOSITE MATERIALS LAB
Introducing Tensors
The magnetic flux density B in volt-sec/m2 and the magnetization H in Amp/m are related through the permeability in H/m by the expression
B = H
For free space, is a scalar with value = 4 10-7 H/m. Thus, the flux density and the magnetization in free space differ in magnitude but not in direction. In some exotic materials, however, the component atoms or molecules have peculiar dipole properties that make these terms differ in both magnitude and direction.
ADVANCED COMPOSITE MATERIALS LAB
Introducing Tensors
In such materials, the scalar permeability is then replaced by the tensor permeability , and we write, in place of the previous equation:
B = H
The permeability is a tensor of rank 2. Remember that B and H are both vectors, but they now differ from one another in both magnitude and direction.
ADVANCED COMPOSITE MATERIALS LAB
Introducing Tensors
Another classical example of the use of tensors in physics is Stress in a material object.
dF = TdA
dF and dA are force and area respectively, both are vectors. Thus, the stress T in the equation must be either a scalar or a tensor. But there are two different types of stress: tensile stress and shear stress. (How can a single denominate number represent both?) We must conclude that stress is a tensor of rank 2 and that the force must be an inner product of stress and area.
ADVANCED COMPOSITE MATERIALS LAB
Introducing Tensors
In summary, notice that in the progression from single number to scalar to vector to tensor, etc., information is being added at every step. The complexity of the physical situation being modeled determines the rank of the tensor representation we must choose.
Their dyad product UV will be:
Notice that, by setting u1v1 = 11, u1v2 = 12, etc., this dyad can be rewritten as
Preliminary Mathematical Considerations
The scalar components ij can be arranged in the familiar configuration of a 3x3 matrix:
Just as a matrix is generally not equal to its transpose, so with dyads it is the case that the dyad product is not commutative.
Preliminary Mathematical Considerations
Using the known rules of matrix multiplication, we can, by extension, write the rules associated with dyad multiplication.
In matrices, the results of pre- and post-multiplication are usually different; i.e., matrix multiplication does not, in general, commute. So it the case of dyad product, it is not commutative. Let the scalar components of dyad M be represented by the 3 3 matrix [ij] i, j = 1, 2, 3; then for scalar ; M = [ij] = [ij] = M Similarly, the product of a dyad UV and a scalar is defined as (UV) = (U)V = (U)V = U(V) = U(V) = (UV).
Preliminary Mathematical Considerations
The inner product of a matrix M and a vector V
Let V (Vi) be a row vector with i = 1, 2, 3
Pre-multiplication of a vector,
Post-multiplication of a vector,
Hence
It is clear that
Tensors of Rank > 2
Tensors of rank 2 result from dyad products of vectors. n an entirely analogous way, tensors of rank 3 arise from triad products, UVW. tensors of rank n arise from n-ad products of vectors, UVW...AB. In three-dimensional space, the number of components in each of these systems is 3n. The rules governing these higher rank objects are defined in the same way as the rules developed above.
ADVANCED COMPOSITE MATERIALS LAB
Important Question
ADVANCED COMPOSITE MATERIALS LAB
Invariant Quantities
A quantity we can measure that do not depend upon our frame of reference is called an invariant. This condition must be met for a mathematical object to be a tensor. For example Temperature is a tensor but Distance is not a tensor. Although both are scalars. Similarly Displacement is a tensor but Position is not a tensor. Although both are vectors.
ADVANCED COMPOSITE MATERIALS LAB
Example
Displacement Vs Position Vector X 2 X 2
O Position Vector OP
X1
O Displacement vector OP
X1
OPOP PQ = PQ
Specific Statements for Tensors per se
We now extend the properties and rules of operation for familiar objects scalars, vectors, and matrices to tensors per se.
1. 2. 3. 4. 5. 6.
All scalars are not tensors, although all tensors of rank 0 are scalars. All vectors are not tensors, although all tensors of rank 1 are vectors. All dyads or matrices are not tensors, although all tensors of rank 2 are dyads or matrices. Tensors can be multiplied by other tensors to form new tensors. The product of a tensor and a scalar (tensor of rank 0) is commutative. The pre-multiplication of a given tensor by another tensor produces a different result from post-multiplication; i.e., tensor multiplication in general is not commutative.
Specific Statements for Tensors per se
7.
The rank of a new tensor formed by the product of two other tensors is the sum of their individual ranks. The inner product of a tensor and a vector or of two tensors is not commutative. The rank of a new tensor formed by the inner product of two other tensors is the sum of their individual ranks minus 2. A tensor of rank n in three-dimensional space has 3n components.
8. 9.
10.
Re-examining Magnetic Permeability
Now we can see why the magnetic permeability must be a rank 2 tensor. We use the tensor form; B = H B and H are vectors (rank 1 tensors). Rank of B must be sum of their individual ranks of and H minus 2. 1 = x + 1 2 x = 2 (where x is the rank of ).
ADVANCED COMPOSITE MATERIALS LAB
Tensor Notations
Summation can be expressed very simply with tensor notations. For example:
Bs = t st Ht
Where scalar components of B are Bs, shown as sum of the products of and H. Summation is occurring over the repeated index, t. This representation has become the standard in the literature.
ADVANCED COMPOSITE MATERIALS LAB
Covariance and Contravariance
At any point P, we can specify three local axes and three local planes determined by these axes. In accordance with strict definitions, the axes must be mutually perpendicular and, by extension, so must the planes. Now, choose three unit vectors at P such that each vector is tangent to one of the axes. Such a triple is usually designated (i, j, k). V = i + j + k
ADVANCED COMPOSITE MATERIALS LAB Rectangular coordinates
Covariance and Contravariance
Now suppose that we had chosen unit vectors perpendicular to each of the planes rather than tangent to each of the coordinate axes. Lets do so and call the resulting triple (i*, j*, k*). Again, any vector V at P can be written as: V = *i* + *j* + *k*
Covariance and Contravariance
Our representation is satisfactory provided we ensure that: i + j + k = *i* + *j* + *k* It is apparent from geometry that the two unit vector triples comprise the same set; i.e., that i = i* j = j* k = k*. Will This approach work in all cases?? NO
ADVANCED COMPOSITE MATERIALS LAB
Covariance and Contravariance
To understand why the answer is NO, lets modify our Cartesian system If the axes are no longer mutually orthogonal for example, so that they meet at 60. (i, j, k) and (i*, j*, k*) are now two different sets of unit vectors.
i and i* now meet at an angle of 60, as do j and j*, and k and k*. Now they specify different sets of directions.
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# How do you solve sqrt(4x+1) - sqrt(x-2)=3?
Aug 11, 2015
$\left\{\begin{matrix}x = 6 \\ x = 2\end{matrix}\right.$
#### Explanation:
Start by writing the conditions that a value of $x$ must satisfy in order to be a valid solution to this equation.
You have to do that because you're dealing with radical terms, which means that the expressions under the radical signs must be positive for real numbers.
• $4 x + 1 \ge 0 \implies x \ge - \frac{1}{4}$
• $x - 2 \ge 0 \implies x \ge 2$
Combine these two conditions to get $x \ge 2$.
Now go on to solve the equation. Square both sides to get
${\left(\sqrt{4 x + 1} - \sqrt{x - 2}\right)}^{2} = {3}^{2}$
${\left(\sqrt{4 x + 1}\right)}^{2} - 2 \sqrt{\left(4 x + 1\right) \left(x - 2\right)} + {\left(\sqrt{x - 2}\right)}^{2} = 9$
$4 x + 1 - 2 \sqrt{\left(4 x + 1\right) \left(x - 2\right)} + x - 2 = 9$
This is equivalent to
$2 \sqrt{\left(4 x + 1\right) \left(x - 2\right)} = 5 x - 10$
Square both sides of the equation again to get rid of the last radical term
${\left(2 \sqrt{\left(4 x + 1\right) \left(x - 2\right)}\right)}^{2} = {\left(5 x - 10\right)}^{2}$
$4 \left(4 x + 1\right) \left(x - 2\right) = 25 {x}^{2} - 100 x + 100$
$16 {x}^{2} - 28 x - 8 = 25 {x}^{2} - 100 x + 100$
$9 {x}^{2} = 72 x + 108 = 0$
Simplify this quadratic equation by dividing everything by $9$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}} {x}^{2}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{9}}}} - 8 x + 12 = 0$
You can find the two roots to this quadratic by using the quadratic formula
${x}_{1 , 2} = \frac{- \left(8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \cdot 1 \cdot 12}}{2 \cdot 1}$
${x}_{1 , 2} = \frac{8 \pm \sqrt{16}}{2} = \frac{8 \pm 4}{2} = \left\{\begin{matrix}{x}_{1} = \frac{8 + 4}{2} = 6 \\ {x}_{2} = \frac{8 - 4}{2} = 2\end{matrix}\right.$
Since both ${x}_{1}$ and ${x}_{2}$ satisfy your condition $x \ge 2$, they will both be valid solutions to the original equation.
You can do a quick check to make sure that you got the calculations right
$\sqrt{4 \cdot 6 + 1} - \sqrt{6 - 2} = 3$
$\sqrt{25} - \sqrt{4} = 3$
$5 - 2 = 3 \textcolor{g r e e n}{\sqrt{}}$
and
$\sqrt{4 \cdot 2 + 1} - \sqrt{2 - 2} = 3$
$\sqrt{9} - \sqrt{0} = 3$
$3 - 0 = 3 \textcolor{g r e e n}{\sqrt{}}$
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# What is 2/653 Simplified?
Are you looking to calculate how to simplify the fraction 2/653? In this really simple guide, we'll teach you exactly how to simplify 2/653 and convert it to the lowest form (this is sometimes calling reducing a fraction to the lowest terms).
To start with, the number above the line (2) in a fraction is called a numerator and the number below the line (653) is called the denominator.
So what we want to do here is to simplify the numerator and denominator in 2/653 to their lowest possible values, while keeping the actual fraction the same.
To do this, we use something called the greatest common factor. It's also known as the greatest common divisor and put simply, it's the highest number that divides exactly into two or more numbers.
In our case with 2/653, the greatest common factor is 1. Once we have this, we can divide both the numerator and the denominator by it, and voila, the fraction is simplified:
2/1 = 2
653/1 = 653
2 / 653
As you can see, 2/653 cannot be simplified any further, so the result is the same as we started with. Not very exciting, I know, but hopefully you have at least learned why it cannot be simplified any further!
So there you have it! You now know exactly how to simplify 2/653 to its lowest terms. Hopefully you understood the process and can use the same techniques to simplify other fractions on your own. The complete answer is below:
2/653
## Convert 2/653 to Decimal
Here's a little bonus calculation for you to easily work out the decimal format of the fraction we calculated. All you need to do is divide the numerator by the denominator and you can convert any fraction to decimal:
2 / 653 = 0.0031
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 2/653 Simplified?". VisualFractions.com. Accessed on February 4, 2023. http://visualfractions.com/calculator/simplify-fractions/what-is-2-653-simplified/.
• "What is 2/653 Simplified?". VisualFractions.com, http://visualfractions.com/calculator/simplify-fractions/what-is-2-653-simplified/. Accessed 4 February, 2023.
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# Find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Last updated date: 17th Jun 2024
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Hint: We start solving the problem by finding the total possibilities of getting numbers by fixing each digit in unit place. We then find the sum of all the numbers present in the unit place. Similarly, we multiply 10 for the sum of digits in tenth place, 100 for the sum of digits in tenth place and 1000 for the sum of digits in thousandth place. We then add all these sums to get the required answer.
According to the problem, we need to find the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time.
Let us first fix a number in a unit place and find the total number of words possible due on fixing this number.
We need to arrange the remaining three places with three digits. We know that the number of ways of arranging n objects in n places is $n!$ ways.
So, we get $3!=6$ numbers on fixing the unit place with a particular digit.
Now, let us find the sum of all digits. We get sum as $2+3+4+5=14$.
Now, we get a sum of digits in units place for all the numbers as $14\times 6=84$.
We use the same digits in ten, hundred and thousand places also. So, the sum of those digits will also be 84 but with the multiplication of its place value.
i.e., We multiply the sum of the digits in tenth place with 10, hundredth place with 100 and so on. We then add these sums.
So, we get the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time is $\left( 84\times 1000 \right)+\left( 84\times 100 \right)+\left( 84\times 10 \right)+\left( 84\times 1 \right)$.
$\Rightarrow$ Sum = $84000+8400+840+84$.
$\Rightarrow$ Sum = $93324$.
We have found the sum of all the numbers that can be formed with the digits 2,3,4,5 taken all at a time as 93324.
Note: Alternatively, we can use the formula for the sum of numbers as $\left( n-1 \right)!\times \left( \text{sum of digits} \right)\times \left( 11111......ntimes \right)$. We can also solve this problem by writing all the possible numbers and finding the sum of them which will be time taking and make us confused. We should know that the value of the digits is determined by the place where they were present. We should check whether there is zero in the given digits and whether there are any repetitions present in the numbers. Similarly, we can expect problems to find the sum of numbers formed by these digits with repetition allowed.
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# How do you integrate int 1/(xsqrt(x^2-1) )dx using trigonometric substitution?
Nov 5, 2017
$\int \setminus \frac{1}{x \sqrt{{x}^{2} - 1}} \setminus \mathrm{dx} = a r c \sec x + C$
#### Explanation:
We seek:
$I = \int \setminus \frac{1}{x \sqrt{{x}^{2} - 1}} \setminus \mathrm{dx}$
Let us attempt a substitution of the form:
$\sec \theta = x$
Then differentiating wrt $x$ we have:
$\sec \theta \tan \theta \frac{d \theta}{\mathrm{dx}} = 1$
Substituting into the integral we have:
$I = \int \setminus \frac{1}{\sec \theta \sqrt{{\sec}^{2} \theta - 1}} \setminus \sec \theta \tan \theta \setminus d \theta$
$\setminus \setminus = \int \setminus \frac{1}{\sec \theta \sqrt{{\tan}^{2} \theta}} \setminus \sec \theta \tan \theta \setminus d \theta$
$\setminus \setminus = \int \setminus \frac{1}{\sec \theta \tan \theta} \setminus \sec \theta \tan \theta \setminus d \theta$
$\setminus \setminus = \int \setminus d \theta$
$\setminus \setminus = \theta + C$
Restoring the substitution gives:
$I = a r c \sec x + C$
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# The equation of the tangent to a circle - Higher
A tangent to a circle at point P is a straight line that touches the circle at P. The tangent is perpendicular to the radius which joins the centre of the circle to the point P.
As the tangent is a straight line, the equation of the tangent will be of the form . We can use perpendicular gradients to find the value of , then use the coordinates of P to find the value of in the equation.
### Example
Find the equation of the tangent to the circle at the point (3, -4).
The tangent will have an equation in the form so to find the equation you need to find the values of and . First, find , the gradient of the tangent.
On a diagram, draw the circle and the tangent at the point P (3, -4) and draw the radius from the centre (0, 0) to the point P. The gradient of the radius is given by
The radius that joins the centre of the circle (0, 0) to the point P is at right angles to the tangent, so the gradient of the tangent is the negative reciprocal of the gradient of the radius. This means that the gradient of the tangent, .
Next, find the value of . For the point P, the value of and the value of . Substituting these values, as well as the value of , in to the formula gives:
So the equation of the tangent to the circle at the point (3, -4) is
Question
Find where the tangent at (2, 4) to the circle, , crosses the x-axis.
Sketch a diagram to show the circle and the tangent at the point (2, 4) labelling this P. Draw the radius from the centre of the circle to P. The tangent will have an equation in the form .
First, find , the gradient of the tangent.
The gradient of the tangent is the negative reciprocal of the gradient of the radius. This means that the gradient of the tangent, .
Next, find the value of . For the point P, the value of and the value of . Substituting these values, as well as the value of , in to the formula gives:
So the equation of the tangent .
Finally, the point where the tangent crosses the -axis will have a -coordinate of 0. Substituting this value into the equation for the tangent gives:
So the tangent crosses the -axis at (10, 0).
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# quant question
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31 May 2010, 13:16
hi everyone,
is there any other way to do the following question? i understand the explanation provided by the GMAT club. however i think this is a time-consuming approach. any short cut to this question? thanks!
How many times will the digit 7 be written when listing the integers from 1 to 1000?
• 110
• 111
• 271
• 300
• 304
There are several ways to count the number of times 7 appears between 7 and 997. One way is to consider the number of 7's in single, double, and triple digit numbers separately.
One-digit numbers: 7 is the only one-digit number.
Two-digit numbers: 7 could be the first digit or the second digit. Case 1: 7 is the first digit. There are 9 ways to place 7 as the first digit of a two-digit number. Case 2: There are 10 ways to place the second digit, i.e. 0-9. Remember that we have counted 07 already. Thus, for two-digit numbers we have: numbers that contain a 7.
Three-digit numbers: Use the knowledge from the previous two scenarios: each hundred numbers will contain one 7 in numbers such as 107 or 507 and also 19 other sevens in numbers such as 271 or 237. Thus a total of 20 sevens per each hundred and 200 sevens for 1000. Since we have 700's within the range, that adds another 100 times that a seven will be written for a total of 300 times.
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31 May 2010, 13:43
Please continue the discussion in the proper thread in the GMAT Club Tests Forum.
Don't be lazy - do a search.
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# Area Of Square Using Diagonals
The area of a square can be calculated using its diagonals. Apart from the traditional method of finding the area of a square by squaring the sides, this is one of the most useful methods to compute the area of a square if the diagonal length is given.
## Formula to Calculate Area of a Square Using Diagonal
It is not always mandatory to have the measure of the side for computing a square’s area. If we have the length of the diagonal, then the area can be calculated as:
Area of Square = ½ × d2 Square units
• Here, “d” is the length of any of the diagonal (in a square, diagonals are equal)
### Derivation for Area of Square using Diagonal Formula
The formula to find the area of any square if its diagonals are given can be derived using Pythagoras theorem as explained below:
Consider a square of sides “a” units and diagonal as “d” units.
In the above figure, the square of the side “a” unit, has been divided into two right triangles with the help of diagonal of length “d” units. Thus, the diagonal of the square divides it into two right triangles. Consider any right triangle and apply Pythagoras theorem.
According to Pythagoras theorem, for a right-angled triangle,
Hypotenuse2 = Perpendicular2 + Base2
In the above diagram,
• Perpendicular = a
• Base = a
• Hypotenuse = d
So,
a2 + a2 = d2
⇒ 2a2 = d2
Or, a2 = d2/2
Now, area of a square = a2 = d2/2
So, area of a square using diagonals = ½ × d2 Square units.
### Example Question to Solve Area of Square When Diagonal is Given:
Question: Find the area of a square having a diagonal of length 10 cm.
Solution:
Given,
d = 10 cm
Now, the area of square = ½ × (diagonal)2
So,
A = ½ × 102
Or, area = 50 cm2
Note: Using the diagonal, the perimeter of the square can also be found as explained below.
### Finding Perimeter of Square Using Diagonal
It is known that the area of a square in terms of diagonal = ½ × d2
Now, a × a = ½ × d2
Or, a = √(½ × d2)
As the perimeter of square = 4a,
P = 4 × √(d2/2) units.
### More Topics Related to Squares:
1. unknown
A boy runs across a rectangular field diagonally covering a distance of 60m. If the length of the field is 50m, calculate the width of the field to the nearest meter
1. lavanya
The diagonal of the field divides the rectangular field into two right triangle.
Thus, we diagonal will be considered here as hypotenuse of triangle, equal to 60m and length of the base is 50 m
By using Pythagoras theorem, we know, in a right triangle;
Hypotenuse^2=Base^2+Perpendicular^2
60^2=50^2+width^2
3600 = 2500 + width^2
width^2 = 3600 – 2500 = 1100
width=√1100 m
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Take the aid of What is x percent of y calculator an virtual math device that calculates 20% of 110 easily in addition to a step by step solution detailing how the result 22 arrived.
You are watching: What is 20 percent of 110
20 percent *110
= (20/100)*110
= (20*110)/100
= 2200/100 = 22
Now us have: 20 percent the 110 = 22
Question: What is 20 percent the 110?
We need to determine 20% of 110 now and the procedure explaining it as such
Step 1: In the given instance Output value is 110.
Step 2: allow us think about the unknown worth as x.
Step 3: think about the output worth of 110 = 100%.
Step 4: In the exact same way, x = 20%.
Step 5: On dividing the pair of an easy equations we acquired the equation together under
110 = 100% (1).
x = 20% (2).
(110%)/(x%) = 100/20
Step 6: reciprocal of both the sides outcomes in the following equation
x%/110% = 20/100
Step 7: simplifying the above obtained equation further will call what is 20% that 110
x = 22%
Therefore, 20% that 110 is 22
## What is 20 Percent of 110?
20 percent *110
= (20/100)*110
= (20*110)/100
= 2200/100 = 22
Now we have: 20 percent the 110 = 22
Question: What is 20 percent that 110?
We need to determine 20% of 110 now and also the procedure explaining it as such
Step 1: In the given case Output worth is 110.
Step 2: let us consider the unknown value as x.
Step 3: think about the output worth of 110 = 100%.
Step 4: In the same way, x = 20%.
Step 5: On dividing the pair of an easy equations we acquired the equation together under
110 = 100% (1).
x = 20% (2).
(110%)/(x%) = 100/20
Step 6: mutual of both the sides outcomes in the following equation
x%/110% = 20/100
Step 7: simple the above obtained equation additional will call what is 20% that 110
x = 22%
Therefore, 20% of 110 is 22
### Frequently Asked concerns on What is 20 percent of 110?
1. Just how do ns calculate percent of a total?
To calculate percentages, begin by composing the number you desire to turn into a percentage over the full value so you end up with a fraction. Then, revolve the portion into a decimal by separating the top number by the bottom number. Finally, multiply the decimal by 100 to find the percentage.
2. What is 20 percent the 110?
20 percent the 110 is 22.
See more: Who Is Faster Quicksilver Or Flash, Quicksilver Vs The Flash
3. How to calculation 20 percent the 110?
Multiply 20/100 with 110 = (20/100)*110 = (20*110)/100 = 22.
### surrounding Results
20% ofResult
110 22
110.01 22.002
110.02 22.004
110.03 22.006
110.04 22.008
110.05 22.01
110.06 22.012
110.07 22.014
110.08 22.016
110.09 22.018
110.1 22.02
110.11 22.022
110.12 22.024
110.13 22.026
110.14 22.028
110.15 22.03
110.16 22.032
110.17 22.034
110.18 22.036
110.19 22.038
110.2 22.04
110.21 22.042
110.22 22.044
110.23 22.046
110.24 22.048
20% of result
110.25 22.05
110.26 22.052
110.27 22.054
110.28 22.056
110.29 22.058
110.3 22.06
110.31 22.062
110.32 22.064
110.33 22.066
110.34 22.068
110.35 22.07
110.36 22.072
110.37 22.074
110.38 22.076
110.39 22.078
110.4 22.08
110.41 22.082
110.42 22.084
110.43 22.086
110.44 22.088
110.45 22.09
110.46 22.092
110.47 22.094
110.48 22.096
110.49 22.098
20% of an outcome
110.5 22.1
110.51 22.102
110.52 22.104
110.53 22.106
110.54 22.108
110.55 22.11
110.56 22.112
110.57 22.114
110.58 22.116
110.59 22.118
110.6 22.12
110.61 22.122
110.62 22.124
110.63 22.126
110.64 22.128
110.65 22.13
110.66 22.132
110.67 22.134
110.68 22.136
110.69 22.138
110.7 22.14
110.71 22.142
110.72 22.144
110.73 22.146
110.74 22.148
20% of result
110.75 22.15
110.76 22.152
110.77 22.154
110.78 22.156
110.79 22.158
110.8 22.16
110.81 22.162
110.82 22.164
110.83 22.166
110.84 22.168
110.85 22.17
110.86 22.172
110.87 22.174
110.88 22.176
110.89 22.178
110.9 22.18
110.91 22.182
110.92 22.184
110.93 22.186
110.94 22.188
110.95 22.19
110.96 22.192
110.97 22.194
110.98 22.196
110.99 22.198
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# What is 1 divided by 3/481
If you have the whole number 1 and you want to divide it by the fraction 3/481 then you have found the perfect article. In this quick math lesson, we'll show you how you can divide any whole number by a fraction. If dividing numbers by fractions is your jam, read on my friend!
Want to quickly learn or show students how to divide a whole number by a fraction? Play this very quick and fun video now!
Now, remember kids, the number above the fraction like is called the numerator, and the number below it is called the denominator. We'll be using these terms throughout the guide. Pretty simple stuff, but it's always nice to do a quick term recap.
Let's put our whole number and fraction side by side so we can visualize the problem we're trying to solve:
1 ÷ 3 / 481
The trick to working out 1 divided by 3/481 is similar to the method we use to work out dividing a fraction by a whole number.
All we need to do here is multiply the whole number by the numerator and make that number the new numerator. The old numerator then becomes the new denominator. Let's write this down visually:
1 x 481 / 3 = 481 / 3
So, the answer to the question "what is 1 divided by 3/481?" is:
481 / 3
Sometimes, after calculating the answer we can simplify the resulting fraction down to lower terms. In this example though 481/3 is already in it's lowest possible form.
If you made it this far you must really love your fractions and dividing whole numbers by them. Hopefully this simple guide was easy for you to follow along and you can now go forth and divide more whole numbers by as many fractions as your heart desires.
## Convert 1 divided by 3/481 to Decimal
One last little calculation before you go. It's common to want to express your result as a decimal and, to do that, all you need to do is divide your numerator by your denominator:
481 / 3 = 160.3333
If you found this content useful in your research, please do us a great favor and use the tool below to make sure you properly reference us wherever you use it. We really appreciate your support!
• "What is 1 divided by 3/481". VisualFractions.com. Accessed on February 22, 2024. http://visualfractions.com/calculator/whole-divided-by-fraction/what-is-1-divided-by-3-481/.
• "What is 1 divided by 3/481". VisualFractions.com, http://visualfractions.com/calculator/whole-divided-by-fraction/what-is-1-divided-by-3-481/. Accessed 22 February, 2024.
## Whole Number Divided by Fraction
Enter a whole number, numerator, denominator
÷
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# Age
Problems on ages basically use formulae from Algebra and do not require any special formula. It is like a question type of algebra.
### Sample Problems
Question 1 : A’s age after 15 years would be equal to 5 times his age 5 years ago. Find his age 3 years hence.
Solution : Let A’s present age be ‘n’ years.
According to the question,
n + 15 = 5 (n – 5)
=> n + 15 = 5 n – 25
=> 4n = 40
=> n = 10
=> A’s present age = 10 years
Therefore, A’s age 3 years hence = 10 + 3 = 13 years
Question 2 : The product of the ages of A and B is 240. If twice the age of B is more than A’s age by 4 years, what was B’s age 2 years ago?
Solution : Let A’s present age be x years. Then, B’s present age = 240 / x years
So, according to question
2 (240 / x ) – x = 4
=> 480 – x2 = 4 x
=> x2 + 4 x – 480 = 0
=> (x + 24) (x – 20) = 0
=> x = 20
=> B’s present age = 240 / 20 = 12 years
Thus, B’s age 2 years ago = 12 – 2 = 10 years
Question 3 : The present age of a mother is 3 years more than three times the age of her daughter. Three years hence, mother’s age will be 10 years more than twice the age of the daughter. Find the present age of the mother.
Solution : Let the daughter’s present age be ‘n’ years.
=> Mother’s present age = (3n + 3) years
So, according to the question
(3n + 3 + 3) = 2 (n + 3) + 10
=> 3n + 6 = 2n + 16
=> n = 10
Hence, mother’s present age = (3n + 3) = ((3 x 10) + 3) years = 33 years
Question 4 : The ratio of present ages of A and B is 6 : 7. Five years hence, this ratio would become 7 : 8. Find the present age of A and B.
Solution : Let the common ratio be ‘n’.
=> A’s present age = 6 n years
=> B’s present age = 7 n years
So, according to the question
(6 n + 5) / (7 n + 5) = 7 / 8
=> 48 n + 40 = 49 n + 35
=> n = 5
Thus, A’s present age = 6 n = 30 years and B’s present age = 7 n = 35 years
Please write comments if you have any doubts related to the topic discussed above, or if you are facing difficulty in any question or if you would like to discuss a question other than those mentioned above.
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Every number is the product of a unique set of prime factors, a group of prime numbers (including repeats) that, when multiplied together, equals that number. You can find those prime factors for a given number, by using a process called decomposition.
An easy way to decompose a number is to make a factorization tree. Here’s how:
1. Find two numbers that multiply to equal the original number; write them as numbers that branch off the original one.
2. If either number is prime, circle it and end that branch.
3. Continue branching off non-prime numbers into two factors; whenever a branch reaches a prime number, circle it and close the branch.
When every branch ends in a circled number, you’re finished — just gather up the circled numbers.
Sample question
1. Decompose the number 48 into its prime factors.
48 = 2 x 2 x 2 x 2 x 3. Begin making a factorization tree by finding two numbers that multiply to equal 48:
Continue making branches of the tree by doing the same for 6 and 8:
Circle the prime numbers and close those branches. At this point, the only open branch is 4. Break it down into 2 and 2:
Every branch ends in a circled number, so you’re finished. The prime factors are 2, 2, 2, 2, and 3.
Practice questions
1. Decompose 18 into its prime factors.
2. Decompose 42 into its prime factors.
3. Decompose 81 into its prime factors.
4. Decompose 120 into its prime factors.
Following are the answers to the practice questions:
1. 18 = 2 x 3 x 3. Here’s one possible factoring tree:
2. 42 = 2 x 3 x 7. Here’s one possible factoring tree:
3. 81 = 3 x 3 x 3 x 3. Here’s one possible factoring tree:
4. 120 = 2 x 2 x 2 x 3 x 5. Here’s one possible factoring tree:
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The K5 Learning Blog urges parents to be pro-active in helping their children reach their full academic potential.
K5 Learning
provides an online reading and math program for kindergarten to grade 5 students.
# Smarter Ways to Subtract
When we were at school, we really only learned the one traditional way to subtract numbers. This is where you subtract from left to right. We called it carrying or borrowing, and it’s now called re-grouping. You may remember this?
Let’s say we are subtracting 156 from 360.
We start from the right, and immediately have to borrow.
Then you subtract the tens column.
And finally, the hundreds column – the left column.
However, there are easier ways to help our kids learn to subtract. Here are some of those methods. We call them mental math, Singapore method and counting up.
## Mental Math
In mental math you subtract from left to right. You start with the hundreds. So, 360 – 100.
Then the tens, so 260 – 50.
And finally the ones column: 210 – 6.
## Singapore Method
In the Singapore method, you don’t have to carry over the number. In our example, 6 is greater than 0. So to avoid carrying, we’ll subtract 6 from both numbers.
Subtract 6 from the top line:
Subtract 6 from the bottom line:
And then subtract the sums of those numbers, which is now a straight subtraction with no carrying:
## Counting Up
This is a good for example for when you need to give change.
In this method, you start from the lower number and count up to the top number.
First, let’s add up to tens. Adding 4 to 1.56 brings it up to 1.60.
Then, add up to 100s. Adding 50 to 1.60 brings you up to 2.00.
Then, keep counting up. Adding 1 to 2, brings us up to 3.00.
And finally, let’s add the .6 to 3 to bring it to 3.60.
To work out the difference between 1.56 and 3.60, you now add the numbers in orange.
0.60 + 0.4 + 1 + 0.04 = 2.04
If you are looking for practice sheets for subtracting larger numbers in columns, check out our grade 4 subtraction worksheets:
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# NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers
NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers
### NCERT Solutions for Class 8 Maths Chapter 16 Playing with Numbers Exercise 16.1
Find the values of the letters in each of the following and give reasons for the steps involved.
Ex 16.1 Class 8 Maths Question 1.
Solution:
Ex 16.1 Class 8 Maths Question 2.
Solution:
Ex 16.1 Class 8 Maths Question 3.
Solution:
Ex 16.1 Class 8 Maths Question 4.
Solution:
Ex 16.1 Class 8 Maths Question 5.
Solution:
3 × B = B
⇒ B = D
3 × A = CA
⇒ 3 × 5 = 15
Thus A = 5 and C = 1
Hence A = 5, B = 0 and C = 1
Ex 16.1 Class 8 Maths Question 6.
Solution:
5 × B = B
⇒ B = 0 or 5
5 × A = CA
5 × 5 = 25
Only possible when B = 0
Thus A = 5 and C = 2
Hence A = 5, B = 0 and C = 2
Ex 16.1 Class 8 Maths Question 7.
Solution:
B × 6 = B
6 × 4 = 24 → B = 4 and 2 is carried to
6 × A = BB
⇒ 6 × 7 = 42 + 2 (carried on) = 44
Thus B = 7
Hence A = 7 and B = 4
Ex 16.1 Class 8 Maths Question 8.
Solution:
1 + B = 0
1 + 9 = 10 → unit digit is 0 and 1 is carried to A
+ 1 +1 (carried on) = B = 9
A + 2 = 9 ⇒ A = 9 – 2 = 7
Hence A = 7 and B = 9
Ex 16.1 Class 8 Maths Question 9.
Solution:
B + 1 = 8 ⇒ B = 8 – 1 = 7
A + B = a number with unit digit 1
A + B = 11
⇒ A + 7 = 11
⇒ A = 11 – 7 = 4 (1 Carried to)
Now 1 carried on + 2 + A = B
3 + A = 7
⇒ A = 7 – 3 = 4
Hence A = 4, B = 7
Ex 16.1 Class 8 Maths Question 10.
Solution:
9 = A + B
9 = 1 + 8 or 2 + 7 or 3 + 6 or 4 + 5 or 8 + 1 or 7 + 2 or 6 + 3 or 5 + 4 or 0 + 9 or 9 + 0
Now 0 is required at unit place
2 + A = 10
⇒ A = 10 – 2 = 8
B = 9 – 8 = 1
1 + 6 + 1 (carried on) = A = 8
Hence A = 8 and B = 1
## SabDekho
The Complete Educational Website
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# What is a limit of a function
1. Jul 23, 2014
### Greg Bernhardt
Definition/Summary
Limits are a mathematical tool which is used to define the 'limiting value' of a function i.e. the value a function seems to approach when it's argument(s) approach a particular value. Although, the argument of the function can be taken to approach any value, limits are helpful in cases where the argument approaches a value where the function is not defined or becomes exceedingly large.
While defining a limit, we say that the argument 'tends to' a value. For example,
$$\lim_{x \to c}f(x) = m$$
is said: "As x tends to c, the function f(x) tends to m". This statement however makes no assertion of what the value of f(c) would be. Rather, it means that as x becomes exceedingly close to 'c', f(x) becomes exceedingly close to m.
If, however, the function is defined and continuous at 'c', then:
$$\lim_{x \to c}f(x) = f(c)$$
(See explanation)
Equations
Identities of limits:
$$\lim_{x \to c}f(x) + g(x) = \lim_{x \to c}f(x) + \lim_{x \to c}g(x)$$
$$\lim_{x \to c}f(x) \cdot g(x) = \lim_{x \to c}f(x) \cdot \lim_{x \to c}g(x)$$
$$\lim_{x \to c}\frac{f(x)}{g(x)} = \frac{\lim_{x \to c}f(x)}{\lim_{x \to c}g(x)}$$
$$\lim_{x \to c}\lambda f(x) = \lambda\,\lim_{x \to c} f(x)$$
Extended explanation
1. Informal approach to understanding limits
Let there be a function f(x) such that:
$$f(x) = \frac{x^2 - 4}{x - 2}$$
This function can be simplified to:
$$f(x) = \frac{(x - 2)(x + 2)}{x - 2} = x + 2$$
However, when (x - 2) is cancelled in both the numerator and the denominator, x - 2 ≠ 0 or x ≠ 2. This is because $\frac{0}{0}$ is an indeterminate form. Hence, the function f(x) = x + 2 at points other than x = 2.
But, the function is still defined at points very close to 2. As x gets closer and closer to 2, f(x) gets closer and closer to 4. So, it is said that f(x) tends to 4 as x tends to 2. This is written as:
$$\lim_{x \to 2} f(x) = 4$$
This function, however is not defined at x = 2 i.e. f(2) does not exist at all and hence,
$$\lim_{x \to 2} f(x) \ne f(2)$$
and hence, the function is said to be non-continuous at x = 2. This can be understood as there being a gap on the line plot of this function.
However, consider the function g(x) = x + 2. It is defined at all points, and hence is continuous everywhere. Other than for their behavior at x = 2, f(x) and g(x) are identical functions.
Even g(x) can be made to be continuous by describing it as:
$$g(x) = \begin{cases} x + 2, & x \ne 2 \\ 0, & x = 2 \end{cases}$$
Here, g(2) = 0, and hence:
$$\lim_{x \to 2} g(x) \ne g(2)$$
and hence the function is not continuous at x = 2. The following function however, is continuous at x = 2:
$$g(x) = \begin{cases} x + 2, & x \ne 2 \\ 4, & x = 2 \end{cases}$$
2. Tending to infinity
Consider the function f(x) such that:
$$f(x) = \frac{1}{x}$$
As 'x' gets close to zero, the value of f(x) becomes large. At x = 0, f(x) is not defined. However, f(x) 'tends to' infinity, as 'x' tends to zero:
$$\lim_{x \to 0} f(x) = \infty$$
3. Right hand and Left hand limits
In a graph of a function, when we approach a particular value, we can do it either from the right side, or from the left side of the graph. Mathematically, when we say that, 'x tends to c', it can mean either that x is infinitesimally larger than c or it is infinitesimally smaller than c'.
$$\lim{x \to c}f(x)$$
Here, let:
$$x = c + h$$
and
$$as~~x \to c~~;h \to 0$$
As such, 'x' is infinitesimally larger than 'x'. On a graph, 'x' lies on the right of x, and such such a limit is called the 'Right Hand Limit'(RHL). Whereas, if we take, x = c - h, it is called the 'Left Hand Limit'(LHL). It is denoted as follows:
$$RHL:~~~\lim_{x \to c^+} f(x)~~=~~\lim_{h \to 0} f(c + h)$$
$$LHL:~~~\lim_{x \to c^-} f(x)~~=~~\lim_{h \to 0} f(c - h)$$
For a continuous function, both these values are equal to each other and also to f(c).
4. List of standard limits
In addition to the identities given in the equation box below, the following standard limits are helpful in evaluating limits of compound functions. The following listed limits are commonly used and are hence enlisted for easy reference:
$$\lim_{x \to c} x = c$$
$$\lim_{x \to 0} \frac{1}{x} = \infty$$
$$\lim_{x \to \infty} x = \infty$$
$$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$$
$$\lim_{x \to 0} \frac{\tan(x)}{x} = 1$$
$$\lim_{x \to 0} \frac{1 - \cos(x)}{x} = 0$$
$$\lim_{x \to a} \frac{x^n - a^n}{x - a} = na^{n - 1}$$
$$\lim_{x \to a} \frac{\sin(x - a)}{x - a} = 1$$
$$\lim_{x \to a} \frac{\tan(x - a)}{x - a} = 1$$
$$\lim_{x \to 0} \frac{log_e(1 + x)}{x} = 1$$
$$\lim_{x \to 0} \frac{a^x - 1}{x} = log_e a$$
$$\lim_{x \to 0} \frac{e^x - 1}{x} = 1$$
* This entry is from our old Library feature. If you know who wrote it, please let us know so we can attribute a writer. Thanks!
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# How do you integrate (x^(1/3))/(((x^(1/3))-1))?
Jan 7, 2017
$x + \frac{3}{2} {x}^{\frac{2}{3}} + 3 {x}^{\frac{1}{3}} + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid + C$
#### Explanation:
$I = \int {x}^{\frac{1}{3}} / \left({x}^{\frac{1}{3}} - 1\right) \mathrm{dx}$
Let $u = {x}^{\frac{1}{3}} - 1$. This implies that ${x}^{\frac{1}{3}} = u + 1$ and that $x = {\left(u + 1\right)}^{3}$, a fact we use to show that $\mathrm{dx} = 3 {\left(u + 1\right)}^{2} \mathrm{du}$.
Substituting in what we know, this becomes:
$I = \int \frac{u + 1}{u} 3 {\left(u + 1\right)}^{2} \mathrm{du} = 3 \int {\left(u + 1\right)}^{3} / u \mathrm{du}$
Expanding, then dividing:
$I = 3 \int \frac{{u}^{3} + 3 {u}^{2} + 3 u + 1}{u} \mathrm{du} = 3 \int \left({u}^{2} + 3 u + 3 + \frac{1}{u}\right) \mathrm{du}$
Integrating term by term:
$I = 3 \left({u}^{3} / 3 + \frac{3}{2} {u}^{2} + 3 u + \ln \left\mid u \right\mid\right)$
$I = {u}^{3} + \frac{9}{2} {u}^{2} + 9 u + 3 \ln \left\mid u \right\mid$
Using $u = {x}^{\frac{1}{3}} - 1$:
$I = {\left({x}^{\frac{1}{3}} - 1\right)}^{3} + \frac{9}{2} {\left({x}^{\frac{1}{3}} - 1\right)}^{2} + 9 \left({x}^{\frac{1}{3}} - 1\right) + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid$
If you wish, you can expand all of these terms and combine for the simplified answer of:
$I = \left(x - 3 {x}^{\frac{2}{3}} + 3 {x}^{\frac{1}{3}} - 1\right) + \frac{9}{2} \left({x}^{\frac{2}{3}} - 2 {x}^{\frac{1}{3}} + 1\right) + 9 {x}^{\frac{1}{3}} - 9 + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid$
Continue with combining like terms, and have the $- 9 , 1 ,$ and $\frac{9}{2}$ absorb into the constant of integration:
$I = x + \frac{3}{2} {x}^{\frac{2}{3}} + 3 {x}^{\frac{1}{3}} + 3 \ln \left\mid {x}^{\frac{1}{3}} - 1 \right\mid + C$
Jan 8, 2017
I also got $x + 3 {x}^{\text{1/3" + 3/2x^"2/3" + 3ln|x^"1/3}} - 1 | + C$.
Another way to do it is:
int x^"1/3"/(x^"1/3" - 1)dx
= int cancel((x^"1/3" - 1)/(x^"1/3" - 1))^(1) + 1/(x^"1/3" - 1)dx
$= \int 1 + \frac{1}{{x}^{\text{1/3}} - 1} \mathrm{dx}$
For this, let $u = {x}^{\text{1/3}}$ so that $\mathrm{du} = \frac{1}{3} {x}^{- \text{2/3}} \mathrm{dx}$, or $\mathrm{dx} = 3 {x}^{\text{2/3}} \mathrm{du} = 3 {u}^{2} \mathrm{du}$. Then:
$\implies 3 \int {u}^{2} \mathrm{du} + 3 \int \frac{{u}^{2}}{u - 1} \mathrm{du}$
Dividing the second integrand gives:
$\frac{\left({u}^{2} - 1\right) + 1}{u - 1} = u + 1 + \frac{1}{u - 1}$
so, this overall gives:
$\implies 3 \int {u}^{2} \mathrm{du} + 3 \int 1 \mathrm{du} + 3 \int u \mathrm{du} + 3 \int \frac{1}{u - 1} \mathrm{du}$
$= {u}^{3} + 3 u + \frac{3}{2} {u}^{2} + 3 \ln | u - 1 |$
$= \textcolor{b l u e}{x + 3 {x}^{\text{1/3" + 3/2x^"2/3" + 3ln|x^"1/3}} - 1 | + C}$
Jan 8, 2017
Making $y = {x}^{\frac{1}{3}}$ we have $\mathrm{dy} = \frac{1}{3} {x}^{- \frac{2}{3}} \mathrm{dx} = \frac{1}{3} \frac{\mathrm{dx}}{y} ^ 2$ so
${x}^{\frac{1}{3}} / \left({x}^{\frac{1}{3}} - 1\right) \mathrm{dx} = 3 {y}^{3} / \left(y - 1\right) \mathrm{dy} = 3 \left(\frac{{y}^{3} - 1}{y - 1} + \frac{1}{y - 1}\right) \mathrm{dy} = 3 \left(1 + y + {y}^{2} + \frac{1}{y - 1}\right) \mathrm{dy}$
so
$\int {x}^{\frac{1}{3}} / \left({x}^{\frac{1}{3}} - 1\right) \mathrm{dx} = 3 \int \left(1 + y + {y}^{2} + \frac{1}{y - 1}\right) \mathrm{dy} = 3 y + \frac{3}{2} {y}^{2} + {y}^{3} + \log \left(\left\mid y - 1 \right\mid\right) + C$ or
$\int {x}^{\frac{1}{3}} / \left({x}^{\frac{1}{3}} - 1\right) \mathrm{dx} = 3 {x}^{\frac{1}{3}} + \frac{3}{2} {x}^{\frac{2}{3}} + x + 3 \log \left(\left\mid {x}^{\frac{1}{3}} - 1 \right\mid\right) + C$
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# Ratio and Proportion
Real life applications of ratio and proportion are numerous! The concept occurs in many places in mathematics
When you prepare recipes, paint your house, or repair gears in a large machine or in a car transmission, you use ratios and proportions.
Say a recipe to make brownie requires 4 cups of flour for 6 persons
You may want to know how much flower to put for 24 persons
As a former math teacher, I used to tell my students to set it up as you see below:
You just cannot go wrong when you take this first step:
The setup you see above can be translated into the following proportion:
4 cups / x = 6 / 24
The lessons below are geared toward setting up problems like these and solving them
Say you need 4 gallons of paint to paint 1000 square feet of area in your house.
You may want to know how much paint, you will need for 1500 square feet
Try setting up that one yourself as shown above!
Probably one of the best applications we can find is that of gears in a car transmission.
A gear look like a circle and it has teeth all around it.
Gear theory in a vehicle transmission is a complex study. We won't discuss this here in details, or I am going to have to read again my old mechanics books.
Having said that, I will only show you the forest. You will need to do your own research to see all the trees.
My goal is to introduce you to an interesting application of the topic.
In sum, transmissions contain several combinations of large and small gears.
Say for instance a small gear (20 teeth) drives a large gear(40 teeth).
The large gear will turn at half the speed of the small gear.
20 / 4 = 1 / 2
However, this situation increases the turning force (or torque) of the large gear.
In general, the larger the gear the bigger the torque.
Therefore, you can see that knowing the ratio of gears, can help us determine the speed and how much torque each gear will deploy.
This concept is important when putting a transmission together. The situation we just describe may for instance set the proper gear ratio for moving a load.
However, at cruising speed, gears may have the same ratio so that the amount of torque that enter the transmission equal the amount of torque that goes out.
Automotive engineers do lots of gear ratios calculation before they can put a transmission together.
This concludes our introduction. Dig deeper by reading the following lessons:
The following lessons about ratio and proportion follow a logical order, so try your best to learn them in order
Ratios
Some formal definitions of ratio are given to include continued ratio. Learn how to write ratios given some real life situations
Proportions
Leanr what a proportion is and how to set it up. Learn also about fourth proportional and how to find equivalent proportions
Solving proportions
Shows you how to set up a proportion and solve for x. Some special techniques to quickly simplify a proportion are also introduced
Proportion word problems
Solve some carefully chosen real life proportion word problems.
Scale drawings
Learn how you can apply proportions to scale drawing
What is the golden ratio
Lastly, challenge yourself. Learn what the golden ratio is and how to derive a formula
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Question
# Trace the following central conics.$3{(2x - 3y + 4)^2} + 2{(3x + 2y - 5)^2} = 78.$
Hint: Since this equation is similar to that of ellipse hence we will compare this equation with
the standard equation of ellipse.
The standard equation of ellipse is:
$\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1$
Now comparing this equation with our given equation we see that we have an ellipse
$3{(2x - 3y + 4)^2} + 2{(3x + 2y - 5)^2} = 78.$
Now multiplying and dividing LHS by $13$ we have
$\dfrac{{3 \times 13{{(2x - 3y + 4)}^2}}}{{{{(\sqrt {13} )}^2}}} + \dfrac{{2 \times 13{{(3x + 2y - 5)}^2}}}{{{{(\sqrt {13} )}^2}}} = 78$
Now let $X = \dfrac{{2x - 3y + 4}}{{\sqrt {13} }},Y = \dfrac{{3x + 2y - 5}}{{\sqrt {13} }}$
on substituting it in our equation we have
$39{X^2} + 26{Y^2} = 78$
Now dividing both sides by $78$ we get
$\dfrac{{{X^2}}}{2} + \dfrac{{{Y^2}}}{3} = 1$
on comparing this equation with the standard equation of ellipse we get
$a = \sqrt 2 ,b = \sqrt 3$
Note: While attempting question on conic sections and especially locus questions we should always compare the given equation in the question with standard equations we know of various conics because the equation in the question is always the modified equation of any of the conics and hence by comparing it and knowing of which conic it is we can further continue by then converting it to the standard form and we finally arrive to solution by this method
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# Teacher Notes
### Why use this resource?
This resource presents diagrams that offer an interesting and concrete way for students to think about infinite geometric series. Prior knowledge of geometric sequences is not required and this task could be used to introduce geometric sequences by linking the scale factor from one shape to the next to the idea of a common ratio. Working with diagrammatic representations means that no formulae are required and with some basic geometry students can find the sum of series in the form of $\dfrac{1}{n}+\dfrac{1}{n^2} + \dfrac{1}{n^3}+ \dotsb$.
### Possible approaches
The warm-up could be used as a stimulus to let students explore the mathematics they see in the diagram, or it could be made more focused by using the questions in the ‘Things you might think about’ toggle. This could help students to appreciate how infinite sums can have a finite answer.
Once students have attempted the problem, it is likely that some talking points will have arisen. One of which is likely to be whether the patterns in the diagrams continue infinitely or not. Discussing these ideas and sharing students’ own diagrams could be a way to bring out the language and ideas that surround geometric sequences and series.
### Key questions
• What is the link between each term of the sequence?
• How do your answers change if you assume this sequence continues forever?
• Do we know anything about the value of this sum? What must it be bigger than? What must it be smaller than?
• Are there other ways these sums could be represented?
### Possible support
There are images contained in the solution section that may help students to notice how the shaded sequences can be thought of as a simple fraction of the whole square.
### Possible extension
At the end of the solution two questions are given as possible ways to extend the problem.
• Can you use these infinite sums to find the total of others? For example, can you find the total sum of $\dfrac{1}{2} + \dfrac{1}{10} + \dfrac{1}{50} + \dfrac{1}{250} + \dotsb$?
• Is it possible that other sums of fractions will sum to the same totals? For example, can you find another infinite sum that equals $\dfrac{1}{4}$?
If exploring the second question, students may wish to look at A puzzling pentagon.
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# Approximating square roots using binomial expansion.
Through the binomial expansion of $(1 - 2x)^\frac{1}{2}$, I am required to find an approximation of $\sqrt2$.
Binomial expansion
$(1 + x)^n = 1 + \frac{n}{1}x + \frac{n(n-1)}{1*2}x^2 + ...$
Thus, the expansion of $(1 - 2x)^\frac{1}{2}$: $$= 1 - x -\frac{1}{2}x^2 - \frac{1}{2}x^3 + ...$$
The suggested way, is to choose a value for $x$ so that $(1-2x)$ has the form $2*$'a perfect square'. This can be done by taking $x = 0.01$. Thus, $(1 - 2x)=(1-2*0.01) = 0.98 = 2*0.7^2$
And $$(1 - 2x)^\frac{1}{2} = 0.98^\frac{1}{2} = 0.7\sqrt2$$ Which is equal to the previously established expansion, so we can now go ahead and find $\sqrt2$. The problem I am facing, is that there was no mention of how the value of $x=0.01$ was arrived at.
Is there an easy way to determine an appropriate value for $x$?
We want to (manually) approximate $$\sqrt{2}$$ by using the first few terms of the binomial series expansion of \begin{align*} \sqrt{1-2x}&= \sum_{n=0}^\infty \binom{\frac{1}{2}}{n}(-2x)^n\qquad\qquad\qquad\qquad |x|<\frac{1}{2}\\ &= 1-x-\frac{1}{2}x^2-\frac{1}{2}x^3+\cdots\tag{1} \end{align*} Here we look for a way to determine appropriate values of $$x$$ using the binomial expansion.
In order to apply (1) we are looking for a number $$y$$ with \begin{align*} \sqrt{1-2x}&=\sqrt{2y^2}=y\sqrt{2}\tag{2}\\ \color{blue}{\sqrt{2}}&\color{blue}{=\frac{1}{y}\sqrt{1-2x}} \end{align*}
We see it is convenient to choose $$y$$ to be a square number which can be easily factored out from the root. We obtain from (2) \begin{align*} 1-2x&=2y^2\\ \color{blue}{x}&\color{blue}{=\frac{1}{2}-y^2}\tag{3} \end{align*}
When looking for an appropriate $$y$$ which fulfills (3) there are some aspects to consider:
• We have to respect the radius of convergence $$|x|<\frac{1}{2}$$.
• Since we want to calculate an approximation of $$\sqrt{2}$$ by hand we should take $$y\in\mathbb{Q}$$ with rather small numbers as numerator and denominator.
• Last but not least: We want to find a value $$x$$ which provides a good approximation for $$\sqrt{2}$$.
We will see it's not hard to find values which have these properties.
We see in (1) a good approximation is given if $$x$$ is close to $$0$$. If $$x$$ is close to zero we will also fulfill the convergence condition. $$x$$ close to zero means that in (3) we have to choose $$y$$ so that $$y^2$$ is close to $$\frac{1}{2}$$. We have already (1) and (3) appropriately considered. Now we want to find small natural numbers $$a,b$$ so that
\begin{align*} y^2=\frac{a^2}{b^2}\approx \frac{1}{2} \end{align*} This can be done easily. When going through small numbers of $$a$$ and $$b$$ whose squares are apart by a factor $$2$$ we might quickly come to $$100$$ and $$49$$. These are two small squares and we have $$2\cdot 49=98$$ close to $$100$$. That's all.
Now it's time to harvest. We choose $$y^2=\frac{49}{100}$$ resp. $$\color{blue}{y=\frac{7}{10}}$$. We obtain for $$x$$ from (3) \begin{align*} x=\frac{1}{2}-y^2=\frac{1}{2}-\frac{49}{100}=\frac{1}{100} \end{align*} We have now a nice value $$\color{blue}{x=\frac{1}{100}}$$ and we finally get from (1) the approximation: \begin{align*} \color{blue}{\sqrt{2}} \approx \frac{10}{7}\left(1- 10^{-2}-\frac{1}{2}\cdot 10^{-4}-\frac{1}{2}\cdot 10^{-6}\right)\color{blue}{=1.414\,213\,5}71\ldots \end{align*}
We have $$\color{blue}{\sqrt{2}=1.414\,213\,5}62\ldots$$ with an approximation error $$\approx 9.055\times 10^{-9}$$. This result is quite impressive when considering that we have used just four terms of the binomial series.
Note: In a section about binomial series expansion in Journey through Genius by W. Dunham the author cites Newton: Extraction of roots are much shortened by this theorem, indicating how valuable this technique was for Newton.
Personally, I wouldn't have done it that way. So here is how I would've done it:
## Method 1:
$$\sqrt2=\sqrt{1+1}=1+\frac12-\frac18+\dots\approx1+\frac12-\frac18=\frac{11}8=1.375$$
which is much clearer to me, since it avoids having to take decimals raised to powers and gives you something you can easily do by hand.
$$1.375^2=1.890625$$
Obviously it approaches the correct value as you take more terms.
## Method 2:
This is called fixed-point iteration/Newton's method, and it basically goes like this:
$$x=\sqrt2\implies x^2=2$$
$$2x^2=2+x^2$$
Divide both sides by $2x$ and we get
$$x=\frac{2+x^2}{2x}$$
Now, interestingly, I'm going to call the $x$'s on the left $x_{n+1}$ and the $x$'s on the right $x_n$, so
$$x_{n+1}=\frac{2+(x_n)^2}{2x_n}$$
and with $x_0\approx\sqrt2$, we will then have $x=\lim_{n\to\infty}x_n$. For example, with $x_0=1$,
$x_0=1$
$x_1=\frac{2+1^2}{2(1)}=\frac32=1.5$
$x_2=\frac{2+(3/2)^2}{2(3/2)}=\frac{17}{12}=1.41666\dots$
$x_3=\dots=\frac{577}{408}=1.414215686$
And one can quickly check that $(x_3)^2=2.000006007\dots$, which is pretty much the square root of $2$.
It is often easier to begin with a slightly different problem to simplify slightly. Stay with me and you will see why.
Instead of finding the square root of 2, look at the square root of 200 and note that:
200 = 14² + 4
This is part of a general approach that says if N = a² + b, where b << a, then we may write:
$$\sqrt N = a(1 + \frac b {a^2} ) =a[1 + \frac b {2a^2} - \frac {b^2} {8a^4} + \frac {b^3} {16a^6} - ... ] = a + \frac b {2a} - \frac {b^2} {8a^3} + ...$$
Now, if we use N = 200, a = 14, b = 4, we get the first two terms give:
$$\sqrt 200 = 14\frac 1 7$$
That is $$\sqrt 2 = 1.4142857...$$ which is correct to 5 significant figures. By including the next term, we get $$\sqrt 200 = 14 + \frac 1 7 - \frac 1 {1372}$$. This gives the superb result:
$$\sqrt 2 = 1.414212827...$$, which is correct to 6 significant figures.
APPENDIX - Another example of $$\sqrt{13}$$
Using the same approach and only the first two terms, we note that 36² = 1296. Thus, $$\sqrt{1300} = 36\frac 1 {18} =$$. This gives:
$$\sqrt{13} = 3.60555$$, correct to 6 significant figures.
In fact you can take any two numbers which can be added to get 2 (not nesserly 0.01 but at least you should know the root of one of them
So for example $\sqrt{2} = {(1+1)^{1/2}}$
Know all what you need is to expand it using bio theorem and for 2 terms you ll get 1.5
$1+(0.5)\cdot 1= 1.5$ Which is not too bad approximation for $\sqrt{2} = 1.4142135624$
• I have edited your answer to use MathJax, but it is still in need of improvement. What do you mean when you say $\sqrt{2} = \frac{(1+1)^1}{2}$? This is not an equality... Also, what is "bio theorem"? Do you mean the binomial theorem? Your answer is altogether hard to understand. Commented Apr 8, 2018 at 21:58
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# Relationship between Zeroes and Coefficients of a Polynomial
Polynomials are algebraic expressions with constants and variables that can be linear i.e. the highest power o the variable is one, quadratic and others. The zeros of the polynomials are the values of the variable (say x) that on substituting in the polynomial give the answer as zero. While the coefficients of a polynomial are the constants that are multiplied by the variables of the polynomial. There is a relation between the Zeroes of a Polynomial and the Coefficients of a Polynomial which is widely used in solving problems in algebra.
In this article, we will learn about the Zeroes of a Polynomial, the Coefficients of a Polynomial, and their relation in detail.
## Zeroes of a Polynomial
For any polynomial f(x) zeros of the polynomial are defined as the values of the x which substituting in f(x) results in the zero value of the polynomial. They are also called the solution of the polynomial. Mathematically we define the zero of the polynomial as for any polynomial f(x) a is the zero of the polynomial if
f(a) = 0
For example, if the given polynomial is f(x) = x+3 then -3 is the zero of the polynomial as,
f(-3) = -3+3 =0
The number of zeroes of the polynomial depends on the degree of the polynomial, i.e. for a linear polynomial (polynomial with degree 1) we have 1 zero, for a quadratic polynomial (polynomial with degree 2) we have 2 zeros, and so on.
## Coefficients of a Polynomial
A polynomial is a function of x and its powers. It can be linear, quadratic, and others. A coefficient of a polynomial is the numerical value related to each x and its power in the polynomial.
For example, in the linear polynomial ax + b the coefficient of x is ‘a’ similarly, in px2 + qx + r the coefficient of x is ‘p’ and the coefficient of x is ‘q’
The coefficient of a polynomial can be any real number, fractions, or decimals and they can also be imaginary numbers.
## Relationship between Zeros and Coefficients of a Polynomial
We know that zeros of any polynomial are the points where the graph of the polynomial cuts the x-axis. These zeros can also be found using the coefficients of different terms in a polynomial. Let’s look at the relationship between the zeros and coefficients of the polynomials for various types of polynomials.
The following image shows the relationship between Zeros and Coefficients of a Polynomial.
## Linear Polynomial
A linear polynomial, in general, is defined by,
y = ax + b
We know, for zeros we need to find the points at which y = 0. Solving this general equation for y = 0.
y = ax + b
⇒ 0 = ax + b
⇒ x = -b/a
This gives us the relationship between zero and the coefficient of a linear polynomial.
In general for a linear equation y = ax + b, a ≠0, the graph of ax + b is a straight line that cuts the x-axis at (-b/a, 0)
Example: Find the zeros of the linear polynomial.
y = 4x + 2
Solution:
Given Equation y = 4x + 2
Here,
• a = 4
• b = 2
So, by the formula mentioned above the zero will occur at (-b/a, 0) that is (-2/4, 0) or (-1/2, 0)
Let’s verify this zero using Graphical Method.
Given Equation
y = 4x + 2
In intercept Form
x/(-1/2) + y/(2) = 1
Now we know the intercepts on the x and y-axis.
Quadratic polynomials are polynomials with a degree of 2. We can find the zeroes of the quadratic polynomial using various methods such as, and along with
• Factorization Method
As Quadratic Polynomial has the highest degree of 2, there exist 2 zeros of the quadratic polynomial.
The relationship between the zeros of the quadratic polynomial and the coefficient of the quadratic polynomial is,
For any polynomial P(x) = ax2 + bx + c if the zeroes of the quadratic polynomial are α, and β then,
• Sum of the zeroes (α + β) = – Coefficient of x / Coefficient of x2 = -b/a
• Product of the zeroes (αβ) = Constant term / Coefficient of x2 = c/a
Let’s check the relationship between zeros and coefficients of a quadratic polynomial with the help of an example.
Example: Find the zeros of the polynomial, P(x) = 2x2 – 8x + 6
Solution:
P(x) = 2x2 -8x + 6
⇒ P(x) = 2x2 – 6x – 2x + 6
⇒ P(x) = 2x(x – 3) -2(x – 3)
⇒ P(x) = 2(x – 1)(x – 3)
So the zeroes of the polynomial are,
x – 1 = 0
⇒ x = 1
And x – 3 = 0
⇒ x = 3
• Sum of Zeros = 1 + 3 = 4
• Product of Zeros = 1 Ă— 3 = 3
Using the relationship as discussed above.
Given equation,
2x2 -8x + 6 = 0 comparing with ax2 + bx + c = 0
a = 2, b = -8, and c = 6
• Sum of Roots = -b/a = -(-8)/2 = 8/2 = 4
• Production of the roots = c/a = 6/2 = 3
Thus, the relationship between the zeros of the quadratic polynomial and the coefficient of the quadratic polynomial holds true.
## Cubic Polynomial
A cubic polynomial is a polynomial of degree 3 and since it has its highest degree as 3, there exist three zeros of a cubic polynomial. Let’s suppose the zeros of the polynomials ax3 + bx2 + cx + d = 0 are p, q, and r, the relationship between the zeros and polynomials and the coefficient of the polynomial will be given as
Given Cubic Polynomials,
ax3 + bx2 + cx + d = 0
which has roots x = p, q and r
• Sum of the Zeroes (p + q+ r) = – Coefficient of x2/ coefficient of x3 = -b/a
• Sum of the product of the Zeroes (pq + qr + pr) = Coefficient of x/Coefficient of x3 = c/a
• Product of the zeroes (pqr) = – Constant Term/Coefficient of x3 = -d/a
## Examples of Relationship Between Zeroes and Coefficients of Polynomials
Example 1: Find the sum of the roots and the product of the roots of the polynomial x3 -2x2 – x + 2.
Solution:
Given Polynomial,
x3 -2x2 – x + 2
comparing with ax3 + bx2 + cx + d = 0
a = 1, b= -2, c = -1, and d = 2
Sum of the roots (p + q+ r) = – Coefficient of x2/ coefficient of x3
= -b/a
= -(-2)/1 = 2
Product of the roots (pqr) = – Constant Term/Coefficient of x3
= -d/a
= -2/1 = -2
Example 2: Find the sum and product of the zeros of the quadratic polynomial 6x2 + 18 = 0
Solution:
Given Polynomial 6x2 + 18 = 0
It can be also written as, 6x2 + 0x + 18 = 0
Comparing with ax2 + bx + c = 0
a = 6, b = 0, and c = 18
Sum of Zeroes = – Coefficient of x/ Coefficient of x2
= -b/a
= -0/6
= 0
Product of the Zeroes = Constant term / Coefficient of x2
= c/a
= 18/6
= 3
Example 3: For the given polynomial ax2 + bx + 1 = 0. Its roots are -1 and 3. Find the values of a and b.
Solution:
Let m and n be the roots of the quadratic equation ax2 + bx + 1 = 0
Here,
• m = -1
• n = 3
We know that,
m + n = -b/a
⇒ -1 + 3 = -b/a
⇒ -b/a = 2…(i)
And m.n = c/a
⇒ (-1)(3) = 1/a
⇒ -3 = 1/a
⇒ a = -1/3…(ii)
from (i) we get,
-b/a = 2
⇒ b = -2a
⇒ b = -2(-1/3) = 2/3
## FAQs on Relationship Between Zeroes and Coefficients of Polynomials
### Q1: What does the Degree of a Polynomial mean?
The degree of the polynomial is the highest power of the independent variable it tells us how many time the curve cuts the x-axis as linear polynomial in x cut the x-axis once, whereas the quadratic polynomial in x cut the x-axis twice, etc.
### Q2: What is the Relationship Between Coefficient of Polynomials and Roots of Polynomials?
Their is a definite relationship between coefficient of polynomials and roots of polynomials for a quadratic polynomial ax2 + bx + c = 0 if the zeroes of the polynomial are p, and q then,
• Sum of the roots (p+q) = -b/a
• Product of the roots(pq) = c/a
### Q3: What Is the Relationship Between Zeroes and Coefficients of a Cubic Polynomial?
For the cubic polynomial, ax3 + bx2 + cx + d =0. The relationship between zeroes and coefficients of a cubic polynomial is,
• Sum of Zeroes = – Coefficient of x2/ coefficient of x3 = = -b/a
• Sum of product of Zeroes = – Coefficient of x/Coefficient of x3 = c/a
• Product of Zeroes = – Constant term/Coefficient of x = -d/a
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# Difference between revisions of "2005 AIME I Problems/Problem 5"
## Problem
Robert has 4 indistinguishable gold coins and 4 indistinguishable silver coins. Each coin has an engraving of one face on one side, but not on the other. He wants to stack the eight coins on a table into a single stack so that no two adjacent coins are face to face. Find the number of possible distinguishable arrangements of the 8 coins.
## Solution 1
There are two separate parts to this problem: one is the color (gold vs silver), and the other is the orientation.
There are ${8\choose4} = 70$ ways to position the gold coins in the stack of 8 coins, which determines the positions of the silver coins.
Create a string of letters H and T to denote the orientation of the top of the coin. To avoid making two faces touch, we cannot have the arrangement HT. Thus, all possible configurations must be a string of tails followed by a string of heads, since after the first H no more tails can appear. The first H can occur in a maximum of eight times different positions, and then there is also the possibility that it doesn’t occur at all, for $9$ total configurations. Thus, the answer is $70 \cdot 9 = \boxed{630}$.
## Solution 2
We can imagine the $8$ coins as a string of $0\text{'s}$ and $1\text{'s}$. Because no $2$ adjacent coins can have $2$ faces touching, subsequent to changing from $0$ to $1$, the numbers following $1$ must be $1\text{'s}$; therefore, the number of possible permutations if all the coins are indistinguishable is $9$ (there are $8$ possible places to change from $0$ to $1$ and there is the possibility that there no change occurs). There are $\binom 8 4$ possibilities of what coins are gold and what coins are silver, so the solution is $\boxed{9\cdot \binom 8 4=630}$.
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# Unit 9: First Order Dierential Equations with Applications
First Order Dierential Equations By a dierential equation, we mean an equation which involves the derivative of some unknown functon, say y , and the task at hand is to nd an explicit expression for y . We will study two types of rst order dierential equation, namely separable, and linear dierential equations ( shortened to D.E.s for the obvious reason)). Each will have its own prescribed method of solution.
Denition 9.1. By a separable dierential equation we will mean an equation which may be put into one of the following forms: dy f (x) dy g (y ) dy = or = or = f (x)g (y ) dx g (y ) dx f (x) dx To solve a dierential equation is to nd all possible solutions ( remember C in integration?). Strictly speaking we have already dealt with some D.E.s, as the following example will show. We rst outline some basic steps for solving such a problem.
Steps For Solving a separable Dierential Equation: Assuming that our D.E. is given in terms of being clear:
dy dx
## with the generalization
(1) Recognize the problem as a separable D.E. (2) Seperate the equation into the form (terms not involving x)(dy ) = (terms not involving y)(dx) (3) Integrate both sides, adding one arbitrary constant, say C, to the x-side. This is done since adding an arbitrary constant to both sides after integrating is equivalent to adding just a single arbiratary constant to one side. (4) Solve for y if possible. If we can solve for y we get what is called the general solution, and if we can not solve for y we get what is called an implicit solution.
dy dx
Example 1. Solve:
= x2
Solution: First we recognize that we have a separable D.E. of either of the rst two forms listed above (with g(y)=1), and so proceed: dy dy y = x2 dx = =
x3 3
x2 dx +C
## So we have in this case found the general solution, x3 +C y= 3 2
Example 2. Solve:
dy dx
4x 3y 3
Solution: We rst recognize this as a separable D.E., and then move on: 3y 3 dy 3y 3 dy 3 y4 3 y4
4
= 4xdx = 4xdx
2
## step (2) Begin step(3) step(3) and step (4)
= 4x +C 2 = 2x2 + C =4 2x + 4 C 3 2 3 =8 x2 + 4 C 3 3 =8 x2 + C 3 =
4 2
y4 y4 y4 y
8 2 x 3
+C
4
## So once again we have a general solution, namely y =
dy Example 3. Solve: ey dx = 4x3 + 2x
8 2 x 3
+C
Solution: We rst recognize this as a separable D.E., and then move on: ey dy ey dy ey ln(ey ) y = (4x3 + 2x)dx = (4x3 + 2x)dx = x4 + x2 + C = ln(x4 + x2 + C ) = ln(x4 + x2 + C ) step (2) Begin step(3) step(3) and step (4)
## So once again we have a general solution, namely y = ln(x4 + x2 + C )
Example 4. Solve:
dy dx
4x3 3x y +sin y
Solution:We rst recognize this as a separable D.E., and then move on: (y + sin y )dy (y + sin y )dy
y2 2
= (4x3 3x)dx
step (2)
## = (4x3 3x)dx Begin step(3) = x4 3 x2 + C 2 = x4 3 x2 + C 2 step(3) and step (4)
cos y
1 y 2 cos y 2
Here, we see that we can not solve explicitly for y and so we have an implicit solution, namely 1 y 2 cos y = x4 3 x2 + C . 2 2
Side Conditions and D.E.s Recall that we used side conditions with integration to allow the constant of integration, C, to be found. Similarily we may use such conditions to solve for C here.
## Example 5. Solve: Solve:
dy dx
y x2
given that y ( 1 ) = e. 2
) y( 1 2
Solution: We solve the D.E. as usual, and then use the side condition = e, to solve for C.:
1 dy y
= =
1 dx x2
step (2)
1 dy y
x2 dx Begin step(3)
x 1 1
ln(y ) =
+C
## step(3) and step (4)
1 ln(y ) = x +C
eln(y) y We now use the fact that y ( 1 ) = e to get: 2 e e3 C Substituting C = 3 into (*) we get: y y y
= e x +C = e x eC
1
(*)
= e2 eC = eC =3 multiplying by e2
= e x e3 = e x +3 = e3 x
1 1 1
So here we have the specic solution, y = e3 x We now move on to the second type of D.E. mentioned above, linear differential equations.
Denition 9.2. By a linear dierential equation we will mean a D.E. which can be put into the following form: dy + p(x)y = q (x) dx where p(x) and q (x) are expressions that may or may not involve x, but very importantly they do not involve y .
The approach we learn here for solving linear D.E.s is much dierent from the approach we took to separable D.E.s. In order to give our method 5
## we need the following denition.
Denition 9.3. Given a linear D.E. dy + p(x)y = q (x) dx we dene the integrating factor (x) of the D.E. by: (x) = e Note: when calculating zero.
p(x)dx
## p(x)dx here we set the integration constant C to
The use of the integration factor is illustrated below, but often these steps are skipped ( see next theorem). We begin with a linear D.E.: dy + p(x)y = q (x) dx e
p(x)dx dy dx
+e
p(x)dx
p(x)y
= q (x)e = q (x)e =
## p(x)dx p(x)dx p(x)dx
- multiply both sides by (x) -recognize the equivalence of the left hand sides -integrating both sides
p(x)dx
d (ye p(x)dx ) dx
ye
p(x)dx
q (x)e
dx
= e
p(x)dx
q (x)e
## dx -multiply both sides by (x)
p(x)dx
1 q (x)(x)dx -use the fact that (x) = e y = ( x) So we have the general solution of the problem. We now be state the above result:
dy dx
+ p(x)y = q (x)
p(x)dx
## , the general solution is given by:
y= Example 6. Solve:
dy dx 2 +x y=
1 (x)
q (x)(x)dx
cos(5x) x2
Solution: By the previous theorem we need only nd (x), and then apply the formula. x) 2 and q (x) = cos(5 so we have p(x) = x x2 2 1 p(x)dx = x dx = 2 x dx = 2 ln(x), so the integrating factor is given by 2 (x) = e p(x)dx = e2 ln(x) = eln(x ) = x2 Substituting into the general solution formula we then have: 1 q (x)(x)dx y = ( x) = = =
1 x2 1 x2 1 1 ( x2 5 cos(5x) 2 x dx x2
## 1 C = 5x 2 sin(5x) + x2 ) So we have the general solution:y =
sin(5x) 5x2
C ) x2
Example 7. Solve:
dy dx
2y = xe3x
Solution: We need only nd (x), p(x) and q (x), and then apply the formula. p(x) = 2 and q (x) = xe3x so we have p(x)dx = So the integrating factor is given by (x) = e
p(x)dx
2dx = 2x
= e2x
= =
1 (x) 1 e2x
## q (x)(x)dx e2x xe3x dx xex dx
= e2x
= e2x (xex ex + C ) (use integration by parts to get this) = xe3x e3x + Ce2x = (x 1)e3x + Ce2x So we have the general solution:y = (x 1)e3x + Ce2x We now introduce a subtopic of dierential equations, exponential growth and decay. Exponential functions are very useful when it comes to modeling many real world phenomena. Any such model will involve a dierential equation of the form dy = ky (which is separable), we will provide an easier method dt of solving such an equation. Theorem 9.5. The solution to the dierential equation y = y0 at t = 0 is given by y = y0 ekt Example 8. Suppose it is known that the cells of a given bacterial culture divide every 3.5 hours. If there are 1000 cells in a dish to begin with, how many will there be after 12 hours? Solution: Recognizing that this is an exponential growth problem, we let y = the number of cells present at time =t and so y = y0 ekt (since this is an exponential growth problem), and y0 = 500 (this is given) The only variable left to determine is k , and since the information provided 8
dy dt
= ky , subject to
that the bacterial culture divides every 3.5 hours has not been used yet, it makes sense that this is what we should use to nd k , indeed: The number of cells doubles every 3.5 hours, so after 3.5 hours there should be 1000 cells 1000 = 500ek(3.5) 2 = ek(3.5) ln(2) = (3.5)k ln(2) 0.198042 3.5
k=
Substituting this into the solution formula y = y0 ekt we have: y = 500e(0.198042)t Therefore after 12 hours there wil be 500e(0.198042)12 5383 cells in the dish. Denition 9.6. The Half-life of a radioactive element is the time required for half of the radioactive nuclei present in a sample to decay. Amazingly enough the half-life of an element is totally independent of the number of nuclei present initially. This can be easily seen: Let y0 be the number of radioactive nuclei present initially, then the number y of nuclei present at time t will be given by: y = y0 ekt Since we are looking for half-life we wish to know when 1 y0 ekt = y0 2 cancelling y0 s we have: ekt = 9 1 2
## 1 kt = ln( ) 2 kt = ln(2) ln(2) k
t=
Thus the half-life depends only on k and not on y0 at all, the formula above is worth noting for future use: Half lif e =
ln(2) k
Example 9. The radioactive lifetime of plutonium-210 is extremely short and so is measured in days rather than years. The number of atoms remaining after t days with an initial amount of y0 radioactive atoms is given by: y = y0 e(4.9510 Find the half-life of plutonium-210. Solution: Using the formula above we have that Half lif e = =
ln(2) k ln(2) 4.95103
3 )t
140 Therefore the half-life of plutonium-210 is approximately 140 days. Example 10. Scientists who do carbon-14 dating use a value of 5700 years for its half-life. Find the age of an object that has been excavated and found to have 90 Solution: Using the equation y = y0 ekt we see that we must nd two things: (a) the value of k 90 9 (b) the value of t for which y0 ekt = 100 y0 , i.e. when ekt = 10
10
## (a) Find the value of k: Using the half-life equation we have k = =
ln(2) half lif e ln(2) 5700
4 )t
9 10
## = 0.9 (1.2 104 )t = ln(0.9) t =
ln(0.9) (1.2104 )
11
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# How do you number a coordinate grid?
## How do you number a coordinate grid?
The numbers on a coordinate grid are used to locate points. Each point can be identified by an ordered pair of numbers; that is, a number on the x-axis called an x-coordinate, and a number on the y-axis called a y-coordinate. Ordered pairs are written in parentheses (x-coordinate, y-coordinate).
## How do you pair coordinates?
An ordered pair contains the coordinates of one point in the coordinate system. A point is named by its ordered pair of the form of (x, y). The first number corresponds to the x-coordinate and the second to the y-coordinate. To graph a point, you draw a dot at the coordinates that corresponds to the ordered pair.
How are the 4 areas of a coordinate grid labeled?
The origin is at 0 on the x-axis and 0 on the y-axis. The intersecting x- and y-axes divide the coordinate plane into four sections. These four sections are called quadrants. Quadrants are named using the Roman numerals I, II, III, and IV beginning with the top right quadrant and moving counter clockwise.
### How do you read a grid coordinate?
When giving a four-figure grid reference, you should always give the eastings number first and the northings number second, very much like when giving the reading of a graph in school, where you give the x coordinate first followed by the y.
### How do you find the grid coordinates on a map?
Make sure you use the correct scale. First, locate the grid square in which the point (for example, Point A, Figure 2) is located (the point should already be plotted on the map). The number of the vertical grid line on the left (west) side of the grid square is the first and second digits of the coordinates.
What is the six digit grid coordinate for 100 meters?
On the bottom scale, the 100 meter mark nearest the vertical grid line provides the third digit, 5. On the vertical scale, the 100 meter mark nearest Point A provides the sixth digit, 3. Therefore, the six-digit grid coordinate is 115813.
#### Do you have to use imaginary lines to find coordinates?
You do not have to use imaginary lines; you can find the exact coordinates using a Coordinate Scale and Protractor. This device has two coordinating scales, 1:25,000 meters and 1:50,000 meters. Make sure you use the correct scale.
#### Why are the coordinates on my Garmin not working?
When given coordinates for a GPS position, you may find the format does not match the device’s format or that the coordinates have too many numbers. If the formatting of the coordinates does not match the format on the Garmin Device, confirm the format of the provided coordinates. Maps will often provide the position format they use.
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# Question Video: Finding the Quotient of Two Complex Numbers in Polar Form Mathematics
Given that π§β = 20 (cos (π/2) + π sin (π/2)) and π§β = 4 (cos (π/6) + π sin (π/6)), find π§β/π§β in polar form.
01:52
### Video Transcript
Given that π§ one equals 20 times cos π by two plus π sin π by two and π§ two equals four times cos π by six plus π sin π by six, find π§ one over π§ two in polar form.
We have two complex numbers, both given in polar form, and we have to find their quotient, also in polar form. Writing numbers in polar form makes it very easy to find their product or quotient. To find their product, weβd simply have to multiply their moduluses together to find the modulus of the product and add their arguments together to find the argument of the product.
But as we want to find the quotient, weβll have to divide one modulus by the other and subtract one argument from the other. So letβs write this as 20 times cos π by two plus π sin π by two over four times cos π by six plus π sin π by six. And we want to express this in polar form, so that means π times cos π plus π sin π.
We find the value of π, which is the modulus of our answer, by dividing the modulus in the numerator, 20, by the modulus in the denominator, four. This gives us a modulus of five.
Now we simply have to find the argument of our answer, π. We do this by finding the argument in the numerator and subtracting the argument in the denominator. So π is equal to π by two minus π by six, which is π by three.
So our answer is five times cos π by three plus π sin π by three. And notice how having all three numbers written in polar form made finding this quotient very easy.
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Part 9: Geometry and Measurement
# 9.1: Angles
Notes
Angles measure the rotation in degrees (denoted by a small circle: º) about a fixed point. For example, if you’re facing north and turn 90º to the right, you’ll be facing east. Another 90º turn to the right (or 180º total) and you’ll be facing south. Then, another 90º turn to the right (or 270º total) and you’ll be facing west. Then, another 90º turn to the right (or 360º total) and you’ll be back facing north again.
##### Lines
An angle can also measure between two lines that meet at a point (called the vertex). The angle measures how much one line would have to rotate about the vertex to line up with the other line.
Four angles are formed when two lines cross. Each pair of opposite angles is equal:
Angles a and c are opposite angles, as are angles b and d:
• angle a = angle c
• angle b = angle d
Also, angles a and b add to 180º and are called supplementary angles. Other supplementary angles here are angles b and c, angles c and d, and angles d and a.
Eight angles are formed when one line crosses two parallel lines. Each pair of corresponding angles is equal.
Angles a and e are corresponding angles, as are angles b and f, angles c and g, and angles d and h. The two shorter lines are parallel, so:
• angle a = angle c = angle e = angle g
• angle b = angle d = angle f = angle h
##### Polygons
Polygons are closed shapes made from (straight, non-crossing) line segments. For example, a polygon made from three line segments is a triangle, a polygon made from four line segments is a quadrilateral, a polygon made from five line segments is a pentagon, and so on. The interior angles in a polygon add up to (n – 2) x 180º, where n is the number of sides. So, the interior angles of a triangle sum to 180º, the interior angles of a quadrilateral sum to 360,º the interior angles of a pentagon sum to 540,º and so on.
If the interior angles of a polygon are all equal (and the side lengths are also all equal), then the polygon is called regular. A regular triangle has 60º interior angles and is called an equilateral triangle. A regular quadrilateral has 90º interior angles and is called a square.
Other types of triangle are:
• an isosceles triangle has (at least) two sides that are the same length
• a scalene triangle has all different side lengths
Other types of quadrilateral are:
• a rectangle has equal (90º) angles
• a rhombus (or diamond) has equal side lengths
• a parallelogram has opposite sides that are parallel
##### Right angles
An angle of 90º is called a right angle and a triangle with a right angle is called a right triangle. The opposite side of the right angle in a right triangle is called the hypotenuse. The side lengths of a right triangle are related according to the Pythagorean theorem: the square of the hypotenuse is equal to the sum of squares of the other sides. For example:
Pythagorean theorem: 52 = 32 + 42 (correct since 25 = 9 + 16).
The Pythagorean theorem can be used to calculate the length of one unknown side of a right triangle. For example:
• If the hypotenuse above was unknown, x = √(32 + 42) = √25 = 5 cm.
• If the left side length was unknown, x = √(52 – 42) = √9 = 3 cm.
• If the bottom side length was unknown, x = √(52 – 32) = √16 = 4 cm.
The video below works through some examples of calculating angles and applying the Pythagorean theorem.
Practice Exercises
Do the following exercises to practice calculating angles and applying the Pythagorean theorem.
1. Consider the following diagram in which the two lines sloping upwards to the right are parallel.
## License
Mathematics For Elementary Teachers Copyright © 2023 by Iain Pardoe is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License, except where otherwise noted.
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# Introduction to Differential Equations
## Section1.7Numerical methods: Euler’s method
Note: 1 lecture, can safely be skipped, §2.4 in [EP], §8.1 in [BD]
Unless $$f(x,y)$$ is of a special form, it is generally very hard if not impossible to get a nice formula for the solution of the problem
\begin{equation*} y' = f(x,y), \qquad y(x_0) = y_0 . \end{equation*}
If the equation can be solved in closed form, we should do that. But what if we have an equation that cannot be solved in closed form? What if we want to find the value of the solution at some particular $$x\text{?}$$ Or perhaps we want to produce a graph of the solution to inspect the behavior. In this section we will learn about the basics of numerical approximation of solutions.
The simplest method for approximating a solution is Euler’s method 1 . It works as follows: Take $$x_0$$ and compute the slope $$k = f(x_0,y_0)\text{.}$$ The slope is the change in $$y$$ per unit change in $$x\text{.}$$ Follow the line for an interval of length $$h$$ on the $$x$$-axis. Hence if $$y = y_0$$ at $$x_0\text{,}$$ then we say that $$y_1$$ (the approximate value of $$y$$ at $$x_1 = x_0 + h$$) is $$y_1 = y_0 + h k\text{.}$$ Rinse, repeat! Let $$k = f(x_1,y_1)\text{,}$$ and then compute $$x_2 = x_1 + h\text{,}$$ and $$y_2 = y_1 + h k\text{.}$$ Now compute $$x_3$$ and $$y_3$$ using $$x_2$$ and $$y_2\text{,}$$ etc. Consider the equation $$y' = \nicefrac{y^2}{3}\text{,}$$ $$y(0)=1\text{,}$$ and $$h=1\text{.}$$ Then $$x_0=0$$ and $$y_0 = 1\text{.}$$ We compute
\begin{equation*} \begin{aligned} & x_1 = x_0 + h = 0 + 1 = 1, & & y_1 = y_0 + h \, f(x_0,y_0) = 1 + 1 \cdot \nicefrac{1}{3} = \nicefrac{4}{3} \approx 1.333,\\ & x_2 = x_1 + h = 1 + 1 = 2, & & y_2 = y_1 + h \, f(x_1,y_1) = \nicefrac{4}{3} + 1 \cdot \frac{{(\nicefrac{4}{3})}^2}{3} = \nicefrac{52}{27} \approx 1.926. \end{aligned} \end{equation*}
We then draw an approximate graph of the solution by connecting the points $$(x_0,y_0)\text{,}$$ $$(x_1,y_1)\text{,}$$ $$(x_2,y_2)\text{,....}$$ For the first two steps of the method see Figure 1.16.
More abstractly, for any $$i=0,1,2,3,\ldots\text{,}$$ we compute
\begin{equation*} x_{i+1} = x_i + h , \qquad y_{i+1} = y_i + h\, f(x_i,y_i) . \end{equation*}
The line segments we get are an approximate graph of the solution. Generally it is not exactly the solution. See Figure 1.17 for the plot of the real solution and the approximation.
We continue with the equation $$y' = \nicefrac{y^2}{3}\text{,}$$ $$y(0)=1\text{.}$$ Let us try to approximate $$y(2)$$ using Euler’s method. In Figures Figure 1.16 and Figure 1.17 we have graphically approximated $$y(2)$$ with step size 1. With step size 1, we have $$y(2) \approx 1.926\text{.}$$ The real answer is 3. We are approximately 1.074 off. Let us halve the step size. Computing $$y_4$$ with $$h=0.5\text{,}$$ we find that $$y(2) \approx 2.209\text{,}$$ so an error of about 0.791. Table 1.7.1 gives the values computed for various parameters.
The difference between the actual solution and the approximate solution is called the error. We usually talk about just the size of the error and we do not care much about its sign. The point is, we usually do not know the real solution, so we only have a vague understanding of the error. If we knew the error exactly ... what is the point of doing the approximation?
Notice that except for the first few times, every time we halved the interval the error approximately halved. This halving of the error is a general feature of Euler’s method as it is a first order method. There exists an improved Euler method, see the exercises, which is a second order method. A second order method reduces the error to approximately one quarter every time we halve the interval. The meaning of “second” order is the squaring in $$\nicefrac{1}{4} = \nicefrac{1}{2} \times \nicefrac{1}{2} = {(\nicefrac{1}{2})}^2\text{.}$$
To get the error to be within 0.1 of the answer we had to already do 64 steps. To get it to within 0.01 we would have to halve another three or four times, meaning doing 512 to 1024 steps. That is quite a bit to do by hand. The improved Euler method from the exercises should quarter the error every time we halve the interval, so we would have to approximately do half as many “halvings” to get the same error. This reduction can be a big deal. With 10 halvings (starting at $$h=1$$) we have 1024 steps, whereas with 5 halvings we only have to do 32 steps, assuming that the error was comparable to start with. A computer may not care about this difference for a problem this simple, but suppose each step would take a second to compute (the function may be substantially more difficult to compute than $$\nicefrac{y^2}{3}$$). Then the difference is 32 seconds versus about 17 minutes. We are not being altogether fair, a second order method would probably double the time to do each step. Even so, it is 1 minute versus 17 minutes. Next, suppose that we have to repeat such a calculation for different parameters a thousand times. You get the idea.
Note that in practice we do not know how large the error is! How do we know what is the right step size? Well, essentially we keep halving the interval, and if we are lucky, we can estimate the error from a few of these calculations and the assumption that the error goes down by a factor of one half each time (if we are using standard Euler).
Exericse: In the table above, suppose you do not know the error. Take the approximate values of the function in the last two lines, assume that the error goes down by a factor of 2. Can you estimate the error in the last time from this? Does it (approximately) agree with the table? Now do it for the first two rows. Does this agree with the table?
Let us talk a little bit more about the example $$y' = \nicefrac{y^2}{3}\text{,}$$ $$y(0) = 1\text{.}$$ Suppose that instead of the value $$y(2)$$ we wish to find $$y(3)\text{.}$$ The results of this effort are listed in Table 1.7.2 for successive halvings of $$h\text{.}$$ What is going on here? Well, you should solve the equation exactly and you will notice that the solution does not exist at $$x=3\text{.}$$ In fact, the solution goes to infinity when you approach $$x=3\text{.}$$
Another case where things go bad is if the solution oscillates wildly near some point. The solution may exist at all points, but even a much better numerical method than Euler would need an insanely small step size to approximate the solution with reasonable precision. And computers might not be able to easily handle such a small step size.
In real applications we would not use a simple method such as Euler’s. The simplest method that would probably be used in a real application is the standard Runge–Kutta method (see exercises). That is a fourth order method, meaning that if we halve the interval, the error generally goes down by a factor of 16 (it is fourth order as $$\nicefrac{1}{16} = \nicefrac{1}{2} \times \nicefrac{1}{2} \times \nicefrac{1}{2} \times \nicefrac{1}{2}$$).
Choosing the right method to use and the right step size can be very tricky. There are several competing factors to consider.
• Computational time: Each step takes computer time. Even if the function $$f$$ is simple to compute, we do it many times over. Large step size means faster computation, but perhaps not the right precision.
• Roundoff errors: Computers only compute with a certain number of significant digits. Errors introduced by rounding numbers off during our computations become noticeable when the step size becomes too small relative to the quantities we are working with. So reducing step size may in fact make errors worse. There is a certain optimum step size such that the precision increases as we approach it, but then starts getting worse as we make our step size smaller still. Trouble is: this optimum may be hard to find.
• Stability: Certain equations may be numerically unstable. What may happen is that the numbers never seem to stabilize no matter how many times we halve the interval. We may need a ridiculously small interval size, which may not be practical due to roundoff errors or computational time considerations. Such problems are sometimes called stiff. In the worst case, the numerical computations might be giving us bogus numbers that look like a correct answer. Just because the numbers seem to have stabilized after successive halving, does not mean that we must have the right answer.
We have seen just the beginnings of the challenges that appear in real applications. Numerical approximation of solutions to differential equations is an active research area for engineers and mathematicians. For example, the general purpose method used for the ODE solver in Matlab and Octave (as of this writing) is a method that appeared in the literature only in the 1980s.
### Subsection1.7.1Exercises
#### Exercise1.7.2.
Consider $$\dfrac{dx}{dt} = {(2t-x)}^2\text{,}$$ $$x(0)=2\text{.}$$ Use Euler’s method with step size $$h=0.5$$ to approximate $$x(1)\text{.}$$
#### Exercise1.7.3.
Consider $$\dfrac{dx}{dt} = t-x\text{,}$$ $$x(0)=1\text{.}$$
1. Use Euler’s method with step sizes $$h = 1, \nicefrac{1}{2}, \nicefrac{1}{4}, \nicefrac{1}{8}$$ to approximate $$x(1)\text{.}$$
2. Solve the equation exactly.
3. Describe what happens to the errors for each $$h$$ you used. That is, find the factor by which the error changed each time you halved the interval.
#### Exercise1.7.4.
Approximate the value of $$e$$ by looking at the initial value problem $$y'=y$$ with $$y(0)=1$$ and approximating $$y(1)$$ using Euler’s method with a step size of $$0.2\text{.}$$
#### Exercise1.7.5.
Example of numerical instability: Take $$y' = -5y\text{,}$$ $$y(0) = 1\text{.}$$ We know that the solution should decay to zero as $$x$$ grows. Using Euler’s method, start with $$h=1$$ and compute $$y_1, y_2, y_3, y_4$$ to try to approximate $$y(4)\text{.}$$ What happened? Now halve the interval. Keep halving the interval and approximating $$y(4)$$ until the numbers you are getting start to stabilize (that is, until they start going towards zero). Note: You might want to use a calculator.
The simplest method used in practice is the Runge–Kutta method. Consider $$\frac{dy}{dx}=f(x,y)\text{,}$$ $$y(x_0) = y_0\text{,}$$ and a step size $$h\text{.}$$ Everything is the same as in Euler’s method, except the computation of $$y_{i+1}$$ and $$x_{i+1}\text{.}$$
\begin{equation*} \begin{aligned} & k_1 = f(x_i,y_i) , & & \\ & k_2 = f\bigl(x_i + \nicefrac{h}{2},y_i + k_1 (\nicefrac{h}{2})\bigr) , & & x_{i+1} = x_i + h , \\ & k_3 = f\bigl(x_i + \nicefrac{h}{2},y_i + k_2 (\nicefrac{h}{2})\bigr) , & & y_{i+1} = y_i + \frac{k_1 + 2k_2 + 2k_3 + k_4}{6}\,h , \\ & k_4 = f(x_i + h,y_i + k_3 h) . \end{aligned} \end{equation*}
#### Exercise1.7.6.
Consider $$\dfrac{dy}{dx} = yx^2\text{,}$$ $$y(0)=1\text{.}$$
1. Use Runge–Kutta (see above) with step sizes $$h=1$$ and $$h=\nicefrac{1}{2}$$ to approximate $$y(1)\text{.}$$
2. Use Euler’s method with $$h=1$$ and $$h=\nicefrac{1}{2}\text{.}$$
3. Solve exactly, find the exact value of $$y(1)\text{,}$$ and compare.
#### Exercise1.7.7.
Let $$x' = \sin(xt)\text{,}$$ and $$x(0)=1\text{.}$$ Approximate $$x(1)$$ using Euler’s method with step sizes 1, 0.5, 0.25. Use a calculator and compute up to 4 decimal digits.
Answer.
Approximately: 1.0000, 1.2397, 1.3829
#### Exercise1.7.8.
Let $$x' = 2t\text{,}$$ and $$x(0)=0\text{.}$$
1. Approximate $$x(4)$$ using Euler’s method with step sizes 4, 2, and 1.
2. Solve exactly, and compute the errors.
3. Compute the factor by which the errors changed.
Answer.
a) 0, 8, 12 b) $$x(4) = 16\text{,}$$ so errors are: 16, 8, 4. c) Factors are 0.5, 0.5, 0.5.
#### Exercise1.7.9.
Let $$x' = x e^{xt+1}\text{,}$$ and $$x(0)=0\text{.}$$
1. Approximate $$x(4)$$ using Euler’s method with step sizes 4, 2, and 1.
2. Guess an exact solution based on part a) and compute the errors.
Answer.
a) 0, 0, 0 b) $$x=0$$ is a solution so errors are: 0, 0, 0.
There is a simple way to improve Euler’s method to make it a second order method by doing just one extra step. Consider $$\frac{dy}{dx}=f(x,y)\text{,}$$ $$y(x_0) = y_0\text{,}$$ and a step size $$h\text{.}$$ What we do is to pretend we compute the next step as in Euler, that is, we start with $$(x_i,y_i)\text{,}$$ we compute a slope $$k_1 = f(x_i,y_i)\text{,}$$ and then look at the point $$(x_i+h,y_i + k_1h)\text{.}$$ Instead of letting our new point be $$(x_i+h,y_i + k_1h)\text{,}$$ we compute the slope at that point, call it $$k_2\text{,}$$ and then take the average of $$k_1$$ and $$k_2\text{,}$$ hoping that the average is going to be closer to the actual slope on the interval from $$x_i$$ to $$x_i+h\text{.}$$ And we are correct, if we halve the step, the error should go down by a factor of $$2^2 = 4\text{.}$$ To summarize, the setup is the same as for regular Euler, except the computation of $$y_{i+1}$$ and $$x_{i+1}\text{.}$$
\begin{equation*} \begin{aligned} & k_1 = f(x_i,y_i) , & & x_{i+1} = x_i + h , \\ & k_2 = f(x_i + h,y_i + k_1h) , & & y_{i+1} = y_i + \frac{k_1+k_2}{2}\,h . \end{aligned} \end{equation*}
#### Exercise1.7.10.
Consider $$\dfrac{dy}{dx} = x+y\text{,}$$ $$y(0)=1\text{.}$$
1. Use the improved Euler’s method (see above) with step sizes $$h=\nicefrac{1}{4}$$ and $$h=\nicefrac{1}{8}$$ to approximate $$y(1)\text{.}$$
2. Use Euler’s method with $$h=\nicefrac{1}{4}$$ and $$h=\nicefrac{1}{8}\text{.}$$
3. Solve exactly, find the exact value of $$y(1)\text{.}$$
4. Compute the errors, and the factors by which the errors changed.
Answer.
a) Improved Euler: $$y(1) \approx 3.3897$$ for $$h=\nicefrac{1}{4}\text{,}$$ $$y(1) \approx 3.4237$$ for $$h=\nicefrac{1}{8}\text{,}$$ b) Standard Euler: $$y(1) \approx 2.8828$$ for $$h=\nicefrac{1}{4}\text{,}$$ $$y(1) \approx 3.1316$$ for $$h=\nicefrac{1}{8}\text{,}$$ c) $$y = 2e^x-x-1\text{,}$$ so $$y(2)$$ is approximately $$3.4366\text{.}$$ d) Approximate errors for improved Euler: $$0.046852$$ for $$h=\nicefrac{1}{4}\text{,}$$ and $$0.012881$$ for $$h=\nicefrac{1}{8}\text{.}$$ For standard Euler: $$0.55375$$ for $$h=\nicefrac{1}{4}\text{,}$$ and $$0.30499$$ for $$h=\nicefrac{1}{8}\text{.}$$ Factor is approximately $$0.27$$ for improved Euler, and $$0.55$$ for standard Euler.
Named after the Swiss mathematician Leonhard Paul Euler 2 (1707–1783). The correct pronunciation of the name sounds more like “oiler.”
en.wikipedia.org/wiki/Euler
For a higher quality printout use the PDF version: https://www.jirka.org/diffyqs/diffyqs.pdf
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Lesson 18 Section 3
# RATIO AND PROPORTIONMixed ratio
Back to Section 1
We have seen that
means
Three times 6 plus half of 6.
Three times 6 is 18; half of 6 is 3; therefore, three and a half times 6
is 21.
In other words, the ratio of 21 to 6 is:
21 is three and a half times 6.
That is called a mixed ratio.
6. What is a mixed ratio? 25 to 10 A ratio in which the larger number is not an exact multiple of the smaller. The larger number will be a multiple of the smaller, plus a part of it.
Example 1. What ratio has 25 to 10?
Answer. 25 is composed of two 10's, plus a remainder of 5.
25 = 20 + 5.
The remainder 5 is a part of 10, namely half. Therefore we say,
"25 is two and a half times 10."
Two times 10 is 20; half of 10 is 5; 20 plus 5 is 25.
We always say that a larger number is so many times a smaller number. 25 is two and a half times 10.
Example 2. What ratio has 14 to 4 -- that is, 14 is how many times 4?
Answer. Again, we can decompose 14 into a multiple of 4 plus a remainder:
14 = 12 + 2.
14 is made up of three 4's with remainder 2, which is half of 4. Therefore we say,
"14 is three and a half times 4."
Again, we say that a larger number is so many times a smaller. And when the first term is larger, the word "times" will immediately precede the second term. "14 is . . . times 4"
Example 3. What ratio has 50 to 40?
Answer. 50 is one and a quarter times 40.
For, 50 = 40 + 10.
50 contains 40 one time with remainder 10, which is a quarter of 40.
What is most important is that we see that we can always express in words the relationship -- the ratio -- of any two natural numbers.
Example 4. What ratio has 7 to 3?
7 = 6 + 1,
then
"7 is two and a third times 3."
The remainder 1 is a third of 3.
Example 5. What ratio has 8 to 3?
8 = 6 + 2,
then
"8 is two and two thirds times 3."
The remainder 2 is two thirds of 3.
Example 6. In a survey there were 5 Yes's for every 2 No's. There were 406 No's. How many Yes's were there?
Solution 1. The ratio of Yes's to No's was 5 to 2. What ratio has 5 to 2?
"5 is two and a half times 2."
5 = 4 + 1.
The number of Yes's, then, is two and half times the number of No's -- 406.
Two times 406 is 812. Half of 406 is 203. 812 + 203 = 1,015.
Solution 2. Proportionally,
Yes'sNo's = 52 = ? 406
406 = 203 × 2. Therefore, the missing term is 203 × 5.
200 × 5 + 3 × 5 = 1000 + 15 = 1,015.
Example 7. 8 out of 50 students got A on an exam. Assuming that same ratio of the part to the whole:
a) How many would have got A if 250 students had taken the exam?
Solution. 8 50 = ? 250
250 is five times 50. Therefore, five times 8 -- 40 -- students would have got A.
b) How many would have got A if 75 students had taken the exam?
Solution. 8 50 = ? 75
What number times 50 is 75? That is, what ratio has 75 to 50?
75 = 50 + 25 = 50 + Half of 50.
75 is one and a half times 50. Therefore, one and a half times 8 -- 12 -- students would have got A.
Example 8. If 6 workers can paint 4 rooms in 5 hours, how long will it take 15 workers to paint 14 rooms?
Solution. We must find out how many rooms 15 workers could paint in ONE hour. Why? Because that will tell us the number of rooms per hour. (Lesson 11.) It will then be a simple matter to know how many hours will be needed to paint 14 rooms.
(This is the standard procedure in what is called a "jobs" problem.)
Now, 6 workers can paint 4 rooms in 5 hours. How many rooms could 15 workers paint in 5 hours?
15 workers are two and half times 6 workers. (15 = 12 + 3. Compare Example 1.) Therefore in 5 hours they could paint two and a half times as many rooms.
Two and a half times 4 = Two times 4 + Half of 4 = 8 + 2 = 10 rooms.
15 workers can paint 10 rooms, then, in 5 hours. This implies that they can paint 2 rooms per hour. Therefore to paint 14 rooms will require 7 hours.
At this point, please "turn" the page and do some Problems.
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Circular motion and gravitation
# Circular motion
If an object moves on a circle, we call that object's motion circular motion. A ball on a string that swing in a circle, motion of the moon around the earth are examples of circular motion. We can also consider the motion of an object on a curved path as circular motion, if the path is an arc of a circle, because the object is in circular motion over that part of the circle.
Since the direction of motion of an object in circular motion changes continuously, the direction of velocity is different at different times (or points). At a given point on the path, the direction of velocity is tangent to the circle at that point.
#### Uniform circular motion
If the speed of an object in circular motion is constant, we call that uniform circular motion. Remember that, speed is the magnitude of velocity, although speed is constant, velocity is not constant, because the direction of velocity is changing. In this page, we focus our attention on uniform circular motion.
#### Centripetal force
According to Newton's first law, every object that is in motion continues its motion on a straight line when there is no external force acting on. Circular motion is not in a straight line. So, in order to make an object's motion circular (non-straight line), there must be a net external force applied on the object. That net external force is called centripetal force, $\vec F_c$. To keep an object in a circle, the applied force should point to the center of the circle at each and every points. So, the direction of the centripetal force is towards the center of the circle. Moon revolves around the earth in a nearly circular orbit, the force of gravity on the moon by the earth is the centripetal force of the moon.
#### Acceleration in uniform circular motion
Since there is a net force called centripetal force is acting on an object in circular motion, there must be an acceleration of the object according to Newton's second law. That acceleration, we call centripetal acceleration, $\vec a_c$. Since the direction of acceleration is the direction of the net force on the object, direction of centripetal acceleration is also towards the center of the circle.
Note that an object in uniform circular motion has an acceleration although the speed is constant, because velocity of the object is not constant.
Magnitude of the centripetal acceleration is (for derivation, refer the textbook),
$a_c=\dfrac{v^2}{r}$
where $v$ is the speed of the object and $r$ is the radius of the circle.
If $m$ is the mass of the object, then the centripetal force is,
$F_c=ma_c=\dfrac{mv^2}{r}$ [by Newton's second law]
#### Period of circular motion
If an object, continues the motion on a circle, it repeats the motion in certain time interval, that interval is the period of the motion. Period is the time taken by the object to complete one circle. In one period of time, the object travels a distance equal to the circumference of the circle, that is $2\pi r$.Therefore, the period, $T$ of circular motion is,
$T=\dfrac{2\pi r}{v}$
#### Unbanked and banked high way curves
There are two types of highway curves (roads), unbanked or horizontal curve and banked curve. Banked curve is advantageous during the snowy season as it provides better control against vehicles slide off.
#### Unbanked curve:
On an unbanked or horizontal curve, a car or truck follows circular path (or part of a circle). Since the vehicle is in circular motion, there must be an external force acting on it. On a horizontal road, the friction force on the tires by the road provide the required centripetal force. If there is no friction, a car cannot follow a horizontal curve as you know that it is very difficult to make a turn on a flat icy pavement. Therefore, on a banked curve,
$F_c=F_{fr}$
Now, the question is what type of friction provides the centripetal force on an unbanked flat road? is it static or kinetic ? Since friction is the centripetal force on a horizontal curve, the friction force's direction should be towards the center of the circle. But there is no motion of the vehicle in that direction, so the friction is static.
Depends upon the friction between the road and the tires, there is a maximum speed, above which a car cannot follow a curve or make a turn on a flat road. This maximum speed,$v_{max}$ can be obtained by equating centripetal force to the maximum available force of friction:
$\dfrac{mv_{max}^2}{r}=\mu_sF_n$.
If $m$ is the mass of the vehicle, then the normal force on the car, is $F_n=mg$. Substituting this in the above equation, and solving for $v_{max}$, we get,
$v_{max}=\sqrt{\mu_s rg}$
This is the maximum speed at which a car can safely turn on a flat road. If there is ice on the road, then $\mu_s$, decreases that reduces the safe speed. The maximum speed is independent of the mass of the vehicle.
#### Banked curve
A banked curve is tilted above the horizontal by an angle, usually towards the inside of the curve. The angle of the tilt is called banking angle. If you make a circle out of the curved path of a car, it is a horizontal circle. So, for the car to follow the curve, there must be a horizontal force acting on it towards the center of the circle, that is the centripetal force. How can we have a horizontal force on the car so that it can follow the curve?. If you look at the normal force on the car, it is not vertical. That means, there is a horizontal component of the normal force. This component of the normal force acts as the required centripetal force for the car to follow the curve.
Let us take the $x$ axis towards the center of the circle and the $y$ axis vertical. From the geometry of the figure, you can see that the normal force is at an angle $\theta$ from the vertical. Finding the $x$ and the $y$ components of the normal force, we have
$F_{nx}=F_n \sin\theta$ and
$F_{ny}=F_n \cos\theta$
Since the $x$ component of the normal force is the centripetal force, we can write
$F_n \sin\theta=\dfrac{mv^2}{r}$
From the figure, normal force does not balance the force of gravity, $mg$, but the $y$ component of the normal force does. Therefore,
$F_n \cos\theta=mg$.
Dividing the previous equation by this equation, we get,
$\tan \theta = \dfrac{v^2}{rg}$
Solving for $v$,
$v=\sqrt{rg\tan\theta}$
This $v$ is called the design speed or the safe speed to drive on a banked curve. Design speed is independent of the mass of the vehicle.
Note that, when deriving the above equation we ignore the friction force on the tires by the road.
#### Law of universal gravitation:
The law of universal gravitation was published by Isaac Newton in 1687. It states that
"every particle in the universe attracts every other particle with a force, which is proportional to the product of their masses and inversely proportional to the square of the distance between them."
If there are two objects, they attract each other with a force called gravitational force. The magnitude of the gravitational force between two objects with masses $m_1$ and $m_2$ is
$F_g=G\dfrac{m_1m_2}{r^2}$
where $r$ is the distance between the two objects, it is measured from the center to center of the objects; and $G=6.67\times10^{-11}Nm^2/kg^2$ is a constant, called, the gravitational constant. Gravitational force is always attractive. It decreases as the distance between the object increases as $1/r^2$.
In the figure, you see that, object 1 is exerting a gravitational force on object 1 and the object 2 is exerting an equal amount of gravitational force on object 2, but in the opposite direction, according to Newton's third law.
#### Acceleration due to gravity on a planet:
You already learned that near the surface of the earth, acceleration due to gravity, $g$ is $9.8 m/s^2$. Acceleration due to gravity decreases with altitude.
Now, we are going to derive an equation for acceleration due to gravity on a planet or on a celestial object. Let $M_p$ is the mass of the planet or the celestial object. Assume that you want to find the acceleration due to gravity at a point P, that is at a distance $r$ from the planet (this the distance from the center of the planet). To do that, we imagine that there is a test object of mass, $m$ at P.
Let $g$ is the acceleration due to gravity of the object at the point P. So, the force of gravity on the test object is
$F_g=mg$
You can also find the force of gravity, i.e., the gravitational force on the object by the planet using the law of universal gravitation,
$F_g=G\dfrac{mM_p}{r^2}$.
Now, equate the two equations above and solve for $g$, you will get,
$g=\dfrac{GM_p}{r^2}$.
Acceleration due to gravity depends on the planet's mass and the distance from its center. It decreases with distance as $1/r^2$, an inverse square law. On the surface of the moon, $g$ is about one-sixth that of the earth, because of its smaller mass.
### Satellites
Satellites revolve around the earth in nearly circular orbits. Man made satellites are used for various applications from weather forecasting to navigation. Since a satellite is in circular motion, there must be a centripetal force acting on the satellite. That centripetal force is actually the gravitational force on the satellite by the earth.
If $m$ is the mass of a satellite that is revolving around the earth, then the gravitational force on the satellite by the earth is
$F_g=G\dfrac{M_E\,m}{r^2}$.
Since this is the centripetal force, we have
$\dfrac{mv^2}{r}=G\dfrac{M_E\,m}{r^2}$.
where $r$ is the radius of the orbit, i.e., the distance of the satellite from the center of the earth.
Solving for $v$,we get,
$v=\sqrt{\dfrac{GM_E}{r}}$.
In the above equation, there is no $m$, the mass of the satellite, so the speed of a satellite is independent of its mass. Speed depends only on the distance $r$. It varies as $1/r^2$, an inverse square law, closer the satellite to the earth, faster it moves.
#### Geosynchronous satellites:
Satellites that appear to stay at the same spot above the earth are called a geosynchronous satellites. For a satellite to be geosynchronous, it should meet two conditions: (i) the orbital period of the satellite should be same as the earth's rotational period, and (ii) the orbit should be exactly aligned with the equator of the earth. So, the period of a geosynchronous satellite is 1 day, the earth's rotational period. Since a geosynchronous satellite appears to be fixed at one spot above the earth, they are used for satellite television broadcasts and communication purposes. Based on the period, a geosynchronous satellite orbits the earth at a height of approximately 36,000 km with a speed of about 2800 km/h.
#### Weightlessness:
We all feel weight due to a normal force acting on us. When there is no normal force, we feel weightlessness. You can realize this by standing in an elevator. When the elevator accelerates downward, you feel less weight, and when it accelerates upward you feel more. The weight that you feel is the normal force (also called apparent weight) exerted on you by the surface of the elevator. As you already learned in "forces", the normal force on a person in an elevator is
$F_{n}-mg=ma$
where $a$ is the acceleration of the elevator (or the person).
If the elevator cable suddenly breaks, then the elevator and the person are in free fall. Since the person is in free fall, her/his acceleration, is $a=-g$. If you substitute this in the above equation, you get $F_{n}=0$, and the person feels weightlessness. An astronomer in a spacecraft is also in free fall along with the spacecraft as gravity is the only force acting on the astronomer or the spacecraft, so he/she feels weightlessness. Note that one feel weightless due to the absence of normal force. Although apparent weight is zero during free fall, it doesn't mean that the actual weight of the person is zero, that is still $mg$.
#### Planetary motion:
All the planets including earth revolve around the sun. There are three laws developed by Kepler’s that describe the motion of planets around the sun. These laws were developed in the 1600’s, even before the invention of telescope and also about 50 years before the Newton’s laws.
#### Kepler’s first law:
Each planet revolves around the Sun in an elliptical path with the Sun at one focus.
#### Kepler’s second law:
An imaginary line drawn from the Sun to a planet sweeps out equal areas in equal times.
In the figure, the shaded areas 1 and 2, are the same if the time taken by the planet to move from A to B is same as that from C to D.
#### Kepler’s third law:
The ratio of the squares of the periods of any two planets revolving around the sun is equal to the ratio of the cubes of their mean distances from the sun.
$\bigg(\dfrac{T_1}{T_2}\bigg)^2=\bigg(\dfrac{s_1}{s_2}\bigg)^3$ where $T_1$ and $T_2$ are the periods of planets 1 and 2 and $s_1$ and $s_2$ are their distances from the sun. For circular orbits, $s=r$, the radius of the orbit.
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# One step equations using multiplication and division.
## Presentation on theme: "One step equations using multiplication and division."— Presentation transcript:
One step equations using multiplication and division
ONE STEP EQUATIONS What you do to one side of the equation must also be done to the other side to keep it balanced. An equation is like a balance scale because it shows that two quantities are equal.
One way to solve an equation is to use inverse operations. An inverse operation is an operation that “undoes” another operation. Operation Inverse Operation AdditionSubtraction Addition Multiplication Division
1) Solve. -5t = 60 To get the variable by itself, which number needs to be moved? -5 To move the -5, you have to do the opposite operation. What operation will we use? division
1) Solve -5t = 60 -5 -5 t = -12 -5(-12) = 60 1.Draw “the river” to separate the equation into 2 sides 2.Divide both sides by -5 3.Check your answer
2) Solve 15 = 6n 6 6 2.5 = n 15 = 6(2.5) 1.Draw “the river” 2.Divide both sides by 6 3.Check your answer
3) Solve -3v = -129 1.v = -126 2.v = -43 3.v = 43 4.v = 126 Answer Now
4) Solve You don’t like fractions? Let’s get rid of them! “Clear the fraction” by multiplying both sides of the equation by the denominator.
4) Solve 4 · · 4 x = -48 1.Draw “the river” 2.Clear the fraction – multiply both sides by 4 3.Check your answer
4) Solve -15 · · · (-15) x = 450 1.Draw “the river” 2.Clear the fraction multiply both sides by -15 3.Check your answer
Practice: Worksheet
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Class 9 Maths
# Statistics
## Exercise 14.4
Question 1. The following number of goals were scored by a team in a series of 10 matches:
2, 3, 4, 5, 0, 1, 3, 3, 4, 3
Find the mean, median and mode of these scores.
Answer: Mean = Sum of observations ÷ Number of observations
= (2 + 3 + 4 + 5 + 0 + 1 + 3 + 3 + 4 + 3) ÷ 10
=28 ÷ 10 = 2.8
Question 2: In a mathematics test given to 15 students, the following marks (out of 100) are recorded:
41, 39, 48, 52, 46, 62, 54, 40, 96, 52, 98, 40, 42, 52, 60
Find the mean, median and mode of this data.
Answer: Mean = Sum of observations ÷ Number of observations
= (39 + 40 + 40 + 41 + 42 + 46 + 48 + 52 + 52 + 52 + 54 + 60 + 62 + 96 + 98) ÷ 15
=(792)/(15)=52.8
As there are odd number of observations so, Median =(n+1)/2=(15+1)/2=9
8th term = 52
Mode = The term with most frequency = 52
Question 3: The following observations have been arranged in ascending order. If the median of the data is 63, find the value of x.
29, 32, 48, 50, x, x + 2, 72, 78, 84, 95
Answer: Mean = 630 = (29 + 32 + 48 + 50 + x + x + 2 + 72 + 78 + 84 + 95) ÷ 10
Or, 630=490+2x
Or, 2x=630–490=140
Or, x=70
Question 4: Find the mode of 14, 25, 14, 28, 18, 17, 18, 14, 23, 22, 14, 18.
Answer: Mode = term with most frequency = 14
Question 5: Find the mean salary of 60 workers of a factory from the following table:
Salary (in Rs.)Number of workers
300016
400012
500010
60008
70006
80004
90003
100001
Total60
Salary (in Rs.)(xi)No. of workers (fi)fixi
30001648000
40001248000
50001050000
6000848000
7000642000
8000432000
9000327000
10000110000
TotalΣf_i = 60Σf_i\x_i = 305000
Me\an\=(Σf_i\x_i)/(Σf_i)=305000/60=5083.33
Question 6: Give one example of a situation in which
(i) the mean is an appropriate measure of central tendency.
Answer: When data does not have extreme values then mean is an appropriate measure of central tendency.
(ii) the mean is not an appropriate measure of central tendency but the median is an appropriate measure of central tendency
Answer: When data has extreme values then median is an appropriate measure of central tendency.
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## Precalculus (6th Edition) Blitzer
In order to plot the graph of $f\left( x \right)=\frac{{{x}^{2}}-x-6}{x+1}$, substitute y in place of $f\left( x \right)$ and evaluate the value of the y variable for each value of the x variable. The function $f\left( x \right)=\frac{{{x}^{2}}-x-6}{x+1}$ is undefined at $x=-1$. Therefore, $x=-1$ is the vertical asymptote of the function. To find the value of the y-intercept, substitute $x=0$ as given below: \begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( 0 \right)}^{2}}-0-6}{0+1} \\ & =-\frac{6}{1} \\ & =-6 \end{align} To find the value of the y-intercept, substitute $x=3$ as given below: \begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( 3 \right)}^{2}}-3-6}{3+1} \\ & =0 \end{align} To find the value of the y-intercept, substitute $x=-2$ as given below: \begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( -2 \right)}^{2}}-\left( -2 \right)-6}{-2+1} \\ & =\frac{4+2-6}{-1} \\ & =0 \end{align} To find the value of the y-intercept, substitute $x=-3$ as given below: \begin{align} & y=\frac{{{x}^{2}}-x-6}{x+1} \\ & =\frac{{{\left( -3 \right)}^{2}}-\left( -3 \right)-6}{-3+1} \\ & =\frac{9+3-6}{-2} \\ & =-\frac{6}{2} \\ & =-3 \end{align} Therefore, plot the intercepts $\left( 0,-6 \right),\left( 3,0 \right)\text{,}\left( -2,0 \right)\text{, and }\left( -3,-3 \right)$ and join them with a free hand and draw a vertical asymptote line $x=-1$ in order to get the graph of the equation $f\left( x \right)=\frac{{{x}^{2}}-x-6}{x+1}$ as given below: Note that there is a break in the graph of the function at $x=-1$; it changes its position at this point.
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# Distributing Decimals: How to Multiply with a 1-digit Whole Number
The distributive property is a fundamental principle in mathematics that states that multiplication distributes over addition. When applied to decimals, it provides a systematic approach to multiplying them by a 1-digit whole number. Let's delve into this method.
Distributive Property Formula:
$$a \times (b + c) = (a \times b) + (a \times c)$$
## Multiplying Decimals by a 1-digit Whole Number Using the Distributive Property
### Example 1:
Multiply $$2.3$$ by $$4$$.
Solution Process:
1. Break $$2.3$$ into $$2 + 0.3$$.
2. Apply the distributive property: $$4 \times (2 + 0.3) = (4 \times 2) + (4 \times 0.3)$$.
$$4 \times 2 = 8$$ and $$4 \times 0.3 = 1.2$$. Summing these gives $$9.2$$.
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### Example 2:
Multiply $$3.5$$ by $$3$$.
Solution Process:
1. Break $$3.5$$ into $$3 + 0.5$$.
2. Apply the distributive property: $$3 \times (3 + 0.5) = (3 \times 3) + (3 \times 0.5)$$.
$$3 \times 3 = 9$$ and $$3 \times 0.5 = 1.5$$. Summing these gives $$10.5$$.
The distributive property offers a structured approach to multiplying decimals by 1-digit whole numbers. By breaking down the decimal into its whole number and fractional parts, you can simplify the multiplication process. This method not only reinforces your understanding of decimals but also strengthens your foundational knowledge of multiplication. With practice, you’ll find that using the distributive property makes decimal multiplication a breeze!
### Practice Questions:
1. Multiply $$1.4$$ by $$5$$ using the distributive property.
2. Determine the product of $$2.6$$ and $$6$$ using the distributive property.
3. Multiply $$0.7$$ by $$8$$ using the distributive property.
4. Determine the product of $$3.8$$ and $$7$$ using the distributive property.
5. Multiply $$4.9$$ by $$2$$ using the distributive property.
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1. $$7$$
2. $$15.6$$
3. $$5.6$$
4. $$26.6$$
5. $$9.8$$
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# Make a Number Line to Add — Let's Practise!
Do you want to learn faster and more easily?
Then why not use our learning videos, and practice for school with learning games.
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## Make a Number Line to Add — Let's Practise! exercise
Would you like to apply the knowledge you’ve learnt? You can review and practice it with the tasks for the video Make a Number Line to Add — Let's Practise!.
• ### Show the steps of using a number line to solve a problem.
Hints
Try making your own number line to solve 17 + 22. What did you do first?
Think about the problem step by step. Before we count forwards by the second number, what do we need to do?
Solution
The steps to using a number line to solve a problem are:
1. Draw a number line.
2. Put the larger number on the left.
3. Count forwards by small amounts of the second number.
4. Solve the equation.
• ### Show how to add using the number line.
Hints
Remember, put the larger number on the left.
Partition 16 into tens and ones. First add the tens, then the ones.
What numbers will go on the number line when we add 10 and 6?
Solution
This is how we use a number line to add.
The larger number 76 is on the left. The smaller number 16 is partitioned into smaller amounts (+10 and +6).
76 + 16 = 92.
• ### Use the number line to add the numbers and find the answer.
Hints
Remember, first we put the larger number on the left. Which number is larger?
Once the larger number is on the left, count forwards by small amounts of the second number. Fill in the blanks as you count on.
Solution
53 + 32 = 85.
We start at the larger number, 53. From there, we add the smaller number 32 by small amounts (10 + 10 + 10 + 2) until the numbers are added to make 85.
• ### Use the number line to solve the equation.
Hints
Remember, put the larger number on the left.
To fill in the numbers on the top, think about how to partition 12 into tens and ones.
The parts of 12 are filled in on the number line. Can you place the sums below?
Solution
64 + 12 = 76.
First, we put the larger number 64 on the left. Then, we partition 12 into smaller amounts (10 + 2). 64 + 10 = 74, 74 + 2 = 76.
• ### Complete the number line.
Hints
Remember, when we add 17, we are adding 10 and then 7.
Add 10 first and write the total on the number line, then add 7.
Solution
• The number line starts with the larger number (40) on the left.
• We count forwards by small amounts of the second number.
• 17 has been partitioned into 10 and 7.
• 40 +10 = 50, 50 + 7 = 57.
The answer to 40 + 17 = 57
• ### Create your own number lines to solve.
Hints
Remember to partition the second number into small amounts. For example: 24 + 14. 14 is the smaller number. We can partition it into +10 and +4.
Remember, you have solved the problem once you have added the entire second number.
Solution
By making our own number lines, we can solve the equations:
24 + 14 = 38
45 + 18 = 63
58 + 13 = 71
|
# RD Sharma Solutions For Class 12 Maths Exercise 17.1 Chapter 17 Increasing and Decreasing Functions
RD Sharma Solutions for Class 12 Maths Exercise 17.1 Chapter 17 Increasing and Decreasing Functions are provided here. Strictly increasing, strictly decreasing and verification of increasing or decreasing functions are covered under this exercise. By practising RD Sharma Solutions for Class 12 Maths, students can improve their performance and work on their strengths and weaknesses.
The PDF of Chapter 17 Exercise 17.1 of RD Sharma Solutions are provided here. These solutions help them to a great extent. A set of experts at BYJU’S formulated these solutions to guide the students. Some of the important topics of this exercise are listed below.
• The solution of rational algebraic inequations
• Strictly increasing functions
• Strictly decreasing functions
• Monotonic functions
• Monotonically increasing function
• Monotonically decreasing functions
## RD Sharma Solutions Class 12 Maths Chapter 17 Increasing and Decreasing Functions Exercise 17.1:
### Access another exercise of RD Sharma Solutions For Class 12 Chapter 17 – Increasing and Decreasing Functions
Exercise 17.2 Solutions
### Access answers to Maths RD Sharma Solutions For Class 12 Chapter 17 – Increasing and Decreasing Functions Exercise 17.1
Exercise 17.1 Page No: 17.10
1. Prove that the function f(x) = loge x is increasing on (0, ∞).
Solution:
Let x1, x2 ∈ (0, ∞)
We have, x1 < x2
⇒ loge x1 < loge x2
⇒ f (x1) < f (x2)
So, f(x) is increasing in (0, ∞)
2. Prove that the function f(x) = loga x is increasing on (0, ∞) if a > 1 and decreasing on (0, ∞), if 0 < a < 1.
Solution:
3. Prove that f(x) = ax + b, where a, b are constants and a > 0 is an increasing function on R.
Solution:
Given,
f (x) = ax + b, a > 0
Let x1, x2 ∈ R and x1 > x2
⇒ ax1 > ax2 for some a > 0
⇒ ax1 + b> ax2 + b for some b
⇒ f (x1) > f(x2)
Hence, x1 > x2 ⇒ f(x1) > f(x2)
So, f(x) is increasing function of R
4. Prove that f(x) = ax + b, where a, b are constants and a < 0 is a decreasing function on R.
Solution:
Given,
f (x) = ax + b, a < 0
Let x1, x2 ∈ R and x1 > x2
⇒ ax1 < ax2 for some a > 0
⇒ ax1 + b < ax2 + b for some b
⇒ f (x1) < f(x2)
Hence, x1 > x2⇒ f(x1) < f(x2)
So, f(x) is decreasing function of R
|
Home | | Maths 5th Std | Using letters
# Using letters
Symbols are frequently used in mathematical writing. The use of symbols makes the writing very short. Letters can be used like symbols to make our writing short and simple.
Using letters
Symbols are frequently used in mathematical writing. The use of symbols makes the writing very short. For example, using symbols, division of 63 by 9 gives us ‘7’ can be written in short as “63 ÷ 9 = 7”. It is also easier to grasp.
Letters can be used like symbols to make our writing short and simple
While adding, subtracting or carrying out other operations on numbers, you must have discovered many properties of the operations.
For example, what properties do you see in sums like (7 + 3), (3 + 7)?
The sum of any two numbers and the sum obtained by reversing the order of the two numbers and the sum obtained by reversing the order of the two numbers is the same.
Now see how much easier and faster it is to write this property using letters.
Let us use a and b to represent any two numbers. Their sum will be ‘a + b’
Changing the order of those numbers will make the addition as ‘b + a’.
Therefore, the rule will be, for all values of ‘a’ and ‘b’
(a + b) = (b + a).
Let us see two more examples
* Multiplying any number by 1 gives the number itself. It the number is replaced by an alphabet ‘a‘, then the above statement can be represented as a × 1 = a.
* Given two unequal numbers, the division of the first by the second is not the same as the division of the second by the first.
In short, if a and b are two different numbers, then (a ÷ b) is not equal to (b ÷ a)
Take the valve of ‘a’ as 6 and the value ‘b’ as 2 and verify the above property by yourself.
Activity
Use a letter for “any number” and write the following properties in short.
i. The sum of a number and zero is the number itself
ii. The product of any two numbers and the product obtained after changing the order of those numbers is the same.
iii. The product of a number and zero is zero
iv. Write the following properties in words
i. n−0=n
ii. m÷1=m
Tags : Algebra | Term 3 Chapter 4 | 5th Maths , 5th Maths : Term 3 Unit 4 : Algebra
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Writing Numbers in Words
Start Practice
## How to Write Numbers in Words
You can write numbers with either digits or words.
1 or One
Whenever you use a number less than one hundred in a sentence, write it as a word.
For example:
I bought three tomatoes.
Kobe Bryant scored sixty goals in his last game.
### Writing Numbers up to 100
Here's how we write the multiples of Tens and Ones in words:
đ Most two-digit numbers are a combination of these words.
How do we write 30? đ€
30 doesn't have any Ones. So we just write:
thirty
Start by dividing it into Tens and Ones:
60 + 5
Next we turn each place value into a word, and add a hyphen (-).
sixty-five
Always add a hyphen (-) when writing numbers with both tens and ones.
Let's try one more. How do we write 91?
91 has both Tens and Ones.
90 + 1
đ So we write each term in words with a hyphen between them.
ninety-one
### Writing 11-19
11 through 19 are special. They don't follow the pattern we just learned. đ
You'll have to memorize them:
11 = eleven
12 = twelve
13 = thirteen
14 = fourteen
15 = fifteen
16 = sixteen
17 = seventeen
18 = eighteen
19 = nineteen
Tip: thirteen through nineteen end in teen. That's how old teenagers are.
### Writing Numbers up to 1,000
Now that youâre in 3rd grade, youâll learn to write numbers up to 1,000! đȘ
Letâs start by adding a column to our chart:
When we have a 3-digit number, we have to write the Hundreds place out, too!
The easy part is we donât have to add a hyphen!
Letâs try it:
675
Start by writing the number out in Hundreds, Tens, and Ones.
600 + 70 + 5
Great! Now, letâs find the word for each number:
Six hundred seventy-five
We write every digit and its value, and add a hyphen only between the Tens and Ones.
Letâs try one more:
383
First, expand the number:
300 + 80 + 3
Now, write the word for each number:
Three hundred eighty-three
Great job. đ€
Now, complete the practice. đș You'll really be able to write words after!Â
It's an important life skill. Do your best!
Start Practice
Complete the practice to earn 1 Create Credit
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Class 7 Maths Chapter 3 Exercise 3.2 Pdf Notes NCERT Solutions
Class 7 Maths Chapter 3 Data Handling Exercise 3.2 pdf notes:-
Exercise 3.2 Class 7 maths Chapter 3 Pdf Notes:-
Ncert Solution for Class 7 Maths Chapter 3 Data Handling Exercise 3.2 Tips:-
ARITHMETIC MEAN
The most common representative value of a group of data is the arithmetic mean or the
mean. To understand this in a better way, let us look at the following example:
Two vessels contain 20 litres and 60 litres of milk respectively. What is the amount that
each vessel would have, if both share the milk equally? When we ask this question we are
seeking the arithmetic mean.
MODE
As we have said Mean is not the only measure of central tendency or the only form of
representative value. For different requirements from a data, other measures of central
tendencies are used.
Look at the following example
To find out the weekly demand for different sizes of shirt, a shopkeeper kept records of sales
of sizes 90 cm, 95 cm, 100 cm, 105 cm, 110 cm. Following is the record for a week:
Size (in inches) 90 cm 95 cm 100 cm 105 cm 110 cm Total
Number of Shirts Sold 8 22 32 37 6 105
If he found the mean number of shirts sold, do you think that he would be able to
decide which shirt sizes to keep in stock?
Mean of total shirts sold =
Total number of shirts sold
Number of different sizes of shirts = = 105
5
21
Should he obtain 21 shirts of each size? If he does so, will he be able to cater to the
needs of the customers?
The shopkeeper, on looking at the record, decides to procure shirts of sizes 95 cm,
100 cm, 105 cm. He decided to postpone the procurement of the shirts of other sizes
because of their small number of buyers.
Look at another example
The owner of a readymade dress shop says, “The most popular size of dress I sell is the
size 90 cm.
Observe that here also, the owner is concerned about the number
of shirts of different sizes sold. She is however looking at the shirt size
that is sold the most. This is another representative value for the data.
The highest occuring event is the sale of size 90 cm.This representative
value is called the mode of the data.
The mode of a set of observations is the observation that occurs
most often.
EXAMPLE 4 Find the mode of the given set of numbers: 1, 1, 2, 4, 3, 2, 1, 2, 2, 4
SOLUTION Arranging the numbers with same values together, we get
1, 1, 1, 2, 2, 2, 2, 3, 4, 4
Mode of this data is 2 because it occurs more frequently than other observations.
3.6.1 Mode of Large Data
Putting the same observations together and counting them is not easy if the number of
observations is large. In such cases we tabulate the data. Tabulation can begin by putting
tally marks and finding the frequency, as you did in your previous class.
Test Paper Of Class 8th
• Maths 8th Class
• Science 8th class
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• Test Paper Of Class 7th
• Maths 7th Class
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|
# Eureka Math Precalculus Module 2 Lesson 18 Answer Key
## Engage NY Eureka Math Precalculus Module 2 Lesson 18 Answer Key
### Eureka Math Precalculus Module 2 Lesson 18 Example Answer Key
Example 1: Vectors and Translation Maps in ℝ3
Translate by the vector v = 〈1, 3, 5〉 by applying the translation map Tv to the following objects in ℝ3. A sketch of the original object and the vector is shown. Sketch the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x + 1 \\ y + 3 \\ z + 5 \end{array}\right]$$
a. The point A(2, – 2, 4)
The new point will be (2 + 1, – 2 + 3, 4 + 5) or (3, 1, 9).
b. The plane 2x + 3y – z = 0
The new plane will be 2(x – 1) + 3(y – 3) – (z – 5) = 0, which is equivalent to 2x + 3y – z = 6.
c. The sphere (x – 1)2 + (y – 3)2 + z2 = 9
The new sphere will be ((x – 1) – 1)2 + ((y – 3) – 3)2 + (z – 5)2 = 9, which is equivalent to (x – 2)2 + (y – 6)2 + (z – 5)2 = 9.
Example 2: What is the Magnitude of a Vector in ℝ3?
a. Find a general formula for ‖v‖2.
‖v‖2 = $$x^{2} + y^{2} + z^{2}$$
b. Solve this equation for ‖v‖ to find the magnitude of the vector.
‖v‖ = $$\sqrt{x^{2} + y^{2} + z^{2}}$$
### Eureka Math Precalculus Module 2 Lesson 18 Exercise Answer Key
Opening Exercise
Write each vector described below in component form and find its magnitude. Draw an arrow originating from (0, 0) to represent each vector’s magnitude and direction.
a. Translate 3 units right and 4 units down.
u = 〈3, – 4〉, ‖u‖ = 5
b. Translate 6 units left.
v = 〈 – 6, 0〉, ‖v‖ = 6
c. Translate 2 units left and 2 units up.
w = 〈 – 2, 2〉, ‖w‖ = 2$$\sqrt{2}$$
d. Translate 5 units right and 7 units up.
t = 〈5, 7〉, ‖t‖ = $$\sqrt{25 + 49}$$ = $$\sqrt{74}$$
Exercises 1–3
Exercise 1.
Write a translation map defined by each vector from the opening.
Consider the vector v = 〈 – 2, 5〉 and its associated translation map:
$$T_{\mathrm{v}}\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x – 2 \\ y + 5 \end{array}\right]$$
Question 2.
Suppose we apply the translation map Tv to each point on the circle (x + 4)2 + (y – 3)2 = 25.
a. What is the radius and center of the original circle?
The radius is 5 and the center is ( – 4, 3).
b. Show that the image points satisfy the equation of another circle.
Suppose that Tv $$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x{\prime} \\ y{\prime} \end{array}\right]$$. Then x’ = x – 2 and y’ = y + 5, so x = x’ + 2 and y = y’ – 5.
Since (x + 4)2 + (y – 3)2 = 25, we have
(x + 4)2 + (y – 3)2 = 25
((x’ + 2) + 4)2 + ((y’ – 5) – 3)2 = 25
(x’ + 6)2 + (y’ – 8)2 = 25
So, the image points $$\left[\begin{array}{l} x{\prime} \\ y{\prime} \end{array}\right]$$ lie on a circle.
c. What are the center and radius of this image circle?
The radius of 5 remains unchanged. The new center is ( – 6, 8).
Exercise 3.
Suppose we apply the translation map Tv to each point on the line 2x – 3y = 10.
a. What are the slope and y – intercept of the original line?
The slope is $$\frac{2}{3}$$ and the y – intercept is (0, – $$\frac{10}{3}$$).
b. Show that the image points satisfy the equation of another line.
Suppose that Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x^{\prime} \\ y^{\prime} \end{array}\right]$$. Then x’ = x – 2 and y’ = y + 5, so x = x’ + 2 and y = y’ – 5.
Since 2x – 3y = 10, we have
2(x’ + 2) – 3(y’ – 5) = 10
2x’ + 4 – 3y’ + 15 = 10
2x’ – 3y’ = – 9
c. What are the slope and y – intercept of this image line?
The slope is $$\frac{2}{3}$$ and the y – intercept is (0, 3).
Exercise 4.
Given the sphere (x + 3)2 + (y – 1)2 + (z – 3)2 = 10.
a. What are its center and radius?
The center is ( – 3, 1, 3), and the radius is √10.
b. Write a vector and its associated translation map that would take this sphere to its image centered at the origin.
We need to translate the center from the point ( – 3, 1, 3) to the point (0, 0, 0). This represents a translation of 3 units in the x direction, – 1 unit in the y direction, and – 3 units in the z direction.
The vector is v = 〈3, – 1, – 3〉, and the translation map would be
Tv$$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x + 3 \\ y – 1 \\ z – 3 \end{array}\right]$$
Exercises 5–8
Exercise 5.
Which vector has greater magnitude, v = 〈0, 5, – 4〉 or u = 〈3, – 4, 4〉? Show work to support your answer.
‖v‖ = $$\sqrt{0^{2} + 5^{2} + ( – 4)^{2}}$$ = $$\sqrt{41}$$
‖u‖ = $$\sqrt{3^{2} + ( – 4)^{2} + 4^{2}}$$ = $$\sqrt{41}$$
These vectors have equal magnitude.
Exercise 6.
Explain why vectors can have equal magnitude but not be the same vector.
A vector has both a magnitude and a direction. If you graphed the vectors from Exercise 5, you can see they point in different directions so they cannot be the same even though they have equal magnitude.
Exercise 7.
Vector arithmetic in ℝ3 is analogous to vector arithmetic in ℝ2. Complete the graphic organizer to illustrate these ideas.
Exercise 8.
Given v = 〈2, 0, – 4〉 and u = 〈 – 1, 5, 3〉.
a. Calculate the following.
i. v + u
v + u = 〈2 + ( – 1), 0 + 5, – 4 + 3〉 = 〈1, 5, – 1〉
ii. 2v – u
2v – u = 〈2∙2 – ( – 1), 2∙0 – 5, 2∙( – 4) – 3〉 = 〈5, – 5, – 11〉
iii. ‖u‖
‖u‖ = $$\sqrt{( – 1)^{2} + 5^{2} + 3^{2}}$$ = $$\sqrt{35}$$
b. Suppose the point (1, 3, 5) is translated by v and then by u. Determine a vector w that would return the point back to its original location (1, 3, 5).
From part (a), we have v + u = 〈1, 5, – 1〉. The point will be translated from (1, 3, 5) to (2, 8, 4)
since 1 + 1 = 2, 5 + 3 = 8, and 5 – 1 = 4. The vector that will return this point to its original location will be the opposite of v + u.
– (v + u) = 〈 – 1, – 5, 1〉
And this vector will translate (2, 8, 4) back to (1, 3, 5) because 2 – 1 = 1, 8 – 5 = 3, and 4 + 1 = 5.
### Eureka Math Precalculus Module 2 Lesson 18 Problem Set Answer Key
Question 1.
Myishia says that when applying the translation map Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 1 \\ y – 2 \end{array}\right]$$ to a set of points given by an equation relating x and y, we should replace every x that is in the equation by x + 1, and y by y – 2. For example, the equation of the parabola y = x2 would become y – 2 = (x + 1)2. Is she correct? Explain your answer.
No, she is not correct. What she did translates the points on the parabola y = x2 in the opposite direction—one unit to the left and 2 units upward, which is not the geometric effect of Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 1 \\ y – 2 \end{array}\right]$$. In order to have the correct translation based on Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 1 \\ y – 2 \end{array}\right]$$, we need to set x’ = x + 1 and y’ = y – 2, which is equivalent to x = x’ – 1 and y = y’ + 2.
This process gives the transformed equation y’ + 2 = (x’ – 1)2, which we write as y + 2 = (x – 1)2.
Question 2.
Given the vector v = 〈 – 1, 3〉, find the image of the line x + y = 1 under the translation map Tv. Graph the original line and its image, and explain the geometric effect of the map Tv on the line.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x – 1 \\ y + 3 \end{array}\right]$$, (x – ( – 1)) + (y – (3)) = 1, x + 1 + y – 3 = 1, x + y = 3
Every point on the line is shifted one unit left and three units upward. The slopes of the lines remain – 1.
Question 3.
Given the vector v = 〈2, 1〉, find the image of the parabola y – 1 = x2 under the translation map Tv. Draw a graph of the original parabola and its image, and explain the geometric effect of the map Tv on the parabola. Find the vertex and x – intercepts of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 2 \\ y + 1 \end{array}\right]$$, y – 2 = (x – 2)2
Every point on the parabola is shifted two units to the right and one unit upward.
The vertex is (2, 2), and there are no x – intercepts.
Question 4.
Given the vector v = 〈3, 2〉, find the image of the graph of y + 1 = (x + 1)3 under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the x – intercepts of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 3 \\ y + 2 \end{array}\right]$$, y – 1 = (x – 2)3
Every point on the curve is shifted three units to the right and two units upward.
The x – intercept is 1.
Question 5.
Given the vector v = 〈3, – 3〉, find the image of the graph of y + 2 = $$\sqrt{x + 1}$$ under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the x – intercepts of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 3 \\ y – 3 \end{array}\right]$$, y + 5 = $$\sqrt{x – 2}$$
Every point on the curve is shifted three units to the right and three units downward.
The x – intercept is 27.
Question 6.
Given the vector v = 〈 – 1, – 2〉, find the image of the graph of y = $$\sqrt{9 – x^{2}}$$ under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the x – intercepts of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x – 1 \\ y – 2 \end{array}\right]$$, y + 2 = $$\sqrt{9 – (x + 1)^{2}}$$
Every point on the semicircle is shifted one unit to the left and two units downward.
x – intercepts: – 1 + $$\sqrt{5}$$ and – 1 – $$\sqrt{5}$$.
Question 7.
Given the vector v = 〈1, 3〉, find the image of the graph of y = $$\frac{1}{x + 2}$$ + 1 under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the equations of the asymptotes of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 1 \\ y + 3 \end{array}\right]$$, y = $$\frac{1}{x + 1}$$ + 4
Every point on the curve is shifted one unit to the right and three units upward.
The new vertical asymptote is x = – 1, the new horizontal asymptote is y = 4.
Question 8.
Given the vector v = 〈 – 1, 2〉, find the image of the graph of y = |x + 2| + 1 under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the x – intercepts of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x – 1 \\ y + 2 \end{array}\right]$$, y = |x + 3| + 3
Every point on the graph is shifted one unit to the left and two units upward.
There are no x – intercepts.
Question 9.
Given the vector v = 〈1, – 2〉, find the image of the graph of y = 2x under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the x – intercepts of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 1 \\ y – 2 \end{array}\right]$$, y = 2x – 1 – 2
Every point on the curve is shifted one unit to the right and two units downward.
x – intercept is 2.
Question 10.
Given the vector v = 〈 – 1, 3〉, find the image of the graph of y = log2x, under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the x – intercepts of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x – 1 \\ y + 3 \end{array}\right]$$, y = log2(x + 1) + 3
Every point on the curve is shifted one unit to the left and three units upward.
x – intercept: – $$\frac{7}{8}$$.
Question 11.
Given the vector v = 〈2, – 3〉, find the image of the graph of $$\frac{x^{2}}{4} + \frac{y^{2}}{16}$$ = 1 under the translation map Tv. Draw the original graph and its image, and explain the geometric effect of the map Tv on the graph. Find the new center, major and minor axis of the graph of the image.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 2 \\ y – 3 \end{array}\right]$$, $$\frac{(x – 2)^{2}}{4} + \frac{(y + 3)^{2}}{16}$$ = 1
Every point on the ellipse is shifted two units to the right and three units downward.
The new center is (2, – 3), the major axis is 4, and the minor axis is 2.
Question 12.
Given the vector v, find the image of the given point P under the translation map Tv. Graph P and its image.
a. v = 〈3, 2, 1〉, P = $$\left[\begin{array}{l} 1 \\ 2 \\ 3 \end{array}\right]$$
Tv$$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x + 3 \\ y + 2 \\ z + 1 \end{array}\right]$$, the new image: $$\left[\begin{array}{l} 4 \\ 4 \\ 4 \end{array}\right]$$
b. v = 〈 – 2, 1, – 1〉, P = $$\left[\begin{array}{c} 2 \\ – 1 \\ – 4 \end{array}\right]$$
Tv $$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x – 2 \\ y + 1 \\ z – 1 \end{array}\right]$$, the new image: $$\left[\begin{array}{c} 0 \\ 0 \\ – 5 \end{array}\right]$$
Question 13.
Given the vector v, find the image of the given plane under the translation map Tv. Sketch the original vector and its image.
a. v = 〈2, – 1, 3〉, 3x – 2y – z = 0
Tv $$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x + 2 \\ y – 1 \\ z + 3 \end{array}\right]$$, 3x – 2y – z = 5
b. v = 〈 – 1, 2, – 1〉, 2x – y + z = 1
Tv$$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x – 1 \\ y + 2 \\ z – 1 \end{array}\right]$$, 2x – y + z = – 4
Question 14.
Given the vector v, find the image of the given sphere under the translation map Tv. Sketch the original sphere and its image.
a. v = 〈 – 1, 2, 3〉, x2 + y2 + z2 = 1
Tv$$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x – 1 \\ y + 2 \\ z + 3 \end{array}\right]$$, (x + 1)2 + (y – 2)2 + (z – 3)2 = 1
b. v = 〈 – 3, – 2, 1〉, (x + 2)2 + (y – 3)2 + (z + 1)2 = 1
Tv$$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x – 3 \\ y – 2 \\ z + 1 \end{array}\right]$$, (x + 5)2 + (y – 1)2 + (z)2 = 1
Question 15.
Find a vector v and translation map Tv that will translate the line x – y = 1 to the line x – y = – 3. Sketch the original vector and its image.
Answers vary. For example, v = $$\left[\begin{array}{l} 0 \\ 4 \end{array}\right]$$ and Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 0 \\ y + 4 \end{array}\right]$$.
Question 16.
Find a vector v and translation map Tv that will translate the parabola y = x2 + 4x + 1 to the parabola y = x2
Because y = x2 + 4x + 1 can be written as y + 3 = (x + 2)2, v = $$\left[\begin{array}{l} 2 \\ 3 \end{array}\right]$$ and Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x + 2 \\ y + 3 \end{array}\right]$$.
Question 17.
Find a vector v and translation map Tv that will translate the circle with equation x2 + y2 – 4x + 2y – 4 = 0 to the circle with equation (x + 3)2 + (y – 4)2 = 9
Because x2 – 4x + y2 + 2y = 4 can be written as (x – 2)2 + (y + 1)2 = 9, v = $$\left[\begin{array}{c} – 5 \\ 5 \end{array}\right]$$ and Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x – 5 \\ y + 5 \end{array}\right]$$.
Question 18.
Find a vector v and translation map Tv that will translate the graph of y = $$\sqrt{x – 3}$$ + 2 to the graph of y = $$\sqrt{x + 2}$$ – 3.
v = $$\left[\begin{array}{l} – 5 \\ – 5 \end{array}\right]$$ and Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left[\begin{array}{l} x – 5 \\ y – 5 \end{array}\right]$$
Question 19.
Find a vector v and translation map Tv that will translate the sphere (x + 2)2 + (y – 3)2 + (z + 1)2 = 1 to the sphere (x – 3)2 + (y + 1)2 + (z + 2)2 = 1
v = $$\left[\begin{array}{c} 5 \\ – 4 \\ – 1 \end{array}\right]$$ and Tv$$\left(\left[\begin{array}{l} x \\ y \\ z \end{array}\right]\right) = \left[\begin{array}{l} x + 5 \\ y – 4 \\ z – 1 \end{array}\right]$$
Question 20.
Given vectors u = 〈2, – 1, 3〉, v = 〈2, 0, – 2〉, and w = 〈 – 3, 6, 0〉, find the following.
a. 3u + v + w
〈5, 3, 7〉
b. w – 2v – u
〈 – 9, 7, 1〉
c. 3(2u – $$\frac{1}{2}$$v) – $$\frac{1}{3}$$w
〈10, – 8, 21〉
d. – 2u – 3(5v – 3w)
〈 – 61, 56, 24〉
e. ‖u‖, ‖v‖, and ‖w‖
‖u‖ = $$\sqrt{14}$$, ‖v‖ = 2$$\sqrt{2}$$, ‖w‖ = $$\sqrt{45}$$
f. Show that 2‖v‖ = ‖2v‖.
2‖v‖ = 2 (2$$\sqrt{2}$$) = 4$$\sqrt{2}$$, ‖2v‖ = ‖〈4, 0, – 4〉‖ = $$\sqrt{32}$$ = 4$$\sqrt{2}$$
g. Show that ‖u + v‖ ≠ ‖u‖ + ‖v‖.
‖u + v‖ = ‖〈4, – 1, 1〉‖ = $$\sqrt{18}$$, ‖u‖ + ‖v‖ = $$\sqrt{14}$$ + 2$$\sqrt{2}$$
h. Show that ‖v – w‖ ≠ ‖v‖ – ‖w‖.
‖v – w‖ = ‖〈5, – 6, – 2〉‖ = $$\sqrt{65}$$, ‖v‖ – ‖w‖ = 2$$\sqrt{2}$$ – $$\sqrt{45}$$
i. $$\frac{1}{\|\mathrm{u}\|} \mathrm{u}$$ and $$\left\|\frac{1}{\|\mathrm{u}\|} \mathrm{u}\right\|$$
### Eureka Math Precalculus Module 2 Lesson 18 Exit Ticket Answer Key
Question 1.
Given the vector v = 〈 – 2, 1〉, find the image of the line 3x – 2y = 2 under the translation map Tv. Graph the original line and its image, and explain the geometric effect of the map Tv.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left(\begin{array}{l} x – 2 \\ y + 1 \end{array}\right)$$; image of the original line: 3(x + 2) – 2(y – 1) = 2.
3x + 6 – 2y + 2 = 2, 3x – 2y = – 6
Every point on the line is shifted 2 units to the left and 1 unit upward.
Question 2.
Given the vector v = 〈 – 1, 2〉, find the image of the circle (x – 2)2 + (y + 1)2 = 4 under the translation map Tv. Graph the original circle and its image, and explain the geometric effect of the map Tv.
Tv$$\left(\left[\begin{array}{l} x \\ y \end{array}\right]\right) = \left(\begin{array}{l} x – 1 \\ y + 2 \end{array}\right)$$; image of original circle: (x – 1)2 + (y – 1)2 = 4.
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# Parametric Differentiation: Summary
by Batool Akmal
My Notes
• Required.
Learning Material 2
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DLM Parametric Differentiation Calculus Akmal.pdf
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Transcript
00:01 Once again, just to recap over everything that we've done and we do it more formally now.
00:08 If you have a function of x defined as f of t or any function that includes t and a function of y defined by a different function g of t, you can bring them together firstly by differentiating each part separately. So the x, we have to differentiate with respect to t so we’ll get dx/dt.
00:27 Then the y, we have to differentiate with respect to t as well to get dy of dt. Then to get your dy/dx, you flip dx/dt to give you dt/dx, so that you can cancel the dt’s and you end up with dy/dx.
00:44 So, both methods that I mentioned earlier are now here. You can see that you could either learn dy/dx as dy/dt divided by dx/dt or if you prefer to multiply, you'd have to remember that it's dy/dt left as it is and then you flip the other fraction. So, dx/dt, you flip to make it into dt/dx so you can end up with dy/dx.
01:09 Let’s look at another example now just to consolidate the ideas. We are looking for dy/dx.
01:17 Again, you can see that you have two equations. You can see that this is a parametric equation because you have not only x's and y's but a third variable, t. Don’t ever worry about what this letter is.
01:29 It could be p, or q, or z, or anything. But as long as you spot a third variable, you have a third parameter that’s holding it together. Let’s try and differentiate this using the rules we've just learned.
01:43 We have y = t² - 5 and x = t³ + 5. Remember what we said. Firstly, spot what kind of equation it is.
01:56 I know that you know a lot of different methods and lots of different types of equations.
02:00 You can question all of them. Is it chain rule or product rule or quotient rule, implicit differentiation? Hopefully, when you come to the part of thinking about parametric equations, you'd recognize that this is a parametric equation. So, in order to differentiate it, we're going to have to do this separately. We'll do dy/dt here. So, that just gives me 2t and the constant 5 just disappears. We have dx/dt. When you differentiate that, that gives you 3t².
02:34 Bring the power to the front and decrease the power by 1 and then the 5 just disappears.
02:39 Now, our definition was this. Dy/dx is dy/dt multiplied by dt/dx or if you wanted to do it the other way, you could do dy/dt divided by dx/dt. Both of them are the same thing. If I’m using this form of it, I obviously need to flip this before. So, dx/dt needs to become dt/dx. You can't just do that to one side of the equation. We have to do it to this side as well. So, this is 3t²/1.
03:17 So, you can write it as 1/3t². We just flipped the entire equation. Put it together here in the formula. Dy/dt is 2t. Dt/dx is 1/3t². Nice and straightforward. We can now just simplify this a little bit to say that dy/dx. You can cancel this one t with one of them is 2/3t.
03:48 Moving on then. We now might wonder how to find the second differential of a parametric equation.
03:56 It’s not as straightforward as just differentiating a function again. So, when you're doing the second differential, we have to use the next part of this definition. In order to do d²y/dx², we essentially need to differentiate dy/dx again as we do with any other differentiation. If you have to do a second differential of any function, you just differentiate it again using the same rules.
04:24 However, because this time you have t’s in your equation, you can’t just dy/dx with respect to x.
04:31 So, we use this definition. We use d²y/dx². We differentiate dy/dx which we've already found in the first part with respect to t because that's important. Then to make up for that dt that we've just done there, we divide it by dx/dt or we multiply it with dx/dt flipped, so by dt/dx.
04:56 It looks fairly complicated but it really isn't when you start to use this with numbers or you start to apply it to real questions. So, let’s have a look at an example.
The lecture Parametric Differentiation: Summary by Batool Akmal is from the course Parametric Differentiation.
### Included Quiz Questions
1. dy/dx = (3t² + 5) / (2t - 1)
2. dy/dx = (3t² + 5) / (2t)
3. dy/dx = (3t + 5) / (2t)
4. dy/dx = (t² + 5) / (2t + 1)
5. dy/dx = (t² - 5) / (2t + 1)
1. dy/dx = t - 2
2. dy/dx = 2t - 2
3. dy/dx = 2t - 4
4. dy/dx = t²/2
5. dy/dx = 4t - 8
1. d(dy/dx) /dt . dt/dx
2. (dy/dx)²
3. 2dy/dx
4. d²y/dt² . (dt/dx)²
5. d(dy/dx) / dt . dx/dt
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Friday, March 24, 2023
## Grade 12 Probability
#### 2. Identities
The addition rule (also called the sum rule) for any 2 events, A and B is
P(A or B) = P(A) + P(B) - P(A and B)
This rule relates the probabilities of 2 events with the probabilities of their union and
intersection.
The addition rule for 2 mutually exclusive events is
P(A or B) = P(A) + P(B)
This rule is a special case of the previous rule. Because the events are mutually exclusive,
P(A and B) = 0.
The complementary rule is
P(not A) = 1 - P(A)
This rule is a special case of the previous rule. Since A and (not A) are complementary,
P(A or (not A)) = 1.
The product rule for independent events A and B is:
P(A and B) = P(A) x P(B)
If two events A and B are dependent then:
QUESTION
Write down which of the following events are dependent and which are independent:
1. The student council chooses a head student and then a deputy head student.
2. A bag contains blue marbles and red marbles. You take a red marble out of the bag and then throw it back in again before you take another marble out of the bag.
SOLUTION
Step 1: Ask the question: Did the available choices change for the second event
because of the first event?
1. Yes, because after selecting the head student there are fewer council members available to choose for the deputy head student position. Therefore, the two events are dependent.
2. No, because when you throw the first marble back into the bag, there are the
same number and color composition of choices for the second marble. Therefore
the two events are independent.
QUESTION
A bag contains 3 yellow and 4 black beads. We remove a random bead from the bag,
record its colour and put it back into the bag. We then remove another random bead
from the bag and record its colour.
1. What is the probability that the first bead is yellow?
2. What is the probability that the second bead is black?
3. What is the probability that the first bead is yellow and the second bead is black?
4. Are the first bead being yellow and the second bead being black independent
events?
SOLUTION
Step 1: Probability of a yellow bead first
Since there is a total of 7 beads, of which 3 are yellow, the probability of getting a
P(first bead yellow) =3/7
Step 2: Probability of a black bead second
The problem states that the first bead is placed back into the bag before we take the
second bead. This means that when we draw the second bead, there are again a total
of 7 beads in the bag, of which 4 are black. Therefore the probability of drawing a
P(second bead black) =4/7
Step 3: Probability of yellow first and black second
When drawing two beads from the bag, there are 4 possibilities. We can get
• a yellow bead and then another yellow bead;
• a yellow bead and then a black bead;
• a black bead and then a yellow bead;
• a black bead and then another black bead.
We want to know the probability of the second outcome, where we have to get a yellow bead first. Since there are 3 yellow beads and 7 beads in total, there are ways to get a yellow bead first. Now we put the first bead back, so there are again 3 yellow beads and 4 black beads in the bag. Therefore there are 4/7 ways to get a black bead second if the first bead was yellow. This means that there are 3/7
QUESTION
In the previous example, we picked a random bead and put it back into the bag before continuing. This is called sampling with replacement. In this worked example, we will follow the same process, except that we will not put the first bead back into the bag. This is called sampling without replacement. So, from a bag with 3 red and 5 green beads, we remove a random bead and record its colour. Then, without putting back the first bead, we remove another random bead from the bag and record its colour.
1. What is the probability that the first bead is red?
2. What is the probability that the second bead is green?
3. What is the probability that the first bead is red and the second bead is green?
4. Are the first bead being red and the second bead being green independent
events?
SOLUTION
Step 1: Count the number of outcomes
We will examine the number ways in which we can get the different possible outcomes
when removing 2 beads. The possible outcomes are
• a red bead and then another red bead (RR);
• a red bead and then a green bead (RG);
• a green bead and then a red bead (GR);
• a green bead and then another green bead (GG).
For the first outcome, we have to get a red bead first. Since there are 3 red beads and
8 beads in total, there are 3/8 ways to get a red bead first. After we have taken out a red bead, there are now 2 red beads and 5 green beads left. Therefore there are 2/7 ways to get a red bead second if the first bead was also red. This means that there are
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## Thinking Mathematically (6th Edition)
$4011_{seven}$
$(5\times 3)_{10}=(15)_{10}=2\times 7+1=(21)_{7}$ Record the $1$; carry the $2$. $\begin{array}{llll} & 5 & 4 & 3_{seven}\\ \times & & & 5_{seven}\\ -- & -- & -- & --\\ & & & 1_{seven} \end{array}$ $(5\times 4+2)_{10}=(22)_{10}=3\times 7+1=(31)_{7}$ Record the $1$; carry the $3$. $\begin{array}{llll} & 5 & 4 & 3_{seven}\\ \times & & & 5_{seven}\\ -- & -- & -- & --\\ & & 1 & 1_{seven} \end{array}$ $(5\times 5+3)_{10}=(28)_{10}=4\times 7+0=(40)_{7}$ Record the $40.$ Result: $4011_{seven}$
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# Lesson 36 Quadratic Trigonometric Equations Pre Calculus 212022
• Slides: 21
Lesson 36 – Quadratic Trigonometric Equations Pre. Calculus 2/1/2022 Pre. Calculus 1
FAST FIVE Solve each equation by factoring n n Solve the quadratic equation x 2 + 2 x = 15 algebraically Solve the quadratic equation x 2 + 2 x = 15 graphically 1. (k+1)(k-5) = 0 2. (a+1)(a+2) = 0 3. (4 k+5)(k+1) = 0 4. (2 m+3)(4 m+3) = 0 5. x 2 – 11 x + 19 = -5 6. n 2 + 7 n + 15 = 5 7. n 2 – 10 n + 22 = -2 8. n 2 + 3 n – 12 = 6 9. 6 n 2 – 18 n – 18 = 6 10. 7 r 2 – 14 r = – 7 2/1/2022 Pre. Calculus 2
(A) Prerequisite Skill: Factoring with Trig n Factor the following trig expressions: 2/1/2022 Pre. Calculus 3
(B) Solving Quadratic Trigonometric Equations n We will outline a process by which we come up with the solution to a trigonometric equation ANALYTICALLY using ALGEBRA n We will outline a process by which we come up with the solution to a trigonometric equation GRAPHICALLY using ALGEBRA n it is important you understand WHY we carry out these steps, rather than simply memorizing them and simply repeating them on a test of quiz 2/1/2022 Pre. Calculus 4
(B) Solving Quadratic Trigonometric Equations 2/1/2022 Pre. Calculus 5
(B) Solving Quadratic Trigonometric Equations n Solve sin 2 x – 1 = 0 on the domain 0 < x < 4π. n Your first solution will be analytical. n We will VERIFY with a graphic solution 2/1/2022 Pre. Calculus 6
(B) Solving Quadratic Trigonometric Equations n Solve the equation 2 cos 2 x – 1 = -cosx on the domain 0 < x < 4π. n Your first solution will be analytical. n We will VERIFY with a graphic solution 2/1/2022 Pre. Calculus 7
(B) Solving Quadratic Trigonometric Equations n Solve the equation 8 sin 2 x + 13 sinx = 4 – 4 sin 2 x on the domain - π < x < 3π. n Your first solution will be analytical. n We will VERIFY with a graphic solution 2/1/2022 Pre. Calculus 8
(B) Solving Quadratic Trigonometric Equations n Solve 2 cos 2(θ) = 1 if 0° < θ < 360 ° 2/1/2022 Pre. Calculus 9
(B) Solving Quadratic Trigonometric Equations n Solve 2 cos 2(θ) = 1 if 0° < θ < 360 ° 2/1/2022 Pre. Calculus 10
(B) Solving Quadratic Trigonometric Equations n Solve 2 cos 2(θ – 45°) = 1 if 0° < θ < 360 ° 2/1/2022 Pre. Calculus 11
(B) Solving Quadratic Trigonometric Equations n Solve 2 cos 2(θ + π/3) = 1 if -2π < θ < 5π 2/1/2022 Pre. Calculus 12
(B) Solving Quadratic Trigonometric Equations n Solve cos 2(x) + 2 cos(x) = 0 for 0 < x < 2π 2/1/2022 Pre. Calculus 13
(B) Solving Quadratic Trigonometric Equations n Solve cos 2(x) + 2 cos(x) = 0 for 0 < x < 2π 2/1/2022 Pre. Calculus 14
(B) Solving Quadratic Trigonometric Equations n Solve cos 2(2 x) + 2 cos(2 x) = 0 for 0 < x < 2π 2/1/2022 Pre. Calculus 15
(B) Solving Quadratic Trigonometric Equations n Solve 2 cos 2(x) - 3 cos(x) + 1 = 0 for 0 < x < 2π 2/1/2022 Pre. Calculus 16
(B) Solving Quadratic Trigonometric Equations n Solve 2 cos 2(x) - 3 cos(x) + 1 = 0 for 0 < x < 2π 2/1/2022 Pre. Calculus 17
(B) Solving Quadratic Trigonometric Equations n Solve 2 cos 2(2 x - 2π/3) - 3 cos(2 x - 2π/3) + 1 = 0 for xεR 2/1/2022 Pre. Calculus 18
(C) Further Examples n Solve the following without a calculator 2/1/2022 Pre. Calculus 19
(C) Further Examples n Solve the following algebraically, without a GRAPHICAL approach 2/1/2022 Pre. Calculus 20
(D) Homework n Nelson Textbook, Chap 6. 6 n http: //mrsantowski. tripod. com/2010 Math. SLY 1/Ass essments/Nelson. S 66 p 541. pdf 2/1/2022 Pre. Calculus 21
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# The dimensional formula of modulus of elasticity is(A) $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$(B) $\left[ {{M^0}L{T^{ - 2}}} \right]$(C) $\left[ {ML{T^{ - 2}}} \right]$(D) $\left[ {M{L^2}{T^{ - 2}}} \right]$
Last updated date: 22nd Mar 2024
Total views: 33.6k
Views today: 1.33k
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Hint When a body is deformed by an external force, the internal restoring forces will oppose this force and restore the original shape of the object. This restoring force developed per unit area is called stress. The ratio of change in dimension to the original dimension is called strain.
Complete Step by step solution
According to Hooke’s law within the elastic limit, stress is proportional to strain
Stress $\propto$strain
$\Rightarrow \dfrac{{stress}}{{strain}} = const$
This constant is called the modulus of elasticity.
Stress can be defined as the force per unit area, i.e.
Stress = $\dfrac{F}{a}$
We know that force, $F = ma$
The dimensional formula for force is $F = m \times a = \left[ {ML{T^{ - 2}}} \right]$
The area is given by length $\times$breadth
The dimensional formula for the area can be written as, $a = l \times b = \left[ {{M^0}{L^2}{T^0}} \right]$
The dimensional formula for stress can be written as,
Stress $= \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{M^0}{L^2}{T^0}} \right]}}$
The dimensional formula for linear stress can be written as,
Stress $= \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
The strain is a dimensionless quantity.
Therefore, the dimensional formula for modulus of elasticity can be written as,
Modulus of elasticity $= \dfrac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]}}{{\left[ {{M^0}{L^0}{T^0}} \right]}} = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
The answer is: Option (A): $\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$
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# Five-number Summary And Box-and-whisker Plot! Trivia Quiz
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Quizzes Created: 1 | Total Attempts: 116
Questions: 10 | Attempts: 116
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How well do you know the five-number summary and box-and-whisker plot? Do you know how to construct this box, the importance of each line, and what it means for the data? Try taking up the trivia quiz below and get to see what facts you are yet to learn about the whole process and how to be effective. All the best!
• 1.
### Determine the 5 number summary for the set of data: 5, 20, 15, 25, 0, 10, 15, 5, 25, 30, 20
• A.
5, 15, 10, 25, 20
• B.
0, 7.5, 17.5, 22.5, 30
• C.
0, 5, 15, 25, 30
C. 0, 5, 15, 25, 30
Explanation
The 5 number summary provides a summary of the distribution of the data. It includes the minimum value (0), the first quartile (5), the median (15), the third quartile (25), and the maximum value (30). These values give an indication of the spread and central tendency of the data set.
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• 2.
### Determine the Interquartile range for the following set of data: 5, 20, 15, 25, 0, 10, 15, 5, 25, 30, 20
• A.
20
• B.
25
• C.
15
A. 20
Explanation
The interquartile range is a measure of the spread of data and is calculated by finding the difference between the upper quartile and the lower quartile. In this case, the upper quartile is 25, which is the third largest value in the data set, and the lower quartile is 5, which is the third smallest value. Therefore, the interquartile range is 25 - 5 = 20.
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• 3.
### Based on the box and whisker plot below, what is the median of the data?
• A.
30
• B.
45
• C.
20
A. 30
Explanation
The median is the middle value in a set of data when arranged in order. In this case, the box and whisker plot shows that the median is located at the middle of the box, which is at the value of 30. Therefore, the median of the data is 30.
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• 4.
### Based on the box and whisker plot below, what is the IQR of the data?
• A.
42
• B.
25
• C.
30
B. 25
Explanation
The IQR, or interquartile range, is a measure of statistical dispersion, specifically the range between the first quartile (Q1) and the third quartile (Q3) of the data. In this case, the box and whisker plot is not provided, so it is not possible to determine the IQR or provide an explanation.
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• 5.
• A.
42
• B.
25
• C.
30
A. 42
• 6.
### What is the 5 number summary of the following data? 100, 150, 50, 30, 90, 50
• A.
30, 50, 50, 100, 150
• B.
30, 50, 90, 100, 150
• C.
30, 50, 75, 100, 150
C. 30, 50, 75, 100, 150
Explanation
The 5 number summary consists of the minimum value, the first quartile (Q1), the median (Q2), the third quartile (Q3), and the maximum value of a dataset. In this case, the minimum value is 30, the first quartile is 50, the median is 75, the third quartile is 100, and the maximum value is 150.
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• 7.
### What is the interquartile range of the following data? 100, 150, 50, 30, 90, 50
• A.
50
• B.
75
• C.
100
B. 75
Explanation
The interquartile range is a measure of the spread of data and is calculated by subtracting the value of the lower quartile from the value of the upper quartile. In this case, the lower quartile is 50 and the upper quartile is 100. Therefore, the interquartile range is 100 - 50 = 50.
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• 8.
### What percent of the data listed below is higher than the upper quartile?88, 53, 72, 67, 48, 63, 25, 34, 46, 55, 72, 77
• A.
25%
• B.
50%
• C.
75%
A. 25%
Explanation
The upper quartile is the median of the upper half of the data. To find the upper quartile, we first need to arrange the data in ascending order: 25, 34, 46, 48, 53, 55, 63, 67, 72, 72, 77, 88. The upper half of the data is: 63, 67, 72, 72, 77, 88. The median of this upper half is the upper quartile. In this case, the upper quartile is 72. There are 6 data points higher than 72 in the given data set. Since there are a total of 12 data points, the percentage of data higher than the upper quartile is (6/12) * 100 = 50%. Therefore, the given answer of 25% is incorrect.
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• 9.
### What percent of the data listed below is higher than the lower quartile?88, 53, 72, 67, 48, 63, 25, 34, 46, 55, 72, 77
• A.
25%
• B.
50%
• C.
75%
C. 75%
Explanation
The lower quartile is the median of the lower half of the data set. In this case, the lower quartile is 48. To determine the percentage of data higher than the lower quartile, we need to count the number of values that are greater than 48. Out of the 12 values listed, 9 values (88, 53, 72, 67, 63, 55, 72, 77) are higher than 48. Therefore, the percentage of data higher than the lower quartile is 75%.
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• 10.
### What percent of the data listed below is higher than the median?88, 53, 72, 67, 48, 63, 25, 34, 46, 55, 72, 77
• A.
25%
• B.
50%
• C.
75%
B. 50%
Explanation
The median is the middle value when the data is arranged in ascending order. In this case, the median is 63. To determine the percentage of data higher than the median, we count the number of values greater than 63, which is 6 (88, 72, 67, 72, 77, 55). Since there are a total of 12 values, the percentage of data higher than the median is 6/12 = 50%.
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# Math Word Problems About Bees
Humans have practiced apiculture -- beekeeping -- for more than 5000 years, and some of the world's oldest artwork depicts scenes of humans attending hives or collecting honey. Much of bee life is organized with a mathematical precision not commonly seen in other social organisms. And as with all enterprises and sciences, there is plenty of math involved in the production of honey and the study of bee colonies. With this in mind, here are seven math word problems inspired by bees.
## (1) New Bees -- Easy
The new queen bee of a certain hive lays lots of eggs every day, but only a small fraction of them develop into healthy worker bees and drones. Yet, as time goes on, a larger and larger proportion of her eggs become bees. Suppose during her first month on the job she produces 1000 new bees, and every month after that she produces 250 more bees than the previous month. How many new bees will she have produced after 3 years on the job?
Solution: Since 3 years is 36 months, on the 36th month she will produce 1000 + 35*250 = 9750 new bees. The total number of new bees produced over the 36 months is the sum 1000 + 1250 + 1500 + 1750 + ... + 9750. To simplify this calculation, we can rewrite it as
(1000 + 1000 + 1000 + ... + 1000) + (0 + 250 + 500 + 750 + ... + 8750)
= 36*1000 + (250 + 500 + 750 + ... + 8750)
= 36000 + 250*(1 + 2 + 3 + ... + 35)
The sum 1 + 2 + 3 + ... + 35 is the 35th triangular number. The nth triangular number T(n) can be computed with the formula T(n) = n(n+1)/2. Plugging in n = 35 gives us T(35) = 630. Going back to the original computation we get
36000 + 250*630 = 193500.
Therefore, she produces new 193500 bees during her first 3 years.
## (2) The Dying Queen -- Easy
The queen of a hive is on her last legs. For the past several months she has not been able to lay as many eggs, in fact, her output has been decreasing at a non-constant rate. Today she produced 8 fewer eggs than yesterday, tomorrow she will produce 9 fewer eggs than today, the day after she will produce 10 fewer eggs than tomorrow, etc. If today she produced 100 eggs, how long will it be until she produces no eggs? And how many eggs will she have laid between today and her last day?
Solution: To solve this problem we start with 100 and successively subtract larger amounts. This gives us the sequence
• 100
• 100 - 9 = 91
• 91 - 10 = 81
• 81 - 11 = 70
• 70 - 12 = 58
• 58 - 13 = 45
• 45 - 14 = 31
• 31 - 15 = 16
• 16 - 16 = 0
The dying queen has seven more days of egg laying ahead. In this time she will lay a total of 100 + 91 + 81+ 70 + 58 + 45 + 31 + 16 = 492 eggs.
## (3) Honeycomb Cell Game -- Medium
Cindy and Lindy are naughty worker bees who would rather play games than collect nectar and pollen. They like to play a guessing game with 19 hexagonal honeycomb cells called "Find the Rock." One of them hides a tiny rock in one of the hexagonal cells, and the other tries to guess where it is with hints based on the previous guess.
The hint the seeker receives is always of the same form: The hider does a clockwise dance if the rock is in a cell adjacent to the one guessed, a counterclockwise dance if the rock is not adjacent to the guessed cell. The hider does a side-to-side dance when the seeker guesses correctly. Cindy's and Lindy's game board is shown below.
Cindy's and Lindy's honeycomb game board with 19 cells.
Suppose Lindy is the hider, Cindy is the seeker, and Cindy's first guess is the center cell. If Lindy does a clockwise dance on Cindy's first guess, what is the maximum number of total guesses Cindy has to make to find the rock, assuming Cindy guesses with optimal strategy after the first guess? (The total is counting the very first guess and the following guesses, including the last guess when she finds the rock.)
Solution: In this problem we need to analyze the worst case scenarios to determine the maximum number of guesses when Cindy's first guess at the center cell elicits a clockwise dance. Cindy makes the best choices available to her and "worst case scenarios" are when Cindy's failure guess the location of the rock is due to bad luck, not to poor strategy.
Lindy's clockwise dancing after the first guess indicates that the rock is hidden in one of the six cells surrounding the first guess. Cindy's second guess will naturally be one of these six cells. The six cells are equivalent in terms of their relative positions, so there is no further strategy for Cindy to apply other than to pick one at random. There are three cases for what happens next.
Case I: Lindy does a side-to-side dance indicating Cindy found the rock. This is not the worst case scenario, however, so we need to look at Cases II and III.
Case II: Lindy does a clockwise dance, indicating the rock is adjacent to Cindy's second guess. This leaves only two possibilities for where the rock is. These two cells are equivalent in terms of their relative positions to the previous guesses and there is no strategy to apply except to guess one of them. The worst case scenario in Case II is that it takes her two more guesses after the second, for a total of four guesses.
Case III: Lindy does a counter clockwise dance, indicating the rock is not adjacent to her second guess. This leaves only three possibilities for where the rock is. These three cells are connected in a chain as in the figure below.
In Case III, Cindy's best strategy is for her third guess to be a cell at one of the ends of this chain. This results in one of three sub cases:
• Case IIIa: Cindy's third guess is right.
• Case IIIb: Cindy's third guess is wrong and Lindy does a clockwise dance. This means the rock is in the middle of the three cells and Cindy's fourth guess will be the last.
• Case IIIc: Cindy's third guess is wrong and Lindy does a counterclockwise dance. This means the rock is in the cell at the other end of the chain, in which case Cindy's fourth guess will still be her last.
All cases and sub-cases considered, if Cindy plays with optimal strategy at every turn, she will need to make at most four guesses to find the rock.
## (4) Nectar and Pollen Proportions -- Medium
Beetrice is the hive's accountant bee and she likes to do annual summaries of all the nectar and pollen collected (by weight). This year, the colony collected 50% more pollen than nectar. Last year, the colony collected 40% more pollen than nectar. Also this year, the colony's nectar collection exceeded last year's nectar collection by 25%. By how much did this year's pollen collection exceed last year's pollen collection?
Solution: Let's say that last year the colony collected X amount of nectar. Then from the information in the problem last year's pollen collection was 1.4X. This is because a 40% increase is computed by multiplying by 1.4.
This year's nectar collection was 1.25X, 25% more than last year's. This year's pollen collection was 1.5(1.25X) = 1.875X.
To compute the increase in pollen from last year to this year, we compute the ratio 1.875X to 1.4X. This works out to 1.875/1.4 ≈ 1.3393. Therefore, this year's pollen collection exceeded last year's by about 33.93%.
## (5) Finding a Shorter Path -- Hard
Buzz has to collect pollen from eight tulips before she can stop working for the day. In each of the four cardinal directions there is one tulip 10 meters from the hive. Also in each of the four cardinal directions there is a tulip 20 meters away from the hive. Starting and ending at the hive, can Buzz visit all eight of these tulips with a path that is less than 134 meters long? Assume the hive is level with the tops of the tulips so that there is no vertical distance to figure in.
Solution: There is more than one path that hits all eight tulips and measures less than 134 meters. Here is one possibility that works out to 60 + 50*sqrt(2) ≈ 130.7 meters.
In the diagram above, H is the hive and the Vs are the tulips. Each square represents 10 meters. You can slightly alter this path to create another with a total length of 60 + 20*sqrt(5) + 20*sqrt(2) meters, which is approximately 133 meters.
## (6) Five Hives Riddle -- Hard
Finn is a beekeeper with five small hives he calls, conveniently, A, B, C, D, and E. He can't remember how many are in each hive so he asks the queen of each hive. Instead of giving Finn a straight answer, they collude to make him work out the answer with a math puzzle.
• The queen of Hive A tells Finn there are 244 bees total in Hives A, B, and C.
• The queen of Hive B tells Finn ther are 159 bees total in Hives B and E
• The queen of Hive C tells Finn there are 297 bees total in Hives C, D, and E
• The queen of Hive D tells Finn there are 226 bees total in Hives A and D.
• The queen of Hive E tells Finn the product of the populations of Hives A and E is 10290.
How many bees are in each hive?
Solution: With a little algebra, we can get the population of Hive C from the first four clues. Notice that
(A+B+C) - (B+E) + (C+D+E) - (A+D) = 2C
which works out to
244 - 159 + 297 - 226 = 2C
156 = 2C
78 = C
Now that Hive C is worked out, the next step is to find the sizes of Hives A and E. The first four equations can be manipulated in a different way to give us
(A+B+C) - (B+E) - (C+D+E) + (A+D) = 2A - 2E
which works out to
244 - 159 - 297 + 226 = 2A - 2E
14 = 2A - 2E
7 = A - E
Now we know that A - E equals 7. We also know that A times E equals 10290. This means we need to find a factor pair for 10290 with a difference of 7. This can be done with trial and error or with the quadratic formula. Using the latter method we get
A*E = 10290
A*(A - 7) = 10290
A^2 - 7A - 10290 = 0
A = [7 ± sqrt(49 + 41160)]/2
A = 105 or -98
Since the population can't be negative, we get A = 105. This also gives us E = 98. Finally, working backward from the second and fourth clues we get B = 61 and D = 121.
## (7) Honeypots -- Hard
Liesl's large honeypots store 2.5 times as much honey as her medium pots, which hold 4 quarts more than her small honeypots. Two large pots and one small pot together can hold as much honey as N medium pots and N small pots, where N is a positive whole number. What is the value of N and how many quarts can Liesl's various honeypots hold?
Solution: Let the small pot's capacity be X. The first statement tells us the medium pot's capacity is X + 4, and the large pot's is 2.5X + 10. The second statement tells us
2(2.5X + 10) + X = N(X + 4) + NX
This simplifies to
6X + 20 = 2NX + 4N
3X + 10 = NX + 2N
NX + 2N - 3X - 10 = 0
(NX + 2N - 3X - 6) - 4 = 0
(N - 3)(X + 2) - 4 = 0
(N - 3)(X + 2) = 4
Since we are given that N is an integer, we can start plugging in N = 4, N = 5, N = 6, etc. into the equation (N - 3)(X + 2) = 4 to see what X values they return.
• N = 4 gives us X = 2
• N = 5 gives us X = 0
• N = 6 gives us X = -2/3
Going higher than N = 6 gives more negative values of X. The only value of N that returns a positive value of X is N = 4. Therefore, the capacities of the pots are 2 quarts, 6 quarts, and 15 quarts.
Bee photographs courtesy of Pixabay public domain stock.
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# 4th Grade MEAP Math Practice Test Questions
Preparing your student for the 4th Grade MEAP Math test? Help your students build MEAP Math test skills by following common 4th Grade MEAP Math questions. Practicing common math questions is the best way to help your students improve their Math skills and prepare for the test. Here, we provide a step-by-step guide to solve 10 common MEAP Math practice problems covering the most important math concepts on the 4th Grade MEAP Math test.
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## 10 Sample 4th Grade MEAP Math Practice Questions
1- What is the value of A in the equation $$64 ÷ A = 8$$
A. 2
B. 4
C. 6
D. 8
2- Jason’s favorite sports team has won 0.62 of its games this season. How can Jason express this decimal as a fraction?
A. $$\frac{6}{2}$$
B. $$\frac{62}{10}$$
C. $$\frac{62}{100}$$
D. $$\frac{6.2}{10}$$
3- Use the models below to answer the question.
Which statement about the models is true?
A. Each shows the same fraction because they are the same size.
B. Each shows a different fraction because they are different shapes.
C. Each shows the same fraction because they both have 3 sections shaded.
D. Each shows a different fraction because they both have 3 shaded sections but a different number of total sections.
4- Sophia flew 2,448 miles from Los Angeles to New York City. What is the number of miles Sophia flew rounded to the nearest thousand?
A. 2000
B. 2400
C. 2500
D. 3000
5- Write $$\frac{124}{1000}$$ as a decimal number.
A. 1.24
B. 0.124
C. 12.4
D. 0.0124
6- Erik made 12 cups of juice. He drinks 3 cups of juice each day. How many days will Erik take to drink all of the juice he made?
A. 2 days
B. 4 days
C. 8 days
D. 9 days
7- Jason has prepared $$\frac{4}{10}$$ of his assignment. Which decimal represents the part of the assignment Jason has prepared?
A. 4.10
B. 4.01
C. 0.4
D. 0.04
8- Emma described a number using these clues.
• 3 digits of the number are 4, 7, and 9
• The value of the digit 4 is $$(4 × 10)$$
• The value of the digit 7 is $$(7 × 1000)$$
• The value of the digit 9 is $$(9 × 10000)$$
Which number could fit Emma’s description?
A. 9,724.04
B. 90,734.40
C. 97,040.04
D. 98,740.70
9- There are 18 boxes and each box contains 26 pencils. How many pencils are in the boxed in total?
A. 108
B. 208
C. 468
D. Not here
10- Emily and Ava were working on a group project last week. They completed $$\frac{7}{10}$$ of their project on Tuesday and the rest on Wednesday.
Ava completed $$\frac{3}{10}$$ of their project on Tuesday. What fraction of the group project did Emily complete on Tuesday?
A. $$\frac{2}{10}$$
B. $$\frac{3}{10}$$
C. $$\frac{4}{10}$$
D. $$\frac{5}{10}$$
## Best 4th Grade MEAP Math Prep Resource for 2022
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1- D
$$A = 64 ÷ 8$$
$$A=8$$
2- C
0.62 is equal to $$\frac{62}{100}$$
3- D
the model for the first fraction is divided into 6 equal parts. We shade 3_6 to show the same amount as 1_2. The model for the second fraction is divided into 8 equal parts. We shade 3_ 8 that it shows these two models are different fractions.
4- A
When rounding to the nearest thousand, you will need to look at the last three digits. If the last three digits are 449 or less round to the next number that is smaller than the number given and end with three zeros.
On the other hand, If the last three digits are 500 or more, round to the next number bigger than the given number and end with three zeros.
5- B
$$\frac{124}{1000}$$ is equal to 0.124
6- B
$$12 ÷ 3 = 4$$
7- C
$$\frac{4}{10}=0.4$$
8- C
9- C
$$18 × 26 = 468$$
10- C
$$\frac{7}{10}-\frac{3}{10}=\frac{4}{10}$$
## The Best Books to Ace the 4th Grade MEAP Math Test
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# Prime Factor Chart
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Prime number is a number which has no other factors other than one and itself. Examples of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29……..
Prime factors of a given number is expressing the given number as a product of all prime numbers.
Example 1: Find the prime factors of the number 24?
Solution: Given is a number 24.
To find its prime factors we divide the given number with the smallest prime number first.
24 ÷ 2 = 12; so, 24 = 12 * 2
Now 12 can be further divided by the prime number 2.
So here we have 12 ÷ 2 = 6. 24 = 2 * 2 * 6.
The number 6 can be further divided by 2.
So here we have 6 ÷ 2 = 3.
The number 3 itself is a prime number. So it cannot be further divided by another prime number.
Hence the number 24 = 2 * 2 * 2 * 3.
Example 2: Find the prime factors of the number 15?
Solution: Given is a number 15.
To find its prime factors we divide the given number with the smallest prime number first.
Here 15 cannot be divided by 2, so the next prime number 3 can be used for factorization.
So here we have 15 ÷ 3 = 5.
The number 5 itself is a prime number. So it cannot be further divided by another prime number.
Hence the number 15 = 3 * 5.
Prime number is a number which has no other factors other than one and itself. Examples of prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29……..
Prime factors of a given number is expressing the given number as a product of all prime numbers.
Example 1: Find the prime factors of the number 24?
Solution: Given is a number 24.
To find its prime factors we divide the given number with the smallest prime number first.
24 ÷ 2 = 12; so, 24 = 12 * 2
Now 12 can be further divided by the prime number 2.
So here we have 12 ÷ 2 = 6. 24 = 2 * 2 * 6.
The number 6 can be further divided by 2.
So here we have 6 ÷ 2 = 3.
The number 3 itself is a prime number. So it cannot be further divided by another prime number.
Hence the number 24 = 2 * 2 * 2 * 3.
Example 2: Find the prime factors of the number 15?
Solution: Given is a number 15.
To find its prime factors we divide the given number with the smallest prime number first.
Here 15 cannot be divided by 2, so the next prime number 3 can be used for factorization.
So here we have 15 ÷ 3 = 5.
The number 5 itself is a prime number. So it cannot be further divided by another prime number.
Hence the number 15 = 3 * 5.
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# Video: CBSE Class X • Pack 5 • 2014 • Question 19
CBSE Class X • Pack 5 • 2014 • Question 19
05:23
### Video Transcript
Two ships are at sea on either side of a lighthouse, such that the ships and the lighthouse are all lying on the same straight line. The angles of depression of the two ships, as observed from the top of the lighthouse, are 60 degrees and 45 degrees. If the height of the lighthouse is 200 metres, find the distance between the two ships. Take root three is equal to 1.73.
Let’s begin by drawing a sketch of the ships and lighthouse. Remember, a sketch does not need to be to scale, but it should be roughly in proportion so that we can use it to figure out how best to answer the question. Remember, the angle of depression of a ship from the lighthouse is the angle between the horizontal and the line of sight of the ship from the top of the lighthouse.
Since the sea is parallel to the horizontal, we can include the angle between the line of sight and the sea. Alternate angles are equal, so it’s 60 degrees. Similarly, the angle between the line of sight of the top of the lighthouse from the second ship and the sea is 45 degrees.
We can assume that the lighthouse is perpendicular to the sea. And in doing so, we can see that we have two right-angled triangles, for which we know the measure of an additional angle and the length of one of the sides. We can use right angle trigonometry to calculate missing lengths in these triangles.
Let’s consider the triangle made by the first ship and the lighthouse. Labelling the triangle, we can see that the side representing the line of sight is the hypotenuse. That’s the longest side of the triangle, and it can be found by looking for the side directly opposite the right angle.
The opposite side is the side representing the lighthouse; it’s the side opposite the given angle. The adjacent side is the other side. This time, it’s the one next to the included angle, and it’s the distance between the lighthouse and the first ship.
Since we know the length of the opposite side and we’re looking to calculate the length of the adjacent, we use the tangent ratio. Tan of 𝜃 is equal to opposite over adjacent. Substituting what we know into this formula, and we get tan of 60 is equal to 200 divided by 𝑥 one. Tan of 60 is equal to root three, so our equation becomes root three is equal to 200 over 𝑥 one. And to solve, we first multiply both sides by 𝑥 one. That gives us root three 𝑥 one is equal to 200. Next, we divide through by root three, and we get that 𝑥 one is equal to 200 over root three.
Now, at this point, it’s sensible to rationalise the denominator. And we do so by multiplying by the numerator and the denominator of this fraction by root three. This just creates an equivalent fraction. But in doing so, we end up with just three on the denominator, which is a rational number. We get 𝑥 one is equal to 200 root three over three. The distance between the first ship and the lighthouse is 200 root three over three metres.
We’re now going to repeat this process to find the distance between the second ship and the lighthouse. Once again, we know the length of the opposite side. And we’re looking to calculate the length of the adjacent, so we’ll use the tangent ratio. This time, we get tan of 45 is equal to 200 over 𝑥 two. Tan of 45 is equal to one, so we get one is equal to 200 over 𝑥 two. We can solve this equation by multiplying both sides by 𝑥 two. And in doing so, we can see that the distance between the lighthouse and the second ship is 200 metres.
The total distance between the two ships is given by the sum of the distance between this first ship and the lighthouse and the distance between the second ship and the lighthouse. That’s 200 root three over three plus 200. Now we were told to take root three is equal to 1.73. We’ll replace root three with 1.73, and we’ll calculate 200 multiplied by 1.73 over three. 200 can be written as two multiplied by 100, and 100 multiplied by 1.73 is 173. Two multiplied by 173 is 346. So our calculation becomes 346 divided by three.
Now the digits three, four, and six when added together do not make a multiple of three. So we’re not gonna get a particularly nice number when we divide 346 by three. Let’s use the bus stop method and see what we get. Three divided by three is one. Four divided by three is one remainder one. And 16 divided by three is five remainder one. 10 divided by three is three remainder one, and 10 divided by three again is three remainder one.
We can see that this three is going to reoccur, so 346 divided by three is 115.3 recurring. We can replace 200 root three over three with 115.3 recurring in our calculation, to give us an answer of 315.3 recurring. Correct to one decimal place, the distance between the two ships is 315.3 metres.
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Module 4: Discrete Random Variables
# Poisson Distribution
Barbara Illowsky & OpenStax et al.
There are two main characteristics of a Poisson experiment.
1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate and independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages.
2. The Poisson distribution may be used to approximate the binomial if the probability of success is “small” (such as 0.01) and the number of trials is “large” (such as 1,000). You will verify the relationship in the homework exercises. n is the number of trials, and p is the probability of a “success.”
The random variable $X=$ the number of occurrences in the interval of interest.
### Example
The average number of loaves of bread put on a shelf in a bakery in a half-hour period is 12. Of interest is the number of loaves of bread put on the shelf in five minutes. The time interval of interest is five minutes. What is the probability that the number of loaves, selected randomly, put on the shelf in five minutes is three?
Let $X=$ the number of loaves of bread put on the shelf in five minutes. If the average number of loaves put on the shelf in 30 minutes (half-hour) is 12, then the average number of loaves put on the shelf in five minutes is $left(frac{5}{30}right)left(12right)=2$ loaves of bread.
The probability question asks you to find $P(x=3)$.
## Notation for the Poisson: $P=$ Poisson Probability Distribution Function
$X{sim}P(mu)$
Read this as “X is a random variable with a Poisson distribution.” The parameter is $mu$ (or $lambda$); $mu$ (or $lambda$)$=$ the mean for the interval of interest.
### Example
Leah’s answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
Let $X=$ the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or $frac{1}{4}$ hour.)
$x=0,1,2,3,...$
If Leah receives, on the average, six telephone calls in two hours, and there are eight 15-minute intervals in two hours, then Leah receives $left(frac{1}{8}right)left(6right)=0.75$ calls in 15 minutes, on average. So, $mu=0.75$ for this problem.
$X{sim}P(0.75)$
Find $P(x>1)$. $P(x>1)=0.1734$ (calculator or computer)
• Press 1 – and then press 2nd DISTR.
• Arrow down to poissoncdf. Press ENTER.
• Enter (.75,1).
• The result is $P(x>1)=0.1734$.
Note: The TI calculators use $lambda$ (lambda) for the mean.
The probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734:
$P(x>1)=1−text{poissoncdf}(0.75,1)$
The graph of $X{sim}P(0.75)$ is:
The y-axis contains the probability of x where $X=$ the number of calls in 15 minutes.
### Example
According to Baydin, an email management company, an email user gets, on average, 147 emails per day. Let $X=$ the number of emails an email user receives per day. The discrete random variable X takes on the values $x=$0, 1, 2 …. The random variable has a Poisson distribution: $X{sim}P(147)$. The mean is 147 emails.
1. What is the probability that an email user receives exactly 160 emails per day?
2. What is the probability that an email user receives at most 160 emails per day?
3. What is the standard deviation?
1. $P(x=160)=text{poissonpdf}(147, 160){approx}0.0180$
2. $P(xleq160)=text{poissoncdf}(147, 160){approx}0.8666$
3. Standard Deviation$=sigma=sqrt{mu}=sqrt{144}approx12.1244$
## Review
A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is “small” (less than or equal to 0.05) and the number of trials is “large” (greater than or equal to 20).
## Formula Review
$X{sim}P(mu)$ means that X has a Poisson probability distribution where $X=$ the number of occurrences in the interval of interest.
X takes on the values $x=$0, 1, 2, 3, …
The mean μ is typically given.
The variance is $σ^{2}=mu$, and the standard deviation is $sigma=sqrt{mu}$.
When P(μ) is used to approximate a binomial distribution, $mu=np$ where n represents the number of independent trials and represents the probability of success in a single trial.
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# Positive Definite Negative Definite
## The Discriminant and the Quadratic Graph
The discriminant of the quadratic equation $ax^2+bx+c=0$ is $\Delta = b^2-4ac$.
We used the discriminant to determine the number of real roots of the quadratic equation. If they exist, these roots correspond to $x$-intercepts of the quadratic $y = ax^2+bx+c$. The discriminant tells us about the relationship between a quadratic function and the $x$-axis.
Case 1: $y=x^2-2x+2$
\begin{align} \displaystyle a &= 1, b=-2, c=2 \\ \Delta &= b^2-4ac \\ &= (-2)^2 - 4 \times 1 \times 2 \\ &= -4 \lt 0 \\ \end{align}
The graph does not cut the $x$-axis and is always above the $x$-axis, which means $y=x^2-2x+2 \gt 0$.
This case is Positive Definite because $x^2-2x+2$ is always positive for all real $x$.
Case 2: $y=x^2-2x+1$
\begin{align} \displaystyle a &= 1, b=-2, c=1 \\ \Delta &= b^2-4ac \\ &= (-2)^2 - 4 \times 1 \times 1 \\ &= 0 \\ \end{align}
The graph touches the $x$-axis the graph is mostly above the $x$-axis, which means $y=x^2-2x+1 \ge 0$.
This case is Not Positive Definite because $x^2-2x+1$ is not always positive for all real $x$.
Case 3: $y=x^2-2x$
\begin{align} \displaystyle a &= 1, b=-2, c=0 \\ \Delta &= b^2-4ac \\ &= (-2)^2 - 4 \times 1 \times 0 \\ &= 4 \gt 0 \\ \end{align}
The graph cuts the $x$-axis twice and and the graph is above or below the $x$-axis depending on $x$ values.
This case is Not Positive Definite because $x^2-2x$ is not always positive for all real $x$.
Case 4: $y=-x^2+2x-2$
\begin{align} \displaystyle a &= -1, b=2, c=-2 \\ \Delta &= b^2-4ac \\ &= 2^2 - 4 \times (-1) \times (-2) \\ &= -4 \lt 0 \\ \end{align}
The graph does not cut the $x$-axis and is always below the $x$-axis, which means $y=-x^2+2x-2 \lt 0$.
This case is Negative Definite because $-x^2+2x-2$ is always negative for all real $x$.
Case 5: $y=-x^2+2x-1$
\begin{align} \displaystyle a &= -1, b=2, c=-1 \\ \Delta &= b^2-4ac \\ &= 2^2 - 4 \times (-1) \times (-1) \\ &= 0 \\ \end{align}
The graph touches the $x$-axis the graph is mostly below the $x$-axis, which means $y=-x^2+2x-1 \le 0$.
This case is Not Positive Definite because $x^2-2x+1$ is not always negative for all real $x$.
Case 6: $y=-x^2+2x$
\begin{align} \displaystyle a &= -1, b=2, c=0 \\ \Delta &= b^2-4ac \\ &= 2^2 - 4 \times (-1) \times 0 \\ &= 4 \gt 0 \\ \end{align}
The graph cuts the $x$-axis twice and and the graph is above or below the $x$-axis depending on $x$ values.
This case is Not Negative Definite because $-x^2+2x$ is not always negative for all real $x$.
## Positive Definite
Quadratics which are positive for all real values of $x$ like Case 1: $y=x^2-2x+2$
\begin{align} \displaystyle ax^2+bx+c &\gt 0 \\ a &\gt 0 \\ b^2 - 4ac &\lt 0 \\ \end{align}
## Negative Definite
Quadratics which are negative for all real values of $x$ like Case 4: $y=-x^2+2x-2$
\begin{align} \displaystyle ax^2+bx+c &\lt 0 \\ a &\lt 0 \\ b^2 - 4ac &\lt 0 \\ \end{align}
### Example 1
Use the discriminant to determine the relationship between the graph of $y=2x^2-4x+5$ and the $x$-axis.
### Example 2
Use the discriminant to determine the relationship between the graph of $y=-x^2+6x-9$ and the $x$-axis.
### Example 3
Use the discriminant to determine the relationship between the graph of $y=-x^2+2x-4$ and the $x$-axis.
### Example 4
Use the discriminant to determine the relationship between the graph of $y=x^2+x-2$ and the $x$-axis.
### Example 5
Use the discriminant to determine the relationship between the graph of $y=4x^2-4x+1$ and the $x$-axis.
### Example 6
Use the discriminant to determine the relationship between the graph of $y=-2x^2+3x+1$ and the $x$-axis.
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# What Is The Cube Root Of 75?
• Cube root ∛75 cannot be reduced, because it already is in its simplest form.
• All radicals are now simplified. The radicand no longer has any cubed factors.
## Determine The Cubed Root Of 75?
• The cubed root of seventy-five ∛75 = 4.2171633265087
## How To Calculate Cube Roots
• The process of cubing is similar to squaring, only that the number is multiplied three times instead of two. The exponent used for cubes is 3, which is also denoted by the superscript³. Examples are 4³ = 4*4*4 = 64 or 8³ = 8*8*8 = 512.
• The cubic function is a one-to-one function. Why is this so? This is because cubing a negative number results in an answer different to that of cubing it's positive counterpart. This is because when three negative numbers are multiplied together, two of the negatives are cancelled but one remains, so the result is also negative. 7³ = 7*7*7 = 343 and (-7)³ = (-7)*(-7)*(-7) = -343. In the same way as a perfect square, a perfect cube or cube number is an integer that results from cubing another integer. 343 and -343 are examples of perfect cubes.
## Mathematical Information About Numbers 7 5
• About Number 7. Seven is a prime number. It is the lowest natural number that cannot be represented as the sum of the squares of three integers. The corresponding cyclic number is 142857. You can use this feature to calculate the result of the division of natural numbers by 7 without a calculator quickly. A seven-sided shape is a heptagon. One rule for divisibility by 7 leads to a simple algorithm to test the rest loose divisibility of a natural number by 7: Take away the last digit, double it and subtract them from the rest of the digits. If the difference is negative, then you're leaving the minus sign. If the result has more than one digit, so you repeat steps 1 through fourth. Eventually results are 7 or 0, then the number is divisible by 7 and not otherwise.
• About Number 5. Integers with a last digit as a zero or a five in the decimal system are divisible by five. Five is a prime number. All odd multiples of five border again with the five (all even with zero). The fifth number of the Fibonacci sequence is a five. Five is also the smallest prime number that is the sum of all other primes which are smaller than themselves. The Five is a Fermat prime: 5 = 2 ^ {2 ^ 1} +1 and the smallest Wilson prime. Number five is a bell number (sequence A000110 in OEIS). There are exactly five platonic bodies. There are exactly five tetrominoes.
## What is a cube root?
In arithmetic and algebra, the cube of a number n is its third power: the result of the number multiplied by itself twice: n³ = n * n * n. It is also the number multiplied by its square: n³ = n * n².
This is also the volume formula for a geometric cube with sides of length n, giving rise to the name. The inverse operation of finding a number whose cube is n is called extracting the cube root of n. It determines the side of the cube of a given volume. It is also n raised to the one-third power.
Both cube and cube root are odd functions: (-n)³ = -(n³). The cube of a number or any other mathematical expression is denoted by a superscript 3, for example 2³ = 8 or (x + 1)³.
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Help with Understanding Geometry
Geometry is easier than you think. Plane geometry is based on concepts that on their own are extremely simple; if you can draw a line, you can do geometry. So, whether you need some help with homework or you’d just like to understand geometry a bit more, here’s a guide to understanding it.
Firstly, lets go right back to basics. Draw two lines on a sheet of paper; one going vertically down the left hand side and the other horizontally along the bottom. With a ruler, put a small mark every centimetre on both lines. Already, you have in front of you the basis of all geometry. With geometry, using nothing more than these two lines you can describe any location on the page.
For example, we can describe the exact position of a dot on the page (technically now a two-dimensional ‘plane’) with just two numbers. These numbers are called co-ordinates (“co” means “together”, since we always need both numbers together), and are written in brackets like this: (4, 7).
This simply tells us where our dot is in relation to our lines. To find our dot, you could put a pen at the place where the two lines cross, called the ‘origin’, and follow the numbers like instructions. (4, 7) therefore means we move the pen 4 units to the right, then 7 units towards the top of the page. Simple! We just located a dot, technically called a ‘point’, using just two numbers. In fact, we could even use negative numbers to describe a point that is to the left of the vertical line or below the horizontal line. For example, (5, -7) means ‘five units right and seven units down’.
The two lines we drew are called ‘axes’ – a horizontal axis and a vertical axis. To make it quicker to write down, we can give them a single letter instead. The standard convention is to call the horizontal axis ‘x’ and the vertical one ‘y’. If you’re wondering where ‘z’ has got to, we’ll mention this later.
Now we have a system for describing locations in the form of points we can describe more complex objects. A point is known as zero-dimensional because it has no size; it is just the name given to a place precisely at the co-ordinates we want. If we draw two points and join them up we now have a one-dimensional line segment. It’s only dimension is length (of course, the pencil line has width so we can see it but in theory it has no width).
We can also define lines by an equation that tells us what the co-ordinates will be. For example, we know that (4, 7) means x=4 and y=7, meaning ‘four along and seven up’. Now I can describe a line as y=4. This is a simple way of saying the line contains the points (0, 4), (1, 4), (2, 4) and so on. Whatever x is, y will always be 4 and if we join these points up we get a straight line that goes through the y-axis right on the number four. We can do the same for the x-axis: x=6 will be a vertical line that goes through the x-axis on the sixth mark.
The same concept can be used in geometry to describe any line, it just depends how adventurous you want to get with the equation! For example, y=2x simply means whatever number you go across, you go twice as many up. The points (1, 2), (2, 4) and (3, 6) would all be on this line. There is no limit to how complex a line can be, for example, y=17x + 5 simply means whatever x value you choose, the corresponding y value is 17 times larger and has five added.
Lines are infinitely long because we can choose any value for x. This means that any two lines on the same plane will either be parallel (equal distance from each other all the way along) or will eventually meet each other at some point.
If the equation for a line has numbers being squared (x squared will be called x^2 due to Helium’s typesetting limitations) then the line will be a curve. If you try and draw y=x^2 you might not think it looks that way if you only put a dot on (1, 1), (2, 4), (3, 9) and so on. If, however, we take half units into account, for example by putting a dot on (1.5, 2.25), (2.5, 6.25) and so on, it looks a bit more curvy. If we put a dot on every 10th of a unit it would look even more like a curve. If it was possible to put an infinite number of dots on the page the line would be a perfect curve. In practice, though, it’s fine to put in only a few dots and sketch the curve in yourself.
In case you’re wondering, a circle can be drawn by plotting y^2 + x^2 = the radius of the circle squared. Try plotting y^2 + x^2 = 25 and you’ll get a circle with a radius of five units.
There’s no limit to how far we can extend geometry – we can add a third dimension to describe points above or below the page, which would look like (2, 5, 9), and the third axis would be called ‘z’. Or, we could make even more complex line equations like x cubed, x divided by a number, or even a number divided by x. But, as we have just seen, geometry is built on principles that are really quite simple!
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# Sum (Sigma) Notation
Hello. This post will be on sigma notation. Instead of having , you can use sigma notation such as to save space, ink and paper.
When it comes to Sigma notation, one must have a keen eye in identifying patterns in numbers.
Finite Sums
To illustrate finite sums/series, an example will be shown.
In a finite sum/series there are a finite number of terms.
In the sigma (sum) notation we start from substituting to get 1 as the first term. The next case is when to get 2, then to get 3 all the way to the upper limit of .
Another version of a finite sum is where the upper limit is an arbitrary positive integer. As an example we can have:
A more generalized version of a finite sum is where we add unknown integers from to (where ) such as:
Sometimes we deal with the sum of variables with increasing numeric subscripts. Consider this example:
Infinite Sums
In an infinite sum/series, there is an infinite (never-ending) number of terms. As an example we can have:
The sum above computes to positive infinity and can be considered an infinite sum.
Some Properties
There are a lot of properties when it comes to sigma/summation notation. Here are a couple.
Sum Formulas
This section will briefly look at sum formulas. A more extensive look at sum formulas can be found in a university level Calculus course which focuses on Sequences and Series.
Here are a few sum formulas.
Examples
Example One
The sum can be represented by . Using the third formula from the previous section, this sum is equal to
Example Two
If we add even numbers starting from 2 up to 50 we have:
.
Factoring out a 2 from each term we have:
In sigma notation, the above can be represented as:
This sum evaluates to:
Example Three (Fractions)
There are cases where we add fractions. Consider this example.
From term to term, the denominator increases by one. The sum above can be represented as:
Example Four (Alternating Sums)
This example is a more involved example (more for university calculus students).
When we think of sums adding is usually involved. However we can “add” negative numbers. There are cases where the sign is alternating. One such example is:
When we add positive numbers from 1 to 20 the was needed. Since we have alternating sums we need the part in the sum. Remember that . Since every second term is negative, we have in the exponent of .
One can notice that we have two sums within this alternating signs sum. We have odd numbers and even numbers. Given a positive whole number an odd number is the form of (or ) and an even number is of the form of . Using these ideas in mind the sum can be separated into two.
The sum of odd numbers () can be represented as:
The sum of (negative) even numbers () can be represented as:
The original sum can be split into two sums. One sum has the positive odd numbers from 1 to 20 and the other sum has negative numbers from 1 to 20.
Using summation properties the above can be simplified and evaluated.
The sum computes to -10. This makes sense since the first pair of 1 – 2 is -1, the next pair of 3 – 4 is -1. This happens eight more times so we add (-1) ten times to get -10.
Practice Problems
Here are some practice problems. The summation properties and formulas will come in handy. Hints and solutions are provided in the next sections.
1) Find the sum of the numbers from 1 to 15 (including 1 and 15).
2) Find the sum of the numbers from 1 to 500 (including 1 and 500).
3) Evaluate the sum using summation notation, properties and/or formulas.
4) Find the sum of using summation notation, properties and/or formulas.
5) Evaluate using summation notation and properties.
6) Is greater than ?
7) Evaluate .
8) Evaluate .
Hints To Practice Problems
1) Use the formula .
2) Use the formula .
3) Notice how each term in the sum is a squared number.
4) Each term in the sum is a cubed number.
5) This is an alternating signs case.
6) The first sum can be simplified and evaluated. The second sum is similar to the alternating signs example.
7) Use properties and formulas to evaluate the sum.
8) Use properties and formulas to evaluate the sum.
Solutions To Practice Problems
1) 120
2) 125250
3) 650
4) 1296
5)
The original sum can be split into two sums. One sum has negative odd numbers and the other sum has positive even numbers. Here are the calculations.
6) The first sum computes to:
The second sum can be represented as:
The above can be simplified and evaluated as follows.
7)
8) This question is similar to number seven.
Notes
There are more topics related to sigma notation, series and other types of sums which are not covered here. (i.e Arithmetic series, geometric series, Taylor series, etc.)
|
# Polynomial And Synthetic Division
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Are you ready to test your knowledge of polynomial division? Do you know all there is to know about synthetic division? Get ready to test your knowledge!
NOTE:
You will probably want to have a piece of scrap paper and a pencil nearby to write with!
Questions and Answers
• 1.
### Which of the following is a polynomial term?
• A.
365
• B.
5x + 6874
• C.
6874
• D.
2
Correct Answer
B. 5x + 6874
Explanation
A polynomial contains two or more terms.
Rate this question:
• 2.
### After the dividend is divided by the divisor, the left- over part is the
Correct Answer
remainder
Explanation
After the dividend is divided by the divisor, the leftover part is known as the remainder. The remainder represents the amount that is left over after the division process is complete. It is the difference between the dividend and the product of the divisor and quotient. The remainder can be expressed as a whole number or a fraction, depending on the type of division being performed.
Rate this question:
• 3.
### After the expression 5x + 35 is divided by 5, what is the remainder?
Correct Answer
x + 7 , x+7
Explanation
When the expression 5x + 35 is divided by 5, the remainder is 0. This means that the expression is divisible by 5 without any remainder. Therefore, the remainder is x + 7, which is equivalent to 0 when divided by 5.
Rate this question:
• 4.
### What is the Greatest Common Factor in the following expression? 32x2-80x3+16x
• A.
16x
• B.
16
• C.
8x
• D.
2
Correct Answer
A. 16x
Explanation
The greatest common factor (GCF) is the largest number or term that divides evenly into all the terms in an expression. In this case, the GCF of 32x^2, -80x^3, and 16x is 16x. This is because 16x is a common factor of all the terms, and there is no larger factor that can divide evenly into all the terms.
Rate this question:
• 5.
### Divide x2-9x-10 by (x+1)
Correct Answer
x-10, x - 10, x- 10 , X- 10
Explanation
The given expression is x^2-9x-10 and we need to divide it by (x+1). To divide, we can use long division method. By dividing x^2-9x-10 by (x+1), we get x-10 as the quotient and the remainder is 0. Therefore, the correct answer is x-10, x - 10, x- 10 , X- 10.
Rate this question:
• 6.
### What is the remainder when dividing 2x3 – 9x2 + 15 by 2x – 5?
• A.
3
• B.
2x-5
• C.
-5
• D.
-10
Correct Answer
D. -10
Explanation
When dividing 2x^3 - 9x^2 + 15 by 2x - 5 using long division, we get a quotient of x^2 - 2x - 3 and a remainder of -10. Therefore, the remainder when dividing the polynomial by 2x - 5 is -10.
Rate this question:
• 7.
• 8.
### How would you check to make sure that your answer is correct?
• A.
The coeffiecients will all add up to be 37.
• B.
When multiplied times the divisor, the product is the original expression
• C.
There is no way to check this kind of math. (duh!)
• D.
Ask your neighbor in the seat next to you.
Correct Answer
B. When multiplied times the divisor, the product is the original expression
Explanation
To check if the answer is correct, you can multiply the coefficients by the divisor and see if the product matches the original expression.
Rate this question:
• 9.
### Divide x2+9x+14 by x+7.
Correct Answer
x+2, x+ 2, x + 2
Explanation
The given expression x^2 + 9x + 14 can be factored as (x + 2)(x + 7). When we divide this expression by x + 7, we are left with (x + 2) as the quotient. Therefore, the correct answer is x + 2.
Rate this question:
• 10.
### Factor out x2 + 9x + 14 note: type in the answer with NO spaces and using ( & )
Correct Answer
(x+7)(x+2), (x+2)(x+7)
Explanation
The given expression, x^2 + 9x + 14, can be factored as (x+7)(x+2) or (x+2)(x+7). Both of these factorizations are equivalent and correct.
Rate this question:
• 11.
• 12.
### CHALLENGE: Divide 3x3 – 5x2 + 10x – 3 by 3x + 1note: enter the answer with no spaces, the exponents as large numbers (ex- 3x3 = 3x3), and the remainder as a quantity. {ex- (-45/3x+8)}
Correct Answer
x2-2x+4+(-7/3x+1)
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## Intermediate Algebra (12th Edition)
$\left[ -\dfrac{11}{2},-\dfrac{1}{2} \right]$
$\bf{\text{Solution Outline:}}$ To solve the given inequality, $|-2x-6| \le 5 ,$ use the definition of absolute value inequalities. Use the properties of inequalities to isolate the variable. For the interval notation, use a parenthesis for the symbols $\lt$ or $\gt.$ Use a bracket for the symbols $\le$ or $\ge.$ For graphing inequalities, use a hollowed dot for the symbols $\lt$ or $\gt.$ Use a solid dot for the symbols $\le$ or $\ge.$ $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|\lt c$ implies $-c\lt x\lt c$ (or $|x|\le c$ implies $-c\le x\le c$), the inequality above is equivalent to \begin{array}{l}\require{cancel} -5 \le -2x-6 \le 5 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -5+6 \le -2x-6+6 \le 5+6 \\\\ 1 \le -2x \le 11 .\end{array} Dividing by a negative number (and consequently reversing the sign), the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{1}{-2} \ge \dfrac{-2x}{-2} \ge \dfrac{11}{-2} \\\\ -\dfrac{1}{2} \ge x \ge -\dfrac{11}{2} \\\\ -\dfrac{11}{2} \le x \le -\dfrac{1}{2} .\end{array} In interval notation, the solution set is $\left[ -\dfrac{11}{2},-\dfrac{1}{2} \right] .$ The colored graph is the graph of the solution set.
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Course: Calculus, all content (2017 edition)>Unit 4
Lesson 8: Riemann sums
Comparing areas of Riemann sums worksheet
Practice ordering the areas of a left, midpoint, and right Riemann sum from smallest to largest.
Problem 1
Let $A$ denote the area of the shaded region shown below.
We can approximate the exact area $A$ using the following Riemann sums. $L\left(6\right)$ is the left-hand rule with $6$ equal subdivisions. $M\left(6\right)$ is the midpoint rule with $6$ equal subdivisions. $R\left(6\right)$ is the right-hand rule with $6$ equal subdivisions.
Put these three approximations in order from smallest to largest.
Problem 2
Let $A$ denote the area of the shaded region shown below.
We can approximate the exact area $A$ using the following Riemann sums. $L\left(6\right)$ is the left-hand rule with $6$ equal subdivisions. $M\left(6\right)$ is the midpoint rule with $6$ equal subdivisions. $R\left(6\right)$ is the right-hand rule with $6$ equal subdivisions.
Put these three approximations in order from smallest to largest.
Problem 3
Let $A$ denote the area of the shaded region shown below.
We can approximate the exact area $A$ using the following Riemann sums. $L\left(6\right)$ is the left-hand rule with $6$ equal subdivisions. $M\left(6\right)$ is the midpoint rule with $6$ equal subdivisions. $R\left(6\right)$ is the right-hand rule with $6$ equal subdivisions.
Put these three approximations in order from smallest to largest.
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• So, it depends on whether the function is increasing or decreasing?
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UPSKILL MATH PLUS
Learn Mathematics through our AI based learning portal with the support of our Academic Experts!
The natural numbers are $$1$$, $$2$$, $$3$$, $$4$$, …
We need to find the value of $$1 + 2 + 3 + 4 + …. + n$$.
Consider the identity $$(x + 1)^{k + 1} - x^{k + 1}$$.
Since the sum of all the numbers is $$1$$, put $$k = 1$$ in the above identity.
$$(x + 1)^{1 + 1} - x^{1 + 1} = (x + 1)^2 - x^2$$
$$= x^2 + 2x + 1 - x^2$$
$$= 2x + 1$$
$$(x + 1)^2 - x^2 = 2x + 1$$ - - - - (I)
Now, substitute $$x = 1, 2, 3, … n$$ in equation (I).
When $$x = 1$$, $$2^2 - 1^2 = 2(1) + 1$$
When $$x = 2$$, $$3^2 - 2^2 = 2(2) + 1$$
When $$x = 3$$, $$4^2 - 3^2 = 2(3) + 1$$
$$\vdots$$ $$\vdots$$ $$\vdots$$
When $$x = n - 1$$, $$n^2 - (n - 1)^2 = 2(n - 1) + 1$$
When $$x = n$$, $$(n + 1)^2 - n^2 = 2(n) + 1$$
Add all the above equations of $$x$$ values.
$$2^2 - 1^2 + 3^2 - 2^2 + 4^2 - 3^2 + \cdots + n^2 - (n - 1)^2 + (n + 1)^2 - n^2 = 2(1) + 1 + 2(2) + 1 + 2(3) + 1 + \cdots + 2(n - 1) + 1 + 2(n) + 1$$
$$2^2 + 3^2 + 4^2 + \cdots + n^2 + (n + 1)^2 - (1^2 + 2^2 + 3^2 + \cdots + (n - 1)^2 + n^2) = 2(1) + 2(2) + 2(3) + \cdots + 2(n - 1) + 2(n) + (1 + 1 + 1 + ... n \ times )$$
By cancelling the same terms with opposite signs on the LHS, we get:
$$(n + 1)^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n$$
$$n^2 + 2n + 1^2 - 1^2 = 2(1 + 2 + 3 + … + (n - 1) + n) + n$$
$$n^2 + 2n = 2(1 + 2 + 3 + … + (n - 1) + n) + n$$
$$n^2 + 2n - n = 2(1 + 2 + 3 + … + (n - 1) + n)$$
$$n^2 + n = 2(1 + 2 + 3 + … + (n - 1) + n)$$
$\frac{{n}^{2}+n}{2}=1+2+3+...+\left(n-1\right)+n$
Therefore, $1+2+3+...+\left(n-1\right)+n=\frac{n\left(n+1\right)}{2}$.
Sum of first $$n$$ natural numbers $$=$$ $\frac{n\left(n+1\right)}{2}$
Important!
1. The sum of the first $$n$$ natural numbers is also called a Triangular Number because they form triangle shapes.
2. The sum of squares of the first $$n$$ natural numbers are also called Square Pyramidal Numbers because they form pyramid shapes with a square base.
|
# Change of Sign Method.
Pure Mathematics 2 Katie Ruck
Solution of Equations by Numerical Methods
## Change of Sign Method
If we let f(x)=3x³-0.5x²-0.5x-1, in order to solve this equation and determine its roots, it is necessary for it to be written in the form 3x³-0.5x²-0.5x-1=0.
The root of the equation f(x)=0 is indicated where y=f(x) crosses the x-axis.
Roots of the equation 3x³-0.5x²-0.5x-1=0, will be found to a three decimal place accuracy.
Having illustrated the equation graphically using Autograph, it is evident that the equation has only root that lies between the interval [0,1].
Before proceeding, it is necessary to check that there is a sign change in the above interval:
f(0) = 0-0-0-1= -1
f(1) = 3-0.5-0.5-1 =1
The method that will be used for the numerical solution of the equation is the decimal search method.
I will first take the increments in x of size 0.1 within the interval [0,1], calculating the value of the function 3x³-0.5x²-0.5x-1 for each one, until a change of sign is found.
The above table illustrates that there is a sign change. It can therefore be understood that the root lies in the interval [0.8,0.9].
Having narrowed down the interval, I will continue with the decimal search, but now using increments of 0.01 within the interval [0.8,0.9].
This indicates that the root lies in the new interval [0.83,0.84].
I will continue the process using the increments of 0.001 within the interval [0.83,0.84].
The root therefore lies in the interval [0.838,0.839]. I shall now use increments of 0.0001 in the interval [0.838,0.839].
Therefore the root lies between 0.8389 and 0.8390. To three decimal places, the root = 0.839.
## Error Bounds
A change of sign method such as the one used, provides bounds within which a root lies so that the maximum possible error in a result is known.
When x = 0.8385, f(0.8385) = -0.00218772512
When x = 0.8395, f(0.8395) = 0.00280856463
The error bounds of the root 0.839 are 0.839 ± 0.0005.
However, I am able to say that I have a more accurate solution, as I know that the root lies in the interval [0.8389,0.8390].
## Failure of the Change of Sign Method
There are a number of situations that can cause problems for change of sign methods. For example, let y=f(x)=x³-2x²-x+2.63.
This curve is shown graphically below.
With this example I shall use the decimal search method to find a change of sign and so investigate the roots of the previous equation. Integers will be used as the x-values.
From the graph it is evident that the equation x³-2x²-x+2.63=0 has more than one root, i.e. roots that lie in the intervals [-2,-1] and [1,2]. However, the above table illustrates that the method used only detects one such root that lies in the interval [-2,-1].
In this case, using the method of decimal search has caused an incorrect conclusion to be reached. This is because the curve touches the x-axis between x=1 and x=2, therefore there is no change of sign and consequently all change of sign methods are doomed to failure.
## Fixed Point Iteration Method
Let y=f(x)=x3+2x2-4x-4.58. The graph is shown below:
The roots of the equation can be found where f(x)=0. From the graph, it is evident that the roots of the equation lie in the intervals [-3,-2],[-1,0] and [1,2].
Using a sign change search verifies that the intervals within which the roots ...
|
# 10.1 Conics and Calculus. Each conic section (or simply conic) can be described as the intersection of a plane and a double-napped cone. CircleParabolaEllipse.
## Presentation on theme: "10.1 Conics and Calculus. Each conic section (or simply conic) can be described as the intersection of a plane and a double-napped cone. CircleParabolaEllipse."— Presentation transcript:
10.1 Conics and Calculus
Each conic section (or simply conic) can be described as the intersection of a plane and a double-napped cone. CircleParabolaEllipse Hyperbola Conic Sections General second-degree equation Each of the conics can be defined as a collection of points satisfying a certain geometric property,
A circle can be defined as the collection of all points (x, y) that are equidistant from a fixed point (h, k). This definition easily produces the standard equation of a circle Circle
A parabola is the set of all points (x, y) that are equidistant from a fixed line called the directrix and a fixed point called the focus not on the line. The midpoint between the focus and the directrix is the vertex, and the line passing through the focus and the vertex is the axis of the parabola. The standard form of the equation of a parabola with vertex (h,k) isParabolas Vertical axis Horizontal axis
Find the focus of the parabola given by Example
An ellipse is the set of all points (x, y) the sum of whose distances from two distinct fixed points called foci is constant. The line through the foci intersects the ellipse at two points, called the vertices. Ellipses The chord joining the vertices is the major axis, and its midpoint is the center of the ellipse. The chord perpendicular to the major axis at the center is the minor axis of the ellipse.
The standard form of an ellipse with the center (h, k) and major and minor axes of length 2a and 2b, where a > b, is The foci lie on the major axis, c units from the center, with Ellipses Horizontal major axis Vertical major axis
Example Find the center, vertices and the foci of the ellipse given by
Find the center, vertices, and foci of the ellipse given by 4x 2 + y 2 – 8x + 4y – 8 = 0. Solution: 4x 2 – 8x + y 2 + 4y = 8 4(x 2 – 2x + 1) + (y 2 + 4y + 4) = 8 + 4 + 4 4(x – 1) 2 + (y + 2) 2 = 16 Example h = 1, k = –2, a = 4, b = 2, and c = Center: (1, –2) Vertices: (1, –6) and (1, 2)
To measure the ovalness of an ellipse, we define the eccentricity e to be the ratio Eccentricity Notice that 0 < c < a, and thus 0 < e < 1. For an ellipse that is nearly circular, e is very small, and for an elongated ellipse, e is close to 1. If e > 1, we will have a hyperbola.
A hyperbola is the set of all points (x, y) for which the absolute value of the difference between the distances from two distinct fixed points called foci is constant. The line through the two foci intersects a hyperbola at two points called the vertices. The line segment connecting the vertices is the transverse axis, and the midpoint of the transverse axis is the center of the hyperbola. A hyperbola has two separate branches, and has two asymptotes that intersect at the center. Hyperbolas
The standard form of a hyperbola with center (h, k) and transverse axis of length 2a is The foci lie on the transverse axis, c units from the center, with Hyperbola Horizontal Transverse Axis Vertical Transverse Axis The asymptotes pass through the vertices of a rectangle of dimensions 2a by 2b, with its center at (h, k).
Sketch the graph of the hyperbola whose equation is 4x 2 – y 2 = 16. Solution: h = 0, k = 0, a = 2, b = 4, c = The transverse axis is horizontal and the vertices occur at (–2, 0) and (2, 0). The ends of the conjugate axis occur at (0, –4) and (0, 4). Example
Download ppt "10.1 Conics and Calculus. Each conic section (or simply conic) can be described as the intersection of a plane and a double-napped cone. CircleParabolaEllipse."
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710 minus 70 percent
This is where you will learn how to calculate seven hundred ten minus seventy percent (710 minus 70 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 710 minus 70 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 710 of something.
710
(100%)
70 percent means 70 per hundred, so for each hundred in 710, you want to subtract 70. Thus, you divide 710 by 100 and then multiply the quotient by 70 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(710 ÷ 100) × 70
= 497
We made a pink square that we put on top of the image shown above to illustrate how much 70 percent is of the total 710:
The dark blue not covered up by the pink is 710 minus 70 percent. Thus, we simply subtract the 497 from 710 to get the answer:
710 - 497
= 213
The explanation and illustrations above are the educational way of calculating 710 minus 70 percent. You can also, of course, use formulas to calculate 710 minus 70%.
Below we show you two formulas that you can use to calculate 710 minus 70 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
710 - ((710 × 70/100))
710 - 497
= 213
Formula 2
Number × (1 - (Percent/100))
710 × (1 - (70/100))
710 × 0.3
= 213
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
711 minus 70 percent
Here is the next percent tutorial on our list that may be of interest.
|
## Intermediate Algebra (12th Edition)
$6\sqrt{2} \text{ units}$
$\bf{\text{Solution Outline:}}$ Use the Distance Formula to find the distance between the given points $\left( \sqrt{2},\sqrt{6} \right)$ and $\left( -2\sqrt{2},4\sqrt{6} \right)$. $\bf{\text{Solution Details:}}$ With the given points, then $x_1= \sqrt{2} ,$ $x_2= -2\sqrt{2} ,$ $y_1= \sqrt{6} ,$ and $y_2= 4\sqrt{6} .$ Using the Distance Formula which is given by $d=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2} ,$ then \begin{array}{l}\require{cancel} d=\sqrt{(\sqrt{2}-(-2\sqrt{2}))^2+(\sqrt{6}-4\sqrt{6})^2} \\\\ d=\sqrt{(\sqrt{2}+2\sqrt{2})^2+(\sqrt{6}-4\sqrt{6})^2} \\\\ d=\sqrt{(3\sqrt{2})^2+(-3\sqrt{6})^2} \\\\ d=\sqrt{9(2)+9(6)} \\\\ d=\sqrt{18+54} \\\\ d=\sqrt{72} \\\\ d=\sqrt{36\cdot2} \\\\ d=\sqrt{(6)^2\cdot2} \\\\ d=6\sqrt{2} .\end{array} Hence, the distance is $6\sqrt{2} \text{ units} .$
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# Use Angle Properties in a Triangle
In this worksheet, students will use the key fact, that angles in a triangle always add to 180°, to find the value of unknown angles in triangles and identify triangles accurately.
Key stage: KS 4
Year: GCSE
GCSE Subjects: Maths
GCSE Boards: Eduqas, AQA, Pearson Edexcel, OCR,
Curriculum topic: Geometry and Measures, Basic Geometry
Curriculum subtopic: Properties and Constructions Angles
Popular topics: Geometry worksheets
Difficulty level:
#### Worksheet Overview
Oh yes, the humble triangle, we take it for granted.
Its shape has been used by Egyptian kings, musicians and in warning signs, to name just a few of its many uses!
There are lots of properties (facts) about triangles that people have learned over the centuries, and now it is your turn.
Special Triangles
From left to right:
An equilateral has 3 equal sides
An isosceles has 2 equal sides
scalene has no equal sides.
We need to look out for the markings on these triangles, as these tell us which type of triangle we are looking at.
We also need to remember the rules about angles in each type of triangle:
All three angles in an equilateral are equal.
The two angles at the base of an isosceles are equal.
All three angles in a scalene are different.
Finally, any triangle containing a 90 degree angle is called a right-angled triangle.
Right-angled triangles can be scalene or isosceles, but not equilateral.
A Key Triangle Fact:
Angles in a triangle always add up to 180°.
Using this knowledge, we can find missing angles in a triangle.
Let's look at how we can do this now.
e.g. What are the value of angles a and b in the triangle below?
From the markings on these triangles, we can see this is an isosceles triangle.
As a result, we know that the base angles must be the same.
Therefore, b = 70º
Finally, angles in a triangle add to 180º, so angle a = 180 - 70 - 70 = 40º
e.g. What is the value of angle c in the diagram below?
Here, we have been asked to find an angle that is external to the triangle.
We need to put two rules we know together here - if we extend the base of the triangle we have a straight line.
What facts do we know about straight lines?
The angles on a straight line will also add up to 180°, so we can use this fact to find the missing exterior angle.
So angle c = 180 - 45 = 135º
Right then, let's put what we know into action now!
In this activity, we will use the key fact, that angles in a triangle always add to 180°, to find the value of unknown angles in triangles, identify triangles accurately and solve problems involving triangles.
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# Geometry: Constructions
## Contents
page 1 of 2
Page 1
Page 2
#### Angles
An angle is a geometric figure consisting of two rays with a common endpoint. It looks like this:
Figure %: Angle ABC
The common endpoint is called the vertex of the angle; in this case the vertex is point A, which is a part of the ray AB as well as the ray AC. The angle can be called either angle CAB or angle BAC. The only rule in naming an angle is that the vertex must always be the middle "initial" of the angle. The symbol for an angle is this:
Figure %: The symbol for angle ABC
#### Measuring Angles
Long ago people wanted to measure angles, so numbers were arbitrarily assigned to determine the size of angles. Under this arbitrary numbering system, one complete rotation around a point is equal to a 360 degree rotation. (There is another unit of measure for angles besides degrees called radians, in which one full rotation is equal to 2Π radians; in this text we will use degrees as our default unit for measuring angles.) Two angles with the same measure are called congruent angles. Congruence in angles is symbolized by a small arc drawn in the region between rays. Congruent angles are drawn with the same number of such arcs between their rays. An angle's measure determines how it is classified.
#### Zero Angles
An angle with a measure of zero degrees is called a zero angle. If this is hard to visualize, consider two rays that form some angle greater than zero degrees, like the rays in the . Then picture one of the rays rotating toward the other ray until they both lie in the same line. The angle they create has been shrunk from its original measure to zero degrees. The angle that is now formed has a measure of zero degrees.
Figure %: A zero angle
#### Right Angles
An angle with a measure of 90 degrees is called a right angle. A right angle is symbolized with a square drawn in the corner of the angle.
Figure %: A right angle
#### Straight Angles
An angle with a measure of 180 degrees is called a straight angle. It looks just like a line. Don't mix up straight angles with zero angles.
Figure %: A straight angle
#### Acute and Obtuse angles
Another way to classify angles by their measures is to consider whether the angle's measure is greater or less than 90 degrees. If an angle measures less than 90 degrees, it is called an acute angle. If it measures more than 90 degrees, it is called an obtuse angle. Right angles are neither acute nor obtuse. They're just right.
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You are reading an older version of this FlexBook® textbook: CK-12 Geometry - Basic Go to the latest version.
# 3.1: Lines and Angles
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Define parallel lines, skew lines, and perpendicular planes.
• Understand the Parallel Line Postulate and the Perpendicular Line Postulate.
• Identify angles made by two lines and a transversal.
## Review Queue
1. What is the equation of a line with slope -2 and $y-$intercept 3?
2. What is the slope of the line that passes through (3, 2) and (5, -6)?
3. Find the $y-$intercept of the line from #2. Write the equation too.
4. Define parallel in your own words.
Know What? To the right is a partial map of Washington DC. The streets are designed on a grid system, where lettered streets, A through $Z$ run east to west and numbered streets, $1^{st}$ to $30^{th}$ run north to south. Every state also has its own street that runs diagonally through the city.
Which streets are parallel? Which streets are perpendicular? How do you know?
If you are having trouble viewing this map, look at the interactive map: http://www.travelguide.tv/washington/map.html
## Defining Parallel and Skew
Parallel: Two or more lines that lie in the same plane and never intersect.
To show that lines are parallel, arrows are used.
Label It Say It
$\overleftrightarrow{AB} || \overleftrightarrow{MN}$ Line $AB$ is parallel to line $MN$
$l || m$ Line $l$ is parallel to line $m$.
Lines must be marked parallel with the arrows in order to say they are parallel. Just because two lines LOOK parallel, does not mean that they are.
Recall the definition of perpendicular from Chapter 1. Two lines are perpendicular when they intersect to form a $90^\circ$ angle. Below $l \perp \overline{AB}$.
In the definitions of parallel and perpendicular, the word “line,” is used. Line segments, rays and planes can also be parallel or perpendicular.
The image to the left shows two parallel planes, with a third blue plane that is perpendicular to both of them.
An example of parallel planes could be the top of a table and the floor. The legs would be in perpendicular planes to the table top and the floor.
Skew lines: Lines that are in different planes and never intersect.
In the cube:
$\overline{AB}$ and $\overline{FH}$ are skew
$\overline{AC}$ and $\overline{EF}$ are skew
Example 1: Using the cube above, list:
(a) A pair of parallel planes
(b) A pair of perpendicular planes
(c) A pair of skew lines.
Solution: Remember, you only need to use three points to label a plane. Below are answers, but there are other possibilities too.
(a) Planes $ABC$ and $EFG$
(b) Planes $ABC$ and $CDH$
(c) $\overline{BD}$ and $\overline{CG}$
## Parallel Line Postulate
Parallel Postulate: For any line and a point not on the line, there is one line parallel to this line through the point.
There are infinitely many lines that go through $A$, but only one that is parallel to $l$.
Investigation 3-1: Patty Paper and Parallel Lines
Tools Needed: Patty paper, pencil, ruler
1. Get a piece of patty paper (a translucent square piece of paper). Draw a line and a point above the line.
2. Fold up the paper so that the line is over the point. Crease the paper and unfold.
3. Are the lines parallel?
Yes. This investigation duplicates the line we drew in #1 over the point. This means that there is only one parallel line through this point.
## Perpendicular Line Postulate
Perpendicular Line Postulate: For any line and a point not on the line, there is one line perpendicular to this line passing through the point.
There are infinitely many lines that pass through $A$, but only one that is perpendicular to $l$.
Investigation 3-2: Perpendicular Line Construction; through a Point NOT on the Line
Tools Needed: Pencil, paper, ruler, compass
1. Draw a horizontal line and a point above that line. Label the line $l$ and the point $A$.
2. Take the compass and put the pointer on $A$. Open the compass so that it reaches past line $l$. Draw an arc that intersects the line twice.
3. Move the pointer to one of the arc intersections. Widen the compass a little and draw an arc below the line. Repeat this on the other side so that the two arc marks intersect.
4. Take your straightedge and draw a line from point $A$ to the arc intersections below the line. This line is perpendicular to $l$ and passes through $A$.
To see a demonstration of this construction, go to:
Investigation 3-3: Perpendicular Line Construction; through a Point on the Line
Tools Needed: Pencil, paper, ruler, compass
1. Draw a horizontal line and a point on that line. Label the line $l$ and the point $A$.
2. Take the compass and put the pointer on $A$. Open the compass so that it reaches out horizontally along the line. Draw two arcs that intersect the line on either side of the point.
3. Move the pointer to one of the arc intersections. Widen the compass a little and draw an arc above or below the line. Repeat this on the other side so that the two arc marks intersect.
4. Take your straightedge and draw a line from point $A$ to the arc intersections above the line. This line is perpendicular to $l$ and passes through $A$.
To see a demonstration of this construction, go to:
Example 2: Construct a perpendicular line through the point below.
Solution: Even though the point is below the line, the construction is the same as Investigation 3-2. However, draw the arc marks in step 3 above the line.
## Angles and Transversals
Transversal: A line that intersects two other lines.
The area between $l$ and $m$ is the interior.
The area outside $l$ and $m$ is the exterior.
Looking at $t, l$ and $m$, there are 8 angles formed. They are labeled below.
There are 8 linear pairs and 4 vertical angle pairs.
An example of a linear pair would be $\angle 1$ and $\angle 2$.
An example of vertical angles would be $\angle 5$ and $\angle 8$.
Example 3: List all the other linear pairs and vertical angle pairs in the picture above.
Solution:
Linear Pairs: $\angle 2$ and $\angle 4, \ \angle 3$ and $\angle 4, \ \angle 1$ and $\angle 3, \ \angle 5$ and $\angle 6, \ \angle 6$ and $\angle 8, \ \angle 7$ and $\angle 8, \ \angle 5$ and $\angle 7$
Vertical Angles: $\angle 1$ and $\angle 4, \ \angle 2$ and $\angle 3, \ \angle 6$ and $\angle 7$
There are also 4 new angle relationships.
Corresponding Angles: Two angles that are on the same side of the transversal and the two different lines. Imagine sliding the four angles formed with line $l$ down to line $m$. The angles which match up are corresponding.
Above, $\angle 2$ and $\angle 6$ are corresponding angles.
Alternate Interior Angles: Two angles that are on the interior of $l$ and $m$, but on opposite sides of the transversal.
Above, $\angle 3$ and $\angle 5$ are alternate exterior angles.
Alternate Exterior Angles: Two angles that are on the exterior of $l$ and $m$, but on opposite sides of the transversal.
Above, $\angle 2$ and $\angle 7$ are alternate exterior angles.
Same Side Interior Angles: Two angles that are on the same side of the transversal and on the interior of the two lines.
Above, $\angle 3$ and $\angle 5$ are same side interior angles.
Example 4: Using the picture above, list all the other pairs of each of the newly defined angle relationships.
Solution:
Corresponding Angles: $\angle 3$ and $\angle 7, \ \angle 1$ and $\angle 5, \ \angle 4$ and $\angle 8$
Alternate Interior Angles: $\angle 4$ and $\angle 5$
Alternate Exterior Angles: $\angle 2$ and $\angle 7$
Same Side Interior Angles: $\angle 4$ and $\angle 6$
Example 5: For the picture below, determine:
(a) A corresponding angle to $\angle 3$?
(b) An alternate interior angle to $\angle 7$?
(c) An alternate exterior angle to $\angle 4$?
Solution:
(a) $\angle 1$
(b) $\angle 2$
(c) $\angle 5$
Know What? Revisited For Washington DC, all of the lettered and numbered streets are parallel. The lettered streets are perpendicular to the numbered streets. We do not have enough information about the state-named streets to say if they are parallel or perpendicular.
## Review Questions
• Questions 1-3 use the definitions of parallel, perpendicular, and skew lines.
• Question 4 asks about the Parallel Line Postulate and the Perpendicular Line Postulate.
• Questions 5-9 use the definitions learned in this section and are similar to Example 1.
• Questions 10-20 are similar to Examples 4 and 5.
• Question 21 is similar to Example 2 and Investigation 3-2.
• Questions 22-30 are Algebra I review.
1. Which of the following is the best example of parallel lines?
2. Lamp Post and a Sidewalk
3. Longitude on a Globe
4. Stonehenge (the stone structure in Scotland)
2. Which of the following is the best example of perpendicular lines?
1. Latitude on a Globe
2. Opposite Sides of a Picture Frame
3. Fence Posts
4. Adjacent Sides of a Picture Frame
3. Which of the following is the best example of skew lines?
1. Roof of a Home
2. Northbound Freeway and an Eastbound Overpass
3. Longitude on a Globe
4. The Golden Gate Bridge
4. Writing What is the difference between the Parallel Line Postulate and the Perpendicular Line Postulate? How are they similar?
Use the figure below to answer questions 5-9. The two pentagons are parallel and all of the rectangular sides are perpendicular to both of them.
1. Find two pairs of skew lines.
2. List a pair of parallel lines.
3. List a pair of perpendicular lines.
4. For $\overline{AB}$, how many perpendicular lines would pass through point $V$? Name this line.
5. For $\overline{XY}$, how many parallel lines would pass through point $D$? Name this line.
For questions 10-16, use the picture below.
1. What is the corresponding angle to $\angle 4$?
2. What is the alternate interior angle with $\angle 5$?
3. What is the corresponding angle to $\angle 8$?
4. What is the alternate exterior angle with $\angle 7$?
5. What is the alternate interior angle with $\angle 4$?
6. What is the same side interior angle with $\angle 3$?
7. What is the corresponding angle to $\angle 1$?
Use the picture below for questions 17-20.
1. If $m\angle 2 = 55^\circ$, what other angles do you know?
2. If $m\angle 5 = 123^\circ$, what other angles do you know?
3. If $t \perp l$, is $t \perp m$? Why or why not?
4. Is $l || m$? Why or why not?
5. Construction Draw a line and a point not on the line. Construct a perpendicular line to the one your drew.
Algebra Review Find the slope of the line between the two points, $\frac{y_2-y_1}{x_2-x_1}$.
1. (-3, 2) and (-2, 1)
2. (5, -9) and (0, 1)
3. (2, -7) and (5, 2)
4. (8, 2) and (-1, 5)
5. Find the equation of the line from #22. Recall that the equation of a line is $y = mx + b$, where $m$ is the slope and $b$ is the $y-$intercept.
6. Find the equation of the line from #23.
7. Find the equation of the line from #24.
8. Is the line $y=-x+3$ parallel to the line in #26? How do you know?
9. Is the line $y=-x+3$ perpendicular to the line in #26? How do you know?
1. $y = -2x + 3$
2. $m = \frac{-6-2}{5-3} = \frac{-8}{2} = -4$
3. ${\;} \ \ \ 2 = -4(3) + b\!\\{\;} \ \ \ 2 = -12 + b\!\\-14 = b\!\\{\;} \ \ \ y = -4x + 14$
4. Something like: Two lines that never touch or intersect and in the same plane. If we do not say “in the same plane,” this definition could include skew lines.
8 , 9 , 10
Feb 22, 2012
Oct 28, 2015
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# Find HCF and LCM of 510 and 92 by applying the prime factorization method.
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Hint: Break 92 and 510 in prime factors. Then take the lowest power of each of the prime factors to compute HCF. Similarly, to find LCM take the highest power of each of the prime factors.
Given 510 and 92,
92 can be factorized as $2 \times 2 \times 23$.
510 can be factorized as $2 \times 3 \times 5 \times 17$.
After prime factoring the numbers given, let us list down all the prime factors.
Then take the lowest/least power of each of the prime factors. This is the method to compute HCF. The lowest power of the prime factor 2 is 1, which is to say we have a 2 in both the numbers.
Hence HCF is 2.
Next, take the highest power of each of the prime factors. This is the method to compute LCM. The highest power of the prime factors 2, 3, 5, 17 and 23 are 2, 1, 1, 1 and 1 respectively.
Therefore LCM is ${2^2} \times 3 \times 5 \times 17 \times 23 = 23460$
Hence the HCF and LCM of 510 and 92 are 2 and 23460 respectively.
Note: LCM of 92 and 510 , is the smallest number divisible by 92 and 510 both. Therefore we have to take maximum power of all factors to calculate LCM.
Whereas HCF of 92 and 510 is the greatest number that divides both 92 and 510.
For 2 numbers always the product of LCM and HCF is equal to the Product of the given two numbers.
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## Enter fraction(s) below:
15 20
##### First, evaluate the numerator and denominator
The numerator is 15
The denominator is 20
##### Reduce our fraction
GCF(15 and 20) = 5
(1/5) x 15 = 3
(1/5) x 20 = 4
3 4
3/4
#### You have 2 free calculationss remaining
##### Convert this fraction number to a mixed number.
Let's find the whole number first.
The formula for this is:
Numerator/Denominator) in whole numbers
##### Using the fraction you entered of 15/20, let's evaluate:
Whole Number = 15 / 20 = 0.75
Whole Number = Integer(0.75)
Whole Number = 0
##### Find the fractional portion of the mixed number
(Numerator - Denominator x Whole Number) Denominator
##### Using the values we entered and calculated, we get:
Fraction Portion = (15 - 20 x 0) 20
Fraction Portion = 15 - 0 20
Fraction Portion = 15 20
##### Combine both pieces for the mixed number:
Mixed Number = Whole Number & fraction Portion
Mixed Number = 0 & 15/20
Whole Number = 0
3/4
#### You have 2 free calculationss remaining
3/4
##### How does the Fractions and Mixed Numbers Calculator work?
Free Fractions and Mixed Numbers Calculator - Given (improper fractions, proper fraction, mixed numbers, or whole numbers), this performs the following operations:
* Subtraction (Subtracting)
* Positive Difference (Absolute Value of the Difference)
* Multiplication (Multiplying)
* Division (Dividing: complex fraction division is included)
* Compare Fractions
* Simplifying of proper and improper fractions as well as mixed numbers. Fractions will be reduced down as far as possible (Reducing Fractions).
* Reciprocal of a Fraction
* Find all fractions between two fractions
* reduce a fraction
This calculator has 2 inputs.
### What 8 formulas are used for the Fractions and Mixed Numbers Calculator?
a/c + b/c = (a + b)/c
a/c - b/c = (a - b)/c
a/c * b/c = ab/c * c
Reciprocal of a/b = b/a
For more math formulas, check out our Formula Dossier
### What 10 concepts are covered in the Fractions and Mixed Numbers Calculator?
math operation involving the sum of elements
denominator
The bottom portion of a fraction. For a/b, b is the denominator
division
separate a number into parts
fraction
how many parts of a certain size exist
a/b where a is the numerator and b is the denominator
mixed number
a number consisting of an integer and a proper fraction.
multiplication
math operation involving the product of elements
numerator
the number above the line in a common fraction
reciprocal
a number which when multiplied by x yields the multiplicative identity, 1
simplify
reducing the expression/fraction/problem in a simpler form.
subtraction
math operation involving the difference of elements
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9310 minus 35 percent
This is where you will learn how to calculate nine thousand three hundred ten minus thirty-five percent (9310 minus 35 percent). We will first explain and illustrate with pictures so you get a complete understanding of what 9310 minus 35 percent means, and then we will give you the formula at the very end.
We start by showing you the image below of a dark blue box that contains 9310 of something.
9310
(100%)
35 percent means 35 per hundred, so for each hundred in 9310, you want to subtract 35. Thus, you divide 9310 by 100 and then multiply the quotient by 35 to find out how much to subtract. Here is the math to calculate how much we should subtract:
(9310 ÷ 100) × 35
= 3258.5
We made a pink square that we put on top of the image shown above to illustrate how much 35 percent is of the total 9310:
The dark blue not covered up by the pink is 9310 minus 35 percent. Thus, we simply subtract the 3258.5 from 9310 to get the answer:
9310 - 3258.5
= 6051.5
The explanation and illustrations above are the educational way of calculating 9310 minus 35 percent. You can also, of course, use formulas to calculate 9310 minus 35%.
Below we show you two formulas that you can use to calculate 9310 minus 35 percent and similar problems in the future.
Formula 1
Number - ((Number × Percent/100))
9310 - ((9310 × 35/100))
9310 - 3258.5
= 6051.5
Formula 2
Number × (1 - (Percent/100))
9310 × (1 - (35/100))
9310 × 0.65
= 6051.5
Number Minus Percent
Go here if you need to calculate any other number minus any other percent.
9320 minus 35 percent
Here is the next percent tutorial on our list that may be of interest.
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# Percentage Difference, Percentage Error, Percentage Change
They are very similar ...
They all show a difference between two values as a percentage of one (or both) values
• Use Percentage Change when comparing an Old Value to a New Value
• Use Percentage Error when comparing an Approximate Value to an Exact Value
• Use Percentage Difference when both values mean the same kind of thing (one value is not obviously older or better than the other).
(Refer to those links for more details)
## How to Calculate
Step 1: Subtract one value from the other
Step 2: Then divide by ... what?
• Percentage Change: Divide by the Old Value
• Percentage Error: Divide by the Exact Value
• Percentage Difference: Divide by the Average of The Two Values
Step 3: Is the answer negative?
• Percentage Change: a positive value is an increase, a negative value is a decrease.
• Percentage Error: ignore a minus sign (just leave it off), unless you want to know if the error is under or over the exact value
• Percentage Difference: ignore a minus sign, because neither value is more important, so being "above" or "below" does not make sense.
Step 4: Convert this into a percentage (multiply by 100 and add a % sign)
## The Formulas
(Note: the "|" symbols mean absolute value, so negatives become positive.)
Percent Change = New Value - Old Value × 100% |Old Value|
Example: There were 200 customers yesterday, and 240 today:
240 - 200 × 100% = (40/200) × 100% = 20% |200|
A 20% increase.
Percent Error = |Approximate Value - Exact Value| × 100% |Exact Value|
Example: I thought 70 people would turn up to the concert, but in fact 80 did!
|70 - 80| × 100% = (10/80) × 100% = 12.5% |80|
I was in error by 12.5%
(Without using the absolute value, the error is -12.5%, meaning I under-estimated the value)
Percentage Difference = | First Value - Second Value | × 100% (First Value + Second Value)/2
Example: "Best Shoes" gets 200 customers, and "Cheap Shoes" gets 240 customers:
| 240 - 200 | × 100% = |40/220| × 100% = 18.18...% (200+240)/2
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Key
```Intro to Analysis of Algorithms
CSE 4081
VERSION
UG
Fall 2015
Exam 3
Points 40
Time 70 min
[Students ignore these key words: Foundations, Problem solving, Algorithm
understanding.]
Q1a. LP
Minimize (-6x1 -2x2 -4x3)
Such that: 2x1 +2x2 +6x3 ≤ 60
-4x1 -4x2 -10x3 ≥ -48
8x1 +2x2 +4x3 ≤ 72
where: x1, x2, x3 ≥ 0
[7+1+2]
Standard Form
Max: 6x1 + 2x2 + 4x3
Constraints:
2x1 + 2x2 + 6x3 <= 60
4x1 + 4x2 + 10x3 <= 48
8x1 + 2x2 + 4x3 <= 72
Where x1, x2, x3 >= 0
Equivalent Slack Form:
z = 0 + 6x1 + 2x2 + 4x3
Linear Equations:
x4 = 60 – (2x1 + 2x2 + 6x3)
x5 = 48 – (4x1 + 4x2 + 10x3)
x6 = 72 – (8x1 + 2x2 + 4x3)
Constraints:
xi >= 0, i = 1, 2, 3, 4, 5, 6.
Sets: N = {x1, x2, x3}, B = {x4, x5, x6}
Entering variables xe in pivot, let’s say xe = x1. So x6 is the leaving variable (the most
constraining basic variable).
Rewrite:𝑥6 = 72 − 8𝑥1 − 2𝑥2 − 4𝑥3
1
1
1
To: 𝑥1 = 9 − 9 𝑥6 − 4 𝑥2 − 2 𝑥3
Then: z = 0 + 6x1 + 2x2 + 4x3 will be
1
1
1
2
1
𝑧 = 0 + 6 (9 − 𝑥6 − 𝑥2 − 𝑥3 ) + 2𝑥2 + 4𝑥3 = 54 − 𝑥6 + 𝑥2 + 𝑥3
9
4
2
3
2
And:
1
1
1
𝑥1 = 9 − 𝑥6 − 𝑥2 − 𝑥3
9
4
2
1
1
1
2
3
𝑥4 = 60 − 2 (9 − 𝑥6 − 𝑥2 − 𝑥3 ) − 2𝑥2 − 6𝑥3 = 42 + 𝑥6 − 𝑥2 − 5𝑥3
9
4
2
9
2
1
1
1
4
𝑥5 = 48 − 4 (9 − 𝑥6 − 𝑥2 − 𝑥3 ) − 4𝑥2 − 10𝑥3 = 12 + 𝑥6 − 3𝑥2 − 8𝑥3
9
4
2
9
Then x2 is chosen to be the entering variable, then the leaving variable is x5.
1
4
8
Similarly, 𝑥2 = 4 − 3 𝑥5 + 27 𝑥6 − 3 𝑥3
2
1
2
1
1
4
8
So, 𝑧 = 54 − 3 𝑥6 + 2 𝑥2 + 𝑥3 = 54 − 3 𝑥6 + 2 (4 − 3 𝑥5 + 27 𝑥6 − 3 𝑥3 ) + 𝑥3
1
16
1
= 56 − 𝑥5 − 𝑥6 − 𝑥3
6
27
3
Q1b. Is LP in NP-class?
Yes, it is in P-class and hence in NP-class. Polynomial algorithms do exist (Kahachien’s,
Karmarkar’s, and other interior point algorithms).
Q1c. Write a dual LP for the above problem.
Check LP slide number 33.
Q2a. Create the output 3SAT problem for the following SAT problem instance.
[4+6]
Input SAT-clause
Output 3SAT clause
(~a, b)
(~a, b, z1), (~a, b, ~z1)
T: z1=do-not-care T/F
(a, c, d) T
(a, c, d)
T
(g)
(g, z2, z3), (g, z2, ~z3), (g, ~z2, z3), (g, ~z2, ~z3)
z2=T, z3=T: show that clauses=T, T, T, F
z2=T, z3=F: show that clauses=T, T, F, T
z2=F, z3=T: show that clauses=T, F, T, T
z2=F, z3=F: show that clauses=F, T, T, T
All 4 3-clauses cannot be made T.
T
F
(a, c, b, e, f, ~g)
T
(b, ~c, d, e, ~f)
T
(a, c, z4), (~z4, b, z5), (~z5, e, z6), (~z6, f, ~g)
T: z4, z5, z6: do-not-care
(b, ~c, z9), (~z9, d, z10), (~z10, e, ~f)
T: z9, z10 = do-not-care
Q2b. Below is a set of variable assignments for the above source SAT problem:
(a=T, b=T, c=T, e=T, d=F, e=F, f=F, g=F).
Try to use the new 3SAT variables’ assignments to show that the T/F remains the same
for each row. That is, if the source-side clause is T with the above assignments, then all
corresponding target side clauses in the same row can be made T; and if the source-side
clause is F with the above assignments, then all corresponding target side clauses cannot
be made T with respective new variables’ assignments.
Use above values for old variables, and T+F values for each of new variables in each
row. Show that when left column (source-SAT) is T, so is right column (target-3SAT), but
when left column is F, then each assignments on right column is F.
Q3a. Is Strongly-connected-component finding problem (SCC) an NP-hard problem?
[2+8]
No, the algorithm we have is polynomial. Complexity is not on any of my slides, but DFTspanning tree creation is polynomial, I mentioned it many a times.
Q3b. Find SCC in the following directed graph. Show results of intermediate steps, start
with node v1.
v1
v3
v2
v5
v4
v6
From v1 do DFS, then we have: v1(6)-v4(5)-v6(4)-v2(3)-v3(2)-v5(1)
Provide the trees.
Note, there could be many trees, even when started with v1.
No points be deducted if one starts with different node.
Tree may like this:
Create Gr
DFS on Gr, start with node v1 since v1 has the highest number. Then we have:
v1-v2-v6-v4-v3, and v5
v1 tree may like this:
Q4a. What action will you take if the graph below is a communication network, and you
want to avoid the network getting disconnected when any node dies. For any algorithm you
run, you must show the intermediate trees and graphs, start your algorithm with node v1.
(Note that the action asked of you above is subjective, based on the result of your
algorithm.)
[8+2]
Key:
Articulation Points Finding Alg:
Only one AP (v3).
Must show the DFT spanning tree with the (num, low) numbers and tree. Justify, why v3 is
found to be AP.
Action: by-pass v3 by adding an arc in such a way that there is no AP in the graph, e.g.
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Horizontal Translations or Phase Shifts
## Shift right or left along the x-axis.
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Practice Horizontal Translations or Phase Shifts
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Horizontal Translations or Phase Shifts
You are working on a graphing project in your math class, where you are supposed to graph several functions. Things seem to be going well, until you realize that there is a bold, vertical line three units to the left of where you placed your "y" axis! As it turns out, you've accidentally shifted your entire graph. You didn't notice that your instructor had placed a bold line where the "y" axis was supposed to be. And now, all of the points for your graph of the cosine function are three points farther to the right than they are supposed to be along the "x" axis.
You might be able to keep all of your work, if you can find a way to rewrite the equation so that it takes into account the change in your graph.
Can you think of a way to rewrite the function so that the graph is correct the way you plotted it?
### Horizontal Translations
Horizontal translations involve placing a constant inside the argument of the trig function being plotted. If we return to the example of the parabola, y=x2\begin{align*}y = x^2\end{align*}, what change would you make to the equation to have it move to the right or left? Many students guess that if you move the graph vertically by adding to the y\begin{align*}y-\end{align*}value, then we should add to the x\begin{align*}x-\end{align*}value in order to translate horizontally. This is correct, but the graph itself behaves in the opposite way than what you may think.
Here is the graph of y=(x+2)2\begin{align*}y = (x + 2)^2\end{align*}.
Notice that adding 2 to the x\begin{align*}x-\end{align*}value shifted the graph 2 units to the left, or in the negative direction.
To compare, the graph y=(x2)2\begin{align*}y = (x - 2)^2\end{align*} moves the graph 2 units to the right or in the positive direction.
We will use the letter C\begin{align*}C\end{align*} to represent the horizontal shift value. Therefore, subtracting C\begin{align*}C\end{align*} from the x\begin{align*}x-\end{align*}value will shift the graph to the right and adding C\begin{align*}C\end{align*} will shift the graph C\begin{align*}C\end{align*} units to the left.
Adding to our previous equations, we now have y=D±sin(x±C)\begin{align*}y=D \pm \sin (x \pm C)\end{align*} and y=D±cos(x±C)\begin{align*}y=D \pm \cos (x \pm C)\end{align*} where D\begin{align*}D\end{align*} is the vertical translation and C\begin{align*}C\end{align*} is the opposite sign of the horizontal shift.
#### Sketch the graph
Sketch y=sin(xπ2)\begin{align*}y=\sin \left( x - \frac{\pi}{2} \right)\end{align*}
This is a sine wave that has been translated π2\begin{align*}\frac{\pi}{2}\end{align*} units to the right.
Horizontal translations are also referred to as phase shifts. Two waves that are identical, but have been moved horizontally are said to be “out of phase” with each other. Remember that cosine and sine are really the same waves with this phase variation.
y=sinx\begin{align*}y = \sin x\end{align*} can be thought of as a cosine wave shifted horizontally to the right by π2\begin{align*}\frac{\pi}{2}\end{align*} radians.
Alternatively, we could also think of cosine as a sine wave that has been shifted π2\begin{align*}\frac{\pi}{2}\end{align*} radians to the left.
#### Sketch the graph
Draw a sketch of y=1+cos(xπ)\begin{align*}y = 1 + \cos (x - \pi)\end{align*}
This is a cosine curve that has been translated up 1 unit and π\begin{align*}\pi\end{align*} units to the right. It may help you to use the quadrant angles to draw these sketches. Plot the points of y=cosx\begin{align*}y = \cos x\end{align*} at 0,π2,π,3π2,2π\begin{align*}0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\end{align*} (as well as the negatives), and then translate those points before drawing the translated curve. The blue curve below is the final answer.
#### Graph the function
Graph y=2+sin(x+3π2)\begin{align*}y=-2+\sin \left(x+\frac{3\pi}{2}\right)\end{align*}
This is a sine curve that has been translated 2 units down and moved 3π2\begin{align*}\frac{3\pi}{2}\end{align*} radians to the left. Again, start with the quadrant angles on y=sinx\begin{align*}y = \sin x\end{align*} and translate them down 2 units.
Then, take that result and shift it 3π2\begin{align*}\frac{3\pi}{2}\end{align*} to the left. The blue graph is the final answer.
### Examples
#### Example 1
Earlier, you were asked if you can think of a way to rewrite the function so that the graph is correct the way you plotted it.
As you've now seen by reading this Concept, it is possible to shift an entire graph to the left or the right by changing the argument of the graph.
So in this case, you can keep your graph by changing the function to y=cos(x3)\begin{align*}y = \cos(x - 3)\end{align*}
#### Example 2
Draw a sketch of y=3+cos(xπ2)\begin{align*}y = 3 + \cos (x - \frac{\pi}{2})\end{align*}
As we've seen, the 3 shifts the graph vertically 3 units, while the π2\begin{align*}-\frac{\pi}{2}\end{align*} shifts the graph to the right by π2\begin{align*}\frac{\pi}{2}\end{align*} units.
#### Example 3
Draw a sketch of y=sin(x+π4)\begin{align*}y = \sin (x + \frac{\pi}{4})\end{align*}
The π4\begin{align*}\frac{\pi}{4}\end{align*} shifts the graph to the left by π4\begin{align*}\frac{\pi}{4}\end{align*}.
#### Example 4
Draw a sketch of y=2+cos(x+2π)\begin{align*}y = 2 + \cos (x + 2\pi)\end{align*}
The 2 added to the function shifts the graph up by 2 units, and the 2π\begin{align*}2\pi\end{align*} added in the argument of the function brings the function back to where it started, so the cosine graph isn't shifted horizontally at all.
### Review
Graph each of the following functions.
1. y=cos(xπ2)\begin{align*}y=\cos(x-\frac{\pi}{2})\end{align*}
2. y=sin(x+3π2)\begin{align*}y=\sin(x+\frac{3\pi}{2})\end{align*}
3. y=cos(x+π4)\begin{align*}y=\cos(x+\frac{\pi}{4})\end{align*}
4. y=cos(x3π4)\begin{align*}y=\cos(x-\frac{3\pi}{4})\end{align*}
5. y=1+cos(xπ4)\begin{align*}y=-1+\cos(x-\frac{\pi}{4})\end{align*}
6. y=1+sin(x+π2)\begin{align*}y=1+\sin(x+\frac{\pi}{2})\end{align*}
7. y=2+cos(x+π4)\begin{align*}y=-2+\cos(x+\frac{\pi}{4})\end{align*}
8. y=3+cos(x3π2)\begin{align*}y=3+\cos(x-\frac{3\pi}{2})\end{align*}
9. y=4+sec(xπ4)\begin{align*}y=-4+\sec(x-\frac{\pi}{4})\end{align*}
10. y=3+csc(xπ2)\begin{align*}y=3+\csc(x-\frac{\pi}{2})\end{align*}
11. y=2+tan(x+π4)\begin{align*}y=2+\tan(x+\frac{\pi}{4})\end{align*}
12. y=3+cot(x3π2)\begin{align*}y=-3+\cot(x-\frac{3\pi}{2})\end{align*}
13. y=1+cos(x3π4)\begin{align*}y=1+\cos(x-\frac{3\pi}{4})\end{align*}
14. \begin{align*}y=5+\sec(x+\frac{\pi}{2})\end{align*}
15. \begin{align*}y=-1+\csc(x+\frac{\pi}{4})\end{align*}
16. \begin{align*}y=3+\tan(x-\frac{3\pi}{2})\end{align*}
To see the Review answers, open this PDF file and look for section 2.13.
### My Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
Amplitude
The amplitude of a wave is one-half of the difference between the minimum and maximum values of the wave, it can be related to the radius of a circle.
Horizontal shift
A horizontal shift is the result of adding a constant term to the function inside the parentheses. A positive term results in a shift to the left and a negative term in a shift to the right.
periodic function
A periodic function is a function with a predictable repeating pattern. Sine waves and cosine waves are periodic functions.
Phase Shift
A phase shift is a horizontal translation or shift.
sinusoidal axis
The sinusoidal axis is the neutral horizontal line that lies between the crests and the troughs of the graph of a sine or cosine function.
sinusoidal function
A sinusoidal function is a sine or cosine wave.
sinusoidal functions
A sinusoidal function is a sine or cosine wave.
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# Euler's Number
Euler's number (not to be confused with Euler's constant, which is something else altogether) is so-called because it was the Swiss mathematician and physicist Leonhard Euler (1707 - 1783) who gave it the label e, although it was not Euler that discovered it. The number is thought to have been discovered by another Swiss mathematician, Jacob Bernoulli (1655-1705), while he was working on the solution to a problem involving compound interest. So what was it exactly that Bernoulli found? Well, suppose we wanted to borrow one hundred pounds (£100) - or whatever currency you prefer to work with - for a period of ten years. Let's assume a rate of interest of ten percent (10%) per year.
If we apply simple interest (interest that is not compounded at all), we would just pay ten percent interest each year on the original loan (called the principal), which over ten years would come to one hundred per cent of the original loan, or one hundred pounds (we would of course have to pay back the original one hundred pounds as well). However, the interest is going to be compounded. This means that at regular intervals during the period of the loan, the amount of interest currently owed is added to the original loan. Each time this happens, the total amount owed increases, and we must thereafter pay the agreed rate of interest on the total amount outstanding. This means we pay more interest (as a percentage of the original loan) as time goes by.
Let's see how this works. We'll assume for the moment that the interest on our ten year loan is compounded annually. The table below shows how the amount of interest paid each year increases. We used a spreadsheet to do the calculations. By the end of the ten-year period, the total amount of money that must be repaid is £259.37, or the original sum of one hundred pounds multiplied by 2.5937.
Annually Compounded Interest
YearPrincipal + interestInterest (10% p.a.)Total owed
1£100.00£10.00£110.00
2£110.00£11.00£121.00
3£121.00£12.10£133.10
4£133.10£13.31£146.41
5£146.41£14.64£161.05
6£161.05£16.11£177.16
7£177.16£17.72£194.87
8£194.87£19.49£214.36
9£214.36£21.44£235.79
10£235.79£23.58£259.37
There is actually a standard formula for calculating the total amount A to be repaid:
A = P × (1 + r ) nt n
In this formula, P is the principal, r is the annual rate of interest expressed as a decimal, t is the number of years over which the money is borrowed or invested, and n is the number of times per year the interest is compounded. If we apply the formula to our loan (which you will remember is one hundred pounds for ten years at a rate of ten percent), we get the following result:
A = 100 × (1 + 0.1 ) (1 × 10) = 259.37 1
This is the same result we achieved previously. Obviously, we will pay significantly more interest on a loan (or accumulate more interest on an investment) if the interest is compounded. What happens, though, if we compound the interest more frequently? Suppose that instead of compounding the interest once a year, we compound it once a month. If we do that, the value of n becomes twelve (12), and our formula becomes:
A = 100 × (1 + 0.1 ) (12 × 10) = 270.70 12
The total has increased by just over ten pounds. A significant amount, but perhaps not as much as we might have expected given the fact that we are compounding the interest twelve times as often as before. Under these terms, the amount we must repay at the end of the ten-year period is £270.70 - the original sum of one hundred pounds multiplied by 2.7070. Let's make things even more interesting (no pun intended). Instead of compounding the interest monthly, we'll compound it daily. This means we will take a value for n of three hundred and sixty-five (365). For the purposes of this exercise, we will ignore the fact that there would normally be at least two leap years during the period of the loan. Our formula now becomes:
A = 100 × (1 + 0.1 ) (365 × 10) = 271.79 365
So, even though we are now compounding the interest every day, the total amount to be repaid is only one pound and nine pence (£1.09) more than the amount we would have to repay if interest was compounded monthly. This amount is the original sum multiplied by 2.7179. You can probably see by now that, as the frequency with which we compound the interest increases, the value of this multiplier seems to be converging on some fixed point. To confirm this, let's apply the formula one more time. This time we'll set the interest to be compounded every second of every day for ten years! To find the value for n, we must multiply the number of days by the number of seconds in each day (for the sake of accuracy, this time we will take leap years into account and add a quarter of a day to each year):
n = 365.25 × 24 × 60 × 60 = 31,557,600
Plugging this value into our formula, we get:
A = 100 × (1 + 0.1 ) (31,557,600 × 10) = 271.83 31,557,600
After all that, the total to be repaid is only four pence more than we got when we were compounding interest every day. It seems like our multiplier is converging on a value of 2.7183. In fact, if you do the above calculation on a calculator you should get a figure for the multiplier of 2.7182818241521875836961248899295 (the exact figure will depend on how many decimal places your calculator will display). Jacob Bernoulli must have recognised the significance of this number - at least in part. The full significance would not become apparent until much later. Bernoulli realised that the number (we'll call it e) approaches a finite limit as n approaches infinity. This limit can be expressed as follows:
e = lim (1 + 1 ) n n→∞ n
There have been many attempts to calculate the value of this number, which we know today as Euler's number or just e. Its exact value cannot be calculated because it is an irrational number (a number that cannot be expressed as the quotient of two integers, and whose decimal representation does not terminate or repeat). Like the mathematical constant Pi (π), it is also a transcendental number, which means that it cannot be a root of a non-zero polynomial equation with rational coefficients. At the time of writing, the value of e is reported to have been calculated to an accuracy of over one trillion digits. Fortunately, we can usually get by with a much less accurate approximation! The known value of e to fifteen decimal places is:
e = 2.718281828459045
From the above, we can derive a formula involving e that gives us the limit (i.e. the maximum value possible) of the total amount to be repaid against a sum of money borrowed for a given period of time, at a specific rate of interest compounded over that period. The formula is:
A = Pe rt
In this formula, as before, P is the principal, r is the annual rate of interest expressed as a decimal, and t is the number of years over which the money is borrowed. Let's suppose that we borrow three hundred and fifty pounds (£350) over seven years, at an annual interest rate of 7.5%, continuously compounded. To discover how much we are going to have to repay at the end of the seven years, we could carry out the following calculation:
A = Pe rt = 350 × 2.7183 (0.075 × 7) = 591.66
So Euler's number, or e, can be very useful for calculations involving compound interest. It turns out, however, that this seemingly innocuous number is also of vital importance in many areas of science, engineering and mathematics. Let's turn our thoughts now to exponential functions. You have no doubt come across functions such as y = x 2. This function takes the variable x as its input, and outputs the value of x 2. Here we have the variable x being raised to the power of the constant value two (2). The variable is the base, and the constant value is the exponent. In an exponential function, the positions are reversed. Suppose we have the exponential function y = 2 x. Here, the constant value is the base, and the variable is the exponent. Consider the following illustration, which shows the graphs of the functions y = 2 x and y = 3x.
The graphs of the functions y = 2x and y = 3x
Both of these exponential functions have similar curves, but whereas the curve for y = 2 x starts off slightly ahead of the curve for y = 3 x, the two curves change places at x = 0. From that point onwards, as the value of x increases, the curve for y = 3 x climbs more steeply. It is interesting to look at how quickly these functions are growing. We'll start by looking at the function y = 2 x. The function that describes the growth rate of this function is y = ln 2 · 2 x. Here are the graphs of these two functions side by side:
The graphs of the functions y = 2 x and y = ln 2 · 2 x
As you can see, the graph of function y = ln 2 · 2 x follows a similar path to that of y = 2 x. However, it gradually falls behind as the value of y = 2 x increases. Now let's see what happens when we carry out the same exercise for the function y = 3 x. The function that describes the growth rate of this function is y = ln 3 · 3 x. Here are the graphs of the two functions:
The graphs of the functions y = 3 x and y = ln 3 · 3 x
Again, you can see that the graph of function y = ln 3 · 3 x follows a similar path to the graph of function y = 3 x. The graphs are closer together, but they still diverge. This time, the growth rate is slightly ahead of the function itself. You might be starting to suspect that there is a constant number c such that the graph of the function y = c x and the graph of the function y = ln c · c x follow exactly the same path. You might feel that if this number does indeed exist, it lies somewhere between two and three - maybe a little bit closer to three than two. Given what you have learned so far, you might even hazard a guess that the number you are looking for is e. Let's test this theory. Here are the graphs of the functions y = e x and y = ln e · e x.
The graphs of the functions y = e x and y = ln e · e x
As you can see, the graphs are indeed identical. The function y = e x is known as the exponential function, and the function y = ln e · e x describes its growth rate. A function that describes the rate at which another function is changing is called a derivative function. If you already have a working knowledge of differential calculus, you will be familiar with the concept of functions and their derivatives. If not, don't worry too much about the details. It is enough to understand that for any differentiable function ƒ(x), the derivative function ƒ′(x) will give the instantaneous rate of change of that function for any value of x. The exponential function is unique in that its derivative function is the exponential function itself (ln e · e x = e x). In other words, the instantaneous rate of change of the exponential function for any value of x matches its output for that value of x.
The instantaneous rate of change (or derivative) of a function at some point P on the graph of the function can be found by measuring the slope of the graph at that point, i.e. the slope of the line that is tangent to the graph at point P. For the exponential function e x, the slope of the graph at point P will always be equal to the y coordinate of point P. This property is unique to the exponential function, and is why mathematicians, scientists and engineers find the exponential function so useful.
The slope of the tangent at e x = 1 is one (1)
The exponential function e x - sometimes written exp(x) - is of particular interest in the branch of mathematics known as calculus. This is because we use calculus to study things that change in a non-linear way. We have already looked at how the value of a loan or investment grows when the rate of interest is compounded. We can find many more examples of things that change in a non-linear way throughout the natural world. Examples include population growth, the spread of a disease, and the decay of a radioactive material. These are all quantities that increase or decrease at a rate that is proportional to their current value. The exponential function is particularly useful for describing such phenomena, because as we have seen, its growth rate always matches its current value.
The inverse function of the exponential function e x is the natural logarithm, ln(x) or loge(x). You might wonder why we use natural logarithms, which are to the base e, rather than common logarithms, which are to the base ten (log10). After all, it might seem more logical to use a number base with which we are familiar. On the other hand, the natural logarithm is directly related to the exponential function, so there is reason to suspect that it might be just as useful. Let's have a look at another graph to try and get a feel for what makes the natural logarithm so special. The following illustration shows the graph of the logarithmic function y = ln(x). By definition, the natural logarithm of e is one, and we have indicated this on the graph. We have also included the tangent to the graph at the point (1, 0). If you look closely at the tangent, you can see that it also passes through the point (0, -1), so the slope of the graph (remember "rise over run") must be one (1).
The graph of the logarithmic function ln(x)
It turns out that the slope of the graph of ln(x) will be 1/x for any value of x. As we have seen, this means that the instantaneous rate of change (or derivative) of the function ln(x) will also be 1/x for any value of x. This property is unique to the natural logarithm function, and can often greatly simplify calculations involving logarithmic quantities. There is another property of natural logarithms that is rather interesting, especially if you are studying integral calculus (the implications will be discussed in the section dealing with that subject). Consider the graph of the function y = 1/x (the derivative of the natural logarithm function):
The graph of the function y = 1/x
Note that we have shaded the area under the graph between x = 1 and x = e. If you were able to measure this area accurately, you would find that it is exactly one (1), which is the natural logarithm of e. In fact, the natural logarithm of any positive number n will be equal to the area under the graph between x = 1 and x = n. If n is one, the area will be zero - this makes perfect sense, since e 0 = 1. For any value of n greater than zero but less than one, the value of ln(n) is considered to be negative.
We have already seen the following definition of Euler's number, which allows us to calculate the value of e by plugging in a suitably large value for n:
e = lim (1 + 1 ) n n→∞ n
There are a number of other ways to calculate the value of e. The following definition can also be useful in some situations:
e = lim (1 + x) 1/x x→0
One of the more frequently used methods involves the representation of e as the sum of an infinite series, as shown here:
e = 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + · · · 1! 2! 3! 4! 5! 6! 7!
In this representation, we start with one, and then add the reciprocals of successive factorials. The factorial n! is simply the product of all positive integers less than or equal to n. For example, the factorial 5! is a shorthand way of writing 1 × 2 × 3 × 4 × 5, which is equal to one hundred and twenty (120). The number of terms used for the calculation will determine the accuracy of the resulting approximation. You should be able to see that each successive fraction represents a significantly smaller increment than its predecessor.
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Horizontal and Vertical Lines Worksheet
📆 1 Jan 1970
🔖 Line Category
📂 Gallery Type
Printable Blank Number Line Worksheet
One point perspective
Adding Subtracting Decimals Common Core Sheets
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# A man deposited Rs.10000 in a bank with an interest rate of 5% simple interest annually. Find the amount in 15th year since he deposited the amount and also calculate the amount after 20 years.
Last updated date: 24th Jul 2024
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Hint: Initially find the interest of the first year by using the given data. And by using the interest we can also find the amount that has been deposited in the 15th year and this helps to calculate the amount after 20 years.
Here from the data it is given that a man deposited Rs.10000 in a bank with an interest rate of 5% simple interest annually.
So, it is known that in simple interest, the interest remains the same in all years.
This means if we find interest in the first year then the interest will be the same for all years.
Now let us find interest for first year
$I = PTR$
Here I is the interest, P= Rs.10000, R=5% and t=1(for one year)
$\begin{gathered} \Rightarrow I = 10000 \times 5\% \times 1 \\ \Rightarrow I = 10000 \times \dfrac{5}{{100}} \\ \Rightarrow I = 500 \\ \end{gathered}$
So interest of first year =Rs.500
Therefore interest for all years is also Rs.500 as it is simple interest used by the person.
Hence,
Amount in 1st year=Rs.10000
Amount in 2nd year = Amount in 1st year +Interest = Rs.10000 + Rs.500 =Rs.10500
Amount in 3rd year = Amount in 2nd year + Interest = Rs10500 + Rs.500 =Rs.11000
Hence our series is
Rs.10000, Rs.10500, Rs.11000 …
If we observe the series it is of AP
Where a=10000 and d=500 (interest is 500 which is common for all years)
Or else d=a2-a1 which implies d=10500-10000=500
There d=500
Now here we need to find amount in 15th year, as the series are in AP let us use nth term of AP where n=15
So, nth term of AP
$\Rightarrow {a_n} = a + (n - 1)d$
Amount in 15th year = $a + (n - 1)d$
=10000 + (15-1)500
=10000+14(500)
=17000
Therefore amount in 15th year is Rs.17000
And also here we need to find amount after 20 year
i.e. we need to calculate ${a_n}$ value where n=21
Amount after 20 year = ${a_n}$
= $a + (n - 1)d$
=10000 + (20-1)500
=10000+19(500)
=20,000
Therefore amount after 20 years is Rs. 20,000.
Note: As the person has taken simple interest, so for every year the interest will be the same .So find interest for 1st year which will be the same for all other years. Later adding interest to the previous year amount we get the present year amount. As interest is common the amount will be in AP series. Make a note that the series is of AP so we have used the nth term of AP formula to find amount for 15th year and amount after 20 years.
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# Graphs of Simple Function
Here we will learn about the graphs of simple functions that is multiples of different numbers.
The example will help us to understand the graph of multiples of different numbers.
1. Draw the graph of the function y = 3x.
2. From the graph, find the value of y, when
(a) x = 4
(b) x = 5
Solution:
1. The given function is y = 3x.
For some different values of x, the corresponding values of y are given below.
x 0 1 2 3 y = 3x 0 3 6 9
On a graph paper plot the points O (0, 0), A (1, 3), B (2, 6) and C (3, 9) and join them successively to obtain the required graph.
2. Reading off from the graph of simple function:
(a) On the x-axis, take the point L at x = 4.
Draw LP ⊥ x-axis, meeting the graph at P.
Clearly, PL = 12 units.
Therefore, x = 4 ⇒ y = 12.
(b) On the x-axis, take the point M at x = 5.
Draw MQ ⊥ x-axis, meeting the graph at Q.
Clearly MQ = 15 units
Therefore, x = 5 ⇒ y = 15
Related Concepts:
Coordinate Graph
Signs of Coordinates
Find the Coordinates of a Point
Coordinates of a Point in a Plane
Plot Points on Co-ordinate Graph
Graph of Linear Equation
Simultaneous Equations Graphically
Graph of Perimeter vs. Length of the Side of a Square
Graph of Area vs. Side of a Square
Graph of Simple Interest vs. Number of Years
Graph of Distance vs. Time
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# 2020 CIME I Problems/Problem 7
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Problem 7
For every positive integer $n$, define $$f(n)=\frac{n}{1 \cdot 3 \cdot 5 \cdots (2n+1)}.$$ Suppose that the sum $f(1)+f(2)+\cdots+f(2020)$ can be expressed as $\frac{p}{q}$ for relatively prime integers $p$ and $q$. Find the remainder when $p$ is divided by $1000$.
## Solution
Let $S(n) = f(1)+f(2)+\cdots+f(n)$. We claim that $$S(n) = \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!}.$$ We show this using induction. Suppose this is true for some $n = k$. Then, it must be true for $k+1$. The base case when $k = 1$ is trivial. Then, we have that \begin{align*} S(k+1) &= f(k) + S(k) \\ {} &= \frac{n}{(2n+1)!!} + \frac{1}{2} - \frac{1}{2 \cdot (2n+1)!!} \\ {} &= \frac{1}{2} - \frac{1}{2 \cdot (2n+3)!!}. \end{align*} Hence, this completes the induction therefore proving the claim. So, the numerator $p$ is $\frac{1}{2}(4041!! - 1)$.
We proceed using Euler's Theorem combined with Chinese Remainder Theorem. It is obvious that $4041!! \equiv 0 \pmod{125}$ so $p \equiv 62\pmod{125}$. Also, instead of considering $p$ modulo $8$, we consider it modulo $16$. Then, we get $$4041!! \equiv 3^{253} \cdot 5^{253} \cdot 7^{253} \cdot 9^{252} \cdot 11^{252} \cdot 13^{252} \cdot 15^{252} \equiv 3 \cdot 5 \cdot 7 \cdot 9 \equiv 1\pmod{16},$$ by Euler's Totient Theorem as $\phi(16) = 8$. This implies that $\frac{1}{2}(4041!! - 1) \equiv 0\pmod{8}$, so $p \equiv 0\pmod{8}$. Solving the system of congruences, we get $p \equiv \boxed{312}\pmod{1000}$. ~rocketsri (based off of official solution)
## See also
2020 CIME I (Problems • Answer Key • Resources) Preceded byProblem 6 Followed byProblem 8 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All CIME Problems and Solutions
The problems on this page are copyrighted by the MAC's Christmas Mathematics Competitions.
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Friday, April 26, 2019 0
Why should it? Well, try this:
• 5 ÷ 10 = 0.5
• 5 ÷ 1 = 5
• 5 ÷ 0.1 = 50
• 5 ÷ 0.01 = 500
• 5 ÷ 0.001 = 5000
As we divide by smaller and smaller numbers the result gets ever bigger. Logically, then, as the divisor tends to (ie gets closer to) zero, so the result tends to infinity. But this is not the same as saying that division by zero actually is infinity, is it? What about drawing a graph with an asymptote?
Here’s the graph of
We can see that as x tends to 2, so the graph gets closer to the line x = 2 (the line is called an asymptote), with values tending towards , but again, it doesn’t actually happen – there is no point on the graph (2, ).
Let’s look at it another way. As an example, 12 ÷ 4 means “how many 4s make 12? Answer: 3. So 12 ÷ 0 must mean “how many zeroes make twelve?” Answer: impossible. Even an infinity of zeroes still makes zero. All we can say is that if we divide by a number which gets ever closer to zero, the result gets bigger and bigger – for ever!
Of course, dividing into zero is another matter. 0 ÷ x = 0, whatever the value of x. But this leads us to another knotty problem – what is the value of 0 ÷ 0? We know that:
is undefined.
So could be any of these? Let’s pose the problem another way. Suppose . It would then follow that 0 = k × 0, but that means that k could take any value. So we must conclude that is also undefined.
#### A well known fallacy
Division by zero is at the heart of this fallacious argument which proves that 2=1!
Suppose we have two numbers a and b, such that a = b
Then:
• ab = b2
• aba2 = b2a2
• a(ba) = (b + a)(ba)
• a = b + a
• a = 2a (Since b = a)
• So 1 = 2.
The fallacy lies after the third line of working where I divided both sides by (b – a); since a = b this means ba = 0, and that makes everything go horribly wrong! Effectively, it’s like saying 2 × 0 = 3 × 0, so divide both sides by 0 to get 2 = 3.
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I have forgotten
• https://me.yahoo.com
# Variable Cross Section
Time of emptying a tank of variable cross-section through an orifice
## Overview
Previously the time of emptying of geometrical tanks (i.e., rectangular, hemispherical and circular) was discussed. But, sometimes, we come across tanks, which have variable cross-section. In such cases, there are two variables instead of one, as in the case of tanks of uniform cross-section. Since a single relation cannot be derived for different cross-sections, it is therefore essential that such problems should be solved from the first principles i.e., from the equation.
This can be best understood from the following examples.
Example:
[metric]
##### Example - Time of emptying a tank of variable cross-section
Problem
A rectangular tank of 20m12m at the top and 10m6m at the bottom is 3m deep as shown in figure. There is an orifice of 450mm diameter at the bottom of the tank. Determine the time taken to empty the tank completely, if coefficient of discharge is 0.64.
Workings
Given,
• Top length = 20m
• Top width = 12m
• Bottom length = 10m
• Bottom width = 6m
• Depth of water = 3m
• Diameter of orifice, = 450mm = 0.45m
• = 0.64
We know that the area of the orifice,
First of all, let us consider a small strip of water of thickness (dh) at a height (h) from the bottom of the tank. From the geometry of the figure, we find that the length of the strip of water,
and the breadth of the strip of water,
So, the area of the strip,
Now let us use the general equation for the time to empty a tank,
The total time required to empty the tank may be found by integrating the above equation between the limits 3 and 0 (because initial head of water is 3m and final head of water is 0m).
Solution
Time taken to empty the tank = 14 min 20 s
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Continued Fractions
Get Continued Fractions essential facts below. View Videos or join the Continued Fractions discussion. Add Continued Fractions to your PopFlock.com topic list for future reference or share this resource on social media.
Continued Fractions
${\displaystyle a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{\ddots +{\cfrac {1}{a_{n}}}}}}}}}}$
A finite continued fraction, where ${\displaystyle n}$ is a non-negative integer, ${\displaystyle a_{0}}$ is an integer, and ${\displaystyle a_{i}}$ is a positive integer, for ${\displaystyle i=1,\ldots ,n}$.
In mathematics, a continued fraction is an expression obtained through an iterative process of representing a number as the sum of its integer part and the reciprocal of another number, then writing this other number as the sum of its integer part and another reciprocal, and so on.[1] In a finite continued fraction (or terminated continued fraction), the iteration/recursion is terminated after finitely many steps by using an integer in lieu of another continued fraction. In contrast, an infinite continued fraction is an infinite expression. In either case, all integers in the sequence, other than the first, must be positive. The integers ${\displaystyle a_{i}}$ are called the coefficients or terms of the continued fraction.[2]
Continued fractions have a number of remarkable properties related to the Euclidean algorithm for integers or real numbers. Every rational number has two closely related expressions as a finite continued fraction, whose coefficients ai can be determined by applying the Euclidean algorithm to ${\displaystyle (p,q)}$. The numerical value of an infinite continued fraction is irrational; it is defined from its infinite sequence of integers as the limit of a sequence of values for finite continued fractions. Each finite continued fraction of the sequence is obtained by using a finite prefix of the infinite continued fraction's defining sequence of integers. Moreover, every irrational number ${\displaystyle \alpha }$ is the value of a unique infinite continued fraction, whose coefficients can be found using the non-terminating version of the Euclidean algorithm applied to the incommensurable values ${\displaystyle \alpha }$ and 1. This way of expressing real numbers (rational and irrational) is called their continued fraction representation.
It is generally assumed that the numerator of all of the fractions is 1. If arbitrary values and/or functions are used in place of one or more of the numerators or the integers in the denominators, the resulting expression is a generalized continued fraction. When it is necessary to distinguish the first form from generalized continued fractions, the former may be called a simple or regular continued fraction, or said to be in canonical form.
The term continued fraction may also refer to representations of rational functions, arising in their analytic theory. For this use of the term, see Padé approximation and Chebyshev rational functions.
## Motivation and notation
Consider, for example, the rational number , which is around 4.4624. As a first approximation, start with 4, which is the integer part; . The fractional part is the reciprocal of which is about 2.1628. Use the integer part, 2, as an approximation for the reciprocal to obtain a second approximation of . The remaining fractional part, , is the reciprocal of , and is around 6.1429. Use 6 as an approximation for this to obtain as an approximation for and , about 4.4615, as the third approximation; . Finally, the fractional part, , is the reciprocal of 7, so its approximation in this scheme, 7, is exact and produces the exact expression for .
The expression is called the continued fraction representation of . This can be represented by the abbreviated notation = [4; 2, 6, 7]. (It is customary to replace only the first comma by a semicolon.) Some older textbooks use all commas in the (n + 1)-tuple, for example, [4, 2, 6, 7].[3][4]
If the starting number is rational, then this process exactly parallels the Euclidean algorithm. In particular, it must terminate and produce a finite continued fraction representation of the number. If the starting number is irrational, then the process continues indefinitely. This produces a sequence of approximations, all of which are rational numbers, and these converge to the starting number as a limit. This is the (infinite) continued fraction representation of the number. Examples of continued fraction representations of irrational numbers are:
• = [4;2,1,3,1,2,8,2,1,3,1,2,8,...] (sequence in the OEIS). The pattern repeats indefinitely with a period of 6.
• e = [2;1,2,1,1,4,1,1,6,1,1,8,...] (sequence in the OEIS). The pattern repeats indefinitely with a period of 3 except that 2 is added to one of the terms in each cycle.
• ? = [3;7,15,1,292,1,1,1,2,1,3,1,...] (sequence in the OEIS). No pattern has ever been found in this representation.
• ? = [1;1,1,1,1,1,1,1,1,1,1,1,...] (sequence in the OEIS). The golden ratio, the irrational number that is the "most difficult" to approximate rationally. See: A property of the golden ratio ?.
Continued fractions are, in some ways, more "mathematically natural" representations of a real number than other representations such as decimal representations, and they have several desirable properties:
• The continued fraction representation for a rational number is finite and only rational numbers have finite representations. In contrast, the decimal representation of a rational number may be finite, for example , or infinite with a repeating cycle, for example
• Every rational number has an essentially unique continued fraction representation. Each rational can be represented in exactly two ways, since [a0;a1,... an-1,an] = [a0;a1,... an-1,(an-1),1]. Usually the first, shorter one is chosen as the canonical representation.
• The continued fraction representation of an irrational number is unique.
• The real numbers whose continued fraction eventually repeats are precisely the quadratic irrationals.[5] For example, the repeating continued fraction is the golden ratio, and the repeating continued fraction is the square root of 2. In contrast, the decimal representations of quadratic irrationals are apparently random. The square roots of all (positive) integers, that are not perfect squares, are quadratic irrationals, hence are unique periodic continued fractions.
• The successive approximations generated in finding the continued fraction representation of a number, that is, by truncating the continued fraction representation, are in a certain sense (described below) the "best possible".
## Basic formula
A continued fraction is an expression of the form
${\displaystyle a_{0}+{\cfrac {b_{1}}{a_{1}+{\cfrac {b_{2}}{a_{2}+{\cfrac {b_{3}}{a_{3}+{_{\ddots }}}}}}}}}$
where ai and bi can be any complex numbers. Usually they are required to be integers. If bi = 1 for all i the expression is called a simple continued fraction. If the expression contains a finite number of terms, it is called a finite continued fraction. If the expression contains an infinite number of terms, it is called an infinite continued fraction.[6]
Thus, all of the following illustrate valid finite simple continued fractions:
Examples of finite simple continued fractions
Formula Numeric Remarks
${\displaystyle \ a_{0}}$ ${\displaystyle \ 2}$ All integers are a degenerate case
${\displaystyle \ a_{0}+{\cfrac {1}{a_{1}}}}$ ${\displaystyle \ 2+{\cfrac {1}{3}}}$ Simplest possible fractional form
${\displaystyle \ a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}}}}}}$ ${\displaystyle \ -3+{\cfrac {1}{2+{\cfrac {1}{18}}}}}$ First integer may be negative
${\displaystyle \ a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}}}}}}}}$ ${\displaystyle \ {\cfrac {1}{15+{\cfrac {1}{1+{\cfrac {1}{102}}}}}}}$ First integer may be zero
## Calculating continued fraction representations
Consider a real number r. Let ${\displaystyle i=\lfloor r\rfloor }$ be the integer part of r and let ${\displaystyle f=r-i}$ be the fractional part of r. Then the continued fraction representation of r is ${\displaystyle [i;a_{1},a_{2},\ldots ]}$, where ${\displaystyle [a_{1};a_{2},\ldots ]}$ is the continued fraction representation of ${\displaystyle 1/f}$.
To calculate a continued fraction representation of a number r, write down the integer part (technically the floor) of r. Subtract this integer part from r. If the difference is 0, stop; otherwise find the reciprocal of the difference and repeat. The procedure will halt if and only if r is rational. This process can be efficiently implemented using the Euclidean algorithm when the number is rational. The table below shows an implementation of this procedure for the number 3.245, resulting in the continued fraction expansion [3; 4,12,4].
Find the continued fraction for ${\displaystyle 3.245={\frac {649}{200}}}$
Step Real
Number
Integer
part
Fractional
part
Simplified Reciprocal
of f
1 ${\displaystyle r={\frac {649}{200}}}$ ${\displaystyle i=3}$ ${\displaystyle f={\frac {649}{200}}-3}$ ${\displaystyle ={\frac {49}{200}}}$ ${\displaystyle {\frac {1}{f}}={\frac {200}{49}}}$
2 ${\displaystyle r={\frac {200}{49}}}$ ${\displaystyle i=4}$ ${\displaystyle f={\frac {200}{49}}-4}$ ${\displaystyle ={\frac {4}{49}}}$ ${\displaystyle {\frac {1}{f}}={\frac {49}{4}}}$
3 ${\displaystyle r={\frac {49}{4}}}$ ${\displaystyle i=12}$ ${\displaystyle f={\frac {49}{4}}-12}$ ${\displaystyle ={\frac {1}{4}}}$ ${\displaystyle {\frac {1}{f}}={\frac {4}{1}}}$
4 ${\displaystyle r=4}$ ${\displaystyle i=4}$ ${\displaystyle f=4-4}$ ${\displaystyle =0}$ STOP
Continued fraction form for ${\displaystyle 3.245={\frac {649}{200}}=[3;4,12,4]}$
= 3 +
## Notations
The integers ${\displaystyle a_{0}}$, ${\displaystyle a_{1}}$ etc., are called the coefficients or terms of the continued fraction.[2] One can abbreviate the continued fraction
${\displaystyle x=a_{0}+{\cfrac {1}{a_{1}+{\cfrac {1}{a_{2}+{\cfrac {1}{a_{3}}}}}}}}$
in the notation of Carl Friedrich Gauss
${\displaystyle x=a_{0}+{\underset {i=1}{\overset {3}{\mathrm {K} }}}~{\frac {1}{a_{i}}}}$
or as
${\displaystyle x=[a_{0};a_{1},a_{2},a_{3}]}$,
or in the notation of Pringsheim as
${\displaystyle x=a_{0}+{\frac {1\mid }{\mid a_{1}}}+{\frac {1\mid }{\mid a_{2}}}+{\frac {1\mid }{\mid a_{3}}},}$
or in another related notation as
${\displaystyle x=a_{0}+{1 \over a_{1}+{}}{1 \over a_{2}+{}}{1 \over a_{3}{}}.}$
Sometimes angle brackets are used, like this:
${\displaystyle x=\left\langle a_{0};a_{1},a_{2},a_{3}\right\rangle .}$
The semicolon in the square and angle bracket notations is sometimes replaced by a comma.[3][4]
One may also define infinite simple continued fractions as limits:
${\displaystyle [a_{0};a_{1},a_{2},a_{3},\,\ldots ]=\lim _{n\to \infty }[a_{0};a_{1},a_{2},\,\ldots ,a_{n}].}$
This limit exists for any choice of ${\displaystyle a_{0}}$ and positive integers ${\displaystyle a_{1},a_{2},\ldots }$[7][8]
## Finite continued fractions
Every finite continued fraction represents a rational number, and every rational number can be represented in precisely two different ways as a finite continued fraction, with the conditions that the first coefficient is an integer and the other coefficients are positive integers. These two representations agree except in their final terms. In the longer representation the final term in the continued fraction is 1; the shorter representation drops the final 1, but increases the new final term by 1. The final element in the short representation is therefore always greater than 1, if present. In symbols:
[a0; a1, a2, ..., an - 1, an, 1] = [a0; a1, a2, ..., an - 1, an + 1].
[a0; 1] = [a0 + 1].
## Of reciprocals
The continued fraction representations of a positive rational number and its reciprocal are identical except for a shift one place left or right depending on whether the number is less than or greater than one respectively. In other words, the numbers represented by ${\displaystyle [a_{0};a_{1},a_{2},\ldots ,a_{n}]}$ and ${\displaystyle [0;a_{0},a_{1},\ldots ,a_{n}]}$ are reciprocals.
For instance if ${\displaystyle a}$ is an integer and ${\displaystyle x<1}$ then
${\displaystyle x=0+{\frac {1}{a+{\frac {1}{b}}}}}$ and ${\displaystyle {\frac {1}{x}}=a+{\frac {1}{b}}}$.
If ${\displaystyle x>1}$ then
${\displaystyle x=a+{\frac {1}{b}}}$ and ${\displaystyle {\frac {1}{x}}=0+{\frac {1}{a+{\frac {1}{b}}}}}$.
The last number that generates the remainder of the continued fraction is the same for both ${\displaystyle x}$ and its reciprocal.
For example,
${\displaystyle 2.25={\frac {9}{4}}=[2;4]}$ and ${\displaystyle {\frac {1}{2.25}}={\frac {4}{9}}=[0;2,4]}$.
## Infinite continued fractions and convergents
Every infinite continued fraction is irrational, and every irrational number can be represented in precisely one way as an infinite continued fraction.
An infinite continued fraction representation for an irrational number is useful because its initial segments provide rational approximations to the number. These rational numbers are called the convergents of the continued fraction.[9][10] The larger a term is in the continued fraction, the closer the corresponding convergent is to the irrational number being approximated. Numbers like ? have occasional large terms in their continued fraction, which makes them easy to approximate with rational numbers. Other numbers like e have only small terms early in their continued fraction, which makes them more difficult to approximate rationally. The golden ratio ? has terms equal to 1 everywhere--the smallest values possible--which makes ? the most difficult number to approximate rationally. In this sense, therefore, it is the "most irrational" of all irrational numbers. Even-numbered convergents are smaller than the original number, while odd-numbered ones are larger.
For a continued fraction [a0; a1, a2, ...], the first four convergents (numbered 0 through 3) are
, , , .
The numerator of the third convergent is formed by multiplying the numerator of the second convergent by the third coefficient, and adding the numerator of the first convergent. The denominators are formed similarly. Therefore, each convergent can be expressed explicitly in terms of the continued fraction as the ratio of certain multivariate polynomials called continuants.
If successive convergents are found, with numerators h1, h2, ... and denominators k1, k2, ... then the relevant recursive relation is:
hn = anhn - 1 + hn - 2,
kn = ankn - 1 + kn - 2.
The successive convergents are given by the formula
= .
Thus to incorporate a new term into a rational approximation, only the two previous convergents are necessary. The initial "convergents" (required for the first two terms) are 0/1 and 1/0. For example, here are the convergents for [0;1,5,2,2].
n an hn kn -2 -1 0 1 2 3 4 0 1 5 2 2 0 1 0 1 5 11 27 1 0 1 1 6 13 32
When using the Babylonian method to generate successive approximations to the square root of an integer, if one starts with the lowest integer as first approximant, the rationals generated all appear in the list of convergents for the continued fraction. Specifically, the approximants will appear on the convergents list in positions 0, 1, 3, 7, 15, ... , 2k-1, ... For example, the continued fraction expansion for is [1;1,2,1,2,1,2,1,2,...]. Comparing the convergents with the approximants derived from the Babylonian method:
n an hn kn -2 -1 0 1 2 3 4 5 6 7 1 1 2 1 2 1 2 1 0 1 1 2 5 7 19 26 71 97 1 0 1 1 3 4 11 15 41 56
x0 = 1 =
x1 = (1 + ) = = 2
x2 = (2 + ) =
x3 = ( + ) =
### Properties
A Baire space is a topological space on infinite sequences of natural numbers. The infinite continued fraction provides a homeomorphism from the Baire space to the space of irrational real numbers (with the subspace topology inherited from the usual topology on the reals). The infinite continued fraction also provides a map between the quadratic irrationals and the dyadic rationals, and from other irrationals to the set of infinite strings of binary numbers (i.e. the Cantor set); this map is called the Minkowski question mark function. The mapping has interesting self-similar fractal properties; these are given by the modular group, which is the subgroup of Möbius transformations having integer values in the transform. Roughly speaking, continued fraction convergents can be taken to be Möbius transformations acting on the (hyperbolic) upper half-plane; this is what leads to the fractal self-symmetry.
### Some useful theorems
If ${\displaystyle a_{0}}$, ${\displaystyle a_{1}}$, ${\displaystyle a_{2}}$, ${\displaystyle \ldots }$ is an infinite sequence of positive integers, define the sequences ${\displaystyle h_{n}}$ and ${\displaystyle k_{n}}$ recursively:
${\displaystyle h_{n}=a_{n}h_{n-1}+h_{n-2}}$ ${\displaystyle h_{-1}=1}$ ${\displaystyle h_{-2}=0}$ ${\displaystyle k_{n}=a_{n}k_{n-1}+k_{n-2}}$ ${\displaystyle k_{-1}=0}$ ${\displaystyle k_{-2}=1}$
Theorem 1. For any positive real number ${\displaystyle z}$
${\displaystyle \left[a_{0};a_{1},\,\dots ,a_{n-1},z\right]={\frac {zh_{n-1}+h_{n-2}}{zk_{n-1}+k_{n-2}}}.}$
Theorem 2. The convergents of [${\displaystyle a_{0}}$; ${\displaystyle a_{1}}$, ${\displaystyle a_{2}}$, ${\displaystyle \ldots }$] are given by
${\displaystyle \left[a_{0};a_{1},\,\dots ,a_{n}\right]={\frac {h_{n}}{k_{n}}}.}$
Theorem 3. If the ${\displaystyle n}$th convergent to a continued fraction is ${\displaystyle h_{n}}$/${\displaystyle k_{n}}$, then
${\displaystyle k_{n}h_{n-1}-k_{n-1}h_{n}=(-1)^{n}.}$
Corollary 1: Each convergent is in its lowest terms (for if ${\displaystyle h_{n}}$ and ${\displaystyle k_{n}}$ had a nontrivial common divisor it would divide ${\displaystyle k_{n}h_{n-1}-k_{n-1}h_{n}}$, which is impossible).
Corollary 2: The difference between successive convergents is a fraction whose numerator is unity:
${\displaystyle {\frac {h_{n}}{k_{n}}}-{\frac {h_{n-1}}{k_{n-1}}}={\frac {h_{n}k_{n-1}-k_{n}h_{n-1}}{k_{n}k_{n-1}}}={\frac {(-1)^{n+1}}{k_{n}k_{n-1}}}.}$
Corollary 3: The continued fraction is equivalent to a series of alternating terms:
${\displaystyle a_{0}+\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{k_{n}k_{n+1}}}.}$
Corollary 4: The matrix
${\displaystyle {\begin{bmatrix}h_{n}&h_{n-1}\\k_{n}&k_{n-1}\end{bmatrix}}}$
has determinant plus or minus one, and thus belongs to the group of ${\displaystyle 2\times 2}$ unimodular matrices ${\displaystyle \mathrm {GL} (2,\mathbb {Z} )}$.
Theorem 4. Each (${\displaystyle s}$th) convergent is nearer to a subsequent (${\displaystyle n}$th) convergent than any preceding (${\displaystyle r}$th) convergent is. In symbols, if the ${\displaystyle n}$th convergent is taken to be ${\displaystyle [a_{0};a_{1},\ldots ,a_{n}]=x_{n}}$, then
${\displaystyle \left|x_{r}-x_{n}\right|>\left|x_{s}-x_{n}\right|}$
for all ${\displaystyle r.
Corollary 1: The even convergents (before the ${\displaystyle n}$th) continually increase, but are always less than ${\displaystyle x_{n}}$.
Corollary 2: The odd convergents (before the ${\displaystyle n}$th) continually decrease, but are always greater than ${\displaystyle x_{n}}$.
Theorem 5.
${\displaystyle {\frac {1}{k_{n}(k_{n+1}+k_{n})}}<\left|x-{\frac {h_{n}}{k_{n}}}\right|<{\frac {1}{k_{n}k_{n+1}}}.}$
Corollary 1: A convergent is nearer to the limit of the continued fraction than any fraction whose denominator is less than that of the convergent.
Corollary 2: A convergent obtained by terminating the continued fraction just before a large term is a close approximation to the limit of the continued fraction.
## Semiconvergents
If
${\displaystyle {\frac {h_{n-1}}{k_{n-1}}},{\frac {h_{n}}{k_{n}}}}$
are consecutive convergents, then any fractions of the form
${\displaystyle {\frac {h_{n-1}+mh_{n}}{k_{n-1}+mk_{n}}},}$
where ${\displaystyle m}$ is an integer such that ${\displaystyle 0\leq m\leq a_{n+1}}$, are called semiconvergents, secondary convergents, or intermediate fractions. The ${\displaystyle (m+1)}$-st semiconvergent equals the mediant of the ${\displaystyle m}$-th one and the convergent ${\displaystyle {\tfrac {h_{n}}{k_{n}}}}$. Sometimes the term is taken to mean that being a semiconvergent excludes the possibility of being a convergent (i.e., ${\displaystyle 0), rather than that a convergent is a kind of semiconvergent.
It follows that semiconvergents represent a monotonic sequence of fractions between the convergents ${\displaystyle {\tfrac {h_{n-1}}{k_{n-1}}}}$ (corresponding to ${\displaystyle m=0}$) and ${\displaystyle {\tfrac {h_{n+1}}{k_{n+1}}}}$ (corresponding to ${\displaystyle m=a_{n+1}}$). The consecutive semiconvergents ${\displaystyle {\tfrac {a}{b}}}$ and ${\displaystyle {\tfrac {c}{d}}}$ satisfy the property ${\displaystyle ad-bc=\pm 1}$.
If a rational approximation ${\displaystyle {\tfrac {p}{q}}}$ to a real number ${\displaystyle x}$ is such that the value ${\displaystyle \left|x-{\tfrac {p}{q}}\right|}$ is smaller than that of any approximation with a smaller denominator, then ${\displaystyle {\tfrac {p}{q}}}$ is a semiconvergent of the continued fraction expansion of ${\displaystyle x}$. The converse is not true, however.
## Best rational approximations
One can choose to define a best rational approximation to a real number x as a rational number , d > 0, that is closer to x than any approximation with a smaller or equal denominator. The simple continued fraction for x can be used to generate all of the best rational approximations for x by applying these three rules:
1. Truncate the continued fraction, and reduce its last term by a chosen amount (possibly zero).
2. The reduced term cannot have less than half its original value.
3. If the final term is even, half its value is admissible only if the corresponding semiconvergent is better than the previous convergent. (See below.)
For example, 0.84375 has continued fraction [0;1,5,2,2]. Here are all of its best rational approximations.
Continued fraction Rational approximation Decimal equivalent Error [0;1] [0;1,3] [0;1,4] [0;1,5] [0;1,5,2] [0;1,5,2,1] [0;1,5,2,2] 1 1 0.75 0.8 ~0.83333 ~0.84615 ~0.84211 0.84375 +18.519% -11.111% -5.1852% -1.2346% +0.28490% -0.19493% 0%
The strictly monotonic increase in the denominators as additional terms are included permits an algorithm to impose a limit, either on size of denominator or closeness of approximation.
The "half rule" mentioned above requires that when ak is even, the halved term ak/2 is admissible if and only if |x - [a0 ; a1, ..., ak - 1]| > |x - [a0 ; a1, ..., ak - 1, ak/2]|[11] This is equivalent[11] to:[12]
[ak; ak - 1, ..., a1] > [ak; ak + 1, ...].
The convergents to x are "best approximations" in a much stronger sense than the one defined above. Namely, n/d is a convergent for x if and only if |dx - n| has the smallest value among the analogous expressions for all rational approximations m/c with c d; that is, we have |dx - n| < |cx - m| so long as c < d. (Note also that |dkx - nk| -> 0 as k -> ?.)
### Best rational within an interval
A rational that falls within the interval (x, y), for 0 < x < y, can be found with the continued fractions for x and y. When both x and y are irrational and
x = [a0; a1, a2, ..., ak - 1, ak, ak + 1, ...]
y = [a0; a1, a2, ..., ak - 1, bk, bk + 1, ...]
where x and y have identical continued fraction expansions up through ak-1, a rational that falls within the interval (x, y) is given by the finite continued fraction,
z(x,y) = [a0; a1, a2, ..., ak - 1, min(ak, bk) + 1]
This rational will be best in the sense that no other rational in (x, y) will have a smaller numerator or a smaller denominator.[]
If x is rational, it will have two continued fraction representations that are finite, x1 and x2, and similarly a rational y will have two representations, y1 and y2. The coefficients beyond the last in any of these representations should be interpreted as +?; and the best rational will be one of z(x1, y1), z(x1, y2), z(x2, y1), or z(x2, y2).
For example, the decimal representation 3.1416 could be rounded from any number in the interval [3.14155, 3.14165). The continued fraction representations of 3.14155 and 3.14165 are
3.14155 = [3; 7, 15, 2, 7, 1, 4, 1, 1] = [3; 7, 15, 2, 7, 1, 4, 2]
3.14165 = [3; 7, 16, 1, 3, 4, 2, 3, 1] = [3; 7, 16, 1, 3, 4, 2, 4]
and the best rational between these two is
[3; 7, 16] = = 3.1415929....
Thus, is the best rational number corresponding to the rounded decimal number 3.1416, in the sense that no other rational number that would be rounded to 3.1416 will have a smaller numerator or a smaller denominator.
### Interval for a convergent
A rational number, which can be expressed as finite continued fraction in two ways,
z = [a0; a1, ..., ak - 1, ak, 1] = [a0; a1, ..., ak - 1, ak + 1]
will be one of the convergents for the continued fraction expansion of a number, if and only if the number is strictly between
x = [a0; a1, ..., ak - 1, ak, 2] and
y = [a0; a1, ..., ak - 1, ak + 2]
The numbers x and y are formed by incrementing the last coefficient in the two representations for z. It is the case that x < y when k is even, and x > y when k is odd.
For example, the number has the continued fraction representations
= [3; 7, 15, 1] = [3; 7, 16]
and thus is a convergent of any number strictly between
[3; 7, 15, 2] = ? 3.1415525 [3; 7, 17] = ? 3.1416667
## Comparison
Consider x = [a0; a1, ...] and y = [b0; b1, ...]. If k is the smallest index for which ak is unequal to bk then x < y if (-1)k(ak - bk) < 0 and y < x otherwise.
If there is no such k, but one expansion is shorter than the other, say x = [a0; a1, ..., an] and y = [b0; b1, ..., bn, bn + 1, ...] with ai = bi for 0 i n, then x < y if n is even and y < x if n is odd.
## Continued fraction expansions of ?
To calculate the convergents of ? we may set a0 = ??? = 3, define u1 = ? 7.0625 and a1 = ?u1? = 7, u2 = ? 15.9966 and a2 = ?u2? = 15, u3 = ? 1.0034. Continuing like this, one can determine the infinite continued fraction of ? as
[3;7,15,1,292,1,1,...] (sequence in the OEIS).
The fourth convergent of ? is [3;7,15,1] = = 3.14159292035..., sometimes called Milü, which is fairly close to the true value of ?.
Let us suppose that the quotients found are, as above, [3;7,15,1]. The following is a rule by which we can write down at once the convergent fractions which result from these quotients without developing the continued fraction.
The first quotient, supposed divided by unity, will give the first fraction, which will be too small, namely, . Then, multiplying the numerator and denominator of this fraction by the second quotient and adding unity to the numerator, we shall have the second fraction, , which will be too large. Multiplying in like manner the numerator and denominator of this fraction by the third quotient, and adding to the numerator the numerator of the preceding fraction, and to the denominator the denominator of the preceding fraction, we shall have the third fraction, which will be too small. Thus, the third quotient being 15, we have for our numerator , and for our denominator, . The third convergent, therefore, is . We proceed in the same manner for the fourth convergent. The fourth quotient being 1, we say 333 times 1 is 333, and this plus 22, the numerator of the fraction preceding, is 355; similarly, 106 times 1 is 106, and this plus 7 is 113. In this manner, by employing the four quotients [3;7,15,1], we obtain the four fractions:
, , , , ....
The following Maple code will generate continued fraction expansions of Pi
To sum up, the pattern is ${\displaystyle Numerator_{i}=Numerator_{(i-1)}\cdot Quotient_{i}+Numerator_{(i-2)}}$ ${\displaystyle Denominator_{i}=Denominator_{(i-1)}\cdot Quotient_{i}+Denominator_{(i-2)}}$
These convergents are alternately smaller and larger than the true value of ?, and approach nearer and nearer to ?. The difference between a given convergent and ? is less than the reciprocal of the product of the denominators of that convergent and the next convergent. For example, the fraction is greater than ?, but - ? is less than = (in fact, - ? is just more than = ).
The demonstration of the foregoing properties is deduced from the fact that if we seek the difference between one of the convergent fractions and the next adjacent to it we shall obtain a fraction of which the numerator is always unity and the denominator the product of the two denominators. Thus the difference between and is , in excess; between and , , in deficit; between and , , in excess; and so on. The result being, that by employing this series of differences we can express in another and very simple manner the fractions with which we are here concerned, by means of a second series of fractions of which the numerators are all unity and the denominators successively be the product of every two adjacent denominators. Instead of the fractions written above, we have thus the series:
+ - + - ...
The first term, as we see, is the first fraction; the first and second together give the second fraction, ; the first, the second and the third give the third fraction , and so on with the rest; the result being that the series entire is equivalent to the original value.
## Generalized continued fraction
A generalized continued fraction is an expression of the form
${\displaystyle x=b_{0}+{\cfrac {a_{1}}{b_{1}+{\cfrac {a_{2}}{b_{2}+{\cfrac {a_{3}}{b_{3}+{\cfrac {a_{4}}{b_{4}+\ddots \,}}}}}}}}}$
where the an (n > 0) are the partial numerators, the bn are the partial denominators, and the leading term b0 is called the integer part of the continued fraction.
To illustrate the use of generalized continued fractions, consider the following example. The sequence of partial denominators of the simple continued fraction of ? does not show any obvious pattern:
${\displaystyle \pi =[3;7,15,1,292,1,1,1,2,1,3,1,\ldots ]}$
or
${\displaystyle \pi =3+{\cfrac {1}{7+{\cfrac {1}{15+{\cfrac {1}{1+{\cfrac {1}{292+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{1+{\cfrac {1}{2+{\cfrac {1}{1+{\cfrac {1}{3+{\cfrac {1}{1+\ddots }}}}}}}}}}}}}}}}}}}}}}}$
However, several generalized continued fractions for ? have a perfectly regular structure, such as:
${\displaystyle \pi ={\cfrac {4}{1+{\cfrac {1^{2}}{2+{\cfrac {3^{2}}{2+{\cfrac {5^{2}}{2+{\cfrac {7^{2}}{2+{\cfrac {9^{2}}{2+\ddots }}}}}}}}}}}}={\cfrac {4}{1+{\cfrac {1^{2}}{3+{\cfrac {2^{2}}{5+{\cfrac {3^{2}}{7+{\cfrac {4^{2}}{9+\ddots }}}}}}}}}}=3+{\cfrac {1^{2}}{6+{\cfrac {3^{2}}{6+{\cfrac {5^{2}}{6+{\cfrac {7^{2}}{6+{\cfrac {9^{2}}{6+\ddots }}}}}}}}}}}$
${\displaystyle \displaystyle \pi =2+{\cfrac {2}{1+{\cfrac {1}{1/2+{\cfrac {1}{1/3+{\cfrac {1}{1/4+\ddots }}}}}}}}=2+{\cfrac {2}{1+{\cfrac {1\cdot 2}{1+{\cfrac {2\cdot 3}{1+{\cfrac {3\cdot 4}{1+\ddots }}}}}}}}}$
${\displaystyle \displaystyle \pi =2+{\cfrac {4}{3+{\cfrac {1\cdot 3}{4+{\cfrac {3\cdot 5}{4+{\cfrac {5\cdot 7}{4+\ddots }}}}}}}}}$
The first two of these are special cases of the arctangent function with ? = 4 arctan (1).
${\displaystyle \pi =3+{\cfrac {1^{3}}{6+{\cfrac {1^{3}+2^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}}{6+{\cfrac {1^{3}+2^{3}+3^{3}+4^{3}+5^{3}+6^{3}+7^{3}+8^{3}}{6+\ddots }}}}}}}}}}}$
The continued fraction of ${\displaystyle \pi }$ above consisting of cubes uses the Nilakantha series and an exploit from Leonhard Euler.[13]
## Other continued fraction expansions
### Periodic continued fractions
The numbers with periodic continued fraction expansion are precisely the irrational solutions of quadratic equations with rational coefficients; rational solutions have finite continued fraction expansions as previously stated. The simplest examples are the golden ratio ? = [1;1,1,1,1,1,...] and = [1;2,2,2,2,...], while = [3;1,2,1,6,1,2,1,6...] and = [6;2,12,2,12,2,12...]. All irrational square roots of integers have a special form for the period; a symmetrical string, like the empty string (for ) or 1,2,1 (for ), followed by the double of the leading integer.
### A property of the golden ratio ?
Because the continued fraction expansion for ? doesn't use any integers greater than 1, ? is one of the most "difficult" real numbers to approximate with rational numbers. Hurwitz's theorem[14] states that any irrational number k can be approximated by infinitely many rational with
${\displaystyle \left|k-{m \over n}\right|<{1 \over n^{2}{\sqrt {5}}}.}$
While virtually all real numbers k will eventually have infinitely many convergents whose distance from k is significantly smaller than this limit, the convergents for ? (i.e., the numbers , , , , etc.) consistently "toe the boundary", keeping a distance of almost exactly ${\displaystyle {\scriptstyle {1 \over n^{2}{\sqrt {5}}}}}$ away from ?, thus never producing an approximation nearly as impressive as, for example, for ?. It can also be shown that every real number of the form , where a, b, c, and d are integers such that a d - b c = ±1, shares this property with the golden ratio ?; and that all other real numbers can be more closely approximated.
### Regular patterns in continued fractions
While there is no discernable pattern in the simple continued fraction expansion of ?, there is one for e, the base of the natural logarithm:
${\displaystyle e=e^{1}=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,1,1,12,1,1,\dots ],}$
which is a special case of this general expression for positive integer n:
${\displaystyle e^{1/n}=[1;n-1,1,1,3n-1,1,1,5n-1,1,1,7n-1,1,1,\dots ]\,\!.}$
Another, more complex pattern appears in this continued fraction expansion for positive odd n:
${\displaystyle e^{2/n}=\left[1;{\frac {n-1}{2}},6n,{\frac {5n-1}{2}},1,1,{\frac {7n-1}{2}},18n,{\frac {11n-1}{2}},1,1,{\frac {13n-1}{2}},30n,{\frac {17n-1}{2}},1,1,\dots \right]\,\!,}$
with a special case for n = 1:
${\displaystyle e^{2}=[7;2,1,1,3,18,5,1,1,6,30,8,1,1,9,42,11,1,1,12,54,14,1,1\dots ,3k,12k+6,3k+2,1,1\dots ]\,\!.}$
Other continued fractions of this sort are
${\displaystyle \tanh(1/n)=[0;n,3n,5n,7n,9n,11n,13n,15n,17n,19n,\dots ]}$
where n is a positive integer; also, for integer n:
${\displaystyle \tan(1/n)=[0;n-1,1,3n-2,1,5n-2,1,7n-2,1,9n-2,1,\dots ]\,\!,}$
with a special case for n = 1:
${\displaystyle \tan(1)=[1;1,1,3,1,5,1,7,1,9,1,11,1,13,1,15,1,17,1,19,1,\dots ]\,\!.}$
If In(x) is the modified, or hyperbolic, Bessel function of the first kind, we may define a function on the rationals by
${\displaystyle S(p/q)={\frac {I_{p/q}(2/q)}{I_{1+p/q}(2/q)}},}$
which is defined for all rational numbers, with p and q in lowest terms. Then for all nonnegative rationals, we have
${\displaystyle S(p/q)=[p+q;p+2q,p+3q,p+4q,\dots ],}$
with similar formulas for negative rationals; in particular we have
${\displaystyle S(0)=S(0/1)=[1;2,3,4,5,6,7,\dots ].}$
Many of the formulas can be proved using Gauss's continued fraction.
### Typical continued fractions
Most irrational numbers do not have any periodic or regular behavior in their continued fraction expansion. Nevertheless, Khinchin proved that for almost all real numbers x, the ai (for i = 1, 2, 3, ...) have an astonishing property: their geometric mean tends to a constant (known as Khinchin's constant, K ? 2.6854520010...) independent of the value of x. Paul Lévy showed that the nth root of the denominator of the nth convergent of the continued fraction expansion of almost all real numbers approaches an asymptotic limit, approximately 3.27582, which is known as Lévy's constant. Lochs' theorem states that nth convergent of the continued fraction expansion of almost all real numbers determines the number to an average accuracy of just over n decimal places.
## Applications
### Square roots
Generalized continued fractions are used in a method for computing square roots.
The identity
leads via recursion to the generalized continued fraction for any square root:[15]
### Pell's equation
Continued fractions play an essential role in the solution of Pell's equation. For example, for positive integers p and q, and non-square n, it is true that if p2 - nq2 = ±1, then is a convergent of the regular continued fraction for . The converse holds if the period of the regular continued fraction for is 1, and in general the period describes which convergents give solutions to Pell's equation.[16]
### Dynamical systems
Continued fractions also play a role in the study of dynamical systems, where they tie together the Farey fractions which are seen in the Mandelbrot set with Minkowski's question mark function and the modular group Gamma.
The backwards shift operator for continued fractions is the map h(x) = 1/x - ?1/x? called the Gauss map, which lops off digits of a continued fraction expansion: h([0; a1, a2, a3, ...]) = [0; a2, a3, ...]. The transfer operator of this map is called the Gauss-Kuzmin-Wirsing operator. The distribution of the digits in continued fractions is given by the zero'th eigenvector of this operator, and is called the Gauss-Kuzmin distribution.
### Eigenvalues and eigenvectors
The Lanczos algorithm uses a continued fraction expansion to iteratively approximate the eigenvalues and eigenvectors of a large sparse matrix.[17]
### Networking applications
Continued fractions have also been used in modelling optimization problems for wireless network virtualization to find a route between a source and a destination.[18]
## Examples of rational and irrational numbers
Number r 0 1 2 3 4 5 6 7 8 9 10
123 ar 123
ra 123
12.3 ar 12 3 3
ra 12
1.23 ar 1 4 2 1 7
ra 1
0.123 ar 0 8 7 1 2 5
ra 0
ϕ =
ar 1 1 1 1 1 1 1 1 1 1 1
ra 1 2
ϕ =
ar −2 2 1 1 1 1 1 1 1 1 1
ra −2
ar 1 2 2 2 2 2 2 2 2 2 2
ra 1
ar 0 1 2 2 2 2 2 2 2 2 2
ra 0 1
ar 1 1 2 1 2 1 2 1 2 1 2
ra 1 2
ar 0 1 1 2 1 2 1 2 1 2 1
ra 0 1
ar 0 1 6 2 6 2 6 2 6 2 6
ra 0 1
ar 1 3 1 5 1 1 4 1 1 8 1
ra 1
e ar 2 1 2 1 1 4 1 1 6 1 1
ra 2 3
π ar 3 7 15 1 292 1 1 1 2 1 3
ra 3
Number r 0 1 2 3 4 5 6 7 8 9 10
ra: rational approximant obtained by expanding continued fraction up to ar
## History
• 300 BCE Euclid's Elements contains an algorithm for the greatest common divisor which generates a continued fraction as a by-product
• 499 The Aryabhatiya contains the solution of indeterminate equations using continued fractions
• 1572 Rafael Bombelli, L'Algebra Opera - method for the extraction of square roots which is related to continued fractions
• 1613 Pietro Cataldi, Trattato del modo brevissimo di trovar la radice quadra delli numeri - first notation for continued fractions
Cataldi represented a continued fraction as ${\displaystyle a_{0}}$ & ${\displaystyle {\frac {n_{1}}{d_{1}\cdot }}}$ & ${\displaystyle {\frac {n_{2}}{d_{2}\cdot }}}$ & ${\displaystyle {\frac {n_{3}}{d_{3}\cdot }}}$ with the dots indicating where the following fractions went.
## Notes
1. ^ "Continued fraction - mathematics".
2. ^ a b Pettofrezzo & Byrkit (1970, p. 150)
3. ^ a b Long (1972, p. 173)
4. ^ a b Pettofrezzo & Byrkit (1970, p. 152)
5. ^
6. ^ Collins, Darren C. "Continued Fractions" (PDF). MIT Undergraduate Journal of Mathematics. Archived from the original (PDF) on 2001-11-20.
7. ^ Long (1972, p. 183)
8. ^ Pettofrezzo & Byrkit (1970, p. 158)
9. ^ Long (1972, p. 177)
10. ^ Pettofrezzo & Byrkit (1970, pp. 162-163)
11. ^ a b M. Thill (2008), "A more precise rounding algorithm for rational numbers", Computing, 82: 189-198, doi:10.1007/s00607-008-0006-7
12. ^ Shoemake, Ken (1995), "I.4: Rational Approximation", in Paeth, Alan W. (ed.), Graphic Gems V, San Diego, California: Academic Press, pp. 25-31, ISBN 0-12-543455-3
13. ^ Foster, Tony (June 22, 2015). "Theorem of the Day: Theorem no. 203" (PDF). Robin Whitty. Retrieved 2015.
14. ^ Theorem 193: Hardy, G.H.; Wright, E.M. (1979). An Introduction to the Theory of Numbers (Fifth ed.). Oxford.
15. ^ Ben Thurston, "Estimating square roots, generalized continued fraction expression for every square root", The Ben Paul Thurston Blog
16. ^ Niven, Ivan; Zuckerman, Herbert S.; Montgomery, Hugh L. (1991). An introduction to the theory of numbers (Fifth ed.). New York: Wiley. ISBN 0-471-62546-9.
17. ^ Martin, Richard M. (2004), Electronic Structure: Basic Theory and Practical Methods, Cambridge University Press, p. 557, ISBN 9781139643658.
18. ^ Afifi, Haitham; et al. (April 2018). "MARVELO: Wireless Virtual Network Embedding for Overlay Graphs with Loops". 2018 IEEE Wireless Communications and Networking Conference (WCNC).
19. ^ Sandifer, Ed (February 2006). "How Euler Did It: Who proved e is irrational?" (PDF). MAA Online.
20. ^
21. ^ Wolfram, Stephen (2002). A New Kind of Science. Wolfram Media, Inc. p. 915. ISBN 1-57955-008-8.
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# How to Find the Area of a Triangle
The area of a triangle is found using the formula:
In this formula, b is the length of the base of the triangle and h is the height of the triangle. The image below shows what we mean by the lengths of the bases and the height:
# A Real Example of How to Find the Area of a Triangle
#### An Example Question
What is the area of a triangle with a base of 5 cm and a height of 3 cm, as shown below?
Step 1
Area = ½bh
Don't forget: ½bh = ½ × b × h
Step 2
Substitute the length of the base and the height into the formula. In our example, b = 5 and h = 3.
Area = ½ × 5 × 3 = 7.5 cm2
Don't forget: ½ × a number = 0.5 × a number = a number ÷ 2.
The area of the triangle with a base of 5 cm and a height of 3 cm is 7.5 cm2.
# Another Real Example of How to Find the Area of a Triangle
The slider below shows another real example of how to find the area of a triangle:
# How to Find the Area of a Triangle Using Trigonometry
The area of a triangle can also be found using trigonometry.
The area is found using the formula:
In the formula, a and b are lengths of two sides of the triangle and C is the angle between them. The image below shows what we mean by the two sides and the angle between them:
Read more about how to find the area of a triangle using trigonometry
Geometry Lessons
##### Interactive Test
show
Here's a second test on finding the area of a triangle.
Here's a third test on finding the area of a triangle.
# What Is a Triangle?
A triangle is a flat shape with three straight sides and three corners (also called vertices).
There are different types of triangle:
# What's in a Name?
A triangle has three corners and so three angles.
Just as a tricycle has three wheels, and a tripod has three legs, a shape with three angles in called a triangle.
# Areas of Triangles and Rectangles
The area of a triangle is ½bh.
The area of a rectangle is bh.
The area of the triangle is half that of the rectangle with the same base length and height.
Can you see from the image of a right-angled triangle below that this is the case?
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