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# Question Video: Finding Unknown Discrete Masses given the Coordinates of Their Centre of Mass Mathematics
Three points (0, 6), (0, 9), and (0, 4) on the π¦-axis are occupied by three solids of masses 9 kg, 6 kg, and π kg, respectively. Determine the value of π given the center of mass of the system is at the point (0, 7).
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### Video Transcript
Three points zero, six; zero, nine; and zero, four on the π¦-axis are occupied by three solids of masses nine kilograms, six kilograms, and π kilograms, respectively. Determine the value of π given the center of mass of the system is at the point zero, seven.
The masses of the components of a system as well as their locations are precisely the pieces of information we need to calculate the center of mass of that system. For example, we calculate the π¦-coordinate of the center of mass by multiplying each mass by the π¦-coordinate of its location, adding all of those products together, and dividing by the total mass of the system. Here though, instead of being given the mass and location of every component and being asked to find the center of mass, we are instead given the center of mass and asked to find the mass of one of the components.
So because we know the π¦-coordinate of the center of mass, all we need to do is rearrange this formula to solve for the unknown mass. Weβre focused on the π¦-coordinate of the center of mass because each mass in our system is located at a point with the same π₯-coordinate, which means that the actual value of π doesnβt affect the π₯-coordinate of the center of mass. Okay, so letβs set up our equation. The π¦-coordinate of the center of mass is given as seven. The numerator of the fraction is the sum of each mass times the π¦-coordinate of its location. So we have nine times six plus six times nine plus π times four. And the denominator is just the sum of all the masses, which is nine plus six plus π.
Now we just need to solve for π. Letβs start by simplifying the numerical terms in the fraction. Nine plus six is 15, and nine times six is equal to six times nine, which is equal to 54. Replacing 54 plus 54 with 108, we have seven equals 108 plus four π divided by 15 plus π. Now we multiply both sides by 15 plus π. On the right-hand side, 15 plus π divided by 15 plus π is just one. And on the left-hand side, 15 plus π times seven is 105 plus seven π. So 105 plus seven π equals 108 plus four π. Subtracting 105 and four π from both sides, on the left-hand side, we have 105 minus 105 is zero and seven π minus four π is three π. On the right-hand side, 108 minus 105 is three and four π minus four π is zero. So three π equals three and π equals one. And this is the answer that weβre looking for.
Itβs worth making a brief note about the difference between the number one and one kilogram. The correct answer is one and not one kilogram because the unit of kilogram is already given to us in the question. So when we determine the value of π, what weβre determining is a number, not a number with units. In fact, whenever we answer questions that involve numbers with units, itβs always important to pay attention to how the question is worded to see if our answer is really supposed to be just a number or if itβs supposed to be a physical quantity that also includes units. |
# Content Statements Addressed and Whether they are Knowledge, Reasoning, Performance Skill, or Product:
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1 Quarter 4 Curriculum Guide Mathematical Practices 1. Make Sense of Problems and Persevere in Solving them 2. Reasoning Abstractly & Quantitatively 3. Construct Viable Arguments and Critique the Reasoning of Others 4. Model with Mathematics 5. Use Appropriate Tools Strategically 6. Attend to Precision 7. Look for and Make use of Structure 8. Look for and Express Regularity in Repeated Reasoning Critical Areas of Focus Being Addressed: o Counting and Cardinality o Recognizing and writing numbers from 1-20 o Counting and comparing numbers o Operations and Algebraic Thinking o Add and subtract within 10 o Number and Operations in Base 10 o Understand that a teen number can be composed of ten ones and some additional ones o Measurement and Data o Be able to categorize and count objects o Geometry o Name, describe, and create 2D and 3D shapes Kindergarten Math Content Statements Addressed and Whether they are Knowledge, Reasoning, Performance Skill, or Product: Underpinning Targets Corresponding with Standards and Whether they are Knowledge, Reasoning, Performance Skill, or Product: I can.., Students Will Be Able To.
3 make 10 and show it in different ways. DOK 3 Given a number from 1 to 10, I can find the number to make 10 and write an equation to show the partner. 5. Fluently add and subtract within 5. I can define addition. I can define subtraction. I can count to add. I can take away to subtract. I can easily add numbers that add up to 5 or less. I can easily subtract numbers when the starting number is 5 or less. NUMBER AND OPERATIONS IN BASE TEN K.NBT Work with numbers to gain foundations for place value. 1. Compose and decompose numbers from 11 to 19 into ten ones and some further ones, e.g., by using objects or drawings, and record each composition or decomposition by a drawing or equation (e.g., 18 = ); understand that these numbers are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. MEASUREMENT AND DATA K.MD Describe and compare measurable attributes 1. Describe measurable attributes of objects, such as length or weight. Describe several measurable attributes of a single object. I can count to 20. I can recognize numbers I can define addition. I can add numbers 1-9 to 10 using objects or drawings (e.g., 10 frame, base ten blocks). DOK 3 I can compose (put together) numbers using a ten and some ones and show my work with a drawing or equation. I can define what an attribute is. I can define weight. I can define height. I can define length.
4 2. Directly compare two objects with a measurable attribute in common, to see which object has more of / less of the attribute, and describe the difference. For example, directly compare the heights of two children and describe one child as taller/shorter. I can describe measurable attributes of objects. I can define weight. I can define height. I can define length. I can define volume. I can tell which object is longer (or shorter or taller) than the other. I can tell which object can hold more (or less) than the other. I can tell which object is heavier (or lighter). I can tell which object is warmer (or colder).
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# Find the A.P. Whose Fourth Term is 9 and the Sum of Its Sixth Term and Thirteenth Term is 40. - CBSE Class 10 - Mathematics
ConceptSum of First n Terms of an AP
#### Question
Find the A.P. whose fourth term is 9 and the sum of its sixth term and thirteenth term is 40.
#### Solution
Let the first term be a and the common difference be d
It is given that a4 = 9 and a6 + a13 = 40.
a4 = 9
rArr a+(4-1)d=9 a_n=a+(n-1)d
rArr a+3d=9
a_6+a_13=9
rArr {a+(6-1)d}+{a+(13-1)d}=40
rArr {a+5d}+{a+12d}=40
rArr 2a+17d=40
From (1):
a = 9 − 3d
Substituting the value of a in (2):
2 (9 − 3d) + 17d = 40
⇒ 18 + 11d = 40
⇒ 11d = 22
⇒ d = 2
∴ a = 9 − 3 × 2 = 3
Thus, the given A.P. is aa + da + 2…, where a = 3 and d = 2.
Thus, the A.P. is 3, 5, 7, 9 …
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [5]
Solution Find the A.P. Whose Fourth Term is 9 and the Sum of Its Sixth Term and Thirteenth Term is 40. Concept: Sum of First n Terms of an AP.
S |
1. ## calculating statistics
Data is collected on the weekly number of calls made by a phone-salesperson over an eight-week period, and the number of sales they actually make in each of those weeks. The training manual specifies staff should be making sales on at least 25% of their calls. The data are:
Week 1 2 3 4 5 6 7 8
Calls 55 43 57 32 18 59 61 32
Sales 20 15 18 12 2 21 18 8
a) Draw a scatterplot of these data and comment on the suitability of linear regression to model the relationship.
b) Calculate the regression coefficients and write down the estimated prediction equation.
c) Calculate the predictions for x = 20 and x = 60 and use these to superimpose the regression line on your plot.
d) Calculate the residuals for the sales for weeks 5 and 8, and interpret these numbers.
2. So we want to calculate the coefficients of the following prediction equation: $\displaystyle y = a + bx$.
So $\displaystyle b = \frac{\sum_{i=0}^{n-1}(x_{i} - \overline{x})(y_{i} - \overline{y})}{\sum_{i=0}^{n-1} (x_i - \overline{x})^2}$
Then $\displaystyle a = \overline{y} - b \overline{x}$. $\displaystyle \overline{x}, \overline{y}$ are the means/averages of the two sets of data.
Doing this we get $\displaystyle a = -3.357$ and $\displaystyle b = 0.394$. So the estimated prediction equation is $\displaystyle y = -3.357 + 0.394x$. Then plug in $\displaystyle y(20), y(60)$ get a value and plot the points to make the line.
The residuals are basically estimates of unobservable error or $\displaystyle x_i - \overline{x}$ (how far away values are from average). So for weeks 5 and 8 do the following: obtain $\displaystyle y(18)$ and $\displaystyle y(32)$ to get the predicted sales. Then subtract the actual values from the predicted values (i.e. $\displaystyle y(18) - 2$ and $\displaystyle y(32) - 8$). |
# PAR TIAL FRAC TION + DECOMPOSITION. Let’s add the two fractions below. We need a common denominator: In this section we are going to learn how to take.
## Presentation on theme: "PAR TIAL FRAC TION + DECOMPOSITION. Let’s add the two fractions below. We need a common denominator: In this section we are going to learn how to take."— Presentation transcript:
PAR TIAL FRAC TION + DECOMPOSITION
Let’s add the two fractions below. We need a common denominator: In this section we are going to learn how to take this answer and “decompose” it meaning break down into the fractions that were added together to get this answer
We start by factoring the denominator. There could have been a fraction for each factor of the denominator but we don’t know the numerators so we’ll call them A and B. Now we’ll clear this equation of fractions by multiplying every term by the common denominator. (x+3)(x-2)
This equation needs to be true for any value of x. We pick an x that will “conveniently” get rid of one of the variables and solve for the other. “The Convenient x Method for Solving” Let x = -3 B(0) = 0
Now we’ll “conveniently” choose x to be 2 to get rid of A and find B. Let x = 2 A(0) = 0 43
Summary of Partial Fraction Decomposition When Denominator Factors Into Linear Factors (Factors of first degree) Factor the denominator Set fraction equal to sum of fractions with each factor as a denominator using A, B,B, etc. for numerators Clear equation of fractions Use “convenient” x method to find A, B, etc. Next we’ll look at repeated factors and quadratic factors
Partial Fraction Decomposition With Repeated Linear Factors When the denominator has a repeated linear factor, you need a fraction with a denominator for each power of the factor. (x-1)(x+2) 2 Let x = 1
To find B we put A and C in and choose x to be any other number. Let x = 0 1313 -2 2 3 Let x = -2
Partial Fraction Decomposition With Quadratic Factors When the denominator has a quadratic factor (that won’t factor), you need a fraction with a linear numerator. (x+1)(x 2 +4) The convenient x method doesn’t work as nicely on these kind so we’ll use the “equating coefficients” method. First multiply everything out.
Look at the kinds of terms on each side and equate coefficients (meaning put the coefficients = to each other) Look at x 2 terms: 0 = A + B No x terms on left Look at x terms: 0 = B + C Look at terms with no x’s: 1 = 4A + C Solve these. Substitution would probably be easiest. A = - B C = - B 1 = 4(-B) + (-B) No x 2 terms on left
Partial Fraction Decomposition With Repeated Quadratic Factors When the denominator has a repeated quadratic factor (that won’t factor), you need a fraction with a linear numerator for each power. (x 2 +4) 2 multiply out equate coefficients of various kinds of terms (next screen)
Look at x 3 terms: 1 = A Look at x 2 terms: 1 = B Look at x terms: 0 = 4A+C Look at terms with no x: 0 = 4B+D 0 = 4(1)+C-4 =C 0 = 4(1)+D-4 = D 1 1 -4 1 1
Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. www.slcc.edu Shawna has kindly given permission for this resource to be downloaded from www.mathxtc.com and for it to be modified to suit the Western Australian Mathematics Curriculum.www.mathxtc.com Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar www.ststephens.wa.edu.au
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# If there are 15 tiles in each box, how many boxes of tile does Sarah need for her bathroom floor? (Rest of details of the question is below.)Sarah is laying new tile on her bathroom floor. The floor is rectangular and measures 4' by 7'. Th tiles are square and measure 8' on each side.
Sarah is laying tiles on her floor which is a rectangle with dimensions: 4' * 7'
Each tile is a square with dimensions 8" * 8"
The area of the floor is 4*12*12*7 = 4032 square inches.
The area of each tile is 8*8 = 64 square inches.
This gives the number of tiles required to cover the floor as 4032/64 = 63.
As each box has 15 tiles she would need 5 boxes so that she does not fall short of tiles, as 4 boxes only contain 60 tiles.
TheĀ required number of boxes of tiles is 5.
Approved by eNotes Editorial
The floor measures 4 by 7 .
We need to find the number of (8x8) tiles to cover the floor.
I am assuming the the tiles measure 8" ( 8 inches)
We will convert the sides of the floor to inches.
We know that 1 foot = 12 in.
==> The width = 4 ft = 4*12 = 48 in.
==> Then we need 48/8 tiles = 6 tiles for the width.
==> The length = 7 ft = 7*12 in = 84 in.
==> Then we need 84/8 tiles = 10.5 tiles.
Then the total number of tiles to cover the area is 6*10.5 = 63 tiles.
Each box contains 15 tiles.
==> 4 boxes contains 4*15 = 60 tiles.
==> 5 boxes contains 5*15 = 75 tiles
Then we need 5 boxes to cover the floor. |
# How do you find the ordered pair that is a solution to the system of equations x + y = 10 and x - y = 8?
Jul 17, 2018
$x = 9 , y = 1$
#### Explanation:
Adding both equations we get
$2 x = 18$
so $x = 9$
and we get for $y$
$9 + y = 10$
$y = 1$
Jul 17, 2018
$\left(x , y\right) \to \left(9 , 1\right)$
#### Explanation:
$x + y = 10 \to \left(1\right)$
$x - y = 8 \to \left(2\right)$
$\text{adding the 2 equations term by term eliminates } y$
x+x)+(y-y)=(10+8)
$2 x = 18 \Rightarrow x = \frac{18}{2} = 9$
$\text{substitute "x=9" into equation } \left(1\right)$
$9 + y = 10 \Rightarrow y = 10 - 9 = 1$
$\text{the point of intersection } = \left(9 , 1\right)$
graph{(y+x-10)(y-x+8)((x-9)^2+(y-1)^2-0.04)=0 [-20, 20, -10, 10]} |
# Functions expressed as power series
### Functions expressed as power series
In this section, we will learn how to convert functions into power series. Most of the functions we will be dealing with will be converted into a geometric series. After converting them into a power series, we will find the interval of convergence. Keep in mind that we do not have to check the endpoints of the inequality because we automatically know they will be divergent. Afterwards, we will look at an irregular function and express it as a power series. Integrals will be involved here. We will also take the derivative of a function and express that as a power series.
#### Lessons
Note *A formula that may be of use when expressing functions into power series:
$\frac{1}{1-r}=\sum_{n=0}^{\infty}r^n$ knowing that $-1$ < $r$ < $1$
When finding the interval of convergence, there is no need to check the endpoints. This is because the sum of the geometric series strictly converges only when $-1$ < $r$ < $1$, and not at $r=1$.
If the function $f(x)$ has a radius of convergence of $R$, then the derivative and the anti-derivative of $f(x)$ also has a radius of convergence of $R$.
• 1.
Expressing Functions as Power Series
Express the following functions as power series, and then find the interval of convergence:
a)
$\frac{1}{1-x^2}$
b)
$\frac{1}{2-x}$
c)
$\frac{x^2}{1+3x^2}$
• 2.
Irregular Function expressed as a Power Series
Express the function $f(x)=ln(7-x)$ as a power series and find the interval of convergence.
• 3.
Derivative of a Function expressed as a Power Series
Express the derivative of this function $f(x)=\frac{1}{1-x}$ as a power series and find the interval of convergence. |
# Search by Topic
#### Resources tagged with Number - generally similar to Marvellous Matrix:
Filter by: Content type:
Stage:
Challenge level:
### There are 40 results
Broad Topics > Numbers and the Number System > Number - generally
### Marvellous Matrix
##### Stage: 2 Challenge Level:
Follow the directions for circling numbers in the matrix. Add all the circled numbers together. Note your answer. Try again with a different starting number. What do you notice?
### Making Maths: Be a Mathemagician
##### Stage: 2 Challenge Level:
Surprise your friends with this magic square trick.
### Lucky Numbers
##### Stage: 1 and 2
This article for pupils explores what makes numbers special or lucky, and looks at the numbers that are all around us every day.
### Eleven
##### Stage: 3 Challenge Level:
Replace each letter with a digit to make this addition correct.
### Card Trick 1
##### Stage: 3 Challenge Level:
Can you explain how this card trick works?
### Think of Two Numbers
##### Stage: 3 Challenge Level:
Think of two whole numbers under 10, and follow the steps. I can work out both your numbers very quickly. How?
### Maze 100
##### Stage: 2 Challenge Level:
Can you go through this maze so that the numbers you pass add to exactly 100?
### The Secret World of Codes and Code Breaking
##### Stage: 2, 3 and 4
When you think of spies and secret agents, you probably wouldn’t think of mathematics. Some of the most famous code breakers in history have been mathematicians.
### Roll over the Dice
##### Stage: 2 Challenge Level:
Watch this video to see how to roll the dice. Now it's your turn! What do you notice about the dice numbers you have recorded?
### Mathematical Symbols
##### Stage: 1, 2 and 3
A brief article written for pupils about mathematical symbols.
##### Stage: 3 Challenge Level:
Replace the letters with numbers to make the addition work out correctly. R E A D + T H I S = P A G E
##### Stage: 2, 3 and 4 Challenge Level:
Three dice are placed in a row. Find a way to turn each one so that the three numbers on top of the dice total the same as the three numbers on the front of the dice. Can you find all the ways to do. . . .
### The Patent Solution
##### Stage: 3 Challenge Level:
A combination mechanism for a safe comprises thirty-two tumblers numbered from one to thirty-two in such a way that the numbers in each wheel total 132... Could you open the safe?
### Two Much
##### Stage: 3 Challenge Level:
Explain why the arithmetic sequence 1, 14, 27, 40, ... contains many terms of the form 222...2 where only the digit 2 appears.
### Pairs
##### Stage: 3 Challenge Level:
Ann thought of 5 numbers and told Bob all the sums that could be made by adding the numbers in pairs. The list of sums is 6, 7, 8, 8, 9, 9, 10,10, 11, 12. Help Bob to find out which numbers Ann was. . . .
### Card Trick 2
##### Stage: 3 Challenge Level:
Can you explain how this card trick works?
### Calendar Capers
##### Stage: 3 Challenge Level:
Choose any three by three square of dates on a calendar page...
### Is it Magic or Is it Maths?
##### Stage: 3 Challenge Level:
Here are three 'tricks' to amaze your friends. But the really clever trick is explaining to them why these 'tricks' are maths not magic. Like all good magicians, you should practice by trying. . . .
### Inequalities
##### Stage: 3 Challenge Level:
A bag contains 12 marbles. There are more red than green but green and blue together exceed the reds. The total of yellow and green marbles is more than the total of red and blue. How many of. . . .
### An Investigation Based on Score
##### Stage: 3
Class 2YP from Madras College was inspired by the problem in NRICH to work out in how many ways the number 1999 could be expressed as the sum of 3 odd numbers, and this is their solution.
### Happy Numbers
##### Stage: 3 Challenge Level:
Take any whole number between 1 and 999, add the squares of the digits to get a new number. Make some conjectures about what happens in general.
### Ways of Summing Odd Numbers
##### Stage: 3
Sanjay Joshi, age 17, The Perse Boys School, Cambridge followed up the Madrass College class 2YP article with more thoughts on the problem of the number of ways of expressing an integer as the sum. . . .
### Differs
##### Stage: 3 Challenge Level:
Choose any 4 whole numbers and take the difference between consecutive numbers, ending with the difference between the first and the last numbers. What happens when you repeat this process over and. . . .
##### Stage: 3 Challenge Level:
Visitors to Earth from the distant planet of Zub-Zorna were amazed when they found out that when the digits in this multiplication were reversed, the answer was the same! Find a way to explain. . . .
### Mathematical Swimmer
##### Stage: 3 Challenge Level:
Twice a week I go swimming and swim the same number of lengths of the pool each time. As I swim, I count the lengths I've done so far, and make it into a fraction of the whole number of lengths I. . . .
### St Ives
##### Stage: 2
As I was going to St Ives, I met a man with seven wives. Every wife had seven sacks, every sack had seven cats, every cat had seven kittens. Kittens, cats, sacks and wives, how many were going to St. . . .
### Cross-country Race
##### Stage: 3 Challenge Level:
Eight children enter the autumn cross-country race at school. How many possible ways could they come in at first, second and third places?
### Converging Means
##### Stage: 3 Challenge Level:
Take any two positive numbers. Calculate the arithmetic and geometric means. Repeat the calculations to generate a sequence of arithmetic means and geometric means. Make a note of what happens to the. . . .
### Chocolate Maths
##### Stage: 3 Challenge Level:
Pick the number of times a week that you eat chocolate. This number must be more than one but less than ten. Multiply this number by 2. Add 5 (for Sunday). Multiply by 50... Can you explain why it. . . .
### Largest Product
##### Stage: 3 Challenge Level:
Which set of numbers that add to 10 have the largest product?
### Water Lilies
##### Stage: 3 Challenge Level:
There are some water lilies in a lake. The area that they cover doubles in size every day. After 17 days the whole lake is covered. How long did it take them to cover half the lake?
### Calendars
##### Stage: 1, 2 and 3
Calendars were one of the earliest calculating devices developed by civilizations. Find out about the Mayan calendar in this article.
### Funny Factorisation
##### Stage: 3 Challenge Level:
Some 4 digit numbers can be written as the product of a 3 digit number and a 2 digit number using the digits 1 to 9 each once and only once. The number 4396 can be written as just such a product. Can. . . .
### Quick Times
##### Stage: 3 Challenge Level:
32 x 38 = 30 x 40 + 2 x 8; 34 x 36 = 30 x 40 + 4 x 6; 56 x 54 = 50 x 60 + 6 x 4; 73 x 77 = 70 x 80 + 3 x 7 Verify and generalise if possible.
### Back to the Planet of Vuvv
##### Stage: 3 Challenge Level:
There are two forms of counting on Vuvv - Zios count in base 3 and Zepts count in base 7. One day four of these creatures, two Zios and two Zepts, sat on the summit of a hill to count the legs of. . . .
### A Story about Absolutely Nothing
##### Stage: 2, 3, 4 and 5
This article for the young and old talks about the origins of our number system and the important role zero has to play in it.
### Do Unto Caesar
##### Stage: 3 Challenge Level:
At the beginning of the night three poker players; Alan, Bernie and Craig had money in the ratios 7 : 6 : 5. At the end of the night the ratio was 6 : 5 : 4. One of them won \$1 200. What were the. . . . |
## Implicit Differentiation
In many examples, especially the ones derived from differential equations, the variables involved are not linked to each other in an explicit way. Most of the time, they are linked through an implicit formula, like F(x,y) =0. Once x is fixed, we may find y through numerical computations. (By some fancy theorems, we may formally show that y may indeed be seen as a function of x over a certain interval). The question becomes what is the derivative , at least at a certain a point? The method of implicit differentiation answers this concern. Let us illustrate this through the following example.
Example. Find the equation of the tangent line to the ellipse
25 x2 + y2 = 109
at the point (2,3). One way is to find y as a function of x from the above equation, then differentiate to find the slope of the tangent line. We will leave it to the reader to do the details of the calculations. Here, we will use a different method. In the above equation, consider y as a function of x:
25 x2 + y(x)2 = 109,
and use the techniques of differentiation, to get
From this, we obtain
which implies that at the point (2,3). So the equation of the tangent line is given by
or equivalently
3y + 50 x = 109.
You may wonder why bother if this is just a different way of finding the derivative? Consider the following example! It can be very hard or in fact impossible to solve explicitly for y as a function of x.
Example. Find y' if
This is a wonderful example of an implicit relation between xand y. So how do we find y'? Let us differentiate the above equation with respect to x where y is considered to be a function of x. We get
Easy algebraic manipulations give
We can also find higher derivatives of y such as y'' in this manner. We only have to differentiate the above result. Of course the calculations get little more messy.
Exercise 1. Find y' if xy3 + x2y2 + 3x2 - 6 = 1.
Exercise 2. Prove that an equation of the tangent line to the graph of the hyperbola
at the point P(x0,y0) is
Exercise 3. Show that if a normal line to each point on an ellipse passes through the center of an ellipse, then the ellipse is a circle.
[Back] [Next]
[Trigonometry] [Calculus]
[Geometry] [Algebra] [Differential Equations]
[Complex Variables] [Matrix Algebra] |
# Transformations of the Reciprocal Function
## Key Questions
An element(point, line, plane or any other geometric, complex or whatever figure) is said to undergo a transformation when ever one or more of its properties are changed.
#### Explanation:
A transformation just a rule; its more like a function. It takes an object and returns that object's image.
Transformations are done using: functions, matrices, complex numbers etc.
What we call object can be a point, a line etc. The basic fact about all objects is that object haves properties.
For example: The point $A \left(3 , 2\right)$ has only the property of position(in a Cartesian coordinate system).
Once you change point $A$'s position to let's say $B \left(6 , 4\right)$ by a particular procedure, we say you have transformed point $A$ to be point $B$.
And in which case the object is $A \left(x , y\right)$ and the image is $B \left(6 , 4\right)$
Our transformation could be the matrix: $\left(\left.\begin{matrix}2 \text{ "0 \\ 2" } 0\end{matrix}\right.\right)$
Proof : Because $\left(\left.\begin{matrix}2 \text{ "0 \\ 0" } 2\end{matrix}\right.\right) \times \left(\left.\begin{matrix}3 \\ 2\end{matrix}\right.\right) = \left(\left.\begin{matrix}6 \\ 4\end{matrix}\right.\right)$
The basic reciprocal function is $\frac{1}{x}$
#### Explanation:
The graph looks like:
graph{1/x [-10, 10, -5, 5]}
• The reciprocal function is:
$f \left(x\right) = \frac{1}{x}$
It's graph is as following:
This is an example of asymptote.
Since $x$ can take all values except $0$ for $f \left(x\right)$ to be defined,
Domain: $R - \left\{0\right\}$, i.e., all real numbers except 0.
Range: $R - \left\{0\right\}$, i.e., all real numbers except 0. |
# When 3 dice are rolled find the probability of getting a sum of at least 15 list the favorable outcomes?
Contents
= 91/216.
## What is the probability of rolling two dice and getting a sum of at least 7?
Probabilities for the two dice
Total Number of combinations Probability
4 3 8.33%
5 4 11.11%
6 5 13.89%
7 6 16.67%
## What is the probability of rolling two dice and getting a sum of at least 10?
When you consider the sum being 10, there are only 3 combinations. So, the probability of getting a 10 would be 3/36 = 1/12.
## What is the probability of rolling 3 dice?
Two (6-sided) dice roll probability table
Roll a… Probability
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
## What is the probability of getting at least one 6 when 3 dice are rolled?
Question: What is the probability of getting at least one six in a single throw of three unbiased dice? Answer: The probability of getting either 1 or 2 or 3 or 4 or 5 when one dice is thrown is 5/6 x 5/6 x 5/6 for 3 dices = 125/216. This is the probability of getting at lease one 6 when 3 dices are thrown.
THIS IS IMPORTANT: Was The Boondocks on BET?
## What is the probability of getting a sum of 7 with two dice?
For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.
## What is the chance of throwing at least 7?
Now, we need at least 7. So, the sum of the numbers on the dice must be greater than or equal to 7. Here, we can clearly see that the total number of favourable outcomes is 21. Thus, the chance of throwing at least 7, in a single throw, using two dice is \$dfrac{text{7}}{text{12}}\$ . |
# How to calculate 133 divided by 20 using long division?
133 ÷ 20
=
6.65
Division is a fundamental arithmetic operation where we calculate how many times a number (divisor or denominator) can fit into another number (dividend or numerator). In this case, we are dividing 133 (the dividend) by 20 (the divisor).
There are three distinct methods to convey the same information: in decimal, fractional, and percentage formats:
• 133 divided by 20 in decimal = 6.65
• 133 divided by 20 in fraction = 133/20
• 133 divided by 20 in percentage = 665%
## What is the Quotient and Remainder of 133 divided by 20?
The quotient is calculated by dividing the dividend by the divisor, and the remainder is what's left over if the division doesn't result in a whole number.
The quotient of 133 divided by 20 is 6, and the remainder is 13. Thus,
### 133 ÷ 20 = 6 R 13
When you divide One Hundred And Thirty Three by Twenty, the quotient is Six, and the remainder is Thirteen.
## Verdict
The division of 133 by 20 results in a quotient of 6 and a remainder of 13, meaning 20 goes into 133 Six times with 13 left over. Understanding this division process is crucial in both basic arithmetic and real-life applications where division is used, such as in financial calculations, data analysis, and everyday problem-solving.
## Random Division Problems?
No worries, we got your back! Tell us what are you brainstorming with and we will bring correct answers to you.
Start Now
### How do we differentiate between divisor and dividend?
A dividend is a number we divide, while a divisor is a number by which we divide. Divisor comes on second, followed by the dividend that we write first.
For instance, if you have 12 candies and want to distribute them among 3 children, the equation will be 12 ÷ 3. You will put 12 first because this is the number being divided. So here, 12 is a dividend. On the other hand, 3 is written after 12, and it is the number with which we are dividing 12. Hence, 3 is a divisor.
### Which formula is used to find a divisor?
There are two formulas used to find a divisor.
The first one is: Divisor = Dividend ÷ Quotient. This formula is used to find a divisor when the remainder is 0.
Second is: Divisor = (Dividend – Remainder) /Quotient. This formula is used when the remainder is not 0.
### Is there a possibility of a number having the same divisor?
Yes, there is. Every number can be divided by itself, leaving 1 as the quotient. So, it would not be wrong to say that all the numbers can have the same divisors.
Let’s take the example of 5. If we divide 5 by 5 (5 ÷ 5), then 5 will be the divisor of 5. And ultimately, 1 will be the quotient.
### What is the difference between a divisor and a factor?
A divisor is a number with which we can divide any number. However, a factor is different from a divisor. It is the number that can be divided with another number leaving no remainder. All factors are divisors, but not all divisors are factors.
### Is it possible to do division by repeated subtraction?
Fortunately yes. You can do division by repeated subtraction. In repeated subtraction, we continuously subtract a number from a bigger number. It continues until we get the 0 or any other number less than the actual number as a remainder.
However, it can be a lengthy process, so we can use division as a shortcut.
### Can I check the remainder and the quotient in a division problem?
Yes, you can quickly check the remainder and quotient in a division problem by using this relationship:
Dividend = Divisor x Quotient + Remainder |
# Irrational Numbers - Math Education Page
```lESSON
Irrational Numbers
In Lesson 2 you learned how to show that any
terminating or repeating decimal can be converted to a fraction. In other words, you know
how to show that terminating or repeating decimals are rational numbers.
2.
Start the factorization of 990 by writing
990 = 3 · 330. Do you get the same prime
factors?
3.
Start the factorization of 990 a third way.
Do you get the same prime factors?
If a decimal is neither repeating nor terminating, it represents an irrational number (one
that is not rational).
Each whole number greater than 1 has only
one prime factorization. Find it for the following numbers:
For example, the number
O.OlOl!OlllOllllOlllllO ... ,
created by inserting one, two, three, ... l 's
between the O's, never ends or repeats.
Therefore it cannot be written as a fraction,
because if it were, it would have to terminate
or repeat.
4.
12
5.
345
1.
Create an irrational number that is
a. greater than l and less than 1.1;
b. greater than 1.11 and Jess than 1.12.
While most numbers we deal with every day
are rational, and even though there is an infinite number of rational numbers, mathematicians have proved that most real numbers are
irrational.
6.
7.
v
6789
Find the prime factorization of several perfect squares. Try to find one having an odd
number of prime factors.
Take the numbers 6 and 8. We have
6 = 2 · 3 and 8 = 2 3
Six has two prime factors, an even number.
Eight has three prime factors, an odd number.
When we square them, we get:
62 = (2 . 3) 2 = 22 • 32
82
= (23)2 = 26
{2 and [.3 are familiar examples of irrational
numbers. They cannot be written as a fraction
having whole number numerators and denominators. In order to prove this, we will need to
review prime factorization.
8. . _ Explain why any perfect square must
have an even number of prime factors.
Every whole number can be written as a product of prime factors.
This section explains why {2 is not a rational
number. The way we are going to do this is
to show that if it were, it would lead to an
impossible situation. This is called proof by
lAhiaUIJ~''-'• 990 = 99 • j 0
=9·11·2·5
=2·3·3·5·11
9. . _ Explain why any number that is equal
to twice a perfect square must have an odd
number of prime factors.
Note that 990 has a total of five prime factors. (Three is counted twice since it appears
twice.)
44o&
Chapter 11 Interpreting Ratios
If p and q were nonzero whole numbers and
E=f2
q
It would follow that (~)' =
17. Does the line y = f2x pass through any
lattice points?
p'
= 2
q
2l
10. Explain each step in the previous
calculations.
11. Explain why p 2 must have an even number
of prime factors.
12. Explain why 2l must have an odd number
of prime factors.
13. Explain why p 2 cannot equal
15. ..,_. Show why the method does not work
to prove that {4 is irrational.
16. Does the decimal expansion of {2 terminate or repeat?
({2)'
2
p2 =
14. ..,_.Use the same method to show that {3
is irrational.
18. ..,_. Do all lines through the origin eventually pass through a lattice point?
Discuss.
19.1;ii#ii§l!in is probably the world's most
famous irrational number. Find out about
its history.
2l.
We conclude that there can be no whole numbers p and q such that {2 = p/q, and therefore
{2 is irrati on a!.
SUM FRACTIONS
20. Find two lowest-term fractions having
different denominators whose sum is 8/9.
11.4 Irrational Numbers
COMPARING COUPONS
21. Which is a better deal, 15% off the
purchase price, or \$1 off every \$5 spent?
Make a graph that shows how much you
save with each discount, for various purchases from \$1 to \$20. Write about your
conclusions.
4074
``` |
Prove thati) the product of 4 consequtive positive integers plus 1 must be a perfect squareand therefore show thatii) for any positive integer n, n^4+2n^3+2n^2+2n+1 is not a perfect square.
i) The product of 4 consecutive integers + 1 can be written as
`n(n+1)(n+2)(n+3) +1`
Multiplying this out we get
`(n^2+n)(n^2 + 5n+6) + 1 `
`= n^4 + ([5n^3],[n^3]) + ([6n^2],[5n^2]) + 6n + 1`
` ` `= n^4 + 6n^3 + 11n^2 + 6n+1`.
If this can be factorised into two identical factors then the factors will be of the form
`n^2 + an + b`
Since the unit term in the expression is 1 we have that `b=1` . Multiplying out the factors we get
`(n^2+an+1)(n^2+an+1) `
`= n^4 + ([an^3],[an^3]) + ([n^2],[a^2n^2],[n^2]) + ([an],[an]) +1 ` `= n^4 + 2an^3 + (a^2+2)n^2 + 2an +1`
Matching the coefficient in `n^3` to that in the expression of interest we have that `a` must equal 3. Checking the other coefficients, they match up: `(a^2+2) = 11` , `2a = 6`. Therefore, the expression is a perfect square and can be written as `(n^2+3n+1)^2` .
ii) This second expression `n^4 + 2n^3 + 2n^2 + 2n+1` can only be a perfect square if it is consistent with the equation above, namely for some `a` it equals
`n^4 + 2an^3 + (a^2+2)n^2 + 2an +1` (since again `b=1` here)
Matching up the coefficients in `n^3` we require then that `a=1`. However with this value for `a` the coefficients in `n^2` don't match: `(a^2 + 2) = 3 != 2`.
Therefore this second expression cannot be a perfect square for positive integers `n`.
Answer: i) the expression is a perfect square ii) the expression isn't a perfect square |
" ">
# A gulab jamun, contains sugar syrup up to about 30% of its volume. Find approximately how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with length 5 cm and diameter 2.8 cm (see figure)."
Given:
A gulab jamun contains sugar syrup up to about 30% of its volume.
To do:
We have to find how much syrup would be found in 45 gulab jamuns, each shaped like a cylinder with two hemispherical ends with a length 5 cm and a diameter of 2.8 cm
Solution:
The volume of one piece of gulab jamun
$=$ Volume of the cylindrical portion $+$ Volume of two hemispherical ends
Radius of each hemispherical portion $= \frac{2.8}{2}$
$= 1.4\ cm$
Volume of one hemispherical end $=\frac{2}{3} \pi r^{3}$
$=\frac{2}{3} \times \frac{22}{7}(1.4)^{3}$
$=\frac{2}{3} \times \frac{22}{7} \times (1.4)^3$
$=\frac{2 \times 22 \times 2 \times 14 \times 14}{3 \times 10 \times 10 \times 10}$
$=5.74 \mathrm{~cm}^{3}$
Volume of both hemispherical ends $= 2 \times 5.74$
$= 11.48\ cm^3$
Height of the cylindrical portion $=$ Total height $-$ Radius of both hemispherical ends
$= 5-2(1.4)\ cm$
$= 5-2.8$
$= 2.2\ cm$
The radius of the cylindrical portion $= 1.4\ cm$
The volume of the cylindrical portion of gulab jamun $= \pi r^2h$
$= \frac{22}{7} \times (1.4)^2 \times 2.2$
$= 22\times2\times1.4\times2.2$
$= 13.55\ cm^3$
The total volume of one gulab jamun $=$ Volume of the two hemispherical ends $+$ Volume of the cylindrical portion
$= 11.48+ 13.55$
$= 25.03\ cm^3$
The volume of sugar syrup $= 30 \%$ of the volume of gulab jamun
$= \frac{30}{100} \times 25.03$
$= 7.50\ cm^3$
Therefore,
The volume of sugar syrup in 45 gulab jamuns
$= 45 \times$ (volume of sugar syrup in one gulab jamun)
$= 45 \times 7.50$
$= 337.5\ cm^3$
$= 338\ cm^3$
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» » Systems of Inequalities Word Problems
Systems of InequalitiesWord Problems
As you get further into Algebra 1, you will find that the real world problems become more complex. They have more questions to be answered and require more steps to find the solution.
When you get into systems of inequalities, this is especially true because you are dealing with two inequalities.
But... don't let that intimidate you! You have all the skills that you need to solve these problems. Take one step at a time and think about what you need in order to answer the question. Read through my example very carefully, and study how I performed each step. Pay careful attention to the key words (highlighted words) and how each inequality was written based on the problem.
Then complete the practice problems. You can do it!
Example 1
For most real world problems, it will be easiest to graph the inequalities using the x and y intercepts. Make sure that you scale your grid so that both inequalities can be graphed on the same grid.
Steps for Solving a System of Inequalities Word Problem
• Read the problem and highlight important information.
• Identify the variables.
• Find one piece of information in the problem that you can use to write an inequality.
• Find a different piece of information that you can use to write a second inequality.
• Graph both inequalities on a grid. Make sure you use appropriate boundary lines and shade the correct half plane for each inequality.
• Identify the intersection of the two inequalities and answer the questions that pertain to the problem.
Now it's time to move onto the practice problems. You can do it!
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# Module 13 - Math 1A - Course Guide
## Module Overview:
In this module students learn about probability. Students will learn about number sentences. Students will group by 10's and use a see saw to determine which animals are heaviest. Students will practice addition facts to 14 and count to 49.
## Module Materials:
• Pencils
• Crayons or colored pencils
• Notebook
• Scissors
• Glue
• Highlighter
• Markers
• Blank paper
Module Objectives:
Lesson # Lesson Title Objective(s)
1 Probability
1. Determine probability of rolling dice; using terms more; less; and equally likely.
2. Complete a probability chart.
2 Number Sentences
1. Write number sentences from addition and subtraction word problems.
3 Group by 10's
1. Group by tens when counting over 19.
2. Balance a scale by moving weights from one side to the other.
4 Fourteen: Addition Facts & Count to 49
1. Practice the addition facts of 14.
2. Count to 49.
Module Key Words:
Key Words
probability
tens
grouping
Module Assignments:
Lesson # Lesson Title Page # Assignment Title
1 Probability 2 Likely Assignment
2 Number Sentences 2 Math Word Problems Assignment
3 Group by 10's 3 Counting Dimes Assignment
## Learning Coach Notes:
Lesson # Lesson Title Notes
1 Probability Extend this lesson more by getting a penny and showing your students heads and tails. Ask them what they think the probability is of flipping heads or tails on a penny. Get a piece of paper and have your student write heads on one side and tails on the other. Have your student flip the coin and record what side it lands on. Have them flip the coin 10 times recording the outcome. Discuss the findings.
2 Number Sentences In the Word Problem Activity, challenge your student to write their answers to each word problem using a complete sentence.
3 Group by 10's Extend this activity by collecting dimes around your house and having your student count to see how much money you have in dimes by counting by tens.
4 Fourteen: Addition Facts & Count to 49 Either purchase at a dollar store or make your own using index cards addition flashcards. Make it part of your daily math routine with your student to review the addition facts daily by showing them the cards and saying the answers. For every answer they get correct, they get the card. Make it their goal to collect all of the cards. When they have reached a point where they know all of the facts, make it a speed contest, time how long it takes them to go through all of the facts, challenge them to beat their time each day.
## Module Guiding Questions:
When a student starts a lesson ask them questions to check for prior knowledge and understanding and to review concepts being taught. At the end of the lesson ask the questions again to see if their answer changes.
Lesson Title Question
Probability
1. What is probability?
Number Sentences
1. What is a number sentence?
Group by 10's
1. Can you count by 10's?
2. How do you measure weight?
Fourteen: Addition Facts & Count to 49
1. Can you count to 49?
2. What numbers can you add together to make 14?
## Module Video Questions:
When a student watches a video take time to ask them questions about what they watched. Suggested questions for the videos in this module are listed here. Suggestion: Have the student watch the entire video first all the way through. Then have them watch the video a second time, as they watch it pause the video and ask the questions.
Lesson Title Video Question
Probability The Probability Tree
1. What does probability mean?
2. Describe how probability was demonstrated in this video in your Math notebook.
Number Sentences Number Sentences
1. In your Math notebook, write down the steps Beatrice used to solve the word problem.
Group by 10's Balancing Act
1. How is a seesaw like a balance scale?
## Module Suggested Read Aloud Books:
Take time to read to your student or have them read aloud to you. Read a different book each day. While reading the book point out concepts being taught. You may purchase these books or find them at your local library. Suggested things to discuss while reading the book:
• What is the main idea?
• What are three things new you learned?
• How does this book relate to what you are learning about?
# Book Author Lexile Level
1 For Good Measure: The Ways We Say How Much, How Far, How Heavy, How Big, How Old Ken Robbins
2 The 100-Pound Problem: Weight Jennifer Dussling 540L
3 It's Probably Penny Loreen Leedy 530L
4 Bad Luck Brad: Probability Gail Herman 490L
## Module Outing:
Take some time to apply what your student is learning to the real world. Suggested outings are below.
# Outing |
# Cosine Ratio Problems
In these lessons, we will learn how to find the angles and sides using the cosine ratio and how to solve word problems using the cosine ratio.
Related Topics: More Trigonometry Lessons
### Hints on solving trigonometry problems:
• If no diagram is given, draw one yourself.
• Mark the right angles in the diagram.
• Show the sizes of the other angles and the lengths of any lines that are known
• Mark the angles or sides you have to calculate.
• Consider whether you need to create right triangles by drawing extra lines. For example, divide an isosceles triangle into two congruent right triangles.
• Decide whether you will need Pythagoras theorem, sine, cosine or tangent.
• Check that your answer is reasonable. The hypotenuse is the longest side in a right triangle.
The following figure shows how to use SOHCAHTOA to decide whether to use sine, cosine or tangent in a given problem. Scroll down the page for examples and solutions.
Example:
Calculate the value of cos * θ* in the following triangle.
Solution:
Use Pythagoras’ theorem to evaluate the length of PR.
### Find Missing Sides
Geometry - Math worksheet - Find the missing length of a triangle using the cosine function
This video shows you how to find the missing length of a triangle using the cosine function. We are given a right triangle and two side lengths and a missing length. We show a right triangle and label the hypotenuse, and two legs the opposite side and adjacent side according to their relation to theta.
How to find the length of a side of a triangle using cosine?
Step 1: Figure out which trigonometric ratio to use
Step 2: Evaluate
### Find Missing Angles
Trigonometry : using SOHCAHTOA to find missing angles.
Trig Ratios - SOH - CAH - TOA - Find missing Angle - Using Inverse Sine, Cosine
### Cosine Word Problems
Trigonometric Functions To Find Unknown Sides of Right Triangles,
This video uses information about the length of the hypotenuse of a right triangle as well as a trig function to find the length of a missing side.
Example:
A 60 ft ladder is leaning against a wall so that the top of the ladder makes a 15-degree angle with the wall. How high up the wall does the ladder reach?
How to Use Cosine to Solve a Word Problem?
Example:
A ramp is pulled out of the back of truck. There is a 38 degrees angle between the ramp and the pavement.If the distance from the end of the ramp to to the back of the truck is 10 feet. How long is the ramp?
Step 1: Find the values of the givens
Step 2: Substitute the values into the cosine ratio
Step 3: Solve for the missing side
Step 4: Write the units
Trigonometry Word Problems
Examples:
1. An airplane over the Pacific sights an atoll at a 20° angle of depression. If the plane is 435 m above water, how many kilometers is it from a point 435 m directly above the center of the atoll?
2. To find the height of a pole, a surveyor moves 70 feet away from the base of the pole and then, with a transit 2 feet tall, measures the angle of elevation to the top of the poleto be 30°. What is the height of the pole to the nearest foot?
3. A ladder 16 feet long makes an angle of 35° with the ground as it leans against a store. To the nearest hundredth, how far up the store does the ladder reach?
Try the free Mathway calculator and problem solver below to practice various math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
# Math Challenge: How To Find Horizontal Asymptotes In Calculus
The study of calculus is a common challenge for all students. Being able to identify and define the concept of horizontal asymptotes is a key skill in calculus that allows students to better understand and apply their skills in problem-solving. In this article, we will discuss the process of finding horizontal asymptotes in calculus. We will examine the various definitions, methods, and approaches of identifying these asymptotes, giving readers a complete overview of this important mathematical concept.
## I. Introduction to Horizontal Asymptotes in Calculus
Horizontal Asymptotes are an important concept to understand in Calculus. An asymptote is a line that the graph of a function approaches, but never quite reaches. Horizontal asymptotes are lines on the x-axis that the graph of a function approaches either from the left or from the right, but never crosses.
There are two types of horizontal asymptotes:
• A horizontal asymptote at y = b – is called the “y = b” asymptote.
• A horizontal asymptote at x = ± ∞ – is called an “end behavior” asymptote.
In order to determine whether or not a function has a horizontal asymptote and to determine the value of the horizontal asymptote, calculus techniques such as limits are needed. Once the values have been determined, then the domain and range of the graph of the function can be determined by looking at the horizontal asymptote and the location of the turning points (or the intercepts) of the graph.
## II. Definition and Examples of Horizontal Asymptotes
What Are Horizontal Asymptotes?
A horizontal asymptote is a straight line on a graph that a curve approaches but does not intersect. Horizontal asymptotes are the lines that a function approaches, but never meets. They are also referred to as limit asymptotes or simply “asymptotes.” In some cases, if the asymptote does not intersect the curve, the limits of the line may be referred to as oblique asymptotes.
Where Do Horizontal Asymptotes Occur?
Horizontal asymptotes occur when the terms of the equation increase steadily or when they increase or decrease without bound. The equation may include polynomials or infinity symbols (∞).
• If the highest power of the equation is even, there will always be a horizontal asymptote at y = 0.
• If the highest power of the equation is odd, the constant coefficient divided by the highest coefficient will give you the value of y for the horizontal asymptote.
• Horizontal asymptotes can be calculated for fractions by looking at the dimensions of the highest power in the numerator and denominator.
• If the horiztonal asymptote equals either the numerator or denominator, no equation exists because the terms continue to increase or decrease without bound.
Examples of Horizontal Asymptotes
• y = 3x + 5 – The highest powers of the equation is 1, so the horizontal asymptote is y = 5.
• y = x4 + 2x3 +3x2 – 15 – The highest powers of the equation is 4, so there is an asymptote of y = 0.
• y = 0.2x2 + 4x – 9 – The highest powers of the equation is 2, so the horizontal asymptote is y = -9.
## III. Analyzing Horizontal Asymptotes in Calculus
Horizontal asymptotes are lines that a graph approaches but never touches. Identifying a horizontal asymptote is an important skill in calculus when analyzing a graph. This section will explain what constitutes a horizontal asymptote and provide steps on how to find a horizontal asymptote.
What Is A Horizontal Asymptote? A horizontal asymptote is a line in the shape of y=c, where c is a constant. The value of c is determined by the degree of the numerator and denominator of the rational function being analyzed. For example, in the case of a rational function f(x)=p(x)/q(x) where the degree of p is less than the degree of q, the horizontal asymptote will be the line y=0.
Finding A Horizontal Asymptote:
• Determine the degree of the numerator and denominator.
• If the degree of the denominator is higher than the degree of the numerator, the horizontal asymptote will be y=0.
• If the degrees are the same, the asymptote will be y=c where c is the leading coefficient of the numerator divided by the leading coefficient of the denominator.
• If the degree of the numerator is higher than the denominator, there will be no horizontal asymptote.
In summary, a horizontal asymptote can be found by comparing the degree of the numerator and denominator of a rational function. If the degree of the numerator is higher than the denominator, there will be no horizontal asymptote. If the degree of the denominator is higher than the degree of the denominator, the horizontal asymptote will be y=0. For the same degree, the horizontal asymptote will be determined by the leading coefficients of the numerator and denominator.
## IV. Solving Horizontal Asymptote Problems
In order to correctly identify an equation’s horizontal asymptote, it is important to understand the concept of limits. Limits measure how the y-value of a function behaves as the x-value approaches a specific point. If the limit of the function is a finite number, the horizontal asymptote exists and is equal to the limit. If the limit is infinite, then there is no horizontal asymptote.
Once you understand limits and the purpose of horizontal asymptotes, you can start to solve equations. To solve a horizontal asymptote equation, you need to:
• Identify the degree of the equation
• Calculate the highest degree term’s y-intercept by setting x to 0
• Identify the direction of the asymptote. If the highest degree term’s y-intercept is 0, the asymptote will be either horizontal or vertical.
Once you have identified the equation’s degree, y-intercept, and asymptote direction, you can start to assess the equation’s behavior. If the degree is even and the y-intercept is 0, the equation has a horizontal asymptote at y=0. If the degree is even and the y-intercept is greater than 0, the equation will have a horizontal asymptote at y=y-intercept. Similarly, if the degree is odd and the y-intercept is 0, the equation will have a vertical asymptote at x=0. If the degree is odd and the y-intercept is greater than 0, the equation will have a vertical asymptote at x=-y-intercept.
## V. Conclusion
In conclusion, it is clear that the presented system of random sampling provides an accurate summary of the data. The accuracy of the system can be accounted for by its unique methods of calculation, its valid results, and its reliable sample estimates.
The system also provides a streamlined process for calculating the sample size, which makes it a valuable tool for any data analysis project. Moreover, the system is highly adaptive, as it can easily be tailored to the needs of different research groups. |
## Welcome to the Math Hub Blog
The Math Hub is a place for learning and sharing expertise about the use of adaptive technology to increase math achievement.
Join the conversation!
## the math hub blog by scholastic/tom snyder productions
### Winter Math Activity: Warm Up to Mitten Math
Teachers are always looking for ways to motivate students, especially at this time of year when it seems just a bit more challenging to crawl out of bed and head to school. So, it’s important to find ways to help students have fun in math class. Last year, I created “Mitten Math”, an activity that I used to help my 5th grade students practice long division. Now, if you’ve ever taught long division, you know that it’s not the most exciting topic to teach, and likewise, it’s not always fun for students to practice this skill.
In this activity, students practice four challenging long division problems in a unique way. Students receive the following materials: Mitten Math Recording Sheet, 8 Math Mittens (made from construction paper with holes punched in the “wrists”), and yarn. Four of the mittens have a division problem written on them; two of the mittens have a quotient written on them and two of the mittens are left blank. The students’ goal is to pair the problem mitten with its correct quotient mitten. Two of the quotients are provided to students, and two of the quotients must be filled after the problem is completed. All work is completed on the mitten math sheet, and once a pair is found, students use the yarn to tie the pair (problem and quotient) of mittens together.
A few things to think about...first, problems need to be selected so that students aren’t able to easily recognize the quotient by looking at the problem. This is part of the reason why I left two of the mittens blank. Students aren’t able to match all four pairs because they don’t know the value of the missing quotients. Second, this activity is great for differentiation. I created a different set of mittens for each of my students. You could do the same and base the problems on students’ ability levels. Or, you could make just a few sets of mittens based on different levels and group students accordingly.
The activity is relatively simple and the “fun” part doesn’t take up much class time, but it motivated students to work hard to solve challenging division problems and seemed more effective than giving them just another worksheet. Have fun!
## What's the Math Hub?
The Math Hub is a place for sharing expertise on math education and the use of adaptive technology to increase student achievement. We invite you to enhance our conversation by submitting your own comments.
Bloggers are compensated by Scholastic. The opinions expressed by the authors on this blog should not be taken to reflect the opinions of Scholastic or Tom Snyder Productions.
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# 2010 AMC 12B Problems/Problem 14
## Problem 14
Let $a$, $b$, $c$, $d$, and $e$ be positive integers with $a+b+c+d+e=2010$ and let $M$ be the largest of the sum $a+b$, $b+c$, $c+d$ and $d+e$. What is the smallest possible value of $M$?
$\textbf{(A)}\ 670 \qquad \textbf{(B)}\ 671 \qquad \textbf{(C)}\ 802 \qquad \textbf{(D)}\ 803 \qquad \textbf{(E)}\ 804$
## Solution 1
We want to try make $a+b$, $b+c$, $c+d$, and $d+e$ as close as possible so that $M$, the maximum of these, is smallest.
Notice that $2010=670+670+670$. In order to express $2010$ as a sum of $5$ numbers, we must split up some of these numbers. There are two ways to do this (while keeping the sum of two numbers as close as possible): $2010=670+1+670+1+668$ or $2010=670+1+669+1+669$. We see that in both cases, the value of $M$ is $671$, so the answer is $671 \Rightarrow \boxed{B}$.
## Solution 2
First, note that, simply by pigeonhole, at least one of a, b, c, d, e is greater than or equal to $\frac{2010}{5}=402,$ so none of C, D, or E can be the answer. Thus, the answer is A or B. We will show that A is unattainable, leaving us with B as the only possible answer.
Assume WLOG that $d+e$ is the largest sum. So $d+e=670,$ meaning $a+b+c=2010-670=1340.$ Because we let $d+e=M,$ we must have $a+b \leq M=670$ and $b+c \leq M=670.$ Adding these inequalities gives $a+2b+c \leq 1340.$ But we just showed that $a+b+c=1340,$ which means that $b=0,$ a contradiction because we are told that all the variables are positive.
Therefore, the answer is $\boxed{B}.$
### Solution 3
Since $a + b \le M$, $d + e \le M$, and $c < b + c \le M$, we have that $2010 = a + b + c + d + e < 3M$. Hence, $M > 670$, or $M \ge 671$.
For the values $(a,b,c,d,e) = (669,1,670,1,669)$, $M = 671$, so the smallest possible value of $M$ is $\boxed{671}$. The answer is (B). |
## Wednesday, September 9, 2015
### geometric mean and sieve
We usually think of the mean of two numbers a and c as their average b = 1/2(+ c). Another way to think of b is that it is the number that makes a, b, c an arithmetic sequence, so b is more properly called the arithmetic mean of a and c. Similarly, we can find the geometric mean of two numbers, a and c. This time, b will be the number that makes a, b, c into a geometric sequence. The geometric mean of two numbers a and c is given by b = sqrt (* c).
Here is a construction for the geometric mean of two numbers that uses a parabola in an interesting way:
1. Graph the parabola y = x^2.
2. Plot the points a and c on the y axis: let A be (0,a) and let C be (0,c).
3. Draw a line through A parallel to the x axis - the point where it touches the parabola's positive arm is A'. Do the same for C - the point where it touches the parabola's negative arm is C'.
4. Draw the line through A' C'. The point where this line crosses the y axis is B = (0, b), where b is the geometric mean of a, c.
This construction of the geometric mean can be used to create a geometric prime sieve. Usually, we visualize the sieve of Eratosthenes on a number chart. Starting with 2, w cross out all the multiples of 2 (except for 2 itself): these numbers are clearly not prime. Then we continue with the first number that we did not cross out, 3, and cross out all its multiples (except for 3 itself)... if we continued indefinitely we will cross out all composites, leaving only primes behind (we have woven a sieve that only lets the primes fall through). Practically, if we are hunting for primes less than n, we only need to go as far as crossing out multiples of primes up to sqrt(n).
Here is how we can construct a similar prime sieve using a parabola - the geometric mean construction can help you see why we are able to hit all the composites.
1. Graph the parabola y = x^2.
2. Plot the points on the parabola obtained from x = -2, -3, -4, ...
3. Plot x = 2 on the parabola. Draw lines from this point on the parabola to each point drawn in step 2. The y-intercepts of these lines are the multiples of 2.
4. Plot x = 3 on the parabola. Draw lines from this point on the parabola to each point drawn in step 2. The y-intercepts of these lines are the multiples of 3.
5. Continue the process for x values that are not among the multiples found in previous steps.The numbers on the y axis, greater than 1 that are not touched by the constructed lines are the primes left behind by the sieve. |
Notes on Scale Drawing and Bearing | Grade 7 > Compulsory Maths > Scale Drawing and Bearing | KULLABS.COM
## Scale Drawing and Bearing (adsbygoogle = window.adsbygoogle || []).push({});
• Note
• Things to remember
• Videos
• Exercise
• Quiz
#### Scale Drawing :
As we know that we cannot draw the fixed shape and size of very small or very big figure on the sheet of paper . It is very difficult to show distances lengths or breath which are too big to draw full size . To draw the figure in the figure in the paper we can just reduce their actual size and for a small figure , we should enlarge their actual size in drawing by taking a scale . A scale drawing is defined as a drawing that shows the size of an object into the form which may be reduction or enlargement .
Scale Factor:
The scale of the drawing is the ratio of the size of the picture or figure to the same size of the objects. For example :
• A distance of 7Km.
Scale : 1cm to 1km ( or ,1:100000)
Scale drawing .
The scale of a drawing is called scale factor . It is used to break (reduce) or increase (enlarge ) a figure according to the scale ratio.
Bearing:
You can see the adjoining compass and learn the different direction . Whereas in the compass NOS shows North-South and EOW show East - West direction . In the angle between N and E is 90o. Whereas the direction NE lies between N and E . The angle between N and NE is 45o. O is taken as the point of reference and ON as a base to find the direction of any place in degrees measuring clockwise from the baseline ( NO ) . The direction is usually written using three digits for example :
• The direction of NE from O is 045o. It is called bearing of NE from O.
• The direction of E from O is 090o. It is called bearing of E and O.
• Whereas the direction of SW from O is 225o. It is called bearing of SW from O and so on.
The direction in which one object or place is sighted may be specified by giving the angle in degrees that the direction makes with the north line is called the bearing if the object. Usually, the bearing of an object or place is written in three digits . So , it is also called three digit bearing.
• A scale drawing is defined as a drawing that shows the size of an object into the form which may be reduction or enlargement .
• The direction in which on object or place is sighted may be specified by giving the angle in degrees that the direction makes with the north line is called the bearing if the object.
• Bearing help shows the direction of the place .
.
### Very Short Questions
Solution:
Here, the scale 1:500000 means, 1 cm represents 500000 cm = $$\frac{500000}{100}$$ m = 5000 m = 5 km
Now, 1 cm represents 5 km
$$\therefore$$ 3.2 represents 3.2 × 5 km = 16 km
So, the distance between two places is 16 km.
Solution:
1. The bearings of A from O is 065°.
2. The bearings of B from O is (360° - 140°) = 220°.
3. The bearings of C from O is (360° - 30°) = 330°.
Solution:
Here, bearing of A from B is 075°
Since, BN//AN, $$\angle$$NBA and $$\angle$$N1AB are co-interior angles.
$$\therefore$$ $$\angle$$ N1AB = 180° - 75° [Co-interior angles are supplementary.]
= 105°
$$\therefore$$ Bearing of B from A = 360° - 105° = 255°.
0%
15.5 m
12.2 m
14.4 m
13.3 m
6.4 cm
3.6 cm
4.8 cm
5.3 cm
11.3 m
12.5 m
13.4 m
14.3 m
• ### The scale of a drawing is also known as the ______ .
scale drawing
scale measure
scale factor
scale mode
3.6 cm
2.4 cm
1.4 cm
4.5 cm
## ASK ANY QUESTION ON Scale Drawing and Bearing
No discussion on this note yet. Be first to comment on this note |
# What is a Linear Function? - Definition & Examples
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Lesson Transcript
Instructor: Yuanxin (Amy) Yang Alcocer
Amy has a master's degree in secondary education and has taught math at a public charter high school.
In this lesson, you will learn the key identifying mark of linear functions. You'll also learn how to graph them with just two points and how you can use the graph to easily find your answers.
## Definition of a Linear Function
A linear function is any function that graphs to a straight line. What this means mathematically is that the function has either one or two variables with no exponents or powers. If the function has more variables, the variables must be constants or known variables for the function to remain a linear function.
## Identifying Linear Functions
To identify linear functions, you can create a checklist of several items the function must meet.
1. The first item the function must satisfy is that it must have either one or two real variables. If another variable is present, it must be a known variable or constant. For example, the function C = 2 * pi * r is a linear function because only the C and r are real variables, with the pi being a constant.
2. The second item is that none of the variables can have an exponent or power to them. They cannot be squared, cubed, or anything else. All variables must be in the numerator.
3. The third item is that the function must graph to a straight line. Any kind of a curve disqualifies the function.
So, linear functions all have some kind of straight line when graphed. The line could be going up and down, left and right, or slanted but the line is always straight. It doesn't matter where on the graph the function is plotted as long as the line comes out straight.
## Working with Linear Functions
If you know that a function is linear, you can plot the graph using just two points. If you are unsure, you can use three or four points to double check.
To figure out your points to plot them, set up a T-chart and start plugging in values for one of the variables. Plug the values into the function to calculate the other variable and note it on the T-chart. When you have filled the T-chart, go ahead and plot the points on the graph. Then, take a ruler and make a straight line through them. All linear functions will have points that are lined up nicely.
Let's try graphing y = 2x.
First, we fill up a T-chart. I'm going to do four points so you can see how the points line up and how, if you know that the function is linear, just two points are sufficient.
x y
0 0
1 2
2 4
3 6
Now, that I've filled up my T-chart, I can go ahead and plot them. Let's see what we get.
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# Properties of Definite Integrals
An integral that has a limit is known as a definite integral. It has an upper limit and a lower limit. It is represented as
f(x) = F(b) − F(a)
There are many properties regarding definite integral. We will discuss each property one by one with proof.
## Properties of Definite Integrals
Various properties of the definite integrals are added below,
### Property 1: f(x) dx = f(y) dy
Proof:
f(x) dx…….(1)
Suppose x = y
dx = dy
Putting this in equation (1)
f(y) dy
### Property 2: f(x) dx = – f(x) dx
Proof:
f(x) dx = F(b) – F(a)……..(1)
f(x) dx = F(a) – F(b)………. (2)
From (1) and (2)
We can derive f(x) dx = – f(x) dx
### Property 3:f(x) dx = f(x) dx + f(x) dx
Proof:
f(x) dx = F(b) – F(a) ………..(1)
f(x) dx = F(p) – F(a) ………..(2)
f(x) dx = F(b) – F(p) ………..(3)
From (2) and (3)
f(x) dx +f(x) dx = F(p) – F(a) + F(b) – F(p)
f(x) dx + f(x) dx = F(b) – F(a) = f(x) dx
Hence, it is Proved.
### Property 4.1: f(x) dx = f(a + b – x) dx
Proof:
Suppose
a + b – x = y…………(1)
-dx = dy
From (1) you can see
when x = a
y = a + b – a
y = b
and when x = b
y = a + b – b
y = a
Replacing by these values he integration on right side becomes f(y)dy
From property 1 and property 2 you can say that
f(x) dx = f(a + b – x) dx
### Property 4.2: If the value of a is given as 0 then property 4.1 can be written as
f(x) dx = f(b – x) dx
### Property 5: f(x) dx = f(x) dx + f(2a – x) dx
Proof:
We can write f(x) dx as
f(x) dx = f(x) dx + f(x) dx ………….. (1)
I = I1 + I
(from property 3)
Suppose 2a – x = y
-dx = dy
Also when x = 0
y = 2a, and when x = a
y = 2a – a = a
So, f(2a – x)dx can be written as
f(y) dy = I2
Replacing equation (1) with the value of I2 we get
f(x) dx = f(x) dx + f(2a – x) dx
### = 0; if f(2a – x) = -f(x)
Proof:
From property 5 we can write f(x) dx as
f(x) dx =f(x) dx + f(2a – x) dx ………….(1)
Part 1: If f(2a – x) = f(x)
Then equation (1) can be written as
f(x) dx =f(x) dx + f(x) dx
This can be further written as
f(x) dx = 2 f(x) dx
Part 2: If f(2a – x) = -f(x)
Then equation (1) can be written as
f(x) dx=f(x) dx – f(x) dx
This can be further written as
f(x) dx= 0
### = 0; if a function is odd i.e. f(-x) = -f(x)
Proof:
From property 3 we can write
f(x) dx as
f(x) dx = f(x) dx + f(x) dx ………(1)
Suppose
f(x) dx = I1 ……(2)
Now, assume x = -y
So, dx = -dy
And also when x = -a then
y= -(-a) = a
and when x = 0 then, y = 0
Putting these values in equation (2) we get
I1 f(-y)dy
Using property 2, I1 can be written as
I1 f(-y)dy
and using property 1 I1 can be written as
I1 f(-x)dx
Putting value of I1 in equation (1), we get
f(x) dx = f(-x) dx +f(x) dx ……….(3)
Part 1: When f(-x) = f(x)
Then equation(3) becomes
f(x) dx = f(x) dx + f(x) dx
f(x) dx = 2f(x) dx
Part 2: When f(-x) = -f(x)
Then equation 3 becomes
f(x) dx = –f(x) dx +f(x) d
f(x)dx = 0
## Example on Properties of Definite Integrals
Example 1: I = x(1 – x)99 dx
Solution:
Using property 4.2 he given question can be written as
(1 – x) [1 – (1 – x)]99 dx
(1 – x) [1 – 1 + x]99 dx
(1 – x)x99 dx
= 1/100 – 1/101
= 1 / 10100
Example 2: I = cos(x) log
Solution:
f(x) = cos(x) log
f(-x) = cos(-x) log
f(-x) = -cos(x) log
f(-x) = -f(x)
Hence the function is odd. So, Using property
f(x)dx = 0; if a function is odd i.e. f(-x) = -f(x)
cos(x) log = 0
Example 3: I = [x] dx
Solution:
0 dx + 1 dx + 2 dx + 3 dx + 4 dx [using Property 3]
= 0 + [x]21 + 2[x]32 + 3[x]43 + 4[x]54
= 0 + (2 – 1) + 2(3 – 2) + 3(4 – 3) + 4(5 – 4)
= 0 + 1 + 2 + 3 + 4
= 10
Example 4: I = |x| dx
Solution:
(-x) dx + (x) dx [using Property 3]
= -[x2/2]0-1 + [x2/2]2
= -[0/2 – 1/2] + [4/2 – 0]
= 1/2 + 2
= 5/2 |
# Factor by grouping solver with steps
Keep reading to understand more about Factor by grouping solver with steps and how to use it. Our website can solve math word problems.
## The Best Factor by grouping solver with steps
Factor by grouping solver with steps can be found online or in math books. There are a number of ways to solve quadratic equations, but one of the most reliable methods is to factor the equation. This involves breaking down the equation into its component parts, which can then be solved individually. For example, if the equation is x2+5x+6=0, it can be rewritten as (x+3)(x+2)=0. From here, it is a simple matter of solving each individual term and finding the value of x that makes both terms equal to zero. While it may take a bit of practice to become proficient at factoring equations, it is a valuable skill to have in your mathematical toolkit.
Solving by square roots Solving by square roots Solving by square roots Solving by square Solving by square Solving Solving by Solving Solving Solving Solving Solvingsolving solving Equation Assume the given equation is of the form: ax^2 + bx + c = 0. Then, the solution to the equation can be found using the following steps: 1) Determine the value of a, b, and c. 2) Find the discriminant, which is equal to b^2 - 4ac. 3) If the discriminant is negative, then there are no real solutions to the equation. 4) If the discriminant is equal to zero, then there is one real solution to the equation. 5) If the discriminant is positive, then there are two real solutions to the equation. 6) Use the quadratic formula to find the value of x that solves the equation. The quadratic formula is as follows: x = (-b +/-sqrt(b^2-4ac))/2a.
Any problem, no matter how complex, can be solved if you break it down into smaller, more manageable pieces. The first step is to identify the goal, or what you want to achieve. Once you have a clear goal in mind, you can start to break the problem down into smaller steps that will lead you to your goal. It is important to be as specific as possible when identifying these steps, and to create a timeline for each one. Otherwise, it will be easy to get overwhelmed and lost in the process. Finally, once you have a plan in place, it is important to stick with it and see it through to the end. Only then can you achieve your goal and move on to the next problem.
A rational function is a function that can be written in the form of a ratio of two polynomial functions. In other words, it is a fraction whose numerator and denominator are both polynomials. Solving a rational function means finding the points at which the function equals zero. This can be done by setting the numerator and denominator equal to zero and solving for x. However, this will only give you the x-intercepts of the function. To find the y-intercepts, you will need to plug in 0 for x and solve for y. The points at which the numerator and denominator are both equal to zero are called the zeros of the function. These points are important because they can help you to graph the function. To find the zeros of a rational function, set the numerator and denominator equal to zero and solve for x. This will give you the x-intercepts of the function. To find the y-intercepts, plug in 0 for x and solve for y. The points at which the numerator and denominator are both zero are called the zeros of the function. These points can help you to graph the function.
## Math solver you can trust
I don't personally use this app because I no longer attend school, however I help my nephew with his homework and I recommended that he tries the app for help. It's a very impressive application that solves math problems and shows the steps as well. So, it's not necessarily cheating if it helps you learn too. In the real world, people use search engines and calculators. I know complex math problems teach critical thinking skills but let's be honest, most people will never use them. It's time to stop teaching kids the "manual" way and teach them to use these apps because this, and some artificial intelligence is the future.
Helen Howard
This is a great app. Basically, it scans the question, gives you the answer, tells you how to do it, and all for free! It doesn't end there either. You can use this app from the very basics and to even some of the most complex math there is. Granted, there are a few bugs, particularly with the camera, and a few with the calculation process (to be exact it sometimes says it can't calculate a question but it can). Don't get me wrong, I still love the app. Math teachers beware!
Giana Ramirez
Help with math app Pre cal identities solver How to solve natural log equations Hard math equations Indicated variable solver |
Unit 5
Numbers 10 20 and Counting to 100
Kindergarten
Math
Description: Students explore numbers 10-20. They apply their skill with and understanding of numbers within 10 to teen numbers, which are decomposed as 10 ones and some ones. For example, 12 is 2 more than 10. The number 10 is special and is the anchor that will eventually become the ten unit in the place value system if future grades.
Standards:
Counting and Cardinality K.CC.1 Count to 100 by ones and by tens. K.CC.2 Count forward beginning from a given number within the known sequence (instead of having to begin at 1). K.CC.3 Write numbers from 0 20. Represent a number of objects with a written numeral 0 20 (with 0 representing a count of no objects). K.CC.4 Understand the relationship between numbers and quantities; connect counting to cardinality. a. When counting objects in standard order, say the number names as they relate to each object in the group, demonstrating one-to-one correspondence. b. Understand that the last number name said tells the number of objects counted. The number of objects is the same regardless of their arrangement or the order in which they were counted. c. Understand that each successive number name refers to a quantity that is one larger. K.CC. 5 K.CC.5: Count to answer How many? questions. a. Count objects up to 20, arranged in a line, a rectangular array, or a circle. b. Count objects up to 10 in a scattered configuration. c. When given a number from 1-20, count out that many object. Numbers and Operations in Base Ten K.NBT.1 Gain understanding of place value. a. Understand that the numbers 1119 are composed of ten ones and one, two, three, four, five, six, seven, eight, or nine ones. b. Compose and decompose numbers 11 to 19 using place value (e.g., by using objects or drawings).c. Record each composition or decomposition using a drawing or equation (e.g., 18 is one ten and eight ones, 18 = 1 ten + 8 ones, 18 = 10 + 8).
Enduring Understandings:
· Students will count to 100 by ones and tens. They will count objects to 20 and represent that number of objects with a written numeral from 1-20.
· Students will use comparative vocabulary to describe items in two sets between 120.
· Students will compare two objects with a measurable attribute in common to see which object has more of/less of the attribute and describe the difference. They will be introduced to solving addition and subtraction word problems and to adding and subtracting within 10 using objects fingers and/or drawings.
· Students will decompose numbers up to 20 into partners in multiple ways (e.g. 12 = 10 + 2 and 12 = 7 + 5). They will begin to find a number that makes 20 when given any number from 1 19.
· Students will gain an understanding of place value by composing and decomposing teen numbers in addition sentences.
· Students will record each composition or decomposition using a drawing or equation (e.g., 18 is one ten and eight ones, 18 = 1 ten + 8 ones, 18 = 10 + 8).
Essential Questions:
• What is base ten and how can it be used?
• What are different ways to represent a number?
• How do we count? Why do we count?
• Is there more than one way to count?
• Why is it important for me to think in numbers?
• How do I show my thinking in different ways?
• How can I compare numbers?
• How can I use concrete objects to add and subtract in a story problem? |
GreeneMath.com - Interval Notation Test #2
# In this Section:
In this section, we review how to write the solution for a linear inequality in one variable using interval notation. Interval notation allows us to notate a specific interval or range of numeric values. Typically we see this when solving a linear inequality in one variable. We may see a result of: x > 3. In this case, any value that is larger than 3 is a solution to that particular inequality. This is notated with interval notation as: (3, ∞). Additionally, we review how to display intervals using a number line. This process closely resembles interval notation. We use a bracket at the number to include it or a parenthesis to not include it. Some courses will use a filled in circle for inclusion and an open circle for exclusion. We shade areas to visually represent the solutions to the inequality. Lastly, we will review how to write a solution using set builder notation. As an example, x > 3 would be displayed in set builder notation as: {x|x > 3}. This is read as “the set of all x, such that x is greater than 3”. This essentially just tells us that x can take on any real number that is larger than 3.
Sections:
# In this Section:
In this section, we review how to write the solution for a linear inequality in one variable using interval notation. Interval notation allows us to notate a specific interval or range of numeric values. Typically we see this when solving a linear inequality in one variable. We may see a result of: x > 3. In this case, any value that is larger than 3 is a solution to that particular inequality. This is notated with interval notation as: (3, ∞). Additionally, we review how to display intervals using a number line. This process closely resembles interval notation. We use a bracket at the number to include it or a parenthesis to not include it. Some courses will use a filled in circle for inclusion and an open circle for exclusion. We shade areas to visually represent the solutions to the inequality. Lastly, we will review how to write a solution using set builder notation. As an example, x > 3 would be displayed in set builder notation as: {x|x > 3}. This is read as “the set of all x, such that x is greater than 3”. This essentially just tells us that x can take on any real number that is larger than 3. |
### A2 Ch 6 Polynomials Notesx
```Algebra 2
Warm Up
A monomial is an expression that is either a
real number, a variable or a product of real
numbers and variables.
A polynomial is a monomial or the sum of
monomials.
The exponent of the variable in a term
determines the degree of that term.
Standard form of a polynomial has the
variable in descending order by degree.
The degree of a polynomial is the greatest
degree of any term in the polynomial
Write each polynomial in standard form and
classify it by degree.
You can write a polynomial as a product of its linear factors
You can sometimes use the GCF to help factor a
polynomial. The GCF will contain variables
common to all terms, as well as numbers
If a linear factor of a polynomial is repeated,
the zero is repeated. A repeated zero is called
a multiple zero. A multiple zero has
multiplicity equal to the number of times the
zero occurs.
page 323 (1-11, 17-35)odd
you do NOT need to graph the functions.
Two people per worksheet.
Take turns at each step, first partner decides
what you multiply the divisor by, second
partner agrees and does the multiplication,
first partner agrees and does the subtraction,
then switch for next term.
You may do the work on the worksheet,
paper or the white board. If you use the white
board you must have me check EACH answer
as you complete it.
Warm Up:
1. Write a polynomial function in standard
form with zeros at -1, 2 and 5.
2. Use long division to divide:
3.
Use long division to divide
Solve for all three roots
Homework:
page 330 (227-33) odd
page 336 (13 – 31) odd,
Solve these equations:
1. x3 + 125 = 0
2. x4 + 3x2 – 28 = 0
To find all the roots of a polynomial:
◦ determine the possible rational roots using the
rational root theorem (ao/an)
◦ Use synthetic division to test the possible rational
roots until one divides evenly
◦ Write the factored form and solve for all roots
Use the quadratic formula if necessary
You may need to use synthetic division more than once
Warm Up
Find the polynomial equation in standard
form that has roots at -5, -4 and 3
Find f(-2) for f(x) = x4 – 2x3 +4x2 + x + 1
using synthetic division
Solve x4 – 100 = 0
Practice Problem:
List all the possible rational roots of
◦ 3x3 + x2 – 15x – 5 = 0
Use synthetic division to determine which of these is
a root
Factor and solve for the rest of the roots of the
equation.
A third degree polynomial has roots 2 and
√3. Write the polynomial in standard form.
Homework
p 345 (11-23) odd
A selection of items in which order does not
matter is called a combination
homework
p 354 (1-29) odd
Warm Up
Find the zeros of the function by finding the
possible rational roots and using synthetic
division.
multiply each and write in standard form:
(x + y)2
(x + y)3
(x + y)4
Notice that each set of coefficients matches a
row of Pascal’s Triangle
Each row of Pascal’s Triangle contains
coefficients for the expansion of (a+b)n
For example, when n = 6 you can find the
coefficients for the expansion of (a+b)6 in the
7th row of the triangle.
Use Pascal’s Triangle to expand (a+b)6
If the terms of the polynomial have
coefficients other than 1, you can still base
the expansion on the triangle.
Warm up
To find a particular term of a binomial
expansion you do not need to calculate the
entire polynomial!
Ex: Find the 5th term of (x – 4)8
Find the 4th term of (x – 3)8
Chapter 6 Test this Thursday (5th) or Friday
(4th)
Homework: Complete practice test
``` |
# Basic math angle facts; angles in a triangle, on a straight line and around a point.
There are 3 main angle facts to get to grips with in your math exams:
1) Angles on a straight line add up to 180⁰ (also known as angles in a half circle)
2) Angle around a point add up to 360⁰ (also known as angles in a full circle)
3) Angles in a triangle add up to 180⁰
So to solve a basic angle fact question you will first have to decide which angle fact to use. Then you need to add up the given angles and subtract this answer either 180⁰ or 360⁰ (depending on the angle fact you have used).
Let’s take a look at 3 examples that involve using these 3 basic angle facts that you get in many math tests.
Example 1
Work out the missing angle x.
The angle fact you will need to use is angles around a point add up to 360⁰ (since the angles join up to make a full circle).
Next add up the angles that you are given. Make sure that you include the right angle in your total (a right angle is 90⁰):
64 + 90 = 154⁰
Next subtract this answer from 360⁰ to give you the size of angle x:
360 – 154 = 206⁰
x = 206⁰
You can see this angle is correct as the angle you were finding is a reflex angle (larger than 180⁰).
Example 2
Work out the missing angle y.
The angle fact you will need to use is angles on a straight line add up to 180⁰ (since the angles form a half circle).
Next add up the angles that you are given:
46 + 35 = 81⁰
Next subtract this answer from 180⁰ to give you the size of angle y:
180 – 81 = 99⁰
y = 99⁰
Example 3
Work out the size of angle z.
Since the shape is triangle, the angle fact is clearly angles in a triangle add up to 180⁰.
Like the last two angle questions first add up all of the angles in the triangle:
128 + 22 = 150⁰
Next subtract this answer from 180⁰ to give you the size of angle z:
180 – 150 = 30⁰
y = 30⁰
As you can see with these 3 examples the hardest part of the question is identifying the correct angle fact to use. Harder questions will involve applying more than one of the 3 angle facts listed above.
## 1 comment
Joan Whetzel 5 years ago
Great descriptions and diagrams. I always love reading your math articles. They're informative and understandable for non-math people like me.
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Before us begin, let"s recognize the an interpretation of the square root. The symbol square root is created as and is one integral part of mathematics. Once you recognize the basics of recognize the square root of a number, you can solve any kind of square root-related problem. In this short lesson, we will learn around the square root that 48 and find it with various methods like long department method. Let united state see what the square root of 48 is.
You are watching: Simplify square root of negative 48
Square source of 4848 = 6.928Square of 48: 482 = 2304
1 What Is the Square root of 48? 2 Is Square source of 48 Rational or Irrational? 3 How to uncover the Square root of 48? 4 FAQs ~ above Square root of 48
The square source of a number is the number that gets multiplied to itself to provide the product. The square root of 48 is 6.92820323. Square root of 48 in the radical form is to express as 48 and in exponent form, it is expressed as 481/2. The square root of 48 rounded to 5 decimal locations is 6.9282...
48 is no a perfect square. We can discover the square source of 48 by the approximation method. For the accurate value, we have the right to use the long division method. In the approximation method, we discover square number close come 48. We watch that 36 and 49 are the perfect square number close to 48. The square root of 36 is 6 and the square source of 49 is 7. As such the square root of 48 must lie between 6 and 7 and much more close come 7 as 48 is closer to 49. This an approach only provides us an approximate answer. To understand the exact value we can use the long division method and find a an ext accurate decimal value for 48.
Simplified Radical kind of Square root of 48
48 is a composite number. When we discover square source of any number, we take one number from each pair the the same numbers native its prime factorization and we main point them. The administrate of 48 is 2 × 3 × 2 × 2 × 2 which has actually 1 pair of the very same number. It can likewise be written as 48 = 24 × 31. Thus, the easiest radical form of 48 is 43 itself.
Let united state follow the steps to find the square root of 48 by lengthy division.
Step 1: starting from the right, we will pair up the digits by putting a bar over them.Step 2: discover a number which, once multiplied to itself, offers the product less than or equal to 48. So, the number is 4. Placing the divisor together 6, we gain the quotient together 6 and the remainder 36.Step 3: Now twin the divisor. Guess the largest possible digit i m sorry will additionally become the new digit in the quotient, together that when the new divisor is multiply to the new quotient the product is less than or equal to the dividend. Divide and write the remainder.
Explore Square roots utilizing illustrations and interactive examples
Important Notes:
In exponent kind square source of 48 is expressed together 481/2.The real roots the 48 are 6.928.
Challenging Questions:
What is the worth of 48?Determine the square source of -48.
Example 2: Help Ross simplify the square source of 48 to its shortest radical form.
Solution
We should express 48 as the product of its prime determinants as 48 = 2 × 3 × 2 × 2 × 2.Therefore, √48 = (2 × 2 × 2 × 2 × 2) = 43.Thus, 4√3 is in the shortest radical form.
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# ISEE Lower Level Math : Geometry
## Example Questions
### Example Question #101 : Geometry
What is the perimeter of a rectangle with a width of and a length of ?
Explanation:
The perimeter of a rectangle is equal to the sum of all its sides. The formula for finding the perimeter of a rectangle is
The length of the rectangle is eight, and the width is five; .
Therefore, the perimeter of the rectangle is .
### Example Question #7 : How To Find The Perimeter Of A Rectangle
Mr. Barker is building a rectangular fence. His yard has an area of feet, and the one side of the fence he's already built is feet long.
What will the perimeter of his fence be when he is finished building it?
Explanation:
The perimeter is calculated by adding up all the sides of the rectangle—in this case,
So the perimeter is feet.
### Example Question #1 : How To Find The Perimeter Of A Rectangle
A rectangle has sides 10 cm and 4 cm. What is its perimeter?
Explanation:
A rectangle has two sides of congruent, or equal, sides. Therefore, there are two 10 cm sides and two 4 cm sides. Perimeters is the sum of all the sides, so you must add up the four sides. The answer is 28 cm.
### Example Question #1 : How To Find The Perimeter Of A Rectangle
If a rectangle has a width of 6.5 inches and a length of 9 inches, what is its perimeter?
Explanation:
The perimeter of a rectangle is found by adding together all four sides. Two sides will be equal to the length, and two sides will be equal to the width.
Since the width is 6.5 inches and the length is 9 inches, the perimeter would be:
### Example Question #1 : How To Find The Perimeter Of A Rectangle
If Margaret is buying a tablecloth for a table that is 4 feet by 2 feet, what should be the area of the tablecloth?
Explanation:
We will need to find the area of the rectangle by multiplying the length by the width:
Use the given dimensions:
### Example Question #81 : Plane Geometry
If the perimeter of a rectangle is equal to 54 inches, what are possible values for the width and length?
Width of 10 inches and length of 16 inches
Width of 10 inches and length of 17 inches
None of these
Width of 11 inches and length of 17 inches
Width of 10 inches and length of 18 inches
Width of 10 inches and length of 17 inches
Explanation:
The perimeter of a rectangle is equal to:
The only answer choice that provides dimensions for which the perimeter of a rectangle would be 54 inches is when there is a width of 10 inches a length of 17 inches.
Therefore, a width of 10 inches and length of 17 inches is the correct answer.
### Example Question #81 : Plane Geometry
If the width of a rectangle is 2 inches and the length is 6 inches, what is the perimeter in inches?
Explanation:
The perimeter of a rectangle can be found by adding together all the sides. The formula is:
Use the given values for the length and the width to solve.
16 inches is the correct answer.
### Example Question #13 : How To Find The Perimeter Of A Rectangle
A rectangle has a width of 5 inches and an area of 35 inches. What is the length, in inches?
Explanation:
The area of a rectangle is equal to the length times the width.
Here, we know that the area is 35, and the width is 5. We can fill in part of the equation with these values.
The only number that equals 35 when multiplied by 5 is 7.
7 is the correct answer.
### Example Question #71 : Rectangles
If the width of a rectangle is 3 inches and the length is 5 inches, what is the perimeter?
Explanation:
The perimeter of a rectangle can be found by adding together all the sides. This would give us:
Therefore, 16 inches is the correct answer.
### Example Question #15 : How To Find The Perimeter Of A Rectangle
Marcell wants to put a new fence around the perimeter of his garden. Marcell's rectangular garden is 30 feet long and 15 feet wide. How many feet of fence does Marcell need? |
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# If a spring has a period $T$, and is cut into the $n$ equal parts, then the period of each part will be:A. $T\sqrt n$B. $\dfrac{T}{{\sqrt n }}$C. $nT$D. $T$
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Hint: The spring constant is always in the linear range. At the lower gravity, the period did not change, The system’s equilibrium position gets different. Hence the period fully depends on the spring constant and mass of the object.
Complete step by step solution:
The hooke's law execute a simple harmonic motion as follows
The period $T$ , mass $m$ , and spring of spring constant $K$ is given by the equation as follows,
$T = 2\pi \sqrt {\dfrac{m}{k}}$ ………(1)
After the spring is cut into four equal parts, let $K$ be the spring constant of little spring. Hence all parts are equal we resume all small springs following Hooke's Law and have equal spring constant.
Where,
$K$ is known as the proportionality constant and which is also called a spring constant. The spring constant $K$ depends on the factors that are listed below,
Nature of the material of the spring,
Cross-Sectional Area of the spring,
Length of the spring.
For a spring,
$T = 2\pi \sqrt {\dfrac{m}{k}}$
For every piece of spring, the constant is $nk$
$T' = 2\pi \sqrt {\dfrac{m}{{nk}}}$
Rearrange the above equation and simplify the equation we get,
$T' = 2\pi \sqrt {\dfrac{m}{k} \times } \dfrac{1}{{\sqrt n }}$
Hence the period of each part is, $T' = \dfrac{T}{{\sqrt n }}$
So, the correct answer is, Option (B)
Note:
The spring constant and mass help to determine the period.
The equilibrium position of the system is determined by the gravitational force.
The equilibrium position does not depend on gravity.
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# The Law of Cosines Study Guide
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## The Law of Cosines
When we know the angles of a triangle and one of the sides, we can figure out the lengths of the other sides with the Law of Sines. In this lesson, we see how a new theorem, the Law of Cosines, enables us to find the third length of a triangle when we know the lengths of the other two sides and the angle between them. For example, suppose we know the lengths A and B of the triangle in Figure 18.1 and the measure θ of the angle between them. We would like to find the length C of the third side.
### Law of Cosines
C2 = A2 + B2 – 2AB·cosθ
We can prove this with vectors (if the proof is confusing, feel free to skip to the examples on page 175). To make things easier, rotate the triangle so that the angle θ is in standard position, as shown in Figure 18.2.
Let the origin be the right endpoint of the side with length B. If we look at the side of length B as a vector, then its components are (–B,0), because it goes a distance of B to the left and does not go up or down at all. The side with length A goes up at an angle of θ , so as a vector, its components are (A· cos(θ),A · sin(θ)). The sum of these two vectors is thus (A· cos(θ) – B,A· sin(θ)), as illustrated in Figure 18.3.
The side with length C goes from the origin (0,0) to the point (A· cos(θ) – B,A· sin(θ)). This means that it is the hypotenuse of a triangle with legs of length | A · cos(θ) – B | and A · sin(θ), as shown in Figure 18.4. Because we don't know if A · cos(θ) or B is bigger, we use absolute values to describe the length of the bottom of this triangle. When we square this length, it won't matter whether A · cos(θ) – B was positive or negative.
We can now find the length C with the Pythagorean theorem:
C2 = (A· sin(θ))2 + (A· cos(θ) – B)2
C2 = A2· sin2(θ) + A2 · cos2(θ) – 2AB · cos(θ) + B2
We can factor the A2out of the first two terms on the right.
C2 = A2(sin2(θ) + cos2(θ)) – 2AB · cos(θ) + B2
If we use sin2(θ) + cos2(θ) = 1, proved in Lesson 6, we end up with the Law of Cosines:
C2 = A2 – 2AB · cos(θ) + B2
Usually, this is written as C2 = A2 + B2 – 2AB · cos(θ). We can use this to quickly find the lengths of triangles.
#### Example 1
Find the length x in Figure 18.5.
We can use the Law of Cosines with A = 8, B = 11, C = x, and θ = 39°:
C2 = A2 + B2 – 2AB · cos(θ)
x2 = 82 + 112 – 2(8)(11) · cos(39°) ≈ 48.22
x ≈ √48.22 ≈ 6.9
The length is x ≈ 6.9 inches.
#### Example 2
Find the length x in Figure 18.6.
We use the Law of Cosines with A = 40, B = 72, C = x, and θ = 108°.
C2 = A2 + B2 – 2AB · cos(θ)
x2 = 402 + 722 – 2(40)(72) · cos(108°) ≈ 8,563.94
x ≈ √8,563.94 ≈ 92.5 cm
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# The mystery of 495 explained
In yesterday’s post, we chose a three digit positive integer and we arranged the digits in ascending and descending order forming the largest and the smallest integer that can be formed using the three digits. We subtracted the smaller integer from the larger, and repeated the process several times. We ended up with 495. We tried other integers and repeated the process, and they all ended up with 495. In this post, we are going to discuss the reason behind our observation.
Let’s take the first example in yesterday’s post. We chose 592. Arranging the numbers in descending and ascending order, we got 259 and 952. Subtracting the numbers, we got 693.
Let us examine closely what happens when we subtract the numbers with rearranged digits (examine the figure above):
1. In the ones’ place, the subtrahend is always greater than the minuend (Why?).
2. In the tens’ place, the difference is always 9 (Why?).
3. In the hundreds’ place, we always subtract 1 from the hundred’s digit of the minuend, before subtracting the hundreds’ digit.
The generalization of the subtraction process is shown below. If the digits of the numbers are represented by x, y, and z, where x>y>z (strictly speaking, at most two numbers can be equal), then the largest number that we can form from this digits is xyz and the smallest is zyx. If abc is the difference, then the following relationships hold (can you see why?):
c = 10 + zx
b = 9
a = x – 1 – z
From the equations above, we can also conclude that a + c = 9. This means that the difference is of the form a9c where a + c = 9. This only gives us ten numbers: 099, 198, 297, 396, 495, 594, 693, 792, 891, and 990. Notice that these numbers are all multiples of 99, and, manually, we can check that all these numbers lead to 495. Note that we have already shown the integers in red led to 495 in the previous post, so we only have to verify the remaining four integers.
How about other numbers?
We only have 10 numbers above, so how do we show that any 3-digit number chosen will end up with 495?
In general, any 3-digit number with digits xyz, where x>y>z, is of the form 100x + 10y + z. The smaller number is of the form 100z + 10y + x. Subtracting, we have 99(xz) which means that the difference of any two digit numbers satisfying the conditions above is a multiple of 99. But all the three digit numbers that are multiples of 99 are the ten numbers listed above. Therefore, repeating the process above will eventually lead to 495 for any three digit number. That explains the mystery behind 495. |
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You are reading an older version of this FlexBook® textbook: CK-12 Algebra I Concepts - Honors Go to the latest version.
# 2.6: Equations with Decimals, Fractions and Parentheses
Difficulty Level: At Grade Created by: CK-12
Pens are $9 per dozen and pencils are$6 per dozen. Janet need to buy a half dozen of each for school. How much is the total cost of her purchase?
### Guidance
Recall that the distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. Here, you will use the distributive property with linear equations and fractions. The same rules apply. You have to multiply what is on the outside of the parentheses by what is on the inside of the parentheses and this is your first step in solving equations with one variable where there are parentheses. After you remove parentheses, you then solve the equation by combining like terms, moving constants to one side of the equals sign, variables to the other side of the equals sign, and finally isolating the variable to find the solution.
#### Example A
Solve: \begin{align*}\frac{2}{5}(d+4) = 6\end{align*}.
Find the LCD for 5, 5, and 1. Since it is 5, multiply the last number by \begin{align*}\frac{5}{5}\end{align*}, to get the same denominator.
Since all of the denominators are the same, the equation becomes:
Therefore \begin{align*}d = 11\end{align*}.
#### Example B
Solve: \begin{align*}\frac{1}{4}(3x+7) =2\end{align*}.
Find the LCD for 4, 4, and 1. Since it is 4, multiply the last number by \begin{align*}\frac{4}{4}\end{align*}, to get the same denominator.
Since all of the denominators are the same, the equation becomes:
Therefore \begin{align*}x = -5\end{align*}.
#### Example C
Solve: \begin{align*}\frac{1}{3}(x-2) = -\frac{2}{3}(2x+4)\end{align*}.
Since all of the denominators are the same, the equation becomes:
Therefore \begin{align*}x=\frac{-6}{5}\end{align*}.
#### Concept Problem Revisited
Pens are $9 per dozen and pencils are$6 per dozen. Janet need to buy a half dozen of each for school. How much is the total cost of her purchase?
First you should write down what you know:
Let \begin{align*}x =\end{align*} total cost
Cost of pens: $9/dozen Cost of pencils:$6/dozen
Janet needs one half dozen of each.
The total cost would therefore be:
Therefore Janet would need \$7.50 to buy these supplies.
### Vocabulary
Distributive Property
The distributive property is a mathematical way of grouping terms. It states that the product of a number and a sum is equal to the sum of the individual products of the number and the addends. For example, in the expression: \begin{align*}{\color{red}\frac{2}{3}} ({\color{blue}x + 5})\end{align*}, the distributive property states that the product of a number \begin{align*}({\color{red}\frac{2}{3}})\end{align*} and a sum \begin{align*}({\color{blue}x + 5})\end{align*} is equal to the sum of the individual products of the number \begin{align*}({\color{red}\frac{2}{3}})\end{align*} and the addends \begin{align*}({\color{blue}x}\end{align*} and \begin{align*}{\color{blue}5})\end{align*}.
### Guided Practice
1. Solve for x: \begin{align*}\frac{1}{2}(5x+3)=1\end{align*}.
2. Solve for x: \begin{align*}\frac{2}{3}(9x-6)=2\end{align*}.
3. Solve for x: \begin{align*}\frac{2}{3}(3x+9)=\frac{1}{4}(2x+5)\end{align*}.
1.
Find the LCD for 2, 2, and 1. Since it is 2, multiply the last number by \begin{align*}\frac{2}{2}\end{align*}, to get the same denominator.
Since all of the denominators are the same, the equation becomes:
Therefore \begin{align*}x=\frac{-1}{5}\end{align*}.
2.
Find the LCD for 3, 3, and 1. Since it is 3, multiply the last number by \begin{align*}\frac{3}{3}\end{align*}, to get the same denominator.
Since all of the denominators are the same, the equation becomes:
Therefore \begin{align*}x=1\end{align*}.
3.
Find the LCD for 3, 3, and 4, 4. Since it is 12, multiply the first two fractions by \begin{align*}\frac{4}{4}\end{align*} and the second two fractions by \begin{align*}\frac{3}{3}\end{align*}, to get the same denominator.
Since all of the denominators are the same, the equation becomes:
Therefore \begin{align*}x=\frac{-57}{18}\end{align*}.
### Practice
Solve for the variable in each of the following problems.
1. \begin{align*}\frac{1}{2} (x+5)=6\end{align*}
2. \begin{align*}\frac{1}{4}(g+2)=8\end{align*}
3. \begin{align*}0.4(b+2)=2\end{align*}
4. \begin{align*}0.5(r-12)=4\end{align*}
5. \begin{align*}\frac{1}{4}(x-16)=7\end{align*}
Solve for the variable in each of the following problems.
1. \begin{align*}26.5-k=0.5(50-k)\end{align*}
2. \begin{align*}2(1.5c+4)=-1\end{align*}
3. \begin{align*}-\frac{1}{2}(3x-5)=7\end{align*}
4. \begin{align*}0.35+0.10(m-1)=5.45\end{align*}
5. \begin{align*}\frac{1}{4}+\frac{2}{3}(t+1)=\frac{1}{2}\end{align*}
Solve for the variable in each of the following problems.
1. \begin{align*}\frac{1}{2}x-3 (x+4)=\frac{2}{3}\end{align*}
2. \begin{align*}-\frac{5}{8}x+x=\frac{1}{8}\end{align*}
3. \begin{align*}0.4(12-d)=18\end{align*}
4. \begin{align*}0.25(x+3)=0.4(x-5)\end{align*}
5. \begin{align*}\frac{2}{3}(t-2)=\frac{3}{4}(t+2)\end{align*}
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# What is 8 Out of 9 As a Percentage?
8 Out of 9 As a Percentage is a very popular question. The answer is 888.89%.
Checkout this video:
## 8 Out of 9 As a Percentage
8 Out of 9 is equal to 88.89%. This can be useful when you want to know how much of something there is compared to the whole. For example, if you have 8 out of 9 people in your group that are going to the movies, that means that 8/9 of your group is going to the movies.
### 8 Out of 9 As a Percentage of What?
To calculate 8 out of 9 as a percentage, you first need to determine what 8 out of 9 represents a portion of.
For example, if 8 out of 9 is a portion of a total number of items, then 8 out of 9 as a percentage would be 8/9 x 100 = 88.89%.
But if 8 out of 9 is a portion of a total number of people, then 8 out of 9 as a percentage would be 8/(9-1) x 100 = 800%.
So the answer to “What is 8 out of 9 As a Percentage?” depends on what exactly 8 out of 9 represents a portion of.
### How to Calculate 8 Out of 9 As a Percentage
To calculate 8 out of 9 as a percentage, you would first divide 8 by 9 to get 0.888888889. Then, you would multiply that decimal by 100 to turn it into a percentage. Therefore, 8 out of 9 as a percentage would be 88.889%.
## Examples
To convert a decimal to a percentage, multiply the decimal by 100 and include the % sign.
### Example 1
To convert 8 out of 9 into a percent, simply divide 8 by 9 to get .8888889, and then multiply by 100 to get 88.88889%.
### Example 2
If you want to know what 8 out of 9 is as a percentage, you need to first understand how to calculate percentages. A percentage is a way of expressing a number as a fraction of 100. So, if we want to calculate the percentage of 8 out of 9, we need to first turn 8 out of 9 into a fraction. To do this, we can use what is called the “fraction strips method.”
First, let’s draw 9 strips. Then, we’ll color in 8 of them:
Now, let’s count how many total strips there are. There are 9 strips in total. So, the fraction 8 out of 9 can be simplified to 8/9.
To calculate the percentage of 8 out of 9, we need to take the fraction 8/9 and multiply it by 100. So, our equation will look like this:
8/9 x 100 =
To solve this equation, we can use a calculator or mental math. If we use mental math, we can easily see that 8 goes into 90 ten times (90/8 = 10 with no remainder). So, our answer will be 900%. But let’s double check this by using a calculator just to be sure:
As you can see from the calculator screenshot above, 900% is indeed the answer!
## What is 8 Out of 9 As a Percentage?
If you have a fraction, such as 8/9, and you want to know what that is as a percentage, there are a couple of steps that you need to follow. To convert a fraction to a percentage, you need to first multiply the fraction by 100. So, 8/9 multiplied by 100 is equal to 800/9. To simplify this fraction, you can divide both the numerator and the denominator by 9, which will give you the answer as 88.89%.
## 8 Out of 9 As a Percentage of What?
8 out of 9 as a percentage is 89%.
## How to Calculate 8 Out of 9 As a Percentage
To calculate 8 out of 9 as a percentage, you need to divide 8 by 9 and then multiply the result by 100. Dividing 8 by 9 gives you 0.888888889, and multiplying by 100 gives you 88.88888889%. Therefore, 8 out of 9 as a percentage is 88.88888889%.
## Examples
8 out of 9 can be expressed as a percentage in two different ways. The first way is to divide 8 by 9 to get .8888888888888888888888888888889 and then multiply that by 100 to get 88.888888888888888888888888888889%. The other way is to simply take 8 out of 9 and multiply it by 100 which also gives you 88.888888888888888888888888888889%.
### Example 1
To calculate 8 out of 9 as a percentage, we first divide 8 by 9 to get 0.888888889. We then multiply this decimal by 100 to turn it into a percentage. This gives us 88.88888889%.
### Example 2
Here is an example of how to convert 8 out of 9 to a percentage.
To turn a fraction into a percent, multiply the top number of the fraction by 100 and then divide by the bottom number of the fraction. In this particular case, that would mean you multiply 8 by 100 and then divide by 9.
If you multiply 8 by 100, you get 800. Then, if you divide 800 by 9, you get 88 with a remainder of 8/9. This can also be written as 88.88% with an infinite number of decimal places after the decimal point since 9 does not go evenly into 800. |
# Calculating with Fractions
Before we begin calculating with fractions, lets refresh some terms.
### A Mixed Number is a mixture of whole numbers and fractions.
If two or more fractions have the same denominator, they are known as 'Like Fractions'.
so 1/4 and 3/4 are like fractions, as are 3/25, 9/25, 23/25 etc
### A Like Fraction has the same denominator (bottom number).
Think about adding two like fractions, say, one quarter plus one quarter, you probably know that the answer is two quarters because you can easily imagine having two quarters of a cake and putting them together - you have two quarters or one half.
But how can we write a calculation for this?
If you look carefully, you will see that we have added the numerators, 1 + 1 = 2, but the denominator has remained the same, 4. Lets try another:
We have added the numerators again, 1 + 2 = 3, and the denominator has again remained the same, 4.
### To add like fractions (same denominator) you just add the numerators and keep the denominator the same.
Now think about adding two fractions that have different denominators, say, one quarter plus one eighth, you cant really visualise this and get a precise answer because they are different quantities - quarters and eighths.
The only way we can add these fractions is if the denominators are the same - so lets change the quarter into eighths. To make the denominator an 8 we multiply it by 2 because 4 x 2 = 8, and remember that we must do the same to the top and bottom. So we also multiply 1 x 2 = 2. Changing the fraction from 1/4 to 2/8.
Now both fractions have the same denominator so we can simply add the numerators.
Lets try a more difficult example, say one half plus one third,
The problem we now have is that the denominators are not the same and we cannot change one denominator into the other easily. The solution is to find a denominator that is common to both numbers - a common denominator. The easiest way to find a common denominator is to multiply the denominators, ie 2 x 3 = 6,
Now we change both fractions so that their denominators are 6.
(basically we multiply each fraction by the other fractions denominator).
and
Both fractions now have the same denominator so we can add the numerators. |
# Coffee is draining from a conical filter into a cylindrical coffee pot of radius 4 inches at the rate of 20 cubic inches per minute. How fast is the level in the pot rising when the coffee in the cone is 5 inches deep. How fast is the level in the cone falling then?
The aim of this question is to use the geometric formulas of volume of different shapes for the solution of word problems.
The volume of the cone-shaped body is given by:
$V \ = \ \dfrac{ 1 }{ 3 } \pi r^2 h$
Where h is the depth of the cone.
The volume of the cylindrical-shaped body is given by:
$V \ = \ \pi r^2 h$
Where h is the depth of the coffee pot.
Part (a) – The volume of the cylindrical-shaped coffee pot is given by the following formula:
$V \ = \ \pi r^2 h$
Differentiating both sides:
$\dfrac{ dV }{ dt } \ = \ \pi r^2 \dfrac{ dh }{ dt }$
Since the rate of rise of volume of the cylindrical coffee pot $\dfrac{ dV }{ dt }$ has to be same as the rate of fall of volume in the conical filter, we can say that:
$\dfrac{ dV }{ dt } \ = \ 20 \ in^3/min$
Also, given that $r \ = \ 4 \ inches$, the above equation becomes:
$10 \ = \ \pi ( 4 )^2 \dfrac{ dh }{ dt }$
$\Rightarrow 20 \ = \ 16 \pi \dfrac{ dh }{ dt }$
$\Rightarrow \dfrac{ dh }{ dt } \ = \ \dfrac{ 20 }{ 16 \pi } \ = \ \dfrac{ 5 }{ 4 \pi }$
Part (b) – Given that the radius r’ of the cone is 3 inches at the maximum height h’ of 6 inches, we can deduce following relationship between r’ and h’:
$\dfrac{ r’ }{ h’ } \ = \ \dfrac{ 3 }{ 6 } \ = \ \dfrac{ 1 }{ 2 }$
$\Rightarrow r’ \ = \ \dfrac{ 1 }{ 2 } h’$
Differentiating both sides:
$\Rightarrow \dfrac{ r’ }{ t } \ = \ \dfrac{ 1 }{ 2 } \dfrac{ h’ }{ t }$
The volume of the cone-shaped conical filter is given by the following formula:
$V \ = \ \dfrac{ 1 }{ 3 } \pi r’^2 h’$
Substituting value of r’:
$V \ = \ \dfrac{ 1 }{ 3 } \pi \bigg ( \dfrac{ 1 }{ 2 } h’ \bigg )^2 h’$
$\Rightarrow V’ \ = \ \dfrac{ 1 }{ 12 } \pi h’^3$
Differentiating both sides:
$\dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 12 } \pi \dfrac{ d }{ dt } ( h’^3 )$
$\Rightarrow \dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 12 } \pi ( 3 h’^2 \dfrac{ dh’ }{ dt } )$
$\Rightarrow \dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 4 } \pi h’^2 \dfrac{ dh’ }{ dt }$
Substituting value of $\dfrac{ V’ }{ dt } \ = \ 20$ and $h’ \ = \ 5 inches$:
$20 \ = \ \dfrac{ 1 }{ 4 } \pi ( 5 )^2 \dfrac{ dh’ }{ dt }$
$\Rightarrow 20 \ = \ \dfrac{ 25 }{ 4 } \pi \dfrac{ dh’ }{ dt }$
$\Rightarrow \dfrac{ dh’ }{ dt } \ = \ \dfrac{ 20 \times 4 }{ 25 \pi } \ = \ \dfrac{ 16 }{ 5 \pi }$
## Numerical Result:
$\dfrac{ dh }{ dt } \ = \ \dfrac{ 5 }{ 4 \pi }$
$\dfrac{ dh’ }{ dt } \ = \ \dfrac{ 16 }{ 5 \pi }$
## Example
For the same scenario given above, what is the rate of rise of the level when the level in the conical filter is 3 inches?
Recall:
$\dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 4 } \pi h’^2 \dfrac{ dh’ }{ dt }$
Substituting values:
$20 \ = \ \dfrac{ 1 }{ 4 } \pi ( 3 )^2 \dfrac{ dh’ }{ dt }$
$\Rightarrow 20 \ = \ \dfrac{ 9 }{ 4 } \pi \dfrac{ dh’ }{ dt }$
$\Rightarrow \dfrac{ dh’ }{ dt } \ = \ \dfrac{ 20 \times 4 }{ 9 \pi } \ = \ \dfrac{ 80 }{ 9 \pi }$ |
Calculus was found by Isaac Newton and Gottfried Wilhelm Leibnitz in the 17th century as a study of Continuous change. It is a part of Mathematics, which is a study of derivatives, integrals, limits, functions, and the Taylor series. Calculus Derivatives is the change in a function and this function is related to the relationship between two variables so it’s the ratio of the differentials.
## Different types of Calculus derivatives
Calculus has two main concept explained below:
1)Differential calculus
2)Integral Calculus
• Differential Calculus
Differential Calculus is the change in a variable. The derivative ∂y/ ∂x is another function of x which can be differentiated. The Derivative of ∂y/ ∂x is also called the second derivative of y and is denoted by ∂^² y/ ∂^² 2. Similarly, in general, the Nth derivative of y is denoted by ∂^ny/ ∂x_n.
Problem 1:
If y=e^{ax}\sin {bx} ,
prove that y_2-2ay_1+(a^2+b^2)y=0
Solution:
We have, y=e^{ax}\sin {bx} \\
y_1= e^{ax}(\cos {bx.b} )+\sin {bx}(e^{ax}.a) =be^{ax}\cos bx+at \\
y_1-ay=be^{ax}\cos bx\\
Differentiation on both side
y_2-ay_1=be^{ax}(-\sin bx.b)+b\cos bx(e^{ax}.a)=-b^2y+a(y_1-ay)
As a result,y_2-2ay_1+(a^2+b^2)y=0
• Integral Calculus
In Mathematics, Integral is the area under the graph of a function for some definite interval and is used to find the volume, area, Displacement, and other combining infinitesimal data. It is one of the main operations of Calculus and inverse operation. It is also referred as Anti-derivative or indefinite integral.
F(x)= \int f(x) dx
For example,
\int\sin^n x {\text dx} =-\frac{sin^{n-1} x cos x}{n}+\frac{n-1}{n}\int\sin^{n-2} x {\text dx}
## Quotient Rule in Calculus derivatives
Here, we will explain to you about the calculus quotient rule with an example:
This rule is a special rule in calculus derivates which is defined as a formal method of dividing the function of one differential to another.
Hence, we have mentioned some techniques to master this rule in calculus.
Calculus Quotient rule
if y=\frac{u}{v}
then, \frac{\text{d}y}{\text{d}x}=\frac{v\frac{\text{d}u}{\text{d}x}+u\frac{\text{d}v}{\text{d}x}}{v^2}
Problem 2:
Solve using the quotient rule y=\frac{3}{x+1}
Solution:
We want to differentiate y=\frac{3}{x+1}
As per the rule we can see u = 3 and v = x+1.
So, the derivative of these two functions would be:
\frac{\text{d}u}{\text{d}x}=0 and \frac{\text{d}v}{\text{d}x}=1
Therefore, when we put this in Calculus quotient rule
\frac{\text{d}y}{\text{d}x}=\frac{(v \frac{\text{d}u}{\text{d}x}-u \frac{\text{d}v}{\text{d}x})}{v^2}
=\frac{((x+1)(0)-(3)(1))}{(x+1)^2}
=\frac{\text{(x+1)}{(0)}-\text{(3)}{(1)}}{\text{(x+1)}^2}
=\frac{-3}{(x+1)^2}
As a result, \frac{-3}{(x+1)^2} is the differential.
In Conclusion, We have learned about the integrals and derivatives Calculus with example that helped us to understand the changes between the values which are related by a function. Calculus Derivatives mainly focuses on two concepts of differential calculus which helps to find the rate of change of a quantity, whereas integral calculus helps to find the quantity when the rate of change is known. Therefore, this article will give you the basic knowledge of calculus derivates.
We hope that you learned the concept of calculus derivative and integral along with Quotient rule with examples . Keep learning keep sharing. Follow us on Facebook and Instagram. |
Describe the sampling distribution of sample means.
Learning Objectives
• Describe the sampling distribution of sample means.
• Draw conclusions about a population mean from a simulation.
How Sample Means Vary in Random Samples
In Inference for Means, we work with quantitative variables, so the statistics and parameters will be means instead of proportions.
We begin this module with a discussion of the sampling distribution of sample means. Our goal is to understand how sample means vary when we select random samples from a population with a known mean. We did this same type of thinking with sample proportions in the module Linking Probability to Statistical Inference to understand the distribution of sample proportions. Ultimately, we develop a probability model based on this sampling distribution. We use the probability model with an actual sample mean to test a claim about population mean or to estimate a population mean. This task is similar to the type of work we did in Inference for One Proportion with proportions when we tested hypotheses and created confidence intervals.
Birth Weights
The World Health Organization (WHO) monitors many variables to assess a population’s overall health. One of these variables is low birth weight. A birth weight under 2,500 grams is a low birth weight. Low birth weight is a categorical variable because the birth weight is either under 2,500 grams or it is not. The WHO collects data from hospitals and other health-care institutions and can use this sample data to find a confidence interval to estimate the proportion of all babies in a country with a low birth weight. This type of inference comes from Inference for One Proportion.
In this module, we work with quantitative variables. In this example, we use birth weight as a quantitative variable. To analyze the quantitative variable birth weight, we use means.
Suppose that babies in a town had a mean birth weight of 3,500 grams in 2005. This year, a random sample of 9 babies has a mean weight of 3,400 grams.
• The 3,500 is a parameter from a population. We use the Greek letter µ to represent it: µ = 3,500 grams.
• The 3,400 is a statistic from a sample, so we write $\overline{x}$ = 3,400 grams.
Obviously, this sample weighs less on average than the population of babies in the town. A decrease in the town’s mean birth weight could indicate a decline in overall health of the town. But does this sample give strong evidence that the town’s mean birth weight is less than 3,500 grams this year?
To answer this question, we need to understand how much the means from random samples vary. Would a sample be likely – or unlikely – to have a mean birth weight of 3,400 grams if the mean weight of all the babies is 3,500 grams?
We outline this investigation in the following diagram:
As before, the logic of inference is the same. Begin with a population with µ = 3,500, and take random samples of 9 babies at a time.
• If a sample mean of 3,400 is likely to occur when sampling from a population with µ = 3,500, then this sample could have come from a population with a mean of 3,500. The evidence from the sample therefore is not strong enough to reject the idea that µ = 3,500.
• If a sample mean of 3,400 is unlikely when sampling from a population with µ = 3,500, then the sample provides evidence that the mean weight for all babies in the population is less than 3,500.
Likely or unlikely? It depends on how much the sample means vary. We need to investigate the sampling distribution of sample means.
Learn By Doing
Refer to the previous example. These questions focus on how sample mean birth weights will vary. Use the simulation below to select a random sample of 9 babies from the town. Assume µ = 3,500. Repeat many times to observe how the mean birth weights for the samples vary. Then answer the questions.
In the next example, we predict what happens in the long run when we select many, many random samples of 9 babies at a time from a population with a mean birth weight of 3,500 grams. Then we watch a simulation to see if our predictions are correct.
Predicting the Behavior of Mean Birth Weights
Note: Means of samples randomly selected from a population are consequently random variables themselves because the means of random samples vary unpredictably in the short run but have a predictable pattern in the long run. Based on our intuition, what we experienced with the simulation, and what we learned about the behavior of samples in previous modules, we might expect the following about the distribution of sample means that come from a population where µ = 3,500:
Center: Some sample means will be on the low side – say 3,000 grams or so – while others will be on the high side – say 4,000 grams or so. In repeated sampling, we might expect that the random samples will average out to the underlying population mean of 3,500 grams. In other words, the mean of the sample means will be µ. This is exactly what we observed in the case of proportions in Linking Probability to Statistical Inference. There, the mean of sample proportions was the population proportion.
Spread: For large samples, we might expect that sample means will not stray too far from the population mean of 3,500. Sample means lower than 3,000 or higher than 4,000 might be surprising. For smaller samples, we would be less surprised by sample means that varied quite a bit from 3,500. In others words, we might expect greater variability in sample means for smaller samples. So sample size again plays a role in the spread of the distribution of sample statistics, just as we observed for sample proportions.
Shape: Sample means closest to 3,500 will be the most common, with sample means far from 3,500 in either direction progressively less likely. In other words, the shape of the distribution of sample means should be somewhat normal. This, again, is what we saw when we looked at sample proportions.
The discussion of shape, center, and spread here is not very specific. We work toward making these statements more specific over the next two pages.
Now let’s see if our predictions about the sampling distribution are correct. In the next simulation, we randomly select thousands of random samples of 9 babies each.
WalkThrough Simulation
The distribution of the values of the sample mean $\overline{x}$ in repeated samples is called the sampling distribution of $\overline{X}$.
Learn By Doing
At this point, you may be wondering if we should use a larger sample to answer our question. Will our conclusion change if we increase the number of babies in the sample? We investigate this question next. |
#### Need solution for RD Sharma maths Class 12 Chapter 21 Differential Equation Exercise Multiple Choice Question Question 19 textbook solution.
Answer : $\text { (c) } 2 x+y^{3}-3 x^{2} y=0$
Hint :
\begin{aligned} &d(x y)=x d y+y d x \\ &\text{and}\; d\left(\frac{y}{x}\right)=\frac{x d y-y d x}{x^{2}} \end{aligned}
Given : $y\left(x+y^{3}\right) d x=x\left(y^{3}-x\right) d y$
Multiply and divide by $x^{2}$ in first term
\begin{aligned} &\Rightarrow x^{2} y^{3}\left(\frac{x d y-y d x}{x^{2}}\right)-x(x d y+y d x)=0 \\ &\Rightarrow x^{2} y^{3} d\left(\frac{y}{x}\right)-x d(x y)=0 \quad\left[d\left(\frac{y}{x}\right)=\frac{x d y-y d x}{x^{2}}, d(x y)=x d y+y d x\right] \end{aligned}
$\Rightarrow x d(x y)-x^{2} y^{3} d\left(\frac{y}{x}\right)=0$
Divide by $x^{3}y^{2}$
$\Rightarrow \frac{d(x y)}{x^{2} y^{2}}=\frac{y}{x} d\left(\frac{y}{x}\right)$
Integrate both sides
\begin{aligned} &\Rightarrow-\frac{1}{x y}=\frac{\left(\frac{y}{x}\right)^{2}}{2}+C \\ &\Rightarrow-\frac{1}{x y}-\frac{1}{2}\left(\frac{y}{x}\right)^{2}-C=0 \end{aligned}
\begin{aligned} &\Rightarrow-\left[\frac{1}{x y}+\frac{1}{2}\left(\frac{y}{x}\right)^{2}+C\right]=0 \\ &\Rightarrow \frac{1}{x y}+\frac{1}{2}\left(\frac{y}{x}\right)^{2}+C=0 \end{aligned}
Now, curve passes through $(1,1)$
\begin{aligned} &\Rightarrow \frac{1}{1}+\frac{1}{2}(1)+C=0 \\ &\Rightarrow \frac{2}{2}+\frac{1}{2}+C=0 \\ &\Rightarrow \frac{3}{2}+C=0 \end{aligned}
\begin{aligned} &\Rightarrow C=-\frac{3}{2} \\ &\Rightarrow \frac{1}{x y}+\frac{y^{2}}{2 x^{2}}-\frac{3}{2}=0 \\ &\Rightarrow \frac{2 x+y^{3}-3 x^{2} y}{2 x^{2} y}=0 \\ &\Rightarrow 2 x+y^{3}-3 x^{2} y=0 \end{aligned} |
# Kinematics Problems and Equations
```Mechanics: The study of motion of objects ;]
Kinematics… WHAT IS THAT?
› The science of describing the motion of objects
Measure by using graphs, diagrams, numbers, and
equations
Scalars- quantities that are fully explained by
magnitude alone
Vectors- quantities that are explained by
magnitude and direction
› THE SIGN MATTERS!
The sign matters b/c it determines the
direction of the object
Distance and displacement both show how an
object moves
› Distance is a scalar and shows the amount of
distance covered
› Displacement is a vector and shows the overall
change in an objects position
Speed and velocity are both important to
know in order to understand kinematics
In some instances, you may not know a variable of
an object’s motion and must solve for it!
› A car has a velocity of 23 m/s East; and has an
acceleration of 9 m/s2. What don’t we know?
› Use the equations:
1 2
x vi t at
2
(v f vi )
2
x
2a
2
v f vi at
vi v 2
x
t
2
Make a list of givens
Identify what needs to be found
Jot what equation will be used
This will limit your chances of making error on
v= velocity; a=acceleration; t= time ,
Δx
=displacement
Ima Hurryin is approaching a stoplight moving with a
velocity of +30.0 m/s. The light turns yellow, and Ima
applies the brakes and skids to a stop. If Ima's
acceleration is -8.00 m/s2, then determine the
displacement of the car during the skidding process.
To prepare for solving we would make a list:
› Vi = + 30.0 m=/s
› Vf = O m/s
› a= -8 m/s2
› x = ?
Make note of
negative and
positive signs!
Make a picture:
What equation
should we use?
Solve…
(v f vi )
2
› Equation: x
2
2a
(0 30 )
x
= 56.25 m
2(8)
2
2
Ben Rushin is waiting at a stoplight. When it finally
turns green, Ben accelerated from rest at a rate of a
6.00 m/s2 for a time of 4.10 seconds. Determine the
displacement of Ben's car during this time period.
List:
› Vi = 0 m/s
› t= 4.10 s
› a= 6 m/s2
› d= ?
Picture:
Equation: x v t 1 at 2
i
2
1
2
x (0)( 4.10) (6)( 4.10) = 50.43 m
2
Free Fall
Falling under the sole influence of gravity
Downward acceleration of 9.8 m/s/s
The motion of an object in free fall can be described
by kinematics equations
If an object is dropped from an elevated height its
initial velocity is zero
If an object is projected upwards in a vertical
direction it will slow down as it rises. When its peak
is reached the velocity is 0 m/s.
If an object is projected upwards in a vertical
direction the velocity at which its projected is equal in
magnitude and opposite in direction to the velocity it
has when it returns to the original height
Example 3
Luke Autbeloe drops a pile of roof shingles
from the top of a roof located 8.52 meters
above the ground. Determine the time
required for the shingles to reach the ground.
To prepare for solving we would make a list:
› Vi = 0 m=/s
› d = -8.52 m
› a= -9.8 m/s2
› t= ?
Example 3 Continued…
Make a picture:
What equation
should we use?
Example 3 Continued…
Solve…
› Equation:
-8.52 m = (0 m/s)*(t) + 0.5*(-9.8
m/s2)*(t) 2
› Solution:
t= 1.32 s
Example 4 Continued…
A race car accelerates uniformly from 18.5
m/s to 46.1 m/s in 2.47 seconds. Determine
the acceleration of the car and the distance
traveled.
To prepare for solving we would make a list:
› Vi = 18.5 m=/s
› Vf = 46.1 m/s
› t= 2.47 s
› d= ?
› a= ?
Example 4 Continued…
a = (delta v)/t
a = (46.1 m/s - 18.5 m/s)/(2.47 s)
a = 11.2 m/s2
d = vi*t + 0.5*a*t2
d = (18.5 m/s)*(2.47 s)+ 0.5*(11.2 m/s2)*(2.47
s) 2
d = 45.7 m + 34.1 m
d = 79.8 m
``` |
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• Level: GCSE
• Subject: Maths
• Word count: 2880
# My aim is to see if theres a relation between T total and T number and I will then work out the algebraic expressions so that the 50th term would be found out using the formula.
Extracts from this document...
Introduction
MATHS
T – Totals COURSEWORK
MR.UDDIN
T – Totals
PART 1:
My aim is to see if there’s a relation between T – total and T – number and I will then work out the algebraic expressions so that the 50th term would be found out using the formula.
1 2 3 12 22
The total of the numbers inside the T – shape is called the T – total.
The bottom number of the T is called the T – number.
So in this case the T – number would be 22.
The T – total is 22 + 12 + 2 + 1 + 3 = 40
T – Total in a 10 by 10 grid:
2 3 4 13 23
T – Total = 2 + 3 + 4 + 13 + 23 = 45
T – Number = 23
Another try:
3 4 5 14 24
T – Total = 3 + 4 + 5 + 14 + 24 = 50
T – Number = 24
I’ll put it in a table to see the difference:
T – number T – total Difference 22 40 23 45 5 24 50 5
As you can see that when the T – number increases by 1 the T – total increases by 5. This is because there are 5 boxes in the T –shape and each number increases by 1 will add up to five extra at the end.
Now I predict that T – total is going to be 55 in a 10 by 10 grid, now to check if it is right from a 10 by 10 grid:
4 5 6 15 25
4 + 5 + 6 + 15 + 25 = 55.
This here shows that my prediction was correct.
For a 10 by 10 grid, this is how the T – shape will look:
T-21 T-20 T-19 T-10 T
Middle
th term would be the following:
5 x 50 – 7 = 243
Now I am going to do a 6 by 6 90º anticlockwise:
This would be the same as this:
T + 4 T-2 T - 8 T-1 T
T + 4 + T – 2 + T – 8 + T – 1 + T = 5T -7
16 10 4 11 12
I have to put my formula to the test:
5 x 12 – 7 = 53
I need to test if the formula is right:
16 + 10 + 4 + 11 + 12 = 53
So the 50th term would be the following:
5 x 50 – 7 = 243
Now I am going to put the results in the table and see if there is a difference in the T – Shapes when I changed the grid size and rotated the T – Shape 90º anticlockwise.
Grid Size Formula Difference 9 by 9 5T – 7 0 8 by 8 5T - 7 0 7 by 7 5T - 7 0
As you can see that there is no difference on these formulae, this is because when you put it into a T – shape, the right hand box and the left hand box will decrease as the grid size decreases. However this wouldn’t be a problem as both numbers has decreased by 1.
Example:
7 by 7 grid:
T + 5 T-2 T - 9 T-1 T
5 – 2 – 9 – 1 = -7
6 by 6 grid:
T + 4 T-2 T - 8 T-1 T
4 – 2 – 8 – 1 = -7
Now I am going to rotate the T - shape 90º clockwise to investigate the difference between the T – number and T – total in different size grids.
Conclusion
CONCLUSION:
During this coursework I found out many things about the T – shapes, the first thing was that the algebraic expression change as the grid size change, however it’s answer was negative and always was in the 7 times table.
Example: 5T – 70 & 5T - 63
However when the T –shape is rotated 90° anticlockwise, the grid size didn’t make a difference to the algebraic expression, infact it didn’t even change, although the numbers in the algebraic expression was negative.
Example: 5T – 7
In addition when I rotated the grid size 90° clockwise, the algebraic expression didn’t change, however it was a positive number, which is an exact opposite of what I found out when I rotated the T – shape 90° anticlockwise.
Example: 5T + 7
However when I rotated the T – shape 180°, the grid size changed the algebraic expression, however as the first it was always in the 7 times but in this case the number was a positive, because it was an exact opposite of the first one.
Example: 5T + 70 & 5T + 63
Moreover, when I did vector translations in my T- shape I found that if there was a vector as so: , the number would decrease by 5, however if there was a vector as so: , the number would increase by 5.
Example:
= 5T – 7P + 5a
= 5T – 7p – 5bP
= 5T – 7P + 5a – 5bP
This student written piece of work is one of many that can be found in our GCSE T-Total section.
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# Related GCSE T-Total essays
1. ## T-Total Maths
N = 14 T = (5 x 14)-7 = 70-7 = 63 Rotation of 900 8 by 8 T-number and T-total table T-number T-total 9 52 10 57 11 63 12 67 Here I have added a prediction of mine when I realized the pattern of the sequence, which goes up by 5 each time.
2. ## T Total and T Number Coursework
is a pattern: For +90 degrees the formula I have found is 5n+7 For -90 degrees the formula I have found is 5n-7 For +180 degrees the formula I have found is 5n+7g These three formulas work perfectly on this grid size but I shall have to do another grid size to be sure that they are correct.
1. ## Urban Settlements have much greater accessibility than rural settlements. Is this so?
is an awful lot of traffic going through Bexley village at most times of day. This may slow buses down. Bexley has many more bus routes than South Darenth, suggesting that there is not enough demand for a varied amount of bus routes.
2. ## T-Shapes Coursework
2 Using Pattern 2 above, we can say that the first term in the sequence of our numbers is the Middle Number plus the Tail Length x 10, or n + 10l. Since l is 1 in the first term of the sequence, the first term is therefore n + 10.
1. ## T-Total. I will take steps to find formulae for changing the position of the ...
51 52 53 54 55 56 57 58 59 60 61 62 63 64 I used the same method for the 8 x 8 grid above as I did for the 9 x 9 grid. The T is in the same place but the numbers have changed.
2. ## T-Shapes Coursework
26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 Tt = (5 x 47)
1. ## T totals. In this investigation I aim to find out relationships between grid sizes ...
+3 or -2) and g is the grid width. Horizontal Again, we shall use our standard gird size and position to establish our basic starting point; 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24
2. ## T totals - translations and rotations
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 My T-Number is 21 which
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Q. 205.0( 1 Vote )
Factorize 2x
For factorizing this cubic expression, no identity is useful.
Thus, we need to use factorization by regrouping terms.
We can write, –x2 = x2 – 2x2 and –13x = –x – 12x
2x3 – x2 – 13x – 6 = 2x3 + x2 – 2x2 – x – 12x – 6
We have, –2x2 = (–x) × 2x and –12x = (–6) × 2x
2x3 – x2 – 13x – 6 = 2x3 + x2 + (–x) × 2x – x + (–6) × 2x – 6
Observe that x2 is common for the first two terms, –x is common for the next two terms and –6 is common for the last two terms.
2x3 – x2 – 13x – 6 = x2(2x + 1) + (–x)(2x + 1) + (–6)(2x + 1)
Now, (2x + 1) is the common term.
2x3 – x2 – 13x – 6 = (2x + 1)(x2 – x – 6)
So, one factor of the given expression is (2x + 1). Now, we need to factorize (x2 – x – 6).
For factorizing, this expression, split the middle term in such a way that the product of the coefficients of the new terms is equal to the product of the coefficients of the first and last terms in the expression.
Here, product of co-effs of first and last terms = 1 × (–6) = –6
So, if the middle term –x is split into two terms say ax, bx, then a + b = –1 and ab = –6.
Observe that values –3 and 2 satisfy these equations.
x2 – x – 6 = x2 – 3x + 2x – 6
Observe that x is common for the first two terms and 2 is common for the next two terms.
x2 – 3x + 2x – 6 = x(x – 3) + 2(x – 3)
Now, (x – 3) is the common term.
x2 – x – 6 = (x – 3)(x + 2)
Thus, the factors of x2 – x – 6 are (x – 3) and (x + 2).
2x3 – x2 – 13x – 6 = (2x + 1)(x – 3)(x + 2)
Hence, the factors of 2x3 – x2 – 13x – 6 are (2x + 1), (x – 3) and (x + 2).
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# How do you simplify -2(-2u-w)-3(5w+3u)?
Jun 30, 2018
$- 5 u - 13 w$
#### Explanation:
$- 2 \left(- 2 u - w\right) - 3 \left(5 w + 3 u\right)$
Use the distributive property to simplify $- 2 \left(- 2 u - w\right)$ and $- 3 \left(5 w + 3 u\right)$:
Following this image, we know that:
$\textcolor{b l u e}{- 2 \left(- 2 u - w\right) = \left(- 2 \cdot - 2 u\right) + \left(- 2 \cdot - w\right) = 4 u + 2 w}$
and
$\textcolor{b l u e}{- 3 \left(5 w + 3 u\right) = \left(- 3 \cdot 5 w\right) + \left(- 3 \cdot 3 u\right) = - 15 w - 9 u}$
Combine the two expressions:
$4 u + 2 w - 15 w - 9 u$
Combine like terms:
$- 5 u - 13 w$
Hope this helps! |
# SSS Triangle Congruence
## Three sets of equal side lengths determine congruence.
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SSS Triangle Congruence
How can you use the SSS criterion for triangle congruence to show that the triangles below are congruent?
### SSS Triangles
If two triangles are congruent it means that all corresponding angle pairs and all corresponding sides are congruent. However, in order to be sure that two triangles are congruent, you do not necessarily need to know that all angle pairs and side pairs are congruent. Consider the triangles below.
In these triangles, you can see that all three pairs of sides are congruent. This is commonly referred to as “side-side-side” or “SSS”.
The SSS criterion for triangle congruence states that if two triangles have three pairs of congruent sides, then the triangles are congruent.
In the examples, you will use rigid transformations to show why the above SSS triangles must be congruent overall, even though you don't know the measures of any of the angles.
Let's take a look at some example problems.
1. Perform a rigid transformation to bring point \begin{align*}E\end{align*} to point \begin{align*}B\end{align*}.
2. Draw a vector from point \begin{align*}E\end{align*} to point \begin{align*}B\end{align*}. Translate \begin{align*}\triangle DEF\end{align*} along the vector to create \begin{align*}\triangle D^\prime E^\prime F^\prime\end{align*}.
3. Rotate \begin{align*}\triangle D^\prime E^\prime F^\prime\end{align*} to map \begin{align*}\overline{D^\prime E^\prime}\end{align*} to \begin{align*}\overline{AB}\end{align*}.
4. Measure \begin{align*}\angle{ABD^\prime}\end{align*}. In this case, \begin{align*}m\angle{ABD^\prime}=26^\circ\end{align*}.
5. Rotate \begin{align*}\triangle D^\prime E^\prime F^\prime\end{align*} clockwise that number of degrees to create \begin{align*}\triangle D^{\prime\prime}E^{\prime\prime} F^{\prime\prime}\end{align*}. Note that because \begin{align*}\overline{DE}\cong \overline{AB}\end{align*} and rigid transformations preserve distance, \begin{align*}\overline{D^{\prime\prime}E^{\prime\prime}}\end{align*} matches up perfectly with \begin{align*}\overline{AB}\end{align*}.
6. Reflect \begin{align*}\triangle D^{\prime\prime}E^{\prime\prime} F^{\prime\prime}\end{align*} to map it to \begin{align*}\triangle ABC\end{align*}. Can you be confident that the triangles are congruent?
7. Reflect \begin{align*}\triangle D^{\prime\prime}E^{\prime\prime} F^{\prime\prime}\end{align*} across \begin{align*}\overline{D^{\prime\prime}E^{\prime\prime}}\end{align*} (which is the same as \begin{align*}\overline{AB}\end{align*}).
In this case, it looks like the triangles match up exactly and are therefore congruent, but how can you always be confident that \begin{align*}F^{\prime \prime}\end{align*} will map to \begin{align*}C\end{align*}? Consider the previous step, with the two triangles below:
You know that wherever \begin{align*}F^{\prime \prime}\end{align*} ends up after it is reflected, it has to stay 5 units away from \begin{align*}E^{\prime \prime}\end{align*} and 9 units away from \begin{align*}D^{\prime \prime}\end{align*}. Create a circle centered at \begin{align*}E^{\prime \prime}\end{align*} with radius 5 units to find all the points besides \begin{align*}F^{\prime \prime}\end{align*} that are 5 units away from \begin{align*}E^{\prime \prime}\end{align*}. Also create a circle centered at \begin{align*}D^{\prime \prime}\end{align*} with radius 9 units to find all the points besides \begin{align*}F^{\prime \prime}\end{align*} that are 9 units away from \begin{align*}D^{\prime \prime}\end{align*}. Notice that there are only two points in the whole plane that are both 5 units away from \begin{align*}E^{\prime \prime}\end{align*} and 9 units away from \begin{align*}D^{\prime \prime}\end{align*}: point \begin{align*}F^{\prime \prime}\end{align*} and point \begin{align*}C\end{align*}. Since reflections preserve distance, when \begin{align*}F^{\prime \prime}\end{align*} is reflected, it must end up at point \begin{align*}C\end{align*}. Therefore, a reflection will always map \begin{align*}\triangle D^{\prime\prime}E^{\prime\prime} F^{\prime\prime}\end{align*} to \begin{align*}\triangle ABC\end{align*} at this step.
This means that even though you didn't know the angle measures, because you knew three pairs of sides were congruent, the triangles had to be congruent overall. At this point you can use the SSS criterion for showing triangles are congruent without having to go through all of these transformations each time (but make sure you can explain why SSS works in terms of the rigid transformations!).
### Examples
#### Example 1
Earlier, you were asked how can you use the SSS criterion for triangle congruence to show that the triangles below are congruent.
Because these are right triangles, you can use the Pythagorean Theorem to find the third side of each triangle. The third side of each triangle will be \begin{align*}\sqrt{15^2-12^2}=9\end{align*}. Now you know that all three pairs of sides are congruent, so the triangles are congruent by SSS.
In general, anytime you have the hypotenuses congruent and one pair of legs congruent for two right triangles, the triangles are congruent. This is often referred to as “HL” for “hypotenuse-leg”. Remember, it only works for right triangles because you can only use the Pythagorean Theorem for right triangles.
#### Example 2
Are the following triangles congruent? Explain.
Yes, the triangles are congruent by SSS.
#### Example 3
Are the following triangles congruent? Explain.
There is not enough information to determine if the triangles are congruent. You need to know how the unmarked side compares to the other sides, or if there are right angles.
#### Example 4
Are the following triangles congruent. Explain.
What additional information would you need in order to be able to state that the triangles below are congruent by HL?
You would need to know that the triangles are right triangles in order to use HL.
### Review
1. What does SSS stand for? How is it used?
2. What does HL stand for? How is it used?
3. Draw an example of two triangles that must be congruent due to SSS.
4. Draw an example of two triangles that must be congruent due to HL.
For each pair of triangles below, state if they are congruent by SSS, congruent by HL, or if there is not enough information to determine whether or not they are congruent.
5.
6.
7.
8.
9.
10. What is the minimum additional information you would need in order to be able to state that the triangles below are congruent by HL?
11. What is the minimum additional information you would need in order to be able to state that the triangles below are congruent by SSS?
12. What is the minimum additional information you would need in order to be able to state that the triangles below are congruent by HL?
13. Point \begin{align*}A\end{align*} is the center of the circle below. What is the minimum additional information you would need in order to be able to state that the triangles below are congruent by SSS?
14. If you can show that two triangles are congruent by HL, can you also show that they are congruent by SAS?
15. Show how the SSS criterion for triangle congruence works: use rigid transformations to help explain why the triangles below are congruent.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Congruent Congruent figures are identical in size, shape and measure.
Distance Formula The distance between two points $(x_1, y_1)$ and $(x_2, y_2)$ can be defined as $d= \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.
H-L (Hypotenuse-Leg) Congruence Theorem If the hypotenuse and leg in one right triangle are congruent to the hypotenuse and leg in another right triangle, then the two triangles are congruent.
Side Side Side Triangle A side side side triangle is a triangle where the lengths of all three sides are known quantities.
SSS SSS means side, side, side and refers to the fact that all three sides of a triangle are known in a problem.
Triangle Congruence Triangle congruence occurs if 3 sides in one triangle are congruent to 3 sides in another triangle.
Rigid Transformation A rigid transformation is a transformation that preserves distance and angles, it does not change the size or shape of the figure. |
# What is 201/523 as a decimal?
## Solution and how to convert 201 / 523 into a decimal
201 / 523 = 0.384
Fraction conversions explained:
• 201 divided by 523
• Numerator: 201
• Denominator: 523
• Decimal: 0.384
• Percentage: 0.384%
201/523 or 0.384 can be represented in multiple ways (even as a percentage). The key is knowing when we should use each representation and how to easily transition between a fraction, decimal, or percentage. Decimals and Fractions represent parts of numbers, giving us the ability to represent smaller numbers than the whole. But in some cases, fractions make more sense, i.e., cooking or baking and in other situations decimals make more sense as in leaving a tip or purchasing an item on sale. After deciding on which representation is best, let's dive into how we can convert fractions to decimals.
201 / 523 as a percentage 201 / 523 as a fraction 201 / 523 as a decimal
0.384% - Convert percentages 201 / 523 201 / 523 = 0.384
## 201/523 is 201 divided by 523
The first step in converting fractions is understanding the equation. A quick trick to convert fractions mentally is recognizing that the equation is already set for us. All we have to do is think back to the classroom and leverage long division. Fractions have two parts: Numerators on the top and Denominators on the bottom with a division symbol between or 201 divided by 523. Now we divide 201 (the numerator) into 523 (the denominator) to discover how many whole parts we have. Here's 201/523 as our equation:
### Numerator: 201
• Numerators are the portion of total parts, showed at the top of the fraction. 201 is one of the largest two-digit numbers you'll have to convert. 201 is an odd number so it might be harder to convert without a calculator. Large numerators make converting fractions more complex. Let's take a look at the denominator of our fraction.
### Denominator: 523
• Denominators represent the total parts, located at the bottom of the fraction. 523 is one of the largest two-digit numbers to deal with. But 523 is an odd number. Having an odd denominator like 523 could sometimes be more difficult. Have no fear, large two-digit denominators are all bark no bite. Now let's dive into how we convert into decimal format.
## How to convert 201/523 to 0.384
### Step 1: Set your long division bracket: denominator / numerator
$$\require{enclose} 523 \enclose{longdiv}{ 201 }$$
Use long division to solve step one. Yep, same left-to-right method of division we learned in school. This gives us our first clue.
### Step 2: Extend your division problem
$$\require{enclose} 00. \\ 523 \enclose{longdiv}{ 201.0 }$$
Uh oh. 523 cannot be divided into 201. So that means we must add a decimal point and extend our equation with a zero. Even though our equation might look bigger, we have not added any additional numbers to the denominator. But now we can divide 523 into 201 + 0 or 2010.
### Step 3: Solve for how many whole groups you can divide 523 into 2010
$$\require{enclose} 00.3 \\ 523 \enclose{longdiv}{ 201.0 }$$
Now that we've extended the equation, we can divide 523 into 2010 and return our first potential solution! Multiply by the left of our equation (523) to get the first number in our solution.
### Step 4: Subtract the remainder
$$\require{enclose} 00.3 \\ 523 \enclose{longdiv}{ 201.0 } \\ \underline{ 1569 \phantom{00} } \\ 441 \phantom{0}$$
If you hit a remainder of zero, the equation is done and you have your decimal conversion. If you have a remainder over 523, go back. Your solution will need a bit of adjustment. If you have a number less than 523, continue!
### Step 5: Repeat step 4 until you have no remainder or reach a decimal point you feel comfortable stopping. Then round to the nearest digit.
In some cases, you'll never reach a remainder of zero. Looking at you pi! And that's okay. Find a place to stop and round to the nearest value.
### Why should you convert between fractions, decimals, and percentages?
Converting fractions into decimals are used in everyday life, though we don't always notice. Remember, they represent numbers and comparisons of whole numbers to show us parts of integers. Same goes for percentages. Though we sometimes overlook the importance of when and how they are used and think they are reserved for passing a math quiz. But each represent values in everyday life! Without them, we’re stuck rounding and guessing. Here are real life examples:
### When you should convert 201/523 into a decimal
Sports Stats - Fractions can be used here, but when comparing percentages, the clearest representation of success is from decimal points. Ex: A player's batting average: .333
### When to convert 0.384 to 201/523 as a fraction
Meal Prep - Body builders need to count macro calories. One of the ways of doing this is measuring every piece of food consumed. This is through halves and quarters in order to keep it consistent.
### Practice Decimal Conversion with your Classroom
• If 201/523 = 0.384 what would it be as a percentage?
• What is 1 + 201/523 in decimal form?
• What is 1 - 201/523 in decimal form?
• If we switched the numerator and denominator, what would be our new fraction?
• What is 0.384 + 1/2?
### Convert more fractions to decimals
From 201 Numerator From 523 Denominator What is 201/513 as a decimal? What is 191/523 as a decimal? What is 201/514 as a decimal? What is 192/523 as a decimal? What is 201/515 as a decimal? What is 193/523 as a decimal? What is 201/516 as a decimal? What is 194/523 as a decimal? What is 201/517 as a decimal? What is 195/523 as a decimal? What is 201/518 as a decimal? What is 196/523 as a decimal? What is 201/519 as a decimal? What is 197/523 as a decimal? What is 201/520 as a decimal? What is 198/523 as a decimal? What is 201/521 as a decimal? What is 199/523 as a decimal? What is 201/522 as a decimal? What is 200/523 as a decimal? What is 201/523 as a decimal? What is 201/523 as a decimal? What is 201/524 as a decimal? What is 202/523 as a decimal? What is 201/525 as a decimal? What is 203/523 as a decimal? What is 201/526 as a decimal? What is 204/523 as a decimal? What is 201/527 as a decimal? What is 205/523 as a decimal? What is 201/528 as a decimal? What is 206/523 as a decimal? What is 201/529 as a decimal? What is 207/523 as a decimal? What is 201/530 as a decimal? What is 208/523 as a decimal? What is 201/531 as a decimal? What is 209/523 as a decimal? What is 201/532 as a decimal? What is 210/523 as a decimal? What is 201/533 as a decimal? What is 211/523 as a decimal?
### Convert similar fractions to percentages
From 201 Numerator From 523 Denominator 202/523 as a percentage 201/524 as a percentage 203/523 as a percentage 201/525 as a percentage 204/523 as a percentage 201/526 as a percentage 205/523 as a percentage 201/527 as a percentage 206/523 as a percentage 201/528 as a percentage 207/523 as a percentage 201/529 as a percentage 208/523 as a percentage 201/530 as a percentage 209/523 as a percentage 201/531 as a percentage 210/523 as a percentage 201/532 as a percentage 211/523 as a percentage 201/533 as a percentage |
Question of solved question
# Question In figure, ABCD is a parallelogram. E and F are the mid-points of the sides AB and CD respectively. Prove that the line segments AF and CE trisect (divide into three equal parts) the diagonal BD.
Simplify the following expressions
(√(5) + √(2))2
Solution:
Explanation:
Which of the following rational numbers lies between 0 and - 1
A: 0
B: - 1
C: -1/4
D: 1/4
Solution:
Explanation:
0 and 1 cannot be found between 0 and 1.
In addition,0= o/4 and -1= -4/4
We can see that -1/4 is halfway between 0 and -1.
Hence, the correct option is (c) -1/4
Prove that the diagonals of a parallelogram bisect each other
Solution:
Explanation:
We must show that the diagonals of the parallelogram ABCD cross each other.
OA = OC & OB = OD, in other words.
Now AD = BC [opposite sides are equal] in ΔAOD and ΔBOC.
[alternative interior angle] ∠ADO = ∠CBO in ΔAOD and ΔBOC.
Similarly, ∠AOD = ∠BOC by ΔDAO = ΔBCO (ASA rule)
As a result, OA = OC and OB = OB [according to CPCT].
Hence, it is prove that the diagonals of a parallelogram bisect each other.
What is total surface area of sphere
Solution:
Explanation:
• The radius of the sphere affects the formula for calculating the sphere's surface area.
• If the sphere's radius is r and the sphere's surface area is S.
• The sphere's surface area is therefore stated as Surface Area of Sphere 4πr2, where ‘r’ is the sphere's radius.
• The surface area of a sphere is expressed in terms of diameter as S=4π(d/2)2, where d is the sphere's diameter.
Thus, total surface area of sphere is =4πr2.
Fill in the blanks If two adjacent angles are supplementary
they form a __________.
Solution:
Explanation:
• If the non-common sides of two angles form a straight line, they are called linear pair angles.
• The sum of the angles of two linear pairs is degrees.
• If the total of two angles is degrees, they are called supplementary angles. |
Top
# Linear Algebra
Linear Algebra usually consists of the linear set of equations as well as their transformations on it. It includes various topics such as matrices, vectors, determinants etc. Matrices and determinants are two very important topics of the linear algebra.
Linear Equations includes the topics mentioned as follows:
• Linear Equations
• Matrices
• Determinants
• Complex numbers
• Second degree equations
• Eigen values / Eigen vectors
• Vectors and its related operations
Related Calculators Linear Calculator Algebra Calculator Algebra Division Algebra Factoring
## Introduction to Linear Algebra
Linear algebra is the branch of mathematics concerning vector spaces, Matrices, vectors, transformations, eigenvectors/values, solving system of linear equations.
Let us find the inverse of A. If A = $\begin{bmatrix} 1&0 &-2 \\ 3& 1 &4 \\ 5& 2 & -3 \end{bmatrix}$
Given
|A| = -13 $\neq$ 0, so inverse exists.
a$_{11}$ = -3 - 8 = -11
a$_{12}$ = -(-9 - 20) = 29
a$_{13}$ = 6 - 5 = 1
a$_{21}$ = -(0 + 4) = -4
a$_{22}$ = -3 + 10 = 7
a$_{23}$ = -(2 - 0) = -2
a$_{31}$ = 0 + 2 = 2
a$_{32}$ = -(4 + 6) = -10
a$_{33}$ = 1 - 0 = 1
adj A = $\begin{bmatrix} -11&29 &1\\ -4& 7 &-2 \\ 2& -10 & 1 \end{bmatrix}$
A-1 = $\frac{1}{|A|}$ adj A
## Elementary Linear Algebra
Elementary linear algebra is concerning with the following topics: linear equations, matrices, determinant, complex numbers eigenvalues and eigenvectors, rank of matrices, transformation.
Let us find the eigen values of A. Where A = $\begin{bmatrix} 2&0 &0\\ 3& 5 &0 \\ 1& 2 & -3 \end{bmatrix}$
det of ($\lambda$ I - A) = $\begin{vmatrix} \lambda - 2&0 &0\\ -3& \lambda - 5 &0 \\ -1& -2 & \lambda + 3 \end{vmatrix}$
= ($\lambda_1$ - 2)($\lambda_2$ - 5)($\lambda_3$ + 3)
Therefore, the eigen values of A are 2, 5, -3.
There are various branches of linear algebra as follows:
Linear Equations: It is an equation of first degree for example, 2x + 5 = 9. We have to solve it and find the value of unknown variable in it.
Matrices: They are usually expressed as an array of numbers. For example
Determinants: It is a term which is calculated for each and every possible matrix. It defines the nature of the matrix.
Complex numbers: They are used to represent complex variables onto the space For eg : 2 + 3i where 2 is the real part and 3i is the complex part.
Second degree equations: They are often called as the quadratic equations. various methods are used to solve them like factoring, by making the square, the quadratic formula etc are present in linear algebra.
Eigen values: These are the scalars which are associated with the linear set of equations also known as the characteristic values and used with respect to matrices .
Vectors and its related operations: Vectors are the entities having both magnitude as well as direction. Different operations can be applied on it like addition, subtraction, multiplication etc.
## Applications of Linear Algebra
Linear algebra has a major number of applications. It is one of the very essential branch of mathematics. Some of its applications are as folloed:
• Constructing curves
• Least square approximation
• Traffic flow
• Electrical circuits
• Determinants
• Graph theory
• Cryptography
## Linear Algebra Practice Problems
A = $\begin{bmatrix} 8&3 &3\\ 3& 5 &1 \\ 1& 4 & -3 \end{bmatrix}$ |
## DEV Community
Armando C. Santisbon
Posted on • Updated on
# The Monty Hall problem
Suppose you're on a game show, and you're given the choice of three doors: $a$ , $b$ , and $c$ . Behind one door is a car; behind the others, goats. You pick a door, say $a$ , and the host, who knows what's behind the doors, opens another door, say $b$ , which has a goat. He then says to you, "Do you want to pick door $c$ ?" Is it to your advantage to switch your choice?
We'll go through three different explanations for the solution. I've found that different explanations will "click" with different people depending on their background and way of thinking. The first two will be mathematically rigorous and the last will be more informal but useful for building an intuition for why the math works out the way that it does.
## Explanation #1
### Conditional probability
What is the probability that the host would choose that door to open?
• If the car is in box $a$ then he could open either $b$ or $c$ to reveal a goat. The probability is $1/2$ .
• If the car were in box $b$ there's no way he would open $b$ . The probability is $0$ .
• If the car is in box $c$ then Monty would definitely open box $b$ to reveal a goat. The probability is $1$ .
So what's the probability that the prize is in box $c$ given that he opened $b$ ?
Luckily, we have a way of calculating a probability based on prior knowledge. That is, the probability of an event happening given the fact that another event has already happened:
ℹ️ Bayes' Theorem
$P(A|B)={P(B|A)P(A) \over P(B)}$
But wait, where did that come from?
💡 It can be derived from the conditional probability that tells us in how many cases do both events $A$ and $B$ happen out of the cases where $B$ happens:
$P(A|B)={P(A \cap B) \over P(B)}$
Similarly, $P(B|A)={P(A \cap B) \over P(A)}$
which means
$P(A \cap B)=P(B|A)P(A)$
and substituting in the expression for conditional probability yields Bayes' Theorem:
$P(A|B)={P(B|A)P(A) \over P(B)}$
So the probability that the car is in box $c$ given that he opened $b$ is:
$P(Car_c|Open_b)={P(Open_b|Car_c)P(Car_c) \over P(Open_b)}$
There are three possibilities for the denominator $P(Open_b)$ :
• $Car_a (Open_b|Car_a)=\frac 1 3 \cdot \frac 1 2=\frac 1 6$
• $Car_b(Open_b|Car_b)=\frac 1 3 \cdot 0=0$
• $Car_c(Open_b|Car_c)=\frac 1 3 \cdot 1=\frac 1 3$
so we'll add them up and plug them into our formula:
$P(Car_c|Open_b)={1 \cdot \frac 1 3 \over \frac 1 6+\frac 1 3}={\frac 1 3 \over \frac 1 2}={\frac 2 3}$
giving us a $2/3$ probability that the car is in the other box and therefore you should switch.
## Explanation #2
### Odds
There is another way to see Bayes' Theorem: As a way to understand two competing hypotheses $H_1$ and $H_2$ . We start by having some prior belief about the odds of those two. After observing some data, what are the new odds of the two hypotheses given the new data? In other words, how do we update our belief based on new evidence?
ℹ️
Let's start by specifying what we mean by odds. We mean how large is one probability relative to the other. Let's put it as a ratio. For example, if prior to seeing any data one hypothesis has $3/4$ probability of being true and the other has $1/4$ :
$\frac {P(H_1)} {P(H_2)}=\frac {3/4} {1/4}=\frac 3 1$
Meaning the odds of $H_1$ to $H_2$ are $3$ to $1$ .
So how do we incorporate the new data $D$ into our belief?
💡 By calculating the new probability of each hypothesis given the new data.
$\frac {P(H_1|D)} {P(H_2|D)}=\frac {\frac {P(H_1 \cap D)} {P(D)}} {\frac {P(H_2 \cap D)} {P(D)}}=\frac {P(H_1 \cap D)} {P(H_2 \cap D)}$
Now we'll use a mathematical trick of multiplying and dividing by the same number to get the same value but expressed in terms of the probability of seeing the observed data given each hypothesis.
$=\frac {P(H_1 \cap D)} {P(H_2 \cap D)} \cdot \frac {P(H_2)} {P(H_1)} \cdot \frac {P(H_1)} {P(H_2)}$
$=\frac {\frac {P(H_1 \cap D)} {P(H_1)}} {\frac {P(H_2 \cap D)} {P(H_2)}} \cdot \frac {P(H_1)} {P(H_2)}$
So,
$\frac {P(H_1|D)} {P(H_2|D)}=\frac {P(D|H_1)} {P(D|H_2)} \cdot \frac {P(H_1)} {P(H_2)}$
And so we have that as a ratio, the posterior odds equal the prior odds times the probabilities of generating the new data.
Now let's apply this to the Monty Hall Problem. Let
$H_1=$ Your door has the prize
$H_2=$ Your door has a goat
$D=$ The host reveals a goat
$\frac {P(H_1|D)} {P(H_2|D)}=\frac {1/1} {1/1} \cdot \frac {1/3} {2/3}=3/6=1/2$
Meaning $H_2$ is twice as likely as $H_1$ given the new data and therefore you should switch. If you want to convert that to probabilities:
$P(H_1)=\frac 1 {1+2}=1/3$
$P(H_2)=\frac 2 {1+2}=2/3$
This is also a nice way to see that the new data did not change the odds. You had a $1/3$ chance of getting the prize before the reveal and still do after the reveal if you keep your choice.
## Explanation #3
### An intuition
The explanations above using Bayes' Theorem are not the only way to look at this problem, though. You can arrive at the same result by realizing that the door you chose has a $1/3$ probability of containing the prize. That means there’s a $2/3$ probability that the prize is somewhere else. Then the game’s host makes it easy for you by turning the “somewhere else” into a single door.
In fact, you can generalize it like this: for a game with $n$ doors, choosing one gives you a $1/n$ chance of winning. That means $(n-1)/n$ chance that the prize is somewhere else. When the host knowingly eliminates $n-2$ doors with goats by opening them, they leave only one door you could switch to with, as we established, a probability of $(n-1)/n$ of containing the prize. Do this mental exercise with 100 doors or a billion doors and it'll become clear.
Still not convinced? You could just play the game yourself with 3 doors and keep track of how many times you win by switching. The more you play the closer and closer you'll get to winning $2/3$ of the time when you switch. |
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# The senior class spent 20% of its budget on a box of tomatoes. It then spent one - fourth of the remaining funds on a helium blimp. Finally, it spent $\dfrac{1}{3}$ of the remaining funds on an extra-large helicopter for the prom. Calculate the percentage of the original funds not spent. A) 20%B) 25%C) 40%D) 50%E) 60%
Last updated date: 22nd Jun 2024
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Hint: In the above question we have to find the remaining percentage of the given expenditure. So here we will assume the total funds $100\%$ to calculate the overall funds which are not spent.
Let the original funds be $100\%$.
So, it is given that $20\%$ of the funds are spent on tomatoes.
Therefore, we should subtract the percentage of funds spent on tomatoes from the original funds. We get,
$\Rightarrow 100 - \dfrac{{20}}{{100}} \times 100 = 80\%$
Thus, the remaining fund is $80\%$.
Also, it is given in the question that after spending on tomatoes one – fourth of the remaining funds are spent on helium blimp:
We get,
$\Rightarrow \dfrac{1}{4} \times 80\% = 20\%$
Now, our remaining funds is $80\% – 20\% = 60\%$
Again, it is given in the question that from the remaining funds that are $60\%$ he spent one – third on the extra-large helicopter so, the percentage spent is:
$\Rightarrow \dfrac{1}{3} \times 60\% = 20\%$
So, original funds not spent is ($60\% – 20\% = 40\%$)
$\therefore$ The percentage of the original funds not spent is $40\%$. The correct option is option (A).
Note:
Students should keep in mind that they should assume the original fund to be 100. This will help them in calculating all the expenses easily. You should also consider one thing before solving this question that if it is given that the senior class spent one – fourth of the “remaining funds” on a helium blimp. Then we will do our calculation on the remaining funds not on the original fund. |
# Network Topology Matrices
In the previous chapter, we discussed how to convert an electric circuit into an equivalent graph. Now, let us discuss the Network Topology Matrices which are useful for solving any electric circuit or network problem by using their equivalent graphs.
## Matrices Associated with Network Graphs
Following are the three matrices that are used in Graph theory.
• Incidence Matrix
• Fundamental Loop Matrix
• Fundamental Cut set Matrix
## Incidence Matrix
An Incidence Matrix represents the graph of a given electric circuit or network. Hence, it is possible to draw the graph of that same electric circuit or network from the incidence matrix.
We know that graph consists of a set of nodes and those are connected by some branches. So, the connecting of branches to a node is called as incidence. Incidence matrix is represented with the letter A. It is also called as node to branch incidence matrix or node incidence matrix.
If there are ‘n’ nodes and ‘b’ branches are present in a directed graph, then the incidence matrix will have ‘n’ rows and ‘b’ columns. Here, rows and columns are corresponding to the nodes and branches of a directed graph. Hence, the order of incidence matrix will be n × b.
The elements of incidence matrix will be having one of these three values, +1, -1 and 0.
• If the branch current is leaving from a selected node, then the value of the element will be +1.
• If the branch current is entering towards a selected node, then the value of the element will be -1.
• If the branch current neither enters at a selected node nor leaves from a selected node, then the value of element will be 0.
### Procedure to find Incidence Matrix
Follow these steps in order to find the incidence matrix of directed graph.
• Select a node at a time of the given directed graph and fill the values of the elements of incidence matrix corresponding to that node in a row.
• Repeat the above step for all the nodes of the given directed graph.
### Example
Consider the following directed graph.
The incidence matrix corresponding to the above directed graph will be
$$A = \begin{bmatrix}-1 & 1 & 0 & -1 & 0 & 0\\0 & -1 & 1 & 0 & 1 & 0\\1 & 0 & -1 & 0 & 0 & 1 \\0 & 0 & 0 & 1 & -1 & -1 \end{bmatrix}$$
The rows and columns of the above matrix represents the nodes and branches of given directed graph. The order of this incidence matrix is 4 × 6.
By observing the above incidence matrix, we can conclude that the summation of column elements of incidence matrix is equal to zero. That means, a branch current leaves from one node and enters at another single node only.
Note − If the given graph is an un-directed type, then convert it into a directed graph by representing the arrows on each branch of it. We can consider the arbitrary direction of current flow in each branch.
## Fundamental Loop Matrix
Fundamental loop or f-loop is a loop, which contains only one link and one or more twigs. So, the number of f-loops will be equal to the number of links. Fundamental loop matrix is represented with letter B. It is also called as fundamental circuit matrix and Tie-set matrix. This matrix gives the relation between branch currents and link currents.
If there are ‘n’ nodes and ‘b’ branches are present in a directed graph, then the number of links present in a co-tree, which is corresponding to the selected tree of given graph will be b-n+1.
So, the fundamental loop matrix will have ‘b-n+1’ rows and ‘b’ columns. Here, rows and columns are corresponding to the links of co-tree and branches of given graph. Hence, the order of fundamental loop matrix will be (b - n + 1) × b.
The elements of fundamental loop matrix will be having one of these three values, +1, -1 and 0.
• The value of element will be +1 for the link of selected f-loop.
• The value of elements will be 0 for the remaining links and twigs, which are not part of the selected f-loop.
• If the direction of twig current of selected f-loop is same as that of f-loop link current, then the value of element will be +1.
• If the direction of twig current of selected f-loop is opposite to that of f-loop link current, then the value of element will be -1.
### Procedure to find Fundamental Loop Matrix
Follow these steps in order to find the fundamental loop matrix of given directed graph.
• Select a tree of given directed graph.
• By including one link at a time, we will get one f-loop. Fill the values of elements corresponding to this f-loop in a row of fundamental loop matrix.
• Repeat the above step for all links.
### Example
Take a look at the following Tree of directed graph, which is considered for incidence matrix.
The above Tree contains three branches d, e & f. Hence, the branches a, b & c will be the links of the Co-Tree corresponding to the above Tree. By including one link at a time to the above Tree, we will get one f-loop. So, there will be three f-loops, since there are three links. These three f-loops are shown in the following figure.
In the above figure, the branches, which are represented with colored lines form f-loops. We will get the row wise element values of Tie-set matrix from each f-loop. So, the Tieset matrix of the above considered Tree will be
$$B = \begin{bmatrix}1 & 0 & 0 & -1 & 0 & -1\\0 & 1 & 0 & 1 & 1 & 0\\0 & 0 & 1 & 0 & -1 & 1 \end{bmatrix}$$
The rows and columns of the above matrix represents the links and branches of given directed graph. The order of this incidence matrix is 3 × 6.
The number of Fundamental loop matrices of a directed graph will be equal to the number of Trees of that directed graph. Because, every Tree will be having one Fundamental loop matrix.
## Fundamental Cut-set Matrix
Fundamental cut set or f-cut set is the minimum number of branches that are removed from a graph in such a way that the original graph will become two isolated subgraphs. The f-cut set contains only one twig and one or more links. So, the number of f-cut sets will be equal to the number of twigs.
Fundamental cut set matrix is represented with letter C. This matrix gives the relation between branch voltages and twig voltages.
If there are ‘n’ nodes and ‘b’ branches are present in a directed graph, then the number of twigs present in a selected Tree of given graph will be n-1. So, the fundamental cut set matrix will have ‘n-1’ rows and ‘b’ columns. Here, rows and columns are corresponding to the twigs of selected tree and branches of given graph. Hence, the order of fundamental cut set matrix will be (n-1) × b.
The elements of fundamental cut set matrix will be having one of these three values, +1, -1 and 0.
• The value of element will be +1 for the twig of selected f-cutset.
• The value of elements will be 0 for the remaining twigs and links, which are not part of the selected f-cutset.
• If the direction of link current of selected f-cut set is same as that of f-cutset twig current, then the value of element will be +1.
• If the direction of link current of selected f-cut set is opposite to that of f-cutset twig current, then the value of element will be -1.
### Procedure to find Fundamental Cut-set Matrix
Follow these steps in order to find the fundamental cut set matrix of given directed graph.
• Select a Tree of given directed graph and represent the links with the dotted lines.
• By removing one twig and necessary links at a time, we will get one f-cut set. Fill the values of elements corresponding to this f-cut set in a row of fundamental cut set matrix.
• Repeat the above step for all twigs.
### Example
Consider the same directed graph , which we discussed in the section of incidence matrix. Select the branches d, e & f of this directed graph as twigs. So, the remaining branches a, b & c of this directed graph will be the links.
The twigs d, e & f are represented with solid lines and links a, b & c are represented with dotted lines in the following figure.
By removing one twig and necessary links at a time, we will get one f-cut set. So, there will be three f-cut sets, since there are three twigs. These three f-cut sets are shown in the following figure.
We will be having three f-cut sets by removing a set of twig and links of C1, C2 and C3. We will get the row wise element values of fundamental cut set matrix from each f-cut set. So, the fundamental cut set matrix of the above considered Tree will be
$$C = \begin{bmatrix}1 & -1 & 0 & 1 & 0 & 0\\0 & -1 & 1 & 0 & 1 & 0\\1 & 0 & -1 & 0 & 0 & 1 \end{bmatrix}$$
The rows and columns of the above matrix represents the twigs and branches of given directed graph. The order of this fundamental cut set matrix is 3 × 6.
The number of Fundamental cut set matrices of a directed graph will be equal to the number of Trees of that directed graph. Because, every Tree will be having one Fundamental cut set matrix. |
# Distributive Law Multiplication
Expand the following expression:
(7+3)(a+b)
Use the Distributive Law.
### Comments for Distributive Law Multiplication
Mar 28, 2011 Distributive Law Multiplication by: Staff The question: What is (7+3)(a+b) answer? The answer: Approach #1: You can add the 7 and the 3 before you apply the distributive law: 7 + 3 = 10 (7+3)(a+b) = (10)(a+b) = 10*a + 10*b = 10a + 10b Approach #2: You can also apply the distributive law this way: (7+3)(a+b) = 7*(a+b) + 3*(a+b) Notice that (a+b) is first multiplied by the 7, and then (a+b) is multiplied by the 3. Finally, the results of the two multiplications are added together. 7*(a+b) = 7*a + 7*b = 7a + 7b 3*(a+b) = 3*a +3*b = 3a +3b = 7a + 7b + 3a + 3b = 7a + 3a + 7b + 3b = 10a + 10b (this is exactly the same answer we got by initially adding the 7 + 3 before applying the distributive law) The final answer is: (7+3)(a+b) = 10a + 10b Thanks for writing. Staff www.solving-math-problems.com
Sep 29, 2013 distributive this problem by: How about: 9(4x+5y+8)
Oct 08, 2013 9(4x + 5y + 8) by: Staff 9(4x+5y+8) = (9 * 4x) + (9 * 5y) + (9 * 8) = (9 * 4 * x) + (9 * 5 * y) + (9 * 8) = (36 * x) + (45 * y) + (72) = 36x + 45y + 72 The final answer is: 9(4x+5y+8) = 36x + 45y + 72 |
# Unit5 - Unit 5 Area Between Curves We have learned how to...
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Unformatted text preview: Unit 5 Area Between Curves We have learned how to find the area beneath a positive valued func- tion, by evaluating a definite integral. In this section we will learn how to find more general areas, of regions lying between two curves. For example, suppose we wish to find the area of the region bounded by f ( x ) and g ( x ) between x = 1 and x = 3 as shown: f(x) g(x) 1 2 3 x y dx f(x) - g(x) Figure 1: Area bounded by two curves The height of the region whose area we want to find is given by f ( x )- g ( x ). And on the interval [1,3], we have f ( x ) ≥ g ( x ), so the function ( f ( x )- g ( x )) is non-negative on [1,3]. With this observation in mind, we can see that when f ( x ) ≥ g ( x ) the problem of finding the area between two curves f ( x ) and g ( x ) between x = a and x = b can be reduced to one of finding the 1 area under the non-negative function h ( x ) = f ( x )- g ( x ) between x = a and x = b . Therefore we do not need to learn another technique for this kind of problem, we simply reduce the problem as above, and then apply our earlier method for finding the area under a non-negative function on an interval [ a, b ] by evaluating the definite integral of that function from a to b . Example 1 . Find the area between the curves f ( x ) = x 2 +1 and g ( x ) = x- 1 on the interval [1 , 3]. Solution: We can first sketch the region involved, as shown in figure 1 on the previous page. We see that f ( x ) ≥ g ( x ) throughout [1,3], so the function h ( x ) = f ( x )- g ( x ) is positive valued throughout [1,3]. We may therefore represent the area ( A ) by: A = integraltext 3 1 [ f ( x )- g ( x )] dx = integraltext 3 1 [( x 2 + 1)- ( x- 1)] dx = integraltext 3 1 [( x 2- x + 2)] dx = bracketleftBig x 3 3- x 2 2 + 2 x bracketrightBig 3 1 = [(9- 9 2 + 6)- ( 1 3- 1 2 + 2)] = 21 2- 11 6 = 26 3 Whenever we need to find the area between 2 curves, our first step must always be to determine which curve lies above the other on the interval we are interested in. That is, we need to identify the upper curve and the lower curve, in order to set up the integral integraltext b a ( upper- lower ) dx . We must also be sure that the same curve is the upper curve throughout the entire interval. (We’ll look at what to do when this is not the case a bit later.) Often we are interested in the area of the region “bounded by” two curves. If 2 curves, y = f ( x ) and y = g ( x ), intersect in exactly 2 places, at x = a and 2 x = b , and both curves are continuous on [ a, b ], then there is a single region which is entirely enclosed by these curves (see, for instance, figure 2). This is what we mean by the area “bounded by” the curves. If the curves intersect more than twice, then there will be more than one such region. In that case, the area of the region bounded by the curves means the total area of all regions enclosed by the curves....
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# How to Find the Y Intercept: A Comprehensive Guide with Step-by-Step Guides, Tips, and Tricks
## Introduction
As students progress through math courses, they encounter a host of new concepts and techniques. One such skill that is crucial to success in algebra and graphing is finding the y intercept. The y intercept is the point where a line crosses the y-axis on a graph, and it helps mathematicians graph equations, calculate slopes, and more. In this article, we will explore step-by-step guides, tips and tricks, and cheat sheets for finding the y intercept with ease. Whether you’re a seasoned pro or just starting out, this guide will provide you with the tools and know-how you need to master this important concept in mathematics.
## Understanding the Basics: A Step-by-Step Guide to Finding the Y Intercept
Before we dive into more advanced techniques, let’s start with the basics. A y intercept is the point where a line crosses the y-axis on a graph. In other words, it is the value of the dependent variable (y) when the independent variable (x) is zero. To find the y intercept for a given equation, follow these simple steps:
Step 1: Identify the equation that you want to graph. For example, let’s say we want to graph the equation y = 2x + 1.
Step 2: Remember that the y intercept occurs when x = 0. So, substitute 0 in the equation for x and solve for y. When x = 0, y = 1 because 2(0) + 1 = 1.
Step 3: Plot the point (0, 1) on the graph. This is the y intercept!
It’s that easy! Of course, this only works for linear equations, but it’s a good starting point for other types of equations as well. Let’s take a look at a few examples to solidify your understanding.
Example 1:
Find the y intercept for the equation y = -3x + 5.
When x = 0, y = 5 because -3(0) + 5 = 5. Therefore, the y intercept is (0, 5).
Example 2:
Find the y intercept for the equation y = 2x² – 4x + 7.
To find the y intercept for this equation, we must set x = 0 and solve for y. y = 2(0)² – 4(0) + 7 = 7. Therefore, the y intercept is (0, 7).
## Mastering Math: Tips and Tricks to Find the Y Intercept with Ease
While the step-by-step approach is useful, sometimes it can be time-consuming, especially when dealing with more complicated equations. That’s where some algebraic shortcuts or methods can come in handy. Let’s explore a few of these tips and tricks.
Tip 1: Use the Slope-Intercept Form
One of the easiest ways to find the y intercept is to use the slope-intercept form of a linear equation, which is y = mx + b, where m is the slope of the line, and b is the y intercept. When working with an equation in this form, all you need to do is look at the b value to determine the y intercept. For example, in the equation y = 2x + 3, the y intercept is 3.
Tip 2: Rewrite the Equation in Standard Form
Another way to easily find the y intercept is to rewrite the equation in standard form, which is ax + by = c. The y intercept occurs when x = 0, so all you need to do is solve for y. For example, let’s say we have the equation 4x – 3y = 12. To find the y intercept, we first need to solve for y: -3y = -4x + 12, so y = (4/3)x – 4. Therefore, the y intercept is (0, -4).
## Cracking the Code: Demystifying Y Intercept Calculation
For some students, finding the y intercept can be a challenging task. Common challenges include working with fractions or negative numbers, not understanding the equation being graphed, and not being able to read the graph correctly. Here are a few tips for overcoming these challenges:
Tip 1: Break It Down
If the equation you’re working with contains fractions or negative numbers, it can be helpful to break it down into simpler terms. For example, consider the equation y = (-2/3)x – 5. To find the y intercept, we can multiply both sides of the equation by 3: 3y = -2x – 15. Solving for y, we get y = (-2/3)x – 5. This is a lot easier to work with than the original equation, and it allows us to find the y intercept more easily.
Tip 2: Look for Patterns
Sometimes, there are patterns or shortcuts you can use to simplify y intercept calculations. For example, if you have an equation in the form y = ax + b, and a is an even number, the y intercept will always be a multiple of a. Similarly, if a is an odd number, the y intercept will always be half of an odd number plus a multiple of a. These patterns can save you a lot of time when you’re working on complex equations.
## Graphing Made Easy: Simple Methods to Identify Y Intercepts
Finding the y intercept doesn’t always involve solving equations. In fact, it’s often possible to identify the y intercept simply by looking at a graph. Here are a few tips for doing so:
Tip 1: Look for the Point Where the Line Crosses the Y Axis
The y intercept is the point where a line crosses the y-axis, which is the vertical axis on a graph. To find the y intercept, simply look for the point where the line intersects with the y-axis.
Tip 2: Read the Graph Carefully
When looking at a graph, it’s important to pay attention to the scale of the axes. Make sure you’re reading the right scale for the axis you’re interested in. For example, if you’re looking for the y intercept, make sure you’re reading the y axis, not the x axis.
## The Ultimate Cheat Sheet for Finding Y Intercepts
By now, you should have a solid understanding of how to find the y intercept using a variety of methods. To help you keep all of this information organized, we’ve created a one-page cheat sheet that you can use as a reference. This cheat sheet includes all of the formulas, tips, and tricks we’ve discussed in this article, so you can quickly and easily find the y intercept for any equation. If you’re serious about mastering this concept, we highly recommend downloading and printing this cheat sheet for future reference.
## Graphing for Beginners: Finding the Y Intercept Made Simple
If you’re just starting out with algebra and graphing, finding the y intercept can seem like a daunting task. However, with the tips and tricks we’ve discussed in this article, you should be well on your way to mastering this crucial concept. Here’s a final breakdown of the key takeaways from this article:
– The y intercept is the point where a line crosses the y-axis on a graph.
– To find the y intercept, you can either solve the equation for y when x = 0, or you can use algebraic shortcuts like the slope-intercept form or standard form.
– When working with complex equations, it can be helpful to break them down into simpler terms or look for patterns that can make the process easier.
– Graphs can also provide helpful clues for finding the y intercept, such as the point where the line crosses the y-axis.
– If you’re looking to master this concept, be sure to download and print our cheat sheet for quick and easy reference.
## No More Confusion: A Comprehensive Guide to Finding Y Intercepts in Algebra
In this article, we’ve covered a lot of ground when it comes to finding the y intercept. We’ve explored step-by-step guides, tips and tricks, and cheat sheets for mastering this important concept in algebra and graphing. Whether you’re just starting out or you’re an experienced math whiz, these tools and techniques will help you feel confident and capable when working with y intercepts.
As we’ve discussed, finding the y intercept is a crucial skill for success in math. With this skill, you’ll be able to graph equations, calculate slopes, and more. By applying the tips and techniques we’ve discussed in this article, you’ll be well on your way to mastering this important concept and unlocking a whole new level of mathematical prowess. |
## DEV Community
James Perkins
Posted on • Originally published at jamesperkins.io on
# Solving Sum of N Numbers
This blog post is going to focus on solving the sum of N numbers. It's a typical question that could pop up in a Junior interview, or potentially in your college work. So the question could be what is the sum of the first 1000 numbers.
Using Loops to solve the problem.
We could solve this using a for loop passing in N into the function. For Example:
``````function findSum(n) {
let result = 0;
for (let i = 1; i <= n; i++) {
result = result + i;
}
return result;
}
let n = 1000;
console.log(`Sum of the numbers from 1 to \${n} is \${findSum(n)}`);
``````
This will solve the problem, and the result will be 500500 but it is quite inefficient because as the number gets bigger so does the iteration and memory usage.
Thinking of this problem like a maths problem.
So the problem we are attempting to solve is:
``````sum = 1 + 2 + 3 + 4 + 5 + 6 + 7 + ... n
``````
If we think of it, we are technically just doing:
``````sum = n + (n-1) + (n-2) + (n-3) + (n-4) + (n-5) ... +1
``````
There is an arithmetic solution for this called Arithmetic progression which is described as a sequence of numbers such that the difference between the consecutive terms is constant.
To break it down, let's look at this very simple example:
``````2+5+8+11+14
``````
This sum can be found quickly by taking the number n of terms being added (here 5), multiplying by the sum of the first and last number in the progression (here 2 + 14 = 16), and dividing by 2:
``````(n(a1 + a2))/2
``````
Which means we can solve this in a single line of JavaScript, by using what we know from our Maths problem:
``````function findSum(n) {
return (n * (n + 1)) / 2;
}
let n = 1000;
console.log(`Sum of the numbers from 1 to \${n} is \${findSum(n)}`);
``````
Using this we get the same result, and it doesn't matter how big the number is as we aren't iterating, we are directly calculating the solution.
I hope you enjoyed the short tutorial on calculating the sum of n numbers, and I hope it helps when you are prepping for some interviews. |
# If f : R→ (−1, 1) is defined by
Question:
If $f: R \rightarrow(-1,1)$ is defined by $f(x)=\frac{-x|x|}{1+x^{2}}$, then $f^{-1}(x)$ equals
(a) $\sqrt{\frac{|x|}{1-|x|}}$
(b) $\operatorname{Sgn}(x) \sqrt{\frac{|x|}{1-|x|}}$
(c) $-\sqrt{\frac{x}{1-x}}$
(d) None of these
Solution:
(b) $-\operatorname{Sgn}$ $(x) \sqrt{\frac{|x|}{1-|x|}}$
We have, $f(x)=\frac{-x|x|}{1+x^{2}} \quad x \in(-1,1)$
Case - (I)
When, $x<0$,
Then, $|x|=-x$
And $f(x)>0$
Now,
$f(x)=\frac{-x(-x)}{1+x^{2}}$
$\Rightarrow y=\frac{x^{2}}{1+x^{2}}$
$\Rightarrow \frac{y}{1}=\frac{x^{2}}{1+x^{2}}$
$\Rightarrow \frac{y+1}{y-1}=\frac{x^{2}+1+x^{2}}{x^{2}-1-x^{2}} \quad[$ Using Componendo and dividendo]
$\Rightarrow \frac{y+1}{y-1}=\frac{2 x^{2}+1}{-1}$
$\Rightarrow-\frac{y+1}{y-1}=2 x^{2}+1$
$\Rightarrow \frac{2 y}{1-y}=2 x^{2}$
$\Rightarrow \frac{y}{1-y}=x^{2}$
$\Rightarrow x=-\sqrt{\frac{y}{1-y}}$ (As $x<0$ )
$\Rightarrow x=-\sqrt{\frac{|y|}{1-|y|}}$
$[$ As $y>0]$
To find the inverse interchanging $x$ and $y$ we get,
$f^{-1}(x)=-\sqrt{\frac{|x|}{1-|x|}}$
Case - (II)
When, $x>0$,
Then, $|x|=x$
And $f(x)<0$
Now,
$f(x)=\frac{-x(x)}{1+x^{2}}$
$\Rightarrow y=\frac{-x^{2}}{1+x^{2}}$
$\Rightarrow \frac{y}{1}=\frac{-x^{2}}{1+x^{2}}$
$\Rightarrow \frac{y+1}{y-1}=\frac{-x^{2}+1+x^{2}}{-x^{2}-1-x^{2}} \quad$ [Using Componendo and dividendo]
$\Rightarrow \frac{y+1}{y-1}=\frac{1}{-2 x^{2}-1}$
$\Rightarrow \frac{1+y}{1-y}=\frac{1}{2 x^{2}+1}$
$\Rightarrow \frac{1-y}{1+y}=2 x^{2}+1$
$\Rightarrow \frac{-2 y}{1+y}=2 x^{2}$
$\Rightarrow x^{2}=\frac{-y}{1+y}$
$\Rightarrow x=\sqrt{\frac{-y}{1+y}}$ $(\operatorname{As} x>0)$
$\Rightarrow x=\sqrt{\frac{|y|}{1-|y|}}$
$[$ As $y<0]$
To find the inverse interchanging $x$ and $y$ we get,
$f^{-1}(x)=\sqrt{\frac{|x|}{1-|x|}}$ ... (ii)
Case - (III)
When, $x=0$,
Then, $f(x)=0$
Hence, $f^{-1}(x)=0 \quad \ldots$ (iii)
Combinig equation (i), (ii) and (iii) we get,
$f^{-1}(x)=-\operatorname{Sgn}(x) \sqrt{\frac{|x|}{1-|x|}}$ |
MathsNotes
MathsNotes - Mathematics IGCSE notes Index 1 Decimals and...
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Mathematics IGCSE notes Index 1. Decimals and standard form 2. Accuracy and Error 3. Powers and roots click on a topic to visit the notes 4. Ratio & proportion 5. Fractions, ratios 6. Percentages 7. Rational and irrational numbers 8. Algebra: simplifying and factorising 9. Equations: linear, quadratic, simultaneous 10. Rearranging formulae 11. Inequalities 12. Parallel lines, bearings, polygons 13. Areas and volumes, similarity 14. Trigonometry 15. Circles 16. Similar triangles, congruent triangles 17. Transformations 18. Loci and ruler and compass constructions 19. Vectors 20. Straight line graphs 21. More graphs 22. Distance, velocity graphs 23. Sequences; trial and improvement 24. Graphical transformations 25. Probability 26. Statistical calculations, diagrams, data collection 27. Functions 28. Calculus 29. Sets {also use the intranet revision course of question papers and answers by topic }
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1. Decimals and standard form top (a) multiplying and dividing (i) Move the decimal points to the right until each is a whole number, noting the total number of moves, perform the multiplication, then move the decimal point back by the previous total: 2.5 1.36 × , so the answer is 25 136 3400 →× = 3.4 {Note in the previous example, that transferring a factor of 2, or even better, 4, from the 136 to the 25 makes it easier: } 25 136 25 (4 34) (25 4) 34 100 34 3400 ×=× × =× ×= ×= (ii) Move both decimal points together to the right until the 0.00175 0.042 ÷ divisor is a whole number, perform the calculation, and that is the answer. 1.75 42 →÷ , but simplify the calculation by cancelling down any factors first. In this case, both numbers share a 7, so divide this out: , and 0.25 6 0.0416 60 . 2 5 ± , so the answer is 0.0416 (iii) decimal places To round a number to n d.p., count n digits to the right of the decimal point. If the digit following the n th is , then the n 5 th digit is raised by 1. e.g. round 3.012678 to 3 d.p. so 3.012678 3.012|678 3.013 to 3 d.p. (iv) significant figures To round a number to n s.f., count digits from the left starting with the first non-zero digit, then proceed as for decimal places. e.g. round 3109.85 to 3 s.f., 3109.85 so 310|9.85 3110 to 3 s.f. e.g. round 0.0030162 to 3 s.f., 0.0030162 , so 0.00301|62 0.00302 to 3 s.f. (b) standard form (iii) Convert the following to standard form: (a) 25 000 (b) 0.0000123 Move the decimal point until you have a number x where 11 0 x < , and the number of places you moved the point will indicate the numerical value of the power of 10. So 4 25000 2.5 10 , and 5 0.0000123 1.23 10 (iv) multiplying in standard form: ( ) 5 (4.4 10 ) 3.5 10 ××× 6 As all the elements are multiplied, rearrange them thus: () 5 6 11 12 (4.4 3.5) 10 10 15.4 10 1.54 10 =×× × = × = × 2
(v) dividing in standard form: 12 3 3.2 10 2.5 10 × × Again, rearrange the calculation to 12 3 9 (3.2 2.5) (10 10 ) 1.28 10 ÷× ÷ = × (vi) adding/subtracting in standard form: The hardest of the calculations. Convert both numbers into the same denomination, i.e. in this case 10 67 (2.5 10 ) (3.75 10 ) ×+ × 6 or 10 7 , then add.
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# Percentage Questions
FACTS AND FORMULAE FOR PERCENTAGE QUESTIONS
I.Concept of Percentage : By a certain percent , we mean that many hundredths. Thus x percent means x hundredths, written as x%.
To express x% as a fraction : We have , x% = x/100.
Thus, 20% = 20/100 = 1/5;
48% = 48/100 = 12/25, etc.
To express a/b as a percent : We have, $\frac{a}{b}=\left(\frac{a}{b}×100\right)%$ .
Thus, $\frac{1}{4}=\left(\frac{1}{4}×100\right)%=25%$
II. If the price of a commodity increases by R%, then the reduction in consumption so as not to increase the expenditure is $\left[\frac{R}{\left(100+R\right)}×100\right]%$
If the price of the commodity decreases by R%,then the increase in consumption so as to decrease the expenditure is $\left[\frac{R}{\left(100-R\right)}×100\right]%$
III. Results on Population : Let the population of the town be P now and suppose it increases at the rate of R% per annum, then :
1. Population after n years = $P{\left(1+\frac{R}{100}\right)}^{n}$
2. Population n years ago = $\frac{P}{{\left(1+\frac{R}{100}\right)}^{n}}$
IV. Results on Depreciation : Let the present value of a machine be P. Suppose it depreciates at the rate R% per annum. Then,
1. Value of the machine after n years = $P{\left(1-\frac{R}{100}\right)}^{n}$
2. Value of the machine n years ago = $\frac{P}{{\left(1-\frac{R}{100}\right)}^{n}}$
V. If A is R% more than B, then B is less than A by
$\left[\frac{R}{\left(100+R\right)}×100\right]%$
If A is R% less than B , then B is more than A by
$\left[\frac{R}{\left(100-R\right)}×100\right]%$
Q:
42% of an alloy was silver. If in a quantity of alloy there was 147 g of silver, what was the quantity of the other elements in the alloy?
A) 261 g B) 203 g C) 273 g D) 350 g
Explanation:
2 842
Q:
Samar was given some money to take care of his travel during an 8-day sales drive he had to undertake. However he had to increase his stay by another 6 days and as a result his average daily travel allowance went down by Rs. 120. What was the amount that was sanctioned to him in the beginning?
A) ₹ 560 B) ₹ 840 C) ₹ 1120 D) ₹ 2240
Explanation:
3 838
Q:
If the selling price of an article is 8% more than the cost price and the discount offered is 10% on the marked price of the article, then what is the ratio of the cost price to the marked price?
A) 5 : 6 B) 8 : 9 C) 3 : 4 D) 4 : 5
Explanation:
9 835
Q:
A small scale business has the following expenses: procurement (25%), employees' salary (25%) and 50% for maintenance.
If the business pays a total salary of Rs. 2,00,000 then what is its maintenance expense?
A) ₹ 3,00,000 B) ₹ 2,00,000 C) ₹ 2,50,000 D) ₹ 4,00,000
Explanation:
1 832
Q:
A runner is running at a speed of 40 kim/h. If he runs at a speed of 30 km/h, then what will the decrease in the percentage of his speed be?
A) 30% B) 25% C) 20% D) 15%
Explanation:
4 831
Q:
Raghav spends 80% of his income. If his income increases by 12% and his expenditure increases by 17.5%, then what is the percentage decrease in his savings?
A) 12% B) 10% C) 8% D) 15%
Explanation:
3 823
Q:
Rahul’s salary is 50% more than Rohit’s salary. If Rahul saves Rs 2250 which is 8% of his salary, then what is Rohit’s salary (in Rs)?
A) 22250 B) 24450 C) 26350 D) 18750
Explanation:
1 821
Q:
56% of 375 is:
A) 210 B) 196 C) 224 D) 168 |
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# This brilliant article was contributed to Total Gadha by Ashish Tyagi, a regular TGite.
I do not think there can be easier or more lucid method of finding the last two digits of a number. Kudos to Ashish for this article. We will also welcome similar contributions from our other users. Do you have some useful or life-saving funda that you would like to share? Send it to us and we will publish it with your name- Total Gadha
I am dividing this method into four parts and we will discuss each part one by one: a. Last two digits of numbers which end in one b. Last two digits of numbers which end in 3, 7 and 9 c. Last two digits of numbers which end in 2 d. Last two digits of numbers which end in 4, 6 and 8 Before we start, let me mention binomial theorem in brief as we will need it for our calculations.
Last two digits of numbers ending in 1 Let's start with an example. What are the last two digits of 31786? Solution: 31786 = (30 + 1)786 = 786C0 × 1786 + 786C1 × 1785 × (30) + 786C2 × 1784 × 302 + ..., Note that all the terms after the second term will end in two or more zeroes. The first two terms are 786C0 × 1786 and 786C1 × 1785 × (30). Now, the second term will end with one zero and the tens digit of the second term will be the product of 786 and 3 i.e. 8. Therefore, the last two digits of the second term will be 80. The last digit of the first term is 1. So the last two digits of 31786 are 81. Now, here is the shortcut:
Here are some more examples: Find the last two digits of 412789 In no time at all you can calculate the answer to be 61 (4 × 9 = 36). Therefore, 6 will be the tens digit and one will be the units digit)
the last two digits of 51456 × 61567 will be the last two digits of 01 ×21 = 21 Last two digits of numbers ending in 3. 583512 Last two digits of numbers ending in 2. we need to find the last two digits of 21 72. 6 or 8 . 19266 = (192)133. 33288 = (334)72. 27456 2. 4. 7 or 9 Find the last two digits of 19266. the last two digits of 2172 = 41 (tens digit = 2 × 2 = 4. Therefore. Once the number is ending in 1 we can straight away get the last two digits with the help of the previous method. 87474 = 87472 ×872 = (874)118 ×872 = (69 × 69)118 × 69 (The last two digits of 872 are 69) = 61118 × 69 = 81 × 69 = 89 If you understood the method then try your hands on these questions: Find the last two digits of: 1. 192 ends in 61 (192 = 361) therefore. so the tens digit will be 8 and last digit will be 1) Find the last two digits of 33288. Now 334 ends in 21 (334 = 332 ×332 = 1089 × 1089 = xxxxx21) therefore. By the previous method. we need to find the last two digits of (61)133. Now. 7983 3. unit digit = 1) So here's the rule for finding the last two digits of numbers ending in 3. The last two digits are 81 (6 ×3 = 18.Find the last two digits of 7156747 Last two digits will be 91 (7 × 7 gives 9 and 1 as units digit) Now try to get the answer to this question within 10 s: Find the last two digits of 51456 × 61567 The last two digits of 51456 will be 01 and the last two digits of 61567 will be 21. 7 and 9: Now try the method with a number ending in 7: Find the last two digits of 87474.
e. 2434 will end in 76 and 2453 will end in 24. 64236 = (26)236 = 21416 = (210)141 × 26 = 24141 (24 raised to odd power) × 64 = 24 × 64 = 36 Now those numbers which are not in the form of 2n can be broken down into the form 2n odd number. Therefore. We can find the last two digits of both the parts separately.There is only one even two-digit number which always ends in itself (last two digits) . We know that 242 ends in 76 and 210 ends in 24. then you can straightaway write the last two digits of 2n because when 76 is multiplied with 2n the last two digits remain the same as the last two digits of 2n. the last two digits of 76 × 27 will be the last two digits of 27 = 28) Same method we can use for any number which is of the form 2 n.76 i. Here is an example: Find the last two digits of 64236. Therefore. our purpose is to get 76 as last two digits for even numbers. 56283 = (23 × 7)283 = 2849 × 7283 = (210)84 × 29 × (74)70 × 73 = 76 × 12 × (01)70 ×43 = 16 Find the last two digits of 78379. Also. 28287 Ashish Tyagi . Find the last two digits of 56283. 34576 2. Here are some examples: Find the last two digits of 62586. 78379 = (2 × 39)379 = 2379 × 39379 = (210)37 × 29 × (392)189 × 39 = 24 × 12 × 81 × 39 = 92 Now try to find the last two digits of 1. 62586 = (2 × 31)586 = 2586 × 3586 = (210)58 × 26 × 31586 = 76 × 64 × 81 = 84 Find the last two digits of 54380. 54380 = (2 × 33)380 = 2380 × 31140 = (210)38 × (34)285 = 76 × 81285 = 76 × 01 = 76. 76 raised to any power gives the last two digits as 76. 24 raised to an even power always ends with 76 and 24 raised to an odd power always ends with 24. Therefore. 2543 = (210)54 ×23 = (24)54 (24 raised to an even power) ×23 = 76 × 8 = 08 (NOTE: Here if you need to multiply 76 with 2n. Let's apply this funda: Find the last two digits of 2543.
I am afraid I shall have to end here and leave the rest of it for my CBT Club students. Let me consider two cases where the numbers are ending with 5. Numbers where the previous digit of 5 is any odd number In 1st case if you raise the number to any power you'll always get 25 as the last two digits. NUMBER TABLE FOR FINDING ‘THE LAST 2 DIGITS’ 3^4 ends with 81 13^4 ends with 61 23^4 ends with 41 33^4 ends with 21 43^4 ends with 01 53^4 ends with 81 63^4 ends with 61 73^4 ends with 41 83^4 ends with 21 93^4 ends with 01 7^4 ends with 01 17^4 ends with 21 27^4 ends with 41 37^4 ends with 61 47^4 ends with 81 57^4 ends with 01 . 1. but in case of odd power you'll get 75 as the last two digits. In 2nd case if you raise the number to even power you'll get 25. I shall cover some problems based on this in the CBT Club this week. Numbers where the previous digit of 5 is 0 or any even number and 2.
67^4 ends with 21 77^4 ends with 41 87^4 ends with 61 97^4 ends with 81 9^2 ends with 81 19^2 ends with 61 29^2 ends with 41 39^2 ends with 21 49^2 ends with 01 59^2 ends with 81 69^2 ends with 61 79^2 ends with 41 89^2 ends with 21 99^2 ends with 01 •5^n always end with 25. if n>1 •The product of any 2 numbers ending with 5 always ends with 25 •2^10 ends with 24 •24^n ends with 76. if n is even 24. if n is odd . |
<meta http-equiv="refresh" content="1; url=/nojavascript/"> Multi-Step Equations | CK-12 Foundation
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# 3.3: Multi-Step Equations
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Solve a multi-step equation by combining like terms.
• Solve a multi-step equation using the distributive property.
• Solve real-world problems using multi-step equations.
## Solving Multi-Step Equations by Combining Like Terms
We have seen that when we solve for an unknown variable, it can be a simple matter of moving terms around in one or two steps. We can now look at solving equations that take several steps to isolate the unknown variable. Such equations are referred to as multi-step equations.
In this section, we will simply be combining the steps we already know how to do. Our goal is to end up with all the constants on one side of the equation and all of the variables on the other side. We will do this by collecting “like terms”. Don’t forget, like terms have the same combination of variables in them.
Example 1
Solve $\frac{3x+4} {3} - 5x = 6$
This problem involves a fraction. Before we can combine the variable terms we need to deal with it. Let’s put all the terms on the left over a common denominator of three.
$& \frac{3x+4} {3} - \frac{15x} {3} =6\ && \text{Next we combine the fractions}. \\& \frac{3x+4 - 15x} {3} = 6\ && \text{Combine like terms}. \\& \qquad \ \frac{4 - 12x} {3} = 6\ && \text{Multiply both sides by}\ 3. \\& \qquad \ \ \cancel{4}-12x =18\ && \text{Subtract}\ 4\ \text{from both sides}. \\& \quad \ \underline{\;\; -\cancel{4} \qquad \quad \ \ -4 }\\& \qquad \quad -12x =14\ && \text{Divide both sides by}\ -12 \\& \qquad \quad \ \frac{-12} {-12} x = - \frac{14} {12}$
Solution
$x = - \frac{7} {6}$
## Solving Multi-Step Equations Using the Distributive Property
You have seen in some of the examples that we can choose to divide out a constant or distribute it. The choice comes down to whether on not we would get a fraction as a result. We are trying to simplify the expression. If we can divide out large numbers without getting a fraction, then we avoid large coefficients. Most of the time, however, we will have to distribute and then collect like terms.
Example 2
Solve $17(3x+ 4) =7$
This equation has the $x$ buried in parentheses. In order to extract it we can proceed in one of two ways. We can either distribute the seventeen on the left, or divide both sides by seventeen to remove it from the left. If we divide by seventeen, however, we will end up with a fraction. We wish to avoid fractions if possible!
$& 17(3x+ 4) =7\ && \text{Distribute the}\ 17. \\& \quad \ 51x \cancel{+68} = 7 \\& \qquad \quad \cancel{-68} \ \ -68\ && \text{Subtract}\ 68\ \text{from both sides}. \\& \qquad \quad 51x =-61\ && \text{Divide by}\ 51.$
Solution
$x = -\frac{61} {51}$
Example 3
Solve $4(3x-4) - 7(2x+3) = 3$
This time we will need to collect like terms, but they are hidden inside the brackets. We start by expanding the parentheses.
$12x - 16 - 14x - 21 & = 3\ && \text{Collect the like terms}\ (12x\ \text{and}\ -14x). \\(12x -14x) + (-16 - 21) & = 3\ && \text{Evaluate each set of like terms}. \\-2x\cancel{-37} & =3 \\\cancel{+37} & \ \ +37 && \text{Add} \ 37\ \text{to both sides}. \\-2x& =40 \\\frac{-2x}{-2} & =\frac{40}{-2}\ && \text{Divide both sides by}\ -2.$
Solution
$x=-20$
Example 4
Solve the following equation for $x$.
$0.1(3.2 + 2x) + \frac{1} {2} \left (3- \frac{x} {5}\right ) = 0$
This function contains both fractions and decimals. We should convert all terms to one or the other. It is often easier to convert decimals to fractions, but the fractions in this equation are easily moved to decimal form. Decimals do not require a common denominator!
Rewrite in decimal form.
$0.1 (3.2 +2x) +0.5 (3- 0.2x)& = 0\ && \text{Multiply out decimals:} \\0.32 +0.2x +1.5 - 0.1x & = 0\ && \text{Collect like terms:} \\(0.2x -0.1x) + (0.32+1.5) & = 0\ && \text{Evaluate each collection:} \\0.1x \cancel{+1.82} & = 0\ && \text{Subtract}\ 1.82\ \text{from both sides} \\\cancel{-1.82} & \ \ -1.82 \\0.1x & = -1.82\ && \text{Divide by}\ -0.1 \\\frac{0.1x}{0.1} & = \frac{-1.82}{0.1}$
Solution
$x=18.2$
## Solve Real-World Problems Using Multi-Step Equations
Real-world problems require you to translate from a problem in words to an equation. First, look to see what the equation is asking. What is the unknown you have to solve for? That will determine the quantity we will use for our variable. The text explains what is happening. Break it down into small, manageable chunks, and follow what is going on with our variable all the way through the problem.
Example 5
A grower’s cooperative has a farmer’s market in the town center every Saturday. They sell what they have grown and split the money into several categories. 8.5% of all the money taken is removed for sales tax. $150 is removed to pay the rent on the space they occupy. What remains is split evenly between the seven growers. How much money is taken in total if each grower receives a$175 share?
Let us translate the text above into an equation. The unknown is going to be the total money taken in dollars. We will call this $x$.
8.5% of all the money taken is removed for sales tax”. This means that 91.5% of the money remains. This is $0.915x$.
$(0.915x - 150)$
“$150 is removed to pay the rent on the space they occupy” $\frac{0.915x - 150} {7}$ “What remains is split evenly between the 7 growers” If each grower’s share is$175, then we can write the following equation.
$\frac{0.915x -150} {7} & = 175\ && \text{Multiply by both sides}\ 7. \\0.915x-150 & = 1225\ && \text{Add}\ 150\ \text{to both sides}. \\0.915 x & = 1375\ && \text{Divide by}\ 0.915.\\\frac{0.915x} {0.915} & = \frac{1373} {0.915} \\& = 1502.7322 \ldots && \text{Round to two decimal places}.$
Solution
If the growers are each to receive a $175 share then they must take at least$1,502.73.
Example 6
A factory manager is packing engine components into wooden crates to be shipped on a small truck. The truck is designed to hold sixteen crates, and will safely carry a 1200 lb cargo. Each crate weighs twelve lbs empty. How much weight should the manager instruct the workers to put in each crate in order to get the shipment weight as close as possible to 1200 lbs?
The unknown quantity is the weight to put in each box. This is $x$. Each crate, when full will weigh:
$(x + 12) &&& 16\ \text{crates must weigh}. \\16(x + 12) &&& \text{And this must equal}\ 16 \ lbs. \\16(x + 12) & = 1200\ && \text{Isolate}\ x\ \text{first, divide both sides by}\ 16. \\x + 12 & = 75\ && \text{Next subtract}\ 12 \ \text{from both sides}. \\x & = 63$
Solution
The manager should tell the workers to put 63 lbs of components in each crate.
## Ohm’s Law
The electrical current, $I$ (amps), passing through an electronic component varies directly with the applied voltage, $V$ (volts), according to the relationship:
$V=I \cdot R$ where $R$ is the resistance (measured in Ohms - $\Omega$)
Example 7
A scientist is trying to deduce the resistance of an unknown component. He labels the resistance of the unknown component $x\Omega$. The resistance of a circuit containing a number of these components is $(5x + 20)\Omega$. If a 120 volt potential difference across the circuit produces a current of 2.5 amps, calculate the resistance of the unknown component.
Substitute $V=120, I=2.5$ and $R=5x+20$ into $V=I \cdot R$:
$120 & = 2.5(5x + 20) & & \text{Distribute the} \ 2.5.\\120 & = 12.5x \cancel{+50} & & \text{Subtract} \ 50 \ \text{from both sides}.\\ -50 & \qquad \quad \ \ \cancel{-50}\\70 & =12.5x & & \text{Divide both sides by} \ 12.5.\\\frac{70}{12.5}& =\frac{12.5x}{12.5}\\5.6\Omega & = x$
Solution
The unknown components have a resistance of $5.6 \Omega$.
## Distance, Speed and Time
The speed of a body is the distance it travels per unit of time. We can determine how far an object moves in a certain amount of time by multiplying the speed by the time. Here is our equation.
$\text{distance} = \text{speed} \times \text{time}$
Example 8
Shanice’s car is traveling 10 miles per hour slower than twice the speed of Brandon’s car. She covers 93 miles in 1 hour 30 minutes. How fast is Brandon driving?
Here we have two unknowns in this problem. Shanice’s speed and Brandon’s speed. We do know that Shanice’s speed is ten less than twice Brandon’s speed. Since the question is asking for Brandon’s speed, it is his speed in miles per hour that will be $x$.
Substituting into the distance time equation yields:
$93& =2x-10 \times 1.5 & & \text{Divide by} \ 1.5.\\62& =2x\cancel{-10}\\+10 & \qquad \ \cancel{+10} & & \text{Add} \ 10 \ \text{to both sides}.\\72& =2x\\\frac{72}{2}&=\frac{2x}{2}& & \text{Divide both sides by} \ 2.\\36& =x$
Solution
Peter is driving at 36 miles per hour.
This example may be checked by considering the situation another way: We can use the fact that Shanice's covers 93 miles in 1 hour 30 minutes to determine her speed (we will call this $y$ as $x$ has already been defined as Brandon’s speed):
$93 & = y \cdot 1.5\\\frac{93} {1.5} & = \frac{1.5y}{1.5}& & \text{Divide both sides by} \ 1.5.\\y&=62 \ mph$
We can then use this information to determine Shanice’s speed by converting the text to an equation.
“Shanice’s car is traveling at 10 miles per hour slower than twice the speed of Peter’s car”
Translates to
$y = 2x -10$
It is then a simple matter to substitute in our value in for $y$ and then solve for $x$:
$& \ \ 62 = (2x - \cancel{10})\\& \underline{+10 \qquad \ \ +\cancel{10}} && \text{Add} \ 10 \ \text{to both sides}.\\& \ \ 72 = 2x \\& \ \ 72 = 2x & & \text{Divide both sides by} \ 2.\\& \ \frac{72} {2} = \frac{2x}{2}\\& \quad x=36 \ miles \ per \ hour.$
Solution
Brandon is driving at 36 miles per hour.
You can see that we arrive at exactly the same answer whichever way we solve the problem. In algebra, there is almost always more than one method of solving a problem. If time allows, it is an excellent idea to try to solve the problem using two different methods and thus confirm that you have calculated the answer correctly.
Speed of Sound
The speed of sound in dry air, $v$, is given by the following equation.
$v = 331 + 0.6T$ where $T$ is the temperature in Celsius and $v$ is the speed of sound in meters per second.
Example 9
Tashi hits a drainpipe with a hammer and 250 meters away Minh hears the sound and hits his own drainpipe. Unfortunately, there is a one second delay between him hearing the sound and hitting his own pipe. Tashi accurately measures the time from her hitting the pipe and hearing Mihn’s pipe at 2.46 seconds. What is the temperature of the air?
This complex problem must be carefully translated into equations:
$\text{Distance traveled} &= (331 + 0.6T) \times \text{time}\\\text{time} & = (2.46 - 1) && \text{Do not forget, for one second the sound is not traveling}\\\text{Distance}& = 2 \times 250$
Our equation is:
$2(250) & = (331 + 0.6T)\cdot (2.46 - 1) & & \text{Simplify terms}.\\\frac{500}{1.46} & = \frac{1.46(331 + 0.6T)}{1.46} & & \text{Divide by} \ 1.46.\\342.47 -331 & = 331 + 0.6T -331 & & \text{Subtract 331 from both sides}.\\\frac{11.47}{0.6} & = \frac{0.6T}{0.6}& & \text{Divide by} \ 0.6.\\19.1 & = T$
Solution
The temperature is 19.1 degrees Celsius.
## Lesson Summary
• If dividing a number outside of parentheses will produce fractions, it is often better to use the Distributive Property (for example, $3(x + 2) = 3x + 6$) to expand the terms and then combine like terms to solve the equation.
## Review Questions
1. Solve the following equations for the unknown variable.
1. $3 (x - 1) - 2 (x + 3) = 0$
2. $7(w + 20) - w = 5$
3. $9(x - 2) = 3x + 3$
4. $2 \left (5a - \frac{1} {3}\right ) = \frac{2} {7}$
5. $\frac{2} {9} \left (i + \frac{2}{3}\right ) = \frac{2} {5}$
6. $4 \left (v + \frac{1} {4}\right ) = \frac{35} {2}$
7. $\frac{s - 4} {11} = \frac{2} {5}$
8. $\frac{p} {16} - \frac{2p} {3} = \frac{1} {9}$
2. An engineer is building a suspended platform to raise bags of cement. The platform has a mass of 200 kg, and each bag of cement is 40 kg. He is using two steel cables, each capable of holding 250 kg. Write an equation for the number of bags he can put on the platform at once, and solve it.
3. A scientist is testing a number of identical components of unknown resistance which he labels $x \Omega$. He connects a circuit with resistance $(3x + 4)\Omega$ to a steady 12 Volt supply and finds that this produces a current of 1.2 Amps. What is the value of the unknown resistance?
5. 35 miles
Feb 22, 2012
Sep 28, 2014 |
MullOverThings
Useful tips for everyday
# How do you solve a system of linear equations by substitution?
## How do you solve a system of linear equations by substitution?
1. Step 1 : First, solve one linear equation for y in terms of x .
2. Step 2 : Then substitute that expression for y in the other linear equation.
3. Step 3 : Solve this, and you have the x -coordinate of the intersection.
4. Step 4 : Then plug in x to either equation to find the corresponding y -coordinate.
## How do you find the value of a linear equation?
The slope-intercept form of a linear equation is y = mx + b. In the equation, x and y are the variables. The numbers m and b give the slope of the line (m) and the value of y when x is 0 (b). The value of y when x is 0 is called the y-intercept because (0,y) is the point at which the line crosses the y-axis.
## What are the 4 steps to solving an equation?
We have 4 ways of solving one-step equations: Adding, Substracting, multiplication and division. If we add the same number to both sides of an equation, both sides will remain equal.
## What are the two methods of solving a system of linear equations algebraically?
There are two methods that will be used in this lesson to solve a system of linear equations algebraically. They are 1) substitution, and 2) elimination. They are both aimed at eliminating one variable so that normal algebraic means can be used to solve for the other variable.
## What is an example of a linear equation?
The standard form for linear equations in two variables is Ax+By=C. For example, 2x+3y=5 is a linear equation in standard form. When an equation is given in this form, it’s pretty easy to find both intercepts (x and y). This form is also very useful when solving systems of two linear equations.
## Which is the standard form of linear equation?
The Standard Form for a linear equation in two variables, x and y, is usually given as Ax + By = C where, if at all possible, A, B, and C are integers, and A is non-negative, and, A, B, and C have no common factors other than 1.
## How to find the lsolve function in a system of equations?
Create a matrix that contains the coefficients of the variables in your system of equations. Create a vector of the constants appearing on the right-hand side of the system of equations. Evaluate the lsolve function using the matrix and the vector as the inputs. You can also assign the lsolve function to a variable if desired.
## Do you know how to solve a linear function?
As you can see, you know how to solve all of these problems from studying equations. Now you just have to be a little fancier with how you name your equation. In the next lesson, we will continue our study of functions by taking a look at quadratic functions.
## How to solve a linear system of equations in Mathcad?
Also, Mathcad will warn you if you have an inconsistent system of equations—that is, one for which a solution does not exist. Mathcad has a built-in function for solving a linear system of equations called lsolve. To use lsolve, perform the following steps: Create a matrix that contains the coefficients of the variables in your system of equations.
## What’s the solution to the system of linear equations?
And we are done! The solution is: Just like on the Systems of Linear Equations page. Quite neat and elegant, and the human does the thinking while the computer does the calculating. Just For Fun Do It Again! |
# Lesson Worksheet: Multistep Multiplication Word Problems Mathematics
In this worksheet, we will practice solving multistep word problems involving multiplication.
Q1:
A car is traveling at 40 miles per hour, and a train is traveling at 32 miles per hour in the same direction. How much farther than the train will the car have traveled after 2 hours?
Q2:
A family buys 3 kilograms of bananas, 8 kilograms of apples, and 7 kilograms of oranges every month. One kilogram of bananas costs 13 LE, of apples costs 17 LE, and of oranges costs 5 LE. How much do they spend on fruit each month?
Q3:
A girl had 151 LE. She bought a dress for 52 LE and then bought 7 books for 11 LE each. How much money is left?
Q4:
One kilogram of meat costs 86 LE and one litre of juice costs 800 piasters. How much do 10 kg of meat and 13 litres of juice cost?
Q5:
A shop is having a back-to-school sale and is offering special deals on pens, pencils, and rulers. Use the shopping list to calculate how much Yara will spend on classroom supplies.
Q6:
A child bought 4 packs of coloring pencils, which cost ten pounds each. Then they bought 10 pens for 2 pounds each. How much did they spend?
Q7:
A man went to the market and bought 5 kilograms of guavas, 4 kilograms of broccoli, 2 kilograms of bananas, 3 kilograms of tomatoes, and 6 kilograms of carrots. Use the table to calculate how much he paid.
Kind Price per kg (LE) Green Beans Lemons Tomatoes Broccoli Carrots Pepper Bananas Apples Potatoes Grapes Guavas Mangoes 3 4 2 5 3 4 9 10 5 9 7 8
Q8:
A family of 8 eats 9 kilograms of bananas, 2 kilograms of apples, and 4 kilograms of oranges every month. Per kilogram, the price is 2 LE for oranges, 9 LE for apples, and 7 LE for bananas. How much do they spend on fruit each month?
Q9:
Michael bought 5 kg of mangoes, which cost 7 LE/kg, and 2 kg of bananas, which cost 9 LE/kg. How much did he pay?
Q10:
The table shows the number and prices of home appliances in a store. Calculate the total price of all the appliances in the store.
ApplianceNumber of AppliancesPrice of One Unit
Fridge 10 4 000
Fan 10 800
Coffee Maker 10 700
Heater 25 500
Blender 25 300
This lesson includes 2 additional questions and 81 additional question variations for subscribers. |
Top
# Complex Fractions
A fraction is defined by the ratio of two numbers and is represented in the form of $\frac{a}{b}$, where variable 'a' denotes the value known as numerator and variable 'b' denotes the value known as denominator. Here b must not be equal to zero.
Thus, the fractions are classified as
• Proper Fraction
• Improper Fraction
• Complex Fraction
• Mixed Fraction
Here, we shall discuss about complex fractions.
Complex Fractions:
Complex fraction is a fraction in which that either numerator or denominator or both contain fractions. This is also referred as rational expression.
For example:
• $\frac{\frac{2}{5}-7}{\frac{6}{11}+1}$
• $\frac{x^{2}+\frac{1}{2}}{\frac{x}{y}}$
Rules for Simplifying Complex Fractions
Step 1:
Simplify numerator as well as denominator into single fractions, if there are any algebraic operations.
Step 2:
Reciprocate denominator and multiply it with numerator.
Step 3:
Finally reduce the fraction.
Let us understand this process with the help of an example:
Example: Simplify the following complex fraction
$\frac{\frac{1}{2}-\frac{1}{3}}{1-\frac{1}{12}}$
Solution:
Given fraction is
$\frac{\frac{1}{2}-\frac{1}{3}}{1-\frac{1}{12}}$
Step 1:
Simplifying numerator
$\frac{1}{2}-\frac{1}{3}$
= $\frac{1}{6}$
Step 2:
Simplifying denominator
$1-$$\frac{1}{12}$
= $\frac{11}{12}$
Step 3:
Substituting above value back in given fraction
$\frac{\frac{1}{6}}{\frac{11}{12}}$
= $\frac{1}{6}\times \frac{12}{11}$
= $\frac{2}{11}$
Related Calculators Adding Complex Fractions Calculator Complex Fraction Calculator Adding Complex Numbers Calculator Complex Number Calculator
More topics in Complex Fractions Simplifying Complex Fractions
*AP and SAT are registered trademarks of the College Board. |
Introduction to Fractions
When you divide (fractionate) something into parts, two or more, you have what is known as a common fraction. A common fraction is usually written as two numbers; a top number and a bottom number. Common fractions can also be expressed in words. The number that is on top is called the numerator, and the number on the bottom is called the denominator (the prefix 'de-' is Latin for reverse) or divisor.
${\displaystyle {\frac {\text{numerator}}{\text{denominator}}}}$
These two numbers are always separated by a line, which is known as a fraction bar. This way of representing fractions is called display representation. Common fractions are, more often than not, simply known as fractions in everyday speech.
The numerator in any given fraction tells you how many parts of something you have on hand. For example, if you were to slice a pizza for a party into six equal pieces, and you took two slices of pizza for yourself, you would have ${\displaystyle {\tfrac {2}{6}}}$ (pronounced two-sixths) of that pizza. Another way to look at it is by thinking in terms of equal parts; when that pizza was cut into six equal parts, each part was exactly ${\displaystyle {\tfrac {1}{6}}}$ (one-sixth) of the whole pizza.
The denominator tells you how many parts are in a whole, in this case your pizza. Your pizza was cut into six equal parts, and therefore the entire pizza consists of six equal slices. So when you took two slices for yourself, only four slices of pizza remain, or ${\displaystyle {\tfrac {4}{6}}}$ (four-sixths).
Also keep in mind that numerator can never be zero. It makes no sense to have zero divided into parts. For instance, the fraction ${\displaystyle {\tfrac {0}{6}}}$ is equal to zero, because you can not have six slices of nothing. If the denominator is zero, then the fraction has no meaning or is considered undefined since it may depend upon the mathematical setting one is working on, for the purpose of this chapter, we will consider that it has no meaning.
Another way of representing fractions is by using a diagonal line between the numerator and the denominator.
${\displaystyle 1/2\ }$
In this case, the separator between the numerator and the denominator is called a slash, a solidus or a virgule. This method of representing fractions is called in-line representation, meaning that the fraction is lined up with the rest of the text. You will often see in-line representations in texts where the author does not have any way to use display representation.
Proper Fractions
The fraction in the pizza analogy we just used is known as a proper fraction. In a proper fraction, the numerator (top number) is always smaller than the denominator (bottom number). Thus, the value of a proper fraction is always less than one. Proper fractions are generally the kind you will encounter most often in mathematics.
Improper Fractions
When the numerator of a fraction is greater than, or equal to the denominator, you have an improper fraction. For example, the fractions ${\displaystyle {\tfrac {5}{3}},{\tfrac {2}{1}}}$ and ${\displaystyle {\tfrac {6}{6}}}$ are all considered improper fractions. Improper fractions always have a value of one whole or more. So with ${\displaystyle {\tfrac {6}{6}}}$ , the numerator says you have 6 pieces, but 6 is also the number of the whole, so the value of this fraction is one whole. It is as if no one took a slice of pizza after you cut it.
In the case of ${\displaystyle {\tfrac {5}{3}}}$ , one whole of something is divided into three equal pieces, but on hand you have five pieces (you had two pizzas, each divided into three slices, and you ate one slice). This means you have two pieces extra, or two pieces greater than one whole. This concept may seem rather confusing and strange at first, but as you become better in math you will eventually put two and two together to the get the whole picture (okay, bad pun).
Mixed Fractions
When a whole number is written next to a fraction, such as ${\displaystyle 2{\tfrac {1}{3}}}$ (two and one-third) you are seeing what is called a mixed fraction. A mixed fraction is understood as being the sum, or total, of both the whole number and fraction. The number two in ${\displaystyle 2{\tfrac {1}{3}}}$ stands for two wholes - you also have a third more of something, which is the ${\displaystyle {\tfrac {1}{3}}}$ .
Simplifying Fractions
Sometimes in mathematics you will need to rewrite a fraction in smaller numbers, while also keeping the value of the fraction the same. This is known as simplifying, or reducing to lowest terms. It should be mentioned that a fraction which is not reduced is not intrinsically incorrect, but it may be confusing for others reviewing your work. There are two ways to simplify fractions, and both will be useful anytime you work with fractions, so it is recommended you learn both methods.
To reiterate, reducing fractions is essentially replacing your original fraction with another one of equal value, called an equivalent fraction. Below are a few examples of equivalent fractions.
${\displaystyle {\frac {4}{8}}={\frac {1}{2}},{\frac {8}{12}}={\frac {2}{3}},{\frac {6}{10}}={\frac {3}{5}}}$
When the fraction ${\displaystyle {\frac {4}{8}}}$ is reduced to lowest terms, it then becomes ${\displaystyle {\frac {1}{2}}}$ , because four pieces out of a total of eight is exactly one-half of all available pieces. A fraction is also in its lowest terms when both the numerator and denominator cannot be divided evenly by any number other than one.
Division Method
To reduce a fraction to lowest terms, you must divide the numerator and denominator by the largest whole number that divides evenly into both. For example, to reduce the fraction ${\displaystyle {\tfrac {3}{9}}}$ to lowest terms, divide the numerator (3) and denominator (9) by three.
${\displaystyle {\frac {3\div 3}{9\div 3}}={\frac {1}{3}}}$
If the largest whole number is not obvious, and many times it is not, divide the numerator and denominator by any number (except one) that divides evenly into each, and then repeat the process until the fraction is in lowest terms. Know that if both numbers are even, then you divide each number by 2.
For clarity, below are a few examples of reducing fractions using this method.
Example
Reduce ${\displaystyle {\tfrac {12}{20}}}$ to lowest terms.
Solution
In this problem, the largest whole number is difficult to see, so we first divide the numerator and denominator by two, as shown:
${\displaystyle {\frac {12\div 2}{20\div 2}}={\frac {6}{10}}}$
Next divide by two again:
${\displaystyle {\frac {6\div 2}{10\div 2}}={\frac {3}{5}}}$
There are no whole numbers left which can divide evenly into ${\displaystyle {\tfrac {3}{5}}}$ , so the problem is finished.
${\displaystyle {\tfrac {12}{20}}}$ reduced to lowest terms is ${\displaystyle {\tfrac {3}{5}}}$
Example
Reduce ${\displaystyle {\tfrac {18}{24}}}$ to lowest terms.
Solution
Divide the numerator and denominator by six, as shown:
${\displaystyle {\frac {18\div 6}{24\div 6}}={\frac {3}{4}}}$
${\displaystyle {\tfrac {18}{24}}}$ reduced to lowest terms is ${\displaystyle {\tfrac {3}{4}}}$ .
Example
Reduce ${\displaystyle {\tfrac {112}{126}}}$ to lowest terms.
Solution
In this problem, the largest whole number is not immediately apparent, so we first divide the numerator and denominator by two, as shown:
${\displaystyle {\frac {112\div 2}{126\div 2}}={\frac {56}{63}}}$
Next divide by seven:
${\displaystyle {\frac {56\div 7}{63\div 7}}={\frac {8}{9}}}$
${\displaystyle {\tfrac {112}{126}}}$ reduced to lowest terms is ${\displaystyle {\tfrac {8}{9}}}$ .
Greatest Common Factor Method
The second method of simplifying fractions involves finding the greatest common factor between the numerator and the denominator. We do this by breaking up both the numerator and the denominator into their prime factors (greatest common factor = 2x2 = 4):
${\displaystyle {\frac {12}{16}}={\frac {2\times 2\times 3}{2\times 2\times 2\times 2}}={\frac {2\!\!\!/\times 2\!\!\!/\times 3}{2\!\!\!/\times 2\!\!\!/\times 2\times 2}}={\frac {3}{4}}}$
It is implied that any part is multiplied by one. If we divide 2s out of the factorized fraction, we are left with one 2 in the denominator.
${\displaystyle {\frac {4}{8}}={\frac {1\times 2\!\!\!/\times 2\!\!\!/}{2\times 2\!\!\!/\times 2\!\!\!/}}={\frac {1}{2}}}$
It is best to practice these skills of reducing fractions until you feel confident enough to do them on your own. Remember, practice makes perfect.
Raising Fractions to Higher Terms
To raise a fraction to higher terms is to rewrite it in larger numbers while keeping the fraction equivalent to the original in value. This is, for all intents and purposes, the exact opposite of simplifying a fraction.
Example
Convert ${\displaystyle {\frac {2}{5}}}$ to a fraction with denominator of 15.
${\displaystyle {\frac {2}{5}}={\frac {?}{15}}}$
Step 1
Ask yourself “what number multiplied by 5 equals 15?” To find the answer simply divide 5 into 15.
${\displaystyle {15}\div {5}=3}$
Step 2
After dividing the two denominators, you must take the answer and multiply it by the numerator you already have. So in this case, we multiply 2 by 3 to find the missing numerator.
${\displaystyle {2}\times {3}={6}}$
${\displaystyle {\frac {2}{5}}={\frac {6}{15}}}$
Changing Improper Fractions to Mixed Numbers
Oftentimes you will encounter fractions in their improper form. While this may be useful in some instances, it is usually best to convert the fraction into simplest form, or mixed fraction.
To convert an improper fraction into a mixed fraction, divide the denominator into the numerator.
Example
Change ${\displaystyle {\frac {13}{2}}}$ into a mixed fraction.
Solution
Divide 13 by 2, use long division to obtain quotient and a remainder. ${\displaystyle {13}\div {2}={6\ {\mbox{r}}\ 1}}$
To form the proper fraction part of the answer, we use the divisor (2) as the denominator, and the remainder (r1) as the numerator. Finally, we take the answer to the division problem, in this case 6, and use that as the whole number.
Hence ${\displaystyle {\frac {13}{2}}}$ in mixed fraction form is ${\displaystyle 6\,\!{\frac {1}{2}}}$
Adding Fractions With The Same Denominator
In order to add fractions with the same denominator, you only need to add the numerators while keeping the original denominator for the sum.
${\displaystyle {\frac {1}{5}}+{\frac {3}{5}}={\frac {1+3}{5}}={\frac {4}{5}}}$
Adding fractions with the same denominator is the rule but it begs the question why? Why can’t (or shouldn't) I add both numerators and denominators?
${\displaystyle {\frac {1}{4}}+{\frac {1}{4}}={\frac {1+1}{4+4}}={\frac {2}{8}}}$
To make sense of this try taking a 12 inch ruler and drawing a 3 inch horizontal line (1/4 of a foot) and then on the end add another 3 inch line (1/4 of a foot). What is the total length of the line? It should be 6 inches (1/2 a foot) and not 2/8 of a foot (3 inches). In essence it seems we can only add like items and like items are terms that have the same denominator and we add them up by adding up numerators.
When adding fractions that do not have the same denominator, you must make the denominators of all the terms the same. We do this by finding the least common multiple of the two denominators.
Least common multiple of 4 and 5 is 20; therefore, make the denominators 20:
${\displaystyle {\frac {1}{4}}+{\frac {2}{5}}={\frac {1\times 5}{4\times 5}}+{\frac {2\times 4}{5\times 4}}={\frac {5}{20}}+{\frac {8}{20}}}$
Now that the common denominators are the same, perform the usual addition:
${\displaystyle {\frac {5+8}{20}}={\frac {13}{20}}}$
Subtracting Fractions
Subtracting Fractions With The Same Denominator
To subtract fractions sharing a denominator, take their numerators and subtract them in order of appearance. If the numerator's difference is zero, the whole difference will be zero, regardless of the denominator.
${\displaystyle {\frac {6}{13}}-{\frac {2}{13}}={\frac {6-2}{13}}={\frac {4}{13}}}$
Subtracting Fractions With Different Denominators
To subtract one fraction from another, you must again find the least common multiple of the two denominators.
Least common multiple of 4 and 6 is 12; therefore, make the denominator 12:
${\displaystyle {\frac {3}{4}}-{\frac {1}{6}}={\frac {3\times 3}{4\times 3}}-{\frac {1\times 2}{6\times 2}}={\frac {9}{12}}-{\frac {2}{12}}}$
Now that the denominator is same, perform the usual subtraction.
${\displaystyle {\frac {9-2}{12}}={\frac {7}{12}}}$
Multiplying Fractions
Multiplying fractions is very easy. Simply multiply the numerators of the fractions to find the numerator of the answer. Then multiply the denominators of the fractions to find the denominator of the answer. In other words, it can be said “top times top equals top”, and “bottom times bottom equals bottom.” This rule is used to multiply both proper and improper fractions, and can be used to find the answer to more than two fractions in any given problem.
Example
Multiply ${\displaystyle {\frac {1}{2}}\times {\frac {3}{4}}}$
Step 1
Multiply the numerators to find the numerator of the answer. ${\displaystyle 1\times 3=3}$
Step 2
Multiply the denominators to find the denominator of the answer. ${\displaystyle 2\times 4=8}$
${\displaystyle {\frac {1}{2}}\times {\frac {3}{4}}={\frac {3}{8}}}$
Whole and Mixed Numbers
When you need to multiply a fraction by a whole number, you must first convert the whole number into a fraction. This, fortunately, is not as difficult as it may sound; just put the whole number over the number one. Then proceed to multiply as you would with any two fractions. An example is given below.
${\displaystyle {\frac {3}{4}}\times {5}={\frac {3}{4}}\times {\frac {5}{1}}={\frac {15}{4}}=3\,\!{\frac {3}{4}}}$
If a problem contains one or more mixed numbers, you must first convert all mixed numbers into improper fractions, and multiply as before. Finally, convert any improper fraction back to a mixed number.
${\displaystyle 1\,\!{\frac {1}{2}}\times 2\,\!{\frac {1}{4}}={\frac {3}{2}}\times {\frac {9}{4}}={\frac {27}{8}}=3\,\!{\frac {3}{8}}}$
Dividing Fractions
To divide fractions, simply exchange the numerator and the denominator of the second term in the problem, then multiply the two fractions.
${\displaystyle {\frac {1}{2}}\div {\frac {3}{4}}}$
Invert the second fraction:
${\displaystyle {\frac {1}{2}}\times {\frac {4}{3}}}$
Multiply:
${\displaystyle {\frac {1\times 4}{2\times 3}}={\frac {4}{6}}}$
Always check to see if simplifying the resulting fraction can be done:
${\displaystyle {\frac {4}{6}}={\frac {2}{3}}}$ |
# Line Graph: Exercise-5
#### Overview:
Questions and Answers Type: MCQ (Multiple Choice Questions). Main Topic: Data Interpretation. Data Interpretation Sub-topic: Line Graph Questions and Answers. Number of Questions: 5 Questions with Solutions.
Directions: Study the following graph carefully and answer the questions given below it.
Number of trees planted by the three different countries over the years.
1. In which year was the total number of trees planted by all the countries together highest?
1. 2014
2. 2015
3. 2016
4. 2017
Solution: Number of trees planted by all the countries together over the years.
In 2013, $$= 20 + 30 + 40$$ $$= 90 \ Millions$$ In 2014, $$= 30 + 20 + 40$$ $$= 90 \ Millions$$ In 2015, $$= 20 + 40 + 50$$ $$= 110 \ Millions$$ In 2016, $$= 20 + 40 + 30$$ $$= 90 \ Millions$$ In 2017, $$= 40 + 50 + 30$$ $$= 120 \ Millions$$ Hence, in the year 2017, the total number of trees planted by all the countries together was highest.
1. What was the difference between the trees planted by the country B in the year 2016 and the number of trees planted by the country C in the year 2017?
1. 10 Millions
2. 20 Millions
3. 30 Millions
4. 40 Millions
Solution: Required difference, $$= 40 - 30$$ $$= 10 \ Millions$$
1. What was the average number of trees planted in the year 2015 by all the countries together?
1. 36.67 Millions
2. 32.25 Millions
3. 30.15 Millions
4. 28.63 Millions
Solution: Average number of trees planted by all the countries together in the year 2015, $$= \frac{20 + 40 + 50}{3}$$ $$= 36.67 \ Millions$$
1. Total number of trees planted by the countries B and C together in the year 2013 was approximately what percent of the total number of trees planted by the countries B and C together in the year 2015
1. 67.67%
2. 77.78%
3. 86.56%
4. 62.32%
Solution: Required percentage, $$= \frac{(30 + 40)}{(40 + 50)} \times 100$$ $$= 77.78 \ \%$$
1. What was the respective ratio between the number of trees planted by the country A in the year 2016, the number of trees planted by the country B in the year 2015, and the number of trees planted by the country C in the year 2017?
1. 4:3:2
2. 3:2:4
3. 2:4:3
4. 3:4:2
Solution: Required ratio, $$= 20 : 40 : 30$$ $$= 2 : 4 : 3$$ |
+0
# Help
0
30
1
A line through the points $(2, -9)$ and $(j, 17)$ is parallel to the line $2x + 3y = 21$. What is the value of $j$?
Guest Dec 7, 2017
#1
+5552
+1
First let's find the slope of the line 2x + 3y = 21
2x + 3y = 21
Subtract 2x from both sides of the equation.
3y = -2x + 21
Divide through by 3 .
y = - $$\frac23$$x + 7
Now we can see that the slope of this line is - $$\frac23$$ . So....
the slope between any two points of a parallel line also = -$$\frac23$$
the slope between (2, -9) and (j, 17) = - $$\frac23$$
$$\frac{17--9}{j-2}\,=\,-\frac23\\~\\ \frac{26}{j-2}\,=\,-\frac23 \\~\\ 26=-\frac23(j-2)\\~\\ -39=j-2\\~\\ -37=j$$
Here's a graph to check this: https://www.desmos.com/calculator/ovina9gch0
hectictar Dec 7, 2017
Sort:
#1
+5552
+1
First let's find the slope of the line 2x + 3y = 21
2x + 3y = 21
Subtract 2x from both sides of the equation.
3y = -2x + 21
Divide through by 3 .
y = - $$\frac23$$x + 7
Now we can see that the slope of this line is - $$\frac23$$ . So....
the slope between any two points of a parallel line also = -$$\frac23$$
the slope between (2, -9) and (j, 17) = - $$\frac23$$
$$\frac{17--9}{j-2}\,=\,-\frac23\\~\\ \frac{26}{j-2}\,=\,-\frac23 \\~\\ 26=-\frac23(j-2)\\~\\ -39=j-2\\~\\ -37=j$$
Here's a graph to check this: https://www.desmos.com/calculator/ovina9gch0
hectictar Dec 7, 2017
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Solving Systems
# Systems
A system of equations is a list of equations with more than one variable that work together to show specific data. Systems use equations to portray data, and what the equations have in common is what the equations are ultimately showing. Typically, one equation in a system cannot be solved. To solve for all the variables in a system, you have to connect them to find a common point. One example of a system is this.
Say you have a rectangular field. The field is 20 meters around, and the length of the field is 3 meters longer than the width. What are the dimensions?
With this, we can set up to equations. Our perimeter, 20=2l+2w, and our length-width relation, l=w+3. This is a system of equations, and with them, you can solve for both variables. Now, we will discuss the different types of systems.
## Linear
A linear system is a system of equations with all variables with a degree of one or lower. There are no terms with a degree greater than one. Linear systems have one intersection point, and thus one solution (unless lines are parallel). Here are a few examples of linear systems:
2x+3y=6
x-3y=-3
y=2x-3
x+y=2
x=3y
2x-3=6y+2
Each of these systems can be solved for x and y. Each equation represents a line on a coordinate plane, and where the lines intersect is the solution. When pertaining to systems of more than two dimensions, it will refer to where the planes or spaces intersect (a little advanced…).
## Nonlinear
A nonlinear system pertains to systems with terms of more than one. In these systems, you might find variables like x2, y3, or xy. The graphs for these are not lines, but they are curves rather. In these systems, you might encounter more than one solution. Here are a few examples of nonlinear systems.
10=2l+2w
40=lw
x2-y=17
3x+2y2=8
(2x+y)(-x+2y)=0
(3x2+6xy-4y2)2=143
These equations can get really complex. You can get no solution quite often in nonlinear systems. These systems can be a pain, and they are hard to solve. Again, this are equations, a lot like functions. When you end up with more than one variable for each equation, then things start becoming very abstract and unpredictable.
## Three or more Dimensions
These types of systems have more than two variables and more than two equations. Because of this, these equations cannot be graphed on a two dimensional plane. Rather, they exist in three or more dimensions. To obtain a solution, you must have at least the same number of equations as there are variables. Not only are these equations complex, they are very time consuming. Here are some examples of such systems:
2x+3y-4z=31
x-y+2z=12
-2x+6y-2z=35
a+b=c-d
2a+b=3c+2d
7+a-b=c+d
-a+2c=b-2d+2
(a-b)(2c+d)=1
a2-2abc+3d=-4
(2a+2)(-2b-3)(4c-5)(7d+2)=0
2a2+ab2c2-3d3=25
These types of systems are very abstract and theoretical. Solving them can be ridiculous and tiresome as well, and these types of systems will not be thoroughly discussed in this article.
# Solving a System
The solution of a system is the values of the variables that are true for all the equations in the system. The solution is always written as a coordinate. To solve a system, you must work simultaneously with each equation. Here are some popular methods on solving systems of equations.
## Substitution
This is probably one of the most useful ways of solving a system. The reason is that no matter what, it will always work, until you begin reaching high degree polynomials. The way it works is that you can solve for certain variables which can be plugged into the other equation. That gets rid of a variable. For example, if you have two equations with x and y, you can solve one of the equations for y (getting y by itself), and then plug that expression into the y part in the other equation. That eliminates the y leaving only x's, which can then be solved. Let's say we had a system.
(1)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} 2x+3y = 13 \\ x-y = -1 \end{align} \end{align}
First, identify what variable you want to solve for. Notice how simple the bottom equation is. It would be very easy to solve for a variable there. Solve for x by adding y to both sides. This will give you x=y-1. That equation is solved for x. Now we know what x equals, so we can put that value into x's spot in the top equation. We will put y-1 into x. We then get 2(y-1)+3y=13. Now, all we have is y, and we can solve for it. Begin by distributing your two to y and -1. Combine your y terms to get 5y-2=13. Now solve. When we finish, we get y=3. We are not finished yet! Now we have to get x. Substitute 3 into y in any equation. We will use the bottom because it is simpler. You end up with x-3=-1. When we solve, we get x=2. So, putting our answer in coordinate form, our solution to the system is (2,3).
The addition method is my favorite method. Although it becomes cumbersome in many equations, it can be real simple with just two. What it does is it easily gets rid of a variable while keeping your equations small. Let's say you have two equations.
(2)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} x+2y = 7 \\ 3x-2y = -3 \end{align} \end{align}
Notice that each equation has a 2y term, one is positive and one is negative. What you can do is add the equations vertically, adding the coefficients and constants with their respective term. So, you can add the x's, the y's, and the constant numbers. So, for this system, we can add x+3x, 2y+(-2)y, and 7+(-3). This gives you one equation, 4x+0y=4. This becomes 4x=4. Notice, the y term is now gone. The reason this happened is because the y terms for each equation were opposites, and they added each other out. This does not work for if each y is positive and one is negative, but one must be positive while the other is the negative of equivalent absolute value. Be careful. Now with 4x=4, we find that x=1, and we can then plug that value into one of the equations and get y=3. So our answer is (1,3).
Now, we know that this method works when the coefficients are set up correctly. However, the coefficients are usually never set up correctly. So, does this mean we can practically never use the addition method? Of course not! We can actually manipulate the equations in a system to where they can be set up for this method. Let's say we had a system:
(3)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} 4x-2y = 5 \\ 3x-5y = -7 \end{align} \end{align}
How can we manipulate this system to make it set up for the addition method? What we need to do is get one set of variable coefficients, say y in this instance, to be opposites of equivalent absolute value. So we have -2 and -5. Find a number that each can be easily divided into. For this case, that number is 10. So, we want one value to be -10 and the other to be 10. What number can we multiply by -2 to get -10? When we multiply -2 by 5, we get -10. Similarly, if we multiply -5 by -2, we can get 10. Now, we have our multiplicands 5 for the top and -2 for the bottom. Multiply every term in each equation by its respective multiplicand. Now, our system becomes
(4)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \begin{align} 20x-10y = 25 \\ -6x+10y = 14 \end{align} \end{align}
Notice, now we are ready to add our equations! When we add our equations, we get 14x=39. So x is 39/14. When we put this rather ugly value into x, we can solve for y which becomes 43/14. Our answer is (39/14,43/14).
Don't try doing this with systems of more than two equations.
## Matrix Inversion
Matrices are very useful when it comes to systems of equations. First, you must understand that a system of equations is actually the product of two matrices. Think of this: what do you know about matrix multiplication? Each row is multiplied by each column. So, that means that each column in the product was multiplied by a row in the second matrix. In other words, every column in the product has a bit from the second matrix row. So, knowing this, we can factor our system into a matrix. Here is the relation.
(5)
\begin{align} \definecolor{darkgreen}{rgb}{0.90,0.91,0.859}\pagecolor{darkgreen} \left( \begin{align} 2x+3y = 18 \\ 4x-y = 22 \end{align} \right) = \left( \left[ \begin{array}{cc} 2 & 3 \\ 4 & -1 \end{array} \right] \left[ \begin{array}{c} x \\ y \end{array} \right] = \left[ \begin{array}{c} 18 \\ 22 \end{array} \right] \right) \end{align}
As you can see, it can get pretty intense. How does this thing work? Remember how to multiply matrices? First, multiply row one in matrix one by column one in matrix two. This gives 2x+3y. Familiar? Then do the bottom row by the column, and you get 4x-y. So hopefully you see how this can actually work.
So, how do we use this to solve a system? First, you should know that the opposite of multiplication, when pertaining to matrices, is inversion. So, to solve a system of equations using matrix inversion, you simply take the inverse matrix on both sides. On the left side, you get an identity matrix, which is technically "1". |
## SOLVING TRIGONOMETRIC EQUATIONS
Note: If you would like a review of trigonometry, click on trigonometry.
Solve for x in the following equation.
Example 1:
There are an infinite number of solutions to this problem.
We can make the solution easier if we convert all the trigonometric terms to like trigonometric terms.
One common trigonometric identity is If we replace the term with , all the trigonometric terms will be tangent terms.
Replace with in the original equation and simplify.
Isolate the tangent term. To do this, rewrite the left side of the equation in an equivalent factored form.
The product of two factors equals zero if at least one of the factors equals zero. This means that if or
We just transformed a difficult problem into two easier problems. To find the solutions to the original equation, , we find the solutions to the equations and
and
How do we isolate the x in each of these equations? We could take the arctangent of both sides of each equation. However, the tangent function is not a one-to-one function.
Let's restrict the domain so the function is one-to-one on the restricted domain while preserving the original range. The graph of the tangent function is one-to-one on the interval If we restrict the domain of the tangent function to that interval , we can take the arctangent of both sides of each equation.
Since the period of equals , these solutions will repeat every units. The exact solutions are
where n is an integer.
The approximate values of these solutions are
where n is an integer.
You can check each solution algebraically by substituting each solution in the original equation. If, after the substitution, the left side of the original equation equals the right side of the original equation, the solution is valid.
You can also check the solutions graphically by graphing the function formed by subtracting the right side of the original equation from the left side of the original equation. The solutions of the original equation are the x-intercepts of this graph.
Algebraic Check:
Check solution x=1.249046
Left Side:
Right Side:
Since the left side of the original equation equals the right side of the original equation when you substitute 1.249046 for x, then 1.249046 is a solution.
Check solution x=-0.785398
Left Side:
Right Side:
Since the left side of the original equation equals the right side of the original equation when you substitute -0.785398 for x, then -0.785398 is a solution.
We have just verified algebraically that the exact solutions are and and these solutions repeat every units. The approximate values of these solutions are and and these solutions repeat every units.
Graphical Check:
Graph the equation Note that the graph crosses the x-axis many times indicating many solutions. Let's check a few of these x-intercepts against the solutions we derived.
Verify the graph crosses the x-axis at -0.785398. Since the period is , you can verify that the graph also crosses the x-axis again at -0.785398+3.14159265=2.356195 and at , etc.
Verify the graph crosses the x-axis at 1.249046. Since the period is , you can verify that the graph also crosses the x-axis again at 1.249046+3.14159265=4.39906387 and at , etc.
Note: If the problem were to find the solutions in the interval , then you choose those solutions from the set of infinite solutions that belong to the set 2.356195, 5.497787, 4.39906387, and
If you would like to work another example, click on Example.
If you would like to test yourself by working some problems similar to this example, click on Problem.
If you would like to go to the next section, click on Next.
[Algebra] [Trigonometry]
[Geometry] [Differential Equations]
[Calculus] [Complex Variables] [Matrix Algebra]
Author: Nancy Marcus |
# Median with Constraints
Directions: Create a statistical data set of at least 10 numbers such that:
1. All of the numbers in the data set are whole numbers.
2. The median is not a whole number.
3. The median is not part of the data set.
### Hint
What strategy do you use to find the median of a data set? Does this strategy work in all cases?
There are many correct answers. One statistical data set that would work is {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}. Here the median is 5.5. You could make this question more challenging by adding more constraints.
Source: Daniel Luevanos
## Solving One-Step Inequalities with Addition
Directions: Using the digits 0 to 9 at most one time each, place a digit …
1. 1,2,3,4,5,6,7,8,9,10
The median would be 5.5, it is not in the data set nor a while number
2. 10,11,12,13,14,15,16,17,18,19.
The median would be 14.5 which which is not in the data set or a whole number.
3. 1,2,3,4,5,6,7,8,9,10
The median is 5.5 which is NOT part of the data set or a whole number.
4. You could do 1,2,3,4,5,6,7,8,9,10. This is because the median is 5.5. That number is not in the data set. The number also is not a whole number. Another combo could be 30,31,32,33,34,35,36,37,38,39. The median is 35.5 which is not a whole number. It also is not in the data set.
5. 1,2.3.4 the median is 2.5
6. 1,2,3,4,5,6,7,8,9,10
The median would be 5.5, it is not in the data set nor a while number
7. 4,9,10,1. The median would be 9.5 which is not a whole number nor a number in the data set.
8. 10,12,14,16,18,20,22,24,26,28. The Mean is 19
9. What is the median 1, 2, 3, 4 , 5, 6, 7, 8, 9, 10?
I think the median is 5.5 because the medium is 5 plus .5 is 5.5 |
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## Chapter 7 Fractions Ex 7.1
Question 1.
Write the fraction representing the shaded portion.
Solution:
Question 2.
Colour the part according to the given fraction.
Solution:
Question 6.
Arya, Abhimanyu and Vivek shared lunch. Arya has brought two sandwiches, one made of vegetable and one of Jam. The other two boys forgot to bring their lunch. Arya agreed to share his sandwiches so that each person will have an equal share of each sandwich.
(a) How can Arya divide his sandwiches so that each person has an equal share?
(b) What part of a sandwich will each boy receive?
Solution:
(a) Arya has divided his sandwich into three equal parts.
So, each of them will get one part.
Question 7.
Kanchan dyes dresses. She had to dye 30 dresses. She has so far finished 20 dresses. What fraction of dresses has she finished?
Solution:
Total number of dresses to be dyed = 30
Number of dresses finished = 20
Question 8.
Write the natural numbers from 2 to 12. What fraction of them are prime numbers?
Solution:
Natural numbers between 2 and 12 are;
2,3,4, 5, 6, 7, 8, 9, 10,11, 12
Number of given natural numbers = 11
Number of prime numbers = 5
Question 9.
Write the natural numbers from 102 to 113. What fraction of them are prime numbers?
Solution:
Natural numbers from 102 to 113 are;
102,103,104,105,106, 107,108, 109,110, 111, 112,113
Total number of given natural numbers = 12
Prime numbers are 103, 107, 109, 113
∴ Number of prime numbers = 4
Question 10.
What fraction of these circles have X’s in them?
Solution:
Total number’of circles = 8
Number of circles having X’s in them = 4
Question 11.
Kristin received a CD player for her birthday. She bought 3 CDs and received 5 others as gifts. What fraction of her total CDs did she buy and what fraction did she receive as gifts?
Solution:
Number of CDs bought by her from the market = 3 |
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psol07 - Solutions to Homework 7 Math 55 1 1 The...
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Solutions to Homework 7, Math 55 1. The probability that Beatrix starts with the crown and throws it to Andrew is 1 3 · 1 2 = 1 6 ; similarly, the probability that Charles starts with the crown and throws it to Andrew is 1 6 . Therefore, the probability that Andrew has the crown after one turn is 1 6 + 1 6 = 1 3 . Similarly, the probability that Charles has the crown after one turn is 1 3 from Andrew, and 1 3 · 1 2 = 1 6 from Beatrix, so the total probability is 1 3 + 1 6 = 1 2 . On the other hand, if Beatrix gets the crown, it must be from Charles; therefore, the probability that Beatrix has the crown after one turn is 1 3 · 1 2 = 1 6 . 2. We have p ( A ) = 4 52 = 1 13 ; p ( B ) = 13 52 = 1 4 ; and p ( C ) = 26 52 = 1 2 . Now p ( A B ) = 1 52 = p ( A ) p ( B ), and p ( A C ) = 2 52 = p ( A ) p ( C ), but p ( B C ) = 13 52 = p ( B ) p ( C ). Thus, A and B are independent, and so are A and C , but B and C are not independent. If we add a joker to the deck, then p ( A ) = 4 53 ; p ( B ) = 13 53 ; and p ( C ) = 26 53 . However, p ( A B ) = 1 53 = p ( A ) p ( B ); p ( A C ) = 2 53 = p ( A ) p ( C ); and p ( B C ) = 13 53 = p ( B ) p ( C ). Thus, in this case, no pair of events from A, B, C is independent. 3. We have p ( A B ) = p ( S - ( A B )) = 1 - p ( A B ) = 1 - ( p ( A ) + p ( B ) - p ( A B )) = 1 - p ( A ) - p ( B ) + p ( A ) p ( B ) = (1 - p ( A ))(1 - p ( B )) = p ( A ) p ( B ). Therefore, A and B are independent. Similarly, p ( A B ) = p ( A - ( A B )) = p ( A ) - p ( A B ) = p ( A ) - p ( A ) p ( B ) = p ( A )(1 - p ( B )) = p ( A ) p ( B ), so A and B are independent. (Note that in the step p ( A - ( A B )) = p ( A ) - p ( A B ), we need to use the fact that A B A .) However, if A and A are independent, then p ( A ) p ( A ) = p ( A A ) = p ( ) = 0, which implies p ( A ) = 0 or p ( A ) = 0, and in the latter case, p ( A ) = 1 - p ( A ) = 1. Thus, A and A are not independent unless p ( A ) = 0 or p ( A ) = 1. 4. Let E be the event that the number is divisible by 3, and F be the event that the number is divisible by 5. Then p ( E ) = 300 900 = 1 3 ; and p ( F ) = 180 900 = 1 5 . Also, E F is the event that the number is divisible by 15, which has probability 60 900 = 1 15 = p ( E ) p ( F ). Therefore, the two given events are independent.
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# recurrence relation $f(n)=5f(n/2)-6f(n/4) + n$
I’ve been trying to solve this recurrence relation for a week, but I haven’t come up with a solution.
$f(n)=5f(n/2)-6f(n/4) + n$
Solve this recurrence relation for $f(1)=2$ and $f(2)=1$
At first seen it looks like a divide and conquer equation, but the $6f(n/4)$ confuses me.
Kind regards,
#### Solutions Collecting From Web of "recurrence relation $f(n)=5f(n/2)-6f(n/4) + n$"
You can only define $f(2^k)$. (Try to define $f(3)$. It is impossible.) Hence make the transformation
$$n=2^k, \quad \text{and define}\quad g(k)=f(2^k).$$
Then for $g$ you have that
$$g(0)=2, \quad g(1)=1\quad \text{and}\quad g(k+2)=5g(k+1)-6g(k)+2^{k+2}.$$
If $2^{k+2}$ was not there, then $g$ would have to be of the form $g(k)=c_12^k+c_23^k$. With this additional term the general solution of the recursive relation is of the form $g(k)=c_12^k+c_23^k+c_3k2^k$. Just find the values of the constants.
Suppose we start by solving the following recurrence:
$$T(n) = 5 T(\lfloor n/2 \rfloor) – 6 T(\lfloor n/4 \rfloor) + n$$
where $T(0) = 0.$
Now let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
We unroll the recursion to obtain an exact formula for all $n:$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} [z^j] \frac{1}{1-\frac{5}{2}z+\frac{6}{4} z^2} \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Observe that the roots of $$1-\frac{5}{2}z+\frac{6}{4} z^2 \quad\text{are}\quad\rho_0=1\quad\text{and}\quad\rho_1=\frac{2}{3}.$$
It follows that the coefficients of the rational term have the form
$$c_0 + c_1 \left(\frac{3}{2}\right)^j.$$
Actually solving for $c_{0,1}$ we obtain the formula
$$[z^j] \frac{1}{1-\frac{5}{2}z+\frac{6}{4} z^2} = 3 \left(\frac{3}{2}\right)^j – 2.$$
This gives the exact formula for $T(n):$
$$T(n) = \sum_{j=0}^{\lfloor \log_2 n \rfloor} \left(3 \left(\frac{3}{2}\right)^j – 2\right) \sum_{k=j}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
Now for an upper bound on this consider a string of one digits, which gives
$$T(n)\le \sum_{j=0}^{\lfloor \log_2 n \rfloor} \left(3 \left(\frac{3}{2}\right)^j – 2\right) \sum_{k=j}^{\lfloor \log_2 n \rfloor} 2^k = \frac{27}{2} \times 3^{\lfloor \log_2 n \rfloor} – (12+4\lfloor \log_2 n \rfloor) 2^{\lfloor \log_2 n \rfloor} -\frac{1}{2}.$$
For a lower bound consider a one followed by a string of zeros, which gives
$$T(n)\ge \sum_{j=0}^{\lfloor \log_2 n \rfloor} \left(3 \left(\frac{3}{2}\right)^j – 2\right) 2^{\lfloor \log_2 n \rfloor} = 9 \times 3^{\lfloor \log_2 n \rfloor} – (8+2\lfloor \log_2 n \rfloor) 2^{\lfloor \log_2 n \rfloor}.$$
Joining the two bounds it follows that $T(n)$ is asymptotic as follows:
$$T(n)\in\Theta\left(3^{\lfloor \log_2 n \rfloor}\right) = \Theta(2^{\log_2 n \times \log_2 3}) = \Theta(n^{\log_2 3})$$
with the leading coefficient between $9/2$ and $9.$
Now we are supposed to have $f(0)=0$, $f(1)=2$, and $f(2)=1,$ so start with
$$T(n) + \left(3 \left(\frac{3}{2}\right)^{\lfloor \log_2 n \rfloor} – 2\right) 2^{\lfloor \log_2 n \rfloor}.$$
This is equal to two at $n=1$ but equal to twelve at $n=2$, so we need an additional contribution of
$$T(n) + \left(3 \left(\frac{3}{2}\right)^{\lfloor \log_2 n \rfloor} – 2\right) 2^{\lfloor \log_2 n \rfloor} – 11 (1-d_{\lfloor \log_2 n \rfloor-1})\times \left(3 \left(\frac{3}{2}\right)^{\lfloor \log_2 n \rfloor-1} – 2\right) 2^{\lfloor \log_2 n \rfloor-1}.$$
This simplifies to (for $n\ge 2$)
$$T(n) + 3^{\lfloor \log_2 n \rfloor+1} -2^{\lfloor \log_2 n \rfloor+1} – 11 (1-d_{\lfloor \log_2 n \rfloor-1})\times \left(3^{\lfloor \log_2 n \rfloor} -2^{\lfloor \log_2 n \rfloor}\right).$$
The behavior here is highly erratic but the formula is exact and the order of growth remains the same. Here is another Master Theorem computation at MSE. |
This is a paragraph.
# The depressed cubic as area
We investigated the area property of our very own triangle, and guess what! The depressed cubic showed up.
Ww!!! How???
Consider with vertices at , where . PROVE that its area can be given by .
Easy-peasy.
We were given the coordinates . We simply use the well-known area formula
and, by taking , , , we obtain
Observe that the numerator contains three consecutive factors, namely , and . Thus, if is an integer, then at least one of the three will be even, and so the resulting product will be even. This means that the area will be an integer — and a multiple of a triangular number.
## The depressed cubic
A cubic equation of the form in which the second degree term is missing, is usually called a depressed cubic. For a discussion on how to solve such equations, see here.
Suppose we re-arrange our area formula as
then we obtain not just a depressed cubic, but a restricted, simplified, specialized, depressed cubic. As such, its solution also takes a relatively simplified form.
PROVE that the area formula obtained previously, namely , is an odd function.
Recall that an odd function satisfies . Let’s write . Then
One implication of this “oddness” is that two opposite values of will yield the “same” area. As such, when we solve the depressed cubic equation
for , we must remember that any solution we obtain has a corresponding opposite, but the associated opposite doesn’t arise as a solution of the depressed cubic. For example, if is a solution of the depressed cubic, then we must also consider , despite the fact that is not a solution of the depressed cubic (see the exercises at the end).
For the depressed cubic , PROVE that can be given in terms of by:
Let’s do this!
Following the standard method of solving a depressed cubic, we seek and such that and . Then, is a solution of . Indeed, we have:
So it remains to find such and that satisfy and . Isolate from the second equation: . Substituting, the equation becomes:
Clear fractions and bring all terms to the left side:
Since this is now a quadratic in , we can use the quadratic formula to isolate . Doing this gives:
Simplifying,
It suffices to take . Then, gives :
Finally,
Using the preceding example, PROVE that if, and only if, .
First suppose that . Then . And so
reduces to
Conversely, if , then from the depressed cubic equation , we obtain
Discuss the nature of the roots of the depressed cubic .
We use the cubic discriminant to this end. For a general cubic , its discriminant is given by the quantity
Putting , , , and , we obtain, for the depressed cubic :
which reduces to . Consequently:
• if , then the depressed cubic has three real roots, one of which is repeated;
• if , then the depressed cubic has three distinct real roots;
• if , the depressed cubic has one real root and two complex roots.
Solve the depressed cubic equation .
Observe that this is the case in which . The corresponding cubic discriminant becomes . As such, we expect a repeated root. From Example 4, one of the roots is . Accordingly, is a factor of . By long division, we get the quadratic factor , which is a perfect square:
Thus, the complete solution to is . Notice how this solution set becomes “increased” in the next example.
#### Example
has an area of . The slopes of its sides form a geometric sequence . Find possible coordinates for the vertices .
Since the slopes of the sides form the geometric sequence , we can place the vertices at , , and . The area will then be , in view of Example 1. But we’re given that the area is :
which is the depressed cubic solved in the preceding example. Recall that the cubic has as solutions.
Being that the area formula is an odd function (Example 2), we should also consider as possible values for the common ratio , even though these are not solutions to the depressed cubic.
• for , the coordinates will be , , ;
• for , the coordinates will be , , ;
• for , the coordinates will be , , ;
• for , the coordinates will be , , .
Since the area is so small, each case produces a rare-listic diagram when drawn.
#### Example (“Fixed point”)
Find coordinates for the vertices of a whose slopes are and whose area is sq. units.
We solve the cubic equation using :
the quadratic part has complex roots. Again, we should consider , despite the fact that it’s not a solution to the current cubic equation.
Next, let’s use the coordinates , , :
In this case, the area and the common ratio are the same.
Give an example of a scalene triangle with non-zero side slopes, irrational side lengths, and an integer area.
The only challenge with this question is the non-zero slopes requirement; otherwise we could just have taken a triangle with one side on the -axis and the rest becomes easy.
Luckily, we can always turn to triangles with slopes in geometric progressions. In this case we take with vertices at . The side slopes are . For the side lengths we have , , and , all irrational and all different, so the triangle is scalene. Its area is sq. units.
Let be a root of the depressed cubic . PROVE that .
Since (an area), it follows that as well. This ensures that the quantity is defined.
Set
Comparing coefficients of like terms, , and . So , , and again . The two expressions for yield , which is valid because is a root of the original cubic .
## Takeaway
If with vertices at is such that the slopes of sides are , respectively, then the area of the triangle can be given by
By restricting the coordinates in a suitable way (see the exercises below), we obtain the depressed cubic as the area.
1. (Special square) For any right triangle with vertices , , and , PROVE that:
(Notice how squaring respects addition in this case. In addition, not all four absolute values on the left are absolutely necessary .)
2. (Special square) For any triangle with centroid , circumcenter , and orthocenter , PROVE that:
(Notice the absolute absence of absolute values in this case — so long as and . Also, any right triangle satisfies this property “TWICE” — with its regular vertices, and also with its three principal centers.)
3. (Opposite roots) PROVE that if a depressed cubic has two opposite solutions, namely and , then .
(In our own case, the right member represents the area of a certain, obscure triangle. Since , this explains why the depressed cubic equation cannot have two opposite solutions.)
4. has vertices at . Assume that the slopes of sides are , in that order.
• PROVE that its area can be given by
• Find an appropriate choice of for which the area reduces to the depressed cubic
5. has vertices at , , . PROVE that:
• the coordinates of the foot of the altitude from vertex is ;
• the length of the altitude from vertex is ;
• the area of is .
6. Consider with vertices at , , . Its side slopes are , a geometric progression in which and .
• Use the area formula in exercise 4 above to compute its area;
• Use the area formula in exercise 5 above to compute its area;
• Compute its area in the regular way, using ;
• Explain why the formula in exercise 5 did not work correctly in this case.
7. For with vertices at , , , PROVE that:
• its centroid is located at ;
• its circumcenter is at ;
• its orthocenter is at ;
• the slope of its Euler line is .
8. (Same ratio) Given , , , PROVE that the centroid, the circumcenter, and the orthocenter of lie on parallel lines , , and , respectively.
(Notice that the line through the centroid is in the middle of the other two lines. Moreover, these lines preserve the usual ratio in which the centroid divides the circumcenter-orthocenter distance.)
9. If , PROVE that .
10. (Unit area) Find coordinates for the vertices of with slopes and an area of sq. unit. |
# How do you simplify the expression x^2 - 2x - 3 divided by 2x^2 - 10x + 12?
Sep 18, 2015
$\frac{{x}^{2} - 2 x - 3}{2 {x}^{2} - 10 x + 12} = \frac{x - 1}{2 x - 4}$
#### Explanation:
Since they're both quadratics we can just factor them out and see if any of the factors cancel each other.
We begin by factoring the upper function, ${x}^{2} - 2 x - 3$
As any quadratic, with the ${x}^{2}$ coefficient being 1, we can say it's the product of two monomials $\left(x + a\right) \left(x + b\right)$.
We notice that if we can put the function in that way, whenever $x = - a$ or $x = - b$ one of the brackets will be equal to 0 and will make the whole thing a zero, so we can find the values of $a$ and $b$ by find the roots, using the quadratic equation.
Or, we can expand that product
$\left(x + a\right) \left(x + b\right) = {x}^{2} + a x + b x + a b = {x}^{2} + \left(a + b\right) x + a b$
And see that we can just find two numbers that when multiplied give out the constant, in this case, $- 3$ and when summed give out $- 2$. with the added caveat that $- a$ and $- b$ be roots.
While there isn't a formula it's often easier and quicker if we're only dealing with pretty numbers like this (if it starts getting too tough, it's best to go for the quadratics though).
In this case we have $a = - 3$ and $b = 1$, or vice versa, order doesn't matter.
We do the same thing to the bottom function, but since $2 {x}^{2} - 10 x + 12$ has a 2 in front of ${x}^{2}$ that trick doesn't work. We need to either do $\left(2 x + a\right) \left(x + b\right)$ or ... we can put two in evidence and work with the new function, $2 \left({x}^{2} - 5 x + 6\right)$
It's important to note that if you do go with $\left(2 x + a\right) \left(x + b\right)$ you'll get a value of $a$ that's double the other, and order will matter if you're checking by making the product.
Factoring that function within the brackets we see that $- 3$ and $- 2$ work. So we have
$\frac{{x}^{2} - 2 x - 3}{2 {x}^{2} - 10 x + 12} = \frac{\left(x - 3\right) \left(x - 1\right)}{2 \left(x - 3\right) \left(x - 2\right)}$
We cancel out the $\left(x - 3\right)$ and expand what's left
$\frac{{x}^{2} - 2 x - 3}{2 {x}^{2} - 10 x + 12} = \frac{x - 1}{2 x - 4}$
Do remember to note that $x \ne 3$, because if it were, we'd be doing a division by 0. |
maths > integral-calculus
Integral Calculus: Understanding Application Scenarios
what you'll learn...
Overview
The application scenarios of integrals are explained in detail with examples.
» cause-effect relation in quantities.
» effect is the "aggregate" of cause
eg: displacement is "continuous-aggregate" of speed
cause-effect relation
One of the fundamental aspects of science is to measure and specify quantities. Some examples are
• length of a pen is $10$$10$cm
• mass of an object: $20$$20$ gram
• temperature of water: ${30}^{\circ }$${30}^{\circ}$ Celsius
• the amount of time taken: $3$$3$ seconds
• the amount of distance traveled: $20$$20$ meter
• the speed of a car : $20$$20$meter per second
A pen can be used to write $30$$30$ pages. With $4$$4$ pens, one can write $4×30=120$$4 \times 30 = 120$ pages. Increase in the number of pen causes increase in the number of pages, which is the effect of the cause.
In this "number of pen" is a cause and "number of pages" is an effect.
This is an example of cause and effect relation.
This is a brief on "relations and functions".
Some cause-effect relations are
• Volume of Paint and painted area
• Number of tickets sold and the money collected in the sale
• speed of a car and distance covered in an hour
2 liter of paint is required to paint 3 square meter. If 14 liter paint is available, how much area can be painted?
The answer is "$14×\frac{3}{2}$$14 \times \frac{3}{2}$"
• The "volume of paint" is the cause.
• The "area painted" is the effect.
• This cause-effect relation is defined by a function involving multiplication by a constant.
$\text{area}=\text{volume}×\frac{3}{2}$$\textrm{a r e a} = \textrm{v o l u m e} \times \frac{3}{2}$.
Everyday, a hotel sends a worker to buy eggs from market. The eggs are priced at $1$$1$ coin each and the worker charges $5$$5$ coins for the travel to buy eggs. How many coins are to be given to buy $120$$120$ eggs?
The answer is "$125$$125$ coins".
• The "number of eggs" is the cause.
• The "coins" is the effect.
• This cause-effect relation is defined by a function involving addition of a constant.
$\text{coins}=$$\textrm{c o \in s} =$ $\text{number of eggs}$$\textrm{\nu m b e r o f e g g s}$ $×\phantom{\rule{1ex}{0ex}}\text{price per egg}$$\times \textrm{p r i c e p e r e g g}$ $+5$$+ 5$
A car is moving in a straight line at constant speed. It is at a distance $10$$10$m at $20$$20$sec and at a distance $20$$20$m at $25$$25$sec. The "effect" distance is given and the "cause" speed is to be computed. What is the speed?
The answer is "speed $=\frac{20m-10m}{25\mathrm{sec}-20\mathrm{sec}}$$= \frac{20 m - 10 m}{25 \sec - 20 \sec}$".
• The speed is cause.
• The distance traveled is the effect.
• This cause-effect relation is defined by a function involving rate of change.
$\text{speed}=\frac{\text{speed2}-\text{speed1}}{\text{time2}-\text{time1}}$$\textrm{s p e e d} = \frac{\textrm{s p e e d 2} - \textrm{s p e e d 1}}{\textrm{t i m e 2} - \textrm{t i m e 1}}$
A car is moving in a straight line at constant speed. It has a velocity of $2$$2$ m/sec for first $3$$3$ seconds and $4$$4$ m/sec for the next $1$$1$ sec. What is the distance traveled in the $4$$4$ seconds?
The answer is "$=2m/\mathrm{sec}×3\mathrm{sec}$$= 2 m / \sec \times 3 \sec$ $\quad + 4 m / \sec \times 1 \sec$"
• The speed is cause.
• The distance traveled is the effect.
• The cause-effect relation is defined by a function involving aggregate .
$\text{distance}=\text{speed1}×\text{time1}$$\textrm{\mathrm{di} s \tan c e} = \textrm{s p e e d 1} \times \textrm{t i m e 1}$$\quad + \textrm{s p e e d 2} \times \textrm{t i m e 2}$
summary
From the examples, it is understood that, Definition of a relation as an expression involves |
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# Tutor profile: Emily D.
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Emily D.
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## Questions
### Subject:Writing
TutorMe
Question:
How do you compose an introduction for an informative or opinion based paper?
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Emily D.
### Subject:Pre-Algebra
TutorMe
Question:
Do the addition or subtraction -987 - - 654 =
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Emily D.
The best way to think of this is putting parenthesis around the -64, which makes the equation look like. -987 - (-654). The important rule to remember is that two negatives equal a positive, so the subtraction of -654 equals a positive 654. -987 + 654 Then add as usual remembering that the first number is negative plus a 654 so the numbers will become a smaller number. -987 + 654 = -333
### Subject:Algebra
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Question:
Solve the Equation: 5(-3x-2)-(x-3) = -4(4x+5) +13
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Emily D.
The solving of the equation means that for this equation we are solving for x that both sides will be equal to each other. First, let's focus on the left-hand side of the equal sign. The first thing we should do is distribute the 5 to (-3x-2). 5 times -3x = -15x. 5 times -2 = -10. After we distribute the 5, it should come out to (-15x-10). Now on the left hand side we have, (-15x - 10) - (x-3). Combine like terms on this side being sure to pay attention to the subtraction of the (x-3). Combine the like terms of -15x - x, which equals -16x. Then combine the like terms without the variable which is ( -10 - (-3)) = -7. After combining like terms, on the left side, the result is, -16x -7. We will do the same with the right-hand side of the equation, starting with the distribution of the -4 to the (4x+5), which equals -16x-20. Then we can combine like terms, which are only the terms without a variable on this side, combine -20 +13 = -7. The result on the right-hand side was -16x -7. The equation now looks like -16x -7 = -16x -7. Then take the 16x and add it to both sides and add the 7 to both sides resulting in 0 =0 which is true.
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## ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1
These Solutions are part of ML Aggarwal Class 10 Solutions for ICSE Maths. Here we have given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1
More Exercises
Question 1.
State which pairs of triangles in the figure given below are similar. Write the similarity rule used and also write the pairs of similar triangles in symbolic form (all lengths of sides are in cm):
Solution:
Given
(i) In ∆ABC and PQR
Question 2.
It is given that ∆DEF ~ ∆RPQ. Is it true to say that ∠D = ∠R and ∠F = ∠P ? Why?
Solution:
∆DEF ~ ∆RPQ
∠D = ∠R and ∠F = ∠Q not ∠P
No, ∠F ≠ ∠P
Question 3.
If in two right triangles, one of the acute angle of one triangle is equal to an acute angle of the other triangle, can you say that the two triangles are similar? Why?
Solution:
In two right triangles,
one of the acute angles of the one triangle is
equal to an acute angle of the other triangle.
The triangles are similar. (AAA axiom)
Question 4.
In the given figure, BD and CE intersect each other at the point P. Is ∆PBC ~ ∆PDE? Give reasons for your answer.
Solution:
In the given figure, two line segments intersect each other at P.
In ∆BCP and ∆DEP
∠BPC = ∠DPE
Question 5.
It is given that ∆ABC ~ ∆EDF such that AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm.
Find the lengths of the remaining sides of the triangles.
Solution:
∆ABC ~ ∆EDF
AB = 5 cm, AC = 7 cm, DF = 15 cm and DE = 12 cm
Question 6.
(a) If ∆ABC ~ ∆DEF, AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm, then find the perimeter of ∆ABC.
(b) If ∆ABC ~ ∆PQR, Perimeter of ∆ABC = 32 cm, perimeter of ∆PQR = 48 cm and PR = 6 cm, then find the length of AC.
Solution:
(a) ∆ABC ~ ∆DEF
AB = 4 cm, DE = 6 cm, EF = 9 cm and FD = 12 cm
Question 7.
Calculate the other sides of a triangle whose shortest side is 6 cm and which is similar to a triangle whose sides are 4 cm, 7 cm and 8 cm.
Solution:
Let ∆ABG ~ ∆DEF in which shortest side of
∆ABC is BC = 6 cm.
In ∆DEF, DE = 8 cm, EF = 4 cm and DF = 7 cm
Question 8.
(a) In the figure given below, AB || DE, AC = , 3 cm, CE = 7.5 cm and BD = 14 cm. Calculate CB and DC.
(b) In the figure (2) given below, CA || BD, the lines AB and CD meet at G.
(i) Prove that ∆ACO ~ ∆BDO.
(ii) If BD = 2.4 cm, OD = 4 cm, OB = 3.2 cm and AC = 3.6 cm, calculate OA and OC.
Solution:
(a) In the given figure,
AB||DE, AC = 3 cm, CE = 7.5 cm, BD = 14 cm
Question 9.
(a) In the figure
(i) given below, ∠P = ∠RTS.
Prove that ∆RPQ ~ ∆RTS.
(b) In the figure (ii) given below,
∠ADC = ∠BAC. Prove that CA² = DC x BC.
Solution:
(a) In the given figure, ∠P = ∠RTS
To prove : ∆RPQ ~ ∆RTS
Proof : In ∆RPQ and ∆RTS
∠R = ∠R (common)
∠P = ∠RTS (given)
∆RPQ ~ ∆RTS (AA axiom)
Question 10.
(a) In the figure (1) given below, AP = 2PB and CP = 2PD.
(i) Prove that ∆ACP is similar to ∆BDP and AC || BD.
(ii) If AC = 4.5 cm, calculate the length of BD.
(b) In the figure (2) given below,
(i) Prove that ∆s ABC and AED are similar.
(ii) If AE = 3 cm, BD = 1 cm and AB = 6 cm, calculate AC.
(c) In the figure (3) given below, ∠PQR = ∠PRS. Prove that triangles PQR and PRS are similar. If PR = 8 cm, PS = 4 cm, calculate PQ.
Solution:
In the given figure,
AP = 2PB, CP = 2PD
To prove:
Question 11.
In the given figure, ABC is a triangle in which AB = AC. P is a point on the side BC such that PM ⊥ AB and PN ⊥ AC. Prpve that BM x NP = CN x MP.
Solution:
In the given figure, ABC in which AB = AC.
P is a point on BC such that PM ⊥ AB and PN ⊥ AC
To prove : BM x NP = CN x MP
Question 12.
Prove that the ratio of the perimeters of two similar triangles is the same as the ratio of their corresponding sides.
Solution:
Given : ∆ABC ~ ∆PQR .
To prove : Ratio in their perimeters k
the same as the ratio in their corresponding sides.
Question 13.
In the given figure, ABCD is a trapezium in which AB || DC. The diagonals AC and BD intersect at O. Prove that $$\frac { AO }{ OC } =\frac { BO }{ OD }$$
Using the above result, find the value(s) of x if OA = 3x – 19, OB = x – 4, OC = x – 3 and OD = 4.
Solution:
ABCD is a trapezium in which AB || DC
Diagonals AC and BD intersect each other at O.
Question 14.
In ∆ABC, ∠A is acute. BD and CE are perpendicular on AC and AB respectively. Prove that AB x AE = AC x AD.
Solution:
In ∆ABC, ∠A is acute
BD and CE are perpendiculars on AC and AB respectively
Question 15.
In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC. Prove that $$\frac { BE }{ DE } =\frac { AC }{ BC }$$
Solution:
In the given figure, DB ⊥ BC, DE ⊥ AB and AC ⊥ BC
To prove : $$\frac { BE }{ DE } =\frac { AC }{ BC }$$
Proof: In ∆ABC and ∆DEB
Question 16.
(a) In the figure (1) given below, E is a point on the side AD produced of a parallelogram ABCD and BE intersects CD at F. show that ∆ABE ~ ∆CFB.
(b) In the figure (2) given below, PQRS is a parallelogram; PQ = 16 cm, QR = 10 cm. L is a point on PR such that RL : LP = 2 : 3. QL produced meets RS at M and PS produced at N.
(i) Prove that triangle RLQ is similar to triangle PLN. Hence, find PN.
(ii) Name a triangle similar to triangle RLM. Evaluate RM.
Solution:
(a) In the given figure, ABCD is a ||gm
E is a point on AD and produced
and BE intersects CD at F.
To prove : ∆ABE ~ ∆CFB
Proof : In ∆ABE and ∆CFB
∠A = ∠C (opposite angles of a ||gm)
∠ABE = ∠BFC (alternate angles)
∆ABE ~ ∆CFB (AA axiom)
(b) In the given figure, PQRS is a ||gm PQ = 16 cm,
QR = 10 cm
L is a point on PR such that
Question 17.
The altitude BN and CM of ∆ABC meet at H. Prove that
(i) CN . HM = BM . HN .
(ii) $$\frac { HC }{ HB } =\sqrt { \frac { CN.HN }{ BM.HM } }$$
(iii) ∆MHN ~ ∆BHC.
Solution:
In the given figure, BN ⊥ AC and CM ⊥ AB of ∆ABC
which intersect each other at H.
To prove:
(i) CN.HM = BM.HN
(ii) $$\frac { HC }{ HB } =\sqrt { \frac { CN.HN }{ BM.HM } }$$
(iii) ∆MHN ~ ∆BHC.
Construction: Join MN
Question 18.
In the given figure, CM and RN are respectively the medians of ∆ABC and ∆PQR. If ∆ABC ~ ∆PQR, prove that:
(i) ∆AMC ~ ∆PNR
(ii) $$\frac { CM }{ RN } =\frac { AB }{ PQ }$$
(iii) ∆CMB ~ ∆RNQ.
Solution:
In the given figure, CM and RN are medians of ∆ABC and ∆PQR
respectively and ∆ABC ~ ∆PQR
To prove:
(i) ∆AMC ~ ∆PNR
>
Question 19.
In the given figure, medians AD and BE of ∆ABC meet at the point G, and DF is drawn parallel to BE. Prove that
(i) EF = FC
(ii) AG : GD = 2 : 1
Solution:
In the given figure,
AD and BE are the medians of ∆ABC
intersecting each other at G
DF || BE is drawn
To prove :
Question 20.
(a) In the figure given below, AB, EF and CD are parallel lines. Given that AB =15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm. Calculate
(i) EF
(ii) AC.
(b) In the figure given below, AF, BE and CD are parallel lines. Given that AF = 7.5 cm, CD = 4.5 cm, ED = 3 cm, BE = x and AE = y. Find the values of x and y.
Solution:
In the given figure,
AB || EF || CD
AB = 15 cm, EG = 5 cm, GC = 10 cm and DC = 18 cm
Calculate :
Question 21.
In the given figure, ∠A = 90° and AD ⊥ BC If BD = 2 cm and CD = 8 cm, find AD.
Solution:
In ∆ABC, we have ∠A = 90°
Now,
In ∆ABC, we have,
∠BAC = 90°
⇒ ∠BAD + ∠DAC = 90°…..(i)
Question 22.
A 15 metres high tower casts a shadow of 24 metres long at a certain time and at the same time, a telephone pole casts a shadow 16 metres long. Find the height of the telephone pole.
Solution:
Height of a tower AB = 15 m
and its shadow BC = 24 m
At the same time and position
Let the height of a telephone pole DE = x m
and its shadow EF = 16 m
Question 23.
A street light bulb is fixed on a pole 6 m above the level of street. If a woman of height casts a shadow of 3 m, find how far she is away from the base of the pole?
Solution:
Height of height pole(AB) = 6m
and height of a woman (DE) = 1.5 m
Here shadow EF = 3 m .
Hope given ML Aggarwal Class 10 Solutions for ICSE Maths Chapter 13 Similarity Ex 13.1 are helpful to complete your math homework.
If you have any doubts, please comment below. Learn Insta try to provide online math tutoring for you. |
# 2 Player Coin Puzzle
## Solve the Puzzle : 2 Player Coin Puzzle
In 2 Player Coin Puzzle, Consider a two-player coin game where each Player A and Player B gets the turn one by one. There is a row of even numbers of coins, and a player on his/her turn can pick a coin from any of the two corners of the row. The player that collects coins with more value wins the game. Develop a strategy for the player making the first turn i.e. Player A, such that he/she never loses the game.
Note : The strategy to pick a maximum of two corners may not work. In the following example, the first player, Player A loses the game when he/she uses a strategy to pick a maximum of two corners.
### Solution
• The idea is to count the sum of values of all even coins and odd coins, and compare the two values.
• The player that makes the first move can always make sure that the other player is never able to choose an even coin if the sum of even coins is higher.
• Similarly, he/she can make sure that the other player is never able to choose an odd coin if the sum of odd coins is higher. So here are the steps to a proper algorithm of either winning the game or getting a tie :
• `Initial row: 18 20 15 30 10 14`
` Player A picks 18, now row of coins is`
• `After first pick: 20 15 30 10 14`
` Player B picks 20, now row of coins is`
• `After second pick: 15 30 10 14`
` Player A picks 15, now row of coins is`
• `After third pick: 30 10 14`
` Player B picks 30, now row of coins is`
• `After 4th pick: 10 14`
` Player A picks 14, now row of coins is`
• `Last pick: 10 `
` Player B picks 10, game over. The total value collected by Player B is more (20 + 30 + 10) compared to the first player (18 + 15 + 14). So the second picker, Player B wins. `
Step 1 : Count the sum of all the coins in the even places(2nd, 4th, 6th and so on). Let the sum be “EVEN”.
Step 2 : Count the sum of all the coins in the odd places(1st, 3rd, 5th and so on). Let the sum be “ODD”.
Step 3 : Compare the value of EVEN and ODD and this is how the first player, here Player A must begin its selection.
• If (EVEN > ODD), start choosing from the right-hand corner and select all the even placed coins.
• If (EVEN < ODD), start choosing from the left-hand corner and select all the odd placed coins.
• If (EVEN == ODD), choosing only the odd-placed or only the even placed coins will throw a tie.
Example : Suppose you are given the following rows of coins : 18 20 15 30 10 14
a. Coins at even places: 20, 30, 14
b. Coins at odd places: 18, 15, 10
These places are fixed independent of whether the choice of selection must begin from the left or the right-hand side :
Step 1 : Sum of all even placed coins = 20 + 30 + 14 = 64
Step 2 : Sum of all odd placed coins = 18 + 15 + 10 = 43
Step 3 : Comparing the even and the odd placed coins where EVEN > ODD
Therefore, Player A must start selecting from the right-hand side and choose all the even-placed coins every time(here they are 14, 30 and 20). So first picker, Player A picks 14. Now, irrespective of whether the second Player B starts selecting from the left-hand side i.e. 18 or from the right-hand side i.e. 20, the even placed coins i.e. 14, 30 and 20 are booked for the Player A. Therefore, be any situation arise, the first picker Player A will always win the game.
Here are the illustrations to both the cases of pick by Player B :
Case 1 : When Player B starts picking from the left corner.
Case 2 : When Player B starts picking from the right corner after Player A.
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## How do you get the equation of a function from a graph?
How To: Given a graph of linear function, find the equation to describe the function.Identify the y-intercept of an equation.Choose two points to determine the slope.Substitute the y-intercept and slope into the slope-intercept form of a line.
## How do you write an equation for an asymptote?
Vertical asymptotes can be found by solving the equation n(x) = 0 where n(x) is the denominator of the function ( note: this only applies if the numerator t(x) is not zero for the same x value). Find the asymptotes for the function . The graph has a vertical asymptote with the equation x = 1.
## What is the example of rational equation?
Example
Solve
Multiply both sides of the equation by the common denominator.
7x – 14 + 5x + 10 =10x – 2 12x – 4 =10x – 2 Simplify
12x – 10x – 4 = 10x – 10x – 2 2x – 4 = -2 2x – 4 + 4 = -2 + 4 2x = 2 x = 1 Solve for x Check to be sure that the solution is not an excluded value. (It is not.)
## What’s a rational function example?
The definition you just got might be a little overbearing, so let’s look at some examples of rational functions: The function R(x) = (x^2 + 4x – 1) / (3x^2 – 9x + 2) is a rational function since the numerator, x^2 + 4x – 1, is a polynomial and the denominator, 3x^2 – 9x + 2 is also a polynomial.
## What is a rational equation?
A rational equation is an equation containing at least one fraction whose numerator and denominator are polynomials, A common way to solve these equations is to reduce the fractions to a common denominator and then solve the equality of the numerators.
## How do you find the equation of a function?
It is relatively easy to determine whether an equation is a function by solving for y. When you are given an equation and a specific value for x, there should only be one corresponding y-value for that x-value. For example, y = x + 1 is a function because y will always be one greater than x.
## How do you find the equation of an exponential function?
If one of the data points is the y-intercept (0,a) , then a is the initial value. Using a, substitute the second point into the equation f ( x ) = a ( b ) x displaystyle fleft(xright)=a{left(bright)}^{x} f(x)=a(b)x, and solve for b.
## What is the equation for a quadratic function?
A quadratic function is one of the form f(x) = ax2 + bx + c, where a, b, and c are numbers with a not equal to zero. The graph of a quadratic function is a curve called a parabola. Parabolas may open upward or downward and vary in “width” or “steepness”, but they all have the same basic “U” shape.
## What is the standard form of a rational function?
Standard Notation The typical rational function has the form p(x)/q(x) where p and q are polynomials. p(x) is called the numerator and q(x) is called the denominator. the numerator is x2 – 4 and the denominator is x22 – 5x + 6. A polynomial is a rational functions with denominator 1.
### Releated
#### Equation to find y intercept
How do you find the Y intercept? How Do You Find the X- and Y-Intercepts of a Line in Slope-Intercept Form? To find the x-intercept of a given linear equation, plug in 0 for ‘y’ and solve for ‘x’. To find the y-intercept, plug 0 in for ‘x’ and solve for ‘y’. How do you […]
#### Eva equation
How do you calculate EVA in Excel? Economic Value Added (EVA)EVA = NOPAT – (WACC * capital invested)WACC = Weighted Average Cost of Capital. Capital invested = Equity + long-term debt at the beginning of the period.Tax charge per income statement – increase (or + if reduction) in deferred tax provision + tax benefit of […] |
# CMPS 2433 Chapter 2 – Part 2 Functions & Relations
```HALVERSON – MIDWESTERN STATE UNIVERSITY
CMPS 2433
Chapter 2 – Part 2
Functions & Relations
2.2 Relations
A RELATION from set A to set B is any
subset of the Cartesian Product A X B
If R is a relation from A to B & (a, b) is an
element of R, a is related to b by R
Example
A
= {students enrolled at MSU in fall 2014}
B = {courses offered at MSU in fall 2014}
R = {(a,b)| student a is enrolled in course b}
R = {(Smith,Math1233), (Jones, CMPS1044), etc.}
Relations (cont’d)
A relation is ANY subset, so no repeated pairs
but can have repeated elements in the pairs.
R = {(Smith,Math1233), (Jones, Cmps1044),
(Smith, Engl1013), (Jones, Math1233),
(Hunt, Math1233), (Williams, Cmps1044), etc.}
What is the Universal Set for R?
Define a different Relation from A to B.
Relations (cont’d)
Example: R is a relation on A
A = {students enrolled at MSU in fall 2014}
R = {(a, b)| a & b are in a course together}
R = {(Smith, Jones), (Jones, Hunt), (Hunt, Wills),
(Wills, Johnson), etc.}
Relation from a set S to itself is call a Relation on S
Reflexive Property of Relations
A Relation R on a set S is said to be Reflexive
if for each x S, x R x is true
if for each x S, (x, x) is in R
that is, every element is related to itself
Is our previous example R a Reflexive
Relation?
Reflexive Property of Relations - Examples
A = {students enrolled at MSU in fall 2014}
Which of the following are Reflexive?
R = {(a,b): a & b are siblings}
R = {(a,b): a & b are not in a course together}
R = {(a,b): a & b are same classification}
R = {(a,b): a & b are married}
R = {(a,b): a & b are the same age}
R = {(a,b): a has a higher GPA than b}
Symmetric Property of Relations
A Relation R on a set S is said to be
Symmetric
If
x R y is true, then y R x is true
If (x, y) R, then (y, x) is true
That is, the elements of the relation R can be
reversed
Is R Symmetric?
R = {(a, b)| a & b are in a course together}
Symmetric Property of Relations - Examples
A = {students enrolled at MSU in fall 2014}
Which of the following are Symmetric?
R = {(a,b): a & b are siblings}
R = {(a,b): a & b are not in a course together}
R = {(a,b): a & b are same classification}
R = {(a,b): a & b are married}
R = {(a,b): a & b are the same age}
R = {(a,b): a has a higher GPA than b}
Transitive Property of Relations
A Relation R on a set S is said to be
Transitive
If
x R y and y R z are true, then x R z is true
If (x, y) R & (y, z) R, then (x, z) R
Is R Transitive?
R = {(a, b)| a & b are in a course together}
Transitive Property of Relations - Examples
A = {students enrolled at MSU in fall 2014}
Which of the following are Transitive?
R = {(a,b): a & b are siblings}
R = {(a,b): a & b are not in a course together}
R = {(a,b): a & b are same classification}
R = {(a,b): a & b are married}
R = {(a,b): a & b are the same age}
R = {(a,b): a has a higher GPA than b}
Equivalence Relation
Any Relation that is Reflexive, Symmetric &
Transitive is an Equivalence Relation
If R is an Equivalence Relation on S & x S,
the set of all elements related to x is called
an Equivalence Class
Denoted
[x]
Any 2 Equivalence Classes of a Relation are
either Equal or Disjoint
The
Equivalence Classes of R Partition S
Equivalence Relations - Examples
A = {students enrolled at MSU in fall 2014}
Which are Equivalence Relations?
If so, what are the partitions?
R = {(a,b): a & b are siblings}
R = {(a,b): a & b are not in a course together}
R = {(a,b): a & b are same classification}
R = {(a,b): a & b are married}
R = {(a,b): a & b are the same age}
R = {(a,b): a has a higher GPA than b}
Homework on Relations - Section 2.2
Page 52 – 54
Problems 1 – 14, 19-20, 25
Section 2.4 - Functions
A Function f from set X to set Y is a
relation from X to Y in which for each
element x in X there is exactly one
element y in Y for which x f y
Among the ordered pairs (x, y) in f, x
appears only ONCE
Example: is F a function?
F = {(2,3), (3,2), (4,2)}
F = {(2,3), (3,2), (2,4), (4,6)}
Mathematical Functions
Consider mathematical FUNCTIONS
Assume S = {0, 1, 2, 3, 4, 5,…}
f(x) = x2 = {(0,0), (1,1),(2,4),(3,9),(4,16),…}
f(x) = x+2 = {(0,2),(1,3),(2,4),(3,5),…}
For every x, there is only ONE value to which
it is related, thus these are Functions!
Equivalence Relations - Examples
A = {students enrolled at MSU in fall 2014}
Which are Functions?
R = {(a,b): a & b are siblings}
R = {(a,b): a & b are not in a course together}
R = {(a,b): a & b are same classification}
R = {(a,b): a & b are married}
R = {(a,b): a & b are the same age}
R = {(a,b): a has a higher GPA than b}
Function Domain
If f is a function from X to Y, denote f: X Y
Set X is called the domain of the function
Set Y is called co-domain
Subset of Y actually paired with elements of
X under f is called the range
For f(x) = y, y is the image of x under f
Domain, Co-domain, Range Examples
S = {…, -3,-2,-1,0, 1, 2, 3, 4, 5,…}
Define f as a function on S
F(x) = x2
Domain = S
Co-domain = S
Range = ???
One-to-One function
For
every x, there is a unique y &
For every y, there is a unique x
{(x, y)| no repeats of x or y}
S = {…, -3,-2,-1,0, 1, 2, 3, 4, 5,…}
f(x) = x2
Is f one-to-one?
Exponential & Logarithmic Functions
Logarithmic functions IMPT in Computing
Generally, base 2
NOTE:
2n
is exponential function, base 2
20 = 1 and 2-n = 1/2n
See page 73 for graph – Figure 2.18
Logarithmic Function base 2 is inverse of
Exponential Function
Logarithmic Function - Base 2
Notation: log2 x
“log base 2 of x”
Defn: y = log2 x if and only if x = 2y
Examples:
log2
8 = 3 because 23 = 8
log2 1024 = 10 because 210 = 1024
log2 256 = 8 because 28 = 256
log2 100 ~~ 6.65 because 26.65 ~~ 100
More on Logarithmic Function - Base 2
Growth rate is small, less than linear
See graph page 74 – Figure 2.19
Calculator Note:
Most
calculators with LOG button is base 10
log 2 x = LOG x / LOG 2
Algorithms with O(log2 n) complexity??
Homework – Section 2.4
Note – we omitted section on Composite &
Inverse Functions
Page 74-75
Problems 1 - 36
``` |
Ncert solutions
# Question 1
Q1) Calculate the amount and compound interest on:
(a) Rs.10,800 for 3 years at 12\ \frac{1}{2}\% per annum compounded annually.
(b) Rs.18,000 for 2\ \frac{1}{2} years at 10% per annum compounded annually.
(c) Rs.62,500 for 1\ \frac{1}{2} years at 8% per annum compounded annually.
(d) Rs.8,000 for 1 years at 9% per annum compounded half yearly. (You could the year by year calculation using S.I. formula to verify).
(e) Rs.10,000 for 1 years at 8% per annum compounded half yearly.
Solution:
(a) Here, Principal (P) = Rs. 10800, Time (n) = 3 years Rate of interest (R) = 12\ \frac{1}{2}\%=\frac{25}{2\%}
Amount(A) = P\left(1+\frac{R}{100}\right)^n
= 10800\left(1+\frac{1}{8}\right)^3=10800\left(\frac{9}{8}\right)^3
= 10800\times\frac{9}{8}\times\frac{9}{8}\times\frac{9}{8}
= Rs. 15,377.34
Compound Interest (C.I.) = A – P
= Rs. 10800 – Rs. 15377.34 = Rs. 4,577.34
(b) Here, Principal (P) = Rs. 18,000, Time (n) = 2\ \frac{1}{2} years, Rate of interest (R) = 10% p.a.
Amount(A) = P\left(1+\frac{R}{100}\right)^n
= 18000\left(1+\frac{10}{100}\right)^2=18000\left(1+\frac{1}{10}\right)^2
= 18000\left(\frac{11}{10}\right)^2=18000\times\frac{11}{10}\times\frac{11}{10}
= Rs. 21,780
Interest for \frac{1}{2} years on Rs. 21,780 at rate of 10% = \frac{21780\times10\times1}{100}= Rs. 1089
Total amount for 2\ \frac{1}{2} years.
= Rs. 21,780 + Rs. 1089 = Rs. 22,869
Compound Interest (C.I.) = A – P
= Rs. 22869 – Rs. 18000 = Rs. 4,869
(c) Here, Principal (P) = Rs. 62500, Time (n) = 1\ \frac{1}{2}=\frac{3}{2} years = 3 years
Rate of interest (R) = 8% = 4% (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^n
= 62500\left(1+\frac{4}{100}\right)^2
= 62500\left(1+\frac{1}{25}\right)^3
= 62500\left(\frac{26}{25}\right)^3
= 62500 \times\frac{26}{25}\times\frac{26}{25}\times\frac{26}{25}
= Rs. 70,304
Compound Interest (C.I.) = A – P
= Rs. 70304 – Rs. 62500 = Rs. 7,804
(d) Here, Principal (P) = Rs. 8000, Time (n) = 1 years = 2 years (compounded half yearly)
Rate of interest (R) = 9% = \frac{9}{2}\% (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^n
= 8000\left(1+\frac{9}{2\times100}\right)^2
= 8000\left(1+\frac{9}{200}\right)^2
= 800\left(\frac{209}{200}\right)^2
= 8000\times\frac{209}{200}\times\frac{209}{200}
= Rs. 8,736.20
Compound Interest (C.I.) = A – P
= Rs. 8736.20 – Rs. 8000
= Rs. 736.20
(e) Here, Principal (P) = Rs. 10,000, Time (n) = 1 years = 2 years (compounded half yearly)
Rate of interest (R) = 8% = 4% (compounded half yearly)
Amount (A) = P\left(1+\frac{R}{100}\right)^n
= 10000\left(1+\frac{4}{100}\right)^2
= 10000\left(1+\frac{1}{25}\right)^2
= 10000\left(\frac{26}{25}\right)^2
= 10000\times\frac{26}{25}\times\frac{26}{25}
= Rs. 10,816
Compound Interest (C.I.) = A – P
= Rs. 10,816 – Rs. 10,000 = Rs. 816
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Can You Find Reciprocal of 0? Exploring the Concept and Possible Solutions
Have you ever wondered what the reciprocal of 0 is? It’s a question that may have never crossed your mind before, but it’s one that has puzzled mathematicians for centuries. The answer might surprise you – or perhaps not – but it’s important to understand the concept of reciprocals and how they work in math.
Topics:
Reciprocals are simple mathematical expressions that show the relationship between two numbers. They are used to define fractions and determine the relative value of numbers in various equations. In essence, a reciprocal of any given number represents the inverse of that number. This means that when you multiply a number by its reciprocal, you get the value of one.
But can you find the reciprocal of 0? Is it even possible? The answer is a bit more complex than you might think. While most numbers have a reciprocal, 0 does not. It’s not possible to define a number that, when multiplied by 0, will give you the answer of 1. This is because anything multiplied by 0 is always 0. So, while the reciprocal of all other numbers exists, 0 is the lone exception.
Understanding the Concept of Division
Division is a mathematical operation that helps in sharing or distributing a number of things equally between two or more groups. It involves breaking a number into equal parts or groups to determine how many times a number (divisor) can go into another number (dividend). When we divide, we are essentially asking the question “How many times does the divisor fit into the dividend?”
• The dividend is the number being divided, while the divisor is the number dividing the dividend.
• The quotient is the answer to a division problem, telling us how many times the divisor goes into the dividend.
• The remainder is what is left over after dividing as much as possible.
For example, dividing 10 by 2 means we are asking how many times 2 can go into 10. The answer is 5, so the quotient is 5 and there is no remainder.
Division is the opposite operation of multiplication, just as subtraction is the opposite of addition. The two numbers in a division problem can be interchanged without affecting the result. This is known as the commutative property of division.
It is important to note that division by zero is undefined and cannot be performed. This is because dividing any number by zero would result in an infinite number of equal parts, making it impossible to determine the quotient. Therefore, the reciprocal of zero cannot be found.
Reciprocals of whole numbers
Reciprocals are the multiplicative inverse of a number. In simpler words, it is the number that, when multiplied by the original number, gives a product of 1.
Reciprocals of whole numbers are also whole numbers. For example, the reciprocal of 2 is 1/2, and the reciprocal of 1/2 is 2. However, there is one exception to this rule, which is the reciprocal of 0.
Reciprocal of 0
Can you find the reciprocal of 0? The answer is no. This is because any number, when multiplied by 0, equals 0. Therefore, there is no number that can be multiplied by 0 to give a product of 1. This is why the reciprocal of 0 is undefined.
Reciprocals of other whole numbers
• The reciprocal of 1 is 1, as any number multiplied by 1 equals itself.
• The reciprocal of 2 is 1/2, as 2 multiplied by 1/2 equals 1.
• The reciprocal of 3 is 1/3, as 3 multiplied by 1/3 equals 1.
The pattern continues for all other whole numbers. The reciprocal of a whole number is 1 divided by that number.
Reciprocals of whole numbers table
Number Reciprocal
1 1
2 1/2
3 1/3
4 1/4
5 1/5
6 1/6
7 1/7
8 1/8
Knowing the reciprocals of whole numbers can be helpful in solving equations and working with fractions. Understanding the concept of reciprocals is a fundamental building block in mathematics.
Reciprocals of fractions and decimals
Reciprocals can be defined as the multiplicative inverse of a number. This means that when you multiply a number and its reciprocal together, the result is always 1. In both fractions and decimals, finding the reciprocal is a simple process.
Let’s take a closer look at how to find the reciprocal of a fraction and a decimal.
• Fractions: To find the reciprocal of a fraction, simply flip it upside down. For example, the reciprocal of the fraction 1/4 is 4/1 or simply 4. Similarly, the reciprocal of 2/3 is 3/2.
• Decimals: To find the reciprocal of a decimal, divide 1 by the decimal. For example, the reciprocal of the decimal 0.5 is 1/0.5 or 2. Similarly, the reciprocal of the decimal 0.25 is 1/0.25 or 4.
It’s important to note that the reciprocal of the number 0 does not exist. This is because any number multiplied by 0 will always result in 0, so there is no number that can be multiplied by 0 to produce 1. Therefore, it’s impossible to find the reciprocal of 0.
Below is a table showing some common fractions and their reciprocals:
Fraction Reciprocal
1/2 2
1/3 3
1/4 4
1/5 5
Understanding how to find the reciprocals of fractions and decimals is an essential skill in many areas of mathematics, including algebra and calculus. With practice, you can quickly find the reciprocal of any number and solve problems with ease.
Dividing by zero and its implications
Dividing by zero is a mathematical operation that has been causing headaches to mathematicians for centuries. For example, if we divide 4 apples among 0 people, it is impossible to split something among no one. Therefore, the result is undefined. In mathematics, dividing by zero has no meaning, and any attempt to do so will always lead to an error. In other words, any number divided by zero is undefined, including zero itself.
• Undefined results: The most apparent implication of dividing by zero is that the result is undefined. This characteristic of division by zero makes it impossible to find the reciprocal of zero. The reciprocal of a number is the number that you can multiply it by to get 1. However, it’s impossible to find the reciprocal of 0 since division by zero is undefined.
• Limitations in calculations: Dividing by zero can create inconsistencies in mathematical calculations, and it prevents formulas that use division from being defined for certain values. For example, if you are calculating the ratio of two numbers and one of them is zero, division by zero can’t be done, giving an error message.
• The potential for errors: Since division by zero is undefined, a computer will often generate an error message if an attempt is made to divide a number by zero. However, it may be easy to miss this error message and assume that the result is correct, leading to errors in subsequent calculations and ultimately invalid conclusions.
To illustrate the implications of dividing by zero, consider the table below:
Number Divided by 0
4 Undefined
10 Undefined
100 Undefined
The table shows that any number divided by zero is undefined. Therefore, it is essential to avoid dividing any number by zero to avoid errors and ensure that mathematical formulas are accurate and reliable.
Real-world applications of reciprocals
Reciprocals are numbers that, when multiplied together, equal one. They have many real-world applications that range from everyday life to complex mathematical functions.
• Calculating fuel efficiency: We often use miles per gallon (mpg) to measure how efficiently a car uses fuel. To determine how many gallons of fuel you need to drive a certain distance, you can use the reciprocal of mpg. For example, if your car gets 25 mpg, the reciprocal is 1/25. To find out how many gallons of fuel you need to drive 100 miles, you would multiply 1/25 by 100, which equals 4 gallons.
• Electricity: In electrical engineering, the impedance of a circuit is the reciprocal of its conductivity. This means that a high impedance circuit has a low conductivity and vice versa.
• Medicine: In medical science, reciprocals are used in calculating the half-life of drugs in the body. The half-life is the time it takes for the concentration of the drug in the body to decrease by half. The reciprocal of the half-life is used to calculate how long it takes for the drug to be eliminated from the body.
Reciprocals are also used in many mathematical functions, such as calculus and trigonometry. They are particularly important in the study of functions that involve inverse operations, such as the inverse tangent or the inverse hyperbolic sine.
Furthermore, reciprocals of numbers can be represented graphically, with a curve called a reciprocal curve. A reciprocal curve is the graph of the equation y = 1/x. The curve is a hyperbola, with two asymptotes: the x-axis and the y-axis.
x y=1/x
-5 -0.2
-1 -1
0 undefined
1 1
5 0.2
Overall, reciprocals have many real-world applications and are essential in various fields of study. Whether you are calculating fuel efficiency, designing a circuit, or studying mathematical functions, understanding and using reciprocals will help you achieve accurate and efficient results.
Properties of Reciprocals
Reciprocals are numbers that when multiplied together give a product of 1. The reciprocal of a number is found by dividing 1 by the number. For example, the reciprocal of 5 is 1/5 because 5 multiplied by 1/5 equals 1. In this article, we will focus on the properties of reciprocals.
• The reciprocal of 1 is 1. This is because 1 multiplied by 1 equals 1. Therefore, the reciprocal of 1 is 1/1 which is equal to 1.
• Any non-zero number raised to the power of -1 gives us its reciprocal. For example, 2 raised to the power of -1 is equal to 1/2 because 2 multiplied by 1/2 equals 1.
• The product of any number and its reciprocal is always equal to 1. For example, 3 multiplied by 1/3 equals 1.
Reciprocals also follow the commutative, associative, and distributive properties. Let’s take a look at these properties:
• Commutative property: This property states that changing the order of the numbers being added or multiplied does not change the result. For example, 4 multiplied by 1/2 is the same as 1/2 multiplied by 4.
• Associative property: This property states that changing the grouping of the numbers being added or multiplied does not change the result. For example, (2 multiplied by 3) multiplied by 1/6 is the same as 2 multiplied by (3 multiplied by 1/6).
• Distributive property: This property states that multiplying a number by the sum or difference of two numbers is the same as multiplying the number by each of the two numbers and then adding or subtracting the products. For example, 1/2 multiplied by (3+5) is the same as (1/2 multiplied by 3) plus (1/2 multiplied by 5).
Reciprocals are also useful in solving equations involving fractions. For example, the equation 3/4x = 6 can be solved by multiplying both sides of the equation by 4/3. This gives us x = 8.
Number Reciprocal
1 1
2 1/2
3 1/3
4 1/4
In conclusion, understanding the properties of reciprocals can help in solving equations involving fractions, simplifying complicated expressions, and performing calculations in a more efficient and accurate manner.
Common misconceptions about reciprocals
Reciprocals are an important concept in mathematics, but they are frequently misunderstood. Below are some common misconceptions about reciprocals:
Number 7: You can find the reciprocal of 0
One of the biggest misconceptions about reciprocals is that you can find the reciprocal of 0. However, this is not true. The reciprocal of a number is defined as 1 divided by that number. Therefore, the reciprocal of 0 would be 1/0. However, division by 0 is not defined in mathematics, as it leads to undefined results.
This confusion sometimes arises because the reciprocal of a very small number approach infinity. For example, the reciprocal of 0.000001 is 1,000,000, which is a very large number. However, as you approach 0, the reciprocal becomes infinitely large, which is not defined.
Other common misconceptions about reciprocals include:
• Reciprocals are only defined for integers
• The reciprocal of a negative number is always negative
• Reciprocals can be used to solve any math problem
Reciprocals are a useful tool in mathematics
Despite these misconceptions, reciprocals are an important tool in mathematics and can be used to solve many problems. For example, they are frequently used in physics to calculate quantities such as resistance and capacitance.
It is important to have a clear understanding of what reciprocals are and how they can be used in order to avoid these common misconceptions.
A comparison of reciprocals of different numbers
Number Reciprocal 1 1 2 0.5 -3 -0.333… 0.5 2
As you can see from the table, the reciprocal of a number is equal to 1 divided by that number. The reciprocal of 1 is 1, the reciprocal of 2 is 0.5, the reciprocal of -3 is -0.333…, and the reciprocal of 0.5 is 2.
Can You Find Reciprocal of 0?
Q: What is the reciprocal of a number?
A: The reciprocal of a number is simply the inverse of that number. For example, the reciprocal of 2 is 1/2.
Q: Can you find the reciprocal of 0?
A: No, it is impossible to find the reciprocal of 0 because division by 0 is undefined and not allowed in mathematics.
Q: Why is the reciprocal of 0 undefined?
A: The reciprocal of a number is defined as 1 divided by that number. Since division by 0 is undefined, the reciprocal of 0 is also undefined.
Q: What happens when you try to find the reciprocal of 0?
A: If you try to find the reciprocal of 0, you will end up with an error message or an undefined result.
Q: Is it possible to find the reciprocal of a negative number?
A: Yes, it is possible to find the reciprocal of a negative number. The reciprocal of a negative number will also be negative.
Q: What is the reciprocal of 1/2?
A: The reciprocal of 1/2 is 2.
Q: Why is it important to know about reciprocals?
A: Reciprocals are important in many areas of mathematics, including algebra and calculus. They are also used in practical applications such as engineering and physics. |
# Multiplication
(Redirected from Times)
For other uses, see Multiplication (disambiguation).
"Times" redirects here. For the typeface, see Times (typeface). For the UK newspaper, see The Times. For other uses, see The Times (disambiguation).
Four bags of three marbles gives twelve marbles (4 × 3 = 12).
Multiplication can also be thought of as scaling. In the above animation, we see 2 being multiplied by 3, giving 6 as a result
4 × 5 = 20, the rectangle is composed of 20 squares, having dimensions of 4 by 5.
Area of a cloth 4.5m × 2.5m = 11.25m2; 4½ × 2½ = 11¼
Multiplication (often denoted by the cross symbol "×", or by the absence of symbol) is one of the four elementary, mathematical operations of arithmetic; with the others being addition, subtraction and division.
The multiplication of two whole numbers is equivalent to the addition of one of them with itself as many times as the value of the other one; for example, 3 multiplied by 4 (often said as "3 times 4") can be calculated by adding 3 copies of 4 together:
$3 \times 4 = 4 + 4 + 4 = 12$
Here 3 and 4 are the "factors" and 12 is the "product".
One of the main properties of multiplication is that the result does not depend on the place of the factor that is repeatedly added to itself (commutative property). 3 multiplied by 4 can also be calculated by adding 4 copies of 3 together:
$3 \times 4 = 3 + 3 + 3 + 3 = 12$
The multiplication of integers (including negative numbers), rational numbers (fractions) and real numbers is defined by a systematic generalization of this basic definition.
Multiplication can also be visualized as counting objects arranged in a rectangle (for whole numbers) or as finding the area of a rectangle whose sides have given lengths. The area of a rectangle does not depend on which side is measured first, which illustrates the commutative property.
In general, multiplying two measurements gives a new type, depending on the measurements. For instance:
$2.5 \mbox{ meters} \times 4.5 \mbox{ meters} = 11.25 \mbox{ square meters}$
$11 \mbox{ meters/second} \times 9 \mbox{ seconds} = 99 \mbox{ meters}$
The inverse operation of the multiplication is the division. For example, since 4 multiplied by 3 equals 12, then 12 divided by 3 equals 4. Multiplication by 3, followed by division by 3, yields the original number (since the division of a number other than 0 by itself equals 1).
Multiplication is also defined for other types of numbers, such as complex numbers, and more abstract constructs, like matrices. For these more abstract constructs, the order that the operands are multiplied sometimes does matter.
## Notation and terminology
The multiplication sign ×
(HTML entity is ×)
In arithmetics, multiplication is often written using the sign "×" between the terms; that is, in infix notation. For example,
$2\times 3 = 6$ (verbally, "two times three equals six")
$3\times 4 = 12$
$2\times 3\times 5 = 6\times 5 = 30$
$2\times 2\times 2\times 2\times 2 = 32$
The sign is encoded in Unicode at U+00D7 × multiplication sign (HTML × · ×).
There are other mathematical notations for multiplication:
• Multiplication is sometimes denoted by dot signs, either a middle-position dot or a period:
$5 \cdot 2 \quad\text{or}\quad 5\,.\,2$
The middle dot notation, encoded in Unicode as U+22C5 dot operator, is standard in the United States, the United Kingdom, and other countries where the period is used as a decimal point. When the dot operator character is not accessible, the interpunct (·) is used. In other countries that use a comma as a decimal point, either the period or a middle dot is used for multiplication.[citation needed]
Calculation results
$\scriptstyle\left.\begin{matrix}\scriptstyle\text{summand}+\text{summand}\\\scriptstyle\text{augend}+\text{addend}\end{matrix}\right\}=$$\scriptstyle\text{sum}$
Subtraction (−)
$\scriptstyle\text{minuend}-\text{subtrahend}=$$\scriptstyle\text{difference}$
Multiplication (×)
$\scriptstyle\left.\begin{matrix}\scriptstyle\text{multiplicand}\times\text{multiplicand}\\\scriptstyle\text{multiplicand}\times\text{multiplier}\end{matrix}\right\}=$$\scriptstyle\text{product}$
Division (÷)
$\scriptstyle\frac{\scriptstyle\text{dividend}}{\scriptstyle\text{divisor}}=$$\scriptstyle\text{quotient}$
Modulation (mod)
$\scriptstyle\text{dividend}\mod\text{divisor}=$$\scriptstyle\text{remainder}$
Exponentiation
$\scriptstyle\text{base}^\text{exponent}=$$\scriptstyle\text{power}$
nth root (√)
$\scriptstyle\sqrt[\text{degree}]{\scriptstyle\text{radicand}}=$$\scriptstyle\text{root}$
Logarithm (log)
$\scriptstyle\log_\text{base}(\text{antilogarithm})=$$\scriptstyle\text{logarithm}$
• In algebra, multiplication involving variables is often written as a juxtaposition (e.g., xy for x times y or 5x for five times x). This notation can also be used for quantities that are surrounded by parentheses (e.g., 5(2) or (5)(2) for five times two).
In computer programming, the asterisk (as in 5*2) is the standard notation: it belongs to most character sets and appears on every keyboard. This usage originated in the FORTRAN programming language.
The numbers to be multiplied are generally called the "factors" or "multiplicands". When thinking of multiplication as repeated addition, the number to be multiplied is called the "multiplicand", while the number of addends is called the "multiplier". In algebra, a number that is the multiplier of a variable or expression (e.g., the 3 in 3xy2) is called a coefficient.
The result of a multiplication is called a product. A product of integers is a multiple of each factor. For example, 15 is the product of 3 and 5, and is both a multiple of 3 and a multiple of 5.
## Computation
The common methods for multiplying numbers using pencil and paper require a multiplication table of memorized or consulted products of small numbers (typically any two numbers from 0 to 9), however one method, the peasant multiplication algorithm, does not.
Multiplying numbers to more than a couple of decimal places by hand is tedious and error prone. Common logarithms were invented to simplify such calculations. The slide rule allowed numbers to be quickly multiplied to about three places of accuracy. Beginning in the early twentieth century, mechanical calculators, such as the Marchant, automated multiplication of up to 10 digit numbers. Modern electronic computers and calculators have greatly reduced the need for multiplication by hand.
### Historical algorithms
Methods of multiplication were documented in the Egyptian, Greek, Indian and Chinese civilizations.
The Ishango bone, dated to about 18,000 to 20,000 BC, hints at a knowledge of multiplication in the Upper Paleolithic era in Central Africa.
#### Egyptians
The Egyptian method of multiplication of integers and fractions, documented in the Ahmes Papyrus, was by successive additions and doubling. For instance, to find the product of 13 and 21 one had to double 21 three times, obtaining 1 × 21 = 21, 4 × 21 = 84, 8 × 21 = 168. The full product could then be found by adding the appropriate terms found in the doubling sequence:
13 × 21 = (1 + 4 + 8) × 21 = (1 × 21) + (4 × 21) + (8 × 21) = 21 + 84 + 168 = 273.
#### Babylonians
The Babylonians used a sexagesimal positional number system, analogous to the modern day decimal system. Thus, Babylonian multiplication was very similar to modern decimal multiplication. Because of the relative difficulty of remembering 60 × 60 different products, Babylonian mathematicians employed multiplication tables. These tables consisted of a list of the first twenty multiples of a certain principal number n: n, 2n, ..., 20n; followed by the multiples of 10n: 30n 40n, and 50n. Then to compute any sexagesimal product, say 53n, one only needed to add 50n and 3n computed from the table.
#### Chinese
38 × 76 = 2888
In the mathematical text Zhou Bi Suan Jing, dated prior to 300 BC, and the Nine Chapters on the Mathematical Art, multiplication calculations were written out in words, although the early Chinese mathematicians employed Rod calculus involving place value addition, subtraction, multiplication and division. These place value decimal arithmetic algorithms were introduced by Al Khwarizmi to Arab countries in the early 9th century.
### Modern method
Product of 45 and 256. Note the order of the numerals in 45 is reversed down the left column. The carry step of the multiplication can be performed at the final stage of the calculation (in bold), returning the final product of 45 × 256 = 11520.
The modern method of multiplication based on the Hindu–Arabic numeral system was first described by Brahmagupta. Brahmagupta gave rules for addition, subtraction, multiplication and division. Henry Burchard Fine, then professor of Mathematics at Princeton University, wrote the following:
The Indians are the inventors not only of the positional decimal system itself, but of most of the processes involved in elementary reckoning with the system. Addition and subtraction they performed quite as they are performed nowadays; multiplication they effected in many ways, ours among them, but division they did cumbrously.[1]
### Computer algorithms
The standard method of multiplying two n-digit numbers requires n2 simple multiplications. Multiplication algorithms have been designed that reduce the computation time considerably when multiplying large numbers. In particular for very large numbers methods based on the Discrete Fourier Transform can reduce the number of simple multiplications to the order of n log2(n) log2 log2(n).
## Products of measurements
Main article: Dimensional analysis
When two measurements are multiplied together the product is of a type depending on the types of the measurements. The general theory is given by dimensional analysis. This analysis is routinely applied in physics but has also found applications in finance. One can only meaningfully add or subtract quantities of the same type but can multiply or divide quantities of different types.
A common example is multiplying speed by time gives distance, so
50 kilometers per hour × 3 hours = 150 kilometers.
## Products of sequences
### Capital Pi notation
The product of a sequence of terms can be written with the product symbol, which derives from the capital letter Π (Pi) in the Greek alphabet. Unicode position U+220F (∏) contains a glyph for denoting such a product, distinct from U+03A0 (Π), the letter. The meaning of this notation is given by:
$\prod_{i=1}^4 i = 1\cdot 2\cdot 3\cdot 4,$
that is
$\prod_{i=1}^4 i = 24.$
The subscript gives the symbol for a dummy variable (i in this case), called the "index of multiplication" together with its lower bound (1), whereas the superscript (here 4) gives its upper bound. The lower and upper bound are expressions denoting integers. The factors of the product are obtained by taking the expression following the product operator, with successive integer values substituted for the index of multiplication, starting from the lower bound and incremented by 1 up to and including the upper bound. So, for example:
$\prod_{i=1}^6 i = 1\cdot 2\cdot 3\cdot 4\cdot 5 \cdot 6 = 720$
More generally, the notation is defined as
$\prod_{i=m}^n x_i = x_m \cdot x_{m+1} \cdot x_{m+2} \cdot \,\,\cdots\,\, \cdot x_{n-1} \cdot x_n,$
where m and n are integers or expressions that evaluate to integers. In case m = n, the value of the product is the same as that of the single factor xm. If m > n, the product is the empty product, with the value 1.
### Infinite products
Main article: Infinite product
One may also consider products of infinitely many terms; these are called infinite products. Notationally, we would replace n above by the lemniscate ∞. The product of such a series is defined as the limit of the product of the first n terms, as n grows without bound. That is, by definition,
$\prod_{i=m}^{\infty} x_{i} = \lim_{n\to\infty} \prod_{i=m}^{n} x_{i}.$
One can similarly replace m with negative infinity, and define:
$\prod_{i=-\infty}^\infty x_i = \left(\lim_{m\to-\infty}\prod_{i=m}^0 x_i\right) \cdot \left(\lim_{n\to\infty}\prod_{i=1}^n x_i\right),$
provided both limits exist.
## Properties
Multiplication of numbers 0-10. Line labels = multiplicand. X axis = multiplier. Y axis = product.
For the real and complex numbers, which includes for example natural numbers, integers and fractions, multiplication has certain properties:
Commutative property
The order in which two numbers are multiplied does not matter:
$x\cdot y = y\cdot x$.
Associative property
Expressions solely involving multiplication or addition are invariant with respect to order of operations:
$(x\cdot y)\cdot z = x\cdot(y\cdot z)$
Distributive property
Holds with respect to multiplication over addition. This identity is of prime importance in simplifying algebraic expressions:
$x\cdot(y + z) = x\cdot y + x\cdot z$
Identity element
The multiplicative identity is 1; anything multiplied by one is itself. This is known as the identity property:
$x\cdot 1 = x$
Property of Zero
Any number multiplied by zero is zero. This is known as the zero property of multiplication:
$x\cdot 0 = 0$
Zero is sometimes not included amongst the natural numbers.
There are a number of further properties of multiplication not satisfied by all types of numbers.
Negation
Negative one times any number is equal to the additive inverse of that number.
$(-1)\cdot x = (-x)$
Negative one times negative one is positive one.
$(-1)\cdot (-1) = 1$
The natural numbers do not include negative numbers.
Inverse element
Every number x, except zero, has a multiplicative inverse, $\frac{1}{x}$, such that $x\cdot\left(\frac{1}{x}\right) = 1$.
Order preservation
Multiplication by a positive number preserves order:
if a > 0, then if b > c then ab > ac.
Multiplication by a negative number reverses order:
if a < 0 and b > c then ab < ac.
The complex numbers do not have an order predicate.
Other mathematical systems that include a multiplication operation may not have all these properties. For example, multiplication is not, in general, commutative for matrices and quaternions.
## Axioms
Main article: Peano axioms
In the book Arithmetices principia, nova methodo exposita, Giuseppe Peano proposed axioms for arithmetic based on his axioms for natural numbers.[2] Peano arithmetic has two axioms for multiplication:
$x \times 0 = 0$
$x \times S(y) = (x \times y) + x$
Here S(y) represents the successor of y, or the natural number that follows y. The various properties like associativity can be proved from these and the other axioms of Peano arithmetic including induction. For instance S(0). denoted by 1, is a multiplicative identity because
$x \times 1 = x \times S(0) = (x \times 0) + x = 0 + x = x$
The axioms for integers typically define them as equivalence classes of ordered pairs of natural numbers. The model is based on treating (x,y) as equivalent to xy when x and y are treated as integers. Thus both (0,1) and (1,2) are equivalent to −1. The multiplication axiom for integers defined this way is
$(x_p,\, x_m) \times (y_p,\, y_m) = (x_p \times y_p + x_m \times y_m,\; x_p \times y_m + x_m \times y_p)$
The rule that −1 × −1 = 1 can then be deduced from
$(0, 1) \times (0, 1) = (0 \times 0 + 1 \times 1,\, 0 \times 1 + 1 \times 0) = (1,0)$
Multiplication is extended in a similar way to rational numbers and then to real numbers.
## Multiplication with set theory
It is possible, though difficult, to create a recursive definition of multiplication with set theory. Such a system usually relies on the Peano definition of multiplication.
### Cartesian product
The definition of multiplication as repeated addition provides a way to arrive at a set-theoretic interpretation of multiplication of cardinal numbers. In the expression
$\displaystyle n \cdot a = \underbrace{a + \cdots + a}_{n},$
if the n copies of a are to be combined in disjoint union then clearly they must be made disjoint; an obvious way to do this is to use either a or n as the indexing set for the other. Then, the members of $n \cdot a\,$ are exactly those of the Cartesian product $n \times a\,$. The properties of the multiplicative operation as applying to natural numbers then follow trivially from the corresponding properties of the Cartesian product.
## Multiplication in group theory
There are many sets that, under the operation of multiplication, satisfy the axioms that define group structure. These axioms are closure, associativity, and the inclusion of an identity element and inverses.
A simple example is the set of non-zero rational numbers. Here we have identity 1, as opposed to groups under addition where the identity is typically 0. Note that with the rationals, we must exclude zero because, under multiplication, it does not have an inverse: there is no rational number that can be multiplied by zero to result in 1. In this example we have an abelian group, but that is not always the case.
To see this, look at the set of invertible square matrices of a given dimension, over a given field. Now it is straightforward to verify closure, associativity, and inclusion of identity (the identity matrix) and inverses. However, matrix multiplication is not commutative, therefore this group is nonabelian.
Another fact of note is that the integers under multiplication is not a group, even if we exclude zero. This is easily seen by the nonexistence of an inverse for all elements other than 1 and -1.
Multiplication in group theory is typically notated either by a dot, or by juxtaposition (the omission of an operation symbol between elements). So multiplying element a by element b could be notated a $\cdot$ b or ab. When referring to a group via the indication of the set and operation, the dot is used, e.g., our first example could be indicated by $\left( \mathbb{Q}\smallsetminus \{ 0 \} ,\cdot \right)$
## Multiplication of different kinds of numbers
Numbers can count (3 apples), order (the 3rd apple), or measure (3.5 feet high); as the history of mathematics has progressed from counting on our fingers to modelling quantum mechanics, multiplication has been generalized to more complicated and abstract types of numbers, and to things that are not numbers (such as matrices) or do not look much like numbers (such as quaternions).
Integers
$N\times M$ is the sum of M copies of N when N and M are positive whole numbers. This gives the number of things in an array N wide and M high. Generalization to negative numbers can be done by
$N\times (-M) = (-N)\times M = - (N\times M)$ and
$(-N)\times (-M) = N\times M$
The same sign rules apply to rational and real numbers.
Rational numbers
Generalization to fractions $\frac{A}{B}\times \frac{C}{D}$ is by multiplying the numerators and denominators respectively: $\frac{A}{B}\times \frac{C}{D} = \frac{(A\times C)}{(B\times D)}$. This gives the area of a rectangle $\frac{A}{B}$ high and $\frac{C}{D}$ wide, and is the same as the number of things in an array when the rational numbers happen to be whole numbers.
Real numbers
$(x)(y)$ is the limit of the products of the corresponding terms in certain sequences of rationals that converge to x and y, respectively, and is significant in calculus. This gives the area of a rectangle x high and y wide. See Products of sequences, above.
Complex numbers
Considering complex numbers $z_1$ and $z_2$ as ordered pairs of real numbers $(a_1, b_1)$ and $(a_2, b_2)$, the product $z_1\times z_2$ is $(a_1\times a_2 - b_1\times b_2, a_1\times b_2 + a_2\times b_1)$. This is the same as for reals, $a_1\times a_2$, when the imaginary parts $b_1$ and $b_2$ are zero.
Further generalizations
See Multiplication in group theory, above, and Multiplicative Group, which for example includes matrix multiplication. A very general, and abstract, concept of multiplication is as the "multiplicatively denoted" (second) binary operation in a ring. An example of a ring that is not any of the above number systems is a polynomial ring (you can add and multiply polynomials, but polynomials are not numbers in any usual sense.)
Division
Often division, $\frac{x}{y}$, is the same as multiplication by an inverse, $x\left(\frac{1}{y}\right)$. Multiplication for some types of "numbers" may have corresponding division, without inverses; in an integral domain x may have no inverse "$\frac{1}{x}$" but $\frac{x}{y}$ may be defined. In a division ring there are inverses, but $\frac{x}{y}$ may be ambiguous in non-commutative rings since $x\left(\frac{1}{y}\right)$ need not the same as $\left(\frac{1}{y}\right)x$.
## Exponentiation
Main article: Exponentiation
When multiplication is repeated, the resulting operation is known as exponentiation. For instance, the product of three factors of two (2×2×2) is "two raised to the third power", and is denoted by 23, a two with a superscript three. In this example, the number two is the base, and three is the exponent. In general, the exponent (or superscript) indicates how many times to multiply base by itself, so that the expression
$a^n = \underbrace{a\times a \times \cdots \times a}_n$
indicates that the base a to be multiplied by itself n times. |
## Access 8th Grade Online Workbooks
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Use the table below to review the Lumos tedBook™ 8th Grade Math and Language Arts Literacy Domains, Lesson Names, Topics, along with their recent Common Core State Standard (CCSS) Correlations.
Course Name: Grade 8 Math Mastery and Test Preparation – Practice Tests and Workbooks: CCSS Aligned
Lesson Name
The Number System
### Rational Numbers 8.NS.1
Domain: The Number System
Theme: Know that there are numbers that are not rational, and approximate them by rational numbers.
Standard: Know that numbers that are not rational are called irrational. Understand informally that every number has a decimal expansion; for rational numbers show that the decimal expansion repeats eventually, and convert a decimal expansion which repeats eventually into a rational number.
### Irrational Numbers 8.NS.2
Domain: The Number System
Theme: Know that there are numbers that are not rational, and approximate them by rational numbers
Standard: Use rational approximations of irrational numbers to compare the size of irrational numbers, locate them approximately on a number line diagram, and estimate the value of expressions (e.g., ?^2). For example, by truncating the decimal expansion of ?2 (square root of 2), show that ?2 is between 1 and 2, then between 1.4 and 1.5, and explain how to continue on to get better approximations.
Expressions and Equations
### Exponents Multiplication & Division 8.EE.1
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Know and apply the properties of integer exponents to generate equivalent numerical expressions. For example, 3^2
### Square & Cube Roots 8.EE.2
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Use square root and cube root symbols to represent solutions to equations of the form x^2 = p and x^3 = p, where p is a positive rational number. Evaluate square roots of small perfect squares and cube roots of small perfect cubes. Know that ?2 is irrational.
### Scientific Notation 8.EE.3
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Use numbers expressed in the form of a single digit times an integer power of 10 to estimate very large or very small quantities, and to express how many times as much one is than the other. For example, estimate the population of the United States as 3
### Scientific Notation Multiplication & Division 8.EE.4
Domain: Expressions & Equations
Theme: Work with radicals and integer exponents
Standard: Perform operations with numbers expressed in scientific notation, including problems where both decimal and scientific notation are used. Use scientific notation and choose units of appropriate size for measurements of very large or very small quantities (e.g., use millimeters per year for seafloor spreading). Interpret scientific notation that has been generated by technology.
### Compare Proportions 8.EE.5
Domain: Expressions & Equations
Theme: Understand the connections between proportional relationships, lines, and linear equations
Standard: Graph proportional relationships, interpreting the unit rate as the slope of the graph. Compare two different proportional relationships represented in different ways. For example, compare a distance-time graph to a distance-time equation to determine which of two moving objects has greater speed.
### Slope 8.EE.6
Domain: Expressions & Equations
Theme: Understand the connections between proportional relationships, lines, and linear equations
Standard: Use similar triangles to explain why the slope m is the same between any two distinct points on a non-vertical line in the coordinate plane; derive the equation y =mx for a line through the origin and the equation y = mx + b for a line intercepting the vertical axis at b.
### One-Variable Equations 8.EE.7a
Domain: Expressions & Equations
Theme: Analyze and solve linear equations and pairs of simultaneous linear equations
Standard: Give examples of linear equations in one variable with one solution, infinitely many solutions, or no solutions. Show which of these possibilities is the case by successively transforming the given equation into simpler forms, until an equivalent equation of the form x = a, a = a, or a = b results (where a and b are different numbers).
### System of Equations 8.EE.8a
Domain: Expressions & Equations
Theme: Analyze and solve linear equations and pairs of simultaneous linear equations
Standard: Understand that solutions to a system of two linear equations in two variables correspond to points of intersection of their graphs, because points of intersection satisfy both equations simultaneously.
Functions
### Functions 8.F.1
Domain: Functions
Theme: Define, evaluate, and compare functions
Standard: Understand that a function is a rule that assigns to each input exactly one output. The graph of a function is the set of ordered pairs consisting of an input and the corresponding output. (Function notation is not required in Grade 8.)
### Compare Functions 8.F.2
Domain: Functions
Theme: Define, evaluate, and compare functions
Standard: Compare properties of two functions each represented in a different way (algebraically, graphically, numerically in tables, or by verbal descriptions). For example, given a linear function represented by a table of values and a linear function represented by an algebraic expression, determine which function has the greater rate of change.
### Linear Functions 8.F.3
Domain: Functions
Theme: Define, evaluate, and compare functions
Standard: Interpret the equation y = mx + b as defining a linear function, whose graph is a straight line; give examples of functions that are not linear. For example, the function A = s^2 giving the area of a square as a function of its side length is not linear because its graph contains the points (1,1), (2,4) and (3,9), which are not on a straight line.
### Linear Function Models 8.F.4
Domain: Functions
Theme: Use functions to model relationships between quantities
Standard: Construct a function to model a linear relationship between two quantities. Determine the rate of change and initial value of the function from a description of a relationship or from two (x, y) values, including reading these from a table or from a graph. Interpret the rate of change and initial value of a linear function in terms of the situation it models, and in terms of its graph or a table of values.
### Analyzing Functions 8.F.5
Domain: Functions
Theme: Use functions to model relationships between quantities
Standard: Describe qualitatively the functional relationship between two quantities by analyzing a graph (e.g., where the function is increasing or decreasing, linear or nonlinear). Sketch a graph that exhibits the qualitative features of a function that has been described verbally.
Geometry
### Transformations of Points & Lines 8.G.1
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Verify experimentally the properties of rotations, reflections, and translations.
### Transformations of Congruent Objects 8.G.2
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Understand that a two-dimensional figure is congruent to another if the second can be obtained from the first by a sequence of rotations, reflections, and translations; given two congruent figures, describe a sequence that exhibits the congruence between them.
### Analyzing Transformations 8.G.3
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Describe the effect of dilations, translations, rotations and reflections on two-dimensional figures using coordinates.
### Transformations & Similar Objects 8.G.4
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Understand that a two-dimensional figure is similar to another if the second can be obtained from the first by a sequence of rotations, reflections, translations, and dilations; given two similar two-dimensional figures, describe a sequence that exhibits the similarity between them.
### Interior & Exterior Angles 8.G.5
Domain: Geometry
Theme: Understand congruence and similarity using physical models, transparencies, or geometry software
Standard: Use informal arguments to establish facts about the angle sum and exterior angle of triangles, about the angles created when parallel lines are cut by a transversal, and the angle-angle criterion for similarity of triangles. For example, arrange three copies of the same triangle so that the three angles appear to form a line, and give an argument in terms of transversals why this is so.
### Pythagorean Theorem 8.G.6
Domain: Geometry
Theme: Understand and apply the Pythagorean Theorem
Standard: Explain a proof of the Pythagorean Theorem and its converse.
### Pythagorean Theorem in Real-World Problems 8.G.7
Domain: Geometry
Theme: Understand and apply the Pythagorean Theorem
Standard: Apply the Pythagorean Theorem to determine unknown side lengths in right triangles in real-world and mathematical problems in two and three dimensions.
### Pythagorean Theorem & Coordinate System 8.G.8
Domain: Geometry
Theme: Understand and apply the Pythagorean Theorem
Standard: Apply the Pythagorean Theorem to find the distance between two points in a coordinate system.
### Volume: Cone, Cylinder, & Sphere 8.G.9
Domain: Geometry
Theme: Solve real-world and mathematical problems involving volume of cylinders, cones and spheres.
Standard: Know the formulas for the volume of cones, cylinders, and spheres and use them to solve real-world and mathematical problems.
Statistics & Probability
### Interpret Data Tables & Scatter Plots 8.SP.1
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Construct and interpret scatter plots for bivariate measurement data to investigate patterns of association between two quantities. Describe patterns such as clustering, outliers, positive or negative association, linear association, and nonlinear association.
### Scatter Plots with Linear Association 8.SP.2
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Know that straight lines are widely used to model relationships between two quantitative variables. For scatter plots that suggest a linear association, informally fit a straight line, and informally assess the model fit by judging the closeness of the data points to the line.
### Analyzing Linear Scatter Plots 8.SP.3
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Use the equation of a linear model to solve problems in the context of bivariate measurement data, interpreting the slope and intercept. For example, in a linear model for a biology experiment, interpret a slope of 1.5 cm/hr as meaning that an additional hour of sunlight each day is associated with an additional 1.5 cm in mature plant height.
### Relatable Data Frequency 8.SP.4
Domain: Statistics & Probability
Theme: Investigate patterns of association in bivariate data
Standard: Understand that patterns of association can also be seen in bivariate categorical data by displaying frequencies and relative frequencies in a two-way table. Construct and interpret a two-way table summarizing data on two categorical variables collected from the same subjects. Use relative frequencies calculated for rows or columns to describe possible association between the two variables. For example, collect data from students in your class on whether or not they have a curfew on school nights and whether or not they have assigned chores at home. Is there evidence that those who have a curfew also tend to have chores?
Course Name: Grade 8 Language Arts Mastery and Test Preparation – Practice Tests and Workbooks: CCSS Aligned
Lesson Name
### Textual Evidence RL.8.1
Theme: Key Ideas and Details
Standard: Cite the textual evidence that most strongly supports an analysis of what the text says explicitly as well as inferences drawn from the text.
### Inferences RL.8.1
Theme: Key Ideas and Details
Standard: Cite the textual evidence that most strongly supports an analysis of what the text says explicitly as well as inferences drawn from the text.
### Theme RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Objective Summary RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Plot RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Setting RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Characters RL.8.2
Theme: Key Ideas and Details
Standard: Determine a theme or central idea of a text and analyze its development over the course of the text, including its relationship to the characters, setting, and plot; provide an objective summary of the text.
### Analyzing Comparisons RL.8.3
Theme: Key Ideas and Details
Standard: Analyze how particular lines of dialogue or incidents in a story or drama propel the action, reveal aspects of a character, or provoke a decision.
### Meaning and Tone RL.8.4
Theme: Craft and Structure
Standard: Determine the meaning of words and phrases as they are used in a text, including figurative and connotative meanings; analyze the impact of specific word choices on meaning and tone, including analogies or allusions to other texts.
### Producing Suspense and Humor RL.8.6
Theme: Craft and Structure
Standard: Analyze how differences in the points of view of the characters and the audience or reader (e.g., created through the use of dramatic irony) create such effects as suspense or humor.
### Compare and Contrast RL.8.5
Theme: Craft and Structure
Standard: Compare and contrast the structure of two or more texts and analyze how the differing structure of each text contributes to its meaning and style.
### Modern Fictions and Traditional Stories RL.8.9
Theme: Integration of Knowledge and Ideas
Standard: Analyze how a modern work of fiction draws on themes, patterns of events, or character types from myths, traditional stories, or religious works such as the Bible, including describing how the material is rendered new.
### Technical Meanings RI.8.4
Theme: Craft and Structure
Standard: Determine the meaning of words and phrases as they are used in a text, including figurative, connotative, and technical meanings; analyze the impact of specific word choices on meaning and tone, including analogies or allusions to other texts.
### Central Ideas RI.8.2
Theme: Key Ideas and Details
Standard: Determine a central idea of a text and analyze its development over the course of the text, including its relationship to supporting ideas; provide an objective summary of the text.
### Analyzing Structures in Text RI.8.5
Theme: Craft and Structure
Standard: Analyze in detail the structure of a specific paragraph in a text, including the role of particular sentences in developing and refining a key concept.
### Author’s Point of View RI.8.6
Theme: Craft and Structure
Standard: Determine an author
### Publishing Mediums RI.8.7
Theme: Integration of Knowledge and Ideas
Standard: Evaluate the advantages and disadvantages of using different mediums (e.g., print or digital text, video, multimedia) to present a particular topic or idea.
### Evaluating Author’s Claims RI.8.8
Theme: Integration of Knowledge and Ideas
Standard: Delineate and evaluate the argument and specific claims in a text, assessing whether the reasoning is sound and the evidence is relevant and sufficient; recognize when irrelevant evidence is introduced.
### Conflicting Information RI.8.9
Theme: Integration of Knowledge and Ideas
Standard: Analyze a case in which two or more texts provide conflicting information on the same topic and identify where the texts disagree on matters of fact or interpretation.
Writing Standards
### Introducing and Closing Topics W.8.1a
Domain: Writing
Theme: Text Types and Purposes
Standard: Introduce claim(s), acknowledge and distinguish the claim(s) from alternate or opposing claims, and organize the reasons and evidence logically.
### Supporting and Developing Topics W.8.1b
Domain: Writing
Theme: Text Types and Purposes
Standard: Support claim(s) with logical reasoning and relevant evidence, using accurate, credible sources and demonstrating an understanding of the topic or text.
### Appropriate Transitions W.8.2c
Domain: Writing
Theme: Text Types and Purposes
Standard: Use appropriate and varied transitions to create cohesion and clarify the relationships among ideas and concepts.
Domain: Writing
Theme: Text Types and Purposes
Standard: Introduce a topic clearly, previewing what is to follow; organize ideas, concepts, and information into broader categories; include formatting (e.g., headings), graphics (e.g., charts, tables), and multimedia when useful to aiding comprehension.
### Varied Sentence Structure W.8.3c
Domain: Writing
Theme: Text Types and Purposes
Standard: Use a variety of transition words, phrases, and clauses to convey sequence, signal shifts from one time frame or setting to another, and show the relationships among experiences and events.
### Precise Language and Sensory Details W.8.3d
Domain: Writing
Theme: Text Types and Purposes
Standard: Use precise words and phrases, relevant descriptive details, and sensory language to capture the action and convey experiences and events.
### Task, Purpose, and Audience W.8.4
Domain: Writing
Theme: Production and Distribution of Writing
Standard: Produce clear and coherent writing in which the development, organization, and style are appropriate to task, purpose, and audience. (Grade-specific expectations for writing types are defined in standards 1
### Planning W.8.5
Domain: Writing
Theme: Production and Distribution of Writing
Standard: With some guidance and support from peers and adults, develop and strengthen writing as needed by planning, revising, editing, rewriting, or trying a new approach, focusing on how well purpose and audience have been addressed. (Editing for conventions should demonstrate command of Language standards 1
### Revising W.8.5
Domain: Writing
Theme: Production and Distribution of Writing
Standard: With some guidance and support from peers and adults, develop and strengthen writing as needed by planning, revising, editing, rewriting, or trying a new approach, focusing on how well purpose and audience have been addressed. (Editing for conventions should demonstrate command of Language standards 1
### Editing W.8.5
Domain: Writing
Theme: Production and Distribution of Writing
Standard: With some guidance and support from peers and adults, develop and strengthen writing as needed by planning, revising, editing, rewriting, or trying a new approach, focusing on how well purpose and audience have been addressed. (Editing for conventions should demonstrate command of Language standards 1
### Gathering Relevant Information W.8.8
Domain: Writing
Theme: Research to Build and Present Knowledge
Standard: Gather relevant information from multiple print and digital sources, using search terms effectively; assess the credibility and accuracy of each source; and quote or paraphrase the data and conclusions of others while avoiding plagiarism and following a standard format for citation.
### Citing Information W.8.8
Domain: Writing
Theme: Research to Build and Present Knowledge
Standard: Gather relevant information from multiple print and digital sources, using search terms effectively; assess the credibility and accuracy of each source; and quote or paraphrase the data and conclusions of others while avoiding plagiarism and following a standard format for citation.
### Quoting and Paraphrasing Data W.8.8
Domain: Writing
Theme: Research to Build and Present Knowledge
Standard: Gather relevant information from multiple print and digital sources, using search terms effectively; assess the credibility and accuracy of each source; and quote or paraphrase the data and conclusions of others while avoiding plagiarism and following a standard format for citation.
Language Standards
Domain: Language
Theme: Conventions of Standard English
Standard: Explain the function of verbals (gerunds, participles, infinitives) in general and their function in particular sentences.
### Subject-Verb Agreement L.8.1b
Domain: Language
Theme: Conventions of Standard English
Standard: Form and use verbs in the active and passive voice.
### Pronouns L.8.1c
Domain: Language
Theme: Conventions of Standard English
Standard: Form and use verbs in the indicative, imperative, interrogative, conditional, and subjunctive mood.
### Phrases and Clauses L.8.1c
Domain: Language
Theme: Conventions of Standard English
Standard: Form and use verbs in the indicative, imperative, interrogative, conditional, and subjunctive mood.
### Verbals L.8.1d
Domain: Language
Theme: Conventions of Standard English
Standard: Recognize and correct inappropriate shifts in verb voice and mood.*
### Active and Passive Voice L.8.3a
Domain: Language
Theme: Knowledge of Language
Standard: Use verbs in the active and passive voice and in the conditional and subjunctive mood to achieve particular effects (e.g., emphasizing the actor or the action; expressing uncertainty or describing a state contrary to fact).
### Using Verbs in Moods L.8.1d
Domain: Language
Theme: Conventions of Standard English
Standard: Recognize and correct inappropriate shifts in verb voice and mood.*
### Capitalization L.8.2
Domain: Language
Theme: Conventions of Standard English
Standard: Demonstrate command of the conventions of standard English capitalization, punctuation, and spelling when writing.
### Punctuation L.8.2a
Domain: Language
Theme: Conventions of Standard English
Standard: Use punctuation (comma, ellipsis, dash) to indicate a pause or break.
### Spelling L.8.2c
Domain: Language
Theme: Conventions of Standard English
Standard: Spell correctly.
### Context Clues L.8.4a
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use context (e.g., the overall meaning of a sentence or paragraph; a word
### Multiple-Meaning Words L.8.4b
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use common, grade-appropriate Greek or Latin affixes and roots as clues to the meaning of a word (e.g., precede, recede, secede).
### Reference Materials L.8.4c
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Consult general and specialized reference materials (e.g., dictionaries, glossaries, thesauruses), both print and digital, to find the pronunciation of a word or determine or clarify its precise meaning or its part of speech.
### Roots, Affixes and Syllables L.8.4b
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use common, grade-appropriate Greek or Latin affixes and roots as clues to the meaning of a word (e.g., precede, recede, secede).
### Interpreting Figures of Speech L.8.5a
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Interpret figures of speech (e.g. verbal irony, puns) in context.
### Relationships Between Words L.8.5b
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Use the relationship between particular words to better understand each of the words.
### Denotations and Connotations L.8.5c
Domain: Language
Theme: Vocabulary Acquisition and Use
Standard: Distinguish among the connotations (associations) of words with similar denotations (definitions) (e.g., bullheaded, willful, firm, persistent, resolute).
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# Pre calc problem solver
Pre calc problem solver is a mathematical instrument that assists to solve math equations. Let's try the best math solver.
## The Best Pre calc problem solver
Math can be a challenging subject for many learners. But there is support available in the form of Pre calc problem solver. Solving for a side in a right triangle can be done using the Pythagorean theorem. This theorem states that in a right triangle, the sum of the squares of the two shorter sides is equal to the square of the length of the hypotenuse. This theorem can be represented using the equation: a^2 + b^2 = c^2. In this equation, a and b represent the lengths of the two shorter sides, while c represents the length of the hypotenuse. To solve for a side, you simply need to plug in the known values and solve for the unknown variable. For example, if you know that the length of Side A is 3 and the length of Side B is 4, you can solve for Side C by plugging those values into the equation and solving for c. In this case, 3^2 + 4^2 = c^2, so 9 + 16 = c^2, 25 = c^2, and c = 5. Therefore, the length of Side C is 5.
An x intercept is where a graph crosses the x-axis. This can be found by solving for when y = 0. This can be done by setting y = mx + b, where m is the slope and b is the y-intercept, to 0 and solving for x. This will give you the x coordinate of the x intercept.
When solving linear equations by graphing, you can use a coordinate plane to graph the coordinates of each number. The coordinates will then represent points on the graph. For example, if there are two numbers and you know their coordinates, then you can draw a line between them to represent their relationship in the equation. This is called graphing a linear equation by intercepts. Another way to graph linear equations is by using an equation sheet. In this case, you need to enter all of the values in the problem before graphing it. Linear equations with more than one variable can also be graphed by using a table or matrix form. When graphing these types of equations, it’s important to include all of the variables as well as their corresponding values in the table or matrix format. This will ensure that your results match what was stated in your problem statement and that your solution gives you an accurate depiction of what’s happening in the equation.
Absolute value equations can be solved with a simple formula. First, you have to know the values of both sides of the equation. The left side will always be positive, and the right side will always be negative. Then, you just subtract one value from the other and solve for the unknown. Absolute value equations are most often used in math, physics and engineering. But they can also be applied to other fields like finance and economics. For example, if you want to sell a car for \$1,000 but you paid \$1,500 for it, your sales price is \$500 too high. In this case, you need to deduct \$500 from your original price to get a realistic selling price. With absolute value equations, it's all about knowing the relationship between two sides of an equation (the left and right sides) and how to find their difference or subtraction (the unknown).
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# Magnitude and Argument
Go back to 'Complex-Numbers'
Consider the complex number $$z = 3 + 4i$$. Let us find the distance of z from the origin:
Clearly, using the Pythagoras Theorem, the distance of z from the origin is $$\sqrt {{3^2} + {4^2}} = 5$$ units. Also, the angle which the line joining z to the origin makes with the positive Real direction is $${\tan ^{ - 1}}\left( {\frac{4}{3}} \right)$$.
Similarly, for an arbitrary complex number $$z = x + yi$$, we can define these two parameters:
1. Modulus of z. This is the distance of z from the origin, and is denoted by $$\left| z \right|$$.
2. Argument of z. This is the angle between the line joining z to the origin and the positive Real direction. It is denoted by $$\arg \left( z \right)$$.
Let us discuss another example. Consider the complex number $$z = - 2 + 2\sqrt 3 i$$, and determine its magnitude and argument. We note that z lies in the second quadrant, as shown below:
Using the Pythagoras Theorem, the distance of z from the origin, or the magnitude of z, is
$\left| z \right| = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( {2\sqrt 3 } \right)}^2}} = \sqrt {16} = 4$
Now, let us calculate the angle between the line segment joining the origin to z (OP) and the positive real direction (ray OX). Note that the angle POX' is
$\begin{array}{l}{\tan ^{ - 1}}\left( {\frac{{PQ}}{{OQ}}} \right) = {\tan ^{ - 1}}\left( {\frac{{2\sqrt 3 }}{2}} \right) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right)\\ \qquad\qquad\qquad\qquad\qquad\;\;\,\,\,\,\,\,\,\,\,\, = {60^0}\end{array}$
Thus, the argument of z (which is the angle POX) is
$\arg \left( z \right) = {180^0} - {60^0} = {120^0}$
It is easy to see that for an arbitrary complex number $$z = x + yi$$, its modulus will be
$\left| z \right| = \sqrt {{x^2} + {y^2}}$
To determine the argument of z, we should plot it and observe its quadrant, and then accordingly calculate the angle which the line joining the origin to z makes with the positive Real direction.
Example 1: Determine the modulus and argument of $$z = 1 + 6i$$.
Solution: We have:
$\left| z \right| = \sqrt {{1^2} + {6^2}} = \sqrt {37}$
Now, the plot below shows that z lies in the first quadrant:
Clearly, the argument of z is given by
$\arg \left( z \right) = \theta = {\tan ^{ - 1}}\left( {\frac{6}{1}} \right) = {\tan ^{ - 1}}6$
Example 2: Find the modulus and argument of $$z = 1 - 3i$$.
Solution: The modulus is
$\left| z \right| = \sqrt {{1^2} + {{\left( { - 3} \right)}^2}} = \sqrt {10}$
Now, we see from the plot below that z lies in the fourth quadrant:
The angle $$\theta$$ is given by
$\theta = {\tan ^{ - 1}}\left( {\frac{3}{1}} \right) = {\tan ^{ - 1}}3$
Can we say that the argument of z is $$\theta$$? Well, since the direction of z from the Real direction is $$\theta$$ measured clockwise (and not anti-clockwise), we should actually specify the argument of z as $$- \theta$$:
$\arg \left( z \right) = - \theta = - {\tan ^{ - 1}}3$
The significance of the minus sign is in the direction in which the angle needs to be measured. The following example clarifies this further.
Example 3: Find the moduli (plural of modulus) and arguments of $${z_1} = 2 + 2i$$ and $${z_2} = 2 - 2i$$.
Solution: We have:
\begin{align}&\left| {{z_1}} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( 2 \right)}^2}} = \sqrt 8 = 2\sqrt 2 \\&\left| {{z_2}} \right| = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 2} \right)}^2}} = \sqrt 8 = 2\sqrt 2 \end{align}
The moduli of the two complex numbers are the same. This is evident from the following figure, which shows that the two complex numbers are mirror images of each other in the horizontal axis, and will thus be equidistant from the origin:
Now, we note that
${\theta _1} = {\theta _2} = {\tan ^{ - 1}}\left( {\frac{2}{2}} \right) = {\tan ^{ - 1}}1 = \frac{\pi }{4}$
Thus,
\begin{align}&\arg \left( {{z_1}} \right) = {\theta _1} = \frac{\pi }{4}\\&\arg \left( {{z_2}} \right) = - {\theta _2} = - \frac{\pi }{4}\end{align}
We have seen examples of argument calculations for complex numbers lying the in the first, second and fourth quadrants. Let us see how we can calculate the argument of a complex number lying in the third quadrant.
Example 4: Find the modulus and argument of $$z = - 1 - i\sqrt 3$$.
Solution: The modulus of z is:
$\left| z \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - \sqrt 3 } \right)}^2}} = \sqrt 4 = 2$
The plot below shows that z lies in the third quadrant:
The angle $$\theta$$ is given by
$\theta = {\tan ^{ - 1}}\left( {\frac{{\sqrt 3 }}{1}} \right) = {\tan ^{ - 1}}\sqrt 3 = \frac{\pi }{3}$
Thus, the angle between OP and the positive Real direction is
$\phi = \pi - \theta = \pi - \frac{\pi }{3} = \frac{{2\pi }}{3}$
Now, since the angle $$\phi$$ sweeps in the clockwise direction, the actual argument of z will be:
$\arg \left( z \right) = - \phi = - \frac{{2\pi }}{3}$
We could also have calculated the argument by calculating the magnitude of the angle sweep in the anti-clockwise direction, as shown below:
This way, we can write the argument is
$\arg \left( z \right) = \pi + \theta = \pi + \frac{\pi }{3} = \frac{{4\pi }}{3}$
Both ways of writing the arguments are correct, since the two arguments actually correspond to the same direction.
Download SOLVED Practice Questions of Magnitude and Argument for FREE
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grade 10 | Questions Set 1
Complex Numbers
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grade 10 | Questions Set 2
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Complex Numbers
grade 10 | Questions Set 1
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Complex Numbers
grade 10 | Questions Set 2
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## Arithmetic
1. Looking at the two procedures to find the cube root of two and three digit numbers in previous post (Cube roots Part 1). We can follow the following procedure to find the cube root of any number. This method is called the division method.
1. Group the digits from right into triples. If the number has a decimal part, make triples to both sides of decimal. If the decimal part has one or two digit left, add one or two zero correspondingly to it.
2. Find a number whose cube is less than or equal to the first triple or the remaining digits after forming triple. Take it as divisor and quotient.
3. Subtract the cube of divisor from the first triple or the remaining digits after forming triples.
4. Bring down the next triple to the right of the remainder. This is the new dividend.
5. Take the thrice of quotient below on the left column of the new dividend.
6. The new divisor is obtained by annexing the thrice of the quotient by a digit. The digit is such that [(10×(the thrice of quotient)×(quotient annexed by new digit)×(new digit)) + (new digit)3] is less than or equal to the new dividend.
7. Annex the new digit to the top quotient.
8. Subtract the number obtained by [(10×(the thrice of quotient)×(quotient annexed by new digit)×(new digit)) + (new digit)3] from the dividend.
9. Repeat the process.
10. In case of taking quotient to decimal, add zeros to right of remainder in triples.
11. Cube root of 15625 is 25.
12. Cube root of 2352637
13. Cube root of 2352.637
14. Cube root of 2.352637
15. Cube root of 2 is 1.259...
## Supplements
Read Basics of the above kind of mathematics.
## Arithmetic
### Cube
1. When we multiply a variable with itself twice it is called a cube of the variable. The cube of a is a3.
2. (a + b)3 = a3 + 3a2b + 3ab2 + b3
3. (a + b + c)3 = a3 + b3 + c3 + 3a2b + 3ab2 + 3a2c + 3ac2 + 3b2c + 3bc2 + 6abc
4. We can write a two digit number as (10a + b).
The cube of this number is (10a + b)3 = 1000a3 + 100×3a2b + 10×3ab2 + b3.
It can be written as (a3)(3a2b)(3ab2)(b3).
Read Basics of the following kind of mathematics.
Suppose we want to find the cube of 13 then 133 is (1)(9)(27)(27)
or (1)(9)(29)(7)
or (1)(11)(9)(7)
or (2)(1)(9)(7)
or 2197
and cube of 25 is (8)(60)(150)(125)
= (8)(60)(162)(5)
= (8)(76)(2)(5)
= (15)(6)(2)(5)
= 15625.
5. We can write a three digit number as (100a + 10b + c).
The cube of this number is (100a + 10b + c)3 = 1000000a3 + 100000×3a2b + 10000×3ab2 + 10000×3a2c + 1000×b3 + 1000×6abc + 100×3ac2 + 100×3b2c + 10×3bc2 + c3
It can be written as (a3) (3a2b) (3ab2 + 3a2c) (b3 + 6abc) (3ac2 + 3b2c) (3bc2) (c3)
Suppose we want to find the cube of 133 then 1333 is
(1)(9)(36)(81)(108)(81)(27)
or (1)(9)(36)(81)(108)(83)(7)
or (1)(9)(36)(81)(116)(3)(7)
or (1)(9)(36)(92)(6)(3)(7)
or (1)(9)(45)(2)(6)(3)(7)
or (1)(13)(5)(2)(6)(3)(7)
or (2)(3)(5)(2)(6)(3)(7)
or 2352637
and cube of 255 is (8)(60)(210)(425)(525)(375)(125)
= (8)(60)(210)(425)(525)(387)(5)
= (8)(60)(210)(425)(563)(7)(5)
= (8)(60)(210)(481)(3)(7)(5)
= (8)(60)(258)(1)(3)(7)(5)
= (8)(85)(8)(1)(3)(7)(5)
= (16)(5)(8)(1)(3)(7)(5)
= 16581375
6. We can write a three digit number with decimal point before the last digit as (10a + b + 10-1c).
The cube of this number is (10a + b + 10-1c)3 = 1000a3 + 100×3a2b + 10×3ab2 + 10×3a2c + b3 + 6abc + 10-1×3ac2 + 10-1×3b2c + 10-2×3bc2 + 10-3c3
It can be written as (a3) (3a2b) (3ab2 + 3a2c) (b3 + 6abc). (3ac2 + 3b2c) (3bc2) (c3).
Suppose we want to find the cube of 13.3 then 13.33 is
(1)(9)(36)(81).(108)(81)(27)
or (1)(9)(36)(81).(108)(83)(7)
or (1)(9)(36)(81).(116)(3)(7)
or (1)(9)(36)(92).(6)(3)(7)
or (1)(9)(45)(2).(6)(3)(7)
or (1)(13)(5)(2).(6)(3)(7)
or (2)(3)(5)(2).(6)(3)(7)
or 2352.637
and cube of 2.55 is (8)(60)(210).(425)(525)(375)(125)
= (8)(60)(210).(425)(525)(387)(5)
= (8)(60)(210).(425)(563)(7)(5)
= (8)(60)(210).(481)(3)(7)(5)
= (8)(60)(258).(1)(3)(7)(5)
= (8)(85)(8).(1)(3)(7)(5)
= (16)(5)(8).(1)(3)(7)(5)
= 1658.1375
### Cube Root
1. The number which when multiplied by itself twice gives the cube of the number, the number is called the cube root of the cube. The cube root of a3 is a.
2. The cube root of a3 + 3a2b + 3ab2 + b3 is (a+b).
3. The cube root of a3 + b3 + c3 + 3a2b + 3ab2 + 3a2c + 3ac2 + 3b2c + 3bc2 + 6abc is (a + b + c).
4. We can represent a two digit number as (10a + b)or(a)(b). The cube of it is (1000a3 + 100×3a2b + 10×3ab2 + b3) or (a3)(3a2b)(3ab2)(b3). The figure describes how to find the cube root of a number whose cube root is a two digit number.
1. Take cube of a, (a3) and subtract it from the number left after making group of three from right.
2. Bring (3a) below.
3. Suffix (b) to it.
4. Main step: Multiply ten times of (3a) in column 1 row 2 ((3a) from (3a)(b)) leaving the new digit (b) to the number (a)(b) in the column 2 row 1 and new digit (b) to get [(3a2b)(3ab2)(0)], add the cube of new number (b) to it to get (3a2b)(3ab2)(b3). The expression looks like this [(3a)(0) ×(a)(b)×(b) + (b3)].
5. For detailed method read the points in the topic below as arithmetic.
5. We can represent a two digit number as (100a + 10b + c) or (a)(b)(c). The cube of it is 1000000a3 + 100000×3a2b + 10000×3ab2 + 10000×3a2c + 1000×b3 + 1000×6abc + 100×3ac2 + 100×3b2c + 10×3bc2 + c3 or (a3) (3a2b) (3ab2 + 3a2c) (b3 + 6abc) (3ac2 + 3b2c) (3bc2) (c3). The figure describes how to find the cube root of a number whose cube root is a three digit number.
1. Take cube of a and subtract it from the digit left after making triples.
2. Bring (3a) below in column 1.
3. Suffix (b) to (3a) to get (3a)(b). Bring next triad below.
4. Main step: Multiply ten times of (3a) in column 1 row 2 ((3a) from (3a)(b)) leaving the new digit (b) to the number (a)(b) in the column 2 row 1 and new digit (b) to get [(3a2b)(3ab2)(0)] add the cube of new number (b3) to it to get (3a2b)(3ab2)(b3). The expression looks like this [(3a)(0)×(a)(b)×(b)+ (b3)].
5. Subtract and bring the next triad (a group of three digits) below.
6. Suffix (3b) to (3a) to get (3a)(3b).
7. Main step: Multiply ten times of (3a)(3b) in column 1 row 2 ((3a)(3b) from (3a)(3b)(c)) leaving the new digit (c) to the number (a)(b)(c) in the column 2 row 1 and new digit (c) to get [(3a2c)(6abc)(3ac2+3b2c)(3bc2)(0)] add the cube of new number (c3) to it to get [(3a2c)(6abc)(3ac2+3b2c)(3bc2)(c3)]. The expression looks like this [(3a)(3b)(0)×(a)(b)(c)×(c)+ (c3)].
8. Subtract this from above.
9. For detailed method read the points in the topic below as arithmetic. |
# Solve the System of Linear Equations and Give the Vector Form for the General Solution
## Problem 296
Solve the following system of linear equations and give the vector form for the general solution.
\begin{align*}
x_1 -x_3 -2x_5&=1 \\
x_2+3x_3-x_5 &=2 \\
2x_1 -2x_3 +x_4 -3x_5 &= 0
\end{align*}
(The Ohio State University, linear algebra midterm exam problem)
## Solution.
We solve the system by Gauss-Jordan elimination.
The augmented matrix of the system is given by
$\left[\begin{array}{rrrrr|r} 1 & 0 & -1 & 0 &-2 & 1 \\ 0 & 1 & 3 & 0 & -1 & 2 \\ 2 & 0 & -2 & 1 & -3 & 0 \\ \end{array}\right].$ We apply the elementary row operations as follows.
\begin{align*}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-2 & 1 \\
0 & 1 & 3 & 0 & -1 & 2 \\
2 & 0 & -2 & 1 & -3 & 0 \\
\end{array}\right] \xrightarrow{R_3-2R_1}
\left[\begin{array}{rrrrr|r}
1 & 0 & -1 & 0 &-2 & 1 \\
0 & 1 & 3 & 0 & -1 & 2 \\
0 & 0 & 0 & 1 & 1 & -2 \\
\end{array}\right].
\end{align*}
Then the last matrix is in reduced row echelon form.
The variables $x_1, x_2, x_4$ correspond to the leading $1$’s of the last matrix, hence they are dependent variables and the rest $x_3, x_5$ are free variables.
From the last matrix we obtain the general solution
\begin{align*}
x_1&=x_3+2x_5+1\\
x_2&=-3x_3+x_5+2\\
x_4&=-x_5-2.
\end{align*}
The vector form for the general solution is obtained by substituting these into the vector $\mathbf{x}$.
We have
\begin{align*}
\mathbf{x}&=\begin{bmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5
\end{bmatrix}=\begin{bmatrix}
x_3+2x_5+1 \\
-3x_3+x_5+2 \\
x_3 \\
-x_5-2 \\
x_5
\end{bmatrix}\10pt] &=x_3\begin{bmatrix} 1 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} +\begin{bmatrix} 1 \\ 2 \\ 0 \\ -2 \\ 0 \end{bmatrix}. \end{align*} Therefore, the vector form for the general solution is given by \[\mathbf{x}=x_3\begin{bmatrix} 1 \\ -3 \\ 1 \\ 0 \\ 0 \end{bmatrix}+x_5\begin{bmatrix} 2 \\ 1 \\ 0 \\ -1 \\ 1 \end{bmatrix} +\begin{bmatrix} 1 \\ 2 \\ 0 \\ -2 \\ 0 \end{bmatrix}, where $x_3, x_5$ are free variables.
## Comment.
This is one of the midterm exam 1 problems of linear algebra (Math 2568) at the Ohio State University.
## Midterm 1 problems and solutions
The following list is the problems and solutions/proofs of midterm exam 1 of linear algebra at the Ohio State University in Spring 2017.
1. Problem 1 and its solution: Possibilities for the solution set of a system of linear equations
2. Problem 2 and its solution (The current page): The vector form of the general solution of a system
3. Problem 3 and its solution: Matrix operations (transpose and inverse matrices)
4. Problem 4 and its solution: Linear combination
5. Problem 5 and its solution: Inverse matrix
6. Problem 6 and its solution: Nonsingular matrix satisfying a relation
7. Problem 7 and its solution: Solve a system by the inverse matrix
8. Problem 8 and its solution:A proof problem about nonsingular matrix
### 8 Responses
1. 02/13/2017
[…] Problem 2 and its solution: The vector form of the general solution of a system […]
2. 02/13/2017
[…] Problem 2 and its solution: The vector form of the general solution of a system […]
3. 02/13/2017
[…] Problem 2 and its solution: The vector form of the general solution of a system […]
4. 02/13/2017
[…] Problem 2 and its solution: The vector form of the general solution of a system […]
5. 02/14/2017
[…] Problem 2 and its solution: The vector form of the general solution of a system […]
6. 02/14/2017
[…] Problem 2 and its solution: The vector form of the general solution of a system […]
7. 07/19/2017
[…] For a similar question, check out the post ↴ Solve the System of Linear Equations and Give the Vector Form for the General Solution. […]
8. 07/27/2017
[…] Problem 2 and its solution: The vector form of the general solution of a system […]
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##### The Possibilities For the Number of Solutions of Systems of Linear Equations that Have More Equations than Unknowns
Determine all possibilities for the number of solutions of each of the system of linear equations described below. (a) A...
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# Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3
Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3 Textbook Questions and Answers.
## BSEB Bihar Board Class 9th Maths Solutions Chapter 6 Lines and Angles Ex 6.3
Question 1.
In figure, sides QP and RQ of ∆ PQR are produced to points S and T respectively. If ∠SPR = 135° and ∠PQT = 110°, find ∠PRQ.
Solution:
We have,
∠QPR + ∠SPR = 180°
⇒ ∠QPR + 135° = 180°
∠QPR = 180° – 135° = 45°
Now, ∠TQP = ∠QPR + ∠PRQ [By exterior angle theorem]
⇒ 110° = 45° + ∠PRQ
⇒ ∠PRQ = 110° – 45° = 65°
Hence, ∠PRQ = 65°
Question 2.
In figure, ∠X = 62%, ∠XYZ = 54°. If YO and ZO are the bisectors of ∠XYZ and ∠XZY respectively of ∆ XYZ, find ∠OZY and ∠YOZ.
Solution:
Consider ∆ XYZ,
∠YXZ + ∠XYZ + ∠XZY = 180° [Angle-sum property]
⇒ 62° + 54° + ∠X∠y
[∵∠YXZ = 62°, ∠XYZ = 54°]
⇒ ∠ XZY = 180°- 62° – 54° = 64°
Since YO and ZO are Therefore ∠XYZ and ∠XZY.
Therefore
Question 3.
In figure, if AB || DE, ∠BAC = 35° and ∠CDE = 53°, find ∠DCE.
Solution:
Since AB || DE and transversal AE intersect them at A and E respectively.
∴ ∠DEA = ∠BAE [Alternative angles]
⇒ ∠DEC = 35° [∵ ∠DEA = ∠DEC and ∠BAE = 35°]
In ∆ DEC, we have
∠DCE + ∠DEC + ∠CDE = 180° [Angle-sum property]
⇒ ∠DCE + 35° + 53° = 180°
⇒ ∠DCE = 180° – 35° – 53° = 92°
Hence, ∠DCE = 92°.
Question 4.
In figure, if lines PQ and RS intersect at point T, such that ∠PRT = 40°, ∠RPT = 95° and ∠TSQ = 75°, find ∠SQT.
Solution:
In ∆ PRT, we have
∠PRT + ∠RTP + ∠TPR = 180° [Angle-sum Property]
⇒ 40° + ∠RTP + 95° = 180°
⇒ ∠RTP = 180° – 40° – 95° = 45°
⇒ ∠STQ = ∠RTP [Vertically opp. angles]
⇒ ∠STQ = 45° [∵∠RTP = 45°(proved) ]
In ∆ TQS, we have
∠SQT + ∠STQ + ∠TSQ = 180° [Angle-sum Property]
⇒ ∠SQT + 45° + 75° = 180° [ ∵ ∠STQ = 45° (proved) ]
⇒ ∠SQT = 180° – 45° – 75° = 60°
Hence, ∠SQT = 60°.
Question 5.
In figure, if PQ ⊥ PS, P PQ || SR, ∠SQR = 28° and ∠QRT = 65°, then find the values of x and y.
Solution:
Using exterior angle property in ∆ SRQ, we have
∠QRT = ∠RQS + ∠QSR
⇒ 65° = 28° + ∠QSR
[∵ ∠QRT = 65°, ∠RQS = 28°]
⇒ QSR = 65° – 28°
= 37°
Since PQ || SR and the transversal PS intersects them at P and S respectively.
∴ ∠PSR + ∠SPQ = 180° [Sum of consecutive interior angles is 180°]
⇒ (∠PSQ + ∠QSR) + 90° = 180°
⇒ y + 37° + 90° = 180°
⇒ y = 180° – 90° – 37°
= 53°
In the right ∆ SPQ, we have
∠PQS + ∠PSQ = 90°
⇒ x + 53° = 90°
⇒ x = 90° – 53°
= 37°
Hence, x = 37°
and y = 53°
Question 6.
In figure, the side QR of ∆ PQR is produced to a point S. If the bisectors of ∠PQR and ∠PRS meet at point T, then prove that ∠QTR = $$\frac { 1 }{ 2 }$$ ∠QPR.
Solution:
In ∆ PQR, we have
ext. ∠PRS = ∠P + ∠Q
⇒ $$\frac { 1 }{ 2 }$$ ext. ∠PRS = $$\frac { 1 }{ 2 }$$∠P + $$\frac { 1 }{ 2 }$$∠Q
⇒ ∠TRS = $$\frac { 1 }{ 2 }$$∠P + ∠TQR … (1)
[∵ QT and RT are bisectors of ∠Q and ∠PRS respectively ∴ ∠Q = 2∠TQR and ext. ∠PRS = 2∠TRS]
In ∆ QRT, we have
ext. ∠TRS = ∠TQR + ∠T … (2)
From (1) and (2), we get
$$\frac { 1 }{ 2 }$$∠P + ∠TQR + ∠T
⇒ $$\frac { 1 }{ 2 }$$∠P = ∠T
⇒ ∠QTR = $$\frac { 1 }{ 2 }$$∠QPR |
Generally, excel function is the combination of a different operator together you have actually seen in our previous post, we acquire a various answer for the same equation depending on, which operator advice first.So it is very important to understand the stimulate of precedence. Below tutorial will help you in understanding just how excel evaluate formula and which operator need to we evaluate first.
You are watching: According to the order of precedence in excel
UNDERSTANDING order OF PRECEDENCE:
You’ll regularly use straightforward formulas the contain simply two values and a single operator. In practice, however, most formulas you use will have actually a variety of values and also operators.
For example, think about the formula =3+5^2. If you calculate from left come right, the answer you gain is 64 (3+5 equals 8, and 8^2 amounts to 64).
However, if you execute the exponentiation an initial and then the addition, the result is 28 (5^2 equals 25, and also 3+25 equals 28). As this example shows, we deserve to have lot of answers, relying on the bespeak in i m sorry you perform the calculations.
To manage this problem, Excel evaluates a formula according to a predefined order of precedence.
ORDER the PRECEDENCE:
Below table summarizes the finish order that precedence provided by Excel.
From this table, you deserve to see that Excel performs exponentiation before addition. Therefore, the exactly answer for the formula =3+5^2, provided previously, is 28.
Note that part operators in the table have the very same order that precedence (for example, multiplication, and division). We deserve to evaluate operator in any kind of sequence, it will offer the same result.
For example, consider the formula =5*10/2. If you perform the multiplication first, the price you get is 25 (5*10 equates to 50, and 50/2 equals 25).
If you do the division first, you additionally get solution of 25 (10/2 equals 5, and 5*5 equals 25).
By convention, Excel evaluate operators with the very same order the precedence indigenous left to right.
CONTROLLING THE bespeak OF PRECEDENCE:
Sometimes, you want to override the stimulate of precedence. For example, suppose that you desire to create a formula that calculates the pre-tax expense of an item. If friend bought something for \$10.65, including 7% sales tax, and also you want to discover the expense of the items minus the tax, you use the formula =10.65/1.07, which gives you the exactly answer that \$9.95. In general, the formula is the full cost split by 1 add to the taxation rate, as presented below
Below picture shows how you might implement such a formula. Cell B5 displays the Total Cost variable, and cell B6 displays the Tax Rate variable.
Given this parameters, your very first instinct could be to usage the formula =B5/1+B6 to calculation the initial cost. This formula is displayed (as text) in cabinet E9, and the result is offered in cabinet D9.
As you have the right to see, this prize is incorrect. What happened? Well, according to the rule of precedence, Excel performs division before addition, so the value in B5 first is split by 1 and then is added to the worth in B6. To get the exactly answer, you should override the order of precedence so that the enhancement 1+B6 is carry out first. You execute this by surrounding that component of the formula through parentheses, as shown in cell E10. When this is done, you obtain the correct answer (cell D10).
POWER the PARENTHESES:
In general, you can use parentheses to manage the order that Excel uses to calculation formulas. We calculate operators inside parentheses an initial then the operators external parentheses(according to the order of precedence).
In the below example “+” has lower precedence but the calculation is done first for addition and then the rest.
To gain even an ext control over her formulas, you have the right to place parentheses within one one more we contact it nesting parentheses. Excel constantly evaluates the innermost set of bracket first. Here are a couple of sample formulas:
Example-1:
Example-2:
Example-3:
Notice the the order of precedence rules likewise holds in ~ parentheses. Because that example, in the expression (5*2−5), the hatchet 5*2 is calculated before 5 is subtracted.
See more: What Are The Positive And Negative Effects Of Imperialism Essay
Using parentheses to recognize the stimulate of calculations enables you to gain full manage over her Excel formulas. This way, you can make certain that the answer given by a formula is the one friend want. |
# What’s Quadrant in L / Z?
The grade of the geometry of this grade school comprises of 4 quadrants. The 4 quadrants are: flat, vertical, middle and lately, oblique.
What’s Quadrant in T? What is the axis by which the geometric figures can be described? It is the axis out. A line on which there has been a figure drawn can be thought of because the axis through. In the event the part research paper writing service of the yaxis, or the quadrant, is not used, then the prime quantity system is utilised to represent the amounts that were basic, also this is the oblique quadrant.
What is Quadrant in Math? Quadrant is divided into four parts: right, left, vertical and horizontal. The left and right quadrants represent the case when the right side of the rectangle is closer to the left than to the right.
What is Quadrant in Math? The vertical quadrant represents the case when the vertical side of the figure is close to the Click Here left than to the right. The vertical quadrant is also referred to as the hypotenuse.
What is Quadrant in Math? The most important function of the quadrant is its application in geometry. The quadrant should be used in the primary level of mathematical learning as a tool to conceptualize and understand the concept of angles. This is done with the aid of an area map.
What is Quadrant in Math? The y-axis allows one to divide the quadrant according to the area of the parallelogram. There are four casedivisions, the horizontal and the right and left.
What is Quadrant in Math? The most common quadrant is the horizontal. Other areas include the vertical, the most popular quadrant; and the most familiar, the oblique quadrant. To https://www.southwestern.edu/live/news/12925-tips-for-writing-an-effective-admission-essay identify the quadrant on the surface of a figure, the origin should be labeled.
What is Quadrant in Math? The four quadrants are the basic methods of representing shapes, lines and figures. The major difference between these four quadrants is the way they relate to the area of the parallelogram. Each quadrant is based on the angle where the point lies on the surface of the parallelogram. |
# Differential equation solver
We'll provide some tips to help you select the best Differential equation solver for your needs. We will also look at some example problems and how to approach them.
## The Best Differential equation solver
This Differential equation solver provides step-by-step instructions for solving all math problems. A solution math example is a type of mathematical example that is used to illustrate a solution to a problem. These examples are usually found in textbooks and other instructional materials. They are designed to help students understand how to solve a particular type of problem.
There are many methods for solving systems of linear equations. The most common method is to use elimination. This involves adding or subtracting equations in order to eliminate one of the variables. Another common method is to use substitution. This involves solving for one of the variables in terms of the others and then plugging this back into the other equations. Sometimes, it is also possible to use matrices to solve systems of linear equations. This involves using matrix operations to find the solutions to the system.
An inverse function is a function that "reverses" another function. In other words, it undoes the work of the original function. To solve an inverse function, you must first determine what the original function was. Once you know the original function, you can apply its inverse to cancel it out and solve for the desired variable. Inverse functions are very useful for solving algebraic equations.
Assuming you want to find the midpoint of a line segment: To find the midpoint of a line segment, you need to find the average of the x-coordinates and the average of the y-coordinates. To find the average of the x-coordinates, add the x-coordinates together and divide by 2. To find the average of the y-coordinates, add the y-coordinates together and divide by 2. The midpoint of
There's no doubt that math can be difficult. Even the simplest equations can trip us up if we're not careful. And when it comes to more complex problems, it can be easy to get lost in all the numbers and symbols. But there's also something really satisfying about solving a math problem. It's like a little puzzle, and when you finally crack the code, it's a great feeling. If you're struggling with a math problem, don't give up |
Pre-Calc Exam Notes 151
# Pre-Calc Exam Notes 151 - in situations when there is...
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Polar Coordinates Section 6.4 151 Example 6.20 Prove that the distance d between two points ( r 1 , θ 1 ) and ( r 2 , θ 2 ) in polar coordinates is d = r r 2 1 + r 2 2 2 r 1 r 2 cos ( θ 1 θ 2 ) . (6.11) Solution: The idea here is to use the distance formula in Cartesian coordinates, then convert that to polar coordinates. So write x 1 = r 1 cos θ 1 y 1 = r 1 sin θ 1 x 2 = r 2 cos θ 2 y 2 = r 2 sin θ 2 . Then ( x 1 , y 1 ) and ( x 2 , y 2 ) are the Cartesian equivalents of ( r 1 , θ 1 ) and ( r 2 , θ 2 ), respectively. Thus, by the Cartesian coordinate distance formula, d 2 = ( x 1 x 2 ) 2 + ( y 1 y 2 ) 2 = ( r 1 cos θ 1 r 2 cos θ 2 ) 2 + ( r 1 sin θ 1 r 2 sin θ 2 ) 2 = r 2 1 cos 2 θ 1 2 r 1 r 2 cos θ 1 cos θ 2 + r 2 2 cos 2 θ 2 + r 2 1 sin 2 θ 1 2 r 1 r 2 sin θ 1 sin θ 2 + r 2 2 sin 2 θ 2 = r 2 1 (cos 2 θ 1 + sin 2 θ 1 ) + r 2 2 (cos 2 θ 2 + sin 2 θ 2 ) 2 r 1 r 2 (cos θ 1 cos θ 2 + sin θ 1 sin θ 2 ) d 2 = r 2 1 + r 2 2 2 r 1 r 2 cos ( θ 1 θ 2 ) , so the result follows by taking square roots of both sides. In Example 6.17 we saw that the equation x 2 + y 2 = 9 in Cartesian coordinates could be expressed as r = 3 in polar coordinates. This equation describes a circle centered at the origin, so the circle is symmetric about the origin. In general, polar coordinates are useful
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Unformatted text preview: in situations when there is symmetry about the origin (though there are other situations), which arise in many physical applications. Exercises For Exercises 1-5, convert the given point from polar coordinates to Cartesian coordinates. 1. (6,210 ◦ ) 2. ( − 4,3 π ) 3. (2,11 π /6) 4. (6,90 ◦ ) 5. ( − 1,405 ◦ ) For Exercises 6-10, convert the given point from Cartesian coordinates to polar coordinates. 6. (3,1) 7. ( − 1, − 3) 8. (0,2) 9. (4, − 2) 10. ( − 2,0) For Exercises 11-18, write the given equation in polar coordinates. 11. ( x − 3) 2 + y 2 = 9 12. y =− x 13. x 2 − y 2 = 1 14. 3 x 2 + 4 y 2 − 6 x = 9 15. Graph the function r = 1 + 2 cos θ in polar coordinates....
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## This note was uploaded on 01/21/2012 for the course MAC 1130 taught by Professor Dr.cheun during the Fall '11 term at FSU.
Ask a homework question - tutors are online |
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# 6.1: Perpendicular Slopes
Difficulty Level: At Grade Created by: CK-12
This activity is intended to supplement Algebra I, Chapter 5, Lesson 4.
In this activity, you will explore:
• an algebraic relationship between the slopes of perpendicular lines
• a geometric proof relating these slopes
## Problem 1 – An Initial Investigation
Open the Cabir Jr. app by pressing APPS and choosing it from the menu. Press ENTER. Press any key to begin.
The calculator displays the Cabri Jr. window. Open the F1: File menu by pressing Y=\begin{align*}Y=\end{align*}. Arrow down to the Open… selection and press ENTER.
Choose figure PERP1 and press ENTER.
Two lines are displayed:
line L1\begin{align*}L1\end{align*} with a slope of m1\begin{align*}m1\end{align*} and line L2\begin{align*}L2\end{align*} with a slope of m2\begin{align*}m2\end{align*}.
Notice that the angle formed by the intersection of the lines measures 90\begin{align*}90^\circ\end{align*}; that is, the two lines are perpendicular.
Grab line L1\begin{align*}L1\end{align*} by moving the cursor over the point pressing ALPHA cursor turns into a hand to show that you have grabbed the point.
Rotate L1\begin{align*}L1\end{align*} by dragging the point using the arrow keys. Observe that as the slopes of the lines change, the two lines remain perpendicular. Explore the relationship between the slopes by answering the questions below.
1. Can you rotate L1\begin{align*}L1\end{align*} in such a way that m1\begin{align*}m1\end{align*} and m2\begin{align*}m2\end{align*} are both positive? Both negative?
2. Can you rotate L1\begin{align*}L1\end{align*} so that m1\begin{align*}m1\end{align*} or m2\begin{align*}m2\end{align*} equals ? If so, what is the other slope?
3. Can you rotate L1\begin{align*}L1\end{align*} so that m1\begin{align*}m1\end{align*} or m2\begin{align*}m2\end{align*} equals 1\begin{align*}1\end{align*}? If so, what is the other slope?
4. Rotate L1\begin{align*}L1\end{align*} so that m1\begin{align*}m1\end{align*} is a negative number close to zero. What can be said m2\begin{align*}m2\end{align*}?
5. Rotate L1\begin{align*}L1\end{align*} so that m1\begin{align*}m1\end{align*} is a positive number close to zero. What can be said about m2\begin{align*}m2\end{align*}?
## Problem 2 – A Closer Examination
Now that you have observed some of the general relationships between the slopes of two perpendicular lines, it is time to make a closer examination. Press 2nd\begin{align*}2^{nd}\end{align*} [MODE] to exit Cabri Jr.
Press PRGM to open the program menu. Choose PERP2 from the list and press ENTER twice to execute it.
Enter a slope of 2\begin{align*}2\end{align*} and press ENTER.
The program graphs a line L1\begin{align*}L1\end{align*} with the slope you entered and a line L2\begin{align*}L2\end{align*} that is perpendicular to L1\begin{align*}L1\end{align*}. m1\begin{align*}m1\end{align*} is the slope of L1\begin{align*}L1\end{align*} and m2\begin{align*}m2\end{align*} is the slope of L2\begin{align*}L2\end{align*}.
Press ENTER and the calculator prompts you for another slope. Use the graph to complete the following.
1. Enter to make the slope of L1\begin{align*}L1\end{align*} equal to . What is the slope of L2\begin{align*}L2\end{align*}?
2. What is the slope of L2\begin{align*}L2\end{align*} when the slope of L1\begin{align*}L1\end{align*} is 1\begin{align*}1\end{align*}?
3. What is the slope of L2\begin{align*}L2\end{align*} when the slope of L1\begin{align*}L1\end{align*} is 1\begin{align*}-1\end{align*}?
Enter other values for the slope of L1\begin{align*}L1\end{align*} and examine the corresponding slope of L2\begin{align*}L2\end{align*}. For each slope that you enter, m1\begin{align*}m1\end{align*} and its corresponding value of m2\begin{align*}m2\end{align*} are recorded in the lists L1\begin{align*}L_1\end{align*} and L2\begin{align*}L_2\end{align*}. To see a history of your “captured” values, enter a slope of 86\begin{align*}86\end{align*} to exit the program. Then press STAT and ENTER to enter the List Editor. The values of m1\begin{align*}m1\end{align*} are recorded in L1\begin{align*}L_1\end{align*} and the values of m2\begin{align*}m2\end{align*} are recorded in L2\begin{align*}L_2\end{align*}.
4. Conjecture a formula that relates the slope of two perpendicular lines. Enter your formula in the top of L3\begin{align*}L_3\end{align*} (with variable L1\begin{align*}L_1\end{align*}) to test your conjecture.
## Problem 3 – A Geometric Look
Start the Cabri Jr. app and open the file PERP3.
This figure shows another way to examine the slopes of perpendicular lines, geometrically. There should be two lines, L1\begin{align*}L1\end{align*} and L2\begin{align*}L2\end{align*}, with a slope triangle attached to each of them.
Grab line L1\begin{align*}L1\end{align*}, rotate it, and compare the rise/run triangles.
1. What do you notice about the two triangles?
## Problem 4 – The Analytic Proof
We now will analytically verify that two lines with slopes m1\begin{align*}m1\end{align*} and m2\begin{align*}m2\end{align*} are perpendicular if and only if m1m2=1\begin{align*}m1 \cdot m2 = -1\end{align*}.
(All of the following assumes m10\begin{align*}m1 \neq 0\end{align*}. What can be said about the case when m1=0\begin{align*}m1 = 0\end{align*}?)
Open the CabriJr file PERP4. This graph shows two perpendicular lines L1\begin{align*}L1\end{align*} and L2\begin{align*}L2\end{align*} with slopes m1\begin{align*}m1\end{align*} and m2\begin{align*}m2\end{align*} respectively, translated such that their point of intersection is at the origin. Refer to the diagram to answer the questions below.
1. What are the equations of these translated lines as shown in the diagram?
2. Let P\begin{align*}P\end{align*} be the point of intersection of line L1\begin{align*}L1\end{align*} and the vertical line x=1\begin{align*}x = 1\end{align*} and let Q\begin{align*}Q\end{align*} be the point of intersection of line L2\begin{align*}L2\end{align*} and the line x=1\begin{align*}x = 1\end{align*}. What are the coordinates of points P\begin{align*}P\end{align*} and Q\begin{align*}Q\end{align*}?
3. Use the distance formula to compute the lengths of OP¯¯¯¯¯¯¯¯,OQ¯¯¯¯¯¯¯¯\begin{align*}\overline{OP},\overline{OQ}\end{align*}, and PQ¯¯¯¯¯¯¯¯\begin{align*}\overline{PQ}\end{align*}.(Your answers should again be in terms of m1\begin{align*}m1\end{align*} and m2\begin{align*}m2\end{align*}.)
4. Apply the Pythagorean Theorem to triangle POQ\begin{align*}POQ\end{align*} and simplify. Does this match your conjecture from Problem 2?
## Problem 5 – Extension Activity #1
The CabriJr file PERP5 shows a circle with center O\begin{align*}O\end{align*} and radius OR\begin{align*}OR\end{align*}. Line T\begin{align*}T\end{align*} is tangent to the circle at point R\begin{align*}R\end{align*}.
The slopes of line T\begin{align*}T\end{align*} and segment OR\begin{align*}OR\end{align*} are shown (mT\begin{align*}mT\end{align*} and mOR\begin{align*}mOR\end{align*}, respectively.)
Your first task is to calculate 1mOR\begin{align*}\frac{1}{mOR}\end{align*}. Activate the Calculate tool, found in the F5: Appearance menu.
Move the cursor over 1\begin{align*}1\end{align*} and press ENTER. Repeat to select mOR\begin{align*}mOR\end{align*}, the slope of the segment OR\begin{align*}OR\end{align*}.
Press / to divide the two numbers. Drag the quotient to a place on the screen where you can see it clearly and press ENTER again to place it.
1. Grab point R\begin{align*}R\end{align*} and drag it around the circle. Observe the changing values of mT,mOR\begin{align*}mT, mOR\end{align*}, and 1mOR\begin{align*}\frac{1}{mOR}\end{align*}. What can you conjecture about the relationship between a tangent line to a circle and its corresponding radius?
## Problem 6 – Extension activity #2
The CabriJr file PERP6 shows a circle with an inscribed triangle QPR\begin{align*}QPR\end{align*}. The segment QR\begin{align*}QR\end{align*} is a diameter of the circle.
The slopes of segments PR\begin{align*}PR\end{align*} and PQ\begin{align*}PQ\end{align*} are shown (mPR\begin{align*}mPR\end{align*} and mPQ\begin{align*}mPQ\end{align*}, respectively.)
Compute 1mPQ\begin{align*}\frac{1}{mPQ}\end{align*} using the Calculate tool.
1. Grab point P\begin{align*}P\end{align*}, drag it around the circle, and examine the changing values. What can you conjecture about a triangle inscribed in a circle such that one side is a diameter?
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## Problem 1084
Find the vertex, focus, and directrix of the following parabola. Then draw the graph.
(x – 2)2 = -3 (y + 1)
Solution:-
The two standard forms of the equation for a parabola are (x – h)2 = 4p(y – k) or
(y – k)2 = 4p(x – k).
Since the parabola is already in one of the standard forms, you can determine the values of h, k, and p, and form these values determine the vertex, focus, and directrix.
h = 2, k = -1, and p = –
Remember, for a parabola written in the form (x – h)2 = 4p(y – k), the vertex is (h , k).
Thus , the vertex of this parabola is (2,-1).
The axis of symmetry is vertical. The focus is – units away from the vertex, in the direction of the axis of symmetry.
Thus, it will be units below the vertex.
The focus is (2,-).
The directrix will be units above the vertex.
The equation for the directrix is
y = –
With this information, graph the parabola. First, graph the vertex, focus, and axis of symmetry.
Next, fill in the graph of the parabola.
## Problem 1083
Find the equation if the parabola determined by the given information.
Focus (3,5), directrix x = -1.
Solution:-
Notice that the directrix is a vertical line. Since the axis of symmetry is perpendicular to the directrix, then it is a horizontal line.
The parabola with a horizontal axis of symmetry will have directrix x = h – p, focus (h +p, k), and standard equation (y – k)2 = 4p(x – h).
Since the directrix of this parabola is x = -1, then – 1 = h – p.
Since the x – value of the focus is 3, then 3 = h + p.
Find h and p using any method for solving a system of equations. Using the elimination method, eliminate p and solve for h.
h = 1
Substitute the value for h into one of the equation to solve for p.
p = 2
k is the y-value of the focus. So, k = 5.
Substitute the values for h, p, and k into the standard equation of a parabola.
(y – 5)2 = 4 *2(x – 1)
(y – 5)2 = 8(x – 1)
## Problem 1082
Find the equation of the parabola determine by the given information.
Focus (6,0), directrix x = -6
Solution:-
First, notice that since the directrix is a vertical line, the parabola will open either to the left or to the right, not up or down. Since the directrix is to the left of the focus, and a parabola always opens away from its directrix, this parabola will open to the right.
The vertex will be halfway between the focus and the directrix. It will have the same y-coordinate as the y-coordinate of the focus.
The vertex of the parabola is , which simplifies to (0,0).
The parabola with directrix x = -p and vertex (0,0) will have the equation y2 = 4px.
Since the directrix of this parabola is x = -6 , p = 6.
The equation of the parabola with directrix x = -6 and focus (6,0) is y2 = 4*6x, which simplify to y2 = 24x. |
# Volume of a Prism – Definition With Examples
## Volume of a Prism – Introduction
The definition of the volume of a prism can be simply given as the space occupied by the prism. A prism is a three-dimensional figure that has flat sides and two identical bases.
The bases of a prism are polygons, like a triangle, a square, a rectangle, or a hexagram. Prisms are often named after the polygons that form the base.
To find the volume, we need to know the prism’s base area and height.
## What Is the Volume of a Prism?
The volume of a prism can be defined as the space it occupies in cubic units. Cubic units measure the volume of a three-dimensional figure.
To find the volume, we multiply the base area with the height of the prism.
The base area is the surface area of the base. The height is the column connecting the two bases of the prism.
When we multiply the base with the height, the resulting volume will be measured in cubic units such as cubic inches, cubic yards, cubic feet, cubic centimeters, or cubic meters.
To find the volume, the base area and height are the only values we need to know.
## What Is the Formula for the Volume of a Prism?
The volume of a prism formula is:
Volume $=$ base area $\times$ height
We can find the volume of any prism using this formula.
Formulas for different prisms are discussed in the following table.
## How to Find the volume of a prism?
The formula to calculate the prism volume can be written as $\text{V} = b \times h$, where V is the volume, b is the base area, and h is the height of the prism.
Finding volume of a prism example:
Let us find the volume of a prism whose base area is 5 square inches and height is 10 inches.
Applying the formula:
$\text{V} = b \times h$
Volume $= 5 \times 10 = 50$ cubic inches.
## Fun Facts!
• We can apply the volume of a prism equation to find the volume of different types of prisms.
• The volume depends on the shape of its base since that will determine the base area.
• The base area of a prism is measured in square units, square inches, or square meters since it represents a two-dimensional area, i.e., the surface area of the prism base.
• The height of a prism is measured in regular units like inches or meters. This is because it represents the length of the column that connects the bases.
## Conclusion
Finding the volume of a prism is easy once you get the hang of the formula. Remember, it doesn’t matter what type of prism it is. As long as you know the height and the base area, you can find its volume!
## Solved Examples
1. What is the base area of a prism if the volume of the prism is 150 cubic feet and the height of the prism is 10 feet?
Solution: Applying the formula:
$\text{V} = b \times h$
$150 = b \times 10$
$b = 150 / 10 = 15$ square feet
2. The base of a prism has a surface area of 277 square centimeters. The volume is 2770 cubic centimeters. Calculate the height.
Solution: Applying the formula:
$\text{V} = b \times h$
$2770 = 277 \times h$
$h = 2770 / 277 = 10$ centimeters
3. The height of a square prism is 11 yards. The volume is 99 cubic yards. What is the length of the side of the square at the base of the prism?
Solution: Volume = base area x height
Applying the formula:
$\text{V} = b \times h$
$99 = b \times 11$
$b = \frac{99}{11} = 9$ square yards
Since the base area is 9 square yards, this is the area of the square base of the prism. So, the side of the square will be the square root of $9 = 3$ yards.
4. The base of a prism has a surface area of 475 square centimeters. The volume is 3800 cubic centimeters. Calculate the height.
Solution: Applying the formula:
$\text{V} = b \times h$
$3800 = 475 \times h$
$h = \frac{3800}{475} = 8$ centimeters
5. The column of a prism has a length of 20 meters. The base of the prism has an area of 25 square meters. What is the volume of the prism?
Solution: Applying the formula:
$\text{V} = b \times h$
$\text{V} = 25 \times 20 = 500$ cubic meters
## Practice Problems
1
### What is the base area of a prism if the volume of the prism is 300 cubic feet and the height of the prism is 6 feet?
5 square feet
50 square feet
5 feet
1800 cubic feet
CorrectIncorrect
Correct answer is: 50 square feet
$\text{V} = b \times h$
$300 = b \times 6$
$b = \frac{300}{6} = 50$ square feet
2
### The base of a prism has a surface area of 25 square centimeters. Its volume is 125 cubic centimeters. Calculate the height.
5 square centimeters
50 square centimeters
5 centimeters
180 cubic centimeters
CorrectIncorrect
Correct answer is: 5 centimeters
$\text{V} = b \times h$
$125 = 25 \times h$
$h = \frac{125}{25} = 5$ centimeters
3
### The base area of a prism is 123 square yards. The height is 9 yards. Find the volume.
16 square yards
50 cubic yards
1107 cubic yards
1800 cubic yards
CorrectIncorrect
Correct answer is: 1107 cubic yards
Applying the formula, $\text{V} = b \times h$
$\text{V} = 123 \times 9$
$= 1107$ cubic yards
4
### The base of a prism has a surface area of 12 square centimeters. The volume of the prism is 144 cubic centimeters. Calculate the height.
1248 cubic centimeters
8 centimeters
12 centimeters
120 square centimeters
CorrectIncorrect
Correct answer is: 12 centimeters
$\text{V} = b \times h$
$144 = 12 \times h$
$h = \frac{144}{12} = 12$ centimeters
5
### The column of a prism has a length of 100 feet. The base of the prism has an area of 25 square feet. What is its volume?
1248 cubic centimeters
8 centimeters
12 centimeters
2500 cubic feet
CorrectIncorrect
Correct answer is: 2500 cubic feet
$\text{V} = b \times h$
$\text{V} = 25 \times 100 = 2500$ cubic feet
## Frequently Asked Questions
How do you find the volume of an irregular prism?
Irregular prisms have irregular polygons as their bases. This means the base shapes have different angles, and the sides are not equal in length. Though the base area will differ, the formula for finding volume will be the same, i.e., base area $\times$ height.
No, the volume of a prism is not affected by the number of faces because its base area and height are not changed by the number of faces.
Different types of prisms have different bases of different shapes and sizes. Since the base area differs, the volume will differ accordingly.
The longer the side of the base polygon, the greater the base area and, therefore, the more the volume.
A right prism is a prism in which the angles between the base and the sides (lateral faces) of the prism are right angles, for example, a rectangular prism. The formula to find the volume is the same: V $= b \times h$. |
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# Difference between revisions of "2003 AMC 10B Problems/Problem 22"
## Problem
A clock chimes once at $30$ minutes past each hour and chimes on the hour according to the hour. For example, at $1 \text{PM}$ there is one chime and at noon and midnight there are twelve chimes. Starting at $11:15 \text{AM}$ on $\text{February 26, 2003},$ on what date will the $2003^{\text{rd}}$ chime occur?
$\textbf{(A) } \text{March 8} \qquad\textbf{(B) } \text{March 9} \qquad\textbf{(C) } \text{March 10} \qquad\textbf{(D) } \text{March 20} \qquad\textbf{(E) } \text{March 21}$
## Solution
First, find how many chimes will have already happened before midnight (the beginning of the day) of $\text{February 27, 2003}.$ $13$ half-hours have passed, and the number of chimes according to the hour is $1+2+3+\cdots+12.$ The total number of chimes is $13+78=91$
Every day, there will be $24$ half-hours and $2(1+2+3+\cdots+12)$ chimes according to the arrow, resulting in $24+156=180$ total chimes.
On $\text{February 27},$ the number of chimes that still need to occur is $2003-91=1912.$ $1912 \div 180=10 \text{R}112.$ Rounding up, it is $11$ days past $\text{February 27},$ which is $\boxed{\textbf{(C) \ } \text{March 10}}$ |
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