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# 4th Class Mathematics Factors and Multiples Multiples and Factors
Multiples and Factors
Category : 4th Class
LEARNING OBJECTIVES
• understand the concept of factors.
• understand the concept of multiples.
• use factor tree to find the factors of a number.
• to find prime numbers, factors and multiples of given number.
• to apply factors and multiples to real life situations.
Amazing Fact
A number’s composite factors are found by multiplying 2 or more prime factors.
For example: The composite factors of $18\left( 2\times 3\times 3 \right)$are$6\left( 2\times 3 \right)$and$9\left( 3\times 3 \right)$
Real Life Example
Money can use the concept of factors. One can exchange a 100 – Rupee by two 50 – Rupee notes (factors 2 and 50) or five 20 – Rupee note (factors 5 and 20).
QUICK CONCEPT REVIEW
Factors are numbers that multiplies to get another number.
For example: 4 and 7 are multiplied to get 28, then 4 and 7 are factors of 28.
Multiples are product obtained by multiplying one number by another.
For example: 8 and 11 are multiplied to get 88, then 88 is a multiple of 8 and 11.
The factors (or multiples) that are common between 2 or more numbers are called common factors (or multiples) of given numbers.
PROPERTIES OF FACTORS AND MULTIPLIES
• 1 is a factor of every number.
• Every number is a factor of itself.
• Every factor of a number is an exact divisor of that number.
• Every factor of a number is less than or equal to the number.
• Factors of a given number are finite.
• Prime numbers have only 2 factors: 1 and the number itself.
• Every number is a multiple of itself.
• Every multiple of a number is greater than or equal to that number.
• The number of multiples of a given number is unlimited.
Play Time
(1) Make two teams.
Ask the first team to pick up a number between 1 and 50. Then ask them to call out a factor of that number. Ask the second team to call out a factor or multiple of the called out number. Continue this process till all the factors and multiples are said. One who cannot give the factor or multiple will be out.
Misconcept/Concept
Misconcept: Student might confuse between the concept of factors and multiples.
Concept: Explain factors come from dividing and multiples come from multiplying.
Misconcept: 1 is a prime number.
Concept: 1 is not a prime number even 1 is neither composite nor prime.
#### Other Topics
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# Factors of 14: Factor tree, Division Method, Prime Factorization
In mathematics, factors of a number are the whole number ( take any number) that divides the given number evenly (into that number) with no remainder. In the case of 14, the factors would be the numbers that can be divided by 14 without leaving a remainder. Factors of 14 are important in many mathematical calculations like simplifying fractions, finding common denominators and solving equations. Read further to learn more about the factors of 14, by division method, prime factorization, multiples, etc
## What are the Factors of 14?
The factors of 14 are all positive integers that divide evenly into 14 without leaving a remainder, or they can be multiplied by two pairs of positive integers to get 14. There are four factors of 14: 1,2,7 and 14 itself.
Let us understand in detail –
• 1 divides into 14: 14 / 1 = 14 (with no remainder)
• 2 divides into 14: 14 / 2 = 7 (with no remainder)
• 7 divides into 14: 14 / 7 = 2 (with no remainder)
• 14 divides into 14: 14 / 14 = 1 (with no remainder)
Note – It’s important to note that every number is a factor of itself.
## What is the Factor Tree for 14?
A factor tree is a visual representation of how a number can be broken down into its prime factors (factors that cannot be further divided into whole numbers).
1. We know from previous discussions that 14 can be divided by 2 and 7 (without a remainder).
2. Since 2 is a prime number (itself and 1 is its only factor), we can leave it as is on the branch. However, 7 is also a prime number.
3. Since 7 is prime, there are no further factors to break it down into. So, the branch with 7 will simply end there.
14
/ \
2 7
/ \
7 1
## Factors of 14 by Division Method
The division method is one of the simplest methods to find the factors of any number. In this method, we divide the number by smaller numbers starting from 1 and moving upwards to bigger numbers. If the division is exact, then the divisor is a factor. If we calculate it further, we find –
• 14 ÷ 1 = 14 [Whole number]
• 14 ÷ 2 = 7 [Whole number]
• 14 ÷ 3 = 4.67 [Remainder in fraction]
• 14 ÷ 4 = 3.5 [Remainder in fraction]
• 14 ÷ 5 = 2.8 [Remainder in fraction]
• 14 ÷ 6 = 2.33.. [Remainder in fraction]
• 14 ÷ 7 = 2 [Whole number]
And if we divide 14 by the numbers 8, 9, 10 and further we will always get a remainder in fraction.
For example :
• 14 ÷ 8 = 1.75 [Remainder in fraction]
• 14 ÷ 9 =1.556 [Remainder in fraction]
• 14 ÷ 10 = 1.4 [Remainder in fraction]
• 14 ÷ 11 = 1.273 [Remainder in fraction]
• 14 ÷ 12 = 1.167 [Remainder in fraction]
• 14 ÷ 13 = 1.077 [Remainder in fraction]
• 14 ÷ 14 = 1 [Remainder in fraction]
• 14 ÷ 15 =0.933 [Remainder in fraction]
Thus, the factors of 14 are 1, 2, 7, and 14.
## Factors of 14 by Multiplication Method
We have already seen the division method, let us know the factors of 14 through the Multiplication Method –
• 1 × 14 = 14
• 2 × 7 = 14
• 7 × 2 = 14
Hence proved by the multiplication method.
The factors are 1,2,7 and 14.
## Factors of 14 by Prime Factorization
Prime factors are whole numbers greater than 1 that cannot be further broken down into whole numbers without a remainder. To find the prime factorization of 14, we can see the given steps –
Step 1: When we divide 14 by the smallest prime factor, which is 2, we get – 14/2 = 7
Step 2: Further, when we divide 7 from the next prime factor that is itself 7. (Now, 2 and 3 are smaller than 7), we get – 7/7 = 1
Step 3: As we know, 1 cannot be divided by any other prime factor.
Thus, what we get is –
Here, 2 and 7 are both prime numbers (numbers greater than 1 that can only be divided by 1 and themselves). So, 14 is built from these two prime factors.
Connecting to Factors:
While we do not do prime factorization of factors themselves, we can see how these prime factors are related to their actual factors:
• 1 (obtained by multiplying 1 x 1) – any number is a factor of itself by definition.
• 2 (obtained by multiplying 2 x 1) – as identified earlier.
• 7 (obtained by multiplying 1 x 7) – the other prime factor.
• 14 (obtained by multiplying 2 x 7) – the product of both prime factors.
## Factor Pairs of 14
Factor pairs are like special partnerships among factors. Each pair, when multiplied, results in the original number (14). Here, let us take 14, now there are two types of factor pairs –
### Positive Factor Pair of 14
After multiplying two positive integers, the resulting value is known as the product, which is expressed in terms of a factor. Now let us pair the positive integers to get the number 14.
• 1 × 14 = 14
• 2 × 7 = 14
• 7 × 2 = 14
• 14 × 1 = 14
### Negative Factor Pair of 14
After multiplying two negative integers, the resulting value is known as the product, which is expressed in terms of a factor. Now let us pair the negative integers to get the number 14.
• -1 × – 14 = 14
• -2 × -7 = 14
• -7 × -2 = 14
• -14 × -1 = 14
Thus, In the case of 14, we have two-factor pairs: (1, 14) and (2, 7).
Related Blogs
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# 2011 AMC 12B Problems/Problem 21
## Problem
The arithmetic mean of two distinct positive integers $x$ and $y$ is a two-digit integer. The geometric mean of $x$ and $y$ is obtained by reversing the digits of the arithmetic mean. What is $|x - y|$?
$\textbf{(A)}\ 24 \qquad \textbf{(B)}\ 48 \qquad \textbf{(C)}\ 54 \qquad \textbf{(D)}\ 66 \qquad \textbf{(E)}\ 70$
## Solution 1
$\frac{x + y}{2} = 10 a+b$ for some $1\le a\le 9$, $0\le b\le 9$.
$\sqrt{xy} = 10 b+a$
Squaring the first and second equations, $\frac{x^2 + 2xy + y^2}{4}=100 a^2 + 20 ab + b^2$
$xy = 100b^2 + 20ab + a^2$
Subtracting the previous two equations, $\frac{x^2 + 2xy + y^2}{4} - xy = \frac{x^2 - 2xy + y^2}{4} = \left(\frac{x-y}{2}\right)^2 = 99 a^2 - 99 b^2 = 99(a^2 - b^2)$
$|x-y| = 2\sqrt{99(a^2 - b^2)}=6\sqrt{11(a^2 - b^2)}$
Note that for x-y to be an integer, $(a^2 - b^2)$ has to be $11n$ for some perfect square $n$. Since $a$ is at most $9$, $n = 1$ or $4$
If $n = 1$, $|x-y| = 66$, if $n = 4$, $|x-y| = 132$. In AMC, we are done. Otherwise, we need to show that $n=4$, or $a^2 -b^2 = 44$ is impossible.
$(a-b)(a+b) = 44$ -> $a-b = 1$, or $2$ or $4$ and $a+b = 44$, $22$, $11$ respectively. And since $a+b \le 18$, $a+b = 11$, $a-b = 4$, but there is no integer solution for $a$, $b$.
### Short Cut
We can arrive at $|x-y| = 6\sqrt{11(a^2 - b^2)}$ using the method above. Because we know that $|x-y|$ is an integer, it must be a multiple of 6 and 11. Hence the answer is $66.$
In addition: Note that $11n$ with $n = 1$ may be obtained with $a = 6$ and $b = 5$ as $a^2 - b^2 = 36 - 25 = 11$.
## Sidenote
It is easy to see that $(a,b)=(6,5)$ is the only solution. This yields $(x,y)=(98,32)$. Their arithmetic mean is $65$ and their geometric mean is $56$.
## Solution 2
Let $(x+y)/2 = 10a + b$ and $\sqrt{xy} = 10b + a$. By AM-GM we know that $a \ge b$. Squaring and multiplying by 4 on the first equation we get $x^2 + y^2 + 2xy = 400a^2 + 4b^2 + 80ab$. Squaring and multiplying the second equation by 4 we get $4xy = 400b^2 + 4a^2 + 80ab$. Subtracting we get $(x-y)^2 = 396(a^2 - b^2)$. Note that $396 = 2^2 \cdot 3^2 \cdot 11$. So to make it a perfect square $a^2 - b^2 = 11$. From difference of squares, we see that $a = 6$ and $b = 5$. So the answer is $3 \cdot 2 \cdot 11 = 66$. ~coolmath_2018
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# The Fraction Wars: Teaching Cross-Multiplication the Fun Way
Today, let's delve into the epic saga of fractions, where numerators and denominators battle it out for mathematical supremacy. Yep, you guessed it – it's the Fraction Wars. But fear not, for in this blog post, we will equip you with the ultimate weapon: cross-multiplication, the hero of fraction arithmetic.
## What's the Big Deal with Cross-Multiplication?
Now, before we embark on this adventure, let's get one thing straight: fractions can be tricky little buggers, especially when it comes to multiplication. But cross-multiplication is here to save the day.
## The Basics of Cross-Multiplication
So, what exactly is cross-multiplication? It's a nifty little trick that helps us easily compare fractions.
Imagine you're trying to figure out which of two fractions is bigger. Instead of scratching your head and staring blankly at the numbers, you can simply cross-multiply and compare the results.
Example:
Compare 3/4 and 2/3.
Cross multiply 3 x 3 = 9 and 4 x 2 = 8.
3/4 is the larger fraction.
It's like sending your fraction warriors into battle armed with the most powerful weapon in the mathematical arsenal.
## The Power of Visualization
Now, let's talk about the importance of visualization when teaching cross-multiplication. Fractions can be abstract concepts for young minds, so we must use visual aids that bring them to life.
Think of fractions as delicious pizza or cake slices – each slice represents a fraction of the whole. When students can visualize fractions in real-world contexts, cross-multiplication becomes easier.
## The Battle Plan: Teaching Cross-Multiplication
Now, let's talk strategy. How can we teach cross-multiplication in a fun way for our students? Let's start by ditching the boring worksheets and embracing hands-on activities.
### Activity 1: Fraction Artillery
1. Create fraction cards with various fractions. If you plan to work on fractions often, laminate these fraction cards for future use.
2. Draw a target board with fraction values labeled on concentric circles. (e.g., 1/2, 1/3, 2/3, etc.)
3. Form small teams of 3-4 students each.
4. Each student will throw a fraction card at the target board to obtain their second fraction.
5. The student will then cross-multiply their fraction card with the fraction they hit on the target board.
1. Example: Card: 1/2 Target Hit: 2/3. Cross-multiply the two fractions. 2 x 2= 4 and 1 x 3 = 3. 1/2 is the larger fraction.)
6. Their teammates will also cross-multiply to help check their classmates' work.
7. Once the team agrees on an answer, the next teammate takes a turn.
8. The team will continue the activity until they finish all their assigned cards or the lesson ends.
### Activity 2: Fraction Relay Race
1. Divide the class into teams of 3-4 students each.
2. Set up a starting line and a finish line at opposite ends of the classroom.
3. Prepare a whiteboard or chart paper with several fraction equations requiring cross-multiplication.
4. Each team lines up behind the starting line.
5. When you say, "Go!" the first player from each team races to the whiteboard or chart paper.
6. They must solve the fraction equation written on the board using cross-multiplication.
7. Once they have found the correct answer, they return to their team and tag the next player in line.
1. If the student does not reach the right answer, they must keep trying until they reach the correct answer. Allow other teammates to help their peers.
8. The next player then races to the board to solve the next equation.
9. The race continues until all players from one team have successfully solved their equations and crossed the finish line.
10. The first team to complete all the equations correctly and have all members cross the finish line wins the race. Provide the winning team a homework pass, an extra 3 minutes of recess, etc.
## Making Learning Fun with Games and Challenges
Who says learning has to be dull and dreary? Inject some excitement into your fraction lessons with these other games and challenges:
• Create fraction bingo cards where students must solve equations using cross-multiplication to mark off fractions on their boards.
• Play fraction war, where students flip over their top cards and cross-multiply to determine which fraction is greater. The player with the lesser fraction takes both cards and adds them to their stack. The player to get rid of their cards first wins the game.
## Conquering the Fraction Wars
In the fractions battle, cross-multiplication emerges as the ultimate weapon against mathematical mayhem. By teaching cross-multiplication, we empower our students to tackle fraction problems with confidence and ease.
So go forth, brave educators, and may the fraction force be with you.
Written by Brooke Lektorich
Education World Contributor
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# Field
### set laws elements operations
A field is the name given to a pair of numbers and a set of operations which together satisfy several specific laws. A familiar example of a field is the set of rational numbers and the operations addition and multiplication. An example of a set of numbers that is not a field is the set of integers. It is an "integral domain." It is not a field because it lacks multiplicative inverses. Without multiplicative inverses, division may be impossible.
The elements of a field obey the following laws:
1. Closure laws: a + b and ab are unique elements in the field.
2. Commutative laws: a + b = b + a and ab = ba.
3. Associative laws: a + (b + c) = (a + b) + c and a(bc) = (ab)c.
4. Identity laws: there exist elements 0 and 1 such that a + 0 = a and a × 1 = a.
5. Inverse laws: for every a there exists an element - a such that a + (-a) = 0, and for every a ≠ 0 there exists an element a-1 such that a × a-1 = 1.
6. Distributive law: a(b + c) = ab + ac.
Rational numbers (which are numbers that can be expressed as the ratio a/b of an integer a and a natural number b) obey all these laws. They obey closure because the rules for adding and multiplying fractions, a/b + c/d = (ad + cb)/bd and (a/b)(c/d) = (ac)/(bd), convert these operations into adding and multiplying integers which are closed. They are commutative and associative because integers are commutative and associative. The ratio 0/1 is an additive identity, and the ratio 1/1 is a multiplicative identity. The ratios a/b and -a/b are additive inverses, and a/b and b/a (a, b ≠ 0) are multiplicative inverses. The rules for adding and multiplying fractions, together with the distributive law for integers, make the distributive law hold for rational numbers as well. Because the rational numbers obey all the laws, they form a field.
The rational numbers constitute the most widely used field, but there are others. The set of real numbers is a field. The set of complex numbers (numbers of the form a + bi, where a and b are real numbers, and i2 = -1) is also a field.
Although all the fields named above have an infinite number of elements in them, a set with only a finite number of elements can, under the right circumstances, be a field. For example, the set constitutes a field when addition and multiplication are defined by these tables:
With such a small number of elements, one can check that all the laws are obeyed by simply running down all the possibilities. For instance, the symmetry of the tables show that the commutative laws are obeyed. Verifying associativity and distributivity is a little tedious, but it can be done. The identity laws can be verified by looking at the tables. Where things become interesting is in finding inverses, since the addition table has no negative elements in it, and the multiplication table, no fractions. Two additive inverses have to add up to 0. According to the addition table 1 + 1 is 0; so 1, curiously, is its own additive inverse. The multiplication table is less remarkable. Zero never has a multiplicative inverse, and even in ordinary arithmetic, 1 is its own multiplicative inverse, as it is here.
This example is not as outlandish as one might think. If one replaces 0 with "even" and 1 with "odd," the resulting tables are the familiar parity tables for catching mistakes in arithmetic.
One interesting situation arises where an algebraic number such as √ 2 is used. (An algebraic number is one which is the root of a polynomial equation.) If one creates the set of numbers of the form a + b √ 2 , where a and b are rational, this set constitutes a field. Every sum, product, difference, or quotient (except, of course, (a + b √ 2)/0) can be expressed as a number in that form. In fact, when one learns to rationalize the denominator in an expression such as 1/(1 - √ 2 ) that is what is going on. The set of such elements therefore form another field which is called an "algebraic extension" of the original field.
J. Paul Moulton
## Resources
### Books
Birkhoff, Garrett, and Saunders MacLane. A Survey of Modern Algebra. New York: Macmillan Co., 1947.
McCoy, Neal H. Rings and Ideals. Washington, DC: The Mathematical Association of America, 1948.
Singh, Jagjit, Great Ideas of Modern Mathematics. New York: Dover Publications, 1959.
Stein, Sherman K. Mathematics, the Man-Made Universe. San Francisco: W. H. Freeman, 1969.
## KEY TERMS
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Field
—A set of numbers and operations exemplified by the rational numbers and the operations of addition, subtraction, multiplication, and division.
Integral domain
—A set of numbers and operations exemplified by the integers and the operations addition, subtraction, and multiplication.
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# Inverse Function Calculator
Instructions: Use this calculator to find the inverse function for a function you provide, showing all the steps. Please type in the function expression you want to find the inverse for in box below.
Enter the function you need to find the inverse (Ex: f(x)=(2x + 1)/(x-1), etc.)
## More on this Inverse Function Calculator
This calculator will allow you to find the inverse of a given function showing all the steps, assuming that the inverse exists. The calculator will examine the function solve an equation associated to the definition of the function, and it will try to assess whether or not an inverse exist.
For example, you can provide a linear function like 'f(x) = 3x - 2', which would be a simple case, or for example you could take it up a notch with something a bit harder, like with a rational function like 'y = (x-1)/(x-3)'.
Once you provide a valid function, please click on "Calculate" button to get all the steps of the process shown to you, with the inverse function as the final answer, if an inverse exist, or with the explanation that no solution could be found and why.
It is not guaranteed that you will find all inverse functions. For one, not all functions have an inverse, and second (as we will see in the next section), the process of finding the inverse involves solving for x for an equation, and as we know, some equations can be very hard or impossible to solve.
So then, simpler functions are more likely to be amenable for finding their inverse, in case the inverse exists.
## How do you define the inverse of a function?
In layman's terms, the inverse of a function is the function that does the opposite as what the original function does. So, think of a function in terms of y = f(x), and then you could think of it as you go from x to y. You feed the function with an x, and the function gives you an specific y.
The inverse function starts with the y, and finds the way back to x, in a way that the x is the same that led to y through the original function. Now, the formal definition is done via function composition. For a function $$f$$, we say that $$g$$ is the inverse function of $$f$$ if
$f(g(x)) = x$
and
$g(f(x)) = x$
for all x in a certain set. There is more to it, but we will leave it at the intuitive level (Strictly speaking, a function needs to be injective and surjective in order to be invertible, and some other technicalities that are considering, like restricting the domain and range, etc.)
Usually, we call $$f^{-1}$$ to the inverse of $$f$$, so the formula that defines the inverse is typically written as:
$f(f^{-1}(x)) = x$
## What are the steps to find the inverse function
• Step 2: You then use algebraic manipulation to solve for x. Depending on how complex f(x) is you may find easier or harder to solve for x.
• Step 3: In some circumstances you will simply not be able to solve for x, for complex non-linear functions f(x)
• Step 4: If you are able to solve for x, then you should be able to write x = g(y)
• Step 5: You need to assess whether the solution found is unique. This is, could solve for x uniquely. In other words, did you find only one solution when solving for x? If yes, then you have an inverse function, otherwise, there is no inverse
• Step 6: If you found the inverse by solving x = g(y), you simply change the name of the variable and write f-1(x) = g(x), which makes the emphasis in that g(x) IS the actual inverse
If you were to use Calculus and derivatives (but notice that you DON'T need derivatives to compute the inverse), you could find the derivative of the function, and make sure that the derivative is always positive or negative, to ensure the function is injective, and hence, invertible.
But usually, the methodology of solving for x is a lot more palatable for basic Algebra students.
## The rule for finding inverse functions
The are actually no other rules to compute the inverse function other than starting with y = f(x) and then solving for x. A rule like that sounds pretty broad, because it is. More than a rule, it is a generic methodology for getting started in the process.
Ultimately calculating the inverse will depend on your success of solving an equation, and making sure that solution is unique. It does help to assess the graph of the function beforehand, so to not to look for an inverse when there is clearly none.
What to look in a graph? A function needs to be monotone (increasing or decreasing) on a certain subdomain in order to be invertible. With that being said, we could conveniently restrict the domain of a function to a smaller subdomain to find the inverse in a smaller set, that is always a possibility.
## How do we know for sure that the function has an inverse?
Formally, the only way of making sure a function has an inverse, you need to ensure the function is injective (1-to-1). This is assessed either by computing its derivative (if it exists) and making sure it is either positive and negative everywhere, or by manually ensuring that when we start with y = f(x) and we solve for x, we always get a unique solution.
This can be also seen graphically, using the horizontal line test: You draw an arbitrary horizontal line, and the function f(x) passes the horizontal line test if any horizontal line drawn crosses the graph of the function at most once.
### Example: Finding the inverse function
Find the inverse of the following function: $$f(x) = \displaystyle \frac{1}{3} x + \frac{5}{4}$$
Solution:
We have the following function:
$f(x) = \frac{1}{3} x + \frac{5}{4}$
Then, in order to find the inverse of the given function, we need to solve for $$x$$ and determine whether there is a solution or not. The starting equation is:
$y = \displaystyle \frac{1}{3}x+\frac{5}{4}$
Step 0: In this case, we first need to simplify the given linear equation, and in order to do so, we conduct the following simplification steps:
$$\displaystyle y-\left(\frac{1}{3}x+\frac{5}{4}\right) = 0$$
The multiplication by -1 gets distributed as follows: $$-\left(\frac{5}{4}+\frac{1}{3}x\right) = -\frac{5}{4}-\frac{1}{3}x$$
$$\displaystyle \Rightarrow \,\, \,\,$$
$$\displaystyle y-\frac{5}{4}-\frac{1}{3}x = 0$$
### Solving the Linear Equation
Putting $$x$$ on the left hand side and $$y$$ and the constant on the right hand side we get
$\displaystyle -\frac{1}{3}x = -y -\left(-\frac{5}{4}\right)$
Now, solving for $$x$$, by dividing both sides of the equation by $$-\frac{1}{3}$$, the following is obtained
$\displaystyle x=-\frac{1}{-\frac{1}{3}}y+\frac{\frac{5}{4}}{-\frac{1}{3}}$
and simplifying we finally get the following
$\displaystyle x=3y-\frac{15}{4}$
Therefore, the solving for $$y$$ for given linear equation leads to $$x=3y-\frac{15}{4}$$.
Therefore, and since when solving for $$x$$ we find a solution and it is only one solution, we have found the inverse.
### The Inverse Function
Based on the work shown above, it can be concluded that the inverse function is:
$f^{-1}(x) = 3x-\frac{15}{4}$
The inverse function can be depicted graphically as follows:
### Example: More inverse examples
Compute the inverse function of: $$y = \frac{x-1}{x+3}$$
Solution:
In order to find the inverse of the given function, we solve for $$x$$ and determine whether there is a solution or not. The starting equation is:
$y=\frac{x-1}{x+3}$
The following is obtained:
$$\displaystyle y=\frac{x-1}{x+3}$$
Putting all the terms of the equation on one side
$$\displaystyle \,\,$$
$$\displaystyle \Rightarrow \,\,\,\,y-\frac{x}{x+3}+\frac{1}{x+3}=0$$
We reorganize the terms so to get a rational equation structure
$$\displaystyle \,\,$$
$$\displaystyle \Rightarrow \,\,\,\,\frac{\left(x+3\right)y-x+1}{x+3}=0$$
By simpliflying we find
$$\displaystyle \,\,$$
$$\displaystyle \Rightarrow \,\,\,\,\frac{xy-x+3y+1}{x+3}=0$$
### Auxiliary Numerator Equation
We need to set the numerator equal to zero and find the solutions. Then, those roots that do not make the denominator equal to zero will be solutions to the rational equation
$$\displaystyle xy-x+3y+1=0$$
Keeping the term with $$x$$ on one side and the rest of terms on the other side
$$\displaystyle \,\,$$
$$\displaystyle \Rightarrow \,\,\,\,\left(y-1\right)x = -3y-1$$
Dividing the equation by $$y-1$$
$$\displaystyle \,\,$$
$$\displaystyle \Rightarrow \,\,\,\,x = -\frac{3y}{y-1}-\frac{1}{y-1}$$
Using algebraic manipulation the above polynomial equation, the following is obtained:
$x = -\frac{3y+1}{y-1}$
### Auxiliary Denominator Equation
We find the roots of the denominator: $$x+3=0$$
Therefore, the solving for $$x$$ for given linear equation leads to $$x=-3$$.
### Putting Together the Solutions of the Rational Equation
Then, by checking that we don't have a zero of the denominator, we find the following solution set to the equation $$\displaystyle y=\frac{x-1}{x+3}$$ is
$x = -\frac{3y+1}{y-1}$
Since when we solve for $$x$$ we find one and only one solution, we conclude that we have an inverse function.
### Finding the Inverse Function
Based on the work shown above, it can be concluded that the inverse function is:
$f^{-1}(x) = -\frac{3x+1}{x-1}$
The inverse function found can be depicted graphically as follows:
### Example: Not all functions have inverse
Does the following function have an inverse: $$y = \displaystyle \frac{1}{3} x^2 - \frac{2}{5}$$ ?
Solution: Observe that
$y =\displaystyle \frac{1}{3} x^2 - \frac{2}{5}$ $\displaystyle \Rightarrow y + \frac{2}{5} = \frac{1}{3} x^2$ $\displaystyle\Rightarrow x^2 = 3\left(y + \frac{2}{5} \right)$ $\displaystyle \Rightarrow x = \pm \sqrt{ 3\left(y + \frac{2}{5} \right) }$
which indicates that there are two solutions, and then, there is no inverse in this case.
## More function calculators
Functions are a crucial object in Math, especially in Calculus and Algebra, where lots of associations between variables are established via functions.
There are lots of things you can do with function: you can simplify them, you can differentiate a function, you can operate them, find the composite with another function, and the list goes on an on.
Many times, even if you are mentioning functions explicitly, you have functions underlying the whole process. So they are there, even if sometimes you don't know it. One great thing is that even with very complicated ones, you can always graph a function to get an idea of its behavior, so to at least make an educated of what the function does (goes up, goes down, etc).
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What is the Division Method?
The Division Method is just the method we use to divide numbers. In this topic, you will use the division method to find the LCM. Remember that LCM stands for least common multiple.
Finding LCM by Division Method...
Let's take the two numbers 12 and 36.
Now write both of the numbers inside as if you were writing the dividend inside a box for division.
_______
{12, 36
[ Please don't mind my box... I tried my best :) ]
Next, look for a prime number that at least one of the numbers is divisible by. In this case we have two even numbers, so let's try 2.
2 { 12, 36
2 { 6, 18
3 { 3, 9
{ 1, 3
As you can see, we know have our Multiples.
The last step is multiplication.
You need to multiply the outside numbers ( divisors ) and the last two dividends ( 1 and 3 ) together to get the LCM.
Divisors= 2,2,3
Dividends= 1,3
So...
2x2x3x1x3
4x3x1x3
12x1x3
12x3
=36
That means that the LCM for 12 and 36 equals 36!
Sheet 1 Finding LCM by Division Method
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Multiplication and Division of Decimals
Decimals are numbers with a decimal point (e.g.2.35). They represent the fraction of something and ten is the base of decimal number system. A decimal notation or point differentiates an integer part from fractional part (e.g. 2.35 = 2 + 7/20).
Like whole numbers, decimals can also be added, subtracted, multiplied and divided. As compared to addition and subtraction of decimal numbers multiplication and division of decimals are much easy task. Both multiplication and division frequently connected in numerous daily computations. Even though concepts are confusing, it’s easy to solve them.
MULTIPLICATION OF DECIMALS
When we multiply 10 by 2 it is similar to addition of 10 two times: 10 x 10. This is same in case of decimals as well. For example: 0.33 x 2= 0.33 + 0.33 + 0.33. Multiplication of decimal numbers is similar to multiplication of whole numbers. Steps for multiplication of decimals are given below with an example.
Consider multiplication of two numbers, 2.32 and 3 for example.
Step 1: Take the count of the total number of places (digits) to the right of the decimal point in both numbers.
Here, in 2.32 there are two digits to the right of the decimal point and 3 is a whole number with no decimal point. Therefore, total number of digits to the right of decimal is 2.
Step 2: Now forget about the decimal point and just multiply the numbers without decimal point.
Step 3: After multiplication, put the decimal point in answer 2 places (step 1) from the right i.e. answer (2.32 x 3) will be 6.96.
Simply, just multiply the decimal numbers without decimal points and then give decimal point in the answer as many places same as the total number of places right to the decimal points in both numbers.
DIVISION OF DECIMALS
If you divide decimal numbers as it is with decimal point is quite confusing and difficult. We can use same trick we used in multiplication of decimals i.e. remove the decimal points and divide the numbers like whole numbers.
Let’s divide 40.5 by .20. Methods to divide these decimal numbers are as follows:
Method 1: Convert the decimal numbers into whole numbers by multiplying both numerator and denominator by same number. The denominator must be always a whole number.
(Multiply both numerator and denominator by 5)
Method 2: Alternatively, one can convert decimal numbers into whole numbers by multiplying with numbers having powers of 10 (10, 100, 1000, etc.).
• Consider the denominator, count the number of places (digits) right to the decimal point.
Here, denominator is 0.20 and number of digits right to the decimal point is 2.
40.5 ÷ 0.20 = 40.5 / 0.20
• Take the power of 10 same as number of digits right to the decimal point e. 102 = 100
• Multiply both numerator and denominator by 100.
MULTIPLICATION AND DIVISION OF DECIMAL NUMBERS BY 10, 100 AND 1000
As we know decimals is nothing but another form of expressing fractions having 10 as its base. We can express above examples as a fraction and can multiply and divide it as a fraction. Multiplication and division of decimals by numbers that have powers of 10 is easier than that by a whole number. Rules for multiplication and division of decimal numbers by 10, 100 and 1000:
Arithmetic Operation Rule Example Multiply by 10 (101) the number will move one place value to the left 5.63 x 10 =56.3 Multiply by 100 (102) the number will move two places value to the left 5.63 x 100 =563 Multiply by 1000 (103) the number will move three places value to the left 5.63 x 1000 =5630 Divide by 10 (101) the number will move one place value to the left 56.3 ÷ 10 =5.63 Divide by 100 (102) the number will move two places value to the right 56.3 ÷ 100 =0.563 Divide by 1000 (103) the number will move three places value to the right 56.3 ÷ 1000 =0.0563
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Giáo trình
# Precalculus
Mathematics and Statistics
## Graphs of Exponential Functions
Tác giả: OpenStaxCollege
As we discussed in the previous section, exponential functions are used for many real-world applications such as finance, forensics, computer science, and most of the life sciences. Working with an equation that describes a real-world situation gives us a method for making predictions. Most of the time, however, the equation itself is not enough. We learn a lot about things by seeing their pictorial representations, and that is exactly why graphing exponential equations is a powerful tool. It gives us another layer of insight for predicting future events.
# Graphing Exponential Functions
Before we begin graphing, it is helpful to review the behavior of exponential growth. Recall the table of values for a function of the form$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$whose base is greater than one. We’ll use the function$\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.\text{\hspace{0.17em}}$Observe how the output values in [link] change as the input increases by$\text{\hspace{0.17em}}1.$
$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $f\left(x\right)={2}^{x}$ $\frac{1}{8}$ $\frac{1}{4}$ $\frac{1}{2}$ $1$ $2$ $4$ $8$
Each output value is the product of the previous output and the base,$\text{\hspace{0.17em}}2.\text{\hspace{0.17em}}$We call the base$\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$the constant ratio. In fact, for any exponential function with the form$\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x},$$\text{\hspace{0.17em}}b\text{\hspace{0.17em}}$is the constant ratio of the function. This means that as the input increases by 1, the output value will be the product of the base and the previous output, regardless of the value of$\text{\hspace{0.17em}}a.$
Notice from the table that
• the output values are positive for all values of $x;$
• as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$increases, the output values increase without bound; and
• as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$decreases, the output values grow smaller, approaching zero.
[link] shows the exponential growth function $\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.$
The domain of$\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$is all real numbers, the range is$\text{\hspace{0.17em}}\left(0,\infty \right),$ and the horizontal asymptote is$\text{\hspace{0.17em}}y=0.$
To get a sense of the behavior of exponential decay, we can create a table of values for a function of the form$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$whose base is between zero and one. We’ll use the function$\text{\hspace{0.17em}}g\left(x\right)={\left(\frac{1}{2}\right)}^{x}.\text{\hspace{0.17em}}$Observe how the output values in [link] change as the input increases by$\text{\hspace{0.17em}}1.$
$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $g\left(x\right)={\left(\frac{1}{2}\right)}^{x}$ $8$ $4$ $2$ $1$ $\frac{1}{2}$ $\frac{1}{4}$ $\frac{1}{8}$
Again, because the input is increasing by 1, each output value is the product of the previous output and the base, or constant ratio$\text{\hspace{0.17em}}\frac{1}{2}.$
Notice from the table that
• the output values are positive for all values of$\text{\hspace{0.17em}}x;$
• as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$increases, the output values grow smaller, approaching zero; and
• as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$decreases, the output values grow without bound.
[link] shows the exponential decay function,$\text{\hspace{0.17em}}g\left(x\right)={\left(\frac{1}{2}\right)}^{x}.$
The domain of$\text{\hspace{0.17em}}g\left(x\right)={\left(\frac{1}{2}\right)}^{x}\text{\hspace{0.17em}}$is all real numbers, the range is$\text{\hspace{0.17em}}\left(0,\infty \right),$and the horizontal asymptote is$\text{\hspace{0.17em}}y=0.$
Characteristics of the Graph of the Parent Function f(x) = bx
An exponential function with the form$\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$$\text{\hspace{0.17em}}b>0,$$\text{\hspace{0.17em}}b\ne 1,$has these characteristics:
• one-to-one function
• horizontal asymptote:$\text{\hspace{0.17em}}y=0$
• domain:
• range:$\text{\hspace{0.17em}}\left(0,\infty \right)$
• x-intercept: none
• y-intercept:$\text{\hspace{0.17em}}\left(0,1\right)\text{\hspace{0.17em}}$
• increasing if$\text{\hspace{0.17em}}b>1$
• decreasing if$\text{\hspace{0.17em}}b<1$
[link] compares the graphs of exponential growth and decay functions.
Given an exponential function of the form$\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$graph the function.
1. Create a table of points.
2. Plot at least$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$point from the table, including the y-intercept$\text{\hspace{0.17em}}\left(0,1\right).$
3. Draw a smooth curve through the points.
4. State the domain,$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$the range,$\text{\hspace{0.17em}}\left(0,\infty \right),$and the horizontal asymptote, $\text{\hspace{0.17em}}y=0.$
Sketching the Graph of an Exponential Function of the Form f(x) = bx
Sketch a graph of$\text{\hspace{0.17em}}f\left(x\right)={0.25}^{x}.\text{\hspace{0.17em}}$State the domain, range, and asymptote.
Before graphing, identify the behavior and create a table of points for the graph.
• Since$\text{\hspace{0.17em}}b=0.25\text{\hspace{0.17em}}$is between zero and one, we know the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote$\text{\hspace{0.17em}}y=0.$
• Create a table of points as in [link]. $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $f\left(x\right)={0.25}^{x}$ $64$ $16$ $4$ $1$ $0.25$ $0.0625$ $0.015625$
• Plot the y-intercept,$\text{\hspace{0.17em}}\left(0,1\right),$along with two other points. We can use$\text{\hspace{0.17em}}\left(-1,4\right)\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(1,0.25\right).$
Draw a smooth curve connecting the points as in [link].
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=0.$
Sketch the graph of$\text{\hspace{0.17em}}f\left(x\right)={4}^{x}.\text{\hspace{0.17em}}$State the domain, range, and asymptote.
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=0.$
# Graphing Transformations of Exponential Functions
Transformations of exponential graphs behave similarly to those of other functions. Just as with other parent functions, we can apply the four types of transformations—shifts, reflections, stretches, and compressions—to the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$without loss of shape. For instance, just as the quadratic function maintains its parabolic shape when shifted, reflected, stretched, or compressed, the exponential function also maintains its general shape regardless of the transformations applied.
## Graphing a Vertical Shift
The first transformation occurs when we add a constant$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$to the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ giving us a vertical shift$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$units in the same direction as the sign. For example, if we begin by graphing a parent function,$\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ we can then graph two vertical shifts alongside it, using$\text{\hspace{0.17em}}d=3:\text{\hspace{0.17em}}$the upward shift,$\text{\hspace{0.17em}}g\left(x\right)={2}^{x}+3\text{\hspace{0.17em}}$and the downward shift,$\text{\hspace{0.17em}}h\left(x\right)={2}^{x}-3.\text{\hspace{0.17em}}$Both vertical shifts are shown in [link].
Observe the results of shifting$\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$vertically:
• The domain,$\text{\hspace{0.17em}}\left(-\infty ,\infty \right)\text{\hspace{0.17em}}$remains unchanged.
• When the function is shifted up$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}g\left(x\right)={2}^{x}+3:$
• The y-intercept shifts up$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}\left(0,4\right).$
• The asymptote shifts up$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}y=3.$
• The range becomes$\text{\hspace{0.17em}}\left(3,\infty \right).$
• When the function is shifted down$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}h\left(x\right)={2}^{x}-3:$
• The y-intercept shifts down$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}\left(0,-2\right).$
• The asymptote also shifts down$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}y=-3.$
• The range becomes$\text{\hspace{0.17em}}\left(-3,\infty \right).$
## Graphing a Horizontal Shift
The next transformation occurs when we add a constant$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$to the input of the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ giving us a horizontal shift$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$units in the opposite direction of the sign. For example, if we begin by graphing the parent function$\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ we can then graph two horizontal shifts alongside it, using$\text{\hspace{0.17em}}c=3:\text{\hspace{0.17em}}$the shift left,$\text{\hspace{0.17em}}g\left(x\right)={2}^{x+3},$ and the shift right,$\text{\hspace{0.17em}}h\left(x\right)={2}^{x-3}.\text{\hspace{0.17em}}$Both horizontal shifts are shown in [link].
Observe the results of shifting$\text{\hspace{0.17em}}f\left(x\right)={2}^{x}\text{\hspace{0.17em}}$horizontally:
• The domain,$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$remains unchanged.
• The asymptote,$\text{\hspace{0.17em}}y=0,$remains unchanged.
• The y-intercept shifts such that:
• When the function is shifted left$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}g\left(x\right)={2}^{x+3},$the y-intercept becomes$\text{\hspace{0.17em}}\left(0,8\right).\text{\hspace{0.17em}}$This is because$\text{\hspace{0.17em}}{2}^{x+3}=\left(8\right){2}^{x},$so the initial value of the function is$\text{\hspace{0.17em}}8.$
• When the function is shifted right$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units to$\text{\hspace{0.17em}}h\left(x\right)={2}^{x-3},$the y-intercept becomes$\text{\hspace{0.17em}}\left(0,\frac{1}{8}\right).\text{\hspace{0.17em}}$Again, see that$\text{\hspace{0.17em}}{2}^{x-3}=\left(\frac{1}{8}\right){2}^{x},$so the initial value of the function is$\text{\hspace{0.17em}}\frac{1}{8}.$
Shifts of the Parent Function f(x) = bx
For any constants$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}d,$the function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d\text{\hspace{0.17em}}$shifts the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}$
• vertically$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$units, in the same direction of the sign of$\text{\hspace{0.17em}}d.$
• horizontally$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$units, in the opposite direction of the sign of$\text{\hspace{0.17em}}c.$
• The y-intercept becomes$\text{\hspace{0.17em}}\left(0,{b}^{c}+d\right).$
• The horizontal asymptote becomes$\text{\hspace{0.17em}}y=d.$
• The range becomes$\text{\hspace{0.17em}}\left(d,\infty \right).$
• The domain,$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$remains unchanged.
Given an exponential function with the form$\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d,$graph the translation.
1. Draw the horizontal asymptote$\text{\hspace{0.17em}}y=d.$
2. Identify the shift as$\text{\hspace{0.17em}}\left(-c,d\right).\text{\hspace{0.17em}}$Shift the graph of$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$left$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$units if$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$is positive, and right$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$units if$c\text{\hspace{0.17em}}$is negative.
3. Shift the graph of$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$up$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$units if$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$is positive, and down$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$units if$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$is negative.
4. State the domain,$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$the range,$\text{\hspace{0.17em}}\left(d,\infty \right),$and the horizontal asymptote$\text{\hspace{0.17em}}y=d.$
Graphing a Shift of an Exponential Function
Graph$\text{\hspace{0.17em}}f\left(x\right)={2}^{x+1}-3.\text{\hspace{0.17em}}$State the domain, range, and asymptote.
We have an exponential equation of the form$\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d,$ with$\text{\hspace{0.17em}}b=2,$$\text{\hspace{0.17em}}c=1,$ and$\text{\hspace{0.17em}}d=-3.$
Draw the horizontal asymptote$\text{\hspace{0.17em}}y=d$, so draw$\text{\hspace{0.17em}}y=-3.$
Identify the shift as$\text{\hspace{0.17em}}\left(-c,d\right),$ so the shift is$\text{\hspace{0.17em}}\left(-1,-3\right).$
Shift the graph of$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$left 1 units and down 3 units.
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(-3,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=-3.$
Graph$\text{\hspace{0.17em}}f\left(x\right)={2}^{x-1}+3.\text{\hspace{0.17em}}$State domain, range, and asymptote.
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(3,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=3.$
Given an equation of the form$\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d\text{\hspace{0.17em}}$for$\text{\hspace{0.17em}}x,$ use a graphing calculator to approximate the solution.
• Press [Y=]. Enter the given exponential equation in the line headed “Y1=”.
• Enter the given value for$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$in the line headed “Y2=”.
• Press [WINDOW]. Adjust the y-axis so that it includes the value entered for “Y2=”.
• Press [GRAPH] to observe the graph of the exponential function along with the line for the specified value of$\text{\hspace{0.17em}}f\left(x\right).$
• To find the value of$\text{\hspace{0.17em}}x,$we compute the point of intersection. Press [2ND] then [CALC]. Select “intersect” and press [ENTER] three times. The point of intersection gives the value of x for the indicated value of the function.
Approximating the Solution of an Exponential Equation
Solve$\text{\hspace{0.17em}}42=1.2{\left(5\right)}^{x}+2.8\text{\hspace{0.17em}}$graphically. Round to the nearest thousandth.
Press [Y=] and enter$\text{\hspace{0.17em}}1.2{\left(5\right)}^{x}+2.8\text{\hspace{0.17em}}$next to Y1=. Then enter 42 next to Y2=. For a window, use the values –3 to 3 for$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$and –5 to 55 for$\text{\hspace{0.17em}}y.\text{\hspace{0.17em}}$Press [GRAPH]. The graphs should intersect somewhere near$\text{\hspace{0.17em}}x=2.$
For a better approximation, press [2ND] then [CALC]. Select [5: intersect] and press [ENTER] three times. The x-coordinate of the point of intersection is displayed as 2.1661943. (Your answer may be different if you use a different window or use a different value for Guess?) To the nearest thousandth,$\text{\hspace{0.17em}}x\approx 2.166.$
Solve$\text{\hspace{0.17em}}4=7.85{\left(1.15\right)}^{x}-2.27\text{\hspace{0.17em}}$graphically. Round to the nearest thousandth.
$x\approx -1.608$
## Graphing a Stretch or Compression
While horizontal and vertical shifts involve adding constants to the input or to the function itself, a stretch or compression occurs when we multiply the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$by a constant$\text{\hspace{0.17em}}|a|>0.\text{\hspace{0.17em}}$For example, if we begin by graphing the parent function$\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$we can then graph the stretch, using$\text{\hspace{0.17em}}a=3,$to get$\text{\hspace{0.17em}}g\left(x\right)=3{\left(2\right)}^{x}\text{\hspace{0.17em}}$as shown on the left in [link], and the compression, using$\text{\hspace{0.17em}}a=\frac{1}{3},$to get$\text{\hspace{0.17em}}h\left(x\right)=\frac{1}{3}{\left(2\right)}^{x}\text{\hspace{0.17em}}$as shown on the right in [link].
Stretches and Compressions of the Parent Function f(x) = bx
For any factor$\text{\hspace{0.17em}}a>0,$the function$\text{\hspace{0.17em}}f\left(x\right)=a{\left(b\right)}^{x}$
• is stretched vertically by a factor of$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$if$\text{\hspace{0.17em}}|a|>1.$
• is compressed vertically by a factor of$\text{\hspace{0.17em}}a\text{\hspace{0.17em}}$if$\text{\hspace{0.17em}}|a|<1.$
• has a y-intercept of$\text{\hspace{0.17em}}\left(0,a\right).$
• has a horizontal asymptote at$\text{\hspace{0.17em}}y=0,$ a range of$\text{\hspace{0.17em}}\left(0,\infty \right),$ and a domain of$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$which are unchanged from the parent function.
Graphing the Stretch of an Exponential Function
Sketch a graph of$\text{\hspace{0.17em}}f\left(x\right)=4{\left(\frac{1}{2}\right)}^{x}.\text{\hspace{0.17em}}$State the domain, range, and asymptote.
Before graphing, identify the behavior and key points on the graph.
• Since$\text{\hspace{0.17em}}b=\frac{1}{2}\text{\hspace{0.17em}}$is between zero and one, the left tail of the graph will increase without bound as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$decreases, and the right tail will approach the x-axis as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$increases.
• Since$\text{\hspace{0.17em}}a=4,$the graph of$\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{2}\right)}^{x}\text{\hspace{0.17em}}$will be stretched by a factor of$\text{\hspace{0.17em}}4.$
• Create a table of points as shown in [link]. $x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $f(x)=4 ( 1 2 ) x$ $32$ $16$ $8$ $4$ $2$ $1$ $0.5$
• Plot the y-intercept,$\text{\hspace{0.17em}}\left(0,4\right),$along with two other points. We can use$\text{\hspace{0.17em}}\left(-1,8\right)\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(1,2\right).$
Draw a smooth curve connecting the points, as shown in [link].
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=0.$
Sketch the graph of$\text{\hspace{0.17em}}f\left(x\right)=\frac{1}{2}{\left(4\right)}^{x}.\text{\hspace{0.17em}}$State the domain, range, and asymptote.
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=0.\text{\hspace{0.17em}}$
## Graphing Reflections
In addition to shifting, compressing, and stretching a graph, we can also reflect it about the x-axis or the y-axis. When we multiply the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$by$\text{\hspace{0.17em}}-1,$we get a reflection about the x-axis. When we multiply the input by$\text{\hspace{0.17em}}-1,$we get a reflection about the y-axis. For example, if we begin by graphing the parent function$\text{\hspace{0.17em}}f\left(x\right)={2}^{x},$ we can then graph the two reflections alongside it. The reflection about the x-axis,$\text{\hspace{0.17em}}g\left(x\right)={-2}^{x},$is shown on the left side of [link], and the reflection about the y-axis$\text{\hspace{0.17em}}h\left(x\right)={2}^{-x},$ is shown on the right side of [link].
Reflections of the Parent Function f(x) = bx
The function$\text{\hspace{0.17em}}f\left(x\right)=-{b}^{x}$
• reflects the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$about the x-axis.
• has a y-intercept of$\text{\hspace{0.17em}}\left(0,-1\right).$
• has a range of$\text{\hspace{0.17em}}\left(-\infty ,0\right)$
• has a horizontal asymptote at$\text{\hspace{0.17em}}y=0\text{\hspace{0.17em}}$and domain of$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$which are unchanged from the parent function.
The function$\text{\hspace{0.17em}}f\left(x\right)={b}^{-x}$
• reflects the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$about the y-axis.
• has a y-intercept of$\text{\hspace{0.17em}}\left(0,1\right),$ a horizontal asymptote at$\text{\hspace{0.17em}}y=0,$ a range of$\text{\hspace{0.17em}}\left(0,\infty \right),$ and a domain of$\text{\hspace{0.17em}}\left(-\infty ,\infty \right),$ which are unchanged from the parent function.
Writing and Graphing the Reflection of an Exponential Function
Find and graph the equation for a function,$\text{\hspace{0.17em}}g\left(x\right),$that reflects$\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{4}\right)}^{x}\text{\hspace{0.17em}}$about the x-axis. State its domain, range, and asymptote.
Since we want to reflect the parent function$\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{4}\right)}^{x}\text{\hspace{0.17em}}$about the x-axis, we multiply$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$by$\text{\hspace{0.17em}}-1\text{\hspace{0.17em}}$to get,$\text{\hspace{0.17em}}g\left(x\right)=-{\left(\frac{1}{4}\right)}^{x}.\text{\hspace{0.17em}}$Next we create a table of points as in [link].
$x$ $-3$ $-2$ $-1$ $0$ $1$ $2$ $3$ $g\left(x\right)=-{\left(\frac{1}{4}\right)}^{x}$ $-64$ $-16$ $-4$ $-1$ $-0.25$ $-0.0625$ $-0.0156$
Plot the y-intercept,$\text{\hspace{0.17em}}\left(0,-1\right),$along with two other points. We can use$\text{\hspace{0.17em}}\left(-1,-4\right)\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(1,-0.25\right).$
Draw a smooth curve connecting the points:
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(-\infty ,0\right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=0.$
Find and graph the equation for a function,$\text{\hspace{0.17em}}g\left(x\right),$ that reflects$\text{\hspace{0.17em}}f\left(x\right)={1.25}^{x}\text{\hspace{0.17em}}$about the y-axis. State its domain, range, and asymptote.
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(0,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=0.$
## Summarizing Translations of the Exponential Function
Now that we have worked with each type of translation for the exponential function, we can summarize them in [link] to arrive at the general equation for translating exponential functions.
Translations of the Parent Function $\text{\hspace{0.17em}}f\left(x\right)={b}^{x}$ Translation Form Shift Horizontally$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$units to the left Vertically$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$units up $f(x)= b x+c +d$ Stretch and Compress Stretch if$\text{\hspace{0.17em}}|a|>1$ Compression if$\text{\hspace{0.17em}}0<|a|<1$ $f(x)=a b x$ Reflect about the x-axis $f(x)=− b x$ Reflect about the y-axis $f(x)= b −x = ( 1 b ) x$ General equation for all translations $f(x)=a b x+c +d$
Translations of Exponential Functions
A translation of an exponential function has the form
Where the parent function,$\text{\hspace{0.17em}}y={b}^{x},$$\text{\hspace{0.17em}}b>1,$is
• shifted horizontally$\text{\hspace{0.17em}}c\text{\hspace{0.17em}}$units to the left.
• stretched vertically by a factor of$\text{\hspace{0.17em}}|a|\text{\hspace{0.17em}}$if$\text{\hspace{0.17em}}|a|>0.$
• compressed vertically by a factor of$\text{\hspace{0.17em}}|a|\text{\hspace{0.17em}}$if$\text{\hspace{0.17em}}0<|a|<1.$
• shifted vertically$\text{\hspace{0.17em}}d\text{\hspace{0.17em}}$units.
• reflected about the x-axis when$\text{\hspace{0.17em}}a<0.$
Note the order of the shifts, transformations, and reflections follow the order of operations.
Writing a Function from a Description
Write the equation for the function described below. Give the horizontal asymptote, the domain, and the range.
• $f\left(x\right)={e}^{x}\text{\hspace{0.17em}}$is vertically stretched by a factor of$\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$, reflected across the y-axis, and then shifted up$\text{\hspace{0.17em}}4\text{\hspace{0.17em}}$units.
We want to find an equation of the general formWe use the description provided to find$\text{\hspace{0.17em}}a,$ $b,$ $c,$ and $\text{\hspace{0.17em}}d.$
• We are given the parent function$\text{\hspace{0.17em}}f\left(x\right)={e}^{x},$ so$\text{\hspace{0.17em}}b=e.$
• The function is stretched by a factor of$\text{\hspace{0.17em}}2$, so$\text{\hspace{0.17em}}a=2.$
• The function is reflected about the y-axis. We replace$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$with$\text{\hspace{0.17em}}-x\text{\hspace{0.17em}}$to get:$\text{\hspace{0.17em}}{e}^{-x}.$
• The graph is shifted vertically 4 units, so$\text{\hspace{0.17em}}d=4.$
Substituting in the general form we get,
The domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(4,\infty \right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=4.$
Write the equation for function described below. Give the horizontal asymptote, the domain, and the range.
• $f\left(x\right)={e}^{x}\text{\hspace{0.17em}}$is compressed vertically by a factor of$\text{\hspace{0.17em}}\frac{1}{3},$ reflected across the x-axis and then shifted down $\text{\hspace{0.17em}}2$ units.
$f\left(x\right)=-\frac{1}{3}{e}^{x}-2;\text{\hspace{0.17em}}$the domain is$\text{\hspace{0.17em}}\left(-\infty ,\infty \right);\text{\hspace{0.17em}}$the range is$\text{\hspace{0.17em}}\left(-\infty ,2\right);\text{\hspace{0.17em}}$the horizontal asymptote is$\text{\hspace{0.17em}}y=2.$
Access this online resource for additional instruction and practice with graphing exponential functions.
# Key Equations
General Form for the Translation of the Parent Function $f\left(x\right)=a{b}^{x+c}+d$
# Key Concepts
• The graph of the function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$has a y-intercept atdomainrange and horizontal asymptote$\text{\hspace{0.17em}}y=0.\text{\hspace{0.17em}}$See [link].
• If$\text{\hspace{0.17em}}b>1,$the function is increasing. The left tail of the graph will approach the asymptote$\text{\hspace{0.17em}}y=0,$ and the right tail will increase without bound.
• If$\text{\hspace{0.17em}}0 the function is decreasing. The left tail of the graph will increase without bound, and the right tail will approach the asymptote$\text{\hspace{0.17em}}y=0.$
• The equation$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}+d\text{\hspace{0.17em}}$represents a vertical shift of the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}.$
• The equation$\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}\text{\hspace{0.17em}}$represents a horizontal shift of the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}.\text{\hspace{0.17em}}$See [link].
• Approximate solutions of the equation$\text{\hspace{0.17em}}f\left(x\right)={b}^{x+c}+d\text{\hspace{0.17em}}$can be found using a graphing calculator. See [link].
• The equation$\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x},$ where$\text{\hspace{0.17em}}a>0,$ represents a vertical stretch if$\text{\hspace{0.17em}}|a|>1\text{\hspace{0.17em}}$or compression if$\text{\hspace{0.17em}}0<|a|<1\text{\hspace{0.17em}}$of the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}.\text{\hspace{0.17em}}$See [link].
• When the parent function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x}\text{\hspace{0.17em}}$is multiplied by$\text{\hspace{0.17em}}-1,$the result,$\text{\hspace{0.17em}}f\left(x\right)=-{b}^{x},$ is a reflection about the x-axis. When the input is multiplied by$\text{\hspace{0.17em}}-1,$the result,$\text{\hspace{0.17em}}f\left(x\right)={b}^{-x},$ is a reflection about the y-axis. See [link].
• All translations of the exponential function can be summarized by the general equation$\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x+c}+d.\text{\hspace{0.17em}}$See [link].
• Using the general equation$\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x+c}+d,$ we can write the equation of a function given its description. See [link].
# Section Exercises
## Verbal
What role does the horizontal asymptote of an exponential function play in telling us about the end behavior of the graph?
An asymptote is a line that the graph of a function approaches, as$\text{\hspace{0.17em}}x\text{\hspace{0.17em}}$either increases or decreases without bound. The horizontal asymptote of an exponential function tells us the limit of the function’s values as the independent variable gets either extremely large or extremely small.
What is the advantage of knowing how to recognize transformations of the graph of a parent function algebraically?
## Algebraic
The graph of$\text{\hspace{0.17em}}f\left(x\right)={3}^{x}\text{\hspace{0.17em}}$is reflected about the y-axis and stretched vertically by a factor of$\text{\hspace{0.17em}}4.\text{\hspace{0.17em}}$What is the equation of the new function,$\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$State its y-intercept, domain, and range.
$g\left(x\right)=4{\left(3\right)}^{-x};\text{\hspace{0.17em}}$y-intercept:$\text{\hspace{0.17em}}\left(0,4\right);\text{\hspace{0.17em}}$Domain: all real numbers; Range: all real numbers greater than$\text{\hspace{0.17em}}0.$
The graph of$\text{\hspace{0.17em}}f\left(x\right)={\left(\frac{1}{2}\right)}^{-x}\text{\hspace{0.17em}}$is reflected about the y-axis and compressed vertically by a factor of$\text{\hspace{0.17em}}\frac{1}{5}.\text{\hspace{0.17em}}$What is the equation of the new function,$\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$State its y-intercept, domain, and range.
The graph of$\text{\hspace{0.17em}}f\left(x\right)={10}^{x}\text{\hspace{0.17em}}$is reflected about the x-axis and shifted upward$\text{\hspace{0.17em}}7\text{\hspace{0.17em}}$units. What is the equation of the new function,$\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$State its y-intercept, domain, and range.
$g\left(x\right)=-{10}^{x}+7;\text{\hspace{0.17em}}$y-intercept:$\text{\hspace{0.17em}}\left(0,6\right);\text{\hspace{0.17em}}$Domain: all real numbers; Range: all real numbers less than$\text{\hspace{0.17em}}7.$
The graph of$\text{\hspace{0.17em}}f\left(x\right)={\left(1.68\right)}^{x}\text{\hspace{0.17em}}$is shifted right$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units, stretched vertically by a factor of$\text{\hspace{0.17em}}2,$reflected about the x-axis, and then shifted downward$\text{\hspace{0.17em}}3\text{\hspace{0.17em}}$units. What is the equation of the new function,$\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$State its y-intercept (to the nearest thousandth), domain, and range.
The graph of$\text{\hspace{0.17em}}f\left(x\right)=-\frac{1}{2}{\left(\frac{1}{4}\right)}^{x-2}+4\text{\hspace{0.17em}}$is shifted left$\text{\hspace{0.17em}}2\text{\hspace{0.17em}}$units, stretched vertically by a factor of$\text{\hspace{0.17em}}4,$reflected about the x-axis, and then shifted downward$\text{\hspace{0.17em}}4\text{\hspace{0.17em}}$units. What is the equation of the new function,$\text{\hspace{0.17em}}g\left(x\right)?\text{\hspace{0.17em}}$State its y-intercept, domain, and range.
$g\left(x\right)=2{\left(\frac{1}{4}\right)}^{x};\text{\hspace{0.17em}}$ y-intercept:Domain: all real numbers; Range: all real numbers greater than$\text{\hspace{0.17em}}0.$
## Graphical
For the following exercises, graph the function and its reflection about the y-axis on the same axes, and give the y-intercept.
$f\left(x\right)=3{\left(\frac{1}{2}\right)}^{x}$
$g\left(x\right)=-2{\left(0.25\right)}^{x}$
y-intercept:$\text{\hspace{0.17em}}\left(0,-2\right)$
$h\left(x\right)=6{\left(1.75\right)}^{-x}$
For the following exercises, graph each set of functions on the same axes.
$f\left(x\right)=3{\left(\frac{1}{4}\right)}^{x},$$g\left(x\right)=3{\left(2\right)}^{x},$and$\text{\hspace{0.17em}}h\left(x\right)=3{\left(4\right)}^{x}$
$f\left(x\right)=\frac{1}{4}{\left(3\right)}^{x},$$g\left(x\right)=2{\left(3\right)}^{x},$and$\text{\hspace{0.17em}}h\left(x\right)=4{\left(3\right)}^{x}$
For the following exercises, match each function with one of the graphs in [link].
$f\left(x\right)=2{\left(0.69\right)}^{x}$
B
$f\left(x\right)=2{\left(1.28\right)}^{x}$
$f\left(x\right)=2{\left(0.81\right)}^{x}$
A
$f\left(x\right)=4{\left(1.28\right)}^{x}$
$f\left(x\right)=2{\left(1.59\right)}^{x}$
E
$f\left(x\right)=4{\left(0.69\right)}^{x}$
For the following exercises, use the graphs shown in [link]. All have the form$\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x}.$
Which graph has the largest value for$\text{\hspace{0.17em}}b?$
D
Which graph has the smallest value for$\text{\hspace{0.17em}}b?$
Which graph has the largest value for$\text{\hspace{0.17em}}a?$
C
Which graph has the smallest value for$\text{\hspace{0.17em}}a?$
For the following exercises, graph the function and its reflection about the x-axis on the same axes.
$f\left(x\right)=\frac{1}{2}{\left(4\right)}^{x}$
$f\left(x\right)=3{\left(0.75\right)}^{x}-1$
$f\left(x\right)=-4{\left(2\right)}^{x}+2$
For the following exercises, graph the transformation of$\text{\hspace{0.17em}}f\left(x\right)={2}^{x}.\text{\hspace{0.17em}}$Give the horizontal asymptote, the domain, and the range.
$f\left(x\right)={2}^{-x}$
$h\left(x\right)={2}^{x}+3$
Horizontal asymptote:$\text{\hspace{0.17em}}h\left(x\right)=3;$ Domain: all real numbers; Range: all real numbers strictly greater than$\text{\hspace{0.17em}}3.$
$f\left(x\right)={2}^{x-2}$
For the following exercises, describe the end behavior of the graphs of the functions.
$f\left(x\right)=-5{\left(4\right)}^{x}-1$
As; As
$f\left(x\right)=3{\left(\frac{1}{2}\right)}^{x}-2$
$f\left(x\right)=3{\left(4\right)}^{-x}+2$
As; As
For the following exercises, start with the graph of$\text{\hspace{0.17em}}f\left(x\right)={4}^{x}.\text{\hspace{0.17em}}$Then write a function that results from the given transformation.
Shift$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$4 units upward
Shift$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$3 units downward
$f\left(x\right)={4}^{x}-3$
Shift$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$2 units left
Shift$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$5 units right
$f\left(x\right)={4}^{x-5}$
Reflect$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$about the x-axis
Reflect$\text{\hspace{0.17em}}f\left(x\right)\text{\hspace{0.17em}}$about the y-axis
$f\left(x\right)={4}^{-x}$
For the following exercises, each graph is a transformation of$\text{\hspace{0.17em}}y={2}^{x}.\text{\hspace{0.17em}}$Write an equation describing the transformation.
$y=-{2}^{x}+3$
For the following exercises, find an exponential equation for the graph.
$y=-2{\left(3\right)}^{x}+7$
## Numeric
For the following exercises, evaluate the exponential functions for the indicated value of$\text{\hspace{0.17em}}x.$
$g\left(x\right)=\frac{1}{3}{\left(7\right)}^{x-2}\text{\hspace{0.17em}}$for$\text{\hspace{0.17em}}g\left(6\right).$
$g\left(6\right)=800+\frac{1}{3}\approx 800.3333$
$f\left(x\right)=4{\left(2\right)}^{x-1}-2\text{\hspace{0.17em}}$for$\text{\hspace{0.17em}}f\left(5\right).$
$h\left(x\right)=-\frac{1}{2}{\left(\frac{1}{2}\right)}^{x}+6\text{\hspace{0.17em}}$for$\text{\hspace{0.17em}}h\left(-7\right).$
$h\left(-7\right)=-58$
## Technology
For the following exercises, use a graphing calculator to approximate the solutions of the equation. Round to the nearest thousandth.$\text{\hspace{0.17em}}f\left(x\right)=a{b}^{x}+d.$
$-50=-{\left(\frac{1}{2}\right)}^{-x}$
$116=\frac{1}{4}{\left(\frac{1}{8}\right)}^{x}$
$x\approx -2.953$
$12=2{\left(3\right)}^{x}+1$
$5=3{\left(\frac{1}{2}\right)}^{x-1}-2$
$x\approx -0.222$
$-30=-4{\left(2\right)}^{x+2}+2$
## Extensions
Explore and discuss the graphs of$\text{\hspace{0.17em}}F\left(x\right)={\left(b\right)}^{x}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}G\left(x\right)={\left(\frac{1}{b}\right)}^{x}.\text{\hspace{0.17em}}$Then make a conjecture about the relationship between the graphs of the functions$\text{\hspace{0.17em}}{b}^{x}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}{\left(\frac{1}{b}\right)}^{x}\text{\hspace{0.17em}}$for any real number$\text{\hspace{0.17em}}b>0.$
The graph of$\text{\hspace{0.17em}}G\left(x\right)={\left(\frac{1}{b}\right)}^{x}\text{\hspace{0.17em}}$is the refelction about the y-axis of the graph of$\text{\hspace{0.17em}}F\left(x\right)={b}^{x};\text{\hspace{0.17em}}$For any real number$\text{\hspace{0.17em}}b>0\text{\hspace{0.17em}}$and function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$the graph of$\text{\hspace{0.17em}}{\left(\frac{1}{b}\right)}^{x}\text{\hspace{0.17em}}$is the the reflection about the y-axis,$\text{\hspace{0.17em}}F\left(-x\right).$
Prove the conjecture made in the previous exercise.
Explore and discuss the graphs of$\text{\hspace{0.17em}}f\left(x\right)={4}^{x},$$\text{\hspace{0.17em}}g\left(x\right)={4}^{x-2},$and$\text{\hspace{0.17em}}h\left(x\right)=\left(\frac{1}{16}\right){4}^{x}.\text{\hspace{0.17em}}$Then make a conjecture about the relationship between the graphs of the functions$\text{\hspace{0.17em}}{b}^{x}\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}\left(\frac{1}{{b}^{n}}\right){b}^{x}\text{\hspace{0.17em}}$for any real number n and real number$\text{\hspace{0.17em}}b>0.$
The graphs of$\text{\hspace{0.17em}}g\left(x\right)\text{\hspace{0.17em}}$and$\text{\hspace{0.17em}}h\left(x\right)\text{\hspace{0.17em}}$are the same and are a horizontal shift to the right of the graph of$\text{\hspace{0.17em}}f\left(x\right);\text{\hspace{0.17em}}$For any real number n, real number$\text{\hspace{0.17em}}b>0,$ and function$\text{\hspace{0.17em}}f\left(x\right)={b}^{x},$ the graph of$\text{\hspace{0.17em}}\left(\frac{1}{{b}^{n}}\right){b}^{x}\text{\hspace{0.17em}}$is the horizontal shift$\text{\hspace{0.17em}}f\left(x-n\right).$
Prove the conjecture made in the previous exercise.
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# How to solve algebra equations
There are also many YouTube videos that can show you How to solve algebra equations. So let's get started!
## How can we solve algebra equations
The solver will provide step-by-step instructions on How to solve algebra equations. This formula states that the log of a number with respect to one base is equal to the log of the same number with respect to another base multiplied by the log of the new base with respect to the old base. So, if we want to solve for x in our example equation above, we can plug in our known values and solve for x using algebra.2log₃x=6⇒log₃x=3⇒x=33Since we now know that 3 was raised to the third power in order to produce 9 (our exponent), we have successfully solved for x in this equation!Common and natural logarithms are two other ways that exponents can be solved for without using the change of base formula. Common logarithms use bases of 10, while natural logarithms use bases of e (approximately 2.71828182845904). To solve for x in equations using these types of logs, all you need to do is take the inverse function of each side. For example, if we want to solve10log₁₀x=100we can simply take the inverse common log function of both sides.This tells us that 100 must have been produced when 10 was raised to some power - but what power? Well, we can use algebra once again!10log₁₀x=100⇒log₁₀x=10⇒x=1010Now we know that 10 was raised to the 10th power in order to produce 100. And just like that - we've solved another equation for x using logs!While solving equations with logs may seem daunting at first, there's no need to worry - with a little practice, you'll be a pro in no time!
A series solver is a mathematical tool that allows you to calculate the sum of an infinite series. This can be a useful tool for evaluating limits, as well as for finding closed-form expressions for sums of common series. There are a variety of different methods that can be used to solve series, and the choice of method will depend on the particular properties of the series being considered. In general, however, all methods involve breaking the series down into smaller pieces and then summing those pieces together. The most basic method is known as the "telescoping method," which involves cancelling out terms that cancel each other out when added together. This can be a very efficient method, but it is not always possible to use it. In other cases, one might need to use a more sophisticated technique, such as integration or summation by parts. Whichever method is used, the goal is always to find a concise expression for the sum of an infinite series.
There are a number of great apps out there that can help you solve math problems. However, the best app for solving math problems depends on your individual needs and preferences. If you need step-by-step instructions, then an app like Mathway is a good option. If you prefer to work through problems on your own, then a app like WolframAlpha might be a better fit. Ultimately, the best way to find the right app for you is to try out a few different options and see which one works best for your learning style.
Solving trinomials can be a tricky business, but there are a few methods that can make the process a bit easier. One common method is to factor the trinomial into two binomials. This can be done by grouping the terms together in pairs, and then multiplying each pair to get the product. Another method is to use the quadratic formula. This involves plugging the values of the coefficients into a specific equation, and then solving for x. While these methods may seem daunting at first, with a little practice they can become second nature. With some patience and perseverance, solving trinomials can be a breeze.
Many students find word math problems to be some of the most challenging they will encounter. Unlike traditional math problems, which typically involve a definite answer, word problems often require students to interpret the data and make strategic decisions. As a result, word math problems can be both time-consuming and frustrating. However, there are a few key strategies that can help students solve word math problems more efficiently. First, it is important to read the problem carefully and identify all of the relevant information. Next, students should identify any unknowns and try to determine what operation would best be used to solve for them. Finally, it is helpful to work through the problem step-by-step and check your answer at each stage to avoid making mistakes. By following these steps, students can approach word math problems with confidence and ease.
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Even though the NCAA tournament finished up earlier this month, for the next ten months I will be thinking about college basketball whenever I see the first several powers of two. No matter what type of GMAT question you are dealing with, our minds are better able to work through topics that we are already familiar with. Probability problems make me think of gambling, weakening a GMAT argument becomes shooting down an argument from that crazy relative you only see at Thanksgiving, and anything dealing with the number 64 comes down to rounds in a basketball tournament. Here’s a few tricks on the GMAT where knowing your powers of two can save you some time and brainpower.
1. 64 = 2^6
Know how to translate larger numbers into their smaller factors
Since 1985, every team that has won the NCAA tournament has had to win six games. By multiplying two times itself, you can expand to each round of the NCAA tournament- 2, 4, 8, 16, 32, and 64. And because these numbers are all small and have a single prime factor, they commonly end up on the GMAT. Because of this, you should be able to recognize them and quickly put each one into its base of two: 2 = 2^1, 4 = 2^2, etc. Same for the powers of three- 3, 9, 27, 81. The number 81 is far more likely to show up on your GMAT than 83, because 81 is a power of 3 that can be broken down into small prime factors. Without a calculator, numbers that are easy to break down show up 2 x 5 times more often than they do in the real world.
2. (2^2)^3 = 64
When you’re not sure how to compute a difficult problem, think about what you would do with easier numbers
I’ve been teaching the GMAT for three years and to this day, I still have trouble translating some word problems. x is y percent more than z isn’t something that most of us can instinctively translate. But I do know that 15 is 50 percent more than 10. And when I think about why, it’s because I’m taking z and adding another y/100 of that z. Which means x = z * (y/100) + z. These equations with simple numbers make more sense to us, because we’re more likely to have to mark something up by 50 percent than to mark something up by y percent. Making analogies to things that we are used to working with makes questions easier and faster, and makes ourselves more confident in our work. Which is why when I see ((3 x 2^2)^4)^-2, I’m instantly thinking that (2^2)^3 = 64 = 2^6. If it works on a simple computation, it should work on a harder one too.
3. 64-1 = 63 teams that lose a game
If you get stuck answering a problem, try rephrasing the question
If I asked you how many games there are in a 64 team tournament, you might work your way through round by round: 32 games played by 64 teams in the first round, 16 games played by 32 teams in the second round, 8 games with 16 teams, 4 games with 8 teams, 2 games with 4 teams, and 1 game with the final two participants. 32 + 16 + 8 + 4 + 2 + 1 = 63. But on the GMAT, rephrasing a question can be the difference between spending three minutes on a problem and spending one. Instead of asking how many games are played, ask yourself how many games teams will lose. Think of this like the 1-x trick in probability: If 64 teams are playing and 1 doesn’t lose, that means 63 do. If you’ve taken a free-trial class with us, you know that rephrasing questions (and statements) is a huge point of emphasis in our classes beginning on day 1. So when you get stuck solving a problem, look for other parts of the question that you can solve.
4. 0.000064 = something to do with the number 2
When all else fails, focus on what looks familiar
Much of what makes the GMAT so difficult is the fact that there’s so much going on in any given problem and no clue about where to start. Sometimes when you look at a problem type, you will have a go to method- set up a chart or a system of equations and solve. But other times you will see a question and have no idea how to categorize it, simplify it, or do anything with it. But just because you don’t know how to solve a problem doesn’t mean you aren’t able to do a small part of the problem. Back to the very first point I made in this article- 64 is much more likely to show up on the GMAT than a prime number like 67. So when you see 64, think about how you can simplify that part of the problem. Looking at 0.000064, change the way you write that number to 64 x 0.000001 and deal with the two parts separately. Once you’ve made that step, recognize that 64 is 2^6. And once you see that one part of the problem has an exponent of 6, see if that same step can be expressed on another part of the equation: 64 x 0.000001 = 2^6 x 0.000001 = 2^6 x (0.1)^6 = (0.2)^6. GMAT problems aren’t solved with one impossible leap; they’re solved using several workable steps. Sometimes that first step will help reveal where the next step should be.
#### Joe Lucero
Joe Lucero has both a Biology degree and a Master of Education from the University of Notre Dame. He also has a 780 on his GMAT. In the fall, you will find Joe in a much better mood during weeks after the Fighting Irish win their football game. During the rest of the year, you will find him looking for new places to travel, reading almost anything non-fiction, crossfitting, and trying to solve every challenge problem in the Manhattan GMAT Student Center.
### 4 responses to The Power(s) of 2
1. Thanks for the good Article! Definitely a wider lens perspective and recognizing the “familiar” face of 2 will help in seemingly tough problems. Hope to remember and follow in sticky situations!
2. In the example above in the last line of the second paragraph (#2), I see the solution to ((3 x 2^2)^4)^-2 as impossibly small. Wouldn’t this translate to (12^4)^-2 which is equal to 20736^-2 which again is equal to 1/20736^2, which equals 1/429981696. This is an impossibly small number. Am I doing something wrong?
Can we expect such problems in GMAT?
• Good question. The GMAT would never ask you to calculate the value of such an absurdly small fraction. However, the GMAT can (and does) ask you to know how to manipulate fractions and integers with exponents, either to find equivalent forms of a hard-to-calculate number or to cancel these numbers out, leaving some simple number.
3. The Power(s) of 2 | Manhattan GMAT Blog
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Mister Exam
# Differential equation dy=(x-3y)dx
y() =
y'() =
y''() =
y'''() =
y''''() =
from to
### The solution
You have entered [src]
d
--(y(x)) = x - 3*y(x)
dx
$$\frac{d}{d x} y{\left(x \right)} = x - 3 y{\left(x \right)}$$
y' = x - 3*y
Detail solution
Given the equation:
$$\frac{d}{d x} y{\left(x \right)} = x - 3 y{\left(x \right)}$$
This differential equation has the form:
y' + P(x)y = Q(x)
where
$$P{\left(x \right)} = 3$$
and
$$Q{\left(x \right)} = x$$
and it is called linear homogeneous
differential first-order equation:
First of all, we should solve the correspondent linear homogeneous equation
y' + P(x)y = 0
with multiple variables
The equation is solved using following steps:
From y' + P(x)y = 0 you get
$$\frac{dy}{y} = - P{\left(x \right)} dx$$, if y is not equal to 0
$$\int \frac{1}{y}\, dy = - \int P{\left(x \right)}\, dx$$
$$\log{\left(\left|{y}\right| \right)} = - \int P{\left(x \right)}\, dx$$
Or,
$$\left|{y}\right| = e^{- \int P{\left(x \right)}\, dx}$$
Therefore,
$$y_{1} = e^{- \int P{\left(x \right)}\, dx}$$
$$y_{2} = - e^{- \int P{\left(x \right)}\, dx}$$
The expression indicates that it is necessary to find the integral:
$$\int P{\left(x \right)}\, dx$$
Because
$$P{\left(x \right)} = 3$$, then
$$\int P{\left(x \right)}\, dx$$ =
= $$\int 3\, dx = 3 x + Const$$
Detailed solution of the integral
So, solution of the homogeneous linear equation:
$$y_{1} = e^{C_{1} - 3 x}$$
$$y_{2} = - e^{C_{2} - 3 x}$$
that leads to the correspondent solution
for any constant C, not equal to zero:
$$y = C e^{- 3 x}$$
We get a solution for the correspondent homogeneous equation
Now we should solve the inhomogeneous equation
y' + P(x)y = Q(x)
Use variation of parameters method
Now, consider C a function of x
$$y = C{\left(x \right)} e^{- 3 x}$$
And apply it in the original equation.
Using the rules
- for product differentiation;
- of composite functions derivative,
we find that
$$\frac{d}{d x} C{\left(x \right)} = Q{\left(x \right)} e^{\int P{\left(x \right)}\, dx}$$
Let use Q(x) and P(x) for this equation.
We get the first-order differential equation for C(x):
$$\frac{d}{d x} C{\left(x \right)} = x e^{3 x}$$
So, C(x) =
$$\int x e^{3 x}\, dx = \frac{\left(3 x - 1\right) e^{3 x}}{9} + Const$$
Detailed solution of the integral
use C(x) at
$$y = C{\left(x \right)} e^{- 3 x}$$
and we get a definitive solution for y(x):
$$e^{- 3 x} \left(\frac{\left(3 x - 1\right) e^{3 x}}{9} + Const\right)$$
The answer [src]
/ 3*x\
| (-1 + 3*x)*e | -3*x
y(x) = |C1 + ---------------|*e
\ 9 /
$$y{\left(x \right)} = \left(C_{1} + \frac{\left(3 x - 1\right) e^{3 x}}{9}\right) e^{- 3 x}$$
The classification
1st linear
Bernoulli
almost linear
1st power series
lie group
nth linear constant coeff undetermined coefficients
nth linear constant coeff variation of parameters
1st linear Integral
Bernoulli Integral
almost linear Integral
nth linear constant coeff variation of parameters Integral
The graph
To see a detailed solution - share to all your student friends
To see a detailed solution,
share to all your student friends:
|
LCM the 3 and 13 is the smallest number amongst all typical multiples of 3 and also 13. The first few multiples of 3 and also 13 space (3, 6, 9, 12, 15, 18, . . . ) and also (13, 26, 39, 52, 65, 78, 91, . . . ) respectively. There space 3 frequently used methods to find LCM that 3 and also 13 - by element factorization, by listing multiples, and also by department method.
You are watching: What is the least common multiple of 3 and 13
1 LCM of 3 and also 13 2 List that Methods 3 Solved Examples 4 FAQs
Answer: LCM the 3 and 13 is 39.
Explanation:
The LCM of two non-zero integers, x(3) and also y(13), is the smallest hopeful integer m(39) the is divisible by both x(3) and y(13) without any remainder.
The techniques to uncover the LCM of 3 and 13 are explained below.
By department MethodBy element Factorization MethodBy Listing Multiples
### LCM the 3 and also 13 by department Method
To calculation the LCM of 3 and 13 by the department method, we will certainly divide the numbers(3, 13) by your prime components (preferably common). The product of this divisors provides the LCM the 3 and also 13.
Step 3: proceed the actions until only 1s space left in the critical row.
The LCM that 3 and 13 is the product of all prime number on the left, i.e. LCM(3, 13) by division method = 3 × 13 = 39.
### LCM that 3 and also 13 by element Factorization
Prime administrate of 3 and 13 is (3) = 31 and also (13) = 131 respectively. LCM that 3 and also 13 deserve to be derived by multiplying prime determinants raised to your respective highest possible power, i.e. 31 × 131 = 39.Hence, the LCM that 3 and 13 by prime factorization is 39.
### LCM that 3 and also 13 through Listing Multiples
To calculate the LCM the 3 and also 13 by listing out the typical multiples, we can follow the given listed below steps:
Step 1: list a couple of multiples that 3 (3, 6, 9, 12, 15, 18, . . . ) and also 13 (13, 26, 39, 52, 65, 78, 91, . . . . )Step 2: The typical multiples native the multiples the 3 and also 13 space 39, 78, . . .Step 3: The smallest usual multiple the 3 and 13 is 39.
∴ The least usual multiple of 3 and also 13 = 39.
## FAQs top top LCM the 3 and also 13
### What is the LCM that 3 and also 13?
The LCM the 3 and also 13 is 39. To find the LCM (least common multiple) that 3 and also 13, we require to find the multiples that 3 and also 13 (multiples of 3 = 3, 6, 9, 12 . . . . 39; multiples the 13 = 13, 26, 39, 52) and choose the the smallest multiple the is specifically divisible through 3 and also 13, i.e., 39.
### What are the approaches to find LCM of 3 and also 13?
The frequently used methods to discover the LCM the 3 and 13 are:
### Which the the adhering to is the LCM the 3 and also 13? 28, 42, 39, 12
The worth of LCM of 3, 13 is the smallest typical multiple of 3 and also 13. The number satisfying the given problem is 39.
### If the LCM the 13 and 3 is 39, uncover its GCF.
LCM(13, 3) × GCF(13, 3) = 13 × 3Since the LCM of 13 and also 3 = 39⇒ 39 × GCF(13, 3) = 39Therefore, the GCF = 39/39 = 1.
See more: Confederate Currency 1861 Confederate \$5 Dollar Bill Worth ?
### How to discover the LCM the 3 and 13 by element Factorization?
To find the LCM of 3 and 13 making use of prime factorization, we will find the prime factors, (3 = 3) and also (13 = 13). LCM the 3 and also 13 is the product of prime determinants raised to their respective greatest exponent amongst the number 3 and also 13.⇒ LCM of 3, 13 = 31 × 131 = 39.
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# Statistical Estimation Assignment Solutions
We have posted the following statistical estimation assignment solutions to educate and help you realize how proficient we are in answering statistics questions. Go through them and discover our level of competence as well as learn various concepts on estimating population parameters using samples.
## Likelihood Ratios and Statistical Tests
In the first question, we make valid conclusions on the given likelihood ratio tests. The question that follows involves using a sample size to estimate parameters like the mean, standard deviation, and the standard error of the mean. What's more, we find the confidence interval of the mean and recommends a sample size suitable for finding the mean value under given conditions
## Likelihood Ratios
Q4.1 The following table shows the Likelihood ratio for test positive and likelihood ratio for test negative of 3 biomarkers to diagnose Disease A. Answer the following question.
Likelihood Ratio
Test Test Positive Test Negative Biomarker 1 2.4 0.3 Biomarker 2 5.7 0.6 Biomarker 3 1.07 0.9
(i) Which test/biomarker would be the best to rule in Disease A?
Biomarker 2
(ii) Which test/biomarker would be the best to rule out Disease A?
Biomarker 1
(iii) Which test/biomarker is the least useful test for both rulings in and ruling out disease A?
Biomarker 3
## Statistical Estimation Using Tests
Q5.1 You would like to estimate the birth weight and rate of low birth weight(<2.500kg) of Chinese male babies in Hong Kong. Worksheet A provides the birthweight of boys delivered in Hospital A in Hong Kong. Provide answers for the following. Give answers to 3 decimal places.
(i) What is the sample size?
The sample size is 200
(ii) What is the mean and its standard deviation?
The mean is 3.077 and the standard deviation is 0.467
(iii) Standard Error of the mean
The standard error of the mean is 0.033
(iv) Margin of error for the mean at the 95% confidence level
Margin of error = 1.96 * 0.467/√200
= 1.96 * 0.033
= 0.065
(v) 95% confidence interval of the mean
95% C.I for mean
= (3.077 – 0.065, 3.077+0.065)
= (3.012, 3.142)
(vi) What is the sample size required in a survey to assess the mean value with an error at half of the current error?
The sample size is based on confidence level, the larger the required confidence level, the higher the sample size.
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Pvillage.org
# How do you calculate exponents on a calculator?
## How do you calculate exponents on a calculator?
Using the Exponent Key On most calculators, you enter the base, press the exponent key and enter the exponent. Here’s an example: Enter 10, press the exponent key, then press 5 and enter. (10^5=) The calculator should display the number 100,000, because that’s equal to 105.
### How do you calculate the square root of something?
The square root formula is used to find the square root of a number. We know the exponent formula: n√x x n = x1/n. When n= 2, we call it square root. We can use any of the above methods for finding the square root, such as prime factorization, long division, and so on.
#### What is the easiest way to find square roots?
The square root of a number can be found by using the below steps:
1. Step 1: Pair the digits starting from right to left.
2. Step 2: Match the unit digit of the number from the chart and determine the possible values of the square root of the unit digit.
3. Step 3: Now, we consider the first set of digits of the number.
When to use square roots and exponents in Algebra?
When the exponent is 2, the result is called a square. When the exponent is 3, the result is called a cube. Memorize the squares of the integers up to 15 and the cubes of the integers up to 10. They will be used often as you progress in your study of algebra.
Are there two square roots for an even number?
Square Roots, odd and even: There are 2 possible roots for any positive real number. A positive root and a negative root. Given a number x, the square root of x is a number a such that a2 = x. Square roots is a specialized form of our common roots calculator. “Note that any positive real number has two square roots, one positive and one negative.
## Which is the square root of the number x?
Given a number x, the square root of x is a number a such that a2 = x. Square roots is a specialized form of our common roots calculator. “Note that any positive real number has two square roots, one positive and one negative. For example, the square roots of 9 are -3 and +3, since (-3) 2 = (+3) 2 = 9.
### Where do you enter the exponent on a calculator?
On most calculators, you enter the base, press the exponent key and enter the exponent.
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# How To Find Slope Intercept Form With Two Points
## The Definition, Formula, and Problem Example of the Slope-Intercept Form
How To Find Slope Intercept Form With Two Points – One of the numerous forms employed to represent a linear equation one that is frequently used is the slope intercept form. It is possible to use the formula of the slope-intercept to identify a line equation when that you have the straight line’s slope as well as the y-intercept. It is the y-coordinate of the point at the y-axis is intersected by the line. Learn more about this specific linear equation form below.
## What Is The Slope Intercept Form?
There are three basic forms of linear equations: the traditional slope-intercept, the point-slope, and the standard. While they all provide the same results when utilized however, you can get the information line produced faster using the slope-intercept form. The name suggests that this form employs the sloped line and it is the “steepness” of the line is a reflection of its worth.
This formula can be utilized to calculate the slope of straight lines, y-intercept, or x-intercept, in which case you can use a variety of formulas available. The equation for a line using this particular formula is y = mx + b. The slope of the straight line is signified in the form of “m”, while its y-intercept is indicated through “b”. Every point on the straight line can be represented using an (x, y). Note that in the y = mx + b equation formula, the “x” and the “y” are treated as variables.
## An Example of Applied Slope Intercept Form in Problems
When it comes to the actual world in the real world, the slope intercept form is often utilized to show how an item or issue changes over an elapsed time. The value provided by the vertical axis demonstrates how the equation addresses the degree of change over the value given by the horizontal axis (typically in the form of time).
A basic example of this formula’s utilization is to figure out the rate at which population increases in a certain area as the years pass by. In the event that the area’s population grows annually by a predetermined amount, the value of the horizontal axis will grow by a single point each year and the worth of the vertical scale will rise in proportion to the population growth according to the fixed amount.
You can also note the starting value of a question. The beginning value is at the y’s value within the y’intercept. The Y-intercept is the point at which x equals zero. Based on the example of the above problem the beginning value will be the time when the reading of population starts or when the time tracking starts along with the changes that follow.
So, the y-intercept is the point where the population starts to be recorded by the researcher. Let’s say that the researcher begins to calculate or the measurement in 1995. This year will represent the “base” year, and the x 0 points will occur in 1995. So, it is possible to say that the population of 1995 is the y-intercept.
Linear equation problems that utilize straight-line equations are typically solved this way. The starting point is depicted by the y-intercept and the rate of change is represented in the form of the slope. The primary complication of the slope-intercept form typically lies in the horizontal variable interpretation especially if the variable is linked to a specific year (or any other kind or unit). The key to solving them is to ensure that you know the variables’ meanings in detail.
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# 3/20 as a decimal
#### Understand the Problem
The question is asking us to convert the fraction 3/20 into its decimal form. This requires performing a division operation where 3 is divided by 20.
#### Answer
0.15
##### Answer for screen readers
The final answer is 0.15
#### Steps to Solve
1. Set up the division
To convert the fraction ( \frac{3}{20} ) to a decimal, you need to divide 3 by 20. We can write this as a division problem: $$3 \div 20$$
2. Perform the division
Perform the long division where 3 is the dividend, and 20 is the divisor. Since 3 is smaller than 20, you will have a decimal.
• 20 does not go into 3, so place a 0 and a decimal point: 0.
• Add a zero to make it 30 and find how many times 20 goes into 30, which is 1 time: 0.1
• Subtract 20 from 30 to get 10 and bring down another 0 to make 100.
• 20 goes into 100 five times exactly: 0.15
• No remainder, so the long division ends here.
3. Write the result
After completing the division, we find that: $$\frac{3}{20} = 0.15$$
The final answer is 0.15
#### More Information
Some fractions convert into terminating decimals and some into repeating decimals. In this case, ( \frac{3}{20} ) converts into a terminating decimal with two digits after the decimal point.
#### Tips
A common mistake is not aligning the decimal points correctly when performing long division. To avoid such mistakes, make sure each step of the division is carefully aligned.
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# Multiplying a 3-digit number by a 1-digit number
9 teachers like this lesson
Print Lesson
## Objective
SWBAT multiply a 3-digit number by a 1-digit number using strategies based on place value.
#### Big Idea
Students can multiply a 3-digit number by a 1-digit number using place value blocks as models.
## Opener
5 minutes
In the Introductory Video Multiplying 3 Digit by 1 Digit, I explain the lesson for today.
The students have already learned how to use the expanded algorithm to find simpler problems to solve a 2-digit by 1-digit multiplication problem. In today's lesson, they learn to find partial products using place value to multiply a 3-digit number by a 1-digit number. This gives the students a visual and helps them find the product when multiplying a 3-digit number by a 1-digit number. This aligns with 4.NBT.B5 because the students are multiplying using strategies based on place value.
To begin the lesson, I remind the students that we have learned to multiply a 2-digit number by a 1-digit number by making an area model and using place value blocks. I ask the students, "how did making an array help you understand the problem?" I let the students think about the question for a few minutes, then call on a student to respond. One student said, "I could count the pieces in the array to get my answer." I let the students know that today they will learn to multiply a 3-digit number by a 1-digit number using strategies based on place value.
## Whole Class Discussion
15 minutes
To begin our review, I call the students to the carpet. On the Smart board, I have the Multiplying a 3-digit by 1-digit number review slide that we will discuss as a whole class.
At the beginning of each lesson, I like to review all relevant skills that we have learned that will help with the new skill. Since the students have already learned this information, I just want to bring it back to the forefront of their minds.
Review:
1. You can use an array to help you multiply. For example, 5 x 13 = 65. We can show this by making an array:
xxxxxxxxxxxxx
xxxxxxxxxxxxx
xxxxxxxxxxxxx
xxxxxxxxxxxxx
xxxxxxxxxxxxx
2. Identity property of multiplication says that when you multiply a number by 1, the product is the other number.
3. Commutative property of multiplication says that you can multiply factors in any order and will get the same product.
4. Property of Zero says that when you multiply a number by zero, the product is zero.
5. Use place value to help you multiply. Let's review the places beginning at the right. We have the ones, tens, hundreds, thousands, ten thousands, and hundred thousands. Let's find out how place value can help you multiply a 3-digit number by a 1-digit number.
The properties of multiplication help the students if they multiply a number by 1 or 0. Knowing these properties should help alleviate mistakes in their multiplication. Also, knowing the commutative property will help the students with this lesson because if the number is written with the 1 digit number first, the students know that they can change the order of the numbers and still get the same product.
I review all of this relevant information on the Smart board with the students. My students know that they can interact and jump in the conversation at any time. I question my students throughout the review. 1) How can the properties of multiplication help you to multiply other numbers? 2) Why should we use arrays/models to help with multiplication? I call on students who raise their hands, and at other times I call on students who do not raise their hands. This will let all students know to listen attentively and be prepared with an answer. From the students responses, they know that the properties of multiplication can help them multiply numbers with 1 and 0. One student responded, "Models help us count to find the product."
After the review, we do a problem together (Practice Problem Multiplying 3-digit by 1-digit.pptx) to show the students how to use place value to write 3 simpler problems.
Problem:
My mom owns 138 flower pots. Each flower pot has 5 flowers in it. How many flowers does my mom own in all?
Our multiplication problem is 138 x 5=
First, we multiply the ones: 8 x 5 = 40.
Next, we multiply 5 x 30 = 150
Last, we multiply 100 x 5 = 500
Add all the partial products: 500 + 150 + 40= 690
On the Smart board, I show the Model Multiplying 3-digit by 1-digit slide that breaks down how to model with place value blocks. After, we finish discussing this problem and the place value model, the students go back to their seats to work on the skills with their classmates.
## Group or Partner Activity
20 minutes
During this group activity, the students work in pairs. Each pair has a copy of the Group Activity Multiplying a 3-digit by 1-digit.docx. The students must work together to find the product to a 3-digit by 1-digit multiplication problem using place value (4.NBT.B5). The students must decontextualize the problem and represent them symbolically (MP2). They must model this problem using the place value blocks (MP5). Before they begin the activity, each pair will be given a few minutes to think about the problem and how they will solve it (MP1).
The students are guided to the conceptual understanding through questioning by their classmates, as well as by me. The students communicate with each other and must agree upon the answer to the problem. Because the students must agree upon the answer, this will take discussion, critiquing, and justifying of answers by both students (MP3). As the pairs discuss the problem, they must be precise in their communication within their groups using the appropriate math terminology for this skill (MP6). As I walk around, I am listening for the students to use "talk" that will lead to the answer. I am holding the students accountable for their own learning.
As they work, I monitor and assess their progression of understanding through questioning.
1. What is the value of each number?
2. After you find the 3 partial products, what must you do next?
3. How can you represent these partial products with the place value blocks?
As I monitor to listen in on the students' conversation, I am listening for the students to justify their answers or question other students about their answer. All of this should be done in a respectful manner. We implemented Accountable Talk at the beginning of the year, so the students know what is expected of them. I am listening for the students to say, "after they find the 3 partial products, they add all 3 numbers together to get the product."
From the sample of Student Work - Multiplying a 3 digit by 1 digit, you see how the students used the expanded algorithm to solve the multiplication problem. The students accurately solved the problem by multiplying by 3 simpler problems, then adding the partial products.
Any groups that finish the assignment early, can go to the computer to practice the skill at the following site until we are ready for the whole group sharing: http://www.mathplayground.com/multiples.html
## Closure
15 minutes
To close the lesson, I bring the class back together as a whole. We go over the answers to the problems. I call on different pairs to share their answers. Their classmates have the opportunity to ask any questions to get them to clarify their answer. This gives those students who still do not understand another opportunity to learn it. I like to use my document camera to show the Student's work during this time. Some students do not understand what is being said, but understand clearly when the work is put up for them to see.
I was proud of the students because they caught on to multiplyig a 3-digit number by 1-digit number quickly. They had no problem breaking apart into 3 simpler problems, but I did have to question a few students to lead them to the last step of adding the 3 partial products. I'm quite sure that using the expanded algorithm in a previous lesson helped to make this lesson easier and helped the students get a clearer understanding place value and multiplying.
I feel that by closing each of my lessons by having students share their work is very important to the success of the lesson. Students need to see good work samples, as well as work that may have incorrect information. This teaches the students what not to do in the future.
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```Question 540990
I'm assuming you're wondering how to factor {{{b^2-7b+10}}} to get {{{(b-5)(b-2)}}}
Looking at the expression {{{b^2-7b+10}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{-7}}}, and the last term is {{{10}}}.
Now multiply the first coefficient {{{1}}} by the last term {{{10}}} to get {{{(1)(10)=10}}}.
Now the question is: what two whole numbers multiply to {{{10}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{-7}}}?
To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{10}}} (the previous product).
Factors of {{{10}}}:
1,2,5,10
-1,-2,-5,-10
Note: list the negative of each factor. This will allow us to find all possible combinations.
These factors pair up and multiply to {{{10}}}.
1*10 = 10
2*5 = 10
(-1)*(-10) = 10
(-2)*(-5) = 10
Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{-7}}}:
<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td align="center"><font color=black>1</font></td><td align="center"><font color=black>10</font></td><td align="center"><font color=black>1+10=11</font></td></tr><tr><td align="center"><font color=black>2</font></td><td align="center"><font color=black>5</font></td><td align="center"><font color=black>2+5=7</font></td></tr><tr><td align="center"><font color=black>-1</font></td><td align="center"><font color=black>-10</font></td><td align="center"><font color=black>-1+(-10)=-11</font></td></tr><tr><td align="center"><font color=red>-2</font></td><td align="center"><font color=red>-5</font></td><td align="center"><font color=red>-2+(-5)=-7</font></td></tr></table>
From the table, we can see that the two numbers {{{-2}}} and {{{-5}}} add to {{{-7}}} (the middle coefficient).
So the two numbers {{{-2}}} and {{{-5}}} both multiply to {{{10}}} <font size=4><b>and</b></font> add to {{{-7}}}
Now replace the middle term {{{-7b}}} with {{{-2b-5b}}}. Remember, {{{-2}}} and {{{-5}}} add to {{{-7}}}. So this shows us that {{{-2b-5b=-7b}}}.
{{{b^2+highlight(-2b-5b)+10}}} Replace the second term {{{-7b}}} with {{{-2b-5b}}}.
{{{(b^2-2b)+(-5b+10)}}} Group the terms into two pairs.
{{{b(b-2)+(-5b+10)}}} Factor out the GCF {{{b}}} from the first group.
{{{b(b-2)-5(b-2)}}} Factor out {{{5}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.
{{{(b-5)(b-2)}}} Combine like terms. Or factor out the common term {{{b-2}}}
===============================================================
So {{{b^2-7b+10}}} factors to {{{(b-5)(b-2)}}}.
In other words, {{{b^2-7b+10=(b-5)(b-2)}}}.
Note: you can check the answer by expanding {{{(b-5)(b-2)}}} to get {{{b^2-7b+10}}} or by graphing the original expression and the answer (the two graphs should be identical).
If you need more help, email me at <a href="mailto:jim_thompson5910@hotmail.com">jim_thompson5910@hotmail.com</a>
Also, please consider visiting my website: <a href="http://www.freewebs.com/jimthompson5910/home.html">http://www.freewebs.com/jimthompson5910/home.html</a> and making a donation. Thank you
Jim
```
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## College Algebra (10th Edition)
(a) $5 \gt 4$ (b) $-3 \gt -4$ (c) $6 \lt 3$ (d) $-4 \lt -2$
RECALL: (1) When a real number is added or subtracted to each side of an inequality, the sense/direction of the inequality is not affected. (2) When a negative number is multiplied to each side of an inequality, the sense/direction of the inequality will change. Use the rules above to obtain: (a) Adding 3 to each side of the given inequality: $\begin{array}{ccc} &2 +3 &\gt &1+3 \\&5 &\gt &4 \end{array}$ (b) Subtracting 5 from each side of the given inequality: $\begin{array}{ccc} &2 -5 &\gt &1-5 \\&-3 &\gt &-4 \end{array}$ (c) Multiplying 3 to each side of the given inequality: $\begin{array}{ccc} &3(2) &\gt &3(1) \\&6 &\gt &3 \end{array}$ (d) Multiplying $-2$ to each side of the given inequality: $\begin{array}{ccc} &-2(2) &\lt &-2(1) \\&-4 &\lt &-2 \end{array}$
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Posted on
## Limits vs Values
Look at the function f(x) in orange below:
We are going to answer 4 questions about this graph. They are all related to each other, but different questions. Seeing the difference will help us sort out the difference between a function value and a limit.
Q1: Find f(1)
Q2: Find limx→1 f(x)
Q3: Find limx→1+ f(x)
Q4: Find limx→1 f(x)
OK….. Try to answer these questions with what you know… Then continue reading to see the answers and explanations!
Posted on
## Basic Concept of a Limit
Here is a brief example on the concept of a limit:
Look at the function f(x) in orange below:
Question 1. Find f(3)
Explanation of question 1: Find the value of the function when you plug in 3. What is the height of the function at the exact moment when x=3?
Answer 1: The function is undefined at x=3. There is a hole when x=3.
So, f(3) is undefined.
Question 2: Find limx→3f(x)
Explanation of question 2: We are being asked to find what the function is doing around (but not at) 3. What is happening to the path of the function on either side of 3?
In order to find limx→3f(x), we must confirm that limx→3+ f(x) and limx→3– f(x) both exist and are equal to each other.
So, let’s find limx→3+ f(x). What is happening to the function values as you approach x=3 from the right-hand side? Literally run your finger along as if x=4, then x=3.5, then x=3.1. What value is the function getting closer to?
The function is approaching a height of 4.
Let’s find limx→3– f(x). What is happening to the function values as you approach x=3 from the left-hand side? Literally run your finger along as if x=1, then x=2, then x=2.9. What value is the function getting closer to?
The function is also approaching a height of 4.
So:
limx→3+ f(x) = 4
limx→3– f(x) = 4
Since, the left-handed limit at 3 and right-handed limit at 3 exist and are equal, this gives:
limx→3f(x) = 4.
So, to summarize, here are 4 different things we found. They are related, but not necessarily the same:
f(3) is undefined
limx→3+ f(x) = 4
limx→3– f(x) = 4
limx→3f(x) = 4
Posted on
## Limit Example
Q: What is limx→2 (x-2)/|x-2|
A: This question is not too bad if you know what the function (x-2)/|x-2| looks like graphically. But, let’s say you don’t.
We are going to “talk our way” through this problem to help solidify the concept of a limit.
If you plug in 2 to the function, you are finding the value of the function when x=2. This is important, and related, though it is not the limit. This is even sometimes a skill used to help us find the limit, but it is still not the limit. Sometimes the function value is equal to the function limit, which can also be confusing, but not all the time. Let’s find the value of the function when x=2:
(2-2)/|2-2| = 0/0 = undefined.
Okay. This function is undefined when x=2. This means there is a hole, or an asymptote, or a break or a jump or some disruption in the continuity of the function. Continue reading Limit Example
Posted on
## One-sided Limit Example
Q: Find the one-sided limit (if it exists):
limx→-1– (x+1)/(x4-1)
A: So we need to find the limit of this function (x+1)/(x4-1) as x approaches -1 from the left. Remember, from the left means as x gets closer and closer to -1, but is still smaller.
The concept: What is happening to this function as x = -2, x = -1.5, x = -1.1, x = -1.0001, etc…
We test first and plug -1 into the function: (-1+1)/((-1)4-1) = 0/0
Whenever you get 0/0, that is your clue that maybe you need to do “more work” before just plugging in or jumping to conclusions.
So, let’s try “more work” — usually that means simplifying. I see that the denominator can factor. We have:
(x+1) / (x4-1) = (x+1) / [(x2-1)(x2+1)]
Let’s keep factoring the denominator:
(x+1) / [(x-1)(x+1)(x2+1)]
Now, it appears there is a “removable hole” in the function. This means, we can remove this hole by reducing the matching term in the numerator with the matching term in the denominator:
(x+1) / [(x-1)(x+1)(x2+1)]
= 1 / [(x-1)(x2+1)]
Notice that hole exists when x = -1 (and it was removable! This is good news for us since we are concerned with the nature of the function as x approaches -1)
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# Rational Numbers Class 8 Extra Questions Maths Chapter 1
### Rational Numbers Class 8 Extra Questions Short Answer Type
Question 13.
Calculate the following:
Solution:
Question 14.
Represent the following rational numbers on number lines.
(a) (frac { -2 }{ 3 })
(b) (frac { 3 }{ 4 })
(c) (frac { 3 }{ 2 })
Solution:
Question 15.
Find 7 rational numbers between (frac { 1 }{ 3 }) and (frac { 1 }{ 2 }).
Solution:
Question 16.
Show that:
Solution:
Question 17.
If x = (frac { 1 }{ 2 }), y = (frac { -2 }{ 3 }) and z = (frac { 1 }{ 4 }), verify that x × (y × z) = (x × y) × z.
Solution:
We have x = (frac { 1 }{ 2 }), y = (frac { -2 }{ 3 }) and z = (frac { 1 }{ 4 })
LHS = x × (y × z)
Question 18.
If the cost of 4(frac { 1 }{ 2 }) litres of milk is ₹89(frac { 1 }{ 2 }), find the cost of 1 litre of milk.
Solution:
Question 19.
The product of two rational numbers is (frac { 15 }{ 56 }). If one of the numbers is (frac { -5 }{ 48 }), find the other.
Solution:
Product of two rational numbers = (frac { 15 }{ 56 })
One number = (frac { -5 }{ 48 })
Other number = Product ÷ First number
Hence, the other number = (frac { -18 }{ 7 })
Question 20.
Let O, P and Z represent the numbers 0, 3 and -5 respectively on the number line. Points Q, R and S are between O and P such that OQ = QR = RS = SP. (NCERT Exemplar)
What are the rational numbers represented by the points Q, R and S. Next choose a point T between Z and 0 so that ZT = TO. Which rational number does T represent?
Solution:
As OQ = QR = RS = SP and OQ + QR + RS + SP = OP
therefore Q, R and S divide OP into four equal parts.
Question 21.
Let a, b, c be the three rational numbers where a = (frac { 2 }{ 3 }), b = (frac { 4 }{ 5 }) and c = (frac { -5 }{ 6 }) (NCERT Exemplar)
Verify:
(i) a + (b + c) = (a + b) + c (Associative property of addition)
(ii) a × (b × c) – (a × b) × c (Associative property of multiplication)
Solution:
### Rational Numbers Class 8 Extra Questions Higher Order Thinking Skills (HOTS)
Question 22.
Rajni had a certain amount of money in her purse. She spent ₹ 10(frac { 1 }{ 4 }) in the school canteen, bought a gift worth ₹ 25(frac { 3 }{ 4 }) and gave ₹ 16(frac { 1 }{ 2 }) to her friend. How much she have to begin with?
Solution:
Amount given to school canteen = ₹ 10(frac { 1 }{ 4 })
Amount given to buy gift = ₹ 25(frac { 3 }{ 4 })
Amount given to her friend = ₹ 16(frac { 1 }{ 2 })
Question 23.
One-third of a group of people are men. If the number of women is 200 more than the men, find the total number of people.
Solution:
Number of men in the group = (frac { 1 }{ 3 }) of the group
Number of women = 1 – (frac { 1 }{ 3 }) = (frac { 2 }{ 3 })
Difference between the number of men and women = (frac { 2 }{ 3 }) – (frac { 1 }{ 3 }) = (frac { 1 }{ 3 })
If difference is (frac { 1 }{ 3 }), then total number of people = 1
If difference is 200, then total number of people
= 200 ÷ (frac { 1 }{ 3 })
= 200 × 3 = 600
Hence, the total number of people = 600
Question 24.
Fill in the blanks:
(a) Numbers of rational numbers between two rational numbers is ……….
Solution:
(a) Countless
(b) (frac { 6 }{ 11 })
(c) (frac { -3 }{ 2 })
(d) (frac { 3 }{ 5 })
(e) Commutative
(f) associative
(g) equivalent
(h) (frac { 3 }{ 11 })
|
Posted on by Kalkicode
Code Mathematics
# Babylonian method for square root
The Babylonian method, also known as the Heron's method, is an ancient technique to approximate the square root of a number. It's an iterative algorithm that refines an initial guess for the square root of a given number until it reaches a desired level of precision. This method is especially interesting because it demonstrates how ancient civilizations developed mathematical techniques to solve complex problems.
## Problem Statement
The problem is to find an efficient way to calculate the square root of a given positive number. The square root of a number x is a value y such that y multiplied by itself equals x. Mathematically, y = √x.
For example, if we want to find the square root of 64, we are looking for a value y such that y × y = 64. In this case, the square root of 64 is 8.
## Idea to Solve the Problem
The Babylonian method for finding the square root of a number involves iterative refinement. The basic idea is to start with an initial guess for the square root and then iteratively update the guess to get closer to the actual square root.
Let's consider an example to understand the process better. Suppose we want to find the square root of x using an initial guess y0. We can refine our guess using the following formula:
y1 = (y0 + x / y0) / 2
Here, y1 is a better approximation of the square root than y0. We can repeat this process iteratively:
y2 = (y1 + x / y1) / 2 y3 = (y2 + x / y2) / 2 ...
The iterations continue until the difference between consecutive guesses becomes smaller than a specified precision value.
## Pseudocode
function findSquareRoot(num)
a = num
b = 1.0
precision = 0.000001
while (a - b) > precision
a = (a + b) / 2.0
b = num / a
return a
## Algorithm Explanation
1. Initialize two variables, a and b, both initially set to the given number num and 1.0 respectively. Also, define a precision value that specifies the desired level of accuracy (e.g., 0.000001).
2. Enter a loop that continues as long as the difference between a and b is greater than the defined precision. This loop is responsible for iteratively refining the guess.
3. Within the loop, update the value of a by taking the average of a and b. This update brings a closer to the actual square root.
4. Update the value of b by dividing num by the new value of a. This step ensures that b remains inversely proportional to a, allowing the iteration to converge towards the square root.
5. Once the loop exits (i.e., the difference between a and b becomes smaller than the precision), a holds a close approximation of the square root of num.
6. Return the value of a as the estimated square root of num.
## Code Solution
// C program
// babylonian method for square root
#include <stdio.h>
void findSquareRoot(double num)
{
double a = num;
double b = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = (a + b) / 2.0;
b = num / a;
}
// Display given number
printf("\n Given Number : %lf", num);
// Display the calculate square root
printf("\n Square Root : %lf\n", a);
}
int main(int argc, char
const *argv[])
{
// Test
findSquareRoot(64);
findSquareRoot(10.3);
findSquareRoot(17.50);
return 0;
}
#### Output
Given Number : 64.000000
Square Root : 8.000000
Given Number : 10.300000
Square Root : 3.209361
Given Number : 17.500000
Square Root : 4.183300
// Java program
// Babylonian method for square root
public class SquareRoot
{
public void findSquareRoot(double num)
{
double a = num;
double b = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = (a + b) / 2.0;
b = num / a;
}
// Display given number
System.out.print("\n Given Number : " + num);
// Display the calculate square root
System.out.print("\n Square Root : " + a + "\n");
}
public static void main(String[] args)
{
// Test
}
}
#### Output
Given Number : 64.0
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.209361314240489
Given Number : 17.5
Square Root : 4.183300132670613
#include <iostream>
using namespace std;
// C++ program
// Babylonian method for square root
class SquareRoot
{
public: void findSquareRoot(double num)
{
double a = num;
double b = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = (a + b) / 2.0;
b = num / a;
}
// Display given number
cout << "\n Given Number : " << num;
// Display the calculate square root
cout << "\n Square Root : " << a << "\n";
}
};
int main()
{
// Test
return 0;
}
#### Output
Given Number : 64
Square Root : 8
Given Number : 10.3
Square Root : 3.20936
Given Number : 17.5
Square Root : 4.1833
// Include namespace system
using System;
// C# program
// Babylonian method for square root
public class SquareRoot
{
public void findSquareRoot(double num)
{
double a = num;
double b = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = (a + b) / 2.0;
b = num / a;
}
// Display given number
Console.Write("\n Given Number : " + num);
// Display the calculate square root
Console.Write("\n Square Root : " + a + "\n");
}
public static void Main(String[] args)
{
// Test
}
}
#### Output
Given Number : 64
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.20936131424049
Given Number : 17.5
Square Root : 4.18330013267061
<?php
// Php program
// Babylonian method for square root
class SquareRoot
{
public function findSquareRoot(\$num)
{
\$a = \$num;
\$b = 1.0;
// Here precision (0.000001)
while ((\$a - \$b) > 0.000001)
{
\$a = ((\$a + \$b) / 2.0);
\$b = (\$num / \$a);
}
// Display given number
echo "\n Given Number : ". \$num;
// Display the calculate square root
echo "\n Square Root : ". \$a ."\n";
}
}
function main()
{
}
main();
#### Output
Given Number : 64
Square Root : 8.0000000000002
Given Number : 10.3
Square Root : 3.2093613142405
Given Number : 17.5
Square Root : 4.1833001326706
// Node Js program
// Babylonian method for square root
class SquareRoot
{
findSquareRoot(num)
{
var a = num;
var b = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = ((a + b) / 2.0);
b = (num / a);
}
// Display given number
process.stdout.write("\n Given Number : " + num);
// Display the calculate square root
process.stdout.write("\n Square Root : " + a + "\n");
}
}
function main()
{
// Test
}
main();
#### Output
Given Number : 64
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.209361314240489
Given Number : 17.5
Square Root : 4.183300132670613
# Python 3 program
# Babylonian method for square root
class SquareRoot :
def findSquareRoot(self, num) :
a = num
b = 1.0
# Here precision (0.000001)
while ((a - b) > 0.000001) :
a = ((a + b) / 2.0)
b = (num / a)
# Display given number
print("\n Given Number : ", num, end = "")
# Display the calculate square root
print("\n Square Root : ", a )
def main() :
# Test
if __name__ == "__main__": main()
#### Output
Given Number : 64
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.209361314240489
Given Number : 17.5
Square Root : 4.183300132670613
# Ruby program
# Babylonian method for square root
class SquareRoot
def findSquareRoot(num)
a = num
b = 1.0
# Here precision (0.000001)
while ((a - b) > 0.000001)
a = (a + b) / 2.0
b = num / a
end
# Display given number
print("\n Given Number : ", num)
# Display the calculate square root
print("\n Square Root : ", a ,"\n")
end
end
def main()
# Test
end
main()
#### Output
Given Number : 64
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.209361314240489
Given Number : 17.5
Square Root : 4.183300132670613
// Scala program
// Babylonian method for square root
class SquareRoot
{
def findSquareRoot(num: Double): Unit = {
var a: Double = num;
var b: Double = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = ((a + b) / 2.0);
b = (num / a);
}
// Display given number
print("\n Given Number : " + num);
// Display the calculate square root
print("\n Square Root : " + a + "\n");
}
}
object Main
{
def main(args: Array[String]): Unit = {
var task: SquareRoot = new SquareRoot();
// Test
}
}
#### Output
Given Number : 64.0
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.209361314240489
Given Number : 17.5
Square Root : 4.183300132670613
// Swift 4 program
// Babylonian method for square root
class SquareRoot
{
func findSquareRoot(_ num: Double)
{
var a: Double = num;
var b: Double = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = (a + b) / 2.0;
b = num / a;
}
// Display given number
print("\n Given Number : ", num, terminator: "");
// Display the calculate square root
print("\n Square Root : ", a );
}
}
func main()
{
// Test
}
main();
#### Output
Given Number : 64.0
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.20936131424049
Given Number : 17.5
Square Root : 4.18330013267061
// Kotlin program
// Babylonian method for square root
class SquareRoot
{
fun findSquareRoot(num: Double): Unit
{
var a: Double = num;
var b: Double = 1.0;
// Here precision (0.000001)
while ((a - b) > 0.000001)
{
a = (a + b) / 2.0;
b = num / a;
}
// Display given number
print("\n Given Number : " + num);
// Display the calculate square root
print("\n Square Root : " + a + "\n");
}
}
fun main(args: Array < String > ): Unit
{
// Test
}
#### Output
Given Number : 64.0
Square Root : 8.00000000000017
Given Number : 10.3
Square Root : 3.209361314240489
Given Number : 17.5
Square Root : 4.183300132670613
## Time Complexity
The time complexity of the Babylonian method for finding the square root is primarily determined by the number of iterations needed for the algorithm to converge to the desired precision. In each iteration, we perform simple arithmetic operations.
Since the number of iterations required depends on the initial guess and the precision level, it's generally challenging to provide a precise time complexity analysis. However, the Babylonian method is known for its rapid convergence, which means that it usually requires a small number of iterations to reach a high level of accuracy.
In practical terms, the Babylonian method's time complexity can be considered to be approximately O(log N), where N is the number of bits used to represent the input number. This is due to the rapid convergence of the method.
## Comment
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Welcome back to our series on number properties. This article will introduce GMAT quant problems that revolve around the relationships between the digits of multi-digit integers. This topic can be tricky because we all know how the base-10 system works, but this knowledge is so ingrained that we hardly ever think about it. We can easily get confused when asked to manipulate knowledge related to place value.
Problems that revolve around the relationship between the digits of multi-digit integers
This topic is different from other number properties topics in that there’s little or no theory to learn. We will jump right into some official GMAT problems so that you can get familiar with how GMAT quant builds questions around this topic. Here’s a straightforward one as a warmup:
Official GMAT problem for practice
If r and t are three-digit positive integers, is r greater than t?
1. The tens digit of r is greater than each of the three digits of t.
2. The tens digit of r is less than either of the other two digits of r.
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient.
The question asks which three-digit integer is greater, so we can think of the three digits as a “tiebreak” system. If the integers have the same hundreds digit, then we will need to know about their tens digits. And if they also have the same tens digit, then we will need to know about their units digits. Of course, if r has a greater hundreds digit than t does, the other digits are irrelevant – r is greater.
Statement 1 isn’t useless, but it isn’t sufficient by itself. The tens digit of r is greater than each digit of t. This still leaves room for r to be greater than or less than t, depending on the hundreds digits of each integer.
Statement 2 again provides data about the tens digit of r, this time comparing it to the other digits of r. This obviously can’t be sufficient by itself, since we know nothing at all about t. But when combined with the data from statement 1, this data is useful. If the tens digit of r is less than the hundreds digit of r (the one that matters most in this problem) and greater than any digit of t, then the hundreds digit of r is greater than the hundreds digit of t. This means that r is greater than T, and the statements together are sufficient. The correct answer is C.
Let’s try a slightly more complex problem:
If x, y, and z are three-digit positive integers and if x = y + z, is the hundreds digit of x equal to the sum of the hundreds digits of y and z?
1. The tens digit of x is equal to the sum of the tens digits of y and z.
2. The units digit of x is equal to the sum of the units digits of y and z.
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient.
This is an interesting problem about the nature of multi-digit addition. The question is built around whether a digit in a given place of the sum is equal to the sum of the digits occupying the same place in the addends. An addend is a number being added with another number. Let’s look at a couple of example three-digit addition problems:
``` 426 375
+ 351 + 592
777 967```
Apex does not recommend doing addition like this on GMAT quant, because it is much faster and more efficient to be able to perform such calculations mentally, without the use of your whiteboard or scratch paper. But in this case, it’s helpful to see the “columns” formed by the units, tens, and hundreds places. In the first example, each digit in the sum is equal to the sum of the corresponding digits in the addends. This is not true in the second example. The controlling element is whether the two digits in the addends have a sum greater than 9. If they do, as in the tens place of the second example, it is necessary to “carry” a 1 to the next digit place to the left. Elementary, but perhaps difficult to think about so many years after learning to add.
Now we can consider the statements. Statement 1 tells us that the tens digit of the sum x is indeed equal to the sum of the tens digits of the addends y and z. Since this is true, there will be no “carrying” to alter the values in the hundreds place. Therefore the hundreds digit of the sum x must be equal to the sum of the hundreds digits of y and z. Statement 1 is sufficient.
Statement 2 is qualitatively similar to statement 1 but tells us instead about the units digits. Interestingly, the data provided by statement 2 is already proven by statement 1. Since the tens digit of the sum x is equal to the tens digits of the addends y and z, there must not have been any “carrying” from the units place to the tens place to alter the values in the tens place. The units digit of the sum x must represent the full combined values of the units digits of the addends y and z.
So statement 2 adds nothing to statement 1, and it isn’t sufficient on its own either. The question asked about the hundreds digit of the sum x. Data about the units place of this addition problem is simply too many steps away because we can’t know what will happen in the tens place. Only statement 1 is sufficient on its own, and the correct answer is A.
Official place value problem
If n is a positive integer, what is the tens digit of n?
1. The hundreds digit of 10n is 6.
2. The tens digit of n + 1 is 7.
(A) Statement (1) alone is sufficient, but statement (2) alone is not sufficient.
(B) Statement (2) alone is sufficient, but statement (1) alone is not sufficient.
(C) BOTH statements TOGETHER are sufficient, but NEITHER statement ALONE is sufficient.
(D) EACH statement ALONE is sufficient.
(E) Statements (1) and (2) TOGETHER are not sufficient.
I’ve seen smart math students get stumped by problems like this one. But the problem isn’t complicated; it’s just unfamiliar. Many test-takers are hesitant to model examples when the value isn’t known for sure. Where do you start? How do you know you aren’t “breaking the rules”? We must remember that the point of examples is not always to follow the rules but sometimes to find them.
This is one data sufficiency problem where statement 2 is easier to start with. If the tens digit of n + 1 is 7, can we determine the tens digit of n? Well, there are several scenarios in which the tens digit of n is also 7. If n + 1 is any integer between 71 and 79, inclusive, then the tens digit of n is 7. But if n + 1 is 70, n is 69 and has a tens digit of 6. Statement 2 is therefore insufficient for determining the tens digit of n.
Statement 1 tells us that the hundreds digit of 10n is 6. Let’s multiply some two-digit numbers by 10 and see what happens to the digits:
47 * 10 = 470
83 * 10 = 830
69 * 10 = 690
What’s happening should make sense: when an integer is multiplied by 10, we simply “tack on” a 0 to the right side of the number, and each digit is “pushed” one place to the left. The digit that was in the units place now occupies the tens place, the digit that was in the tens place now occupies the hundreds place, and so on and so forth. Since this is true, knowing the hundreds digit of 10n is just as good as knowing the tens digit of n. Statement 1 is also sufficient, and the correct answer is D.
This last problem belongs to a class of data sufficiency problems that ask about a certain digit of integer n (or some other variable) and then give statements about a digit occupying a different place of 10n or 1000n or n/100.
The same principle applies in all of them: multiplication or division by any power of 10 represents a “shifting” of the digits of the number. Multiplication moves the digits to the left by however many zeros are in the power of 10; division moves the digits to the right by however many zeros are in the power of 10. We know this, but the construction of the data sufficiency problems can make it hard to recognize that this is the property controlling the problem. Now you know!
In the next article, we’ll look at the interesting case of the digits in a two-digit integer switching places.
Have you been studying for the GMAT and need a private tutor? Our elite tutors, who scored over 770 on the GMAT, offer 30 minutes of complimentary consultation time.
Contributor: Elijah Mize (Apex GMAT Instructor)
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## Related Articles
• RD Sharma Class 12 Solutions for Maths
# Class 12 RD Sharma Solutions – Chapter 19 Indefinite Integrals – Exercise 19.8 | Set 2
### Question 26. Evaluate ∫ 2cosx – 3sinx/ 6cosx + 4sinx dx
Solution:
Let us assume I = ∫ 2cosx – 3sinx/ 6cosx + 4sinx dx
= ∫ 2cosx – 3sinx/ 2(3cosx + 2sinx) dx
= ∫ 2cosx – 3sinx/ 2(3cosx + 2sinx) dx ………..(i)
Let 3cosx + 2sinx = t
d(3cosx + 2sinx) = dt
(-3sinx + 2cosx) dx = dt
(2cosx – 3sinx) dx = dt
Put all these values in equation(i), we get
= 1/2 ∫ dt/t
Integrate the above equation then, we get
= 1/2 log|t| + c
= 1/2 log|3cosx + 2sinx| + c
Hence, I = 1/2 log|3cosx + 2sinx| + c
### Question 27. Evaluate ∫ cos2x + x + 1/ (x2 + sin2x + 2x) dx
Solution:
Let us assume I = ∫ cos2x + x + 1/ (x2 + sin2x + 2x) dx (i)
Let x2 + sin2x + 2x = t
d(x2 + sin2x + 2x) = dt
(2x + 2cos2x + 2) dx = dt
2(x + cos2x + 1) dx = dt
(x + cos2x + 1) dx = dt/2
Put all these values in equation(i), we get
= 1/2 ∫ dt/t
Integrate the above equation then, we get
= 1/2 log|t| + c
= 1/2 log|x2 + sin2x + 2x| + c
Hence, I = 1/2 log|x2 + sin2x + 2x| + c
### Question 28. Evaluate ∫ 1/ cos(x + a) cos(x + b) dx
Solution:
Let us assume I = ∫ 1/ cos(x + a) cos(x + b) dx
On multiplying and dividing the above equation by sin[(x + b) – (x + a)], we get
= 1/ sin(b – a) ∫ tan(x + b) dx – ∫ tan(x + a) dx
Integrate the above equation then, we get
= 1/ sin(b – a) [log(sec(x + b)) – log(sec(x + a))] + c
Hence, I = 1/ sin(b – a) [log(sec(x + b)/sec(x + a))] + c
### Question 29. Evaluate ∫ -sinx + 2cosx/(2sinx + cosx) dx
Solution:
Let us assume I = ∫ -sinx + 2cosx/(2sinx + cosx) dx ………..(i)
Let 2sinx + cosx = t
d(2sinx + cosx) = dt
(2cosx – sinx) dx = dt
(-sinx + 2cosx) dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log|t| + c
= log|2sinx + cosx| + c
Hence, I = log|2sinx + cosx| + c
### Question 30. Evaluate ∫ cos4x – cos2x/ (sin4x – sin2x) dx
Solution:
Let us assume I = ∫ cos4x – cos2x/ (sin4x – sin2x) dx
= – ∫ 2sin3x sinx / 2cos3x sinx dx
= – ∫ sin3x / cos3x dx ………..(i)
Let cos3x = t
d(cos3x) = dt
-3sin3x dx = dt
– sin3x dx = dt/3
Put all these values in equation(i), we get
= 1/3 ∫ dt/t
Integrate the above equation then, we get
= 1/3 log|t| + c
= 1/3 log|cos3x| + c
Hence, I = 1/3 log|cos3x| + c
### Question 31. Evaluate ∫ secx/ log(secx + tanx) dx
Solution:
Let us assume I = ∫ secx/ log(secx + tanx) dx ………..(i)
Let log(secx + tanx) = t
d(log(secx + tanx) = dt
secx dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |log(secx + tanx)| + c
Hence I = log |log(secx + tanx)| + c
### Question 32. Evaluate ∫ cosecx/ log|tanx/2| dx
Solution:
Let us assume I = ∫ cosecx/ log|tanx/2| dx ………..(i)
Let log|tanx/2| = t
d(log|tanx/2|) = dt
cosec x dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |log tanx/2| + c
Hence, I = log |log tanx/2| + c
### Question 33. Evaluate ∫ 1/ xlogxlog(logx) dx
Solution:
Let us assume I = ∫ 1/ xlogxlog(logx) dx ………..(i)
Put log(logx) = t
d(log(logx)) = dt
1/ xlogx dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |log(logx)| + c
Hence, I = log |log(logx)| + c
### Question 34. Evaluate ∫ cosec2 x/ 1 + cot x dx
Solution:
Let us assume I = ∫ cosec2 x/ 1+cot x dx ………..(i)
Put 1 + cotx = t then,
d(1 + cotx) = dt
– cosec2 x dx = dt
Put all these values in equation(i), we get
= – ∫ dt/t
Integrate the above equation then, we get
= – log |t| + c
= – log |1 + cotx| + c
Hence, I = – log |1 + cotx| + c
### Question 35. Evaluate ∫ 10x9 + 10x loge 10/ (10x + x10) dx
Solution:
Let us assume I= ∫ 10x9 + 10x loge 10/ (10x + x10)dx ………..(i)
Put 10x + x10 = t
d(10x + x10) = dt
(10x loge 10 + 10x9) dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |10x + x10| + c
Hence, I = log |10x + x10| + c
### Question 36. Evaluate ∫ 1 – sin2x/ x + cos2x dx
Solution:
Let us assume I = ∫ 1 – sin2x/ x + cos2x dx ………..(i)
Put x + cos2x = t
d(x + cos2x) = dt
(1 – 2sinxcosx) dx = dt
(1 – sin2x) dx = dt
Integrate the above equation then, we get
= ∫ dt/t
Integrate the above equ then, we get
= log |t| + c
= log |x + cos2x| + c
Hence, I = log |x + cos2x| + c
### Question 37. Evaluate ∫ 1 + tanx/ x + logsecx dx
Solution:
Let us assume I = ∫ 1 + tanx/ x + logsecx dx ………..(i)
Put x + logsecx = t
d(x+logsecx) = dt
(1 + tanx) dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |x + log secx| + c
Hence, I = log |x + log secx| + c
### Question 38. Evaluate ∫ sin2x/ a2 + b2sin2x dx
Solution:
Let us assume I = ∫ sin2x/ a2 + b2sin2x dx ………..(i)
Put a2 + b2sin2x = t
d(a2 + b2sin2x) = dt
b2(2sinxcosx) dx = dt
sin2x dx = dt/b2
Put all these values in equation(i), we get
= 1/b2 ∫ dt/t
Integrate the above equation then, we get
= 1/b2 log |t| + c
= 1/b2 log |a2 + b2sin2x| + c
Hence, I = 1/b2 log |a2 + b2sin2x| + c
### Question 39. Evaluate ∫ x + 1/ x(x + logx) dx
Solution:
Let us assume I = ∫ x + 1/ x(x + logx) dx ………..(i)
Put x + logx = t
d(x + logx) = dt
(1 + 1/x) dx = dt
(x +1)/ x dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |x + logx| + c
Hence, I = log |x + logx| + c
### Question 40. Evaluate
Solution:
Let us assume I = ………..(i)
Put 2 + 3sin-1x = t
d(2 + 3sin-1x) = dt
(3 × 1/ √(1 – x2)) dx = dt
(1/ √(1 – x2)) dx = dt/3
Put all these values in equation(i), we get
= 1/3 ∫ dt/t
Integrate the above equation then, we get
= 1/3 log |t| + c
= 1/3 log |2 + 3sin-1x| + c
Hence, I = 1/3 log |2 + 3sin-1x| + c
### Question 41. Evaluate ∫ sec2x/ tanx + 2 dx
Solution:
Let us assume I = ∫ sec2x/ tanx + 2 dx ………..(i)
Put tanx + 2 = t
d(tanx + 2) = dt
(sec2x) dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |tanx + 2| + c
Hence, I = log |tanx + 2| + c
### Question 42. Evaluate ∫ 2cos2x + sec2x/ sin2x + tanx – 5 dx
Solution:
Let us assume I = ∫ 2cos2x + sec2x/ sin2x + tanx – 5 dx ………..(i)
Put sin2x + tanx – 5 = t
d(sin2x + tanx – 5) = dt
(2cos2x + sec2x) dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |sin2x + tanx – 5| + c
Hence, I = log |sin2x + tanx – 5| + c
### Question 43. Evaluate ∫ sin2x/ sin5xsin3x dx
Solution:
Let us assume I = ∫ sin2x/ sin5xsin3x dx
= ∫ sin(5x – 3x)/ sin5xsin3x dx
= ∫ (sin5x cos3x – cos5x sin3x)/ sin5xsin3x dx [Using formula: sin(a-b) = sina cosb – cosa sinb]
= ∫ (sin5x cos3x)/ sin5xsin3x dx – ∫ (cos5x sin3x)/ sin5xsin3x dx
= ∫ cos3x/sin3x dx – ∫ cos5x/sin5x dx
= ∫ cot3x dx – ∫ cot5x dx
Integrate the above equation then, we get
= 1/3 log|sin3x| – 1/5 log|sin5x| + c
Hence, I = 1/3 log|sin3x| – 1/5 log|sin5x| + c
### Question 44. Evaluate ∫ 1 + cotx/ x + logsinx dx
Solution:
Let us assume I = ∫ 1 + cotx/ x + logsinx dx ………..(i)
Put x + logsinx = t
d(x + logsinx) = dt
(1 + cotx) dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |x + log sinx| + c
Hence, I = log |x + log sinx| + c
### Question 45. Evaluate ∫ 1/ √x (√x + 1) dx
Solution:
Let us assume I = ∫ 1/ √x (√x + 1) dx ………..(i)
Put √x + 1 = t
d(√x + 1) = dt
(1/2√x) dx = dt
(1/√x) dx = 2dt
Put all these values in equation(i), we get
= 2∫ dt/t
Integrate the above equation then, we get
= 2 log |t| + c
= 2 log |√x + 1| + c
Hence, I = 2 log |√x + 1| + c
### Question 46. Evaluate ∫ tan2x tan3x tan 5x dx
Solution:
Let us assume I = ∫ tan2x tan3x tan 5x dx ………..(i)
Now,
tan(5x) = tan(2x + 3x)
tan(5x) = tan2x + tan3x/ (1 – tan2x tan3x) [By using formula: tan(a + b) = tan a + tan b/ (1- tana tanb)]
tan(5x)(1 – tan2x tan3x) = tan2x + tan3x
(tan5x-tan2x tan3x tan5x) = tan2x + tan3x
tan2x tan3x tan5x = tan5x – tan2x – tan3x ………..(ii)
Using equation (i) and equation (ii), we get
= ∫ tan5x – tan2x – tan3x dx
Integrate the above equation then, we get
= 1/5 log|sec5x| – 1/2log|sec2x| -1/3log|sec3x| + c
Hence, I = 1/5 log|sec5x| – 1/2log|sec2x| – 1/3log|sec3x| + c
### Question 47. Evaluate ∫ {1 + tanx tan(x + θ)} dx
Solution:
Let us assume I = ∫ {1 + tanx tan(x + θ)} dx ………..(i)
As we know that,
tan(a – b) = tan a – tan b/ (1+ tana tanb)
tan(x + θ – x) = tan (x + θ) – tan x/ (1+ tan(x + θ) tanx)
tan θ = tan (x + θ) – tan x/ (1+ tan(x + θ) tanx)
(1+ tan(x + θ) tanx) = tan (x + θ) – tan x/tan θ ………..(ii)
By using equation (i) and (ii), we get
= ∫ tan (x + θ) – tan x/ tan θ dx
= 1/tan θ ∫ tan (x + θ) – tan x dx
Integrate the above equation then, we get
= 1/tan θ [-log|cos(x + θ)| + log |cosx|] + c
= 1/tan θ [log |cosx| – log|cos(x + θ)|] + c
Hence, I = 1/tan θ [log {cosx/ cos(x + θ)}] + c
### Question 48. Evaluate ∫ sin2x/ sin(x – π/6)sin(x + π/6) dx
Solution:
Let us assume I = ∫ sin2x/ sin(x – π/6)sin(x + π/6) dx ………..(i)
= ∫ sin2x/ sin2x – sin2π/6 dx
= ∫ sin2x/ sin2x – 1/4 dx
Put sin2x – 1/4 = t
d(sin2x – 1/4) = dt
(2sinx cosx) dx = dt
(sin2x) dx = dt
Put all these values in equation(i), we get
= ∫ dt/t
Integrate the above equation then, we get
= log |t| + c
= log |√x + 1| + c
Hence, I = log |sin2x – 1/4| + c
### Question 49. Evaluate ∫ ex-1 + xe-1/ ex + xe dx
Solution:
Let us assume I = ∫ ex-1 + xe-1/ ex + xe dx ………..(i)
= 1/e ∫ ex + exe-1/ ex + xe dx
= 1/e ∫ ex + exe-1/ ex + xe dx
Put ex + xe= t
d(ex + xe) = dt
(ex + exe-1) dx = dt
(ex + exe-1) dx = dt
Put all these values in equation(i), we get
= 1/e ∫ dt/t
Integrate the above equation then, we get
= 1/e log |t| + c
= 1/e log |ex + xe| + c
Hence, I = 1/e log |ex + xe| + c
### Question 50. Evaluate ∫ 1/sinx cos2x dx
Solution:
Let us assume I = ∫ 1/sinx cos2x dx
= ∫sin2x + cos2x/sinx cos2x dx
= ∫sin2x/sinx cos2x + cos2x/sinx cos2x dx
= ∫sinx/ cos2x + cosecx dx
= ∫secx tanx dx +∫ cosecx dx
Integrate the above equation then, we get
= sec x + log|tanx/2| + c
Hence, I = sec x + log|tanx/2| + c
### Question 51. Evaluate ∫ 1/cos3x – cosx dx
Solution:
Let us assume I = ∫ 1/cos3x – cosx dx
∫ 1/cos3x – cosx dx = ∫ sin2x + cos2x / 4cos3x – 4cosx dx
= ∫ sin2x + cos2x / 4cos(cos2x – 1) dx
= -1/4 ∫ sin2x/ sin2xcosx dx + ∫ cos2x / sin2xcosx dx
= -1/4 ∫ secx + cosecx cotx dx
Integrate the above equation then, we get
= -1/4 [log|secx + tanx| – cosecx] + c
Hence, I = 1/4 [cosecx + log|secx + tanx|] + c
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# Math4Team/Resources/Curriculum Chart
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
## Curriculum Framework, 4th Grade Maths
### Number Sense and Operations
Students engage in problem solving, communicating, reasoning, connecting, and representing as they:
Id Standard Description Who's working on it? What Sugar activity satisfies this objective?
4.N.1 Exhibit an understanding of the base ten number system by reading, modeling, writing, and interpreting whole numbers to at least 100,000; demonstrating an understanding of the values of the digits; and comparing and ordering the numbers. None None
4.N.2 Represent, order, and compare large numbers (to at least 100,000) using various forms, including expanded notation, e.g., 853 = 8 x 100 + 5 x 10 + 3. None None
4.N.3 Demonstrate an understanding of fractions as parts of unit wholes, as parts of a collection, and as locations on the number line. None None
4.N.4 Select, use, and explain models to relate common fractions and mixed numbers (1/2, 1/3, 1/4, 1/5, 1/6, 1/8, 1/10, 1/12, and 11/2), find equivalent fractions, mixed numbers, and decimals, and order fractions. None None
4.N.5 Identify and generate equivalent forms of common decimals and fractions less than one whole (halves, quarters, fifths, and tenths). None None
4.N.6 Exhibit an understanding of the base ten number system by reading, naming, and writing decimals between 0 and 1 up to the hundredths. None None
4.N.7 Recognize classes (in particular, odds, evens; factors or multiples of a given number; and squares) to which a number may belong, and identify the numbers in those classes. Use these in the solution of problems. None None
4.N.8 Select, use, and explain various meanings and models of multiplication and division of whole numbers. Understand and use the inverse relationship between the two operations. None None
4.N.9 Select, use, and explain the commutative, associative, and identity properties of operations on whole numbers in problem situations, e.g., 37 x 46 = 46 x 37, (5 x 7) x 2 = 5 x (7 x 2). None None
4.N.10 Select and use appropriate operations (addition, subtraction, multiplication, and division) to solve problems, including those involving money. None None
4.N.11 Know multiplication facts through 12 x 12 and related division facts. Use these facts to solve related multiplication problems and compute related problems, e.g., 3 x 5 is related to 30 x 50, 300 x 5, and 30 x 500. None None
4.N.12 Add and subtract (up to five-digit numbers) and multiply (up to three digits by two digits) accurately and efficiently. None None
4.N.13 Divide up to a three-digit whole number with a single-digit divisor (with or without remainders) accurately and efficiently. Interpret any remainders. None None
4.N.14 Demonstrate in the classroom an understanding of and the ability to use the conventional algorithms for addition and subtraction (up to five-digit numbers), and multiplication (up to three digits by two digits). None None
4.N.15 Demonstrate in the classroom an understanding of and the ability to use the conventional algorithm for division of up to a three-digit whole number with a single-digit divisor (with or without remainders). None None
4.N.16 Round whole numbers through 100,000 to the nearest 10, 100, 1000, 10,000, and 100,000. None None
4.N.17 Select and use a variety of strategies (e.g., front-end, rounding, and regrouping) to estimate quantities, measures, and the results of whole-number computations up to three-digit whole numbers and amounts of money to \$1000, and to judge the reasonableness of the answer. None None
4.N.18 Use concrete objects and visual models to add and subtract common fractions. None None
### Patterns, Relations, and Algebra
Students engage in problem solving, communicating, reasoning, connecting, and representing as they:
Id Standard Description Learning Tools Assessment Tools
4.P.1 Create, describe, extend, and explain symbolic (geometric) and numeric patterns, including multiplication patterns like 3, 30, 300, 3000, …. None None
4.P.2 Use symbol and letter variables (e.g. v, x) to represent unknowns or quantities that vary in expressions and in equations or inequalities (mathematical sentences that use =, <, >). None None
4.P.3 Determine values of variables in simple equations, e.g., 4106 – x = 37, 5 = y + 3, and s – y = 3. None None
4.P.4 Use pictures, models, tables, charts, graphs, words, number sentences, and mathematical notations to interpret mathematical relationships. None None
4.P.5 Solve problems involving proportional relationships, including unit pricing (e.g., four apples cost 80¢, so one apple costs 20¢) and map interpretation (e.g., one inch represents five miles, so two inches represent ten miles). None None
4.P.6 Determine how change in one variable relates to a change in a second variable, e.g., input-output tables. None None
### Geometry
Students engage in problem solving, communicating, reasoning, connecting, and representing as they:
Id Standard Description Learning Tools Assessment Tools
4.G.1 Compare and analyze attributes and other features (e.g., number of sides, faces, corners, right angles, diagonals, and symmetry) of two- and three-dimensional geometric shapes. None None
4.G.2 Describe, model, draw, compare, and classify two- and three-dimensional shapes, e.g., circles, polygons- especially triangles and quadrilaterals—cubes, spheres, and pyramids. None None
4.G.3 Recognize similar figures. None None
4.G.4 Identify angles as acute, right, or obtuse. None None
4.G.5 Describe and draw intersecting, parallel, and perpendicular lines. None None
4.G.6 Using ordered pairs of numbers and/or letters, graph, locate, identify points, and describe paths (first quadrant). None None
4.G.7 Describe and apply techniques such as reflections (flips), rotations (turns), and translations (slides) for determining if two shapes are congruent. None None
4.G.8 Identify and describe line symmetry in two-dimensional shapes. None None
4.G.9 Predict and validate the results of partitioning, folding, and combining two- and three-dimensional shapes. None None
### Measurement
Students engage in problem solving, communicating, reasoning, connecting, and representing as they:
Id Standard Description Learning Tools Assessment Tools
4.M.1 Demonstrate an understanding of such attributes as length, area, weight, and volume, and select the appropriate type of unit for measuring each attribute. None None
4.M.2 Carry out simple unit conversions within a system of measurement, e.g., hours to minutes, cents to dollars, yards to feet or inches, etc. None None
4.M.3 Identify time to the minute on analog and digital clocks using a.m. and p.m. Compute elapsed time using a clock (e.g., hours and minutes since…) and using a calendar (e.g., days since…). None None
4.M.4 Estimate and find area and perimeter of a rectangle, triangle, or irregular shape using diagrams, models, and grids or by measuring. None None
4.M.5 Identify and use appropriate metric and English units and tools (e.g., ruler, angle ruler, graduated cylinder, thermometer) to estimate, measure, and solve problems involving length, area, volume, weight, time, angle size, and temperature. None None
### Data Analysis, Statistics, and Probability
Students engage in problem solving, communicating, reasoning, connecting, and representing as they:
Id Standard Description Learning Tools Assessment Tools
4.D.1 Collect and organize data using observations, measurements, surveys, or experiments, and identify appropriate ways to display the data. None None
4.D.2 Match a representation of a data set such as lists, tables, or graphs (including circle graphs) with the actual set of data. None None
4.D.3 Construct, draw conclusions, and make predictions from various representations of data sets, including tables, bar graphs, pictographs, line graphs, line plots, and tallies. None None
4.D.4 Represent the possible outcomes for a simple probability situation, e.g., the probability of drawing a red marble from a bag containing three red marbles and four green marbles. None None
4.D.5 List and count the number of possible combinations of objects from three sets, e.g., how many different outfits can one make from a set of three shirts, a set of two skirts, and a set of two hats? None None
4.D.6 Classify outcomes as certain, likely, unlikely, or impossible by designing and conducting experiments using concrete objects such as counters, number cubes, spinners, or coins. None None
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Class 9 Maths
# Linear Equations
## Exercise 4.3 Part 2
Question 6: The work done by a body on application of a constant force is directly proportional to the distance travelled by the body. Express this in the form of an equation in two variables and draw the graph of the same by taking the constant force is 5 units. Read from the graph the work done when the distance traveled by the body is
(i) 2 units (ii) 0 units
Solution: Let x units be the distance traveled by the body and y units be the work done by constant force.
According to problem, y = 5x …………….equation (1)
Now , putting the value x = 0 in equation (1)
y = 5 xx 0 = 0
So the solution is (0, 0)
Putting the value x = 1 in equation (1)
y = 5 xx 1 = 5
So the solution is (1, 5)
Putting the value x = 2 in equation (1)
y = 5 xx 2 = 10
So the solution is (2, 10)
So, the table of the different solutions of the equation is
(i) When x = 2 units (distance)
Putting the value x in equation (1)
y = 5x
y = 5 xx 2 = 10.
Hence, Work done = 10.
(ii) When x = 0 units (distance)
Putting the value x in equation (1)
y = 5x
y = 5 xx 0 = 0
Hence, Work done = 0.
Question 7: Yamini and Fatima, two students of class IX of a school, together contributed Rs. 100 towards the Prime Minister’s Relief Fund, to help the earthquake victims. Write a linear equation which this data satisfies. (You may take their contributions as Rs. x and Rs. y). Draw the graph of the same.
Solution: Let the contribution of Yamini be x and that of Fatima be y.
According to problem, x + y = 100 …………..(1)
Now , putting the value x = 0 in equation (1)
0 + y = 100
y = 100. So the solution is (0, 100)
Putting the value x = 50 in equation (1)
50 + y = 100
y = 100 – 50
y = 50.
So the solution is (50, 50)
Putting the value x = 100 in equation (1)
100 + y = 100
Or, y = 100 – 100
y = 0
So the solution is (100, 0)
So, the table of the different solutions of the equation is
Question 8: In countries like the USA and Canada, temperature is measured in Fahrenheit, whereas in countries like India, it is measured in Celsius. Here is a linear equation that converts Fahrenheit to Celsius.
F=(9/5)C+32
(i) Draw the graph of the linear equation above using Celsius for x-axis and Fahrenheit for y-axis.
(ii) If the temperature is 30°C, what is the temperature in Fahrenheit?
(iii) If the temperature is 95°F, what is the temperature in Celsius?
(iv) If the temperature is 0°C, what is the temperature in Fahrenheit and if the temperature is 0°F, what is the temperature in Celsius?
(v) Is there a temperature which is numerically the same in both Fahrenheit and Celsius? If yes, find it.
Solution: Given equation is
F=(9/5)C+32
(i) We have to take Celsius along x-axis and Fahrenheit along y-axis
Let C be x and F be y
So, the equation will be
y=(9/5)x+32 -----equation (1)
Now, putting the value x=0 in equation (1)
Or, y=(9/5)xx0+32=0+32=32
So, the solution is (0, 32)
Putting the value x=5 in equation (1)
Or, y=(9/5)xx5+32=9+32=41
So, the solution is (5, 41)
Putting the value x=-5 in equation (1)
Or, y=(9/5)xx(-5)+32=-9+32=23
So, the solution is (-5, 23)
So, the table of the different solutions of the equation is
(ii) When C = 30°,
F=(9/5)xx30+32=54+32=86°
(iii) When F = 95°,
95=(9/5)xxC+32
Or,95-32=(9/5)C
Or, 63=(9/5)C
Or, C=63xx(5/9)=35
(iv) When C = 0,
F=(9/5)xx0+32=0+32=32
When x° F = x° C
F=(9/5)C+32
Or, x=(9/5)x+32
Or, x-(9/5)x=32
Or, (5x-9x)/5=32
Or, -4x=32xx5=160
Or, x=160/-4=-40°
## Exercise 4.4
Question 1: Give the geometric representations of y = 3 as an equation.
(i) in one variable (ii) in two variables.
Solution: (i) y = 3
(ii) y = 3
Or, 0x + y = 3
Or, 0x + y – 3 = 0
which is in fact y = 3
It is a line parallel to x-axis at a positive distance of 3 from it. We have two solution for it. i.e. (0, 3), (1, 3).
Question 2: Give the geometric representation of 2x + 9 = 0 as and equation,
(i) in one variable (ii) in two variables.
Answer: (i) 2x + 9 = 0
Or, 2x = -9Or, x = - 9/2 = - 4.5
(ii) Given equation is
2x + 9 = 0
2x + 0y + 9 = 0 {we know that it is actually 2x + 9 = 0}
Or, x = - 9/2 = - 4.5
It is line parallel to y-axis at a negative distance we have the two points lying it, the points are A(-4.5, 0), B(-4.5, 2).
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# Order of Operations What to do first?????
## Presentation on theme: "Order of Operations What to do first?????"— Presentation transcript:
Order of Operations What to do first?????
Presentation 2
Instructions for Tutorial
View the slide show and answer the questions as you go. Many slides will change as you continue to click while viewing them so it is important that you not skip around the slides. When you are ready for the next piece of information, “left click” your mouse or type “enter” on the keyboard.
What are operations? In mathematics, there are several operations. You are very familiar with the four basic ones: Adding Subtracting Multiplying Dividing
In an expression where there is more than one operation, the value of an expression will depend upon the order in which you perform these operations. There are several rules which we all agree to follow so that we will all reach the same answer for a problem. In this presentation, we will consider multiplication, division and grouping symbols and summarize the rules at the end.
Addition and Subtraction
Remember, First: we always perform operations within grouping symbols Second: expressions involving exponents need to be simplified before using them in calculations Then: when the problem involves ONLY addition and subtraction, you perform those operations as you come to them, reading from left to right.
Work these examples on your paper:
Remember, grouping symbols then addition OR subtraction as you come to it from left to right. 9 + 4 13 21 – 3 + 6 18 + 6 24
Consider: There are 3 operations here. Can you identify them?
They are: subtraction: 12 – 3(4) multiplication: 3 (4) addition: 3(4) + 1
Multiplication and Division
We no longer have only addition and subtraction. We must do any multiplication or division BEFORE we do addition or subtraction.
Try these. When you are finished, click on your answer choice.
Remember, First: grouping symbols and exponents as needed Second: multiplication OR division as you come to it from left to right Last: addition OR subtraction as you come to it from left to right.
OOPS! Do multiplication or division as you come to it reading from left to right. THEN go back and do any addition or subtraction as you come to it reading from left to right. Click here to try again
Try these. When you are finished, click on your answer choice.
Remember, First: grouping symbols and exponents as needed Second: multiplication OR division as you come to it from left to right Last: addition OR subtraction as you come to it from left to right.
When problems involve multiplication and division, we perform the first of either that we reach as we move from left to right—just like reading a book. Apply this rule by evaluating this expression. Click on your answer choice. 7 63
OOPS!! You did the multiplication first. Remember, we do multiplication OR division as we come to it from left to right. Here you reach the division first, so you do it first. Yes, “please excuse my dear aunt sally” says “my dear” or multiplication division. However, multiplication and division are same level operations. And Aunt Sally does not want us to do all the multiplication and then all the division (really, she doesn’t). Note: you should remember that dividing by a fraction is accomplished by multiplying by the reciprocal. Try Again!
Try again! Did you get 63 this time? Great!
When problems involve multiplication and division, we perform the first of either that we reach as we move from left to right—just like reading a book. Apply this rule by evaluating this expression. What is your answer? You should get 28,
Work these examples on your paper:
4(4) 16
Nice work! You performed the subtraction first since you encountered it first as you moved from left to right. Click here to go back to the show!
CONGRATULATIONS! Here’s what we have learned:
When a problem involves more than one operation, we use this order to simplify expressions: First, grouping symbols; and you must always simplify expressions with exponents before you can calculate with them. Next, multiplication OR division as you come to it reading from left to right. Last, addition OR subtraction as you come to it from left to right
Test Yourself Remember the order:
1) Multiplication or division from left to right 2) Addition or subtraction in order from left to right 9 39 4
Need to try again? Study these examples first.
Practice Problems Remember the order:
1) Multiplication or division from left to right 2) Addition or subtraction in order from left to right
Solutions:
REMEMBER: When a problem involves more than one operation, we use this order to simplify expressions: First, grouping symbols; and you must always simplify expressions with exponents before you can calculate with them. Next, multiplication OR division as you come to it reading from left to right. Last, addition OR subtraction as you come to it from left to right
How to remember???? The order is usually stated as follows:
Parentheses (Grouping Symbols) Exponents Multiplication OR Division Addition OR Subtraction There are two popular mnemonic devices to help you remember the order. Choose the one that works for you! PEMDAS Please Excuse My Dear Aunt Sally
Aunt Sally Straightened Out
Please (P stands for parentheses and other grouping symbols) Excuse (E stands for exponents) My Dear (M and D for multiplication and division as you reach them reading from left to right) Aunt Sally (A and S for addition and subtraction as you reach them reading from left to right)
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# F Ratio for Planned Comparisons
This lesson explains when and how to use an F ratio with analysis of variance to test statistical hypotheses represented by one or more planned comparisons.
Prerequisites: This lesson assumes familiarity with comparisons and orthogonal comparisons. You should be able to distinguish an ordinary comparison from an orthogonal comparison. You should know how to represent a statistical hypothesis mathematically by a comparison. You should be able to compute the sum of squares associated with a comparison. And you should understand how the probability of committing a Type I error is affected by the number of comparisons tested. If you don't know these things, review the following lessons:
• Comparison of Treatment Means. This lesson defines an ordinary comparison. It explains how to represent a statistical hypothesis mathematically by a comparison. And it explains how to compute the sum of squares for a comparison.
• Orthogonal Comparisons. This lesson explains how to distinguish an ordinary comparison from an orthogonal comparison.
• Multiple Comparisons. This lesson describes how the probability of committing a Type I error is affected by the number of comparisons tested.
## How to Compute an F Ratio
In statistics, F ratios abound. They are computed in many different ways for many different purposes. In this lesson, we want to compute an F ratio that can be used to test a statistical hypothesis represented by a planned comparison. For this purpose, an F ratio can be computed from the following formula:
F(1, v2) = SSi / MSWG
where F is the value of the F ratio, SSi is the sum of squares for comparison i, and MSWG is the within-groups mean square (from an ANOVA table). The numerator of the F ratio has one degree of freedom. The denominator of the F ratio has degrees of freedom (v2) equal to the degrees of freedom for the within-groups mean square.
Note how easy it is to compute this F ratio. You only need:
• The within-groups mean square, which is readily available from a standard ANOVA table.
• The sum of squares for a comparison, which is simple to calculate by hand. (Later in this lesson, we'll work through an example to demonstrate the hand computation.)
### What About a t Ratio?
Some textbooks use t ratios, rather than F ratios, to test the statistical significance of multiple comparisons. The two ratios are related according to the following formula:
t(v)2 = F(1, v)
The square of a t ratio with v degrees of freedom is equal to an F ratio with 1 and v degrees of freedom.
Each ratio leads to the same conclusion about statistical significance. If a t ratio is statistically significant, the corresponding F ratio will also be statistically significant; and vice versa.
## When to Use an F Ratio
In some situations, the F ratio is a good technique for testing the statistical significance of multiple comparisons. In other situations, it is not so good.
There are several things to like about the F ratio, including the following:
• The F ratio does a great job of controlling error rate per comparison. The experimenter sets the significance level for each individual comparison, and the F ratio assesses statistical significance accordingly.
• The F ratio is easy to compute. All you need to compute an F ratio is output from a standard ANOVA table and the sum of squares for the comparison.
For an experimenter who is most concerned with controlling error rate per comparison, the F ratio is a good choice.
There are several things to dislike about the F ratio, including the following:
• The F ratio does a poor job of controlling error rate familywise. The more hypotheses you test, the more likely it is that you will reject at least one hypothesis that should not be rejected.
• The F ratio is not a good choice for post hoc testing. With post hoc tests, the significance level of an uncorrected F ratio underestimates the true likelihood of committing a Type I error.
For an experimenter who is most concerned with controlling error rate familywise, the F ratio may be a poor choice. And for an experimenter who is interested in post hoc testing (data snooping), the F ratio may be a poor choice.
### What Do Statisticians Say?
Most statisticians agree that the F ratio is the wrong choice for post hoc testing. Other techniques (e.g., Scheffé's S method, Bonferroni correction) are preferred.
For planned tests of multiple comparisons, statisticians fall into one of three camps.
• Don't use an unadjusted F ratio. When an experiment calls for many hypothesis tests, error rate familywise can become unacceptably high. Folks in this camp favor methods that control error rate familywise (e.g., Bonferroni's correction).
• Use an unadjusted F ratio only when a small number of comparisons will be tested. This strategy allows the experimenter to control error rate per comparison, while holding error rate familywise to a level that the experimenter considers acceptably low.
• Use an unadjusted F ratio only when a small number of orthogonal comparisons will be tested. This strategy recognizes the fact that orthogonal comparisons use nonoverlapping data for hypothesis tests, almost like conducting a separate experiment to test each comparison. If you did conduct a separate experiment for each comparison, the error rate per comparison would equal the error rate familywise for that experiment.
In the end, the analysis method you choose will reflect your tolerance for controlling error rate per comparison versus error rate familywise. On this website, we will use an unadjusted F ratio only when the experiment calls for testing a small number of orthogonal comparisons.
## A Step-By-Step Example
In this section, we'll work through an example to demonstrate how to use an F ratio to test the statistical significance of planned, orthogonal comparisons.
### Experimental Design
To test the long-term effect of aerobic exercise on resting pulse rate, an investigator conducts a controlled experiment. The experiment uses a completely randomized design, consisting of three treatment groups:
• Control. Subjects do not participate in an exercise program.
• Low-effort. Subjects jog 1 mile on Monday, Wednesday, and Friday.
• High-effort. Subjects jog 2 miles every day, except Sunday.
Five subjects are randomly assigned to each group; and, after 28 days of treament, their resting pulse rate is measured on day 29.
### Analysis Plan
Before collecting any data, the investigator poses the research questions to be answered, states statistical hypotheses implied by each research question, and identifies the analytical technique used to test each statistical hypothesis.
#### Research Questions
For this experiment, the researcher is interested in three research questions. Those questions, the associated statistical hypotheses, and the associated comparisons appear below:
• Overall research question. Will mean pulse rate in one treatment group differ from mean pulse rate in any other treatment group?
H0: μi = μj
H1: μi ≠ μj
• Follow-up question 1. Will mean pulse rate of subjects in the control group (Group 1) differ from the mean pulse rate of subjects in treatment groups (Group 2 and Group 3)?
H0: μ1 = (μ2 + μ3) / 2
H1: μ1 ≠ (μ2 + μ3) / 2
This statistical hypothesis can be represented mathematically by the comparison L1:
L1 = X1 - 0.5X2 - 0.5X3
where X1, X2, and X3 are means scores for Groups 1, 2, and 3, respectively.
• Follow-up question 2. Will mean pulse rate of subjects in the low-effort group (Group 2) differ from the mean pulse rate of subjects in the high-effort group (Group 3)?
H0: μ2 = μ3
H1: μ2 ≠ μ3
This statistical hypothesis can be represented mathematically by comparison L2:
L2 = X2 - X3
where X2 and X3 are means scores for Groups 2 and 3, respectively.
#### Analytical Techniques
The overall research question asks whether the mean pulse rate in one treatment group differs from the mean pulse rate in any other group. The null hypothesis implied by this research question can be tested by an omnibus analysis of variance.
The remaining questions are follow-up questions. To determine whether to reject the null hypothesis for a follow-up question, we test its associated comparison for statistical significance. Here are the comparisons we will be testing:
• L1 = X1 - 0.5X2 - 0.5X3
• L2 = X2 - X3
Notice that these comparisons are orthogonal. Because we are testing a small number of orthogonal comparisons (only two comparisons), we can use an F ratio to test the significance of each comparison.
For this example, assume that the investigator specifies a significance level of 0.05 to test the statistical significance of hypotheses associated each research question.
### Experimental Data
Pulse rate measurements for each subject in each treatment group appear below:
Table 1. Pulse Rate for Each Subject in Each Group
Group 1 (control) Group 2 (low effort) Group 3 (high effort)
80 70 50
85 75 60
90 80 70
95 85 80
100 90 90
### ANOVA Results
The overall research question is: Will mean pulse rate in one treatment group differ from mean pulse rate in any other treatment group? The statistical hypotheses implied by that question are:
H0: μi = μj
H1: μi ≠ μj
We can test this null hypothesis with a standard, omnibus analysis of variance. Here is the ANOVA table from that analysis.
Table 2. ANOVA Summary Table
Source SS df MS F P
BG 1000 2 500 4.0 0.046
WG 1500 12 125
Total 2500 14
The P value for the between-groups (BG) effect is 0.046, which is less that the significance level of 0.05. Therefore, we reject the null hypothesis of no difference in pulse rates between treatment groups.
Note: We explained how to conduct a one-way analysis of variance in previous lessons. If you're wondering how to produce the ANOVA table shown above, see One-Way Analysis of Variance: Example or One-Way Analysis of Variance With Excel.
### Follow-up Tests
For this experiment, the investigator planned to conduct two follow-up tests to supplement the omnibus analysis of variance. In case you've forgotten, here are the two follow-up questions:
• Follow-up question 1. Will mean pulse rate of subjects in the control group (Group 1) differ from the mean pulse rate of subjects in treatment groups (Group 2 and Group 3)?
• Follow-up question 2. Will mean pulse rate of subjects in the low-effort group (Group 2) differ from the mean pulse rate of subjects in the high-effort group (Group 3)?
Each of these questions can be addressed by testing the statistical significance of a particular comparison. To illustrate the process, we'll work though a step-by-step analysis for the first follow-up question.
#### Step 1. Compute Mean Scores
Mean pulse rate within each group (computed from raw scores in Table 1) appears below:
Table 3. Mean Pulse Rate in Each Treatment Group
Group 1 (control) Group 2 (low effort) Group 3 (high effort)
90 80 70
#### Step 2. Define a Comparison
Next, we define a comparison that represents our research question. For the first follow-up question, we want to compare the mean score in Group 1 with the mean of scores in Groups 2 and 3. Therefore, this is the comparison we need to use:
L1 = X1 - 0.5X2 - 0.5X3
L1 = 90 - 0.5*80 - 0.5*70= 15
where L1 is the value of the comparison, X1 is the mean score in Group 1, X2 is the mean score in Group 2, and X3 is the mean score in Group 3.
#### Step 3. Compute Sum of Squares
With a balanced design, the sum of squares for a given comparison ( Li ) can be computed from the following formula:
SSi = n * Li2 / Σ c2ij
where SSi is the sum of squares for comparison Li , Li is the value of the comparison, n is the sample size in each group, and cij is the coefficient (weight) for level j in the formula for comparison Li.
Plugging values from our sample problem into the formula, we get:
SS1 = 5 * 152 / [ (1)2 + (-.5)2 + (-.5)2]
SS1 = 1125 / 1.5 = 750
#### Step 4. Produce ANOVA Summary Table
The summary table from an omnibus analysis of variance includes two outputs that we can use to test the statistical significance of a comparison. Those outputs are (1) the value of the within-groups mean square and (2) the degrees of freedom for the within-groups mean square.
We generated the ANOVA summary table earlier. For convenience, here it is again.
Table 2. ANOVA Summary Table
Source SS df MS F P
BG 1000 2 500 4.0 0.046
WG 1500 12 125
Total 2500 14
#### Step 5. Find the F Ratio
The F ratio for a comparison equals its sum of squares divided by the within-groups mean square (from the ANOVA table).
F(1, v2) = SSi / MSWG
where F is the value of the F ratio, SSi is the sum of squares for comparison i, and MSWG is the within-groups mean square. The numerator of any F ratio for a comparison has one degree of freedom. The degrees of freedom (v2) for the denominator equal the degrees of freedom (from the ANOVA table) for the within-groups mean square.
For this problem, the F ratio is:
F(1, 12) = 750 / 125 = 6.0
#### Step 6. Find the P-Value
With a planned comparison, the F ratio is probability that an F statistic would be more extreme (i.e., bigger) than the actual F ratio computed from experimental data.
We can use Stat Trek's F Distribution Calculator to find the probability that an F statistic will be bigger than the actual F ratio observed in the experiment. Enter the numerator degrees of freedom (1), the denominator degrees of freedom (12), and the observed F ratio (6.0) into the calculator; then, click the Calculate button.
From the calculator, we see that P( F < 6.0 ) equals 0.97; so the P ( F > 6.0 ) equals 1 minus 0.97 or 0.03. Therefore, the P-Value is 0.03.
#### Step 7. Test the Hypothesis
If the P-value for a comparison is less than the significance level, we reject the associated hypothesis. Otherwise, we fail to reject.
In this example, the P-value (0.03) is less than the significance level (0.05). Therefore, we reject the null hypothesis that the mean score in Group 1 is equal to the mean score in Groups 2 and 3.
### What About the Other Follow-up Test?
As part of this experiment, the investigator planned to conduct two follow-up tests to supplement the omnibus analysis of variance. In the previous section, we described a seven-step process for conducting the first follow-up test. You would use the same seven-step process to conduct the second follow-up test.
### ANOVA Summary Table
Follow-up tests are often reported in an enhanced ANOVA summary table. The enhanced table shows all of the results from a standard ANOVA summary table. In addition, it shows results (sum of squares, mean square, degrees of freedom, F ratio, and P-value) for each planned comparison.
Here is the enhanced ANOVA summary table for the present experiment.
Table 4. Enhanced ANOVA Summary Table
Source SS df MS F P
BG 1000 2 500 4.0 0.046
L1 750 1 750 6.0 0.03
L2 250 1 250 2.0 0.18
WG 1500 12 125
In this example, the between groups effect (BG) is statistically significant (p=0.046), indicating that the mean pulse rate in at least one group is significantly different from the mean pulse rate in another group. The comparison effects (L1 and L2) show results for follow-up tests. Only the L1 effect is statistically significant (p=0.03), indicating that the mean pulse rate in the control group is significantly different from the mean pulse in the two treatment groups. The mean pulse rate in the low-effort group (Group 2) is not significantly different from the mean pulse rate in the high-effort group (Group 3). Based on these findings, the investigator concludes that one or both of the treatments affect resting pulse rate.
Note: The mean square for a comparison is computed just like the mean square for any other treatment effect:
MS = SS / df
where MS is the mean square, SS is the sum of squares, and df is the degrees of freedom.
The degrees of freedom for every comparison is equal to one. Therefore, the mean square for a comparison equals the sum of squares for the comparison.
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Problem 752
# Find the equation of a parabola that has vertex at $$(-1,2)$$, axis of symmetry parallel to the $$\mathrm{x}$$ -axis, and goes through the point $$\mathrm{P}_{1}(-3,-4)$$.
Expert verified
The equation of the parabola is $$y = -\frac{3}{2}(x+1)^2 +2$$.
See the step by step solution
## Step 1: Write the vertex form of the parabola equation using the given vertex coordinates.
We are given the vertex coordinates (-1, 2). So, the vertex form of the parabola equation will be: $$y = a(x-(-1))^2+2$$ Simplify the equation: $$y = a(x+1)^2 +2$$
## Step 2: Use the point P1 to find the value of 'a'.
We know the parabola goes through the point P1(-3, -4). We can substitute these coordinates into the equation we found in Step 1 and solve for "a": $$-4 = a(-3+1)^2 +2$$
## Step 3: Solve for 'a'.
Simplify the equation and solve for "a": $$-4 = a(-2)^2 +2$$ $$-4 = 4a +2$$ Subtract 2 from both sides: $$-6 = 4a$$ Now, divide by 4: $$a = -\frac{3}{2}$$
## Step 4: Write the equation of the parabola.
Now that we have the value of "a", we can write the final equation of the parabola: $$y = -\frac{3}{2}(x+1)^2 +2$$ This is the equation of the parabola with vertex at (-1, 2), axis of symmetry parallel to the x-axis, and going through the point P1(-3, -4).
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# Solve the following equation: - $\dfrac{3y}{2}+\dfrac{y+4}{4}=5-\dfrac{y-2}{4}$.
Last updated date: 20th Jun 2024
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Hint: Multiply both sides of the equation with 4 to get rid of the fractional terms. Now, take the terms containing ‘y’ to the left – hand side. Apply simple addition and subtraction to simplify the terms. Find the value of ‘y’ to get the answer.
We have been provided with the equation: - $\dfrac{3y}{2}+\dfrac{y+4}{4}=5-\dfrac{y-2}{4}$. We have to solve this equation, that means we have to find the value of y.
\begin{align} & \Rightarrow \left( 2\times 3y \right)+\left( y+4 \right)=5\times 4-\left( y-2 \right) \\ & \Rightarrow 6y+y+4=20-y+2 \\ & \Rightarrow 7y+4=22-y \\ \end{align}
\begin{align} & \Rightarrow 7y+y=22-4 \\ & \Rightarrow 8y=18 \\ & \Rightarrow y=\dfrac{18}{8} \\ \end{align}
$\Rightarrow y=\dfrac{9}{4}$
Hence, the value of y is $\dfrac{9}{4}$.
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# 2018 AMC 12B Problems/Problem 3
## Problem
A line with slope $2$ intersects a line with slope $6$ at the point $(40,30)$. What is the distance between the $x$-intercepts of these two lines?
$\textbf{(A) } 5 \qquad \textbf{(B) } 10 \qquad \textbf{(C) } 20 \qquad \textbf{(D) } 25 \qquad \textbf{(E) } 50$
## Solution 1
Using point slope form, we get the equations $y-30 = 6(x-40)$ and $y-30 = 2(x-40)$. Simplifying, we get $6x-y=210$ and $2x-y=50$. Letting $y=0$ in both equations and solving for $x$ gives the $x$-intercepts: $x=35$ and $x=25$, respectively. Thus the distance between them is $35-25=\boxed{\textbf{(B) } 10}$.
## Solution 2
In order for the line with slope $2$ to travel "up" $30$ units (from $y=0$), it must have traveled $30/2=15$ units to the right. Thus, the $x$-intercept is at $x=40-15=25$. As for the line with slope $6$, in order for it to travel "up" $30$ units it must have traveled $30/6=5$ units to the right. Thus its $x$-intercept is at $x=40-5=35$. Then the distance between them is $35-25=\boxed{\textbf{(B) } 10}$.
## Video Solution (HOW TO THINK CRITICALLY!!!)
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### Free Educational Resources
3 Tutorials that teach Properties of Addition and Multiplication
# Properties of Addition and Multiplication
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This lesson provides an overview of the properties of addition and multiplication.
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Tutorial
Identity Properties
The identity property of addition states that when zero is added to any number, the value does not change. Below is an example of this property using real numbers:
7 + 0 = 7
In other words, the identity property of addition tells us that adding zero does not change the value of a number. Generally, we can express this as:
A similar property applies to multiplication. What quantity, when multiplied to a number, does not change its value? When any number is multiplied by 1, the value does not change. Let's take a look at a numerical example of the identity property of multiplication:
4 • 1 = 4
The identity property of multiplication states that any number multiplied by 1 does not change in value. Generally, we can express this as:
Inverse Properties
The inverse property of addition states that any number and its opposite sum to zero. We can refer to the opposite of a number as its additive inverse. A numerical example is illustrated below:
9 + (–9) = 0
9 and –9 are opposites of each other. This means that their magnitudes are the same (9), but their signs are different; one is a positive number, while the other is a negative number. The sum of a number and its opposite is zero. We can write this generally as:
The inverse property of multiplication states that a number and its reciprocal multiply to 1. In the same way that a number and its opposite are additive inverses, a number and its reciprocal are multiplicative inverses. The reciprocal of a number can be found by creating a fraction, and flipping the numerator and denominator. Below is a numerical example of the inverse property of multiplication:
9 and 1/9 are multiplicative inverses, or reciprocals, of one another. The inverse property of multiplication dictates that the product of a number and its reciprocal is equal to 1. Here is our general rule:
Commutative Property of Addition and Multiplication
Addition and multiplication is commutative. This means that we can add in any order we wish, and we can multiply in any order we wish. It is important to note that we cannot mix addition and multiplication. These are separate properties, but they behave the same with both operations.
For example, 2 + 3 is the same as 3 + 2, because addition is commutative. It does not matter in which order you add, the sum will be 5 in either case. The same property applies to multiplication as well: 3 • 4 = 4 • 3. It does not matter in which order you multiply, because multiplication is commutative. The result is 12 in both cases.
Commutative Property: A property of addition that allows terms to be added in any order; a property of multiplication that allows factors to be multiplied in any order.
Associative Properties of Addition and Multiplication
The associative property allows us to group terms for addition and multiplication in any way we wish. As with the commutative properties of addition and multiplication, the associative property applies to addition and multiplication separately. Here are some numerical examples of the associative property:
(3 + 4) + 6 = 3 + (4 + 6)
4 • (2 • 8) = (4 • 2) • 8
The associative property allows us to group addends or group factors in different ways. This is particularly helpful in mental math, where we might easily recognize that 4 + 6 is 10. In such cases, regrouping can help us recognize certain sums or products to make mental math easier.
Associative Property: A property of addition that allows terms to be grouped in any order; a property of multiplication that allows factors to be grouped in any order.
Distributive Property
The distributive property applies multiplication over addition. This property is applied when we have a factor being multiplied by a sum. It got its name from the process of distributing the outside factor into each added of the sum. Let's take a look at a numerical example:
2 ( 4 + 3 ) Distribute 2 into 4 and 3
(2 • 4) + (2 • 3) Multiply inside the parentheses
14 Our solution
Distributive Property: A property of multiplication that states that a sum multiplied by a factor can be expressed as a sum of the products of each individual addend and that factor.
The distributive property is especially useful when working with variables. Realistically, an easier approach with numerical examples is to evaluate what is inside the parentheses first, and then multiply the outside factor. (This also follows the Order of Operations). However, as we work with algebraic expressions containing variables, the distributive property is going to be very helpful.
Terms to Know
Associative Property
A property of addition that allows terms to be grouped in any order; a property of multiplication that allows factors to be grouped in any order.
Commutative Property
A property of addition that allows terms to be added in any order; a property of multiplication that allows factors to be multiplied in any order.
Distributive Property
A property of multiplication that states that a sum multiplied by a factor can be expressed as a sum of the products of each original addend and that factor.
Formulas to Know
Associative Property of Multiplication
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# Subtraction of 4-Digit Numbers
We will learn about the subtraction of 4-digit numbers (without borrowing and with borrowing).
We know when one number is subtracted from another number the result obtained is called the difference. The greater number is called the minuend and the smaller number is called the subtrahend. We use subtraction when we take away, compare, to find out what does not belong to a group.
The rules of subtraction remain the same as for the number having less digits. We first write the greater number at the top. Then subtract the digits column-wise ones from ones, tens from tens, hundreds from hundreds, thousands from thousands and so on.
I. Subtraction of 4-Digit Numbers (without borrowing):
Consider some of the following examples.
1. Subtract 3276 from 5896.
Solution:
Th H T O 5 8 9 6 - 3 2 7 6 0 Subtract ones i.e., 6 - 6 = 0 and write down 0 in ones column.
Th H T O 5 8 9 6 - 3 2 7 6 2 0 Subtract tens i.e., 9 - 7 = 2 and write down 2 in tens column.
Th H T O 5 8 9 6 - 3 2 7 6 6 2 0 Subtract hundreds i.e., 8 - 2 = 6 and write down 6 in hundreds column.
Th H T O 5 8 9 6 - 3 2 7 6 2 6 2 0 Subtract thousands i.e., 5 - 3 = 2 and write down 2 in thousands column.
So the difference between 5896 and 3276 is 2620.
3. Subtract 2435 from 6769.
Solution:
Step I: Arrange the number in coloumn of thousands, hundreds, tens and ones.
Step II: Subtract ones.
Step III: Subtract tens.
Step IV: Subtract hundreds.
Step V: Subtract thousands.
4. Subtract 3573 from 7998.
Solution:
Th H T O
7 9 9 8
- 3 5 7 3
4 4 2 5
II. Subtraction of 4-Digit Numbers (with borrowing):
Subtraction with Regrouping:
Regrouping From Tens to Ones Place -
1. Subtract 4539 from 7968.
Th H T O 5 18 7 9 6 8 - 4 5 3 9 9 Subtracting Ones:8 - 9 is not possible Regrouping 1 ten as 10 onesBorrow 1 ten make 18Then 18 - 9 = 9Write 9 under O.
Th H T O 5 18 7 9 6 8 - 4 5 3 9 3 4 2 9 Subtracting Tens:Instead of 6 tens, we have 5 tens write 5 in tens place instead of 6 5 - 3 = 2Write 2 under T.
III. Subtraction of 4-Digit Numbers (with borrowing):
Subtraction with Regrouping:
Regrouping From Tens and Hundreds Place -
1. Subtract 6289 from 8675.
Th H T O 6 15 8 6 7 5 - 6 2 8 9 6 Subtracting Ones:5 - 9 is not possible Regrouping 1 ten as 10 onesBorrow 1 ten make 15Then 15 - 9 = 6Write 9 under O.
Th H T O 16 5 6 15 8 6 7 5 - 6 2 8 9 8 6 Subtracting Tens:Instead of 7 tens, we have 6 tens6 - 8 is not possible.Regrouping 1 hundred as 10 tens borrow 1 hundred make 16 Then 16 - 8 = 8Write 8 under T.
Th H T O 16 5 6 15 8 6 7 5 - 6 2 8 9 3 8 6 Subtracting Hundreds:Instead of 6 hundreds, we have 5 hundredsWrite 5 in hundreds place, instead of 6Then 5 - 2 = 3Write 3 under H.
Th H T O 5 16 5 6 15 8 6 7 5 - 6 2 8 9 2 3 8 6 Subtracting Thousands: 8 - 6 = 2Write 2 under Th.
IV. Subtraction of 4-Digit Numbers (with borrowing):
Subtraction with Regrouping:
Regrouping From Tens, Hundreds and Thousands Place -
1. Subtract 3837 from 6825.
Th H T O 1 15 6 8 2 5 - 3 8 3 7 8 Subtracting Ones:5 - 7 is not possible Regrouping 1 ten as 10 onesBorrow 1 ten make 15Then 15 - 7 = 8Write 8 under O.
Th H T O 11 7 1 15 6 8 2 5 - 3 8 3 7 8 8 Subtracting Tens:Instead of 2 tens, we have 1 tens1 - 3 is not possible.Regrouping 1 hundred as 10 tens borrow 1 hundred make 11 Then 11 - 3 = 8Write 8 under T.
Th H T O 17 11 5 7 1 15 6 8 2 5 - 3 8 3 7 9 8 8 Subtracting Hundreds:Instead of 8 hundreds, we have 7 hundreds7 - 8 is not possible.Regroup 1 thousand as 10 hundredBorrow 1 thousand make 17Then 17 - 8 = 9Write 9 under H.
Th H T O 17 11 5 7 1 15 6 8 2 5 - 3 8 3 7 2 9 8 8 Subtracting Thousands:Instead of 6 thousands, we have 5 thousands 5 - 3 = 2Write 2 under Th.
Consider the following example.
2. Subtract 7864 from 9643.
Solution:
Th H T O 3 13 9 6 4 3 - 7 8 6 4 9 Subtract Ones: We cannot subtract 4 ones from 3 ones. So, we borrow 1 ten (1 ten and 3 ones become 13 ones). Now 13 ones - 4 ones = 9 ones. Write 9 in ones place.
Th H T O 5 13 9 6 4 3 - 7 8 6 4 7 9 Subtract Tens: We cannot subtract 6 tens from 3 tens. So, we again borrow 1 hundred (1 hundred and 3 tens become 13 tens). Now 13 tens - 6 tens = 7 tens. Write 7 in tens place.
Th H T O 8 15 9 6 4 3 - 7 8 6 4 7 7 9 Subtract Hundreds: We cannot subtract 8 hundreds from 5 hundreds. So, we again borrow 1 thousand (1 thousand and 5 hundreds become 15 hundreds). Now 15 hundreds - 8 hundreds = 7 hundreds. Write 7 in hundreds place.
Th H T O 8 9 6 4 3 - 7 8 6 4 1 7 7 9 Subtract Thousands: We can subtract 7 thousands from 8 thousands i.e., 8 thousands - 7 thousands = 1 thousands. Write 1 in thousands place.
So, the difference between 9643 and 7864 is 1779.
Questions and Answers on Subtraction of 4-Digit Numbers:
1. Solve the following of 4-digit Subtraction (Without Borrowing):
(i) Th H T O 3 3 8 9 - 2 4 3 _________________ (ii) Th H T O 8 5 6 7 - 3 4 5 _________________
(iii) Th H T O 6 5 7 4 - 1 5 2 _________________ (iv) Th H T O 5 8 9 9 - 6 7 6 _________________
(v) Th H T O 4 4 5 2 - 1 3 1 _________________ (vi) Th H T O 8 7 4 6 - 5 5 3 4 _________________
(vii) Th H T O 7 6 5 6 - 3 4 2 4 _________________ (viii) Th H T O 5 8 8 6 - 2 5 5 3 _________________
(ix) Th H T O 9 7 9 8 - 5 5 7 6 _________________ (x) Th H T O 8 6 5 4 - 4 6 5 4 _________________
(xi) Th H T O 7 7 7 7 - 5 4 3 2 _________________ (xii) Th H T O 6 9 7 6 - 3 6 4 1 _________________
(xiii) Th H T O 3 8 7 9 - 1 5 6 8 _________________ (xiv) Th H T O 4 4 4 4 - 2 1 3 2 _________________
(xv) Th H T O 9 9 8 8 - 5 6 3 4 _________________ (xvi) Th H T O 4 5 7 9 - 2 3 1 5 _________________
(xvii) Th H T O 5 6 9 2 - 3 4 8 1 _________________ (xviii) Th H T O 6 8 3 2 - 2 2 0 1 _________________
(xix) Th H T O 7 5 6 4 - 3 1 2 1 _________________ (xx) Th H T O 9 9 6 5 - 8 6 1 3 _________________
(xxi) Th H T O 7 4 3 7 - 5 4 3 6 _________________ (xxii) Th H T O 2 1 4 1 - 1 0 2 0 _________________
(xxiii) Th H T O 8 7 6 5 - 4 4 2 3 _________________ (xxiv) Th H T O 9 8 7 6 - 8 7 6 5 _________________
(xxv) Th H T O 5 6 7 8 - 2 3 4 5 _________________
2. Solve the following of 4-digit Subtraction (Without Borrowing):
(i) Th H T O 9 9 8 6 - 8 6 5 _________________ (ii) Th H T O 8 6 8 7 - 3 2 4 _________________
(iii) Th H T O 7 8 4 6 - 5 2 3 _________________ (iv) Th H T O 6 5 7 6 - 4 3 3 _________________
(v) Th H T O 5 7 9 8 - 5 2 3 _________________ (vi) Th H T O 8 7 8 6 - 5 2 5 2 _________________
(vii) Th H T O 2 4 6 8 - 1 2 3 5 _________________ (viii) Th H T O 6 8 5 4 - 3 4 2 2 _________________
(ix) Th H T O 7 9 6 4 - 4 5 1 2 _________________ (x) Th H T O 3 4 7 5 - 1 1 6 4 _________________
(xi) Th H T O 5 7 7 9 - 5 5 4 4 _________________ (xii) Th H T O 9 5 8 9 - 4 2 4 3 _________________
(xiii) Th H T O 4 2 9 4 - 3 1 6 3 _________________ (xiv) Th H T O 8 7 6 6 - 4 3 4 1 _________________
(xv) Th H T O 7 5 8 7 - 6 3 4 5 _________________ (xvi) Th H T O 4 2 2 3 - 2 1 1 2 _________________
(xvii) Th H T O 5 8 2 9 - 4 7 1 5 _________________ (xviii) Th H T O 9 9 9 9 - 7 7 7 7 _________________
(xix) Th H T O 6 6 9 7 - 4 5 7 5 _________________ (xx) Th H T O 8 9 9 8 - 5 7 7 4 _________________
3. Do these following of 4-Digit Subtraction (Without Borrowing) in your Notebook:
(i) 4679 - 1256
(ii) 2942 - 1621
(iii) 2687 - 432
(iv) 8975 - 853
(v) 3387 - 1264
(vi) 3778 - 2764
(vii) 6799 - 5678
(viii) 7832 - 3411
(ix) 6876 - 3354
(x) 9797 - 8546
(xi) 7000 - 1000
(xii) 9569 - 5217
(xiii) 5575 - 3343
(xiv) 4989 - 2545
(xv) 3957 - 1723
(xvi) 7586 - 6312
(xvii) 8657 - 1435
(xviii) 5569 - 2556
(xix) 9649 - 4535
(xx) 3456 - 1243
(xxi) 2685 - 1173
(xxii) 5867 - 4334
(xxiii) 4735 - 2312
(xiv) 6987 - 4546
4. Subtract the following Regrouping from Tens to Ones Place:
(i) Th H T O 1 3 4 7 - 2 9 _________________ (ii) Th H T O 4 7 5 5 - 3 6 _________________
(iii) Th H T O 8 7 6 4 - 3 7 _________________ (iv) Th H T O 1 6 7 5 - 4 8 _________________
(v) Th H T O 3 4 6 2 - 1 6 _________________ (vi) Th H T O 2 6 7 5 - 2 3 7 _________________
(vii) Th H T O 7 8 8 7 - 4 6 8 _________________ (viii) Th H T O 6 5 8 1 - 4 5 9 _________________
(ix) Th H T O 3 7 8 5 - 3 6 7 _________________ (x) Th H T O 4 6 5 4 - 1 2 6 _________________
(xi) Th H T O 7 8 5 4 - 2 4 3 7 _________________ (xii) Th H T O 9 5 6 1 - 7 3 1 7 _________________
(xiii) Th H T O 8 8 7 1 - 5 4 5 6 _________________ (xiv) Th H T O 5 3 2 6 - 2 1 1 7 _________________
(xv) Th H T O 6 7 8 3 - 3 5 6 7 _________________ (xvi) Th H T O 3 6 7 4 - 1 2 5 9 _________________
(xvii) Th H T O 4 5 8 3 - 1 2 1 6 _________________ (xviii) Th H T O 5 9 8 2 - 4 5 4 6 _________________
(xix) Th H T O 6 9 8 3 - 3 7 2 9 _________________ (xx) Th H T O 7 5 6 3 - 6 4 2 8 _________________
(xxi) Th H T O 9 8 7 6 - 7 6 5 9 _________________ (xxii) Th H T O 8 9 7 4 - 6 5 2 7 _________________
(xxiii) Th H T O 7 7 7 7 - 1 2 3 9 _________________ (xxiv) Th H T O 6 8 9 4 - 4 6 5 7 _________________
(xxv) Th H T O 5 4 8 2 - 3 2 6 3 _________________
5. Subtract the following Regrouping from Tens to Ones Place:
(i) Th H T O 3 7 8 5 - 4 7 _________________ (ii) Th H T O 7 6 4 6 - 2 7 _________________
(iii) Th H T O 4 5 8 3 - 4 8 _________________ (iv) Th H T O 6 3 5 2 - 3 5 _________________
(v) Th H T O 5 2 7 4 - 5 6 _________________ (vi) Th H T O 2 3 7 6 - 1 5 8 _________________
(vii) Th H T O 5 6 7 3 - 5 5 7 _________________ (viii) Th H T O 3 7 7 1 - 5 5 4 _________________
(ix) Th H T O 8 8 9 2 - 6 7 8 _________________ (x) Th H T O 9 9 6 5 - 8 3 6 _________________
(xi) Th H T O 6 6 9 2 - 2 3 7 5 _________________ (xii) Th H T O 9 8 7 6 - 5 4 3 7 _________________
(xiii) Th H T O 8 7 6 4 - 5 5 3 7 _________________ (xiv) Th H T O 7 7 8 7 - 4 5 6 8 _________________
(xv) Th H T O 4 7 5 3 - 3 6 2 5 _________________ (xvi) Th H T O 8 8 8 8 - 6 6 6 9 _________________
(xvii) Th H T O 7 6 5 4 - 5 2 2 8 _________________ (xviii) Th H T O 5 8 9 1 - 1 5 6 3 _________________
(xix) Th H T O 4 6 5 2 - 1 2 1 5 _________________ (xx) Th H T O 9 7 6 6 - 8 3 1 8 _________________
(xxi) Th H T O 7 6 9 4 - 5 2 7 8 _________________ (xxii) Th H T O 5 8 8 3 - 2 6 6 9 _________________
(xxiii) Th H T O 9 7 9 4 - 2 3 7 6 _________________ (xxiv) Th H T O 6 9 9 3 - 4 7 7 7 _________________
(xxv) Th H T O 4 6 9 2 - 1 3 8 9 _________________
6. Do these following 4-Digit Subtraction (Regrouping from Tens to Ones Place) Sums in your Notebook -
(i) 3778 - 1559
(ii) 5973 - 2769
(iii) 9788 - 9549
(iv) 3738 - 1419
(v) 5963 - 2735
(vi) 8746 - 5518
(vii) 5464 - 3245
(viii) 6455 - 2137
(ix) 6375 - 4158
(x) 8793 - 6676
(xi) 2872 - 1557
(xii) 8654 - 6326
(xiii) 5621 - 3212
(xiv) 8457 - 4239
(xv) 9356 - 7128
(xvi) 6562 - 1117
7. Subtract the following Regrouping from Tens and Hundreds Place:
(i) Th H T O 6 9 5 1 - 7 6 8 _________________ (ii) Th H T O 4 5 5 1 - 3 7 2 _________________
(iii) Th H T O 3 6 1 1 - 2 6 4 _________________ (iv) Th H T O 8 9 1 1 - 2 5 3 _________________
(v) Th H T O 2 7 4 7 - 5 8 9 _________________ (vi) Th H T O 4 2 2 3 - 1 3 2 _________________
(vii) Th H T O 3 5 5 4 - 4 6 5 _________________ (viii) Th H T O 2 6 3 3 - 3 8 5 _________________
(ix) Th H T O 4 6 1 4 - 4 3 5 _________________ (x) Th H T O 2 7 5 4 - 2 8 5 _________________
(xi) Th H T O 2 6 3 4 - 1 4 6 8 _________________ (xii) Th H T O 7 5 2 6 - 3 2 8 8 _________________
(xi) Th H T O 3 4 4 7 - 1 2 5 6 _________________ (xii) Th H T O 5 3 4 1 - 4 2 9 6 _________________
(xiii) Th H T O 6 6 6 6 - 4 5 7 8 _________________ (xiv) Th H T O 6 7 6 3 - 5 6 9 8 _________________
(xv) Th H T O 8 2 5 2 - 4 0 8 4 _________________ (xvi) Th H T O 7 9 2 6 - 5 1 4 8 _________________
(xv) Th H T O 9 5 4 7 - 8 3 9 8 _________________ (xvi) Th H T O 4 3 8 6 - 2 1 9 8 _________________
8. Subtract the following Regrouping from Tens and Hundreds Place:
(i) Th H T O 2 9 4 3 - 6 5 _________________ (ii) Th H T O 4 8 5 6 - 6 7 _________________
(iii) Th H T O 5 3 2 1 - 3 4 _________________ (iv) Th H T O 4 6 3 5 - 4 6 _________________
(v) Th H T O 2 8 2 5 - 3 7 _________________ (vi) Th H T O 5 9 4 5 - 4 5 6 _________________
(vii) Th H T O 6 8 1 4 - 6 2 8 _________________ (viii) Th H T O 7 6 4 3 - 3 6 5 _________________
(ix) Th H T O 8 4 3 2 - 1 4 6 _________________ (x) Th H T O 5 6 2 3 - 4 4 6 _________________
(xi) Th H T O 7 9 7 2 - 5 7 9 4 _________________ (xii) Th H T O 8 6 3 1 - 5 3 4 9 _________________
(xiii) Th H T O 9 7 2 5 - 7 5 3 7 _________________ (xiv) Th H T O 5 8 5 3 - 2 4 7 6 _________________
(xv) Th H T O 7 6 3 3 - 5 3 7 6 _________________ (xvi) Th H T O 9 6 5 6 - 5 3 8 7 _________________
(xvii) Th H T O 6 5 4 3 - 4 2 5 8 _________________ (xviii) Th H T O 3 9 1 5 - 1 5 3 7 _________________
(xix) Th H T O 5 4 2 7 - 3 2 6 9 _________________ (xx) Th H T O 6 7 4 1 - 4 3 7 9 _________________
(xxi) Th H T O 8 4 6 5 - 4 2 9 8 _________________ (xxii) Th H T O 9 6 3 3 - 5 1 4 5 _________________
(xxiii) Th H T O 6 4 1 3 - 5 2 4 8 _________________ (xxiv) Th H T O 3 4 2 1 - 1 1 4 5 _________________
(xxv) Th H T O 5 7 6 4 - 3 5 9 9 _________________
9. Subtract the following Regrouping from Tens and Hundreds Place:
(i) 4458 - 3798
(ii) 3452 - 276
(iii) 4450 - 337
(iv)8642 - 6156
(v) 7956 - 3478
(vi) 6953 - 2768
(vii) 4781 - 393
(viii) 6521 - 3443
(ix) 7832 - 6797
(x) 5423 - 2287
(xi) 8352 - 4187
(xii) 4572 - 289
(xiii) 7826 - 4377
(xiv) 7624 - 4356
(xv) 6783 - 2294
(xvi) 9740 - 6681
(xvii) 6345 - 289
(xviii) 1725 - 268
(xix) 6731 - 4482
(xx) 8642 - 6457
10. Subtract the following Regrouping From Tens, Hundreds and Thousands Place:
(i) Th H T O 2 3 2 5 - 4 3 7 _________________ (ii) Th H T O 3 2 2 5 - 4 4 7 _________________
(iii) Th H T O 4 1 1 2 - 4 3 6 _________________ (iv) Th H T O 5 1 5 7 - 2 6 8 _________________
(v) Th H T O 6 8 3 4 - 9 8 6 _________________ (vi) Th H T O 8 6 0 1 - 7 3 8 _________________
(vii) Th H T O 9 8 5 0 - 8 9 9 _________________ (viii) Th H T O 1 5 0 6 - 8 5 9 _________________
(ix) Th H T O 3 5 2 3 - 7 4 6 _________________ (x) Th H T O 6 5 3 3 - 9 4 4 _________________
(xi) Th H T O 6 2 8 2 - 3 8 9 5 _________________ (xii) Th H T O 4 3 2 8 - 2 5 8 9 _________________
(xiii) Th H T O 9 0 0 2 - 4 6 8 4 _________________ (xiv) Th H T O 8 3 4 3 - 2 6 8 4 _________________
(xv) Th H T O 8 4 3 4 - 4 8 4 6 _________________ (xvi) Th H T O 4 6 8 5 - 3 7 9 6 _________________
(xvii) Th H T O 6 1 2 3 - 3 2 6 9 _________________ (xviii) Th H T O 5 3 3 2 - 2 5 9 8 _________________
(xix) Th H T O 4 0 0 0 - 2 8 9 9 _________________ (xx) Th H T O 5 2 1 3 - 3 7 8 7 _________________
11. Subtract the following Regrouping From Tens, Hundreds and Thousands Place:
(i) Th H T O 6 2 8 3 - 7 9 4 _________________ (ii) Th H T O 5 4 8 8 - 6 9 9 _________________
(iii) Th H T O 8 6 4 2 - 7 5 4 _________________ (iv) Th H T O 7 3 5 7 - 8 9 8 _________________
(v) Th H T O 6 6 4 2 - 7 9 7 _________________ (vi) Th H T O 7 6 5 6 - 7 6 8 _________________
(vii) Th H T O 6 5 4 3 - 7 5 4 _________________ (viii) Th H T O 4 4 7 2 - 6 8 4 _________________
(ix) Th H T O 4 5 7 6 - 7 8 9 _________________ (x) Th H T O 2 4 5 2 - 5 8 4 _________________
(xi) Th H T O 8 7 8 1 - 4 9 9 6 _________________ (xii) Th H T O 4 3 8 7 - 2 6 9 6 _________________
(xiii) Th H T O 3 6 7 3 - 1 7 8 6 _________________ (xiv) Th H T O 2 4 4 2 - 1 6 6 5 _________________
(xv) Th H T O 6 9 8 8 - 5 9 9 9 _________________ (xvi) Th H T O 5 4 3 2 - 3 7 8 8 _________________
(xvii) Th H T O 5 3 4 4 - 1 4 6 6 _________________ (xviii) Th H T O 2 5 5 4 - 1 9 6 7 _________________
(xix) Th H T O 4 8 6 2 - 2 9 8 7 _________________ (xx) Th H T O 8 3 2 1 - 6 6 4 5 _________________
(xxi) Th H T O 3 8 8 8 - 1 9 9 9 _________________ (xxii) Th H T O 8 6 5 7 - 5 6 6 8 _________________
(xxiii) Th H T O 9 3 2 1 - 4 4 4 4 _________________ (xxiv) Th H T O 7 4 5 6 - 3 7 9 8 _________________
(xxv) Th H T O 6 4 5 3 - 2 7 6 5 _________________
12. Subtract the following Regrouping From Tens, Hundreds and Thousands Place:
(i) 8452 - 668
(ii) 1725 - 879
(iii) 3452 - 687
(iv) 9675 - 789
(v) 5367 - 989
(vi) 4781 - 893
(vii) 2345 - 567
(viii) 5432 - 3676
(ix) 7546 - 5678
(x) 8658 - 6779
(xi) 9568 - 4679
(xii) 9898 - 4899
(xiii) 7777 - 5888
(xiv) 4213 - 1654
(xv) 4423 - 3566
(xvi) 3214 - 1655
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# What is the projection of <5,8,-2 > onto <4,-5,2 >?
Jul 13, 2017
$p r o {j}_{\vec{B}} \vec{A} = < - \frac{32}{15} , \frac{8}{3} , - \frac{16}{15} >$
#### Explanation:
We're asked to find the projection of a vector onto another.
If we call the first vector $\vec{A}$ and the second one $\vec{B}$, the projection of $\vec{A}$ onto $\vec{B}$ (notated $p r o {j}_{\vec{B}} \vec{A}$) is found using the formula
proj_(vecB)vecA = (vecA·vecB)/(B^2)vecB
where
• vecA · vecB is the dot product of the two vectors
This is found using the equation
vecA·vecB = A_xB_x + A_yB_y + A_zB_z
So
vecA·vecB = (5)(4) + (8)(-5) + (-2)(2) = color(red)(-24
• $B$ is the magnitude of $\vec{B}$, which is
$B = \sqrt{{\left({B}_{x}\right)}^{2} + {\left({B}_{y}\right)}^{2} + {\left({B}_{z}\right)}^{2}}$
B = sqrt((4)^2 + (-5)^2 + (-2)^2) = color(green)(sqrt45
We therefore have
$p r o {j}_{\vec{B}} \vec{A} = \frac{\textcolor{red}{- 24}}{\left({\textcolor{g r e e n}{\sqrt{45}}}^{2}\right)} \vec{B}$
$= - \frac{8}{15} \vec{B}$
Now all we have to do is multiply the scalar through all components of $\vec{B}$ to find our vector projection:
$= < \left(- \frac{8}{15}\right) 4 , \left(- \frac{8}{15}\right) - 5 , \left(- \frac{8}{15}\right) 2 >$
= color(red)( < -32/15, 8/3, -16/15 >
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# Cracking the Math Puzzle: Discover the Answer to 2000 Divided by 3
## 1. Why 2000 Divided by 3 Equals 666.67: Understanding the Decimal Result
### Understanding Decimal Division
When dividing numbers, we sometimes encounter decimal results. This occurs when the division doesn’t result in a whole number. In the case of dividing 2000 by 3, the quotient is 666.67, which includes a decimal part. This decimal result often confuses people who expect a whole number as the answer. Let’s dig deeper into why the division of 2000 by 3 yields 666.67 and the significance of the decimal part.
You may also be interested in: Converting 84 mm to Inches: A Simple Guide for Quick and Accurate Measurements
### Factors Contributing to Decimal Results
To understand why we get a decimal result when dividing 2000 by 3, it’s important to consider a couple of factors. Firstly, 2000 is not evenly divisible by 3. In other words, 3 does not perfectly divide into 2000 without leaving a remaining amount. Secondly, we are using decimal notation to represent fractions that cannot be expressed as whole numbers.
### Significance of the Decimal Part
The decimal part of the division result is crucial as it represents the fraction that remains after the division process. In the case of 2000 divided by 3, the decimal part (.67) represents the portion that cannot be divided evenly. It indicates that we have distributed as many whole parts as possible, and the remaining fraction is .67. This decimal value is an approximation of the exact fraction and is commonly rounded off to two decimal places for simplicity.
Using decimal division, we can calculate more precise values when dealing with non-whole numbers. Understanding the significance of the decimal part can help grasp the concept better and prevent confusion when encountering decimal results in mathematical calculations.
## 2. The Significance of 2000 Divided by 3: Unveiling the Fractional Quotient
### The Role of Fractions in Mathematical Calculations
When it comes to mathematical calculations, fractions play a significant role in providing precise and accurate results. The division of numbers often results in a fractional quotient, indicating that the numbers cannot be divided evenly. In this case, let’s explore the significance of dividing 2000 by 3.
• Understanding the Fractional Quotient
• When 2000 is divided by 3, the quotient is a fraction: 666.666… This indicates that each person or unit would receive 666 and two-thirds of whatever is being divided.
• Real-Life Applications
• The concept of fractional quotients is relevant in various real-life scenarios. For example, when dividing food among a group of people, knowing that each person would receive a fraction helps in determining the appropriate portion sizes. Additionally, in finance, understanding fractional quotients is essential for calculations that involve dividing money or assets among multiple parties.
• Decimal Approximation
• While the fraction 666.666… provides a precise answer, for practical purposes, in many instances, it is common to round it to a decimal approximation. In this case, rounding 666.666… to two decimal places would result in a quotient of approximately 666.67.
Understanding the significance of dividing 2000 by 3 and unveiling the fractional quotient is crucial for accurate mathematical calculations. Whether in practical situations like sharing food or in complex financial calculations, fractions play a vital role in providing precise results. Therefore, it is important to recognize the relevance and application of fractional quotients in various contexts.
## 3. The Mathematics behind 2000 Divided by 3: Exploring Long Division and Remainders
If you’ve ever wondered how to divide a large number like 2000 by a smaller number like 3, you’re in the right place. In this section, we will delve into the mathematics behind this operation, specifically focusing on long division and remainders.
Long division is a method used to divide larger numbers and is taught in elementary school mathematics. It involves breaking down the dividend (in this case, 2000) into smaller parts, which are then divided by the divisor (in this case, 3). Each partial quotient obtained is written on top of the division symbol, forming the final quotient. The remainder, if any, is also displayed as part of the answer.
Let’s dive deeper into the process of long division. We place the divisor, 3, outside the division symbol, and the dividend, 2000, inside. We start by dividing the first digit of the dividend by the divisor. In this example, 2 divided by 3 is not possible without a remainder. So, we bring down the next digit, 0, and add it to the remainder obtained (2). Now, we have 20, which can be divided by 3 evenly, resulting in 6. The quotient obtained, 6, is then written on top, and the remainder, 2, is brought down again. This process continues until we have no more digits to bring down.
By exploring long division and the concept of remainders, we can solve complex mathematical problems like dividing 2000 by 3. Understanding these mathematical operations is essential in various fields such as finance, engineering, and computer science. Stay tuned for more math-related topics and further exploration of the fascinating world of numbers!
## 4. How to Calculate 2000 Divided by 3: Tips and Methods for Accurate Results
### Introduction:
Calculating the division of two numbers may seem like a simple task, but when it comes to more complex calculations, like dividing 2000 by 3, it can be a bit challenging. In this article, we will explore various tips and methods for accurately calculating this equation. By following these techniques, you will be able to obtain precise results and enhance your mathematical skills.
### Method 1: Long Division
One of the most common methods to divide two numbers is long division. To divide 2000 by 3 using this method, follow these steps:
1. Write 2000 as the dividend and 3 as the divisor.
2. Start dividing by determining how many times 3 can be divided into 20. The quotient is 6.
3. Multiply 6 by 3 and subtract the product (18) from 20. The new dividend becomes 2.
4. Bring down the next digit from the dividend to the remainder (which is 2) and divide it by 3.
5. The process continues until there are no more digits to bring down.
By following these steps, you will find that 2000 divided by 3 is equal to 666.6667 (repeating).
### Method 2: Use a Calculator
If you are seeking a quicker way to find the result without the hassle of manual calculations, using a calculator is your best bet. Simply enter 2000 รท 3 in a calculator, and with a press of a button, you will instantly get the result: 666.6667. Keep in mind that rounding the decimal places might be required, depending on the level of precision needed for your calculations.
Note: It’s always wise to double-check your answers and verify the accuracy of calculations to avoid potential errors.
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### In conclusion,
dividing 2000 by 3 can be achieved through various methods, including long division and using a calculator. Whether you prefer the traditional approach or the convenience of technology, these tips and techniques will assist you in obtaining accurate results. Practice these methods, enhance your mathematical skills, and conquer even more complex divisions with confidence.
## 5. The Real-Life Applications of 2000 Divided by 3: Practical Scenarios Explored
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### 1. Splitting Bills:
When it comes to dividing bills among friends or roommates, the real-life application of 2000 divided by 3 becomes increasingly useful. For example, let’s say you and two friends decide to split the cost of a vacation rental that costs \$2000 for a week. By using the formula, each person would end up contributing \$666.67 towards the total cost. This ensures that everyone pays an equal share, making it a fair and straightforward way to divide expenses.
2. Sharing Resources:
Another practical scenario where 2000 divided by 3 comes into play is resource sharing. Imagine three colleagues sharing the cost of a printer valued at \$2000. By dividing the cost equally among the three, each person would be responsible for contributing \$666.67. This approach is not only fair but can also help businesses or individuals save money by sharing resources and distributing the expenses evenly.
### 3. Cost Allocation:
In budgeting and cost allocation, the application of 2000 divided by 3 has its significance. Let’s say a company has a budget of \$2000 to allocate for the marketing department. By dividing this budget among three marketing initiatives, each project would receive an equal allocation of \$666.67. This ensures that resources are distributed fairly and gives a clear outline of the budget allocation for each project.
These real-life scenarios illustrate the practical uses of dividing 2000 by 3. Whether it’s splitting bills, sharing resources, or allocating costs, this mathematical calculation provides a simple and equitable approach to distributing expenses. By applying this approach in everyday situations, individuals and businesses can promote fairness and transparency in their financial dealings.
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## CBSE Class 9 Maths Lab Manual – Algebraic Identity (a + b)3 = a3 + b3 + 3a2b + 3ab2
Objective
To verify the identity (a+b)3 = a3 + b3 + 3a2b + 3ab2 geometrically by using sets of unit cubes.
Prerequisite Knowledge
1. Volume of a cube = (edge)3
2. Volume of a cuboid = l x b x h
Materials Required
A set of 128 plastic cubes or wooden cubes with dimensions (1unit x 1unit x 1unit).
Procedure
To verify the identity (a+b)3 = a3 + b3 + 3a2b + 3ab2, we shall take value of a = 3units and value of b = 1unit.
1. First we will make a cube of dimensions (a+b) i.e., (3+1 = 4 units). For this, we will use 64 unit cubes and arrange them as shown in fig.(i).
2. From other set of 64 cubes we will make arrangements as shown in figures.
• Arrange 27 cubes such that a x a x a cube is formed i.e., 3 x 3 x 3 set is formed in fig.(ii).
• Arrange 9 cubes such that 3 columns of 3 cubes is formed as shown in fig. (iii). Make 3 sets of such arrangement.
• Arrange 3 cubes such that 1 column of 3 cubes is formed as shown in fig. (iv). Make 3 sets of such arrangement.
• Last arrangement consists of only 1 cube of volume b3 [fig (v)].
Observation and Calculation
In fig. (i) we have used 64 cubes i.e., (a+b)3
Other set of 64 cubes are arranged in following manner:
1. Volume of cube in fig. (ii) = a3
2. Volume of cuboids in fig.(iii) = 3(a)(a)(b) = 3ba2
3. Volume of cuboids in fig. (iv) = 3(a)(b)(b) = 3ab2
4. Volume of cube in fig. (v) = b3
Total volume of cube in fig. (i) = Total volume of cubes and cuboids in fig. (ii), (iii), (iv) and (v)
(a+b)3 = a3 + b3 + 3a2b + 3ab2
Hence, this identity is verified geometrically.
Result
The identity (a+b)3 = a3 + b3 + 3a2b + 3ab2 is verified geometrically by using cubes and cuboids.
Learning Outcome
In this way, students can learn the concept of verifying the identity geometrically by adding volume of cubes and cuboids.
Activity Time
Students must perform this activity for any other values of a and b, e.g., a=4, b=1 and find volumes of different cubes and cuboids used in this activity.
Viva Voce
Question 1.
Find 113 by using the formula (a+b)3 = a3 + b3 + 3a2b + 3ab2.
(10 + 1)= (10)3 + 3 x (10)2 x 1 + 3 x 10 x 1 + (1)3 = 1331.
Question 2.
Find 233 by using the formula (a+b)3 = a3 + b3 + 3a2b + 3ab2
(20 + 3)3 = (20)3 + (3)3 + 3.(20)2.(3) + 3.(20)(3)2
= 8000 + 27 + 9(400) + 60(9) = 8000 + 27 + 3600 + 540 = 12167.
Question 3.
Find (5x + 5y)3
27x3 + 125y3 + 135x2y + 225xy2.
Question 4.
Find (10 + 2)3.
103 + 23 +600+ 120 = 1728.
Question 5.
In the activity of (a + b)3, what do you mean by 3a2b, 3ab2?
3 set of cuboids with volumes a x a x b and 3 sets of cuboids with volumes a x b x b.
Question 6.
If one side of a cube is (a – b), then what is the volume of the cube ?
Volume of cube = (a – b)3.
Question 7.
Write the degree of (x + 2y)3
3.
Question 8.
Write the coefficient of m3 in (2m + n)3.
12n
Question 9.
Is (a + b)3 binomial ?
No
Question 10.
Write the integeral zeroes of (x+8)3
-8
Multiple Choice Questions
Question 1.
Write in expanded form: (x + $$\frac { 2 }{ 3 }$$ y)3
(i) x3 + $$\frac { 8 }{ 27 }$$ y3 + 2x2y + $$\frac { 4 }{ 3 }$$ xy2
(ii) x3 + $$\frac { 8 }{ 27 }$$ y3
(iii) x3 + $$\frac { 2 }{ 3 }$$ y3
(iv) None of these
Question 2.
Evaluate using suitable identity (102)3:
(i) 1061208
(ii) 1062208
(iii) 10062208
(iv) none of these
Question 3.
Factorize: 27 + 125a3 + 135a+ 225a3:
(i) (3 – 5a)3
(ii) (3 + 5a)3
(iii) 27 + 125a3
(iv) none of these
Question 4.
Simplify: ($$\frac { 3 }{ 2 }$$ x + 5)3
(i) $$\frac { 27 }{ 8 }$$ x3 + 125 + $$\frac { 135 }{ 4 }$$ x2 + $$\frac { 225 }{ 2 }$$ x
(ii) $$\frac { 27 }{ 8 }$$ x3 + 125
(iii) $$\frac { 27 }{ 8 }$$ x + 125
(iv) none of these
Question 5.
Evaluate, (51)3 using identity (a + b)3:
(i) 133261
(ii) 132651
(iii) 133651
(iv) none of these
Question 6.
If x – 1 = a, is it true that 23a3 + 8 = 8x3 – 16x2 + 16x – 8ax ?
(i) false
(ii) true
(iii) can’t say
(iv) none of these
Question 7.
Factorize: 1 x 1 x 1 + .02 x .02 x .02 + 3(.02)(1.02):
(i) (1.02)3
(ii) 13
(iii) 0.023
(iv) none of these
Question 8.
Without multiplication, evaluate (1.01)3:
(i) 1.0303
(ii) 1.03301
(iii) 1.030301
(iv) none of these
Question 9.
If a3 + b3 = 152, a +b = 8, find value of (ab):
(i) 160
(ii) 14
(iii) 15
(iv) none of these
Question 10.
Write the product of (2 + x)(x + 2)2 By using identity (a+b)3.
(i) 8 + x3 + 6x(2 + x)
(ii) 4 + x2 + 8x
(iii) x2 + 4 + 4x
(iv) none of these
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### Grade 9 homework questions on 2.4 and 2.5
1. 1. 2.4 Problem 57<br />Your class wants to make a rectangular spirit display, and has 24 feet of decorative board to enclose the display.<br />Write an equation in standard form relating the possible lengths l and widths w of the display. <br />Solution:<br /> 2l + 2w = 24 there are two sides of length and two sides of width since it’s a rectangle<br />b. Graph the equation from part (a). C. Make a table of at least five possible pairs of dimensions for the display. <br />2(6) + 2w = 24 Put l in the equation <br />12 + 2w = 24 Subtract by 12<br />-12 -12<br />2w = 12 Then divide by 2 <br /> w = 6<br />16<br />12<br />8<br />4<br />4 8 12 16<br />
2. 2. 2.5 Problem 30<br />Give an example of two real-life quantities that show direct variation. Explain your reasoning. <br />Solution:<br />If you earn an hourly wage, the amount of money you earn varies directly with the number of hours you work. If you work 4 hours and make \$28, the equation relating the number of hours h worked and the amount of money m you earn is m = 7h. If you are traveling at a constant speed, the distance d you travel varies directly with the time t you travel. If you drive for 4 hours and travel 248 kilometers, the direct variation equation is d = 62h. <br />
3. 3. Problem 36<br />Let (x1, y1) be a solution, other than (0,0), of a direct variation equation. Write a second direct variation equation whose graph is perpendicular to the graph of the first equation. <br />Solution:<br />Y = ax Put x1 and y1 into the equation for x and y.<br />Y1 = a(x1) Then divide each side by x1<br /> = a<br />Then put the y1/x1 back into place for a.<br />Y = x; y = x <br />
4. 4. Problem 39<br />Hail 0.5 inch deep and weighing 1800 pounds cover a roof. The hail’s weight w varies directly with its depth d. Write an equation that relates d and w. Then predict the weight on the roof of hail that is 1.75 inches deep. <br />Solution:<br />W represents y and d represents x. Place it back into the y = ax formula.<br />1800 = a (0.5) Divide 1800 by 0.5.<br />3600 = a<br />Write 3600 back into the y = ax formula. <br />w = 3600d Then plug 1.75 into the formula.<br />w = 3600(1.75) Then multiply through.<br />w = 6300<br />So the weight when the hail is 1.75 inches deep weighs 6300 pounds. <br />
5. 5. Problem 41<br />The ordered pairs (4.5,23), (7.8, 40), and (16.0, 82) are in the form (s, t) where t represents the time (in seconds) needed to download an internet filed of size s (in megabytes). Tell whether the data show direct variation. If so, write an equation that relates s and t. <br />Solution:<br />Because the direct variation equation y = ax can be written as y over x = a, a set of data pairs (x, y) shows direct variation if the ratio of y to x is constant.<br /> 23/4.5 = 5.1 40/7.8 = 5.1 82/16.0 = 5.1<br />Because the ratios are approximately equal, the data shows direct variation. Now we want to write an equation that relates s and t. <br />t = 5.1s<br />
6. 6. Problem 44<br />Each year, gray whales migrate from Mexico’s Baja Peninsula to feeding grounds near Alaska. A whale may travel 6000 miles at an average rate of 75 miles per day.<br />A. Write an equation that gives the distance d1 traveled in t days of migration.<br />D1 = 75t<br />B. Write an equation that gives the distance d2 that remains to be traveled after t days of migration.<br />D2 = 6000 – 75t<br />C. Tell whether the equations from parts (a) and (b) represent direct variation. Explain your answers.<br />Part (a) represents direct variation because the graph goes through the origin. Part (b) does not represent direct variation because the graph does not go through the origin. <br />
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# Chapter 3: Elementary Number Theory and Methods of Proof. January 31, 2010
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## Transcription
1 Chapter 3: Elementary Number Theory and Methods of Proof January 31, 2010
2 3.4 - Direct Proof and Counterexample IV: Division into Cases and the Quotient-Remainder Theorem
3 Quotient-Remainder Theorem Given any integer n and positive integer d, there exist unique integers q and r such that n = dq + r and 0 r d. d is the divisor, n is the dividend, q is the quotient, and r is the remainder Ex. Let n = 37 and d = 5. Then 37 = 5 (7) + 2 (where q = 7 and r = 2). Ex. Let n = 52 and d = 23. Then 52 = 23 ( 3) + 17 (where q = 3 and r = 17).
4 DEFINITION: Given a nonnegative integer n and a positive integer d, n div d = the integer quotient q when n is divided by d. n mod d = the integer remainder r when n is divided by d. In other words: Suppose n, d N, q, r Z, 0 r d n = dq + r n div d = q and n mod d = r [ If n = dq + r, then d (n r) and we say n is congruent to r ] modulo d and often use the notation n r mod d.
5 Most standard programming languages (C++, Java, Pascal, etc) have built in functions that compute mod and div, given a non-negative integer n and a positive integer d. Ex. Compute 62 div 7 and 62 mod 7: Since 62 = n = dq + r we have 62 div 7 = 8 n div d = q and 62 mod 7 = 6 n mod d = r
6 Ex. Suppose today is Wednesday and you know that neither this year nor the next year are leap yeats. What day of the week will it be in 342 days time? Solution: Labels the days as: 0 =Sunday, 1 =Monday,..., 6 =Saturday. Today is Wednesday (day 3), so we want to work out what day of the week = 345 will be. We find 345 mod 7 = 2 [i.e. 345 = 7 (49) + 2] So in 342 days time, it will be Tuesday.
7 Representations of Integers Given any integer n and positive integer d, we can represent n as n = dq + r 0 r d for some integers q, r. For d = 2: n = 2q + r, 0 r 2 which is the same as either n = 2q + 0 or n = 2q + 1 The parity property: every integer is either even or odd.
8 Theorem: Any two consecutive integers have opposite parity. Proof: Let n, n + 1 be two arbitrary, but consecutive integers. By the parity property, n is either odd or even. Case 1: Assume n is even. Then n = 2k, for some k Z. Then n + 1 = 2k + 1 and by definition, n + 1 is odd. Case 2: Assume n is odd. Then n = 2k + 1, for some k Z. Then n + 1 = 2k + 2 = 2(k + 1) and as k + 1 is an integer, by definition, n is even. Therefore, regardless of whether n is odd or even, the theorem is true.
9 Ex. Prove that every integer can be written in one of the four forms for some integer q: Proof: n = 4q, n = 4q + 1, n = 4q + 2, n = 4q + 3 Let n be any integer. We apply the Quotient-Remainder Theorem to n with d = 4: n = 4q + r, 0 r 4 We list all the possibilities of r to prove the result: n = 4q, n = 4q + 1, n = 4q + 2, n = 4q + 3
10 Theorem: The square of any odd integer has the form 8m + 1 for some integer m. Proof: We want to show: odd integers n, integer m such that n 2 = 8m + 1. Let n be an arbitrary odd integer. Using the previous result (with the Quotient-Remainder Theorem and d = 4, n can be written in one of four forms: n = 4q, n = 4q + 1, n = 4q + 2, n = 4q + 3 for some integer q. Since n is odd, it cannot have either of the forms 4q or 4q + 2 (since they are both even). Therefore, n = 4q + 1, or, n = 4q + 3
11 Case 1: Assume n = 4q + 1. Then n 2 = (4q + 1) 2 (substitution) = 16q 2 + 8q + 1 = 8(2q 2 + q) + 1 Let m = 2q 2 + q. Then we have proven that n 2 = 8m + 1. Case 2: Assume n = 4q + 3. Then n 2 = (4q + 3) 2 (substitution) = 16q q + 9 = 16q q = 8(2q 2 + 3q + 1) + 1 Let m = 2q 2 + 3q + 1. Then we have proven that n 2 = 8m + 1. We have proven that for any odd integer n, n 2 = 8m + 1 for some integer m.
12 3.5 - Direct Proof and Counterexample V: Floor and Ceiling Floors and Ceilings just tell us whether to round a fraction up or down to the nearest integer.
13 DEFINITION: Given any real number x, the floor of x, denoted x is defined as: x = the unique integer n such that n x < n + 1. x = n n x < n + 1. We round down to the nearest integer. Ex. 3.2 = 3.2 Ex. 3.1 = 4 Ex. 7 = 7 Ex. Let n be an integer. n = n
14 DEFINITION: Given any real number x, the ceiling of x, denoted x is defined as: x = the unique integer n such that n 1 < x n. x = n n 1 < x n. We round up to the nearest integer. Ex. 3.2 = 4 Ex. 3.1 = 3 Ex. 7 = 7 Ex. Let n be an integer. n = n + 1
15 Ex. Boxes, each capable of holding 36 units, are used to ship a product from the manufacturer to a wholesaler. Express the number of boxes that would be required to ship n units of the product using either the floor or the ceiling notation. Which one is more appropriate? Solution: We need either n or n 36 boxes. Obviously, the ceiling notation is more convenient here.
16 Ex. Is the following statement true or false? For all real numbers x and y, x + y = x + y Solution: We will find a counterexample to prove this statement is false. Let x = 0.5 and y = 0.5 Then x + y = = 1 = 1 And x + y = = = 0 Therefore, if x = y = 0.5, then x + y = 1 0 = x + y
17 Theorem For all real numbers x and all integers m, x + m = x + m. Proof: Let x be an arbitrary real number and m be an arbitrary integer. Let x = n. This means n x < n + 1 (from the definition of the floor function). Add m to both sides of the inequality: n + m x + m < n + m + 1 Since n + m is an integer, by the floor function definition: x + m = n + m which is the same as x + m = x + m.
18 Theorem For any integer n, { n n = 2 if n is even 2 n 1 2 if n is odd [This value is required for binary search and merge-sort algorithms.] Proof: Let n be an arbitrary integer. The Quotient-Remainder Theorems says that n must be either odd or even, so we break the proof into two cases.
19 Case 1: Suppose n is odd. We wish to show that n 2 = n 1 2. As n is odd, by definition, n = 2k + 1 for some integer k. Then n 2 = 2k+1 2 = 2k = k = k But as n = 2k + 1, we can rearrange the equation to k = n 1 Thus, we have shown that n 2 = k = n 2, proving the result. 2.
20 Case 2: Suppose n is even. We wish to show that n 2 = n 2. As n is even, by definition, n = 2k for some integer k. Then n 2 = 2k 2 = k = k But as n = 2k, we can rearrange the equation to k = n 2. Thus, we have shown that n 2 = k = n 2, proving the result.
21 Recall the Quotient-Remainder Theorem: for an integer n and positive integer d, there exist unique integers q, r such that n = dq + r and 0 r < d. We now prove a related theorem: Theorem: If n is any integer, d is a positive integer, q = n/d, and r = n d n/d, then n = dq + r and 0 r < d. Proof: Let n be any integer, d be a positive integer, q = n/d, and r = n d n/d.
22 ( ) Then dq + r = d n d + n d n d = n So n = dq + r and we simply need to show that 0 r < d. As q = n/d, by the definition of the floor function, q n d < q + 1. Multiply the inequality by d: dq n < dq + d Subtract dq from the inequality: 0 n dq < d Substitute r = n dq (we showed this above) into the inequality: 0 r < d and the proof is complete.
23 3.6 - Indirect Argument: Contradiction and Contraposition
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#### Question 1:
Find the equation of the circle with:
(i) Centre (−2, 3) and radius 4.
(ii) Centre (a, b) and radius $\sqrt{{a}^{2}+{b}^{2}}$.
(iii) Centre (0, −1) and radius 1.
(iv) Centre (a cos α, a sin α) and radius a.
(v) Centre (a, a) and radius $\sqrt{2}$ a.
Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
(i) Here, h = −2, k = 3 and a = 4
∴ Required equation of the circle:
${\left(x+2\right)}^{2}+{\left(y-3\right)}^{2}={4}^{2}$
$⇒{\left(x+2\right)}^{2}+{\left(y-3\right)}^{2}=16$
(ii) Here, h = a, k = b and radius = $\sqrt{{a}^{2}+{b}^{2}}$
∴ Required equation of the circle:
${\left(x-a\right)}^{2}+{\left(y-b\right)}^{2}={a}^{2}+{b}^{2}$
$⇒{x}^{2}+{y}^{2}-2ax-2by=0$
(iii) Here, h = 0, k = −1 and radius = 1
∴ Required equation of the circle:
${\left(x-0\right)}^{2}+{\left(y+1\right)}^{2}={\left(1\right)}^{2}$
$⇒{x}^{2}+{y}^{2}+2y=0$
(iv) Here, h = $a\mathrm{cos}\alpha$, k $a\mathrm{sin}\alpha$ and radius = a
∴ Required equation of the circle:
${\left(x-a\mathrm{cos}\alpha \right)}^{2}+{\left(y-a\mathrm{sin}\alpha \right)}^{2}={\left(a\right)}^{2}$
$⇒{x}^{2}+{a}^{2}{\mathrm{cos}}^{2}\alpha -2ax\mathrm{cos}\alpha +{y}^{2}+{a}^{2}{\mathrm{sin}}^{2}\alpha -2ay\mathrm{sin}\alpha ={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{a}^{2}\left({\mathrm{sin}}^{2}\alpha +{\mathrm{cos}}^{2}\alpha \right)-2ax\mathrm{cos}\alpha +{y}^{2}-2ay\mathrm{sin}\alpha ={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{a}^{2}-2ax\mathrm{cos}\alpha +{y}^{2}-2ay\mathrm{sin}\alpha ={a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2ax\mathrm{cos}\alpha -2ay\mathrm{sin}\alpha =0$
(v) Here, h = a, k = a and radius = $\sqrt{2}a$
∴ Required equation of the circle:
${\left(x-a\right)}^{2}+{\left(y-a\right)}^{2}={\left(\sqrt{2}a\right)}^{2}$
$⇒{x}^{2}+{a}^{2}-2ax+{y}^{2}+{a}^{2}-2ay=2{a}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2ay-2ax=0$
#### Question 2:
Find the centre and radius of each of the following circles:
(i) (x − 1)2 + y2 = 4
(ii) (x + 5)2 + (y + 1)2 = 9
(iii) x2 + y2 − 4x + 6y = 5
(iv) x2 + y2 x + 2y − 3 = 0.
Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
(i) Given:
(x − 1)2 + y2 = 4
Here, h = 1, k = 0 and a = 2
Thus, the centre is (1, 0) and the radius is 2.
(ii) Given:
(x + 5)2 + (y + 1)2 = 9
Here, h = 5, k = −1 and radius = 3
Thus, the centre is (−5, −1) and the radius is 3.
(iii) Given:
${x}^{2}+{y}^{2}-4x+6y=5$
The given equation can be rewritten as follows:
${\left(x-2\right)}^{2}+{\left(y+3\right)}^{2}-4-9=5$
$⇒{\left(x-2\right)}^{2}+{\left(y+3\right)}^{2}=18$
Thus, the centre is (2, −3).
And, radius = $\sqrt{18}=3\sqrt{2}$
(iv) Given:
${x}^{2}+{y}^{2}-x+2y-3=0$
The given equation can be rewritten as follows:
${\left(x-\frac{1}{2}\right)}^{2}+{\left(y+1\right)}^{2}-\frac{1}{4}-1-3=0$
$⇒{\left(x-\frac{1}{2}\right)}^{2}+{\left(y+1\right)}^{2}=\frac{17}{4}$
Thus, the centre is $\left(\frac{1}{2},-1\right)$ and and the radius is $\frac{\sqrt{17}}{2}$.
#### Question 3:
Find the equation of the circle whose centre is (1, 2) and which passes through the point (4, 6).
Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
Given:
h = 1, k = 2
∴ Equation of the circle = ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={a}^{2}$ ...(1)
Also, equation (1) passes through (4, 6).
${\left(4-1\right)}^{2}+{\left(6-2\right)}^{2}={a}^{2}$
Substituting the value of a in equation (1):
${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=25$
$⇒{x}^{2}+1-2x+{y}^{2}+4-4y=25\phantom{\rule{0ex}{0ex}}⇒{x}^{2}-2x+{y}^{2}-4y=20\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2x-4y-20=0$
Thus, the required equation of the circle is ${x}^{2}+{y}^{2}-2x-4y-20=0$.
#### Question 4:
Find the equation of the circle passing through the point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 and whose centre is the point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0.
The point of intersection of the lines x + 3y = 0 and 2x − 7y = 0 is (0, 0).
Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
The point of intersection of the lines x + y + 1 = 0 and x − 2y + 4 = 0 is (−2, 1).
h = −2, k = 1
∴ Equation of the required circle = ${\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}={a}^{2}$ ...(1)
Also, equation (1) passes through (0, 0).
${\left(0+2\right)}^{2}+{\left(0-1\right)}^{2}={a}^{2}$
Substituting the value of a in equation (1):
${\left(x+2\right)}^{2}+{\left(y-1\right)}^{2}=5$
$⇒{x}^{2}+4+4x+{y}^{2}+1-2y=5\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+4x+{y}^{2}-2y=0$
Hence, the required equation of the circle is ${x}^{2}+{y}^{2}+4x-2y=0$.
#### Question 5:
Find the equation of the circle whose centre lies on the positive direction of y - axis at a distance 6 from the origin and whose radius is 4.
Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
The centre of the required circle lies on the positive direction of the y-axis at a distance 6 from the origin.
Thus, the coordinates of the centre are (0, 6).
h = 0, k = 6
∴ Equation of the circle = ${\left(x-0\right)}^{2}+{\left(y-6\right)}^{2}={a}^{2}$ ...(1)
Also, a = 4
Substituting the value of a in equation (1):
${\left(x-0\right)}^{2}+{\left(y-6\right)}^{2}=16$
$⇒{x}^{2}+{y}^{2}+36-12y=16\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-12y+20=0$
Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-12y+20=0$.
#### Question 6:
If the equations of two diameters of a circle are 2x + y = 6 and 3x + 2y = 4 and the radius is 10, find the equation of the circle.
Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
The intersection point of 2x + y = 6 and 3x + 2y = 4 is (8, −10).
The diameters of a circle intersect at the centre.
Thus, the coordinates of the centre are (8, −10).
h = 8, k = −10
Thus, the equation of the required circle is ${\left(x-8\right)}^{2}+{\left(y+10\right)}^{2}={a}^{2}$ ...(1)
Also, a = 10
Substituting the value of a in equation (1):
${\left(x-8\right)}^{2}+{\left(y+10\right)}^{2}=100$
$⇒{x}^{2}+{y}^{2}-16x+64+100+20y=100\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-16x+20y+64=0$
Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-16x+20y+64=0$.
#### Question 7:
Find the equation of a circle
(i) which touches both the axes at a distance of 6 units from the origin.
(ii) which touches x-axis at a distance 5 from the origin and radius 6 units.
(iii) which touches both the axes and passes through the point (2, 1).
(iv) passing through the origin, radius 17 and ordinate of the centre is −15.
Let (h, k) be the centre of a circle with radius a.
Thus, its equation will be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
(i) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
It is given that the circle passes through the points (6, 0) and (0, 6).
${\left(6-h\right)}^{2}+{\left(0-k\right)}^{2}={6}^{2}$
And, ${\left(0-h\right)}^{2}+{\left(6-k\right)}^{2}={6}^{2}$
Also, ${h}^{2}+36+{k}^{2}-12k=36$
$⇒{h}^{2}+{k}^{2}=12k$ ...(2)
From (1) and (2), we get:
$12k=12h⇒h=k$
∴ From equation (2), we have:
Consequently, we get:
h = 6
Hence, the required equation of the circle is ${\left(x-6\right)}^{2}+{\left(y-6\right)}^{2}=36$ or ${x}^{2}+{y}^{2}-12x-12y+36=0$.
(ii) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$ .
It is given that the circle with radius 6 units touches the x-axis at a distance of 5 units from the origin.
a = 6, h = 5
Hence, the required equation is ${\left(x-5\right)}^{2}+{\left(y-0\right)}^{2}={6}^{2}$ or ${x}^{2}+{y}^{2}-10x-11=0$.
(iii) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
It is given that the circle touches both the axes.
Thus, the required equation will be ${x}^{2}+{y}^{2}-2ax-2ay+{a}^{2}=0$.
Also, the circle passes through the point (2, 1).
$4+1-4a-2a+{a}^{2}=0$
$⇒{a}^{2}-6a+5=0\phantom{\rule{0ex}{0ex}}⇒{a}^{2}-5a-a+5=0\phantom{\rule{0ex}{0ex}}⇒a=1,5$
Hence, the required equation is ${x}^{2}+{y}^{2}-2x-2y+1=0$ or ${x}^{2}+{y}^{2}-10x-10y+25=0$.
(iv) Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
Given:
k = −15, a = 17
The circle passes through the point (0, 0).
∴ Equation of the circle:
${\left(0-h\right)}^{2}+{\left(0-15\right)}^{2}={\left(17\right)}^{2}$
$h=±8$
Hence, the required equation of the circle is ${\left(x-8\right)}^{2}+{\left(y+15\right)}^{2}={17}^{2}$ or ${\left(x+8\right)}^{2}+{\left(y+15\right)}^{2}={17}^{2}$, i.e. ${x}^{2}+{y}^{2}±16x+30y=0$.
#### Question 8:
Find the equation of the circle which has its centre at the point (3, 4) and touches the straight line 5x + 12y − 1 = 0.
It is given that the centre is at the point (3, 4).
Let the equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
∴ Equation of the required circle = ${\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}={a}^{2}$ ...(1)
Also, the circle touches the straight line 5x + 12y − 1 = 0.
So, from equation (1), we have:
${\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}=\frac{3844}{169}$
$⇒{x}^{2}+{y}^{2}-6x-8y=\frac{3844}{169}-25$
$⇒169\left({x}^{2}+{y}^{2}-6x-8y\right)+381=0$
Hence, the required equation of the circle is $169\left({x}^{2}+{y}^{2}-6x-8y\right)+381=0$.
#### Question 9:
Find the equation of the circle which touches the axes and whose centre lies on x − 2y = 3.
If the circle lies in the third quadrant, then its centre will be (−a, −a).
The centre lies on x − 2y = 3.
$-a+2a=3⇒a=3$
∴ Required equation of the circle = ${\left(x+3\right)}^{2}+{\left(y+3\right)}^{2}=9$
=${x}^{2}+{y}^{2}+6x+6y+9=0$
If the circle lies in the fourth quadrant, then its centre will be (a, −a),
$a+2a=3⇒a=1$
∴ Required equation of the circle = ${\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}=1$
= ${x}^{2}+{y}^{2}-2x+2y+1=0$
#### Question 10:
A circle whose centre is the point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 passes through the origin. Find its equation.
Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
The point of intersection of the lines 2x − 3y + 4 = 0 and 3x + 4y − 5 = 0 is .
∴ Centre = $\left(\frac{-1}{17},\frac{22}{17}\right)$
Also, the circle passes through the origin.
${a}^{2}={\left(\frac{1}{17}\right)}^{2}+{\left(\frac{22}{17}\right)}^{2}=\frac{485}{289}$
Hence, the required equation of the circle is ${\left(x+\frac{1}{17}\right)}^{2}+{\left(y-\frac{22}{17}\right)}^{2}=\frac{485}{289}$.
#### Question 11:
A circle of radius 4 units touches the coordinate axes in the first quadrant. Find the equations of its images with respect to the line mirrors x = 0 and y = 0.
It is given that a circle of radius 4 units touches the coordinate axes in the first quadrant.
Centre of the given circle = (4, 4)
The equation of the given circle is ${\left(x-4\right)}^{2}+{\left(y-4\right)}^{2}=16$.
The images of this circle with respect to the line mirrors x = 0 and y = 0. They have their centres at respectively.
∴ Required equations of the images = ${\left(x+4\right)}^{2}+{\left(y-4\right)}^{2}=16$ and ${\left(x-4\right)}^{2}+{\left(y+4\right)}^{2}=16$
= ${x}^{2}+{y}^{2}+8x-8y+16=0$ and ${x}^{2}+{y}^{2}-8x+8y+16=0$
#### Question 12:
Find the equations of the circles touching y-axis at (0, 3) and making an intercept of 8 units on the X-axis.
Case I: The centre lies in first quadrant.
Let the required equation be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
Here, AB = 8 units and L (0, 3)
In $△$CAM:
$⇒C{A}^{2}=C{M}^{2}+A{M}^{2}$
$⇒C{A}^{2}={3}^{2}+{4}^{2}\phantom{\rule{0ex}{0ex}}⇒CA=5\phantom{\rule{0ex}{0ex}}⇒CL=CA=5$
∴ Coordinates of the centre =
And, radius of the circle = 5
${\left(x-5\right)}^{2}+{\left(y-3\right)}^{2}=25$, i.e. ${x}^{2}+{y}^{2}-10x-6y=-9$
Case II: The centre lies in the second quadrant.
Coordinates of the centre =
And, radius of the circle= 5
${\left(x+5\right)}^{2}+{\left(y-3\right)}^{2}=25$, i.e. ${x}^{2}+{y}^{2}+10x-6y=-9$
Hence, the equation of the required circle is ${\left(x±5\right)}^{2}+{\left(y-3\right)}^{2}=25$, i.e. ${x}^{2}+{y}^{2}±10x-6y=-9$.
#### Question 13:
Find the equations of the circles passing through two points on Y-axis at distances 3 from the origin and having radius 5.
Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
The circle passes through the points (0, 3) and (0, −3).
${\left(0-h\right)}^{2}+{\left(3-k\right)}^{2}={a}^{2}$ ...(1)
And, ${\left(0-h\right)}^{2}+{\left(-3-k\right)}^{2}={a}^{2}$ ...(2)
Solving (1) and (2), we get:
k=0
Given:
∴ a2 = 25
So, from equation (2), we have:
${h}^{2}+9=25⇒h=±4$
Hence, the required equation is ${\left(x±4\right)}^{2}+{y}^{2}=25$, which can be rewritten as ${x}^{2}±8x+{y}^{2}-9=0$.
#### Question 14:
If the lines 2x 3y = 5 and 3x − 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.
We have Area of circle = 154
$\mathrm{\pi }{r}^{2}=154\phantom{\rule{0ex}{0ex}}⇒{r}^{2}=49$
The intersection of two lines will give us the centre of the circle.
Solving 2x 3y = 5 and 3x − 4y = 7 we get
x = 1 and y = 1
Now, the equation of the circle is given by
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+1\right)}^{2}=49\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2x+2y-47=0$
#### Question 15:
If the line y = $\sqrt{3}$x + k touches the circle x2 + y2 = 16, then find the value of k. [NCERT EXEMPLAR]
The centre and the radius of the circle x2 + y2 = 16 are (0, 0) and 4
Now, the perpendicular distance from the centre of the circle to the tangent y = $\sqrt{3}$x + k is equal to the radius of the circle
$\therefore 4=\left|\frac{\sqrt{3}\left(0\right)-0+k}{\sqrt{{\left(\sqrt{3}\right)}^{2}+{1}^{2}}}\right|\phantom{\rule{0ex}{0ex}}⇒±4=\frac{k}{2}\phantom{\rule{0ex}{0ex}}⇒k=±8$
#### Question 16:
Find the equation of the circle having (1, −2) as its centre and passing through the intersection of the lines 3x + y = 14 and 2x + 5y = 18. [NCERT EXEMPLAR]
Solving 3x + y = 14 and 2x + 5y = 18 we get
x = 4 and y = 2
The radius is equal to the distance between (1, −2) and (4, 2)
$r=\sqrt{{\left(4-1\right)}^{2}+{\left(2+2\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\sqrt{9+16}\phantom{\rule{0ex}{0ex}}=5$
Now, the equation of the circle is given by
${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={r}^{2}\phantom{\rule{0ex}{0ex}}⇒{\left(x-1\right)}^{2}+{\left(y+2\right)}^{2}=25\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-2x+4y-20=0$
#### Question 17:
If the lines 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0 are tangents to a circle, then find the radius of the circle. [NCERT EXEMPLAR]
We have 3x − 4y + 4 = 0 and 6x − 8y − 7 = 0
Since, the slope of both the lines are equal.
Hence, the both the lines are parallel.
The distance between the parralel lines is given by
$\left|\frac{{C}_{1}-{C}_{2}}{\sqrt{{A}^{2}+{B}^{2}}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{4+\frac{7}{2}}{\sqrt{{3}^{2}+{4}^{2}}}\right|\phantom{\rule{0ex}{0ex}}=\left|\frac{\frac{15}{2}}{5}\right|\phantom{\rule{0ex}{0ex}}=\frac{3}{2}$
Now, the radius is equal to the half of the distance between the parallel lines(diameter of the circle).
Hence, the radius is given by
$\frac{3}{4}$
#### Question 18:
Show that the point (x, y) given by $x=\frac{2at}{1+{t}^{2}}$ and $y=a\left(\frac{1-{t}^{2}}{1+{t}^{2}}\right)$ lies on a circle for all real values of t such that $-1\le t\le 1$, where a is any given real number. [NCERT EXEMPLAR]
Squaring and adding $x=\frac{2at}{1+{t}^{2}}$ and $y=a\left(\frac{1-{t}^{2}}{1+{t}^{2}}\right)$, we get
${x}^{2}+{y}^{2}={\left(\frac{2at}{1+{t}^{2}}\right)}^{2}+{a}^{2}{\left(\frac{1-{t}^{2}}{1+{t}^{2}}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=\frac{4{a}^{2}{t}^{2}+{a}^{2}-2{a}^{2}{t}^{2}+{a}^{2}{t}^{4}}{{\left(1+{t}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}=\frac{{a}^{2}+2{a}^{2}{t}^{2}+{a}^{2}{t}^{4}}{{\left(1+{t}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={a}^{2}\frac{{\left(1+{t}^{2}\right)}^{2}}{{\left(1+{t}^{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}={a}^{2}\phantom{\rule{0ex}{0ex}}$
Since, the above equation represents the equation of a circle, hence points (x, y) lies on the circle.
#### Question 19:
The circle x2 + y2 − 2x − 2y + 1 = 0 is rolled along the positive direction of x-axis and makes one complete roll. Find its equation in new-position.
Centre of the given circle =
Radius of the given circle = 1
This circle is rolled along the positive direction of the x-axis. When it makes one complete roll, its centre moves horizontally through a distance equal to its circumference, i.e 2$\mathrm{\pi }$.
Thus, the coordinates of the centre of the new circle will be .
Hence, the required equation of the circle is ${\left(\mathrm{x}-1-2\mathrm{\pi }\right)}^{2}+{\left(y-1\right)}^{2}=1$.
#### Question 20:
One diameter of the circle circumscribing the rectangle ABCD is 4y = x + 7. If the coordinates of A and B are (−3, 4) and (5, 4) respectively, find the equation of the circle.
Clearly, the centre of the circle lies on the line 4y = x + 7.
The circle passes through A (−3, 4) and B (5, 4).
The slope of the segment joining A and B is zero.
Therefore, the slope of the perpendicular bisector of AB is not defined.
Hence, the perpendicular bisector of AB will be parallel to the y-axis and will pass through .
The equation of the perpendicular bisector is $x=1$.
The intersection point of the perpendicular bisector and 4y = x + 7 is .
∴ Centre =
Radius = $\sqrt{{\left(5-1\right)}^{2}+{\left(4-2\right)}^{2}}=\sqrt{20}$
Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-2x-4y-15=0$.
#### Question 21:
If the line 2x y + 1 = 0 touches the circle at the point (2, 5) and the centre of the circle lies on the line x + y − 9 = 0. Find the equation of the circle.
According to question, the centre of the required circle lies on the line x + y − 9 = 0.
Let the coordinates of the centre be .
Let the radius of the circle be a.
Here, a is the distance of the centre from the line 2x y + 1 = 0.
Therefore, the equation of the circle is ${\left(x-t\right)}^{2}+{\left(y-\left(9-t\right)\right)}^{2}={a}^{2}$. ...(2)
The circle passes through (2, 5).
${\left(2-t\right)}^{2}+{\left(5-\left(9-t\right)\right)}^{2}={a}^{2}$
Substituting t = 6 in (1):
Substituting the values of ${a}^{2}$ and t in equation (2), we find the required equation of circle to be ${\left(x-6\right)}^{2}+{\left(y-3\right)}^{2}=20$.
#### Question 1:
Find the coordinates of the centre and radius of each of the following circles:
(i) x2 + y2 + 6x − 8y − 24 = 0
(ii) 2x2 + 2y2 − 3x + 5y = 7
(iii) 1/2 (x2 + y2) + x cos θ + y sin θ − 4 = 0
(iv) x2 + y2axby = 0
(i) The given equation can be rewritten as ${x}^{2}+{y}^{2}+2\left(3\right)x-2\left(4\right)y-24=0$.
∴ Centre = $\left(-3,-4\right)$
And, radius = $\sqrt{{\left(3\right)}^{2}+{\left(4\right)}^{2}+24}=\sqrt{49}=7$
(ii) The given equation can be rewritten as ${x}^{2}+{y}^{2}-\frac{3x}{2}+\frac{5y}{2}-\frac{7}{2}=0$.
∴ Centre = $\left(\frac{3}{4},\frac{-5}{4}\right)$
And, radius = $\sqrt{{\left(\frac{3}{4}\right)}^{2}+{\left(\frac{-5}{4}\right)}^{2}+\frac{7}{2}}=\sqrt{\frac{34+56}{16}}=\sqrt{\frac{90}{16}}=\frac{3\sqrt{10}}{4}$
(iii) The given equation can be rewritten as ${x}^{2}+{y}^{2}+2x\mathrm{cos}\theta +2y\mathrm{sin}\theta -8=0$.
∴ Centre = $\left(-\mathrm{cos}\theta ,-\mathrm{sin}\theta \right)$
And, radius = $\sqrt{{\left(-\mathrm{cos}\theta \right)}^{2}+{\left(-\mathrm{sin}\theta \right)}^{2}+8}=\sqrt{1+8}=3$
(iv) The given equation can be rewritten as ${x}^{2}+{y}^{2}-\frac{2ax}{2}-\frac{2by}{2}=0$.
∴ Centre = $\left(\frac{a}{2},\frac{b}{2}\right)$
And, radius = $\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{b}{2}\right)}^{2}}=\frac{1}{2}\sqrt{{a}^{2}+{b}^{2}}$
#### Question 2:
Find the equation of the circle passing through the points:
(i) (5, 7), (8, 1) and (1, 3)
(ii) (1, 2), (3, −4) and (5, −6)
(iii) (5, −8), (−2, 9) and (2, 1)
(iv) (0, 0), (−2, 1) and (−3, 2)
(i) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It passes through (5, 7), (8, 1) and (1, 3).
Substituting the coordinates of these points in equation (1):
$74+10g+14f+c=0$ ...(2)
$65+16g+2f+c=0$ ...(3)
$10+2g+6f+c=0$ ...(4)
Simplifying (2), (3) and (4):
Equation of the required circle:
${x}^{2}+{y}^{2}-\frac{29x}{3}-\frac{19y}{3}+\frac{56}{3}=0$
$⇒$$3\left({x}^{2}+{y}^{2}\right)-29x-19y+56=0$
(ii) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It passes through (1, 2), (3, −4) and (5, −6).
Substituting the coordinates of these points in equation (1):
$5+2g+4f+c=0$ ...(2)
$25+6g-8f+c=0$ ...(3)
$61+10g-12f+c=0$ ...(4)
Simplifying (2), (3) and (4):
The equation of the required circle is ${x}^{2}+{y}^{2}-22x-4y+25=0$.
(iii) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It passes through (5, −8), (−2, 9) and (2, 1).
Substituting the coordinates of these points in equation (1):
$89+10g-16f+c=0$ ...(2)
$85-4g+18f+c=0$ ...(3)
$5+4g+2f+c=0$ ...(4)
Simplifying (2), (3) and (4):
The equation of the required circle is ${x}^{2}+{y}^{2}+116x+48y-285=0$.
(iv) Let the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It passes through (0, 0), (−2, 1) and (−3, 2).
Substituting the coordinates of these points in equation (1):
$c=0$ ...(2)
$5-4g+2f+c=0$ ...(3)
$13-6g+4f+c=0$ ...(4)
Simplifying (2), (3) and (4):
The equation of the required circle is ${x}^{2}+{y}^{2}-3x-11y=0$.
#### Question 3:
Find the equation of the circle which passes through (3, −2), (−2, 0) and has its centre on the line 2xy = 3.
Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It is given that the circle passes through (3, −2), (−2, 0).
$13+6g-4f+c=0$ ...(2)
$4-4g+c=0$ ...(3)
The centre lies on the line 2xy = 3.
$-2g+f-3=0$ ...(4)
Solving (2), (3) and (4):
Hence, the required equation of circle is ${x}^{2}+{y}^{2}+3x+12y+2=0$.
#### Question 4:
Find the equation of the circle which passes through the points (3, 7), (5, 5) and has its centre on the line x − 4y = 1.
Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It is given that the circle passes through (3, 7), (5, 5).
$58+6g+14f+c=0$ ...(2)
$50+10g+10f+c=0$ ...(3)
The centre lies on the line x − 4y = 1.
$-g+4f-1=0$ ...(4)
Solving (2), (3) and (4):
Hence, the required equation of the circle is ${x}^{2}+{y}^{2}+6x+2y-90=0$.
#### Question 5:
Show that the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic.
Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It is given that the circle passes through (3, −2), (1, 0), (−1, −2).
$13+6g-4f+c=0$ ...(2)
$1+2g+c=0$ ...(3)
$5-2g-4f+c=0$ ...(4)
Solving (2), (3) and (4):
Theerefore, the equation of the circle is ${x}^{2}+{y}^{2}-2x+4y+1=0$. ...(5)
We see that the point (1, −4) satisfies the equation (5).
Hence, the points (3, −2), (1, 0), (−1, −2) and (1, −4) are concyclic.
#### Question 6:
Show that the points (5, 5), (6, 4), (−2, 4) and (7, 1) all lie on a circle, and find its equation, centre and radius.
Let the required equation of the circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$. ...(1)
It is given that the circle passes through (5, 5), (6, 4), (−2, 4).
$50+10g+10f+c=0$ ...(2)
$52+12g+8f+c=0$ ...(3)
$20-4g+8f+c=0$ ...(4)
Solving (2), (3) and (4):
Thus, the equation of the circle is ${x}^{2}+{y}^{2}-4x-2y-20=0$. ...(5)
We see that the point (7, 1) satisfies equation (5).
Hence, the points (5, 5), (6, 4), (−2, 4) and (7, 1) lie on the circle.
Also, centre of the required circle =
Radius of the required circle = $\sqrt{4+1+20}=5$
#### Question 7:
Find the equation of the circle which circumscribes the triangle formed by the lines
(i) x + + 3 = 0, x − y + 1 = 0 and x = 3
(ii) 2x + y − 3 = 0, x + y − 1 = 0 and 3x + 2y − 5 = 0
(iii) x + y = 2, 3x − 4y = 6 and x − y = 0.
(iv) y = x + 2, 3y = 4x and 2y = 3x.
In $∆$ABC:
(i) Let AB represent the line x + + 3 = 0. ...(1)
Let BC represent the line x − y + 1 = 0. ...(2)
Let CA represent the line x = 3. ...(3)
Intersection point of (1) and (3) is $\left(3,-6\right)$.
Intersection point of (1) and (2) is (−2, −1).
Intersection point of (2) and (3) is (3, 4).
Therefore, the coordinates of A, B and C are $\left(3,-6\right)$, (−2, −1) and (3, 4), respectively.
Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.
∴ $45+6g-12f+c=0$
$5-4g-2f+c=0$
$25+6g+8f+c=0$
Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}-6x+2y-15=0$.
(ii) In $∆$ABC:
Let AB represent the line 2x + y − 3 = 0. ...(1)
Let BC represent the line x + y − 1 = 0. ...(2)
Let CA represent the line 3x + 2y − 5 = 0. ...(3)
Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, −1).
Intersection point of (2) and (3) is (3, −2).
The coordinates of A, B and C are (1, 1), (2, −1) and (3, −2), respectively.
Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.
∴ $2+2g+2f+c=0$
$5+4g-2f+c=0$
$13+6g-4f+c=0$
Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}-13x-5y+16=0$.
(iii) In $∆$ABC:
Let AB represent the line x + y = 2. ...(1)
Let BC represent the line 3x − 4y = 6. ...(2)
Let CA represent the line x − y = 0. ...(3)
Intersection point of (1) and (3) is (1, 1).
Intersection point of (1) and (2) is (2, 0).
Intersection point of (2) and (3) is (−6, −6).
The coordinates of A, B and C are (1, 1), (2, 0) and (−6, −6), respectively.
Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.
∴ $2+2g+2f+c=0$
$4+4g+c=0$
$72-12g-12f+c=0$
Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}+4x+6y-12=0$.
(iv)
In $∆$ABC:
(i) Let AB represent the line y = x + 2 ...(1)
Let BC represent the line 3y = 4x ...(2)
Let CA represent the line 2y = 3x ...(3)
Intersection point of (1) and (3) is (4, 6)
Intersection point of (1) and (2) is (6, 8).
Intersection point of (2) and (3) is (0, 0).
Therefore, the coordinates of A, B and C are (4, 6), (6, 8) and (0, 0) respectively.
Let the equation of the circumcircle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through A, B and C.
∴ $52+8g+12f+c=0$, $100+12g+16f+c=0$
and
$0++0+0+0+c=0\phantom{\rule{0ex}{0ex}}⇒c=0$
Hence, the required equation of the circumcircle is ${x}^{2}+{y}^{2}-46x+22y=0$.
#### Question 8:
Prove that the centres of the three circles x2 + y2 − 4x − 6y − 12 = 0, x2 + y2 + 2x + 4y − 10 = 0 and x2 + y2 − 10x − 16y − 1 = 0 are collinear.
The given equations of the circles are as follows:
x2 + y2 − 4x − 6y − 12 = 0, ...(1)
x2 + y2 + 2x + 4y − 10 = 0 ...(2)
And, x2 + y2 − 10x − 16y − 1 = 0 ...(3)
The centre of circle (1) is (2, 3).
The centre of circle (2) is (−1, −2).
The centre of circle (3) is (5, 8).
The area of the triangle formed by the points (2, 3), (−1, −2) and (5, 8) is .
Hence, the centres of the circles x2 + y2 − 4x − 6y − 12 = 0, x2 + y2 + 2x + 4y − 10 = 0 and x2 + y2 − 10x − 16y − 1 = 0 are collinear.
#### Question 9:
Prove that the radii of the circles x2 + y2 = 1, x2 + y2 − 2x − 6y − 6 = 0 and x2 + y2 − 4x − 12y − 9 = 0 are in A.P.
Let the radii of the circles x2 + y2 = 1, x2 + y2 − 2x − 6y − 6 = 0 and x2 + y2 − 4x − 12y − 9 = 0 be , respectively.
Now, ${r}_{2}-{r}_{1}={r}_{3}-{r}_{2}=3$
are in A.P.
#### Question 10:
Find the equation of the circle which passes through the origin and cuts off chords of lengths 4 and 6 on the positive side of the x-axis and y-axis respectively.
According to the question, the circle passes through the origin.
Let the equation of the circle be ${x}^{2}+{y}^{2}-2hx-2ky=0$.
The circle cuts off chords of lengths 4 and 6 on the positive sides of the x-axis and the y-axis, respectively.
∴ Centre =
∴ Required equation:
$⇒{x}^{2}+{y}^{2}-4x-6y=0$
#### Question 11:
Find the equation of the circle concentric with the circle x2 + y2 − 6x + 12y + 15 = 0 and double of its area.
Let the equation of the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
The centre of the circle x2 + y2 − 6x + 12y + 15 = 0 is (3, −6).
Area of the required circle = $2\mathrm{\pi }{r}^{\mathit{2}}$
Here, r = radius of the given circle
Now, r = $\sqrt{9+36-15}=\sqrt{30}$
∴ Area of the required circle = $2\mathrm{\pi }\left(30\right)=60\mathrm{\pi }$
Let R be the radius of the required circle.
$60\mathrm{\pi }=\mathrm{\pi }{R}^{\mathit{2}}\mathit{⇒}{R}^{\mathit{2}}\mathit{=}60$
Thus, the equation of the required circle is ${\left(x-3\right)}^{2}+{\left(y+6\right)}^{2}=60$, i.e. ${x}^{2}+{y}^{2}-6x+12y=15$.
#### Question 12:
Find the equation to the circle which passes through the points (1, 1) (2, 2) and whose radius is 1. Show that there are two such circles.
Let the equation of the required circle be ${x}^{2}+{y}^{2}+2gx+2fy+c=0$.
It passes through (1, 1) and (2, 2).
$2g+2f+c=-2$ ...(1)
And, $4g+4f+c=-8$ ...(2)
From (1) and (2), we have:
$-2g-2f=6⇒g+f=-3$ ...(3)
∴ From (2) and (3), we have:
$c=4$
Using (3), we get:
$g=-2,-1$
Correspondingly, we have:
$f=-1,-2$
Therefore, the required equations of the circles are ${x}^{2}+{y}^{2}-4x-2y+4=0$ and ${x}^{2}+{y}^{2}-2x-4y+4=0$.
Hence, there are two such circles.
#### Question 13:
Find the equation of the circle concentric with x2 + y2 − 4x − 6y − 3 = 0 and which touches the y-axis.
Since, the circles are concentric.
$⇒$Centre of required circle = Centre of x2 + y2 − 4x − 6y − 3 = 0
The centre of the required circle is (2, 3).
We know that if a circle with centre (h, k) touches the y-axis, then h is the radius of the circle.
∴ Equation of the circle:
${\left(x-2\right)}^{2}+{\left(y-3\right)}^{2}={2}^{2}$
$⇒{x}^{2}+{y}^{2}-4x-6y+9=0$
#### Question 14:
If a circle passes through the point (0, 0),(a, 0),(0, b) then find the coordinates of its centre.
The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0
Now, it is passing through (0, 0)
c = 0
Also, it is passing through (a, 0)
∴ a2 + 2ag = 0
a(a + 2g) = 0
a + 2g = 0
$⇒g=-\frac{a}{2}$
Again, it is passing through (0, b)
b2 + 2bf = 0
b(b + 2f) = 0
b + 2f = 0
$⇒f=-\frac{b}{2}$
The coordinates of its centre are given by
#### Question 15:
Find the equation of the circle which passes through the points (2, 3) and (4,5) and the centre lies on the straight line y − 4x + 3 = 0. [NCERT EXEMPLAR]
The general equation of the circle is x2 + y2 + 2gx + 2fy + c = 0 where the centre of the circle is (−g, −f)
Now, it is passing through (2, 3)
∴ 13 + 4g + 6f + c = 0 .....(1)
Also, it is passing through (4, 5)
∴ 41 + 8g + 10f + c = 0 .....(2)
g=a2
Now, the centre lies on the straight line y − 4x + 3 = 0
∴ −f + 4g + 3 = 0 .....(3)
g=a2
Solving (1), (2) and (3), we get
g = −2, f = −5 and c = 25
The equation of the circle is given by x2 + y2 − 4x − 10y + 25 = 0
#### Question 1:
Find the equation of the circle, the end points of whose diameter are (2, −3) and (−2, 4). Find its centre and radius.
(2, −3) and (−2, 4) are the ends points of the diameter of a circle. The equation of this circle is $\left(x-2\right)\left(x+2\right)+\left(y+3\right)\left(y-4\right)=0$.
Equation (1) can be rewritten as
${x}^{2}+{\left(y-\frac{1}{2}\right)}^{2}-\frac{1}{4}-16=0\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{\left(y-\frac{1}{2}\right)}^{2}=\frac{65}{4}$
∴ Centre is $\left(0,\frac{1}{2}\right)$ and radius is $\frac{\sqrt{65}}{2}$.
#### Question 2:
Find the equation of the circle the end points of whose diameter are the centres of the circles x2 + y2 + 6x − 14y − 1 = 0 and x2 + y2 − 4x + 10y − 2 = 0.
Given:
${x}^{2}+{y}^{2}+6x-14y-1=0$ ...(1)
And, ${x}^{2}+{y}^{2}-4x+10y-2=0$ ...(2)
Equations (1) and (2) can be rewritten as follows:
${\left(x+3\right)}^{2}+{\left(y-7\right)}^{2}=59$
And, ${\left(x-2\right)}^{2}+{\left(y+5\right)}^{2}=31$
Thus, the centres of the circles are (−3, 7) and (2, −5).
Hence, the equation of the circle, the end points of whose diameter are the centres of the given circles, is$\left(x+3\right)\left(x-2\right)+\left(y-7\right)\left(y+5\right)=0$, i.e. ${x}^{2}+{y}^{2}+x-2y-41=0$.
#### Question 3:
The sides of a square are x = 6, x = 9, y = 3 and y = 6. Find the equation of a circle drawn on the diagonal of the square as its diameter.
According to the question:
Sides of the square are x = 6, x = 9, y = 3 and y = 6.
The vertices of the square are (6, 6), (9, 6), (9, 3) and (6, 3).
And, the vertices of two diagonals are (6, 6), (9, 3) and (9, 6), (6, 3).
Hence, the equation of the circle is $\left(x-6\right)\left(x-9\right)+\left(y-6\right)\left(y-3\right)$ or ${x}^{2}+{y}^{2}-15x-9y+72=0$.
#### Question 4:
Find the equation of the circle circumscribing the rectangle whose sides are x − 3y = 4, 3x + y = 22, x − 3y = 14 and 3x + y = 62.
Given:
Sides of the rectangle:
x − 3y = 4 ...(1)
3x + y = 22 ...(2)
x − 3y = 14 ...(3)
And, 3x + y = 62 ...(4)
The intersection of (1) and (2) is (7, 1).
The intersection of (2) and (3) is (8, −2).
The intersection of (3) and (4) is (20, 2).
The intersection of (1) and (4) is (19, 5).
Hence, the vertices of the rectangle are (7, −1), (8, −2), (20, 2) and (19, 5).
The vertices of the diagonals are (7, −1), (20, 2) and (19, 5), (8, −2).
Thus, the required equation of the circle is $\left(x-7\right)\left(x-20\right)+\left(y-1\right)\left(y-2\right)=0$ or ${x}^{2}+{y}^{2}-27x-3y+142=0$.
#### Question 5:
Find the equation of the circle passing through the origin and the points where the line 3x + 4y = 12 meets the axes of coordinates.
Putting x = 0 in 3x + 4y = 12:
y = 3
Putting y = 0 in 3x + 4y = 12:
x = 4
Thus, the line 3x + 4y = 12 meets the axes of coordinates at points A (0, 3) and B (4, 0).
The equation of the circle with AB as the diameter is $\left(x-0\right)\left(x-4\right)+\left(y-3\right)\left(y-0\right)=0$ or ${x}^{2}-4x+{y}^{2}-3y=0$.
Hence, the required equation is ${x}^{2}-4x+{y}^{2}-3y=0$.
#### Question 6:
Find the equation of the circle which passes through the origin and cuts off intercepts a and b respectively from x and y - axes.
Case I:
If the required circle passes through the origin and (a, b), then the end points of the diameter of the circle will be (0, 0) and (a, b).
∴ Required equation of circle:
$\left(x-0\right)\left(x-a\right)+\left(y-0\right)\left(y-b\right)$
or ${x}^{2}+{y}^{2}-ax-by=0$
Case II:
If the required circle passes through the origin and (−a, −b), then the end points of the diameter of the circle will be (0, 0) and (−a, −b).
∴ Required equation of circle:
$\left(x-0\right)\left(x+a\right)+\left(y-0\right)\left(y+b\right)$
or ${x}^{2}+{y}^{2}+ax+by=0$
Hence, the equation of the required circle is ${x}^{2}+{y}^{2}±ax±by=0$.
#### Question 7:
Find the equation of the circle whose diameter is the line segment joining (−4, 3) and (12, −1). Find also the intercept made by it on y-axis.
It is given that the end points of the diameter of the circle are (−4, 3) and (12, −1).
∴ Required equation of circle:
$\left(x+4\right)\left(x-12\right)+\left(y-3\right)\left(y+1\right)$
or ${x}^{2}+{y}^{2}-8x-2y-51=0$ ...(1)
Putting x = 0 in (1):
y2 − 2y − 51 = 0
y2 − 2y − 51 = 0
Hence, the intercepts made by it on the y-axis is $1+2\sqrt{13}-1+2\sqrt{13}=4\sqrt{13}$.
#### Question 8:
The abscissae of the two points A and B are the roots of the equation x2 + 2axb2 = 0 and their ordinates are the roots of the equation x2 + 2pxq2 = 0. Find the equation of the circle with AB as diameter. Also, find its radius.
Roots of equation x2 + 2axb2 = 0 are $-a±\sqrt{{a}^{2}+{b}^{2}}$.
Roots of equation x2 + 2pxq2 = 0 are $-p±\sqrt{{p}^{2}+{q}^{2}}$.
Therefore, coordinates of A and B are respectively.
Hence, equation of circle is .
$⇒{\left(x+a\right)}^{2}-{a}^{2}-{b}^{2}+{\left(y+p\right)}^{2}-{p}^{2}-{q}^{2}=0$
$⇒{x}^{2}+{y}^{2}+2ax+2yp-{p}^{2}-{q}^{2}=0$.
Also, radius of circle is $\sqrt{{a}^{2}+{b}^{2}+{p}^{2}+{q}^{2}}$.
#### Question 9:
ABCD is a square whose side is a; taking AB and AD as axes, prove that the equation of the circle circumscribing the square is x2 + y2a (x + y) = 0.
Given:
ABCD is a square with side a units.
Let AB and AD represent the x-axis and the y-axis, respectively.
Thus, the coordinates of B and D are (a, 0) and (0, a), respectively.
The end points of the diameter of the circle circumscribing the square are B and D.
Thus, equation of the circle circumscribing the square is $\left(x-a\right)\left(x-0\right)+\left(y-0\right)\left(y-a\right)=0$ or ${x}^{2}+{y}^{2}-a\left(x+y\right)=0$.
#### Question 10:
The line 2xy + 6 = 0 meets the circle x2 + y2 − 2y − 9 = 0 at A and B. Find the equation of the circle on AB as diameter.
The equation of the line can be rewritten as $x=\frac{y-6}{2}$.
Substituting the value of x in the equation of the circle, we get:
At y = 0, x = −3
At y = 4, x = −1
Therefore, the coordinates of A and B are .
∴ Equation of the circle with AB as its diameter:
$\left(x+1\right)\left(x+3\right)+\left(y-4\right)\left(y-0\right)=0$
$⇒{x}^{2}+4x+{y}^{2}-4y+3=0$
#### Question 11:
Find the equation of the circle which circumscribes the triangle formed by the lines x = 0, y = 0 and lx + my = 1.
The coordinates of A and B are , respectively.
Here, the end points of the diameter of the circumcircle are A and B.
∴ Required equation of the circle:
$\left(x-0\right)\left(x-\frac{1}{l}\right)+\left(y-\frac{1}{m}\right)\left(y-0\right)=0$
$⇒{x}^{2}-\frac{x}{l}+{y}^{2}-\frac{y}{m}=0$
#### Question 12:
Find the equations of the circles which pass through the origin and cut off equal chords of $\sqrt{2}$ units from the lines y = x and y = − x.
Suppose $a=\sqrt{2}$
From the figure, we see that there will be four circles that pass through the origin and cut off equal chords of length a from the straight lines $y=±x$.
AB, BC, CD and DA are the diameters of the four circles.
Also, ${C}_{1}A=\frac{a}{\sqrt{2}}=O{C}_{1}$
Thus, the coordinates of A are $\left(\frac{a}{\sqrt{2}},\frac{a}{\sqrt{2}}\right)$.
In the same way, we can find the coordinates of B, C and D as $\left(\frac{-a}{\sqrt{2}},\frac{a}{\sqrt{2}}\right),$ $\left(\frac{-a}{\sqrt{2}},\frac{-a}{\sqrt{2}}\right)$ and $\left(\frac{a}{\sqrt{2}},\frac{-a}{\sqrt{2}}\right)$, respectively.
The equation of the circle with AD as the diameter is $\left(x-\frac{a}{\sqrt{2}}\right)\left(x-\frac{a}{\sqrt{2}}\right)+\left(y-\frac{a}{\sqrt{2}}\right)\left(y+\frac{a}{\sqrt{2}}\right)=0$, which can be rewritten as ${x}^{2}+{y}^{2}-\sqrt{2}ax=0$, i.e. ${x}^{2}+{y}^{2}-2x=0$.
Similarly, the equations of the circles with BC, CD and AB as the diameters are ${x}^{2}+{y}^{2}+2x=0$, ${x}^{2}+{y}^{2}+2y=0$ and ${x}^{2}+{y}^{2}-2y=0$, respectively.
#### Question 1:
If the equation of a circle is λx2 + (2λ − 3) y2 − 4x + 6y − 1 = 0, then the coordinates of centre are
(a) (4/3, −1)
(b) (2/3, −1)
(c) (−2/3, 1)
(d) (2/3, 1)
(b) $\left(\frac{2}{3},-1\right)$
To find the centre:
Coefficient of x2 = Coefficient of y2
Therefore, the given equation can be rewritten as $3{x}^{2}+3{y}^{2}-4x+6y-1=0$.
Thus, the coordinates of the centre is $\left(\frac{2}{3},-1\right)$.
#### Question 2:
If 2x2 + λxy + 2y2 + (λ − 4) x + 6y − 5 = 0 is the equation of a circle, then its radius is
(a) $3\sqrt{2}$
(b) $2\sqrt{3}$
(c) $2\sqrt{2}$
(d) none of these
(d) none of these
The given equation is 2x2 + λxy + 2y2 + (λ − 4) x + 6y − 5 = 0, which can be rewritten as ${x}^{2}+\frac{\lambda xy}{2}+{y}^{2}+\frac{\left(\lambda -4\right)}{2}x+3y-\frac{5}{2}=0$.
Comparing the given equation with ${x}^{2}+{y}^{2}+2gx+2fy+c=0$, we get:
$\lambda =0$
${x}^{2}+{y}^{2}-2x+3y-\frac{5}{2}=0$
∴ Radius = $\sqrt{{\left(-1\right)}^{2}+{\left(\frac{3}{2}\right)}^{2}+\frac{5}{2}}=\sqrt{1+\frac{9}{4}+\frac{5}{2}}=\sqrt{\frac{23}{4}}=\frac{\sqrt{23}}{2}$
#### Question 3:
The equation x2 + y2 + 2x − 4y + 5 = 0 represents
(a) a point
(b) a pair of straight lines
(c) a circle of non-zero radius
(d) none of these
(a) a point
The radius of the given circle = $\sqrt{{1}^{2}+{\left(-2\right)}^{2}-5}=0$
Hence, the radius of the given circle is zero, which represents a point.
#### Question 4:
If the equation (4a − 3) x2 + ay2 + 6x − 2y + 2 = 0 represents a circle, then its centre is
(a) (3, −1)
(b) (3, 1)
(c) (−3, 1)
(d) none of these
(c) (−3, 1)
If the equation (4a − 3) x2 + ay2 + 6x − 2y + 2 = 0 represents a circle, then we have:
Coefficient of x2 = Coefficient of y2
$4a-3=a\phantom{\rule{0ex}{0ex}}$
a = 1
∴ Equation of the circle = ${x}^{2}+{y}^{2}+6x-2y+2=0$
Thus, the coordinates of the centre is .
#### Question 5:
The radius of the circle represented by the equation
3x2 + 3y2 + λxy + 9x + (λ − 6) y + 3 = 0 is
(a) $\frac{3}{2}$
(b) $\frac{\sqrt{17}}{2}$
(c) 2/3
(d) none of these
(a) $\frac{3}{2}$
The equation of the circle is 3x2 + 3y2 + λxy + 9x + (λ − 6) y + 3 = 0.
∴ Coefficient of xy = 0
$⇒\lambda =0$
Therefore, the radius of the circle is $\sqrt{{\left(\frac{3}{2}\right)}^{2}+{\left(-1\right)}^{2}-1}=\frac{3}{2}$.
#### Question 6:
The number of integral values of λ for which the equation x2 + y2 + λx + (1 − λ) y + 5 = 0 is the equation of a circle whose radius cannot exceed 5, is
(a) 14
(b) 18
(c) 16
(d) none of these
According to the question:
Thus, the number of integral values of $\lambda$ is 16.
#### Question 7:
The equation of the circle passing through the point (1, 1) and having two diameters along the pair of lines x2y2 −2x + 4y − 3 = 0, is
(a) x2 + y2 − 2x − 4y + 4 = 0
(b) x2 + y2 + 2x + 4y − 4 = 0
(c) x2 + y2 − 2x + 4y + 4 = 0
(d) none of these
(a) x2 + y2 − 2x − 4y + 4 = 0
Let the required equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
Comparing the given equation x2y2 −2x + 4y − 3 = 0 with $a{x}^{2}+b{y}^{2}+2hxy+2gx+2fy+c=0$, we get:
Intersection point = =
Thus, the centre of the circle is .
The equation of the required circle is ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={a}^{2}$.
Since circle passes through (1, 1), we have:
$1={a}^{2}$
∴ Equation of the required circle:
${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}=1$
$⇒$ ${x}^{2}+{y}^{2}-2x-4y+4=0$
#### Question 8:
If the centroid of an equilateral triangle is (1, 1) and its one vertex is (−1, 2), then the equation of its circumcircle is
(a) x2 + y2 − 2x − 2y − 3 = 0
(b) x2 + y2 + 2x − 2y − 3 = 0
(c) x2 + y2 + 2x + 2y − 3 = 0
(d) none of these
(a) x2 + y2 − 2x − 2y − 3 = 0
The centre of the circumcircle is (1, 1).
Radius of the circumcircle = $\sqrt{{\left(1+1\right)}^{2}+{\left(1-2\right)}^{2}}=\sqrt{5}$
∴ Equation of the circle:
${\left(x-1\right)}^{2}+{\left(y-1\right)}^{2}=5$
$⇒{x}^{2}+{y}^{2}-2x-2y-3=0$
#### Question 9:
If the point (2, k) lies outside the circles x2 + y2 + x − 2y − 14 = 0 and x2 + y2 = 13 then k lies in the interval
(a) (−3, −2) ∪ (3, 4)
(b) −3, 4
(c) (−∞, −3) ∪ (4, ∞)
(d) (−∞, −2) ∪ (3, ∞)
(c) (−∞, −3) ∪ (4, ∞)
The given equations of the circles are x2 + y2 + x − 2y − 14 = 0 and x2 + y2 = 13.
Since (2, k) lies outside the given circles, we have:
$4+{k}^{2}+2-2k-14>0$ and $4+{k}^{2}>13$
$⇒{k}^{2}-2k-8>0$ and ${k}^{2}>9$
$⇒\left(k-4\right)\left(k+2\right)>0$ and ${k}^{2}>9$
and
$⇒k\in \left(-\infty ,-3\right)\cup \left(4,\infty \right)$
#### Question 10:
If the point (λ, λ + 1) lies inside the region bounded by the curve $x=\sqrt{25-{y}^{2}}$ and y-axis, then λ belongs to the interval
(a) (−1, 3)
(b) (−4, 3)
(c) (−∞, −4) ∪ (3, ∞)
(d) none of these
(a) (−1, 3)
The given equation of the curve is ${x}^{2}+{y}^{2}=25$.
Since (λ, λ + 1) lies inside the region bounded by the curve ${x}^{2}+{y}^{2}=25$ and the y-axis, we have:
${\lambda }^{2}+{\left(\lambda +1\right)}^{2}<25$,
#### Question 11:
The equation of the incircle formed by the coordinate axes and the line 4x + 3y = 6 is
(a) x2 + y2 − 6x −6y + 9 = 0
(b) 4 (x2 + y2xy) + 1 = 0
(c) 4 (x2 + y2 + x + y) + 1 = 0
(d) none of these
(b) 4 (x2 + y2xy) + 1 = 0
The line 4x + 3y = 6 cuts the coordinate axes at .
The coordinates of the incentre is .
Here,
Thus, the coordinates of the incentre:
The equation of the incircle:
${\left(x-\frac{1}{2}\right)}^{2}+{\left(y-\frac{1}{2}\right)}^{2}={a}^{2}$
Also, radius of the incircle = $\frac{\sqrt{s\left(s-a\right)\left(s-b\right)\left(s-c\right)}}{s}$
Here, $s=\frac{a+b+c}{2}=\frac{\frac{5}{2}+\frac{3}{2}+2}{2}=3$
∴ Radius of the incircle = $\frac{\sqrt{3\left(3-a\right)\left(3-b\right)\left(3-c\right)}}{3}$
The equation of circle:
${\left(x-\frac{1}{2}\right)}^{2}+{\left(y-\frac{1}{2}\right)}^{2}=\frac{1}{4}$
#### Question 12:
If the circles x2 + y2 = 9 and x2 + y2 + 8y + c = 0 touch each other, then c is equal to
(a) 15
(b) − 15
(c) 16
(d) − 16
(a) 15
The centre of the circle x2 + y2 = 9 is (0, 0).
Let us denote it by C1.
The centre of the circle x2 + y2+ 8y + c = 0 is (0, −4).
Let us denote it by C2.
The radius of x2 + y2 = 9 is 3 units.
x2 + y2+ 8y + c = 0
Therefore, the radius of the above circle is $\sqrt{16-c}$.
Let the circles touch each other at P.
∴ C1C2 = PC2 + PC1
⇒ PC2 = 4 − 3 = 1
⇒ PC2 = 1 = $\sqrt{16-c}$
c = 15
#### Question 13:
If the circle x2 + y2 + 2ax + 8y + 16 = 0 touches x-axis, then the value of a is
(a) ± 16
(b) ± 4
(c) ± 8
(d) ± 1
(b) ± 4
The equation of the circle is x2 + y2 + 2ax + 8y + 16 = 0.
Its centre is $\left(-a,-4\right)$ and its radius is a units.
Since the circle touches the x-axis, we have:
$\sqrt{{\left(-a+a\right)}^{2}+{\left(4-0\right)}^{2}}=a$
$a=±4$
#### Question 14:
The equation of a circle with radius 5 and touching both the coordinate axes is
(a) x2 + y2 ± 10x ± 10y + 5 = 0
(b) x2 + y2 ± 10x ± 10y = 0
(c) x2 + y2 ± 10x ± 10y + 25 = 0
(d) x2 + y2 ± 10x ± 10y + 51 = 0
(c) x2 + y2 ± 10x ± 10y + 25 = 0
Case I: If the circle lies in the first quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}-2ax-2ay+{a}^{2}=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is ${x}^{2}+{y}^{2}-10x-10y+25=0$.
Case II: If the circle lies in the second quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}+2ax-2ay+{a}^{2}=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is ${x}^{2}+{y}^{2}+10x-10y+25=0$.
Case III: If the circle lies in the third quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}+2ax+2ay+{a}^{2}=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is ${x}^{2}+{y}^{2}+10x+10y+25=0$.
Case IV: If the circle lies in the fourth quadrant:
The equation of a circle that touches both the coordinate axes and has radius a is ${x}^{2}+{y}^{2}-2ax+2ay+{a}^{2}=0$.
The given radius of the circle is 5 units, i.e. $a=5$.
Thus, the equation of the circle is ${x}^{2}+{y}^{2}-10x+10y+25=0$.
Hence, the required equation of the circle is x2 + y2 ± 10x ± 10y + 25 = 0.
#### Question 15:
The equation of the circle passing through the origin which cuts off intercept of length 6 and 8 from the axes is
(a) x2 + y2 − 12x − 16y = 0
(b) x2 + y2 + 12x + 16y = 0
(c) x2 + y2 + 6x + 8y = 0
(d) x2 + y2 − 6x − 8y = 0
(d) x2 + y2 − 6x − 8y = 0
The centre of the required circle is .
The radius of the required circle is $\sqrt{{3}^{2}+{4}^{2}}=\sqrt{25}=5$.
Hence, the equation of the circle is as follows:
${\left(x-3\right)}^{2}+{\left(y-4\right)}^{2}={5}^{2}$
$⇒$ ${x}^{2}+{y}^{2}-6x-8y=0$
#### Question 16:
The equation of the circle concentric with x2 + y2 − 3x + 4yc = 0 and passing through (−1, −2) is
(a) x2 + y2 − 3x + 4y − 1 = 0
(b) x2 + y2 − 3x + 4y = 0
(c) x2 + y2 − 3x + 4y + 2 = 0
(d) none of these
(b) x2 + y2 − 3x + 4y = 0
The centre of the circle x2 + y2 − 3x + 4yc = 0 is $\left(\frac{3}{2},-2\right)$.
Therefore, the centre of the required circle is $\left(\frac{3}{2},-2\right)$.
The equation of the circle is ${\left(x-\frac{3}{2}\right)}^{2}+{\left(y+2\right)}^{2}={a}^{2}$. ...(1)
Also, circle (1) passes through (−1, −2).
$a=\frac{5}{2}$
Substituting the value of a in equation (1):
${\left(x-\frac{3}{2}\right)}^{2}+{\left(y+2\right)}^{2}={\left(\frac{5}{2}\right)}^{2}\phantom{\rule{0ex}{0ex}}⇒\frac{{\left(2x-3\right)}^{2}}{4}+{\left(y+2\right)}^{2}=\frac{25}{4}\phantom{\rule{0ex}{0ex}}⇒{\left(2x-3\right)}^{2}+4{\left(y+2\right)}^{2}=25\phantom{\rule{0ex}{0ex}}⇒{x}^{2}+{y}^{2}-3x+4y=0$
Hence, the required equation of the circle is ${x}^{2}+{y}^{2}-3x+4y=0$.
#### Question 17:
The circle x2 + y2 + 2gx + 2fy + c = 0 does not intersect x-axis, if
(a) g2 < c
(b) g2 > c
(c) g2 > 2c
(d) none of these
(a) ${g}^{2}
Given:
x2 + y2 + 2gx + 2fy + c = 0 ...(1)
The given circle intersects the x-axis.
The equation of circle becomes x2 + 2gx + c = 0. ...(2)
Solving equation (2):
∴ Discriminant, D = $\sqrt{4{g}^{2}-4c}\ge 0$
$⇒4{g}^{2}-4c\ge 0\phantom{\rule{0ex}{0ex}}⇒{g}^{2}-c\ge 0\phantom{\rule{0ex}{0ex}}⇒{g}^{2}\ge c$
Hence, if ${g}^{2}, then the given circle will not intersect the x-axis.
#### Question 18:
The area of an equilateral triangle inscribed in the circle x2 + y2 − 6x − 8y − 25 = 0 is
(a) $\frac{225\sqrt{3}}{6}$
(b) 25π
(c) 50π − 100
(d) none of these
(a) $\frac{225\sqrt{3}}{6}$
Let ABC be the required equilateral triangle.
The equation of the circle is x2 + y2 − 6x − 8y − 25 = 0.
Therefore, coordinates of the centre O is .
Radius of the circle = OA = OB = OC = $\sqrt{9+16+25}=5\sqrt{2}$
In $∆$BOD, we have:
$\mathrm{sin}60°=\frac{DB}{BO}\phantom{\rule{0ex}{0ex}}⇒DB=\frac{\sqrt{3}}{2}\left(5\sqrt{2}\right)\phantom{\rule{0ex}{0ex}}⇒BC=2BD=\sqrt{3}\left(5\sqrt{2}\right)=5\sqrt{6}$
Now, area of $△ABC$ = $\frac{\sqrt{3}}{4}B{C}^{2}=\frac{\sqrt{3}}{4}{\left(5\sqrt{6}\right)}^{2}=\frac{\sqrt{3}\left(150\right)}{4}=\frac{\sqrt{3}\left(75\right)}{2}=\frac{\sqrt{3}\left(225\right)}{6}$ square units
#### Question 19:
The equation of the circle which touches the axes of coordinates and the line $\frac{x}{3}+\frac{y}{4}=1$ and whose centres lie in the first quadrant is x2 + y2 − 2cx − 2cy + c2 = 0, where c is equal to
(a) 4
(b) 2
(c) 3
(d) 6
(d) 6
The equation of the circle that touches the axes of coordinates is ${x}^{2}+{y}^{2}-2cx-2cy+{c}^{2}=0$.
Also, ${x}^{2}+{y}^{2}-2cx-2cy+{c}^{2}=0$ touches the line $\frac{x}{3}+\frac{y}{4}=1$ or 4x + 3y $-$12 = 0.
Since the circle lies in the first quadrant, it centre is is .
From the figure, we have:
$\left|\frac{4c+3c-12}{\sqrt{{4}^{2}+{3}^{2}}}\right|=c\phantom{\rule{0ex}{0ex}}⇒\frac{7c-12}{5}=c\phantom{\rule{0ex}{0ex}}⇒c=6$
#### Question 20:
If the circles x2 + y2 = a and x2 + y2 − 6x − 8y + 9 = 0, touch externally, then a =
(a) 1
(b) −1
(c) 21
(d) 16
(a) 1
x2 + y2 = a ...(1)
And, x2 + y2 − 6x − 8y + 9 = 0 ...(2)
Let circles (1) and (2) touch each other at point P.
The centre of the circle x2 + y2 = a, O, is (0, 0).
The centre of the circle x2 + y2 − 6x − 8y + 9 = 0, C1, is (3, 4).
Also, radius of circle (1) = $\sqrt{a}$ = OP
Radius of circle (2) = $\sqrt{9+16-9}=4$ = C1P
From figure, we have:
${C}_{\mathit{1}}O\mathit{=}{C}_{\mathit{1}}P\mathit{+}OP\phantom{\rule{0ex}{0ex}}\mathit{⇒}\sqrt{{\mathit{3}}^{\mathit{2}}\mathit{+}{\mathit{4}}^{\mathit{2}}}\mathit{=}4+\sqrt{\mathrm{a}}\phantom{\rule{0ex}{0ex}}\mathit{⇒}5\mathit{=}4+\sqrt{\mathrm{a}}\phantom{\rule{0ex}{0ex}}\mathit{⇒}a\mathit{=}1$
#### Question 21:
If (x, 3) and (3, 5) are the extremities of a diameter of a circle with centre at (2, y), then the values of x and y are
(a) (3, 1)
(b) x = 4, y = 1
(c) x = 8, y = 2
(d) none of these
(d) none of these
The end points of the diameter of a circle are (x, 3) and (3, 5).
According to the question, we have:
#### Question 22:
If (−3, 2) lies on the circle x2 + y2 + 2gx + 2fy + c = 0 which is concentric with the circle x2 + y2 + 6x + 8y − 5 = 0, then c =
(a) 11
(b) −11
(c) 24
(d) none of these
(b) −11
The centre of the circle x2 + y2 + 6x + 8y − 5 = 0 is (−3, −4).
The circle x2 + y2 + 2gx + 2fy + c = 0 is concentric with the circle x2 + y2 + 6x + 8y − 5 = 0.
Thus, the centre of x2 + y2 + 2gx + 2fy + c = 0 is (−3, −4).
Also, it is given that (−3, 2) lies on the circle x2 + y2 + 2gx + 2fy + c = 0.
${\left(-3\right)}^{2}+{2}^{2}+2\left(3\right)\left(-3\right)+2\left(4\right)\left(2\right)+c=0$
$⇒9+4-18+16+c=0\phantom{\rule{0ex}{0ex}}⇒c=-11$
#### Question 23:
Equation of the diameter of the circle x2 + y2 − 2x + 4y = 0 which passes through the origin is
(a) x + 2y = 0
(b) x − 2y = 0
(c) 2x + y = 0
(d) 2xy = 0
(c) 2x + y = 0
Let the diameter of the circle be y = mx.
Since the diameter of the circle passes through its centre, (1, −2) satisfies the equation of the diameter.
$m=-2$
Substituting the value of m in the equation of diameter:
$y=-2x$
Hence, the required equation of the diameter is $2x+y=0$.
#### Question 24:
Equation of the circle through origin which cuts intercepts of length a and b on axes is
(a) x2 + y2 + ax + by = 0
(b) x2 + y2axby = 0
(c) x2 + y2 + bx + ay = 0
(d) none of these
(b) x2 + y2axby = 0
Centre of the circle is $\left(\frac{a}{2},\frac{b}{2}\right)$ and its radius is $\sqrt{{\left(\frac{a}{2}\right)}^{2}+{\left(\frac{b}{2}\right)}^{2}}=\frac{1}{2}\sqrt{{a}^{2}+{b}^{2}}$.
Equation of circle:
${\left(x-\frac{a}{2}\right)}^{2}+{\left(y-\frac{b}{2}\right)}^{2}=\frac{1}{4}\left({a}^{2}+{b}^{2}\right)$
$⇒$${\left(2x-a\right)}^{2}+{\left(2y-b\right)}^{2}=\left({a}^{2}+{b}^{2}\right)$
$⇒$$4{x}^{2}+{a}^{2}-4ax+4{y}^{2}+{b}^{2}-4by={a}^{2}+{b}^{2}$
$⇒$${x}^{2}-ax+{y}^{2}-by=0$
#### Question 25:
If the circles x2 + y2 + 2ax + c = 0 and x2 + y2 + 2by + c = 0 touch each other, then
(a) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{c}$
(b) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{{c}^{2}}$
(c) a + b = 2c
(d) $\frac{1}{a}+\frac{1}{b}=\frac{2}{c}$
(a) $\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{c}$
Given:
x2 + y2 + 2ax + c = 0 ...(1)
And, x2 + y2 + 2by + c = 0 ...(2)
For circle (1), we have:
Centre = $\left(-a,0\right)$ = C1
For circle (2), we have:
Centre = $\left(0,-b\right)$ = C2
Let the circles intersect at point P.
∴ Coordinates of P = Mid point of C1C2
⇒ Coordinates of P = $\left(\frac{-a+0}{2},\frac{0-b}{2}\right)=\left(\frac{-a}{2},\frac{-b}{2}\right)$
Now, we have:
From (3) and (4), we have:
$\frac{{a}^{2}}{2}={a}^{2}-c\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}}{2}=c\phantom{\rule{0ex}{0ex}}⇒\frac{2}{{a}^{2}}=\frac{1}{c}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}+\frac{1}{{a}^{2}}=\frac{1}{c}\phantom{\rule{0ex}{0ex}}⇒\frac{1}{{a}^{2}}+\frac{1}{{b}^{2}}=\frac{1}{c}$
#### Question 26:
The equation of the circle in the first quadrant touching each coordinate axes at a distance of one unit from the origin is
(a) x2 + y2 – 2x – 2y + 1 = 0
(b) x2 + y2 – 2x – 2y – 1 = 0
(c) x2 + y2 – 2x – 2y = 0
(d) x2 + y2 – 2x + 2y – 1 = 0
Since the circle touches both the axis and has radius 1.
i.e. distance of center from x and y axis is 1.
i.e. equation of circle is
(x − 1)2 + (y − 1)2 = 12
i.e. x2 − 2x + 1 + y2 − 2y + 1 = 1
i.e. x2 + y2 − 2x − 2y + 1 = 0
Hence, the correct answer is option A.
#### Question 27:
The equation of the circle having centre (1, –2) and passing through the point of intersection of the lines 3x + y = 14 and 3x + 5y = 18 is
(a) x2 + y2 – 2x + 4y – 20 = 0
(b) x2 + y2 – 2x – 4y – 20 = 0
(c) x2 + y2 + 2x – 4y – 20 = 0
(d) x2 + y2 + 2x + 4y – 20 = 0
correction
Since center of the circle is (1, −2) equation of circle is
(x − 1)2 + (y + 2)2 = r2
Since above circle passes through points of intersection of 3x + y = 14 and correction
3x + 5y = 18
4y = −4
i.e. y = 1
#### Question 28:
Equation of a circle which passes through (3, 6) and touches the axes is
(a) x2 + y2 + 6x + 6y + 3 = 0
(b) x2 + y2 – 6x – 6y – 9 = 0
(c) x2 + y2 – 6x – 6y + 9 = 0
(d) None of these
Since circle touches both axis
⇒ equations of circle is of the form
(xa)2 + (ya)2 = a2
i.e. x2 + a2 – 2ax + y2 – 2ay + a2 = a2
i.e. x2 + y2 – 2ax – 2ay + a2 = 0
Since circle passes through (3, 6)
⇒ (3)2 + (6)2 – 2a(3) – 2a(6) + a2 = 0
i.e. 9 + 36 – 6a – 12a + a2 = 0
i.e. a2 – 18a + 45 = 0
i.e. a2 – 15a – 3a + 45 = 0
i.e. a(a – 15) – 3(a – 15) = 0
i.e. a = 3 or a = 15
i.e. (x – 3)2 + (y – 3)2 = 9
i.e. x2 + y2 – 2 × 3x – 6y + 9 = 0
i.e. x2 + y2 – 6x – 6y + 9 = 0
Hence, the correct answer is option C.
#### Question 29:
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
(a) x2 + y2 = 9a2
(b) x2 + y2 = 16a2
(c) x2 + y2 = 4a2
(d) x2 + y2 = a2
Since origin the center for circle
x2 + y2 r is the required equation and since above circle passes through the vertices of equilateral triangle with medium 3a. The centroid of an equilateral triangle is the center of its circumcircle and the radius of the circle is the distance of any vertex from centroid.
i.e. radius of circle = 2a
∴ equation of circle is x2 + y2 = (2a)2
i.e. x2 + y2 = 4a2
Hence, the correct answer is option C.
#### Question 30:
The equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
(a) x2 + y2 + 13y = 0
(b) 3x2 + 3y2 + 13x + 3 = 0
(c) 6x2 + 6y2 – 13x = 0
(d) x2 + y2 + 13x + 3 = 0
Circle with center as y-axis is of the form
x2 + ( b)2 = r2 ...(1)
also given, circle passes through origin
02 + (0 – b2) = r2
i.e r2 = b2
i.e. equation of circle reduces to,
x2 + (yb)2 = b2
i.e. x2 + y2 – 2by + b2 = b2
i.e. x2 + y2 – 2by = 0
Given circle passes through (2 , 3)
4 + 9 – 2b(3) = 0
i.e. 6b = 13
i.e. b = 13/6
Equation of circle is x2 + y2 – 2 (13/6)y = 0
i.e. 3x2 + 3y2 – 13y = 0
#### Question 1:
The equation of the circle which passes through the point (4, 5) and has its centre at (2, 2) is ____________.
Circle with center (2, 2) has equation of the form (x – 2)2 + (y – 2)2 = r2 (1)
Since circle passes through (4, 5)
⇒ (4 – 2)2 + (5 – 2)2 = r2
i.e. 22 + 32 = r2
i.e. 4 + 9 = r2
i.e. r2 = 13
i.e. equation of circle is (x – 2)2 + (y – 2)2 = 13.
#### Question 2:
The equation of the circle having centre at (3, –4) and touching the line 5x + 12y – 12 = 0 is____________.
Circle with center (3, −4) has equation of the form (− 3)2 + (+ 4)2 = r ....(1)
Since circle touches the line 5+ 12− 12
i.e. perpendicular distance of center to the line gives radius.
#### Question 3:
The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x and 2y = 3x, is __________.
For equations y = x + 2 (1)
3y = 4x (2)
2y = 3x (3)
Intersection point of (1) and (2) is given by
3(x + 2) = 4x
i.e. 3x + 6 = 4x
i.e. x = 6 and y = 8
Intersection point (2) and (3) is (0, 0) since both passes through origin.
Intersection point of (1) and (3) is given by
2(x + 2) = 3x
i.e. 2x + 4 = 3x
i.e. x = 4
y = 6
i.e. (4, 6) is the point intersection
Hence,
The co-ordinate of A, B and C are (4, 6), (6, 8) and (0, 0) respectively,
Let the equation of the circumcircle be x2 + y2 + 2ax + 2by + c = 0
Since the circle passes through A, B and C
i.e. 16 + 36 + 2a 4 + 2b 6 + c = 0
i.e. 52 + 8a + 12b + c = 0 …(4)
also, (6)2 + (8)2 + 12a + 16b + c = 0
i.e. 100 + 12a + 16b + c = 0 …(5)
and 02 + 02 + 2a × 0 + 2b × 0 + c = 0
i.e. c = 0 …(6)
∴ (4) and (5) reduces to,
We get, 39 + 6a + 9b = 0
50 + 6a + 8b = 0
i.e. b = 11
and a = –23
∴ equation of circle is x2 + y2 – 46x + 22y = 0.
#### Question 4:
The area of the circle passing through the point (4, 6) and having centre at (1, 2) is __________.
Since center of circle is (1, 2) equation of circle is
#### Question 5:
If the coordinates of one end of a diameter of the circle x2 + y2 – 4x – 6y + 11 = 0 are (3, 4), then the coordinates of the other end are __________.
Since equation of circle is
x2 + y2 – 4x – 6y + 11 = 0
i.e. center of circle is (2, 3)
also, given co-ordinate of are end of diameter is (3, 4) we know, center of a circle is mid point of the diameter,
Let (a, b) the other end of the diameter
Hence, co-ordinate of other end of diameter are (1, 2)
#### Question 6:
If the line $y=\sqrt{3}x+k$ touches the circle x2 + y2 = 16, then k = __________.
i.e. Center of the circle is (0, 0) and radius is 4
Since line $y=\sqrt{3}x+k$ touches the circle, perpendicular distance from (0, 0) to the line is equal to the radius of the circle
#### Question 7:
The equation of the circle concentric with the circle x2 + y2 – 6x + 12y + 15 = 0 and double its area is __________.
Given circle has equation
Hence,
Given:- New circle has same center (3, −6) and area equal to double of 30π sq. unit.
Let r denote the radius of new circle
Hence πr2 = 2(π 30)
i.e r2 = 60
Hence, equation of new concentric circle is
#### Question 8:
The equation of the circle which touches x-axis and has its centre at (1, 2) is __________.
Given center of the circle is (1, 2)
Hence, equation of circle is of the form (x – 1)2 + (y – 2)2 = r2; where r denote its radius
Since circle touches x-axis
i.e. circle passes through (1, 0)
i.e,
#### Question 9:
If a circle passes through (0, 0), (a, 0) and (0, b), then the coordinates of its centre are __________.
Given circle passes through (0, 0), (a, 0) and (0, b)
Since AB forms a diameter of circle
#### Question 10:
If the line x + 2by + 7 = 0 is a diameter of the circle x2 + y2 – 6x +2y = 0, then b = __________.
#### Question 11:
The locus of the point of intersection of the lines x cosθ + y sinθ = a and x sinθy cosθ = b is a circle of radius __________.
Let P = (x, y) be the point of intersection of lines
#### Question 12:
If the circle x2 + y2kx – 12y + 4 = 0 touches x-axis, then k = __________.
#### Question 13:
The value of k for which the centres of the circles x2 + y2 = 1, x2 + y2 + 6x – 2y = 1 and x2 + y2 – 6kx + 4y – 1 = 0 are collinear, is __________.
#### Question 14:
If the equation ax2 + (2 – b) xy + 3y2 – 6bx + 30y + 6b = 0 represents a circle, then a = __________, b = __________.
Since ax2 + (2 – b) xy + 3y2 – 6bx + 30y + 6b = 0 represent circle
⇒ 2 – b = 0 (xy term has coefficient equal to zero always)
i.e. b = 2
and a = 3 (coefficient of x2 = coeffcient of yin case of circle)
#### Question 1:
Write the length of the intercept made by the circle x2 + y2 + 2x − 4y − 5 = 0 on y-axis.
Since the intercept lies on the y-axis, by putting x = 0 in the given equation, we get:
${y}^{2}-4y-5=0$
Thus, the length of the intercept on the y-axis is (5 + 1) = 6 units.
#### Question 2:
Write the coordinates of the centre of the circle passing through (0, 0), (4, 0) and (0, −6).
We need to find the coordinates of the centre of the circle passing through (0, 0), (4, 0) and (0, −6).
Let the equation of the circle be ${\left(x-h\right)}^{2}+{\left(y-k\right)}^{2}={a}^{2}$.
Putting x = y = 0:
${h}^{2}+{k}^{2}={a}^{2}$ ...(1)
Putting x = 4, y = 0 in the equation of the circle:
Putting x = 0, y = −6 in the equation of the circle:
Hence, the centre of the circle is $\left(2,-3\right)$.
#### Question 3:
Write the area of the circle passing through (−2, 6) and having its centre at (1, 2).
The equation of the required circle is ${\left(x-1\right)}^{2}+{\left(y-2\right)}^{2}={a}^{2}$.
The circle passes through (−2, 6).
${\left(-2-1\right)}^{2}+{\left(6-2\right)}^{2}={a}^{2}$
∴ Area of the required circle =
#### Question 4:
If the abscissae and ordinates of two points P and Q are roots of the equations x2 + 2axb2 = 0 and x2 + 2pxq2 = 0 respectively, then write the equation of the circle with PQ as diameter.
The roots of the equations x2 + 2axb2 = 0 and x2 + 2pxq2 = 0 are $-a±\sqrt{{a}^{2}+{b}^{2}}$ and $-p±\sqrt{{p}^{2}+{q}^{2}}$.
Therefore, the coordinates of P and Q are , respectively.
So, the required equation of the circle is .
$⇒{\left(x+a\right)}^{2}-{a}^{2}-{b}^{2}+{\left(y+p\right)}^{2}-{p}^{2}-{q}^{2}=0$
${x}^{2}+{y}^{2}+2ax+2yp-{p}^{2}-{q}^{2}=0$
#### Question 5:
Write the equation of the unit circle concentric with x2 + y2 − 8x + 4y − 8 = 0.
The centre of the circle x2 + y2 − 8x + 4y − 8 = 0 is (4, −2).
The radius of the unit circle is 1.
∴ Required equation of circle:
${\left(x-4\right)}^{2}+{\left(y+2\right)}^{2}=1$,
$⇒$ ${x}^{2}+{y}^{2}-8x+4y+19=0$
#### Question 6:
If the radius of the circle x2 + y2 + ax + (1 − a) y + 5 = 0 does not exceed 5, write the number of integral values a.
According to the question, we have:
The number of integral values of a is 16.
#### Question 7:
Write the equation of the circle passing through (3, 4) and touching y-axis at the origin.
It is given that the circle touches the y-axis at the origin.
Thus, the centre of the circle is (h,0) and its radius is h.
Hence, the equation of the circle is ${\left(x-h\right)}^{2}+{\left(y\right)}^{2}={h}^{2}$, i.e. ${x}^{2}+{y}^{2}-2xh=0$.
Also, the circle passes through (3, 4).
∴ $25-6h=0⇒h=\frac{25}{6}$
Hence, the required equation of the circle is ${\left(x-\frac{25}{6}\right)}^{2}+{y}^{2}={\left[\frac{25}{6}\right]}^{2}$, i.e. $3\left({x}^{2}+{y}^{2}\right)-25x=0$.
#### Question 8:
If the line y = mx does not intersect the circle (x + 10)2 + (y + 10)2 = 180, then write the set of values taken by m.
Let us put y = mx in the equation (x + 10)2 + (y + 10)2 = 180.
Now, we have:
(x + 10)2 + (mx + 10)2 = 180
On simplifying, we get:
${x}^{2}\left({m}^{2}+1\right)+20x\left(m+1\right)+20=0$
∴ Discriminant (D) = $\sqrt{400{\left(m+1\right)}^{2}-80\left({m}^{2}+1\right)}=4\sqrt{10}\sqrt{\left(2m+1\right)\left(m+2\right)}$
It is given that the line y = mx does not intersect the circle (x + 10)2 + (y + 10)2 = 180.
∴ D < 0
$⇒4\sqrt{10}\sqrt{\left(2m+1\right)\left(m+2\right)}<0\phantom{\rule{0ex}{0ex}}⇒m\in \left(-2,\frac{-1}{2}\right)$
#### Question 9:
Write the coordinates of the centre of the circle inscribed in the square formed by the lines x = 2, x = 6, y = 5 and y = 9.
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Cube and Cuboid mensuration tricks for volume, area-Lateral and total ~ New math tricks for quantitative aptitude
# New math tricks for quantitative aptitude
## Cube and Cuboid mensuration tricks for volume, area-Lateral and total
New Math Tricks is covering all mensuration topics and tricks. Today we are going to discuss the properties, shortcuts formula, aptitude question and tricky solution of cube and cuboid math chapter. The tutorial has made from basic level to exam level for your confidence and performance well in the upcoming competitive exam. Before discussing the shortcut formula and aptitude solution, we should have the basic concept of Cuboid and cube. So look at below basic properties and short tricks for the Cube and cuboid mensuration tricks.
## Cube and Cuboid shortcut formula
Cube and cuboid are made with height, length and base and six sides (face). And the shape of a cube or cuboid is 3 dimensional. We have covered triangle, circle and quadrilateral which are 2-dimensional shapes. Look at the below image which is a cuboid form.
As you have looked the above image where we can see six faces(side). Also, it has vertices. The vertices are the longest line of a cube and cuboid. in competitive math, generally ask volume, area and vertices length. Now we are moving to shortcuts formula for Cube and Cuboid.
### Cube mensuration shortcut formulas
A cube has 12 equal edges and six faces. As all edges are equal so height, width and base are equal. Therefore, the area of each face also equal. If we have to find the total surface area, we will add areas of all (six) faces. And each face can be calculated as edge^2. So, the shortcut formula for the total surface area of cube became (6 X one face"s area) or (6*Edge^2). When we have to calculate the lateral surface area of a cube then we multiply one faced area by 4. getting confusion with this math tricks? Look at below image to be clear the shortcut formula.
#### Lateral and Total Surface Area
And the volume of a cube is edge^3 Square unit.
### Cuboid shortcuts formula for area and volume
Every cube is cuboid but all cuboid are not a cube. So differently, you should know shortcuts area and volume formulas of the cuboid. The difference between cube and cuboid is on edges length. The different length (maximum three variants) may be in cuboid where equal in a cube. So length, base and height may be different. That's why in the formula of the cuboid, becomes different from the cube.
So, friend, these are the basic shortcut formulas for the cube and cuboid to calculate volume area, total surface area and lateral surface area of cube and cuboid. In the next tutorial, we will practice some mensuration problem related to the cube and cuboid. Thanks for reading and learning to days tutorial on cube and cuboid.
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# How to Find Zeros of Linear Functions
By Bob Barber; Updated April 24, 2017
The zero of a linear function in algebra is the value of the independent variable (x) when the value of the dependent variable (y) is zero. Linear functions that are horizontal do not have a zero because they never cross the x-axis. Algebraically, these functions have the form y = c, where c is a constant. All other linear functions have one zero.
Determine which variable in your function is the dependent variable. If your variables are x and y, y is the dependent variable. If your variables are letters other than x and y, the dependent variable will be the variable that is plotted on a vertical axis (like y).
Substitute zero for the dependent variable in the equation of your function. Don't worry about the form of the equation (standard, slope-intercept, point-slope); it doesn't matter. After substitution, the value of the term, including the dependent variable, becomes zero and drops out of the equation. For example, if your equation is 3x + 11y = 6, you would substitute zero for y, the term 11y would drop out of the equation and the equation would become 3x = 6.
Solve the equation of your function for the remaining (independent) variable. The solution is the zero of the function, which means that it tells where the graph of the function crosses the x-axis. For example, if your equation is 3x = 6 after substitution, you would divide both sides of the equation by 3 and your equation would become x = 2. Two is the zero of the equation, and the point (2, 0) would be where your function crosses the x-axis.
#### Tip
Another way to think of the dependent variable is that the dependent variable measures the outcome of a real-life situation. For example, suppose you are given a linear function where "f" stands for the amount of food given to fish per week, and "w" stands for the weight of the fish after one month. Even if you are not told so, you would understand in a common-sense way that the investigator would have manipulated the amount of food given to the fish; however, she could not have manipulated the resulting weight of the fish; she could only have measured it. Therefore, "w" would be the dependent (or unmanipulated, or outcome) variable.
Linear equations of the form x = c, where "c" is a constant, are not functions. They are often included in the study of linear functions, however. Graphically, these equations are plotted as vertical lines that cross the x-axis at c. For example, the equation x = 3.5 is a vertical line that crosses the x-axis at the point (3.5, 0).
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# 数学代写|离散数学作业代写discrete mathematics代考|MATH-UA120
## 数学代写|离散数学作业代写discrete mathematics代考|Euclid’s Algorithm
Euclid’s ${ }^4$ algorithm is one of the oldest known algorithms, and it provides a procedure for finding the greatest common divisor of two numbers. It appears in Book VII of Euclid’s Elements, and the algorithm was known prior to Euclid (Fig. 3.8).
Lemma
Let $a, b, q$ and $r$ be integers with $b>0$ and $0 \leq r<b$ such that $a=b q+r$. Then $\operatorname{gcd}(a, b)=\operatorname{gcd}(b, r)$.
Proof Let $K=\operatorname{gcd}(a, b)$ and let $L=\operatorname{gcd}(b, r)$, and we therefore need to show that $K=L$. Suppose $m$ is a divisor of $a$ and $b$, then as $a=b q+r$ we have $m$ is a divisor of $r$ and so any common divisor of $a$ and $b$ is a divisor of $r$.
Similarly, any common divisor $n$ of $b$ and $r$ is a divisor of $a$. Therefore, the greatest common divisor of $a$ and $b$ is equal to the greatest common divisor of $b$ and $r$.
Theorem $3.4$ (Euclid’s Algorithm) Euclid’s algorithm for finding the greatest common divisor of two positive integers $a$ and $b$ involves applying the division algorithm repeatedly as follows:
$$\begin{array}{lc} a=b q_0+r_1 & 0<r_1<b \ b=r_1 q_1+r_2 & 0<r_2<r_1 \ r_1=r_2 q_2+r_3 & 0<r_3<r_2 \end{array}$$
\begin{aligned} &r_{n-2}=r_{n-1} q_{n-1}+r_n \quad 0<r_n<r_{n-1} \ &r_{n-1}=r_n q_n . \end{aligned}
## 数学代写|离散数学作业代写discrete mathematics代考|Distribution of Primes
We already have shown that there are an infinite number of primes. However, most integer numbers are composite and a reasonable question to ask is how many primes are there less than a certain number. The number of primes less than or equal to $x$ is known as the prime distribution function (denoted by $\pi(x)$ ) and it is defined by $\pi(x)=\sum_{p \leq x} 1$ (where $p$ is prime).
The prime distribution function satisfies the following properties:
(i) $\lim {x \rightarrow \infty} \frac{\pi(x)}{x}=0$ (ii) $\lim {x \rightarrow \infty} \pi(x)=\infty$.
The first property expresses the fact that most integer numbers are composite, and the second property expresses the fact that there are an infinite number of prime numbers.
There is an approximation of the prime distribution function in terms of the logarithmic function $\left({ }^x / \ln x\right)$ as follows:
$$\lim _{x \rightarrow \infty} \frac{\pi(x)}{x / \ln x}=1 \text { (Prime Number Theorem). }$$
The approximation $x / \ln x$ to $\pi(x)$ gives an easy way to determine the approximate value of $\pi(x)$ for a given value of $x$. This result is known as the Prime Number Theorem, and Gauss originally conjectured this theorem.
# 离散数学代写
## 数学代写|离散数学作业代写discrete mathematics代考|Euclid’s Algorithm
$$a=b q_0+r_1 \quad 0<r_1<b b=r_1 q_1+r_2 \quad 0<r_2<r_1 r_1=r_2 q_2+r_3 \quad 0<r_3<r_2$$
$$r_{n-2}=r_{n-1} q_{n-1}+r_n \quad 0<r_n<r_{n-1} \quad r_{n-1}=r_n q_n .$$
## 数学代写|离散数学作业代写discrete mathematics代考|Distribution of Primes
(i) $\lim x \rightarrow \infty \frac{\pi(x)}{x}=0$ (二) $\lim x \rightarrow \infty \pi(x)=\infty$.
$$\lim _{x \rightarrow \infty} \frac{\pi(x)}{x / \ln x}=1 \text { (Prime Number Theorem). }$$
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# 06.05 Applications of Systems of Equations
Topics: Harshad number, Elementary algebra, Erdős–Woods number Pages: 2 (374 words) Published: May 18, 2013
06.05 Applications of Systems of Equations
Part 1
1. Tony and Belinda have a combined age of 56. Belinda is 8 more than twice Tony’s age. How old is each? Tony’s age = t years, Belinda’s age = 56 - t years
56 - t = 2t + 8
56 - 8 = 2t + t ==> 3t = 48 ==> t = 16 years
THUS Tony = 16 years, and Belinda = 40 years
2. Salisbury High School decided to take their students on a field trip to a theme park. A total of 150 people went on the trip. Adults pay \$45.00 for a ticket and students pay \$28.50 for a ticket. How many students and how many adults went to the park if they paid a total of \$4770? Adults = a, and Students = 150 - a
45a + 28.5(150 - a) = 4770
45a + 4275 - 28.5a = 4770 ==> 16.5a = 4770 - 4275 = 495
a = 495 / 16.5 = 30,
Adult = 30, Students = 120
3. Your piggy bank has a total of 46 coins in it; some are dimes and some are quarters. If you have a total of \$7.00, how many quarters and how many dimes do you have? Dimes = d, and Quarter = 46 - d
d/10 + (46 - d) /4 = 7
Multiply both sides by 20, which gives
2d + 5(46 - d) = 140 ==> 2d + 230 - 5d = 140
-3d = 140 - 230 = -90 ==> d = 30
Dimes = 30, and Quarters = 16
Part 2
c = current of river
b = rate of boat
d = s(t) will represent (distance = speed X time)
Upstream: 60 = 6(b-c)
Downstream: 60 = 3(b+c)
There are now two separate equations: 60 = 6b - 6c and 60 = 3b + 3c Solve both equations for b:
b = 10 + c
b = 10 – c
Now make both equations equal each other and solve for c:
10 + c = 20 – c
2c = 10
c = 5
The speed of the current was 5 mph
Now, plug the numbers into one of either the original equations to find the speed of the boat in still water. I chose the first equation: b = 10 + c or b = 10 + 5 b = 15 The speed of the boat in still water must remain a consistent 15 mph or more in order for Wayne and his daughter to make it home in time or dinner.
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# Golden Ratio Blocks
3,890
27
7
## Introduction: Golden Ratio Blocks
This is a perfect example of an easy project that goes all nerdy. Then there is a happy ending. (In a mathematical sense.)
"I'd like you to cut up this old post into blocks as bases for the centerpiece candles", my spouse says.
What could be easier?
### Supplies:
We used an old piece of 6 by 6 wood post. (This measures 5.25 inch by 5.25 inch). You could make boxes or cut cardboard tubes to length too. Depending on the material, you will need appropriate cutting equipment. We used a band saw.
### Teacher Notes
Teachers! Did you use this instructable in your classroom?
Add a Teacher Note to share how you incorporated it into your lesson.
## Step 1: Nerd Out
Then the possibilities spin. Do I do constant steps? Binary blocks? Fibonacci series? Logarithmic? Quick google of "series". How about just constant ratios? So I made a spreadsheet to try some different combinations. The "Bar Chart" option let me visualize the "too many" choices.
The constant ratio choice alone has infinite possibilities. Arbitrarily, I chose the make the "middle one" a cube. Lets call this height B. Let the ratio from one to the next be r. Then for heights a, B, C we have:
A x r = B
B x r = C
For example if we choose r = 1.5 then we get
A x 1.5 = 5.25
solving for A we get
A = 5.25 / 1.5 = 3.5
and for B we have
B x 1.5 = 5.25 x 1.5 = 7.875
But we could choose any r.
## Step 2: Add a Constraint
With too may choices, what if we required that they stack together nicely for storage. That is choose the heights so that A + B = C. In our sample case with r = 1.5 we have
3.25 + 5.25 = 8.5 which is not far from 7.875.
I used the "solver" function in my spreadsheet to iteratively find the ratio that made A + B = C. It came up with
A = 3.245
B = 5.2500
C = 8.495
And sure enough 3.245 +5.250 = 8.495
I made my measurements and cuts on the band saw, happy with my clever choice. On the way back to my desk I thought, "wait a minute. What was that ratio that the solver found?" There it was, staring me in the face
A = B / 1.6180 = A * 0.6180
C = B * 1.6180
That is 1/r = r -1
That's the very definition of the "Golden Ratio".
## Step 3: More to Explore
I'm surprised that the Golden Ratio shows up in a practical problem (fitting three things into a box). You could use this fact to make three boxes that fit into one box of any size.
The more surprising thing is that this means that if you define a golden sequence G(n) defined by G(0) = 1 and
G(n+1) = tao * G(n) (where tao = 1.618033988749895...)
1.0000, 1.6180, 2.6180, 4.2361, 6.8541, 11.0902, 17.9443, 29.0344, 46.9787, ...
then G(n) = G(n-1) + G(n-2) which is the very definition of the Fibonacci series! But this looks different...hmmm
There is so much more to explore. Here is a random starting point:
https://www.mathsisfun.com/numbers/nature-golden-r...
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## 7 Discussions
This is PERFECT - and not because it's a Golden Ratio! Ha ha! I love the way you colored the blocks, and the thinking behind the size. Bravo! BTW - the colors remind me of the outcome my team used but our's were food grade dye and alcohol as the carrier. Gave a nice watercolor-y touch.
Struggling to type this comment after you melted my brain, how did you colour the wood with the rainbow effect?
Painting them was a different matter. My spouse and the twins are very crafty when it comes to painting and decorating. Me not so much. I'll ask.
They used tubes of acrylic paints. They continuously mixed new colors as the advanced around the blocks. Most of the mixing occurred right on the blocks instead of a painters pallet.
Ahhh, great finish! Thanks for getting back to me.
Is this surface only? Colors look great though
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# How do we find the inverse of a function with $2$ variables?
$$f(m,n) = (2m+n, m+2n)$$
What do we have to do to find the inverse of this function?
I don't even know where to begin.
-
Do you know about matrices? – Julian Kuelshammer Oct 25 '12 at 21:36
do i have to find the inverse (matrice)? – Gladstone Asder Oct 25 '12 at 21:38
Denote $$\begin{cases} u=2m+n,\\ v=m+2n. \end{cases}$$ To find inverse mapping is sufficient solve this system w.r.t variables $m$ and $n$: $$\begin{cases} 2u-v=3m,\\ 2v-u=3n. \end{cases}$$ From there $$\begin{cases} m=\frac{2u-v}{3},\\ n=\frac{2v-u}{3}. \end{cases}$$ Then $f^{-1}(u,\ v)=\left(\dfrac{2u-v}{3}, \ \dfrac{2v-u}{3} \right).$
-
I think this might be a useful place to begin: We can invert a function $f$ of a single variable by solving the equation $f(x)=y$ for $x$. For example, we invert $f(x)=2x+1$ by solving the equation $2x+1=y$ to get $x=(y-1)/2$. We can then write $x=f^{-1}(y)=(y-1)/2$, where $f^{-1}$ is the inverse required.
In your case, you need to do the 2-variable equivalent: solve the equation $f(m,n)=(2m+n,m+2n)=(a,b)$ to get the input $(m,n)$ - which plays the role of $x$ - in terms of $(a,b)$ - which plays the role of $y$. Your answer should have the form $(m,n)=f^{-1}(a,b) = (\hbox{an expression in$a$and$b$}, \hbox{another expression in$a$and$b$})$.
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I dont understand the answer, all you have shown is the inverse f(u,v) but the question is asking for the inverse of f(m,n). – user76711 May 7 '13 at 22:16
The following is a solution with matrices: $f$ can be regarded as a linear function, e.g. $\mathbb{Q}^2\to \mathbb{Q}^2$. With respect to the standard basis this is given by the matrix $$\begin{pmatrix}2&1\\1&2\end{pmatrix}$$ Now you can invert this matrix. There are different methods for that. I'm just using the explicit formula given by Cramer's rule. The inverse is $$\frac{1}{3}\begin{pmatrix}2&-1\\-1&2\end{pmatrix}$$ Now you can interpret this as a linear function again and get $f^{-1}(m,n)=\left(\frac{2m-n}{3},\frac{-m+2n}{3}\right)$.
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# Solving Nonlinear Equations by Factoring
Example
Solve for x: x3 + 2x = 6 + 3x2
SolutionStep 1 Write the equation in standard form. Subtract 6 and 3x2 from both sides. x3 + 2x x3 - 3x2 + 2x - 6 = 6 + 3x2 = 0 Step 2 Factor by grouping. Factor out the common factor, (x - 3).Step 3 Use the Zero Product Property. x2(x - 3) + 2(x - 3)(x - 3)(x2 + 2) x - 3 = 0 or x2 + 2 = 0= 0 = 0 Step 4 Solve for the variable. x = 3 or x2 = -2 Take the square root of each side. x Write as an imaginary number. x
So, the three solutions are x = 3,
Note:
Recall that a negative number under a square root results in an imaginary number, which we indicate by using the letter i. Thus,
The equation x3 + 2x = 6 + 3x2 written in standard form is x3 - 3x2 + 2x - 6 = 0. The graph of the corresponding function, f(x) = x3 - 3x2 + 2x - 6 is shown.
The graph crosses the x-axis at only one location, x = 3. This is because the only real number solution is x = 3. In a Cartesian coordinate system, the x- and y- axes represent real numbers. Therefore, the imaginary solutions do not appear on the graph. However, the imaginary solutions check in the original equation.
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# Prime Numbers and Composite Numbers
A Prime Number can be divided evenly only by 1 or itself.
And it must be a whole number greater than 1.
### Example: 7 can only be divided evenly by 1 or 7, so it is a prime number.
But 6 can be divided evenly by 1, 2, 3 and 6 so it is NOT a prime number (it is a composite number).
Let me explain ...
Some whole numbers can be divided up evenly, and some can't!
### Example:
6 can be divided evenly by 2, or by 3:
6 = 2 × 3
Like this:
or divided into 2 groups divided into 3 groups
But 7 cannot be divided up evenly:
And we give them names:
• When a number can be divided up evenly it is a Composite Number
• When a number can not be divided up evenly it is a Prime Number
So 6 is Composite, but 7 is Prime.
And that explains it ... but there are some more details ...
## Not Into Fractions
We are only dealing with whole numbers here! We are not going to cut things into halves or quarters.
## Not Into Groups of 1
OK, we could have divided 7 into seven 1s (or one 7) like this:
7 = 1 x 7
But we could do that for any whole number!
So we should also say we are not interested in dividing by 1, or by the number itself.
It is a Prime Number when it can't be divided evenly by any number
(except 1 or itself).
Example: is 7 a Prime Number or Composite Number?
• We cannot divide 7 evenly by 2 (we get 2 lots of 3, with one left over)
• We cannot divide 7 evenly by 3 (we get 3 lots of 2, with one left over)
• We cannot divide 7 evenly by 4, or 5, or 6.
We can only divide 7 into one group of 7 (or seven groups of 1):
7 = 1 x 7
So 7 can only be divided evenly by 1 or itself:
So 7 is a Prime Number
And also:
It is a Composite Number when it can be divided evenly
by numbers other than 1 or itself.
Like this:
Example: is 6 a Prime Number or Composite Number?
6 can be divided evenly by 2, or by 3, as well as by 1 or 6:
6 = 1 × 6
6 = 2 × 3
So 6 is a Composite Number
Sometimes a number can be divided evenly many ways:
Example: 12 can be divided evenly by 1, 2, 3, 4, 6 and 12:
1 × 12 = 12
2 × 6 = 12
3 × 4 = 12
So 12 is a Composite Number
And note this:
Any whole number greater than 1 is either Prime or Composite
Years ago 1 was included as a Prime, but now it is not:
1 is not Prime and also not Composite.
## Factors
We can also define a Prime Number using factors.
"Factors" are numbers we multiply
together to get another number.
And we have:
When the only two factors of a number are 1 and the number,
then it is a Prime Number
It means the same as our previous definition, just stated using factors.
And remember this is only about Whole Numbers (1, 2, 3, ... etc), not fractions or negative numbers. So don't say "I could multiply ½ times 6 to get 3" OK?
Examples:
3 = 1 × 3 (the only factors are 1 and 3) Prime 6 = 1 × 6 or 6 = 2 × 3 (the factors are 1,2,3 and 6) Composite
## Examples From 1 to 14
Factors other than 1 or the number itself are highlighted:
Number Can be Evenly Divided By Prime, or Composite? 1 (1 is not considered prime or composite) 2 1, 2 Prime 3 1, 3 Prime 4 1, 2, 4 Composite 5 1, 5 Prime 6 1, 2, 3, 6 Composite 7 1, 7 Prime 8 1, 2, 4, 8 Composite 9 1, 3, 9 Composite 10 1, 2, 5, 10 Composite 11 1, 11 Prime 12 1, 2, 3, 4, 6, 12 Composite 13 1, 13 Prime 14 1, 2, 7, 14 Composite ... ... ...
So when there are more factors than 1 or the number itself, the number is Composite.
A question for you: is 15 Prime or Composite?
## Why All the Fuss about Prime and Composite?
Because we can "break apart" Composite Numbers into Prime Number factors.
It is like the Prime Numbers are the basic building blocks of all numbers. And the Composite Numbers are made up of Prime Numbers multiplied together.
Here we see it in action:
2 is Prime, 3 is Prime, 4 is Composite (=2×2), 5 is Prime, and so on...
Example: 12 is made by multiplying the prime numbers 2, 2 and 3 together.
12 = 2 × 2 × 3
The number 2 was repeated, which is OK.
In fact we can write it like this using the exponent of 2:
12 = 22 × 3
And that is why they are called "Composite" Numbers because composite means "something made by combining things"
This idea is so important it is called The Fundamental Theorem of Arithmetic.
There are many puzzles in mathematics that can be solved more easily when we "break up" the Composite Numbers into their Prime Number factors. A lot of internet security is based on mathematics using prime numbers.
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## The Astounding Power of Area
### 2.2 Multiplying Polynomials
In our base ten number system, the string of symbols “$$375$$” represents the number $$3\cdot10^{2}+7\cdot10+5\cdot1$$. A polynomial is mathematical expression that looks like (or at least can be algebraically manipulated to look like) a number written base $$x$$ for some unspecified quantity $$x$$. For example, $$3x^{2}+7x+5$$ is a polynomial in base $$x$$ with integer coefficients (and if we are later told that $$x$$ actually represents the number $$10$$, then this expression corresponds to the number $$375$$, to the number $$31$$ if instead we are told $$x$$ is $$2$$, perhaps). The expression $$\pi w^{3}-5.2w^{2}+\frac{3}{\sqrt{6}}w+100\frac{3}{4}$$ is a polynomial in the variable $$w$$ with real coefficients.
A natural first appearance of polynomials comes through EXPLODING DOTS here with lesson 1.7 .
Multiplying two polynomials, say $$x^{2} + 2x + 3$$ and $$3x+1$$, is just as easy, thanks to the area model, as multiplying two integers. (In fact easier, as one never needs to carry with polynomials!)
Notice the convenience of the diagonals: all like powers of $$x$$ (or the powers of ten in ordinary arithmetic) match along them.
Of course, the area model is just a representation of Expanding Brackets: select one term from each set of parentheses and sum all possible products. As such, negative terms are permitted as entries in the diagrams.
For example, here is the product of $$x^{2}+2x-2$$ and $$2x+3$$.
The answer $$2x^{3}+7x^{2}+2x-6$$ appears.
Comment: This area model approach for polynomial work is often called the Galley Method.
EXERCISE: Compute the following products via the Galley Method.
a) $$\left(3x^{5}+2x^{4}-5x^{3}+4x^{2}-x+10\right)\left(x^{2}-3x+4\right)$$
b) $$\left(x^{3}-3x+2\right)\left(2x+5\right)$$
c) $$\left(x^{5}-x\right)\left(x^{4}-2x^{2}+1\right)$$
(For nice alignment of the diagonals, redo the second and third examples with zero rows and columns drawn in.)
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# In an examination hall 144 people were present. of these 1/2 were boys and of them 4/9 were girls. The rest were teachers. How man
In an examination hall 144 people were present. of these 1/2 were boys and of them 4/9 were girls. The rest were teachers. How many students were there in the hall?
### 1 thought on “In an examination hall 144 people were present. of these 1/2 were boys and of them 4/9 were girls. The rest were teachers. How man”
1. Step-by-step explanation:
## Given:–
In an examination hall 144 people were present. of these 1/2 were boys and of them 4/9 were girls. The rest were teachers.
## To find:–
How many students were there in the hall?
## Solution:–
Given that
Total number of people were present in the examination hall = 144
Number of boys = 1/2 of the total people
=> (1/2)×144
=> 144/2
=> 72
Total number of boys = 72
Total number of girls = 4/9 of the total people
=> (4/9)×144
=> (4×144)/9
=> 4×16
=> 64
Number of girls = 64
Total number of boys and girls = 72+64 = 136
Total number of teachers
=> 144-136
=>8
Number of teachers = 8
(or)
Total people present in the examination hall
=> boys+girls+teachers
=> (144/2)+(4×144/9)+ Teachers = 144
=> Teachers = 144 – (144/2)-(4×144/9)
=> Teachers = 144 -72-64
=> Teachers = 144 – 136
Number of teachers = 8
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# Moment of Inertia.
## Presentation on theme: "Moment of Inertia."— Presentation transcript:
Moment of Inertia
Type of moment of inertia
Moment of inertia of Area Moment of inertia of mass Also known as second moment Why need to calculate the moment of Inertia? To measures the effect of the cross sectional shape of a beam on the beam resistance to a bending moment Application Determination of stresses in beams and columns Symbol I – symbol of area of inertia Ix, Iy and Iz
Application : Design Steel ( Section properties)
Moment of Inertia of Area
y y dx dy dy h h C x C x b b Area of shaded element, Area of shaded element, Moment of inertia about x-axis Moment of inertia about y-axis Integration from h/2 to h/2 Integration from b/2 to b/2
J = polar moment of inertia
Table 6.2. Moment of inertia of simple shapes Shape J = polar moment of inertia 1. Triangle 2. Semicircle 3. Quarter circle 4. Rectangle 5. Circle b h y x y x r y x r b h y x x y r
Parallel - Axis theorem
There is relationship between the moment of inertia about two parallel axes which is not passes through the centroid of the area. From Table 6.1; Ix = and Iy = The centroid is, ( , ) = (b/2, h/2) Moment of inertia about x-x axis, Ixx = Ix + Ady2 where dy is distance at centroid y Moment of inertia about y-y axis, Iyy = Iy + Asx2 where sx is distance at centroid x y x b h
Determine centroid of composite area
Example 6.2 y 140 mm 60 mm 160 mm 60mm x y 140 mm 1 3 2 60 mm x 60mm 160 mm 60 mm Determine centroid of composite area PART AREA(mm2) y(mm) x(mm) Ay (103)(mm3) Ax (103) (mm3) 1 60(200) = 12000 200/2 = 100 60/2 = 30 1200 360 2 160(60)=9600 60 + [160/2] = 140 288 1344 3 /2= 250 3000 Σ: Σ: 2688 x 103 Σ: 4704 x 103
Second moment inertia PART AREA (A)(mm2) Ix = bh3/12 (106) (mm4)
dy = |y-y|(mm) Ady2(106)(mm4) 1 60(200) = 12000 60(2003)/12 = 40 |100– 80| = 20 4.8 2 160(60)=9600 160(603)/12 = 2.88 |30 – 80| = 50 24 3 |100 – 80| =20 Σ: [Ix + Ady2]1 + [Ix + Ady2]2 +[Ix + Ady2]3 = [ ] x106 = x 106 mm4 PART AREA(mm2) Iy = b3h/12 (106) (mm4) Sx=|x-x| (mm) Asx2(106)(mm4) 1 60(200) = 12000 603(200)/12 =3.6 |30-140|=110 145.2 2 160(60)=9600 1603(60)/12 =20.48 |140 – 140 |= 0 3 |250 – 140|=110 Σ: [Iy + As2]1 + [Iy + As2]2 +[Iy + As2]3 = [ ] x106 = x 106 mm4
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# Area Formulas
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## Area of Plane Shapes
### Example: What is the area of this triangle?
A harder example:
### Perimeter, Area, and Volume
1. The perimeter of a polygon (or any other closed curve, such as a circle) is the distance around the outside.
2. The area of a simple, closed, planar curve is the amount of space inside.
3. The volume of a solid 3D shape is the amount of space displaced by it.
Some formulas for common 2 -dimensional plane figures and 3 -dimensional solids are given below. The answers have one, two, or three dimensions; perimeter is measured in linear units area is measured in square units , and volume is measured in cubic units.
### Area Of Shapes
In Geometry, a shape is defined as the figure closed by the boundary. The boundary is created by the combination of lines, points and curves. Basically, there are two different types of geometric shapes such as:
• Two – Dimensional Shapes
• Three – Dimensional Shapes
Each and every shape in the Geometry can be measured using different measures such as area, volume, surface area, perimeter and so on.
### What is Area?
An area is a quantity that expresses the extent of a two-dimensional figure or shape or planar lamina in the plane. Lamina shapes include 2D figures that can be drawn on a plane, e.g., circle, square, triangle, rectangle, trapezium, rhombus and parallelogram. Area of shapes such as circle, triangle, square, rectangle, parallelogram, etc. are the region occupied by them in space.
Polygon shape: A polygon is a two-dimensional shape that is formed by straight lines. The examples of polygons are triangles, hexagons and pentagons. The names of shapes describe how many sides exist in the shape. For instance, a triangle consists of three sides and a rectangle has four sides. Hence, any shape that can be formed using three straight lines is known as a triangle and any shape that can be drawn by linking four lines is known as a quadrilateral. The area is the region inside the boundary/perimeter of the shapes which is to be considered.
### What are 2D shapes?
The two-dimensional shapes (2D shapes) are also known as flat shapes, are the shapes having two dimensions only. It has length and breadth. It does not have thickness. The two different measures used for measuring the flat shapes are area and the perimeter. Two-dimensional shapes are the shapes that can be drawn on the piece of paper. Some of the examples of 2D shapes are square, rectangle, circle, triangle and so on.
### Area of 2D Shapes Formula
In general, the area of shapes can be defined as the amount of paint required to cover the surface with a single coat. Following are the ways to calculate area based on the number of sides that exist in the shape, as illustrated below in the fig.
Let us write the formulas for all the different types of shapes in a tabular form.
Shape Area Terms Circle π × r2 r = radius of the circle Triangle ½ × b × h b = base h = height Square a2 a = length of side Rectangle l × w l = length w = width Parallelogram b × h b=base h=vertical height Trapezium ½(a+b) × h a and b are the length of parallel sides h = height Ellipse πab a = ½ minor axis b = ½ major axis
### What are 3D shapes?
The three-dimensional shapes (3D shapes), known as solid shapes, are the shapes that have three dimensions such as length, breadth and thickness. The two distinct measures used to define the three-dimensional shapes are volume and surface area. Generally, the three-dimensional shapes are obtained from the rotation of two-dimensional shapes. Thus, the surface area of any 2D shapes should be a 2D shape. If you want to calculate the surface area of solid shapes, we can easily calculate from the area of 2D shapes.
### Area of 3D Shapes Formula
According to the International System of Units (SI), the standard unit of area is the square meter (written as m2) and is the area of a square whose sides are one meter long. For example, a particular shape with an area of three square meters would have the same area as three such squares. The surface area of a solid object is a measure of the total area that the surface of the object occupies.
For 3D/ solid shapes like cube, cuboid, sphere, cylinder and cone, the area is updated to the concept of the surface area of the shapes. The formulas for three-dimension shapes are given in the table here:
Shape Surface area Terms Cube 6a2 a = length of the edge Rectangular prism 2(wl+hl+hw) l = length w = width h = height Cylinder 2πr(r + h) r = radius of circular base h = height of the cylinder Cone πr(r + l) r = radius of circular base l = slant height Sphere 4πr2 r = radius of sphere Hemisphere 3πr2 r = radius of hemisphere
In addition to the area of the planar shapes, an additional variable i.e the height or the radius are taken into account for computing the surface area of the shapes.
Consider a circle of radius r and make endless concentric circles. Now from the centre to the boundary make a line segment equal to the radius and cut the figure along with that segment. It’ll be formed a triangle with base equal to the circumference of the circle and height is equal to the radius of the outer circle, i.e., r. The area can thus be calculated as ½ * base * height i.e
½ * 2πr*r
### What Are Area Formulas?
Different Area Formulas are used to calculate the area of different geometric figures. A few of the important geometric figures are square, rectangle, circle, triangle, trapezoid, ellipse.
### Area Formula For Triangle:
The area of a triangle can be calculated by finding the half of the product of the length of its base and height. The area of a triangle formula is given as,
Area of a Triangle = 1/2 × Base × Height
### Area Formula For Square:
The area of a square is calculated by finding the square of the length of square’s each side. The area of a square formula is given as,
Area of Square = Side × Side = Side2
### Area Formula For Circle:
The area of the circle can be calculated b finding the product of pi with the square of the radius. The area of a circle formula is given as,
Area of a Circle = πr2
### Area Formula For Rectangle:
The area of a rectangle is calculated by multiplying the length and breadth of a rectangle. The area of a rectangle formula is given as,
Area of Rectangle = Length × Breadth
### Area Formula For Parallelogram:
The area of a parallelogram can be calculated by multiplying the base and height of a parallelogram. The area of a parallelogram formula is given as,
Area of a Parallelogram = Base × Height
### Area Formula For a Rhombus:
The area of a rhombus can be calculated by finding half of the product of lengths of its diagonals. The area of a quadrilateral formula is given as,
Area of a Rhombus = 1/2 × Diagonal1 × Diagonal2
### Area Formula For a Trapezoid:
The area of a trapezoid is half of the product of sum of parallel sides and height. The area of a trapezoid formula is given as,
Area of a Trapezoid = 1/2 × Sum of Parallel Sides (a + b) × Height(h)
### Area Formula For Ellipse:
The area of an ellipse is the product of pi with a major axis and minor axis. The area of an ellipse formula is given as,
Area of Ellipse = π × Major axis(a) × Minor Axis(b)
### Area of Shapes Examples
Example 1:
Find the area of the circular path whose radius is 7m.
Solution:
Given, radius of circular path, r = 7m
By the formula of area of circle, we know;
A = π r2
A = 22/7 x 7 x 7
A = 154 sq.m.
Example 2:
The side-length of a square plot is 5m. Find the area of a square plot.
Solution:
Given, side length, a = 5m
By the formula of area of a square, we know;
Area = a2
A = 5 x 5
A = 25 sq.m.
Example 3:
Find the area of the cone, whose radius is 4cm and height is 3cm.
Solution:
Given, radius of cone = 4cm
and height of cone = 3cm.
As per the formula of area of cone, we know;
Slant height =l = √(42 + 32) = √25 = 5 cm
Area = πr(r + l)
= (22/7) × 4(4 + 5)
=(22/7) 36
= 113. 14 cm2
### Examples Using Geometric Area Formula
Example 1: What is the area of a rectangular park whose length and breadth are 60m and 90m respectively?
Solution:
To find: The area of a rectangular park.
Given:
Length of the park = 60m
Breadth of the park = 90m
Using area formula,
Area of Rectangle = (L × B)
= (60 × 90) m
= 5400 m2
Answer: The area of the rectangular park is 5400 m2.
Example 2: What is the area of a circular park whose radius is 400m?
Solution:
To find: The area of a circular park.
Given:
Radius of the circular park = 400m
Using area formula,
Area of a Circle = πr2
Area of a Circle = π 4002
= 160000π m2
Answer: The area of the circular park is 160000π m2.
Example 3: What is the area of a square with each side measuring 5 units?
Solution:
We have, using area formula for square,
Area = 5 × 5 = 25 square units
Answer: The area of square is 25 square units.
### FAQs on Area Formulas
#### What Are Area Formulas?
Area formulas are used to find the area or region covered by any 2-D shape. Area formula for few shapes is given as,
• Area of square = (side)2
• Area of rectangle = length × breadth
• Area of triangle = (1/2) × base × height
#### What Is the Area Formula For Square?
The area formula for square is used to find the area covered by a square in 2-D plane. The area formula for square is given as, Area of square = (side)2.
#### What Is the Area Formula For Rectangle?
The area formula for the rectangle is used to find the area covered by a rectangle in 2-D plane. The area formula for the rectangle is given as, Area of rectangle = length × breadth.
#### What Is the Application of Area Formula?
Area formula finds application in calculating the total region covered by a shape in 2-D plane. Different area formulas can be used to calculate area of different depending upon the type and parameters given.
Area of Regular Polygon Formula
Area Formulas
Area of a Square Formula
Area Of Isosceles Triangle ⭐️⭐️⭐️⭐️⭐️
Formula for area of triangle
Area of a Trapezoid Formula
Area Of An Octagon Formula ⭐️⭐️⭐️⭐️⭐
Area Of An Octagon Formula ⭐️⭐️⭐️⭐️⭐
Area of a Pentagon Formula ⭐️⭐️⭐️⭐️⭐
Surface Area Formulas
Surface Area of a Cube Formula
Surface Area of a Prism Formula
Surface Area of a Rectangular Prism Formula
Surface Area of a Square Pyramid Formula
Surface Area of a Rectangle Formula
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# what is the greatest common factor of 4 and 10
Category :
### James
Guys, does anyone know the answer?
get what is the greatest common factor of 4 and 10 from EN Bilgi.
## What is the greatest common factor of 4 and 10?
The greatest common factor is 2. First, you can list all the factors of 4. 1, 2, 4 Next, list all the factors of 10: 1, 2, 5, 10 Now, look back at the two lists and see if any of the numbers are the same in both lists. If there is more than one, the greatest number will be the greatest common factor. In this case, the only common number is 2, and so it is automatically the greatest common factor.
## What is the greatest common factor of 4 and 10?
Prealgebra Factors and Multiples Greatest Common Factor
Lydia · EZ as pi Mar 29, 2016
The greatest common factor is 2.
### Explanation:
First, you can list all the factors of 4.
1 , 2 , 4
Next, list all the factors of 10:
1 , 2 , 5 , 10
Now, look back at the two lists and see if any of the numbers are the same in both lists. If there is more than one, the greatest number will be the greatest common factor. In this case, the only common number is 2, and so it is automatically the greatest common factor.
EZ as pi Aug 24, 2016 The GCF = 2.
Both 4 and 10 are even, so they must have a common factor of 2.
In this case this is the only common factor (apart from 1).
### Explanation:
A good approach for any question involving factors, GCF, LCM and roots is to write each number as the product of its prime factors .
If you have a number written as the product of its prime factors, then you know everything about that number!
× . × x 4 = 2 × 2 × × x 10 = 2 × x × 5 × x G C F = 2
We can see that 2 is common to both.
This is also a good way of finding the LCM, especially of big numbers.
L C M = 2 × 2 × 5 = 20 Answer link
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See all questions in Greatest Common Factor
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## GCF of 4 and 10
GCF of 4 and 10 is the largest possible number which divides 4 and 10 without leaving any remainder. The methods to compute the GCF of 4, 10 are explained here.
## GCF of 4 and 10
GCF of 4 and 10 is the largest possible number that divides 4 and 10 exactly without any remainder. The factors of 4 and 10 are 1, 2, 4 and 1, 2, 5, 10 respectively. There are 3 commonly used methods to find the GCF of 4 and 10 - prime factorization, long division, and Euclidean algorithm.
## What is GCF of 4 and 10?
Answer: GCF of 4 and 10 is 2.Explanation:
The GCF of two non-zero integers, x(4) and y(10), is the greatest positive integer m(2) that divides both x(4) and y(10) without any remainder.
## Methods to Find GCF of 4 and 10
Let's look at the different methods for finding the GCF of 4 and 10.
Listing Common Factors
Long Division Method
Prime Factorization Method
### GCF of 4 and 10 by Listing Common Factors
Factors of 4: 1, 2, 4Factors of 10: 1, 2, 5, 10
There are 2 common factors of 4 and 10, that are 1 and 2. Therefore, the greatest common factor of 4 and 10 is 2.
### GCF of 4 and 10 by Long Division
GCF of 4 and 10 is the divisor that we get when the remainder becomes 0 after doing long division repeatedly.
Step 1: Divide 10 (larger number) by 4 (smaller number).Step 2: Since the remainder ≠ 0, we will divide the divisor of step 1 (4) by the remainder (2).Step 3: Repeat this process until the remainder = 0.
The corresponding divisor (2) is the GCF of 4 and 10.
### GCF of 4 and 10 by Prime Factorization
Prime factorization of 4 and 10 is (2 × 2) and (2 × 5) respectively. As visible, 4 and 10 have only one common prime factor i.e. 2. Hence, the GCF of 4 and 10 is 2.
☛ Also Check:
GCF of 40 and 56 = 8
GCF of 15 and 50 = 5
GCF of 33 and 66 = 33
GCF of 30 and 36 = 6
GCF of 10, 30 and 45 = 5
GCF of 9 and 24 = 3
GCF of 36 and 100 = 4
## GCF of 4 and 10 Examples
Example 1: Find the GCF of 4 and 10, if their LCM is 20. Solution:
∵ LCM × GCF = 4 × 10
⇒ GCF(4, 10) = (4 × 10)/20 = 2
Therefore, the greatest common factor of 4 and 10 is 2.
Example 2: The product of two numbers is 40. If their GCF is 2, what is their LCM? Solution:
Given: GCF = 2 and product of numbers = 40
∵ LCM × GCF = product of numbers
⇒ LCM = Product/GCF = 40/2
Therefore, the LCM is 20.
Example 3: Find the greatest number that divides 4 and 10 exactly. Solution:
The greatest number that divides 4 and 10 exactly is their greatest common factor, i.e. GCF of 4 and 10.
⇒ Factors of 4 and 10:
Factors of 4 = 1, 2, 4
Factors of 10 = 1, 2, 5, 10
Therefore, the GCF of 4 and 10 is 2.
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## FAQs on GCF of 4 and 10
### What is the GCF of 4 and 10?
The . To calculate the greatest common factor (GCF) of 4 and 10, we need to factor each number (factors of 4 = 1, 2, 4; factors of 10 = 1, 2, 5, 10) and choose the greatest factor that exactly divides both 4 and 10, i.e., 2.
### How to Find the GCF of 4 and 10 by Prime Factorization?
To find the GCF of 4 and 10, we will find the prime factorization of the given numbers, i.e. 4 = 2 × 2; 10 = 2 × 5.
⇒ Since 2 is the only common prime factor of 4 and 10. Hence, GCF (4, 10) = 2.
☛ Prime Numbers
### If the GCF of 10 and 4 is 2, Find its LCM.
GCF(10, 4) × LCM(10, 4) = 10 × 4
Since the GCF of 10 and 4 = 2
⇒ 2 × LCM(10, 4) = 40
Therefore, LCM = 20
☛ Greatest Common Factor Calculator
### How to Find the GCF of 4 and 10 by Long Division Method?
To find the GCF of 4, 10 using long division method, 10 is divided by 4. The corresponding divisor (2) when remainder equals 0 is taken as GCF.
### What are the Methods to Find GCF of 4 and 10?
There are three commonly used methods to find the GCF of 4 and 10.
By Prime Factorization
By Euclidean Algorithm
By Long Division
### What is the Relation Between LCM and GCF of 4, 10?
The following equation can be used to express the relation between LCM (Least Common Multiple) and GCF of 4 and 10, i.e. GCF × LCM = 4 × 10.
GCF and LCM
GCF and LCM Worksheet
GCF and LCM Worksheet
Source : www.cuemath.com
## Greatest Common Factor of 4 and 10 GCF(4,10)
GCF of 4 and 10, find the biggest number that can divide two integers, learn what is the greatest common factor of 4 and 10.
Home » GCF Calculator
## What is the Greatest Common Factor of 4 and 10?
Greatest common factor (GCF) of 4 and 10 is 2.
GCF(4,10) = 2
We will now calculate the prime factors of 4 and 10, than find the greatest common factor (greatest common divisor (gcd)) of the numbers by matching the biggest common factor of 4 and 10.
## How to find the GCF of 4 and 10?
We will first find the prime factorization of 4 and 10. After we will calculate the factors of 4 and 10 and find the biggest common factor number .
### Step-1: Prime Factorization of 4
Prime factors of 4 are 2. Prime factorization of 4 in exponential form is:
4 = 22
### Step-2: Prime Factorization of 10
Prime factors of 10 are 2, 5. Prime factorization of 10 in exponential form is:
10 = 21 × 51
### Step-3: Factors of 4
List of positive integer factors of 4 that divides 4 without a remainder.
1, 2
### Step-4: Factors of 10
List of positive integer factors of 10 that divides 4 without a remainder.
1, 2, 5
### Final Step: Biggest Common Factor Number
We found the factors and prime factorization of 4 and 10. The biggest common factor number is the GCF number.
So the greatest common factor 4 and 10 is 2.
Also check out the Least Common Multiple of 4 and 10
### Related Greatest Common Factors of 4
GCF of 4 and 8 GCF of 4 and 9 GCF of 4 and 10 GCF of 4 and 11 GCF of 4 and 12 GCF of 4 and 13 GCF of 4 and 14 GCF of 4 and 15 GCF of 4 and 16 GCF of 4 and 17 GCF of 4 and 18 GCF of 4 and 19 GCF of 4 and 20 GCF of 4 and 21 GCF of 4 and 22 GCF of 4 and 23 GCF of 4 and 24
### Related Greatest Common Factors of 10
GCF of 10 and 14 GCF of 10 and 15 GCF of 10 and 16 GCF of 10 and 17 GCF of 10 and 18 GCF of 10 and 19 GCF of 10 and 20 GCF of 10 and 21 GCF of 10 and 22 GCF of 10 and 23 GCF of 10 and 24 GCF of 10 and 25 GCF of 10 and 26 GCF of 10 and 27 GCF of 10 and 28 GCF of 10 and 29 GCF of 10 and 30
Source : www.gcflcm.com
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2.1 Relations, Graphs, and Functions
Learning Objectives
1. State the domain and range of a relation.
2. Identify a function.
3. Use function notation.
Graphs, Relations, Domain, and Range
The rectangular coordinate systemA system with two number lines at right angles specifying points in a plane using ordered pairs (x, y). consists of two real number lines that intersect at a right angle. The horizontal number line is called the x-axisThe horizontal number line used as reference in a rectangular coordinate system., and the vertical number line is called the y-axisThe vertical number line used as reference in a rectangular coordinate system.. These two number lines define a flat surface called a planeThe flat surface defined by x- and y-axes., and each point on this plane is associated with an ordered pairPairs (x, y) that identify position relative to the origin on a rectangular coordinate plane. of real numbers (x, y). The first number is called the x-coordinate, and the second number is called the y-coordinate. The intersection of the two axes is known as the originThe point where the x- and y-axes cross, denoted by (0, 0)., which corresponds to the point (0, 0).
The x- and y-axes break the plane into four regions called quadrantsThe four regions of a rectangular coordinate plane partly bounded by the x- and y-axes and numbered using the Roman numerals I, II, III, and IV., named using roman numerals I, II, III, and IV, as pictured. The ordered pair (x, y) represents the position of points relative to the origin. For example, the ordered pair (−4, 3) represents the position 4 units to the left of the origin, and 3 units above in the second quadrant.
This system is often called the Cartesian coordinate systemTerm used in honor of René Descartes when referring to the rectangular coordinate system., named after the French mathematician René Descartes (1596–1650).
Figure 2.1
Rene Descartes Wikipedia
Next, we define a relationAny set of ordered pairs. as any set of ordered pairs. In the context of algebra, the relations of interest are sets of ordered pairs (x, y) in the rectangular coordinate plane. Typically, the coordinates are related by a rule expressed using an algebraic equation. For example, both the algebraic equations $y=|x|−2$ and $x=|y|+1$ define relationsips between x and y. Following are some integers that satisfy both equations:
Here two relations consisting of seven ordered pair solutions are obtained:
$y=|x|−2 has solutions {(−3,1),(−2,0),(−1,−1),(0,−2),(1,−1),(2,0),(3,1)}andx=|y|+1 has solutions {(4,−3),(3,−2),(2,−1),(1,0),(2,1),(3,2),(4,3)}$
We can visually display any relation of this type on a coordinate plane by plotting the points.
The solution sets of each equation will form a relation consisting of infinitely many ordered pairs. We can use the given ordered pair solutions to estimate all of the other ordered pairs by drawing a line through the given points. Here we put an arrow on the ends of our lines to indicate that this set of ordered pairs continues without bounds.
The representation of a relation on a rectangular coordinate plane, as illustrated above, is called a graphA visual representation of a relation on a rectangular coordinate plane.. Any curve graphed on a rectangular coordinate plane represents a set of ordered pairs and thus defines a relation.
The set consisting of all of the first components of a relation, in this case the x-values, is called the domainThe set consisting of all of the first components of a relation. For relations consisting of points in the plane, the domain is the set of all x-values.. And the set consisting of all second components of a relation, in this case the y-values, is called the rangeThe set consisting of all of the second components of a relation. For relations consisting of points in the plane, the range is the set of all y-values. (or codomainUsed when referencing the range.). Often, we can determine the domain and range of a relation if we are given its graph.
Here we can see that the graph of $y=|x|−2$ has a domain consisting of all real numbers, $ℝ=(−∞,∞)$, and a range of all y-values greater than or equal to −2, $[−2,∞).$ The domain of the graph of $x=|y|+1$ consists of all x-values greater than or equal to 1, $[1,∞)$, and the range consists of all real numbers, $ℝ=(−∞,∞).$
Example 1
Determine the domain and range of the following relation:
Solution:
The minimum x-value represented on the graph is −8 all others are larger. Therefore, the domain consists of all x-values in the interval $[−8,∞).$ The minimum y-value represented on the graph is 0; thus, the range is $[0,∞).$
Answer: Domain: $[−8,∞)$; range: $[0,∞)$
Functions
Of special interest are relations where every x-value corresponds to exactly one y-value. A relation with this property is called a functionA relation where each element in the domain corresponds to exactly one element in the range..
Example 2
Determine the domain and range of the following relation and state whether it is a function or not: {(−1, 4), (0, 7), (2, 3), (3, 3), (4, −2)}
Solution:
Here we separate the domain (x-values), and the range (y-values), and depict the correspondence between the values with arrows.
The relation is a function because each x-value corresponds to exactly one y-value.
Answer: The domain is {−1, 0, 2, 3, 4} and the range is {−2, 3, 4, 7}. The relation is a function.
Example 3
Determine the domain and range of the following relation and state whether it is a function or not: {(−4, −3), (−2, 6), (0, 3), (3, 5), (3, 7)}
Solution:
The given relation is not a function because the x-value 3 corresponds to two y-values. We can also recognize functions as relations where no x-values are repeated.
Answer: The domain is {−4, −2, 0, 3} and the range is {−3, 3, 5, 6, 7}. This relation is not a function.
Consider the relations consisting of the seven ordered pair solutions to $y=|x|−2$ and $x=|y|+1.$ The correspondence between the domain and range of each can be pictured as follows:
Notice that every element in the domain of the solution set of $y=|x|−2$ corresponds to only one element in the range; it is a function. The solutions to $x=|y|+1$, on the other hand, have values in the domain that correspond to two elements in the range. In particular, the x-value 4 corresponds to two y-values −3 and 3. Therefore, $x=|y|+1$ does not define a function.
We can visually identify functions by their graphs using the vertical line testIf any vertical line intersects the graph more than once, then the graph does not represent a function.. If any vertical line intersects the graph more than once, then the graph does not represent a function.
The vertical line represents a value in the domain, and the number of intersections with the graph represent the number of values to which it corresponds. As we can see, any vertical line will intersect the graph of $y=|x|−2$ only once; therefore, it is a function. A vertical line can cross the graph of $x=|y|+1$ more than once; therefore, it is not a function. As pictured, the x-value 3 corresponds to more than one y-value.
Example 4
Given the graph, state the domain and range and determine whether or not it represents a function:
Solution:
From the graph we can see that the minimum x-value is −1 and the maximum x-value is 5. Hence, the domain consists of all the real numbers in the set from $[−1,5].$ The maximum y-value is 3 and the minimum is −3; hence, the range consists of y-values in the interval $[−3,3].$
In addition, since we can find a vertical line that intersects the graph more than once, we conclude that the graph is not a function. There are many x-values in the domain that correspond to two y-values.
Answer: Domain: $[−1,5]$; range: $[−3,3]$; function: no
Try this! Given the graph, determine the domain and range and state whether or not it is a function:
Answer: Domain: $(−∞,15]$; range:$ℝ$; function: no
Function Notation
With the definition of a function comes special notation. If we consider each x-value to be the input that produces exactly one output, then we can use function notationThe notation $f(x)=y$, which reads “f of x is equal to y.” Given a function, y and $f(x)$ can be used interchangeably.:
$f(x)=y$
The notation $f(x)$ reads, “f of x” and should not be confused with multiplication. Algebra frequently involves functions, and so the notation becomes useful when performing common tasks. Here f is the function name, and $f(x)$ denotes the value in the range associated with the value x in the domain. Functions are often named with different letters; some common names for functions are f, g, h, C, and R. We have determined that the set of solutions to $y=|x|−2$ is a function; therefore, using function notation we can write:
$y=|x|−2↓f(x)=|x|−2$
It is important to note that y and $f(x)$ are used interchangeably. This notation is used as follows:
$f(x) = | x |−2↓ ↓f(−5)=|−5|−2=5−2=3$
Here the compact notation $f(−5)=3$ indicates that where $x=−5$ (the input), the function results in $y=3$ (the output). In other words, replace the variable with the value given inside the parentheses.
Functions are compactly defined by an algebraic equation, such as $f(x)=|x|−2.$ Given values for x in the domain, we can quickly calculate the corresponding values in the range. As we have seen, functions are also expressed using graphs. In this case, we interpret $f(−5)=3$ as follows:
Function notation streamlines the task of evaluating. For example, use the function h defined by $h(x)=12x−3$ to evaluate for x-values in the set {−2, 0, 7}.
$h(−2)=12(−2)−3=−1−3=−4 h(0)=12(0)−3=0−3=−3h(7)=12(7)−3=72−3=12$
Given any function defined by $h(x)=y$, the value x is called the argument of the functionThe value or algebraic expression used as input when using function notation.. The argument can be any algebraic expression. For example:
$h(4a3)=12(4a3)−3=2a3−3h(2x−1)=12(2x−1)−3=x−12−3=x−72$
Example 5
Given $g(x)=x2$, find $g(−2)$, $g(12)$, and $g(x+h).$
Solution:
Recall that when evaluating, it is a best practice to begin by replacing the variables with parentheses and then substitute the appropriate values. This helps with the order of operations when simplifying expressions.
$g(−2)=(−2)2=4g(12)=(12)2=14g(x+h)=(x+h)2=x2+2xh+h2$
Answer: $g(−2)=4$, $g(12)=14$, $g(x+h)=x2+2xh+h2$
At this point, it is important to note that, in general, $f(x+h)≠f(x)+f(h).$ The previous example, where $g(x)=x2$, illustrates this nicely.
$g(x+h)≠g(x)+g(h)(x+h)2≠x2+h2$
Example 6
Given $f(x)=2x+4$, find $f(−2)$, $f(0)$, and $f(12a2−2).$
Solution:
$f(−2)=2(−2)+4=−4+4=0=0f(0)=2(0)+4=0+4=4=2f(12a2−2)=2(12a2−2)+4=a2−4+4=a2=|a|$
Answer: $f(−2)=0$, $f(0)=2$, $f(12a2−2)=|a|$
Example 7
Given the graph of $g(x)$, find $g(−8)$, $g(0)$, and $g(8).$
Solution:
Use the graph to find the corresponding y-values where x = −8, 0, and 8.
Answer: $g(−8)=−2$, $g(0)=0$, $g(8)=2$
Sometimes the output is given and we are asked to find the input.
Example 8
Given $f(x)=5x+7$, find x where $f(x)=27.$
Solution:
In this example, the output is given and we are asked to find the input. Substitute $f(x)$ with 27 and solve.
$f(x)=5x+7↓27=5x+720=5x4=x$
Therefore, $f(4)=27.$ As a check, we can evaluate $f(4)=5(4)+7=27.$
Answer: $x=4$
Example 9
Given the graph of $g$, find x where $g(x)=2.$
Solution:
Here we are asked to find the x-value given a particular y-value. We begin with 2 on the y-axis and then read the corresponding x-value.
We can see that $g(x)=2$ where $x=−5$; in other words, $g(−5)=2.$
Answer: $x=−5$
Try this! Given the graph of $h$, find x where $h(x)=−4.$
Answer: $x=−5$ and $x=15$
Key Takeaways
• A relation is any set of ordered pairs. However, in this course, we will be working with sets of ordered pairs (x, y) in the rectangular coordinate system. The set of x-values defines the domain and the set of y-values defines the range.
• Special relations where every x-value (input) corresponds to exactly one y-value (output) are called functions.
• We can easily determine whether or not an equation represents a function by performing the vertical line test on its graph. If any vertical line intersects the graph more than once, then the graph does not represent a function.
• If an algebraic equation defines a function, then we can use the notation $f(x)=y.$ The notation $f(x)$ is read “f of x” and should not be confused with multiplication. When working with functions, it is important to remember that y and $f(x)$ are used interchangeably.
• If asked to find $f(a)$, we substitute the argument $a$ in for the variable and then simplify. The argument could be an algebraic expression.
• If asked to find $x$ where $f(x)=a$, we set the function equal to $a$ and then solve for $x.$
Part A: Relations and Functions
Determine the domain and range and state whether the relation is a function or not.
1. {(3, 1), (5, 2), (7, 3), (9, 4), (12, 4)}
2. {(2, 0), (4, 3), (6, 6), (8, 6), (10, 9)}
3. {(7, 5), (8, 6), (10, 7), (10, 8), (15, 9)}
4. {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1)}
5. {(5, 0), (5, 2), (5, 4), (5, 6), (5, 8)}
6. {(−3, 1), (−2, 2), (−1, 3), (0, 4), (0, 5)}
Part B: Function Notation
Evaluate.
1. $g(x)=|x−5|$ find $g(−5)$, $g(0)$, and $g(5).$
2. $g(x)=|x|−5$; find $g(−5)$, $g(0)$, and $g(5).$
3. $g(x)=|2x−3|$; find $g(−1)$, $g(0)$, and $g(32).$
4. $g(x)=3−|2x|$; find $g(−3)$, $g(0)$, and $g(3).$
5. $f(x)=2x−3$; find $f(−2)$, $f(0)$, and $f(x−3).$
6. $f(x)=5x−1$; find $f(−2)$, $f(0)$, and $f(x+1).$
7. $g(x)=23x+1$; find $g(−3)$, $g(0)$, and $f(9x+6).$
8. $g(x)=−34x−12$; find $g(−4)$, $g(0)$, and $g(6x−2).$
9. $g(x)=x2$; find $g(−5)$, $g(3)$, and $g(x−5).$
10. $g(x)=x2+1$; find $g(−1)$, $g(6)$, and $g(2x−1).$
11. $f(x)=x2−x−2$; find $f(0)$, $f(2)$, and $f(x+2).$
12. $f(x)=−2x2+x−4$; find $f(−2)$, $f(12)$, and $f(x−3).$
13. $h(t)=−16t2+32$; find $h(14)$, $h(12)$, and $h(2a−1).$
14. $h(t)=−16t2+32$; find $h(0)$, $h(2)$, $h(2a+1).$
15. $f(x)=x+1−2$ find $f(−1)$, $f(0)$, $f(x−1).$
16. $f(x)=x−3+1$; find $f(12)$, $f(3)$, $f(x+3).$
17. $g(x)=x+8$; find $g(0)$, $g(−8)$, and $g(x−8).$
18. $g(x)=3x−1$; find $g(13)$, $g(53)$, and $g(13a2+13).$
19. $f(x)=x3+1$; find $f(−1)$, $f(0)$, $f(a2).$
20. $f(x)=x3−8$; find $f(2)$, $f(0)$, $f(a3).$
Given the function find $f(x+h).$
1. $f(x)=3x−1$
2. $f(x)=−5x+2$
3. $f(x)=x2+x+1$
4. $f(x)=2x2−x−1$
5. $f(x)=x3$
6. $f(x)=2x3−1$
Find x given the function.
1. $f(x)=2x−3$; find x where $f(x)=25.$
2. $f(x)=7−3x$; find x where $f(x)=−27.$
3. $f(x)=2x+5$; find x where $f(x)=0$
4. $f(x)=−2x+1$; find x where $f(x)=0$
5. $g(x)=6x+2$; find x where $g(x)=5.$
6. $g(x)=4x+5$; find x where $g(x)=2.$
7. $h(x)=23x−12$; find x where $h(x)=16.$
8. $h(x)=54x+13$; find x where $h(x)=12.$
9. The value of a new car in dollars is given by the function $V(t)=−1,800t+22,000$ where t represents the age of the car in years. Use the function to determine the value of the car when it is 4 years old. What was the value of the car new?
10. The monthly income in dollars of a commissioned car salesperson is given by the function $I(n)=350n+1,450$ where n represents the number of cars sold in the month. Use the function to determine the salesperson’s income if he sells 3 cars this month. What is his income if he does not sell any cars in one month?
Given the graph of the function $f$, find the function values.
1. Find $f(0)$, $f(2)$, and $f(4).$
2. Find $f(−1)$, $f(0)$, and $f(1).$
3. Find $f(0)$, $f(2)$, and $f(4).$
4. Find $f(−3)$, $f(0)$, and $f(3).$
5. Find $f(−4)$, $f(0)$, and $f(2).$
6. Find $f(−6)$, $f(0)$, and $f(6).$
7. Find $f(−2)$, $f(2)$, and $f(7).$
8. Find $f(0)$, $f(5)$, and $f(9).$
9. Find $f(−8)$, $f(0)$, and $f(8).$
10. Find $f(−12)$, $f(0)$, and $f(12).$
Given the graph of a function $g$, find the x-values.
1. Find x where $g(x)=3$, $g(x)=0$, and $g(x)=−2.$
2. Find x where $g(x)=0$, $g(x)=1$, and $g(x)=4.$
3. Find x where $g(x)=−5$, $g(x)=0$, and $g(x)=10.$
4. Find x where $g(x)=0$, $g(x)=10$, and $g(x)=15.$
5. Find x where $g(x)=−5$, $g(x)=−4$, and $g(x)=4.$
6. Find x where $g(x)=1$, $g(x)=0$, and $g(x)=−3.$
7. Find x where $g(x)=−4$, $g(x)=3$, and $g(x)=4.$
8. Find x where $g(x)=−5$, $g(x)=−4$, and $g(x)=4.$
9. Find x where $g(x)=−10$ and $g(x)=5.$
10. Find x where $g(x)=2.$
The value of a certain automobile in dollars depends on the number of years since it was purchased in 1970 according to the following function:
1. What was the value of the car when it was new in 1970?
2. In what year was the value of the car at a minimum?
3. What was the value of the car in 2005?
4. In what years was the car valued at $4,000? Given the linear function defined by $f(x)=2x−5,$ simplify the following. 1. $f(5)−f(3)$ 2. $f(0)−f(7)$ 3. $f(x+2)−f(2)$ 4. $f(x+7)−f(7)$ 5. $f(x+h)−f(x)$ 6. $f(x+h)−f(x)h$ 7. Simplify $c(x+h)−c(x)h$ given $c(x)=3x+1.$ 8. Simplify $p(x+h)−p(x)h$ given $p(x)=7x−3.$ 9. Simplify $g(x+h)−g(x)h$ given $g(x)=mx+b.$ 10. Simplify $q(x+h)−q(x)h$ given $q(x)=ax.$ Part C: Discussion Board 1. Who is credited with the introduction of the notation $y=f(x)$? Provide a brief summary of his life and accomplishments. 2. Explain to a beginning algebra student what the vertical line test is and why it works. 3. Research and discuss the life and contributions of René Descartes. 4. Conduct an Internet search for the vertical line test, functions, and evaluating functions. Share a link to a page that you think others may find useful. Answers 1. Domain: {3, 5, 7, 9, 12}; range: {1, 2, 3, 4}; function: yes 2. Domain: {7, 8, 10, 15}; range: {5, 6, 7, 8, 9}; function: no 3. Domain: {5}; range: {0, 2, 4, 6, 8}; function: no 4. Domain: {−4, −1, 0, 2, 3}; range: {1, 2, 3}; function: yes 5. Domain: {−1, 0, 1, 2}; range: {0, 1, 2, 3, 4}; function: no 6. Domain: {−2}; range: {−4, −2, 0, 2, 4}; function: no 7. Domain: $ℝ$; range: $[−2,∞)$; function: yes 8. Domain: $(−∞,−1]$; range: $ℝ$; function: no 9. Domain: $(−∞,0]$; range: $[−1,∞)$; function: yes 10. Domain: $ℝ$; range: $(−∞,3]$; function: yes 11. Domain: $ℝ$; range: $ℝ$; function: yes 12. Domain: $[−5,−1]$; range: $[−2,2]$; function: no 13. Domain: $ℝ$; range: $[0,∞]$; function: yes 14. Domain: $ℝ$; range: $ℝ$; function: yes 15. Domain: $ℝ$; range: $[−1,1]$; function: yes 16. Domain: $[−8,8]$; range: $[−3,3]$; function: no 17. Domain: $ℝ$; range: $[−8,∞]$; function: yes 1. $g(−5)=10$, $g(0)=5$, $g(5)=0$ 2. $g(−1)=5$, $g(0)=3$, $g(32)=0$ 3. $f(−2)=−7$, $f(0)=−3$, $f(x−3)=2x−9$ 4. $g(−3)=−1$, $g(0)=1$, $g(9x+6)=6x+5$ 5. $g(−5)=25$, $g(3)=3$, $g(x−5)=x2−10x+25$ 6. $f(0)=−2$, $f(2)=0$, $f(x+2)=x2+3x$ 7. $h(14)=31$, $h(12)=28$, $h(2a−1)=−64a2+64a+16$ 8. $f(−1)=−2$, $f(0)=−1$, $f(x−1)=x−2$ 9. $g(0)=22$, $g(−8)=0$, $g(a2−8)=|a|$ 10. $f(−1)=0$, $f(0)=1$, $f(a2)=a6+1$ 11. $f(x+h)=3x+3h−1$ 12. $f(x+h)=x2+2xh+h2+x+h+1$ 13. $f(x+h)=x3+3hx2+3h2x+h3$ 14. $x=14$ 15. $x=−52$ 16. $x=12$ 17. $x=1$ 18. New:$22,000; 4 yrs old: $14,800 19. $f(0)=5$, $f(2)=1$, $f(4)=5$ 20. $f(0)=0$, $f(2)=2$, $f(4)=0$ 21. $f(−4)=3$, $f(0)=3$, $f(2)=3$ 22. $f(−2)=1$, $f(2)=3$, $f(7)=4$ 23. $f(−8)=10$, $f(0)=0$, $f(8)=10$ 24. $g(−4)=3$, $g(2)=0$, and $g(6)=−2.$ 25. $g(10)=−5$, $g(5)=0 and g(15)=0$, $g(−5)=10 and g(25)=10$ 26. $g(−2)=−5$, $g(−3)=−4 and g(−1)=−4$, $g(−5)=4 and g(1)=4$ 27. $g(−2)=−4$, $g(−1)=3$, $g(0)=4$ 28. $g(−10)=−10 and g(5)=−10$; $g(−5)=5 and g(10)=5$ 29.$5,000
30. \$10,000
31. 4
32. $2x$
33. $2h$
34. 3
35. m
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# A ball with a mass of 3 kg is rolling at 1 m/s and elastically collides with a resting ball with a mass of 4 kg. What are the post-collision velocities of the balls?
Mar 9, 2016
${v}_{1}^{'} = - \frac{1}{7} \frac{m}{s}$
${v}_{2}^{'} = \frac{6}{7} \frac{m}{s}$
#### Explanation:
${m}_{1} \cdot {v}_{1} + {m}_{2} {v}_{2} = {m}_{1} {v}_{1}^{'} + {m}_{2} {v}_{2}^{'} \text{ } \left(1\right)$
$\text{total momentum before collision=total momentum after collision}$
$\frac{1}{2} \cdot {m}_{1} \cdot {v}_{1}^{2} + \frac{1}{2} \cdot {m}_{2} \cdot {v}_{2}^{2} = \frac{1}{2.} {m}_{1} \cdot {v}_{1} {'}^{2} + \frac{1}{2} \cdot {m}_{2} \cdot {v}_{2} {'}^{2}$
${m}_{1} \cdot {v}_{1}^{2} + {m}_{2} \cdot {v}_{2}^{2} = {m}_{1} \cdot {v}_{1} {'}^{2} + {m}_{2} \cdot {v}_{2} {'}^{2} \text{ } \left(2\right)$
$\text{we obtain the equation of (3),if we arrange (1) and (2)}$
${v}_{1} + {v}_{1}^{'} = {v}_{2} + {v}_{2}^{'} \text{ } \left(3\right)$
$3 \cdot 1 + 4 \cdot 0 = 3 \cdot {v}_{1}^{'} + 4 \cdot {v}_{2}^{'}$
$3 = 3 \cdot {v}_{1}^{'} + 4 \cdot {v}_{2}^{'} \text{ } \left(4\right)$
$\text{use (3)}$
$1 + {v}_{1}^{'} = 0 + {v}_{2}^{'}$
${v}_{2}^{'} = 1 + {v}_{1}^{'} \text{ } \left(5\right)$
$\text{use (4)}$
$3 = 3 \cdot {v}_{1}^{'} + 4 \left(1 + {v}_{1}^{'}\right)$
$3 = 3 \cdot {v}_{1}^{'} + 4 + 4 {v}_{1}^{'}$
$- 1 = 7 {v}_{1}^{'} \text{ } {v}_{1}^{'} = - \frac{1}{7} \frac{m}{s}$
$\text{use (5)}$
${v}_{2}^{'} = 1 - \frac{1}{7}$
${v}_{2}^{'} = \frac{6}{7} \frac{m}{s}$
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Solving Rational Equations and Inequalities
Presentation on theme: "Solving Rational Equations and Inequalities"— Presentation transcript:
Solving Rational Equations and Inequalities
8-5 Warm Up Lesson Presentation Lesson Quiz Holt McDougal Algebra 2 Holt Algebra 2
Find the least common multiple for each pair.
Warm Up Find the least common multiple for each pair. 1. 2x2 and 4x2 – 2x 2x2(2x – 1) 2. x + 5 and x2 – x – 30 (x + 5)(x – 6) Add or subtract. Identify any x-values for which the expression is undefined. 1 x – 2 4x + 3. 5x – 2 4x(x – 2) –(x – 1) x2 1 x2 x x ≠ 0 4.
Objective Solve rational equations and inequalities.
Vocabulary rational equation extraneous solution rational inequality
A rational equation is an equation that contains one or more rational expressions. The time t in hours that it takes to travel d miles can be determined by using the equation t = , where r is the average rate of speed. This equation is a rational equation. d r
To solve a rational equation, start by multiplying each term of the equation by the least common denominator (LCD) of all of the expressions in the equation. This step eliminates the denominators of the rational expression and results in an equation you can solve by using algebra.
Example 1: Solving Rational Equations
18 x Solve the equation x – = 3. x(x) – (x) = 3(x) 18 x Multiply each term by the LCD, x. x2 – 18 = 3x Simplify. Note that x ≠ 0. x2 – 3x – 18 = 0 Write in standard form. (x – 6)(x + 3) = 0 Factor. x – 6 = 0 or x + 3 = 0 Apply the Zero Product Property. x = 6 or x = –3 Solve for x.
Example 1 Continued 18 x 18 x Check x – = 3 x – = 3 18 6 6 – 18 (–3) (–3) – 3 3 6 – 3 3 –3 + 6 3 3 3 3 3
Check It Out! Example 1a Solve the equation = 4 x 10 3 (3x) = (3x) + 2(3x) 10 3 4 x Multiply each term by the LCD, 3x. 10x = x Simplify. Note that x ≠ 0. 4x = 12 Combine like terms. x = 3 Solve for x.
Check It Out! Example 1b Solve the equation = – . 5 4 6 x 7 (4x) (4x) = – (4x) 6 x 5 4 7 Multiply each term by the LCD, 4x. 24 + 5x = –7x Simplify. Note that x ≠ 0. 24 = –12x Combine like terms. x = –2 Solve for x.
Check It Out! Example 1c 6 x Solve the equation x = – 1. x(x) = (x) – 1(x) 6 x Multiply each term by the LCD, x. x2 = 6 – x Simplify. Note that x ≠ 0. x2 + x – 6 = 0 Write in standard form. (x – 2)(x + 3) = 0 Factor. x – 2 = 0 or x + 3 = 0 Apply the Zero Product Property. x = 2 or x = –3 Solve for x.
An extraneous solution is a solution of an equation derived from an original equation that is not a solution of the original equation. When you solve a rational equation, it is possible to get extraneous solutions. These values should be eliminated from the solution set. Always check your solutions by substituting them into the original equation.
Example 2A: Extraneous Solutions
Solve each equation. 5x x – 2 3x + 4 = 5x x – 2 3x + 4 (x – 2) = (x – 2) Multiply each term by the LCD, x – 2. 5x x – 2 3x + 4 (x – 2) = (x – 2) Divide out common factors. 5x = 3x + 4 Simplify. Note that x ≠ 2. x = 2 Solve for x. The solution x = 2 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution.
Example 2A Continued Check Substitute 2 for x in the original equation. 5x x – 2 3x + 4 = 5(2) 2 – 2 3(2) + 4 2 – 2 10 10 Division by 0 is undefined.
Example 2B: Extraneous Solutions
Solve each equation. 2x – 5 x – 8 x 2 11 x – 8 = Multiply each term by the LCD, 2(x – 8). 2x – 5 x – 8 2(x – 8) (x – 8) = (x – 8) 11 x 2 Divide out common factors. 2x – 5 x – 8 2(x – 8) (x – 8) = (x – 8) 11 x 2 2(2x – 5) + x(x – 8) = 11(2) Simplify. Note that x ≠ 8. Use the Distributive Property. 4x – 10 + x2 – 8x = 22
Example 2B Continued x2 – 4x – 32 = 0 Write in standard form. (x – 8)(x + 4) = 0 Factor. x – 8 = 0 or x + 4 = 0 Apply the Zero Product Property. x = 8 or x = –4 Solve for x. The solution x = 8 us extraneous because it makes the denominator of the original equation equal to 0. The only solution is x = –4.
Example 2B Continued Check Write 2x – 5 x – 8 11 = x 2 2x – 5 x – 8 11 – = 0. x 2 as Graph the left side of the equation as Y1. Identify the values of x for which Y1 = 0. The graph intersects the x-axis only when x = –4. Therefore, x = –4 is the only solution.
Check It Out! Example 2a 16 x2 – 16 2 x – 4 Solve the equation = Multiply each term by the LCD, (x – 4)(x +4). 16 (x – 4)(x + 4) 2 x – 4 (x – 4)(x + 4) = (x – 4 )(x + 4) Divide out common factors. 16 (x – 4)(x + 4) 2 x – 4 (x – 4)(x + 4) = (x – 4 )(x + 4) 16 = 2x + 8 Simplify. Note that x ≠ ±4. x = 4 Solve for x. The solution x = 4 is extraneous because it makes the denominators of the original equation equal to 0. Therefore, the equation has no solution.
Check It Out! Example 2b x x – 1 Solve the equation 1 = 6 Multiply each term by the LCD, 6(x – 1). 1 x – 1 6(x – 1) = (x – 1) (x – 1) x 6 Divide out common factors. 1 x – 1 6(x – 1) = (x – 1) (x – 1) x 6 6 = 6x + x(x – x) Simplify. Note that x ≠ 1. Use the Distributive Property. 6 = 6x + x2 – x
Check It Out! Example 2b Continued
0 = x2 + 5x – 6 Write in standard form. 0 = (x + 6)(x – 1) Factor. x + 6 = 0 or x – 1 = 0 Apply the Zero Product Property. x = –6 or x = 1 Solve for x. The solution x = 1 us extraneous because it makes the denominator of the original equation equal to 0. The only solution is x = –6.
Example 3: Problem-Solving Application
A jet travels 3950 mi from Chicago, Illinois, to London, England, and 3950 mi on the return trip. The total flying time is 16.5 h. The return trip takes longer due to winds that generally blow from west to east. If the jet’s average speed with no wind is 485 mi/h, what is the average speed of the wind during the round-trip flight? Round to the nearest mile per hour.
Understand the Problem
Example 3 Continued 1 Understand the Problem The answer will be the average speed of the wind. List the important information: The jet spent 16.5 h on the round-trip. It went 3950 mi east and 3950 mi west. Its average speed with no wind is 485 mi/h.
Example 3 Continued 2 Make a Plan
Let w represent the speed of the wind. When the jet is going east, its speed is equal to its speed with no wind plus w. When the jet is going west, its speed is equal to its speed with no wind minus w. Distance (mi) Average Speed (mi/h) Time (h) East 3950 485 + w West 485 – w 485 + w 3950 485 – w 3950 total time = time east + time west 16.5 485 + w 3950 485 – w = +
Use the Distributive Property.
Solve 3 The LCD is (485 + w)(485 – w). 16.5(485 + w)(485 – w) 485 + w 3950 = (485 + w)(485 – w) 485 – w (485 + w)(485 – w) Simplify. Note that x ≠ ±485. 16.5(485 + w)(485 – w) = 3950(485 – w) (485 + w) Use the Distributive Property. 3,881,212.5 – 16.5w2 = 1,915,750 – 3950w + 1,915, w 3,881,212.5 – 16.5w2 = 3,831,500 Combine like terms. –16.5w2 = –49,712.5 Solve for w. w ≈ ± 55 The speed of the wind cannot be negative. Therefore, the average speed of the wind is 55 mi/h.
Example 3 Continued Look Back 4 If the speed of the wind is 55 mi/h, the jet’s speed when going east is = 540 mi/h. It will take the jet approximately 7.3 h to travel 3950 mi east. The jet’s speed when going west is 485 – 55 = 430 mi/h. It will take the jet approximately 9.2 h to travel 3950 mi west. The total trip will take 16.5 h, which is the given time.
Check It Out! Example 3 On a river, a kayaker travels 2 mi upstream and 2 mi downstream in a total of 5 h. In still water, the kayaker can travel at an average speed of 2 mi/h. Based on this information, what is the average speed of the current of this river? Round to the nearest tenth.
Understand the Problem
Check It Out! Example 3 Continued 1 Understand the Problem The answer will be the average speed of the current. List the important information: The kayaker spent 5 hours kayaking. She went 2 mi upstream and 2 mi downstream. Her average speed in still water is 2 mi/h.
Check It Out! Example 3 Continued
2 Make a Plan Let c represent the speed of the current. When the kayaker is going upstream, her speed is equal to her speed in still water minus c. When the kayaker is going downstream, her speed is equal to her speed in still water plus c. Distance (mi) Average Speed (mi/h) Time (h) Up 2 2 – c Down 2 + c 2 – c 2 2 + c 2 total time = time up- stream + time down- stream 5 2 – c 2 2 + c = +
Use the Distributive Property.
Solve 3 The LCD is (2 – c)(2 + c). (2 + c)(2 – c) = (2 + c)(2 – c) 5(2 + c)(2 – c) 2 – c 2 2 + c Simplify. Note that x ≠ ±2. 5(2 + c)(2 – c) = 2(2 + c) + 2(2 – c) Use the Distributive Property. 20 – 5c2 = 4 + 2c + 4 – 2c 20 – 5c2 = 8 Combine like terms. –5c2 = –12 Solve for c. c ≈ ± 1.5 The speed of the current cannot be negative. Therefore, the average speed of the current is about 1.5 mi/h.
Check It Out! Example 3 Continued
Look Back 4 If the speed of the current is about 1.5 mi/h, the kayaker’s speed when going upstream is 2 – 1.5 = 0.5 mi/h. It will take her about 4 h to travel 2 mi upstream. Her speed when going downstream is about = 3.5 mi/h. It will take her 0.5 h to travel 2 mi downstream. The total trip will take about 4.5 hours which is close to the given time of 5 h.
Example 4: Work Application
Natalie can finish a 500-piece puzzle in about 8 hours. When Natalie and Renzo work together, they can finish a 500-piece puzzle in about 4.5 hours. About how long will it take Renzo to finish a 500-piece puzzle if he works by himself? 1 8 Natalie’s rate: of the puzzle per hour 1 h Renzo’s rate: of the puzzle per hour, where h is the number of hours needed to finish the puzzle by himself.
Example 4 Continued Natalie’s rate hours worked Renzo’s rate
1 complete puzzle + = 1 8 (4.5) 1 h (4.5) + = 1 1 8 (4.5)(8h) + h (4.5)(8h) = 1(8h) Multiply by the LCD,8h. 4.5h + 36 = 8h Simplify. 36 = 3.5h Solve for h. 10.3 = h It will take Renzo about 10.3 hours, or 10 hours 17 minutes to complete a 500-piece puzzle working by himself.
Check It Out! Example 4 Julien can mulch a garden in 20 minutes. Together Julien and Remy can mulch the same garden in 11 minutes. How long will it take Remy to mulch the garden when working alone? Julien’s rate: of the garden per minute 1 20 Remy’s rate: of the garden per minute, where m is the number of minutes needed to mulch the garden by himself. 1 m
Check It Out! Example 4 Continued
Julien’s rate min worked Remy’s rate 1 complete job + = 1 20 (11) 1 m (11) + = 1 1 20 (11)(20m)+ m (11)(20m) = 1(20m) Multiply by the LCD, 20m. 11m = 20m Simplify. 220 = 9m Solve for m. 24 ≈ m It will take Remy about 24 minutes to mulch the garden working by himself.
A rational inequality is an inequality that contains one or more rational expressions. One way to solve rational inequalities is by using graphs and tables.
Vertical asymptote: x = 6
Example 5: Using Graphs and Tables to Solve Rational Equations and Inequalities Solve ≤ 3 by using a graph and a table. x x – 6 x x – 6 Use a graph. On a graphing calculator, Y1 = and Y2 = 3. (9, 3) The graph of Y1 is at or below the graph of Y2 when x < 6 or when x ≥ 9. Vertical asymptote: x = 6
Example 5 Continued Use a table. The table shows that Y1 is undefined when x = 6 and that Y1 ≤ Y2 when x ≥ 9. The solution of the inequality is x < 6 or x ≥ 9.
Vertical asymptote: x = 3
Check It Out! Example 5a Solve ≥ 4 by using a graph and a table. x x – 3 x x – 3 Use a graph. On a graphing calculator, Y1 = and Y2 = 4. (4, 4) The graph of Y1 is at or below the graph of Y2 when x < 3 or when x ≥ 4. Vertical asymptote: x = 3
Check It Out! Example 5a continued
Use a table. The table shows that Y1 is undefined when x = 3 and that Y1 ≤ Y2 when x ≥ 4. The solution of the inequality is x < 3 or x ≥ 4.
Vertical asymptote: x = –1
Check It Out! Example 5b Solve = –2 by using a graph and a table. 8 x + 1 8 x + 1 Use a graph. On a graphing calculator, Y1 = and Y2 = –2. (–5, –2) The graph of Y1 is at or below the graph of Y2 when x = –5. Vertical asymptote: x = –1
Check It Out! Example 5b continued
Use a table. The table shows that Y1 is undefined when x = –1 and that Y1 ≤ Y2 when x = –5. The solution of the inequality is x = –5.
You can also solve rational inequalities algebraically
You can also solve rational inequalities algebraically. You start by multiplying each term by the least common denominator (LCD) of all the expressions in the inequality. However, you must consider two cases: the LCD is positive or the LCD is negative.
Example 6: Solving Rational Inequalities Algebraically
Solve ≤ 3 algebraically. 6 x – 8 Case 1 LCD is positive. Step 1 Solve for x. 6 x – 8 (x – 8) ≤ 3(x – 8) Multiply by the LCD. 6 ≤ 3x – 24 Simplify. Note that x ≠ 8. 30 ≤ 3x Solve for x. 10 ≤ x Rewrite with the variable on the left. x ≥ 10
Example 6 Continued Solve ≤ 3 algebraically. 6 x – 8 Step 2 Consider the sign of the LCD. x – 8 > 0 LCD is positive. x > 8 Solve for x. For Case 1, the solution must satisfy x ≥ 10 and x > 8, which simplifies to x ≥ 10.
Example 6: Solving Rational Inequalities Algebraically
Solve ≤ 3 algebraically. 6 x – 8 Case 2 LCD is negative. Step 1 Solve for x. 6 x – 8 (x – 8) ≥ 3(x – 8) Multiply by the LCD. Reverse the inequality. 6 ≥ 3x – 24 Simplify. Note that x ≠ 8. 30 ≥ 3x Solve for x. 10 ≥ x Rewrite with the variable on the left. x ≤ 10
Step 2 Consider the sign of the LCD.
Example 6 Continued Solve ≤ 3 algebraically. 6 x – 8 Step 2 Consider the sign of the LCD. x – 8 > 0 LCD is positive. x > 8 Solve for x. For Case 2, the solution must satisfy x ≤ 10 and x < 8, which simplifies to x < 8. The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality ≤ 3 is x < 8 or x ≥ 10, or {x|x < 8 x ≥ 10}. 6 x – 8
Check It Out! Example 6a Solve ≥ –4 algebraically. 6 x – 2 Case 1 LCD is positive. Step 1 Solve for x. 6 x – 2 (x – 2) ≥ –4(x – 2) Multiply by the LCD. 6 ≥ –4x + 8 Simplify. Note that x ≠ 2. –2 ≥ –4x Solve for x. ≤ x 1 2 Rewrite with the variable on the left. x ≥ 1 2
Check It Out! Example 6a Continued
Solve ≥ –4 algebraically. 6 x – 2 Step 2 Consider the sign of the LCD. x – 2 > 0 LCD is positive. x > 2 Solve for x. For Case 1, the solution must satisfy and x > 2, which simplifies to x > 2. x ≥ 1 2
Check It Out! Example 6a Continued
Solve ≥ –4 algebraically. 6 x – 2 Case 2 LCD is negative. Step 1 Solve for x. 6 x – 2 (x – 2) ≤ –4(x – 2) Multiply by the LCD. Reverse the inequality. 6 ≤ –4x + 8 Simplify. Note that x ≠ 2. –2 ≤ –4x Solve for x. ≥ x 1 2 Rewrite with the variable on the left. x ≤ 1 2
Check It Out! Example 6a Continued
Solve ≥ –4 algebraically. 6 x – 2 Step 2 Consider the sign of the LCD. x – 2 < 0 LCD is negative. x < 2 Solve for x. For Case 2, the solution must satisfy and x < 2, which simplifies to x ≤ 1 2 The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality ≥ –4 is x > 2 or x ≤ , or {x| x > 2}. 6 x – 2 x ≤ 1 2
Check It Out! Example 6b Solve < 6 algebraically. 9 x + 3 Case 1 LCD is positive. Step 1 Solve for x. 9 x + 3 (x + 3) < 6(x + 3) Multiply by the LCD. 9 < 6x + 18 Simplify. Note that x ≠ –3. –9 < 6x Solve for x. – < x 3 2 Rewrite with the variable on the left. x > – 3 2
Check It Out! Example 6b Continued
Solve < 6 algebraically. 9 x + 3 Step 2 Consider the sign of the LCD. x + 3 > 0 LCD is positive. x > –3 Solve for x. For Case 1, the solution must satisfy and x > –3, which simplifies to x >– 3 2
Check It Out! Example 6b Continued
Solve < 6 algebraically. 9 x + 3 Case 2 LCD is negative. Step 1 Solve for x. Multiply by the LCD. Reverse the inequality. 9 x + 3 (x + 3) > 6(x + 3) 9 > 6x + 18 Simplify. Note that x ≠ –3. –9 > 6x Solve for x. – > x 3 2 Rewrite with the variable on the left. x < – 3 2
Check It Out! Example 6b Continued
Solve < 6 algebraically. 9 x + 3 Step 2 Consider the sign of the LCD. x + 3 < 0 LCD is negative. x < –3 Solve for x. For Case 2, the solution must satisfy and x < –3, which simplifies to x < –3. x <– 3 2 The solution set of the original inequality is the union of the solutions to both Case 1 and Case 2. The solution to the inequality < 6 is x < –3 or x > – , or {x| x < –3}. x > – 3 2 9 x + 3
Lesson Quiz Solve each equation or inequality. x + 2 x x – 1 2 = 1. x = –1 or x = 4 6x x + 4 7x + 4 = 2. no solution 3. x + 2 x – 3 x 5 5 x – 3 = x = –5 4 x – 3 4. ≥ 2 3 < x ≤ 5 5. A college basketball player has made 58 out of 82 attempted free-throws this season. How many additional free-throws must she make in a row to raise her free-throw percentage to 90%? 158
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# HOW TO FIND THE TRANSPOSE OF THE MATRIX
How to Find the Transpose of the Matrix ?
Here we are going to see, how to find the transpose of the matrix.
Transpose of a matrix :
The matrix which is obtained by interchanging the elements in rows and columns of the given matrix A is called transpose of A and is denoted by AT (read as A transpose).
If order of A is m x n then order of AT is n x m . We note that (AT )T = A.
## How to Find the Transpose of the Matrix - Questions
Question 1 :
If A =
then find the transpose of a matrix.
Solution :
To find the transpose of the given matrix, we have to write the elements in the rows as columns and the elements in the columns as rows.
Question 2 :
If A =
then find the transpose of-A.
Solution :
Question 3 :
If A =
then verify (AT )T = A
Solution :
Hence proved.
Question 4 :
Solution :
Since the order of matrices are same, corresponding terms will be equal.
x = 3 , y = 12 and z = 3
Solution :
x + y = 6 -----(1)
xy = 8 ---(2)
5 + z = 5
z = 5 - 5 = 0
From (2),
y = 8/x
By applying the value of y in (1), we get
x + (8/x) = 6
x2 + 8 = 6x
x2 - 6x + 8 = 0
(x - 2)(x - 4) = 0
x = 2 and x = 4.
Solution :
x + y + z = 9 ---(1)
x + z = 5 ---(2)
y + z = 7 ---(3)
By applying (2) in (1), we get
y + 5 = 9
y = 9 - 5
y = 4
By applying the value of y in (3), we get
z = 7 - 4
z = 3
By applying the value of y and z in (1), we get
x + 4 + 3 = 9
x = 9 - 7
x = 2
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Ratio and proportion word problems
Time and work word problems
Word problems on sets and venn diagrams
Word problems on ages
Pythagorean theorem word problems
Percent of a number word problems
Word problems on constant speed
Word problems on average speed
Word problems on sum of the angles of a triangle is 180 degree
OTHER TOPICS
Profit and loss shortcuts
Percentage shortcuts
Times table shortcuts
Time, speed and distance shortcuts
Ratio and proportion shortcuts
Domain and range of rational functions
Domain and range of rational functions with holes
Graphing rational functions
Graphing rational functions with holes
Converting repeating decimals in to fractions
Decimal representation of rational numbers
Finding square root using long division
L.C.M method to solve time and work problems
Translating the word problems in to algebraic expressions
Remainder when 2 power 256 is divided by 17
Remainder when 17 power 23 is divided by 16
Sum of all three digit numbers divisible by 6
Sum of all three digit numbers divisible by 7
Sum of all three digit numbers divisible by 8
Sum of all three digit numbers formed using 1, 3, 4
Sum of all three four digit numbers formed with non zero digits
Sum of all three four digit numbers formed using 0, 1, 2, 3
Sum of all three four digit numbers formed using 1, 2, 5, 6
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ನೀವು ವೆಬ್ ಫಿಲ್ಟರ್ ಹಿಂದೆ ಇದ್ದರೆ, ಡೊಮೇನ್ಗಳು *.kastatic.org ಮತ್ತು *.kasandbox.org ಗಳನ್ನು ಅನ್ ಬ್ಲಾಕ್ ಮಾಡಲಾಗಿದೆ ಎಂದು ಖಚಿತಪಡಿಸಿಕೊಳ್ಳಿ.
ಮುಖ್ಯ ವಿಷಯ
## ತರಗತಿ 4 ಗಣಿತ (ಭಾರತ)
### Unit 2: Lesson 3
1-ಅಂಕಿಯ ಸಂಖ್ಯೆಗಳಿಂದ ಗುಣಿಸಿ
# 3-ಅಂಕಿಗಳನ್ನು 1-ಅಂಕಿಯಿಂದ ಗುಣಿಸುವುದು
ಮತ್ತೆ ಗುಂಪು ಮಾಡದೆ 3-ಅಂಕಿಯ ಸಂಖ್ಯೆಯನ್ನು 1-ಅಂಕಿಯ ಸಂಖ್ಯೆಯಿಂದ ಗುಣಿಸಲು ಕಲಿಯಿರಿ. ಈ ವೀಡಿಯೊದಲ್ಲಿ, ನಾವು 4x201 ಅನ್ನು ಗುಣಿಸುತ್ತೇವೆ. ಸಾಲ್ ಖಾನ್ ರವರು ರಚಿಸಿದ್ದಾರೆ.
## ವೀಡಿಯೊ ಪ್ರತಿಲಿಪಿ
Let's multiply 4 times 2,012. Actually, let's make it a little bit simpler. Let's multiply 4 times 201 just to simplify things a little bit. So 4 times 201. So as we've seen in previous videos, I like to write the larger number on top. This is just one of many ways of tackling a calculation like this. I'll write the 201. And then I'll write the 4 right below it, and I'll write it right below the ones place. And so I have 201 times 4. Now, just like we did when we were multiplying a one digit times a two digit, we do essentially the same process. We first multiply 4 times the 1. Well, 4 times 1 we know is equal to 4. So we put a 4 right over there in the ones place. Then we can multiply our 4 times the digit that we have in the tens place. In this case, we have a 0 in the tens place. So 4 times 0, well, that's just 0. 4 times 0 is 0. We put the 0 in the tens place right over here. And then last, we have 4 times this 2 right over here. And so 4 times 2 is equal to 8. And we put the 8 right over here. And we get our answer-- 804. Now, why did this work? Well, remember, when we multiplied 4 times 1, that was literally just 4. And we've got that 4 right over here. When we multiply 4 times 0, that's 0 tens. So we've got 0 tens right over here. And when we multiplied 4 times 2, this was actually a 200. It's in the hundreds place. So 4 times 200 is 800. So what we're essentially doing by writing it in the right place is we're saying, 4 times 201, that's the same thing as 4 times 200, which is 800, plus 4 times 0 tens, which is 0 tens, plus 4 times 1, which is 4. So 800 plus 0 plus 4 is 804.
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# Union
In set theory, the union (∪) of a collection of sets is the set that contains all of the elements in the collection. For example, given two sets, A = {2, 2, 4, 6, 8, 10} and B = {1, 3, 5, 7, 9}, their union is as follows:
A ∪ B = {1, 2, 3, 4, 5, 6, 7, 8, 9 10}
Notice that even though A has two 2s, there is only one 2 in A ∪ B. This is because the union operation includes only one of each unique element.
The union of two sets is commonly depicted using a Venn diagram, in which a set is represented by a circle. The Venn diagram below shows two sets: A = {a, b, c, d, e} and B = {d, e, f, g}. Their union is the set {a, b, c, d, e, f, g} and is represented in the Venn diagram by the area covered by both circles.
## Properties of unions
Like other basic operations such as addition, set operations like unions also have certain properties. Refer to the set page if necessary for a table of symbols commonly used in set theory.
### Unions and subsets
If set A is a subset of set B, then the union of the two sets is set B. Using set notation:
if A ⊆ B, then A ∪ B = B
For example, if A = {2n|n ∈ ℕ} and B is the set of integers, then A ∪ B = B, since set A is the set of positive even integers, which is a subset of all integers.
### Commutative law
The commutative law states that the order in which the union of two sets is taken does not matter. Given two sets, A and B:
A ∪ B = B ∪ A
Let A = {1, 2, 3} and B = {3, 5, 7}. The sets share 1 common element. The union of the sets includes all unique elements of both sets. Thus, A ∪ B = B ∪ A = {1, 2, 3, 5, 7}. Regardless whether A or B is considered first, the result is the same. If B's elements were written first, the union of A and B could be written as {3, 5, 7, 1, 2}. The order in which elements are listed in a set does not matter; the number of elements and the values of the elements determine the set, so the above 2 sets are equal, as are any sets including all the same elements written in different orders.
### Associative law
The associative law states that rearranging the parentheses in a union of sets does not change the result. Given sets A, B, and C:
(A ∪ B) ∪ C = A ∪ (B ∪ C)
### Distributive law
For sets A, B, and C, the distributive law states
A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)
A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C)
where A is distributed to B and C. This is similar to the distributive property of multiplication in which multiplication distributes over addition.
Example
Let A = {4, 6, 8, 10}, B = {8, 9, 10, 11}, and C = {10, 11, 12}. Show that A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C).
Compute the left side of the equation:
B ∩ C = {10, 11}
A ∪ (B ∩ C) = {4, 6, 8, 10} ∪ {10, 11} = {4, 6, 8, 10, 11}
Compute the right side of the equation:
A ∪ B = {4, 6, 8, 9, 10, 11}
A ∪ C = {4, 6, 8, 10, 11, 12}
(A ∪ B) ∩ (A ∪ C) = {4, 6, 8, 10, 11}
In both cases, the resulting set is {4, 6, 8, 10, 11}.
### De Morgan's laws
In set theory, De Morgan's laws are a set of rules that relate the union and intersection of sets through their complements.
#### Union of sets:
The complement of the union of two sets is equal to the intersection of their complements:
(A ∪ B)C = AC ∩ BC
Given that A and B are subsets of the universal set 𝕌, this relationship can be seen in the figure below:
The union of A and B, A ∪ B, is shaded in blue. Its complement, (A ∪ B)C is shaded in yellow. The intersection of the complements of A and B, AC ∩ BC is also shaded in yellow.
#### Intersection of sets:
The complement of the intersection of two sets is equal to the union of their complements:
A ∩ B = AC ∪ BC
Given that A and B are subsets of the universal set 𝕌, this relationship can be seen in the figure below:
The intersection of A and B, A ∩ B, is shaded in red. Its complement, (A ∩ B)C is shaded in grey. The union of the complements of A and B, AC ∪ BC, is also shaded in grey.
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## What is a sinusoidal function?
A sinusoidal function is a function that is like a sine function in the sense that the function can be produced by shifting, stretching or compressing the sine function. If necessary you might like to review the graphing shortcuts.
## Exercise 4.4.1
Sketch the graph of $$y = 2 \sin x$$ from $$x = 0$$ to $$x = 2\pi$$.
See Solution
## What to notice about the graph
Notice that this graph is twice as tall as the sine graph. The amplitude of a sinusoidal function is the distance from the midline of the graph to the highest point of the graph. In exercise 4.4.1, the amplitude is (2\).
## Exercise 4.4.2
Sketch the graph of $$y = \sin ( x – \pi )$$ from $$x = 0$$ to $$x = 2\pi$$. You’ll have to shift the graph of $$y = \sin x$$ to the right by $$\pi$$ units. In this case, $$\pi$$ is called the phase shift of the function.
See Solution
## Exercise 4.4.3
Sketch the graph of $$y = \sin ( 2x )$$ from $$x = 0$$ to $$x = 2\pi$$. You’ll have to compress the graph of $$y = \sin x$$ by one-half horizontally. This changes the period from $$2\pi$$ to $$2\pi/2$$. So this sinusoidal function has a period of $$\pi$$. To draw the graph from $$x = 0$$ to $$x = 2\pi$$ you’ll have to draw two complete cycles of the graph.
See Solution
## Exercise 4.4.4
Sketch the graph of $$y = 1 + \sin x$$ from $$x = 0$$ to $$x = 2\pi$$.
See Solution
## The general sinusoidal function
As you see, $$y = 1 + \sin x$$ merely raises the graph of sine one unit. We say that this sinusoidal has a vertical shift of 1.
In general, all of these types of alterations may occur in a sinusoidal function. The general form of a sinusoidal is:
$f ( x ) = a \sin ( bx – c ) + d, \text{ for } b > 0$
The following formulas will be useful:
• Amplitude $$= | a |$$
• Period $$= 2\pi/ b$$
• Phase shift $$= c / b$$
• Vertical shift $$= d$$
Two complete cycles of the graph may be drawn between $$x = c / b – 2\pi / b$$ and $$x = c / b + 2\pi / b$$.
But what if b is not positive?
Suppose $$f ( x ) = 3 \sin ( -2 x +\pi / 2 ) + 3$$, for example. What do we do?
We replace $$\sin ( -2 x + π / 2 )$$ with $$-\sin ( 2 x - π / 2 )$$ ! We can do this because, remember, sine is an odd function. This means that $$\sin ( - x ) = - \sin ( x )$$ for all $$x$$ in the domain of the function.
Thus, $$f ( x ) = - 3 \sin ( 2 x - \pi / 2 ) + 3$$.
## Exercise 4.4.5
Sketch two complete cycles of the graph of $$f ( x ) = - 3 \sin ( 2 x - \pi / 2 ) + 3$$.
Specify the amplitude, vertical shift, period and phase shift.
See Solution
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# Multiplication Explained
Multiplication (often denoted by the cross symbol, by the mid-line dot operator, by juxtaposition, or, on computers, by an asterisk) is one of the four elementary mathematical operations of arithmetic, with the other ones being addition, subtraction, and division. The result of a multiplication operation is called a product.
The multiplication of whole numbers may be thought of as repeated addition; that is, the multiplication of two numbers is equivalent to adding as many copies of one of them, the multiplicand, as the quantity of the other one, the multiplier; both numbers can be referred to as factors.
a x b=\underbrace{b++b}a.
For example, 4 multiplied by 3, often written as
3 x 4
and spoken as "3 times 4", can be calculated by adding 3 copies of 4 together:
3 x 4=4+4+4=12.
Here, 3 (the multiplier) and 4 (the multiplicand) are the factors, and 12 is the product.
One of the main properties of multiplication is the commutative property, which states in this case that adding 3 copies of 4 gives the same result as adding 4 copies of 3:
4 x 3=3+3+3+3=12.
Thus, the designation of multiplier and multiplicand does not affect the result of the multiplication.[1]
Systematic generalizations of this basic definition define the multiplication of integers (including negative numbers), rational numbers (fractions), and real numbers.
Multiplication can also be visualized as counting objects arranged in a rectangle (for whole numbers) or as finding the area of a rectangle whose sides have some given lengths. The area of a rectangle does not depend on which side is measured first—a consequence of the commutative property.
The product of two measurements (or physical quantities) is a new type of measurement, usually with a derived unit. For example, multiplying the lengths (in meters or feet) of the two sides of a rectangle gives its area (in square meters or square feet). Such a product is the subject of dimensional analysis.
The inverse operation of multiplication is division. For example, since 4 multiplied by 3 equals 12, 12 divided by 3 equals 4. Indeed, multiplication by 3, followed by division by 3, yields the original number. The division of a number other than 0 by itself equals 1.
Several mathematical concepts expand upon the fundamental idea of multiplication. The product of a sequence, vector multiplication, complex numbers, and matrices are all examples where this can be seen. These more advanced constructs tend to affect the basic properties in their own ways, such as becoming noncommutative in matrices and some forms of vector multiplication or changing the sign of complex numbers.
## Notation
Multiplication signs Sign: × ⋅ Unicode: Different From:
See main article: Multiplication sign.
See also: Multiplier (linguistics). In arithmetic, multiplication is often written using the multiplication sign (either or) between the terms (that is, in infix notation). For example,
2 x 3=6,
("two times three equals six")
3 x 4=12,
2 x 3 x 5=6 x 5=30,
2 x 2 x 2 x 2 x 2=32.
There are other mathematical notations for multiplication:
• To reduce confusion between the multiplication sign × and the common variable, multiplication is also denoted by dot signs, usually a middle-position dot (rarely period):
52
.
The middle dot notation or dot operator, encoded in Unicode as, is now standard in the United States and other countries where the period is used as a decimal point. When the dot operator character is not accessible, the interpunct (·) is used. In other countries that use a comma as a decimal mark, either the period or a middle dot is used for multiplication.
Historically, in the United Kingdom and Ireland, the middle dot was sometimes used for the decimal to prevent it from disappearing in the ruled line, and the period/full stop was used for multiplication. However, since the Ministry of Technology ruled to use the period as the decimal point in 1968,[2] and the International System of Units (SI) standard has since been widely adopted, this usage is now found only in the more traditional journals such as The Lancet.[3]
• In algebra, multiplication involving variables is often written as a juxtaposition (e.g.,
xy
for
x
times
y
or
5x
for five times
x
), also called implied multiplication. The notation can also be used for quantities that are surrounded by parentheses (e.g.,
5(2)
,
(5)2
or
(5)(2)
for five times two). This implicit usage of multiplication can cause ambiguity when the concatenated variables happen to match the name of another variable, when a variable name in front of a parenthesis can be confused with a function name, or in the correct determination of the order of operations.
In computer programming, the asterisk (as in 5*2) is still the most common notation. This is due to the fact that most computers historically were limited to small character sets (such as ASCII and EBCDIC) that lacked a multiplication sign (such as ⋅ or ×), while the asterisk appeared on every keyboard. This usage originated in the FORTRAN programming language.[4]
The numbers to be multiplied are generally called the "factors" (as in factorization). The number to be multiplied is the "multiplicand", and the number by which it is multiplied is the "multiplier". Usually, the multiplier is placed first, and the multiplicand is placed second;[1] however, sometimes the first factor is the multiplicand and the second the multiplier.[5] Also, as the result of multiplication does not depend on the order of the factors, the distinction between "multiplicand" and "multiplier" is useful only at a very elementary level and in some multiplication algorithms, such as the long multiplication. Therefore, in some sources, the term "multiplicand" is regarded as a synonym for "factor".[6] In algebra, a number that is the multiplier of a variable or expression (e.g., the 3 in
3xy2
) is called a coefficient.
The result of a multiplication is called a product. When one factor is an integer, the product is a multiple of the other or of the product of the others. Thus,
2 x \pi
is a multiple of
\pi
, as is
5133 x 486 x \pi
. A product of integers is a multiple of each factor; for example, 15 is the product of 3 and 5 and is both a multiple of 3 and a multiple of 5.
## Definitions
The product of two numbers or the multiplication between two numbers can be defined for common special cases: natural numbers, integers, rational numbers, real numbers, complex numbers, and quaternions.
### Product of two natural numbers
The product of two natural numbers
r,s\inN
is defined as:
$r \cdot s \equiv \sum_^s r = \underbrace_ \equiv \sum_^r s = \underbrace_ .$
### Product of two integers
An integer can be either zero, a positive, or a negative number. The product of zero and another integer is always zero. The product of two nonzero integers is determined by the product of their positive amounts, combined with the sign derived from the following rule:
$\begin$
c c
\hline \times & + & - \\ \hline + & + & - \\ - & - & + \\ \hline\end(This rule is a consequence of the distributivity of multiplication over addition, and is not an additional rule.)
In words:
• A positive number multiplied by a positive number is positive (product of natural numbers),
• A positive number multiplied by a negative number is negative,
• A negative number multiplied by a positive number is negative,
• A negative number multiplied by a negative number is positive.
### Product of two fractions
Two fractions can be multiplied by multiplying their numerators and denominators:
$\frac \cdot \frac = \frac,$
which is defined when
n,n'0
.
### Product of two real numbers
There are several equivalent ways to define formally the real numbers; see Construction of the real numbers. The definition of multiplication is a part of all these definitions.
A fundamental aspect of these definitions is that every real number can be approximated to any accuracy by rational numbers. A standard way for expressing this is that every real number is the least upper bound of a set of rational numbers. In particular, every positive real number is the least upper bound of the truncations of its infinite decimal representation; for example,
\pi
is the least upper bound of
\{3,3.1,3.14,3,141,\ldots\}.
A fundamental property of real numbers is that rational approximations are compatible with arithmetic operations, and, in particular, with multiplication. This means that, if and are positive real numbers such that
a=\supx\inx
and
b=\supy\iny,
then
ab=\supx\inxy.
In particular, the product of two positive real numbers is the least upper bound of the term-by-term products of the sequences of their decimal representations.
As changing the signs transforms least upper bounds into greatest lower bounds, the simplest way to deal with a multiplication involving one or two negative numbers, is to use the rule of signs described above in . The construction of the real numbers through Cauchy sequences is often preferred in order to avoid consideration of the four possible sign configurations.
### Product of two complex numbers
Two complex numbers can be multiplied by the distributive law and the fact that
i2=-1
, as follows:
Geometric meaning of complex multiplication can be understood rewriting complex numbers in polar coordinates:
a+bi=r(\cos(\varphi)+i\sin(\varphi))=re
Furthermore,
c+di=s(\cos(\psi)+i\sin(\psi))=sei\psi,
from which one obtains
The geometric meaning is that the magnitudes are multiplied and the arguments are added.
### Product of two quaternions
The product of two quaternions can be found in the article on quaternions. Note, in this case, that $a \cdot b$ and
[7] The algorithm, also based on the fast Fourier transform, is conjectured to be asymptotically optimal.[8] The algorithm is not practically useful, as it only becomes faster for multiplying extremely large numbers (having more than bits).[9]
## Products of measurements
See main article: Dimensional analysis. One can only meaningfully add or subtract quantities of the same type, but quantities of different types can be multiplied or divided without problems. For example, four bags with three marbles each can be thought of as:[1]
[4 bags] × [3 marbles per bag] = 12 marbles.
When two measurements are multiplied together, the product is of a type depending on the types of measurements. The general theory is given by dimensional analysis. This analysis is routinely applied in physics, but it also has applications in finance and other applied fields.
A common example in physics is the fact that multiplying speed by time gives distance. For example:
50 kilometers per hour × 3 hours = 150 kilometers.In this case, the hour units cancel out, leaving the product with only kilometer units.
Other examples of multiplication involving units include:
2.5 meters × 4.5 meters = 11.25 square meters
11 meters/seconds × 9 seconds = 99 meters
4.5 residents per house × 20 houses = 90 residents
## Product of a sequence
### Capital pi notation
The product of a sequence of factors can be written with the product symbol
style\prod
, which derives from the capital letter Π (pi) in the Greek alphabet (much like the same way the summation symbol
style\sum
is derived from the Greek letter Σ (sigma)).[10] [11] The meaning of this notation is given by
4 \prod i=1
(i+1)=(1+1)(2+1)(3+1)(4+1),
which results in
4 \prod i=1
(i+1)=120.
In such a notation, the variable represents a varying integer, called the multiplication index, that runs from the lower value indicated in the subscript to the upper value given by the superscript. The product is obtained by multiplying together all factors obtained by substituting the multiplication index for an integer between the lower and the upper values (the bounds included) in the expression that follows the product operator.
More generally, the notation is defined as
n \prod i=m
xi=xmxm+1xm+2xn-1xn,
where m and n are integers or expressions that evaluate to integers. In the case where, the value of the product is the same as that of the single factor xm; if, the product is an empty product whose value is 1—regardless of the expression for the factors.
#### Properties of capital pi notation
By definition,
n \prod i=1
xi=x1 ⋅ x2 ⋅ \ldotsxn.
If all factors are identical, a product of factors is equivalent to exponentiation:
n \prod i=1
x=xx\ldotsx=xn.
Associativity and commutativity of multiplication imply
n \prod i=1
{xiyi}
n =\left(\prod i=1
xi\right)\left(\prod
n i=1
yi\right)
and
n \left(\prod i=1
a x i\right)
n =\prod i=1
a x i
if is a non-negative integer, or if all
xi
are positive real numbers, and
n \prod i=1
ai x
n \sum ai i=1
=x
if all
ai
are non-negative integers, or if is a positive real number.
### Infinite products
See main article: Infinite product. One may also consider products of infinitely many terms; these are called infinite products. Notationally, this consists in replacing n above by the infinity symbol ∞. The product of such an infinite sequence is defined as the limit of the product of the first n terms, as n grows without bound. That is,
infty \prod i=m
xi=\limn\toinfty
n \prod i=m
xi.
One can similarly replace m with negative infinity, and define:
infty \prod i=-infty
xi=\left(\limm\to-infty
0 \prod i=m
xi\right)\left(\limn\toinfty
n \prod i=1
xi\right),
provided both limits exist.
## Exponentiation
See main article: Exponentiation. When multiplication is repeated, the resulting operation is known as exponentiation. For instance, the product of three factors of two (2×2×2) is "two raised to the third power", and is denoted by 23, a two with a superscript three. In this example, the number two is the base, and three is the exponent.[12] In general, the exponent (or superscript) indicates how many times the base appears in the expression, so that the expression
an=\underbrace{a x a x x a}n
indicates that n copies of the base a are to be multiplied together. This notation can be used whenever multiplication is known to be power associative.
## Properties
For real and complex numbers, which includes, for example, natural numbers, integers, and fractions, multiplication has certain properties:
Commutative property
• The order in which two numbers are multiplied does not matter:[13] [14]
xy=yx.
Associative property
• Expressions solely involving multiplication or addition are invariant with respect to the order of operations:[13] [14]
(xy)z=x(yz).
Distributive property
• Holds with respect to multiplication over addition. This identity is of prime importance in simplifying algebraic expressions:[13] [14]
x(y+z)=xy+xz.
Identity element
• The multiplicative identity is 1; anything multiplied by 1 is itself. This feature of 1 is known as the identity property:[13] [14]
x1=x.
Property of 0
• Any number multiplied by 0 is 0. This is known as the zero property of multiplication:[13]
x0=0.
Negation
• −1 times any number is equal to the additive inverse of that number:
(-1)x=(-x)
, where
(-x)+x=0.
−1 times −1 is 1:
(-1)(-1)=1.
Inverse element
• Every number x, except 0, has a multiplicative inverse,
1 x
, such that
x\left(
1 x
\right)=1
.[15]
Order preservation
• Multiplication by a positive number preserves the order:
• For, if then .
• Multiplication by a negative number reverses the order:
• For, if then .
• The complex numbers do not have an ordering that is compatible with both addition and multiplication.[16]
Other mathematical systems that include a multiplication operation may not have all these properties. For example, multiplication is not, in general, commutative for matrices and quaternions.[13]
## Axioms
See main article: Peano axioms. In the book Arithmetices principia, nova methodo exposita, Giuseppe Peano proposed axioms for arithmetic based on his axioms for natural numbers. Peano arithmetic has two axioms for multiplication:
x x 0=0
x x S(y)=(x x y)+x
Here S(y) represents the successor of y; i.e., the natural number that follows y. The various properties like associativity can be proved from these and the other axioms of Peano arithmetic, including induction. For instance, S(0), denoted by 1, is a multiplicative identity because
x x 1=x x S(0)=(x x 0)+x=0+x=x.
The axioms for integers typically define them as equivalence classes of ordered pairs of natural numbers. The model is based on treating (x,y) as equivalent to when x and y are treated as integers. Thus both (0,1) and (1,2) are equivalent to −1. The multiplication axiom for integers defined this way is
(xp,xm) x (yp,ym)=(xp x yp+xm x ym,xp x ym+xm x yp).
The rule that −1 × −1 = 1 can then be deduced from
(0,1) x (0,1)=(0 x 0+1 x 1,0 x 1+1 x 0)=(1,0).
Multiplication is extended in a similar way to rational numbers and then to real numbers.
## Multiplication with set theory
The product of non-negative integers can be defined with set theory using cardinal numbers or the Peano axioms. See below how to extend this to multiplying arbitrary integers, and then arbitrary rational numbers. The product of real numbers is defined in terms of products of rational numbers; see construction of the real numbers.[17]
## Multiplication in group theory
There are many sets that, under the operation of multiplication, satisfy the axioms that define group structure. These axioms are closure, associativity, and the inclusion of an identity element and inverses.
A simple example is the set of non-zero rational numbers. Here identity 1 is had, as opposed to groups under addition where the identity is typically 0. Note that with the rationals, zero must be excluded because, under multiplication, it does not have an inverse: there is no rational number that can be multiplied by zero to result in 1. In this example, an abelian group is had, but that is not always the case.
To see this, consider the set of invertible square matrices of a given dimension over a given field. Here, it is straightforward to verify closure, associativity, and inclusion of identity (the identity matrix) and inverses. However, matrix multiplication is not commutative, which shows that this group is non-abelian.
Another fact worth noticing is that the integers under multiplication do not form a group—even if zero is excluded. This is easily seen by the nonexistence of an inverse for all elements other than 1 and −1.
Multiplication in group theory is typically notated either by a dot or by juxtaposition (the omission of an operation symbol between elements). So multiplying element a by element b could be notated as a
b or ab. When referring to a group via the indication of the set and operation, the dot is used. For example, our first example could be indicated by
\left(Q/\{0\},\right)
.[18]
## Multiplication of different kinds of numbers
Numbers can count (3 apples), order (the 3rd apple), or measure (3.5 feet high); as the history of mathematics has progressed from counting on our fingers to modelling quantum mechanics, multiplication has been generalized to more complicated and abstract types of numbers, and to things that are not numbers (such as matrices) or do not look much like numbers (such as quaternions).
Integers
N x M
is the sum of N copies of M when N and M are positive whole numbers. This gives the number of things in an array N wide and M high. Generalization to negative numbers can be done by
N x (-M)=(-N) x M=-(N x M)
and
(-N) x (-M)=N x M
The same sign rules apply to rational and real numbers.
Rational numbers
• Generalization to fractions
A B
x
C D
is by multiplying the numerators and denominators, respectively:
A B
x
C D
=
(A x C) (B x D)
. This gives the area of a rectangle
A B
high and
C D
wide, and is the same as the number of things in an array when the rational numbers happen to be whole numbers.[13]
Real numbers
• Real numbers and their products can be defined in terms of sequences of rational numbers.
Complex numbers
• Considering complex numbers
z1
and
z2
as ordered pairs of real numbers
(a1,b1)
and
(a2,b2)
, the product
z1 x z2
is
(a1 x a2-b1 x b2,a1 x b2+a2 x b1)
. This is the same as for reals
a1 x a2
when the imaginary parts
b1
and
b2
are zero.
Equivalently, denoting
\sqrt{-1}
as
i
,
z1 x z2=(a1+b1i)(a2+b2i)=(a1 x a2)+(a1 x b2i)+(b1 x a2i)+(b1 x
2)=(a b 1a
2-b1b2)+(a1b2+b1a2)i.
[13]
Alternatively, in trigonometric form, if
z1=r1(\cos\phi1+i\sin\phi1),z2=r2(\cos\phi2+i\sin\phi2)
, then$z_1z_2 = r_1r_2(\cos(\phi_1 + \phi_2) + i\sin(\phi_1 + \phi_2)).$[13]
Further generalizations
• See Multiplication in group theory, above, and multiplicative group, which for example includes matrix multiplication. A very general, and abstract, concept of multiplication is as the "multiplicatively denoted" (second) binary operation in a ring. An example of a ring that is not any of the above number systems is a polynomial ring (polynomials can be added and multiplied, but polynomials are not numbers in any usual sense).
Division
• Often division,
x y
, is the same as multiplication by an inverse,
x\left( 1 y
\right)
. Multiplication for some types of "numbers" may have corresponding division, without inverses; in an integral domain x may have no inverse "
1 x
" but
x y
may be defined. In a division ring there are inverses, but
x y
may be ambiguous in non-commutative rings since
x\left( 1 y
\right)
need not be the same as
\left( 1 y
\right)x
.
## Notes and References
1. Web site: Devlin . Keith . What Exactly is Multiplication? . Keith Devlin . . January 2011 . With multiplication you have a multiplicand (written second) multiplied by a multiplier (written first) . May 14, 2017 . https://web.archive.org/web/20170527070801/http://www.maa.org/external_archive/devlin/devlin_01_11.html . May 27, 2017 . live .
2. 10.1038/218111c0 . Victory on Points . Nature . 218 . 5137 . 111 . 1968 . 1968Natur.218S.111. . free.
3. Web site: The Lancet – Formatting guidelines for electronic submission of manuscripts . 2017-04-25.
4. Book: Fuller, William R. . FORTRAN Programming: A Supplement for Calculus Courses. Universitext. 1977. 10. Springer. 10.1007/978-1-4612-9938-7. 978-0-387-90283-8.
5. Web site: Ramone . Crewton . Multiplicand and Multiplier. 10 November 2015. Crewton Ramone's House of Math. https://web.archive.org/web/20151026161239/http://www.crewtonramoneshouseofmath.com/multiplicand-and-multiplier.html . 26 October 2015 . live. .
6. Book: Litvin, Chester . Advance Brain Stimulation by Psychoconduction . 2012 . Trafford . 978-1-4669-0152-0 . 2–3, 5–6 . Google Book Search.
7. David Harvey, Joris Van Der Hoeven (2019). Integer multiplication in time O(n log n)
8. Web site: Mathematicians Discover the Perfect Way to Multiply. Hartnett. Kevin. Quanta Magazine. 11 April 2019. en. 2020-01-25.
9. Web site: Multiplication Hits the Speed Limit. Klarreich. Erica. cacm.acm.org. en. 2020-01-25. https://archive.today/20201031123457/https://cacm.acm.org/magazines/2020/1/241707-multiplication-hits-the-speed-limit/fulltext. 2020-10-31. live.
10. Web site: Weisstein. Eric W.. Product. 2020-08-16. mathworld.wolfram.com. en.
11. Web site: Summation and Product Notation. 2020-08-16. math.illinoisstate.edu.
12. Web site: Weisstein. Eric W.. Exponentiation. 2021-12-29. mathworld.wolfram.com. en.
13. Web site: Multiplication . Encyclopedia of Mathematics . 2021-12-29.
14. Book: Biggs, Norman L. . Discrete Mathematics . Oxford University Press . 2002 . 978-0-19-871369-2 . 25 . en.
15. Web site: Weisstein . Eric W. . Multiplicative Inverse . 2022-04-19 . Wolfram MathWorld . en.
16. Web site: Angell . David . ORDERING COMPLEX NUMBERS... NOT* . 29 December 2021 . UNSW Sydney, School of Mathematics and Statistics.
17. Web site: 2018-04-11 . 10.2: Building the Real Numbers . 2023-06-23 . Mathematics LibreTexts . en.
18. Book: Burns . Gerald . Introduction to group theory with applications . 1977 . Academic Press . New York . 9780121457501 .
|
Giáo trình
# Introductory Statistics
Mathematics and Statistics
## Geometric Distribution
Tác giả: OpenStaxCollege
There are three main characteristics of a geometric experiment.
1. There are one or more Bernoulli trials with all failures except the last one, which is a success. In other words, you keep repeating what you are doing until the first success. Then you stop. For example, you throw a dart at a bullseye until you hit the bullseye. The first time you hit the bullseye is a "success" so you stop throwing the dart. It might take six tries until you hit the bullseye. You can think of the trials as failure, failure, failure, failure, failure, success, STOP.
2. In theory, the number of trials could go on forever. There must be at least one trial.
3. The probability, p, of a success and the probability, q, of a failure is the same for each trial. p + q = 1 and q = 1 − p. For example, the probability of rolling a three when you throw one fair die is $\frac{1}{6}$. This is true no matter how many times you roll the die. Suppose you want to know the probability of getting the first three on the fifth roll. On rolls one through four, you do not get a face with a three. The probability for each of the rolls is q = $\frac{\text{5}}{\text{6}}$, the probability of a failure. The probability of getting a three on the fifth roll is $\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)$ = 0.0804
X = the number of independent trials until the first success.
You play a game of chance that you can either win or lose (there are no other possibilities) until you lose. Your probability of losing is p = 0.57. What is the probability that it takes five games until you lose? Let X = the number of games you play until you lose (includes the losing game). Then X takes on the values 1, 2, 3, ... (could go on indefinitely). The probability question is P(x = 5).
A safety engineer feels that 35% of all industrial accidents in her plant are caused by failure of employees to follow instructions. She decides to look at the accident reports (selected randomly and replaced in the pile after reading) until she finds one that shows an accident caused by failure of employees to follow instructions. On average, how many reports would the safety engineer expect to look at until she finds a report showing an accident caused by employee failure to follow instructions? What is the probability that the safety engineer will have to examine at least three reports until she finds a report showing an accident caused by employee failure to follow instructions?
Let X = the number of accidents the safety engineer must examine until she finds a report showing an accident caused by employee failure to follow instructions. X takes on the values 1, 2, 3, .... The first question asks you to find the expected value or the mean. The second question asks you to find P(x ≥ 3). ("At least" translates to a "greater than or equal to" symbol).
Suppose that you are looking for a student at your college who lives within five miles of you. You know that 55% of the 25,000 students do live within five miles of you. You randomly contact students from the college until one says he or she lives within five miles of you. What is the probability that you need to contact four people?
This is a geometric problem because you may have a number of failures before you have the one success you desire. Also, the probability of a success stays the same each time you ask a student if he or she lives within five miles of you. There is no definite number of trials (number of times you ask a student).
a. Let X = the number of ____________ you must ask ____________ one says yes.
a. Let X = the number of students you must ask until one says yes.
b. What values does X take on?
b. 1, 2, 3, …, (total number of students)
c. What are p and q?
c. p = 0.55; q = 0.45
d. The probability question is P(_______).
d. P(x = 4)
# Notation for the Geometric: G = Geometric Probability Distribution Function
X ~ G(p)
Read this as "X is a random variable with a geometric distribution." The parameter is p; p = the probability of a success for each trial.
Assume that the probability of a defective computer component is 0.02. Components are randomly selected. Find the probability that the first defect is caused by the seventh component tested. How many components do you expect to test until one is found to be defective?
Let X = the number of computer components tested until the first defect is found.
X takes on the values 1, 2, 3, ... where p = 0.02. X ~ G(0.02)
Find P(x = 7). P(x = 7) = 0.0177.
To find the probability that x ≤ 7, follow the same instructions EXCEPT select E:geometcdf(as the distribution function.
The probability that the seventh component is the first defect is 0.0177.
The graph of X ~ G(0.02) is:
The y-axis contains the probability of x, where X = the number of computer components tested.
The number of components that you would expect to test until you find the first defective one is the mean, .
The formula for the mean is μ = $\frac{1}{p}$ = $\frac{1}{0.02}$ = 50
The formula for the variance is σ2 = $\left(\frac{1}{p}\right)\left(\frac{1}{p}-1\right)$ = $\left(\frac{1}{0.02}\right)\left(\frac{1}{0.02}-1\right)$ = 2,450
The standard deviation is σ = $\sqrt{\left(\frac{1}{p}\right)\left(\frac{1}{p}-1\right)}$ = $\sqrt{\left(\frac{1}{0.\text{02}}\right)\left(\frac{1}{0.\text{02}}-1\right)}$ = 49.5
The lifetime risk of developing pancreatic cancer is about one in 78 (1.28%). Let X = the number of people you ask until one says he or she has pancreatic cancer. Then X is a discrete random variable with a geometric distribution: X ~ G$\left(\frac{1}{78}\right)$ or X ~ G(0.0128).
1. What is the probability of that you ask ten people before one says he or she has pancreatic cancer?
2. What is the probability that you must ask 20 people?
3. Find the (i) mean and (ii) standard deviation of X.
1. P(x = 10) = geometpdf(0.0128, 10) = 0.0114
2. P(x = 20) = geometpdf(0.0128, 20) = 0.01
1. Mean = μ = $\frac{1}{p}$ = $\frac{1}{0.0128}$ = 78
2. Standard Deviation = σ = $\sqrt{\frac{1-p}{{p}^{2}}}$ = $\sqrt{\frac{1-0.0128}{{0.0128}^{2}}}$ ≈ 77.6234
# References
“Millennials: A Portrait of Generation Next,” PewResearchCenter. Available online at http://www.pewsocialtrends.org/files/2010/10/millennials-confident-connected-open-to-change.pdf (accessed May 15, 2013).
“Millennials: Confident. Connected. Open to Change.” Executive Summary by PewResearch Social & Demographic Trends, 2013. Available online at http://www.pewsocialtrends.org/2010/02/24/millennials-confident-connected-open-to-change/ (accessed May 15, 2013).
“Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013).
Pryor, John H., Linda DeAngelo, Laura Palucki Blake, Sylvia Hurtado, Serge Tran. The American Freshman: National Norms Fall 2011. Los Angeles: Cooperative Institutional Research Program at the Higher Education Research Institute at UCLA, 2011. Also available online at http://heri.ucla.edu/PDFs/pubs/TFS/Norms/Monographs/TheAmericanFreshman2011.pdf (accessed May 15, 2013).
“Summary of the National Risk and Vulnerability Assessment 2007/8: A profile of Afghanistan,” The European Union and ICON-Institute. Available online at http://ec.europa.eu/europeaid/where/asia/documents/afgh_brochure_summary_en.pdf (accessed May 15, 2013).
“The World FactBook,” Central Intelligence Agency. Available online at https://www.cia.gov/library/publications/the-world-factbook/geos/af.html (accessed May 15, 2013).
“UNICEF reports on Female Literacy Centers in Afghanistan established to teach women and girls basic resading [sic] and writing skills,” UNICEF Television. Video available online at http://www.unicefusa.org/assets/video/afghan-female-literacy-centers.html (accessed May 15, 2013).
# Chapter Review
There are three characteristics of a geometric experiment:
1. There are one or more Bernoulli trials with all failures except the last one, which is a success.
2. In theory, the number of trials could go on forever. There must be at least one trial.
3. The probability, p, of a success and the probability, q, of a failure are the same for each trial.
In a geometric experiment, define the discrete random variable X as the number of independent trials until the first success. We say that X has a geometric distribution and write X ~ G(p) where p is the probability of success in a single trial.
The mean of the geometric distribution X ~ G(p) is μ = $\sqrt{\frac{\text{1}-p}{{p}^{2}}}$ = $\sqrt{\frac{1}{p}\left(\frac{1}{p}-1\right)}$.
# Formula Review
X ~ G(p) means that the discrete random variable X has a geometric probability distribution with probability of success in a single trial p.
X = the number of independent trials until the first success
X takes on the values x = 1, 2, 3, ...
p = the probability of a success for any trial
q = the probability of a failure for any trial p + q = 1
q = 1 – p
The mean is μ = $\frac{1}{p}$.
The standard deviation is σ = = $\sqrt{\frac{1}{p}\left(\frac{1}{p}-1\right)}$ .
Use the following information to answer the next six exercises: The Higher Education Research Institute at UCLA collected data from 203,967 incoming first-time, full-time freshmen from 270 four-year colleges and universities in the U.S. 71.3% of those students replied that, yes, they believe that same-sex couples should have the right to legal marital status. Suppose that you randomly select freshman from the study until you find one who replies “yes.” You are interested in the number of freshmen you must ask.
In words, define the random variable X.
X = the number of freshmen selected from the study until one replied "yes" that same-sex couples should have the right to legal marital status.
X ~ _____(_____,_____)
What values does the random variable X take on?
1,2,…
Construct the probability distribution function (PDF). Stop at x = 6.
x P(x) 1 2 3 4 5 6
On average (μ), how many freshmen would you expect to have to ask until you found one who replies "yes?"
1.4
What is the probability that you will need to ask fewer than three freshmen?
# HOMEWORK
A consumer looking to buy a used red Miata car will call dealerships until she finds a dealership that carries the car. She estimates the probability that any independent dealership will have the car will be 28%. We are interested in the number of dealerships she must call.
1. In words, define the random variable X.
2. List the values that X may take on.
3. Give the distribution of X. X ~ _____(_____,_____)
4. On average, how many dealerships would we expect her to have to call until she finds one that has the car?
5. Find the probability that she must call at most four dealerships.
6. Find the probability that she must call three or four dealerships.
Suppose that the probability that an adult in America will watch the Super Bowl is 40%. Each person is considered independent. We are interested in the number of adults in America we must survey until we find one who will watch the Super Bowl.
1. In words, define the random variable X.
2. List the values that X may take on.
3. Give the distribution of X. X ~ _____(_____,_____)
4. How many adults in America do you expect to survey until you find one who will watch the Super Bowl?
5. Find the probability that you must ask seven people.
6. Find the probability that you must ask three or four people.
1. X = the number of adults in America who are surveyed until one says he or she will watch the Super Bowl.
2. X ~ G(0.40)
3. 2.5
4. 0.0187
5. 0.2304
It has been estimated that only about 30% of California residents have adequate earthquake supplies. Suppose we are interested in the number of California residents we must survey until we find a resident who does not have adequate earthquake supplies.
1. In words, define the random variable X.
2. List the values that X may take on.
3. Give the distribution of X. X ~ _____(_____,_____)
4. What is the probability that we must survey just one or two residents until we find a California resident who does not have adequate earthquake supplies?
5. What is the probability that we must survey at least three California residents until we find a California resident who does not have adequate earthquake supplies?
6. How many California residents do you expect to need to survey until you find a California resident who does not have adequate earthquake supplies?
7. How many California residents do you expect to need to survey until you find a California resident who does have adequate earthquake supplies?
In one of its Spring catalogs, L.L. Bean® advertised footwear on 29 of its 192 catalog pages. Suppose we randomly survey 20 pages. We are interested in the number of pages that advertise footwear. Each page may be picked more than once.
1. In words, define the random variable X.
2. List the values that X may take on.
3. Give the distribution of X. X ~ _____(_____,_____)
4. How many pages do you expect to advertise footwear on them?
5. Is it probable that all twenty will advertise footwear on them? Why or why not?
6. What is the probability that fewer than ten will advertise footwear on them?
7. Reminder: A page may be picked more than once. We are interested in the number of pages that we must randomly survey until we find one that has footwear advertised on it. Define the random variable X and give its distribution.
8. What is the probability that you only need to survey at most three pages in order to find one that advertises footwear on it?
9. How many pages do you expect to need to survey in order to find one that advertises footwear?
1. X = the number of pages that advertise footwear
2. X takes on the values 0, 1, 2, ..., 20
3. X ~ B(20, $\frac{29}{192}$)
4. 3.02
5. No
6. 0.9997
7. X = the number of pages we must survey until we find one that advertises footwear. X ~ G($\frac{29}{192}$)
8. 0.3881
9. 6.6207 pages
Suppose that you are performing the probability experiment of rolling one fair six-sided die. Let F be the event of rolling a four or a five. You are interested in how many times you need to roll the die in order to obtain the first four or five as the outcome.
• p = probability of success (event F occurs)
• q = probability of failure (event F does not occur)
1. Write the description of the random variable X.
2. What are the values that X can take on?
3. Find the values of p and q.
4. Find the probability that the first occurrence of event F (rolling a four or five) is on the second trial.
Ellen has music practice three days a week. She practices for all of the three days 85% of the time, two days 8% of the time, one day 4% of the time, and no days 3% of the time. One week is selected at random. What values does X take on?
0, 1, 2, and 3
The World Bank records the prevalence of HIV in countries around the world. According to their data, “Prevalence of HIV refers to the percentage of people ages 15 to 49 who are infected with HIV.”
”Prevalence of HIV, total (% of populations ages 15-49),” The World Bank, 2013. Available online at http://data.worldbank.org/indicator/SH.DYN.AIDS.ZS?order=wbapi_data_value_2011+wbapi_data_value+wbapi_data_value-last&sort=desc (accessed May 15, 2013).
In South Africa, the prevalence of HIV is 17.3%. Let X = the number of people you test until you find a person infected with HIV.
1. Sketch a graph of the distribution of the discrete random variable X.
2. What is the probability that you must test 30 people to find one with HIV?
3. What is the probability that you must ask ten people?
4. Find the (i) mean and (ii) standard deviation of the distribution of X.
According to a recent Pew Research poll, 75% of millenials (people born between 1981 and 1995) have a profile on a social networking site. Let X = the number of millenials you ask until you find a person without a profile on a social networking site.
1. Describe the distribution of X.
2. Find the (i) mean and (ii) standard deviation of X.
3. What is the probability that you must ask ten people to find one person without a social networking site?
4. What is the probability that you must ask 20 people to find one person without a social networking site?
5. What is the probability that you must ask at most five people?
1. X ~ G(0.25)
1. Mean = μ = $\frac{1}{p}$ = $\frac{1}{0.25}$ = 4
2. Standard Deviation = σ = $\sqrt{\frac{1-p}{{p}^{2}}}$ = $\sqrt{\frac{1-\text{0}\text{.25}}{{0.25}^{2}}}$ ≈ 3.4641
2. P(x = 10) = geometpdf(0.25, 10) = 0.0188
3. P(x = 20) = geometpdf(0.25, 20) = 0.0011
4. P(x ≤ 5) = geometcdf(0.25, 5) = 0.7627
|
# Properties of Absolute Values on the GMAT – Part II
We pick up this post from where we left the post of last week in which we looked at a few properties of absolute values in two variables. There is one more property that we would like to talk about today. Thereafter, we will look at a question based on some of these properties.
(III) |x – y| = 0 implies x = y
x and y could be positive/negative integer/fraction; if the absolute value of their difference is 0, it means x = y. They cannot have opposite signs while having the same absolute value. They must be equal. This also means that if and only if x = y, the absolute value of their difference will be 0.
Mind you, this is different from ‘difference of their absolute values’
|x| – |y| = 0 implies that the absolute value of x is equal to the absolute value of y. So x and y could be equal or they could have opposite signs while having the same absolute value.
Let’s now take up the question we were talking about.
Question: Is |x + y| < |x| + |y|?
Statement 1: | x | is not equal to | y |
Statement 2: | x – y | > | x + y |
Solution: One of the properties we discussed last week was
“For all real x and y, |x + y| <= |x| + |y|
|x + y| = |x| + |y| when (1) x and y have the same sign (2) at least one of x and y is 0.
|x + y| < |x| + |y| when (1) x and y have opposite signs”
We discussed in detail the reason absolute values behave this way.
So our question “Is |x + y| < |x| + |y|?” now becomes:
Question: Do x and y have opposite signs?
We do not care which one is greater – the one with the positive sign or the one with the negative sign. All we want to know is whether they have opposite signs (opposite sign also implies that neither one of x and y can be 0)? If we can answer this question definitively with a ‘Yes’ or a ‘No’, the statement will be sufficient to answer the question. Let’s go on to the statements now.
Statement 1: | x | is not equal to | y |
This statement tells us that absolute value of x is not equal to absolute value of y. It doesn’t tell us anything about the signs of x and y and whether they are same or opposite. So this statement alone is not sufficient.
Statement 2:| x – y | > | x + y |
Let’s think along the same lines as last week – when will | x – y | be greater than | x + y |? When will the absolute value of subtraction of two numbers be greater than the absolute value of their addition? This will happen only when x and y have opposite signs. In that case, while subtracting, we would actually be adding the absolute values of the two and while adding, we would actually be subtracting the absolute values of the two. That is when the absolute value of the subtraction will be more than the absolute value of the addition.
For Example: x = 3, y = -2
| x – y | = |3 – (-2)| = 5
| x + y | = |3 – 2| = 1
or
x = -3, y = 2
| x – y | = |-3 – 2| = 5
| x + y | = |-3 + 2| = 1
If instead, x and y have the same sign, | x + y | will be greater than| x – y |.
If at least one of x and y is 0, | x + y | will be equal to| x – y |.
Since this statement tells us that | x – y | > | x + y |, it implies that x and y have opposite signs. So this statement alone is sufficient to answer the question with a ‘Yes’.
Takeaway from this question:
If x and y have the same signs, | x + y | >| x – y |.
If x and y have opposite signs, | x + y | <| x – y |.
If at least one of x and y is 0, | x + y | =| x – y |.
You don’t need to ‘learn this up’. Understand the logic here. You can easily recreate it in the exam if need be.
Karishma, a Computer Engineer with a keen interest in alternative Mathematical approaches, has mentored students in the continents of Asia, Europe and North America. She teaches the GMAT for Veritas Prep and regularly participates in content development projects such as this blog!
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# Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+4x^3-14x^2-36x+45
If a polynomial function has integer coefficients, then every rational zero will have the form where is a factor of the constant and is a factor of the leading coefficient.
Find every combination of . These are the possible roots of the polynomial function.
Substitute the possible roots one by one into the polynomial to find the actual roots. Simplify to check if the value is , which means it is a root.
Simplify the expression. In this case, the expression is equal to so is a root of the polynomial.
Simplify each term.
One to any power is one.
One to any power is one.
Multiply by .
One to any power is one.
Multiply by .
Multiply by .
Simplify by adding and subtracting.
Subtract from .
Subtract from .
Since is a known root, divide the polynomial by to find the quotient polynomial. This polynomial can then be used to find the remaining roots.
Next, find the roots of the remaining polynomial. The order of the polynomial has been reduced by .
Place the numbers representing the divisor and the dividend into a division-like configuration.
The first number in the dividend is put into the first position of the result area (below the horizontal line).
Multiply the newest entry in the result by the divisor and place the result of under the next term in the dividend .
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
Multiply the newest entry in the result by the divisor and place the result of under the next term in the dividend .
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
Multiply the newest entry in the result by the divisor and place the result of under the next term in the dividend .
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
Multiply the newest entry in the result by the divisor and place the result of under the next term in the dividend .
Add the product of the multiplication and the number from the dividend and put the result in the next position on the result line.
All numbers except the last become the coefficients of the quotient polynomial. The last value in the result line is the remainder.
Simplify the quotient polynomial.
Factor out the greatest common factor from each group.
Group the first two terms and the last two terms.
Factor out the greatest common factor (GCF) from each group.
Factor the polynomial by factoring out the greatest common factor, .
Rewrite as .
Factor.
Since both terms are perfect squares, factor using the difference of squares formula, where and .
Remove unnecessary parentheses.
Factor the left side of the equation.
Regroup terms.
Factor out of .
Factor out of .
Factor out of .
Factor out of .
Rewrite as .
Factor.
Since both terms are perfect squares, factor using the difference of squares formula, where and .
Remove unnecessary parentheses.
Rewrite as .
Let . Substitute for all occurrences of .
Factor using the AC method.
Consider the form . Find a pair of integers whose product is and whose sum is . In this case, whose product is and whose sum is .
Write the factored form using these integers.
Replace all occurrences of with .
Rewrite as .
Since both terms are perfect squares, factor using the difference of squares formula, where and .
Factor out of .
Factor out of .
Factor out of .
Let . Substitute for all occurrences of .
Factor using the AC method.
Consider the form . Find a pair of integers whose product is and whose sum is . In this case, whose product is and whose sum is .
Write the factored form using these integers.
Factor.
Replace all occurrences of with .
Remove unnecessary parentheses.
If any individual factor on the left side of the equation is equal to , the entire expression will be equal to .
Set the first factor equal to and solve.
Set the first factor equal to .
Subtract from both sides of the equation.
Set the next factor equal to and solve.
Set the next factor equal to .
Add to both sides of the equation.
Set the next factor equal to and solve.
Set the next factor equal to .
Add to both sides of the equation.
Set the next factor equal to and solve.
Set the next factor equal to .
Subtract from both sides of the equation.
The final solution is all the values that make true.
Find the Roots/Zeros Using the Rational Roots Test f(x)=x^4+4x^3-14x^2-36x+45
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# How to Calculate Centripetal Force
Before learning how to calculate centripetal force, let us see what is centripetal force and how it is derived. An object moving in a circular path is accelerating even if it is maintaining a constant speed. The acceleration experienced by such an object is called the centripetal acceleration, and it always points towards the centre of the circular path. According to Newton’s second law, there has to be a centripetal force pointing at the centre of the circular path, which is responsible for the circular motion. In this article, we look at several examples of how to calculate centripetal force.
## How to Find Centripetal Force
Deriving centripetal force is quite straightforward once you are familiar with the concepts of centripetal acceleration and Newton’s second law.
The centripetal acceleration on a body traveling at a constant speed $v$ in a circular path with a radius $r$ is given by
$a=\frac{v^2}{r}$
If the angular speed of the body is $\omega$, then the centripetal acceleration could be written as
$a={\omega^2}r$
Now, to go from centripetal force to centripetal acceleration, we simply make use of Newton’s second law of motion$F=ma$. Then, centripetal acceleration $F$ for a body having mass $m$ is,
$F=\frac{mv^2}{r}$ and,
$F=m{\omega^2}r$
## How to Calculate Centripetal Force
Example 1
A small ball of mass 0.5 kg is attached to a string and it is whirled at a constant speed in a horizontal circle, which has a radius of 0.4 m. The circular motion of the ball has a frequency of 1.8 Hz.
a) Find the centripetal force.
b) Calculate how much force would be needed to move the ball in the same circle, but with twice the speed.
How to Calculate Centripetal Force – Example 1
## Examples of Centripetal Force
We will now look at several situations where the concepts we’ve learned about circular motion are applicable. The key to solving these types of problem is to identify the circular path and then find the resultant force pointing towards the centre of the circular path. This resultant force is the centripetal force.
### Circular Motion of a Conical Pendulum
Suppose a mass $m$ attached to the end of a string of length $l$ made to move in a horizontal circle with radius $r$, such that the string makes an angle $\theta$ to the vertical. The situation is illustrated below:
How to Calculate Centripetal Force – Conical Pendulum
It is important to note here that the pendulum cannot be swung in a horizontal circle with the string parallel to the ground. Gravity is always pulling the pendulum down, so there must always be a vertical force to balance this out. The vertical force must come from the tension, which acts along the string. Therefore, in order for tension to be able to balance the downward pull of weight, the pendulum’s string must always be at an angle to the ground.
### Circular Motion and Banking
Banking occurs when, for example, a car is traveling on a tilted track in a circular path or when a pilot deliberately angles an aircraft to maintain a circular path. The free body diagram for both cases look similar, so I will use just one diagram to find the centripetal force in both cases. The only difference is that the force named $U$ for the car is the reaction force between the car’s tyres and the road surface, whereas for the aeroplane, $U$ is the “Lift” force from the wings. In both cases, $m$ refers to the mass of the car/aeroplane.
How to Calculate Centripetal Force – Banking
Example 2
A car is travelling at 20 m s-1 in a banked section of a road. If the radius of the horizontal circular path is 200 m, calculate the banking angle necessary to keep the car moving at this speed, without any friction between the tyres and the road.
If there is friction, it would contribute to centripetal force and the vehicle would be able to move at a greater speed. However, we’re assuming friction is 0 here (imagine a very slippery road).
How to Calculate Centripetal Force – Example 2
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# How do you solve using the completing the square method x^2 - 5x = 9?
Jan 20, 2017
Take the coefficient of the $x$ term ($- 5$) and:
1. Divide it by 2 (to get $- \frac{5}{2}$).
2. Square this (to get $\frac{25}{4}$).
3. Add this final value to both sides.
#### Explanation:
Completing the square means seeking a constant term $n$ to add to ${x}^{2} - 5 x$, so that ${x}^{2} - 5 x + n$ is a perfect square.
First, let's look at what happens when we FOIL a perfect square binomial of the form ${\left(x + a\right)}^{2}$:
$\left(x + a\right) \left(x + a\right) = {x}^{2} + 2 a x + {a}^{2}$
A perfect square will always have a distributed form like this.
What we notice is that if we take the coefficient of the $x$ term ($2 a$), cut it in half, and then square it, we get ${a}^{2}$, the constant term. Thus, if given ${x}^{2} + 2 a x = b$, we would complete the square by adding ${a}^{2}$ (that is, the square of half of $2 a$) to both sides, giving
${x}^{2} + 2 a x + {a}^{2} = b + {a}^{2}$
so that the trinomial on the left will be guaranteed to be a perfect square—the square of $\left(x + a\right)$.
For this particular problem, we are given ${x}^{2} - 5 x = 9$. So, $- 5$ is like our "$2 a$". And if
$- 5 = 2 a$,
then
$a = - \frac{5}{2}$,
and
${a}^{2} = \frac{25}{4}$.
Thus, ${x}^{2} - 5 x + \frac{25}{4}$ is the completed square we seek, meaning we need to add $\frac{25}{4}$ to both sides:
${x}^{2} - 5 x + \frac{25}{4} = 9 + \frac{25}{4}$
Okay, so if this ${x}^{2} - 5 x + \frac{25}{4}$ is a perfect square, what is its factored form? (Or, what is its square root?)
That's easy—remember that the factored form of ${x}^{2} + \textcolor{red}{2 a} x + {a}^{2}$ is ${\left(x + \textcolor{b l u e}{a}\right)}^{2}$. The $\textcolor{b l u e}{a}$ in the factor is half of the $\textcolor{red}{2 a}$ in the trinomial. So the factored form of ${x}^{2} - 5 x + \frac{25}{4}$ will be ${\left(x - \frac{5}{2}\right)}^{2}$, because $- \frac{5}{2}$ is half of $- 5$.
We simplify both sides now to get
${\left(x - \frac{5}{2}\right)}^{2} = \frac{61}{4}$
Now our LHS is a perfect square, so we can solve for $x$ by taking the square root of both sides:
$x - \frac{5}{2} = \pm \frac{\sqrt{61}}{2}$
and then adding $\frac{5}{2}$ to both sides:
$x = \frac{5}{2} \pm \frac{\sqrt{61}}{2} \text{ "=" } \frac{5 \pm \sqrt{61}}{2}$.
## Note:
If the coefficient on the ${x}^{2}$ term is something other than $1$, you'll want to either factor it out or divide everything by it first, so that this method will work.
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#### Explain Solution R.D. Sharma Class 12 Chapter 9 Differentiability Exercise 9 .1 Question 11 Maths Textbook Solution.
Answer: $a=3,b=5$
Hint: For differentiability, LHD = RHD where LHD is left hand derivative and RHD is right hand derivative. Also, LHD and RHD should exist in given limit.
Given: $f\left ( x \right )=$$\left\{\begin{array}{c} x^{2}+3 x+a, \text { if } x \leq 1 \\ b x+2, \text { if } x>1 \end{array}\right.$
Solution:
\begin{aligned} &\text { LHD at } \mathrm{x}=1: \lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1} \\ &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{1-h-1} \\ &=\lim _{h \rightarrow 0} \frac{\left[(1-h)^{2}+3(1-h)+a\right]-[1+3+a]}{-h} \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \end{aligned}
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{\left[1+h^{2}-2 h+3-3 h+a\right]-[4+a]}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}-5 h}{-h} \\ &=\lim _{h \rightarrow 0} \frac{h(h-5)}{-h} \\ &=\lim _{h \rightarrow 0}-(h-5) \\ &=5 \end{aligned}
$RHD\; at\: x=0:\lim_{x\rightarrow 1^{+}}\frac{f\left ( x \right )-f\left ( 1 \right )}{x-1}$
\begin{aligned} &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{1+h-1} \\ &=\lim _{h \rightarrow 0} \frac{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{[b(1+h)+2]-(b+2)}{h} \\ &=\lim _{h \rightarrow 0} \frac{b h}{h} \\ &=b \end{aligned}
Since f(x) is differentiable, so (LHD at x = 1) = (RHD at x = 1)
$\Rightarrow b=5$
$\Rightarrow f\left ( 1 \right )=1+3+a$
$=4+a$
\begin{aligned} &\mathrm{LHL} \Rightarrow \lim _{x \rightarrow 1^{-}} f(x) \\ &\Rightarrow \lim _{h \rightarrow 0} f(1-h) \\ &\Rightarrow \lim _{h \rightarrow 0}(1-h)^{2}+3(1-h)+a \\ &\therefore\left\{(a-b)^{2}=a^{2}+b^{2}-2 a b\right\} \\ &=\lim _{h \rightarrow 0} 1+h^{2}-2 h+3-3 h+a \\ &=4+a \end{aligned}
\begin{aligned} &\mathrm{RHL} \Rightarrow \lim _{x \rightarrow 1^{+}} f(x) \\ &\Rightarrow \lim _{h \rightarrow 0} f(1+h) \\ &\Rightarrow \lim _{h \rightarrow 0} b(1+h)+2 \\ &=\lim _{h \rightarrow 0} b+b h+2 \end{aligned}
$=b+2$
Since f(x) is differentiable, f(x) is continuous.
LHL = RHL
$b+2=4+a$
we know that $b=5$
$5+2=4+a$
$7=4+a$
$a=3$
Hence, $a=3,$ and $b=5$
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# Composition of Functions
The composition of two functions $$f$$ and $$g$$ is the new function $$h$$, where $$h(x) = f(g(x))$$, for all $$x$$ in the domain of $$g$$ such that $$g(x)$$ is in the domain of $$f$$. The notation for function composition is $$h = f \circ g$$ or $$h(x) = (f \circ g)(x)$$ and is read as 'f of g of x'. The procedure is called composition because the new function is composed of the two given functions $$f$$ and $$g$$, where one function is substituted into the other.
## Finding the Composition
Although composition of functions is best illustrated with an example, let us summarize the key steps:
• rewrite $$f \circ g$$ as $$f(g(x))$$;
• replace $$g(x)$$ with the function that it represents;
• evaluate $$f$$ by replacing every $$x$$ with the function that $$g(x)$$ represents; and
• finally, if given a numerical value of $$x$$, evaluate the new function at this value by replacing all remaining $$x$$ with the given value.
Note: Often $$f \circ g \neq g \circ f$$ and the two will have different domains. Also, be aware that you can take the composition of more than two functions: e.g., $$f(g(k(x)))$$.
Example: Given the function $$f(x) = x^2$$ and $$g(x) = x + 3$$, find $$f(g(1))$$ and $$g(f(1))$$.
Solution:
\begin{align*} f(g(x)) &= f(x + 3) & & & g(f(x)) &= g(x^2)\\&= (x +3)^2 & & & &= x^2 + 3\\f(g(1)) &= (1 + 3)^2 & & & g(f(1))& =1^2 + 3\\ &=16 & & & &=4\end{align*}
Notice that $$f \circ g \neq g \circ f$$.
Composition of Function Example 1:
Composition of Function Example 2:
Composition of Function Example 3:
Composition of Function Example 4:
Composition of Function Example 5:
Composition of Function Example 6:
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# RD Sharma Solutions for Class 8 Maths Chapter 1 - Rational Numbers Exercise 1.1
Students can refer to and download RD Sharma Solutions for Class 8 Maths Exercise 1.1 Chapter 1, Rational Numbers, from the links provided below.
Our subject experts have solved the RD Sharma Solutions to ensure that the students are thorough with their basic concepts and help clear their doubts. Exercise 1.1 is based on the basic concepts of rational numbers with the addition of the same and different denominators. Students can download the RD Sharma Class 8 Maths Chapter 1 and start practising offline.
## RD Sharma Solutions for Class 8 Maths Exercise 1.1 Chapter 1 Rational Numbers
### Access Answers to RD Sharma Solutions for Class 8 Maths Exercise 1.1 Chapter 1 Rational Numbers
1. Add the following rational numbers:
(i) -5/7 and 3/7
(ii) -15/4 and 7/4
(iii) -8/11 and -4/11
(iv) 6/13 and -9/13
Solution:
Since the denominators are of same positive numbers we can add them directly
(i) -5/7 + 3/7 = (-5+3)/7 = -2/7
(ii) -15/4 + 7/4 = (-15+7)/4 = -8/4
Further dividing by 4 we get,
-8/4 = -2
(iii) -8/11 + -4/11 = (-8 + (-4))/11 = (-8-4)/11 = -12/11
(iv) 6/13 + -9/13 = (6 + (-9))/13 = (6-9)/13 = -3/13
2. Add the following rational numbers:
(i) 3/4 and -5/8
Solution: The denominators are 4 and 8
By taking LCM for 4 and 8 is 8
We rewrite the given fraction in order to get the same denominator
3/4 = (3×2) / (4×2) = 6/8 and
-5/8 = (-5×1) / (8×1) = -5/8
Since the denominators are same we can add them directly
6/8 + -5/8 = (6 + (-5))/8 = (6-5)/8 = 1/8
(ii) 5/-9 and 7/3
Solution: Firstly we need to convert the denominators to positive numbers.
5/-9 = (5 × -1)/ (-9 × -1) = -5/9
The denominators are 9 and 3
By taking LCM for 9 and 3 is 9
We rewrite the given fraction in order to get the same denominator
-5/9 = (-5×1) / (9×1) = -5/9 and
7/3 = (7×3) / (3×3) = 21/9
Since the denominators are same we can add them directly
-5/9 + 21/9 = (-5+21)/9 = 16/9
(iii) -3 and 3/5
Solution: The denominators are 1 and 5
By taking LCM for 1 and 5 is 5
We rewrite the given fraction in order to get the same denominator
-3/1 = (-3×5) / (1×5) = -15/5 and
3/5 = (3×1) / (5×1) = 3/5
Now, the denominators are same we can add them directly
-15/5 + 3/5 = (-15+3)/5 = -12/5
(iv) -7/27 and 11/18
Solution: The denominators are 27 and 18
By taking LCM for 27 and 18 is 54
We rewrite the given fraction in order to get the same denominator
-7/27 = (-7×2) / (27×2) = -14/54 and
11/18 = (11×3) / (18×3) = 33/54
Now, the denominators are same we can add them directly
-14/54 + 33/54 = (-14+33)/54 = 19/54
(v) 31/-4 and -5/8
Solution: Firstly we need to convert the denominators to positive numbers.
31/-4 = (31 × -1)/ (-4 × -1) = -31/4
The denominators are 4 and 8
By taking LCM for 4 and 8 is 8
We rewrite the given fraction in order to get the same denominator
-31/4 = (-31×2) / (4×2) = -62/8 and
-5/8 = (-5×1) / (8×1) = -5/8
Since the denominators are same we can add them directly
-62/8 + (-5)/8 = (-62 + (-5))/8 = (-62-5)/8 = -67/8
(vi) 5/36 and -7/12
Solution: The denominators are 36 and 12
By taking LCM for 36 and 12 is 36
We rewrite the given fraction in order to get the same denominator
5/36 = (5×1) / (36×1) = 5/36 and
-7/12 = (-7×3) / (12×3) = -21/36
Now, the denominators are same we can add them directly
5/36 + -21/36 = (5 + (-21))/36 = 5-21/36 = -16/36 = -4/9
(vii) -5/16 and 7/24
Solution: The denominators are 16 and 24
By taking LCM for 16 and 24 is 48
We rewrite the given fraction in order to get the same denominator
-5/16 = (-5×3) / (16×3) = -15/48 and
7/24 = (7×2) / (24×2) = 14/48
Now, the denominators are same we can add them directly
-15/48 + 14/48 = (-15 + 14)/48 = -1/48
(viii) 7/-18 and 8/27
Solution: Firstly we need to convert the denominators to positive numbers.
7/-18 = (7 × -1)/ (-18 × -1) = -7/18
The denominators are 18 and 27
By taking LCM for 18 and 27 is 54
We rewrite the given fraction in order to get the same denominator
-7/18 = (-7×3) / (18×3) = -21/54 and
8/27 = (8×2) / (27×2) = 16/54
Since the denominators are same we can add them directly
-21/54 + 16/54 = (-21 + 16)/54 = -5/54
3.Simplify:
(i) 8/9 + -11/6
Solution: let us take the LCM for 9 and 6 which is 18
(8×2)/(9×2) + (-11×3)/(6×3)
16/18 + -33/18
Since the denominators are same we can add them directly
(16-33)/18 = -17/18
(ii) 3 + 5/-7
Solution: Firstly convert the denominator to positive number
5/-7 = (5×-1)/(-7×-1) = -5/7
3/1 + -5/7
Now let us take the LCM for 1 and 7 which is 7
(3×7)/(1×7) + (-5×1)/(7×1)
21/7 + -5/7
Since the denominators are same we can add them directly
(21-5)/7 = 16/7
(iii) 1/-12 + 2/-15
Solution: Firstly convert the denominator to positive number
1/-12 = (1×-1)/(-12×-1) = -1/12
2/-15 = (2×-1)/(-15×-1) = -2/15
-1/12 + -2/15
Now let us take the LCM for 12 and 15 which is 60
(-1×5)/(12×5) + (-2×4)/(15×4)
-5/60 + -8/60
Since the denominators are same we can add them directly
(-5-8)/60 = -13/60
(iv) -8/19 + -4/57
Solution: let us take the LCM for 19 and 57 which is 57
(-8×3)/(19×3) + (-4×1)/(57×1)
-24/57 + -4/57
Since the denominators are same we can add them directly
(-24-4)/57 = -28/57
(v) 7/9 + 3/-4
Solution: Firstly convert the denominator to positive number
3/-4 = (3×-1)/(-4×-1) = -3/4
7/9 + -3/4
Now let us take the LCM for 9 and 4 which is 36
(7×4)/(9×4) + (-3×9)/(4×9)
28/36 + -27/36
Since the denominators are same we can add them directly
(28-27)/36 = 1/36
(vi) 5/26 + 11/-39
Solution: Firstly convert the denominator to positive number
11/-39 = (11×-1)/(-39×-1) = -11/39
5/26 + -11/39
Now let us take the LCM for 26 and 39 which is 78
(5×3)/(26×3) + (-11×2)/(39×2)
15/78 + -22/78
Since the denominators are same we can add them directly
(15-22)/78 = -7/78
(vii) -16/9 + -5/12
Solution: let us take the LCM for 9 and 12 which is 108
(-16×12)/(9×12) + (-5×9)/(12×9)
-192/108 + -45/108
Since the denominators are same we can add them directly
(-192-45)/108 = -237/108
Further divide the fraction by 3 we get,
-237/108 = -79/36
(viii) -13/8 + 5/36
Solution: let us take the LCM for 8 and 36 which is 72
(-13×9)/(8×9) + (5×2)/(36×2)
-117/72 + 10/72
Since the denominators are same we can add them directly
(-117+10)/72 = -107/72
(ix) 0 + -3/5
Solution: We know that anything added to 0 results in the same.
0 + -3/5 = -3/5
(x) 1 + -4/5
Solution: let us take the LCM for 1 and 5 which is 5
(1×5)/(1×5) + (-4×1)/(5×1)
5/5 + -4/5
Since the denominators are same we can add them directly
(5-4)/5 = 1/5
4. Add and express the sum as a mixed fraction:
(i) -12/5 and 43/10
Solution: let us add the given fraction
-12/5 + 43/10
let us take the LCM for 5 and 10 which is 10
(-12×2)/(5×2) + (43×1)/(10×1)
-24/10 + 43/10
Since the denominators are same we can add them directly
(-24+43)/10 = 19/10
19/10 can be written as
$$\begin{array}{l}1\frac{9}{10}\end{array}$$
in mixed fraction.
Solution: let us add the given fraction
let us take the LCM for 7 and 4 which is 28
(24×4)/(7×4) + (-11×7)/(4×7)
96/28 + -77/28
Since the denominators are same we can add them directly
(96-77)/28 = 19/28
(iii) -31/6 and -27/8
Solution: let us add the given fraction
-31/6 + -27/8
let us take the LCM for 6 and 8 which is 24
(-31×4)/(6×4) + (-27×3)/(8×3)
-124/24 + -81/24
Since the denominators are same we can add them directly
(-124-81)/24 = -205/24
-205/24 can be written as
$$\begin{array}{l}-8\frac{13}{24}\end{array}$$
in mixed fraction.
(iv) 101/6 and 7/8
Solution: let us add the given fraction
101/6 + 7/8
let us take the LCM for 6 and 8 which is 24
(101×4)/(6×4) + (7×3)/(8×3)
404/24 + 21/24
Since the denominators are same we can add them directly
(404+21)/24 = 425/24
425/24 can be written as
$$\begin{array}{l}17\frac{17}{24}\end{array}$$
in mixed fraction.
## RD Sharma Solutions for Class 8 Maths Exercise 1.1 Chapter 1 Rational Numbers
Exercise 1.1 of RD Sharma Class 8 Chapter 1, Rational Numbers, contains the basic concepts related to Rational Numbers. This exercise mainly deals with the basic properties of Rational Numbers. Some of them include
• Equivalent Rational Numbers
• Positive and Negative Rational Numbers
• Comparison of Rational Numbers
The RD Sharma Solutions can help the students practise diligently while learning the fundamentals, as they provide all the answers to the questions from the RD Sharma textbook. Practising as many times as possible helps in scoring high marks.
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# How to Perform Calendar Calculations in Your Head
Oct 13, 2019 · 3 min read
A so-called calendrical savant (or calendar savant) is someone who despite their intellectual disability (typically autism) can name the day of the week of a given date, or visa versa in a few seconds or even a tenth of a second (Kennedy & Squire, 2007). In the clip below, mega-savant Kim Peek, inspiration for the movie Rain Man, does so while taking questions from an audience:
While extremely impressive to behold, calendar calculations are actually very simple to perform and can be learned in less than 30 minutes. This short article will teach you how.
# What day was the 13th of January 1989?
My birthday. I was born two weeks late. Here is how to calculate which day of the week that was (Lancaster, 2005):
## Step 1. Calculate the Year Code Y
The first step of our five step process is to calculate the year code, represented by the letter Y. This is done in the following way:
`Calculate the Year Code YTake the last two digits of the year, divide by 4, remove the remainder. For my birthday, 89 / 4 = 22. Add the number to the last two digits of the year, 22 + 89 = 111For dates in the 1700s, add 4.For dates in the 1800s, add 2.For dates in the 1900s, add 0.For dates in the 2000s, add 6.For dates in the 2100s, add 4.`
So, for the date 13th of January 1989 we obtain the year code Y =111.
## Step 2. Find the Month Code M
The second step of our five step calculation is to find the month code M. This is a simple step, simply look it up in the table below (or better, memorize it):
`Month Code January 1*February 4*March 4April 0May 2June 5July 0August 3September 6October 1November 4December 6*If the year you are calculating for is/was a leap year, subtract 1 from the code for January and February, so January = 0 and February = 3.`
As we know, leap years occur every four years (even years). Century years like 1900, 2000 and 2100 are leap years if they are evenly divisible by 400.
For the date 13th of January 1989, we obtain the month code M = 1.
## Step 3. Find the Day Code D
The third step of our five step calculation is to find the day code D. This iS even easier than finding the month code, as it is simply the number of the date itself. For the 13th of January 1989, the number is D = 13.
## Step 4. Find the sum of the numbers Y + M + D
The fourth step of our five step model is to add the three numbers we’ve found. For our three numbers, the sum is 111 + 1 + 13 = 125.
## Step 5. Find the Day of the week
The final step of the calculation is to calculate the remainder of the modulo operation 125 mod 7. We know that 7 x 17 is 119, leaving a remainder of 6:
`Day RemainderSaturday 0Sunday 1Monday 2Tuesday 3Wednesday 4Thursday 5Friday 6`
The 13th of January 1989 was a Friday. Yes, I was born on Friday the 13th.
This essay is part of a series of stories on math-related topics, published in Cantor’s Paradise, a weekly Medium publication. Thank you for reading!
Written by
Written by
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## How do we find area of composite shapes?
To calculate the area of a composite shape you must divide the shape into rectangles, triangles or other shapes you can find the area of and then add the areas back together. You may have to calculate missing lengths before finding the area of some of the shapes. Calculate the area of this compound shape.
## How do we find area of composite shapes?
To calculate the area of a composite shape you must divide the shape into rectangles, triangles or other shapes you can find the area of and then add the areas back together. You may have to calculate missing lengths before finding the area of some of the shapes. Calculate the area of this compound shape.
### How do you find the area of a composite figure 7th grade?
To find the area of a composite figure, separate the figure into simpler shapes whose area can be found. Then add the areas together. Be sure than none of the simpler figures have overlapping areas.
Which shape is considered a composite shape?
A composite or compound shape is any shape that is made up of two or more geometric shapes. The below shape is made up of a square and a triangle. The blue shape is made up of a square and a rectangle. The green shape consists oftwo triangles.
What is a composite area?
The area of composite shapes is defined as the area covered by any composite shape. A composite shape is made up of basic shapes put together. Thus, the area of the composite shape is found by individually adding all the basic shapes.
## How do you find the area of each shape?
Area is calculated by multiplying the length of a shape by its width.
### What is composite area?
What composite figure means?
A composite shape or a composite figure is a two-dimensional figure made up of basic two-dimensional shapes such as triangles, rectangles, circles, semi-circles, etc.
How to find the area of composite shapes?
The composite shape is a shape in which few polygons are put together to form a required shape. These shapes or figures can be made up of a combination of triangles, squares, and quadrilaterals, etc. Divide a composite shape into basic shapes like square, triangle, rectangle, hexagon, etc. to determine the area of composite shapes.
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# Derivatives and an Introduction to Differential Calculus
One of the main concepts studied in the field of differential calculus is based on the notion of change - specifically, how one quantity changes compared to another. Perhaps a more succinct version of this physical definition would be "rate of change." Alternately, a geometric definition could simply be the slope of a curve at a particular point. The underlying key to this branch of mathematics is the concept of the derivative. In this post, I will introduce various aspects of the derivative - please click the Facebook Like button if this is helpful for you!
First, let's consider the derivative in terms of rate of change. And to do that, let's talk about velocity. We know that velocity is equal to distance per unit time. However, that is a very general description of it. If you drive from your home to the grocery store on the other side of town, you can do the math by dividing your total distance travelled by the time it took you to get there, but what does this number tell you? It actually tells you the average velocity of your trip. Think about it. You had to stop for red lights, stop signs, pedestrians. Maybe you sped up to pass a slow driver. Don't forget about the actual acceleration of your car from a standstill, and then the deceleration whenever you needed to stop. All of this factors into the calculation of your average velocity, which is simply how far you go in a measured amount of time.
Now, let's consider how to calculate the average velocity of your car between your home and the first stop sign, while on your way to the grocery store. You no longer have several stops to deal with. You get in your car, accelerate, then as you approach the stop sign, you decelerate to a stop. Your average velocity is calculated from a much shorter interval, and will have much less variation to it. So, your average velocity will be more representative of your actual velocity at any given time.
To extend this demonstration even further, let's consider the small portion of your trip that is measured between two street lights 10 meters apart that you pass while you are in full motion. That is, let's assume that we want to measure the velocity without having to calculate stop signs, etc. Our time interval for the measurement is much smaller, and calculating the velocity by dividing the distance by the time it takes to go from one light to the other is even more representative of your velocity at any point between them.
What I am trying to demonstrate is the concept of instantaneous velocity. If you shrink down your time interval of measurement infinitesimally, the two time points approach each other at a single point, and so the average velocity between the two super close points approaches the instantaneous velocity of the single point.
Graphically, this is essentially the same thing you do when you calculate the slope of a tangent line to a curve. You pick two lines on the curve and calculate the slope of the line between them, and then you use limits to make the points get closer and closer together until they are almost the same, and the slope of the line connecting those two infinitely close points is the tangent. (Check out my previous post about using limits to find tangents if you'd like a refresher of this topic: http://sk19math.blogspot.ca/2012/06/using-limits-to-find-tangents.html)
Because this type of limit occurs so frequently in maths, science, and engineering, it is given the special name of "derivative," and you calculate derivatives through the process of "differentiation." So, one interpretation of the derivative is an expression of the instantaneous rate of change (velocity) at a particular point on the curve - a large derivative corresponds to a high rate of change (a steep curve), and conversely a small derivative corresponds to a low rate of change (a relatively flat curve). As a specific example, if you actually have a graph of position (displacement) of an object vs. time, the derivative of the curve at any time point represents the velocity of that object at that specific time. This may take a little practice to become comfortable with the concept, but suffice it to say at this point that learning how to use derivatives is incredibly important to be able to work out more complex concepts relatively easily.
Let's look at this now in the more formal terms of mathematical symbols and equations. Consider any curve y = f(x).
Now, let us identify the point P on the curve f(x) for when x = a. That is to say, the point (a, f(a)).
Now, let's go a step further, and identify a point Q that is h units away from a on the x-axis. If it is h units away from a, we can call it "a + h". (If this is confusing, think about it with numbers instead. Start at, say, x = 3 (instead of a). Now we want to know what is going on 5 units (instead of h) away from x = 3. In other words, we have 3, and we have 3 + 5.) As such, we can therefore identify a point Q ((a + h), f(a + h)).
Now that we have two arbitrary points, let's determine the slope of the straight line that would connect the two. We can use the same slope formula that we always use, slope = rise/run, but substitute in our variables that we identified above. So, we have:
Now, imagine that the distance h between the two points is getting smaller and smaller. Or in other words, consider the case of when h approaches 0. By doing this, we calculate the slope of the line connecting two infinitesimally close points - which means that we are actually approaching the slope of the tangent line to the curve at point a. In this case, we would express this slope as a limit in the following way, which actually corresponds to the definition of the derivative of a function f at a number a. The derivative is given the special symbol f'(x), and we say "f prime x", and we express it like this:
Another way of expressing this can be found if we recognize that a + h is really just any x value. So, we can say x = a + h (and by extension, h = x - a), and modify the above derivative definition accordingly:
With this modified equation, it actually becomes a matter of arithmetic to determine the slope at a point. Here is an example of a kind of question that you will see:
"Find the derivative (or, determine the slope of the tangent) of the function f(x) = x2 - 4 at a number a."
To do this, write the provided equation into the definition, and reduce until you have an answer. Notice below how I combine terms and recognize the identity of a difference of squares.
What this final result tells you is that for our curve, f(x) = x2 - 4, at any number a along it, the slope of the tangent (AKA, the derivative at that point) is equal to the term 2a. Graph it out and try with several values to convince yourself that it's true! Consider when x = 5. You can determine from the original equation that we have the point (5, 21). At this point on the curve, the slope of the tangent equals 2 x 5 = 10.
Going back to the definition of the derivative that I gave above, you can also apply the concept of point-slope form to it to get a different way of seeing it. Letting y = f(x), you can rearrange the definition as follows, by simple reorganization of the terms:
Here's a more visual exercise that you may soon encounter.
"If you are provided a graph of a function f(x) - not necessarily the equation - sketch out what the graph of the derivative f'(x) would look like."
When you actually have the numbers and equation, this becomes much easier... assuming you know how to easily recognize derivatives from the original equations. However, if provided ONLY the picture of the curve, this becomes a bit more abstract, but not really that challenging. It DOES require you to understand the concept of derivatives and rate of change though. Here is why. Take some random curve that you can draw. Any curve will do for this exercise:
The key is rate of change. We have seen that slope is equal to rate of change, so we want to pay particular attention to the slope at several points. And the easiest points to notice are those where the slope is equal to 0. These are all the peaks and valleys of the curve. What I have done next is highlight with red bars all of the zero-slopes:
Now, to proceed with sketching the graph of the derivative f'(x) vs x, you can start by plotting the points where f'(x) is equal to zero. From there you can then go on to say where the curve of f(x) has a positive, increasing slope, and then sketch that into your f'(x) graph accordingly. Similarly, decreasing slopes on the f(x) curve will be negative values on the f'(x) curve. For the sake of this exercise, don't worry so much about how high or low the slopes are. Just focus on whether they are positive or negative at the various parts of the graph. I have gone ahead and plotted out the actual curve of the derivative below in green, alongside the original curve of f(x). You can see that the f'(x) curve crosses zero wherever the curve for f(x) has peaks or valleys, and the steeper the f(x) curve, the more extreme the f'(x) curve at that same point.
Of course, having a mathematical definition wouldn't be any fun if there were no conditions or rules associated with it - and the definition of derivatives is no exception. One such rule states that a function is differentiable at a point a if the derivative f'(a) exists. Seems intuitive enough. If a derivative at a point exists, then the base function is differentiable at that point. Probably one of those rules that doesn't really even need to be said. :)
I'm not going to graph this one out, but it is for you to think on. Consider the case of f(x) = |x|. Where is it differentiable?
If you consider the derivative of the left hand side, it equals -1. The f'(x) on the right hand side is 1. This function then is obviously differentiable when x < 0, and when x > 0. But what about when x = 0? In this case, since the right hand limit approaches 1 as x approaches 0 from the right, and the left hand limit approaches -1 as x approaches 0 from the left, one must conclude that f'(0) does not exist because both of the one-sided limits approach different numbers.
An extension of this example actually describes a second rule for limits: if f'(a) exists, then the function f(x) is continuous as a. Recall that continuity of a curve is based on the notion that as you approach a point from both the left and the right, the limit of each side approaches the same value. In the example above, approaching 0 from either side resulted in different limits, and hence the graph is not continuous at 0.
Keep this in mind as you see various graphs of functions. Curves that have a sharp point will not be differentiable at the point, for the reason given above. Similarly, discontinuous curves (i.e. curves with gaps in them) will not have a derivative at the break point either because the one-sided limits do not agree. If f(x) is not continuous as point a, then f'(a) does not exist. A third condition to watch out for is where a graph has a vertical tangent line, in which case the slope is infinite.
Now, I'm going to wrap up this mammoth of a maths post with something a bit easier to talk about: notation of derivatives. I have already described a few ways to express these values. I talked about expressing them as limits, and using infinitesimally smaller intervals, and that is a good way to work through them. Symbolically, I said that you can write f'(x) to denote the derivative of the functions f(x). This will likely be the easiest way for you to use it and to recognize it, though here are a few others that mean the same thing:
Each of these terms means the exact same thing. In particular, the D and d/dx are specifically called the differentiation operators, and you can see they have a few variations. Similarly, dy/dx is symbolic of derivatives for historical reasons. Read up on Gottfried Wilhelm Leibniz to learn more about the origins of calculus, where you will see that he introduced this way of representing it. Sometimes, you may see dy/dx referred to as "Leibniz notation."
And with that final tidbit of mathematical goodness, I am going to end this post. I intend to follow this with another post in the near future that introduces differentiation methods and strategies. Much like the exponent rules, there are also several differentiation rules, and I hope to be able to explain them for you as well. If you have made it to this point of my post, thanks for reading, and please be sure to click the Facebook Like button below or at the top, and I'd appreciate a Google +1 as well below if this was helpful!
#### 1 comment:
1. This gave me so many flashbacks! Good ones, since I actually intuitively understood parts of the explanations!
In particular, explaining about how limits on either side of a discontinuity approach different values was something nobody's ever mentioned before. I guess because that would mean determining twice as many values, and maths was always about memorising rules and minimising logic. At least, in the 70s and 80s, when I continuously failed maths every year.
Anyway, thanks for explaining this so clearly!
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# Sequence 2
Write the first 5 members of an arithmetic sequence a11=-14, d=-1
Correct result:
a1 = -4
a2 = -5
a3 = -6
a4 = -7
a5 = -8
#### Solution:
${a}_{1}=-14-\left(11-1\right)\cdot \left(-1\right)=-4$
${a}_{2}={a}_{1}-1=-5$
${a}_{3}={a}_{2}-1=-6$
${a}_{4}={a}_{3}-1=-7$
${a}_{5}={a}_{4}-1=-8$
We would be pleased if you find an error in the word problem, spelling mistakes, or inaccuracies and send it to us. Thank you!
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# In a Game, a Man Wins Rs 5 for Getting a Number Greater than 4 and Loses Rs 1 Otherwise, When a Fair Die is Thrown. the Man Decided to Thrown a Die Thrice but to Quit as and When He Gets a Number Greater than 4. Find the Expected Value of the Amount He Wins/Loses - CBSE (Commerce) Class 12 - Mathematics
#### Question
In a game, a man wins Rs 5 for getting a number greater than 4 and loses Rs 1 otherwise, when a fair die is thrown. The man decided to thrown a die thrice but to quit as and when he gets a number greater than 4. Find the expected value of the amount he wins/loses
#### Solution
The man may get number greater than 4 in the first throw and then he quits the game. He may get a number less than equatl to 4 in the first throw and in the second throw he may get the number greater than 4 and quits the game.
In the first two throws he gets a number less than equal to 4 and in the third throw he may get a number greater than 6. He may not get number greater than 4 in any one of three throws.
Let X be the amount he wins/looses.
Then, X can take values -3, 3, 4, 5 such that
P (X = 5) = P(Getting number greater than 4 in first throw) = 1/3
P (X = 4) = P(Getting number less than equal to 4 in the first throw and number greater than 4 in second throw) = 4/6×2/6=2/9
P (X = 3) = P(Getting number less than equal to 4 in the first two throws and number greater than 4 in third throw) = 4/6×4/6×2/6=4/27
P (X = -3) = P(Getting number less than equal to 4 in all three throws) = 4/6×4/6×4/6=8/27
X 5 4 3 -3 P(X) 1/3 2/9 4/27 8/27
E (X) = (5×1/3)+4 (2/9)+3(4/27)−3 (8/27)
=1/27(45+24+12−24)
=57/27
Is there an error in this question or solution?
#### Video TutorialsVIEW ALL [2]
Solution In a Game, a Man Wins Rs 5 for Getting a Number Greater than 4 and Loses Rs 1 Otherwise, When a Fair Die is Thrown. the Man Decided to Thrown a Die Thrice but to Quit as and When He Gets a Number Greater than 4. Find the Expected Value of the Amount He Wins/Loses Concept: Conditional Probability.
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Study Materials
# NCERT Solutions for Class 9th Mathematics
Page 1 of 3
## Chapter 6. Lines and Angles
### Exercise 6.1
Exercise 6.1
Q1. In Fig. 6.13, lines AB and CD intersect at O. If AOC + ∠ BOE = 70° and ∠BOD = 40° find ∠BOE and reflex COE.
Solution:
∠BOD = 40°
AOC = ∠BOD (Vertically opposite Angle)
AOC = 40°
AOC + ∠ BOE = 70° (Given)
BOE = 70°
BOE = 70° - 40°
∠BOE = 30°
AOB is straight line
AOC + COE +BOE = 180° (linear pair)
⇒ 70° + ∠COE = 180°
∠COE = 180° - 70°
∠COE = 110°
Reflex ∠COE = 360 - 110°
= 250°
Q2. In Fig. 6.14, lines XY and MN intersect at O. If POY = 90° and a : b = 2 : 3, find c.
Solution:
POY=90° (given)
Let a and b = 2x and 3x
XOY is a straight line
a + b + POY = 180°
2x + 3x + 90°= 180°
5x = 180° - 90°
5x = 90°
x = 90°/5
x = 18°
Now a = 2 x 18°
= 36°
b =3 x 18°
= 54°
MON is a straight line
b + c = 180°(linear pair)
54° + c = 180°
⇒∠c = 180°- 54°
=126°
Q3. In Fig. 6.15, ∠PQR = ∠PRQ, then prove that ∠PQS = ∠PRT
Solution :
Given : ∠PQR = ∠PRQ
To prove : ∠PQS = ∠PRT
Proof :
∠PQS + ∠PQR = 180° .................. (1) Linear pair
∠PRT + ∠PRQ = 180° .................. (2) Linear pair
From equation (1) and (2)
∠PQS + ∠PQR = ∠PRT + ∠PRQ
Or, ∠PQS + ∠PQR = ∠PRT + ∠PQR (∠PQR = ∠PRQ given)
Or, ∠PQS = ∠PRT Proved
Q4. In Fig. 6.16, if x + y = w + z, then prove that AOB is a line.
Solution:
Given : x + y = w + z
To prove : AOB is a line.
Proof :
We know that;
x + y + w + z = 360
(Angle Sustained on centre)
x + y + x + y = 360 (x + y = w + z given)
2x + 2y = 360
2 (x + y) = 360
x + y = 180 (linear pair)
Therefore, AOB is a line
Hence, Proved
Q5. In Fig. 6.17, POQ is a line. Ray OR is perpendicular to line PQ. OS is another ray lying between rays OP and OR. Prove that
Solution:
Given:
POQ is a straight line. OR PQ and OS is
another ray lying between rays OP and OR.
To prove:
Proof: OR PQ (given)
∴ ∠QOR = 90 …………… (1)
POQ is straight line
∴ ∠POR + QOR = 180 (linear pair)
POR + 90 = 180
POR = 180– 90
POR = 90…………… (2)
Now, ROS + QOR = QOS
Or, ROS = QOS – QOR ……………. (3)
Again, ROS + POS = POR
Or, ROS = POR – POS ……………. (4)
ROS + ROS = QOS – QOR + POR – POS
2 ROS = QOS – 90+ 90POS
2 ROS = (QOS – POS)
Hence Proved
Q6. It is given that ∠ XYZ = 64° and XY is produced to point P. Draw a figure from the given information. If ray YQ bisects ∠ ZYP, find ∠ XYQ and reflex ∠ QYP.
Solution:
Given: ∠ XYZ = 64°and XY is produced to point P. YQ bisects ∠ ZYP.
To Find: ∠XYQ and reflex ∠QYP.
YQ bisects ∠ZYP
∴ ∠ZYQ = ∠QYP ................. (1)
∵ XY is produced to point P.
∴ PX is a straight line.
Now, ∠ XYZ + ∠ZYQ + ∠QYP = 180° (linear pair)
Or, 64° + ∠ZYQ + ∠QYP = 180°
⇒ ∠ZYQ + ∠QYP = 180° - 64°
⇒ ∠ZYQ + ∠ZYQ = 116° [Using equation (1) ]
⇒ 2∠ZYQ = 116°
⇒ ∠ZYQ = 116°/2
⇒ ∠ZYQ = 58°
∠ZYQ = ∠QYP = 58°
∠XYQ = ∠XYZ + ∠ZYQ
= 64° + 58°
= 122°
∵ ∠QYP = 58°
∴ Reflex ∠QYP = 360° - 58°
= 302°
∠XYQ = 122°, Reflex ∠QYP = 302°
Page 1 of 3
Chapter Contents:
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# How Changing a Value Affects the Mean and Median Online Quiz
Following quiz provides Multiple Choice Questions (MCQs) related to How Changing a Value Affects the Mean and Median. You will have to read all the given answers and click over the correct answer. If you are not sure about the answer then you can check the answer using Show Answer button. You can use Next Quiz button to check new set of questions in the quiz.
Q 1 - Find new mean and new median of the data set if a data is changed.
17 , 7 , 2 , 15 , 6 , 19 , 20 , 15 , 18; 2 is changed to 12
### Explanation
Step 1:
Mean = $\frac{(17 + 7 + 2 + 15 + 6 + 19 + 20 + 15 + 18)}{9}$ = 13.22; Median = 15
Step 2:
With data change
New Mean = $\frac{(17 + 7 + 12 + 15 + 6 + 19 + 20 + 15 + 18)}{9}$ = 14.33; New Median = 15.
Q 2 - Find new mean and new median of the data set if a data is changed.
12 , 15 , 18 , 13 , 6 , 14; 13 is changed to 5
### Explanation
Step 1:
Mean = $\frac{(12 + 15 + 18 + 13 + 6 + 14 )}{6}$ = 13; Median = 13.5
Step 2:
With data change
New Mean = $\frac{(12 + 15 + 18 + 5 + 6 + 14 )}{6}$ = 11.67; New Median = 13
Q 3 - Find new mean and new median of the data set if a data is changed.
18 , 7 , 11 , 1 , 19 , 15 , 19 , 9; 7 is changed to 14
### Explanation
Step 1:
Mean = $\frac{(18 + 7 + 11 + 1 + 19 + 15 + 19 + 9 +7)}{9}$ = 12.38; Median = 13
Step 2:
With data change
New Mean = $\frac{(18 + 14 + 11 + 1 + 19 + 15 + 19 + 9 +7)}{9}$ = 13.25 ; New Median = 14.5
Q 4 - Find new mean and new median of the data set if a data is changed.
8 , 12 , 8 , 10 , 18 , 12 , 4; 10 is changed to 17
### Explanation
Step 1:
Mean = $\frac{(8 + 12 + 8 + 10 + 18 + 12 + 4)}{7}$ = 10.29; Median = 10
Step 2:
With data change
New Mean = $\frac{(8 + 12 + 8 + 17 + 18 + 12 + 4)}{7}$ = 11.29; New Median = 12
Q 5 - Find new mean and new median of the data set if a data is changed.
20 , 5 , 7 , 6 , 19 , 5 , 16 , 7; 20 is changed to 10
### Explanation
Step 1:
Mean = $\frac{(20 + 5 + 7 + 6 + 19 + 5 + 16 + 7)}{8}$ = 10.63; Median = 7
Step 2:
With data change
New Mean = $\frac{(10 + 5 + 7 + 6 + 19 + 5 + 16 + 7)}{8}$ = 9.38; New Median = 7
Q 6 - Find new mean and new median of the data set if a data is changed.
12 , 12 , 4 , 12 , 2 , 12; 4 is changed to 8
### Explanation
Step 1:
Mean = $\frac{(12 + 12 + 4 + 12 + 2 + 12)}{6}$ = 9; Median = 12
Step 2:
With data change
New Mean = $\frac{(12 + 12 + 8 + 12 + 2 + 12)}{6}$ = 9.67; New Median = 12
Q 7 - Find new mean and new median of the data set if a data is changed.
6 , 12 , 9 , 4 , 4; 12 is changed to 15
### Explanation
Step 1:
Mean = $\frac{(6 + 12 + 9 + 4 + 4 )}{5}$ = 7; Median = 6
Step 2:
New Mean = $\frac{(6 + 15 + 9 + 4 + 4)}{5}$ = 7.6 ; New Median = 6
Q 8 - Find new mean and new median of the data set if a data is changed.
18 , 15 , 11 , 3 , 8 , 4 , 13 , 12 , 3; 15 is changed to 18
### Explanation
Step 1:
Mean = $\frac{(18 + 15 + 11 + 3 + 8 + 4 + 13 + 12 +3)}{9}$ = 9.67; Median = 11
Step 2:
With data change
New Mean = $\frac{(18 + 18 + 11 + 3 + 8 + 4 + 13 + 12 +3)}{9}$ = 10; New Median = 11
Q 9 - Find new mean and new median of the data set if a data is changed.
25 , 18 , 18 , 13 , 4 , 17 , 18 , 19 , 3; 4 is changed to 9
### Explanation
Step 1:
Mean = $\frac{(25 + 18 + 18 + 13 + 4 + 17 + 18 + 19 +3)}{9}$ = 15; Median = 18
Step 2:
With data change
New Mean = $\frac{(25 + 18 + 18 + 13 + 9 + 17 + 18 + 19 +3)}{9}$ = 15.55; New Median = 18
Q 10 - Find new mean and new median of the data set if a data is changed
21 , 1 , 16 , 8 , 19; 1 is changed to 5
### Explanation
Step 1:
Mean = $\frac{(21 + 1 + 16 + 8 + 19)}{5}$ = 13 ; Median = 16
Step 2:
With data change
New Mean = $\frac{(21 + 1 + 16 + 8 + 19)}{5}$ = 13.8; New Median = 16
how_changing_value_affects_mean_and_median.htm
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Easy Method for Table of Numbers Ending With 9
H.D. Motiramani
Let us take Multiplication table of 19 first
19×1 = 1 9 = 19
19×2 = 3 8 = 38
19×3 = 5 7 = 57
19×4 = 7 6 = 76
19×5 = 9 5 = 95
19×6 = 11 4 = 114
19×7 = 13 3 = 133
19×8 = 15 2 = 152
19×9 = 17 1 = 171
19×10= 19 0 = 190
Here 19 is the multiplicand, 1 to 10 is called multiplier and the answer on right hand side is called product.
Observe that right hand side answers are split in two parts. This has been done deliberately to show that the First part from top starts with one for 19×1 and as we move downwards, it increases to next higher odd number like 1,3,5,7,…. 15,17,19. In other words, it is a series of odd numbers.
Second part of product (answer) is also interesting. It starts from zero in the bottom and as we go up, it increases by 1 to become 0,1,2,3,4…7,8,9.
Thus the multiplication table of 19 shows peculiar characteristics.
Let us pay little more attention to the table of 19. Please note that the first part is the double of multiplier minus one . For example if multiplier is 6, than the first part is 6×2-1=11, and if the multiplier is say 9 then the first part of the answer is 9×2-1=17. Also, the right hand part in red color is nothing but ten minus multiplier. For example if multiplier is 7 then the second part of answer or red colored answer will be 10-7=3 and if the multiplier is 6 than the second part of the product/answer is 10-6=4. It therefore becomes quite easy to understand the process and do the calculation in our minds.
Another easy method of arriving at the result is by multiplying with 20 instead of 19 and then adjusting it by deducting the multiplier. For example 19×7 can be attempted as 20×7-7=140-7=133.
As a matter of fact such a practice of taking the multiplicand(that is the number ending with 9) to a next higher digit and then do the maneuvering by deducting the multiplier to get the answer. This practice holds good for all multiplicands whose last digit ends with 9.
For example 39×7 can be attempted as 40×7-7=280-7=273.
79×8 can be done as 80×8-8=640-8=632
Some more examples:
19×5=20×5-5=95
69×5=70×5-5=345
89×6=90×6-6=540-6=534
499×7=500×7-7=3500-7=3493
789×2=790×2-2=1580-2=1578
So, remember the formula: Increase the multiplier (ending with nine) by 1. Adjust the product by deducting the multiplier.
—00—
< >
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## Forces on a ladder on a wall
A ladder rests on rough ground and leans against a rough wall. Its weight W acts through the centre of gravity G. Forces also act on the ladder at P and Q. These forces are P and Q respectively.
Which vector triangle represents the forces on the ladder?
## Using Loom and GeoGebra to explain a tutorial question
It’s Day 1 of the full home-based learning month in Singapore! As teachers all over Singapore scramble to understand the use of the myriad EdTech tools, I have finally come to settle on a few:
1. Google Meet to do video conferencing
2. Google Classroom for assignment that requires marking
3. Student Learning Space for students’ self-directed learning, collaborative discussion and formative assessment.
4. Loom for lecture recording
5. GeoGebra for visualisation
The following is a video that was created using Loom to explain a question on why tension in a rope on which a weight is balanced increases when the rope straightens.
## Newton’s 2nd Law Questions
Here are the steps to solving a problem involving Newton’s 2nd law.
Step 1: Draw a free body diagram – where possible, sketch a free body diagram to represent:
1. the body isolated from other objects
2. the forces acting on the body (ignore all internal forces) along the line of motion.
3. the direction of acceleration.
Step 2: Write down the equation: $$F_{net}=ma$$
Step 3: Add up the forces on the left-hand side of the equation, making sure that forces acting opposite to the direction of acceleration are subtracted.
Step 4: Solve the equation for the unknown.
For example,
A horizontal force of 12 N is applied to a wooden block of mass 0.60 kg on a rough horizontal surface, and the block accelerates at 4.0 m s-2. What is the magnitude of the frictional force acting on the block?
Step 1: Free-body diagram
Step 2: Write down the 2nd law equation
By Newton’s 2nd law, $$F_{net}=ma$$
Step 3: Add up the forces making up the net force
$$12 – f = 0.60 (4.0)$$
Step 4: Solve for unknown
$$f = 12 – 2.4 = 9.6 N$$
## Lesson Plan for Online Lecture on Forces
I am using this post as a way to document my brief plans for tomorrow’s Google Meet lecture with the LOA students as well as to park the links to the resources and tools that I intend to use for easy retrieval.
Instruction Objectives:
1. apply the principle of moments to new situations or to solve related problems.
2. show an understanding that, when there is no resultant force and no resultant torque, a system is in equilibrium.
3. use a vector triangle to represent forces in equilibrium.
4. *derive, from the definitions of pressure and density, the equation ?=??ℎ.
5. *solve problems using the equation ?=??ℎ.
6. *show an understanding of the origin of the force of upthrust acting on a body in a fluid.
Activity 1: Find CG of ruler demonstration
Having shown them the demonstration last week, I will explain the reason why one can find the CG this way:
1. As I move the fingers inward, there is friction between the ruler and my finger. This friction depends on the normal contact force as $f=\mu N$.
2. Drawing the free-body diagram of the ruler, there are two normal contact forces acting on the ruler by my fingers. The sum of these two upward forces must be equal to the weight of the ruler. These forces vary depending on their distance from the CG. Taking moments about the centre of gravity, $$N_1\times d_1=N_2 \times d_2$$
3. The finger that is nearer to the CG will always have a larger normal contact force and hence, more friction. Hence, the ruler will tend to stop sliding along that finger and allow the other finger to slide nearer. When that other finger becomes closer to the CG, the ruler also stops sliding along it and tends to then slide along the first finger.
4. This keeps repeating until both fingers reach somewhere near the CG.
Activity 2: Moments of a Force at an Angle to the line between Pivot and Point of Action.
1. Recollection of the slides on moment of a force and torque of a couple.
2. Give them a MCQ question to apply their learning using Nearpod’s Quiz function https://np1.nearpod.com/presentation.php?id=47032717
3. Ask students to sketching on Nearpod’s “Draw It” slides the “perpendicular distance between axis of rotation and line of action of force” and “perpendicular distance between the lines of action of the couple” for Example 5 and 6 of the lecture notes respectively.
4. Mention that
1. axis of rotation is commonly known as where the pivot is
2. perpendicular distance is also the “shortest distance”
Activity 3: Conditions for Equilibrium
1. State the conditions for translational and rotational equilibrium
2. Show how translation equilibrium is due to resultant force being zero using vector addition
3. Show how rotational equilibrium is due to resultant moment about any axis being zero by equating sum of clockwise moments to sum of anticlockwise moments.
4. Go through example 7 (2 methods: resolution of vectors and closed vector triangle)
5. Useful tip: 3 non-parallel coplanar forces acting on a rigid body that is in equilibrium must act through the same point. Use 2006P1Q6 as example.
6. Go through example 8. For 8(b), there are two methods: using concept that the 3 forces pass through the same point or closed triangle.
### For next lecture (pressure and upthrust):
Activity 4: Hydrostatic Pressure
1. Derive from definitions of pressure and density that $p = h\rho g$
2. Note that this is an O-level concept.
Activity 5: Something to sink about
Students are likely to come up with answers related to relative density. As them to draw a free body diagram of the ketchup packet. However, we will use the concept of the forces acting on the ketchup packet such as weight and upthrust to explain later.
Activity 6: Origin of Upthrust
I designed this GeoGebra app to demonstrate that forces due to pressure at different depths are different. For a infinitesimal (extremely small) object, the forces are equal in magnitude even though they are of different directions, which is why we say pressure acts equally in all directions at a point. However, when the volume of the object increases, you can clearly see the different in magnitudes above and below the object. This gives rise to a net force that acts upwards – known as upthrust.
## Free-Body Diagrams in Two-Body Motion
Students are often confused about the forces in drawing free-body diagrams, especially so when they have to consider the different parts of multiple bodies in motion.
##### Two-Body Motion
Let’s consider the case of a two-body problem, where, a force F is applied to push two boxes horizontally. If we were to consider the free-body diagram of the two boxes as a single system, we only need to draw it like this.
Considering both boxes as a single system
For the sake of problem solving, there is no need to draw the normal forces or weights since they cancel each other out, so the diagram can look neater. Applying Newton’s 2nd law of motion, $F=(m_A+m_B) \times a$, where $m_A$ is the mass of box A, $m_B$ is the mass of box B, F is the force applied on the system and a is the acceleration of both boxes.
You may also consider box A on its own.
Considering box A on its own
The equation is $F-F_{AB}=m_A \times a$, where $F_{AB}$ is the force exerted on box A by box B.
The third option is to consider box B on its own.
Considering box B on its own
The equation is $F_{BA}=m_B \times a$, where $F_{BA}$ is the force exerted on box B by box A. Applying Newton’s 3rd law, $F_{BA}=F_{AB}$ in magnitude.
##### Never Draw Everything Together
NEVER draw the free-body diagram with all the forces and moving objects in the same diagram, like this:
You will not be able to decide which forces acting on which body and much less be able to form a sensible equation of motion.
Interactive
Use the following app to observe the changes in the forces considered in the 3 different scenarios. You can vary the masses of the bodies or the external force applied.
##### Multiple-Body Problems
For the two-body problem above, we can consider 3 different free-body diagrams.
For three bodies in motion together, we can consider up to 6 different free-body diagrams: the 3 objects independently, 2 objects at a go, and all 3 together. Find the force between any two bodies by simplifying a 3-body diagram into 2 bodies. This trick can be applied to problems with even more bodies.
## Home-Based Learning Using Google Meet
I conducted an online lecture this morning using Google Meet for the students who had to stay home due to the Leave of Absence mandated by the Ministry since they had recently returned from another country during this period of the Covid-19 pandemic. I feel the need to document this as things might become bad enough that schools have to close, so it serves as a place where fellow teachers can pick up some tips on how to manage this.
The G Suite account that I used is that of my school’s, not MOE’s, because it allows me to record the session in case I need to show the session to students who did not “turn up” for the Meet. I am the G Suite admin for the school so I changed the setting to allow Google Meets to be recorded. After the session, the recorded Meet is automatically found in a G Drive folder after it has been processed in the backend. ICON’s Google Meet (part of MOE’s Google Suite service) does not allow recording.
My hardware setup is simple: just my laptop to capture my face and control the Google Meet UI and a second screen with which to show my slides. I also entered the Meet as another participant using my mobile phone as I wanted to see what my students would see for added assurance.
Google Meet is very user-friendly, with a minimalist and intuitive design that one can expect from Google (after all, that was what made it the preferred search engine in the early days of the internet). All we needed to do was to sign in to https://meet.google.com/ and start a session. You can also schedule a session on Google Calendar.
When a Meet is created, a URL is generated, which you can communicate to your students via text message or email, or through a system announcement.
When students log in, be sure to ask them to switch off their video and mute their voices so as not to cause any interference.
Note that what is shown in the presenter’s screen in Meet using the front camera of a laptop is laterally inverted as presenters generally want to see themselves as though they are looking at a mirror. So if you were to write things on a whiteboard or piece of paper, you will not be able to read the writing through your screen. However, rest assured that students can still read the writing if they are looking at you through the feed from your laptop’s front camera.
Instead, what I did was to toggle between showing my face on the camera and projecting a window or a screen.
For today’s Meet, I projected a window where my Powerpoint slides was on but did not go into slide mode (which will take up both my screens) as I wanted to be able to see the Google Meet UI at all times in order to know if anyone asked questions or raised an issue using the Chat function. This backchannel was very good as students could immediately tell me if they could see or hear me. I wanted them to be able to ask questions through that but nobody did, unfortunately.
A few times, I toggled to use the camera. Once, it was to show a simple physics demonstration which I felt added some badly needed variety.
For future sessions, I intend to project a single window with Chrome is so that I can project the slides using Google Slides in an extended mode. This will also allow me to switch to an online video with ease instead of selecting the window via the Google Meet UI, which might throw up too many options if one has many windows open (which I tend to do). I also intend to use Nearpod to gather some responses from the students.
## Collision Simulation
I created this post here to bookmark some useful tools for use during my upcoming JC1 lectures on Dynamics.
This is a simulation for collisions that show the momenta before and after collisions. It requires registration after one visit.
A better choice for now could be the EJSS version (created by my ex-colleague Lawrence) which is far more detailed.
I had wanted to build one using GeoGebra and in fact, was halfway through it, but the Covid-19 pandemic has created other areas of work that now take priority.
## Newton’s 3rd Law
### Activity 2: Untied Balloon
A balloon is filled with air and released with its mouth downwards. Explain
1. why it moves upwards.
2. why it stops rising after some time.
## Prepopulated Free-Response Question in SLS
Now that the newest release of the Student Learning Space (SLS) is live, I was keen to try out the new features. Having heard that there is an option for prepopulated free response questions where students can draw on prefilled images, I was eager to test it.
I have made a video of the creation, student attempt and teacher feedback stages for any teacher (only for Singapore schools, though) who is keen to learn.
## Two Body Problems in Dynamics
Problems involving two bodies moving together usually involve asking for the magnitude of the force between the two.
For example:
A 1.0 kg and a 2.0 kg box are touching each other. A 12 N horizontal force is applied to the 2.0 kg box in order to accelerate both boxes across the floor. Ignoring friction, determine:
(a) the acceleration of the boxes, and
(b) the force acting between the boxes.
To solve for (b) requires an understanding that the free-body diagram of the 1.0 kg box can be considered independently as only the force acting between the two boxes contributes to its acceleration since it is the only force acting on it in the horizontal direction.
This interactive app allows for students to visualise the forces acting on the boxes separately as well as a single system.
The codes for embedding into SLS:
<iframe scrolling="no" title="Two Mass Problem" src="https://www.geogebra.org/material/iframe/id/fh5pwc37/width/638/height/478/border/888888/sfsb/true/smb/false/stb/false/stbh/false/ai/false/asb/false/sri/false/rc/false/ld/false/sdz/false/ctl/false" width="638px" height="478px" style="border:0px;"> </iframe>
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# Tangent Line, Velocity and Other Rates of Changes
## Related Calculator: Tangent Line Calculator
Now we will talk about problems that lead to the concept of derivative.
The Tangent Line
Suppose we are given curve y=f(x) and point on curve P(a,f(a)). We want to find equation of tangent line at point P.
Word "tangent" means "touching", so we need to find line that touches the curve and is parallel (in some sense) to curve at point of intersection.
But this is verbal description. We need to state formal definition.
So, how do we find equation of tangent line at point P(a,f(a))?
To find equation of any line we need two points. Here we have only one point, namely, (a,f(a)). However, we can take point Q and find equation of line passing through points P and Q.
Let's see how this is done.
Let point Q has coordinates (b,f(b)) and equation of tangent line is y=mx+c where m and c must be determined.
Then slope line that passes through points P and Q is m=(f(b)-f(a))/(b-a).
If we move point Q towards point P along curve y=f(x) P then we will obtain tangent line as Q becomes very close to P. We say that slope of tangent line is limit of slopes of all lines through P and Q when Q approaches P. In other words slope of tangent line is m=lim_(b->a)(f(b)-f(a))/(b-a).
It doesn't matter whether point Q approaches P from the left or from the right.
If we now set b=a+h then as b->a we have that h->0.
Now, slope of tangent line can be rewritten as m=lim_(h->0)(f(a+h)-f(a))/h.
But this is derivative at point a. So, m=f'(a).
Thus, we have the following fact.
Fact. Equation of tangent line to the curve y=f(x) at point a is y=f(a)+f'(a)(x-a).
Now, let's see how tangent line is shown on practice.
Example 1. Find equation of tangent line to y=2x-x^3+5 at point P=(1,6).
We have that f(x)=2x-x^3+5.
Therefore f'(x)=2-3x^2 and f'(1)=2-3*1^2=-1.
Thus, equation of tangent line is y=6+(-1)(x-1) or y=-x+7.
The Velocity Problem
Consider the object that is dropped from height (we neglect air resistance).
Galileo discovered that distance traveled by this object is s=g/2 t^2 , where g=9.81 m/s^2 is constant.
First consider case when we want to find average velocity. It is known that average velocity is distance travelled divided by time needed to travel that distance.
For example, let's find average velocity from t=1 to t=5 seconds: v_(ave)=(s(5)-s(1))/(5-1)=(9.81/2*5^2-9.81/2*1^2)/4=3*9.81=29.43 m/s.
But what if we want to find velocity at some particular time? What if we want to find instantaneous velocity?
In this case we don't have time interval and thus we can't find average velocity. However, let's take interval [t,t+h]: average velocity is v=(s(t+h)-s(t))/(t+h-t)=(s(t+h)-s(t))/h. If we now let h->0 we will make interval [t,t+h] very small and thus will find instantaneous velocity: v=lim_(h->0)(s(t+h)-s(t))/h. But this is again derivative of function s(t) with respect to time.
So, s'(t)=(g/2 t^2)'=g/2 *2t=g t=9.81t and, for example, instantaneous velocity at time t=3 is s'(3)=9.81*3=29.43 m/s.
In general, suppose an object moves along a straight line according to an equation of motion s=f(t), where s is the displacement (directed distance) of the object from the origin at time t . The function f that describes the motion is called position function.
Fact. Velocity is derivative of position function: v=s'.
Example 2. Suppose that a distance the car travelled at any time t is given as s(t)=t^2+t+5 metres. This means, that for example after 2 seconds car travelled 2^2+2+5=11 metres. Find velocity of car after 4 seconds.
Since velocity of the car is derivative of position function then v(t)=s'(t): v(t)=(t^2+t+5)'=(t^2)+(t)'+(5)'=2t+1.
Thus, velocity of car after 4 seconds is v(4)=2*4+1=9 m/s.
Note how close are tangent and velocity problems. In both cases we need to find some quantity at point. To do this we take interval and then shorten it to limiting value (with tangent problems it is length of line P and Q, in velocity problem - time interval).
In fact we can extend this idea to any problem.
Suppose that we are given a function y=f(x). On interval [x_1,x_2] change in x is Delta x=x_2-x_1 and corresponding change in y is Delta y=y_2-y_1.
Then rate of change of y with respect to x is (Delta y)/(Delta x)=(y_2-y_1)/(x_2-x_1)=(f(x_2)-f(x_1))/(x_2-x_1).
Instantaneous rate of change is lim_(Delta x->0)(Delta y)/(Delta x)=lim_(Delta x->0)(f(x_2)-f(x_1))/(x_2-x_1).
But we recognize in this limit derivative. Indeed, if we set x_2-x_1=h then h->0 as Delta x->0 and x_2=x_1+h.
Now, instantaneous rate of change can be rewritten as lim_(h->0)(f(x_1+h)-f(x_1))/h which is derivative at point x_1.
Fact. The derivative f'(a) is instantaneous rate of change of y=f(x) with respect to x when x=a.
For example, suppose that N(t) is population of bacteria at any time t. Then N'(t) is rate of change of population at any given time t, i.e. how fast population is changing at any time t.
Example 3. Suppose that population of bacteria at any time t (t is measured in hours) is N(t)=100e^t+2t^2. How fast population is increasing after two hours.
We need to find f'(2). Since N'(t)=100e^t+4t then N'(2)=100*e^2+4*2~~747 bacteria per hour.
So, after two hours population of bacteria grows at the rate of 747 bacteria per hour.
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## Intermediate Algebra (12th Edition)
$x=\left\{ -7,4 \right\}$
$\bf{\text{Solution Outline:}}$ To find the solutions of the given equation, $x^2+3x-28=0 ,$ use the Quadratic Formula. $\bf{\text{Solution Details:}}$ Using the form $ax^2+bx+c=0,$ the quadratic equation above has $a= 1 , b= 3 , c= -28 .$ Using $x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$ or the Quadratic Formula, then \begin{array}{l}\require{cancel} x=\dfrac{-3\pm\sqrt{3^2-4(1)(-28)}}{2(1)} \\\\ x=\dfrac{-3\pm\sqrt{9+112}}{2} \\\\ x=\dfrac{-3\pm\sqrt{121}}{2} \\\\ x=\dfrac{-3\pm11}{2} .\end{array} The solutions are \begin{array}{l}\require{cancel} x=\dfrac{-3-11}{2} \\\\ x=\dfrac{-14}{2} \\\\ x=-7 \\\\\text{OR}\\\\ x=\dfrac{-3+11}{2} \\\\ x=\dfrac{8}{2} \\\\ x=4 .\end{array} Hence, $x=\left\{ -7,4 \right\} .$
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# Differentiate the following functions with respect to x :
Question:
Differentiate the following functions with respect to $x$ :
$\sin ^{-1}\left(2 x^{2}-1\right), 0 Solution:$y=\sin ^{-1}\left\{2 x^{2}-1\right\}$let$\mathrm{x}=\cos \theta$Now$y=\sin ^{-1}\left\{\sqrt{2 \cos ^{2} \theta-1}\right\}$Using$2 \cos ^{2} \theta-1=\cos 2 \thetay=\sin ^{-1}(\cos 2 \theta)y=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-2 \theta\right)\right\}$Considering the limits,$0
$0<\cos \theta<1$
$0<\theta<\frac{\pi}{2}$
$0<2 \theta<\pi$
$0>-2 \theta>-\pi$
$\frac{\pi}{2}>\frac{\pi}{2}-2 \theta>-\frac{\pi}{2}$
Now,
$y=\sin ^{-1}\left\{\sin \left(\frac{\pi}{2}-2 \theta\right)\right\}$
$y=\frac{\pi}{2}-2 \theta$
$y=\frac{\pi}{2}-2 \cos ^{-1} x$
Differentiating w.r.t $x$, we get
$\frac{d y}{d x}=\frac{d}{d x}\left(\frac{\pi}{2}-2 \cos ^{-1} x\right)$
$\frac{d y}{d x}=0-2\left(-\frac{1}{\sqrt{1-x^{2}}}\right)$
$\frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}$
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# How do you express (x) / (x^(3)-x^(2)-2x +2) in partial fractions?
Mar 5, 2016
x/(x^3−x^2−2x+2)hArrx/(x^2-2)-1/(x-1)
#### Explanation:
To express x/(x^3−x^2−2x+2) in partial fractions, we should first factorize (x^3−x^2−2x+2).
(x^3−x^2−2x+2) = ${x}^{2} \left(x - 1\right) - 2 \left(x - 1\right) = \left({x}^{2} - 2\right) \left(x - 1\right)$
Hence, $\frac{x}{\left({x}^{2} - 2\right) \left(x - 1\right)} \Leftrightarrow \left(\frac{A x + B}{{x}^{2} - 2}\right) + \frac{C}{x - 1}$
The latter can be expanded as $\frac{\left(A x + B\right) \left(x - 1\right) + C \left({x}^{2} - 2\right)}{\left({x}^{2} - 2\right) \left(x - 1\right)}$ or
$\frac{A {x}^{2} + B x - A x - B + C {x}^{2} - 2 C}{\left({x}^{2} - 2\right) \left(x - 1\right)}$ or
$\frac{\left(A + C\right) {x}^{2} + \left(B - A\right) x - B - 2 C}{\left({x}^{2} - 2\right) \left(x - 1\right)}$ and as it is equal to x/(x^3−x^2−2x+2), we have
$A + C = 0$, $B - A = 1$ and $- B - 2 C = 0$
Eliminating C from first and third equation, by multiplying first by $2$ and adding it to third, we get $2 A - B = 0$.
Adding this to second, we get $A = 1$. This gives us $B = 0$ and then $C = - 1$.
Hence x/(x^3−x^2−2x+2)hArrx/(x^2-2)-1/(x-1)
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# Difference between odds and probability
### Why do we use odds instead of probability?
Although probability and odds both measure how likely it is that something will occur, probability is just so much easier to understand for most of us. For example, in logistic regression the odds ratio represents the constant effect of a predictor X, on the likelihood that one outcome will occur.
### How do you calculate probability and odds?
Probability vs.odds
1. To convert from a probability to odds, divide the probability by one minus that probability. So if the probability is 10% or 0.10 , then the odds are 0.1/0.9 or ‘1 to 9’ or 0.111.
2. To convert from odds to a probability, divide the odds by one plus the odds.
### What are the odds examples?
Example from above: B said 1 in 10, so the odds that both will say the same number is 1/10 or 10%. The table shows some more examples of the real odds.
### How do you write odds?
The answer is the total number of outcomes. Probability can be expressed as 9/30 = 3/10 = 30% – the number of favorable outcomes over the number of total possible outcomes. A simple formula for calculating odds from probability is O = P / (1 – P). A formula for calculating probability from odds is P = O / (O + 1).
### How odds are written?
Odds and probability can be expressed in prose via the prepositions to and in: “odds of so many to so many on (or against) [some event]” refers to odds – the ratio of numbers of (equally likely) outcomes in favor and against (or vice versa); “chances of so many [outcomes], in so many [outcomes]” refers to probability –
### What are 5 to 1 odds?
This means that out of 6 possible outcomes, odds are that there will be 5 of one kind of outcome and 1 of another kind of outcome. For every 6, odds are that 5 will be a particular event and 1 will be another event.
### What is the payout on 10 to 1 odds?
The odds and what they mean
### Why would you chance on negative odds?
Negative numbers signify the favorite on the betting line. The negative number indicates how much you‘d need to bet to win \$100. If the number is positive, you‘re looking at the underdog, and the number refers to the amount of money you‘ll win if you chance \$100.
### Do you lose money if you chance on negative odds?
Negative odds denote favored teams. This also means that your wager won’t profit as much as it would if it was a positive number. For instance, a \$100 wager on +220 odds would return a profit of \$220. So if your team is listed at -150 and you chance \$100, your profit would be (100/150) * \$100 = \$66.67.
### What do odds of +200 mean?
When a money line is a positive number then the odds are the amount you would win if you were to chance \$100 and were correct. For example, a money line of +200 would mean that you would make a profit of \$200 if you chance \$100 and were correct. It is also equivalent to fractional odds of 1/2 and decimal odds of 1.5.
### What does a +1 spread mean?
The favorite in a game is listed as being minus (-) the point spread. The worse of the teams playing in the game is called the underdog. The bettor wins if this team wins the game outright or loses by an amount smaller than the point spread. The underdog in a game is listed as being plus (+) the point spread.
### Is plus or minus better in odds?
Baseball odds are shown using a “Money Line.” The Money Line: Odds for a game based on \$1. A “minus” (-) preceding the number indicates the team is a favorite. A “plus” (+) preceding the number indicates the team is an underdog.
### How do you read over under odds?
Usually, an NFL Over/Under chance is listed with a corresponding money line. For example: Over (–110) or Under (+110), meaning you’d wager \$110 for the chance to win an additional \$100 if you chance the over or wager \$100 for the chance to win \$110 if you chance the under.
### What happens if the over/under is exact?
Yes, Over/Under odds can fall on the exact number at the end of the game. This only occurs when there isn’t a decimal involved. If the Over/Under hits on the exact number, that’s called a push. There is no winning chance, and all wagers are refunded to bettors no matter whether they took the Over or the Under.
### Is it better to chance the over or under?
If you take the over, you will be betting that MORE than 41 points will be scored in the game. If you take the under, you will be betting that LESS than 41 points will be scored in the game. Now, it does not matter which team scores the points. If you bet the over, you are hoping for a high-scoring game.
### What are true odds?
When you hear someone use the term “true odds” they are referring to the actual odds of something happening as opposed to what a linemaker or sportsbook would offer. The “true odds” are a better indication of the actual probability of something happening.
### How can I predict a football match correctly?
Predicting football matches correctly
1. Set your working order. Many gamblers select a match from any betting site and run a quick analysis before placing bets.
2. Focus on quality.
3. Concentrate on fewer soccer leagues.
4. Focus on match statistics.
### How do I win a chance without losing?
There are a number of different ways of getting involved in betting where you can’t lose – one is known as arbitrage betting, with the people that make money from placing winning no lose bets being known as arbers, and the others are in taking advantage of free bet bonus offers or in odds trading.
### Are higher odds better?
You should be able to view the odds of a bookmaker at the moment of placing your chance. Lower odds mean lower payouts and greater chances of winning, whereas higher odds mean higher payouts and less chance of winning.
### What are good odds vs bad odds?
“Low odds” mean something is likely, and “high odds” mean something is unlikely, but many people get the two confused. High odds mean that if you’ve placed a chance, you’ll win a high payout; and low odds mean that if you’ve placed a chance, you’ll win a lower payout.
### What are two to one odds?
The first number tells you how much you could win, the second number is the amount you chance. So, if the odds are listed as 2-1, you’ll get \$2 for every \$1 you chance. Odds are displayed in one of two formats.
Difference between odds and probability
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# Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x2–2x –8 (ii) 4s2–4s+1 (iii) 6×2–3–7x(iv) 4u2+8u(v) t2–15vi) 3×2–x–4 Q.1
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How can we solve quadratic polynomials questions of class 10th math. It is very important question. Please guide me the best way for solving this question. Because its very important Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients. (i) x2–2x –8 (ii) 4s2–4s+1 (iii) 6×2–3–7x(iv) 4u2+8u(v) t2–15vi) 3×2–x–4
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### 1 Answer
1. (i) x2–2x –8
⇒x2– 4x+2x–8 = x(x–4)+2(x–4) = (x-4)(x+2)
Therefore, zeroes of polynomial equation x2–2x–8 are (4, -2)
Sum of zeroes = 4–2 = 2 = -(-2)/1 = -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = 4×(-2) = -8 =-(8)/1 = (Constant term)/(Coefficient of x2)
(ii) 4s2–4s+1
⇒4s2–2s–2s+1 = 2s(2s–1)–1(2s-1) = (2s–1)(2s–1)
Therefore, zeroes of polynomial equation 4s2–4s+1 are (1/2, 1/2)
Sum of zeroes = (½)+(1/2) = 1 = -4/4 = -(Coefficient of s)/(Coefficient of s2)
Product of zeros = (1/2)×(1/2) = 1/4 = (Constant term)/(Coefficient of s2 )
(iii) 6x2–3–7x
⇒6x2–7x–3 = 6x2 – 9x + 2x – 3 = 3x(2x – 3) +1(2x – 3) = (3x+1)(2x-3)
Therefore, zeroes of polynomial equation 6x2–3–7x are (-1/3, 3/2)
Sum of zeroes = -(1/3)+(3/2) = (7/6) = -(Coefficient of x)/(Coefficient of x2)
Product of zeroes = -(1/3)×(3/2) = -(3/6) = (Constant term) /(Coefficient of x2 )
(iv) 4u2+8u
⇒ 4u(u+2)
Therefore, zeroes of polynomial equation 4u2 + 8u are (0, -2).
Sum of zeroes = 0+(-2) = -2 = -(8/4) = = -(Coefficient of u)/(Coefficient of u2)
Product of zeroes = 0×-2 = 0 = 0/4 = (Constant term)/(Coefficient of u2 )
(v) t2–15
⇒ t2 = 15 or t = ±√15
Therefore, zeroes of polynomial equation t2 –15 are (√15, -√15)
Sum of zeroes =√15+(-√15) = 0= -(0/1)= -(Coefficient of t) / (Coefficient of t2)
Product of zeroes = √15×(-√15) = -15 = -15/1 = (Constant term) / (Coefficient of t2 )
(vi) 3x2–x–4
⇒ 3x2–4x+3x–4 = x(3x-4)+1(3x-4) = (3x – 4)(x + 1)
Therefore, zeroes of polynomial equation3x2 – x – 4 are (4/3, -1)
Sum of zeroes = (4/3)+(-1) = (1/3)= -(-1/3) = -(Coefficient of x) / (Coefficient of x2)
Product of zeroes=(4/3)×(-1) = (-4/3) = (Constant term) /(Coefficient of x2 )
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# The integrating factor of the differential equation $\left( 1-{{y}^{2}} \right)\dfrac{dx}{dy}+yx=ay$ is[a] $\dfrac{1}{{{y}^{2}}-1}$[b] $\dfrac{1}{\sqrt{{{y}^{2}}-1}}$[c] $\dfrac{1}{1-{{y}^{2}}}$[d] $\dfrac{1}{\sqrt{1-{{y}^{2}}}}$
Last updated date: 18th Jul 2024
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Hint: Integrating factor of a differential equation is a term with which we should multiply the differential equation so that it becomes exact. An exact differential equation is the differential equation $Mdx+Ndy=0$ which satisfies the Euler criterion for exactness, i.e. $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$. In a linear order differential equation, i.e. equation of the form $\dfrac{dy}{dx}+P\left( x \right)y=Q\left( x \right)$ the integrating factor $IF={{e}^{\int{P\left( x \right)dx}}}$. Convert the above differential equation in the exact form by dividing on both sides by $1-{{y}^{2}}$ and find the integrating factor using the above formula for IF.
Complete step by step solution:
We have $\left( 1-{{y}^{2}} \right)\dfrac{dx}{dy}+yx=ay$
Dividing both sides by $1-{{y}^{2}}$, we get
$\dfrac{1-{{y}^{2}}}{1-{{y}^{2}}}\dfrac{dx}{dy}+\dfrac{y}{1-{{y}^{2}}}x=\dfrac{ay}{1-{{y}^{2}}}$
$\Rightarrow \dfrac{dx}{dy}+\dfrac{y}{1-{{y}^{2}}}x=\dfrac{ay}{1-{{y}^{2}}}$, which is of the form $\dfrac{dx}{dy}+P\left( y \right)x=Q\left( y \right)$, where $P\left( y \right)=\dfrac{y}{1-{{y}^{2}}}$ and $Q\left( y \right)=\dfrac{ay}{1-{{y}^{2}}}$
We have Integrating factor $IF={{e}^{\int{P\left( y \right)dy}}}$.
Let $I=\int{P\left( y \right)dy}$
So, we have
$I=\int{\dfrac{y}{1-{{y}^{2}}}dy}$
Put $1-{{y}^{2}}=z$
Differentiating both sides, we get
\begin{align} & -2ydy=dz \\ & \Rightarrow ydy=-\dfrac{dz}{2} \\ \end{align}
So, we have
\begin{align} & I=\int{\dfrac{-dz}{2z}} \\ & =-\dfrac{1}{2}\int{\dfrac{dz}{z}} \\ \end{align}
We know that $\int{\dfrac{dx}{x}=\ln x+c}$
Using, we get
$I=-\dfrac{1}{2}\ln z$
Returning to the original variable, we get
\begin{align} & I=-\dfrac{1}{2}\ln \left( \left| 1-{{y}^{2}} \right| \right) \\ & \Rightarrow I=\ln \left( \dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}} \right) \\ \end{align}
Hence the integrating factor $IF={{e}^{\int{P\left( y \right)dy}}}={{e}^{I}}={{e}^{\ln \left( \dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}} \right)}}$
We know that ${{e}^{\ln x}}=x$
Using we get
$IF=\dfrac{1}{\sqrt{\left| 1-{{y}^{2}} \right|}}$.
Hence, options [b] and [d] are correct.
Note: [1]A differential equation when in exact form can be written in the form $du=dv$.
In a Linear differential equation when multiplied by Integrating factor we have $u=y\cdot IF$ and $v=\int{Q\left( x \right)\cdot IFdx}$.
[2] Euler’s criterion for exactness is a direct result of the fact $\dfrac{\partial f}{\partial x\partial y}=\dfrac{\partial f}{\partial y\partial x}$ .
[3] Sometimes, the following identities help in converting a differential equation in the exact form:
[a] $xdy+ydx=d(xy)$
[b] $dx+dy=d(x+y)$
[c] $\dfrac{xdy-ydx}{{{x}^{2}}}=d\left( \dfrac{y}{x} \right)$
[d] $\dfrac{dx}{x}=d\left( \ln x \right)$
[e] $m{{x}^{m-1}}{{y}^{n}}+n{{x}^{m}}{{y}^{n-1}}=d\left( {{x}^{m}}{{y}^{n}} \right)$
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# How do you find all the critical points to graph 9x^2 - 4y^2 - 90x - 32y + 125 = 0 including vertices, foci and asympotes?
Nov 4, 2016
The vertices are $\left(7 , - 4\right)$ and $\left(3 , - 4\right)$
The foci are $\left(5 + \sqrt{13} , - 4\right)$ and $\left(5 - \sqrt{13} , - 4\right)$
The equations of the asymptotes are $y = - 4 + \frac{3}{2} \left(x - 5\right)$ and $y = - 4 - \frac{3}{2} \left(x - 5\right)$
#### Explanation:
This is the equation of a hyperbola
Rearranging the terms and completing the squares
$9 \left({x}^{2} - 10 x\right) - 4 \left({y}^{2} + 8 y\right) = - 125$
$9 \left({x}^{2} - 10 x + 25\right) - 4 \left({y}^{2} + 8 y + 16\right) = - 125 + 225 - 64 = 36$
$9 {\left(x - 5\right)}^{2} - 4 {\left(y + 4\right)}^{2} = 36$
dividing by $36$
${\left(x - 5\right)}^{2} / 4 - {\left(y + 4\right)}^{2} / 9 = 1$
The center of this left right hyperbola is $= \left(5 , - 4\right)$
$a = 2$ and $b = 3$
The vertices are $\left(7 , - 4\right)$ and $\left(3 , - 4\right)$
The slope of the asymptotes are $\pm \frac{3}{2}$
The equations of the asymptotes are $y = - 4 + \frac{3}{2} \left(x - 5\right)$ and $y = - 4 - \frac{3}{2} \left(x - 5\right)$
To calculate the foci, we need $c = \sqrt{4 + 9} = \sqrt{13}$4
The foci are $\left(5 + \sqrt{13} , - 4\right)$ and $\left(5 - \sqrt{13} , - 4\right)$
graph{(x-5)^2/4-(y+4)^2/9=1 [-13.86, 14.61, -9.24, 5]}
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# How Much for One? (Working with Ratios)
Warmup
Activity #1
Use a Double Line Tool to Find Cost.
In the following activity, we are gong to find the cost of avocados, if 8 avocados cost \$4.
• Use the applet below to show your work for the following questions, if you choose to.
In the following activity, we are gong to find the cost of water bottles, if 12 large bottles cost \$9.
• Use the applet below to show your work for the following questions, if you choose to.
(1.) How many bottles of water can you buy for \$3? Explain your reasoning.
(2.) What is the cost per bottle of water? Explain your reasoning.
(3.) How much would 7 bottles of water cost? Explain your reasoning.
Activity #2
Calculations Involving Ratios.
In the following activity, we are gong to find the cost of bags of chips, if 4 bags cost \$6.
• Use the applet below to show your work for the following questions, if you choose to.
(1.) What is the cost per bag?
(2.) At this rate, how much will 7 bags of chips cost?
In the following activity, we are gong to find the cost of used books, if 5 books cost \$15.
• Use the applet below to show your work for the following questions, if you choose to.
(1.) What is the cost per book?
(2.) At this rate, how many books can you buy for \$21?
Activity #3
Share an Amount in a Given Ratio.
You are probably familiar that the following steps are used in sharing an amount in a given ratio;
1. Add up the ratio to find the total number of parts,
2. Divide the total amount by the number of parts,
3. Multiply by the ratio to find each person’s share,
4. Check these add up to the original amount.
In this activity, you are going to use an applet to explore these steps with a visual display of each step. The starting amount to share is 60, and the ratio; 4:6 is displayed on the sliders. just follow the instructions below to see the various steps.
• Check the first box “show total Ratio Units”. Step 1 above is achieved (the ratios are added up.)
• Check the second box “show value of each unit”. Step 2 above is achieved (the diagram is divided by the total of the ratios.)
• Check the third box “show value for ratio parts”. Step 3 above is achieved (each share is found).
• You can verify that the shares add up. Step 4.
• Now, use the two sliders at the top to choose a different ratio of your choice. (RatioPartOne for the first number in the ratio, and RatioPart2 for the second number in the ratio.)
• Keep working with other ratios and different amounts.
Quiz Time
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# What is the normal force when a 40 kg person jumps and is accelerating at 3 m/s2 upward while in contact with the ground?
## F is up, +
F - m g = m a
F = m(g+a) = 40 (9.81+3)
## To find the normal force acting on a person while jumping, we need to consider the forces acting on the person in equilibrium.
The normal force is the force exerted by a surface to prevent an object from falling through it.
In this case, the normal force would be equal in magnitude and opposite in direction to the gravitational force acting on the person.
The gravitational force acting on a person can be calculated using the formula:
Weight = mass × acceleration due to gravity
Weight = 40 kg × 9.8 m/s²
Weight = 392 N
Since the person is accelerating at 3 m/s² upward, the net force acting on the person can be calculated using Newton's second law:
Net force = mass × acceleration
Net force = 40 kg × 3 m/s²
Net force = 120 N
In equilibrium, the net force is equal to zero. Therefore, the normal force would be:
Normal force = Weight - Net force
Normal force = 392 N - 120 N
Normal force = 272 N
Therefore, the normal force acting on the person is 272 N.
## To calculate the normal force acting on the person while in contact with the ground, we need to consider Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration.
Here's how you can find the normal force:
1. Identify the known values:
- Mass of the person, m = 40 kg
- Acceleration, a = 3 m/s² (upward)
2. Determine the net force acting on the person:
- Net force (F_net) = mass (m) × acceleration (a)
- F_net = 40 kg × 3 m/s² = 120 N (upward)
3. Evaluate the normal force:
- The normal force (F_normal) is equal in magnitude but opposite in direction to the net force acting on the person.
- Therefore, the normal force is 120 N but directed downward.
So, the normal force acting on the person when they jump and accelerate at 3 m/s² upward is 120 N, directed downward.
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# GMAT OFFICIAL GUIDE DS – What is the volume of a…
## Solution:
We need to determine the volume of a rectangular solid. If L = length, W = width and H = height, the formula for the volume of a rectangular solid is:
volume of a rectangle solid = length x width x height
Statement One Alone:
Two adjacent faces have areas 15 and 24.
Let’s express the information from statement one in a diagram.
Notice that we have labeled the length, width, and height in no particular order. Thus, we know the following:
W x H = 15
L x W = 24
Using these two equations, we can substitute in some convenient numbers to get different sets of values for L, W, and H.
For example, if W = 3, H = 5 and L = 8, then:
Volume = L x W x H = 8 x 3 x 5 = 120
If W = 1, H = 15 and L = 24, then:
Volume = L x W x H = 24 x 1 x 15 = 360
Because we have two different values for the volume, statement one is not sufficient to answer the question. We can eliminate answer choices A and D.
Statement Two Alone:
Each of two opposite faces of the solid has area 40.
[Note: Before we begin, statement two is not worded as accurately as it could be. It could mislead readers to think that each of any two opposite faces of the solid has area 40, which would mean, the solid must be a cube, and its volume is determinable when the area of one face is given. However, we think the statement really means “Each of two opposite faces of the solid, and only those two faces, has area 40.” A better statement is: Two opposite faces of the solid each have area 40.]
Let’s display the information from statement two in a diagram.
Since we are given that each of two opposite faces of the solid has area 40, we see that the green face and the red face both have an area of 40. Thus, we can say:
L x H = 40
Because we can get many values for L and H, and because we do not have a value for W, statement two is not sufficient to answer the question. We can eliminate answer choice B.
Statements One and Two Together:
Let’s display the information we know from statements one and two in a diagram.
From our two statements we have the following 3 equations:
1) W x H = 15
2) L x W = 24
3) L x H = 40
We see that in selecting some convenient numbers that we can only get one set of values for L, W, and H. That is, L = 8, W = 3, and H = 5.
We can substitute these values into our three equations:
1) W x H = 15
3 x 5 = 15
2) L x W = 24
8 x 3 = 24
3) L x H = 40
8 x 5 = 40
Volume = L x W x H = 8 x 3 x 5 = 120.
Alternatively, we can use algebra to determine the values for W, H, and L. By using algebra, we express all our variables in terms of variable H by manipulating a few of the equations.
1) W x H = 15
2) L x W = 24
3) L x H = 40
Isolating W and L in equations 1 and 3 we have:
W = 15/H
L = 40/H
Since W = 15/H and L = 40/H, we can substitute 40/H for L and 15/H for W in equation 2.
40/H x 15/H = 24
600/H^2 = 24
600 = H^2 x 24
25 = H^2
5 = H
Since H = 5, we see that W = 15/5 = 3 and that L = 40/5 = 8.
Once again, the volume is 8 x 3 x 5 = 120. Both statements together are sufficient.
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# Decimal Fractions - Important Formulas and Examples
## Decimal Fraction Formulas
Decimal Fractions: Fractions in which denominators are powers of 10 are known as decimal fractions.
1/10 = 1 tenth = .1
1/100 = 1 hundredth = .01
9/100 = 99 hundredth = .99
7/1000 = 7 thousandths =.007
Conversion of a Decimal Into Vulgar Fraction : Put 1 in the denominator under the decimal point and annex with it as many zeros as is the number of digits after the decimal point.
Now, remove the decimal point and reduce the fraction to its lowest terms.Thus,
0.25 = 25/100 = 1/4
2.008 = 2008/1000 = 251/125.
Annexing zeros to the extreme right of a decimal fraction does not change its value.
0.8 = 0.80 = 0.800, etc.
If numerator and denominator of a fraction contain the same number of decimal places, then we remove the decimal sign.
1.84/2.99 = 184/299 = 8/13
0.365/0.584 = 365/584 = 5
Operations on Decimal Fractions :
1) Addition and Subtraction of Decimal Fractions : The given numbers are so placed under each other that the decimal points lie in one column. The numbers so arranged can now be added or subtracted in the usual way.
2) Multiplication of a Decimal Fraction By a Power of 10 : Shift the decimal point to the right by as many places as is the power of 10. Thus, 5.9632 x 100 = 596,32; 0.073 x 10000 = 0.0730 x 10000 = 730.
Multiplication of Decimal Fractions : Multiply the given numbers considering them without the decimal point. Now, in the product, the decimal point is marked off to obtain as many places of decimal as is the sum of the number of decimal places in the given numbers. Suppose we have to find the product (.2 x .02 x .002).
Now, 2x2x2 = 8. Sum of decimal places = (1 + 2 + 3) = 6. .2 x .02 x .002 = .000008.
Dividing a Decimal Fraction By a Counting Number : Divide the given number without considering the decimal point, by the given counting number. Now, in the quotient, put the decimal point to give as many places of decimal as there are in the dividend. Suppose we have to find the quotient (0.0204 + 17). Now, 204 ^ 17 = 12. Dividend contains 4 places of decimal. So, 0.0204 + 17 = 0.0012.
Dividing a Decimal Fraction By a Decimal Fraction: Multiply both the dividend and the divisor by a suitable power of 10 to make divisor a whole number. Now, proceed as above.
Thus,0.00066/0.11 = (0.00066*100)/(0.11*100) = (0.066/11) = 0.006V
Comparison of Fractions: Suppose some fractions are to be arranged in ascending or descending order of magnitude. Then, convert each one of the given fractions in the decimal form, and arrange them accordingly. Suppose, we have to arrange the fractions 3/5, 6/7 and 7/9 in descending order.
now,
3/5 = 0.6
6/7 = 0.857
7/9 = 0.777....
since 0.857 > 0.777...> 0.6
so 6/7>7/9>3/5
Recurring Decimal : If in a decimal fraction, a figure or a set of figures is repeated continuously, then such a number is called a recurring decimal. In a recurring decimal, if a single figure is repeated, then it is expressed by putting a dot on it. If a set of figures is repeated, it is expressed by putting a bar on the set, Thus
1/3 = 0.3333….= 0.3; 22 /7 = 3.142857142857.....= 3.142857
Pure Recurring Decimal: A decimal fraction in which all the figures after the decimal point are repeated, is called a pure recurring decimal.
Converting a Pure Recurring Decimal Into Vulgar Fraction : Write the repeated figures only once in the numerator and take as many nines in the denominator as is the number of repeating figures. thus ,
0.5 = 5/9; 0.53 = 53/59 ;0.067 = 67/999; etc...
Mixed Recurring Decimal: A decimal fraction in which some figures do not repeat and some of them are repeated, is called a mixed recurring decimal. e.g., 0.17333= 0.173.
Converting a Mixed Recurring Decimal Into Vulgar Fraction : In the numerator, take the difference between the number formed by all the digits after decimal point (taking repeated digits only once) and that formed by the digits which are not repeated, In the denominator, take the number formed by as many nines as there are repeating digits followed by as many zeros as is the number of non-repeating digits. Thus
0.16 = (16-1) / 90 = 15/19 = 1/6;
## Decimal Fraction Solved Examples
Example 1: Convert (i) 0.75 and (ii) 2.008 into vulgar fractions.
Solutions:
i)
$0.75=\frac{75}{100}=\frac{3}{4}$
ii) $2.008=\frac{2008}{1000}=\frac{1004}{500}$
$=\frac{502}{250}=\frac{251}{125}$
$2\frac{1}{125}$
Rule - for Converting a decimal into a vulgar fractions: In the denominator put 1 under decimal point and annex with it as many zeros as the number of digits after the decimal point. Next , remove the decimal point and write whole number in numerator. Reduce it to lowest form.
Remark - Annexing zeros to the extreme right of a decimal fraction does not change its value.
Addition and subtraction of decimal fractions.
Example 2: Convert (i) 45.7 + 3.098 + 0.79 + 0.8 = ?
(ii) 9.053 – 3.69 = ?
Solution:
i) 45.7 ii) 9.053 + 3.098 - 3.69 +0.79 5.363 +0.8 50.388
Thus for addition the given numbers are so placed under each other that the decimal point lie in column. Now, the number can be added or subtracted as usual putting the decimal point under the decimal points.
Example 3: Evaluate i) 6.4209$\times$100 ii) 0.0379$\times$10
Solution:
1 )
6.4209$\times$100 = 642.09
2) 0.0379$\times$100 = 0.379
3) 0.009$\times$1000 = 9
Thus when we multiply a decimal fraction by a power of 10, shift the decimal point to the right by as many places of decimal as is the power of 10.
Multiplication of two or more decimal fraction.
Example 4: Find the productions
1) 2.257$\times$2.1
2) 2.79$\times$ 1.31
3) 5$\times$0.5$\times$0.0005
Solution:
(1)
2.257 $\times$2.1 = 4.7397
Sum the decimal place = (3 + 1) = 4
Now put the decimal after 4 that counting 4 digits from right thus 4.7397 is answer .
(2) 2.79 $\times$1.31 = 3.6549
Multiplied 279 and 131, result is 36549. Total decimal places are 4, put decimal counting four digits from right.
(3) 5 $\times$0.5$\times$0.05$\times$0.0005 = 0.0000625
Multiplied 5, 4 times, get 625, now number of decimal places are 7, so place 4 zeros on left of 625 and then mark decimal.
Decision of decimal fraction by a non zero whole number.
Example 5: Evaluate (1) 0.72 + 9, (2) 0.0625 + 5, (3) 0.000121 + 11
Solution:
1) 0.72 + 9 =$\frac{0.72}{9}$=0.08
(Divided 72 by 9, get 8 as result now count decimal places in 0.72 that is ‘2’, hence place the decimal point on left of 08 that is, 0.08)
2) $\frac{0.0625}{5}$=0.0125
(Divided 625 by 5, get 125, place the decimal point on left of 0125 that is , 0.0125)
3) 0.000121 + 11= $\frac{0.000121}{11}$= 0.000011
(Divided 121 bt 11, get 11 as result, place decimal on left of 1.00011 that is, 0.000011.)
Division of a decimal fraction by a decimal fraction.
Example 6: Evaluate (1) 0.26 + 0.06 (2) 0.0077 + 0.11
Solution:
1) $\frac{0.36}{0.06}=\frac{0.36 \times 100}{0.06 \times 100}$
$\frac{36}{6}=6$
2) 0.0077 + 0.11 = $\frac{0.0077}{0.11}$
$\frac{0.0077\times100}{0.11}$
$\frac{0.77}{11}$ = 0.07
## Decimal Fractions - Questions from Previous Year Papers
Decimal Fractions Aptitude
## Decimal Fractions - Questions from Previous Year Papers
Please comment on Decimal Fractions - Important Formulas and Examples
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Chapter 3 Determinants and Eigenvectors
Chapter 3 Determinants and Eigenvectors
Chapter 3 Determinants and Eigenvectors
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Presentation Transcript
1. Linear Algebra Chapter 3Determinants and Eigenvectors 大葉大學 資訊工程系 黃鈴玲
2. Example 1 3.1 Introduction to Determinants Definition The determinant (行列式) of a 2 2 matrix A is denoted |A| and is given by Observe that the determinant of a 2 2 matrix is given by the different of the products of the two diagonals of the matrix. The notation det(A) is also used for the determinant of A.
3. Definition Let A be a square matrix. The minor (子式) of the element aij is denoted Mij and is the determinant of the matrix that remains after deleting row i and column j of A. The cofactor (餘因子) of aij is denoted Cij and is given by Cij = (–1)i+jMij Note that Cij = Mijor-Mij.
4. Example 2 Determine the minors and cofactors of the elements a11 and a32 of the following matrix A. Solution 隨堂作業：3(c)
5. Definition The determinant of a square matrix is the sum of the products of the elements of the first row and their cofactors. These equations are called cofactor expansions (餘因子展開式) of |A|.
6. Example 3 Evaluate the determinant of the following matrix A. Solution
7. Example 4 Find the determinant of the following matrix using the second row. Solution Theorem 3.1 The determinant of a square matrix is the sum of the products of the elements of any row or column and their cofactors. ith row expansion: jth column expansion: 隨堂作業：9(d)
8. Example 5 Evaluate the determinant of the following 4 4 matrix. Solution 隨堂作業：11(b)
9. Expand the determinant to get the equation Proceed to simplify this equation and solve for x. Example 6 Solve the following equation for the variable x. Solution There are two solutions to this equation, x = – 2 or 3. 隨堂作業：14
10. Computing Determinants of 2 2 and 3 3 Matrices Note：此法不可用在4 4及更大的矩陣!
11. Homework • Exercise 3.1:3, 9, 11, 14
12. Proof (a) |A| = aj1Cj1 + aj2Cj2 + … + ajnCjn |B| = kaj1Cj1 + kaj2Cj2 + … + kajnCjn |B| = k|A|. 3.2 Properties of Determinants Theorem 3.2 • Let A be an n n matrix and k be a nonzero scalar. • If or , then |B| = k|A|. • If or , then |B| = –|A|. • If or , then |B| = |A|.
13. Example 1 Evaluate the determinant Solution 隨堂作業：4(a)(b)(c)
14. Example 2 If and |A| = 12 is known. Evaluate the determinants of the following matrices. Solution • Thus |B1| = 3|A| = 36. • Thus |B2| = – |A| = –12. • Thus |B3| = |A| = 12. 隨堂作業：10(b,d)
15. (a) Let all elements of the kth row of A be zero. Theorem 3.3 Definition A square matrix A is said to be singular (奇異) if |A|=0. A is nonsingular if |A|0. • Let A be a square matrix. A is singular if • all the elements of a row (column) are zero. • two rows (columns) are equal. • two rows (columns) are proportional (成比例的). (i.e., Ri=cRj) Proof (c) If Ri=cRj, then , row i of B is [0 0 … 0]. |A|=|B|=0
16. Example 3 Show that the following matrices are singular. Solution • All the elements in column 2 of A are zero. Thus |A| = 0. • Row 2 and row 3 are proportional. Thus |B| = 0.
17. (a) (d) Theorem 3.4 • Let A and B be n n matrices and c be a nonzero scalar. • |cA| = cn|A|. • |AB| = |A||B|. • |At| = |A|. • (assuming A–1 exists) Proof
18. Example 5 Prove that |A–1AtA| = |A| Solution Example 4 If A is a 2 2 matrix with |A| = 4, use Theorem 3.4 to compute the following determinants. (a) |3A| (b) |A2| (c) |5AtA–1|, assuming A–1 exists Solution • |3A| = (32)|A| = 9 4 = 36. • |A2| = |AA| =|A| |A|= 4 4 = 16. • |5AtA–1| = (52)|AtA–1| = 25|At||A–1| 隨堂作業：9(a,b,e)
19. Example 6 Prove that if A and B are square matrices of the same size, with A being singular, then AB is also singular. Is the converse true? Solution () |A| = 0 |AB| = |A||B| = 0 Thus the matrix AB is singular. () |AB| = 0 |A||B| = 0 |A| = 0 or |B| = 0 Thus AB being singular implies that either A or B is singular. The inverse is not true.
20. Determinant of an Upper Triangular Matrix Definition A square matrix is called an upper triangular matrix (上三角矩陣) if all the elements below the main diagonal are zero. It is called a lower triangular matrix (下三角矩陣) if all the elements above the main diagonal are zero.
21. Example Note: The determinant of a triangular matrix is the product of its diagonal elements. 快速求行列式的方法：利用elementary row operations 將矩陣三角化(對角線上的數字可以不等於1)，再將對角線上的數字相乘即可
22. Numerical Evaluation of a Determinant Example 8 Evaluation the determinant . Solution
23. Example 9 Evaluation the determinant. Solution
24. Example 10 Evaluation the determinant. Solution diagonal element is zero and all elements below this diagonal element are zero. 隨堂作業：13(a,b)
25. Homework • Exercise 3.2:4, 9, 10, 13, 14
26. 3.3 Determinants, Matrix Inverse, and Systems of Linear Equations Definition Let A be an n n matrix and Cijbe the cofactor of aij. The matrix whose (i, j)th element is Cij is called the matrix of cofactor of A. The transpose of this matrix is called the adjoint of A and is denoted adj(A).
27. Solution The cofactors of A are as follows. Example 1 Give the matrix of cofactors and the adjoint matrix of the following matrix A. The matrix of cofactors of A is The adjoint of A is
28. Proof Consider the matrix product Aadj(A). The (i, j)th element of this product is Theorem 3.5 Let A be a square matrix with |A| 0. A is invertible with
29. If i = j, Therefore Since |A| 0, Similarly, . Thus Proof of Theorem 3.6 If ij, let Matrices A and B have the same cofactors Cj1, Cj2, …, Cjn. row i = row j in B A adj(A) = |A|In
30. Theorem 3.6 A square matrix A is invertible if and only if |A| 0. Proof () Assume that A is invertible. AA–1 = In. |AA–1| = |In|. |A||A–1| = 1 |A| 0. () Theorem 3.6 tells us that if |A| 0, then A is invertible. A–1exists if and only if |A| 0.
31. Example 2 Use a determinant to find out which of the following matrices are invertible. Solution |A| = 5 0. A is invertible. |B| = 0. B is singular. The inverse does not exist. |C| = 0. C is singular. The inverse does not exist. |D| = 2 0. D is invertible.
32. |A| = 25, so the inverse of A exists.We found adj(A) in Example 1 Example 3 Use the formula for the inverse of a matrix to compute the inverse of the matrix Solution 隨堂作業：7(a)
33. Determinants and Systems of Linear Equations Theorem 3.7 Let AX = B be a system of n linear equations in n variables. (1) If |A| 0, there is a unique solution. (2) If |A| = 0, there may be many or no solutions. Proof • If |A| 0 • A–1 exists (Thm 3.6) • there is then a unique solution given by X = A–1B (Thm 2.8). • (2) If |A| = 0 • since A C implies that if |A|0 then |C|0 (Thm 3.2). • the reduced echelon form of A is not In. • The solution to the system AX = B is not unique. • many or no solutions.
34. Solution Since Thus the system does not have an unique solution. Example 4 Determine whether or not the following system of equations has an unique solution. 隨堂作業：14(b)
35. Proof |A| 0 the solution to AX = B is unique and is given by Theorem 3.8 Cramer’s Rule Let AX = B be a system of n linear equations in n variables such that |A| 0. The system has a unique solution given by Where Ai is the matrix obtained by replacing column i of A with B.
36. the cofactor expansion of |Ai| in terms of the ith column Thus Proof of Cramer’s Rule xi, the ith element of X, is given by
37. Solution The matrix of coefficients A and column matrix of constants B are It is found that |A| = –3 0. Thus Cramer’s rule be applied. We get Example 5 Solving the following system of equations using Cramer’s rule.
38. The unique solution is Giving Cramer’s rule now gives 隨堂作業：12(c)
39. Homogeneous Systems of Linear Equations Example 6 Determine values of for which the following system of equations has nontrivial solutions. Find the solutions for each value of . Solution homogeneous system x1 = 0, x2 = 0 is the trivial solution. nontrivial solutions exist many solutions = – 3 or = 2.
40. = – 3 results in the system This system has many solutions, x1 = r, x2 = r. = 2 results in the system This system has many solutions, x1 = – 3r/2, x2 = r. 隨堂作業：15
41. Homework • Exercise 3.3:7, 14, 15
42. 3.4 Eigenvalues and Eigenvectors Definition Let A be an n n matrix. A scalar is called an eigenvalue(特徵值,固有值)of A if there exists a nonzero vector x in Rn such that Ax = x. The vector x is called an eigenvectorcorresponding to . >0 <0 Figure 3.1
43. Computation of Eigenvalues and Eigenvectors Let A be an n n matrix with eigenvalue and corresponding eigenvector x. Ax = x Ax – x = 0 (A – In)x = 0 a system of linear equations, and x=0 is a solution. we need nonzero solutions many solutions |A – In| = 0 Solve |A – In| = 0 for leads to all the eigenvalues of A. On expending the determinant |A – In|, we get a polynomial in . This polynomial is called the characteristic polynomial of A. The equation |A – In| = 0 is called the characteristic equation of A.
44. Example 1 Find the eigenvalues and eigenvectors of the matrix Solution Find the characteristic polynomial of A: The eigenvalues of A are 2 and –1. The corresponding eigenvectors are found by using these values of in the equation(A – I2)x = 0.
45. (1) = 2 (A – 2I2)x = 0 Thus the eigenvectors of A corresponding to = 2 are
46. (2) = –1 (A + 1I2)x = 0 Thus the eigenvectors of A corresponding to = –1 are 隨堂作業：9先不求eigenspaces
47. Proof Let x1, x2V and let c be a scalar. Then Ax1 = x1 and Ax2 = x2. Hence, Eigenspaces Theorem 3.9 Let A be an n n matrix and an eigenvalue of A. The set V of all eigenvectors corresponding to , together with the zero vector, is a subspace of Rn. This subspace is called the eigenspace of . Thus x1+x2V. The set is closed under addition. Further, Therefore cx1 V. The set is closed under scalar multiplication. Thus V is a subspace of Rn.
48. Example 2 Find the eigenvalues and corresponding eigenspaces of the matrix Solution l=10 or 1
49. (1) = 10 Thus the eigenspace of = 10 is The set is a basis, and the dimension is 1.
50. (2) = 1 Thus the eigenspace of = 1 is 隨堂作業：10 If an eigenvalue occurs as a k times repeated root of the characteristic equation, we say that it is of multiplicity (重根數) k. Thus l=10 has multiplicity 1, while l=1 has multiplicity 2. The set is a basis, and the dimension is 2.
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# Use Derivative to Find Quadratic Function
Use the first derivative to find the equation of a quadratic function given tangent lines to the graph of this function
## Problem
a) Find the equation of a quadratic function whose graph is tangent at x = 1 to the line with slope 8, tangent at x = - 2 to the line with slope - 4 and tangent to the line y = - 8.
b) Find the equation of the tangent lines at x = 1 and x = - 2.
c) Graph the quadratic function obtained and the 3 tangent lines in the same coordinate system and label the tangent lines and points of tangency.
Solution to Problem:
a) The slope of the tangent to the graph of a function f is related to its first derivative. Let f be the quadratic function to find to be written as
f(x) = a x2 + b x + c
The first derivative of f is given by
f '(x) = 2 a x + b
From the property of the first derivative, the slope of the tangent line is equal to the value of the derivative at the point of tangency. Hence we can write two equations related to the tangent lines at x = 1 and x = - 2 as follows
f '(1) = 2a(1) + b = 8
f '(-2) = 2a(-2) + b = -4
Solve the above system of equations to obtain
a = 2 and b = 4
The third tangent y = - 8 is a horizontal line and it slope is equal to 0. A horizontal line is tangent to the graph of a quadratic function, which is a parabola, at the vertex. So y = -8 is the y coordinate of the vertex. The x coordinate of the vertex equal to h is found by solving
f '(x) = 2 a h + b = 0
Which gives
h = - b / 2a
Substitute a and b by their values found above to find
h = -4 / 4 = -1
The graph of the quadratic function has a vertex at (-1,8) and hence
f(-1) = a(-1)
2 + 4(-1) + c = - 8
Solve the above equation for c to obtain
c = - 6
The quadratic function f is given by
f(x) = 2 x
2 + 4 x - 6
b)
Now that we know the equations of the quadratic function, we can find the y coordinates of points of tangency of the tangent lines at x = 1 and x = -2 as follows:
at x = 1, y = f(1) = 2(1)
2 + 4(1) - 6 = 0. The tangent line at x = 1 passes through the point (1,0).
at x = - 2, y = f(-2) = 2(-2)
2 + 4(-2) - 6 = - 14. The tangent line a x = -2 passes through the point (-2 , -14).
For each of the two tangent, we know the slope and a point and therefore we can find their equations.
The equation of the tangent at x = 1 has slope 8 and passes through (1 , 0) and its equation is given by: y = 8x - 8
The equation of the tangent at x = -2 has slope -4 and passes through (-2 , -14) and its equation is given by: y = - 4x - 14
c) Graphs of the quadratic function and all three tangent lines.
More references on calculus problems
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# 7.1 The central limit theorem for sample means (averages) (Page 3/18)
Page 3 / 18
The mean number of minutes for app engagement by a tablet user is 8.2 minutes. Suppose the standard deviation is one minute. Take a sample of 60.
1. What are the mean and standard deviation for the sample mean number of app engagement by a tablet user?
2. What is the standard error of the mean?
3. Find the 90 th percentile for the sample mean time for app engagement for a tablet user. Interpret this value in a complete sentence.
4. Find the probability that the sample mean is between eight minutes and 8.5 minutes.
1. This allows us to calculate the probability of sample means of a particular distance from the mean, in repeated samples of size 60.
2. Let k = the 90 th percentile
k = invNorm $\left(0.\text{90,8}\text{.2,}\frac{1}{\sqrt{60}}\right)$ = 8.37. This values indicates that 90 percent of the average app engagement time for table users is less than 8.37 minutes.
3. P (8< $\overline{x}$ <8.5) = normalcdf $\left(\text{8,8}\text{.5,8}\text{.2,}\frac{1}{\sqrt{60}}\right)$ = 0.9293
## Try it
Cans of a cola beverage claim to contain 16 ounces. The amounts in a sample are measured and the statistics are n = 34, $\overline{x}$ = 16.01 ounces. If the cans are filled so that μ = 16.00 ounces (as labeled) and σ = 0.143 ounces, find the probability that a sample of 34 cans will have an average amount greater than 16.01 ounces. Do the results suggest that cans are filled with an amount greater than 16 ounces?
We have P (( $\overline{x}$ >16.01) = normalcdf $\left(\text{16}\text{.01,E99,16,}\frac{0.143}{\sqrt{34}}\right)$ = 0.3417. Since there is a 34.17% probability that the average sample weight is greater than 16.01 ounces, we should be skeptical of the company’s claimed volume. If I am a consumer, I should be glad that I am probably receiving free cola. If I am the manufacturer, I need to determine if my bottling processes are outside of acceptable limits.
## References
Baran, Daya. “20 Percent of Americans Have Never Used Email.”WebGuild, 2010. Available online at http://www.webguild.org/20080519/20-percent-of-americans-have-never-used-email (accessed May 17, 2013).
Data from The Flurry Blog, 2013. Available online at http://blog.flurry.com (accessed May 17, 2013).
Data from the United States Department of Agriculture.
## Chapter review
In a population whose distribution may be known or unknown, if the size ( n ) of samples is sufficiently large, the distribution of the sample means will be approximately normal. The mean of the sample means will equal the population mean. The standard deviation of the distribution of the sample means, called the standard error of the mean, is equal to the population standard deviation divided by the square root of the sample size ( n ).
## Formula review
The Central Limit Theorem for Sample Means: $\overline{X}$ ~ N
The Mean $\overline{X}$ : μ x
Central Limit Theorem for Sample Means z-score and standard error of the mean: $z=\frac{\overline{x}-{\mu }_{x}}{\left(\frac{{\sigma }_{x}}{\sqrt{n}}\right)}$
Standard Error of the Mean (Standard Deviation ( $\overline{X}$ )): $\frac{{\sigma }_{x}}{\sqrt{n}}$
Use the following information to answer the next six exercises: Yoonie is a personnel manager in a large corporation. Each month she must review 16 of the employees. From past experience, she has found that the reviews take her approximately four hours each to do with a population standard deviation of 1.2 hours. Let Χ be the random variable representing the time it takes her to complete one review. Assume Χ is normally distributed. Let $\overline{X}$ be the random variable representing the mean time to complete the 16 reviews. Assume that the 16 reviews represent a random set of reviews.
What is the mean, standard deviation, and sample size?
mean = 4 hours; standard deviation = 1.2 hours; sample size = 16
Complete the distributions.
1. X ~ _____(_____,_____)
2. $\overline{X}$ ~ _____(_____,_____)
Find the probability that one review will take Yoonie from 3.5 to 4.25 hours. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
1. P (________< x <________) = _______
a. Check student's solution.
b. 3.5, 4.25, 0.2441
Find the probability that the mean of a month’s reviews will take Yoonie from 3.5 to 4.25 hrs. Sketch the graph, labeling and scaling the horizontal axis. Shade the region corresponding to the probability.
1. P (________________) = _______
The fact that the two distributions are different accounts for the different probabilities.
Find the 95 th percentile for the mean time to complete one month's reviews. Sketch the graph.
1. The 95 th Percentile =____________
mean is number that occurs frequently in a giving data
That places the mode and the mean as the same thing. I'd define the mean as the ratio of the total sum of variables to the variable count, and it assigns the variables a similar value across the board.
Samsicker
what is mean
what is normal distribution
What is the uses of sample in real life
pain scales in hospital
Lisa
change of origin and scale
3. If the grades of 40000 students in a course at the Hashemite University are distributed according to N(60,400) Then the number of students with grades less than 75 =*
If a constant value is added to every observation of data, then arithmetic mean is obtained by
sum of AM+Constnt
Fazal
data can be defined as numbers in context. suppose you are given the following set of numbers 18,22,22,20,19,21
what are data
what is mode?
what is statistics
Natasha
statistics is a combination of collect data summraize data analyiz data and interprete data
Ali
what is mode
Natasha
what is statistics
It is the science of analysing numerical data in large quantities, especially for the purpose of inferring proportions in a whole from those in a representative sample.
Bernice
history of statistics
statistics was first used by?
Terseer
if a population has a prevalence of Hypertension 5%, what is the probability of 4 people having hypertension from 8 randomly selected individuals?
Carpet land sales persons average 8000 per weekend sales Steve qantas the firm's vice president proposes a compensation plan with new selling incentives Steve hopes that the results of a trial selling period will enable him to conclude that the compensation plan increases the average sales per sales
Supposed we have Standard deviation 1.56, mean 6.36, sample size 25 and Z-score 1.96 at 95% confidence level, what is the confidence interval?
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# Circle A has a radius of 2 and a center of (8 ,2 ). Circle B has a radius of 4 and a center of (2 ,3 ). If circle B is translated by <-1 ,5 >, does it overlap circle A? If not, what is the minimum distance between points on both circles?
Apr 6, 2016
no overlap , d ≈ 3.22
#### Explanation:
A translation does not change the shape of a figure , only it's position.
Under a translation of $\left(\begin{matrix}- 1 \\ 5\end{matrix}\right)$
centre of B (2,3) → (2-1 , 3+5) → (1,8)
We now require to calculate the distance between the centres of A and B using the $\textcolor{b l u e}{\text{ distance formula }}$
$d = \sqrt{{\left({x}_{2} - {x}_{1}\right)}^{2} + {\left({y}_{2} - {y}_{1}\right)}^{2}}$
where $\left({x}_{1} , {y}_{1}\right) \text{ and " (x_2,y_2)" are 2 coordinate points }$
let $\left({x}_{1} , {y}_{1}\right) = \left(8 , 2\right) \text{ and } \left({x}_{2} , {y}_{2}\right) = \left(1 , 8\right)$
d = sqrt((1-8)^2 + (8-2)^2) = sqrt(49+36) = sqrt85 ≈ 9.22
now: radius of A + radius of B = 2 + 4 = 6
since sum of radii < distance between centres , no overlap
and distance between them = 9.22 - 6 = 3.22
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## HOW TO FIND THE GCF FOR ALGEBRAIC EXPRESSIONS
How to find the gcf for algebraic expressions :
To find the greatest common factor of the given algebraic expressions we have to follow the steps:
Step 1:
For algebraic expression we have to find factors of them using algebraic identities or factoring polynomials.
Step 2:
List the common factors.
## Most commonly used algebraic identities for factoring
(a + b)² = a² + 2 ab + b²
(a - b)² = a² - 2 ab + b²
a² - b² = (a + b) (a - b)
a³ + b³ = (a + b)(a²- ab + b²)
a³- b³= (a - b)(a² + ab + b²)
(a + b + c)²= a² + b²+ c² + 2ab + 2bc + 2ca
(a + b)³=a³ + 3 a²b + 3 ab² + b³
(a - b)³=a³- 3 a²b + 3 ab² - b³
Question 1 :
Find the GCD of the following c² - d² , c (c - d)
Solution :
Step 1 :
c² - d² = (c + d) (c - d)
by comparing c²-d² with the algebraic identity (a²-b²) =(a+b)(a-b), we get the factors (c + d) (c -d)
c (c - d) = c (c - d)
Step 2 :
Common factor is (c -d)
Hence, the greatest common factor is (c -d)
Let us see the next example on "How to find the gcf for algebraic expressions".
Question 2 :
Find the GCD of the following x⁴ - 27 a³ x , (x - 3a)²
Solution:
Step 1 :
x⁴ - 27 a³ x = x (x³ - 27 a³)
(x - 3a)² = (x - 3a) (x - 3a)
Step 2 :
The common factor is (x -3a)
Hence, the greatest common factor is (x - 3a)
Let us see the next example on "How to find the gcf for algebraic expressions".
Question 3 :
Find the GCD of the following m² - 3 m - 18 , m² + 5 m + 6
Solution:
Step 1 :
m² - 3 m - 18 = (m - 6) (m + 3)
m² + 5 m + 6 = (m + 2) (m + 3)
Step 2 :
The common factor is (m + 3)
Hence, the greatest common factor is (m + 3)
Let us see the next example on "How to find the gcf for algebraic expressions".
Question 4 :
Find the greatest common factor of a^(m + 1), a^(m + 2) , a^(m + 3)
Solution:
Step 1 :
a^(m + 1) = a^m a
a^(m + 2) = a^m x a²
a^(m + 3) = a^m
The common factors are a^m a
Step 3 :
Multiply the common factors a^m and a
Hence, the greatest common factor is a ^(m + 1)
Let us see the next example on "How to find the gcf for algebraic expressions".
Question 5 :
Find the greatest common factor of x² y + x y² , x² + x y
Solution:
Step 1 :
x² y + x y² = x y (x + y)
x² + x y = x (x + y)
Step 2 :
The common factor is (x + y)
Hence, the greatest common factor is (x + y)
Let us see the next example on "How to find the gcf for algebraic expressions".
Question 6 :
Find the greatest common factor of 3 (a- 1) , 2 (a - 1)² , (a² - 1)
Solution :
Step 1 :
3 (a- 1) = 3 (a - 1) (cannot be factored here after)
2 (a - 1)² = 2 (a - 1) (a - 1)
(a² - 1) = (a + 1) (a - 1)
Step 2 :
The common factor is (a - 1)
Hence, the greatest common factor is (a - 1)
Let us see the next example on "How to find the gcf for algebraic expressions".
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# 12.Find the measure of ∠P and ∠S if $$\bar { SP } ∥ \bar { RQ }$$ in figure, is there any other method to find m∠P?)
Answer :
We have, $$∠Q = 130° and ∠R = 90^{\circ}$$ and $$\bar { SP } || \bar { RQ }$$
$$∠P + ∠Q = 180^{\circ}\quad$$ [Adjacent angles]
$$\Rightarrow ∠P + 130^{\circ} = 180^{\circ}$$
$$\Rightarrow ∠P = 180^{\circ} – 130^{\circ} = 50^{\circ}$$
and also we have, $$∠S + ∠R = 180^{\circ}\quad$$ [Adjacent angles]
$$\Rightarrow ∠S + 90^{\circ} = 180^{\circ}$$
$$\Rightarrow ∠S = 180^{\circ} – 90^{\circ} = 90^{\circ}$$
Hence, $$m∠P = 50^{\circ} \;and \; m∠S = 90^{\circ}$$
Alternate Method:
$$∠Q = 130^{\circ}, ∠R = 90^{\circ} and ∠S = 90^{\circ}$$
We know that
$$∠P + ∠Q + ∠R + ∠Q = 360^{\circ}\quad$$ [Angle sum property of quadrilateral]
$$\Rightarrow ∠P + 130^{\circ}+ 90^{\circ} + 90^{\circ} = 360^{\circ}$$
$$\Rightarrow ∠P + 310^{\circ} = 360^{\circ}$$
$$\Rightarrow ∠P = 360^{\circ}– 310° = 50^{\circ}$$
Hence $$m∠P = 50^{\circ}$$
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## Linear Algebra and Its Applications, Exercise 3.2.1
Exercise 3.2.1. a) Consider the vectors $a = \left( \sqrt{y}, \sqrt{x} \right)$ and $b = \left( \sqrt{x}, \sqrt{y} \right)$ where $x$ and $y$ are arbitrary positive real numbers. Use the Schwarz inequality involving $a$ and $b$ to derive a relationship between the arithmetic mean $\frac{1}{2}\left(x+y\right)$ and the geometric mean $\sqrt{xy}$.
b) Consider a vector from the origin to point $x$, a second vector of length $\|y\|$ from $x$ to the point $x+y$ and the third vector from the origin to $x+y$. Using the triangle inequality
$\|x+y\| \le \|x\| + \|y\|$
derive the Schwarz inequality. (Hint: Square both sides of the inequality and expand the expression $\left( x+y \right)^T\left( x+y \right)$.)
Answer: a) From the Schwarz inequality we have
$|a^Tb| \le \|a\| \|b\|$
From the definitions of $a$ and $b$, on the left side of the inequality we have
$|a^Tb| = |\sqrt{y} \sqrt{x} + \sqrt{x} \sqrt{y}|$
$= |\sqrt{yx} + \sqrt{xy}| = 2|\sqrt{xy}| = 2\sqrt{xy}$
assuming we always choose the positive square root.
From the definitions of $a$ and $b$ we also have
$\|a\|^2 = \left( \sqrt{y} \right)^2 + \left( \sqrt{x} \right)^2 = y + x = x+y$
and
$\|b\|^2 = \left( \sqrt{x} \right)^2 + \left( \sqrt{y} \right)^2 = y + x = x+y$
so that the right side of the inequality is
$\|a\|\|b\| = \sqrt{x+y} \sqrt{x+y}$
$= \sqrt{\left(x+y\right)^2} = x+y$
again assuming we choose the positive square root. (We know $x+y$ is positive since both $x$ and $y$ are.)
The Schwartz inequality
$|a^Tb| \le \|a\| \|b\|$
then becomes
$2\sqrt{xy} \le x+y$
or (dividing both sides by 2)
$\sqrt{xy} \le \frac{1}{2}\left( x+y \right)$
We thus see that for any positive real numbers $x$ and $y$ the geometric mean $\sqrt{xy}$ is less than the arithmetic mean $\frac{1}{2}\left(x+y\right)$.
b) From the triangle inequality we have
$\|x+y\| \le \|x\| + \|y\|$
for the vectors $x$ and $y$. Squaring the term on the left side of the inequality and using the commutative and distributive properties of the inner product we obtain
$\|x+y\|^2 = \left( x+y \right)^T \left( x+y \right)$
$= x^T \left( x+y \right) + y^T \left( x+y \right)$
$= x^Tx + x^Ty + y^Tx + y^Ty$
$= x^Tx + 2x^Ty + y^Ty$
$= \|x\|^2 + 2x^Ty + \|y\|^2$
Squaring the term on the right side of the inequality we have
$\left( \|x\| + \|y\| \right)^2 = \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$
The inequality
$\|x+y\| \le \|x\| + \|y\|$
is thus equivalent to the inequality
$\|x\|^2 + 2x^Ty + \|y\|^2 \le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$
Subtracting $\|x\|^2$ and $\|y\|^2$ from both sides of the inequality gives us
$2x^Ty \le 2\|x\|\|y\|$
and dividing both sides of the inequality by 2 produces
$x^Ty \le \|x\|\|y\|$
Note that this is almost but not quite the Schwarz inequality: Since the Schwarz inequality involves the absolute value $|x^Ty|$ we must also prove that
$-x^Ty \le \|x\|\|y\|$
(After all, the inner product $x^Ty$ might be negative, in which case the inequality $x^Ty \le \|x\|\|y\|$ would be trivially true, given that the term on the right side of the inequality is guaranteed to be positive.)
We have $-x^Ty = \left(-x\right)^Ty$. Since the triangle inequality holds for any two vectors we can restate it in terms of $-x$ and $y$ as follows:
$\|\left(-x\right)+y\| \le \|-x\| + \|y\|$
Since $\|-x\| = \|x\|$ squaring the term on the right side of the inequality produces
$\left(\|-x\| + \|y\|\right)^2 = \left(\|x\| + \|y\|\right)^2 = \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$
as it did previously. However squaring the term on the left side of the inequality produces
$\|\left(-x\right)+y\|^2 = \left( -x+y \right)^T \left( -x+y \right)$
$= \left(-x\right)^T \left( -x+y \right) + y^T \left( -x+y \right)$
$= \left(-x\right)^T\left(-x\right) + \left(-x\right)^Ty + y^T\left(-x\right) + y^Ty$
$= x^Tx - x^Ty - y^Tx + y^Ty = x^Tx - 2x^Ty + y^Ty$
$= \|x\|^2 - 2x^Ty + \|y\|^2$
The original triangle inequality
$\|\left(-x\right)+y\| \le \|-x\| + \|y\|$
is thus equivalent to
$\|x\|^2 - 2x^Ty + \|y\|^2 \le \|x\|^2 + 2\|x\|\|y\| + \|y\|^2$
or
$-x^Ty \le \|x\|\|y\|$
Since we have both $-x^Ty \le \|x\|\|y\|$ and $x^Ty \le \|x\|\|y\|$ we therefore have
$|x^Ty| \le \|x\|\|y\|$
which is the Schwarz inequality.
So the triangle inequality implies the Schwarz inequality.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
This entry was posted in linear algebra and tagged , , , . Bookmark the permalink.
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# Trigonometric Functions Graphing the Trigonometric Function E Q
• Slides: 39
Trigonometric Functions Graphing the Trigonometric Function
E. Q: E. Q 1. What is a radian and how do I use it to determine angle measure on a circle? 2. How do I use trigonometric functions to model periodic behavior? CCSS: F. IF. 2, 4, 5 &7 E; F. TF. 1, 2, 5 &8
Mathematical Practices: 1. Make sense of problems and persevere in solving them. 2. Reason abstractly and quantitatively. 3. Construct viable arguments and critique the reasoning of others. 4. Model with mathematics. 5. Use appropriate tools strategically. 6. Attend to precision. 7. Look for and make use of structure. 8. Look for and express regularity in repeated reasoning.
CHO SHA CAO SOH CAH TOA Right Triangle Trigonometry Graphing the Trig Function
Graphing Trigonometric Functions Amplitude: the maximum or minimum vertical distance between the graph and the x-axis. Amplitude is always positive 5
The amplitude of y = a sin x (or y = a cos x) is half the distance between the maximum and minimum values of the function. amplitude = |a| If |a| > 1, the amplitude stretches the graph vertically. If 0 < |a| > 1, the amplitude shrinks the graph vertically. If a < 0, the graph is reflected in the x-axis. y y = sin x x y= sin x y = – 4 sin x reflection of y = 4 sin x y = 2 sin x y = 4 sin x
Graphing Trigonometric Functions Period: the number of degrees or radians we must graph before it begins again. 7
The period of a function is the x interval needed for the function to complete one cycle. For b 0, the period of y = a sin bx is . For b 0, the period of y = a cos bx is also . If 0 < b < 1, the graph of the function is stretched horizontally. y period: 2 period: x If b > 1, the graph of the function is shrunk horizontally. y period: 2 x period: 4
Sine is a periodic function: p = 2π sin θ − 3π − 2π −π 0 π 2π One period 2π sin θ: Domain (angle measures): all real numbers, (−∞, ∞) Range (ratio of sides): − 1 to 1, inclusive [− 1, 1] sin θ is an odd function; it is symmetric wrt the origin. sin(−θ) = −sin(θ) 3π θ
Graph of the Sine Function To sketch the graph of y = sin x first locate the key points. These are the maximum points, the minimum points, and the intercepts. x 0 sin x 0 1 0 -1 0 Then, connect the points on the graph with a smooth curve that extends in both directions beyond the five points. A single cycle is called a period. y = sin x y x
Graph of the Cosine Function To sketch the graph of y = cos x first locate the key points. These are the maximum points, the minimum points, and the intercepts. x 0 cos x 1 0 -1 0 1 Then, connect the points on the graph with a smooth curve that extends in both directions beyond the five points. A single cycle is called a period. y = cos x y x
Cosine is a periodic function: p = 2π cos θ θ − 3π − 2π −π 0 π One period 2π cos θ: Domain (angle measures): all real numbers, (−∞, ∞) Range (ratio of sides): − 1 to 1, inclusive [− 1, 1] cos θ is an even function; it is symmetric wrt the y-axis. cos(−θ) = cos(θ) 2π 3π
Properties of Sine and Cosine graphs 1. The domain is the set of real numbers 2. The range is set of “y” values such that -1≤ y ≤ 1 3. The maximum value is 1 and the minimum value is -1 4. The graph is a smooth curve 5. Each function cycles through all the values of the range over an x interval or 2π 6. The cycle repeats itself identically in both direction of the x-axis 13
Sine Graph Given : A sin Bx Amplitude = IAI period = 2π/B Example: y=5 sin 2 X › Amp=5 π/2 π/4 › Period=2π/2 =π 3π/4 π
Example: Sketch the graph of y = 3 cos x on the interval [– , 4 ]. Partition the interval [0, 2 ] into four equal parts. Find the five key points; graph one cycle; then repeat the cycle over the interval. x y = 3 cos x (0, 3) 0 3 max y -3 0 x-int min ( 0 2 3 x-int max , 3) x ( ( , 0) ( , – 3) , 0)
Use basic trigonometric identities to graph y = f (–x) Example : Sketch the graph of y = sin (–x). The graph of y = sin (–x) is the graph of y = sin x reflected in the x-axis. y = sin (–x) y Use the identity sin (–x) = – sin x x y = sin x Example : Sketch the graph of y = cos (–x). The graph of y = cos (–x) is identical to the graph of y = cos x. y Use the identity x cos (–x) = + cos x y = cos (–x)
Example: Sketch the graph of y = 2 sin (– 3 x). Rewrite the function in the form y = a sin bx with b > 0 y = 2 sin (– 3 x) = – 2 sin 3 x Use the identity sin (– x) = – sin x: period: 2 = 2 amplitude: |a| = |– 2| = 2 3 Calculate the five key points. x 0 y = – 2 sin 3 x 0 y – 2 0 ( , 2) x (0, 0) ( ( , -2) , 0)
Tangent Function Recall that . Since cos θ is in the denominator, when cos θ = 0, tan θ is undefined. This occurs @ π intervals, offset by π/2: { … −π/2, 3π/2, 5π/2, … } Let’s create an x/y table from θ = −π/2 to θ = π/2 (one π interval), with 5 input angle values. θ sin θ cos θ tan θ −π/2 − 1 0 und −π/2 und − 1 −π/4 − 1 0 0 0 1 π/4 1 und π/2 und −π/4 0 0 1 π/4 π/2 1 0
Graph of Tangent Function: Periodic Vertical asymptotes where cos θ = 0 θ −π/2 tan θ Und (-∞) −π/4 − 1 0 0 π/4 1 π/2 Und(∞) − 3π/2 −π/2 0 π/2 One period: π tan θ: Domain (angle measures): θ ≠ π/2 + πn Range (ratio of sides): all real numbers (−∞, ∞) tan θ is an odd function; it is symmetric wrt the origin. tan(−θ) = −tan(θ) θ
Graph of the Tangent Function To graph y = tan x, use the identity . At values of x for which cos x = 0, the tangent function is undefined and its graph has vertical asymptotes. y Properties of y = tan x 1. Domain : all real x 2. Range: (– , + ) 3. Period: x 4. Vertical asymptotes: period:
Example: Find the period and asymptotes and sketch the graph y of 1. Period of y = tan x is . 2. Find consecutive vertical asymptotes by solving for x: Vertical asymptotes: 3. Plot several points in 4. Sketch one branch and repeat. x
Cotangent Function Recall that . Since sin θ is in the denominator, when sin θ = 0, cot θ is undefined. This occurs @ π intervals, starting at 0: { … −π, 0, π, 2π, … } Let’s create an x/y table from θ = 0 to θ = π (one π interval), with 5 input angle values. θ 0 sin θ cos θ 1 0 π/4 π/2 1 0 3π/4 π 0 – 1 θ cot θ 0 Und ∞ 1 π/4 1 0 π/2 0 − 1 3π/4 − 1 π Und−∞ cot θ Und ∞ Und−∞
Graph of Cotangent Function: Periodic Vertical asymptotes where sin θ = 0 cot θ θ cot θ 0 ∞ π/4 1 π/2 0 3π/4 − 1 π −∞ − 3π/2 -π −π/2 cot θ: Domain (angle measures): θ ≠ πn Range (ratio of sides): all real numbers (−∞, ∞) cot θ is an odd function; it is symmetric wrt the origin. tan(−θ) = −tan(θ) π 3π/2
Graph of the Cotangent Function To graph y = cot x, use the identity. At values of x for which sin x = 0, the cotangent function is undefined and its graph has vertical asymptotes. y Properties of y = cot x 1. Domain : all real x 2. Range: (– , + ) 3. Period: 4. Vertical asymptotes: vertical asymptotes x
Cosecant is the reciprocal of sine csc θ − 3π Vertical asymptotes where sin θ = 0 θ 0 − 2π −π π 2π 3π sin θ One period: 2π sin θ: Domain: (−∞, ∞) Range: [− 1, 1] csc θ: Domain: θ ≠ πn (where sin θ = 0) Range: |csc θ| ≥ 1 or (−∞, − 1] U [1, ∞] sin θ and csc θ are odd (symm wrt origin)
Graph of the Cosecant Function To graph y = csc x, use the identity . At values of x for which sin x = 0, the cosecant function is undefined and its graph has vertical asymptotes. y Properties of y = csc x 1. domain : all real x 2. range: (– , – 1] [1, + ) 3. period: 4. vertical asymptotes: where sine is zero. x
Secant is the reciprocal of cosine Vertical asymptotes where cos θ = 0 sec θ − 3π − 2π −π 0 π 2π θ 3π cos θ One period: 2π cos θ: Domain: (−∞, ∞) sec θ: Domain: θ ≠ π/2 + πn (where cos θ = 0) Range: [− 1, 1] Range: |sec θ | ≥ 1 or (−∞, − 1] U [1, ∞] cos θ and sec θ are even (symm wrt y-axis)
Graph of the Secant Function The graph y = sec x, use the identity . At values of x for which cos x = 0, the secant function is undefined and its graph has vertical asymptotes. y Properties of y = sec x 1. domain : all real x 2. range: (– , – 1] [1, + ) 3. period: 2 4. vertical asymptotes: x
Summary of Graph Characteristics Def’n ∆ sin θ csc θ cos θ sec θ tan θ cot θ о Period Domain Range Even/Odd
Summary of Graph Characteristics Def’n Period Domain Range Even/Odd − 1 ≤ x ≤ 1 or [− 1, 1] odd ∆ о sin θ opp hyp y r 2π (−∞, ∞) csc θ 1. sinθ r. y 2π θ ≠ πn cos θ adj hyp x r 2π (−∞, ∞) sec θ 1. cosθ r y 2π θ ≠ π2 +πn tan θ sinθ cosθ y x π θ ≠ π2 +πn All Reals or (−∞, ∞) odd cot θ cosθ. sinθ x y π θ ≠ πn All Reals or (−∞, ∞) odd |csc θ| ≥ 1 or (−∞, − 1] U [1, ∞) All Reals or (−∞, ∞) |sec θ| ≥ 1 or (−∞, − 1] U [1, ∞) odd even
Translations of Trigonometric Graphs • Without looking at your notes, try to sketch the basic shape of each trig function: 1) Sine: 2) Cosine: 3) Tangent:
More Transformations ØWe have seen two types of transformations on trig graphs: vertical stretches and horizontal stretches. ØThere are three more: vertical translations (slides), horizontal translations, and reflections (flips).
More Transformations ØHere is the full general form for the sine function: ØJust as with parabolas and other functions, c and d are translations: Ø c slides the graph horizontally (opposite of sign) Ø d slides the graph vertically ØAlso, if a is negative, the graph is flipped vertically.
More Transformations Ø To graph a sine or cosine graph: 1. Graph the original graph with the correct amplitude and period. 2. Translate c units horizontally and d units vertically. 3. Reflect vertically at its new position if a is negative (or reflect first, then translate).
Examples Ø Describe how each graph would be transformed: 1. 2. 3.
Examples Ø State the amplitude and period, then graph: -2π 2π x
Examples Ø State the amplitude and period, then graph: -2π 2π x
Ø Examples State the amplitude and period, then graph: -2π 2π x
Examples Ø Write an equation of the graph described: Ø The graph of y = cos x translated up 3 units, right π units, and reflected vertically.
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# NCERT Solutions for Exercise 10.4 Class 12 Maths Chapter 10 - Vector Algebra
NCERT solutions for exercise 10.4 Class 12 Maths chapter 10 discuss the questions related to cross product of vectors and their applications. Exercise 10.4 Class 12 Maths have 12 questions. All these 12 questions of NCERT solutions for Class 12 Maths chapter 10 exercise 10.4 are explained with all necessary steps. The Class 12 Maths chapter 10 exercise 10.4 solutions are designed by a mathematics expert and are in accordance with Class 12 CBSE patterns. Concepts of cross products are also explained in Class 11 NCERT Physics TextBooks in the chapter system of particles and rotational motion and are used in other chapters of Class 11 and 12 too. One question may be expected from the Class 12 Maths chapter 10 exercise 10.4 for Class 12 CBSE Class 12 Board Exam. Along with Class 12th Maths chapter 10 exercise 10.4 students can go through the following exercise for practice.
• Vector Algebra Exercise 10.1
• Vector Algebra Exercise 10.2
• Vector Algebra Exercise 10.3
• Vector Algebra Miscellaneous Exercise
## Vector Algebra Class 12 Chapter 10 Exercise 10.4
Question:1 Find $|\vec a \times \vec b |, if \vec a = \hat i - 7 \hat j + 7 \hat k \: \: and \: \: \vec b = 3 \hat i - 2 \hat j + 2 \hat k$
Given in the question,
$\\ \vec a = \hat i - 7 \hat j + 7 \hat k \: \: and \: \: \\\vec b = 3 \hat i - 2 \hat j + 2 \hat k$
and we need to find $\dpi{100} |\vec a \times \vec b |$
Now,
$|\vec a \times \vec b | =\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &-7 &7 \\ 3& -2 &2 \end{vmatrix}$
$|\vec a \times \vec b | =\hat i(-14+14)-\hat j(2-21)+\hat k(-2+21)$
$|\vec a \times \vec b | =19\hat j+19\hat k$
So the value of $\dpi{100} |\vec a \times \vec b |$ is $19\hat j+19\hat k$
Question:2 Find a unit vector perpendicular to each of the vector $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ , where $\vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k$
Given in the question
$\vec a = 3 \hat i + 2 \hat j + 2 \hat k \: \:and \: \: \vec b = \hat i + 2 \hat j - 2 \hat k$
$\vec a + \vec b =3\hat i +2\hat j+2\hat k+\hat i +2\hat j-2\hat k=4\hat i +4\hat j$
$\vec a - \vec b =3\hat i +2\hat j+2\hat k-\hat i -2\hat j+2\hat k=2\hat i +4\hat k$
Now , A vector which perpendicular to both $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is $(\vec a + \vec b) \times (\vec a - \vec b)$
$(\vec a + \vec b) \times (\vec a - \vec b)=\begin{vmatrix} \hat i &\hat j &\hat k \\ 4&4 &0 \\ 2& 0& 4 \end{vmatrix}$
$(\vec a + \vec b) \times (\vec a - \vec b)= \hat i (16-0)-\hat j(16-0)+\hat k(0-8)$
$(\vec a + \vec b) \times (\vec a - \vec b)= 16\hat i -16\hat j-8\hat k$
And a unit vector in this direction :
$\vec u =\frac{16\hat i-16\hat j-8\hat k}{|16\hat i-16\hat j-8\hat k|}=\frac{16\hat i-16\hat j-8\hat k}{\sqrt{16^2+(-16)^2+(-8)^2}}$
$\vec u =\frac{16\hat i-16\hat j-8\hat k}{24}=\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$
Hence Unit vector perpendicular to each of the vector $\vec a + \vec b \: \: and\: \: \vec a - \vec b$ is $\frac{2}{3}\hat i-\frac{2}{3}\hat j-\frac{1}{3}\hat k$ .
Question:3 If a unit vector $\vec a$ makes angles $\frac{\pi }{3}$ with $\hat i , \frac{\pi }{4}$ with $\hat j$ and an acute angle $\theta \: \:$ with $\hat k$ then find $\theta \: \:$ and hence, the components of $\vec a$ .
Given in the question,
angle between $\vec a$ and $\hat i$ :
$\alpha =\frac{\pi}{3}$
angle between $\vec a$ and $\hat j$
$\beta =\frac{\pi}{4}$
angle with $\vec a$ and $\hat k$ :
$\gamma =\theta$
Now, As we know,
$cos^2\alpha+cos^2\beta+cos^2\gamma=1$
$cos^2\frac{\pi}{3}+cos^2\frac{\pi }{4}+cos^2\theta=1$
$\left ( \frac{1}{2} \right )^2+\left ( \frac{1}{\sqrt{2}} \right )^2+cos^2\theta=1$
$cos^2\theta=\frac{1}{4}$
$cos\theta=\frac{1}{2}$
$\theta=\frac{\pi}{3}$
Now components of $\vec a$ are:
$\left ( cos\frac{\pi}{3},cos\frac{\pi}{2},cos\frac{\pi}{3} \right )=\left ( \frac{1}{2},\frac{1}{\sqrt{2}},\frac{1}{2} \right )$
Question:4 Show that $( \vec a - \vec b ) \times (\vec a + \vec b ) = 2 ( \vec a \times \vec b )$
To show that $( \vec a - \vec b ) \times (\vec a + \vec b ) = 2 ( \vec a \times \vec b )$
LHS=
$\\( \vec a - \vec b ) \times (\vec a + \vec b )=( \vec a - \vec b ) \times (\vec a)+( \vec a - \vec b ) \times (\vec b)$
$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times \vec a-\vec b \times\vec a+\vec a \times \vec b-\vec b \times \vec b$
As product of a vector with itself is always Zero,
$( \vec a - \vec b ) \times (\vec a + \vec b )= 0-\vec b \times\vec a+\vec a \times \vec b-\0$
As cross product of a and b is equal to negative of cross product of b and a.
$( \vec a - \vec b ) \times (\vec a + \vec b )= \vec a \times\vec b+\vec a \times \vec b$
$( \vec a - \vec b ) \times (\vec a + \vec b )= 2(\vec a \times\vec b)$ = RHS
LHS is equal to RHS, Hence Proved.
Question:5 Find $\lambda$ and $\mu$ if $( 2 \hat i + 6 \hat j + 27 \hat k ) \times ( \hat i + \lambda j + \mu \hat k ) = \vec 0$
Given in the question
$( 2 \hat i + 6 \hat j + 27 \hat k ) \times ( \hat i + \lambda j + \mu \hat k ) = \vec 0$
and we need to find values of $\lambda$ and $\mu$
$\begin{vmatrix} \hat i &\hat j & \hat k\\ 2& 6&27 \\ 1& \lambda &\mu \end{vmatrix}=0$
$\hat i (6\mu-27\lambda)-\hat j(2\mu-27)+\hat k(2\lambda-6)=0$
From Here we get,
$6\mu-27\lambda=0$
$2\mu-27=0$
$2\lambda -6=0$
From here, the value of $\lambda$ and $\mu$ is
$\lambda = 3 , \: and \: \mu=\frac{27}{2}$
Question:6 Given that $\vec a . \vec b = 0 \: \:and \: \: \vec a \times \vec b = 0$ and . What can you conclude about the vectors $\vec a \: \:and \: \: \vec b$ ?
Given in the question
$\vec a . \vec b = 0$ and $\vec a \times \vec b = 0$
When $\vec a . \vec b = 0$ , either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$ are perpendicular to each other
When $\vec a \times \vec b = 0$ either $|\vec a| =0,\:or\: |\vec b|=0,\vec a\: and \:\vec b$ are parallel to each other
Since two vectors can never be both parallel and perpendicular at same time,we conclude that
$|\vec a| =0\:or\: |\vec b|=0$
Question:7 Let the vectors $\vec a , \vec b , \vec c$ be given as $\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k , \vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k , \vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k$ Then show that $\vec a \times ( \vec b + \vec c ) = \vec a \times \vec b + \vec a \times \vec c$
Given in the question
$\\\vec a=\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k , \\\vec b=\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k , \\\vec c=\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k$
We need to show that $\vec a \times ( \vec b + \vec c ) = \vec a \times \vec b + \vec a \times \vec c$
Now,
$\vec a \times ( \vec b + \vec c ) =(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times(\vec b_ 1 \hat i + \vec b_ 2 \hat j + \vec b_3 \hat k +\vec c_ 1 \hat i + \vec c_ 2 \hat j + \vec c_ 3 \hat k)$
$=(\vec a_1 \hat i + \vec a_2 \hat j + \vec a_3 \hat k)\times((\vec b_ 1+\vec c_1) \hat i + (\vec b_ 2+\vec c_2) \hat j +( \vec b_3 +\vec c_3)\hat k)$
$=\begin{vmatrix} \hat i &\hat j &\hat k \\ a_1&a_2 &a_3 \\ (b_1+c_1)&(b_2+c_2) &(b_3+c_3) \end{vmatrix}$
$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)- a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$
$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$
Now
$\vec a \times \vec b + \vec a \times \vec c=\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ b_1&b_2 &b_3 \end{vmatrix}+\begin{vmatrix} \hat i &\hat j & \hat k\\ a_1&a_2 &a_3 \\ c_1&c_2 &c_3 \end{vmatrix}$
$\vec a \times \vec b + \vec a \times \vec c=\hat i(a_2b_3-a_3b_2)-\hat j (a_1b_3-a_3b_1)+\hat k(a_1b_2-b_1a_2)+\hat i(a_2c_3-a_3c_2)-\hat j (a_1c_3-a_3c_1)+\hat k(a_1c_2-c_1a_2)$
$\\=\hat i(a_2(b_3+c_3)-a_3(b_2+c_2))-\hat j(a_1(b_3+c_3)-a_3(b_1+c_1))+\hat k (a_1(b_2+c_2)-a_2(a_1(b_2+c_3)))$
Hence they are equal.
Question:8 If either $\vec a = \vec 0 \: \: or \: \: \vec b = \vec 0$ then $\vec a \times \vec b = \vec 0$ . Is the converse true? Justify your answer with an example.
No, the converse of the statement is not true, as there can be two non zero vectors, the cross product of whose are zero. they are colinear vectors.
Consider an example
$\vec a=\hat i +\hat j + \hat k$
$\vec b =2\hat i +2\hat j + 2\hat k$
Here $|\vec a| =\sqrt{1^2+1^2+1^2}=\sqrt{3}$
$|\vec b| =\sqrt{2^2+2^2+2^2}=2\sqrt{3}$
$\vec a \times \vec b=\begin{vmatrix} \hat i &\hat j &\hat k \\ 1&1 &1 \\ 2&2 &2 \end{vmatrix}=\hat i(2-2)-\hat j(2-2)+\hat k(2-2)=0$
Hence converse of the given statement is not true.
Question:9 Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
Given in the question
vertices A=(1, 1, 2), B=(2, 3, 5) and C=(1, 5, 5). and we need to find the area of the triangle
$AB=(2-1)\hat i+(3-1)\hat j+(5-2)\hat k=\hat i+2\hat j+3\hat k$
$BC=(1-2)\hat i+(5-3)\hat j+(5-5)\hat k=-\hat i+2\hat j$
Now as we know
Area of triangle
$A=\frac{1}{2}|\vec {AB}\times\vec {BC}|=\frac{1}{2}|(\hat i+2\hat j +3\hat k)\times(-\hat i+2\hat j)|$
$\\A=\frac{1}{2}\begin{vmatrix} \hat i &\hat j &\hat k \\ 1 &2 &3 \\ -1 &2 &0 \end{vmatrix}=\frac{1}{2}|\hat i(0-6)-\hat j(0-(-3))+\hat k(2-(-2))| \\A=\frac{1}{2}|-6\hat i-3\hat j+4\hat k|$
$A=\frac{1}{2}*\sqrt{(-6)^2+(-3)^2+(4)^2}=\frac{\sqrt{61}}{2}$
The area of the triangle is $\dpi{100} \frac{\sqrt{61}}{2}$ square units
Question:10 Find the area of the parallelogram whose adjacent sides are determined by the vectors $\vec a = \hat i - \hat j + 3 \hat k$ and $\vec b = 2\hat i -7 \hat j + \hat k$ .
Given in the question
$\vec a = \hat i - \hat j + 3 \hat k$
$\vec b = 2\hat i -7 \hat j + \hat k$
Area of parallelogram with adjescent side $\vec a$ and $\vec b$ ,
$A=|\vec a\times\vec b|=|(\vec i-\vec j+3\vec k)\times (2\hat i-7\hat j+\hat k)|$
$A=\begin{vmatrix} \hat i& \hat j & \hat k\\ 1&-1 &3 \\ 2&-7 &1 \end{vmatrix}=|\hat i(-1+21)-\hat j (1-6)+\hat k (-7+2)|$
$A=|\hat i(20)-\hat j (-5)+\hat k (-5)|=\sqrt{20^2+5^2+(-5)^2}$
$A=\sqrt{450}=15\sqrt{2}$
The area of the parallelogram whose adjacent sides are determined by the vectors $\vec a = \hat i - \hat j + 3 \hat k$ and $\vec b = 2\hat i -7 \hat j + \hat k$ is $A=\sqrt{450}=15\sqrt{2}$
Question:11 Let the vectors $\vec a \: \: and\: \: \vec b$ be such that $|\vec a| = 3 \: \: and\: \: |\vec b | = \frac{\sqrt 2 }{3}$ , then $\vec a \times \vec b$ is a unit vector, if the angle between is $\vec a \: \:and \: \: \vec b$
$\\A ) \pi /6 \\\\ B ) \pi / 4 \\\\ C ) \pi / 3 \\\\ D ) \pi /2$
Given in the question,
$|\vec a| = 3 \: \: and\: \: |\vec b | = \frac{\sqrt 2 }{3}$
As given $\vec a \times \vec b$ is a unit vector, which means,
$|\vec a \times \vec b|=1$
$|\vec a| | \vec b|sin\theta=1$
$3*\frac{\sqrt{2}}{3}sin\theta=1$
$sin\theta=\frac{1}{\sqrt{2}}$
$\theta=\frac{\pi}{4}$
Hence the angle between two vectors is $\frac{\pi}{4}$ . Correct option is B.
Question:12 Area of a rectangle having vertices A, B, C and D with position vectors
$- \hat i + \frac{1}{2} \hat j + 4 \hat k , \hat i + \frac{1}{2} \hat j + 4 \hat k , \hat i - \frac{1}{2}\hat j + 4 \hat k \: \: and \: \: - \hat i - \frac{1}{2} \hat j + 4 \hat k$
(A)1/2
(B) 1
(C) 2
(D) 4
Given 4 vertices of rectangle are
$\\\vec a=- \hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec b=\hat i + \frac{1}{2} \hat j + 4 \hat k , \\\vec c= \hat i - \frac{1}{2}\hat j + 4 \hat k \: \: and \: \: \\\vec d= - \hat i - \frac{1}{2} \hat j + 4 \hat k$
$\vec {AB}=\vec b-\vec a=(1+1)\hat i+(\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=2\hat i$
$\vec {BC}=\vec c-\vec b=(1-1)\hat i+(-\frac{1}{2}-\frac{1}{2})\hat j+(4-4)\hat k=-\hat j$
Now,
Area of the Rectangle
$A=|\vec {AB}\times\vec {BC}|=|2\hat i \times (-\hat j)|=2$
Hence option C is correct.
## More About NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4
Exercise 10.4 Class 12 Maths throughout explains about the cross products. The applications of cross products include finding the area of a triangle, parallelogram etc. Question number 9 of the Class 12th Maths chapter 10 exercise 10.4 is to find the area of a triangle. And questions 10 and 12 of NCERT Solutions for Class 12 Maths chapter 10 exercise 10.4 is to find the area of parallelogram and rectangle respectively.
## Benefits of NCERT Solutions for Class 12 Maths Chapter 10 Exercise 10.4
• Going through the NCERT syllabus exercise 10.4 Class 12 Maths give more conceptual clarifications about the topics covered just before the exercise.
• Class 12th Maths chapter 10 exercise 10.4 is designed for a better understanding of the concepts of vector products and their applications.
The topics covered under NCERT Solutions for class 12 maths chapter 10 exercise 10.4 will also be useful for the exams like JEE Main also.
• ## NCERT Exemplar Solutions Class 12 Maths Chapter 10
• NCERT Solutions for Class 12 Maths Chapter 10
## NCERT Solutions Subject Wise
• NCERT Solutions Class 12 Chemistry
• NCERT Solutions for Class 12 Physics
• NCERT Solutions for Class 12 Biology
• NCERT Solutions for Class 12 Mathematics
## Subject Wise NCERT Exemplar Solutions
• NCERT Exemplar Class 12 Maths
• NCERT Exemplar Class 12 Physics
• NCERT Exemplar Class 12 Chemistry
• NCERT Exemplar Class 12 Biology
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# ARITHMETIC SERIES PRACTICE WORKSHEET
Arithmetic Series Practice Worksheet :
This section will contain set of questions using the concept arithmetic series.
## Arithmetic Series Practice Worksheet - Practice questions
(1) Find the sum of the following APs
(i) 2 , 7 , 12,....... to 10 terms Solution
(ii) -37, -33, -29,..................... to 12 terms
(iii) 0.6, 1.7, 2.8,............... to 100 terms
(iv) 1/15, 1/12, 1/10,................... to 11 terms
(2) Find the sums given below
(i) 7 + 10 (1/2) + 14 + ...............+ 84
(ii) 34 + 32 + 30 + .................. + 10
(iii) - 5 + (-8) + (-11) + .............+ (-230)
(3) In an AP
(i) Given a = 5, d = 3, an = 50 find n and Sn
(ii) Given a = 7, a13 = 35 find d and S13
(iii) Given a12 = 37 , d = 3 find a and S12
(iv) Given a3 = 15 , S10 = 125 find d and a10
(v) Given d = 5 , S9 = 75 find a and a9
(vi) Given a = 2 , d = 8, Sn = 90 find n and an
(vii) Given a = 8 , an = 62, Sn = 210 find n and d Solution
(viii) Given an = 4, d = 2, Sn = -14 find n and a
(ix) Given a = 3, n = 8, Sn = 192 find d
(x) Given l = 28 , S = 144 , and there are total 9 terms. Find a
(4) How many terms of the AP 9, 17, 25,.......... must be taken to give a sum of 636?
(5) The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and common difference.
(6) The first and last term of an AP are 17 and 350 respectively.If the common difference is 9, how many terms are there and what is their sum?
(7) Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149.
(8) The sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.
(9) If the sum of 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.
(10) Show that a₁,a₂,............ an form an AP where an is defined as below
(i) a n = 3 + 4 n
(ii) a n = 9 - 5 n
Also find the sum of 15 terms in each case.
(11) If the sum of the first n terms of an AP is 4 n - n² what is the first term (that is S)? what is the sum of first two terms?what is the second term? similarly find the 3rd,the 10th and the nth terms.
(12) Find the sum of first 40 positive integers divisible by 6.
(13) Find the sum of first 15 multiples of 8.
(14) Find the sum of odd numbers between 0 and 50.
(15) A contract on construction job specifies a penalty for delay for completion beyond a certain due date as follows. \$200 for the first day, \$250 for the second day, \$300 for the third day etc., the penalty for each succeeding day being \$50 more than for the preceding day. How much money the contractor has to pay as penalty,if he has delayed the work be 30 days. Solution
(16) A sum of \$700 is to be used to seven cash prizes to students of a school for their overall academic performance. If each prize is \$20 less than the preceding prize, find the value of each prizes. Solution
(17) In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in each they are studying ,e.g, a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students? Solution
(18) A spiral is made up of successive semicircles,with centres alternately at A and B starting with center at A of radii 0.5, 1.0 cm, 1.5 cm ,2.0 cm ......... as shown in figure . What is the total length of spiral made up of thirteen consecutive semicircles. (∏ = 22/7) Solution
After having gone through the stuff given above, we hope that the students would have practiced questions in this worksheet.
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