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A Summer 2015 Mathematics A To Z: quintile
Quintile.
Why is there statistics?
There are many reasons statistics got organized as a field of study mostly in the late 19th and early 20th century. Mostly they reflect wanting to be able to say something about big collections of data. People can only keep track of so much information at once. Even if we could keep track of more information, we’re usually interested in relationships between pieces of data. When there’s enough data there are so many possible relationships that we can’t see what’s interesting.
One of the things statistics gives us is a way of representing lots of data with fewer numbers. We trust there’ll be few enough numbers we can understand them all simultaneously, and so understand something about the whole data.
Quintiles are one of the tools we have. They’re a lesser tool, I admit, but that makes them sound more exotic. They’re descriptions of how the values of a set of data are distributed. Distributions are interesting. They tell us what kinds of values are likely and which are rare. They tell us also how variable the data is, or how reliably we are measuring data. These are things we often want to know: what is normal for the thing we’re measuring, and what’s a normal range?
We get quintiles from imagining the data set placed in ascending order. There’s some value that one-fifth of the data points are smaller than, and four-fifths are greater than. That’s your first quintile. Suppose we had the values 269, 444, 525, 745, and 1284 as our data set. The first quintile would be the arithmetic mean of the 269 and 444, that is, 356.5.
The second quintile is some value that two-fifths of your data points are smaller than, and that three-fifths are greater than. With that data set we started with that would be the mean of 444 and 525, or 484.5.
The third quintile is a value that three-fifths of the data set is less than, and two-fifths greater than; in this case, that’s 635.
And the fourth quintile is a value that four-fifths of the data set is less than, and one-fifth greater than. That’s the mean of 745 and 1284, or 1014.5.
From looking at the quintiles we can say … well, not much, because this is a silly made-up problem that demonstrates how quintiles are calculated rather instead of why we’d want to do anything with them. At least the numbers come from real data. They’re the word counts of my first five A-to-Z definitions. But the existence of the quintiles at 365.5, 484.5, 635, and 1014.5, along with the minimum and maximum data points at 269 and 1284, tells us something. Mostly that numbers are bunched up in the three and four hundreds, but there could be some weird high numbers. If we had a bigger data set the results would be less obvious.
If the calculating of quintiles sounds much like the way we work out the median, that’s because it is. The median is the value that half the data is less than, and half the data is greater than. There are other ways of breaking down distributions. The first quartile is the value one-quarter of the data is less than. The second quartile a value two-quarters of the data is less than (so, yes, that’s the median all over again). The third quartile is a value three-quarters of the data is less than.
Percentiles are another variation on this. The (say) 26th percentile is a value that 26 percent — 26 hundredths — of the data is less than. The 72nd percentile a value greater than 72 percent of the data.
Are quintiles useful? Well, that’s a loaded question. They are used less than quartiles are. And I’m not sure knowing them is better than looking at a spreadsheet’s plot of the data. A plot of the data with the quintiles, or quartiles if you prefer, drawn in is better than either separately. But these are among the tools we have to tell what data values are likely, and how tightly bunched-up they are.
Author: Joseph Nebus
I was born 198 years to the day after Johnny Appleseed. The differences between us do not end there. He/him.
12 thoughts on “A Summer 2015 Mathematics A To Z: quintile”
1. Aw, sorry. Don’t worry about it, and maybe give it a try again when you’re rested.
If I’d had better organization I’d have put some drawings together. Ways to visualize data should be done in pictures, where possible.
Liked by 1 person
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# Fractional Arithmetic
### Rational Numbers
The set of all fractions of natural numbers is the set of rational numbers. The natural numbers are a subset of the rational numbers. The natural numbers are in the so-called rational numbers as improper fractions (m: 1).
### Greatest common divisor (gcd)
The gcd is the greatest natural number can be divided by without a remainder of two integers.
### Least common multiple (lcm)
The least common multiple of two integers m and n is the smallest natural number which is both multiple of m as well as a multiple of n. The smallest possible common denominator (the so-called. Main denominator) of two fractions is the lcm.
### Special fractions
A fraction with a zero numerator has the value zero.
$\frac{0}{a}=0$
A fraction with a denominator of value one has the value of the numerator.
$\frac{a}{1}=a$
Are numerator and denominator in a fraction equal to the fraction has the value one.
$\frac{a}{a}=1$
A division by zero is undefined.
$\frac{a}{0}=\text{not defined}$
### Sign of fractions
Fractures with the same sign in numerator and denominator have positive sign.
$\frac{+a}{+b}=\frac{-a}{-b}=\frac{a}{b}$
The quotient of two numbers with unequal sign is negative. It follows that the signs of the numerator and denominator can be interchanged.
$\frac{+a}{-b}=\frac{-a}{+b}=-\frac{a}{b}$
A sign before the fraction can be placed in either the numerator or the denominator.
$-\frac{a}{b}=\frac{-a}{b}=\frac{a}{-b}$
## Fractions
As a fraction is called the quotient of two numbers this is the division problem m divided by n, commonly written with a fraction bar. The value above the fraction bar is called numerator and the value below is the denominator.
General fraction with the numbers m and n $m : n = m n$ here is m the numerator and n the denominator of the fraction.
#### Examples for fractions
$7:8=\frac{7}{8}$
Fraction with the numerator 7 and the denominator 8
$\frac{a{x}^{2}+by}{ax}$
General example with a sum in the numerator of the fraction
### Reducing
Reducing of a fraction means the dividing of the numerator and the denominator by the same factor unequal zero. The value of the fraction is unchanged by this reducing. Reducing with the greatest common divisor of numerator and denominator results in a irreducible fraction
General reducing of a fraction: $a⋅c b⋅c = a b ⋅ c c = a b ⋅1 = a b$
Are in the numerator and / or denominator sums then the common factor must be present in all summands and must divided in each summand respectively: $a⋅c+b⋅c x⋅c+y⋅c = ca+b cx+y = c c ⋅ a+b x+y = a+b x+y$
#### Examples for the reducing of fractions
In 12 and 16 is the common factor 4 and with this factor the fraction can be reduced. 4 is the greatest common divisor of 12 and 16. Separation of the fraction in a product gives the faktor 4/4 and this is 1.
$\frac{a{x}^{2}+axy}{ax}=\frac{ax\left(x+y\right)}{ax}$ $=\frac{ax}{ax}\cdot \frac{x+y}{1}=x+y$
In the sum is the common factor a⋅x and can be reduced.
The expanding of a fraction is the opposite of reducing. The multiplying of the numerator and denominator by the same factor does not change the value of the fraction it is the same as the multiplication of the total fraction with one.
The addition of fractions is done by expanding the fraction so that they get the same denominator.
Addition of fractions in general: $a b + c d = a⋅d b⋅d + c⋅b d⋅b = a⋅d+c⋅b b⋅d$
#### Example for the addition of fractions
$\frac{1}{2}+\frac{2}{3}$
Expanding the fraction to the main denominator (LCM) 6 and summarized. Each denominator must be multiplied by a factor to the main denominator. In order not to change the value of the fraction, the factor is multiplied also in the numerator. The multiplication of a factor in the numerator and denominator of the fraction is referred to expanding the fraction.
### Calculator for numerical addition of two fractions
The calculator can also be used for subtracting, when the numerator of a fraction is input with a negative sign.
+
+
### Subtraction
The subtraction of fractions is analogous to the addition.
General subtraction of two fractions: $a b - c d = a⋅d b⋅d - c⋅b d⋅b = a⋅d-c⋅b b⋅d$
#### Example of the subtraction of fractions
$\frac{1}{2}-\frac{2}{3}$
Expanding the fractions to the main denominator 6 analogous to addition and summarized under consideration of the sign
#### Example of the addition / subtraction of fractions in steps
$\frac{4}{9}+\frac{2}{-15}$
In this example all steps are explained step by step.
$=\frac{4}{9}-\frac{2}{15}$
Step 1: Negative signs in the numerator or denominator can be moved in front of the fraction. It is - time - gives + and - times + gives -. The minus in the denominator of the second fraction is multiplied plus in front of the fraction and + times - results in -.
Step 2: Determine the least common multiple of the denominators.
Multiple of 9 are 9; 18; 27; 36; 45
Multiple of 15 are 15; 30; 45
The LCM is 45. That is so both fractions must be adapted so that the denominator is 45. For this purpose, the first break with 45/9 = 5 is extended and the second break with 45/15 = 3 extended.
Step 3: The fractions are now brought to the common denominator and can be written to a common fraction bar.
$=\frac{14}{45}$
Step 4: Evaluate the numerators returns the result. It remains to be examined whether the fracture has a common divisor and can be shortened.
Divisors of 14 are 1; 2; 7; 14
Divisors of 45 are 1; 3; 5; 9; 15; 45
The greatest common divisor is 1. That is so the fraction can not be further reduced. Otherwise you would numerator and denominator divide by the gcd.
### Multiplication
The multiplication of fractions carried by the numerator and denominator are multiplied respectively.
General multiplying two fractions: $a b ⋅ c d = a⋅c b⋅d$
#### Example for the multiplication of fractures
$\frac{1}{2}\cdot \frac{2}{3}$ $=\frac{2}{6}=\frac{1}{3}$
Multiply the numerator and denominator and then reducing the fraction
### Division
The division of fractions is carried out by the first fraction is multiplied by the reciprocal value of the second.
General division of two fractions: $a b : c d = a b ⋅ d c = a b c d = a⋅d b⋅c$
#### Example of the Division of fractions with main fraction bar
$a+b x + 1 x2 1 + 1 x$
In this example all steps are explained step by step.
$= xa+b x2 + 1 x2 1 + 1 x$
Step 1: The fractions in the numerator are brought to the common denominator. That means the first fraction is expanded with x.
$= xa+b x2 + 1 x2 x x + 1 x$
Step 2: The fractions in the denominator are brought to the common denominator.
$= xa+b+1 x2 x+1 x$
Step 3: Now the fractions in the numerator and the denominator in the fraction can be written on their common denominator.
$= xa+b+1x x2x+1$
Step 4: Execution of the division.
$= xa+b+1 xx+1$
Step 5: Reducing by x.
### Power
A fraction is raised to the power in the way that numerator and denominator potentiated seperatly.
$a b p = ap bp$
### Roots
The root of a fraction is obtained by dividing the roots of the numerator and denominator of the fraction.
$a b = a b$
#### Example of a fraction with roots
$\sqrt{\frac{a{x}^{2}}{\left(a-b\right)}}=\frac{\sqrt{a{x}^{2}}}{\sqrt{\left(a-b\right)}}=\frac{\sqrt{a}x}{\sqrt{\left(a-b\right)}}$
The root is applied to the numerator and denominator. |
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# Subtracting Mixed Number
## Subtracting Mixed Number
Numbers are the basic need for the mathematical process. We are going to learn
about how to perform mathematical operations on the mixed numbers. Here we
When we talk about subtracting mixed numbers, we will first convert the mixed
numbers in the form of improper fraction numbers. Once the mixed numbers are
converted in the form of improper fraction numbers, we say that the denominators
of the two numbers need to be same.
It means to do the addition or subtraction on the improper fraction numbers is
only possible when we have two fraction numbers as like fractions. So we check if
the subtrahend and the minuend are like fractions or not.
If they are unlike fraction numbers, then we come to the conclusion that the
fraction numbers must be converted into their equivalent form such that the
denominators of the two fractions become same.
## Know More About :- How To Graph On A Coordinate Plane
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For this we will first find the LCM of the two denominators and then the fraction
will be converted in the form of equivalent fraction such that the denominator
becomes equal to the LCM.
Now if becomes the simple problem of subtraction and can be solved by simply
subtracting the numerator of the subtrahend from the numerator of the minuend.
Now we will see how will we perform the operation of Subtracting Mixed Numbers.
Let us take the two numbers as follows :
Subtract 3 from 5 2/3
This expression can be written as 5 2/3 3
First we will convert both the mixed fraction numbers in the form of improper
fraction numbers. So we proceed as follows : ( 5 * 3 + 2 ) / 3 - ( 4 * 3 + 1 ) / 4
= 17 / 3 - 13 / 4
Now both the fraction numbers are in the form of improper fractions. More over
the two fractions are unlike fraction numbers.
So we will first find the LCM of the two denominators 3 and 4. So we will get 12 as
the LCM.
Now we write 17 / 3 as an equivalent fraction with the denominator 12 , for this
the numerator and the denominator will be multiplied by 4 and we get : ( 17 * 4 )
/ 12 and another fraction will be written as ( 13 * 3 ) / 12
= ( 68 / 12) ( 39 / 12 )
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= ( 68 39 ) / 12
= 29 / 12
Now we observe that the HCF of the numerator and the denominator is 1, so we
say that the result is in its lowest form. But the resultant fraction is improper
fraction, so it can be converted into the mixed form as follows:
Divide 29 by 12 , we get 2 as the quotient and 5 as the remainder. So we write
the fraction 29 / 12 as 2 whole 5 /12 Ans.
In this way we can find the difference between the two fraction numbers and get
the result.
We can learn how to plot Different Types of Graphs, with the given data which can
help us to predict the conclusions by the help of online math help given by math
online tutors. We can also get sample papers for physics CBSE board available on
the CBSE website to get the idea of the examination pattern.
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Presentation |
# The Difference Between Vertical Asymptotes and Removable Discontinuities
Asymptotes are a vital part of graphing a rational function. They help show the direction the graph moves in and where it is defined and undefined.
Removable discontinuities and vertical asymptotes are undefined areas of a rational function. Both bring shape and direction to the graph of a rational function. Vertical asymptotes are invisible or ghost lines that show where a rational function is not allowed. A removable discontinuity is a hole along the curve of a function in a rational function graph. It is an undefined point instead of a line. The discontinuity is not as stark as the vertical asymptote, but it is still undefined at that particular point.
As we continue, we’ll break down how to find each one so that the concepts are a bit clearer. We will dive deeper into finding vertical asymptotes and how to find removable discontinuities. We’ll start with removable discontinuities.
## What is a removable discontinuity?
Both removable discontinuities and vertical asymptotes are undefined parts of a rational function. They are both areas on the graph where the function cannot exist since they create a zero in the denominator. Because of the nature of a removable discontinuity, they are less severe than a vertical asymptote.
A removable discontinuity is a puddle that the graph merely jumps over. It appears as a hole in the graph. On the other hand, a vertical asymptote is the same as an invisible wall. The graph is not allowed to pass.
The first step in our process is to see if the rational function has any removable discontinuities. We do this by finding out if there are any common factors in the numerator and denominator.
For example, if we have the following rational function.
The first step that we have to take is to reduce this function. We have to see if there are any common factors in the numerator and denominator.
When we factor the rational function, we find a common factor in the numerator and denominator. The common factor shows us where the removable discontinuity is.
Our removable discontinuity exists at a point (x, y). To find this point, we take the common factor out from our function and set it equal to zero. Our common factor is x+6. Once we set it equal to zero, the result is x=-6.
Now to find the y-coordinate of our removable discontinuity.
To find the y-coordinate, we plug the x back into the reduced function. The result is y = -1/12.
The discontinuity exists at the point (-6, -1/12). When we graph this rational function, there will be a hole at that one point.
Remember that removable discontinuity won’t always exist in a rational function. Removable discontinuities or holes only exist when there is a common factor in the numerator and denominator.
## What are vertical asymptotes?
After reducing the rational function, we can find the vertical asymptotes.
Keep in mind that there cannot be a zero in the denominator of any rational functions. The denominator is where we will look to find the vertical asymptotes.
Let’s take the example from above. Remember that the vertical asymptote shows where our function is undefined, so we must find the values that create a zero in the denominator. Once we have reduced the function, we can find the vertical asymptotes.
Our denominator contains x-6. We set x-6 equal to zero and solve.
We find that our vertical asymptote is at x=6. The function, shown in red, is being influenced by the vertical asymptote. You can’t see it here, but the function approaches the asymptote but never crosses it.
## Here are a few more examples.
The first example is for a function that doesn’t have any common factors in the numerator and denominator. Since the function cannot be reduced, there are no removable discontinuities.
Our next step is to find the vertical asymptotes. Looking at the denominator, we find the function:
Vertical asymptotes exist where the denominator is equal to zero. If we look at this function, we see that this function can never be zero. If we put in any negative number because we squared it, it will always be positive. That means that no value can create a zero in the denominator.
Therefore, there are no vertical asymptotes.
This function has no removable discontinuities and no vertical asymptotes.
For the following example, we see the function:
With this function, there may be a possibility to find common factors. But once we factor the polynomials, we find that the numerator and denominator do not share a common factor.
Now we can move on to find the vertical asymptotes. Concentrating on the denominator, we find that we have 2x-3 and x+2.
Once we set each of these equations to zero, we find x=3/2 and x=-2. These two x-values are the vertical lines that will guide our rational function. The vertical asymptotes are at these x-values.
For the last example, we see this function:
We immediately see that this function can be reduced. The common factor in the numerator and denominator is the x. Since it’s just x, we can set the x equal to zero. That’s the x-coordinate of our removable discontinuity.
To find the y-coordinate, we put 0 back into our reduced function and solve.
Our removable discontinuity is at (0,4).
Now to find the vertical asymptote. Looking at the denominator, we see that we have x-2 and x+2.
Setting both of these equal to zero, we find that we have vertical asymptotes at x=2 and x=-2.
Hopefully, this has been helpful. Keep practicing, and don’t give up!
Danielle
Just a nerd who loves math. Trying to help the world one problem at a time. |
# Video: Using Bayesβ Rule to Find the Conditional Probability of an Event
Suppose that π΄ and π΅ are events with probabilities π(π΄) = 0.63 and π(π΅) = 0.77. Given that π(π΅|π΄) = 0.88, find π(π΄|π΅).
01:40
### Video Transcript
Suppose that π΄ and π΅ are events with probabilities: probability of π΄ is 0.63 and probability of π΅ is 0.77. Given that the probability of π΅ given π΄ is equal to 0.88, find the probability of π΄ given π΅.
We can find the probability of π΄ given π΅ using a formula. We can use the formula the probability of π΄ given π΅ is equal to the probability of π΅ given π΄ times the probability of π΄ divided by the probability of π΅. Or we could use the formula the probability of π΄ given π΅ is equal to the probability of π΄ and π΅ divided by the probability of π΅.
Notice, however, in this second formula, we have the probability of π΄ and π΅. In a sample space, here will be event π΄ and here will be event π΅. Where π΄ and π΅ overlap would be the intersection of π΄ and π΅. So the probability of π΄ and π΅ would be inside of here.
But looking at what weβre given, we are given the probability of π΄, the probability of π΅, and the probability of π΅ given that π΄ has already happened. And all three of those are found here in this formula. So this is what weβll use.
The probability of π΅ given π΄ is 0.88. The probability of π΄ is 0.63. And the probability of π΅ is 0.77. Multiplying on the numerator, we get 0.5544. And now we need to divide by 0.77. And we get 0.72. So this means that the probability of π΄ happening given that π΅ has already happened is 0.72. |
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# ZingPath: Concepts of Function
Searching for
## Concepts of Function
Learn in a way your textbook can't show you.
Explore the full path to learning Concepts of Function
### Lesson Focus
#### Determining Whether a Relation is also a Function
Algebra-1
You will use various representations of relations to show what makes a relation a function.
### Now You Know
After completing this tutorial, you will be able to complete the following:
• Determine whether a relation is a function or not when the relation is described by a verbal rule, an equation, a table or a Venn diagram.
• Determine whether a relation is a function or not when the relation is represented on a graph
### Everything You'll Have Covered
Function terminology explained
A function is a dependency between the elements of two sets. The first set, the independent variable, is called the domain and is also referred to as the input variable. The second set, the dependent variable, is called the range and is also referred to as the output variable.
Here is an example. You go to a cell phone provider and are told the plan you want is \$25.00 a month plus \$0.15 cents for each text message. You tend to send eight or less text messages a month. The number of text messages is the independent variable; the dependent variable is the total cost per month. This function is represented below using a graph, equation, table, and mapping diagram.
Relation = (number of test messages, total cost)
y = 25.00 + 0.15x
Relation as a Function
A relation is a function where every input has an output and each input has only one output. A relation is a function in which every element in the domain is mapped to an element in the range and each element in the domain is mapped to only one element in the range. No element in the domain is left unmapped and no element in the domain is mapped to more than one element in the range.
The above example represents a function because all the criteria are met.
Expressing functions
Functions can be expressed using a verbal rule, an equation, a graph, and with a table or a mapping diagram. An example follows:
Verbal rule: B = {(x, y): 3x = y, where x and y are positive integers and x is less than or equal to 4}
Equation: f(x) = 3x with Domain = {1, 2, 3, 4}
Table:
Graph:
Mapping diagram:
### Tutorial Details
Approximate Time 20 Minutes Pre-requisite Concepts Learners should be familiar with the basic properties of sets, basic understanding of relations, graphing relations Course Algebra-1 Type of Tutorial Skills Application Key Vocabulary functions, graphs of functions, relation |
Question 89fd2
2 Answers
Feb 1, 2018
The middle number is : $x = 18$
Explanation:
Given that the three numbers are consecutive.
If the middle number is $' x '$,
the first number will be $' x - 1 '$
and the last number will be $' x + 1 '$
Given:
$\left(x - 1\right) + x + \left(x + 1\right) = 54$
$\implies x - 1 + x + x + 1 = 54$
$\implies 3 x = 54$
$\implies x = \frac{54}{3} = 18$
So the three consecutive numbers that sum up to 54 are : 17, 18 , and 19.
And, the middle number is $x = 18$
Feb 1, 2018
$\text{middle number } = 18$
Explanation:
$\text{assuming the numbers to be whole then}$
$\text{the middle number is } x$
$\text{the number before it is } x - 1$
$\text{the number after it is } x + 1$
rArr"sum "=xcancel(-1)+x+xcancel(+1) =54#
$\Rightarrow 3 x = 54$
$\text{divide both sides by 3}$
$\frac{\cancel{3} x}{\cancel{3}} = \frac{54}{3}$
$\Rightarrow x = 18 \leftarrow \textcolor{red}{\text{middle number}}$
$\text{the 3 numbers are "17,18" and } 19$ |
# What is the trick for comparing fractions?
## What is the trick for comparing fractions?
Instead of rewriting them in terms of a common denominator, the fastest way to compare fractions is to convert them into decimal numbers. After you do that, you can then put the fractions you’re comparing in ascending or descending order simply by ordering them in terms of their decimal representations.
What is the greatest fraction?
👉The fraction with the biggest numerator is the greatest. That’s right! When written in order from least to greatest, you have 6/12 < 8/12 < 9/12.
How do you compare fractions using a number line?
Comparing fractions on a number line is very similar to finding equivalent fractions on a number line. The children begin by finding the least common multiple of the two fractions and creating two lines with that many spaces. Once the place the fractions on the number line it is easy to see which fraction is greater.
### What is comparing and ordering of fractions?
How to Compare and Order Fractions Method 1 of 3: Converting Fractions to Decimals. List the fractions you are ordering in one column. Next to each fraction, write an equal sign. Method 2 of 3: Finding a Common Denominator. List the fractions you are ordering in one column. Method 3 of 3: Reasoning About Fraction Size. Compare and order unit fractions.
How can you use number line to compare two numbers?
Here are two ways to compare numbers: Count. The number you get to first is smaller. Use a number line. The greater, or higher number is always further along the line.
What is the part of a fraction above the line?
Fractions only have two main parts: the numerator and the denominator. The numerator is the number above the line in the fraction. The denominator is the number below the line in the fraction. The denominator tells you how many pieces the whole is broken up into. |
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### Feb. 18, 2014
1. 1. February 18, 2014 Today: Exponents, Exponents, Exponents; (Exponents3) Test Scores Posted Khan Topics Posted/Alt.Khan here Class Work
2. 2. Vocabulary & Formulas Section of Notebook
3. 3. Exponents are the mathematician's shorthand. In general, the format for using exponents is: (base)exponent where the exponent tells you how many of the base are being multiplied together.
4. 4. A Summary of Exponent Rules 1. 3. 4. 6. 8. 2. Also Known as the power of a power rule. 5. 7.
5. 5. Common Errors with Exponents
6. 6. Understanding Exponent Rules Solve:
7. 7. Practice Problem Section of Notebook
8. 8. Understanding Exponent Laws:
9. 9. Understanding Exponent Laws:
10. 10. Understanding Exponent Laws:
11. 11. Understanding Exponent Laws:
12. 12. Understanding Exponent Laws: Practice Problems 1. 72 2. (-8)2 3. (-9) 3 4. -24 5. -43
13. 13. Exponent Laws Simplify to lowest terms:
14. 14. Exponent Laws 1. p2 • p4 • q3 • q5 = p 6q 8 Only exponents of the same bases can be added. Power of a Power Property To find the power of a power, multiply the exponents.
15. 15. Exponents: Zero and One as Exponents This is what the algebra text said about 0 as an exponent: “When you have a 0 as an exponent, your answer will always be 1.The only exception is 00, which is undefined”. But this is incorrect. What they meant to say is: When you have a 0 as an exponent, your answer will always almost always be 1. The only exception One exception is 00, which is undefined. But there are others: 40 = 1 50000 = 1 - (-5)0 = -1 50 = 1 (-5)0 = 1 -(5)0 = -1
16. 16. Negative and Zero Exponents Take a look at the following problems and see if you can find the pattern. The expression a-n is the reciprocal of an Examples:
17. 17. Negative and Zero Exponents Example 1 Example 2 Since 2/3 is in parenthesis, we must apply the power of a quotient property and raise both the 2 and 3 to the negative 2 power. First take the reciprocal to get rid of the negative exponent. Then raise (3/2) to the second power.
18. 18. Negative and Zero Exponents Example 3 Step 1: Step 2: Step 3:
19. 19. Negative and Zero Exponents Example 4: Step 1:
20. 20. Negative and Zero Exponents Step 2: Step 3: Step 4:
21. 21. Negative and Zero Exponents Step 5: Step 6-7:
22. 22. Practice Problems |
# Motion of a Cycloid
With today's modern computers, there are many ways to model the motion of a cycloid. As a cycloid is the locus of a point on a circle that is rolling along a line, a graphing program would be the easiest way to generate one. To simplify the graphing procedure , we will examine the motion of the point seperately in the horizontal and vertical plane. By breaking up the motion into two components, it will be easier to visualize and the results will be able to be graphed parametrically.
The first motion that we should consider is the motion in the horizontal plane. The center of the circle will be moving forward at a constant rate t. The rotation of the circle will also affect the motion of the point. Before the circle begins rotating, the point is directly in line with the center, below it. As the circle begins to rotate, the point falls behind the center until it has rotated one fourth the way around, it then begins catching upto to the center, until it is even with it after one half of a rotation. As sine is a function that begins at zero, increases and then decreases, the second part of the motion in the horizontal plane is sin[t]. As both of these motions will depend upon the radius, the function that will define horizontal motion is r(t-sin[t]).
The point will also be moving in the vertical direction as the circle rotates. The point will start below the center, as the rotation begins it will get closer to the center, until it reaches the height of the center one fourth of the way through a revolution. A function that begins at its maximum and decreases is cosine. To start at floor level, the function needs more terms that just the cosine term. The function that defines the motion in the vertical direction is r(1-cos[t]).
To achieve one complete revolution, t needs to begin at 0 and stop at two Pi. The parametric equations that define the motion of a cycloid are
### x=r(t-sin[t]) y=r(1-cos[t])
where t measures the angle of rotation of the cycloid. Many different graphing programs can be used to graph the results of this parametirc equation, and Mathematica 3.0 was chosen for this investigation. The following lines, when entered into Mathematica, will produce a cycloid where the circle has undergone two complete rotations.
```r = 1;
g1=ParametricPlot[{(r t - r Sin[t]),r(1 - Cos[t])},{t,0,4Pi}]```
The following graphs show the cycloid generated by the equations above. The green point on the circle is the point whose locus is being used to generate the cycloid.
```
```
Each time the circle and point have been drawn, t has increased by a Pi/3. It is interesting to note that the majority of the horizontal motion occurs from 2Pi/3 to 4Pi/3. This is easier to see in the graph below that has multiple circles on the same graph.
As the radius is increased, two results occur. The first result is that the highest point on the cycloid increases, it is the diameter of the circle. The other result is that the distance in the horizontal direction that is required to complete a revolution increases. The distance is the circumference of the circle. The following graphs illustrate the point. The red cycloid has a radius of 0.5, the purple cycloid has a radius of 1, the aqua cycloid has a radius of 2 and the green cycloid has a radius of 4.
```
``` |
Types of vectors and algebraic operations
Equal vectors: Vectors with the same length and direction, and must represent the same quantity (such as force or velocity).
Unit vector: For this vector, the length is always 1.For a vector 𝑎 ⃗ , a unit vector is in the
the same direction as 𝑎 ⃗ and is given by: ${a} = \frac{\overrightarrow{a}}{\vert \overrightarrow{a}\vert }$
$\begin{matrix} a_{x}=(x_{B}-x_{A}) & \\ a_{y}=(y_{B}-y_{A}) & \\ a_{z}=(z_{B}-z_{A}) & \\ \end{matrix}$
${a} = \frac{a_{x}{i} + a_{y}{j} + a_{z}{k}}{\sqrt{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}}=\frac{1}{\sqrt{a_{x}^{2}+a_{y}^{2}+a_{z}^{2}}} (\begin{matrix} a_{x} & \\ a_{y} & \\ a_{z} & \\ \end{matrix} )$
Zero vector: When the initial and terminal points of the vector coincide, we obtain a zero vector or null vector and it is denoted as $\overrightarrow{0}$. For example, $\overrightarrow{AA}$ or $\overrightarrow{PP}$ represents a zero vector.
Parallel vectors:
$\overrightarrow{a}$ is parallel to $\overrightarrow{b} ⇔ \overrightarrow{a}= k\overrightarrow{b}k\epsilon R$
$\overrightarrow{a}= (\begin{matrix} 6 & \\ 9 & \\ 3 & \\ \end{matrix} ) \overrightarrow{b}= (\begin{matrix} 2 & \\ 3 & \\ 1 & \\ \end{matrix} ) (\begin{matrix} 6 & \\ 9 & \\ 3 & \\ \end{matrix} ) = 3(\begin{matrix} 2 & \\ 3 & \\ 1 & \\ \end{matrix} ) \to \overrightarrow{a} || \overrightarrow{b}$
Coinitial vectors:
Two or more vectors that have the same initial points are called coinitial vectors. Eg. $\overrightarrow{AB }$ and $\overrightarrow{AC}$ are coinitial vectors.
Collinear vectors:
Points are collinear if they lie on the same line. A, B and C are collinear ⇔ $\overrightarrow{AB}=k\overrightarrow{AC}$ for some scalar k. This means that they have one common point and in the same direction.
Example: Show that P(0, 2, 4), Q(10, 0, 0) and R(5, 1, 2) are collinear.
$\overrightarrow{PR}=(\begin{matrix} 5 & \\ -1 & \\ -2 & \\ \end{matrix} ), \overrightarrow{QR}= (\begin{matrix} -5 & \\ 1 & \\ 2 & \\ \end{matrix} ) \overrightarrow{QR}= -1\times \overrightarrow{PR}$
$\overrightarrow{QR}$ and $\overrightarrow{PR}$ have a common direction and a common point. Therefore P, Q, and R are collinear.
Negative of a vector:
Negative of a vector or Inverse vectors have the same length, but opposite direction.
The sum of two or more vectors = resultant vector. The resultant vector can replace the vectors from which it is obtained. It is completely canceled out by adding it to its inverse, which is then called the equilibrant.
Triangle Law:
Let $\overrightarrow{AB}$ represent a vector a and $\overrightarrow{BC}$ represent a vector b. Then $\overrightarrow{AC}$ represents the vector c in the triangle shown below:
Parallelogram law:
Arrange the tail of the vector to tail in the correct direction and draw to scale. Now sketch two identical vectors as the originals to form a parallelogram. Draw in the diagonal of the parallelogram. This is your answer called a resultant. Measure the resultant and find the angle.
Vector multiplication with scalar: Let $\overrightarrow{a}$ and $\overrightarrow{b}$be two vectors and g and h as the scalar quantity. Then we can assume that:
• $g\overrightarrow{a}+h\overrightarrow{a} =(g+h)\overrightarrow{a}$
• $g(h\overrightarrow{a})=(gh)\overrightarrow{a}$
• $g(\overrightarrow{a} +\overrightarrow{b})=g\overrightarrow{a} +g\overrightarrow{b}$
Properties of vectors:
1. Commutative Property: A+B = B+A
2. Associative Property: (A+B)+C = A+(B+C)
3. Zero Property: A+(-B) = 0, iff, A has the same magnitude to B and pointing in the opposite direction.
4. Subtraction: A – B = A + (-B)
5. Multiplication: 3 x A = 3A
Components of a vector: In three dimensions, the vector components of vector A are three perpendicular vectors AxAyand Az that are parallel to the x, y and z axes, respectively, and add together vectorially so that
A = Ax + Ay + Az
In this case: Ax = A cos Qx, Ay = A Sin Qy and Az = A Tan Qz
The position vector of a point that divides a line segment in a given ratio:
Let the point on line segment AB :A = (2, 7, 8) B = ( 2, 3, 12)
X divides [AB]in the ratio means $\overrightarrow{AX}:\overrightarrow{XB}=a :b$
Example: P divides [AB] internally in ratio 1:3. Find P
$\overrightarrow{AP}:\overrightarrow{PB}=1:3 \to \overrightarrow{AP}=\frac{1}{4} \overrightarrow{AB}$
$\overrightarrow{AP}=\frac{1}{4} \overrightarrow{AB}$
$(\begin{matrix} x-2 & \\ y-7 & \\ z-8 & \\ \end{matrix} )=\frac{1}{4}(\begin{matrix} 0 & \\ -4 & \\ 4 & \\ \end{matrix} )$
point P is (2, 6, 9)
Example: X divide [AB] externally in ratio 2:1, or X divide [AB] in ratio –2:1. Find Q
$\overrightarrow{AQ}:\overrightarrow{QB}=-2:1$
$\overrightarrow{BQ}=\overrightarrow{AB}$
$(\begin{matrix} x-2 & \\ y-3 & \\ z-12 & \\ \end{matrix} )=(\begin{matrix} 0 & \\ -4 & \\ 4 & \\ \end{matrix} )$
point Q is (2,– 1,16) |
# Basketball Statistics – Math in Sports
Photo by: jc.winkler
## Math in Sports: Basketball Statistics
Math can be found in every area of life including sports. Children will enjoy using math to find their own statistics for the sports they play. Basketball is a popular winter sport for children of all ages. Finding common basketball statistics for themselves and their teammates is a fun way to reinforce math concepts.
## Total Points Per Game
Children love to know how many points they scored in a game. Keeping track of how many 2 point shots, 3 point shots, and free throws your child scored provides an opportunity to practice either addition or both multiplication and addition.
• A child could simply add up all the points they made throughout the game.
• If you want to make it a little harder, you could have them multiply the number of each type of shot by the value of the shot and then add the three totals together. (9 two pointers would be 9 x 2 = 18, 4 three pointers would be 4 x 3 = 12, and 5 free throws made would be 5 x 1 = 5. Then add 18 + 12 + 5 to get 35 total points for that game.)
• He could also add up all the points in all the games for a season to find out how many points he scored for the entire season!
## Free Throw Percentage
Finding the percentage of free throws a child makes is another interesting basketball statistic. To get this percentage, your child would divide the number of free throws he made by the number he attempted. (If he made 6 free throw shots but he attempted 16 shots, the answer would be 0.375.) Then move the decimal in the answer two places to the right to get a percentage. (0.375 becomes 37.5%.)
Children will love being able to tell their friends and family their own basketball statistics from game to game. They may even overlook the fact that they are doing math to get those statistics. In our next post, we will discuss some football statistics that use math.
Do you use your children’s activities to teach math? |
# Restriction of a Linear Transformation on the x-z Plane is a Linear Transformation
## Problem 428
Let $T:\R^3 \to \R^3$ be a linear transformation and suppose that its matrix representation with respect to the standard basis is given by the matrix
$A=\begin{bmatrix} 1 & 0 & 2 \\ 0 &3 &0 \\ 4 & 0 & 5 \end{bmatrix}.$
(a) Prove that the linear transformation $T$ sends points on the $x$-$z$ plane to points on the $x$-$z$ plane.
(b) Prove that the restriction of $T$ on the $x$-$z$ plane is a linear transformation.
(c) Find the matrix representation of the linear transformation obtained in part (b) with respect to the standard basis
$\left\{\, \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} \,\right\}$ of the $x$-$z$ plane.
## Proof.
### (a) Prove that the linear transformation $T$ sends points on the $x$-$z$ plane to points on the $x$-$z$ plane.
Each point on the $x$-$z$ plane is of the form
$\begin{bmatrix} x \\ 0 \\ z \end{bmatrix}$ for some $x, z \in \R$.
We have
\begin{align*}
T\left(\,\begin{bmatrix}
x \\
0 \\
z
\end{bmatrix}\,\right)&=A\begin{bmatrix}
x \\
0 \\
z
\end{bmatrix}\6pt] &=\begin{bmatrix} 1 & 0 & 2 \\ 0 &3 &0 \\ 4 & 0 & 5 \end{bmatrix}\begin{bmatrix} x \\ 0 \\ z \end{bmatrix}\\[6pt] &=\begin{bmatrix} x+2z \\ 0 \\ 4x+5z \end{bmatrix}. \end{align*} Since the y-coordinate of the last vector is 0, and thus the output vector lies in the x-z plane. ### (b) Prove that the restriction of T on the x-z plane is a linear transformation. Let V be the x-z plane in \R^3. Then V is a subspace of the vector space \R^3. In part (a), we showed that the restriction of T on V is given by the formula \begin{align*} T\left(\,\begin{bmatrix} x \\ 0 \\ z \end{bmatrix}\,\right)=\begin{bmatrix} x+2z \\ 0 \\ 4x+5z \end{bmatrix}. \tag{*} \end{align*} We abuse the notation and write this restriction as T: V\to V. (The precise notation is T|_{V}:V\to V.) Let \[\begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix}, \begin{bmatrix} x_2 \\ 0 \\ z_2 \end{bmatrix} be arbitrary vectors in $V$ and let $r\in \R$ be an arbitrary real number.
Then we have
\begin{align*}
T\left(\,\begin{bmatrix}
x_1 \\
0 \\
z_1
\end{bmatrix}+\begin{bmatrix}
x_2 \\
0 \\
z_2
\end{bmatrix}\,\right)&=T\left(\,\begin{bmatrix}
x_1+x_2 \\
0 \\
z_1 +z_2
\end{bmatrix}\,\right)\6pt] &=\begin{bmatrix} (x_1+x_2)+2(z_1+z_2) \\ 0 \\ 4(x_1+x_2)+5(z_1+z_2)\\ \end{bmatrix}\\ &=\begin{bmatrix} x_1+2z_1 \\ 0 \\ 4x_1+5z_1 \end{bmatrix}+\begin{bmatrix} x_2+2z_2 \\ 0 \\ 4x_2+5z_2 \end{bmatrix}\\[6pt] &=T\left(\,\begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix}\,\right)+T\left(\,\begin{bmatrix} x_2 \\ 0 \\ z_2 \end{bmatrix}\,\right) \end{align*} and \begin{align*} T\left(\,r\begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix}\,\right)&=T\left(\,\begin{bmatrix} rx_1 \\ 0 \\ rz_1 \end{bmatrix}\,\right)\\[6pt] &=\begin{bmatrix} (rx_1)+2(rz_1) \\ 0 \\ 4(rx_1)+5(z_1) \end{bmatrix}\\[6pt] &=r\begin{bmatrix} x_1+2z_1 \\ 0 \\ 4x_1+5z_1 \end{bmatrix}\\[6pt] &=r T\left(\, \begin{bmatrix} x_1 \\ 0 \\ z_1 \end{bmatrix} \,\right). \end{align*} It follows that the restriction T:V\to V is a linear transformation. ### (c) Find the matrix representation of the linear transformation obtained in part (b) with respect to the standard basis Let B=\{\mathbf{u}, \mathbf{v}\} be the standard basis of the x-z plane: \[\mathbf{u}=\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, \mathbf{v}=\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}. It follows from formula (*) that we have
\begin{align*}
T(\mathbf{u})=T\left(\, \begin{bmatrix}
1 \\
0 \\
0
\end{bmatrix} \,\right)=\begin{bmatrix}
1 \\
0 \\
4
\end{bmatrix}=\mathbf{u}+4\mathbf{v}\end{align*}
and
\begin{align*}
T(\mathbf{v})=T\left(\, \begin{bmatrix}
0 \\
0 \\
1
\end{bmatrix}\,\right)=\begin{bmatrix}
2 \\
0 \\
5
\end{bmatrix} =2\mathbf{u}+5\mathbf{v}.
\end{align*}
Thus the coordinate vectors with respect to the basis $B$ are
$[T(\mathbf{u})]_B=\begin{bmatrix} 1\\ 4 \end{bmatrix}_{B} , [T(\mathbf{v})]_B=\begin{bmatrix} 2 \\ 5 \end{bmatrix}_{B},$ and the matrix representation of the linear transformation $T:V\to V$ with respect to the standard basis $B=\{\mathbf{u}, \mathbf{v}\}$ is
$\left[\, [T(\mathbf{u})]_B, [T(\mathbf{v})]_B \,\right] =\begin{bmatrix} 1 & 2\\ 4& 5 \end{bmatrix}.$
Let $W_1, W_2$ be subspaces of a vector space $V$. Then prove that $W_1 \cup W_2$ is a subspace of... |
# Focus
As we get further into algebra, we will start to deal with problems involving focus. But what exactly is a focus? How is this concept related to algebra, and what can it teach us about math? Let''s find out:
## What is a focus of a circle?
The concept of a focus is related to conic sections and circles. A circle has one focus, and it exists at its center. Take a look:
As we can see, the focus of this circle exists not only at its center but also at the origin of the plane. We might say that a circle is determined by its focus, and that it is the set of all points in a plane at a given distance from its focus.
Fun fact: The plural of focus is "foci."
## The focus of a parabola
As we might recall, a parabola is a type of conic section. Although we see parabolas in two dimensions on a plane, it is equivalent to the cross-section of a three-dimensional cone. The focus exists at the center of the conic section just as it exists at the center of a circle. We might also say that the focus represents the "point" of a conic section since that also exists at its center.
But what exactly does the focus of a parabola look like? Take a look:
As we can see, there is also a directrix on this graph that takes the form of a straight line. The vertex is the highest or lowest point of the line, depending on the equation. In this case, the vertex is at the lowest point of the line -- also known as the "trough." These two terms are related to the focus because the distance between the vertex and directrix is equal to the distance between the vertex and the focus.
## The foci of an ellipse
Another type of conic section is called an "ellipse." This type of conic section has not one but two foci. Let''s see what this looks like:
As we can see, an ellipse is not a perfect circle, but rather a circle that has been stretched or compressed into more of an oval shape. If we look at any point on the ellipse, we see that the sum of the distances to each focus remains constant.
## The foci of a parabola
A hyperbola also has two foci. This conic section represents the cross-section of two inverted cones, and it has two characteristic bows or arms. Let''s see what the foci of a parabola look like:
If we look at any point on either of the two arms, we can see that the difference between the distances to each focus is constant.
## Topics related to the Focus
Focus of a Parabola
Ellipse
Finding the Equation of a Parabola given Focus and Directrix
## Flashcards covering the Focus
Algebra II Flashcards
College Algebra Flashcards
## Practice tests covering the Focus
Algebra II Diagnostic Tests
College Algebra Diagnostic Tests
## Pair your student with a tutor who can clearly explain the focus of conic sections
Sometimes, students need to have difficult concepts like foci explained in new ways. A 1-on-1 tutoring session outside of class is an excellent option, as tutors can use examples geared towards your student''s hobbies. For example, your student might benefit from having a parabola explained in the context of a soccer ball flying through the air. Tutors can also cater to your student''s learning style, which is very useful if your student prefers verbal teaching methods to visual representations. Tutors can also personalize lessons based on your student''s ability level. Reach out to our Educational Directors to learn more about the benefits of tutoring. Varsity Tutors will pair your student with an outstanding tutor. |
# The diameter of a roller is 72 cm and its length is 120 cm. It takes 200 complete revolutions to move over to level playground. Find the area of the playground.
Given:
Diameter of the roller$=72\ cm$
Length of the roller$=120\ cm$.
Number of revolutions taken to level the playground$=200$.
To do:
We have to find the area of the playground.
Solution:
Radius of the roller$=\frac{72}{2}\ cm=36\ cm$.
We know that,
Curved surface area of a cylinder of radius r and height h is $2 \pi rh$.
Therefore,
Area covered in 1 revolution$=$Curved surface area of the roller
Area of the playground$=$Number of revolutions$\times$Curved surface area of the roller
$=200\times2\times3.14\times36\times120\ cm^2$
$=5425920\ cm^2$
$=\frac{5425920}{10000}\ m^2$
$=542.592\ m^2$
The area of the playground is $542.592\ m^2$.
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### Course: 5th grade (Eureka Math/EngageNY)>Unit 3
Lesson 4: Topic D: Further applications
# Solving for the missing fraction
Learn how to add and subtract fractions with unlike denominators. Watch examples, practice finding common denominators, and apply these skills to solve problems.
## Want to join the conversation?
• what if the missing fraction is on the right side?
• Well you just add the fractions on the left side
• What is the difference between BEDMAS, BODMAS and PEMDAS?
• B - brackets
E - exponents
D - division
M - multiplication
S - subtraction
• where did he get the 8 at?
• He got the 8 from the 16. Because 16 divided by 2 is 8.
• What is 2/5 + blank equals 7/10.
• If we had a simpler problem such as 3 + blank = 8, we would do 8-3 = 5.
Similarly, for 2/5 + blank = 7/10, we would do 7/10 - 2/5.
We can use 10 as a common denominator because 5 divides evenly into 10.
Multiplying top and bottom of 2/5 each by 2 gives 4/10, so 2/5 = 4/10.
So the blank is 7/10 - 2/5 = 7/10 - 4/10 = 3/10. The answer is 3/10.
May Jesus richly bless you today!
• what if there are 2 fractions missing?
• If there are two missing fractions, then generally you will need to be given a second equation in order to have enough information to find the two missing fractions.
• you have to get common denominator
• i don't really get this i need some help why do we have to change the common denominator
• We need to change it so it's easier to subtract or add.
1/2 + 3/12 is harder because the denominator is not the same. If the denominator was the same, it would be easier. In that case, 1/2 + 3/12 would be 6/12 + 3/12
which is 9/12.
Since making the dinomenator the same, it makes answering the question easier.
Sorry if my grammar is bad, I just started English. (used to go to French immersion)
Hope this helps!
• I honestly still don't get it but what if there is not a question that has two numbers like what I mean is like what if there is a question like -- + 3\4 = 1\2?? |
# How do you evaluate \frac { ( 3.00\times 10^ { 6} ) ( 2.0\times 10^ { - 3} ) } { 5.0\times 10^ { - 2} }?
Dec 10, 2016
$120 , 000$ or $1.2 \times {10}^{5}$
#### Explanation:
First, rearrange the terms in the numerator:
$\frac{3.00 \times 2.0 \times {10}^{6} \times {10}^{-} 3}{5.00 \times {10}^{-} 2}$
Using the rule for exponents where $\textcolor{red}{{x}^{a} \times {x}^{b} = {x}^{a + b}}$ we can simplify the numerator to:
$\frac{6.00 \times {10}^{6 - 3}}{5.0 \times {10}^{-} 2}$
$\frac{6.00 \times {10}^{3}}{5.0 \times {10}^{-} 2}$
Now using the rule for exponents where $\textcolor{red}{{x}^{a} / {x}^{b} = {x}^{a - b}}$ we can simplify the fraction to:
$\frac{6.00 \times {10}^{\left(3 - - 2\right)}}{5.0}$
$\frac{6.00 \times {10}^{5}}{5.00}$
Expanding the numerator gives us:
$\frac{600000}{5}$
$120000$
Or in scientific notation form:
$1.2 \times {10}^{5}$
Dec 10, 2016
$1.2 \cdot {10}^{5}$
#### Explanation:
((3.00 * 10^6)(2.0 * 10^-3))/(5.0 * 10^-2
multiply out brackets:
$\frac{6.0 \cdot {10}^{3}}{5.0 \cdot {10}^{-} 2}$
divide 6 by 5:
$\frac{1.2 \cdot {10}^{3}}{10} ^ - 2$
and then the powers of 10:
$1.2 \cdot {10}^{5}$
(reason:) law of indices:
$\left({a}^{m} / {a}^{n} = {a}^{m - n}\right)$
$1.2$ is between 1 and 10, so we do not need to change this or the power of 10.
final answer: $1.2 \cdot {10}^{5}$ |
# Differentiate the function f(x)=(3x+14)/(3x^2+7x)
neela | Student
f(x)=(3x+14)/(3x^2+7x).
We use u(x)/v(x) = {(u'(x)v(x)-u(x)v'(x)}/(v(x))^2 to differentiate the given expression.
f'(x) = {(3x+4)'(3x^2+7x) -(3x+14)(3x^2+7x)'}/(3x^2+7x)62.
f(x) = {3(3x^2+7x)-(3x+14)(6x+7)}/(3x^2+7x)^2.
f'(x) = {9x^2+21x-18x^2-21x-84x-98}/(3x^2+7x)^2.
f'(x) = {-9x^2-84x-98}/(3x^2+7x)^2.
f'(x) = -(9x^2+84x+98)/(3x^2+7x)^2.
giorgiana1976 | Student
Since the given function is a quotient, we'll apply the quotient rule to find it's first derivative:
(u/v)' = (u'*v - u*v')/v^2 (*)
We'll put u = 3x+14 => u' = 3
We'll put v = 3x^2 + 7x => v' = 6x + 7
We'll substitute u,v,u',v' in the formula (*):
f'(x) = [3(3x^2 + 7x) - (3x+14)(6x + 7)]/(3x^2+7x)^2
We'll remove the brackets:
f'(x) = (9x^2 + 21x - 18x^2 - 21x - 84x - 98)/(3x^2+7x)^2
We'll combine and eliminate like terms:
f'(x) = -(9x^2 + 84x + 98)/(3x^2+7x)^2 |
# Algebra II : Mean, Standard Deviation, and Normal Distribution
## Example Questions
### Example Question #1 : Mean, Standard Deviation, And Normal Distribution
The mean of the following numbers is 13. Solve for x.
18, 15, 10, 8, x
15
13
14
12
14
Explanation:
where n equals the number of events.
### Example Question #2 : Mean, Standard Deviation, And Normal Distribution
On 3 exams, Willow has an average score of 95 points. Her lowest score was an 90. If this score is dropped, what is her new average?
98
100
97.5
96
97.5
Explanation:
To answer this question, you must first calculate Willow's total points from the 3 exams. The total points will be equal to 3 times the original average:
95*3=285
Now, if the 90 is removed, his new total will be:
285-90=195
With the score dropped, her new total number of exams is 2. Therefore, the new average is:
### Example Question #3 : Mean, Standard Deviation, And Normal Distribution
The ages of the students at Rosa Parks High School are normally distributed with a mean of 15.5 and a standard deviation of 0.6 years.
What is the proportion of students that are older than 16.2 years old?
5%
0%
1%
2.5%
2.5%
Explanation:
This question relates to the 689599.7 rule of normal distribution. We know that 95% of the data are within 2 standard deviations from the mean.
In this case this means that 95% of the students are between
15.520.6 and 15.5+20.5
151.2 and 15+1.2
13.8 and 16.2
Therefore we have 5% of the students that are outside of this range. Since the normal distribution is symmetric, the proportion of students below 13.8 is the same as the proportion of students above 16.2.
Thus the right answer is one half of 5%, or 2.5%.
2.5% of the students are older than 16.2 years old.
### Example Question #4 : Mean, Standard Deviation, And Normal Distribution
The ages of the students at Central Universityl are normally distributed with a mean of 23.5 and a standard deviation of 2.5 years.
What is the proportion of students that are younger than 18.5 years old?
1%
0%
5%
2.5%
2.5%
Explanation:
This question relates to the 689599.7 rule of normal distribution. We know that 95% of the data are within 2 standard deviations from the mean.
In this case this means that 95% of the students are between
23.52⋅2.5 and 23.5+2⋅2.5
23.5-5 and 23.5+5
18.5 and 28.5
Therefore we have 5% of the students that are outside of this range. Since the normal distribution is symmetric, the proportion of students below 18.5 is the same as the proportion of students above 28.5.
Thus the right answer is half of 5% or 2.5%.
### Example Question #5 : Mean, Standard Deviation, And Normal Distribution
Find the standard deviation of the following set of numbers:
Explanation:
To begin, we must remember the formula for standard deviation:
where is the standard deviation, N is the number of values in our set, is the value we're currently evaluating in the summation, and is the mean of our set of numbers. All the summation part of the equation means is that we subtract our mean from each number in the set, square that value, and then add all of those values for each number together. So before we find the values that will be added together, we must first find our mean for the set of numbers:
Now we can determine the value for our summation for each number in the set:
Looking at our equation for standard variation, now all we must do is sum all of the values above, divide by N, and take the square root:
### Example Question #5 : Mean, Standard Deviation, And Normal Distribution
The formula for standard deviation is the following:
.
Where,
,
Five students took a test and recieved the following grades: , , , , . What is the standard deviation of the test grades to the nearest decimal place?
Explanation:
The first step in solving for standrd deviation is to find the mean of the data set.
For this problem:
.
Now we can evaluate the summation:
Now we can rewrite the standard deviation expression:
There are 5 data points, so n = 5
### Example Question #6 : Mean, Standard Deviation, And Normal Distribution
In a normal distribution, what percentage is covered within one standard deviation?
50%
68%
95%
34%
99.7%
68%
Explanation:
By drawing a bell curve, the middle line is 50%. One standard deviation left and right of the middle line is 34% each. That means one standard deviation within is 68%
### Example Question #7 : Mean, Standard Deviation, And Normal Distribution
If the mean is and the standard deviation is , which of the following is NOT within two standard deviations?
Explanation:
Two standard deviations means to double the standard deviation value.
This means the range is to
Only is not in the range.
### Example Question #8 : Mean, Standard Deviation, And Normal Distribution
Ibram was paid $5000 for editing a 2200-page encyclopedia. What was his rate earned per page? Possible Answers:$0.44
$1.27$2.27
$0.56 Correct answer:$2.27
Explanation:
This question is really just asking for the average dollar per page. This could be easily calculated:
However, you might also think of this as a rate problem:
D=RT
Using the information from the question, we can see that D=5000 and T=2200.
This comes out to what we solved above:
### Example Question #3 : Mean, Standard Deviation, And Normal Distribution
Standard Deviation can be calculated from what statistical term?
Range
Median
Variance
Mode
Quartile |
# 基本的な1次変換の問題
## ■問題
(1)点$\text{P}\left(x,y\right)$ を原点を中心に$60°$ 回転させ点$\text{Q}\left(u,v\right)$ に移す1次変換を表す行列 $A$ を求めよ． 三角関数は計算すること．
(2) $\text{Q}\left(u,v\right)$$y$ 軸に関して対称な点 $\text{R}\left(r,s\right)に$ 移す.1次変換を表す行列 $B$ を求めよ．
(3)点$\text{P}\left(x,y\right)$ を点$\text{R}\left(r,s\right)$ に移す1次変換を表す行列$C$ を求めよ．
## ■答
• (1)
$\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$
• (2)
$\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$
• (3)
$\left(\begin{array}{cc}-\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$
## ■解き方
(1)
$A=\left(\begin{array}{cc}\mathrm{cos}60°& -\mathrm{sin}60°\\ \mathrm{sin}60°& \mathrm{cos}60°\end{array}\right)$$=\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$
$\left(\begin{array}{c}u\\ v\end{array}\right)=\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$ ･･････(i)
(2)
$r=-u=-1·u+0·v$
$s=v=0·u+1·v$
$\left(\begin{array}{c}r\\ s\end{array}\right)=\left(\begin{array}{c}-u\\ v\end{array}\right)=\left(\begin{array}{c}-1\cdot u+0\cdot v\\ 0\cdot u+1\cdot v\end{array}\right)$$=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{c}u\\ v\end{array}\right)$ ･･････(ii)
よって
$B=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)$
(3)
(i)，(ii)より
$\left(\begin{array}{c}r\\ s\end{array}\right)=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)$
よって
$C=BA$$=\left(\begin{array}{cc}-1& 0\\ 0& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{2}& -\frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$$=\left(\begin{array}{cc}-\frac{1}{2}& \frac{\sqrt{3}}{2}\\ \frac{\sqrt{3}}{2}& \frac{1}{2}\end{array}\right)$
ホーム>>カテゴリー分類>>行列>>線形代数>>線形代数に関する問題>>線形写像・1次変換>>基本的な1次変換の問題 |
# Angles in a circle
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0/2
##### Intros
###### Lessons
1. Terms related to circles
• Diameter
• Circumference
• Central angle
• minor arc
• major arc
• Inscribed angle
• Chord
• perpendicular bisector
• Tangent
• point of tangency
2. What are Inscribed angles and Central angles?
0/6
##### Examples
###### Lessons
1. In the following diagram, the radius is 24 cm and $\angle$BDC is 75°.
1. Find $\angle$BAC.
2. Find the chord BC.
2. You are setting up a display board for the science fair. You have two spotlights: one projects through an angle of 20°; and one projects through an angle of 40°. Where is the best place to put the display board? Show your answer in a diagram.
1. Given $\angle$BAE = 44.5° and $\angle$ADC = 64.27°.
Find.
1. $\angle$BCD
2. $\angle$AEB
3. $\angle$CED
###### Topic Notes
In a circle, chords, angles, inscribed angles and arc length all have special relationships with each other. This lesson focuses on exploring the relationships among inscribed angles in a circle as well as those of inscribed angle and central angle with the same arc. We will make use of the relationships to solve related questions in this lesson.
## What is a central angle
Previously, when you worked with angles in a triangle, you were mainly working with central angles. These are angles that had their vertices at the center of the circle. However, in this lesson, we won't be working with central angles.
## What is an inscribed angle
If we're not working with central angles, then what angles in a circle will we be focusing on? We'll be tackling inscribed angles.
Inscribed angles are angles that can be anywhere on the circle's circumference. While we're not solely working with central angles anymore, you'll learn that the central angle of a circle will still come into play when we're going through the example problems in this lesson.
## Inscribed angle theorem
There are several inscribed angle theorems that we'll be learning today. One of them is the angle in the center theorem. This tells us that an inscribed angle $\theta$ is half of the central angle 2?. This is seen illustrated below.
## Angles subtended on the same arc theorem
Another theorem that we'll learn is the angles subtended by the same arc theorem. Given that the endpoints are the same, you'll realize that from the below illustration, it doesn't matter where the inscribed angle $\theta$ is. It will remain the same. As you can also see, this is because they share the same arc (the orange line), hence the name of this theorem.
Question 1:
In the following diagram, the radius is 24 cm and angle BDC is 75°.
a) Find angle BAC.
Solution:
Angle BDC is the central angle, which equals 75 degrees. Angle BAC is the inscribed angle. According to the inscribed angle theorem, angle BAC equals half of angle BDC. Conversely, you can think about the relationship between angle BDC and angle BAC as the central angle always being doubled that of the inscribed angle.
Central angle = inscribed angle x 2
$75$°$=2\theta$
$\theta=37.5$°
b) Find the chord BC
Solution:
Let us draw a bisector in triangle DBC to cut it in half
Now, look at one of the halves cut from triangle DBC
X is half of the chord BC that we are looking for. So solve for x using sin (remember the principles of SohCahToa?)
$BC=2x$
$\sin 37.5$°$=\frac{x}{24}$
$x \cong 14.6$
$BC=2x=29.2cm$
Question 2:
Given angle BAE = 44.5° and angle ADC = 64.27°.
Find angle BCD.
Solution:
Angle BCD and angle BAE are inscribed angles on the same arc. So, angle BCD = BAE = 44.5°. This is based off the angles subtended by the same arc theorem.
Still not sure about the theorems? This online demonstration can show you the proof through you dragging the lines in the circle that form central and inscribed angles.
Next up, you'll be dealing with problems that'll require you to find the arcs of a circle and the area of a sector in circles. You'll also expand on the central and inscribed angles that you learned here, as well as move on to doing proofs on the inscribed angles. |
# Evaluate the following integrals:
Question:
Evaluate the following integrals:
$\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x$
Solution:
Given I $=\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x$
Integral is of form $\int \frac{p x+q}{\sqrt{a x^{2}+b x+c}} d x$
Writing numerator as $\mathrm{px}+\mathrm{q}=\lambda\left\{\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}\right)\right\}+\mu$
$\Rightarrow p x+q=\lambda(2 a x+b)+\mu$
$\Rightarrow 2 x+3=\lambda(2 x+4)+\mu$
$\therefore \lambda=1 / 2$ and $\mu=-1$
Let $2 x+3=2 x+4-1$ and split,
$\Rightarrow \int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=\int\left(\frac{2 x+4}{\sqrt{x^{2}+4 x+5}}-\frac{1}{\sqrt{x^{2}+4 x+5}}\right) d x$
$=2 \int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{1}{\sqrt{x^{2}+4 x+5}} d x$
Consider $\int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x$
Let $\mathrm{u}=\mathrm{x}^{2}+4 \mathrm{x}+5 \rightarrow \mathrm{dx}=\frac{1}{2 \mathrm{x}+4} \mathrm{du}$
$\Rightarrow \int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{1}{2 \sqrt{u}} d u$
$=\frac{1}{2} \int \frac{1}{\sqrt{u}} d u$
We know that $\int x^{n} d x=\frac{x^{n+1}}{n+1}+c$
$\Rightarrow \frac{1}{2} \int \frac{1}{\sqrt{u}} d u=\frac{1}{2}(2 \sqrt{u})$
$=\sqrt{u}=\sqrt{x^{2}+4 x+5}$
Consider $\int \frac{1}{\sqrt{x^{2}+4 x+5}} d x$
$\Rightarrow \int \frac{1}{\sqrt{x^{2}+4 x+5}} d x=\int \frac{1}{\sqrt{(x+2)^{2}+1}} d x$
Let $u=x+2 \rightarrow d x=d u$
$\Rightarrow \int \frac{1}{\sqrt{(x+2)^{2}+1}} d x=\int \frac{1}{\sqrt{u^{2}+1}} d u$
We know that $\int \frac{1}{\sqrt{x^{2}+1}} d x=\sinh ^{-1} x+c$
$\Rightarrow \int \frac{1}{\sqrt{u^{2}+1}} d u=\sinh ^{-1}(u)$
$=\sinh ^{-1}(x+2)$
Then,
$\Rightarrow \int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=2 \int \frac{x+2}{\sqrt{x^{2}+4 x+5}} d x-\int \frac{1}{\sqrt{x^{2}+4 x+5}} d x$
$=2 \sqrt{x^{2}+4 x+5}-\sinh ^{-1}(x+2)+c$
$\therefore I=\int \frac{2 x+3}{\sqrt{x^{2}+4 x+5}} d x=2 \sqrt{x^{2}+4 x+5}-\sinh ^{-1}(x+2)+c$ |
# How do you integrate int (6(5 - x))/( (x - 7)(4 - x)) using partial fractions?
Apr 13, 2017
$\int \frac{6 \left(5 - x\right)}{\left(x - 7\right) \left(4 - x\right)} \mathrm{dx} = 4 \ln | x - 7 | + 2 \ln | x - 4 | + C$
#### Explanation:
By multiplying the numerator and the denominator by $- 1$,
$\frac{6 \left(x - 5\right)}{\left(x - 7\right) \left(x - 4\right)} = \frac{A}{x - 7} + \frac{B}{x - 4} = \frac{A \left(x - 4\right) + B \left(x - 7\right)}{\left(x - 7\right) \left(x - 4\right)}$
By matching the numerators,
$A \left(x - 4\right) + B \left(x - 7\right) = 6 \left(x - 5\right)$
To find $A$, set $x = 7 R i g h t a r r o w 3 A = 12 R i g h t a r r o w A = 4$
To find $B$, set $x = 4 R i g h t a r r o w - 3 B = - 6 R i g h t a r r o w B = 2$
So, we have
int(6(5-x))/((x-7)(4-x))dx =int(4/(x-7)+2/(x-4) )dx
By Log Rule,
$= 4 \ln | x - 7 | + 2 \ln | x - 4 | + C$
I hope that this was clear. |
Write-up #10
Question #3
## Parametric Curves
If x and y are given funtions
### x=f(t) y=g(t)
over an interval of t values, the the set of points (f(t), g(t)) defined by these equations is a parametric curve in the coordinate plane. The equations are parametric equations for the curve, and we say the curve has been parametrized. The variable t is the parameter of the curve, and its domain I = [a,b] = [tMin, tMax] is the parameter interval. The point (f(a), g(a)) is the initial point of the curve, and (f(b), g(b)) is the terminal point of the curve.
Consider
### x = a cos(t) y = b sin(t)
for for various values of a and b.
## We want to investigate the curve when a < b, a= b, and a> b.Our exploration begins when a < b.
Consider the graph when a=1 and b = 3. The associated equations are
x = 1 cos(t)
y = 3 sin(t)
The graph is an ELLIPSE crossing the x axis at -1 and 1 and the y axis at -3 and 3.
What might you expect when a = -4 and b = -2?
As expected, the ellipse intercepts the x axis at -4 and 4 and the yaxis at -2 and 2. Notice, however, the ellipse has the x axis as it's major axis whereas in the previous example the major axis was the y axis. Recall from your study of the conic sections , particularly the ellipse,
that you can associate with the major axis. This translates into our knowledge of parametric equations.
For example, we have our beginning equations
If we divide each side of the equations by a and b respectively, we obtain the following:
.
Now if we square both sides of each equation and add each side of the equations,
we obtain the following:
.
This equation can be simplified further using the trigonometric identity, .
Thus the equation becomes
which is the equation for an ellipse.
Recall from above that when a<b, we get two different cases where the major axis can be the x-axis or y-axis. What case produces the x-axis as its major axis? and y-axis as its major axis?
Let's try several more examples to see if there is a pattern. Remember a<b.
Consider the two cases:
a=-4 and b=1 (pink)
a=-4 and b=9 (torquoise). The graph suggest the following ellipses:
In both cases, a<b. The key to identifying the major axis of the ellipse is looking at the absolute value of a and b. Notiçe when |a|>|b|, the x-axis is the major axis. When |a|<|b| the y-axis is the major axis.
Now, consider the graph when a=-3 and b=3.
As expected, the ellipse crosses the x-axis at -3 and 3 and the y-axis at - 3 and 3. Yet, in this case we get a circle which can also be considered a "special case" of an ellipse.
Notice .
If we consider our equations for an ellipse, since , our ellipse will be a circle.
Furthermore, when |a|=|b|, the equation will be an ellipse.
What if we consider the case where either a or b is equal to zero?
Let a=p;2 and b=0 , equations are
x = -2 cos (t)
y = 0 sin (t)
where .
The graph is a straight line whose endpoints are -2 and 2. For every t, the output is an ordered pair (x,y) which in this case is (-2 cos(t),0). Thus, a line segment is expected.
Another perspective to answer the question, "What happens if either a or b is equal to zero?" is to investigate the parametric equations
x = a cos (t)
y = b sin (t)
and their respective graphs as b tends to zero and a is any number.
Letting a=-2 and b tend to zero, consider the following graph where
b=1/2 (pink)
b=1/4 (turquoise)
b=1/8 (green)
b=1/16 (red)
b=1/32 (royal blue)
Notice, as b tends to zero, the ellipse becomes thinner and thinner. If you let b=0, the ellipse will become a segment.
## Our next journey investigates when a = b.
Consider when a=b=3, we have the equations
x=3 cos(t)
y=3 sin(t)
and the graph
is a CIRCLE whose radius is a=b, in this case a=b=3. You can also consider this circle as a "special case" of an ellipse where the x- and y-intercepts are the same. Hence, the distance from the origin to each vertice is equal and a circle is formed.
Consider when a=b=-6.
As predicted, the graph is a circle whose radius is |-6 | = 6. Remember, the length of a radius is always positive. Thus, in the parametriç equations
when a=b, the graph will always be a circle with radius=|a|=|b|.
## Our last investigation centers around when a>b?
Consider when a=5 and b=3. The associated equations are
x = 5 cos(t)
y = 3 sin(t)
and the graph is as follows:
The graph is an ellipse with x-intercepts at and y-intercepts at with |a|>|b|. Using our previous conclusion the graph should have the x-axis as its major axis. This is a true statement.
Consider the case when a=4 and b=-2. The associated equations are
x = 4 cos (t)
y = -2 sin (t)
and the graph is as follows:
Consider the case when both a and b are both negative and a > b.
Let a=-4 and b=-8. So,
x=-4 cos (t)
y=-8 sin (t)
and the graph (pink) is
Also, consider several other cases where a=-5 and b=-9 (turquoise) and a=-3 and b=-6 (green).
Notice the y-axis is the major axis. Recall from our previous discussion, when |a| < |b|, the major axis is the y-axis.
## Summary
Overall, this investigation explores the connection between parametric equations as it relates to the equation of an ellipse and its associated graph. The assertion that the parametric equations
x = a cos (t)
y = b sin (t)
will always produce an ellipse was shown algebraically. The process involved squaring both sides of each equation, adding the equations together, and using a common trigonometric identity. This proved essential in making the connection. We also explored different cases involving a<b, a=b, and a>b. Several conclusions drawn were when
a) |a| > |b|, the major axis lies on the x-axis
b) |a| < |b|, the major axis lies on the y-axis
c) a=b, the ellipse will be a circle.
This exploration would prove most beneficial to a secondary mathematics course when studying ellipses. Why? Parametric equations are not dealt with heavily in the traditional curriculum. This investigation could be used after ellipses have been discussed and in conjunction with an introduction into the concept of parametric equations. Hence, students can sample a taste of an entirely new concept, parametric equations, along with a connection to an old concept, ellipses. This mathematical activity allows students a new perspective on ellipses providing them with a different way to see ellipses. As a mathematics educator, one of my goals is to provide students with alternative ways of viewing a mathematical concept. Furthermore, another goal is to provide opportunities for students to make connections between mathematics concepts themselves. This exploration fulfills both goals. In addition, the use of technology is utilized. The technology used provides students with new mathematical learning experiences while at the same time the learning is enhanced using technology.
Return |
# Area of a Room: Basic Practice
5 teachers like this lesson
Print Lesson
## Objective
SWBAT express multiple ways to create rooms of a given area.
#### Big Idea
Creating rooms with different dimensions but the same area improves understanding of multiplicative principles.
## Guided Review/ Practice Finding Area and Perimeter of Rectangles
25 minutes
Prior to taking students outside for the hands-on part of this lesson, either teach students about or remind them of the very direct relationship between finding the area of a rectangle and using arrays in multiplication. If the idea of perimeter has not yet been introduced, explain the difference between area and perimeter. One tried and true classic is to compare the area of the classroom to the floor space covered by tile or carpet, and the perimeter to the distance a number line or border would cover if it went all around the tops of the classroom's walls.
Here is an example that differentiates between area and perimeter and shows how to find the area and perimeter of a rectangle: Examples of Area and Perimeter On Level and Examples of Area and Perimeter Enrich.
These student practice pages (Student Practice Find Area and Perimeter Extra SupportStudent Practice Find Area and Perimeter On Level, and Student Practice Find Area and Perimeter Enrich) can either be printed, projected, or accessed by students on individual computers.
## Measuring (Imaginary) Rooms and Reasoning About Size
43 minutes
If you have leveled your students, you may want to break them into groups based on the type of practice they need. Ideally, each child should have a meter or yard stick for this activity. Whatever they have, make sure they are using consistent units, either yards or meters.
Give the students some of the following questions/prompts and circulate around the group(s) to ask them to explain their thinking and to assist them in kindly critiquing the reasoning of others. Also prompt them to recall the basic facts, addition and subtraction, involved in answering these problems.
What is one way to make a room with a perimeter of 12? What is the area? What is another way to make a room with a perimeter of 12? Is the area the same? Why or why not?
* Would this room be large enough to be used as a... (classroom, bedroom, closet, and so on).
Make a rectalinear room with a perimeter of 24. What is the area? What would be another way to make a room with a perimeter of 24? What is the area of this second room? If one of these rooms needed to be used as a classroom for a small group tutoring room, which room would be better suited to this purpose? Why?
Here is an example of students creating a room: Area Outside
When asked how many kids will fit in classroom, the students choose an interesting way to determine how much space is needed for each child.
This video clip shows a student critiquing the reasoning of another student. The students are building a room for an (imaginary) orphanage and have the idea that if they can sit in the room, then it is a big enough space. This child draws the line at a nursery that has extremely small dimensions.
Here is one more example of a child explaining his reasoning.
## Wrap-up
4 minutes
Have the students think (silently) of something new they discovered or were able to explain today. Ask them to either write this concept down to share with an adult at home or have them share with one or more other students using an inside-outside line.
An inside-outside line is my tiny classroom version of the Inside Outside Circle. |
# ACT Math : How to find the area of a trapezoid
## Example Questions
### Example Question #1 : How To Find The Area Of A Trapezoid
What is the area of this regular trapezoid?
32
45
20
26
32
Explanation:
To solve this question, you must divide the trapezoid into a rectangle and two right triangles. Using the Pythagorean Theorem, you would calculate the height of the triangle which is 4. The dimensions of the rectangle are 5 and 4, hence the area will be 20. The base of the triangle is 3 and the height of the triangle is 4. The area of one triangle is 6. Hence the total area will be 20+6+6=32. If you forget to split the shape into a rectangle and TWO triangles, or if you add the dimensions of the trapezoid, you could arrive at 26 as your answer.
### Example Question #1 : How To Find The Area Of A Trapezoid
What is the area of the trapezoid above if a = 2, b = 6, and h = 4?
16
32
64
8
24
16
Explanation:
Area of a Trapezoid = ½(a+b)*h
= ½ (2+6) * 4
= ½ (8) * 4
= 4 * 4 = 16
### Example Question #1 : How To Find The Area Of A Trapezoid
Find the area of a trapezoid if the height is , and the small and large bases are and , respectively.
Explanation:
Write the formula to find the area of a trapezoid.
Substitute the givens and evaluate the area.
### Example Question #1 : How To Find The Area Of A Trapezoid
Trapezoid has an area of . If height and , what is the measure of ?
Explanation:
The formula for the area of a trapezoid is:
We have here the height and one of the bases, plus the area, and we are being asked to find the length of base . Plug in known values and solve.
Thus,
### Example Question #1 : How To Find The Area Of A Trapezoid
Find the area of a trapezoid given bases of length 6 and 7 and height of 2. |
# Maharashtra Board Class 6 Maths Solutions Chapter 8 Divisibility Practice Set 22
## Maharashtra State Board Class 6 Maths Solutions 8 Divisibility Practice Set 22
Question 1.
There are some flowering trees in a garden. Each tree bears many flowers with the same number printed on it. Three children took a basket each to pick flowers. Each basket has one of the numbers, 3, 4 or 9 on it. Each child picks those flowers which have numbers divisible by the number on his or her basket. If He / She takes only 1 flower from each tree. Can you tell which numbers the flowers in each basket will have?
Solution:
Each child will have flowers bearing the following numbers:
Girl with basket number 3: 111, 369, 435, 249, 666, 450, 960, 432, 999, 72, 336, 90, 123, 108
Boy with basket number 4: 356, 220, 432, 960, 72, 336, 108
Girl with basket number 9: 369, 666, 450, 432, 999, 72, 90, 108
#### Maharashtra Board Class 6 Maths Chapter 8 Divisibility Practice Set 22 Intext Questions and Activities
Question 1.
Read the numbers given below. Which of these numbers are divisible by 2, by 5, or by 10? Write them in the empty boxes. 125,364,475,750,800,628,206,508,7009,5345,8710. (Textbook pg. no. 43)
Divisible by 2 Divisible by 5 Divisible by 10
Solution:
Divisible by 2 Divisible by 5 Divisible by 10 364,750, 800, 628, 206, 508, 8710 125,475, 750, 800, 5345, 8710 750, 800, 8710
Question 2.
Complete the following table: (Textbook pg. no. 43)
Number Sum of digits in the number Is the sum divisible by 3? Is the given number divisible by 3? 63 6 + 3 = 9 ✓ ✓ 872 17 X X 91 552 9336 4527
Solution:
Number Sum of digits in the number Is the sum divisible by 3? Is the given number divisible by 3? 63 6 + 3 = 9 ✓ ✓ 872 8 + 7 + 2 = 17 X X 91 9 + 1 = 10 X X 552 5 + 5 + 2 = 12 ✓ ✓ 9336 9 + 3 + 3 + 6 = 21 ✓ ✓ 4527 4 + 5 + 2 + 7 = 18 ✓ ✓
Question 3.
Complete the following table: (Textbook pg. no. 44)
Number Divide the number by 4. Is it completely divisible? The number formed by the digits in the tens and units places. Is this number divisible by 4? 992 ✓ 92 ✓ 7314 6448 8116 7773 3024
Solution:
Number Divide the number by 4. Is it completely divisible? The number formed by the digits in the tens and units places. Is this number divisible by 4? 992 ✓ 92 ✓ 7314 X 14 X 6448 ✓ 48 ✓ 8116 ✓ 16 ✓ 7773 X 73 X 3024 ✓ 24 ✓
Question 4.
Complete the following table: (Textbook pg. no. 44)
Number Divide the number by 9. Is it completely divisible? Sum of the digits in the number. Is the sum divisible by 9? 1980 ✓ 1 + 9 + 8 + 0 = 18 ✓ 2999 X 2 + 9 + 9 + 9 = 29 X 5004 13389 7578 69993
Solution:
Number Divide the number by 9. Is it completely divisible? Sum of the digits in the number. Is the sum divisible by 9? 1980 ✓ 1 + 9 + 8 + 0 = 18 ✓ 2999 X 2 + 9 + 9 + 9 = 29 X 5004 ✓ 5 + 0 + 0 + 4 = 9 ✓ 13389 X 1 + 3 + 3 + 8 + 9 = 24 X 7578 ✓ 7 + 5 + 7 + 8 = 27 ✓ 69993 ✓ 6 + 9 + 9 + 9 + 3 = 36 ✓ |
800score.comhttp://www.800score.com/forum/ GMAT Fractionshttp://www.800score.com/forum/viewtopic.php?f=3&t=74 Page 1 of 1
Author: questioner [ Tue Nov 02, 2010 3:22 am ] Post subject: GMAT Fractions A very similar question is http://800score.com/forum/viewtopic.php?t=128.Which of the following fractions is the largest?A. 11/14B. 4/5C. 9/14D. 7/10E. 5/6(E) One way to solve this type of problem is to find the lowest common denominator for all of the fractions.To do this, first factor all of the denominators:11/14: 14 factors into 2 and 74/5: 5 is prime17/21: 21 factors into 3 and 729/35: 35 factors into 5 and 75/6: 6 factors into 2 and 3.Then, to get a common denominator, we make sure that each of the factors is represented exactly once (since no factor is raised to a power in any denominator):2 × 3 × 5 × 7 = 210.Therefore, 210 is the lowest common denominator. Now we can express all the fractions in terms of the common denominator of 210 to see which is the largest:11/14 × 15/15 = 165/2104/5 × 42/42 = 168/2109/14 × 15/15 = 135/2107/10 × 21/21 = 147/2105/6 × 35/35 = 175/210.Therefore, 5/6 is the largest (175/210). The correct answer is choice (E).-------------I don't understand how you arrived at 15/15 , 42/42, 15/15 , 21/21, 35/35.Can you please explain further.
Author: Gennadiy [ Tue Nov 02, 2010 3:29 am ] Post subject: Re: math (test 1, question 13): fractions Once we find that the common denominator is 210, we need to convert all the fractions. Let us convert the fraction 11/14 first.210 / 14 = 15. Therefore wee need to multiple the denominator by 15 to get 210. In order to keep the fraction the same we need to multiply the numerator by 15 as well.11/14 = (11 × 15)/(14 × 15) = 165/210.The rest of the fractions are converted the same way.
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# Improper Integral Calculator
Provide the function with the required limits and the calculator will let you know whether it is convergent or divergent over the range, with the steps shown.
An online improper integral calculator is specially designed to measure the integral with defined limits. You can also determine whether the given function is convergent or divergent by using a convergent or divergent integral calculator. Before we start using this free calculator, let us discuss the basic concept of improper integral.
## What is Improper Integral?
In the context of calculus, an improper integral is a type of integration that determines the area between a curve. This kind of integral has an upper limit and a lower limit. An improper integral can be considered as a type of definite integral. An improper integral is said to be a reversal process of differentiation. Subjecting to an online improper integral calculator is one of the key methods that are best described to solve an improper integral.
### Types of Improper Integrals:
Depending upon the limits we use, there exist two types of improper integral. Type 1 (Integration Over an Infinite Domain): In type one, we classify those improper integral that contain upper and lower limits as infinity. We must keep in mind that infinity is a never-ending process and can’t be considered as a number. Let us suppose that we have a function f(x) which is defined for the interval [a, ∞). Now, if we consider to integrate over a finite domain, the limits become: $${\int\limits_a^\infty {f\left( x \right)dx} }={ \lim\limits_{n \to \infty } \int\limits_a^n {f\left( x \right)dx} .}$$ If the function is defined for the interval(-∞, b], then the integral becomes: $${\int\limits_{ – \infty }^b {f\left( x \right)dx} }={ \lim\limits_{n \to – \infty } \int\limits_n^b {f\left( x \right)dx} .}$$ It should be remembered that if the limits are finite and result in a number, the improper integral is convergent. But if limits are not a number, then the given integral is divergent. Now, let us discuss the case in which our improper integral has two infinite limits. In this situation, we choose an arbitrary point and break the integral at that particular point. After doing so, we get two integrals having one of the two limits as infinite. $${\int\limits_{ – \infty }^\infty {f\left( x \right)dx} } = {\int\limits_{ – \infty }^c {f\left( x \right)dx} }+{ \int\limits_c^\infty {f\left( x \right)dx} .}$$ These kind of integrals can easily be evaluated with the help of free online improper integral calculator. Type 2(Improper Integrals With Infinite Discontinuity): These integrals have undefined integrands at one or more points of integration. Let f(x) is a function that is discontinuous at x = b and is continuous in the interval [a, b). $${\int\limits_a^b {f\left( x \right)dx} }={ \lim\limits_{\tau \to 0 + } \int\limits_a^{b – \tau } {f\left( x \right)dx} .}$$ Like above, we consider that our function is continuous at the interval (a, b] and discontinuous at x = a. $${\int\limits_a^b {f\left( x \right)dx} }={ \lim\limits_{\tau \to 0 + } \int\limits_{a + \tau }^b {f\left( x \right)dx} .}$$ Now if the function is continuous at the interval (a, c] (c, b] with a discontinuity at x = c. $${\int\limits_a^b {f\left( x \right)dx} } = {\int\limits_a^c {f\left( x \right)dx} }+{ \int\limits_c^b {f\left( x \right)dx} ,}$$
### How to Evaluate the Improper Integral?
We can approach various clever methods to solve an improper integral. Let us solve some examples to understand the concept more clearly. Example # 01: Evaluate the improper integral given: $$\int\limits_{0}^\infty \frac{1}{x} dx$$ Solution: Your input is: $$\int\limits_{0}^{\infty} \frac{1}{x}\, dx$$ First, we need to calculate the definite integral: $$\int \frac{1}{x}\, dx = \log{\left(x \right)}$$ (for steps, see Integral Calculator). $$\left(\log{\left(x \right)}\right)|_{x=0}=- f i n$$ $$\lim_{x \to \infty}\left(\log{\left(x \right)}\right)=\infty$$ $$\int\limits_{0}^{\infty} \frac{1}{x}\, dx = \left(\left(\log{\left(x \right)}\right)|_{x=0} \right) - \left(\lim_{x \to \infty}\left(\log{\left(x \right)}\right(\right) = \infty$$ $$\int\limits_{0}^{\infty} \frac{1}{x}\, dx=\infty$$ Since the value of the integral is not a finite number, so the integral is divergent. Moreover, the integral convergence calculator is the best option to obtain more precise results. Example # 02: Evaluate the improper integral: $$\int\limits_{-1}^{\infty} \frac{1}{x^{2}}\, dx$$ Solution: As the given input is: $$\int\limits_{-1}^{\infty} \frac{1}{x^{2}}\, dx$$ So, we have to solve for the indefinite integral first: $$\int \frac{1}{x^{2}}\, dx = - \frac{1}{x}$$ (for steps, see Integral Calculator). $$\left(- \frac{1}{x}\right)|_{x=-1}=1.0$$ $$\lim_{x \to \infty}\left(- \frac{1}{x}\right)=0$$ $$\int\limits_{-1}^{\infty} \frac{1}{x^{2}}\, dx = \left(\left(- \frac{1}{x}\right)|_{x=-1}\right) - \left(\lim_{x \to \infty}\left(- \frac{1}{x}\right)\right) = -1.0$$ $$\int\limits_{-1}^{\infty} \frac{1}{x^{2}}\, dx=-1.0$$ As we have got a finite number, the given integral is said to be convergent. Also, you can clear your doubts by feeding the same function in integral of convergence calculator. Example # 03: Evaluate the improper integral given below: $$\int\limits_{-\infty}^{\infty} \left(2 x^{2} - 2\right)\, dx$$ Solution: First, we have to determine the indefinite integral: $$\int \left(2 x^{2} - 2\right)\, dx = \frac{2 x \left(x^{2} - 3\right)}{3}$$ (for steps, see Integral Calculator). $$\lim_{x \to -∞}\left(\frac{2 x \left(x^{2} - 3\right)}{3}\right)=-\infty$$ $$\lim_{x \to \infty}\left(\frac{2 x \left(x^{2} - 3\right)}{3}\right)=\infty$$ $$\int\limits_{-\infty}^{\infty} \left(2 x^{2} - 2\right)\, dx = \left(\lim_{x \to -∞}\left(\frac{2 x \left(x^{2} - 3\right)}{3}\right)\right) - \left(\lim_{x \to \infty}\left(\frac{2 x \left(x^{2} -3\right)}{3}\right) \right) = \infty$$ $$\int\limits_{-\infty}^{\infty} \left(2 x^{2} - 2\right)\, dx=\infty$$ As the integral is not a finite number so it is said to be divergent. For verification, you can use our free online improper integrals calculator.
### How Improper Integral Calculator Works?
We can adopt different ways to evaluate the improper integral. However, the best option is to use our free divergent or convergent integral calculator. Let us see what we need to do while using an improper integral calculator. Input:
• Select the variable w.r.t which you wish to determine the integral
• Select your desired limits for integration
• Click ‘calculate’
Output: Our improper integral solver calculates:
• The definite or indefinite integral
• Apply limits to determine whether the integral in convergent or divergent
• Shows step-by-step calculations carried out.
## FAQ’s:
### How do you know if an integral is improper?
If an integral has either upper, lower or both limits as infinite, you can say that this is an improper integral.
### Can you split an improper integral?
Yes, splitting an improper integral at 0 is a little bit easier but you can also split it at any number you want.
### Is 0 convergent or divergent?
Whenever you add terms of the sequence that get closer and closer to 0, we can say that the sum is always converging at some finite value. That is why if the terms get small and small enough, we say that the integral does not diverge.
### What do you mean by convergence in real life?
In real life, we should know about the convergence theory, also known as catch-up effect which states that “poorer economies tend to grow at a faster rate than more developed economies”.
## Conclusion:
Determining the area under a curve with the help of an improper integral is a suitable approach as it enables you to understand the period in which the integral gives some value. You can not compute an improper integral using a normal Riemann integral. However, the use of an online improper integral calculator makes it easy to determine whether the given function is convergent or divergent for the limits defined.
## Reference:
From the source of Wikipedia: Convergence of the integral, Types of integrals, Improper Riemann integrals, and Lebesgue integrals, Cauchy principal value, Multivariable improper integrals. From the source of khan academy: Improper integrals, Divergent improper integral. Fro the source opf lumen learning: Basic Integration Principles, Properties, Integration By Parts, Trigonometric Integrals, Trigonometric Substitution, The Method of Partial Fractions, Integration Using Tables and Computers, Approximate Integration, Numerical Integration. |
# 7.3 Unit circle (Page 7/11)
Page 7 / 11
Access these online resources for additional instruction and practice with sine and cosine functions.
## Key equations
Cosine $\mathrm{cos}\text{\hspace{0.17em}}t=x$ Sine $\mathrm{sin}\text{\hspace{0.17em}}t=y$ Pythagorean Identity ${\mathrm{cos}}^{2}t+{\mathrm{sin}}^{2}t=1$
## Key concepts
• Finding the function values for the sine and cosine begins with drawing a unit circle, which is centered at the origin and has a radius of 1 unit.
• Using the unit circle, the sine of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the y -value of the endpoint on the unit circle of an arc of length $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ whereas the cosine of an angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ equals the x -value of the endpoint. See [link] .
• The sine and cosine values are most directly determined when the corresponding point on the unit circle falls on an axis. See [link] .
• When the sine or cosine is known, we can use the Pythagorean Identity to find the other. The Pythagorean Identity is also useful for determining the sines and cosines of special angles. See [link] .
• Calculators and graphing software are helpful for finding sines and cosines if the proper procedure for entering information is known. See [link] .
• The domain of the sine and cosine functions is all real numbers.
• The range of both the sine and cosine functions is $\text{\hspace{0.17em}}\left[-1,1\right].$
• The sine and cosine of an angle have the same absolute value as the sine and cosine of its reference angle.
• The signs of the sine and cosine are determined from the x - and y -values in the quadrant of the original angle.
• An angle’s reference angle is the size angle, $\text{\hspace{0.17em}}t,$ formed by the terminal side of the angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and the horizontal axis. See [link] .
• Reference angles can be used to find the sine and cosine of the original angle. See [link] .
• Reference angles can also be used to find the coordinates of a point on a circle. See [link] .
## Verbal
Describe the unit circle.
The unit circle is a circle of radius 1 centered at the origin.
What do the x- and y- coordinates of the points on the unit circle represent?
Discuss the difference between a coterminal angle and a reference angle.
Coterminal angles are angles that share the same terminal side. A reference angle is the size of the smallest acute angle, $\text{\hspace{0.17em}}t,$ formed by the terminal side of the angle $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ and the horizontal axis.
Explain how the cosine of an angle in the second quadrant differs from the cosine of its reference angle in the unit circle.
Explain how the sine of an angle in the second quadrant differs from the sine of its reference angle in the unit circle.
The sine values are equal.
## Algebraic
For the following exercises, use the given sign of the sine and cosine functions to find the quadrant in which the terminal point determined by $\text{\hspace{0.17em}}t\text{\hspace{0.17em}}$ lies.
$\text{sin}\left(t\right)<0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\text{cos}\left(t\right)<0$
$\text{sin}\left(t\right)>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)>0$
I
$\text{sin}\left(t\right)>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)<0$
$\text{sin}\left(t\right)>0\text{\hspace{0.17em}}$ and $\text{\hspace{0.17em}}\mathrm{cos}\left(t\right)>0$
IV
For the following exercises, find the exact value of each trigonometric function.
$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{2}$
$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{3}$
$\frac{\sqrt{3}}{2}$
$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{2}$
$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{3}$
$\frac{1}{2}$
$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{4}$
$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{4}$
$\frac{\sqrt{2}}{2}$
$\mathrm{sin}\text{\hspace{0.17em}}\frac{\pi }{6}$
$\mathrm{sin}\text{\hspace{0.17em}}\pi$
0
$\mathrm{sin}\text{\hspace{0.17em}}\frac{3\pi }{2}$
$\mathrm{cos}\text{\hspace{0.17em}}\pi$
-1
$\mathrm{cos}\text{\hspace{0.17em}}0$
$\mathrm{cos}\text{\hspace{0.17em}}\frac{\pi }{6}$
$\frac{\sqrt{3}}{2}$
By the definition, is such that 0!=1.why?
(1+cosA+IsinA)(1+cosB+isinB)/(cos@+isin@)(cos$+isin$)
hatdog
Mark
how we can draw three triangles of distinctly different shapes. All the angles will be cutt off each triangle and placed side by side with vertices touching
bsc F. y algebra and trigonometry pepper 2
given that x= 3/5 find sin 3x
4
DB
remove any signs and collect terms of -2(8a-3b-c)
-16a+6b+2c
Will
Joeval
(x2-2x+8)-4(x2-3x+5)
sorry
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
x²-2x+9-4x²+12x-20 -3x²+10x+11
Miranda
(X2-2X+8)-4(X2-3X+5)=0 ?
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
The anwser is imaginary number if you want to know The anwser of the expression you must arrange The expression and use quadratic formula To find the answer
master
Y
master
master
Soo sorry (5±Root11* i)/3
master
Mukhtar
2x²-6x+1=0
Ife
explain and give four example of hyperbolic function
What is the correct rational algebraic expression of the given "a fraction whose denominator is 10 more than the numerator y?
y/y+10
Mr
Find nth derivative of eax sin (bx + c).
Find area common to the parabola y2 = 4ax and x2 = 4ay.
Anurag
y2=4ax= y=4ax/2. y=2ax
akash
A rectangular garden is 25ft wide. if its area is 1125ft, what is the length of the garden
to find the length I divide the area by the wide wich means 1125ft/25ft=45
Miranda
thanks
Jhovie
What do you call a relation where each element in the domain is related to only one value in the range by some rules?
A banana.
Yaona
a function
Daniel
a function
emmanuel
given 4cot thither +3=0and 0°<thither <180° use a sketch to determine the value of the following a)cos thither
what are you up to?
nothing up todat yet
Miranda
hi
jai
hello
jai
Miranda Drice
jai
aap konsi country se ho
jai
which language is that
Miranda
I am living in india
jai
good
Miranda
what is the formula for calculating algebraic
I think the formula for calculating algebraic is the statement of the equality of two expression stimulate by a set of addition, multiplication, soustraction, division, raising to a power and extraction of Root. U believe by having those in the equation you will be in measure to calculate it
Miranda |
# Bayes theorem¶
The reverse probability page has a game, that we analyzed by simulation, and then by reflection.
The game is:
• I have two boxes; BOX4 with 4 red balls and 1 green ball, and BOX2 with two red balls and three green balls.
• I offer you one of these two boxes, with a 30% chance that I give you BOX4, and 70% chance I give you BOX2.
• You draw a ball at random from the box, and you get a red ball.
• What is the probability that I gave you BOX4?
We found by simulation, and later by reflection, that the probability is about 0.462.
The logic we discovered was:
• We want the proportion of “red” trials that came from BOX4.
• Calculate the proportion of trials that are both BOX4 and red, and divide by the overall proportion of red trials.
Here is a Sankey diagram of that calculation:
We found the proportion of red trials that are both BOX4 and red is (the proportion of BOX4 trials) multiplied by (the proportion of BOX4 trials that are red.
In our case above, the value we need is the probability of red trials from BOX4 (red AND BOX4) divided by the probability of red from either box. In our case this is 0.24 / (0.24 + 0.28) = 0.4615.
The logic above is a fundamental rule in probability called Bayes theorem.
In this page, we relate the logic above to the usual way of describing Bayes theorem.
First we need some notation.
The probability that I give you BOX4 on any one trial is 0.3.
I could write this out long-hand as “probability of BOX4”, but this will soon become inconvenient, so I will write the “probability of BOX4” in a much more compact form as $$P(B4)$$. As you remember $$P(B4)$$ is 0.3:
$P(B4) = 0.3$
Read this as “the probability of BOX4 is 0.3”.
Similarly:
$P(B2) = 0.7$
The probability of getting a red ball, given that I am drawing from BOX4, is 4/5 = 0.8. Again, we could write that out longhand as “probability of red given BOX4, or:
$P(Red\ \mathrm{given}\ BOX4)$
but I want to make this more compact. First I will shorten $$\mathrm{red}$$ to $$R$$, and then I will write “given” with the bar: $$\mid$$.
$P(R \mid B4) = 0.8$
Read this as “the probability of drawing a red ball given I have BOX4 is 0.8”.
Similarly:
$P(R \mid B2) = 0.4$
“The probability of red given BOX2 is 0.4”.
We follow the logic above, with this notation. Here is the logic again:
1. We want the proportion of “red” trials that came from BOX4.
2. Calculate the proportion of trials that are both BOX4 and red, and divide by the overall proportion of red trials.
We can express the first statement by saying that we are trying to find $$P(B4 \mid R)$$ — “the probability of BOX4 given I have drawn a red ball”.
Now we need some more notation. We need the idea of the probability of getting both BOX4 and red. This is the probability for any one trial that we will end up with a red from BOX4, as opposed to a red from BOX2 or a green from either box.
We could write this as and $$P(B4\ \mathrm{and}\ R)$$, but in fact, we usually shorten the and to this symbol: $$\cap$$:
$P(B4 \cap R)$
Read this as the “probability we got BOX4 and then a red”.
We have already found that that we get this probability of BOX4 and red by multiplying the probability of BOX4 (0.3) by the probability of — getting a red ball, given BOX4 (0.8). In our notation, this multiplication is
$P(B4 \cap R) = P(B4) * P(R \mid B4)$
“The probability of BOX4 and red is the probability of BOX4 multiplied by the probability of — red given BOX4”.
Remember too, from the reverse probability page that we found $$P(R)$$ by adding the probabilities of the two different ways we can get a red ball: $$P(R) = P(R \mid B4) + P(R \mid B2)$$.
Let’s put all that on the Sankey diagram:
Putting the first and second statements together into one, we get:
$P(B4 \mid R) = \frac{P(B4) P(R \mid B4)}{P(R)}$
This is Bayes theorem, although it is usually written with the multiplication in the other order:
$P(B4 \mid R) = \frac{P(R \mid B4) P(B4)}{P(R)}$
See Bayes theorem on Wikipedia for more detail. |
The new training field of the Gryffindor Quidditch team is in the shape of a triangle, with small towers in its corners $A,B,C$. The angles at $A$, $B$, $C$ are respectively $50^{\circ}$, $50^{\circ}$, $80^{\circ}$. George Weasley is standing in a point $G$ somewhere in the middle of the field, and notes that $\angle GAB=30^{\circ}$ and that $\angle ABG=10^{\circ}$.
Question: How large is the angle $\angle BGC$?
(The answer to this question will be an integer. A good solution will clearly explain the reason why an integer number shows up here.)
This puzzle continues and is closely related to the following puzzles: $*$ A triangular Quidditch field $*$ The adventitious 18-gon $*$ https://puzzling.stackexchange.com/questions/30174/another-adventitious-triangle-for-ivo-beckers
• Aaagh. Feel like I'm right there but can't quite get it – Kevin Apr 11 '16 at 17:52
The angle is
$70^{\circ}$
For this, we will need the law of sines:
(Law of Sines) In any triangle $ABC$ it holds that
$$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=d_{ABC}$$
where $d_{ABC}$ is a constant, the diameter of $ABC$'s circumcenter.
I provide an auxiliary image, drawn to scale, to help us navigate the answer. We will use $a$ to denote the side of $ABC$ opposite $A$, and similarly for $b$ and $c$.
We have $\angle BAG = 30^{\circ}$ so $d_{ABG}=\frac{r}{\sin 30^{\circ}} = 2r$. Using the law of sines on $ABG$ once again, we find that $c = 2r \sin 140^{\circ} = 2r \sin 40^{\circ}$ ($40^{\circ}$ and $140^{\circ}$ are supplementary).
Now, the angle at $C$ is $80^{\circ}$, so $d_{ABC} = \frac{c}{\sin 80^{\circ}} = \frac{2r \sin 40^{\circ}}{\sin 80^{\circ}}$. By the double angle formula $\sin 2\theta = 2 \sin\theta\cos\theta$, it follows that $d_{ABC} = \frac{r}{\cos 40^{\circ}}$.
We use the law of sines once again on $ABC$. Since the angle at $A$ is $50^{\circ}$, we find that$$\frac{a}{\sin 50^{\circ}} = \frac{r}{\cos 40^{\circ}}$$
Since $40^{\circ}$ and $50^{\circ}$ are complementary, this implies $a = r$, so that $BCG$ is isoceles. Finally, $\angle ABG = 10^{\circ}$ and the angle at $B$ is $50^{\circ}$, so $\angle CBG = 40^{\circ}$ and thus $\angle BGC = 70^{\circ}$.
EDIT: In the spirit of the previous problems, perhaps I should mention that this too is a problem inscribed in an $18$-gon (or perhaps more appropriately, a $36$-gon, as f'' observed).
• And if you expand to a 36-gon, CG becomes a diagonal as well. – f'' Apr 11 '16 at 19:32
• @f'' That's correct. Put it another way, CG intercepts the corresponding side of the $18$-gon right down the middle. – Fimpellizieri Apr 11 '16 at 19:37
• Here's another way to inscribe it into an 18-gon: i.stack.imgur.com/z2YKj.png – f'' Apr 11 '16 at 19:47
Brute-forcing with the law of sines works, but here's a different approach:
Let's make a new point $D$ such that $BCD$ is an equilateral triangle. Then draw the line $DA$, as in the picture:
$\angle ACD = 140^{\circ}$, so as $\triangle ACD$ is isosceles $\angle CAD = 20^{\circ} = \angle CAG$. Therefore $A$, $G$ and $D$ all lie in the same line.
$\angle GBD = 40^{\circ} + 60^{\circ} = 100^{\circ}$, and $\angle GDB = 60^{\circ} - 20^{\circ} = 40^{\circ}$, so $\angle DGB = 180^{\circ} - 100^{\circ} - 40^{\circ} = 40^{\circ}$. Therefore $\triangle BGD$ is an isosceles triangle, and $BG = BD = BC$, which implies $\triangle BCG$ is also an isosceles triangle.
From the fact that $\angle GBC = 40^{\circ}$, the base angles of the isosceles triangle $\triangle BCG$ each equal $70^{\circ}$.
• +1 for not using trig on a problem which can be solved using only elementary geometry. That technique of erecting an equilateral triangle on a segment is useful, isn't it. – Rosie F Jun 17 '18 at 19:04 |
## Intermediate Algebra (12th Edition)
$x=\left\{ 2,18 \right\}$
$\bf{\text{Solution Outline:}}$ To solve the given equation, $|5-0.5x|=4 ,$ use the definition of absolute value equality. Then use the properties of equality to isolate the variable. $\bf{\text{Solution Details:}}$ Since for any $c\gt0$, $|x|=c$ implies $x=c \text{ or } x=-c,$ the equation above is equivalent to \begin{array}{l}\require{cancel} 5-0.5x=4 \\\\\text{OR}\\\\ 5-0.5x=-4 .\end{array} Solving each equation results to \begin{array}{l}\require{cancel} 5-0.5x=4 \\\\ -0.5x=4-5 \\\\ -0.5x=-1 \\\\ 10(-0.5x)=10(-1) \\\\ -5x=-10 \\\\ x=\dfrac{-10}{-5} \\\\ x=2 \\\\\text{OR}\\\\ 5-0.5x=-4 \\\\ -0.5x=-4-5 \\\\ -0.5x=-9 \\\\ 10(-0.5x)=10(-9) \\\\ -5x=-90 \\\\ x=\dfrac{-90}{-5} \\\\ x=18 .\end{array} Hence, $x=\left\{ 2,18 \right\} .$ |
# How do you solve 3= \frac { 5x - 4} { 3} + \frac { 3x - 1} { 5}?
##### 2 Answers
Jan 16, 2018
$x = 2$
#### Explanation:
$3 = \frac{5 x - 4}{3} + \frac{3 x - 1}{5}$
Multiply all terms by the LCM of $3$ and $5$ which is $15$.
$3 \times 15 = 15 \times \frac{5 x - 4}{3} + 15 \times \frac{3 x - 1}{5}$
$45 = 5 \cancel{15} \times \frac{5 x - 4}{1 \cancel{3}} + 3 \cancel{15} \times \frac{3 x - 1}{1 \cancel{5}}$
$45 = 5 \left(5 x - 4\right) + 3 \left(3 x - 1\right)$
Open the brackets and simplify.
$45 = 25 x - 20 + 9 x - 3$
$45 = 25 x + 9 x - 20 - 3$
$45 = \left(25 x + 9 x\right) - \left(20 + 3\right)$
$45 = 34 x - 23$
Add $23$ to both sides.
$45 + 23 = 34 x - 23 + 23$
$68 = 34 x$
Divide both sides by $34$.
$\frac{68}{34} = \frac{34 x}{34}$
$\frac{2 \cancel{68}}{1 \cancel{34}} = \frac{\cancel{34} x}{\cancel{34}}$
$2 = x$ or $x = 2$
Jan 16, 2018
Please see the step process below;
#### Explanation:
$3 = \setminus \frac{5 x - 4}{3} + \setminus \frac{3 x - 1}{5}$
First Step: Multiply via the LCM, in this case the LCM $= 15$
$15 \left(\frac{3}{1}\right) = 15 \left(\frac{5 x - 4}{3}\right) + 15 \left(\frac{3 x - 1}{5}\right)$
Second Step: Simplify
$15 \left(\frac{3}{1}\right) = {\cancel{15}}^{5} \left(\frac{5 x - 4}{\cancel{3}}\right) + {\cancel{15}}^{3} \left(\frac{3 x - 1}{\cancel{5}}\right)$
$15 \left(3\right) = 5 \left(5 x - 4\right) + 3 \left(3 x - 1\right)$
$45 = 25 x - 20 + 9 x - 3$
Third Step: Collecting like terms
$45 = 25 x + 9 x - 20 - 3$
$45 = 34 x - 23$
$45 + 23 = 34 x$
$68 = 34 x$
Divide both sides by $34$
$\frac{68}{34} = \frac{34 x}{34}$
$\frac{68}{34} = \frac{\cancel{34} x}{\cancel{34}}$
$\frac{68}{34} = x$
$x = 2$ |
# Four Forces of Flight
Learn about the four forces of flight.
You’ve probably seen an airplane flying at some point. But have you ever wondered how an aircraft flies? The answer is easy - with a little physics! Flight is all about forces and movement, which can be explained using physics.
Let’s start with the forces. There are four forces that act on things that fly. These are weightliftthrust, and drag. Each of these plays a key role in keeping an aircraft in the air and moving forward.
The first of the four forces exerted on aircraft is weight. The weight of an object is the force on the object due to gravity. Certain objects in space, including planets like the Earth, exert a force that attracts objects toward itself. In the case of the Earth, “toward itself” means “down toward the ground.” The force exerted on a body due to gravity can be expressed using the equation:
#### F = mg
Where F is the force in (N), m is the mass of the object in kg and g is the acceleration due to gravity. When doing this calculation, it is best to use the unit for gravity in N/kg:
#### g = 9.81 N/kg
In terms of the four forces acting upon an aircraft, weight is measured as the F in the above equation. However, we usually use the symbol W when specifically talking about weight. Substituting W for F above we get:
#### W = mg
From this equation, we can see that when we talk about ‘weight’ we are actually talking about how much force is acting on a mass due to gravity. This force, as mentioned above, also has a direction. We could call it “down”.
When we stand on the ground, we push down on the Earth and the Earth pushes up against our feet with the same amount of force in the opposite direction. This is an example of Newton’s Third Law). Newton’s Third Law states that for every action, there is an equal and opposite reaction. We usually abbreviate this upward force of the Earth as Fn.
Image - Text Version
Shown is a cartoon drawing of a group of three people.
To the left is a young Caucasian man sitting in a wheelchair and reading a book. To his right is a young brown-skinned woman sitting cross-legged on a pillow on the floor and reading a book. Behind them is a tall brown-skinned man standing up and holding a ball. On the right side of the picture there are two large arrows pointing at each other. At the top is a red arrow pointing downwards beside the text "Weight DOWN = Action". Below this is a green arrow of the same shape and size pointing upwards beside the text "Normal Force UP = Reaction".
Any type of flying machine experiences weight. This weight is always in the direction of the Earth, no matter which way the aircraft is travelling. It is very important to know the weight of an aircraft before flight. Too much weight can cause an aircraft to fly poorly. Heavy aircraft may need higher takeoff speeds and longer runways. They also may not be able to fly as far or as high.
Image - Text Version
Shown, from left to right, are photographs of a hot air balloon, a large commercial airliner and a fighter jet. Directional arrows and text are overlaid on the photo.
The green hot air balloon appears to be floating in a clear blue sky. Below the balloon is a red arrow pointing downward beside the word "weight", indicating that weight is a downward force, and the direction of the force is perpendicular to the ground.
The large twin-engined airliner, appears to be flying in a level position and travelling from left to right. Below the centre of the aircraft is a red arrow pointing downward beside the word "weight".
The fighter jet appears to be flying almost straight up into the sky. Below the rear part of the aircraft is once again a red arrow pointing downward with the word "weight" beside it.
If an aircraft is being pulled down toward the Earth by gravity and its own mass, how does it stay in the air? The answer is the second force, lift. Lift refers to the force that an object needs to overcome its weight
Lift is an upward force caused by air moving over a wing. A wing or blade, such as that of a propeller, rotor, or turbine as seen in cross-section has a special shape called an airfoil.
As a wing moves through air, the air splits and flows both over and under the wing. The difference in the movement of the air on top of the wing and below the wing generates lift. There are two explanations for the causes of lift: deflection and differences in pressure.
Image - Text Version
Shown is a line drawing of how an airfoil redirects air as it moves.
The airfoil is a long teardrop shape with a thicker, rounded end to the left and a longer, thinner end to the right. It is slightly bulged on the top compared to the bottom.
The airfoil is horizontal in the middle of the picture, with the rounded edge pointing slightly upwards towards the left and the pointed edge pointing slightly downwards towards the right.
Above the airfoil is a blue arrow that points to the left to indicate the direction of motion.
There are five thin parallel blue lines, representing the airflow, that go across the image horizontally. Two blue lines go below the airfoil shape, rising up slightly towards the left and then sloping down, following the shape of the bottom of the airfoil. One blue line connects to the middle of each side of the airfoil. The other two blue lines follow shape of the top of the airfoil. The lines get closer to each other above the thickest part of the airfoil and then return to their normal spacing once they are past the airfoil shape.
Below the middle of the airfoil is a red arrow that is pointed downwards. Beside this is text indicates that air is pushed downward by the underside of the wing. This is the action.
Above the middle of the airfoil is a green arrow that is the same size and shape as the red arrow pointing upwards. Beside this is text that indicates when the wing is pushed upwards there is lift. This is the reaction.
#### Deflection
As air passes along a wing, some of the air is directed downward. This is called deflection. Once again Newton’s Third Law is in action. Here, the ACTION is air pushing downward under the wing, and the REACTION is the wing moving upwards.
When the leading edge of the wing points upward, such as when the aircraft is climbing, it creates a positive angle of attack. Angle of attack is the angle between the and the direction of motion. Since air is being deflected downward by the wing, there is lift.
The opposite is also true. When the wings point downward (a negative angle of attack), there is less lift and the aircraft goes down.
Image - Text Version
Shown is a line drawing with labels indicating the parts of an airfoil.
The airfoil is a long teardrop shape with a thicker, rounded end to the left and a longer, thinner end to the right. It is slightly bulged on the top compared to the bottom.
The airfoil is horizontal in the middle of the picture, with the rounded edge pointing slightly upwards towards the left and the pointed edge pointing slightly downwards towards the right.
Below the airfoil is a blue arrow that points to the left to indicate the direction of motion.
The rounded end of the airfoil is labelled as the leading edge.
The pointed end of the airfoil is labelled as the trailing edge.
A blue line extends from just past the leading edge to the trailing edge. This line is labelled as the chord line.
The angle between the direction of motion and the chord line is called the angle of attack. When the angle is upward, this is known as a positive angle of attack, which is the orientation of the airfoil shown in this diagram.
#### Pressure Differences
Lift can also result from differences in pressure. These differences occur above and below the wing as air moves past the wing.
Air pressure is measured by dividing the force of the air molecules by the area that the air molecules are in. When air moves over a wing, the layer of air is squeezed into a smaller area. As a result, the speed of the air increases and the pressure of the air decreases. The opposite occurs below the wing. The air is squeezed less, resulting in slower moving air that has higher pressure.
Image - Text Version
Shown is another interpretation of how air moves around an airfoil. Above and below the airfoil shape are four thin blue parallel lines that follow the shape of the airfoil. Two go over the airfoil and two go under.
Between the airfoil and the first blue line above it, the space is shaded pink. This is to indicate that there is less force from air. There is also a small red arrow pointing downwards toward the airfoil. Beside the arrow is the phrase "fast air = low pressure".
Between the airfoil and the first blue line below it, the space is shaded pale blue. This is to indicate that there is more force from air. There is also a large green arrow pointing upwards toward the airfoil. Below the arrow is the phrase "slow air = high pressure".
Lift can be explained using Bernoulli’s Principle. It states that “as the speed of a moving increases, the pressure within the fluid decreases.” Since the force pushing up from the high pressure air is greater than the force pushing down from the low pressure air, there is lift in an upward direction.
Early flew in hot air balloons. These lighter than air (LTA) vehicles could easily go up and down, but once in the air, they were at the mercy of the wind. A pilot had no way to steer the balloon.
Not long after they were invented, people began to think of ways to make balloons go in the direction they wanted. To accomplish this, they needed a way to push the balloon forward. This pushing is known as thrust. Like lift, thrust is another type of reaction force that can be explained using Newton’s Third Law.
#### Propellers
In 1784, Jean-Pierre Blanchard attached a hand-powered propeller to a balloon, which is the first recorded use of propulsion by a hot air balloon. People tried many other forms of propulsion in the 1700s and early 1800s, and it wasn’t until 1852 that Henri Giffard created an airship which used an engine to turn a propeller.
Propellers are rotating blades which may be found at the front or back of an aircraft. If they are on the front, they are called tractors. If they are at the back, they are called pushers.
A propeller consists of two or more blades connected together by a hub.
Early propellers were turned by hand, pedalled by foot, or powered by steam engines. Today, propellers are powered by either internal combustion engines or jet engines (see below).
Image - Text Version
Shown is a photograph of the front end of a shiny silver World War II fighter plane sitting on the ground.
The propeller is connected to the front of the aircraft at the hub, which is painted a deep yellow. The location of the hub is labelled. Attached to the hub are four flat paddle-shaped propeller blades. These are painted black and have yellow tips. One of the blades is labelled. The engine, which is connected to the hub, is also labelled.
Each of the blades of a propeller is shaped like an airfoil. When they turn, they act as spinning wings. As the propeller turns, it pulls slow air towards itself and pushes fast air out behind itself. This generates a force directly behind the propeller - the action - that pushes the aircraft forward - the reaction.
Image - Text Version
Shown is a photograph of a vintage World War II US navy fighter plane, in flight. It appears to be travelling to the left. Directional arrows and text are overlaid on the photo.
Four small parallel blue arrows point towards the propeller. Above these arrows is text identifying this as "slower air."
Four larger parallel pink arrows point from the propeller back towards the rear of the plane. Above these arrows in text identifying this as "faster air."
Pointing backwards, away from the propeller and below the aircraft is a red arrow and the word "Action". Pointing forwards, away from the propeller and below the aircraft is a green arrow and the word "Reaction". The two arrows are the same size and shape.
#### Rotors
Rather than propellers, helicopters use a set of rotary wings called the rotor. A rotor is made up of two or more rotor blades. Helicopters typically have two rotors. These are the main rotor, which is located at the top of the aircraft and the tail rotor, which is located at the back of the aircraft.
Image - Text Version
Shown is a photograph of a bright red search and rescue helicopter with a white stripe and black rotors. A person is suspended on a wire directly below the helicopter. Directional arrows and text are overlaid on the photo.
Above the helicopter is a label indicating that the main rotor is located above the body of the aircraft. Another arrow points to one of the blades indicating that it is a rotor blade. At the back end of the helicopter is an arrow pointing to another smaller rotor embedded in the tail. The label indicates that this is the tail rotor.
Unlike a propeller, a rotor produces both lift and thrust. In order to fly in a particular direction, the pilot changes the of the rotor blades. This makes the rotor tip in a given direction. The helicopter will then move in that direction.
Rotors allow helicopters to take off and land vertically, as well as hover. This makes them useful for search and rescue, firefighting, and medical transport.
Image - Text Version
Shown is a photograph of a bright red search and rescue helicopter with a white stripe and black rotors. A person is suspended on a wire directly below the helicopter. Directional arrows and text are overlaid on the photo.
Pointing upward from two of the main rotor blades are two green arrows and the text "Reaction = Lift and Thrust".
Pointing downward from two of the main rotor blades, and directly below the green arrows are two red arrows, each of which is beside the text "Action."
#### Jet Engines
Many modern aircraft have replaced engines turning propellers with jet engines. These engines create thrust by:
1. pulling air into the engine,
2. mixing the air with fuel,
3. igniting the fuel/air mixture, and
4. pushing the hot air out of the back of the engine at high speed.
As with the propeller, the jet engine pushes out air at a higher speed than the air entering the engine. This causes the aircraft to move forwards.
Image - Text Version
Shown is an illustration of a cross section of a tubojet engine.
The engine looks like a tube that narrows, then widens, then narrows again.
Labelled as step 1, cool air, identified with blue arrows, is pulled into the engine on the left side of the image.
Next, the air passes through a section that looks like seveal striped lines. Labelled as step 2, this is where the air is mixed with fuel. The interior of the engine appears blue to indicate that the air is cold in this region.
Next the air moves into the narrow part of the engine. Here the air and fuel mixture is ignited. Labelled as step 3, this is where the air and fuel mixture is ignited. The interior of the engine appears as yellow to indicate that the air is heated in this region. A red directional arrow points further into the engine.
Finally, the air passes through several more striped lines. Labelled as step 4, this is where a turbine pushes hot air out the back of the engine at high speed.
The fourth and final force of flight is called drag. Another term for drag is air resistance. Like other fluids, air can resist, or try to stop the movement of an object through it. This is similar to how water behaves when you try to walk or swim through it. The same is true for aircraft. Air resists the movement of aircraft through it. This resistance counteracts thrust and slows down forward motion.
There are two main types of drag: parasite drag and lift-induced drag.
#### Parasite Drag
Form drag is drag that is caused by the shape of an object travelling through a fluid. Some shapes, such as the airfoil shape, move fairly smoothly through air. The air moves neatly above and below the shape without creating a lot of behind it. However, other shapes do not move smoothly through air. Shapes like the sphere and the flat plate create a lot of turbulence behind them. This turbulence slows down their movement.
Image - Text Version
Shown are four drawings used to describe how air flows around objects of different shapes. The air flow is represented by thin blue lines with small arrows that are point to the right. In each diagram, the lines initially come towards the shape in parallel.
The top left drawing shows how the air would move around a thin rectangular object in a vertical position. The lines indicate that some air would be blocked by the object, but that most air move over and below the object. Directly behind the object are a number of curved lines. These indicate turbulent air, which occurs directly behind the object.
The top right drawing shows how the air would move around a circular object. The lines indicate that some air would be blocked by the object, but that most air move over and below the object. Directly behind the object are a few curved lines indicating turbulent air. There are fewer turbulence lines than in the drawing of the rectangular object.
The bottom left drawing shows how the air would move around an airfoil. The lines indicate that almost all air would move over and below the object. As the airfoil is parallel to the direction of airflow, there are no turbulence lines.
The bottom right drawing shows how the air would move around a thin rectangular object in a horizontal position. The lines indicate that almost all of the air would move over and below the object. The airflow lines are almost undisturbed by the object and there is no turbulence.
Early aircraft such as the Curtiss 1911 Model D had a lot of form drag, especially from vertical parts such as wing struts. Over time, advances in aerodynamics and materials has led to much more streamlined designs, such as the SR-71 Blackbird.
Image - Text Version
Shown are two photographs of aircraft in flight.
The photo on the left is of an early biplane. This early aircraft looks a lot like a box kite. It is made from wood and fabric with wires connecting the two sets of parallel wings. A pilot sits in the open near the engine.
The photo on the right is of a modern stealth aircraft. It is very flat and angular, with a long flat fuselage and a single set of sleek wings.
Surface friction drag occurs whenever an object moves through a fluid. The roughness of the surface affects how much the fluid slows down the movement of the object. This is because the rough spots cause turbulence. Surface drag is actually a form of. To reduce surface friction drag, aircraft are designed to be as smooth as possible.
Image - Text Version
Shown is a photograph of vintage World War II fighter aircraft in flight. The aircraft has a shiny metallic finish. It is so smooth that the runway and grass below is reflected on its side!
#### Lift-Induced Drag
The other main type of drag is lift-induced drag. This type of drag is a result of lift. The greater the lift, the greater the lift-induced drag.
When a wing is roughly parallel to the airflow the air tends to flow smoothly past the wing. As you increase the angle of attack, though, you begin to get more unstable air behind the wing. This is due to the shape of the wing becoming more like the flat plate mentioned under form drag.
Image - Text Version
Shown are two drawings of how air flows around airfoils with different angles of attack.
In each drawing, the air flow is represented by four thin blue lines which initially come towards the airfoil in parallel from the left side of the drawing.
The top drawing shows how the air would move around an airfoil with a small angle of attack. Above the trailing edge of the airfoil are a few curved lines. These indicate turbulence, which is fairly minimal in this airfoil position.
The bottom drawing shows how the air would move around an airfoil with a large angle of attack. In this case, the airfoil acts al lot like the vertical rectangle shape in the diagram about form drag. Directly behind the airfoil are many curved lines indicating that extensive turbulence is generated.
There comes a point where the angle of attack becomes so great that the wing is no longer able to generate lift. This is known as the critical angle of attack. At this point, the aircraft stalls. Many modern aircraft have warning systems that alert the pilot if the aircraft is about to stall.
So to summarize, there are four forces that keep an aircraft in the air and moving forward: weight, lift, thrust, and drag. But if you think about it, that means an aircraft is falling, rising, moving forward, and being pulled back - all at the same time! Scientific innovations over the centuries have allowed us to keep these four forces in balance so that we can fly aircraft from one place to another.
Fly8MA.com Ep. 5: How Airplanes Fly
This video (5:36) from Fly8MA, a private pilot education company, describes with visuals the four forces that act upon an airplane in flight.
Thinking Captain: How Do Planes Fly?
This video (2:14) provides an easy and quick breakdown on how airplanes stay in the sky and the four forces of flight.
No One Can Explain Why Planes Fly In The Air
This article (2020) in Scientific American by Ed Regis delves into the idea that, while we can clearly engineer aircraft that fly through the air, we don’t have a precise mathematical model that explains lift.
Physics Challenges For Green Aviation
This article (2020) by Brian Tillotson in Physics World tackles some of the future-looking ideas for aircraft design to help reduce fuel usage due to drag and other environmentally problematic aspects of flight.
U.S. Space and Rocket Center: Four Forces of Flight with Paper Airplanes
This video (9:27) examines the four forces of flight through both visual aids and the construction and flight of paper airplanes. |
# How to simplify this?
## $\frac{{a}^{2} - {b}^{2}}{a - b} ^ 2$
Mar 26, 2016
$\frac{a + b}{a - b}$
#### Explanation:
Given the expression:
$\textcolor{w h i t e}{\text{XXX}} \frac{{a}^{2} - {b}^{2}}{{\left(a - b\right)}^{2}}$
Recall that
$\textcolor{w h i t e}{\text{XXX}} \left({a}^{2} - {b}^{2}\right) = \left(a + b\right) \left(a - b\right)$
Therefore
$\textcolor{w h i t e}{\text{XXX}} \frac{{a}^{2} - {b}^{2}}{{\left(a - b\right)}^{2}} = \frac{\left(a + b\right) \cancel{\left(a - b\right)}}{\left(a - b\right) \cancel{\left(a - b\right)}} = \frac{a + b}{a - b}$ |
# Lesson 4Practice Solving Equations and Representing Situations with Equations
Let's solve equations by doing the same to each side.
### Learning Targets:
• I can explain why different equations can describe the same situation.
• I can solve equations that have whole numbers, fractions, and decimals.
## 4.1Number Talk: Subtracting From Five
Find the value of each expression mentally.
## 4.2Row Game: Solving Equations Practice
Solve the equations in one column. Your partner will work on the other column.
Check in with your partner after you finish each row. Your answers in each row should be the same. If your answers aren’t the same, work together to find the error and correct it.
column A column B
## 4.3Choosing Equations to Match Situations
• Circle all of the equations that describe each situation. If you get stuck, draw a diagram.
• Find the solution for each situation.
1. Clare has 8 fewer books than Mai. If Mai has 26 books, how many books does Clare have?
___________
2. A coach formed teams of 8 from all the players in a soccer league. There are 14 teams. How many players are in the league?
___________
3. Kiran scored 223 more points in a computer game than Tyler. If Kiran scored 409 points, how many points did Tyler score?
___________
4. Mai ran 27 miles last week, which was three times as far as Jada ran. How far did Jada run?
___________
### Are you ready for more?
Mai’s mother was 28 when Mai was born. Mai is now 12 years old. In how many years will Mai’s mother be twice Mai’s age? How old will they be then?
## Lesson 4 Summary
Suppose a scientist has 13.68 liters of acid and needs 16.05 liters for an experiment. How many more liters of acid does she need for the experiment?
• We can represent this situation with the equation:
• When working with hangers, we saw that the solution can be found by subtracting 13.68 from each side. This gives us some new equations that also represent the situation:
• Finding a solution in this way leads to a variable on one side of the equal sign and a number on the other. We can easily read the solution—in this case, 2.37—from an equation with a letter on one side and a number on the other. We often write solutions in this way.
Let’s say a food pantry takes a 54-pound bag of rice and splits it into portions that each weigh of a pound. How many portions can they make from this bag?
• We can represent this situation with the equation:
• We can find the value of by dividing each side by . This gives us some new equations that represent the same situation:
• The solution is 72 portions.
## Lesson 4 Practice Problems
1. Select all the equations that describe each situation and then find the solution.
1. Kiran’s backpack weighs 3 pounds less than Clare’s backpack. Clare’s backpack weighs 14 pounds. How much does Kiran’s backpack weigh?
2. Each notebook contains 60 sheets of paper. Andre has 5 notebooks. How many sheets of paper do Andre’s notebooks contain?
2. Solve each equation.
3. For each equation, draw a tape diagram that represents the equation.
4. Find each product.
5. For a science experiment, students need to find 25% of 60 grams. Jada says, “I can find this by calculating of 60.” Andre says, “25% of 60 means .” Lin says both of their methods work. Do you agree with Lin? Explain your reasoning. |
# NCERT Solutions for Chapter 11 Algebra Class 6 Maths
Chapter Name Algebra NCERT Solutions Class CBSE Class 6 Textbook Name Maths Related Readings NCERT Solutions for Class 6NCERT Solutions for Class 6 Maths Summary of Algebra
Exercise 11.1
Question 1: Find the rule, which gives the number of matchsticks required to make the following matchsticks patterns. Use a variable to write the rule.
(a) A pattern of letter T as
(b) A pattern of letter Z as
(c) A pattern of letter U as
(d) A pattern of letter V as
(e) A pattern of letter E as
(f) A pattern of letter S as
(g) A pattern of letter A as
Solution
(a) Pattern of letter = 2n (as two matchstick used in each letter)
(b) Pattern of letter = 3n (as three matchstick used in each letter)
(c) Pattern of letter = 3n (as three matchstick used in each letter)
(d) Pattern of letter = 2n (as two matchstick used in each letter)
(e) Pattern of letter = 5n (as five matchstick used in each letter)
(f) Pattern of letter = 5n (as five matchstick used in each letter)
(g) Pattern of letter = 6n (as six matchstick used in each letter)
Question 2: We already know the rule for the pattern of letter L, C and F. Some of the letters from Q.1 (given above) give us the same rule as that given by L. Which are these? Why does this happen?
Solution
The letter ‘T’ and ‘V’ that has pattern 2n, since 2 matchsticks are used in all these letters.
Question 3: Cadets are marching in a parade. There are 5 cadets in a row. What is the rule, which gives the number of cadets, given the number of rows? (Use n for the number of rows)
Solution
Number of rows = n
Cadets in each row = 5
Therefore, total number of cadets = 5n
Question 4: If there are 50 mangoes in a box, how will you write the total number of mangoes in terms of the number of boxes? (Use b for the number of boxes)
Solution
Number of boxes = b
Number of mangoes in each box = 50
Therefore, total number of mangoes = 50b
Question 5: The teacher distributes 5 pencils per student. Can you tell how many pencils are needed, given the number of students? (Use s for the number of students)
Solution
Number of students = s
Number of pencils to each student = 5
Therefore, total number of pencils needed are = 5s
Question 6: A bird flies 1 kilometer in one minute. Can you express the distance covered by the bird in terms of its flying time in minutes? (Use t for flying time in minutes)
Solution
Time taken by bird = t minutes
Speed of bird = 1 km per minute
Therefore, Distance covered by bird = speed × time = 1 × t = t km
Question 7: Radha is drawing a dot Rangoli (a beautiful pattern of lines joining dots with chalk powder as in figure). She has 8 dots in a row. How many dots will her Rangoli have for r rows? How many dots are there if there are 8 rows? If there are 10 rows?
Solution
Number of dots in each row = 8 dots
Number of rows = r
Therefore, number of dots = 8r
When there are 8 rows, then number of dots = 8 × 8 = 64 dots
When there are 10 rows, then number of dots = 8 × 10 = 80 dots
Question 8: Leela is Radha’s younger sister. Leela is 4 years younger than Radha. Can you write Leela’s age in terms of Radha’s age? Take Radha’s age to be x years.
Solution
Therefore, Leela’s age = (x – 4) years
Question 9: Mother has made laddus. She gives some laddus to guests and family members; still 5 laddus remain. If the number of laddus mother gave away is l, how many laddus did she make?
Solution
Number of laddus gave away = l
Number of laddus remaining = 5
Total number of laddus = (l + 5)
Question 10: Oranges are to be transferred from larger boxes into smaller boxes. When a large box is emptied, the oranges from it fill two smaller boxes and still 10 oranges remain outside. If the number of oranges in a small box are taken to be x, what is the number of oranges in the larger box?
Solution
Number of oranges in one box = x
Number of boxes = 2
Therefore, total number of oranges in boxes = 2x
Remaining oranges = 10
Thus, number of oranges = 2x + 10
Question 11: (a) Look at the following matchstick pattern of squares. The squares are not separate. Two neighbouring squares have a common matchstick. Observe the patterns and find the rule that gives the number of matchsticks in terms of the number of squares. (Hint: If you remove the vertical stick at the end, you will get a pattern of Cs.)
(b) Figs. Below gives a matchstick pattern of triangles. As in Exercise 11 (a) above find the general rule that gives the number of matchsticks in terms of the number of triangles.
Solution
If we remove 1 from each then they makes table of 3, i.e., 3, 6, 9, 12... So the required equation = 3x + 1 , where x is number of squares
If we remove 1 from each then they makes table of 2, i.e., 2, 4, 6, 8... So the required equation = 2x + 1 , where x is number of triangles..
Exercise 11.2
Question 1: The side of an equilateral triangle is shown by l. Express the perimeter of the equilateral triangle using l.
Solution
Side of equilateral triangle = l
Therefore, Perimeter of equilateral triangle = 3 × side = 3l
Question 2: The side of a regular hexagon is denoted by l. Express the perimeter of the hexagon using l. (Hint: A regular hexagon has all its six sides in length)
Solution
Side of hexagon = l
Therefore, Perimeter of Hexagon = 6 × side = 6l
Question 3: A cube is a three-dimensional figure. It has six faces and all of them are identical squares. The length of an edge of the cube is given by l. find the formula for the total length of the edges of a cube.
Solution
Length of one edge of cube = l
Number of edges in a cube = 12
Therefore, total length = 12 × l = 12l
Question 4: The diameter of a circle is a line, which joins two points on the circle and also passes through the centre of the circle. (In the adjoining figure AB is a diameter of the circle; C is its centre). Express the diameter of the circle (d) in terms of its radius (r).
Solution
Since, length of diameter is double the length of radius.
Therefore, d = 2r
Question 5: To find sum of three numbers 14, 27 and 13. We can have two ways.
(a) We may first add 14 and 27 to get 41 and then add 13 to it to get the total sum 54, or
(b) We may add 27 and 13 to get 40 and then add 14 to get the sum 54. Thus
(14 + 27) + 13 = 14 + (27 + 13)
This can be done for any three numbers. This property is known as the associativity of addition of numbers. Express this property which we have already studied in the chapter on Whole Numbers, in a general way, by using variables a, b and c.
Solution
(a + b) + c = a + (b + c)
Exercise 11.3
Question 1: Make up as many expressions with numbers (no variables) as you can from three numbers 5, 7 and 8. Every number should be used not more than once. Use only addition, subtraction and multiplication. (Hint: Three possible expressions are 5 + (8 – 7), 5 – (8 – 7), (5 x 8) + 7 make the other expressions)
Solution
(a) (8 × 5) – 7
(b) (8 + 5) – 7
(c) (8 × 7) – 5
(d) (8 + 7) – 5
(e) 5 × (7 + 8)
(f) 5 + (7 x 8)
(g) 5 + (8 – 7)
(h) 5 – (7 + 8)
Question 2: Which out of the following are expressions with numbers only:
(a) y + 3
(b) (7 × 20) – 8z
(c) 5 (21 – 7) + 7 × 2
(d) 5
(e) 3x
(f) 5 – 5n
(g) (7 × 20) – (5 × 10) – 45 + p
Solution
(c) and (d)
Question 3: Identify the operations (addition, subtraction, division, multiplication) in forming the following expressions and tell how the expressions have been formed:
(a) z + 1, z – 1, y + 17, y – 17
(b) 17y, y/17, 5z
(c) 2y + 17, 2y – 17
(d) 7m, - 7m + 3, - 7m – 3
Solution
z – 1→ Subtraction
y – 17→ Subtraction
(b) 17y→ Multiplication
y/17→ Division
5z→ Multiplication
(c) 2y + 17→ Multiplication and Addition
2y – 17→ Multiplication and Subtraction
(d) 7m→ Multiplication
- 7m + 3→ Multiplication and Addition
- 7m – 3→ Multiplication and Subtraction
Question 4: Give expressions for the following cases:
(b) 7 subtracted from p.
(c) p multiplied by 7.
(d) p divided by 7.
(e) 7 subtracted from -m.
(f) -p multiplied by 5.
(g) -p divided by 5.
(h) p multiplied by -5.
Solution
(a) p +7
(b) p – 7
(c) 7p
(d) p/7
(e) -m-7
(f) -5p
(g) –p/5
(h) -5p
Question 5: Give expression in the following cases:
(b) 11 subtracted from 2m
(c) 5 times y to which 3 is added.
(d) 5 times y from which 3 is subtracted.
(e) y is multiplied by -8.
(f) y is multiplied by -8 and then 5 is added to the result.
(g) y is multiplied by 5 and result is subtracted from 16.
(h) y is multiplied by -5 and the result is added to 16.
Solution
(a) 2m + 11
(b) 2m – 11
(b) 5y + 3
(d) 5y – 3
(e) – 8y
(f) – 8y + 5
(g) 16 – 5y
(h) – 5y + 16
Question 6: (a) From expressions using t and 4. Use not more than one number operation. Every expression must have t in it.
(b) Form expressions using y, 2 and 7. Every expression must have y in it. Use only two number operations. These should be different.
Solution
(a) t + 4, t – 4, 4 – t, 4t, t/4, 4/t
(b) 2y + 7, 2y – 7, 7y = 2, 7y – 2 and so on.
Exercise 11.4
(a) Take Sarita’s present age to be y years.
(i) What will be her age 5 years from now?
(ii) What was her age 3 years back?
(iii) Sarita’s grandfather is 6 times her age. What is the age of her grandfather?
(iv) Grandmother is 2 years younger than grandfather. What is grandmother’s age?
(v) Sarita’s father’s age is 5 years more than 3 times Sarita’s age. What is her father’s age?
(b) The length of a rectangular hall is 4 meters less than 3 times the breadth of the hall. What is the length, if the breadth is b meters?
(c) A rectangular box has height h cm. Its length is 5 times the height and breadth is 10 cm less than the length. Express the length and the breadth of the box in terms of the height.
(d) Meena, Beena and Leena are climbing the steps to the hill top. Meena is at step s, Beena is 8 steps ahead and Leena 7 steps behind. Where are Beena and Meena? The total number of steps to the hill top is 10 less than 4 times what Meena has reached. Express the total number of steps using s.
(e) A bus travels at v km per hour. It is going from Daspur to Beespur. After the bus has travelled 5 hours. Beespur is still 20 km away. What id the distance from Daspur to Beespur? Express it using v.
Solution
(a) (i) y + 5
(ii) y - 3
(iii) 6y
(iv) 6y - 2
(v) 3y + 5
(b) Length = 3b and Breadth = (3b – 4) meters
(c) Height of the box = h cm
Length of the box = 5 times the height = 5h cm
Breadth of the box = 10 cm less than length = (5h – 10) cm
(d) Meena’s position = s
Beena’s position = 8 steps ahead = s + 8
Leena’s position = 7 steps behind = s - 7
Total number of steps = 4s - 10
(e) Speed of the bus = v km/h
Distance travelled in 5 hours = 5v km
Remaining distance = 20 km
Therefore, total distance = (5v + 20) km
Question 2: Change the following statements using expressions into statements in ordinary language. (For example, given Salim scores r runs in a cricket match, nalin scores (r + 15) runs. In ordinary language – Nalin scores 15 runs more than Salim).
(a) A note book costs ₹ p. A book costs ₹ 3 . p
(b) Tony puts q marbles on the table. He has 8q marbles in his box.
(c) Our class has n students. The school has 20n students.
(d )Jaggu is z years old. His uncle is 4z years old and his aunt is (4z – 3) years old.
(e) In an arrangement of dots there are r rows. Each row contains 5 dots.
Solution
(a) A book cost 3 times the cost of a notebook.
(b) The number of marbles in box is 8 times the marble on the table.
(c) Total number of students in the school is 20 times that in our class.
(d) Jaggu’s uncle’s age is 4 times the age of Jaggu. Jaggu’s aunt is 3 years younger than his uncle.
(e) The total number of dots is 5 times the number of rows.
Question 3: (a) Given, Munnu’s age to be x years. Can you guess what (x – 2) may show? (Hint: Think of Munnu’s younger brother). Can you guess what (x + 4) may show? What (3x + 4) may show?
(b) Given Sara’s age today to be y years. Think of her age in the future or in the past. What will the following expression indicate? Y + 7, y – 3, 4 ½, y – 2 ½
(c) Given, n students in the class like football, what may 2n show? What may n/2 show? (Hint: Think of games other than football).
Solution
(a) Munnu’s age = x years
His younger brother is 2 years younger than him = (x – 2) years
His elder brother’s age is 4 years more than his age = (x + 4) years
His father is 7 year’s more than thrice of his age = (3x + 7) years
(b) Her age in past = (y – 3), (y – 2 ½)
Her age in future = (y + 7), (y + 4 ½)
(c) Number of students like hockey is twice the students liking football, i.e., 2n
Number of students like tennis is half the students like football, i.e. n/2.
Exercise 11.5
Question 1: State which of the following are equations (with a variable). Given reason for your answer. Identify the variable from the equations with a variable.
(a) 17 = x + 7
(b) (t – 7) > 5
(c) 4/2 = 2
(d) (7 × 3) – 19 = 8
(e) 5 × 4 – 8 = 2x
(f) x – 2 = 0
(g) 2m < 30
(h) 2n + 1 = 11
(i) 7 = (11 × 5) – (12 × 4)
(j) 7 = (11 × 2) + p
(k) 20 = 5y
(l) 3q/2 < 5
(m) z + 12 > 24
(n) 20 – (10 – 5) = 3 × 5
(o) 7 – x = 5
Solution
(a) It is an equation of variable as both the sides are equal. The variable is x.
(b) It is not an equation as L.H.S. is greater than R.H.S.
(c) It is an equation with no variable. But it is a false equation.
(d) It is an equation with no variable. But it is a false equation.
(e) It is an equation of variable as both the sides are equal. The variable is x.
(f) ) It is an equation of variable x.
(g) It is not an equation as L.H.S. is less than R.H.S.
(h) It is an equation of variable as both the sides are equal. The variable is n.
(i) It is an equation with no variable as its both sides are equal.
(j) It is an equation of variable p.
(k) It is an equation of variable y.
(l) It is not an equation as L.H.S. is less than R.H.S.
(m) It is not an equation as L.H.S. is greater than R.H.S.
(n) It is an equation with no variable.
(o) It is an equation of variable x.
Question 2: Complete the entries of the third column of the table:
S. no. Equation Value of Variable Equation Satisfied Yes/No (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) 10y = 80 10y = 80 10y = 80 4l = 20 4l = 20 4l = 20 b + 5 = 9 b + 5 = 9 b + 5 = 9 h – 8 = 5 h – 8 = 5 h – 8 = 5 p + 3 = 1 p + 3 = 1 p + 3 = 1 p + 3 = 1 p + 3 = 1 y = 10 y = 8 y = 5 l = 20 l = 80 l = 5 b = 5 b = 9 b = 4 h = 13 h =8 h = 0 p = 3 p = 1 p = 0 p = -1 p = -2
Solution
S. No. Equation Value of variable Equ. Satisfied yes/No Sol. of L.H.S. (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) (k) (l) (m) (n) (o) (p) (q) 10y = 80 10y = 80 10y = 80 4l = 20 4l = 20 4l = 20 b + 5 = 9 b + 5 = 9 b + 5 = 9 h – 8 = 5 h – 8 = 5 h – 8 = 5 p + 3 = 1 p + 3 = 1 p + 3 = 1 p + 3 = 1 p + 3 = 1 y = 10 y = 8 y = 5 l = 20 l = 80 l = 5 b = 5 b = 9 b = 4 h = 13 h = 8 h = 0 p = 3 p = 1 p = 0 p = -1 p = -2 No Yes No No No Yes No Yes Yes Yes No No No No No No Yes 10 × 10 = 100 10 × 8 = 80 10 × 5 = 50 4 × 20 = 80 4 × 80 = 320 4 × 5 = 20 5 + 5 = 10 9 + 5 = 14 4 + 5 = 9 13 – 8 = 5 8 – 8 = 0 0 – 8 = -8 3 + 3 = 6 1 + 3 = 4 0 + 3 = 3 -1 + 3 = 2 -2 + 3 = 1
Question 3: Pick out the solution from the values given in the bracket next to each equation. Show that the other values do not satisfy the equation.
(a) 5m = 60 (10, 5, 12, 15)
(b) n + 12 = 20 (12, 8, 20, 0)
(c) p – 5 = 5 (0, 10, 5, –5)
(d) q/2 = 7 (7, 2, 10, 14)
(e) r – 4 = 0 (4, –4, 8, 0)
(f) x + 4 = 2 (–2, 0, 2, 4)
Solution
(a) 5m = 60
Putting the given values in L.H.S.,
5 × 10 = 50
∵ L.H.S. ≠ R.H.S.
∴ M = 10 is not the solution.
5 × 5 = 25
∵ L.H.S. ≠ R.H.S.
∴ M = 5 is not the solution.
5 × 12 = 60
∵ L.H.S. = R.H.S.
∴ M = 12 is a solution.
5 × 15 = 75
∵ L.H.S. ≠ R.H.S.
∴ M = 15 is not the solution.
(b) n + 12 = 20
Putting the given values in L.H.S.,
12 + 12 = 24
∵ L.H.S. ≠ R.H.S.
∴ n = 12 is not the solution.
8 + 12 = 20
∵ L.H.S. = R.H.S.
∴ n = 8 is a solution.
20 + 12 = 32
∵ L.H.S. ≠ R.H.S.
∴ n = 20 is not the solution.
0 + 12 = 12
∵ L.H.S. ≠ R.H.S.
∴ n = 0 is not the solution.
(c) p – 5 = 5
Putting the given values in L.H.S.,
0 – 5 = –5
∵ L.H.S. ≠ R.H.S.
∴ p = 0 is not the solution.
10 – 5 = 5
∵ L.H.S. = R.H.S.
∴ p = 10 is a solution.
5 – 5 = 0
∵ L.H.S. ≠ R.H.S.
∴ p = 5 is not the solution.
–5 – 5 = –10
∵ L.H.S. ≠ R.H.S.
∴ p = - 5 is not the solution.
(d) q/2 = 7
Putting the given values in L.H.S.,
7/2
∵ L.H.S. ≠ R.H.S.
∴ q = 7 is not the solution.
2/2 = 1
∵ L.H.S. ≠ R.H.S.
∴ q = 2 is not the solution.
10/2 = 5
∵ L.H.S. ≠ R.H.S.
∴ q = 10 is not the solution.
14/2 = 7
∵ L.H.S. = R.H.S.
∴ q = 14 is a solution.
(e) r – 4 = 0
Putting the given values in L.H.S.,
4 – 4 = 0
∵ L.H.S. = R.H.S.
∴ r = 4 is a solution.
–4 – 4 = –8
∵ L.H.S. ≠ R.H.S.
∴ r = - 4 is not the solution.
8 – 4 = 4
∵ L.H.S. ≠ R.H.S.
∴ r = 8 is not the solution.
0 – 4 = –4
∵ L.H.S. ≠ R.H.S.
∴ r = 0 is not the solution.
(f) x + 4 = 2
Putting the given values in L.H.S.,
–2 + 4 = 2
∵ L.H.S. = R.H.S.
∴ X = - 2 is a solution.
0 + 4 = 4
∵ L.H.S. ≠ R.H.S.
∴ X = 0 is not the solution.
2 + 4 = 6
∵ L.H.S. ≠ R.H.S.
∴ X = 2 is not the solution.
4 + 4 = 8
∵ L.H.S. ≠ R.H.S.
∴ X = 4 is not the solution.
Question 4: (a) Complete the table and by inspection of the table find the solution to the equation m + 10 = 16
m 1 2 3 4 5 6 7 8 9 10 …. …. …. m + 10 …. …. …. …. …. …. …. …. …. …. …. …. ….
(b) Complete the table and by inspection of the table find the solution to the equation 5t = 35
t 3 4 5 6 7 8 9 10 11 …. …. …. …. 5t …. …. …. …. …. …. …. …. …. …. …. …. ….
(c) Complete the table and by inspection of the table find the solution to the equation z/3 = 4
(d) Complete the table and by inspection of the table find the solution to the equation m – 7 =3
m 5 6 7 8 9 10 11 12 13 …. …. m - 7 …. …. …. …. …. …. …. …. …. …. ….
Solution
(a)
m 1 2 3 4 5 6 7 8 9 10 11 12 13 m + 10 11 12 13 14 15 16 17 18 19 20 21 22 23
∵ At m = 16, m + 10 = 16
∴ m = 6 is the solution.
(b)
t 3 4 5 6 7 8 9 10 11 12 13 14 15 16 5t 15 20 25 30 35 40 45 50 55 60 65 70 75 80
∵ At t = 7, 5t = 35
∴ t = 7 is the solution.
(c)
∵ At z = 12, z/3 = 4
∴ z = 12 is the solution.
(d)
m 5 6 7 8 9 10 11 12 13 14 15 m - 7 -2 -1 0 1 2 3 4 5 6 7 8
∵ At m = 10, m – 7 = 3
∴ m = 10 is the solution. |
# 6th Class Mathematics Algebra ALGEBRA
ALGEBRA
Category : 6th Class
Learning Objective
• To understand algebraic expression, variables, constants, like and unlike terms, coefficient of term.
• To understand types of algebraic expression (Monomial, binomial trinonial and polynomials)
• To learn about the algebraic equation and how to solve it.
• To learn how to add, subtract and multiply and divide the polynomials.
The branch of mathematics which deals with numbers is called Arithmetic. Algebra can be considered as generalization of arithmetic, where we use letters in place of numbers, which allows to write rules and form in general way
VARIABLE
A symbol which takes various numerical values is called a variable.
CONSTANT
A symbol which takes a fixed numerical value is called a constant.
ALGEBRAIC EXPRESSIONS
When variables and constants are combined with the help of mathematical operations of addition, subtraction, multiplication and division, we get an algebraic expression. For example, 3x + 7, 15y - 23 are algebraic expression. 3x+7 is an algebraic expression in variable x, it is obtained by multiplying the variable x by constant 3 and then adding 7 to the product.
TERMS
Look at the expression (3x + 7). This is formed by first forming 3x as product of 3 and x and then adding 7 to the product. Similarly $(10{{x}^{2}}+15)$ can be formed by first forming $10{{x}^{2}}$ as product of 10, x and x and adding 15 to it. Such parts of an expression which are formed first and then added are called terms. Consider $(9{{y}^{2}}-8x),$ here we can say that $9{{y}^{2}}$ and - 8x are two terms of given expressions.
FACTORS OF A TERM
Now we know that an expression consist of terms. $(9{{y}^{2}}-8x)$ has two terms $9{{y}^{2}}$ and (-8x). The term $9{{y}^{2}}$ is a product of 9, y and y. Here we say that 9, y and y are factors of term$9{{y}^{2}}$. A term is represented as product of its factors.
For term (-8.x), -8 and x are factors.
COEFFICIENTS OF A TERM
We know that any term of an expression can be expressed as product of its factors. These factors are numeric or variables. The numerical factor is called numerical coefficient or coefficient of the term.
In $9{{y}^{2}}$, 9 is the coefficient of the term. It is also called coefficient ofy2. In - lO^z2, -10 is the coefficient of ${{y}^{2}}{{z}^{2}}$.
If the coefficient of any term is + 1, we omit it.
$1{{y}^{2}}$ can be written as y2, 1xy is written as xy coefficient (-1) is indicated by minus (-) sign, $(-1){{y}^{2}}$ is written as $-{{y}^{2}},\,(-1)\,{{y}^{2}}{{z}^{2}}$ as $-{{y}^{2}}-{{z}^{2}}$ etc.
as -y2, (-1) y^z2 as - y2 - z2 etc.
For example: In the following expressions identify the terms, factors and coefficients.
$(4x+3y),\,3{{x}^{2}}-4x,\,3{{p}^{2}}q+7pq\,-8p{{q}^{2}}$
Expression Terms Factors Coefficients $4x+3y$ $4x$ $3y$ $4,\,x$ $3,y$ 4 3 $3x{{y}^{2}}-4x$ $3x{{y}^{2}}$ $-4x$ $3,\,x,\,y,\,y$$-4,\,x$ 3 -4 $3{{p}^{2}}q+7pq$$-8p{{q}^{2}}$ $3{{p}^{2}}q$ $7pq$ $-8p{{q}^{2}}$ $3p,\,p,q$$7,\,p,\,q$ $-8p,\,q,\,q$ 3 7 -8
LIKE AND UNLIKE TERMS
In any algebraic expression, terms which have same variable(s) factor are called like terms. Terms which have different variable(s) factors are called unlike terms.
For example of expression $3{{y}^{2}}+2x-2{{y}^{2}}+5,$ in this expression factors of $3{{y}^{2}}$ are 3, y and y, factors of $-2{{y}^{2}}$ are -2, y and y. Thus their variables factors are same so $3{{y}^{2}}$ and $-2{{y}^{2}}$ are like terms whereas $3{{y}^{2}}$ and 2x are unlike terms because their variable factors are different, similarly 2x and 5 are also unlike terms.
TYPES OF ALGEBRAIC EXPRESSIONS
(i) An algebraic expression that has only one term is called a monomial.
For example:
$x,\,4{{x}^{2}}y,\,-3{{p}^{2}}{{q}^{2}}$etc.
(ii) An algebraic expression that contains two unlike terms is called a binomial.
For example:
$x+y,\,3x+4y,\,x-10,\,-y-5$ etc.
(iii) An algebraic expression that contains three unlike terms is called a trinomial.
For example:
$a+b+5,\,{{x}^{2}}-{{y}^{2}}+6,\,\,{{x}^{2}}y+x{{y}^{2}}+xy$ etc.
$zy+10-y$ is not a trinomial as 2y and -y are like terms.
(iv) An algebraic expression that contains more than three unlike terms is called a polynomial.
For example:
${{x}^{3}}+4{{x}^{2}}+7x+3y+5,\,x{{y}^{2}}+7y+2x+3$ etc.
The degree of a Polynomial is the greatest of the exponents (indices) of its various terms.
For example:
(i) $2+x+{{x}^{2}}+{{x}^{3}}$ is a polynomial of degree 3
(ii) $3+2x+5{{x}^{2}}+{{x}^{4}}$ is a polynomial of degree 4
ALGEBRAIC EQUATIONS
An equation is a mathematical statement equating two quantities, e.g. 3x + 4 = 6x + 1, 7x + 2 = 11 + x, 12a + 9 = 8a + 4 etc.
SOLUTION OF EQUATION
It is the value of the unknown that balances an equation e.g.
(i) $x+7=15$
$\therefore \,\,x=8$
(ii) $2x+3=7$
$2x=7-3$
$2x=4\Rightarrow \,x=\frac{4}{2}=2$
Addition (or subtraction) is possible even if terms are like. Addition (or subtraction) of two unlike terms is not possible.
For example:
$2x+3x=(2+3)\,x=5x;$
$5xy+7xy=(5+7)\,xy=12xy$
$8a-3a=(8-3)a=5a$ etc.
$2{{x}^{2}}+4x+3-3{{x}^{2}}-5x+7$
$=(2-3){{x}^{2}}+(4-3)x+(3+7)$
$=-{{x}^{2}}+x+10$
Hence, the sum (or difference) of several like terms is another like term whose coefficient is the sum (or difference) of the coefficients of several like terms.
MULTIPLICATION OF POLYNOMIALS
To multiply one polynomial with the other, (i) multiply each term of the polynomial by each term of the other, and (ii) then add the terms thus obtained.
For example:
(i)$x(a+b)\,=xa+xb$
[Distributive law of multiplication]
(ii) $(x+a)\,(x+b)\,=x(x+b)\,+a(x+b)$
$={{x}^{2}}+xb+xa\,+ab$
$={{x}^{2}}+x(b+a)+ab$.
$={{x}^{2}}+x(a+b)\,+ab$
(iii) $(3x+2)\,(x+4)$
$3x(x+4)\,+2(x+4)$
$=3{{x}^{2}}+12x+2x+8$
$=3{{x}^{2}}+14x+8$
#### Other Topics
##### Notes - Algebra
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# What is the derivative of (lnx)^(sinx)?
Jan 9, 2016
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln \left(x\right)\right)}^{\sin} \left(x\right) \left\{\sin \frac{x}{x \ln \left(x\right)} + \cos \left(x\right) \ln \left(\ln \left(x\right)\right)\right\}$
#### Explanation:
Let $y = {\left(\ln \left(x\right)\right)}^{\sin} \left(x\right)$
To find the derivative of such a problem we need to take logarithms on both the sides.
$\ln \left(y\right) = \ln \left({\left(\ln \left(x\right)\right)}^{\sin} \left(x\right)\right)$
This step is done to move the exponent to the front of the equation.
as color(brown)(ln(a^n) = n*ln(a)
$\ln \left(y\right) = \sin \left(x\right) \cdot \ln \left(\ln \left(x\right)\right)$
Now let us differentiate both sides with respect to $x$
We shall use product rule color(blue)((uv)'=uv'+vu'
(1/y)dy/dx = sin(x)d/dx(ln(ln(x))) + ln(ln(x))d/dx(sin(x)
$\textcolor{b l u e}{\text{Derivative of ln(ln(x)) to be done using chain rule}}$
(1/y)dy/dx = sin(x)(1/ln(x))*d/dx(ln(x)+ln(ln(x))(cos(x))
$\left(\frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \sin \left(x\right) \left(\frac{1}{\ln} \left(x\right)\right) \cdot \frac{1}{x} + \cos \left(x\right) \ln \left(\ln \left(x\right)\right)$
$\left(\frac{1}{y}\right) \frac{\mathrm{dy}}{\mathrm{dx}} = \sin \frac{x}{x \ln \left(x\right)} + \cos \left(x\right) \ln \left(\ln \left(x\right)\right)$
$\frac{\mathrm{dy}}{\mathrm{dx}} = y \left\{\sin \frac{x}{x \ln \left(x\right)} + \cos \left(x\right) \ln \left(\ln \left(x\right)\right)\right\}$
$\frac{\mathrm{dy}}{\mathrm{dx}} = {\left(\ln \left(x\right)\right)}^{\sin} \left(x\right) \left\{\sin \frac{x}{x \ln \left(x\right)} + \cos \left(x\right) \ln \left(\ln \left(x\right)\right)\right\}$ |
# 7: Correlation - What it really means
7: Correlation - What it really means
## Overview
#### Case Study: Baseball
Clark is a big Boston Red Sox fan. Noticing that the Red Sox had a slightly better home record than away, he was interested to see if there is an association between batting averages at away (y) verses home (x) games in the 2018 regular season. Clark included stats from players who played at least 3 away and 3 home games (One player had a batting average of .000 for home games). He got his data from: mlb.com/stats. Clark used Minitab to create a scatterplot of the data.
From his scatterplot, he was surprised to see that the higher batting averages at home were not always higher away. He also noticed two of the two top hitters: Mookie Betts and JD Martinez had data up in the upper right hand corner.
In looking at the data, Clark could not decide what to conclude. He decided to run a correlation in Minitab. He got the following output: r = 0.280, p-value = 0.261. What should he conclude about the batting averages home vs away?
Clark is off to a great start. He realizes that he is working with two quantitative variables (home batting average and away batting average). Let’s take a closer look at the statistical methods he used to come up with his results.
## Objectives
Upon completion of this lesson, you should be able to:
• Use a scatterplot to appropriately graph two quantitative variables
• Apply a correlation technique to two quantitative variables
• Define the difference between a correlation and a covariance
• Identify the magnitude, direction, and linearity of a correlation coefficient.
# 7.1 - Scatterplots
7.1 - Scatterplots
With Clark’s interest in investigating the relationship between two quantitative variables, he correctly started with a scatterplot.
Scatterplot
A graphical representation of two quantitative variables where the explanatory variable is on the x-axis and the response variable is on the y-axis.
When we look at the scatterplot, keep in mind the following questions:
1. What is the direction of the relationship?
2. Is the relationship linear or nonlinear?
3. Is the relationship weak, moderate, or strong?
4. Are there any outliers or extreme values?
We describe the direction of the relationship as positive or negative. A positive relationship means that as the value of the explanatory variable increases, the value of the response variable increases, in general. A negative relationship implies that as the value of the explanatory variable increases, the value of the response variable tends to decrease.
Looking at Clark’s scatterplot, we see a positive direction. As home batting average is increasing in value (moving toward the right on the X axis) away batting average is also increasing in value (moving up the Y axis).
While Clark does not have a lot of data, it is possible to see that the pattern of the data suggests a linear (straight line) pattern. There is no discernable curve to the pattern. While this is not as obvious as we might like it to be, for real-world data, this is pretty typical.
Finally, we want to as about the magnitude of the data. We do this by looking at the degree of the slope (think running up a hill). The slope of the red line indicates a slope that is closer to a zero line (no hill running) then a steep slope (hard hill running).
Now that Clark has taken a closer look at describing his data, let’s take a closer look at analyzing the relationship between home and away batting averages!
# 7.2 - Correlation
7.2 - Correlation
## The Correlation Coefficient
If we want to provide a measure of the strength of the linear relationship between home and away batting averages (two quantitative variables), a good way is to report the correlation coefficient between them.
The sample correlation coefficient is typically denoted as $$r$$. It is also known as Pearson’s $$r$$. The population correlation coefficient is generally denoted as $$\rho$$, pronounced “rho.”
Sample Correlation Coefficient
The sample correlation coefficient, $$r$$, is calculated using the following formula:
$$r=\dfrac{\sum (x_i-\bar{x})(y_i-\bar{y}) }{\sqrt{\sum (x_i-\bar{x})^2}\sqrt{\sum (y_i-\bar{y})^2}}$$
If you have a solid foundation of the material covered in this course up to this point you should notice that the term $$x-\bar{x}$$ (and also $$y-\bar{y}) are simple deviation scores. As you (should) know, the deviation score is the starting point to calculate the variance of a variable. Thus a correlation coefficient is simply the co-variance of two variables! The advantage of the correlation coefficient is that the denominator provides a standardization of the value of the correlation coefficient because it divides the covariance by the product of the standard deviations of the two variables. #### Properties of the Correlation Coefficient, r To summarize, some important properties of the correlation coefficient, r : 1. \(-1\le r\le 1$$, i.e. $$r$$ takes values between -1 and +1, inclusive.
2. The sign of the correlation provides the direction of the linear relationship. The sign indicates whether the two variables are positively or negatively related.
3. A correlation of 0 means there is no linear relationship.
4. There are no units attached to $$r$$.
5. As the magnitude of $$r$$ approaches 1, the stronger the linear relationship.
6. As the magnitude of $$r$$ approaches 0, the weaker the linear relationship.
7. If we fit the simple linear regression model between Y and X, then $$r$$ has the same sign as $$\beta_1$$, which is the coefficient of X in the linear regression equation. -- more on this later.
8. The correlation value would be the same regardless of which variable we defined as X and Y.
# 7.3 - Visualizing Correlation
7.3 - Visualizing Correlation
The following four graphs illustrate four possible situations for the values of r. Pay particular attention to graph (d) which shows a strong relationship between y and x but where r = 0. Note that no linear relationship does not imply no relationship exists!
a) $$r > 0$$
b) $$r < 0$$
c) $$r = 0$$
d) $$r=0$$
With a correlation coefficient, Clark can now make a conclusion about the association between home and away batting averages. With a scatterplot that suggests linearity, Clark can conclude the correlation statistic (r=0.280) is not statistically significant (p>.05). This indicates a player’s home batting average is not associated with how well a player bats away.
Let’s take another look at Clark’s data. Now that we have a good understanding of correlation, we can take a closer look at another measure of association for quantitative variables, the covariance. Let’s revisit the covariance formula.
The variance is a measure of how much any one observation deviates from its mean. In Clark’s example, the variance of the home batting average would be a single player’s batting average minus the mean batting average.
$$\text { Variance }=\dfrac{\sum\left(x_{i}-\bar{x}\right)^{2}}{N-1}$$
Now, the covariance is simply a measure of the variance of two variables. In Clark’s example, the covariance would be the product of a player’s home batting average minus the mean of all home batting averages times that same player’s away batting average minus the mean of all away batting averages.
$$Cov(x,y)=\dfrac{\sum(x_i-\bar{x})(y_i-\bar{y})}{N-1}$$
Now, if you look back at the formula for the correlation, you can see that the correlation is simply the covariance divided by the product of the standard deviations of the two variables. In other words, just as we standardized the variance to create the standard deviation, we standardize the covariance to create the correlation.
$$=\dfrac{\sum(x_i-\bar{x})(y_i-\bar{y})}{(N-1)s_xs_y}$$
Now, why do we even bother talking about the covariance when the correlation is much easier to understand (just like the standard deviation is much easier to understand)? The reason is that as statistical techniques become more advanced the covariance of two variables can tell us a lot about how well our statistics are doing. While we will not get too far into this, it is important to understand this foundational idea of covariance if you progress further into statistics.
To summarize, the correlation is used to test if a significant relationship exists between two quantitative variables, but these variables must be linearly related, which we can see through using a scatterplot before conducting a correlation. While you may likely never see nor use the covariance, it is an important diagnostic tool for more advanced statistical techniques.
# 7.4 - Summary
7.4 - Summary
#### Case-Study: Baseball
Clark correctly identified his data as two quantitative variables. His question about these two variables was simple, is there any relationship between them. By examining the magnitude, direction, and linearity of the scatterplot, we were able to describe the relationship between home and away batting average. Next, the correlation coefficient allowed Clark to conclude that home and away batting averages are not related. So we may not know what to expect from the Red Sox on the road!
[1] Link ↥ Has Tooltip/Popover Toggleable Visibility |
Question
# The midpoint P of the line segment joining the points A (-10, 4) and B (-2, 0) lies on the line segment joining the points C (-9, -4) and D (-4, y). Find the ratio in which P divides CD and also, find the value of y.
Hint: Find the coordinates of the midpoint P of the line segment AB. Then use the section formula of line segment CD for the abscissa of point P to find the ratio in which P divides CD. Use the section formula of line segment CD for the ordinate of point P to find y.
From section formula, if P (x, y) divides the line segment joining $C({x_3},{y_3})$ and $D({x_4},{y_4})$ in the ratio m:n, then:
$x = \dfrac{{m{x_4} + n{x_3}}}{{m + n}};{\text{ }}y = \dfrac{{m{y_4} + n{y_3}}}{{m + n}}{\text{ }}.........{\text{(2)}}$
Substituting equation (1) in equation (2) and using coordinates of C and D, we get:
$- 6 = \dfrac{{m( - 4) + n( - 9)}}{{m + n}}{\text{ }}..........{\text{(3)}}$
$2 = \dfrac{{m(y) + n( - 4)}}{{m + n}}{\text{ }}...........{\text{(4)}}$
Simplifying equation (3) to get the ratio in which P divided CD, we get:
$- 6 = \dfrac{{ - 4m - 9n}}{{m + n}}{\text{ }}$
$- 6(m + n) = - 4m - 9n$
$- 6m - 6n = - 4m - 9n$
$- 6m + 4m = - 9n + 6n$
$- 2m = - 3n$
$\dfrac{m}{n}{\text{ = }}\dfrac{3}{2}{\text{ }}..........{\text{(5)}}$
Simplifying equation (4) to obtain the value of y, we get:
$2 = \dfrac{{my - 4n}}{{m + n}}$
$2(m + n) = my - 4n$
$2m + 2n = my - 4n$
Gathering terms containing m on RHS and terms containing n on LHS, we get:
$4n + 2n = my - 2m$
$6n = m(y - 2)$
Divide both sides by n, to get:
$6 = \dfrac{m}{n}(y - 2)$
Substituting equation (5) in the above equation, we get:
$6 = \dfrac{3}{2}(y - 2)$
Multiply both sides by $\dfrac{2}{3}$ and simplify.
$\dfrac{2}{3} \times 6 = y - 2$
$4 = y - 2$
$y = 6$
Hence, the value of y is 6
Therefore, P divides CD in the ratio 3:2 and the value of y is 6.
Note: The possibility for mistake is writing the section formula for points $C({x_3},{y_3})$ and $D({x_4},{y_4})$ wrongly as $x = \dfrac{{m{x_3} + n{x_4}}}{{m + n}};{\text{ }}y = \dfrac{{m{y_3} + n{y_4}}}{{m + n}}$ instead of $x = \dfrac{{m{x_4} + n{x_3}}}{{m + n}};{\text{ }}y = \dfrac{{m{y_4} + n{y_3}}}{{m + n}}$ . You might also think, it is impossible to find three variables from two equations but you are just finding the ratio between m and n and then the value of y, which requires only two equations. |
# Converting Fractions to Percents
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Lesson Transcript
Instructor: Jennifer Beddoe
There are times when it is important to know how to convert a fraction to a percent. Fortunately, the steps to perform this conversion are not difficult. This lesson will show you those steps and give some examples.
## Let's Make a Deal
Have you ever been at the mall shopping for new clothes, and you see a sign - '1/3 off everything on this rack'? Then you turn and see another sign - 'Everything with a red dot is 35% off'. Your budget is tight, so the sale will really help out. But which sale is a better deal? Fortunately, you know how to convert a fraction to a percent, so you can easily tell where to begin looking for the best deal.
## Definitions
A fraction is a number that represents a part of a whole number. Numbers such as 1/2 and 35/72 are fractions.
A percent is a ratio expressed as a fraction of 100. Examples of percents are 25% and 85%
## How to Convert Fractions to Percents
There are two ways to convert a fraction to a percent. Both take a little math skills, but are not difficult at all.
The first method works best with a calculator, but can be done without one. The steps are as follows:
1. Divide the top number by the bottom number.
2. Multiply that result by 100.
For example -
Convert 3/5 to a percent.
3 divided by 5 is 0.6.
0.6 * 100 is 60.
So, 3/5 is 60%.
The second method is to change the fraction so that the denominator is 100. This works because percent means per 100, so if the fraction has 100 as the denominator, the numerator will be the percent.
The steps for converting fractions to percents using this method are:
1. Change your fraction so that the denominator is 100. Remember that whatever you do to change the denominator, you must do the same to the numerator.
2. The new numerator will be your percent.
Let's convert the same example as before, using this new method.
Convert 3/5 to a percent.
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## College Algebra (10th Edition)
$$\sum_{k=1}^{n}(2k+1) = n^2+2n$$
RECALL: For any constant $c$, (1) $$\sum_{k=1}^{n}(k+c) = \sum_{k=1}^{n}k + \sum_{k=1}^{n}c$$ (2) $$\sum_{i=1}^{n}c = nc$$ (3) $$\sum_{k=1}^{n}k = \dfrac{n(n+1)}{2}$$ (4) $$\sum_{k=1}^{n}ck= c\sum_{k=1}^{n}k$$ Using rule (1) above gives: $$\sum_{k=1}^n(2k + 1) = \sum_{k=1}^{n}2k + \sum_{k=1}^{n}1$$ Using rule (4) above gives: $$2\sum_{k=1}^{n}k + \sum_{k=1}^{n}1$$ Use rule (3) and rule (2) above, respectively, to obtain: $$2\sum_{k=1}^{n}k + \sum_{k=1}^{n}1 \\= 2\left(\dfrac{n(n+1)}{2}\right) + 1(n) \\= n(n+1) + n \\=n^2+n+n \\= n^2+2n$$ Thus, $$\sum_{k=1}^{n}(2k+1) = n^2+2n$$ |
# CIRKILL - Editorial
EASY
Math
### PROBLEM
You are given N points in 2D space. You select a set of 3 points, say A, and another point, say B. What is the probability that
• You can draw a unique circle that goes through each point in A
• B lies inside the circle
### QUICK EXPLANATION
It should be apparent that a you can brute force all possible choices for points in A. Then, iterate over all the possible choices for B. Now, all we need to check is
• Can you draw a unique circle that passes through the 3 points in A
• If you can, then does B lie inside this circle
To check whether you can draw a unique circle that passes through the 3 points in A, you just need to verify that the 3 points in A are not colinear. This can be done by finding the area of the triangle that passes through the 3 points, and check that this area is non-zero.
This particular check can be done without performing any divisions. As we will see in the rest of this editorial, the solution proposed needs no divisions and avoids any issues with precision.
### EXPLANATION
Given three points that are not collinear, we wish to find the equation of the circle that passes through these 3 points. The general form of the equation of a circle in co-ordinate geometry is
```(x2 + y2) + Ax + By + C = 0
```
Since we already know the co-ordinate of three points that this circle passes through, we already have 3 equations for 3 unknowns A, B and C.
```(x12 + y12) + Ax1 + By1 + C = 0
(x22 + y22) + Ax2 + By2 + C = 0
(x32 + y32) + Ax3 + By3 + C = 0
```
We can use Cramer’s Rule to find the values of A, B and C.
```A =
| (x12+y12) y1 1 |
- | (x22+y22) y2 1 |
| (x32+y32) y3 1 |
-----------------------
| x1 y1 1 |
| x2 y2 1 |
| x3 y3 1 |
```
```B =
| (x12+y12) x1 1 |
| (x22+y22) x2 1 |
| (x32+y32) x3 1 |
-----------------------
| x1 y1 1 |
| x2 y2 1 |
| x3 y3 1 |
```
```C =
| (x12+y12) x1 y1 |
- | (x22+y22) x2 y2 |
| (x32+y32) x3 y3 |
-----------------------
| x1 y1 1 |
| x2 y2 1 |
| x3 y3 1 |
```
substituting these values in the equation above, gives us the following determinant, which is the equation of the circle that passes through 3 points
```| (x2 + y2) x y 1 |
| (x12+y12) x1 y1 1 | = 0
| (x22+y22) x2 y2 1 |
| (x32+y32) x3 y3 1 |
```
Now, we can put the co-ordinates of chosen point B in the above determinant and perform only integer calculations to find whether the point is inside the circle or not. If the result is 0, the point is on the circle. If the result is of the same sign as the sign of the term (x2 + y2) in the above equation, then the point is outside the circle.
Lastly, I request you to read kuruma’s wonderful post about his experience solving this problem. His approach involved actually finding the co-ordinates of the center of the circle. This works of course, but one has to be very careful to obtain a numerically stable answer.
### SETTER’S SOLUTION
Can be found here.
### TESTER’S SOLUTION
Can be found here.
7 Likes
but how to generate the no of cases for a given input of co-ordinates???
There is a direct formula to find the circumcenter of a triangle, then you can equate the radius with the co-ordinates of ASH to find whether or not he comes inside the circle ! simple approach but it lacked arithmetic precision, had to check +/- 10^-6 with radius to get AC !
I solved this by finding the equation of 2 perpendicular bisectors by getting slope of perpendiculars and then equation of line (this line passes through the mid-point of the side).then get the co-ordinates of the circumcentre by finding point of intersection of the 2 lines get radius. Then check distance of remaining points from circumcentre to see if it was less than or equal to radius. I was using Python so did not have problem with precision but it might cause problem in other languages. Here is [link to my solution.](point of intersection)
1 Like
I tried this problem a lot of times and got WA , I think my approach is right but there is some precision issue, I am writing here my approach and my solution link , please anybody go through my solution and it will nice for me if some one can detect a problem.
Approach :
Select a point (i), and select another three point(j, k, l) and check that the points can form a unique circle or not.
If they can form , than check that point i (arr*[0], arr*1) is in the circle or not, if it is in than dead counter will be increased , every time increase the total case counter , than the final answer will be (dead/total) .
I followed the line intersection approach as given in the above link…
While computing the values of Center’s X coordinate and Center’s Y coordinate I was doing instead of these
double x = (B2C1 - B1C2)/det
double y = (A1C2 - A2C1)/det
I was just doing this
double x = (B2C1 - B1C2)
and
double y = (A1C2 - A2C1)
and suppose the 4th Point is
t1,t2
then the radius is computed by
((t1 * det) - x)^2 + ((t1 * det -y)^2
I dint need to divide the entire numerator by det because while finding another radius it the numerator will again be divided by det again… so I can totally eradicate divion by this process
i used basic Geometry to check whether a triangle can be formed or not, then found the center of the circle which passes through these 3 points (i=0, i<n-2, i++ then, j=i+1, j<n-1 ,j++ then k=j+1, k<n, k++ ) then the 4th for loop of l(l!=i or j or k)… let distance b/w l and centre is d… if d<=r(radius) then its counts as Ash catchem’s death
but wa due to something !?!
@problem setter, does the answer depend on the way the points are given ?!?
Thank you @gamabunta for mentioning my post and also for teaching me a new and very cool method with the usage of Cramer’s rule!!! I hope I can keep learning a lot more!
3 Likes
see this also…same approach lyk kuruma http://www.codechef.com/viewsolution/2334403
In case when x1=x2 (in is_circleformed), notice that x will be nan , i.e not a number (same problem happens whenever the denominator becomes 0 in any of the other case . m1,m2) …so it is giving wa…and also double gave me WA, precision issues i guess, long double worked…
Hello @srinu634, double is enough you can refer to my explanation to see its so…
@upendra1234, your approach is obviously correct, but the way you did those things in terms of precision or not, could make all the difference in AC or WA status… For some precision discussion, check my solution and my non official editorial for this problem |
1. Chapter 4 Class 12 Determinants
2. Serial order wise
Transcript
Step 2 Calculate |A| |A| = 2111−2−103−5 = 2 −2−13−5 – 1 1−10−5 + 1 1−203 = 2 (10 + 3 ) – 1 (– 5 + 0) + 1 (3 – 0) = 2 (13) – 1 (– 5 ) + 21 (3) = 34 Thus, |A| ≠ 0 ∴ The system of equation is consistent & has a unique solution Now, AX = B X = A-1 B Step 3 Calculate X = A-1 B A-1 = 1|A| adj (A) adj (A) = A11 A12 A13 A21 A22 A23 A31 A32 A33′ = A11 A21 A31 A12 A22 A23 A13 A32 A33 A = 2111−2−103−5 M11 = −2−13−5 = 15 + 3 = 13 M12 = 1−10−5 = – 5 + 0 = – 5 M13 = 1−203 = 3 + 0 = 3 M21 = 113−5 = – 5 – 3 = – 8 M22 = 210−5 = – 10 + 0 = – 10 M23 = 2103 = 6 + 0 = 6 M31 = 11−2−1 = – 1 + 2 = 1 M32 = 211−1 = – 2 – 1 = – 3 M33 = 211−2 = – 4 – 1 = – 5 A11 = ( – 1)1+1 . M11 = ( –1 )2 . (13) = 13 A12 = ( – 1)1+2 . M12 = ( –1 )3 . (– 5) = 5 A13 = ( – 1)1+3 . M13 = ( –1 )4 . (3) = 3 A21 = ( – 1)2+1 . M21 = ( –1 )3 . ( – 8) = 8 A22 = ( – 1)2+2 . M22 = ( –1 )4 . ( – 10) = – 10 A23 = ( – 1)2+3 . M23 = ( –1 )5 . (6) = – 6 A31 = ( – 1)3+1 . M31 = ( –1 )4 . (1) = 1 A32 = ( – 1)3+2 . M32 = ( –1 )5 . (– 3) = 3 A33 = ( – 1)3+3 . M33 = ( –1 )6 . (– 5) = – 5 Thus, adj (A) = 13815−1033−6−5 Now, A-1 = 1|A| adj A Putting values = 134 13815−1033−6−5 Also, X = A-1 B Putting values 𝑥𝑦𝑧 = 134 13815−1033−6−5 1 329 𝑥𝑦𝑧 = 134 13 1+8 32+1 95 1+ −10 32+3 93 1+ −6 32+ −5 9 𝑥𝑦𝑧 = 134 13+12+95−15+273−9−45 = 134 3417−51 𝑥𝑦𝑧 = 1 12 −32 Hence x = 1, y = 𝟏𝟐 & z = −𝟑𝟐 |
Licchavi Lyceum
ll
# NCERT Solutions For Class 10 Math Chapter 1 Real Numbers Ex. 1.1
Question 1. Use Euclid’s Division Algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii) 867 and 255
Solution:
(i) 135 and 225
Step 1: Apply the Division Algorithm
Divide 225 by 135:
$$225 = 135 \times 1 + 90$$
Here, 225 divided by 135 gives a quotient of 1 and a remainder of 90.
Step 2: Replace and Repeat
Now, replace 225 with 135 and 135 with 90, then repeat the division process:
Divide 135 by 90:
$$135 = 90 \times 1 + 45$$
Here, 135 divided by 90 gives a quotient of 1 and a remainder of 45.
Step 3: Replace and Repeat Again
Now, replace 135 with 90 and 90 with 45, then repeat the division process:
Divide 90 by 45:
$$90 = 45 \times 2 + 0$$
Here, 90 divided by 45 gives a quotient of 2 and a remainder of 0. Since the remainder is now 0, the divisor at this step, which is 45, is the HCF of 135 and 225. Therefore, the HCF of 135 and 225 is 45.
(ii) 196 and 38220
Step 1: Apply the Division Algorithm
Divide 38220 by 196:
$$38220 = 196 \times 195 + 0$$
Here, 38220 divided by 196 gives a quotient of 195 and a remainder of 0.
Since the remainder is now 0, the divisor at this step, which is 196, is the HCF of 196 and 38220. Therefore, the HCF of 196 and 38220 is 196.
(iii) 867 and 255
Step 1: Apply the Division Algorithm
Divide 867 by 255:
$$867 = 255 \times 3 + 102$$
Here, 867 divided by 255 gives a quotient of 3 and a remainder of 102.
Step 2: Replace and Repeat
Now, replace 867 with 255 and 255 with 102, then repeat the division process:
Divide 255 by 102:
$$255 = 102 \times 2 + 51$$
Here, 255 divided by 102 gives a quotient of 2 and a remainder of 51.
Step 3: Replace and Repeat Again
Now, replace 255 with 102 and 102 with 51, then repeat the division process:
Divide 102 by 51:
$$102 = 51 \times 2 + 0$$
Here, 102 divided by 51 gives a quotient of 2 and a remainder of 0. Since the remainder is now 0, the divisor at this step, which is 51, is the HCF of 867 and 255. Therefore, the HCF of 867 and 255 is 51. |
# Direct Proportion Formula
Here we will learn about direct proportion formulas, including what the proportion formulas are and how to interpret them.
There are also direct proportion formula worksheets based on Edexcel, AQA and OCR exam questions, along with further guidance on where to go next if you’re still stuck.
## What is the direct proportion formula?
The direct proportion formula is an algebraic formula which represents the directly proportional relationship between two variables.
Direct proportion is one type of a proportionality relationship. As one value increases, so does the other value.
The proportionality symbol is . If the variables x and y are directly proportional then we can write the relationship using the proportionality symbol.
y\propto x
We can also write this as a formula using the constant of proportionality, k.
k is a constant value that links the two variables.
y=kx
Proportional relationships can also be represented by graphs.
The following is a type of direct proportion graph.
It is a straight line graph going through the origin. It has the equation y=kx.
Direct proportion formulas can involve powers and roots. If the variables were x and y, and y is directly proportional to x^{2}, we can write the relationship using the proportionality symbol.
y\propto x^2
We can also write this as a formula using the constant of proportionality k.
y=kx^2
The graph of this relationship would be quadratic and have the shape of a parabola.
## How to recognise a direct proportion formula
In order to recognise a direct proportion formula:
1. Look for an equation in the form y=kx^n.
2. Check there are no other terms in the equation.
## Direct proportion formula examples
### Example 1: recognise the direct proportion formula
Which of these equations indicate that y\propto x?
A B C D
y=x+5 \hspace{1cm} y=\frac{5}{x} \hspace{1cm} y=1-5x \hspace{1cm} y=5x
1. Look for an equation in the form y=kx^n.
We are looking for a direct proportion equation. We are looking for an equation with the variable, x^n, being multiplied by a number.
We can eliminate equation B as this has a \frac{1}{x} term, this would be inverse proportion.
2Check there are no other terms in the equation.
For direct proportion there can be no addition or subtraction involved in the equation.
So we can eliminate equations A and C.
The only equation which only has an x^n -term being multiplied by a number is equation D. Here the power of the x is 1.
y=5x indicates that y \propto x.
### Example 2: recognise the direct proportion formula
Which of these equations indicate that a direct proportion?
A B C D
y=\frac{4}{x} \hspace{1cm} y=4x^2 \hspace{1cm} y=4x-3 \hspace{1cm} y=4x+2
Look for an equation in the form y=kx^n.
Check there are no other terms in the equation.
## How to construct a direct proportion formula
In order to construct a direct proportion formula:
1. Write down the general proportion formula.
2. Find the constant \textbf{k} , the constant of proportionality.
3. Write down the equation.
## Constructing direct proportion formula examples
### Example 3: constructing a direct proportion equation
y is directly proportional to x.
When y=35, \ x=5.
Find a formula for y in terms of x.
Write down the general proportion formula.
Find the constant of proportionality.
Write down the equation.
### Example 4: constructing a direct proportion equation
y is directly proportional to x.
When y=4, \ x=20.
Find a formula for y in terms of x.
Write down the general proportion formula.
Find the constant of proportionality.
Write down the equation.
### Example 5: constructing a direct proportion equation involving a power or root
y is directly proportional to \sqrt{x}.
When y=40, \ x=4.
Find a formula for y in terms of x.
Write down the general proportion formula.
Find the constant of proportionality.
Write down the equation.
### Common misconceptions
• Confusing direct proportion and inverse proportion
You will need to learn which formula is for which type of proportionality.
Direct proportionInverse proportion
y\propto x^n y\propto \frac{1}{x^n}
y=kx^n y=\frac{k}{x^n}
• Not recognising and applying any powers or roots
Most direct proportional formulas just involve x, but they can involve powers such as x^{2} or roots such as \sqrt{x} . \sqrt{x} is the same as x^{\frac{1}{2}} .
• Assuming the constant of proportionality can only be an integer
The constant of proportionality can be an integer (a whole number), but they can also be decimals or fractions.
For example,
y=\frac{x}{2}
Even though there is a fraction, the constant of proportionality is \frac{1}{2} or 0.5.
y=\frac{x}{2}=\frac{1}{2} \times x=0.5 \times x
## Related lessons on direct proportion formula
Direct proportion formula is part of our series of lessons to support revision on proportion. You may find it helpful to start with the main proportion lesson for a summary of what to expect, or use the step by step guides below for further detail on individual topics. Other lessons in this series include:
### Practice direct proportion formula questions
1. Which of these equations does NOT indicate y \propto x?
y=2x
y=x+2
y=\frac{x}{2}
y=20x
For direct proportion there can be no addition or subtraction involved in the equation. The incorrect equation is y=x+2.
2. Which of these equations indicate y \propto x?
y=x+5
y=5x
y=\frac{5}{x}
y=5x^2
For direct proportion there can be no addition or subtraction involved in the equation. The correct equation is y=5x.
3. Which of these equations indicates direct proportion?
y=4-10x
y=10x+4
y=\frac{10}{x^2}
y=10x^3
For direct proportion there can be no addition or subtraction involved in the equation. The correct equation is y=10x^3.
4. y is directly proportional to x.
When y=24, \ x=8.
Find a formula for y in terms of x.
y=4x
y=\frac{3}{x}
y=\frac{x}{4}
y=3x
Write down the relationship between y and x and substitute the given values to find k, the constant of proportionality.
\begin{aligned} y&\propto x \\\\ y&= kx \\\\ 24 &= k\times 8 \\\\ k&=24\div 8=3 \end{aligned}
So, the equation is
y=3x.
5. y is directly proportional to x.
When y=10, \ x=5.
Find a formula for y in terms of x.
y=50x
y=2x
y=\frac{x}{2}
y=\frac{2}{x}
Write down the relationship between y and x and substitute the given values to find k, the constant of proportionality.
\begin{aligned} y&\propto x \\\\ y&= kx \\\\ 10&= k\times 5 \\\\ k&=10\div 5=2 \end{aligned}
So, the equation is
y=2x.
6. y is directly proportional to x^{2}.
When y=18, \ x=3.
Find a formula for y in terms of x^{2}.
y=6x
y=2x^2
y=6x^2
y=3x
Write down the relationship between y and x and substitute the given values to find k, the constant of proportionality.
\begin{aligned} y&\propto x^2 \\\\ y&= kx^2 \\\\ 18&= k\times 3^2 \\\\ 18&= 9k \\\\ k&=18\div 3=2 \end{aligned}
So, the equation is
y=2x^2.
### Direct proportion formula GCSE questions
1. y is directly proportional to x and k is a constant.
Identify the correct equation.
Equation A – y=\frac{k}{x}
Equation B – y=kx
Equation C – y=k-x
Equation D – y=\frac{x}{k}
(1 mark)
Equation B
(1)
2. y is proportional to x.
When y=36, \ x=12.
(a) Find the equation for y in terms of x.
(b) Find the value of x when y=50.
(4 marks)
(a)
\begin{aligned} y&=kx \\\\ 36&=k \times 12 \\\\ k&=36 \div 12=3 \end{aligned}
(1)
y=3x
(1)
(b)
50=3x \ so \ x=50 \div 3
(1)
x=\frac{50}{3}=16\frac{2}{3}
(1)
3. m is directly proportional to p .
When p=5, \ m=20.
Find m when p=3.
(3 marks)
m=kp \ or \ 20=k \times 5
(1)
k=4 \ or \ m=4p
(1)
m=12
(1)
## Learning checklist
You have now learned how to:
• Recognise a direct proportion formula
• Construct a direct proportion equation
• Model situations by translating them into algebraic formulae
## Still stuck?
Prepare your KS4 students for maths GCSEs success with Third Space Learning. Weekly online one to one GCSE maths revision lessons delivered by expert maths tutors.
Find out more about our GCSE maths tuition programme. |
If your student has trouble with percentages, it is essential to troubleshoot the problem early, as future math concepts build upon prior knowledge. Learning the basics of percentages may begin as early as third grade and should play an important role through eighth grade, according to The National Council of Teachers of Mathematics. A student needs to understand the meaning of percent, its visual representation and its relationship to decimals and fractions.
## Understand the Term
Knowing that the "cent" part of the word "percent" means "100" can act as a starting point for understanding. Khan Academy recommends associating the 100 years in a century with this term. The “century” becomes the whole, and the “100 years” represents the parts of the whole. In other words, the word “percent” means "per 100." In addition, an NCTM Illuminations activity suggests that you relate percents to everyday events. A teacher could ask, "What does it mean to score 100 percent on a spelling test?" or "What does it mean to have 50 percent of a candy bar?" or "If 4 percent of 100 parking places should be available for people with disabilities, what does that mean? How many spaces would that be?” Questions like these can assess where students need to begin.
## Create Grids
By using grids of 100 squares to demonstrate percents, teachers can demonstrate the “parts” and the “whole.’ If the students color 15 small parts out of 100, they can visualize 15 percent. If they color in all 100 parts, then they have colored 100 percent of the grid or an entire large square. Christopher Scaptura and other mathematics instructors who collaborated at George Mason University, propose using the 10-by-10 grid as an artwork assignment. The students can devise their own designs by color and then compute the percentage of each color. The artwork engages the students and promotes understanding.
## Understand Percents Over 100 Percent
Often, a figure like 200 percent confuses students, because they might assume the value means 200 times more. By using two large squares, each divided into 100 parts, students can see what percents over 100 means visually. For example, filling in 100 parts of the first large square and 25 parts of the second square will equal 125 percent. If a student thinks the answer should be 125 out of 200, remind him that percent refers only to parts out of 100. Once a student fills in all 200 smaller parts, he will realize that he has filled in two large wholes. Therefore, 200 percent refers to two large squares, not 200.
## Apply the Concepts
Viewing an interactive visual model allows students to compare percents to other concepts. One Illuminations model allows students to experiment with percents, fractions and decimals. At first, the student can view the numerator and denominator 1/1 converted to 100 percent, a 1.0 decimal or one purple rectangle. As the student makes changes, moving the numerator to 2/1 or 200 percent, she will see two rectangles and a decimal of 2.0. If she moves to one-half, she will see half a rectangle and 50 percent or 0.5. Such experimentation can engage a student and encourage an interest in mathematics. |
# How do you evaluate the expression: 28 ÷ 2 ÷ 2 ÷ 2?
Jul 17, 2016
$3 \frac{1}{2}$ by a variety of methods, but take care!
#### Explanation:
As there is only one operation (all division) we work from left to right.
$28 \div 2 \div 2 \div 2$
=$14 \div 2 \div 2$
=$7 \div 2$
= $3 \frac{1}{2}$
However, it is important to note that division is not commutative.
We also CANNOT do the following:
$28 \div 2 \div \left(2 \div 2\right)$
This will give $14 \div 1 = 14$
Dividing by a divide should make a multiplication!
= $28 \div 2 \div \left(2 \times 2\right) = 14 \div 4 = 3 \frac{1}{2}$
We CAN do the following:
$28 \div \left(2 \times 2 \times 2\right) = 28 \div 8 = 3 \frac{1}{2}$
We could also write the expression as a fraction:
$\frac{28}{2 \times 2 \times 2} = \frac{7}{2} = 3 \frac{1}{2}$ |
Set Theory and Logic Supplementary Materials Math 103: Contemporary Mathematics with Applications. A. Calini, E. Jurisich, S.
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1 Set Theory and Logic Supplementary Materials Math 103: Contemporary Mathematics with Applications A. Calini, E. Jurisich, S. Shields c2008
2 2
3 Chapter 1 Set Theory 1.1 Basic definitions and notation A set is a collection of objects. For example, a deck of cards, every student enrolled in Math 103, the collection of all even integers, these are all examples of sets of things. Each object in a set is an element of that set. The two of diamonds is an element of the set consisting of a deck of cards, one particular student is an element of the set of all students enrolled in Math 103, the number 4 is an element of the set of even integers. We often use capital letters such as A to denote sets, and lower case letters such as a to denote the elements. Definition 1. Given a set A, ifu is an element of A we write If the element u is not in the set A we write u A. u/ A. Some sets that you may have encountered in mathematics courses before are: The integers Z The even integers 2Z The set of rational numbers Q The set of real numbers R. We can now practice using our element notation: Example We have 4 2Z. 3
4 4 CHAPTER 1. SET THEORY Example Z, Example / 2Z. Example / Q So far, we have been defining sets by describing them in words. We can also specify some sets by listing their elements. For example, define the set T by writing T = {a, b, c, d, e}. When defining a set by listing, always use the brackets {, }. Another set that we can define by listing is the set of natural numbers N = {0, 1, 2, 3, 4, }, where we have indicated a general pattern (hopefully easily regognized!) by writing. Many sets cannot be listed so easily (or at all for that matter), and in many of these cases it is convenient to use a rule to specify a set. For example, suppose we want to define a set S that consists of all real numbers between 1 and 1, inclusive. We use the notation S = {x x R and 1 x 1}. We read the above as S equals the set of all x such that x is a real number and x is greater than or equal to 1, and less than or equal to 1. What happens if someone specifies a set by a rule like x is a negative integer greater than 1000? What should we do? There are no numbers that are negative and greater than We allow examples of rules of this kind, and make the following definition: Definition 2. The empty set is the set with no elements, and is denoted by the symbol φ, or by {}. Thus, the above set {x x Z,x<0 and x>1000} = {}= φ. Definition 3. Two sets are equal if they have exactly the same elements, denoted If A and B are not equal, we write A = B. A = B. Example Let T = {a, b, c, d, e} and let R = {e, d, a, c, b}. We can check that T and R have exactly the same elements, so T = R. Example Let S = {x x Z and x 0}, and let A = {3n n Z}. We can see that S = A because A consists of all integer multiples of 3, hence 3 A but 3 / S. Thisshows S = A.
5 1.2. SUBSETS 5 As we have seen from our examples, sets may contain a finite number of elements, or an infinite number of elements. Examples of finite sets include T from Example 1.1.5, and also the set of students enrolled in Math 103. Examples of infinite sets are Z and R. Definition 4. If a set S is finite, we let n(s) denote the number of elements in S. Example Let T be as in Example 1.1.5, then n(t ) = Subsets One important relation between sets is the idea of a subset. Given sets A and B, wesay B is a subset of A if every element of B is also an element of A. We denote this as Example {2, 4, 6} 2Z. B A. Example Let A = {a, b, c, d, e}, and B = {a, e} then B A. Example Let s list all subsets of A from Example that have four elements: {a, b, c, d}, {a, b, c, e}, {a, b, d, e}, {a, c, d, e}, {b, c, d, e}. For any set A, since every element of A is in A we have A A. This says that a set is always a subset of itself. We also consider the empty set to be a subset of any set A, φ A. Let S = {a, b, c, d}, let slistall subsets of the set S = {a, b, c, d}. To organize our work, we will list them by size. Table 1.1: Subsets of S number of elements subsets 0 φ 1 {a}, {b}, {c}, {d} 2 {a, b}, {a, c}, {a, d}, {b, c}, {b, d}{c, d} 3 {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d} 4 {a, b, c, d} We have listed all of the subets of S. Notice that there are 16 of them. In fact, one can prove the following theorem by using methods of counting covered later in this course. Theorem Let S be a set having N elements. Then there are 2 N subsets of S.
6 6 CHAPTER 1. SET THEORY 1.3 Union, Intersection, and Complement Let U be a set. Given two subsets A and B of U we define the union of A and B to be the subset of U that contains all elements that are in A, or in B, or possibly in both. The union of A and B is denoted A B. In our rule notation A B = {x U x A or x B, or both}. Example Let U = {1, 2, 3, 10}. Let S = {2, 4, 6, 8, 10}, T = {5, 6, 7, 8}. Then S T = {2, 4, 5, 6, 7, 8, 10}. We often use what is known as a Venn diagram to illustrate sets. In a Venn diagram circles are used to represent subsets of a set U (denoted by a large rectangle). Here is a Venn diagram illustrating A B. Figure 1.1: A B We have the following facts about the union: 1. A φ = A 2. A A = A 3. A B = B A 4. (A B) C = A (B C) We define the intersection of subsets A and B of U to be the subset of U that contains all of the elements that are in both A and B. The intersection of A and B is denoted A B. We have A B = {x U x A and x B}. Here is a Venn diagram illustrating the intersection:
7 1.3. UNION, INTERSECTION, AND COMPLEMENT 7 Figure 1.2: A B Example If U, S and T are given as in Example above, then S T = {6, 8}. We have the following facts about the intersection: 1. A φ = φ 2. A A = A 3. A B = B A 4. (A B) C = A (B C) Given the two operations, we can apply them in combination, as long as we remember to use parenthesis to indicate in what order the operations should be performed. Example Let U = {a, b, c, d, e, f, g}, lets = {a, e},h = {a, b, c, d},k = {a, c, e, f}. Then (S H) K = {a} K = {a, c, e, f}, S (H K) =S {a, b, c, d, e, f} = {a, e}, (S H) K = {a}. Notice that (S H) K = S (H K). It is important to always use parenthesis in the appropriate place when working with three or more sets, statements like A B C D do not have one interpretation so do not actually specify a set. The exception is when all operations are the same, as in properties (4) of intersection and union. For example A B C =(A B) C = A (B C).
8 8 CHAPTER 1. SET THEORY Example Venn diagrams illustrating the sets (A B) C, and A (B C). Figure 1.3: (A B) C A (B C) Given A a subset of U, thecomplement of A is the subset of U consisting of all elements not in A. The complement of A is denoted A. Figure 1.4: A Example: Let U = { 4, 3, 2, 1, 0, 1, 2, 3, 4, },letn = {0, 1, 2, 3 }.Then N = { 1, 2, 3, 4, }. The complement satisfies the following rules: 1. (A ) = A 2. U = φ and φ = U 3. A A = U 4. A A = φ
9 1.3. UNION, INTERSECTION, AND COMPLEMENT 9 Example Let D be the set of a standard deck of cards. Let R be the subset of red cards, let F be the subset of face cards. (The face cards include all suits of K, Q, J.) Find the following sets: (R F ), R F, R F. The set R F = {A,A, 2, 2, K,K,K,K,Q,Q,J,J } i.e. consists of all cards that are either red, or black face cards. The complement of R F consists of the cards not listed above and is (R F ) = {A,A, 2, 2,, 10, 10 }. The set R is the set of black cards, the set F is the set of non-face cards (of any suit), so the intersection is the set of black non-face cards: R F = {A,A, 2, 2,, 10, 10 }. This is the same set as (R F ). Now let s find R F the union of the black cards and the non-face cards. R F = {A,A, 2, 2, 10, 10,A,A, 2, 2, J,J,Q,Q,K,K }. We see that R F is not equal to (R F ). Theorem De Morgan s Laws: Given two sets A, B U, (A B) = A B (A B) = A B Example Fill in the Venn diagrams for (A B), and for A B. Figure 1.5: (A B) A B
10 10 CHAPTER 1. SET THEORY If we are interested in elements of a set A that are not contained in a set B, we can write this set as A B. This concept comes up so often we define the difference of two sets A and B: A B = A B, Figure 1.6: A B For example, if S is the set of all juices in the supermarket, and T is the set of all foodstuffs in the supermarket with added sugar, then S T is the set of all juices in the market without added sugar. 1.4 Cardinality and Survey Problems If a set S is finite, recall that n(s) denotes the number of elements in S. Example Let D denote a standard deck of cards. n(d) = 52. Theorem If A and B are both finite sets, then n(a B) =n(a)+n(b) n(a B) To see how this theorem works, lets consider our set D. Let the set of all red cards be denoted R, and let the set of face cards be denoted F. How many elements are in R F? We can count them as listed in Example 1.3.5, or we can use the formula. The intersection consists of the six red face cards: {K,K,Q,Q,J,J }. Using the formula gives n(r F )=n(r)+n(f ) n(r F ) = = 32. What we should not do is simply add the number of red cards to the number of face cards, if we do that we have counted the red face cards twice. We can use our formula for the number of elements to analyze surveys. Example Suppose Walter s online music store conducts a customer survey to determine the preferences of its customers. Customers are asked what type of music they
11 1.4. CARDINALITY AND SURVEY PROBLEMS 11 like. They may choose from the following categories: Pop (P), Jazz (J), Classical (C), and none of the above (N). Of 100 customers some of the results are as follows: 44 like Classical 27 like all three 15 like only Pop 10 like Jazz and Classical, but not Pop How many like Classical but not Jazz? We can fill in the Venn diagram below to keep track of the numbers. There are n(c) = 44 total that like Classical, and n(c J) = = 37 that like both Jazz and Classical, so = 7 like Classical but not Jazz. Example Let s look at some more survey results from Example 1.4.3: 78 customers like Jazz or Pop (or possibly both). 19 customers marked None of the above when asked what they like. 12 like Jazz and Pop, but not classsical. How many like only Jazz? To answer this, let s fill in more of the diagram:
12 12 CHAPTER 1. SET THEORY We have n(c) = 44, n((p J C) ) = 19. If we let j be the number of surveyed customers who like only Jazz, then because there are 100 surveyed customers, we see j = 100. Solving for j gives j = 10. How many like Pop and Classical, but not Jazz? We know that n(p J) = 78. Using the diagram, the number who like Pop and Classical, but not Jazz is = 4.
13 1.5. CARTESIAN PRODUCTS Cartesian Products You may recall the Cartesian plane R 2 which is the set of all points in the plane. This set consists of ordered pairs of numbers (x, y) wherex and y are real numbers. The point (1, 2) is not the same as (2, 1). We use round brackets (, ) to denote ordered pairs, reserving the brackets {, } for sets. We can make a more general definition involving ordered pairs: Given two sets A, B we define the Cartesian product to be Example A B = {(a, b) a A and b B}. {2, 3, 4} {7, 9, 10} = {(2, 7), (2, 9), (2, 10), (3, 7), (3, 9), (3, 10), (4, 7), (4, 9), (4, 10)} Theorem If A and B are two finite sets, then the number of elements in the Cartesian product A B is given by n(a B) =n(a) n(b). Example If we roll two dice, and create a set of all possible results. How many elements are there? We can think of the possible results of rolling dice as a set of ordered pairs. Let D1 denote the set of possible results of rolling the first die D1 ={1, 2, 6}, and let D2 denote the set of possible results of rolling the second die, D2 ={1, 2, 6}. There are 6 6 = 36 possible results from rolling the pair: (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Example Write out the subset of D1 D2 that represents cases where the sum of the numbers showing is either 7 or 11. How many elements are in this set? The subset we are looking for is {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1)} {(5, 6), (6, 5)} = {(1, 6), (2, 5), (3, 4), (4, 3), (5, 2), (6, 1), (5, 6), (6, 5)}. and the number of elements is = 8.
14 14 CHAPTER 1. SET THEORY 1.6 Excercises 1. Let U = {1, 2, 3, 4, 5,, 10} A = {2, 4, 6, 8, 10} B = {3, 6, 9} C = {1, 2, 3, 8, 9, 10} perform the indicated operations (a) A B (b) A B (c) A C (d) (A C) (e) (A B) C (f) (A B) A 2. Determine if the following statements are true or false. Here A represents any set. (a) φ A (b) A A (c) (A ) = A 3. Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and A = {1, 3, 5, 7, 9} and B = {1, 4, 5, 9}. (a) Find A B (b) Find A B (c) Use a Venn diagram to represent these sets. 4. Let U be the set of integers. Let A = {3, 6},B = {3, 8, 10, 12} and C = {6, 8, 10}. Perform the indicated operations. (a) B C (b) (A C ) (c) (A C) B (d) B C (e) C B (f) B C (g) A C 5. Use Venn diagrams to verify DeMorgan s laws. 6. Represent the following sets with a Venn diagram (a) (B C) A
15 1.6. EXCERCISES 15 (b) (A B) C 7. Denote the set A = {x x Z and x<3} by the listing method. 8. A proper subset of a set A is one that is not equal to the set A itself. If a set has 6 elements, how many proper subsets does it have? 9. Describe the shaded region using,,, : a) b) c) d) e) f)
16 16 CHAPTER 1. SET THEORY 10. One hundred students were surveyed and asked if they are currently taking math (M), English (E) and/or History (H) The survey findings are summarized here: Table 1.2: Survey Results n(m) = 45 n(m E) = 15 n(e) = 41 n(m H) = 18 n(h) = 40 n(m E H) =7 n[(m E) (M H) (E H)] = 36 (a) Use a Venn diagram to represent this data. (b) How many students are only taking math? 11. Ninety people at a Superbowl party were surveyed to see what they ate while watching the game. The following data was collected: 48 had nachos. 39 had wings. 35 had a potato skins. 20 had both wings and potato skins. 19 had both potato skins and nachos. 22 had both wings and nachos. 10 had nachos, wings and potato skins. (a) Use a Venn diagram to represent this data. (b) How many had nothing? 12. In Example how many pairs sum to an even number, or one greater than 9? 13. Let D be a standard deck of cards, let S = {A,K,Q,J }. a) List all the subsets of S that contain both the A and K. How many subsets of this type are there? Discuss why this is the same number as all the subsets of {Q,J }. b) How many subsets of D contain both the A and K? 14. Answer the following True or False. (a) {1, 2, 3} is a subset of {3, 2, 1, 4}. (b) {3, 2, 1, 4} is a subset of {1, 2, 3}.
17 1.6. EXCERCISES 17 (c) The empty set is a subset of every set. (d) 1 is an element of {3, 2, 1, 4}. (e) {1} is an element of {3, 2, 1, 4}. (f) {1} is a proper subset of {3, 2, 1, 4} (g) {3, 2, 1, 4} = {1, 2, 3, 4} (h) (0, 1/2) is an element of Q Z (i) (0, 1/2) is an element of Z Q (j) ( 7/8, 0) is an element of Q Q (k) ( 7/8, 0) is an element of Z Z
18 18 CHAPTER 1. SET THEORY 1.7 Solutions to exercises 1. a) {6}, b){2, 3, 4, 6, 8, 9, 10}, c){1, 3, 9}, d){1, 3, 4, 5, 6, 7, 9}, e){2, 3, 8, 9, 10}, f)a. 2. a) T b) F c) T. 3. a) {1, 3, 4, 5, 7, 9}, b){1, 5, 9} 4. a) {3, 12}, b){6}, c){3, 8, 10}, d){3, 12}, e){6}, f) all integers except 6, g) {(3, 6), (3, 8), (3, 10), (6, 6), (6, 8), (6, 10)} {, 4, 3, 2, 1, 0, 1, 2} a) (A B) (B A) b)(a B) (A B) c)b A d) A (A B C) e) [(A B) (A C) (B A)] [A B C] f)[(a B) (A B)] (B C) 10. b) b) = a) {A,K }, {A,K,Q }, {A,K,J }, {A,K,Q,J }. b) a) True, b) False, c) True, d) True, e) False, f) True, g) True, h) False, i) True, j) True, k) False
19 Chapter 2 Logic 2.1 Statements and Connectives Symbolic logic studies some parts and relationships of the natural language by representing them with symbols. The main ingredients of symbolic logic are statements and connectives. A statement is an assertion that can be either true or false. Examples. The following sentences: It is sunny today; Ms. W. will have a broader audience next month; I did not join the club; are statements, while questions (e.g. How s the weather?), interjections (e.g. Cool!) and incomplete sentences (e.g. If I could...) are not considered to be statements unless rephrased appropriately. Simple statements do not contain other statements as their parts. (All of the examples above are simple statements.) We typically represent simple statements using lower-case letters p, q, r,...; for example s= Your bicycle is slick; c= I like its color. Connectives join simple statements into more complex statements, called compound statements. The most common connectives and their symbols are: and/but = ; or = ; if..., then =. Example. Your bicycleisslick andi likeitscolor= s c. 19
20 20 CHAPTER 2. LOGIC The operation not = turns a single statement into its negation and it is not a connective. The symbols representing statements, connectives, and the negation operation form our dictionary. Parentheses are used for punctuation. Simple statements. p p is true (Assertion) p p is false (Negation) Connectives and compound statements. Notes. p q p and q (Conjunction) p q either p or q, or both (Disjunction) p q if p then q (Conditional) 1. The connective or, in logic, has an inclusive meaning. For example, Bob will play tennis or go to the movies is interpreted as follows: Bob will either play tennis, or go to the movies, or do both. 2. The connective but has an identical role as the connective and, thus the same symbol is used for both. For example, Your bicycle is slick, but I don t like its color is written symbolically as b c. Parentheses. The use of parentheses is important and needs particular attention. Suppose we want to convert the following compound statement into symbolic form: If I do a web search for pages containing the terms termites or cattle, then I will search for pages containing global warming. First we identify the simple statements present in these expression and assign letters to each of them (we can rephrase them slightly without modifying their meaning): t=i search for pages containing termites; c=i search for pages containing cattle; g=i search for pages containing global warming. Connectives and punctuation help with splitting compound statements into simple ones. For example, the particle then splits the compound statement into two parts: the first part
21 2.1. STATEMENTS AND CONNECTIVES 21 is the disjunction t c, and the second is the simple statement g. The symbolic form of the compound statement is then (t c) g. Note. If we had accidentally skipped the parenthesis, we would have created the compound statement t c g, which could be read as: I search for pages containing termites or if I search for pages containing cattle, then I search for those containing global warming. This has a rather different meaning from the original statement! Exercises 1. Convert the following compound statements into symbolic statements, by assigning symbols to each simple statements (for example, f= my favorite dish has lots of anchovies ) and using the appropriate connectives: (a) Jim is a lawyer, yet he is not a crook. (b) Although our professor is young, he is knowledgeable. (c) My favorite dish has lots of anchovies or is not spicy and also it comes in a large portion. (d) My favorite dish has lots of anchovies and it comes in a large portion or it is spicy and it also comes in a large portion. (e) If you do not attend class, then either you read a book or you will not pass the exam. (f) I am doing a web search for pages containing the terms global warming, but not for pages containing both termites and cattle. (g) I am doing a web search for pages containing the terms global warming, but not for pages containing the word termites and not for pages containing the word cattle. 2. Convert the following symbolic statement into words if s=the sunroof is extra, r=the radial tires are included, w=power windows are optional. (a) (s w) r. (b) r [s ( w)]. (c) ( r) [( s) ( w)]. (d) (w s).
22 22 CHAPTER 2. LOGIC 2.2 Truth Values and Truth Tables Every logical statement, simple or compound, is either true or false. We say that the truth value of a statement is true (represented by the letter T) when the statement is true, and false (represented by the letter F) when the statement is false. The truth value of a compound statement can always be deduced from the truth values of the simple statements that compose it. Example. If p=i play the piano is false, and q=i study logic is true, then the conjunction p q =I play the piano and study logic is a false statement. A truth table summarizes all possible truth values of a statement. For example, p can only either be true (T) or false (F), so its truth table (the simplest of all) is: p T F The next simplest truth table is the truth table for the negation, whose truth values are always the opposite as those of the original statement: p p T F F T The truth tables for the conjunction and the disjunction are shown next. An easy way to remember them is to note that the statement p q (conjunction) is true only when p and q are both true; while the statement p q (disjunction) is false only when p and q are both false. (Conjunction) p q p q T T T T F F F T F F F F p q p q T T T T F T F T T F F F (Disjunction) Example. Construct the truth table for the compound statement (p q) p. We will first break down this statement in components of increasing complexity: the simple statements p and q, the disjunction p q, its negation (p q), and finally the statement (p q) p. We will create one column for each of these components: p q p q (p q) (p q) p T T T F F T F F
23 2.2. TRUTH VALUES AND TRUTH TABLES 23 and fill them out according to the Basic Rules 1. The negation reverses truth values. (So the values in the fourth column are the opposite as the values in the third column.) 2. The only case in which a conjunction of two statements is true is when both statements are true. 3. The only case in which a disjunction of two statements is false is when both statements are false. At the end, we obtain the completed truth table: Exercises p q p q (p q) (p q) p T T T F F T F T F F F T T F F F F F T F 1. Fill in the missing values in the following truth table: p q p q p q ( p) q [p q] [( p) q] T T F F F T T T F F T T F T T F F T T F F T T T 2. Construct truth tables for the following compound statements (a) p q. (b) ( p) p. (c) (p q). (d) ( p) q. (e) (p q) p. 3. If p is a true statement and q is a false statement, what are the truth values of the following statements? (For such problems, you need only construct one row of the truth table.) (a) (p q). (b) (p q).
24 24 CHAPTER 2. LOGIC Big Truth Tables Compound statements may contain several simple statements; in order to figure out how many columns you need in the truth table: Break the compound statement into building blocks of increasing complexity, starting with a column for each letter, and ending with the compound statement itself. We illustrate the procedure for the following compound statement containing three simple statements (the most difficult case we will encounter): (p r) (q r). If we break the statement into increasingly complex pieces, we will need one column for each of the following (in order): p, q, r (the simplest building blocks) r p r and q r (p r) (q r) (the most complex block) So that the truth table will contain the seven columns: p q r r p r q r (p r) (q r) Next, we fill in the first three columns. Since each of the 3 statements p, q, r is either true (T) or false (F), there are 2 3 = 8 possibilities. (In fact, r can be true or false for each of the 4 possible pairs of truth values associated with p and q.) Here is the truth table with the first three columns filled in p q r r p r q r (p r) (q r) T T T T F T F T T F F T T T F T F F F T F F F F In order to fill in the remaining columns, we follow the Basic Rules Here is a partially completed truth table, complete the rest on your own.
25 2.3. CONDITIONAL STATEMENTS AND THEIR TRUTH TABLES Exercises p q r r p ( r) q r (p r) (q r) T T T F F T T T F T F T F T T F F T F F T T T T T T T T T F F T T F F T F T F T T F F F T F F 1. Construct the truth tables for the following compound statements: (a) p (q r). (b) (p ( q)] r. (c) (r p) q. (d) (p q) (p r). 2.3 Conditional statements and their truth tables A compound statement of the form If p then q, written symbolically as p q, is called a conditional statement; p is called the antecedent, and q is called the consequent of the conditional statement. Example. Consider the conditional statement: If M is a human being, then M is mortal. In this statement, p= M is a human being is the antecedent, and q= M is mortal is the consequent. The truth table for a conditional statement is shown below: p q p q T T T T F F F T T F F T and it is best justified by looking at the special case in which the conditional statement is in the form of a promise, such as If you deliver this pizza by 7PM, then I will give you a \$5 bonus.
26 26 CHAPTER 2. LOGIC For this conditional statement, the antecendent is p=you deliver this pizza by 7PM and the consequent is q=i will give you a \$5 bonus.clearly, if you deliver the pizza by the stated time (p = T ) and I give you \$5 (q = T ), then the conditional is a true promise (i.e. p q is true). It is also clear that, if you deliver the pizza by 7PM (p = T ), but I do not give you \$5 (q = F ), then the promise was broken (i.e. p q is false). Now, suppose you do not deliver the pizza by 7PM (p = F ), then, whatever my decision is (to give you or not give you the \$5), then my original promise is still standing, (i.e. p q is true) Exercises 1. Given the conditional statement If your course average is better than 98% then you will earn an A in the class, (a) Identify the antecedent and the consequent, and write the statement in symbolic form. (b) Examine each of the four possible scenarios described in the truth table (p and q both true, p=t and q=f, etc.) and explain why the corresponding truth value for this conditional statement makes sense. 2. Construct a truth table for each of the following compound statements. (a) (p q) q. (b) ( p) (p q). (c) [p ( q r)] (q p). 2.4 Tautologies and Contradictions Anything that happens, happens. Anything that in happening causes something else to happen, causes something else to happen. Anything that in happening happens again, happens again. Though not necessarily in that order. (From Mostly Harmless, by Douglas Adams.) A tautology is a statement that is always true. The expression A is A (often attributed to Aristotle), is one of the most common tautologies. Consider these examples: A quote from a student: The main idea behind data compression is to compress data. A quote from George W. Bush concerning Native American tribes: Tribal sovereignty means that, it s sovereign. Both statements are certainly true, but... rather pointless. The way we check whether a statement is a tautology is by using truth tables. Let us look at the following paragraph taken from Through the Looking Glass by Lewis Carrol:
27 2.4. TAUTOLOGIES AND CONTRADICTIONS 27 You are sad, the Knight said in an anxious tone: let me sing you a song to comfort you... Everybody that hears me sing it either it brings the tears into their eyes, or else Or else what? said Alice, for the Knight had made a sudden pause. Or else it doesn t, you know. The statement in bold face can be written in symbolic form as p p, wherep =It brings tears into their eyes. The truth table for this statement is p p p p T F T F T T which says that any statement of the form p p is a tautology. Now fill out the following truth table for the statement p (p q) and show that this statement is a tautology: p q p q (p q) p (p q) T T T F F T F F As opposed to tautologies, contradictions are statements that are always false. The expression A and A is a contradiction. I don t believe in reincarnation, but I did in my past life. (Anonymous) Another example of a contradiction is [p ( p)] (show why). For more practice, complete the following truth table for the statement p (p q) and show that this statement is a contradiction Exercises p q p q (p q) p (p q) T T T F F T T F F F F 1. Show that the statement (p q) [( p) q] is a tautology by constructing its truth table. 2. Use a truth table to check whether the following are tautologies, contradictions, or neither.
28 28 CHAPTER 2. LOGIC (a) p ( p). (b) p p. (c) p (p q). (d) q [p p]. (e) q (p p). (f) [(p q) q]. 3. (a) If a proposition is neither a tautology nor a contradiction, what can be said about its truth table? (b) If A and B are two (possibly compound) statements such that A B is a contradiction, what can you say about A and B? (c) If A and B are two (possibly compound) statements such that A B is a tautology, what can you say about A and B? 2.5 Logical Equivalence Two statements are said to be logically equivalent when they have the same logical content. The simplest example of two logically equivalent statements is that of any statement p and its double negation ( p). In fact, if p is true (or false) then so is its double negation and viceversa. It is easy to check whether two statements are logically equivalent by using truth tables Two statements are logically equivalent when their truth tables are identical. Example. Any two statements of the form p q and ( p) q are logically equivalent. To show this we construct the following truth table (note that we duplicated a column for q since it makes it easier to fill out the last column): p q p q p q ( p) q T T T F T T T F F F F F F T T T T T F F T T F T The third and last columns (set in boldface) are identical, showing that the two statements have identical truth values regardless of their contents, and thus are logically equivalent. Note that this implies that, for example, the statements If the price is right, I will buy this and The price is not right or I will buy this have the same logical content.
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# 2002 Indonesia MO Problems/Problem 4
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
Given a triangle $ABC$ with $AC > BC$. On the circumcircle of triangle $ABC$ there exists point $D$, which is the midpoint of arc $AB$ that contains $C$. Let $E$ be a point on $AC$ such that $DE$ is perpendicular to $AC$. Prove that $AE = EC + CB$.
## Solution
We use the method of phantom points.
Draw $AC$ and $BC$, and extend line $AC$ past $C$ to a point $B'$ such that $BC = B'C$. Draw point $E'$ at the midpoint of $AB'$, and $D'$ at the intersection of the perpendicular to $AC$ from $E'$ and the perpendicular bisector of $AB$.
Since $D'B = D'B', CB = CB', D'C = D'C$, we have $\triangle D'BC \cong \triangle D'B'C$ by side-side-side similarity. Then $\angle D'AC = \angle D'AE' = \angle D'B'E' = \angle D'B'C = \angle D'BC$, so $ADCB$ is cyclic.
In particular, since we have $AD' = B'D' = BD'$, we know that $D'$ must be the midpoint of the arc of the circumcircle of $\triangle ABC$ that contains point $C$, and since $D'$ was on the perpendicular to $AC$ from $E'$, we must have that $E'$ is the foot of the perpendicular of $D'$ to $AC$. But this uniquely identifies $D' = D, E' = E$, and we are done. |
Problem with alternate solution — Equation of plane through point and containing intersection line of two planes [Stewart P $803, 12.5.37$]
$37.$ Find an equation of the plane that passes through the point $(1, -2, 1)$
and contains the line of intersection of the planes $x + y - z = 2$ and $2x - y + 3z = 1$. $\bbox[3px,border:2px solid grey]{\text{ Official solution : }}$My modified picture illustrates that the red line of intersection belongs to both planes. A normal vector of each plane $\perp$ to this red line of intersection. Thus, the direction vector of the red line = $\mathbb{n_1} \times \mathbb{n_2} = (1, 1, -1) \times (2, -1, 3) = \color{#C154C1}{(2, -5, -3)}$
For want of a normal vector to the requested plane, require another direction vector on this requested plane. Thus need the vector containing any point on the red line of intersection to the given point $(-1, 2, 1)$ in the plane. WLOG, set $\color{#FF4F00}{x = 0}$: $\begin{cases} x + y - z = 2 \\ 2x - y + 3z = 1 \end{cases} \implies \begin{cases} y - z = 2 \\ -y + 3z = 1 \end{cases}$ $\implies (\color{#FF4F00}{x}, y, z) = (\color{#FF4F00}{0}, 7/2, 3/2).$
Thus another vector parallel to the plane is $(-1, 2, 1) - (\color{#FF4F00}{0}, 7/2, 3/2) = \color{#C154C1}{(-1, -3/2, -1/2)}$.
In toto, a normal vector to the plane $= \color{#C154C1}{(2, -5, -3) \times (-1, -3/2, -1/2)} = 2(-1, 2, -4).$ $... \blacksquare$
$\bbox[3px,border:2px solid grey]{\text{ My solution : }}$ ♦ Any vector on the line of intersection of the two planes produces one vector $\parallel$ to the requested plane.
♠ The other vector $\parallel$ the requested plane is any vector connecting any point on this line of intersection with the given point $(-1, 2, 1)$ in the plane.
♦ Subtract the two equations of the plane to produce their line of intersection: $3x + 2z = 3$. For a direction vector of this line, subtract any two vectors on it. WLOG, put $z = 0 \implies \color{brown}{(1, 0, 0)}$ and put $z = 3 \implies \color{brown}{(-1, 0, 3)}$. Thus direction vector $= (-1, 0, 3) - (1, 0, 0) = (-2, 0, 3)$.
♠ Observe the given point $(-1, 2, 1)$ isn't on this line (but is given to be on the plane). Thus either $\color{brown}{(1, 0, 0)} - (-1, 2, 1) = (2, -2, -1)$ or $\color{brown}{(-1, 0, 3)} - (-1, 2, 1)$ are on the plane.
In toto, a normal vector to the plane = $(-2, 0, 3) \times (2, -2, -1) = 2(3, 2, 2) \neq k(-1, 2, -4)... \blacksquare$
Since the solution's and my normal vectors differ already, what's wrong with my solution?
Let's call the given point $P$. Then there is a typo in your question: at the beginning you specify $P$ as $(1,-2,1)$ and later you change it to $(-1,2,1)$. Let's assume $P =(-1,2,1)$.
Your mistake happens in the third step where you subtract the equations. The point $(1,0,0)$ is not on the line of intersection of the two planes. To see this note that if it were it would satisfy the equation of the first plane, $x+y-z=2$ but $1+0-0\neq 2$.
My guess of what happened: After you subtracted the equations the variable $y$ got eliminated. You assumed that this meant that $y=0$. But this is not the case: any point $(x,y,z)$ on the red line is specified by three coordinates satisfying both $x+y-z=2$ and $2x-y+3z =1$.
• Yep. $3x + 2z = 3$ in $\mathbb R^3$ defines a plane containing the red line rather than the line itself. – epimorphic May 21 '15 at 19:29 |
#### Chapter 6 Elementary Functions
Section 6.2 Linear Functions and Polynomials
# 6.2.5 Absolute Value Functions
In Module 2 the absolute value of a real number $x$ was introduced in the following way:
$|x|=\left\{\begin{array}{cc}\hfill x& \text{for}\mathrm{ }x\ge 0\hfill \\ \hfill -x& \text{for}\mathrm{ }x<0 .\hfill \end{array}$
In the context of this module, the absolute value can be regarded as a function. This results in the absolute value function:
$b:\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & ℝ\hfill \\ \hfill x& \hfill ⟼\hfill & |x| .\hfill \end{array}$
##### Exercise 6.2.6
What is the range ${W}_{b}$ of the absolute value function $b$?
Due to the definition by cases
$b\left(x\right)=|x|=\left\{\begin{array}{cc}\hfill x& \text{for}\mathrm{ }x\ge 0\hfill \\ \hfill -x& \text{for}\mathrm{ }x<0 ,\hfill \end{array}$
the absolute value function is an example of a piecewise defined function. If absolute values are defined according to different cases, then it is also said that the absolute value is resolved. Then, the graph of the absolute value function $b$ looks as follows:
One property of the graph of the absolute value function, which most of the more general functions involving absolute values have in common, is the kink at $x=0$. The absolute value function $b$ defined above is only the simplest case of a function involving an absolute value. More complicated examples of functions can be constructed, involving one or several absolute values, e.g.
$f:\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & ℝ\hfill \\ \hfill x& \hfill ⟼\hfill & |2x-1| .\hfill \end{array}$
For such functions, it is a relevant task to get an idea of how the graph of the function looks. To do this, we use the piecewise definition of the absolute value, and the approach is similar to the one for the solution of absolute value equations and inequalities. Here, we demonstrate this approach for the example of the function $f$ defined above:
##### Example 6.2.7
Consider the function
$f:\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & ℝ\hfill \\ \hfill x& \hfill ⟼\hfill & |2x-1|\hfill \end{array}.$
What does the graph look like?
We calculate:
$f\left(x\right)=|2x-1|=\left\{\begin{array}{cc}\hfill 2x-1& \text{for} 2x-1\ge 0\hfill \\ \hfill -\left(2x-1\right)& \text{for} 2x-1<0\hfill \end{array}=\left\{\begin{array}{cc}\hfill 2x-1& \text{for} x\ge \frac{1}{2}\hfill \\ \hfill -2x+1& \text{for} x<\frac{1}{2} .\hfill \end{array}$
Thus, we obtain a piecewise defined function whose graph is an increasing line with the slope $2$ and the $y$-intercept $-1$ for $x$ in the region $x\ge \frac{1}{2}$ and a decreasing line with the slope $-2$ and the $y$-intercept $1$ for $x$ in the region $x<\frac{1}{2}$. With this information we can draw the graph of $f$:
##### Info 6.2.8
CAUTION! If absolute values are resolved as in the calculation in the example above, two important calculation rules have to be observed:
1. The regions for the cases are defined by inequalities for the entire expression between the absolute value bars, here $2x-1\ge 0$ and $2x-1<0$, and not only by $x\ge 0$ and $x<0$. It is always that way if absolute values are resolved.
2. For the case $<0$ the entire expression gets a minus sign. Here, care has to be taken that the expression is bracketed. In the example above, we therefore have $-\left(2x-1\right)=-2x+1$ and not $-2x-1$. This is always the case if absolute values are resolved.
##### Exercise 6.2.9
Sketch the graph of the function
$\alpha :\mathrm{ }\left\{\begin{array}{ccc}\hfill ℝ& \hfill \to \hfill & ℝ\hfill \\ \hfill x& \hfill ⟼\hfill & |-8x+1|-1 .\hfill \end{array}$
Moreover, specify its range ${W}_{\alpha }$. |
# 1 Connected simple graphs on four vertices
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1 1 Connected simple graphs on four vertices Here we briefly answer Exercise 3.3 of the previous notes. Please come to office hours if you have any questions about this proof. Theorem 1.1. There are exactly six simple connected graphs with only four vertices. They are listed in Figure 1. Figure 1: An exhaustive and irredundant list. We order the graphs by number of edges and then lexicographically by degree sequence. The proof is arranged around first, the number of edges and second, the idea of the degree sequence. We begin with a few observations. Suppose that G is simple, connected, and V (G) = 4. It follows that the degree sequence of G has length four. Also, all degrees are at most three (as G is simple) and at least 1 (as G is connected so it has no isolated vertices). If G is to be connected, then E(G) 3. As G is simple, E(G) 6. We deal with the possible values of E(G), namely 3, 4, 5, and 6 in turn. The handshaking lemma implies that the sum of the degrees is even. From all of this it follows that the lexicographically first possible degreee sequence is (1, 1, 1, 1). However any graph with this degree sequence has only two edges. V (G) = {x, y, z, w}. Then x is incident to only one edge. Say the other endpoint of this edge is y (as the other possibilities are similar). Then neither x nor y has any other edge incident and we conclude that G is not connected. The next possible degree sequence for G is (2, 2, 1, 1). We take V (G) = {x, y, z, w}. Suppose that the degrees of x, y, z, and w are non-increasing. Thus both z and w are degree one vertices, ie leaves. If z and w are connected to each other then G is not connected, a contradiction. If both z and w are connected to x, say, then y must have a loop, a contradiction. We conclude that x and y share an edge and G is isomorphic to the first graph of Figure /09/12 1
2 Next we have (3, 1, 1, 1). If G is to be simple, then x is connected to each of the other three vertices exactly once. We obtain the third graph of Figure 1. We now increase the number of allowed edges by one the sequences must sum to eight instead of six. The next degree sequence is (2, 2, 2, 2). Prove to your own satisfaction that any graph with degree sequence (2, 2, 2,..., 2) must be a union of cycles. It will follow that G is isomorphic to C 4. Next is (3, 2, 2, 1). Again, x is connected to each of the other three vertices exactly once. This accounts for three edges. We must add exactly one more. Check that regardless of which two vertices we connect (y to z, z to w, or w to y) we obtain the forth graph. We now deal with E(G) = 5 or 6. The next sequence is (3, 3, 2, 2). Clearly x is connected to y, z, and w. Likewise y is connected to x, z, and w. This gives the correct degree sequence and we have the fifth graph. The sequence (3, 3, 3, 1) does not correspond to a simple graph. Finally we have the sequence (3, 3, 3, 3) which gives the complete graph K 4. We are done. 2 After all of that it is quite tempting to rely on degree sequences as an infallable measure of isomorphism. However, that would be a mistake, as we shall now see. In our first example, Figure 2, we have two connected simple graphs, each with five vertices. You should check that the graphs have identical degree sequences. Figure 2: A pair of five vertex graphs, both connected and simple. Both have the same degree sequence. Are they isomorphic? However, the graphs are not isomorphic. To prove this, notice that the graph on the left has a triangle, while the graph on the right has no triangles. Similarly, in Figure 3 below, we have two connected simple graphs, each with six vertices, each being 3-regular. It follows that they have identical degree sequences. Again, the graph on the left has a triangle; the graph on the right does not. (Check!). It is easy to see what is going on: the two graphs are different because one has some kind of path that the other graph does not have. We formalize this notion as follows. 2005/09/12 2
3 Figure 3: The graphs C 3 I and K 3,3. Both are connected, simple, and three-regular. Are they isomorphic? Definition 2.1. Suppose that G is a graph. A sequence of edges e 1, e 2,..., e i, e i+1,..., e n is a walk in G of length n if there are vertices v 0, v 1,..., v n so that for all i the edge e i connects v i 1 to v i. This is somewhat formal if G is simple it is equivalent to think of a walk as just a sequence of adjacent vertices. It all cases it is ok to think of a walk as a way of tracing your pencil over the edges of the graph, starting at v 0, ending at v n, without lifting the pencil. If we never repeat an edge in the walk then we call it a trail. If we never repeat a vertex in the walk (except for possibly allowing v 0 = v n ) then we call it a path. If in fact v 0 = v n then we call the walk, trail, or path closed. Another name for a closed path is a cycle. So what we were calling a triangle above now has another name it is a cycle of length three. Counting cycles in G gives isomorphism invariants. For example, if G has 3 triangles and H has 4 then G is not isomorphic to H. Here is a simple lemma which we will need below: Lemma 2.2. If G contains a closed walk of odd length, then G contains a cycle of odd length. Proof. We induct on the length, n, of the given closed walk W = (v 0, v 1,..., v n ). If n = 1 then G contains a loop and we are done. Suppose now that n 3. If W is a path we are done. So suppose that v i = v j where i < j < n. If j i is odd then the walk W = (v i, v i+1,..., v j ) is a closed walk of odd length, shorter than n. If j i is even then the walk W = (v 0, v 1,..., v i 1, v i, v j+1, v j+2,..., v n ) is a closed walk of odd length, shorter than n. By induction, the proof is complete. 2005/09/12 3
4 Definition 2.3. We say that a graph is Eulerian if there is a closed trail which vists every edge of the graph exactly once. Definition 2.4. We say that a graph is Hamiltonian if there is a closed path walk which vists every vertex of the graph exactly once. We will discuss these in greater detail next week. 3 Making small examples There are many ways of creating new graphs from old. Here we concentrate on making G smaller. Definition 3.1. If G is a graph then any graph appearing in G is called a subgraph of G. For example, if e is an edge of G then we can delete e from G to form the graph G e. If v is a vertex of G then we can delete v and all edges incident to v to form G v. Every subgraph of G is obtained by deleting some sequence of edges and vertices of G. If H is a subgraph of G we write H G. Exercise 3.2. Suppose that G is a simple graph on n vertices. Show that G appears as a subgraph of K n, ie G K n. 4 Return to connectedness Recall that a graph G is disconnected if there is a partition V (G) = A B so that no edge of E(G) connects a vertex of A to a vertex of B. If there is no such partition, we call G connected. We now use paths to give a characterization of connected graphs. Theorem 4.1. A graph G is connected if and only if for every pair of vertices v and w there is a path in G from v to w. Proof. We begin with the forward direction. Fix a vertex v V (G). We define: A v = {w V (G) v and w are connected by a path in G}. Let B v = V (G) A v. We will show that B v is empty and thus the conclusion holds. Suppose, for a contradiction, that B v is not empty. As G is connected there is an edge e connecting some w A v to some u B v. There is path P from v to w which does not mention u (because u/ A v ). So append the edge e to form a longer path P e, connecting v to u. It follows that u A v! This is a contradiction. To prove the reverse direction: Fix a partition V (G) = A B. We must find an edge e running from a vertex of A to a vertex of B. To this end, choose any v A and w B. By assumption there is a path from v to w. Label the vertices of the path 2005/09/12 4
5 v 0, v 1, v 2,..., v n so that v = v 0, w = v n, and v i and v i+1 are consecutive vertices of the path. Then there is a vertex with smallest possible label, say v k, with v k B. Now, v k v, as v A. So k > 0. Thus v k 1 is in A, v k B, and v k 1 and v k are connected by an edge. It is easy to adapt the proof just give to show that: G is connected if and only if there is a walk connecting every pair of vertices of G. Disconnected graphs break naturally into smaller pieces: Suppose H is a connected subgraph of G. Suppose also that all subgraphs H G containing H are disconnected. Then we call H a component of G. Put another way: if you draw a disconnected graph and then draw circles around each of the pieces then you have circled the components. There another two common definitions relating to connectedness. Definition 4.2. If G is connected and G e is disconnected then we call the edge e a bridge. Definition 4.3. If G is connected and G v is disconnected then we call the vertex v a cut-vertex. For example, every edge of the path graph P n is a bridge but no edge of the cycle C n is. All of the vertices of P n having degree two are cut vertices. A leaf is never a cut vertex. A very important class of graphs are the trees: a simple connected graph G is a tree if every edge is a bridge. (Equivalently, if every non-leaf vertex is a cut vertex.) The idea of a bridge or cut vertex can be generalized to sets of edges and sets of vertices. We will develop such extensions later in the course. 5 Making large examples Now we may focus on making big examples. Definition 5.1. Suppose that G is a graph. Choose a point v which is not a vertex of G. Form the graph G by adding v to the vertex set of G and by adding edges between v and every vertex of G. We call G the cone on G. Note that the graph G naturally appears as a subgraph of the cone G. example, the cone on a cycle is called a wheel. As an Definition 5.2. Suppose that G and H are graphs. We form the product of G and H, called G H, taking V (G H) = V (G) V (H) and E(G H) = E(G) E(H). The product of two edges is a square: {v, w} {a, b} gives the four edges {(v, a), (w, a)}, {(w, a), (w, b)}, {(w, b), (v, b)}, and {(v, b), (v, a)} in the product graph. 2005/09/12 5
6 The simplest example of this is taking the product of a graph G with the interval: the simple graph I with exactly two vertices and exactly one edge. For example Q 2 = I I is the square; which is identical to the 4-cycle C 4. Q 3 = I C 4 is the cube, and Q 4 = I cube is the hypercube or tesseract. The hypercube is often called the 4-cube. The ordinary cube Q 3 would then be called the 3-cube, and so on. See Figure 4. Figure 4: The interval, the square, and the cube. Draw the tesseract as an exercise. In general we take an n-cube to be the graph Q n = I I... I: the product of the interval with itself n times. The vertices of the n-cube are in bijection with the set of binary strings of length n. Two vertices are adjacent in the cube if and only if the corresponding binary strings differ in exactly one position. Our last family of examples are the bipartite graphs. Here is the definition: Definition 5.3. A graph G is bipartite if we can partition the vertex set V = A B so that A B =, no edge of G has both endpoints in A and no edge of G has both endpoints in B. The complete bipartite graph K m,n has A = m, has B = n, and has an edge {a, b} for all a A and b B. So, for example, bipartite graphs never have loops. Note that the graph on the right hand side of Figure 3 is a copy of K 3,3. Exercise 5.4. Show that K m,n has exactly m n edges. Exercise 5.5. Give a direct proof that the n-cube is a bipartite graph. (Hint: Add up all the ones in the binary string, mod 2.) We end this section with a nice characterization of bipartite graphs: Theorem 5.6. A graph is bipartite if and only if all cycles in the graph have even length. Proof. Here is just a hint of the proof. Suppose, to begin with, that G is bipartite. Let v 0, v 1,..., v n = v 0 be a cycle. We want to show that n is even. Suppose that v 0 is in A. Now prove by induction that v i A if and only if i is even. Since v n = v 0 A it will follow that n is even. 2005/09/12 6
7 Suppose now that all cycles have even length. Fix v V (G) to be a basepoint. Define A V (G) to be the set of all those vertices which can be reached by some path of even length, starting at v 0. Define B V (G) to be the set of all those vertices which can be reached by some path of odd length, starting at v 0. We leave it to the reader to show that A B = V (E) and A B =. We also leave it to the reader to show that there is no edge between vertices of A and there is no edge between vertices of B. Theorem 4.1 and Lemma 2.2 should be quite useful! 6 Matrices To investigate the path structure of a graph there are two kinds of matrices which suggest themselves. The first is the adjacency matrix, constructed as follows. Label all of the vertices of G so that V (G) = {v 1, v 2,..., v n }. Let A(G) be the matrix where the entry a ij equals the number of edges connecting v i to v j. We make several observations about A(G): We have deg(v i ) = j a ij. A(G) is non-negative and symmetric. A(G) records enough information to recover the graph G. The ij th entry of the matrix A k (G), the k th power of A(G), records the number of walks of length exactly k between v i and v j. For example, if G is connected and has an odd cycle then there is a k so that all entries of A k (G) are positive. Exercise 6.1. Write down the adjacency matrix for the complete bipartite graph K 3,3 on vertices {a 1, a 2, a 3, b 1, b 2, b 3 }. What do you notice about A(K 3,3 )? Describe the powers of A(K 3,3 ). Suppose that G is any bipartite graph. Show that A(G) is a block anti-diagonal matrix. Describe qualititively what the powers of A(G) look like. We now turn to the incidence matrix of G. Label all of the vertices of G so that V (G) = {v 1, v 2,..., v n }. Label all fo the edges of G so that E(G) = {e 1, e 2,..., e m }. Let B(G) be the matrix where the entry b ij is a one if v i is an endpoint of e j. We make several observations about A(G): We have deg(v i ) = j b ij and 2 = i b ij. B(G) is non-negative, but usually is not square. B(G) records enough information to recover the graph G. 2005/09/12 7
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# Triangle Formula
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Last updated date: 15th Aug 2024
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## What is the Triangle Formula?
### What is a Triangle?
A polygon with three sides and three vertices is a triangle. In geometry, it is one of the fundamental topics of geometry. A triangle with vertices A, B, and C, is represented as △ ABC. In Euclidean geometry, any three non-collinear points determine a unique triangle and a unique plane at the same time. In a triangle, three angles are there. Each angle is formed when any two sides of the triangle meet at a common point, known as the vertex.
### Perimeter of Triangle Formula
The perimeter of any polygon is the sum of the lengths of the edges.
In the triangle,
Perimeter = The sum of the three sides
### Area of a Triangle
The triangle’s area is the total region that is enclosed by the three sides of any particular triangle. It is equal to half of the height of the basic periods. Therefore, we have to know the base and height of it to find the field of a tri-sided polygon.
Let us find out the area of different types of triangles.
• Area of Isosceles Triangle
In a triangle of isosceles, two sides are equal in length. Also equal to each other are the two angles opposite to the two equal sides.
• In case base and height are given, we use the following formula:
• A = ½ × height × base
• If three sides are given :
• A = ½[√(a2b2/4) × b]
• Using 2 sides of the triangle and an angle between them :
• A = ½ × b × c × sin(α)
• Using two angles between two sides and their length :
• A = [c2 × sin(β) × sin(α)/ 2 × sin(2π−α−β)]
• Area of Scalene Triangle Formula
A scalene triangle is a type of triangle in which there are different side dimensions on all three sides. The three angles are therefore different from each other due to this.
A = ½ × height × base
• Area of Equilateral Triangle Formula
There are all three sides of an equilateral triangle equal to each other. As a consequence, all the inner angles are equal degrees, i.e. each angle is 60°.
A = (√3)/4 × side2
Where,
A is the area of the triangle.
a is the length of the triangle.
b is the base of the triangle.
c is the third side of the triangle.
h is the height of the triangle.
α and β are the angles between two sides.
### Solved Example
Determine a triangle area with a base of 12cm and a height of 10cm.
Solution:
Area of a triangle = ½ × height × base
= ½ × 12 × 10
= 6 × 10
= 60 cm2
### Conclusion
Since a triangle is a three-sided polygon, therefore to find the perimeter of a triangle we have to find the sum of the three sides. Similarly to find the area of a triangle, we must first know about the lengths of the sides of the triangle.
## FAQs on Triangle Formula
1. What is the Formula for Finding a Triangle's Perimeter?
Solution: Perimeter of a triangle = a + b + c
Where c, b, and a are the three sides of a triangle.
2. How can you Find the Third Side of a Given Two Sides of a Triangle?
Solution: We can use the Pythagoras theorem to find the third side if two sides are given. It states that the square of the side on which the hypotenuse is located is equal to the sum of the square areas on the other two sides. |
# Set
Sets are one of the most important and fundamental concepts in modern mathematics. Basic set theory, having only been invented at the end of the 19th century, is now a ubiquitous part of mathematics education, being introduced as early as elementary school . This article gives a brief and basic introduction to what mathematicians call "intuitive" or "naive" set theory; for a more detailed account see Naive set theory. For a rigorous axiomatic treatment of sets see Axiomatic set theory.
## Introduction
Informally, a set is just a collection of objects considered as a whole. The objects of a set are called elements or members. The elements of a set can be anything: numbers, people, letters of the alphabet, other sets, etc. Sets will usually be denoted by capital letters, A, B, C, etc. Two sets A and B are said to be equal, written A = B, if they have the same members.
## Ways of describing a set
A set may be described by words, for example:
A = the first three natural numbers greater than zero
B = the colors red, white, blue, and green
Another way to describe a set is to list its elements between curly braces, for example:
C = {1, 2, 3}
D = {red, white, blue, green}
Even though two sets may be described differently, they still may be identical as sets. For example, for the sets described above, A = C and B = D, since they have precisely the same members.
It makes no difference in what order the elements are listed, or whether there are repetitions in the list. For example, the three sets: {2, 4}, {4, 2}, and {2, 2, 4, 2} are identical, since again, they all have the same members.
## The number of members in a set
Each of the sets described above, have a definite number of members, for example the set A has three members, while the set B has four members.
A set can also have zero members. Such a set is called the empty set (or the null set) and is denoted by the symbol ∅. For example, as of 2004, the set A of all people living on the moon, has zero members, and thus, A = ∅. Like the number zero, seemingly trivial, the empty set turns out to be quite important in mathematics.
A set can also have an infinite number of members. For example the set of natural numbers is infinite.
For more information on infinity and the size of sets see Cardinal number.
## Subsets
If every member of the set A is also a member of the set B, then A is said to be a subset of B, written AB. If A is subset of B, and A is not equal to B, then A is called a proper subset of B, written A ⊂ B.
A is a subset of B
Examples:
• The set of all men is a proper subset of the set of all people.
• The set of all natural numbers is a proper subset of all integers.
• {1, 3} ⊂ {1, 2, 3, 4}
• {1, 2, 3, 4} ⊆ {1, 2, 3, 4}
The empty set is a subset of every set and every set is a subset of itself:
• ∅ ⊆ A
• AA
## Unions
There are several ways to construct new sets from existing ones. Two sets can be "added" together. The union of A and B, denoted by A ∪ B, is the set of all things which are members of either A or B.
The union of A and B
Examples:
• {1, 2} ∪ {red, white} = {1, 2, red, white}
• {1, 2, green} ∪ {red, white, green} = {1, 2, red, white, green}
• {1, 2} ∪ {1, 2} = {1, 2}
Some basic properties of unions:
• A ∪ B = B ∪ A
• A ⊆ A ∪ B
• A ∪ A = A
• A ∪ ∅ = A
## Intersections
A new set can also be constructed by determining which members two sets have "in common". The intersection of A and B, denoted by A ∩ B, is the set of all things which are members of both A and B. If A ∩ B = ∅, then A and B are said to be disjoint.
The intersection of A and B
Examples:
• {1, 2} ∩ {red, white} = ∅
• {1, 2, green} ∩ {red, white, green} = {green}
• {1, 2} ∩ {1, 2} = {1, 2}
Some basic properties of intersections:
• A ∩ B = B ∩ A
• A ∩ B ⊆ A
• A ∩ A = A
• A ∩ ∅ = ∅
## Complements
Two sets can also be "subtracted". The relative complement of A in B, denoted by B − A, is the set of all things which are members of B, but not members of A.
In certain settings all sets under discussion are considered to be subsets of a given universal set U. In such cases, U − A, is called the absolute complement or simply complement of A, and is denoted by A′.
The relative complement
of A in B
The complement of A in U
Examples:
• {1, 2} − {red, white} = {1, 2}
• {1, 2, green} − {red, white, green} = {1, 2}
• {1, 2} − {1, 2} = ∅
• If U is the set of integers, then the complement of the even integers is the odd integers
Some basic properties of complements:
• A ∪ A′ = U
• A ∩ A′ = ∅
• (A′)′ = A
• A − B = A ∩ B′ |
# How do you solve 5x+7=27?
Feb 3, 2016
$5 x + 7 = 27$
$\rightarrow 5 x = 27 - 7$
$\rightarrow 5 x = 20$
$\rightarrow x = \frac{20}{5} = 4$
Feb 3, 2016
Subtract $7$ from both sides of the equation
then divide both sides by $5$ to isolate $x$
giving $x = 4$
#### Explanation:
If
$\textcolor{w h i t e}{\text{XXX}} 5 x + 7 = 27$
then
$\textcolor{w h i t e}{\text{XXX}} 5 x + 7 \textcolor{red}{- 7} = 27 \textcolor{red}{- 7}$
which simplifies as
$\textcolor{w h i t e}{\text{XXX}} 5 x = 20$
If
$\textcolor{w h i t e}{\text{XXX}} 5 x = 20$
then
$\textcolor{w h i t e}{\text{XXX}} \left(5 x\right) \textcolor{red}{\div 5} = 20 \textcolor{red}{\div 5}$
which simplifies as
$\textcolor{w h i t e}{\text{XXX}} x = 4$ |
How do you solve log_2 x = log_4 (x+6)?
3 Answers
Using the change of base rule we have that
logx/log2=log(x+6)/(log2^2) => 2*logx=log(x+6)=> x^2=x+6=> x^2-x-6=0=> (x+2)(x-3)=0=> x_1=3,x_2=-2
Only $x = 3$ is acceptable because $x > 0$
Nov 14, 2015
We have to use the properties of logarithms.
Explanation:
${\log}_{2} x = {\log}_{4} \left(x + 6\right)$
$\implies \ln \frac{x}{\ln} 2 = \ln \frac{x + 6}{\ln} 4$
$\implies \ln \frac{x}{\ln} 2 = \ln \frac{x + 6}{\ln} \left({2}^{2}\right)$
$\implies \ln \frac{x}{\cancel{\ln 2}} = \ln \frac{x + 6}{2 \cdot \cancel{\ln 2}}$
$\implies 2 \ln x = \ln \left(x + 6\right)$
$\implies \ln \left({x}^{2}\right) = \ln \left(x + 6\right)$
$\implies {x}^{2} = x + 6$
$\implies {x}^{2} - x - 6 = 0$
which is thus reduced to a quadratic equation which can be solved by the quadratic formula to get the two roots: $x = - 2 \mathmr{and} x = 3$
For $x = 3$, the arguments within the logarithms in the original equation come out to be positive, thus $x = 3$ is a valid solution.
But for $x = - 2$, one of the logarithmic argument is $- 2$ , giving rise to ${\log}_{2} \left(- 2\right)$ which is completely meaningless. Thus $x = - 2$ is discarded and cannot be considered as a solution.
Nov 14, 2015
Make both sides have a base of $4$.
Explanation:
Change to 4^(log_2(x))=4^(log_4(x+6).
Notice that on the right side, the $4$ and ${\log}_{4}$ will cancel, leaving just $\left(x + 6\right)$.
${4}^{{\log}_{2} \left(x\right)} = x + 6$
Change the $4$ to ${2}^{2}$.
${\left({2}^{2}\right)}^{{\log}_{2} \left(x\right)} = x + 6$
${2}^{2 {\log}_{2} \left(x\right)} = x + 6$
Remember this logarithm rule.
${2}^{{\log}_{2} \left({x}^{2}\right)} = x + 6$
Like before, the $2$ and ${\log}_{2}$ will cancel.
${x}^{2} = x + 6$
${x}^{2} - x - 6 = 0$
$\left(x - 3\right) \left(x + 2\right) = 0$
x=3 or cancel(x=-2
Remember that in ${\log}_{a} \left(b\right) , b > 0$. |
# How do you graph r^2=3sin2θ?
Dec 12, 2016
#### Explanation:
The relation between polar coordinates $\left(r , \theta\right)$ and Cartesian coordinates $\left(x , y\right)$ is $x = r \cos \theta$, $y = r \sin \theta$ and ${r}^{2} = {x}^{2} + {y}^{2}$.
We can use this to convert equation in polar coordinates to an equation with Cartesian coordinates.
${r}^{2} = 3 \sin 2 \theta$
$\Leftrightarrow {r}^{2} = 3 \times 2 \sin \theta \cos \theta$
or ${r}^{2} \times {r}^{2} = 6 \times r \sin \theta \times r \cos \theta$
or ${\left({x}^{2} + {y}^{2}\right)}^{2} = 6 x y$
Note that
(a) As $6 x y$ is a complete square, it is positive and hence curve can lie only in first and third quadrant.
(b) Further as maximum value of ${r}^{2} = 3 \sin 2 \theta$, maximum possible value for ${r}^{2}$ is $3$ and so $r$ cannot be more than $\sqrt{3} = 1.732 \ldots .$
(c) As replacing $x$ and $y$ with each other does not change the equation, it is symmetric along $x = y$.
Now we can put different values of $x$ to get $y$ (both less than $\sqrt{3}$) and draw the graph.
The function appears as follows.
graph{((x^2+y^2)^2-6xy)(x-y)=0 [-5, 5, -2.5, 2.5]} |
# 2019 AIME II Problems/Problem 4
## Problem
A standard six-sided fair die is rolled four times. The probability that the product of all four numbers rolled is a perfect square is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
## Solution 1
Notice that, other than the number 5, the remaining numbers 1, 2, 3, 4, 6 are only divisible by 2 and/or 3. We can do some cases on the number of 5's rolled (note that there are $6^4 = 1296$ outcomes).
Case 1 (easy): Four 5's are rolled. This has probability $\frac{1}{6^4}$ of occurring.
Case 2: Two 5's are rolled.
Case 3: No 5's are rolled.
To find the number of outcomes for the latter two cases, we will use recursion. Consider a 5-sided die with faces numbered 1, 2, 3, 4, 6. For $n \ge 1$, let $a_n$ equal the number of outcomes after rolling the die $n$ times, with the property that the product is a square. Thus, $a_1 = 2$ as 1 and 4 are the only possibilities.
To find $a_{n+1}$ given $a_n$ (where $n \ge 1$), we observe that if the first $n$ rolls multiply to a perfect square, then the last roll must be 1 or 4. This gives $2a_n$ outcomes. Otherwise, the first $n$ rolls do not multiply to a perfect square ($5^n - a_n$ outcomes). In this case, we claim that the last roll is uniquely determined (either 2, 3, or 6). If the product of the first $n$ rolls is $2^x 3^y$ where $x$ and $y$ are not both even, then we observe that if $x$ and $y$ are both odd, then the last roll must be 6; if only $x$ is odd, the last roll must be 2, and if only $y$ is odd, the last roll must be 3. Thus, we have $5^n - a_n$ outcomes in this case, and $a_{n+1} = 2a_n + (5^n - a_n) = 5^n + a_n$.
Computing $a_2$, $a_3$, $a_4$ gives $a_2 = 7$, $a_3 = 32$, and $a_4 = 157$. Thus for Case 3, there are 157 outcomes. For case 2, we multiply $a_2$ by $\binom{4}{2} = 6$ to distribute the two 5's among four rolls. Thus the probability is
$$\frac{1 + 6 \cdot 7 + 157}{6^4} = \frac{200}{6^4} = \frac{25}{162} \implies m+n = \boxed{187}$$
-scrabbler94
## Solution 2
We can solve this without finding the amount of cases. Notice when we roll a 1 or a 4, it does not affect whether or not the product is a square number. We have a 1/3 chance of rolling either a 1 or 4. We have a 2/3 chance of rolling a 2,3,5 or 6. Let's call rolling 1 or 4 rolling a dud (a perfect square).
Probability of rolling 4 duds: $\left(\frac{1}{3}\right)^4$
Probability of rolling 3 duds: $4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3}$
Probability of rolling 2 duds: $6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2$
Probability of rolling 1 dud: $4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3$
Probability of rolling 0 duds: $\left(\frac{2}{3}\right)^4$
Now we will find the probability of a square product given we have rolled each amount of duds
Probability of getting a square product given 4 duds: 1
Probability of getting a square product given 3 duds: 0 (you will have 1 non-dud and that's never going to be square)
Probability of getting a square product given 2 duds: $\frac{1}{4}$ (as long as our two non-duds are the same, our product will be square)
Probability of getting a square product given 1 dud: $\frac{3!}{4^3}$ = $\frac{3}{32}$ (the only way to have a square product is rolling a 2,3 and 6. There are 3! ways of doing that and a total of $4^3$ ways to roll 3 non-duds).
Probability of getting a square product given 0 duds: $\frac{40}{4^4}$= $\frac{5}{32}$ (We can have any two non-duds twice. For example, 2,2,5,5. There are $\binom{4}{2} = 6$ ways of choosing which two non-duds to use and $\binom{4}{2} = 6$ ways of choosing how to arrange those 4 numbers. That gives us 6*6=36 combinations. We can also have 2,2,2,2 or 3,3,3,3 or 5,5,5,5 or 6,6,6,6. This gives us a total of 40 combinations).
We multiply each probability of rolling k duds with the probability of getting a square product given k duds and then sum all the values. $$\left(\frac{1}{3}\right)^4 * 1 + 4 * \left(\frac{1}{3}\right)^3 * \frac{2}{3} * 0 + 6 * \left(\frac{1}{3}\right)^2 * \left(\frac{2}{3}\right)^2 * \frac{1}{4} + 4 * \frac{1}{3} * \left(\frac{2}{3}\right)^3 * \frac{3}{32} + \left(\frac{2}{3}\right)^4 * \frac{5}{32} = \frac{25}{162}.$$
$25+162$ = $\boxed{187}$
-dnaidu (silverlizard)
## Solution 3
Note that rolling a 1/4 will not affect whether or not the product is a perfect square. This means that in order for the product to be a perfect square, all non 1/4 numbers rolled must come in pairs, with the only exception being the triplet 2,3, 6. Now we can do casework:
If there are four 1/4's, then there are $2^4=16$ combinations. If there are three 1/4's, then there are 0 combinations, because the fourth number isn't a square. If there are two 1/4's, there are $2^2=4$ ways to choose the two 1/4's, 4 ways to choose the remaining pair of numbers, and $\frac{4!}{2!2!}=6$ ways to arrange, so there are $4\cdot 4\cdot 6=96$ combinations for this case. If there is one 1/4, then there are 2 ways to choose whether it is a 1 or 4, and the remaining three numbers must be 2, 3, and 6, so there are $4!$ ways to order, meaning there are $2\cdot 4!=48$ combinations for this case. Our final case is if there are no 1/4's, in which case we must have two pairs. If the two pairs are of different numbers, then there $\binom{4}{2}$ to choose the numbers and $\frac{4!}{2!2!}=6$ ways to arrange them, so $6\cdot 6=36$. If all four numbers are the same there are $4$ combinations, so there are $4+36=40$ combinations for this case.
Hence there are $16+0+96+48+40=200$ combinations where the product of the dice is a perfect square, and there are $6^4=1296$ total combinations, so the desired probability is $\frac{200}{1296}=\frac{25}{162}$, yielding an answer of $25+162=\boxed{187}$.
-Stormersyle
## Solution 4 (Casework)
Another way to solve this problem is to do casework on all the perfect squares from $1^2$ to $36^2$, and how many ways they can be ordered $1^2$- $1,1,1,1$- $1$ way. $2^2$- $4,1,1,1$ or $2,2,1,1$- $\binom{4}{2}+4=10$ ways. $3^2$- $3,3,1,1$- $\binom{4}{2}=6$ ways. $4^2$- $4,4,1,1$, $2,2,2,2$, or $2,2,4,1$- $\binom{4}{2}+1+12=19$ ways. $5^2$- $5,5,1,1$- $\binom{4}{2}=6$ ways. $6^2$- $6,6,1,1$, $1,2,3,6$, $2,3,2,3$, $3,3,4,1$- $2*\binom{4}{2}+4!+12=48$ ways. $7^2$- Since there is a prime greater than 6 in its prime factorization there are $0$ ways. $8^2$- $4,4,4,1$ or $2,4,2,4$- $\binom{4}{2}+4=10$ ways. $9^2$- $3,3,3,3$- $1$ way. $10^2$- $2,2,5,5$ or $1,4,5,5$- $6+12=18$ ways. $11^2$- $0$ ways for the same reason as $7^2$. $12^2$- $6,6,2,2$, $4,4,3,3$, $2,3,4,6$, or $1,4,6,6$- $2*\binom{4}{2}+4!+12=48$ ways. $13^2$- $0$ ways. $14^2$- $0$ ways. $15^2$- $3,3,5,5$- $\binom{4}{2}=6$ ways. $16^2$- $4,4,4,4$- $1$ way. $17^2$- $0$ ways. $18^2$- $3,3,6,6$- $\binom{4}{2}=6$ ways. $19^2$- $0$ ways. $20^2$- $4,4,5,5$- $\binom{4}{2}=6$ ways. $21^2$- $0$ ways. $22^2$- $0$ ways. $23^2$- $0$ ways. $24^2$-$4,4,6,6$- $\binom{4}{2}=6$ ways. $25^2$- $5,5,5,5$- $1$ way. $26^2$- $0$ ways. $27^2$- $0$ ways. $28^2$- $0$ ways. $29^2$- $0$ ways. $30^2$- $5,5,6,6$- $\binom{4}{2}$ ways. $31^2$- $0$ ways. $32^2$- $0$ ways. $33^2$- $0$ ways. $34^2$- $0$ ways. $35^2$- $0$ ways. $36^2$- $6,6,6,6$- $1$ way.
There are $6^4=1296$ ways that the dice can land. Summing up the ways, it is easy to see that there are $200$ ways. This results in a probability of $\frac{200}{1296}=\frac{25}{162}\implies\boxed{187}$ -superninja2000
## Solution 5 (Recursion)
We can do recursion on the number of rolls to find the number of ways we can get $4$ rolls to multiply to a square.
After $n$ rolls, let us say that the product is $p = 2^a3^b5^c$.
Define the following:
$A_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is odd, and $b$, $c$ are even
$B_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is odd, and $a$, $c$ are even
$C_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is odd, and $a$, $b$ are even
$D_{n} =$ the number of ways to have a product after $n$ rolls where $c$ is even, and $a$, $b$ are odd
$E_{n} =$ the number of ways to have a product after $n$ rolls where $b$ is even, and $a$, $c$ are odd
$F_{n} =$ the number of ways to have a product after $n$ rolls where $a$ is even, and $b$, $c$ are odd
$G_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all odd
$S_{n} =$ the number of ways to have a product after $n$ rolls where $a, b,$ and $c$ are all even (square!)
We have the following equations after considering the possible values of the nth roll:
$$A_{n} = S_{n-1}+B_{n-1}+D_{n-1}+E_{n-1}+2A_{n-1}$$
$$B_{n} = A_{n-1}+D_{n-1}+F_{n-1}+S_{n-1}+2B_{n-1}$$
$$C_{n} = S_{n-1}+E_{n-1}+F_{n-1}+G_{n-1}+2C_{n-1}$$
$$D_{n} = S_{n-1}+A_{n-1}+B_{n-1}+G_{n-1}+2D_{n-1}$$
$$E_{n} = A_{n-1}+C_{n-1}+F_{n-1}+G_{n-1}+2E_{n-1}$$
$$F_{n} = B_{n-1}+E_{n-1}+C_{n-1}+G_{n-1}+2F_{n-1}$$
$$G_{n} = C_{n-1}+D_{n-1}+F_{n-1}+E_{n-1}+2G_{n-1}$$
$$S_{n} = A_{n-1}+C_{n-1}+B_{n-1}+D_{n-1}+2S_{n-1}$$
We have the following values after considering the possible values of the 1st roll:
$$A_1 = B_1 = C_1 = D_1 = 1; E_1 = F_1 = G_1 = 0; S_1 = 2$$
After applying recursion twice, we get:
$$A_2 = B_2 = D_2 = 6, C_2 = 4, E_2 = F_2 = G_2 = 2, S_2 = 8$$
$$A_3 = B_3 = D_3 = 34, C_3 = 22, E_3 = F_3 = G_3 = 18, S_3 = 38$$
Finally, we have $S_4 = 200$, $\frac{m}{n} = \frac{200}{1296} = \frac{25}{162}$meaning our answer is $\boxed{187}$.
## Solution 6
Consider all the distinct "fundamental" groups of integers from $1$ to $6$ whose product is a perfect square. A "fundamental" group is one that cannot be broken into two smaller groups that each have a perfect square product. For example, $\{2,2\}$ is a fundamental group, while $\{3,3,4\}$ is not, because it can be broken up into $\{3,3\}$ and $\{4\}$.
$1$ and $4$ are already perfect squares, so they each form a "fundamental" group and cannot belong in any other group. Pairs of the other $4$ numbers ($\{2,2\}$,$\{3,3\}$, etc. ) form "fundamental" groups as well. The last "fundamental" group is $\{2,3,6\}$. It can be easily seen that no more groups exist.
Thus, we have the "fundamental" groups $\{1\}$,$\{4\}$,$\{2,2\}$,$\{3,3\}$,$\{5,5\}$,$\{6,6\}$, and $\{2,3,6\}$.
We now consider the ways to use these groups to form a sequence of $4$ numbers whose product is a perfect square. To form a set, we can simply select zero to two groups of size $2$ or $3$ and fill in any remaining spots with $1$s and $4$s. We can do this in one of $5$ ways: Using only $1$s and $4$s, using one group of size $2$, using one group of size $3$, using two different groups of size $2$, and using the same group of size $2$ twice.
If we only use $1$s and $4$s, each of the $4$ slots can be filled with one of the $2$ numbers, so there are $2^4=16$ possibilities.
If we use one group of size $2$, there are $4$ options for the group to use, $\binom{4}{2}$ ways to place the two numbers (since they are identical), and $2^2$ ways to fill in the remaining slots with $1$s and $4$s, so there are $4\cdot\binom{4}{2}\cdot2^2=96$ possibilities.
If we use one group of size $3$, there is only $1$ option for the group to use, $4\cdot3\cdot2$ ways to place the three numbers (since they are distinct), and $2$ ways to fill in the remaining slot, so there are $4\cdot3\cdot2\cdot2=48$ possibilities.
If we use two different groups of size $2$, there are $\binom{4}{2}$ options for the groups to use and $\binom{4}{2}$ ways to place the four numbers (since there are $2$ groups of identical numbers, and one group's placement uniquely determines the other's), so there are $\binom{4}{2}\cdot\binom{4}{2}=36$ possibilities.
If we use the same group of size $2$ twice, there are $4$ options for the group to use and $1$ way to place the four numbers (since they are all identical), so there are $4=4$ possibilities.
This gives us a total of $16+96+48+36+4=200$ possibilities, and since there are $6^4=1296$ total sequences that can be rolled, the probability is equal to $\frac{200}{1296}=\frac{25}{162}$, so the answer is $25+162=\boxed{187}$. ~emerald_block
## Solution 7(Generating functions)
Let's look at the prime factorization of some of these rolls:
01 => 2^0*3^0*5^0
02 => 2^1*3^0*5^0
03 => 2^0*3^1*5^0
04 => 2^2*3^0*5^0
05 => 2^0*3^0*5^1
06 => 2^1*3^1*5^0
Now, using multi-variable generating functions, we get:
f(x,y,z)=1+x+y+x^2+z+xy
| |
\| |/
'v'
f(x,y,z)=1+x+y+z+x^2+xy
=(for our purposes)
=2+x+y+z+xy
Let's square that!
4+2x+2y+2z+2xy+2x+x^2+xy+xz+x^2y+2y+xy+y^2+yz+xy^2+2z+xz+yz+z^2+xyz+2xy+x^2y+xy^2+xyz+x^2y^2
Combining like terms . . .
4+4x+4y+4z+x^2+y^2+z^2+6xy+2xz+2yz+2xyz+2x^2y+2xy^2+x^2y^2
Since we only want the parity of each of the exponents, we can collapse it again.
8+6x+6y+4z+6xy+2xz+2yz+2xyz
Last simplification: factor out a factor of two and save it for later.
4+3x+3y+2z+3xy+xz+yz+xyz
Let's take a better approach this time.
___ term :4*4+3*3+3*3+2*2+3*3+1*1+1*1+1*1=16+09+09+04+09+01+01+01=50
__x term :4*3+3*4+3*3+2*1+3*3+1*2+1*1+1*1=12+12+09+02+09+02+01+01=48
__y term :4*3+3*3+3*4+2*1+3*3+1*1+1*2+1*1=12+09+12+02+09+01+02+01=48
__z term :4*2+3*1+3*1+2*4+3*1+1*3+1*3+1*3=08+03+03+08+03+03+03+03=34
_xy term :4*3+3*3+3*3+2*1+3*4+1*1+1*1+1*2=12+09+09+02+12+01+01+02=48
_xz term :4*1+3*2+3*1+2*3+3*1+1*4+1*3+1*3=04+06+03+06+03+04+03+03=32
_yz term :4*1+3*1+3*2+2*3+3*1+1*3+1*4+1*3=04+03+06+06+03+03+04+03=32
xyz term :4*1+3*1+3*1+2*3+3*2+1*3+1*3+1+4=04+03+03+06+06+03+03+04=32
Result: 50+48x+48y+34z+48xy+32xz+32yz+32xyz
I know we could use vectors and dot products to make it look neater but come on. It already looks neat enough. Also, we didn't need the other parts, but it just looks nicer. Now let's stick back the two that turned into a four.
200+192x+192y+136z+192xy+128xz+128yz+128xyz
We seek the constant term which is 200. 200/1296=100/648=50/324=25/162, 25+162=187
~ Afly (talk)
## Solution 8
There are a total of $2^4=1296$ possible die rolls.
We use casework:
Case 1: All 4 numbers are the same. There are obviously $6$ ways.
Case 2: Sets of 2 different numbers.
A set of two different numbers is basically $(x,x,y,y)$ . There are a total of $\frac{4!}{2!\cdot 2!}=6$ ways to arrange the numbers.
By listing these cases, we quickly see a pattern:
$(1,1,2,2)$
$(1,1,3,3)$
$...$
$(1,1,6,6)$
$(2,2,3,3)$
$...$
$(2,2,6,6)$
$...$
$(5,5,6,6)$
There are a total of $5+4+3+2+1=15$ cases. Multiplying this by $6$ yields $15\cdot 6=90$ ways.
Case 3: Sets of numbers in the form of $(x,x,1,4)$
A special case must be made for the number $4$ because $4$ itself is a perfect square.
$(1,1,1,4)$ $4$
$(2,2,1,4)$ $12$
$(3,3,1,4)$ $12$
$(4,4,1,4)$ $4$
$(5,5,1,4)$ $12$
$(6,6,1,4)$ $12$
Summing these up yields a total of $4+12+12+4+12+12=56$ ways.
Case 4: Sets with all 4 numbers different
Note that the sets
$(1,2,3,6)$
$(2,3,4,6)$
Multiply to perfect square. The total of these cases are $24+24=48$.
Adding all these cases together yields $6+90+56+48=200$ ways that the product of the values of the die can be a perfect square.
Therefore the probability is
$\frac{200}{1296}=\frac{25}{162}$
$m+n = 25+162 = \boxed{187}$
-elchen3_f |
# Reflection
Reflection is one of the non deformative transformation that any solid object can be subjected to. It involves determining the mirror image of the object accross a point or a line in 2D and accross a point, a line or a plane or in 3D. In this transformation any changes applied to a coordinate point can be directly applied to the equations of objects as well.
## Reflections in 2D
Reflections in 2D can happen in the following ways:
1. Reflection with respect to coordinate axes or the origin: These are the simplest form of reflection that can happen in 2D. Following summerises these.
Reflection Type Equation Form Matrix Form Accross x axis $$x' = x$$ $$y' = -y$$ $$\begin{bmatrix} 1 & 0 \\0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x' \\ y' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} 1 & 0 \\0 & -1 \end{bmatrix} = \begin{bmatrix} x' & y' \end{bmatrix}$$ Accross y axis $$x' = -x$$ $$y' = y$$ $$\begin{bmatrix} -1 & 0 \\0 & 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x' \\ y' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} -1 & 0 \\0 & 1 \end{bmatrix} = \begin{bmatrix} x' & y' \end{bmatrix}$$ Accross origin (0,0)(Similar to rotation by 180° accross origin (0,0)) $$x' = -x$$ $$y' = -y$$ $$\begin{bmatrix} -1 & 0 \\0 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x' \\ y' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y \end{bmatrix} \begin{bmatrix} -1 & 0 \\0 & -1 \end{bmatrix} = \begin{bmatrix} x' & y' \end{bmatrix}$$
2. Reflection with respect to lines $$x=constant$$ or $$y=constant$$ or arbitrary point $$(o_x,o_y)$$: Following summerises these.
Reflection Type Equation Form Matrix Form Accross line x=c(c=constant) $$x' = 2c - x$$ $$y' = y$$ $$\begin{bmatrix} - 1 & 0 & 2c \\0 & 1 & 0 \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0\\0 & 1 & 0 \\ 2c & 0 & 1 \end{bmatrix} = \begin{bmatrix} x' & y' & 1\end{bmatrix}$$ Accross line y=c(c=constant) $$x' = x$$ $$y' = 2c-y$$ $$\begin{bmatrix} 1 & 0 & 0 \\0 & -1 & 2c \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\0 & -1 & 0 \\ 0 & 2c & 1 \end{bmatrix} = \begin{bmatrix} x' & y' & 1 \end{bmatrix}$$ Accross point $$(o_x,o_y)$$(Similar to rotation by 180° accross point $$(o_x,o_y)$$) $$x' = 2o_x - x$$ $$y' = 2o_y - y$$ $$\begin{bmatrix} -1 & 0 & 2o_x \\0 & -1 & 2o_y \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0\\0 & -1 & 0 \\ 2o_x & 2o_y & 1 \end{bmatrix} = \begin{bmatrix} x' & y' & 1 \end{bmatrix}$$ Please note that when $$c, o_x,o_y$$ are 0 these equations are similar to reflection with respect to coordinate axes or the origin
3. Reflection with respect to line $$y=mx$$ making and angle $$\theta$$ with positive direction of x axis:
Reflection Based On Equation Form Matrix Form Angle $$\theta$$ $$x' = x cos 2\theta + y sin 2\theta$$ $$y' = x sin 2\theta - y cos 2\theta$$ $$\begin{bmatrix} cos 2\theta & sin 2\theta \\sin 2\theta & -cos 2\theta\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} x' \\ y'\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} cos 2\theta & sin 2\theta \\sin 2\theta & -cos 2\theta\end{bmatrix} = \begin{bmatrix} x' & y'\end{bmatrix}$$ Slope m (m=$$\tan\theta$$) $$x' = x (\frac{1-m^2}{1+m^2}) + y (\frac{2m}{1+m^2})$$ $$y' = x (\frac{2m}{1+m^2}) - y (\frac{1-m^2}{1+m^2})$$ $$\begin{bmatrix} \frac{1-m^2}{1+m^2} & \frac{2m}{1+m^2} \\\frac{2m}{1+m^2} & -\frac{1-m^2}{1+m^2}\end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} x' \\ y'\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} \frac{1-m^2}{1+m^2} & \frac{2m}{1+m^2} \\\frac{2m}{1+m^2} & -\frac{1-m^2}{1+m^2}\end{bmatrix} = \begin{bmatrix} x' & y'\end{bmatrix}$$
4. Reflection with respect to line $$ax + by=0$$:
Reflection Based On Equation Form Matrix Form When a and b are not unit vectors i.e. $$a^2 + b^2 \neq 1$$ $$x' = x (\frac{b^2-a^2}{a^2+b^2}) + y (\frac{-2ab}{a^2+b^2})$$ $$y' = x (\frac{-2ab}{a^2+b^2}) + y (\frac{a^2-b^2}{a^2+b^2})$$ $$\begin{bmatrix} \frac{b^2-a^2}{a^2+b^2} & \frac{-2ab}{a^2+b^2} \\\frac{-2ab}{a^2+b^2} & \frac{a^2-b^2}{a^2+b^2} \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} x' \\ y'\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} \frac{b^2-a^2}{a^2+b^2} & \frac{-2ab}{a^2+b^2} \\\frac{-2ab}{a^2+b^2} & \frac{a^2-b^2}{a^2+b^2} \end{bmatrix} = \begin{bmatrix} x' & y'\end{bmatrix}$$ When a and b are unit vectors i.e. $$a^2 + b^2 =1$$ $$x' = x (1-2a^2) + y (-2ab)$$ $$y' = x (-2ab) + y (1-2b^2)$$ $$\begin{bmatrix} 1-2a^2 & -2ab \\-2ab & 1-2b^2 \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} x' \\ y'\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} 1-2a^2 & -2ab \\-2ab & 1-2b^2 \end{bmatrix} = \begin{bmatrix} x' & y'\end{bmatrix}$$ Please note that when a and b are unit vectors i.e. $$a^2 + b^2 =1$$, then $$1-2a^2 = b^2-a^2$$ and $$1-2b^2 = a^2-b^2$$
5. Reflection with respect to line $$ax=by$$:
Reflection Based On Equation Form Matrix Form When a and b are not unit vectors i.e. $$a^2 + b^2 \neq 1$$ $$x' = x (\frac{b^2-a^2}{a^2+b^2}) + y (\frac{2ab}{a^2+b^2})$$ $$y' = x (\frac{2ab}{a^2+b^2}) + y (\frac{a^2-b^2}{a^2+b^2})$$ $$\begin{bmatrix} \frac{b^2-a^2}{a^2+b^2} & \frac{2ab}{a^2+b^2} \\\frac{2ab}{a^2+b^2} & \frac{a^2-b^2}{a^2+b^2} \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} x' \\ y'\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} \frac{b^2-a^2}{a^2+b^2} & \frac{2ab}{a^2+b^2} \\\frac{2ab}{a^2+b^2} & \frac{a^2-b^2}{a^2+b^2} \end{bmatrix} = \begin{bmatrix} x' & y'\end{bmatrix}$$ When a and b are unit vectors i.e. $$a^2 + b^2 =1$$ $$x' = x (1-2a^2) + y (2ab)$$ $$y' = x (2ab) + y (1-2b^2)$$ $$\begin{bmatrix} 1-2a^2 & 2ab \\2ab & 1-2b^2 \end{bmatrix} \begin{bmatrix} x \\ y\end{bmatrix} = \begin{bmatrix} x' \\ y'\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y\end{bmatrix} \begin{bmatrix} 1-2a^2 & 2ab \\2ab & 1-2b^2 \end{bmatrix} = \begin{bmatrix} x' & y'\end{bmatrix}$$ Please note that when a and b are unit vectors i.e. $$a^2 + b^2 =1$$, then $$1-2a^2 = b^2-a^2$$ and $$1-2b^2 = a^2-b^2$$
6. Reflection with respect to line making and angle $$\theta$$ with positive direction of x axis with y intercept=$$c_y$$ and x intercept=$$c_x$$:
Reflection Based On Equation Form Matrix Form Shifting origin to $$(c_x,0)$$ $$x' = (x-c_x) cos 2\theta + y sin 2\theta + c_x$$ $$y' = (x-c_x) sin 2\theta - y cos 2\theta$$ $$\begin{bmatrix} cos 2\theta & sin 2\theta & -c_x cos 2\theta + cx\\sin 2\theta & -cos 2\theta & -c_x sin 2\theta\\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1\end{bmatrix} \begin{bmatrix} cos 2\theta & sin 2\theta & 0\\sin 2\theta & -cos 2\theta & 0\\-c_x cos 2\theta + cx & -c_x sin 2\theta & 1\end{bmatrix} = \begin{bmatrix} x' & y' & 1\end{bmatrix}$$ Shifting origin to $$(0,c_y)$$ $$x' = x cos 2\theta + (y-c_y) sin 2\theta$$ $$y' = x sin 2\theta - (y-c_y) cos 2\theta + c_y$$ $$\begin{bmatrix} cos 2\theta & sin 2\theta & -c_y sin 2\theta\\sin 2\theta & -cos 2\theta & c_y cos 2\theta + c_y\\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1\end{bmatrix} \begin{bmatrix} cos 2\theta & sin 2\theta & 0\\sin 2\theta & -cos 2\theta & 0\\-c_y sin 2\theta & c_y cos 2\theta + c_y & 1\end{bmatrix} = \begin{bmatrix} x' & y' & 1\end{bmatrix}$$
7. Reflection with respect to line making and angle $$\theta$$ with positive direction of x axis at a distance c from the origin:
Equation Form Matrix Form $$x' = x cos 2\theta + y sin 2\theta - 2c sin\theta$$ $$y' = x sin 2\theta - y cos 2\theta + 2c cos\theta$$ $$\begin{bmatrix} cos 2\theta & sin 2\theta & -2c sin \theta\\sin 2\theta & -cos 2\theta & 2c cos \theta\\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1\end{bmatrix} \begin{bmatrix} cos 2\theta & sin 2\theta & 0\\sin 2\theta & -cos 2\theta & 0\\ -2c sin \theta & 2c cos \theta & 1\end{bmatrix} = \begin{bmatrix} x' & y' & 1\end{bmatrix}$$
8. Reflection with respect to line $$ax+by+c=0$$ when a and b are unit vectors i.e. $$a^2 + b^2 =1$$:
Equation Form Matrix Form $$x' = x (1-2a^2) + y (-2ab) - 2ac$$ $$y' = x (-2ab) + y (1-2b^2) - 2bc$$ $$\begin{bmatrix} 1-2a^2 & -2ab & -2ac\\-2ab & 1-2b^2 & -2bc \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1\end{bmatrix} \begin{bmatrix} 1-2a^2 & -2ab & 0\\-2ab & 1-2b^2 & 0\\ -2ac & -2bc & 1\end{bmatrix} = \begin{bmatrix} x' & y' & 1\end{bmatrix}$$ Please note that since a and b are unit vectors i.e. $$a^2 + b^2 =1$$, therefore $$1-2a^2 = b^2-a^2$$ and $$1-2b^2 = a^2-b^2$$
9. Reflection with respect to line $$by=ax + c$$ when a and b are unit vectors i.e. $$a^2 + b^2 =1$$:
Equation Form Matrix Form $$x' = x (1-2a^2) + y (2ab) - 2ac$$ $$y' = x (2ab) + y (1-2b^2) + 2bc$$ $$\begin{bmatrix} 1-2a^2 & 2ab & -2ac\\2ab & 1-2b^2 & 2bc \\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & 1\end{bmatrix} \begin{bmatrix} 1-2a^2 & 2ab & 0\\2ab & 1-2b^2 & 0\\ -2ac & 2bc & 1\end{bmatrix} = \begin{bmatrix} x' & y' & 1\end{bmatrix}$$ Please note that since a and b are unit vectors i.e. $$a^2 + b^2 =1$$, therefore $$1-2a^2 = b^2-a^2$$ and $$1-2b^2 = a^2-b^2$$
10. Reflection with respect to any line $$ax + by + c=0$$:
Equation Form Matrix Form $$x' = x - 2D \hat{a}$$ $$y' = y - 2D \hat{b}$$ $$\begin{bmatrix} x' \\ y' \end{bmatrix} = \begin{bmatrix} x \\ y \end{bmatrix} - 2D \begin{bmatrix} \hat{a} \\ \hat{b}\end{bmatrix}$$ Please note that $$\hat{a},\hat{b}$$ are unit vectors corresponding to coefficients a and b and D is the distance from point to line.
11. Reflection of a point C with respect to any line having direction ratio $$\vec{A}$$ making a vector $$\vec{B}$$ with any point on the line:
When $$\vec{B} \perp \vec{A}$$ $$\vec{P}= \vec{C} - 2 \vec{B}$$ When $$\vec{B}$$ not $$\perp \vec{A}$$ $$\vec{P}= \vec{C} - 2 (\frac{\vec{A} \times (\vec{B} \times \vec{A})}{\vert \vec{A} \vert ^2})$$ Please note that $$\vec{C}$$ is the position vector of point C and $$\vec{P}$$ is position vector of reflected point.
The reflection types 3,4 and 5 are different ways of representing reflection accross a line passing through origin which is not parallel or perpendicular to any coordinate axes.
The reflection types 6,7,8 and 9 are different ways of representing reflection accross a line not passing through origin which is not parallel or perpendicular to any coordinate axes.
## Reflections in 3D
Reflections in 3D can happen in the following ways:
1. Reflection with respect to a coordinate plane or a coordinate axis (i.e. 2 coordinate planes) or with respect to origin (0,0,0) (3 coordinate planes): These are the simplest form of reflection that can happen in 3D. Following summerises these.
Reflection Type Equation Form Matrix Form Accross xy plane $$x' = x$$ $$y' = y$$ $$z' = -z$$ $$\begin{bmatrix} 1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & -1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z\end{bmatrix} \begin{bmatrix} 1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -1\end{bmatrix} = \begin{bmatrix} x' & y' & z' \end{bmatrix}$$ Accross yz plane $$x' = -x$$ $$y' = y$$ $$z' = z$$ $$\begin{bmatrix} -1 & 0 & 0\\0 & 1 & 0\\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\0 & 1 & 0\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} x' & y' & z' \end{bmatrix}$$ Accross zx plane $$x' = x$$ $$y' = -y$$ $$z' = z$$ $$\begin{bmatrix} 1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & -1 & 0\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} x' & y' & z' \end{bmatrix}$$ Accross xy & yz plane (y Axis) $$x' = -x$$ $$y' = y$$ $$z' = -z$$ $$\begin{bmatrix} -1 & 0 & 0 \\0 & 1 & 0 \\0 & 0 & -1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z\end{bmatrix} \begin{bmatrix} -1 & 0 & 0\\0 & 1 & 0\\0 & 0 & -1\end{bmatrix} = \begin{bmatrix} x' & y' & z' \end{bmatrix}$$ Accross yz & zx plane (z Axis) $$x' = -x$$ $$y' = -y$$ $$z' = z$$ $$\begin{bmatrix} -1 & 0 & 0\\0 & -1 & 0\\0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\0 & -1 & 0\\0 & 0 & 1\end{bmatrix} = \begin{bmatrix} x' & y' & z' \end{bmatrix}$$ Accross xy & zx plane (x Axis) $$x' = x$$ $$y' = -y$$ $$z' = -z$$ $$\begin{bmatrix} 1 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \\0 & -1 & 0\\0 & 0 & -1\end{bmatrix} = \begin{bmatrix} x' & y' & z' \end{bmatrix}$$ Accross origin (0,0,0) (all three coordinate planes) $$x' = -x$$ $$y' = -y$$ $$z' = -z$$ $$\begin{bmatrix} -1 & 0 & 0\\0 & -1 & 0\\0 & 0 & -1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 \\0 & -1 & 0\\0 & 0 & -1\end{bmatrix} = \begin{bmatrix} x' & y' & z' \end{bmatrix}$$
2. Reflection with respect to planes $$x=constant$$ or $$y=constant$$ or $$z=constant$$ or arbitrary point $$(o_x,o_y,o_z)$$: Following summerises these.
Reflection Type Equation Form Matrix Form Accross plane x=c(c=constant) $$x' = 2c - x$$ $$y' = y$$ $$z' = z$$ $$\begin{bmatrix} - 1 & 0 & 0 & 2c \\0 & 1 & 0 & 0 \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 2c & 0 & 0 & 1\end{bmatrix} = \begin{bmatrix} x' & y' & z' & 1\end{bmatrix}$$ Accross plane y=c(c=constant) $$x' = x$$ $$y' = 2c-y$$ $$z' = z$$ $$\begin{bmatrix} 1 & 0 & 0 & 0\\0 & -1 & 0 & 2c \\0 & 0 & 1 & 0 \\0 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\0 & -1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 2c & 0 & 1\end{bmatrix} = \begin{bmatrix} x' & y' & z' & 1\end{bmatrix}$$ Accross plane z=c(c=constant) $$x' = x$$ $$y' = y$$ $$z' = 2c-z$$ $$\begin{bmatrix} 1 & 0 & 0 & 0 \\0 & 1 & 0 & 0 \\0 & 0 & -1 & 2c \\0 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z & 1 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 & 0\\0 & 1 & 0 & 0 \\ 0 & 0 & -1 & 0 \\ 0 & 0 & 2c & 1\end{bmatrix} = \begin{bmatrix} x' & y' & z' & 1\end{bmatrix}$$ Accross point $$(o_x,o_y,o_z)$$ $$x' = 2o_x - x$$ $$y' = 2o_y - y$$ $$z' = 2o_z - z$$ $$\begin{bmatrix} -1 & 0 & 0 & 2o_x \\0 & -1 & 0 & 2o_y \\0 & 0 & -1 & 2o_z \\0 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z & 1 \end{bmatrix} \begin{bmatrix} -1 & 0 & 0 & 0\\0 & -1 & 0 & 0 \\0 & 0 & -1 & 0\\ 2o_x & 2o_y & 2o_z & 1 \end{bmatrix} = \begin{bmatrix} x' & y' & z' & 1 \end{bmatrix}$$ Please note that when $$c, o_x,o_y,o_z$$ are 0 these equations are similar to reflection with respect to coordinate planes or the origin
3. Reflection with respect to plane $$ax + by + cz=0$$ (i.e. a plane passing through origin) when a , b and c are unit vectors i.e. $$a^2 + b^2 + c^2 =1$$:
Equation Form Matrix Form $$x' = x (1-2a^2) + y (-2ab) + z (-2ac)$$ $$y' = x (-2ab) + y (1-2b^2) + z(-2bc)$$ $$z' = x (-2ac) + y (-2bc) + z(1-2c^2)$$ $$\begin{bmatrix} 1-2a^2 & -2ab & -2ac\\-2ab & 1-2b^2 & -2bc \\-2ac & -2bc & 1-2c^2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z'\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z\end{bmatrix} \begin{bmatrix} 1-2a^2 & -2ab & -2ac\\-2ab & 1-2b^2 & -2bc \\-2ac & -2bc & 1-2c^2 \end{bmatrix} = \begin{bmatrix} x' & y' & z'\end{bmatrix}$$
4. Reflection with respect to plane $$ax + by + cz +d=0$$ (i.e. a plane not passing through origin) when a , b and c are unit vectors i.e. $$a^2 + b^2 + c^2 =1$$:
Equation Form Matrix Form $$x' = x (1-2a^2) + y (-2ab) + z (-2ac) - 2ad$$ $$y' = x (-2ab) + y (1-2b^2) + z(-2bc) - 2bd$$ $$z' = x (-2ac) + y (-2bc) + z(1-2c^2) - 2cd$$ $$\begin{bmatrix} 1-2a^2 & -2ab & -2ac & -2ad\\-2ab & 1-2b^2 & -2bc & -2bd \\-2ac & -2bc & 1-2c^2 & -2cd\\0 & 0 & 0 & 1\end{bmatrix} \begin{bmatrix} x \\ y \\ z \\ 1\end{bmatrix} = \begin{bmatrix} x' \\ y' \\ z' \\ 1\end{bmatrix}\hspace{.5cm} OR \hspace{.5cm} \begin{bmatrix} x & y & z & 1\end{bmatrix} \begin{bmatrix} 1-2a^2 & -2ab & -2ac & 0 \\-2ab & 1-2b^2 & -2bc & 0\\-2ac & -2bc & 1-2c^2 &0 \\-2ad & -2bd & -2cd & 1\end{bmatrix} = \begin{bmatrix} x' & y' & z' & 1\end{bmatrix}$$
5. Reflection with respect to any plane $$ax + by + cz +d=0$$:
Equation Form Matrix Form $$x' = x - 2D \hat{a}$$ $$y' = y - 2D \hat{b}$$ $$z' = z - 2D \hat{c}$$ $$\begin{bmatrix} x' \\ y' \\ z' \end{bmatrix} = \begin{bmatrix} x \\ y \\ z \end{bmatrix} - 2D \begin{bmatrix} \hat{a} \\ \hat{b} \\ \hat{c} \end{bmatrix}$$ Please note that $$\hat{a},\hat{b},\hat{c}$$ are unit vectors corresponding to coefficients a, b and c and D is the distance from point to plane.
6. Reflection of a point C with respect to any line having direction ratio $$\vec{A}$$ making a vector $$\vec{B}$$ with any point on the line:
When $$\vec{B} \perp \vec{A}$$ $$\vec{P}= \vec{C} - 2 \vec{B}$$ When $$\vec{B}$$ not $$\perp \vec{A}$$ $$\vec{P}= \vec{C} - 2 (\frac{\vec{A} \times (\vec{B} \times \vec{A})}{\vert \vec{A} \vert ^2})$$ Please note that $$\vec{C}$$ is the position vector of point C and $$\vec{P}$$ is position vector of reflected point. |
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# Triangle Inequality Theorem
## Sum of the lengths of any two sides of a triangle is greater than the third side.
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Triangle Inequality Theorem
What if you had to determine whether the three lengths 5, 7 and 10 make a triangle? After completing this Concept, you'll be able to use the Triangle Inequality Theorem to determine if any three side lengths make a triangle.
### Guidance
Can any three lengths make a triangle? The answer is no. There are limits on what the lengths can be. For example, the lengths 1, 2, 3 cannot make a triangle because , so they would all lie on the same line. The lengths 4, 5, 10 also cannot make a triangle because .
The arc marks show that the two sides would never meet to form a triangle. The Triangle Inequality Theorem states that the sum of the lengths of any two sides of a triangle must be greater than the length of the third.
#### Example A
Do the lengths 4, 11, 8 make a triangle?
To solve this problem, check to make sure that the smaller two numbers add up to be greater than the biggest number. and so yes these lengths make a triangle.
#### Example B
Find the length of the third side of a triangle if the other two sides are 10 and 6.
The Triangle Inequality Theorem can also help you find the range of the third side. The two given sides are 6 and 10, so the third side, , can either be the shortest side or the longest side. For example could be 5 because . It could also be 15 because . Therefore, the range of values for is .
Notice the range is no less than 4, and not equal to 4. The third side could be 4.1 because . For the same reason, cannot be greater than 16, but it could 15.9, .
#### Example C
The base of an isosceles triangle has length 24. What can you say about the length of each leg?
To solve this problem, remember that an isosceles triangle has two congruent sides (the legs). We have to make sure that the sum of the lengths of the legs is greater than 24. In other words, if is the length of a leg:
Each leg must have a length greater than 12.
Watch this video for help with the Examples above.
#### Concept Problem Revisited
The three lengths 5, 7, and 10 do make a triangle. The sum of the lengths of any two sides is greater than the length of the third.
### Vocabulary
An isosceles triangle is a triangle with two congruent sides. The congruent sides are called the legs and the third side is called the base. The Triangle Inequality Theorem states that to make a triangle, two sides must add up to be greater than the third side.
### Guided Practice
Do the lengths below make a triangle?
1. 4.1, 3.5, 7.5
2. 4, 4, 8
3. 6, 7, 8
Use the Triangle Inequality Theorem. Test to see if the smaller two numbers add up to be greater than the largest number.
1. . Yes this is a triangle because .
2. . No this is not a triangle because two lengths cannot equal the third.
3. . Yes this is a triangle because .
### Practice
Determine if the sets of lengths below can make a triangle. If not, state why.
1. 6, 6, 13
2. 1, 2, 3
3. 7, 8, 10
4. 5, 4, 3
5. 23, 56, 85
6. 30, 40, 50
7. 7, 8, 14
8. 7, 8, 15
9. 7, 8, 14.99
If two lengths of the sides of a triangle are given, determine the range of the length of the third side.
1. 8 and 9
2. 4 and 15
3. 20 and 32
4. 2 and 5
5. 10 and 8
6. and
7. The legs of an isosceles triangle have a length of 12 each. What can you say about the length of the base?
### Vocabulary Language: English
Triangle Inequality Theorem
Triangle Inequality Theorem
The Triangle Inequality Theorem states that in order to make a triangle, two sides must add up to be greater than the third side. |
# AP CALCULUS 1005: Secants and Tangents. Objectives SWBAT determine the tangent line by finding the limit of the secant lines of a function. SW use both.
## Presentation on theme: "AP CALCULUS 1005: Secants and Tangents. Objectives SWBAT determine the tangent line by finding the limit of the secant lines of a function. SW use both."— Presentation transcript:
AP CALCULUS 1005: Secants and Tangents
Objectives SWBAT determine the tangent line by finding the limit of the secant lines of a function. SW use both informal and precise mathematical language to describe the tangent line of a function
Average Rates of Change The AVERAGE SPEED (average rate of change) of a quantity over a period of time is the amount of change divided by the time it takes. In general, the average rate of change of a function over an interval is the amount of change divided by the length of the interval. Therefore, the average rate of change can be thought of as the slope of a secant line to a curve.
Average Rate of Change Known Formula Average Rate of Change : (in function notation) ax Slope of a secant
Slope of a Tangent Calculus I -The study of ___________________________________ Slope : (in function notation) ax Move x closer to a Rates of change
A. THE DERIVATIVE (AT A POINT) WORDS: Layman’s description: The DERIVATIVE is a __________________ function. Built on the ________________ formula. The derivative (slope of a tangent line) is the limit of the slopes of the secants as the two points are brought infinitely close together Rate of change slope
Let T(t) be the temperature in Dallas( in o F) t hours after midnight on June 2, 2001. The graph and table shows values of this function recorded every two hours,. What is the meaning of the secant line (units)? Estimate the value of the rate of change at t = 10. t T 073 2 73 470 669 8 72 10 81 Estimate the rate of change tangent line As the points get closer together the ROC approach ROC at 10
EX: THE DERIVATIVE AT A POINT EX 1: at a = 4 Notation: Words: f(a) = 8
Equation of the Tangent To write the equation of a line you need: a) b) Point- Slope Form: point slope (4,8) m = 6
There is local linearity When the secant lines get very near a point it acts like a tangent line
Normal to a Curve The normal line to a curve at a point is the line perpendicular to the tangent at the point. Therefore, the slope of the normal line is the negative reciprocal of the slope of the tangent line. EX 1(cont): at a = 4 Find the equation of the NORMAL to the curve
EX: THE DERIVATIVE AT A POINT EX 2: at x = -2 Method: y = 1 Mantra:
At a Joint Point SKIP Piece Wise Defined Functions: The function must be CONTINUOUS Derivative from the LEFT and RIGHT must be equal. The existence of a derivative indicates a smooth curve; therefore,
Last Update: 08/12/10
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# How do I solve this ?
## Given p-2 is one of the roots of the quadratic equation ${x}^{2}$ - m$x$ + 9 $- {p}^{2}$ = 0 . Express p in terms of m.
Aug 27, 2017
$p = \frac{2 m + 13}{m + 4}$
#### Explanation:
$p - 2$ is a root of $f \left(x\right) = {x}^{2} - m x + 9 - {p}^{2}$.
So ${\left(p - 2\right)}^{2} - m \left(p - 2\right) + 9 - {p}^{2} = 0$.
Let's expand the parentheses:
$R i g h t a r r o w {p}^{2} - 4 p + 4 - m p + 2 m + 9 - {p}^{2} = 0$
Then, let's simplify the equation:
$R i g h t a r r o w {p}^{2} - {p}^{2} - 4 p - m p + 2 m + 9 + 4 = 0$
$R i g h t a r r o w - p \left(4 + m\right) + 2 m + 13 = 0$
Now, let's solve for $p$:
$R i g h t a r r o w - p \left(4 + m\right) = - \left(2 m + 13\right)$
$R i g h t a r r o w - p = - \frac{2 m + 13}{4 + m}$
$\therefore p = \frac{2 m + 13}{m + 4}$ |
0 0
# How to solve the system of equations by graphing
3x+y=-4
x+2y=2
y= -3x -4
y= -1/2x + 1
Hi Tanesha,
Let's begin by putting both equations into slope-intercept form (y = mx + b):
3x + y = -4 becomes y = -3x - 4 (slope (m) = -3 and y-intercept (b) = -4)
x + 2y = 2 becomes 2y = -x + 2 and then y= -0.5x + 1 (slope (m) = -0.5 and y-intercept = 1)
We now have 1 point (the y-intercept, the y value when x = 0) for each line:
For line, y = -3x - 4, point is (0,-4)
For line, y = -0.5x + 1, point is (0,1)
Now let's get another point for each line - the x-intercept (the x value when y = 0) by plugging in 0 for y and solving for x in each equation:
For the line, y = -3x - 4, 0 = -3x - 4 gives a value of -4/3 for x, so point is (-4/3,0)
For the line, y = -0.5x + 1, 0 = -0.5x + 1 gives a value of 2 for x, so point is (2,0)
Now we have 2 points for each line, so we can graph each line by connecting the 2 points (x and y intercepts) for each line.
Using the slope of each line, we can plot additional points for each line and then observe the point at which the lines intersect (the point which is common to both lines):
For line, y = -3x - 4, additional points can be plotted either by going down 3 units and over 1 unit to the right (slope is -3/1) from the previous point or by going up 3 units and over 1 unit to the left (slope is 3/-1) from the previous point. In either, case we are using using the slope (m) of -3 to plot additional points.
For line, y = -0.5x + 1, additional points can be plotted either by going up 1 unit and over 1 unit to the right (slope is 1/1) from the previous point or by going down 1 unit and over 1 unit to the left (slope is -1/-1) from the previous point. In either case, we are using a slope (m) of 1 to plot additional points.
The point of intersection of these 2 lines, the common point on both lines, gives us our solution. The common point (our solution) is the point (-2,2).
We can verify that this is our solution by plugging in -2 for x and seeing if we get a y value of 2 in each line equation:
y = -3x - 4, y = -3(-2) - 4, y = 6 - 4, y = 2
y = -0.5x + 1, y = -0.5(-2) + 1, y = 1 + 1, y = 2
Both equations yield a value of 2 for y when plugging in the value of -2 for x, so our intersection point (our olution) of (-2,2) is correct.
To summarize our graphing procedure:
1. Put both line equations into slope intercept form (y = mx + b)
2. Plot the the y-intercept (b) for each line.
3. Find the x intercept for each line by setting y to zero and solving for x in the slop-intercept equation for each line.
4. Connect the 2 points (x and y intercept points) for each line.
5. Plot additional points for each line as needed (using slope (m) of each line) to determine at which point the lines intersect. The intersection point is the solution.
6. Check the solution by plugging in the x value of the intersection point in each line equation (slope-intercept form) and confirming the y value of the intersection point in each case.
I trust the above procedure will clarify for you how to graphically solve any system of 2 line equations.
Thanks for submitting this question.
Regards, Jordan.
y = -3x - 4
pick a number as close as possible to the origin. let's pick 0
plug in 0 to x:
y = -3(0) - 4 = -4
so, we have our first point (0, -4)
we can find another point by picking another number to be plugged in to x or we can use the given slope of the equation.
Since the slope is -3, we can represent this as -3/1. We know the slope is the change in y over the change in x. So, from the point (0, -4) move your pencil to the right one unit and down three units. This location will be another point. Now, just use a ruler to connect these points.
Another way to get another point by using the slope is adding the change in y to the y coordinate of the first point and adding the change in x to the x coordinate of the first point. So, we have (0 + 1, -4 + (-3)) = (1, -7), this is the second point.
Do the same for the other equation. |
## When would we use rounding in real life?
We round numbers because the estimate (rounded number) is used when we can’t find an exact WHOLE number or is used to make a number continuing on forever (0.333333333333333333333333333333333333333333 etc.) a whole number.
## Why do we use rounding?
Rounding is a way of simplifying numbers to make them easier to understand or work with. Rounding can be used when an exact number isn’t needed, and an approximate answer will do. For example, if you want to round a number to the nearest ten, the round off digit is the number in the tens’ place.
## What is the rule for rounding?
Rules for Rounding Here’s the general rule for rounding: If the number you are rounding is followed by 5, 6, 7, 8, or 9, round the number up. Example: 38 rounded to the nearest ten is 40. If the number you are rounding is followed by 0, 1, 2, 3, or 4, round the number down.
## Why do we round off 5 examples?
Rounding numbers means shortening the number of digits to the right of the decimal point to make the number easier to work with. If the decimal digit that you are truncating (removing) is 5 or more then you add 1 to the last digit that you keep. This is called rounding up.
## How do you round off to the nearest 1000?
The rule for rounding to the nearest ten thousand is to look at the last four digits. If the last four digits are 5,000 or greater, then we round our ten thousands digit up, and if it is less than 5,000, then we keep our ten thousands digit the same. For example, 5,765 rounds up to 10,000. 43,567 rounds down to 40,000.
## Where do you see decimals in real life?
We use decimals every day while dealing with money, weight, length etc. Decimal numbers are used in situations where more precision is required than the whole numbers can provide. For example, when we calculate our weight on the weighing machine, we do not always find the weight equal to a whole number on the scale.
## Where do you see percentages in real life?
Percentages are used widely and in many different areas. For example, discounts in shops, bank interest rates, rates of inflation and many statistics in the media are expressed as percentages. Percentages are important for understanding the financial aspects of everyday life.
## Where can you find fractions in everyday life?
Here are some examples of fractions in real life: Eating at a restaurant: Think about a time you go to a restaurant with friends and the waitress brings a single bill. To divide the total amongst the friends, you use fractions. Shopping: Think about the time you went shopping for a new school bag.
## Why is there a need for us to know how do you divide decimals mentally?
When you multiply and divide with decimals you can use mental strategies so that you can solve problems in your head. When you multiply a decimal by 10, 100, 1,000, 10,000 or other powers of ten, you can just move the decimal to the right for multiplication or the left for division to form your answer.
## How do we multiply quickly by 10 or 100 mentally?
When multiplying or dividing by 10 or 100, or any power of ten, for that matter, move the decimal point the same number of places as there are zeros in the power of ten factor.
point zero one
left
## How will you multiply decimals by 10 100 and 1000?
Follow this simple method when multiplying and dividing numbers with decimals: When we multiply by 10, 100 and 1000 we shift the digits to the left. One place left for 10, two places left for 100 and three places left for 1000.
## What happens when you multiply by 10 100 or 1000?
When we multiply whole numbers by 10, 100, 1000, we simply rewrite the numbers and put some extra zeros at the end. Rules for the multiplication by 10, 100 and 1000. If we multiply a whole number by a 10, then we write one zero at the end. If we multiply a whole number by a 1000, then we write three zeros at the end.
## What is the decimal of 100 100?
Example Values
Percent Decimal Fraction
80% 0.8 4/5
90% 0.9 9/10
99% 0.99 99/100
100% 1
## What happens when you divide a 2 digit number by 10 100 or 1000?
Therefore, when we divide by 100, the two digits in the ONES and the TENS place form the remainder while the remaining digits form the quotient. 4. Following this method, when we divide by 1000, the remainder will have 3 digits.
## What is the rule for multiplying by 10?
When multiplying whole numbers by 10, simply add a 0 to the end of the number, and you will have your answer. So, 5 * 10 is a 5 with a 0 at the end: 50. 3 * 10 is a 3 with a 0 at the end: 30.
## What can you divide to get 100?
Let n = the unknown number. Therefore, we divide 10,000 (ten thousand) by 100 to get or to produce 100.
## What is the rule for dividing by 10?
Here’s the rule for dividing by 10: move the decimal point one place to the left. Place value is the value of a digit based on its location in the number. Beginning with a decimal point and moving left, we have the ones, tens, hundreds, thousands, ten thousands, hundred thousands, and millions.
## Why do you add a zero when multiplying by 10?
Should you just add a zero when multiplying by 10? In this example, 9.5 x 10 isn’t 9.50 because simply inserting a zero on the end gives exactly the same value.
8
## Which is the same as dividing a number by 1 000?
To divide a number by 1000 we move each digit in that number three place value columns to the right. Dividing a whole number that ends in three zeros by 1000 has the same effect as removing the three zeros. This trick only works for whole numbers that end in three zeros (or numbers that are multiples of 1000).
10%
## How do you work out 5 divided by 100?
5/100 = 120 = 0.05.
## How do you work out 1000 divided by 8?
Answer to math problems Solution Steps
1. Math answers to division of fraction 1000/8. 1000 8 = 125.
2. Homework answers: (1000/2) + 20 = 520. 1000 divide by half plus 40.
3. Homework answers: (1000/2) + 40 = 540. 1000/8 divided by 2.
4. Answer: (1000/8) รท 2 = 62.5.
5. What is Numerator / Denominator.
6. Fraction to decimal conversion steps.
## What does 15 divided by 3 look like?
15 divided by 3 is equal to 5. If you have trouble with division, you can think of it like this.
## How do you solve 8 divided by 3?
The number 8 divided by 3 is 2 with a remainder of 2 (8 / 3 = 2 R.
## How do you work out 20 divided by 8?
What is 20 Divided by 8?
1. 20 divided by 8 in decimal = 2.5.
2. 20 divided by 8 in fraction = 20/8.
3. 20 divided by 8 in percentage = 250% |
# Question Video: Verifying the Outliers of a Data Set Mathematics • 8th Grade
The numbers of matches won by 12 teams in the national league are 11, 5, 6, 6, 9, 10, 19, 14, 11, 9, 9, and 6. Is it true or false that 19 is an outlier of the data?
02:27
### Video Transcript
The numbers of matches won by 12 teams in the national league are 11, five, six, six, nine, 10, 19, 14, 11, nine, nine, and six. Is it true or false that 19 is an outlier of the data?
To identify whether 19 is an outlier or not, we’ll need the interquartile range. And to do that, we’ll have to identify quartile one and quartile three. This means our first step is to put the data in order of size. Now, we have our 12 data points in size order.
We know that the median will come in the middle of these 12 data points and that the median is quartile two. 𝑄 one is the middle of the lower half of the data. Since there are six data points below the median, 𝑄 one will be located between the third and fourth. And similarly, 𝑄 three is the middle of the upper half of the data. There are six points above quartile two. And that means 𝑄 three will be located in the middle of those. It will be between the ninth and 10th value.
Because the third and the fourth value is six, we would call quartile one six. And because the ninth and 10th values are the same, quartile three is equal to 11. The interquartile range equals 𝑄 three minus 𝑄 one. For us, that’s 11 minus six. And so, we have an IQR of five. To find out if 19 is, in fact, an outlier, we’ll use the 1.5 times IQR rule. This rule tells us that a value is an outlier if it’s greater than 𝑄 three plus 1.5 times the IQR or less than 𝑄 one minus 1.5 times the IQR.
Since we’re looking at a data point that’s above 𝑄 three, we’ll look for the greater-than option. And that means we want to know is 19 greater than the quartile three plus 1.5 times the interquartile range? The IQR is five. 𝑄 three is 11. 1.5 times five is 7.5, plus 11 equals 18.5. 19 is greater than 18.5. And so, we can say it’s a true statement that 19 is an outlier of this data set.
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# How would you find a balanced equation if acetylene gas ("C"_2"H"_2) undergoes combustion to form carbon dioxide and water vapor?
Jun 20, 2016
One way is to "balance by inspection".
#### Explanation:
The unbalanced equation is
$\text{C"_2"H"_2 + "O"_2 → "CO"_2 + "H"_2"O}$
Let's start with the most complicated formula, ${\text{C"_2"H}}_{2}$. We put a 1 in front of it.
$\textcolor{red}{1} \text{C"_2"H"_2 + "O"_2 → "CO"_2+ "H"_2"O}$
We have fixed 2 $\text{C}$ atoms on the left, so we need 2 $\text{C}$ atoms on the right. We put a 2 in front of ${\text{CO}}_{2}$.
$\textcolor{red}{1} \text{C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + "H"_2"O}$
We have also fixed 2 $\text{H}$ atoms in the ${\text{C"_2"H}}_{2}$, so we need 2 $\text{H}$ atoms on the right. We put a 1 in front of the $\text{H"_2"O}$.
$\textcolor{red}{1} \text{C"_2"H"_2 + "O"_2 → color(orange)(2)"CO"_2 + color(blue)(1)"H"_2"O}$
Now we have fixed 5 $\text{O}$ atoms on the right, so we need 5 $\text{O}$ atoms on the left.
Oops! We would need to use 2½ molecules of ${\text{O}}_{2}$.
To avoid fractions, we multiply all coefficients by 2 and get
$\textcolor{red}{2} \text{C"_2"H"_2 + "O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O}$
We now have 10 $\text{O}$ atoms on the right, so we need 10 $\text{O}$ atoms on the left. We can now put a 5 in front of the ${\text{O}}_{2}$.
$\textcolor{red}{2} \text{C"_2"H"_2 + color(orange)(5)"O"_2 → color(red)(4)"CO"_2 + color(blue)(2)"H"_2"O}$
The equation should now be balanced. Let's check.
$\text{Atom"color(white)(m) "On the left"color(white)(m) "On the right}$
stackrel(—————————————)(color(white)(m)"C"color(white)(mmmmm)4color(white)(mmmmmm) 4)
$\textcolor{w h i t e}{m} \text{H} \textcolor{w h i t e}{m m m m m} 4 \textcolor{w h i t e}{m m m m m m} 4$
$\textcolor{w h i t e}{m} \text{O} \textcolor{w h i t e}{m m m m l} 10 \textcolor{w h i t e}{m m m m m l l} 10$
The equation is balanced! |
On the campus of
The Five Triangle Puzzle: Composing and Decomposing
Beginning in first grade and continuing through the grades, the Common Core Math Standards emphasize composing and decomposing shapes. One way to give students experience with composing shapes is through put-together puzzles. Actually, working at put-together puzzles involves lots of composing and decomposing. Sometimes what we compose is a solution; sometimes it’s not.
While the tangram puzzle is an obvious, well known example of such a puzzle, there are others that students will find interesting. One of my favorites is a puzzle made up of five triangles. In this post I want to tell you about this puzzle; I will give you a master so you can make a Five Triangle Puzzle for yourself. I’m then going to leave it for you to solve the puzzle, and finally, I’m going to come back to it in a later post to talk about the solutions. I’m hoping in the mean time to hear from you about how you thought about the puzzle and to hear about your solutions and anything you might have learned by working it.
I first saw this puzzle when a colleague dropped it on my desk, probably 20 years ago. He simply commented that he had found one solution, but was sure there was at least one other one. I was hooked; I couldn’t leave it alone until I’d explored all the possibilities for solutions and found them or demonstrated that they were not possible.
Let me show you the pieces in this short video and then I’ll finish up by telling you about the challenge of the puzzle.
The puzzle consists of 5 triangles. They are right scalene triangles; actually, they are 30-60-90 triangles. The triangles are two sizes. There are two that are smaller, and three larger. You can see that when two of the same size are put together, the form an equilateral triangle. They are related in that the hypotenuse of the smaller triangle is equal to the long leg of the larger triangle. This will be important to know as you try to put the triangles together form other triangles. Read on to see the challenges posed by the puzzle.
The first challenge is to simply put all five of the triangles together to form one large triangle. The second challenge is to compose a second, different triangle using all five triangles. The third challenge is to determine whether there might be yet one more triangle that can be formed using all of the pieces.
I’m hoping you will accept the challenge posed by this puzzle and that you will let us know about your experience.
2 Responses to The Five Triangle Puzzle: Composing and Decomposing
1. […] The Five Triangle Puzzle was the subject of a post back on February 11. I’m hopeful that some of you will have downloaded the pieces and solved the puzzle. The challenge was to put all five pieces together to form a triangle and then to determine if there were other triangles that could be formed using all five pieces. […] |
# Spring constant $K$
A mass $$m_1$$ with initial velocity $$V_0$$ collides with a spring attached to mass $$m_2$$ initially resting on a frictionless surface according to the figure below. Considering the spring constant $$K$$ and the negligible mass, do you ask if:
a) Determine the maximum spring compression.
b) If long time after collision both objects travel in the same direction, determine the velocities $$V_1$$ and $$V_2$$ of masses $$m_1$$ and $$m_2$$ respectively.
$$Attemp:$$ This is similar to 2017 F=ma Problem 24.
(a) We use conservation of energy and momentum. We have an inelastic collision when the ball hits the block and spring configuration. Since the spring is massless we can write a conservation of momentum equation without worrying too much about the spring yet. $$m_1v_1=(m_1+m_2)v’\implies v’=\frac{m_1v_1}{m_1+m_2}$$Next we use conservation of energy. We write the equation $$\frac{1}{2}m_1v_1^2=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)v’^2$$ substituting $$v’=\frac{m_1v_1}{m_1+m_2}$$ we get $$\frac{1}{2}m_1v_1^2=\frac{1}{2}kx^2+\frac{1}{2}(m_1+m_2)\left(\frac{m_1v_1}{m_1+m_2}\right)^2$$ simplifying and multiplying both sides by two we have $$m_1v_1^2=kx^2+\frac{m_1^2v_1^2}{m_1+m_2}$$ subtracting both sides and simplifying the equation for only $$x$$ on one side we have $$x=\sqrt{\frac{m_1v_1^2\left(1-\frac{m_1}{m_1+m_2}\right)}{k}}$$ $$x=v_1\sqrt{\frac{m\left(\frac{m_1+m_2}{m_1+m_2}-\frac{m_1}{m_1+m_2}\right)}{k}}$$ $$x=v_1\sqrt{\frac{m_1m_2}{k(m_1+m_2)}}$$
How can I solve a b?
• Why do you assume that the velocities of both the masses will be the same after the collision? Commented Dec 21, 2019 at 15:56
• @sudeep5221 Excuse me! See image above Commented Dec 21, 2019 at 16:06
The result for part a is correct, but the reasoning is not perfect, therefore you have questions about part b. So let's describe what is happening. As soon as the object of mass $$m_1$$ hits the spring, it will start slowing down, and object of mass $$m_2$$ will start accelerating. At any point after the collision, we call $$v_1$$ the velocity of mass $$m_1$$ and $$v_2$$ will be the velocity of mass $$m_2$$. You correctly pointed out that momentum is conserved: $$m_1v_0=m_1v_1+m_2v_2$$ As long as $$v_1>v_2$$ the mass $$m_1$$ will get closer to $$m_2$$. Since $$v_1$$ is decreasing and $$v_2$$ is increasing, at some point they will have the same velocity $$v$$. After this point the spring is still compressed, so $$m_1$$ will still slow down even more and $$m_2$$ will continue accelerating, but the distance between them is increasing. So the minimum distance is when they have the same velocities. Then your calculations seem correct.
Now what is happening afterwards? Like I've mentioned, $$m_1$$ will continue slowing down and $$m_2$$ will continue accelerating, while the spring returns to the original length. At this point there is no more force acting on either objects, so , if the spring does not get stuck to mass $$m_1$$, they will keep moving with whatever velocities they have at that point. You can still write the same equation for conservation of momentum as above, and you can write the conservation of energy, which will look like the one for elastic collision: $$\frac 12 m_1v_0^2=\frac 12 m_1v_1^2+\frac12m_2v_2^2$$ Notice that the spring is uncompressed, so the elastic energy is still the same as before collision. You now have two equations with two unknowns, which I assume you already know how to solve.
Let me know if something is unclear. |
# HOW TO FIND THE UNIT DIGIT IN THE PRODUCT EXPRESSION WITH EXPONENTS
How to Find the Unit Digit in the Product Expression with Exponents ?
In this section, we will see how to find the unit digit in the product expression with exponents.
To identify the unit digit of a number with some power, we must be aware of cyclicity.
## Cyclicity of Numbers
Cyclicity of any number is about the last digit and how they appear in a certain defined manner.
Example 1 :
Let us consider the values of 2n, where n = 1, 2, 3, ...........
2 = 2
22 = 4
23 = 8
24 = 16
25 = 32
26 = 64
In the above calculations of 2n
We get unit digit 2 in the result of 2n, when n = 1.
Again we get 2 in the unit digit of 2n, when n = 5.
That is, in the fifth term.
So, the cyclicity of 2 is 4.
Example 2 :
Let us consider the values of 3n, where n = 1, 2, 3, ...........
31 = 3
32 = 9
33 = 27
34 = 81
35 = 243
36 = 729
In the above calculations of 3n
We get unit digit 3 in the result of 3n, when n = 1.
Again we get 3 in the unit digit of 3n, when n = 5.
That is, in the fifth term.
So, the cyclicity of 3 is 4.
In the same way, we can get cyclicity of others numbers as shown below.
## Cyclicity of 4 and 5
Cyclicity of 441 = 442 = 1643 = 6444 = 256The cyclicity of 4 is 2. Cyclicity of 551 = 552 = 2553 = 125The cyclicity of 5 is 1.
## Cyclicity of 6 and 7
Cyclicity of 661 = 662 = 3663 = 216The cyclicity of 6 is 1. Cyclicity of 771 = 772 = 4973 = 34374 = 240175 = 16807The cyclicity of 7 is 4.
## Cyclicity of 8, 9 and 10
Cyclicity of 881 = 882 = 6483 = 51284 = 409685 = 32768Cyclicity of 8 is 4. Cyclicity of 991 = 992 = 8193 = 729Cyclicity of 9 is 2.Cyclicity of 10101 = 10102 = 100Cyclicity of 10 is 1.
## Cyclicity of Numbers - Summary
Number12345678910 Cyclicity of a number1442114421
## Find the Unit Digit in the Product Expression with Exponents - Examples
Example 1 :
Find the unit digit in the product :
(3547)153 x (251)72
Solution :
In (3547)153, unit digit is 7.
The cyclicity of 7 is 4. Dividing 153 by 4, we get 1 as remainder.
71 = 7
So, the unit digit of 7153 is 7.
In 25172, unit digit is 1.
Because 1 has the cyclicity 1, the unit digit of 25172 is 1.
By multiplying the unit digits, we get
7 x 1 = 7
Therefore, the unit digit of the expression
(3547)153 x (251)72 is 7.
Example 2 :
Find unit digit in the product :
(6374)1793 x (625)317 x (341)491
Solution :
In (6374)1793, unit digit is 4.
The cyclicity of 4 is 2. Dividing 1793 by 4, we get 1 as remainder.
41 = 4
So, the unit digit of (6374)1793 is 4.
In (625)317, unit digit is 5.
Since 5 has the cyclicity 1, the unit digit of (625)317 is 5.
In (341)491, unit digit is 1.
Since 1 has the cyclicity 1, the unit digit of (341)491 is 1.
By multiplying the unit digits, we get
4 x 5 x 1 = 20
The unit digit of 20 is '0'.
Therefore, the unit digit of the expression
(6374)1793 x (625)317 x (341)491 is 0
After having gone through the stuff given above, we hope that the students would understood how to find the unit digit in the product expression with exponent.
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# 14.7: Directional Derivatives and the Gradient
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Learning Objectives
• Determine the directional derivative in a given direction for a function of two variables.
• Determine the gradient vector of a given real-valued function.
• Explain the significance of the gradient vector with regard to direction of change along a surface.
• Use the gradient to find the tangent to a level curve of a given function.
• Calculate directional derivatives and gradients in three dimensions.
A function $$z=f(x,y)$$ has two partial derivatives: $$∂z/∂x$$ and $$∂z/∂y$$. These derivatives correspond to each of the independent variables and can be interpreted as instantaneous rates of change (that is, as slopes of a tangent line). For example, $$∂z/∂x$$ represents the slope of a tangent line passing through a given point on the surface defined by $$z=f(x,y),$$ assuming the tangent line is parallel to the $$x$$-axis. Similarly, $$∂z/∂y$$ represents the slope of the tangent line parallel to the $$y$$-axis. Now we consider the possibility of a tangent line parallel to neither axis.
## Directional Derivatives
We start with the graph of a surface defined by the equation $$z=f(x,y)$$. Given a point $$(a,b)$$ in the domain of $$f$$, we choose a direction to travel from that point. We measure the direction using an angle $$θ$$, which is measured counterclockwise in the $$xy$$-plane, starting at zero from the positive $$x$$-axis (Figure $$\PageIndex{1}$$). The distance we travel is $$h$$ and the direction we travel is given by the unit vector $$\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}.$$ Therefore, the $$z$$-coordinate of the second point on the graph is given by $$z=f(a+h\cos θ,b+h\sin θ).$$
We can calculate the slope of the secant line by dividing the difference in $$z$$-values by the length of the line segment connecting the two points in the domain. The length of the line segment is $$h$$. Therefore, the slope of the secant line is
$m_{sec}=\dfrac{f(a+h\cos θ,b+h\sin θ)−f(a,b)}{h}$
To find the slope of the tangent line in the same direction, we take the limit as $$h$$ approaches zero.
Definition: Directional Derivatives
Suppose $$z=f(x,y)$$ is a function of two variables with a domain of $$D$$. Let $$(a,b)∈D$$ and define $$\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}$$. Then the directional derivative of $$f$$ in the direction of $$\vecs u$$ is given by
$D_{\vecs u}f(a,b)=\lim_{h→0}\dfrac{f(a+h \cos θ,b+h\sin θ)−f(a,b)}{h} \label{DD}$
provided the limit exists.
Equation \ref{DD} provides a formal definition of the directional derivative that can be used in many cases to calculate a directional derivative.
Note that since the point $$(a, b)$$ is chosen randomly from the domain $$D$$ of the function $$f$$, we can use this definition to find the directional derivative as a function of $$x$$ and $$y$$.
That is,
$D_{\vecs u}f(x,y)=\lim_{h→0}\dfrac{f(x+h \cos θ,y+h\sin θ)−f(x,y)}{h} \label{DDxy}$
Example $$\PageIndex{1}$$: Finding a Directional Derivative from the Definition
Let $$θ=\arccos(3/5).$$ Find the directional derivative $$D_{\vecs u}f(x,y)$$ of $$f(x,y)=x^2−xy+3y^2$$ in the direction of $$\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}$$.
Then determine $$D_{\vecs u}f(−1,2)$$.
Solution
First of all, since $$\cos θ=3/5$$ and $$θ$$ is acute, this implies
$\sin θ=\sqrt{1−\left(\dfrac{3}{5}\right)^2}=\sqrt{\dfrac{16}{25}}=\dfrac{4}{5}. \nonumber$
Using $$f(x,y)=x^2−xy+3y^2,$$ we first calculate $$f(x+h\cos θ,y+h\sin θ)$$:
\begin{align*} f(x+h\cos θ,y+h\sin θ) &=(x+h\cos θ)^2−(x+h\cos θ)(y+h\sin θ)+3(y+h\sin θ)^2 \\ &=x^2+2xh\cos θ+h^2\cos^2 θ−xy−xh\sin θ−yh\cos θ−h^2\sin θ\cos θ+3y^2+6yh\sin θ+3h^2\sin^2 θ \\ &=x^2+2xh(\frac{3}{5})+\frac{9h^2}{25}−xy−\frac{4xh}{5}−\frac{3yh}{5}−\frac{12h^2}{25}+3y^2+6yh(\frac{4}{5})+3h^2(\frac{16}{25})\\ &=x^2−xy+3y^2+\frac{2xh}{5}+\frac{9h^2}{5}+\frac{21yh}{5}. \end{align*}
We substitute this expression into Equation \ref{DD} with $$a = x$$ and $$b = y$$:
\begin{align*} D_{\vecs u}f(x,y) &=\lim_{h→0}\frac{f(x+h\cos θ,y+h\sin θ)−f(x,y)}{h}\\ &=\lim_{h→0}\frac{(x^2−xy+3y^2+\frac{2xh}{5}+\frac{9h^2}{5}+\frac{21yh}{5})−(x^2−xy+3y^2)}{h}\\ &=\lim_{h→0}\frac{\frac{2xh}{5}+\frac{9h^2}{5}+\frac{21yh}{5}}{h}\\ &=\lim_{h→0}\frac{2x}{5}+\frac{9h}{5}+\frac{21y}{5}\\ &=\frac{2x+21y}{5}. \end{align*}
To calculate $$D_{\vecs u}f(−1,2),$$ we substitute $$x=−1$$ and $$y=2$$ into this answer (Figure $$\PageIndex{2}$$):
$D_{\vecs u}f(−1,2)=\dfrac{2(−1)+21(2)}{5}=\dfrac{−2+42}{5}=8. \nonumber$
An easier approach to calculating directional derivatives that involves partial derivatives is outlined in the following theorem.
Directional Derivative of a Function of Two Variables
Let $$z=f(x,y)$$ be a function of two variables $$x$$ and $$y$$, and assume that $$f_x$$ and $$f_y$$ exist. Then the directional derivative of $$f$$ in the direction of $$\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}$$ is given by
$D_{\vecs u}f(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ. \label{DD2v}$
Proof
Applying the definition of a directional derivative stated above in Equation \ref{DD}, the directional derivative of $$f$$ in the direction of $$\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}$$ at a point $$(x_0, y_0)$$ in the domain of $$f$$ can be written
$D_{\vecs u}f((x_0, y_0))=\lim_{t→0}\dfrac{f(x_0+t \cos θ,y_0+t\sin θ)−f(x_0,y_0)}{t}.$
Let $$x=x_0+t\cos θ$$ and $$y=y_0+t\sin θ,$$ and define $$g(t)=f(x,y)$$. Since $$f_x$$ and $$f_y$$ both exist, we can use the chain rule for functions of two variables to calculate $$g′(t)$$:
$g′(t)=\dfrac{∂f}{∂x}\dfrac{dx}{dt}+\dfrac{∂f}{∂y}\dfrac{dy}{dt}=f_x(x,y)\cos θ+f_y(x,y)\sin θ.$
If $$t=0,$$ then $$x=x_0$$ and $$y=y_0,$$ so
$g′(0)=f_x(x_0,y_0)\cos θ+f_y(x_0,y_0)\sin θ$
By the definition of $$g′(t),$$ it is also true that
$g′(0)=\lim_{t→0}\dfrac{g(t)−g(0)}{t}=\lim_{t→0}\dfrac{f(x_0+t\cos θ,y_0+t\sin θ)−f(x_0,y_0)}{t}.$
Therefore, $$D_{\vecs u}f(x_0,y_0)=f_x(x_0,y_0)\cos θ+f_y(x_0,y_0)\sin θ$$.
Since the point $$(x_0,y_0)$$ is an arbitrary point from the domain of $$f$$, this result holds for all points in the domain of $$f$$ for which the partials $$f_x$$ and $$f_y$$ exist.
Therefore, $D_{\vecs u}f(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ.$
Example $$\PageIndex{2}$$: Finding a Directional Derivative: Alternative Method
Let $$θ=\arccos (3/5).$$ Find the directional derivative $$D_{\vecs u}f(x,y)$$ of $$f(x,y)=x^2−xy+3y^2$$ in the direction of $$\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}$$.
Then determine $$D_{\vecs u}f(−1,2)$$.
Solution
First, we must calculate the partial derivatives of $$f$$:
\begin{align*}f_x(x,y) &=2x−y \\ f_y(x,y) &=−x+6y, \end{align*}
Then we use Equation \ref{DD2v} with $$θ=\arccos (3/5)$$:
\begin{align*} D_{\vecs u}f(x,y) &=f_x(x,y)\cos θ+f_y(x,y)\sin θ \\ &=(2x−y)\dfrac{3}{5}+(−x+6y)\dfrac{4}{5} \\ &=\dfrac{6x}{5}−\dfrac{3y}{5}−\dfrac{4x}{5}+\dfrac{24y}{5}\\ &=\dfrac{2x+21y}{5}. \end{align*}
To calculate $$D_{\vecs u}f(−1,2),$$ let $$x=−1$$ and $$y=2$$:
$D_{\vecs u}f(−1,2)=\dfrac{2(−1)+21(2)}{5}=\dfrac{−2+42}{5}=8.\nonumber$
This is the same answer obtained in Example $$\PageIndex{1}$$.
Exercise $$\PageIndex{1}$$:
Find the directional derivative $$D_{\vecs u}f(x,y)$$ of $$f(x,y)=3x^2y−4xy^3+3y^2−4x$$ in the direction of $$\vecs u=(\cos \dfrac{π}{3})\,\hat{\mathbf i}+(\sin \dfrac{π}{3})\,\hat{\mathbf j}$$ using Equation \ref{DD2v}.
What is $$D_{\vecs u} f(3,4)$$?
Hint
Calculate the partial derivatives and determine the value of $$θ$$.
$$D_{\vecs u}f(x,y)=\dfrac{(6xy−4y^3−4)(1)}{2}+\dfrac{(3x^2−12xy^2+6y)\sqrt{3}}{2}$$
$$D_{\vecs u}f(3,4)=\dfrac{72−256−4}{2}+\dfrac{(27−576+24)\sqrt{3}}{2}=−94−\dfrac{525\sqrt{3}}{2}$$
If the vector that is given for the direction of the derivative is not a unit vector, then it is only necessary to divide by the norm of the vector. For example, if we wished to find the directional derivative of the function in Example $$\PageIndex{2}$$ in the direction of the vector $$⟨−5,12⟩$$, we would first divide by its magnitude to get $$\vecs u$$. This gives us $$\vecs u=⟨−\frac{5}{13},\frac{12}{13}⟩$$.
Then
\begin{align*} D_{\vecs u}f(x,y) &=f_x(x,y)\cos θ+f_y(x,y)\sin θ \\ &=−\dfrac{5}{13}(2x−y)+\dfrac{12}{13}(−x+6y) \\ &=−\dfrac{22}{13}x+\dfrac{17}{13}y \end{align*}
The right-hand side of Equation \ref{DD2v} is equal to $$f_x(x,y)\cos θ+f_y(x,y)\sin θ$$, which can be written as the dot product of two vectors. Define the first vector as $$\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}$$ and the second vector as $$\vecs u=(\cos θ)\,\hat{\mathbf i}+(\sin θ)\,\hat{\mathbf j}$$. Then the right-hand side of the equation can be written as the dot product of these two vectors:
$D_{\vecs u}f(x,y)=\vecs ∇f(x,y)⋅\vecs u. \label{gradDirDer}$
The first vector in Equation \ref{gradDirDer} has a special name: the gradient of the function $$f$$. The symbol $$∇$$ is called nabla and the vector $$\vecs ∇f$$ is read “del $$f$$.”
Let $$z=f(x,y)$$ be a function of $$x$$ and $$y$$ such that $$f_x$$ and $$f_y$$ exist. The vector $$\vecs ∇f(x,y)$$ is called the gradient of $$f$$ and is defined as
$\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}. \label{grad}$
The vector $$\vecs ∇f(x,y)$$ is also written as “grad $$f$$.”
Example $$\PageIndex{3}$$: Finding Gradients
Find the gradient $$\vecs ∇f(x,y)$$ of each of the following functions:
1. $$f(x,y)=x^2−xy+3y^2$$
2. $$f(x,y)=\sin 3 x \cos 3y$$
Solution
For both parts a. and b., we first calculate the partial derivatives $$f_x$$ and $$f_y$$, then use Equation \ref{grad}.
a. $$f_x(x,y)=2x−y$$ and $$f_y(x,y)=−x+6y$$, so
\begin{align*} \vecs ∇f(x,y) &=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}\\ &=(2x−y)\,\hat{\mathbf i}+(−x+6y)\,\hat{\mathbf j}.\end{align*}
b. $$f_x(x,y)=3\cos 3x \cos 3y$$ and $$f_y(x,y)=−3\sin 3x \sin 3y$$, so
\begin{align*} \vecs ∇f(x,y) &=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j} \\ &=(3\cos 3x \cos 3y)\,\hat{\mathbf i}−(3\sin 3x \sin 3y)\,\hat{\mathbf j}. \end{align*}
Exercise $$\PageIndex{2}$$
Find the gradient $$\vecs ∇f(x,y)$$ of $$f(x,y)=\dfrac{x^2−3y^2}{2x+y}$$.
Hint
Calculate the partial derivatives, then use Equation \ref{grad}.
$$\vecs ∇f(x,y)=\dfrac{2x^2+2xy+6y^2}{(2x+y)^2}\,\hat{\mathbf i}−\dfrac{x^2+12xy+3y^2}{(2x+y)^2}\,\hat{\mathbf j}$$
The gradient has some important properties. We have already seen one formula that uses the gradient: the formula for the directional derivative. Recall from The Dot Product that if the angle between two vectors $$\vecs a$$ and $$\vecs b$$ is $$φ$$, then $$\vecs a⋅\vecs b=‖\vecs a‖‖\vecs b‖\cos φ.$$ Therefore, if the angle between $$\vecs ∇f(x_0,y_0)$$ and $$\vecs u=(cosθ)\,\hat{\mathbf i}+(sinθ)\,\hat{\mathbf j}$$ is $$φ$$, we have
$D_{\vecs u}f(x_0,y_0)=\vecs ∇f(x_0,y_0)⋅\vecs u=\|\vecs ∇f(x_0,y_0)\|‖\vecs u‖\cos φ=\|\vecs ∇f(x_0,y_0)\|\cos φ.$
The $$‖\vecs u‖$$ disappears because $$\vecs u$$ is a unit vector. Therefore, the directional derivative is equal to the magnitude of the gradient evaluated at $$(x_0,y_0)$$ multiplied by $$\cos φ$$. Recall that $$\cos φ$$ ranges from $$−1$$ to $$1$$.
If $$φ=0,$$ then $$\cos φ=1$$ and $$\vecs ∇f(x_0,y_0)$$ and $$\vecs u$$ both point in the same direction.
If $$φ=π$$, then $$\cos φ=−1$$ and $$\vecs ∇f(x_0,y_0)$$ and $$\vecs u$$ point in opposite directions.
In the first case, the value of $$D_{\vecs u}f(x_0,y_0)$$ is maximized; in the second case, the value of $$D_{\vecs u}f(x_0,y_0)$$ is minimized.
We can also see that if $$\vecs ∇f(x_0,y_0)=\vecs 0$$, then
$D_{\vecs u}f(x_0,y_0)=\vecs ∇f(x_0,y_0)⋅\vecs u=0$
for any vector $$\vecs u$$. These three cases are outlined in the following theorem.
Properties of the Gradient
Suppose the function $$z=f(x,y)$$ is differentiable at $$(x_0,y_0)$$ (Figure $$\PageIndex{3}$$).
1. If $$\vecs ∇f(x_0,y_0)=\vecs 0$$, then $$D_{\vecs u}f(x_0,y_0)=0$$ for any unit vector $$\vecs u$$.
2. If $$\vecs ∇f(x_0,y_0)≠0$$, then $$D_{\vecs u}f(x_0,y_0)$$ is maximized when $$\vecs u$$ points in the same direction as $$\vecs ∇f(x_0,y_0)$$. The maximum value of $$D_{\vecs u}f(x_0,y_0)$$ is $$\|\vecs ∇f(x_0,y_0)\|$$.
3. If $$\vecs ∇f(x_0,y_0)≠0$$, then $$D_{\vecs u}f(x_0,y_0)$$ is minimized when $$\vecs u$$ points in the opposite direction from $$\vecs ∇f(x_0,y_0)$$. The minimum value of $$D_{\vecs u}f(x_0,y_0)$$ is $$−\|\vecs ∇f(x_0,y_0)\|$$.
Example $$\PageIndex{4}$$: Finding a Maximum Directional Derivative
Find the direction for which the directional derivative of $$f(x,y)=3x^2−4xy+2y^2$$ at $$(−2,3)$$ is a maximum. What is the maximum value?
Solution:
The maximum value of the directional derivative occurs when $$\vecs ∇f$$ and the unit vector point in the same direction. Therefore, we start by calculating $$\vecs ∇f(x,y$$):
$f_x(x,y)=6x−4y \; \text{and}\; f_y(x,y)=−4x+4y \nonumber$
so
$\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}=(6x−4y)\,\hat{\mathbf i}+(−4x+4y)\,\hat{\mathbf j}. \nonumber$
Next, we evaluate the gradient at $$(−2,3)$$:
$\vecs ∇f(−2,3)=(6(−2)−4(3))\,\hat{\mathbf i}+(−4(−2)+4(3))\,\hat{\mathbf j}=−24\,\hat{\mathbf i}+20\,\hat{\mathbf j}. \nonumber$
We need to find a unit vector that points in the same direction as $$\vecs ∇f(−2,3),$$ so the next step is to divide $$\vecs ∇f(−2,3)$$ by its magnitude, which is $$\sqrt{(−24)^2+(20)^2}=\sqrt{976}=4\sqrt{61}$$. Therefore,
$\dfrac{\vecs ∇f(−2,3)}{\|\vecs ∇f(−2,3)\|}=\dfrac{−24}{4\sqrt{61}}i+\dfrac{20}{4\sqrt{61}}j=−\dfrac{6\sqrt{61}}{61}\,\hat{\mathbf i}+\dfrac{5\sqrt{61}}{61}\,\hat{\mathbf j}. \nonumber$
This is the unit vector that points in the same direction as $$\vecs ∇f(−2,3).$$ To find the angle corresponding to this unit vector, we solve the equations
$\cos θ=\dfrac{−6\sqrt{61}}{61}\; \text{and}\; \sin θ=\dfrac{5\sqrt{61}}{61} \nonumber$
for $$θ$$. Since cosine is negative and sine is positive, the angle must be in the second quadrant. Therefore, $$θ=π−\arcsin((5\sqrt{61})/61)≈2.45$$ rad.
The maximum value of the directional derivative at $$(−2,3)$$ is $$\|\vecs ∇f(−2,3)\|=4\sqrt{61}$$ (Figure $$\PageIndex{4}$$).
Exercise $$\PageIndex{3}$$
Find the direction for which the directional derivative of $$g(x,y)=4x−xy+2y^2$$ at $$(−2,3)$$ is a maximum. What is the maximum value?
Hint
Evaluate the gradient of $$g$$ at point $$(−2,3)$$.
The gradient of $$g$$ at $$(−2,3)$$ is $$\vecs ∇g(−2,3)=\,\hat{\mathbf i}+14\,\hat{\mathbf j}$$. The unit vector that points in the same direction as $$\vecs ∇g(−2,3)$$ is
$\dfrac{\vecs ∇g(−2,3)}{\|\vecs ∇g(−2,3)\|}=\dfrac{1}{\sqrt{197}}\,\hat{\mathbf i}+\dfrac{14}{\sqrt{197}}\,\hat{\mathbf j}=\dfrac{\sqrt{197}}{197}\,\hat{\mathbf i}+\dfrac{14\sqrt{197}}{197}\,\hat{\mathbf j},\nonumber$
which gives an angle of $$θ=\arcsin ((14\sqrt{197})/197)≈1.499$$ rad.
The maximum value of the directional derivative is $$\|\vecs ∇g(−2,3)\|=\sqrt{197}$$.
Figure $$\PageIndex{5}$$ shows a portion of the graph of the function $$f(x,y)=3+\sin x \sin y$$. Given a point $$(a,b)$$ in the domain of $$f$$, the maximum value of the gradient at that point is given by $$\|\vecs ∇f(a,b)\|$$. This would equal the rate of greatest ascent if the surface represented a topographical map. If we went in the opposite direction, it would be the rate of greatest descent.
When using a topographical map, the steepest slope is always in the direction where the contour lines are closest together (Figure $$\PageIndex{6}$$). This is analogous to the contour map of a function, assuming the level curves are obtained for equally spaced values throughout the range of that function.
## Gradients and Level Curves
Recall that if a curve is defined parametrically by the function pair $$(x(t),y(t)),$$ then the vector $$x′(t)\,\hat{\mathbf i}+y′(t)\,\hat{\mathbf j}$$ is tangent to the curve for every value of $$t$$ in the domain. Now let’s assume $$z=f(x,y)$$ is a differentiable function of $$x$$ and $$y$$, and $$(x_0,y_0)$$ is in its domain. Let’s suppose further that $$x_0=x(t_0)$$ and $$y_0=y(t_0)$$ for some value of $$t$$, and consider the level curve $$f(x,y)=k$$. Define $$g(t)=f(x(t),y(t))$$ and calculate $$g′(t)$$ on the level curve. By the chain Rule,
$g′(t)=f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t).$
But $$g′(t)=0$$ because $$g(t)=k$$ for all $$t$$. Therefore, on the one hand,
$f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t)=0;$
on the other hand,
$f_x(x(t),y(t))x′(t)+f_y(x(t),y(t))y′(t)=\vecs ∇f(x,y)⋅⟨x′(t),y′(t)⟩.$
Therefore,
$\vecs ∇f(x,y)⋅⟨x′(t),y′(t)⟩=0.$
Thus, the dot product of these vectors is equal to zero, which implies they are orthogonal. However, the second vector is tangent to the level curve, which implies the gradient must be normal to the level curve, which gives rise to the following theorem.
Gradient Is Normal to the Level Curve
Suppose the function $$z=f(x,y)$$ has continuous first-order partial derivatives in an open disk centered at a point $$(x_0,y_0)$$. If $$\vecs ∇f(x_0,y_0)≠0$$, then $$\vecs ∇f(x_0,y_0)$$ is normal to the level curve of $$f$$ at $$(x_0,y_0).$$
We can use this theorem to find tangent and normal vectors to level curves of a function.
Example $$\PageIndex{5}$$: Finding Tangents to Level Curves
For the function $$f(x,y)=2x^2−3xy+8y^2+2x−4y+4,$$ find a tangent vector to the level curve at point $$(−2,1)$$. Graph the level curve corresponding to $$f(x,y)=18$$ and draw in $$\vecs ∇f(−2,1)$$ and a tangent vector.
Solution:
First, we must calculate $$\vecs ∇f(x,y):$$
$f_x(x,y)=4x−3y+2 \;\text{and}\; f_y=−3x+16y−4 \;\text{so}\; \vecs ∇f(x,y)=(4x−3y+2)\,\hat{\mathbf i}+(−3x+16y−4)\,\hat{\mathbf j}.\nonumber$
Next, we evaluate $$\vecs ∇f(x,y)$$ at $$(−2,1):$$
$\vecs ∇f(−2,1)=(4(−2)−3(1)+2)\,\hat{\mathbf i}+(−3(−2)+16(1)−4)\,\hat{\mathbf j}=−9\,\hat{\mathbf i}+18\,\hat{\mathbf j}.\nonumber$
This vector is orthogonal to the curve at point $$(−2,1)$$. We can obtain a tangent vector by reversing the components and multiplying either one by $$−1$$. Thus, for example, $$−18\,\hat{\mathbf i}−9\,\hat{\mathbf j}$$ is a tangent vector (Figure $$\PageIndex{7}$$).
Exercise $$\PageIndex{4}$$
For the function $$f(x,y)=x^2−2xy+5y^2+3x−2y+4$$, find the tangent to the level curve at point $$(1,1)$$. Draw the graph of the level curve corresponding to $$f(x,y)=8$$ and draw $$\vecs ∇f(1,1)$$ and a tangent vector.
Hint
Calculate the gradient at point $$(1,1)$$.
$$\vecs ∇f(x,y)=(2x−2y+3)\,\hat{\mathbf i}+(−2x+10y−2)\,\hat{\mathbf j}$$
$$\vecs ∇f(1,1)=3\,\hat{\mathbf i}+6\,\hat{\mathbf j}$$
Tangent vector: $$6\,\hat{\mathbf i}−3\,\hat{\mathbf j}$$ or $$−6\,\hat{\mathbf i}+3\,\hat{\mathbf j}$$
## Three-Dimensional Gradients and Directional Derivatives
The definition of a gradient can be extended to functions of more than two variables.
Definition: Gradients in 3D
Let $$w=f(x, y, z)$$ be a function of three variables such that $$f_x, \, f_y$$, and $$f_z$$ exist. The vector $$\vecs \nabla f(x,y,z)$$ is called the gradient of $$f$$ and is defined as
$\vecs ∇f(x,y,z)=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k}.$
$$\vecs ∇f(x,y,z)$$ can also be written as grad $$f(x,y,z).$$
Calculating the gradient of a function in three variables is very similar to calculating the gradient of a function in two variables. First, we calculate the partial derivatives $$f_x, \, f_y,$$ and $$f_z$$, and then we use Equation \ref{gradDirDer}.
Example $$\PageIndex{6}$$: Finding Gradients in Three Dimensions
Find the gradient $$\vecs ∇f(x,y,z)$$ of each of the following functions:
1. $$f(x,y,z)=5x^2−2xy+y^2−4yz+z^2+3xz$$
2. $$f(x,y,z)=e^{−2z}\sin 2x \cos 2y$$
Solution:
For both parts a. and b., we first calculate the partial derivatives $$f_x,f_y,$$ and $$f_z$$, then use Equation \ref{gradDirDer}.
a. $$f_z(x,y,z)=10x−2y+3z$$, $$f_y(x,y,z)=−2x+2y−4z$$, and $$f_z(x,y,z)=3x−4y+2z$$, so
\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(10x−2y+3z)\,\hat{\mathbf i}+(−2x+2y−4z)\,\hat{\mathbf j}+(−4x+3y+2z)\,\hat{\mathbf k}. \end{align*}
b. $$f_x(x,y,z) =−2e^{−2z}\cos 2x \cos 2y$$, $$f_y(x,y,z)=−2e^{−2z} \sin 2x \sin 2y$$, and $$f_z(x,y,z)=−2e^{−2z}\sin 2x \cos 2y$$, so
\begin{align*} \vecs ∇f(x,y,z) &=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k} \\ &=(2e^{−2z}\cos 2x \cos 2y)\,\hat{\mathbf i}+(−2e^{−2z})\,\hat{\mathbf j}+(−2e^{−2z})\,\hat{\mathbf k} \\ &=2e^{−2z}(\cos 2x \cos 2y \,\hat{\mathbf i}−\sin 2x \sin 2y\,\hat{\mathbf j}−\sin 2x \cos2y\,\hat{\mathbf k}). \end{align*}
Exercise $$\PageIndex{5}$$:
Find the gradient $$\vecs ∇f(x,y,z)$$ of $$f(x,y,z)=\dfrac{x^2−3y^2+z^2}{2x+y−4z.}$$
$\vecs ∇f(x,y,z)=\dfrac{2x^2+2xy+6y^2−8xz−2z^2}{(2x+y−4z)^2}\,\hat{\mathbf i}−\dfrac{x^2+12xy+3y^2−24yz+z^2}{(2x+y−4z)^2}\,\hat{\mathbf j}+\dfrac{4x^2−12y^2−4z^2+4xz+2yz}{(2x+y−4z)^2}\,\hat{\mathbf k}\nonumber$
The directional derivative can also be generalized to functions of three variables. To determine a direction in three dimensions, a vector with three components is needed. This vector is a unit vector, and the components of the unit vector are called directional cosines. Given a three-dimensional unit vector $$\vecs u$$ in standard form (i.e., the initial point is at the origin), this vector forms three different angles with the positive $$x$$-, $$y$$-, and $$z$$-axes. Let’s call these angles $$α,β,$$ and $$γ$$. Then the directional cosines are given by $$\cos α,\cos β,$$ and $$\cos γ$$. These are the components of the unit vector $$\vecs u$$; since $$\vecs u$$ is a unit vector, it is true that $$\cos^2 α+\cos^2 β+\cos^2 γ=1.$$
Definition: Directional Derivative of a Function of Three variables
Suppose $$w=f(x,y,z)$$ is a function of three variables with a domain of $$D$$. Let $$(x_0,y_0,z_0)∈D$$ and let $$\vecs u=\cos α\,\hat{\mathbf i}+\cos β\,\hat{\mathbf j}+\cos γ\,\hat{\mathbf k}$$ be a unit vector. Then, the directional derivative of $$f$$ in the direction of $$u$$ is given by
$D_{\vecs u}f(x_0,y_0,z_0)=\lim_{t→0}\dfrac{f(x_0+t \cos α,y_0+t\cos β,z_0+t\cos γ)−f(x_0,y_0,z_0)}{t}$
provided the limit exists.
We can calculate the directional derivative of a function of three variables by using the gradient, leading to a formula that is analogous to Equation \ref{DD2v}.
Directional Derivative of a Function of Three Variables
Let $$f(x,y,z)$$ be a differentiable function of three variables and let $$\vecs u=\cos α\,\hat{\mathbf i}+\cos β\,\hat{\mathbf j}+\cos γ\,\hat{\mathbf k}$$ be a unit vector. Then, the directional derivative of $$f$$ in the direction of $$\vecs u$$ is given by
$D_{\vecs u}f(x,y,z)=\vecs ∇f(x,y,z)⋅\vecs u=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_z(x,y,z)\cos γ. \label{DDv3}$
The three angles $$α,β,$$ and $$γ$$ determine the unit vector $$\vecs u$$. In practice, we can use an arbitrary (nonunit) vector, then divide by its magnitude to obtain a unit vector in the desired direction.
Example $$\PageIndex{7}$$: Finding a Directional Derivative in Three Dimensions
Calculate $$D_{\vecs v}f(1,−2,3)$$ in the direction of $$v=−\,\hat{\mathbf i}+2\,\hat{\mathbf j}+2\,\hat{\mathbf k}$$ for the function
$f(x,y,z)=5x^2−2xy+y^2−4yz+z^2+3xz. \nonumber$
Solution:
First, we find the magnitude of $$v$$:
$‖\vecs v‖=\sqrt{(−1)^2+(2)^2+(2)^2}=\sqrt{9}=3. \nonumber$
Therefore, $$\dfrac{\vecs v}{‖\vecs v‖}=\dfrac{−\hat{\mathbf i}+2\,\hat{\mathbf j}+2\,\hat{\mathbf k}}{3}=−\dfrac{1}{3}\,\hat{\mathbf i}+\dfrac{2}{3}\,\hat{\mathbf j}+\dfrac{2}{3}\,\hat{\mathbf k}$$ is a unit vector in the direction of $$\vecs v$$, so $$\cos α=−\dfrac{1}{3},\cos β=\dfrac{2}{3},$$ and $$\cos γ=\dfrac{2}{3}$$. Next, we calculate the partial derivatives of $$f$$:
\begin{align*} f_x(x,y,z) &=10x−2y+3z \\ f_y(x,y,z) &=−2x+2y−4z \\ f_z(x,y,z) &=−4y+2z+3x, \end{align*}
then substitute them into Equation \ref{DDv3}:
\begin{align*} D_{\vecs v}f(x,y,z) &=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_z(x,y,z)\cos γ \\ &=(10x−2y+3z)(−\dfrac{1}{3})+(−2x+2y−4z)(\dfrac{2}{3})+(−4y+2z+3x)(\dfrac{2}{3}) \\ &=−\dfrac{10x}{3}+\dfrac{2y}{3}−\dfrac{3z}{3}−\dfrac{4x}{3}+\dfrac{4y}{3}−\dfrac{8z}{3}−\dfrac{8y}{3}+\dfrac{4z}{3}+\dfrac{6x}{3} \\ &=−\dfrac{8x}{3}−\dfrac{2y}{3}−\dfrac{7z}{3}. \end{align*}
Last, to find $$D_{\vecs v}f(1,−2,3),$$ we substitute $$x=1,\, y=−2$$, and $$z=3:$$
\begin{align*} D_{\vecs v}f(1,−2,3) &=−\dfrac{8(1)}{3}−\dfrac{2(−2)}{3}−\dfrac{7(3)}{3} \\ &=−\dfrac{8}{3}+\dfrac{4}{3}−\dfrac{21}{3} \\ &=−\dfrac{25}{3}. \end{align*}
Exercise $$\PageIndex{6}$$:
Calculate $$D_{\vecs v}f(x,y,z)$$ and $$D_{\vecs v}f(0,−2,5)$$ in the direction of $$\vecs v=−3\,\hat{\mathbf i}+12\,\hat{\mathbf j}−4\,\hat{\mathbf k}$$ for the function
$f(x,y,z)=3x^2+xy−2y^2+4yz−z^2+2xz.\nonumber$
Hint
First, divide $$\vecs v$$ by its magnitude, calculate the partial derivatives of $$f$$, then use Equation \ref{DDv3}.
$$D_{\vecs v}f(x,y,z)=−\dfrac{3}{13}(6x+y+2z)+\dfrac{12}{13}(x−4y+4z)−\dfrac{4}{13}(2x+4y−2z)$$
$$D_{\vecs v}f(0,−2,5)=\dfrac{384}{13}$$
## Summary
• A directional derivative represents a rate of change of a function in any given direction.
• The gradient can be used in a formula to calculate the directional derivative.
• The gradient indicates the direction of greatest change of a function of more than one variable.
### Key Equations
• directional derivative (two dimensions) $D_{\vecs u}f(a,b)=\lim_{h→0}\dfrac{f(a+h\cos θ,b+h\sin θ)−f(a,b)}{h} \nonumber$ or $D_{\vecs u}f(x,y)=f_x(x,y)\cos θ+f_y(x,y)\sin θ\nonumber$
• gradient (two dimensions) $\vecs ∇f(x,y)=f_x(x,y)\,\hat{\mathbf i}+f_y(x,y)\,\hat{\mathbf j}\nonumber$
• gradient (three dimensions) $\vecs ∇f(x,y,z)=f_x(x,y,z)\,\hat{\mathbf i}+f_y(x,y,z)\,\hat{\mathbf j}+f_z(x,y,z)\,\hat{\mathbf k}\nonumber$
• directional derivative (three dimensions) $D_{\vecs u}f(x,y,z)=\vecs ∇f(x,y,z)⋅\vecs u=f_x(x,y,z)\cos α+f_y(x,y,z)\cos β+f_x(x,y,z)\cos γ\nonumber$
### Glossary
directional derivative
the derivative of a function in the direction of a given unit vector
the gradient of the function $$f(x,y)$$ is defined to be $$\vecs ∇f(x,y)=(∂f/∂x)\,\hat{\mathbf i}+(∂f/∂y)\,\hat{\mathbf j},$$ which can be generalized to a function of any number of independent variables |
Lines Prepared by Doron Shahar. Warm-up: Class Notes Page 25 Today we are starting to learn about lines. In honor of that, here are several one-liners:
Presentation on theme: "Lines Prepared by Doron Shahar. Warm-up: Class Notes Page 25 Today we are starting to learn about lines. In honor of that, here are several one-liners:"— Presentation transcript:
Lines Prepared by Doron Shahar
Warm-up: Class Notes Page 25 Today we are starting to learn about lines. In honor of that, here are several one-liners: I wondered why the Frisbee was getting bigger, and then it hit me. I had a dream I was eating a giant marshmallow, and when I woke up my pillow was missing! Right now I'm having amnesia and deja vu at the same time! I think I've forgotten this before? How’s that new line I introduced you to? Ohh, she goes on forever. Prepared by Doron Shahar
Find points on a graph (−2,3) (2,1) x coordinate first y coordinate second (x, y) Prepared by Doron Shahar
Finding intercepts on a graph (0,2) (4,0) x-intercept y-intercept where line touches y-axis where line touches x-axis Prepared by Doron Shahar
Find slope from a graph Rise Run Prepared by Doron Shahar
The slope of a line passing through two distinct points (x 1, y 1 ) and (x 2, y 2 ) is given by the formula____________ Exception: The slope is undefined if x 1 =x 2. Slope formula Extra 1: Find the slope of the line passing through the points (−2, −3) and (−4, 5). (x 1, y 1 ) and (x 2, y 2 ) Prepared by Doron Shahar
Slope-intercept form The slope-intercept form of the equation of a line with slope m and y-intercept (0, b) is given by the formula _____________ 2.3.4 Find the slope-intercept form of the equation of the line with y-intercept (0, −8) and having slope m=2/5. Solution Prepared by Doron Shahar
Extra 2: Slope-intercept form Find the slope-intercept form of the equation of the line displayed in the graph below. (0, 3) y-intercept (2, 4) (x 1, y 1 ) (x 2, y 2 ) Solution Prepared by Doron Shahar
2.3.4 Sketch a graph y-intercept: Sketch a graph of the line y=2/5x −8. Slope: (0, −8) +( ) y-intercept 2 5 Two points define a line. Prepared by Doron Shahar
2.3.4 Identifying intercepts Find the y-intercept and the x-intercept of the line y=2/5x−8. To find the x-intercept, set y=0 and solve for x. Setting y equal to zero Solution To find the y-intercept, set x=0 and solve for y. Setting x equal to zero Solution y-intercept: x-intercept: Prepared by Doron Shahar
Point-slope form The point-slope form of the equation of a line passing through the point (x 1, y 1 ) and having slope m is given by the formula ______________________ Extra 3: Find the point-slope form of the equation of the line passing through the point (− 3,2) with slope m= −2. Solution (x 1, y 1 ) Prepared by Doron Shahar
Extra 4: Point-Slope form Find the point-slope form of the equation of the line passing through the points (−1, 3) and (1, 4). (x 1, y 1 ) and (x 2, y 2 ) First find the slope. Solution Prepared by Doron Shahar
2.3.2: Point-slope form Find the point-slope form of the equation of the line displayed in the graph below. (−4, 5) (−2, −3) First find the slope. (x 1, y 1 ) (x 2, y 2 ) Solution Prepared by Doron Shahar
Extra 3: Sketch a graph Point: Sketch a graph of the line y−2= −2(x+3). Slope: (−3, 2) Point 2 1 Two points define a line. Prepared by Doron Shahar
2.3.2 Identifying Intercepts Find the y-intercept and the x-intercept of the line y−5= −4(x+4). To find the x-intercept, set y=0 and solve for x. Setting y equal to zero Solution To find the y-intercept, set x=0 and solve for y. Setting x equal to zero Solution y-intercept: x-intercept: Prepared by Doron Shahar
Point-slope to Slope-intercept form 2.3.2 Write the line y−5=−4(x+4) in slope-intercept form. Starting Equation Distribute right hand side Slope-intercept form Prepared by Doron Shahar
Standard form of a line The standard from of the equation of a line is given by the formula __________________, where A and B are not both zero. 2.3.1 Write x/3 − 4y +1 =0 in standard form such that A, B, C are integers. Subtract 1 from both sides Multiply both sides by 3 Prepared by Doron Shahar
2.3.1 Identifying Intercepts Find the y-intercept and the x-intercept of the line x−12y= −3. To find the x-intercept, set y=0 and solve for x. Setting y equal to zero Solution To find the y-intercept, set x=0 and solve for y. Setting x equal to zero Solution y-intercept: x-intercept: Prepared by Doron Shahar
2.3.1 Sketch a graph x-intercept: Sketch a graph of the line x−12y= −3. (−3, 0) Two points define a line. y-interceptx-intercept y-intercept: (0, 1/4) Prepared by Doron Shahar
Standard form Every line can be written in standard form. This is not true of the other two forms. In particular, verticals lines can only be written in standard form, because their slope is undefined. It is difficult to get the equation of a line directly into standard form given a graph or “usual” information. Instead, you will need to be able to get the equations of lines into and out of this form. Prepared by Doron Shahar
Point-slope to Standard form To get the equation of a line from point-slope form into standard form, first convert it into slope-intercept form as we have already learned. Then proceed as on the next slide to convert the equation into standard form. Prepared by Doron Shahar
Slope-intercept to Standard form 2.3.4 Write the line y=2/5x−8 in standard form such that A, B, and C are integers. Starting Equation Multiply both sides by 5 Distribute the 5 Alternative Standard form Prepared by Doron Shahar
Standard to Slope-intercept to form 2.3.1 Write the line x−12y = −3 in slope-intercept form. Starting Equation Subtract x from both sides Divide by −12 Slope-intercept form Prepared by Doron Shahar
Standard to Slope-intercept form One reason for converting the equation of a line from Standard form to slope-intercept form is to find the slope of the line. It also helps if you are trying to graph the line. 2.3.1 Find the slope of the x−12y = −3. Slope-intercept form Slope: Prepared by Doron Shahar
Turtle Sign language for turtle Prepared by Doron Shahar
Horizontal lines 2.3.5 Write the equation of the horizontal line passing through the point (−9/2, 15/2). A horizontal line passing through the point (a, b) is given by the equation ________. The slope of a horizontal line is _________. Solution Prepared by Doron Shahar
Extra 5: Horizontal lines Find the equation of the horizontal line displayed below. Solution (2, 3) (a, b) Prepared by Doron Shahar
2.3.5 Sketch a graph Sketch a graph of the line y=15/2. Slope: (−9/2, 15/2) Prepared by Doron Shahar
2.3.5 Identifying intercepts Find the y-intercept and the x-intercept of the line y=15/2. To find the x-intercept, set y=0 and solve for x. Setting y equal to zero To find the y-intercept, set x=0 and solve for y. Setting x equal to zero Solution y-intercept: There is no x-intercept! ¡PROBLEMA! Prepared by Doron Shahar
Vertical lines 2.3.3 Write the equation of the vertical line with x-intercept (1/3, 0). Solution A vertical line passing through the point (a, b) is given by the equation ________. The slope of a vertical line is _____________. Prepared by Doron Shahar
Extra 6: Vertical lines Find the equation of the vertical line displayed below. Solution (2, 3) (a, b) Prepared by Doron Shahar
2.3.3 Sketch a graph Sketch a graph of the line x=1/3. Slope: (1/3, 0) Prepared by Doron Shahar
2.3.3 Identifying intercepts Find the y-intercept and the x-intercept of the line x=1/3. To find the x-intercept, set y=0 and solve for x. Setting y equal to zero To find the y-intercept, set x=0 and solve for y. Setting x equal to zero Solution x-intercept: There is no y-intercept! ¡PROBLEMA! Prepared by Doron Shahar
Horizontal lines vs. Vertical lines Horizontal LinesVerticals lines Equation Slope x-intercept y-intercept Exception: y=0 Exception: y=0 Prepared by Doron Shahar
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Intervals and Interval Notation - Function Basics - High School Algebra I Unlocked (2016)
## High School Algebra I Unlocked (2016)
### Lesson 7.2. Intervals and Interval Notation
Often the domain and range of a function will be expressed in interval notation. An interval is a set of real numbers between, and at times including, two numbers. Consider the following examples that express the domain of a function using interval notation:
The use of brackets ([ ]) in interval notation indicates that the value of the endpoint is included in the interval; graphically, the endpoints will be filled circles. Conversely, the use of parentheses (( )) in interval notation indicates that the value of the endpoint is not included in the interval; graphically, the endpoints will be unfilled circles.
Consider the graph of g(x). In this function, the least x-value is −4, the greatest x-value is 3, and all of the points on the graph are depicted as filled circles. Therefore, the domain of g(x) includes all real numbers between −4 and 3, including both −4 and 3, and is written in interval notation as [−4, 3]. You can also use interval notation to express the range, or all possible y-values, of a function. Let’s try a couple questions that focus on interval notation.
EXAMPLE
Express the range of g(x) in interval notation.
In the graph of g(x), the lowest y−value is −3, which occurs at point (−4, −3), and the greatest y-value is 2, which occurs at point (1, 2). Since both of these points are filled-in circles, and therefore included in the range, the function g(x) has a range of [−3, 2].
EXAMPLE
Express the domain and range of h(x) in interval notation.
Since h(x) has x-values that range from −4 to 3, but point (−4, −3) has an unfilled endpoint, the domain of h(x) is (−4, 3]. Similarly, h(x) has y-values that range from −3 to 2, but point (−4, −3) has an unfilled endpoint, signifying that the range of h(x) is (−3, 2].
Thus, the function h(x) has a domain of (−4, 3] and a range of (−3, 2].
But what if the function is discontinuous at various points? How would it be represented in interval notation? To express a domain that is true over multiple intervals, you use the union symbol: ∪. To understand how this works, look at the graph below.
Let’s use this graph to determine the domain and range of m(x), and state them in interval notation. Based on the graph, you can determine that m(x) is has x-values that range from x = −10 to x = −8, where x = −8 is not included in the domain. Therefore, you would express this interval as [−10, −8). Then find the next interval, which exists from x = −8 to x = −2; again, x = −8 is not included. Thus, this interval is expressed as (−8, −2]. Repeat the process for the next interval, which goes from x = −2 to x = −1, where −1 is not included. This interval is expressed as [−2, −1). Finally, find the last interval, which goes from x = −1 to x = 5, where x = −1 is not included. This interval is expressed as (−1, 5].
Now you need to join all of the intervals together with the union symbol to express the domain of m(x). Therefore, the domain of m(x) is [−10, −8) ∪ (−8, −2] ∪ [−2, −1) ∪ (−1, 5]. Since −2 is included in both the intervals (−8, −2] and [−2, −1), the domain can also be expressed as [−10, −8) ∪ (−8, −1) ∪ (−1, 5].
But what if the function goes on forever? Is that even possible? Of course it is! Consider the linear function ƒ(x) = x + 6:
Unlike the previous graphs, g(x) and h(x), the graph of ƒ(x) doesn’t have any endpoints. No worries—you can still define the domain and range of ƒ(x) using the infinity sign (∞) in your interval notation to represent intervals that extend indefinitely in one or both directions.
Since the function ƒ(x) = x + 6 extends infinitely in both directions, across all x- and y-values, both the domain and range of ƒ(x) would be represented by the interval (−∞, ∞).
Notice that the infinity
sign is not associated with
brackets. Since the infinity
symbol does not represent
a specific number, it will
always be accompanied
by a parenthesis.
|
Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 12
# Two Parallel Wires Carry Equal Currents of 10 a Along the Same Direction and Are Separated by a Distance of 2.0 Cm. Find the Magnetic Field at a Point Which is 2.0 Cm Away from Each of These Wires. - Physics
Short Note
Two parallel wires carry equal currents of 10 A along the same direction and are separated by a distance of 2.0 cm. Find the magnetic field at a point which is 2.0 cm away from each of these wires.
#### Solution
Given:
Magnitude of currents, I1 = I2 = 10 A
Separation of the point from the wires, d = 2 cm
Thus, the magnetic field due to current in the wire is given by
$B_1 = B_2 = \frac{\mu_0 I}{2\pi d}$
In the figure, dotted circle shows the magnetic field lines due to current carrying wire placed in a plane perpendicular to the plane of the paper.
From the figure, we can see that $∆ P I_1 I_2$ is an equilateral triangle.
$\angle I_1 P I_2 = {60}^\circ$
Angle between the magnetic fields due to current in the wire, θ = 60°
∴ Required magnetic field at P
$B_{net} = \sqrt{{B_1}^2 + {B_2}^2 + 2 B_1 B_2 \cos\theta}$
$= \sqrt{\left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)^2 + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)^2 + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right) + \left( \frac{2 \times {10}^{- 7} \times 10}{2 \times {10}^{- 2}} \right)\cos60^\circ}$
$= \sqrt{( {10}^{- 4} ) + ( {10}^{- 4} )^2 + 2( {10}^{- 4} )( {10}^{- 4} ) \times \frac{1}{2}}$
$= \sqrt{3} \times {10}^{- 4} T$
$= 1 . 732 \times {10}^{- 4} T$
Is there an error in this question or solution?
#### APPEARS IN
HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 13 Magnetic Field due to a Current
Q 10 | Page 250 |
April 12, 2024
This article offers a beginner-friendly guide on how to calculate and interpret mean values in statistics. It outlines essential tools for calculating mean, explains common mistakes, and offers step-by-step instructions for computing mean from frequency tables. It also provides examples and tips on how to accurately interpret mean values from different fields.
Introduction
Mean or average is a fundamental concept in statistics that is used to measure central tendency. It’s an essential metric that gives us an idea of what a typical value of a data set looks like. For instance, if we want to know the average age of a group of individuals or the average salary of employees in a company, we need to calculate the mean.
This article is a beginner-friendly guide on how to calculate the mean, interpret its values, and avoid common mistakes. Whether you’re a student, a researcher, or simply curious about statistics, this guide will provide you with everything you need to know to get started.
Part 1: How to Calculate the Mean: A Step-by-Step Guide
To calculate the mean, we first need to understand its definition. In statistics, mean is defined as the sum of all the values in a data set divided by the total number of values.
The mathematical formula for calculating mean is as follows:
$\Large&space;\bar{x}=\frac{\sum_{i=1}^{n}x_i}{n}$
Here, $\Large&space;\bar{x}}$ represents the mean value, $\Large&space;\sum$ represents the sum of all values, $\Large&space;i$ represents each value in the data set, and $\Large&space;n$ represents the total number of values.
Here is a step-by-step guide for computing mean:
Step 1: Add up all the values in the data set.
Step 2: Count the number of values in the data set.
Step 3: Divide the sum of all values by the total number of values.
Let’s take an example to understand this better. Suppose we have the following data set: 2, 4, 8, 10, 12, 15.
Step 1: Add up all the values in the data set.
2 + 4 + 8 + 10 + 12 + 15 = 51
Step 2: Count the number of values in the data set.
There are six values in the data set.
Step 3: Divide the sum of all values by the total number of values.
51/6 = 8.5
Therefore, the mean of this data set is 8.5.
Here are some sample problems for readers to practice.
1. Calculate the mean of the following data set: 6, 12, 18, 24.
2. Calculate the mean of the following data set: 1.2, 3.4, 5.6, 7.8, 9.10.
Part 2: Essential Tools for Computing Mean
There are different methods to calculate mean depending on the type of data set and the tools or software at your disposal. Here are some essential tools for computing mean:
– Microsoft Excel: Microsoft Excel is a popular tool used for analyzing data. It offers a built-in function for calculating mean. You can find the function in the ‘Formulas’ tab under the ‘Math & Trig’ category. Simply select the range of cells containing the values and apply the function.
– Python: Python is a programming language commonly used for data analysis. It has built-in libraries like Numpy and Pandas that make calculating mean simple. In Python, you can calculate mean by importing the Numpy or Pandas library and using their built-in functions.
– Calculator: If you don’t have access to Excel or Python, you can use a calculator to compute the mean. Simply add up all the values and divide by the total number of values.
Each of these tools has its pros and cons, and it is up to you to decide which one works best for your needs. Here are some easy-to-follow steps for computing mean using each tool.
Microsoft Excel:
Step 1: Enter the values in a column or row in Excel.
Step 2: Click on an empty cell where you want to display the mean.
Step 3: Type “=AVERAGE(” followed by the range of cells containing the values. For instance, if your values are in cells A1 to A6, you would type “=AVERAGE(A1:A6)”.
Step 4: Press Enter. Excel will display the mean value.
Python:
Step 1: Open your Python environment (e.g., Jupyter notebook).
Step 2: Import the necessary libraries (e.g., Numpy).
Step 3: Enter the values in the form of a list or an array.
Step 4: Apply the appropriate function (e.g., np.mean()) to the list or array.
Step 5: Display the output.
Calculator:
Step 1: Add up all the values in the data set.
Step 2: Count the number of values in the data set.
Step 3: Divide the sum of all values by the total number of values.
Here are some sample problems for readers to practice.
1. Calculate the mean of the following data set using Microsoft Excel: 2, 8, 12, 16, 20.
2. Calculate the mean of the following data set using Python: 5.6, 7.8, 9.0, 3.4, 1.2.
Part 3: Common Mistakes When Computing the Mean
Computing the mean may seem like a simple task, but it’s easy to make mistakes if you’re not careful. Here are some common mistakes to avoid when calculating the mean.
Mistake 1: Forgetting to add one or more values in the data set.
Explanation: To calculate the mean, you need to include all values in the data set. If you forget to add one or more values, the mean will be incorrect.
Solution: Before calculating the mean, double-check that all values in the data set are included.
Mistake 2: Using the wrong formula.
Explanation: There are different types of means, such as arithmetic mean, geometric mean, and harmonic mean. Each mean requires a different formula, and using the wrong formula will give you incorrect results.
Solution: Make sure you’re using the correct formula for the type of mean you want to calculate.
Mistake 3: Dividing by the wrong number.
Explanation: To compute the mean, you need to divide the sum of all values by the total number of values. If you divide by the wrong number, your result will be inaccurate.
Solution: Always double-check that you’re dividing by the correct number.
By avoiding these mistakes, you can ensure that your mean calculations are accurate.
Part 4: Calculating the Mean from a Frequency Table
A frequency table is a table that shows how often each value or group of values occurs in a data set. To calculate the mean from a frequency table, you need to follow these steps:
Step 1: Multiply each value by its frequency.
Step 2: Add up all the products from Step 1.
Step 3: Divide the sum of products by the total number of values.
Let’s take an example to understand this better. Suppose we have the following frequency table:
| Value | Frequency |
| —– | ——— |
| 10 | 3 |
| 20 | 5 |
| 30 | 2 |
Step 1: Multiply each value by its frequency.
(10 x 3) + (20 x 5) + (30 x 2) = 130
Step 2: Add up all the products from Step 1.
130/10 = 13
Therefore, the mean of this data set is 13.
Here are some example problems for readers to practice.
1. Calculate the mean from the following frequency table:
| Value | Frequency |
| —– | ——— |
| 5 | 2 |
| 10 | 4 |
| 15 | 6 |
| 20 | 3 |
2. Calculate the mean from the following frequency table:
| Value | Frequency |
| —– | ——— |
| 0 | 5 |
| 5 | 7 |
| 10 | 3 |
| 15 | 1 |
| 20 | 2 |
Part 5: How to Interpret Mean Values
Mean values can be used in different fields, such as education, medicine, and economics, to name a few. In each field, mean values convey different information. Here are some tips on how to accurately interpret mean values.
Tip 1: Check for outliers.
An outlier is a value that is significantly different from the other values in the data set. Outliers can affect the mean value and provide an inaccurate representation of the data. Therefore, it’s important to check for outliers before interpreting the mean value.
Tip 2: Consider the context.
Mean values can only be interpreted accurately if you understand the context behind the data. For instance, the mean age of employees in a company may be higher than the national average due to the nature of the industry. Therefore, it’s essential to consider the context behind the data before making any conclusions.
Tip 3: Look for patterns.
Mean values can reveal patterns in the data that may not be visible otherwise. For instance, if the mean test scores of students have been decreasing over the years, it may indicate a need for educational reform.
By following these tips, you can accurately interpret mean values from different fields.
Part 6: Fun Examples Using Mean
Mean values aren’t just limited to academic fields; they can be used in creative and fun ways too! Here are some examples of how mean is used in different fields.
Example 1: Sports
In sports, mean values can be used to measure player performance. For instance, the mean batting average (the number of hits divided by the number of at-bats) of a baseball player can provide insight into their overall performance.
Example 2: Animal Kingdom
In the animal kingdom, mean values can be used to measure the size of different animal species. For instance, the mean weight of African elephants is approximately 6 tons, while the mean weight of Asian elephants is around 4-5 tons.
Example 3: Food
In the food industry, mean values can be used to determine the popularity of different dishes. For example, the mean number of times a dish is ordered in a restaurant can indicate its popularity among customers.
These are just a few examples of how mean values can be used creatively.
Conclusion
In conclusion, understanding how to compute and interpret mean values is essential in statistics. We’ve covered the definition of mean, how to calculate it, and different tools for computing mean. We’ve also discussed common mistakes to avoid, how to calculate mean from a frequency table, how to interpret mean values, and fun examples of how mean is used. By following this guide, readers should feel more confident in their ability to compute and interpret mean values. Remember to always double-check your calculations and consider the context behind the data before making any conclusions. |
# Parasitic number
An n-parasitic number (in base 10) is a positive natural number which, when multiplied by n, results in movement of the last digit of its decimal representation to its front. Here n is itself a single-digit positive natural number. In other words, the decimal representation undergoes a right circular shift by one place. For example:
4 × 128205 = 512820, so 128205 is 4-parasitic.
Most mathematicians do not allow leading zeros to be used, and that is a commonly followed convention.
So even though 4 × 25641 = 102564, the number 25641 is not 4-parasitic.
## Derivation
An n-parasitic number can be derived by starting with a digit k (which should be equal to n or greater) in the rightmost (units) place, and working up one digit at a time. For example, for n = 4 and k = 7
4 × 7 = 28
4 × 87 = 348
4 × 487 = 1948
4 × 9487 = 37948
4 × 79487 = 317948
4 × 179487 = 717948.
So 179487 is a 4-parasitic number with units digit 7. Others are 179487179487, 179487179487179487, etc.
Notice that the repeating decimal
${\displaystyle x=0.179487179487179487\ldots =0.{\overline {179487}}{\mbox{ has }}4x=0.{\overline {717948}}={\frac {7.{\overline {179487}}}{10}}.}$
Thus
${\displaystyle 4x={\frac {7+x}{10}}{\mbox{ so }}x={\frac {7}{39}}.}$
In general, an n-parasitic number can be found as follows. Pick a one digit integer k such that kn, and take the period of the repeating decimal k/(10n−1). This will be ${\displaystyle {\frac {k}{10n-1}}(10^{m}-1)}$ where m is the length of the period; i.e. the multiplicative order of 10 modulo (10n − 1).
For another example, if n = 2, then 10n − 1 = 19 and the repeating decimal for 1/19 is
${\displaystyle {\frac {1}{19}}=0.{\overline {052631578947368421}}.}$
So that for 2/19 is double that:
${\displaystyle {\frac {2}{19}}=0.{\overline {105263157894736842}}.}$
The length m of this period is 18, the same as the order of 10 modulo 19, so 2 × (1018 − 1)/19 = 105263157894736842.
105263157894736842 × 2 = 210526315789473684, which is the result of moving the last digit of 105263157894736842 to the front.
The step-by-step derivation algorithm depicted above is a great core technique but will not find all n-parasitic numbers. It will get stuck in an infinite loop when the derived number equals the derivation source. An example of this occurs when n = 5 and k = 5. The 42-digit n-parasitic number to be derived is 102040816326530612244897959183673469387755. Check the steps in Table One below. The algorithm begins building from right to left until it reaches step 15—then the infinite loop occurs. Lines 16 and 17 are pictured to show that nothing changes. There is a fix for this problem, and when applied, the algorithm will not only find all n-parasitic numbers in base ten, it will find them in base 8 and base 16 as well. Look at line 15 in Table Two. The fix, when this condition is identified and the n-parasitic number has not been found, is simply to not shift the product from the multiplication, but use it as is, and append n (in this case 5) to the end. After 42 steps, the proper parasitic number will be found.
### Table One
1. 5 × 5 = 25 − Shift = 55 2. 5 × 55 = 275 − Shift = 755 3. 5 × 755 = 3775 − Shift = 7755 4. 5 × 7755 = 38775 − Shift = 87755 5. 5 × 87755 = 438775 − Shift = 387755 6. 5 × 387755 = 1938775 − Shift = 9387755 7. 5 × 9387755 = 46938775 − Shift = 69387755 8. 5 × 69387755 = 346938775 − Shift = 469387755 9. 5 × 469387755 = 2346938775 − Shift = 3469387755 10. 5 × 3469387755 = 17346938775 − Shift = 73469387755 11. 5 × 73469387755 = 367346938775 − Shift = 673469387755 12. 5 × 673469387755 = 3367346938775 − Shift = 3673469387755 13. 5 × 3673469387755 = 18367346938775 − Shift = 83673469387755 14. 5 × 83673469387755 = 418367346938775 − Shift = 183673469387755 15. 5 × 183673469387755 = 918367346938775 − Shift = 183673469387755 16. 5 × 183673469387755 = 918367346938775 − Shift = 183673469387755 17. 5 × 183673469387755 = 918367346938775 − Shift = 183673469387755
### Table Two
1. 5 × 5 = 25 − Shift = 55 2. 5 × 55 = 275 − Shift = 755 3. 5 × 755 = 3775 − Shift = 7755 4. 5 × 7755 = 38775 − Shift = 87755 5. 5 × 87755 = 438775 − Shift = 387755 6. 5 × 387755 = 1938775 − Shift = 9387755 7. 5 × 9387755 = 46938775 − Shift = 69387755 8. 5 × 69387755 = 346938775 − Shift = 469387755 9. 5 × 469387755 = 2346938775 − Shift = 3469387755 10. 5 × 3469387755 = 17346938775 − Shift = 73469387755 11. 5 × 73469387755 = 367346938775 − Shift = 673469387755 12. 5 × 673469387755 = 3367346938775 − Shift = 3673469387755 13. 5 × 3673469387755 = 18367346938775 − Shift = 83673469387755 14. 5 × 83673469387755 = 418367346938775 − Shift = 183673469387755 15. 5 × 183673469387755 = 918367346938775 − Shift = 9183673469387755 16. 5 × 9183673469387755 = 45918367346938775 − Shift = 59183673469387755 17. 5 × 59183673469387755 = 295918367346938775 − Shift = 959183673469387755
There is one more condition to be aware of when working with this algorithm, leading zeros must not be lost. When the shift number is created it may contain a leading zero which is positionally important and must be carried into and through the next step. Calculators and computer math methods will remove leading zeros. Look at Table Three below displaying the derivation steps for n = 4 and k = 4. The Shift number created in step 4, 02564, has a leading zero which is fed into step 5 creating a leading zero product. The resulting Shift is fed into Step 6 which displays a product proving the 4-parasitic number ending in 4 is 102564.
### Table Three
1. 4 × 4 = 16 − Shift = 64 2. 4 × 64 = 256 − Shift = 564 3. 4 × 564 = 2256 − Shift = 2564 4. 4 × 2564 = 10256 − Shift = 02564 5. 4 × 02564 = 010256 − Shift = 102564 6. 4 × 102564 = 410256 − Shift = 102564
## Smallest n-parasitic numbers
The smallest n-parasitic numbers are also known as Dyson numbers, after a puzzle concerning these numbers posed by Freeman Dyson.[1][2][3] They are: (leading zeros are not allowed) (sequence A092697 in the OEIS)
n Smallest n-parasitic number Digits Period of
1 1 1 1/9
2 105263157894736842 18 2/19
3 1034482758620689655172413793 28 3/29
4 102564 6 4/39
5 142857 6 7/49 = 1/7
6 1016949152542372881355932203389830508474576271186440677966 58 6/59
7 1014492753623188405797 22 7/69
8 1012658227848 13 8/79
9 10112359550561797752808988764044943820224719 44 9/89
## General note
In general, if we relax the rules to allow a leading zero, then there are 9 n-parasitic numbers for each n. Otherwise only if kn then the numbers do not start with zero and hence fit the actual definition.
Other n-parasitic integers can be built by concatenation. For example, since 179487 is a 4-parasitic number, so are 179487179487, 179487179487179487 etc.
## Other bases
In duodecimal system, the smallest n-parasitic numbers are: (using inverted two and three for ten and eleven, respectively) (leading zeros are not allowed)
n Smallest n-parasitic number Digits Period of
1 1 1 1/Ɛ
2 10631694842 Ɛ 2/1Ɛ
3 2497 4 7/2Ɛ = 1/5
4 10309236ᘔ88206164719544 4/3Ɛ
5 1025355ᘔ9433073ᘔ458409919Ɛ715 25 5/4Ɛ
6 1020408142854ᘔ997732650ᘔ18346916306 6/5Ɛ
7 101899Ɛ864406Ɛ33ᘔᘔ15423913745949305255Ɛ17 35 7/6Ɛ
8 131ᘔ8ᘔ 6 /7Ɛ = 2/17
9 101419648634459Ɛ9384Ɛ26Ɛ533040547216ᘔ1155Ɛ3Ɛ12978ᘔ399 45 9/8Ɛ
(10) 14Ɛ36429ᘔ7085792 14 12/9Ɛ = 2/15
Ɛ (11) 1011235930336ᘔ53909ᘔ873Ɛ325819Ɛ9975055Ɛ54ᘔ3145ᘔ42694157078404491Ɛ 55 Ɛ/ᘔƐ
## Strict definition
In strict definition, least number m beginning with 1 such that the quotient m/n is obtained merely by shifting the leftmost digit 1 of m to the right end are
1, 105263157894736842, 1034482758620689655172413793, 102564, 102040816326530612244897959183673469387755, 1016949152542372881355932203389830508474576271186440677966, 1014492753623188405797, 1012658227848, 10112359550561797752808988764044943820224719, 10, 100917431192660550458715596330275229357798165137614678899082568807339449541284403669724770642201834862385321, 100840336134453781512605042016806722689075630252, ... (sequence A128857 in the OEIS)
They are the period of n/(10n − 1), also the period of the decadic integer -n/(10n − 1).
Number of digits of them are
1, 18, 28, 6, 42, 58, 22, 13, 44, 2, 108, 48, 21, 46, 148, 13, 78, 178, 6, 99, 18, 8, 228, 7, 41, 6, 268, 15, 272, 66, 34, 28, 138, 112, 116, 179, 5, 378, 388, 18, 204, 418, 6, 219, 32, 48, 66, 239, 81, 498, ... (sequence A128858 in the OEIS) |
# Decimal Place Value Chart: Definition, How to Write, and Examples
Mathematics is the study of numbers, shapes, and patterns. It includes various complex and simple arithmetic topics that help people in their daily life routine. In Maths, Numbers play a major role and they can be of different types like Real Numbers, Whole Numbers, Natural Numbers, Decimal Numbers, Rational Numbers, etc. Today, we are going to discuss one of the main topics of Decimal Numbers. In Decimals, identifying the Decimal Place Values is a fundamental topic and everyone should know the techniques clearly. So, here we will be discussing elaborately the topic of Decimal Place Values Chart.
Let’s get into it.
## What is a Decimal in Math?
In algebra, a decimal number can be represented as a number whose whole number part and the fractional part is divided by a decimal point. The dot in a decimal number is called a decimal point. The digits following the decimal point show a value smaller than one.
### What is the Place Value of Decimals?
Place value is a positional notation system where the position of a digit in a number, determines its value. The place value for decimal numbers is arranged exactly the identical form of treating whole numbers, but in this case, it is reverse. On the basis of the preceding exponential of 10, the place value in decimals can be decided.
### Decimal Place Value Chart
On the place value chart, the numbers on the left of the decimal point are multiplied with increasing positive powers of 10, whereas the digits on the right of the decimal point are multiplied with increasing negative powers of 10 from left to right.
• The first digit after the decimal represents the tenths place.
• The second digit after the decimal represents the hundredths place.
• The third digit after the decimal represents the thousands place.
• The rest of the digits proceed to fill in the place values until there are no digits left.
### How to write the place value of decimals for the number 132.76?
• The place of 6 in the decimal 132.76 is 6/100
• The place of 7 in the decimal 132.76 is 7/10
• The place of 2 in the decimal 132.76 is 2
• The place of 3 in the decimal 132.76 is 30
• The place of 1 in the decimal 132.76 is 100.
#### Examples:
1. Write the place value of digit 7 in the following decimal number: 5.47?
The number 7 is in the place of hundredths, and its place value is 7 x 10 -2 = 7/100 = 0.07.
2. Identify the place value of the 6 in the given number: 689.87?
Given number is 689.87
The place of 6 in the decimal 689.87 is 600 or 6 hundreds.
3. Write the following numbers in the decimal place value chart.
(i) 4532.079
(ii) 490.7042
Solutions:
(i) 4532.079
4532.079 in the decimal place value chart.
(ii) 490.7042
490.7042 in the decimal place value chart. |
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Get in touch with us
Sep 10, 2022
## Key Concepts
• Prove that a quadrilateral is a parallelogram.
• Find the relation between the angles of a parallelogram.
• Use properties of sides, and diagonals to identify a parallelogram.
### Parallelogram
A simple quadrilateral in which the opposite sides are of equal length and parallel is called a parallelogram.
### Real-life examples of a parallelogram
1. Dockland office building in Hamburg, Germany.
1. An eraser
1. A striped pole
1. A solar panel
#### Theorem
If one pair of opposite sides of a quadrilateral is parallel and congruent, then the quadrilateral is a parallelogram.
Given: WX=ZY
To prove: WXYZ is a parallelogram.
Proof: Now, in △XZY and △ZXW,
WX = ZY
∠WXZ = ∠XZY [Alternate interior angles]
XZ = ZX [Reflexive property]
So, △XZY ≅ △ZX by Side-Angle-Side congruence criterion
If two triangles are congruent, their corresponding sides are equal.
Hence, WZ=XY
Therefore,
WXYZ is a parallelogram.
#### Theorem
If an angle of a quadrilateral is supplementary to both of its consecutive angles, then the quadrilateral is a parallelogram
Given: ∠A+∠B = 180° and ∠A+∠D = 180°
To prove: ABCD is a parallelogram.
Proof:
Given: ∠A and ∠B are supplementary.
If two lines are cut by a transversal, then the consecutive interior angles are supplementary, and the lines are parallel.
Given: ∠A and ∠D are supplementary.
We know that if the consecutive interior angles are supplementary, then the lines are parallel.
So, AB ∥ CD …(2)
From (1) and (2), we get AB∥CD and AD∥BC
Hence,
ABCD is a parallelogram.
#### Theorem
If both the pairs of opposite sides of a quadrilateral are congruent, then the quadrilateral is a parallelogram
Given: PQ=RS and PS=QR
To prove: PQRS is a parallelogram.
Proof:
Now, in △PQS and △RSQ,
PQ = RS
PS = RQ
QS = SQ [Reflexive property]
So, △PQS ≅ △RSQ by Side-Side-Side congruence criterion
If two triangles are congruent, their corresponding angles are equal.
Hence, ∠PQS = ∠RSQ and ∠QSP = ∠SQR
We know that if two lines are cut by a transversal, then the alternate interior angles are congruent, and the lines are parallel.
So, PQ ∥ SR and PS ∥ QR
Therefore,
PQRS is a parallelogram.
#### Theorem
If the diagonals of a quadrilateral bisect each other, then the quadrilateral is a parallelogram
Given: LP=PN and OP=PM
To prove: LMNO is a parallelogram.
Proof:
Now, in
△OPN and △MPL, OP=MP
∠OPN=∠MPL [Vertically opposite angles]
PN=PL
So,
△OPN ≅ △MPL by Side-Angle-Side congruence criterion.
If two triangles are congruent, their corresponding sides are equal.
Hence,
ON = LM …(1)
And, in △LPO and △NPM,
LP = NP
∠LPO = ∠NPM [Vertically opposite angles]
PO = PM
So, △LPO ≅ △NPM by Side-Angle-Side congruence criterion.
If two triangles are congruent, their corresponding sides are equal.
Hence,
LO = MN …(2)
From (1) and (2), we get
ON = LM and LO = MN.
∴LMNO is a parallelogram.
#### Theorem
If both pairs of opposite angles of a quadrilateral are congruent, then the quadrilateral is a parallelogram
Given: ∠A=∠C; ∠B=∠D
To prove: ABCD is a parallelogram.
Proof:
We know that the sum of angles of a quadrilateral is 360°.
So,
∠A+∠B+∠C+∠D=360°
∠A+∠B+∠A+∠B=360°
2(∠A+∠B) =360°
∠A+∠B=180°
If two lines are cut by a transversal, then the consecutive interior angles are supplementary, and the lines are parallel.
Since ∠A and ∠B are supplementary,
Or we can write
∠A+∠B+∠C+∠D=360°
∠A+∠D+∠A+∠D=360°
2(∠A+∠D) =360°
∠A+∠D=180°
Since the angles ∠A and ∠D are supplementary, so,
AB ∥ CD …(2)
From (1) and (2), we get AB ∥ CD and AD ∥ BC
So, ABCD is a parallelogram.
## Exercise
• For what values of x and y is the given quadrilateral a parallelogram?
• For what values of w and z is the given figure a parallelogram?
• Is the figure below a parallelogram?
• For what values of x and y is the given quadrilateral a parallelogram?
• Is the figure below a parallelogram?
### What we have learned
• A quadrilateral whose two pairs of opposite sides are parallel is called a parallelogram.
• If a quadrilateral is a parallelogram, then its opposite sides are congruent.
• If a quadrilateral is a parallelogram, then its consecutive angles are supplementary.
• If a quadrilateral is a parallelogram, then opposite angles are congruent.
• If a quadrilateral is a parallelogram, then its diagonals bisect each other.
#### Addition and Multiplication Using Counters & Bar-Diagrams
Introduction: We can find the solution to the word problem by solving it. Here, in this topic, we can use 3 methods to find the solution. 1. Add using counters 2. Use factors to get the product 3. Write equations to find the unknown. Addition Equation: 8+8+8 =? Multiplication equation: 3×8=? Example 1: Andrew has […]
#### Dilation: Definitions, Characteristics, and Similarities
Understanding Dilation A dilation is a transformation that produces an image that is of the same shape and different sizes. Dilation that creates a larger image is called enlargement. Describing Dilation Dilation of Scale Factor 2 The following figure undergoes a dilation with a scale factor of 2 giving an image A’ (2, 4), B’ […]
#### How to Write and Interpret Numerical Expressions?
Write numerical expressions What is the Meaning of Numerical Expression? A numerical expression is a combination of numbers and integers using basic operations such as addition, subtraction, multiplication, or division. The word PEMDAS stands for: P → Parentheses E → Exponents M → Multiplication D → Division A → Addition S → Subtraction Some examples […] |
# Lesson 8
Ratios and Rates With Fractions
## 8.1: Number Talk: Dividing by Half as Much (5 minutes)
### Warm-up
The purpose of this number talk is to elicit strategies students have for reasoning about division. In the previous lesson, students learned an algorithm for dividing a fraction by a fraction. Later in this lesson, students will need to be able to divide a fraction by a fraction to solve problems about rates in contexts.
### Launch
Reveal one problem at a time. Give students 30 seconds of quiet think time for each problem and ask them to give a signal when they have an answer and a strategy. Keep all previous problems displayed throughout the talk. Follow with a whole-class discussion.
Representation: Internalize Comprehension. To support working memory, provide students with sticky notes or mini whiteboards.
Supports accessibility for: Memory; Organization
### Student Facing
Find each quotient mentally.
$$743 \div 10$$
$$743 \div 5$$
$$99 \div 3$$
$$99 \div \frac32$$
### Anticipated Misconceptions
The divisor for the second problem is half as much as the divisor in the first problem. Some students may think that the quotient will be half as much as well, instead of double. Remind the students of the two meanings of division they have learned and encourage them to draw a diagram to represent the problem.
If students get stuck on the last problem, help them see that previous problems can be used to figure out an answer to this last one. Since $$\frac32$$ is half of 3 the answer is going to be double the previous answer.
### Activity Synthesis
Ask students to share their strategies for each problem. Record and display their explanations for all to see. To involve more students in the conversation, consider asking:
• “Who can restate ___’s reasoning in a different way?”
• “Did anyone have the same strategy but would explain it differently?”
• “Did anyone solve the problem in a different way?”
• “Does anyone want to add on to _____’s strategy?”
• “Do you agree or disagree? Why?”
Speaking: MLR8 Discussion Supports.: Display sentence frames to support students when they explain their strategy. For example, "First, I _____ because . . ." or "I noticed _____ so I . . . ." Some students may benefit from the opportunity to rehearse what they will say with a partner before they share with the whole class.
Design Principle(s): Optimize output (for explanation)
## 8.2: Using an Algorithm to Divide Fractions (15 minutes)
### Activity
This activity allows students to practice using the algorithm from earlier to solve division problems that involve a wider variety of fractions. Students can use any method of reasoning and are not expected to use the algorithm. As they encounter problems with less-friendly numbers, however, they notice that it becomes more challenging to use diagrams or other concrete strategies, and more efficient to use the algorithm. As they work through the activity, students choose their method.
Monitor the strategies students use and identify those with different strategies— including those who may not have used the algorithm—so they can share later.
### Launch
Keep students in groups of 2. Give students 5–7 minutes of quiet work time, followed by 2–3 minutes to discuss their responses with a partner.
Representation: Internalize Comprehension. Activate or supply background knowledge. Provide students with access to blank tape diagrams. Encourage students to attempt more than one strategy for at least one of the problems.
Supports accessibility for: Visual-spatial processing; Organization
### Student Facing
Calculate each quotient. Show your thinking and be prepared to explain your reasoning.
1. $$\frac 89 \div 4$$
2. $$\frac 34 \div \frac 12$$
3. $$3 \frac13 \div \frac29$$
4. $$\frac92 \div \frac 38$$
5. $$6 \frac 25 \div 3$$
6. After biking $$5 \frac 12$$ miles, Jada has traveled $$\frac 23$$ of the length of her trip. How long (in miles) is the entire length of her trip? Write an equation to represent the situation, and then find the answer.
### Student Facing
#### Are you ready for more?
Suppose you have a pint of grape juice and a pint of milk. You pour 1 tablespoon of the grape juice into the milk and mix it up. Then you pour 1 tablespoon of this mixture back into the grape juice. Which liquid is more contaminated?
### Activity Synthesis
Select previously identified students to share their responses. Sequence their presentations so that students with the more concrete strategies (e.g., drawing pictures) share before those with more abstract strategies. Students using the algorithm should share last. Find opportunities to connect the different methods. For example, point out where the multiplication by a denominator and division by a numerator are visible in a tape diagram.
Conversing, Representing: MLR7 Compare and Connect. As students consider the different strategies, invite them to make connections between the various representations and approaches. Ask, “What do each of the strategies have in common?”, “How are the strategies different?” and “Which strategy is more efficient? Why?” Listen for and amplify observations that include mathematical language and reasoning.
Design Principle(s): Maximize meta-awareness; Optimize output (for comparison)
## 8.3: A Train is Traveling at . . . (10 minutes)
### Activity
The purpose of this activity is to review different strategies for working with ratios and to prepare students to use these strategies with ratios involving fractions. The activity also foreshadows percentages by asking about the distance traveled in 100 minutes.
Monitor for different strategies like these:
• divide $$\frac{15}{2}\div 6$$ to find the distance traveled in 1 minute, and then multiply it by 100.
• draw a double number line.
• create a table of equivalent ratios.
Depending on their prior learning, students might lean towards the first strategy.
### Launch
Give students 3 minutes of quiet work time. Encourage them to find more than one strategy if they have time. Follow with whole-class discussion around the various strategies they used.
### Student Facing
A train is traveling at a constant speed and goes 7.5 kilometers in 6 minutes. At that rate:
1. How far does the train go in 1 minute?
2. How far does the train go in 100 minutes?
### Anticipated Misconceptions
Students might calculate the unit rate as $$6\div \frac{15}{2}$$. Ask students what this number would mean in this problem? (This number means that it takes $$\frac45$$ of a minute to travel 1 kilometer.) In this case, students should be encouraged to create a table or a double number line, since it will help them make sense of the meaning of the numbers.
### Activity Synthesis
Select students to share the strategies they used. To the extent possible, there should be one student per strategy listed. If no students come up with one or more representations, create them so that students can compare and contrast.
• Divide ($$\frac{15}{2}\div 6$$) to find the number of kilometers traveled in 1 minute, and multiply by 100
• Double Number Line
• Table
Display strategies for all to see throughout the discussion.
Help students connect the strategies by asking:
• Was there a place in your solution where you calculated $$\frac{15}{2}\div 6$$?
• How can we see this value being used in the double number line? Table?
Engagement: Develop Effort and Persistence. Encourage and support opportunities for peer interactions. Prior to the whole-class discussion, invite students to share their work with a partner. Display sentence frames to support student conversation such as: “First, I _____ because...”, “I noticed _____ so I...”, “Why did you...?”, “I agree/disagree because….”
Supports accessibility for: Language; Social-emotional skills
Speaking: MLR7 Compare and Connect. Use this routine when students present the strategies they used to determine the distance traveled. Ask students to consider what is the same and what is different about each approach. Draw students' attention to the different ways the unit rate and total distance can be seen in each representation (i.e., tape diagram, double number line and table). Listen for and amplify students' correct use of the term “unit rate.” These exchanges can strengthen students' mathematical language use as they reason to make sense of strategies used to calculate unit rates and distance traveled.
Design Principle(s): Maximize meta-awareness
## 8.4: Comparing Running Speeds (10 minutes)
### Activity
The purpose of this activity is to provide another context that leads students to calculate a unit rate from a ratio of fractions. This work is based on students’ work in grade 6 on dividing fractions.
Students notice and wonder about two statements and use what they wonder to create questions that are collected for all to see. Each student picks a question secretly and calculates the answer, then shares the answer with their partner. The partner tries to guess the question. Most of the time in this activity should be spent on students engaging in partner discussion.
### Launch
Arrange students into groups of 2. Display the two statements for all to see. Ask students to write down what they notice and wonder, and then use what they wonder to come up with questions that can be answered using the given information. Create a list of questions and display for all to see. Here are suggested questions to listen for:
• Who ran faster, Noah or Lin?
• How far would Lin run in 1 hour?
• How far did Noah run in 1 hour?
• How long would it take Lin to run 1 mile at that rate?
• How long would it take Noah to run 1 mile at that rate?
Action and Expression: Internalize Executive Functions. Chunk this task into more manageable parts to support students who benefit from support with organization and problem solving. For example, present one question at a time and monitor students to ensure they are making progress throughout the activity.
Supports accessibility for: Organization; Attention
Speaking: MLR8 Discussion Supports. Provide sentence frames to students to support their explanations of their process for finding the answer to their selected question. For example, “First, I _____. Then, I ____.”
Design Principle(s): Support sense-making, Optimize output (for explanation)
### Student Facing
Lin ran $$2 \frac34$$ miles in $$\frac25$$ of an hour. Noah ran $$8 \frac23$$ miles in $$\frac43$$ of an hour.
1. Pick one of the questions that was displayed, but don’t tell anyone which question you picked. Find the answer to the question.
2. When you and your partner are both done, share the answer you got (do not share the question) and ask your partner to guess which question you answered. If your partner can’t guess, explain the process you used to answer the question.
### Student Facing
#### Are you ready for more?
Nothing can go faster than the speed of light, which is 299,792,458 meters per second. Which of these are possible?
1. Traveling a billion meters in 5 seconds.
2. Traveling a meter in 2.5 nanoseconds. (A nanosecond is a billionth of a second.)
3. Traveling a parsec in a year. (A parsec is about 3.26 light years and a light year is the distance light can travel in a year.)
### Anticipated Misconceptions
The warm-up was intended to remind students of some strategies for dividing fractions by fractions, but students may need additional support working with the numbers in this task.
Students might have a hard time guessing their partner's question given only the answer. Ask their partners to share the process they used to calculate the solution, they might leave out numbers and describe in general the steps they took to find the answer first. If their partner is still unable to guess the question, have them share the specific number they used. If they need additional support to guess the question, have their partner show them their work on paper (without sharing the question they answered) and see if this helps them figure out the question.
### Activity Synthesis
After both partners have a chance to guess each other’s question, ask a few different students to share their strategies for guessing which question their partner answered.
## Lesson Synthesis
### Lesson Synthesis
In this lesson, we worked with ratios of fractions.
• “What are strategies we can use to find solutions to ratio problems that involve fractions?” (double number line, tables, calculating unit rate)
• “How are those strategies different from and similar to ways we previously solved ratio problems that didn't involve fractions?” (They are structurally the same, but the arithmetic might take more time.)
## Student Lesson Summary
### Student Facing
There are 12 inches in a foot, so we can say that for every 1 foot, there are 12 inches, or the ratio of feet to inches is $$1:12$$. We can find the unit rates by dividing the numbers in the ratio:
$$1\div 12 = \frac{1}{12}$$
so there is $$\frac{1}{12}$$ foot per inch.
$$12 \div 1 = 12$$
so there are 12 inches per foot.
The numbers in a ratio can be fractions, and we calculate the unit rates the same way: by dividing the numbers in the ratio. For example, if someone runs $$\frac34$$ mile in $$\frac{11}{2}$$ minutes, the ratio of minutes to miles is $$\frac{11}{2}:\frac34$$.
$$\frac{11}{2} \div \frac34 = \frac{22}{3}$$, so the person’s
pace is $$\frac{22}{3}$$ minutes per mile.
$$\frac34 \div \frac{11}{2} = \frac{3}{22}$$, so the person’s
speed is $$\frac{3}{22}$$ miles per minute. |
# Chapter 6: linear equations and the greatest common factor (1)
Time：2020-12-5
### Theorem of linear equation
Likeax+byThe smallest positive integer of is equal togcd(a,b)
We use Euclidean algorithm to construct the appropriatexAndyIn other words, the equation will be describedax + by= gcd(a,b)Integer solutionxAndyMethods. Because of each numberax+bycovergcd(a,b)to be divisible by,ax + byThe smallest positive integer value of is exactlygcd(a,b)
Solving equations by Euclidean algorithmax+by=gcd(a,b)
For example, try to solve the problem22x+60y=gcd(22,60)
In the first step, Euclidean algorithm is used to calculate the greatest common factor
60 = 2·22 +16
22 = 1·16 +6
16=2·6+4
6=1·4+2
4=2·2+0
This shows thatgcd(22,60)=2This is an obvious fact that there is no need for Euclidean algorithm. However, it is very important to use Euclidean algorithm because we use the middle quotient and remainder to solve the equation. First, rewrite the first equation into
16 = a-2b, where a = 60, B = 22
Next, replace the value in the second equation with this value16, get
b=1·16+6=1·(a-2b)+6
Rearrange the equation6Move to one side
6=b-(a-2b)=-a+3b
Now the value of16And6Put in the next equation16=2·6+4
a-2b=16=2·6+4=2(-a+3b)+4
The result of transfer
4=(a-2b)-2(-a+3b)=3a-8b
Finally, use the equation6=1·4+2have to
-a+3b=6=1·4+2=1·(3a-8b)+2
Rearrange the equation to get the desired solution
-4a+11b=2 |
# The Formula
$$\sum\limits_{n=1}^{\infty}a_{n} \text{ diverges if } \lim\limits_{n\rightarrow\infty}a_{n} \text{ fails to exists or does not equal zero.}$$
Note: If the limit equals zero that does not mean that the series converges . It means that it could converge. If the limit equals zero, more tests have to be performed to prove whether or not the series converges. However, if the limit is not zero, we know the series diverges. This is why the test is also known as the divergence test.
The Nth term test is often the first test that you should use to rule out the series diverges before you go any further.
Series Limit Conclusion $$\sum\limits_{n=0}^{\infty}\frac{n}{2n+1}$$ $$\lim\limits_{n\rightarrow\infty}\frac{n}{2n+1}=\frac{1}{2}$$ diverges since $$\lim\limits_{n\rightarrow\infty}\frac{n}{2n+1}\neq 0$$ $$\sum\limits_{n=0}^{\infty}\frac{n!+1}{3n!-5}$$ $$\lim\limits_{n\rightarrow\infty}\frac{n!+1}{3n!-5}=\frac{1}{3}$$ diverges since $$\lim\limits_{n\rightarrow\infty}\frac{n!+1}{3n!-5}\neq0$$ $$\sum\limits_{n=0}^{\infty}\frac{n}{ln(n)}$$ $$\lim\limits_{n\rightarrow\infty}\frac{n}{ln(n)}=\infty$$ L'Hopitals Rule diverges since $$\sum\limits_{n=0}^{\infty}\frac{n}{ln(n)}$$ $$\sum\limits_{n=0}^{\infty}n \cdot sin(\frac{1}{n})$$ $$\lim\limits_{n\rightarrow\infty}n \cdot sin(\frac{1}{n})=1$$ L'Hopitals Rule diverges since $$\lim\limits_{n\rightarrow\infty}n \cdot sin(\frac{1}{n})\neq0$$ $$\sum\limits_{n=0}^{\infty}\frac{1}{n}$$ $$\lim\limits_{n\rightarrow\infty}\frac{1}{n}=0$$ could possibly converge
In fact, in the last example we found that $$\lim\limits_{n\rightarrow\infty}\frac{1}{n}=0$$, however, we know that the series $$\sum\limits_{n=0}^{\infty}\frac{1}{n}$$ is a P-series that diverges.
On the other hand, consider the series $$\sum\limits_{n=0}^{\infty}\frac{1}{n^2}$$ the $$\lim\limits_{n\rightarrow\infty}\frac{1}{n^2}=0$$ . In this case, we have a P-series with p=2 so this series converges. |
# Word Problems That Involve The Mean (II) Worksheet
Related Topics & Worksheets:
Mean Worksheet 1
Mean Worksheet 3
More Statistics Worksheets
Objective: I know how to solve word problems that involve the mean.
In statistics, mode, median and mean are typical values to represent a pool of numerical observations. They are calculated from the pool of observations.
Mode is the most common value among the given observations. For example, a person who sells ice creams might want to know which flavor is the most popular.
Median is the middle value, dividing the number of data into 2 halves. In other words, 50% of the observations is below the median and 50% of the observations is above the median.
Mean is the average of all the values. For example, a teacher may want to know the average marks of a test in his class.
Fill in all the gaps, then press "Check" to check your answers. Use the "Hint" button to get a free letter if an answer is giving you trouble. You can also click on the "[?]" button to get a clue. Note that you will lose points if you ask for hints or clues!
1) The mean of 15 numbers is 20. The mean of another 10 numbers is 15. What is the mean of all the 25 numbers?
Mean =
2) There are 20 students in a drama class. The mean age of 12 students is 18 years. The mean age of the remaining 8 students of the drama class is 23 years. What is the mean age of all the students in the drama class?
Mean =
3) The mean mass of 6 boxes is 45 kg. One of the boxes with a mass of 50 kg is removed. Calculate the new mean mass of the remaining boxes.
Mean mass = kg
4) The following is a set of numbers: 2, 5, 6, 13, 5, 21, 15, 5, 19, 22, 8
When a number is x is added to the set, the new mean is 12. Calculate the value of x.
x =
5) Given that a set of numbers = {5, 7, 8, 12, 6, 14, 11}.
When a number x is added to the above set, the new mean is 10. Calculate the value of x.
x =
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.
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UnFOILing is a method for factoring a trinomial into two binomials. When you multiply two binomials together, you use the FOIL method, multiplying the First, then the Outer, then the Inner, and finally the Last terms of the two binomials into a trinomial. But when you need to factor a trinomial, you unFOIL by determining the factor pairs for a and c, the correct signs to place inside the two binomials, and what combination of factor pairs of a and c results in b.
The key to unFOILing is being organized:
• Be sure you have an expression in the form:
• Write the terms in the order of decreasing powers.
• Remember how to assign the correct signs in each binomial:
• The signs are both positive, if c is positive and b is positive.
• The signs are both negative, if c is positive and b is negative.
• One sign is positive and one negative, if c is negative; which binomial is positive and which one is negative depends on whether b is positive or negative and how you arranged the factors.
Example:
1. Determine all the ways you can multiply two numbers to get a.
You can get these numbers from the prime factorization of a. Sometimes, writing out the list of ways to multiply is a big help. In this example, a is 24, and the list of ways you can multiply two numbers to get 24 is:
1 × 24, 2 × 12, 3 × 8, or 4 × 6.
2. Determine all the ways you can multiply two numbers to get c.
In this example, c is 45, and you can multiply the following numbers to get 45:
1 × 45, 3 × 15, or 5 × 9.
Ignore the sign at this point. You don't need to worry about signs until Step 3.
3. Look at the sign of c and your lists from Steps 1 and 2 to see if you want a sum or difference.
If c is positive, find a value from your Step 1 list and another from your Step 2 list such that the sum of their product and the product of the two remaining numbers in those steps results in b.
If c is negative, find a value from your Step 1 list and another from your Step 2 list such that the difference of their product and the product of two remaining numbers from those steps results in b.
For the trinomial
c is negative, so you want a difference of 34 between products.
4. Choose a product from Step 1 and a product from Step 2 that result in the correct sum or difference determined in Step 3.
Because you determined in Step 3 that you want a difference of 34 between products, use 4 × 6 from a and 5 × 9 from c.
The product of 4 and 5 is 20. The product of 6 and 9 is 54. The difference of these products is 34.
5. Arrange your choices as binomials so the results are those you want.
(4x 9)(6x 5)
6. Place the signs to give the desired results.
(4x – 9)(6x + 5)
7. FOIL the two binomials to check your work.
If the binomials are correct, you'll end up with the original problem when you FOIL them. |
# 2001 AIME II Problems/Problem 10
## Problem
How many positive integer multiples of $1001$ can be expressed in the form $10^{j} - 10^{i}$, where $i$ and $j$ are integers and $0\leq i < j \leq 99$?
## Solution 1
The prime factorization of $1001 = 7\times 11\times 13$. We have $7\times 11\times 13\times k = 10^j - 10^i = 10^i(10^{j - i} - 1)$. Since $\text{gcd}\,(10^i = 2^i \times 5^i, 7 \times 11 \times 13) = 1$, we require that $1001 = 10^3 + 1 | 10^{j-i} - 1$. From the factorization $10^6 - 1 = (10^3 + 1)(10^{3} - 1)$, we see that $j-i = 6$ works; also, $a-b | a^n - b^n$ implies that $10^{6} - 1 | 10^{6k} - 1$, and so any $\boxed{j-i \equiv 0 \pmod{6}}$ will work.
To show that no other possibilities work, suppose $j-i \equiv a \pmod{6},\ 1 \le a \le 5$, and let $j-i-a = 6k$. Then we can write $10^{j-i} - 1 = 10^{a} (10^{6k} - 1) + (10^{a} - 1)$, and we can easily verify that $10^6 - 1 \nmid 10^a - 1$ for $1 \le a \le 5$.
If $j - i = 6, j\leq 99$, then we can have solutions of $10^6 - 10^0, 10^7 - 10^1, \dots\implies 94$ ways. If $j - i = 12$, we can have the solutions of $10^{12} - 10^{0},\dots\implies 94 - 6 = 88$, and so forth. Therefore, the answer is $94 + 88 + 82 + \dots + 4\implies 16\left(\dfrac{98}{2}\right) = \boxed{784}$.
## Solution 2
Observation: We see that there is a pattern with $10^k \pmod{1001}$. $$10^0 \equiv 1 \pmod{1001}$$ $$10^1 \equiv 10 \pmod{1001}$$ $$10^2 \equiv 100 \pmod{1001}$$ $$10^3 \equiv -1 \pmod{1001}$$ $$10^4 \equiv -10 \pmod{1001}$$ $$10^5 \equiv -100 \pmod{1001}$$ $$10^6 \equiv 1 \pmod{1001}$$ $$10^7 \equiv 10 \pmod{1001}$$ $$10^8 \equiv 100 \pmod{1001}$$
So, this pattern repeats every 6.
Also, $10^j-10^i \equiv 0 \pmod{1001}$, so $10^j \equiv 10^i \pmod{1001}$, and thus, $$j \equiv i \pmod{6}$$. Continue with the 2nd paragraph of solution 1, and we get the answer of $\boxed{784}$
-AlexLikeMath
## Solution 3
Note that $1001=7\cdot 11\cdot 13,$ and note that $10^3 \equiv \pmod{p}$ for prime $p | 1001$; therefore, the order of 10 modulo $7,11$, and $13$ must divide 6. A quick check on 7 reveals that it is indeed 6. Therefore we note that $i-j=6k$ for some natural number k. From here, we note that for $j=0,1,2,3,$ we have 16 options and we have 15,14,...,1 option(s) for the next 90 numbers (6 each), so our total is $4\cdot 16 + 6 \cdot \frac{15 \cdot 16}{2} = \boxed{784}$.
~Dhillonr25 |
# Partial Fraction Decomposition
I am really stuck on how to do this. It is a step I need to do for an inductive proof. I have $$\frac{1}{n(n-1)}$$
Do I set it up like this: $\frac{ A}{n} + \frac{B}{n-1}$ ? $$1= A(n-1) + B(n)$$ How do I solve for $A$ and $B$
You have the fraction $$\frac{1}{n(n-1)}$$ And you want it to be $$\frac{1}{n(n-1)} =\frac{A}{n}+\frac{B}{n-1}$$ Now you expand it and get $$\frac{A(n-1)+Bn}{n(n-1)}=\frac{1}{n(n-1)}$$ Now you compare the coefficients, which give you that $$A(n-1) + Bn=0\cdot n + 1$$ 2 polynomials are the same iff all their coefficients are the same. $$(A+B) \cdot n =0 \cdot n$$ and $$-A = 1$$ This gives $B=1$ and $A=-1$ hence $$\frac{1}{n(n-1)} = \frac{1}{n-1}-\frac{1}{n}$$
Instead of the coefficient comparism you can $k+1$ values for the polynomial where $k$ is the degree of the polynomial. Showing that this works to is a result of polynom interpolation, that works at least for $\mathbb{R}$.
• Do the same as @Dominic suggested till you get $$A(n-1)+Bn=1$$ Now set $n=1$ above so $A(1-1)+B=1$ so $B=1$. Set $n=0$ above so $A(0-1)+0=1$ so $A=-1$. – mrs Apr 3 '13 at 18:39 |
Cosine Rule - Formula, Revision and worksheets. | Maths Made Easy
# Cosine Rule – Formula, Revision and worksheets.
Navigate topic
## What you need to know
The cosine rule is an equations that help us find missing side-lengths and angles in any triangle.
It is expressed expressed according to the triangle on the right.
The cosine rule is
$\textcolor{limegreen}{a}^2=\textcolor{blue}{b}^2+\textcolor{red}{c}^2-2\textcolor{blue}{b}\textcolor{red}{c}\cos \textcolor{limegreen}{A}$
In this topic, we’ll go through examples of cosine rule questions.
## Example: Cosine rule to find a length
Use the cosine rule to find the side-length marked $x$ to $1$dp
First we match up the information in the question to the letters in the formula.
$a=x$, $A = 44\degree$, $b=5$ and $c=7$.
So, we’re ready to substitute the values into the formula.
$x^2=5^2+7^2-(2\times5\times7\times\cos(44))=25+49-70\cos(44)$
Taking the square root of both sides, and putting it into the calculator, we get
$x=\sqrt{25+49-70\cos(44)}=4.9\text{ (1dp)}$.
### Example Questions
Firstly, we need appropriately label the sides of this triangle. Firstly, we set $a=x$, and therefore we get that $A=19$, since it is the angle opposite. It doesn’t matter how we label the other two sides, so here we’ll let $b=86$ and $c=65$.
Now, subbing these values into the cosine rule equation, we get
$x^2=86^2+65^2-(2\times86\times65\times\cos(19))=7,396+4,225-11,180\cos(19)$
Then, taking the square root, and putting it into the calculator, we get
$x=\sqrt{7,396+4,225-11,180\cos(19)}=32\text{cm (2sf)}$
As always, we must label our triangle. Firstly, assign the thing we’re looking for to be $a=x$, and therefore make the side opposite to it is $A=6$. Then, it doesn’t matter how we choose the other two sides, so we will let $b=5$ and $c=7$.
Here, we will use the rearranged version of the formula that looks like
$\cos A=\dfrac{b^2+c^2-a^2}{2bc},$
So, subbing these values into the equation, we get
$\cos x=\dfrac{5^2+7^2-6^2}{2\times5\times7}=\dfrac{25+49-36}{70}$
Taking $\cos^{-1}$ of both sides, and putting it into a calculator, we get
$x=\cos^{-1}\left(\dfrac{25+49-36}{70}\right)=57.1\degree\text{ (3sf)}$.
### Learning resources you may be interested in
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361 Do You See a Pattern?
When 2^361 is divided by 361, the remainder is 116, not 2. That means that 361 is definitely a composite number. Its factors are listed at the end of this post.
361 isn’t used as often, but it is just as special as some of the numbers in the table below:
The pattern can also be seen along the diagonals in this ordinary multiplication table:
This pattern could be very helpful to students who are learning to multiply. I have seen plenty of students who knew 7 x 7 = 49, but couldn’t remember what 6 x 8 is.
Years after I learned the multiplication facts, I learned how to multiply binomials in an algebra class. I learned about the difference of two squares. In the example below one of the squares is n² and the other square is 1² which is equal to 1. I learned that the equation
is true for ALL numbers, but nobody pointed out any practical examples to make it more meaningful. The table at the top of the page contains twelve practical examples. Let’s see how you do applying it to products of a few larger numbers.
Sometimes we find easy ways to remember certain products like
We can use those products to help us remember other products easily by applying the difference of two squares. Try these: (Yes, you can easily do them without a calculator!)
• 13 x 13 = 169. How much is 12 x 14?
• 14 x 14 = 196. How much is 13 x 15?
• 20 x 20 = 400. How much is 19 x 21?
• If you know that the first ten powers of 2 are 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, then it’s easy to remember that 16 x 16 = 256. How much is 15 x 17?
• What if we go in the opposite direction…It isn’t too hard to multiply 18 x 20 in your head to get 360. How much is 19 x 19?
• 22 x 20 = 440 was also easy to find. How much is 21 x 21?
• 30 x 30 = 900. How much is 29 x 31?
• 100 x 100 = 10,000. How much is 99 x 101?
Did you figure out what 361 has to do with this pattern? It is a perfect square just like 1, 4, 9, 16, and 25. Here is its factoring information:
• 361 is a composite number.
• Prime factorization: 361 = 19^2
• The exponent in the prime factorization is 2. Adding one we get (2 + 1) = 3. Therefore 361 has exactly 3 factors.
• Factors of 361: 1, 19, 361
• Factor pairs: 361 = 1 x 361 or 19 x 19
• 361 is a perfect square. √361 = 19
One thought on “361 Do You See a Pattern?”
1. Fascinating! I have never heard of this before.
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# 2013 AMC 10A Problems/Problem 13
## Problem
How many three-digit numbers are not divisible by $5$, have digits that sum to less than $20$, and have the first digit equal to the third digit?
$\textbf{(A)}\ 52 \qquad\textbf{(B)}\ 60 \qquad\textbf{(C)}\ 66 \qquad\textbf{(D)}\ 68 \qquad\textbf{(E)}\ 70$
## Solution
We use a casework approach to solve the problem. These three digit numbers are of the form $\overline{xyx}$.($\overline{abc}$ denotes the number $100a+10b+c$). We see that $x\neq 0$ and $x\neq 5$, as $x=0$ does not yield a three-digit integer and $x=5$ yields a number divisible by 5.
The second condition is that the sum $2x+y<20$. When $x$ is $1$, $2$, $3$, or $4$, $y$ can be any digit from $0$ to $9$, as $2x<10$. This yields $10(4) = 40$ numbers.
When $x=6$, we see that $12+y<20$ so $y<8$. This yields $8$ more numbers.
When $x=7$, $14+y<20$ so $y<6$. This yields $6$ more numbers.
When $x=8$, $16+y<20$ so $y<4$. This yields $4$ more numbers.
When $x=9$, $18+y<20$ so $y<2$. This yields $2$ more numbers.
Summing, we get $40 + 8 + 6 + 4 + 2 = \boxed{\textbf{(B) }60}$
~savannahsolver |
# How do you solve 2x + 3 = 63?
Jan 25, 2017
See the entire solution process below:
#### Explanation:
First, subtract $\textcolor{red}{3}$ from each side of the equation to isolate the $x$ term while keeping the equation balanced:
$2 x + 3 - \textcolor{red}{3} = 63 - \textcolor{red}{3}$
$2 x + 0 = 60$
$2 x = 60$
Now, divide each side of the equation by $\textcolor{red}{2}$ to solve for $x$ while keeping the equation balanced:
$\frac{2 x}{\textcolor{red}{2}} = \frac{60}{\textcolor{red}{2}}$
$\frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{2}}} x}{\cancel{\textcolor{red}{2}}} = 30$
$x = 30$ |
# Inverse Sine
The inverse sine function (also called arcsine) is the inverse of sine function. Since sine of an angle (sine function) is equal to ratio of opposite side and hypotenuse, thus sine inverse of same ratio will give the measure of the angle. Let’s say θ is the angle, then:
sin θ = (Opposite side to θ/Hypotenuse)
Or θ = Sin-1 (Opposite side to θ/Hypotenuse)
In the same way, Inverse Cosine and Inverse Tan can be determined.
Fact: If the ratio of opposite side of angle θ and hypotenuse equals to 1, then, θ = sin-1 (1) = sin-1 (sin 90) = 90 degrees
The function of inverse or inverse function is used to determine the angle measure, with the use of basic trigonometric ratios from the right triangle. Mostly, the inverse sine function is represented using sin-1. It does not mean that sine is not raised to the negative power.
## What is Sine Function?
In a right-angle triangle, a sine function of an angle θ is equal to the opposite side to θ divided by hypotenuse.
Sin θ = Opposite side/Hypotenuse
This is the basic formula for sine function. See the figure below.
Sin θ = Opposite / Hypotenuse
## What is Inverse Sine Function?
In the above explanation, we learned about sine function, where we can determine the sin of any angle if the opposite side and hypotenuse is known to us.
What if we have to find just the measure of angle θ? The inverse sine function or Sin-1 takes the ratio, Opposite Side / Hypotenuse Side and produces angle θ. It is also written as arcsin.
Sin inverse is denoted by sin-1 or arcsin
θ = Sin-1 (Opposite side/Hypotenuse)
Let us see an example of inverse of sine function.
Example: In a triangle, ABC, AB= 4.9m, BC=4.0 m, CA=2.8 m and angle B = 35°.
Solution:
• Sin 35° = Opposite / Hypotenuse
• Sin 35° = 2.8 / 4.9
• Sin 35° = 0.57°
So, Sin-1 (Opposite / Hypotenuse) = 35°
Sin-1 (0.57) = 35°
## Inverse Sine Formula
Let us consider if we want to find the depth(d) of the seabed from the bottom of the ship and the following two parameters are given:
• The angle which the cable makes with the seabed.
• The cable’s length.
The Sine function will help to find the distance/depth d of the ship from the sea bed by the following method:
If the angle is 39° and the cable’s length is 40 m.
• Sin 39° = Opposite / Hypotenuse
• Sin 39° = d / 40
• d = Sin 39° × 40
• d = 0.6293 × 40
• d = 25.172 cm
Therefore, the depth d is 25.17 cm.
Summary of Formula:
The formula for the trigonometric sine function is given by:
sin (θ) Opposite Side/ Hypotenuse
The inverse sine function formula or the arcsin formula is given as:
sin-1 (Opposite side/ hypotenuse) = θ
## Graph of Inverse Sine Function
Arcsine trigonometric function is the sine function is shown as sin-1 a and is shown by the below graph.
## Inverse Sine Derivative
The derivative of inverse sine function is given by: d/dx Sin-1x= 1 / √(1-x2)
Let us proof this equation.
Proof: f(x) =sin(x) and g(x) = sin-1x
If we differentiate g(x) with respect to x, we get;
g'(x) = 1/f'(g(x)) = 1/cos(sin-1x) …………….(1)
Now we know, the inverse of sine function, y = Sin-1x
or sin y = x …………(2)
Hence we can write, the derivative as;
g'(x) = 1/cos y (from eq.1)
Since, we know, by trigonometric identities;
cos2 y + sin2 y = 1
Therefore, we can write;
cos y = √(1-sin2 y) = √(1 – x2) (since sin y = x)
Now, again putting this value in the derivative;
g'(x) = 1/cos y = 1/√(1 – x2)
Thus,
d/dx(sin-1x) = 1/√(1 – x2)
Hence, proved.
## Inverse Sine Table
θ Sin-1 or Arcsin(θ) (in Radian) Sin-1 or Arcsin(θ) (in Degree) -1 -π/2 -90° -√3/2 -π/3 -60° -√2/2 -π/4 -45° -1/2 -π/6 -30° 0 0 0° 1/2 π/6 30° √2/2 π/4 45° √3/2 π/3 60° 1 π/2 90°
## Solved Examples
Now, if the angle is not given and we want to calculate it, then we use the Inverse functions and the question will be asked in the following way:
Problem: What is the angle Sin = Opposite / Hypotenuse, has?
Solution: Let’s take the measurement from above example only.
• Distance d = 25.17 cm
• Cable’s length = 40 cm.
We want to find the angle “α ”
Step 1: Find the sin α°
• Sin α° = Opposite / Hypotenuse
• Sin α° = 25.17 / 40
• Sin α° = 0.6293
Step 2: Now, for which angle sin α° = 0.6293
Let’s find it out with Inverse sin:
α° = Sin-1 / (0.6293)
α° = 38.1°
Did you know: Sin and Sin-1 are vice-versa.
Example: Sin 30° = 0.5 and Sin-1 (0.5) = 30°
Did you know: Sin and Sin-1 are vice-versa. Example: Sin 30° = 0.5 and Sin-1 (0.5) = 30°
Need to practice more on the trigonometric functions, Download BYJUS -The learning app.
## Frequently Asked Questions – FAQs
Q1
### What is inverse sine?
Inverse sine is one of the trigonometric functions which is used to find the measure of angle in a right triangle. Suppose, α is the angle between hypotenuse and its adjacent side. Then, the measure of angle α is given by;
α = sin-1 (opposite side of α/hypotenuse)
Where sin-1 represents the sine inverse function.
Q2
### What is the value of sin inverse of 1?
Let α is an angle equal to Sin-1 (1)
α = sin-1(1)
We know that, sin 90 degrees = 1
Therefore, α = sin-1 (sin 90) = 90 degrees
Q3
### What is arcsin?
Arcsine is the inverse of sine function. It is used to evaluate the angle whose sine value is equal to the ratio of its opposite side and hypotenuse. Therefore, if we know the length of opposite side and hypotenuse, then we can find the measure of angle.
Q4
### What is the difference between sin-1 and 1/sin x?
Sin-1 is the inverse of sine function. -1 here do not represent the exponent. Arcsin α means the arc whose sine is α. Whereas 1/sin x shows the reciprocal of sine function, which is also equal to cosecant function.
Q5
### Is sine inverse equal to cosec function?
Sine inverse or arcsine is the inverse of sine function which returns the value of angle for which sine function is equal to opposite side and hypotenuse ratio. It generates the value of angle. But cosec function is the reciprocal of sine function and is not equal to sine inverse.
Test your knowledge on Inverse sine |
# probability selecting marbles
can someone solve this example?
An urn contains 2 Red marbles, 3 White marbles and 4 Blue marbles. You reach in and draw out 3 marbles at random (without replacement). What is the probability that you will get one marble of each color? What is the answer if you draw out the marbles one at a time with replacement?
There are $9$ marbles in the urn, so there are $\binom93$ different sets of $3$ marbles that you could draw without replacement. How many contain one marble of each color? To build such a set, you could pick either of the $2$ red marbles, any one of the $3$ white marbles, and any one of the $4$ blue marbles, so there are $2\cdot3\cdot4$ such sets. Each of the $\binom93$ sets is equally likely to be drawn, and $2\cdot3\cdot4$ of them are ‘successes’, so the probability of success is
$$\frac{2\cdot3\cdot4}{\binom93}=\frac{24}{84}=\frac27\;.$$
If you draw with replacement, however, you can potentially draw the same marble twice, and you also have to take into account the order of the draws. On each of your $3$ draws you can get any of the $9$ marbles, so there are $9^3$ possible sequences of $3$ marbles that you can draw, and they’re all equally likely. How many of them contain one marble of each color? As in the first problem, there are $2\cdot3\cdot4=24$ different sets of $3$ marbles that will work, but each of them can be drawn in several different orders to give several different successful sequences of draws. In fact $3$ different objects can be arranged in $3!=6$ different orders, so each of those $24$ sets of $3$ marbles of $3$ different colors can be drawn in $6$ different orders. Thus, there are actually $6\cdot24$ successful sequences of $3$ draws. The final probability of success when you draw with replacement is therefore ... ?
We do the "with replacement" problem. The event "we got a red, a white, and a blue" can happen in various orders.
We find the probability of red then white then blue. This is $\dfrac{2}{9}\cdot \dfrac{3}{9}\cdot \dfrac{4}{9}$. For since we are drawing with replacement, the results on the three picks are independent.
We can end up having obtained a red, a white, and a blue in various other ways, such as blue then red then white. This has the same probability as red then white then blue. In total there are $3!$ ways we can end up having drawn one of each colour. So the required probability is $$3!\left(\frac{2}{9}\cdot\frac{3}{9}\cdot \frac{4}{9} \right).$$
The same probabilistic approach works for sampling without replacement. The probability of red then white then blue is $\dfrac{2}{9}\cdot \dfrac{3}{8}\cdot \dfrac{4}{7}$.
For blue then red then white, we get $\dfrac{4}{9}\cdot \dfrac{2}{8}\cdot \dfrac{3}{7}$. The denominators do not change, and the numerators are permuted. So the required probability is $$3!\left(\frac{2}{9}\cdot\frac{3}{8}\cdot \frac{4}{7} \right).$$
$$n(s)=\binom93$$ $$n(A)= \binom 31 \binom 21 \binom41$$ $$P(A)=n(A)/n(s)$$ |
# USING THE DISTANCE FORMULA WORKSHEET
## About "Using the distance formula worksheet"
Using the distance formula worksheet :
Worksheet given in this section is much useful to the students who would like to practice problems on finding distance between two points using distance formula.
The distance formula is a formula for computing the distance between two points.
Let A(x, y) and B(x, y) be the two points in a coordinate plane as shown below.
Then, the distance between A and B is
## Using the distance formula worksheet - Problems
Problem 1 :
Find the lengths of the segments AB, AC and AD. Say whether any of the segments have the same length.
Problem 2 :
On the map shown below, the city blocks are 340 feet apart east-west and 480 feet apart north-south.
(i) Find the walking distance between A and B.
(ii) What would the distance be if a diagonal street existed between the two points ?
## Using the distance formula worksheet - Solution
Problem 1 :
Find the lengths of the segments AB, AC and AD. Say whether any of the segments have the same length.
Use the distance formula to find the length of each segment.
Length of AB :
(x, y) = A(-1, 1)
(x, y) = B(-4, 3)
Then, we have
AB = √[(-4 + 1)² + (3 - 1)²]
AB = √[(-3)² + (2)²]
AB = √[9 + 4]
AB = √13
Length of AC :
(x, y) = A(-1, 1)
(x, y) = C(3, 2)
Then, we have
AC = √[(3 + 1)² + (2 - 1)²]
AC = √[(4)² + (1)²]
AC = √[16 + 1]
AC = √17
Length of AD :
(x, y) = A(-1, 1)
(x, y) = D(2, -1)
Then, we have
AD = √[(2 + 1)² + (-1 - 1)²]
AD = √[(3)² + (-2)²]
AD = √[9 + 4]
AD = √13
So, AB and AD have the same lengths, but AC has different length.
Note :
Segments that have the same length are called congruent segments. For instance, in the above example, segments AB and AD are congruent, because each has a length of √13 units.
There is a special symbol, ≅ for indicating congruence.
So, we have
AB ≅ AD
Problem 2 :
On the map shown below, the city blocks are 340 feet apart east-west and 480 feet apart north-south.
(i) Find the walking distance between A and B.
(ii) What would the distance be if a diagonal street existed between the two points ?
Solution :
Solution (i) :
To walk from A to B, we would have to walk five blocks east and three blocks north.
5 blocks 340 feet/block = 1700 feet
3 blocks 480 feet/block = 1440 feet
So, the walking distance is
1700 + 1440 = 3140 feet
Solution (ii) :
To find the diagonal distance between A and B, use the distance formula.
(x, y) = A(-680, -480)
(x, y) = B(1020, 960)
Then, we have
AB = √[(1020 + 680)² + (960 + 480)²]
AB = √[(1700)² + (1440)²]
AB = √4,963,600
AB 2228 feet
So, the diagonal distance would be about 2228 feet, that is 912 feet less than walking distance.
It has been illustrated in the picture given below.
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# How do you solve using the completing the square method 0=x^2-3x-6?
Aug 14, 2017
$x = \frac{3}{2} \pm {\left(33\right)}^{\frac{1}{2}} / 2$
#### Explanation:
We need to add a number to create the constant
${\left(\frac{b}{2}\right)}^{2}$ which will make the perfect square ${\left(x - \left(\frac{b}{2}\right)\right)}^{2}$
Since b = - 3 then ${\left(\frac{b}{2}\right)}^{2} = \frac{9}{4}$
But we already have - 6 = - 24/4 so we need to add 33/4
And we also need to subtract 33/4 to keep the equation true
This results in
${\left(x - \left(\frac{3}{2}\right)\right)}^{2} = \frac{33}{4}$
Taking the root and adding 3/2 to both sides
gives the result
$x = \frac{3}{2} \pm {\left(33\right)}^{\frac{1}{2}} / 2$ |
Linear Equations in One Variable with Formula, Rules and Solved Examples
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Linear equations in one variable are the first basic topic that a candidate must know to solve questions of Algebra. Algebra is one of the most important topics in the Quantitative Aptitude section of any government examination. Hence, it is vital to be clear about the basics of linear equations before starting your preparations.
In this article, we will describe linear equations in one variable along with the various rules and concepts associated with it. To help you understand these concepts better, we have listed a few solved sample questions on linear equations in one variable.
What are Linear Equations In One Variable?
Linear equation in one variable is an algebraic equation of equality that involves only one variable, the greatest exponent of which is 1. The equality which contains one or more literal numbers is called an equation. For every linear equation in one variable, there is a single root or solution. This root/solution is a value that when replaced with the variable makes the equation true.
Example: Consider the linear equation in one variable: $$4x+6=14$$. Can you tell what is the number that should replace ‘x’ so that the given equation results in a true statement?
The linear equation will result in a true statement when the value of ‘x’ is 2.
$$4*2+6=8+6=14$$
The linear equation will not result in a true statement if the value of ‘x’ is taken other than 2. Thus, the given equation 4x + 6 = 14 results in a true statement only when ‘x’ is replaced by 2. In this case, we call the number 2, the root or the solution of the equation 4x + 6 = 14. The questions in examinations are usually about solving an equation, that is, finding the root or the value of the variable for which the equation is true.
Standard Form of Linear Equations in One Variable
Linear equations in one variable are those sorts of equations where there is only one variable along with this there is only one solution to the given equation.
Standard form of linear equations in one variable is written as ax+b=0, (a is not equal to zero) where x is the one variable, while a and b are any real numbers.
Therefore, the formula of the linear equations in one variable is formulated as ax + b = 0, such that ‘a’ and ‘b’ are not equivalent to zero. When a graph is drawn for such an equation it seems to be a straight line either horizontally or vertically.
Linear Equations in One Variable Formula
A linear equation is a straightforward approach to represent a mathematical statement. Moreover, solving a linear equation comprises a collection of simple methods. For this, we separate the variables on one side of the equation and the constants on another side of the given equation and conclude the final value of the unknown quantity/elements.
Important formulas of linear equations in one variable are:
Term Linear equations in one variable Formula Distributive Property x(y+z)=xy+xz Commutative Property x+y=y+x (Addition)x.y=y.x (Multiplication) Associative Property x+(y+z)=(x+y)+z (Addition)x.(y.z)=(x.y).z (Multiplication) Identity Property x+0=x (Additive)x.1=x (Multiplicative) Inverse Property x+(-x)=0 (Additive)x.(1/x)=1 (Multiplicative)
How to solve Linear Equations in One Variable?
We can solve linear equations in one variable by following the below steps:
• Step 1 – Clear the equation of fractions (or decimals) by multiplying both sides by the LCM of the denominators (or by a power of 10 in the case of decimals).
• Step 2 – Remove group symbols from both sides, if any, and simplify the equation.
• Step 3 – Bring all variable terms on one side and all the constants on the other. Combine the like terms.
• Step 4 – Divide the terms into both sides by the numerical coefficient of the variable.
Let us understand how to solve linear equations in one variable with the help of the following example:
Example: Consider the following equation and solve for ‘x’ $$\frac{x}{4}=28-\frac{x}{3}$$
Step 1. Clear the fraction by multiplying both sides by the LCM of denominators (4 and 3), which is 12.
$$12\frac{x}{4}=12*28-12\frac{x}{3}$$
Step 2. Simplify the equation
$$3x=336-4x$$
Step 3. Isolate all variable terms on one side and all the constants on the other and combine the like terms.
$$3x+4x=336$$
$$7x=336$$
Step 4. Divide the terms into both sides by the numerical coefficient of the variable, which is 7.
$$x=48$$
When you’ve finished with linear equations in one variable, you can read about Linear Equations in Two variables.
Linear Equation in One Variable vs Non-linear Equations
So far we have seen the linear equations in one variable definition and how to solve them. Now let us also learn about non-linear equations. As even non-linear equations have multiple applications in geometry, trigonometry, as well in calculus.
The following table shows the comparison of linear equations in one variable vs non-linear equations:
Linear Equations in One Variable Non-linear Equations A linear equation in one variable is a single-degree equation that is represented by a line on a coordinate plane. A non-linear equation in contrast to a linear one is represented by a curve on the coordinate axis. Linear equation in one variable is of the form ax+b=0. A nonlinear equation is of a higher degree. The equation of curves such as parabola, circle, ellipse, hyperbola are examples of non-linear equations. Examples of linear equations in one variable:x = 8, 6x + 11 = 19, 2x + 4y = 22 Examples of non-linear equations:$$x^2 + y^2= 16$$, $$\frac{x^2}{8}+\frac{y^2}{25}=1$$
Rules for solving Linear Equations in One Variable
In these types of linear equations in one variable questions, a problem (or situation) will be given in verbal language involving certain numbers. You will have to find the solution to the problem by translating words into a linear equation.
There are some specific rules for solving linear equations in one variable which are mentioned as follows:
• Adding the same number (positive or negative) to both sides of an equation, then the resulting equation will be equivalent to the original equation.
• Multiplying or dividing both sides of an equation by the same non-zero number, then the resulting equation will be equivalent to the original equation.
• Transposing a term from one side to the other side of the equation, then the resulting equation will be equivalent to the original equation, but remember that the sign of the transposed term changes.
Linear Equations In One Variable Examples
The following are a few examples of questions that can come in examinations, related to Linear Equations in one variable. The questions can be asked in the following two types:
Example 1. The sum of the two numbers is 45 and their ratio is 2 : 3. Find the numbers.
Solution: Let x be one of the numbers. Then, the other number is 45-x.
Since the two numbers are in the ratio 2 : 3, we have
$$\frac{x}{45-x}=\frac{2}{3}$$
$$3x=2(45-x)$$
$$3x=90-2x$$
$$5x=90$$
$$x=\frac{90}{5}$$
$$x=18$$
Thus, one number is 18 and the other number is 45-18=27.
Example 2. Solve the following equations for $$x :$$ $$4(x+5)=14x+50$$
Solution: $$4x+20=14x+50$$ [Removing the brackets]
$$4x-14x=50-20$$ [Transposing like terms to same side]
$$-10x=30$$[Dividing both sides by -10]
$$x= -3$$
Example 3. Solve the following equations for $$x :$$ $$2(x-1)-3(x-2)=4(x-5)$$
Solution: $$2x-2-3x+6=4x-20$$
$$2x-3x-4x=2-6-20$$
$$-5x= -24$$
$$x=245$$
Example 4. Solve the following equations for $$x :$$ $$\frac{3x}{8}+\frac{1}{4}=\frac{x-3}{3}$$
Solution: $$24(\frac{3x}{8}+\frac{1}{4})=24\frac{x-3}{3}$$
$$9x+6=8x-24$$
$$9x-8x= -6-24$$
$$x = -30$$
Example 5. Solve the following equations for $$x :$$ $$\frac{5}{x}=\frac{9}{x-8}$$
Solution: Multiplying throughout by $$x(x – 8)$$, LCM of x and (x – 4), we get
$$5(x-8)=9x$$
$$5x-40=9x$$
$$5x-9x=40$$
$$-4x=40$$
$$x= -10$$
Example 6. Solve the following equations for $$x :$$ $$\frac{x+b}{a+b}=\frac{x-b}{a-b}$$
Solution: Multiplying throughout by $$(a + b)(a – b)$$, LCM of the denominators
$$x(a-b)+b(a-b)=a(x-b)+b(x-b)$$
$$xa-xb+ba-b^2=ax-ab+bx-b^2$$
$$ax-bx-ax-bx= -ab-b^2-ab+b^2$$
$$- 2bx= -2ab$$
$$x=a$$
Example 7. A woman is five times as old as her daughter. In two years’ time, she will be four times as old as her daughter. Find their present ages.
Solution: Let the present age of the daughter be $$x$$ years.
Then, the present age of woman = $$5x$$ years
After two years,
Daughter’s age = $$(x+2)$$ years
Woman’s age = $$(5x+ 2)$$years
According to the question,
$$5x+2=4(x+2)$$
$$5x+2=4x+8$$
$$5x-4x=8-2$$
$$x=6$$
Thus, the daughter’s present age is 6 years and the woman’s present age is (56) years, i.e., 30 years
Example 8. A number consists of two digits whose sum is 9. If 9 is subtracted from the number its digits are interchanged. Find the number.
Solution: Let the unit digit of the number be x.
Then, the tens digit =9-x
Therefore, Number =$$10(9-x)+x$$
$$=90-10x+x$$
$$=90-9x$$
Number with interchanged digits =10 x+9-x
$$=9x+9$$
According to the question,
$$(90-9x)-9=9x+9$$
$$81-9x=9x+9$$
$$9x+9x=81-9$$
$$18x=72$$
$$x=4$$
Therefore, the units digit $$=4$$ and the tens digit $$=9-4=5$$
Hence, the number is $$54$$.
Example 9. Rahul has 3 boxes of different vegetables. Box A weighs 2.50 kg more than box B and box C weighs 10.25 kg more than box B. The total weight of the three boxes is 48.75 kg. How much does box A weighs?
Solution:
Let the weight of box B be x kg.
Then,
• Weight of box A = $$(x+2.50)$$kg
• Weight of box C = $$(x+10.25)$$kg
According to the question,
$$(x+2.50)+x+(x+10.25)=48.75$$
$$x+x+x=48.75-2.50-10.25$$
$$3x=36$$
$$x=12$$
Therefore, Weight of box A =(x+2.50) kg=14.50 kg.
Example 10. The distance between the two stations is 340 km. two trains start simultaneously from these stations on parallel tracks to cross each other. The speed of one of them is greater than that of the other by 5 km/h. If the distance between the two trains after 2 hours of their start is 30 km, find the speed of each train.
Solution: Let the speed of the train starting from A be $$x$$ km/h.
Therefore, The speed of the second train starting from B will be $$(x + 5)$$ km/h.
The first train reaches C and the second train reaches D in 2 hours.
According to the question,
$$x2+30+(x+5)2=340$$
$$2x+30+2x+10=340$$
$$2x+2x=340-40$$
$$4x=300$$
$$x=75$$
Therefore, The speeds of the trains are 75 km/h and 80 km/h respectively.
Exams where Linear Equations In One Variable is Part of Syllabus
Linear Equations in one variable are a frequent occurrence in almost all mathematics. It is a fundamental topic in Algebra and Quantitative Aptitude. In almost all government examinations, these two topics are an important section. In the following exams, Algebra is an important topic in the Quantitative Aptitude section.
We hope you found this article useful, informative and helpful in clearing all your doubts regarding linear equations in one variable and how to solve questions regarding the same.
If you are checking Linear Equations In One Variable article, also check the related maths articles: Analytic Geometry Differential Equations Factoring Quadratic Equations Pair Of Linear Equations Simultaneous Equations Algebraic Equation
Linear Equations in One Variable FAQs
Q.1 What is a linear equation in one variable?
Ans.1 Linear equation in one variable is an algebraic equation of equality that involves only one variable, the greatest exponent of which is 1. It is represented as ax + b = 0, where a is not equal to 0.
Q.2 What is the formula for linear equations in one variable?
Ans.2 The formula for linear equations in one variable is ax + b = 0, where a is not equal to 0. It is also known as the standard form of linear equation in one variable.
Q.3 Can a linear equation include more than one variable?
Ans.3 A linear equation can contain more than one variable. Such equations are known as linear equations in two variables or linear equations in three variables depending on the number of variables.
Q.4 Can I get more than one answer for a linear equation in one variable?
Ans.4 No, there is only one solution (or root) to a linear equation in one variable.
Q.5 What is the power of the variable in a linear equation in one variable?
Ans.5 In a one-variable linear equation, the power of the variable is one. |
# a. 2x = sqr root 12x+72 b. sqr root x+5 = 5 - sqr root x c. |2x| = -|x+6| d. 5 = |x+4|+|x-1| e. |x-2| = 4 -|x-3|Please Help!!!
giorgiana1976 | Student
a) Supposing that you want to solve:
2x = sqr root (12x+72)
First you have to check for what x values, the square roots exists. To check, you have to solve the inequality:
12x+72>=0
If you divide the inequality with 4, you'll have:
3x+18>=0
3x>=-18
x>=-18/3
x>=-6
That means that, solving the equation, you have to keep all x values which belong to the interval set by the condition of existence of the square root.
Now, let's solve it:
(2x)^2=(sqr root 12x+72)^2
4x^2=12x+72
4x^2-12x-72=0
x^2-3x-18=0
x1=[3+sqrt(9+72)]/2
x1=(3+9)/2
x1=6
x2=(3-9)/2
x2=-3
Though both solutions belongs to the interval, we have to verify them into the equation.
If we put x=-3 in equation, we'll have:
-6=sqrt(-36+72)
-6=sqrt36
-6=6, which is not true.
b) sqr root (x+5) = 5 - sqr root x
x+5 = 25-10sqrt x+x
20-10sqrt x=0
2-sqrt x=0
sqrt x = 2
x=4
c) |2x| = -|x+6|
In order to solve this equation, first let's see for what value of x, 2x>=0
|2x|=2x for x>=0
|2x|=-2x, for x<0.
|x+6|=x+6, for x>=-6
|x+6|=-x-6, for x<6
From these condition, occure 3 cases:
1) x belongs to (-inf., -6)
-2x=-(-x-6)
-2x=x+6
-3x=6
x=-2 which is not in the interval (-inf., -6).
2) x belongs to [-6,0)
-2x=-(x+6)
-2x+x=-6
-x=-6
x=6 which belongs to the interval [-6,0).
3) x belongs to [0,+inf.)
2x=-x-6
3x=-6
x=-2, which is not i the interval [0,+inf.).
The only solution of the equation is x=6.
d) 5= |x+4|+|x-1|
In order to solve this equation, first let's see for what value of x, |x+4|>0
|x+4|=x+4, for x>=-4
|x+4|=-x-4, for x<-4
Now, let's see, for what values of x,|x-1|>0.
|x-1|=x-1, for x>=1
|x-1|=-x+1, for x<1
From these condition, occure 3 cases:
1)x belongs to (-inf., -4)
5=-x-4-x+1
8=-2x
x=-4 which is not in the interval.
2) x belongs to [-4,1)
5=x+4-x+1
5=5, for any value from [-4,1).
3) x belongs to [1,+inf.)
5=x+4+x-1
2=2x
x=1, which belongs to [1, +inf.)
e) |x-2| = 4 -|x-3|
In order to solve this equation, first let's see for what value of x, |x-2|>0
|x-2|=x-2, for x>=2
|x-2|=-x+2, for x<2
Now, let's see, for what values of x,|x-3|>0.
|x-3|=x-3, for x>=3
|x-3|=-x+3, for x<3
From these condition, occure 3 cases:
1) x belongs to (-inf., 2)
-x+2=4+x-3
2x=1
x=1/2 which belongs to (-inf., 2).
2)x belongs to [2,3)
x-2=4+x-3
-2=1, is not true!
3) x belongs to [3, +inf.)
x-2=4-x+3
2x=9
x=4.5which belongs to [3, +inf.)
The solutions of the equation are:
x={0.5,4.5}
neela | Student
a)
2x = sqrt12x+72. Hope sqrt 12x means (sqrt12)x and not sqrt(12x). So,
2x-sqr12 x = 72. Or
(2-sqrt12)x = 72 Or
x = 72/(2-sqrt12) = 72(2+sqrt12)/(2-12) , after rationalising the denominator,
= 72(2+2sqrt3)/(-10)
=-7.2(2+2sqrt3)
b) sqrtx+5 = 5-sqrtx. Sqrtx+5 and sqrt(x+5) are different. Hope you do not mean sqrt(x+5) on LHS.
sqrtx=-sqrtx . Adding sqrtx to both sides,
2sqrtx = 0.
sqrtx = 0 . Or x = 0.
c)
|2x| = -|x+6|
Case (i) x<-6
|2x| = -(6-x) when x<-6 Or
-2x = 6-x. Or
-2x = 6. Or
-2x+x =-6.
-x = -6
So x = 6 and x<-6 which is a contradiction.
case(2) when 6< x <0
-2x = -(x+6). Or
-2x+x = -6. Or
-x = -6. Or x = 6 and x<0 is a contradiction.
Case (iii) x>0
2x = -(x+6) Or
2x+x = -6. Or
3x=-6 Or x =-2 and x>0 is a contradiction.
So there is no solution
d) 5 = |x+4|+|x-1|
When x>=1, 5 = x+4+x-1 Or 5 = 2x+3. Or x= (5-3)/2 = 1 is a contradiction as x>=1 and x=1. So x =1 is a solution.
When 0 = < x < 1, 5 = x+4 +(1-x). Or
5=5+2x . Or x = 0. Is a solution.
When -4 =< x<0, 5 = x+4+1-x, 5 =5 is an identity.So -4<=x< 0 is a solution.
When x <4, 5 = 4-x+(1-x) . Or 5 = 5-2x Or x=0 is a contradiction as x<-4 and x=5/2 are inconsistent.
So x=1 or (-4<x<=0) are the solution
e)
|x-2| = 4-|x-3|
case (i) x>=3
x-2 =4-(x-3). Or 2x = 4+3+2 or x =9/2 is consistent with x>=3. So x= 4.5 is a solution.
Case(ii) 2<x<3
x-2 = 4-(3-x) . Or x -2 = 3+x Or -2 = 3 is inconsistent result. So 2<x<3 cannot be a solution.
When x<2,
2-x =4-(3-x). Or
2-x=4+x-3. Or
2-4+3 = 2x Or
1 = 2x . Or x = 1/2 is consistent with x<2. So x=1/2 is a solution. |
# Perspectives on Proportions and Ratios - PowerPoint PPT Presentation
Perspectives on Proportions and Ratios
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Perspectives on Proportions and Ratios
## Perspectives on Proportions and Ratios
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##### Presentation Transcript
1. Perspectives on Proportions and Ratios Presented by Lyn Jones and Katreena Daniels
2. To get you thinking • Two students are measuring the height of the plants their class is growing. Plant A is 6 counters high. Plant B is 9 counters high. • When they measure the plants using paper clips they find that Plant A is 4 paper clips high. • What is the height of Plant B in paper clips ? With thanks to nzmaths
3. Consider… • Scott thinks Plant B is 7 paper clips high. • Wendy thinks Plant B is 6 paper clips high. • Who is correct? • What is the possible reasoning behind each of their answers? With thanks to nzmaths
4. Wendy is correct, Plant B is 6 paper clips high. Scott’s reasoning: To find Plant B’s height you add 3 to the height of Plant A: 4 + 3 = 7. Wendy’s reasoning: Plant B is one and a half times taller than Plant A: 4 x 1.5 = 6. The ratio of heights will remain constant. 6:9 is equivalent to 4:6. 3 counters are the same height as 2 paper clips. There are 3 lots of 3 counters in plant B, therefore 3 x 2 = 6 paper clips. With thanks to nzmaths
5. Use ratio tables to identify the multiplicative relationships between the numbers involved. With thanks to nzmaths
6. Use double-number lines to help visualise the relationships between the numbers. Tips on teaching with the dual number line: All the relationships are multiply or divide (not add or subtract). The same operations are done on both sides. The directions of the arrows are important. Acknowledge that there are several different correct ways to solve ratio problems. With thanks to nzmaths
7. Key Idea The key to proportional thinking is being able to see combinations of factors within numbers.
8. What’s the Difference? The ratio focuses on the relationship of the split of the whole into parts whilst the proportion shows the 'gap'- or 'space between' which needs to be maintained - almost like it provides the track or parallel lines the figures must travel along to maintain their relationship.
9. How full is the jug? How many different ways can you express the mathematical concepts depicted in this picture? Who is right?
10. How full is the jug? • How many different ways can you express the mathematical concepts depicted in this picture? • A fraction=1/2 • A percentage=50% • A decimal= .5 • A ratio? • Why/ Why not? • Who was right?
11. Key Ideas- Ratios A ratio is a way of writing how much of something there is compared to another thing. Ratios can compare a part to the whole.It can also be expressed as a fraction. Ratios can compare parts to parts. Part-to-whole and part-to-part ratios compare two quantities of the same thing. 50 % or .5 compares part to whole while 50:50 compares part-to-part where there are two quantities of the same thing Ratios can also be a rate, where a comparison is made between quantities of different things e.g. New minimum wage \$13.75 per hour. A proportion expresses the relationship between two ratios. A proportion is a statement of the equality between two rations.
12. Common Misconceptions – Do you see what I see? Teacher • Teaching fractions of a circle- cutting / shapes referring back to the whole. Give a part and ask to draw the whole exercises. • Teaching numerator and denominator. Key=equal parts and irregular fractions. Give a part and ask to draw the whole exercises. • Ordering fractions- smallest to biggest and vise versa. • Always presenting equal parts as models – PAT mathsgliches • Reading, Reading, Reading- FIO-Unpack the reading. Rewrite the problem into user friendly words. Use materials. • A percentage is expressed as ?/100 – use of variety of visuals and materials. Student perspective: • Fractions are part of a circle. Stumped when drawing equal parts of rectangles, squares, diamonds, many other shapes • How may parts are coloured in. Stumped with irregular fractions. • Two is always the biggest number. • I remember this … it goes backwards • I just have to use visual clues to sort this out. • I am not a Level 26 reader yet? • Its out of a 100
13. Common Misconceptions – Do you see what I see? Teacher • Decimals add up to ten. Use of Cuisenaire rods to measure and portray decimals. • Proportions taught with a bias to percentages only. Cards to show the relationship between fractions, percentages and decimals. Student therefore understands the value of quantity of the part changes not the proportion of the whole and the inter relationships between the expressions of the proportions i.e. ½= 50%= .5 of the whole. Student perspective: • Its out of a 10. Stumped when adding and subtracting decimals. • Always expressed as percentages.
14. Progressions Reproduced thanks to Charlotte Wilkinson
15. Level One Unpacking & Attacking • Provide materials (any level) • Remind there are multiple possible answers and strategies to solve the problem • Encourage looking for patterns Possible Challenges • Colour perceptions (colour blindness) • Reading abilities • Reluctance to look for more than one solution • Once a colour has been used as a fraction e.g. half, then a block to seeing it as a third of a larger whole (different colour) With thanks Charlotte Wilkinson, Wilkie Way
16. Level Two Unpacking & Attacking • Recap on finding ½ and ¼ of sets mentally • Provide materials (any level), e.g. complete a hexagon using: poi = trapezium (1/2), rhombus = rakau (1/3), triangle = waiata (1/6) • Check understanding of relationship between number of students and materials mentioned in problem Possible Challenges • Reading abilities & language/experience barriers • Remind there are multiple possible strategies to solve the problem • Confusion or missing relationship between materials and number of students • Balking at larger numbers and not seeing relationship between them. 60 being double 30 or breaking down 204 into 60+60+60+24. With thanks Figure It Out, Number Sense & Algebraic Thinking, Book 1 L2-3
17. Level Three Unpacking & Attacking • Model/suggest double number lines, ratio tables • Order of problem helps guide towards finding solution • Need to search for common factors and equivalent fractions • . Possible Challenges • Reading abilities • Knowledge of time as a measure • Requires some knowledge of inverse relationships • Understanding of equivalent fractions • Knowledge of multiplication factors, e.g. looking for common factor of both 30 and 45 (easier to work with 3 rather than 5min intervals to simplify fraction work) With thanks Figure It Out, Proportional Reasoning, Book 2 L3-4
18. Level Four • The results: • Ratio: 5:3 • Fractions: 5/8 and 3/8 • Rachel: \$500 ÷8= \$62.50, \$62.5x5=\$312.50 • Hemi: \$500÷8=\$62.50, \$62.50x3=\$187.50 • Check: \$187.50 +\$312.50 =\$500 ✓ Paul Mounsey Year 10 Proportions and Ratios -HGHS
19. NZ Maths on Ratios and Proportions Big Ideas: Ratios allow us to compare the relative sizes of two quantities. Background points for teaching An understanding of ratios involves understanding the following: Ratios can compare a part to the whole. An example of a part to a whole ratio is the number of females in a class to the number of students in the class. If there are 8 females in the class of 20 students the ratio of girls to students can be expressed 8:20 (females to students). Because this ratio is relating a part to a whole it can also be expressed as a fraction (8/20) or as a percentage (40%). Ratios can compare parts to parts. An example of a part to a part ratio is where the number of females in a class is compared to the number of males. If there are 8 females in a class of 20 the ratio of females to males is 8:12. It is important when using ratios to clearly state what the comparison is made in relation to. One of the most common uses of part-to-part ratios are odds. The odds of an event happening is a ratio of the number of ways an event can happen to the number of ways it cannot happen. Ratios can also be a rate. Part-to-whole and part-to-part ratios compare two quantities of the same thing. Rates on the other hand are examples of ratios where a comparison is made between quantities of different things. In rates the measuring units are different for the quantities being compared and the rate is expressed as one quantity per the other quantity. For example the value of food can be expressed as price per kilogram, fuel efficiency can be expressed as litres per 100 km. A proportion expresses the relationship between two ratios. A proportion is a statement of the equality between two rations. For example if it takes 10 balls of wool to make 15 beanies, 6 beanies will take 4 balls of wool. In this example the ratio of 2:3 (balls to beanies) can be applied to each situation. Solving proportional problems involves applying a known ratio to situations that are proportionally related and finding one of the measures when the other is given. For example, in the beanie situation the ratio of 2:3 (balls to beanies) can be applied to the problem where you want to find out how many balls of wool are needed to make 33 beanies. 2:3 = ?:33, ? = 22
20. Where to from here? Where to go ...when we don’t know. • Book 7 The Numeracy Professional Development Projects documents • www.nzmaths.co.nz • http://www.ncwilkinsons.com/wilkieway/pages/educational_resources.php • http://www.primaryresources.co.uk/maths/mathsB7.htm • http://www.oxfordowl.co.uk/maths/any-questions/ |
# Cayley’s Formula – How many trees can you make from n labelled vertices?
Tee Jet Whaw
Suppose you have 2 vertices (points) – 1 red and 1 blue – how many ways can you join them together? There’s only way to do it, which is to draw a line between the 2 points.
Now, what if you have 3 vertices – 1 red, 1 blue and 1 green – how many ways can you join these all together?
Let’s start by drawing 2 lines as shown below:
Here we see that red and blue are joined, and red and green are joined. So, we can move the points around and create a different graph – for example joining red and green, and green and blue to get a different tree:
There is one final way to arrange the graph, that is blue joining to green and blue joining to red, which gives a total of 3 different trees that can be made from 3 labelled (meaning differently coloured) vertices.
For 4 and beyond, things get a little trickier. Have a go for yourself and see how many trees you can create using 4 different vertices (red, blue, green and yellow). And if you’re up for a real challenge, try to come up with a formula for the number of tress that can be made from n labelled vertices.
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After a little bit of thought, you will hopefully see that there are in fact 16 different trees – 12 which have the points joined in a loop (in different orders as with the case of 3 points), and a further 4 that can be formed by using a central point (4 choices as they are 4 different colours).
For the general case of n vertices, there are a total of nn-2 different trees that can be made. You can check this result for the above cases with n = 2, nn-2 = 20 = 1. For n = 3, nn-2 = 31 = 3 and for n = 4, nn-2 = 42 = 16.
There are many ways to prove the general result – known as Cayley’s Formula – and here is one example using double counting (don’t worry if you can’t follow all of the steps, this is quite advanced!):
To start, suppose there are F(n) undirected trees (this is what we want to know). We can then produce directed trees with numbered edges (meaning the order in which you traverse the graph is given) as follows:
Step 1: For each undirected tree, choose a root (starting point) so that the tree automatically becomes directed. There are n choices for this root (since we have n points).
Step 2: Number the (n-1) edges (if there are n vertices which are all connected as a tree then we know there are (n-1) of them – you can check this with our examples above for small n). There are (n-1)! ways to do this numbering.
Therefore, in total we have F(n) x n x (n-1)! = F(n) x n! directed trees with numbered edges that can be constructed out of n points.
Meanwhile, we can also produce a directed tree with numbered edges by adding edges to n separate vertices:
1. Note that if we’ve added k edges to the empty graph, we’re left with (n-k) trees since adding an edge decreases the number of trees by 1 (as two of the original number must now be joined).
2. When we add an edge each time, we can choose any of the points to be our starting point, and the root of any other tree to be our end point. This gives n starting points and (n-k-1) end points to choose from each time.
Overall, cycling through the values of k from 0 to n-1 we therefore have (n(n-1)) x … x (n(1)) = nn-1 x (n-1)! = n(n-2) x n! directed trees with numbered edges.
Since both methods are counting the same thing (the number of directed tress with numbered edges) we can equate the formulas to get F(n) x n! = nn-2 x n! and so F(n) = n(n-2). |
# Math Snap
## The formula to compute the direct labor cost per employee is: Direct labor hours times total direct labor cost Direct labor hours times direct labor cost per hour Total labor hours times labor cost per hour Total labor hours times direct labor co'st per hour
#### STEP 1
Assumptions 1. We need to find the formula to compute the direct labor cost per employee. 2. The direct labor cost per employee is typically calculated based on the number of hours worked by the employee and the cost per hour of labor.
#### STEP 2
To find the direct labor cost per employee, we need to identify the correct formula from the given options.
#### STEP 3
Let's analyze each option one by one to determine which one correctly calculates the direct labor cost per employee.
#### STEP 4
Option 1: Direct labor hours times total direct labor cost This option suggests multiplying the direct labor hours by the total direct labor cost. However, this would not give us the cost per employee, as it does not take into account the cost per hour of labor.
#### STEP 5
Option 2: Direct labor hours times direct labor cost per hour This option suggests multiplying the direct labor hours by the direct labor cost per hour. This correctly calculates the labor cost for the hours worked by an individual employee.
#### STEP 6
Option 3: Total labor hours times labor cost per hour This option suggests multiplying the total labor hours (for all employees) by the labor cost per hour. This would give the total labor cost for all employees, not the cost per employee.
#### STEP 7
Option 4: Total labor hours times direct labor cost per hour This option is similar to Option 3 and would also give the total labor cost for all employees, not the cost per employee.
#### STEP 8
Based on the analysis, the correct formula to compute the direct labor cost per employee is: $\text{Direct labor hours} \times \text{Direct labor cost per hour}$ This corresponds to Option 2.
##### SOLUTION
Therefore, the correct answer is: Option 2: Direct labor hours times direct labor cost per hour |
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