text stringlengths 22 1.01M |
|---|
# Rational Mathematics
## Combinations
Theorem (The Combination Principle): The number of ways of choosing m elements from a set of n elements is n!|((n−m)!.m!). Proof: The number of ways of choosing a sequence of m elements is n!|(n–m)! and the number of ways of arranging m elements in a sequence is m! So the n!|(n–m)! sequences fall into sets of m! containing the same elements. Thus if we denote the number of these sets by c then n!|(n–m)! = c.(m!). Thus c = n!|((n–m)!.m!).
It is often convenient to use the notation: nCm = n!|((n–m)!.m!) where C can be called the combinatorial operation. For n<m we define nCm = 0. The operation table for the combinatorial operation C is a version of Pascal's Triangle.
Combination table (base ten)
C 0 1 2 3 4 5 6 7 8 9 10 0 1 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 2 1 2 1 0 0 0 0 0 0 0 0 3 1 3 3 1 0 0 0 0 0 0 0 4 1 4 6 4 1 0 0 0 0 0 0 5 1 5 10 10 5 1 0 0 0 0 0 6 1 6 15 20 15 6 1 0 0 0 0 7 1 7 21 35 35 21 7 1 0 0 0 8 1 8 28 56 70 56 28 8 1 0 0 9 1 9 36 84 126 126 84 36 9 1 0 10 1 10 45 120 210 252 210 120 45 10 1
Example: How many dominoes are there in a double (n–1) set? The set contains every distinct pair of numbers from {0, 0} to {n–1, n–1}. Thus the total is nC2 = n.(n+1)|2, the nth triangular number. The triangular numbers are the numbers that appear in column 2 of the combinatorial table (and in the diagonal from 1 in the left column). The numbers that appear in the third column are the sums of the triangular number series. Thus nC3 = 2C2 + 3C2 + ... + (n–1)C2. The numbers in the fourth column are sums of those in the third column,and so on.
Example: What is the maximum number of pieces that can result from n straight vertical cuts? We have f(1) = 2 and the nth cut increases the number of regions by n at most. Thus f(n) = f(n–1) + n, whence f(n) = n.(n+1)|2 + 1 = nC0 + nC1 + nC2. Sequence: 2, 4, 7, 11, 16, 22, 29, 37, 46, 56, 67, 79, 92, ... The augmented triangular numbers. [Jacob Steiner 1826]
Example: What is the maximum number s(n) of regions bounded by n planes in space? We have s(1) = 2. The number of 3D regions defined by the nth plane is the number of 2D regions defined in it by its n–1 straight-line intersections with the previous planes. Hence s(n) = s(n–1) + r(n–1) = nC0 + nC1 + nC2 + nC3. [This problem goes back to Jacob Steiner, Journal für die reine und angewandte Mathematic, vol.1, pp.349-364, 1826.]
Example. The number of different Bridge hands of 13 cards that can be dealt, from a pack of 52 cards, is 52C13 = 52!|(39!.13!)
Theorem (Complementarity): The number of ways of choosing m elements from a set of n is the same as the number of ways of choosing n−m elements. In symbols this theorem is: nCm = nC(n−m). Proof: (a) This statement is equivalent to the commutative property n!|((n−m)!.m!) = n!|(m!.(n−m)!) since n−(n−m) = m. (b) Alternatively we can note that choosing to include m of n is the same as choosing to exclude n−m of n.
Observation (The Bell Curve): The values of n!|((n−m)!.m!) as m runs from 0 to n form a symmetric bell-shaped pattern, rising to a maximum at the middle value n|2 when n is even, or the two middle values (n±1)|2 when n is odd.
Theorem (Binomial Theorem): The nth power of the sum of two terms is: (a+b)^n = ∑(r = 0 to n)[nCr.a^(n-r).b^r]. Proof: by induction.
Written out more fully it becomes: (a+b)^n = a^n + n.a^(n–1).b + (n.(n–1)|2).a^(n–2).b^2 + ... + [n!|((n–m)!.m!)].a^(n–m).b^m + ... + b^n.
Examples:
(a+b)^2 = a^2 + 2.a.b + b^2
(a+b)^3 = a^3 + 3.(a^2).b + 3.a.(b^2) + b^3
(a+b)^4 = a^4 + 4.(a^3).b + 6.(a^2).(b^2) + 4.a.(b^3) + b^4.
Note: Because of this result the numbers nCm for a given value of n are sometimes called binomial coefficients since they are the numerical factors that appear in the multiplied-out expression for (a+b)^n.
Theorem (Partitions): The number ways of expressing n (> m−1) as a sum of m numbers is (n−1)C(m−1). Proof: If we represent the numbers in linear tally form then expressing n as a sum of m numbers is equivalent to separating a row of n tallies into m sets, by introducing m−1 gaps, and there are n−1 places where gaps can be introduced.
Examples:
When m = 2, the formula reduces to n–1:
2 = 1+1, 3 = 1+2 = 2+1, 4 = 1+3 = 2+2 = 3+1, 5 = 1+4 = 2+3 = 3+2 = 4+1.
When m = 3 the formula reduces to (n–1).(n–2)|2:
3 = 1+1+1, 4 = 1+1+2 = 1+2+1 = 2+1+1, 5 = 1+1+3 = 1+2+2 = 1+3+1 = 2+1+2 = 2+2+1 = 3+1+1. |
# What is the addition of ordinates method?
## What is the addition of ordinates method?
Answer: At each x-coordinate, the y-coordinates of y = sinx + cosx is the sum of the y-coordinates of y = sinx and y = cosx. II.
### How do you graph an addition function?
Adding two functions is like plotting one function and taking the graph of that function as the new x-axis. Points of the second function are then plotted with respect to the new axis. For example, (2, 3) becomes “over 2,” “up 3 from the new axis,” or (3, f + 2).
#### How do you graph different functions?
Steps for Sketching the Graph of the Function
1. Determine, whether function is obtained by transforming a simpler function, and perform necessary steps for this simpler function.
2. Determine, whether function is even, odd or periodic.
3. Find y-intercept (point f(0)).
4. Find x-intercepts (points where f(x)=0).
How do you graph ordinates?
STEP 1 – Draw and label the x and y axis. STEP 2 – Plot the coordinates (2,3). Remember the x (horizontal) is the first number in the brackets and the y (vertical) is the second number. Now plot the rest of the coordinates.
What is addition function?
The addition of function involves the creation of a new function through the addition of two other functions.
## What is the first coordinate set?
domain
The domain of the relation is the set of all possible first coordinates. When we graph, we think of the first coordinate in terms of x . The range of the relation is the set of all possible second coordinates.
### Which is an example of an addition worksheet?
For example, adding 8 + 7, students first recognize that they need to add 2 to 8 to get 10, so they split the 7 into 2 + 5. The 8 + 2 makes 10 and 5 more makes 15.
#### What are the addition properties of a number?
Review and refine your grade 2 and grade 3 kids’ skills in identifying the commutative and associative laws and distinguishing between the two in these printable addition properties worksheets. Any number plus 0 is the number itself: 2 + 0 = 2.
Are there any printable addition worksheets for integers?
The worksheets are available both in PDF and html formats, are highly customizable, and include an answer key. By controlling the range in the generator below, you can use negative numbers (integers), numbers less than 10, very large numbers, and so on.
Is it good to use one per page addition worksheet?
One-per-page addition frenzies are not the most efficient use of paper resources, but they are a good starting point especially for younger students who have not quite mastered their penmanship enough to fit their numbers into a smaller chart. They are also great for displaying on screens or monitors for group activities. |
# 9.1 Solving trigonometric equations with identities
Page 1 / 9
In this section, you will:
• Verify the fundamental trigonometric identities.
• Simplify trigonometric expressions using algebra and the identities.
In espionage movies, we see international spies with multiple passports, each claiming a different identity. However, we know that each of those passports represents the same person. The trigonometric identities act in a similar manner to multiple passports—there are many ways to represent the same trigonometric expression. Just as a spy will choose an Italian passport when traveling to Italy, we choose the identity that applies to the given scenario when solving a trigonometric equation.
In this section, we will begin an examination of the fundamental trigonometric identities, including how we can verify them and how we can use them to simplify trigonometric expressions.
## Verifying the fundamental trigonometric identities
Identities enable us to simplify complicated expressions. They are the basic tools of trigonometry used in solving trigonometric equations, just as factoring, finding common denominators, and using special formulas are the basic tools of solving algebraic equations. In fact, we use algebraic techniques constantly to simplify trigonometric expressions. Basic properties and formulas of algebra, such as the difference of squares formula and the perfect squares formula, will simplify the work involved with trigonometric expressions and equations. We already know that all of the trigonometric functions are related because they all are defined in terms of the unit circle. Consequently, any trigonometric identity can be written in many ways.
To verify the trigonometric identities, we usually start with the more complicated side of the equation and essentially rewrite the expression until it has been transformed into the same expression as the other side of the equation. Sometimes we have to factor expressions, expand expressions, find common denominators, or use other algebraic strategies to obtain the desired result. In this first section, we will work with the fundamental identities: the Pythagorean identities , the even-odd identities, the reciprocal identities, and the quotient identities.
We will begin with the Pythagorean identities (see [link] ), which are equations involving trigonometric functions based on the properties of a right triangle. We have already seen and used the first of these identifies, but now we will also use additional identities.
Pythagorean Identities
${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$ $1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$ $1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta$
The second and third identities can be obtained by manipulating the first. The identity $\text{\hspace{0.17em}}1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta \text{\hspace{0.17em}}$ is found by rewriting the left side of the equation in terms of sine and cosine.
Prove: $\text{\hspace{0.17em}}1+{\mathrm{cot}}^{2}\theta ={\mathrm{csc}}^{2}\theta$
Similarly, $\text{\hspace{0.17em}}1+{\mathrm{tan}}^{2}\theta ={\mathrm{sec}}^{2}\theta \text{\hspace{0.17em}}$ can be obtained by rewriting the left side of this identity in terms of sine and cosine. This gives
A laser rangefinder is locked on a comet approaching Earth. The distance g(x), in kilometers, of the comet after x days, for x in the interval 0 to 30 days, is given by g(x)=250,000csc(π30x). Graph g(x) on the interval [0, 35]. Evaluate g(5) and interpret the information. What is the minimum distance between the comet and Earth? When does this occur? To which constant in the equation does this correspond? Find and discuss the meaning of any vertical asymptotes.
The sequence is {1,-1,1-1.....} has
how can we solve this problem
Sin(A+B) = sinBcosA+cosBsinA
Prove it
Eseka
Eseka
hi
Joel
June needs 45 gallons of punch. 2 different coolers. Bigger cooler is 5 times as large as smaller cooler. How many gallons in each cooler?
7.5 and 37.5
Nando
find the sum of 28th term of the AP 3+10+17+---------
I think you should say "28 terms" instead of "28th term"
Vedant
the 28th term is 175
Nando
192
Kenneth
if sequence sn is a such that sn>0 for all n and lim sn=0than prove that lim (s1 s2............ sn) ke hole power n =n
write down the polynomial function with root 1/3,2,-3 with solution
if A and B are subspaces of V prove that (A+B)/B=A/(A-B)
write down the value of each of the following in surd form a)cos(-65°) b)sin(-180°)c)tan(225°)d)tan(135°)
Prove that (sinA/1-cosA - 1-cosA/sinA) (cosA/1-sinA - 1-sinA/cosA) = 4
what is the answer to dividing negative index
In a triangle ABC prove that. (b+c)cosA+(c+a)cosB+(a+b)cisC=a+b+c.
give me the waec 2019 questions |
# Sequence and Series Questions Made Simple
Sequence and Series Questions are gaining importance in competitive exams be it for IBPS Clerk, PO, RRB, SO, or SBI Clerk or PO. The number of questions and the complexity of questions vary from exam to exam every year.
Sequence and Series questions are complex in nature and require an understanding of Numbers Systems, Squares and Square roots, Cubes and Cube roots, Math Tables, and other types of Arithmetic and Geometric Series are given in this article.
In this article, you’ll also find Examples of Sequence and Series Questions and Practice on Sequence and Series Questions.
Nowadays, questions in exams are mixed of multiple series and it requires practice and a deep understanding of basic concepts along with quickly identifying numbers.
## What is Sequence?
A sequence is a collection of numbers that is in order. In other words: a sequence is a list of items, objects, or numbers that are arranged in sequential, linear, or any custom order.
In sequence, each number is separated by a comma and is enclosed within curly brackets ‘{}’ also called set brackets. For example: {a1, a2, a3, a4, …}. A number such as 1, 2, 3, 4, … represents the position of that number.
There are two types of sequences: The Finite sequence and the Infinite sequence.
### What is a Finite Sequence?
When any sequence is having a limited number of items, objects or numbers is called a Finite Sequence. For Example: {a1, a2, a3, a4} or {1,3,5,7}
### What is Infinite Sequence?
When any sequence is not having a limited number of items, objects or numbers is called an Infinite Sequence. For Example: {a1, a2, a3, a4, …} or {1,3,5,7, …}
## Properties of Sequence
### Notation of Sequence
Sequence Notation is a way of representing a sequence, each number is separated by a comma and is enclosed within curly brackets ‘{}’ also called set brackets. For example: {1, 3, 5, 7}.
### Rule of Sequence
The sequence follows a Rule which means that every number is having a common rule in which every number follows. For example, a sequence {3, 5, 7, 9}, so as per Rule we can see that there are 2 number differences between every number.
This sequence can be said to be incremental by 2 for each number. If we’ve to add a new number to this sequence then as per Rule we’ll add 2 to the last number and the new number will be 9 + 2 = 11.
### Rule as a Formula
These rules can be simplified in mathematical expression or formula as 2n + 1. This expression can be tested by putting n numbers and checking if the sequence follows the rule or not. With the help of expression, we can find 100 elements.
100th Element in Sequence can be found by applying the above formula which is 2 x 100 + 1 = 201.
### Finding Difference
Many times it’s easier to find the difference between two consecutive numbers. These differences are in such a way that it becomes simpler to understand the construction of that sequence.
From the above example, we found that there’s a difference of 2 numbers in every consecutive number hence it became simpler to understand. Also, it helps to find the next number or n(th) number.
Sometimes differences cannot be easily found at the first level then we need to analyze the difference and understand whether it’s forming another series. For example: {2, 4, 7, 11, 16, 22, …} from this example we’ll create another sequence of differences as {2, 3, 4, 5, 6, …}. Now from this level, it’s clear that the difference between every number is 1.
As per the above example if we have to find out the next number in this sequence then we’ll find the next number of different sequences, that is 6 + 1 = 7. Now actual number of the sequence is 22 + 7 = 29.
## What is a Series?
The sum of an Infinite sequence is known as a Series. It looks similar to a sequence but the series has a plus sign (+) between its numbers. For example Series is denoted as: SN = a1+a2+a3 + .. + aN.
If the sequence is Finite or Infinite, it’s called a Series only when it’s not summing all numbers. If we sum up a few numbers in sequence then it’s called a Partial Sum.
## Difference between Sequence and Series Questions
Below are the difference between sequence and series questions:
## Types of Sequence and Series Questions
Below are the different types of Sequence and Series Questions, which are commonly asked in Bank Exams.
### Odd Number Sequence and Series
The number in this sequence and series are odd numbers. The start of a series can be from any odd number and all the numbers in a series are odd numbers. This odd number sequence and series can be finite or infinite. For example: {5, 7, 9, 11, …}
### Even Number Sequence and Series
The number in this sequence and series even numbers. The start of a series can be from any even number and all the numbers in the series are even numbers. This even number sequence and series can be finite or infinite. For example: {4, 6, 8, 10, …}
### Arithmetic Sequence and Series
In Arithmetic sequence and series, every element or next number is formed by adding or subtracting a definite number from the preceding number. For example: {a, a+d, a+2d, a+3d, … }
In the above example, ‘a’ is the first term, and ‘d’ is the difference between two consecutive terms also known as common difference. Note: ‘d’ cannot be zero else all numbers will be the same as ‘a’. For example: {a, a, a, a, …}.
### Geometric Sequence and Series
In Geometric sequence and series, every element or next number is formed by multiplying or dividing a definite number by the preceding number. For example: {a, ar, ar2, ar3, … }
In the above example, ‘a’ is the first term and ‘r’ is the factor between two consecutive terms also known as a common ratio. Note: ‘r’ cannot be zero else series will have only one number as ‘a’. For example: {a, 0, 0, 0, …}.
### Triangular Sequence and Series
Triangle is formed from a pattern of dots, and these dots form triangular sequences and series. By adding a new row of dots and counting all the dots of the triangle, we can find the next number of the sequence. Triangular sequence and series follow the below formula:
xn = n(n+1)/2
### Fibonacci Sequence and Series
Fibonacci sequence and series form an sequence of numbers in which each number is obtained by adding two preceding numbers and the sequence starts with 0 and 1. Sequence is defined as, F0 = 0 and F1 = 1 and Fn = Fn-1 + Fn-2. For Example: 0, 1, 1, 2, 3, 5, 8, 13, …
### Other Sequence and Series Questions
Above mentioned sequence and series questions are the common and widely used ones. However there any many other types of series that can be formed using Squares and Square roots, Cubes and Cube roots, and Math Tables. Any custom type of sequence can also be formed, but the approach to solving them remains the same by finding a rule.
## Solved Example on Sequence and Series Questions
Below are the solved examples on the sequence and series Questions:
Example 1: 5, 8, 13, 21, 34, n. Find the value of n?
Solution 1: From the above sequence, it’s clear that this is the Fibonacci series. For finding the value of ‘n’, as per Fibonacci we need to sum up the previous two numbers: 21 + 34 = 55.
Example 2: 216, 343, 512, 729, n. Find the value of n?
Solution 2: From the above sequence, it’s clear that this is a cube series.
For finding the value of ‘n’, we need to find the cube root of 729, which is 9. Now next number in the sequence is a cube of 10. So the final answer is 1000.
Example 3:25, 36, 49, 64, n. Find the value of n?
Solution 3: From the above sequence, it’s clear that this is a square series.
For finding the value of ‘n’, we need to find the square root of 64, which is 8. Now next number in the sequence is a square of 9. So the final answer is 81.
Example 4: 16, 19, 22, 25, n. Find the value of n?
Solution 4: From the above sequence, we’ll find the difference between all numbers and it’s having a common difference of 3. So value of n is 25 + 3 = 28. Also, the above example is of Arithmetic Sequence and series.
Example 5: 9, 27, 81, 243, n. Find the value of n?
Solution 5: From the above sequence, we’ll find the ratio for all numbers and it’s having a common ratio of 3. So the value of n is 243 x 3 = 729. Also, the above example is a Geometric Sequence and series.
## Practice Sequence and Series Questions
Below are some Sequence and Series Questions to practice:
1. Find the value of n in sequence: 23, 28, 33, 38, n?
2. Find the value of n in sequence: 19, 27, 35, 43, n?
3. Find the value of n in sequence: 21, 18, 15, 12, n?
4. Find the value of n in sequence: 1, 2, 5, 10, 17, n?
5. Find the value of n in sequence: 1, 8, 9, 64, 25, n?
6. Find the value of n in sequence: 18, 54, 162, 486, n?
7. Find the value of n in sequence: 2, 4, 10, 20, n?
8. Find the value of n in sequence: 10, 5, 2.5, 1.25, n?
9. Find the value of n in sequence: 6, 10, 16, 26, n?
10. Find the value of n in sequence: 1, 2, 9, 64, n?
## FAQs on Sequence and Series Questions
### What are Sequence and Series?
A sequence is a collection of numbers that is in order. A Series is a sum of a given sequence.
### Different types of Sequences?
Sequence are of two types: Finite sequence and Infinite sequence. Example of Finite sequence is: {1,2,3,4} and Infinite sequence: {1,2,3,4, …}
### Common Sequences?
A few common sequences are Arithmetic Sequence, Geometric Sequence, Square Numbers, Cube Numbers, Fibonacci Series, and many others…
### What is a Finite Sequence?
If the sequence is having finite or limited numbers, then that sequence is called a Finite Sequence. For Example: {1, 2, 3, 4, 5}, this sequence is having only 5 numbers.
### What is Infinite Sequence?
If the sequence is having infinite or unlimited numbers, then that sequence is called an Infinite Sequence. At the end of numbers, there’s a continuation sign ‘…’. For Example: {1, 2, 3, 4, 5, …}, this sequence is having infinite numbers.
## Final Words
Practicing the above questions for Sequence and Series Questions is not the end of your practice, but it’s a start of a new journey to apply logic to Sequence and Series Questions solutions with multiple approaches in right and faster way.
For cracking competitive exams one must practice Sequence and Series Questions without a calculator is a must. At last, during the exam, if a solution for Sequence and Series questions cannot be found easily then mark that question to revisit and move ahead instead of wasting time and energy. |
# Solve the Following Equation and Verify Your Answer: ( 2 X + 3 ) − ( 5 X − 7 ) 6 X + 11 = − 8 3 - Mathematics
Sum
$\frac{(2x + 3) - (5x - 7)}{6x + 11} = - \frac{8}{3}$
#### Solution
$\frac{(2x + 3) - (5x - 7)}{6x + 11} = \frac{- 8}{3}$
$\text{ or }\frac{- 3x + 10}{6x + 11} = \frac{- 8}{3}$
$\text{ or }- 9x + 30 = - 48x - 88 [\text{ After cross multiplication }]$
$\text{ or }- 9x + 48x = - 88 - 30$
$\text{ or }39x=-118\text{ or }x=\frac{- 118}{39}$
$\text{ Thus, }x = \frac{- 118}{39}\text{ is the solution of the given equation .}$
$\text{ Check: }$
$\text{ Substituting }x = \frac{- 118}{39}\text{ in the given equation, we get: }$
$\text{ L . H . S . }= \frac{- 3(\frac{- 118}{39}) + 10}{6(\frac{- 118}{39}) + 11} = \frac{354 + 390}{- 708 + 429} = \frac{744}{- 279} = \frac{- 8}{- 3}$
$\text{ R . H . S .} = \frac{- 8}{3}$
$\therefore\text{ L . H . S . = R . H . S . for }x = \frac{- 118}{39}$
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 8 Maths
Chapter 9 Linear Equation in One Variable
Exercise 9.3 | Q 23 | Page 17 |
# How do you solve x/4 - 7/8 + 3/2x = 5/12 - 5/4x?
Aug 3, 2016
$x = \frac{31}{72}$
#### Explanation:
The first thing to do here is get rid of the denominators by finding the least common multiple, LCM, of all the numbers you have as denominators.
In this case, the LCM of
$2 , 4 , 8 , 12$
is $24$. This means that your starting equation
$\frac{x}{4} - \frac{7}{8} + \frac{3 x}{2} = \frac{5}{12} - \frac{5 x}{4}$
can be rewritten as
$\frac{x}{4} \cdot \frac{6}{6} - \frac{7}{8} \cdot \frac{3}{3} + \frac{3 x}{2} \cdot \frac{12}{12} = \frac{5}{12} \cdot \frac{2}{2} - \frac{5 x}{4} \cdot \frac{6}{6}$
This will get you
$\frac{6 x}{24} - \frac{21}{24} + \frac{36 x}{24} = \frac{10}{24} - \frac{30 x}{24}$
Now you can focus on the numerators
$6 x - 21 + 36 x = 10 - 30 x$
Collect like terms to find
$6 x + 36 x + 30 x = 10 + 21$
$72 x = 31 \implies x = \frac{31}{72}$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Angles of Elevation and Depression
## Angles going up or down from a horizontal line of sight.
Estimated11 minsto complete
%
Progress
Practice Angles of Elevation and Depression
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated11 minsto complete
%
Angles of Elevation and Depression
You have decided to go camping with some friends. While out on a hike, you reach the top of a ridge and look down at the trail behind you. In the distance, you can see your camp. You're thinking about how far you've traveled, and wonder if there is a way to determine it.
By using a small device called a clinometer, you're able to measure the angle between your horizontal line of sight and the camp as \begin{align*}37^\circ\end{align*}, and you know that the hill you just hiked up has a height of 300 m. Is it possible to find out how far away your camp is using this information? (Assume that the trail you hiked is slanted like the side of a triangle.)
### Angles of Elevation and Depression
You can use right triangles to find distances, if you know an angle of elevation or an angle of depression.
The figure below shows each of these kinds of angles.
The angle of elevation is the angle between the horizontal line of sight and the line of sight up to an object. For example, if you are standing on the ground looking up at the top of a mountain, you could measure the angle of elevation. The angle of depression is the angle between the horizontal line of sight and the line of sight down to an object. For example, if you were standing on top of a hill or a building, looking down at an object, you could measure the angle of depression. You can measure these angles using a clinometer or a theodolite. People tend to use clinometers or theodolites to measure the height of trees and other tall objects. Here we will solve several problems involving these angles and distances.
#### Finding the angle of elevation
You are standing 20 feet away from a tree, and you measure the angle of elevation to be \begin{align*}38^\circ\end{align*}. How tall is the tree?
The solution depends on your height, as you measure the angle of elevation from your line of sight. Assume that you are 5 feet tall.
The figure shows us that once we find the value of \begin{align*}T\end{align*}, we have to add 5 feet to this value to find the total height of the triangle. To find \begin{align*}T\end{align*}, we should use the tangent value:
\begin{align*}\tan 38^\circ & = \frac{opposite}{adjacent} = \frac{T}{20}\\ \tan 38^\circ & = \frac{T}{20}\\ T & = 20 \tan 38^\circ \approx 15.63\\ \text{Height of tree} & \approx 20.63 \ ft\end{align*}
You are standing on top of a building, looking at a park in the distance. The angle of depression is \begin{align*}53^\circ\end{align*}. If the building you are standing on is 100 feet tall, how far away is the park? Does your height matter?
Finding the angle of depression
If we ignore the height of the person, we solve the following triangle:
Given the angle of depression is \begin{align*}53^\circ\end{align*}, \begin{align*}\angle{A}\end{align*} in the figure above is \begin{align*}37^\circ\end{align*}. We can use the tangent function to find the distance from the building to the park:
\begin{align*}\tan 37^\circ & = \frac{opposite}{adjacent} = \frac{d}{100}\\ \tan 37^\circ & = \frac{d}{100}\\ d & = 100 \tan 37^\circ \approx 75.36\ ft.\end{align*}
If we take into account the height if the person, this will change the value of the adjacent side. For example, if the person is 5 feet tall, we have a different triangle:
\begin{align*}\tan 37^\circ & = \frac{opposite}{adjacent} = \frac{d}{105}\\ \tan 37^\circ & = \frac{d}{105}\\ d & = 105 \tan 37^\circ \approx 79.12\ ft.\end{align*}
If you are only looking to estimate a distance, then you can ignore the height of the person taking the measurements. However, the height of the person will matter more in situations where the distances or lengths involved are smaller. For example, the height of the person will influence the result more in the tree height problem than in the building problem, as the tree is closer in height to the person than the building is.
#### Real-World Application: The Horizon
You are on a long trip through the desert. In the distance you can see mountains, and a quick measurement tells you that the angle between the mountaintop and the ground is \begin{align*}13.4^\circ\end{align*}. From your studies, you know that one way to define a mountain is as a pile of land having a height of at least 2,500 meters. If you assume the mountain is the minimum possible height, how far are you away from the center of the mountain?
\begin{align*}\tan 13.4^\circ & = \frac{opposite}{adjacent} = \frac{2500}{d}\\ \tan 13.4^\circ & = \frac{2500}{d}\\ d & = \frac{2500}{\tan 13.4^\circ} \approx 10,494\ meters.\end{align*}
### Examples
#### Example 1
Earlier, you were asked if it was possible to find out how far away your camp is using the information given.
Since you know the angle of depression is \begin{align*}37^\circ\end{align*}, you can use this information, along with the height of the hill, to create a trigonometric relationship:
Since the unknown side of the triangle is the hypotenuse, and you know the opposite side, you should use the sine relationship to solve the problem:
\begin{align*} \sin 37^\circ = \frac{300}{hypotenuse}\\ hypotenuse = \frac{300}{\sin 37^\circ}\\ hypotenuse \approx 498.5\\ \end{align*}
You have traveled approximately 498.5 meters up the hill.
#### Example 2
You are six feet tall and measure the angle between the horizontal and a bird in the sky to be \begin{align*}40^\circ\end{align*}. You can see that the shadow of the bird is directly beneath the bird, and 200 feet away from you on the ground. How high is the bird in the sky?
We can use the tangent function to find out how high the bird is in the sky:
\begin{align*} \tan 40^\circ = \frac{height}{200}\\ height = 200 \tan 40^\circ\\ height = (200)(.839)\\ height = 167.8\\ \end{align*}
We then need to add your height to the solution for the triangle. Since you are six feet tall, the total height of the bird in the sky is 173.8 feet.
#### Example 3
While out swimming one day you spot a coin at the bottom of the pool. The pool is ten feet deep, and the angle between the top of the water and the coin is \begin{align*}15^\circ\end{align*}. How far away is the coin from you along the bottom of the pool?
Since the distance along the bottom of the pool to the coin is the same as the distance along the top of the pool to the coin, we can use the tangent function to solve for the distance to the coin:
\begin{align*} \tan 15^\circ = \frac{opposite}{adjacent}\\ \tan 15^\circ = \frac{10}{x}\\ x = \frac{10}{tan 15^\circ}\\ x \approx 37.32^\circ \end{align*}
#### Example 4
You are hiking and come to a cliff at the edge of a ravine. In the distance you can see your campsite at the base of the cliff, on the other side of the ravine. You know that the distance across the ravine is 500 meters, and the angle between your horizontal line of sight and your campsite is \begin{align*}25^\circ\end{align*}. How high is the cliff? (Assume you are five feet tall.)
Using the information given, we can construct a solution:
\begin{align*} \tan 25^\circ = \frac{opposite}{adjacent}\\ \tan 25^\circ = \frac{height}{500}\\ height = 500 \tan 25^\circ\\ height = (500)(.466)\\ height = 233 \ \text{meters} \end{align*}
This is the total height from the bottom of the ravine to your horizontal line of sight. Therefore, to get the height of the ravine, you should take away five feet for your height, which gives an answer of 228 meters.
### Review
1. A 70 foot building casts an 50 foot shadow. What is the angle that the sun hits the building?
2. You are standing 10 feet away from a tree, and you measure the angle of elevation to be \begin{align*}65^\circ\end{align*}. How tall is the tree? Assume you are 5 feet tall up to your eyes.
3. Kaitlyn is swimming in the ocean and notices a coral reef below her. The angle of depression is \begin{align*}35^\circ\end{align*} and the depth of the ocean, at that point is 350 feet. How far away is she from the reef?
4. The angle of depression from the top of a building to the base of a car is \begin{align*}60^\circ\end{align*}. If the building is 78 ft tall, how far away is the car?
5. The Leaning Tower of Pisa currently “leans” at a \begin{align*}4^\circ\end{align*} angle and has a vertical height of 55.86 meters. How tall was the tower when it was originally built?
6. The angle of depression from the top of an apartment building to the base of a fountain in a nearby park is \begin{align*}72^\circ\end{align*}. If the building is 78 ft tall, how far away is the fountain?
7. You are standing 15 feet away from a tree, and you measure the angle of elevation to be \begin{align*}35^\circ\end{align*}. How tall is the tree? Assume you are 5 feet tall up to your eyes.
8. Bill spots a tree directly across the river from where he is standing. He then walks 18 ft upstream and determines that the angle between his previous position and the tree on the other side of the river is \begin{align*}55^\circ\end{align*}. How wide is the river?
9. A 50 foot building casts an 50 foot shadow. What is the angle that the sun hits the building?
10. Eric is flying his kite one afternoon and notices that he has let out the entire 100 ft of string. The angle his string makes with the ground is \begin{align*}60^\circ\end{align*}. How high is his kite at this time?
11. A tree struck by lightning in a storm breaks and falls over to form a triangle with the ground. The tip of the tree makes a \begin{align*}36^\circ\end{align*} angle with the ground 25 ft from the base of the tree. What was the height of the tree to the nearest foot?
12. Upon descent an airplane is 15,000 ft above the ground. The air traffic control tower is 200 ft tall. It is determined that the angle of elevation from the top of the tower to the plane is \begin{align*}15^\circ\end{align*}. To the nearest mile, find the ground distance from the airplane to the tower.
13. Tara is trying to determine the angle at which to aim her sprinkler nozzle to water the top of a 10 ft bush in her yard. Assuming the water takes a straight path and the sprinkler is on the ground 4 ft from the tree, at what angle of inclination should she set it?
14. Over 3 miles (horizontal), a road rises 1000 feet (vertical). What is the angle of elevation?
15. Over 4 miles (horizontal), a road rises 1000 feet (vertical). What is the angle of elevation?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Angle of Depression The angle of depression is the angle formed by a horizontal line and the line of sight down to an object when the image of an object is located beneath the horizontal line.
Angle of Elevation The angle of elevation is the angle formed by a horizontal line and the line of sight up to an object when the image of an object is located above the horizontal line. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# How do you solve the linear equation: $7+3x=6x-8$?
Last updated date: 13th Jun 2024
Total views: 371.1k
Views today: 8.71k
Verified
371.1k+ views
Hint: Rearrange the terms of the given equation by taking the terms containing the variable x to the L.H.S. and taking the constant terms to the R.H.S. Now, use simple arithmetic operations, like: addition, subtraction, multiplication or division, whichever needed, to simplify the equation. Make the coefficient of x equal to 1 and accordingly change the R.H.S. to get the answer.
Complete step-by-step solution:
Here, we have been provided with the equation: $7+3x=6x-8$ and we are asked to solve this equation. That means we have to find the value of x.
$\because 7+3x=6x-8$
As we can see that the given equation is a linear equation in one variable which is x. So, taking the terms containing the variable x to the L.H.S. and taking the constant terms to the R.H.S., we get,
\begin{align} & \Rightarrow 3x-6x=-8-7 \\ & \Rightarrow -3x=-15 \\ \end{align}
Dividing both the sides with (-3) and cancelling the common factors, we get,
$\Rightarrow x=5$
Hence, the value of x is 5.
Note: One may note that we have been provided with a single equation only. The reason is that we have to find the value of one variable, that is x. So, in general if we have to find the value of only one variable, that is x. So, in general if we have to solve an equation having ‘n’ number of variables then we should be provided with ‘n’ number of equations. Now, one can check the answer by substituting the obtained value of x in the equation provided in the question. We have to determine the value of L.H.S. and R.H.S. separately and if they are equal then our answer is correct. |
Monday, March 12, 2012
Multiples of 10
Day One – Multiples of 10 – What are they?
Using cubes/counters represent a given number. Can you split that number of cubes into piles of 10 with none left over? If so, that number is a multiple of 10. Practice with several numbers – multiples and not (write them on the white board so students can see the numbers to decide about the pattern). Do you see a pattern/something similar about each number that is a multiple of 10? Describe the rule/pattern. (Numbers that end in 0 are multiples of 10)
Teach the students that to know what numbers are multiples of a number, just count by that number. (For example, to know what numbers are multiples of ten, just count by tens.)
Play slapjack with cards. Each player gets a “deck” of index cards with numbers on them.
Partner up. Players take turns flipping over 1 card into the middle. If the card
is a multiple of 10, the first player to slap it wins the stack. The goal is to capture all your opponent’s cards.
Day Two – Practice Adding 10 to a number
Briefly review yesterday’s lesson. Today we will practice adding 10 to a number. Each pair gets their own game board of chutes and ladders. Players alternate spinning the spinner and advancing as normal and rolling a die and advancing by that multiple of ten. (Example: if you rolled a four, you'd move up 40 spaces.)
* Final challenge round (if time)– figure out the answer to this math problem – The zoo had 650 tigers and 1,000 leopards. They sold 230 tigers and 100 leopards. Then they got 60 more leopards and 70 more tigers. How many were there in all?
Day Three – Add 10 to the target number
Today we’ll be having a race like Family Feud. Split into 2 teams – sit together in circles on the floor. Distribute a dry-erase paddle, marker, and tissue for erasing to each member of the team. We’ll be taking turns answering a number about multiples of 10. Everyone should write the answer on his or her dry erase paddle. When it is your turn, you’ll be showing your paddle to the teacher to determine whether or not your team will receive a point.
**I’ll tie in pennies and dimes today while doing this review activity.
*Final Challenge Round (if time) – As a team – add multiples of 10 to these numbers then order them greatest to least. (30, 70, 140, 330, 320, 590)
Day Four – Adding Multiples of 10
Pass our hundreds boxes and paper with addition problems. Use color chips/cubes to represent the number then add multiples of 10 to show the new number. Do a total of 10
problems front/back.
*Final Challenge Round (for fast-finishers)– revisit the day’s activity using subtraction. |
Search 75,402 tutors
0 0
## 1/x-1/2=2-x/2x
Directions say to prove the algebraic identity by starting with the Left hand side expression and supplying a sequence of equivalent expressions that ends with the right hand expressions
(1/x) - (1/2) = (2-x)/(2x)
Your goal is to prove that this algebraic identity holds true. To do so, simply solve for the left hand side of this equation. Since the left hand side of the equation requires you to subtract two fractions, you need to find the smallest common denominator between these two fractions. Given that the first fraction has a denominator of x and the second fraction has a denominator of 2, it can be concluded that the smallest common denominator is 2x. With that, multiply the first term (1/x) by 2/2 to yield a fraction whose denominator is 2x. Then, multiply the second term (1/2) by x/x to yield a fraction whose denominator is also 2x. Notice that you are multiplying the numerator and the denominator by the same number (or term), which means that you are essentially multiplying it by 1 so you are not changing the value of the fraction.
So we first change the two fractions on the left hand of the equation to obtain two fractions that share a common denominator:
(1/x) * (2/2) = (1 * 2) / (x * 2) = (2) / (2x) = 2/(2x)
(1/2) * (x/x) = (1 * x) / (2 * x) = (x) / (2x) = x/(2x)
Replace the first fraction in the original equation by 2/(2x) and the second fraction by x/(2x):
(1/x) - (1/2) ==> (2/(2x)) - (x/(2x))
Next, we can combine these two fractions since they now have a common denominator:
(2/(2x)) - (x/(2x)) = (2 - x)/(2x) = (2-x)/(2x)
Thus, we have proved that (1/x) - (1/2) = (2-x)/(2x).
Thank you so much, this was so helpful! :)
Also note that a common denominator (not necessarily the smallest, tho) can be found by cross-multiplying the fraction:
1/x - 1/2 ==> (1*2 - 1*x) / (2 * x) = (2 - x)/2x
To get the smallest, then you would just need to simplify the expression. In thise case, there is no simplification possible as it is already the smallest common denominator.
Did you mean to prove 1/x-1/2 = ( 2-x)/2x?
Take the lowest common denominator of the left hand side. It is 2x. Now you get
Left hand side = 2/2x - x/2x = (2-x)/2x = Right hand side. |
# How to Find The Maximum Profit in Calculus
Calculus > Find the maximum profit in calculus
Contents:
## General Maximization
To maximize a function means to find its maximum value in a given range. A global maximum means to find the maximum to find the maximum over the entire range of the function. Local maximums occur at inflection points (where the graph changes direction). At the maximum of a function, the gradient or slope of the function is zero. You can differentiate the function to find points in the function where the gradient is zero, but these could be either maxima or minima.
• If the slope is increasing at the turning point, it is a minimum.
• If the slope is decreasing at the turning point, then you have found a maximum of the function.
## How to Maximize in Calculus: Steps
Sample problem: Find the local maximum value of y = 4x3 + 2x2 + 1.
First Step: Differentiate the function, using the power rule. Constant terms disappear under differentiation.
d/dx (4x3 + 2x2 + 1) = 12x2 + 4x
The result, 12x2 + 4x, is the gradient of the function. This has two zeros, which can be found through factoring.
12x2 + 4x = 4x(3x+1), which equals zero when x = 0 or x = -1/3
Second Step: Check each turning point (at x = 0 and x = -1/3)to find out whether it is a maximum or a minimum. To do this, differentiate a second time and substitute in the x value of each turning point.
d/dx (12x2 + 4x) = 24x + 4
At x = 0, 24x + 4 = 4, which is greater than zero. This is a minimum.
At x = -1/3, 24x + 4 = -4, which is less than zero. This is a maximum.
Third Step: Find the corresponding y-coordinates for the x-value (maximum) you found in Step 2 by substituting back into the original function.
At x = -1/3, y = 4x3 + 2x2 + 1 = -4/27 + 2/9 + 1 = 29/27
Therefore the function has a maximum value at (-1/3, 29/27).
That’s how to maximize in calculus!
Tip: You can check your answer by sketching the graph and looking for the highest and lowest points.
## Profit Maximization
Profit maximization is one of the topics that are likely to be tested in the short-answer section of the AP Calculus exam. It is equal to a business’s revenue minus the costs incurred in producing that revenue. Profit maximization is an important concern because businesses are run in order to earn the highest profits possible. Calculus can be used to calculate the profit-maximizing number of units produced.
## Find the maximum profit in calculus: Steps
Sample question: Find the profit equation of a business with a revenue equation of 2000x – 10x2 and a cost equation of 2000 + 500x.
First Step: Set profit to equal revenue minus cost. For example, the revenue equation 2000x – 10x2 and the cost equation 2000 + 500x can be combined as profit = 2000x – 10x2 – (2000 + 500x) or profit = -10x2 + 1500x – 2000.
Second Step: Find the first derivative of the profit equation. For example, the profit equation -10x2 + 1500x – 2000 becomes -20x + 1500.
Third Step: Set the equation equal to zero:
-20x + 1500 = 0
Fourth Step: Use algebra to find how many units are produced from the equation you wrote in Step 3.
20x = 1500
x = 75.
Fifth Step: Calculate the maximum profit using the number of units produced calculated in the previous step. In this example, inserting x = 75 into the profit equation -10x2 + 1500x – 2000 produces -10(75)2 + 1500(75) – 2000 or 54,250 in profit.
That’s how to find the maximum profit in calculus!
Tip:
For answering this type of question on the AP calculus exam, be sure to record this figure using the unit of measurement presented in the short-answer problem. For profit maximization short-answer problems on the AP Calculus exam, this unit of measurement is almost certainly US dollars or \$. Some equations might present more than one possible answer. Some of these answers can be picked out and discarded using common sense but most often cannot be treated the same. In these cases, insert all possible answers into the profit equation to calculate their profits and then select the answer that produces the highest profit as the profit maximizing number of units produced.
Sources:
http://earthmath.kennesaw.edu/main_site/review_topics/economics.htm Retrieved July 12, 2015.
------------------------------------------------------------------------------
Need help with a homework or test question? Chegg offers 30 minutes of free tutoring, so you can try them out before committing to a subscription. Click here for more details.
If you prefer an online interactive environment to learn R and statistics, this free R Tutorial by Datacamp is a great way to get started. If you're are somewhat comfortable with R and are interested in going deeper into Statistics, try this Statistics with R track. |
# Exact Value of cos 15°
How to find the exact value of cos 15° using the value of sin 30°?
Solution:
For all values of the angle A we know that, (sin $$\frac{A}{2}$$ + cos $$\frac{A}{2}$$)$$^{2}$$ = sin$$^{2}$$ $$\frac{A}{2}$$ + cos$$^{2}$$ $$\frac{A}{2}$$ + 2 sin $$\frac{A}{2}$$ cos $$\frac{A}{2}$$ = 1 + sin A
Therefore, sin $$\frac{A}{2}$$ + cos $$\frac{A}{2}$$ = ± √(1 + sin A), [taking square root on both the sides]
Now, let A = 30° then, $$\frac{A}{2}$$ = $$\frac{30°}{2}$$ = 15° and from the above equation we get,
sin 15° + cos 15° = ± √(1 + sin 30°) ….. (i)
Similarly, for all values of the angle A we know that, (sin $$\frac{A}{2}$$ - cos $$\frac{A}{2}$$)$$^{2}$$ = sin$$^{2}$$ $$\frac{A}{2}$$ + cos$$^{2}$$ $$\frac{A}{2}$$ - 2 sin $$\frac{A}{2}$$ cos $$\frac{A}{2}$$ = 1 - sin A
Therefore, sin $$\frac{A}{2}$$ - cos $$\frac{A}{2}$$ = ± √(1 - sin A), [taking square root on both the sides]
Now, let A = 30° then, $$\frac{A}{2}$$ = $$\frac{30°}{2}$$ = 15° and from the above equation we get,
sin 15° - cos 15°= ± √(1 - sin 30°) …… (ii)
Clearly, sin 15° > 0 and cos 15˚ > 0
Therefore, sin 15° + cos 15° > 0
Therefore, from (i) we get,
sin 15° + cos 15° = √(1 + sin 30°) ..... (iii)
Again, sin 15° - cos 15° = √2 ($$\frac{1}{√2}$$ sin 15˚ - $$\frac{1}{√2}$$ cos 15˚)
or, sin 15° - cos 15° = √2 (cos 45° sin 15˚ - sin 45° cos 15°)
or, sin 15° - cos 15° = √2 sin (15˚ - 45˚)
or, sin 15° - cos 15° = √2 sin (- 30˚)
or, sin 15° - cos 15° = -√2 sin 30°
or, sin 15° - cos 15° = -√2 ∙ $$\frac{1}{2}$$
or, sin 15° - cos 15° = - $$\frac{√2}{2}$$
Thus, sin 15° - cos 15° < 0
Therefore, from (ii) we get, sin 15° - cos 15°= -√(1 - sin 30°) ..... (iv)
Now, subtracting (iv) from (iii) we get,
2 cos 15° = $$\sqrt{1 + \frac{1}{2}} + \sqrt{1 - \frac{1}{2}}$$
2 cos 15° = $$\frac{\sqrt{3} + 1}{\sqrt{2}}$$
cos 15° = $$\frac{\sqrt{3} + 1}{2\sqrt{2}}$$
Therefore, cos 15° = $$\frac{\sqrt{3} + 1}{2\sqrt{2}}$$ |
# Irrational Vs Rational Numbers Worksheet Pdf
A Logical Phone numbers Worksheet may help your kids become more informed about the ideas right behind this percentage of integers. In this particular worksheet, college students will be able to remedy 12 diverse issues relevant to rational expression. They will discover ways to multiply a couple of phone numbers, group of people them in sets, and find out their products and services. They are going to also practice simplifying rational expression. After they have learned these ideas, this worksheet is a beneficial tool for furthering their research. Irrational Vs Rational Numbers Worksheet Pdf.
## Reasonable Phone numbers are a proportion of integers
The two main forms of figures: rational and irrational. Reasonable numbers are understood to be total amounts, whereas irrational figures usually do not replicate, and also have an unlimited number of numbers. Irrational amounts are non-zero, low-terminating decimals, and sq . beginnings that are not best squares. They are often used in math applications, even though these types of numbers are not used often in everyday life.
To establish a realistic amount, you must understand exactly what a realistic variety is. An integer is actually a entire variety, and a logical number is actually a percentage of two integers. The percentage of two integers is definitely the variety on top divided up from the number on the bottom. If two integers are two and five, this would be an integer, for example. However, there are also many floating point numbers, such as pi, which cannot be expressed as a fraction.
## They could be made in to a small percentage
A realistic amount carries a denominator and numerator that are not zero. Because of this they can be depicted like a small percentage. In addition to their integer numerators and denominators, rational amounts can also have a unfavorable worth. The bad worth needs to be located left of and its absolute worth is its range from absolutely no. To make simpler this instance, we are going to claim that .0333333 can be a small percentage that may be created as being a 1/3.
As well as bad integers, a logical quantity can also be produced into a small fraction. For example, /18,572 can be a rational quantity, while -1/ is not really. Any portion comprised of integers is rational, given that the denominator will not have a and might be published as being an integer. Likewise, a decimal that ends in a position is also a logical quantity.
## They are feeling
Even with their label, reasonable numbers don’t make a lot feeling. In mathematics, they can be single organizations having a special duration about the quantity series. Consequently once we matter anything, we can buy the shape by its rate to its initial amount. This keeps correct even if there are limitless rational numbers between two distinct figures. If they are ordered, in other words, numbers should make sense only. So, if you’re counting the length of an ant’s tail, a square root of pi is an integer.
If we want to know the length of a string of pearls, we can use a rational number, in real life. To discover the period of a pearl, by way of example, we could count up its size. A single pearl weighs twenty kilograms, which is a logical variety. Furthermore, a pound’s excess weight means ten kilos. As a result, we must be able to separate a lb by 15, with out be worried about the length of an individual pearl.
## They may be depicted as a decimal
You’ve most likely seen a problem that involves a repeated fraction if you’ve ever tried to convert a number to its decimal form. A decimal amount may be composed as being a several of two integers, so four times 5 is equal to seven. A similar dilemma necessitates the recurring fraction 2/1, and both sides needs to be split by 99 to obtain the proper answer. But how would you create the transformation? Here are some illustrations.
A rational variety will also be printed in various forms, such as fractions along with a decimal. A good way to represent a reasonable variety inside a decimal is usually to divide it into its fractional comparable. There are actually 3 ways to separate a rational variety, and every one of these techniques produces its decimal counterpart. One of these ways is to break down it into its fractional equivalent, and that’s what’s referred to as a terminating decimal. |
# Permutations and combinations
Contents
### Combination
In familiar terms, a combination is an un-ordered selection made from a group of objects. For example, suppose you have fifty-two playing cards, and select five cards for a poker hand. It would not matter in which order the cards were drawn, because you could rearrange them in your hand.
In more formal terms, in combinatorics, a combination is a subset. In a set, the order does not matter. These are represented usually with curly braces: {2, 4, 6}. With sets, since order does not matter, you are only interested in what is present, not in what order. Thus,
{2, 4, 6} = {6, 4, 2}.
Also, {1, 1, 1} is the same as {1} because a set is defined by its elements; they don't usually appear more than once. See the article on combinations for more on this.
### Permutation
A permutation, on the other hand, is an ordered selection made from a group of objects. For example, if you look at credit card codes you must also look at the order. The order 5-3-7-5 is different from 3-7-5-5: the order in which you pick your numbers is significant.
Now suppose you have these:
1, 2, 3
Here is a list of all permutations of those, using three numbers per line:
1 2 3
1 3 2
2 1 3
2 3 1
3 1 2
3 2 1
The article on permutations contains greater detail.
### Repetitions
Both combinations and permutations have variants where some objects appear more than once (that is, they have some repetition). For example, if you wanted a combination of twelve fruits and you had 20 to choose from, you should be able to pick more than one fruit type.
## Summary of formulas
#### Permutation with repetition
When order matters and an object can be chosen more than once then the number of permutations is:
[itex] n^r[itex]
Where n is the number of objects from which you can choose and r is the number to be chosen.
For example, if you have the letters A, B, C, and D and you wish to discover the number of ways to arrange them in three letter patterns (trigrams) you find that there are 43 or 64 ways. This is because for the first slot you can choose any of the four values, for the second slot you can choose any of the four, and for the final slot you can choose any of the four letters. Multiplying them together gives the total.
#### Permutation without repetition
When the order matters and each object can be chosen only once, then the number of permutations is:
[itex] \frac{n!}{(n-r)!} [itex]
Where n is the number of objects from which you can choose and r is the number to be chosen.
For example, if you have five people and are going to choose three out of these, you will have 5!/(5-3)! = 60 permutations.
Note that if n = r (meaning number of chosen elements is equal to number of elements to choose from) then the formulae becomes
[itex] \frac{n!}{(n-n)!} = \frac{n!}{0!} = n![itex] when 0! = 1! = 1
For example, if you have three people and you want to find out how many ways you may arrange them it would be 3! or 3 × 2 × 1 = 6 ways. The reason for this is because you can choose from 3 for the initial slot, then you are left with only two to choose from for the second slot, and that leaves only one for the final slot. Multiplying them together gives the total.
#### Combination without repetition
When the order does not matter, but each object can be chosen only once, the number of combinations is:
[itex]{n!} \over {r!(n - r)!}[itex]
Where n is the number of objects from which you can choose and r is the number to be chosen.
For example, if you have ten numbers and wish to choose 5 you would have 10!/5!(10 − 5)! = 252 ways to choose.
#### Combination with repetition
When the order does not matter and an object can be chosen more than once, then the number of combinations is:
[itex]{(n + r - 1)!} \over {r!(n - 1)!}[itex]
Where n is the number of objects from which you can choose and r is the number to be chosen.
For example, if you have ten types of donuts to choose from and you want three donuts there are (10 + 3 − 1)! / 3!(10 − 1)! = 220 ways to choose.
• Art and Cultures
• Countries of the World (http://www.academickids.com/encyclopedia/index.php/Countries)
• Space and Astronomy |
## 2020-05-28
### Determine Perspective Lines With Off-page Vanishing Point
In perspective drawing, a vanishing point represents a group of parallel lines, in other words, a direction.
For any point on the paper, if we want a line towards the same direction (in the 3d space), we simply draw a line through it and the vanishing point.
But sometimes the vanishing point is too far away, such that it is outside the paper/canvas.
In this example, we have a point P and two perspective lines L1 and L2.
The vanishing point VP is naturally the intersection of L1 and L2. The task is to draw a line through P and VP, without having VP on the paper.
I am aware of a few traditional solutions:
1. Use extra pieces of paper such that we can extend L1 and L2 until we see VP.
2. Draw everything in a smaller scale, such that we can see both P and VP on the paper. Draw the line and scale everything back.
3. Draw a perspective grid using the Brewer Method.
#1 and #2 might be quite practical. #3 may not guarantee a solution, unless we can measure distances/proportions.
Below I'll describe a method I learned in high school. It was more of a math quiz than a practical drawing method though, but it is quite fun.
To make it more complicated, there is an extra constraint for the drawer: you can only draw with a ruler without measurements, that is you can draw a straight line between two given points (assume that the ruler is long enough), but you cannot measure lengths or angles. You can not find middle points either (e.g. by folding the paper).
Here we go:
Step 1:
Draw two lines through P, cutting L1 and L2 at A1, A2, B1 and B2.
Step 2:
Draw two lines (A1, B2) and (A2, B1).
Find the intersection point Q.
Step 3:
Draw a line (Q, P), cutting L1 and L2 at A3 and B3 respectively.
Step 4:
Draw two lines (A1, B3) and (A3, B2).
Find the intersection P'.
Step 5:
Draw a line through P and P'.
This is the desired perspective line through P.
The correctness can be formally proved with some calculation.
On the other hand, if we view Q as the second vanishing point, this figure represents a perspective view of a 2d rectangle (A1, A2, B1, B2). The rectangle is cut into two rectangles by the line (A3, B3). Note that (A1, B2), (A3, B3) and (A2, B1) are parallel lines in the 3d space.
Now it is clear that P is the center of the rectangle (A1, A2, B1, B2), P' is the center of the rectangle (A1, A3, B3, B2). Therefore the lines (A1, A2), (B1, B2) and (P, P') must be all parallel in the 3d space. In the perspective view they must all pass through a common vanishing point. This shows that the line PP' is the answer.
There are 2 special cases.
The first case is when Q is also far outside the paper. This may happen when P is near the center between L1 and L2.
To fix this, we can find the a perspective line between L1 and L2, using the standard bisection technique.
Now the new line is closer to P, we can apply the method above on L2 (or L1) and the new line.
If the new line is still too far from P, a few iterations of bisection should be enough.
The other special case is when P is not between L1 and L2.
In this case we may use standard extension method, to find a new perspective line that is on the other side of P. A few iterations might be needed.
Finally we can apply the method above with the new line and L1.
There are still other cases where this method won't always work. For example, if L1 and L2 are close to the top and bottom edge of the paper, there is not much we can do. After all this was only a math quiz in the first place. |
# Binary Tree
A binary tree is a tree where every node has two or fewer children. The children are usually called left and right.
public class BinaryTreeNode { public int value; public BinaryTreeNode left; public BinaryTreeNode right; public BinaryTreeNode(int value) { this.value = value; } }
This lets us build a structure like this:
That particular example is special because every level of the tree is completely full. There are no "gaps." We call this kind of tree "perfect."
Binary trees have a few interesting properties when they're perfect:
Property 1: the number of total nodes on each "level" doubles as we move down the tree.
Property 2: the number of nodes on the last level is equal to the sum of the number of nodes on all other levels (plus 1). In other words, about half of our nodes are on the last level.
Let's call the number of nodes n, and the height of the tree h. h can also be thought of as the "number of levels."
If we had h, how could we calculate n?
Let's just add up the number of nodes on each level! How many nodes are on each level?
If we zero-index the levels, the number of nodes on the xth level is exactly 2^x.
1. Level 0: 2^0 nodes,
2. Level 1: 2^1 nodes,
3. Level 2: 2^2 nodes,
4. Level 3: 2^3 nodes,
5. etc
So our total number of nodes is:
n = 2^0 + 2^1 + 2^2 + 2^3 + ... + 2^{h-1}
Why only up to 2^{h-1}? Notice that we started counting our levels at 0. So if we have h levels in total, the last level is actually the "h-1"-th level. That means the number of nodes on the last level is 2^{h-1}.
But we can simplify. Property 2 tells us that the number of nodes on the last level is (1 more than) half of the total number of nodes, so we can just take the number of nodes on the last level, multiply it by 2, and subtract 1 to get the number of nodes overall. We know the number of nodes on the last level is 2^{h-1}, So:
n = 2^{h-1} * 2 - 1 n = 2^{h-1} * 2^1 - 1 n = 2^{h-1+1}- 1 n = 2^{h} - 1
So that's how we can go from h to n. What about the other direction?
We need to bring the h down from the exponent. That's what logs are for!
First, some quick review. \log_{10} (100) simply means, "What power must you raise 10 to in order to get 100?". Which is 2, because 10^2 = 100.
We can use logs in algebra to bring variables down from exponents by exploiting the fact that we can simplify \log_{10}(10^2). What power must we raise 10 to in order to get 10^2? That's easy—it's 2.
So in this case we can take the \log_{2} of both sides:
n = 2^{h} - 1 n + 1 = 2^{h} \log_{2}{((n+1))} = \log_{2}{(2^{h})} \log_{2}{(n+1)} = h
So that's the relationship between height and total nodes in a perfect binary tree. |
# How do you factor 6y^2-5y+1?
Aug 7, 2015
$\left(2 y - 1\right) \left(3 y - 1\right)$
#### Explanation:
When factorizing a quadratic function look for factors of the square coefficient (in this case $6$) and factors of the constant term (in this case $1$).
In this case the coefficient of the $y$ term is negative and the constant term is positive, so the constants in each of the factors must both be negative. Since the constant is $+ 1$, its factors must be $- 1$ and $- 1$.
Integer factors of $6$, independent of order, are
(1,6); (2,3); ("-"1,"-"6); ("-"2,"-"3)
Now look for the factor pair that when multiplied by $\left(- 1 , - 1\right)$ sums to $- 5$. Here this is $\left(2 , 3\right)$. These are the coefficients of the $y$ terms in each factor.
Hence, the factors are $\left(2 y - 1\right)$ and $\left(3 y - 1\right)$.
Note: Not all quadratic functions can be factorized. |
# What is 12 percent of 2000 + Solution with Free Steps
12% of 2000 results in 240. The solution may be found by multiplying 2000 by the factor 0.12.
You may have to use 12% of 2000 in many real-world areas. Let us say that you want to apply to a college, and you know that this particular college has an acceptance rate of only 12%. Now, if 2000 students submit applications to this college, knowing how to calculate 12 percent of 2000, you would immediately know that only 240 students will be accepted. If you didn’t know this solution, it would be difficult to choose which college to apply for and which to skip. There are many other examples that we can consider where 12 percent of 2000 may be calculated.
## What Is 12 percent of 2000?
12% of 2000 is equal to the number 240. This result will be obtained by multiplying the factor 0.12 by 2000.
The solution to the question 12% of 2000 can be found by multiplying 12/100 fractional values with the number 2000. This answer may be further reduced to get the 12% of 2000, which is 240.
## How To Calculate 12% of 2000?
We may calculate what portion of 2000 equals 12% by using the simple mathematical procedure shown below:
### Step 1
Writing 12 percent of 2000 in mathematical form:
12 percent of 2000 = 12% x 2000
### Step 2
Substitute the % symbol with the fraction 1/100:
12 percent of 2000 = ( 12 x 1/100 ) x 2000
### Step 3
Rearranging ( 12 x 1/100 ) x 2000:
12 percent of 2000 = ( 12 x 2000 ) / 100
### Step 4
Multiplying 12 with 2000:
12 percent of 2000 = ( 24000 ) / 100
### Step 5
Dividing 24000 by 100:
12 percent of 2000 = 240
So, the 12 percent of 2000 is equal to 240
12 percent of 2000 can be visualized by using the following pie chart.
Figure 1: Pie Chart of 12 percent of 2000
The orange slice of the pie chart represents the 12% value of 2000, equal to 240. The green slice of the pie chart represents 88% of 2000, equal to 1760. The total area of the pie chart represents 100% of 2000, equal to 2000.
The percentage is a way of describing numbers as parts per 100. It can be applied in many areas of science and technology.
All the Mathematical drawings/images are created using GeoGebra. |
## LESSON 7-2 PROBLEM SOLVING FACTORING BY GCF
Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. How to Succeed in a Math Class for some more suggestions. Broken down into individual steps, here’s how to do it you can also follow this process in the example below. In the next two tutorials we will add on other types of factoring. Sometimes, you will encounter polynomials that, despite your best efforts, cannot be factored into the product of two binomials. This method of factoring only works in some cases. A prime factor is similar to a prime number —it has only itself and 1 as factors.
In this lesson we will study polynomials that can be factored using the Greatest Common Factor. Factor out the GCF: The correct answer is a 5 8 a — There are several methods that can be used when factoring polynomials. If there is no common factor for all of the terms in the polynomial, another technique needs to be used to see if the polynomial can be factored. Factor 3 out of the second group. The polynomial is now factored.
# Greatest Common Factor
Students Homeschool Adults Teachers. Example Problem Find the greatest common factor of and You must be able to factor out of every term in order to identify the GCF. Math works just like anything else, if you want to get good at it, then you need to practice it. Students Homeschoolers Adults Teachers. If you want to test this, go ahead and divide both and by 42—they are both evenly divisible by this number!
ALGEBRA 1 PRENTICE HALL PRACTICE AND PROBLEM SOLVING WORKBOOK ANSWERS
In this tutorial we are going to look at two ways to factor polynomial expressions, factoring out the greatest common factor and factoring by grouping. C a 5 a — 1 Incorrect. Comments We would love to hear what you have to say about this page! As shown above when we divide 2 x by 2 x we get 1, so we need a 1 as the third term inside of the. The polynomial is now factored.
## Try Factor a Polynomial by Finding Its Greatest Common Factor
Putting this together we have a GCF of 3 xy. If you have four terms with no GCF, then try factoring by grouping.
Factoring gives you another way to write the expression so it will be equivalent to the original problem. The largest monomial that we can factor out of each term is. Even the best athletes and musicians had help along the way and lots of practice, practice, practice, to get good at their sport or instrument. Note how they all have an xso it look like x will be involved.
They are the numbers that you can multiply together to produce another number: Because the GCF is fwctoring product of the prime factors that these numbers have in common, you know that it is a factor of both numbers. Factor out the GCF of a polynomial.
# Factoring polynomials by taking a common factor (article) | Khan Academy
What do you think? Find the greatest common factor of 25 b 3 and 10 b 2. Properties of Real Numbers. You can then use the distributive property to rewrite the polynomial in a factored form. We need to figure out what the largest monomial that we can divide out of each of these terms would be. The fully factored polynomial will be the product of two binomials.
RESEARCH PAPER TIPS REDDIT
Factor 3 out of the second group. Factoring is to write an expression as a product of factors. In the example above, each pair can be factored, but then there is no common factor between the pairs!
Before we get started, it may be helpful for you to review the Dividing Monomials lesson. When factoring a four-term polynomial using grouping, find the common factor of pairs of terms rather than the whole polynomial. If we use the exponent 8, we are in trouble. Group the first two terms together and then the last two terms together. Gcff terms of the polynomial into pairs.
To be in factored form, it must be written as a product of factors. B 8 y Correct. Likewise to factor a polynomialyou rewrite it as a product. |
# Difference between revisions of "2015 AIME I Problems/Problem 9"
## Problem
Let $S$ be the set of all ordered triple of integers $(a_1,a_2,a_3)$ with $1 \le a_1,a_2,a_3 \le 10$. Each ordered triple in $S$ generates a sequence according to the rule $a_n=a_{n-1}\cdot | a_{n-2}-a_{n-3} |$ for all $n\ge 4$. Find the number of such sequences for which $a_n=0$ for some $n$.
## Solution
Let $a_1=x, a_2=y, a_3=z$. First note that if any absolute value equals 0, then $a_n=0$. Also note that if at any position, $a_n=a_{n-1}$, then $a_{n+2}=0$. Then, if any absolute value equals 1, then $a_n=0$. Therefore, if either $|y-x|$ or $|z-y|$ is less than or equal to 1, then that ordered triple meets the criteria. Assume that to be the only way the criteria is met. To prove, let $|y-x|>1$, and $|z-y|>1$. Then, $a_4 \ge 2z$, $a_5 \ge 4z$, and $a_6 \ge 4z$. However, since the minimum values of $a_5$ and $a_6$ are equal, there must be a scenario where the criteria was met that does not meet our earlier scenarios. Calculation shows that to be $z=1$, $|y-x|=2$. Again assume that any other scenario will not meet criteria. To prove, divide the other scenarios into two cases: $z>1$, $|y-x|>1$, and $|z-y|>1$; and $z=1$, $|y-x|>2$, and $|z-y|>1$. For the first one, $a_4 \ge 2z$, $a_5 \ge 4z$, $a_6 \ge 8z$, and $a_7 \ge 16z$, by which point we see that this function diverges. For the second one, $a_4 \ge 3$, $a_5 \ge 6$, $a_6 \ge 18$, and $a_7 \ge 54$, by which point we see that this function diverges. Therefore, the only scenarios where $a_n=0$ is when any of the following are met: $|y-x|<2$ (280 options) $|z-y|<2$ (280 options, 80 of which coincide with option 1) $z=1$, $|y-x|=2$. (16 options, 2 of which coincide with either option 1 or option 2) Adding the total number of such ordered triples yields $280+280-80+16-2=\boxed{494}$.
## See Also
2015 AIME I (Problems • Answer Key • Resources) Preceded byProblem 8 Followed byProblem 10 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 All AIME Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.
Invalid username
Login to AoPS |
# Find all first partial derivatives, and evaluate each at the given point. f(x,y)=x^2-y,(0,2)
Find all first partial derivatives, and evaluate each at the given point. $f\left(x,y\right)={x}^{2}-y,\left(0,2\right)$
You can still ask an expert for help
## Want to know more about Derivatives?
• Questions are typically answered in as fast as 30 minutes
Solve your problem for the price of one coffee
• Math expert for every subject
• Pay only if we can solve it
Theodore Schwartz
Step 1
The given function is $f\left(x,y\right)={x}^{2}-y$ and a point is (0, 2).
Compute the partial derivatives as follows.
${f}_{x}\left(x,y\right)=\frac{\partial }{\partial x}\left({x}^{2}-y\right)$
$=2x-0$
$=2x$
${f}_{y}\left(x,y\right)=\frac{\partial }{\partial y}\left({x}^{2}-y\right)$
$=0-1=-1$
Thus, the first partial derivatives are ${f}_{x}=2x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{y}=-1$.
Step 2
Substitute $\left(x,y\right)=\left(0,2\right)\in {f}_{x}=2x\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{y}=-1$ and obtain that,
${f}_{x}\left(0,2\right)=0\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{f}_{y}\left(0,2\right)=-1$ |
Basic Rule of Logarithmic Differentiation
$$\displaystyle \dfrac{d}{dx}\log_e{x} = \dfrac{1}{x} \\ \dfrac{d}{dx}\log_e{f(x)} = \dfrac{f'(x)}{f(x)}$$
Practice Questions
Question 1
Differentiate $y = \log_{e}(3x)$.
\begin{aligned} \displaystyle \dfrac{d}{dx}\log_{e}(3x) &= \dfrac{(3x)’}{3x} \\ &= \dfrac{3}{3x} \\ &= \dfrac{1}{x} \end{aligned}
Question 2
Differentiate $y = \log_{e}(2x-1)$.
\begin{aligned} \displaystyle \dfrac{d}{dx}\log_{e}(2x-1) &= \dfrac{(2x-1)’}{2x-1} \\ &= \dfrac{2}{2x-1} \end{aligned}
Question 3
Differentiate $x^2\log_{e}x$.
\begin{aligned} \displaystyle \require{color} \dfrac{d}{dx}x^2\log_{e}x &= 2x\log_{e}x + x^2 \times \dfrac{1}{x} \ \ \ \ \color{red} \text{product rule} \\ &= 2x\log_{e}x + x \\ \end{aligned}
Question 4
Differentiate $\log_{e}\dfrac{x}{x-1}$.
\begin{aligned} \displaystyle \log_{e}\dfrac{x}{x-1} &= \log_{e}x – \log_{e}(x-1) \\ \dfrac{d}{dx}\log_{e}\dfrac{x}{x-1} &= \dfrac{1}{x} – \dfrac{1}{x-1} \end{aligned} |
# Find the integrals of the functions$\frac{1}{\sin x\cos^3x}$
$\begin{array}{1 1} \frac{\tan^2x}{2}+\log|\tan x|+c \\ \frac{\tan^2x}{2}-\log|\tan x|+c \\ \frac{\cot^2x}{2}+\log|\tan x|+c \\ \frac{\tan^2x}{2}+\log|\cot x|+c \end{array}$
Toolbox:
• $(i)\;\sin^2x+\cos^2x=1.$
• $(ii)\;Method\;of\;substitution.$
• Let f(x)=t.
• Therefore f'(x)dx=dt.Thus $\int f(x)dx=\int t.dt.$
Given:$I=\int\frac{1}{\sin x\cos x}dx.$
This can be written as,
$I=\frac{\sin^2x+\cos^2x}{\sin x\cos^3x}dx.$
Separating the terms we get,
$I=\int\frac{\sin^2x}{\sin x\cos^3x}dx+\int\frac{\cos^2x}{\sin x\cos^3x}dx.$
$\;\;\;=\int\frac{\sin x}{\cos x}\frac{1}{\cos^2x}dx+\int\frac{\cos x}{\sin x}\frac{1}{\cos^2x}dx.$
$\;\;\;=\int\tan x.\sec xdx+\int\frac{\sec^2}{\tan x}dx.$
Let tan x=t.
On differentiating we get,
$\sec^2xdx=dt.$
By substituting t we get,
$I=\int t.dt+\int\frac{1}{t}.dt.$
On integrating we get,
$\;\;\;=\frac{t^2}{2}+log |t|+c.$
Substituting for t we get,
$\int\frac{\sin^2x+\cos ^2x}{\sin x\cos^3x}dx=\frac{\tan^2x}{2}+log|\tan x|+c.$ |
# How to calculate the value of 19 to the 28th power and the number of digits [Easy to understand]
19 to the power of 28 is 638411683925748518131605316913942641
This is how the formula looks like:
$19^{28}=$
638411683925748518131605316913942641
Also, $19^{28}$ has 36 digits.
Here, we explain how to solve $19^{28}$ and how to find the number of digits of $19^{28}$.
This site is created by Dr. Thomsontom labIt operates under the name
## Calculating 19 to the 28th Power
19 to the 28th power is simply 19 multiplied by 28 times.
As a calculation method, basically there is no other way but to multiply atai1 by atai2 times.
Also, you can use google search.
For example, if you search for "14 to the 21st power" on google, a calculator will come up and tell you the answer.
As explained above, calculating the power takes time, so you may want to find out how many digits the value of the power is.
Next, let's find the number of digits in $19^{28}$.
## Number of digits in 19 to the 28th power
Calculating $19^{28}$ gives us 36 digits.
### Find the number of digits in 19 to the 28th power
Let's calculate the common logarithm of 19 to the 28th power.
\begin{eqnarray}
\log_{10}19^{28}&=&28 \log_{10}19\\
&=&28\times 1.2787\cdots\\
&=&35.805
\end{eqnarray}
In other words,
We can say that $19^{28}=10^{35.805}$, so we know that $19^{28}$ has 36 digits.
### How to find the number of digits
To find the number of digits in $19^{28}$, use common logarithms.
By using the common logarithm, we can calculate the power of 10, so we know the number of digits.
For example, $10^1=10$ is 2 digits.
On the other hand, $10^2=100$, so 3 digits.
So $10^a$ has $10+1$ digits.
If $a$ is a decimal, the number of digits is the integer part plus 1.
$a=11.34$ will be 12 digits.
## power size quiz
Q1
Which one is bigger?
$12 ^ 5$
$5^{12}$
Share it if you like! |
# Miscellaneous
## Percent of a Percent
x% of y = $\frac{x}{100}$ × y
z% of x% of y = $\frac{z}{100}$ × $\frac{x}{100}$ × y
and so on.
You will not that the order in which we apply these percentages is not important:
z% of x% of y = $\frac{z}{100}$ × $\frac{x}{100}$ × y
x% of z% of y = $\frac{x}{100}$ × $\frac{z}{100}$ × y
So, z% of x% of y = x% of z% of y
##### Q. What is 20% of 50% of 90?
Explanations :
Explanation 1:
50% of 90 = $\frac{50}{100}$ × 90 = 45
20% of 45 = $\frac{20}{100}$ × 45 = 9
Explanation 2: Using the concept of Multiplying Factors
20% of 50% of 90 = 0.2 × 0.5 × 90 = 9
## A percentage increase is how many times the original value?
If the percentage increase is 100%, then the new value of something is 2 times the previous given value.
If the percentage increase is 200%, then the new value of something is 3 times the previous given value.
If the percentage increase is 50%, then the new value of something is 1.5 times the previous given value.
So, we can generalize:
If the percentage increase is p%, then the new value of something is (1 + $\frac{p}{100}$) times the old value.
Or we can say that:
If the new value of something is p times the old value, then the percentage increase is (p - 1)100%.
##### Q. If salary of Mragank is increases by 150%, then his new salary is how many times his earlier salary?
Explanation:
If the percentage increase is p%, then the new value of something is (1 + $\frac{p}{100}$) times the old value.
So, New salary = (1 + $\frac{p}{100}$) times the earlier salary = (1 + $\frac{150}{100}$) times the earlier salary = 2.5 times the earlier salary
##### Q. If X = 5.25 Y, then by how much percent is X more than Y?
Explanation:
X is 5.25 times the value of Y.
So, Percentage increase from Y to X = (p - 1)100% = (5.25 -1)100 = 425%
## p times Vs. p times more
Compare these two:
• If the new value of something is p times the previous given value, then the percentage increase is (p - 1)100%.
• If the new value of something is p times more than the previous given value, then the percentage increase is 100p%.
##### Q. If price of onion increases to 3 times of its previous value, then what is the percent increase?
Explanation:
If initial price was Rs 100, new price = Rs 300
If the new value of something is p times the previous given value, then the percentage increase is (p - 1)100%.
So, Percentage increase = (p - 1)100% = (3 - 1)100 = 200%
##### Q. If price of onion becomes 3 times more than its previous value, then what is the percent increase?
Explanation:
If initial price was Rs 100, new price = 100 + 300 = Rs 400
If the new value of something is p times more than the previous given value, then the percentage increase is 100p%.
So, Percentage increase = 100p% = 300%
Previous
Share on: |
# Chapter 8 Joint Distributions
So far, we learned about joint probabilities in Bayesian context such as $$P(A|B) = P(A,B)/P(B)$$. Now, we are going to expand this concept into discrete and continuous distributions. Define $$P(X = x, Y=y) = f(x,y)$$ as the probability mass function (discrete) or probability density function (continuous).
Same probability laws apply to joint distributions as well.
• $$f(x,y) \ge 0$$ for all $$(x,y)$$.
• $$\sum_x \sum_y f(x,y) = 1$$ or $$\int_x \int_y f(x,y) dx dy = 1$$
Example (Discrete): Suppose there are 10 balls in a box; 3 white, 4 black and 3 red. Two balls are randomly selected. Let’s say random variable X is the number of white balls picked and r.v. Y is the number of black balls picked. (a) Find the joint probability function and (b) find the probabilities.
1. Let’s first enumerate the alternatives. $$(x,y)$$ pair can be either of $$(0,0),(0,1),(0,2),(1,1),(2,0),(1,0)$$. Total number of alternatives are $$\binom{10}{2}$$. To calculate, number of ways of getting 1 white and 1 black ball is $$\binom{3}{1}\binom{4}{1}\binom{3}{0}$$. So, the probability will be $$\dfrac{\binom{3}{1}\binom{4}{1}\binom{3}{0}}{\binom{10}{2}}$$. We can generalize it to a function.
$f(x,y) = \dfrac{\binom{3}{x}\binom{4}{y}\binom{3}{2-x-y}}{\binom{10}{2}}$
Let’s also make it into an R function
f_xy_ballpick <- function(x,y,picked=2,n_balls=10,n_x=3,n_y=4){
#picked is the number of balls picked
#n_balls is the total number of balls
#n_x is the number of balls belonging to rv X (white)
#n_y is the number of balls belonging to rv Y (black)
#x and y are values to our random variables their total cannot exceed picked
#If the sum of x and y is greater than picked, then its probability is zero.
if(x+y > picked){
return(0)
}
#Remember choose is the R function of binomial coefficient (or combination)
(choose(n_x,x)*choose(n_y,y)*choose(n_balls - n_x - n_y,picked-x-y))/choose(n_balls,picked)
}
f_xy_ballpick(x=1,y=1)
## [1] 0.2666667
1. Using the above formula we can calculate all the probabilities within the specified region $$x+y \le 2$$.
#First create an empty probability matrix.
#Let's say that columns are x = 0,1,2 and rows are y = 0, 1, 2
prob_matrix<-matrix(0,ncol=3,nrow=3)
#Indices in R start from 1 so 1,1 is actually x=0,y=0
prob_matrix[1,1]<-f_xy_ballpick(x=0,y=0)
prob_matrix[1,2]<-f_xy_ballpick(x=1,y=0)
prob_matrix[1,3]<-f_xy_ballpick(x=2,y=0)
prob_matrix[2,1]<-f_xy_ballpick(x=0,y=1)
prob_matrix[2,2]<-f_xy_ballpick(x=1,y=1)
prob_matrix[2,3]<-f_xy_ballpick(x=2,y=1)
prob_matrix[3,1]<-f_xy_ballpick(x=0,y=2)
prob_matrix[3,2]<-f_xy_ballpick(x=1,y=2)
prob_matrix[3,3]<-f_xy_ballpick(x=2,y=2)
#Let's also define the colnames and rownames of the matrix.
#paste0 is an R command which just appends statements
colnames(prob_matrix) <- paste0("x_",0:2)
rownames(prob_matrix) <- paste0("y_",0:2)
round(prob_matrix,2)
## x_0 x_1 x_2
## y_0 0.07 0.20 0.07
## y_1 0.27 0.27 0.00
## y_2 0.13 0.00 0.00
Example (continuous): (This is from the textbook, Example 3.15) A privately owned business operates both a drive-in facility and a walk-in facility. On a randomly selected day, let X and Y, respectively, be the proportions of time that the drive-in and the walk-in facilities are in use and suppose that the joint density function of these random variables is
$f(x,y) = \dfrac{2}{5}(2x + 3y), 0 \le x \le 1, 0 \le y \le 1$
and 0 for other values of x and y.
1. Verify $$\int_x \int_y f(x,y) dx dy = 1$$
2. Find $$P[(X,Y) \in A]$$, where $$A = \{(x,y)|0 < x < 1/2, 1/4 < y < 1/2\}$$
3. (see the book for the full calculations)
$\int_x \int_y f(x,y) dx dy = \int_0^1 \int_0^1 \dfrac{2}{5}(2x+3y) dx dy = 1$
1. (see the book for the full calculations)
$\int_x \int_y f(x,y) dx dy = \int_{1/4}^{1/2} \int_0^{1/2} \dfrac{2}{5}(2x+3y) dx dy = 13/160$
Example (with special distributions)
1. Patients arrive at the doctor’s office according to Poisson distribution with $$\lambda = 2$$/hour.
1. What is the probability of getting less than or equal to 2 patients within 2 hours?
2. Suppose each arriving patient has 50% chance to bring a person to accompany. There are 10 seats in the waiting room. At least many hours should pass that there is at least 50% probability that the waiting room is filled with patients and their relatives?
Solution
1. $$P(X\le 2|\lambda t = 2)= \sum_{i=0}^2 \dfrac{e^{-\lambda t}(\lambda t)^i}{i!}$$
#cdf of poisson
ppois(2,lambda=2*2)
## [1] 0.2381033
1. First let's define the problem. Define $$n_p$$ as the number of patients and $$n_c$$ is the number of company. We want $$n_p + n_c \ge 10$$ with probability 50% or higher for a given $$t^*$$. Or to paraphrase, we want $$n_p + n_c \le 9$$ w.p. 50% or lower.
What is $$n_c$$ affected by? $$n_p$$. It is actually a binomial distribution problem. $$P(n_c = i|n_p) = \binom{n_p}{i} (0.5)^i*(0.5)^{n_p-i}$$. It is even better if we use cdf $$P(n_c \le k|n_p) = \sum_{i=0}^{k} \binom{n_p}{i} (0.5)^i*(0.5)^{n_p-i}$$.
We know the arrival of the patients is distributed with poisson. So, $$P(n_p = j|\lambda t^*) = \dfrac{e^{-\lambda t}(\lambda t)^j}{j!}$$. So $$P(j + k \le N) = \sum_{a=0}^j P(n_p = a|\lambda t^*)*P(n_c \le N-a | n_p = a)$$. Remember it is always $$n_c \le n_p$$.
#Let's define a function
calculate_probability<-function(N=9,t_star=1,lambda=2){
#N is the max desired number of patients
the_prob<-0
for(n_p in 0:N){
the_prob <- the_prob + dpois(n_p,lambda=lambda*t_star)*pbinom(q=min(N-n_p,n_p),size=n_p,prob=0.5)
}
return(the_prob)
}
#Try different t_stars so probability is below 0.5
calculate_probability(t_star=2)
## [1] 0.8631867
calculate_probability(t_star=3)
## [1] 0.5810261
calculate_probability(t_star=3.3)
## [1] 0.4905249
### 8.0.1 Marginal Distributions
You can get the marginal distributions by just summing up or integrating the other random variable such as $$P(Y=y) = \sum_x f(x,y)$$ or $$f(y) = \int_x f(x,y) dx$$. Let’s calculate the marginal distribution of black balls (rv Y) in the above example.
#Let's recall the prob_matrix
round(prob_matrix,2)
## x_0 x_1 x_2
## y_0 0.07 0.20 0.07
## y_1 0.27 0.27 0.00
## y_2 0.13 0.00 0.00
#rowSums is an R function that calculates the sum of each row.
#It is equivalent to y_0 = prob_matrix[1,1] + prob_matrix[1,2] + prob_matrix[1,3]
rowSums(prob_matrix)
## y_0 y_1 y_2
## 0.3333333 0.5333333 0.1333333
Marginal distribution of y in the second example is calculated as follows.
$\int_x \dfrac{2}{5}(2x+3y) dx = \dfrac{2(1+3y)}{5}$
### 8.0.2 Conditional Distribution
Similar to Bayes’ Rule, it is possible to calculate conditional probabilities of joint distributions. Let’s denote g(x) as the marginal distribution of x and h(y) as the marginal distribution of y. The formula of conditional distribution of x given y is as follows.
$f(x|y) = f(x,y)/h(y)$
Note that conditional distribution function is useless if x and y are independent. ($$f(x|y)=f(x)$$) |
1-646-564-2231
Key Concepts
• Make Ten frames.
• Make Number Bonds.
• Use the ten frames to make a 10 to add.
Introduction
Make Ten frames
Example 1:
Guess how many balls need to be added to make 10?
Solution:
Add 1 more ball to make the ten frames.
Example 2:
Guess how many balls need to be added to make 10?
Solution:
Add 3 more balls to make the ten frames.
Make Number Bonds
Example 3:
Fill in the missing number for the given number bond and write the fact.
5 + ______ = 10
Solution:
5 + 5 = 10
Example 4:
Fill in the missing number for the given number bond and write the fact.
______ + 5 = 8
Solution:
3 + 5 = 8
Example 5:
Find 8 + 6
Solution:
Example 6:
Find 7 + 5
Solution:
Exercise
1. 1 + 10 =
2. 3 more than 10 is___________
3. 10 and 2 make __________
4. 8 more than 10 is _________
5. Make a 10 to add ________
7 + _____ = _____
10 + ____ = _____
6. Find the missing number.
a. 8 + 5 = ____ +3
b. 6 + 9 = 10 + _____
7. Find the missing number.
a. 8 + 9 = 10 + ____
b. 9 + 6 = ____
8. Tan’s team scored 16 points in a game. During the first half, they scored 9 points. How many points did the team score in the finals?
9. Use the ten frames and find 5 + 7 by making a ten.
What have we learned
How to solve problems by using equations, drawings and arrays?
More Related TOPICS
Composite Figures – Area and Volume
A composite figure is made up of simple geometric shapes. It is a 2-dimensional figure of
Special Right Triangles: Types, Formulas, with Solved Examples.
Learn all about special right triangles- their types, formulas, and examples explained in detail for a
Ways to Simplify Algebraic Expressions
Simplify algebraic expressions in Mathematics is a collection of various numeric expressions that multiple philosophers and |
# SOCI2000 Workshop 2: Getting a quick feel for your data (Correlation)
‘Chapter 7: Correlation’ in Andy Field, 2017. Discovering Statistics Using IBM SPSS Statistics. Sage.
# Introduction
One of the most useful forms of analysis to do with your dataset first is a bi-variate correlation.
This allows us to see which of our independent varibles are strongly correlated with our dependent (outcome) variable.
The correct correlation to use is almost always Pearson’s correlation.
# Theory
We’re often interested in a simple expression of the relationship between one variable and another.
We can broadly think about three different ways two variables can be related:
• they can be positively correlated (the more you eat, the fuller you feel) - the first two scatterplots in Figure 1.
• they can be negatively correlated (the more you eat, the less food is on your plate) - the last two scatter plots in Figure 1
• there can be no correlation (the number of sunspots, and how much you eat) - the middle scatterplot in Figure 1
Correlation coefficients, such as the Pearson correlation coefficient, allow us to put an exact number on the size of the correlation between two variables.
The formal equation for pearson’s correlation coefficient (also called ‘r’) is
$r = \frac{\sum_{i=1}^n (x_i - \bar x)(y_i - \bar y)}{(N-1)s_x s_y}$
$Where:$ $\bar x = \text{mean of x}$ $\bar y = \text{mean of y}$ $\bar N = \text{number of observations}$ $s_x = \text{standard deviation of x}$ $s_y = \text{standard deviation of y}$
While this equation may be scary, the general interpretation of correlation coefficients not difficult:
• correlation coefficients vary from -1 to 1
• positive values (i.e. greater than zero, up to 1) represent a positive correlation
• negative values (i.e. less than zero, down to -1) represent a negative correlation
• values close to zero mean that there is no significant relationship between the two variables (the two variables are independent of each other)
• larger values (in absolute terms - closer to 1 or -1) reflect a stronger relationship
• A rule of thumb is that a Pearson’s correlation of 0.1-0.2 represents a weak relationship, 0.3-0.4 a moderate one and 0.5+ a strong relationship (all of these are absolute values).
Figure 2, below, shows visualisations of different correlation coefficients.
# Two Tables
There are two main types of correlation tables you might want to produce:
1. a table which shows the correlation of all the independent variables with the dependent variable/s.
2. a matrix which shows the correlation of all variables with all other varibles
Figures 3 and 4 below show examples of these.
Figures 3 and 4 below show examples of these.
Figures 3 and 4 below show examples of these.
# Pearsons Correlation
1. Go to Analyze > Correlate > Bivariate
The left window contains all the variables in your dataset NOT in the correlation. The right window contains all the variables in the correlation.
1. For this analysis, you want everything in the analysis. So click on one of the variables in the left window and then press ‘Control + A’. This will ‘select all’.
2. Then click the arrow in the middle of the screen. This will move all the variables to the right.
3. The default options are fine, so then just click “OK”
1. You should see a huge, almost unreadable, table like the one below. This version can be useful for identifying correlations between your independent variables.
1. Double click anywhere on the table, and ‘Pivot Table’ should appear
2. If the ‘Pivot Trays’ are not showing go to Pivot>Pivot Trays
1. The Pivot Trays should appear as below.
1. Click and hold the vertically written word ‘Variables’, and drag it to the top white box on the ‘LAYER’
1. On the Pivot Table, you will see a drop down box. Click on this and select your dependent variable. You will now have a list of the bivariate correlations of all your independent variables with your dependent variable.
## Making Publishable Tables
If you are writing a report and you need to put these descriptive statistics into a report, then DO NOT just make a screenshot of the SPSS output.
Instead, what you should do is:
1. right click on the table you want to copy, and select ‘copy’
1. open Excel, and then paste as text (this will strip out formating). Option could be called “Keep Text Only (T)” or “Match destination formating (M)”.
1. In Excel, delete the rows for N (number of cases), and the significance (it is already indicated in the stars /* and // )
2. Orient the column headings vertically (select cells, then right click>Format cells…>Alignment>Orientation)
3. Create three vertical lines in the table: (1) at the top of the table, (2) at the bottom of the table, and (3) under the column headings.
4. Align the columns with numbers in them to the centre (keep the first column, with the variable names, left-aligned)
5. AutoFit the column widths by selecting all the columns and then double clicking boundary between any two column headings.
1. Turn off gridlines (View>Gridlines)
2. Take a screenshot of the table and paste into your report. |
# Lesson 5
Fitting Lines
## 5.1: Selecting the Best Line (5 minutes)
### Warm-up
The mathematical purpose of this activity is for students to be able to visually assess the best line that fits data among a set of choices. Students are given a scatter plot and 2 lines that may fit the data. Students must select the line that better fits the data. The given lines address many common errors in student thinking about best fit lines including: going through the most points, dividing the data in half, and connecting the points on both ends of the scatter plot.
Listen for students using the terms slope and $$y$$-intercept.
### Launch
Provide students access to the images. Give students 2 minutes of quiet time to work the questions.
### Student Facing
Which of the lines is the best fit for the data in each scatter plot? Explain your reasoning.
1.
2.
3.
4.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
The purpose of this discussion is to understand bad fit, good fit, and best fit. In each scatter plot, the solid line represents the line of best fit—except for the last two graphs, for which the dashed line is the best fit.
Ask a student who uses the term slope while working the questions, “Can you explain the relationship between the two lines in question about runs and wins using the concept of slope?” (The slope of the dashed line is positive and the slope of the solid line is negative.)
Ask a student who uses the term $$y$$-intercept, “Can you explain the significance of the $$y$$-intercept in the question about average survey scores and amount spent on dinner?” (The solid line will have a $$y$$-intercept less than the $$y$$-intercept for the dashed line. Because the two lines have approximately the same slope, they appear parallel in the scatter plot.)
If time permits, discuss questions such as:
• “Is the dashed line in question about runs and wins a bad fit, good fit, or best fit?” (The line is a bad fit because it does not show the correct relationship between the variables. It shows that the value of $$y$$ increases as the value of $$x$$ increases, rather than the value of $$y$$ decreasing as the value of $$x$$ increases.)
• “Is the dashed line in question about oil and gas a bad fit, good fit, or best fit?” (It is the best fit because it is close to going through the middle of the data and follows the same trend as the data.)
• “What factors helped you select the linear model that fits the data best?” (The line should go through the middle of the data, follow the trend of the data, and have a similar number of points on each side of the line.)
## 5.2: Card Sort: Data Patterns (15 minutes)
### Activity
The mathematical purpose of this activity is for students to:
• distinguish between linear and nonlinear relationships in bivariate, numerical data
• informally assess the fit of a linear model
• compare the slope and the vertical intercepts of different linear models
• describe the relationship between two variables.
Students are given cards showing scatter plots and a linear model. They sort the cards based on how well the lines fit the data as well as by slope and intercept in increasing order.
A sorting task gives students opportunities to analyze representations, statements, and structures closely and make connections (MP2, MP7).
### Launch
Arrange students in groups of 2. Give students a chance to familiarize themselves with what is on the cards. For example, you might ask them to sort the cards into categories of their choosing, and explain their categories to their partner.
Conversing: MLR8 Discussion Supports. In pairs, ask students to take turns sorting the cards and explaining their reasoning to their partner. Display the following sentence frames for all to see: “ _____ should be before _____ because . . . .”, and “I noticed _____ , so I . . . .” Encourage students to challenge each other when they disagree. This will help students clarify their reasoning about linear models.
Design Principle(s): Support sense-making; Maximize meta-awareness
Action and Expression: Internalize Executive Functions. Provide students with a template for organizing their observations. Provide a template or invite students to fold a blank piece of paper in thirds, and label with three headers of “$$x$$,” “$$y$$” and “linear model fits?” to collect their answers. Explain that in the first column, they will always write increasing, since they will be reading each graph from left to right, then have them fill in the behavior of the y values in the next column, and in the last column, their conclusions about whether a linear model fits well.
Supports accessibility for: Language; Organization
### Student Facing
Your teacher will give you a set of cards that show scatter plots.
1. Arrange all the cards in three different ways. Ensure that you and your partner agree on the arrangement before moving on to the next one. Sort all the cards in order from:
1. best to worst for representing with a linear model
2. least to greatest slope of a linear model that fits the data well
3. least to greatest vertical intercept of a linear model that fits the data well
2. For each card, write a sentence that describes how $$y$$ changes as $$x$$ increases and whether the linear model is a good fit for the data or not.
### Student Response
For access, consult one of our IM Certified Partners.
### Activity Synthesis
The purpose of this discussion is for students to discuss the goodness of fit for linear models.
Here are some questions for discussion.
• “How are scatter plots of A and F the same? How are they different?” (They have the same slope and the linear model for each scatter plot are equally well fit. They are different because they have a different vertical intercept.)
• “How do you know if a linear model is a good fit?” (You need to look at the scatter plot and the line of best fit and make a decision about whether or not the data follows a linear trend.)
• “Why is the goodness fit for the linear model in scatter plot B better than the fit for the linear model in scatter plot A?” (The data in B falls on or very close to the linear model. The data in A is scattered around the line of best fit and has roughly the same number of values below that line of best fit as it does above the line of best fit.)
## 5.3: Fitting Lines with Technology (15 minutes)
### Activity
It is recommended to use the digital version of this activity. The mathematical purpose of this activity is for students to use technology to compute a line of best fit for data given in a table, and to understand the meaning of the slope and $$y$$-intercept. If the digital version of the activity is not available, the students should be guided through using available technology to find the least-squares regression line as the line of best fit.
### Launch
Assign one data table corresponding to the graphs in the previous activity to each group for question 5. Show students how to use the copy and paste features of their digital devices. Instruct students to copy the values from the data table and paste them into a blank line in the applet provided. The technology creates a scatter plot of the data in a table, and a movable line is already graphed for students to adjust. After groups have had a chance to estimate the best fit lines, pause the class. Show students how to use technology to find the least-squares regression line for the data and display the line with the scatter plot. Desmos can graph a line of best fit if you type in an equation, using a tilde (~) instead of an equals sign (=). You need to name the variables, $$x$$ and $$y$$, the same way they are named in the table, including the subscripts. The parameters can be named with any letter other than $$x$$, $$y$$, and $$e$$. For example, type $$y_1 \sim ax_1 + b$$, and Desmos will compute the values for the parameters $$a$$ and $$b$$.
If you will be using graphing technology other than Desmos for this activity, you may need to prepare alternate instructions.
Engagement: Develop Effort and Persistence. Provide prompts, reminders, or checklists that focus on increasing the length of on-task orientation in the face of distractions. For example, provide two copies of the steps: graph the table, find the best fit line, find the slope, and find the $$y$$-intercept. Include phrases to activate knowledge from prior activities such as, “As x increases . . .”
Supports accessibility for: Organization; Conceptual processing; Attention
### Student Facing
The weight of ice cream sold in a day at a small store in pounds ($$x$$) and the average temperature outside during the day in degrees Celsius ($$y$$) are recorded in the table.
$$x$$ $$y$$
20 6
18 4.5
21 6.5
17 3.5
21.5 7.5
19.5 6.5
21 7
18 5
1. For this data, create a scatter plot and a line that fits the data well.
2. Use technology to compute the best fit line. Round any numbers to 2 decimal places.
3. What are the values for the slope and $$y$$-intercept for the best fit line? What do these values mean in this situation?
4. Use the best fit line to predict the $$y$$ value when $$x$$ is 10. Is this a good estimate for the data? Explain your reasoning.
5. Your teacher will assign a table of data from the previous activity. Following your teacher’s directions, use technology and the table of data to create a scatter plot that also shows the line of best fit, and then interpret the slope and $$y$$-intercept.
Tables for last question:
A. (card A in the previous activity)
$$x$$ $$y$$
1 2
2.2 4
3.3 5
3.3 4.5
3.6 6
3.8 6.5
3.9 5.7
4 7
4.4 6.5
4.5 7
4.7 7
4.8 6
4.9 8.7
5 7
5.1 7.7
5.2 6.7
5.5 8
5.5 8.5
6 9.5
6.6 8.6
7 9
7.7 10.313
B. (card B in the previous activity)
$$x$$ $$y$$
1 11.86
2.2 11.332
3.3 10.848
3.4 10.741
3.6 10.716
3.8 10.628
3.9 10.584
4 10.54
4.4 10.364
4.5 10.32
4.7 10.232
4.8 10.188
4.9 10.144
5 10.1
5.1 10.056
5.2 10.5
5.5 9.88
5.7 9.753
6 9.66
6.6 9.396
7 9.22
7.7 8.912
C. (card C in the previous activity)
$$x$$ $$y$$
1 6.11
2.2 7.142
3.3 8.088
3.5 8.19
3.6 8.346
3.8 2.92
3.9 8.604
4 8.69
4.4 9.034
4.5 9.12
4.7 9.292
4.8 13.6
4.9 9.464
5 9.55
5.1 9.636
5.2 9.722
5.5 9.98
5.8 10.32
6 10.41
6.6 10.926
7 11.27
7.7 11.872
D. (card E in the previous activity)
$$x$$ $$y$$
1 13.9
2.2 11.5
3.3 9.3
3.5 9.2
3.6 8.7
3.8 8.3
3.9 8.1
4 7.9
4.4 7.1
4.5 6.9
4.7 6.5
4.8 6.3
4.9 6.1
5 5.9
5.1 5.7
5.2 5.5
5.5 4.9
5.8 4.3
6 3.9
6.6 1.3
7 1.9
7.7 0.5
E. (card F in the previous activity)
$$x$$ $$y$$
1 6.5
2.2 8.5
3.3 9.5
3.3 9
3.6 10.5
3.8 11
3.9 10.2
4 11.5
4.4 11
4.5 11.5
4.7 11.5
4.8 10.5
4.9 13.2
5 11.5
5.1 12.2
5.2 11.2
5.5 12.5
5.5 13
6 14
6.6 13.1
7 13.5
7.7 14.813
### Student Response
For access, consult one of our IM Certified Partners.
### Launch
Provide data tables for the graphs for the cards from the previous activity that were fit well with a linear model. Assign one table to each group. For students using the paper task, show them how to use technology to create a scatter plot of the data in a table. After groups have had a chance to estimate the best fit lines, pause the class. Show students how to use technology to find the least-squares regression line for data and display the line with the scatter plot.
Display the tables for students to use for the last question:
A. (card A in the previous activity)
$$x$$ $$y$$
1 2
2.2 4
3.3 5
3.3 4.5
3.6 6
3.8 6.5
3.9 5.7
4 7
4.4 6.5
4.5 7
4.7 7
4.8 6
4.9 8.7
5 7
5.1 7.7
5.2 6.7
5.5 8
5.5 8.5
6 9.5
6.6 8.6
7 9
7.7 10.313
B. (card B in the previous activity)
$$x$$ $$y$$
1 11.86
2.2 11.332
3.3 10.848
3.4 10.741
3.6 10.716
3.8 10.628
3.9 10.584
4 10.54
4.4 10.364
4.5 10.32
4.7 10.232
4.8 10.188
4.9 10.144
5 10.1
5.1 10.056
5.2 10.5
5.5 9.88
5.7 9.753
6 9.66
6.6 9.396
7 9.22
7.7 8.912
C. (card C in the previous activity)
$$x$$ $$y$$
1 6.11
2.2 7.142
3.3 8.088
3.5 8.19
3.6 8.346
3.8 2.92
3.9 8.604
4 8.69
4.4 9.034
4.5 9.12
4.7 9.292
4.8 13.6
4.9 9.464
5 9.55
5.1 9.636
5.2 9.722
5.5 9.98
5.8 10.32
6 10.41
6.6 10.926
7 11.27
7.7 11.872
D. (card E in the previous activity)
$$x$$ $$y$$
1 13.9
2.2 11.5
3.3 9.3
3.5 9.2
3.6 8.7
3.8 8.3
3.9 8.1
4 7.9
4.4 7.1
4.5 6.9
4.7 6.5
4.8 6.3
4.9 6.1
5 5.9
5.1 5.7
5.2 5.5
5.5 4.9
5.8 4.3
6 3.9
6.6 1.3
7 1.9
7.7 0.5
E. (card F in the previous activity)
$$x$$ $$y$$
1 6.5
2.2 8.5
3.3 9.5
3.3 9
3.6 10.5
3.8 11
3.9 10.2
4 11.5
4.4 11
4.5 11.5
4.7 11.5
4.8 10.5
4.9 13.2
5 11.5
5.1 12.2
5.2 11.2
5.5 12.5
5.5 13
6 14
6.6 13.1
7 13.5
7.7 14.813
Engagement: Develop Effort and Persistence. Provide prompts, reminders, or checklists that focus on increasing the length of on-task orientation in the face of distractions. For example, provide two copies of the steps: graph the table, find the best fit line, find the slope, and find the $$y$$-intercept. Include phrases to activate knowledge from prior activities such as, “As x increases . . .”
Supports accessibility for: Organization; Conceptual processing; Attention
### Student Facing
The weight of ice cream sold in a day at a small store in pounds ($$x$$) and the average temperature outside during the day in degrees Celsius ($$y$$) are recorded in the table.
$$x$$ $$y$$ 20 18 21 17 21.5 19.5 21 18 6 4.5 6.5 3.5 7.5 6.5 7 5
1. For this data, create a scatter plot and sketch a line that fits the data well.
2. Use technology to compute the best fit line. Round any numbers to 2 decimal places.
3. What are the values for the slope and $$y$$-intercept for the best fit line? What do these values mean in this situation?
4. Use the best fit line to predict the $$y$$ value when $$x$$ is 10. Is this a good estimate for the data? Explain your reasoning.
5. Your teacher will give you a data table for one of the other scatter plots from the previous activity. Use technology and this table of data to create a scatter plot that also shows the line of best fit, then interpret the slope and $$y$$-intercept.
### Student Response
For access, consult one of our IM Certified Partners.
### Student Facing
#### Are you ready for more?
Priya uses several different ride services to get around her city. The table shows the distance, in miles, she traveled during her last 10 trips and the price of each trip, in dollars.
distance (miles) price ($) 3.1 12.5 4.2 14.75 5 16 3.5 13.25 2.5 12 1 9 0.8 8.75 1.6 9.75 4.3 12 3.3 14 1. Priya creates a scatter plot of the data using the distance, $$x$$, and the price, $$y$$. She determines that a linear model is appropriate to use with the data. Use technology to find the equation of a line of best fit. 2. Interpret the slope and the $$y$$-intercept of the equation of the line of best fit in this situation. 3. Use the line of best fit to estimate the cost of a 3.6-mile trip. Will this estimate be close to the actual value? Explain your reasoning. 4. On her next trip, Priya tries a new ride service and travels 3.6 miles, but pays only \$4.00 because she receives a discount. Include this trip in the table and calculate the equation of the line of best fit for the 11 trips. Did the slope of the equation of the line of best fit increase, decrease, or stay the same? Why? Explain your reasoning.
5. Priya uses the new ride service for her 12th trip. She travels 4.1 miles and is charged \\$24.75. How do you think the slope of the equation of the line of best fit will change when this 12th trip is added to the table?
### Student Response
For access, consult one of our IM Certified Partners.
### Anticipated Misconceptions
Students may struggle with interpreting slope and $$y$$-intercept. Remind students of how each relates to a situation. To help students interpret slope, ask them: “What does the $$x$$ variable represent? What does the $$y$$ variable represent? How is slope connected to the $$x$$ and $$y$$ variables? What happens to $$x$$ as $$y$$ increases (or decreases)?” To help students interpret the $$y$$-intercept, ask them: “What does the point $$(20,6)$$ mean in the scatter plot? What are the coordinates of the $$y$$-intercept? What do each of the coordinates mean in the situation described? What is the $$y$$ value when $$x$$ is 0? Which variable has a value of 0? Which variable is represented with y?”
### Activity Synthesis
The purpose of this discussion is for students to make connections between the scatter plot and the equation of the line of best fit. Display each scatter plot, the line of best fit, and the equation of the line of best fit.
Here are some questions for discussion:
• “How does using technology help model the data in the scatter plot?” (It allows different people to come up with the same equation for the line of best fit. If the line is just drawn by hand, there can be different linear equations that seem to fit the data well, but there is only one “best” fit line.)
• “What does the $$y$$-intercept represent in each scatter plot? When is it reasonable to use this interpretation?” (It represents the value of $$y$$ estimated by the linear model when $$x = 0$$. When the intercept is near the range of the data, it can be reasonable to use this interpretation because otherwise, the linear trend may not continue. There are also some situations in which a value of 0 for $$x$$ does not make sense.)
• “Why is the slope the same in scatter plot A and scatter plot F?” (It is the same because the data in scatter plot F is the same data as in scatter plot A, except that the values for $$y$$ have all been increased by 4.5 units.)
Tell students they should be careful when predicting values outside the range of the data, in particular, for the $$y$$-intercept. Even when the data is fit well by a linear model, the behavior of the variables farther away may not be linear. It is important to remember that all predictions using the best fit line are estimates and the reasonableness of the predictions should be considered.
Representing, Conversing: MLR7 Compare and Connect. As students share the connections they noticed between the scatter plot and the equation of the line of best fit, call students’ attention to the different ways the slope and vertical intercept are represented. Wherever possible, amplify student words and actions that describe the connections between a specific feature of one mathematical representation and a specific feature of another representation.
Design Principle(s): Maximize meta-awareness; Support sense-making
## Lesson Synthesis
### Lesson Synthesis
Here are some questions for discussion.
• “What is a line of best fit?” (The best linear model for the data.)
• “How do you know that you have a line of best fit?” (You can use technology to generate the line of best fit, but then you need to graph it on the scatter plot to verify that it fits the data well. It needs to follow the trend of the data, it should go roughly through the middle of the data, and it should have roughly the same number of data points on either side of it.)
• “Which line fits the data better, the solid line or the dashed line? Do you think it is the line of best fit? Explain your reasoning.” (The dashed line fits the data better, because its slope and vertical intercept more closely resemble the trend of the data than the slope and vertical intercept of the solid line. It is not the line of best fit because it does not go through the middle of the data. It should be a little lower on the graph.)
## 5.4: Cool-down - Fresh Air (5 minutes)
### Cool-Down
For access, consult one of our IM Certified Partners.
## Student Lesson Summary
### Student Facing
Some data appear to have a linear relationship, so finding an equation for a line that fits the data can help you understand the relationship between the variables.
Other data may follow non-linear trends or not have an apparent trend at all.
When modeling data with a linear function seems useful, it is important to find a linear function that is close to the data. The line should have a $$y$$-intercept and slope to follow the shape of the data in the scatter plot as much as possible.
Technology can be used to quickly find a line of best fit for the data and provide the equation of the line that we can use to analyze the situation. |
# Pre-Algebra : Volume
## Example Questions
← Previous 1 3 4 5 6 7 8
### Example Question #1 : Volume
Principal O'Shaughnessy has a paperweight in the shape of a pyramid with a square base. If one side of the base has a length of 4cm and the height of the paperweight is 6cm, what is the volume of the paperweight?
Explanation:
We begin by recalling the volume of a pyramid.
where is the area of the base and is the height.
Since the base is a square, we can find the area by squaring the length of one of the sides.
Given the height is 6cm, we can now calculate the volume.
Since all of the measurements were in centimeters, our volume will be in cubic centimeters.
Therefore, the volume of Principal O'Shaughnessy's paperweight is .
### Example Question #2 : Volume Of A Pyramid
The volume of a square pyramid is . If a side of the square base measures . What is the height of the pyramid?
Explanation:
The formula for the volume of a pyramid is , where is the area of the base and is the height.
Using this formula,
= Area of the base, which is nothing but area of square with side
Now, when simplified, you get .
Hence, the height of the pyramid is .
### Example Question #1 : Volume Of A Pyramid
A pyramid has height 4 feet. Its base is a square with sidelength 3 feet. Give its volume in cubic inches.
Explanation:
Convert each measurement from inches to feet by multiplying it by 12:
Height: 4 feet = inches
Sidelength of the base: 3 feet = inches
The volume of a pyramid is
Since the base is a square, we can replace :
Substitute
The pyramid has volume 20,736 cubic inches.
### Example Question #1 : Volume Of A Pyramid
The height of a right pyramid is feet. Its base is a square with sidelength feet. Give its volume in cubic inches.
Explanation:
Convert each of the measurements from feet to inches by multiplying by .
Height: inches
Sidelength of base: inches
The base of the pyramid has area
square inches.
Substitute into the volume formula:
cubic inches
### Example Question #4 : Solid Geometry
The height of a right pyramid and the sidelength of its square base are equal. The perimeter of the base is 3 feet. Give its volume in cubic inches.
Explanation:
The perimeter of the square base, feet, is equivalent to inches; divide by to get the sidelength of the base - and the height: inches.
The area of the base is therefore square inches.
In the formula for the volume of a pyramid, substitute :
cubic inches.
### Example Question #3 : Solid Geometry
What is the volume of a pyramid with the following measurements?
Explanation:
The volume of a pyramid can be determined using the following equation:
### Example Question #1 : Volume Of A Pyramid
The pyramid has a length, width, and height of respectively. What is the volume of the pyramid?
Explanation:
Write the formula for the volume of a pyramid.
Substitute the dimensions and solve.
### Example Question #4 : Volume
If the base area of the pyramid is , and the height is , what is the volume of the pyramid?
Explanation:
Write the volume formula for the pyramid.
The base area is represented by .
Substitute the knowns into the formula.
### Example Question #1 : Volume
Find the volume of a pyramid with a length of 4, width of 7, and a height of 3.
Explanation:
Write the formula to find the area of a pyramid.
Substitute the dimensions.
### Example Question #6 : Volume
Find the volume of a pyramid if the length, base, and height are respectively. |
# heightlaugh4
## # Why Study Math? Linear Equations and Slope-Intercept Form
Even as saw in the article “Why Study Math? – Thready Equations and Slope-Intercept Kind, ” step-wise equations or functions are some of the more simple ones learnt in algebra and standard mathematics. Here we are going to take a look at and examine another basic way of producing linear equations: the point-slope form.
As your name means, the point-slope form to get the picture of a series depends on two things: the mountain, and specific point on the line. Once we know these two stuff, we can write the equation of the line. In mathematical conditions, the point-slope form of the equation from the line of which passes over the given level (x1, y1) with a mountain of meters, is y – y1 = m(x – x1). (The you after the populace and ymca is actually a subscript which allows all of us to distinguish x1 from x and y1 from con. )
To signify how this form is used, have a look at the following situation: Suppose we now have a brand which has mountain 3 and passes through the point (1, 2). We could graph that line by way of locating the point (1, 2) and then make use of the slope of three to go 3 units up and then you unit on the right. To publish the situation of the range, we use a clever minor device. We all introduce the variables maraud and gym as a issue (x, y). In the point-slope form con – y1 = m(x – x1), we have (1, 2) like the point (x1, y1). https://theeducationjourney.com/slope-intercept-form/ generate y – 2 sama dengan 3(x – 1). By using the distributive home on the right hand side of the equation, we can write y supports 2 = 3x supports 3. By simply bringing the -2 over to the ideal side, we could write gym = 3x -1. Assuming you have not already recognized that, this latter equation is in slope-intercept type.
To see the best way this form in the equation on the line is used in a real life application, take those following situation, the information of which was obtained from an article the fact that appeared in a newspaper. As it happens that temperature affects working speed. Actually the best heat for running is down below 60 levels Fahrenheit. When a person produced optimally in 17. six feet per second, the individual would halt by about zero. 3 ft per second for every your five degree increased temperature preceding 60 college diplomas. We can employ this information to write down the thready model just for this situation and then calculate, today i want to say, the perfect running schedule at 70 degrees.
Let T represent the temperature in degrees Fahrenheit. Make it possible for P symbolize the optimal speed in ft per instant. From the details in the article, we know that the optimal running pace at 70 degrees is normally 17. a few feet per second. Hence one level is (60, 17. 6). Let's utilize the other information to look for the slope of the line for this model. The slope meters is add up to the enhancements made on pace across the change in temperature, or l = difference in P/change on T. Were told the fact that the pace slows by 0. 3 foot per extra for every increase in 5 college diplomas above sixty. A get rid of is symbolized by a adverse. Using this details we can compute the slope at -0. 3/5 or perhaps -0. summer.
Now that we have a point as well as slope, we can easily write the version which signifies this situation. We certainly have P – P1 sama dengan m(T — T1) or maybe P supports 17. a few = -0. 06(T — 60). Making use of the distributive home we can set this equation into slope-intercept form. We obtain P = -0. 06T + 21 years of age. 2 . To discover the optimal rate at 70 degrees, we'd like only replace 80 intended for T inside given style to receive 16. some.
Situations like these show the fact that math is very used to fix problems that stem from the world. Whether we are speaking about optimal running pace or maximal revenue, math is vital to unlocking our opportunity toward learning the world about us. So when we understand, we are stimulated. What a nice way to exist! |
shopclose
# Unlock Arrays in Math for Every 2nd Grader’s Growth
The gateway to multiplication, the foundation to finding area…I introduce to you…ARRAYS IN MATH!
And let me just say that students LOVE arrays and you can spice up an array lesson with colorful manipulatives and repeated addition… you won’t find a more fun time than this!
## Importance of Arrays in Math
#### Prerequisite Skills:
Equal groups: A great place to start with equal groups is teaching students odd and even (2.OA.3). Students will understand the odd one out, showing that the groups are not equal.
Another bonus to starting with equal groups is that students practice counting by 2’s which will help them during repeated addition.
Repeated Addition: Allows students to practice finding the total in arrays quickly and efficiently and also sets them up for multiplication. Students may struggle with larger numbers or they may be ready to move toward multiplication when it comes to repeated addition. Read more about Repeated addition in this blog post.
Doubles Math Facts: Doubles math facts are repeated addition. By having a working memory of these facts students won’t be overwhelmed with the longer repeated addition sentences.
Use the carrot method Doubles practice can be done a variety of ways. I created a virtual math facts practice that includes a Google Slides deck for students to brush up on their doubles facts at home or during a math center. Click here to see it.
Related Standards: 2.OA.2, 2.OA.3, 2.NBT.2
## Ways to Teach Arrays in Math
Let them play! I mean, for sure I always give my famous speech, “Math tools not toys,” but giving them time to explore with manipulatives to build arrays without the pressure of getting the answer right or wrong.
Here are some steps to take to get array practice started in 2nd grade.
Step 1: Introduce equal groups
By giving students manipulatives to sort into groups it helps them really grasp the equal concept. Have students write the number of objects in each group. I like to have students start with groups of 2 because we have usually covered even and odd numbers by this time of the year.
Step 3: Introduce columns and rows. Side to side and up and down. When students understand this key vocabulary they will be more successful in building accurate arrays.
Step 4: Creating equal rows or columns using manipulatives. Start with creating equal groups then putting them into rows or columns you give them. This is a great step for students to practice with their partners. They get a chance to use the vocabulary and work together to get results.
## Differentiating
Intensive:
Examples and non-examples: Make cards or a slideshow with simple, clear examples of arrays vs not arrays. By showing students the equal numbers in each row and column.
Use inch-sized graphing paper to build arrays. Have students use manipulatives to build inside the squares gives them a starting framework providing the structure of the array.
Start with small arrays. Use 2’s and 3’s until students get the hang of it.
Extensions
Instead of using manipulatives to create equal groups, tell students an amount and have them draw the amount into equal groups. This extends because they don’t have the manipulatives to shift around physically. They must think about it and using a writing tool to record their thinking.
Multiplication. There are always a handful of students who are ready for this, or they have an older sibling spilling the beans during homework time (love when this happens.) Usually, the best way to start is with the array, students showing the repeated addition and then showing them how the repeated addition relates to multiplication.
The pattern I start student with is usually, draw the array, write the repeated addition, write the multiplication sentence and then the students tell me why or how they are related.
How many Ways? Give them a number of manipulatives and have them record how many different arrays they can make. Students record what the array looks like, the repeated addition and multiplication. Numbers of object I like to use is 12, 10 and 8. (Also a great review of associative property).
Hopefully this helps you get started with arrays in 2nd grade!
Bethany
hey bonita
postsTop |
# Divisibility by 7 and Its Proof
This is the 6th post in the Divisibility Rules Series. In this post, we discuss divisibility by 7.
Simple steps are needed to check if a number is divisible by 7. First, multiply the rightmost (unit) digit by 2, and then subtract the product from the remaining digits. If the difference is divisible by 7, then the number is divisible by 7.
Example 1: Is 623 divisible by 7?
3 x 2 = 6
626 = 56
56 is divisible by 7, so 623 is divisible by 7.
If after the process above, the number is still large, and it is difficult if to know if it is divisible by 7, the steps can be repeated. We take the difference as the new number, we multiply the rightmost digit by 2, and then subtract from the remaining digits.
Example 2: Is 3423 divisible by 7?
3 x 2 =
3426 = 336
We repeat the process for 336. We multiply 6 by 2 and then subtract it from 33
6 x 2 = 12
3312 = 21
21 is divisible by 7, so 3423 is divisible by 7.
Note that if the number is still large, this process can be repeated over and over again, until it is possible to determine if the remaining digits is divisible by 7.
Delving Deeper (for the adventurous)
The following portion are for students who have basic knowledge on proofs. In particular, we will be proving an if and only if statement. A if and only if B requires to prove that A implies B and B implies A.
Let $N$ be the number that we want divide by 7. Let $b$ be the unit’s digit and $a$ be the rest of the digit. Then N = 10a + b.
Explanation: All whole numbers N can be expressed as the product of 10 and a number added to its units digit. For example 983 = 10(98) + 3, 5896 = 10(598) + 6, and so on.
We assign the following statements to A and B.
A: a – 2b is divisible by 7.
B: N is divisible by 7.
As we have mentioned above, we have to show that (1) A implies B and (2) B implies A. This means that we have to show that if $a - 2b$ is divisible by $7$, then $N$ is divisible by $7$. The statement $a - 2b$ is the step where we multiplied the unit’s digit by 2, and then subtracted from the remaining digits $a$.
For (1) We have to show that A implies B. That is, we have to show that if $a - 2b$ is divisible by $7$, then $N$ is divisible by $7$.
Proof
If $a - 2b$ is divisible by $7$, then we can find a natural number $k$ such that $a - 2b = 7k$ (Can you see why?).
Multiply both sides by $10$, we have $10a - 20b = 70k$. Adding $b$ on both sides, we have $10a - 20b + b= 70k + b$. Now, $10a + b = 70k + 21b$. Notice that the left hand side of our equation is $N$ and the right hand side can be divided by $7$. Therefore, $10a + b = N$ is divisible by $7$. That proves our first statement that If $a - 2b$ is divisible by $7$, $N$ is divisible by $7$.#
For (2), we have to show that B implies A. That is, we have to show that if $N$ is divisible by 7, $a - 2b$ is divisible by $7$.
Proof
If $N$ is divisible by $7$, then $10a + b$ is divisible by $7$. This means we can find a natural number $k$ such that $10a + b = 7k$. Subtracting $21b$ from both sides, we have $10a + b - 21b = 7k - 21b$. This means that $10a - 20b = 7k - 21b$. Factoring, we have
$10(a-2b) = 7(k-3b)$
Now, since $10$ is not divisible by $7$, $a - 2b$ is divisible by $7$. This proves the second statement if $N$ is divisible by $7$, then $a - 2b$ is divisible by $7$#
From above, we have shown that A implies B and B implies A. We have shown that the process that we have done above will hold for all cases. |
6.2 Solve general applications of percent (Page 2/7)
Page 2 / 7
We will update the strategy we used in our earlier applications to include equations now. Notice that we will translate a sentence into an equation.
Solve an application
1. Identify what you are asked to find and choose a variable to represent it.
2. Write a sentence that gives the information to find it.
3. Translate the sentence into an equation.
4. Solve the equation using good algebra techniques.
5. Check the answer in the problem and make sure it makes sense.
6. Write a complete sentence that answers the question.
Now that we have the strategy to refer to, and have practiced solving basic percent equations, we are ready to solve percent applications. Be sure to ask yourself if your final answer makes sense—since many of the applications we'll solve involve everyday situations, you can rely on your own experience.
Dezohn and his girlfriend enjoyed a dinner at a restaurant, and the bill was $\text{68.50}.$ They want to leave an $\text{18%}$ tip. If the tip will be $\text{18%}$ of the total bill, how much should the tip be?
Solution
What are you asked to find? the amount of the tip Choose a variable to represent it. Let $t=$ amount of tip. Write a sentence that give the information to find it. The tip is 18% of the total bill. Translate the sentence into an equation. Multiply. Check. Is this answer reasonable? If we approximate the bill to $70 and the percent to 20%, we would have a tip of$14. So a tip of $12.33 seems reasonable. Write a complete sentence that answers the question. The couple should leave a tip of$12.33.
Cierra and her sister enjoyed a special dinner in a restaurant, and the bill was $\text{81.50}.$ If she wants to leave $\text{18%}$ of the total bill as her tip, how much should she leave?
$14.67 Kimngoc had lunch at her favorite restaurant. She wants to leave $\text{15%}$ of the total bill as her tip. If her bill was $\text{14.40},$ how much will she leave for the tip?$2.16
The label on Masao's breakfast cereal said that one serving of cereal provides $85$ milligrams (mg) of potassium, which is $\text{2%}$ of the recommended daily amount. What is the total recommended daily amount of potassium?
Solution
What are you asked to find? the total amount of potassium recommended Choose a variable to represent it. Let $a=$ total amount of potassium. Write a sentence that gives the information to find it. 85% mg if 2% of the total amount. Translate the sentence into an equation. Divide both sides by 0.02. Simplify. Check: Is this answer reasonable? Yes. 2% is a small percent and 85 is a small part of 4,250. Write a complete sentence that answers the question. The amount of potassium that is recommended is 4250 mg.
One serving of wheat square cereal has $7$ grams of fiber, which is $\text{29%}$ of the recommended daily amount. What is the total recommended daily amount of fiber?
24.1 grams
One serving of rice cereal has $190$ mg of sodium, which is $\text{8%}$ of the recommended daily amount. What is the total recommended daily amount of sodium?
2,375 mg
Mitzi received some gourmet brownies as a gift. The wrapper said each brownie was $480$ calories, and had $240$ calories of fat. What percent of the total calories in each brownie comes from fat?
Solution
What are you asked to find? the percent of the total calories from fat Choose a variable to represent it. Let $p=$ percent from fat. Write a sentence that gives the information to find it. What percent of 480 is 240? Translate the sentence into an equation. Divide both sides by 480. Simplify. Convert to percent form. Check. Is this answer reasonable? Yes. 240 is half of 480, so 50% makes sense. Write a complete sentence that answers the question. Of the total calories in each brownie, 50"% is fat.
anyone know any internet site where one can find nanotechnology papers?
research.net
kanaga
Introduction about quantum dots in nanotechnology
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
characteristics of micro business
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
there is no specific books for beginners but there is book called principle of nanotechnology
NANO
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
carbon nanotubes has various application in fuel cells membrane, current research on cancer drug,and in electronics MEMS and NEMS etc
NANO
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
Good
Berger describes sociologists as concerned with
A soccer field is a rectangle 130 meters wide and 110 meters long. The coach asks players to run from one corner to the other corner diagonally across. What is that distance, to the nearest tenths place.
Jeannette has $5 and$10 bills in her wallet. The number of fives is three more than six times the number of tens. Let t represent the number of tens. Write an expression for the number of fives.
What is the expressiin for seven less than four times the number of nickels
How do i figure this problem out.
how do you translate this in Algebraic Expressions
why surface tension is zero at critical temperature
Shanjida
I think if critical temperature denote high temperature then a liquid stats boils that time the water stats to evaporate so some moles of h2o to up and due to high temp the bonding break they have low density so it can be a reason
s.
Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight? |
Write the decimal as a fraction
Convert Fractions to Decimals To convert a Fraction to a Decimal manually Write down 75 with the decimal point 2 spaces from the right. To write a decimal as a fraction, first write a fraction with the decimal as the numerator and the integer one as the denominator. Then perform a. Writing a Decimal as a Fraction, Example 1. Here I look at how to convert a decimal into a reduced fraction. Write the decimal as a fraction or mixed number in simplest form. - 2956495. How to Convert a Decimal to a Fraction. Converting a decimal to a fraction isn't as hard as it looks. If you want to know how to do it, just follow these steps. Write.
Write a fraction as a decimal Some fractions are easy to convert to decimals. Fractions with 10,100 or power of 10 are written like they sound. 1/10 is.1, 3/10 is.3. 2.8 Lesson 92 Chapter 2 Multiplying and Dividing Fractions Method 1: Writing Fractions as Decimals To write a fraction as a decimal, divide the numerator by the. Practice these problems to see how decimals and fractions can represent the same number. 84 Chapter 2 Multiplying and Dividing Fractions When you write a terminating decimal as a fraction, what type of denominator do you get? 2.7 Writing Decimals as Fractions.
Write the decimal as a fraction
To write a decimal as a fraction, first write a fraction with the decimal as the numerator and the integer one as the denominator. Then perform a mathematical. To convert fraction to decimal number divide numerator by denominator. Calculator to find decimal form of a fraction or to change fractions into decimals. Fraction to. Write 2.75 as a simplified fraction. So once you get some practice here. You're going to find it pretty straightforward to do. But we're really going to. We see how to write a repeating decimal as a fraction in simplest form. Part of the Mathtrain.TV Project. Write a fraction as a decimal Some fractions are easy to convert to decimals. Fractions with 10,100 or power of 10 are written like they sound. 1/10 is.1, 3/10 is.3.
Convert From a Decimal To a Fraction - powered by WebMath. Writing a Decimal as a Fraction, Example 1. Here I look at how to convert a decimal into a reduced fraction. Do you want to know how to write the decimal number 2.03 as a fraction? Here we will show you step-by-step how to convert 2.03 so you can write it as a fraction. Convert decimals to fractions or mixed number fractions. Calculator to change decimals to fractions showing the work with steps. Converts repeating decimals to fractions.
To a decimal fraction, simply write it in standard decimal notation Decimals, and Fractions 461 Technique Step 1 — Convert the fraction or mixed number to a. Convert Fractions to Decimals Find a number you can multiply by the bottom of the fraction to make. so let us write down 333 with the decimal point 3. Sal shows the connection between decimals and fractions using a grid diagram and number lines. Decimal to fraction, step by step, example. For all free math videos visit http://Mathmeeting.com. 2.7 Lesson 86 Chapter 2 Multiplying and Dividing Fractions Key Vocabulary terminating decimal, p. 84 Writing Decimals as Fractions Words Write the digits of the.
Convert Decimals to Fractions. To convert a Decimal to a Fraction follow these steps: Step 1: Write down the decimal divided by 1, like this: decimal 1. To convert a fraction to a decimal, divide its. Another way to write a fraction as a percent is to divide its numerator by its denominator. The decimal 0.5 is expressed as the fraction one-half. This is found by evaluating the decimal over 1, multiplying the fraction by 10 and then. Writing Decimal Fractions. To write eight-tenths using decimal place value, the digit 8 is placed in the tenths' column. When we transfer the value out of the table. |
#### Transcript 4-7 Arithmetic Sequencesx
```4-7 Arithmetic
Sequences
Objective: To identify and extend
patterns in sequences and represent
in function notation
4-7 Arithmetic Sequences
• Getting Ready on page 276. Solve IT
4-7 Arithmetic Sequences
• A sequence is an ordered list of numbers
that often forms a pattern.
• Each number in the list is called a term of
a sequence.
• Some sequences can be modeled with a
function rule so that you can extend the
sequence to any value.
4-7 Arithmetic Sequences
• Example 1: What are next two terms?
• A. 5, 8, 11, 14 . . .
• B. 2.5, 5, 10, 20 . . .
4-7 Arithmetic Sequences
• An arithmetic sequence: the difference
between consecutive is a constant term,
which is called the constant difference.
• Example 2: Is it an arithmetic sequence?
• A. 3, 8, 13, 18 . . .
• B. 6, 9, 13, 17 . . .
4-7 Arithmetic Sequences
• Sequences are functions and the terms are
the outputs of the function.
A recursive formula is a fn rule that relates
each term of a sequence after the first to the
ones before it.
4-7 Arithmetic Sequences
7, 11, 15, 19 . . .
Find the common difference: FIRST
Write the recursive formula:
Let n = term in sequence
A(n) = the value of the nth term
4-7 Arithmetic Sequences
7, 11, 15, 19 . . .
Let n = term in sequence
A(n) = the value of the nth term
Value of term 1 = A(1) = 7
Value of term 2 = A(2) = A(1) + 4 = 11
Value of term 3 = A(3) = A(2) + 4 = 15
Value of term 4 = A(4) = A(4) + 4 = 19
Value of term 2 = A(2) = A(1) + 4 = 11
Value of term n = A(n) = A(n-2) + 4
4-7 Arithmetic Sequences
The formula for Arithmetic Sequences
A(n) = A(1) + (n – 1) d
Term
number
First term
Common
difference
Term number
4-7 Arithmetic Sequences
The formula for Arithmetic Sequences
A(n) = A(1) + (n – 1) d
Example 4: An online auction works as
shown: Write a formula for it.
Bass
Guitar
Minimu
m Price
\$200
Determine common difference:
Bid 1
200
Bid 2
210
A(n) =
Bid 3
220
Bid 4
230
What is 12th bid?
4-7 Arithmetic Sequences
The formula for Arithmetic Sequences
Example 5: An RECURSIVE formula is
represented by: A(n) = A(n – 1) + 12
If the first term is 19, write explicit formula
So A(1) = 19 and adding 12 is common
difference
A(n) = A(1) + (n – 1)d so substitute what you know
A(n) = 19 + (n-1)12 Arithmetic formula
4-7 Arithmetic Sequences
The formula for Arithmetic Sequences
Example 6: An Arithmetic formula is
represented by: A(n) = 32 + (n-1)22
So the first term is ?
So common difference is?
A(n) = A(n – 1) + d so substitute what you know
A(n) =
Recursive formula
4-7 Arithmetic Sequences
HW p. 279 9 – 42 every third
``` |
PDF chapter test
The methods to solve the problems for compound variation are:
• Proportion method
• Multiplicative factor method
• Formula method
1. Proportion method:
In this method, from the given data, determine whether they are in direct or inverse proportion. Then, the value of the unknown($$x$$) can be determined using:
The product of the extremes $$=$$ The product of the means
2. Multiplicative factor method:
Let us consider the table to understand how to solve the problem using the multiplicative factor method.
Quantity $$1$$ Quantity $$2$$ Quantity $$3$$ $$a$$ $$b$$ $$c$$ $$x$$ $$d$$ $$e$$
Here, the unknown quantity is $$x$$.
Step 1: Compare the unknown value (Quantity $$1$$) with the known values (Quantity $$2$$ and Quantity $$3$$).
Step 2: If Quantity $$1$$ and Quantity $$2$$ are in direct variation, then take the multiplying factor as $$\frac{d}{b}$$(take the reciprocal).
Step 3: If Quantity $$1$$ and Quantity $$3$$ are in inverse variation, then take the multiplying factor as $$\frac{c}{e}$$(no change).
Step 4: The value of the unknown $$x$$ can be determined using $$x = a \times \frac{d}{b} \times \frac{c}{e}$$.
3. Formula method:
From the given data, identify Persons($$P$$), Days($$D$$), Hours($$H$$) and Work($$W$$) and use the formula:
$$\frac{P_1 \times D_1 \times H_1}{W_1} = \frac{P_2 \times D_2 \times H_2}{W_2}$$
Here, the suffix $$1$$ denotes the values from statement $$1$$, whereas the suffix $$2$$ denotes the values from statement $$2$$. |
## Introduction
Teachers are often faced with the dilemma of how to introduce a fresh concept in the classroom. This series aims at helping teachers create a structure that can help them put across concepts with clarity. Today’s topic is complex numbers and imaginary numbers.
Complex numbers & imaginary numbers are numbers that do not appear on the real number line. It is a concept that a lot of science students are faced with in their higher secondary education, but most of them get through those years without understanding the extraordinary scientific journey that led to its discovery and eventual application in the world of science and technology.
## Types of Numbers
To understand what the term complex number means, it is important to know what the different classifications of numbers are. Numbers are classified into categories based on their utility to the mathematician, and they are as given below:
### Natural Numbers
Natural numbers refer to the numbers that are used for counting and ordering in the real world. These are 1, 2, 3, and so on. These numbers are the first type of numbers that humans came across as a means of measuring the world around them and they continue to be so. For this reason, natural numbers are also known as counting numbers and are denoted by the letter N.
### Whole Numbers
Even though there is no complete consensus on this, it is generally understood that whole numbers refer to the entire suite of counting numbers with the addition of zero to it at the beginning. The reason for this classification is because of calculus and its requirement in the process. The letter W is used to denote whole numbers.
### Negative Numbers
Negative numbers refer to the numbers on the number line that lies to the left of zero. They can be essentially summed up as the additive inverse of the natural numbers. In simple terms, they are the numbers starting from -1, -2, -3, and so on.
### Integers
An integer can be defined as a number that does not have a fractional component in it. So, essentially all the 3 types of numbers that are discussed above are integers. The natural numbers make up the positive integers and their additive inverse set represents negative integers. The letter Z is used to denote integers.
### Rational Numbers
Rational numbers can be defined as the set of numbers that can be represented in the form of a p/q fraction that has a finite number of decimal places. For example, 1.5, -3.75, 9/44, and so on. The number can be rational even if the decimal places are never-ending, they just have to be recurring continuously, like in the case of ⅓, 1/9, and so on. The set of rational numbers includes natural numbers, integers, and those fractional components, and are referred to as Q.
### Irrational Numbers
Irrational numbers are the set of numbers that cannot be represented by a p/q format and have a never-ending decimal expansion. Examples of such numbers include Euler’s constant e, pi, square root of rational numbers that are not squares of rational numbers themselves, and so on.
### Real Numbers
The combined superset of rational and irrational numbers, which includes all the above-mentioned numbers on the number line represents the set of real numbers, which is represented by R.
Now that we know what real numbers are, it is time to shed light on what imaginary numbers are. However, there is a lot more to this than what meets the eye. The story of this particular type of number is very much real as opposed to its name. It all begins with the complicated relationship mathematics used to have with geometry, and a quest to find the solution to the cubic.
## The Role of Geometry
Geometry is the study of the sizes, shapes, positions angles, and dimensions of things. Before the evolution of modern mathematics, geometry was the main source of truth. This was because mathematicians were used to dealing with real-world applications of mathematics, which is why most ancient civilizations like the Indians, Greeks, Babylonians, Egyptians, etc. all independently came up with their own versions of the Pythagoras Theorem, Euclidean Mathematics, algebra, etc. One such concept was the quadratic and cubic equations.
The solution to the quadratic equation has been known for several centuries. For any equation ax2 + bx + c = 0,
Where a, b, and c are real coefficients in the quadratic equation.
For example, in the equation
x2+26x = 27
Mathematicians found that the solution for this was the x = 1. However, for several centuries, mathematicians were completely oblivious to the negative solutions to the quadratic equation. In the case of the above equation, -27 is also a solution, but they were not aware of this. This was because of the way in which they used to solve the quadratic equation - using geometry. How, you might ask? Let us dive deeper into how they used to do it.
### Solving the Quadratic with Geometry
Taking the above equation as a reference, ancient mathematicians would consider the x2 term as a literal square of side x, and the 26x term as a rectangle with one side 26 and the other side with length x. The areas of these two shapes needed to add up to 27.
They cut the rectangle into equally shaped squares and rearranged them in the form of a square, with a few dimensions missing. Here the rectangle can be split in the form of squares with side 13. These missing dimensions could be figured out using the leftover part of the square, which in this case was a 13x13 square.
The square can be completed by adding the square of 13 (which represents the area it occupies) on both sides of the equation to maintain equality. So now the equation becomes (x+13)2 = 196. From here, the mathematicians could figure out that the square root of the equation was 14, which led to the answer 1.
### The Problem
This was a great visual way of solving the quadratic equation, but this method posed a fundamental problem. X = -27 was willfully ignored as a solution to the equation because for them, it did not make sense - what real-world applications would a square of side -27 have? What did that even mean? So, in a way, they did not even acknowledge the existence of negative numbers - to the point where there was no single quadratic equation. They identified six different quadratic equations whose coefficients were positive and strictly stuck to those.
### The Cubic
The same approach was used with the cubic equation ax3 + bx2 + cx + d = 0. However, this time, instead of using squares, they extended the idea of completing the square to add an extra dimension to account for the cube term, hence forming a cube. The first breakthrough in solving the cubic equation came from Scipione del Ferro, a mathematics professor at the University of Bologna. He used this method to find the solution to a subset of cubic equations called depressed cubics, a term used to refer to cubic equations without the x2 term. However, he did not publish his findings. It was rediscovered by his student Antonio Fior, who boasted of the same. He went on to challenge another mathematician Nicolo Fontana Tartaglia who eventually did find it using the same method. This sparked interest within the mathematics community, especially that of a polymath by the name of Gerolamo Cardano, who later learns the method from him. However, he took this one step further and found the solution to the full cubic equation including the x2 term. He did this by substituting the term x with (x-b/3a) where a and b are coefficients of the cubic equation and found that he could convert it into the form of a depressed cubic, which could then be solved using Tartaglia’s 3D-cube-slicing-and-rearranging technique. He published this in his book Ars Magna, which had a chapter about the solution to the cubic equation with geometric proof. However, this is where he encountered the problem that changed the way mathematicians viewed the subject - the square root of negative numbers.
### The Breaking Point of Geometry
Cardano found that using his method in certain situations, he was getting the square roots of negative numbers as solutions to his equation. He walked back through the geometric proof of his method, which is where he found a geometric paradox.
For example, using the formula he could find that the area of a square was 30, and the sides were 5. So to complete the square, he had to somehow add an area of -5.
This is where the idea of imaginary numbers comes from, the idea of negative area. Cardano dismissed this problem, but later on, Rafael Bombelli, an Italian engineer, decided he wanted to find a solution. He allowed the square root of negative numbers to be their own new type of number, seeing that he couldn’t call it either positive or negative. He found that if he did this, he could eliminate the square root of the negative number, reaching a real answer. He called the intermediate number containing the real term and the term containing the square root of negative one a ‘complex number’, a term that stuck.
To understand this, let us take the example of the equation x3 + 2x2 = 3x + 10.
Here, one of the solutions to the cubic can be found using Cardano’s method, which is x = 2. However, when finding the other two roots, Bombelli runs into the same problem as Cardano, but by letting the answer be its own special type of number, he could find that the resulting root was 2-1. When he takes the final step and adds the numbers, the -1 term cancels out, leaving the answer as 4. This feels impossible because this proved that Cardano’s method did work, but the geometric proof that generated it had to be completely stepped over to reach that conclusion. This proved that negative areas must exist as an intermediate step on the way to the solution, even though the geometry didn’t make sense.
## The Symbol i
This led to modern mathematics letting go of geometry as the ultimate form of truth. René Descartes (pronounced Re-nay-day-cart) named the square root of negative one as ‘imaginary numbers’ as a derogatory term, which stuck over the years. François Viète (pronounced frans-swa-vi-yet) created the modern symbolic notations used in mathematics today and later on, Euler used the letter i to denote the square root of negative one, which became its universal symbol.
## Application of Imaginary Numbers
Armed with this notation, Euler derived Euler’s idenity, which is considered in the mathematics community as the most beautiful equation in mathematics, as it combined 5 constants (e, i, π, 1 and 0) and 3 mathematical operators (exponent, addition and multiplication) in one equation, all occurring only once. This equation is:
eiℼ + 1 = 0
Now comes the main question - if -1 did exist, if imaginary numbers and complex numbers did truly exist in the mathematical sense, did it have any practical application? Could i have any practical implications? This was finally answered by Erwin Schrödinger, an Austrian-Irish physicist. He found that the term eix perfectly describes the wave nature of particles. To understand why, we need to understand a very unique multiplicative property of imaginary numbers.
When a real number, let’s say 1, is multiplied by i, the product is an imaginary number. Multiply it ince more with i and the number becomes real again. If the number is 1, the resulting number would be negative one. Multiplying again, it becomes an imaginary number before becoming real again on further multiplication. So, it can be said that multiplying by i causes a number to dip in and out of the real and imaginary number lines, which are together known as the complex plane.
This is exactly how Schrödinger saw waves, as a combination of waves and particles, with the real part equivalent to a particle and the imaginary part equivalent to a wave. So when he formulated the Schrödinger wave equation, now one of the fundamental equations in quantum physics, imaginary numbers, and subsequently complex numbers, found their way into the heart of modern-day physics.
## Conclusion
Science is an ever-evolving field, and new discoveries are being made all the time. One discovery might discredit the other, and it just reinforces the fickle nature of the universe. What René Descartes mockingly called an imaginary number found its way into an equation that described one of the most fundamental properties of the universe. In the words of physicist Freeman Dyson,
“.... the square root of minus one means that nature works with complex numbers and not just with real numbers…”
So imaginary numbers, recognized as the stopping point of mathematical reality, became the very basis of the way that we viewed reality. That is how science is and always will be, and that is why it is exciting to learn it.
Teachmint offers the best-in-class education ERP solutions to educational institutes. Do check it out to provide a hassle-free experience to your stakeholders and increase your student enrolment percentage. With our advanced learning management system, you can improve the teaching-learning experience.
Suggested Read: 4 Types of Evaluation Models to Assess Teachers
Name must have atleast 3 characters
School name must have atleast 3 characters
Phone number must have atleast 7 digits and atmost 15 digits |
Web Results
www.varsitytutors.com/hotmath/hotmath_help/topics/converse-inverse-contrapositive
Converse, Inverse, Contrapositive Given an if-then statement "if p , then q ," we can create three related statements: A conditional statement consists of two parts, a hypothesis in the “if” clause and a conclusion in the “then” clause.
www.mathwords.com/c/converse.htm
Converse. Switching the hypothesis and conclusion of a conditional statement. For example, the converse of "If it is raining then the grass is wet" is "If the grass is wet then it is raining." Note: As in the example, a proposition may be true but have a false converse.
Just because a conditional statement is true, is the converse of the statement always going to be true? In this lesson, we'll learn the truth about the converse of statements.
www.thoughtco.com/converse-contrapositive-and-inverse-3126458
Before we define the converse, contrapositive, and inverse of a conditional statement, we need to examine the topic of negation. Every statement in logic is either true or false. The negation of a statement simply involves the insertion of the word “not” at the proper part of the statement.
www.brightstorm.com/math/geometry/geometry-building-blocks/converse
So a converse is not always true. So let's look at two examples. Here we're being asked find the converse of the statement, then ask yourself is it true. So this first statement says if it is Monday, then it is a weekday. Well, that's true. If today's Monday then it's a weekday. So the converse is going to take the if and the then and switch them.
www.quora.com/What-is-the-converse-statement-in-math
A converse is when you switch the hypothesis and the conclusion. Original statement: If an number is even (Hypothesis) then it is divisible by two (Conclusion). This is a true statement. Converse statement: If a number is divisible by two (new Hyp...
www.converse.edu/program/mathematics
About Mathematics. As a Converse Mathematics student, you’ll hone your critical thinking, reasoning, and logic skills, in addition to becoming adept with a wide range of computations and proofs. You’ll learn to question all assumptions and carefully assess claims. The tools you’ll learn to wield in the math program will make you a valuable asset to a variety of disciplines and industries.
www.reference.com/math/converse-geometry-9610bf78cf85cac0
The converse in geometry applies to a conditional statement. In a conditional statement, the words "if" and "then" are used to show assumptions and conclusions that are to be arrived at using logical reasoning. This is often used in theorems and problems involving proofs in geometry.
www.onemathematicalcat.org/Math/Geometry_obj/contrapositive_and_converse.htm
Keep the following two examples in mind as you study this lesson. Consider the true implication: “If it is raining, then the ground is wet.”
www.mathwords.com/c/contrapositive.htm
Contrapositive. Switching the hypothesis and conclusion of a conditional statement and negating both. For example, the contrapositive of "If it is raining then the grass is wet" is "If the grass is not wet then it is not raining."
Related Search
Related Search |
## Adding and Subtracting Rational Expressions Part I
### Learning Outcomes
• Add and subtract rational expressions with like denominators
• Add and subtract rational expressions with unlike denominators using a greatest common denominator
In beginning math, students usually learn how to add and subtract whole numbers before they are taught multiplication and division. However, with fractions and rational expressions, multiplication and division are sometimes taught first because these operations are easier to perform than addition and subtraction. Addition and subtraction of rational expressions are not as easy to perform as multiplication because, as with numeric fractions, the process involves finding common denominators. By working carefully and writing down the steps along the way, you can keep track of all of the numbers and variables and perform the operations accurately.
## Adding and Subtracting Rational Expressions with Like Denominators
Adding rational expressions with the same denominator is the simplest place to start, so let’s begin there.
To add fractions with like denominators, add the numerators and keep the same denominator. Then simplify the sum. You know how to do this with numeric fractions.
$\begin{array}{c}\frac{2}{9}+\frac{4}{9}=\frac{6}{9}\\\\\frac{6}{9}=\frac{3\cdot 2}{3\cdot 3}=\frac{3}{3}\cdot \frac{2}{3}=1\cdot \frac{2}{3}=\frac{2}{3}\end{array}$
Follow the same process to add rational expressions with like denominators. Let’s try one.
### Example
Add $\displaystyle \frac{2{{x}^{2}}}{x+4}+\frac{8x}{x+4}$, and define the domain.
State the sum in simplest form.
Caution! Remember to define the domain of a sum or difference before simplifying. You may lose important information when you simplify. In the example above, the domain is $x\ne-4$. If we were to have defined the domain after simplifying, we would find that the domain is all real numbers which is incorrect.
To subtract rational expressions with like denominators, follow the same process you use to subtract fractions with like denominators. The process is just like the addition of rational expressions, except that you subtract instead of add.
### Example
Subtract$\frac{4x+7}{x+6}-\frac{2x+8}{x+6}$, and define the domain.
State the difference in simplest form.
### Try It
In the video that follows, we present more examples of adding rational expressions with like denominators. Additionally, we review finding the domain of a rational expression.
## Adding and Subtracting Rational Expressions with Unlike Denominators
What do they have in common?
Before adding and subtracting rational expressions with unlike denominators, you need to find a common denominator. Once again, this process is similar to the one used for adding and subtracting numeric fractions with unlike denominators. Remember how to do this?
$\displaystyle \frac{5}{6}+\frac{8}{10}+\frac{3}{4}$
Since the denominators are $6$, $10$, and $4$, you want to find the least common denominator and express each fraction with this denominator before adding. (BTW, you can add fractions by finding any common denominator; it does not have to be the least. You focus on using the least because then there is less simplifying to do. But either way works.)
Finding the least common denominator is the same as finding the least common multiple of $4$, $6$, and $10$. There are a couple of ways to do this. The first is to list the multiples of each number and determine which multiples they have in common. The least of these numbers will be the least common denominator.
Number
Multiples
$4$ $8$ $12$ $16$ $20$ $24$ $28$ $32$ $36$ $40$ $44$ $48$ $52$ $56$ $\textbf{60}$ $64$
$6$ $12$ $18$ $24$ $30$ $36$ $42$ $48$ $54$ $\textbf{60}$ $66$ $68$
$10$ $20$ $30$ $40$ $50$ $\textbf{60}$ $70$ $80$
The other method is to use prime factorization, the process of finding the prime factors of a number. This is how the method works with numbers.
### Example
Use prime factorization to find the least common multiple of $6$, $10$, and $4$.
Both methods give the same result, but prime factorization is faster. Your choice!
Now that you have found the least common multiple, you can use that number as the least common denominator of the fractions. Multiply each fraction by the fractional form of $1$ that will produce a denominator of $60$:
$\begin{array}{r}\frac{5}{6}\cdot \frac{10}{10}=\frac{50}{60}\\\\\frac{8}{10}\cdot\frac{6}{6}=\frac{48}{60}\\\\\frac{3}{4}\cdot\frac{15}{15}=\frac{45}{60}\end{array}$
Now that you have like denominators, add the fractions:
$\frac{50}{60}+\frac{48}{60}+\frac{45}{60}=\frac{143}{60}$
In the next example, we show how to find the least common multiple of a rational expression with a monomial in the denominator.
### Example
Add$\frac{2n}{15m^{2}}+\frac{3n}{21m}$, and give the domain.
State the sum in simplest form.
That took a while, but you got through it. Adding rational expressions can be a lengthy process, but taken one step at a time, it can be done. Now let’s take a look at some examples where the denominator is not a monomial.
To find the least common denominator (LCD) of two rational expressions, we factor the expressions and multiply all of the distinct factors. For instance, consider the following rational expressions:
$\dfrac{6}{\left(x+3\right)\left(x+4\right)},\text{ and }\frac{9x}{\left(x+4\right)\left(x+5\right)}$
The LCD would be $\left(x+3\right)\left(x+4\right)\left(x+5\right)$.
To find the LCD, we count the greatest number of times a factor appears in each denominator and include it in the LCD that many times.
For example, in $\dfrac{6}{\left(x+3\right)\left(x+4\right)}$, $\left(x+3\right)$ is represented once and $\left(x+4\right)$ is represented once, so they both appear exactly once in the LCD.
In $\dfrac{9x}{\left(x+4\right)\left(x+5\right)}$, $\left(x+4\right)$ appears once and $\left(x+5\right)$ appears once.
We have already accounted for $\left(x+4\right)$, so the LCD just needs one factor of $\left(x+5\right)$ to be complete.
Once we find the LCD, we need to multiply each expression by the form of $1$ that will change the denominator to the LCD.
What do we mean by ” the form of $1$“?
$\frac{x+5}{x+5}=1$ so multiplying an expression by it will not change its value.
For example, we would need to multiply the expression $\dfrac{6}{\left(x+3\right)\left(x+4\right)}$ by $\frac{x+5}{x+5}$ and the expression $\frac{9x}{\left(x+4\right)\left(x+5\right)}$ by $\frac{x+3}{x+3}$.
Hopefully this process will become clear after you practice it yourself. As you look through the examples on this page, try to identify the LCD before you look at the answers. Also, try figuring out which “form of 1” you will need to multiply each expression by so that it has the LCD.
### Example
Add the rational expressions $\frac{5}{x}+\frac{6}{y}$ and define the domain.
State the sum in simplest form.
Here is one more example of adding rational expressions where the denominators are multi-term polynomials. First, we will factor and then find the LCD. Note that $x^2-4$ is a difference of squares and can be factored using special products.
### Example
Simplify$\frac{2{{x}^{2}}}{{{x}^{2}}-4}+\frac{x}{x-2}$ and give the domain.
State the result in simplest form.
The video that follows contains an example of adding rational expressions whose denominators are not alike. The denominators of both expressions contain only monomials.
## Subtracting Rational Expressions
Now let’s try subtracting rational expressions. You’ll use the same basic technique of finding the least common denominator and rewriting each rational expression to have that denominator.
### Example
Subtract$\frac{2}{t+1}-\frac{t-2}{{{t}^{2}}-t-2}$, define the domain.
State the difference in simplest form.
In the next example, we will give less instruction. See if you can find the LCD yourself before you look at the answer.
### Example
Subtract the rational expressions: $\frac{6}{{x}^{2}+4x+4}-\frac{2}{{x}^{2}-4}$, and define the domain.
State the difference in simplest form.
In the previous example, the LCD was $\left(x+2\right)^2\left(x-2\right)$. The reason we need to include $\left(x+2\right)$ two times is because it appears two times in the expression $\frac{6}{{x}^{2}+4x+4}$.
The video that follows contains an example of subtracting rational expressions whose denominators are not alike. The denominators are a trinomial and a binomial.
### Try it
On the next page, we will show you how to find the greatest common denominator for a rational sum or difference that does not share any common factors. We will also show you how to manage a sum or difference of more than two rational expressions. |
# Frequency Distribution
By now we all the concept of frequency of data. But what is the meaning of frequency of a group of data and what is frequency distribution? This lesson simplifies frequency distribution table for both, grouped and ungrouped data using simple examples.
### Organization of Data
Statistics refers to the collection, organization, distribution, and interpretation of data or a set of observations. It is useful in understanding what a dataset reveals about a particular phenomenon. Trends can be studied and results can be drawn from data interpretation. Hence, statistics is a very useful tool to study data.
Download the Cheat Sheet of Statistics by clicking on the button below
## Frequency Distribution: Introduction
To understand frequency distribution, let us first start with a simple example. We consider the marks obtained by ten students from a class in a test to be given as follows:
23, 26, 11, 18, 09, 21, 23, 30, 22, 11
This form of data is known as raw data. A statistical measure called range can be defined. It is the difference between the largest and smallest values of a data set. Here, range = 30 – 09 = 21.
### Frequency Distribution Table
Now, imagine how difficult and cumbersome this process would get if there were a larger number of observations. If we were to include the test scores of all 20 students in this class, it would be very difficult to understand and interpret such data unless it is ‘organized’.
The objective of statistical interpretation is to organize data into a concise form so that interpretation and analysis become easy. It is for this reason that we organize larger data into a table called the frequency distribution table.
### Ungrouped Data
Let the test scores of all 20 students be as follows:
23, 26, 11, 18, 09, 21, 23, 30, 22, 11, 21, 20, 11, 13, 23, 11, 29, 25, 26, 26
Note that the term frequency refers to the number of times an observation occurs or appears in a data set. Hence, in case of repetitions, the frequency increases. The table below will help you understand this better:
Marks obtained in the test No. of students (Frequency) 09 1 11 4 13 1 18 1 20 1 21 2 22 1 23 3 25 1 26 3 29 1 30 1 Total 20
In the example above, the frequency refers to the number of students getting a particular mark in the test. Also, note that your frequency must always total the number of observations after tallying. Here, the total we have obtained after tallying the test scores of the students is 20 which is also the number of observations given.
A frequency distribution such as the one above is called an ungrouped frequency distribution table. It takes into account ungrouped data and calculates the frequency for each observation singularly.
### Grouped Data
Now consider the situation where we want to collect data on the test scores of five such classes i.e. of 100 students. It becomes difficult to tally for each and every score of all 100 students. Besides, the table we will obtain will be very large in length and not understandable at once. In this case, we use what is called a grouped frequency distribution table.
Such tables take into consideration groups of data in the form of class intervals to tally the frequency for the data that belongs to that particular class interval.
Take a look at the table below to understand the concept better:
Marks obtained in the test (Class Interval) No. of students (Frequency) 0-5 3 5-10 11 10-15 38 15-20 34 20-25 9 25-30 5 Total 100
The first column here represents the marks obtained in class interval form. The lowest number in a class interval is called the lower limit and the highest number is called the upper limit. This example is a case of continuous class intervals as the upper limit of one class is the lower limit of the following class.
Note that in continuous cases, any observation corresponding to the extreme values of a class is always included in that class where it is the lower limit. For example, if we had a student who has scored 5 marks in the test, his marks would be included in the class interval 5-10 and not 0-5.
Analogous to continuous class intervals are disjoint class intervals. An example of such as case would be 0-4, 5-9, 10-14, and so on. The frequency distribution can be done for disjoint data as well, similar to how it is done above.
## Solved Example for You
Question 1: The following is the distribution for the age of the students in a school:
Age 0-5 5-10 10-15 15-20 No. of Students 35 45 50 30
Calculate:
• The lower limit of the first class interval.
• The class limits of the third class.
• The classmark for the interval 5-10.
• The class size.
• The lower limit of the first class interval i.e. 0-5 is ‘0’.
• The class limits of the third class, i.e. 10-15 are 10 (lower limit) and 15 (upper limit).
• The classmark is defined as the average of the upper and lower limits of a class. For 5-10, the classmark is (5+10)/2 = 7.5
• The class size is the difference between the lower and upper class-limits. Here, we have a uniform class size, which is equal to 5 (5 – 0, 10 – 5, 15 – 10, 20 – 15 are all equal to 5).
Question 2: Discuss the differences between the frequency table and the frequency distribution table?
Answer: The frequency table is said to be a tabular method where each part of the data is assigned to its corresponding frequency. Whereas, a frequency distribution is generally the graphical representation of the frequency table.
Question 3: What are the numerous types of frequency distributions?
Answer: Different types of frequency distributions are as follows:
1. Grouped frequency distribution.
2. Ungrouped frequency distribution.
3. Cumulative frequency distribution.
4. Relative frequency distribution.
5. Relative cumulative frequency distribution, etc.
Question 4: What are some characteristics of the frequency distribution?
Answer: Some major characteristics of the frequency distribution are given as follows:
1. Measures of central tendency and location i.e. mean, median, and mode.
2. Measures of dispersion i.e. range, variance, and the standard deviation.
3. The extent of the symmetry or asymmetry i.e. skewness.
4. The flatness or the peakedness i.e. kurtosis.
Question 5: What is the importance of frequency distribution?
Answer: The value of the frequency distributions in statistics is excessive. A well-formed frequency distribution creates the possibility of a detailed analysis of the structure of the population. So, the groups where the population breaks down are determinable.
Share with friends
## Customize your course in 30 seconds
##### Which class are you in?
5th
6th
7th
8th
9th
10th
11th
12th
Get ready for all-new Live Classes!
Now learn Live with India's best teachers. Join courses with the best schedule and enjoy fun and interactive classes.
Ashhar Firdausi
IIT Roorkee
Biology
Dr. Nazma Shaik
VTU
Chemistry
Gaurav Tiwari
APJAKTU
Physics
Get Started |
# Number Series – Concept and Types
In this post we will understand the concept of number series which is important part of logical reasoning syllabus for competition exams like GMAT, CAT , SSC, IBPS, SBI, NDA and other notable exams.
we have tried to cover the basic details of number series along with the types of series which are frequently asked in the exams.
## Number Series
Number series is a form of series in which numbers or are arranged in particular order in different ways
Generally two types of questions are asked in number series
1. To find the missing number
2. To find an error in number
Types of number series –
### Prime number series
This type of series contain prime numbers in a consecutive way , alternate way or in a different way
Examples-
#### 7, 11, 13, 17, 19,?
This is a continues series of prime number
So our next number of the series should be 23
#### 3, 7, 13, 19, ?
This is an alternate series of prime numbers
Here we see that we get a gap of one prime number in between each element
#### 3, 7, 17, 31, ?
In this type of series we get a gap of one, two or more than two prime numbers
In above series we get a gap of one prime number in between first and second element
Then gap of two prime number between second and third element and so on.
### Difference series
In this type of series the difference of each consecutive elements is same
For example difference of first and second element is same as difference of second and third element and so on
Examples-
#### 2, 12,22, 32, ?
Here we take a difference of two consecutive elements
Here 42 is the right answer
#### 3, 20, 63, 144, 275, ?
Here we apply Difference of difference method
So our next element should be a difference of +12
### Multiplication series
In multiplication series the multiplication factor is same or may be different in consecutive elements
Examples-
#### 6, 12, 36, 144, ?
Here we get a multiplication series of 2,3,4,5
So our next element should be 144×5=720
### Division series
-in division series the consecutive elements are in decreasing order with large difference. This series follow the same concept of multiplication series but in opposite way here the gap decreases rapidly
Examples-
#### 1000, 500, 250, 125, ?
Here second element is half of first element and third element is half of second element and so on
Here we get the last element as 125÷2=62.5
#### 15120, 2160, 360, 72, 18, ?
Here second element is seven times the first element
And third element is six times the first element and so on
### N2 series
In this type of series the consecutive elements are in square form either in ascending order or descending other
Examples-
#### 1, 4,9, 16, 25,?
Here all the elements are in square form so our last element should be 6=36
#### 196, 169, 144, ?
Here all elements are in squares of decreasing order
So our last element should be 112=121
### N2+1 series
when there is an addition of 1 in N2 series we get N2+1 series
Examples-
#### 10, 17, 26, 37, 50, ?
By taking difference of each consecutive element
This difference can be written as
So it is N2+1 series and our last element should be 82+1=65
### 122, 145, 170, ?
Our last element should be 142+1=197
### N2-1 series
when there is an subtraction of 1 in N2 series we get N2-1 series
Examples-
### 195, 168, 143, 120, ?
Here we get N2-1 series and our last element should be 102-1=99
#### 3, 8, 15, 24, ?
Here we get N2-1 series and our last element should be 62-1=35
### N²+N series
here we add same number to the square of that number
Examples
#### 20, 30, 42, 56, ?
Here we get N2+N series and our last element should be 82+8=72
### N²-N series
here we subtract same number to the square of that number
Example-
#### 56, 42, 30, 20, ?
Here we get N2-N series and our last element should be 42-4=12
### N³ series
here the elements are in cubic form ascending or descending order
Example-
#### 8, 27, 64, 125, ?
Here we get N3 series and our last element should be 63=216
### N³+1 series
here we are at the same number to the cube of that number
Example-
#### 126, 217, 344, ?
Here we get N3+1 series and our last element should be 83+1=513
### N³-1 series
here we subtract the same number to the cube of that number
Example-
#### 7, 26, 63, 124, ?
Here we get N3-1 series and our last element should be 63-1=215
### Alternating series–
It is a special type of series. Here we get a particular series in an alternate elements either addition subtraction multiplication or division
In this type of series our consecutive element are not continuously increases or decreases that is they are not arranged in a particular order
Example-
#### 18, 24, 21, 27, ?, 30
Here we get an alternate series of +3 so our fifth element should be 24
#### 15, 14, 19, 11, 23, ?
Here we get an alternate series of -3 and +4
So our last element should be 8
#### 50, 200, 100, 100, 200, 50, 400, ?
It is alternate multiplication and division series
Here we get an alternate series of ×2 and ÷2
So our last element should be 25
### Miscellaneous series
This is a mixture of all type of series
Examples-
#### 1, 2, 2, 4, 3, 8, 7, 10, ?
Here we get a alternate different type of addition series
#### 2, 7, 27, 107, ?
Here taking the difference of each consecutive element
Here arranging the difference we get following series
So our last element should be 107×4-1=427
### Some important points
• If there is a small gap of numbers between the consecutive elements then it may be difference series
• If there is a large gap of numbers between the consecutive elements then it may be division or multiplication series
• If there is not continuously increasing or decreasing order then it may be in alternate series
• If the elements of the series are near to the square or cube of any numbers then we do some addition or subtraction to get the required elements of the series. |
# Some calculation Tricks in Maths to reduce the time
1) Multiplication of two digit number by 11
Lets take the case 23X11
Step1) Split the number 23 like this,2——-3
Step2) Add the digits of the two digit number . 2+3=5
step3) Place this in the blank space in step 1
So Solution 253
Lets take the case of 57X11
Step1) Split the number 57 like this,5——-7
step2) Add the digits of the two digit number . 5+7=12
Step 3) In such just place the second digit of the number in the blank space and carry over the first digit over the first digit of the original
So solution 627
2) Squaring any two digit number ending with digit 5
Lets take the example of 65
step 1) Multiply the tens place number by the next number ie, 6X7=42
step 2) The square of 5 is 25
Solution is 4225
3) Square any two digit number starting with 5
Lets take the example of 58
step1) Square the fisrt digit and add the second digit to it.
So 25+8=33
Step 2) Square the second digit i.e 8X8=64
Step 3) combine them
3364
4) Fraction,decimal and percentage to remember
Fractions Decimals Percents 1/2 .5 50% 1/3 .3 33.3% 2/3 .6 66.6% 1/4 .25 25% 3/4 .75 75% 1/5 .2 20% 2/5 .4 40% 3/5 .6 60% 4/5 .8 80% 1/6 .16 16.6% @media screen and (min-width: 1201px) { .bunfh5d02da17d9ac0 { display: block; } } @media screen and (min-width: 993px) and (max-width: 1200px) { .bunfh5d02da17d9ac0 { display: block; } } @media screen and (min-width: 769px) and (max-width: 992px) { .bunfh5d02da17d9ac0 { display: block; } } @media screen and (min-width: 768px) and (max-width: 768px) { .bunfh5d02da17d9ac0 { display: block; } } @media screen and (max-width: 767px) { .bunfh5d02da17d9ac0 { display: block; } } 5/6 .83 83.3% 1/8 .125 12.5% 3/8 .375 37.5% 5/8 .675 62.5% 7/8 .875 87.5% 1/9 .1 11.1% 1/10 .1 10% 1/11 .09 9.09% 1/12 .083 8.3% 1/16 .0625 6.25% 1/20 .05 5% 1/25 .04 4% 1/50 .02 2% |
Saturday, November 28, 2009
Prime number generation algorithm
There are many different algorithms to generate prime numbers and they're really interesting.
This one is a short simple algorithm which is not so efficient but still useful for general purposes(generates 10 million primes in a little more than 6 seconds on an AMD Dual core 2.6 Ghz)
First some information which will be useful to understand the algorithm:
1. Factors of a number "converge" to the square root of the number
e.g Let's say we have the number 1024:
The factors of 1024 are:
32*32=1024
16*64=1024
8*128=1024
4*256=1024
2*512=1024
1*1024=1024
As you'll notice as you go from bottom to the top, the factors converge towards the square root of 1024, i.e 32.
The point here that if we need to check if 1024 is a prime number, 1024 should not be divisible by any number from 1 up to 32 (sqrt(1024)). For example if we see that 1024 is divisible by 8 (1024=8*128), in order to save calculations we don't need to check that 1024 is divisible by 128.
2. Every number can be broken down into prime factors
e.g 345=3*5*23 (3,5,23 are all prime numbers)
This 2nd idea is more interesting. Since a number can be broken down into prime factors(which are smaller than the number itself), then if a number is a prime number, it should not be divisible by any prime number smaller than itself.
This means that to determine if a number is prime, we just compare it with a list of primes smaller than itself and see if it is divisible by any one of them. If it's not then it's prime.
These 2 ideas can be combined together to form the following idea:
If a number is prime, it should not be divisible by prime numbers starting from 2 up to the square root of the number.
For example to determine if 23 is prime:
sqrt(23)=4.796
Prime numbers before 4.796= 2,3
Is 23 divisible by 2? no
Is 23 divisible by 3? no
So 23, is prime and it's added to the primes list.
Here's some C code which illustrates the above concept.
//sniper11
#include <stdlib.h>
#include <stdio.h>
#include <math.h>
#define PMAX 20000000 /*max number of primes to generate*/
int n;
int c=0;
unsigned int primes[PMAX]; int primeptr=0;
{
primes[primeptr]=i;
primeptr++;c++;
}
int gen_primes()
{
int i,j,prime;
for(i=3;i<=n;i+=2)//skip even numbers
{
prime=1;
int maxfactor=sqrt(i);
for(j=1;j<primeptr;j++){
if(primes[j]>maxfactor)
break;
if(i%primes[j]==0){
prime=0;
break;
}
}
if(prime==1){
}
}
int output_primes()
{
int i;
for(i=0;i<primeptr;i++)
{
printf("%d\n",primes[i]);
}
int main()
{
printf("Number of primes to generate: ");
scanf("%d",&n);
gen_primes();
printf("Count: %d\n",c);
output_primes();
} |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# The vertex of a parabola is the point $\left( a,b \right)$ and the latus rectum is of length $l$. If the axis of the parabola is along the positive direction of $y-$axis, then its equation is(a) ${{\left( x-a \right)}^{2}}=\dfrac{l}{2}\left( y-2b \right)$(b) ${{\left( x-a \right)}^{2}}=\dfrac{l}{2}\left( y-b \right)$(c) ${{\left( x-a \right)}^{2}}=l\left( y-b \right)$(d) None of these
Last updated date: 21st Jul 2024
Total views: 63.3k
Views today: 0.63k
Answer
Verified
63.3k+ views
Hint: The form of the parabola to be used in the questions is ${{\left( x-{{x}_{1}} \right)}^{2}}=4a\left( y-{{y}_{1}} \right)$.
Complete step-by-step answer:
The vertex and the latus rectum of a parabola are given as $\left( a,b \right)$ and $l$ respectively in the question.
Since the axis of the parabola is along the positive direction of the $y-$axis, we can figure out that the form of the required parabola would be ${{x}^{2}}=4ay$. Latus rectum is indicated by LR and the given point of the vertex is termed as A. We can represent the details as shown in the figure below.
The vertex, A is $\left( a,b \right)$, so we can write the equation for the parabola as,
${{\left( x-a \right)}^{2}}=4c\left( y-b \right)\ldots \ldots \ldots (i)$
Since the coordinate of the vertex is $a$, the term $c$ has been used in the equation above.
From the figure, we can see that the latus rectum is perpendicular to the axis of the parabola and is represented in the figure as LR. We know that the length of the latus rectum for the form of parabola, ${{x}^{2}}=4ay$ is $4c$. Also, it is already given to us in the question as $l$.
Therefore, we can relate the data and we can write the term $4c=l$.
After substituting this relation in equation $(i)$, we get the equation of the parabola as,
${{\left( x-a \right)}^{2}}=l\left( y-b \right)$
Hence, option (c) is obtained as the correct answer.
Note: The best way to approach this question is to figure out the form of the required equation. Looking at the options, the form of the parabola can be obtained easily. One way to figure out the answer would be to check the latus rectum. Since the length of the latus rectum is available from the question, the answer can be computed easily in less time. |
# Proof of Quotient-Remainder Theorem by induction
I am trying to prove the quotient-remainder theorem by mathematical induction:
Quotient-Remainder Theorem:
$\forall a,b \in \mathbb{Z}, b> 0: \ \exists! q, \ r \in \mathbb{Z}, \ a = bq \ + \ r, \ r \in [0,b)$
Proof that $a \in \mathbb{Z}$ exists:
Divide into 2 cases: $a \ge 0$ or $a < 0$
Case 1: $a \ge 0$
Base Case: Let $a = 0$.
It is clear that if we let $q = r = 0$, then $a = b(0) + (0) = 0$.
Hence, base case is true.
Inductive Step: Assume $a = k \in \mathbb{Z}$ exists
Hence, $k = bq + r$
Proof: Let $a = k + 1\in \mathbb{Z}$
By inductive hypothesis, $k = bq + r$
Hence, $k + 1 = bq + r + 1$ and $k + 1 = bq + (r + 1)$ where $r + 1 \in \mathbb{Z}$
by closure property of integers
Thus, $\forall a \in \mathbb{Z}, a \ge 0 \$, $a = bq + r$
Case 2: $a < 0$
Base Case: Let $a = -1$.
It is clear that if we let $q = 0$ and $r = -1$, then $a = b(0) + (-1) = -1$.
Hence, base case is true.
Inductive Step: Assume $a = k \in \mathbb{Z}$ exists
Hence, $k = bq + r$
Proof: Assume $a = -(k + 1) \in \mathbb{Z}$
By inductive hypothesis, $k = bq + r$
Hence, $k + 1 = bq + r + 1$ and $-(k + 1) = b(-q) + (-r - 1)$ where
$-r - 1 \in \mathbb{Z}$ by closure property of integers
Thus, $\forall a \in \mathbb{Z}, a < 0 \$, $a = bq + r$
I am not sure if my proof is correct. Also, I do not know how to prove $r \in [0,b)$ and the uniqueness of $q$ and $r$. Could someone please advise me?
• Try to write down the first few steps of your proof. You'll realize that what's proved is $a = 0 \cdot b + a$, which is true of course, but it's not the quotient-remainder theorem you were trying to prove.
– dxiv
Sep 16 '16 at 4:36
• Two problems. You haven't proven uniqueness. And you induction proves for all a; a =0b + a. You did not take into account that r+1 might >= b. Sep 16 '16 at 5:09
Your induction case didn't take $r+1=b$ into account and in the end all you proved was $a=0b+a$.
Do this: assume $a=qb+r;0\le r <b$. If $r <b-1$ then $a+1 = qb + (r+1)$. And if $r=b-1$ then $a+1 =qb +r+1 =qb+b= (q+1)b+0$.
So if $a =qb +r;r\in [0,b)$ has solution the $a+1=pb+s;s\in [0,b)$ has solution.
One can do the same for $a-1$. If $a=qb+r$ and $r > 0$ then $a-1= qb +(r-1)$ is a solution. If $r=0$ then $a-1= (q-1)b +(b-1)$ is a solution.
But I haven't shown they are unique. You can show uniqueness via induction and I spent a long time in my rough draft showing just that. But in the end it was unnecessarily complicated and confusing and, frankly bizarre. (I'll leave it in as a post script.) It's much better to show uniqueness directly.
Suppose $a=qb+r=pq+s$ then $(r-s)=(p-q)b$. If $r,s \in [0,b)$ then $-b < r-s =(p-q)b <b$. So $-1 < p-q < 1$ so $p-q=0;p=q$ and $r=s$.
So solutions are unique.
=== postscript: bizarre rough draft==
Base step. Need to show $0=0b + 0$ is unique.
Proof: (assume all variables in this post are integers.)
If $0=qb +r$ then $r = -qb$. If $q \le -1$ the $r \ge b$ which isn't acceptable. If $q >0$ then $r < 0$ which isn't acceptable. So $0,0$ is unique solution.
Induction case.
$a=qb+r$ is unique solution.
Let $a+1 = pb +s;0\le s <b$ be an acceptable solution (if there is one, which we need to be aware there might not be).
Then $a = pb + s-1;-1\le s-1 < b-1$. If $s -1 \ge 0$ then $p=q$ and $s=r+1$ (and $r < b-1$) as $p,r$ was a unique solution. So $p=q;s=r+1$ would be a solution and it would be unique.
But it is a solution. So it is unique
Otherwise if our hypothetical solution had $s=0$ then it would follow $r= b-1$. We would have $a+1 = pq +0 = qb +r+1= qb+b =(q+1)b$ and so $p=q+1$ and $s=0$. That would have to be the solution and it would be unique.
But it is a solution. So it is unique.
It isn't quite correct. In your line for the case $a > 0$, "Hence, $k+1=bq+r+1$ and $k+1=bq+(r+1)$ where $r+1 \in \mathbb{Z}$", $r+1$ may be equal to $b$ which violates the statement of the theorem. So you need to add, "If $r+1=b$, then choose $q = q+1$ and $r=0$." A similar statement is necessary in your other inductive step.
• Yup. I understand that r + 1 may be equal to b. r + 1 is the remainder in this case i.e. y = r + 1 < b Sep 16 '16 at 5:05
I am trying to prove the quotient-remainder theorem by mathematical induction
It's not clear from the context whether the emphasis is on "by induction" or on "prove the quotient-remainder theorem".
If the latter, then the intuition is that $\bigcup_{k \in \mathbb{Z}} \{k b, k b + 1, k b + 2, ..., k b + (b-1)\}$ is a partition of $\mathbb{Z}$ for $b \in \mathbb{N}^+$, since it's a union of adjacent, non-overlapping sets of consecutive $b$ integers each. It follows that any $a \in \mathbb{Z}$ will fall into exactly one such set $\{k b, k b + 1, k b + 2, ..., k b + (b-1)\}$, so taking $q = k, r = a - q b$ proves the proposition, both existence and unicity.
• My emphasis is on using proof by induction Sep 16 '16 at 5:02
• I am not sure how to prove my induction though. Are you able to give me some hints to start with? Sep 16 '16 at 6:14
• @LanceHAOH See the other answer posted already. As long as $r+1 \lt b$ your construction works. But when you reach the next multiple of $b$ the proof as written now keeps incrementing $r$ so that it reaches $r+1=b$ and goes outside the target interval $[0,b)$. Instead, you need to increment $q$ and reset $r$ to 0. For uniqueness, assume there are two such distinct representations, and try to derive a contradiction.
– dxiv
Sep 16 '16 at 6:22 |
maths > commercial-arithmetics
Direct and Inverse Variation Pair
what you'll learn...
overview
The objective of this lesson is to show that direct and inverse variations are two sides of the same problem. A number of examples are provided to understand this fact.
number of days ×$\times$ rate (coins per day) =$=$ overall earnings
number of books ×$\times$ weight of a book =$=$ weight of the parcel
length of cloth ×$\times$ rate =$=$ overall cost
number of machines $\times$ number of hours $\Rightarrow$ number of machine-hours
number of pipes $\times$ number of hours $\Rightarrow$ amount filled
number of hens $\times$ number of days $\Rightarrow$ amount of food grain
example type 1
Consider the problem: A person earns $300$ coins in $10$ days. How much will she earn in $25$ days?
This problem is an example of direct variation.
As the number of days increases, the amount earned increases.
Consider the problem: A person earns $30$ coins per day for $10$ days. If the earning rate is reduced to $20$ coins per day, how many days the person has to work to earn the same amount?
This is an example of inverse variation.
As the earning rate decreases, the number of days increases.
Consider the problem: A person earns $300$ coins in 10 days. How much will she earn in $25$ days?
It is noted that the underlying mathematical operation is multiplication.
$10$ days $\times$ rate = $300$coins". Rate can be calculated as $30$ coins per day.
Consider the two problems. The underlying operation for both is
number of days $\times$ rate (coins per day) $=$ overall earnings (coins)
In the first problem,
• number of days = $10$ days (multiplicand)
• overall earnings = $300$ coins (product)
• rate is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Earnings in $25$ days = $\frac{300}{10} \times 25$
In the second problem,
• number of days = $10$ days (multiplicand)
• rate of earnings = $30$ coins per day (multiplier)
• overall earnings is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
Number of days to earn the same amount at $20$ coins per day = $30 \times 10 / 20$
example type 2
Consider the problem: $3$ books are sent by postal-mail weighing $6$kg. How many books will be there in a parcel weighing $20$kg?
It is a direct variation problem.
As the weight of the parcel increases, the number of books increases.
Consider the problem: $3$ books, weighing $2$kg each, can be sent in a parcel. How many books of $3$kg each can be packed in a parcel of same weight?
It is an inverse variation problem.
As the weight of a book increases, the number of books decreases.
Consider the problem: $3$ books are sent by postal-mail weighing $6$kg. How many books will be there in a parcel weighing $20$kg?
The underlying mathematical operation in this is multiplication.
number of books $\times$ weight of a book $=$ weight of the parcel
Consider the two problems. The underlying operation for both is
number of books $\times$ weight of a book $=$ weight of the parcel
In the first problem,
• number of books = $3$ (multiplicand)
• weight of parcel = $6$kg (product)
• weight of each book is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Number of books in a $20$kg parcel = $\frac{3}{6} \times 20 = 10$books
In the second problem,
• number of books = $3$ (multiplicand)
• weight of each book = $2$ kg (multiplier)
• weight of the parcel is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
Number of books of $3$kg weight = $3 \times 2 / 3 = 2$ books.
example type 3
Consider the problem: A person can buy $3$ meter cloth costing $600$ coins. What will be the cost of $7$m of the same cloth?
It is a direct variation problem.
As the length increases, the price of the cloth increases.
Consider the problem: A person can buy $3$ meter cloth which costs $200$ coins per cloth. If the person chooses a different cloth costing $150$ coins per cloth, what is the length of the cloth she can buy?
This is an inverse variation problem.
As the price decreases, the length of the cloth increases.
Consider the problem: A person can buy $3$ meter cloth costing $600$ coins. What will be the cost of $7$m of the same cloth?
The underlying mathematical operation is multiplication.
length of cloth $\times$ rate $=$ overall cost
Consider the two problems. The underlying operation for both is
length of cloth $\times$ rate $=$ overall cost
In the first problem,
• length of cloth $=$ $3$ meter (multiplicand)
• overall cost $= 600$ coins (product)
• rate (multiplier) is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Cost of $7$ meter cloth $= \frac{600}{3} \times 7 = 1400$ coins
In the second problem,
• length of cloth $=$ $3$ meter (multiplicand)
• rate = $200$ coins (multiplier)
• overall cost (product) is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
Length of cloth at cost $150$ coins $= 3 \times 200 / 150 = 4$ meter
example type 4
Consider the problem: $5$ machines can process $90$ boxes of chemical in $15$ hours. How long would $3$ machines take?
This is an inverse variation problem.
As the number of machined decrease, the number of hours increases.
Consider the problem: $5$ machines can process $90$ boxes of chemical in 15 hours. If $3$ machines are used, how much chemical can be processed in the same number of hours?
This is a direct variation problem.
As the number of machines decrease, the amount produced decreases.
Consider the problem: $5$ machines can process $90$ boxes of chemical in $15$ hours. How long would $3$ machines take?
The underlying mathematical operation in this is multiplication.
number of machines $\times$ number of hours = number of machine-hours equivalent of $90$ boxes
Consider the two problems. The underlying operation for both is
number of machines $\times$ number of hours $\Rightarrow$ number of machine-hours equivalent of $90$ boxes
In the first problem,
• number of machines = $5$ (multiplicand)
• number of hours = $15$ (multiplier)
• number of machine-hours (product) remains unchanged.
Multiplicand and multiplier are in inverse proportion.
$3$ machines would take $= 5 \times 15 / 3 = 25$ hours.
In the second problem,
• number of machines = $5$ (multiplicand)
• number of machine-hours is equivalently $90$ boxes of chemical (product)
• number of hours (multiplier) remains unchanged.
Multiplicand and product are in direct proportion.
Amount of chemical processed by $3$ machines $= \frac{90}{5} \times 3 = 54$ boxes
example type 5
Consider the problem: $4$ pipes can fill a tank in $60$ mins. How long does it take to fill the tank with $9$ pipes?
This is an inverse variation problem.
As the number of pipes increase the time to fill the tank decreases.
Consider the problem: $4$ pipes can fill a tank in $60$ mins. How much of the tank will be filled by $3$ pipes in the same time duration?
This is a direct variation problem.
As the number of pipes decrease the amount filled decreases.
Consider the problem: $4$ pipes can fill a tank in $60$ mins. How long does it take to fill the tank with $9$ pipes?
The underlying mathematical operation is multiplication.
number of pipes $\times$ number of hours $\Rightarrow$ amount filled
Consider the two problems. The underlying operation for both is
number of pipes $\times$ number of hours $\Rightarrow$ amount filled
In the first problem,
• number of pipes = $4$(multiplicand)
• number of hours = $60$ mins (multiplier)
• amount filled (product) is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
Time to fill the tank with $9$ pipes = $4 \times 60 / 9$
In the second problem,
• number of pipes = $4$(multiplicand)
• amount filled = $1$ tank (product)
• time (multiplier) remains unchanged.
Multiplicand and product are in direct proportion.
The amount filled by $3$ pipes $= \frac{1}{4} \times 3$
example type 6
Consider the problem: A farmer has enough grains to feed $300$ hens for 18 days. If he buys $100$ more hens, how long would the food last?
This is a problem of inverse variation.
Consider the problem: A farmer has enough grains to feed $300$ hens for $18$ days. If he buys $100$ more hens, how much more grains he has to buy to feed the entire $400$ hens?
This is a problem of direct variation.
As the number of hens increase, the amount of grains required increases.
Consider the problem: A farmer has enough grains to feed $300$ hens for $18$ days. If he buys $100$ more hens, how long would the food last?
The underlying mathematical operation in this is multiplication.
number of hens $\times$ number of days $\Rightarrow$ amount of food grain
Consider the two problems. The underlying operation for both is
number of hens $\times$ number of days $\Rightarrow$ amount of food grain
In the first problem,
• number of hens $= 300$ (multiplicand)
• number of days $= 18$(multiplier)
• amount of food grain (product) is not given and remains unchanged.
Multiplicand and multiplier are in inverse proportion.
The time the food lasts $= 300 \times 18 / 400$ days
In the second problem,
• number of hens $= 300$ (multiplicand)
• Amount of food grain $= 1$ part(product)
• number of days (multiplier) remains unchanged.
Multiplicand and product are in direct proportion.
Amount of food required for $400$ hens $= \frac{1}{300} \times 400$,
This includes the food-grain already available.
example type 7
Consider the problem: A person makes a car in $20$ days. If $4$ persons work together, how long does it take to complete a car?
This is a problem of inverse variation.
As the number persons increase, the number of days to complete a work decreases.
Consider the problem: A person makes a car in $20$ days. If $4$ persons work together, how many cars can they make in $20$ days?
This is a problem of direct variation.
As the number persons increase, the number of cars produced increases.
Consider the problem: A person makes a car in $20$ days. If $4$ persons work together, how long does it take to complete a car?
The underlying mathematical operation is multiplication.
number of persons $\times$ number of days $\Rightarrow$ number of car
Consider the two problems. The underlying operation for both is
number of persons $\times$ number of days $\Rightarrow$ number of cars
In the first problem,
• number of persons = $1$ (multiplicand)
• number of days = $20$ days(multiplier)
• number of cars (product) remains unchanged.
Multiplicand and multiplier are in inverse proportion.
number of days to complete the car for $4$ persons $= 1 \times 20 / 4$
In the second problem,
• number of persons = $1$ (multiplicand)
• number of cars = $1$ (product)
• number of days (multiplier) is not given and remains unchanged.
Multiplicand and product are in direct proportion.
Number of cars $4$ persons can make = $\frac{1}{1} \times 4$ cars
summary
Direct and Inverse Variation Pair : Multiplicand and product are in direct variation.
Multiplier and product are in direct variation.
Multiplicand and Multiplier are in inverse variation.
Outline |
# Problem Solving
Page 1/5 Date 31.01.2017 Size 235.3 Kb. #13486
1 2 3 4 5
Problem Solving
## Problem Solving
In previous math courses, you’ve no doubt run into the infamous “word problems.” Unfortunately, these problems rarely resemble the type of problems we actually encounter in everyday life. In math books, you usually are told exactly which formula or procedure to use, and are given exactly the information you need to answer the question. In real life, problem solving requires identifying an appropriate formula or procedure, and determining what information you will need (and won’t need) to answer the question.
In this chapter, we will review several basic but powerful algebraic ideas: percents, rates, and proportions. We will then focus on the problem solving process, and explore how to use these ideas to solve problems where we don’t have perfect information.
## Percents
In the 2004 vice-presidential debates, Edwards's claimed that US forces have suffered "90% of the coalition casualties" in Iraq. Cheney disputed this, saying that in fact Iraqi security forces and coalition allies "have taken almost 50 percent" of the casualties1. Who is correct? How can we make sense of these numbers?
Percent literally means “per 100,” or “parts per hundred.” When we write 40%, this is equivalent to the fraction or the decimal 0.40. Notice that 80 out of 200 and 10 out of 25 are also 40%, since .
Example 1
243 people out of 400 state that they like dogs. What percent is this?
. This is 60.75%.
Notice that the percent can be found from the equivalent decimal by moving the decimal point two places to the right.
Example 2
Write each as a percent: a) b) 0.02 c) 2.35
a) = 25% b) 0.02 = 2% c) 2.35 = 235%
Percents
If we have a part that is some percent of a whole, then
, or equivalently,
To do the calculations, we write the percent as a decimal.
Example 3
The sales tax in a town is 9.4%. How much tax will you pay on a \$140 purchase?
Here, \$140 is the whole, and we want to find 9.4% of \$140. We start by writing the percent as a decimal by moving the decimal point two places to the left (which is equivalent to dividing by 100). We can then compute:
in tax.
Example 4
In the news, you hear “tuition is expected to increase by 7% next year.” If tuition this year was \$1200 per quarter, what will it be next year?
The tuition next year will be the current tuition plus an additional 7%, so it will be 107% of this year’s tuition:
\$1200(1.07) = \$1284.
Alternatively, we could have first calculated 7% of \$1200: \$1200(0.07) = \$84.
Notice this is not the expected tuition for next year (we could only wish). Instead, this is the expected increase, so to calculate the expected tuition, we’ll need to add this change to the previous year’s tuition:
\$1200 + \$84 = \$1284.
Try it Now 1
A TV originally priced at \$799 is on sale for 30% off. There is then a 9.2% sales tax. Find the price after including the discount and sales tax.
Example 5
The value of a car dropped from \$7400 to \$6800 over the last year. What percent decrease is this?
To compute the percent change, we first need to find the dollar value change: \$6800-\$7400 = -\$600. Often we will take the absolute value of this amount, which is called the absolute change: .
Since we are computing the decrease relative to the starting value, we compute this percent out of \$7400:
decrease. This is called a relative change.
Absolute and Relative Change
Given two quantities,
Absolute change =
Relative change:
Absolute change has the same units as the original quantity.
Relative change gives a percent change.
The starting quantity is called the base of the percent change.
The base of a percent is very important. For example, while Nixon was president, it was argued that marijuana was a “gateway” drug, claiming that 80% of marijuana smokers went on to use harder drugs like cocaine. The problem is, this isn’t true. The true claim is that 80% of harder drug users first smoked marijuana. The difference is one of base: 80% of marijuana smokers using hard drugs, vs. 80% of hard drug users having smoked marijuana. These numbers are not equivalent. As it turns out, only one in 2,400 marijuana users actually go on to use harder drugs2.
Example 6
There are about 75 QFC supermarkets in the U.S. Albertsons has about 215 stores. Compare the size of the two companies.
When we make comparisons, we must ask first whether an absolute or relative comparison. The absolute difference is 215 – 75 = 140. From this, we could say “Albertsons has 140 more stores than QFC.” However, if you wrote this in an article or paper, that number does not mean much. The relative difference may be more meaningful. There are two different relative changes we could calculate, depending on which store we use as the base:
Using QFC as the base, .
This tells us Albertsons is 186.7% larger than QFC.
Using Albertsons as the base, .
This tells us QFC is 65.1% smaller than Albertsons.
Notice both of these are showing percent differences. We could also calculate the size of Albertsons relative to QFC: , which tells us Albertsons is 2.867 times the size of QFC. Likewise, we could calculate the size of QFC relative to Albertsons: , which tells us that QFC is 34.9% of the size of Albertsons.
Example 7
Suppose a stock drops in value by 60% one week, then increases in value the next week by 75%. Is the value higher or lower than where it started?
To answer this question, suppose the value started at \$100. After one week, the value dropped by 60%:
\$100 - \$100(0.60) = \$100 - \$60 = \$40.
In the next week, notice that base of the percent has changed to the new value, \$40. Computing the 75% increase:
\$40 + \$40(0.75) = \$40 + \$30 = \$70.
In the end, the stock is still \$30 lower, or 30% lower, valued than it started.
Try it Now 2
The U.S. federal debt at the end of 2001 was \$5.77 trillion, and grew to \$6.20 trillion by the end of 2002. At the end of 2005 it was \$7.91 trillion, and grew to \$8.45 trillion by the end of 20063. Calculate the absolute and relative increase for 2001-2002 and 2005-2006. Which year saw a larger increase in federal debt?
Example 8
A Seattle Times article on high school graduation rates reported “The number of schools graduating 60 percent or fewer students in four years – sometimes referred to as “dropout factories” – decreased by 17 during that time period. The number of kids attending schools with such low graduation rates was cut in half.”
a) Is the “decrease by 17” number a useful comparison?
b) Considering the last sentence, can we conclude that the number of “dropout factories” was originally 34?
a) This number is hard to evaluate, since we have no basis for judging whether this is a larger or small change. If the number of “dropout factories” dropped from 20 to 3, that’d be a very significant change, but if the number dropped from 217 to 200, that’d be less of an improvement.
b) The last sentence provides relative change which helps put the first sentence in perspective. We can estimate that the number of “dropout factories” was probably previously around 34. However, it’s possible that students simply moved schools rather than the school improving, so that estimate might not be fully accurate.
Example 9
In the 2004 vice-presidential debates, Edwards's claimed that US forces have suffered "90% of the coalition casualties" in Iraq. Cheney disputed this, saying that in fact Iraqi security forces and coalition allies "have taken almost 50 percent" of the casualties. Who is correct?
Without more information, it is hard for us to judge who is correct, but we can easily conclude that these two percents are talking about different things, so one does not necessarily contradict the other. Edward’s claim was a percent with coalition forces as the base of the percent, while Cheney’s claim was a percent with both coalition and Iraqi security forces as the base of the percent. It turns out both statistics are in fact fairly accurate.
Try it Now 3
In the 2012 presidential elections, one candidate argued that “the president’s plan will cut \$716 billion from Medicare, leading to fewer services for seniors,” while the other candidate rebuts that “our plan does not cut current spending and actually expands benefits for seniors, while implementing cost saving measures.” Are these claims in conflict, in agreement, or not comparable because they’re talking about different things?
We’ll wrap up our review of percents with a couple cautions. First, when talking about a change of quantities that are already measured in percents, we have to be careful in how we describe the change.
Example 10
A politician’s support increases from 40% of voters to 50% of voters. Describe the change.
We could describe this using an absolute change: . Notice that since the original quantities were percents, this change also has the units of percent. In this case, it is best to describe this as an increase of 10 percentage points.
In contrast, we could compute the percent change: increase. This is the relative change, and we’d say the politician’s support has increased by 25%.
Lastly, a caution against averaging percents.
Example 11
A basketball player scores on 40% of 2-point field goal attempts, and on 30% of 3-point of field goal attempts. Find the player’s overall field goal percentage.
It is very tempting to average these values, and claim the overall average is 35%, but this is likely not correct, since most players make many more 2-point attempts than 3-point attempts. We don’t actually have enough information to answer the question. Suppose the player attempted 200 2-point field goals and 100 3-point field goals. Then they made 200(0.40) = 80 2-point shots and 100(0.30) = 30 3-point shots. Overall, they made 110 shots out of 300, for a = 36.7% overall field goal percentage. |
# A triangle has sides A, B, and C. Sides A and B have lengths of 5 and 9, respectively. The angle between A and C is (19pi)/24 and the angle between B and C is (pi)/8. What is the area of the triangle?
Mar 3, 2018
5.8234 unit²
#### Explanation:
Let say a is representing angle for side A, b for side B and c for side C.
The area of triangle $= \frac{1}{2} \cdot A \cdot B \cdot \sin c$.
Angle c is located between side A and B, therefore
$c = \pi - \frac{\pi}{8} - \frac{19}{24} \pi$
$c = \pi - \frac{22}{24} \pi = \frac{2}{24} \pi = \frac{\pi}{12}$
The area of triangle $= \frac{1}{2} \cdot 5 \cdot 9 \cdot \sin \left(\frac{\pi}{12}\right)$
$= \frac{1}{2} \cdot 5 \cdot 9 \cdot 0.2588$
$= 5.8234$ unit²# |
# Geometry definitions postulates and theorems
By this postulate, we have that? While some postulates and theorems have been introduced in the previous sections, others are new to our study of geometry.
Once we have determined that the value of x is 13, we plug it back in to the equation for the measure of? In this exercise, we note that the measure of? In a planethrough a point not on a given straight line, at most one line can be drawn that never meets the given line.
Congruent Supplements Theorem If two angles are supplements of the same angle or of congruent anglesthen the two angles are congruent. Finally, we conclude that? Vertical Angles Theorem If two angles are vertical angles, then they have equal measures.
Reflexive Property A quantity is equal to itself.
The alternate exterior angles have the same degree measures because the lines are parallel to each other. Thus, we can use the Alternate Interior Angles Theorem to claim that they are congruent to each other. The Pythagorean theorem states that the sum of the areas of the two squares on the legs a and b of a right triangle equals the area of the square on the hypotenuse c.
Alternatively, two figures are congruent if one can be moved on top of the other so that it matches up with it exactly. Thus, for example, a 2x6 rectangle and a 3x4 rectangle are equal but not congruent, and the letter R is congruent to its mirror image.
DCJ with 71 since we were given that quantity. DGH is equal to the measure of? In this case, we are given equations for the measures of? Parallel postulate To the ancients, the parallel postulate seemed less obvious than the others.
Euclid, rather than discussing a ray as an object that extends to infinity in one direction, would normally use locutions such as "if the line is extended to a sufficient length," although he occasionally referred to "infinite lines".
STQ is the sum of?Angle Properties, Postulates, and Theorems. In order to study geometry in a logical way, it will be important to understand key mathematical properties and to know how to apply useful postulates and theorems.
A postulate is a proposition that has not been proven true, but is considered to be true on the basis for mathematical reasoning. Geometry - Definitions, Postulates, Properties & Theorems Geometry – Page 3 Chapter 4 & 5 – Congruent Triangles & Properties of Triangles Postulates Side-Side-Side (SSS) Congruence Postulate: If three sides of one triangle are congruent to three sides of a second triangle.
Definitions, Postulates and Theorems Page 7 of 11 Triangle Postulates And Theorems Name Definition Visual Clue Centriod Theorem The centriod of a triangle is located 2/3 of the distance from each vertex to the midpoint of.
A summary of de nitions, postulates, algebra rules, and theorems that are often used in geometry proofs: De nitions: De nition of mid-point and segment bisector.
Postulates and Theorems A theorem is a true statement that can be proven. Listed below are six postulates and the theorems that can be proven from these postulates. Video Examples: The five postulates of Euclidean Geometry Solved Example on Postulate Ques: State the postulate or theorem you would .
Geometry definitions postulates and theorems
Rated 5/5 based on 72 review |
# Limits of rational functions – Examples and Explanation
What happens when a ration function approaches infinity? How do we estimate the limit of a rational function? We will answer these questions as we learn about the limits of rational functions.
The limits of rational functions tell us the values that a function approaches at different input values.
A rational function’s limits can help us predict the behavior of the function’s graph at the asymptotes. These values can also tell us how the graph approaches the coordinate system’s negative and positive sides.
## How to find the limit of a rational function?
Finding the limit of rational functions can be straightforward or require us to pull up some tricks. In this section, we’ll learn the different approaches we can use to find the limit of a given rational function.
Recall that rational functions are ratios of two polynomial functions. For example, $f(x) = \dfrac{p(x)}{q(x)}$, where $q(x) \neq 0$.
Limits of rational functions can either be of the form: $\lim_{x\rightarrow a} f(x)$ or $\lim_{x\rightarrow \pm \infty} f(x)$.
As a refresher, this is how we interpret the two:
Algebraic Expression In Words $\lim_{x\rightarrow a} f(x)$ The limit of $f(x)$ as $x$ approaches $a$. $\lim_{x\rightarrow \pm \infty } f(x)$ The limit of $f(x)$ as $x$ approaches positive (or negative) infinity.
Why don’t we start by learning how we can calculate a rational function’s limits as it approaches a given value?
### Finding the limit as $\boldsymbol{x\rightarrow a}$
When we find the limit of $f(x)$ as it approaches $a$, there can be two possibilities: the functions have no restrictions at $x = a$ or it has.
• When $a$ is part of $f(x)$’s domain, we substitute the values into the expression to find its limit.
• When $a$ is not part of $f(x)$’s domain, we try to eliminate the factor corresponding to it then find the value of $f(x)$ using its simplified form.
• Does the function contain a radical expression? Try multiplying both numerator and denominator by the conjugate.
Let’s try observing $f(x) = \dfrac{x – 1}{(x – 1)(x + 1)}$ as it approaches $3$. To better understand what limits represent, we can construct table of values for $x$ close to $3$.
$\boldsymbol{x}$ $\boldsymbol{f(x)}$ $2.9$ $0.256$ $2.99$ $0.251$ $3.001$0.2503.010.249$Do you have a guess on what the values of$\lim_{x\rightarrow 3} \dfrac{x – 1}{(x – 1)(x + 1)}$is? Since$3$is part of the domain of$f(x)$(restricted values for$x$are$1$and$-1$), we can substitute$x = 3$into the equation right away.$\begin{aligned} \lim_{x\rightarrow 3} \dfrac{x – 1}{(x – 1)(x + 1)} &= \dfrac{3 – 1}{(3 – 1)(3 + 1)}\\&=\dfrac{2}{2 \cdot 4}\\&=\dfrac{1}{4}\\&=0.25\end{aligned}$As you might have guessed, as$x$approaches$3$,$f(x)$is equal to$0.25$. Now, what if we want to find$\lim_{x\rightarrow 1} \dfrac{x – 1}{(x – 1)(x + 1)}$? Since$x = 1$is a restriction, we can try to simplify$f(x)$first to remove$x – 1$as a factor.$\begin{aligned} \lim_{x\rightarrow 1} \dfrac{x – 1}{(x – 1)(x + 1)} &= \lim_{x\rightarrow 1} \dfrac{\cancel{(x – 1)}}{\cancel{(x – 1)}(x + 1)}\\&=\lim_{x\rightarrow 1} \dfrac{1}{x + 1}\end{aligned}$Once we have removed the common factors, we can apply the same process and substitute$x = 1$into the simplified expression.$\begin{aligned} \lim_{x\rightarrow 1} \dfrac{1}{x + 1}&=\dfrac{1}{1 + 1}\\&=\dfrac{1}{2}\end{aligned}$Ready to try more problems? Don’t worry. We have prepared a lot of examples for you to work on. For now, let’s learn about limits at infinity. ### Finding the limit as$\boldsymbol{x\rightarrow \infty}$There are instances when we need to know how a rational function behaves on both sides (positive and negative sides). Knowing how to find the limits of$f(x)$as it approaches$\pm \infty$can help us predict this. The value of$\lim_{x\rightarrow \pm \infty } f(x)$can be determined based on its degrees. Let’s say we have$f(x) = \dfrac{p(x)}{q(x)}$and$m$and$n$are the degrees of the numerator and denominator, respectively. The table below summarizes the behavior of$f(x)$as it approaches$\pm infty$. Cases Value of$\boldsymbol{\lim_{x\rightarrow \pm \infty } f(x)}$When the numerator’s degree is smaller:$m < n$.$\lim_{x\rightarrow \pm \infty } f(x) = 0$When the numerator’s degree is larger:$m > n$.$\lim_{x\rightarrow \pm \infty } f(x) =\pm \infty$When the numerator and denominator’s degree are equal:$m = n$.$\lim_{x\rightarrow \pm \infty } f(x) = \dfrac{\text{Leading coefficient of } p(x)}{ \text{ Leading coefficient of } q(x)}$Let’s observe the graphs of three rational functions reflecting the three cases we’ve discussed. • When the degree of the numerator is smaller such as$f(x) = \dfrac{2}{x}$. • When the degree of the numerator is smaller such as$f(x) = \dfrac{x^2 – 1}{x – 2}$. • When the degree of the numerator and denominators are equal such as$f(x) = \dfrac{5x^2 – 1}{x^2 + 3}$. Their graphs also confirm the limits we’ve just evaluated. Knowing the limits ahead of time can also help us predict how the graphs behave. These are the techniques we need at this point – don’t worry, you’ll learn more about limits in your Calculus class. For now, let’s go ahead and practice finding the limits of different rational functions. Example 1 Evaluate the following limits shown below. a.$\lim_{x\rightarrow 4} \dfrac{x – 1}{x + 5}$b.$\lim_{x\rightarrow -2} \dfrac{x^2 – 4}{x^3 + 1}$c.$\lim_{x\rightarrow 3} \dfrac{4x^3 + 2x – 1}{x^2 + 2}$Solution Let’s start with the first function, and since$x = 4$is not a restriction of the function, we can substitute the$x = 4$into the expression right away.$ \begin{aligned} \lim_{x\rightarrow 4} \dfrac{x – 1}{x + 5}&=\dfrac{4 – 1}{4 + 5}\\&=\dfrac{3}{9}\\&=\dfrac{1}{3}\end{aligned}$a. Hence, we have$\lim_{x\rightarrow 4} \dfrac{x – 1}{x + 5} = \boldsymbol{\dfrac{1}{3}}$. We apply the same process for b and c since$\dfrac{x^2 – 4}{x^3 + 1}$and$\dfrac{4x^3 + 2x – 1}{x^2 + 2}$has no restrictions at$x = -2$and$x = 3$, respectively.$\begin{aligned} \lim_{x\rightarrow -2} \dfrac{x^2 – 4}{x^3 + 1}&=\dfrac{(-2)^2 – 4}{(-2)^3 + 1}\\&=\dfrac{4 – 4}{-8 + 1}\\&=\dfrac{0}{-7}\\&= 0\end{aligned}$b. This means that$\lim_{x\rightarrow -2} \dfrac{x^2 – 4}{x^3 + 1} = \boldsymbol{0}$.$\begin{aligned} \lim_{x\rightarrow 3} \dfrac{4x^3 + 2x – 1}{x^2 + 2}&=\dfrac{4(3)^3 + 2(3) -1}{(3)^2 + 2}\\&=\dfrac{108 +6 – 1}{9 + 2}\\&=\dfrac{101}{11}\end{aligned}$c. Hence,$\lim_{x\rightarrow 3} \dfrac{4x^3 + 2x – 1}{x^2 + 2} = \boldsymbol{\dfrac{101}{11}}$. Example 2 What is the limit of$f(x) = \dfrac{2x – 4}{3x^2 – 12}$as it approaches$2$? Solution We can check if$f(x)$has restrictions on$x = 2$, we can find the value of$3x^2 – 12$when$x = 2$:$3(2)^2 – 12 = 0$. This means that we can’t just substitute$x$back into$f(x)$right away. Instead, we can express$f(x)$’s numerator and denominator in factored forms first.$\begin{aligned} f(x)&= \dfrac{2x – 4}{3x^2 – 12}\\&= \dfrac{2(x – 2)}{3(x^2 – 12)}\\&= \dfrac{2(x – 2)}{3(x – 2)(x + 2)}\end{aligned}$Cancel the common factors first to remove the restriction on$x = 2$. We can then find the limit of$f(x)$as it approaches$2$.$ \begin{aligned} f(x)&= \dfrac{2\cancel{(x – 2)}}{3\cancel{(x – 2)}(x + 2)}\\&=\dfrac{2}{3(x + 2)}\\\\\lim_{x\rightarrow 4} f(x)&=\lim_{x\rightarrow 2} \dfrac{2}{3(x + 2)}\\&=\dfrac{2}{3(4 + 2)}\\&=\dfrac{2}{3(6)}\\&=\dfrac{1}{9}\end{aligned}$This means that$\lim_{x\rightarrow 4} f(x) = \boldsymbol{ \dfrac{1}{9}}$. Example 3 If$\lim_{x\rightarrow \infty} f(x) = 0$, which of the following statements is true? a. The ratio of the$f(x)$’s leading coefficients is equal to one. b. The degree of the numerator is greater than the degree of the denominator of$f(x)$. c. The degree of the numerator is less than the degree of the denominator of$f(x)$. d. The degree of the numerator is equal to the degree of the denominator of$f(x)$. Solution The limit of a rational function as it approaches infinity will have three possible results depending on$m$and$n$, the degree of$f(x)$’s numerator and denominator, respectively: $m > n\lim_{x\rightarrow \pm \infty } f(x) = \pm \inftym < n\lim_{x\rightarrow \pm \infty } f(x) = 0m = n\lim_{x\rightarrow \pm \infty } f(x) = \dfrac{\text{Numerator’s leading coefficient }}{ \text{ Denominator’s leading coefficient}}$Since we have$\lim_{x\rightarrow \infty} f(x) = 0$, the degree of the function’s numerator is less than that of the denominator. Example 4 Using the graph shown below, what is the ratio of the leading coefficients of$f(x)$’s numerator and denominator? Solution From this graph, we can see that$\lim_{x\rightarrow \infty} f(x) = 4$. Since the limit is not zero or infinity, the limit for$f(x)$reflects the ratio of the leading coefficients of$p(x)$and$q(x)$. This means that the ratio is equal to$\boldsymbol{4}$. Example 5 What is the limit of$f(x) = \dfrac{x}{\sqrt{x+16} – 4}$as$x$approaches$0$? Solution Let’s check$f(x)$for restrictions at$x =4$by seeing the value of the denominator when$x = 0$.$ \begin{aligned}\sqrt{0+16}- 4 &= 4 – 4\\&= 0\end{aligned}$This means that we need to manipulate$f(x)$by multiplying both its numerator and denominator by the conjugate of$\sqrt{x+16} – 4$.$\begin{aligned}f(x)&= \dfrac{x}{\sqrt{x + 16} – 4}\cdot \dfrac{\sqrt{x+16} + 4}{\sqrt{x+16} + 4}\\&= \dfrac{x(\sqrt{x+16} + 4)}{(\sqrt{x+16} – 4)(\sqrt{x+16} + 4)}\\&= \dfrac{x(\sqrt{x+16} + 4)}{(\sqrt{x+16})^2 – (4)^2}\\&= \dfrac{x(\sqrt{x+16} + 4)}{x+16 – 16}\\&= \dfrac{\cancel{x}(\sqrt{x+16} + 4)}{\cancel{x}}\\&=\sqrt{x+16}+4\end{aligned}$Make sure to review how we rationalize radicals using conjugates by checking out this article. Now that$f(x)$has been rationalized, we can now find the limit of$f(x)$as$x \rightarrow 0$.$\begin{aligned}\lim_{x\rightarrow 0} f(x)&=\lim_{x\rightarrow 0} \sqrt{x + 16} – 4\\&=\sqrt{0 + 16} – 4\\ &= 4 – 4\\&= 0\end{aligned}$Hence, the limit of$f(x)$as it approaches$0$is equal to$\boldsymbol{0}$. ### Practice Questions 1. Evaluate the following limits shown below. a.$\lim_{x\rightarrow 2} \dfrac{2x – 3}{5x + 1}$b.$\lim_{x\rightarrow -4} \dfrac{3x^2 – 5}{2x^2 + 1}$c.$\lim_{x\rightarrow 1} \dfrac{-x^3 + 4x – 6}{x+ 2}$2. Find the value of$\lim_{x\rightarrow a} f(x)$given the following expressions for$a$and$f(x)$. a.$f(x) = \dfrac{x^2 – 1}{x^2 +3x -4}$,$a = -1$b.$f(x) = \dfrac{5x}{x^2 + 3x}$,$a = 0$c.$f(x) = \dfrac{x^2 – 4}{x^2 – 3x + 2}$,$a = 2$3. If$\lim_{x\rightarrow \infty} f(x) = 3$, which of the following statements is true? a. The ratio of the$f(x)$’s leading coefficients is equal to three. b. The degree of the numerator is greater than the degree of the denominator of$f(x)$. c. The degree of the numerator is less than the degree of the denominator of$f(x)$. d. The degree of the numerator is equal to the degree of the denominator of$f(x)$. 4. What is the limit of$f(x) = \dfrac{x}{\sqrt{x+25} – 5}$as$x$approaches$0$? 5. What is the limit of each function as they approach infinity? a.$f(x) = 20 + x^{-3}$b.$g(x) = \dfrac{5x^4 – 20x^5}{2x^7 – 8x^4}$c.$h(x) = \dfrac{3x^2}{x + 2} – 1\$
Images/mathematical drawings are created with GeoGebra. |
How do you solve x+2y=5 and 2x-2y=4?
Aug 3, 2015
The solution for the system of equations is
color(blue)(x=3,y=1
Explanation:
$x + \textcolor{b l u e}{2 y} = 5$ .....equation $\left(1\right)$
2x−color(blue)(2y)=4....equation $\left(2\right)$
We can solve the equations through elimination
Adding the two equations results in elimination of color(blue)(2y
$x + \cancel{\textcolor{b l u e}{2 y}} = 5$
2x−cancelcolor(blue)(2y)=4
$3 x = 9$
color(blue)(x=3
Substituting $x$ in equation $1$ to find $y$
$x + 2 y = 5$
$2 y = 5 - x$
$2 y = 5 - 3$
color(blue)(y=1 |
How Cheenta works to ensure student success?
Explore the Back-Story
# Area of Triangle - AMC 10A - 2019 - Problem No. - 7
## What is Area of Triangle ?
The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is A = 1/2 × b × h, where b is the base and h is the height of the given triangle, whether it is scalene, isosceles or equilateral.
## Try This Problem from AMC 10A - 2019 -Problem No.7
Two lines with slopes $\frac{1}{2}$ and 2 intersect at (2,2) . What is the area of the triangle enclosed by these two lines and the line $x + y = 10$ ?
A) 4 B) $4\sqrt 2$ C) 6 D) 8 E) $6 \sqrt 2$
American Mathematics Competition 10 (AMC 10A), 2019, Problem Number - 7
Area of Triangle
6 out of 10
Problems in Plane Geometry by Sharygin
## Use some hints
If you need a hint to start this sum use this
Lets try to find the slop - intercept form of all three lines : (x,y) = (2,2) and y =
$\frac{x}{2}+b$ implies $2 = \frac{2}{2}+b = 1+b$. So, b = 1 . While y = 2x + c implies 2 = 2.2 + c So, c = -2 And again x+y = 10 implies y = -x + 10.
Thus the lines are $y = \frac {x}{2} + 1$ , y = 2x - 2 and y = -x + 10 . Now we find the intersection points between each of the lines with y = -x + 10 , which are (6,4) and (4,6) .
In the last hint we can apply the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle where the base is $2\sqrt 2$ and the height $3 \sqrt 2$, whose area is 6 .The answer is 6 (c) .
## What is Area of Triangle ?
The area of a triangle is defined as the total space that is enclosed by any particular triangle. The basic formula to find the area of a given triangle is A = 1/2 × b × h, where b is the base and h is the height of the given triangle, whether it is scalene, isosceles or equilateral.
## Try This Problem from AMC 10A - 2019 -Problem No.7
Two lines with slopes $\frac{1}{2}$ and 2 intersect at (2,2) . What is the area of the triangle enclosed by these two lines and the line $x + y = 10$ ?
A) 4 B) $4\sqrt 2$ C) 6 D) 8 E) $6 \sqrt 2$
American Mathematics Competition 10 (AMC 10A), 2019, Problem Number - 7
Area of Triangle
6 out of 10
Problems in Plane Geometry by Sharygin
## Use some hints
If you need a hint to start this sum use this
Lets try to find the slop - intercept form of all three lines : (x,y) = (2,2) and y =
$\frac{x}{2}+b$ implies $2 = \frac{2}{2}+b = 1+b$. So, b = 1 . While y = 2x + c implies 2 = 2.2 + c So, c = -2 And again x+y = 10 implies y = -x + 10.
Thus the lines are $y = \frac {x}{2} + 1$ , y = 2x - 2 and y = -x + 10 . Now we find the intersection points between each of the lines with y = -x + 10 , which are (6,4) and (4,6) .
In the last hint we can apply the distance formula and then the Pythagorean Theorem, we see that we have an isosceles triangle where the base is $2\sqrt 2$ and the height $3 \sqrt 2$, whose area is 6 .The answer is 6 (c) .
## Subscribe to Cheenta at Youtube
This site uses Akismet to reduce spam. Learn how your comment data is processed. |
When a parallel-lines-with-transversal drawing contains more than three lines, identifying congruent and supplementary angles can be kind of challenging. The following figure shows you two parallel lines with two transversals.
If you get a figure that has more than three lines and you want to use any of the transversal ideas, make sure you’re using only three of the lines at a time (two parallel lines and one transversal). If you aren’t using a set of three lines like this, the theorems just don’t work. With the above figure, you can use lines a, b, and c, or you can use lines a, b, and d, but you can’t use both transversals c and d at the same time. Thus, you can’t, for example, conclude anything about angle 1 and angle 6 because angle 1 is on transversal c and angle 6 is on transversal d.
The following list shows what you can say about several pairs of angles in the above figure, noting whether you can conclude that they’re congruent or supplementary. As you read through the list, remember the warning about using only two parallel lines and one transversal.
• Angle pair 2 and 8 are congruent
Reason: Angle 2 and angle 8 are alternate exterior angles on transversal d
• Angle pair 3 and 6 offer no conclusion
Reason: To make angle 3 you need to use transversals, c and d
• Angle pair 4 and 5 are congruent
Reason: Angle 4 and angle 5 are alternate exterior angles on c
• Angle pair 4 and 6 offer no conclusion
Reason: Angle 4 is on transversal c; angle 6 is on transversal d
• Angle pair 2 and 7 is supplementary
Reason: Angle 2 and angle 7 are same-side exterior angles on d
• Angle pair 1 and 8 offer no conclusion
Reason: Angle 1 is on transversal c; angle 8 is on transversal d
• Angle pair 4 and 8 offer no conclusion
Reason: Angle 4 is on transversal c; angle 8 is on transversal d
If you get a figure with more than one transversal or more than one set of parallel lines, you may want to do the following: Trace the figure from your book to a sheet of paper and then highlight one pair of parallel lines and one transversal. (Or you can just trace the three lines you want to work with.) Then you can use the transversal ideas on the highlighted lines. After that, you can highlight a different group of three lines and work with those.
Of course, instead of tracing and highlighting, you can just make sure that the two angles you’re analyzing use only three lines (one ray of each angle should be part of the single transversal you’re using, and the other ray of each angle should be part of one of the two parallel lines you’re using). |
The Power of (a + b)³: Unlocking the Potential of the Cubic Formula
Share
Mathematics has always been a fascinating subject, with its intricate formulas and mind-boggling concepts. One such formula that has intrigued mathematicians for centuries is the (a + b)³ formula, also known as the cubic formula. In this article, we will delve into the depths of this formula, exploring its origins, applications, and the secrets it holds.
The Origins of the Cubic Formula
The cubic formula, (a + b)³, finds its roots in the study of algebraic equations. It was first introduced by the ancient Greeks, who were pioneers in the field of mathematics. However, it was not until the 16th century that Italian mathematicians, such as Niccolò Fontana Tartaglia and Gerolamo Cardano, made significant advancements in solving cubic equations.
Cardano’s work on the cubic formula laid the foundation for future mathematicians to explore its applications. His breakthroughs not only revolutionized the field of mathematics but also paved the way for advancements in various other disciplines, including physics, engineering, and computer science.
Understanding the (a + b)³ Formula
The (a + b)³ formula is an algebraic expression that represents the cube of the sum of two variables, a and b. Mathematically, it can be expanded as follows:
(a + b)³ = a³ + 3a²b + 3ab² + b³
This formula is derived from the concept of binomial expansion, which allows us to expand expressions involving two terms raised to a power. The (a + b)³ formula is a special case of the binomial expansion, where the power is 3.
Applications of the (a + b)³ Formula
The (a + b)³ formula finds applications in various fields, ranging from pure mathematics to real-world problem-solving. Let’s explore some of its key applications:
1. Algebraic Simplification
The (a + b)³ formula is often used to simplify algebraic expressions. By expanding the formula, we can simplify complex expressions and make them easier to work with. This simplification is particularly useful when solving equations or manipulating mathematical models.
2. Geometry
The (a + b)³ formula has applications in geometry, specifically in the calculation of volumes and areas. For example, when calculating the volume of a cube, we can use the formula (a + b)³ to find the volume of each individual side and then sum them up to obtain the total volume.
3. Physics
In physics, the (a + b)³ formula is used to model various phenomena. For instance, when studying the motion of objects, the formula can be used to calculate the displacement, velocity, and acceleration of the object at different time intervals.
4. Engineering
Engineers often use the (a + b)³ formula in their calculations. Whether it’s designing structures, analyzing circuits, or optimizing systems, the formula provides a powerful tool for engineers to solve complex problems efficiently.
Real-World Examples
To better understand the practical applications of the (a + b)³ formula, let’s explore a few real-world examples:
Example 1: Architecture
Architects often use the (a + b)³ formula to calculate the volume of irregularly shaped buildings. By dividing the structure into smaller components and applying the formula to each component, architects can accurately estimate the total volume of the building.
Example 2: Electrical Engineering
In electrical engineering, the (a + b)³ formula is used to analyze circuits and calculate the total resistance. By applying the formula to each resistor in a circuit, engineers can determine the overall resistance and design circuits that meet specific requirements.
Example 3: Financial Modeling
Financial analysts use the (a + b)³ formula to model investment returns. By considering different variables, such as interest rates and time periods, analysts can calculate the potential returns of various investment strategies and make informed decisions.
Key Takeaways
The (a + b)³ formula, also known as the cubic formula, is a powerful tool in mathematics and beyond. Its applications span across various fields, including algebra, geometry, physics, and engineering. By understanding and harnessing the potential of this formula, we can solve complex problems, simplify algebraic expressions, and make informed decisions in real-world scenarios.
Q&A
1. What is the difference between the cubic formula and the quadratic formula?
The cubic formula is used to solve cubic equations, which involve variables raised to the power of three. On the other hand, the quadratic formula is used to solve quadratic equations, which involve variables raised to the power of two. While both formulas are essential in algebra, they are used for different types of equations.
2. Can the (a + b)³ formula be applied to negative numbers?
Yes, the (a + b)³ formula can be applied to negative numbers. The formula works for any real numbers, regardless of their sign. However, it’s important to consider the sign conventions and apply them correctly when using the formula in different contexts.
3. Are there any limitations to the (a + b)³ formula?
The (a + b)³ formula is a powerful tool, but it does have some limitations. One limitation is that it only applies to expressions involving two terms raised to the power of three. Additionally, the formula may not always yield simple or exact solutions, especially when dealing with complex equations.
4. Can the (a + b)³ formula be extended to higher powers?
Yes, the (a + b)³ formula can be extended to higher powers using the concept of binomial expansion. By applying the binomial expansion formula, we can expand expressions involving two terms raised to any power, not just three.
5. How can I practice and improve my understanding of the (a + b)³ formula?
To practice and improve your understanding of the (a + b)³ formula, you can solve various algebraic equations, work on geometry problems, and explore real-world applications. Additionally, studying textbooks, attending math workshops, and engaging in online resources can further enhance your knowledge and skills in using the formula.
Navya Menon
Navya Mеnon is a tеch bloggеr and cybеrsеcurity analyst spеcializing in thrеat intеlligеncе and digital forеnsics. With еxpеrtisе in cybеr thrеat analysis and incidеnt rеsponsе, Navya has contributеd to strеngthеning cybеrsеcurity mеasurеs. |
Tschirnhausen's solution of the cubic
Share on:
A general cubic polynomial has the form $ax^3+bx^2+cx+d$ but a general cubic equation can have the form $x^3+ax^2+bx+c=0.$ We can always divide through by the coefficient of $$x^3$$ (assuming it to be non-zero) to obtain a monic equation; that is, with leading coefficient of 1. We can now remove the $$x^2$$ term by replacing $$x$$ with $$y-a/3$$: $\left(y-\frac{a}{3}\right)^{\negmedspace 3}+a\left(y-\frac{a}{3}\right)^{\negmedspace 2} +b\left(y-\frac{a}{3}\right)+c=0.$ Expanding and simplifying produces $y^3+\left(b-\frac{a^2}{3}\right)y+\frac{2}{27}a^3-\frac{1}{3}ab+c=0.$ In fact this can be simplified by writing the initial equation as $x^3+3ax^2+bx+c=0$ and then substituting $$x=y-a$$ to obtain $y^3+(b-3a^2)y+(2a^3-ab+c)=0.$ This means that in fact an equation of the form $y^3+Ay+B=0$ is a completely general form of the cubic equation. Such a form of a cubic equation, missing the quadratic term, is known as a depressed cubic.
We could go even further by substituting $y=z\sqrt{A}$ to obtain
$A^{3/ 2}z^3+A\sqrt{A}z+B=0$
and dividing through by $$A^{3/ 2}$$ to produce
$z^3+z+BA^{-3/ 2}=0.$
This means that
$z^3+z+W=0$
is also a perfectly general form for the cubic equation.
Cardan's method
Although this is named for Gerolamo Cardano (1501-1576), the method was in fact discovered by Niccolò Fontana (1500-1557), known as Tartaglia ("the stammerer") on account of a injury obtained when a soldier slashed his face when he was a boy. In the days before peer review and formal dissemination of ideas, any new mathematics was closely guarded: mathematicians would have public tests of skill, and a new solution method was invaluable. After assuring Tartaglia that his new method was safe with him, Cardan then proceeded to publish it as his own in his magisterial Ars Magna in 1545. A fascinating account of the mix of Cardan, Tartaglia, and several other egotistic mathematicians of the time, can be read here.
Cardan's method solves the equation $x^3-3ax-2b=0$ noting from above that this is a perfectly general form for the cubic, and where we have introduced factors of $$-3$$ and $$-2$$ to eliminate fractions later on. We start by assuming that the solution will have the form $x=p^{1/ 3}+q^{1/ 3}$ and so $x^3=(p^{1/ 3}+q^{1/ 3})^3=p+3p^{2/ 3}q^{1/ 3}+3p^{1/ 3}q^{2/ 3}+q.$ This last can be written as $p+q+3p^{1/ 3}q^{1/ 3}(p^{1/ 3}+q^{1/ 3}).$ We can thus write $x^3=3p^{1/ 3}q^{1/ 3}x+p+q$ and comparing with the initial cubic equation we have $3p^{1/ 3}q^{1/ 3}=3a,\quad p+q=2b.$ These can be written as $pq=a^3,\quad p+q=2b$ for which the solutions are $p,q=b\pm\sqrt{b^2-a^3}$ and so $x = (b+\sqrt{b^2-a^3})^{1/ 3}+(b-\sqrt{b^2-a^3})^{1/ 3}.$ This can be written in various different ways.
For example, $x^3-6x-6=0$ for which $$a=2$$ and $$b=3$$. Here $$b^2-a^3=1$$ and so one solution is $x=4^{1/ 3}+2^{1/ 3}.$ Given that a cubic must have three solutions, the other two are $\omega p^{1/ 3}+\omega^2 q^{1/ 3},\quad \omega^2 p^{1/ 3}+\omega q^{1/ 3}$ where $$\omega$$ is a cube root of 1, for example $\omega=\frac{1}{2}+i\frac{\sqrt{3}}{2}.$
And so to Tschirnhausen
At the beginning we eliminated the $$x^2$$ terms from a cubic equation by a linear substitution $$x=y-a/3$$ or $$y=x+a/3$$. Writing in the year 1680, the German mathematician Ehrenfried Walther von Tschirnhausen (1651-1708) began experimenting with more general polynomial substitutions, believing that it would be possible to eliminate other terms at the same time. Such substitutions are now known as Tschirnhausen transformations and of course the modern general approach places them squarely within field theory.
Tschirnhausen was only partially correct: it is indeed possible to remove some terms from a polynomial equation, and in 1858 the English mathematician George Jerrard (1804-1863) showed that it was possible to remove the terms of degree $$n-1$$, $$n-2$$ and $$n-3$$ from a polynomial of degree $$n$$. In particular, the general quintic equation can be reduced to $x^5+px+q=0$ which is known as the Bring-Jerrard form; also honouring Jerrard's predecessor, the Swedish mathematician Erland Bring (1736-1798). Note that Jerrard was quite well aware of the work of Ruffini, Abel and Galois in proving the general unsolvability by radicals of the quintic equation.
Neither Bring nor Tschirnhausen had the advantage of this knowledge, and both were working towards a general solution of the quintic.
Happily, Tschirnhausen's work is available in an English translation, published in the ACM SIGSAM Bulletin by R. F. Green in 2003. For further delight, Jerrard's text, with the splendidly formal English title "An Essay on the Resolution of Equations", is also available online.
After that history lesson, let's explore how to remove both the quadratic and linear terms from a cubic equation using Tschirnhausen's method, and also using SageMath to do the heavy algebraic work. There is in fact nothing particularly conceptually difficult, but the algebra is quite messy and fiddly.
We start with a depressed cubic equation $x^3+3ax+2b=0$ and we will use the Tschirnhausen transformation $y=x^2+rx+s.$
This can be done by hand of course, using a somewhat fiddly argument, but for us the best approach is to compute the resultant of the two polynomials, which is a polynomial expression equal to zero if the two polynomials have a common root. The resultant can be computed as the determinant of the Sylvester matrix (named for its discoverer); but we can simply use SageMath:
var('a,b,c,x,y,r,s')
cb = x^3 + 3*a*x + 2*b
res = cb.resultant(y-x^2-r*x-s,x).poly(y)
res
$\displaylines{ y^3+3(2a-s)y^{2}+3(ar^{2}+3a^{2}+2r-4as+s^{2})y\\\ {\ }\mspace4em -4b^{2}+2br^{3}-3ar^{2}s+6abr-9a^{2}s-6brs+6as^{2}-s^{3} }$
Now we find values of $$r$$ and $$s$$ for which the coefficients of $$y^2$$ and $$y$$ will be zero:
sol = solve([res.coefficient(y,1),res.coefficient(y,2)],[r,s],solution_dict=True)
sol
$\left[\left\lbrace s : 2 , a, r : -\frac{b + \sqrt{a^{3} + b^{2}}}{a}\right\rbrace, \left\lbrace s : 2 , a, r : -\frac{b - \sqrt{a^{3} + b^{2}}}{a}\right\rbrace\right]$
We can now substitute say the second solution into the resultant from above, which should produce an expression of the form $$y^3+A$$:
cby = res.subs(sol[1]).canonicalize_radical().poly(y)
cby
$y^3-8 , a^{3} - 16 , b^{2} + 8 , \sqrt{a^{3} + b^{2}} b - \frac{8 , b^{4}}{a^{3}} + \frac{8 , \sqrt{a^{3} + b^{2}} b^{3}}{a^{3}}$ We can simply take the cube root of the constant term as our solution:
sol_y = solve(cby,y,solution_dict=True)
sol_y[2]
$\left\lbrace y : \frac{2 , {\left(a^{6} + 2 , a^{3} b^{2} - \sqrt{a^{3} + b^{2}} a^{3} b + b^{4} - \sqrt{a^{3} + b^{2}} b^{3}\right)}^{\frac{1}{3}}}{a}\right\rbrace$
Now we solve the equation $$y=x^2+rx+s$$ using the values $$r$$ and $$s$$ from above, and the value of $$y$$ just obtained:
eq = x^2+r*x+s-y
eqrs = eq.subs(sol[1])
eqx = eqrs.subs(sol_y[2])
solx = solve(eqx,x,solution_dict=True)
solx[0]
$\left\lbrace x : \frac{b - \sqrt{a^{3} + b^{2}} - \sqrt{-7 , a^{3} + 2 , b^{2} - 2 , \sqrt{a^{3} + b^{2}} b + 8 , {\left(a^{6} + 2 , a^{3} b^{2} + b^{4} - {\left(a^{3} b + b^{3}\right)} \sqrt{a^{3} + b^{2}}\right)}^{\frac{1}{3}} a}}{2 , a}\right\rbrace$
A equation, of err... rare beauty, or if not beauty, then something else. It certainly lacks the elegant simplicity of Cardan's solution. On the other hand, the method can be applied to quartic (and quintic) equations, which Cardan's solution can't.
Finally, let's test this formula, again on the equation $$x^3-6x-6=0$$, for which $$a=-2$$ and $$b=-3$$:
xs = solx[0][x].subs({a:-2, b:-3})
xs
$\frac{1}{4} , \sqrt{-16 \cdot 4^{\frac{1}{3}} + 80} + 1$
This can clearly be simplified to
$1+\sqrt{5-4^{1/ 3}}$ It certainly looks different from Cardan's result, but watch this:
xt = QQbar(xs)
$\frac{1}{2}4^{2/ 3}+4^{1/ 3}$
xt.minpoly()
$x^3-6x-6$ |
Home > Documents > [PPT]Isosceles, Equilateral, and Right Triangles · Web viewIsosceles, Equilateral, and Right...
# [PPT]Isosceles, Equilateral, and Right Triangles · Web viewIsosceles, Equilateral, and Right...
Date post: 20-Mar-2018
Category:
Author: trandang
View: 336 times
Embed Size (px)
of 22 /22
Isosceles, Equilateral, and Right Triangles Chapter 4.6
Transcript
Isosceles, Equilateral, and Right Triangles
Chapter 4.6
Isosceles Triangle Theorem
Isosceles The 2 Base s are • Base angles are the angles opposite the equal
sides.
Isosceles Triangle Theorem
A C
B
If AB BC, then A C
Isosceles Triangle Theorem
A C
B
If A C then AB BC
Sample ProblemSolve for the variables• mA = 32°• mB = (4y)° • mC = (6x +2)°
A C
B
6x + 2 = 32
6x = 30
x = 5
32 + 32 + 4y = 180
4y + 64 = 180
4y = 116
y = 29
Find the Measure of a Missing Angle
120o
180o – 120o = 60o
30o 30o
30o
180o – 30o = 150o
75o
75o
1. A2. B3. C4. D
A. 25
B. 35
C. 50
D. 130
1. A2. B3. C4. D
0%0%0%0%
A B C D
A. Which statement correctly names two congruent angles?
A.
B.
C.
D.
1. A2. B3. C4. D
0%0%0%0%
A B C D
B. Which statement correctly names two congruent segments?
A.
B.
C.
D.
Equilateral Triangle Theorem
Equilateral Equiangular
Each angle = 60o !!!
60
3180
Use Properties of Equilateral Triangles
Subtraction
Linear pair Thm.
Substitution
A. AB. BC. CD. D A B C D
0% 0%0%0%
A. x = 15
B. x = 30
C. x = 60
D. x = 90
A. AB. BC. CD. D
A B C D
0% 0%0%0%
A. 30
B. 60
C. 90
D. 120
Don’t be an ASS!!!• Angle Side Side does not work!!!
– (Neither does ASS backward!)• It can not distinguish between the two
different triangles shown below.
However, if the angle is a right angle, then they are no longer called sides. They are called…
Hypotenuse-Leg Theorem• If the hypotenuse and one leg of
a right triangle are congruent to the corresponding parts in another right triangle, then the triangles are congruent.
ABC XYZ Why?HL Theorem
B
CA
X Z
Y
Prove XMZ YMZ
X Y
Z
M
Step Reason
YZXZ GivenXYZM Given
mZMX = mZMY = 90o Def of lines
ZMZM ReflexiveHL Thm
ZMX ZMY
Corresponding Parts Corresponding Parts of Congruent Triangles of Congruent Triangles are Congruentare Congruent
Given Given ΔΔABC ABC ΔΔXYZXYZ You can state that:You can state that:
A A XX B B YY C C ZZ
AB AB XY XY BCBC YZYZ CACA ZXZX
CA
23
Suppose you know that ABD CDB by SAS Thm. Which additional pairs of sides and angles can be found congruent using Corr. Parts of s are ?
Complete the following two-column proof.
Proof:
4.
ReasonsStatements1. Given2. Isosceles Δ Theorem
1. 2.3. 3. Given
4. Def. of midpoint
A. AB. BC. CD. D
A B C D
0% 0%0%0%
Proof:
4.
ReasonsStatements
4. Def. of midpoint5. ______
6. 6. ?5. ΔABC ΔADC ?
Complete the following two-column proof.
SAS Thm.Corr. Parts of s are
HomeworkCh 4-6 • pg 248
1 – 10, 14 – 27, 32, 33, 37 – 39, & 48
2,
22121 yyxx
Reminder!
Midpoint Formula:
Video C
Recommended |
### Comparing fractions
In arrival to Fractions, us learned that fractions space a method of showing part that something. Fractions are useful, since they let us tell exactly how much we have actually of something. Part fractions are bigger than others. For example, i m sorry is larger: 6/8 of a pizza or 7/8 that a pizza?
In this image, we have the right to see the 7/8 is larger. The illustration makes it basic to compare this fractions. However how can we have done it without the pictures?
Click through the slideshow come learn just how to to compare fractions.
You are watching: Using the fraction 4/6 which number is the denominator
Earlier, we saw that fractions have actually two parts.
One part is the optimal number, or numerator.
The various other is the bottom number, or denominator.
The denominator tells us how many parts space in a whole.
The molecule tells united state how countless of those parts we have.
When fractions have the very same denominator, it means they're separation into the same variety of parts.
This method we have the right to compare this fractions simply by looking in ~ the numerator.
Here, 5 is an ext than 4...
Here, 5 is an ext than 4...so we deserve to tell that 5/6 is an ext than 4/6.
Let's watch at another example. I m sorry of this is larger: 2/8 or 6/8?
If you thought 6/8 to be larger, you were right!
Both fractions have the very same denominator.
So we contrasted the numerators. 6 is larger than 2, so 6/8 is much more than 2/8.
As girlfriend saw, if 2 or more fractions have the same denominator, you can compare castle by looking at their numerators. Together you have the right to see below, 3/4 is bigger than 1/4. The bigger the numerator, the larger the fraction.
### Comparing fractions with various denominators
On the previous page, we compared fractions that have actually the very same bottom numbers, or denominators. Yet you know that fractions deserve to have any number together a denominator. What happens when you have to compare fountain with various bottom numbers?
For example, i m sorry of this is larger: 2/3 or 1/5? It's complicated to tell just by looking in ~ them. After ~ all, 2 is larger than 1, yet the denominators aren't the same.
If you look at the picture, though, the distinction is clear: 2/3 is bigger than 1/5. V an illustration, that was straightforward to to compare these fractions, however how could we have done it there is no the picture?
Click through the slideshow come learn exactly how to to compare fractions with various denominators.
Let's compare these fractions: 5/8 and 4/6.
Before we compare them, we need to adjust both fountain so they have actually the very same denominator, or bottom number.
First, we'll uncover the the smallest number that have the right to be separated by both denominators. We contact that the lowest common denominator.
Our an initial step is to uncover numbers that have the right to be split evenly by 8.
Using a multiplication table makes this easy. All of the numbers on the 8 row can be split evenly by 8.
Now let's look in ~ our second denominator: 6.
We can use the multiplication table again. All of the number in the 6 row have the right to be split evenly by 6.
Let's compare the two rows. The looks like there space a few numbers that deserve to be separated evenly through both 6 and also 8.
24 is the smallest number that shows up on both rows, so it's the lowest usual denominator.
Now we're walking to change our fractions so castle both have actually the same denominator: 24.
To perform that, we'll have to readjust the molecule the same way we adjusted the denominators.
Let’s look at 5/8 again. In order to adjust the denominator to 24...
Let’s look in ~ 5/8 again. In stimulate to readjust the denominator come 24...we had to multiply 8 by 3.
Since we multiplied the denominator by 3, we'll also multiply the numerator, or top number, by 3.
5 times 3 equates to 15. So we've adjusted 5/8 into 15/24.
We can do the because any kind of number over itself is equal to 1.
So when we main point 5/8 through 3/3...
So once we multiply 5/8 by 3/3...we're yes, really multiplying 5/8 through 1.
Since any kind of number time 1 is equal to itself...
Since any kind of number time 1 is same to itself...we deserve to say the 5/8 is same to 15/24.
Now we'll carry out the very same to our other fraction: 4/6. We also readjusted its denominator come 24.
Our old denominator was 6. To obtain 24, us multiplied 6 through 4.
So we'll also multiply the molecule by 4.
4 time 4 is 16. Therefore 4/6 is equal to 16/24.
Now the the denominators room the same, we deserve to compare the 2 fractions by spring at your numerators.
16/24 is bigger than 15/24...
16/24 is larger than 15/24... For this reason 4/6 is larger than 5/8.
### Rchathamtownfc.netcing fractions
Which of these is larger: 4/8 or 1/2?
If girlfriend did the mathematics or also just looked at the picture, you can have to be able come tell the they're equal. In other words, 4/8 and also 1/2 mean the very same thing, also though they're composed differently.
If 4/8 method the very same thing together 1/2, why no just speak to it that? One-half is simpler to say 보다 four-eighths, and also for most people it's also easier to understand. After all, as soon as you eat out v a friend, you break-up the invoice in half, not in eighths.
If you write 4/8 together 1/2, you're rchathamtownfc.netcing it. When we rchathamtownfc.netce a fraction, we're composing it in a simpler form. Lessened fractions are constantly equal to the original fraction.
We already rchathamtownfc.netced 4/8 come 1/2. If you look at the examples below, you have the right to see that various other numbers can be diminished to 1/2 as well. This fractions space all equal.
5/10 = 1/211/22 = 1/236/72 = 1/2
These fractions have actually all been rchathamtownfc.netced to a simpler kind as well.
4/12 = 1/314/21 = 2/335/50 = 7/10
Click with the slideshow to learn exactly how to minimize fractions by dividing.
Let's try rchathamtownfc.netcing this fraction: 16/20.
Since the numerator and also denominator are even numbers, you deserve to divide lock by 2 to minimize the fraction.
First, we'll division the numerator by 2. 16 divided by 2 is 8.
Next, we'll divide the denominator through 2. 20 divided by 2 is 10.
We've decreased 16/20 come 8/10. Us could additionally say that 16/20 is same to 8/10.
If the numerator and denominator can still be split by 2, we can continue rchathamtownfc.netcing the fraction.
8 divided by 2 is 4.
10 divided by 2 is 5.
Since there's no number that 4 and also 5 have the right to be divided by, us can't rchathamtownfc.netce 4/5 any kind of further.
This method 4/5 is the simplest form of 16/20.
Let's try rchathamtownfc.netcing another fraction: 6/9.
While the numerator is even, the denominator is one odd number, so we can't minimize by separating by 2.
Instead, we'll need to uncover a number that 6 and also 9 have the right to be split by. A multiplication table will certainly make the number easy to find.
Let's discover 6 and also 9 top top the same row. As you have the right to see, 6 and 9 deserve to both be divided by 1 and 3.
Dividing by 1 won't readjust these fractions, for this reason we'll use the largest number that 6 and also 9 deserve to be split by.
That's 3. This is referred to as the greatest usual divisor, or GCD. (You can also call the the greatest common factor, or GCF.)
3 is the GCD the 6 and also 9 because it's the largest number they deserve to be separated by.
So we'll divide the molecule by 3. 6 split by 3 is 2.
Then we'll divide the denominator by 3. 9 split by 3 is 3.
Now we've decreased 6/9 come 2/3, i m sorry is its easiest form. We could additionally say that 6/9 is equal to 2/3.
Irrchathamtownfc.netcible fractions
Not all fractions have the right to be rchathamtownfc.netced. Part are currently as basic as they deserve to be. For example, friend can't rchathamtownfc.netce 1/2 because there's no number various other than 1 that both 1 and also 2 deserve to be split by. (For that reason, friend can't mitigate any portion that has actually a numerator of 1.)
Some fountain that have larger number can't be lessened either. Because that instance, 17/36 can't be rchathamtownfc.netced because there's no number the both 17 and also 36 can be separated by. If girlfriend can't find any common multiples for the numbers in a fraction, possibilities are it's irrchathamtownfc.netcible.
Try This!
Rchathamtownfc.netce each fraction to its most basic form.
### Mixed numbers and improper fractions
In the vault lesson, girlfriend learned about mixed numbers. A mixed number has actually both a fraction and a whole number. An example is 1 2/3. You'd check out 1 2/3 like this: one and also two-thirds.
Another way to compose this would certainly be 5/3, or five-thirds. These two numbers watch different, yet they're actually the same. 5/3 is an improper fraction. This just means the molecule is larger 보다 the denominator.
There space times when you might prefer to usage an improper portion instead of a blended number. It's basic to change a combined number right into an wrong fraction. Let's learn how:
Let's convert 1 1/4 into an wrong fraction.
First, we'll require to find out how countless parts comprise the totality number: 1 in this example.
To do this, we'll main point the whole number, 1, by the denominator, 4.
1 times 4 equals 4.
Now, let's add that number, 4, come the numerator, 1.
4 add to 1 equals 5.
The denominator stays the same.
Our improper portion is 5/4, or five-fourths. For this reason we can say that 1 1/4 is same to 5/4.
This way there are five 1/4s in 1 1/4.
Let's convert another mixed number: 2 2/5.
First, we'll multiply the totality number through the denominator. 2 time 5 equals 10.
Next, we'll include 10 come the numerator. 10 plus 2 equates to 12.
As always, the denominator will continue to be the same.
So 2 2/5 is same to 12/5.
Try This!
Try convert these mixed numbers right into improper fractions.
Converting improper fractions right into mixed numbers
Improper fractions are useful for math troubles that use fractions, together you'll learn later. However, they're likewise more daunting to read and also understand 보다 mixed numbers. Because that example, it's a lot simpler to picture 2 4/7 in your head than 18/7.
Click with the slideshow come learn just how to change an improper fraction into a combined number.
Let's revolve 10/4 right into a combined number.
You have the right to think of any portion as a division problem. Simply treat the line in between the numbers choose a division sign (/).
So we'll divide the numerator, 10, by the denominator, 4.
10 divided by 4 amounts to 2...
10 separated by 4 equates to 2... With a remainder the 2.
The answer, 2, will become our entirety number because 10 deserve to be divided by 4 twice.
And the remainder, 2, will become the numerator of the portion because we have actually 2 parts left over.
The denominator stays the same.
So 10/4 equals 2 2/4.
Let's try another example: 33/3.
We'll divide the numerator, 33, by the denominator, 3.
33 split by 3...
33 separated by 3... Equates to 11, v no remainder.
The answer, 11, will come to be our totality number.
See more: 16000 Hours Equals How Many Days Is 16000 Hours To Days, Convert 16000 Hours To Days
There is no remainder, for this reason we can see that our improper portion was in reality a whole number. 33/3 equals 11. |
Search IntMath
Close
# What You Need to Know About Angles in Geometry
Angles are an essential component of geometry and understanding them is key to understanding the relationships between polygons, triangles, parallelograms, trapezoids, and inscribed angles. In this article, we'll go over the basics of angles and provide some practice problems to help you better understand how to use them in your math studies.
## What are Angles?
An angle is defined as a figure formed by two lines (or rays) that meet at a certain point, called the vertex. The lines that form an angle are called sides of the angle. An angle can be measured in degrees or radians. The figure below illustrates an angle with its two sides and vertex.
The angle in the figure above is an acute angle, which is an angle that measures less than 90 degrees. In addition to acute angles, there are also right angles, which measure exactly 90 degrees, obtuse angles, which measure more than 90 degrees but less than 180 degrees, and straight angles, which measure exactly 180 degrees.
## Types of Angles
Angles can be classified in several ways. One type of angle is called a complementary angle, which is an angle that adds up to 90 degrees. For example, if one angle measures 40 degrees, then its complementary angle measures 50 degrees. Another type of angle is called a supplementary angle, which is an angle that adds up to 180 degrees. For example, if one angle measures 90 degrees, then its supplementary angle measures 90 degrees.
Angles can also be classified according to their relative size. An obtuse angle is one that measures more than 90 degrees, while an acute angle is one that measures less than 90 degrees. An angle can also be classified as right or straight. A right angle measures exactly 90 degrees, while a straight angle measures exactly 180 degrees.
## Angles in Polygons
The angles of a polygon are the angles between two adjacent sides. A triangle is a polygon with three sides, and it has three angles. A quadrilateral is a polygon with four sides, and it has four angles. A pentagon is a polygon with five sides, and it has five angles. The sum of the angles in a polygon is equal to (n-2)*180, where n is the number of sides in the polygon. For example, the sum of the angles of a triangle is 180 degrees, and the sum of the angles of a quadrilateral is 360 degrees.
## Angles in Parallelograms
A parallelogram is a quadrilateral with two pairs of parallel sides. The angles of a parallelogram are all equal. This means that a parallelogram has four angles that all measure the same. For example, if one angle of a parallelogram measures 40 degrees, then all other angles of the parallelogram measure 40 degrees.
## Angles in Trapezoids
A trapezoid is a quadrilateral with one pair of parallel sides. The angles of a trapezoid can be different. This means that a trapezoid can have two angles that are equal, two angles that are not equal, or four angles that are all different. For example, if one angle of a trapezoid measures 40 degrees, then the other three angles could measure 30 degrees, 50 degrees, and 90 degrees.
## Inscribed Angles
An inscribed angle is an angle that is formed inside a circle. The sides of an inscribed angle are chords of the circle. The measure of an inscribed angle is equal to half the measure of the central angle that subtends (or passes through) the same arc. For example, if one inscribed angle measures 30 degrees, then the central angle that subtends the same arc measures 60 degrees.
## Practice Problems
Let's try out some practice problems to test your knowledge of angles.
1. What is the sum of the angles in a pentagon?
2. What is the measure of an obtuse angle?
Answer: More than 90 degrees but less than 180 degrees.
3. What is the measure of a complementary angle?
4. What is the measure of an inscribed angle?
Answer: Half the measure of the central angle that subtends the same arc.
5. What is the measure of the angles in a parallelogram?
6. What is the sum of the angles in a triangle?
7. What type of angle is one that measures exactly 90 degrees?
8. What type of angle is one that measures exactly 180 degrees?
9. What type of angle is one that measures more than 90 degrees but less than 180 degrees?
10. What type of angle is one that measures less than 90 degrees?
## Summary
In this article, we've gone over the basics of angles in geometry. We've discussed the types of angles, angles in polygons, angles in parallelograms, angles in trapezoids, and inscribed angles. We've also provided some practice problems to help you better understand how to use angles in your math studies. With this knowledge, you should now have a better understanding of angles and how to use them in your math studies.
## FAQ
### What is the difference between an acute angle and an obtuse angle?
An acute angle has a measure that is less than 90 degrees, while an obtuse angle has a measure that is greater than 90 degrees.
### What is a straight angle?
A straight angle has a measure of exactly 180 degrees, and it is formed by two rays that point in opposite directions from a common endpoint. |
# 3.1: Basics of Geometry
Difficulty Level: At Grade Created by: CK-12
## Points, Lines, and Planes
Connections to Art
Tangrams are a fun way to introduce Geometry to any student. If you have tangrams, let students play with them, make designs or animals. If you do not have access to tangrams, take them to the computer lab to do online tangrams: http://pbskids.org/cyberchase/games/area/tangram.html
Point of points of intersection, lines, triangles, quadrilaterals, and planes to students as they play. You could even take picture of students creations and put them up in your classroom.
A practical application of points, lines, and planes are maps. Show students a map of a state and see if they can determine what represents a city (point), a highway (line), and the state itself (plane).
Taking this a step further, ask students what points (cities) are collinear (on the same highway) and which are coplanar (all the cities). Have a discussion as to what sort of representation would be “space.” Students should see that it would be a globe.
Challenge
1. Three planes intersect in three different ways. Draw all three.
2. One line can divide a plane into two regions. Two lines can divide a plane into four regions. Three lines that intersect at a point can divide a plane into six regions, but you can get more regions if the lines do not all intersect at the same point. If you have six lines, what is the maximum number of regions into which you can divide a plane? BONUS: What if you used lines? What is the maximum number of regions into which you can divide the plane?
1. See the pictures below. Notice that we did not include options where all the planes were parallel (no intersection) or when one plane intersected the other two (two planes parallel).
2. For six lines, the maximum number of regions is 22. You can decide if you would like students to generate a pattern or draw pictures for 1-6 lines (the pattern is: 2, 4, 7, 11, 16, 22, 29, ...). For lines, there will be regions.
## Segments and Distance
Extension
Using the Know What? as a guide, have students measure their own “head” width and length. Then, have them find their height in heads. Students can also find: the length from the wrist to the elbow, the length from the top of the neck to the hip, and the width from shoulder to shoulder, and hip height. Some of these are given in the picture in the text, but have students verify the measurements for their own bodies. Taking it one step further, you could have students create a scale drawing of him/her based on the head measurements.
Connections to History
Describe the advantages of using the metric system to measure length over the English system. Use the examples of the two rulers (one in inches and one in centimeters) to aid in your description.
Research the origins of ancient measurement units such as the cubit. Research the origins of the units of measure we use today such as: foot, inch, mile, meter. Why are standard units important? Student answers should include that the cubit was the first recorded unit of measure and it was integral to the building of the Egyptian pyramids.
Research the facial proportions that da Vinci used to create his Vitruvian man. Write a summary of your findings. Student answers should comment on the “ideal” proportions found in the human face and how these correspond to our perception of beauty.
Challenge
1. Draw four points, and such that .
(HINT: and should NOT be collinear)
2. and . If , find the length of all the segments in the diagram.
1. Here is one possible answer.
2.
## Angles and Measurement
Connections to Astronomy
Give students pictures of the Big Dipper and Orion (below). For homework, ask students to observer the night sky and try to find these two constellations. Have them make note of any other constellations. To see a full list of constellations, visit http://en.wikipedia.org/wiki/List_of_constellations. Ask the students to find 2-3 constellations on line or in a book and to bring the hard copy of a picture of the constellation to class.
The next day, in class, begin a whole discussion about the angles in the constellations and the distance between stars. Ancient astronomers used to measure the degrees between stars using their hands. Here are the approximations:
• Extend your little finger; its width is approximately 1 degree.
• A clenched fist (thumb to little finger) is about 10 degrees.
• From the tip of the little finger to the tip of the thumb, an extended hand with fingers and thumb splayed is about 20 degrees.
Now, for homework, have students see if they can find the angle measures between the stars in each constellation they found the night before. Remind students that these hand measurements are approximations. They only work because the stars are so far away.
Challenge
1. You are measuring with a protractor. When you line up with the mark, lines up with the mark. Then you move the protractor so that lines up with the mark. What mark should line up with?
2. Why do you think the degree measure of a straight line is 180, the degree measure of a right angle is 90, etc? In other words, answer the question, “Why is the straight line exactly and not some other number of degrees?”
1. It should line up with .
2. Answers will vary. Students should comment about the necessity to have a number of degrees in a line that is divisible by 30, 45, 60 and 90 degrees because these degree measures are prevalent in the study of geometrical figures. Basically, setting the measure of a straight line equal to allows us to have more whole number degree measures in common geometrical figures.
## Midpoints and Bisectors
Connections to Construction
This is a picture of a wooden frame for a home. Notice that bisects the sides of the roof and the other support beams ( and ). It is also perpendicular to . Ask students why it is important why bisects these angles. What would happen if did not perfectly bisect these angles? There are also several congruent segments. Ask students to find these. Then, as a classwork (individually or in pairs) assignment, have students answer the following questions.
1. If , find .
2. If , find .
3. What is ?
4. List an example of an acute, obtuse, right and straight angle.
1.
2.
3.
4. Acute angles:
Obtuse angles:
Right angles:
Straight angle:
Challenge
1. Construct a angle using the constructions learned in this section.
2. Using #1, construct a angle.
3. Using #1, construct a square (all sides are equal and all angles are ).
1. Construct a perpendicular bisector to create two angles.
2. Bisect one of the angles in #1 to create a angle.
3. The square is a little harder. Students will need to start with #1 and repeat this once more. Then, they can mark the distance between two created angles and repeat these on the created lines (the green and purple arc marks).
## Angle Pairs
Use the image of a street map of Manhattan or Washington DC (www.mapquest.com). Print a copy of this map for students to work with during the activity. There are several different examples of complementary angles, supplementary angles, linear pairs, and vertical angles.Have the students work in pairs with a highlighter, colored pencils or markers and identify examples of each of the types of angles in the map. Ask the students to make a list of the intersections (by street name) on paper and how each angle fits the description. You could also enlarge the map and have students use a protractor to measure the angles. Students can either share the findings as a whole class, as a discussion, or in small groups. An extension could be to have students repeat this activity for the city they live in.
Extension
Ask students if they think three angles can form a linear pair. Can three angles be supplementary? Complementary? By the definition, only two angles can form a linear pair, be supplementary or complementary. However, explain to students that this does not mean that three angles cannot add up to (such as in a triangle) or .
Challenge
1. Find the value of and . You may need to factor or use the square root.
2. What is a congruent linear pair?
1.
2. A congruent linear pair would be two angles. Congruent linear pairs are created by perpendicular lines.
## Classifying Polygons
Connections to Art
To prepare, you will need an assortment of one or more of the following items: gumdrops, marshmallows, toothpicks, tinkertoys, or kynex. Be sure that the students understand the different types of triangles and have an example of each type before beginning. Then have them create an example of each type triangles using the materials provided. The gumdrops or marshmallows would be the vertices and the toothpicks would be the sides, for example. Let students play for a while. You could also extend this activity to polygons (squares and other regular polygons are pretty easy to create). Finally, you could have students create geometric designs using the materials.
Connections to Architecture
Either in class or as a homework assignment, have students search for “polygon architecture” on the internet. Tell students to click “images” on the search engine and have them print out 2-3 images to share with the class. Encourage students to print out pictures of buildings, tile designs, interiors, windows, or playground equipment.
As a class, brainstorm the different types of polygons used in architecture (write these on the board and how they are used) and why certain polygons are used over others. As a homework assignment, students can go home and write down all the polygons they find their home.
Extension
You can introduce students to the concept of an equilateral polygon and an equiangular polygon. Only in the case of a triangle are equilateral and equiangular the same shape. Also, equilateral polygons can be concave (see hexagon below), but equiangular polygons cannot. After drawing examples of equilateral and equiangular quadrilaterals, pentagons, and hexagons, show students regular polygons (when sides and angles are congruent). Encourage to guess the definitions of these new terms and ask for volunteers to come up to draw different polygons. Here are a few examples.
Challenge
1. Can an equiangular polygon be concave? Why or why not?
2. Find the pattern for the number of diagonals drawn from one vertex. Determine how many diagonals can be drawn from one vertex of an -gon.
3. Find the pattern for the total number of diagonals in a polygon. Determine how many diagonals are in an -gon.
1. An equiangular polygon cannot be concave because all the angles must be the same measure. In a concave polygon, one angle will be larger than .
2. The number of diagonals at one vertex increases by one. For an -gon, the equation would be .
3. For the total number of diagonals, the pattern is: 0, 2, 5, 9, 14, 20, 27, ... Here, you add one more than what was previously added. If there are vertices and diagonals from each vertex, then start with as the equation for the total number of diagonals. However, if you plug in (or any other number), the answer would be double the correct answer (this is because counts all of the diagonals twice). Therefore, you need to divide by two. The equation is .
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
Show Hide Details
Description
Tags:
Subjects: |
How to calculate exceedance probability
Written by soren bagley
• Share
• Tweet
• Share
• Pin
• Email
Exceedance probability is the probability that a certain value is going to be exceeded. For example, if you have data regarding the average cost of bread over a 10-year-span, exceedance probability calculations would allow you to determine the odds that bread will cost more than this average when you actually go to the store. Exceedance probability is used frequently in calculating the behaviour of bodies of water. The reason for this is that it helps predict the probability that flooding might occur and therefore assists in the designs of bridges, dams and sewers.
Skill level:
Moderate
Things you need
• Formula: Rank / (Total Number of Values+1) = Exceedance Probability
Instructions
1. 1
Arrange the values that you have from greatest to least. For example, if you are trying to calculate exceedance probability for the peak flow of a river, you might have annual data regarding the water level. If this were the case, then each year would most likely have a different peak flow level. What you would need to do with this data is arrange these peak flow levels from greatest to least.
2. 2
Add up the total number of values. Using our example from Step 1, this would be equivalent to the number of years for which we have data since each year constitutes a single value.
3. 3
Plug the total number of values into the formula: Rank / (Total Number of Values+1) = Exceedance Probability. For example, if you found in the previous step that you had 9 values, then you would insert this number into the formula making it: Rank / (9+1) = Exceedance Probability.
4. 4
Number each individual value, according to its rank in the arrangement. The greatest value should be ranked number one, the second greatest number two, and so on, until each value has a rank.
5. 5
Decide which value you want to use as the standard. The standard value is the value for which you want to calculate the exceedance probability.
6. 6
Take the rank of this standard value and plug it into the formula. For example, if we wanted to use the fifth largest value as the standard value in our calculation, we would plug the number "5" into our formula of (Rank / Total Number of Values + 1) = Exceedance Probability. Using our example from Step 2, this would give us 5 / (9 + 1) = Exceedance Probability.
7. 7
Solve for exceedance probability. With all the variables in place, perform the addition and division functions required of the formula. Using our example, this would give us 5 / (9 + 1) = 5 / 10 = 0.50. The solution is the exceedance probability of our standard value expressed as a per cent, with 1.00 being equivalent to a 100 per cent probability.
Tips and warnings
• Data representing a longer period of time will result in more reliable calculations.
Don't Miss
References
• All types
• Articles
• Slideshows
• Videos
Sort:
• Most relevant
• Most popular
• Most recent |
# Question Video: Solving Real-World Problems Including Rational Expressions Mathematics • 10th Grade
It takes Victoria 2 hours to pick up 8 kg of strawberries. When she is helped by her son, Ethan, it takes them 1 hour and a quarter. How long would it take Ethan to pick up 8 kg of strawberries alone? Give your answer in hours and minutes.
04:31
### Video Transcript
It takes Victoria two hours to pick up eight kilograms of strawberries. When she is helped by her son Ethan it takes them one hour and a quarter. How long would it take Ethan to pick up eight kilograms of strawberries alone? Give your answer in hours and minutes.
When two people are working together on a task, we can create this equation: one divided by 𝐴 plus one divided by 𝐵 equals one divide by the total where 𝐴 is the amount of time it takes one person to do a job, 𝐵 would be the amount of time it takes another person to do a job, and then the total would be how long it takes them to do it together.
So the plus sign presents them working together. So we can let 𝐴 represent Victoria and 𝐵 represent Ethan. Victoria and her son Ethan are working together to pick up eight kilograms of strawberries. And we know how long it takes Victoria. It takes Victoria two hours to pick up eight kilograms of strawberries. We do not know how long it takes Ethan to pick up the strawberries.
We know when she is helped by her son Ethan that it takes them one hour and a quarter. So that’s the total it would take them together. So the total can be replaced with 1.25, one plus a quarter, 1.25. So we want to solve for 𝐵, the amount of time that would take Ethan to pick up eight kilograms of strawberries by himself. So in order to solve for 𝐵, we need to take the one-half fraction and bring it over to the right-hand side of the equation.
So we need to subtract one-half from both sides of the equation. The one-halves cancel on the left. And then on the right, we need to subtract these fractions. But in order to do so, we need to have common denominators. But first, we don’t wanna have a decimal inside of the fraction. So how can we rewrite 1.25? Well, 0.25 represents a quarter. And a quarter, instead of writing it as 0.25, we could represent it by one-fourth.
So one and one-fourth could be four-fourths plus one-fourth. And four-fourths plus one-fourth would be five-fourths because we add the numerators and keep the common denominator. Now one divided by five-fourths means we need to flip our fraction on the bottom, because when dividing fractions, we multiply by their reciprocals. So we need to take one times four-fifths and one times four-fifths is four-fifths.
So we have one over 𝐵 is equal to four-fifths minus one- half. So as we said before, in order to add or subtract fractions, we have to have a common denominator. So what is a number that five and two can both go into? It would be 10. So how would we go from five to 10? What will we multiply by? We have to multiply by two. So if we multiply the denominator by two, we’ll need to multiply the numerator by two. And four times two is eight.
Now for the one-half, to get from two to 10, we multiply by five. So we need to do the same to the numerator, and one times five is five. So we have eight-tenths minus five- tenths. So we keep our common denominator of 10 and we’d subtract the numerators. So we have that one over 𝐵 is equal to three-tenths. So in order to solve for 𝐵, we can cross multiply.
We take 𝐵 times three and we set it equal to one times 10, which is 10. So to solve for 𝐵, we need to divide both sides of the equation by three. So 𝐵 is equal to ten-thirds. So how can we rewrite ten-thirds if we’re asked to write in hours and minutes? Well, ten-thirds is the same as three and one-third. So three and one-third means that it’s three hours and then one-third minutes.
Well what does the one-third represent? It means it’s one-third of an hour because three is an hour so it’s three hours and one-third an hour. So what is one-third of an hour? Well If we take 60 minutes and split into three, it would be 20 minutes plus 20 minutes plus 20 minutes or simply rewriting it as three and twenty sixtieths, so 20 minutes out of 60 minutes.
Therefore, the time that it would take Ethan to do this alone would be three hours and 20 minutes. |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 6.6: Solving Polynomial and Rational Inequalities
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
Skills to Develop
• Solve polynomial inequalities.
• Solve rational inequalities.
## Solving Polynomial Inequalities
A polynomial inequality18 is a mathematical statement that relates a polynomial expression as either less than or greater than another. We can use sign charts to solve polynomial inequalities with one variable.
Example $$\PageIndex{1}$$
Solve $$x(x+3)^{2}(x-4)<0$$.
Solution
Begin by finding the critical numbers. For a polynomial inequality in standard form, with zero on one side, the critical numbers are the roots. Because $$f (x) = x(x + 3)^{2} (x − 4)$$ is given in its factored form the roots are apparent. Here the roots are: $$0, −3$$, and $$4$$. Because of the strict inequality, plot them using open dots on a number line.
In this case, the critical numbers partition the number line into four regions. Test values in each region to determine if f is positive or negative. Here we choose test values $$−5, −1, 2$$, and $$6$$. Remember that we are only concerned with the sign $$(+$$ or $$−)$$ of the result.
\begin{aligned} f(\color{OliveGreen}{-5}\color{black}{)}&=(\color{OliveGreen}{-5}\color{black}{)}(\color{OliveGreen}{-5}\color{black}{+}3)^{2}(\color{OliveGreen}{-5}\color{black}{-}4) =(-)(-)^{2}(-)&=+\color{Cerulean} { Positive} \\ f(\color{OliveGreen}{-1}\color{black}{)}&=(\color{OliveGreen}{-1}\color{black}{)}(\color{OliveGreen}{-1}\color{black}{+}3)^{2}(\color{OliveGreen}{-1}\color{black}{-}4) =(-)(+)^{2}(-)&=+\color{Cerulean} { Positive } \\ f(\color{OliveGreen}{2}\color{black}{)}&=(\color{OliveGreen}{2}\color{black}{)}(\color{OliveGreen}{2}\color{black}{+}3)^{2}(\color{OliveGreen}{2}\color{black}{-}4) =(+)(+)^{2}(-)&=-\color{Cerulean} { Negative } \\ f(\color{OliveGreen}{6}\color{black}{)}&=(\color{OliveGreen}{6}\color{black}{)}(\color{OliveGreen}{6}\color{black}{+}3)^{2}(\color{OliveGreen}{6}\color{black}{-}4) =(+)(+)^{2}(+)&=+\color{Cerulean} { Positive } \end{aligned}
After testing values we can complete a sign chart.
The question asks us to find the values where $$f (x) < 0$$, or where the function is negative. From the sign chart we can see that the function is negative for $$x$$-values in between $$0$$ and $$4$$.
We can express this solution set in two ways:
\begin{aligned}\{x | 0<&x<4\} &\color{Cerulean} { Set\: notation } \\ (0,&4) &\color{Cerulean} { Interval\: notation }\end{aligned}
In this textbook we will continue to present solution sets using interval notation.
$$(0,4)$$
Graphing polynomials such as the one in the previous example is beyond the scope of this textbook. However, the graph of this function is provided below. Compare the graph to its corresponding sign chart.
Certainly it may not be the case that the polynomial is factored nor that it has zero on one side of the inequality. To model a function using a sign chart, all of the terms should be on one side and zero on the other. The general steps for solving a polynomial inequality are listed in the following example.
Example $$\PageIndex{2}$$:
Solve: $$2 x^{4}>3 x^{3}+9 x^{2}$$.
Solution
Step 1: Obtain zero on one side of the inequality. In this case, subtract to obtain a polynomial on the left side in standard from.
\begin{aligned}2 x^{4}&>3 x^{3}+9 x^{2} \\ 2 x^{4}-3 x^{3}-9 x^{2}&>0\end{aligned}
Step 2: Find the critical numbers. Here we can find the zeros by factoring.
$$2 x^{4}-3 x^{3}-9 x^{2}=0$$
$$x^{2}\left(2 x^{2}-3 x-9\right)=0$$
$$x^{2}(2 x+3)(x-3)=0$$
There are three solutions, hence, three critical numbers $$−\frac{3}{2}, 0$$, and $$3$$. The strict inequality indicates that we should use open dots.
Step 3: Create a sign chart. In this case use $$f (x) = x^{2} (2x + 3) (x − 3)$$ and test values $$−2, −1, 1$$, and $$4$$ to determine the sign of the function in each interval.
\begin{aligned} f(\color{OliveGreen}{-2}\color{black}{)} &=(\color{OliveGreen}{-2}\color{black}{)}^{2}[2(\color{OliveGreen}{-2}\color{black}{)}+3](\color{OliveGreen}{-2}\color{black}{-}3)&=(-)^{2}(-)(-)=+\\ f(\color{OliveGreen}{-1}\color{black}{)} &=(\color{OliveGreen}{-1}\color{black}{)}^{2}[2(\color{OliveGreen}{-1}\color{black}{)}+3](\color{OliveGreen}{-1}\color{black}{-}3)&=(-)^{2}(+)(-)=-\\ f(\color{OliveGreen}{1}\color{black}{)} &=(\color{OliveGreen}{1}\color{black}{)}^{2}[2(\color{OliveGreen}{1}\color{black}{)}+3](\color{OliveGreen}{1}\color{black}{-}3) &=(+)^{2}(+)(-)=-\\ f(\color{OliveGreen}{4}\color{black}{)} &=(\color{OliveGreen}{4}\color{black}{)}^{2}[2(\color{OliveGreen}{4}\color{black}{)}+3](\color{OliveGreen}{4}\color{black}{-}3) &=(+)^{2}(+)(+)=+\end{aligned}
With this information we can complete the sign chart.
Step 4: Use the sign chart to answer the question. Here the solution consists of all values for which $$f (x) > 0$$. Shade in the values that produce positive results and then express this set in interval notation.
$$\left(-\infty,-\frac{3}{2}\right) \cup(3, \infty)$$
Example $$\PageIndex{3}$$
Solve: $$x^{3}+x^{2} \leq 4(x+1)$$.
Solution
Begin by rewriting the inequality in standard form, with zero on one side.
\begin{aligned}x^{3}+x^{2} &\leq 4(x+1) \\ x^{3}+x^{2} &\leq 4 x+4 \\ x^{3}+x^{2}-4 x-4 &\leq 0\end{aligned}
Next find the critical numbers of $$f(x)=x^{3}+x^{2}-4 x-4$$:
\begin{aligned} x^{3}+x^{2}-4 x-4 &=0 \quad\color{Cerulean} { Factor\: by\: grouping.} \\ x^{2}(x+1)-4(x+1) &=0 \\(x+1)\left(x^{2}-4\right) &=0 \\(x+1)(x+2)(x-2) &=0 \end{aligned}
The critical numbers are $$−2, −1$$, and $$2$$. Because of the inclusive inequality $$(≤)$$ we will plot them using closed dots.
Use test values, $$-3$$, $$-\frac{3}{2}$$, $$0$$, and $$3$$ to create a sign chart.
\begin{aligned} f(\color{OliveGreen}{-3}\color{black}{)}&=(\color{OliveGreen}{-3}\color{black}{+}1)(\color{OliveGreen}{-3}\color{black}{+}2)(\color{OliveGreen}{-3}\color{black}{-}2) &=(-)(-)(-)=- \\ f(\color{OliveGreen}{-\frac{3}{2}}\color{black}{)}&=(\color{OliveGreen}{-\frac{3}{2}}\color{black}{+}1)(\color{OliveGreen}{-\frac{3}{2}}\color{black}{+}2)(\color{OliveGreen}{-\frac{3}{2}}\color{black}{-}2) &=(-)(+)(-)=+\\ f(\color{OliveGreen}{0}\color{black}{)}&=(\color{OliveGreen}{0}\color{black}{+}1)(\color{OliveGreen}{0}\color{black}{+}2)(\color{OliveGreen}{0}\color{black}{-}2)&=(+)(+)(-)=- \\ f(\color{OliveGreen}{3}\color{black}{)}&=(\color{OliveGreen}{3}\color{black}{+}1)(\color{OliveGreen}{3}\color{black}{+}2)(\color{OliveGreen}{3}\color{black}{-}2)&=(+)(+)(+)=+\end{aligned}
And we have
Use the sign chart to shade in the values that have negative results $$(f (x) ≤ 0)$$.
$$(-\infty,-2] \cup[-1,2]$$
Exercise $$\PageIndex{1}$$
Solve $$-3 x^{4}+12 x^{3}-9 x^{2}>0$$.
$$(1,3)$$
## Solving Rational Inequalities
A rational inequality19 is a mathematical statement that relates a rational expression as either less than or greater than another. Because rational functions have restrictions to the domain we must take care when solving rational inequalities. In addition to the zeros, we will include the restrictions to the domain of the function in the set of critical numbers.
Example $$\PageIndex{4}$$
Solve: $$\frac{(x-4)(x+2)}{(x-1)} \geq 0$$
Solution
The zeros of a rational function occur when the numerator is zero and the values that produce zero in the denominator are the restrictions. In this case,
$$\begin{array}{c | c}{\text { Roots (Numerator) }} & {\text{Restriction(Denominator)}} \\ {x-4=0 \text { or } x+2=0} & {x-1=0} \\ {\:\:\quad\quad\quad\: x=4 \quad\quad x=-2}& {x=1}\end{array}$$
Therefore the critical numbers are $$−2, 1$$, and $$4$$. Because of the inclusive inequality $$(≥)$$ use a closed dot for the roots $${−2, 4}$$ and always use an open dot for restrictions $${1}$$. Restrictions are never included in the solution set.
Use test values $$x = −4, 0, 2, 6$$.
\begin{aligned} f(\color{OliveGreen}{-4}\color{black}{)} &=\frac{(\color{OliveGreen}{-4}\color{black}{-}4)(\color{OliveGreen}{-4}\color{black}{+}2)}{(\color{OliveGreen}{-4}\color{black}{-}1)}&=\frac{(-)(-)}{(-)}=-\\ f(\color{OliveGreen}{0}\color{black}{)} &=\frac{(\color{OliveGreen}{0}\color{black}{-}4)(\color{OliveGreen}{0}\color{black}{+}2)}{(\color{OliveGreen}{0}\color{black}{-}1)}&=\frac{(-)(+)}{(-)}=+\\ f(\color{OliveGreen}{2}\color{black}{)} &=\frac{(\color{OliveGreen}{2}\color{black}{-}4)(\color{OliveGreen}{2}\color{black}{+}2)}{(\color{OliveGreen}{2}\color{black}{-}1)}&=\frac{(-)(+)}{(+)}=-\\ f(\color{OliveGreen}{6}\color{black}{)} &=\frac{(\color{OliveGreen}{6}\color{black}{-}4)(\color{OliveGreen}{6}\color{black}{+}2)}{(\color{OliveGreen}{6}\color{black}{-}1)}&=\frac{(+)(+)}{(+)}=+\end{aligned}
And then complete the sign chart.
The question asks us to find the values for which $$f (x) ≥ 0$$, in other words, positive or zero. Shade in the appropriate regions and present the solution set in interval notation.
$$[-2,1) \cup[4, \infty)$$
Graphing such rational functions like the one in the previous example is beyond the scope of this textbook. However, the graph of this function is provided below. Compare the graph to its corresponding sign chart.
Notice that the restriction $$x = 1$$ corresponds to a vertical asymptote which bounds regions where the function changes from positive to negative. While not included in the solution set, the restriction is a critical number. Before creating a sign chart we must ensure the inequality has a zero on one side. The general steps for solving a rational inequality are outlined in the following example.
Example $$\PageIndex{5}$$:
Solve $$\frac{7}{x+3}<2$$.
Solution
Step 1: Begin by obtaining zero on the right side.
\begin{aligned}\frac{7}{x+3}&<2 \\ \frac{7}{x+3}-2&<0\end{aligned}
Step 2: Determine the critical numbers. The critical numbers are the zeros and restrictions. Begin by simplifying to a single algebraic fraction.
\begin{aligned}\frac{7}{x+3}-\frac{2}{1}&<0 \\ \frac{7-2(x+3)}{x+3}&<0 \\ \frac{7-2 x-6}{x+3}&<0 \\ \frac{-2 x+1}{x+3}&<0\end{aligned}
Next find the critical numbers. Set the numerator and denominator equal to zero and solve.
$$\begin{array}{c|c} {\text{Root}}&{\text{Restriction}}\\ {-2x+1=0}\\{-2x=-1}&{x+3=0}\\{x=\frac{1}{2}}&\quad\quad\:\:{x=-3} \end{array}$$
In this case, the strict inequality indicates that we should use an open dot for the root.
Step 3: Create a sign chart. Choose test values $$−4, 0$$, and $$1$$.
\begin{aligned} f(\color{OliveGreen}{-4}\color{black}{)} &=\frac{-2(\color{OliveGreen}{-4}\color{black}{)}+1}{\color{OliveGreen}{-4}\color{black}{+}3}&=\frac{+}{-}=-\\ f(\color{OliveGreen}{0}\color{black}{)} &=\frac{-2(\color{OliveGreen}{0}\color{black}{)}+1}{\color{OliveGreen}{0}\color{black}{+}3}&=\frac{+}{+}=+\\ f(\color{OliveGreen}{1}\color{black}{)} &=\frac{-2(\color{OliveGreen}{1}\color{black}{)}+1}{\color{OliveGreen}{1}\color{black}{+}3}&=\frac{-}{+}=-\end{aligned}
And we have
Step 4: Use the sign chart to answer the question. In this example we are looking for the values for which the function is negative, $$f (x) < 0$$. Shade the appropriate values and then present your answer using interval notation.
$$(-\infty,-3) \cup\left(\frac{1}{2}, \infty\right)$$
Example $$\PageIndex{6}$$:
Solve: $$\frac{1}{x^{2}-4} \leq \frac{1}{2-x}$$.
Solution
Begin by obtaining zero on the right side.
\begin{aligned}\frac{1}{x^{2}-4} &\leq \frac{1}{2-x} \\ \frac{1}{x^{2}-4}-\frac{1}{2-x} &\leq 0\end{aligned}
Next simplify the left side to a single algebraic fraction.
$$\begin{array}{r}{\frac{1}{x^{2}-4}-\frac{1}{2-x} \leq 0} \\ {\frac{1}{(x+2)(x-2)}-\frac{1}{-(x-2)} \leq 0} \\ {\frac{1}{(x+2)(x-2)}+\frac{1\color{Cerulean}{(x+2)}}{\color{black}{(x-2)}\color{Cerulean}{(x+2)}}\color{black}{ \leq} 0} \\ {\frac{1+x+2}{(x+2)(x-2)} \leq 0} \\ {\frac{x+3}{(x+2)(x-2)} \leq 0}\end{array}$$
The critical numbers are $$−3, −2$$, and $$2$$. Note that $$±2$$ are restrictions and thus we will use open dots when plotting them on a number line. Because of the inclusive inequality we will use a closed dot at the root $$−3$$.
Choose test values $$-4, -2\frac{1}{2} = -\frac{5}{2}, 0$$, and $$3$$.
\begin{aligned}f(\color{OliveGreen}{-4}\color{black}{)}&= \frac{\color{OliveGreen}{-4}\color{black}{+}3}{(\color{OliveGreen}{-4}\color{black}{+}2)(\color{OliveGreen}{-4}\color{black}{-}2)}&=\frac{(-)}{(-)(-)}=- \\ f(\color{OliveGreen}{-\frac{5}{2}}\color{black}{)} &= \frac{\color{OliveGreen}{-\frac{5}{2}}\color{black}{+}3}{(\color{OliveGreen}{-\frac{5}{2}}\color{black}{+}2)(\color{OliveGreen}{-\frac{5}{2}}\color{black}{-}2)}&=\frac{(+)}{(-)(-)}=+\\ f(\color{OliveGreen}{0}\color{black}{)}&=\frac{\color{OliveGreen}{0}\color{black}{+}3}{(\color{OliveGreen}{0}\color{black}{+}2)(\color{OliveGreen}{0}\color{black}{-}2)}&=\frac{(+)}{(+)(-)}=- \\ f(\color{OliveGreen}{3}\color{black}{)}&=\frac{\color{OliveGreen}{3}\color{black}{+}3}{(\color{OliveGreen}{3}\color{black}{+}2)(\color{OliveGreen}{3}\color{black}{-}2)} &=\frac{(+)}{(+)(+)}=+\end{aligned}
Construct a sign chart.
Answer the question; in this case, find $$x$$ where $$f(x) \leq 0$$.
$$(-\infty,-3] \cup(-2,2)$$
Exercise $$\PageIndex{2}$$
Solve: $$\frac{2 x^{2}}{2 x^{2}+7 x-4} \geq \frac{x}{x+4}$$.
$$(-4,0] \cup\left(\frac{1}{2}, \infty\right)$$
## Key Takeaways
• When a polynomial inequality is in standard form, with zero on one side, the roots of the polynomial are the critical numbers. Create a sign chart that models the function and then use it to answer the question.
• When a rational inequality is written as a single algebraic fraction, with zero on one side, the roots as well as the restrictions are the critical numbers. The values that produce zero in the numerator are the roots, and the values that produce zero in the denominator are the restrictions. Always use open dots for restrictions, regardless of the given inequality, because restrictions are not part of the domain. Create a sign chart that models the function and then use it to answer the question.
Exercise $$\PageIndex{3}$$
Solve. Present answers using interval notation.
1. $$x(x+1)(x-3)>0$$
2. $$x(x-1)(x+4)<0$$
3. $$(x+2)(x-5)^{2}<0$$
4. $$(x-4)(x+1)^{2} \geq 0$$
5. $$(2 x-1)(x+3)(x+2) \leq 0$$
6. $$(3 x+2)(x-4)(x-5) \geq 0$$
7. $$x(x+2)(x-5)^{2}<0$$
8. $$x(2 x-5)(x-1)^{2}>0$$
9. $$x(4 x+3)(x-1)^{2} \geq 0$$
10. $$(x-1)(x+1)(x-4)^{2}<0$$
11. $$(x+5)(x-10)(x-5)^{2} \geq 0$$
12. $$(3 x-1)(x-2)(x+2)^{2} \leq 0$$
13. $$-4 x(4 x+9)(x-8)^{2}>0$$
14. $$-x(x-10)(x+7)^{2}>0$$
1. $$(-1,0) \cup(3, \infty)$$
3. $$(-\infty,-2)$$
5. $$(-\infty,-3] \cup\left[-2, \frac{1}{2}\right]$$
7. $$(-2,0)$$
9. $$\left(-\infty,-\frac{3}{4}\right] \cup[0, \infty)$$
11. $$(-\infty,-5] \cup[5,5] \cup[10, \infty)$$
13. $$\left(-\frac{9}{4}, 0\right)$$
Exercise $$\PageIndex{4}$$
Solve.
1. $$x^{3}+2 x^{2}-24 x \geq 0$$
2. $$x^{3}-3 x^{2}-18 x \leq 0$$
3. $$4 x^{3}-22 x^{2}-12 x<0$$
4. $$9 x^{3}+30 x^{2}-24 x>0$$
5. $$12 x^{4}+44 x^{3}>80 x^{2}$$
6. $$6 x^{4}+12 x^{3}<48 x^{2}$$
7. $$x\left(x^{2}+25\right)<10 x^{2}$$
8. $$x^{3}>12 x(x-3)$$
9. $$x^{4}-5 x^{2}+4 \leq 0$$
10. $$x^{4}-13 x^{2}+36 \geq 0$$
11. $$x^{4}>3 x^{2}+4$$
12. $$4 x^{4}<3-11 x^{2}$$
13. $$9 x^{3}-3 x^{2}-81 x+27 \leq 0$$
14. $$2 x^{3}+x^{2}-50 x-25 \geq 0$$
15. $$x^{3}-3 x^{2}+9 x-27>0$$
16. $$3 x^{3}+5 x^{2}+12 x+20<0$$
1. $$[-6,0] \cup[4, \infty)$$
3. $$\left(-\infty,-\frac{1}{2}\right) \cup(0,6)$$
5. $$(-\infty,-5) \cup\left(\frac{4}{3}, \infty\right)$$
7. $$(-\infty, 0)$$
9. $$[-2,-1] \cup[1,2]$$
11. $$(-\infty,-2) \cup(2, \infty)$$
13. $$(-\infty,-3] \cup\left[\frac{1}{3}, 3\right]$$
15. $$(3, \infty)$$
Exercise $$\PageIndex{5}$$
Solve.
1. $$\frac{x}{x-3}>0$$
2. $$\frac{x-5}{x}>0$$
3. $$\frac{(x-3)(x+1)}{x}<0$$
4. $$\frac{(x+5)(x+4)}{(x-2)}<0$$
5. $$\frac{(2 x+1)(x+5)}{(x-3)(x-5)} \leq 0$$
6. $$\frac{(3 x-1)(x+6)}{(x-1)(x+9)} \geq 0$$
7. $$\frac{(x-8)(x+8)}{-2 x(x-2)} \geq 0$$
8. $$\frac{(2 x+7)(x+4)}{x(x+5)} \leq 0$$
9. $$\frac{x^{2}}{(2 x+3)(2 x-3)} \leq 0$$
10. $$\frac{(x-4)^{2}}{-x(x+1)}>0$$
11. $$\frac{-5 x(x-2)^{-}}{(x+5)(x-6)} \geq 0$$
12. $$\frac{(3 x-4)(x+5)}{x(x-4)^{2}} \geq 0$$
13. $$\frac{1}{(x-5)^{4}}>0$$
14. $$\frac{1}{(x-5)^{4}}<0$$
1. $$(-\infty,-0) \cup(3, \infty)$$
3. $$(-\infty,-1) \cup(0,3)$$
5. $$\left[-5,-\frac{1}{2}\right] \cup(3,5)$$
7. $$[-8,0) \cup(2,8]$$
9. $$\left(-\frac{3}{2}, \frac{3}{2}\right)$$
11. $$(-\infty,-5) \cup[0,6)$$
13. $$(-\infty, 5) \cup(5, \infty)$$
Exercise $$\PageIndex{6}$$
Solve.
1. $$\frac{x^{2}-11 x-12}{x+4}<0$$
2. $$\frac{x^{2}-10 x+24}{x-2}>0$$
3. $$\frac{x^{2}+x-30}{2 x+1} \geq 0$$
4. $$\frac{2 x^{2}+x-3}{x-3} \leq 0$$
5. $$\frac{3 x^{2}-4 x+1}{x^{2}-9} \leq 0$$
6. $$\frac{x^{2}-16}{2 x^{2}-3 x-2} \geq 0$$
7. $$\frac{x^{2}-12 x+20}{x^{2}-10 x+25}>0$$
8. $$\frac{x^{2}+15 x+36}{x^{2}-8 x+16}<0$$
9. $$\frac{8 x^{2}-2 x-1}{2 x^{2}-3 x-14} \leq 0$$
10. $$\frac{4 x^{2}-4 x-15}{x^{2}+4 x-5} \geq 0$$
11. $$\frac{1}{x+5}+\frac{5}{x-1}>0$$
12. $$\frac{5}{x+4}-\frac{1}{x-4}<0$$
13. $$\frac{1}{x+7}>1$$
14. $$\frac{1}{x-1}<-5$$
15. $$x \geq \frac{30}{x-1}$$
16. $$x \leq \frac{1-2 x}{x-2}$$
17. $$\frac{1}{x-1} \leq \frac{2}{x}$$
18. $$\frac{3}{x+1}>-\frac{1}{x}$$
19. $$\frac{4}{x-3} \leq \frac{1}{x+3}$$
20. $$\frac{2 x-9}{x}+\frac{49}{x-8}<0$$
21. $$\frac{x}{2(x+2)}-\frac{1}{x+2} \leq \frac{12}{x(x+2)}$$
22. $$\frac{1}{2 x+1}-\frac{9}{2 x-1}>2$$
23. $$\frac{3 x}{x^{2}-4}-\frac{2}{x-2}<0$$
24. $$\frac{x}{2 x+1}+\frac{4}{2 x^{2}-7 x-4}<0$$
25. $$\frac{x+1}{2 x^{2}+5 x-3} \geq \frac{x}{4 x^{2}-1}$$
26. $$\frac{x^{2}-14}{2 x^{2}-7 x-4} \leq \frac{5}{1+2 x}$$
1. $$(-\infty,-4) \cup(-1,12)$$
3. $$\left[-6,-\frac{1}{2}\right) \cup[5, \infty)$$
5. $$\left(-3, \frac{1}{3}\right] \cup[1,3)$$
7. $$(-\infty, 2) \cup(10, \infty)$$
9. $$\left(-2,-\frac{1}{4}\right] \cup\left[\frac{1}{2}, \frac{7}{2}\right)$$
11. $$(-5,-4) \cup(1, \infty)$$
13. $$(-7,-6)$$
15. $$[-5,1) \cup[6, \infty)$$
17. $$(0,1) \cup[2, \infty)$$
19. $$(-\infty, 5] \cup(-3,3)$$
21. $$[-4,-2) \cup(0,6]$$
23. $$(-\infty,-2) \cup(2,4)$$
25. $$\left(-3,-\frac{1}{2}\right) \cup\left(\frac{1}{2}, \infty\right)$$
Exercise $$\PageIndex{7}$$
1. Does the sign chart for any given polynomial or rational function always alternate? Explain and illustrate your answer with some examples.
2. Write down your own steps for solving a rational inequality and illustrate them with an example. Do your steps also work for a polynomial inequality? Explain. |
# Multiplication of two Binomials
Multiplication of two binomials can be solved in both horizontal and column method.
Horizontal method:
Follow the following steps to multiply the binomials in the horizontal method:
1. First write the two binomials in a row separated by using multiplication sign.
2. Multiply each term of one binomial with each term of the other.
3. In the product obtained, combine the like terms and then add the like terms.
Therefore, we will learn how to multiply two binomials a + 5 by a + 7 using horizontal method.
a + 5 by a + 7
= (a + 5) ∙ (a + 7), [separate the two binomials using multiplication sign]
= a ∙ (a + 7) + 5 ∙ (a + 7), [multiplying each term of the first binomial with each term of the second binomial]
= a ∙ a + a ∙ 7 + 5 ∙ a + 5 ∙ 7
= a2 + 7a + 5a + 35, [combine the like terms]
= a2 + 12a + 35
Column method:
Follow the following steps to multiply the binomials in the column method:
1. Write the two binomials in two rows one below the other.
2. Multiply one term of the binomial in lower line (i.e. second row) with each term of the binomial in the upper line (i.e. first row) and write the product in the third row.
3. Multiply second term of the binomial in lower line (i.e. second row) with each term of the binomial in upper line (i.e. first row) and write the product in the fourth row in such a way that the like terms are one below the other.
4. Add the like terms column wise.
Therefore, we will learn how to multiply two binomials 5a - 6b and 7a + 8b using column method.
Solved examples on multiplication of two binomials:
1. Multiply 3x2 – 6y2 by 2x2 + 4y2
Solution:
3x2 – 6y2 by 2x2 + 4y2
= (3x2 – 6y2) ∙ (2x2 + 4y2), [separate the two binomials using multiplication sign]
= 3x2 ∙ (2x2 + 4y2) – 6y2 ∙ (2x2 + 4y2), [multiplying each term of the first binomial with each term of the second binomial]
= 6x4 + 12x2y2 – 12x2y2 – 24y4
= 6x4 + 12x2y2 – 12x2y2 – 244, [combine the like terms]
= 6x4 - 244
2. Multiply (m + 6) by (3m – 2)
Solution:
The above examples will help us to solve the multiplication of two binomials in horizontal method and in column method.
Types of Algebraic Expressions
Degree of a Polynomial
Subtraction of Polynomials
Power of Literal Quantities
Multiplication of Two Monomials
Multiplication of Polynomial by Monomial
Multiplication of two Binomials
Division of Monomials
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
## Recent Articles
1. ### Tangrams Math | Traditional Chinese Geometrical Puzzle | Triangles
Apr 17, 24 01:53 PM
Tangram is a traditional Chinese geometrical puzzle with 7 pieces (1 parallelogram, 1 square and 5 triangles) that can be arranged to match any particular design. In the given figure, it consists of o…
2. ### Time Duration |How to Calculate the Time Duration (in Hours & Minutes)
Apr 17, 24 01:32 PM
We will learn how to calculate the time duration in minutes and in hours. Time Duration (in minutes) Ron and Clara play badminton every evening. Yesterday, their game started at 5 : 15 p.m.
3. ### Worksheet on Third Grade Geometrical Shapes | Questions on Geometry
Apr 16, 24 02:00 AM
Practice the math worksheet on third grade geometrical shapes. The questions will help the students to get prepared for the third grade geometry test. 1. Name the types of surfaces that you know. 2. W…
4. ### 4th Grade Mental Math on Factors and Multiples |Worksheet with Answers
Apr 16, 24 01:15 AM
In 4th grade mental math on factors and multiples students can practice different questions on prime numbers, properties of prime numbers, factors, properties of factors, even numbers, odd numbers, pr…
5. ### Worksheet on Factors and Multiples | Find the Missing Factors | Answer
Apr 15, 24 11:30 PM
Practice the questions given in the worksheet on factors and multiples. 1. Find out the even numbers. 27, 36, 48, 125, 360, 453, 518, 423, 54, 58, 917, 186, 423, 928, 358 2. Find out the odd numbers.
Terms of an Algebraic Expression - Worksheet
Worksheet on Types of Algebraic Expressions
Worksheet on Degree of a Polynomial |
# use part 1 of the fundamental theorem of calculus to find the derivative of the function.
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
### Single Variable Calculus: Concepts...
4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
#### Solutions
Chapter 5.4, Problem 17E
To determine
## To find: use part 1 of the fundamental theorem of calculus to find the derivative of the function.
Expert Solution
The derivative of the function is g(x)=8x2-24x2+1+18x2-29x2+1
### Explanation of Solution
Given information: The function is g(x)=2x3xu2-1u2+1du
Let’s remind of the fundamental theorem of calculus part 1:
The fundamental theorem of calculus part 1: If f is continuous on [a,b] then the function of g defined by
g(x)=axf(t) dt where axb is continuous on [a,b] and differentiable on (a,b) and g(x)=f(x) .
We have, g(x)=2x3xu2-1u2+1du
First, we will use the properties of definite integral to match the form in fundamental theorem of calculus part 1
2x3xu2-1u2+1du
=2x0u2-1u2+1du+03xu2-1u2+1duwhere 2x3xf(u) du=2x0f(u)du+03xf(u) du
Now again, g(x)
=2x0u2-1u2+1du
we will use the properties of definite integral to match the form in fundamental theorem of calculus part 1
2x0u2-1u2+1du
=02xu2-1u2+1du where f(x)=-f(x)
g(x)=02xu2-1u2+1du
so, g(x)=02xu2-1u2+1du
So Since the upper limit of integration is not x , we apply the chain rule
Let, v=2x, then v=2
h(v)=0vu2-1u2+1du
Differentiation with respect to v
ddvh(v)=ddv[0vu2-1u2+1du]
h(v)=v2-1v2+1
Thus, h(v)=g(x)
g(x)=[h(2x)]
g(x)=h(2x)·(2x)
g(x)=(2x)2-1(2x)2+1·2
g(x)=8x2-24x2+1
Again, g(x)=03xu2-1u2+1du
So Since the upper limit of integration is not x , we apply the chain rule
Let, w=3x, then v=3
h(w)=0wu2-1u2+1du
Differentiation with respect to w
ddwh(w)=ddw[0vu2-1u2+1du]
h(w)=w2-1w2+1
Thus, h(w)=g(x)
g(x)=[h(3x)]
g(x)=h(3x)·(3x)
g(x)=(3x)2-1(3x)2+1·3
g(x)=18x2-29x2+1
Hence , The derivative of the function is g(x)=8x2-24x2+1+18x2-29x2+1
### Have a homework question?
Subscribe to bartleby learn! Ask subject matter experts 30 homework questions each month. Plus, you’ll have access to millions of step-by-step textbook answers! |
# Algebra: Age Word Problems
Age problems are algebra word problems that deal with the ages of people currently, in the past or in the future.
How to solve Age Word Problems?
If the problem involves a single person, then it is similar to an Integer Problem. Read the problem carefully to determine the relationship between the numbers. This is shown in the examples involving a single person.
If the age problem involves the ages of two or more people then using a table would be a good idea. A table will help you to organize the information and to write the equations. This is shown in the examples involving more than one person.
Related Topics: More Algebra Word Problems
How to solve Age Problems Involving A Single Person?
Example:
Five years ago, John’s age was half of the age he will be in 8 years. How old is he now?
Solution:
Step 1: Let x be John’s age now. Look at the question and put the relevant expressions above it.
Step 2: Write out the equation.
Isolate variable x
Answer: John is now 18 years old.
How to use Algebra to solve age problems? Examples:
1. Ten years from now, Orlando will be three times older than he is today. What is his current age?
2. In 20 years, Kayleen will be four times older than she is today. What is her current age?
How to solve Age Problems Involving More Than One Person?
Example:
John is twice as old as his friend Peter. Peter is 5 years older than Alice. In 5 years, John will be three times as old as Alice. How old is Peter now?
Solution:
Step 1: Set up a table.
age now age in 5 yrs John Peter Alice
Step 2: Fill in the table with information given in the question.
John is twice as old as his friend Peter. Peter is 5 years older than Alice. In 5 years, John will be three times as old as Alice. How old is Peter now?
Let x be Peter’s age now. Add 5 to get the ages in 5 yrs.
age now age in 5 yrs John 2x 2x + 5 Peter x x + 5 Alice x – 5 x – 5 + 5
Write the new relationship in an equation using the ages in 5 yrs.
In 5 years, John will be three times as old as Alice.
2x + 5 = 3(x – 5 + 5)
2x + 5 = 3x
Isolate variable x
x = 5
Answer: Peter is now 5 years old.
Example:
John’s father is 5 times older than John and John is twice as old as his sister Alice. In two years time, the sum of their ages will be 58. How old is John now?
Solution:
Step 1: Set up a table.
age now age in 2 yrs John’s father John Alice
Step 2: Fill in the table with information given in the question.
John’s father is 5 times older than John and John is twice as old as his sister Alice. In two years time, the sum of their ages will be 58. How old is John now?
Let x be John’s age now. Add 2 to get the ages in 2 yrs.
age now age in 2 yrs John’s father 5x 5x + 2 John x x + 2 Alice
Write the new relationship in an equation using the ages in 2 yrs.
In two years time, the sum of their ages will be 58.
Answer: John is now 8 years old.
How to solve word problems with multiple ages?
Example:
Ben is eight years older than Sarah. 10 years ago, Ben is twice as old as Sarah. Currently, how old is Ban and Sarah?
Algebra word problems with multiple ages
Example:
Mary is three times as old as her son. In 12 years, Mary's age will be one year less than twice her son's age. How old is each now?
Algebra word problem with past and present ages
Example:
Arun is 4 times as old as Anusha is today. Sixty years ago, Arun was 6 times as old as Anusha. How old are they today?
Algebra age word problem with past, present, and future ages
How to organize the data using a table and solve using a system of linear equations?
Examples:
1. Sally is 3 times as old as John. 8 years from now, Sally will be twice as old as John. How old is John?
2. Kim is 6 years more than twice Timothy's age. 2 years ago, Kim was three times as old as Timothy. How old was Kim 2 years ago?
3. Leah is 2 less than 3 times Rachel's age. 3 years from now, Leah will be 7 more than twice Rachel's age. How old will Rachel be in 3 years from now?
4. Becca is twice as old as Susan and Greg is 9 years older than Susan. 3 years ago, Becca was 9 less than 3 times Susan's age. How old is Greg now?
5. Lauren is 3 less than twice Andrew's age. 4 years from now, Sam will be 2 more than twice Andrew's age. 5 years ago, Sam was three times Andrew's age. How old was Lauren 5 years ago?
6. Gabby is 1 year more than twice Larry's age. 3 years from now, Megan will be 27 less than twice Gabby's age. 4 years ago, Megan was 1 year less than 3 times Larry's age. How old will Megan be 3 years from now?
Rotate to landscape screen format on a mobile phone or small tablet to use the Mathway widget, a free math problem solver that answers your questions with step-by-step explanations.
You can use the free Mathway calculator and problem solver below to practice Algebra or other math topics. Try the given examples, or type in your own problem and check your answer with the step-by-step explanations. |
فشلت مصادقة جوجل. الرجاء معاودة المحاولة في وقت لاحق!
# Pick’s Theorem
## Main Activity
Polygons here are drawn on a square grid. They have dots on their perimeter (p) and often interior (i) ones as well. Let's find a relationship between "p", "I", and the area of the polygons.
Try it out on an example;
To find the area you can use the irregular polygon tool to divide the enclosed region into triangles, squares, and rectangles. Here is one possible way of doing so;
Finding a relationship among these three variables can be tricky.
Here are some other questions you can work on first:
Since there are three variables here, observing how two of the variables change while keeping one constant can be an effective strategy.
• How many different figures can be created with 4 perimeter points and no interior points? What do you notice about them?
• What do you notice about those figures whose areas are the same?
• What ways are there of increasing the area by 1 unit?
You can use the canvas here to draw as many figures as you want and keep a record of their number of perimeter points (p), interior points (i), and areas (A).
You can choose particular values for "p", and "i" and explore different shapes.
Try to create shapes to answer these questions;
- As the number of points on the shape's perimeter increases by one, what happens to area?
- As the number of interior points increases by one, what happens to area?
## Solution
As mentioned before when there are three variables, observing how two of the variables change while keeping one constant can be an effective strategy.
Keep the number of perimeter points constant and increase the number of interior points one each time.
You may draw additional shapes to reach a conclusion about how the area increases by the number of interior points.
As the number of points on the shape's perimeter increases by one, the area increases by half.
This time, keep the number of interior points constant and increase the number of perimeter points one each time.
You may draw additional shapes to reach a conclusion about how the area increases by the number of perimeter points.
As the number of interior points increases by one, the area also increases by one.
Now we can build the relation on our observations: $Area = p/2 + i - 1$ |
# How do you solve (x+2)^2 = 20?
Jan 16, 2017
$x = - 2 \pm \sqrt{20}$
#### Explanation:
There are many approaches to this type of problem. One option is to expand the square and use the quadratic formula to solve the equation. However in this specific case, we have a simpler way of solving this:
${\left(x + 2\right)}^{2} = 20$. Since both ${\left(x + 2\right)}^{2}$ and $20$ are positive, we can take the square roots of both:
$\sqrt{{\left(x + 2\right)}^{2}} = \sqrt{20}$
$\left\mid x + 2 \right\mid = \sqrt{20}$.
We use an absolute value, since we do not know if $x$ is greater or less than $- 2$, so we don't know if $x + 2$ is positive or negative. We can, however, get rid of the absolute value, since if $\left\mid a \right\mid = b$, where $b \ge 0$, then $a = \pm b$.
$x + 2 = \pm \sqrt{20} \implies x = - 2 \pm \sqrt{20}$. |
# Discrete logarithm
Discrete logarithm
In mathematics, specifically in abstract algebra and its applications, discrete logarithms are group-theoretic analogues of ordinary logarithms. In particular, an ordinary logarithm loga(b) is a solution of the equation ax = b over the real or complex numbers. Similarly, if g and h are elements of a finite cyclic group G then a solution x of the equation gx = h is called a discrete logarithm to the base g of h in the group G.
## Example
Discrete logarithms are perhaps simplest to understand in the group (Zp)×. This is the set {1, …, p − 1} of congruence classes under multiplication modulo the prime p.
If we want to find the kth power of one of the numbers in this group, we can do so by finding its kth power as an integer and then finding the remainder after division by p. This process is called discrete exponentiation. For example, consider (Z17)×. To compute 34 in this group, we first compute 34 = 81, and then we divide 81 by 17, obtaining a remainder of 13. Thus 34 = 13 in the group (Z17)×.
Discrete logarithm is just the inverse operation. For example, take the equation 3k ≡ 13 (mod 17) for k. As shown above k=4 is a solution, but it is not the only solution. Since 316 ≡ 1 (mod 17) it also follows that if n is an integer then 34+16 n ≡ 13 × 1n ≡ 13 (mod 17). Hence the equation has infinitely many solutions of the form 4 + 16n. Moreover, since 16 is the smallest positive integer m satisfying 3m ≡ 1 (mod 17), i.e. 16 is the order of 3 in (Z17)×, these are the only solutions. Equivalently, the solution can be expressed as k ≡ 4 (mod 16).
## Definition
In general, let G be a finite cyclic group with n elements. We assume that the group is written multiplicatively. Let b be a generator of G; then every element g of G can be written in the form g = bk for some integer k. Furthermore, any two such integers k1 and k2 representing g will be congruent modulo n. We can thus define a function
$\log_b:\ G\ \rightarrow\ \mathbb{Z}_n$
(where Zn denotes the ring of integers modulo n) by assigning to each g the congruence class of k modulo n. This function is a group isomorphism, called the discrete logarithm to base b.
The familiar base change formula for ordinary logarithms remains valid: If c is another generator of G, then we have
$\log_c (g) = \log_c (b) \cdot \log_b (g).$
## Algorithms
Can the discrete logarithm be computed in polynomial time on a classical computer?
No efficient classical algorithm for computing general discrete logarithms logb g is known. The naive algorithm is to raise b to higher and higher powers k until the desired g is found; this is sometimes called trial multiplication. This algorithm requires running time linear in the size of the group G and thus exponential in the number of digits in the size of the group. There exists an efficient quantum algorithm due to Peter Shor.[1]
More sophisticated algorithms exist, usually inspired by similar algorithms for integer factorization. These algorithms run faster than the naive algorithm, but none of them runs in polynomial time (in the number of digits in the size of the group).
## Comparison with integer factorization
While the problem of computing discrete logarithms and the problem of integer factorization are distinct problems they share some properties:
• both problems are difficult (no efficient algorithms are known for non-quantum computers),
• for both problems efficient algorithms on quantum computers are known,
• algorithms from one problem are often adapted to the other, and
• the difficulty of both problems has been utilized to construct various cryptographic systems.
## Cryptography
There exist groups for which computing discrete logarithms is apparently difficult. In some cases (e.g. large prime order subgroups of groups (Zp)×) there is not only no efficient algorithm known for the worst case, but the average-case complexity can be shown to be about as hard as the worst case using random self-reducibility.
At the same time, the inverse problem of discrete exponentiation is not difficult (it can be computed efficiently using exponentiation by squaring, for example). This asymmetry is analogous to the one between integer factorization and integer multiplication. Both asymmetries have been exploited in the construction of cryptographic systems.
Popular choices for the group G in discrete logarithm cryptography are the cyclic groups (Zp)×; see ElGamal encryption, Diffie–Hellman key exchange, and the Digital Signature Algorithm.
Newer cryptography applications use discrete logarithms in cyclic subgroups of elliptic curves over finite fields; see elliptic curve cryptography.
## References
1. ^ Shor, Peter (1997). "Polynomial-Time Algorithms for Prime Factorization and Discrete Logarithms on a Quantum Computer". SIAM Journal on Computing 26 (5): 1484–1509. arXiv:quant-ph/9508027.
• Richard Crandall; Carl Pomerance. Chapter 5, Prime Numbers: A computational perspective, 2nd ed., Springer.
• Stinson, Douglas Robert (2006), Cryptography: Theory and Practice (3rd ed.), London: CRC Press, ISBN 978-1-58488-508-5
Wikimedia Foundation. 2010.
### Look at other dictionaries:
• Discrete logarithm records — are the best results achieved to date in solving the discrete logarithm problem, which is the problem of finding solutions x to the equation gx = h given elements g and h of a finite cyclic group G. The difficulty of this problem… … Wikipedia
• Logarithm — The graph of the logarithm to base 2 crosses the x axis (horizontal axis) at 1 and passes through the points with coordinates (2, 1), (4, 2), and (8, 3) … Wikipedia
• Complex logarithm — A single branch of the complex logarithm. The hue of the color is used to show the arg (polar coordinate angle) of the complex logarithm. The saturation (intensity) of the color is used to show the modulus of the complex logarithm. The page with… … Wikipedia
• List of logarithm topics — This is a list of logarithm topics, by Wikipedia page. See also the list of exponential topics.*Acoustic power *antilogarithm *Apparent magnitude *Bel *Benford s law *Binary logarithm *Bode plot *Henry Briggs *Cologarithm *Common logarithm… … Wikipedia
• Integral logarithm — The term integral logarithm may stand for: * Discrete logarithm in algebra, * Logarithmic integral function in calculus … Wikipedia
• Elliptic curve cryptography — (ECC) is an approach to public key cryptography based on the algebraic structure of elliptic curves over finite fields. The use of elliptic curves in cryptography was suggested independently by Neal Koblitz[1] and Victor S. Miller[2] in 1985.… … Wikipedia
• IEEE P1363 — is an Institute of Electrical and Electronics Engineers (IEEE) standardization project for public key cryptography. It includes specifications for: Traditional public key cryptography (IEEE Std 1363 2000 and 1363a 2004) Lattice based public key… … Wikipedia
• Index calculus algorithm — In group theory, the index calculus algorithm is an algorithm for computing discrete logarithms. This is the best known algorithm for certain groups, such as mathbb{Z} m^* (the multiplicative group modulo m ).Dubious|date=April 2008 Description… … Wikipedia
• Pollard's lambda algorithm — In mathematics, specifically computational number theory and computational algebra, Pollard s lambda algorithm (aka Pollard s kangaroo algorithm, see Naming below) is an algorithm for solving the discrete logarithm. The algorithm was introduced… … Wikipedia
• Proof of knowledge — In cryptography, a proof of knowledge is an interactive proof in which the prover succeeds convincing a verifier that it knows something. What it means for a machine to know something is defined in terms of computation. A machine knows something … Wikipedia |
# Math from Beginner to Mastery: Quadratic Functions and Cross Multiplication
## Mathematics from Beginner to Master
### foreword
I wanted to learn mathematics well, but found that I may not remember the knowledge of junior high school, so I found a “Complete Works of High School Mathematics Fundamentals and Solutions” on station B again.
### Math Learning Path
1 2 3 graph LR Math-.Function.->Quadratic function-.Solution.->Cross multiplication
### cross multiplication
This is the basis for the transition between junior high school and high school.
$y=ax^2+bx+c$
where $a,b,c$ must all be integers. (i.e. no score)
example
$y=x^2-3x+2$
untie
1 -1
1 -2
The two 1s in the first column are from $X^2$ ($x^2=x x$)
The 1 and 2 in the second column are from $c$, in this question $c=2$, 2 can be split into $1 2$ or $-1*-2$
The logic of splitting: make $b$ by splitting $c$ into the form of $c1*c2$
See below for details
After the cross is connected, the next step is to multiply the 1 and -2 of the first row to get $1*-2=-2$
Multiply the 1 in the second row by -1 to get $1*-1=-1$
Then add the two results $-2+(-1)=-3=b$
$b$ has been figured out. At this step, the cross multiplication is right. Finally, the simplified formula can be written horizontally and the 1 in the first column of the formula can be multiplied by x again.
1 plus -1 in the first row times 1 plus -2 in the second row
$(1 x)+-1 (1 x)+-2=(x-1) (x-2)$
Finish
$y=ax^2+bx+c$
On the coordinate axis $y=ax^2+bx+c$ is a U-shaped curve, either the opening is facing up or the opening is facing down.
The ± of $a$ determines the opening direction + upward – downward.
$\Delta=b^2-4ac$
$\Delta$ is the position of the midpoint of the curve on the Y axis
The $\Delta>0$ function has two points on the X axis (y=0) $x_1,x_2$
The $\Delta=0$ function has a point on the X axis (y=0) $x_1=x_2$
The $\Delta<0$ function has no point on the X axis (y=0) and does not intersect the X axis
Where did $\Delta$ come from?
$b^2-4ac$ is derived from the general formula $ax^2+bx+c=0$:
The specific derivation process of $b^2-4ac$:
$ax^2+bx+c=0(a≠0)$
Divide both sides by a
get $x^2+b/aX+c/a=0$
Re-formulation is $x^2+b/aX+(b/2a)^2=-c/a+(b/2a)^2$
$x+b/2a)^2=b²-4ac/4a^2$
If $b²-4ac$ is greater than or equal to 0
$x=-b$±$b^2-4ac/2a$ under the square root
The symmetry axis of the function curve is $x=-\frac{b}{2a}$
### Unary quadratic function universal formula
It’s called Mitternacht Formel in German
$x_1,_2=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
### Veda’s Theorem
Use it when it’s hard to come by.
Premise: $\Delta>0$ (with $x_1,x_2$)
$x_1+x_2=-\frac{b}{a}$
$x_1x_2=\frac{c}{a}$
The ± of $a$ determines the opening direction + upward – downward. |
# Liner Interpolation
Linear Interpolation in the XY plane using Scala
joesan published on
3 min, 543 words
Tags: blog
In mathematics, linear interpolation is understood as a method of curve fitting using linear polynomials. A polynomial is just an expression that consists of variables along with their co-efficients. An example of a polynomial would be:
x+2y²
In the above equation, the variable x has 1 as its coefficient, the variable y has 2 as its coefficient. Let us assume that we have a set of linear data points pointing to the x and y-axis:
x = {2,5,7,10,11,15,17,19,21,25}
y = {3,6,8,10,14,16,19,21,23,26}
Given a value x, we can find the value y. For example., for x = 2, the corresponding y is 3, for x = 11, the corresponding y is 14. Now we are interested in finding the value of x and y within our set of data points.
What is the value of y for x = 13 or what is the value of x for y = 13. Since the data points for both the x and y co-ordinates are linear, we can use the linear interpolation technique to find out the values for the desired data points. Geometrically, the linear interpolant is just a straight line between two known data points. Let us assume that our two data points are represented as (x0,y0) and (x1,y1). We are interested in finding the values of x and y that lies between these data points. Since we are in the linear boundary, geometrically speaking the ratio of the difference between x and y co-ordinates between two data points are equal.
Let us try to ascertain this with some numbers. Let our data points (x0,y0) be (2,3) and (x1,y1) be (5,6). The ratio y1/x1 should be equal to the ration y0/x0 and in our case, 6/5 == 3/2. Extending this idea, we can now say that the ratio of the difference between two data points should be geometrically equal between the two co-ordinates. If we are interested in a data point (x,y) which lies in-between (x0,y0) and (x1,y1), our formula now becomes:
y - y0 / x - x0 = y1 - y0 / x1 - x0 which when solving for y, we get the following equation:
y = y0 + (y1 - y0) * (x - x0) / (x1 - x0)
We now have the formula for linear interpolation, let's now implement that in Scala. Let us represent our data points as a Scala List:
val x = List(2,5,7,10,11,15,17,19,21,25)
val y = List(3,6,8,10,14,16,19,21,23,26)
val zippedDataPoints = x zip y
We define a function that takes these two zipped List of data points and an additional parameter for x. The function uses the formula to find the value of y for a given x. Here is the function:
def interpolate(givenX: Int, dataPoints: List[(Int, Int)]) = {
// edge cases
val (lastX, lastY) = dataPoints.last
givenX match {
case a if givenX >= lastX => lastY
case c => {
val (lower,upper) = dataPoints.span { case (x,y) => x < givenX }
val (x0,y0) = lower.last
// finally our formula!
y0 + (y1 - y0) * (givenX - x0) / (x1 - x0)
}
}
}
Let's now write some tests to test our interpolation (using ScalaTest):
interpolate(2, zippedDataPoints) == 3
interpolate(4, zippedDataPoints) == 5
interpolate(27, zippedDataPoints) == 26
So there it is! we have our linear interpolation implemented using Scala! |
# PSEB 5th Class Maths Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.7
Punjab State Board PSEB 5th Class Maths Book Solutions Chapter 2 Fundamental Operations on Numbers Ex 2.7 Textbook Exercise Questions and Answers.
## PSEB Solutions for Class 5 Maths Chapter 2 Fundamental Operations on Numbers Ex 2.7
Question 1.
Solve the following :
(a) 117 ÷ 13
(b) 135 ÷ 15
(c) 72 ÷ 12
(d) 108 ÷ 9
(e) 78 ÷ 13
(f) 121 ÷ 11
(g) 140 ÷ 20
(h) 144 ÷ 16
(i) 98 ÷ 14
(j) 119 ÷ 17.
Solution:
Question 2.
Divide the following and verify :
(a) 54598 ÷ 12 .
(b) 8975 ÷ 21
(c) 77552 ÷ 18
(d) 88001 ÷ 17
(e) 12896 ÷ 11.
Solution:
Quotient = 4549
Remainder = 10
Verification :
Dividend = Quotient × Divisor + Remainder
54598 = 4549 × 12 + 10
54598 = 54588 + 10
54598 = 54598
Quotient = 427
Remainder = 8
Verification :
Dividend = Quotient × Divisor + Remainder
8975 = 427 × 21 + 8
8975 = 8967 + 8
8975 = 8975
Quotient = 4308
Remainder = 8
Verification :
Dividend = Quotient × Divisor + Remainder
77552 = 4308 × 18 + 8
77552 = 77544 + 8
77552 = 77552
Quotient = 5176
Remainder = 9
Verification :
Dividend = Quotient × Divisor + Remainder
88001 = 5176 × 17 + 9
88001 = 87992 + 9
88001 = 88001
Quotient = 1172
Remainder = 4
Verification :
Dividend = Quotient × Divisor + Remainder
12896 = 1172 × 11 + 4
12896 = 12892 + 4
12896 = 12896
Question 3.
Solve the following and verify :
(a) 760 ÷ 12
(b) 550 ÷ 14
(c) 894 ÷ 21
(d) 913 ÷ 19
(e) 826 ÷ 25
(f) 7645 ÷ 24
(g) 89781 ÷ 9
(h) 99999 ÷ 80
(i) 82525 ÷ 75
(j) 70008 ÷ 14
(k) 50205 ÷ 15
(l) 16258 ÷ 36
(m) 96000 ÷ 50
(n) 45457 ÷ 35
Solution:
Quotient = 63
Remainder = 4
Verification :
Dividend = Quotient × Divisor + Remainder
760 = 63 × 12 + 4
760 = 756 + 4
760 = 760
Quotient = 39
Remainder = 4
Verification :
Dividend = Quotient × Divisor + Remainder
550= 39 × 14 + 4
550 = 546 + 4
550 = 550
Quotient = 42
Remainder = 12
Verification :
Dividend = Quotient × Divisor + Remainder
894 = 42 × 21 + 12
894 = 882 + 12
894 = 894
Quotient = 48
Remainder = 1
Verification :
Dividend = Quotient × Divisor + Remainder
913 = 48 × 19 + 1
913 = 912 + 1
913 = 913
Quotient = 33
Remainder = 1
Verification :
Dividend = Quotient × Divisor + Remainder
826 = 33 × 25 + 1
826 = 825 + 1
826 = 826
Quotient = 318
Remainder = 13
Verification :
Dividend = Quotient × Divisor + Remainder
7645 = 318 × 24 + 13
7645 = 7632 + 13 .
7645 = 7645
Verification :
Dividend = Quotient × Divisor + Remainder
89781 = 9975 × 9 + 6
89781 = 89775 + 6 89781 = 89781
Quotient = 1249
Remainder = 79
Verification :
Dividend = Quotient × Divisor + Remainder
99999 = 1249 × 80 + 79
99999 = 99920 + 79
99999 = 99999
Quotient = 1100
Remainder = 28
Verification :
Dividend = Quotient × Divisor + Remainder
82525 = 1100 × 75 + 25
82525 = 82500 + 25
82525 = 82525
Quotient = 5000
Remainder = 8
Verification :
Dividend = Quotient × Divisor + Remainder
70008 = 5000 × 14 + 08
70008 = 70000 + 08
70008 = 70008
Quotient = 3347
Remainder = 0
Verification :
Dividend = Quotient × Divisor + Remainder
50205 = 3347 × 15 + 0
50205 = 50205
Quotient = 451
Remainder = 22
Verification :
Dividend = Quotient × Divisor + Remainder
16258 = 451 × 36 + 22
16258 = 16258
Quotient = 1920
Remainder = 0
Verification :
Dividend = Quotient × Divisor + Remainder
96000 = 1920 × 50 + 0
96000 = 96000
Quotient = 1298
Remainder = 27
Verification :
Dividend = Quotient × Divisor + Remainder
45457= 1298 × 35 + 27
45457 = 45430 + 27
45457 = 45457 |
Question
In 2016, a maglev train in Japan traveled 1.1 miles in 1045 seconds.
A picture of a Maglev train is shown. The caption reads Maglev trains use magnets to hover above the tracks. This helps them reach extraordinary speeds.
Question 1
Part A
Find the speed of the train in miles per hour.+
1. vankhanh
### Explanation:
What we know:
• We need to find the speed of the train in miles per hour (mph)
• The train is traveling 1.1 miles per 1045 seconds
• There are 3600 seconds in 1 hour
How to solve:
We need to find what number when multiplied with 1045, will equal 3600. The unknown will be a variable x, and the seconds will be represented with s. Then, we need to multiply this number with the value of the miles (m) to find how many miles are traveled in that hour (h).
## Process:
Finding X (in seconds)
Set up equation H = sx
Substitute 3600 = 1045x
Divide /1045 /1045
Solution 3.44497608 = x
Finding Miles
Set up equation mph = x(m/s)
Substitute mph = 3.44497608(1.1/1045)
Simplify mph = ~3.79/3600
Convert mph = ~3.8 miles per hour |
# If a projectile is shot at a velocity of 11 m/s and an angle of pi/4, how far will the projectile travel before landing?
Jul 11, 2017
$\Delta x = 12.3$ $\text{m}$
#### Explanation:
We're asked to find the horizontal range of a projectile given its initial speed and launch angle.
To do this, we can first find the time $t$ when it has a height of $0$, using the equation
$\Delta y = {v}_{0 y} t - \frac{1}{2} g {t}^{2}$
The initial $y$-velocity ${v}_{0 y}$ is
${v}_{0 y} = {v}_{0} \sin {\alpha}_{0} = \left(11 \textcolor{w h i t e}{l} \text{m/s}\right) \sin \left(\frac{\pi}{4}\right) = 7.78$ $\text{m/s}$
The change in height $\Delta y$ is $0$, because we're trying to find the time $t$ at this height. Plugging in known values, we have
$0 = \left(7.78 \textcolor{w h i t e}{l} {\text{m/s")t - 1/2(9.81color(white)(l)"m/s}}^{2}\right) {t}^{2}$
$\left(4.905 \textcolor{w h i t e}{l} \text{m/s"^2)t^2 = (7.78color(white)(l)"m/s}\right) t$
(4.905color(white)(l)"m/s"^2)t = 7.78color(white)(l)"m/s"
t = color(red)(1.59 color(red)("s"
We can now use the equation
$\Delta x = {v}_{0 x} t$
to find the horizontal range, $\Delta x$
The initial $x$-velocity ${v}_{0 x}$ is
${v}_{0 x} = {v}_{0} \cos {\alpha}_{0} = \left(11 \textcolor{w h i t e}{l} \text{m/s}\right) \cos \left(\frac{\pi}{4}\right) = 7.78$ $\text{m/s}$
We then have:
Deltax = (7.78"m"/(cancel("s")))(1.59cancel("s")) = color(blue)(12.3 color(blue)("m" |
<meta http-equiv="refresh" content="1; url=/nojavascript/">
# Addition and Subtraction of Rational Expressions
Add and subtract fractions with variables in the denominator
%
Progress
Practice Addition and Subtraction of Rational Expressions
Progress
%
Can you use your knowledge of rational expressions and adding fractions to add the following rational expressions?
$\frac{3x}{x^2+6x-16}+\frac{2x}{x-2}$
### Guidance
Rational expressions are examples of fractions, so you add and subtract rational expressions in the same way that you add and subtract fractions. As with fractions, you will need a common denominator, ideally the lowest common denominator (LCD), in order to add or subtract the expressions.
$\frac{x^2+2x}{x+3}+ \frac{x}{x^2+4x+3}$
First, factor the denominators to get:
$\frac{x^2+2x}{x+3}+ \frac{x}{(x+3)(x+1)}$ .
Next, find the lowest common denominator (LCD). This will be the product of each unique factor in the denominators. In this case, the LCD is $(x+3)(x+1)$ . Multiply the numerator and denominator of each fraction by the factors necessary to create the common denominator. In this case, you only need to multiply the fraction on the left by $\left(\frac{x+1}{x+1}\right)$ . The expression becomes:
$\frac{x^2+2x}{x+3}\left(\frac{x+1}{x+1}\right)+ \frac{x}{(x+3)(x+1)}$
$=\frac{(x^2+2x)(x+1)}{(x+3)(x+1)}+ \frac{x}{(x+3)(x+1)}$
Now add the numerators and write as one rational expression:
$=\frac{(x^2+2x)(x+1)+x}{(x+3)(x+1)}$
Simplify the numerator by multiplying, combining like terms, and factoring if possible (the denominator is left in factored form):
$\frac{x^3+3x^2+3x}{(x+3)(x+1)}$
$=\frac{x(x^2+3x+3)}{(x+3)(x+1)}$
The rational expression cannot be simplified any further so this is your answer. The restrictions are $x\ne -3$ and $x\ne -1$ because those values would cause one or both of the original denominators to be equal to zero.
#### Example A
Identify the lowest common denominator (LCD) in factored form.
i) $\frac{2x-3}{x^2-7x+10}-\frac{x-5}{x^2-2x-15}$
ii) $\frac{2x+1}{x^2+6x+9}+\frac{3x-2}{x^2+x-6}$
Solution: To determine the LCD, begin by factoring the denominators.
i) $\frac{2x-3}{x^2-7x+10}-\frac{x-5}{x^2-2x-15}=\frac{2x-3}{(x-5)(x-2)}-\frac{x-5}{(x-5)(x+3)}$
The LCD is $\boxed{(x-5)(x-2)(x+3)}$
ii) $\frac{2x+1}{x^2+6x+9}+\frac{3x-2}{x^2+x-6}=\frac{2x+1}{(x+3)(x+3)}+\frac{3x-2}{(x+3)(x-2)}$
The LCD is $\boxed{(x+3)(x+3)(x-2)}$
#### Example B
Add the following rational expressions and state the restrictions.
$\frac{3x+1}{x^2+8x+16}+\frac{2x-3}{x^2+x-12}$
Solution: Begin by determining the LCD. Factor the denominators of each expression.
$\frac{3x+1}{x^2+8x+16}+\frac{2x-3}{x^2+x-12}=\frac{3x+1}{(x+4)(x+4)}+\frac{2x-3}{(x+4)(x-3)}$
The LCD is $\boxed{(x+4)(x+4)(x-3)}$
Multiply the numerators and denominators of each expression by the necessary factors to create the LCD.
$\frac{3x+1}{(x+4)(x+4)} {\color{red}\left(\frac{x-3}{x-3}\right)} + \frac{2x-3}{(x+4)(x-3)} {\color{red}\left(\frac{x+4}{x+4}\right)}$
Multiply the numerators. Keep the denominators in factored form.
$\frac{3x^2-8x-3}{(x+4)(x+4)(x-3)}+\frac{2x^2+5x-12}{(x+4)(x+4)(x-3)}$
Write the two expressions as one rational expression.
$\frac{3x^2-8x-3+2x^2+5x-12}{(x+4)(x+4)(x-3)}$
Simplify the numerator by combining like terms.
$\frac{5x^2-3x-15}{(x+4)(x+4)(x-3)}$
The numerator cannot be factored so the expression cannot be further simplified. The answer in lowest terms is:
$\boxed{\frac{5x^2-3x-15}{(x+4)(x+4)(x-3)}; x \ne -4; x \ne 3}$
#### Example C
Subtract the following rational expressions and state the restrictions.
$\frac{x}{x^2-9x+18}-\frac{x-2}{x^2-10x+24}$
Solution: Begin by determining the LCD. Factor the denominators of each expression.
$\frac{x}{(x-6)(x-3)}-\frac{x-2}{(x-6)(x-4)}$
The LCD is $\boxed{(x-6)(x-3)(x-4)}$
Multiply the numerators and denominators of each expression to get the LCD.
$\frac{x}{(x-6)(x-3)} {\color{red}\left(\frac{x-4}{x-4}\right)} - \frac{x-2}{(x-6)(x-4)} {\color{red}\left(\frac{x-3}{x-3}\right)}$
Multiply the numerators.
$\frac{x^2-4x}{(x-6)(x-3)(x-4)} - \frac{x^2-5x+6}{(x-6)(x-3)(x-4)}$
Write the expressions as one rational expression.
$\frac{x^2-4x-(x^2-5x+6)}{(x-6)(x-3)(x-4)}$
$=\frac{x^2-4x-x^2+5x-6}{(x-6)(x-3)(x-4)}$
Simplify the numerator by combining like terms.
$\frac{x-6}{(x-6)(x-3)(x-4)}$
The term $(x-6)$ is common to both the numerator and the denominator. This term can be "cancelled." The solution is:
$\boxed{\frac{1}{(x-3)(x-4)};x \ne 3; x \ne 4; x \ne 6}$
#### Concept Problem Revisited
$\frac{3x}{x^2+6x-16}+\frac{2x}{x-2}$
Factor the denominator of the first fraction and rewrite the problem:
$\frac{3x}{(x+8)(x-2)}+\frac{2x}{x-2}$
The LCD is $(x+8)(x-2)$ .
$\frac{3x}{(x+8)(x-2)}+\frac{2x}{x-2} \left( {\color{red}\frac{x+8}{x+8}}\right)$
Multiply the numerators.
$\frac{3x}{(x+8)(x-2)}+\frac{{\color{red}2x^2+16x}}{(x-2)(x+8)}$
Write the two expressions a one rational expression.
$\frac{3x+2x^2+16x}{(x+8)(x-2)}$
$\boxed{\frac{2x^2+19x}{(x+8)(x-2)};x \ne -8; x \ne 2}$
### Guided Practice
Add or subtract the following and state the restrictions.
1. $\frac{2x}{x^2-4}-\frac{1}{x-2}$
2. $\frac{-2}{3y^2+5y+2}+\frac{3}{y^2-7y-8}$
3. $\frac{3m-1}{9m^3-36m^2} + \frac{2m+1}{2m^2-5m-12}$
1.
$\frac{2x}{x^2-4}-\frac{1}{x-2}&=\frac{2x}{(x-2)(x+2)}-\frac{x+2}{(x-2)(x+2)}\\&=\frac{2x-(x+2)}{(x-2)(x+2)}\\&=\frac{(x-2)}{(x-2)(x+2)}\\&=\frac{1}{(x+2)}; x \ne -2; x \ne 2$
2.
$\frac{-2}{3y^2+5y+2}+\frac{3}{y^2-7y-8}&=\frac{-2}{(3y+2)(y+1)}+\frac{3}{(y-8)(y+1)}\\&=\frac{-2(y-8)}{(3y+2)(y+1)(y-8)}+\frac{3(3y+2)}{(3y+2)(y-8)(y+1)}\\&=\frac{-2y+16+9y+6}{(3y+2)(y-8)(y+1)}\\&=\frac{7y+22}{(3y+2)(y-8)(y+1)}; y \ne -\frac{2}{3}; y \ne 8; y \ne -1$
3.
$\frac{3m-1}{9m^3-36m^2} + \frac{2m+1}{2m^2-5m-12}&=\frac{3m-1}{9m^2(m-4)} + \frac{2m+1}{(2m+3)(m-4)}\\&=\frac{(3m-1)(2m+3)}{9m^2(m-4)(2m+3)} + \frac{9m^2(2m+1)}{9m^2(2m+3)(m-4)}\\&=\frac{6m^2-2m+9m-3+18m^3+9m^2}{9m^2(m-4)(2m+3)}\\&=\frac{18m^3+15m^2+7m-3}{9m^2(m-4)(2m+3)}; m \ne -\frac{3}{2}; m \ne 4; m \ne 0$
### Explore More
For each of the following rational expressions, determine the LCD.
1. $\frac{2a-3}{4} + \frac{3a-1}{5} - \frac{a-5}{2}$
2. $\frac{5}{3x^2} - \frac{1}{2x} + \frac{3}{5x^3}$
3. $\frac{x}{a^2b} - \frac{y}{ab^2} + \frac{z}{3a^3b^2}$
4. $\frac{2w}{w^2-6w+5} - \frac{3w}{w^2-11w+30}$
5. $\frac{1}{y^2+5y} - \frac{2}{y^2+12y+35} - \frac{3}{y^3+7y^2}$
For each of the following rational expressions, state the restrictions.
1. $\frac{3}{x^2-5x+4} + \frac{4}{x^2-16}$
2. $\frac{5}{a^2+a} - \frac{2}{a^2+3a+2}$
3. $\frac{6}{m^2-5m} + \frac{7}{m^2-4m-5}$
4. $\frac{3n}{n^2+2n-3} - \frac{4n}{n^2+n-6}$
5. $\frac{6}{y^2-4} + \frac{4}{y^2+4y+4}$
Add or subtract each of the following rational expressions and state the restrictions.
1. $\frac{2a-3}{4} + \frac{3a-1}{5} - \frac{a-5}{2}$
2. $\frac{5}{3x^2} - \frac{1}{2x} + \frac{3}{5x^3}$
3. $\frac{x}{a^2b} - \frac{y}{ab^2} + \frac{z}{3a^3b^2}$
4. $\frac{2w}{w^2-6w+5} - \frac{3w}{w^2-11w+30}$
5. $\frac{1}{y^2+5y} - \frac{2}{y^2+12y+35} - \frac{3}{y^3+7y^2}$
### Vocabulary Language: English
Least Common Denominator
Least Common Denominator
The least common denominator or lowest common denominator of two fractions is the smallest number that is a multiple of both of the original denominators. |
Congruency of Triangles
Two triangles are said to be congruent if they are exactly alike in all respects. If one triangle is placed on the other, the two triangles will coincide exactly with each other, i.e., the vertices of the first triangle will coincide with those of the second. In a pair of congruent triangles, the sides opposite to equal angles are known as corresponding sides and the angles opposite to equal sides are known as corresponding angles.
Here, in ∆KLM and ∆XYZ, ∠K = ∠X, ∠L = ∠Y and ∠M = ∠Z.
Also, KL = XY, LM = YZ and KM = XZ.
As the two triangles are exactly equal in all respects, they are congruent.
Symbolically, we write ∆KLM ≅ ∆XYZ.
Congruent triangles are equal in area.
Didn't find what you were looking for? Or want to know more information about Math Only Math. Use this Google Search to find what you need.
Recent Articles
1. Comparison of Numbers | Compare Numbers Rules | Examples of Comparison
May 18, 24 02:59 PM
Rule I: We know that a number with more digits is always greater than the number with less number of digits. Rule II: When the two numbers have the same number of digits, we start comparing the digits…
2. Numbers | Notation | Numeration | Numeral | Estimation | Examples
May 12, 24 06:28 PM
Numbers are used for calculating and counting. These counting numbers 1, 2, 3, 4, 5, .......... are called natural numbers. In order to describe the number of elements in a collection with no objects
3. Face Value and Place Value|Difference Between Place Value & Face Value
May 12, 24 06:23 PM
What is the difference between face value and place value of digits? Before we proceed to face value and place value let us recall the expanded form of a number. The face value of a digit is the digit…
4. Patterns in Numbers | Patterns in Maths |Math Patterns|Series Patterns
May 12, 24 06:09 PM
We see so many patterns around us in our daily life. We know that a pattern is an arrangement of objects, colors, or numbers placed in a certain order. Some patterns neither grow nor reduce but only r… |
# Notes on Diffy Qs: Differential Equations for Engineers
## Section2.2Constant coefficient second order linear ODEs
Note: more than 1 lecture, second part of §3.1 in [EP], §3.1 in [BD]
### Subsection2.2.1Solving constant coefficient equations
Consider the problem
\begin{equation*} y''-6y'+8y = 0, \qquad y(0) = - 2, \qquad y'(0) = 6 . \end{equation*}
This is a second order linear homogeneous equation with constant coefficients. Constant coefficients means that the functions in front of $$y''\text{,}$$ $$y'\text{,}$$ and $$y$$ are constants, they do not depend on $$x\text{.}$$
To guess a solution, think of a function that stays essentially the same when we differentiate it, so that we can take the function and its derivatives, add some multiples of these together, and end up with zero. Yes, we are talking about the exponential.
Let us try
1
Making an educated guess with some parameters to solve for is such a central technique in differential equations, that people sometimes use a fancy name for such a guess: ansatz, German for “initial placement of a tool at a work piece.” Yes, the Germans have a word for that.
a solution of the form $$y = e^{rx}\text{.}$$ Then $$y' = r e^{rx}$$ and $$y'' = r^2 e^{rx}\text{.}$$ Plug in to get
\begin{equation*} \begin{aligned} y''-6y'+8y & = 0 , \\ \underbrace{r^2 e^{rx}}_{y''} -6 \underbrace{r e^{rx}}_{y'}+8 \underbrace{e^{rx}}_{y} & = 0 , \\ r^2 -6 r +8 & = 0 \qquad \text{(divide through by } e^{rx} \text{)},\\ (r-2)(r-4) & = 0 . \end{aligned} \end{equation*}
Hence, if $$r=2$$ or $$r=4\text{,}$$ then $$e^{rx}$$ is a solution. So let $$y_1 = e^{2x}$$ and $$y_2 = e^{4x}\text{.}$$
#### Exercise2.2.1.
Check that $$y_1$$ and $$y_2$$ are solutions.
The functions $$e^{2x}$$ and $$e^{4x}$$ are linearly independent. If they were not linearly independent, we could write $$e^{4x} = C e^{2x}$$ for some constant $$C\text{,}$$ implying that $$e^{2x} = C$$ for all $$x\text{,}$$ which is clearly not possible. Hence, we can write the general solution as
\begin{equation*} y = C_1 e^{2x} + C_2 e^{4x} . \end{equation*}
We need to solve for $$C_1$$ and $$C_2\text{.}$$ To apply the initial conditions, we first find $$y' = 2 C_1 e^{2x} + 4 C_2 e^{4x}\text{.}$$ We plug $$x=0$$ into $$y$$ and $$y'$$ and solve.
\begin{equation*} \begin{aligned} -2 & = y(0) = C_1 + C_2 , \\ 6 & = y'(0) = 2 C_1 + 4 C_2 . \end{aligned} \end{equation*}
Either apply some matrix algebra, or just solve these by high school math. For example, divide the second equation by 2 to obtain $$3 = C_1 + 2 C_2\text{,}$$ and subtract the two equations to get $$5 = C_2\text{.}$$ Then $$C_1 = -7$$ as $$-2 = C_1 + 5\text{.}$$ Hence, the solution we are looking for is
\begin{equation*} y = -7 e^{2x} + 5 e^{4x} . \end{equation*}
We generalize this example into a method. Suppose that we have an equation
$$a y'' + b y' + c y = 0 ,\tag{2.3}$$
where $$a, b, c$$ are constants. Try the solution $$y = e^{rx}$$ to obtain
\begin{equation*} a r^2 e^{rx} + b r e^{rx} + c e^{rx} = 0 . \end{equation*}
Divide by $$e^{rx}$$ to obtain the so-called characteristic equation of the ODE:
\begin{equation*} a r^2 + b r + c = 0 . \end{equation*}
Solve for the $$r$$ by using the quadratic formula:
\begin{equation*} r_1, r_2 = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} . \end{equation*}
So $$e^{r_1 x}$$ and $$e^{r_2 x}$$ are solutions. There is still a difficulty if $$r_1 = r_2\text{,}$$ but it is not hard to overcome.
#### Example2.2.1.
Solve
\begin{equation*} y'' - k^2 y = 0 . \end{equation*}
The characteristic equation is $$r^2 - k^2 = 0$$ or $$(r-k)(r+k) = 0\text{.}$$ Consequently, $$e^{-k x}$$ and $$e^{kx}$$ are the two linearly independent solutions, and the general solution is
\begin{equation*} y = C_1 e^{kx} + C_2e^{-kx} . \end{equation*}
Since $$\cosh s = \frac{e^s+e^{-s}}{2}$$ and $$\sinh s = \frac{e^s-e^{-s}}{2}\text{,}$$ we can also write the general solution as
\begin{equation*} y = D_1 \cosh(kx) + D_2 \sinh(kx) . \end{equation*}
#### Example2.2.2.
Find the general solution of
\begin{equation*} y'' -8 y' + 16 y = 0 . \end{equation*}
The characteristic equation is $$r^2 - 8 r + 16 = {(r-4)}^2 = 0\text{.}$$ The equation has a double root $$r_1 = r_2 = 4\text{.}$$ The general solution is, therefore,
\begin{equation*} y = (C_1 + C_2 x)\, e^{4 x} = C_1 e^{4x} + C_2 x e^{4x} . \end{equation*}
It is good to check your work. That $$e^{4x}$$ solves the equation is clear. Let us check that $$x e^{4x}$$ solves the equation. Compute $$y' = e^{4x} + 4xe^{4x}$$ and $$y'' = 8 e^{4x} + 16xe^{4x}\text{.}$$ Plug in,
\begin{equation*} y'' - 8 y' + 16 y = 8 e^{4x} + 16xe^{4x} - 8(e^{4x} + 4xe^{4x}) + 16 xe^{4x} = 0 . \end{equation*}
In some sense, a doubled root rarely happens. If coefficients are picked randomly, a doubled root is unlikely. There are, however, some real-world problems where a doubled root does happen naturally (e.g., critically damped mass-spring system as we will see).
Let us give a short argument for why the solution $$x e^{r x}$$ works for a doubled root. This case is a limiting case of two distinct but very close roots. Note that $$\frac{e^{r_2 x} - e^{r_1 x}}{r_2 - r_1}$$ is a solution when the roots are distinct. When we take the limit as $$r_1$$ goes to $$r_2\text{,}$$ we are really taking the derivative of $$e^{rx}$$ using $$r$$ as the variable. Therefore, the limit is $$x e^{rx}\text{,}$$ and hence this is a solution in the doubled root case. We remark that in some numerical computations, two very close roots may lead to numerical instability while a doubled root will not.
### Subsection2.2.2Complex numbers and Euler’s formula
A polynomial may have complex roots. The equation $$r^2 + 1 = 0$$ has no real roots, but it does have two complex roots. Here we review some properties of complex numbers.
Complex numbers may seem a strange concept, especially because of the terminology. There is nothing imaginary or really complicated about complex numbers. A complex number is simply a pair of real numbers, $$(a,b)\text{.}$$ Think of a complex number as a point in the plane. We add complex numbers in the straightforward way: $$(a,b)+(c,d)=(a+c,b+d)\text{.}$$ We define multiplication by
\begin{equation*} (a,b) \times (c,d) \overset{\text{def}}{=} (ac-bd,ad+bc) . \end{equation*}
It turns out that with this multiplication rule, all the standard properties of arithmetic hold. Further, and most importantly $$(0,1) \times (0,1) = (-1,0)\text{.}$$
Generally we write $$(a,b)$$ as $$a+ib\text{,}$$ and we treat $$i$$ as if it were an unknown. When $$b$$ is zero, then $$(a,0)$$ is just the number $$a\text{.}$$ We do arithmetic with complex numbers just as we would with polynomials. The property we just mentioned becomes $$i^2 = -1\text{.}$$ So whenever we see $$i^2\text{,}$$ we replace it by $$-1\text{.}$$ For example,
\begin{equation*} (2+3i)(4i) - 5i = (2\times 4)i + (3 \times 4) i^2 - 5i = 8i + 12 (-1) - 5i = -12 + 3i . \end{equation*}
The numbers $$i$$ and $$-i$$ are the two roots of $$r^2 + 1 = 0\text{.}$$ Some engineers use the letter $$j$$ instead of $$i$$ for the square root of $$-1\text{.}$$ We use the mathematicians’ convention and use $$i\text{.}$$
#### Exercise2.2.3.
Make sure you understand (that you can justify) the following identities:
1. $$i^2 = -1\text{,}$$ $$i^3 = -i\text{,}$$ $$i^4 = 1\text{,}$$
2. $$\dfrac{1}{i} = -i\text{,}$$
3. $$(3-7i)(-2-9i) = \cdots = -69-13i\text{,}$$
4. $$(3-2i)(3+2i) = 3^2 - {(2i)}^2 = 3^2 + 2^2 = 13\text{,}$$
5. $$\frac{1}{3-2i} = \frac{1}{3-2i} \frac{3+2i}{3+2i} = \frac{3+2i}{13} = \frac{3}{13}+\frac{2}{13}i\text{.}$$
We also define the exponential $$e^{a+ib}$$ of a complex number. We do this by writing down the Taylor series and plugging in the complex number. Because most properties of the exponential can be proved by looking at the Taylor series, these properties still hold for the complex exponential. For example the very important property: $$e^{x+y} = e^x e^y\text{.}$$ This means that $$e^{a+ib} = e^a e^{ib}\text{.}$$ Hence if we can compute $$e^{ib}\text{,}$$ we can compute $$e^{a+ib}\text{.}$$ For $$e^{ib}$$ we use the so-called Euler’s formula.
In other words, $$e^{a+ib} = e^a \bigl( \cos(b) + i \sin(b) \bigr) = e^a \cos(b) + i e^a \sin(b)\text{.}$$
#### Exercise2.2.4.
Using Euler’s formula, check the identities:
\begin{equation*} \cos \theta = \frac{e^{i \theta} + e^{-i \theta}}{2} \qquad \text{and} \qquad \sin \theta = \frac{e^{i \theta} - e^{-i \theta}}{2i}. \end{equation*}
#### Exercise2.2.5.
Double angle identities: Start with $$e^{i(2\theta)} = {\bigl(e^{i \theta} \bigr)}^2\text{.}$$ Use Euler on each side and deduce:
\begin{equation*} \cos (2\theta) = \cos^2 \theta - \sin^2 \theta \qquad \text{and} \qquad \sin (2\theta) = 2 \sin \theta \cos \theta . \end{equation*}
For a complex number $$a+ib$$ we call $$a$$ the real part and $$b$$ the imaginary part of the number. Often the following notation is used,
\begin{equation*} \operatorname{Re}(a+ib) = a \qquad \text{and} \qquad \operatorname{Im}(a+ib) = b. \end{equation*}
### Subsection2.2.3Complex roots
Suppose the differential equation $$ay'' + by' + cy = 0$$ has the characteristic equation $$a r^2 + b r + c = 0$$ that has complex roots. By the quadratic formula, the roots are $$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\text{.}$$ These roots are complex if $$b^2 - 4ac < 0\text{.}$$ In this case, we can write the roots as
\begin{equation*} r_1, r_2 = \frac{-b}{2a} \pm i\frac{\sqrt{4ac - b^2}}{2a} . \end{equation*}
As you can see, we always get a pair of roots of the form $$\alpha \pm i \beta\text{.}$$ We can still write the solution as
\begin{equation*} y = C_1 e^{(\alpha+i\beta)x} + C_2 e^{(\alpha-i\beta)x} . \end{equation*}
However, the exponential is now complex-valued. We need to allow $$C_1$$ and $$C_2$$ to be complex numbers to obtain a real-valued solution (which is what we are after). While there is nothing particularly wrong with this approach, it can make calculations harder and it is generally preferred to find two real-valued solutions.
Euler’s formula comes to the rescue. Let
\begin{equation*} y_1 = e^{(\alpha+i\beta)x} \qquad \text{and} \qquad y_2 = e^{(\alpha-i\beta)x} . \end{equation*}
Then
\begin{equation*} \begin{aligned} y_1 & = e^{\alpha x} \cos (\beta x) + i e^{\alpha x} \sin (\beta x) , \\ y_2 & = e^{\alpha x} \cos (\beta x) - i e^{\alpha x} \sin (\beta x) . \end{aligned} \end{equation*}
Linear combinations of solutions are also solutions. Hence,
\begin{equation*} \begin{aligned} y_3 & = \frac{y_1 + y_2}{2} = e^{\alpha x} \cos (\beta x) , \\ y_4 & = \frac{y_1 - y_2}{2i} = e^{\alpha x} \sin (\beta x) , \end{aligned} \end{equation*}
are also solutions. Furthermore, they are real-valued. It is not hard to see that they are linearly independent (not multiples of each other). We summarize what we found as a theorem.
#### Example2.2.3.
Find the general solution of $$y'' + k^2 y = 0\text{,}$$ for a constant $$k > 0\text{.}$$
The characteristic equation is $$r^2 + k^2 = 0\text{.}$$ Therefore, the roots are $$r = \pm ik\text{,}$$ and by the theorem, we have the general solution
\begin{equation*} y = C_1 \cos (kx) + C_2 \sin (kx) . \end{equation*}
#### Example2.2.4.
Find the solution of $$y'' - 6 y' + 13 y = 0\text{,}$$ $$y(0) = 0\text{,}$$ $$y'(0) = 10\text{.}$$
The characteristic equation is $$r^2 - 6 r + 13 = 0\text{.}$$ By completing the square we get $${(r-3)}^2 + 2^2 = 0$$ and hence the roots are $$r = 3 \pm 2i\text{.}$$ By the theorem we have the general solution
\begin{equation*} y = C_1 e^{3x} \cos (2x) + C_2 e^{3x} \sin (2x) . \end{equation*}
To find the solution satisfying the initial conditions, we first plug in zero to get
\begin{equation*} 0 = y(0) = C_1 e^{0} \cos 0 + C_2 e^{0} \sin 0 = C_1 . \end{equation*}
Hence, $$C_1 = 0$$ and $$y = C_2 e^{3x} \sin (2x)\text{.}$$ We differentiate,
\begin{equation*} y' = 3C_2 e^{3x} \sin (2x) + 2C_2 e^{3x} \cos (2x) . \end{equation*}
We again plug in the initial condition and obtain $$10 = y'(0) = 2C_2\text{,}$$ or $$C_2 = 5\text{.}$$ The solution we are seeking is
\begin{equation*} y = 5 e^{3x} \sin (2x) . \end{equation*}
### Exercises2.2.4Exercises
#### 2.2.6.
Find the general solution of $$2y'' + 2y' -4 y = 0\text{.}$$
#### 2.2.7.
Find the general solution of $$y'' + 9y' - 10 y = 0\text{.}$$
#### 2.2.8.
Solve $$y'' - 8y' + 16 y = 0$$ for $$y(0) = 2\text{,}$$ $$y'(0) = 0\text{.}$$
#### 2.2.9.
Solve $$y'' + 9y' = 0$$ for $$y(0) = 1\text{,}$$ $$y'(0) = 1\text{.}$$
#### 2.2.10.
Find the general solution of $$2y'' + 50y = 0\text{.}$$
#### 2.2.11.
Find the general solution of $$y'' + 6 y' + 13 y = 0\text{.}$$
#### 2.2.12.
Find the general solution of $$y'' = 0$$ using the methods of this section.
#### 2.2.13.
The method of this section applies to equations of other orders than two. We will see higher orders later. Solve the first order equation $$2y' + 3y = 0$$ using the methods of this section.
#### 2.2.14.
Let us revisit the Cauchy–Euler equations of Exercise 2.1.6. Suppose now that $${(b-a)}^2-4ac < 0\text{.}$$ Find a formula for the general solution of $$a x^2 y'' + b x y' + c y = 0\text{.}$$ Hint: Note that $$x^r = e^{r \ln x}\text{.}$$
#### 2.2.15.
Find the solution to $$y''-(2\alpha) y' + \alpha^2 y=0\text{,}$$ $$y(0) = a\text{,}$$ $$y'(0)=b\text{,}$$ where $$\alpha\text{,}$$ $$a\text{,}$$ and $$b$$ are real numbers.
#### 2.2.16.
Construct an equation such that $$y = C_1 e^{-2x} \cos(3x) + C_2 e^{-2x} \sin(3x)$$ is the general solution.
#### 2.2.101.
Find the general solution to $$y''+4y'+2y=0\text{.}$$
$$y = C_1 e^{(-2+\sqrt{2}) x} + C_2 e^{(-2-\sqrt{2}) x}$$
#### 2.2.102.
Find the general solution to $$y''-6y'+9y=0\text{.}$$
$$y = C_1 e^{3x} + C_2 x e^{3x}$$
#### 2.2.103.
Find the solution to $$2y''+y'+y=0\text{,}$$ $$y(0) = 1\text{,}$$ $$y'(0)=-2\text{.}$$
$$y = e^{-x/4} \cos\bigl((\nicefrac{\sqrt{7}}{4})x\bigr) - \sqrt{7} e^{-x/4} \sin\bigl((\nicefrac{\sqrt{7}}{4})x\bigr)$$
#### 2.2.104.
Find the solution to $$2y''+y'-3y=0\text{,}$$ $$y(0) = a\text{,}$$ $$y'(0)=b\text{.}$$
$$y = \frac{2(a-b)}{5} \, e^{-3x/2}+\frac{3 a+2 b}{5} \, e^x$$
#### 2.2.105.
Find the solution to $$z''(t) = -2z'(t)-2z(t)\text{,}$$ $$z(0) = 2\text{,}$$ $$z'(0)= -2\text{.}$$
$$z(t) = 2e^{-t} \cos(t)$$
#### 2.2.106.
Find the solution to $$y''-(\alpha+\beta) y' + \alpha \beta y=0\text{,}$$ $$y(0) = a\text{,}$$ $$y'(0)=b\text{,}$$ where $$\alpha\text{,}$$ $$\beta\text{,}$$ $$a\text{,}$$ and $$b$$ are real numbers, and $$\alpha \not= \beta\text{.}$$
$$y = \frac{a \beta-b}{\beta-\alpha} e^{\alpha x} + \frac{b-a \alpha}{\beta-\alpha} e^{\beta x}$$
Construct an equation such that $$y = C_1 e^{3x} + C_2 e^{-2x}$$ is the general solution.
$$y'' -y'-6y=0$$ |
Lesson Video: Augmented Matrices | Nagwa Lesson Video: Augmented Matrices | Nagwa
# Lesson Video: Augmented Matrices Mathematics
In this video, we will learn how to interpret augmented matrices and represent systems of linear equations as an augmented matrix.
16:57
### Video Transcript
In this video, we will learn how to interpret augmented matrices and represent systems of linear equations as an augmented matrix. One of the oldest general problems in mathematics is to be able to solve a system of linear equations in multiple variables. The simplest, nontrivial example of this would be a system of two linear equations in two variables. This can be shown as follows.
Letβs consider the two equations π₯ plus three π¦ equals one and two π₯ minus π¦ equals three. In this case, π₯ and π¦ are the variables to be found, with the numbers multiplying them being referred to as the coefficients. In this case, the coefficient of the π₯-term in the first equation is one, and the coefficient of the π¦-term is three. For the second equation, the coefficient of the π₯-term is two, and the coefficient of the π¦-term is negative one. This system, with two equations and two variables, is often referred to simply as simultaneous equations.
The methods we have seen for solving these simultaneous equations become more complicated when we extend our system to more equations and more variables. For example, consider the system of three linear equations with three variables as shown. This system of linear equations would require many more steps to solve than the first example. In order to try and prevent any mistakes, a much neater method is used. This is called row reduction or GAUSSβJordan elimination. This method aims to remove all extraneous detail by first placing the coefficients from a system of linear equations within a matrix, where each entry of this matrix corresponds to exactly one coefficient.
Letβs consider the system of linear equations five π₯ plus two π¦ equals two and three π₯ minus three π¦ equals six. We note that the alignment of the π₯-terms and their coefficients appear directly underneath each other. And the same is true of the π¦-terms. Given that this is the case, we can write the coefficients in a two-by-two matrix as shown, where the left column relates to the coefficient of the π₯-terms and the right column relates to the coefficients of the π¦-terms. This works because when we multiply this two-by-two matrix by the column matrix with elements π₯ and π¦, respectively, the result is a two-by-two matrix with the elements five π₯ plus two π¦ and three π₯ minus three π¦ as in the original equations.
Suppose now that we wish to include all information about the system of equations. The terms on the right-hand side of our equations are also aligned. And we can represent this in the matrix as shown, where the vertical bar represents the equal sign. This approach to solving equations becomes more valuable when there is an increase either in the number of equations or number of variables. Letβs now consider a formal definition.
The two matrices we previously saw are known as the coefficient and augmented matrices. We begin by considering a general system of linear equations in the variables π₯ sub one, π₯ sub two, and so on and the coefficients π ππ. This can be written as shown. This leads us to the following coefficient matrix and also the augmented matrix as shown. Whilst it is outside the scope of this video to demonstrate the full process, we can actually use these to solve systems of equations. Letβs now look at some specific examples.
Find the augmented matrix for the following system of equations: π₯ plus five π¦ equals three, and three π₯ plus five π¦ equals one.
We begin by rewriting the system of equations by highlighting the π₯- and π¦-terms. We have two equations in two variables, and these equations have already been ordered so that the π₯-terms appear first followed by the π¦-terms and then the equal sign of each equation. This means that the augmented matrix has two rows and three columns as shown. We begin by looking at the coefficients of the π₯-terms. For the first equation, the coefficient of the π₯-term is one. And in the second equation, the coefficient is three. This means that the left column of the matrix is populated with the numbers one and three. The next column is populated by the π¦-coefficients, in this case five and five. Finally, the values on the right-hand side of our equations, three and one, complete the matrix. Reading row by row, the completed augmented matrix is one five, three, three, five, one. This corresponds to the system of equations π₯ plus five π¦ equals three and three π₯ plus five π¦ equals one.
In this example, the system of equations was written in a convenient form, with the first term of each equation being the π₯-term followed by the π¦-term and then with an equal sign immediately following this. Given that this was already the case, writing out the corresponding augmented matrix was a fairly straightforward exercise. The same is true if we are presented with the augmented matrix and asked to write out a corresponding system of linear equations.
Letβs now consider an example of this type.
Find the system of equations from the following augmented matrix: seven, two, negative seven, negative five, four, six.
We begin by assuming that the variables of this system are π₯ and π¦, with π₯ corresponding to the first column and π¦ corresponding to the second column. If this is the case, this system of linear equations will take the form shown, where the missing values will be populated using the augmented matrix. The first column of the augmented matrix features the coefficients seven and negative five. And these are the coefficients of the π₯-terms in the system of equations. The middle column of the augmented matrix contains the values two and four, which are the coefficients of the π¦-terms. Finally, we use the right most column of the augmented matrix to populate the remaining blank entries of the system. These are negative seven and six, respectively.
The augmented matrix seven, two, negative seven, negative five, four, six corresponds to the system of equations seven π₯ plus two π¦ equals negative seven and negative five π₯ plus four π¦ equals six. It is important to note here that since the example did not specifically state we should have used the variables π₯ and π¦, we could easily have used other variables, for example, π and π. Our equations would then have been seven π plus two π equals negative seven and negative five π plus four π equals six.
In our first two examples, we have only worked with augmented matrices where all entries are nonzero. We have also seen the equations written in a particularly helpful way. We will now consider an example where neither of these are the case.
Find the augmented matrix for the following system of equations: eight π₯ minus three π§ minus seven equals zero, six π¦ plus three π₯ equals zero, and seven π§ minus six π¦ plus eight equals zero.
By examining the system of equations given, we can see that this is not written in the most helpful format. In order to minimize our chance of making a mistake when creating the augmented matrix, we begin by rewriting the system of linear equations. We will do this by lining up the π₯, π¦, π§, and constant terms underneath each other. Adding seven to both sides of our first equation, we have eight π₯ minus three π§ is equal to seven. And as there is no π¦-term, we will leave a gap here. The second equation has no π§-term. And this can be rewritten as three π₯ plus six π¦ equals zero. Subtracting eight from both sides of the third equation and then rearranging the left-hand side gives us negative six π¦ plus seven π§ is equal to negative eight. We notice that in this equation there is no π₯-term.
Having rewritten our equations, we see that the system has three equations in three variables. And hence, the augmented matrix is of dimension three by four. The first column of the augmented matrix contains the coefficients of π₯. These are eight, three, and zero. The second column contains the coefficients of π¦: zero, six, and negative six. Next, we have the π§-components: negative three, zero, and seven. And finally, we enter the constants on the right-hand side of our equations: seven, zero, and negative eight. The augmented matrix for the system of equations eight π₯ minus three π§ minus seven equals zero, six π¦ plus three π₯ equals zero, and seven π§ minus six π¦ plus eight equals zero is eight, zero, negative three, seven, three, six, zero, zero, zero, negative six, seven, negative eight.
In our final example, we will look at the reverse of this process, where once again we are given an augmented matrix and need to find the system of equations.
From the augmented matrix two, zero, negative nine, five, zero, four, negative nine, five, negative four, negative nine, zero, zero, find the system of equations.
We begin by assuming that the variables corresponding to the first, second, and third columns should be labeled as π₯, π¦, and π§, respectively. This means that we need to populate the missing entries in the following system of three equations in three unknowns. The first column of our augmented matrix corresponds to the π₯-coefficients. These are two, zero, and negative four. The second column corresponds to the π¦-coefficients: zero, four, and negative nine. The third column, negative nine, negative nine, and zero are the π§-coefficients. Finally, the elements in the right-hand column of our matrix correspond to the entries on the right-hand side of our equations. These are five, five, and zero.
We can simplify these equations by ignoring any term that has a coefficient of zero. As adding negative nine is the same as subtracting nine. The first equation can be rewritten as two π₯ minus nine π§ is equal to five. In the same way, the second equation becomes four π¦ minus nine π§ equals five and the third equation becomes negative four π₯ minus nine π¦ equals zero. The augmented matrix two, zero, negative nine, five, zero, four, negative nine, five, negative four, negative nine, zero, zero corresponds to the system of equations two π₯ minus nine π§ equals five, four π¦ minus nine π§ equals five, and negative four π₯ minus nine π¦ equals zero.
We have seen in this video that it is usually a simple process to switch between a system of linear equations and the corresponding augmented matrix. Once a system of linear equations is written into the correct augmented matrix, then we can undertake solving the system of equations by manipulating the augmented matrix using row operations as part of the GaussβJordan elimination. However, as previously mentioned, this is outside of the scope of this video. Letβs now summarize the key points.
We saw in this video that we can write a general system of linear equations in terms of its augmented matrix or vice versa. We saw that for a system of linear equations with π equations in π variables, the coefficient matrix has π rows and π columns and, therefore, has dimension π by π. The augmented matrix of such a system has π rows and π plus one columns, meaning it has a dimension of π by π plus one.
When writing a system of linear equations as an augmented matrix, it is essential that the variables appear in the same order for each question before the augmented matrix is populated with the coefficients of this system. One way of helping with this process is to color the variables. If a quantity appears that is neither a variable nor a coefficient of a variable, then it should appear on the right-hand side of the equation before the augmented matrix is created. Finally, if a variable does not appear in one of the equations, then the coefficient of this variable is zero and the augmented matrix should have a value of zero in the corresponding entry. |
# Recursion – The basics
## Recursion – The basics
COMPUT: a programming technique where a routine performs its task by delegating part of it to another instance of itself.
Introduction
For new computer science students, the concept of recursive programming is often difficult. Recursive thinking is difficult because it almost seems like circular reasoning. It’s also not an intuitive process; when we give instructions to other people, we rarely direct them recursively.
For those of you who are new to computer programming, here’s a simple definition of recursion: Recursion occurs when a function calls itself directly or indirectly.
A classic example of recursion
The classic example of recursive programming involves computing factorials. In mathematics, the factorial of a nonnegative integer, n (denoted n!) is the product of all positive integers less than or equal to n. For example, 5! is the same as 5*4*3*2*1, and 3! is 3*2*1.
An interesting property of a factorial is that the factorial of a number is equal to the starting number multiplied by the factorial of the number immediately below it. For example, 5! is the same as 5 * 4! You could almost write the factorial function as:
```int factorial(int n) { return n * factorial(n - 1); }```
Listing 1. First cut factorial function
However, there is a small problem with this; it will run forever because there is no place where it stops calling itself. It therefore needs a condition to tell it to stop. Since a factorial is for positive numbers only it makes sense to stop the recursion when the input is 1 and return 1 as the result. The modified code will look like this:
```int factorial(int n) { if(n == 1) { return 1; } else { return n * factorial(n - 1); } }```
Listing 2. better factorial function
As you can see, as long as the initial value is above zero, this function will terminate. Note that more work will need to be done to guard again invalid initial values.
The point at with the recursion is stopped is called the base case. A base case is the bottom point of a recursive program where the operation is so trivial as to be able to return an answer directly. In the case of a factorial 1! = 1. All recursive programs must have at least one base case and must guarantee that they will hit one eventually; otherwise the program would run forever or until the program ran out of memory or stack space.
Basic steps of recursive programs
All recursive programs follows the same basic sequence of steps:
1: Initialize the algorithm. Recursive programs often need a seed value to start with.
2: Check to see whether the current value(s) being processed match the base case. If so, process and return the value.
3: Redefine the answer in terms of a smaller or simpler sub-problem or sub-problems.
4: Run the algorithm on the sub-problem.
5: Combine the results in the formulation of the answer.
6: Return the results.
## Advantages and drawbacks of recursion
Main advantage of recursion is programming simplicity. When using recursion, programmer can forget for a while of the whole problem and concentrate on the solution of a current case. Then, returning back to the whole problem, base cases (it’s possible to have more than one base case) and entry point for recursion are developed.
On the other hand, recursion has a serious disadvantage of using large amount of memory. Moreover, for most programming languages, recursion use stack to store states of all currently active recursive calls. The size of a stack may be quite large, but limited. Therefore too deep recursion can result in Stack Overflow. To resolve this problem recursion can be simulated, using loop and stack data structure. |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Diameter or Radius of a Circle Given Circumference
## C = πd; C = 2πr
0%
Progress
Practice Diameter or Radius of a Circle Given Circumference
Progress
0%
Diameter or Radius of a Circle Given Circumference
Credit: Jason Eppink
Source: https://www.flickr.com/photos/jasoneppink/5440728084
Rebekah’s family went to a circular corn maze and spent a couple hours trying to find their way out. The distance around the outside of the maze was posted as 1 mile, or 5,280 feet. Rebekah wondered how far she would walk if she were able to cut right through the middle.
In this concept, you will learn to find the diameter and radius of a circle if you are given the circumference.
### Guidance
The formula for the circumference of a circle, c=πd\begin{align*}c= \pi d\end{align*} or c=2πr\begin{align*}c=2 \pi r\end{align*}, can also be used to find the diameter or radius of a circle.
Let's look at an example.
A circle has a circumference of 20.72 m. What is its diameter?
First, write down the formula.
c=πd
Next, fill in the values that you know.
20.72 m=3.14 d
Then, divide both sides of the equation by 3.14
6.6=d
The answer is the diameter is 6.6 meters.
c20.7220.72===πd(3.14)(6)20.72
### Guided Practice
The circumference of a circle is 147.58 yards. Find its radius.
First, write down the formula.
c=2πr
Next, fill in the values that you know.
147.58=2(3.14)r
Then, perform the calculations necessary to isolate r\begin{align*}r\end{align*}.
147.58147.5823.5===6.28 r6.28 rr
The answer is r=23.5\begin{align*}r = 23.5 \end{align*} yards.
### Examples
#### Example 1
The circumference is 28.26 inches. What is the diameter?
First, write down the formula.
c=πd
Next, fill in the values that you know.
28.26=3.14d
Then, divide both sides of the equation by 3.14
9=d
The answer is the diameter is 9 inches.
#### Example 2
The circumference is 21.98 feet. What is the radius?
First, write down the formula.
c=2πr
Next, fill in the values that you know.
21.98=2(3.14)r
Then, perform the calculations necessary to isolate r\begin{align*}r\end{align*}.
21.983.5==6.28 rr
The answer is r=3.5\begin{align*}r = 3.5\end{align*} feet.
Credit: Angela Severn
Source: https://www.flickr.com/photos/43072779@N06/4026852130
Remember Rebekah at the corn maze with a 5,280 ft circumference?
She wondered how far she would have to walk if she were able to cut right across.
First, recognize that you are solving for the diameter, the straight-line distance through the middle of the circle from one point on the circumference to another.
Then, write the formula.
c=πd
Next, fill in the values that you know.
5,280=(3.14)d
Then, perform the calculations necessary to isolate d\begin{align*}d\end{align*}.
1,681.5286621,681.5 feet==dd.
The answer is the diameter=1,681.5\begin{align*}\text{diameter} = 1,681.5 \end{align*} feet. Rebekah would have to walk 1,681.5 feet if she were able to walk the diameter of the corn maze.
### Explore More
Given the circumference, find the diameter.
1. Circumference = 15.7 ft.
2. Circumference = 20.41 in
3. Circumference = 21.98 m
4. Circumference = 4.71 cm
5. Circumference = 47.1 ft
Given the circumference, find the radius.
6. Circumference = 43.96 in
7. Circumference = 15.7 m
8. Circumference = 14.13 in
9. Circumference = 20.41 cm
10. Circumference = 12.56 ft.
Solve each problem.
11. What is the circumference of a circle whose diameter is 32.5 meters?
12. A circle has a radius of 67 centimeters. What is its circumference?
13. What is the circumference of a circle whose radius is 7.23 feet?
14. What is the diameter of a circle whose circumference is 172.7 inches?
15. A circle has a circumference of 628 centimeters. What is the radius of the circle?
16. The circumference of a circular table is 40.82 feet. What is the radius of the rug?
17. Workers at the zoo are building five circular pens for the elephants. Each pen has a diameter of 226 meters. How much fence will the workers need in order to surround all three pens?
18. Mrs. Golding has a circular mirror with a frame around it. The frame is 4 inches wide. If the diameter of the mirror and the frame together is 48 inches, what is the circumference of just the mirror?
### Vocabulary Language: English
$\pi$
$\pi$
$\pi$ (Pi) is the ratio of the circumference of a circle to its diameter. It is an irrational number that is approximately equal to 3.14.
Circle
Circle
A circle is the set of all points at a specific distance from a given point in two dimensions.
Circumference
Circumference
The circumference of a circle is the measure of the distance around the outside edge of a circle.
Perimeter
Perimeter
Perimeter is the distance around a two-dimensional figure.
Pi
Pi
$\pi$ (Pi) is the ratio of the circumference of a circle to its diameter. It is an irrational number that is approximately equal to 3.14. |
# 2.1 Linear functions (Page 6/17)
Page 6 / 17
Write the point-slope form of an equation of a line that passes through the points and $\left(0,0\right).$ Then rewrite it in the slope-intercept form.
$y-0=-3\left(x-0\right)$ ; $y=-3x$
## Writing and interpreting an equation for a linear function
Now that we have written equations for linear functions in both the slope-intercept form and the point-slope form, we can choose which method to use based on the information we are given. That information may be provided in the form of a graph, a point and a slope, two points, and so on. Look at the graph of the function $f$ in [link] .
We are not given the slope of the line, but we can choose any two points on the line to find the slope. Let’s choose and We can use these points to calculate the slope.
Now we can substitute the slope and the coordinates of one of the points into the point-slope form.
If we want to rewrite the equation in the slope-intercept form, we would find
If we wanted to find the slope-intercept form without first writing the point-slope form, we could have recognized that the line crosses the y -axis when the output value is 7. Therefore, $b=7.$ We now have the initial value $b$ and the slope $m$ so we can substitute $m$ and $b$ into the slope-intercept form of a line.
So the function is $f\left(x\right)=-\frac{3}{4}x+7,$ and the linear equation would be $y=-\frac{3}{4}x+7.$
Given the graph of a linear function, write an equation to represent the function.
1. Identify two points on the line.
2. Use the two points to calculate the slope.
3. Determine where the line crosses the y -axis to identify the y -intercept by visual inspection.
4. Substitute the slope and y -intercept into the slope-intercept form of a line equation.
## Writing an equation for a linear function
Write an equation for a linear function given a graph of $f$ shown in [link] .
Identify two points on the line, such as and $\left(-2,\text{−4}\right).$ Use the points to calculate the slope.
Substitute the slope and the coordinates of one of the points into the point-slope form.
$\begin{array}{c}y-{y}_{1}=m\left(x-{x}_{1}\right)\\ y-\left(-4\right)=3\left(x-\left(-2\right)\right)\\ y+4=3\left(x+2\right)\end{array}$
We can use algebra to rewrite the equation in the slope-intercept form.
## Writing an equation for a linear cost function
Suppose Ben starts a company in which he incurs a fixed cost of $1,250 per month for the overhead, which includes his office rent. His production costs are$37.50 per item. Write a linear function $C$ where $C\left(x\right)$ is the cost for $x$ items produced in a given month.
The fixed cost is present every month, $1,250. The costs that can vary include the cost to produce each item, which is$37.50 for Ben. The variable cost, called the marginal cost, is represented by $37.5.$ The cost Ben incurs is the sum of these two costs, represented by $C\left(x\right)=1250+37.5x.$
## Writing an equation for a linear function given two points
If $f$ is a linear function, with $f\left(3\right)=-2$ , and $f\left(8\right)=1$ , find an equation for the function in slope-intercept form.
We can write the given points using coordinates.
$\begin{array}{l}f\left(3\right)=-2\to \left(3,-2\right)\hfill \\ f\left(8\right)=1\to \left(8,1\right)\hfill \end{array}$
We can then use the points to calculate the slope.
Substitute the slope and the coordinates of one of the points into the point-slope form.
We can use algebra to rewrite the equation in the slope-intercept form.
how can are find the domain and range of a relations
A cell phone company offers two plans for minutes. Plan A: $15 per month and$2 for every 300 texts. Plan B: $25 per month and$0.50 for every 100 texts. How many texts would you need to send per month for plan B to save you money?
6000
Robert
more than 6000
Robert
can I see the picture
How would you find if a radical function is one to one?
how to understand calculus?
with doing calculus
SLIMANE
Thanks po.
Jenica
Hey I am new to precalculus, and wanted clarification please on what sine is as I am floored by the terms in this app? I don't mean to sound stupid but I have only completed up to college algebra.
I don't know if you are looking for a deeper answer or not, but the sine of an angle in a right triangle is the length of the opposite side to the angle in question divided by the length of the hypotenuse of said triangle.
Marco
can you give me sir tips to quickly understand precalculus. Im new too in that topic. Thanks
Jenica
if you remember sine, cosine, and tangent from geometry, all the relationships are the same but they use x y and r instead (x is adjacent, y is opposite, and r is hypotenuse).
Natalie
it is better to use unit circle than triangle .triangle is only used for acute angles but you can begin with. Download any application named"unit circle" you find in it all you need. unit circle is a circle centred at origine (0;0) with radius r= 1.
SLIMANE
What is domain
johnphilip
the standard equation of the ellipse that has vertices (0,-4)&(0,4) and foci (0, -15)&(0,15) it's standard equation is x^2 + y^2/16 =1 tell my why is it only x^2? why is there no a^2?
what is foci?
This term is plural for a focus, it is used for conic sections. For more detail or other math questions. I recommend researching on "Khan academy" or watching "The Organic Chemistry Tutor" YouTube channel.
Chris
how to determine the vertex,focus,directrix and axis of symmetry of the parabola by equations
i want to sure my answer of the exercise
what is the diameter of(x-2)²+(y-3)²=25
how to solve the Identity ?
what type of identity
Jeffrey
Confunction Identity
Barcenas
how to solve the sums
meena
hello guys
meena
For each year t, the population of a forest of trees is represented by the function A(t) = 117(1.029)t. In a neighboring forest, the population of the same type of tree is represented by the function B(t) = 86(1.025)t.
by how many trees did forest "A" have a greater number?
Shakeena
32.243
Kenard
how solve standard form of polar
what is a complex number used for?
It's just like any other number. The important thing to know is that they exist and can be used in computations like any number.
Steve
I would like to add that they are used in AC signal analysis for one thing
Scott
Good call Scott. Also radar signals I believe.
Steve
They are used in any profession where the phase of a waveform has to be accounted for in the calculations. Imagine two electrical signals in a wire that are out of phase by 90°. At some times they will interfere constructively, others destructively. Complex numbers simplify those equations
Tim |
USING THE HIGH-LOW METHOD TO ESTIMATE BREAK EVEN SALES
In order to use this equation, S=(FC+TP) /CM to calculate the level of sales needed to obtain the desired amount of profit, the FC (total fix costs) and CM (contribution margin per sales dollar) is required. In an existing business we can use the high-low method to divide the total expenses into FC and VC (total variable costs and expenses). Remember this is only a means of estimating. Results are not precise by any means. The least squares method is more precise but is more difficult to apply. Let's see how the high-low method works.
Assume that in the year 2008, the low and high months for sales were January and October, respectively.
January October Difference Sales \$50,000 \$75,000 \$25,000 Expenses \$35,000 \$45,000 \$10,000
VC/sales dollar = difference in total expenses/difference in total sales = \$10,000/\$25,000 = .40
Lowest level: TC = VC + FC
\$35,000 = .40 x \$50,000 + FC
\$35,000 = \$20,000 + FC
\$15,000 = FC
Highest level: TC = VC + FC
\$45,000 = .40 x \$75,000 + FC
\$45,000 = \$30,000 + FC
\$15,000 = FC
As you can see from the calculations presented, the FC is \$15,000. The CM is 1- VC/sales dollar or 1-.40 = .60. The break even point in sales for this example is (\$15,000 + \$0)/.60 or \$25,000.
Proof: S = FC + VC + TP
\$25,000 = \$15,000 + (.40 x \$25,000) + \$0
\$25,000 = \$15,000 + \$10,000 = \$25,000
If we desired \$3,000 in profits, our calculation would be (\$15,000 + \$3,000)/.60 or \$30,000
Proof: S = FC + VC + TP
\$30,000 = \$15,000 + (.40 x \$30,000) + \$3,000
\$30,000 = \$15,000 + \$12,000 + \$3,000 = \$30,000
Break even analysis also known as cost-volume-profit analysis is very useful for setting prices of products and services, making decisions about adding product lines, expanding capacity and many other day to day financial challenges. The skillful use of this tool will significantly improve the management of your business. |
# Proportions and Ratios
### Definition of Ratio
A ratio is a relationship between two values. For instance, a ratio of 1 pencil to 3 pens would imply that there are three times as many pens as pencils. For each pencil there are 3 pens, and this is expressed in a couple ways, like this: 1:3, or as a fraction like 1/3. There do not have to be exactly 1 pencil and 3 pens, but some multiple of them. We could just as easily have 2 pencils and 6 pens, 10 pencils and 30 pens, or even half a pencil and one-and-a-half pens! In fact, that is how we will use ratios -- to represent the relationship between two numbers.
### Definition of Proportion
A proportion can be used to solve problems involving ratios. If we are told that the ratio of wheels to cars is 4:1, and that we have 12 wheels in stock at the factory, how can we find the number of cars we can equip? A simple proportion will do perfectly. We know that 4:1 is our ratio, and the number of cars that match with those 12 wheels must follow the 4:1 ratio. We can setup the problem like this, where x is our missing number of cars:
$$\frac{4}{1}=\frac{12}{x}$$
To solve a proportion like this, we will use a procedure called cross-multiplication. This process involves multiplying the two extremes and then comparing that product with the product of the means. An extreme is the first number (4), and the last number (x), and a mean is the 1 or the 12.
To multiply the extremes we just do $$4 * x = 4x$$. The product of the means is $$1 * 12 = 12$$. The process is very simple if you remember it as cross-multiplying, because you multiply diagonally across the equal sign.
You should then take the two products, 12 and 4x, and put them on opposite sides of an equation like this: $$12 = 4x$$. Solve for x by dividing each side by 4 and you discover that $$x = 3$$. Reading back over the problem we remember that x stood for the number of cars possible with 12 tires, and that is our answer.
It is possible to have many variations of proportions, and one you might see is a double-variable proportion. It looks something like this, but it easy to solve.
$$\frac{16}{x}=\frac{x}{1}$$
Using the same process as the first time, we cross multiply to get $$16 * 1 = x * x$$. That can be simplified to $$16 = x^2$$, which means x equals the square root of 16, which is 4 (or -4). You've now completed this lesson, so feel free to browse other pages of this site or search for more lessons on proportions.
## Ratios and Proportions Calculator
Use the tool below to convert between fractions and decimal, or to take a given ratio expression and solve for the unknown value. |
# GRE Algebra | Applications
Last Updated : 07 Jun, 2019
Algebra and its application cover a large chunk of the quantitative reasoning part of the GRE test paper. This is due to the fact that algebra has a good number of topics under its head. The topics which form a part of algebra as follows:
1. Expanding expressions
2. Basic equations
3. Systems of equations
5. Equations with exponents
6. Equations with fractions
7. Equations with square roots
8. Equations with absolute values
9. The coordinate plane
10. Equations of lines
12. Simplifying algebraic expressions
The application of algebra is in word problems. The most important and preliminary step of solving any word problem is in understanding the textual description and translating it into an equivalent numerical equation.
The word problems mainly can be divided into given topics:
1. Average and Mixture Word Problems
2. Distance, Rate, Time Word Problems
3. Work Word Problems
4. Word Problems involving simultaneous equations and inequalities.
A few examples to show how to translate word problems into equations:
1. The fourth power of y is added to cubes of x,
` y^4+ x^3`
2. Smita salary is x and is increased by 10% every year,
`x = x + x/10`
Some useful tricks and shortcuts which will help while solving problems in algebra:
FOIL Method:
In binomial problems like the one given below it is at times difficult to know which term to multiply with which term. F.O.I.L can be used in this case:
`(x^4 + y^3)(x^5 + y^2)`
```F: multiply the first term
O: multiply the outer term
I: multiply the inner term
L: multiply the last term ```
Examples:
Ques-1: Given two equations. If x and y satisfy the system of the equation shown, what is the value of x-y ?
```7x + 3y = 12
3x + 7y = 6 ```
Options:
```(a) 2/3
(b) 3/2
(c) 1
(d) 4
(e) 6 ```
Explanation:
Solving the above two equations,
```7x + 3y = 12 ------ (1)
3x + 7y = 6 --------(2)
Equation (1) X 3
Equation (2) X 7
Solving we get,
x = 231/70
y = 3/20
x-y = 210/140 = 3/2 ```
So, option (b) is correct.
Ques-2: A mixture of 160 gallons of wine and water contains 25% water. How much water must be added to the mixture in order to increase the percentage of water to 40% of the new mixture ?
Options:
```(a) 40 gals
(b) 50 gals
(c) 80 gals
(d) 33 gals ```
Explanation:
Initially in the 160 galloons mixture, 25% water was present which makes ratio of wine to water as 120:40 or 3:1.
Now we need to add W gallons more water so that the percentage of water in the mixture is 40%.
That means the water would become 40 + W , while the wine remains at 120.
120 gallons of wine corresponds to 60% of the mixture.
Let M be the total mixture:
```60% of M = 120
M = 200 gallons
W = 40% of 200 = 80 gallons ```
Initially, 40 gallons of water was added.
Hence more 80 – 40 = 40 gallons of water has to be added.
So, option (a) is correct.
Ques-3: The average age of a group of men is increased by 6 when a man ages 26 years is replaced by a new person of age 56 years. How many men are there in the group ?
Options:
```(a) 3
(b) 4
(c) 5
(d) 6 ```
Explanation:
Let the number of men be X and the total age of all men be T.
When a person aged 26 years is replaced by a person aged 56 years, the total age of the group goes up by 30 years.
This leads to an increase in the average age by 6 years.
Hence,
```T/(X+30) - T/X = 6
Solving for X, we get
X = 5 ```
The number of men in the group is 5.
So, option (c) is correct.
Ques-4: Nishu and Archana can do a piece of work in 10 days and Nishu alone can do it in 12 days . In how many days can Archana do it alone ?
Option:
```(a) 3 60 days
(b) 3 30 days
(c) 3 50 days
(d) 3 45 days ```
Explanation:
Nishu and Archana can do a piece of work in 10 days.
In 1 day, Nishu and Archana can do (1/10)th part of the work.
Nishu alone can do the work in 12 days.
In 1 day, Nishu can do (1/12)th part of the work.
So in 1 day, Archana can do,
`1/10 - 1/12 = (1/60) th part of the work`
So, Archana can do the work alone in 60 days.
So, option (a) is correct.
Previous
Next |
Section 8.1 - Determining Optimum Area And Perimeter
Approved & Edited by ProProfs Editorial Team
The editorial team at ProProfs Quizzes consists of a select group of subject experts, trivia writers, and quiz masters who have authored over 10,000 quizzes taken by more than 100 million users. This team includes our in-house seasoned quiz moderators and subject matter experts. Our editorial experts, spread across the world, are rigorously trained using our comprehensive guidelines to ensure that you receive the highest quality quizzes.
| By Seixeiroda
S
Seixeiroda
Community Contributor
Quizzes Created: 41 | Total Attempts: 23,734
Questions: 9 | Attempts: 518
Settings
Complete the following questions.
• 1.
Each rectangle has a perimeter of 24 units. Which one has the greatest area?
• A.
Diagram a
• B.
Diagram b
• C.
Diagram c
• D.
Diagram d
D. Diagram d
Explanation
Among the given options, diagram d has the greatest area because it has the longest side length among all the rectangles. Since the perimeter of each rectangle is the same, having a longer side length means that diagram d has a larger area.
Rate this question:
• 2.
What is the maximum area of a rectangle with a perimeter of 60km?
• A.
200 km^2
• B.
225 km^2
• C.
240 km^2
• D.
360 km^2
B. 225 km^2
Explanation
To find the maximum area of a rectangle with a given perimeter, we need to consider the case where the rectangle is a square, as a square has the maximum area for a given perimeter. In this case, the perimeter is 60km, so each side of the square would be 15km. Therefore, the area of the square is 15km multiplied by 15km, which equals 225 km^2.
Rate this question:
• 3.
Sylvia is fencing a rectangular rose garden. The hardware store sells fencing for \$22.50/m. Her family has \$250 to spend. What dimensions should Sylvia use to build a garden with the greatest area?
• A.
Length = 2.75m, width = 2.75m
• B.
Length = 9m, width = 2m
• C.
Length = 2.25m, width = 3.25m
• D.
Length = 3m, width = 3m
A. Length = 2.75m, width = 2.75m
Explanation
To maximize the area of the garden, Sylvia should use the dimensions of length = 2.75m and width = 2.75m. The area of a rectangle is calculated by multiplying its length and width together. In this case, the area would be 2.75m x 2.75m = 7.5625 square meters. By choosing these dimensions, Sylvia can achieve the greatest area possible within her budget and the available fencing materials.
Rate this question:
• 4.
Jordan is making a paddleball court. The court consists of a wall outlined by 40m of paint. What dimensions will maximize the area of the paddleball court?
• A.
Length = 20 m, width = 20 m
• B.
Length = 13.33 m, width = 13.33 m
• C.
Length = 10 m, width = 10 m
• D.
Length = 10 m, width = 20 m
D. Length = 10 m, width = 20 m
Explanation
To maximize the area of the paddleball court, the dimensions should be such that the perimeter is maximized. Since the court consists of a wall outlined by 40m of paint, the perimeter of the court should be 40m. The dimensions that satisfy this condition are a length of 10m and a width of 20m. This results in a perimeter of 40m and hence maximizes the area of the court.
Rate this question:
• 5.
Each rectangle has an area of 49 square units. Which one has the greatest perimeter?
• A.
Diagram a
• B.
Diagram b
• C.
Diagram c
• D.
Diagram d
C. Diagram c
Explanation
The area of a rectangle is determined by multiplying its length and width. Since all the rectangles have an area of 49 square units, it means that they all have the same length and width. The perimeter of a rectangle is calculated by adding up the lengths of all its sides. In this case, since all the rectangles have the same length and width, the rectangle with the greatest perimeter would be the one with the longest sides. Looking at the diagrams, it can be observed that diagram c has the longest sides compared to the other rectangles, therefore it has the greatest perimeter.
Rate this question:
• 6.
Raquel is making a quilt. She has 540cm of fabric to border the quilt. What is the greatest possible area for the quilt?
• A.
11 664 cm^2
• B.
18 225 cm^2
• C.
72 900 cm^2
• D.
291 600 cm^2
B. 18 225 cm^2
Explanation
To find the greatest possible area for the quilt, we need to determine the dimensions that will maximize the area. Since Raquel has 540cm of fabric to border the quilt, we can assume that the border will be the same width on all sides. Let's call the width of the border x cm. This means that the dimensions of the quilt will be (540-2x) cm by (540-2x) cm. The area of the quilt can be calculated by multiplying these dimensions, which gives us (540-2x)^2 cm^2. To maximize the area, we need to find the maximum value of this expression. This can be done by taking the derivative and setting it equal to zero. However, since we have a multiple-choice question, we can simply plug in the given answer options and see which one gives the maximum area. By plugging in the dimensions for each answer option, we find that the greatest possible area is 18,225 cm^2.
Rate this question:
• 7.
What is the minimum perimeter of a rectangle with an area of 625 mm^2
• A.
100 mm
• B.
125 mm
• C.
156.25 mm
• D.
312.5 mm
A. 100 mm
Explanation
The minimum perimeter of a rectangle can be achieved when the length and width of the rectangle are equal. In this case, the area of the rectangle is 625 mm^2, so both the length and width would be the square root of 625, which is 25 mm. The perimeter is calculated by adding all four sides of the rectangle, which would be 25 + 25 + 25 + 25 = 100 mm. Therefore, the minimum perimeter of a rectangle with an area of 625 mm^2 is 100 mm.
Rate this question:
• 8.
Brandon is making a rectangular pen for his pigs. The hardware store sells chicken wire for \$8.50/m. Brandon has \$75 to spend. What dimensions should he use to build a pen with the greatest area?
• A.
Length = 4.4 m, width = 4.4 m
• B.
Length = 3.1 m, width = 1.3 m
• C.
Length = 8.8 m, width = 8.8 m
• D.
Length = 2.2 m, width = 2.2 m
D. Length = 2.2 m, width = 2.2 m
Explanation
To maximize the area of the rectangular pen, Brandon should use the dimensions length = 2.2 m and width = 2.2 m. The area of a rectangle is calculated by multiplying its length and width. By using the same value for both the length and width, Brandon will create a square-shaped pen, which will have the largest area possible.
Rate this question:
• 9.
Kelly is making an area to keep her dog outside. She has 25m of fencing. The area will be against a garage as shown. What dimensions will maximize the area of the dog run?
• A.
Length = 6.25 m, width = 12.5 m
• B.
Length = 5 m, width = 5 m
• C.
Length = 8.33 m, width = 8.34 m
• D.
Length = 7.5 m, width = 10 m
A. Length = 6.25 m, width = 12.5 m
Explanation
The correct answer is length = 6.25 m, width = 12.5 m. This is because to maximize the area of the dog run, the length and width should be equal and half of the total fencing. In this case, the total fencing is 25m, so half of that is 12.5m. Therefore, the length and width should both be 12.5m/2 = 6.25m.
Rate this question:
Quiz Review Timeline +
Our quizzes are rigorously reviewed, monitored and continuously updated by our expert board to maintain accuracy, relevance, and timeliness.
• Current Version
• Mar 20, 2023
Quiz Edited by
ProProfs Editorial Team
• Dec 06, 2008
Quiz Created by
Seixeiroda
Related Topics |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# Ram walks to school from home every morning and it takes him 30 minutes. Once on his way he realized that he had forgotten his physics book at home. If he continued walking to school at the same speed, he would be there 12 minutes before the bell. But he went back home for the physics book and reached school 6 minutes after the bell. Assuming he had walked all the way with his usual speed, the fraction of the way to his school he covered till the moment he turned back is _______.
Last updated date: 13th Jun 2024
Total views: 53.7k
Views today: 0.53k
Verified
53.7k+ views
Hint: The time required to go back home and make it to school is the time of the bell plus 6 minute. The extra distance he would have travelled had he not gone back would have been total distance minus distance already covered and the time taken to travel the distance would be time of the bell minus 12 minutes.
Formula used: In this solution we will be using the following formulae;
\[v = \dfrac{d}{t}\] where \[v\] is the speed of an object, \[d\] is the distance travelled, and \[t\] is the time taken to cover such distance.
Complete Step-by-Step Solution:
Let the distance Ram travel be a distance \[x\], but the total distance be \[d\].
Now at point \[x\] if Ram had continued, he would have gotten to school 12 minute before the bell. Let the time of the bell be \[{t_1}\], then the time he would have taken to get to school will be \[{t_1} - 12\]. But the distance remaining was \[d - x\]. Hence,
\[{t_1} - 12 = \dfrac{{d - x}}{v}\] where \[v\] is the speed of his movement.
\[ \Rightarrow {t_1} = \dfrac{{d - x}}{v} + 12\]
However, after going home, he finally got to school 6 minutes after the bell, i.e. \[{t_1} + 6\]. But the total distance travelled is \[x + d\] (because he went back by \[x\] and travelled the full distance \[d\]). Hence,
\[{t_1} + 6 = \dfrac{{x + d}}{v}\]
\[ \Rightarrow {t_1} = \dfrac{{x + d}}{v} - 6\]
Hence, by equation
\[\dfrac{{d - x}}{v} + 12 = \dfrac{{x + d}}{v} - 6\]
By collecting like terms, we have
\[\dfrac{{d - x}}{v} - \dfrac{{x + d}}{v} = - 6 - 12\]
Hence, by performing subtraction on both sides, we get
\[ - \dfrac{{2x}}{v} = - 18\]
\[ \Rightarrow x = 9v\]
Since, usually it takes him 30 minutes to travel the distance \[d\], then,
\[d = 30v\]
Then the ratio of the distance travelled till he came back to the total distance is
\[\dfrac{x}{d} = \dfrac{{9v}}{{30v}} = \dfrac{3}{{10}}\]
Note: For understanding, observe how the quantity used was speed and not velocity. The velocity in general would have been inappropriate to find the distance since it neglects any interval of space travelled twice. Hence, the velocity will neglect the distance \[x\] and only consider the distance \[d - x\]. |
# How do you find a standard form equation for the line with A (-1,4); Slope: 2/5?
Jul 19, 2017
$y - \frac{2}{5} x = 4 \frac{2}{5}$ which leads to
$2 x - 5 y = - 22$
#### Explanation:
The standard form of a line is expressed in the following form
$A x + B y = C$, where $A , B \mathmr{and} C$ are integers
To find this form use the formula $\left(y - {y}_{1}\right) = m \left(x - {x}_{1}\right)$
substitute in the point and slope (m)
$\left(y - 4\right) = \frac{2}{5} \left(x - \left(- 1\right)\right) \text{ }$ or $\text{ } \left(y - 4\right) = \frac{2}{5} \left(x + 1\right)$
$y - 4 = \frac{2}{5} x + \frac{2}{5}$
$y = \frac{2}{5} x + \frac{2}{5} + 4$
$y = \frac{2}{5} x + 4 \frac{2}{5} \text{ }$ now put into standard form
Multiply by $5$ to clear the denominators
$5 y - \frac{\cancel{5} \times 2}{\cancel{5}} x = \cancel{5} \times \frac{22}{\cancel{5}}$
$5 y - 2 x = 22$
$\Rightarrow 2 x - 5 y = - 22$ |
# NCERT Solutions Class 9 Mathematics Solutions for Heron’s Formula - Exercise 12.2 in Chapter 12 - Heron’s Formula
Question 7 Heron’s Formula - Exercise 12.2
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm
and sides 6 cm each is to be made of three different shades as shown in Fig. 12.17. How much
paper of each shade has been used in it?
As the kite is in the shape of a square, its area will be
\begin{array}{l} A=(1 / 2) \times(\text { diagonal })^{2} \\ \Rightarrow \text { Area of the kite }=(3 / 2) \times 32 \times 32=512 \mathrm{cm}^{2} \end{array}
\begin{aligned} &\Rightarrow 512 / 2 \mathrm{cm}^{2}=256 \mathrm{cm}^{2}\\ &\text { So, the total area of the paper that is required in each shade }=256 \mathrm{cm}^{2} \end{aligned}
For the triangle section (III),
The sides are given as 6 cm, 6 cm and 8 cm
Now, the semi perimeter of this isosceles triangle = (6 + 6 + 8)/2 cm = 10 cm
By using Heron's formula, the area of the III triangular piece will be
= √[s (s-a) (s-b) (s-c)]
\begin{array}{l} =\sqrt{10}(10-6)(10-6)(10-8) \mathrm{cm}^{2} \\ =\sqrt{10 \times 4 \times 4 \times 2 \mathrm{cm}^{2}} \\ =8 \mathrm{v} 6 \mathrm{cm}^{2} \end{array}
Related Questions
Lido
Courses
Teachers
Book a Demo with us
Syllabus
Maths
CBSE
Maths
ICSE
Science
CBSE
Science
ICSE
English
CBSE
English
ICSE
Coding
Terms & Policies
Selina Question Bank
Maths
Physics
Biology
Allied Question Bank
Chemistry
Connect with us on social media! |
# Ordinal Numbers 1st To 10th
A vast array of sets can be listed using ordinal numbers as a tool. These numbers can be used as a tool to generalize ordinal figures.
## 1st
One of the basic concepts of math is the ordinal number. It is a number which indicates where an object is within a set of. The ordinal number is typically the number that ranges from one to twenty. Although ordinal numbers serve various purposes, they are most commonly utilized to represent the order of the items in a list.
You can use charts as well as words and numbers to represent ordinal numbers. They can also be used to indicate how a group is organised.
The majority of ordinal numbers are classified into one or one or. Transfinite ordinals can be depicted using lowercase Greek letters, while finite ordinals are represented by Arabic numbers.
Based on the axioms that govern selection, every set should include at least one ordinal or two. The first person in the class, for example, would receive the highest grade. The winner of the contest was the student with the highest score.
## Combinational ordinal figures
The ordinal numbers that are compound, meaning they could have multiple numbers, are also known as. They are generated by multiplying an ordinal number by its final digit. These numbers are most commonly utilized for ranking and dating purposes. They don’t have a distinct ending for each digit, like cardinal number.
To denote the order of elements within the collection, ordinal numbers can be used. They may be used as names of items within the collection. There are two kinds of ordinal numbers: regular and suppletive.
Regular ordinals are created by prefixing a cardinal number with the suffix -u. Then, the number has to be entered in words followed by a colon added. There are additional suffixes. The suffix “nd” is used to indicate numbers that end in two numbers. The suffix “th” can indicate numbers that end with between 4 and 9.
By affixing words with the -u, or–ie suffix creates suffixtive ordinals. The suffix, which can be used to count, is a bit longer than the standard.
## Limit of ordinal magnitude
Limit ordinal values that do not reach zero are ordinal numbers. Limit ordinal numbers come with one disadvantage: there is no maximum element that they can use. You can make them by joining sets, but without maximum element.
Transfinite recursion definitions also employ restricted ordinal numbers. According to the von Neumann model, every infinite cardinal also acts as an order limit.
A limit ordinal equals the sum of the other ordinals below. Limit ordinal number can be calculated using arithmetic or in a series natural numbers.
Data is organized by ordinal number. They are used to explain the numerical location of objects. They are often utilized in Arithmetic, set theory and in other settings. Despite being in the same category however, they are not considered to be natural numbers.
A well-ordered set is utilized to construct the von Neumann model. Assume that fy fy is one of the subfunctions of the function g’ which is described as a single function. If fy is only one subfunction (ii) the function g’ must meet the criteria.
A limit ordinal that is of the Church-Kleene kind is also referred to as the Church-Kleene ordinal. The Church-Kleene oral defines an appropriate limit as a well organized collection of smaller ordinals. It also has a nonzero ordinal.
## Stories with examples of ordinal numbers
Ordinal numbers are commonly used to show the order of entities or objects. They are critical to organize, count, as well as ranking reasons. They can also be used to indicate the order in which items are placed and also the place of objects.
The ordinal number is typically identified by the letter “th”. Sometimes though it is possible that the “nd” letter can be used in place of “th”. The titles of books often include ordinal numbers.
While ordinal numbers are typically written in lists but they are also expressed in words. They can also be found in acronyms and numbers. According to research, these numbers are more comprehensible than cardinal ones.
Ordinary numbers come in three distinct flavors. It is possible to learn more about them by engaging in games, practice, and engaging in various other pursuits. Learning about them is an important part of improving your math skills. Coloring exercises are a great easy and fun method to increase your proficiency. A simple marking sheet can be used to track your results. |
# Modulus and Conjugate of a Complex Number
View Notes
## What is a Complex Number?
In mathematics, a complex number is said to be a number which can be expressed in a + bi form, where a and b are real numbers and i is the imaginary unit.
Here, a is named as the real part of the number and b is referred to as the imaginary part of a number.
The following table provides a representation of the complex numbers.
Complex Number Standard Form (a + bi) Explanation 5i + 7 7 + 5i Real part is 7 and imaginary part is 5 2i 0 + 2i Here real part is 0 and imaginary part is 2 -3 – 5i -3 + (-5)i Here the real part is -3 and the imaginary part is -5
### What is the Modulus of Complex Number?
Modulus of complex number defined as | z | where if z = a + bi is a complex number. The modulus of the complex number will be defined as follows:
| Z | =a + bi
1. | z | =0 then it indicates a=b=0
2. | -z | = | z |
Imagine z1 and z2 are two complex numbers, then
1. | z1.z2 | = | z1 | | z2 |
2. | z1 + z2 | ≤ | z1 | + | z2 |
3. | z1/ z2 | = | z1 | / | z2 |
### Modulus of a Complex Number
There seems to be a method to get a sense of how large these numbers are. We consider the conjugate complex and multiply it by the complex number specified in (1). Therefore we describe the product $z\overline{z}$ as the square of a complex number's Absolute value or modulus. So let's write $z\overline{z}$ = |z|2.
As per the explanation, $z\overline{z}$ provides a calculation of the absolute value or magnitude of the complex number. When you learn about the Argand Plane, the exact explanation for that concept will become apparent.
Therefore, |z|2 = (a2 + b2) [Using (1)]
Hence, |z| = √(a2+b2) …(2)
The equation above is the modulus or absolute value of the complex number z.
### Conjugate of a Complex Number
The complex conjugate of a complex number is the number with the same real part and the imaginary part equal in magnitude, but are opposite in terms of their signs.
Complex conjugates are responsible for finding polynomial roots. According to the complex conjugate root theorem, if a complex number in one variable with real coefficients is a root to a polynomial, so is its conjugate.
### How to Find Conjugate of a Complex Number
If you are wondering how to find the conjugate of a complex number, then go through this. Each complex number has a relationship with another complex number known as its complex conjugate. You can find the conjugate complex by merely changing the symbol of the imaginary part of the Complex numbers.
For example:
We alter the sign of the imaginary part to find the complex conjugate of 4 + 7i. So the complex conjugate is 4 − 7i.
Example:
We alter the sign of the imaginary part to find the complex conjugate of 1−3i. So the complex conjugate is 1 + 3i.
Example:
We alter the sign of the imaginary component to find the complex conjugate of −4 − 3i. So the complex conjugate is −4 + 3i.
It is to be noted that the conjugate complex has a very peculiar property.
If we multiply a complex number by its complex conjugate, think about what will happen.
Let’s take this example:
Multiply (4 + 7i) by (4 − 7i): (4 + 7i)(4 − 7i) = 16 − 28i + 28i − 16 + 49i2 = 65
We find the answer to this is a strictly real number; there is no imaginary part. It often occurs when a complex number is multiplied by its conjugate, the consequence is a real number.
### Modulus of the Sum of Two Complex Numbers
To add two complex numbers of the x plus iy form, we have to add the real parts and the imaginary parts individually.
Let z = a + ib reflect a complex number. Module of z , referred to as z, is defined as the real number (a2 + b2)1/2 z = (a2 + b2)1/2
### Conjugate of Complex Number Class 11
Numerical: Evaluate the modulus of (3-4i)
z = (a2 + b2)1/2 = (32 + 42)1/2 = 5
Let z = a + ib reflect a complex number. Z conjugate is the complex number a - ib, i.e., = a - ib.
Z * = Z
Or Z–1 = / Z (Useful to find a complex number in reverse)
### Properties of Complex Numbers
Properties of complex numbers are mentioned below:
1. In case of a and b are real numbers and a + ib = 0 then a = 0, b = 0
Proof:
According to the property,
a + ib = 0 = 0 + i ∙ 0,
Therefore, we conclude that, x = 0 and y = 0.
2. For any three the set complex numbers z1, z2 and z3 satisfies the commutative, associative and distributive laws.
(i) z1 + z2 = z2 + z1 (Commutative law for addition)
(ii) z1 ∙ z2 = z2 ∙ z1 (Commutative law for multiplication)
(iii) (z1 + z2) + z3 = z1 + (z2 + z3) (Associative law for addition)
(iv) (z1z2)z3 = z1(z2z3) (Associative law for multiplication)
(v) z1(z1 + z3) = z1z2 + z1z3 (Distributive law)
3. The sum of two complex conjugate numbers is real.
Proof:
Let, z = a + ib (a, b are real numbers) be a complex number. Then, a conjugate of z is $\overline{z}$ = a - ib.
Now, z + $\overline{z}$ = a + ib + a - ib = 2a, which is real.
4. The product of two complex conjugate numbers is real.
Proof:
Let, z = a + ib (a, b are real numbers) be a complex number. Then, a conjugate of z is $\overline{z}$ = a - ib.
z ∙ $\overline{z}$ = (a + ib)(a - ib) = a2 - i2b2 = a2 + b2, (Since i2 = -1), which is real.
Hence, $z\overline{z}$ −−√ = a2+b2−−−−−−√
Therefore, |z| = $\overline{z}$ −−√
Thus, the modulus of any complex number is equal to the positive square root of the product of the complex number and its conjugate complex number.
5. When the sum of two complex numbers is real, and the product of two complex numbers is also natural, then the complex numbers are conjugated.
Proof:
According to the property,
z1 + z2 = a+ ib + c + id = (a + c) + i(b + d) is real.
Therefore, b + d = 0
⇒ d = -b
And,
z1z2 = (a + ib)(c + id) = (a + ib)(c +id) = (ac– bd) + i(ad + bc) is real.
Therefore, ad + bc = 0
⇒ -ab + bc = 0, (Since, d = -b)
⇒ b(c - a) = 0
⇒ c = a (Since, b ≠ 0)
Hence, z2 = c + id = a + i(-b) = a - ib = $\overline{z1}$
Therefore, we conclude that z1 and z2 are conjugate to each other.
|z1 + z2| ≤ |z1| + |z2|, for two complex numbers z1 and z2.
FAQ (Frequently Asked Questions)
Q1. What is a Complex Conjugate Number?
Ans. Complex numbers are called a complex conjugate of each other when in two complex numbers, the sign of the imaginary part is differing. Complex numbers in the binomial form are depicted as (a + ib). It invites you to work with the "+" symbol. What about if we turn it to a minus sign? Let z = a + ib represent a complex number. We describe another complex number ¯z such that ¯z = a – ib. We are calling ¯ z or even the complex number acquired by altering the symbol of the imaginary part (positive to negative or vice versa), as the conjugate of z. Let's just find the product z¯z = (a + ib)×(a – ib).
Thus, z¯z = {a2 -i(ab) + i(ab) + b2 } = (a2 + b2 ) …(1)
If a and b are big numbers, then the sum in (1) becomes more significant. And one can use this equation to determine a complex number's value. |
Education.com
Try
Brainzy
Try
Plus
# The Disc Method for Volumes of Solids for AP Calculus
based on 2 ratings
By — McGraw-Hill Professional
Updated on Oct 24, 2011
Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus
The volume of a solid of revolution using discs: Revolving about the x-axis:
See Figure 12.4-5.
Revolving about a line y = k:
Revolving about a line x = h:
See Figure 12.4-6.
### Example 1
Find the volume of the solid generated by revolving about the x-axis the region bounded by the graph of , the x-axis, and the line x = 5.
Step 1. Draw a sketch. See Figure 12.4-7.
Step 2. Determine the radius of a disc from a cross section.
Step 3. Set up an integral.
Step 4. Evaluate the integral.
Verify your result with a calculator.
### Example 2
Find the volume of the solid generated by revolving about the x -axis the region bounded by the graph of , the x-axis, and the y-axis.
Step 1. Draw a sketch. See Figure 12.4-8.
Step 2. Determine the radius from a cross section.
Step 3. Set up an integral.
Step 4. Evaluate the integral.
Thus the volume of the solid is .
Verify your result with a calculator.
### Example 3
Find the volume of the solid generated by revolving about the y-axis the region in the first quadrant bounded by the graph of y = x2, the y-axis, and the line y = 6.
Step 1. Draw a sketch. See Figure 12.4-9.
Step 2. Determine the radius from a cross section.
is the part of the curve involved in the region.
Step 3. Set up an integral.
Step 4. Evaluate the integral.
The volume of the solid is 18.
Verify your result with a calculator.
### Example 4
Using a calculator, find the volume of the solid generated by revolving about the line y = 8 the region bounded by the graph of y = x 2 + 4, the line y = 8.
Step 1. Draw a sketch. See Figure 12.4-10.
Step 2. Determine the radius from a cross section.
r = 8 - y = 8 - (x2 + 4) = 4 - x2
Step 3. Set up an integral.
To find the intersection points, set
Step 4. Evaluate the integral.
Thus the volume of the solid is .
Verify your result with a calculator.
### Example 5
Using a calculator, find the volume of the solid generated by revolving about the line y = –3 the region bounded by the graph of y = ex, the y-axis, the lines x = ln 2 and y = – 3.
Step 1. Draw a sketch. See Figure 12.4-11.
Step 2. Determine the radius from a cross section.
r = y – (–3) = y + 3 = e x +3
Step 3. Set up an integral.
Step 4. Evaluate the integral.
The volume of the solid is approximately 13.7383.
Practice problems for these concepts can be found at: Areas and Volumes Practice Problems for AP Calculus
150 Characters allowed
### Related Questions
#### Q:
See More Questions
Top Worksheet Slideshows |
Quadratic Formula - PHA Math Central
```Do Now:
WRITE THE EQUATION
FORMULA FROM
MEMORY!!!!
Take out HW-hand in
Imaginary Numbers!
We will
Investigate how to find
the equation of a
Formula WORKSHEET –
odds only!!!
Intro:
equation is:
y = ax2 + bx + c
The coefficients are: a , b, c
The
variables are: y, x
The
ROOTS (or
solutions) of a
polynomial are
its x-intercepts
The
x-intercepts
occur where y =
0.
Roots
Example: Find the
roots: y = x2 + x - 6
Solution: Factoring:
y = (x + 3)(x - 2)
0 = (x + 3)(x - 2)
The roots are:
x = -3; x = 2
Roots
After centuries of
work,
mathematicians
realized that as long
as you know the
coefficients, you can
find the roots of the
doesn’t factor!
y ax 2 bx c, a 0
b b 2 4ac
x
2a
Watch this:
eature=related
(another time…)
http://vihart.com/doodling/
Extra Credit if you create a video/song/anything
creative!! You have 2 weeks!
Solve: y = 5x 8x 3
a 5, b 8, c 3
2
8 64 60
x
10
2
b b 4ac
8
4
x
x
2a
10
2
(8) (8) 4(5)(3)
8 2
x
x
2(5)
10
8 2
x
10
8 2 10
x
1
10
10
82 6 3
x
10 10 5
Roots
2
y 5 35 8 35 3
y 5 9 25 24 5 3
45
24 3
y
2
25
5
y 5(1) 8(1) 3
9
24
15
y
y 583
5
5
5
y0
y0
Plug in your
If you’re right,
you’ll get y = 0.
2
Solve : y 2x 7x 4
DO NOW ** WRITE THE
EQUATION FOR THE
MEMORY then solve the
Class Work:
AS PROMISED….
Imaginary numbers!!
WE will:
Check out the world of
IMAGINARY NUMBERS!!!
WORKSHEET – 1-10!!!
7
49
32
Solve : y 2x 7x 4 x
4
a 2, b 7, c 4
7 81
x
2
b b 4ac
4
x
7 9 2 1
2a
x
x
4
4 2
2
(7) (7) 4(2)(4)
16
x
x
4
2(2)
4
2
Remember: All the terms must be on one
side BEFORE you use the quadratic
formula.
•Example: Solve 3m2 - 8 = 10m
•Solution: 3m2 - 10m - 8 = 0
•a = 3, b = -10, c = -8
Free Fallin’
The acceleration of
gravity (g) for objects in
free fall at the earth's
surface is 9.8 m/s2.
Galileo found that all
things fall at the same
rate.
Free Fall
The rate of falling
increases by 9.8 m/s
every second.
Height = ½ gt2
For example:
½ (9.8 )12 = 4.9 m
½(9.8)22 = 19.6 m
½ (9.8)32 = 44.1 m
½ (9.8)42 = 78.4 m
Free Fall
A ball thrown horizontally
will fall at the same rate as
a ball dropped directly.
Free Fall
A ball thrown into the air
will slow down, stop, and
then begin to fall with the
acceleration due to gravity.
When it passes the
thrower, it will be traveling
at the same rate at which it
was thrown.
Free Fall
An object thrown upward at an angle to the ground follows a curved
path called a parabola.
Air Resistance
• In air…
– A stone falls faster than a
feather
• Air resistance affects
stone less
• In a vacuum
– A stone and a feather will fall
at the same speed.
Air Resistance
• Free Fall
– A person in free fall reaches a
terminal velocity of around 54
m/s
– With a parachute, terminal
velocity is only 6.3 m/s
• Allows a safe
landing
The initial velocity is the coefficient for the middle
term, and the initial height is the constant term.
The coefficient on the leading term comes from the
force of gravity. This coefficient is negative, since
gravity pulls downward, and the value will either be
"4.9" (if your units are "meters") or "16" (if your
units are "feet"). In general, the format is:
s(t) = –gt2 + v0t + h0
...where "g" here is the "4.9" or the "16" derived
from the value of the force of gravity (technically,
it's half of the force of gravity, but you probably
don't need to know that right now), "v0" ("veenaught", or "vee-sub-zero") is the initial velocity,
and "h0" ("aitch-naught", or "aitch-sub-zero") is the
initial height.
An object is launched at 19.6 meters per second
(m/s) from a 58.8-meter tall platform. The equation
for the object's height s at time t seconds after
launch is s (t) = –4.9t2 + 19.6t + 58.8, where s is in
meters. When does the object strike the ground?
What do we know vs. What do we need to know? (3
min)
I'll set s equal to zero, and solve:
0 = –4.9t2 + 19.6t + 58.8
0 = t2 – 4t – 12
0 = (t – 6)(t + 2)
Then t = 6 or t = –2. The second solution is from two
seconds before launch, which doesn't make sense in this
context. (It makes sense on the graph, because the line
crosses the x-axis at –2, but negative time won't work in
this word problem.) So "t = –2" is an extraneous solution,
and I'll ignore it.
The object strikes the ground six seconds after launch.
An object in launched directly upward at 64 feet per
second (ft/s) from a platform 80 feet high. What
will be the object's maximum height? When will it
attain this height?
Think Pair Share (3 min)
What do I know?
What do I need to find out??
The initial height is 80 feet above ground and the
initial speed is 64 ft/s. Since my units are "feet",
then the number for gravity will be 16, and my
equation is:
s(t) = –16t2 + 64t + 80
Yes?
They want me to find the maximum height. For a
negative quadratic like this, the maximum will be at
the vertex of the upside-down parabola. So they
really want me to find the vertex. From our past
experience with graphing, I know how to find the
vertex;
in this case, the vertex is at (2, 144):
x = –b/2a = –(64)/2(–16) = –64/–32 = 2
Y = s(2) = –16(2)2 + 64(2) + 80 = –16(4) + 128 + 80
= 208 – 64 = 144
But what does this vertex tell me? According to my
equation, I'm plugging in time values and extracting
height values, so the input "2" must be the time and
the output "144" must be the height.
It takes two seconds to reach the maximum height of
144 feet.
2 4 84
x
6
2 88
x
6
2
2 4 • 22
b b 4ac
x
x
6
2a
2
(2) (2) 4(3)(7) x 2 2 22
x
6
1 22
2(3)
x
3
Solve:
3x2 = 7 - 2x
Solution: 3x2 + 2x - 7 =
0
a = 3, b = 2, c = -7
``` |
# 2.3.2 Standard Form, PT3 Focus Practice
Question 6:
(a) Round off 0.05079 correct to three significant figures.
(b) Find the value of 5 × 107 + 7.2 × 105 and state its answer in standard form.
Solution:
(a) 0.05079 = 0.0508 (correct to three significant figures)
(b) 5 × 107 + 7.2 × 105 = 500 × 105 + 7.2 × 105
= (500 + 7.2) × 105
= 507.2 × 105
= 5.072 × 107
Question 7:
(a) Calculate the value of 70.2 – 3.22 × 8.4 and round off its answer correct to three significant figures.
(b) Express $\frac{840}{0.000021}$ as a number in standard form.
Solution:
(a) 70.2 – 3.22 × 8.4 = 70.2 – 27.048
= 43.152 = 43.2 (correct to three significant figures)
(b)
$\begin{array}{l}\frac{840}{0.000021}=\frac{8.4×{10}^{2}}{2.1×{10}^{-5}}\\ \text{}=\frac{8.4}{2.1}×{10}^{2-\left(-5\right)}\\ \text{}=4×{10}^{7}\end{array}$
Question 8:
Calculate the value of 7 × (2 × 10-2)3– 4.3 × 10-5 and state its answer in standard form.
Solution:
7 × (2 × 10-2)3 – 4.3 × 10-5= 7 × 23 × (10-2)3 – 4.3 × 10-5
= 56 × 10-6 – 4.3 × 10-5
= 5.6 × 10-5 – 4.3 × 10-5
= 1.3 × 10-5
Question 9:
$3.17×{10}^{-8}-1.20×{10}^{-9}=$
Solution:
$\begin{array}{l}3.17×{10}^{-8}-1.20×{10}^{-9}\\ =3.17×{10}^{-8}-\left(0.120×{10}^{1}×{10}^{-9}\right)\\ =3.17×{10}^{-8}-0.120×{10}^{-8}\\ =\left(3.17-0.120\right)×{10}^{-8}\\ =3.05×{10}^{-8}\end{array}$
Question 10:
Solution:
$\frac{0.096}{{\left(2×{10}^{3}\right)}^{3}}=\frac{0.096}{8×{10}^{9}}=1.2×{10}^{-11}$ |
+
# Special Products of Polynomials
(2)
• (0)
• (0)
• (0)
• (1)
• (1)
Author: Abby S
##### Description:
To teach kids about solving special polynomials because they occur a lot in everyday life.
You can learn how to solve special polynomials!
(more)
Sophia’s self-paced online courses are a great way to save time and money as you earn credits eligible for transfer to over 2,000 colleges and universities.*
No credit card required
28 Sophia partners guarantee credit transfer.
281 Institutions have accepted or given pre-approval for credit transfer.
* The American Council on Education's College Credit Recommendation Service (ACE Credit®) has evaluated and recommended college credit for 25 of Sophia’s online courses. More than 2,000 colleges and universities consider ACE CREDIT recommendations in determining the applicability to their course and degree programs.
Tutorial
## Special Products of Polynomials
When you learn how to recognize the special product polynomials quickly and easily, you can solve them a lot faster
(x-2)(x+2)--- are there any patterns?
(x+5)2--- any patterns here?
In the first example, you can cancel out the last two number using the equation for a sum and difference pattern: a2 - b2
In the next two examples you can solve easily using the equation for a squared binomial pattern : a2-2ab+b2
Not, let's work out some equations:
1. Write out the sum and difference pattern: a2-b2
(x-2)(x+2)= x2-22
Then you solve the equation:
=x2-4
Now, Let's solve a squared binomial:
1. Write out the square of a binomial pattern: a2+2ab+b2
=(x+4)2
= x2+2(4)(x)+42
Then, solve the equation:
=x2+8x+16
Here is one more square of a binomial equation:
(2x-5)2
Write out the square of a binomial pattern: a2+2ab+b2
Substitute the equation numbers into the model equation:
4x2+2(2x)(-5)+25
Solve the equation:
=4x2-20x+25
As long as you follow the model equation for sum and difference patterns, a2-b2, and the model equation for square of a binomial pattern, a2+2ab+b2, it is extremely easy! |
# Find the angle to intersection of the following curves:
Question:
Find the angle to intersection of the following curves:
$x^{2}+y^{2}=2 x$ and $y^{2}=x$
Solution:
Given:
Curves $x^{2}+y^{2}=2 x \ldots(1)$
$\& y^{2}=x \ldots(2)$
Solving $(1) \&(2)$, we get
Substituting $y^{2}=x$ in $x^{2}+y^{2}=2 x$
$\Rightarrow x^{2}+x=2 x$
$\Rightarrow x^{2}-x=0$
$\Rightarrow x(x-1)=0$
$\Rightarrow x=0$ or $(x-1)=0$
$\Rightarrow x=0$ or $x=1$
Substituting $x=0$ or $x=1$ in $y^{2}=x$, we get,
when $x=0$
$\Rightarrow y^{2}=0$
$\Rightarrow y=0$
when $x=1$
$\Rightarrow y^{2}=1$
$\Rightarrow y=1$
The point of intersection of two curves are $(0,0) \&(1,1)$
Now, Differentiating curves (1) \& (2) w.r.t $x$, we get
$\Rightarrow x^{2}+y^{2}=2 x$
$\Rightarrow 2 x+2 y \cdot \frac{d y}{d x}=2$
$\Rightarrow x+y \cdot \frac{d y}{d x}=1$
$\Rightarrow y \cdot \frac{d y}{d x}=1-x$
$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{1-\mathrm{x}}{\mathrm{y}} \ldots(3)$
$\Rightarrow y^{2}=x$
$\Rightarrow 2 y \cdot \frac{d y}{d x}=1$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 y} \ldots(4)$
At $(1,1)$ in equation $(3)$, we get
$\Rightarrow \frac{d y}{d x}=\frac{1-x}{y}$
$\Rightarrow \frac{d y}{d x}=\frac{1-1}{1}$
$\Rightarrow \mathrm{m}_{1}=0$
At $(1,1)$ in equation $(4)$, we get
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 y}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2 \times 1}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{2}$
$\Rightarrow m_{2}=\frac{1}{2}$
when $m_{1}=0 \& m_{2}=\frac{1}{2}$
$\Rightarrow \tan \theta=\left|\frac{0-\frac{1}{2}}{1+0 \times \frac{1}{2}}\right|$
$\Rightarrow \tan \theta=\left|\frac{\frac{-1}{2}}{1+0}\right|$
$\Rightarrow \tan \theta=\left|\frac{-1}{2}\right|$
$\Rightarrow \tan \theta=\frac{1}{2}$
$\Rightarrow \theta=\tan ^{-1}\left(\frac{1}{2}\right)$
$\Rightarrow \theta \cong 26.56$ |
# Resposta rápida: Are Kites opposite angles congruent?
Contents
In a kite, ONE pair of opposite angles is congruent AND ONE diagonal is the perpendicular bisector of the other diagonal (diagonals are always perpendicular but only ONE diagonal is bisected by the other).
## Are the opposite angles of a kite equal?
The two interior angles of a kite that are on opposite sides of the symmetry axis are equal.
## Which angles in a kite are equal?
By definition, a kite is a polygon with four total sides (quadrilateral). The sum of the interior angles of any quadrilateral must equal: degrees degrees degrees. Additionally, kites must have two sets of equivalent adjacent sides & one set of congruent opposite angles.
## Can a kite have 3 congruent angles?
Kite properties include (1) two pairs of consecutive, congruent sides, (2) congruent non-vertex angles and (3) perpendicular diagonals. Other important polygon properties to be familiar with include trapezoid properties, parallelogram properties, rhombus properties, and rectangle and square properties.
## Are the vertex angles of a kite congruent?
The angles between the congruent sides are called vertex angles. … If we draw the diagonal through the vertex angles, we would have two congruent triangles. Theorem: The non-vertex angles of a kite are congruent.
ЭТО ИНТЕРЕСНО: Pergunta frequente: Do you pee when you skydive?
## Is every kite a rhombus?
For example, kites, parallelograms, rectangles, rhombuses, squares, and trapezoids are all quadrilaterals. Kite: A quadrilateral with two pairs of adjacent sides that are equal in length; a kite is a rhombus if all side lengths are equal.
## Can a kite have a right angle?
Thus the right kite is a convex quadrilateral and has two opposite right angles. If there are exactly two right angles, each must be between sides of different lengths. All right kites are bicentric quadrilaterals (quadrilaterals with both a circumcircle and an incircle), since all kites have an incircle.
## What does congruent mean?
Congruent means same shape and same size. So congruent has to do with comparing two figures, and equivalent means two expressions are equal. So to say two line segments are congruent relates to the measures of the two lines are equal.
## Does a rhombus have right angles?
A square has two pairs of parallel sides, four right angles, and all four sides are equal. It is also a rectangle and a parallelogram. A rhombus is defined as a parallelogram with four equal sides. … No, because a rhombus does not have to have 4 right angles.
## Why is a rectangle not a kite?
A kite and a rectangle cannot be the same at any time. The reasons are: Two pairs of adjacent sides are equal in a kite, but not so in a rectangle. Two diagonals intersect at right angles in a kite, but not so in a rectangle.
## Are all sides of a kite congruent?
Geometry For Dummies, 2nd Edition
A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means that one side can’t be used in both pairs). … The opposite angles at the endpoints of the cross diagonal are congruent (angle J and angle L).
ЭТО ИНТЕРЕСНО: Você perguntou: What happens to the kite when the wind falls?
## Are opposite angles of a trapezoid congruent?
Opposite sides of an isosceles trapezoid are the same length (congruent). The angles on either side of the bases are the same size/measure (congruent). The diagonals (not show here) are congruent.
## Does a kite diagonals bisect each other?
If two distinct pairs of consecutive sides of the quadrilateral are congruent, then it’s a kite. If one of the diagonals bisects the other diagonal at a perpendicular angle, it’s a kite.
360°
## What are the vertex angles of a kite?
The vertex angles of a kite are the angles formed by two congruent sides. The non-vertex angles are the angles formed by two sides that are not congruent. The two non-vertex angles are always congruent.
two |
Co-ordinate Geometry , Distance Formula , Section Formula , Centroid , Orthocentre , Incentre , Circumcentre of Triangle
Coordinate Geometry is the unification of algebra and geometry in which algebra is used in the study of geometrical relations and geometrical figures are represented by means of equations.
The most popular coordinate system is the rectangular cartesian system.
Coordinates of a point are the real variables associated in an order to describe its location in space.
Here we consider the space to be two-dimensional.
Through a point O, referred to as the origin, we take two mutually perpendicular lines XOX’ and YOY’ and call them x and y axes respectively.
The position of a point is completely determined with reference to these axes by means of an ordered pair of real numbers (x, y) called the coordinates of P where |x| and |y| are the distances of the point P from the y-axis and the x – axis respectively.
x is called the x-coordinate or the abscissa of P and y is called the y-coordinate or the ordinate of the point P.
Distance between two points:
Let A and B be two given points, whose coordinates are given by A(x1, y1) and B(x2, y2) respectively.
Then $\displaystyle AB = \sqrt{(x_1 – x_2)^2 + (y_1-y_2)^2}$
Section formula:
Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) internally in the given ratio λ1 : λ2 are
$\displaystyle ( \frac{\lambda_1 x_2 + \lambda_2 x_1}{\lambda_1 + \lambda_2} , \frac{\lambda_1 y_2 + \lambda_2 y_1}{\lambda_1 + \lambda_2} )$
Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2, y2) externally in the ratio of λ1 : λ2 are
$\displaystyle ( \frac{\lambda_1 x_2 – \lambda_2 x_1}{\lambda_1 – \lambda_2} , \frac{\lambda_1 y_2 – \lambda_2 y_1}{\lambda_1 – \lambda_2} )$
In both the cases,λ12 is positive.
Notes:
⋄ If the ratio, in which a given line segment is divided, is to be determined, then sometimes, for convenience (instead of taking the ratio λ1 : λ2), we take the ratio k : 1. If the value of k turns out to be positive, it is an internal division otherwise it is an external division.
⋄ The coordinates of the mid-point of the line-segment joining (x1, y1) and (x2, y2) are
$\displaystyle ( \frac{x_1 + x_2}{2} , \frac{y_1 + y_2}{2} )$
Definitions Related to Co-Ordinate Geometry
Centroid: The point of concurrency of the medians of a triangle is called the centroid of the triangle. The centroid of a triangle divides each median in the ratio 2 :1.
The coordinates are given by
$\displaystyle ( \frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3} )$
Orthocentre: The point of concurrency of the altitudes of a triangle is called the orthocentre of the triangle.
The co-ordinates of the orthocentre of the triangle A(x1, y1), B(x2, y2), C(x3,y3) are
$\displaystyle ( \frac{x_1 tanA + x_2 tanB + x_3 tanC}{tanA + tanB + tanC} , \frac{y_1 tanA + y_2 tanB + y_3 tanC}{tanA + tanB + tanC} )$
Incentre : The point of concurrency of the internal bisectors of the angles of a triangle is called the incentre of the triangle. The coordinates of the incentre are given by
$\displaystyle ( \frac{a x_1 + b x_2 + c x_3}{a + b + c} , \frac{a y_1 + b y_2 + c y_3}{a + b + c} )$
where a, b and c are length of the sides BC, CA and AB respectively
Circumcentre : The point of concurrency of the perpendicular bisectors of the sides of a triangle is called circumcentre of the triangle.
The coordinates of the circumcentre of the triangle with vertices A(x1, y1), B(x2, y2), C(x3, y3) are
$\displaystyle ( \frac{x_1 sin2A + x_2 sin2B + x_3 sin2C}{sin2A + sin2B + sin2C} , \frac{y_1 sin2A + y_2 sin2B + y_3 sin2C}{sin2A + sin2B + sin2C} )$
⋄ If the triangle is right angle then circumcentre is the mid point of the hypotenuse
Note:
⋄ Centriod G, orthocentre H and Circumcentre P of a ΔABC are collinear and G Divides HP in the ratio 2 :1. i.e. HG:GP =2:1. Also, AH = 2PD.
Example : If midpoints of the sides of a triangle are (0, 4), (6, 4) and (6, 0), then find the vertices of triangle, centroid and circumcentre of triangle.
Solution: Let points A (x1, y1), B (x2, y2) and C (x3, y3) be vertices of ΔABC.
x1 + x3 = 0 , y1 + y3 = 8
x2 + x3 = 12 , y2 + y3 = 8
x1 + x2 = 12, y1 + y2 = 0
Solving we get A (0, 0), B (12, 0) and C (0, 8)
Hence ΔABC is right angled triangle.
∠A = π/2
Circumcentre is midpoint of hypotenuse which is (6, 4) itself and centroid
$\displaystyle (\frac{x_1 + x_2 + x_3}{3} , \frac{y_1 + y_2 + y_3}{3} )$
= (4 , 8/3)
Illustration : Prove that the incentre of the triangle whose vertices are given by A(x1,y1),B(x2,y2), C(x3, y3) is
$\displaystyle ( \frac{a x_1 + b x_2 + c x_3}{a + b + c} , \frac{a y_1 + b y_2 + c y_3}{a + b + c} )$
where a, b, and c are the sides opposite to the angles A, B and C respectively.
Solution:
By geometry, we know that
$\displaystyle \frac{BD}{DC} = \frac{AB}{AC}$ (since AD bisects ∠A)
If the lengths of the sides AB, BC and AC are c, a and b respectively, then
$\displaystyle \frac{BD}{DC} = \frac{AB}{AC} = \frac{c}{b}$
Coordinates of D are
$\displaystyle (\frac{b x_2 + c x_3}{b + c} , \frac{b y_2 + c y_3}{b + c} )$
Since
$\displaystyle \frac{BD}{DC} = \frac{c}{b}$
$\displaystyle BD = \frac{a c}{b+c}$
IB bisects ∠B. Hence
$\displaystyle \frac{ID}{IA} = \frac{BD}{BA} = \frac{a}{c+b}$
Let the coordinates of I be (x , y)
Then
$\displaystyle ( x = \frac{a x_1 + b x_2 + c x_3}{a + b + c} , y = \frac{a y_1 + b y_2 + c y_3}{a + b + c} )$
(using section formula) |
# An isosceles triangle has sides A, B, and C with sides B and C being equal in length. If side A goes from (7 ,5 ) to (1 ,2 ) and the triangle's area is 36 , what are the possible coordinates of the triangle's third corner?
Nov 5, 2016
One possible coordinate would be $\left(- .8 , 13.1\right)$ and $\left(8.8 , - 6.1\right)$
#### Explanation:
One possible corner would be opposite to side length A if side A is the base of the isosceles triangle. We can find the length of A using the distance formula:
${A}^{2} = {6}^{2} + {3}^{2}$
$A = 3 \sqrt{5}$
The area of the triangle can be found using the formula:
$A r e a = \frac{1}{2} b h$ where b= base length and h= height
$36 = \left(\frac{1}{2}\right) \left(3 \sqrt{5}\right) \left(h\right)$
$h = \frac{24}{\sqrt{5}}$
Now we need to find the coordinates of a point that is $\frac{24}{\sqrt{5}}$ units away from the point (4,3.5) (which is the midpoint of the given line segment A) and perpendicular to line segment A.
The two possible points for the third corner are at $\left(- .8 , 13.1\right)$ and $\left(8.8 , - 6.1\right)$. I found this by using the distance formula and setting it equal to $\frac{24}{\sqrt{5}}$, and I also used the fact that the slope of the line perpendicular to A is -2. |
# Lesson 12Solutions to Linear Equations
Let’s think about what it means to be a solution to a linear equation with two variables in it.
### Learning Targets:
• I know that the graph of an equation is a visual representation of all the solutions to the equation.
• I understand what the solution to an equation in two variables is.
## 12.1Estimate Area
Which figure has the largest shaded region?
## 12.2Apples and Oranges
At the corner produce market, apples cost $1 each and oranges cost$2 each.
1. Find the cost of:
1. 6 apples and 3 oranges
2. 4 apples and 4 oranges
3. 5 apples and 4 oranges
4. 8 apples and 2 oranges
2. Noah has $10 to spend at the produce market. Can he buy 7 apples and 2 oranges? Explain or show your reasoning. 3. What combinations of apples and oranges can Noah buy if he spends all of his$10?
## 12.3Solutions and Everything Else
You have two numbers. If you double the first number and add it to the second number, the sum is 10.
1. Let represent the first number and let represent the second number. Write an equation showing the relationship between , , and 10.
2. Draw and label a set of - and -axes. Plot at least five points on this coordinate plane that make the statement and your equation true. What do you notice about the points you have plotted?
3. List ten points that do not make the statement true. Using a different color, plot each point in the same coordinate plane. What do you notice about these points compared to your first set of points?
## Lesson 12 Summary
Think of all the rectangles whose perimeters are 8 units. If represents the width and represents the length, then expresses the relationship between the width and length for all such rectangles.
For example, the width and length could be 1 and 3, since or the width and length could be 2.75 and 1.25, since .
We could find many other possible pairs of width and length, , that make the equation true—that is, pairs that when substituted into the equation make the left side and the right side equal.
A solution to an equation with two variables is any pair of values that make the equation true.
We can think of the pairs of numbers that are solutions of an equation as points on the coordinate plane. Here is a line created by all the points that are solutions to . Every point on the line represents a rectangle whose perimeter is 8 units. All points not on the line are not solutions to .
## Glossary Terms
solution to an equation with two variables
A solution to an equation with two variables is a pair of values of the variables that make the equation true.
For example, one possible solution to the equation is . Substituting 6 for and 0 for makes this equation true because .
## Lesson 12 Practice Problems
1. Select all of the ordered pairs that are solutions to the linear equation .
2. The graph shows a linear relationship between and .
represents the number of comic books Priya buys at the store, all at the same price, and represents the amount of money (in dollars) Priya has after buying the comic books.
1. Find and interpret the - and -intercepts of this line.
2. Find and interpret the slope of this line.
3. Find an equation for this line.
4. If Priya buys 3 comics, how much money will she have remaining?
3. Match each equation with its three solutions.
1. , ,
2. , ,
3. , ,
4. , ,
5. , ,
4. A container of fuel dispenses fuel at the rate of 5 gallons per second. If represents the amount of fuel remaining in the container, and represents the number of seconds that have passed since the fuel started dispensing, then and satisfy a linear relationship.
In the coordinate plane, will the slope of the line representing that relationship have a positive, negative, or zero slope? Explain how you know.
5. A sandwich store charges a delivery fee to bring lunch to an office building. One office pays $33 for 4 turkey sandwiches. Another office pays$61 for 8 turkey sandwiches. How much does each turkey sandwich add to the cost of the delivery? Explain how you know. |
Centre of the Circle Coincides with the Origin
We will learn how to form the equation of a circle when the centre of the circle coincides with the origin.
The equation of a circle with centre at (h, k) and radius equal to a, is (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$.
When the centre of the circle coincides with the origin i.e., h = k = 0.
Then the equation (x - h)$$^{2}$$ + (y - k)$$^{2}$$ = a$$^{2}$$ becomes x$$^{2}$$ + y$$^{2}$$ = a$$^{2}$$
Solved examples on the central form of the equation of a circle whose centre coincides with the origin:
1. Find the equation of the circle whose centre coincides with the origin and radius is √5 units.
Solution:
The equation of the circle whose centre coincides with the origin and radius is √5 units is x$$^{2}$$ + y$$^{2}$$ = (√5)$$^{2}$$
⇒ x$$^{2}$$ + y$$^{2}$$ = 5
⇒ x$$^{2}$$ + y$$^{2}$$ - 5 = 0.
2. Find the equation of the circle whose centre coincides with the origin and radius is 10 units.
Solution:
The equation of the circle whose centre coincides with the origin and radius is 10 units is x$$^{2}$$ + y$$^{2}$$ = (10)$$^{2}$$
x$$^{2}$$ + y$$^{2}$$ = 100
x$$^{2}$$ + y$$^{2}$$ - 100 = 0.
3. Find the equation of the circle whose centre coincides with the origin and radius is 2√3 units.
Solution:
The equation of the circle whose centre coincides with the origin and radius is 2√3 units is x$$^{2}$$ + y$$^{2}$$ = (2√3)$$^{2}$$
x$$^{2}$$ + y$$^{2}$$ = 12
x$$^{2}$$ + y$$^{2}$$ - 12 = 0.
4. Find the equation of the circle whose centre coincides with the origin and radius is 13 units.
Solution:
The equation of the circle whose centre coincides with the origin and radius is 13 units is x$$^{2}$$ + y$$^{2}$$ = (13)$$^{2}$$
x$$^{2}$$ + y$$^{2}$$ = 169
x$$^{2}$$ + y$$^{2}$$ - 169 = 0
5. Find the equation of the circle whose centre coincides with the origin and radius is 1 unit.
Solution:
The equation of the circle whose centre coincides with the origin and radius is 1 unit is x$$^{2}$$ + y$$^{2}$$ = (1)$$^{2}$$
x$$^{2}$$ + y$$^{2}$$ = 1
x$$^{2}$$ + y$$^{2}$$ - 1 = 0 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.