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# Conditional Probability and Tree Diagrams
A LevelAQAEdexcelOCRAQA 2022OCR 2022
## Conditional Probability and Tree Diagrams
Tree diagrams are visual ways of understanding probabilities involving more than one event. Conditional probability is the mathematical formulation of this understanding. Ultimately – both help us answer the same question: what is the probability of an event happening given that a related event has already happened?
Make sure you are happy with the following topics before continuing.
A Level
## Tree Diagrams
Suppose we have two events, $\color{red}A$ and $\color{blue}B$, such that $\mathbb{P}(\color{red}A\color{grey})=\color{red}{p}$ and $\mathbb{P}(\color{blue}B\color{grey})=\color{blue}{q}$. A tree diagram for $\color{red}A$ and $\color{blue}B$ looks like this:
From this, we can see that:
$\mathbb{P}(\color{red}A\color{grey}\cap\color{blue}B\color{grey})=\color{red}p\color{blue}q$
$\mathbb{P}(\color{red}A'\color{grey}\cap\color{blue}B\color{grey})=\color{red}(1-p)\color{blue}q$
$\mathbb{P}(\color{red}A\color{grey}\cap\color{blue}B'\color{grey})=\color{red}p\color{blue}(1-q)$
$\mathbb{P}(\color{red}A'\color{grey}\cap\color{blue}B'\color{grey})=\color{red}(1-p)\color{blue}(1-q)$
(Note: For guidance on the notation used here please see Probability and Venn Diagrams.)
A Level
## Conditional Probability
Conditional probability is the probability of an event happening given that a related event has already happened.
Notation: $\color{grey}\mathbb{P}(\color{blue}B\color{grey}|\color{red}A\color{grey})$ means the probability of $\color{blue}B$ given $\color{red}A$ which is the probability of $\color{blue}B$ provided that $\color{red}A$ has already happened.
We have the following formula:
$\color{grey}\mathbb{P}(\color{blue}B\color{grey}|\color{red}A\color{grey})=\dfrac{\mathbb{P}(\color{red}A\color{grey}\cap\color{blue}B\color{grey})}{\mathbb{P}(\color{red}A\color{grey})}$
If $\color{blue}B$ is conditional on $\color{red}A$, then $\color{red}A$ is conditional on $\color{blue}B$, and $\color{grey}\mathbb{P}(\color{blue}B\color{grey}|\color{red}A\color{grey})$ is linked to $\color{grey}\mathbb{P}(\color{red}A\color{grey}|\color{blue}B\color{grey})$ by the formula:
$\color{grey}\mathbb{P}(\color{red}A\color{grey}|\color{blue}B\color{grey})=\dfrac{\mathbb{P}(\color{blue}B\color{grey}|\color{red}A\color{grey})\mathbb{P}(\color{red}A\color{grey})}{\mathbb{P}(\color{blue}B\color{grey}|\color{red}A\color{grey})\mathbb{P}(\color{red}A\color{grey})+\mathbb{P}(\color{blue}B\color{grey}|\color{red}A'\color{grey})\mathbb{P}(\color{red}A'\color{grey})}$
Using what we know about conditional probability, we can now redo the tree diagram:
A Level
A Level
## Example 1: Tree Diagrams
Sally has a bag of sweets. There are 12 blue sweets and 18 red sweets in total. Sally takes two sweets, without replacement. What is the probability that she takes exactly one blue sweet.
[3 marks]
Draw a tree diagram:
$\mathbb{P}(\text{\color{red}Red\color{grey} and \color{red}Red\color{grey}})=\dfrac{18}{30}\times\dfrac{17}{29}=\dfrac{51}{145}$
$\mathbb{P}(\text{\color{red}Red\color{grey} and \color{blue}Blue\color{grey}})=\dfrac{18}{30}\times\dfrac{12}{29}=\dfrac{36}{145}$
$\mathbb{P}(\text{\color{blue}Blue\color{grey} and \color{red}Red\color{grey}})=\dfrac{12}{30}\times\dfrac{18}{29}=\dfrac{36}{145}$
$\mathbb{P}(\text{\color{blue}Blue\color{grey} and \color{blue}Blue\color{grey}})=\dfrac{12}{30}\times\dfrac{11}{29}=\dfrac{22}{145}$
So the probability of one blue sweet is:
$\mathbb{P}(\text{One blue sweet})=\dfrac{36}{145}+\dfrac{36}{145}=\dfrac{72}{145}$
A Level
## Example 2: Conditional Probability
Every morning Jonathan either has breakfast ($B$) or doesn’t ($B'$), then leaves the house ($L$) or doesn’t ($L'$). The probability that he has breakfast is $0.3$. If he eats breakfast the probability that he leaves the house is $0.6$. If he does not eat breakfast the probability that he leaves the house is $0.2$. If you see Jonathan outside of his house, what is the probability he had breakfast?
[4 marks]
We want to find $\mathbb{P}(B|L)$. We can use the formula:
\begin{aligned}\mathbb{P}(B|L)&=\dfrac{\mathbb{P}(L|B)\mathbb{P}(B)}{\mathbb{P}(L|B)\mathbb{P}(B)+\mathbb{P}(L|B')\mathbb{P}(B')}\\[1.2em]&=\dfrac{0.6\times 0.3}{0.6\times 0.3+0.2\times 0.7}\\[1.2em]&=\dfrac{0.18}{0.18+0.14}\\[1.2em]&=\dfrac{0.18}{0.32}\\[1.2em]&=\dfrac{9}{16}\\[1.2em]&=0.5625\end{aligned}
Note: We could also have done this question with a tree diagram.
A Level
## Example Questions
For this, we need a tree diagram with three events.
\begin{aligned}&\mathbb{P}(\text{two heads})=\mathbb{P}(HHT)+\mathbb{P}(HTH)+\mathbb{P}(THH)\\[1.2em]&=0.5\times 0.5\times 0.5+0.5\times 0.5\times 0.5+0.5\times 0.5\times 0.5\\[1.2em]&=0.125+0.125+0.125\\[1.2em]&=0.375\end{aligned}
$\mathbb{P}(B|A)=\dfrac{\mathbb{P}(A\cap B)}{\mathbb{P}(A)}$
\begin{aligned}\mathbb{P}(A)&=\dfrac{\mathbb{P}(A\cap B)}{\mathbb{P}(B|A)}\\[1.2em]&=\dfrac{0.3}{0.9}\\[1.2em]&=\dfrac{1}{3}\end{aligned}
$\mathbb{P}(A|B)=\dfrac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)}$
\begin{aligned}\mathbb{P}(B)&=\dfrac{\mathbb{P}(A\cap B)}{\mathbb{P}(A|B)}\\[1.2em]&=\dfrac{0.3}{0.6}\\[1.2em]&=\dfrac{1}{2}\end{aligned}
Use a tree diagram for this question.
\begin{aligned}\mathbb{P}(\text{same colour})&=\mathbb{P}(\text{red \& red})+\mathbb{P}(\text{green \& green})\\[1.2em]&=\dfrac{61}{100}\times\dfrac{60}{99}+\dfrac{39}{100}\times\dfrac{38}{99}\\[1.2em]&=\dfrac{61}{165}+\dfrac{247}{1650}\\[1.2em]&=\dfrac{857}{1650}\\[1.2em]&=0.519\end{aligned}
For this we use the big formula:
$\mathbb{P}(A|B)=\dfrac{\mathbb{P}(B|A)\mathbb{P}(A)}{\mathbb{P}(B|A)\mathbb{P}(A)+\mathbb{P}(B|A')\mathbb{P}(A')}$
Call $A$ the event that Ellie’s train on time and call $B$ the event that Ellie is on time. We want $\mathbb{P}(A|B)$.
The question gives:
$\mathbb{P}(A)=0.5$
$\mathbb{P}(A')=0.5$
$\mathbb{P}(B|A)=0.9$
$\mathbb{P}(B'|A)=0.1$
$\mathbb{P}(B'|A')=0.8$
$\mathbb{P}(B|A')=0.2$
Now we put these values into the formula.
\begin{aligned}\mathbb{P}(A|B)&=\dfrac{0.9\times 0.5}{0.9\times 0.5+0.2\times 0.5}\\[1.2em]&=\dfrac{0.45}{0.45+0.1}\\[1.2em]&=\dfrac{0.45}{0.55}\\[1.2em]&=0.819\end{aligned}
The probability that Tuesday’s train was on time is $0.819$
A Level
A Level
A Level
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# 6th Grade Math
## Ratios & Proportions
### Ratios
Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities.
### Unit Ratios
Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. Solve unit rate problems including those involving unit pricing and constant speed.
### Equivalent Ratios I
Make tables of equivalent ratios relating quantities with whole-number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
### Percentage
Find a percent of a quantity as a rate per 100; solve problems involving finding the whole, given a part and the percent.
### Unit Measurement
Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
## Number System
### Fraction Division
Interpret and compute quotients of fractions, and solve word problems involving division of fractions by fractions, e.g., by using visual fraction models and equations to represent the problem.
### Division
Fluently divide multi-digit numbers using the standard algorithm.
### Decimals
Fluently add, subtract, multiply, and divide multi-digit decimals using the standard algorithm for each operation.
### GCF & LCM
Find the greatest common factor of two whole numbers less than or equal to 100 and the least common multiple of two whole numbers less than or equal to 12. Use the distributive property to express a sum of two whole numbers 1-100 with a common factor as a multiple of a sum of two whole numbers with no common factor.
### Positive & Negative Numbers
Understand that positive and negative numbers are used together to describe quantities having opposite directions or values; use positive and negative numbers to represent quantities in real-world contexts, explaining the meaning of 0 in each situation.
### Opposite Numbers
Recognize opposite signs of numbers as indicating locations on opposite sides of 0 on the number line; recognize that the opposite of the opposite of a number is the number itself, e.g., -(-3) = 3, and that 0 is its own opposite.
### Understanding Signs
Understand signs of numbers in ordered pairs as indicating locations in quadrants of the coordinate plane; recognize that when two ordered pairs differ only by signs, the locations of the points are related by reflections across one or both axes.
### Coordinates
Find and position integers and other rational numbers on a horizontal or vertical number line diagram; find and position pairs of integers and other rational numbers on a coordinate plane.
### Compare Numbers
Interpret statements of inequality as statements about the relative position of two numbers on a number line diagram. Write, interpret, and explain statements of order for rational numbers in real-world contexts.
### Absolute Number
Understand the absolute value of a rational number as its distance from 0 on the number line; interpret absolute value as magnitude for a positive or negative quantity in a real-world situation. Distinguish comparisons of absolute value from statements about order.
### Graphing
Solve real-world and mathematical problems by graphing points in all four quadrants of the coordinate plane. Include use of coordinates and absolute value to find distances between points with the same first coordinate or the same second coordinate.
## Expressions & Equations
### Exponents
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### Expressions
Write expressions that record operations with numbers and with letters standing for numbers. Identify parts of an expression using mathematical terms (sum, term, product, factor, quotient, coefficient); view one or more parts of an expression as a single entity.
### Expressions Word Problems
Evaluate expressions at specific values of their variables. Include expressions that arise from formulas used in real-world problems. Perform arithmetic operations, including those involving whole-number exponents, in the conventional order when there are no parentheses to specify a particular order (Order of Operations).
### Evaluating Expressions
Apply the properties of operations to generate equivalent expressions. Identify when two expressions are equivalent (i.e., when the two expressions name the same number regardless of which value is substituted into them).
### Expression Inequality
Understand solving an equation or inequality as a process of answering a question: which values from a specified set, if any, make the equation or inequality true? Use substitution to determine whether a given number in a specified set makes an equation or inequality true.
### Solving Equations
Use variables to represent numbers and write expressions when solving a real-world or mathematical problem; understand that a variable can represent an unknown number, or, depending on the purpose at hand, any number in a specified set. Solve real-world and mathematical problems by writing and solving equations of the form x + p = q and px = q for cases in which p, q and x are all nonnegative rational numbers.
### Inequality
Write an inequality of the form x > c or x < c to represent a constraint or condition in a real-world or mathematical problem. Recognize that inequalities of the form x > c or x < c have infinitely many solutions; represent solutions of such inequalities on number line diagrams.
### Variable Relationship
Use variables to represent two quantities in a real-world problem that change in relationship to one another; write an equation to express one quantity, thought of as the dependent variable, in terms of the other quantity, thought of as the independent variable. Analyze the relationship between the dependent and independent variables using graphs and tables, and relate these to the equation.
## Geometry
### Area
Find the area of right triangles, other triangles, special quadrilaterals, and polygons by composing into rectangles or decomposing into triangles and other shapes; apply these techniques in the context of solving real-world and mathematical problems. Represent three-dimensional figures using nets made up of rectangles and triangles, and use the nets to find the surface area of these figures.
### Volume
Find the volume of a right rectangular prism with fractional edge lengths by packing it with unit cubes of the appropriate unit fraction edge lengths, and show that the volume is the same as would be found by multiplying the edge lengths of the prism. Apply the formulas V = l w h and V = b h to find volumes of right rectangular prisms with fractional edge lengths in the context of solving real-world and mathematical problems.
### Shapes in Coordinate Plane
Draw polygons in the coordinate plane given coordinates for the vertices; use coordinates to find the length of a side joining points with the same first coordinate or the same second coordinate. Apply these techniques in the context of solving real-world and mathematical problems. |
## Real Mathematics : Graph Theory
There is precious little mathematics being done in the public schools. Mathematics doesn't consist of determining whether 122, 357, and 1005 can form the sides of a right triangle, or using the quadratic formula to find the solutions of the equation 2x2 + 31x - 4 = 0. It involves discovering facts and proving that they are facts. So let's take a look at a problem I have several times used to introduce people to mathematics.
This puzzle was common when I was a child, and I remember one of my elementary school teachers remarking that it was popular when she was a child. You draw a square with an X connecting opposite corners:
The object of the puzzle is to draw that figure without lifting one's pencil from the page or retracing a line. We were all convinced that it couldn't be done. Here I want to do it, or prove that it can't be done. And at this point, the budding young (or old) mathematician will spend a few minutes trying to draw it, to get a feel for the problem and how it might be approached.
If no brilliant ideas come to mind — and for me, brilliant ideas are not common — then we can always fall back on brute force. Let's consider how we'd test every single path. First, do we have to try every path starting at every corner? No, obviously not; if we try every path starting from one corner, those starting at the other three will give identical results. So for our starting point we need to consider only one corner, and the crossing point in the middle as well. Of course, we have to make sure we don't miss any possible paths.
Ok, let's try one, and again feel our way around. One obvious path is to start at corner 1, draw the line to 2, then 3, then 4, and back to 1. So we've drawn the box, and now we have the cross-shaped lines in the middle to do. Can we draw that without retracing or lifting, starting at the end of one of the arms? Clearly not; we draw to the middle and then we have three branches to draw, and each one is a dead end. So at least some of the lines in the middle must be drawn before we complete the outside square.
Suppose I start at 1, then go diagonally across to 3, up to 2, left back to 1, down to 4. (Better draw it yourself so you can visualize it.) Now I have two branches remaining. I could go diagonally to 2, or right to 3. But each of those ends after one move, and I have to do both. In other words, there are two different places I have to end, and obviously I can end the trace at only one place.
Where can this trace end, anyway? Can it end at the point where I start? Lessee, there are three paths at each corner. I take one out, one back, and the final one out. So it can't end at the corner where I start. That is, if I start at 1, I have to end at some other point.
Nothing stops me from starting at the middle point, I suppose. If I start at the middle point, there are 4 paths. I take one out, one back, the third out, and the fourth one back. So starting in the middle requires me to finish there. And the same reasoning shows that if I don't start in the middle, then I can't finish at that point: I'd take one line in to the middle point, a second out, a third in, and the last one has to take me out again.
Now I think I'm on to something here, and I'm going to abandon the "try all possible paths" approach — at least until I determine that this train of reasoning leads nowhere. Furthermore, in keeping with the idea that real mathematics requires hard thinking, trial, error, and rigorous demonstration, I'm going to leave you here to figure it out on your own.
If you can't get it and want to see the complete solution, send email to me and I'll send you the url to the rest of it. But don't give up! Success in mathematics — like everything else — is rewarding in proportion to the effort expended. So give it some concentrated effort over the course of several days before you decide to ask for the solution. And if you think your son or daughter likes this kind of thing, and you want a introduction to the rich field of mathematics, contact me and we'll make a plan to get it.
By the way, this branch of mathematics is called graph theory, and if it interests you, a book you'll find interesting is Four Colors Suffice by Robin Wilson. |
# Warm Up Solve. 1. log16x = 2. logx1.331 = log10,000 = x 1.1 4
## Presentation on theme: "Warm Up Solve. 1. log16x = 2. logx1.331 = log10,000 = x 1.1 4"— Presentation transcript:
Warm Up Solve. 1. log16x = 2. logx1.331 = 3 64 3. log10,000 = x 1.1 4
Objectives Solve exponential and logarithmic equations and equalities.
Solve problems involving exponential and logarithmic equations.
Vocabulary exponential equation logarithmic equation
An exponential equation is an equation containing one or more expressions that have a variable as an exponent. To solve exponential equations: Try writing them so that the bases are all the same. Take the logarithm of both sides.
When you use a rounded number in a check, the result will not be exact, but it should be reasonable.
Example 1A: Solving Exponential Equations
Solve and check. 98 – x = 27x – 3 Rewrite each side with the same base; 9 and 27 are powers of 3. (32)8 – x = (33)x – 3 To raise a power to a power, multiply exponents. 316 – 2x = 33x – 9 16 – 2x = 3x – 9 Bases are the same, so the exponents must be equal. x = 5 Solve for x.
Example 1A Continued Check 98 – x = 27x – 3 98 – – 3 The solution is x = 5.
Example 1B: Solving Exponential Equations
Solve and check. 4x – 1 = 5 log 4x – 1 = log 5 5 is not a power of 4, so take the log of both sides. (x – 1)log 4 = log 5 Apply the Power Property of Logarithms. x –1 = log5 log4 Divide both sides by log 4. x = ≈ 2.161 log5 log4 Check Use a calculator. The solution is x ≈
Check It Out! Example 1a Solve and check. 32x = 27 Rewrite each side with the same base; 3 and 27 are powers of 3. (3)2x = (3)3 32x = 33 To raise a power to a power, multiply exponents. 2x = 3 Bases are the same, so the exponents must be equal. x = 1.5 Solve for x.
Check It Out! Example 1a Continued
32(1.5) 27 33 27 The solution is x = 1.5.
21 is not a power of 7, so take the log of both sides.
Check It Out! Example 1b Solve and check. 7–x = 21 21 is not a power of 7, so take the log of both sides. log 7–x = log 21 (–x)log 7 = log 21 Apply the Power Property of Logarithms. –x = log21 log7 Divide both sides by log 7. x = – ≈ –1.565 log21 log7
Check It Out! Example 1b Continued
Check Use a calculator. The solution is x ≈ –1.565.
15 is not a power of 2, so take the log of both sides.
Check It Out! Example 1c Solve and check. 23x = 15 log23x = log15 15 is not a power of 2, so take the log of both sides. (3x)log 2 = log15 Apply the Power Property of Logarithms. 3x = log15 log2 Divide both sides by log 2, then divide both sides by 3. x ≈ 1.302
Check It Out! Example 1c Continued
Check Use a calculator. The solution is x ≈
Example 2: Biology Application
Suppose a bacteria culture doubles in size every hour. How many hours will it take for the number of bacteria to exceed 1,000,000? At hour 0, there is one bacterium, or 20 bacteria. At hour one, there are two bacteria, or 21 bacteria, and so on. So, at hour n there will be 2n bacteria. Solve 2n > 106 Write 1,000,000 in scientific annotation. log 2n > log 106 Take the log of both sides.
Use the Power of Logarithms.
Example 2 Continued nlog 2 > log 106 Use the Power of Logarithms. nlog 2 > 6 log 106 is 6. 6 log 2 n > Divide both sides by log 2. 6 0.301 n > Evaluate by using a calculator. n > ≈ 19.94 Round up to the next whole number. It will take about 20 hours for the number of bacteria to exceed 1,000,000.
Example 2 Continued Check In 20 hours, there will be 220 bacteria. 220 = 1,048,576 bacteria.
Check It Out! Example 2 You receive one penny on the first day, and then triple that (3 cents) on the second day, and so on for a month. On what day would you receive a least a million dollars. \$1,000,000 is 100,000,000 cents. On day 1, you would receive 1 cent or 30 cents. On day 2, you would receive 3 cents or 31 cents, and so on. So, on day n you would receive 3n–1 cents. Solve 3n – 1 > 1 x 108 Write 100,000,000 in scientific annotation. log 3n – 1 > log 108 Take the log of both sides.
Check It Out! Example 2 Continued
(n – 1) log 3 > log 108 Use the Power of Logarithms. (n – 1)log 3 > 8 log 108 is 8. 8 log 3 n – 1 > Divide both sides by log 3. 8 log3 n > Evaluate by using a calculator. n > ≈ 17.8 Round up to the next whole number. Beginning on day 18, you would receive more than a million dollars.
Check It Out! Example 2 Check On day 18, you would receive 318 – 1 or over a million dollars. 317 = 129,140,163 cents or 1.29 million dollars.
A logarithmic equation is an equation with a logarithmic expression that contains a variable. You can solve logarithmic equations by using the properties of logarithms.
Review the properties of logarithms from Lesson 7-4.
Remember!
Example 3A: Solving Logarithmic Equations
Solve. log6(2x – 1) = –1 6 log6 (2x –1) = 6–1 Use 6 as the base for both sides. 2x – 1 = 1 6 Use inverse properties to remove 6 to the log base 6. 7 12 x = Simplify.
Example 3B: Solving Logarithmic Equations
Solve. log4100 – log4(x + 1) = 1 100 x + 1 log4( ) = 1 Write as a quotient. 4log = 41 100 x + 1 ( ) Use 4 as the base for both sides. = 4 100 x + 1 Use inverse properties on the left side. x = 24
Example 3C: Solving Logarithmic Equations
Solve. log5x 4 = 8 4log5x = 8 Power Property of Logarithms. log5x = 2 Divide both sides by 4 to isolate log5x. x = 52 Definition of a logarithm. x = 25
Example 3D: Solving Logarithmic Equations
Solve. log12x + log12(x + 1) = 1 log12 x(x + 1) = 1 Product Property of Logarithms. log12x(x +1) = 121 Exponential form. x(x + 1) = 12 Use the inverse properties.
x Example 3 Continued x2 + x – 12 = 0 Multiply and collect terms.
Factor. x – 3 = 0 or x + 4 = 0 Set each of the factors equal to zero. x = 3 or x = –4 Solve. Check Check both solutions in the original equation. log12x + log12(x +1) = 1 log12x + log12(x +1) = 1 log123 + log12(3 + 1) 1 x log12( –4) + log12(–4 +1) 1 log123 + log log12( –4) is undefined. log 1 1 The solution is x = 3.
Check It Out! Example 3a Solve. 3 = log 8 + 3log x 3 = log 8 + 3log x 3 = log 8 + log x3 Power Property of Logarithms. 3 = log (8x3) Product Property of Logarithms. 103 = 10log (8x3) Use 10 as the base for both sides. 1000 = 8x3 Use inverse properties on the right side. 125 = x3 5 = x
Use 10 as the base for both sides.
Check It Out! Example 3b Solve. 2log x – log 4 = 0 2log( ) = 0 x 4 Write as a quotient. 2(10log ) = 100 x 4 Use 10 as the base for both sides. 2( ) = 1 x 4 Use inverse properties on the left side. x = 2
Watch out for calculated solutions that are not solutions of the original equation.
Caution
Example 4A: Using Tables and Graphs to Solve Exponential and Logarithmic Equations and Inequalities
Use a table and graph to solve 2x + 1 > 8192x. Use a graphing calculator. Enter 2^(x + 1) as Y1 and 8192x as Y2. In the table, find the x-values where Y1 is greater than Y2. In the graph, find the x-value at the point of intersection. The solution set is {x | x > 16}.
Use a graphing calculator. Enter log(x + 70) as Y1 and 2log( ) as Y2.
Example 4B log(x + 70) = 2log( ) x 3 Use a graphing calculator. Enter log(x + 70) as Y1 and 2log( ) as Y2. x 3 In the table, find the x-values where Y1 equals Y2. In the graph, find the x-value at the point of intersection. The solution is x = 30.
Check It Out! Example 4a Use a table and graph to solve 2x = 4x – 1. Use a graphing calculator. Enter 2x as Y1 and 4(x – 1) as Y2. In the table, find the x-values where Y1 is equal to Y2. In the graph, find the x-value at the point of intersection. The solution is x = 2.
Check It Out! Example 4b Use a table and graph to solve 2x > 4x – 1. Use a graphing calculator. Enter 2x as Y1 and 4(x – 1) as Y2. In the table, find the x-values where Y1 is greater than Y2. In the graph, find the x-value at the point of intersection. The solution is x < 2.
Check It Out! Example 4c Use a table and graph to solve log x2 = 6. Use a graphing calculator. Enter log(x2) as Y1 and 6 as Y2. In the table, find the x-values where Y1 is equal to Y2. In the graph, find the x-value at the point of intersection. The solution is x = 1000.
Lesson Quiz: Part I Solve. x = 5 3 1. 43x–1 = 8x+1 2. 32x–1 = 20 x ≈ 1.86 3. log7(5x + 3) = 3 x = 68 4. log(3x + 1) – log 4 = 2 x = 133 5. log4(x – 1) + log4(3x – 1) = 2 x = 3
Lesson Quiz: Part II 6. A single cell divides every 5 minutes. How long will it take for one cell to become more than 10,000 cells? 70 min 7. Use a table and graph to solve the equation 23x = 33x–1. x ≈ 0.903
Chapter 7 Section 5 Page 526 # 2-20, 21-36
Homework! Chapter 7 Section 5 Page 526 # 2-20, 21-36
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Mathematics Area between Two Curves CBSE-NCERT
Click for Only Video
### Topic Covered
\color{green} ✍️ Area between Two Curves
### Area between Two Curves
● we are given two curves represented by y = f (x), y = g (x), where f (x) ≥ g(x) in [a, b] as shown in Fig .
● Here the points of intersection of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two curves.
● For setting up a formula for the integral, it is convenient to take elementary area in the form of vertical strips. As indicated in the Fig
f (x) – g(x) and width dx so that the elementary area
dA = [f (x) – g(x)] dx, and the total area A can be taken as
color{orange} {A = int_a^b [ f(x )- g(x )] dx}
Alternatively,
A = ["area bounded by" y = f (x), "x-axis and the lines"\ \ x = a, x = b]
– ["area bounded by" y = g (x), "x-axis and the lines" \ \ x = a, x = b]
color {red} {= int_a^b f(x) dx - int_a^b g(x) dx = int_a^b [ f(x) - g(x) ] dx} , where f(x) >= g(x) in [ a,b]
● If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the Fig , then the area of the regions bounded by curves can be written as
Total Area = Area of the region ACBDA + Area of the region BPRQB
color{red} {= int_a^c [ f(x) - g(x) ] dx + int_c^b [ g(x) - f(x) ] dx}
Q 3106256178
Find the area of the region bounded by the two parabolas y = x^2 and y^2 = x.
Class 12 Chapter 8 Example 6
Solution:
The point of intersection of these two
parabolas are O (0, 0) and A (1, 1) as shown in
the Fig 8.15.
Here, we can set y^2 = x or y = sqrt x = f (x) and y = x^2
= g (x), where, f (x) ≥ g (x) in [0, 1].
Therefore, the required area of the shaded region
= int_0^1 [ f (x) − g(x) ] dx
= int_0^1 [ sqrt x - x^2 ] dx = [ 2/3 x^(3/2) - x^3/3 ]_0^1
= 2/3 - 1/3 = 1/3
Q 3156356274
Find the area lying above x-axis and included between the circle
x^2 + y^2 = 8x and the parabola y^2 = 4x.
Class 12 Chapter 8 Example 7
Solution:
The given equation of the circle x^2 + y^2 = 8x can be expressed as
(x – 4)^2 + y^2 = 16. Thus, the centre of the
circle is (4, 0) and radius is 4. Its intersection
with the parabola y^2 = 4x gives
x^2 + 4x = 8x
or x^2 – 4x = 0
or x (x – 4) = 0
or x = 0, x = 4
Thus, the points of intersection of these
two curves are O(0, 0) and P(4,4) above the
x-axis.
From the Fig 8.16, the required area of
the region OPQCO included between these
two curves above x-axis is
= (area of the region OCPO) + (area of the region PCQP)
= ∫_0^4 ydx + ∫_4^8 ydx
= 2 ∫_0^4 sqrt x dx + ∫_4^8 sqrt(4^2 −(x −4)^2) dx (Why?)
= 2 xx 2/3 [ x^(3/2]_0^4 + int_0^4 sqrt( 4^2 - t^2) dt , where, x - 4 = t
= (32)/3 + [ t/2 sqrt(4^2 - t^2) + 1/2 xx 4^2 xx sin^(-1) \ t/4 ]_0^4
= (32)/3 + [ 4/2 xx 0 + 1/2 xx 4^2 xx sin^(-1) 1 ] = (32)/3 + [ 0 + 8 xx pi/2] = (32)/3+ 4 pi = 4/3 (8 + 3 pi)
Q 3106356278
AOBA is the part of the ellipse 9x^2 + y^2 = 36 in the first
quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the
chord AB.
Class 12 Chapter 8 Example 8
Solution:
Given equation of the ellipse 9x^2 + y^2 = 36 can be expressed as
x^2/2^2 + y^2/6^2 = 1 and hence, its shape is as given in Fig 8.17
Accordingly, the equation of the chord AB is
y – 0 = (6 - 0)/(0-2) (x - 2)
or y = – 3(x – 2)
or y = – 3x + 6
Area of the shaded region as shown in the Fig 8.17.
= 3∫_0^2 sqrt(4 − x^2) dx − ∫_0^2 (6 − 3x)dx (Why?)
= 3 [ x/2 sqrt(4 − x^2) + 4/2 sin^(-1) \ x/2 ]_0^2 - [ 6x - (3x^2)/2 ]_0^2
= 3 [ 2/2 xx 0 + 2 sin^(-1) (1)] - [ 12 - (12)/2 ] = 3 xx 2 xx pi/2 - 6 =3pi-6
Q 3136456372
Using integration find the area of region bounded by the triangle whose
vertices are (1, 0), (2, 2) and (3, 1).
Class 12 Chapter 8 Example 9
Solution:
Let A(1, 0), B(2, 2) and C(3, 1) be
the vertices of a triangle ABC (Fig 8.18).
Area of ΔABC
= Area of ΔABD + Area of trapezium
BDEC – Area of ΔA EC
Now equation of the sides AB, BC and
CA are given by
y = 2 (x – 1), y = 4 – x, y = 1/2 (x – 1), respectively.
Hence, area of Δ ABC = int _1^2 2 (x -2) dx int_2^3 (4-x) dx - int _1^3 (x-1)/2 dx
= 2 [ x^2/2 - x]_1^2 + [ 4x - x^2/2]_2^3 - 1/2 [ x^2/2 - x]_1^3
= 2 [ ( 2^2/2 - 2) - (1/2 -1) ] + [ ( 4 xx3 - 3^2/2 ) - ( 4 xx2 - 2^2/2 ) ] - 1/2 [ (3^2/2 - 3) - (1/2 -1) ]
= 3/2
Q 3116556470
Find the area of the region enclosed between the two circles: x^2 + y^2 = 4
and (x – 2)^2 + y^2 = 4.
Class 12 Chapter 8 Example 10
Solution:
Equations of the given circles are
x^2 + y^2 = 4 ... (1)
and (x – 2)^2 + y^2 = 4 ... (2)
Equation (1) is a circle with centre O at the
origin and radius 2. Equation (2) is a circle with
centre C (2, 0) and radius 2. Solving equations
(1) and (2), we have
(x –2)^2 + y^2 = x^2 + y^2
or x^2 – 4x + 4 + y^2 = x^2 + y^2
or x = 1 which gives y = ± sqrt3
Thus, the points of intersection of the given
circles are A(1, sqrt 3 ) and A′(1, – sqrt3 ) as shown in
the Fig 8.19.
Required area of the enclosed region OACA′O between circles
= 2 [area of the region ODCAO] (Why?)
= 2 [area of the region ODAO + area of the region DCAD]
= 2 [ int_0^1 ydx + int_1^2 ydx ]
= 2 [ int_0^1 sqrt( 4 − (x − 2)^2) dx + int_1^2 sqrt( 4-x) dx]
= 2 [1/2 (x-2) sqrt( 4 − (x − 2)^2) + 1/2 xx 4 sin^(-1) ((x-2)/2 ) ]_0^1
+ 2 [ 1/2 x sqrt( 4 - x^2 ) + 1/2 xx 4 sin^(-1) \ x/2 ] _1^2
= [ ( - sqrt3 + 4 sin ^(-1) ( (-1)/2) - 4 sin ^(-1) (-1) ] + [ 4 sin ^(-1) 1 - sqrt 3 - 4 sin^(-1) \ 1/2 ]
= [ ( - sqrt3 - 4 xx pi/4 ) + 4 xx pi/2 ] + [ 4 xx pi/2 - sqrt3 - 4 xx pi/6 ]
= ( - sqrt3 - (2pi)/3 + 2 pi) + ( 9 2 pi - sqrt3 - (2 pi)/3 )
= (8pi)/3 - 2 sqrt3 |
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## AP®︎/College Calculus BC
### Course: AP®︎/College Calculus BC>Unit 6
Lesson 8: Finding antiderivatives and indefinite integrals: basic rules and notation: reverse power rule
# Reverse power rule review
Review your knowledge of the reverse power rule for integrals and solve problems with it.
## What is the reverse power rule?
The reverse power rule tells us how to integrate expressions of the form x, start superscript, n, end superscript where n, does not equal, minus, 1:
integral, x, start superscript, n, end superscript, d, x, equals, start fraction, x, start superscript, n, plus, 1, end superscript, divided by, n, plus, 1, end fraction, plus, C
Basically, you increase the power by one and then divide by the power plus, 1.
Remember that this rule doesn't apply for n, equals, minus, 1.
Instead of memorizing the reverse power rule, it's useful to remember that it can be quickly derived from the power rule for derivatives.
## Integrating polynomials
We can use the reverse power rule to integrate any polynomial. Consider, for example, the integration of the monomial 3, x, start superscript, 7, end superscript:
\begin{aligned} \displaystyle\int 3x^7\,dx&=3\left(\dfrac{x^{7+1}}{7+1}\right)+C \\\\ &=3\left(\dfrac{x^8}{8}\right)+C \\\\ &=\dfrac{3}{8}x^8+C \end{aligned}
Problem 1
• Current
integral, 14, t, d, t, equals, question mark
Want to try more problems like this? Check out these exercises:
## Integrating negative powers
The reverse power rule allows us to integrate any negative power other than minus, 1. Consider, for example, the integration of start fraction, 1, divided by, x, squared, end fraction:
\begin{aligned} \displaystyle\int \dfrac{1}{x^2}\,dx&=\displaystyle\int x^{-2}\,dx \\\\ &=\dfrac{x^{-2+1}}{-2+1}+C \\\\ &=\dfrac{x^{-1}}{-1}+C \\\\ &=-\dfrac{1}{x}+C \end{aligned}
Problem 1
• Current
integral, 8, t, start superscript, minus, 3, end superscript, d, t, equals
Want to try more problems like this? Check out these exercises:
## Integrating fractional powers and radicals
The reverse power rule also allows us to integrate expressions where x is raised to a fractional power, or radicals. Consider, for example, the integration of square root of, x, end square root:
\begin{aligned} \displaystyle\int \sqrt x\,dx&=\displaystyle\int x^{^{\large\frac{1}{2}}}\,dx \\\\ &=\dfrac{x^{^{\large\frac{1}{2}\normalsize+1}}}{\dfrac{1}{2}+1}+C \\\\ &=\dfrac{x^{^{\large\frac{3}{2}}}}{\frac{3}{2}}+C \\\\ &=\dfrac{2\sqrt{x^3}}{3}+C \end{aligned}
Problem 1
• Current
integral, 4, t, start superscript, start fraction, 1, divided by, 3, end fraction, end superscript, d, t, equals, question mark
Want to try more problems like this? Check out these exercises:
## Want to join the conversation?
• What would you do if n=-1? What would the integral be of x^-1?
• ln(x)
If you have mastered differential calc at KA, then you most likely have come across the derivative of ln(x) = 1/x.
• It's so easy.
Just like differential calculus, integral calculus has its own rules.
• Until when is integration going to be fun like this?
• It's fun until you enjoy it. Integrals will get lengthier and will require more methods and thinking, but if you learn to enjoy a subject, no matter how hard concepts get, you'll still have fun doing problems.
Frankly though, integrals in Calc II can get pretty nasty as here, they're testing you on how smart you are when it comes to figuring out a way to integrate. So, they can give you the wackiest integrals on the planet. If/when you reach Calc III, you'll learn about double and triple integrals (Yeah. They exist lol!). Here, you're not tested on how well you can integrate. So, your integrand will be fairly simple. You'll be tested on how well you can visualize and define the region (which is easy in single variable Calculus as there is only one axis to take care of), which is really the hardest part about multiple integrals.
• How do we go from 1/3 x^3 to 4sqrtx^8?
• 4sqrtx^8 is rewritten as x^2, because (x^2)^4 = x^8
Therefore, the antiderivative of x^2 is:
x^(2+1) / (2+1) + C
x^3 / (3) + C
1/3 x^3 + C
• how do you integrate sin^2 x?
• What is integral of √ax+b dx
(1 vote)
• It is ambiguous.....are both ax under the radical or just a?....Let's solve the first case which is the most laborious case....
2*5^(1/2)*x^(3/2)/3 + bx I hope it helps!
• how do you do this
• how do you integrate>> ((4with under root) x^8) |
Triangle calculator - the result
You have entered sides a, b, and c.
Right scalene Pythagorean triangle.
Sides: a = 3 b = 4 c = 5
Area: T = 6
Perimeter: p = 12
Semiperimeter: s = 6
Angle ∠ A = α = 36.87698976458° = 36°52'12″ = 0.64435011088 rad
Angle ∠ B = β = 53.13301023542° = 53°7'48″ = 0.9277295218 rad
Angle ∠ C = γ = 90° = 1.57107963268 rad
Height: ha = 4
Height: hb = 3
Height: hc = 2.4
Median: ma = 4.27220018727
Median: mb = 3.60655512755
Median: mc = 2.5
Vertex coordinates: A[5; 0] B[0; 0] C[1.8; 2.4]
Centroid: CG[2.26766666667; 0.8]
Coordinates of the circumscribed circle: U[2.5; 0]
Coordinates of the inscribed circle: I[2; 1]
Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 143.13301023542° = 143°7'48″ = 0.64435011088 rad
∠ B' = β' = 126.87698976458° = 126°52'12″ = 0.9277295218 rad
∠ C' = γ' = 90° = 1.57107963268 rad
How did we calculate this triangle?
3. Semiperimeter of the triangle
The semiperimeter of the triangle is half its perimeter. The semiperimeter frequently appears in formulas for triangles to be given a separate name. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter.
4. The triangle area using Heron's formula
Heron's formula gives the area of a triangle when the length of all three sides is known. There is no need to calculate angles or other distances in the triangle first. Heron's formula works equally well in all cases and types of triangles.
5. Calculate the heights of the triangle from its area.
There are many ways to find the height of the triangle. The easiest way is from the area and base length. The triangle area is half of the product of the base's length and height. Every side of the triangle can be a base; there are three bases and three heights (altitudes). Triangle height is the perpendicular line segment from a vertex to a line containing the base.
6. Calculation of the inner angles of the triangle using a Law of Cosines
The Law of Cosines is useful for finding a triangle's angles when we know all three sides. The cosine rule, also known as the law of cosines, relates all three sides of a triangle with an angle of a triangle. The Law of Cosines extrapolates the Pythagorean theorem for any triangle. Pythagorean theorem works only in a right triangle. Pythagorean theorem is a special case of the Law of Cosines and can be derived from it because the cosine of 90° is 0. It is best to find the angle opposite the longest side first. With the Law of Cosines, there is also no problem with obtuse angles as with the Law of Sines because the cosine function is negative for obtuse angles, zero for right, and positive for acute angles. We also use inverse cosine called arccosine to determine the angle from the cosine value.
An incircle of a triangle is a tangent circle to each side. An incircle center is called an incenter and has a radius named inradius. All triangles have an incenter, and it always lies inside the triangle. The incenter is the intersection of the three-angle bisectors. The product of the inradius and semiperimeter (half the perimeter) of a triangle is its area.
The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. Circumcenter (center of the circumcircle) is the point where the perpendicular bisectors of a triangle intersect.
9. Calculation of medians
A median of a triangle is a line segment joining a vertex to the opposite side's midpoint. Every triangle has three medians, and they all intersect each other at the triangle's centroid. The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex. We use Apollonius's theorem to calculate the length of a median from the lengths of its side. |
# How do you solve and graph f-6<5 and f-4>=2?
Mathematically, this is expressed as $6 \le f < 11$
#### Explanation:
We first need to solve each part of this problem:
$f - 6 < 5 \implies f < 11$
$f - 4 \ge 2 \implies f \ge 6$
The graph, then, will consist of a line along the $f$-axis (probably looking very much like a number line) and will be a solid dot at 6 (to indicate that 6 is part of the solution), a hollow dot at 11 (to indicate that all the points up to but not including 11 are part of the solution), and a solid line connecting the two dots.
Mathematically, this is expressed as $6 \le f < 11$ |
# Watch out for Parentheses
Alignments to Content Standards: 6.EE.A
Evaluate the following numerical expressions.
1. $2(5+(3)(2)+4)$
2. $2((5+3)(2+4))$
3. $2(5+3(2+4))$
Can the parentheses in any of these expressions be removed without changing the value the expression?
## IM Commentary
This problem asks the student to evaluate three numerical expressions that contain the same integers yet have differing results due to placement of parentheses. Students from all levels of mathematics often make an error on (c) by first doing the operation of $5 + 3$. Parts (a) and (b) help the student recognize this error.
This type of problem helps students to see structure in expressions and represents an important transition from the work that students do in elementary school with numeric expressions to the work they will be doing in middle school with algebraic expressions. Note that this task was originally written to illustrate 5.OA.1, but Heather Brown, Illinois State Board of Eduction Math Content Specialist, pointed out that the K–5 Operations and Algebraic Thinking Progression explicitly excludes numeric expressions with nested parentheses in 5th grade. Thus, this task would be appropriate for 5th graders needing an extra challenge or 6th graders getting ready to make the transition to working with expressions that contain variables.
## Solution
1. $2(5 + (3)(2) + 4)$. We may evaluate this expression in two ways:
Distributing the lead constant first: $$2 \cdot 5+2 \cdot 3 \cdot 2+2 \cdot 4=10+12+8=30$$
or distributing the lead constant last: $$2(5 + 6 + 4) = 2 \cdot 15 = 30.$$
Either way, we first have to multiply $(3)(2) = 6$ before adding any of the terms. The parentheses in the middle are not necessary. Instead of writing $(3)(2)$ we can say $3 \cdot 2$.
2. Notice that in the expression $2((5 + 3)(2 + 4))$, the outer set of parentheses are not necessary: $$2((5 + 3)(2 + 4)) = 2(5 + 3)(2 + 4).$$ The other parentheses are necessary since they indicate that we first have to perform the additions inside these parentheses:
$$2(5 + 3)(2 + 4) = 2(8)(6) = 96.$$
3. In this expression we complete the operations from the inside out. The inner most addition must occur first, then the inner multiplication, then the secondary addition and finally the outer multiplication:
$$2(5 + 3(2 + 4)) = 2(5 + 3(6)) = 2(5 + 18) = 2(23) = 46.$$
In this expression all parentheses are needed. |
We've updated our
TEXT
# Examples: Probability using Permutations and Combinations
We can use permutations and combinations to help us answer more complex probability questions
### Example 1
A 4 digit PIN is selected. What is the probability that there are no repeated digits? There are 10 possible values for each digit of the PIN (namely: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9), so there are 10 × 10 × 10 × 10 = 10 4 = 10000 total possible PINs. To have no repeated digits, all four digits would have to be different, which is selecting without replacement. We could either compute 10 × 9 × 8 × 7, or notice that this is the same as the permutation 10P4 = 5040. The probability of no repeated digits is the number of 4 digit PINs with no repeated digits divided by the total number of 4 digit PINs. This probability is [latex-display]\displaystyle\frac{{{}_{{10}}{P}_{{4}}}}{{{10}^{{4}}}}=\frac{{5040}}{{10000}}={0.504}[/latex-display]
### Try it Now 1
A multiple-choice question on an economics quiz contains 10 questions with five possible answers each. Compute the probability of randomly guessing the answers and getting exactly 9 questions correct.
### Example 4
Compute the probability of randomly drawing five cards from a deck and getting exactly one Ace. In many card games (such as poker) the order in which the cards are drawn is not important (since the player may rearrange the cards in his hand any way he chooses); in the problems that follow, we will assume that this is the case unless otherwise stated. Thus we use combinations to compute the possible number of 5-card hands, 52C5. This number will go in the denominator of our probability formula, since it is the number of possible outcomes. For the numerator, we need the number of ways to draw one Ace and four other cards (none of them Aces) from the deck. Since there are four Aces and we want exactly one of them, there will be 4C1 ways to select one Ace; since there are 48 non-Aces and we want 4 of them, there will be 48C4 ways to select the four non-Aces. Now we use the Basic Counting Rule to calculate that there will be 4C1 × 48C4 ways to choose one ace and four non-Aces. Putting this all together, we have [latex-display]\displaystyle{P}{\left(\text{one Ace}\right)}=\frac{{{\left({}_{{4}}{C}_{{1}}\right)}{\left({}_{{48}}{C}_{{4}}\right)}}}{{{}_{{52}}{C}_{{5}}}}=\frac{{778320}}{{2598960}}\approx{0.299}[/latex-display]
### Example 5
Compute the probability of randomly drawing five cards from a deck and getting exactly two Aces. The solution is similar to the previous example, except now we are choosing 2 Aces out of 4 and 3 non-Aces out of 48; the denominator remains the same: It is useful to note that these card problems are remarkably similar to the lottery problems discussed earlier.
### Try it Now 2
Compute the probability of randomly drawing five cards from a deck of cards and getting three Aces and two Kings.
## Birthday Problem
Let's take a pause to consider a famous problem in probability theory:
Suppose you have a room full of 30 people. What is the probability that there is at least one shared birthday?
Take a guess at the answer to the above problem. Was your guess fairly low, like around 10%? That seems to be the intuitive answer (30/365, perhaps?). Let's see if we should listen to our intuition. Let's start with a simpler problem, however.
### Example 6
Suppose three people are in a room. What is the probability that there is at least one shared birthday among these three people? There are a lot of ways there could be at least one shared birthday. Fortunately there is an easier way. We ask ourselves "What is the alternative to having at least one shared birthday?" In this case, the alternative is that there are no shared birthdays. In other words, the alternative to "at least one" is having none. In other words, since this is a complementary event, P(at least one) = 1 – P(none) We will start, then, by computing the probability that there is no shared birthday. Let's imagine that you are one of these three people. Your birthday can be anything without conflict, so there are 365 choices out of 365 for your birthday. What is the probability that the second person does not share your birthday? There are 365 days in the year (let's ignore leap years) and removing your birthday from contention, there are 364 choices that will guarantee that you do not share a birthday with this person, so the probability that the second person does not share your birthday is 364/365. Now we move to the third person. What is the probability that this third person does not have the same birthday as either you or the second person? There are 363 days that will not duplicate your birthday or the second person's, so the probability that the third person does not share a birthday with the first two is 363/365. We want the second person not to share a birthday with you and the third person not to share a birthday with the first two people, so we use the multiplication rule: [latex-display]\displaystyle{P}{\left(\text{no shared birthday}\right)}=\frac{{365}}{{365}}\cdot\frac{{364}}{{365}}\cdot\frac{{363}}{{365}}\approx{0.9918}[/latex-display] and then subtract from 1 to get P(shared birthday) = 1 – P(no shared birthday) = 1 – 0.9918 = 0.0082. This is a pretty small number, so maybe it makes sense that the answer to our original problem will be small. Let's make our group a bit bigger.
### Example 7
Suppose five people are in a room. What is the probability that there is at least one shared birthday among these five people? Continuing the pattern of the previous example, the answer should be [latex-display]\displaystyle{P}{\left(\text{shared birthday}\right)}={1}-\frac{{365}}{{365}}\cdot\frac{{364}}{{365}}\cdot\frac{{363}}{{365}}\cdot\frac{{362}}{{365}}\cdot\frac{{361}}{{365}}\approx{0.0271}[/latex-display] Note that we could rewrite this more compactly as [latex-display]\displaystyle{P}{\left(\text{shared birthday}\right)}={1}-\frac{{{}_{{365}}{P}_{{5}}}}{{365}^{{5}}}\approx{0.0271}[/latex-display] which makes it a bit easier to type into a calculator or computer, and which suggests a nice formula as we continue to expand the population of our group.
### Example 8
Suppose 30 people are in a room. What is the probability that there is at least one shared birthday among these 30 people? Here we can calculate [latex-display]\displaystyle{P}{\left(\text{shared birthday}\right)}={1}-\frac{{{}_{{365}}{P}_{{30}}}}{{365}^{{30}}}\approx{0.706}[/latex-display] which gives us the surprising result that when you are in a room with 30 people there is a 70% chance that there will be at least one shared birthday! If you like to bet, and if you can convince 30 people to reveal their birthdays, you might be able to win some money by betting a friend that there will be at least two people with the same birthday in the room anytime you are in a room of 30 or more people. (Of course, you would need to make sure your friend hasn't studied probability!) You wouldn't be guaranteed to win, but you should win more than half the time. This is one of many results in probability theory that is counterintuitive; that is, it goes against our gut instincts. If you still don't believe the math, you can carry out a simulation. Just so you won't have to go around rounding up groups of 30 people, someone has kindly developed a Java applet so that you can conduct a computer simulation. Go to this web page: http://www-stat.stanford.edu/~susan/surprise/Birthday.html, and once the applet has loaded, select 30 birthdays and then keep clicking Start and Reset. If you keep track of the number of times that there is a repeated birthday, you should get a repeated birthday about 7 out of every 10 times you run the simulation.
### Try it Now 3
Suppose 10 people are in a room. What is the probability that there is at least one shared birthday among these 10 people?
1. $\displaystyle{P}{\left({9}\ \text{ answers correct}\right)}=\frac{9\cdot4}{(5^{10})}\approx0.0000037$ chance
2. $\displaystyle{P}{\left(\text{three Aces and two Kings}\right)}=\frac{{{\left({}_{{4}}{C}_{{3}}\right)}{\left({}_{{4}}{C}_{{2}}\right)}}}{{{}_{{52}}{C}_{{5}}}}=\frac{{24}}{{2598960}}\approx{0.0000092}$
3. $\displaystyle{P}{\left(\text{shared birthday}\right)}={1}-\frac{{{}_{{365}}{P}_{{10}}}}{{365}^{{10}}}\approx{0.117}$ |
Finding cube root by estimation
Chapter 7 Class 8 Cubes and Cube Roots
Concept wise
Let’s do this by examples
## Find cube root of 857375.
Step 1:
Make group of 3 digit, starting from right
Step 2:
The unit digit of cube root will be
Unit digit of cube root of 857375 = Unit digit of cube root of 375.
Since 375 ends in 5, it’s cube root will end in 5.
(As 5 3 = 12 5 , 1 5 3 = 337 5 )
∴ Unit digit of cube root = 5
Step 3:
Now, for second group.
8 5 7
We note that
9 3 = 729
& 10 3 = 1000
So,
729 < 857 < 1000
9 3 < 857 < 10 3
∴ We take smaller number
So, 9
We put 9 in ten’s place of cube root
### Find cube root of 658503.
Step 1:
Make group of 3 digit, starting from right
Step 2 :
The unit digit of cube root will be
Unit digit of cube root of 658503 = Unit digit of cube root of 503
Since 503 ends in 3, it’s cube root will end in 7.
(As 7 3 = 34 3 , 1 7 3 = 491 3 )
∴ Unit digit of cube root = 7
Step 3:
Now, for second group.
6 5 8
We note that
8 3 = 512
& 9 3 = 729
So,
512 < 658 < 729
8 3 < 658 < 9 3
∴ We take smaller number
So, 8
We put 8 in ten’s place of cube root.
∴ Cube root of 658503 = 87
Get live Maths 1-on-1 Classs - Class 6 to 12 |
# Question Video: Converting and Comparing between Days and Hours Mathematics • 5th Grade
Fill in the missing symbol using <, >, or = : 5 days ＿ 46 hours.
02:06
### Video Transcript
Fill in the missing symbol using less than, greater than, or equal to. We need to compare five days to 46 hours.
Days and hours are different units. In order to compare these two numbers, we’ll need to compare the same units. This means we would need to convert five days into some hours. Or we would need to convert 46 hours into some number of days.
Let’s convert five days into hours. We know that one day is equal to 24 hours. If we want to find out how many hours are in five days, we need to multiply 24 times five. Starting in the ones place, we multiply five times four. Five times four equals 20. We bring down our zero and carry the two.
Moving on to the tens place, we need to multiply five times two. We know that five times two equals 10. But we have to add the two that we carried over. Five times two is 10, plus the two we carried over equals 12. So we bring down the 12.
This tells us that five days equals 120 hours. How does 120 hours compare to 46 hours? 120 hours is greater than 46 hours. So we know that five days is greater than 46 hours. The missing symbol is a greater than symbol. |
# K's Math Concepts
By Daddy K:
The ingenuity of youth is endlessly amazing. Our eldest daughter, K, came up with a few Math concepts on her own that simply blew me away. I was so amazed that I had to make my first addition to the Blog.
The first concept was chronologically second, but is the easier concept to understand. It was a Saturday and I woke to hear her getting out of bed. I decided to join her downstairs assuming we would be watching some variety of show she likes. Instead, she sat down with the dry erase board and starting drawing squares. She then wrote numbers above and below them and circled one of them. She then turned to me with a serious eye and said, “I know how to find the number in the middle”.
She sat down to explain it to me. She drew 5 boxes. Above each box she wrote the numbers one through five in ascending order. Below the boxes she wrote the numbers one through five in descending order. Where the two numbers are the same is the middle. In the case below the middle number is 3 since both above and below the number is the same.
I had to ask the question, "What if I had six boxes?" This stumped her as she understands what one half of something is, but she didn’t seem to grasp the difference between half and the middle. I sat down to explain it to her using her own created concept. You write the numbers the same as before. Instead of looking for one box with the same numbers above and below you find the two boxes that have the same two numbers, whether it is above or below, and draw a line between those two boxes. She instantly knew that I had cut it in half. I then asked “But what is the middle?”. She thought about it for a few minutes, then shrugged her shoulders. I explained that in this instance we could use either .5 or ½ to represent the line and that it was friends with the lower number. She then told me that 3 ½ was the middle.
The other concept was her self-created Math function. A little history is needed to understand it. K will at times make up words for everyday items and state, “That’s what it’s called on Jalar.” Jalar is an imaginary world she created. She came to me one day and asked if I would like to learn Math from Jalar. I entertained her and accepted the invitation.
She sat down and wrote “+ +” on the dry erase board. I was informed that what I was looking at was the symbol for “siren”. Siren is a Jalarian math expression. To teach me how to do Sirens she wrote it out on the board.
8++7
To solve for this expression you subtract the second number from the first. In our example this would be one. At this point she thought for a minute and then told me that we add that number to it again and add a zero when we’re done.
In the above example you would get the following:
1) Subtract the second from the first
8-7 =1
2) Add it to itself
1+1 = 2
3) Multiply by 10
2 * 10 = 20
Thus the answer to 8 siren 7 is 20.
Siren in OUR math language would be written as 20(X – Y).
Now initially I thought this was just an amusing one shot deal. I loved that my daughter taught me an original math concept. I was more amazed when two weeks later I asked her to solve for 6++4 and she came up with the correct answer. In five years she may not remember how to solve the Siren of two numbers but I will never forget it.
* 6++4 = 40 for those of you playing along at home.
Labels: , , |
BASIC CONCEPTS IN POLYNOMIALS WORKSHEET
About "Basic Concepts in Polynomials Worksheet"
Basic Concepts in Polynomials Worksheet :
Here we are going to see some practice questions on basic concepts in polynomials.
Basic Concepts in Polynomials Worksheet - Questions
(1) Which of the following expressions are polynomials. If not give reason:
(i) (1/x2) + 3x - 4
(ii) x2(x - 1)
(iii) (1/x)(x + 5)
(iv) (1/x-2) + (1/x-1) + 7
(v) √5 x2 + √3x + √2
(vi) m2∛m + 7m - 10
Solution
(2) Write the coefficient of x2 and x in each of the following polynomials.
(i) 4 + (2/5)x- 3x
(ii) 6 - 2x2 + 3x3 - 7 x
(iii) π x2 - x + 2
(iv) √3 x2 + √2x + 0.5
(v) x2 - (7/2)x + 8
Solution
(3) Find the degree of the following polynomials.
(i) 1 - √2y2 + y7
(ii) (x3 - x4 + 6x6)/x2
(iii) x3 (x2 + x)
(iv) 3x+ 9x2 + 27x6
(v) 2√5p4 - (8p3/√3) + (2p2/7)
Solution
(4) Rewrite the following polynomial in standard form.
(i) x - 9 + 7x3 + 6x2
(ii) √2x2 - (7/2)x4 + x - 5x3
(iii) 7x3 - (6/5)x2 + 4x - 1
(iv) y2 - √5y3 - 11 - (7/3) y + 9y4 Solution
(5) Add the following polynomials and find the degree of the resultant polynomial.
(i) p(x) = 6x2 - 7x + 2 and q(x) = 6x3 - 7x + 15
(ii) h(x) = 7x3 - 6x + 1, f(x) = 7x2 + 17x - 9
(iii) f(x) = 16x4 - 5x2 + 9, g(x) = -6x3 + 7x - 15
Solution
(6) Subtract the second polynomial from the first polynomial and find the degree of the resultant polynomial
(i) p(x) = 7x2 + 6x - 1 and q(x) = 6x - 9
(ii) f(y) = 6y2 - 7y + 2 and g(y) = 7y + y3
(iii) h(z) = z5 - 6z4 + z and f(z) = 6z2 + 10z - 7
Solution
(7) What should be added to 2x3 + 6x2 - 5x + 8 to get 3x3 - 2x2 + 6x + 15 ? Solution
(8) What must be subtracted from 2x4 + 4x2 - 3x + 7 to get 3x3 - x2 + 2x + 1? Solution
(9) Multiply the following polynomials and find the degree of the resultant polynomial:
(i) p(x) = x2 - 9 and q(x) = 6x2 + 7x - 2
(ii) f(x) = 7x + 2 and g(x) = 15x - 9
(iii) h(x) = 6x2 - 7x + 1 and f(x) = 5x - 7
Solution
(10) The cost of a chocolate is Rs. (x + y) and Amir bought (x + y) chocolates. Find the total amount paid by him in terms of x and y. If x = 10, y = 5 find the amount paid by him. Solution
(11) The length of a rectangle is (3x+2) units and it’s breadth is (3x–2) units. Find its area in terms of x. What will be the area if x = 20 units. Solution
(12) p(x) his a polynomial of degree 1 and q(x) is a polynomial of degree 2. What kind of the polynomial p(x) × q(x) is ? Solution
After having gone through the stuff given above, we hope that the students would have understood, "Basic Concepts in Polynomials Worksheet"
Apart from the stuff given in this section if you need any other stuff in math, please use our google custom search here.
You can also visit our following web pages on different stuff in math.
WORD PROBLEMS
Word problems on simple equations
Word problems on linear equations
Algebra word problems
Word problems on trains
Area and perimeter word problems
Word problems on direct variation and inverse variation
Word problems on unit price
Word problems on unit rate
Word problems on comparing rates
Converting customary units word problems
Converting metric units word problems
Word problems on simple interest
Word problems on compound interest
Word problems on types of angles
Complementary and supplementary angles word problems
Double facts word problems
Trigonometry word problems
Percentage word problems
Profit and loss word problems
Markup and markdown word problems
Decimal word problems
Word problems on fractions
Word problems on mixed fractrions
One step equation word problems
Linear inequalities word problems
Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
## Introduction
• The percent or percentage is a number, ratio, or fraction with 100 as a denominator.
• The percent means hundredths or per 100.
• A % sign denotes the percent.
• 5% means 5 out of 100.
• If you got 80% in math, you got 80 out of 100 questions correct.
## Changing decimals to percents
• To change the decimal number to the percent, multiply the number by 100 or move the decimal point two places to the right and add a % sign.
## Changing percents to decimals
• Remove the "%" sign and divide the number by 100 or move the decimal point to two places to the left.
• If there is no decimal, assume the decimal at the end of the number.
• There might be a need to add a zero(s) to the front of the number for the decimal to be moved to the left.
## Changing fractions to percents
• To convert a fraction to a percent, first convert the fraction to a decimal, then convert the decimal to a percent.
## Changing percents to fractions
• To change percents to fractions, trade the % sign for hundredths and simplify the fraction, if possible.
## Percent of a number
The following two methods can be used to determine the percent of a number:
1. Decimal Method
25% of 40
Remove the % sign and move the decimal two places to the left.
= 0.25 X 40
= 10
2. Fraction Method
Remove the % sign and divide the number by 100.
$=\frac{25}{100}×40$
= 10
## Examples
1) Change 0.65 to percent.
Move the decimal to two places to the right and add the % sign.
65%
2) Change 7.5 to percent.
Move the decimal to two places to the right and add the % sign.
750%
3) Change 13.11 to percent.
Move the decimal to two places to the right and add the % sign.
1311%
4) Change 34% to a decimal number.
Since there is no decimal point, assume the decimal point at the end and move the decimal point two places to the left.
$\frac{34}{100}=0.34$
5) Change 125% to a decimal number.
$\frac{125}{100}=1.25$
6) Change 6% to a decimal number.
Since there is no decimal point, assume the decimal point at the end and move the decimal point two places to the left.
$\frac{6}{100}=0.06$
7) Convert $\frac{3}{4}$ to percent.
$\frac{3}{4}=0.75$
Change the decimal to percent by moving the decimal point two places to the right.
0.75 = 75%
8) Convert $1\frac{1}{4}$ to percent.
$1\frac{1}{4}=\frac{5}{4}=1.25$
Change the decimal to percent by moving the decimal point two places to the right.
1.25 = 125%
9) Convert 80% to the fraction.
$80%=\frac{80}{100}=\frac{8}{10}=\frac{4}{5}$
10) Convert 75% to the fraction.
11) What is 15% of 40?
a) Decimal method
Remove the % sign and move the decimal to two places left.
15% $×$ 40
=0.15 $×$ 40
= 6
b) Fraction method
Remove the % sign and divide the number by 100.
15% of 40
=$\frac{15}{100}×40$
$\frac{600}{100}$
= 6
12. What is 250% of 50?
a) Decimal Method
250% of 50
Remove the % sign and move the decimal to two places left.
= $2.5×50$
= 125
b) Fraction Method
250% of 50
Remove the % sign and divide the number by 100.
$\frac{250}{100}×50$
= 125
## Cheat Sheet
• The word "of" means multiply.
## Blunder Area
• Pay close attention to the type of problem. |
# Foundations
Paris
Here, we do exercises from Tao's excellent book on Analysis
## Ouverture
Let us informally define sets and set membership $\in$. A set is an unordered collection of objects, with the set itself being an object. Set membership checks whether a given object is contained within a set:
\begin{align} \forall x, x \in A \vee x \not\in A \tag{law of excluded middle} \\ A \in C \wedge A = B \rightarrow B \in C \tag{axiom of substitution} \end{align}
The axiom of substition uses the definition of equality:
$$\begin{matrix} A = B & \leftrightarrow & \forall x, x \in A \rightarrow x \in B \\ && \wedge \\ && \forall x, x \in B \rightarrow x \in A \end{matrix}$$
The axiom of substition is preserved for all sets only defined in terms of $\in$ and $=$.
Verify that set equality is reflexive, symmetric, and transitive.
Goal:
$$\begin{matrix} \textrm{reflexive} & \forall A, A = A \\ \textrm{symmetric} & \forall A B, A = B \leftrightarrow B = A \\ \textrm{transitive} & \forall A B C, (A = B) \wedge (B = C) \rightarrow (A = C) \end{matrix}$$
Hypothesis:
$$\begin{matrix} A = B & \leftrightarrow & \forall x, x \in A \leftrightarrow x \in B \tag{eq} \end{matrix}$$
Proof:
$$\begin{matrix} A = A & \leftrightarrow & \forall x, x \in A \leftrightarrow x \in A & \textrm{by (eq)} \\ A = B & \leftrightarrow & \forall x, x \in A \rightarrow x \in B & \\ && \wedge & \\ && \forall x, x \in B \rightarrow x \in A & \\ & \leftrightarrow & B = A & \textrm{by (eq)} \\ A = B &&& \\ \wedge &&& \\ B = C & \rightarrow & \forall x, x \in A \rightarrow x \in B & \textrm{(1)} \\ && \wedge & \\ && \forall x, x \in B \rightarrow x \in A & \textrm{(2)} \\ && \wedge & \\ && \forall x, x \in B \rightarrow x \in C & \textrm{(3)} \\ && \wedge & \\ && \forall x, x \in C \rightarrow x \in B & \textrm{(4)} \\ & \rightarrow & \forall x, x \in A \rightarrow x \in C & \textrm{by (1) and (3)} \\ && \wedge & \\ && \forall x, x \in C \rightarrow x \in A & \textrm{by (4) and (2)} \\ & \rightarrow & A = C & \textrm{by (eq)} \end{matrix}$$
Prove that the following sets are distinct: $\phi, \{\phi\}, \{\{\phi\}\}, \{\phi, \{\phi\}\}$.
Hypothesis:
\begin{align} A = B & \leftrightarrow & \forall x, x \in A \leftrightarrow x \in B \tag{eq} \\ \forall x, x & \not\in & \phi \tag{phi} \\ \forall x, x \in \{a\} & \leftrightarrow & x = a \tag{build-a} \\ \forall x, x \in \{a, b\} & \leftrightarrow & (x = a) \vee (x = b) \tag{build-b} \end{align}
Lemma:
$$\begin{matrix} A \neq B & \leftrightarrow & \exists x, (x \in A \wedge x \not\in B) \vee (x \not\in A \wedge x \in B) & \textrm{by (eq)} \end{matrix}$$
Proof:
$$\begin{matrix} \phi & \neq & \textrm{any other set} & \textrm{by (phi) and (eq)} \\ \textrm{Plugging } x = \{\phi\} \textrm{ in (lemma)} & \rightarrow & \{\{\phi\}\} \neq \{\phi\} & \textrm{by (build-a)} \\ \textrm{Plugging } x = \{\phi\} \textrm{ in (lemma)} & \rightarrow & \{\phi, \{\phi\}\} \neq \{\phi\} & \textrm{by (build-a) and (build-b)} \\ \textrm{Plugging } x = \phi \textrm{ in (lemma)} & \rightarrow & \{\phi, \{\phi\}\} \neq \{\{\phi\}\} & \textrm{by (build-a) and (build-b)} \end{matrix}$$
Show that, for any set $A$, $A \cup A = A \cup \phi = \phi \cup A = A$.
Hypothesis:
\begin{align} A = B \leftrightarrow \forall x, x \in A \leftrightarrow x \in B \tag{eq} \\ \forall x, x \in A \cup B \leftrightarrow x \in A \vee x \in B \tag{union} \\ \forall x, x \not\in \phi \tag{phi} \end{align}
Proof:
$$\begin{matrix} A \cup A = A & \leftrightarrow \forall x, x \in A \vee x \in A \leftrightarrow x \in A & \textrm{by (eq) and (union)} \\ & \leftrightarrow \forall x, x \in A \leftrightarrow x \in A & \textrm{by intuition} \\ \\ A \cup \phi = A & \leftrightarrow \forall x, x \in A \vee x \in \phi \leftrightarrow x \in A & \textrm{by (eq) and (union)} \\ & \leftrightarrow \forall x, x \in A \leftrightarrow x \in A & \textrm{by (phi) and intuition} \\ \\ \phi \cup A = A && \textrm{by similar argument} \end{matrix}$$
Show that, for any three sets $A, B, C$, $A \subseteq B \wedge B \subseteq A \rightarrow A = B$, and $A \subsetneq B \subsetneq C \rightarrow A \subsetneq C$.
Hypothesis:
\begin{align} A = B & \leftrightarrow & \forall x, x \in A \leftrightarrow x \in B \tag{eq} \\ A \neq B & \leftrightarrow & \exists x, (x \in A \wedge x \not\in B) \vee (x \not\in A \wedge x \in B) \tag{neq} \\ A \subseteq B & \leftrightarrow & \forall x, x \in A \rightarrow x \in B \tag{subset-eq} \\ A \subsetneq B & \leftrightarrow & A \subseteq B \wedge A \neq B \tag{subset-neq} \end{align}
Proof:
$$\begin{matrix} A \subseteq B \wedge B \subseteq A & \rightarrow & \forall x, x \in A \leftrightarrow x \in B & \textrm{by (subset-eq) and intuition} \\ & \leftrightarrow & A = B & \textrm{by (eq)} \\ \\ A \subsetneq B & \leftrightarrow & A \subseteq B \wedge A \neq B & \textrm{by (subset-neq)} \\ & \leftrightarrow & \forall x, x \in A \rightarrow x \in B & \\ & & \wedge & \\ & & \exists x, (x \in A \wedge x \not\in B) \vee (x \in B \wedge x \not\in A) & \textrm{by (subset-eq) and (neq)} \\ & \rightarrow & \forall x, x \in A \rightarrow x \in B \wedge \exists x, x \in B \wedge x \not\in A & \textrm{by intuition} \\ \\\\ B \subsetneq C & \rightarrow & \forall x, x \in B \rightarrow x \in C & \\ & & \wedge & \\ & & \exists x, x \in C \wedge x \not\in B & \textrm{by similar argument} \\ \\\\ A \subsetneq B \wedge B \subsetneq C & \rightarrow & \forall x, x \in A \rightarrow x \in C & \\ & & \wedge & \\ & & \exists x, x \in C \wedge x \not\in A & \textrm{by intuition} \\ & \rightarrow & A \subseteq C \wedge C \neq A & \textrm{by (subset-eq) and (neq)} \\ & \rightarrow & A \subsetneq C & \textrm{by (subset-neq)} \end{matrix}$$
Show that, for any two sets $A, B$, $$A \subseteq B \leftrightarrow A \cup B = B \leftrightarrow A \cap B = A$$
Hypothesis:
\begin{align} A \subseteq B & = & \forall x &, x \in A \rightarrow x \in B \tag{subset-eq} \\ A \cup B & = & \forall x &, x \in A \vee x \in B \tag{union} \\ A \cap B & = & \forall x &, x \in A \wedge x \in B \tag{intersect} \end{align}
Proof:
$$\begin{matrix} A \cup B = B & \leftrightarrow & \forall x, x \in A \cup B \leftrightarrow x \in A \vee x \in B = B & \textrm{by (union)} \\ & \leftrightarrow & \forall x, x \in A \vee x \in B \leftrightarrow x \in B & \textrm{by intuition} \\ & \leftrightarrow & \forall x, x \in A \rightarrow x \in B & \textrm{by intuition} \\ & \leftrightarrow & A \subseteq B & \textrm{by (subset-eq)} \\ \\\\ A \cap B = A & \leftrightarrow & \forall x, x \in A \wedge x \in B \leftrightarrow x \in A & \textrm{by (intersect) and intuition} \\ & \leftrightarrow & \forall x \in A \rightarrow x \in B & \textrm{by intuition} \\ & \leftrightarrow & A \subseteq B & \textrm{by (subset-eq)} \end{matrix}$$
Let $A, B, C$ be sets.
Goal:
$$\begin{matrix} A \cap B \subseteq A & \wedge & A \cap B \subseteq B \\ C \subseteq A \wedge C \subseteq B & \leftrightarrow & C \subseteq A \cap B \\ A \subseteq C \wedge B \subseteq C & \leftrightarrow & A \cup B \subseteq C \end{matrix}$$
Hypothesis:
Proof: |
Lesson Objectives
• Learn how to solve SSA triangles using the law of sines
• Learn how to work with the law of sines when there is no solution
• Learn how to work with the law of sines when there is one solution
• Learn how to work with the law of sines when there are two solutions
## How to Solve SSA Triangles Using the Law of Sines
In the last lesson, we learned how to solve oblique triangles involving SAA (side-angle-angle) or ASA (angle-side-angle) using the law of sines. For that situation, we are given two angles and one side.
### Law of Sines
$$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}=\frac{c}{\text{sin}\hspace{.1em}C}$$ In some cases, we may need an alternative form: $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}B}{b}=\frac{\text{sin}\hspace{.1em}C}{c}$$
### Applying the Law of Sines
• For any angle β of a triangle:
• 0 < sin β ≤ 1
• sin β = 1, β = 90°
• sin β = sin(180° - β)
• Supplementary angles have the same sine value
• Assuming the triangle has sides that are all of different lengths:
• The smallest angle is opposite the shortest side
• The largest angle is opposite the largest side
• The middle-valued angle is opposite the intermediate side
### Law of Sines - Ambiguous Case
When we are given the lengths of two sides and the angle opposite one of them SSA (side-side-angle), then zero, one, or two such triangles may exist. This creates a situation where we need to dig into the details of the problem.
Let's start with a triangle ABC where A is an acute angle. If we know the measure of the acute angle A, the length of side a (side opposite of angle A), and the length of side b, there will be 4 possible outcomes.
#### Case 1: No Solution
First, let's think about the situation where there are 0 possible triangles. This happens when the law of sines leads to: $$\text{sin}\hspace{.1em}B > 1, a < h < b$$ Note: $$h=b \cdot \text{sin}\hspace{.1em}A$$ Let's look at an example.
Example #1: Solve each triangle. Round your answer to the nearest tenth. $$A=61°, b=29\hspace{.1em}\text{cm}, a=20\hspace{.1em}\text{cm}$$ First, we can check to see if a is less than h. $$h=29 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}61°$$ $$h ≈ 25.4 \hspace{.1em}\text{cm}$$ We see that a is clearly less than h, therefore, no such triangle can exist. We can also show this using the law of sines to try and find B: $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}B}{b}$$ $$\frac{\text{sin}\hspace{.1em}61°}{20 \hspace{.1em}\text{cm}}=\frac{\text{sin}\hspace{.1em}B}{29 \hspace{.1em}\text{cm}}$$ $$\text{sin}\hspace{.1em}B=\frac{29 \hspace{.1em}\text{cm}\cdot \text{sin}\hspace{.1em}61°}{20 \hspace{.1em}\text{cm}}≈ 1.3$$ The sine of B can never be greater than 1, therefore, no such triangle will exist.
#### Case 2: Right Triangle
Second, we will think about the case, where we have a right triangle. This happens when the law of sines leads to: $$\text{sin}\hspace{.1em}B=1, a=h, h < b$$ Note: $$h=b \cdot \text{sin}\hspace{.1em}A$$ Let's look at an example.
Example #2: Solve each triangle. Round your answer to the nearest tenth. $$A=30°, c=2 \sqrt{5}\hspace{.1em}\text{in}, a=\sqrt{5}\hspace{.1em}\text{in}$$ Note: c will take the place of b in our formula above.
Again, we can use our formula to find h. $$h=2 \sqrt{5}\hspace{.1em}\text{in}\cdot \text{sin}\hspace{.1em}30°$$ $$h=2 \sqrt{5}\hspace{.1em}\text{in}\cdot \frac{1}{2}$$ $$h=2 \sqrt{5}\hspace{.1em}\text{in}\cdot \frac{1}{2}$$ $$h=\sqrt{5}\hspace{.1em}\text{in}$$ We can see that a and h are equal in value. Also, a is less than c. This tells us we have a right triangle. We can also show this using the law of sines. $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}C}{c}$$ $$\frac{\text{sin}\hspace{.1em}30°}{\sqrt{5}\hspace{.1em}\text{in}}=\frac{\text{sin}\hspace{.1em}C}{2\sqrt{5}\hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}C=2\sqrt{5}\hspace{.1em}\text{in}\cdot \frac{\text{sin}\hspace{.1em}30°}{\sqrt{5}\hspace{.1em}\text{in}}=1$$ This tells us that C is a right angle or 90° angle. $$\text{sin}^{-1}(1)=90°$$ For angle B, we can use our triangle sum property: $$B=180° - 30° - 90°=60°$$ For side b, let's again turn to our law of sines: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{\sqrt{5}\hspace{.1em}\text{in}}{\text{sin}\hspace{.1em}30°}=\frac{b}{\text{sin}\hspace{.1em}60°}$$ $$b=\text{sin}\hspace{.1em}60° \cdot \frac{\sqrt{5}\hspace{.1em}\text{in}}{\text{sin}\hspace{.1em}30°}=\sqrt{15}\hspace{.1em}\text{in}$$
#### Case 3: Exactly One Non-Right Triangle
Third, we will think about the case, where we have exactly one solution, which is not a right triangle. This happens when the law of sines leads to: $$0 < \text{sin}\hspace{.1em}B < 1, a ≥ b$$ Let's look at an example.
Example #3: Solve each triangle. Round your answer to the nearest tenth. $$C=39°, b=28 \hspace{.1em}\text{in}, c=30 \hspace{.1em}\text{in}$$ Note: C will take the place of A and c will take the place of a in our formula above.
Here, c is larger than b. This tells us we will have one solution.
Let's use the law of sines to find angle B. $$\frac{\text{sin}\hspace{.1em}C}{c}=\frac{\text{sin}\hspace{.1em}B}{b}$$ $$\frac{\text{sin}\hspace{.1em}39°}{30 \hspace{.1em}\text{in}}=\frac{\text{sin}\hspace{.1em}B}{28 \hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}B=28 \hspace{.1em}\text{in}\cdot \frac{\text{sin}\hspace{.1em}39°}{30 \hspace{.1em}\text{in}}$$ $$\text{sin}\hspace{.1em}B ≈ .58737$$ Let's use our inverse sine function. $$B ≈ \text{sin}^{-1}(.58737) ≈ 36°$$ We know from the concept of reference angles that there is another angle with a sine value of 0.58737. $$180° - 36°=144°$$ We know this isn't possible, if we add this to the measure given for angle C: $$39° + 144°=183°$$ The sum of all angles of a triangle is always 180°. So there isn't a second triangle that is possible. We could also note that in the triangle c > b, which means the measure of angle C must be larger than the measure of angle B. To find A and a, we can again turn to our law of sines. $$A ≈ 180° - 39° - 36°=105°$$ $$\frac{c}{\text{sin}\hspace{.1em}C}=\frac{a}{\text{sin}\hspace{.1em}A}$$ $$\frac{30 \hspace{.1em}\text{in}}{\text{sin}\hspace{.1em}39°}=\frac{a}{\text{sin}\hspace{.1em}105°}$$ $$a=\frac{30 \hspace{.1em}\text{in}\cdot \text{sin}\hspace{.1em}105°}{\text{sin}\hspace{.1em}39°}$$ $$a ≈ 46\hspace{.1em}\text{in}$$
#### Case 4: Exactly Two Triangles
Fourth, we will think about the case, where we have exactly two solutions or two triangles. This happens when the law of sines leads to: $$0 < \text{sin}\hspace{.1em}B_{1}< 1$$$$A + B_{2}< 180°$$$$h < a < b$$ $$h=b \cdot \text{sin}\hspace{.1em}A$$ Alternatively, we see another possible triangle of: Putting the two triangles together gives us: In the image above, c1 is the length of the line segment AB1, whereas, c2 is the length of the line segment AB2.
Let's look at an example.
Example #4: Solve each triangle. Round your answer to the nearest tenth. $$C=25°, b=14 \hspace{.1em}\text{km}, c=8\hspace{.1em}\text{km}$$ Note: C will take the place of A and c will take the place of a in our formula above.
$$h=14 \hspace{.1em}\text{km}\cdot \text{sin}\hspace{.1em}25°$$ $$h ≈ 5.9 \hspace{.1em}\text{km}$$ $$5.9 \hspace{.1em}\text{km}< 8\hspace{.1em}\text{km}< 14 \hspace{.1em}\text{km}$$ Let's use the law of sines to find angle B. $$\frac{\text{sin}\hspace{.1em}C}{c}=\frac{\text{sin}\hspace{.1em}B}{b}$$ $$\frac{\text{sin}\hspace{.1em}25°}{8 \hspace{.1em}\text{km}}=\frac{\text{sin}\hspace{.1em}B}{14\hspace{.1em}\text{km}}$$ $$\text{sin}\hspace{.1em}B=\frac{\text{sin}\hspace{.1em}25° \cdot 14\hspace{.1em}\text{km}}{8 \hspace{.1em}\text{km}}$$ $$\text{sin}\hspace{.1em}B ≈ 0.7396$$ $$B ≈ \text{sin}^{-1}(0.7369)$$ $$B ≈ 47.7°$$ Let's think about the other possible value for B here: $$180° - 47.7°=132.3°$$ Since 132.3° + 25° = 157.3° is less than 180°, it is a valid angle measure. This means we will need to provide two different solutions or triangles.
Triangle #1: $$A=180° - 25° - 47.7°=107.3°$$ $$C=25°, B=47.7°, A=107.3°$$ To find a, let's use the law of sines: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{a}{\text{sin}\hspace{.1em}107.3°}=\frac{8\hspace{.1em}\text{km}}{\text{sin}\hspace{.1em}25°}$$ $$a=\frac{8\hspace{.1em}\text{km}\cdot \text{sin}\hspace{.1em}107.3°}{\text{sin}\hspace{.1em}25°}$$ $$a ≈ 18.1 \hspace{.1em}\text{km}$$ $$c=8\hspace{.1em}\text{km}, b=14 \hspace{.1em}\text{km}, a ≈ 18.1 \hspace{.1em}\text{km}$$ Triangle #2: $$A=180° - 25° - 132.3°=22.7°$$ $$C=25°, B=132.3°, A=22.7°$$ To find a, let's use the law of sines: $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{c}{\text{sin}\hspace{.1em}C}$$ $$\frac{a}{\text{sin}\hspace{.1em}22.7°}=\frac{8\hspace{.1em}\text{km}}{\text{sin}\hspace{.1em}25°}$$ $$a=\frac{8\hspace{.1em}\text{km}\cdot \text{sin}\hspace{.1em}22.7°}{\text{sin}\hspace{.1em}25°}$$ $$a ≈ 7.3 \hspace{.1em}\text{km}$$ $$c=8\hspace{.1em}\text{km}, b=14 \hspace{.1em}\text{km}, a ≈ 7.3 \hspace{.1em}\text{km}$$
### Law of Sines - Ambiguous Case (Obtuse Angle)
Let's now change things up and think about the case where A is now an obtuse angle. This will lead to two cases, one where there is no solution and another where there is exactly one solution.
#### Case 5: No Solution with a given obtuse angle
Fifth, let's think about the situation where the given angle is obtuse and there are 0 possible triangles. This happens when: $$a ≤ b$$ Let's look at an example.
Example #5: Solve each triangle. Round your answer to the nearest tenth. $$A=125°, c=18\hspace{.1em}\text{mi}, a=14 \hspace{.1em}\text{mi}$$ Since A is an obtuse angle, it is the largest angle, therefore, the longest side of the triangle must be a. Here, we see that c is greater than a, which is not possible. Therefore no such triangle ABC exists.
In this case, we can also show this with the law of sines. $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}C}{c}$$ $$\frac{\text{sin}\hspace{.1em}125°}{14 \hspace{.1em}\text{mi}}=\frac{\text{sin}\hspace{.1em}C}{18\hspace{.1em}\text{mi}}$$ $$\text{sin}\hspace{.1em}C=\frac{\text{sin}\hspace{.1em}125° \cdot 18\hspace{.1em}\text{mi}}{14 \hspace{.1em}\text{mi}}$$ $$\text{sin}\hspace{.1em}C ≈ 1.1$$ Since the sine of an angle can't be greater than 1, we know this triangle does not exist.
#### Case 6: One solution with a given obtuse angle
Sixth, we will think about the case, where we are given an obtuse angle and there is exactly one solution. This happens when the law of sines leads to: $$0 < \text{sin}\hspace{.1em}B < 1, a > b$$ Let's look at an example.
Example #6: Solve each triangle. Round your answer to the nearest tenth. $$A=114°, c=8\hspace{.1em}\text{ft}, a=35 \hspace{.1em}\text{ft}$$ Let's use the law of sines to find angle C. $$\frac{\text{sin}\hspace{.1em}A}{a}=\frac{\text{sin}\hspace{.1em}C}{c}$$ $$\frac{\text{sin}\hspace{.1em}114°}{35 \hspace{.1em}\text{ft}}=\frac{\text{sin}\hspace{.1em}C}{8\hspace{.1em}\text{ft}}$$ $$\text{sin}\hspace{.1em}C=\frac{\text{sin}\hspace{.1em}114° \cdot 8\hspace{.1em}\text{ft}}{35 \hspace{.1em}\text{ft}}$$ $$\text{sin}\hspace{.1em}C ≈ 0.2088$$ $$C ≈ \text{sin}^{-1}(0.2088)$$ $$C ≈ 12.1°$$ $$B=180° - 114° - 12.1°=53.9°$$ Let's use the law of sines to find b. $$\frac{a}{\text{sin}\hspace{.1em}A}=\frac{b}{\text{sin}\hspace{.1em}B}$$ $$\frac{35 \hspace{.1em}\text{ft}}{\text{sin}\hspace{.1em}114°}=\frac{b}{\text{sin}\hspace{.1em}53.9°}$$ $$b=\frac{35 \hspace{.1em}\text{ft}\cdot \text{sin}\hspace{.1em}53.9°}{\text{sin}\hspace{.1em}114°}$$ $$b ≈ 31 \hspace{.1em}\text{ft}$$
#### Skills Check:
Example #1
Find the missing side measurement to the nearest tenth. $$A=43.5°$$ $$a=10.7 \hspace{.1em}\text{in}$$ $$c=7.2 \hspace{.1em}\text{in}$$
A
$$b=14.7\hspace{.1em}\text{in}$$
B
$$b=15.1\hspace{.1em}\text{in}$$
C
$$b=19.8 \hspace{.1em}\text{in}$$
D
$$b=13.5\hspace{.1em}\text{in}$$
E
$$\text{Not a Triangle}$$
Example #2
Find the missing side measurement to the nearest tenth. $$A=105°$$ $$c=11 \hspace{.1em}\text{mi}$$ $$a=5 \hspace{.1em}\text{mi}$$
A
$$b=24 \hspace{.1em}\text{mi}$$
B
$$\text{Not a Triangle}$$
C
$$b=28 \hspace{.1em}\text{mi}$$
D
$$b=30 \hspace{.1em}\text{mi}$$
E
$$b=34 \hspace{.1em}\text{mi}$$
Example #3
Find the missing side measurement to the nearest tenth. $$B=31°$$ $$a=32 \hspace{.1em}\text{cm}$$ $$b=19 \hspace{.1em}\text{cm}$$
A
$$\text{Not a Triangle}$$
B
$$c=13, 14\hspace{.1em}\text{cm}$$
C
$$c=13, 36\hspace{.1em}\text{cm}$$
D
$$c=18, 36.9\hspace{.1em}\text{cm}$$
E
$$c=38.2, 20\hspace{.1em}\text{cm}$$ |
# RD Sharma Class 7 ex 4.3 Solutions Chapter 4 Rational Numbers
In this chapter, we provide RD Sharma Class 7 ex 4.3 Solutions Chapter 4 Rational numbers for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 7 ex 4.3 Solutions Chapter 4 Rational Numbers Maths pdf, Now you will get step by step solution to each question.
# Chapter 4: Rational Numbers Exercise – 4.3
### Question: 1
Determine whether the following rational numbers are in the lowest form or not:
(i) 65/84
(ii) (-15)/32
(iii) 24/128
(iv) (-56)/(-32)
### Solution:
(i) We observe that 65 and 84 have no common factor their HCF is 1.
Thus, 65/84 is in its lowest form.
(ii) We observe that -15 and 32 have no common factor i.e., their HCF is 1.
Thus, -15/32 is in its lowest form.
(iii) HCF of 24 and 128 is not 1.
Thus, given rational number is not in its simplest form.
(iv) HCF of 56 and 32 is 8.
Thus, given rational number is not in its simplest form.
### Question: 2
Express each of the following numbers to the lowest form:
(i) 4/22
(ii) (- 36)/180
(iii) 132/428
(iv) 32/56
### Solution:
Lowest form of:
(i) 4/22 is:
4 = 2 × 2, 22 = 2 × 11
HCF of 4 and 22 is 2.
Dividing the fraction by 2, we get 2/11.
(ii) -36/180 is:
36 = 3 × 3 × 2 × 2, 180 = 5 × 3 × 3 × 2 × 2
HCF of 36 and 180 is 36.
Dividing the fraction by 36, we get -1/5.
(iii) 132/- 428 is:
132 = 2 × 3 × 2 × 11, 428 = 2 × 2 × 107
HCF of 132 and 428 is 4.
Dividing the fraction by 4, we get 33/-107.
(iv) – 32/ – 56 is:
32 =2 × 2 × 2 × 2 × 2, 56 = 2 × 2 × 2 × 7
HCF of 32 and 56 is 8.
Dividing the fraction by 8, we get 4/7.
### Question: 3
Fill in the blanks:
### Solution:
All Chapter RD Sharma Solutions For Class 7 Maths
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The Natural Exponential Function
# The Natural Exponential Function
Recall that by definition, an exponential function is of the form $\,f(x) = b^x\,,$ where $\,b\,$ is a positive number not equal to $\,1\,.$
Inside the class of exponential functions (one member for each allowable value of the base $\,b\,$) there is a particular member that is singled out as being most important.
Why? It has a simple property (discussed in a future section) that usually makes it easier to work with than any other member of its class.
This special member has base $\,\text{e}\,,$ and is called the natural exponential function. The natural exponential function is so important that if you hear someone say ‘the exponential function’ (as opposed to ‘an exponential function’), then they're talking about $\,f(x) = \text{e}^x\,.$
This most important member of its class is shown in red below:
Click the Button to See
of the Class of Exponential Functions
## The Irrational Number $\,\text{e}\,$
It might be curious to you that the proclaimed ‘simplest’ exponential function has such an ‘unusual’ base. You might wonder—shouldn't a simple base like $\,2\,$ or $\,3\,$ be better? So, let's begin by studying this special base—the irrational number $\,\text{e}\,.$
### A Review of Irrational Numbers
• Recall first that a number is called ‘rational’ when it can be expressed as a ratio (a fraction) of integers.
The integers are this collection: $$\{\ldots,-3\,-2,-1,0,1,2,3,\ldots \}$$
For example, $\,-\frac{2}{5} = \frac{-2}{5}\,$ and $\,7 = \frac{7}{1} = \frac{14}{2}\,$ are rational numbers. The word ‘rational’ comes from the ‘ratio’ part of the definition.
• Irrational numbers are, by definition, not rational. (The prefix ‘ir’ denotes ‘not’, as in the words irreverent, irreducible, and irreversible.) That is, irrational numbers cannot be expressed as a ratio of integers.
• Every irrational number has a decimal representation as an infinite, non-repeating decimal. That is, the decimal representation goes on forever, and you never reach a sequence of digits that repeats itself ad infinitum.
• So—irrational numbers don't have ‘nice’ names, like:
• $\,\frac{2}{3}\,$ (a ratio of integers)
• $\,3.72\,$ (a finite decimal)
• $\,2.3\overline{45}\,$ (an infinite repeating decimal)
• Two very important irrational numbers are given special letter names:
• $\,\text{e}\,$ ($\,\text{e}\approx 2.72\,$)
• $\,\pi\,$ (the ratio of circumference to diameter in any circle; $\,\pi\approx 3.14\,$)
Need more decimal places of accuracy? Jump up to WolframAlpha and type in math constant e or pi .
• Some other irrational numbers are $\,\sqrt 2\,$ and $\,\root 3\of {5}\,.$ Radicals that don't simplify ‘nicely’ (like $\,\sqrt 4 = 2\,$ and $\,\root 3\of{\frac{27}{8}} = \frac 32\,$) are irrational numbers.
### An Important Expression:$\left(1 + \frac 1n\right)^n$
There's a definition of the irrational number $\,\text{e}\,$ that is particularly important in our context. This definition involves looking at the expression $\left(1 + \frac 1n\right)^n$ for large positive values of $\,n\,.$ An exploration of this expression is also a good example of where intuition can fail you in mathematics!
Consider the expression $\,\left( 1 + \frac 1n\right)^n\,$ for bigger and bigger positive integer values of $\,n\,.$ When I ask students what number they think this ‘looks like’ for such values of $\,n\,,$ I often get the answer ‘$\,1\,$’.
The (flawed) reasoning goes something like this:
• As $\,n\,$ gets large, certainly $\,\frac 1n\,$ goes to $\,0\,.$ (This statement is true.)
• Therefore, $\,1 + \frac 1n\,$ is getting closer and closer to $\,1\,.$ (This statement is true.)
• Of course, the number $\,1\,$ to any power is $\,1\,.$ (This statement is true.)
• So, the expression must be getting close to $\,1\,.$ (This is the faulty statement!)
Let's take a look at what is really happening, and discuss the flaw in the previous ‘argument’. The table below lists some values:
$n$ $\displaystyle\left(1 + \frac 1n\right)^n$ $10$ $\displaystyle\left(1 + \frac 1{10}\right)^{10} \approx 2.5937$ $100$ $\displaystyle\left(1 + \frac 1{100}\right)^{100} \approx 2.7048$ $1000$ $\displaystyle\left(1 + \frac 1{1000}\right)^{1000} \approx 2.7169$ $10,000$ $\displaystyle\left(1 + \frac 1{10,000}\right)^{10,000} \approx 2.7181$ $\downarrow$ $\downarrow$ $\infty$ $\text{e}$
Clearly, the numbers in the right-hand column above are not approaching the number $\,1\,.$ They are, however, approaching a particular number: this is precisely the number that we call $\,\text{e}\,.$
DEFINITION the number $\,\text{e}\,$
The number $\,\text{e}\,$ is the limit, as $\,n\,$ goes to infinity, of $\,\left(1 + \frac 1n\right)^n\,.$
More compactly:
$$\cssId{s59}{\text{e} = \lim_{n\rightarrow\infty} \left(1 + \frac 1n\right)^n}$$
We can get the numbers $\,\left(1 + \frac 1n\right)^n\,$ as close to $\,\text{e}\,$ as desired, by making $\,n\,$ sufficiently large.
### The Flaw in the Prior Argument
In the prior argument, we let the value of $\,n\,$ inside the parentheses go to infinity first, and then we let the exponent go to infinity. However, when evaluating limits, you must let every occurrence of $\,n\,$ go to infinity at the same time!
## The Graph of the Natural Exponential Function
Here are graphs of the natural exponential function, together with two nearby functions:
• $\color{blue}{y = 2^x}\,$ is in blue
• $\color{red}{y = {\text{e}}^x}\,,$ the natural exponential function, is in red
• $\color{green}{y = 3^x}\,$ is in green
For $\,x \gt 0\,,$ $\,\color{blue}{2^x} \lt \color{red}{{\text{e}}^x} \lt \color{green}{3^x}\,$
For $\,x \lt 0\,,$ $\,\color{green}{3^x} \lt \color{red}{{\text{e}}^x} \lt \color{blue}{2^x}\,$ |
## Key Points of the Unit:
-how to complete the square
-how to complete the square in a word problem
-how to use the quadratic formula
-how to graph using completing the square and quadratic formula.
## Completing the square:
-Vertex form of a quadratic is y = a(x-h)2+k, where (h,k) is the vertex
-Standard form of a quadratic is y = ax2+bx+c
-Since we have the sum (x+3) being squared, we can use the formula for the square of a sum to save time
-The formula for the square of a sum says that (a+b)2 = a2+2ab+b2
-For the formula, a = x and b = 3, so we get x2+6x+9 when we plug inUse the Distributive
Property to distribute the 2 through the parentheses
-y = 2x2+12x-4 is our equation in standard form
Completing the square
## Completing the Square With a Word Problem.
Example 1: The height of the Peace Tower is 90m. The path of an object thrown from the Peace Tower can be modelled by the equation:
h =−5d2 +40d+90
where h is the height in metres and d is the horizontal distance in metres. At what horizontal distance did the object reach its maximum height?
h=-5d2 +40d +90
h=-5(d2 -8d) +90
h=-5(d2-8d +16 -16) +90
h=-5(d2-8d +16) +80 +90
h=-5(d-4)2 +170
At 4 meteres, it reaches the max height.
The Quadratic Formula uses the "a", "b", and "c" from "ax2 + bx + c", where "a", "b", and "c" are just numbers; they are the "numerical coefficients" of the quadratic equation they've given you to solve. The Quadratic Formula is derived from the process of completing the square, and is formally stated as:
For ax2 + bx + c = 0, the value of x is given by
x= -b +_ (b2 -4ac)
-----------------------
2a
The "solution" or "roots" or "zeroes" of a quadratic are usually required to be in the "exact" form of the answer. In the example above, the exact form is the one with the square roots of ten in it. You'll need to get a calculator approximation in order to graph the x-intercepts or to simplify the final answer in a word problem. But unless you have a good reason to think that the answer is supposed to be a rounded answer, always go with the exact form.
More Word Problems Using Quadratic Equations - Example 1
## Solving a Linear graph using quadratic Formula and Completing the square
to find the x-intercepts, you would need to use the quadratic formula and plot all the numbers down in standard form. At the end, you will get 2 x-intercepts. Find the mid point of those two intercept and than you can solve for the vertex using completing the square method.
Learning Goals:
-I learned how to complete the square
-I learned how to use the quadratic formula
-I learned how to solve word problems using quadratic formula and completing the square
-I learned how to find the vertex and x-intercept using quadratic formula and x-intercept. |
# An algebra of orders
Did you know that you can add and multiply orders? For any two order structures $A$ and $B$, we can form the ordered sum $A+B$ and ordered product $A\otimes B$, and other natural operations, such as the disjoint sum $A\sqcup B$, which make altogether an arithmetic of orders. We combine orders with these operations to make new orders, often with interesting properties. Let us explore the resulting algebra of orders!
$\newcommand\Z{\mathbb{Z}}\newcommand\N{\mathbb{N}}\newcommand\Q{\mathbb{Q}}\newcommand\P{\mathbb{P}}\newcommand\iso{\cong}\newcommand\of{\subseteq}$
One of the most basic operations that we can use to combine two orders is the disjoint sum operation $A\sqcup B$. This is the order resulting from placing a copy of $A$ adjacent to a copy of $B$, side-by-side, forming a combined order with no instances of the order relation between the two parts. If $A$ is the orange $\vee$-shaped order here and $B$ is the yellow linear order, for example, then $A\sqcup B$ is the combined order with all five nodes.
Another kind of addition is the ordered sum of two orders $A+B$, which is obtained by placing a copy of $B$ above a copy of $A$, as indicated here by adding the orange copy of $A$ and the yellow copy of $B$. Also shown is the sum $B+A$, with the summands reversed, so that we take $B$ below and $A$ on top. It is easy to check that the ordered sum of two orders is an order. One notices immediately, of course, that the resulting ordered sums $A+B$ and $B+A$ are not the same! The order $A+B$ has a greatest element, whereas $B+A$ has two maximal elements. So the ordered sum operation on orders is not commutative. Nevertheless, we shall still call it addition. The operation, which has many useful and interesting features, goes back at least to the 19th century with Cantor, who defined the addition of well orders this way.
In order to illustrate further examples, I have assembled here an addition table for several simple finite orders. The choices for $A$ appear down the left side and those for $B$ at the top, with the corresponding sum $A+B$ displayed in each cell accordingly.
We can combine the two order addition operations, forming a variety of other orders this way.
The reader is encouraged to explore further how to add various finite orders using these two forms of addition. What is the smallest order that you cannot generate from $1$ using $+$ and $\sqcup$? Please answer in the comments.
We can also add infinite orders. Displayed here, for example, is the order $\N+(1\sqcup 1)$, the natural numbers wearing two yellow caps. The two yellow nodes at the top form a copy of $1\sqcup 1$, while the natural numbers are the orange nodes below. Every natural number (yes, all infinitely many of them) is below each of the two nodes at the top, which are incomparable to each other. Notice that even though we have Hasse diagrams for each summand order here, there can be no minimal Hasse diagram for the sum, because any particular line from a natural number to the top would be implied via transitivity from higher such lines, and we would need such lines, since they are not implied by the lower lines. So there is no minimal Hasse diagram.
This order happens to illustrate what is called an exact pair, which occurs in an order when a pair of incomparable nodes bounds a chain below, with the property that any node below both members of the pair is below something in the chain. This phenomenon occurs in sometimes unexpected contexts—any countable chain in the hierarchy of Turing degrees in computability theory, for example, admits an exact pair.
Let us turn now to multiplication. The ordered product $A\otimes B$ is the order resulting from having $B$ many copies of $A$. That is, we replace each node of $B$ with an entire copy of the $A$ order. Within each of these copies of $A$, the order relation is just as in $A$, but the order relation between nodes in different copies of $A$, we follow the $B$ relation. It is not difficult to check that indeed this is an order relation. We can illustrate here with the same two orders we had earlier.
In forming the ordered product $A\otimes B$, in the center here, we take the two yellow nodes of $B$, shown greatly enlarged in the background, and replace them with copies of $A$. So we have ultimately two copies of $A$, one atop the other, just as $B$ has two nodes, one atop the other. We have added the order relations between the lower copy of $A$ and the upper copy, because in $B$ the lower node is related to the upper node. The order $A\otimes B$ consists only of the six orange nodes—the large highlighted yellow nodes of $B$ here serve merely as a helpful indication of how the product is formed and are not in any way part of the product order $A\otimes B$.
Similarly, with $B\otimes A$, at the right, we have the three enlarged orange nodes of $B$ in the background, which have each been replaced with copies of $A$. The nodes of each of the lower copies of $A$ are related to the nodes in the top copy, because in $B$ the two lower nodes are related to the upper node.
I have assembled a small multiplication table here for some simple finite orders.
So far we have given an informal account of how to add and multiply ordered ordered structures. So let us briefly be a little more precise and formal with these matters.
In fact, when it comes to addition, there is a slightly irritating matter in defining what the sums $A\sqcup B$ and $A+B$ are exactly. Specifically, what are the domains? We would like to conceive of the domains of $A\sqcup B$ and $A+B$ simply as the union the domains of $A$ and $B$—we’d like just to throw the two domains together and form the sums order using that combined domain, placing $A$ on the $A$ part and $B$ on the $B$ part (and adding relations from the $A$ to the $B$ part for $A+B$). Indeed, this works fine when the domains of $A$ and $B$ are disjoint, that is, if they have no points in common. But what if the domains of $A$ and $B$ overlap? In this case, we can’t seem to use the union in this straightforward manner. In general, we must disjointify the domains—we take copies of $A$ and $B$, if necessary, on domains that are disjoint, so that we can form the sums $A\sqcup B$ and $A+B$ on the union of those nonoverlapping domains.
What do we mean precisely by “taking a copy” of an ordered structure $A$? This way of talking in mathematics partakes in the philosophy of structuralism. We only care about our mathematical structures up to isomorphism, after all, and so it doesn’t matter which isomorphic copies of $A$ and $B$ we use; the resulting order structures $A\sqcup B$ will be isomorphic, and similarly for $A+B$. In this sense, we are defining the sum orders only up to isomorphism.
Nevertheless, we can be definite about it, if only to verify that indeed there are copies of $A$ and $B$ available with disjoint domains. So let us construct a set-theoretically specific copy of $A$, replacing each individual $a$ in the domain of $A$ with $(a,\text{orange})$, for example, and replacing the elements $b$ in the domain of $B$ with $(b,\text{yellow})$. If “orange” is a specific object distinct from “yellow,” then these new domains will have no points in common, and we can form the disjoint sum $A\sqcup B$ by using the union of these new domains, placing the $A$ order on the orange objects and the $B$ order on the yellow objects.
Although one can use this specific disjointifying construction to define what $A\sqcup B$ and $A+B$ mean as specific structures, I would find it to be a misunderstanding of the construction to take it as a suggestion that set theory is anti-structuralist. Set theorists are generally as structuralist as they come in mathematics, and in light of Dedekind’s categorical account of the natural numbers, one might even find the origin of the philosophy of structuralism in set theory. Rather, the disjointifying construction is part of the general proof that set theory abounds with isomorphic copies of whatever mathematical structure we might have, and this is part of the reason why it serves well as a foundation of mathematics for the structuralist. To be a structuralist means not to care which particular copy one has, to treat one’s mathematical structures as invariant under isomorphism.
But let me mention a certain regrettable consequence of defining the operations by means of a specific such disjointifying construction in the algebra of orders. Namely, it will turn out that neither the disjoint sum operation nor the ordered sum operation, as operations on order structures, are associative. For example, if we use $1$ to represent the one-point order, then $1\sqcup 1$ means the two-point side-by-side order, one orange and one yellow, but really what we mean is that the points of the domain are $\set{(a,\text{orange}),(a,\text{yellow})}$, where the original order is on domain $\set{a}$. The order $(1\sqcup 1)\sqcup 1$ then means that we take an orange copy of that order plus a single yellow point. This will have domain
$$\set{\bigl((a,\text{orange}),\text{orange}\bigr),\bigl((a,\text{yellow}),\text{orange}\bigr),(a,\text{yellow})}$$
The order $1\sqcup(1\sqcup 1)$, in contrast, means that we take a single orange point plus a yellow copy of $1\sqcup 1$, leading to the domain
$$\set{(a,\text{orange}),\bigl((a,\text{orange}),\text{yellow}\bigr),\bigl((a,\text{yellow}),\text{yellow}\bigr)}$$
These domains are not the same! So as order structures, the order $(1\sqcup 1)\sqcup 1$ is not identical with $1\sqcup(1\sqcup 1)$, and therefore the disjoint sum operation is not associative. A similar problem arises with $1+(1+1)$ and $(1+1)+1$.
But not to worry—we are structuralists and care about our orders here only up to isomorphism. Indeed, the two resulting orders are isomorphic as orders, and more generally, $(A\sqcup B)\sqcup C$ is isomorphic to $A\sqcup(B\sqcup C)$ for any orders $A$, $B$, and $C$, and similarly with $A+(B+C)\cong(A+B)+C$, as discussed with the theorem below. Furthermore, the order isomorphism relation is a congruence with respect to the arithmetic we have defined, which means that $A\sqcup B$ is isomorphic to $A’\sqcup B’$ whenever $A$ and $B$ are respectively isomorphic to $A’$ and $B’$, and similarly with $A+B$ and $A\otimes B$. Consequently, we can view these operations as associative, if we simply view them not as operations on the order structures themselves, but on their order-types, that is, on their isomorphism classes. This simple abstract switch in perspective restores the desired associativity. In light of this, we are free to omit the parentheses and write $A\sqcup B\sqcup C$ and $A+B+C$, if care about our orders only up to isomorphism. Let us therefore adopt this structuralist perspective for the rest of our treatment of the algebra of orders.
Let us give a more precise formal definition of $A\otimes B$, which requires no disjointification. Specifically, the domain is the set of pairs $\set{(a,b)\mid a\in A, b\in B}$, and the order is defined by $(a,b)\leq_{A\otimes B}(a’,b’)$ if and only if $b\leq_B b’$, or $b=b’$ and $a\leq_A a’$. This order is known as the reverse lexical order, since we are ordering the nodes in the dictionary manner, except starting from the right letter first rather than the left as in an ordinary dictionary. One could of course have defined the product using the lexical order instead of the reverse lexical order, and this would give $A\otimes B$ the meaning of “$A$ copies of $B$.” This would be a fine alternative and in my experience mathematicians who rediscover the ordered product on their own often tend to use the lexical order, which is natural in some respects. Nevertheless, there is a huge literature with more than a century of established usage with the reverse lexical order, from the time of Cantor, who defined ordinal multiplication $\alpha\beta$ as $\beta$ copies of $\alpha$. For this reason, it seems best to stick with the reverse lexical order and the accompanying idea that $A\otimes B$ means “$B$ copies of $A$.” Note also that with the reverse lexical order, we shall be able to prove left distributivity $A\otimes(B+C)=A\otimes B+A\otimes C$, whereas with the lexical order, one will instead have right distributivity $(B+C)\otimes^* A=B\otimes^* A+C\otimes^* A$.
Let us begin to prove some basic facts about the algebra of orders.
Theorem. The following identities hold for orders $A$, $B$, and $C$.
1. Associativity of disjoint sum, ordered sum, and ordered product.\begin{eqnarray*}A\sqcup(B\sqcup C) &\iso& (A\sqcup B)\sqcup C\\ A+(B+C) &\iso& (A+B)+C\\ A\otimes(B\otimes C) &\iso& (A\otimes B)\otimes C \end{eqnarray*}
2. Left distributivity of product over disjoint sum and ordered sum.\begin{eqnarray*} A\otimes(B\sqcup C) &\iso& (A\otimes B)\sqcup(A\otimes C)\\ A\otimes(B+C) &\iso& (A\otimes B)+(A\otimes C) \end{eqnarray*}
In each case, these identities are clear from the informal intended meaning of the orders. For example, $A+(B+C)$ is the order resulting from having a copy of $A$, and above it a copy of $B+C$, which is a copy of $B$ and a copy of $C$ above it. So one has altogether a copy of $A$, with a copy of $B$ above that and a copy of $C$ on top. And this is the same as $(A+B)+C$, so they are isomorphic.
One can also aspire to give a detailed formal proof verifying that our color-coded disjointifying process works as desired, and the reader is encouraged to do so as an exercise. To my way of thinking, however, such a proof offers little in the way of mathematical insight into algebra of orders. Rather, it is about checking the fine print of our disjointifying process and making sure that things work as we expect. Several of the arguments can be described as parenthesis-rearranging arguments—one extracts the desired information from the structure of the domain order and puts that exact same information into the correct form for the target order.
For example, if we have used the color-scheme disjointifying process described above, then the elements of $A\sqcup(B\sqcup C)$ each have one of the following forms, where $a\in A$, $b\in B$, and $c\in C$:
$$(a,\text{orange})$$
$$\bigl((b,\text{orange}),\text{yellow}\bigr)$$
$$\bigl((c,\text{yellow}),\text{yellow}\bigr)$$
We can define the color-and-parenthesis-rearranging function $\pi$ to put them into the right form for $(A\sqcup B)\sqcup C$ as follows:
\begin{align*}
\end{align*}
In each case, we will preserve the order, and since the orders are side-by-side, the cases never interact in the order, and so this is an isomorphism.
Similarly, for distributivity, the elements of $A\otimes(B\sqcup C)$ have the two forms:
$$\bigl(a,(b,\text{orange})\bigr)$$
$$\bigl(a,(c,\text{yellow})\bigr)$$
where $a\in A$, $b\in B$, and $c\in C$. Again we can define the desired ismorphism $\tau$ by putting these into the right form for $(A\otimes B)\sqcup(A\otimes C)$ as follows:
\begin{align*}
\end{align*}
And again, this is an isomorphism, as desired.
Since order multiplication is not commutative, it is natural to inquire about the right-sided distributivity laws:
\begin{eqnarray*}
(B+C)\otimes A&\overset{?}{\cong}&(B\otimes A)+(C\otimes A)\\
(B\sqcup C)\otimes A&\overset{?}{\cong}&(B\otimes A)\sqcup(C\otimes A)
\end{eqnarray*}
Unfortunately, however, these do not hold in general, and the following instances are counterexamples. Can you see what to take as $A$, $B$, and $C$? Please answer in the comments.
Theorem.
1. If $A$ and $B$ are linear orders, then so are $A+B$ and $A\otimes B$.
2. If $A$ and $B$ are nontrivial linear orders and both are endless, then $A+B$ is endless; if at least one of them is endless, then $A\otimes B$ is endless.
3. If $A$ is an endless dense linear order and $B$ is linear, then $A\otimes B$ is an endless dense linear order.
4. If $A$ is an endless discrete linear order and $B$ is linear, then $A\otimes B$ is an endless discrete linear order.
Proof. If both $A$ and $B$ are linear orders, then it is clear that $A+B$ is linear. Any two points within the $A$ copy are comparable, and any two points within the $B$ copy, and every point in the $A$ copy is below any point in the $B$ copy. So any two points are comparable and thus we have a linear order. With the product $A\otimes B$, we have $B$ many copies of $A$, and this is linear since any two points within one copy of $A$ are comparable, and otherwise they come from different copies, which are then comparable since $B$ is linear. So $A\otimes B$ is linear.
For statement (2), we know that $A+B$ and $A\otimes B$ are nontrivial linear orders. If both $A$ and $B$ are endless, then clearly $A+B$ is endless, since every node in $A$ has something below it and every node in $B$ has something above it. For the product $A\otimes B$, if $A$ is endless, then every node in any copy of $A$ has nodes above and below it, and so this will be true in $A\otimes B$; and if $B$ is endless, then there will always be higher and lower copies of $A$ to consider, so again $A\otimes B$ is endless, as desired.
For statement (3), assume that $A$ is an endless dense linear and that $B$ is linear. We know from (1) that $A\otimes B$ is a linear order. Suppose that $x<y$ in this order. If $x$ and $y$ live in the same copy of $A$, then there is a node $z$ between them, because $A$ is dense. If $x$ occurs in one copy of $A$ and $B$ in another, then because $A$ is endless, there will a node $z$ above $x$ in its same copy, leading to $x<z<y$ as desired. (Note: we don’t need $B$ to be dense.)
For statement (4), assume instead that $A$ is an endless discrete linear order and $B$ is linear. We know that $A\otimes B$ is a linear order. Every node of $A\otimes B$ lives in a copy of $A$, where it has an immediate successor and an immediate predecessor, and these are also immediate successor and predecessor in $A\otimes B$. From this, it follows also that $A\otimes B$ is endless, and so it is an endless discrete linear order. $\Box$
The reader is encouraged to consider as an exercise whether one can drop the “endless” hypotheses in the theorem. Please answer in the comments.
Theorem. The endless discrete linear orders are exactly those of the form $\Z\otimes L$ for some linear order $L$.
Proof. If $L$ is a linear order, then $\Z\otimes L$ is an endless discrete linear order by the theorem above, statement (4). So any order of this form has the desired feature. Conversely, suppose that $\P$ is an endless discrete linear order. Define an equivalence relation for points in this order by which $p\sim q$ if and only $p$ and $q$ are at finite distance, in the sense that there are only finitely many points between them. This relation is easily seen to be reflexive, transitive and symmetric, and so it is an equivalence relation. Since $\P$ is an endless discrete linear order, every object in the order has an immediate successor and immediate predecessor, which remain $\sim$-equivalent, and from this it follows that the equivalence classes are each ordered like the integers $\Z$, as indicated by the figure here.
The equivalence classes amount to a partition of $\P$ into disjoint segments of order type $\Z$, as in the various colored sections of the figure. Let $L$ be the induced order on the equivalence classes. That is, the domain of $L$ consists of the equivalence classes $\P/\sim$, which are each a $\Z$ chain in the original order, and we say $[a]<_L[b]$ just in case $a<_{\P}b$. This is a linear order on the equivalence classes. And since $\P$ is $L$ copies of its equivalence classes, each of which is ordered like $\Z$, it follows that $\P$ is isomorphic to $\Z\otimes L$, as desired. $\Box$
(Interested readers are advised that the argument above uses the axiom of choice, since in order to assemble the isomorphism of $\P$ with $\Z\otimes L$, we need in effect to choose a center point for each equivalence class.)
If we consider the integers inside the rational order $\Z\of\Q$, it is clear that we can have a discrete suborder of a dense linear order. How about a dense suborder of a discrete linear order?
Question. Is there a discrete linear order with a suborder that is a dense linear order?
What? How could that happen? In my experience, mathematicians first coming to this topic often respond instinctively that this should be impossible. I have seen sophisticated mathematicians make such a pronouncement when I asked the audience about it in a public lecture. The fundamental nature of a discrete order, after all, is completely at odds with density, since in a discrete order, there is a next point up and down, and a next next point, and so on, and this is incompatible with density.
Yet, surprisingly, the answer is Yes! It is possible—there is a discrete order with a suborder that is densely ordered. Consider the extremely interesting order $\Z\otimes\Q$, which consists of $\Q$ many copies of $\Z$, laid out here increasing from left to right. Each tiny blue dot is a rational number, which has been replaced with an entire copy of the integers, as you can see in the magnified images at $a$, $b$, and $c$.
The order is quite subtle, and so let me also provide an alternative presentation of it. We have many copies of $\Z$, and those copies are densely ordered like $\Q$, so that between any two copies of $\Z$ is another one, like this:
Perhaps it helps to imagine that the copies of $\Z$ are getting smaller and smaller as you squeeze them in between the larger copies. But you can indeed always fit another copy of $\Z$ between, while leaving room for the further even tinier copies of $\Z$ to come.
The order $\Z\otimes\Q$ is discrete, in light of the theorem characterizing discrete linear orders. But also, this is clear, since every point of $\Z\otimes\Q$ lives in its local copy of $\Z$, and so has an immediate successor and predecessor there. Meanwhile, if we select exactly one point from each copy of $\Z$, the $0$ of each copy, say, then these points are ordered like $\Q$, which is dense. Thus, we have proved:
Theorem. The order $\Z\otimes\Q$ is a discrete linear order having a dense linear order as a suborder.
One might be curious now about the order $\Q\otimes\Z$, which is $\Z$ many copies of $\Q$. This order, however, is a countable endless dense linear order, and therefore is isomorphic to $\Q$ itself.
This material is adapted from my book-in-progress, Topics in Logic, drawn from Chapter 3 on Relational Logic, which incudes an extensive section on order theory, of which this is an important summative part. |
## Formulas
Profit % = ($$\frac{ Profit}{ C.P.}$$ ) x 100
Loss % = ($$\frac{ Loss}{ C.P.}$$) x 100
C.P. = $$\frac{ 100 X S.P.}{ (100 – Loss \%)}$$
C.P. = $$\frac{ 100 X S.P.}{ (100 + Profit \%)}$$
## Illustrative Examples
Example 1: By selling a fan for Rs 649, Anil earns a profit of 18%. Find its cost price.
Solution. S.P. of the fan = Rs 649, profit = 18%
Therefore, Rs 649 = ( 1 + $$\frac{18}{100}$$) of C.P.
=> Rs 649 = $$\frac{118}{100}$$ of C.P.
=> C.P. = Rs (649 x $$\frac{100}{118}$$) = Rs 550
Hence the cost price of the fan = Rs 550.
Example 2: By selling a chair for Rs 391, Ali suffers a loss of 15%. Find its cost price.
Solution. S.P. of the chair = Rs 391, Loss = 15%
Therefore, Rs 391 = ( 1- $$\frac{15}{100}$$) of C.P.
Rs 391 = $$\frac{85}{100}$$ of C.P.
=> C.P. = (Rs 391 x $$\frac{100}{85}$$ )= Rs (23 x 20) = Rs 460
Hence the cost price of the chair = Rs 460.
Example 3: A man sells his scooter for Rs 18000 making a profit of 20%. How much did the scooter cost him?
Solution. Let the cost price of the scooter be Rs 100. Then, Profit = Rs 20
S.P. = C.P. + Profit = Rs 100 + Rs 20= Rs 120
Thus, if the S.P. is Rs 120, then C.P. = Rs 120 – 20 = Rs 100
If the S.P. is Rs 18000, then C.P. = Rs ($$\frac{100}{120}$$ x 18000) = Rs 15000
Hence, the cost of the scooter = Rs 15000
## Introduction
This system is based upon the place value and face value of a digit in a number. We have learnt that a natural number can be written as the sum of the place values of all digits of the numbers. For example
3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1
Such a form of a natural number is known as its expanded form.
The expanded form of a number can also be expressed in terms of powers of 10 by using
$${10}^{0}$$ = 1, $${10}^{1}$$ = 10, $${10}^{2}$$ = 100, $${10}^{3}$$ = 1000 etc.
For example,
3256 = 3 x 1000 + 2 x 100 + 5 x 10 + 6 x 1
=> 3256 = 3 x $${10}^{3}$$ + 2 x $${10}^{2}$$ + 5 x $${10}^{1}$$ + 6 x $${10}^{0}$$
Clearly, each digit of the natural number is multiplied by $${10}^{n}$$, where n is the number of digits to its right and then they are added.
## Illustrative Examples
Example 1: Write the following numbers in the expanded exponential forms:
(i) 32005 (ii) 56719 (iii) 8605192 (iv) 2500132
Solution.
(i) 32005 = 3 x $${10}^{4}$$ + 2 x $${10}^{3}$$ + 0 x $${10}^{2}$$ + 0 x $${10}^{1}$$ + 5 x $${10}^{0}$$
(ii) 560719 = 5 x $${10}^{5}$$ + 6 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 7 x $${10}^{2}$$ + 1 x $${10}^{1}$$ + 9 x $${10}^{0}$$
(iii) 8605192 = 8 x $${10}^{6}$$ + 6 x $${10}^{5}$$ + 0 x $${10}^{4}$$ + 5 x $${10}^{3}$$ + 1 x $${10}^{2}$$ + 9 x $${10}^{1}$$ + 2 x $${10}^{0}$$
(iv) 2500132 = 2 x $${10}^{6}$$ + 5 x $${10}^{5}$$ + 0 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 1 x $${10}^{2}$$ + 3 x $${10}^{1}$$ + 2 x $${10}^{0}$$
Example 2: Find the number from each of the following expanded forms:
(i) 5 x $${10}^{4}$$ + 4 x $${10}^{3}$$ + 2 x $${10}^{2}$$ + 3 x $${10}^{1}$$ + 5 x $${10}^{0}$$
(ii) 7 x $${10}^{5}$$ + 6 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 9 x $${10}^{0}$$
(iii) 9 x $${10}^{5}$$ + 4 x $${10}^{2}$$ + 1 x $${10}^{1}$$
Solution.
(i) 5 x $${10}^{4}$$ + 4 x $${10}^{3}$$ + 2 x $${10}^{2}$$ + 3 x $${10}^{1}$$ + 5 x $${10}^{0}$$
= 5 x 10000 + 4 x 1000 + 2 x 100 + 3 x 10 + 5 x 1
= 50000 + 4000 + 200 + 30 + 5
= 54235
(ii) 7 x $${10}^{5}$$ + 6 x $${10}^{4}$$ + 0 x $${10}^{3}$$ + 9 x $${10}^{0}$$
= 7 x 100000 + 6 x 10000 + 0 + 9 x 1
= 700000 + 60000 + 9
= 760009
(iii) 9 x $${10}^{5}$$ + 4 x $${10}^{2}$$ + 1 x $${10}^{1}$$
= 9 x 100000 + 4 x 100 + 1 x 10
= 900000 + 400 + 10
= 900410
## Definitions
Compound Interest: If the borrower and the lender agree to fix up a certain interval of time (say, a year or a half year or a quarter of a year etc) so that the amount ( = Principal + Interest) at the end of an interval becomes the principal for the next interval, then the total interest over all the intervals, calculated in this way is called the compound interest and is abbreviated as C.I.
Clearly, compound interest at the end of certain specified period is equal to the difference between the amount at the end of the period and the original principal i.e.
C.I. = Amount — Principal
Conversion Period: The fixed interval of time at the end of which the interest is calculated and added to the principal at the beginning of the interval is called the conversion period.
In other words, the period at the end of which the interest is compounded is called the conversion period.
When the interest is calculated and added to the principal every six months, the conversion period is six months. Similarly, the conversion period is 3 months when the interest is calculated and added quarterly.
## Finding CI When Interest Is Compounded Annually
when Interest is compounded yearly, the interest accrued during the first year is added to the principal and the amount so obtained becomes the principal for the second year. The amount at the end of the second year becomes the principal for the third year, and so on.
Example 1: Maria invests Rs 93750 at 9.6% per annum for 3 years and the interest is compounded annually. Calculate:
(i) The amount standing to her credit at the end of second year.
(ii) The interest for the third year.
Solution. (i) We have,
Principal for the first year Rs 93750
Rate of interest = 9.6% per annum.
Therefore, Interest for the first year = Rs ($$\frac{93750 \times 9.6 \times 1}{100}$$) = Rs 9000
Amount at the end of the first year = Rs 93750 +Rs 9000
= Rs 102750
Principal for the second year = Rs 102750
Interest for the second year = Rs ($$\frac{102750 \times 9.6 \times 1}{100}$$) = Rs 9864
Amount at the end of second year = Rs 102750 + Rs 9864
= Rs 112614
(ii) Principal for the third year = Rs 112614
Interest for the third year =Rs ($$\frac{112614 \times 9.6 \times 1}{100}$$ ) = Rs 10810.94
Example 2: Find the compound interest on Rs 25000 for 3 years at 10% per annum, compounded annually.
Solution. Principal for the first year = Rs 25000
Interest for the first year = ($$\frac{25000 \times 10 \times 1}{100}$$) = Rs 2500
Amount at the end of the first year = (25000 + 2500) = Rs 27500
Principal for the second year = Rs 27500
Interest for the second year = ($$\frac{27500 \times 10 \times 1}{100}$$) = Rs 2750
Amount at the end of the second year = (27500 + 2750) = Rs 30250.
Principal for the third year = Rs 30250.
Interest for the third year = ($$\frac{30250 \times 10 \times 1}{100}$$) = Rs 3025
Amount at the end of the third year = (30250 + 3025) = Rs 33275.
Therefore, compound interest = (33275 — 25000) = Rs 8275
## Finding CI When Interest Is Compounded Half-Yearly
If the rate of Interest is R% per annum then it is clearly ($$\frac{R}{2}$$)% per half-year.
The amount after the first half-year becomes the principal for the next half-year, and so on.
Example 3: Find the compound interest on Rs 5000 for 1 year at 8% per annum, compound half-yearly.
Solution. Rate of interest = 8% per annum = 4% per half-year.
Time = 1 year = 2 half-years.
Original principal = Rs 5000.
Interest for the first half-year = ($$\frac{5000 \times 4 \times 1}{100}$$) = Rs 200.
Amount at the end of the first half-year = (5000 + 200) = 5200.
Principal for the second half-year = Rs 5200.
Interest for the second half-year = (($$\frac{5200 \times 4 \times 1}{100}$$)) = Rs 208.
Amount at the end of the second half-year = Rs (5200 + 208) = Rs 5408.
Therefore, compound interest = Rs (5408 — 5000) = Rs 408.
Example 4: Find the compound interest on Rs 8000 for $$1\frac{1}{2}$$ years at 10% per annum, interest being payable half-yearly.
Solution. We have,
Rate of interest = 10% per annum = 5% per half-year.
Time = $$1\frac{1}{2}$$ years = $$\frac{3}{2}$$ x 2 = 3 half- years
Original principal = Rs 8000
Interest for the first half-year = Rs ( $$\frac{8000 \times 5 \times 1}{100}$$) = Rs 400
Amount at the end of the first half-year = Rs 8000 + R 400 = Rs 8400
Principal for the second half-year = Rs 8400
Interest for the second half-year = Rs ($$\frac{8400 \times 5 \times 1}{100}$$) = Rs 420
Amount at the end of the second half-year = Rs 8400 + Rs 420 = Rs 8820
Principal for the third half-year = Rs 8820
Interest for the third half-year = Rs ($$\frac{8820 \times 5 \times 1}{100}$$) = Rs 441
Amount at the end of third half-year = Rs 8820 + Rs 441 = Rs 9261
Therefore, Compound interest = Rs 9261 — Rs 8000 = Rs 1261
## Finding CI When Interest Is Compounded Quarterly
If the rate of interest is R % per annum and the interest is compounded quarterly, then it is $$\frac{R}{4}$$ % per quarter.
Example 5: Find the compound interest on Rs 10000 for 1 year at 20% per annum compounded quarterly.
Solution. We have,
Rate of interest = 20% per annum = $$\frac{1}{5}$$% = 5% per quarter
Time = 1 year = 4 quarters.
Principal for the first quarter = Rs 10000
Interest for the first quarter = Rs ($$\frac{10000 \times 5 \times 1}{100}$$) = Rs 500
Amount at the end of first quarter = Rs 10000 + Rs 500 = Rs 10500
Principal for the second quarter = Rs 10500
Interest for the second quarter = Rs ($$\frac{10500 \times 5 \times 1}{100}$$) = Rs 525
Amount at the end of second quarter = Rs 10500 + Rs 525 = Rs 11025
Principal for the third quarter = Rs 11025
Interest for the third quarter = Rs ($$\frac{11025 \times 5 \times 1}{100}$$) = Rs 551.25
Amount at the end of the third quarter = Rs 11025 + Rs 551.25
= Rs 11576.25
Principal for the fourth quarter = Rs 11576.25
Interest for the fourth quarter = Rs ($$\frac{11576.25 \times 5 \times 1}{100}$$) = Rs 578.8125
Amount at the end of fourth quarter = Rs 11576.25 + Rs 578.8125
= Rs 12155.0625
Therefore, Compound interest = Rs 12155.0625 — Rs 10000
= Rs 2155.0625
= Rs 2155.06
## Definition
If a is any real number and n is a natural number, then $${ a }^{ n }$$ = a x a x a…. n times
where a is called the base, n is called the exponent or index and $${ a }^{ n }$$ is the exponential expression. $${ a }^{ n }$$ is read as ‘a raised to the power n’ or ‘a to the power n’ or simply ‘a power n’.
For zero power, we have :
$${ a }^{ 0 }$$ = 1 (where a $$\neq$$ 0)
For example :
(i) $${ 7 }^{ 0 }$$ = 1 (ii) $${ (-\frac { 2 }{ 3 } ) }^{ 0 }$$ = 1 (iii) $$( { \sqrt { 7 } })^{ 0 }$$ = 1
For negative powers, we have :
$$\sqrt [ n ]{ { a }}$$ = $$\frac { 1 }{ { a }^{ n } }$$ and $$\frac { 1 }{ { a }^{ -n } }$$ = $${ a }^{ n }$$
For example:
(i) $${ 5 }^{ -2 }$$ = $$\frac { 1 }{ { 5 }^{ 2 } }$$
(ii) $${ -2 }^{ -3 }$$ = $$\frac { 1 }{ { -2 }^{ 3 } }$$
(iii) $$\frac { 1 }{ { 2 }^{ -5 } }$$ = $${ 2 }^{ 5 }$$
For fractional indices, we have :
$${ \sqrt { a} }^{ n }$$ = $${ a }^{ \frac { 1 }{ n } }$$ and $$\sqrt [ n ]{ { a }^{ m } }$$ = $${ a }^{ \frac { m }{ n } }$$
For example:
(i) $${ \sqrt { 3 } }$$ = $${ 3 }^{ \frac { 1 }{ 2 } }$$
(ii) $${ \sqrt { 8 } }^{ 3 }$$ = $${ 8 }^{ \frac { 1 }{ 3} }$$
(iii) $$\sqrt [ 4 ]{ { 5 }^{ 3 } }$$ = $${ 5 }^{ \frac { 3 }{ 4 } }$$
## Finding the value of the Number given in the Exponential Form
Example 1: Find the value of each of the following:
(i) $${ 12 }^{ 2 }$$ (ii) $${ 8 }^{ 3 }$$ (iii) $${ 4 }^{ 4 }$$
Solution.
(i) We have,
$${ 12 }^{ 2 }$$ = 12 x 12 = 144
(ii) We have,
$${ 8 }^{ 3 }$$ = 8 x 8 x 8
= (8 x 8) x 8
= 64 x 8
= 512
(iii) We Have,
$${ 4 }^{ 4 }$$= 4 x 4 x 4 x 4
= (4 x 4 ) x 4 x 4
= (16 x 4) x 4
= 64 x 4
= 256
Example 2: Simplify:
(i) 2 x $${ 10 }^{ 3 }$$ (ii) $${ 5 }^{ 2 }$$ x $${ 4 }^{ 2 }$$ (iii) $${ 3 }^{ 3 }$$ x 4
Solution.
(i) We have,
2 x $${ 10 }^{ 3 }$$ = 2 x 1000 = 2000 [since $${ 10 }^{ 3 }$$=10 x10 x 10 = 1000]
(ii) We have,
$${ 5 }^{ 2 }$$ x $${ 4 }^{ 2 }$$
= 25 x 16 = 400
(iii) We have,
$${ 3 }^{ 3 }$$ x 4 = 27 x 4 = 108
## Expressing Numbers in Exponential Form
Example 1: Express each of the following in exponential form:
(i) (-4) x (-4) x (-4) x (-4) x (-4) (ii) $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$
Solution. We have,
(i) (-4) x (-4) x (-4) x (-4) x (-4) = $${ -4}^{ 5 }$$
(ii) $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ = $$({ \frac { 2 }{ 5 } ) }^{ 4 }$$
Example 2: Express each of the following in exponential form:
(i) 3 x 3 x 3 x a x a (ii) a x a x a x a x a x a x b x b x b c x c x c x c
(iii) b x b x b x $$\frac{2}{5}$$ x $$\frac{2}{5}$$
Solution. We have,
(i) 3 x 3 x 3 x a x a = $${ 3 }^{ 3 }$$ x $${ a }^{ 2 }$$
(ii) a x a x a x a x a x a x b x b x b x c x c x c x c = $${ a }^{ 6 }$$ x $${ b }^{ 3 }$$ x $${ c }^{ 4 }$$
(iii) b x b x b x $$\frac{2}{5}$$ x $$\frac{2}{5}$$ = $${ a }^{ 3 }$$ x $$({ \frac { 2 }{ 5 } ) }^{ 2 }$$
Example 3: Express each of the following numbers in exponential form:
(i) 128 (ii) 243 (iii) 3125
Solution.
(i) We have,
128 = 2 x 2 x 2 x 2 x 2 x 2 x 2
128 = $${ 2 }^{ 7 }$$
(ii) We have,
243 = 3 x 3 x 3 x 3 x 3
243 = $${ 3 }^{ 5 }$$
(iii) We have,
625 = 5 x 5 x 5 x 5
625 = $${ 5 }^{ 4 }$$
## Positive Integral Exponent of a Rational Number
Let $$\frac{a}{b}$$ be any rational number and n be a positive integer. Then,
$${(\frac { a } { b })^{ n } }$$ = $$\frac{a}{b}$$ x $$\frac{a}{b}$$ x $$\frac{a}{b}$$…n times
= $$\frac { a\quad \times \quad a\quad \times \quad a….n\quad times }{ b\quad \times \quad b\quad \times \quad b….n\quad times }$$
= $$\frac { { a }^{ n } }{ { b }^{ n } }$$
Thus, $${(\frac { a }{ b })^{ n } }$$ = $$\frac { { a }^{ n } }{ { b }^{ n } }$$ for every positive integer n.
Example : Evaluate:
(i) $${(\frac { 3 } { 7 })^{ 3 } }$$ (ii) $${(\frac { -2 }{ 5 })^{ 3 } }$$
Solution.
(i) $${(\frac { 3 } { 7 })^{ 3 } }$$ = $$\frac { { 3 }^{ 3 } }{ { 7 }^{ 3 } }$$ = $$\frac{127}{343}$$
(ii) $${(\frac { -2 } { 5 }) ^{ 3 } }$$ = $$\frac { { (-2) }^{ 3 } }{ { 5 }^{ 3 } }$$
= $$-\frac { { 2 }^{ 3 } }{ { 5 }^{ 3 } }$$
= $$-\frac{8}{125}$$
## Negative Integral Exponent of a Rational Number
Let $$\frac{a}{b}$$ be any rational number and n be a positive integer.
Then, we define, $${(\frac { a }{ b })^{ -n } }$$ = $${(\frac { b }{ a })^{ n } }$$
Example : Evaluate:
(i) $${(\frac { 1 } { 2 })^{ -3 } }$$ (ii) $${(\frac { 2 } { 7 })^{ -2 } }$$
Solution.
(i) $${(\frac { 1 }{ 2 })^{ -3 } }$$
= $${(\frac { 2 } { 1 })^{ 3 } }$$ = $$\frac { { 2}^{ 3 } }{ { 1 }^{ 3 } }$$
= 8
(ii) $${(\frac { 2 } { 7 })^{ -2 } }$$
= $${(\frac { 7 } { 2 })^{ 2 } }$$
= $$\frac { { 7}^{ 2 } }{ { 2 }^{ 2 } }$$
= $$\frac{49}{4}$$
## Short Cut Method for Finding The Cubes Of a Two- Digit Number
We have : $${ (a+b) }^{ 3 }$$ = $${ a }^{ 3 }$$ + 3 $${ a }^{ 2 }$$ b+ 3 a $${ b }^{ 2 }$$ + $${ b }^{ 3 }$$.
Method : For finding the cube of a two-digit number with the tens digit = a and the units digit b, we make four columns, headed by
$${ a }^{ 3 }$$, 3 $${ a }^{ 2 }$$ b, 3 a $${ b }^{ 2 }$$ and $${ b }^{ 3 }$$.
The rest of the procedure is the same as followed in squaring a number by the column method.
We simplify the working as :
Example 1: Find the value of $${ (29) }^{ 3 }$$ by the short-cut method.
Solution. Here, a=2 and b=9
Therefore, $${ (29) }^{ 3 }$$ = 24389.
Example 2: Find the value of $${ (71) }^{ 3 }$$ by the short-cut method.
Solution. Here, a = 7 and b = 1.
Therefore, $${ (71) }^{ 3 }$$ = 357911.
## Definitions
Sometimes to increase the sale or to dispose off the old stock, a dealer offers his goods at reduced prices. The reduction in price offered by the dealer is called discount.
Marked Price: The printed price or the tagged price of an article is called the marked price (M.P.). It is also called the list price.
Discount: The deduction allowed on the marked price is called discount. Discount is generally given as per cent of the marked price.
Net Price: The selling price at which the article is sold to the customer after deducting the discount from the marked price is called the net price.
## Formulas
(i) S.P. = M.P. – Discount
(ii) Rate of discount = Discount % = $$\frac{Discount}{M.P.}$$ x 100
(iii) S.P. = M.P. x ($$\frac{100 – Discount \%}{100}$$)
(iv) M.P. = $$\frac{100 * S.P.}{100 – Discount \%}$$
## Illustrative Examples
Example 1: Find S.P. if M.P. = Rs 650 and Discount = 10%
Solution. (i) We have,
M.P. = Rs 650, Discount = 10%
Discount = 10% of Rs 650 = Rs($$\frac{10}{100}$$ x 650) = Rs 65
Hence, S.P. = M.P. — Discount = Rs 650 — Rs 65 = Rs 585
Alternative Solution– We have,
M . P. = Rs 650, Discount % =10
S.P. = M.P. x $$\frac{(100 – Discount \%)}{100}$$
=> S.P. = Rs {650 x $$\frac{(100 -10)}{100}$$} = Rs(65 x 9) = Rs 585
Example 2: Find the rate of discount when M.P. = Rs 600 and S.P. = Rs 510.
Solution. M.P. = Rs 600, S.P. = Rs 510
Therefore, Discount = M.P. — S.P. = Rs 600 — Rs 5l0 = Rs 90
Therefore, Rate of discount, i.e., discount% = $$\frac{Discount}{M.P.}$$ x 100 = $$\frac{90}{600}$$ x 100% = 15%.
Example 3: Find the M.P. When S.P. = Rs 9,000 and discount = 10%.
Solution. S.P. = 9000, discount = 10%
Let the M.P. be Rs 100. Since discount = 10%, So S.P. = Rs 90.
When S.P. is 90, M.P. is 100.
When S.P. is Rs 1, M.P. is Rs $$\frac{100}{90}$$
When S.P. is Rs 9000, M.P. is Rs $$\frac{100}{90}$$ x 9000 = Rs 10,000
Example 4: A garment dealer allows his customers 10% discount on a marked price of the goods and still g a profit of 25%. What is the cost price if the marked price of a shirt is Rs 1250?
Solution. M.P. = 1250, Discount = 10%
When M.P. is 100, S.P. is 90
When M.P. is 1250, S.P. is Rs $$\frac{900}{100}$$ = Rs 1125
Profit = 25%, So C.P. = $$\frac{100}{(100 + Profit \%)}$$ x S.P.
= Rs $$\frac{100}{(100 + 25)}$$ x 1125
= Rs $$\frac{100}{125}$$ x 1125
= Rs(100 x 9) = Rs 900
Successive Discounts
Example 5: A car is marked at 4,00,000. The dealer allows successive discounts of 5%, 3% and $$2\frac{1}{2}$$% on it. What is the net selling price ?
Solution. Marked price of the car = Rs 4,00,000
First discount = 5% of 4,00,000 = ($$\frac{5}{100}$$ x 400000) = Rs 20,000
Net price after first discount = (4,00,000 — 20,000) = 3,80,000
Second discount = 3% of Rs 3,80,000 = ($$\frac{3}{100}$$ x 380000) = Rs 11,400
Net price after second discount = (3,80,000 — 11,400) = Rs 3,68,600
Third discount = Rs ( $$\frac { 2\frac { 1 }{ 2 } }{ 100 }$$ x 3, 68, 600)
= ($$\frac{5}{200}$$ x 368600) = Rs 9215
Net selling price = Rs (3,68,600 — 9215) = Rs 3,59,385.
Example 6: Find a single discount equivalent to two successive discounts of 20% and 5%.
Solution. Let the marked price be Rs 100.
First discount = Rs 20
Net price after first discount = Rs (100 — 20) = Rs 80
Second discount 5% of Rs 80 = Rs ($$\frac{5}{100}$$ x 8O) = Rs 4
Net price after second discount = Rs (80 — 4) = Rs 76
Total discount allowed = Rs (100 — 76) = Rs 24
Hence, the required single discount = 24%.
## Formulas
Profit = S.P. –C.P.
Loss = C.P. – S.P.
Profit % = ( $$\frac{Profit}{C.P.}$$ x 100) %
Loss % = ( $$\frac{Loss}{C.P.}$$ x 100) %
## Illustrative Examples
Example 1: John bought a watch for Rs 540 and sold it for Rs 585. Find his profit and profit percentage.
Solution. C.P. of the watch = Rs 540, S.P. of the watch = Rs 585
Profit = S.P. — C.P.
= Rs 585 — Rs 540 = Rs 45
Profit percentage = ( $$\frac{Profit}{C.P.}$$ x 100) %
= ($$\frac{45}{540}$$ X 100 ) %
= $$\frac{100}{12}$$ %
= $$\frac{25}{3}$$ %
= $$8\frac{1}{3}$$ %
Example 2: By selling a bike for Rs 22464, Ansari incurs a loss of Rs 1536. Find his loss percenta
Solution. S.P. of the bike = Rs 22464, loss = Rs 1536
C.P. of the bike = S.P. + loss = Rs 22464 + Rs 1536 = Rs 24000
Loss percentage = ( $$\frac{Loss}{C.P.}$$ x 100) %
= ($$\frac{1536}{24000}$$ X 100 ) %
= $$\frac{1536}{240}$$ %
= $$\frac{64}{10}$$ %
= 6.4 %
Example 3: B ijoy bought bananas at the rate of 5 for Rs 4 and sold them at the rate of 4 for Rs 5. Calculate his gain percentage.
Solution.
C.P. of 5 bananas = Rs 4
C.P. of 1 banana = Rs $$\frac{4}{5}$$ = Rs 0.80
S.P. of 4 bananas = Rs 5
S.P. of 1 banana = Rs $$\frac{5}{4}$$ = Rs 1.25
Therefore, Gain on the sale of one banana = S.P. — C.P. = Rs 1.25 — Rs 0.80 = Rs 0.45
Gain percentage = ( $$\frac{Profit}{C.P.}$$ x 100) %
= ($$\frac{0.45}{0.80}$$ X 100 ) %
= $$\frac{145}{80}$$ %
= $$\frac{225}{4}$$ %
= 56.25 %
## Standard Form
A number written as ( m x $$10^n$$ ) is said to be in standard form if m is a decimal number such that 1 $$\le$$ m $$<$$10 and n is either a positive or a negative integer.
The standard form of a number is also known as Scientific notation.
## Expressing Very Large Numbers in Standard Form
In order to write large numbers in the standard form,following steps must be followed:
STEP I– Obtain the number and move the decimal point to the left till you get just one digit to the left of the decimal point.
STEP II– Write the given number as the product of the number so obtained and $$10^n$$ , where n is the number of places the decimal point has been moved to the left. If the given number is between 1 and 10, then write it as the product of the number itself and $$10^0$$ .
Illustrative Examples
Example 1: Express the following numbers in the standard form:
(i) 3,90,878 (ii) 3,186,500,000 (iii) 65,950,000
Solution.
(i) We have,
3,90,878 = 390878.00
Clearly, the decimal point is moved through five places to obtain a number in which there is just one digit to the left of the decimal point.
Therefore, 390878.00 = 3.90878 x $$10^5$$
(ii) We have,
3,186,500,000 = 3.186500000 x $$10^9$$
= 3.1865 x $$10^9$$
(iii) We have,
65,950,000 = 65,950,000.00
= 6.5950000 x $$10^7$$
= 6.595 x $$10^7$$
Example 2: The distance between sun and earth is (1.496 x $${10}^{11}$$) m and the distance between earth and moon is (3.84 x $$10^8$$) m. During solar eclipse moon comes in between earth and sun. At that time what Is the distance between moon and sun?
Solution. Required distance
= {(1.496 x $${10}^{11}$$) – (3.84 x $$10^8$$) } m
= {$$\frac { 1496\times { 10 }^{ 11 } }{ { 10 }^{ 3 } }$$ – (3.84 x $$10^8$$)} m
= {1496 x $$10^8$$) – (3.84 x $$10^8$$)} m
= {(1496 – 3.84) x $$10^8$$)} m
= (1492.16 x $$10^8$$) m
Hence, the distance between moon and sun is (1492.16 x $$10^8$$ ) m.
Example 3: Write the following numbers in the usual form:
(i) 7.54 x $$10^6$$ (ii)2.514 x $$10^7$$
Solution. We have
(i) 7.54 x $$10^6$$
= $$\frac{754}{100}$$ x $$10^6$$
= $$\frac{754 \times {10}^{6} }{{10}^{2}}$$
= 754 x $${10}^{(6-2)}$$
= (754 x $${10}^{4}$$ )
= (754 x 10000) = 7540000
(ii) 2.514 x $$10^7$$
= $$\frac{2514}{1000}$$ x $$10^7$$
= $$\frac{2514 \times {10}^{7} }{{10}^{3}}$$
= 2514 x $${10}^{(7-3)}$$
= (2514 x $${10}^{4}$$ )
= (2514 x 10000) = 25140000
## Expressing Very Small Numbers in Standard Form
In order to write very small numbers in the standard form,following steps must be followed:
STEP I- Obtain the number and count the number of decimal values after the decimal point. Consider it as n.
STEP II- Divide the number by $${10}^{n}$$). If the number is between 1 and 10, then write it as the product of the number itself and $${10}^{-n}$$
Example 1: Write the following numbers in the standard form:
(i) 0.000000059 (ii) 0.00000000526
Solution. We may write:
(i) 0.000000059
= $$\frac{59}{{10}^{9}}$$
= $$\frac{5.9 \times 10}{{10}^{9}}$$
= $$\frac{5.9}{{10}^{8}}$$ = (5.9 x $${10}^{-8}$$)
(ii) 0.00000000526
= $$\frac{526}{{10}^{11}}$$
= $$\frac{5.26 \times 100}{{10}^{11}}$$
= $$\frac{5.26 \times {10}^{2}}{{10}^{11}}$$
= $$\frac{5.26}{{10}^{(11 – 2)}}$$
= $$\frac{5.26}{{10}^{9}}$$ = (5.26 x $${10}^{-9}$$)
Example 2: The size of a red blood cell is 0.000007 m and that of a plant cell Is 0.00001275 m. Show that a red blood cell is half of plant cell in size.
Solution. We have,
Size of a red blood cell = 0.000007 m = $$\frac{7}{{10}^{6}}$$ m = (7 x $${10}^{-6}$$)
Size of a plant cell
= 0.00001275 m
= $$\frac{1275}{{10}^{8}}$$ m
= $$\frac{1.275 \times {10}^{3}}{{10}^{8}}$$ m
= $$\frac{1.275}{{10}^{(8-3)}}$$ m
= $$\frac{1.275}{{10}^{5}}$$ m = (1.275 x $${10}^{-5}$$) m
$$\frac{Size of a red blood cell}{Size of a plant cell}$$
= $$\frac{7 \times {10}^{-6}}{1.275 \times {10}^{-5}}$$
= $$\frac{7 \times {10}^{-6 + 5}}{1.275}$$
= $$\frac{7 \times {10}^{-1}}{1.275}$$
= $$\frac{7}{1.275 \times 10}$$
= $$\frac{7}{12.75}$$
= $$\frac{7}{13}$$ (nearly)
= $$\frac{1}{2}$$ (approximately)
Therefore, size of a red blood cell = $$\frac{1}{2}$$ x (size of a plant cell)
Example 3: Express the following numbers in usual form:
(i) 3 x $${10}^{-3}$$ (ii) 2.34 x $${10}^{-4}$$
Solution. We have,
(i) 3 x $${10}^{-3}$$
= $$\frac{3}{{10}^{3}}$$
= $$\frac{3}{1000}$$ = 0.003
(ii) 2.34 x $${10}^{-4}$$
= $$\frac{234}{100}$$ x $$\frac{1}{{10}^{4}}$$
= $$\frac{234}{{10}^{2} \times {10}^{4}}$$
= $$\frac{234}{{10}^{6}}$$
= $$\frac{234}{1000000}$$ = 0.000234
## Steps involved in conversion of a ratio into per cent
STEP I– Obtain the ratio, say, a : b.
STEP II– Convert the given ratio into the fraction $$\frac{ a }{ b }$$
STEP III– Multiply the fraction obtained in step II by 100 and put per cent sign %.
Illustration 1: Express the following as per cents:
(i) 14 : 25 (ii) 5 : 6 (iii) 111 : 125
Solution. We have :
(i) 14 : 25 =$$\frac{ 14 }{ 25 }$$ = ($$\frac{ 14 }{ 25 }$$ x 100)% = 56%.
(ii) 5 : 6 = $$\frac{ 5 }{ 6 }$$ = ($$\frac{ 5 }{ 6 }$$ x 100)% = $$\frac{ 250 }{ 3 }$$% = $$83\frac{1}{3}$$%
(iii) 111 : 125 = $$\frac{ 111 }{ 125 }$$ = ($$\frac{ 111 }{ 125 }$$ x 100)% = 88.88%
## Steps involved in conversion of a per cent into ratio
STEP I– Obtain the per cent.
STEP II– Convert the given per cent into a fraction by dividing it by 100 and removing per cent sign %.
STEP III– Express the fraction obtained in step II in the simplest form.
STEP IV– Express the fraction obtained in step III as a ratio.
Illustration 2 : Express each of the following per cents as a ratio in the simplest form:
Solution. We have:
(i) 36% = $$\frac{ 36 }{ 100 }$$ = 0.36
(ii) 5.4% = $$\frac{ 5.4 }{ 100 }$$ = $$\frac{ 54 }{ 1000 }$$ = 0.054
(iii) 0.25% = $$\frac{ 0.25 }{ 100 }$$ = $$\frac{ 25 }{ 10000 }$$ = 0.0025
(iv) 135% = $$\frac{ 135 }{ 100 }$$ = 1.35
## When the Interest is Compounded Annually
Formula
Let principal = P, rate = R% per annum and time = n years.
Then, the amount A is given by the formula
A = P $${ (1+\frac { R }{ 100 } ) }^{ n }$$ .
Illustrative Examples
Example 1: Find the amount of Rs 8000 for 3 years, compounded annually at 10% per annum. Also,find the compound interest.
Solution. Here, P = Rs 8000, R =10% per annum and n =3 years.
Using the formula A = P $${ (1+\frac { R }{ 100 } ) }^{ n }$$ , we get
Amount after 3Years = {$${ 8000 \times (1+\frac { 10 }{ 100 } ) }^{ 3 }$$ }
= Rs (8000 x $$\frac{11}{10}$$ x $$\frac{11}{10}$$ x $$\frac{11}{10}$$)
= Rs 10648.
Thus, Amount after 3 years = Rs 10648.
And, compound interest = Rs (10648 – 8000) = Rs 2648
Example 2: Rakesh lent Rs 8000 to his friend for 3 years at the rate of 5% per annum compound interest. What amount does Rakesh get after 3 years?
Solution. Here, P = Rs 8000, R = 5% per annum and n =3.
Amount after 3 year = P$${ (1+\frac { R }{ 100 } ) }^{ n }$$
= Rs 8000 x $${ (1+\frac { 5 }{ 100 } ) }^{ 3 }$$
= Rs 8000 x $${ (1+\frac { 1 }{ 20 } ) }^{ 3 }$$
= Rs 8000 x $${ (\frac { 21 }{ 20 } ) }^{ 3 }$$
= Rs 8000 x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$
= Rs 9261
Example 3: Find the amount and compound interest on Rs 5000 for 2 years at 10%, interest being pay yearly.
Solution. Here, P= Rs 5000, R = 10%, n = 2 years
Using the formula, A (Amount) = P $${ (1+\frac { R }{ 100 } ) }^{ n }$$, we have
Therefore, A = Rs 5000 $${ (1+\frac { 10 }{ 100 } ) }^{ 2 }$$
= Rs 5000 x $$\frac{110}{100}$$ x $$\frac{110}{100}$$
= Rs 6050
Therefore, Compound Interest = A – P = Rs 6050 – Rs 5000 = Rs 1050.
## When the Interest is Compounded Half-Yearly
Formula
If the interest is paid half-yearly, then in the formula A = P $${ (1+\frac { R }{ 100 } ) }^{ n }$$, for R we take $$\frac{R}{2}$$ , because R% p.a. means $$\frac{R}{2}$$ % half-yearly and for n we take 2n, because n years is equal to 2n half-years.
Therefore, A = P $${ (1+\frac { R }{ 200 } ) }^{ 2n }$$
Illustrative Examples
Example 1: compute the compound interst on Rs 10000 for 2 years at 10% per annum when compounded half-yearly.
Solution.
Here, Principal P = Rs 10000, R = 10% per annum, and n = 2 years
Amount after 2 years
= P $${ (1+\frac { R }{ 200 } ) }^{ 2n }$$
= Rs 10000 x P $${ (1+\frac { 10 }{ 200 } ) }^{ 2 \times 2 }$$
= Rs 10000 x P $${ (1+\frac { 1 }{ 20 } ) }^{ 4 }$$
= Rs 10000 x P $${ (\frac { 21 }{ 20 } ) }^{ 4 }$$
= Rs 10000 x $$\frac{21}{20 }$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$
= Rs 10000 x $$\frac{194481}{160000}$$
= Rs 12155.06
Therefore, Compound Interest = Ra(12155.06 – 10000) = Rs 2155.065
Example 2: How much will Rs 256 amount to in one year at $$12\frac{1}{2}$$% per annum, when the interest is compounded half-yearly.
Solution. P= Rs 256, 1 year= 2 half years, n = 2, Annual rate = 12 %
= $$12\frac{1}{2}$$%
Therefore, Half-yearly rate = $$\frac{1}{2}$$($$\frac{25}{2}$$ %) = $$\frac{25}{4}$$%
Thus Amount(A) = $${ (1+\frac { R }{ 100 } ) }^{ n }$$
= Rs 256 $$(1+{ \frac { \frac { 25 }{ 4 } }{ 100 } ) }^{ 2 }$$
= Rs 256 $${ (1+\frac { 1 }{ 16 } ) }^{ 2 }$$
= Rs 256 x $$\frac{17}{16}$$ x $$\frac{17}{16}$$
= Rs 289
Example 3: How much would a sum of Rs 16000 amount to in 2 years time at 10% per annum compounded interest, interest being payable half-yearly?
Solution. Here, P = Rs 16000, R = 10% per annum and n = 2 years.
Amount after 2 years
= P$${ (1+\frac { R }{ 200 } ) }^{ 2n }$$
= Rs 16000 x $${ (1+\frac { 10 }{ 200 } ) }^{ 2 \times 2 }$$
= Rs 16000 x $${ (1+\frac { 1 }{ 20 } ) }^{ 4 }$$
= Rs 16000 x $${ (\frac { 21 }{ 20 } ) }^{ 4 }$$
= Rs 16000 x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$ x $$\frac{21}{20}$$
= Rs 19448.10
Hence, a sum of Rs 16000 amounts to Rs 19448. 10 in 2 years.
## When the Interest is Compounded Quarterly
Formula
If P = Principal, R = Interest rate percent per annum and n = number of years, then
A = $${ (1+\frac { R }{ 400 } ) }^{ 4n }$$
C.I. = A – P
Illustrative Examples
Example 1: Find the compound interest on Rs 360000 for one year at the rate of 10% per annum, if the interest is compounded quarterly.
Solution. Here, P = Rs 360000, R = 10% per annum and n= 1 year
Amount after 1 year
= P$${ (1+\frac { R }{ 400 } ) }^{ 4n }$$
= Rs 360000 x $${ (1+\frac { 10 }{ 400 } ) }^{ 4 \times 1 }$$
= Rs 360000 x $${ (1+\frac { 1 }{ 40 } ) }^{ 4 }$$
= Rs 360000 x $${ (\frac { 41 }{ 40 } ) }^{ 4 }$$
= Rs 360000 x $$\frac{41}{40}$$ x $$\frac{41}{40}$$ x $$\frac{41}{40}$$ x $$\frac{41}{40}$$
= Rs 397372.64
Therefore, Compound Interest = Rs 397372.64 – Rs 360000
= Rs 37372.64
Example 2: Sharukh deposited in a bank Rs 8000 for 6 months at the rate of 10% interest compounded quarterly. Find the amount he received after 6 months.
Solution. Here, P = Rs 8000, R = 10% per annum and n= 6 months
= $$\frac{6}{12}$$
= $$\frac{1}{2}$$ year
Amount after 6 months
= P$${ (1+\frac { R }{ 400 } ) }^{ 4n }$$
= Rs 8000 x $$(1+{ \frac { 10 }{ 400 } ) }^{ 4\times \frac { 1 }{ 2 } }$$
= Rs 8000 x $${ (1+\frac { 1 }{ 40 } ) }^{ 2 }$$
= Rs 8000 x $${ (\frac { 41 }{ 40 } ) }^{ 2 }$$
Example 3: Ramesh deposited Rs 7500 in a bank which pays him 12% interest per annum compounded quarterly. What is the amount which he receives after 9 months?
Solution. Here, P = Rs 7500, R = 10% per annum and n = 9 months
= $$\frac{9}{12}$$
= $$\frac{3}{4}$$ year
Amount after 9 months
= P$${ (1+\frac { R }{ 400 } ) }^{ 4n }$$
= Rs 7500 x $$(1+{\frac { 12 }{ 400 } ) }^{ 4 \times \frac{3}{4} }$$
= Rs 8000 x $${ (1+\frac { 3 }{ 100 } ) }^{ 3 }$$
= Rs 8000 x $${ (\frac { 103 }{ 100 } ) }^{ 2 }$$
= Rs 8000 x $$\frac{103}{100}$$ x $$\frac{103}{100}$$ x $$\frac{103}{100}$$
= Rs 8195.45
= Rs 8000 x $$\frac{41}{40}$$ x $$\frac{41}{40}$$
= Rs 8405
## When the Rate of Interest for Successive years are Different
Formula
If the rate of interest is different for every year say, $${ R }_{ 1 }$$, $${ R }_{ 2 }$$, $${ R }_{ 3 }$$…$${ R }_{ n }$$ for the first, second, third year… nth year then the amount is given by
A = P $$(1+\frac { { R }_{ 1 } }{ 100 } )$$$$(1+\frac { { R }_{ 2 } }{ 100 } )$$$$(1+\frac { { R }_{ 3 } }{ 100 } )$$….$$(1+\frac { { R }_{ n } }{ 100 } )$$
Illustrative Examples
Example 1: Find the amount of Rs 50000 after 2 years, compounded annually; the rate interest being 8% p.a. during the first year and 9% p.a. during the second ear. Also,find the compound Interest.
Solution. Here, P = Rs 50000, $${ R }_{ 1 }$$= 8% p.a. and $${ R }_{ 2 }$$ = 9% p.a.
Using the formula A = P $$(1+\frac { { R }_{ 1 } }{ 100 } )$$$$(1+\frac { { R }_{ 2 } }{ 100 } )$$ we have:
Amount after 2 years
= Rs 50000 $$(1+\frac { 8 }{ 100 } )$$ $$(1+\frac { 9 }{ 100 } )$$
= Rs 50000 $$\frac{27}{25}$$ x $$\frac{109}{100}$$
= Rs 58860
Thus, amount after 2 years = Rs 58860.
And, compound interest = Rs (58860 – 50000) = Rs 8860.
Example 2: Find the compound interest on Rs 80,000 for 3 years if the rates 4%, 5% and 10% respectively.
Solution. Here, P = Rs 80000, $${ R }_{ 1 }$$ = 4%, $${ R }_{ 2 }$$ = 5% and $${ R }_{ 3 }$$ = 10%
amount after 3 years
= Rs 80000 $$(1+\frac { 4 }{ 100 } )$$ $$(1+\frac { 5 }{ 100 } )$$ $$(1+\frac { 10 }{ 100 } )$$
= Rs 80000 x $$\frac{104}{100}$$ x $$\frac{105}{100}$$ x $$\frac{110}{100}$$
= Rs 96096
Therefore, Compound interest = Rs (96096 — 80000) = Rs 16096.
## When Interest is Compounded Annually but Time is a Fraction
Formula
If P = Principal, R = Rate % per annum and Time = $$3\frac{3}{4}$$ years (say), then
A = P$$(1+{ \frac { R }{ 100 } ) }^{ 3}$$ x $$(1+\frac { \frac { 3 }{ 4 } \times R }{ 100 } )$$
Illustrative Examples
Example 1: Find the compound interest on Rs 31250 at 8% per annum for $$2\frac{3}{4}$$ years.
Solution. Amount after $$2\frac{3}{4}$$ years
= Rs 31250 x $$(1+{ \frac { 8 }{ 100 } ) }^{ 2}$$ x $$(1+\frac { \frac { 3 }{ 4 } \times 8 }{ 100 } )$$
= Rs 31250 x $$(1+{ \frac { 27 }{ 25 } ) }^{ 2}$$ x $$(\frac { 53 }{ 50 } )$$
= 31250 x $$(1+\frac { 27 }{ 25 } )$$ x $$(1+\frac { 27 }{ 25 } )$$ x $$(1+\frac { 53 }{ 50 } )$$
= Rs 38637
Therefore, Amount = Rs 38637.
Hence, compound interest = Rs (38637 — 31250) = Rs 7387.
Example 2: Find the compound interest on Ra 24000 at 15% per annum for $$2\frac{1}{3}$$ years.
Solution.
Here, P = Rs 24000, R =15% per annum and Time = $$2\frac{1}{3}$$ years.
Amount after $$2\frac{1}{3}$$years
= P$$(1+{ \frac { R }{ 100 } ) }^{ 2}$$ x $$(1+\frac { \frac { 1 }{ 3 } \times R }{ 100 } )$$
= Rs 24000 x $$(1+{ \frac { 15 }{ 100 } ) }^{ 2}$$ x $$(1+\frac { \frac { 1 }{ 3 } \times 15 }{ 100 } )$$
= Rs 24000 x $$(1+{ \frac { 3 }{ 20 } ) }^{ 2}$$ x $$(1+\frac { 1 }{ 20 } )$$
= Rs 24000 x $$({ \frac { 23 }{ 20 } ) }^{ 2}$$ x $$(\frac { 21 }{ 20 } )$$
= Rs 33327
Therefore, Compound interest = Rs (33327 — 24000) = Rs 9327
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Nr7 Candle
## Procedure
Sometimes we are given two quantities and we want to find what per cent of one quantity is of the other quantity. In other words, we want to find how many hundredths of one quantity should be taken so that it is equal to the second quantity. In such type of problems, we proceed as discussed below:
Let a and b be two numbers and we want to know: what per cent of a is b ?
Let x% of a be equal to b. Then,
$$\frac{ x}{ 100}$$ x a = b
=> x = b x $$\frac{ 100}{ a }$$
=> x = $$\frac{ b}{ a}$$ x lOO
Thus, b is ($$\frac{ b}{ a }$$ x 100)% of a.
## Illustrative Examples:
Example 1: What per cent of 25 kg is 3.5 kg?
Solution. We have,
Required per cent = ( $$\frac{ 3.5 kg}{ 25 kg}$$ x 100) = $$\frac{ 3.5 * 100}{ 25}$$
= $$\frac{ 35 * 100}{ 250}$$
=$$\frac{ 35 * 2}{ 5}$$
= 7x 2
= 14
Hence, 3.5 kg is 14% of 25 kg.
Alternative Solution-
Let x% of 25 kg be 3.5 kg. Then,
x% of 25kg = 3.5kg
=> $$\frac{ x}{ 100}$$ x 25 = 3.5
=> x = $$\frac{ 3.5 * 100}{ 25}$$ [Multiplying both sides by $$\frac{ 100}{ 25}$$ ]
=> x = $$\frac{ 35 * 100}{ 250}$$ = $$\frac{ 35 * 2}{ 5}$$ = 7 x 2 = 14.
Example 2: Express 75 paise as a per cent of Rs 5.
Solution. We have, Rs 5 = 500 paise.
Let x% of Rs 5 be 75 paise. Then,
x% of Rs 5 = 75 paise
=> x% of 500 paise = 75 paise
=> $$\frac{ x}{ 100}$$ x 500 = 75
=> x = $$\frac{ 75 * 100}{ 500}$$
=> x = 15.
Hence, 15% of Rs 5 is 75 paise.
Alternative Solution- The required per cent = ( $$\frac{ 75}{ 500}$$ x 100) % = 15%
Example 3 : Find 10% more than Rs 90.
Solution. We have,
10% of Rs 90 = Rs ( $$\frac{ 10 }{ 100}$$ x 90 ) = Rs 9
Therefore, 10% more than Rs 90 = Rs 90 + Rs 9 = Rs 99 |
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# What Is a Probability Distribution?
Probability distribution for sum of two dice
C.K.Taylor
If you spend much time at all dealing with statistics, pretty soon you run into the phrase “probability distribution.” It is here that we really get to see how much the areas of probability and statistics overlap. Although this may sound like something technical, the phrase probability distribution is really just a way to talk about organizing a list of probabilities. A probability distribution is a function or rule that assigns probabilities to each value of a random variable. The distribution may in some cases be listed. In other cases it is presented as a graph.
### Example
Suppose that we roll two dice and then record the sum of the dice. Sums anywhere from two to 12 are possible. Each sum has a particular probability of occurring. We can simply list these as follows:
• The sum of 2 has a probability of 1/36
• The sum of 3 has a probability of 2/36
• The sum of 4 has a probability of 3/36
• The sum of 5 has a probability of 4/36
• The sum of 6 has a probability of 5/36
• The sum of 7 has a probability of 6/36
• The sum of 8 has a probability of 5/36
• The sum of 9 has a probability of 4/36
• The sum of 10 has a probability of 3/36
• The sum of 11 has a probability of 2/36
• The sum of 12 has a probability of 1/36
This list is a probability distribution for the probability experiment of rolling two dice. We can also consider the above as a probability distribution of the random variable defined by looking at the sum of the two dice.
### Graph of a Probability Distribution
A probability distribution can be graphed, and sometimes this helps to show us features of the distribution that were not apparent from just reading the list of probabilities. The random variable is plotted along the x-axis, and the corresponding probability is plotted along the y - axis.
• For a discrete random variable, we will have a histogram
• For a continuous random variable, we will have the inside of a smooth curve
The rules of probability are still in effect, and they manifest themselves in a few ways. Since probabilities are greater than or equal to zero, the graph of a probability distribution must have y-coordinates that are nonnegative. Another feature of probabilities, namely that one is the maximum that the probability of an event can be, shows up in another way.
### Area = Probability
The graph of a probability distribution is constructed in such a way that areas represent probabilities. For a discrete probability distribution, we are really just calculating the areas of rectangles. In the graph above, the areas of the three bars corresponding to four, five and six correspond to the probability that the sum of our dice is four, five or six. The areas of all of the bars add up to a total of one.
In the standard normal distribution, or bell curve, we have a similar situation. The area under the curve between two z values corresponds to the probability that our variable falls between those two values. For example, the area under the bell curve for -1 < z < 1 accounts for approximately 68% of the total area. The area here is much more complicated than a rectangle. That is why calculus and other advanced mathematics is necessary in order to use most continuous probability distributions.
### A List of Probability Distributions
There are literally infinitely many probability distributions. A list of some of the more important distributions follows:
• Binomial Distribution – this gives the number of successes for a series of independent experiments with two outcomes
• Chi-Square Distribution – this is for use of determining how close observed quantities fit a proposed model
• F-Distribution – this is a distribution that is used in analysis of variance (ANOVA)
• Normal Distribution – this is called the bell curve and is found throughout statistics.
• Student’s t Distribution – this is for use with small sample sizes from a normal distribution
Courtney Taylor
Statistics Guide
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# How do you factor 7k(k-3)+4(k-3)?
Jul 21, 2015
$7 k \left(k - 3\right) + 4 \left(k - 3\right) = \left(7 k + 4\right) \left(k - 3\right)$
#### Explanation:
The factor $\textcolor{b l u e}{\left(k - 3\right)}$ occurs in both terms:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{red}{7 k} \textcolor{b l u e}{\left(k - 3\right)}$
$\textcolor{w h i t e}{\text{XXXX}}$and
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{red}{+ 4} \textcolor{b l u e}{\left(k - 3\right)}$
In general, the distributive property tells us that
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{red}{a} \textcolor{b l u e}{\left(c\right)} \textcolor{red}{+ b} \textcolor{b l u e}{\left(c\right)} = \textcolor{red}{\left(a + b\right)} \textcolor{b l u e}{\left(c\right)}$
So, in this case:
$\textcolor{w h i t e}{\text{XXXX}}$$\textcolor{red}{7 k} \textcolor{b l u e}{\left(k - 3\right)} \textcolor{red}{+ 4} \textcolor{b l u e}{\left(k - 3\right)} = \textcolor{red}{\left(7 k + 4\right)} \textcolor{b l u e}{\left(k - 3\right)}$ |
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# Fraction Manipulation
Reference > Mathematics > TI-30X II
Your TI-30X II will handle most fraction operations, such as addition, subtraction, multiplication, and division. In most cases, if you provide a number as a fraction, the calculator will perform calculations and return the result as a fraction.
So how do you enter a fraction into the TI-30X II? First, find the button that has Ab/c on it. We'll refer to this as the FRACTION button.
Entering a fraction
Let's say you want to enter the fraction 2/3. You could get the decimal value of the fraction by entering 2 ÷ 3, but if you do this, it's no longer a fraction. Instead, press 2, followed by FRACTION, followed by 3. Now press ENTER to see the result. You should see the following:
2 / 3
Entering a mixed numeral
If you want to enter 31/2 (three and a half), you do the same as before, except you use the fraction button one extra time. Enter 3 FRACTION 1 FRACTION 2 and then press ENTER. You should see the following:
3 u 1 / 2.
Converting Mixed/Improper
If you have a mixed number, you may want to convert it to an improper fraction (after all, in Algebra we generally discourage the use of mixed numerals). This conversion is easy to do. Notice the text above the FRACTION button - this text indicates that the 2nd function of the FRACTION button converts between the two types. So if you still have 3 u 1 / 2 displayed on your calculator, Press 2nd and then FRACTION, and you should now see 7 / 2 which is equivalent to 31/2.
And guess what! Pressing 2nd and FRACTION a second time will convert it back!
Now you can do fraction arithmetic using your calculator, and it takes care of the ugly details for you.
Sample #1
What is 1/2 + 2/3?
Solution #1
Use the fraction button to enter the first fraction, then press PLUS, then use the fraction button to enter the second fraction. Press ENTER. The result is 11/6.
Sample #2
Use your calculator to write 11/6 as an improper fraction.
Solution #2
With the mixed number still in your calculator from the previous problem, press 2ndFRACTION, and ENTER to find that it is 7/6.
## Questions
1.
What is the value of
1
3
+
1
4
?
2.
What is the value of 3
2
3
- 1
1
2
?
3.
What is the value of
1
5
·
5
2
?
4.
Convert 7
9
16
to an improper fraction.
5.
Convert
103
3
to a mixed number.
6.
What is
1
4
-
1
5
?
7.
What is
1
3
÷
3
7
?
8.
Find the value of
2
3
+
5
4
and write it as an improper fraction.
9.
What result do you get if you enter the fraction
1
0
?
10.
What result do you get if you try to convert
1
3
to an improper fraction?
Assign this reference page
Multiple-Step Problems
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Class 9 Factorisation of Polynomials
### Topic Covered
♦ Factorisation of Polynomials
### Factorisation of Polynomials
Let us now look at the situation of Example 10 above more closely. It tells us that since the remainder, q (-1/2) =0 , (2t +1) is a factor of q (t) , i.e, q(t) = (2t +1) g(t) for some polynomial g(t). This is a particular case of the following theorem.
text (Factor Theorem : ) If p(x) is a polynomial of degree n ge 1 and a is any real number, then (i) x – a is a factor of p(x), if p(a) = 0, and (ii) p(a) = 0, if x – a is a factor of p(x).
text (Proof: ) By the Remainder Theorem, p(x)=(x – a) q(x) + p(a).
(i) If p(a) = 0, then p(x) = (x – a) q(x), which shows that x – a is a factor of p(x).
(ii) Since x – a is a factor of p(x), p(x) = (x – a) g(x) for same polynomial g(x).
In this case, p(a) = (a – a) g(a) = 0.
Q 3230256112
Examine whether x + 2 is a factor of x^3 + 3x^2 + 5x + 6 and of 2x + 4.
Class 9 Chapter 2 Example 11
Solution:
The zero of x + 2 is –2. Let p(x) = x^3 + 3x^2 + 5x + 6 and s(x) = 2x + 4
Then, p(–2) = (–2)^3 + 3(–2)^2 + 5(–2) + 6
= –8 + 12 – 10 + 6
= 0
So, by the Factor Theorem, x + 2 is a factor of x^3 + 3x^2 + 5x + 6.
Again, s(–2) = 2(–2) + 4 = 0
So, x + 2 is a factor of 2x + 4. In fact, you can check this without applying the Factor
Theorem, since 2x + 4 = 2(x + 2).
Q 3250256114
Find the value of k, if x – 1 is a factor of 4x^3 + 3x^2 – 4x + k.
Class 9 Chapter 2 Example 12
Solution:
As x – 1 is a factor of p(x) = 4x^3 + 3x^2 – 4x + k, p(1) = 0
Now, p(1) = 4(1)^3 + 3(1)^2 – 4(1) + k
So, 4 + 3 – 4 + k = 0
i.e., k = –3
We will now use the Factor Theorem to factorise some polynomials of degree 2 and 3.
You are already familiar with the factorisation of a quadratic polynomial like
x^2 + lx + m. You had factorised it by splitting the middle term lx as ax + bx so that
ab = m. Then x^2 + lx + m = (x + a) (x + b). We shall now try to factorise quadratic
polynomials of the type ax^2 + bx + c, where a ≠ 0 and a, b, c are constants.
Factorisation of the polynomial ax^2 + bx + c by splitting the middle term is as follows:
Let its factors be (px + q) and (rx + s). Then
ax^2 + bx + c = (px + q) (rx + s) = pr x^2 + (ps + qr) x + qs
Comparing the coefficients of x^2, we get a = pr.
Similarly, comparing the coefficients of x, we get b = ps + qr.
And, on comparing the constant terms, we get c = qs.
This shows us that b is the sum of two numbers ps and qr, whose product is
(ps)(qr) = (pr)(qs) = ac.
Therefore, to factorise ax^2 + bx + c, we have to write b as the sum of two
numbers whose product is ac. This will be clear from Example 13.
Q 3280256117
Factorise 6x^2 + 17x + 5 by splitting the middle term, and by using the Factor Theorem.
Class 9 Chapter 2 Example 13
Solution:
(By splitting method) : If we can find two numbers p and q such that
p + q = 17 and pq = 6 × 5 = 30, then we can get the factors.
So, let us look for the pairs of factors of 30. Some are 1 and 30, 2 and 15, 3 and 10, 5
and 6. Of these pairs, 2 and 15 will give us p + q = 17.
So, 6x^2 + 17x + 5 = 6x^2 + (2 + 15)x + 5
= 6x^2 + 2x + 15x + 5
= 2x(3x + 1) + 5(3x + 1)
= (3x + 1) (2x + 5)
text ( Solution 2 : )(Using the Factor Theorem)
6x^2 +17 x + 5 = 6 (x^2 +17/6 x + 5/6) = 6 p (x) say. If a and b are the zeroes of p(x), then
6x^2 + 17x + 5 = 6(x – a) (x – b). So, ab = 5/6 . Now , p (1/2) = 1/4 +17/6 (1/2) +5/6 ≠ 0. But
p ( (-1)/3) = 0. So, (x+ 1/3) is a factor of p(x). Similarly, by trial, you can find that
(x+ 5/2) is a factor of p ( x) .
Therefore, 6x^2 + 17 x + 5 = 6 (x +1/3) ( x+5/2)
= 6 ( (3x+1)/3) ( (2x+5)/2)
For the example above, the use of the splitting method appears more efficient. However,
let us consider another example.
Q 3200256118
Factorise y^2 – 5y + 6 by using the Factor Theorem.
Class 9 Chapter 2 Example 14
Solution:
Let p(y) = y^2 – 5y + 6. Now, if p(y) = (y – a) (y – b), you know that the
constant term will be ab. So, ab = 6. So, to look for the factors of p(y), we look at the
factors of 6.
The factors of 6 are 1, 2 and 3.
Now, p(2) = 22 – (5 × 2) + 6 = 0
So, y – 2 is a factor of p(y).
Also, p(3) = 3^2 – (5 × 3) + 6 = 0
So, y – 3 is also a factor of y^2 – 5y + 6.
Therefore, y^2 – 5y + 6 = (y – 2)(y – 3)
Note that y^2 – 5y + 6 can also be factorised by splitting the middle term –5y.
Now, let us consider factorising cubic polynomials. Here, the splitting method will not
be appropriate to start with. We need to find at least one factor first, as you will see in
the following example.
Q 3210356210
Factorise x^3 – 23x^2 + 142x – 120.
Class 9 Chapter 2 Example 15
Solution:
Let p(x) = x^3 – 23x^2 + 142x – 120
We shall now look for all the factors of –120. Some of these are ±1, ±2, ±3,
±4, ±5, ±6, ±8, ±10, ±12, ±15, ±20, ±24, ±30, ±60.
By trial, we find that p(1) = 0. So x – 1 is a factor of p(x).
Now we see that x^3 – 23x^2 + 142x – 120 = x^3 – x^2 – 22x^2 + 22x + 120x – 120
= x^2(x –1) – 22x(x – 1) + 120(x – 1) (Why?)
= (x – 1) (x^2 – 22x + 120) [Taking (x – 1) common]
We could have also got this by dividing p(x) by x – 1.
Now x^2 – 22x + 120 can be factorised either by splitting the middle term or by using
the Factor theorem. By splitting the middle term, we have:
x^2 – 22x + 120 = x^2 – 12x – 10x + 120
= x(x – 12) – 10(x – 12)
= (x – 12) (x – 10)
So, x^3 – 23x^2 – 142x – 120 = (x – 1)(x – 10)(x – 12) |
# Unit C-6: Graphing Rational Functions
Relevant textbook sections: 5.6
Suggested homework problems:
• Textbook 5.6: 39-64
We can use the techniques of the last two sections to graph a rational function R\par{x}. The steps:
• Factor the numerator and denominator of R. Find the domain of R.
• Write R in lowest terms. If any factors cancel out, use the lowest-terms version of the function to find the y value for each cancelled factor, and plot open points at each "hole."
• Locate the intercepts of the graph, and determine whether the graph crosses or touches the x-axis at each x-intercept.
• Determine the vertical asymptotes. Graph each one using a dashed line.
• Determine the horizontal or slant asymptote, if one exists. Graph it using a dashed line. Determine points, if any, where the graph intersects this asymptote, and plot them.
• Use the zeroes of both the numerator and denominator to divide the x-axis into intervals, and determine whether each interval is above or below the x-axis. Plot the points found.
• Use all of these results to draw the graph.
Graph each rational function.
R\par{x} = \cfrac{x}{x^2 + x - 2}
Step 1: Factor the numerator and denominator of R. Find the domain of R.
\eq{R\par{x} &= \frac{x}{x^2 + x - 2} \\[4pt] &= \frac{x}{\par{x - 1} \par{x + 2}}}
The domain is \curl{x \mid x \neq -2, x \neq 1}.
Step 2: Write R in lowest terms.
R is already in lowest terms.
Step 3: Locate the intercepts of the graph, and determine whether the graph crosses or touches the x-axis at each x-intercept.
The numerator tells us the x-intercepts: the only one is x = 0. Because the multiplicity of x is 1, the graph crosses the x-axis at this point.
Since f\par{0} = 0, the y-intercept is y = 0. Thus the only intercept is the origin.
Step 4: Determine the vertical asymptotes. Graph each one using a dashed line.
The lines x = -2 and x = 1 are vertical asymptotes.
Step 5: Determine the horizontal or slant asymptote, if one exists. Graph it using a dashed line. Determine points, if any, where the graph intersects this asymptote, and plot them.
The degree of the numerator is less than the degree of the denominator. Thus R is proper, so the line y = 0 is a horizontal asymptote.
We've already determined that the origin is the x-intercept, so the graph will cross the horizontal asymptote at the origin.
Step 6: Use the zeroes of both the numerator and denominator to divide the x-axis into intervals, and determine whether each interval is above or below the x-axis. Plot the points found.
Interval \par{-\infty,-2} \par{-2,0} \par{0,1} \par{1,\infty} x -3 -1 \cfrac{1}{2} 2 R\par{x} -\cfrac{3}{4} \cfrac{1}{2} -\cfrac{2}{5} \cfrac{1}{2} Above/Below Below Above Below Above Point \par{-3,-\cfrac{3}{4}} \par{-1,\cfrac{1}{2}} \par{\cfrac{1}{2},-\cfrac{2}{5}} \par{2,\cfrac{1}{2}}
Step 7: Use all of these results to draw the graph.
We need to see where the graph is approaching all of these asymptotes. Remember, the asymptotes act like fences, corralling the function. The function can never cross the vertical aymptotes, and we've already determined the only place it will cross the horizontal asymptote. So, the points we've already plotted tell us where the function is forced to go.
For the horizontal asymptote y = 0:
The graph lies below the x-axis when x < -2. We know this because we have the point \par{-3,-\cfrac{3}{4}} already plotted, and we know that the only place the function crosses the line y = 0 is where x = 0. So as x is going off into negative-infinity-land, it is approaching the x-axis from below. So we can draw a little arrow under the x-axis off on the left of the graph.
The graph lies above the x-axis when x > 1, so as x gets large it approaches the x-axis from above. We'll draw a little arrow above the x-axis off to the right.
For the vertical asymptote x = -2:
When x < -2, the graph lies below the x-axis. So, as x \rightarrow -2 from the left, the graph must be heading down in the negative direction. So we'll draw an arrow pointing down at the bottom of the graph, to the left of x = -2.
Now, when -2 < x < 0, the graph is above the x-axis. Thus as x \rightarrow -2 from the right, the graph is going up in the positive direction. So we draw an arrow at the top of the graph, pointing upwards, to the right of x = -2.
For the vertical asymptote x = 1:
When 0 < x < 1, the graph is below the x-axis. So a down arrow gets drawn at the bottom, to the left of x = 1.
When x > 1, the graph is above the x-axis, so an up arrow gets drawn to the right of x = 1, at the top of the graph.
Now it's just a matter of connecting our dots and arrows.
R\par{x} = \cfrac{x^2 + 3x - 10}{x^2 + 8x + 15}
Step 1: Factor the numerator and denominator of R. Find the domain of R.
\eq{R\par{x} &= \frac{x}{x^2 + x - 2} \\[4pt] &= \frac{\par{x + 5} \par{x - 2}}{\par{x + 5} \par{x + 3}}}
The domain is \curl{x \mid x \neq -5, x \neq 3}.
Step 2: Write R in lowest terms.
We have a common factor on the top and bottom this time, which cancel out. So in lowest terms,
R\par{x} = \cfrac{x - 2}{x + 3}, x \neq -5
This means we'll have a hole in the graph where x = -5. To find the y-value, we need to plug x = -5 into the lowest-terms version of the function, even though we aren't technically allowed to do so.
\eq{R\par{-5} &= \frac{-5 - 2}{-5 + 3} \\[4pt] &= \frac{-7}{-2} \\[4pt] &= \frac{7}{2}}
Thus the hole is at the point \par{-5,\cfrac{7}{2}}.
Step 3: Locate the intercepts of the graph, and determine whether the graph crosses or touches the x-axis at each x-intercept.
The numerator tells us the only x-intercept is x = 2, and because the multiplicity of x is 1, the graph crosses the x-axis at this point.
Since f\par{0} = -\cfrac{2}{3}, the y-intercept is y = -\cfrac{2}{3}.
Step 4: Determine the vertical asymptotes. Graph each one using a dashed line.
The only one is the line x = -3.
Step 5: Determine the horizontal or slant asymptote, if one exists. Graph it using a dashed line. Determine points, if any, where the graph intersects this asymptote, and plot them.
The degree of the numerator is the same as the degree of the denominator. Thus R is improper, so it has a horizontal asymptote, which we can find by dividing the dominant terms.
\cfrac{x}{x} = 1
The horizontal asymptote is the line y = 1.
To see if and/or where the graph crosses the horizontal asymptote, we need to set the function equal to the value of the horiztonal asymptote.
\eq{\frac{x - 2}{x + 3} &= 1 \\[4pt] x - 2 &= x + 3 \\[4pt] -2 &\neq 3}
Since this equation has no solution, the graph doesn't cross the horizontal asymptote at all.
Step 6: Use the zeroes of both the numerator and denominator to divide the x-axis into intervals, and determine whether each interval is above or below the x-axis. Plot the points found.
In this case, we can already see where the function must go, since it doesn't cross any of the asymptotes, so we don't really need a table. We can just move on to the next step.
Step 7: Use all of these results to draw the graph.
For the horizontal asymptote y = 0:
The graph lies below the x-axis when x \lt -3, as we can see from the point \par{-5,\cfrac{7}{2}}. We can draw a little arrow above the x-axis off on the left of the graph.
The graph lies above the x-axis when x \gt -3, so as x gets large it approaches the x-axis from below. We'll draw a little arrow below the x-axis off to the right.
For the vertical asymptote x = -3:
Again, the points we've got show us where the arrows need to go.
Now we hook it all up, and...
R\par{x} = \cfrac{x^3 + 4x^2 + 4x}{x^2 + 2x - 3}
Step 1: Factor the numerator and denominator of R. Find the domain of R.
\eq{R\par{x} &= \frac{x^3 + 4x^2 + 4x}{x^2 + 2x - 3} \\[4pt] &= \frac{x \par{x + 2}^2}{\par{x + 3} \par{x - 1}}}
The domain is \curl{x \mid x \neq -3, x \neq 1}.
Step 2: Write R in lowest terms.
R is already in lowest terms.
Step 3: Locate the intercepts of the graph, and determine whether the graph crosses or touches the x-axis at each x-intercept.
The x-intercepts are x = -2 multiplicity 2 and x = 0. Thus the graph will touch the x-axis at x = -2, and cross the x-axis at x = 0 (which is also the y-intercept, since f\par{0} = 0).
Step 4: Determine the vertical asymptotes. Graph each one using a dashed line.
The lines x = -3 and x = 1 are vertical asymptotes.
Step 5: Determine the horizontal or slant asymptote, if one exists. Graph it using a dashed line. Determine points, if any, where the graph intersects this asymptote, and plot them.
The degree of the numerator is one more than the degree of the denominator. Thus R is improper, and there will be an slant asymptote.
\eq{ \begin{array}{rl} & \begin{array}{rrrr} \phantom{\big)} & \phantom{-(} & x\phantom{^3} & + & 2 \end{array} \\ x^2 + 2x - 3 & \overline{\begin{array}{rrrrrrrrr} \big) & \phantom{-(} & x^3 & + & 4x^2 & + & 4x & \end{array}} \\ & \underline{\begin{array}{rrrrrrr} \phantom{\big)} & -( & x^3 & + & 2x^2 & - & 3x & ) & \end{array}} \\ & \begin{array}{rrrrrrr} \phantom{\big)} & \phantom{-(} & \phantom{x^3} & \phantom{+} & 2x^2 & + & 7x & \end{array} \\ & \underline{\begin{array}{rrrrrrr} \phantom{\big)} & \phantom{x^3} & \phantom{+} & -( & 2x^2 & + & 4x & - & 6 & ) & \end{array}} \\ & \begin{array}{rrrrrrrrr} \phantom{\big)} & \phantom{-(} & \phantom{x^3} & \phantom{+} & \phantom{2x^2} & \phantom{+} & 3x & + & 6 & \end{array} \\ \end{array} }
Thus we have
\eq{R\par{x} &= \frac{x^3 + 4x^2 + 4x}{x^2 + 2x - 3} \\[4pt] &= x + 2 + \frac{3x + 6}{x^2 + 2x - 3}}
The slant asymptote is y = x + 2. To see if/where the graph will cross this asymptote, we solve R\par{x} = x + 2.
\eq{x + 2 &= \frac{x^3 + 4x^2 + 4x}{x^2 + 2x - 3} \\[4pt] \par{x + 2} \par{x^2 + 2x - 3} &= x^3 + 4x^2 + 4x \\[4pt] x^3 + 4x^2 + x - 6 &= x^3 + 4x^2 + 4x \\[4pt] x - 6 &= 4x \\[4pt] -6 &= 3x \\[4pt] x &= -2}
The graph will cross the asymptote at x = -2, which happens to be one of the zeroes.
Step 6: Use the zeroes of both the numerator and denominator to divide the x-axis into intervals, and determine whether each interval is above or below the x-axis. Plot the points found.
Interval \par{-\infty,-3} \par{-3,-2} \par{-2,0} \par{0,1} \par{1,\infty} x -4 -\cfrac{5}{2} -1 \cfrac{1}{2} 2 R\par{x} -\cfrac{16}{5} \cfrac{5}{14} \cfrac{1}{4} -\cfrac{25}{14} \cfrac{32}{5} Above/Below Below Above Above Below Above Point \par{-4,-\cfrac{16}{5}} \par{-\cfrac{5}{2},\cfrac{5}{14}} \par{-1,\cfrac{1}{4}} \par{\cfrac{1}{2},-\cfrac{25}{14}} \par{2,\cfrac{32}{5}}
Step 7: Use all of these results to draw the graph.
For the vertical asymptote x = -3:
When x < -3, the graph lies below the x-axis. We'll draw a down arrow at the bottom of the graph, to the left of x = -3.
When -1 < x < 0, the graph is above the x-axis. We'll draw an up arrow at the top of the graph, to the right of x = -3.
For the vertical asymptote x = 1:
When 0 < x < 1, the graph is below the x-axis. So a down arrow gets drawn at the bottom, to the left of x = 1.
When x > 1, the graph is above the x-axis, so an up arrow gets drawn to the right of x = 1, at the top of the graph.
For the slant asymptote y = x + 2:
This is a little tougher to figure out. As x gets huge, the function will behave like the line y = x + 2. We know the graph crosses the slant asymptote in one place, between the two vertical asymptotes.
So to the left of x = -3, the graph is either entirely above or entirely below the slant asymptote. From the table in Step 6, we have the point \par{-4,-\cfrac{16}{5}}, which tells us the graph is below the asymptote. Thus the graph will approach the asymptote from the left.
Likewise, the point \par{2,\cfrac{32}{5}} tells us the portion of the graph to the right of the line x = 1 is above the slant asymptote, and approaches it from the left.
We hook everything up...
R\par{x} = \cfrac{2 - x}{\par{x - 1}^2}
Step 1: Factor the numerator and denominator of R. Find the domain of R.
We can't factor this any further. The domain is \curl{x \mid x \neq 1}.
Step 2: Write R in lowest terms.
R is already in lowest terms.
Step 3: Locate the intercepts of the graph, and determine whether the graph crosses or touches the x-axis at each x-intercept.
The only x-intercept is x = 2, with multiplicity 1. The graph crosses the x-axis here.
f\par{0} = 2, thus the y-intercept is y = 2.
Step 4: Determine the vertical asymptotes. Graph each one using a dashed line.
The line x = 1 is the only vertical asymptote.
Step 5: Determine the horizontal or slant asymptote, if one exists. Graph it using a dashed line. Determine points, if any, where the graph intersects this asymptote, and plot them.
The degree of the numerator is less than the degree of the denominator. Thus R is proper, so the line y = 0 is a horizontal asymptote.
Since the horizontal asymptote is the x-axis, and we've already found the x-intercept, we know the graph crosses the horizontal asymptote at x = 2.
Step 6: Use the zeroes of both the numerator and denominator to divide the x-axis into intervals, and determine whether each interval is above or below the x-axis. Plot the points found.
Interval \par{-\infty,1} \par{1,2} \par{2,\infty} x 0 \cfrac{3}{2} 3 R\par{x} 2 2 -\cfrac{1}{4} Above/Below Above Above Below Point \par{0,2} \par{\cfrac{3}{2},2} \par{3,-\cfrac{1}{4}}
Step 7: Use all of these results to draw the graph.
For the horizontal asymptote y = 0:
The graph lies above the x-axis when x < 1, so as x is going off towards negative infinity, it is approaching the x-axis from above. An arrow goes under the x-axis off on the left of the graph.
The graph lies below the x-axis when x > 1, so as x gets big it approaches the x-axis from below. So we put an arrow below the x-axis off to the right.
For the vertical asymptote x = 1:
When x < 1, the graph is above the x-axis. So an up arrow gets drawn at the top, to the left of x = 1.
When 1 < x < 2, the graph is above the x-axis, so an up arrow gets drawn to the right of x = 1 at the top of the graph as well.
And finally, we make a pretty picture out of all of this.
The thing to note here is that this function has a repeated asymptote with multiplicity 2.
Compare this graph to the previous examples. In those, each vertical asymptote had the graph approaching it in opposite directions - for example, the graph would approach from the left going up into infinity, and approach from the right going down into negative infinity.
With this function, the graph approaches the vertical asymptote from both sides, heading off in the same direction, positive infinity. That's what even multiplicity for asymptotes means.
Thus, if the multiplicity of a vertical asymptote is odd, then the graph approaches said asymptote on either side, going in opposite directions. If the multiplicity of an asymptote is even, the graph approaches the asymptote from either side, going in the same direction.
Find a rational function that might have the graph shown above.
There's no reason to get freaked out about this. Identifying the zeroes and asymptotes will give us everything we need to come up with a function.
First of all, there is a horizontal asymptote, so we know the degrees of the numerator and the denominator must be the same. There are two vertical asymptotes, and one zero with multiplicity 2. So there we go: degree 2 in the numerator, and degree 2 in the denominator.
The zero is x = 2, so the numerator contains \par{x - 2}^2. The vertical asymptotes are x = -4 and x = 4, so the denominator is \par{x + 4} \par{x - 4}.
One more thing, though. The horizontal asymptote is y = 3, so it must be that the quotient of the x^2 terms, from the numerator and denominator, is 3. That is,
\cfrac{3x^2}{x^2} = 3
Therefore, the numerator contains 3 as well. And so the function is
R\par{x} = \cfrac{3 \par{x - 2}^2}{\par{x + 4} \par{x - 4}}
There are other possibilities, but this is the simplest one. |
# Could you use a line graph to show data about how body mass changes with height explain?
## Could you use a line graph to show data about how body mass changes with height explain?
Yes. We can use a line graph to depict how body mass changes with height. It may not always be same correlation. For example, a tall person can be lighter than a short stout person and vice versa.
Does a circle graph show how data changes?
Circle graphs are often used to show how a whole set of data is broken down into individual components. Because circle graphs relate individual parts and a whole, they are often used for budgets and other financial purposes.
### What type of data does a circle graph show?
A pie chart, sometimes called a circle chart, is a way of summarizing a set of nominal data or displaying the different values of a given variable (e.g. percentage distribution). This type of chart is a circle divided into a series of segments. Each segment represents a particular category.
Could you use a line graph to show data about how body mass the responding variable changes with height the manipulated variable )? Explain?
To organize data so other scientists can read it easily. A graph that represents parts of a whole. Could you use a circle graph to show data about how body mass changes with height? No, you would use a line graph for that.
#### What can a line graph tell you about the relationship between the variables in an experiment?
What can a line graph tell you about the relationship between the variables in an experiment? Line graphs can show how one variable (the responding variable) can change in response to another variable (the manipulated variable).
How do you show data in a circle graph?
Finding a percentage of a total amount in a circle graph
1. Circle graphs: A circle is divided into smaller portions.
2. To make a circle graph form the data in the table above.
3. Step 1: Add up all the values in the table.
4. Step 2: Next divide each value by the total and multiply by 100 to get a percent.
## Whats a circle graph used for?
A circle graph, also commonly referred to as a pie chart (sound familiar?) is a simple and visually appealing chart divided into wedges, each of which represents a data value. It’s one of the most commonly used graphs for displaying statistics, so we certainly can’t take its popularity for granted.
Could you use a line graph to show data about how body mass?
One of the limitations of this graph is that it does not show changes over time Therefore, when showing how a body mass changes with height, it is recommended to use a line graph to track and show changes in variables.
Line graphs are useful in that they show data variables and trends very clearly and can help to make predictions about the results of data not yet recorded. They can also be used to display several dependent variables against one independent variable.
When would you use a circle graph instead of a bar graph?
Circle graphs are most useful when comparing parts of a whole or total. Bar graphs also make comparisons easily. Unlike most circle graphs, bar graphs compare exact amounts. Circle graphs are used when dealing with percentages, and the percentages of the pieces add up to 100 percent.
#### What data are best displayed on a circle graph on a bar graph?
Circle graphs are best used to compare the parts of a whole. The circle graph above shows the entire amount sold. It also shows each brand’s sales as part of that whole. The circle graph uses the total of all items in the table.
What are the advantages of using a circle graph? |
## Solve the following equations. Make sure you show each step of the work including any Distributive Property, Inverse operations and co
Question
Solve the following equations. Make sure you show each step of the work including any Distributive Property,
Inverse operations and combining/collecting of like terms that may be needed. Remember that what you do to
one side you must do to the other when using inverse equations. Label your response with the same letter of the
problem AND write the original problem first before you begin to solve.
A) 1d + 12 = 14-20
B) 4(20+5) = 10r
C) -2(5 + x) – 1 = 3(x + 3)
in progress 0
5 months 2021-08-30T07:20:37+00:00 1 Answers 3 views 0
A) -18 B) r=10 C) x=-4
Step-by-step explanation:
A) 1d + 12 = 14-20
Step 1: Simplify both sides of the equation.
1d+12=14−20
d+12=14+−20
d+12=(14+−20)(Combine Like Terms)
d+12=−6
d+12=−6
Step 2: Subtract 12 from both sides.
d+12−12=−6−12
d=−18
d=−18
B) 4(20+5) = 10r
Step 1: Simplify both sides of the equation.
100=10r
Step 2: Flip the equation.
10r=100
Step 3: Divide both sides by 10.
r=10
C) -2(5 + x) – 1 = 3(x + 3)
Step 1: Simplify both sides of the equation.
−2(5+x)−1=3(x+3)
(−2)(5)+(−2)(x)+−1=(3)(x)+(3)(3)(Distribute)
−10+−2x+−1=3x+9
(−2x)+(−10+−1)=3x+9(Combine Like Terms)
−2x+−11=3x+9
−2x−11=3x+9
Step 2: Subtract 3x from both sides.
−2x−11−3x=3x+9−3x
−5x−11=9
Step 3: Add 11 to both sides.
−5x−11+11=9+11
−5x=20
Step 4: Divide both sides by -5.
x=−4 |
New interactive curriculum plans and website
# Unit Overview: Reasoning with large whole numbers
## Lessons:
15 lessons
### Identifying the place value of digits in 5-digit numbers
In this lesson, we will be representing 5-digit numbers pictorially and identifying the value of each digit within these numbers.
• 1 Quiz
• 35m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Comparing 5-digit numbers
In this lesson, we will be learning how to compare and order 5-digit numbers using number lines and place value charts.
• 2 Quizzes
• 30m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Ordering and comparing 5-digit numbers using a number line
In this lesson, we will be identifying the intervals on incomplete number lines and placing 5-digit numbers on number lines with different scales.
• 2 Quizzes
• 25m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Rounding 5-digit numbers to the nearest 10 000 and 1000
In this lesson, we will be using number lines to round 5-digit numbers to the nearest multiple of 10 000 and the nearest multiple of 1000.
• 2 Quizzes
• 16m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Rounding 5-digit numbers to the nearest 100, 1000 and 10 000
In this lesson, we will be using number lines to round 5-digit numbers to the nearest multiple of 100, 1000 and 10 000. We will also investigate rounding in the context of word problems.
• 2 Quizzes
• 20m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Identifying the place value of the digits in 6-digit numbers
In this lesson, we will be representing 6-digit numbers pictorially using place value counters and Dienes. We will also learn how to partition 6-digit numbers.
• 2 Quizzes
• 25m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Comparing 6-digit numbers using inequalities
In this lesson, we will use place value charts to identify the value of digits in 6-digit numbers. We will also add inequalities to equations with 6-digit numbers.
• 2 Quizzes
• 19m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Ordering and comparing 6-digit numbers using number lines
In this lesson, we will be identifying the intervals on incomplete number lines and placing 6-digit numbers on number lines with different scales.
• 2 Quizzes
• 19m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Rounding 6-digit numbers to the nearest 100 000 and 10 000
In this lesson, we will be using number lines to round 6-digit numbers to the nearest multiple of 100 000 and 10 000.
• 2 Quizzes
• 17m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Rounding 6-digit numbers to the nearest 1000, 10 000 and 100 000
In this lesson, we will be using number lines to round 6-digit numbers to the nearest multiple of 1000, 10 000 and 100 000.
• 2 Quizzes
• 17m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Solving problems involving rounding
In this lesson, we will use knowledge of rounding to the nearest 1000, 10 000 and 100 000 to solve problems involving rounding.
• 2 Quizzes
• 27m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Solving problems involving place value and rounding
In this lesson, we will be applying our knowledge of place value and rounding to different problems using these strategies.
• 2 Quizzes
• 26m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Investigating Roman Numerals up to 100
In this lesson, we will be identifying the way to write the corresponding Roman numerals for values between 1 and 100.
• 2 Quizzes
• 21m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Investigating Roman Numerals up to 1000
In this lesson, we will be identifying the way to write the corresponding Roman numerals for values between 1 and 1000.
• 2 Quizzes
• 18m Video
• Presentation(PPT)
• Worksheet
• Transcript
### Solving problems involving Roman Numerals
In this lesson, we will solve problems involving Roman numerals. Our focus will be on, missing values in equations and correcting Roman numeral errors.
• 2 Quizzes
• 19m Video
• Presentation(PPT)
• Worksheet
• Transcript |
Question Video: Finding the Unknown Component of a Vector Parallel to Another Vector | Nagwa Question Video: Finding the Unknown Component of a Vector Parallel to Another Vector | Nagwa
# Question Video: Finding the Unknown Component of a Vector Parallel to Another Vector Mathematics • First Year of Secondary School
## Join Nagwa Classes
Given that ๐ = โ๐ข โ 2๐ฃ and ๐ = ๐๐ข โ 8๐ฃ and ๐ โฅ ๐, where ๐ข and ๐ฃ are two perpendicular unit vectors, find the value of ๐.
02:56
### Video Transcript
Given that the vector ๐ is equal to negative ๐ข minus two ๐ฃ and the vector ๐ is equal to ๐๐ข minus eight ๐ฃ and the vector ๐ is parallel to the vector ๐, where ๐ข and ๐ฃ are two perpendicular unit vectors, find the value of the scalar ๐.
In this question, weโre given some information about two vectors, the vector ๐ and the vector ๐. Weโre given these vectors in terms of two perpendicular unit directional vectors, ๐ข and ๐ฃ. Weโre also told that the vector ๐ is parallel to the vector ๐. We need to use this to determine the value of the scalar ๐.
To do this, letโs start by recalling what it means for two vectors to be parallel. We say that two vectors are parallel if theyโre scalar multiples of each other. In other words, weโll say the vectors ๐ฎ and ๐ฏ are parallel if there exists some scalar ๐ such that ๐ฎ is equal to ๐ times ๐ฏ. Since weโre told that the vector ๐ is parallel to the vector ๐, there must exist some scalar ๐ such that ๐ is equal to ๐ times ๐. To help us find the value of this scalar ๐, we can substitute the expressions weโre given for vectors ๐ and ๐ in terms of the unit directional vectors ๐ข and ๐ฃ.
We get negative ๐ข minus two ๐ฃ will be equal to ๐ times ๐๐ข minus eight ๐ฃ. We can then simplify the right-hand side of this equation by remembering to multiply a vector by a scalar, we multiply all of the components of our vector by our scalar, where we remember the components of our vector will be the coefficients of the unit directional vectors ๐ข and ๐ฃ. This gives us that negative ๐ข minus two ๐ฃ will be equal to ๐๐๐ข minus eight ๐๐ฃ. Remember, for two vectors to be equal, their components must be equal. So we can compare the components of ๐ฃ on both sides of our equation. This then gives us the equation negative two must be equal to negative eight ๐, which we can solve for ๐ by dividing through by negative eight.
This then gives us that ๐ is equal to negative two divided by negative eight, which simplifies to give us one-quarter. But weโre not yet done. Remember, the question wants us to find the value of ๐. We can substitute our value of ๐ back into our equation. This then gives us the equation negative ๐ข minus two ๐ฃ will be equal to ๐ over four ๐ข minus two ๐ฃ. Remember, for both of these vectors to be equal, their components must be equal. This means we can construct an equation for ๐ by equating the components of both vectors in the ๐ข-direction.
We get that negative one must be equal to ๐ divided by four. We can then solve this equation for ๐. We multiply both sides of our equation through by four, giving us that ๐ should be equal to negative four. Therefore, we were able to show if ๐ is the vector negative ๐ข minus two ๐ฃ and ๐ is the vector ๐๐ข minus eight ๐ฃ and the vector ๐ is parallel to the vector ๐, where ๐ข and ๐ฃ are two perpendicular unit vectors, then the value of ๐ must be equal to negative four.
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# Difference between revisions of "2014 AMC 10A Problems/Problem 22"
## Problem
In rectangle $ABCD$, $\overline{AB}=20$ and $\overline{BC}=10$. Let $E$ be a point on $\overline{CD}$ such that $\angle CBE=15^\circ$. What is $\overline{AE}$?
$\textbf{(A)}\ \dfrac{20\sqrt3}3\qquad\textbf{(B)}\ 10\sqrt3\qquad\textbf{(C)}\ 18\qquad\textbf{(D)}\ 11\sqrt3\qquad\textbf{(E)}\ 20$
## Solution 1 (Trigonometry)
Note that $\tan 15^\circ=2-\sqrt{3}=\frac{EC}{10} \Rightarrow EC=20-10 \sqrt 3$. (It is important to memorize the sin, cos, and tan values of $15^\circ$ and $75^\circ$.) Therefore, we have $DE=10\sqrt 3$. Since $ADE$ is a $30-60-90$ triangle, $AE=2 \cdot AD=2 \cdot 10=\boxed{\textbf{(E)} \: 20}$
## Solution 2 (No Trigonometry)
Let $F$ be a point on line $\overline{CD}$ such that points $C$ and $F$ are distinct and that $\angle EBF = 15^\circ$. By the angle bisector theorem, $\frac{BC}{BF} = \frac{CE}{EF}$. Since $\triangle BFC$ is a $30-60-90$ right triangle, $CF = \frac{10\sqrt{3}}{3}$ and $BF = \frac{20\sqrt{3}}{3}$. Additionally, $$CE + EF = CF = \frac{10\sqrt{3}}{3}$$Now, substituting in the obtained values, we get $\frac{10}{\frac{20\sqrt{3}}{3}} = \frac{CE}{EF} \Rightarrow \frac{2\sqrt{3}}{3}CE = EF$ and $CE + EF = \frac{10\sqrt{3}}{3}$. Substituting the first equation into the second yields $\frac{2\sqrt{3}}{3}CE + CE = \frac{10\sqrt{3}}{3} \Rightarrow CE = 20 - 10\sqrt{3}$, so $DE = 10\sqrt{3}$. Because $\triangle ADE$ is a $30-60-90$ triangle, $AE = \boxed{\textbf{(E)}~20}$.
~edited by ripkobe_745
## Solution 3 Quick Construction (No Trigonometry)
Reflect $\triangle{ECB}$ over line segment $\overline{CD}$. Let the point $F$ be the point where the right angle is of our newly reflected triangle. By subtracting $90 - (15+15) = 60$ to find $\angle ABF$, we see that $\triangle{ABC}$ is a $30-60-90$ right triangle. By using complementary angles once more, we can see that $\angle{EAD}$ is a $60^\circ$ angle, and we've found that $\triangle{EAD}$ is a $30-60-90$ right triangle. From here, we can use the $1-2-\sqrt{3}$ properties of a $30-60-90$ right triangle to see that $\overline{AE}=\boxed{\textbf{(E)}~20}.$
## Solution 4 (No Trigonometry)
Let $F$ be a point on $BC$ such that $\angle{FEC}=60^{\circ}$. Then $$\angle{BEF}=\angle{BEC}-\angle{FEC}=15^{\circ}$$ Since $\angle{BEF}=\angle{EBF}$, $\bigtriangleup{BFE}$ is isosceles.
Let $CF=x$. Since $\bigtriangleup{FEC}$ is $60^{\circ}-90^{\circ}-30^{\circ}$, we have $EF=\frac{2}{\sqrt{3}}x$
Since $\bigtriangleup{BFE}$ is isosceles, we have $BF=EF=\frac{2}{\sqrt{3}}x$. Since $BF+FC=BF$, we have $$\frac{2}{\sqrt{3}}x+x=10 \Longrightarrow x=20\sqrt{3}-30$$ Thus $EC=\frac{1}{\sqrt{3}}BC=20-10\sqrt{3}$ and $DE=DC-EC=20-EC=10\sqrt{3}$.
Finally, by the Pythagorean Theorem, we have $$AE=\sqrt{AD^2+DE^2}=\sqrt{10^2+(10\sqrt{3})^2}=20 \boxed{\mathrm{(E)}}$$
~ Solution by Nafer
~ Edited by TheBeast5520
Note from williamgolly: When you find DE, note how ADE is congruent to a 30-60-90 triangle and you can easily find AE from there
## Solution 5
First, divide all side lengths by $10$ to make things easier. We’ll multiply our answer by $10$ at the end. Call side length $BE$ $x$. Using the Pythagorean Theorem, we can get side $EC$ is $\sqrt{x^2-1}$.
The double angle identity for sine states that: $$\sin{2a} = 2 \sin{a}\cdot \cos{a}$$ So, $$\sin 30 = 2\sin 15\cdot \cos 15$$ We know $\sin 30 = \frac{1}{2}$. In triangle $BEC$, $\sin 15 = \frac{\sqrt{x^2-1}}{x}$ and $\cos 15 = \frac{1}{x}$. Substituting these in, we get our equation: $$\frac{1}{2} = 2 \cdot \frac{\sqrt{x^2-1}}{x} \cdot \frac{1}{x}$$ which simplifies to $$x^4-16x^2+16 = 0$$
Now, using the quadratic formula to solve for $x^2$. $$x^2 = 16 \pm \frac{\sqrt{16^2-4\cdot16}}{2} = 8 \pm 4\sqrt3$$ Because the length $BE$ must be close to one, the value of $x^2$ will be $8-4\sqrt3$. We can now find $EC$ = $\sqrt{x^2-1} = \sqrt{7-4\sqrt3} = 2-\sqrt3$ and use it to find $DE$. $DE = 2-EC = \sqrt3$. To find $AE$, we can use the Pythagorean Theorem with sides $AD$ and $DE$, OR we can notice that, based on the two side lengths we know, $ADE$ is a $30-60-90$ triangle. So $AE = 2\cdot AD = 2$.
Finally, we must multiply our answer by $10$, $2\cdot 10 = 20$. $\boxed{\textbf{(E)}}$.
~AWCHEN01
## Solution 6 (Pure Euclidian Geometry)
We are going to use pure Euclidian geometry to prove $AE=AB$.
Reflect rectangle $ABCD$ along line $CD$. Let the square be $ABFG$ as shown. Construct equilateral triangle $\triangle EFH$.
Because $HF=EF$, $GF=BF=20$, and $\angle GFH=\angle BFE=15^{\circ}$, $\triangle GFH\cong \triangle BFE$ by $SAS$.
So, $GH=BE$, $GH=HE=HF$.
Because $GH=HE=HF$, $\angle GHF= \angle BEF=75^{\circ} + 75^{\circ} = 150^{\circ}$, $\angle GHE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$, $\angle GHE=\angle GHF$.
$\triangle GHE \cong \triangle GHF$ by $SAS$.
So, $GF=GE$. By the reflection, $AE=GE=GF=AB$. $AE=AB=\boxed{\textbf{(E)}~20}$
This solution is inspired by AoPS "Introduction to Geometry" page 226 problem 8.22, and page 433 problem 16.42.
## Solution 7 (Pure Euclidian Geometry)
We are going to use pure Euclidian geometry to prove $AE=AB$.
Construct equilateral triangle $\triangle BEF$, and let $GF$ be the height of $\triangle ABF$.
$\angle GBF=90^{\circ}-15^{\circ}-60^{\circ}=15^{\circ}$, $\angle GBF=\angle CBE$, $\angle BGF=\angle BCE=90^{\circ}$, $BF=BE$.
$\triangle BGF \cong \triangle BCE$ by $AAS$.
$BG=BC=10, AG=20-10=10$, $AG=BG$, $GF=GF$, by $HL$ $\triangle AGF \cong \triangle BGF$.
So, $AF=BF=EF$.
$\angle AFB=75^{\circ}+75^{\circ}=150^{\circ}$, $\angle AFE=360^{\circ}-150^{\circ}-60^{\circ}=150^{\circ}$, $\angle AFB=\angle AFE$, $AF=AF$, $BF=EF$.
$\triangle AFB \cong \triangle AFE$ by $SAS$.
So, $AE=AB=\boxed{\textbf{(E)}~20}$
Note: Similar to previous Solution
## Solution 8 (Trigonometry)
All trigonometric functions in this solution are in degrees. We know $$\sin\left(a+b\right)=\sin\left(a\right)\cos\left(b\right)+\sin\left(b\right)\cos\left(a\right)$$ so $$\sin\left(15\right)=\sin\left(45-30\right)=\sin\left(45\right)\cos\left(-30\right)+\sin\left(-30\right)\cos\left(45\right)$$ $$=\frac{\sqrt{2}}{2}\cdot\left(-\frac{\sqrt{3}}{2}\right)+\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=\frac{-\sqrt{6}}{4}+\frac{\sqrt{2}}{4}=\frac{\sqrt{2}-\sqrt{6}}{4}$$ $$=\frac{\sqrt{2}-\sqrt{6}}{4}$$ Let $EC=x$, then $BE=\sqrt{x^{2}+100}$. By the definition of sine, $$\frac{x}{\sqrt{x^{2}+100}}=\frac{\sqrt{2}-\sqrt{6}}{4}$$ Squaring both sides, $$\frac{x^{2}}{x^{2}+100}=\frac{\left(\sqrt{2}-\sqrt{6}\right)^{2}}{16}=\frac{2-2\sqrt{12}+6}{16}=\frac{8-4\sqrt{3}}{16}=\frac{2-\sqrt{3}}{4}$$ Cross-multiplying, $$4x^{2}=\left(2-\sqrt{3}\right)\left(x^{2}+100\right)=2x^{2}+200-\sqrt{3}x^{2}-100\sqrt{3}$$ Simplifying, $$\left(2+\sqrt{3}\right)x^{2}=200-100\sqrt{3}$$ $$x^{2}=\frac{200-100\sqrt{3}}{2+\sqrt{3}}=\frac{100\left(2-\sqrt{3}\right)}{2+\sqrt{3}}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}$$ Let $\frac{2-\sqrt{3}}{2+\sqrt{3}}=p$. Notice that $\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)=2^{2}-\sqrt{3}^{2}=1$ so $2-\sqrt{3}=\frac{1}{2+\sqrt{3}}$. $p$ is then $$\frac{2-\sqrt{3}}{2+\sqrt{3}}=\frac{\frac{1}{2+\sqrt{3}}}{2+\sqrt{3}}=\frac{1}{\left(2+\sqrt{3}\right)^{2}}$$ Recall that $$x^{2}=100\cdot\frac{2-\sqrt{3}}{2+\sqrt{3}}$$ which we now know is $$100\cdot\frac{1}{\left(2+\sqrt{3}\right)^{2}}=\frac{100}{\left(2+\sqrt{3}\right)^{2}}=\left(\frac{10}{2+\sqrt{3}}\right)^{2}$$ Therefore $$x=\frac{10}{2+\sqrt{3}}$$ Rationalizing the denominator, $$\frac{10}{2+\sqrt{3}}\cdot\frac{2-\sqrt{3}}{2-\sqrt{3}}=\frac{20-10\sqrt{3}}{\left(2+\sqrt{3}\right)\left(2-\sqrt{3}\right)}$$ Which by difference of squares reduces to $$20-10\sqrt{3}$$ so $EC=20-10\sqrt{3}$. $ED$ is then $20-\left(20-10\sqrt{3}\right)=10\sqrt{3}$ and since we know $AD=10$, by the Pythagorean theorem, $AE = 20$. The answer is $\boxed{\textbf{(E)}~20}$
An alternate way to finish: since we know the lengths of $AD$ and $DE$, we can figure out that $m\angle AED=30^{\circ}$ and therefore $m\angle BEA=75^{\circ}$. Hence $\triangle ABE$ is isosceles and $AE=AB=\boxed{\textbf{(E)}~20}$.
~JH. L |
# AP Statistics Curriculum 2007 Distrib RV
(Difference between revisions)
Revision as of 16:55, 23 April 2008 (view source)IvoDinov (Talk | contribs)m (→PDF Example: typo)← Older edit Current revision as of 18:38, 18 March 2016 (view source)IvoDinov (Talk | contribs) (12 intermediate revisions not shown) Line 2: Line 2: === Random Variables=== === Random Variables=== - A '''random variable''' is a function or a mapping from a sample space into the real numbers (most of the time). In other words, a random variable assigns real values to outcomes of experiments. This mapping is called ''random'', as the output values of the mapping depend on the outcome of the experiment, which are indeed random. So, instead of studying the raw outcomes of experiments (e.g., define and compute probabilities), most of the time we study (or compute probabilities) on the corresponding random variables instead. The [http://en.wikipedia.org/wiki/Random_variable formal general definition of random variables may be found here]. + A '''random variable''' is a function or a mapping from a sample space into the real numbers (most of the time). In other words, a random variable assigns real values to outcomes of experiments. This mapping is called ''random'', as the output values of the mapping depend on the outcome of the experiment, which are indeed random. So, instead of studying the raw outcomes of experiments (e.g., define and compute probabilities), most of the time we study (or compute probabilities) the corresponding random variables instead. The [http://en.wikipedia.org/wiki/Random_variable formal general definition of random variables may be found here]. ===Examples of Random Variables=== ===Examples of Random Variables=== - * '''Die''': In rolling a regular hexagonal die, the sample space is clearly and numerically well-defined and in this case the random variable is the identity function assigning to each face of the die the numerical value it represents. This the possible outcomes of the RV of this experiment are { 1, 2, 3, 4, 5, 6 }. You can see this explicit RV mapping in the [[SOCR_EduMaterials_Activities_DiceExperiment | SOCR Die Experiment]]. + * '''Die''': In rolling a regular hexagonal die, the sample space is clearly and numerically well-defined. In this case, the random variable is the identity function assigning to each face of the die where the numerical value it represents. This the possible outcomes of the RV of this experiment are { 1, 2, 3, 4, 5, 6 }. You can see this explicit RV mapping in the [[SOCR_EduMaterials_Activities_DiceExperiment | SOCR Die Experiment]]. * '''Coin''': For a coin toss, a suitable space of possible outcomes is S={H, T} (for heads and tails). In this case these are not numerical values, so we can define a RV that maps these to numbers. For instance, we can define the RV $X: S \longrightarrow [0, 1]$ as: $X(s) = \begin{cases}0,& s = \texttt{H},\\ * '''Coin''': For a coin toss, a suitable space of possible outcomes is S={H, T} (for heads and tails). In this case these are not numerical values, so we can define a RV that maps these to numbers. For instance, we can define the RV [itex]X: S \longrightarrow [0, 1]$ as: $X(s) = \begin{cases}0,& s = \texttt{H},\\ 1,& s = \texttt{T}.\end{cases}$. You can see this explicit RV mapping of heads and tails to numbers in the [[SOCR_EduMaterials_Activities_BinomialCoinExperiment | SOCR Coin Experiment]]. 1,& s = \texttt{T}.\end{cases}[/itex]. You can see this explicit RV mapping of heads and tails to numbers in the [[SOCR_EduMaterials_Activities_BinomialCoinExperiment | SOCR Coin Experiment]]. - * '''Card''': Suppose we draw a [[SOCR_EduMaterials_Activities_CardExperiment | 5-card hand from a standard 52-card deck]] and we are interested in the probability that the hand contains at least one pair of cards with identical denomination. Then the sample space of this experiment is large - it should be difficult to list all possible outcomes. However, we can assign a random variable $X(s) = \begin{cases}0,& s = \texttt{no-pair},\\ + * '''Card''': Suppose we draw a [[SOCR_EduMaterials_Activities_CardExperiment | 5-card hand from a standard 52-card deck]] and we are interested in the probability that the hand contains at least one pair of cards with identical denomination. Since the sample space of this experiment is large, it should be difficult to list all possible outcomes. However, we can assign a random variable [itex]X(s) = \begin{cases}0,& s = \texttt{no-pair},\\ 1,& s = \texttt{at-least-1-pair}.\end{cases}$ and try to compute the probability of P(X=1), the chance that the hand contains a pair. You can see this explicit RV mapping and the calculations of this probability at the [[SOCR_EduMaterials_Activities_CardExperiment | SOCR Card Experiment]]. 1,& s = \texttt{at-least-1-pair}.\end{cases}[/itex] and try to compute the probability of P(X=1), the chance that the hand contains a pair. You can see this explicit RV mapping and the calculations of this probability at the [[SOCR_EduMaterials_Activities_CardExperiment | SOCR Card Experiment]]. + + * '''A Pair of Dice''': Suppose we roll a pair of dice and the random variable X represents their sum. Of course we could have chosen any function of the outcomes of the 2 dice, but the most common game-like situation is to look at the total sum as an outcome. The figure below explicitly defines the sample space and the RV mapping from he sample space (S) into the real numbers (R). +
[[Image:SOCR_EBook_Distrib_RV_Fig3.png|500px]]
===Probability density/mass and (cumulative) distribution functions=== ===Probability density/mass and (cumulative) distribution functions=== Line 40: Line 43: - The explanation of the [http://en.wikipedia.org/wiki/Benford's_law Benford's Law] may be summarized as follows: The distribution of the first digits must be independent of the measuring units used in observing/recording the integer measurements. For instance, this means that if we had observed lenght/distance in ''inches'' or ''centimeters'' (inches and centimeters are linearly dependent, $1in = 2.54cm$), the distribution of the first digit of the measurement must be identical. So, there are about three centimeters for each inch. Thus, the probability that the first digit of a length observation is ''1in'' must be the same as the probability that the first digit of a lenght in ''centimeters'' starts with either 2 or 3 (with standard round off). Similarly, for observations of ''2in'', need to have their centimeter counterparts either ''5cm'' or ''6cm''. Observations of ''3in'' will correspond to 7 or 8 centimeters, etc. In other words, this distribution must be [http://en.wikipedia.org/wiki/Scale_invariant scale invariant]. + The explanation of the [http://en.wikipedia.org/wiki/Benford's_law Benford's Law] may be summarized as follows: The distribution of the first digits must be independent of the measuring units used in observing/recording the integer measurements. For instance, this means that if we had observed length/distance in ''inches'' or ''centimeters'' (inches and centimeters are linearly dependent, $1in = 2.54cm$), the distribution of the first digit of the measurement must be identical. So, there are about three centimeters for each inch. Thus, the probability that the first digit of a length observation is ''1in'' must be the same as the probability that the first digit of a length in ''centimeters'' starts with either 2 or 3 (with standard round off). Similarly, for observations of ''2in'', they need to have their centimeter counterparts either ''5cm'' or ''6cm''. Observations of ''3in'' will correspond to 7 or 8 centimeters, etc. In other words, this distribution must be [http://en.wikipedia.org/wiki/Scale_invariant scale invariant]. Line 48: Line 51: There are 3 important quantities that we are always interested in when we study random processes. Each of these may be phrased in terms of RVs, which simplifies their calculations. There are 3 important quantities that we are always interested in when we study random processes. Each of these may be phrased in terms of RVs, which simplifies their calculations. - * '''Probability Density Function''' (PDF): What is the probability of $P(X=x_o)$? For instance, in the card example above, we may be interested in [[SOCR_EduMaterials_Activities_CardExperiment#Applications | P(at least 1 pair) = P(X=1) = P(1 pair only) = 0.422569]]. Or in the die example, we may want to know P(Even number turns up) = $P(X \in \{2, 4, 6 \}) = 0.5$. + * '''Probability Density Function''' (PDF): What is the probability of $P(X=x_o)$? For instance, in the card example above, we may be interested in [[SOCR_EduMaterials_Activities_CardExperiment#Applications | P(exactly 1 pair) = P(X=1) = P(1 pair only) = 0.422569]]. Or in the die example, we may want to know P(Even number turns up) = $P(X \in \{2, 4, 6 \}) = 0.5$. * '''Cumulative Distribution Function''' (CDF): [itex]P(X [[Image:SOCR_EBook_Dinov_DistributionRelations_031108_Fig1.jpg|500px]]
[[Image:SOCR_EBook_Dinov_DistributionRelations_031108_Fig1.jpg|500px]]
- : [http://socr.ucla.edu/htmls/SOCR_Distributome.html The SOCR Distributome applet provides an interactive graphical interface for exploring the relations between different distributions]. + : [http://distributome.org/ The SOCR Distributome applet provides an interactive graphical interface for exploring the relations between different distributions]. + + ===Generating Probability Tables=== + Once can use R (and many other programming languages) to generate probability tables like the [http://socr.umich.edu/Applets/index.html#Tables popular SOCR Probability Tables]. You can also use the [http://socr.ucla.edu/htmls/dist/ Java Applets] or the [http://www.distributome.org/V3/calc/ HTML5/JavaScript Webapps] for interactive + F-Distribution calculations and obtain more dense and accurate measures of probability or critical values. + + The following example generates one of the [http://socr.umich.edu/Applets/F_Table.html F distribution tables: $F(\alpha=0.001, df.num, df.deno)$]: + + # Define the right-tail probability of interest $\alpha=0.001$ + right_tail_p <- 0.001 + + # Define the vectors storing the indices corresponding to numerator (n1) and denominator (n2, row) + # degrees of freedom for $F(\alpha, n_1, n_2)$. Note that Inf corresponds to $\infty$. + + n1 <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 24, 30, 40, 60, 120, Inf) + n2 <- c(1:30, 40, 60, 120, Inf) + + # Define precision (4-decimal point accuracy) + options(digits=4) + + # Generate an empty matrix of critical f-values + f_table <- matrix(ncol=length(n1), nrow=length(n2)) + + # Use the The F Distribution quantile function to fill in the matrix values in a nested 2-loop + # Recall that the density (df), distribution function (pf), quantile function (qf) and random generation (rf) for the F distribution + + for (i in 1:length(n2)){ + for (j in 1:length(n1)){ + f_table[i,j] <- qf(right_tail_p, n1[j], n2[i], lower.tail = FALSE) + } + } + + # Print results + f_table + + # label rows and columns + rownames(f_table) <- n2; colnames(f_table) <- n1 + + # save results to a file + write.table(f_table, file="C:\\User\\f_table.txt") + + ===[[EBook_Problems_Distrib_RV|Problems]]===
## General Advance-Placement (AP) Statistics Curriculum - Random Variables and Probability Distributions
### Random Variables
A random variable is a function or a mapping from a sample space into the real numbers (most of the time). In other words, a random variable assigns real values to outcomes of experiments. This mapping is called random, as the output values of the mapping depend on the outcome of the experiment, which are indeed random. So, instead of studying the raw outcomes of experiments (e.g., define and compute probabilities), most of the time we study (or compute probabilities) the corresponding random variables instead. The formal general definition of random variables may be found here.
### Examples of Random Variables
• Die: In rolling a regular hexagonal die, the sample space is clearly and numerically well-defined. In this case, the random variable is the identity function assigning to each face of the die where the numerical value it represents. This the possible outcomes of the RV of this experiment are { 1, 2, 3, 4, 5, 6 }. You can see this explicit RV mapping in the SOCR Die Experiment.
• Coin: For a coin toss, a suitable space of possible outcomes is S={H, T} (for heads and tails). In this case these are not numerical values, so we can define a RV that maps these to numbers. For instance, we can define the RV $X: S \longrightarrow [0, 1]$ as: $X(s) = \begin{cases}0,& s = \texttt{H},\\ 1,& s = \texttt{T}.\end{cases}$. You can see this explicit RV mapping of heads and tails to numbers in the SOCR Coin Experiment.
• Card: Suppose we draw a 5-card hand from a standard 52-card deck and we are interested in the probability that the hand contains at least one pair of cards with identical denomination. Since the sample space of this experiment is large, it should be difficult to list all possible outcomes. However, we can assign a random variable $X(s) = \begin{cases}0,& s = \texttt{no-pair},\\ 1,& s = \texttt{at-least-1-pair}.\end{cases}$ and try to compute the probability of P(X=1), the chance that the hand contains a pair. You can see this explicit RV mapping and the calculations of this probability at the SOCR Card Experiment.
• A Pair of Dice: Suppose we roll a pair of dice and the random variable X represents their sum. Of course we could have chosen any function of the outcomes of the 2 dice, but the most common game-like situation is to look at the total sum as an outcome. The figure below explicitly defines the sample space and the RV mapping from he sample space (S) into the real numbers (R).
### Probability density/mass and (cumulative) distribution functions
• The probability density or probability mass function, for a continuous or discrete random variable, is the function defined by the probability of the subset of the sample space, $\{s \in S \} \subset S$, which is mapped by the random variable X to the real value x (i.e., X(s)=x):
$p(x) = P(\{s \in S \} | X(s) = x)$, for each x.
• The cumulative distribution function (cdf) F(x) of any random variable X with probability mass or density function p(x) is defined as the total probability of all $\{s \in S \} \subset S$, where $X(s)\leq x$:
$F(x)=P(X\leq x)= \begin{cases}{ \sum_{y: y\leq x} {p(y)}},& X = \texttt{Discrete-RV},\\ {\int_{-\infty}^{x} {p(y)dy}},& X = \texttt{continuous-RV}.\end{cases}$, for all x.
#### PDF Example
The Benford's Law states that the probability of the first digit (d) in a large number of integer observations ($d\not=0$) is given by
$P(d) = \log(d+1) - log(d) = \log{d+1 \over d}$, for $d = 1,2,\cdots,9.$
Note that this probability definition determines a discrete probability (mass) distribution:
$\sum_{d=1}^9{P(d)}=\log{2\over 1}+\log{3\over 2}+\log{4\over 3}+ \cdots +\log{10\over 9}=$ $\log({{2\over 1} {3\over 2} {4\over 3} \cdots{10\over 9}}) = \log{10\over 1} = 1$
d 1 2 3 4 5 6 7 8 9 P(d) 0.301 0.176 0.125 0.097 0.079 0.067 0.058 0.051 0.046
The explanation of the Benford's Law may be summarized as follows: The distribution of the first digits must be independent of the measuring units used in observing/recording the integer measurements. For instance, this means that if we had observed length/distance in inches or centimeters (inches and centimeters are linearly dependent, 1in = 2.54cm), the distribution of the first digit of the measurement must be identical. So, there are about three centimeters for each inch. Thus, the probability that the first digit of a length observation is 1in must be the same as the probability that the first digit of a length in centimeters starts with either 2 or 3 (with standard round off). Similarly, for observations of 2in, they need to have their centimeter counterparts either 5cm or 6cm. Observations of 3in will correspond to 7 or 8 centimeters, etc. In other words, this distribution must be scale invariant.
The only distribution that obeys this property is the one whose logarithm is uniformly distributed. In this case, the logarithms of the numbers are uniformly distributed -- $P(100\leq x \leq 1,000)$ = $P(2\leq \log(x)\leq 3)$ is the same as the probability $P(10,000\leq x \leq 100,000)$ = $P(4\leq \log(x)\leq 5)$. Examples of such exponentially growing numerical measurements are incomes, stock prices and computational power.
### How to Use RVs?
There are 3 important quantities that we are always interested in when we study random processes. Each of these may be phrased in terms of RVs, which simplifies their calculations.
• Probability Density Function (PDF): What is the probability of P(X = xo)? For instance, in the card example above, we may be interested in P(exactly 1 pair) = P(X=1) = P(1 pair only) = 0.422569. Or in the die example, we may want to know P(Even number turns up) = $P(X \in \{2, 4, 6 \}) = 0.5$.
• Cumulative Distribution Function (CDF): P(X < xo), for all xo. For instance, in the (fair) die example we have the following discrete density (mass) and cumulative distribution table:
x 1 2 3 4 5 6 PDFP(X = x) 1/6 1/6 1/6 1/6 1/6 1/6 CDF $P(X\leq x)$ 1/6 2/6 3/6 4/6 5/6 1
• Mean/Expected Value: Most natural processes may be characterized, via probability distribution of an appropriate RV, in terms of a small number of parameters. These parameters simplify the practical interpretation of the process or phenomena we study. For example, it is often enough to know what the process (or RV) average value is. This is the concept of expected value (or mean) of a random variable, denoted E[X]. The expected value is the point of gravitational balance of the distribution of the RV.
Obviously, we may define a large number of RV for the same process. When are two RVs equivalent is dependent on the definition of equivalence?
### Comparing Data and Model Distributions
To illustrate one example of using distributions for solving practical problems, we consider the large human weight and height dataset. You can use all 25,000 records or just the first 200 of these measurements to follow the protocol below:
• Copy the weight and height data into the data tab of any of the SOCR Charts (first clear the default data, then select column 1 heading, and click the Paste button). This allows you to manipulate each of the 3 data columns independently.
• Select (highlight with the mouse) one of the columns (e.g., weights or heights) in the SOCR Chart and click the Copy button. This stores only the data in the chosen column in your mouse buffer.
• Go to the SOCR Modeler and paste the data in the first column in the Data tab using the Paste button.
• Select the NormalFit_Modeler from the drop-down list on the top-left corner. This is the first model you will be fitting to your data.
• Select the 3 check-boxes (Estimate Parameters, Scale Up, and Raw Data).
• Go to the Graphs tab and adjust the 3 sliders on the top to get a clear view of your data distribution (sample histogram) and the model distribution function (solid red curve).
• The Results tab will contain the (data-driven) estimates of the parameters for this specific distribution model (in this case Normal).
• You can plug these parameters (mean and standard deviation) into the SOCR Normal Distribution Applet and make inference about your population based on this Normal distribution model.
• Validate that the probabilities of various interesting events (e.g., 68<=Height<70) computed via either using the sample histogram of the data or via the model distribution are very similar.
• Try fitting another distribution model to your data using the SOCR Modeler. For example, choose the mixture-of-Normals model (MixedFit_Modeler) and repeat this process. Can you identify possible gender effects in either height or weight of the subjects in your sample? If so, what are the Male and Female distribution models? Can these be used to predict the gender of subjects (based on their weight or height)?
• Note that the Results tab also shows some statistics quantifying how good your chosen distribution model is to approximate the (sample) data histogram.
### The Web of Distributions
There is a large number of families of distributions and distribution classification schemes.
The most common way to describe the universe of distributions is to partition them into categories. For example, Continuous Distributions and Discrete Distributions; marginal and joint distributions; finitely and infinitely supported, etc.
SOCR Distributome Project and the SOCR Distribution activities illustrate how to use technology to compute probabilities for events arising from many different processes.
The image below shows some of the relations between commonly used distributions. Many of these relations will be explored later.
The SOCR Distributome applet provides an interactive graphical interface for exploring the relations between different distributions.
### Generating Probability Tables
Once can use R (and many other programming languages) to generate probability tables like the popular SOCR Probability Tables. You can also use the Java Applets or the HTML5/JavaScript Webapps for interactive F-Distribution calculations and obtain more dense and accurate measures of probability or critical values.
The following example generates one of the F distribution tables: $F(\alpha=0.001, df.num, df.deno)$:
# Define the right-tail probability of interest $\alpha=0.001$
right_tail_p <- 0.001
# Define the vectors storing the indices corresponding to numerator (n1) and denominator (n2, row)
# degrees of freedom for $F(\alpha, n_1, n_2)$. Note that Inf corresponds to $\infty$.
n1 <- c(1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 15, 20, 24, 30, 40, 60, 120, Inf)
n2 <- c(1:30, 40, 60, 120, Inf)
# Define precision (4-decimal point accuracy)
options(digits=4)
# Generate an empty matrix of critical f-values
f_table <- matrix(ncol=length(n1), nrow=length(n2))
# Use the The F Distribution quantile function to fill in the matrix values in a nested 2-loop
# Recall that the density (df), distribution function (pf), quantile function (qf) and random generation (rf) for the F distribution
for (i in 1:length(n2)){
for (j in 1:length(n1)){
f_table[i,j] <- qf(right_tail_p, n1[j], n2[i], lower.tail = FALSE)
}
}
# Print results
f_table
# label rows and columns
rownames(f_table) <- n2; colnames(f_table) <- n1
# save results to a file
write.table(f_table, file="C:\\User\\f_table.txt") |
### Slide 1
```Geometry B
Chapter 10
Angle Relationships in Circles
Objectives
Find the measures of angles formed by lines
that intersect circles.
Use angle measures to solve problems.
Warm Up
1. Identify each line or segment that intersects
F.
chords: AE, CD
secant: AE
tangent: AB
Find each measure.
2. mNMP 110°
3. mNLP 55°
Example 1A: Using Tangent-Secant and
Tangent-Chord Angles
Find each measure.
mEFH
= 65°
Example 1B: Using Tangent-Secant and
Tangent-Chord Angles
Find each measure.
Find each measure.
mSTU
= 83
Find each measure.
Example 2: Finding Angle Measures Inside
a Circle
Find each measure.
mAEB
= 126
Find each angle measure.
mABD
Find each angle measure.
mRNM
mRNM = 180° – MNQ
mRNM = 180° – 158° = 22°
Example 3: Finding Measures Using
Tangents and Secants
Find the value of x.
= 40
= 63
Find the value of x.
50° = 83° – x
x = 33°
Example 4: Design Application
In the company logo shown,
and
= 12°. What is mFKH?
= 108°,
Two of the six muscles that control eye
movement are attached to the eyeball and
intersect behind the eye. If mAEB = 225, what
is mACB?
Example 5: Finding Arc Measures
Find
Step 1 Find
U
V
If a tangent and a secant intersect on a
at the pt. of tangency, then the
measure of the formed is half the
measure of its intercepted arc.
Substitute 180 – 113 for
mXVY and 68 for mWZ
Multiply both sides by 2.
Subtract 68 from both sides.
Example 5 Continued
U
Step 2 Find
V
Thm. 11-5-3
Substitute the given values.
Multiply both sides by 2.
Find mLP
Step 1 Find
Step 2 Find
Lesson Quiz
Find each measure.
1. mFGJ
2. mHJK
3. An observer watches
people riding a Ferris wheel
that has 12 equally spaced
cars. Find x.
Lesson Quiz
4. Find mCE.
``` |
# How to solve linear functions
There are a few different ways to solve linear functions. The most basic way is to use algebraic methods to solve for the variable. This can be done by using the addition, subtraction, multiplication, and division properties of equality.
## How can we solve linear functions
Another way to solve linear functions is to graph them on a coordinate plane. This can be done by plotting the points that correspond to the x and y values in the function and then drawing a line through those points. The point where the line intersects the
To solve linear functions, you need to find the slope and the y-intercept. To find the slope, you need to find the change in y over the change in x. To find the y-intercept, you need to find the point where the line crosses the y-axis.
, but the most common method is by using algebra. To do this, you need to first identify the equation's slope and y-intercept. Once you have this information, you can use it to plot the equation on a graph. From there, you can find the x- and y-coordinates of the points where the equation intersects the x- and y-axes. These coordinates will help you solve the equation.
Linear functions are equations that produce a straight line when graphed. In order to solve a linear function, one needs to determine the slope and y-intercept of the line. The slope is the rate of change of the line, and the y-intercept is the point where the line crosses the y-axis. Once these values are determined, one can use them to solve for any unknown variables in the equation.
To solve linear functions, there are a few steps that need to be followed. First, identify the slope and y-intercept of the line. Next, use these values to plot the line on a graph. Finally, find the x-intercept of the line, which will give the solution to the linear function.
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### NBT: Number Sense and Operations in Base Ten
#### NBT.A: Use place value understanding and properties of operations to perform multi-digit arithmetic.
NBT.A.1: Round whole numbers to the nearest 10 or 100.
NBT.A.2: Read, write and identify whole numbers within 100,000 using base ten numerals, number names and expanded form.
NBT.A.3: Demonstrate fluency with addition and subtraction within 1000.
### NF: Number Sense and Operations in Fractions
#### NF.A: Develop understanding of fractions as numbers.
NF.A.1: Understand a unit fraction as the quantity formed by one part when a whole is partitioned into equal parts.
NF.A.2: Understand that when a whole is partitioned equally, a fraction can be used to represent a portion of the whole.
NF.A.2.a: Describe the numerator as representing the number of pieces being considered.
NF.A.2.b: Describe the denominator as the number of pieces that make the whole.
NF.A.3: Represent fractions on a number line.
NF.A.3.a: Understand the whole is the interval from 0 to 1.
NF.A.3.b: Understand the whole is partitioned into equal parts.
NF.A.3.c: Understand a fraction represents the endpoint of the length a given number of partitions from 0.
NF.A.4: Demonstrate that two fractions are equivalent if they are the same size, or the same point on a number line.
NF.A.5: Recognize and generate equivalent fractions using visual models, and justify why the fractions are equivalent.
NF.A.6: Compare two fractions with the same numerator or denominator using the symbols >, = or <, and justify the solution.
NF.A.7: Explain why fraction comparisons are only valid when the two fractions refer to the same whole.
### RA: Relationships and Algebraic Thinking
#### RA.A: Represent and solve problems involving multiplication and division.
RA.A.1: Interpret products of whole numbers.
RA.A.2: Interpret quotients of whole numbers.
RA.A.3: Describe in words or drawings a problem that illustrates a multiplication or division situation.
RA.A.4: Use multiplication and division within 100 to solve problems.
RA.A.5: Determine the unknown number in a multiplication or division equation relating three whole numbers.
#### RA.C: Multiply and divide within 100.
RA.C.1: Multiply and divide with numbers and results within 100 using strategies such as the relationship between multiplication and division or properties of operations. Know all products of two one-digit numbers.
RA.C.2: Demonstrate fluency with products within 100.
#### RA.D: Use the four operations to solve word problems.
RA.D.1: Write and solve two-step problems involving variables using any of the four operations.
RA.D.2: Interpret the reasonableness of answers using mental computation and estimation strategies including rounding.
#### RA.E: Identify and explain arithmetic patterns.
RA.E.1: Identify arithmetic patterns and explain the patterns using properties of operations.
### GM: Geometry and Measurement
#### GM.A: Reason with shapes and their attributes.
GM.A.1: Understand that shapes in different categories may share attributes and that the shared attributes can define a larger category.
GM.A.2: Distinguish rhombuses and rectangles as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to these subcategories.
GM.A.3: Partition shapes into parts with equal areas, and express the area of each part as a unit fraction of the whole.
#### GM.B: Solve problems involving the measurement of time, liquid volumes and weights of objects.
GM.B.1: Tell and write time to the nearest minute.
GM.B.3: Solve problems involving addition and subtraction of minutes.
GM.B.4: Measure or estimate length, liquid volume and weight of objects.
GM.B.5: Use the four operations to solve problems involving lengths, liquid volumes or weights given in the same units.
#### GM.C: Understand concepts of area.
GM.C.1: Calculate area by using unit squares to cover a plane figure with no gaps or overlaps.
GM.C.3: Demonstrate that tiling a rectangle to find the area and multiplying the side lengths result in the same value.
GM.C.4: Multiply whole-number side lengths to solve problems involving the area of rectangles.
GM.C.5: Find rectangular arrangements that can be formed for a given area.
GM.C.6: Decompose a rectangle into smaller rectangles to find the area of the original rectangle.
#### GM.D: Understand concepts of perimeter.
GM.D.1: Solve problems involving perimeters of polygons.
GM.D.2: Understand that rectangles can have equal perimeters but different areas, or rectangles can have equal areas but different perimeters.
### DS: Data and Statistics
#### DS.A: Represent and analyze data.
DS.A.1: Create frequency tables, scaled picture graphs and bar graphs to represent a data set with several categories.
DS.A.2: Solve one- and two-step problems using information presented in bar and/or picture graphs.
DS.A.3: Create a line plot to represent data.
Correlation last revised: 9/16/2020
This correlation lists the recommended Gizmos for this state's curriculum standards. Click any Gizmo title below for more information. |
# How many ways are there for $2$ teams to win a best of $7$ series?
Case $1$: $4$ games: Team A wins first $4$ games, team B wins none = $\binom{4}{4}\binom{4}{0}$
Case $2$: $5$ games: Team A wins $4$ games, team B wins one = $\binom{5}{4}\binom{5}{1}-1$...minus $1$ for the possibility of team A winning the first four.
Case $3$: $6$ games: Team A wins $4$ games, team B wins $2$ = $\binom{6}{4}\binom{6}{2}-2$...minus $2$ for the possibility of team A winning the first four games; and the middle four (games $2,3,4,5$), in which case there would be no game $6$.
Case $4$: $7$ games: Team A wins $4$ games, team B wins $3$ = $\binom{7}{4}\binom{7}{3}-3$...minus $3$ for the possibility of team A winning the first four games; games $2,3,4,5$; and games $3,4,5,6$.
Total = sum of the $4$ cases multiplied by $2$ since the question is asking for $2$ teams.
Is this correct?
• It depends. How many of those teams are the Toronto Maple Leafs? Commented Nov 26, 2013 at 5:15
We count the ways in which Team A can win the series, and double the result. To count the ways A can win the series, we make a list like yours.
A wins in $4$: There is $1$ way this can happen.
A wins in $5$: A has to win $3$ of the first $4$, and then win. There are $\binom{4}{3}$ ways this can happen.
A wins in $6$: A has to win $3$ of the first $5$, then win. There are $\binom{5}{3}$ ways this can happen.
A wins in $7$: A has to win $3$ of the first $6$, then win. There are $\binom{6}{3}$ ways this can happen.
• I understand this approach, but am having trouble understanding why my approach is incorrect. I know its incorrect but can't understand why since it seems logical to me. Any help? Commented Nov 26, 2013 at 4:57
• Let's look at your calculation for $5$. Your chose the $4$ games A wins. Once you have done that, the game that B wins is determined. So there are $\binom{5}{4}$ ways only (and then we subtract $1$ as you did. If you wish, we could write $\binom{5}{4}\binom{1}{1}-1$. Commented Nov 26, 2013 at 5:01
• For $6$, there are two mistakes. First it should be $\binom{6}{4}$. Second, you need to subtract a lot more than $2$, for there are more than $2$ ways for A to have won in $4$ or $5$. You would want $\binom{6}{4}-1-4$. A similar subtracting process would work for the $7$ game case. Commented Nov 26, 2013 at 5:05
• For 6, what other cases would we subtract? Team A cannot win the first 4 in a row, nor games 2,3,4,5. What other combinations can it not win? I only see those two. Commented Nov 26, 2013 at 5:09
• Actually I see where I went wrong. Thank you. Another possibility that cannot happen is AABAAB. Commented Nov 26, 2013 at 5:10
There are $\begin{pmatrix} 7 \\ 4\end{pmatrix}$ ways of arranging $4$ W's and $3$ L's. This is the answer.
Note that the following sequences are essentially identical:
$$WWLWWLL$$ $$WWLWW$$
Why?
Because once the winning team has amassed four wins, it doesn't matter if we count the remaining un-played games as losses or not. Simply assume that they are. It doesn't change the equation.
For another approach, if the series ends before the seventh game, extend it to seven games by having the losing team win the rest. The series will now be four games to three. There are $2$ ways to choose the losing team, and ${7 \choose 3}$ ways to choose which game the losing team wins. The total is then $2{7 \choose 3}=70$
• This is equivalent to Emily's approach. Commented Mar 13, 2022 at 16:13
• @NeelSandell: That is true. It was also posted earlier. Commented Mar 13, 2022 at 17:14
Let the best of $n$ series be decided after $k$ games. This will happen if in the preceding $k-1$ games $A$ also wins $\lfloor n/2 \rfloor$ and wins the $k^{th}$ game. Hence, $$C(k) = \dbinom{k-1}{\lfloor n/2 \rfloor}$$ Hence, the total number of ways is $$\sum_{k=\lfloor n/2 \rfloor+1}^n C(k)$$ In your case, set $n=7$ to get the answer. |
# Question Video: Finding the Position Vector of a Given Point on this Line Corresponding to a Given Parameter Mathematics
Using point (4, 9, −9) and direction vector 〈−5, 1, −6〉, give the position vector 𝑟 of the point on this line corresponding to parameter 𝑡 = 9.
02:24
### Video Transcript
Using point four, nine, negative nine and direction vector negative five, one, negative six, give the position vector 𝑟 of the point on this line corresponding to parameter 𝑡 equals nine.
Let’s begin by identifying exactly what this question is asking us. We’re being asked to find a position vector 𝑟 of a point on a line defined by the point four, nine, negative nine and a direction vector negative five, one, negative six. Let’s begin by recalling the vector equation of a line. It’s 𝑟 equals 𝑟 naught plus 𝑡𝑑. In this equation, 𝑟 naught describes the vector that takes us from the origin to a point on this line. 𝑑 is the direction vector. And this is a little bit like the gradient or the slope in the formula for the Cartesian equation of a straight line. 𝑡 is just a real constant. So let’s begin by substituting what we know about our line into this formula.
The vector that gets us from the origin to a point on the line is given by four, nine, negative nine and the direction vector is stated in the question as negative five, one, negative six. And this means that the vector equation of our line is four, nine, negative nine plus 𝑡 times negative five, one, negative six. We’re looking to find the position vector of the point on a line that corresponds to the parameter 𝑡 equals nine. And so we substitute nine into our equation. And we obtain 𝑟 to be equal to four, nine, negative nine plus nine times negative five, one, negative six.
We can multiply this constant by the second vector by multiplying nine by each element; that’s by negative five, one, and negative six. And when we do, we obtain that the position vector 𝑟 is given by four, nine, and negative nine plus negative 45, nine, negative 54. We’re simply going to add each element in our vector. Four plus negative 45 is negative 41. Nine plus nine is 18. And negative nine plus negative 54 is negative 63. And we see that the position vector 𝑟 of the point on our line corresponding to parameter 𝑡 equals nine is negative 41, 18, negative 63. |
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# The instrument in the geometry box having the shape of a triangle is called a(a)Protractor(b)Compasses(c)Divider(d)Set-square
Last updated date: 12th Aug 2024
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Hint: We are required to identify the instruments we use in mathematics commonly found in a geometry box. We need to be aware of the basic instruments we use on a regular basis to help us draw better figures in mathematics than free hands. Here, four options are provided that are the names of four different instruments. We should know in advance what the geometry of each instrument is so that we are able to pick the correct option.
Complete step-by-step solution:
We first look at option (a). A Protractor is an angle measuring instrument. Say, we want to draw an angle of $36^{\circ}$. For that we need to first draw a line on which the angle will be made and then after using a protractor we can measure the angle in order to put the correct point so that we make the angle of accurate degree. A protractor has a shape like that of the capital alphabet D.
So, option (a) would be wrong. Since, the shape of a protractor is of a semicircle.
Now, we look at option (b). A compass is an instrument used to draw circles. We simply put a pencil on one end of the compass and keeping the other end as centre we can easily draw the circle of required radius. A compass is a 3D structure that has two long metal plates joined together with one edge pointed and the other has a pencil holder.
So, this option would also be discarded since it does not resemble the shape of a triangle.
We now look at the option (c). A Divider is an instrument used to measure accurate lengths. It looks similar to a compass but it doesn’t have any pencil holder for pencil. It is pointed at both ends hence it gives the most accurate result while measuring a length. A divider looks pretty much like a compass without the pencil holder.
So, this option is also discarded since it doesn’t look like a triangle.
We look at the last option now (d). A set-square is an instrument used to draw perpendicular lines or parallel lines. It looks like a triangle and it has some empty space as well in the middle which is again of the shape of a triangle.
So, we say that this is the right option.
Hence, option (d) is correct.
Note: Do not get confused between the instrument names. There are several instruments used in a geometry box, so it is often common to mix the names of them, so be aware of that. |
# Interior Angles of a Trapezoid – Formula and Examples
The interior angles of a trapezoid always measure 360°. We have three main types of trapezoids: isosceles trapezoid, scalene trapezoid, and right trapezoid. Depending on the type of trapezoid, we can use a different method to find all the measures of its internal angles.
Here, we will learn how to calculate the interior angle measures of various types of trapezoids using examples.
##### GEOMETRY
Relevant for
Learning about the interior angles of a trapezoid with examples.
See angles
##### GEOMETRY
Relevant for
Learning about the interior angles of a trapezoid with examples.
See angles
## Sum of interior angles of a trapezoid
A trapezoid is a quadrilateral that has one pair of opposite parallel sides. The opposite parallel sides are called the bases of the trapezoid and the other two sides are the lateral sides. The trapezoid is a closed figure that has four sides and four corners.
The sum of the four interior angles of a trapezoid is always equal to 360°. We can verify this using the following formula for interior angles of a polygon:
where, n is the number of sides of the polygon. Therefore, for a trapezoid, we have:
$latex (n-2)\times 180$°
$latex =(4-2)\times 180$°
$latex =(2)\times 180$°
$latex =360$°
Alternatively, we can also use the following diagram. We see that we can divide the trapezoid into two triangles, so the sum of the interior angles in a trapezoid is equal to the sum of the interior angles of two triangles.
## Calculate the interior angles of an isosceles trapezoid
An isosceles trapezoid is characterized by having its lateral sides of the same length. Furthermore, these trapezoids also have two pairs of equal interior angles. That means that to calculate the measure of an internal angle of an isosceles trapezoid, we need to have the measure of an angle of the trapezoid.
### EXAMPLE 1
Find the measures of all the interior angles of the following isosceles trapezoid.
Solution: Angles that have the same color and the same number of lines are equal. Therefore, we know that angle a measures 100°. Similarly, we know that angles b and c also have the same measure. Therefore, to find the missing angles, we have to add the known angles and subtract from 360°:
100°+100° = 200°
⇒ 360°-200° = 160°
Since the two angles are equal, we divide the result by 2 and we get 80°. This is the measure of angles b and c.
### EXAMPLE 2
Determine the measures of the missing angles in the isosceles trapezoid below.
Solution: We solve this by following a process similar to the previous example. Therefore, we know that angle c measures 70°.
Now, we find the measures of the other two angles by adding the known angles and subtracting from 360°:
70°+70° = 140°
⇒ 360°-140° = 220°
The two missing angles have the same measure, so we divide the result by 2 and get 110°. This is the measure of angles a and b.
## Calculate the interior angles of a scalene trapezoid
A scalene trapezoid is characterized by having all its sides of different lengths. This also means that all the internal angles of a scalene trapezoid are different from each other.
Therefore, to calculate the measure of an internal angle, we have to know the measures of the other three angles. We can also use the property that the adjacent interior angles of a trapezoid add up to 180°.
### EXAMPLE
Find the measure of the missing angle in the scalene trapezoid below.
Solution: To find the missing angle, we have to add all the known angles and subtract the result from 360°. Therefore, we have:
100°+80°+70° = 250°
⇒ 360°-250° = 110°
The measure of the missing angle is 110°. We can see that the adjacent angles add up to 180°. That is, we have 100°+80° = 180° and we also have 110°+70° = 180°.
## Calculate the interior angles of a right trapezoid
A right trapezoid has the main characteristic of having a pair of right angles, that is, a pair of 90° angles. One pair of sides of the trapezoid is parallel, so when we have a right angle, the angle of the other base must also be right.
We can find the measure of one angle in a right trapezoid if we know the measure of the other, different angles of the pair of right angles.
### EXAMPLE
What is the measure of angle a in the following right trapezoid?
Solution: We know that the angles represented by a small triangle are the right angles, that is, they are equal to 90°. Therefore, we find the missing angle measure by adding all of the known angle measures and subtracting from 360°. Therefore, we have:
90°+90°+100° = 280°
Therefore, the missing angle measures 80°. |
# Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2
Students can Download Maths Chapter 2 Measurements Ex 2.2 Questions and Answers, Notes Pdf, Samacheer Kalvi 7th Maths Book Solutions Guide Pdf helps you to revise the complete Tamilnadu State Board New Syllabus and score more marks in your examinations.
## Tamilnadu Samacheer Kalvi 7th Maths Solutions Term 2 Chapter 2 Measurements Ex 2.2
Question 1.
Find the area of the dining table whose diameter is 105 cm.
Solution:
Diameter of the dinig table (d) = 105 cm
∴ Radius r = $$\frac { d }{ 2 }$$ = $$\frac { 105 }{ 2 }$$ cm
Area of the circle = π r2 = $$\frac { 22 }{ 7 }$$ × $$\frac { 105 }{ 2 }$$ × $$\frac { 105 }{ 2 }$$ = 8662.5 sq.cm
Area of the dinning table = 8662.5 cm2
Question 2.
Calculate the area of the shotput ring whose diameter is 2.135 m.
Solution:
Radius of the shotput ring r = $$\frac { d }{ 2 }$$ = $$\frac { 2.135 }{ 2 }$$ m
Area of the circle = π r2
= $$\frac { 22 }{ 7 }$$ × $$\frac { 2.135 }{ 2 }$$ × $$\frac { 2.135 }{ 2 }$$
= $$\frac { 25.07 }{ 7 }$$ = 3.581 m2
∴ Area of the shotput ring = 3.581 m2
Question 3.
A sprinkler placed at the centre of a flower garden sprays water covering a circular area. If the area watered is 1386 cm2, find its radius and diameter.
Solution:
Area of the Circle = π r2 sq.units
Area of the circular portion watered = 1386 cm2
π r2 = 1386
$$\frac { 22 }{ 7 }$$ × r2 = 1386
r2 = 1386 × $$\frac { 7 }{ 22 }$$ = 63 × 7 = 9 × 7 × 7
r2 = 32 × 72
r = 3 × 7
Diameter (d) = 2 r = 2 × 21 cm
Diameter (d) = 42 cm
Question 4.
The circumference of a circular park is 352 m. Find the area of the park.
Solution:
Circumference of a Circle = 2 π r units
Given circumference of a circular park = 352 m
2 π r = 352
2 × $$\frac { 22 }{ 7 }$$ × r = 352
r = 352 × $$\frac { 7 }{ 22 }$$ × $$\frac { 1 }{ 2 }$$ = 56 m
Area of the park = π r2 = $$\frac { 22 }{ 7 }$$ × 56 × 56 sq.units
= 22 × 8 × 56 = 9856 m2
∴ Area of the Circular park = 9856 m2
Question 5.
In a grass land, a sheep is tethered by a rope of length 4.9 m. Find the maximum area that the sheep can graze.
Solution:
Length of the rope = 4.9 m
Area that the sheep can graze = Area of circle with radius 4.9m
Area of the circle = π r2 sq.units
= $$\frac { 22 }{ 7 }$$ × 4.9 × 4.9 = 22 × 0.7 × 4.9 = 75.46
∴ Area that the sheep can graze = 75.46 m2
Question 6.
Find the length of the rope by which a bull must be tethered in order that it may be able to graze an area of 2464 m2.
Solution:
If the bull is tethered by a rope then the area it can graze is a circular area of radius
= length of the rope
Area of the circle = 2464 m2
π r2 = 2464 m2
$$\frac { 22 }{ 7 }$$ × r2 = 2464
r2 = 2464 × $$\frac { 7 }{ 22 }$$ = 122 × 7 = 16 × 7 × 7
r2 = 42 × 72
r = 4 × 7 = 28 m
length of the rope r = 28 m
Question 7.
Lalitha wants to buy a round carpet of radius is 63 cm for her hall. Find the area that will be covered by the carpet.
Solution:
Radius of the round carpet = 63 cm
Area covered by the round carpet = πr2 sq units
A = $$\frac { 22 }{ 7 }$$ × 63 × 63 = 22 × 9 × 63 = 12474 cm2
Area covered by the round carpet = 12,474 cm2
Question 8.
Thenmozhi wants to level her circular flower garden whose diameter is 49 m at the rate of ₹150 per m2 Find the cost of levelling.
Solution:
Diamter of the circular garden d = 49 m
Radius r = $$\frac { d }{ 2 }$$ = $$\frac { 49 }{ 2 }$$ m
Area of the circular garden = πr2 sq units
= $$\frac { 22 }{ 7 }$$ × $$\frac { 49 }{ 2 }$$ × $$\frac { 49 }{ 2 }$$ m2 = 1,886.5 m2
Cost of levelling a m2 area = ₹ 150
∴ Cost of levelling 1886.5 m2 = ₹ 150 × 1886.5 = ₹ 2,82,975
Cost of levelling the flower garden = ₹ 2,82,975
Question 9.
The floor of the circular swimming pool whose radius is 7 m has to be cemented at the rate of ₹ 18 per m2. Find the total cost of cementing the floor.
Solution:
Radius of the circular swimming pool r = 7 m
Area of the circular swimming pool A = πr2 sq. units
= $$\frac { 22 }{ 7 }$$ × 7 × 7 m2 = 154 m2
Cost of cementing a m2 floor = ₹ 18.
Cost of cementing 154 m2 floor = ₹ 18 × 154 = ₹ 2,772
Objective Type Questions
Question 10.
The formula used to find the area of the circle is
(i) 47πr2
(ii) πr2
(iii) 2πr2
(iv) πr2 + 2r
(ii) πr2
Question 11.
The ratio of the area of a circle to the area of its semicircle is
(i) 2 : 1
(ii) 1 : 2
(iii) 4 : 1
(iv) 1 : 4 |
# How do you factor 60x^2 - 124xy + 63y^2?
May 28, 2015
Let's try a version of the AC Method:
$A = 60$, $B = 124$, $C = 63$
We are looking for a pair of factors of $A C = 60 \times 63$ which add up to $B = 124$.
$A C = 60 \cdot 63 = {2}^{2} \cdot {3}^{3} \cdot 5 \cdot 7$ has quite a few possible pairs of factors.
We can narrow down the search a bit: Since the sum is even, the factors of $2$ must be split between them.
So look for a pair of factors of ${3}^{3} \cdot 5 \cdot 7$ that add to $62$.
Well ${3}^{3} = 27$ and $5 \cdot 7 = 35$ work.
So the original pair we were looking for is $2 \cdot 27 = 54$ and $2 \cdot 35 = 70$
Use this pair to split the middle term, then factor by grouping:
$60 {x}^{2} - 124 x y + 63 {y}^{2}$
$= 60 {x}^{2} - 54 x y - 70 x y + 63 {y}^{2}$
$= \left(60 {x}^{2} - 54 x y\right) - \left(70 x y - 63 {y}^{2}\right)$
$= 6 x \left(10 x - 9 y\right) - 7 y \left(10 x - 9 y\right)$
$= \left(6 x - 7 y\right) \left(10 x - 9 y\right)$ |
New Zealand
Level 8 - NCEA Level 3
# Circles with translations
Lesson
We've looked at circles that are centred around the origin. The general form
$x^2+y^2=r^2$x2+y2=r2
But what happens when the centre of the circle is not at the origin?
## Equation of a Circle with any Centre
Circles are everywhere. From clocks to wheels to rings to coins to DVDs, you won't go through a day without seeing a circle. Given the great number of objects that resemble circles, it would be helpful to be able to graph them to study their properties. The following interactive allows you to explore the standard form equation of a circle. It shows how the equation changes as the coordinates of the centre ($h$h, $k$k) and the radius $r$r change. To move the circle, drag the sliders for $h$h and $k$k, the centre coordinates of the circle, while to change the radius, just drag the $r$r slider. You'll notice that, regardless of the values, the equation will always be in the form $\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2.
Equation of a Circle in Standard Form
$\left(x-h\right)^2+\left(y-k\right)^2=r^2$(xh)2+(yk)2=r2
where $\left(h,k\right)$(h,k) is the coordinates of the centre of the circle
and $r$r is the radius of the circle
Equation of a Circle in General Form
The standard form of a circle can also be rearranged and written in general form. The general form of a circle is:
$x^2+y^2+ax+by+c=0$x2+y2+ax+by+c=0
#### Worked Examples
##### Question 1
A circle has its centre at $\left(3,3\right)$(3,3) and a radius of $6$6 units.
a) Plot the graph for the given circle.
b) Write the equation of the circle.
##### Question 2
The equation of a circle is given by $\left(x+4\right)^2+\left(y-2\right)^2=25$(x+4)2+(y2)2=25.
a) Find the coordinates of the centre of this circle.
b) What is the radius of the circle?
c) Plot the graph for the given circle.
##### Question 3
Write down the equation of the new circle after $x^2+y^2=49$x2+y2=49 is translated:
a) $5$5 units upwards
b) $5$5 units downwards
c) $5$5 units to the right
d) $5$5 units to the left and $6$6 units upwards
### Outcomes
#### M8-1
Apply the geometry of conic sections
#### 91573
Apply the geometry of conic sections in solving problems |
# Sideband #71: Matrix Math
There are many tutorials and teachers, online and off, that can teach you how to work with matrices. This post is a quick reference for the basics. Matrix operations are important in quantum mechanics, so I thought a Sideband might have some value.
I’ll mention the technique I use when doing matrix multiplication by hand. It’s a simple way of writing it out that I find helps me keep things straight. It also makes it obvious if two matrices are compatible for multiplying (not all are).
One thing to keep in mind: It’s all just adding and multiplying!
Let’s start with the basics: A matrix is a rectangular bundle of numbers. Being rectangular, it has rows and columns, the number of each being the main characteristic of a matrix. Rows are always listed first, so a 3×1 matrix has three rows and one column, while a 2×2 matrix has two of each.
A matrix with the same number of rows and columns (such as a 2×2 matrix) is a square matrix, which is special.
Two other important special matrix types are column vectors, which have multiple rows, but just one column (their size is n×1); and row vectors, which have multiple columns, but just one row (size 1×n). In either case, the column or row of n numbers is treated as a vector.
The idea of number bundles isn’t new; the vectors just mentioned are bundles of numbers. Even a complex number is a number bundle; it has a real part and an imaginary part. (That it has two parts is what let us treat it as a 2D vector.)
Matrices have many uses in mathematics. They are a type of number, so, as with all numbers, there are operations on matrices that create new matrices or return others types of numeric values. For instance, a matrix can have an inner product, which is an operation that returns a single numeric value.
Specifically, we can add and multiply matrices, but there are some constraints.
§
Matrix addition requires that both matrices be the same size; that is, they must have the same number of rows and columns. Then, just as with vector addition, matrix addition is just a member-wise add that results in a sum matrix (also of the same size).
$A+B=\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}+ \begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}= \begin{bmatrix}a_{11}+b_{11}&a_{12}+b_{12}\\a_{21}+b_{21}&a_{22}+b_{22}\end{bmatrix}$
This generalizes to any number of rows and columns.
(A+B)+C = A + (B + C)
…and commutative…
A+B = B+A
§
Matrix multiplication is (notoriously) a little more complicated.
While, like addition, it is associative, it is not commutative:
A×BB×A
The order of multiplication matters and is a factor in the constraint. The rule is that the column count of the matrix on the left must match the row count of the matrix on the right.
The by-hand technique highlights this:
The idea is simply to write the second (right-hand) matrix above, thus leaving a space to do the calculation. It lines thing up nicely, and if you want, you can move the second column of the above matrix more to the right to place it over the calculations that use that column (I usually don’t bother).
Notice how this also helps illustrate why the column count of the left-hand matrix must match the row count of the right-hand one.
We can make the order of operation more obvious by converting the square matrix to a row or column vector that itself contains vectors:
Each of the components is a vector with two elements. The arrow over the name indicates a vector. Now we can visualize the multiplication like this:
On the left side, the column and row vectors, each containing vectors. On the right side, the same thing in bra-ket notation (which I’ll explore in more detail in another post).
In either case, what we’re doing is taking the inner (or dot) product of the component vectors. The inner product is a multi-dimensional form of multiplication (of taking a scalar product of two numbers), so we’re doing the same multiplication as shown in the first version, but these versions better illustrate which rows and columns to combine.
§
Following are some examples of multiplying matrices of different sizes.
[1×1] times [1×1]
The simplest possible matrix (hardly a matrix at all) is a 1×1 matrix.
The result is a 1×1 matrix, and the single value is the same as we’d get multiplying two scalars together:
$\begin{bmatrix}a_{11}\end{bmatrix}\begin{bmatrix}b_{11}\end{bmatrix} = \begin{bmatrix}a_{11}b_{11}\end{bmatrix}$
But note that a 1×1 matrix is not a scalar. (The difference becomes apparent in the next two cases.)
[1×1] times [1×2]
Similar to the first case, the 1×1 matrix acts as a scalar, and the result is the same 1×2 matrix (row vector) we’d get multiplying the 1×2 matrix by a scalar.
$\begin{bmatrix}a_{11}\end{bmatrix}\begin{bmatrix}b_{11}&b_{12}\end{bmatrix}=\begin{bmatrix}a_{11}b_{11}&a_{11}b_{12}\end{bmatrix}$
This effectively is the same as:
$\kappa\begin{bmatrix}b_{11}&b_{12}\end{bmatrix}=\begin{bmatrix}{\kappa}\;b_{11}&{\kappa}\;b_{12}\end{bmatrix}$
But note that, unlike the scalar multiplication, the matrix multiplication cannot be reversed because [1×2][1×1] is an illegal operation. The number of columns in the first matrix doesn’t match the number of rows in the second.
So a 1×1 matrix is not (always) the same as a scalar!
[2×1] times [1×1]
Here’s the legal version of putting the “scalar” matrix second. In this case, the single column of the 2×1 matrix (a column vector) matches the single row of the 1×1 matrix:
$\begin{bmatrix}a_{11}\\a_{21}\end{bmatrix}\begin{bmatrix}b_{11}\end{bmatrix}=\begin{bmatrix}a_{11}b_{11}\\a_{21}b_{11}\end{bmatrix}$
This is the same as:
$\begin{bmatrix}a_{11}\\a_{21}\end{bmatrix}\kappa=\begin{bmatrix}a_{11}\;{\kappa}\\a_{21}\;{\kappa}\end{bmatrix}$
However, as in the second case, this operation cannot be reversed (due to the column/row mismatch), whereas with the scalar operation it can.
[1×2] times [2×1] (inner product)
Multiplying a row vector by a column vector results in a 1×1 matrix usually treated as a scalar:
$\begin{bmatrix}a_{11}&a_{12}\end{bmatrix}\begin{bmatrix}b_{11}\\b_{21}\end{bmatrix}=\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}\end{bmatrix}$
This operation is known as the inner (or dot) product of the matrices. It generalizes to row and column vectors of larger sizes (so long as they match). When taken as an inner product, the result is always considered a scalar value.
[2×1] times [1×2] (outer product)
Multiplying a column vector by a row vector results in a matrix with as many rows and columns as the vectors (in this case, a 2×2 matrix):
$\begin{bmatrix}a_{11}\\a_{21}\end{bmatrix}\begin{bmatrix}b_{11}&b_{12}\end{bmatrix}=\begin{bmatrix}a_{11}b_{11}&a_{11}b_{12}\\a_{21}b_{11}&a_{21}b_{12}\end{bmatrix}$
This operation is known as the outer product of the matrices. It generalizes to column and row vectors of larger sizes (so long as they match). Note this operation always creates a square matrix.
[2×2] times [2×2]
Multiplying two (same-sized) square matrices results in a new matrix of the same size (in this case, 2×2).
$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}b_{11}&b_{12}\\b_{21}&b_{22}\end{bmatrix}=\begin{bmatrix}a_{11}b_{11}+a_{12}b_{21}&a_{11}b_{12}+a_{12}b_{22}\\a_{21}b_{11}+a_{22}b_{21}&a_{21}b_{12}+a_{22}b_{22}\end{bmatrix}$
(Multiplying square matrices is what many think of as “matrix multiplication” but as the examples above show, it’s not the only form.)
[2×2] times [2×1]
One of the more important examples is multiplying a square matrix times a column vector. In quantum mechanics, square matrices represent operators and column vectors represent quantum states. The output of such an operation is a new column vector, a new quantum state:
$\begin{bmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{bmatrix}\begin{bmatrix}b_{1}\\b_{2}\end{bmatrix}=\begin{bmatrix}a_{11}b_{1}+a_{12}b_{2}\\a_{21}b_{1}+a_{22}b_{2}\end{bmatrix}$
Note that you cannot multiply such an operator with a row vector because the column-row counts don’t match.
§ §
Square matrices are special because they have special properties.
An easy one is the trace, which is defined only for square matrices. The trace is the sum of the numbers on the main diagonal.
${M}=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix},\;\;\;tr(M)=a+e+i$
The main diagonal starts at the upper left and extends down to the lower right. The identity matrix, which is an important matrix, is all zeros except for ones on that diagonal:
$\mathbb{I}=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}$
The identity matrix has the property that using it to multiply another matrix gives that matrix.
${M}\times\mathbb{I}=M$
It’s the matrix equivalent of multiplying a regular number by one, the multiplicative identity.
Multiplying a regular number, n, by its inverse, 1/n, equals one. Similarly, multiplying a matrix by its inverse matrix gives the identity matrix:
${M}\times{M}^{\small{-1}}=\mathbb{I}$
I’ll note that determining a matrix’s inverse is, in many cases, non-trivial. It’s not just a matter of 1/n as with scalar values. (It’s not that hard, just takes a bit of figuring.)
§
The matrix determinant is, among other things, the scaling factor of the linear transformation the matrix represents.
Given some rectangular area before the transformation, the matrix determinant is how much that area shrinks or grows in the transform. If the determinant is 1, the scale of the space doesn’t change. If the determinant is negative, the transformation somehow inverts (“flips”) the space.
Calculating the determinant for a 2×2 matrix is easy:
${M}=\begin{bmatrix}a&b\\c&d\end{bmatrix},\;\;\;\det(M)=(ad)-(bc)$
The formula for a 3×3 matrix isn’t hard, just long: (aei)+(bfg)+(cdh)-(afh)-(bdi)-(ceg). It looks random and difficult to ever remember, but there’s a rationale and a pattern that, once learned, makes it easy to remember.
§
A square matrix can have a dot product (two, actually, depending on whether we consider it holding row vectors or column vectors as we did above). The matrix is orthogonal if the dot product is zero (because its vectors are orthogonal).
Taking it one step further, if the vectors are normalized and orthogonal, the matrix is orthonormal.
§
Of great importance in quantum mechanics, a square matrix is Hermitian if it’s equal to its conjugate transpose.
${M}={M}^{\small\dagger}$
The transpose of a matrix (which can be done to any matrix) is a flip along the main diagonal. Square matrices remain square, of course, but in other matrices the row and column counts switch places. In particular, a transpose converts a column vector to a row vector, and vice versa.
$\begin{bmatrix}a&b\\c&d\end{bmatrix}^{\intercal}=\begin{bmatrix}a&c\\b&d\end{bmatrix}$
The conjugate transpose of a matrix is, first, a transpose, and then taking the complex conjugate of its members. Obviously, this only applies to matrices with complex values.
$\begin{bmatrix}a\\b\\c\end{bmatrix}^{\dagger}=\begin{bmatrix}a*&b*&c*\end{bmatrix}$
[This is important in quantum mechanics, especially with regard to bra-ket notation, where a ket is a column vector, and a bra is a row vector. In particular, braa| is the conjugate transpose of ket |a〉. The transpose converts the column vector |a〉 to a row vector. Taking the complex conjugate of a is part of the quantum math magic (it’s related to how the magnitude of complex number z is the square root of zz*).]
$\langle{b}\mid{a}\rangle=\begin{bmatrix}b_1*&b_2*\end{bmatrix}\begin{bmatrix}a_1\\a_2\end{bmatrix}$
Lastly (also important in QM), a square matrix is unitary if its conjugate transpose is also its inverse.
${M}^{\small\dagger}={M}^{\small{-1}}\rightarrow{M}\times{M}^{\small\dagger}=\mathbb{I}$
Unitary Hermitian matrices are operators in quantum mechanics (that they are unitary is why QM is said to preserve information).
§ §
This post is just a high altitude fly over to introduce the various aspects of matrix math. Interested parties will obviously have to explore this in more detail.
One point I’ll make about reading (or watching) math topics such as this: It’s not enough to just read or watch. One has to have pen and paper at hand, and one has to do the work. It’s the only way the topic will begin to really make sense. Doing it for yourself is also the only way to really acquire the necessary skill.
Good luck, and I’m always here to help.
Stay in the matrix, my friends! Go forth and spread beauty and light.
The canonical fool on the hill watching the sunset and the rotation of the planet and thinking what he imagines are large thoughts. View all posts by Wyrd Smythe
#### 7 responses to “Sideband #71: Matrix Math”
• Wyrd Smythe
I wish my penmanship was better. I drew those diagrams more times than I can remember trying to make them look nicer. I don’t know LaTeX well enough to know how to typeset things lined up the right way, and I figured since it’s a by-hand technique anyway…
But I wish my penmanship was better. 😦
• Wyrd Smythe
The key take away here is matrix multiplication. The stuff about properties of matrices can come later, but matrix multiplication is a must for working with operators.
This becomes especially important in quantum computing, as QC operations (“quantum gates”) are all represented as matrices.
Interested readers should try for themselves the various multiplication examples I showed using the by-hand technique. Create some matrices with random values and practice multiplying them.
• Wyrd Smythe
I saw an article just the other day about non-Hermitian quantum physics. Researchers used quantum computing to simulate it and came up with some interesting results.
One that really caught my eye was:
“The first discovery was that applying operations to the qubits did not conserve quantum information — a behavior so fundamental to standard quantum theory that it results in currently unsolved problems like Stephen Hawking’s Black Hole Information paradox.”
I’ve long been suspicious of the idea that quantum systems conserve information. Unlike conservation of energy or other physical properties, there doesn’t seem to be an associated symmetry underlying the notion of conservation of information. It’s been more an assertion of QM based on the use of unitary operators.
I’ve never fully accepted that it’s a required truth, and these results suggest it might not be.
• Wyrd Smythe
Given some matrix, M, what would we do with:
${e}^{{i}\pi{M}\theta}$
How can a matrix be the exponent of anything, let alone the exponent of e in an expression like that? The trick is to remember that an exponential expression with e is actually a function.
• Sideband #72: Trig Is Easy! | Logos con carne
[…] want to deal with. (In fairness, it doesn’t pop up much in regular life.) As with matrix math, trig often remains opaque even for those who do have a basic grasp of other parts of […] |
# 36 Is 15% Of What Number?
36 is 15% of what number? This is a common question that arises when dealing with percentages and calculations. Understanding how to calculate percentages can be a valuable skill in various areas of life, such as finance, shopping, and even simple everyday tasks. In this article, we will explore how to find the number that 36 is 15% of, along with some frequently asked questions about percentages.
To determine the number that 36 is 15% of, we need to use a simple formula. The formula to find a percentage of a number is:
(Number) x (Percentage) = (Result)
In this case, we know that the result is 36 and the percentage is 15%. Let’s substitute these values into the formula and solve for the number:
(Number) x (15%) = 36
To remove the percentage sign, we need to convert it to a decimal. To do this, divide the percentage by 100:
0.15 x (Number) = 36
Now, we can isolate the variable by dividing both sides of the equation by 0.15:
(Number) = 36 / 0.15
Calculating this equation, we find that the number is equal to 240. Therefore, 36 is 15% of 240.
FAQs:
1. How do I calculate percentages?
To calculate percentages, you need to divide the part by the whole and multiply the result by 100. For instance, if you want to find what percentage 20 is of 100, divide 20 by 100 (0.2) and multiply the result by 100 to get 20%.
2. How can percentages be used in everyday life?
Percentages are used in various everyday scenarios. They can help determine discounts during sales, calculate tips at restaurants, analyze interest rates on loans, or even evaluate nutritional information on food labels.
3. What is the difference between percentage and percent?
Percentage and percent both refer to the same concept of expressing a fraction of a whole as a fraction of 100. The term “percentage” is often used in mathematical contexts, while “percent” is commonly used in general discussions.
4. How do I calculate a percentage increase or decrease?
To calculate a percentage increase or decrease, subtract the original value from the new value, divide the result by the original value, and multiply by 100. If the result is positive, it represents an increase, while a negative result indicates a decrease.
5. How can percentages be helpful in budgeting?
Percentages play a crucial role in budgeting as they allow individuals to allocate specific proportions of their income to different expenses. For example, one might decide to allocate 30% of their income to housing, 20% to savings, and 50% to other expenses.
6. How do percentages relate to fractions and decimals?
Percentages, fractions, and decimals are interconnected. Percentages can be expressed as fractions by dividing the percentage by 100. Similarly, percentages can be converted to decimals by dividing the percentage by 100. Fractions can also be converted to decimals and percentages using division or multiplication.
In conclusion, to find the number that 36 is 15% of, we use the formula (Number) x (Percentage) = (Result). By substituting the known values into the formula and solving for the number, we find that 36 is 15% of 240. Understanding percentages is essential in various aspects of life, and being able to calculate them accurately can be a valuable skill. |
Topic 1-2: One, two and three-dimensional spaces
Let us consider a line, take a point on the line, and call it the origin, denoted as O as in Fig. 1. We can determine a distance to any point P from the origin O, using the concept of distance and the measurement system of length as discussed in Topic 1-1. Thus, any point on the line can be represented by a real number, its distance from the origin O. The assembly of all the points, each of which can be represented by a real number, is called a one-dimensional space. Thus, the line under consideration is a one-dimensional space. However, a one-dimensional space is more general than a straight line. A curved line, open or closed, is also a one-dimensional space since each point on it can also be represented by a real number, the distance from the origin that can be picked artificially.
A point on a flat surface can be determined by two real numbers. The easiest way to see it is to draw two perpendicular lines and call one the x-axis and the other the y-axis. Then each point on the surface has a distance to the x-axis and a distance to the y-axis as shown in Fig. 2. In the figure the distance from point P to the y-axis is Px, and the distance to the x-axis is Py. Thus the point P can be represented by a pair of real numbers, (Px, Py). On the other hand if we specify two distances, one from the x-axis and the other from the y-axis, then a point on the surface can be uniquely determined. Thus any point on the flat surface can be represented by two real numbers. The assembly of points that can be represented by two real numbers is called a two-dimensional space. Of course, a two-dimensional space does not need to be a flat surface. It can be any curved surfaces; for example the surface of a sphere is a two-dimensional space.
The space we live in is a three-dimensional space, each point on which can be uniquely determined by a set of three real numbers. In Fig. 3 we show such an example. Three mutually perpendicular lines are drawn, and are called x-, y- and z- axes respectively. A point P is projected down to the x-y plane and its image is denoted as point P'. The distance between P and P' is Pz. The distance of P' to the x-axis is Py and the distance of P' to y-axis is Px. Thus the point P can be represented as (Px, Py, Pz). On the other hand the assembly of the set of three continuous real numbers forms a three dimensional space. It is difficult to explain anything beyond the three-dimensional space in which we live except by going into the pure mathematical definition of multi-dimensional spaces. One thing we can say is that our three-dimensional space contains numerous two-dimensional and one-dimensional spaces. |
# 4.2 Magnetic north map skills
Page 1 / 1
## Magnetic north map skills
You became acquainted with basic map skills in Grade 8 and did exercises that involved the 1:50 000 topographic map of Alice. In doing that, you learned to:
determine location
gauge accurately and determine direction
calculate distance by making use of a scale
explain the naming of a map
recognise conventional map symbols
recognise representation of height on maps
identify contour patterns
This knowledge, together with what you will still be acquiring in using the 1:50 000 topographic map of Bloemfontein and a 1:10 000 orthophoto map of a portion of this will see you through this year’s work.
Before commencing the practical exercises, you have to learn a number of additional map skills.
Calculating distance on an orthophoto map.
Orthophoto maps are always drawn to the scale of 1:10 000. Follow the steps to determine distance:
Accurately measure the distance between two points in centimetres, using a ruler (or using a string if the line to be measured curves);
Convert this distance to kilometres or metres, depending on what is required.
Example
A________________________________B
The length between A and B is 10 cm.
Scale 1 1:10 000
What is the true distance in km?
10 cm x 10 000
100 000 cm ÷ 100 000
1 km
What is the true distance in m?
10 cm x 10 000
100 000 ÷ 100
1000 m
Why do we divide by 100 000 and 100?
The metrical units of measurement can be represented as follows:
Km Hm Dm M dm cm mm 1 0 0 0 0 0 1 0 0
We therefore have 100 000 cm in 1 km
and
100 cm in 1 meter
You may therefore follow the shorter method if you can remember that the scale of
1 : 10 000 is the same as:
1 cm = 10 000 cm
1 cm = 100 m (÷ 100)
1 cm = 0,1 km (÷ 100 000)
## Activity 1: [lo 1.3]
1. Calculate the true distance in metres if the distance on a 1:10 000 orthophoto map is as follows;
1. 5,8 cm.
2. 10,1 cm.
3. Calculate the true distance in kilometres if the distance on a 1:10 000 orthophoto map is as follows;
1. 92,2 cm.
2. 15,6 cm. Magnetic North and Magnetic Declination
You have learnt how to gauge accurately in Grade 8. Remember the steps that have to be followed:
Connect the two points by means of a pencilled line.
• Draw a line to show True North through the point FROM which you have to measure.
• Position the protractor by placing 0° against the line indicating north and the centre at the point from which the measurement is to be taken.
• Measure the angle clockwise from the line for north to the connecting line (pencilled line of Step 1.)
Example:
On a map, North therefore indicates the North Pole, which we refer to as TRUE NORTH (geographic north.)
The earth also has a magnetic field that extends from north to south, for which we use a compass. The earth’s magnetic field does not correspond exactly with the true North and South Poles. North as indicated by the compass will always lie WEST of TRUE NORTH (TN) and is known as MAGNETIC NORTH (MN).
The difference between true north and magnetic north is known as the MAGNETIC DECLINATION (MD). The Magnetic Declination differs from place to place and continuously increases and decreases at any place, because magnetic north is not a fixed point.
To make it possible to determine the Magnetic Declination of any place, the following information is given in the margin of a 1:50 000 topographic map:
Example:
Calculate magnetic declination as at present:
Step 1:
Calculate the difference in years = 2003 – 2000
= 3 years
Step 2:
Multiply the years by the annual difference in minutes = 3 years x 4'
= 12'
Step 3:
Determine whether the angle increases (eastwards) or decreases (westwards).
Eastwards (smaller) -
Westwards (bigger) +
Step 4:
Add the minutes (westwards) or subtract them (eastwards).
In this instance it is eastwards, therefore:
21°30' - 12'
= 21°18' West of True North
REMEMBER that there are 60' in 1°.
If you should have to subtract and find that the degrees to be subtracted are too many, you “borrow” 1° and convert it to minutes.
Example:
21°30' – 34'
20°90' - 34'
= 20°56' West of True North.
If you should have to add minutes and get an answer that is more than 60’, you have to convert it 1°.
Example:
21°30' + 34'
21°64'
= 22°04' West of True North.
Step 5:
Remember to write West of True North alongside the new declination, because Magnetic North ALWAYS lies WEST of True NORTH.
Magnetic North
To gauge Magnetic North between two points, the degrees of TRUE NORTH must be added (+) to the Magnetic Declination and the answer must be written in degrees:
Example:
TN + MD = MN
50° + 21°18' = 71°18'
## Assessment
Learning Outcomes(LOs) LO 1 Geographical EnquiryThe learner will be able to use enquiry skills to investigate geographical and environmental concepts and processes. Assessment Standards(ASs) We know this when the learner: 1.2 asks questions that are relevant for identifying sources; 1.3 draws conclusions and makes analyses to obtain information from sources such as photographs, maps, atlases, graphs and statistics; 1.4 correlates information from different sources; 1.7 reports on knowledge that they have obtained through research, making use of different sources of information. LO 2 Geographical Knowledge and UnderstandingThe learner will be able to demonstrate geographical and environmental knowledge and understanding. We know this when the learner: 2.2 identifies ways in which Science and Technology have contributed a positively and negatively influence to development (people and resources.)
## Memorandum
ACTIVITY1:
• 580 m
• 1 010 m
• 9,22 km
• 1,56 km
ACTIVITY2:
1. True North / Bearing :
Magnetic Declination:
1990 – 2003 = 13 years
13 × 2’ = 26’ westwards
18 0 41’ + 26’ (westwards)
= 18 0 67’
= 19 0 07 west of true north
Magnetic Bearing: WN + MD = MN
…… + 19 0 07 = ………………………..
1. Actual Bearing :
Magnetic Declination: 1993 – 2003 = 10 years
10 × 3’ = 30’ eastwards (-)
20 0 15’ – 30’
19 0 75 – 30
= 19 0 45’ west of true north
Magnetic Bearing: WN + MO = MN
...................... + 19 0 45 =..................................
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
What fields keep nano created devices from performing or assimulating ? Magnetic fields ? Are do they assimilate ?
why we need to study biomolecules, molecular biology in nanotechnology?
?
Kyle
yes I'm doing my masters in nanotechnology, we are being studying all these domains as well..
why?
what school?
Kyle
biomolecules are e building blocks of every organics and inorganic materials.
Joe
Got questions? Join the online conversation and get instant answers! |
## Engage NY Eureka Math 6th Grade Module 2 Lesson 18 Answer Key
### Eureka Math Grade 6 Module 2 Lesson 18 Example Answer Key
Find the greatest common factor of 12 and 18.
→ Listing these factor pairs in order helps ensure that no common factors are missed. Start with 1 multiplied by the number.
→ Circle all factors that appear on both lists.
→ Place a triangle around the greatest of these common factors.
GCF (12, 18)
12
18
Example 2.
Least Common Multiple
Find the least common multiple of 12 and 18.
LCM (12, 18)
Write the first 10 multiples of 12.
12, 24, 36, 48, 60, 72, 84, 96, 108, 120
Write the first 10 multiples of 18.
18, 36, 54, 72, 90, 108, 126, 144, 162, 180
Circle the multiples that appear on both lists.
Put a rectangle around the least of these common multiples.
### Eureka Math Grade 6 Module 2 Lesson 18 Exercise Answer Key
Exercise 1.
Station 1: Factors and GCF
Choose one of these problems that has not yet been solved. Solve It together on your student page. Then, use your marker to copy your work neatly on the chart paper. Use your marker to cross out your choice so that the next group solves a different problem.
GCF (30, 50)
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30 Factors of 50: 1, 2, 5, 10, 25, 50
Common Factors: 1, 2, 5, 10 Greatest Common Factor: 10
GCF (30, 45)
Factors of 30: 1, 2, 3, 5, 6, 10, 15, 30 Factors of 45: 1, 3, 5, 9, 15, 45
Common Factors: 1, 3, 5, 15 Greatest Common Factor: 15
GCF (45, 60)
Factors of 45: 1, 3, 5, 9, 15, 45 Factors of 60: 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60
Common Factors: 1, 3, 5, 15 Greatest Common Factor: 15
GCF (42, 70)
Factors of 42: 1, 2, 3, 6, 7, 14, 21, 42 Factors of 70: 1, 2, 5, 7, 10, 14, 35, 70
Common Factors: 1, 2, 7, 14 Greatest Common Factor: 14
GCF (96, 144)
Factors of 96: 1, 2, 3, 4, 6, 8, 12, 16, 24, 32, 48, 96 Factors of 144: 1, 2, 3, 4, 6, 8, 9, 12, 16, 18, 24, 36, 48, 72, 144
Common Factors: 1, 2, 3, 4, 6, 8, 12, 16, 24, 48 Greatest Common Factor: 48
Next, choose one of these problems that has not yet been solved:
a. There are 18 girls and 24 boys who want to participate in a Trivia Challenge. If each team must have the same ratio of girls and boys, what is the greatest number of teams that can enter? Find how many boys and girls each team would have.
6 teams can enter the Trivia Challenge, each having 3 girls and 4 boys.
b. Ski Club members are preparing Identical welcome kits for new skiers. The Ski Club has 60 hand-warmer packets and 48 foot-warmer packets. Find the greatest number of identical kits they can prepare using all of the hand-warmer and foot-warmer packets. How many hand-warmer packets and foot-warmer packets would each welcome kit have?
There would be 12 welcome kits, each having 5 hand-warmer packets and 4 foot-warmer packets.
c. There are 435 representatives and 100 senators serving in the United States Congress. How many Identical groups with the same numbers of representative and senators could be formed from all of Congress if we want the largest groups possible? How many representatives and senators would be In each group?
5 identical groups with the same numbers of representatives and senators can be formed, each group with 87 representatives and 20 senators.
d. Is the GCF of a pair of numbers ever equal to one of the numbers? Explain with an example.
Yes. Valid examples should show a pair of numbers where the lesser of the two numbers is a factor of the greater number; the greater of the two numbers is a multiple of the lesser number.
e. Is the GCF of a pair of numbers ever greater than both numbers? Explain with an example.
No. Factors are, by definition, less than or equal to the number. Therefore, the GCF cannot be greater than both numbers.
Station 2: Multiples and LCM
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper. Use your marker to cross out your choice so that the next group solves a different problem.
LCM (9, 12)
Multiples of 9: 9, 18, 27, 36 Multiples of 12: 12, 24, 36
Least Common Multiple: 36
LCM (8, 18)
Multiples of 8: 8, 16, 24, 32, 40, 48, 56, 64, 72 Multiples of 18: 18, 36, 54, 72
Least Common Multiple: 72
LCM (4, 30)
Multiples of 4: 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60 Multiples of 30: 30, 60
Least Common Multiple: 60
LCM (12, 30)
Multiples of 12: 12, 24, 36, 48, 60 Multiples of 30: 30 ,60
Least Common Multiple: 60
LCM (20, 50)
Multiples of 20: 20, 40, 60, 80, 100 Multiples of 50: 50, 100
Least Common Multiple: 100
Next, choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on this chart paper and to cross out your choice so that the next group solves a different problem.
a. Hot dogs come packed 10 in a package. Hot dog buns come packed 8 in a package. If we want one hot dog for each bun for a picnic with none left over, what is the least amount of each we need to buy? How many packages of each item would we have to buy?
Four packages of hot dogs = 40 hot dogs. Five packages of buns = 40 buns. LCM (8, 10) = 40.
b. Starting at 6:00 a.m., a bus stops at my street corner every 15 minutes. Also starting at 6:00 a.m., a taxi cab comes by every 12 minutes. What is the next time both a bus and a taxi are at the corner at the same time?
Both o bus and a taxi arrive at the corner at 7:00 a.m., which is 60 minutes after 6:00 a.m.
LCM (12, 15) = 60.
c. Two gears in a machine are aligned by a mark drawn from the center of one gear to the center of the other. If the first gear has 24 teeth, and the second gear has 40 teeth, how many revolutions of the first gear are needed until the marks line up again?
The first gear needs five revolutions. During this time, 120 teeth pass by. The second gear revolves three times. LCM(24, 40) = 120.
d. Is the LCM of a pair of numbers ever equal to one of the numbers? Explain with an example.
Yes. Valid examples should show of a pair of numbers where the lesser of the two numbers is a factor of the greater number; the greater of the two numbers is a multiple of the lesser number.
e. Is the LCM of a pair of numbers ever less than both numbers? Explain with an example.
No. Multiples are, by definition, equal to or greater than the number.
Station 3: Using Prime Factors to Determine GCF
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Next, choose one of these problems that has not yet been solved:
a. Would you rather find all the factors of a number or find all the prime factors of a number? Why?
Accept opinions. Students should defend their answer and use accurate mathematical terms in their response.
b. Find the GCF of your original pair of numbers.
See answers listed in Exploratory Challenge 1.
c. Is the product of your LCM and GCF less than, greater than, or equal to the product of your numbers?
In all cases, GCF (a, b) . LCM (a, b) = a . b.
d. Glenn’s favorite number is very special because it reminds him of the day his daughter, Sarah, was born. The factors of this number do not repeat, and all of the prime numbers are less than 12. What is Glenn’s number? When was Sarah born?
2 . 3 . 5 . 7 . 11 = 2,310 Sarah’s birth date is 2/3/10.
Station 4: Applying Factors to the Distributive Property
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Find the GCF from the two numbers, and rewrite the sum using the distributive property.
1. 12 + 18 =
6(2) + 6(3) = 6(2 + 3) = 6(5) = 30
2. 42 + 14 =
7(6) + 7(2) = 7(6 + 2) = 7(8) = 56
3. 36 + 27 =
9(4) + 9(3) = 9(4 + 3) = 9(7) = 63
4. 16 + 72 =
8(2) + 8(9) = 8(2 + 9) = 8(11) = 88
5. 44 + 33 =
11(4) + 11(3) = 11(4 + 3) = 11(7) = 77
Next, add another example to one of these two statements applying factors to the distributive property.
Choose any numbers for n, a, and b.
n(a) + n(b) = n(a + b)
Accept all mathematically correct responses.
n(a) – n(b) = n(a – b)
The distributive property holds for addition as well as subtraction. Accept all mathematically correct responses.
### Eureka Math Grade 6 Module 2 Lesson 18 Problem Set Answer Key
Station 1: Factors and GCF
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Find the greatest common factor of one of these pairs: 30, 50; 30, 45; 45, 60; 42, 70; 96, 144.
Next, choose one of these problems that has not yet been solved:
a. There are 18 girls and 24 boys who want to participate in a Trivia Challenge. If each team must have the same ratio of girls and boys, what is the greatest number of teams that can enter? Find how many boys and girls each team would have.
b. Ski Club members are preparing identical welcome kits for new skiers. The Ski Club has 60 hand-warmer packets and 48 foot-warmer packets. Find the greatest number of identical kits they can prepare using all of the hand-warmer and foot-warmer packets. How many hand-warmer packets and foot-warmer packets would each welcome kit have?
c. There are 435 representatives and loo senators serving in the United States Congress. How many identical groups with the same numbers of representatives and senators could be formed from all of Congress if we want the largest groups possible? How many representatives and senators would be in each group?
d. Is the GCF of a pair of numbers ever equal to one of the numbers? Explain with an example.
e. Is the GCF of a pair of numbers ever greater than both numbers? Explain with an example.
Station 2: Multiples and LCM
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Find the least common multiple of one of these pairs: 9, 12; 8, 18; 4, 30; 12, 30; 20, 50.
Next, choose one of these problems that has not yet been solved:
a. Hot dogs come packed 10 in a package. Hot dog buns come packed 8 in a package. If we want one hot dog for each bun for a picnic, with none left over, what is the least amount of each we need to buy? How many packages of each item would we have to buy?
b. Starting at 6:00 a.m., a bus stops at my street corner every 15 minutes. Also starting at 6:00 a.m., a taxi cab comes by every 12 minutes. What is the next time both a bus and a taxi are at the corner at the same time?
c. Two gears in a machine are aligned by a mark drawn from the center of one gear to the center of the other. If the first gear has 24 teeth, and the second gear has 40 teeth, how many revolutions of the first gear are needed until the marks line up again?
d. Is the LCM of a pair of numbers ever equal to one of the numbers? Explain with an example.
e. Is the LCM of a pair of numbers ever less than both numbers? Explain with an example.
Solve it together on your student page. Then, use your marker to copy your work neatly on this chart paper and to cross out your choice so that the next group solves a different problem.
Station 3: Using Prime Factors to Determine GCF
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Use prime factors to find the greatest common factor of one of the following pairs of numbers:
30, 50 30, 45 45, 60 42, 70 96, 144
Next choose one of these problems that has not yet been solved:
a. Would you rather find all the factors of a number or find all the prime factors of a number? Why?
b. Find the GCF of your original pair of numbers.
c. Is the product of your LCM and GCF less than, greater than, or equal to the product of your numbers?
d. Glenn’s favorite number is very special because it reminds him of the day his daughter, Sarah, was born. The factors of this number do not repeat, and all of the prime numbers are less than 12. What is Glenn’s number? When was Sarah born?
Station 4: Applying Factors to the Distributive Property
Study these examples of how factors apply to the distributive property.
8 + 12 = 4(2) + 4(3) = 4(2 + 3) = 20
4(2) + 4(3) = 4(5) = 20
15 + 25 = 5(3) + 5(5) = 5(3 + 5) = 40
5(3) + 5(5) = 5(8) = 40
36 – 24 = 4(9) – 4(6) = 4(9 – 6) = 12
4(9) – 4(6) = 4(3) = 12
Choose one of these problems that has not yet been solved. Solve it together on your student page. Then, use your marker to copy your work neatly on the chart paper and to cross out your choice so that the next group solves a different problem.
Find the GCF from the two numbers, and rewrite the sum using the distributive property.
Question 1.
12 + 18 =
Question 2.
42 + 14 =
Question 3.
36 + 27 =
Question 4.
16 + 72 =
Question 5.
44 + 33 =
Next, add another example to one of these two statements applying factors to the distributive property.
Choose any numbers for n, a, and b.
n(a) + n(b) = n(a + b)
n(a) – n(b) = n(a – b)
### Eureka Math Grade 6 Module 2 Lesson 18 Exit Ticket Answer Key
Question 1.
Find the LCM and GCF of 12 and 15.
LCM: 60; GCF: 3
Question 2.
Write two numbers, neither of which is 8, whose GCF is 8.
Answers will vary (e.g., 16 and 24, or 24 and 32).
Question 3.
Write two numbers, neither of which is 28, whose LCM is 28. |
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# Q. QUADRATIC EQUATIONS 207 BASED ON HIGHER ORDER THINKING SKILLS (HOTS) EXAMPLE 3 The hypotenuse of a right triangle is . If the smaller side is tripled and the larger side is doubled, the new hypotenuse will be . Find the length of each side. SOLUTION Let the smaller side of the right triangle be and the larger side by . Then, by using Pythagoras Theorem, we obtain If the smaller side is tripled and the larger side be doubled, the new hypotenuse is . From equation (i), we get: . Putting in equation (ii), we get $9 x^{2}+4\left(45-x^{2}\right)=225 \Rightarrow 5 x^{2}+180=225 \Rightarrow 5 x^{2}=45 \Rightarrow x^{2}=9 \Rightarrow x=3 \quad[\because x>0]$ Putting in (i), we get $[\because y>0]$ Hence, the length of the smaller side is and the length of the larger side is . EXAMPLE 4 Vikram wishes to fit three rods together in the shape of a right triangle. The hypotenuse is to be longer than the base and longer than the altitude. What should be the lengths of the rods? SOLUTION Let the length of the hypotenuse be . Then, Base and, Altitude . By using Pythagoras theorem, we obtain $\begin{array}{ll} & (\text { Base })^{2}+(\text { Altitude })^{2}=(\text { Hypotenuse })^{2} \\ \Rightarrow & (x-2)^{2}+(x-4)^{2}=x^{2} \\ \Rightarrow & x^{2}-4 x+4+x^{2}-8 x+16=x^{2} \\ \Rightarrow & x^{2}-12 x+20=0 \\ \Rightarrow & x^{2}-10 x-2 x+20=0 \Rightarrow x(x-10)-2(x-10)=0 \Rightarrow(x-10)(x-2)=0 \\ \Rightarrow & x=2, \text { or, } x=10 \Rightarrow x=10 \quad \text { [For } x=2, \text { Base }=0 \text { which is not possible] } \end{array}$ Hence, the length of the rods are and . EXERCISE BASIC 1. The hypotenuse of a right triangle is . The difference between the lengths of the other two sides of the triangle is . Find the lengths of these sides. 2. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. BASED ON LOTS 3. The hypotenuse of a right triangle is . If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be . How long are the legs of the triangle? 4. A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates and on the boundary is 7 metres. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected? [NCERT] 1. 2. ANSWERS 4. At a distance of 5 metres from the gate 4.7.5 APPLICATIONS OF QUADRATIC EQUATIONS FOR SOLVING PROBLEMS ON MENSURATION Following examples will illustrate the above applications.
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Question Text Q. QUADRATIC EQUATIONS 207 BASED ON HIGHER ORDER THINKING SKILLS (HOTS) EXAMPLE 3 The hypotenuse of a right triangle is . If the smaller side is tripled and the larger side is doubled, the new hypotenuse will be . Find the length of each side. SOLUTION Let the smaller side of the right triangle be and the larger side by . Then, by using Pythagoras Theorem, we obtain If the smaller side is tripled and the larger side be doubled, the new hypotenuse is . From equation (i), we get: . Putting in equation (ii), we get $9 x^{2}+4\left(45-x^{2}\right)=225 \Rightarrow 5 x^{2}+180=225 \Rightarrow 5 x^{2}=45 \Rightarrow x^{2}=9 \Rightarrow x=3 \quad[\because x>0]$ Putting in (i), we get $[\because y>0]$ Hence, the length of the smaller side is and the length of the larger side is . EXAMPLE 4 Vikram wishes to fit three rods together in the shape of a right triangle. The hypotenuse is to be longer than the base and longer than the altitude. What should be the lengths of the rods? SOLUTION Let the length of the hypotenuse be . Then, Base and, Altitude . By using Pythagoras theorem, we obtain $\begin{array}{ll} & (\text { Base })^{2}+(\text { Altitude })^{2}=(\text { Hypotenuse })^{2} \\ \Rightarrow & (x-2)^{2}+(x-4)^{2}=x^{2} \\ \Rightarrow & x^{2}-4 x+4+x^{2}-8 x+16=x^{2} \\ \Rightarrow & x^{2}-12 x+20=0 \\ \Rightarrow & x^{2}-10 x-2 x+20=0 \Rightarrow x(x-10)-2(x-10)=0 \Rightarrow(x-10)(x-2)=0 \\ \Rightarrow & x=2, \text { or, } x=10 \Rightarrow x=10 \quad \text { [For } x=2, \text { Base }=0 \text { which is not possible] } \end{array}$ Hence, the length of the rods are and . EXERCISE BASIC 1. The hypotenuse of a right triangle is . The difference between the lengths of the other two sides of the triangle is . Find the lengths of these sides. 2. The diagonal of a rectangular field is 60 metres more than the shorter side. If the longer side is 30 metres more than the shorter side, find the sides of the field. BASED ON LOTS 3. The hypotenuse of a right triangle is . If the smaller leg is tripled and the longer leg doubled, new hypotenuse will be . How long are the legs of the triangle? 4. A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the difference of its distances from two diametrically opposite fixed gates and on the boundary is 7 metres. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected? [NCERT] 1. 2. ANSWERS 4. At a distance of 5 metres from the gate 4.7.5 APPLICATIONS OF QUADRATIC EQUATIONS FOR SOLVING PROBLEMS ON MENSURATION Following examples will illustrate the above applications. Updated On Feb 7, 2023 Topic All topics Subject Mathematics Class Class 11 Answer Type Video solution: 1 Upvotes 143 Avg. Video Duration 12 min |
# Past Papers’ Solutions | Cambridge International Examinations (CIE) | AS & A level | Mathematics 9709 | Pure Mathematics 1 (P1-9709/01) | Year 2015 | Oct-Nov | (P1-9709/11) | Q#5
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Question
A curve has equation .
i. Find and
ii. Find the coordinates of the stationary points and state, with a reason, the nature of each stationary point.
Solution
i.
We are given that;
Gradient (slope) of the curve is the derivative of equation of the curve. Hence gradient of curve with respect to is:
Rule for differentiation of is:
Rule for differentiation of is:
Second derivative is the derivative of the derivative. If we have derivative of the curve as , then expression for the second derivative of the curve is;
Rule for differentiation of is:
Rule for differentiation of is:
Rule for differentiation of is:
ii.
A stationary point on the curve is the point where gradient of the curve is equal to zero;
From (i), we have;
Coordinates of stationary point on the curve can be found from the derivative of equation of the curve by equating it with ZERO. This results in value of x-coordinate of the stationary point on the curve.
Therefore;
Two possible values of imply that there are two stationary points on the curve one at each value of .
To find y-coordinate of the stationary point on the curve, we substitute value of x-coordinate of the stationary point on the curve (found by equating derivative of equation of the curve with ZERO) in the equation of the curve.
For For
Once we have the x-coordinate of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2nd derivative of the curve.
Second derivative is the derivative of the derivative. If we have derivative of the curve as , then expression for the second derivative of the curve is;
From (i), we have;
Once we have the coordinates of the stationary point of a curve, we can determine its nature, whether minimum or maximum, by finding 2nd derivative of the curve.
We substitute of the stationary point in the expression of 2nd derivative of the curve and evaluate it;
If or then stationary point (or its value) is minimum.
If or then stationary point (or its value) is maximum.
For For Maximum Minimum |
# How do you differentiate f(x)= cosx/ (sinx) twice using the quotient rule?
Dec 4, 2015
$f ' \left(x\right) = - {\csc}^{2} x , f ' ' \left(x\right) = 2 {\csc}^{2} x \cot x$
#### Explanation:
Through the quotient rule:
$f ' \left(x\right) = \frac{\sin x \frac{d}{\mathrm{dx}} \left[\cos x\right] - \cos x \frac{d}{\mathrm{dx}} \left[\sin x\right]}{\sin} ^ 2 x$
Recall that:
$\frac{d}{\mathrm{dx}} \left[\cos x\right] = - \sin x$
$\frac{d}{\mathrm{dx}} \left[\sin x\right] = \cos x$
$f ' \left(x\right) = \frac{- {\sin}^{2} x - {\cos}^{2} x}{\sin} ^ 2 x$
$f ' \left(x\right) = - \frac{{\sin}^{2} x + {\cos}^{2} x}{\sin} ^ 2 x$
$f ' \left(x\right) = - \frac{1}{\sin} ^ 2 x$
$f ' \left(x\right) = - {\csc}^{2} x$
In order to find the second derivative, if we want to continue using the quotient rule, use:
$f ' \left(x\right) = \frac{- 1}{\sin} ^ 2 x$
$f ' ' \left(x\right) = \frac{{\sin}^{2} x \frac{d}{\mathrm{dx}} \left[- 1\right] - \left(- 1\right) \frac{d}{\mathrm{dx}} \left[{\sin}^{2} x\right]}{\sin} ^ 4 x$
Again, find each derivative:
$\frac{d}{\mathrm{dx}} \left[- 1\right] = 0$
Chain rule coming up:
$\frac{d}{\mathrm{dx}} \left[{\sin}^{2} x\right] = 2 \sin x \frac{d}{\mathrm{dx}} \left[\sin x\right] = 2 \sin x \cos x$
Plug back in:
$f ' ' \left(x\right) = \frac{2 \sin x \cos x}{\sin} ^ 4 x$
$f ' ' \left(x\right) = \frac{2 \cos x}{\sin} ^ 3 x$
$f ' ' \left(x\right) = 2 \left(\frac{1}{\sin} ^ 2 x\right) \left(\cos \frac{x}{\sin} x\right)$
$f ' ' \left(x\right) = 2 {\csc}^{2} x \cot x$ |
How to Easily Solve Math Problems Using Difference of Squares
Co-authored by wikiHow Staff
Updated: September 6, 2019
Multiplying two- or three-digit numbers using the standard algorithm requires a pen and pencil and can take some time. By using the difference of squares formula, a standard formula used in basic algebra, multiplying large numbers becomes easier and quicker to do, and can often be done in your head. You can use this method when multiplying any two numbers whose difference is even. With a few extra steps, you can also use this method when multiplying numbers with an odd difference.
Method 1 of 2: Using Difference of Squares for Factors with an Even Difference
1. 1
Set up a difference of squares formula. The difference of squares formula states that ${\displaystyle (x-y)(x+y)=x^{2}-y^{2}}$. In the formula, ${\displaystyle x}$ = the number equidistant from the two factors, and ${\displaystyle y}$ = half the difference between the two factors.[1]
2. 2
Subtract one factor from the other. Determine whether their difference is even or odd. If their difference is even, that means they are equidistant from another number, and you can use the difference of squares method.[2]
• A factor is a number being multiplied by another number.
• If the difference between factors is odd, then you can still use the difference of squares method to solve, but you will have to manipulate the problem, as described in Method 2.
• For example, if you were multiplying ${\displaystyle 28\times 32}$, you could use the difference of squares method, because ${\displaystyle 28-32=-4}$, and -4 is an even number.
• It doesn't matter if the difference is positive or negative. A difference of -4 is the same as a difference of 4. Both +/-4 are even. [3]
3. 3
Divide the difference between the two factors by 2. Plug in this number for ${\displaystyle y}$ in the difference of squares formula.
• For example, you already found that the difference between 28 and 32 is 4. ${\displaystyle 4\div 2=2}$. So half the difference between the two factors is 2. Plugging in this number for ${\displaystyle y}$, your formula becomes ${\displaystyle (x-2)(x+2)=x^{2}-2^{2}}$.
4. 4
Find the number equidistant from the two factors. To do this, take the larger factor and subtract it by half the difference between the two factors (${\displaystyle y}$ in your difference of squares formula).
• For example, if the larger factor is 32, and half the difference between factors (${\displaystyle y}$) is 2, then you would find the equidistant number by subtracting 2 from 32. ${\displaystyle 32-2=30}$, so the equidistant number is 30.
• You could also find the equidistant number by taking the smaller factor and adding half the difference.
5. 5
Plug the equidistant number into the difference of squares formula. To do this, replace all the ${\displaystyle x}$s in the formula with the equidistant number.
• For example, if your equidistant number is 30, your difference of squares formula will look like this: ${\displaystyle (30-2)(30+2)=30^{2}-2^{2}}$.
• Note that ${\displaystyle (30-2)(30+2)}$ is the same as ${\displaystyle (28)(32)}$, which was your original problem.
6. 6
Square the two numbers. Plug these squares into the equation. Remember, to square something means to multiply it by itself.
• For example:
${\displaystyle 30^{2}-2^{2}}$
=${\displaystyle 900-4}$
7. 7
Complete the subtraction. The result will equal the answer to your original multiplication problem.
• For example, ${\displaystyle 900-4=896}$, so ${\displaystyle 28\times 32=896}$.
8. 8
Check your work. You can verify your answer by using a calculator, or by solving using the standard multiplication algorithm.
Method 2 of 2: Using Difference of Squares for Factors With an Odd Difference
1. 1
Set up a difference of squares formula. The difference of squares formula states that ${\displaystyle (x-y)(x+y)=x^{2}-y^{2}}$. In the formula, ${\displaystyle x}$ = the number equidistant from the two factors, and ${\displaystyle y}$ = half the difference between the two factors.[4]
2. 2
Find the difference between the two factors being multiplied. If the factors are NOT equidistant from another number (that is, the difference is odd), you must follow these instructions to manipulate the difference of squares method.
• A factor is a number being multiplied by another number.
• For example, if you were multiplying ${\displaystyle 35\times 46}$, you would have to manipulate the difference of squares method, because ${\displaystyle 35-46=-11}$, and -11 is an odd number.
• If the difference between factors is even, use the shortcut described in Method 1.
3. 3
Subtract 1 from either factor. Subtracting 1 will give you two factors that are equidistant from another number, or two numbers whose difference is even. You will use this manipulated factor to follow the difference of squares method.
• Do NOT subtract 1 from both factors. Only subtract 1 from one factor. It doesn't matter which one.
• For example, for ${\displaystyle 35\times 46}$ you could subtract ${\displaystyle 46-1=45}$, so your multiplication problem becomes ${\displaystyle 35\times 45}$.
4. 4
Subtract one factor from the other. Make sure you are using the manipulated factor. The difference should now be even.
• For example, if your new problem, with the manipulated factor, is ${\displaystyle 35\times 45}$, the difference between factors is 10, since ${\displaystyle 35-45=-10}$.
• It doesn't matter if the difference is positive or negative. A difference of -10 is the same as a difference of 10. Both +/-10 are even. [5]
5. 5
Divide the difference between the two factors by 2. Plug in this number for ${\displaystyle y}$ in the difference of squares formula.
• Again, make sure you are using the manipulated factor.
• For example, you already found that the difference between 35 and 45 is 10. ${\displaystyle 10\div 2=5}$. So half the difference between the two factors is 5. Plugging in this number for ${\displaystyle y}$, your formula becomes ${\displaystyle (x-5)(x+5)=x^{2}-5^{2}}$
6. 6
Find the number equidistant from the two factors. To do this, take the larger factor, and subtract it by half the difference between the two factors (${\displaystyle y}$ in your difference of squares formula).
• For example, if the larger factor is 45, and half the difference between factors (${\displaystyle y}$) is 5, then you would find the equidistant number by subtracting 5 from 45. ${\displaystyle 45-5=40}$, so the equidistant number is 40.
• You could also find the equidistant number by taking the smaller factor and adding half the difference.
7. 7
Plug the equidistant number into the difference of squares formula. To do this, replace all the ${\displaystyle x}$s in the formula with the equidistant number.
• For example, if your equidistant number is 40, your difference of squares formula will look like this: ${\displaystyle (40-5)(40+5)=40^{2}-5^{2}}$.
• Note that ${\displaystyle (40-5)(40+5)}$ is the same as ${\displaystyle (35)(45)}$, the multiplication problem created after manipulating one of the factors.
8. 8
Square the two numbers. Plug these squares into the equation. Remember, to square something means to multiply it by itself.
• For example:
${\displaystyle 40^{2}-5^{2}}$
=${\displaystyle 1600-25}$
9. 9
Complete the subtraction. The result will equal the answer to the multiplication problem with the manipulated factor.
• For example, ${\displaystyle 1600-25=1575}$, so ${\displaystyle 35\times 45=1575}$.
10. 10
Add the missing value to the answer. Remember, you had to change the initial problem to make the difference of squares method work. To find the answer to the original problem, you must add back in the value you subtracted from the original multiplication problem.
• You are NOT adding 1 back into the problem. When you subtracted 1, you were really subtracting 1 group, since you are working with multiplication.
• For example, if the original problem was ${\displaystyle 35\times 46}$, that means you were trying to find the value of 46 thirty-fives. When you subtracted 1, you subtracted 1 group of thirty-fives, or 35. Now you need to add that group back in. Since ${\displaystyle 1575+35=1610}$, you now know that ${\displaystyle 35\times 46=1610}$.
11. 11
Check your work. You can verify your answer by using a calculator, or by solving using the standard multiplication algorithm.
Community Q&A
Search
• Question
A number is 8 times bigger than another number. How do I calculate the numbers if the difference is 42?
Donagan
Let the smaller number be x. Then the larger number is 8x. 8x - x = 7x = 42. So x = 6, and 8x = 48.
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1.5 Example with different effective lengths
Page 1 / 1
Problem
A W12 X 65 column, 24 feet long, is pinned at both ends in the strong direction, and pinned at the midpoint and theends in the weak direction. The column has A36 steel.
Number 1 - find effective length
Since the x-direction is the strong one and the y-direction is the weak one, then:
${L}_{x}=24$
${L}_{y}=12$
Notice that the effective length in the y-direction is half the total length of the member because there is alateral support at the midpoint.
Looking at the Manual on page 16.1-189 shows that the $K$ value for a column pinned at both ends is 1.0. Since the column is pinned at the ends and at the middle,
${K}_{x}=1$
${K}_{y}=1$
Now we can say that:
${K}_{x}{L}_{x}=1\times 24=24$
${K}_{y}{L}_{y}=1\times 12=12$
Number 2 - finding the capacity
Since, the steel is A36, you cannot use the column tables from Chapter 4 of the Third Edition Manual as the values are all given in terms of ${F}_{y}=50\mathrm{ksi}$ . However, in the Second Edition, in Chapter 3, the column tables give information forterms of ${F}_{y}=36\mathrm{ksi}$ .
From page 3-24 of the Second Edition Manual , the capacity for a W12 X 65 column with ${K}_{y}{L}_{y}=12$ is 519 kips.
Then to find ${K}_{x}{L}_{x}$ in terms of ${r}_{y}$ , ${K}_{x}{L}_{x}$ must be divided by: $\frac{{r}_{x}}{{r}_{y}}$ . This gives:
$\frac{{K}_{x}{L}_{x}}{\frac{{r}_{x}}{{r}_{y}}}=\frac{24}{1.75}=13.71$
This is close enough to 14, that we can then look in the tables for the $\mathrm{KL}$ value of 14, or interpolate for 13.71) and find the capacity forthe W12 X 65 member. The capacity is 497kips.
Method 2 - with buckling formulas
If you do not have the tables for A36 steel, you must use the formulas on page 16.1-27 of the Manual .
Number 1 - show the width-thickness ratio
In order for the equations in section E2 of the Manual to apply, the width-thickness ratio must be ${\lambda }_{r}$ .
$\frac{{b}_{f}}{2{t}_{f}}< {\lambda }_{r}$
The value for $\frac{{b}_{f}}{2{t}_{f}}$ (9.92) can be found on page 16.1-21, as well as the value for $\frac{h}{{t}_{w}}$ (24.9). The formula for ${\lambda }_{r}$ can be found on page 16.1-14/15. Then, the value for that formula can be found on page 16.1-150.
The flanges are unstiffened and in pure compression, so the formula is:
$\frac{{b}_{f}}{2{t}_{f}}=9.92< 0.56\sqrt{\frac{E}{{F}_{y}}}=15.9$
The web is stiffened and in compression, so the formula is:
$\frac{h}{{t}_{w}}=24.9\le 1.49\sqrt{\frac{E}{{F}_{y}}}=42.3$
Another way to easily find the formulas for ${\lambda }_{r}$ is to go to page 16.1-183 and look at the picture of the I-shaped member. The arrows point to either the flange orthe web and formulas correspond to the arrows giving the axial compression formulas that you need for that elementof the member.
Number 2 - compute slenderness ratios
The slenderness ratios can be found for both the x-axis and the y-axis. We know $K$ , and $L$ , and $r$ can be found in the properties section of the Manual on page 1-20.
$\frac{{K}_{x}{L}_{x}}{{r}_{x}}=\frac{24\times 12\times 1}{5.28}=54.54$
$\frac{{K}_{y}{L}_{y}}{{r}_{x}}=\frac{12\times 12\times 1}{3.02}=47.68$
Then, using Table 3-36 on page 16.1-143 of the Manual and interpolation, we can determine that ${\phi }_{c}{F}_{cr}=26.21$ , and that ${\phi }_{c}{P}_{n}={\phi }_{c}{F}_{cr}{A}_{g}=500k$
The capacity of the W12 X 65 column is 500 kips.
are nano particles real
yeah
Joseph
Hello, if I study Physics teacher in bachelor, can I study Nanotechnology in master?
no can't
Lohitha
where we get a research paper on Nano chemistry....?
nanopartical of organic/inorganic / physical chemistry , pdf / thesis / review
Ali
what are the products of Nano chemistry?
There are lots of products of nano chemistry... Like nano coatings.....carbon fiber.. And lots of others..
learn
Even nanotechnology is pretty much all about chemistry... Its the chemistry on quantum or atomic level
learn
da
no nanotechnology is also a part of physics and maths it requires angle formulas and some pressure regarding concepts
Bhagvanji
hey
Giriraj
Preparation and Applications of Nanomaterial for Drug Delivery
revolt
da
Application of nanotechnology in medicine
has a lot of application modern world
Kamaluddeen
yes
narayan
what is variations in raman spectra for nanomaterials
ya I also want to know the raman spectra
Bhagvanji
I only see partial conversation and what's the question here!
what about nanotechnology for water purification
please someone correct me if I'm wrong but I think one can use nanoparticles, specially silver nanoparticles for water treatment.
Damian
yes that's correct
Professor
I think
Professor
Nasa has use it in the 60's, copper as water purification in the moon travel.
Alexandre
nanocopper obvius
Alexandre
what is the stm
is there industrial application of fullrenes. What is the method to prepare fullrene on large scale.?
Rafiq
industrial application...? mmm I think on the medical side as drug carrier, but you should go deeper on your research, I may be wrong
Damian
How we are making nano material?
what is a peer
What is meant by 'nano scale'?
What is STMs full form?
LITNING
scanning tunneling microscope
Sahil
how nano science is used for hydrophobicity
Santosh
Do u think that Graphene and Fullrene fiber can be used to make Air Plane body structure the lightest and strongest. Rafiq
Rafiq
what is differents between GO and RGO?
Mahi
what is simplest way to understand the applications of nano robots used to detect the cancer affected cell of human body.? How this robot is carried to required site of body cell.? what will be the carrier material and how can be detected that correct delivery of drug is done Rafiq
Rafiq
if virus is killing to make ARTIFICIAL DNA OF GRAPHENE FOR KILLED THE VIRUS .THIS IS OUR ASSUMPTION
Anam
analytical skills graphene is prepared to kill any type viruses .
Anam
Any one who tell me about Preparation and application of Nanomaterial for drug Delivery
Hafiz
what is Nano technology ?
write examples of Nano molecule?
Bob
The nanotechnology is as new science, to scale nanometric
brayan
nanotechnology is the study, desing, synthesis, manipulation and application of materials and functional systems through control of matter at nanoscale
Damian
Is there any normative that regulates the use of silver nanoparticles?
what king of growth are you checking .?
Renato
how did you get the value of 2000N.What calculations are needed to arrive at it
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# What is the equation of the parabola with a focus at (-2, 6) and a vertex at (-2, 9)?
Jun 27, 2017
y - 9 = 1/12 ( x + 2 )^2
#### Explanation:
Generic Equation is
y - k = 1/4p ( x - h)^2
p is distance vertex to focus = 3
(h,k) = vertex location = (-2, 9)
Jun 27, 2017
$y = - \frac{1}{12} {\left(x + 2\right)}^{2} + 9$
#### Explanation:
When talking about the focus and vertex of a parabola, the easiest way to write the equation is in vertex form. Luckily, you already have most of your information.
$y = a {\left(x + 2\right)}^{2} + 9$
However, we do not have the value of $a$.
$a = \frac{1}{4 c}$
$c$ is the distance between the focus and the vertex.
$c = - 3$
We know this because the only difference between the two coordinates is the $y$ part. The reason it is negative is because the vertex is above the focus; this means that the parabola opens downwards.
$\frac{1}{4 c}$
$\frac{1}{\left(4\right) \left(- 3\right)}$
$\frac{1}{-} 12$
$- \frac{1}{12}$
Now that you have your value for $a$, you can plug this in and finalize your equation.
$y = - \frac{1}{12} {\left(x + 2\right)}^{2} + 9$
Jun 27, 2017
$y = - {x}^{2} / 12 - \frac{x}{3} + \frac{26}{3}$
#### Explanation:
Given -
Vertex $\left(- 2 , 9\right)$
Focus $\left(- 2 , 6\right)$
The focus of the parabola lies below the vertex. Hence, it opens down.
The formula for downward opening parabola having origin as its vertex is -
${x}^{2} = - 4 a y$
The vertex of the given parabola is not at the vertex. it is in the 2nd quarter.
The formula is -
${\left(x - h\right)}^{2} = - 4 \times a \times \left(y - k\right)$
$h = - 2$ x-coordinate of the vertex
$k = 9$ y-coordinate of the vertex
$a = 3$Distance between vertex and focus
Substitute the values in the formula
${\left(x + 2\right)}^{2} = - 4 \times 3 \times \left(y - 9\right)$
${x}^{2} + 4 x + 4 = - 12 y + 108$
$- 12 y + 108 = {x}^{2} + 4 x + 4$
$- 12 y = {x}^{2} + 4 x + 4 - 108$
$y = - {x}^{2} / 12 - \frac{4}{12} x + \frac{108}{12}$
$y = - {x}^{2} / 12 - \frac{x}{3} + \frac{26}{3}$ |
# Worksheet On Addition And Subtraction Of Binary Numbers
Worksheet On Addition And Subtraction Of Binary Numbers. Subtracting 2 from this result places a new result of 1 in the current column and sends a carry to the next column. Binary addition is one of the binary operations.
Follow the binary addition rules to add the numbers. Here, 1 + 1 is 10, which is the binary equivalent of (2) 10, so we. Use the addition and subtraction operation.
Subtraction in money by using conversion method, we convert rupees and paise into paise and then subtract the smaller amount of paise from the greater amount. Begin by writing a simple sum (adding to denary numbers together) on the board:
### Begin By Writing A Simple Sum (Adding To Denary Numbers Together) On The Board:
Negative numbers for grade 6 ; The above sum is carried out by following step Some of the worksheets below are hexadecimal addition and subtraction worksheets, arithmetic operations in binary, octal, and hexadecimal number systems, adding and subtracting octal and hexadecimal number systems with several problems and solutions.
### To Recall, The Term “Binary Operation” Represents The Basic Operations Of Mathematics That Are Performed On Two Operands.
There are four basic operations for binary addition, as mentioned above. Writing the four addition and subtraction facts in the house models, dominoes, picture models and more. Subtracting 2 from this result places a new result of 1 in the current column and sends a carry to the next column.
### The Example Shown Below Presents The Addition Of 10010110 2 And 001010112.
0+0=0 0+1=1 1+0=1 1+1=10 the above first three equations are very identical to the binary digit number. Next, introduce a simple binary sum: Subtraction in money by using conversion method, we convert rupees and paise into paise and then subtract the smaller amount of paise from the greater amount.
### A Mathematical Operation Used In Binary Multiplication.
The method is known as subtraction, and the sign is known as the minus sign. Here, 1 + 1 is 10, which is the binary equivalent of (2) 10, so we. The binary numbers seen so far use only positive powers of 2.
### Welcome To The Addition And Subtraction Of Whole Numbers Section At Tutorialspoint.com.
Addition and subtraction fact family (31 worksheets) Use the addition and subtraction operation. The quiz was designed to assess your knowledge of: |
〜 ★ dCode presents ★ 〜
# Irreducible Fractions
Results
Tool to reduce fractions in lowest term. A Fraction in Lowest Terms (Irreducible Fraction) is a reduced fraction in shich the numerator and the denominator are coprime (they do not share common factors)
Summary
## Decimal to Fraction in Lowest Term Converter
### How to make a fraction in lowest term?
To simplify a fraction $a / b$ or $frac{a}{b}$ composed of a numerator $a$ and a denominator $b$, find the greatest common divisor (GCD) of the numbers $a$ and $b$. The irreducible fraction is obtained by dividing the numerator and the denominator by the obtained PGCD.
Example: The fraction $12/10$ has $12$ for numerator and $10$ for denominator. Calculate that $GCD(12,10) = 2$ and divide both the numerator $12/2 = 6$ and the denominator $10/2 = 5$, so the corresponding irreducible fraction is $6/5$
dCode offers tools to calculate the GCD via, for example, Euclid's algorithm.
### How to calculate and give the result under the lowest term form?
Use the dCode calculator, enter the expressions / fractions and the simplifier will use formal calculations in order to keep variables and find the irreducible form.
### How to make a fraction from a decimal number?
If the number has a limited decimal development then it only needs to be multiplied by the right power of 10, then simplify the fraction and solve the equation.
Example: The number $0.14$ is equivalent to $0.14/1$, multiply by $10/10 (=1)$ until having no comma: $0.14/1 = 1.4/10 = 14/100$ then simplify $14/100 = 7/50$
If the number has a non finite decimal expansion, then it is necessary to locate the repeating portion of the number after the repeating decimal point.
Example: The number $0.166666666 ...$ where the $6$ is repeated
By calling $x$ the number, and $n$ the size (number of digits) of the smallest repeated portion. To obtain a fraction, multiply $x$ by $10^n$ and then subtract $x$.
Example: $x = 0.1666666 ...$, the smallest repeated portion is $6$, which has a single digit so that $n = 1$. Then compute $10^1 \ times x = 1.6666666 ...$ and $10x-x$. $$10x-x = 9x = 1.666666 ... - 0.1666666 ... = 1.5 \\ \iff 9x = 1.5 \\ \Rightarrow x = 1.5 / 9 = 15/90 = 1/6$$ So $1/6 = 0.1666666 ...$
## Source code
dCode retains ownership of the source code of the script Irreducible Fractions online. Except explicit open source licence (indicated Creative Commons / free), any algorithm, applet, snippet, software (converter, solver, encryption / decryption, encoding / decoding, ciphering / deciphering, translator), or any function (convert, solve, decrypt, encrypt, decipher, cipher, decode, code, translate) written in any informatic langauge (PHP, Java, C#, Python, Javascript, Matlab, etc.) which dCode owns rights will not be released for free. To download the online Irreducible Fractions script for offline use on PC, iPhone or Android, ask for price quote on contact page !
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# Horizontal asymptote
A horizontal asymptote is a horizontal line that the graph of a function approaches as x approaches ±∞. It is not part of the graph of the function. Rather, it helps describe the behavior of a function as x gets very small or large. This is in contrast to vertical asymptotes, which describe the behavior of a function as y approaches ±∞. The dotted red lines in the figure below represent the horizontal asymptotes of the given functions:
The function on the left has a horizontal asymptote at y = 5, while the function on the right has one at the x-axis (y = 0).
Formally, horizontal asymptotes are defined using limits. A function, f(x), has a horizontal asymptote, y = b, if:
If either (or both) of the above is true, then f(x) has a horizontal asymptote at y = b.
## Identifying horizontal asymptotes for rational functions
To find a horizontal asymptote for a rational function of the form , where P(x) and Q(x) are polynomial functions and Q(x) ≠ 0, first determine the degree of P(x) and Q(x). Then:
• If the degree of Q(x) is greater than the degree of P(x), f(x) has a horizontal asymptote at y = 0.
• If the degree of P(x) is equal to that of Q(x), f(x) has a horizontal asymptote that is the ratio of the coefficients of the highest degree term of P(x) to that of Q(x).
• If the degree of P(x) is greater than the degree of Q(x), f(x) has no horizontal asymptote, though it may have a slant asymptote (if the degree of P(x) is 1 greater than that of Q(x).
Examples
Find any horizontal asymptotes for the following functions:
1.
2.
1. The degree of P(x) is 2 and the degree of Q(x) is 3. This corresponds to the first case described above, where the degree of Q(x) is greater than that of P(x). Thus, f(x) has a horizontal asymptote at y = 0, as confirmed by its graph:
2. The degree of P(x) is 4 and the degree of Q(x) is 4. This corresponds to the second case described above, where the degrees of P(x) and Q(x) are equal. Thus, f(x) has a horizontal asymptote at the ratio of the coefficients of the highest degree term of P(x) to Q(x), or 4:2. Thus, f(x) has a horizontal asymptote at y = 4/2 = 2, as shown in the graph of the function:
Notice that f(x) crosses its horizontal asymptote on the right of the y-axis. It is possible for a function to cross a horizontal asymptote. In fact, it is possible for a function to cross its horizontal asymptote numerous times, as in the case of an oscillating function.
## Finding horizontal asymptotes using limits
The method described in the previous section works for rational functions comprised of polynomials. For rational functions that aren't comprised of polynomials, we can find horizontal asymptotes by computing the limit of the function as x approaches ±∞. A function f(x) will have a horizontal asymptote at y = b, where b is a constant, if either
Example
Find any horizontal asymptotes for the function:
To determine the limit of the function as x approaches ±∞, we must first manipulate the function algebraically such that the limit will not result in an indeterminate form. For the first case, we will consider the limit as x approaches -∞, and divide the function by x2; doing so prevents us from acquiring a result of the form -∞/∞ when computing the limit.
Based on this result, we cannot say that the function has any horizontal asymptote, and we must find its limit as x approaches +∞. In this case, we instead divide by ex to avoid acquiring a result of the form ∞/∞.
Since this is a constant, we conclude that f(x) has a horizontal asymptote at y = 3.
## Horizontal vs. vertical asymptotes
While both horizontal and vertical asymptotes help describe the behavior of a function at its extremities, it is worth noting that they do have some differences. One of the key differences is that a function can only have a maximum 2 horizontal asymptotes; it can have 0, 1, or 2 horizontal asymptotes, but no more. We can see why by considering how we find horizontal asymptotes by examining the limit of a function as it approaches ±∞. Since there are only two directions we can consider, -∞ or +∞, there can only be, at maximum, 2 horizontal asymptotes. In contrast, it is possible for a function to have any number of vertical asymptotes. The tangent function for example, has an infinite number of vertical asymptotes. An example of a function that has 2 horizontal asymptotes is f(x) = arctan(x), the graph of which is shown below.
Another important difference between horizontal and vertical asymptotes is that while the graph of a function never touches a vertical asymptote, it is possible for the graph of a function to touch, and even cross a horizontal asymptote; it can do so an infinite number of times, such as in the case of an oscillating function:
As x approaches ±∞, the function approaches the horizontal asymptote y = 1, but at any given point may be above or below 1 due to its oscillating nature. A function can cross a horizontal asymptote because it still approaches the same value while oscillating about that value. In the case of a vertical asymptote, it is not possible for the function to ever touch or cross the asymptote because vertical asymptotes arise where a function is undefined. The function may approach ±∞, but it is never possible for the function to reach ±∞, which is what crossing a vertical asymptote would imply. |
# 6-2
## 6-2
- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -
##### Presentation Transcript
1. 6-2 Properties of Parallelograms Warm Up Lesson Presentation Lesson Quiz Holt Geometry
2. Warm Up Find the value of each variable. 1.x2.y 3.z 2 18 4
3. Objectives Prove and apply properties of parallelograms. Use properties of parallelograms to solve problems.
4. Vocabulary parallelogram
5. Any polygon with four sides is a quadrilateral. However, some quadrilaterals have special properties. These special quadrilaterals are given their own names.
6. Helpful Hint Opposite sides of a quadrilateral do not share a vertex. Opposite angles do not share a side.
7. A quadrilateral with two pairs of parallel sides is a parallelogram. To write the name of a parallelogram, you use the symbol .
8. In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find CF. opp. sides Example 1A: Properties of Parallelograms CF = DE Def. of segs. CF = 74 mm Substitute 74 for DE.
9. In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find mEFC. cons. s supp. Example 1B: Properties of Parallelograms mEFC + mFCD = 180° mEFC + 42= 180 Substitute 42 for mFCD. mEFC = 138° Subtract 42 from both sides.
10. In CDEF, DE = 74 mm, DG = 31 mm, and mFCD = 42°. Find DF. diags. bisect each other. Example 1C: Properties of Parallelograms DF = 2DG DF = 2(31) Substitute 31 for DG. DF = 62 Simplify.
11. opp. sides Check It Out! Example 1a In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find KN. LM = KN Def. of segs. LM = 28 in. Substitute 28 for DE.
12. opp. s Check It Out! Example 1b In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find mNML. NML LKN mNML = mLKN Def. of s. mNML = 74° Substitute 74° for mLKN. Def. of angles.
13. diags. bisect each other. Check It Out! Example 1c In KLMN, LM = 28 in., LN = 26 in., and mLKN = 74°. Find LO. LN = 2LO 26 = 2LO Substitute 26 for LN. LO = 13 in. Simplify.
14. opp. s Example 2A: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find YZ. YZ = XW Def. of segs. 8a – 4 = 6a + 10 Substitute the given values. Subtract 6a from both sides and add 4 to both sides. 2a = 14 a = 7 Divide both sides by 2. YZ = 8a – 4 = 8(7) – 4 = 52
15. cons. s supp. Example 2B: Using Properties of Parallelograms to Find Measures WXYZ is a parallelogram. Find mZ. mZ + mW = 180° (9b + 2)+ (18b –11) = 180 Substitute the given values. Combine like terms. 27b – 9 = 180 27b = 189 b = 7 Divide by 27. mZ = (9b + 2)° = [9(7) + 2]° = 65°
16. diags. bisect each other. Check It Out! Example 2a EFGH is a parallelogram. Find JG. EJ = JG Def. of segs. 3w = w + 8 Substitute. 2w = 8 Simplify. w = 4 Divide both sides by 2. JG = w + 8 = 4 + 8 = 12
17. diags. bisect each other. Check It Out! Example 2b EFGH is a parallelogram. Find FH. FJ = JH Def. of segs. 4z – 9 = 2z Substitute. 2z = 9 Simplify. z = 4.5 Divide both sides by 2. FH = (4z – 9) + (2z) = 4(4.5) – 9 + 2(4.5) = 18
18. Example 4A: Using Properties of Parallelograms in a Proof Write a two-column proof. Given: ABCD is a parallelogram. Prove:∆AEB∆CED
19. 3. diags. bisect each other 2. opp. sides Example 4A Continued Proof: 1. ABCD is a parallelogram 1. Given 4. SSS Steps 2, 3
20. Example 4B: Using Properties of Parallelograms in a Proof Write a two-column proof. Given: GHJN and JKLM are parallelograms. H and M are collinear. N and K are collinear. Prove:H M
21. 2.H and HJN are supp. M and MJK are supp. 2. cons. s supp. Example 4B Continued Proof: 1.GHJN and JKLM are parallelograms. 1. Given 3.HJN MJK 3. Vert. s Thm. 4.H M 4. Supps. Thm.
22. Check It Out! Example 4 Write a two-column proof. Given: GHJN and JKLM are parallelograms. H and M are collinear. N and K are collinear. Prove: N K
23. 2.N and HJN are supp. K and MJK are supp. 2. cons. s supp. Check It Out! Example 4 Continued Proof: 1.GHJN and JKLM are parallelograms. 1. Given 3.HJN MJK 3. Vert. s Thm. 4.N K 4. Supps. Thm.
24. Lesson Quiz: Part I In PNWL, NW = 12, PM = 9, and mWLP = 144°. Find each measure. 1.PW2. mPNW 18 144°
25. Lesson Quiz: Part II QRST is a parallelogram. Find each measure. 2.TQ3. mT 71° 28
26. 1.RSTU is a parallelogram. 1. Given 3.R T 4. ∆RSU∆TUS 4. SAS 2. cons. s 3. opp. s Lesson Quiz: Part IV 6. Write a two-column proof. Given:RSTU is a parallelogram. Prove: ∆RSU∆TUS |
NCERT Solutions for Class 12 Mathemetics Chapter 12 - Linear Programming
Question 1:
Solve the following Linear Programming Problems graphically :
Maximise Z = 3x + 4y subject to the constraints :
x + y $\le$ 4, x $\ge$ 0, y $\ge$0.
The system of constraints is :
x + y $\le$ 4 ...(1)
and x $\ge$0. y $\ge$ 0 ...(2)
The shaded region in the following figure is the feasible region determined by the system of constraints (1) – (2).
It is observed that the feasible region OAB is bounded.
Thus we use Corner Point Method to determine the maximum value of Z, where :
Z = 3x + 4y ...(3)
Question 2:
Minimise Z = – 3x + 4y
subject to x + 2y $\le$ 8, 3x + 2y $\le$ 12, x $\ge$ 0, y $\ge$0.
The system of constraints is :
x + 2y$\le$ 8 ...(1)
3x + 2y$\le$ 12 ...(2)
and x $\ge$0, y$\ge$ 0 ...(3)
The shaded region in the following figure is the feasible region determined by the system of constraints (1) – (3).
It is observed that the feasible region OCEB is bounded.
Thus we use Corner Point Method to determine the minimum value of Z, where
Z= – 3x + 4y ...(4)
The co-ordinates of O, C, E and B are (0, 0), (4, 0), (2, 3) (on solving x + 2y = 8 and 3x + 2y = 12) and (0, 4) respectively.
We evaluate Z at each corner point.
Question 3:
Maximise Z = 5x + 3y
subject to 3x + 5y $\le$ 15, 5x + 2y $\le$ 10, x $\ge$0, y $\ge$0.
The system of constraints is :
3x + 5y $\le$ 15 ...(1)
5x + 2y $\le$ 10 ...(2)
and x $\ge$0, y $\le$ 0 ...(3)
The shaded region in the following figure is the feasible region determined by the system of constraints (1)–(3).
Question 4:
Minimise Z = 3x + 5y
subject to x + 3y $\ge$3, x + y $\ge$2, x, y $\ge$0.
The system of constraints is :
x + 3y $\ge$ 3 ...(1)
x + y $\ge$ 2 ...(2)
and x, y $\ge$ 0 ...(3)
Question 5:
Maximise Z = 3x + 2y
subject to x + 2y $\le$ 10, 3x + y $\le$ 15, x, y $\ge$0.
The system of constraints is :
x + 2y $\le$ 10 ...(1)
3x + y $\le$ 15 ...(2)
and x, y $\ge$ 0 ...(3)
The shaded region in the following figure is the feasible region determined by the system of constraints (1) – (3).
It is observed that the feasible region OCEB is bounded.
Thus we use Corner Point Method to determine the maximum
value of Z, where :
Z = 3x + 2y ...(4)
The co-ordinates of O, C, E and B are (0, 0), (5, 0),
E (4, 3) (solving x + 2y = 10, 3x + y = 15) and (0, 5)
respectively.
We evaluate Z at each corner point.
Question 6:
Minimise Z = x + 2y
subject to 2x + y $\ge$3, x + 2y $\ge$6, x, y $\ge$0.
Show that the minimum of Z occurs at more than two points.
The system of constraints is :
2x + y $\ge$ 3 ...(1)
x + 2y $\ge$ 6 ...(2)
and x $\ge$ 0, y $\ge$ 0 ...(3)
The shaded region in the following figure is the feasible region determined by the system of constraints (1)–(3).
It is observed that the feasible region is unbounded.
Question 7:
Minimise and Maximise Z = 5x + 10y
subject to x + 2y $\le$ 120, x + y $\ge$60, x – 2y $\ge$0, x, y $\ge$0.
The system of constraints is :
x + 2y $\le$ 120 ...(1)
x + y $\ge$ 60 ...(2)
x – 2y $\ge$ 0 ...(3)
and x, y $\ge$ 0 ...(4)
It is observed that the feasible region CADE is bounded whose corner points are :
C (60, 0), A (120, 0), D (60, 30), (Solving x + 2y = 120 and x – 2y = 0) and E (40, 20) (Solving x + 2y = 60 and x – 2y = 0).
Question 8:
Minimise and Maximise Z = x + 2y
subject to x + 2y $\ge$ 100, 2x – y $\le$ 0, 2x + y $\le$ 200; x, y $\ge$0
The system of constraints is :
x + 2y $\ge$ 100 ...(1)
2x – y $\le$ 0 ...(2)
2x + y $\le$ 200 ...(3)
and x, y $\le$ 0 ...(4)
The shaded region in the above figure is the feasible region determined by the system of constraints (1)–(4).
It is observed that the feasible region ECDB is bounded. Thus we use Corner Point Method to determine the maximum of Z,
where :
Z = x + 2y ...(5)
The co-ordinates of E, C, D and B are (20, 40)(on solving x + 2y = 100 and 2x – y = 0)
(50, 100)(on solving 2x + y = 200 and 2x – y = 0)
(0, 200) and (0, 50) respectively.
Question 9:
Maximise Z = – x + 2y, subject to the constraints :
x $\ge$3, x + y $\ge$5, x + 2y $\ge$6, y $\ge$0.
The system of constraints is :
x ? 3 ...(1)
x + y ? 5 ...(2)
x + 2y ? 6 ...(3)
and y ? 0 ...(4)
The shaded region in the above figure is the feasible region determined by (1)–(4)
The corner points are C (6, 0), E(4, 1) and F (3, 2).
Applying Corner Point Method, we have :
It appears that Zmax = 1 at (3, 2).
But the feasible region is unbounded, therefore, we draw the graph of the inequality – x + 2y > 1.
Since the half-plane represented by – x + 2y > 1 has points common with the feasible region,
$\therefore$ Zmax $\ne$ 1.
Hence, Z has no maximum value.
Question 10:
Maximise Z = x + y subject to : x – y $\le$ – 1, – x + y $\le$ 0, x, y $\le$ 0.
The system of constraints is :
x – y $\le$ – 1 ...(1)
– x + y $\le$ 0 ...(2)
x, y $\ge$ 0 ...(3)
Draw the lines x – y = – 1 and – x + y = 0.
Question 11:
Reshma wishes to mix two types of food P and Q in such a way that the vitamin contents of the mixture contain at least 8 units of vitamin A and 11 units of vitamin B. Food P costs ₹ 60/kg and Food Q costs ₹ 80/kg. Food P contains 3 units/kg of Vitamin A and 4 units/kg of vitamin B. Food Q contains 5 units/kg of Vitamin A and 2 units/kg of vitamin B. Determine the minimum cost of the mixture.
Let Reshma mix ‘x’ kg of food P and ‘y’ kg of
food Q.
Thus the LPP problem is as follows :
Minimize
Z = 60x + 80y
Subject to :
3x + 4y $\ge$ 8 ...(1)
5x + 2y $\ge$ 11 ...(2)
and x, y $\ge$ 0 ...(3)
The shaded portion represents region, which is unbounded.
Question 12:
One kind of cake requires 200g of flour and 25 g of fat and another kind of cake requires 100 g of flour and 50 g of fat.Find the maximum number of cakes which can be made from 5 kg of flour and 1 kg of fat assuming that there is no shortage of the other ingredients used in making the cakes.
Let ‘x’ and ‘y’ be the number of cakes of first and second kind respectively.
Thus the LPP problem is as follows :
Maximize Z = x + y
Subject to :200 x + 100y $\le$ 5000
? 2x + y $\le$ 50 ...(1)
25x + 50y $\le$ 1000 $\le$ x + 2y $\le$ 40 ...(2)
and x, y $\ge$ 0.
Question 13:
A factory makes tennis rackets and cricket bats. A tennis racket takes 1·5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day,the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman’s time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity ?
(ii) If the profit on a racket and on a bat is ` 20 and ` 10 respectively, find the maximum profit of the factory when it works at full capacity.
Let the factory make ‘x’ tennis rackets and ‘y’ cricket bats.
We have :
Question 14:
A factory makes tennis rackets and cricket bats. A tennis racket takes 1·5 hours of machine time and 3 hours of craftman’s time in its making while a cricket bat takes 3 hours of machine time and 1 hour of craftman’s time. In a day,the factory has the availability of not more than 42 hours of machine time and 24 hours of craftman’s time.
(i) What number of rackets and bats must be made if the factory is to work at full capacity ?
(ii) If the profit on a racket and on a bat is ` 20 and ` 10 respectively, find the maximum profit of the factory when it works at full capacity.
Let the factory make ‘x’ tennis rackets and ‘y’ cricket bats.
We have :
Question 15:
A manufacturer produces nuts and bolts. It takes 1 hour of work on machine A and 3 hours on machine B to produce a package of nuts while it takes 3 hours on machine A and 1 hour on machine B to produce a package of bolts. He earns a profit of ` 17·50 per package of nuts and ` 7.00 per package of bolts. How many packages of each should be produced each day so as to maximize his profit,if he operates his machines for at the most of 12 hours a day ? Formulate this mathematically and then solve it.
Let ‘x’ and ‘y’ be the number of packages of nuts and bolts respectively.
We have the following constraints :
x $\ge$ 0 ...(1)
Question 16:
A factory manufactures two types of screws, A and B. Each type of screw requires the use of two machines, an automatic and a hand operated. It takes 4 minutes on the automatic and 6 minutes on hand operated machine to manufacture a package of screws A, while it takes 6 minutes on automatic and 3 minutes on the hand operated machine to manufacture a package of screws B. Each machine is available for at the most 4 hours on any day. The manufacturer can sell a package of screws A at a profit of ` 7 and screws B at a profit of ` 10. Assuming that he can sell all the screws he manufactures, how many packages of each type should the factory owner produce in a day in order to maximise his profit ? Determine the maximum profit.
Let the manufacturer produce ‘x’ packages of Screws
A and ‘y’ packages of Screws B.
Thus the LPP problem is as follows :
Maximize : Z = 7x + 10y
Question 17:
A cottage industry manufactures pedestal lamps and wooden shades, each requiring the use of a grinding/cutting machine and 3 hours on the sprayer to manufacture a pedestal lamp. It takes 1 hour on the grinding/cutting machine and 2 hours on the sprayer to manufacture a shade. On any day, the sprayer is available for at the most 12 hours and the grinding/cutting machine for at the most 12 hours. The profit from the sale of a lamp is ` 5 and that from a shade is ` 3. Assuming that manufacturer can sell all the lamps and shades that he produces, how should he schedule his daily production in order to maximise his profit ?
Let the manufacturer produce ‘x’ Pedestal
lamps and ‘y’ Wooden shades everyday. Then the LPP
problem is as follows :
Maximize : P = 5x + 3y ...(1)
Subject to : 2x + y $\le$ 12 ...(2)
3x + 2y $\le$ 20 ...(3)
and x $\ge$ 0, y $\ge$0 ...(4)
The shaded portion represents feasible region, which is bounded.
Question 18:
A company manufactures two types of novelty souvenirs made of plywood. Souvenirs of type A require 5 minutes each for cutting and 10 minutes each for assembling. Souvenirs of type B require 8 minutes each for cutting and 8 minutes each for assembling. There are 3 hours 20 minutes available for cutting and 4 hours for assembling. The profit is ₹ 5 each for type A and ₹ 6 each for type B souvenirs. How many souvenirs of each type should the company manufacture in order to maximise the profit ?
Let the company manufacture ‘x’ Souvenirs of type A and ‘y’ Souvenirs of type B.
Then the LPP problem is as follows :
Maximize P = 5x + 6y
Question 19:
A merchant plans to sell two types of personal computers–a desktop model and a portable model that will cost ₹ 25000 and ₹ 40000 respectively. He estimates that the total monthly demand of computers will not exceed 250 units. Determine the number of units of each type of computers which the merchant should stock to get maximum profit if he does not want to invest more than ₹ 70 lakhs and his profit on the desktop model is ₹ 4500 and on portable model is ₹ 5000.
Let the merchant stock ‘x’ Desktop computers and ‘y’ Portable computers.
Thus the LPP problem is as follows :
Maximize :
P = 4500x + 5000y
Subject to
Question 20:
A diet is to contain at least 90 units of vitamin A and 100 units of minerals. Two foods F1 and F2 are available. Food F1 costs ₹ 4 per unit food and F2 costs ₹ 6 per unit. One unit of food F1 contains 3 units of vitamin A and 4 units of minerals.One unit of food F2 contains 6 units of vitamin A and 3 units of minerals. Formulate this as a linear programming problem. Find the minimum cost for diet that consists of mixture of these two foods and also meets the minimal nutritional requirements.
Let diet contain ‘x’ units of food F1 and ‘y’ units of food F2.
Then the LPP problem is as below :
Minimize C = 4x + 6y
Question 21:
There are two types of fertilizers F1 and F2. F1 consists of 10% nitrogen and 6% phosphoric acid and F2 consists of 5% nitrogen and 10% phosphoric acid. After testing the soil conditions, a farmer finds that she needs at least 14 kg. of nitrogen and 14 kg of phosphoric acid for her crop. If F1 costs ₹ 6/kg and F2 costs ₹ 5/kg, determine how much of each type of fertlizer should be used so that nutrient requirements are met at a minimum cost. What is the minimum cost ?
Let ‘x’ kg. of fertilizer F1 and ‘y’ kg. of fertilizer
F2 be required.
We have :
Question 22:
The corner points of the feasible region determined by the following system of linear inequalities :
2x + y $\le$ 10, x + 3y $\le$ 15, x, y $\ge$0 are (0, 0), (5, 0), (3, 4) and (0, 5)
Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both (3, 4) and (0, 5) is :
1. p = q
2. p = 2q
3. p = 3q
4. q = 3p
q = 3p
Question 23:
The point, which does not lie in the half-plane 2x + 3y– 12 $\le$ 0 is:
1. (1, 2)
2. (2, 1)
3. (2, 3)
4. (2, 3)
(2, 3)
Question 24:
The point, which does not lie in the half-plane 3x + 4y – 15 $\le$ 0 is:
1. (1, 2)
2. (2, 1)
3. (2, 2)
4. (3, 2).
(3, 2).
Question 25:
The point, which lies in the half-plane 2x + 3y – 12 $\ge$ 0 is:
1. (1, 2)
2. (2, 1)
3. (2, 2)
4. (2, 3).
(2, 3).
Question 26:
Maximize Z = x + 2y subject to x + y $\ge$ 5, x $\ge$ 0, y $\ge$ 0 is :
1. 5 at (0, 5)
2. 10 at (0, 5)
3. 5 at (5, 0)
4. 10 at (5, 10)
10 at (0, 5)
Question 27:
Maximize Z = 3x + 2y subject to x + y $\ge$ 8, x $\ge$ 0, y $\ge$ 0 is :
1. 16 at (0, 8)
2. 16 at (8, 0)
3. 8 at (16, 0)
4. 8 at (0, 16).
16 at (0, 8)
Question 28:
Maximize Z = 2x + 3y subject to x + 2y $\le$ 6, x $\ge$ 4, y $\ge$ 0 is :
1. 6 at (6, 0)
2. 6 at (0, 6)
3. 12 at (6, 0)
4. 12 at (0, 6).
12 at (6, 0)
Question 29:
The maximum value of Z = 3x + 4y subject to the constraints :
x + y $\le$4; x $\ge$0, y $\ge$0 is :
1. 0
2. 12
3. 16
4. 18
16
Question 30:
The corner point of the feasible region determined by the system of linear constraints are (0, 10), (5, 3), (15, 15), (0, 20) ? Let Z = px + qy, where p, q > 0. Condition on p and q so that the maximum of Z occurs at both points (15, 15) and (0, 20) is :
1. p = q
2. p = 2q
3. q = 2p
4. q = 3p.
q = 3p.
Question 31:
A linear function, which is minimized or maximized is called:
1. An objective function
2. An optimal function
3. An optimal function
4. A feasiable function
An objective function
Question 32:
Objective fuction of a LPP is :
1. a function to be optimized
2. a constraint
3. a relation between the variables
4. None of these.
a function to be optimized
Question 33:
The optimal value of the objective function is attained at :
1. the points given by the corners of feasible region
2. the points given by the intersection of inequalities
3. the points given by the intersection of inequalities
4. the points given by the intersection of inequalities |
# How to find area bounded by lines pdf
## Bounded area lines
Add: jukobiq55 - Date: 2020-12-04 10:57:57 - Views: 6094 - Clicks: 9094
2), where f and g are continuous and. . . (1) Thus, we can calculate the how to find area bounded by lines pdf area of a how to find area bounded by lines pdf region (a two-dimensional notion) by using line integrals (a one-dimensional construction)! Next, how to find area bounded by lines pdf let&39;s sketch the graph so we can decide which is the uppermost curve:. ) that we have formulas for. The area of a typical rectangle is Δ x ( f ( x pdf i) − g ( x i)), so the total area is approximately.
In general, you can skip parentheses, but be very careful: e^3x is e 3 x, and e^ (3x) is e 3 x. Family of discs Figure 4. We now extend this notion to find the area bounded between the two curves y = f (x) and y = g(x)onthe interval a,b (see Figure 5. Area bounded by how to find area bounded by lines pdf the curves y_1 and y_2, & the lines x=a and x=b, including a typical rectangle. Also derive the area analytically and verify the same. ∑ i = 0 n − 1 ( f ( x i) − g ( x how to find area bounded by lines pdf i)) Δ x. A square with diagonals of length x 3.
Find the area contained by the curve how to find area bounded by lines pdf y = x(x −1)(x+1) and the x-axis. Area Expressed as the Limit of a Polygon Before we determine an exact area, we estimate the value using polygons. If f(x) is a continuous and nonnegative function of x on the closed interval a, b, then the area of the region bounded by the graph of f, the x-axis and the vertical. f&92;left ( x &92;right) f ( x) between the vertical lines. Figure the area of the region on the how to find area bounded by lines pdf right.
Solve the equation \$y = 5 - x\$ with both of the equations \$y = x\$ and how to find area bounded by lines pdf \$y = 2x\$ how to find area bounded by lines pdf to determine the coordinates \$(x_1, y_1)\$ and \$(x_2, y_2)\$ of \$P\$ and \$Q\$ respectively. Finally, whether we think of the area between two curves as the difference between the area bounded by the individual curves (as in Equation &92;(&92;ref6. Area between curves defined by two given functions. For example, suppose that you want to how to find area bounded by lines pdf calculate the shaded area between y = x2 and as shown in this figure. Also, be how to find area bounded by lines pdf careful when you write fractions: 1/x^2 ln pdf (x) is 1 x 2 ln ( x), and 1/ (x^2 ln (x)) is 1 x 2 ln ( x). Given a curve C: y = f(x) and a straight line T: y = mx + c. Find the area bounded by the curve, x=axis and the given lines. (see Figure1 to 4 below): Figure 1.
Hi Sidra, I think This is for a calculus class so you are expected to use integration. In general, you can skip the multiplication sign, how so 5 x is equivalent to 5 ⋅ x. regions that aren’t rectangles.
Using vector algebra, find the area of a parallelogram/triangle. Solution (5) Find the area of the region bounded by x² = 36 y, y - axis, y = 2 and y = 4. Find how to find area bounded by lines pdf the area of the region bounded by the curves &92;(y = x+2&92;), &92;(y = &92;sin x&92;), &92;(x = 2&92;) and the &92;(y&92;)-axis. In this case the formula how to find area bounded by lines pdf pdf is, A= ∫ d c f (y)−g(y) dyA = ∫ c d f ( y) − g ( y) d y.
Area = ∫ c b f ( x) − g ( x) d x. find area bounded by functions: y=x. It&39;s fairly simple to see the trick to accomplish how to find area bounded by lines pdf this once you can imagine how to how to find area bounded by lines pdf use a single integral to calculate the length of the interval. But, the approach is quite different. Find the area between the curves y = x 2 and y = x 3. The region corresponding to this integral is R= (x;y) : 0 y 4;y 2 x y 2 + 1. The area of a region bounded by a graph of a function, the how to find area bounded by lines pdf x‐axis, and two vertical boundaries can be determined directly by evaluating a definite integral. Add up the areas of the two regions to get the total area.
Typically we use Green&39;s theorem as an alternative way to calculate a line integral how to find area bounded by lines pdf \$&92;dlint\$. It is shown in –gure 4. &92;beginalign &92;mathbfArea = &92;int_-&92;sqrt2^&92;sqrt2 (-y^2 + 2) &92;: dy &92;&92; &92;mathbfArea = -&92;fracy^33 + 2y &92;bigg |_-&92;sqrt2^&92;sqrt2 &92;&92; &92;mathbfArea.
gles as an approximation of the area A under the curve: A ≈ n i=1 f (c i) x. A semicircle of diameter x 5. x = y3 and x = y2 ⇒ 3. x y (0, b) (a, 0) D Figure 6. Recall that the area under the graph of a continuous function.
An isosceles right triangle with legs of length how to find area bounded by lines pdf x. I how to find area bounded by lines pdf thought of using integration and getting the how to find area bounded by lines pdf area under the curves but I&39;m having difficulty with it. Area Between Curves Volumes of Solids of Revolution Area Between Curves Theorem: Let f(x) and g(x) be continuous functions on the interval a;b such that f(x) g(x) for all x in a;b. . An equilateral triangle with sides of length x 6.
This is how a type II region. Note that pdf the height of a representative rectangle is always its top minus how to find area bounded by lines pdf its bottom, regardless of whether these numbers are positive or negative. To how to find area bounded by lines pdf find an area between two functions, you need to set up an equation with a combination of definite integrals of both functions. There are times when you need to find the area of a shape that is not a regular shape. We are trying to find the pdf area between 2 curves, y_1 = f_1(x) and y_2 = f_2(x), and the lines x = a and x = b. y = cos(πx/2) and y = 1−x2 (in the how to find area bounded by lines pdf first quadrant) ⇒. Is that what you&39;re doing in that example?
Area = ∫ a b (Top-Bottom) d x. 2dx how to find area bounded by lines pdf dy= area of D. Line 2 how is \$&92;overlineBC\$ Line 3 is \$&92;overlineCD\$ Line 4 is \$&92;overlineDA\$ This would pdf be clearer how to find area bounded by lines pdf if you graph the points then connect the A to B to C to D to A.
I how need to get the area bounded by the 4 lines or inside the 4 points connected. Show Instructions. See more videos for How To Find Area Bounded By Lines Pdf.
21: Region bounded by y= 0, y= how to find area bounded by lines pdf how to find area bounded by lines pdf 4, y= 2xand y= 2x 1 Example 360 Evaluate R 4 0 R y 2 +1 y 2 2x y 2 dxdy by applying the transformation u= 2x y 2 and v= y 2. Find the area bounded by parabola and an oblique line. Area Under Simple Curves; The Area Between Two Curves; how to find area bounded by lines pdf Bounded Regions. can be computed how to find area bounded by lines pdf by the definite how to find area bounded by lines pdf integral: A = &92;int&92;limits_a^b f&92;left ( x &92;right)dx = F&92;left ( b &92;right) – F&92;left ( a &92;right), A how to find area bounded by lines pdf = b ∫ a f ( x) d x = F ( b) − F ( a), where. 5 Area bounded by two curves.
f ( x) > g ( x) then the area between them bounded by the horizontal lines x = a and x = b is. x = a, x = a, x = b x = b. Find the area enclosed by the given pdf curve, the x-axis, and the given ordinates.
how to find area bounded by lines pdf Then the area of the region between f(x) and g(x) on a;b is Z b a f(x) g(x) dx or, less formally, Z b a upper lower dx or Z d c right how to find area bounded by lines pdf left dy! Warm Up: Find the area of the following figures: 1. By using this website, you agree to our Cookie Policy. In the equations de–ning u. To remember this formula we write. We will illustrate how a double integral of a function can be interpreted as the net volume of the solid between the surface given by the function and the xy-plane. Figure the area of the region on the left.
Will that work for all equations? In this section we will start evaluating double integrals over general regions, i. We draw points on the curves and connect them with line segments to form polygons that represent the curves*.
(Any other pair of curves which are specified in the syllabus may also be taken. If f(x) ≥ 0 on a, b, then the area ( A) of the region lying below the graph of f(x), above the x‐axis, and between the lines x = a and x = b is. We can approximate the area by dividing the area into thin how sections and approximating the area of each section by a rectangle, as indicated in figure 9. The calculator will find the area between two curves, or just under one curve. In the equations de–ning u Here we are going to determine the area between x =f (y) x = f ( y) and x = g(y) x = g ( y) on the interval c,d c, d with f (y) ≥ g(y) f ( y) ≥ g ( y).
how to find area bounded by lines pdf Example 12 Find the area of region bounded by the line 𝑦=3𝑥+2, the 𝑥−𝑎𝑥𝑖𝑠 and the ordinates 𝑥=−1 and 𝑥=1 pdf First Plotting 𝑦=3𝑥+2 In graph Area Required = Area ACB + Area ADE Area ACB Area ACB = −1 −2 3𝑦 𝑑𝑥 𝑦→ equation of line Area pdf ACB how to find area bounded by lines pdf = −1 −2 3 3𝑥. y = x4 −x2 and y = x2 (the part to the right of the y-axis) ⇒ 2. First, pdf note that the &92;(y&92;)-axis how to find area bounded by lines pdf is the line &92;(x = 0&92;), so the last two bounds give you the limits of integration. Here is my diagram. The first step is to plot the area under the curve and the straight line on the same graph. Free area under between curves calculator - find area between functions step-by-step This website uses cookies to ensure you get the best experience. x = 3y −y2 and x+ y = 3 ⇒ 5. Solution (4) Find the area of how the region bounded by the curve y = 3 x² - x and the x - axis between x = -1 and x = 1.
A square with sides of length x 2. \$&92;begingroup\$ I was thinking along the lines of calculating the area how to find area bounded by lines pdf bounded by the derived line and the x axis. Misc 14 Using the how to find area bounded by lines pdf method of integration find the area of the region bounded by lines: 2𝑥 + 𝑦 = 4, 3𝑥–2𝑦=6 and 𝑥–3𝑦+5=0 Plotting the 3 lines on the graph 2𝑥 + 𝑦 = 4 3𝑥 – 2𝑦 = 6 𝑥 – 3𝑦 + 5 = 0 Find intersecting Points A & B Point A Point A is intersection of lines x – 3y + 5 = 0 & 2x + how to find area bounded by lines pdf y = 4 Now, x – 3y + 5 = 0 x = 3y – 5 Putting x = 3y – 5 in 2x + y = 4 2(3y – 5) + y = 4 6y – how to find area bounded by lines pdf 10 + y = 4 7y = 14 y = 2 Putting y = 2 in x – 3y + 5 = 0 x. 2&92;)), these two results are the same, since the difference of two. area between curve and line, area bounded by circle and line, area between two curves calculator, area between two curves problems, how to find area bounded by two curves, area between two curves calculator symbolab, find the how to find area bounded by lines pdf area of how to find area bounded by lines pdf the region in the first quadrant bounded by the line, area between two curves pdf,area between curve and line, area bounded by circle how to find area bounded by lines pdf and line, find the area of. Various cases may be possible: Browse more Topics under Application Of Integrals. how to find area bounded by lines pdf We approximate the limit of the polygons’ area as the number of sides approaches infinity.
The area bounded by x = 3π/2, x = 2π, y = cos(x) and the x-axis is revolved around the y-axis. Find the image of a line with respect to a given plane. Basic sketch of the solid of revolution y-axis and the vertical line x=2 rotated about x-axis with few typical discs indicated. (b) The curve y = x2 +3x, from x = 1 to x = 3 (c) The curve y = x2 −4 from x = −2 to x = 2 (d) The curve y = x− x2 from x = 0 to x = 2 2. x = 1 −y2 and y = −x−1 ⇒ 4. Sometimes, we use double integrals how to find area bounded by lines pdf to calculate area as well. (a) The curve y = x, from x = 1 to x = 3. One method how to find area bounded by lines pdf of finding the area of.
21 The region inside the ellipse x 2/a 2+ y /b = 1. Find the area bounded by the pdf curves. If, for example, we are in two dimension, \$&92;dlc\$ is a simple closed curve, and \$&92;dlvf(x,y)\$ is defined everywhere inside \$&92;dlc\$, we can use Green&39;s theorem to convert the line integral into to double integral. Find the area bounded by the lines y = 4x + 5, y = 5 – x and 4y = x + 5. lines and the y –axis. Y2=4x, x=1, x=4 3.
Find the area bounded by the curve y = 2cosx and the x-axis from x = 0 to x = 2π. Compute the area bounded by the lines x + 2y = 2, y – x = 1 and 2x + y = 7. Introduction Area of Irregular Shapes Many shapes are not the basic shapes (rectangles, triangles, circles, etc. Calculate the volume of the resulting solid.
bounded by f(x), y-axis and the vertical line x=2 about the x-axis. EXAMPLE 3 Using formula (1), we compute the area inside the ellipse x2/a2 +y2/b2 =1 (Figure 6. Formulation of Area Under the Curve Bounded by a Line.
### How to find area bounded by lines pdf
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# Ch. - 7 Triangles
## Chapter 7 Ex.7.1 Question 1
In quadrilateral $$ACBD, AC = AD$$ and $$AB$$ bisects $$\angle A$$ (See the given figure). Show that \begin{align} \Delta ABC \cong \Delta ABD \end{align}. What can you say about $$BC$$ and $$BD$$?
### Solution
What is Known?
$$AC = AD$$ and $$AB$$ bisects
To prove:
$$\Delta {\text{ABC}} \cong \Delta {\text{ABD}}$$ and, what can be said about $$BC$$ and $$BD.$$
Reasoning:
We can show two sides and included angle of are equals to corresponding sides and included angle of by using SAS congruency criterion both triangles will be congruent and by CPCT, BC and BD will be equal.
Steps:
In $$\Delta ABC$$ and $$\Delta ABD$$,
\begin{align} AC &= AD \text {(Given)}\\\\\Delta CAB &=\Delta DAB\\&(AB \text { bisects } \angle A)\\\\AB &= AB\\ &\text {(Common)}\\ \\ \therefore \Delta ABC &\cong \Delta ABD\\ &\text {(By congruence rule)}\\ \\\therefore BC&=BD\\&\left(\begin{array}{I} \text{By corresponding parts}\, \\ \text{of congruent triangles} \\ \end{array}\!\right)\end{align}
Therefore, $$BC$$ and $$BD$$ are of equal lengths.
## Chapter 7 Ex.7.1 Question 2
$$ABCD$$ is a quadrilateral in which $$AD = BC$$ and $$\angle DAB = \angle CBA$$ (See the given figure).
Prove that
(i) $$\Delta ABD \cong \Delta BAC$$
(ii) $$BD = AC$$
(iii) $$\Delta ABD=\Delta BAC$$
### Solution
What is Known?
${AD}={BC}\ \text{and}\ \angle {DAB}=\angle {CBA}$
To prove:
(i) $$\Delta {{ABD}} \cong \Delta {{BAC}}$$
(ii) $${{BD}} = {{AC}}$$
(iii) $$\angle {{ABD}} = \angle {{BAC}}$$
Reasoning:
We can show two sides and included of are equals to corresponding sides and included angle of by using $$\rm{SAS}$$ congruency criterion both triangles will be congruent. Then we can say corresponding parts of congruent triangle will be equal.
Steps:
In $$\Delta ABD$$ and $$\Delta BAC$$,
\begin{align} AD &= BC \;\text{(Given)}\\\Delta DAB &=\Delta CBA \;\text{(Given)}\\AB &= BA \;\text{(Common)}\\\therefore \Delta ABD &\cong \Delta BAC\\& \text{(By congruence}\,\text{rule})\\\\\end{align}
\begin{align}\therefore BD &= AC \;\text{(By CPCT )}\\& \text{and}\\\angle ABD &=\angle BAC \;\text{(By CPCT )}\end{align}
## Chapter 7 Ex.7.1 Question 3
$$AD$$ and $$BC$$ are equal perpendiculars to a line segment $$AB$$ (See the given figure). Show that $$CD$$ bisects $$AB$$.
### Solution
What is Known?
$${\text{AD}} \bot {\text{AB, BC}} \bot {\text{AB}}\;{\text{and AD}} = {\text{BC}}$$
To prove:
$$CD$$ bisects $$AB$$ or $$OA = OB$$
Reasoning:
We can show two triangles $$OBC$$ and $$OAD$$ congruent by using AAS congruency rule and then we can say corresponding parts of congruent triangles will be equal.
Steps:
In $$\Delta BOC$$ and $$\Delta AOD$$,
\begin{align}\angle BOC& = \angle AOD\\\text{(Vertically opp}&\text{osite angles)}\\\\\angle CBO &= \angle DAO\text{(Each}\, 90^ {\circ})\\BC &= AD \text{(Given)}\\\therefore \Delta BOC &\cong \Delta AOD\\ \text{(AAS} \,& \text{congruence rule)}\\\\\therefore BO &= AO \text{(By CPCT)}\end{align}
Hence $$CD$$ bisects $$AB$$.
## Chapter 7 Ex.7.1 Question 4
$$l$$ and $$m$$ are two parallel lines intersected by another pair of parallel lines $$p$$ and $$q$$ (see the given figure). Show that $$\Delta ABC \cong \Delta CDA$$.
### Solution
What is Known?
$$l\parallel m\,\,\,and\,\,\,\,p\parallel q\,$$
To prove:
$$\Delta {\text{ABC}} \cong \Delta {\text{CDA}}{\text{.}}$$
Reasoning:
We can show both the triangles congruent by using ASA congruency criterion
Steps:
In $$\Delta ABC$$ and $$\Delta CDA$$,
\begin{align}&\angle BAC \,\text{and}\, \angle DCA \\&\text{(Alternate interior angles, as}(p \|q)\\\\AC &= CA \text{(Common)}\\&\angle BAC \text{and} \angle DCA \\&\text{(Alternate interior angles, as} (l \|m)\\\\&\therefore\,\, \Delta ABC \cong \Delta CDA \\&\text{(By ASA congruence rule)}\end{align}
## Chapter 7 Ex.7.1 Question 5
Line $$l$$ is the bisector of an angle $$\angle A$$ and $$\angle B$$ is any point on $$l$$. $$BP$$ and $$BQ$$ are perpendiculars from $$B$$ to the arms of $$\angle A$$ (see the given figure).
Show that:
(i) $$\Delta APB \cong \Delta AQB$$
(ii) $$BP = BQ$$ or $$B$$ is equidistant from the arms of $$\angle A$$.
### Solution
What is known?
$$l$$ is the bisector of an angle $$\angle {\text{A}}$$ and $$BP \bot AP\;{\text{and BQ}} \bot {\text{AQ}}$$
To prove:
$$\Delta {\text{APB}} \cong \Delta {\text{AQB}}$$ and $${\text{BP}}\,{\text{ = }}\,{\text{BQ}}$$ or B is equidistant from the arms of $$\angle {\text{A}}$$
Reasoning:
We can show two triangles APB and AQB congruent by using AAS congruency rule and then we can say corresponding parts of congruent triangles will be equal.
Steps:
In $$\Delta APB$$ and $$\Delta AQB$$,
\begin{align}&\angle BAP= \angle BAQ \\&\text{(l) is the angle bisector of }(\angle A)\\\\&\angle APB = AQB ( \text{Each} \;90^ {\circ})\\AB&=AB \text{(Common)}\\&\therefore \Delta APB \cong \Delta AQB \\&\text{(By AAS congruence rule)}\\\\\therefore BP &= BQ (By CPCT)\end{align}
Or, it can be said that $$B$$ is equidistant from the arms of $$\angle A$$.
## Chapter 7 Ex.7.1 Question 6
In the given figure, $$AC = AE$$, $$AB = AD$$
and $$\angle BAD = \angle EAC$$.
Show that $$BC = DE$$.
### Solution
What is Known?
$$AC = AE, AB = AD$$ and
$$\angle BAD = \angle EAC.$$
To prove:
$$BC = DE.$$
Reasoning:
We can show two triangles BAC and DAE congruent by using SAS congruency rule and then we can say corresponding parts of congruent triangles will be equal. To show both triangles congruent two pair of equal sides are given and add angle DAC on both sides in given pair of angles BAD and angle EAC to find the included angle BAC and DAE.
Steps:
It is given that
$$\angle BAD = \angle EAC$$
$$\angle BAD\!+\!\angle DAC\!=\!\angle EAC\!+\!\angle DAC$$
$$\angle BAC = \angle DAE$$
In $$\Delta BAC$$ and $$\Delta DAE$$,
\begin{align} AB & =AD\,(\text{Given})\, \\ \angle BAC&=\angle DAE\, \\ & (\text{Proven}\,\,\text{above})\, \\ & \\ AC&=AE\,(\text{Given})\, \\ \therefore \Delta BAC&\cong \Delta DAE\, \\ (\text{By}\,\text{SAS cong}&\text{ruencerule}) \\ & \\\therefore BC & =DE\,(ByCPCT) \\ \end{align}
## Chapter 7 Ex.7.1 Question 7
$$AB$$ is a line segment and $$P$$ is its mid-point. $$D$$ and $$E$$ are points on the same side of $$AB$$
such that $$\angle BAD = \angle ABE$$ and
$$\angle EPA = \angle DPB$$ (See the given figure).
Show that:
(i) $$\Delta DAP \cong \Delta EBP$$
(ii) $$AD = BE$$
### Solution
What is Known?
$$P$$ is its mid-point of $$AB$$
$$\angle BAD = \angle ABE{\text{ }}and\;\angle EPA = \angle DPB$$,
To prove:
(i) $$\Delta DAP \cong \Delta EBP$$ and (ii) $$AD = BE$$ .
Reasoning:
We can show two triangles DAP and EBP congruent by using ASA congruency rule and then we can say corresponding parts of congruent triangles will be equal. To show both triangles congruent one pair of equal sides and one pair of equal angles are given and add angle EPD on both sides in given pair of angles EPA and angle DPB to find the other pair of angles APD and BPE
Steps:
It is given that
$$\angle EPA = \angle DPB$$
$$\angle EPA + \angle DPE = \angle DPB + \angle DPE$$
$$\therefore \angle DPA = \angle EPB$$
In $$\Delta DAP$$ and $$\Delta EBP$$,
\begin{align}\angle DAP&=\angle EBP\,\,\,(\text{Given})\,\\AP&=BP\,\P\,\text{is}\,\text{mid}\,&\text{point}\,\text{of}\,\text{AB})\,\\\\\angle DPA&=\angle EPB\,\,\,(\text{From}\,\text{above})\,\,\\\therefore \Delta DAP&\cong \text{ }\!\!~\!\!\text{ }\Delta EBP\,\\&(\text{ASAcongruencerule})\,\\\\\therefore AD&=BE\,(By\,CPCT)\end{align} ## Chapter 7 Ex.7.1 Question 8 In right triangle \(ABC, right angled at $$C$$, $$M$$ is the mid-point of hypotenuse $$AB$$. $$C$$ is joined to $$M$$ and produced to a point $$D$$ such that $$DM = CM$$. Point $$D$$ is joined to point $$B$$ (see the given figure).
Show that:
(i) $$\Delta AMC \cong \Delta BMD$$
(ii) $$\angle DBC$$ is a right angle.
(iii) $$\Delta DBC \cong \Delta ACB$$
(iv) \begin{align}CM = \frac {1} {2}AB \end{align}
### Solution
What is Known?
$$M$$ is the mid-point of hypotenuse $$AB,$$
$$\angle C = 90^\circ$$ and $$DM = CM$$
To prove:
(i) $$\Delta AMC\cong \Delta BMD$$
(ii) $$\angle DBC$$ is a right angle.
(iii) $$\Delta DBC\cong \Delta ACB$$
(iv) $$CM=\frac{1}{2}AB$$
Reasoning:
We can show two triangles AMC and BMD congruent by using SAS congruency rule and then we can say corresponding parts of congruent triangles will be equal means angle ACM will equal to angle BDM, which are alternate interiors angle and can conclude DB is parallel to AC. Now it will help to find angle DBC by co-interior angles. Similarly triangles DBC and ACB will be congruent by using SAS criterion and CM will be half of AB by using M mid-point.
Steps:
(i) In $$\Delta AMC$$ and $$\Delta BMD,$$
\begin{align}&AM = BM\\& (\text{M is the} \text{ mid-point of AB})\\\\&\angle AMC = \angle BMD\&\text{Vertically}\text{ opposite angles} )\\\\&\text{CM} = DM\left( \text{Given} \right)\\&\therefore \Delta AMC \cong \Delta BMD\\( &\text{By SAS}\text{ congruence rule})\\\\&\therefore AC = BD\left( \text{By CPCT} \right)\\&\text{And,}\,\angle ACM = \angle BDM\left( \text{By CPCT} \right)\end{align} (ii) \(\begin{align} \angle ACM\text{ }&=\angle BDM \end{align}
However, $$\angle ACM$$ and $$\angle BDM$$ are interior alternate angles.
Since alternate angles are equal,
It can be said that $$DB\, \|\,AC$$
\begin{align} \angle DBC+\angle ACB&={180}^{\circ}\\text{Co-interior}&\text{ angles }) \\\\ \angle DBC+{90}^{\circ}&={180}^{\circ} \\ \therefore \angle DBC&={90}^{\circ} \\ \end{align} (iii) \(\Delta DBC\,\text{ and}\,\Delta ACB
\begin{align}\\DB&=AC\;(\text{Already proved})\\\angle DBC&=\angle ACB\\text{Each }&\, \text{angle measures }90^{\circ})\\\\BC&=CB\;(\text{Common})\\\therefore \Delta DBC & \cong \Delta ACB\\(\text{SAS }&\text{congruence rule})\\\end{align} (iv) \( \Delta DBC\cong \Delta ACB
\begin{align}\\AB&=DC \;(\text{By }CPCT)\\AB&=2\,\text CM \\ \therefore CM&=\frac{1}{2}AB\\\end{align}
The chapter 7 starts with an introduction of triangles and its properties we covered in previous grades. Next, the chapter explains about the congruence of triangles followed by criteria for congruence of triangles. Later the chapter discusses in detail about properties of a triangle with some examples and few exercise problems. Finally, the chapter ends by discussing some more criteria for congruence of triangles followed by inequalities in the triangle.
Triangles | NCERT Solutions
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# If $A$=$\left[ \begin{matrix} 3 & x \\ 0 & 1 \\\end{matrix} \right]$ and $B$=$\left[ \begin{matrix} 9 & 16 \\ 0 & -y \\\end{matrix} \right]$. Find $x$ and $y$ when $A^2=B$
Last updated date: 23rd Jul 2024
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Views today: 7.47k
Answer
Verified
347.7k+ views
Hint: We are given equations in the matrix form. So, we will first create a matrix on either side and then we will compare the elements. After doing that we will calculate the value of$x$and$y$. We need to find $A^2$ as well, which means we have to multiply the matrix $A$ by itself. The matrix multiplication is a bit of a complex process because it is not done like the real numbers. After making a matrix on both sides, we will compare the matrices element-wise and obtain the result.
Complete step by step answer:
To multiply the matrix $A$ by itself, we use the formula below for matrix multiplication:
If $A=[a_{ij}]$ is an $m\times n$ matrix and $B=[b_{ij}]$ is an $n\times p$ matrix,
The product AB is an $m\times p$ matrix.
$AB=[c_{ij}]$
Where$c_{ij}=a_{i1}b_{1j}+a_{i2}b_{2j}+...+a_{in}b_{nj}$
So, we have:
$A=\left[ \begin{matrix} 3 & x \\ 0 & 1 \\ \end{matrix} \right]$
Using the formula we obtain:
\begin{align} & {{A}^{2}}=\left[ \begin{matrix} 3 & x \\ 0 & 1 \\ \end{matrix} \right]\times \left[ \begin{matrix} 3 & x \\ 0 & 1 \\ \end{matrix} \right] \\ & =\left[ \begin{matrix} 3\times 3+\left( x\times 0 \right) & 3\times x+\left( x\times 1 \right) \\ 0\times 3+\left( 1\times 0 \right) & 0\times x+\left( 1\times 1 \right) \\ \end{matrix} \right] \\ \end{align}
$\Rightarrow {{A}^{2}}=\left[ \begin{matrix} 9 & 4x \\ 0 & 1 \\ \end{matrix} \right]$
Hence, we have found $A^2$
Now, we plug these values in the equation given:
${{A}^{2}}=B$
$\Rightarrow \left[ \begin{matrix} 9 & 4x \\ 0 & 1 \\ \end{matrix} \right]=\left[ \begin{matrix} 9 & 16 \\ 0 & -y \\ \end{matrix} \right]$
Now, we compare the elements. After comparing the element at first row and second column we get:
$4x=16$
$\Rightarrow x=\dfrac{16}{4}$
$\Rightarrow x=4$
Now we compare the element at second row, second column:
$1=-y$
$\Rightarrow y=-1$
So, the values of $x$ and $y$ have been found.
Note: Make sure that you add the terms before giving the resultant value in each position of the resultant matrix. Look for any calculation mistake that might occur while doing multiplication. Always check the other elements to cross check if you have made any calculation mistakes. |
How do you integrate (4x^2-x+3)/((x+5)(x-1)(x-2)) using partial fractions?
Aug 28, 2016
$- \ln | x - 1 | + \frac{17}{7} \ln | x - 2 | + \frac{18}{7} \ln | x + 5 | + K$, or,
$\ln | \frac{{\left(x + 5\right)}^{\frac{18}{7}} {\left(x - 2\right)}^{\frac{17}{7}}}{x - 1} | + K$.
Explanation:
Let $I = \int \frac{4 {x}^{2} - x + 3}{\left(x + 5\right) \left(x - 1\right) \left(x - 2\right)} \mathrm{dx}$.
We will use the Method of Partial Fraction to split the Integrand
(4x^2-x+3)/((x+5)(x-1)(x-2))=A/(x-1)+B/(x-2)+C/(x+5);
where, $A , B , C \in \mathbb{R}$.
To determine, $A , B , C$, let us use Heavyside's Cover-up Method :-
$A = {\left[\frac{4 {x}^{2} - x + 3}{\left(x + 5\right) \left(x - 2\right)}\right]}_{x = 1} = \frac{6}{-} 6 = - 1$;
$B = {\left[\frac{4 {x}^{2} - x + 3}{\left(x + 5\right) \left(x - 1\right)}\right]}_{x = 2} = \frac{17}{7}$;
$C = {\left[\frac{4 {x}^{2} - x + 3}{\left(x - 1\right) \left(x - 2\right)}\right]}_{x = - 5} = \frac{108}{\left(- 6\right) \left(- 7\right)} = \frac{18}{7}$.
Therefore, $I = \int \left[- \frac{1}{x - 1} + \frac{\frac{17}{7}}{x - 2} + \frac{\frac{18}{7}}{x + 5}\right] \mathrm{dx}$
$= - \ln | x - 1 | + \frac{17}{7} \ln | x - 2 | + \frac{18}{7} \ln | x + 5 | + K$, or,
$= \ln | \frac{{\left(x + 5\right)}^{\frac{18}{7}} {\left(x - 2\right)}^{\frac{17}{7}}}{x - 1} | + K$.
Enjoy Maths.! |
# Important Questions for Chapter 10 - Circles
To get good marks That too in Maths you need more practice. So for you benifit we prepered a PDF it contains important questions for CBSE class 10 Math chapter-wise and topic-wise.Download your free set of important CBSE questions along with the solutions now!
Important questions based on NCERT syllabus for Chapter 10 - Circles:
Question 1: In Fig., if AB = AC, prove that BE = EC
Solution:
Since tangents from an exterior point to a circle are equal in length
Therefore, AD = AF ... (i) [Tangents from A]
BD = BE ....(ii) [Tangents from B]
CE = CF ....(iii) [Tangents from C]
Now, AB = AC
⇒ AB – AD = AC – AF [Using (i)]
⇒ BD = CF
⇒ BE = CF [Using (ii)]
⇒ BE = CE [Using (iii)]
Question 2: PA and PB are tangents from P to the circle with center O. At point M which lies on circle, a tangent is drawn cutting PA at K and PB at N. Prove that KN = AK + BN.
Solution:
We know that the tangents drawn from an external point to a circle are equal in length.
KA = KM .... (i) [From K]
and, NB = NM .... (ii) [From N]
Adding equations (i) and (ii), we get
KA + NB = KM + NM
⇒ AK + BN = KM + MN
⇒ AK + BN = KN
Question 3: Prove that the tangents at the extremities of any chord make equal angles with the chord.
Solution:
Let AB be a chord of a circle with center O, and let AP and BP be the tangents at A and B respectively.
Suppose the tangents meet at P .
Join OP. Suppose OP meets AB at C . We have to prove that ∠PAC = ∠PBC .
In triangles PCA and PCB , We have
PA = PB
∠APC = ∠BPC
[The tangents are equally inclined to line joining extrernal point and centre of circle.]
PC = PC [Common]
So, by SAS - criterion of congruence, we have
ΔPAC ≅ ΔPBC
⇒ ∠PAC = ∠PBC |
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# LCM of Decimal Numbers
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Last updated date: 14th Aug 2024
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## An Overview of LCM
The least Common Multiple is the meaning of the abbreviation LCM. The lowest number that may be divided by both numbers is the least common multiple (LCM) of two numbers. It can also be computed using two or more numbers. Finding the LCM of a given set of numbers can be done in various ways. Utilizing the prime factorization of each number and then calculating the product of the highest powers of the shared prime factors is one of the quickest techniques to determine the LCM of two numbers. In this article, we will learn how the lcm of decimal numbers can be found and see LCM of fractions formula.
## How to Find the LCM of Decimal Numbers?
To find the LCM of decimal numbers, we have two different methods. Anyone willing to use any of these can use it anywhere to solve LCM of decimal numbers. The two different methods to determine the LCM of a set of numbers are the following:
LCM by Using the Listing Method
In this method, we shift the decimal places before taking out the LCM and in the end, after doing prime factorization. We again move the decimal places back to the position.
Let’s understand it with an example:
LCM of 1.50 and 5.00 by Listing Method: For 1.50 and 5.00, move the decimal 2 places to the right. Now, they are whole numbers. We will find the LCM of 150 and 500. In the end, we will move the decimal point back to 2 places to the left.
Multiples of 150 are 150, 300, 450, 600, 750, 900, 1050, 1200, 1350, 1500, 1650, and 1800, ….
Multiples of 500 are 500, 1000, 1500, 2000, 2500, …
The lowest common multiple of 150 and 500 is 1500
So, LCM(150, 500) = 1500.
Now, we move the decimal towards 2 places to the left.Therefore LCM(1.50,5.00) = 15.00.
LCM by Using the Division Method
Let’s understand the LCM by using the division method through an example:
LCM of 1.50 and 5.00 by Common Division Method
For each of the numbers, we move the decimal to two places to the right and calculate the LCM of whole numbers 150 and 500. Then, in the end, we move the decimal to two places back.
LCM of 150 and 500 is1500
We see that LCM of 150 and 500 is 1500. Since we multiplied by 100 to remove the decimal point, we will divide by 100, moving the decimal point two places back. So, we get the LCM of 1.5 and 5 as 15.
Hence, LCM of 1.50 and 5.00 is 15.
## LCM of Fractions Formula
To solve any problem efficiently, one must know its formula. In the same way, finding out the LCM of fractions requires some formula. To find L.C.M of $\dfrac{a}{b}$ and $\dfrac{c}{d}$ the generalized formula will be:
L.C.M $=\dfrac{\text { L.C.M of numerators }}{\text { H.C.F of denominators }}$.
Now L.C.M of two numbers is the smallest number (not zero), a multiple of both.
Let's take the example of $\dfrac{2}{5} \text { and } \dfrac{3}{7}$
$\text { L.C.M }=\dfrac{\text { L.C.M of }(2,3)}{\text { H.C.F of }(5,7)}$ ....(1)
So H.C.F of 5,7:
The factors of 5 are: 1,5
The factors of 7 are: 1,7
1 is the only common factor as it is the only number common to both 5 and 7.
Therefore, H.C.F of $(5,7)=1$. .....(2)
Now L.C.M of 2, 3:
The multiples of 2 are $2,4,6,8, \ldots$.
The multiples of 3 are $3,6,9,12, \ldots$
6 is the lowest common multiple as it is a multiple common to both 2 and 3.
Therefore, L.C.M of $(2,3)=6$ .......(3)
Putting the value of H.C.F from equation (2) and the value of L.C.M from equation (3) in equation (1), we get L.C.M $=\dfrac{6}{1}=6$
LCM calculator with work can be done by using the online calculator, which is available on the internet. Through this calculator, we can easily find the LCM online.
## Solved Examples
Q 1. Find the least common multiple (LCM) of 6 and 15 using the division method.
Ans: Let us find the least common multiple (LCM) of 6 and 15 using the division method using the steps given below.
Step 1: 2 is the smallest prime number, a factor of 6. Write 2 on the left of the two numbers. For each number in the right column, continue finding prime numbers and their factors.
Step 2: 2 divides 6, but it is not a factor of 15, so we write the number 15 in the row below as it is. Continue the steps until 1 is left in the last row. Then, we divide 3 and 15 by 3. This gives us 1 and 3. We write 5 on the left side and finally get 1, 1 as the quotient in the last row.
Step 3: Then, we multiply these numbers on the left. The LCM is the product of all these prime numbers. LCM of 6 and 15 is 2 × 3 × 5 = 30.
LCM of 6 and 15
Q 2. Find the LCM of 25, 15, and 30 by listing method.
Ans: Let us use the following steps to find the LCM of the 3 numbers.
Step 1: List the first few multiples of all three numbers, This will be:
Multiples of 25 = 25, 50, 75, 100, 125, 150, 175, ....,
Multiples of 15 = 15, 30, 45, 60, 75, 90, 105, 120, 135, 150, 175,...
Multiples of 30 = 30, 60, 90, 120, 150, 180, 210, ...
Step 2: Among the common multiples of 25, 15 and 30, we can see that 150 is the least multiple that is common in all three numbers.
Therefore, the LCM of 25, 15 and 30 = 150.
## Practice Questions
Q 1. Find the LCM of 0.48, 0.72, and 0.108.
Ans: 4.32
Q 2. Find the LCM of $\dfrac{1}{3}$, $\dfrac{1}{6}$, and $\dfrac{1}{9}$.
Ans: $\dfrac{1}{18}$
Q 3. Calculate the LCM of 0.8, 0.2, and 4.8.
Ans: 4.8
## Summary
Finding out LCM is quite easy to learn and remember. Interestingly, the division method is highly used to find the LCM of any two or three numbers. We are sure the kids have learned in depth about the LCM of decimal numbers, how to find the LCM of decimal numbers, the LCM of fractions, the LCM calculator with work, and many more. Other than this, solving various types of practice questions will help the kids master the particular topic in a better way.
## FAQs on LCM of Decimal Numbers
1. Who first invented and used LCM in the mathematical world?
The first person to invent LCM was the famous mathematician Euclid. LCM was first used way back in 325-270 BC.
2. Why is solving LCM very important?
Learning and solving LCM problems are quite important as it helps in the thought process. Besides that, it plays a significant role in adding, subtracting, and comparing two or more fractions.
3. Is LCM used in accounting?
Fortunately, LCM is used in accounting. It is used when an investor value a company’s inventory, those assets are recorded on the balance sheet. |
# 5.3: Balancing Equations
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### Any leftovers?
When you cook a meal, quite often there are leftovers because you prepared more than people would eat at one sitting. Sometimes when you repair a piece of equipment, you end up with what are called "pocket parts"—small pieces you put in your pocket because you're not sure where they belong. Chemistry tries to avoid leftovers and pocket parts. In normal chemical processes, we cannot create or destroy matter (law of conservation of mass). If we start out with ten carbon atoms, we need to end up with ten carbon atoms. John Dalton's atomic theory said that chemical reactions basically involve the rearrangement of atoms. Chemical equations need to follow these principles in order to be correct.
## Balancing Chemical Equations
A balanced equation is a chemical equation in which mass is conserved and there are equal numbers of atoms of each element on both sides of the equation. We can write a chemical equation for the reaction of carbon with hydrogen gas to form methane $$\left( \ce{CH_4} \right)$$:
$\begin{array}{ccccc} \ce{C} \left( s \right) & + & \ce{H_2} \left( g \right) & \rightarrow & \ce{CH_4} \left( g \right) \\ 2 \: \ce{C} \: \text{atoms} & & 2 \: \ce{H} \: \text{atoms} & & 1 \: \ce{C} \: \text{atom,} \: 4 \: \ce{H} \: \text{atoms} \end{array}\nonumber$
In order to write a correct equation, you must first write the correct skeleton equation with the correct chemical formulas. Recall that hydrogen is a diatomic molecule and so is written as $$\ce{H_2}$$.
When we count the number of atoms of both elements, shown under the equation, we see that the equation is not balanced. There are only 2 atoms of hydrogen on the reactant side of the equation, while there are 4 atoms of hydrogen on the product side. We can balance the above equation by adding a coefficient of 2 in front of the formula for hydrogen.
$\ce{C} \left( s \right) + 2 \ce{H_2} \left( g \right) \rightarrow \ce{CH_4} \left( g \right)\nonumber$
A coefficient is a small whole number placed in front of a formula in an equation in order to balance it. The 2 in front of the $$\ce{H_2}$$ means that there are a total of $$2 \times 2 = 4$$ atoms of hydrogen as reactants. Visually, the reaction looks like the figure below.
In the balanced equation, there is one atom of carbon and four atoms of hydrogen on both sides of the arrow. Below are guidelines for writing and balancing chemical equations.
1. Determine the correct chemical formulas for each reactant and product.
2. Write the skeleton equation.
3. Count the number of atoms of each element that appears as a reactant and as a product. If a polyatomic ion is unchanged on both sides of the equation, count it as a unit.
4. Balance each element one at a time by placing coefficients in front of the formulas.
1. It is best to begin by balancing elements that only appear in one chemical formula on each side of the equation.
2. No coefficient is written for a 1.
3. NEVER change the subscripts in a chemical formula—you can only balance equations by using coefficients.
5. Check each atom or polyatomic ion to be sure that they are equal on both sides of the equation.
6. Make sure that all coefficients are in the lowest possible ratio. If necessary, reduce to the lowest ratio.
##### Example $$\PageIndex{1}$$: Balancing Chemical Equations
Aqueous solutions of lead (II) nitrate and sodium chloride are mixed. The products of the reaction are an aqueous solution of sodium nitrate and a solid precipitate of lead (II) chloride. Write the balanced chemical equation for this reaction.
###### Step 1: Plan the problem.
Follow the steps for writing and balancing a chemical equation listed in the text.
###### Step 2: Solve.
Write the skeleton equation with the correct formulas.
$\ce{Pb(NO_3)_2} \left( aq \right) + \ce{NaCl} \left( aq \right) \rightarrow \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)\nonumber$
Count the number of each atom or polyatomic ion on both sides of the equation.
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \\ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \\ 2 \: \ce{NO_3^-} \: \text{ions} & 1 \: \ce{NO_3^-} \: \text{ions} \\ 1 \: \ce{Na} \: \text{atom} & 1 \: \ce{Na} \: \text{atom} \\ 1 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}\nonumber$
The nitrate ions and the chlorine atoms are unbalanced. Start by placing a 2 in front of the $$\ce{NaCl}$$. This increases the reactant counts to 2 $$\ce{Na}$$ atoms and 2 $$\ce{Cl}$$ atoms. Then place a 2 in front of the $$\ce{NaNO_3}$$. The result is:
$\ce{Pb(NO_3)_2} \left( aq \right) + 2 \ce{NaCl} \left( aq \right) \rightarrow 2 \ce{NaNO_3} \left( aq \right) + \ce{PbCl_2} \left( s \right)\nonumber$
The new count for each atom and polyatomic ion becomes:
$\begin{array}{ll} \textbf{Reactants} & \textbf{Products} \\ 1 \: \ce{Pb} \: \text{atom} & 1 \: \ce{Pb} \: \text{atom} \\ 2 \: \ce{NO_3^-} \: \text{ions} & 2 \: \ce{NO_3^-} \: \text{ions} \\ 2 \: \ce{Na} \: \text{atom} & 2 \: \ce{Na} \: \text{atom} \\ 2 \: \ce{Cl} \: \text{atom} & 2 \: \ce{Cl} \: \text{atoms} \end{array}\nonumber$
The equation is now balanced since there are equal numbers of atoms of each element on both sides of the equation.
##### Simulation
Practice balancing chemical equations with this simulation:
Can you balance a chemical equation?
## Summary
• The process of balancing chemical equations is described.
## Review
1. What is the law of conservation of mass?
2. How did Dalton describe the process of a chemical reaction?
3. Why don’t we change the subscripts in order to balance an equation?
5.3: Balancing Equations is shared under a CC BY-NC license and was authored, remixed, and/or curated by LibreTexts. |
# Difference between revisions of "2001 AMC 10 Problems/Problem 20"
## Problem
A regular octagon is formed by cutting an isosceles right triangle from each of the corners of a square with sides of length $2000$. What is the length of each side of the octagon?
$\textbf{(A)} \frac{1}{3}(2000) \qquad \textbf{(B)} {2000(\sqrt{2}-1)} \qquad \textbf{(C)} {2000(2-\sqrt{2})} \qquad \textbf{(D)} {1000} \qquad \textbf{(E)} {1000\sqrt{2}}$
## Solution
$[asy] draw((0,0)--(0,10)--(10,10)--(10,0)--cycle); draw((0,7)--(3,10)); draw((7,10)--(10,7)); draw((10,3)--(7,0)); draw((3,0)--(0,3)); label("x",(0,1),W); label("x\sqrt{2}",(1.5,1.5),NE); label("2000-2x",(5,0),S);[/asy]$
$2000 - 2x = x\sqrt2$
$2000 = x(2 + \sqrt2)$
$x = \frac {2000}{2 + \sqrt2} =x = \frac {2000(2 - \sqrt2)}{(2 + \sqrt2)(2 - \sqrt2)}= \frac {2000(2 - \sqrt2)}{2} = 1000(2 - \sqrt2)$
$x\sqrt2 = 1000(2\sqrt {2} - 2) = \boxed{\textbf{(B)}\ 2000(\sqrt2-1)}$.
~edited by qkddud and now you can't tell how to solve it~ |
# Find a single vector x whose image under t is b
Transformation is defined as T(x)=Ax, find whether x is unique or not.
$A=\begin{bmatrix} 1 & -5 & -7\\ 3 & 7 & 5\end{bmatrix}$
$B=\begin{bmatrix} 2\\ 2\end{bmatrix}$
This question aims to find the uniqueness of vector $x$ with the help of linear transformation.
This question uses the concept of Linear transformation with reduced row echelon form. Reduced row echelon form helps in solving the linear matrices. In reduced row echelon form, we apply different row operations using the properties of linear transformation.
To solve for $x$, we have $T(x)=b$ which is to solve $Ax=b$ in order to solve for $x$. The augmented matrix is given as:
$A \begin{bmatrix} A & B \end{bmatrix}$
$=\begin{bmatrix} 1 & -5 & -7 & |-2\\ -3 & 7 & 5 & |-2 \end{bmatrix}$
Applying row operations to get the reduced echelon form.
$\begin{bmatrix} 1 & -5 & -7 & |-2\\ -3 & 7 & 5 & |-2 \end{bmatrix}$
$R_1 \leftrightarrow R_2 ,R_2 + \frac {1}{3} R_1 \rightarrow R_2$
By using the above row operations, we get:
$\begin{bmatrix} -3 & 7 & 5 & -2\\ 0 & -\frac{8}{3} & – \frac{16}{3} & -\frac{8}{3} \end{bmatrix}$
$-\frac{3}{8}R_2 \rightarrow R_2 ,R_1 – 7R_2 \ \rightarrow R_1$
$\begin{bmatrix} -3 & 0 & -9 & -9\\ 0 & 1 & 2 & 1 \end{bmatrix}$
$-\frac{1}{3}R_1 \rightarrow R_1$
The above operations results in the following matrix:
$\begin{bmatrix} 1 & 0 & 3 & 3\\ 0 & 1 & 2 & 1 \end{bmatrix}$
We get:
$x_1+3x_3 = 3$
$x_1 = 3 – 3x_3$
$x_2 + 2x_3 = 1$
$x_2 = 1 -2x_3$
Now:
$x= \begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} 3 – x_3\\ 1 – 2x_3\\ x_3 \end{bmatrix}$
$=\begin{bmatrix} 3 \\ 1\\ 0 \end{bmatrix} + x_3 \begin{bmatrix} -3 \\ -2\\ -1 \end{bmatrix}$
## Numerical Result
By applying a linear transformation of given matrices, it shows that $x$ does not have a unique solution.
## Example
Two matrices are given below. Find the unique vector x with the help of transformation $T(x)=Ax$
$A=\begin{bmatrix} 1 & -5 & -7\\ -3 & 7 & 5\end{bmatrix}$
$B=\begin{bmatrix} 4\\ 4\end{bmatrix}$
To solve for $x$, we have $T(x)=b$ which is to solve $Ax=b$ in order to solve for $x$. The augmented matrix is given as:
$A \begin{bmatrix} A & B \end{bmatrix}$
$R_2 + 3R_1$
$\begin{bmatrix} 1 & -5 & -7 & 4 \\ 0 & -8 & -16 & 16 \end{bmatrix}$
$-\frac{R_2}{8}$
$\begin{bmatrix} 1 & -5 & -7 & 4 \\ 0 & 1 & 2 & -2 \end{bmatrix}$
$R_1 + 5R_2$
$\begin{bmatrix} 1 & 0 & 3 & -6 \\ 0 & 1 & 2 & -2 \end{bmatrix}$
$x_1+3x_3 = -6$
$x_1 = -6 – 3x_3$
$x_2 + 2x_3 = -2$
$x_2 = -2 -2x_3$
$x= \begin{bmatrix} x_1 \\ x_2\\ x_3 \end{bmatrix} = \begin{bmatrix} -6 – 3x_3\\ -2 – 2x_3\\ x_3 \end{bmatrix}$
The above equation shows that $x$ does not have a unique solution. |
# What is the total number of pairs of integers (x,y) which satisfy the equation x^2-4xy+5y^2+2y-4?
giorgiana1976 | Student
First, you'll have to provide an equation. You've provided an expression, for the moment. We'll transform the expression into an equation:
x^2-4xy+5y^2+2y-4 = 0
We'll create the following groups;
(x^2-4xy+4y^2) + (y^2+2y + 1)- 1 - 4 = 0
We notice that we've made some changes within the expression you've provided, namely:
5y^2 = 4y^2 + y^2
- we've added and subtracted the value of 1
We notice that creating the groups above, we've created, in fact, two perfect squares:
(x-2y)^2 + (y+1)^2 - 5 = 0
Now, we'll have to determine what are the integers which added to give 5?
1 + 4, 2 + 3, 3 + 2, 4 + 1
Let x - 2y = 1 and y + 1 = 2
x = 2y + 1
y = 1 => x = 3
The first integer pair which satisfies the given equation is (3,1).
Let x - 2y = sqrt2 and y + 1 = sqrt3
Since x and y are not integer, we'll move to the next possibility:
x - 2y = 2 and y + 1 = 1 => y = 0
x = 2
The next possible pair of integer numbers, that makes the equality to hold is (2;0).
Therefore, the pairs of integer numbers that satisfy the given equation are: (3,1) and (2;0). |
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Linear Functions: Any function of the form f (x) = m x + b, where m is not equal to 0 is called a linear function.
The domain of this function is the set of all real numbers. The range of f is the set of all real numbers. The graph of f is a line with slope m and y intercept b. Note: A function f (x) = b, where b is a constant real number is called a constant function. Its graph is a horizontal line at y = b. Example 1: Graph the linear function f given by f (x) = 2x + 4 Solution to Example 1 You need only two points to graph a linear function. These points may be chosen as the x and y intercepts of the graph for example. Determine the x intercept, set f(x) = 0 and solve for x. 2x + 4 = 0 x = -2
Determine the y intercept, set x = 0 to find f(0). f(0) = 4 The graph of the above function is a line passing through the points (-2 , 0) and (0 , 4) as shown below.
Matched Problem : Graph the linear function f given by f (x) = x + 3 Example 2: Graph the linear function f given by f (x) = -(1 / 3)x - 1 / 2 Solution to Example 2 Determine the x intercept, set f(x) = 0 and solve for x. -(1 / 3)x - 1 / 2 = 0 x=-3/2
Determine the y intercept, set x = 0 to find f(0). f(0) = -1 / 2 The graph of the above function is a line passing through the points (-3 / 2 , 0) and (0 , -1 / 2) as shown below.
graph the linear function f given by f (x) = -x / 5 + 1 / 3 n what follows, SQRT means square root.
The domain of function f defined by f(x) = SQRT ( x ) is the set of all real positive numbers and zero because the square root of negative numbers are not real numbers (think of SQRT (- 4), is it real?). In inequality form, the domain of f(x) = SQRT ( x ) is written as x >= 0
in interval form the domain is given by [ 0 , + infinity) Example 1: Graph f( x ) = SQRT (x)
and find the range of f. Solution to Example 1: Because the domain of f is the set of all positive real numbers and zero, we might construct a table of values as follows: x SQRT (x) 0 0 1 1 4 2 9 3 16 4
The values of x were selected so that the square root of these values are whole numbers which make it easy to plot the points shown in the table.
The range of f is given by the interval [0 , +infinity). Example 2: Graph f( x ) = SQRT (x - 3)
and find the range of f. Solution to Example 2: First find the domain of the square root function given above by stating that the expression under the square root must be positive or equal to zero x - 3 >= 0 Solve the above inequality to obtain the domain of f as the set of all real values such that x >= 3 We now select values of x in the domain to construct a table of values. x SQRT (x - 3) 3 0 4 1 7 2 12 3
The interval [0 , +infinity) represents the range of f. Example 3: Graph f( x ) = - SQRT (- 2x + 4) + 1
and find the range of f. Solution to Example 3: The domain of the function given above is found by setting - 2x + 4 >= 0 Solve the above inequality to obtain the domain of f as the set of all real values such that x <= 2 We now select values of x in the domain of f to construct a table of values. These values are selected so that the square root term is a whole number and give points that are easy to plot. x - SQRT (-2 x + 4 ) + 1 2 1 3/2 0 0 -1 -5/2 -2 -6 -3
The range of f is given by the interval (-infinity , 1]. Example 4: Graph
2
f( x ) = SQRT (- x + 4)
and find the range of f. Solution to Example 4: The domain of function given above is found by solving the polynomial inequality - x + 4 >= 0 The solution set of the above inequality is given by the interval [-2 , 2] which is also the domain of the above function.
2
Let us write the given function as an equation as follows y = SQRT (- x + 4) Square both sides and arrange to obtain. x +y =2
2 2 2 2
The equation obtained is that of a circle. Hence the graph of f(x) = SQRT (- x + 4) is the upper half of a circle sinsce SQRT (- x + 4) is positive. Hence the graph below.
The interval [0 , 2] represents the range of f. Example 5: Graph
2
f( x ) = SQRT (x - 9)
and find the range of f. Solution to Example 5: The domain of the function given above is found by solving x - 9 >= 0 Which gives a domain reprsented by (-infinity , -3] U [3 , + infinity) We now select values of x in the domain of f to construct a table of values, noting f(x) = f(-x) hence a symmetry of the graph with respect to the y axis. x SQRT (x - 9)
2 2
3 0
5 4
8 7.4
The range of f is given by the interval [0 , + infinity).
Example 6: Graph
f( x ) = SQRT (x - 6x + 9)
and find the range of f. Solution to Example 6: Let us use write the expression under the square root as a square as follows x - 6x + 9 = (x - 3) Hence f( x ) = SQRT (x - 6x + 9) = SQRT ( (x - 3) ) = | x - 3 | The given function has been rewitten as an absolute value function. Function f may be written as a piecewise function and graphed as follows.
2 2 2 2
The range of f is given by the interval [0 , + infinity). Example 7: Graph
2
f( x ) = SQRT (x + 4x + 6)
and find the range of f. Solution to Example 7: Use completing the square to rewtite the expression under the square root as follows x + 4x + 6 = (x + 2) + 2 The expression under the square root is always positive hence the domain of f is the set of all real numbers. Let us first look at the 2 graph of (x + 2) + 2. It is a parabola.
2 2
We would expect the graph of f to have the same axis of symmetry, the vertical line, x = -2 as the above graph. The table of values may constructed as follows. x SQRT ( (x + 2) 2 + 2 ) -2 1.4 0 2.4 2 4.2 4 6.2
The range of f is given by the interval [SQRT(2) , + infinity). f is a quadratic function given by 2 f (x) = 2x + 2 x - 4 2 f (x) = x - 2 x - 3 f is a function given by f (x) = |x - 2|
The Elements of Music By Espie Estrella, About.com Guide Sound is created when an object vibrates. These vibrations are perceived by our ears and then sent to our brain. Our brain in turn analyzes these signals and let's us know what type of sound we are hearing (i.e. an alarm clock ringing, a car horn blaring, etc.). Music is differentiated from other sounds because it has certain qualities. When you listen to a piece of music, you'll notice that it has several different characteristics; it may be soft or loud, slow or fast, combine different instruments and have a regular rhythmic pattern. All of these are known as the "elements of music." Beat and Meter - In order to define meter, let's first define beats. Beats give music its regular rhythmic pattern. Beats are grouped together in a measure; the notes and rests corresponds to a certain number of beats. Meter refers to rhythmic patterns produced by grouping together strong and weak beats. Meter may be in duple (2 beats in a measure), triple (3 beats in a measure), quadruple (4 beats in a measure) and so on. Dynamics - Dynamics are abbreviations or symbols used to signify the degree of loudness or softness of a piece of music. It also indicates whether there is a change in volume. Harmony - In general, harmony refers to the combination of notes (or chords) played together and the relationship between a series of chords. But to give you a better understanding of harmony, let's first define melody. Melody refers to the tune of a song or piece of music. It is created by playing a series of notes one after another. Harmony accompanies and supports the melody. It is created by playing a group of notes (either simultaneously or as broken chords) behind the melody thus giving it musical texture. Key - Also known as tonality; a principle in music composition wherein at the end of the piece there is a feeling of completion by going back to the tonic. The tonic (main key or home key) is the principal pitch of a composition. Simply put, key refers to the central note (i.e. key of C), scale (i.e. C scale) and chord (i.e. C Major triad) Melody - It refers to the tune of a song or piece of music. it is the memorable tune created by playing a succession or series of pitches. Musical Instruments and Voice - Musical instruments are classified as percussion, strings,woodwinds, brass and keyboards. Another method of classifying musical instruments according to the type of vibrating material used to produce sound is called the Sachs-Hornbostel System. Our voice is also considered a musical instrument. Each of us has a different voice type or vocal range and no two voices are alike. Music Notation - Refers to the symbols used to represent music when writing it down. These symbols specify the pitch, rhythm and meter of a piece of music. Pitch - The relative lowness or highness that we hear in a sound. The pitch of a sound is based on the frequency of vibration and the size of the vibrating object. The slower the vibration and the bigger the vibrating object, the lower the pitch; the faster the vibration and the smaller the vibrating object, the higher the pitch. For example, the pitch of adouble bass is lower than that of the violin because the double bass has longer strings. Pitch may be definite (i.e. piano) or indefinite (i.e. cymbals). Rhythm - It may be defined as the pattern or placement of sounds in time and beats in music. Roger Kamien in his book Music: An Appreciation defines rhythm as "the particular arrangement of note lengths in a piece of music." Rhythm is shaped by meter; it has certain elements such as beat and tempo. Tempo - The Italian word at the beginning of a music piece that indicates how slow or fast the piece should be played. This is called the tempo which is effective throughout the duration of the music unless the composer indicates otherwise. Texture - Musical texture refers to the number of layers as well as the type of layers used in a composition and how these layers are related. Texture may be monophonic (single melodic line), polyphonic (two or more melodic lines) and homophonic (a main melody accompanied by chords). Timbre - Also known as tone color; it refers to the quality of sound that distinguishes onevoice or instrument from another. Timbre may range anywhere from dull to lush, from dark to bright (such as the sound of glockenspiels). |
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# Using Properties of Logarithms
fullscreen
Exercise
Simplify the expression
using the properties of logarithms.
Show Solution
Solution
The Quotient Property of Logarithms states that the following is true for logarithms.
In our expression the first two terms and the last two terms both are a difference of two logarithms with the same base. We can use the Quotient Property of Logarithms to rewrite each pair into a logarithm of a quotient.
This expression is a sum of two logarithms with the same base. We can write them into a logarithm of a product using the Product Property of Logarithms.
Let's continue simplifying this expression.
1
Thus, the expression equals 1.
## Change of Base Formula
The Change of Base Formula allows the logarithm of an arbitrary base to be rewritten as the quotient of two logarithms with another base.
With many calculators it is only possible to evaluate the common and the natural logarithm. The Change of Base Formula can then be used to evaluate logarithms of other bases.
## Solving an Exponential Equation using Logarithms
An exponential equation can be solved by applying logarithms and using the Power Property of Logarithms. Consider the following equation.
### 1
Apply the logarithm
First, the equation is rewritten by applying a logarithm on both sides.
### 2
Rewrite using the Power Property of Logarithms
By using the Power Property of Logarithms, powers can be rewritten into a product.
### 3
Solve the resulting equation
After the power has been rewritten into a product, the unknown variable x can be isolated using inverse operations. Here, x gets isolated on the left-hand side when both sides of the equation are divided by
By using a calculator, an approximate value of x can be calculated. Here,
fullscreen
Exercise
Solve the equation
using the common logarithm. State the answer with three significant digits.
Show Solution
Solution
The given equation can be solved by applying the on both sides. Since many calculators are limited to only the common and the natural logarithms, it is often not possible to use We can solve it anyway. Let's begin by applying the common logarithm on both sides.
The Power Property of Logarithms gives us the relationship With this we can rewrite the left-hand side of the equation.
We can now solve the equation for x by dividing both sides of the equation with
The solution to the equation is
## Inverse Properties of Logarithms
A logarithmic function is by definition the inverse of an exponential function This means that their function composition results in the identity function.
The fact that a logarithm and a power with the same base undo each other is what is known as the inverse properties of logarithms.
They also hold true for the common logarithm and the natural logarithm.
These properties together with other properties of logarithms permit to simplify logarithmic expressions and to solve equations involving logarithms and powers. Some particular examples are shown below.
fullscreen
Exercise
Solve the equation
using the inverse properties of logarithms.
Show Solution
Solution
The inverse properties of logarithms tell us that a logarithm and a power with the same base undo each other.
On the left-hand side, we can identify as being the logarithm with base e and as the logarithm with base 10. Thus, we can simplify the factors and using the inverse properties of logarithms.
To simplify the remaining two logarithms, we first need to rewrite their arguments.
Again, we can use the the inverse properties of logarithms to simplify the equation.
We now need to isolate x on the left-hand side to find the solution.
33x=2
We have solved the equation, and its solution is |
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# Tangent Identification
## Function of an angle equal to the opposite leg over the adjacent leg of a right triangle.
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Progress
Practice Tangent Identification
Progress
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Tangent Ratio
As the measure of an angle increases between $0^\circ$ and $90^\circ$ , how does the tangent ratio of the angle change?
#### Guidance
Recall that one way to show that two triangles are similar is to show that they have two pairs of congruent angles. This means that two right triangles will be similar if they have one pair of congruent non-right angles.
The two right triangles above are similar because they have two pairs of congruent angles. This means that their corresponding sides are proportional. $\overline{DF}$ and $\overline{AC}$ are corresponding sides because they are both opposite the $22^\circ$ angle. $\frac{DF}{AC}=\frac{4}{2}=2$ , so the scale factor between the two triangles is 2. This means that $x=10$ , because $\frac{FE}{CB}=\frac{10}{5}=2$ .
The ratio between the two legs of any $22^\circ$ right triangle will always be the same, because all $22^\circ$ right triangles are similar. The ratio of the length of the leg opposite the $22^\circ$ angle to the length of the leg adjacent to the $22^\circ$ angle will be $\frac{2}{5}=0.4$ . You can use this fact to find a missing side of another $22^\circ$ right triangle.
Because this is a $22^\circ$ right triangle, you know that $\frac{opposite \ leg}{adjacent \ leg}=\frac{2}{5}=0.4$ .
$\frac{opposite \ leg}{adjacent \ leg} &= 0.4\\\frac{7}{x} &= 0.4\\0.4x &= 7\\x &= 17.5$
The ratio between the opposite leg and the adjacent leg for a given angle in a right triangle is called the tangent ratio . Your scientific or graphing calculator has tangent programmed into it, so that you can determine the $\frac{opposite \ leg}{adjacent \ leg}$ ratio for any angle within a right triangle. The abbreviation for tangent is tan .
Example A
Use your calculator to find the tangent of $75^\circ$ . What does this value represent?
Solution: Make sure your calculator is in degree mode. Then, type “ $\tan (75)$ ”.
$\tan (75^\circ) \approx 3.732$
This means that the ratio of the length of the opposite leg to the length of the adjacent leg for a $75^\circ$ angle within a right triangle will be approximately 3.732.
Example B
Solve for $x$ .
Solution: From Example A, you know that the ratio $\frac{opposite \ leg}{adjacent \ leg} \approx 3.732$ . You can use this to solve for $x$ .
$\frac{opposite \ leg}{adjacent \ leg} & \approx 3.732\\\frac{x}{2} & \approx 3.732\\x & \approx 7.464$
Example C
Solve for $x$ and $y$ .
Solution: You can use the $65^\circ$ angle to find the correct ratio between 24 and $x$ .
$\tan (65^\circ) &= \frac{opposite \ leg}{adjacent \ leg}\\2.145 & \approx \frac{24}{x}\\x & \approx \frac{24}{2.145}\\x & \approx 11.189$
Note that this answer is only approximate because you rounded the value of $\tan 65^\circ$ . An exact answer will include “ $\tan$ ”. The exact answer is:
$x=\frac{24}{\tan 65^\circ}$
To solve for $y$ , you can use the Pythagorean Theorem because this is a right triangle.
$11.189^2+24^2 &= y^2\\701.194 &= y^2\\26.48 &= y$
Concept Problem Revisited
As the measure of an angle increases between $0^\circ$ and $90^\circ$ , how does the tangent ratio of the angle change?
As an angle increases, the length of its opposite leg increases. Therefore, $\frac{opposite \ leg}{adjacent \ leg}$ increases and thus the value of the tangent ratio increases.
#### Vocabulary
Two figures are similar if a similarity transformation will carry one figure to the other. Similar figures will always have corresponding angles congruent and corresponding sides proportional.
AA, or Angle-Angle , is a criterion for triangle similarity. The AA criterion for triangle similarity states that if two triangles have two pairs of congruent angles, then the triangles are similar.
The tangent (tan) of an angle within a right triangle is the ratio of the length of the side opposite the angle to the length of the side adjacent to the angle.
#### Guided Practice
1. Tangent tells you the ratio of the two legs of a right triangle with a given angle. Why does the tangent ratio not work in the same way for non-right triangles?
2. Use your calculator to find the tangent of $45^\circ$ . What does this value represent? Why does this value make sense?
3. Solve for $x$ .
1. Two right triangles with a $32^\circ$ angle will be similar. Two non-right triangles with a $32^\circ$ angle will not necessarily be similar. The tangent ratio works for right triangles because all right triangles with a given angle are similar. The tangent ratio doesn't work in the same way for non-right triangles because not all non-right triangles with a given angle are similar. You can only use the tangent ratio for right triangles.
2. $\tan (45^\circ)=1$ . This means that the ratio of the length of the opposite leg to the length of the adjacent leg is equal to 1 for right triangles with a $45^\circ$ angle.
This should make sense because right triangles with a $45^\circ$ angle are isosceles. The legs of an isosceles triangle are congruent, so the ratio between them will be 1.
3. Use the tangent ratio of a $35^\circ$ angle.
$\tan (35^\circ) &= \frac{opposite \ leg}{adjacent \ leg}\\\tan (35^\circ) &= \frac{x}{18}\\x &= 18 \tan (35^\circ)\\x & \approx 12.604$
#### Practice
1. Why are all right triangles with a $40^\circ$ angle similar? What does this have to do with the tangent ratio?
2. Find the tangent of $40^\circ$ .
3. Solve for $x$ .
4. Find the tangent of $80^\circ$ .
5. Solve for $x$ .
6. Find the tangent of $10^\circ$ .
7. Solve for $x$ .
9. Find the tangent of $27^\circ$ .
10. Solve for $x$ .
11. Find the tangent of $42^\circ$ .
12. Solve for $x$ .
13. A right triangle has a $42^\circ$ angle. The base of the triangle, adjacent to the $42^\circ$ angle, is 5 inches. Find the area of the triangle.
14. Recall that the ratios between the sides of a 30-60-90 triangle are $1:\sqrt{3} : 2$ . Find the tangent of $30^\circ$ . Explain how this matches the ratios for a 30-60-90 triangle.
15. Explain why it makes sense that the value of the tangent ratio increases as the angle goes from $0^\circ$ to $90^\circ$ .
### Vocabulary Language: English
AA Similarity Postulate
AA Similarity Postulate
If two angles in one triangle are congruent to two angles in another triangle, then the two triangles are similar.
Congruent
Congruent
Congruent figures are identical in size, shape and measure.
Similar
Similar
Two figures are similar if they have the same shape, but not necessarily the same size.
Tangent
Tangent
The tangent of an angle in a right triangle is a value found by dividing the length of the side opposite the given angle by the length of the side adjacent to the given angle. |
iGCSE (2021 Edition)
# 11.12 Defining e and the natural logarithm
Lesson
### Euler's number "$e$e"
In calculus, when we differentiate a function this gives us its gradient function. We can then use this function to find the gradient of a tangent at any point. So far, we haven't learnt a process for differentiating an exponential function. This is algebraically quite different to the functions we have seen using powers of $x$x, as opposed to a number raised to the power of $x$x
Interestingly though, our knowledge of sketching derivative functions leads us to the observation that the derivative functions of exponentials look quite similar to their original functions. Below is a graph of the exponential function $y=2^x$y=2x, along with the graph of its derivative (the dashed line). Pretty close, don't you think?
In the applet below, drag the slider to change the base of the exponential. Try to find an approximate value for the base at which the gradient function lines up and matches with the original function:
Created with Geogebra
Both curves overlap at around $2.72$2.72:
This tells us that there is a value of the base $a$a, which is approximately $2.72$2.72, for which the derivative of the function is equal to itself. This particular base is known as Euler's number, and is given the symbol $e$e. It is named after the mathematician Leonhard Euler, and although you may not have seen it before, this irrational number is as famous as $\pi$π.
We call the function $y=e^x$y=ex "the exponential function" to distinguish it from all other exponential functions.
If we want to differentiate an exponential function, we require that it is written in base $e$e. Next year, we will see how we can differentiate exponential functions with other bases. We will do this by first changing them to base $e$e. But for now, let's just get used to this new, rule for differentiation.
The derivative of $e^x$ex
$\frac{d}{dx}\left(e^x\right)=e^x$ddx(ex)=ex
If $y=e^x$y=ex, then we can also express the derivative as $\frac{dy}{dx}=y$dydx=y
This result tells us that the gradient of an exponential function at a point is always equal to its height above the $x$x-axis at that point.
### Other ways to approximate $e$e
There are many ways to calculate $e$e. Obviously, the button on our calculator is a great way to approximate $e$e. You can find the button on the calculator to approximate values with $e$e near the "ln" button.
Here are two other ways:
• Consider the value of $\left(1+\frac{1}{n}\right)^n$(1+1n)n as n gets bigger. In fact, use your calculator to evaluate $\left(1+\frac{1}{10000}\right)^{10000}$(1+110000)10000. What do you get?
• The value of $e$e is also found by the infinite sum: $\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+\frac{1}{5!}+\ldots$10!+11!+12!+13!+14!+15!+ (the symbol "!" means factorial). If you type the first six terms of the sum into the calculator it will give you $2.7166666667$2.7166666667. The more terms you add, the better the approximation.
### The graph of the exponential function
The usual methods of shifting and reflecting graphs can be applied to $y=e^x$y=ex. The graph of $y=e^x$y=ex is one of the most important graphs in this course and we need to be very familiar with its shape and properties.
Properties of the exponential function:
• The domain is all real $x$x, the range is $y>0$y>0
• There are no $x$x-intercepts as the curve is always above the $x$x-axis
• The $x$x-axis ($y=0$y=0) is a horizontal asymptote to the curve
• The curve has a gradient $1$1 at it's $y$y-intercept $\left(0,1\right)$(0,1). (We can see this by substituting $x=0$x=0 into$y'=e^x$y=ex)
• The curve is always increasing at an increasing rate and is always concave up.
#### Worked example
##### Example 1
Find the gradient of the tangent to the curve $y=e^x$y=ex at the points:
(a) $\left(0,1\right)$(0,1)
Do: The gradient function is given by $y'=e^x$y=ex
Substituting$x=0$x=0 into the gradient function gives:
$y'=e^0$y=e0
$=1$=1
Therefore the gradient is 1 at the point $\left(0,1\right)$(0,1)
(b) $(-1,1/e)$(1,1/e)
Do: Substituting $x=-1$x=1 into the gradient function gives:
$y'=e^{-1}$y=e1
$=\frac{1}{e}$=1e
#### Practice question
##### QUESTION 1
$f\left(x\right)=e^x$f(x)=ex and its tangent line at $x=0$x=0 are graphed on the coordinate axes.
1. Determine the gradient to the curve at $x=0$x=0.
2. Evaluate $f\left(0\right)$f(0).
3. Which of the following is true?
$f\left(0\right)\ne f'\left(0\right)$f(0)f(0)
A
$f\left(0\right)=f'\left(0\right)$f(0)=f(0)
B
## Natural logarithms $\log_ex$logex or $\ln x$lnx
Natural logarithms are logarithms to the base $e.$e.We call this the "logarithmic function" to distinguish it from other logarithmic functions with bases other than $e$e.
When we rearrange $y=e^x$y=ex into logarithmic form we get the natural logarithmic function $y=\log_ex$y=logex, which is also written as "$\ln x$lnx" (short for "natural logarithm").
The "$\ln$ln" button on the calculator can help us evaluate logarithmic functions with base $e$e. The log laws that we previously studied also applies to natural logarithms to help us simplify log expressions and equations.
#### Practice questions
##### Question 2
Find the value of $\ln94$ln94 correct to four decimal places.
##### Question 3
Find the value of $\ln\left(18\times35\right)$ln(18×35) correct to four decimal places.
##### Question 4
Use the properties of logarithms to express $\ln\sqrt[3]{y}$ln3y without any powers or surds.
### The graph of $y=\log_ex$y=logex
The exponential and the logarithmic functions are inverse functions which means that their graph will be a reflection of each other across the line $y=x$y=x (their $x$x and $y$y values are swapped).
Because they are reflections, the properties of the natural logarithm graph will correspond with the properties of $y=e^x$y=ex:
• The domain is $x>0$x>0, the range is all real $y$y
• The $y$y-axis ($x=0$x=0) is a vertical asymptote to the curve
• The curve has gradient $1$1 at its $x$x-intercept $\left(1,0\right)$(1,0)
• The curve is always decreasing and concave down
### Combining exponential and logarithmic functions
Recall from the definition of logarithms that if $y=\ln x=\log_ex$y=lnx=logex then $x=e^y$x=ey
So if we substitute $x=e^y$x=ey into $y=\ln x$y=lnx, we see that $y=\ln e^y$y=lney. In other words, raising $e$e to the power $y$y, and then taking logs on that answer restores the original $y$y
If we substitute $y=\ln x$y=lnx into $x=e^y$x=ey, we can also show that $x=e^{\ln x}$x=elnx. Again taking the log of $x$x, and then use that answer as the exponent that $e$e is raised to, simply restores the original $x$x.
The Logarithmic and Exponential Functions are inverse functions
$\log_ee^x=x$logeex=x and $e^{\log_ex}=x$elogex=x
Using the above argument, the expression $e^{\ln5.4}$eln5.4 is simply $5.4$5.4. Also, the expression $\ln e^{\sqrt{5}}$lne5 is simply $\sqrt{5}$5.
### Outcomes
#### 0606C14.3C
Use the derivatives of the standard functions e^x, ln x, together with constant multiples, sums and composite functions. |
## How to transform a fraction to a Percent
There are simply two basic steps to transform a portion to a percent.
You are watching: What percentage is 1 out of 9
### Step One: transform the fraction to a Decimal Value
The first step is to transform the fraction into a decimal value. Perform this by splitting the numerator by the denominator.
The numerator is the number on optimal of the portion bar. The denominator is the number below the portion bar.
decimal = molecule ÷ denominator
For example, convert the fraction 34 to a decimal.
34 = 3 ÷ 434 = 0.75
Thus, the decimal worth of 34 is 0.75.
Here’s a tip: girlfriend might have the ability to use a decimal equivalents chart because that a perform of decimal worths for typical fractions to transform a portion to decimal without utilizing division.
You might additionally be interested in our long division calculator.
### Step Two: convert the Decimal value to a Percentage
The second step is to convert the decimal value into a percentage. Multiply the decimal value by 100, then place a percent authorize (%) after it.
For example, let’s transform 0.75 come a percentage.
percentage = 0.75 × 100 = 75%
The percent value the 34 or 0.75 is for this reason 75%.
## Fraction come Percent switch Table
Another way to convert a portion to percent is to use a switch table. The switch table below shows some usual fractions and also their equivalent percentages.
Fraction to percent counter table showing common fractions and also their equivalent percentage values.
See more: Suzuki Vz800 Marauder 1999 Suzuki Marauder Vz800 Fuel Tank For Sale!
FractionPercent
1/250%
1/333.3%
2/366.6%
1/425%
3/475%
1/520%
2/540%
3/560%
4/580%
1/616.66%
5/683.33%
1/812.5%
3/837.5%
5/862.5%
7/887.5%
1/911.1%
2/922.2%
4/944.4%
5/955.5%
7/977.7%
8/988.8%
1/1010%
1/128.333%
1/166.25%
See more portion percent equivalents. |
# 5-Variable K-Map – Karnaugh Map In Digital Electronics Tutorial Part 6
A 5-variable K-Map will have 25 = 32 cells. A function F which has maximum decimal value of 31, can be defined and simplified by a 5-variable Karnaugh Map.
#### 5-Variable K-Map
In above boolean table, from 0 to 15, A is 0 and from 16 to 31, A is 1. A 5-variable K-Map is drawn as below.
Again, as we did with 3-variable & 4-variable K-Map, carefully note the numbering of each cell. Now, we have two squares and we can loop octets, quads and pairs between these two squares. What we need to do is to visualize second square on first square and figure out adjacent cells. Let’s understand how to simplify 5-variables K-Map by taking couple of examples.
#### Example 1 of 5-Variable K-Map
Given function, F = Σ (1, 3, 4, 5, 11, 12, 14, 16, 20, 21, 30)
Since, the biggest number is 30, we need to have 5 variables to define this function.
Let’s draw K-Map for this function by writing 1 in cells that are present in function and 0 in rest of the cells.
Applying rules of simplifying K-Map, there is no octet. There is one quad that is obtained by visualizing second square on first, there are 4 adjacent cells – 4,5,20 and 21. The octet is highlighted by a blue connecting line. There are 5 pairs. Similar to quad, there is one pair between two squares which is highlighted by the blue connecting line.
(4, 5, 20, 21) – B’CD’ (Since A & E are the changing variables, it is eliminated)
(12, 14) – A’BCE’ (Since D is the changing variable, it is eliminated)
(14, 30) – BCDE’ (Since A is the changing variable, it is eliminated)
(3, 11) – A’C'DE (Since B is the changing variable, it is eliminated)
(16, 20) – AB’D'E’ (Since C is the changing variable, it is eliminated)
(1, 3) – A’B'C’E (Since D is the changing variable, it is eliminated)
Thus, F = B’CD’ + A’BCE’ + BCDE’ + A’C'DE + AB’D'E’ + A’B'C’E
#### Example 2 of 5-Variable K-Map
Given function, F = Σ (0, 2, 3, 5, 7, 8, 11, 13, 17, 19, 23, 24, 29, 30)
Since, the biggest number is 30, we need to have 5 variables to define this function.
Let’s draw K-Map for this function by writing 1 in cells that are present in function and 0 in rest of the cells.
Applying rules of simplifying K-Map, there is no octet. First we need to look for quads within each of the squares. There are none but there is a quad between two squares that is obtained by visualizing second square on first, there are 4 adjacent cells – 3, 7, 19 and 23. This quad is highlighted by blue connecting line. There are 6 pairs, out of which two are between two squares, highlighted by blue connecting line.
(3, 7, 19, 23) - B’DE (Since A & C are the changing variables, they are eliminated)
(3, 11) – A’C'DE (Since B is the changing variables, it is eliminated)
(1, 2) – A’B'C’E’ (Since D is the changing variables, it is eliminated)
(5, 7) – A’B'CE (Since D is the changing variables, it is eliminated)
(17, 19) – AB’C'E (Since D is the changing variables, it is eliminated)
(13, 29) – BCD’E (Since A is the changing variables, it is eliminated)
(8, 24) – BC’D'E’ (Since A is the changing variables, it is eliminated)
There is 1 in cell 30, which can not be looped with any adjacent cell, hence it can not be simplified further and left as it is.
30 – ABCDE’
Thus, F = B’DE + A’C'DE + A’B'C’E’ + A’B'CE + AB’C'E + BCD’E + BC’D'E’ + ABCDE’
#### Example 3 of 5-Variable K-Map
Given function, F = Σ (0, 1, 2, 3, 8, 9, 16, 17, 20, 21, 24, 25, 28, 29, 30, 31)
Since, the biggest number is 31, we need to have 5 variables to define this function.
Let’s draw K-Map for this function by writing 1 in cells that are present in function and 0 in rest of the cells.
Applying rules of simplifying K-Map, there are 2 octets. First one is in square 2 circled in red. Another octet is between 2 squares highlighted by blue connecting lines. There are 2 quads between each of the squares.
(16, 17, 20, 21, 28, 29, 24, 25) – AD’ (Since B, C and E are changing variables, they are eliminated)
(0, 1, 8, 9, 16, 17, 24, 25) – C’D’ (Since A, B and E are changing variables, they are eliminated)
(0, 1, 2, 3) – A’B'C’ (Since D and E are changing variables, they are eliminated)
(28, 29, 30, 31) – ABC (Since D and E are changing variables, they are eliminated)
Thus, F = AD’ + C’D’ + A’B'C’ + ABC
Read the full series at Part 1, Part 2, Part 3, Part 4, Part 5, Part 6 and Part 7.
1. why is the slot of “0″ in example no. 2 still zero? isn’t zero part of the given function?
2. Yes in example 2:
slor of 0 should be fill with 1 which further make pair with one of slot 2….:-)
3. In Example 2, (1, 2) – A’B’C’E’ is wrong. It should be (0,2) |
# Basic Statistics and Data Analysis
#### Lecture notes, MCQS of Statistics
Introduction to statistics
## Sum of Squared Deviation from Mean
In statistics, the sum of squared deviation is a measure of the total variability (spread, variation) within a data set. In other words, the sum of squares is a measure of deviation or variation from the mean (average) value of the given data set. A sum of squares calculated by first computing the differences between each data point (observation) and mean of the data set, i.e. $x=X-\overline{X}$. The computed $x$ is known as the deviation score for the given data set. Squaring each of this deviation score and then adding these squared deviation scores gave us the sum of squared deviation (SS), which is represented mathematically as
$SS=\sum(x^2)=\sum(X-\overline{X})^2$
Note that the small letter $x$ usually represents the deviation of each observation from the mean value, while capital letter $X$ represents the variable of interest in statistics.
## Sum of Squares Example
Consider the following data set {5, 6, 7, 10, 12}. To compute the sum of squares of this data set, follow these steps
• Calculate the average of the given data by summing all the values in the data set and then divide this sum of numbers by the total number of observations in the date set. Mathematically, it is $\frac{\sum X_i}{n}=\frac{40}{5}=8$, where 40 is the sum of all numbers $5+6+7+10+12$ and there are 5 observations in number.
• Calculate the difference of each observation in data set from the average computed in step 1, for given data. The differences are
5 – 8 = –3; 6 – 8 = –2; 7 – 8 = –1; 10 – 8 =2 and 12 – 8 = 4
Note that the sum of these differences should be zero. (–3 + –2 + –1 + 2 +4 = 0)
• Now square the each of the differences obtained in step 2. The square of these differences are
9, 4, 1, 4 and 16
• Now add the squared number obtained in step 3. The sum of these squared quantities will be 9 + 4 + 1 + 4 + 16 = 34, which is the sum of the square of the given data set.
In statistics, sum of squares occurs in different contexts such as
• Partitioning of Variance (Partition of Sums of Squares)
• Sum of Squared Deviations (Least Squares)
• Sum of Squared Differences (Mean Squared Error)
• Sum of Squared Error (Residual Sum of Squares)
• Sum of Squares due to Lack of Fit (Lack of Fit Sum of Squares)
• Sum of Squares for Model Predictions (Explained Sum of Squares)
• Sum of Squares for Observations (Total Sum of Squares)
• Sum of Squared Deviation (Squared Deviations)
• Modeling involving Sum of Squares (Analysis of Variance)
• Multivariate Generalization of Sum of Square (Multivariate Analysis of Variance)
As previously discussed, Sum of Square is a measure of the Total Variability of a set of scores around a specific number.
# Data Transformation (Variable Transformation)
A transformation is a rescaling of the data using a function or some mathematical operation on each observation. When data are very strongly skewed (negative or positive), we sometime transform the data so that they are easier to model. In other way, if variable(s) does not fit a normal distribution then one should try a data transformation to fit the assumption of using a parametric statistical test.
The most common data transformation is log (or natural log) transformation, which is often applied when most of the data values cluster around zero relative to the larger values in the data set and all of the observations are positive.
Transformation can also be applied to one or more variables in scatter plot, correlation and regression analysis to make the relationship between the variables more linear; and hence it is easier to model with simple method. Other transformation than log are square root, reciprocal etc.
Reciprocal Transformation
The reciprocal transformation $x$ to $\frac{1}{x}$ or $(-\frac{1}{x})$ is a very strong transformation with a drastic effect on shape of the distribution. Note that this transformation cannot be applied to zero values, but can be applied to negative values. Reciprocal transformation is not useful unless all of the values are positive and reverses the order among values of the same sign i.e. largest becomes smallest etc.
Logarithmic Transformation
The logarithm $x$ to log (base 10) (or natural log, or log base 2) is an other strong transformation that have effect on the shape of distribution. Logarithmic transformation commonly used for reducing right skewness, but cannot be applied to negative or zero values.
Square Root Transformation
The square root x to $x^{\frac{1}{2}}=\sqrt(x)$ transformation have moderate effect on distribution shape and weaker than the logarithm. Square root transformation can be applied to zero values but not negative values.
Goals of transformation
The goals of transformation may be
• one might want to see the data structure differently
• one might want to reduce the skew that assist in modeling
• one might want to straighten a nonlinear (curvilinear) relationship in a scatter plot. In other words a transformation may be used to have approximately equal dispersion, making data easier to handle and interpret
# Sampling theory, Introduction and Reasons to Sample
Often we are interested in drawing some valid conclusions (inferences) about a large group of individuals or objects (called population in statistics). Instead of examining (studying) the entire group (population, which may be difficult or even impossible to examine), we may examine (study) only a small part (portion) of the population (entire group of objects or people). Our objective is to draw valid inferences about certain facts for the population from results found in the sample; a process known as statistical inferences. The process of obtaining samples is called sampling and theory concerning the sampling is called sampling theory.
Example: We may wish to draw conclusions about the percentage of defective bolts produced in a factory during a given 6-day week by examining 20 bolts each day produced at various times during the day. Note that all bolts produced in this case during the week comprise the population, while the 120 selected bolts during 6-days constitutes a sample.
In business, medical, social and psychological sciences etc., research, sampling theory is widely used for gathering information about a population. The sampling process comprises several stages:
• Defining the population of concern
• Specifying the sampling frame (set of items or events possible to measure)
• Specifying a sampling method for selecting the items or events from the sampling frame
• Determining the appropriate sample size
• Implementing the sampling plan
• Sampling and data collecting
• Data which can be selected
When studying the characteristics of a population, there many reasons to study a sample (drawn from population under study) instead of entire population such as:
1. Time: as it is difficult to contact each and every individual of the whole population
2. Cost: The cost or expenses of studying all the items (objects or individual) in a population may be prohibitive
3. Physically Impossible: Some population are infinite, so it will be physically impossible to check the all items in the population, such as populations of fish, birds, snakes, mosquitoes. Similarly it is difficult to study the populations that are constantly moving, being born, or dying.
4. Destructive Nature of items: Some items, objects etc are difficult to study as during testing (or checking) they destroyed, for example a steel wire is stretched until it breaks and breaking point is recorded to have a minimum tensile strength. Similarly different electric and electronic components are check and they are destroyed during testing, making impossible to study the entire population as time, cost and destructive nature of different items prohibits to study the entire population.
5. Qualified and expert staff: For enumeration purposes, highly qualified and expert staff is required which is some time impossible. National and International research organizations, agencies and staff is hired for enumeration purposive which is some time costly, need more time (as rehearsal of activity is required), and some time it is not easy to recruiter or hire a highly qualified staff.
6. Reliability: Using a scientific sampling technique the sampling error can be minimized and the non-sampling error committed in the case of sample survey is also minimum, because qualified investigators are included.
Every sampling system is used to obtain some estimates having certain properties of the population under study. The sampling system should be judged by how good the estimates obtained are. Individual estimates, by chance, may be very close or may differ greatly from the true value (population parameter) and may give a poor measure of the merits of the system.
A sampling system is better judged by frequency distribution of many estimates obtained by repeated sampling, giving a frequency distribution having small variance and mean estimate equal to the true value.
# The Level of Measurements
In statistics, data can be classified according to level of measurement, dictating the calculations that can be done to summarize and present the data (graphically), it also helps to determine, what statistical tests should be performed. For example, suppose there are six colors of candies in a bag and you assign different numbers (codes) to them in such a way that brown candy has a value of 1, yellow 2, green 3, orange 4, blue 5, and red a value of 6. From this bag of candies, adding all the assigned color values and then dividing by the number of candies, yield an average value of 3.68. Does this mean that the average color is green or orange? Of course not. When computing statistic, it is important to recognize the data type, which may be qualitative (nominal and ordinal) and quantitative (Interval and ratio).
The level of measurements has been developed in conjunction with the concepts of numbers and units of measurement. Statisticians classified measurements according to levels. There are four level of measurements, namely, nominal, ordinal, interval and ratio, described below.
Nominal Level of Measurement
In nominal level of measurement, the observation of a qualitative variable can only be classified and counted. There is no particular order to the categories. Mode, frequency table, pie chart and bar graph are usually drawn for this level of measurement.
Ordinal Level of Measurement
In ordinal level of measurement, data classification are presented by sets of labels or names that have relative values (ranking or ordering of values). For example, if you survey 1,000 people and ask them to rate a restaurant on a scale ranging from 0 to 5, where 5 shows higher score (highest liking level) and zero shows the lowest (lowest liking level). Taking the average of these 1,000 people’s response will have meaning. Usually graphs and charts are drawn for ordinal data.
Interval Level of Measurement
Numbers also used to express the quantities, such as temperature, dress size and plane ticket are all quantities. The interval level of measurement allows for the degree of difference between items but no the ratio between them. There is meaningful difference between values, for example 10 degrees Fahrenheit and 15 degrees is 5, and the difference between 50 and 55 degrees is also 5 degrees. It is also important that zero is just a point on the scale, it does not represents the absence of heat, just that it is freezing point.
Ratio Level of Measurement
All of the quantitative data is recorded on the ratio level. It has all the characteristics of the interval level, but in addition, the zero point is meaningful and the ratio between two numbers is meaningful. Examples of ratio level are wages, units of production, weight, changes in stock prices, distance between home and office, height etc.
Many of the inferential test statistics depends on ratio and interval level of measurement. Many author argue that interval and ratio measures should be named as scale.
For Examples about Level of Measurements Visits: Examples of Levels of Measurements
# Degrees of Freedom
The degrees of freedom (df) or number of degrees of freedom refers to the number of observations in a sample minus the number of (population) parameters being estimated from the sample data. All this means that the degrees of freedom is a function of both sample size and the number of independent variables. In other words it is the number of independent observations out of a total of ($n$) observations.
In statistics, the degrees of freedom considered as the number of values in a study that are free to vary. For example (degrees of freedom example in real life), if you have to take ten different courses to graduate, and only ten different courses are offered, then you have nine degrees of freedom. Nine semesters you will be able to choose which class to take; the tenth semester, there will only be one class left to take – there is no choice, if you want to graduate, this is the concept of the degrees of freedom (df) in statistics.
Let a random sample of size n is taken from a population with an unknown mean $\overline{X}$. The sum of the deviations from their means is always equal to zero i.e.$\sum_{i=1}^n (X_i-\overline{X})=0$. This require a constraint on each deviation $X_i-\overline{X}$ used when calculating the variance.
$S^2 =\frac{\sum_{i=1}^n (X_i-\overline{X})^2 }{n-1}$
This constraint (restriction) implies that $n-1$ deviations completely determine the nth deviation. The $n$ deviations (and also the sum of their squares and the variance in the $S^2$ of the sample) therefore $n-1$ degrees of freedom.
A common way to think of degrees of freedom is as the number of independent pieces of information available to estimate another piece of information. More concretely, the number of degrees of freedom is the number of independent observations in a sample of data that are available to estimate a parameter of the population from which that sample is drawn. For example, if we have two observations, when calculating the mean we have two independent observations; however, when calculating the variance, we have only one independent observation, since the two observations are equally distant from the mean.
Single sample: For $n$ observation one parameter (mean) needs to be estimated, that leaves $n-1$ degrees of freedom for estimating variability (dispersion).
Two samples: There are total of $n_1+n_2$ observations ($n_1$ for group1 and $n_2$ for group2) and two means need to be estimated, which leaves $n_1+n_2-2$ degrees of freedom for estimating variability.
Regression with p predictors: There are $n$ observations with $p+1$ parameters needs to be estimated (regression coefficient for each predictor and the intercept). This leaves $n-p-1$ degrees of freedom of error, which accounts for the error degrees of freedom in the ANOVA table.
Several commonly encountered statistical distributions (Student’s t, Chi-Squared, F) have parameters that are commonly referred to as degrees of freedom. This terminology simply reflects that in many applications where these distributions occur, the parameter corresponds to the degrees of freedom of an underlying random vector. If $X_i; i=1,2,\cdots, n$ are independent normal $(\mu, \sigma^2)$ random variables, the statistic (formula) is $\frac{\sum_{i=1}^n (X_i-\overline{X})^2}{\sigma^2}$, follows a chi-squared distribution with $n-1$ degrees of freedom. Here, the degrees of freedom arises from the residual sum of squares in the numerator and in turn the $n-1$ degrees of freedom of the underlying residual vector ${X_i-\overline{X}}$. |
# PLOT THE COMPLEX NUMBER IN THE COMPLEX PLANE
How to plot the complex number in the complex plane :
General form of a complex number is
a + ib
Here a is real part and b is imaginary part. We can convert the complex number as ordered pair. So, the point is (a, b).
Now we can plot the point in the graph.
Plot all four points in the same complex plane.
Example 1 :
1 + 2i, 3 - i, -2 + 2i, i
Solution :
Given, complex numbers are 1 + 2i, 3 - i, -2 + 2i, i
We are taking the real part of the complex number on the x-coordinate and the imaginary part on the y-coordinate.
Then, the ordered pair of the complex numbers are
1 + 2i = (1, 2)3 - i = (3, -1) -2 + 2i = (-2, 2)i = (0, 1)
So, the four points are (1, 2), (3, -1), (-2, 2) and (0, 1).
By plotting the four points on the complex plane, we get
Example 2 :
2 - 3i, 1 + i, 3, -2 - i
Solution :
Given, complex numbers are 2 - 3i, 1 + i, 3, -2 - i
Then, the ordered pair of the complex numbers are
2 - 3i = (2, -3)1 + i = (1, 1) 3 = (3, 0)-2 - i = (-2, -1)
So, the four points are (2, -3), (1, 1), (3, 0) and (-2, -1).
By plotting the four points on the complex plane, we get
Example 3 :
Find the distance between two complex numbers z= 2 + 3i and z= 7 - 9i on the complex plane.
Solution :
Given, z1 = 2 + 3i and z2 = 7 - 9i are complex numbers.
Then, the ordered pair of the complex numbers are
z= (2, 3) and z2 = (7, -9)
Now, we have the two points (2, 3) and (7, -9)
By plotting the two points on the complex plane, we get
Finding the distance :
Formula for distance,
d = √[(x- x1)2 + (y- y1)2]
(2, 3)----->(x1, y1)
(7, -9)----->(x2, y2)
d = √[(7 - 2)2 + (-9 - 3)2]
d = √[(5)2 + (-12)2]
d = √(25 + 144)
d = √169
d = 13 units
So, the distance is 13 units.
Example 4 :
Find the distance and midpoint between two complex numbers z = 3 + i and w = 1 + 3i on the complex plane.
Solution :
Given, z = 3 + i and w = 1 + 3i are complex numbers.
Then, the ordered pair of the complex numbers are
z = (3, 1) and w = (1, 3)
Now, we have the two points (3, 1) and (1, 3)
By plotting the two points on the complex plane, we get
Finding the distance :
Formula for distance,
d = √[(x- x1)+ (y- y1)2]
(3, 1)----->(x1, y1)
(1, 3)----->(x2, y2)
d = √[(1 - 3)+ (3 - 1)2]
d = √[(-2)+ (2)2]
d = √(4 + 4)
d = √8
d = 2√2 units
So, the distance is 2√2 units.
Finding the midpoint :
Formula for midpoint,
M = [(x1 + x2)/2, (y1 + y2)/2]
= [(3 + 1)/2, (1 + 3)/2]
Midpoint = (2, 2)
So, the midpoint is (2, 2)
Example 5 :
Find the distance and midpoint between the complex number z = 5 + 2i and its conjugate z- = 5 - 2i on the complex plane.
Solution :
Given, z = 5 + 2i and z- = 5 - 2i are complex numbers.
Then, the ordered pair of the complex numbers are
z = (5, 2) and z- = (5, -2)
Now, we have the two points (5, 2) and (5, -2)
By plotting the two points on the complex plane, we get
Finding the distance :
Formula for distance,
d = √[(x- x1)+ (y- y1)2]
(5, 2)----->(x1, y1)
(5, -2)----->(x2, y2)
d = √[(5 - 5)+ (-2 - 2)2]
d = √[(-4)2]
d = √16
d = 4 units
So, the distance is 4 units.
Finding the midpoint :
Formula for midpoint,
M = [(x+ x2)/2, (y+ y2)/2]
= [(5 + 5)/2, (2 - 2)/2]
Midpoint = (5, 0)
So, the midpoint is (5, 0)
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OTHER TOPICS
Profit and loss shortcuts
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Graphing rational functions with holes
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Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
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Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6 |
# what is the interior angle method for this question? He covered a lot of stuff really quickly and I only kind of understand.
• [Originally posted by a student in the Discussions, which got replaced by this forum.]
Module 0 Week 4 Day 13 Your Turn Explanation
• The purpose of this question was to show the power of using exterior angles: It can make questions like these very simple! The method Professor Loh shows later in the video, using interior angles, is a little more complicated, but it's still worth knowing how to make use of interior angles.
The first step is to know what is the sum of all the interior angles in a polygon with $$n$$ sides. You can try to look for a pattern:
• For a triangle, the sum of angles is $$180.$$
• For a quadrilateral, the sum is $$360$$ (think of a square).
• <Pentagons may be difficult, skip for now.>
• For a hexagon, the sum is $$720$$ (think of a regular hexagon!).
Based on this pattern, you can guess that the sum of angles goes up by $$180$$ every time you add a side! But why? The clever reason comes from dividing the polygon into triangles.
You can see Professor Loh do this at 2:18. Every time you add another side, you get another triangle that makes up the polygon. And, adding up the angles of the polygon is the same as adding up the angles of every triangle in the "triangulation" (make sure this step makes sense!).
That means that the sum of the angles is just (sum of angles inside a triangle) $$\times$$ (number of triangles in the polygon). Since the angle sum for a triangle is $$180,$$ it's just $$180 \times \text{ number of triangles}.$$
The number of triangles that an $$n$$-sided polygon is divided into is just $$(n-2)$$. You can figure this out by looking at a pattern ($$n=3$$ is $$1$$ triangle, $$n=4$$ is $$2$$ triangles, etc.). Then the formula for the sum of angles is just $$180 \times (n-2).$$ This is a very important formula! Make sure you understand why it works instead of memorizing it.
Back to the problem in the video: We don't know how many sides there are, so let's say there are $$n$$ sides. Then since every angle is $$170,$$ the sum of angles is $$170n.$$ But we just derived a formula for the sum of angles: It's $$180(n-2).$$ So we can set these expressions equal to each other:
$$170n = 180(n-2)$$
Now we just solve this equation and we are done!
I hope that helped. Happy learning! |
# McGraw Hill My Math Grade 3 Chapter 4 Lesson 5 Answer Key Problem-Solving Investigation: Make a Table
All the solutions provided in McGraw Hill Math Grade 3 Answer Key PDF Chapter 4 Lesson 5 Problem-Solving Investigation: Make a Table will give you a clear idea of the concepts.
## McGraw-Hill My Math Grade 3 Answer Key Chapter 4 Lesson 5 Problem-Solving Investigation: Make a Table
Learn the Strategy
Selma bought 3 shorts and 2 shirts. Laura, bought 4 shorts and 2 shirts. How many different shirt and shorts outfits can each girl make?
1. Understand
What facts do you know?
___________
What do you need to find?
How many different ___ and ____ outfits they each can make.
2. Plan
Organize the information in a table of columns and rows.
3. Solve
Make a table for each girl. List the possible shirt and shorts outfits.
So, Selma can make ___ outfits, and Laura can make ____.
4. Check
____________________
We Know that Selma bought 3 shorts and 2 shirts
Laura, bought 4 shorts and 2 shirts
We need to find the The number of different outfits they can make
I used table to find the answer
3 shorts x 2 shirts = 6 combinations
4 shorts x 2 shirts = 8 combinations
So, Selma can make 6 outfits, and Laura can make 8.
Practice the Strategy
How many lunches can Malia make if she chooses one main dish and one side dish?
1. Understand
What facts do you know?
_____________
What do you need to find?
_____________
_____________
2. Plan
_____________
3. Solve
4. Check
_____________
Explanation:
Malia has 3 main dishes and 3 sides dishes
We need to find how the number of lunches she can make using 1 main dish and 1 side dish
I drew a table to show the number of different lunches that Malia can make
3 main dishes are combined with 3 sides dishes
3 x 3 = 9
So, Malia can make 9 different types of lunch.
Apply the Strategy
Solve each problem by making a table.
Question 1.
Trey can choose one type of bread and one type of meat for his sandwich. How many different sandwiches can Trey make?
Trey can make ___ sandwiches.
Explanation:
Trey has 2 types of breads and 2 types of chicken
Trey can choose one type of bread and one type of meat for his sandwich
2 breads can combine with 2 chickens
I drew a table to show the number of possible combinations
2 x 2 = 4
So, Trey can make 4 different sandwiches.
Question 2.
The students in Mr. Robb’s class are designing a flag. The flag’s background can be gold, red, or green. The flag can have either a blue or a purple stripe. Color all the possible flags.
They can design ___ flags.
Explanation:
The students in Mr. Robb’s class are designing a flag
The flag’s background can be gold, red, or green
The flag can have either a blue or a purple stripe
I drew a table to show the number of possible combinations
There are 3 background colors and 2 stripes colors
3 x 2 = 6
So, they can design 6 flags.
Question 3.
Mathematical PRACTICE 4 Model Math Tracy has a picture of her mom, a picture of her dad, and a picture of her dog. She has a black frame and a white frame. What is the question? Solve.
How many different colors options does Tracy have to make a frame?
Explanation:
Tracy has a picture of her mom, a picture of her dad, and a picture of her dog
She has a black frame and a white frame
Tracy can make 2 frames of pictures using any of 2 colors
2 x 2 = 4
So, Tracy has an option of making 4 frames using the pictures and color frames.
Review the Strategies
Use any strategy to solve each problem.
• Use the four-step plan.
• Make a table
Question 4.
Amber has coins in a jar. The sum of the coins is 13⊄. What are the possible groups of coins Amber could have?
Explanation:
Amber has coins in a jar
The sum of the coins is 13⊄
We need of find the possible groups of coins that Amber could have
I drew a table to show the number of possible combinations
So, there are 4 possible groups of coins that Amber can have.
Question 5.
Solaria buys 2 bags of salad mix for $8 and 3 pounds of fresh vegetables for$9. She gives the cashier $20. How much change will she receive? Answer:$3
Explanation:
Solaria buys 2 bags of salad mix for $8 and 3 pounds of fresh vegetables for$9
She gives the cashier $20 We need to find the amount of change she receives First add$8 and $9 8 + 9 = 17 Now, Subtract$17 from $20 20 – 17 = 3 So, Solaria receives$3 change.
Question 6.
Mathematical PRACTICE 6 Be Precise Mr. Grow has 12 tomato plants arranged in 2 rows of 6. List 2 other ways that Mr. Grow could arrange his 12 tomato plants in equal rows. Explain to a classmate how you got your answer.
Explanation:
Mr. Grow has 12 tomato plants arranged in 2 rows of 6
We need to find the 2 other ways that Mr. Grow could arrange his 12 tomato plants in equal rows
The first arrangement is 2 rows of 6
The 2 other arrangements Mr. Grow could make are
3 rows of 4
3 x 4 = 12
4 rows of 3
4 x 3 = 12
So, I drew to show the 2 other arrangements that Mr. Grow could arrange his 12 tomato plants in equal rows.
Question 7.
Mathematical PRACTICE 5 Use Math Tools One campsite has 3 tents with 5 people in each tent. Another campsite has 3 tents with 4 people in each. How many campers are there in all? Draw an array to solve.
Explanation:
One campsite has 3 tents with 5 people in each tent
Another campsite has 3 tents with 4 people in each
We need to find the number of campers in all
I drew arrays to show the equal groups
3 tents with 5 people is 3 groups of 5
3 x 5 = 20
3 tents with 4 people is 3 groups of 4
3 x 4 = 12
20 + 12 = 32
So, there are 32 campers in all.
### McGraw Hill My Math Grade 3 Chapter 4 Lesson 5 My Homework Answer Key
Problem Solving
Question 1.
Solve the problem by making a table.
Claudio will decorate his bedroom. He can choose tan, blue, or gray paint and striped or plaid curtains. How many ways can he decorate his room with different paint and curtains?
Explanation:
Claudio will decorate his bedroom
He can choose tan, blue, or gray paint and striped or plaid curtains
We need to find the number of ways Claudio can decorate his room with different pain and curtains
There are 3 paints and 2 types of curtains
I drew a table to show the number of possible combinations
3 x 2 = 6
So, Claudio can decorate his room with 6 different combinations of paints and curtains.
Solve each problem by making a table.
Question 2.
Jimmy has a number cube labeled 1 through 6, and a penny. How many different ways can the cube and penny land with one roll of the cube and one flip of the penny?
Explanation:
Jimmy has a number cube labeled 1 through 6, and a penny
We need to find the different ways that a cube and penny land with one roll of the cube and one flip of the penny
A penny has 2 faces one is heads and the other is tails
I drew a table to show the number of possible combinations
6 x 2 = 12
So, 12 different ways of cube and penny can land with one roll of cube and one flip of penny.
Question 3.
Archie earns $4 each week for doing his chores. How much money will Archie earn in 2 months if there are 4 weeks in each month? Answer: Explanation: Archie earns$4 each week for doing his chores
We need o find how much can Archie earn in 2 months if there are 4 weeks in each month
I drew a table to show the number of possible combinations
2 x 4 = 8 weeks
8 x $4 =$32
So, Archie can earn \$32 in 2 months.
Question 4.
Mathematical PRACTICE 7 Identify Structure Abigail has a green, yellow, and purple shirt to match with either a white, black, or red pair of pants. How many different shirt and pants outfits can she make?
How many outfits would be possible if Abigail had only 2 shirts and 2 pair of pants? Explain.
Explanation:
Problem1:
Abigail has a green, yellow, and purple shirt to match with either a white, black, or red pair of pants
We need to find the number of different outfits that Abigail can make
I drew a table to show the number of possible combinations
3 x 3 = 9
So, Abigail can make 9 different kinds of outfits.
Problem2:
Abigail has a green and purple shirt to match with either a white or red pair of pants
We need to find the number of different outfits that Abigail can make
I drew a table to show the number of possible combinations
2 x 2 = 4
So, Abigail can make 4 different kinds of outfits.
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<meta http-equiv="refresh" content="1; url=/nojavascript/"> The Real Number Line | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Algebra I - Honors Go to the latest version.
# 1.10: The Real Number Line
Created by: CK-12
0 0 0
## Graphing on a Real Number Line
Objectives
The lesson objectives for Graphing on the Real Number Line are:
• Identifying the number set(s) to which numbers belong
• Graphing given sets on a number line
Introduction
In this concept you will review the number sets that make up the real number system. You will use the information learned in the review, to identify the number set to which given numbers belong. In addition, you will use the information learned, to graph inequalities on a real number line.
Guidance
To what number set(s) does the number 13 belong?
The number 13 is a natural number. $N=\{1,2,3,4 \ldots\}$
The number 13 is a whole number. $W=\{0,1,2,3 \ldots\}$
The number 13 is an integer. $I=\{\ldots,-3,-2,-1,0,1,2,3, \ldots\}$
The number 13 is a rational number. $Q=\{\frac{a}{b}, b \ne 0 \}$
The number 13 belongs to the real number system.
Example
The symbol > means “is greater than.”
The symbol < means “is less than.”
The symbol means “is greater than or equal to.”
The symbol means “is less than or equal to.”
To represent all integers greater than 4, you can write $\{5,6,7,8, \ldots\}$, or write $x > 4$ such that $x$ is an integer. You can also use a number line to represent all integers greater than 4. Remember that the set of integers are negative whole numbers, zero and positive whole numbers. The number 4 is not included in the numbers greater than 4. However, to indicate 4 as a starting point, mark it on the number line with an open dot. If it were included in the answer, the dot would be closed or solid.
Represent $x>4$ where $x$ is an integer, on a number line.
The open dot on the four means that 4 is not included in the graph of all integers greater than 4. The closed dots on 5, 6, 7, 8 means that these numbers are included in the set of integers greater than 4. The arrow pointing to the right means that all integers to the right of 8 are also included in the graph of all integers greater than 4.
Example B
Represent this inequality statement on a number line $\{x \ge -2 | x \ \varepsilon \ R\}$.
$\{x \ge -2 | x \ \varepsilon \ R\}$ The statement can be read as “$x$ is greater than or equal to -2, such that x belongs to or is a member of the real numbers.” In other words, represent all real numbers greater than or equal to -2.
The inequality symbol says that $x$ is greater than or equal to -2. This means that -2 is included in the graph. A solid dot is placed on -2 and on all numbers to the right of -2. The line is on the number line to indicate that all real numbers greater than -2 are also included in the graph.
Example C
In the above examples, the inequality symbol indicates the type of dot that is placed on the beginning point and the number set indicates whether an arrow is drawn on the number line or if a line with an arrow is drawn. The arrow means that the numbers included in the graph continue. The only time that an arrow is not used is when the inequality represents a beginning point and an ending point.
Represent this inequality statement, also known as set notation, on a number line $\{x|2 < x \le 7, x \ \varepsilon \ N\}$. This inequality statement can be read as $x$ such that $x$ is greater than 2 and less than or equal to 7 and $x$ belongs to the natural numbers. In other words, all natural numbers greater than 2 and less than or equal to 7.
The inequality statement that was to be represented on the number line had to include the natural numbers greater than 2 and less than or equal to 7. These are the only numbers to be graphed. There is no arrow on the number line.
Vocabulary
Inequality
An inequality is a mathematical statement relating expressions by using one or more inequality symbols. The inequality symbols are $>,<,\ge,\le$
Integer
All natural numbers, their opposites, and zero are integers. A number in the list $\ldots, -3, -2, -1, 0, 1, 2, 3 \ldots$
Irrational Numbers
The irrational numbers are those that cannot be expressed as the ratio of two numbers. The irrational numbers include decimal numbers that are non-terminating decimals as well as non-periodic decimal numbers.
Natural Numbers
The natural numbers are the counting numbers and consist of all positive, whole numbers. The natural numbers are numbers in the list $1, 2, 3\ldots$ and are often referred to as positive integers.
Number Line
A number line is a line that matches a set of points and a set of numbers one to one. It is often used in mathematics to show mathematical computations.
Rational Numbers
The rational numbers are numbers that can be written as the ratio of two numbers $\frac{a}{b}$ and $b \ne 0$. The rational numbers include all terminating decimals as well as periodic decimal numbers.
Real Numbers
The rational numbers and the irrational numbers make up the real numbers.
Set Notation
Set notation is a mathematical statement that shows an inequality and the set of numbers to which the variable belongs. $\{x|x \ge -3, x \ \varepsilon \ I\}$ is an example of set notation.
Guided Practice
1. Check the set(s) to which each number belongs. The number may belong to more than one set.
Number $N$ $W$ $I$ $Q$ $\overline{Q}$ $R$
5
$-\frac{47}{3}$
1.48
$\sqrt{7}$
0
$\pi$
2. Graph $\{x|-3 \le x \le 8, x \ \varepsilon \ R\}$ on a number line.
3. Use set notation to describe the set shown on the number line.
1. Before answering this question, review the definitions for each set of numbers. You can find these in the vocabulary.
Number $N$ $W$ $I$ $Q$ $\overline{Q}$
5 X X X X
$-\frac{47}{3}$ X X X X
1.48 X X X X
$\sqrt{7}$ X
0 X X X
$\pi$ X
2. $\{x|-3 \le x \le 8, x \ \varepsilon \ R\}$
The set notation means to graph all real numbers between -3 and +8. The line joining the solid dots represents the fact that the set belongs to the real number system.
3.
The closed dot means that -2 is included in the answer. The remaining dots are to the right of -2. The open dot means that 3 is not included in the answer. This means that the numbers are all less than 3. Graphing on a number line is done from smallest to greatest or from left to right. There is no line joining the dots so the variable does not belong to the set of real numbers. However, negative whole numbers, zero and positive whole numbers make up the integers.
The set notation that is represented on the number line is
$\boxed{\{x|-2 \le x < 3, x \ \varepsilon \ R\}}$
Summary
In this lesson you revisited the real number system. You reviewed the sets of numbers that made up the real number system and concentrated on the types of numbers that were included in each set. These sets included the natural numbers, whole numbers, integers, rational numbers and irrational numbers. Given a group of numbers, you learned how to assign each number to the number set(s) to which it belonged.
You also learned how to represent a set notation on a number line. You now know the meaning of a closed dot and an open dot when it is graphed on a number line. You also learned how to represent the various number sets on a number line. From a number line graph, you learned how to write the set notation that describes the set shown on the number line.
Problem Set
Describe each set notation in words.
1. $\{x|x > 8, x \ \varepsilon \ R\}$
2. $\{x|x \le -3, x \ \varepsilon \ I\}$
3. $\{x|-4 \le x \le 6, x \ \varepsilon \ R\}$
4. $\{x|5 \le x \le 11, x \ \varepsilon \ W\}$
5. $\{x|x \ge 6, x \ \varepsilon \ N\}$
Represent each graph using set notation
For each of the following situations, use set notations to represent the limits.
1. To ride the new tilt-a whirl at the fairgrounds, a child can be no taller than 4.5 feet.
2. A dance is being held at the community hall to raise money for breast cancer. The dance is only for those people 19 years of age or older.
3. A sled driver in the Alaska Speed Quest must start the race with no less than 10 dogs and no more than 16 dogs.
4. The residents of a small community are planning a skating party at the local lake. In order for the event to take place, the outdoor temperature needs to be above $-6^\circ C$ and not above $-5^\circ C$.
5. Juanita and Hans are planning their wedding supper at a local venue. To book the facility, they must guarantee that at least 100 people will have supper but no more than 225 people will eat.
Represent the following set notations on a number line.
1. $\{x|x>6, x \ \varepsilon \ N\}$
2. $\{x|x\le 8, x \ \varepsilon \ R\}$
3. $\{x|-3\le x < 6, x \ \varepsilon \ I\}$
Describe each set...
1. $\{x|x > 8, x \ \varepsilon \ R\}$ All real numbers greater than 8.
1. $\{x|-4 \le x \le 6, x \ \varepsilon \ R\}$ All real numbers between -4 and 6.
1. $\{x|x \ge 6, x \ \varepsilon \ N\}$ All natural numbers greater than or equal to 6.
Represent each graph...
1. $\{x|< -6
1. $\{x|x \le -8, x \ \varepsilon \ R\}$
1. $\{x|-7 \le x < 1, x \ \varepsilon \ R\}$
For each of the following situations...
1. $\{x|x \le 4.5, x \ \varepsilon \ R\}$
1. $\{x|10 \le x \le 16, x \ \varepsilon \ N\}$
1. $\{x|100 \le x \le 225, x \ \varepsilon \ N\}$
Represent the following...
1. $\{x|x > 6, x \ \varepsilon \ N\}$
1. $\{x|x \le 8, x \ \varepsilon \ R\}$
1. $\{x|-3 \le x < 2, x \ \varepsilon \ I\}$
Summary
In this lesson you revisited the real number system $(R)$. You reviewed the sets of numbers that make up the real number system and concentrated on the types of numbers that were included in each set. These sets included the natural numbers $(N)$, whole numbers $(W)$, integers $(I)$, rational numbers $(Q)$ and irrational numbers $(\overline{Q})$. Given a group of numbers, you learned how to assign each number to the number set(s) to which it belonged. When you assigned a number to its number set(s), you had to be careful not to put it in the wrong set. All whole numbers are integers but not all integers are whole numbers.
You also learned that a mathematical statement that shows an inequality and the set of numbers to which the variable belongs is known as set notation. Set notation can be described in words or can be represented on a number line. A closed dot on the number line means that the number is included in the set notation and an open dot means that the number is not included. A closed dot is the result of the inequality symbol $\le$ or $\ge$. An open dot is the result of the inequality sign < or >. You also learned how to represent the various number sets on a number line.
From a number line graph, you learned how to write the set notation that describes the set shown on the number line. You also learned how to write set notation to represent various real life situations. The set notations that represent real life situations can all be drawn on a number line.
## Comparing Real Numbers
Objectives
The lesson objectives for Comparing Real Numbers are:
• Ordering real numbers from least to greatest
• Representing real numbers on a number line
• Using technology to simplify the process
Introduction
In this concept you will revisit the number sets that make up the real number system. You will also apply the skills you have learned for changing fractions to decimal numbers. In addition, you will learn to order real numbers from least to greatest and to place these numbers on a number line. When placing numbers on a number line, you will learn helpful hints to make the process easier. Finally, you will learn to order the numbers using your TI83 calculator.
Guidance
Order the following real numbers from least to greatest.
$\frac{22}{7},1.234 234 \ldots, - \sqrt{7}, -5, -\frac{21}{4}, 7,-1.666,0,8.32,\frac{\pi}{2},-\pi,-5.38$
As you examine the above numbers, you can see that there are natural numbers, whole numbers, integers, rational numbers and irrational numbers. These numbers, as they are presented here, would be very difficult to order from least to greatest. The simplest way to order the numbers is to express them all in the same form – all fractions or all decimal numbers. Since you all have a calculator, use the calculator to express every number as a decimal number. Watch your signs – don't drop any of the negative signs.
Now that all the numbers are in decimal form, make two lists of decimal numbers – negatives and positives. The most places after the decimal point in the given numbers is 6. The decimal numbers that you determined with your calculator need only have 6 numbers after the decimal point.
Given Pos. Value Positives Given Neg. Value Negatives
$\frac{22}{7}$ 3.142857 $-\sqrt{7}$ -2.645751
1.234 234... 1.234234 -5 -5
7 7 $-\frac{21}{4}$ -5.25
0 0 -1.666 -1.666
8.32 8.32 $-\pi$ -3.141592
$\frac{\pi}{2}$ 1.570796 -5.38 -5.38
The negative numbers with the greatest magnitude go left on the number line since they are the smallest of the numbers. Now arrange the numbers from least to greatest using the numbers you were given in the problem.
$-5.38, -\frac{21}{4}, -5, -\pi,\sqrt{7}, -1.666, 0, 1.234234, \frac{\pi}{2},\frac{22}{7}, 7, 8.32$
Example A
Draw a number line and place these numbers on the line.
${\color{red}\sqrt{\frac{2}{5}}}, {\color{blue}0.6467},{\color{red}-\frac{3}{5}},{\color{red}\frac{1}{8}},{\color{green}0},{\color{red}\sqrt{0.5}},{\color{blue}-2.34},{\color{red}\pi},{\color{red}\frac{2 \pi}{3},{\color{green}-1}},{\color{green}2}$
$\sqrt{\frac{2}{5}}=0.6324 \quad -\frac{3}{5}=-0.6 \quad \frac{1}{8}=0.125 \quad \sqrt{0.5}=0.7071 \quad -\pi=-3.1416 \quad \frac{2 \pi}{3}=2.0944$
Start by placing the ${\color{green}\mathbf{integers}}$ on the line first. Next place the ${\color{blue}\mathbf{decimal \ numbers}}$ on the line.
Use your calculator to convert ${\color{red}\mathbf{the \ remaining \ numbers}}$ to decimal numbers. Place these on the line last.
It is impossible to place decimal numbers in the exact location on the number line. However, place them as close as you can to the appropriate spot on the line. Use your estimating skills when doing an exercise like this one.
Example B
For each given pair of real numbers, find another real number that is between each of the pairs.
i) $-2,1$
ii) $3.5,3.6$
iii) $\frac{1}{2},\frac{1}{3}$
iv) $-\frac{1}{3}, -\frac{1}{4}$
The answers to these will vary.
i) The number must be greater than -2 and less than 1. $\boxed{-2, {\color{blue}0},1}$
ii) The number must be greater than 3.5 and less than 3.6. $\boxed{3.5, {\color{blue}3.54},3.6}$
iii) The number must be greater $\frac{1}{3}$ than and less than $\frac{1}{2}$. Write each fraction with a common denominator. $\frac{1}{2}=\frac{3}{6},\frac{1}{3}=\frac{2}{6}$. If you look at $\frac{2}{6}$ and $\frac{3}{6}$, there is no fraction, with a denominator of 6, between these values. Write the fractions with a larger common denominator. $\frac{1}{2}=\frac{6}{12}, \frac{1}{3}=\frac{4}{12}$. If you look at $\frac{4}{12}$ and $\frac{6}{12}$, the fraction $\frac{5}{12}$ is between them. $\boxed{\frac{1}{3},{\color{blue}\frac{5}{12}},\frac{1}{2}}$
iv) The number must be greater than $-\frac{1}{3}$ and less than $-\frac{1}{4}$. Write each fraction with a common denominator. $-\frac{1}{3}=-\frac{4}{12},-\frac{1}{4}=-\frac{3}{12}$. If you look at $-\frac{3}{12}$ and $-\frac{4}{12}$, there is no fraction, with a denominator of 12, between these values. Write the fractions with a larger common denominator. $-\frac{1}{3}=-\frac{8}{24},-\frac{1}{4}=-\frac{6}{24}$. If you look at $-\frac{6}{24}$ and $-\frac{8}{24}$, the fraction $-\frac{7}{24}$ is between them. $\boxed{-\frac{8}{24}, {\color{blue}-\frac{7}{24}}, -\frac{6}{24}}$
Example C
Order the following fractions from least to greatest.
$\frac{2}{11},\frac{7}{9},\frac{8}{7},\frac{1}{11},\frac{5}{6}$
The fractions do not have a common denominator. This makes it almost impossible to arrange the fractions from least to greatest. To determine the common denominator, may take some time. Let’s use the TI83 to order these fractions.
The fractions were entered into the calculator as division problems. The decimal numbers were entered into List 1.
The calculator has sorted the data from least to greatest.
The data is sorted. The decimal numbers and the corresponding fractions can now be matched from the screen where they were first entered.
$\frac{1}{11},\frac{2}{11},\frac{7}{9},\frac{5}{6},\frac{8}{7}$
Guided Practice
1. Arrange the following numbers in order from least to greatest and place them on a number line.
$-3.78, -\frac{11}{4},-4, \frac{\pi}{2}, -\sqrt{6},-1.888,0,0.2424,\pi,\frac{21}{15},2,1.75$
2. For each given pair of real numbers, find another real number that is between each of the pairs.
i) $-3,-5$
ii) $-3.4,-3.5$
iii) $\frac{1}{5},\frac{3}{10}$
iv) $-\frac{3}{4},-\frac{11}{6}$
3. Use technology to sort the following numbers:
$\sqrt{\frac{3}{5}},\frac{15}{38},-\frac{7}{12},\frac{1}{4},0,\sqrt{8},-\frac{13}{21},-\pi,\frac{3 \pi}{5},-6,3$
1. $-3.78, -\frac{11}{4},-4, \frac{\pi}{2}, -\sqrt{6},-1.888,0,0.2424,\pi,\frac{21}{15},2,1.75$
2. i) $-3,-5$
ii) $-3.4,-3.5$
iii) $\frac{1}{5},\frac{3}{10}$
iv) $-\frac{3}{4},-\frac{11}{6}$
i) The number must be greater than -5 and less than -3. $\boxed{-5,{\color{blue}-4},-3}$
ii) The number must be greater than -3.5 and less than -3.4. $\boxed{-3.5,{\color{blue}-3.45},-3.4}$
iii) The number must be greater than $\frac{1}{5}$ and less than $\frac{3}{10}$. Write each fraction with a common denominator. $\frac{1}{5}=\frac{2}{10}$. If you look at $\frac{2}{10}$ and $\frac{3}{10}$, there is no fraction, with a denominator of 10, between these values. Write the fractions with a larger common denominator. $\frac{1}{5}=\frac{4}{20},\frac{3}{10}=\frac{6}{20}$. If you look at $\frac{4}{20}$ and $\frac{6}{20}$, the fraction $\frac{5}{20}=\frac{1}{4}$ is between them. $\boxed{\frac{1}{5}, \frac{{\color{blue}1}}{{\color{blue}4}}, \frac{3}{10}}$
iv) The number must be greater than $-\frac{3}{4}$and less than $-\frac{11}{16}$. Write each fraction with a common denominator. $-\frac{3}{4}=-\frac{12}{16}$. If you look at $-\frac{12}{16}$ and $-\frac{11}{16}$, there is no fraction, with a denominator of 16 between these values. Write the fractions with a larger common denominator. $-\frac{3}{4}=-\frac{24}{32},-\frac{11}{16}=-\frac{22}{32}$. If you look at $-\frac{24}{32}$ and $-\frac{22}{32}$, the fraction $-\frac{23}{32}$ is between them. $\boxed{-\frac{3}{4},-{\color{blue}\frac{23}{32}},-\frac{11}{16}}$
3. $\sqrt{\frac{3}{5}},\frac{15}{38},-\frac{7}{12},\frac{1}{4},0,\sqrt{8},-\frac{13}{21},-\pi,\frac{3 \pi}{5},-6,3$
The numbers have been sorted. The numbers from least to greatest are:
$-6,-\pi,-\frac{13}{21},-\frac{7}{12},0,\frac{1}{4},\frac{15}{38},\sqrt{\frac{3}{5}},\frac{3 \pi}{5},\sqrt{8},3$
## Summary
In this lesson you learned to order real numbers from least to greatest. To facilitate the process, you learned that changing the numbers to decimal numbers made the ordering less difficult.
You also learned to represent the numbers on an appropriate number line. The given integers could be placed on the line first and then any given decimal numbers were located on the number line. The remaining numbers were changed to decimals and placed on the number line last. Placing the numbers on the number line gave you a visual image of the order of the real numbers.
You were also shown the key strokes to use to order the numbers by using technology. The numbers were entered into the calculator and were recorded as decimal numbers. You then sorted the numbers on the calculator to order them. The one thing that you had to remember was to track the numbers as you entered them into the calculator. This step was necessary to match the given numbers with the corresponding decimal numbers.
To apply the concept of ordering numbers, you then solved problems that asked you to find a real number between a pair of given numbers.
Problem Set
Arrange the following numbers in order from least to greatest and place them on a number line.
1. $\{0.5,0.45,0.65,0.33,0,2,0.75,0.28\}$
2. $\{\frac{1}{2},-2,0,-\frac{1}{3},3,\frac{2}{3},-\frac{1}{2}\}$
3. $\{0.3,-\sqrt{2},1,-0.25,0,1.8,-\frac{\pi}{3}\}$
For each given pair of real numbers, find another real number that is between each of the pairs.
i) $8,10$
ii) $-7.6,-7.5$
iii) $\frac{1}{7},\frac{4}{21}$
iv) $-\frac{3}{5},-\frac{1}{2}$
Use technology to sort the following numbers:
1. $\{-2,\frac{2}{3},0,\frac{3}{8},-\frac{7}{5},\frac{1}{2},4,-3.6\}$
2. $\{\sqrt{10},-1,\frac{7}{12},3,-\frac{5}{4},-\sqrt{7},0,-\frac{2 \pi}{3},-\frac{3}{5}\}$
Arrange the following...
1. $\{\frac{1}{2},-2,0,-\frac{1}{3},1,\frac{2}{3},-\frac{1}{2}\}$
For each given pair...
i) The number must be greater than 8 and less than 10. $\boxed{8,{\color{blue}9},10}$
iii) The number must be greater than $\frac{1}{7}$ and less than $\frac{4}{21}$. Write each fraction with a common denominator. $\frac{1}{7}=\frac{3}{21}$. If you look at $\frac{3}{21}$ and $\frac{4}{21}$, there is no fraction, with a denominator of 21, between these values. Write the fractions with a larger common denominator. $\frac{1}{7}=\frac{6}{42},\frac{4}{21}=\frac{8}{42}$. If you look at $\frac{6}{42}$ and $\frac{8}{42}$, the fraction $\frac{7}{42}=\frac{1}{6}$ is between them. $\boxed{\frac{1}{7},{\color{blue}\frac{1}{6}},\frac{4}{21}}$
Use technology...
1. $\Big \{\sqrt{10},-1,\frac{7}{12},3,-\frac{5}{4},-\sqrt{7},0,-\frac{2 \pi}{3},-\frac{3}{5} \Big \}$ The numbers in order from least to greatest are: $\Big \{-\sqrt{7},-\frac{2 \pi}{3},-\frac{5}{4},-1,-\frac{3}{5},0,\frac{7}{12},3,\frac{7}{12} \Big \}$
Summary
In this lesson you learned to order real numbers from least to greatest. To avoid using the guessing method, numbers that were not integers or were not given as decimal numbers, were changed to decimal numbers. When you did this, you were less likely to make an error when ordering the numbers.
You also learned to represent the numbers on an appropriate number line. The given integers could be placed on the line first and then any given decimal numbers were located on the number line. The remaining numbers were changed to decimals and placed on the number line last. If a number line marled in intervals of 10 is used, the decimal numbers can be placed approximately where they belong. It is impossible to place the decimal numbers exactly where they belong. An approximate location is acceptable. This visual representation of the ordered numbers made the process more meaningful.
You were also shown the key strokes to use to order the numbers by using technology. As you entered the numbers into the calculator, they were converted to decimal numbers and stored in a list on the calculator. The numbers were sorted within seconds when the SORT command was used. The one thing that you had to remember was to track the numbers as you entered them into the calculator. This step was necessary to match the given numbers with the corresponding decimal numbers.
To apply the concept of ordering numbers, you then solved problems that asked you to find a real number between a pair of given numbers. When fractions were used, it was often necessary to write equivalent fractions using a common denominator and not a least common denominator.
## Date Created:
Jan 16, 2013
Jun 04, 2014
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# Factoring Trinomials Section 5.3.
## Presentation on theme: "Factoring Trinomials Section 5.3."— Presentation transcript:
Factoring Trinomials Section 5.3
Factor a Trinomial Using Algebra Tiles Factor 2x2 + 7x + 3: x2 tile
unit tile
Factor a Trinomial Using Algebra Tiles 2x2 + 7x + 3
Place x tiles to form a rectangle. This rectangle represents x2 + 7x + 3.
The factors are (2x + 1) and (x + 3).
Factor a Trinomial Using Algebra Tiles Place 2 x tiles and one unit tile in the top row. Place one x tile and 3 unit tiles in the side column. The factors are (2x + 1) and (x + 3). 2x2 + 7x + 3 = (2x + 1)(x + 3)
Apply Factoring Scuba Diving
A scuba diver uses up her air supply in 60 min when swimming at a speed of 30 m/min. If she swims faster, she will cover more distance each minute, but will also breathe more quickly, depleting her air supply more quickly. If she swims at 30 + x m/min, the distance she can swim before running out of air is given by the formula d = 1800 – 30x – 3x2, where d is the distance she can swim, and x is the speed in m/min.
Express the distance formula in factored form.
Apply Factoring Scuba Diving Express the distance formula in factored form. d = 1800 – 30x – 3x2 Start by looking for a common factor A common factor is –3 Factor out –3 d = –3(x2 + 10x – 600)
Apply Factoring Scuba Diving d = –3(x2 + 10x – 600)
Find two integers whose product is –600 and sum is 10. Factor 1 Factor 2 Product Sum 60 --10 --600 50 Factor 1 Factor 2 Product Sum 60 --10 --600 50 --12 38 Factor 1 Factor 2 Product Sum 60 --10 --600 50 --12 38 40 --15 25 30 --20 10 Factor 1 Factor 2 Product Sum 60 --10 --600 50 --12 38 40 --15 25 Factor 1 Factor 2 Product Sum The integers are 30 and –20 d = --3(x + 30)(x –20)
Apply Factoring Scuba Diving Did You Know?
A scuba diver can make maximum use of the available air by breathing slowly and deeply. Slow, deep breathing allows the lungs to transfer more oxygen from the air to the blood before it is exhaled. Once the air is exhaled, it disappears as bubbles. |
# Cube of negative number
In this post we will learn to find the cube of negative numbers with example.
If you want the revise the basics of cube of any number, click the red link.
## Cube of negative number
The cube of negative number is found by multiplying the number by itself thrice.
When we multiply the negative number thrice by itself we get the negative number.
Hence, the cube of negative number is also a negative number.
The general form of cube of negative number is given as;
\mathtt{( -m)^{3} \Longrightarrow ( -m) \times ( -m) \times ( -m) \ }\\\ \\ \mathtt{( -m)^{3} \Longrightarrow \ -m^{3}}
### Rules of negative sign multiplication
To understand the cube of negative number, just remember the below rule.
(a) Multiplication of two negative sign results in positive sign.
It’s like two negatives becoming positive.
(b) When we multiply three negative signs, the final result is also negative.
From the below image you can see that cube of negative number works in three steps;
(i) Two negative numbers multiply to generate positive number.
(ii) The positive number will them multiply the negative number to generate negative number.
For example;
Find the cube of -5
Multiply -5 by itself three times
\mathtt{\Longrightarrow \ ( -5)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( -5) \times ( -5) \times ( -5)}\\\ \\ \mathtt{\Longrightarrow \ -125}
Hence, on calculation we got the negative number.
Example 02
Find the cube of -11
Solution
Multiply -11 by itself three times.
\mathtt{\Longrightarrow \ ( -11)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( -11) \times ( -11) \times ( -11)}\\\ \\ \mathtt{\Longrightarrow \ -133}
Example 03
Find the cube of -12
Solution
Multiply -12 thrice.
\mathtt{\Longrightarrow \ ( -12)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( -12) \times ( -12) \times ( -12)}\\\ \\ \mathtt{\Longrightarrow \ 144\times ( -12)}\\\ \\ \mathtt{\Longrightarrow \ -1728}
Hence, cube of -12 is -1728.
Example 04
Find the cube of -25
Solution
Multiply -25 thrice.
\mathtt{\Longrightarrow \ ( -25)^{3}}\\\ \\ \mathtt{\Longrightarrow \ ( -25) \times ( -25) \times ( -25)}\\\ \\ \mathtt{\Longrightarrow \ 625\ \times ( -25)}\\\ \\ \mathtt{\Longrightarrow \ -15625}
Hence, cube of -25 is -15625.
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# How Was The Area Formula for a Circle Derived?
You might be knowing what is the area of a circle. While solving mensuration problems, you have used following formulae.
• Rectangle: Perimeter = $2 \left(l + b \right)$ and Area = $lb$, where $l$ and $b$ are length and breadth of a rectangle.
• Circle: Perimeter (Circumference) = $2 \pi r$ and Area = $\pi r^{2}$, where $r$ is the radius of a circle.
Also, you might be knowing how you can derive the formulae for perimeter and area of a rectangle.
But do you know how the area formula for a circle is derived?
## What is the Area of a Circle and How is it Derived?
You might be aware that area or space occupied by a rectangle is equal to the product of its length and breadth, i.e., Area = $lb$.
You can use the formula for area of a rectangle to find the area of a circle also.
To understand this let’s consider a circle of radius $r$ and divide into small sectors as shown in the figure below:
The perimeter (circumference) or the length of the boundary of a circle is $2 \pi r$.
Note: Circumference of a circle is equal to sum of the lengths of the arcs = $\pi r + \pi r = 2 \pi r$.
The circle now can be cut-up to form a rectangle. When the central angle of each sector becomes very small or nearly diminishes, the curves on the bottom and top of the rectangle straighten out to form a straight line with length PI r units while the width of the rectangle is r units.
When you consider a very small portion of the circumference of a circle, it is approximately a straight line. For that reason only, you see the surface of earth as flat (although it is curved), because you are seeing a very small portion of the Earth surface while on the ground.
Thus the area of the circle approximates the area of the rectangle.
And now you can find the area of rectangle obtained as Area = $\pi r \times r = \pi r^{2}$
## Let’s Code With Python
Find the circumference and area of a circle for a given radius.
#Circumference & Area of a Circle
PI = 22/7 #Define PI
C = 2 * PI * r #Calculate Circumference
A = PI * (r ** 2) # Calculate Area
#Print results
print('Circumference of circle of radius ', "{:.2f}".format(r), 'is', "{:.2f}".format(C), 'unit')
print('Area of circle of radius ', "{:.2f}".format(r), 'is', "{:.2f}".format(A), 'sq unit')
## Conclusion
Rectangle and triangle are two basic plane figures in geometry and one can use these two figures as base to find the perimeter and area of other plane figures.
## Practice Problems
(Take $\pi = \frac {22}{7}$)
1. For a circle of radius $2.8 cm$, find
1. Circumference
2. Area
2. For what length of radius are the area and circumference of a circle numerically equal?
3. If the circumference of a circle is $88 inch$, then find the area of a circle.
4. If the area of a circle is $0.385 m^{2}$, then find the circumference of a circle.
5. What will the percentage change in the area of a circle if its circumference increases by $10 \%$?
6. What will the percentage change in the circumference of a circle if its area decreases by $10 \%$? |
+ k, the negative in front of the parenthesis tells us that the parabola is pointed downward. It is a point (h, k) on the parabola. For the horizontal axis of symmetry, the equation will be y = 4b or y = -4b. y = x ² + 4 x. y = ax ² + b x + c a = 1, b = 4 and c = 0 Once you know the x-value of the vertex, you can substitute it in the original function to find … Whew, that was a lot of shuffling numbers around! For a parabola with equation $$y=a{x}^{2}+bx+c$$: The axis of … depending upon the orientation of the parabola. + k, here (h,k) are the points on the x-axis and y-axis respectively. Normally, you would take the coefficient of the x term, divide it by 2, and use the result to complete the square. Step 4. For Parabola equation y, The distance between the Vertex and the Focus, which is measured along the axis of symmetry, is termed as the “, It is very easy to locate the Focus point of Parabola when the Parabola equation is given. View solution. The vertex of the parabola is not the origin. If we identify the vertex of a quadratic, we can just plug it in the formula and get the equation. the focus, directrix, Axis, vertex etc .In this post we will explore all the different variety of methods that can be used to find the vertex of parabola. – 4x + 7, here first you need to apply to complete the square method to write this given equation into vertex form. vertex\: (y-2)=3 (x-5)^2. This algebra 2 video tutorial explains how to find the vertex of a parabola given a quadratic equation in standard form, vertex form, and factored form. Then find the equation of the parabola. The squared part is always positive (for a right-side-up parabola), unless it’s zero. Here h = 0 and k = 0, so … Latest Trends of Information Technology in Education, 8 examples of technology that utilizes the Internet of Things, 6 Examples Of Smart Home Devices That Utilize IoT Technology, 6 Reasons Why Companies Need IoT solutions, How Digital Healthcare is revolutionized with IoT. Vertex Form of Equation. A parabola is the arc a ball makes when you throw it, or the cross-section of a satellite dish. vertex\:3x^2+2x+5y-6=0. To Find The Vertex, Focus And Directrix Of The Parabola. The vertex form of a parabola's equation is generally expressed as: y = a(x-h) 2 +k (h,k) is the vertex as you can see in the picture below If a is positive then the parabola opens upwards like a regular "U". The standard equation of the parabola is of the form ax2 + bx + c = 0. Example 2 Graph of parabola given vertex and a point Find the equation of the parabola whose graph is shown below. View solution. Example 1: Find the focus of the parabola y = 1 8 x 2. A parabola forms an integral part of the conic section geometry( among others like ellipse, hyperbola etc). y^{2}+y-x+1=0. I'm going to assume the OP wanted the vertex and focus of the original tilted parabola since they already had rotated it to a standard form where those values are easier to find. (equal shape and equal size) halves. Its equation takes the form y = (x+h)² + k. where (h, k) is the vertex. We are given constants of the parabola equation x, y, and z. vertex form of parabola: y = a(x – h) 2 + k or y = ax 2 + bx + c. Here, the x-coordinate of the vertex is -b/2a. It lies on the axis of symmetry. The standard form of a parabola is #y=ax^2++bx+c#, where #a!=0#.. The vertex is on the axis of symmetry, so its x -coordinate is − b 2a − b 2 a. The easiest way to find the equation of a parabola is by using your knowledge of a special point, called the vertex, which is located on the parabola itself. The Parabola is a set of all points in a plane which are equidistant from a given point and a given line. The vertex is just (h,k) from the equation. The focus lies on the axis of symmetry of the parabola. A parabola is a curve where any point is at an equal distance from: 1. a fixed point (the focus ), and 2. a fixed straight line (the directrix ) Get a piece of paper, draw a straight line on it, then make a big dot for the focus (not on the line!). If the quadratic is written in the form y = a(x – h)2 + k, then the vertex is the point (h, k). Hence. Each parabola has a vertical line or axis of symmetry that … The "Ask A Scientist" website provides this answer: The general form of a parabola is given by the equation: A * x^2 + B * x + C = y where A, B, and C are arbitrary Real constants. The x-value of the vertex can be found using the formula h=-b/2a. I will not limit myself to parabolas with the axis on the coordinate axis, rather I will focus on a generic parabola, be it rotated or translated. + 5, the value of his 1, and k is 5. Learn about the parts of a parabola. So, the parabola opens to the left. To find the vertex of a parabola with axis of symmetry, factor the quadratic equation and find the point at which the equation crosses the x-axis. Well, it is quite easy. If a > 0 in ax2 + bx + c = 0, then the parabola is opening upwards and if a < 0, then the parabola is opening downwards. The standard form of a parabola equation is . Co-ordinates of Vertex of a parabola can be found form the coefficients a,b,c of the standard form of quadratic. Given the values of a, b and c; our task is to find the coordinates of vertex, focus and the equation of the directrix. Now play around with some measurements until you have another dot that is exactly the same distance from the focus and the straight line. Vertex form: y = (x – h)2 + k. You might have heard your friend asking you how to find the vertex of a parabola from a given equation. helpful. There are two ways to do it, and they're both easy. Real World Math Horror Stories from Real encounters, is the maximum or minimum value of the parabola (see picture below), the axis of symmetry intersects the vertex (see picture below). Vertex ” a mirror-symmetrical, plane curve and typically U-shaped able to find the vertex form next calculate. -Coordinate of vertex depends upon the standard form ) is zero focus of. The depth of the parabola equation a vertical line the equation then join the little dots then. On the parabola the beginning of your parabola calculate the midway point, which will lie directly in the... Of little dots and you will have a parabola can be found form coefficients! Ellipse, hyperbola etc ), hyperbola etc ) makes when you throw it, or maximum. Crosses its axis of symmetry of the parenthetical term terms Substitute the vertex a... The general form of a parabola forms an integral part of the required parabola quadratic, we Substitute vertex... = 2 ( x – h ) step 3: by applying these in! ) 2 +k ax2 + bx + c = 0 # ( -2,1 ) # directrix # x=1.... 4X + 7, here first you need to understand the meaning of parabola in the other direction ( vertex... Then join the little dots, then the graph has a vertex at \ ( ( 2,3 ) \.... ) are respectively the vertex of the parenthetical term & x = uy2 + +. Makes when you throw it, and directrix using the equations given in the 's... Into vertex form thousands of other math skills -b/2a, -v/2u ) of any!. Negative in front of the parabola line or axis of symmetry has a vertical line or axis symmetry... On find the vertex is the point ( h, k ) is a point (,. And today we 're gon na talk about how to determine the coordinate points of the parabola changes direction is! The focus of the parabola 's equation in vertex form of a ; we can find... And z with two x-Intercepts More Numerous than Parabolas with two x-Intercepts More Numerous than Parabolas with x-Intercepts... ( ( 2,3 ) \ ) vertex to standard form we will get the vertex.... Once we have identified what they-coordinate is, there ’ s the beginning of your parabola to and! Information of the parabola equation down U '' to complete the square method to vertex... Left, the graph of a parabola '' and thousands of other math skills Substitute the vertex is (,! Following two steps: vertex is the arc a ball makes when you throw it, and k in following... Related terms better if we understand the meaning of parabola '' and thousands other! Terms related to parabola, basic things which one should know about parabola applying these in... Or lowest point on parabola, also called maximum or minimum a ball makes when you throw it and. Problem on find the focus, and k is 5 and the straight line unless it ’ the. Is always positive ( for a reason solution: method 1: find y-coordinate! When coming from the focus, vertex and focus = a ( x-h ) 2 +k the or... The depth of the parabola is the point from where parabola crosses its axis of that... Formulas to find the vertex of a quadratic function as it approaches its vertex is just (,... Given constants of the parabola and the straight line uy2 + vy + w y-axis... B = 6 steps to find the focus, vertex and focus =.... This number represents a maximum or minimum learn how to find the vertex is ( -7 -10. Explains how to find the equation improve your math knowledge with free questions in find the vertex, and! = uy2 + vy + w for y-axis and x-axis respectively -2,1 ) # directrix x=1! Of these calculations identify the vertex of a parabola '' and thousands of other math skills free questions ! ) 2 +k to graph a parabola forms an integral part of the graph has a vertical line or of. 2 ( x, y ) ordered pairs see the next example problem on find the of! And related terms is just ( h, k ) on the parabola the! = 3x 2 + x – h ) 's equation in vertex form ) better. Is just ( h, k ) on the axis of symmetry we have identified they-coordinate! # x=1 # many important terms related to it and related terms ball makes when you throw it or... ^2=-20 ( y-1 ) parabola-function-vertex-calculator curve and typically U-shaped of parabola, also called maximum or minimum of the.. Depth of the required parabola from y 2 = 4 x explains how to find the y-coordinate the., calculate the midway point, which will lie directly in between the two roots of the form +... We will get the equation – 1 ) and ( 0, -4 ) x-axis. That given line opens upwards like a regular U '': find zeros... Not the origin required parabola is always positive ( for a right-side-up parabola ), unless it ’ s for. Directrix # x=1 # until you have three pairs of points that are ( x 2. By using the tools of calculus and derivatives convert your quadratic equation depth! Will get the equation is y = 8 ’ s the beginning of your parabola as we know the... B, c of the parabola which will lie directly in between the roots... The cross-section of a parabola you find the vertex of the parabolic graph can the... Parabola faces upward, hyperbola etc ) form the coefficients a, b = 6 steps to find vertex! These values in the following table have a parabola can be found form the coefficients,! Form ' for a reason how to find the vertex of a parabola: ( y-3 ) ^2=8 ( x-5 ) vertex\: ( y-3 ) (. Vertex at \ ( -\frac { how to find the vertex of a parabola } { 2a } \ ) )! Think about it example 2 the graph of a parabola is given by order graph! We understand the basic 's of a parabola beginning of your parabola -\frac { b } { }. Sign in front of the vertex, we need to understand the meaning of parabola ( 0, 4 and! In order to graph a parabola is and “ y ” -Coordinate “ k ” gives! Line of symmetry hi guys, I 'm Nancy and today we gon! Location of vertex coordinates ( h, k ) are given to...., also called maximum or minimum “ directrix ” of the required parabola if opens. Whether this number represents a maximum or minimum of a parabola with vertex ( -3, 1 ) Their Owners... Point and a given line that given point is called the “ ”. Vertex be ( 4, 5 ) hence the vertex of a parabola is resting on the parabola or of. That are ( x – 2 and graph the parabola from y 2 4... Is on the axis of symmetry = 0 of calculus and derivatives co-ordinates of vertex of parabola is a,. Or downward by just looking at the equation is y = -4b hi guys, I 'm Nancy and we! Would be better if we understand the basic 's of a quadratic, we given! ( y-3 ) ^2=8 ( x-5 ) vertex\: ( x+3 ) ^2=-20 ( y-1 ) parabola-function-vertex-calculator equation the!, let the given equation into vertex form ) is zero also called maximum or minimum the... Line is called the focus of a parabola forms an integral part of the section. } \ ) the parenthesis tells us that the parabola decreases rapidly including... A negative sign in front, then the how to find the vertex of a parabola has a vertical line or axis of symmetry is going discuss... ( -3, -63 { 3/14 } ) \$ example problem on find range... Quick way as I have to see whether this number represents a maximum or minimum the form ax2 bx! Ellipse, hyperbola etc ) k= -4, so the vertex is shape... We will get the equation of the vertex ’ s zero & Trademark Belongs to Respective! X-Intercepts More Numerous than Parabolas with two x-Intercepts More how to find the vertex of a parabola than Parabolas No! Other math skills approaches its vertex is different for both forms of equations Nancy and today we 're na... Seen parabola has a vertex at \ ( x\ ) is zero about it have a parabola forms integral... Y=X 2 +6x-10 we frequently come across i.e 0, -4 ) graph, the. S coordinates for h and k is 5 y-1 ) parabola-function-vertex-calculator = ax2 + bx + c = 0 plane! A quadratic equation on find the vertex is the arc a makes. Once we have identified what they-coordinate is, there ’ s zero by following, a negative sign front. A satellite dish in vertex form are ( x – 2 and graph parabola! Also called maximum or minimum of a satellite dish of y = 3x 2 x... { 2a } \ ) ( 0, 4 ) and directrix to discuss upward and downward parabola have dot! ), unless it ’ s the beginning how to find the vertex of a parabola your parabola and today we 're gon na talk about to! Things which one should know about parabola straight line of quadratic equation have to do a lot of shuffling around. With diagram 2 + x – h ) forms of equations example 1: the... Have lots of little dots, then the graph of a parabola is a set of all in... Vertex ” be converted into the same form by following, ) are respectively the vertex focus! Crosses its axis of symmetry, the graph opens downwards like an upside down U '' from 2. Are already few answers given to this question to apply to complete square. |
McGraw Hill Math Grade 2 Chapter 3 Test Answer Key
Practice questions available in McGraw Hill Math Grade 2 Answer Key PDF Chapter 3 Test will engage students and is a great way of informal assessment.
McGraw-Hill Math Grade 2 Chapter 3 Test Answer Key
Is the number odd or even? Circle your answer. Use small objects to help.
Question 1.
3 odd even
3 is an odd number because it is not divisible by 2.
Question 2.
14 odd even
14 is an even number because it is divisible by 2 and leaves the remainder 0.
Question 3.
10 odd even
10 is an even number because it is divisible by 2 and leaves the remainder 0.
Question 4.
17 odd even
17 is an odd number because it is not divisible by 2.
Add. Write the sum. Tell if the sum ¡s even or odd. Circle your answer.
Question 5.
9 + 10 = ____
odd even
9 + 10 = 19
So, 19 is an odd number.
Question 6.
11 + 5 = _____
odd even
11 + 5 = 16
16 is an even number.
Use the pictures to count by 2s. Write the numbers.
Question 7.
Count the number by 2s and find the number of cherries
2 + 2 + 2 + 2 + 2 = 10 cherries
Question 8.
Count the bananas by 2s
2 × 7 = 14
Thus there are 14 bananas.
Question 9.
Count by 2s
2 + 2 + 2 + 2 + 2 + 2 = 12
6 × 2 = 12
Use the pictures to count by 5s. Write the numbers.
Question 10.
Count each group by 5s.
5 + 5 + 5 + 5 = 20
Question 11.
Count each group by 5s.
5 + 5 + 5 + 5 + 5 + 5 = 30
Look at the array. Write the sum.
Question 12.
3 + 3 + 3 + 3 = _____ cars
There are 4 rows and 3 columns of cars.
3 + 3 + 3 + 3 = 12 cars
Question 13.
5 + 5 + 5 = ____ stars
There are 3 rows and 5 columns of stars
5 + 5 + 5 = 15 stars
Write a number sentence that tells what the array shows.
Question 14.
____ + ___ + ____ + ____ = ____ airplanes |
Lesson 2- Laws of Indices Objectives To know what indices are To learn the rules of indices Oct 2011INTO Foundation L2.
Presentation on theme: "Lesson 2- Laws of Indices Objectives To know what indices are To learn the rules of indices Oct 2011INTO Foundation L2."— Presentation transcript:
Lesson 2- Laws of Indices Objectives To know what indices are To learn the rules of indices Oct 2011INTO Foundation L2
What are Indices? Indices provide a way of writing numbers in a more compact and convenient form Indices is the plural of Index An Index is often referred to as a power Oct 2011INTO Foundation L2
For example 5 x 5 x 5 = 5 3 2 x 2 x 2 x 2= 2 4 7 x 7 x 7x 7 x 7= 7 5 7 5 & 2 4 are numbers in INDEX FORM Oct 2011INTO Foundation L2 7575
Combining numbers 5 x 5 x 5 x 2 x 2 = 5 3 x 2 4 We can not write this any more simply Can ONLY combine BASE NUMBERS if they are the same Oct 2011INTO Foundation L2
Rule 1 : Multiplication 2 6 x 2 4 = 2 10 2 4 x 2 2 = 2 6 3 5 x 3 7 = 3 12 General Rule Law 1 a m x a n = a m+n Oct 2011INTO Foundation L2
Rule 2 : Division 2 6 ÷ 2 4 = 2 2 2 5 ÷ 2 2 = 2 3 3 5 ÷ 3 7 = 3 -2 General Rule Law 2 a m ÷ a n = a m-n Oct 2011INTO Foundation L2
Rule 3 : Brackets (2 6 ) 2 = 2 6 x 2 6 = 2 12 (3 5 ) 3 = 3 5 x 3 5 x 3 5 = 3 15 General Rule Law 3 (a m ) n = a m x n Oct 2011INTO Foundation L2
Rule 4 : Index of 0 How could you get an answer of 3 0 ? 3 5 ÷ 3 5 = 3 5-5 = 3 0 3 0 =1 General Rule Law 4 a 0 = 1 Oct 2011INTO Foundation L2
Putting them together? 2 5 x 2 3 2 4 x 2 2 = 2 8 2 6 2 6 x 2 4 2 3 = 2 10 2 3 3 5 x 3 7 3 4 = 3 12 3 4 = 2 7 = 3 8 = 2 2 Oct 2011INTO Foundation L2
Works with algebra too! a 5 x a 3 a 4 x a 6 = a 8 a 10 = a -2 a 6 x a 4 = a 10 b 5 x b 7 = b 12 c 5 x c 3 c 4 = c 8 c 4 = c 4 Oct 2011INTO Foundation L2
..and a mixture… 2a 3 x 3a 4 = 6a 7 = 2 x 3 x a 3 x a 4 8a 6 ÷ 4a 4 = 2a 2 = (8 ÷ 4) x (a 6 ÷ a 4 ) 8a 6 4a 4 2 2 Oct 2011INTO Foundation L2 = 2a 2
Fractional indices (Using Law 1) We could write But So Oct 2011INTO Foundation L2
Fractional Indices Similarly General Rule Law 5 Oct 2011INTO Foundation L2
Negative Index Numbers. Simplify the expression below: 5 3 5 7 = 5 - 4 To understand this result fully consider the following: Write the original expression again as a quotient: Expand the numerator and the denominator: Cancel out as many fives as possible: Write as a power of five: Now compare the two results: Oct 2011INTO Foundation L2
Negative Indices The last Index rule a m General Rule Law 6 a -m = 1 Oct 2011INTO Foundation L2
Summary Rule 1 : Multiplication of Indices. a n x a m =……… Rule 2 : Division of Indices. a n a m = ……. Rule 4 : For Powers Of Index Numbers. ( a m ) n = ….. Rule 6 : For negative indices a - m =……. Rule 3 : For Powers Of Index Numbers. a 0 = ….. Rule 5 : For fractional indices a 1/n = n √a Oct 2011INTO Foundation L2
Exercises Section 2- Working with Indices Additional Questions if you get that far! Oct 2011INTO Foundation L2
Travelling to Mars How long would it take a space ship travelling at an average speed of 2.6 × 10 3 km/h to reach Mars 8.32 × 10 7 km away? Oct 2011INTO Foundation L2
Calculations involving standard form Time to reach Mars = 8.32 × 10 7 2.6 × 10 3 = 3.2 × 10 4 hours Rearrange speed = distance time time = distance speed to give This is 8.32 ÷ 2.6 This is 10 7 ÷ 10 3 How long would it take a space ship travelling at an average speed of 2.6 × 10 3 km/h to reach Mars 8.32 × 10 7 km away? Oct 2011INTO Foundation L2
Calculations involving standard form Use your calculator to work out how long 3.2 × 10 4 hours is in years. You can enter 3.2 × 10 4 into your calculator using the EXP key: 3. 2 EXP 4 Divide by 24 to give the equivalent number of days. Divide by 365 to give the equivalent number of years. 3.2 × 10 4 hours is over 3½ years. Oct 2011INTO Foundation L2
Download ppt "Lesson 2- Laws of Indices Objectives To know what indices are To learn the rules of indices Oct 2011INTO Foundation L2."
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Question Video: Verifying Whether Two Given Polygons Are Similar | Nagwa Question Video: Verifying Whether Two Given Polygons Are Similar | Nagwa
# Question Video: Verifying Whether Two Given Polygons Are Similar Mathematics • Second Year of Preparatory School
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Are the two polygons similar?
01:49
### Video Transcript
Are the two polygons similar?
Two polygons are similar if the corresponding angles are equal in measure and their corresponding side lengths are proportional. So we need to find the measure of angle ๐ด and the measure of angle ๐ธ. Each polygon should have all of the angles adding up to be 360 degrees. So to find the measure of angle ๐ด, we need to take 360 degrees and then subtract all of the three angles we already know, and then the remainder will be left for angle ๐ด. So weโre subtracting 103 degrees, 84 degrees, and 95 degrees. So the measure of angle ๐ด is 78 degrees.
Now, in the other polygon, the measure of angle ๐ธ, we would do the exact same thing. So after subtracting the three angles from 360, we get 95 degrees. So the measure of angle ๐ด is equal to the measure of angle ๐น, so theyโre both 78 degrees. Angle ๐ต and angle ๐บ are both 103 degrees, angle ๐ถ and angle ๐ป are both 84 degrees, and then lastly angle ๐ท and angle ๐ธ are both 95 degrees. This means our corresponding angles are equal in measure.
Now, letโs check if the side lengths are proportional. ๐ด๐ต to ๐น๐บ is 20 to 14, ๐ต๐ถ to ๐บ๐ป is just 16 to 11, ๐ถ๐ท to ๐ป๐ธ is 20 to 14, and ๐ท๐ด to ๐ธ๐น is 18.6 to 13. So we need to check if each of these fractions are proportional; do they reduce to be the exact same number? Unfortunately, they are not. These fractions do not reduce to be the exact same fraction; theyโre not proportional. Since the side lengths are not proportional, the polygons are not geometrically similar.
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# What is the solution for a cubic equation? Thanks.....
Mar 20, 2016
See explanation for a demonstration of Cardano's method...
#### Explanation:
There is a general formula and there are several methods.
My personal favourite method, which works well when a cubic equation has one Real root and two non-Real roots is Cardano's method.
Given a cubic equation:
$a {x}^{3} + b {x}^{2} + c x + d = 0$
If the coefficients are integers, then it may be useful to first multiply through by ${3}^{3} {a}^{2}$ to reduce the number of fractions we need to deal with.
$27 {a}^{3} {x}^{3} + 27 {a}^{2} b {x}^{2} + 27 {a}^{2} c x + 27 {a}^{2} d = 0$
${\left(3 a x\right)}^{3} + 3 b {\left(3 a x\right)}^{2} + 9 a c \left(3 a x\right) + 27 {a}^{2} d = 0$
Substitute $t = 3 a x + b$ to get:
${t}^{3} + \left(9 a c - 3 {b}^{2}\right) t + \left(27 {a}^{2} d + 2 {b}^{3} - 9 a b c\right) = 0$
Let $p = 9 a c - 3 {b}^{2}$ and $q = 27 {a}^{2} d + 2 {b}^{3} - 9 a b c$, so we have:
${t}^{3} + p t + q = 0$
Next substitute $t = u + v$ to get:
${u}^{3} + {v}^{3} + \left(3 u v + p\right) \left(u + v\right) + q = 0$
Then add the constraint: $v = - \frac{p}{3 u}$ to make $\left(3 u v + p\right) = 0$
${u}^{3} - {p}^{3} / \left(27 {u}^{3}\right) + q = 0$
Multiply through by $27 {u}^{3}$ and rearrange to get:
$27 {\left({u}^{3}\right)}^{2} + 27 q \left({u}^{3}\right) - {p}^{3} = 0$
Use the quadratic formula to find:
${u}^{3} = \frac{- 27 q \pm \sqrt{729 {q}^{2} + 108 {p}^{3}}}{54}$
$= \frac{- 27 q \pm 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}$
The derivation was symmetric in $u$ and $v$, so one of these roots can be used for ${u}^{3}$ and the other for ${v}^{3}$.
If $81 {q}^{2} + 12 {p}^{3} \ge 0$ then the square root is Real and the Real root of the cubic ${t}^{3} + p t + q = 0$ is:
${t}_{1} = \sqrt[3]{\frac{- 27 q + 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}} + \sqrt[3]{\frac{- 27 q - 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}}$
In this case, the Complex roots are given by:
${t}_{2} = \omega \sqrt[3]{\frac{- 27 q + 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}} + {\omega}^{2} \sqrt[3]{\frac{- 27 q - 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}}$
${t}_{3} = {\omega}^{2} \sqrt[3]{\frac{- 27 q + 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}} + \omega \sqrt[3]{\frac{- 27 q - 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}}$
where $\omega = - \frac{1}{2} + \frac{\sqrt{3}}{2} i$ is the primitive Complex cube root of $1$.
If on the other hand $81 {q}^{2} + 12 {p}^{3} < 0$ then the square roots are pure imaginary, the cube roots are of Complex numbers and all three roots are Real.
In this case to match the appropriate cube root $v$ to $u$, we have to use the relation $v = - \frac{p}{3 u}$ to find:
${t}_{k + 1} = {\omega}^{k} \sqrt[3]{\frac{- 27 q + 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}} - \frac{p}{3 {\omega}^{k} \sqrt[3]{\frac{- 27 q + 3 \sqrt{81 {q}^{2} + 12 {p}^{3}}}{54}}}$
$k = 0 , 1 , 2$
From these vales of $t$ we can find $x$ by $x = \frac{t - b}{3 a}$ |
# Conjugacy in a Group
Conjugate Element: If $a,b \in G$, then $b$ is said to be a conjugate of $a$ in $G$ if there exist an element $x \in G$ such that $b = {x^{ - 1}}ax$.
Symbolically, we shall write $a \sim b$ for this and shall refer to this relation as conjugacy.
Then $b \sim a \Leftrightarrow b = {x^{ - 1}}ax$ for some $x \in G$
Theorem: Conjugacy is an equivalence relation in a group.
Proof:
(i) Reflexivity: Let $a \in G$, then $a = {e^{ - 1}}ae$, hence $a \sim a\,\,\,\forall a \in G$ i.e. the relation of conjugacy is reflexive.
(ii) Symmetric: Let $a \sim b$, so that there exist an element $x \in G$ such that $a = {x^{ - 1}}bx,\,\,\,\,a,b \in G$. Now
Thus $a \sim b = b \sim a$. Hence the relation is symmetric.
(iii) Transitivity: Let there exist two elements $x,y \in G$ such that $a = {x^{ - 1}}bx$ and $b = {y^{ - 1}}cy$ for $a,b,c \in G$. Hence $a \sim b$, $b \sim c$
Where $yx \in G$ and $G$ being the group. Therefore$a \sim b,\,\,b \sim c\,\, \Rightarrow a \sim c$.
Hence relation is transitive.
Thus conjugacy is equivalence relation on$G$.
Conjugate Classes: For $a \in G$, let $C\left( a \right) = \left\{ {x:x \in G\,\,{\text{and}}\,\,a \sim x} \right\}$, $C\left( a \right)$, the equivalence class of $a$ in $G$ under conjugacy relation is usually called the conjugate class of $a$ in $G$. It consists of the set of all distinct elements of the type ${y^{ - 1}}ay$. |
# How do you solve -5x^2=-500?
Jul 26, 2016
$x = 10$
#### Explanation:
This is a good example of how operations can act on both sides of an equation in order to simplify and find the value of a variable.
I'll work this out showing each step:
$- 5 {x}^{2} = - 500$
$\left(\frac{1}{-} 5\right) \left(- 5 {x}^{2}\right) = - 500 \left(\frac{1}{-} 5\right)$
${x}^{2} = 100$
$\sqrt{{x}^{2}} = \sqrt{100}$
$x = 10$
Jul 26, 2016
$x = \pm 10$
#### Explanation:
Given:$\text{ } \textcolor{b r o w n}{- 5 {x}^{2} = - 500}$
To make $- 5 {x}^{2}$ positive multiply both sides by (-1)
$\text{ } \textcolor{b r o w n}{5 {x}^{2} = 500}$
Divide both sides by $\textcolor{b l u e}{5}$ turning the 5 in $5 {x}^{2}$ into 1
$\text{ } \textcolor{b r o w n}{\frac{5}{\textcolor{b l u e}{5}} \times {x}^{2} = \frac{500}{\textcolor{b l u e}{5}}}$
$\text{ } \textcolor{b r o w n}{{x}^{2} = 100}$
Square root both sides
$\text{ } \textcolor{b r o w n}{\sqrt{{x}^{2}} = \sqrt{100}}$
$\text{ } \textcolor{g r e e n}{x = \pm 10}$ |
## Distributive Property of Multiplication…oh, my! Yes, captain. Shields up and ready to fire on the Distributive Property of Multiplication.
Root canal without Novacaine. Pins under my fingernails. Sitting through an insurance seminar. Those all seem preferable to teaching the Distributive Property of Multiplication (DPM) to third graders. Seriously, didn’t my generation learn that sometime in MIDDLE SCHOOL? I know, stop with the whining and just teach it.
Here’s the rub. I have never had to teach this property! I met with my team and we talked about how to teach this. Should we break it down into steps? Should we do only hands on first? Should we use the Go Math book? We decided that we had to start with the concrete. Get the students to use manipulatives to build arrays that could then be broken apart and make the connection to the DPM. So here’s what I did.
I presented the problem/question then highlighted key words.
First I introduced the problem:
How can we break apart the array 3 x 7 to make it easier to solve?
If you have read my previous posts about Lesson Study and bansho, then you know I always start with a question that begins the lesson. We read the question together, we underlined important key words (break apart, easier). I explained to the students that I would give them the bag of foam tiles and a blank paper. They were to use the tiles to build the array first and the paper was to record their solutions with drawing and words and numbers.
I began to walk around the room interested in how the students interpreted “break apart.” Some students literally scattered their tiles around leaving no semblance of an array. Some students broke apart the array by putting the tiles into equal groups (hey at least they were equal groups!). Some students broke apart the array into 2 smaller arrays. I saw that one student was already recording his solution so I went to see what he had done. He had broken the array into 2 smaller arrays, counted the tiles in each array, then added. Not bad! I had the student go up to the board and copy his solution and then explain to the class what he had done. This was an excellent opportunity for me to step in after his explanation (which by the way I wish I had recorded because he used the correct vocabulary and his explanation was worthy of any common core question!) to talk about what it means to “break apart.”
Student used tiles to find a solution, then recorded answer then shared it with classmates.
I like to tell stories, so I told the students this story about my backyard. In my backyard, I had to move a lot of bricks from one location to the other. There were more than 30 bricks. I said that there was no way I could move all the bricks at once. So I decided to move about half of them at a time using a hand dolly. This way I divided or divvied up the work to make the task easier. I kept referring back to this idea of breaking up a hard task into easier tasks.
I guided the students to understand that breaking apart the array meant that after you broke it apart, you still had to have an array! Not groups, no piles, not unequal shapes. So once that was cleared up, I gave them another array to make and break apart: 3 x 8.
Another student found a solution for 3 x 8.
This time, the students understood that they had to end up with 2 smaller arrays made out of the bigger array but keeping the same amount of rows or columns. Most of the students I observed broke apart the array into 2 arrays of 3 x 4. Then they added the two arrays (12 + 12). I didn’t see any multiplication sentences yet, so I had to make sure I encouraged using multiplication sentences when labeling. However, I did find one student who had written multiplication sentences so I picked the student to come up and record the solution while the others continued to work.
After the student explained what she had done, I then asked the students:
Where is the best place to break up an array?
I guided the conversation by asking them which factors were easiest to multiply. They came up with 0, 1, 2, 5 and 10. Great! I kept guiding the conversation by saying pointing out that you’re not going to break up an array by 0 or 1. Since we are working with large arrays, I pointed out that 5 would be a better choice. Once that was established, I asked the students to turn their paper over and take notes. A very quick lecture followed introducing them to the DPM. We brainstormed synonyms for the word distribute and we talked about WHY they had to learn this property.
From there, I gave them another array to break apart: 4 x 9. Most of them began by breaking apart the array at 4 x 5 and 4 x 4. From there I saw them write their multiplication sentences and added the partial products to get the total product. Success! For now at least.
I did quickly explain to them how their 2 multiplication sentences needed to be combined into one multiplication sentence 4 x (5 + 4) so they could see where we were going with this whole process. That’s tomorrow’s lesson. Stay tuned. |
3.3 Limits and Continuity: Algebraic Approach
(This topic appears in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus)
Consider the following limit.
limx→2 x2 - 3x4x + 3 .
If you estimate the limit either numerically or graphically, you will find that
limx→2 x2 - 3x4x + 3 ≈ −0.1818
But, notice that you can obtain this answer by simply substituting x = 2 in the given function:
f(x) = x2 - 3x4x + 3 f(2) = 4 - 68 + 3 = - 211 = -0.181818...
This answer is more accurate than the one coming from numerical or graphical method; in fact, it gives the exact limit.
Q Is that all there is to evaluating limits algebraically: just substitute the number x is approaching in the given expression?
A Not always, but this often does happen, and when it does, the function is continuous at the value of x in question. Recall the definition of continuity from the previous tutorial:
Continuous Functions
The function f(x) is continuous at x = a if
limx→a f(x) exists That is, the left-and right limits exist and agree with each other limx→a f(x) = f(a)
The function f is said to be continuous on its domain if it is continuous at each point in its domain. If f is not continuous at a particular point a, we say that f is discontinuous at a or that f has a discontinuity at a.
Q How do you tell if a function is continuous?
A As we saw in the previous tutorial, we can tell whether a function is continuous by looking at its graph. If the graph breaks at some point in the domain, then f has a discontinuity there. If the function is specified algrabcially, sometimes it is easy to tell whether it is continuous by just looking at the formula:
A closed-form function is any function that can be obtained by combining constants, powers of x, exponential functions, radicals, absolute values, trigonometric functions, logarithms (and some other functions you may never see) into a single mathematical formula by means of the usual arithmetic operations and composition of functions. Examples of closed-form functions are:
3x2 - x +1, (x2 - 1)1/26x-2
e(x2 - 1)1/2/x
(log3(4x2 - ex))2/3
They can be as complicated as you like. The following is not a closed form function.
f(x) = -1 if x < -1 x2 + x if -1 ≤ x ≤ 1 2 - x if 1 < x ≤ 2
The reason for this is that f(x) is not specified by a single mathematical expression. What is nice about closed-form functions is the following.
Continuity of Closed Form Functions
Every closed form function is continuous on its domain.
Thus, the limit of a closed-form function at a point in its domain can be obtained by substitution.
Example
f(x) = x2 - 3x4x + 3
is a closed form function, and x = 2 is in its domain. Therefore we can obtain limx→2 f(x) by substitution:
limx→2 x2 - 3x4x + 3 = f(2) = - 211 ,
as we saw above.
Q limx→0 ex22x-3 Select one does not exist = 0 = -4/3 = -1/3 = 4/3 is undefined = -e/3 none of the above Q lim x→3 2x + 3x + 3 =
Q What if f(x) is a closed form function, but the point x = a is not in the domain of the function?
A Then, you either:
1. First try using simplification or some other technique to replace f(x) by another closed form function which does have x = a in its domain. This allows you to substitute x = a in the new function to obtain the limit, or
2. If the obove method does not work, try evaluating the limit numerically or graphically. Note, however, that this may only give you an estimate of the limit.
Evaluating a Limit Using Simplification
(Similar to Example 2 in Section 3.3 of Applied Calculus, or in Section 10.3 of Finite Mathematics and Applied Calculus )
Let us evaluate
lim x→-2 3x2 + x - 10x + 2
1. Is the function f(x) a closed form function?
Answer: yes, since (3x2 + x-10)/(x+2) is a single mathematical formula.
2. Is the value x = a in the domain of f(x)?
Answer: no, since (3(-2)2+(-2)-10)/((-2)+2) is not defined.
Therefore, we consult the above Question/Answer discussion, and simplify the function, if we can.
3x2 + x - 10
x + 2
= (x+2)(3x-5) (x+2)
= 3x-5.
Since we are now left with a closed form function that is defined when x = -2, we can now evaluate the limit by substitution:
lim x→-2 3x2 + x-10x + 2 = lim x→-2 3x - 5 = 3(-2)-5 = -11.
lim x→-1 (x3+1)(x-1)x2 + 3x + 2 Select one does not exist = + ∞ = - ∞ is undefined = 0 = -2 = -4 = -6 = -8 none of the above
Back in the tutorial on functions from the graphical point of view, we looked at the following function:
f(x)= -1 if -4 ≤ x < -1 x if -1 ≤ x ≤ 1 x2-1 if 1 < x ≤ 2
This time, we are not showing you the graph right away, and ask you to look at the formula instead. Notice:
1. The function f is not closed form. (It is not defined by a single formula.)
2. Inside each of the separate intervals [-4, -1), (-1, 1) and (1, 2] the function is closed form, and is hence continuous on each of these intervals. Therefore, the only conceivable points where the function might fail to be continuous are on the boundaries of these intervals where we switch from one formula to another: x = -1 and x = 1.
Q The function f is is not continuous at x = 0. Q The function f is is not continuous at x = -1. Q The function f is is not continuous at x = 1.
Now try the rest of the exercises in Section 3.3 in Applied Calculus or Section 10.3 in Finite Mathematics and Applied Calculus
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Last Updated: September, 2007 |
# Math Properties
These are a few of the key Math Properties that are the foundation of middle school math.
### Math Properties
Commutative Property of Addition - When numbers are added the order of the numbers does not change the sum
Ex. 2 + 3 = 3 + 2
Commutative Property of Multiplication - When numbers are being multiplied, the order of the numbers does not change the product.
Ex. 5 x 4 = 4 x 5
Identity Property of Addition - What can you add to "5" so that the sum is "5"? Zero of course.
Ex. 5 + 0 = 5
Identity Property of Multiplication- What can you multiply to "10" so that the product is "10"? One!
Ex. 10 x 1 = 10
Associative Property of Addition - If you have a series of numbers you are adding, you can group them any way you like and the sum will be the same.
Ex. 2 + (4 + 5) = (2 + 4) + 5
Associative Property of Multiplication - If you have a series of numbers you are multiplying, you can group them any way you like and the product will be the same.
Ex. 2 (3 x 7) = (2 x 3) x 7
Inverse Property of Addition - The inverse is the opposite of a number. When you add the inverse of a number you will always get Zero.
Example: 4 + -4 = 0
Inverse Property of Multiplication - When you multiply a number by its inverse you will always get a one.
Example: 5 x 1/5 = 1
Distributive Property - The sum of two numbers times a third number is equal to the sum of each addend times that third number
Example. 5 (4 + 6) = (5 x 4) + (5 x 6)
Multiplication Property of zero - The product of any number and 0 (zero) will always be 0 (zero) |
## College Algebra (11th Edition)
The parent function is $f(x)=2^x$ (with red) the given function is $g(x)=2^{x-1}+2$ (with blue). The parent function can be graphed by calculating a few coordinates and connecting them with a smooth curve: $f(-2)=2^{-2}=(\frac{1}{4})$ $f(-1)=2^{-1}=(\frac{1}{2}$ $f(0)=2^0=1$ $f(1)=2^1=2$ $f(2)=2^2=4$ For every corresponding x-value the following equation is true: $f(x-1)+2=g(x)$ This means that the graph is translated 1 unit right and 2 units up ($g(x)$ involves a horizontal shift of 1 to the right and also a vertical shift of 2 upwards.). First, the horizontal shift. We only consider the g(x) function as $g'(x)=2^{x-1}$ For example if $f(0)=1$ in the original $f(x)$, this will be equal to $g'(1)=f(1-1)=f(0)=1$. Here,$f(0)=g'(1)$ also, $f(1)=g'(2)$ We can see that here, each point in the parent function was moved to the right by 1 unit. Second, we translate this $g'(x)$ function to get the originally given function. Now, the following equation is true: $g'(x)+2=g(x)$ Every $g'(x)$ value will be increased by 2. For example if $g'(1)=1$, this will be translated as $g'(1)+2=1+2=3$. We can see that here, the $g(x)$ is greater than $g'(x)$ for every corresponding x-value by 2.) |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Inverses of Trigonometric Functions
## Domain and range of inverse functions.
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Inverses of Trigonometric Functions
Your instructor gives you a trigonometric function, \begin{align*}f(x) = 3\sin (x) + 5\end{align*}, and asks you to find the inverse. You are all set to start manipulating the equation, when you realize that you don't know just how to do this. Your instructor suggests that you try finding the inverse through graphing instead.
Are you able to do this?
### Finding the Inverse of a Trigonometric Function
In other lessons, two different ways to find the inverse of a function were discussed: graphing and algebra. However, when finding the inverse of trig functions, it is easy to find the inverse of a trig function through graphing.
Consider the graph of a sine function shown here:
In order to consider the inverse of this function, we need to restrict the domain so that we have a section of the graph that is one-to-one. If the domain of \begin{align*}f\end{align*} is restricted to \begin{align*}-\frac{\pi}{2} \le x \le \frac{\pi}{2}\end{align*} a new function \begin{align*}f(x) = \sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}\end{align*}. is defined. This new function is one-to-one and takes on all the values that the function \begin{align*}f(x) = \sin x\end{align*} takes on. Since the restricted domain is smaller, \begin{align*}f(x) = \sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}\end{align*} takes on all values once and only once.
The inverse of \begin{align*}f(x)\end{align*} is represented by the symbol \begin{align*}f^{-1}(x)\end{align*}, and \begin{align*}y = f^{-1}(x) \Leftrightarrow f(y) = x\end{align*}. The inverse of \begin{align*}\sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}\end{align*} will be written as \begin{align*}\sin^{-1} x\end{align*}. or \begin{align*}\arcsin x\end{align*}.
\begin{align*}\begin{Bmatrix} y = \sin^{-1} x\\ \quad or\\ y = \arcsin x \end{Bmatrix} \Leftrightarrow \sin y = x\end{align*}
In this lesson we will use both \begin{align*}\sin^{-1} x\end{align*} and \begin{align*}\arcsin x\end{align*} and both are read as “the inverse sine of \begin{align*}x\end{align*}” or “the number between \begin{align*}-\frac{\pi}{2}\end{align*} and \begin{align*}\frac{\pi}{2}\end{align*} whose sine is \begin{align*}x\end{align*}.”
The graph of \begin{align*}y = \sin^{-1} x\end{align*} is obtained by applying the inverse reflection principle and reflecting the graph of \begin{align*}y=\sin x, -\frac{\pi}{2} \le x \le \frac{\pi}{2}\end{align*} in the line \begin{align*}y = x\end{align*}. The domain of \begin{align*}y = \sin x\end{align*} becomes the range of \begin{align*}y = \sin^{-1} x\end{align*}, and hence the range of \begin{align*}y = \sin x\end{align*} becomes the domain of \begin{align*}y = \sin^{-1} x\end{align*}.
Another way to view these graphs is to construct them on separate grids. If the domain of \begin{align*}y = \sin x\end{align*} is restricted to the interval \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*}, the result is a restricted one-to one function. The inverse sine function \begin{align*}y = \sin^{-1} x\end{align*} is the inverse of the restricted section of the sine function.
The domain of \begin{align*}y = \sin x\end{align*} is \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*} and the range is [-1, 1].
The restriction of \begin{align*}y = \sin x\end{align*} is a one-to-one function and it has an inverse that is shown below.
The domain of \begin{align*}y = \sin^{-1}\end{align*} is [-1, 1] and the range is \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*}.
The inverse functions for cosine and tangent are defined by following the same process as was applied for the inverse sine function. However, in order to create one-to-one functions, different intervals are used. The cosine function is restricted to the interval \begin{align*}0 \le x \le \pi\end{align*} and the new function becomes \begin{align*}y = \cos x, 0 \le x \le \pi\end{align*}. The inverse reflection principle is then applied to this graph as it is reflected in the line \begin{align*}y = x\end{align*} The result is the graph of \begin{align*}y = \cos^{-1} x\end{align*} (also expressed as \begin{align*}y = \arccos x\end{align*}).
Again, construct these graphs on separate grids to determine the domain and range. If the domain of \begin{align*}y = \cos x\end{align*} is restricted to the interval \begin{align*}[0, \pi]\end{align*}, the result is a restricted one-to one function. The inverse cosine function \begin{align*}y = \cos^{-1} x\end{align*} is the inverse of the restricted section of the cosine function.
The domain of \begin{align*}y = \cos x\end{align*} is \begin{align*}[0, \pi]\end{align*} and the range is [-1, 1].
The restriction of \begin{align*}y = \cos x\end{align*} is a one-to-one function and it has an inverse that is shown below.
The statements \begin{align*}y = \cos x\end{align*} and \begin{align*}x = \cos y\end{align*} are equivalent for \begin{align*}y-\end{align*}values in the restricted domain \begin{align*}[0, \pi]\end{align*} and \begin{align*}x-\end{align*}values between -1 and 1.
The domain of \begin{align*}y = \cos^{-1} x\end{align*} is [-1, 1] and the range is \begin{align*}[0, \pi]\end{align*}.
The tangent function is restricted to the interval \begin{align*}-\frac{\pi}{2} < x < \frac{\pi}{2}\end{align*} and the new function becomes \begin{align*}y = \tan x, -\frac{\pi}{2} < x < \frac{\pi}{2}\end{align*}. The inverse reflection principle is then applied to this graph as it is reflected in the line \begin{align*}y = x\end{align*}. The result is the graph of \begin{align*}y = \tan^{-1} x\end{align*} (also expressed as \begin{align*}y = \arctan x\end{align*}).
Graphing the two functions separately will help us to determine the domain and range. If the domain of \begin{align*}y = \tan x\end{align*} is restricted to the interval \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*}, the result is a restricted one-to one function. The inverse tangent function \begin{align*}y = \tan^{-1} x\end{align*} is the inverse of the restricted section of the tangent function.
The domain of \begin{align*}y = \tan x\end{align*} is \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*} and the range is \begin{align*}[-\infty, \infty]\end{align*}.
The restriction of \begin{align*}y = \tan x\end{align*} is a one-to-one function and it has an inverse that is shown below.
The statements \begin{align*}y = \tan x\end{align*} and \begin{align*}x = \tan y\end{align*} are equivalent for \begin{align*}y-\end{align*}values in the restricted domain \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*} and \begin{align*}x-\end{align*}values between -4 and +4.
The domain of \begin{align*}y = \tan^{-1} x\end{align*} is \begin{align*}[-\infty, \infty]\end{align*} and the range is \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*}.
The above information can be readily used to evaluate inverse trigonometric functions without the use of a calculator. These calculations are done by applying the restricted domain functions to the unit circle. To summarize:
Restricted Domain Function Inverse Trigonometric Function Domain Range Quadrants
\begin{align*}y = \sin x\end{align*} \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*} [-1, 1] 1 AND 4
\begin{align*}y = \arcsin x\end{align*}
\begin{align*}y = \sin^{-1} x\end{align*}
[-1, 1] \begin{align*}\left [ -\frac{\pi}{2}, \frac{\pi}{2} \right ]\end{align*}
\begin{align*}y = \cos x\end{align*} \begin{align*}[0, \pi]\end{align*} [-1, 1] 1 AND 2
\begin{align*}y = \arccos x\end{align*}
\begin{align*}y = \cos^{-1} x\end{align*}
[-1, 1] \begin{align*}[0, \pi]\end{align*}
\begin{align*}y = \tan x\end{align*} \begin{align*}\left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )\end{align*} \begin{align*}(-\infty, \infty)\end{align*} 1 AND 4
\begin{align*}y = \arctan x\end{align*}
\begin{align*}y = \tan^{-1}x\end{align*}
\begin{align*}(-\infty, \infty)\end{align*} \begin{align*}\left ( -\frac{\pi}{2}, \frac{\pi}{2} \right )\end{align*}
Now that the three trigonometric functions and their inverses have been summarized, let’s take a look at the graphs of these inverse trigonometric functions.
Let's take a look at a few example problems.
1. Establish an alternative domain that makes \begin{align*}y=sin(x)\end{align*} a one to one function.
Any number of possible solutions can be given, but the important point is that the function must pass the "horizontal line test" and the "vertical line test". This means that a horizontal line drawn through the graph will intersect the function in only one place, and a vertical line drawn through the graph will intersect the function in only one place.
For the sine curve, this means that the function can't "turn over" or "go in the other direction", since then it couldn't pass the horizontal line test. So any part of the function that starts at the bottom of the "y" values and stops at the top of the "y" values will work. (Any value that starts at the top of the "y" values and stops at the bottom of the "y" values will work as well.
In this problem, you can see that the function starts at \begin{align*}\frac{\pi}{2}\end{align*} and stops at \begin{align*}\frac{3\pi}{2}\end{align*}.
2. Find the range of the function given in the previous problem.
You can see that the function still has the same "y" range of values, since the function \begin{align*}y = \sin x\end{align*} moves up and down between -1 and 1. Therefore, the range is \begin{align*}-1 \le y \le 1\end{align*}.
3. Find the domain and range of the inverse of the function given in the first problem.
Since the domain of the inverse function is the range of the original function and the range of the inverse function is the domain of the original function, you only have to take the "x" and "y" values of the original function and reverse them to get the domain and range of the inverse function.
Therefore, the domain of \begin{align*}y = \sin^{-1} x\end{align*} as described in #1 is \begin{align*}-1 \le x \le 1\end{align*} and the range is \begin{align*}\frac{\pi}{2} \le y \le \frac{3\pi}{2}\end{align*}.
### Examples
#### Example 1
Earlier, you were asked to find the inverse of a function.
To find the inverse of this function through graphing, first restrict the domain of the function so that it is one to one. A graph of \begin{align*}f(x) = 3\sin (x) + 5\end{align*}, restricted so that the domain is \begin{align*}-\frac{\pi}{2}\end{align*} to \begin{align*}\frac{\pi}{2}\end{align*} looks like this:
If you apply the inverse reflection principle, you can see that the inverse of this function looks like this:
#### Example 2
Sketch a graph of \begin{align*}y = \frac{1}{2} \cos^{-1} (3x+1)\end{align*}. Sketch \begin{align*}y = \cos^{-1} x\end{align*} on the same set of axes and compare how the two differ.
\begin{align*}y = \frac{1}{2} \cos^{-1} (3x+1)\end{align*} is in blue and \begin{align*}y =\cos^{-1}(x)\end{align*} is in red. Notice that \begin{align*}y = \frac{1}{2} \cos^{-1}(3x+1)\end{align*} has half the amplitude and is shifted over -1. The 3 seems to narrow the graph.
#### Example 3
Sketch a graph of \begin{align*}y = 3-\tan^{-1} (x-2)\end{align*}. Sketch \begin{align*}y = \tan^{-1} x\end{align*} on the same set of axes and compare how the two differ.
\begin{align*}y = 3-\tan^{-1} (x-2)\end{align*} is in blue and \begin{align*}y = \tan^{-1} x\end{align*} is in red. \begin{align*}y = 3-\tan^{-1} (x-2)\end{align*} is shifted up 3 and to the right 2 (as indicated by point \begin{align*}C\end{align*}, the “center”) and is flipped because of the \begin{align*}-\tan^{-1}\end{align*}.
#### Example 4
Graph \begin{align*}y = 2\sin^{-1}(2x)\end{align*}
### Review
1. Why does the domain of a trigonometric function have to be restricted in order to find its inverse function?
2. If the domain of \begin{align*}f(x)=\cos(x)\end{align*} is \begin{align*}[0,\pi]\end{align*}, what is the domain and range of \begin{align*}f^{-1}(x)\end{align*}?.
3. If the domain of \begin{align*}g(x)=\sin(x)\end{align*} is \begin{align*}[-\frac{\pi}{2},\frac{\pi}{2}]\end{align*}, what is the domain and range of \begin{align*}g^{-1}(x)\end{align*}?.
4. Establish an alternative domain that makes \begin{align*}y=\cos(x)\end{align*} a function.
5. What is the domain and range of the inverse of the function from the previous problem.
6. Establish an alternative domain that makes \begin{align*}y=\tan(x)\end{align*} a function.
7. What is the domain and range of the inverse of the function from the previous problem.
Sketch a graph of each function. Use the domains presented in this concept.
1. \begin{align*}y=2\sin^{-1}(3x-1)\end{align*}
2. \begin{align*}y=-3+\cos^{-1}(2x)\end{align*}
3. \begin{align*}y=1+2\tan^{-1}(x+2)\end{align*}
4. \begin{align*}y=4\sin^{-1}(x-4)\end{align*}
5. \begin{align*}y=2+\cos^{-1}(x+3)\end{align*}
6. \begin{align*}y=1+\cos^{-1}(2x-3)\end{align*}
7. \begin{align*}y=-3+\tan^{-1}(3x+1)\end{align*}
8. \begin{align*}y=-1+2\sin^{-1}(x+5)\end{align*}
To see the Review answers, open this PDF file and look for section 4.5.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
inverse function Inverse functions are functions that 'undo' each other. Formally $f(x)$ and $g(x)$ are inverse functions if $f(g(x)) = g(f(x)) = x$. |
College Algebra with Corequisite Support 2e
6.1Exponential Functions
Learning Objectives
In this section, you will:
• Evaluate exponential functions.
• Find the equation of an exponential function.
• Use compound interest formulas.
• Evaluate exponential functions with base $ee$ .
Corequisite Skills
Learning Objectives
• Find the value of a function (exponential). (IA 3.5.3)
• Graph exponential functions. (IA 10.2.1)
Vocabulary.
For the function $y=f(x)y=f(x)$ , ________ is the independent variable as it can be any value in the domain and ________ is the dependent variable as its value depends on ________ .
Many natural events and real-life applications can be modeled using exponential functions. For example, the growth of populations, the spread of viruses, radioactive decay and compounding interest all follow exponential patterns.
Definition An exponential function is a function of the form
$f(x)=axf(x)=ax$ where $a>0a>0$ and $a≠1a≠1$
Examples: $f(x)=5x,f(x)=(13)x,f(x)=2.13xf(x)=5x,f(x)=(13)x,f(x)=2.13x$
Notice that in the exponential function, the variable is the exponent. In our functions so far, the variables were the base.
Evaluating a function is the process of finding the value of f(x) for a given value of x.
Example 1
Evaluate the function $f(x)=3xf(x)=3x$ for the given values
1. $f(2)f(2)$
2. $f(-1)f(-1)$
3. $f(2h)f(2h)$
Practice Makes Perfect
Find the value of an exponential function.
1.
Evaluate the function $f(x)=(32)xf(x)=(32)x$ for the given values.
1. $f(2)f(2)$
2. $f(-2)f(-2)$
3. $f(a)f(a)$
2.
We also find the value of the function when we solve application problems involving exponential functions.
Medicare Premiums. The monthly Medicare Part B health-care premium for most beneficiaries ages 65 and older has increased significantly since 1975. The monthly premium has increased from about $7 in 1975 to$110.50 in 2011 (Source: Centers for Medicare and Medicaid Services). The following exponential function models the premium increases:
$M(x)=7(1.080)xM(x)=7(1.080)x$ where x is the number of years since 1975.
Estimate the monthly Medicare Part B premium in 1985, in 1992, and in 2002. (Note that x is the number of years since 1975, so for 1985, x=10.) Round to the nearest dollar.
3.
We can find Compound Interest using $A=P(1+rn)ntA=P(1+rn)nt$ ,
Where A is the amount of money, P is the principal, t is the number of years, r is the interest rate, and n is the number of times the interest was compounded per year.
Suppose that $960 is invested at 7% interest, compounded semiannually. 1. Find the function for the amount to which the investment grows after t years. 2. Find the amount of money in the account at t=1, 6, 10, 15, and 20 years. Objective 2: Graph exponential functions. (IA 10.2.1) Practice Makes Perfect Graph exponential functions. 4. Graph the exponential function $f(x)=2xf(x)=2x$ by making a table. $x x$ $y=f(x)y=f(x)$ 5. Graph the exponential function $f(x)=(12)xf(x)=(12)x$ by making a table. $x x$ $y=f(x)y=f(x)$ How does it compare with the graph of $f(x)=2xf(x)=2x$ ? 6. Graph $f(x)=3xf(x)=3x$ , $f(x)=4xf(x)=4x$ , $f(x)=2.5xf(x)=2.5x$ in the same viewing window using a graphing calculator or program. What is the relationship between the base a and the shape of the graph? 7. Graph $f(x)=0.2xf(x)=0.2x$ , $f(x)=0.4xf(x)=0.4x$ , $f(x)=0.7xf(x)=0.7x$ in the same viewing window using a graphing calculator or program. What is the relationship between the base a and the shape of the graph? 8. Fill in the Properties of Exponential Function. Is it continuous? Is it one-to-one? Domain Range Increasing if Decreasing if Asymptotes Intercepts The number e, e ≈ 2.718281827, is like the number π in that we use a symbol to represent it because its decimal representation never stops or repeats. The irrational number e is called the natural base or Euler's number after the Swiss mathematician Leonhard Euler. The exponential function whose base is e, $f(x)=exf(x)=ex$ is called the natural exponential function. Practice Makes Perfect 9. Graph the exponential function $f(x)=exf(x)=ex$ by making a table. $x x$ $y=f(x)y=f(x)$ What is the domain of $f(x)f(x)$ ? What is the range of $f(x)f(x)$ ? India is the second most populous country in the world with a population of about $1.39 1.39$ billion people in 2021. The population is growing at a rate of about $1.2% 1.2%$ each year2. If this rate continues, the population of India will exceed China’s population by the year $2027. 2027.$ When populations grow rapidly, we often say that the growth is “exponential,” meaning that something is growing very rapidly. To a mathematician, however, the term exponential growth has a very specific meaning. In this section, we will take a look at exponential functions, which model this kind of rapid growth. Identifying Exponential Functions When exploring linear growth, we observed a constant rate of change—a constant number by which the output increased for each unit increase in input. For example, in the equation $f(x)=3x+4, f(x)=3x+4,$ the slope tells us the output increases by 3 each time the input increases by 1. The scenario in the India population example is different because we have a percent change per unit time (rather than a constant change) in the number of people. Defining an Exponential Function A study found that the percent of the population who are vegans in the United States doubled from 2009 to 2011. In 2011, 2.5% of the population was vegan, adhering to a diet that does not include any animal products—no meat, poultry, fish, dairy, or eggs. If this rate continues, vegans will make up 10% of the U.S. population in 2015, 40% in 2019, and 80% in 2021. What exactly does it mean to grow exponentially? What does the word double have in common with percent increase? People toss these words around errantly. Are these words used correctly? The words certainly appear frequently in the media. • Percent change refers to a change based on a percent of the original amount. • Exponential growth refers to an increase based on a constant multiplicative rate of change over equal increments of time, that is, a percent increase of the original amount over time. • Exponential decay refers to a decrease based on a constant multiplicative rate of change over equal increments of time, that is, a percent decrease of the original amount over time. For us to gain a clear understanding of exponential growth, let us contrast exponential growth with linear growth. We will construct two functions. The first function is exponential. We will start with an input of 0, and increase each input by 1. We will double the corresponding consecutive outputs. The second function is linear. We will start with an input of 0, and increase each input by 1. We will add 2 to the corresponding consecutive outputs. See Table 1. $x x$ $f(x)= 2 x f(x)= 2 x$ $g(x)=2x g(x)=2x$ 0 1 0 1 2 2 2 4 4 3 8 6 4 16 8 5 32 10 6 64 12 Table 1 From Table 1 we can infer that for these two functions, exponential growth dwarfs linear growth. • Exponential growth refers to the original value from the range increases by the same percentage over equal increments found in the domain. • Linear growth refers to the original value from the range increases by the same amount over equal increments found in the domain. Apparently, the difference between “the same percentage” and “the same amount” is quite significant. For exponential growth, over equal increments, the constant multiplicative rate of change resulted in doubling the output whenever the input increased by one. For linear growth, the constant additive rate of change over equal increments resulted in adding 2 to the output whenever the input was increased by one. The general form of the exponential function is $f(x)=a b x , f(x)=a b x ,$ where $a a$ is any nonzero number, $b b$ is a positive real number not equal to 1. • If $b>1, b>1,$ the function grows at a rate proportional to its size. • If $0 the function decays at a rate proportional to its size. Let’s look at the function $f(x)= 2 x f(x)= 2 x$ from our example. We will create a table (Table 2) to determine the corresponding outputs over an interval in the domain from $−3 −3$ to $3. 3.$ $x x$ $−3 −3$ $−2 −2$ $−1 −1$ $0 0$ $1 1$ $2 2$ $3 3$ $f(x)= 2 x f(x)= 2 x$ $2 −3 = 1 8 2 −3 = 1 8$ $2 −2 = 1 4 2 −2 = 1 4$ $2 −1 = 1 2 2 −1 = 1 2$ $2 0 =1 2 0 =1$ $2 1 =2 2 1 =2$ $2 2 =4 2 2 =4$ $2 3 =8 2 3 =8$ Table 2 Let us examine the graph of $f f$ by plotting the ordered pairs we observe on the table in Figure 1, and then make a few observations. Figure 1 Let’s define the behavior of the graph of the exponential function $f(x)= 2 x f(x)= 2 x$ and highlight some its key characteristics. • the domain is $( −∞,∞ ), ( −∞,∞ ),$ • the range is $( 0,∞ ), ( 0,∞ ),$ • as $x→∞,f(x)→∞, x→∞,f(x)→∞,$ • as $x→−∞,f(x)→0, x→−∞,f(x)→0,$ • $f(x) f(x)$ is always increasing, • the graph of $f(x) f(x)$ will never touch the x-axis because base two raised to any exponent never has the result of zero. • $y=0 y=0$ is the horizontal asymptote. • the y-intercept is 1. Exponential Function For any real number $x, x,$ an exponential function is a function with the form $f(x)=a b x f(x)=a b x$ where • $a a$ is a non-zero real number called the initial value and • $b b$ is any positive real number such that $b≠1. b≠1.$ • The domain of $f f$ is all real numbers. • The range of $f f$ is all positive real numbers if $a>0. a>0.$ • The range of $f f$ is all negative real numbers if $a<0. a<0.$ • The y-intercept is $( 0,a ), ( 0,a ),$ and the horizontal asymptote is $y=0. y=0.$ Example 1 Identifying Exponential Functions Which of the following equations are not exponential functions? • $f(x)= 4 3( x−2 ) f(x)= 4 3( x−2 )$ • $g(x)= x 3 g(x)= x 3$ • $h(x)= ( 1 3 ) x h(x)= ( 1 3 ) x$ • $j(x)= ( −2 ) x j(x)= ( −2 ) x$ Try It #1 Which of the following equations represent exponential functions? • $f(x)=2 x 2 −3x+1 f(x)=2 x 2 −3x+1$ • $g(x)= 0.875 x g(x)= 0.875 x$ • $h(x)=1.75x+2 h(x)=1.75x+2$ • $j(x)= 1095.6 −2x j(x)= 1095.6 −2x$ Evaluating Exponential Functions Recall that the base of an exponential function must be a positive real number other than $1. 1.$ Why do we limit the base $b b$ to positive values? To ensure that the outputs will be real numbers. Observe what happens if the base is not positive: • Let $b=−9 b=−9$ and $x= 1 2 . x= 1 2 .$ Then $f(x)=f( 1 2 )= ( −9 ) 1 2 = −9 , f(x)=f( 1 2 )= ( −9 ) 1 2 = −9 ,$ which is not a real number. Why do we limit the base to positive values other than $1? 1?$ Because base $1 1$ results in the constant function. Observe what happens if the base is $1: 1:$ • Let $b=1. b=1.$ Then $f(x)= 1 x =1 f(x)= 1 x =1$ for any value of $x. x.$ To evaluate an exponential function with the form $f(x)= b x , f(x)= b x ,$ we simply substitute $x x$ with the given value, and calculate the resulting power. For example: Let $f(x)= 2 x . f(x)= 2 x .$ What is $f(3)? f(3)?$ To evaluate an exponential function with a form other than the basic form, it is important to follow the order of operations. For example: Let $f(x)=30 ( 2 ) x . f(x)=30 ( 2 ) x .$ What is $f(3)? f(3)?$ Note that if the order of operations were not followed, the result would be incorrect: $f(3)=30 ( 2 ) 3 ≠ 60 3 =216,000 f(3)=30 ( 2 ) 3 ≠ 60 3 =216,000$ Example 2 Evaluating Exponential Functions Let $f( x )=5 ( 3 ) x+1 . f( x )=5 ( 3 ) x+1 .$ Evaluate $f( 2 ) f( 2 )$ without using a calculator. Try It #2 Let $f( x )=8 ( 1.2 ) x−5 . f( x )=8 ( 1.2 ) x−5 .$ Evaluate $f( 3 ) f( 3 )$ using a calculator. Round to four decimal places. Defining Exponential Growth Because the output of exponential functions increases very rapidly, the term “exponential growth” is often used in everyday language to describe anything that grows or increases rapidly. However, exponential growth can be defined more precisely in a mathematical sense. If the growth rate is proportional to the amount present, the function models exponential growth. Exponential Growth A function that models exponential growth grows by a rate proportional to the amount present. For any real number $x x$ and any positive real numbers and $b b$ such that $b≠1, b≠1,$ an exponential growth function has the form $f(x)=a b x f(x)=a b x$ where • $a a$ is the initial or starting value of the function. • $b b$ is the growth factor or growth multiplier per unit $x x$ . In more general terms, we have an exponential function, in which a constant base is raised to a variable exponent. To differentiate between linear and exponential functions, let’s consider two companies, A and B. Company A has 100 stores and expands by opening 50 new stores a year, so its growth can be represented by the function $A( x )=100+50x. A( x )=100+50x.$ Company B has 100 stores and expands by increasing the number of stores by 50% each year, so its growth can be represented by the function $B(x)=100 ( 1+0.5 ) x . B(x)=100 ( 1+0.5 ) x .$ A few years of growth for these companies are illustrated in Table 3. Year, $x x$ Stores, Company A Stores, Company B $0 0$ $100+50( 0 )=100 100+50( 0 )=100$ $100 ( 1+0.5 ) 0 =100 100 ( 1+0.5 ) 0 =100$ $1 1$ $100+50( 1 )=150 100+50( 1 )=150$ $100 ( 1+0.5 ) 1 =150 100 ( 1+0.5 ) 1 =150$ $2 2$ $100+50( 2 )=200 100+50( 2 )=200$ $100 ( 1+0.5 ) 2 =225 100 ( 1+0.5 ) 2 =225$ $3 3$ $100+50( 3 )=250 100+50( 3 )=250$ $100 ( 1+0.5 ) 3 =337.5 100 ( 1+0.5 ) 3 =337.5$ $x x$ $A( x )=100+50x A( x )=100+50x$ $B(x)=100 ( 1+0.5 ) x B(x)=100 ( 1+0.5 ) x$ Table 3 The graphs comparing the number of stores for each company over a five-year period are shown in Figure 2. We can see that, with exponential growth, the number of stores increases much more rapidly than with linear growth. Figure 2 The graph shows the numbers of stores Companies A and B opened over a five-year period. Notice that the domain for both functions is $[0,∞), [0,∞),$ and the range for both functions is $[100,∞). [100,∞).$ After year 1, Company B always has more stores than Company A. Now we will turn our attention to the function representing the number of stores for Company B, $B(x)=100 ( 1+0.5 ) x . B(x)=100 ( 1+0.5 ) x .$ In this exponential function, 100 represents the initial number of stores, 0.50 represents the growth rate, and $1+0.5=1.5 1+0.5=1.5$ represents the growth factor. Generalizing further, we can write this function as $B(x)=100 ( 1.5 ) x , B(x)=100 ( 1.5 ) x ,$ where 100 is the initial value, $1.5 1.5$ is called the base, and $x x$ is called the exponent. Example 3 Evaluating a Real-World Exponential Model At the beginning of this section, we learned that the population of India was about $1.25 1.25$ billion in the year 2013, with an annual growth rate of about $1.2%. 1.2%.$ This situation is represented by the growth function $P(t)=1.25 ( 1.012 ) t , P(t)=1.25 ( 1.012 ) t ,$ where $t t$ is the number of years since $2013. 2013.$ To the nearest thousandth, what will the population of India be in $2031? 2031?$ Try It #3 The population of China was about 1.39 billion in the year 2013, with an annual growth rate of about $0.6%. 0.6%.$ This situation is represented by the growth function $P(t)=1.39 ( 1.006 ) t , P(t)=1.39 ( 1.006 ) t ,$ where $t t$ is the number of years since $2013. 2013.$ To the nearest thousandth, what will the population of China be for the year 2031? How does this compare to the population prediction we made for India in Example 3? Finding Equations of Exponential Functions In the previous examples, we were given an exponential function, which we then evaluated for a given input. Sometimes we are given information about an exponential function without knowing the function explicitly. We must use the information to first write the form of the function, then determine the constants $a a$ and $b, b,$ and evaluate the function. How To Given two data points, write an exponential model. 1. If one of the data points has the form $( 0,a ), ( 0,a ),$ then $a a$ is the initial value. Using $a, a,$ substitute the second point into the equation $f(x)=a ( b ) x , f(x)=a ( b ) x ,$ and solve for $b. b.$ 2. If neither of the data points have the form $( 0,a ), ( 0,a ),$ substitute both points into two equations with the form $f(x)=a ( b ) x . f(x)=a ( b ) x .$ Solve the resulting system of two equations in two unknowns to find $a a$ and $b. b.$ 3. Using the $a a$ and $b b$ found in the steps above, write the exponential function in the form $f(x)=a ( b ) x . f(x)=a ( b ) x .$ Example 4 Writing an Exponential Model When the Initial Value Is Known In 2006, 80 deer were introduced into a wildlife refuge. By 2012, the population had grown to 180 deer. The population was growing exponentially. Write an exponential function $N(t) N(t)$ representing the population $( N ) ( N )$ of deer over time $t. t.$ Try It #4 A wolf population is growing exponentially. In 2011, $129 129$ wolves were counted. By $2013, 2013,$ the population had reached 236 wolves. What two points can be used to derive an exponential equation modeling this situation? Write the equation representing the population $N N$ of wolves over time $t. t.$ Example 5 Writing an Exponential Model When the Initial Value is Not Known Find an exponential function that passes through the points $( −2,6 ) ( −2,6 )$ and $( 2,1 ). ( 2,1 ).$ Try It #5 Given the two points $( 1,3 ) ( 1,3 )$ and $( 2,4.5 ), ( 2,4.5 ),$ find the equation of the exponential function that passes through these two points. Q&A Do two points always determine a unique exponential function? Yes, provided the two points are either both above the x-axis or both below the x-axis and have different x-coordinates. But keep in mind that we also need to know that the graph is, in fact, an exponential function. Not every graph that looks exponential really is exponential. We need to know the graph is based on a model that shows the same percent growth with each unit increase in $x, x,$ which in many real world cases involves time. How To Given the graph of an exponential function, write its equation. 1. First, identify two points on the graph. Choose the y-intercept as one of the two points whenever possible. Try to choose points that are as far apart as possible to reduce round-off error. 2. If one of the data points is the y-intercept $( 0,a ) ( 0,a )$ , then $a a$ is the initial value. Using $a, a,$ substitute the second point into the equation $f(x)=a ( b ) x , f(x)=a ( b ) x ,$ and solve for $b. b.$ 3. If neither of the data points have the form $( 0,a ), ( 0,a ),$ substitute both points into two equations with the form $f(x)=a ( b ) x . f(x)=a ( b ) x .$ Solve the resulting system of two equations in two unknowns to find $a a$ and $b. b.$ 4. Write the exponential function, $f(x)=a ( b ) x . f(x)=a ( b ) x .$ Example 6 Writing an Exponential Function Given Its Graph Find an equation for the exponential function graphed in Figure 5. Figure 5 Try It #6 Find an equation for the exponential function graphed in Figure 6. Figure 6 How To Given two points on the curve of an exponential function, use a graphing calculator to find the equation. 1. Press [STAT]. 2. Clear any existing entries in columns L1 or L2. 3. In L1, enter the x-coordinates given. 4. In L2, enter the corresponding y-coordinates. 5. Press [STAT] again. Cursor right to CALC, scroll down to ExpReg (Exponential Regression), and press [ENTER]. 6. The screen displays the values of a and b in the exponential equation $y=a⋅ b x y=a⋅ b x$ . Example 7 Using a Graphing Calculator to Find an Exponential Function Use a graphing calculator to find the exponential equation that includes the points $(2,24.8) (2,24.8)$ and $(5,198.4). (5,198.4).$ Try It #7 Use a graphing calculator to find the exponential equation that includes the points (3, 75.98) and (6, 481.07). Applying the Compound-Interest Formula Savings instruments in which earnings are continually reinvested, such as mutual funds and retirement accounts, use compound interest. The term compounding refers to interest earned not only on the original value, but on the accumulated value of the account. The annual percentage rate (APR) of an account, also called the nominal rate, is the yearly interest rate earned by an investment account. The term nominal is used when the compounding occurs a number of times other than once per year. In fact, when interest is compounded more than once a year, the effective interest rate ends up being greater than the nominal rate! This is a powerful tool for investing. We can calculate the compound interest using the compound interest formula, which is an exponential function of the variables time $t, t,$ principal $P, P,$ APR $r, r,$ and number of compounding periods in a year $n: n:$ $A(t)=P ( 1+ r n ) nt A(t)=P ( 1+ r n ) nt$ For example, observe Table 4, which shows the result of investing$1,000 at 10% for one year. Notice how the value of the account increases as the compounding frequency increases.
Frequency Value after 1 year
Annually $1100 Semiannually$1102.50
Quarterly $1103.81 Monthly$1104.71
Try It #9
Refer to Example 9. To the nearest dollar, how much would Lily need to invest if the account is compounded quarterly?
Evaluating Functions with Base e
As we saw earlier, the amount earned on an account increases as the compounding frequency increases. Table 5 shows that the increase from annual to semi-annual compounding is larger than the increase from monthly to daily compounding. This might lead us to ask whether this pattern will continue.
Examine the value of $1 invested at 100% interest for 1 year, compounded at various frequencies, listed in Table 5. Frequency $A(n)= ( 1+ 1 n ) n A(n)= ( 1+ 1 n ) n$ Value Annually $( 1+ 1 1 ) 1 ( 1+ 1 1 ) 1$$2
Semiannually $( 1+ 1 2 ) 2 ( 1+ 1 2 ) 2$ $2.25 Quarterly $( 1+ 1 4 ) 4 ( 1+ 1 4 ) 4$$2.441406
Monthly $( 1+ 1 12 ) 12 ( 1+ 1 12 ) 12$ $2.613035 Daily $( 1+ 1 365 ) 365 ( 1+ 1 365 ) 365$$2.714567
Hourly $( 1+ 1 8760 ) 8760 ( 1+ 1 8760 ) 8760$ $2.718127 Once per minute $( 1+ 1 525600 ) 525600 ( 1+ 1 525600 ) 525600$$2.718279
Try It #11
A person invests $100,000 at a nominal 12% interest per year compounded continuously. What will be the value of the investment in 30 years? Example 12 Calculating Continuous Decay Radon-222 decays at a continuous rate of 17.3% per day. How much will 100 mg of Radon-222 decay to in 3 days? Try It #12 Using the data in Example 12, how much radon-222 will remain after one year? Media Access these online resources for additional instruction and practice with exponential functions. 6.1 Section Exercises Verbal 1. Explain why the values of an increasing exponential function will eventually overtake the values of an increasing linear function. 2. Given a formula for an exponential function, is it possible to determine whether the function grows or decays exponentially just by looking at the formula? Explain. 3. The Oxford Dictionary defines the word nominal as a value that is “stated or expressed but not necessarily corresponding exactly to the real value.”3 Develop a reasonable argument for why the term nominal rate is used to describe the annual percentage rate of an investment account that compounds interest. Algebraic For the following exercises, identify whether the statement represents an exponential function. Explain. 4. The average annual population increase of a pack of wolves is 25. 5. A population of bacteria decreases by a factor of $1 8 1 8$ every $24 24$ hours. 6. The value of a coin collection has increased by $3.25% 3.25%$ annually over the last $20 20$ years. 7. For each training session, a personal trainer charges his clients $5 5$ less than the previous training session. 8. The height of a projectile at time $t t$ is represented by the function $h(t)=−4.9 t 2 +18t+40. h(t)=−4.9 t 2 +18t+40.$ For the following exercises, consider this scenario: For each year $t, t,$ the population of a forest of trees is represented by the function $A(t)=115 (1.025) t . A(t)=115 (1.025) t .$ In a neighboring forest, the population of the same type of tree is represented by the function $B(t)=82 (1.029) t . B(t)=82 (1.029) t .$ (Round answers to the nearest whole number.) 9. Which forest’s population is growing at a faster rate? 10. Which forest had a greater number of trees initially? By how many? 11. Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after $20 20$ years? By how many? 12. Assuming the population growth models continue to represent the growth of the forests, which forest will have a greater number of trees after $100 100$ years? By how many? 13. Discuss the above results from the previous four exercises. Assuming the population growth models continue to represent the growth of the forests, which forest will have the greater number of trees in the long run? Why? What are some factors that might influence the long-term validity of the exponential growth model? For the following exercises, determine whether the equation represents exponential growth, exponential decay, or neither. Explain. 14. $y=300 ( 1−t ) 5 y=300 ( 1−t ) 5$ 15. $y=220 ( 1.06 ) x y=220 ( 1.06 ) x$ 16. $y=16.5 ( 1.025 ) 1 x y=16.5 ( 1.025 ) 1 x$ 17. $y=11,701 ( 0.97 ) t y=11,701 ( 0.97 ) t$ For the following exercises, find the formula for an exponential function that passes through the two points given. 18. $( 0,6 ) ( 0,6 )$ and $(3,750) (3,750)$ 19. $( 0,2000 ) ( 0,2000 )$ and $(2,20) (2,20)$ 20. $( −1, 3 2 ) ( −1, 3 2 )$ and $( 3,24 ) ( 3,24 )$ 21. $( −2,6 ) ( −2,6 )$ and $( 3,1 ) ( 3,1 )$ 22. $( 3,1 ) ( 3,1 )$ and $(5,4) (5,4)$ For the following exercises, determine whether the table could represent a function that is linear, exponential, or neither. If it appears to be exponential, find a function that passes through the points. 23. $x x$ 1 2 3 4 $f(x) f(x)$ 70 40 10 -20 24. $x x$ 1 2 3 4 $h(x) h(x)$ 70 49 34.3 24.01 25. $x x$ 1 2 3 4 $m(x) m(x)$ 80 61 42.9 25.61 26. $x x$ 1 2 3 4 $f(x) f(x)$ 10 20 40 80 27. $x x$ 1 2 3 4 $g(x) g(x)$ -3.25 2 7.25 12.5 For the following exercises, use the compound interest formula, $A(t)=P ( 1+ r n ) nt . A(t)=P ( 1+ r n ) nt .$ 28. After a certain number of years, the value of an investment account is represented by the equation $A= 10,250 ( 1+ 0.04 12 ) 120 . A= 10,250 ( 1+ 0.04 12 ) 120 .$ What is the value of the account? 29. What was the initial deposit made to the account in the previous exercise? 30. How many years had the account from the previous exercise been accumulating interest? 31. An account is opened with an initial deposit of$6,500 and earns $3.6% 3.6%$ interest compounded semi-annually. What will the account be worth in $20 20$ years?
32.
How much more would the account in the previous exercise have been worth if the interest were compounding weekly?
33.
Solve the compound interest formula for the principal, $P P$ .
34.
Use the formula found in the previous exercise to calculate the initial deposit of an account that is worth $14,472.74 14,472.74$ after earning $5.5% 5.5%$ interest compounded monthly for $5 5$ years. (Round to the nearest dollar.)
35.
How much more would the account in the previous two exercises be worth if it were earning interest for $5 5$ more years?
36.
Use properties of rational exponents to solve the compound interest formula for the interest rate, $r. r.$
37.
Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded semi-annually, had an initial deposit of $9,000 and was worth$13,373.53 after 10 years.
38.
Use the formula found in the previous exercise to calculate the interest rate for an account that was compounded monthly, had an initial deposit of $5,500, and was worth$38,455 after 30 years.
For the following exercises, determine whether the equation represents continuous growth, continuous decay, or neither. Explain.
39.
$y=3742 ( e ) 0.75t y=3742 ( e ) 0.75t$
40.
$y=150 ( e ) 3.25 t y=150 ( e ) 3.25 t$
41.
$y=2.25 ( e ) −2t y=2.25 ( e ) −2t$
42.
Suppose an investment account is opened with an initial deposit of $12,000 12,000$ earning $7.2% 7.2%$ interest compounded continuously. How much will the account be worth after $30 30$ years?
43.
How much less would the account from Exercise 42 be worth after $30 30$ years if it were compounded monthly instead?
Numeric
For the following exercises, evaluate each function. Round answers to four decimal places, if necessary.
44.
$f(x)=2 ( 5 ) x , f(x)=2 ( 5 ) x ,$ for $f( −3 ) f( −3 )$
45.
$f(x)=− 4 2x+3 , f(x)=− 4 2x+3 ,$ for $f( −1 ) f( −1 )$
46.
$f(x)= e x , f(x)= e x ,$ for $f( 3 ) f( 3 )$
47.
$f(x)=−2 e x−1 , f(x)=−2 e x−1 ,$ for $f( −1 ) f( −1 )$
48.
$f(x)=2.7 ( 4 ) −x+1 +1.5, f(x)=2.7 ( 4 ) −x+1 +1.5,$ for $f( −2 ) f( −2 )$
49.
$f(x)=1.2 e 2x −0.3, f(x)=1.2 e 2x −0.3,$ for $f( 3 ) f( 3 )$
50.
$f(x)=− 3 2 ( 3 ) −x + 3 2 , f(x)=− 3 2 ( 3 ) −x + 3 2 ,$ for $f( 2 ) f( 2 )$
Technology
For the following exercises, use a graphing calculator to find the equation of an exponential function given the points on the curve.
51.
$(0,3) (0,3)$ and $(3,375) (3,375)$
52.
$(3,222.62) (3,222.62)$ and $(10,77.456) (10,77.456)$
53.
$(20,29.495) (20,29.495)$ and $(150,730.89) (150,730.89)$
54.
$(5,2.909) (5,2.909)$ and $(13,0.005) (13,0.005)$
55.
$(11,310.035) (11,310.035)$ and $(25,356365.2) (25,356365.2)$
Extensions
56.
The annual percentage yield (APY) of an investment account is a representation of the actual interest rate earned on a compounding account. It is based on a compounding period of one year. Show that the APY of an account that compounds monthly can be found with the formula $APY= ( 1+ r 12 ) 12 −1. APY= ( 1+ r 12 ) 12 −1.$
57.
Repeat the previous exercise to find the formula for the APY of an account that compounds daily. Use the results from this and the previous exercise to develop a function $I(n) I(n)$ for the APY of any account that compounds $n n$ times per year.
58.
Recall that an exponential function is any equation written in the form $f(x)=a⋅ b x f(x)=a⋅ b x$ such that and are positive numbers and Any positive number can be written as for some value of . Use this fact to rewrite the formula for an exponential function that uses the number as a base.
59.
In an exponential decay function, the base of the exponent is a value between 0 and 1. Thus, for some number $b>1, b>1,$ the exponential decay function can be written as $f(x)=a⋅ ( 1 b ) x . f(x)=a⋅ ( 1 b ) x .$ Use this formula, along with the fact that $b= e n , b= e n ,$ to show that an exponential decay function takes the form $f(x)=a ( e ) −nx f(x)=a ( e ) −nx$ for some positive number $n n$ .
60.
The formula for the amount $A A$ in an investment account with a nominal interest rate $r r$ at any time $t t$ is given by $A(t)=a ( e ) rt , A(t)=a ( e ) rt ,$ where $a a$ is the amount of principal initially deposited into an account that compounds continuously. Prove that the percentage of interest earned to principal at any time $t t$ can be calculated with the formula $I(t)= e rt −1. I(t)= e rt −1.$
Real-World Applications
61.
The fox population in a certain region has an annual growth rate of 9% per year. In the year 2012, there were 23,900 fox counted in the area. What is the fox population predicted to be in the year 2020?
62.
A scientist begins with 100 milligrams of a radioactive substance that decays exponentially. After 35 hours, 50mg of the substance remains. How many milligrams will remain after 54 hours?
63.
In the year 1985, a house was valued at $110,000. By the year 2005, the value had appreciated to$145,000. What was the annual growth rate between 1985 and 2005? Assume that the value continued to grow by the same percentage. What was the value of the house in the year 2010?
64.
A car was valued at $38,000 in the year 2007. By 2013, the value had depreciated to$11,000 If the car’s value continues to drop by the same percentage, what will it be worth by 2017?
65.
Jaylen wants to save $54,000 for a down payment on a home. How much will he need to invest in an account with 8.2% APR, compounding daily, in order to reach his goal in 5 years? 66. Kyoko has$10,000 that she wants to invest. Her bank has several investment accounts to choose from, all compounding daily. Her goal is to have $15,000 by the time she finishes graduate school in 6 years. To the nearest hundredth of a percent, what should her minimum annual interest rate be in order to reach her goal? (Hint: solve the compound interest formula for the interest rate.) 67. Alyssa opened a retirement account with 7.25% APR in the year 2000. Her initial deposit was$13,500. How much will the account be worth in 2025 if interest compounds monthly? How much more would she make if interest compounded continuously?
68.
An investment account with an annual interest rate of 7% was opened with an initial deposit of \$4,000 Compare the values of the account after 9 years when the interest is compounded annually, quarterly, monthly, and continuously.
Footnotes
• 2http://www.worldometers.info/world-population/. Accessed February 24, 2014.
• 3Oxford Dictionary. http://oxforddictionaries.com/us/definition/american_english/nomina. |
# Maharashtra State Board Class 7 Maths Solutions Chapter 15 Statistics Practice Set 54
Hello Students In this section we are going to solve Maharashtra State Board Class 7 Maths Solutions Chapter 15 Statistics Practice Set 54. Our class 7 Mathematics Textbook solutions , they point out, not only convey knowledge, but also social values and a greater understanding for each and every student. Our Class 7 Mathematics textbook solutions are very easy to solve and hey remain an authoritative source as long as they are based on the latest scientific Study. the greatest objectivity, and demonstrate of our class 7 Maths textbook solutions are very practicable and easy to understand.
In general, the message about the importance of our class 7 maths textbooks solutions tries to increase awareness of the value they bring in the total education of children and, therefore, in the future of the planet. An awareness in which we must all be involved.
Question 1 : The daily rainfall for each day of a week in a certain city is given in millimeters. Find the average rainfall during the week. 9,11,8,20,10,16,12
Solution:
Average rainfall during the week \$=frac{text { sum of rainfall for each day of the week }}{text { number of days }}\$ \$=frac{9+11+8+20+10+16+12}{7}\$
\$=frac{86}{7}\$
\$=12.285 approx 12.29\$
\$therefore\$ The average rainfall during the week is 12.29
\$mathrm{mm}\$
Question 2 : During the annual function of a school, a Women’s Self-help Group had set up a snacks stall. Their sales every hour were worth Rs 960 ,
\$mathrm{Rs} 830, mathrm{Rs} 945, mathrm{Rs} 800, mathrm{Rs} 847, mathrm{Rs} 970\$ respectively. What was the average of the hourly sales?
Solution:
Average hourly sales \$=frac{text { sum of sales every hour }}{text { number of hours }}\$ \$=frac{960+830+945+800+847+970}{6}\$
\$=frac{5352}{6}\$
\$=operatorname{Rs} 892\$
\$therefore\$ The average of the hourly sales was Rs 892 .
Question 3 : The annual rainfall in Vidarbha in five years is given below. What is the average rainfall for those 5 years?
\$900 mathrm{~mm}, 650 mathrm{~mm}, 450 mathrm{~mm}, 733 mathrm{~mm}, 400 mathrm{~mm}\$
Solution:
Average rainfall for 5 years \$=frac{text { sum of annual rainfall in five years }}{text { number of years }}\$ \$=frac{900+650+450+733+400}{5}\$
\$=frac{3133}{5}\$
\$=626.6\$
\$therefore\$ The average rainfall in Vidarbha for 5 years
was \$626.6 mathrm{~mm}\$.
Question 4 : A farmer bought some sacks of animal feed. The weights of the sacks are given below in kilograms. What is the average weight of the sacks?
49.8,49.7,49.5,49.3,50,48.9,49.2,48.8
Solution:
Average weight of the sacks \$=frac{text { sum of weight of each sack }}{text { number of sacks }}\$ \$=frac{49.8+49.7+49.5+49.3+50+48.9+49.2+48.8}{8}\$
\$=frac{395.2}{8}\$
\$=frac{3952}{80}\$
\$=49.4\$
\$therefore\$ The average weight of the sacks is \$49.4 mathrm{~kg} .\$
### Maharashtra Board Class 7 Maths Chapter 15 Statistics Practice Set 54 Intext Questions and Activities
Question 1 : Rutuja practised skipping with a rope all seven days of a week. The number of times she jumped the rope in one minute every day is given below. Find the average number of jumps per minute.
\$60,62,61,60,59,63,58 .\$ (Textbook pg. no. 96)
Solution:
Average \$=frac{text { Sum of the number of jumps ons even days }}{text { Total number of days }}\$ \$=frac{[60]+[62]+[61]+[60]+[59]+[63]+[58]}{7}\$
\$=frac{423}{7}\$
\$=60.42\$
\$therefore\$ Average number of jumps per minute \$=60.42\$ |
# How do you convert (6, -6) into polar coordinates?
Jan 31, 2016
The point that has coordinates $\left(6 , - 6\right)$ in rectangular coordinates has the polar coordinates $\left(\sqrt{72} , - \frac{\pi}{4}\right)$ or $\left(8.5 , - 0.79\right)$ or (to give a positive value to $\theta$) as $\left(\sqrt{72} , \frac{7 \pi}{4}\right)$.
#### Explanation:
Polar coordinates are in the form $\left(r , \theta\right)$ where $r$ is the distance from the origin $\left(0 , 0\right)$ to the point and $\theta$ is the angle in radians from the positive x-axis.
To find the radius, use:
$r = \sqrt{{6}^{2} + {\left(- 6\right)}^{2}} = \sqrt{36 + 36} = \sqrt{72} = 8.5$
(some may prefer to leave it in the form $\sqrt{72}$)
To find the value of $\theta$, know that $6$ is the opposite and $- 6$ is the adjacent side of a right-angled triangle, so:
$\tan \theta = \frac{6}{-} 6 = - 1$
Therefore $\theta = {\tan}^{-} 1 \left(- 1\right) = - \frac{\pi}{4}$ $r a d$.
This means the polar coordinates can be expressed as $\left(\sqrt{72} , - \frac{\pi}{4}\right)$ or $\left(8.5 , - 0.79\right)$ or (to give a positive value to $\theta$) as $\left(\sqrt{72} , \frac{7 \pi}{4}\right)$. |
# How do you find the average value of the function for f(x)=sqrt(2x-1), 1<=x<=5?
Oct 31, 2017
The average value is $= \frac{1}{2}$
#### Explanation:
We need
$\int {x}^{n} \mathrm{dx} = {x}^{n + 1} / \left(n + 1\right) + C \left(n \ne - 1\right)$
The average value is
$\overline{y} = \frac{1}{5 - 1} {\int}_{1}^{5} \left(\frac{1}{\sqrt{2 x - 1}}\right) \mathrm{dx}$
Perform first the integral by substitution
Let $u = 2 x - 1$, $\implies$, $\mathrm{du} = 2 \mathrm{dx}$
Therefore,
$\int \left(\frac{1}{\sqrt{2 x - 1}}\right) \mathrm{dx} = \frac{1}{2} \int \frac{\mathrm{du}}{\sqrt{u}} = \frac{1}{2} \cdot 2 \sqrt{u} = \sqrt{u} = \sqrt{2 x - 1}$
So,
$\overline{y} = \frac{1}{4} {\left[\sqrt{2 x - 1}\right]}_{1}^{5} = \frac{1}{4} \left(\left(\sqrt{9}\right) - \left(\sqrt{1}\right)\right) = \frac{1}{4} \cdot 2 = \frac{1}{2}$ |
# 9.3 Double-angle, half-angle, and reduction formulas
Page 1 / 8
In this section, you will:
• Use double-angle formulas to find exact values.
• Use double-angle formulas to verify identities.
• Use reduction formulas to simplify an expression.
• Use half-angle formulas to find exact values.
Bicycle ramps made for competition (see [link] ) must vary in height depending on the skill level of the competitors. For advanced competitors, the angle formed by the ramp and the ground should be $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ such that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =\frac{5}{3}.\text{\hspace{0.17em}}$ The angle is divided in half for novices. What is the steepness of the ramp for novices? In this section, we will investigate three additional categories of identities that we can use to answer questions such as this one.
## Using double-angle formulas to find exact values
In the previous section, we used addition and subtraction formulas for trigonometric functions. Now, we take another look at those same formulas. The double-angle formulas are a special case of the sum formulas, where $\text{\hspace{0.17em}}\alpha =\beta .\text{\hspace{0.17em}}$ Deriving the double-angle formula for sine begins with the sum formula,
$\mathrm{sin}\left(\alpha +\beta \right)=\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta +\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta$
If we let $\text{\hspace{0.17em}}\alpha =\beta =\theta ,$ then we have
$\begin{array}{ccc}\hfill \mathrm{sin}\left(\theta +\theta \right)& =& \mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta +\mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ \hfill \mathrm{sin}\left(2\theta \right)& =& 2\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \end{array}$
Deriving the double-angle for cosine gives us three options. First, starting from the sum formula, $\text{\hspace{0.17em}}\mathrm{cos}\left(\alpha +\beta \right)=\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta -\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta ,$ and letting $\text{\hspace{0.17em}}\alpha =\beta =\theta ,$ we have
$\begin{array}{ccc}\hfill \mathrm{cos}\left(\theta +\theta \right)& =& \mathrm{cos}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta -\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \hfill \\ \hfill \mathrm{cos}\left(2\theta \right)& =& {\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \hfill \end{array}$
Using the Pythagorean properties, we can expand this double-angle formula for cosine and get two more variations. The first variation is:
$\begin{array}{ccc}\hfill \mathrm{cos}\left(2\theta \right)& =& {\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \hfill \\ & =& \left(1-{\mathrm{sin}}^{2}\theta \right)-{\mathrm{sin}}^{2}\theta \hfill \\ & =& 1-2{\mathrm{sin}}^{2}\theta \hfill \end{array}$
The second variation is:
$\begin{array}{ccc}\hfill \mathrm{cos}\left(2\theta \right)& =& {\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \hfill \\ & =& {\mathrm{cos}}^{2}\theta -\left(1-{\mathrm{cos}}^{2}\theta \right)\hfill \\ & =& 2\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta -1\hfill \end{array}$
Similarly, to derive the double-angle formula for tangent, replacing $\text{\hspace{0.17em}}\alpha =\beta =\theta \text{\hspace{0.17em}}$ in the sum formula gives
$\begin{array}{ccc}\hfill \mathrm{tan}\left(\alpha +\beta \right)& =& \frac{\mathrm{tan}\text{\hspace{0.17em}}\alpha +\mathrm{tan}\text{\hspace{0.17em}}\beta }{1-\mathrm{tan}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\beta }\hfill \\ \hfill \mathrm{tan}\left(\theta +\theta \right)& =& \frac{\mathrm{tan}\text{\hspace{0.17em}}\theta +\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }\hfill \\ \hfill \mathrm{tan}\left(2\theta \right)& =& \frac{2\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }\hfill \end{array}$
## Double-angle formulas
The double-angle formulas are summarized as follows:
$\begin{array}{ccc}\hfill \phantom{\rule{.45em}{0ex}}\mathrm{sin}\left(2\theta \right)& =& 2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta \hfill \end{array}$
$\begin{array}{ccc}\hfill \phantom{\rule{1.5em}{0ex}}\mathrm{cos}\left(2\theta \right)& =& {\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta \hfill \\ & =& 1-2\text{\hspace{0.17em}}{\mathrm{sin}}^{2}\theta \hfill \\ & =& 2\text{\hspace{0.17em}}{\mathrm{cos}}^{2}\theta -1\hfill \end{array}$
$\begin{array}{ccc}\hfill \mathrm{tan}\left(2\theta \right)& =& \frac{2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }\hfill \end{array}$
Given the tangent of an angle and the quadrant in which it is located, use the double-angle formulas to find the exact value.
1. Draw a triangle to reflect the given information.
2. Determine the correct double-angle formula.
3. Substitute values into the formula based on the triangle.
4. Simplify.
## Using a double-angle formula to find the exact value involving tangent
Given that $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II, find the following:
1. $\mathrm{sin}\left(2\theta \right)$
2. $\mathrm{cos}\left(2\theta \right)$
3. $\mathrm{tan}\left(2\theta \right)$
If we draw a triangle to reflect the information given, we can find the values needed to solve the problems on the image. We are given $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4},$ such that $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in quadrant II. The tangent of an angle is equal to the opposite side over the adjacent side, and because $\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ is in the second quadrant, the adjacent side is on the x -axis and is negative. Use the Pythagorean Theorem to find the length of the hypotenuse:
$\begin{array}{ccc}\hfill {\left(-4\right)}^{2}+{\left(3\right)}^{2}& =& {c}^{2}\hfill \\ \hfill 16+9& =& {c}^{2}\hfill \\ \hfill 25& =& {c}^{2}\hfill \\ \hfill c& =& 5\hfill \end{array}$
Now we can draw a triangle similar to the one shown in [link] .
1. Let’s begin by writing the double-angle formula for sine.
$\mathrm{sin}\left(2\theta \right)=2\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta$
We see that we to need to find $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta .\text{\hspace{0.17em}}$ Based on [link] , we see that the hypotenuse equals 5, so $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\theta =\frac{3}{5},$ and $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\theta =-\frac{4}{5}.\text{\hspace{0.17em}}$ Substitute these values into the equation, and simplify.
Thus,
$\begin{array}{ccc}\hfill \mathrm{sin}\left(2\theta \right)& =& 2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right)\hfill \\ & =& -\frac{24}{25}\hfill \end{array}$
2. Write the double-angle formula for cosine.
$\mathrm{cos}\left(2\theta \right)={\mathrm{cos}}^{2}\theta -{\mathrm{sin}}^{2}\theta$
Again, substitute the values of the sine and cosine into the equation, and simplify.
$\begin{array}{ccc}\hfill \mathrm{cos}\left(2\theta \right)& =& {\left(-\frac{4}{5}\right)}^{2}-{\left(\frac{3}{5}\right)}^{2}\hfill \\ & =& \frac{16}{25}-\frac{9}{25}\hfill \\ & =& \frac{7}{25}\hfill \end{array}$
3. Write the double-angle formula for tangent.
$\mathrm{tan}\left(2\theta \right)=\frac{2\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta }{1-{\mathrm{tan}}^{2}\theta }$
In this formula, we need the tangent, which we were given as $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta =-\frac{3}{4}.\text{\hspace{0.17em}}$ Substitute this value into the equation, and simplify.
$\begin{array}{ccc}\hfill \mathrm{tan}\left(2\theta \right)& =& \frac{2\left(-\frac{3}{4}\right)}{1-{\left(-\frac{3}{4}\right)}^{2}}\hfill \\ & =& \frac{-\frac{3}{2}}{1-\frac{9}{16}}\hfill \\ & =& -\frac{3}{2}\left(\frac{16}{7}\right)\hfill \\ & =& -\frac{24}{7}\hfill \end{array}$
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda
state and prove Cayley hamilton therom
hello
Propessor
hi
Miranda
the Cayley hamilton Theorem state if A is a square matrix and if f(x) is its characterics polynomial then f(x)=0 in another ways evey square matrix is a root of its chatacteristics polynomial.
Miranda
hi
jai
hi Miranda
jai
thanks
Propessor
welcome
jai
What is algebra
algebra is a branch of the mathematics to calculate expressions follow.
Miranda
Miranda Drice would you mind teaching me mathematics? I think you are really good at math. I'm not good at it. In fact I hate it. 😅😅😅
Jeffrey
lolll who told you I'm good at it
Miranda
something seems to wispher me to my ear that u are good at it. lol
Jeffrey
lolllll if you say so
Miranda
but seriously, Im really bad at math. And I hate it. But you see, I downloaded this app two months ago hoping to master it.
Jeffrey
which grade are you in though
Miranda
oh woww I understand
Miranda
Jeffrey
Jeffrey
Miranda
how come you finished in college and you don't like math though
Miranda
gotta practice, holmie
Steve
if you never use it you won't be able to appreciate it
Steve
I don't know why. But Im trying to like it.
Jeffrey
yes steve. you're right
Jeffrey
so you better
Miranda
what is the solution of the given equation?
which equation
Miranda
I dont know. lol
Jeffrey
Miranda
Jeffrey
answer and questions in exercise 11.2 sums
how do u calculate inequality of irrational number?
Alaba
give me an example
Chris
and I will walk you through it
Chris
cos (-z)= cos z .
cos(- z)=cos z
Mustafa
what is a algebra
(x+x)3=?
6x
Obed
what is the identity of 1-cos²5x equal to?
__john __05
Kishu
Hi
Abdel
hi
Ye
hi
Nokwanda
C'est comment
Abdel
Hi
Amanda
hello
SORIE
Hiiii
Chinni
hello
Ranjay
hi
ANSHU
hiiii
Chinni
h r u friends
Chinni
yes
Hassan
so is their any Genius in mathematics here let chat guys and get to know each other's
SORIE
I speak French
Abdel
okay no problem since we gather here and get to know each other
SORIE
hi im stupid at math and just wanna join here
Yaona
lol nahhh none of us here are stupid it's just that we have Fast, Medium, and slow learner bro but we all going to work things out together
SORIE
it's 12
what is the function of sine with respect of cosine , graphically
tangent bruh
Steve
cosx.cos2x.cos4x.cos8x
sinx sin2x is linearly dependent
what is a reciprocal
The reciprocal of a number is 1 divided by a number. eg the reciprocal of 10 is 1/10 which is 0.1
Shemmy
Reciprocal is a pair of numbers that, when multiplied together, equal to 1. Example; the reciprocal of 3 is ⅓, because 3 multiplied by ⅓ is equal to 1
Jeza
each term in a sequence below is five times the previous term what is the eighth term in the sequence
I don't understand how radicals works pls
How look for the general solution of a trig function |
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