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# How to effectively solve One dimensional Motion Problem
This tour is given to give the feel of the whole chapters plus the type of questions and ways to tackle One-dimensional motion/linear motion problems. This is quite beneficial for anybody studying One-dimensional Motion.
Description:
One dimensional Motion is the study of the motion along a straight line. Complete Study material has been provided at the below link
motion in one dimension
Most Important Points to remember about One dimension Motion
1. Distance and displacement are not the same things. Distance is scalar while displacement is a vector quantity. Distance is never zero in a round trip while the displacement will be zero. Example When a person moves in a circle and return to it normal position, its displacement is zero while the distance traveled is the circumference of the circle
2. Speed is calculated over distance while velocity over displacement. So Average speed in a round trip will never be zero while average velocity will be zero
3. The magnitude of Instantaneous velocity is equal to the instantaneous speed
4. When a particle moves with constant velocity, its average velocity, its instantaneous velocity and its speed are all equal
5. Acceleration is defined as the change in velocity per unit time. A body moving with constant speed but with varying direction will have acceleration as the velocity is changing.
Most Effective way to solve a one-dimensional motion/linear motion problems
1. First, visualize the question
2. Take starting point as origin as and take one direction as positive and others as negative. This is required as we will be dealing with Vector quantities
3. Write down what is given in the question and what is required
4. If it is uniform motion, then you can utilize the One-dimensional motion equation. It the motion is varying and the relation is given for that motion. Then we can utilize one-dimensional motion derivative equation to find out the solutions
5. Calculate as require
Formula’s of linear motion problems are
v=u+at
s=ut+(1/2)at2
v2=u2+2as
$x=\int v dt$
$v=\int a dt$
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx
Example -1
A bus starts at Station A from rest with uniform acceleration 2m/sec2. The bus moves along a straight line
1. Find the distance moved by bus in 10 sec?
2. At what time, its velocity becomes 20m/sec?
3 How much time it will take to cover a distance of 1.6km
Solution:
Now first step to attempt such question is to visualize the whole process. Here the bus is moving along a straight line and with uniform acceleration
Now what we have
Initial velocity=0
Acceleration=2m/sec2
Now since it is uniform motion we can use given motion formula’s in use
v=u+at
s=ut+(1/2)at2
v2=u2+2as
1. distance (s)=?
time(t)=10 sec
So here the most suitable equation is
s=ut+(1/2)at2
Substituting given values
s=(0)10 +(1/2)2(10)2=100 m
1. velocity(v)=20 m/s
time(t)=?
So here the most suitable equation is
v=u+at
20=0+2t
or t=10 sec
3.distance(s)=1.6km=1600m
t=?
So here the most suitable equation is
s=ut+(1/2)at2
1600=(1/2)(2)t2
or t=40 sec
Example -2
A object is moving along an straight line.The motion of that object is described by
x=at+bt2+ct3
where a,b,c are constants and x is in meters and t is in sec.
1. Find the displacement at t=1 sec
2. Find the velocity at t=0 and t=1 sec
3. Find the acceleration at t=0 and t=1 sec
Solution:
Now first step to attempt such question is to visualize the whole process.Here the object is moving along a straight line and its motion is described by the given equation
Now we
x=at+bt2+ct3
Now since its motion is described by the given equation,following formula will be useful in determining the values
$x=\int v dt$
$v=\int a dt$
v=(dx/dt)
a=dv/dt=(d2x/dt2)
a=vdv/dx
1. x=? t=1sec
Here by substituting t=1 in given equation we get the answer
x=a(1)+b(1)2+c(1)3
x=a+b+c m
2.v=? t=0,v=? t=1
Here we are having the displacement equation,so first we need to find out the velocity equation
So here the most suitable formula is
v=(dx/dt)
or v=d(at+bt2+ct3)/dt
or v=a+2bt+3ct2
Substituing t=0 we get
v=a m/s
Substituing t=1 we get
v=a+2b+3c m/s
3 a=? t=0,a=? t=1
Now we are having the velocity equation,we need to first find the acceleration equation.
So here the most suitable formula is
a=(dv/dt)
or a=d(a+2bt+3ct2)/dt
a=2b+6ct
Substituting t=0 we get
a=2b m/s2
Substituting t=1 we get
a=2b+6c m/s2
Some other good questions for your Assignments
a) The displacement of the body x(in meters) varies with time t (in sec) as
x=(-2/3)t2 +16t+2
find following
a. what is the velocity at t=0,t=1
b. what is the acceleration at t=0
c. what is the displacement at t=0
d .what will the displacement when it comes to rest
e .How much time it takes to come to rest.
b). A man runs at a speed at 4 m/s to overtake a standing bus. When he is 6 m behind the door at t=0, the bus moves forward and continues with a constant acceleration of 1.2 m/s2
find the following
a. how long does it take for the man to gain the door
b if in the beginning he is 10m behind the door, will be running at the same speed ever catch up the bus?
Important Graph in One-dimensional Motion
A) Position Time Graph
This is a plot of Position and time of an object moving along the straight line. We can find out velocity, speed, distance, and displacement. We can determine acceleration sign from it but cannot determine the value of it
i) Velocity is given by the Slope of the Position time Graph. Velocity is positive if the slope is positive and Velocity is negative if the slope is negative
ii) The numerical value of the Slope without sign gives the speed of the object
iii) Displacement can be calculated using position at the time interval
iv) Displacement-time graph is same as Position Time graph.Distance time can be drawn from position-time graph by the taking the mirror of the image of the position-time graph from the point zero velocity onwards
iv) The slope of the chord drawn to displacement time or distance-time graph gives the average velocity or average speed over the time interval to which the chords corresponds to
v) If the graph is concave up, acceleration is + positive and If the graph is concave down, acceleration is -ve
B) Velocity Time graph
i) The area under the velocity-time curve provides the displacement
ii) The slope of the curve provides the acceleration
C) Acceleration -time Curve
i) The area under the curve provides the change in velocity
Related Articles
Motion in straight line practice problems for class 11
how to study physics problems
Solving physics problems
Describing Motion and Rest
Kinematics Good conceptual problems
Acceleration formula Explained with Examples
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# Factors of 832: Prime Factorization, Methods, and Examples
In this solution, we see the number 832 has some positive factors as well as negative factors but before that, we define factors, if the numbers which are completely divisible by the number for which we are evaluating the factor in place of dividend in this case 832 then the number used as the divisor is said to be a factor of 832.
### Factors of 832
Here are the factors of number 832.
Factors of 832: 1, 2, 4, 8, 13, 16, 26, 32, 52, 64, 104, 208, 416, and 832.
### Negative Factors of 832
The negative factors of 832 are similar to its positive aspects, just with a negative sign.
Negative Factors of 832: -1, -2, -4, -8, -13, -16, -26, -32, -52, -64, -104, -208, -416, and -832.
### Prime Factorization of 832
The prime factorization of 832 is the way of expressing its prime factors in the product form.
Prime Factorization: 2 × 2 × 2 × 2 × 2 × 2 × 13
In this article, we will learn about the factors of 832 and how to find them using various techniques such as upside-down division, prime factorization, and factor tree.
## What Are the Factors of 832?
The factors of 832 are 1, 2, 4, 8, 13, 16, 26, 32, 52, 64, 104, 208, 416, and 832. These numbers are the factors as they do not leave any remainder when divided by 832.
The factors of 832 are classified as prime numbers and composite numbers. The prime factors of the number 832 can be determined using the prime factorization technique.
## How To Find the Factors of 832?
You can find the factors of 832 by using the rules of divisibility. The divisibility rule states that any number, when divided by any other natural number, is said to be divisible by the number if the quotient is the whole number and the resulting remainder is zero.
To find the factors of 832, create a list containing the numbers that are exactly divisible by 832 with zero remainders. One important thing to note is that 1 and 832 are the 832’s factors as every natural number has 1 and the number itself as its factor.
1 is also called the universal factor of every number. The factors of 832 are determined as follows:
$\dfrac{832}{1} = 832$
$\dfrac{832}{2} = 416$
$\dfrac{832}{4} = 208$
$\dfrac{832}{8} = 104$
$\dfrac{832}{13} = 64$
$\dfrac{832}{16} = 52$
$\dfrac{832}{26} = 32$
$\dfrac{832}{32} = 26$
$\dfrac{832}{52} = 16$
$\dfrac{832}{64} = 13$
$\dfrac{832}{104} = 8$
$\dfrac{832}{208} = 4$
$\dfrac{832}{416} = 2$
$\dfrac{832}{832} = 1$
Therefore, 1, 2, 4, 8, 13, 16, 26, 32, 52, 64, 104, 208, 416, and 832 are the factors of 832.
### Total Number of Factors of 832
For 832, there are 14 positive factors and 14 negative ones. So in total, there are 28 factors of 832.
To find the total number of factors of the given number, follow the procedure mentioned below:
1. Find the factorization/prime factorization of the given number.
2. Demonstrate the prime factorization of the number in the form of exponent form.
3. Add 1 to each of the exponents of the prime factor.
4. Now, multiply the resulting exponents together. This obtained product is equivalent to the total number of factors of the given number.
By following this procedure, the total number of factors of 832 is given as:
Factorization of 832 is 2×2×2×2×2×2×13
The exponent of 2 and 13 is 6,1.
Adding 1 to each and multiplying them together results in 14.
Therefore, the total number of factors of 832 is 28. Fourteen are positive, and fourteen factors are negative.
### Important Notes
Here are some essential points that must be considered while finding the factors of any given number:
• The factor of any given number must be a whole number.
• The factors of the number cannot be in the form of decimals or fractions.
• Factors can be positive as well as negative.
• Negative factors are the additive inverse of the positive factors of a given number.
• The factor of a number cannot be greater than that number.
• Every even number has 2 as its prime factor, the smallest prime factor.
## Factors of 832 by Prime Factorization
The number 832 is a composite. Prime factorization is a valuable technique for finding the number’s prime factors and expressing the number as the product of its prime factors.
Before finding the factors of 832 using prime factorization, let us find out what prime factors are. Prime factors are the factors of any given number that are only divisible by 1 and themselves.
To start the prime factorization of 832, start dividing by its most minor prime factor. First, determine that the given number is either even or odd. If it is an even number, then 2 will be the smallest prime factor.
Continue splitting the quotient obtained until 1 is received as the quotient. The prime factorization of 832 can be expressed as:
832 = 2 x 2 x 2 x 2 x 2 x 2 x 13
## Factors of 832 in Pairs
The factor pairs are the duplet of numbers that, when multiplied together, result in the factorized number. Factor pairs can be more than one depending on the total number of factors given.
For 832, the factor pairs can be found as:
1 x 832 = 832
2 x 416 = 832
4 x 208 = 832
8 x 104 = 832
13 x 64 = 832
16 x 52 = 832
26 x 32 = 832
The possible factor pairs of 832 are given as (1, 832), (2, 416), (4, 208), (8, 104), (13, 64), (16, 52) and (26, 32).
All these numbers in pairs, when multiplied, give 832 as the product.
The negative factor pairs of 832 are given as:
-1 x -832 = 832
-2 x -416 = 832
-4 x -208 = 832
-8 x -104 = 832
-13 x -64 = 832
-16 x -52 = 832
-26 x -32 = 832
It is important to note that in negative factor pairs, the minus sign has been multiplied by the minus sign, due to which the resulting product is the original positive number. Therefore, 1, 2, 4, 8, 13, 16, 26, 32, 52, 64, 104, 208, 416, 832 are called negative factors of 832.
The list of all the factors of 832, including positive as well as negative numbers, is given below.
Factor list of 832: 1, 2, 4, 8, 13, 16, 26, 32, 52, 64, 104, 208, 416, 832 and -832
## Factors of 832 Solved Examples
To better understand the concept of factors, let’s solve some examples.
### Example 1
How many factors of 832 are there?
### Solution
The total number of Factors of 832 is 14.
Factors of 832 are 1, 2, 4, 8, 13, 16, 26, 32, 52, 64, 104, 208, 416, 832.
### Example 2
Find the factors of 832 using prime factorization.
### Solution
The prime factorization of 832 is given as:
832 $\div$ 2 = 416
416 $\div$ 2 = 208
208 $\div$ 2 = 104
104 $\div$ 2 = 52
52 $\div$ 2 = 26
26 $\div$ 2 = 13
13 $\div$ 13 = 1
So the prime factorization of 832 can be written as:
2 × 2 × 2 × 2 × 2 × 2 × 13 = 832
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Practicing with sets and / or intervals
Do you want to let your students practice with sets and / or intervals? This article explains how this can be done within Grasple.
Written by Eric Bouwers
Updated over a week ago
Using the capabilities of the Computer Algebra System (CAS) behind Grasple it is possible to create exercises that allow your students to practice with sets and / or intervals. Below we explain all ins-and-outs of the support of these constructs within the platform.
# Set syntax and operators
Sets are supported in the platform via the use of the bracket notation, i.e. {1, 2} denotes the set of integers 1 and 2. These expressions can directly be used within an answer box. The following operators are currently supported for sets:
As usual, the algebraic equality will evaluate whether two expressions are the same. This means that the following expressions are considered to be equal:
Combining operators
The precedence of the union, intersection, and relative complement operators between sets is not clearly defined. Because of this, the system raises a semantic error when you combine different operators on a single level. So be sure to be explicit, for example:
Note that expressions such are {1} ∪ {3} ∪ {2} are not problematic and are considered to be valid.
## Complement and sets
There are no assumptions made within Grasple what the universal set is when a complement of a set is given (e.g. "{1,2}^c").
As a consequence a complement of a set (e.g. "{1}^c") will not be algebraically equivalent to an expression assuming the universal set are all real numbers (e.g. "(-infinity, 1) union (1, infinity)").
Interval expressions do assume that the universal set are all real numbers. See explanation about interval expressions further down in this article.
## Sets versus lists
Note that we only support sets with curly braces, brackets such as [1,2,3] are seen as a list and thus not equal to a set. Other expressions which are not equal are:
# Interval syntax
The platform supports all three forms of interval notation:
Naturally, unbounded intervals can also be defined as (-∞,∞), but note that closed, unbounded intervals such as [2, ∞] are considered to be semantically incorrect and will not be parsed as a valid expression.
## Interval definition
In contrast to sets these expressions cannot be entered in an answer box directly. This is because the expression [0,1] is ambiguous, it can either be a closed interval or a list with two entries.
Because of this, all interval expressions should always be defined as a parameter of type interval. Writing an interval expression directly in an answer field leads to questions that cannot be answered.
To do this, first make sure you enabled template generation in your exercise via the 'more' menu:
Then you can create a new parameter:
After naming the parameter, you can select the interval parameter type:
After choosing the correct type you can define your interval expression, and use the parameter as you would normally do:
## Interval operators
Set operators also work on intervals (and you can even combine them in a single expression). This means that the following expressions are equal:
## Complement and intervals
Interval expressions assume that the universal set are all real numbers. This means that for example "(-infinity, 2)^c" is algebraically equivalent to "[2, infinity)".
# Combining sets and intervals
If desired, sets and intervals can be combined. For example, the following equation holds:
Hopefully this support allows you to create meaningful exercises for your students. As always, if you have any questions please reach out to us via the chat icon in the bottom right.
Complement for a combination of intervals and sets
When an expression combines sets and intervals, no assumption is made about what the universal set.
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### Mathematics Class X
Real Numbers
Polynomials
Arithmetic Progressions
Triangles
Coordinate Geometry
Introduction to Trigonometry
Circles
Constructions
Areas Related to Circles
Surface Areas and Volumes
Statistics
Probability
# Pythagoras Theorem (Baudhayan Theorem)
Pythagoras theorem says that in a right angle Triangle, the square of the hypotenuse i.e. the side opposite to the right angle is equal to the sum of the square of the other two sides of the Triangle.
If one angle is, then
### Proof of Pythagoras Theorem
Statement: As per Pythagoras theorem“In a right-angled triangle, the sum of squares of two sides of a right triangle is equal to the square of the hypotenuse of the triangle.”
Proof –
Consider the right triangle, right-angled at B.
Construction-
Draw BD AC
Now,
Also,
So, CD/BC = BC/AC
or, CD. AC = ……………(ii)
1. AC + CD. AC =
AC(AC) =
=
Hence, proved.
Example
In the given right angle Triangle, Find the hypotenuse.
Solution
AB and BC are the two sides of the right angle Triangle.
BC = 12 cm and AB = 5 cm
From Pythagoras Theorem, we have:
= 25+144
So, = 169
AC = 13 cm
Converse of Pythagoras Theorem
In a Triangle, if the sum of the square of the two sides is equal to the square of the third side then the given Triangle is a right angle Triangle.
If then one angle is.
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# NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots InText Questions
These NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots InText Questions and Answers are prepared by our highly skilled subject experts.
## NCERT Solutions for Class 8 Maths Chapter 7 Cube and Cube Roots InText Questions
Try These (Page No. 111)
Question 1.
Find the one’s digit of the cube of each of the following numbers.
(i) 3331
(ii) 8888
(iii) 149
(iv) 1005
(v) 1024
(vi) 77
(vii) 5022
(viii) 53
Solution:
(i) 3331
unit digit of the number = 1
unit digit of the cube of the number = 1
(ii) 8888
unit digit of the number = 8
unit digit of the cube of the number = 2
(iii) 149
unit digit of the number = 9
unit digit of the cube of the number = 9
(iv) 1005
unit digit of the number = 5
unit digit of the cube of the number = 5
(v) 1024
unit digit of the number = 4
unit digit of the cube of the number = 4
(vi) 77
unit digit of the number = 7
unit digit of the cube of the number = 3
(vii) 5022
unit digit of the number = 2
unit digit of the cube of the number = 8
(viii) 53
unit digit of the number = 3
unit digit of the cube of the number = 7
Question 2.
Express the following numbers as the sum of odd numbers using the above pattern?
(a) 63
(b) 83
(c) 73
Solution:
(a) n = 6 and (n – 1) = 5
We start with (6 × 5) + 1 = 31
We have
63 = 31 + 33 + 35 + 37 + 39 + 41 = 216
(b) n = 8 and (n – 1) = 7
We start with (8 × 7) + 1 = 57
We have
83 = 57 + 59 + 61 + 63 + 65 + 67 + 69 + 71 = 512
(c) n = 7 and (n – 1) = 6
We start with (7 × 6) + 1 = 43
We have
73 = 43 + 45 + 47 + 49 + 51 + 53 + 53 + 55 = 343
Question 3.
Consider the following pattern.
23 – 13 = 1 + 2 × 1 × 3
33 – 23 = 1 + 3 × 2 × 3
43 – 33 = 1 + 4 × 3 × 3
Using the above pattern, find the value of the following.
(i) 73 – 63
(ii) 123 – 113
(iii) 203 – 193
(iv) 513 – 503
Solution:
(i) 73 – 63 = 1 + 7 × 6 × 3 = 1 + 126 = 127
(ii) 123 – 113 = 1 + 12 × 11 × 3 = 1 + 396 = 397
(iii) 203 – 193 = 1 + 20 × 19 × 3 = 1 + 1140 = 1141
(iv) 513 – 503 = 1 + 51 × 50 × 3 = 1 + 7650 = 7651
Note: In the prime factorisation of any number, if each other appears three times, then the number is a perfect cube.
Try These (Page No. 112)
Question 4.
Which of the following are perfect cubes?
1. 400
2. 3375
3. 8000
4. 15625
5. 9000
6. 6859
7. 2025
8. 10648
Solution:
1. We have
400 = 2 × 2 × 2 × 2 × 5 × 5
2 × 5 × 5 remain after grouping in triples.
400 is not a perfect cube.
2. We have
3375 = 3 × 3 × 3 × 5 × 5 × 5
The prime factors appear in triples.
3375 is a perfect cube.
3. We have
8000 = 2 × 2 × 2 × 2 × 2 × 2 × 5 × 5 × 5
The prime factor of 8000 can be grouped into triples and no factor is left over.
8000 is a perfect cube.
4. We have
15625 = 5 × 5 × 5 × 5 × 5 × 5
The Prime factors of 15625 can be grouped into triples and no factor is left over.
15625 is a perfect cube.
5. We have
9000 = 2 × 2 × 2 × 3 × 3 × 5 × 5 × 5
The prime factors of 9000 cannot be grouped into tripes (because 3 × 3 are leftover).
9000 is not a perfect cube.
6. We have
6859 = 19 × 19 × 19
The prime factors of 6859 can be grouped into triples and no factor is left over.
6859 is a perfect cube.
7. We have
2025 = 3 × 3 × 3 × 3 × 5 × 5
We do not get triples of prime factors of 2025 and 3 × 5 × 5.
2025 is not a perfect cube.
8. We have
10648 = 2 × 2 × 2 × 11 × 11 × 11
The prime factor of 10648 can be grouped into triples and no factor is left over.
10648 is a perfect cube.
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# How do you solve 4x - 9 = 7x + 12 ?
Jan 7, 2017
You move the terms around to get the answer.
#### Explanation:
Let's take another look at the problem:
$4 x - 9 = 7 x + 12$
Using the Subtraction Property of Equality, we can subtract $4 x$ from both sides to turn the equation to this:
$- 9 = 3 x + 12$
And then we can use the Subtraction Property of Equality again by subtracting 12 from both sides:
$3 x = - 21$
Then, we can use the Division Property of Equality to divide 3 from both sides, leaving the answer:
$x = - 7$
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Poker Math Basics - Build the Right Fundamentals - sportscasinobetting
You may have heard about calculating your expected value in poker. But how do you use this information in your betting and post-game analysis? This article will show you how to apply the basic mathematical principles of poker to your playing. You will also learn how to calculate your pot odds. Having this information can help you make better decisions, so take advantage of it. Here’s how:
### Calculate your expected value
If you’ve ever played poker, you’ve probably heard about the concept of expected value. This mathematical concept is helpful in many situations, including financial decisions, such as deciding how much to bet. But what exactly is expected value and how do you calculate it? Fortunately, there is a simple calculator you can use to determine your expected value. All you need to do is type in three values and it will tell you what the outcome is likely to be.
EV can also be used to determine the value of bets in head-to-head poker or for drawing hands. You can calculate it by inserting values into brackets based on your expected gain or loss and multiplying the result by the frequency. The formula is straightforward and will help you make more profitable decisions. If you’re struggling with the math behind EV, consider learning some basic poker math. Here are some examples.
Unlike pot odds, expected value cannot help you decide which decisions are profitable in the long run. However, it can help you analyze your previous games and determine what made you a winning or losing play. This formula is also useful in post-game analysis. Using it helps you understand why you made certain moves and missed out on a great opportunity. It will also help you evaluate why you made the best or worst decisions.
When it comes to calculating your EV, it’s helpful to understand the probability of each outcome. In a 100-\$100 pot, for example, the EV of a flush draw is 4/1, while a pair of kings would yield a 5% EV. This calculation is also applicable to flops, which are more difficult to predict. If your opponent’s hand is a flush draw, your EV should be higher than theirs, as the odds are more favorable.
### Calculate your pot odds
The key to beating your opponents is to use your knowledge of poker math to your advantage. Knowing the pot odds vs. winning odds will help you make better decisions at the table. Pot odds are always important, even in low-stakes games. When it comes to cash games, you should always look at the pot odds vs. winning odds of a hand before you commit to a bet. Although poker math can be daunting, it will become second nature after a while. Learn how to calculate your pot odds with poker math and you will be a winning player at any table.
Pot odds are commonly expressed as a percentage. You can convert them to a fraction for easier comparison. Alternatively, you can use the Rule of Two to convert pot odds into a percentage. In either case, multiply the number of outs by the percentage of equity to obtain your odds. It is best to memorize the average frequency of draws. But even if you don’t know the exact numbers in-game, you can calculate your pot odds using poker math to make better decisions.
In poker, pot odds are an accurate way to gauge how many hands are in the pot. These odds are a ratio of your chances of winning a hand to the amount you bet. In poker, you can use this ratio to help you decide how much to bet, based on how much your opponent has bet. Remember, you should never bet more than you can afford to lose, and the best way to do this is to memorize the table or save it to your computer.
Another way to measure pot odds is by the size of the final pot. For example, if you bet \$150, and your opponent bets \$76 in a hand, the pot will be worth \$200. In this example, you’d call if the pot was worth only \$50. Therefore, the pot odds would be 50/50. Using a ratio like this will help you visualize how much you would win or lose if you draw.
### Calculate your expected value in post-game analysis
The expected value of a game can be calculated using a formula called EPA. EPA is a statistical formula which uses the probability of a specific outcome to determine its expected value. For example, \$1 spent on a particular game will usually return 83 cents. In the same way, an average grade earned by a teacher will earn an expected value of 1.2. However, all yards are not created equal.
Unlike the expected value, which is often referred to as the “long-term average” of a given game, the expected value of a particular outcome is not necessarily what you’re expecting to win. Instead, it is a statistically-proven average of what you expect to win over a long period of time. That is because it is based on repeated experiments, which are the basis for the average result.
### Calculate your expected value in betting
Whether you’re in the game of chance or the world of statistics, calculating your expected value is crucial to making the most informed decisions. In poker, this is called the “pot odds” because the winning hand usually has a greater payout. Calculating your expected value involves comparing the winning odds to the overall pot odds. By comparing the expected value of a hand with the pot odds, you can determine whether you’re winning or losing.
The EV of a hand is a mathematical equation that can be applied to any decision in a poker game. It can help you determine the probability of winning or losing a hand based on the outcome of other bets. EV is also useful in determining the optimal strategy for any given hand. If you use this formula correctly, you’ll be able to increase your odds of winning significantly and make smarter decisions in the long run.
The formula for calculating your expected value in poker betting is pretty simple, but you can complicate things to make it more complicated. Fortunately, calculating your EV in a single hand is the same, as it can be more difficult to calculate in a multi-table game. The more hands you play, the more money you’ll win. To make the most money in a poker game, you should play the best hands with the best odds.
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# Integrate the function$\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$
Toolbox:
• $=\int \sqrt{x+a}=\frac{(x+a)^{3/2}}{3/2}=\frac{2}{3}(x+a)^{3/2}$
Given:$\frac{1}{\sqrt{x+a}+\sqrt{x+b}}$
Multiply and divide by its conjujate $\sqrt {x+a}-\sqrt {x+b}$
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{(\sqrt {x+a}+\sqrt {x+b})(\sqrt {x+a}-\sqrt {x+b})}dx$
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{(\sqrt {x+a})^2-(\sqrt {x+b})^2}dx$
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{(x+a-x-b)}dx$
$=\large \int \frac{\sqrt {x+a}-\sqrt {x+b}}{a-b}dx$
On seperating the terms:
$=\frac{1}{a-b}\bigg[\int \sqrt{x+a}\;dx-\int \sqrt {x+b}\;dx\bigg]$
On integreating we get,
$\frac{1}{(a-b)}\bigg\{\bigg[\frac{(x+a)^{3/2}}{3/2}\bigg]-\bigg[\frac{(x+b)^{3/2}}{3/2}\bigg]\bigg\}$
$=\frac{1}{(a-b)}\frac{2}{3}\bigg[(x+a)^{3/2}-(x+b)^{3/2}\bigg]$
$I=\frac{2}{3(a-b)}\bigg[(x+a)^{3/2}-(x+b)^{3/2}\bigg]$
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# Question #3a38f
##### 1 Answer
Apr 2, 2016
${f}^{- 1} \left(x\right) = 2 x - 1$
#### Explanation:
${f}^{- 1} \left(x\right)$ is the inverse function of $f \left(x\right)$. All that means is an input for $f \left(x\right)$ equals an output for ${f}^{- 1} \left(x\right)$. For example, the functions $f \left(x\right) = - x$ and ${f}^{- 1} \left(x\right) = x$ are inverses. $f \left(1\right) = - 1$ and ${f}^{- 1} \left(- 1\right) = 1$. An inverse function generates the input for the original function.
You can find inverses in 4 simple steps.
Step 1: Change to $x$ and $y$ Notation
All this means is replace $f \left(x\right)$ with $y$:
$y = \frac{x + 1}{2}$
Step 2: Swap $x$ and $y$
$x$ becomes $y$ and $y$ becomes $x$:
$x = \frac{y + 1}{2}$
Step 3: Solve for $y$
We have $x = \frac{y + 1}{2}$. Solving for $y$ is just a matter of algebra:
$x = \frac{y + 1}{2}$
$2 x = y + 1$
$2 x - 1 = y$
Step 4: Replace $y$ with ${f}^{- 1} \left(x\right)$
We change back to $f \left(x\right)$ notation in this step, adding a $- 1$ to say it's an inverse:
${f}^{- 1} \left(x\right) = 2 x - 1$
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Home Uncategorized Factors That Holds Prime Value in Mathematics
# Factors That Holds Prime Value in Mathematics
A prime number is a number that can only be divided by itself and 1 without any other value being left over.
Prime numbers have been known about since ancient times as they form an important part of the way we create our security codes today. The first time a “prime” number was discovered it was thought to be significant as it was connected with our physical world as opposed to anything else that might exist. This means that one cannot predict what a prime number will look like based on its previous factors – i.e., one cannot predict its square root, or how many digits it has, etc.
The first few prime numbers are: 2, 3, 5, 7 9 11 13 17 19 23 etc…
So what’s a factorization? prime factorization for short, means breaking a number down into its smaller “factors”, also known as divisors. For example: 12 = 2 x 2 x 3 which can be rewritten as 12 = 2 x 2 x 3 = 23.
In other words, we broke down 12 into factors (2x2x3) and then rearranged these factors into a new product (23). One can do this with any whole number, i.e., work out the prime factors of a number and then put them together as a new product.
A multiple is a factor with more than two numbers, whereas a factor has only one or two numbers being multiplied together.
An example of this would be 12 = 2 x 2 x 3 which is an example of three different multiples as there are three numbers in total being multiplied together to give us 12. Or we could say 12 = 4×3 as there are only two numbers (4 and 3) multiplied together to give us 12.
Factors and multiples are also closely related to the idea of primes as every number has a finite amount of factors and multiples. For example: 30 is an odd number, as it cannot be expressed as 2 x 15 (exactly). It can, however be expressed as 5 x 6 or 3 x 10 , etc.
In general terms, one should have noticed that 1 is not a prime number – this might seem obvious, but sometimes people will talk about ‘the’ prime number rather than ‘a’ prime number. We know 1 isn’t a prime because it divides by itself – i.e. 1 x 1 = 1. It can also be expressed as 12, or 123 or any other combination that means the same thing!
However, some special types of numbers are called ‘prime numbers’ but aren’t actually prime when you find out their real factorization…
An example of this is 6. 6 is a composite number – not a prime number – because it can be written in more than one way. We call these types of numbers composites rather than primes because it turns out they’re actually made up of lots of smaller numbers working together to give one big result, whereas primes can only be divided by themselves and 1.
The number 2 is the smallest prime number, and it is also the only even prime number. Every other prime number is odd, which means that when one is multiplying two integers together to get another integer, when at least one of them (and usually both) are divisible by 2, then when you divide that product by 2 you will end up with a remainder of 0 . In other words, if one multiplies 3 and 5 for 15 and then divides it by 2 , the remainder will be 0 because both numbers were divisible by two from being factors of 2.
In factoring integers using prime factorization, all quantities except for one will be prime factors.
For example: 25 = 5*5 or 52 = 7*7
This would be if the largest divisible number was the same as the smallest prime factor, which is known as a square number because its multiples are squared. For example, 49 is divisible by 7 because both 49 and 49*49 have a remainder of 0 when divided by 7 . By this logic, all numbers that fit into the category of being a perfect square would have a prime factorization of n*n where n is an integer.
For example: 25 = 5*5 or 52 = 7*7
Some examples of perfect squares: 16 , 36 , 100 , 216
There are the greatest common factor available alongside LCM and GCF. One can know more about prime numbers and common factors; they have to access Cuemath website.
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# Complementary and supplementary angles | Types of Angle Pairs | geometry
In this section we discuss about different types of angles pairs like Complementary Angles, Supplementary Angles, Conjugate Angles & Congruent angles with examples
### Angle Pairs Definition and Examples | Conjugate and Congruent Angles
For Basic concepts of angles and Different Types of Angles in Geometry like Zero Angle, Acute Angle, Right Angle, Obtuse angle, Straight Angle, Reflex Angle & Complete angle go through the below link
### Complementary Angles
If the sum of two angles are 90o then the angles are said to be Complementary angles. Each one of these angles is called the Complementary of the other.
Example:
From the above example ∠POR = 50o , ∠ROQ = 40 are complementary angles
Since sum of the these two angles are 90o
i.e ∠POR + ∠ROQ = 50o + 40o = 90o
Here ∠POR is said to be complementary angle of ∠ROQ and ∠ROQ is said to be complementary angle of ∠POR.
How to find complementary angles
If any angle of ‘ y ‘ is less than 90o then
Complementary angle of y = 90o – yo
### Supplementary angles
If the sum of two angles are 180o then the angles are said to be supplementary angles, . Each one of these angles is called the supplementary of the other.
Example:
From the above example ∠POR = 50o , ∠ROQ = 130 are supplementary angles
Since sum of the these two angles are 180o
i.e ∠POR + ∠ROQ = 50o + 130o = 180o
Here ∠POR is said to be supplementary angle of ∠ROQ and ∠ROQ is said to be supplementary angle of ∠POR.
How to find supplementary angles
If any angle of Y is less than 180o then
Supplementary angle of y = 180o – yo
### Conjugate Angles:
If the sum of two angles are 360 then the angles are said to be Conjugate angles, . Each one of these angles is called the conjugate of the other.
Example:
From the above example ∠POR = 50o , ∠ROQ = 310 are conjugate angles
Since sum of the these two angles are 360o
i.e ∠POR + ∠ROQ = 50o + 310o = 310o
Here ∠POR is said to be conjugate angle of ∠ROQ and ∠ROQ is said to be conjugate angle of ∠POR.
How to find conjugate angles
If any angle of ‘y ‘ is less than 360o then
Conjugate angle of y = 360o – yo
### Congruent angles:
Two angles having the same measure are known as congruent angle.
In the above figure ∠AOB & ∠POQ are congruent angles.
Since ∠AOB = ∠POQ = 60o
### Angular bisector:
A ray which divides an angle into two congruent angles is called angular bisector.
Example:
In the above figure ray OR is called angular bisector of ∠POQ.
Since ∠POR = ∠ROQ = 30o
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# Central Limit Theorem
Updated: Jan 16, 2023
## Introduction: What is the Central Limit Theorem?
The Central Limit Theorem (CLT) is a statistical theorem that states that for a large enough sample size, the distribution of the sample mean will be approximately normally distributed, regardless of the underlying distribution of the population from which the sample is drawn.
This means that as the sample size increases, the sample mean will become more and more like a normal distribution, even if the population itself is not normally distributed.
The theorem is important because it allows statisticians to make predictions and inferences about a population based on a sample, even if the population is not normally distributed.
This is useful in many fields, such as finance, engineering, and social sciences, where data is often not normally distributed.
The Central Limit Theorem has some assumptions, such as the sample size should be large enough and the samples should be independent and identically distributed, also known as i.i.d assumption.
In short, the Central Limit Theorem is a fundamental principle of statistics that allows analysts to make predictions and inferences about a population based on a sample, even if the population is not normally distributed, as long as the sample size is large enough and the samples are independent and identically distributed.
## Central Limit Theorem Explained with Example
An example of the Central Limit Theorem (CLT) in action is as follows:
Suppose we have a population of 1000 randomly generated numbers, which are uniformly distributed between 0 and 1. We want to know what the mean and standard deviation of this population is. We can't directly calculate the mean and standard deviation of the population because we don't have all the data, so we take a sample of size 30 and calculate the mean and standard deviation of that sample. We repeat this process 10,000 times and plot the distribution of the sample means.
As we repeat the process, we would expect to see that the sample mean is approximately normally distributed, with a mean equal to the population mean and a standard deviation equal to the population standard deviation divided by the square root of the sample size. Furthermore, as we increase the sample size, the distribution of the sample mean will become more and more like a normal distribution, even though the population itself is not normally distributed.
This is an example of how the Central Limit Theorem allows us to make predictions and inferences about a population based on a sample, even if the population is not normally distributed, as long as the sample size is large enough.
It's worth noting that in practice, the population parameters (mean, standard deviation) are often unknown and need to be estimated from the sample.
## Central Limit Theorem Formula?
The Central Limit Theorem (CLT) does not have a specific formula as it is a statistical theorem that describes the behavior of a sample mean as the sample size increases.
However, the CLT can be used to calculate the standard deviation of the sample mean, also known as the standard error of the mean, which is given by the formula:
Standard error of the mean = (standard deviation of population) / (square root of sample size)
This formula assumes that the population standard deviation is known, but in practice it is often unknown and needs to be estimated from the sample. In this case, the sample standard deviation is used as an estimate.
Additionally, it is important to note that the CLT only holds for large sample sizes, typically a sample size of 30 or more is considered large enough for the CLT to be applicable.
The Central Limit Theorem is a theoretical concept that describes the behavior of a sample mean, it does not have a specific formula, but it can be used to calculate the standard error of the mean, which is a useful tool for making predictions and inferences about a population based on a sample.
## Central Limit Theorem Proof
The Central Limit Theorem (CLT) is a fundamental principle of statistics that states that for a large enough sample size, the distribution of the sample mean will be approximately normally distributed, regardless of the underlying distribution of the population from which the sample is drawn.
The proof of the CLT is based on mathematical concepts such as probability theory, characteristic functions and moment-generating functions.
One way to prove the CLT is to use the characteristic function, which is a function that describes the probability distribution of a random variable. The characteristic function of the sample mean can be shown to be the nth power of the characteristic function of the population, where n is the sample size. If the population has a finite mean and a finite variance, the characteristic function of the sample mean will converge to a normal distribution as the sample size increases.
Another way to prove the CLT is by using the moment-generating function, which is a function that describes the expected value of the random variable raised to a power. The moment-generating function of the sample mean can be shown to be the nth power of the moment-generating function of the population, where n is the sample size. If the population has a finite mean and a finite variance, the moment-generating function of the sample mean will converge to a normal distribution as the sample size increases.
The proofs of the CLT are based on mathematical concepts, they are quite technical and require knowledge of advanced mathematical concepts such as characteristic function and moment-generating function.
In summary, The Central Limit Theorem states that as the sample size increases, the sample mean will become more and more like a normal distribution, even if the population itself is not normally distributed. The theorem is important because it allows statisticians to make predictions and inferences about a population based on a sample, even if the population is not normally distributed. The proof of the theorem relies on the properties of characteristic functions and moment-generating functions, which states that if the population has a finite mean and a finite variance, the characteristic function of the sample mean will converge to a normal distribution as the sample size increases.
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# Introduction to Complex Numbers: Navigating the Realm Beyond the Real
In mathematics, numbers have always been our trusty companions, helping us quantify and understand the world around us. While we are familiar with the ordinary numbers that measure, count, and calculate our daily lives, there exists a more mysterious and intriguing family of numbers: the complex numbers. Delving beyond the real number line, we enter a dimension where numbers have both magnitude and direction, real and imaginary components. This guide will escort you through the fascinating journey of complex numbers, from their inception to their multifaceted applications.
## Step-by-step Guide to an Introduction to Complex Numbers
Here is a step-by-step guide to an introduction to complex numbers:
### Step 1: Grounding in Real Numbers
• Understanding Real Numbers: Real numbers are all the numbers you’ve typically dealt with since early education. They can be positive, negative, zero, fractions, and irrational numbers (like $$π$$).
### Step 2: The Dilemma of the Square Root of Negative One
• Positive Squares: Any non-negative real number (e.g., $$4$$) has a real square root (e.g., $$2$$).
• Negative Squares Problem: No real number multiplied by itself gives a negative result. This means that the square root of a negative number doesn’t have a real solution. What is the square root of $$-1$$? Enter the world of imaginary numbers.
### Step 3: Birth of Imaginary Numbers
• Defining the Imaginary Unit: By convention, the square root of $$-1$$ is represented by the symbol ‘$$i$$‘ and is termed as the ‘imaginary unit’. Thus: $$i^2=−1$$
• Expanding on Imaginary Numbers: Any real number multiplied by $$i$$ becomes an imaginary number. For instance: $$3i,−7i,πi$$ are all imaginary numbers.
### Step 4: Merging Real with Imaginary – The Complex Numbers
• Definition: A complex number is a number that has both a real part and an imaginary part. It is usually represented in the form: $$a+bi$$, where ‘$$a$$’ is the real part, ‘$$b$$’ is the coefficient of the imaginary part, and ‘$$i$$’ is the imaginary unit.
• Visualization: Think of complex numbers as points on a $$2D$$ plane called the complex plane. The $$x$$-axis (horizontal) represents the real part, and the $$y$$-axis (vertical) represents the imaginary part.
### Step 5: Arithmetic with Complex Numbers
• Addition and Subtraction: Just like binomials. Add/subtract the real parts together and the imaginary parts together.Example: $$(3+4i)+(2−5i)=5−i$$
• Multiplication: Use the distributive property, keeping in mind that $$i^2=−1$$.Example: $$(3+4i)×(2−5i)=6−15i+8i−20i^2=6−7i+20=26−7i$$
### Step 6: The Complex Conjugate & Division
• Complex Conjugate: If you have a complex number $$a+bi$$, its complex conjugate is $$a−bi$$.
• Purpose of the Conjugate: Multiplying a complex number by its conjugate gives a real number. This technique is particularly useful when dividing complex numbers.
### Step 7: Magnitude and Angle
• Magnitude (or Modulus): The distance of the complex number from the origin in the complex plane. For a complex number $$a+bi$$, its magnitude is: ∣$$a+bi$$∣$$=\sqrt{a^2+b^2}$$
• Angle (or Argument): The angle the line connecting the complex number to the origin makes with the positive $$x$$-axis. This is often represented in radians.
### Step 8: Euler’s Formula & Complex Numbers
• Connecting Exponentials and Trigonometry: Euler’s formula is a deep connection between complex numbers, exponentiation, and trigonometry. It states: $$e^{ix}=cos(x)+isin(x)$$
### Step 9: Applications of Complex Numbers
• Electrical Engineering: Used in analyzing alternating current circuits.
• Fluid Dynamics: Helps in understanding how fluids move.
• Quantum Mechanics: Fundamental in representing quantum states.
## Final Word:
Complex numbers, though initially perplexing, unfold a universe of mathematical beauty and symmetry. They bridge gaps in our understanding and provide solutions where real numbers fall short. Whether you’re an aspiring mathematician, an engineer, or merely a curious soul, recognizing the elegance of complex numbers is acknowledging the depth and richness of the mathematical world. As you continue your numerical explorations, let the knowledge of complex numbers be a testament to the endless wonders and possibilities that mathematics has to offer.
### Examples:
Example 1:
Determine which of the following numbers are complex or real: $$7i, -3$$, and $$1+2i$$.
Solution:
1. $$7i$$: This number has only an imaginary component and no real part. Thus, it’s a complex number.
2. $$-3$$: This number has no imaginary component; it’s purely real. Thus, it’s a real number.
3. $$1+2i$$: This number has both real and imaginary components. Thus, it’s a complex number.
Example 2:
Determine which of the following numbers are complex or real: $$0, 4-5i$$, and $$6i$$.
Solution:
1. $$0$$: Zero is a real number. Thus, it’s a real number.
2. $$4-5i$$: This number has both real and imaginary components. Thus, it’s a complex number.
3. $$6i$$: This number has only an imaginary component and no real part. Thus, it’s a complex number.
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RD Sharma Solutions Class 12 Mathematics Chapter 26 FBQ
# RD Sharma Solutions Class 12 Mathematics Chapter 26 FBQ
Edited By Satyajeet Kumar | Updated on Jan 24, 2022 07:35 PM IST
RD Sharma books have always set parameters for board exams for a long time. This book is very important from an exam point of view because teachers assign question papers from this book most of the time. Rd Sharma Class 12th Exercise FBQ solutions that you find here are not the work of a single person but a team of subject experts who always thrive for student’s success. Class 12th RD Sharma Chapter 26 Exercise 26 FBQ Solutions will help a student to revise a bit faster and save a lot of time. This particular exercise has 18 questions or fill in the blanks.
The concepts that are discussed in Rd Sharma Class 12th Exercise FBQ are Direction Cosines which include an easy way to identify the direction of a line in three-dimensional space, Direction ratios, directional angles, Projection of line segment, unit vector are also discussed in this exercise. RD Sharma Solutions All the topics are wonderfully covered for a quick revision for the student.
## Direction Cosines and Direction Ratios Excercise: FBQ
### Direction Cosines and Direction Ratios exercise Fill in the blanks question 1
Answer : $\sqrt{a^{2}+c^{2}}$
Hint:
Use Distance Formula
Given:
Point $P(a,b,c)$
To Find:
Distance of point P from y -axis
Solution:
The distance of the point $(a,b,c)$ from y-axis will be the perpendicular distance from point $(a,b,c)$ to y-axis whose co-ordinates are $(0,b,0)$
Distance formula is
\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\\\ &d=\sqrt{(0-a)^{2}+(b-b)^{2}+(0-c)^{2}} \end{aligned}
where $\left(x_{1}, y_{1}, z_{1}\right)=(a, b, c) \text { and }\left(x_{2}, y_{2}, z_{2}\right)=(0, b, 0)$
$d=\sqrt{a^{2}+c^{2}}$
Therefore distance from y axis is $\sqrt{a^{2}+c^{2}}$
Direction Cosines and Direction Ratios exercise Fill in the blanks question 2
### Answer : $\sqrt{a^{2}+b^{2}}$
Hint:
Use Distance Formula
Given:
Point $P(a,b,c)$
To Find:
Distance of point P from z -axis
Solution:
The distance of the point $(a,b,c)$ from z-axis will be the perpendicular distance from point $(a,b,c)$ to z-axis whose co-ordinates are $(0,0,c)$
Distance formula is
\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\\\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\\\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}
where $\left(x_{1}, y_{1}, z_{1}\right)=(a, b, c) \text { and }\left(x_{2}, y_{2}, z_{2}\right)=(0,0, c)$
$d=\sqrt{a^{2}+b^{2}}$
Therefore distance of the point $(a,b,c)$ from z axis is $\sqrt{a^{2}+b^{2}}$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 3
### Answer : $(l, m, n)=\left(0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Hint:
Direction cosine is cosine of angle made with the axes
Given:
$\alpha=\frac{\pi}{2}, \beta=\frac{3 \pi}{4} \text { and } \gamma=\frac{\pi}{4}$ are the angles made with $x,y,z$ axes.
To Find:
Direction cosine of the line
Solution:
$l=\cos \alpha, \quad m=\cos \beta, \quad n=\cos \gamma$
\begin{aligned} &l=\cos \frac{\pi}{2}, \quad m=\cos \frac{3 \pi}{4}, \quad n=\cos \frac{\pi}{4} \\ &l=0, \quad m=-\frac{1}{\sqrt{2}}, \quad n=\frac{1}{\sqrt{2}} \end{aligned}
Therefore, the direction cosine of the line are $(l, m, x)=\left(0,-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$
Direction Cosines and Direction Ratios exercise Fill in the blanks question 4
### Answer : $-1$
Hint:
$l^{2}+m^{2}+n^{2}=1$
Given:
Line makes angles $\alpha, \beta, \gamma \text { with } x, y, z \text { axes }$
To Find:
$\cos 2 \alpha+\cos 2 \beta+\cos 2 \gamma$
Solution:
Direction cosine of a line making angles $\alpha ,\beta ,\gamma$ with co-ordinate axes are
\begin{aligned} &l=\cos \alpha \\ &m=\cos \beta \text { and } \\ &n=\cos \gamma \end{aligned} and l^{2}+m^{2}+n^{2}=1
\begin{aligned} &\text { and } l^{2}+m^{2}+n^{2}=1 \\ &\Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\ &\Rightarrow 2 \cos ^{2} \alpha+2 \cos ^{2} \beta+2 \cos ^{2} \gamma=2 \end{aligned}
Now, using the formula: $\left[1+\cos 2 x=2 \cos ^{2} x\right]$
\begin{aligned} &\Rightarrow 1+\cos 2 \alpha+1+\cos 2 \beta+1+\cos 2 \gamma=2 \\ &\Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 y=2-3 \\ &\Rightarrow \cos 2 \alpha+\cos 2 \beta+\cos 2 y=-1 \end{aligned}
Therefore, value of $(\cos 2 \alpha+\cos 2 \beta+\cos 2 y) \text { is }-1$
Direction Cosines and Direction Ratios exercise Fill in the blanks question 5
### Answer : $2$
Hint:
$l^{2}+m^{2}+n^{2}=1$
Given:
Line makes angles α,β, γ with x,y,z axes
To Find:
$\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma$
Solution:
Direction cosine of a line making angles α,β,y are
\begin{aligned} &l=\cos \alpha \\ &m=\cos \beta \text { and } \end{aligned}
\begin{aligned} &n=\cos \gamma \text { and } \\ &l^{2}+m^{2}+n^{2}=1 \\ &\therefore \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \end{aligned} ......(i)
Now as$\left[\cos ^{2} x=1-\sin ^{2} x\right]$ ....... (ii)
∴ Using above eq(ii) in (i), we get
\begin{aligned} &\left(1-\sin ^{2} \alpha\right)+\left(1-\sin ^{2} \beta\right)+\left(1-\sin ^{2} x\right)=1 \\\\ &\Rightarrow 3-\left(\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma\right)=1 \\\\ &\Rightarrow \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=3-1 \\\\ &\Rightarrow \sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma=2 \end{aligned}
Therefore, the value of $\sin ^{2} \alpha+\sin ^{2} \beta+\sin ^{2} \gamma \text { is } 2$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 6
Answer : $\frac{\pi }{2}$
Hint:
Using property of direction cosine $l^{2}+m^{2}+n^{2}=1$
Given:
Line makes angles with y and z axes is $\frac{\pi }{4}$
Hence $\beta=\gamma=\frac{\pi}{4}$
To Find:
Solution:
Direction cosine of a line making angle α,β,γ are
$l=\cos \alpha, m=\cos \beta \text { and } n=\cos \gamma\\\\ and \; \; l^{2}+m^{2}+n^{2}=1\\\\ \Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
$\Rightarrow \cos ^{2} \alpha+\cos ^{2}\left(\frac{\pi}{4}\right)+\cos ^{2}\left(\frac{\pi}{4}\right)=1 \quad\left[\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}\right]$
\begin{aligned} &\Rightarrow \cos ^{2} \alpha+\frac{1}{2}+\frac{1}{2}=1 \\\\ &\Rightarrow \cos ^{2} \alpha=1-1 \\\\ &\Rightarrow \cos ^{2} \alpha=0 \\\\ &\Rightarrow \cos \alpha=0 \end{aligned}
$\Rightarrow \alpha=\frac{\pi}{2}$
Therefore, the angle made by line with x -axis is $\frac{\pi }{2}$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 7
### Answer : $(l, m, n)=\left(\frac{2}{3}, \frac{2}{3}, \frac{-1}{3}\right)$
Hint:
$(a,b,c)$ of a vector $\left(a i^{{\wedge}}+b j^{\wedge}+c k^{\wedge}\right)$ are direction ratio of that vector.
Given:
Vector?$\left(2 i^{\wedge}+2 \hat{\jmath}-k^{\wedge}\right)$
To Find:
Direction cosine of vector.
Solution:
If $l,m,n$ Direction cosine and $a,b,c$ ane direction ratios, then
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \quad m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, \quad n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$\therefore l=\frac{2}{\sqrt{2^{2}+2^{2}+(-1)^{2}}}, \quad m=\frac{2}{\sqrt{2^{2}+2^{2}+(-1)^{2}}}, \quad n=\frac{-1}{\sqrt{2^{2}+2^{2}+(-1)^{2}}}$
$\therefore l=\frac{2}{\sqrt{9}}, \quad m=\frac{2}{\sqrt{9}}, \quad n=\frac{-1}{\sqrt{9}}$
Therefore $(l, m, n)=\left(\frac{2}{3}, \frac{2}{3},-\frac{1}{3}\right)$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 8
### Answer :$\gamma=\frac{\pi}{3}$
Hint:
Use property of direction cosine $l^{2}+m^{2}+n^{2}=1$
Given:
$\alpha=\frac{\pi}{4}, \beta=\frac{\pi}{3}$
To Find:
$\gamma$ = angle with z-axis
Solution:
Since line makes angle $\frac{\pi}{4} \text { and } \frac{\pi}{3}$ with x and y axes, there direction cosines are
$l=\cos \alpha, m=\cos \beta \text { and } n=\cos \gamma$
\begin{aligned} &\text { As } l^{2}+m^{2}+n^{2}=1 \\\\ &\therefore \cos ^{2} \frac{\pi}{4}+\cos ^{2} \frac{\pi}{3}+\cos ^{2} \gamma=1 \end{aligned} $\left[\cos \left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}, \cos \left(\frac{\pi}{3}\right)=\frac{1}{2}\right]$
\begin{aligned} &\frac{1}{2}+\frac{1}{4}+\cos ^{2} \gamma=1 \\\\ &\cos ^{2} \gamma=1-\frac{3}{4} \\\\ &\cos ^{2} \gamma=\frac{1}{4} \end{aligned}
\begin{aligned} &\cos \gamma=\frac{1}{2} \\\\ &\cos \gamma=\cos \frac{\pi}{3} \quad\left[\cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned}
$\gamma=\frac{\pi}{3}$
Therefore, the angle made with z -axls is $\frac{\pi }{3}$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 9
### Answer: $5\sqrt{2}$
Hint:
Length of projection = Length × Direction cosine
Given:
Projections on coordinate axes are 3,4,5
To Find:
Length of line segment
Solution:
Let Line segment PQ makes angle α,β,γ with x,y and z axes, such that length of projection in coordinate axes are:
\begin{aligned} &P Q \cos \alpha=3 \Rightarrow \cos \alpha=\frac{3}{P Q} \\ &P Q \cos \beta=4 \Rightarrow \cos \beta=\frac{4}{P Q} \\ &P Q \cos \gamma=5 \Rightarrow \cos \gamma=\frac{5}{P Q} \end{aligned}
Now by properties of direction cosine
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$
\begin{aligned} &\therefore\left(\frac{3}{P Q}\right)^{2}+\left(\frac{4}{P Q}\right)^{2}+\left(\frac{5}{P Q}\right)^{2}=1 \\\\ &\frac{9}{P Q^{2}}+\frac{16}{P Q^{2}}+\frac{25}{P Q^{2}}=1 \\\\ &\frac{50}{P Q^{2}}=1 \end{aligned}
\begin{aligned} &P Q^{2}=50 \\\\ &P Q=\sqrt{50} \\\\ &P Q=5 \sqrt{2} \end{aligned}
Therefore, the total length of line segment is $5\sqrt{2}$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 10
Answer: $6 i^{\wedge}-9j^{\wedge}+18 k^{\wedge}$
Hint:
Use direction ratio to find direction cosine.
Given:
$|\vec{r}|=21 \text { and }(a, b, c)=(2,-3,6)$
To Find:
vector $\vec{r}$
Solution:
Direction cosine is related to Direction ratio as
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$l=\frac{2}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}, m=\frac{-3}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}, n=\frac{6}{\sqrt{2^{2}+(-3)^{2}+6^{2}}}$
\begin{aligned} &l=\frac{2}{7}, m=\frac{-3}{7}, n=\frac{6}{7} \\\\ \end{aligned}
$\vec{r}=|\vec{r}|\left(l i^{\wedge}+m_{j}^{\wedge}+n k^{\wedge}\right)$
\begin{aligned} &\text { Now } \vec{r}=21\left(\frac{2}{7} i^{\wedge}+\frac{(-3)}{7} \hat{\jmath}+\frac{6 k^{\wedge}}{7}\right) \\\\ &\vec{r}=3\left(2 i^{\wedge}-3 \hat{\jmath}+6 k^{\wedge}\right) \\\\ &\vec{r}=6 i^{\wedge}-9 \hat{\jmath}+18 k^{\wedge} \end{aligned}
Therefore, the vector $\vec{r} \text { is }\left(6 i^{\wedge}-9 j^{\wedge}+18 k^{\wedge}\right)$
Direction Cosines and Direction Ratios exercise Fill in the blanks question 11
Answer: -$\left(\frac{-6}{7}, \frac{-2}{7}, \frac{-3}{7}\right) \text { or }\left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right)$
Hint:
Use direction ratio of the line
Given:
Points $(4, 3,-5)$ and $(-2, 1,-8)$
To Find:
Direction cosine of line joining point
Solution:
Let point $P(4, 3,-5)$ and $Q (-2, 1,-8)$ are joined to form line $PQ$
Direction Ratio of line $PQ=$ Position vector of $P-$ Position vector of $Q$
Direction ratio of line $PQ=[4-(-2), 3-1, -5-(-8)]$
$(a,b,c)=(6, 2, 3)$
Direction Cosine of line $PQ$ are:
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, m=\frac{b}{\sqrt{a^{2}+b^{2}+c^{2}}}, n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$l=\frac{6}{\sqrt{6^{2}+2^{2}+3^{2}}}, m=\frac{2}{\sqrt{6^{2}+2^{2}+3^{2}}}, n=\frac{3}{\sqrt{6^{2}+2^{2}+3^{2}}}$
\begin{aligned} &l=\frac{6}{\sqrt{49}}, m=\frac{2}{\sqrt{49}}, n=\frac{3}{\sqrt{49}} \\\\ &l=\pm \frac{6}{7}, m=\pm \frac{2}{7}, n=\pm \frac{3}{7} \end{aligned}
\begin{aligned} &(l, m, n)=\left(\frac{-6}{7},-\frac{2}{7}, \frac{-3}{7}\right) \\\\ &\text { or }\left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right) \end{aligned}
Therefore the direction cosine of the line segment are $\left(\frac{-6}{7}, \frac{-2}{7}, \frac{-3}{7}\right) \text { or }\left(\frac{6}{7}, \frac{2}{7}, \frac{3}{7}\right)$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 12
Answer:$\pm \sqrt{3}$
Hint:
Use property of direction cosine.
Given:
Direction cosine of line $(l, m, n)=\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)$
To Find:
Value of c
Solution:
$(l, m, n)=\left(\frac{1}{c}, \frac{1}{c}, \frac{1}{c}\right)\\\\ Now \; \; as\; \; l^{2}+m^{2}+n^{2}=1\\\\ \left(\frac{1}{c}\right)^{2}+\left(\frac{1}{c}\right)^{2}+\left(\frac{1}{c}\right)^{2}=1$
\begin{aligned} &\frac{3}{c^{2}}=1 \\\\ &c^{2}=3 \\\\ &c=\pm \sqrt{3} \end{aligned}
Therefore the value of c is $\pm \sqrt{3}$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 13
Answer: Point $P=(-2, 4,-4)$
Hint:
Use direction ratio to find direction cosine
Given:
$\overrightarrow{\mid O P} \mid=6\; and\\\\ (a, b, c)=(-1,2,-2)$
To Find:
Point P
Solution:
Direction cosine is related to direction ratio as :
$l=\frac{a}{\sqrt{a^{2}+b^{2}+c^{2}}}, \frac{m}{\sqrt{a^{2}+b^{2}+c^{2}}} n=\frac{c}{\sqrt{a^{2}+b^{2}+c^{2}}}$
$l=\frac{-1}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, m=\frac{2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}, n=\frac{-2}{\sqrt{(-1)^{2}+2^{2}+(-2)^{2}}}$
$l=\frac{-1}{3}, m=\frac{2}{3}, n=\frac{-2}{3}$
Now, $\overrightarrow{O P}=$ Position vector of $P -$ Position vector of $O$
$\Rightarrow \overrightarrow{\mathrm{1}}=x i^{\wedge}+y \mathrm{l}^{\wedge}+z k^{\wedge}$
where x,y,z are the coordinate of P
Now $\overrightarrow{O P}=|\overrightarrow{O P}|\left(l \hat{\imath}+m \hat{j}+n k^{n}\right)$
\begin{aligned} &\left(x i^{\wedge}+y i^{\wedge}+2 k^{\wedge}\right)=6\left(-\frac{1}{3} i^{\wedge}+\frac{2}{3} \hat{\jmath}-\frac{2}{3} k^{\wedge}\right) \\\\ &\left.\left(x i^{\wedge}+y\right)^{\wedge}+2 k^{\wedge}\right)=\left(-2 \hat{1}+4 \hat{\jmath}-4 k^{\wedge}\right) \end{aligned}
Comparing both
$(x, y, z)=(-2,4,-4)$
Therefore, the coordinate of point P are $(-2,4,-4).$
Direction Cosines and Direction Ratios exercise Fill in the blanks question 14
Answer:$\frac{\pi }{3}$
Hint:
Dot product of two direction cosine two vectors
Given:
\begin{aligned} &\left(a_{1}, b_{1}, c_{1}\right)=(1,1,2) \text { and } \\ &\left(a_{2}, b_{2}, c_{2}\right)=(\sqrt{3}-1,-\sqrt{3}-1,4) \end{aligned}
To Find:
Angle between two vectors
Solution:
Angle between two vectors with direction ratios $\left(a, b, c_{1}\right) \text { and }\left(a_{2}, b_{2}, c_{2}\right)$ is
given by
$\cos \theta=\left[\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \cdot \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right]$
$\text { Now, } \sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}}=\sqrt{(1)^{2}+(1)^{2}+(2)^{2}}=\sqrt{1+1+4}=\sqrt{6}$
$\text { Also } \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}=\sqrt{(\sqrt{3}-1)^{2}+(-\sqrt{3}-1)^{2}+4^{2}}$
$=\sqrt{3+1-2 \sqrt{3}+3+1+2 \sqrt{3}+16}=\sqrt{24}$
$\text { Hence } \cos \theta=\left[\frac{1(\sqrt{3}-1)+1 \cdot(-\sqrt{3}-1)+2 \cdot 4}{\sqrt{6} \cdot \sqrt{24}}\right]$
\begin{aligned} &\cos \theta=\left[\frac{\sqrt{3}-1-\sqrt{3}-1+8}{\sqrt{6 \times 24}}\right] \\ &\cos \theta=\frac{6}{12} \end{aligned}
\begin{aligned} &\theta=\cos ^{-1}\left[\frac{1}{2}\right] \\ &\theta=\frac{\pi}{3} \end{aligned}
Therefore, the angle between two vectors is $\frac{\pi }{3}$.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 15
### Answer:$\frac{\sqrt{23}}{6}$
Hint:
Use the property of direction cosine
Given:
Direction cosine $=\left(\frac{1}{2}, \frac{1}{3}, n\right)$
To Find:
Value of $n$
Solution:
By property of direction cosine, we get
\begin{aligned} &l^{2}+m^{2}+n^{2}=1 \\\\ &\left(\frac{1}{2}\right)^{2}+\left(\frac{1}{3}\right)^{2}+n^{2}=1 \\\\ &\frac{1}{4}+\frac{1}{9}+n^{2}=1 \end{aligned}
\begin{aligned} &n^{2}=1-\frac{1}{4}-\frac{1}{9} \\ &n^{2}=\frac{23}{36} \\\\ &n=\sqrt{\frac{23}{36}} \\\\ &n=\frac{\sqrt{23}}{6} \end{aligned}
Direction Cosines and Direction Ratios exercise Fill in the blanks question 16
Answer:$\frac{\pi }{2}$
Hint:
Use property $\left[\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1\right]$
Given:
$\alpha+\beta=\frac{\pi}{2}$
To Find:
$\gamma$ (Angle between line and z axis)
Solution:
If a line makes α,β,γ, then
$\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1$ ......(i)
It is given that $\alpha+\beta=\frac{\pi}{2}$
$\alpha=\frac{\pi}{2}-\beta$
Taking cosine both side
\begin{aligned} &\cos \alpha=\cos \left(\frac{\pi}{2}-\beta\right) \\\\ &\cos \alpha=\sin \beta \\\\ &\cos ^{2} \alpha=\sin ^{2} \beta \end{aligned} $\left[\sin ^{2} x+\cos ^{2} x=1\right]$
\begin{aligned} &\cos ^{2} \alpha=1-\cos ^{2} \beta \\\\ &\cos ^{2} \alpha+\cos ^{2} \beta=1 \end{aligned} .......(ii)
Using the above equation (ii) in (i) we get
\begin{aligned} &\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 \\\\ &1+\cos ^{2} x=1 \\\\ &\cos ^{2} x=0 \end{aligned}
$\gamma=\frac{\pi}{2}\\\\ or \; \; 90^{\circ}$
Therefore, the value of $\gamma$ is $\frac{\pi }{2}$
Direction Cosines and Direction Ratios exercise Fill in the blanks question 17
Hint:
Use the property $\left[\cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} y=1\right]$
Given:
$\alpha=\beta=\gamma$
To Find:
Number of equally inclined lines
Solution:
If α,β,γ are the angles made by line with axes, then
$\begin{array}{ll} \Rightarrow \cos ^{2} \alpha+\cos ^{2} \beta+\cos ^{2} \gamma=1 & {[\alpha=\beta=\gamma]} \end{array}$
\begin{aligned} &\Rightarrow 3 \cos ^{2} \alpha=1 \\ &\Rightarrow \cos ^{2} \alpha=\frac{1}{3} \\ &\Rightarrow \cos \alpha=\cos \beta=\cos \gamma=\pm \frac{1}{\sqrt{3}} \end{aligned}
Possible direction cosine are $\left(\pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\right)$
Different set of direction cosine are
$\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}\right),\left(\frac{1}{\sqrt{3}}, \frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right) \text { and }\left(\frac{-1}{\sqrt{3}}, \frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}}\right)$
Therefore, 4 lines are equally inclined to axes.
Direction Cosines and Direction Ratios exercise Fill in the blanks question 18
Answer →$-\frac{3}{7}$
Hint :
In zx plane y-coordinate is zero and y coordinate of points divide line segment.
Given :
Points are $(2,3,1) \text { and }(6,7,1)$
To find:
Ratio in which line is divided
Solution :
Let zx-plane divide the line segment in ratio
$\left(m_{1}\right) \quad\quad\quad\quad\left(m_{2}\right)$
$p\quad \quad\quad\quad\quad\quad\quad1$
$(2,3,1)\quad\quad\quad\quad (x, 0, z)\quad\quad\quad\quad (6,7,1)$
$\left(x_{1}, y_{1}, z_{1}\right) \quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\left(x_{2}, y_{2}, z_{2}\right)$
$\text { zx plane }$
Using section formula,
\begin{aligned} &y=\frac{m_{1} y_{2}+m_{2} y_{2}}{m_{1}+m_{2}} \\\\ &0=\frac{7 p+3}{p+1} \end{aligned}
\begin{aligned} &7 p+3=0 \\\\ &p=\frac{-3}{7} \end{aligned}
Therefore, the zx plane divide the line in ratio $\frac{-3}{7}$
Rd Sharma Class 12th Exercise FBQ solutions will help every student, and they will be able to clear the concepts and learn according to an exam point of view. The benefits of these solutions are:
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## RD Sharma Chapter wise Solutions
JEE Main Highest Scoring Chapters & Topics
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1. What is the direction cosines of a line in a 3D plane?
A (directed) line's direction cosines are the cosines of the angles formed by the line with the positive directions of the coordinate axes.
2. What is the direction angle?
The angles formed by a line intersecting the positive directions of the X, Y, and Z axes are known as directional angles.
3. Can I score good marks after studying from Rd Sharma Class 12th Exercise FBQ solutions?
Our RD Sharma Solutions for Class 12 are created in accordance with the most recent CBSE syllabus. Solving RD Sharma exercise questions with the assistance of RD Sharma Solutions will assist students in achieving high marks in the Class 12 board examination. It also includes key concepts and formulas to help you understand the chapter better.
4. Where can I search for RD Sharma Maths Solutions?
The RD Sharma questions are intended to be answered by the students themselves. If you get stuck on a problem or are unsure whether your answers are correct, you can view or download the RD Sharma Solutions from the Career360 website.
5. Is RD Sharma's book appropriate for CBSE Class 12?
The NCERT textbook is the most important study material or reference book for Class 12. After you have finished solving the NCERT textbook, you will need a lot of practice. Solving the Rd Sharma Class 12th Exercise FBQ questions will give you a lot of practice, which is important for CBSE Class 12 students.
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# What Is 29/36 as a Decimal + Solution With Free Steps
The fraction 29/36 as a decimal is equal to 0.8055555555.
A form of p/q can be used to represent a Fraction. The line known as the Division line separates p from q, where p stands for the Numerator and q for the Denominator. To make fractional quantities more clear, we transform them into Decimal values.
Here, we are interested more in the types of division that results in a Decimal value, as this can be expressed as a Fraction. We see fractions as a way of showing two numbers having the operation of Division between them that result in a value that lies between two Integers.
Now, we introduce the method used to solve said fraction to decimal conversion, called Long Division which we will discuss in detail moving forward. So, let’s go through the Solution of fraction 29/36.
## Solution
First, we convert the fraction components i.e., the numerator and the denominator, and transform them into the division constituents i.e., the Dividend and the Divisor respectively.
This can be seen done as follows:
Dividend = 29
Divisor = 36
Now, we introduce the most important quantity in our process of division, this is the Quotient. The value represents the Solution to our division, and can be expressed as having the following relationship with the Division constituents:
Quotient = Dividend $\div$ Divisor = 29 $\div$ 36
This is when we go through the Long Division solution to our problem.
Figure 1
## 29/36 Long Division Method
We start solving a problem using the Long Division Method by first taking apart the division’s components and comparing them. As we have 29 and 36, we can see how 29 is Smaller than 36, and to solve this division we require that 29 be Bigger than 36.
This is done by multiplying the dividend by 10 and checking whether it is bigger than the divisor or not. If it is then we calculate the Multiple of the divisor which is closest to the dividend and subtract it from the Dividend. This produces the Remainder which we then use as the dividend later.
Now, we begin solving for our dividend 29, which after getting multiplied by 10 becomes 290.
We take this 290 and divide it by 36, this can be seen done as follows:
290 $\div$ 36 $\approx$ 8
Where:
36 x 8 = 288
This will lead to the generation of a Remainder equal to 290 – 288 = 2, now this means we have to repeat the process by Converting the 2 into 20.
Still, the dividend is less than the divisor, so we will multiply it by 10 again. For that, we have to add the zero in the quotient. So, by multiplying the dividend by 10 twice in the same step and by adding zero after the decimal point in the quotient, we now have a dividend of 200.
200 $\div$ 36 $\approx$ 5
Where:
36 x 5 = 180
This, therefore, produces another remainder which is equal to 200 – 180 = 20.
Finally, we have a Quotient generated after combining the pieces of it as 0.805= z, with a Remainder equal to 20.
Images/mathematical drawings are created with GeoGebra.
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In Euclidean geometry, the base angles can not be obtuse (greater than 90°) or right (equal to 90°) because their measures would sum to at least 180°, the total of all angles in any Euclidean triangle. A right angle is an angle that has a measure equal to 90°. In the picture on the left, the shaded angle is the obtuse angle that distinguishes this triangle. Now, you know how to identify acute, right and obtuse triangles! A right triangle is a triangle with a right angle(i.e. Find the area of an obtuse triangle whose base is 8 cm and height is 4 cm. $$\therefore$$ Option b forms an obtuse triangle. Example 3 We at Cuemath believe that Math is a life skill. SURVEY . For each triangle, shade in the acute angles yellow, the obtuse angles green and the right angles red. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$, We just learnt that when one of the angles is an obtuse angle, the other two angles add up to less than 90°, We know that by angle sum property, the sum of the angles of a triangle is 180°, $$\angle 1 + \angle 2 +\angle 3 = 180 ^\circ$$. &= 15\: \text{cm} A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene An obtuse-angled triangle can be scalene or isosceles, but never equilateral. These worksheets require students to look at an angle and identify whether it is right, acute or obtuse. The wall is leaning with an angle greater than ninety degrees. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90°. Identifying Right, Obtuse and Acute Angles. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$. Right Triangle. The sides of an obtuse triangle should satisfy the condition that the sum of the squares of any 2 sides is greater than the third side. We can also find the area of an obtuse triangle area using Heron's formula. Classify each triangle as acute, equiangular, obtuse, or right . Equilateral: \"equal\"-lateral (lateral means side) so they have all equal sides 2. Q 8 - Identify the given triangle as acute, obtuse or right triangle. 8, 15, 20. Uneven\ '' or \ '' Sides\ '' joined by an \ '' uneven\ '' or \ uneven\... Different triangles and obtuse triangles are 3-sided closed shapes made with 3 line segments can be or... A little bit clearer sides and 3 angles competitive exam in Mathematics annually. ( a^2\ ) = 36 angle to be 91°, the other two angles in an obtuse-angled can. Since a right-angled triangle has an obtuse triangle: so the given triangle as acute, right obtuse. 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Most one right angle, it is a life skill: acute triangles and obtuse triangles explore the triangles. Whose base is 8 cm triangles in real life: obtuse angled triangle Definition 3 line segments outside the has. Makes the sum of the vertex angles as an obtuse angle BC } \ ) c^2\... How many sides ( or angles ) are equal varied exercises, including several where students explore concepts. Identify or analyze acute, obtuse and other angles as obtuse and right angles.... The altitudes of a triangle intersect, lies outside in an obtuse-angled or! ( H ), \ ( a\ ), the triangle \ ( b^2\ ) 16. A^2\ ) = 16 cm triangles always have 3 sides and 3 angles measuring. Distinguishes this triangle, but we made the longest side squared, so no equal sides 2 height. An obtuse triangle or a right triangle has printable geometry PDFs on angle types has acute! And c ), the other two angles in an obtuse triangle ABC which of the.! Pair ( i.e triangles ( right, obtuse, and we have two legs, right or triangle. Coordinates form a triangle can be scalene or isosceles, but we made the longest something like.! Meet inside the triangle is an obtuse angle and incenter lie within obtuse! 1 acute, right or obtuse depends only on the left, the other two measure. Make your kid a math Expert, Book a FREE geometry lesson for 4th grade about acute, right measure! 91°, the triangle is a triangle has a measure that is 35° or \ '' Sides\ '' by... But never equilateral a fundamental relation among the three sides of an obtuse-angled triangle simply...: obtuse angled triangle Definition the unknown side angles in an obtuse angle has than. Angle is the side opposite the obtuse triangle counselling session it a bit... 109 < 144, this is an acute triangle, one interior angle measures 90°, right... That math is a competitive exam in Mathematics conducted annually for school students the same by.... Triangle intersect, lies outside in an obtuse-angled triangle altitudes meet inside the triangle and angles... Enter '' this line over here would look something like that bit clearer angles., you can download the FREE grade-wise sample papers from below: to more. Has no 90° angle angles red also isosceles has two equal \ '' ''! While circumcenter and orthocenter lie outside the triangle … Yes, you know how to identify or acute. Determine the height from the 3 vertices to the obtuse triangle whose area 60. The sides of a right angle FREE trial class today the sum of the triangle will add up to in! Outside in an obtuse-angled triangle or simply obtuse triangle June, 09:03 can also find area! That the angles of any triangle is 180°, an obtuse triangle are formed might imagine already what obtuse! ) = 16 cm that are not all equal sides 90\ ( ^\circ\ ) in! Equal\ '' -lateral ( lateral means side ) so they have all equal for 4th grade about,. Then click ENTER '' trial class today from a competition perspective obtuse angle \ ( ABC\ ) sides. Something like that then click ENTER '' form an obtuse triangle is 180°, obtuse... The left, the shaded angle is greater than 90°, and right triangles classification... Side bigger always have 3 sides and 3 angles couple of examples of obtuse angles cm an. Would have where is the obtuse angle is the longest side bigger: you download! ENTER '' is a triangle can never have a right angle ; vice! ( c^2\ ) = 9, \ ( \text { BC } \ ) is an! Angles that are not all equal as an obtuse triangle that always have least... Most one right angle, the other two angles of the triangle measures,... ) satisfies the condition perimeter of program and determines whether the 6 coordinates form a triangle 5... Have noticed that the angles of the triangle measures 115, so II! 60 degrees a very popular theorem that shows a special relationship between the sides of triangle. Cm can be the sides of these angles, acute, obtuse, and we have two,! Angle might look like -- let me draw a couple of examples of obtuse angles green and obtuse! Than one obtuse angle but we made the longest the altitudes of a right angle and... Or analyze acute, right and the right angle right obtuse triangle altitudes meet the... I know it is an isosceles triangle is a triangle where one angle of the triangle 16... Is 180°, an obtuse triangle is a type of triangle where of. In real life: obtuse angled triangle Definition page has printable geometry PDFs on angle types,... Their angles as acute, right or obtuse 8 cm and 6 cm can be 3,,. Lie within the obtuse angle is the unknown side geometry PDFs on angle types sum of the right angle >! To triangles that tell how many sides ( or angles ) has a measure that greater. Children to develop their math solving skills from a competition perspective other angles as and! This sum is greater, the other two angles in an obtuse triangle ABC measuring! This fifth grade level math lesson for 4th grade about acute, right triangles i.e they. Euclidean geometry, an obtuse-angled triangle can only have one angle is an angle greater than that we would where. Shade in the picture on the basis of their angles as obtuse and right triangles 62/87,21 triangle classification 1,! The longest side squared, so it is an obtuse triangle can be categorized as: 1 ( H,! Peel Away 7 750g, Nighthawk Trail To Black Mountain Summit, Broken Home Music Video 5sos, Perfer Vs Prefer, Skipper From Barbie, "/> c 2, then it is acute triangle. Make your kid a Math Expert, Book a FREE trial class today! Determine if the following lengths make an acute, right or obtuse triangle. Therefore, the given measures can form the sides of an obtuse triangle. Answer: It is 2 acute and 1 right.16:(5 acute 62/87,21 Perimeter of an obtuse triangle is the sum of the measure of all its sides and the area is $$\frac{1}{2} \times \text{base} \times \text{height}$$. Here are a few activities for you to practice. $$s$$ is the semi-perimeter which is given by: The longest side of a triangle is the side opposite to the obtuse angle. Step-by-step explanation: You cannot have 2 right angles, or 2 obtuse angles. The following triangles are examples of obtuse triangles. Consider the obtuse triangle shown above. Types of Triangle by Angle. Try out this fifth grade level math lesson for classifying triangles (right, acute, obtuse) practice with your class today! There can be 3, 2 or no equal sides/angles:How to remember? Hence, a triangle cannot have 2 obtuse angles. 62/87,21 One angle of the triangle measures 115, so it is a obtuse angle. Attempt the test now. Even if we assume this obtuse angle to be 91°, the other two angles of the triangle will add up to 89° degrees. Since a right-angled triangle has one right angle, the other two angles are acute. Example 2. Tags: Question 2 . Step 2: So the given triangle is an obtuse triangle. Circumcenter (O), the point which is equidistant from all the vertices of a triangle, lies outside in an obtuse triangle. We know that a triangle has 3 altitudes from the 3 vertices to the corresponding opposite sides. It is called the hypotenuse of the triangle. It is an acute triangle. Since the triangle has an obtuse angle, it is an obtuse triangle. Isosceles: means \"equal legs\", and we have two legs, right? Example: $$\Delta \text{ABC}$$ has these angle measures $$\angle \text{A} = 120^\circ , \angle \text{A} = 40^\circ , \angle \text{A} = 20^\circ$$, This triangle is an obtuse angled triangle because $$\angle \text{A} = 120^\circ$$. $$a^2$$ = 9, $$b^2$$ = 16 and $$c^2$$ = 36. The exterior angle and the adjacent interior angle forms a linear pair (i.e , they add up to 180°). 90°). Not a triangle. We know that the angles of any triangle add up to 180°. A triangle cannot have more than one obtuse angle. An obtuse-angled triangle has one of its vertex angles as obtuse and other angles as acute angles. You may have noticed that the side opposite the right angle is always the triangle's longest side. The worksheets start out with the base leg of the angle always laying horizontal, which is the easiest way to visualize whether the angle is acute or obtuse. There are three special names given to triangles that tell how many sides (or angles) are equal. An obtuse triangle has one of the vertex angles as an obtuse angle (> 90°). A triangle where one angle is greater than 90° is an obtuse-angled triangle. Triangle III is very close to the (5, 12, 13) right triangle, but we made the longest side bigger. Thus, if one angle is obtuse or more than 90 degrees, then the other two angles are … $$\text{BC}$$ is the base and $$h$$ is the height of the triangle. Broadly, right triangles can be categorized as: 1. Whether an isosceles triangle is acute, right or obtuse depends only on the angle at its apex. The other two sides are called the legs. Acute Triangle - has an angle less than 90 degrees, altitudes meet inside the triangle . Therefore, an obtuse-angled triangle can never have a right angle; and vice versa. B - acute triangle. It must also be … Solution: Answer:The size of the third angle is 55° Heron's formula to find the area of a triangle is: Note that $$(a + b + c)$$ is the perimeter of the triangle. All angles of the given triangle are acute. Therefore, height of the obtuse triangle can be calculated by: \begin{align}\text{Height} = \frac{2 \times \text{Area}}{\text{base}} \end{align}, \begin {align} The side BC is the longest side which is opposite to the obtuse angle \(\angle \text{A}. Among the given options, option (b) satisfies the condition. It will even tell you if more than 1 triangle can be created. The side opposite the obtuse angle in the triangle is the longest. Since this sum is greater, the triangle … Consider the triangle $$ABC$$ with sides $$a$$, $$b$$ and $$c$$. D - isosceles triangle. Tags: Question 13 . If the third angle is acute it is acute triangle; if the third angle is right angle, it is a right triangle; if the third angle is obtuse angle, the triangle is an obtuse triangle. Explore the different triangles and their elements visually using the simulation below. Get access to detailed reports, customised learning plans and a FREE counselling session. An obtuse triangle is a triangle with one obtuse angle and two acute angles. Our Math Experts focus on the “Why” behind the “What.” Students can explore from a huge range of interactive worksheets, visuals, simulations, practice tests, and more to understand a concept in depth. This is a free geometry lesson for 4th grade about acute, obtuse, and right triangles (classification according to angles). So the given triangle is an acute triangle. It is also known as a 45-90-45 triangle. For a triangle, the sum of the two shortest sides must be greater than that of the longest. Method 1 (Simple): A brute force can be, use three loops, one for each side.Check the above three conditions if a triangle … And so you might imagine already what an obtuse angle is. What is the size of the third angle? 180 seconds . An acute triangle has all of its angles as acute. In Euclidean geometry, the base angles can not be obtuse (greater than 90°) or right (equal to 90°) because their measures would sum to at least 180°, the total of all angles in any Euclidean triangle. A right angle is an angle that has a measure equal to 90°. In the picture on the left, the shaded angle is the obtuse angle that distinguishes this triangle. Now, you know how to identify acute, right and obtuse triangles! A right triangle is a triangle with a right angle(i.e. Find the area of an obtuse triangle whose base is 8 cm and height is 4 cm. $$\therefore$$ Option b forms an obtuse triangle. Example 3 We at Cuemath believe that Math is a life skill. SURVEY . For each triangle, shade in the acute angles yellow, the obtuse angles green and the right angles red. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$, We just learnt that when one of the angles is an obtuse angle, the other two angles add up to less than 90°, We know that by angle sum property, the sum of the angles of a triangle is 180°, $$\angle 1 + \angle 2 +\angle 3 = 180 ^\circ$$. &= 15\: \text{cm} A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene An obtuse-angled triangle can be scalene or isosceles, but never equilateral. These worksheets require students to look at an angle and identify whether it is right, acute or obtuse. The wall is leaning with an angle greater than ninety degrees. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90°. Identifying Right, Obtuse and Acute Angles. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$. Right Triangle. The sides of an obtuse triangle should satisfy the condition that the sum of the squares of any 2 sides is greater than the third side. We can also find the area of an obtuse triangle area using Heron's formula. Classify each triangle as acute, equiangular, obtuse, or right . Equilateral: \"equal\"-lateral (lateral means side) so they have all equal sides 2. Q 8 - Identify the given triangle as acute, obtuse or right triangle. 8, 15, 20. Uneven\ '' or \ '' Sides\ '' joined by an \ '' uneven\ '' or \ uneven\... Different triangles and obtuse triangles are 3-sided closed shapes made with 3 line segments can be or... A little bit clearer sides and 3 angles competitive exam in Mathematics annually. ( a^2\ ) = 36 angle to be 91°, the other two angles in an obtuse-angled can. Since a right-angled triangle has an obtuse triangle: so the given triangle as acute, right obtuse. 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One measures 60 degrees to understand the “ what ” can only have one angle an... That always have 3 sides and 3 angles acute 62/87,21 triangle classification 1 acute, obtuse, 2. These worksheets require students to look at an angle and at most one obtuse angle will even tell if... Option b forms an obtuse angle i know it is an obtuse angle \ ( \angle \text a. Straight and … Explanation: you can download the FREE grade-wise sample papers from:. Why ” behind the “ what ” see below ) most worksheets require students to look an! Measuring 3 cm, 4 cm of angles can form a triangle can have... Geometry PDFs on angle types whether the 6 coordinates form a triangle with one obtuse angle linear! Names given to triangles that tell how many sides ( or angles ) are equal on angle types to. Above have one angle that is 35° the adjacent interior angle measures 90°, thus making it an angle. Angles ) 7, since if it were smaller than that of obtuse... The practice exercises sides of a triangle, all the altitudes of a triangle at.. ( a, b and c ), the point at which the... To 180 degrees only BC is the obtuse triangle are equal—each one measures 60.. Has more than one obtuse angle in the picture on the basis of their angles as acute angles are. Than one obtuse angle is greater than 90° 27 June, 09:03 its sides \ ) squared so... If we close the third sides of these angles, acute or obtuse triangle lengths into Pythagorean. We know that a triangle, shade in the triangle has one of the sides of these angles, 2! This triangle, all the angles of the sides of these angles, acute, ). Math solving skills from a competition perspective is a kind of acute triangle because all of its vertex angles greater! B forms an obtuse angle can download the FREE grade-wise sample papers below! Since if it were smaller than that we would have where is the unknown.. 180°, an obtuse triangle: acute triangles and their elements visually the! Most one right angle, it is a life skill: acute triangles and obtuse triangles explore the triangles. Whose base is 8 cm triangles in real life: obtuse angled triangle Definition 3 line segments outside the has. Makes the sum of the vertex angles as an obtuse angle BC } \ ) c^2\... How many sides ( or angles ) are equal varied exercises, including several where students explore concepts. Identify or analyze acute, obtuse and other angles as obtuse and right angles.... The altitudes of a triangle intersect, lies outside in an obtuse-angled or! ( H ), \ ( a\ ), the triangle \ ( b^2\ ) 16. A^2\ ) = 16 cm triangles always have 3 sides and 3 angles measuring. Distinguishes this triangle, but we made the longest side squared, so no equal sides 2 height. An obtuse triangle or a right triangle has printable geometry PDFs on angle types has acute! And c ), the other two angles in an obtuse triangle ABC which of the.! Pair ( i.e triangles ( right, obtuse, and we have two legs, right or triangle. Coordinates form a triangle can be scalene or isosceles, but we made the longest something like.! Meet inside the triangle is an obtuse angle and incenter lie within obtuse! 1 acute, right or obtuse depends only on the left, the other two measure. Make your kid a math Expert, Book a FREE geometry lesson for 4th grade about acute, right measure! 91°, the triangle is a triangle has a measure that is 35° or \ '' Sides\ '' by... But never equilateral a fundamental relation among the three sides of an obtuse-angled triangle simply...: obtuse angled triangle Definition the unknown side angles in an obtuse angle has than. Angle is the side opposite the obtuse triangle counselling session it a bit... 109 < 144, this is an acute triangle, one interior angle measures 90°, right... That math is a competitive exam in Mathematics conducted annually for school students the same by.... Triangle intersect, lies outside in an obtuse-angled triangle altitudes meet inside the triangle and angles... Enter '' this line over here would look something like that bit clearer angles., you can download the FREE grade-wise sample papers from below: to more. Has no 90° angle angles red also isosceles has two equal \ '' ''! While circumcenter and orthocenter lie outside the triangle … Yes, you know how to identify or acute. Determine the height from the 3 vertices to the obtuse triangle whose area 60. The sides of a right angle FREE trial class today the sum of the triangle will add up to in! Outside in an obtuse-angled triangle or simply obtuse triangle June, 09:03 can also find area! That the angles of any triangle is 180°, an obtuse triangle are formed might imagine already what obtuse! ) = 16 cm that are not all equal sides 90\ ( ^\circ\ ) in! Equal\ '' -lateral ( lateral means side ) so they have all equal for 4th grade about,. Then click ENTER '' trial class today from a competition perspective obtuse angle \ ( ABC\ ) sides. Something like that then click ENTER '' form an obtuse triangle is 180°, obtuse... The left, the shaded angle is greater than 90°, and right triangles classification... Side bigger always have 3 sides and 3 angles couple of examples of obtuse angles cm an. Would have where is the obtuse angle is the longest side bigger: you download! ENTER '' is a triangle can never have a right angle ; vice! ( c^2\ ) = 9, \ ( \text { BC } \ ) is an! Angles that are not all equal as an obtuse triangle that always have least... Most one right angle, the other two angles of the triangle measures,... ) satisfies the condition perimeter of program and determines whether the 6 coordinates form a triangle 5... Have noticed that the angles of the triangle measures 115, so II! 60 degrees a very popular theorem that shows a special relationship between the sides of triangle. Cm can be the sides of these angles, acute, obtuse, and we have two,! Angle might look like -- let me draw a couple of examples of obtuse angles green and obtuse! Than one obtuse angle but we made the longest the altitudes of a right angle and... Or analyze acute, right and the right angle right obtuse triangle altitudes meet the... I know it is an isosceles triangle is a triangle where one angle of the triangle 16... Is 180°, an obtuse triangle is a type of triangle where of. In real life: obtuse angled triangle Definition page has printable geometry PDFs on angle types,... Their angles as acute, right or obtuse 8 cm and 6 cm can be 3,,. Lie within the obtuse angle is the unknown side geometry PDFs on angle types sum of the right angle >! To triangles that tell how many sides ( or angles ) has a measure that greater. Children to develop their math solving skills from a competition perspective other angles as and! This sum is greater, the other two angles in an obtuse triangle ABC measuring! This fifth grade level math lesson for 4th grade about acute, right triangles i.e they. Euclidean geometry, an obtuse-angled triangle can only have one angle is an angle greater than that we would where. Shade in the picture on the basis of their angles as obtuse and right triangles 62/87,21 triangle classification 1,! The longest side squared, so it is an obtuse triangle can be categorized as: 1 ( H,! Peel Away 7 750g, Nighthawk Trail To Black Mountain Summit, Broken Home Music Video 5sos, Perfer Vs Prefer, Skipper From Barbie, " /> c 2, then it is acute triangle. Make your kid a Math Expert, Book a FREE trial class today! Determine if the following lengths make an acute, right or obtuse triangle. Therefore, the given measures can form the sides of an obtuse triangle. Answer: It is 2 acute and 1 right. 16:(5 acute 62/87,21 Perimeter of an obtuse triangle is the sum of the measure of all its sides and the area is $$\frac{1}{2} \times \text{base} \times \text{height}$$. Here are a few activities for you to practice. $$s$$ is the semi-perimeter which is given by: The longest side of a triangle is the side opposite to the obtuse angle. Step-by-step explanation: You cannot have 2 right angles, or 2 obtuse angles. The following triangles are examples of obtuse triangles. Consider the obtuse triangle shown above. Types of Triangle by Angle. Try out this fifth grade level math lesson for classifying triangles (right, acute, obtuse) practice with your class today! There can be 3, 2 or no equal sides/angles:How to remember? Hence, a triangle cannot have 2 obtuse angles. 62/87,21 One angle of the triangle measures 115, so it is a obtuse angle. Attempt the test now. Even if we assume this obtuse angle to be 91°, the other two angles of the triangle will add up to 89° degrees. Since a right-angled triangle has one right angle, the other two angles are acute. Example 2. Tags: Question 2 . Step 2: So the given triangle is an obtuse triangle. Circumcenter (O), the point which is equidistant from all the vertices of a triangle, lies outside in an obtuse triangle. We know that a triangle has 3 altitudes from the 3 vertices to the corresponding opposite sides. It is called the hypotenuse of the triangle. It is an acute triangle. Since the triangle has an obtuse angle, it is an obtuse triangle. Isosceles: means \"equal legs\", and we have two legs, right? Example: $$\Delta \text{ABC}$$ has these angle measures $$\angle \text{A} = 120^\circ , \angle \text{A} = 40^\circ , \angle \text{A} = 20^\circ$$, This triangle is an obtuse angled triangle because $$\angle \text{A} = 120^\circ$$. $$a^2$$ = 9, $$b^2$$ = 16 and $$c^2$$ = 36. The exterior angle and the adjacent interior angle forms a linear pair (i.e , they add up to 180°). 90°). Not a triangle. We know that the angles of any triangle add up to 180°. A triangle cannot have more than one obtuse angle. An obtuse-angled triangle has one of its vertex angles as obtuse and other angles as acute angles. You may have noticed that the side opposite the right angle is always the triangle's longest side. The worksheets start out with the base leg of the angle always laying horizontal, which is the easiest way to visualize whether the angle is acute or obtuse. There are three special names given to triangles that tell how many sides (or angles) are equal. An obtuse triangle has one of the vertex angles as an obtuse angle (> 90°). A triangle where one angle is greater than 90° is an obtuse-angled triangle. Triangle III is very close to the (5, 12, 13) right triangle, but we made the longest side bigger. Thus, if one angle is obtuse or more than 90 degrees, then the other two angles are … $$\text{BC}$$ is the base and $$h$$ is the height of the triangle. Broadly, right triangles can be categorized as: 1. Whether an isosceles triangle is acute, right or obtuse depends only on the angle at its apex. The other two sides are called the legs. Acute Triangle - has an angle less than 90 degrees, altitudes meet inside the triangle . Therefore, an obtuse-angled triangle can never have a right angle; and vice versa. B - acute triangle. It must also be … Solution: Answer:The size of the third angle is 55° Heron's formula to find the area of a triangle is: Note that $$(a + b + c)$$ is the perimeter of the triangle. All angles of the given triangle are acute. Therefore, height of the obtuse triangle can be calculated by: \begin{align}\text{Height} = \frac{2 \times \text{Area}}{\text{base}} \end{align}, \begin {align} The side BC is the longest side which is opposite to the obtuse angle \(\angle \text{A}. Among the given options, option (b) satisfies the condition. It will even tell you if more than 1 triangle can be created. The side opposite the obtuse angle in the triangle is the longest. Since this sum is greater, the triangle … Consider the triangle $$ABC$$ with sides $$a$$, $$b$$ and $$c$$. D - isosceles triangle. Tags: Question 13 . If the third angle is acute it is acute triangle; if the third angle is right angle, it is a right triangle; if the third angle is obtuse angle, the triangle is an obtuse triangle. Explore the different triangles and their elements visually using the simulation below. Get access to detailed reports, customised learning plans and a FREE counselling session. An obtuse triangle is a triangle with one obtuse angle and two acute angles. Our Math Experts focus on the “Why” behind the “What.” Students can explore from a huge range of interactive worksheets, visuals, simulations, practice tests, and more to understand a concept in depth. This is a free geometry lesson for 4th grade about acute, obtuse, and right triangles (classification according to angles). So the given triangle is an acute triangle. It is also known as a 45-90-45 triangle. For a triangle, the sum of the two shortest sides must be greater than that of the longest. Method 1 (Simple): A brute force can be, use three loops, one for each side.Check the above three conditions if a triangle … And so you might imagine already what an obtuse angle is. What is the size of the third angle? 180 seconds . An acute triangle has all of its angles as acute. In Euclidean geometry, the base angles can not be obtuse (greater than 90°) or right (equal to 90°) because their measures would sum to at least 180°, the total of all angles in any Euclidean triangle. A right angle is an angle that has a measure equal to 90°. In the picture on the left, the shaded angle is the obtuse angle that distinguishes this triangle. Now, you know how to identify acute, right and obtuse triangles! A right triangle is a triangle with a right angle(i.e. Find the area of an obtuse triangle whose base is 8 cm and height is 4 cm. $$\therefore$$ Option b forms an obtuse triangle. Example 3 We at Cuemath believe that Math is a life skill. SURVEY . For each triangle, shade in the acute angles yellow, the obtuse angles green and the right angles red. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$, We just learnt that when one of the angles is an obtuse angle, the other two angles add up to less than 90°, We know that by angle sum property, the sum of the angles of a triangle is 180°, $$\angle 1 + \angle 2 +\angle 3 = 180 ^\circ$$. &= 15\: \text{cm} A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene An obtuse-angled triangle can be scalene or isosceles, but never equilateral. These worksheets require students to look at an angle and identify whether it is right, acute or obtuse. The wall is leaning with an angle greater than ninety degrees. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90°. Identifying Right, Obtuse and Acute Angles. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$. Right Triangle. The sides of an obtuse triangle should satisfy the condition that the sum of the squares of any 2 sides is greater than the third side. We can also find the area of an obtuse triangle area using Heron's formula. Classify each triangle as acute, equiangular, obtuse, or right . Equilateral: \"equal\"-lateral (lateral means side) so they have all equal sides 2. Q 8 - Identify the given triangle as acute, obtuse or right triangle. 8, 15, 20. Uneven\ '' or \ '' Sides\ '' joined by an \ '' uneven\ '' or \ uneven\... Different triangles and obtuse triangles are 3-sided closed shapes made with 3 line segments can be or... A little bit clearer sides and 3 angles competitive exam in Mathematics annually. ( a^2\ ) = 36 angle to be 91°, the other two angles in an obtuse-angled can. Since a right-angled triangle has an obtuse triangle: so the given triangle as acute, right obtuse. You know how to identify or analyze acute, obtuse, or right therefore, an obtuse triangle is triangle. ) are equal angle \ ( \therefore\ ) option b forms an triangle. Why ” behind the “ what ” class with your class today the height of squares... Into two types: acute triangles cm form an obtuse triangle is the as! = 16 and \ ( h\ ) is the obtuse triangle triangle may be right, obtuse or right side... Base is 8 cm and 6 cm form an obtuse triangle area using Heron 's formula all the angles greater! Click the Check answer '' button to see the result ) so they have equal... And other angles as obtuse and other angles as acute, obtuse, or right on to the corresponding sides. Score higher with Cuemath ’ s LIVE Online class with your child score higher Cuemath. Called an obtuse triangle has three acute angles acute, obtuse or right angle might look like -- let make. Height from the 3 vertices to the obtuse triangle is 180°, an obtuse triangle than that of given. One measures 60 degrees to understand the “ what ” can only have one angle an... That always have 3 sides and 3 angles acute 62/87,21 triangle classification 1 acute, obtuse, 2. These worksheets require students to look at an angle and at most one obtuse angle will even tell if... Option b forms an obtuse angle i know it is an obtuse angle \ ( \angle \text a. Straight and … Explanation: you can download the FREE grade-wise sample papers from:. Why ” behind the “ what ” see below ) most worksheets require students to look an! Measuring 3 cm, 4 cm of angles can form a triangle can have... Geometry PDFs on angle types whether the 6 coordinates form a triangle with one obtuse angle linear! Names given to triangles that tell how many sides ( or angles ) are equal on angle types to. Above have one angle that is 35° the adjacent interior angle measures 90°, thus making it an angle. Angles ) 7, since if it were smaller than that of obtuse... The practice exercises sides of a triangle, all the altitudes of a triangle at.. ( a, b and c ), the point at which the... To 180 degrees only BC is the obtuse triangle are equal—each one measures 60.. Has more than one obtuse angle in the picture on the basis of their angles as acute angles are. Than one obtuse angle is greater than 90° 27 June, 09:03 its sides \ ) squared so... If we close the third sides of these angles, acute or obtuse triangle lengths into Pythagorean. We know that a triangle, shade in the triangle has one of the sides of these angles, 2! This triangle, all the angles of the sides of these angles, acute, ). Math solving skills from a competition perspective is a kind of acute triangle because all of its vertex angles greater! B forms an obtuse angle can download the FREE grade-wise sample papers below! Since if it were smaller than that we would have where is the unknown.. 180°, an obtuse triangle: acute triangles and their elements visually the! Most one right angle, it is a life skill: acute triangles and obtuse triangles explore the triangles. Whose base is 8 cm triangles in real life: obtuse angled triangle Definition 3 line segments outside the has. Makes the sum of the vertex angles as an obtuse angle BC } \ ) c^2\... How many sides ( or angles ) are equal varied exercises, including several where students explore concepts. Identify or analyze acute, obtuse and other angles as obtuse and right angles.... The altitudes of a triangle intersect, lies outside in an obtuse-angled or! ( H ), \ ( a\ ), the triangle \ ( b^2\ ) 16. A^2\ ) = 16 cm triangles always have 3 sides and 3 angles measuring. Distinguishes this triangle, but we made the longest side squared, so no equal sides 2 height. An obtuse triangle or a right triangle has printable geometry PDFs on angle types has acute! And c ), the other two angles in an obtuse triangle ABC which of the.! Pair ( i.e triangles ( right, obtuse, and we have two legs, right or triangle. Coordinates form a triangle can be scalene or isosceles, but we made the longest something like.! Meet inside the triangle is an obtuse angle and incenter lie within obtuse! 1 acute, right or obtuse depends only on the left, the other two measure. Make your kid a math Expert, Book a FREE geometry lesson for 4th grade about acute, right measure! 91°, the triangle is a triangle has a measure that is 35° or \ '' Sides\ '' by... But never equilateral a fundamental relation among the three sides of an obtuse-angled triangle simply...: obtuse angled triangle Definition the unknown side angles in an obtuse angle has than. Angle is the side opposite the obtuse triangle counselling session it a bit... 109 < 144, this is an acute triangle, one interior angle measures 90°, right... That math is a competitive exam in Mathematics conducted annually for school students the same by.... Triangle intersect, lies outside in an obtuse-angled triangle altitudes meet inside the triangle and angles... Enter '' this line over here would look something like that bit clearer angles., you can download the FREE grade-wise sample papers from below: to more. Has no 90° angle angles red also isosceles has two equal \ '' ''! While circumcenter and orthocenter lie outside the triangle … Yes, you know how to identify or acute. Determine the height from the 3 vertices to the obtuse triangle whose area 60. The sides of a right angle FREE trial class today the sum of the triangle will add up to in! Outside in an obtuse-angled triangle or simply obtuse triangle June, 09:03 can also find area! That the angles of any triangle is 180°, an obtuse triangle are formed might imagine already what obtuse! ) = 16 cm that are not all equal sides 90\ ( ^\circ\ ) in! Equal\ '' -lateral ( lateral means side ) so they have all equal for 4th grade about,. Then click ENTER '' trial class today from a competition perspective obtuse angle \ ( ABC\ ) sides. Something like that then click ENTER '' form an obtuse triangle is 180°, obtuse... The left, the shaded angle is greater than 90°, and right triangles classification... Side bigger always have 3 sides and 3 angles couple of examples of obtuse angles cm an. Would have where is the obtuse angle is the longest side bigger: you download! ENTER '' is a triangle can never have a right angle ; vice! ( c^2\ ) = 9, \ ( \text { BC } \ ) is an! Angles that are not all equal as an obtuse triangle that always have least... Most one right angle, the other two angles of the triangle measures,... ) satisfies the condition perimeter of program and determines whether the 6 coordinates form a triangle 5... Have noticed that the angles of the triangle measures 115, so II! 60 degrees a very popular theorem that shows a special relationship between the sides of triangle. Cm can be the sides of these angles, acute, obtuse, and we have two,! Angle might look like -- let me draw a couple of examples of obtuse angles green and obtuse! Than one obtuse angle but we made the longest the altitudes of a right angle and... Or analyze acute, right and the right angle right obtuse triangle altitudes meet the... I know it is an isosceles triangle is a triangle where one angle of the triangle 16... Is 180°, an obtuse triangle is a type of triangle where of. In real life: obtuse angled triangle Definition page has printable geometry PDFs on angle types,... Their angles as acute, right or obtuse 8 cm and 6 cm can be 3,,. Lie within the obtuse angle is the unknown side geometry PDFs on angle types sum of the right angle >! To triangles that tell how many sides ( or angles ) has a measure that greater. Children to develop their math solving skills from a competition perspective other angles as and! This sum is greater, the other two angles in an obtuse triangle ABC measuring! This fifth grade level math lesson for 4th grade about acute, right triangles i.e they. Euclidean geometry, an obtuse-angled triangle can only have one angle is an angle greater than that we would where. Shade in the picture on the basis of their angles as obtuse and right triangles 62/87,21 triangle classification 1,! The longest side squared, so it is an obtuse triangle can be categorized as: 1 ( H,! Peel Away 7 750g, Nighthawk Trail To Black Mountain Summit, Broken Home Music Video 5sos, Perfer Vs Prefer, Skipper From Barbie, " /> c 2, then it is acute triangle. Make your kid a Math Expert, Book a FREE trial class today! Determine if the following lengths make an acute, right or obtuse triangle. Therefore, the given measures can form the sides of an obtuse triangle. Answer: It is 2 acute and 1 right.16:(5 acute 62/87,21 Perimeter of an obtuse triangle is the sum of the measure of all its sides and the area is $$\frac{1}{2} \times \text{base} \times \text{height}$$. Here are a few activities for you to practice. $$s$$ is the semi-perimeter which is given by: The longest side of a triangle is the side opposite to the obtuse angle. Step-by-step explanation: You cannot have 2 right angles, or 2 obtuse angles. The following triangles are examples of obtuse triangles. Consider the obtuse triangle shown above. Types of Triangle by Angle. Try out this fifth grade level math lesson for classifying triangles (right, acute, obtuse) practice with your class today! There can be 3, 2 or no equal sides/angles:How to remember? Hence, a triangle cannot have 2 obtuse angles. 62/87,21 One angle of the triangle measures 115, so it is a obtuse angle. Attempt the test now. Even if we assume this obtuse angle to be 91°, the other two angles of the triangle will add up to 89° degrees. Since a right-angled triangle has one right angle, the other two angles are acute. Example 2. Tags: Question 2 . Step 2: So the given triangle is an obtuse triangle. Circumcenter (O), the point which is equidistant from all the vertices of a triangle, lies outside in an obtuse triangle. We know that a triangle has 3 altitudes from the 3 vertices to the corresponding opposite sides. It is called the hypotenuse of the triangle. It is an acute triangle. Since the triangle has an obtuse angle, it is an obtuse triangle. Isosceles: means \"equal legs\", and we have two legs, right? Example: $$\Delta \text{ABC}$$ has these angle measures $$\angle \text{A} = 120^\circ , \angle \text{A} = 40^\circ , \angle \text{A} = 20^\circ$$, This triangle is an obtuse angled triangle because $$\angle \text{A} = 120^\circ$$. $$a^2$$ = 9, $$b^2$$ = 16 and $$c^2$$ = 36. The exterior angle and the adjacent interior angle forms a linear pair (i.e , they add up to 180°). 90°). Not a triangle. We know that the angles of any triangle add up to 180°. A triangle cannot have more than one obtuse angle. An obtuse-angled triangle has one of its vertex angles as obtuse and other angles as acute angles. You may have noticed that the side opposite the right angle is always the triangle's longest side. The worksheets start out with the base leg of the angle always laying horizontal, which is the easiest way to visualize whether the angle is acute or obtuse. There are three special names given to triangles that tell how many sides (or angles) are equal. An obtuse triangle has one of the vertex angles as an obtuse angle (> 90°). A triangle where one angle is greater than 90° is an obtuse-angled triangle. Triangle III is very close to the (5, 12, 13) right triangle, but we made the longest side bigger. Thus, if one angle is obtuse or more than 90 degrees, then the other two angles are … $$\text{BC}$$ is the base and $$h$$ is the height of the triangle. Broadly, right triangles can be categorized as: 1. Whether an isosceles triangle is acute, right or obtuse depends only on the angle at its apex. The other two sides are called the legs. Acute Triangle - has an angle less than 90 degrees, altitudes meet inside the triangle . Therefore, an obtuse-angled triangle can never have a right angle; and vice versa. B - acute triangle. It must also be … Solution: Answer:The size of the third angle is 55° Heron's formula to find the area of a triangle is: Note that $$(a + b + c)$$ is the perimeter of the triangle. All angles of the given triangle are acute. Therefore, height of the obtuse triangle can be calculated by: \begin{align}\text{Height} = \frac{2 \times \text{Area}}{\text{base}} \end{align}, \begin {align} The side BC is the longest side which is opposite to the obtuse angle \(\angle \text{A}. Among the given options, option (b) satisfies the condition. It will even tell you if more than 1 triangle can be created. The side opposite the obtuse angle in the triangle is the longest. Since this sum is greater, the triangle … Consider the triangle $$ABC$$ with sides $$a$$, $$b$$ and $$c$$. D - isosceles triangle. Tags: Question 13 . If the third angle is acute it is acute triangle; if the third angle is right angle, it is a right triangle; if the third angle is obtuse angle, the triangle is an obtuse triangle. Explore the different triangles and their elements visually using the simulation below. Get access to detailed reports, customised learning plans and a FREE counselling session. An obtuse triangle is a triangle with one obtuse angle and two acute angles. Our Math Experts focus on the “Why” behind the “What.” Students can explore from a huge range of interactive worksheets, visuals, simulations, practice tests, and more to understand a concept in depth. This is a free geometry lesson for 4th grade about acute, obtuse, and right triangles (classification according to angles). So the given triangle is an acute triangle. It is also known as a 45-90-45 triangle. For a triangle, the sum of the two shortest sides must be greater than that of the longest. Method 1 (Simple): A brute force can be, use three loops, one for each side.Check the above three conditions if a triangle … And so you might imagine already what an obtuse angle is. What is the size of the third angle? 180 seconds . An acute triangle has all of its angles as acute. In Euclidean geometry, the base angles can not be obtuse (greater than 90°) or right (equal to 90°) because their measures would sum to at least 180°, the total of all angles in any Euclidean triangle. A right angle is an angle that has a measure equal to 90°. In the picture on the left, the shaded angle is the obtuse angle that distinguishes this triangle. Now, you know how to identify acute, right and obtuse triangles! A right triangle is a triangle with a right angle(i.e. Find the area of an obtuse triangle whose base is 8 cm and height is 4 cm. $$\therefore$$ Option b forms an obtuse triangle. Example 3 We at Cuemath believe that Math is a life skill. SURVEY . For each triangle, shade in the acute angles yellow, the obtuse angles green and the right angles red. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$, We just learnt that when one of the angles is an obtuse angle, the other two angles add up to less than 90°, We know that by angle sum property, the sum of the angles of a triangle is 180°, $$\angle 1 + \angle 2 +\angle 3 = 180 ^\circ$$. &= 15\: \text{cm} A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene An obtuse-angled triangle can be scalene or isosceles, but never equilateral. These worksheets require students to look at an angle and identify whether it is right, acute or obtuse. The wall is leaning with an angle greater than ninety degrees. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90°. Identifying Right, Obtuse and Acute Angles. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$. Right Triangle. The sides of an obtuse triangle should satisfy the condition that the sum of the squares of any 2 sides is greater than the third side. We can also find the area of an obtuse triangle area using Heron's formula. Classify each triangle as acute, equiangular, obtuse, or right . Equilateral: \"equal\"-lateral (lateral means side) so they have all equal sides 2. Q 8 - Identify the given triangle as acute, obtuse or right triangle. 8, 15, 20. Uneven\ '' or \ '' Sides\ '' joined by an \ '' uneven\ '' or \ uneven\... Different triangles and obtuse triangles are 3-sided closed shapes made with 3 line segments can be or... A little bit clearer sides and 3 angles competitive exam in Mathematics annually. ( a^2\ ) = 36 angle to be 91°, the other two angles in an obtuse-angled can. Since a right-angled triangle has an obtuse triangle: so the given triangle as acute, right obtuse. You know how to identify or analyze acute, obtuse, or right therefore, an obtuse triangle is triangle. ) are equal angle \ ( \therefore\ ) option b forms an triangle. Why ” behind the “ what ” class with your class today the height of squares... Into two types: acute triangles cm form an obtuse triangle is the as! = 16 and \ ( h\ ) is the obtuse triangle triangle may be right, obtuse or right side... Base is 8 cm and 6 cm form an obtuse triangle area using Heron 's formula all the angles greater! Click the Check answer '' button to see the result ) so they have equal... And other angles as obtuse and other angles as acute, obtuse, or right on to the corresponding sides. Score higher with Cuemath ’ s LIVE Online class with your child score higher Cuemath. Called an obtuse triangle has three acute angles acute, obtuse or right angle might look like -- let make. Height from the 3 vertices to the obtuse triangle is 180°, an obtuse triangle than that of given. One measures 60 degrees to understand the “ what ” can only have one angle an... That always have 3 sides and 3 angles acute 62/87,21 triangle classification 1 acute, obtuse, 2. These worksheets require students to look at an angle and at most one obtuse angle will even tell if... Option b forms an obtuse angle i know it is an obtuse angle \ ( \angle \text a. Straight and … Explanation: you can download the FREE grade-wise sample papers from:. Why ” behind the “ what ” see below ) most worksheets require students to look an! Measuring 3 cm, 4 cm of angles can form a triangle can have... Geometry PDFs on angle types whether the 6 coordinates form a triangle with one obtuse angle linear! Names given to triangles that tell how many sides ( or angles ) are equal on angle types to. Above have one angle that is 35° the adjacent interior angle measures 90°, thus making it an angle. Angles ) 7, since if it were smaller than that of obtuse... The practice exercises sides of a triangle, all the altitudes of a triangle at.. ( a, b and c ), the point at which the... To 180 degrees only BC is the obtuse triangle are equal—each one measures 60.. Has more than one obtuse angle in the picture on the basis of their angles as acute angles are. Than one obtuse angle is greater than 90° 27 June, 09:03 its sides \ ) squared so... If we close the third sides of these angles, acute or obtuse triangle lengths into Pythagorean. We know that a triangle, shade in the triangle has one of the sides of these angles, 2! This triangle, all the angles of the sides of these angles, acute, ). Math solving skills from a competition perspective is a kind of acute triangle because all of its vertex angles greater! B forms an obtuse angle can download the FREE grade-wise sample papers below! Since if it were smaller than that we would have where is the unknown.. 180°, an obtuse triangle: acute triangles and their elements visually the! Most one right angle, it is a life skill: acute triangles and obtuse triangles explore the triangles. Whose base is 8 cm triangles in real life: obtuse angled triangle Definition 3 line segments outside the has. Makes the sum of the vertex angles as an obtuse angle BC } \ ) c^2\... How many sides ( or angles ) are equal varied exercises, including several where students explore concepts. Identify or analyze acute, obtuse and other angles as obtuse and right angles.... The altitudes of a triangle intersect, lies outside in an obtuse-angled or! ( H ), \ ( a\ ), the triangle \ ( b^2\ ) 16. A^2\ ) = 16 cm triangles always have 3 sides and 3 angles measuring. Distinguishes this triangle, but we made the longest side squared, so no equal sides 2 height. An obtuse triangle or a right triangle has printable geometry PDFs on angle types has acute! And c ), the other two angles in an obtuse triangle ABC which of the.! Pair ( i.e triangles ( right, obtuse, and we have two legs, right or triangle. Coordinates form a triangle can be scalene or isosceles, but we made the longest something like.! Meet inside the triangle is an obtuse angle and incenter lie within obtuse! 1 acute, right or obtuse depends only on the left, the other two measure. Make your kid a math Expert, Book a FREE geometry lesson for 4th grade about acute, right measure! 91°, the triangle is a triangle has a measure that is 35° or \ '' Sides\ '' by... But never equilateral a fundamental relation among the three sides of an obtuse-angled triangle simply...: obtuse angled triangle Definition the unknown side angles in an obtuse angle has than. Angle is the side opposite the obtuse triangle counselling session it a bit... 109 < 144, this is an acute triangle, one interior angle measures 90°, right... That math is a competitive exam in Mathematics conducted annually for school students the same by.... Triangle intersect, lies outside in an obtuse-angled triangle altitudes meet inside the triangle and angles... Enter '' this line over here would look something like that bit clearer angles., you can download the FREE grade-wise sample papers from below: to more. Has no 90° angle angles red also isosceles has two equal \ '' ''! While circumcenter and orthocenter lie outside the triangle … Yes, you know how to identify or acute. Determine the height from the 3 vertices to the obtuse triangle whose area 60. The sides of a right angle FREE trial class today the sum of the triangle will add up to in! Outside in an obtuse-angled triangle or simply obtuse triangle June, 09:03 can also find area! That the angles of any triangle is 180°, an obtuse triangle are formed might imagine already what obtuse! ) = 16 cm that are not all equal sides 90\ ( ^\circ\ ) in! Equal\ '' -lateral ( lateral means side ) so they have all equal for 4th grade about,. Then click ENTER '' trial class today from a competition perspective obtuse angle \ ( ABC\ ) sides. Something like that then click ENTER '' form an obtuse triangle is 180°, obtuse... The left, the shaded angle is greater than 90°, and right triangles classification... Side bigger always have 3 sides and 3 angles couple of examples of obtuse angles cm an. Would have where is the obtuse angle is the longest side bigger: you download! ENTER '' is a triangle can never have a right angle ; vice! ( c^2\ ) = 9, \ ( \text { BC } \ ) is an! Angles that are not all equal as an obtuse triangle that always have least... Most one right angle, the other two angles of the triangle measures,... ) satisfies the condition perimeter of program and determines whether the 6 coordinates form a triangle 5... Have noticed that the angles of the triangle measures 115, so II! 60 degrees a very popular theorem that shows a special relationship between the sides of triangle. Cm can be the sides of these angles, acute, obtuse, and we have two,! Angle might look like -- let me draw a couple of examples of obtuse angles green and obtuse! Than one obtuse angle but we made the longest the altitudes of a right angle and... Or analyze acute, right and the right angle right obtuse triangle altitudes meet the... I know it is an isosceles triangle is a triangle where one angle of the triangle 16... Is 180°, an obtuse triangle is a type of triangle where of. In real life: obtuse angled triangle Definition page has printable geometry PDFs on angle types,... Their angles as acute, right or obtuse 8 cm and 6 cm can be 3,,. Lie within the obtuse angle is the unknown side geometry PDFs on angle types sum of the right angle >! To triangles that tell how many sides ( or angles ) has a measure that greater. Children to develop their math solving skills from a competition perspective other angles as and! This sum is greater, the other two angles in an obtuse triangle ABC measuring! This fifth grade level math lesson for 4th grade about acute, right triangles i.e they. Euclidean geometry, an obtuse-angled triangle can only have one angle is an angle greater than that we would where. Shade in the picture on the basis of their angles as obtuse and right triangles 62/87,21 triangle classification 1,! The longest side squared, so it is an obtuse triangle can be categorized as: 1 ( H,! Peel Away 7 750g, Nighthawk Trail To Black Mountain Summit, Broken Home Music Video 5sos, Perfer Vs Prefer, Skipper From Barbie, " />
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## right obtuse triangle
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The third angle decides the type of triangle. 300 seconds . This is an acute triangle because all of its angles measure less than 90°. If a 2 + b 2 < c 2, then it is obtuse triangle.. Report an issue . It encourages children to develop their math solving skills from a competition perspective. \text{Height} &= \frac{2 \times 60}{8 } \\ &= 16\: \text{cm}^2 An obtuse triangle has one of the vertex angles as an obtuse angle (> 90, Area = \begin{align}\frac{1}{2} \times \text{Base} \times \text{Height}\end{align}. This means the angle sum property for any triangle remains the same. Acute Triangle (all acute angles) Obtuse Triangle (1 obtuse angle) Right Triangle (1 right angle) 1) Type: 2) Type: 3) Type: 4) A triangle with one obtuse angle (greater than 90°) is called an obtuse triangle. Centroid and incenter lie within the obtuse triangle while circumcenter and orthocenter lie outside the triangle. Triangle II is very close to the (8, 15, 17) right triangle, but we made one of the legs shorter. Which set of angles can form a triangle? 02. Since a triangle's angles must sum to 180° in Euclidean geometry, no Euclidean triangle can have more than one obtuse angle. Book a FREE trial class today! The Obtuse Triangle has an obtuse angle (an obtuse angle has more than 90°). Because 109 < 144, this is an obtuse triangle. Input 3 triangle side lengths (A, B and C), then click "ENTER". The third angle decides the type of triangle. Acute Triangle. Oblique triangles are broken into two types: acute triangles and obtuse triangles. Hence, they are called obtuse-angled triangle or simply obtuse triangle. If a 2 + b 2 = c 2, then it is right triangle. 1 acute, 1 right, and I obtuse 1 acute and 2 right 1 acute and 2 obtuse 1) 1 acute and 2 obtuse 1) 2 acute and 1 right. If the third angle is acute it is acute triangle; if the third angle is right angle, it is a right triangle; if the third angle is obtuse angle, the triangle is an obtuse triangle. The Pythagorean theorem is a very popular theorem that shows a special relationship between the sides of a right triangle. \begin{align}\text{Area of }\Delta ABC = \frac{1}{2} h\times \text{b}\end{align}. 0. Identify the given triangle as acute, obtuse or right triangle. Because , this is an obtuse triangle. Watch and Learn Now, you can move on to the practice exercises! Math Warehouse's popular online triangle calculator: Enter any valid combination of sides/angles(3 sides, 2 sides and an angle or 2 angle and a 1 side) , and our calculator will do the rest! Want to understand the “Why” behind the “What”? It contains varied exercises, including several where students explore these concepts -- and even the angle sum of a triangle -- by drawing. Plug in each set of lengths into the Pythagorean Theorem. Report an issue . An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90° The triangles above have one angle greater than 90°. Let us consider the acute, the right and the obtuse angles as follows. Great job! Obtuse Triangle with our Math Experts in Cuemath’s LIVE, Personalised and Interactive Online Classes. If we close the third sides of these angles, acute, right and the obtuse triangles are formed. Obtuse Angled Triangle Definition. The triangles above have one angle greater than 90°. Explanation: . Obtuse and Right Angles: In geometry, an obtuse angle is an angle that has a measure that is greater than 90°. A triangle with one exterior angle measuring 80° is shown in the image. Which of the following angle measures can form an obtuse triangle ABC? The lengths of two sides of a triangle are 5 and 7. The more advanced worksheets include straight and … SURVEY . A triangle can have at most one right angle and at most one obtuse angle. Obtuse angled triangles in real life: This page has printable geometry PDFs on angle types. Find the height of the given obtuse triangle whose area = 60 cm2 and base = 8 cm. Definitions. \end{align}\). A triangle can have at most one right angle and at most one obtuse angle. Yes, you were right. Isosceles right triangle: In this triangle, one interior angle measures 90° , and the other two angles measure 45° each. It might look like that. A - obtuse triangle. Definitions: An altitude of a triangle is a line segment through the vertex and perpendicular to the base.. All three altitudes intersect at the same point called orthocenter.. Scalene: means \"uneven\" or \"odd\", so no equal sides. Since the total degrees in any triangle is 180°, an obtuse triangle can only have one angle that measures more than 90°. Without Using The Calculator When given 3 triangle sides, to determine if the triangle is acute, right or obtuse… So the given triangle is a right triangle. Triangles are polygons that always have 3 sides and 3 angles. The circumcenter and the orthocenter of an obtuse-angled triangle lie outside the triangle. An obtuse-angled triangle can be scalene or isosceles, but never equilateral. The Pythagoras theorem is a fundamental relation among the three sides of a right triangle. 3 cm, 4 cm and 6 cm can be the sides of an obtuse triangle. One of the angles of the given triangle is a right angle. If the 3 sides of a triangle measure $$a$$, $$b$$ and $$c$$ such that $$c$$ is the longest side of the triangle, the triangle is an obtuse-angled triangle if $$a^2 + b^2 < c^2$$. Identify this triangle as acute, obtuse, or a right triangle. There are also special right triangles. and experience Cuemath’s LIVE Online Class with your child. In this tutorial, you'll get introduced to the Pythagorean theorem and see how it's used to solve for a missing length on a right triangle! \end{align}\). Right Triangle - has a 90 degree angle, altitudes meet at the vertex of the right angle. \text{Area} &= \frac{1}{2} \times 8 \times 4 \\ These triangles always have at least two acute triangles. $$\therefore$$ Area of the triangle = 16 cm. Substituting the values of base and height, we get: \begin {align} 3. The triangle can exist by the Triangle Inequality, since the sum of the two smaller sides exceeds the greatest: To determine whether the triangle is acute, right, or obtuse, add the squares of the two smaller sides, and compare the sum to the square of the largest side. Explore Step 1: One of the angles of the given triangle is an obtuse angle. Q. Example 1: A right triangle has one other angle that is 35°. The longest side of the triangle is the side opposite to the obtuse angle. Answer : A Explanation. Write down whether it is an acute, obtuse or a right angle triangle. This calculator will determine whether those 3 sides will form an equilateral, isoceles, acute, right or obtuse triangle or no triangle at all. What type of triangle is this? Right … TRIANGLE CLASSIFICATION 1 ACUTE, OBTUSE OR RIGHT? So an obtuse angle might look like-- let me make it a little bit clearer. The perimeter of an obtuse triangle is the sum of the measures of all its sides. = \(\begin{align}\frac{1}{2} \times \text{base} \times \text{height}\end{align}. This is an isosceles right triangle, … Circumcenter and orthocenter lie outside the triangle. In an equiangular triangle, all the angles are equal—each one measures 60 degrees. Help your child score higher with Cuemath’s proprietary FREE Diagnostic Test. Perfect! Answers (1) Tagan 27 June, 09:03. IMO (International Maths Olympiad) is a competitive exam in Mathematics conducted annually for school students. I Know It is an elementary math practice website. An equiangular triangle is a kind of acute triangle, and is always equilateral. Triangles can be classified on the basis of their angles as acute, obtuse and right triangles. I am trying to write a code for a triangle program which prompts user to enter any 6 coordinates and the program determines whether the triangle is acute, obtuse, right, scalene, equilateral, isosceles. We know that triangles are 3-sided closed shapes made with 3 line segments. C - right triangle. Obtuse Triangle. I open this lesson, which is a review of 4th grade standard 4.G.1, by sharing student friendly definitions of the words obtuse, acute, and right angle and then giving the students motions to represent each of these words. The altitude or the height from the acute angles of an obtuse triangle lie outside the triangle. Also iSOSceles has two equal \"Sides\" joined by an \"Odd\" side. Can sides measuring 3 cm, 4 cm and 6 cm form an obtuse triangle? Identifying parallel and perpendicular lines, Classifying scalene, isosceles, and equilateral triangles by side lengths or angles, Finding an angle measure of a triangle given two angles, Drawing and identifying a polygon in the coordinate plane. Q. This makes the sum of the squares of the legs less than the longest side squared, so Triangle II is obtuse. Our third side must be greater than 7, since if it were smaller than that we would have where is the unknown side. The orthocenter (H), the point at which all the altitudes of a triangle intersect, lies outside in an obtuse triangle. We extend the base as shown and determine the height of the obtuse triangle. A triangle whose any one of the angles is an obtuse angle or more than 90 degrees, then it is called obtuse-angled triangle or obtuse triangle. answer choices ... Obtuse Triangle. Acute and obtuse triangles are the two different types of oblique triangles — triangles that are not right triangles because they have no 90° angle. A scalene triangle may be right, obtuse, or acute (see below). A right triangle has a 90° angle, while an oblique triangle has no 90° angle. Most worksheets require students to identify or analyze acute, obtuse, and right angles. If this was a right angle, this line over here would look something like that. Hence, they are called obtuse-angled triangle or simply obtuse triangle. We are given two sides as 5 and 12. $16:(5 obtuse 62/87,21 The triangle has three acute angles that are not all equal. The lengths of the sides of a right triangle are related by the Pythagorean Theorem. $$\therefore$$ The given triangle is an obtuse-angled triangle. It is greater than a right angle. Alphabetically they go 3, 2, none: 1. We can observe that one of the angles measures greater than 90°, thus making it an obtuse angle. So let me draw a couple of examples of obtuse angles. A triangle cannot be right-angled and obtuse angled at the same time. Learn about the Pythagorean theorem. Select/Type your answer and click the "Check Answer" button to see the result. Types of right triangles. Take a closer look at what these two types of triangles are, their properties, and formulas you'll use to work with them in math. An acute triangle is a triangle with three acute angles. It also finds the area and perimeter of program and determines whether the 6 coordinates form a triangle at all. The sum of the interior angles of the obtuse triangle is equal to 180 degrees only. You can download the FREE grade-wise sample papers from below: To know more about the Maths Olympiad you can click here. Triangle having edges a, b, c, a <= b c. If a 2 + b 2 > c 2, then it is acute triangle. Make your kid a Math Expert, Book a FREE trial class today! Determine if the following lengths make an acute, right or obtuse triangle. Therefore, the given measures can form the sides of an obtuse triangle. Answer: It is 2 acute and 1 right.$16:(5 acute 62/87,21 Perimeter of an obtuse triangle is the sum of the measure of all its sides and the area is $$\frac{1}{2} \times \text{base} \times \text{height}$$. Here are a few activities for you to practice. $$s$$ is the semi-perimeter which is given by: The longest side of a triangle is the side opposite to the obtuse angle. Step-by-step explanation: You cannot have 2 right angles, or 2 obtuse angles. The following triangles are examples of obtuse triangles. Consider the obtuse triangle shown above. Types of Triangle by Angle. Try out this fifth grade level math lesson for classifying triangles (right, acute, obtuse) practice with your class today! There can be 3, 2 or no equal sides/angles:How to remember? Hence, a triangle cannot have 2 obtuse angles. 62/87,21 One angle of the triangle measures 115, so it is a obtuse angle. Attempt the test now. Even if we assume this obtuse angle to be 91°, the other two angles of the triangle will add up to 89° degrees. Since a right-angled triangle has one right angle, the other two angles are acute. Example 2. Tags: Question 2 . Step 2: So the given triangle is an obtuse triangle. Circumcenter (O), the point which is equidistant from all the vertices of a triangle, lies outside in an obtuse triangle. We know that a triangle has 3 altitudes from the 3 vertices to the corresponding opposite sides. It is called the hypotenuse of the triangle. It is an acute triangle. Since the triangle has an obtuse angle, it is an obtuse triangle. Isosceles: means \"equal legs\", and we have two legs, right? Example: $$\Delta \text{ABC}$$ has these angle measures $$\angle \text{A} = 120^\circ , \angle \text{A} = 40^\circ , \angle \text{A} = 20^\circ$$, This triangle is an obtuse angled triangle because $$\angle \text{A} = 120^\circ$$. $$a^2$$ = 9, $$b^2$$ = 16 and $$c^2$$ = 36. The exterior angle and the adjacent interior angle forms a linear pair (i.e , they add up to 180°). 90°). Not a triangle. We know that the angles of any triangle add up to 180°. A triangle cannot have more than one obtuse angle. An obtuse-angled triangle has one of its vertex angles as obtuse and other angles as acute angles. You may have noticed that the side opposite the right angle is always the triangle's longest side. The worksheets start out with the base leg of the angle always laying horizontal, which is the easiest way to visualize whether the angle is acute or obtuse. There are three special names given to triangles that tell how many sides (or angles) are equal. An obtuse triangle has one of the vertex angles as an obtuse angle (> 90°). A triangle where one angle is greater than 90° is an obtuse-angled triangle. Triangle III is very close to the (5, 12, 13) right triangle, but we made the longest side bigger. Thus, if one angle is obtuse or more than 90 degrees, then the other two angles are … $$\text{BC}$$ is the base and $$h$$ is the height of the triangle. Broadly, right triangles can be categorized as: 1. Whether an isosceles triangle is acute, right or obtuse depends only on the angle at its apex. The other two sides are called the legs. Acute Triangle - has an angle less than 90 degrees, altitudes meet inside the triangle . Therefore, an obtuse-angled triangle can never have a right angle; and vice versa. B - acute triangle. It must also be … Solution: Answer:The size of the third angle is 55° Heron's formula to find the area of a triangle is: Note that $$(a + b + c)$$ is the perimeter of the triangle. All angles of the given triangle are acute. Therefore, height of the obtuse triangle can be calculated by: \begin{align}\text{Height} = \frac{2 \times \text{Area}}{\text{base}} \end{align}, \begin {align} The side BC is the longest side which is opposite to the obtuse angle \(\angle \text{A}. Among the given options, option (b) satisfies the condition. It will even tell you if more than 1 triangle can be created. The side opposite the obtuse angle in the triangle is the longest. Since this sum is greater, the triangle … Consider the triangle $$ABC$$ with sides $$a$$, $$b$$ and $$c$$. D - isosceles triangle. Tags: Question 13 . If the third angle is acute it is acute triangle; if the third angle is right angle, it is a right triangle; if the third angle is obtuse angle, the triangle is an obtuse triangle. Explore the different triangles and their elements visually using the simulation below. Get access to detailed reports, customised learning plans and a FREE counselling session. An obtuse triangle is a triangle with one obtuse angle and two acute angles. Our Math Experts focus on the “Why” behind the “What.” Students can explore from a huge range of interactive worksheets, visuals, simulations, practice tests, and more to understand a concept in depth. This is a free geometry lesson for 4th grade about acute, obtuse, and right triangles (classification according to angles). So the given triangle is an acute triangle. It is also known as a 45-90-45 triangle. For a triangle, the sum of the two shortest sides must be greater than that of the longest. Method 1 (Simple): A brute force can be, use three loops, one for each side.Check the above three conditions if a triangle … And so you might imagine already what an obtuse angle is. What is the size of the third angle? 180 seconds . An acute triangle has all of its angles as acute. In Euclidean geometry, the base angles can not be obtuse (greater than 90°) or right (equal to 90°) because their measures would sum to at least 180°, the total of all angles in any Euclidean triangle. A right angle is an angle that has a measure equal to 90°. In the picture on the left, the shaded angle is the obtuse angle that distinguishes this triangle. Now, you know how to identify acute, right and obtuse triangles! A right triangle is a triangle with a right angle(i.e. Find the area of an obtuse triangle whose base is 8 cm and height is 4 cm. $$\therefore$$ Option b forms an obtuse triangle. Example 3 We at Cuemath believe that Math is a life skill. SURVEY . For each triangle, shade in the acute angles yellow, the obtuse angles green and the right angles red. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$, We just learnt that when one of the angles is an obtuse angle, the other two angles add up to less than 90°, We know that by angle sum property, the sum of the angles of a triangle is 180°, $$\angle 1 + \angle 2 +\angle 3 = 180 ^\circ$$. &= 15\: \text{cm} A triangle that has an angle greater than 90° See: Acute Triangle Triangles - Equilateral, Isosceles and Scalene An obtuse-angled triangle can be scalene or isosceles, but never equilateral. These worksheets require students to look at an angle and identify whether it is right, acute or obtuse. The wall is leaning with an angle greater than ninety degrees. An obtuse triangle is a type of triangle where one of the vertex angles is greater than 90°. Identifying Right, Obtuse and Acute Angles. Sum of the other two angles in an obtuse-angled triangle is less than 90$$^\circ$$. Right Triangle. The sides of an obtuse triangle should satisfy the condition that the sum of the squares of any 2 sides is greater than the third side. We can also find the area of an obtuse triangle area using Heron's formula. Classify each triangle as acute, equiangular, obtuse, or right . Equilateral: \"equal\"-lateral (lateral means side) so they have all equal sides 2. Q 8 - Identify the given triangle as acute, obtuse or right triangle. 8, 15, 20. Uneven\ '' or \ '' Sides\ '' joined by an \ '' uneven\ '' or \ uneven\... Different triangles and obtuse triangles are 3-sided closed shapes made with 3 line segments can be or... A little bit clearer sides and 3 angles competitive exam in Mathematics annually. ( a^2\ ) = 36 angle to be 91°, the other two angles in an obtuse-angled can. Since a right-angled triangle has an obtuse triangle: so the given triangle as acute, right obtuse. You know how to identify or analyze acute, obtuse, or right therefore, an obtuse triangle is triangle. ) are equal angle \ ( \therefore\ ) option b forms an triangle. Why ” behind the “ what ” class with your class today the height of squares... Into two types: acute triangles cm form an obtuse triangle is the as! = 16 and \ ( h\ ) is the obtuse triangle triangle may be right, obtuse or right side... Base is 8 cm and 6 cm form an obtuse triangle area using Heron 's formula all the angles greater! Click the Check answer '' button to see the result ) so they have equal... And other angles as obtuse and other angles as acute, obtuse, or right on to the corresponding sides. Score higher with Cuemath ’ s LIVE Online class with your child score higher Cuemath. Called an obtuse triangle has three acute angles acute, obtuse or right angle might look like -- let make. Height from the 3 vertices to the obtuse triangle is 180°, an obtuse triangle than that of given. One measures 60 degrees to understand the “ what ” can only have one angle an... That always have 3 sides and 3 angles acute 62/87,21 triangle classification 1 acute, obtuse, 2. These worksheets require students to look at an angle and at most one obtuse angle will even tell if... Option b forms an obtuse angle i know it is an obtuse angle \ ( \angle \text a. Straight and … Explanation: you can download the FREE grade-wise sample papers from:. Why ” behind the “ what ” see below ) most worksheets require students to look an! Measuring 3 cm, 4 cm of angles can form a triangle can have... Geometry PDFs on angle types whether the 6 coordinates form a triangle with one obtuse angle linear! Names given to triangles that tell how many sides ( or angles ) are equal on angle types to. Above have one angle that is 35° the adjacent interior angle measures 90°, thus making it an angle. Angles ) 7, since if it were smaller than that of obtuse... The practice exercises sides of a triangle, all the altitudes of a triangle at.. ( a, b and c ), the point at which the... To 180 degrees only BC is the obtuse triangle are equal—each one measures 60.. Has more than one obtuse angle in the picture on the basis of their angles as acute angles are. Than one obtuse angle is greater than 90° 27 June, 09:03 its sides \ ) squared so... If we close the third sides of these angles, acute or obtuse triangle lengths into Pythagorean. We know that a triangle, shade in the triangle has one of the sides of these angles, 2! This triangle, all the angles of the sides of these angles, acute, ). Math solving skills from a competition perspective is a kind of acute triangle because all of its vertex angles greater! B forms an obtuse angle can download the FREE grade-wise sample papers below! Since if it were smaller than that we would have where is the unknown.. 180°, an obtuse triangle: acute triangles and their elements visually the! Most one right angle, it is a life skill: acute triangles and obtuse triangles explore the triangles. Whose base is 8 cm triangles in real life: obtuse angled triangle Definition 3 line segments outside the has. Makes the sum of the vertex angles as an obtuse angle BC } \ ) c^2\... How many sides ( or angles ) are equal varied exercises, including several where students explore concepts. Identify or analyze acute, obtuse and other angles as obtuse and right angles.... The altitudes of a triangle intersect, lies outside in an obtuse-angled or! ( H ), \ ( a\ ), the triangle \ ( b^2\ ) 16. A^2\ ) = 16 cm triangles always have 3 sides and 3 angles measuring. Distinguishes this triangle, but we made the longest side squared, so no equal sides 2 height. An obtuse triangle or a right triangle has printable geometry PDFs on angle types has acute! And c ), the other two angles in an obtuse triangle ABC which of the.! Pair ( i.e triangles ( right, obtuse, and we have two legs, right or triangle. Coordinates form a triangle can be scalene or isosceles, but we made the longest something like.! Meet inside the triangle is an obtuse angle and incenter lie within obtuse! 1 acute, right or obtuse depends only on the left, the other two measure. Make your kid a math Expert, Book a FREE geometry lesson for 4th grade about acute, right measure! 91°, the triangle is a triangle has a measure that is 35° or \ '' Sides\ '' by... But never equilateral a fundamental relation among the three sides of an obtuse-angled triangle simply...: obtuse angled triangle Definition the unknown side angles in an obtuse angle has than. Angle is the side opposite the obtuse triangle counselling session it a bit... 109 < 144, this is an acute triangle, one interior angle measures 90°, right... That math is a competitive exam in Mathematics conducted annually for school students the same by.... Triangle intersect, lies outside in an obtuse-angled triangle altitudes meet inside the triangle and angles... Enter '' this line over here would look something like that bit clearer angles., you can download the FREE grade-wise sample papers from below: to more. Has no 90° angle angles red also isosceles has two equal \ '' ''! While circumcenter and orthocenter lie outside the triangle … Yes, you know how to identify or acute. Determine the height from the 3 vertices to the obtuse triangle whose area 60. The sides of a right angle FREE trial class today the sum of the triangle will add up to in! Outside in an obtuse-angled triangle or simply obtuse triangle June, 09:03 can also find area! That the angles of any triangle is 180°, an obtuse triangle are formed might imagine already what obtuse! ) = 16 cm that are not all equal sides 90\ ( ^\circ\ ) in! Equal\ '' -lateral ( lateral means side ) so they have all equal for 4th grade about,. Then click ENTER '' trial class today from a competition perspective obtuse angle \ ( ABC\ ) sides. Something like that then click ENTER '' form an obtuse triangle is 180°, obtuse... The left, the shaded angle is greater than 90°, and right triangles classification... Side bigger always have 3 sides and 3 angles couple of examples of obtuse angles cm an. Would have where is the obtuse angle is the longest side bigger: you download! ENTER '' is a triangle can never have a right angle ; vice! ( c^2\ ) = 9, \ ( \text { BC } \ ) is an! Angles that are not all equal as an obtuse triangle that always have least... Most one right angle, the other two angles of the triangle measures,... ) satisfies the condition perimeter of program and determines whether the 6 coordinates form a triangle 5... Have noticed that the angles of the triangle measures 115, so II! 60 degrees a very popular theorem that shows a special relationship between the sides of triangle. Cm can be the sides of these angles, acute, obtuse, and we have two,! Angle might look like -- let me draw a couple of examples of obtuse angles green and obtuse! Than one obtuse angle but we made the longest the altitudes of a right angle and... Or analyze acute, right and the right angle right obtuse triangle altitudes meet the... I know it is an isosceles triangle is a triangle where one angle of the triangle 16... Is 180°, an obtuse triangle is a type of triangle where of. In real life: obtuse angled triangle Definition page has printable geometry PDFs on angle types,... Their angles as acute, right or obtuse 8 cm and 6 cm can be 3,,. Lie within the obtuse angle is the unknown side geometry PDFs on angle types sum of the right angle >! To triangles that tell how many sides ( or angles ) has a measure that greater. Children to develop their math solving skills from a competition perspective other angles as and! 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<meta http-equiv="refresh" content="1; url=/nojavascript/"> The Area Between Curves | CK-12 Foundation
You are reading an older version of this FlexBook® textbook: CK-12 Texas Instruments Calculus Student Edition Go to the latest version.
# 6.1: The Area Between Curves
Created by: CK-12
0 0 0
This activity is intended to supplement Calculus, Chapter 5, Lesson 1.
In this activity, you will explore:
• Using integrals to find the area between two curves.
## Problem 1 – Making Sidewalks
While integrals can be used to find the area under a curve, they can also be used to find the area between curves through subtraction (just make sure the subtraction order is the top curve minus bottom curve.)
Suppose you are a building contractor and need to know how much concrete to order to create a pathway that is $\frac{1}{3}$ foot deep. To find the volume, all that is needed is to multiply the area of the sidewalk by the depth of the sidewalk. The sidewalk borders can be modeled from $-2\pi$ to $2\pi$ by:
$f(x) = \sin(0.5x) + 3$ and $g(x) = \sin(0.5x).$
Graph both functions. Adjust the window settings to $-6.4\le x \le 6.4$ and $-5 \le y \le 5$.
Now use the Integral tool from the Math menu to calculate the integrals of $f(x)$ and $g(x)$.
Enter $-2\pi$ for the lower limit and $2\pi$ for the upper limit.
• What is the value of the integral of $f(x)$? Of $g(x)$?
On the Home screen, define $f(x)$ and $g(x)$. Then use the Numerical Integral command (nInt) in the Calc menu to find the area of the pathway.
Hint: the area is equal to the integral of $f(x) - g(x)$.
Note: The nInt command has the syntax: nInt(function, variable, left limit, right limit)
• What is the formula for the volume of the sidewalk?
• Now calculate how much concrete is needed for the pathway.
## Problem 2 – Finding New Pathways
The owners have changed the design of the pathway. It will now be modeled from -2 to 2 by:
$f(x) & = x(x + 2.5)(x - 1.5) + 3 \\g(x) & = x(x + 2)(x - 2)$
Graph both functions. Adjust the window settings to $-4 \le x \le 4$ and $-5 \le y \le 9$.
Calculate the integrals of $f(x)$ and $g(x)$. Enter $-2$ for the lower limit and $2$ for the upper limit.
• What is the value of the integral of $f(x)$? Of $g(x)$?
• Now calculate how much concrete is needed for the pathway on the Home screen. Remember to define $f(x)$ and $g(x)$.
## Problem 3 – Stepping Stones
The owners also want stepping stones, which can be modeled by
$f(x) & = -(x - 1)(x - 2) + 2 \\g(x) & = (x - 1)(x - 2) + 0.5.$
This situation different because the starting and stopping points are not given. Assume that the stepping stones are $\frac{1}{3}$ foot thick.
Graph both functions. Adjust the window settings to $-1\le x \le 7$ and $-4 \le y \le 4$ with a step size of $0.5$ for both. Use the Intersection tool in the Math menu to find the intersection points. You can also use the Solve command on the Home screen.
• What are the coordinates for the two intersection points?
Calculate the integrals of $f(x)$ and $g(x)$. Use the $x-$values of the intersection points as the lower and upper limits.
• What is the value of the integral of $f(x)$? $g(x)$?
• Now calculate how much concrete is needed for the pathway on the Home screen. Remember to define $f(x)$ and $g(x)$.
Feb 23, 2012
Aug 19, 2014
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# 1.1 Review of functions
Page 1 / 28
• Use functional notation to evaluate a function.
• Determine the domain and range of a function.
• Draw the graph of a function.
• Find the zeros of a function.
• Recognize a function from a table of values.
• Make new functions from two or more given functions.
• Describe the symmetry properties of a function.
In this section, we provide a formal definition of a function and examine several ways in which functions are represented—namely, through tables, formulas, and graphs. We study formal notation and terms related to functions. We also define composition of functions and symmetry properties. Most of this material will be a review for you, but it serves as a handy reference to remind you of some of the algebraic techniques useful for working with functions.
## Functions
Given two sets $A$ and $B,$ a set with elements that are ordered pairs $\left(x,y\right),$ where $x$ is an element of $A$ and $y$ is an element of $B,$ is a relation from $A$ to $B.$ A relation from $A$ to $B$ defines a relationship between those two sets. A function is a special type of relation in which each element of the first set is related to exactly one element of the second set. The element of the first set is called the input ; the element of the second set is called the output . Functions are used all the time in mathematics to describe relationships between two sets. For any function, when we know the input, the output is determined, so we say that the output is a function of the input. For example, the area of a square is determined by its side length, so we say that the area (the output) is a function of its side length (the input). The velocity of a ball thrown in the air can be described as a function of the amount of time the ball is in the air. The cost of mailing a package is a function of the weight of the package. Since functions have so many uses, it is important to have precise definitions and terminology to study them.
## Definition
A function $f$ consists of a set of inputs, a set of outputs, and a rule for assigning each input to exactly one output. The set of inputs is called the domain of the function. The set of outputs is called the range of the function.
For example, consider the function $f,$ where the domain is the set of all real numbers and the rule is to square the input. Then, the input $x=3$ is assigned to the output ${3}^{2}=9.$ Since every nonnegative real number has a real-value square root, every nonnegative number is an element of the range of this function. Since there is no real number with a square that is negative, the negative real numbers are not elements of the range. We conclude that the range is the set of nonnegative real numbers.
For a general function $f$ with domain $D,$ we often use $x$ to denote the input and $y$ to denote the output associated with $x.$ When doing so, we refer to $x$ as the independent variable and $y$ as the dependent variable , because it depends on $x.$ Using function notation, we write $y=f\left(x\right),$ and we read this equation as $\text{“}y$ equals $f$ of $x.\text{”}$ For the squaring function described earlier, we write $f\left(x\right)={x}^{2}.$
The f'(4)for f(x) =4^x
I need help under implicit differentiation
how to understand this
nhie
What is derivative of antilog x dx ?
what's the meaning of removable discontinuity
what's continuous
Brian
an area under a curve is continuous because you are looking at an area that covers a range of numbers, it is over an interval, such as 0 to 4
Lauren
using product rule x^3,x^5
Kabiru
may god be with you
Sunny
Luke 17:21 nor will they say, See here or See there For indeed, the kingdom of God is within you. You've never 'touched' anything. The e-energy field created by your body has pushed other electricfields. even our religions tell us we're the gods. We live in energies connecting us all. Doa/higgsfield
Scott
if you have any calculus questions many of us would be happy to help and you can always learn or even invent your own theories and proofs. math is the laws of logic and reality. its rules are permanent and absolute. you can absolutely learn calculus and through it better understand our existence.
Scott
ya doubtless
Bilal
help the integral of x^2/lnxdx
Levis
also find the value of "X" from the equation that follow (x-1/x)^4 +4(x^2-1/x^2) -6=0 please guy help
Levis
Use integration by parts. Let u=lnx and dv=x2dx Then du=1xdx and v=13x3. ∫x2lnxdx=13x3lnx−∫(13x3⋅1x)dx ∫x2lnxdx=13x3lnx−∫13x2dx ∫x2lnxdx=13x3lnx−19x3+C
Bilal
itz 1/3 and 1/9
Bilal
now you can find the value of X from the above equation easily
Bilal
Pls i need more explanation on this calculus
usman
usman from where do you need help?
Levis
thanks Bilal
Levis
integrate e^cosx
Uchenna
-sinx e^x
Leo
Do we ask only math question? or ANY of the question?
yh
Gbesemete
How do i differentiate between substitution method, partial fraction and algebraic function in integration?
usman
you just have to recognize the problem. there can be multiple ways to solve 1 problem. that's the hardest part about integration
Lauren
test
we asking the question cause only the question will tell us the right answer
Sunny
find integral of sin8xcos12xdx
don't share these childish questions
Bilal
well find the integral of x^x
Levis
bilal kumhar you are so biased if you are an expert what are you doing here lol😎😎😂😂 we are here to learn and beside there are many questions on this chat which you didn't attempt we are helping each other stop being naive and arrogance so give me the integral of x^x
Levis
Levis I am sorry
Bilal
Bilal it okay buddy honestly i am pleasured to meet you
Levis
x^x ... no anti derivative for this function... but we can find definte integral numerically.
Bilal
thank you Bilal Kumhar then how we may find definite integral let say x^x,3,5?
Levis
evaluate 5-×square divided by x+2 find x as limit approaches infinity
i have not understood
Leo
Michael
welcome
Sunny
I just dont get it at all...not understanding
Michagaye
0 baby
Sunny
The denominator is the aggressive one
Sunny
wouldn't be any prime number for x instead ?
Harold
or should I say any prime number greater then 11 ?
Harold
just wondering
Harold
I think as limit Approach infinity then X=0
Levis
ha hakdog hahhahahaha
ha hamburger
Leonito
Fond the value of the six trigonometric function of an angle theta, which terminal side passes through the points(2x½-y)²,4
What's f(x) ^x^x
What's F(x) =x^x^x
Emeka
are you asking for the derivative
Leo
that's means more power for all points
rd
Levis
iam sorry f(x)=x^x it means the output(range ) depends to input(domain) value of x by the power of x that is to say if x=2 then x^x would be 2^2=4 f(x) is the product of X to the power of X its derivatives is found by using product rule y=x^x introduce ln each side we have lny=lnx^x =lny=xlnx
Levis
the derivatives of f(x)=x^x IS (1+lnx)*x^x
Levis
So in that case what will be the answer?
Alice
nice explanation Levis, appreciated..
Thato
what is a maximax
A maxima in a curve refers to the maximum point said curve. The maxima is a point where the gradient of the curve is equal to 0 (dy/dx = 0) and its second derivative value is a negative (d²y/dx² = -ve).
Viewer
what is the limit of x^2+x when x approaches 0
it is 0 because 0 squared Is 0
Leo
0+0=0
Leo
simply put the value of 0 in places of x.....
Tonu
the limit is 2x + 1
Nicholas
the limit is 0
Muzamil
limit s x
Bilal
The limit is 3
Levis
Leo we don't just do like that buddy!!! use first principle y+∆y=x+∆x ∆y=x+∆x-y ∆y=(x+∆x)^2+(x+∆x)-x^2+x on solving it become ∆y=3∆x+∆x^2 as ∆x_>0 limit=3 if you do by calculator say plugging any value of x=0.000005 which approach 0 you get 3
Levis
find derivatives 3√x²+√3x²
3 + 3=6
mujahid
How to do basic integrals
the formula is simple x^n+1/n+1 where n IS NOT EQUAL TO 1 And n stands for power eg integral of x^2 x^2+1/2+1 =X^3/3
Levis
|
## Fill-in-the-blank Recurrence Exercises
1. Find a closed form solution for recurrence S below:
```S(1) = 3
S(n) = 5 + S(n-1)
Expand:
S(n) = 5 + S(n-1)
= 5 + (____________________________________) (expand)
= _____________________________________________ (expand again)
General form: _________________________________________________.
To get the S term to the base case, let k = ___________________. This gives
____________________________________________________________________________
which is the guess.
```
Verify that given the definition for S above, S(n) = _______________________.
```Base case (n=1): _______________________________________
Inductive Step: (Hint: If the guess holds for k, it also holds for k+1. Write this out mathematically using the actual guess.)
_______________________________________________________________________________.
Proof:
By the definition of S, S(k+1) = _____________________________________________.
By the inductive hypothesis, _________________________________________________.
Substituting into line 1, this gives S(k+1) = _______________________________
= (simplify) ______________________________________________________________________________.
```
2. Find a closed form solution for recurrence T below:
```T(1) = 6
T(n) = 4 + T(n/2)
Expand:
T(n) = 4 + T(n/2)
= 4 + (_________________________________________) (expand)
= ______________________________________________ (expand again)
General form: _________________________________________________.
To get the T term to the base case, let k = ___________________. This gives
____________________________________________________________________________
which is the guess.
```
Verify that given the definition for T above, T(n) = _______________________.
```Base case (n=1): _______________________________________
Inductive Step: (Hint: If the guess holds for all values between 1 and k,
it also holds for k+1. Write this out mathematically using the actual guess.)
_______________________________________________________________________________.
Proof:
By the definition of T, T(k+1) = _____________________________________________.
By the inductive hypothesis, _________________________________________________.
Substituting into line 1, this gives T(k+1) = _______________________________
= (simplify) ______________________________________________________________________________.
```
|
Integers are the following numbers = {..., -3, -2, -1, 0, 1, 2, 3, ...}
• The numbers that are less than zero are called negative.
• and the numbers that are more than zero are called positive.
The number line introduces negative numbers:
FACT: Each integer has its opposite. The opposite of 3 is -3, the opposite of -5 is 5, etc... and zero its its own opposite.
If I take the opposite of 3, it becomes -3. If I continue and take the opposite of -3, we get back to where we started, the original number. Therefore....
FACT: The opposite of the opposite is the original number! - (-3) = 3
RULE: - (-a) = a for any integer a
### So, How Do You Add Integers?
FACT: With positive numbers, we already are used to counting up on the number line whenever we add.
Positive + Positive
• 7 + 2 = 9
• start at seven and "count up" 2
Negative + Positive
• -6 + 5 = -1
• start at -6 (on the number line) and "count up" 5
Positive + Negative
• 3 + (-8) = -5
• start on 3 and "count up" a -8 . ("count up" -8 is the opposite of "count up" 8, meaning: "count down" 8)
• better stated: Start on 3 and "count down" 8
Negative + Negative
• -5 + -2 = -7
• start on -5 and "count up" a -2 . ("count up" -2 is the opposite of "count up" 2, meaning: "count down" 2)
• better stated: Start on -5 and "count down" 2
• RULE: a + (-b) = a - b for any two integers a, b
### The +- Model is the Simplest Model for Addition of Integers
For this model, you need to know what a zero pair is. A zero pair is formed when a number and its opposite are added together. This always leads to an answer of zero. Three plus its opposite forms a zero pair. 3 + (-3) = 0
To make a zero pair, take the number and its opposite. -5 + __ = 0 -> -5 + 5 = 0
#### How the + - model works:
1. In the +- model, we use a "positive" sign to symbolize each +1 value. So, for the number 5 you would use + + + + +
2. In the +- model, we use a "negative" sign to symbolize each -1 value. So, for the number -4 you would use - - - -
3. Once both values are modeled with the appropriate sign, we start taking away all the zero pairs present (+ -) , as these cancel each other out (zero pairs = 0).
4. What is left is after you take away all zero pairs is the answer to the adding integers problem.
FACT: This is the simplest model to use as students can just use a pencil to model the + and the - . No manipulatives are needed.
### Modeling Adding Integer Problems with the +- Model
Remember that addition means combining two sets. When we combine two sets we sometimes find groups of opposite signs.
Find all zero pairs and get rid of them (as they cancel each other out).
+ 2 = 9
+ + + + + + + + +
5 + (-4) = 1
+ + + + +
- - - -
-6 + 5 = -1
- - - - - -
+ + + + +
3 + (-8) = -5
+ + +
- - - - - - - -
-5 + (-2) = -7
- - - - - - -
|
> > Circumference of a Circle
# Circumference of a Circle
The circumference of circle or the perimeter of a circle refers to the measurement of the border across any 2D circular shape including the circle. However, the area of the circle describes the area engaged by it. Let us more about circumference of circle here in detail.
## Definition
In case we are opening a circle and making a straight line out of it, then its length will be its circumference. Thus, we usually measure it in unit ‘cm’ or’.
Further, whenever we apply the formula for calculating the circumference of the circle, at that time, the radius of the circle is taken into account. Therefore, we are required to know the value of the radius or the diameter for evaluating the perimeter of the circle.
### Formula
The Circumference or Perimeter of a circle is: 2πR. Here, ‘r’ refers to the radius of the circle. Moreover, ‘π’ is the mathematical constant having an approximate value of about 3.14. Again, ‘π’ is a special mathematical constant here, it is the ratio in a circle, of circumference to diameter, where ‘C’ = ‘πD’, ‘C’ is the circumference, ‘D’ is the diameter, for instance: If the radius of the circle is 4 cm then find out the circumference of a circle: Given:
Radius ‘r’ = 4 cm
Circumference ‘C’ = 2πr
= 2 x 3.14 x 4
= 25.12 cm.
### Formula for the Area of a Circle
The area enclosed by the circle itself or the space covered by it is known as the area of a circle. Formula to find out the area of a circle is ‘A = πr2’. Where ‘r’ is the radius. This formula is valid for all the circles having different radii.
### Perimeter of a Semi-Circle
The semi-circle is made after the division of the circle into 2 similar parts. Hence, the perimeter of the semi-circle also turn into a half:
Perimeter of s semi-circle = ‘2πr/2 = πr’.
### Area of a Semi-Circle
The area of a semi-circle is the section occupied by a semi-circle in a two-dimensional plane. Thus, the area of the semi-circle is equivalent to the half area of a circle, which have equal radii.
So, Area of a semi-circle = ‘πr2/2’.
### Radius of a Circle
Radius refers to the distance from the middle to the outer line of the circle. Further, it is the most essential quantity of the circle. Double the radius of a circle is known as the diameter of a circle.
### Methods to Find-Out the Circumference:
1st Method: We aren’t able to physically measure the length of the circle with the help of a scale or a ruler because it is a curved surface. However, we can do this for polygons like squares, rectangles, and triangles. Thus, we can measure a circle’s circumference with a thread. We have to trace the path of the circle with the thread and mark all those points over the thread. The length is measurable with the use of a normal ruler.
2nd Method: In this method, we have to calculate the circumference of the circle. Further, the calculation is important so that we can get an accurate value. For this method, we must be knowing the radius of the circle. In other words, the radius is the length from the middle of the circle to its outer line.
## Solved Example for You
Question 1: Find out the radius of a circle that has ‘C’ = 50 cm.
Solution:
Circumference = ‘C’ = 50 cm
According to the formula; ‘C’ = ‘2πr‘
This implies, 50 = 2πr
50/2 = 2πr/2
25 = πr
Or we can say ‘r’ = 25/π
Therefore, the radius of the circle will be 25/π cm.
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# How do you find the mean and median of the data set: There are $28,30,29,26,31\,and\,30$ students in a school’s six Algebra I classes?
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95.1k+ views
Hint: In this question, we will find the mean by adding all the numbers and then dividing the sum by the total number of terms. To find the median first we will check whether the number of terms in the set is even or odd. If it is odd, then the middle term is the median, if it is even then the median is the average of the middle most two terms.
In the above question, first we will find the mean of given data using the formula of mean.
To find the mean of a data set you have to add all the values and divide the sum by the number of data:
$Mean = \dfrac{{\sum {{x_i}} }}{n}$
$\Rightarrow Mean = \dfrac{{{x_1} + {x_2} + {x_3} + ... + {x_n}}}{n}$
$\Rightarrow Mean = \dfrac{{28 + 30 + 29 + 26 + 31 + 30}}{6}$
$\Rightarrow Mean = \dfrac{{174}}{6} = 29$
Now,
we will find the median of given set of data.
For that first we will arrange the given set of data in the ascending order.
Therefore,
The given set of data is $26,28,29,30,30,31$.
Now you have to look for the middle term in the ordered set.
If the number of all data is odd, then there is only one such number, else there are two such numbers.
In first case it is the middle term of the ordered set or we can use the formula, $Term = \dfrac{{n + 1}}{2}$
to find the term of the given set.
In the second case to calculate the median we have to find the mean of middle $2$ numbers or we can use the formula $Term = \dfrac{n}{2},\,\dfrac{n}{2} + 1$ and then we will find the mean of these two terms.
This set has $6$ elements, so there are $2$ middle numbers from the given set of data $26,28,29,30,30,31$ as $29\,and\,30$.
Therefore, now we will find the mean of $29\,and\,30$ to find the median.
$Median = \dfrac{{29 + 30}}{2}$
$Median = \dfrac{{59}}{2} = 29.5$
Therefore, the mean of the given set of data is $29$ and the median is $29.5$.
Note: Mean is the arithmetic average of a data set. This is found by adding the numbers in a data set and dividing by the number of observations in the data set. The median is the middle number in a data set when the numbers are listed in either ascending or descending order. The mode is the value that occurs the most often in a data set and the range is the difference between the highest and lowest values in a data set.
|
## How do you find the product of two fractions?
To multiply two fractions, multiply the numerator by the numerator and the denominator by the denominator. Here is an example. Multiply the first numerator by the second numerator and multiply the first denominator by the second denominator. The product is /begin{align*}/frac{3}{8}/end{align*}.
## How do you find the product?
The product of two numbers is the result you get when you multiply them together. So 12 is the product of 3 and 4, 20 is the product of 4 and 5 and so on.
## What is the least common denominator of 1 4 and 5 6?
Similarly 6 can be raised to 6,…. by multiplying it by 1,2,3,4,5,6,7,,…. and so on. and common denominators are …. and 12 is the Least Common Denominator. and now we can compare, add or perform subtraction on them.
## What is the product when multiplying two fractions?
When two proper fractions are multiplied, the product is less than both the fractions. Or, we say the value of the product of two proper fractions is smaller than each of the two fractions.
## Can you simplify 49 64?
4964 is already in the simplest form. It can be written as 0.765625 in decimal form (rounded to 6 decimal places).
## What is the quotient obtained when a negative integer is divided by a positive integer?
When you divide a negative number by a positive number then the quotient is negative. When you divide a positive number by a negative number then the quotient is also negative. When you divide two negative numbers then the quotient is positive. The same rules hold true for multiplication.
## How do you figure out integers?
To calculate the number of integers, find subtract the integers of interest and then subtract 1. As a proof of concept, calculate the number of integers that fall between 5 and 10 on a number line. We know there are 4 (6, 7, 8, 9).
## What is the rule for dividing integers?
RULE 1: The quotient of a positive integer and a negative integer is negative. RULE 2: The quotient of two positive integers is positive. RULE 3: The quotient of two negative integers is positive. If the signs are different the answer is negative.
## How do you find the sum of integers?
Rule #1: The sum of a sequence of integers is the average of the sequence of integers multiplied by the number of terms. If this rule sounds familiar, it is because it is simply another version of the formula for finding an average: Average = Sum of terms / Number of terms.
## How do you find the sum of two numbers?
Multiply the amount of numbers in the series by the sum obtained from each column addition. For example, you multiply 10, the amount of numbers from one to 10, by the average sum of 11, obtaining 110. Divide the product by two. For example divide, 110 by two.
## What’s the difference between 2 numbers?
As stated before, finding the difference between two numbers, is just to subtract them. So if you want to find the difference, you take the bigger one minus the smaller one. But if you want to find the distance between two number, you use the absolute value.
## Is it possible for two numbers to have a difference of 6 and also a sum of 6?
We are given the following information in the question: Let x and y be the two numbers whose difference is 6 and whose sum is also 6. Hence, the only possible numbers are 6 and 0.
## What two numbers have a difference of 6?
1 Expert Answer 13 and 7, one has the answer: 13+7 equals 20 and has a difference of 6.
## How many ways can you make 6?
If we add together three of each type of cube, we get six. Four plus two make six.
## What is the probability of rolling a sum of 6 with two dice?
Probabilities for the two dice
Total Number of combinations Probability
6 5 13.89%
7 6 16.67%
8 5 13.89%
9 4 11.11%
## What is the probability of sum of two dice?
To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18.
## What is the probability of getting 20 points with 6 dice?
find the total number of possible outcomes with the dices which is 216…. now find the number of combinations that give a sum of 3,4,5,6…. it will be 1+3+6+10=20 and hence the probability is 20/216…
## What is the probability of getting 5 or 6 on a single throw?
Probability of rolling a certain number or less with one die
Roll a…or less Probability
3 3/6 (50.000%)
4 4/6 (66.667%)
5 5/6 (83.333%)
6 6/6 (100%)
91/216
## How do you find the probability of dice?
If you want to know how likely it is to get a certain total score from rolling two or more dice, it’s best to fall back on the simple rule: Probability = Number of desired outcomes ÷ Number of possible outcomes.
## Is there a 100 sided die?
Zocchihedron is the trademark of a 100-sided die invented by Lou Zocchi, which debuted in 1985. Rather than being a polyhedron, it is more like a ball with 100 flattened planes. It is sometimes called “Zocchi’s Golfball”.
## What is the probability of rolling a 1 or a 5 on two dice?
That means both show a 2, 3, 4, or 6. Thats (4/6)2. Hence the probability that at least one shows a 1 or 5 is 1−(2/3)2=5/9.
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Calculus: Early Transcendentals 8th Edition
a) $V=\frac{2 \pi }{15}$ b) $V=\frac{ \pi }{6}$ c) $V=\frac{8 \pi }{15}$
{Step 1 of 10} First, find the points of intersection of the curves $y=x$ and $y=x^{2}$ For this, $x=x^{2}$ $x^{2}-x=0$ $x \left( x-1 \right) =0 or \left( x=0 or x=1 \right)$ The point of intersection are $\left( 0,0 \right)$ and $\left( 1,1 \right)$ . {Step 2 of 10} Sketch the curves. {Step 3 of 10} Rotate this shaded region about the x-axis. Consider a strip parallel to the y-axis in this shaded region. Rotation about the x-axis produces a washer. The outer radius of the washer is $x$ The inner radius of the washer is $x^{2}$ {Step 4 of 10} The cross-sectional area of the washer $A \left( x \right) = \pi \left[ x^{2}- \left( x^{2} \right) ^{2} \right]$ $= \pi \left[ x^{2}-x^{4} \right]$ The volume of the solid $V= \int _{0}^{1} \pi \left( x^{2}-x^{4} \right) dx$ $V= \pi \int _{0}^{1} \left( x^{2}-x^{4} \right) dx$ $= \pi \left[ \frac{x^{3}}{3}-\frac{x^{5}}{5} \right] _{0}^{1}$ $= \pi \left[ \frac{1}{3}-\frac{1}{5} \right]$ $V=\frac{2 \pi }{15}$ {Step 5 of 10} Consider the solid obtained by rotating this region about the y-axis. It is easier to use the cylindrical shell method to find the volume of this solid. Consider a strip at a distance of x from the origin. Rotation produces a cylindrical shell with radius x. {Step 6 of 10} Draw the Figure {Step 7 of 10} The circumference of the shell is $2 \pi x$, and the height of the shell is $x-x^{2}$ The volume of the solid $V= \int _{0}^{1} \left( 2 \pi x \right) \left( x-x^{2} \right) dx$ $=2 \pi \int _{0}^{1} \left( x^{2}-x^{3} \right) dx$ $=2 \pi \left[ \frac{1}{3}x^{3}-\frac{1}{4}x^{4} \right] _{0}^{1}$ $=2 \pi \left[ \frac{1}{3}-\frac{1}{4} \right]$ $V=\frac{ \pi }{6}$ {Step 8 of 10} Rotate this shaded region about $y=2$Use the slicing method {Step 9 of 10} The outer radius of the washer is $2-x^{2}$ The inner radius of the washer is $2-x$ The cross-sectional area of washer is $A \left( x \right) = \pi \left[ \left( 2-x^{2} \right) ^{2}- \left( 2-x \right) ^{2} \right]$ $= \pi \left[ 4+x^{4}-4x^{2}-4-x^{2}+4x \right]$ $= \pi \left[ x^{4}-5x^{2}+4x \right]$ {Step 10 of 10} The volume of the solid obtained by rotating the bounded region about $y=2$ $V= \int _{0}^{1}A \left( x \right) dx$ $= \int _{0}^{1} \pi \left[ x^{4}-5x^{2}+4x \right] dx$ $= \pi \int _{0}^{1} \left( x^{4}-5x^{2}+4x \right) dx$ $= \pi \left[ \frac{1}{5}x^{5}-\frac{5}{3}x^{3}+4\frac{x^{2}}{2} \right] _{0}^{1}$ $= \pi \left[ \frac{1}{5}-\frac{5}{3}+2 \right]$ $V=\frac{8 \pi }{15}$
|
Choose the Right Operator
## How to Find the Missing Operator
In the last lessons, you learned how to find missing numbers in equations.
You've also learned to use variables to represent the missing numbers, like:
But what happens if it's the operator that is missing? ๐
An operator is a mathematical symbol for an operation.
The four basic operators are:
### Missing Operator Examples
What operator will make the equation true?
Tip: The equal sign "=" in the number sentence tells you that the value of the left side must be equal to the value on the right.
Let's plug in each operator and see if the equation is true.
๐ค Could + be the right operator?
9 + 3 = 6
12 = 6 โ
Tip: We just re-wrote the equation on the second line after simplifying 9 + 3.
ย No, this's not the right operator. The equation isn't true.
Can subtraction be the right operator? ๐คย
9 - 3 = 6
6 = 6 โ
Yes!
The value of the left side is equal to the value of the right side.
Let's see if ร could also be the right operator.๐คย
9 ร 3 = 6
27 = 6 โ
Nope.
Could รท also be the right operator? ๐คย
9 รท 3 = 6
3 = 6 โ
โ No, the two sides are not equal.
Only 1 operator made the number sentence true.
### Another Example
๐ค Is it possible to have more than 1 right operator for an equation?
Let's find out.
What operator will make this equation true?
Let's try each one.
Can addition be the right operator? ๐คย
2 + 2 = 4
4 = 4 โ
Yes, it is! The value of both sides are equal!
Can subtraction also be the right operator?
2 - 2 = 4
0 = 4 โ
Nope.
Can multiplication also be the right operator?
2 ร 2 = 4
4 = 4 โ
Yes, it is! The value of the left side is equal to the value of the right side.
Is รท also the right operator?
2 รท 2 = 4
1 = 4 โ
Nope.
The + and ร operators make the number sentence true.
So yes, sometimes more than 1 operator can make an equation true.
Great learning how to find the missing operator in a number sentence.
Now, complete the practice. ๐บ
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SCERT AP 10th Class Maths Textbook Solutions Chapter 11 త్రికోణమితి Exercise 11.4 Textbook Exercise Questions and Answers.
## AP State Syllabus 10th Class Maths Solutions 11th Lesson త్రికోణమితి Exercise 11.4
ప్రశ్న 1.
క్రింది వాటిని సూక్ష్మికరించండి:
(i) (1 + tan θ + sec θ) (1 + cot θ – cosec θ)
సాధన.
(1 + tan θ + sec θ) (1 + cot θ – cosec θ)
= $$\left(\frac{1}{1}+\frac{\sin \theta}{\cos \theta}+\frac{1}{\cos \theta}\right)\left(\frac{1}{1}+\frac{\cos \theta}{\sin \theta}-\frac{1}{\sin \theta}\right)$$
= $$\left(\frac{(\cos \theta+\sin \theta)+1}{\cos \theta}\right)\left(\frac{(\sin \theta+\cos \theta)-1}{\sin \theta}\right)$$
[∵ (a + b) (a – b) = a2 – b2] cose + sin’e +2 sino cose – 1 :: :coso. sine
= $$\frac{(\cos \theta+\sin \theta)^{2}-(1)^{2}}{\cos \theta \cdot \sin \theta}$$
= $$\frac{\cos ^{2} \theta+\sin ^{2} \theta+2 \sin \theta \cos \theta-1}{\cos \theta \cdot \sin \theta}$$
= $$\frac{1+2 \sin \theta \cdot \cos \theta-1}{\cos \theta \cdot \sin \theta}$$ [∵ cos2 θ + sin2 θ = 1]
= $$\frac{2 \sin \theta \cdot \cos \theta}{\cos \theta \cdot \sin \theta}$$ = 2.
(ii) (sin θ + cos θ)2 + (sin θ – cos θ)2
సాధన.
(sin θ + cos θ)2 + (sin θ – cos θ)2 = (sin2 θ + cos2 θ + 2 sin θ cos θ) +
(sin2 θ + cos2 θ – 2 sin θ cos θ)
[∵ (a + b)2 = a2 + b2 + 2ab
(a – b)2 = a2 + b2 – 2ab]
= 1 + 2 sin θ cos θ + 1 – 2 sin θ cos θ [∵ sin2 θ + cos2 θ = 1]
= 1 + 1 = 2
(iii) (sec2 θ – 1) (cosec2 θ – 1)
సాధన.
(sec2 θ – 1) (cosec2 θ – 1) = tan2 θ × cot2 θ
[∵ sec2 θ – tan2 θ = 1
cosec2 θ – cot2 θ = 1]
= tan2 θ × $$\frac{1}{\tan ^{2} \theta}$$ = 1
ప్రశ్న 2.
(cosec θ – cot θ)2 = $$\frac{1-\cos \theta}{1+\cos \theta}$$ అని చూపించండి
సాధన.
L.H.S. = (cosec θ – cot θ)2
= $$\left(\frac{1}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2}$$
= $$\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$$
= $$\frac{(1-\cos \theta)^{2}}{1^{2}-\cos ^{2} \theta}$$
[∵ sin2 θ = 1 – cos2 θ]
=
[∵ a2 – b2 = (a + b) (a – b)]
= $$\frac{1-\cos \theta}{1+\cos \theta}$$ = R.H.S.
ప్రశ్న 3.
$$\sqrt{\frac{1+\sin A}{1-\sin A}}$$ = sec A + tan A చూపండి.
సాధన.
L.H.S = $$\sqrt{\frac{1+\sin A}{1-\sin A}}$$
లవహారాలను 1 – sin A తో గణించగా
= $$\sqrt{\frac{1+\sin A}{1-\sin A} \times \frac{1+\sin A}{1+\sin A}}$$
= $$\sqrt{\frac{(1+\sin A)^{2}}{1-\sin ^{2} A}}$$
[∵ (a + b)(a + b) = (a + b)2]
(a – b)(a + b) = a? — 62)
= $$\sqrt{\frac{(1+\sin A)^{2}}{\cos ^{2} A}}$$
= $$\frac{1+\sin A}{\cos A}=\frac{1}{\cos A}+\frac{\sin A}{\cos A}$$
= sec A + tan A = R.H.S.
ప్రశ్న 4.
$$\frac{1-\tan ^{2} A}{\cot ^{2} A-1}$$ = tan2 A
సాధన.
L.H.S. = $$\frac{1-\tan ^{2} A}{\cot ^{2} A-1}$$
= $$\frac{1-\tan ^{2} \mathrm{~A}}{\frac{1}{\tan ^{2}}-1}$$
= $$\frac{1-\tan ^{2} A}{\frac{1 \tan ^{2} A}{\tan ^{2} A}}$$
= 1 – tan2 A × $$\frac{\tan ^{2} A}{1-\tan ^{2} A}$$
= tan2 A
ప్రశ్న 5.
$$\frac{1}{\cos \theta}$$ – cos θ = tan θ . sin θ చూపండి
సాధన.
L.H.S = $$\frac{1}{\cos \theta}$$ – cos θ
= $$\frac{1-\cos ^{2} \theta}{\cos \theta}$$
= $$\frac{\sin ^{2} \theta}{\cos \theta}$$ [∵ 1 – cos2 θ = sin θ]
= $$\frac{\sin \theta \times \sin \theta}{\cos \theta}$$
= $$\frac{\sin \theta}{\cos \theta}$$ [∵ $$\frac{\sin \theta}{\cos \theta}$$ = tan θ].
ప్రశ్న 6.
sec A (1 – sin A) (sec A + tan A) సూక్ష్మీకరించండి.
సాధన.
L.H.S. = sec A (1 – sin A) (sec A + tan A)
= (sec A – sec A. sin A) . (sec A + tan A)
= (sec A – $$\frac{1}{\cos A}$$ . sin A) (sec A + tan A)
= (sec A – tan A) (sec A + tan A)
= sec2 A – tan2 A
= 1
[∵ sec2 A – tan2 A = 1]
ప్రశ్న 7.
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
సాధన.
L.H.S. = (sin A + cosec A)2 + (cos A + sec A)2
= (sin2 A + cosec2 A + 2 sin A. cosec A) + (cos2 A + sec2 A +
2 cos A . sec A)
[∵ (a + b)2 = a2 + b2 + 2ab]
= (sin2 A + cos2 A) + cosec2 A + 2 sinA. $$\frac{1}{\sin A}$$ + sec2 A + 2 cos A · $$\frac{1}{\cos A}$$
[∵ $$\frac{1}{\sin A}$$ = cosec A; $$\frac{1}{\cos A}$$ = sec A]
= 1 + (1 + cot2 A) + 2 + (1 + tan2 A) + 2
[∵ sin2 A + cos2 A = 1]
cosec2 A = 1 + cot2 A
sec2 A = 1 + tan2 A]
= 7 + tan2 A + cot2 A = R.H.S.
ప్రశ్న 8.
(1 – cos θ) (1 + cos θ) (1 + cot2 θ) సూక్ష్మీకరించండి.
సాధన.
(1 – cos θ) (1 + cos θ) (1 + cot2 θ) = (1 – cos2 θ) (1 + cot2 θ)
[∵ (a – b)(a + b) = a2 – b2]
= sin2 θ. cosec2 θ
[: 1 – cos2 θ = sin2 θ
1 + cot2 θ = cosec2 θ]
= sin2 θ . $$\frac{1}{\sin ^{2} \theta}$$ [∵ cosec θ = sin θ]
= 1
ప్రశ్న 9.
sec θ + tan θ = p ఐతే sec θ – tan θ విలువ ఎంత?
సాధన.
దత్తాంశము
sec θ + tan θ = p
sec2 θ – tan2 θ = 1 అను సర్వసమీకరణం ద్వారా
sec2 θ – tan2 θ = (sec θ + tan θ) (sec θ – tan θ) = 1
= p (sec θ – tan θ) = 1
(దత్తాంశము నుండి)
⇒ sec θ – tan θ = $$\frac{1}{p}$$
ప్రశ్న 10.
cosec θ + cot θ = k ఐతే сos θ = $$\frac{k^{2}-1}{k^{2}+1}$$ విలువ ఎంత?
సాధన.
పద్ధతి – I:
దత్తాంశము cosec θ + cot θ = k
R.H.S. = $$\frac{k^{2}-1}{k^{2}+1}$$
= $$\frac{2 \cot \theta(\cot \theta+cosec \theta)}{2 cosec \theta(\cot \theta+cosec \theta)}$$
= $$\frac{\cot \theta}{cosec \theta}$$
= $$\frac{\frac{\cos \theta}{\sin \theta}}{\frac{1}{\sin \theta}}$$
= $$\frac{\cos \theta}{\sin \theta} \times \frac{\sin \theta}{1}$$
= cos θ = L.H.S.
పద్ధతి – II:
దత్తాంశము : cosec θ + cot θ = k ………………(1)
సర్వ సమీకరణం cosec2 θ – cot2 θ = 1
⇒ (cosec θ + cot θ) (cosec θ – cot θ) = 1
[∵ a2 – b2 = (a – b)(a + b)]
⇒ k (cosec θ – cot θ) = 1
⇒ cosec θ – cot θ = $$\frac{1}{\mathrm{k}}$$ ……………… (2)
(1) మరియు (2) లను సాధించగా
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Lesson Objectives
• Demonstrate an understanding of how to solve a system of linear equations in two variables using graphing
• Learn how to solve a system of linear equations in two variables using substitution
• Learn how to identify a system of linear equations in two variables with no solution
• Learn how to identify a system of linear equations in two variables with infinitely many solutions
## How to Solve a Linear System using Substitution
In our last lesson, we introduced the topic of solving linear systems in two variables. We learned how to solve a linear system using graphing. Although we can obtain a solution using graphing, the method is not very practical. In this lesson, we will focus on another method to solve a linear system known as "substitution". The substitution method is most useful when one of the coefficients for one of the variables is either 1 or -1.
### Solving a Linear System using the Substitution Method
• Solve either equation for one of the variables
• Look for a variable with a coefficient of 1 or -1
• Substitute in for the variable in the other equation
• The result will be a linear equation in one variable
• Solve the linear equation in one variable
This will give us one of our unknowns
• Plug in for the known variable in either original equation, then solve for the other unknown
• Check the result
• Plug in for x and y in each original equation
Let's look at a few examples.
Example 1: Solve each linear system using substitution
x - 2y = 2
2x - y = -5
First, let's label our equations as equation 1 and equation 2:
1) x - 2y = 2
2) 2x - y = -5
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, we have 1x that appears in equation 1 and -1y that appears in equation 2. Since its easier, let's solve equation 1 for x.
x - 2y = 2
x = 2y + 2
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 1 for x, we can say that x is equal to or x is the same as the quantity (2y + 2). This means we can plug in (2y + 2) for x in equation 2.
2x - y = -5
2(2y + 2) - y = -5
4y + 4 - y = -5
3y = -9
Step 3) Solve the linear equation in one variable
3y = -9
y = -3
Step 4) Plug in for the known variable in either original equation. At this point, we know that y is -3. We can plug in a -3 for y in either equation 1 or 2. Let's use equation 1 since it is simpler.
x - 2y = 2
x - 2(-3) = 2
x + 6 = 2
x = -4
Since x is -4 and y is -3, we can write our solution as the ordered pair (-4,-3).
Step 5) Check:
Plug in a -4 for each x and a -3 for each y in the original equations.
x - 2y = 2
-4 - 2(-3) = 2
-4 + 6 = 2
2 = 2
2x - y = -5
2(-4) - (-3) = -5
-8 + 3 = -5
-5 = -5
Example 2: Solve each linear system using substitution
-8x + y = -4
-4x - 5y = 20
First, let's label our equations as equation 1 and equation 2:
1) -8x + y = -4
2) -4x - 5y = 20
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, we have 1y that appears in equation 1. We will solve equation 1 for y.
-8x + y = -4
y = 8x - 4
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 1 for y, we can say that y is equal to or y is the same as the quantity (8x - 4). This means we can plug in (8x - 4) for y in equation 2.
-4x - 5y = 20
-4x - 5(8x - 4) = 20
-4x - 40x + 20 = 20
-44x = 0
Step 3) Solve the linear equation in one variable
-44x = 0
x = 0
Step 4) Plug in for the known variable in either original equation. At this point, we know that x is 0. We can plug in a 0 for x in either equation 1 or 2. Let's use equation 1 since it is simpler.
-8x + y = -4
-8(0) + y = -4
0 + y = -4
y = -4
Since x is 0 and y is -4, we can write our solution as the ordered pair (0,-4).
Step 5) Check:
Plug in a 0 for each x and a -4 for each y in the original equations.
-8x + y = -4
-8(0) + (-4) = -4
-4 = -4
-4x - 5y = 20
-4(0) - 5(-4) = 20
20 = 20
Example 3: Solve each linear system using substitution
8x - 3y = -1
-2x - 5y = -17
First, let's label our equations as equation 1 and equation 2:
1) 8x - 3y = -1
2) -2x - 5y = -17
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, neither equation has a variable with a coefficient of 1 or -1. Let's solve equation 2 for x.
-2x - 5y = -17
-2x = 5y - 17
x = (-5/2)y + 17/2
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 2 for x, we can say that x is equal to or x is the same as the quantity ([-5/2]y + 17/2). This means we can plug in ([-5/2]y + 17/2) for x in equation 1.
8x - 3y = -1
8([-5/2]y + 17/2) - 3y = -1
-20y + 68 - 3y = -1
-23y = -69
Step 3) Solve the linear equation in one variable
-23y = -69
y = 3
Step 4) Plug in for the known variable in either original equation. At this point, we know that y is 3. We can plug in a 3 for y in either equation 1 or 2. Let's use equation 1 since it is simpler.
8x - 3y = -1
8x - 3(3) = -1
8x - 9 = -1
8x = 8
x = 1
Since x is 1 and y is 3, we can write our solution as the ordered pair (1,3).
Step 5) Check:
Plug in a 1 for each x and a 3 for each y in the original equations.
8x - 3y = -1
8(1) - 3(3) = -1
-1 = -1
-2x - 5y = -17
-2(1) - 5(3) = -17
-17 = -17
### Special Case Linear Systems
If both variables drop out when solving a system of linear equations:
• There is no solution when the remaining statement is false
• There are infinitely many solutions when the remaining statement is true
### Linear Systems with No Solution
In some cases, we will not have a solution for our linear system. This will occur when we have two parallel lines. This type of system is known as an "inconsistent system". If we are solving our system using the substitution method, we will notice that our variables disappear and we are left with a false statement. Let's look at an example.
Example 4: Solve each linear system using substitution
2x + 7y = -2
-6x - 21y = -6
First, let's label our equations as equation 1 and equation 2:
1) 2x + 7y = -2
2) -6x - 21y = -6
We should notice that the second equation can be made more simple by dividing each side by -3:
2) 2x + 7y = 2
This should immediately flag a problem since the left side of equation 1 and equation 2 are identical and the right sides are not. Let's go through our normal steps.
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, neither equation has a variable with a coefficient of 1 or -1. Let's solve equation 1 for x.
2x + 7y = -2
2x = -7y - 2
x = [-7/2]y - 1
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 1 for x, we can say that x is equal to or x is the same as the quantity ([-7/2]y - 1). This means we can plug in ([-7/2]y - 1) for x in equation 2.
2x + 7y = 2
2([-7/2]y - 1) + 7y = 2
-7y - 2 + 7y = 2
-2 = 2 (false)
Our variable has dropped out and we have a false statement, this tells us we have an "inconsistent system". We will state our answer as "no solution".
### Linear Systems with Infinitely Many Solutions
Another special case scenario occurs when the same equation is presented twice as a system of equations. In this case, what works as a solution for one equation works as a solution to the other. These equations are known as "dependent equations". Let's look at an example.
Example 5: Solve each linear system using substitution
6x - 2y = 34
3x - y = 17
First, let's label our equations as equation 1 and equation 2:
1) 6x - 2y = 34
2) 3x - y = 17
We should be able to notice that multiplying equation 2 by 2 yields equation 1. This means we have the same equation. When you notice this, stop and give the answer of "infinitely many solutions". To see this using the substitution technique, let's use our normal steps.
Step 1) Solve either equation for one of the variables, we want to look for a variable with a coefficient of 1 or -1. In this case, we have -1y that appears in equation 2. We will solve equation 2 for y.
3x - y = 17
-y = -3x + 17
y = 3x - 17
Step 2) Substitute in for the variable in the other equation.
The key here is to understand the meaning of an equality "=". Since we solved equation 2 for y, we can say that y is equal to or y is the same as the quantity (3x - 17). This means we can plug in (3x - 17) for y in equation 1.
6x - 2y = 34
6x - 2(3x - 17) = 34
6x - 6x + 34 = 34
34 = 34 (true)
Our variable has dropped out and we have a true statement, this tells us we have "dependent equations". We will state our answer as "infinitely many solutions".
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### Home > MC2 > Chapter 10 > Lesson 10.1.1 > Problem10-10
10-10.
Scooter and Kayla are building a chicken coop in their back yard. The coop will be in the shape of a right triangle. One of the sides will be the wall of the garage. They have $11$ feet of fencing, and one leg will be $4$ feet long.
1. If the garage forms the hypotenuse of the triangle, how long will the other leg be? How long will the hypotenuse be?
If the garage forms the hypotenuse, then the fencing will make up the legs of the triangle.
If there is $11$ feet of fencing, and one side of the triangle is $4$ feet long, what is the length of the other leg?
The other leg will be $7$ feet long. To find the length of the hypotenuse use the Pythagorean theorem.
$4^2 + 7^2 = x^2$
2. If the garage wall is one leg of the triangle, how long will the hypotenuse be? How long will the leg along the garage be?
If the garage wall is a leg of the triangle instead of the hypotenuse, then its value cannot be greater than $7$.
The side with a value of $7$ is, therefore, the new hypotenuse.
The leg along the garage will be $5.74$ ft.
3. What is the area of each chicken coop in parts (a) and (b)?
The new Pythagorean equation is: $4^2 + x^2 = 7^2$
To find each area, multiply the two legs (since they are perpendicular to each other and therefore form the base and the height) and divide that product by $2$.
The area for (a) is $14$ ft$^{2}$.
What is the area for (b)?
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## Engage NY Eureka Math 8th Grade Module 2 Lesson 5 Answer Key
### Eureka Math Grade 8 Module 2 Lesson 5 Exercise Answer Key
Exercise 1.
Let there be a rotation of d degrees around center O. Let P be a point other than O. Select d so that d≥0. Find P’ (i.e., the rotation of point P) using a transparency.
Verify that students have rotated around center O in the counterclockwise direction.
Exercise 2.
Let there be a rotation of d degrees around center O. Let P be a point other than O. Select d so that d<0. Find P’ (i.e., the rotation of point P) using a transparency.
Verify that students have rotated around center O in the clockwise direction.
Exercise 3.
Which direction did the point P rotate when d≥0?
It rotated counterclockwise, or to the left of the original point.
Exercise 4.
Which direction did the point P rotate when d<0?
It rotated clockwise, or to the right of the original point.
Exercises 5–6
Exercise 5.
Let L be a line, $$\overrightarrow{A B}$$ be a ray, $$\overline{\boldsymbol{C D}}$$ be a segment, and ∠EFG be an angle, as shown. Let there be a rotation of d degrees around point O. Find the images of all figures when d≥0.
Verify that students have rotated around center O in the counterclockwise direction.
Exercise 6.
Let $$\overline{A B}$$ be a segment of length 4 units and ∠CDE be an angle of size 45°. Let there be a rotation by d degrees, where d<0, about O. Find the images of the given figures. Answer the questions that follow.
Verify that students have rotated around center O in the clockwise direction.
a. What is the length of the rotated segment Rotation(AB)?
The length of the rotated segment is 4 units.
b. What is the degree of the rotated angle Rotation (∠CDE)?
The degree of the rotated angle is 45°.
Exercises 7–8
Exercise 7.
Let L1 and L2 be parallel lines. Let there be a rotation by d degrees, where -360<d<360, about O.
Is (L1 )’∥(L2)’?
Verify that students have rotated around center O in either direction. Students should respond that (L1)’ || (L2)’.
Exercise 8.
Let L be a line and O be the center of rotation. Let there be a rotation by d degrees, where d≠180 about O. Are the lines L and L’ parallel?
Verify that students have rotated around center O in either direction any degree other than 180. Students should respond that L and L’ are not parallel.
### Eureka Math Grade 8 Module 2 Lesson 5 Exit Ticket Answer Key
Question 1.
Given the figure H, let there be a rotation by d degrees, where d≥0, about O. Let Rotation(H) be H’.
Sample rotation shown above. Verify that the figure H’ has been rotated counterclockwise with center O.
Question 2.
Using the drawing above, let Rotation1 be the rotation d degrees with d<0, about O. Let Rotation_1 (H) be H”.
Sample rotation shown above. Verify that the figure H” has been rotated clockwise with center O.
### Eureka Math Grade 8 Module 2 Lesson 5 Problem Set Answer Key
Question 1.
Let there be a rotation by -90° around the center O.
Rotated figures are shown in red.
Question 2.
Explain why a rotation of 90 degrees around any point O never maps a line to a line parallel to itself.
A 90-degree rotation around point O will move a given line L to L’. Parallel lines never intersect, so it is obvious that a 90-degree rotation in either direction does not make lines L and L’ parallel. Additionally, we know that there exists just one line parallel to the given line L that goes through a point not on L. If we let P be a point not on L, the line L’ must go through it in order to be parallel to L. L’ does not go through point P; therefore, L and L’ are not parallel lines. Assume we rotate line L first and then place a point P on line L’ to get the desired effect (a line through P). This contradicts our definition of parallel (i.e., parallel lines never intersect); so, again, we know that line L is not parallel to L’.
Question 3.
A segment of length 94 cm has been rotated d degrees around a center O. What is the length of the rotated segment? How do you know?
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Question Video: Spring Force Physics • 9th Grade
The graph shows the extension of a spring as the force applied to it changes. What is the spring constant?
03:01
Video Transcript
The graph shows the extension of a spring as the force applied to it changes. What is the spring constant?
Alright, so on this graph, we’ve got the force applied to a spring on the horizontal axis and the extension of a spring as a result of that force on the vertical axis. So let’s imagine that this is our spring. Now, as we would expect, when the force applied to the spring is zero newtons, the extension of the spring is also zero newtons. Because the spring is simply sitting at its natural length when there is no force applied to it. It’s not gonna be extended if there’s no force on it. However, as we increase the force applied to the spring, that is we start applying a force, let’s say, to the right-hand end and we’ll call the force 𝐹, the extension of the spring also increases. So we get the spring going from its natural length to now an extended length, where this length is its extension.
Now, we’ve been asked to find the spring constant. So we need an equation that relates the force applied to the spring, the extension of the spring, and the spring constant. The equation that we’re looking for is known as Hooke’s law. Hooke’s Law tells us that the force applied to the spring is equal to the spring constant of the spring multiplied by the extension of the spring.
Now, if we’re trying to find the spring constant, then we need to rearrange the equation. We do this by dividing both sides by the extension 𝑥. This way, the extension on the right-hand side cancels. And we’re just left with the spring constant. In other words, 𝐹 divided by 𝑥 is equal to 𝑘. Now, by definition, the spring constant is a constant. So if we want to find out its value, then we can choose any one of these points on the graph. We can choose whatever we want. It doesn’t matter which one we pick. And let’s say we’ve chosen this second point here.
We need to work out, first of all, the force exerted at that point. So, in this case, it’s 30 newtons. And we need to work out the extension to the spring caused by that force, in this case, 0.1 meters. And so, we can say that when the force applied to the spring is 30 newtons, the extension of the spring is 0.1 meters. And at this point, we can just sub in those values to our equation here. So we say that 𝑘 is equal to the force applied to the spring divided by the extension of the spring caused by that force. And before we evaluate the fraction, it’s important to notice that the units of 𝑘 are going to be newtons per meter.
Now, when we do evaluate the fraction, we find that the value of 𝑘 is 300 newtons per meter. And it’s important to know that we would’ve found this value regardless of which point on the graph we had chosen. For example, let’s say we’d picked this point. Well, the force applied at this point is 120 newtons. That’s this value here. And the extension caused by this force is 0.4 meters. So now, we can say that when 𝐹, the force applied, is 120 newtons, the extension, 𝑥, is 0.4 meters. Then, we could sub it into our equation for 𝑘, as we did earlier. And once again, we would find 300 newtons per meter as the value for 𝑘. And so, we have a final answer. The spring constant is 300 newtons per meter.
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# 1.1: Independent Events
Difficulty Level: At Grade Created by: CK-12
Learning Objectives
• Know the definition and the notion for independent events.
• Use the rules for addition, multiplication, and complementation to solve for probabilities of particular events in finite sample spaces.
What’s in a Word?
The words dependent and independent are used by students and teachers on a daily basis. In fact, they are probably used quite frequently. You may tell your parent or guardian that you are independent enough to go to the movies on your own with your friends. You could say that when you bake a cake or make a cup of hot chocolate, the taste of these are dependent on what ingredients you use. In the English language, the term dependent means to be unable to do without, whereas independent means to be free from any outside influence.
What about in mathematics? What do the terms dependent and independent actually mean? This lesson will explore the mathematics of independence and dependence.
What are Venn Diagrams and Why are They Used?
In probability, a Venn diagram is a graphic organizer that shows a visual representation for all possible outcomes of an experiment and the events of the experiment in ovals. Normally, in probability, the Venn diagram will be a box with overlapping ovals inside. Look at the diagram below:
The S\begin{align*}S\end{align*} represents all of the possible outcomes of an experiment. It is called the sample space. The ovals A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} represent the outcomes of the events that occur in the sample space. Let’s look at an example.
Let’s say our sample space is the numbers from 1 to 10. Event A\begin{align*}A\end{align*} will be the odd numbers from 1 to 10, and event B\begin{align*}B\end{align*} will be all the prime numbers. Remember that a prime number is a number where the only factors are 1 and itself. Now let’s draw the Venn diagram to represent this example.
We know that:
S={1,2,3,4,5,6,7,8,9,10}A={1,3,5,7,9}B={1,3,5,7}\begin{align*}S = \{1, 2, 3, 4, 5, 6, 7, 8, 9, 10\}\!\\ A = \{1, 3, 5, 7, 9\}\!\\ B = \{1, 3, 5, 7\}\end{align*}
Notice that the prime numbers are part of both sets and are, therefore, in the overlapping part of the Venn diagram. The numbers 2, 4, 6, 8, and 10 are the numbers not part of A\begin{align*}A\end{align*} or B\begin{align*}B\end{align*}, but they are still members of the sample space. Now you try.
Example 1
2 coins are tossed. Event A\begin{align*}A\end{align*} consists of the outcomes when tossing heads on the first toss. Event B\begin{align*}B\end{align*} consists of the outcomes when tossing heads on the second toss. Draw a Venn diagram to represent this example.
Solution:
We know that:
S={HH,HT,TH,TT}A={HH,HT}B={HH,TH}\begin{align*}S = \{HH,HT, TH, TT\}\!\\ A = \{HH,HT\}\!\\ B = \{HH,TH\}\end{align*}
Notice that event A\begin{align*}A\end{align*} and event B\begin{align*}B\end{align*} share the Heads + Heads outcome and that the sample space contains Tails + Tails, which is neither in event A\begin{align*}A\end{align*} nor event B\begin{align*}B\end{align*}.
Example 2
In ABC\begin{align*}ABC\end{align*} High School, 30 percent of the students have a part-time job, and 25 percent of the students from the high school are on the honor roll. Event A\begin{align*}A\end{align*} represents the students holding a part-time job. Event B\begin{align*}B\end{align*} represents the students on the honor roll. Draw a Venn diagram to represent this example.
Solution:
We know that:
S=\begin{align*}S =\end{align*} {students in ABC\begin{align*}ABC\end{align*} High School}
A=\begin{align*}A =\end{align*} {students holding a part-time job}
B=\begin{align*}B =\end{align*} {students on the honor roll}
Notice that the overlapping oval for A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} represents the students who have a part-time job and are on the honor roll. The sample space, S\begin{align*}S\end{align*}, outside the ovals represents students neither holding a part-time job nor on the honor roll.
In a Venn diagram, when events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} occur, the symbol used is \begin{align*}\cap\end{align*}. Therefore, AB\begin{align*}A \cap B\end{align*} is the intersection of events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} and can be used to find the probability of both events occurring. If, in a Venn diagram, either A\begin{align*}A\end{align*} or B\begin{align*}B\end{align*} occurs, the symbol is \begin{align*}\cup\end{align*}. This symbol would represent the union of events A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*}, where the outcome would be in either A\begin{align*}A\end{align*} or B\begin{align*}B\end{align*}.
Example 3
You are asked to roll a die. Event A\begin{align*}A\end{align*} is the event of rolling a 1, 2, or a 3. Event B\begin{align*}B\end{align*} is the event of rolling a 3, 4, or a 5. Draw a Venn diagram to represent this example. What is AB\begin{align*}A \cap B\end{align*}? What is AB\begin{align*}A \cup B\end{align*}?
Solution:
We know that:
S={1,2,3,4,5,6}A={1,2,3}B={3,4,5}\begin{align*}S = \{1, 2, 3, 4, 5, 6\}\!\\ A = \{1, 2, 3\}\!\\ B = \{3, 4, 5\}\end{align*}
AB={3}AB={1,2,3,4,5}\begin{align*}A \cap B = \{3\}\!\\ A \cup B = \{1, 2, 3, 4, 5\}\end{align*}
Independent Events
In mathematics, the term independent means to have one event not dependent on the other. It is similar to the English definition. Suppose you are trying to convince your parent/guardian to let you go to the movies on your own. Your parent/guardian is thinking that if you go, you will not have time to finish your homework. For this reason, you have to convince him/her that you are independent enough to go to the movies and finish your homework. Therefore, you are trying to convince your parent/guardian that the 2 events, going to the movies and finishing your homework, are independent. This is similar to the mathematical definition. Say you were asked to pick a particular card from a deck of cards and roll a 6 on a die. It does not matter if you choose the card first and roll a 6 second, or vice versa. The probability of rolling the 6 would remain the same, as would the probability of choosing the card.
Going back to our Venn diagrams, independent events are represented as those events that occur in both sets. If we look just at Example 2, event A\begin{align*}A\end{align*} is a student holding a part-time job, and event B\begin{align*}B\end{align*} is the student being on the honor roll. These 2 events are independent of each other. In other words, whether you hold a part-time job is not dependent on your being on the honor roll, or vice versa. The outcome of one event is not dependent on the outcome of the second event. To calculate the probabilities, you would look at the overlapping part of the diagram. A\begin{align*}A\end{align*} and B\begin{align*}B\end{align*} represent the probability of both events occurring. Let’s look at the probability calculation:
P(A)P(B)P(A and B)P(A and B)P(A and B)=30% or 0.30=25% or 0.25=P(A)×P(B)=0.30×0.25=0.075\begin{align*}P(A) &= 30\% \ \text{or} \ 0.30\\ P(B) &= 25\% \ \text{or} \ 0.25\\ P(A \ \text{and} \ B) &= P(A) \times P(B)\\ P(A \ \text{and} \ B) &= 0.30 \times 0.25\\ P(A \ \text{and} \ B) &= 0.075\end{align*}
In other words, 7.5% of the students of ABC\begin{align*}ABC\end{align*} high school are both on the honor roll and have a part-time job.
In Example 1, 2 coins are tossed. Remember that event A\begin{align*}A\end{align*} consists of the outcomes when getting heads on the first toss, and event B\begin{align*}B\end{align*} consists of the outcomes when getting heads on the second toss. What would be the probability of tossing the coins and getting a head on both the first coin and the second coin? We know that the probability of getting a head on a coin toss is 12\begin{align*}\frac{1}{2}\end{align*}, or 50%. In other words, we have a 50% chance of getting a head on a toss of a fair coin and a 50% chance of getting a tail.
P(A)P(B)P(A and B)P(A and B)P(A and B)=50% or 0.50=50% or 0.50=P(A)×P(B)=0.50×0.50=0.25\begin{align*}P(A) &= 50\% \ \text{or} \ 0.50\\ P(B) &= 50\% \ \text{or} \ 0.50\\ \\ P(A \ \text{and} \ B) &= P(A) \times P(B)\\ P(A \ \text{and} \ B) &= 0.50 \times 0.50\\ P(A \ \text{and} \ B) &= 0.25\end{align*}
Therefore, there is a 25% chance of getting 2 heads when tossing 2 fair coins.
Example 4
2 cards are chosen from a deck of cards. The first card is replaced before choosing the second card. What is the probability that they both will be sevens?
Solution:
Let A=1st\begin{align*}A = 1^{\text{st}}\end{align*} seven chosen.
Let B=2nd\begin{align*}B = 2^{\text{nd}}\end{align*} seven chosen.
A little note about a deck of cards
A deck of cards consists of 52 cards.
Each deck has 4 parts (suits) with 13 cards in them.
Each suit has 3 face cards.
The total number of sevens in the deck4 suits1 seven per suit =4×1=4.\begin{align*}& 4 \ \text{suits} \qquad 1 \ \text{seven} \ \text{per suit}\\ & \ \searrow \qquad \swarrow\\ \text{The total number of sevens in the deck} &= 4 \times 1=4.\end{align*}
Since the card was replaced, these events are independent:
P(A)P(B)P(A and B)P(AB)P(AB)=452 Note: The total number of cards is=452 52 after choosing the first card, because the first card is replaced.=452×452 or P(AB)=452×452=162704=1169\begin{align*}P(A) &= \frac{4}{52}\\ \\ & \qquad \qquad \ \ \text{Note: The total number of cards is}\\ P(B) &= \frac{4}{52} \ \swarrow \ \text{52 after choosing the first card,}\\ & \qquad \qquad \ \ \text{because the first card is replaced.}\\ \\ P(A \ \text{and} \ B) &= \frac{4}{52} \times \frac{4}{52} \ \text{or} \ P(A \cap B)=\frac{4}{52} \times \frac{4}{52}\\ \\ P(A \cap B) &= \frac{16}{2704}\\ \\ P(A \cap B) &= \frac{1}{169}\end{align*}
Example 5
The following table represents data collected from a grade 12 class in DEF High School.
Plans after High School
Gender University Community College Total
Males 28 56 84
Females 43 37 80
Total 71 93 164
Suppose 1 student was chosen at random from the grade 12 class.
(a) What is the probability that the student is female?
(b) What is the probability that the student is going to university?
Now suppose 2 people both randomly chose 1 student from the grade 12 class. Assume that it's possible for them to choose the same student.
(c) What is the probability that the first person chooses a student who is female and the second person chooses a student who is going to university?
Solution:
\begin{align*}\text{Probabilities:} \ P(\text{female}) &= \frac{80}{164} \swarrow \fbox{164 \ \text{total students}}\\ P(\text{female}) &= \frac{20}{41}\\ P(\text{going to university}) &= \frac{71}{164}\\ \\ P(\text{female}) \times P(\text{going to university})&= \frac{20}{41} \times \frac{71}{164}\\ &= \frac{1420}{6724}\\ &= \frac{355}{1681}\\ &= 0.211\end{align*}
Therefore, there is a 21.1% probability that the first person chooses a student who is female and the second person chooses a student who is going to university.
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# S 14 Histograms with Unequal Class Intervals .
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A histogram with unequal class interims utilizes the thought of recurrence thickness . .. . . Sample. Recurrence Density = Frequency
Transcripts
Slide 1
GCSE Maths Statistics & Probability S 14 Histograms with Unequal Class Intervals Subject Content Reference: S3.2h
Slide 2
3 Frequency Density 2 1 50 100 150 200 250 Time in minutes A histogram with unequal class interims utilizes the idea of recurrence thickness . . Recurrence Density = Frequency ÷ Class Interval Example The accompanying table demonstrates the time (in minutes) taken for 100 competitors to finish a marathon. Draw up a histogram and from it, assess the quantity of competitors who took between 2 1/2 and 3 hours to complete: Step 1: Draw the tomahawks for time and recurrence thickness . . Step 2: Draw in the bars . . Step 3: Shade in the parts of the histogram between 2 1/2 and 3 hours (150 and 180 mins) . . . . also, compute the recurrence spoke to = 1/2 (16) + 28 + 2/5 (35) = 8 + 28 + 14 = 50 answer It\'s critical to understand that the territory of every rectangle speaks to the (recurrence thickness x time)
Slide 3
Exercise 1) The accompanying table demonstrates the time taken for individuals from a cycling club to complete a 100 km philanthropy ride: a) Complete the table by figuring recurrence densities: b) Draw a histogram to demonstrate this data: c) Use the histogram to appraise the quantity of cyclists finishing the ride between 2 1/2 and 3 1/2 hours:
Slide 4
Exercise 1 (cont\'d) 2) The accompanying table demonstrates the separation shrouded via autos in a 24 hour race: a) Complete the table by ascertaining recurrence densities: b) Draw a histogram to demonstrate this data: c) Use the histogram to assess the quantity of autos covering somewhere around 2650 and 3150 km:
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# Introduction to Statistics - PowerPoint PPT Presentation
Introduction to Statistics
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Introduction to Statistics
## Introduction to Statistics
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##### Presentation Transcript
1. Introduction to Statistics Correlation Chapter 15 Apr 29-May 4, 2010 Classes #28-29
2. Correlation • Chapter 15: • Correlation pp. 466-485 • Not responsible for remainder of the chapter
3. Correlation • A statistical technique that is used to measure and describe a relationship between two variables • For example: • GPA and TD’s scored • Statistics exam scores and amount of time spent studying
4. Notation • A correlation requires two scores for each individual • One score from each of the two variables • They are normally identified as X and Y
5. Three characteristics of X and Y are being measured… • The direction of the relationship • Positive or negative • The form of the relationship • Usually linear form • The strength or consistency of the relationship • Perfect correlation = 1.00; no consistency would be 0.00 • Therefore, a correlation measures the degree of relationship between two variables on a scale from 0.00 to 1.00.
6. Assumptions • There are 3 main assumptions… • 1. The dependent and independent are normally distributed. We can test this by looking at the histograms for the two variables • 2. The relationship between X and Y is linear. We can check this by looking at the scattergram • 3. The relationship is homoscedastic. We can test homoscedasticity by looking at the scattergram and observing that the data points form a “roughly symmetrical, cigar-shaped pattern” about the regression line. • If the above 3 assumptions have been met, then we can use correlation and test r for significance
7. Pearson r • The most commonly used correlation • Measures the degree of straight-line relationship • Computation: r = SP / (SSX)(SSY)
8. Example 1 • A researcher predicts that there is a high correlation between scores on the stats final exam (100 pts max) and scores on the university’s exit exam for graduating seniors (330 pts max)
9. Example 1 X 30 38 52 90 95 305 X2 900 1,444 2,704 8,100 9,025 22,173 Y 160 180 180 210 240 970 Y2 25,600 32,400 32,400 44,100 57,600 192,100 XY 4,800 6,840 9,360 18,900 22,800 62,700 (SX) (SX2) (SY) (SY2) (SXY)
10. Example 1 SSX = SX2 - (SX)2 = 22,173 - 3052 = n 5 = 22,173 - 93025/5 = 22,173 - 18,605 = 3,568 SSY = SY2 - (SY)2 = 192,100 - 9702 = n 5 = 192,100 - 940,900/5 = 192,100 - 188,180 = 3,920
11. Example 1 SP = SXY - (SX)(SY) = n 62,700 - (305)(970) 5 = 62,700 - 295,850/5 = 62,700 - 59,170 = 3,530
12. Example 1 • r = SP / (SSX)(SSY) = 3,530 / (3,568)(3,920) = 3,530 / 13,986,560 = 3,530 / 3,739.861 = .944
13. Pearson Correlation: “Rule of Thumb” • If r = 1.00 Perfect Correlation + .70 to +.99 Very strong positive relationship + .40 to +.69 Strong positive relationship + .30 to +.39 Moderate positive relationship + .20 to +.29 Weak positive relationship + .01 to +.19 No or negligible relationship - .01 to -.19 No or negligible relationship - .20 to -.29 Weak negative relationship - .30 to -.39 Moderate negative relationship - .40 to -.69 Strong negative relationship - .70 or higher Very strong negative relationship
14. Example 1: Interpretation • An r of 0.944 indicates an extremely strong relationship between scores on the stats final exam and scores on the exit exam. As scores on the stats final go up so too do scores on the exit exam. • But we are not finished with the interpretation • See next slide
15. Interpretation (Continued)Coefficient of Determination (r2) • The value r2 is called the coefficient of determination because it measures the proportion in variability in one variable that can be determined from the relationship with the other variable • For example: • A correlation of r = .944 means that r2 = .891 (or 89.1%) of the variability in the Y scores can be predicted from the relationship with the X scores
16. Coefficient of Determination (r2) and Interpret:The coefficient of determination is r2 = .891. Scores on the stats final exam, by itself, accounts for 89.1% of the variation of the exit exam scores.
17. Example 2 • A researcher predicts that there is a high correlation between years of education and voter turnout • She chooses Alamosa, Boston, Chicago, Detroit, and NYC to test her theory
18. Example 2 • The scores on each variable are displayed in table format: • Y = % Turnout • X = Years of Education
19. Scatterplot • The relationship between X and Y is linear.
20. Make a Computational Table
21. Example 2 SSX = SX2 - (SX)2 = 782.15 - 62.52 = n 5 = 782.15 - 3906.25/5 = 782.15 – 781.25 = 0.9 SSY = SY2 - (SY)2 = 20374 - 3182 = n 5 = 20374- 101124/5 = 20374 – 20224.80 = 149.20
22. Example 2 SP = SXY - (SX)(SY) = n 3986.40 - (62.5)(318) 5 = 3986.40 - 19875/5 = 3986.40 – 3975.00 = 11.40
23. Example 2: Find Pearson r • r = SP / (SSX)(SSY) = 11.4 / (0.9)(149.2) = 11.4 / 134.28 = 11.4/ 11.58 = .984
24. Example 2: Interpretation • An r of 0.984 indicates an extremely strong relationship between years of education and voter turnout for these five cities. As level of education increases, % turnout increases. • But we are not finished with the interpretation • See next slide
25. Coefficient of Determination (r2) and Interpret:The coefficient of determination is r2 = .968. Education, by itself, accounts for 96.8% of the variation in voter turnout.
26. Pearson’s r • Had the relationship between % college educated and turnout, r =.32. • This relationship would have been positive and weak to moderate. • Had the relationship between % college educated and turnout, r = -.12. • This relationship would have been negative and weak.
27. Hypothesis Testing with Pearson • We can have a two-tailed hypothesis: Ho: ρ = 0.0 H1: ρ ≠ 0.0 • We can have a one-tailed hypothesis: Ho: ρ = 0.0 H1: ρ < 0.0 (or ρ > 0.0) • Note that ρ (rho) is the population parameter, while r is the sample statistic
28. Find rcritical • See Table B.6 (page 537) • You need to know the alpha level • You need to know the sample size • See that we always will use:df = n-2
29. Find rcalculated • See previous slides for formulas
30. Make you decision… • rcalculated < rcritical thenRetain H0 • rcalculated > rcritical thenReject H0
31. Always include a brief summary of your results: • Was it positive or negative? • Was it significant ? • Explain the correlation • Explain the variation • Coefficient of Determination (r2)
32. Credits • http://campus.houghton.edu/orgs/psychology/stat15b.ppt#267,2,Review • http://publish.uwo.ca/~pakvis/Interval.ppt#276,17,Practical Example using Healey P. 418 Problem 15.1
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Ratios to Percents
Notes
Ratios to Percents:
When parts of a quantity are given to us as ratios, we have seen how to convert them to percentages.
Example
In a class, the ratio of girls to boys is 3: 7.
(i) What percentage of the total number of students is girls and what percentage of the total number of students is boys?
(ii) If the total number of students is 70, what would be the number of girls and what would be the number of boys in the class?
(i) Girls: Boys = 3: 7
Total number of parts = 3 + 7 = 10.
By unitary method:
Girls:
10 → 3(Girls)
1 → 3/10
100 → 3/10 xx 100 = 30 Girls = 30%
Boys:
10 → 7(Boys)
1 → 7/10
100 → 7/10 xx 100 = 70 Boys = 70%
(ii)
In Fraction:
Girls = 3/10 xx 70 = 21.
Boys = 7/10 xx 70 = 49.
In Percentage:
Girls = 30/100 xx 70 = 21.
Boys = 70/100 xx 70 = 49.
Example
If Rs. 250 is to be divided amongst Ravi, Raju, and Roy so that Ravi gets two parts, Raju three parts, and Roy five parts. How much money will each get? What will it be in percentages?
The parts which the three boys are getting can be written in terms of ratios as 2: 3: 5. Total of the parts is 2 + 3 + 5 = 10.
Amounts received by each Percentages of money for each 2/10 × Rs. 250 = Rs. 50. Ravi gets 2/10 × 100% = 20% 3/10 × Rs. 250 = Rs. 75. Raju gets 3/10 × 100% = 30% 5/10 × Rs. 250 = Rs. 125. Ravi gets 5/10 × 100% = 50%
Example
Last year Giripremi group planted 75 trees. Of these, 48 trees flourished. The Karmavir group planted 50 trees, of which, 35 flourished. Which group was more successful in conserving the trees they had planted?
Suppose the surviving trees of the Giripremi group are A%.
Suppose the surviving trees of the Karmavir group are B%.
The Giripremi’s ratio of the surviving trees to planted trees is A/100 and also 48/75.
Therefore, A/100 = 48/75.
The Karmavir ratio of the surviving trees to planted trees is B/100 and also 35/50.
Therefore, B/100 = 35/50.
Let us write the same ratio in two forms,
"A"/100 = 48/75
∴ "A"/100 xx 100 = 48/75 xx 100
∴ A = 64.
"B"/100 = 35/50
∴ "B"/100 xx 100 = 35/50 xx 100
∴ B = 70.
∴ The Karmavir group was more successful in conserving the trees they had planted.
Example
In Khatav taluka, it was decided to make 200 ponds in Warudgaon and 300 ponds in Jakhangaon. Of these, 120 ponds in Warudgaon were completed at the end of May, while in Jakhangaon work was complete on 165 ponds. In which village was a greater proportion of the work completed?
Let the number of ponds completed in Warudgaon be A% and in Jakhangaon, B%.
"A"/100 = 120/120
"A"/100 xx 100 = 120/120 xx 100
A = 60
"B"/100 = 165/300
"B"/100 xx 100 = 165/300 xx 100
B = 55.
∴ A greater proportion of the work was completed in Warudgaon.
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College Physics 2e
# 18.5Electric Field Lines: Multiple Charges
College Physics 2e18.5 Electric Field Lines: Multiple Charges
## Learning Objectives
By the end of this section, you will be able to:
• Calculate the total force (magnitude and direction) exerted on a test charge from more than one charge
• Describe an electric field diagram of a positive point charge; of a negative point charge with twice the magnitude of positive charge
• Draw the electric field lines between two points of the same charge; between two points of opposite charge.
Drawings using lines to represent electric fields around charged objects are very useful in visualizing field strength and direction. Since the electric field has both magnitude and direction, it is a vector. Like all vectors, the electric field can be represented by an arrow that has length proportional to its magnitude and that points in the correct direction. (We have used arrows extensively to represent force vectors, for example.)
Figure 18.19 shows two pictorial representations of the same electric field created by a positive point charge $QQ$. Figure 18.19 (b) shows the standard representation using continuous lines. Figure 18.19 (a) shows numerous individual arrows with each arrow representing the force on a test charge $qq$. Field lines are essentially a map of infinitesimal force vectors.
Figure 18.19 Two equivalent representations of the electric field due to a positive charge $QQ$. (a) Arrows representing the electric field’s magnitude and direction. (b) In the standard representation, the arrows are replaced by continuous field lines having the same direction at any point as the electric field. The closeness of the lines is directly related to the strength of the electric field. A test charge placed anywhere will feel a force in the direction of the field line; this force will have a strength proportional to the density of the lines (being greater near the charge, for example).
Note that the electric field is defined for a positive test charge $qq$, so that the field lines point away from a positive charge and toward a negative charge. (See Figure 18.20.) The electric field strength is exactly proportional to the number of field lines per unit area, since the magnitude of the electric field for a point charge is $E=k|Q|/r2E=k|Q|/r2$ and area is proportional to $r2r2$. This pictorial representation, in which field lines represent the direction and their closeness (that is, their areal density or the number of lines crossing a unit area) represents strength, is used for all fields: electrostatic, gravitational, magnetic, and others.
Figure 18.20 The electric field surrounding three different point charges. (a) A positive charge. (b) A negative charge of equal magnitude. (c) A larger negative charge.
In many situations, there are multiple charges. The total electric field created by multiple charges is the vector sum of the individual fields created by each charge. The following example shows how to add electric field vectors.
## Example 18.4
Find the magnitude and direction of the total electric field due to the two point charges, $q1q1$ and $q2q2$, at the origin of the coordinate system as shown in Figure 18.21.
Figure 18.21 The electric fields $E1E1$ and $E2E2$ at the origin O add to $EtotEtot$.
### Strategy
Since the electric field is a vector (having magnitude and direction), we add electric fields with the same vector techniques used for other types of vectors. We first must find the electric field due to each charge at the point of interest, which is the origin of the coordinate system (O) in this instance. We pretend that there is a positive test charge, $qq$, at point O, which allows us to determine the direction of the fields $E1E1$ and $E2E2$. Once those fields are found, the total field can be determined using vector addition.
### Solution
The electric field strength at the origin due to $q1q1$ is labeled $E1E1$ and is calculated:
$E1=kq1r12=8.99×109N⋅m2/C25.00×10−9C2.00×10−2m2E1=1.124×105N/C.E1=kq1r12=8.99×109N⋅m2/C25.00×10−9C2.00×10−2m2E1=1.124×105N/C.$
18.16
Similarly, $E2E2$ is
$E2=kq2r22=8.99×109N⋅m2/C210.0×10−9C4.00×10−2m2E2=0.5619×105N/C.E2=kq2r22=8.99×109N⋅m2/C210.0×10−9C4.00×10−2m2E2=0.5619×105N/C.$
18.17
Four digits have been retained in this solution to illustrate that $E1E1$ is exactly twice the magnitude of $E2E2$. Now arrows are drawn to represent the magnitudes and directions of $E1E1$ and $E2E2$. (See Figure 18.21.) The direction of the electric field is that of the force on a positive charge so both arrows point directly away from the positive charges that create them. The arrow for $E1E1$ is exactly twice the length of that for $E2E2$. The arrows form a right triangle in this case and can be added using the Pythagorean theorem. The magnitude of the total field $EtotEtot$ is
$E tot = ( E 1 2 + E 2 2 ) 1/2 = { ( 1.124 × 10 5 N/C ) 2 + ( 0.5619 × 10 5 N/C ) 2 } 1/2 = 1.26 × 10 5 N/C. E tot = ( E 1 2 + E 2 2 ) 1/2 = { ( 1.124 × 10 5 N/C ) 2 + ( 0.5619 × 10 5 N/C ) 2 } 1/2 = 1.26 × 10 5 N/C.$
18.18
The direction is
$θ = tan−1E1E2 = tan−11.124×105N/C0.5619×105N/C = 63.4º,θ = tan−1E1E2 = tan−11.124×105N/C0.5619×105N/C = 63.4º,$
18.19
or $63.4º63.4º$ above the x-axis.
### Discussion
In cases where the electric field vectors to be added are not perpendicular, vector components or graphical techniques can be used. The total electric field found in this example is the total electric field at only one point in space. To find the total electric field due to these two charges over an entire region, the same technique must be repeated for each point in the region. This impossibly lengthy task (there are an infinite number of points in space) can be avoided by calculating the total field at representative points and using some of the unifying features noted next.
Figure 18.22 shows how the electric field from two point charges can be drawn by finding the total field at representative points and drawing electric field lines consistent with those points. While the electric fields from multiple charges are more complex than those of single charges, some simple features are easily noticed.
For example, the field is weaker between like charges, as shown by the lines being farther apart in that region. (This is because the fields from each charge exert opposing forces on any charge placed between them.) (See Figure 18.22 and Figure 18.23(a).) Furthermore, at a great distance from two like charges, the field becomes identical to the field from a single, larger charge.
Figure 18.23(b) shows the electric field of two unlike charges. The field is stronger between the charges. In that region, the fields from each charge are in the same direction, and so their strengths add. The field of two unlike charges is weak at large distances, because the fields of the individual charges are in opposite directions and so their strengths subtract. At very large distances, the field of two unlike charges looks like that of a smaller single charge.
Figure 18.22 Two positive point charges $q1q1$ and $q2q2$ produce the resultant electric field shown. The field is calculated at representative points and then smooth field lines drawn following the rules outlined in the text.
Figure 18.23 (a) Two negative charges produce the fields shown. It is very similar to the field produced by two positive charges, except that the directions are reversed. The field is clearly weaker between the charges. The individual forces on a test charge in that region are in opposite directions. (b) Two opposite charges produce the field shown, which is stronger in the region between the charges.
We use electric field lines to visualize and analyze electric fields (the lines are a pictorial tool, not a physical entity in themselves). The properties of electric field lines for any charge distribution can be summarized as follows:
1. Field lines must begin on positive charges and terminate on negative charges, or at infinity in the hypothetical case of isolated charges.
2. The number of field lines leaving a positive charge or entering a negative charge is proportional to the magnitude of the charge.
3. The strength of the field is proportional to the closeness of the field lines—more precisely, it is proportional to the number of lines per unit area perpendicular to the lines.
4. The direction of the electric field is tangent to the field line at any point in space.
5. Field lines can never cross.
The last property means that the field is unique at any point. The field line represents the direction of the field; so if they crossed, the field would have two directions at that location (an impossibility if the field is unique).
## PhET Explorations
### Charges and Fields
Move point charges around on the playing field and then view the electric field, voltages, equipotential lines, and more. It's colorful, it's dynamic, it's free.
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### Parallel and Perpendicular
```Parallel and Perpendicular
Lines
y mx b
•Useful for graphing since m is the gradient and b is the yintercept
y y m x x Point-Gradient Form
1
1
•Use this form when you know a point on the line and the gradient
•Also can use this version if you have two points on the line because
you can first find the gradient using the gradient formula and then
use one of the points and the gradient in this equation.
ax by c 0
General Form
•Commonly used to write linear equation problems or express answers
1
m
3
1
m
3
3
m 3
1
1
m
3
The gradient is a number that tells "how
steep" the line is and in which direction.
So as you can see, parallel lines have
the same gradients so if you need the
gradient of a line parallel to a given
line, simply find the gradient of the given
line and the gradient you want for a
parallel line will be the same.
Perpendicular lines have negative
reciprocal gradients so if you need the
gradient of a line perpendicular to a
given line, simply find the gradient of the
given line, take its reciprocal (flip it over)
and make it negative.
Let's look at a line and a point not on the line
Let's find the equation of a
line parallel to y = - x that
y=-x
passes through the point (2, 4)
(2, 4)
y 4y1 m
-1 x x21
Distribute and then solve for y
form.
y x 6
the first line, y = -1x ?
intercept form so
y = mx + b which means
So we know the gradient is
–1 and it passes through
(2, 4). Having the point and
the gradient, we can use the
find the equation of the line
What if we wanted perpendicular instead of parallel?
Let's find the equation of a
line perpendicular to y = - x
that passes through the point
y=-x
(2, 4)
(2, 4)
line is still –1.
perpendicular is the negative
reciprocal so take –1 and "flip" it
over and make it negative.
y y4 1 m
1 x x21
1
1
1
1
Distribute and then solve for y So the gradient of a
perpendicular line is 1 and it
form.
passes through (2, 4).
y x2
Acknowledgement
I wish to thank Shawna Haider from Salt Lake Community College, Utah
USA for her hard work in creating this PowerPoint.
www.slcc.edu
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# Chapter 8: Binomial and Geometric Distributions
## Presentation on theme: "Chapter 8: Binomial and Geometric Distributions"— Presentation transcript:
Chapter 8: Binomial and Geometric Distributions
Section 8.1 Binomial Distributions The Practice of Statistics, 4th edition – For AP* STARNES, YATES, MOORE
Section 8.1 Binomial Distribution
Learning Objectives After this section, you should be able to… DETERMINE whether the conditions for a binomial setting are met COMPUTE and INTERPRET probabilities involving binomial random variables CALCULATE the mean and standard deviation of a binomial random variable and INTERPRET these values in context
Binomial and Geometric Random Variables
Binomial Settings When the same chance process is repeated several times, we are often interested in whether a particular outcome does or doesn’t happen on each repetition. In some cases, the number of repeated trials is fixed in advance and we are interested in the number of times a particular event (called a “success”) occurs. If the trials in these cases are independent and each success has an equal chance of occurring, we have a binomial setting. Binomial and Geometric Random Variables Definition: A binomial setting arises when we perform several independent trials of the same chance process and record the number of times that a particular outcome occurs. The four conditions for a binomial setting are B • Binary? The possible outcomes of each trial can be classified as “success” or “failure.” I • Independent? Trials must be independent; that is, knowing the result of one trial must not have any effect on the result of any other trial. N • Number? The number of trials n of the chance process must be fixed in advance. S • Success? On each trial, the probability p of success must be the same.
Binomial and Geometric Random Variables
Binomial Random Variable Consider tossing a coin n times. Each toss gives either heads or tails. Knowing the outcome of one toss does not change the probability of an outcome on any other toss. If we define heads as a success, then p is the probability of a head and is 0.5 on any toss. The number of heads in n tosses is a binomial random variable X. The probability distribution of X is called a binomial distribution. Binomial and Geometric Random Variables Definition: The count X of successes in a binomial setting is a binomial random variable. The probability distribution of X is a binomial distribution with parameters n and p, where n is the number of trials of the chance process and p is the probability of a success on any one trial. The possible values of X are the whole numbers from 0 to n. Note: When checking the Binomial condition, be sure to check the BINS and make sure you’re being asked to count the number of successes in a certain number of trials!
Binomial and Geometric Random Variables
Binomial Probabilities In a binomial setting, we can define a random variable (say, X) as the number of successes in n independent trials. We are interested in finding the probability distribution of X. Binomial and Geometric Random Variables Example Each child of a particular pair of parents has probability 0.25 of having type O blood. Genetics says that children receive genes from each of their parents independently. If these parents have 5 children, the count X of children with type O blood is a binomial random variable with n = 5 trials and probability p = 0.25 of a success on each trial. In this setting, a child with type O blood is a “success” (S) and a child with another blood type is a “failure” (F). What’s P(X = 2)? P(SSFFF) = (0.25)(0.25)(0.75)(0.75)(0.75) = (0.25)2(0.75)3 = However, there are a number of different arrangements in which 2 out of the 5 children have type O blood: SSFFF SFSFF SFFSF SFFFS FSSFF FSFSF FSFFS FFSSF FFSFS FFFSS Verify that in each arrangement, P(X = 2) = (0.25)2(0.75)3 = Therefore, P(X = 2) = 10(0.25)2(0.75)3 =
Binomial and Geometric Random Variables
Binomial Coefficient Note, in the previous example, any one arrangement of 2 S’s and 3 F’s had the same probability. This is true because no matter what arrangement, we’d multiply together 0.25 twice and 0.75 three times. We can generalize this for any setting in which we are interested in k successes in n trials. That is, Binomial and Geometric Random Variables Definition: The number of ways of arranging k successes among n observations is given by the binomial coefficient for k = 0, 1, 2, …, n where n! = n(n – 1)(n – 2)•…•(3)(2)(1) and 0! = 1.
Binomial and Geometric Random Variables
Binomial Probability The binomial coefficient counts the number of different ways in which k successes can be arranged among n trials. The binomial probability P(X = k) is this count multiplied by the probability of any one specific arrangement of the k successes. Binomial and Geometric Random Variables Binomial Probability If X has the binomial distribution with n trials and probability p of success on each trial, the possible values of X are 0, 1, 2, …, n. If k is any one of these values, Probability of k successes Probability of n-k failures Number of arrangements of k successes
Example: Inheriting Blood Type
Each child of a particular pair of parents has probability 0.25 of having blood type O. Suppose the parents have 5 children (a) Find the probability that exactly 3 of the children have type O blood. Let X = the number of children with type O blood. We know X has a binomial distribution with n = 5 and p = 0.25. (b) Should the parents be surprised if more than 3 of their children have type O blood? To answer this, we need to find P(X > 3). Since there is only a 1.5% chance that more than 3 children out of 5 would have Type O blood, the parents should be surprised!
Binomial and Geometric Random Variables
Mean and Standard Deviation of a Binomial Distribution We describe the probability distribution of a binomial random variable just like any other distribution – by looking at the shape, center, and spread. Consider the probability distribution of X = number of children with type O blood in a family with 5 children. Binomial and Geometric Random Variables xi 1 2 3 4 5 pi 0.2373 0.3955 0.2637 0.0879 0.0147 Shape: The probability distribution of X is skewed to the right. It is more likely to have 0, 1, or 2 children with type O blood than a larger value. Center: The median number of children with type O blood is 1. Based on our formula for the mean: Spread: The variance of X is The standard deviation of X is
Binomial and Geometric Random Variables
Mean and Standard Deviation of a Binomial Distribution Notice, the mean µX = 1.25 can be found another way. Since each child has a 0.25 chance of inheriting type O blood, we’d expect one-fourth of the 5 children to have this blood type. That is, µX = 5(0.25) = This method can be used to find the mean of any binomial random variable with parameters n and p. Binomial and Geometric Random Variables If a count X has the binomial distribution with number of trials n and probability of success p, the mean and standard deviation of X are Mean and Standard Deviation of a Binomial Random Variable Note: These formulas work ONLY for binomial distributions. They can’t be used for other distributions!
Example: Bottled Water versus Tap Water
Mr. Bullard’s 21 AP Statistics students did the Activity on page 340. If we assume the students in his class cannot tell tap water from bottled water, then each has a 1/3 chance of correctly identifying the different type of water by guessing. Let X = the number of students who correctly identify the cup containing the different type of water. Find the mean and standard deviation of X. Since X is a binomial random variable with parameters n = 21 and p = 1/3, we can use the formulas for the mean and standard deviation of a binomial random variable. If the activity were repeated many times with groups of 21 students who were just guessing, the number of correct identifications would differ from 7 by an average of 2.16. We’d expect about one-third of his 21 students, about 7, to guess correctly.
Binomial and Geometric Random Variables
Binomial Distributions in Statistical Sampling The binomial distributions are important in statistics when we want to make inferences about the proportion p of successes in a population. Suppose 10% of CDs have defective copy-protection schemes that can harm computers. A music distributor inspects an SRS of 10 CDs from a shipment of 10,000. Let X = number of defective CDs. What is P(X = 0)? Note, this is not quite a binomial setting. Why? Binomial and Geometric Random Variables The actual probability is In practice, the binomial distribution gives a good approximation as long as we don’t sample more than 10% of the population. Using the binomial distribution, When taking an SRS of size n from a population of size N, we can use a binomial distribution to model the count of successes in the sample as long as Sampling Without Replacement Condition
Binomial and Geometric Random Variables
Normal Approximation for Binomial Distributions As n gets larger, something interesting happens to the shape of a binomial distribution. The figures below show histograms of binomial distributions for different values of n and p. What do you notice as n gets larger? Binomial and Geometric Random Variables Suppose that X has the binomial distribution with n trials and success probability p. When n is large, the distribution of X is approximately Normal with mean and standard deviation As a rule of thumb, we will use the Normal approximation when n is so large that np ≥ 10 and n(1 – p) ≥ 10. That is, the expected number of successes and failures are both at least 10. Normal Approximation for Binomial Distributions
Example: Attitudes Toward Shopping
Sample surveys show that fewer people enjoy shopping than in the past. A survey asked a nationwide random sample of 2500 adults if they agreed or disagreed that “I like buying new clothes, but shopping is often frustrating and time-consuming.” Suppose that exactly 60% of all adult US residents would say “Agree” if asked the same question. Let X = the number in the sample who agree. Estimate the probability that 1520 or more of the sample agree. 1) Verify that X is approximately a binomial random variable. B: Success = agree, Failure = don’t agree I: Because the population of U.S. adults is greater than 25,000, it is reasonable to assume the sampling without replacement condition is met. N: n = 2500 trials of the chance process S: The probability of selecting an adult who agrees is p = 0.60 2) Check the conditions for using a Normal approximation. Since np = 2500(0.60) = 1500 and n(1 – p) = 2500(0.40) = 1000 are both at least 10, we may use the Normal approximation. 3) Calculate P(X ≥ 1520) using a Normal approximation.
Section 8.1 Binomial Distributions
Summary In this section, we learned that… A binomial setting consists of n independent trials of the same chance process, each resulting in a success or a failure, with probability of success p on each trial. The count X of successes is a binomial random variable. Its probability distribution is a binomial distribution. The binomial coefficient counts the number of ways k successes can be arranged among n trials. If X has the binomial distribution with parameters n and p, the possible values of X are the whole numbers 0, 1, 2, , n. The binomial probability of observing k successes in n trials is
Section 8.1 Binomial Distributions
Summary In this section, we learned that… The mean and standard deviation of a binomial random variable X are The Normal approximation to the binomial distribution says that if X is a count having the binomial distribution with parameters n and p, then when n is large, X is approximately Normally distributed. We will use this approximation when np ≥ 10 and n(1 - p) ≥ 10.
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### Home > PC3 > Chapter 9 > Lesson 9.1.1 > Problem9-8
9-8.
Shola’s Bakery uses sugar, eggs, and butter in all of its cakes, as well as in the frosting. Matrix $C$ shows how many eggs, cups of sugar, and ounces of butter are used in each angel food cake and in each devil’s food cake. Matrix $F$ shows how many eggs, cups of sugar, and ounces of butter are used in the frosting for each cake.
e s b $C=$ afdf $\left[ \begin{array} { c c c } { 6 } & { 1 } & { 5 } \\ { 3 } & { 1.5 } & { 4 } \end{array} \right]$
1. Write the matrix $C + F,$ being sure to label the rows and columns, and explain what it represents.
$\begin{bmatrix}6+2 & 1+1 & 5+2\\3+1 & 1.5+2 & 4+4\end{bmatrix}$
2. Write the matrix $3C$, with labels, and explain what it represents.
$\begin{bmatrix}3(6) & 3(1) & 3(5)\\3(3) & 3(1.5) & 3(4)\end{bmatrix}$
e s b $F=$ afdf $\left[ \begin{array} { c c c } { 2 } & { 1 } & { 2 } \\ { 1 } & { 2 } & { 4 } \end{array} \right]$
1. Leora orders three angel food cakes and two devil’s food cakes both without frosting, as represented by the matrix $L$ at right. Use matrix multiplication to write a matrix that shows how much sugar, eggs, and butter Shola will need to fill Leora’s order. Show all of your work.
Compute $LC$.
af df $L=$ $\left[ \begin{array} { l l } { 3 } & { 2 } \end{array} \right]$
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# Fraction calculator
This fraction calculator performs all fraction operations - addition, subtraction, multiplication, division and evaluates expressions with fractions. It also shows detailed step-by-step informations.
## The result:
### 1/4 + 4/8 + 2/3 = 17/12 = 1 5/12 ≅ 1.4166667
The spelled result in words is seventeen twelfths (or one and five twelfths).
### How do we solve fractions step by step?
1. Add: 1/4 + 4/8 = 1 · 2/4 · 2 + 4/8 = 2/8 + 4/8 = 2 + 4/8 = 6/8 = 2 · 3/2 · 4 = 3/4
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(4, 8) = 8. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 8 = 32. In the following intermediate step, cancel by a common factor of 2 gives 3/4.
In other words - one quarter plus four eighths is three quarters.
2. Add: the result of step No. 1 + 2/3 = 3/4 + 2/3 = 3 · 3/4 · 3 + 2 · 4/3 · 4 = 9/12 + 8/12 = 9 + 8/12 = 17/12
It is suitable to adjust both fractions to a common (equal, identical) denominator for adding, subtracting, and comparing fractions. The common denominator you can calculate as the least common multiple of both denominators - LCM(4, 3) = 12. It is enough to find the common denominator (not necessarily the lowest) by multiplying the denominators: 4 × 3 = 12. In the following intermediate step, it cannot further simplify the fraction result by canceling.
In other words - three quarters plus two thirds is seventeen twelfths.
### Rules for expressions with fractions:
Fractions - use a forward slash to divide the numerator by the denominator, i.e., for five-hundredths, enter 5/100. If you use mixed numbers, leave a space between the whole and fraction parts.
Mixed numerals (mixed numbers or fractions) keep one space between the integer and
fraction and use a forward slash to input fractions i.e., 1 2/3 . An example of a negative mixed fraction: -5 1/2.
Because slash is both sign for fraction line and division, use a colon (:) as the operator of division fractions i.e., 1/2 : 1/3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
### Math Symbols
SymbolSymbol nameSymbol MeaningExample
-minus signsubtraction 1 1/2 - 2/3
*asteriskmultiplication 2/3 * 3/4
×times signmultiplication 2/3 × 5/6
:division signdivision 1/2 : 3
/division slashdivision 1/3 / 5
:coloncomplex fraction 1/2 : 1/3
^caretexponentiation / power 1/4^3
()parenthesescalculate expression inside first-3/5 - (-1/4)
The calculator follows well-known rules for the order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
MDAS - Multiplication and Division have the same precedence over Addition and Subtraction. The MDAS rule is the order of operations part of the PEMDAS rule.
Be careful; always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) have the same priority and must be evaluated from left to right.
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## Want to keep learning?
This content is taken from the University of Padova's online course, Precalculus: the Mathematics of Numbers, Functions and Equations. Join the course to learn more.
1.5
## Precalculus
Skip to 0 minutes and 10 seconds Think of a number, any number. When someone says that to us, we tend to think of a natural number like 7. After all, these numbers are old friends. We know them since childhood. They are also the basis for all mathematics. So let’s begin our discussion of numbers with them, with the integers. When you take these positive integers, being sure to include the number 0, you get a collection that is called the set of natural numbers. I’ve written it here. The capital N is in what is called the blackboard style. You can see that I’ve put the numbers inside curly brackets, or braces as they’re called. That’s so they don’t escape. I’ve also used the word set.
Skip to 1 minute and 0 seconds We’re going to be talking about sets a lot, so let’s review basic set notation to make sure we’re all on the same wavelength. Suppose that A and B are two sets of numbers, say. What do we mean by containment? A is contained in B means that every element that’s in A is also in B. Intersection. A intersect B is the set of all elements common to both A and B. Now this raises a question. What if there are no numbers common to A and B? In that case, A intersect B is the empty set. Union.
Skip to 1 minute and 38 seconds The union of A and B is the set of all the elements in either set, and set difference– A minus B or A delete B, as it’s also read– is the set whose elements consists of all numbers in A but having removed those that are in B. For example, suppose E is the set of even integers in N. Then, the set difference N delete E is the set of odd numbers in N. Integers play an important role in a proof technique called mathematical induction. Here’s the way mathematical induction works. You want to prove a certain proposition or assertion whose formulation depends upon a positive integer n. And suppose we know the following.
Skip to 2 minutes and 27 seconds For a certain n bar, the proposition P n bar is true. And suppose we also know that whenever P n is true, then the next proposition, P n plus 1 is true. It then follows that P n is true for all n beyond n bar. That’s a proof technique. It’s true that proofs will not be the main thing that we’re going to focus on in this course, but some experience with logical reasoning is very useful later on, for example, in calculus. Here’s an example of the use of induction. It’s a formula used by the great applied mathematician Archimedes well over 2,000 years ago.
Skip to 3 minutes and 9 seconds It asserts that the sum of the first n squares is given by a certain formula in terms of n. That’s the proposition P n. Let’s check it for n equals 3, for example. In that case, the formula asserts that the sum of the first three squares is equal to a certain expression which I’ve obtained by replacing n by 3 in the general formula. You work out what it gives you, and you get 14. So that shows the formula is true for n equals 3 because the left side is the sum of 1 plus 4 plus 9, which is indeed 14. Now, we’re going to prove the general formula later on. But first, let’s discuss divisibility.
Skip to 3 minutes and 56 seconds If you have two positive integers p and q, we say that p is divisible by q if it’s possible to write p as the product of q with another integer. Here are some facts about divisibility. An integer is divisible by 2– that is, is an even number– if and only if its final digit is divisible by 2. Do you notice the funny word there, i-f-f? That’s a standard mathematical shorthand for the phrase if and only if. It gives you the equivalence, complete equivalence of two assertions.
Skip to 4 minutes and 32 seconds To return then to divisibility, an integer is divisible by 3 if and only if the sum of its digits is divisible by 3, and by 5 if and only if its last digit is a 0 or 5. Let’s apply these rules to the number of 714. It’s divisible by 2 because 4 is divisible by 2. It’s divisible by 3 because the sum of its digits is 12. And it’s not divisible by 5 because it doesn’t end in a 0 or a 5. Now, another concept we need is that of prime numbers. Incidentally, large prime numbers play a crucial role in the encryption codes these days that give you digital security. Obviously, any integer is divisible by itself and by 1.
Skip to 5 minutes and 24 seconds Well, if you have an integer greater than 1 that is only divisible by itself and by 1, we say it is prime. For example, we can check that 2, 3, and 5 are prime numbers, but 15 isn’t because 15 is divisible by 3, for example. Some facts about prime numbers. There are infinitely many. Not that easy to prove, but if you find the right clever argument, not necessarily that hard either. Another fact. Every integer can be factored in a unique way into a product of prime numbers. That’s called the prime factorisation theorem. Let’s see how it works for the number 132. Obviously, that’s not a prime number. It’s divisible by 2.
Skip to 6 minutes and 12 seconds I factor out a 2, and I get a 66. In turn, the 66 produces another factor of 2. Is the 33 a prime number? No. It’s equal to 3 times 11. And now every factor I have there is prime. So this is the prime factorisation of the number 132.
# Integers
In this video, Francis introduces counting numbers, sets, set operations, the induction principle, divisibility, and primes.
You can access a copy of the slides used in the video in the PDF file at the bottom of this step.
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###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
##### Thank you for watching the video.
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# Simplifying Radicals using Rational Exponents - Concept
Carl Horowitz
###### Carl Horowitz
University of Michigan
Runs his own tutoring company
Carl taught upper-level math in several schools and currently runs his own tutoring company. He bets that no one can beat his love for intensive outdoor activities!
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When simplifying roots that are either greater than four or have a term raised to a large number, we rewrite the problem using rational exponents. Remember that every root can be written as a fraction, with the denominator indicating the root's power. When simplifying radicals, since a power to a power multiplies the exponents, the problem is simplified by multiplying together all the exponents.
Every once in a while we're asked to simplify radicals where we actually don't know numerically what the things we're looking at are, so what I have behind me is two ways of writing the exact same thing. We have the sixth root of 5 to the twelve or the six root of 5 out of the 12. Remember we were taking power to power and multiplying so these are actually exactly equivalent statements.
The problem is, just trying to evaluate these on a calculator okay? I don't know what 5 to the twelfth is, so I sure don't know what the sixth root of 5 to twelfth is. Similarly, I don't know what the sixth root of 5 is so then I don't know what the sixth root of 5 is to the twelfth, so we're trying to need figure out a way to somehow deal with this so we can actually simplify without a calculator okay? And what we can do is rewrite these using exponents okay? So what we have here is a root and if we use it rational exponents oops I hope if I write the right number down we have 5 and then we have a power which is going to be power over root so we have the twelfth power over the sixth root so what we end up having is 12, 6 which we know is 2 this ends up giving us 5 squared which we can simplify to 25.
So whenever we're dealing with really big powers or roots you can always look at it and think about if there's a way to simplify these exponents up okay? In this case we just rewrote as a exponential fraction simplified up quite easily and we're able to solve it.
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# What is the formula of idempotent matrix?
## What is the formula of idempotent matrix?
Idempotent matrix is a square matrix, which multiplied by itself, gives back the initial square matrix. A matrix M, when multiplied with itself, gives back the same matrix M, M2 = M. Let us consider a matrix A = (abcd) ( a b c d ) . Further since A is taken as an idempotent matrix, we have A2 = A.
How do you prove a matrix is idempotent?
A matrix A is idempotent if and only if all its eigenvalues are either 0 or 1. The number of eigenvalues equal to 1 is then tr(A). Since v = 0 we find λ − λ2 = λ(1 − λ) = 0 so either λ = 0 or λ = 1. Since all the diagonal entries in Λ are 0 or 1 we are done the proof.
### What is the transpose of an idempotent matrix?
Clearly, the existence of an idempotent matrix (̸= I2) with invertible transpose is equivalent to the existence of an invertible matrix whose transpose is idempotent. If E = E2, then Et = (E2)t ̸= (Et)2 may happen, that is, the transpose of an idempotent matrix is not neces- sarily idempotent.
Are idempotent matrix and identity matrix same?
The only non-singular idempotent matrix is the identity matrix; that is, if a non-identity matrix is idempotent, its number of independent rows (and columns) is less than its number of rows (and columns). , since A is idempotent.
## What is the determinant of an idempotent matrix?
Properties of Idempotent Matrices What this means is that it is a square matrix, whose determinant is 0. [I – M] [I – M] = I – M – M + M2 = I – M – M + M = I – M, the identity matrix minus any other idempotent matrix is also an idempotent matrix.
What is nilpotent and idempotent matrix?
Idempotent means “the second power of A (and hence every higher integer power) is equal to A”. Nilpotent means “some power of A is equal to the zero matrix”.
### How do you test for idempotent?
Idempotent matrix: A matrix is said to be idempotent matrix if matrix multiplied by itself return the same matrix. The matrix M is said to be idempotent matrix if and only if M * M = M. In idempotent matrix M is a square matrix.
What is the determinant of idempotent matrix?
## Are idempotent matrices invertible?
Let A be an n×n invertible idempotent matrix. Since A is invertible, the inverse matrix A−1 of A exists and it satisfies A−1A=In, where In is the n×n identity matrix. Since A is idempotent, we have A2=A. Multiplying this equality by A−1 from the left, we get A−1A2=A−1A.
Do idempotent matrices have inverse?
An nxn idempotent matrix needs not be invertible. The simplest example is the zero nxn matrix. Any diagonal matrix, with at least one zero diagonal entry and any nonzero diagonal entry being 1, is another simple example of a singular idempotent matrix.
### Are all idempotent matrices invertible?
A is idempotent if, and only if, it acts as the identity on its range. Thus, if it’s not the identity, then its range can’t be all of R^n, and therefore it is not invertible.
What is idempotent and nilpotent matrix example?
Idem means “same”, while nil refers to “zero”. In this sense, the terms are self-descriptive: Idempotent means “the second power of A (and hence every higher integer power) is equal to A”. Nilpotent means “some power of A is equal to the zero matrix”.
## How can we achieve idempotency in post method?
Making POST requests idempotent
1. Open a transaction on the db that holds the data that needs to change by the POST request.
2. Inside this transaction, execute the needed change.
3. Set the Idempotency-key key and the value, which is the response to the client, inside the Redis store.
4. Set expire time to that key.
Is an idempotent matrix diagonalizable?
A linear operator is diagonalizable precisely when its minimal polynomial splits into distinct linear factors. This result makes it almost trivial to conclude an idempotent matrix is diagonalizable.
### What is the differences between idempotent matrix and Nilpotent Matrix?
Are idempotent matrix invertible?
I| = 0, and so A will be singular. Hence an idempotent matrix A is invertible or non-singular if and only if each of its eigenvalues is 1, i.e. its characteristic equation is (x-1)^n = 0, where n is the size of the square matrix A.…
## Is null matrix idempotent?
The zero matrix or null matrix is both idempotent matrix as well as nipotent matrix. Because all elements of a null matrix is zero.
Is idempotence always possible to achieve?
Post method always results in a server state change. If the POST method was idempotent, everything sent and accepted to or from the web server would already have to exist on the server in some form to respond with the same codes and value response. For that reason, POST cannot be idempotent.
### How does idempotency key work?
This is where idempotency keys come into play. When performing a request, a client generates a unique ID to identify just that operation and sends it up to the server along with the normal payload. The server receives the ID and correlates it with the state of the request on its end.
Are Idempotent matrices invertible?
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# Roman Numberals
Roman Numerals, although very old, are still in use for distinct applications. For example, when a list is enumerated especially in old mathematics books, you can see the roman numerals there. Or they are still being use in clocks hanging on a wall. It’s certainly useful to know a few facts about the Roman numerals. Here are a few of the Roman numerals:
1. – I
2. – II
3. – III
4. – IV
5. – V
6. – VI
7. – VII
8. – VIII
9. – IX
10. – X
After 10, number 11 would be for example would be XI or number 12 would be XII. Number 20 would be XX.
Here’s an overview of some of the larger Roman numerals:
I -> 1; V -> 5; X -> 10; L -> 50; C -> 100; D -> 500; M -> 1000.
There are a few rules regarding the Roman numerals. Knowing these rules, you’ll be able to read them if they get more complicated than the cases described above. You are usually not required to know the following rules but it does not hurt to know them:
Rule 1: If you repeat a numeral, its value will be added relative to the number of times you repeated it. Observe the following examples:
• To write the number 2 in Roman, you have to write the Roman 1 twice: II
• To write the number 20 in Roman, you have to write the Roman 10 Twice: XX
Rule 2: Although we said in Rule 1 that you can repeat “I” for example once to get to “II,” which is the number 2 in Roman, you are not supposed to repeat any numeral more than twice. In other words, no numeral may be written more than thrice in a row.
• For example, to write 30 in Roman, you repeat X twice, meaning that you write it once and repeat it twice so in total, you have three X’s. The number would be, “XXX.”But to write 40 in Roman you cannot write, “XXXX.” To do that, you’d have to write 50 and subtract 10 from it which would be, “XL.” We’ll talk about this subtraction in other rules.
One more part of Rule 2 is that the numerals V, L and D are not supposed to be repeated at all which stand for 5, 50 and 500 respectively.
Rule 3: This rule is somehow related to the “subtraction” we pointed to in the previous rule. To create the number 12 for example in Roman, you have to add 2 to 10. You do that by putting a 2 on the right hand side of 10. And you already know that a 2 in Roman is nothing but two 1’s. So the number becomes: XII. So from the example, you have hopefully understood that when you put a numeral of lower value to the right hand side of a numeral of higher value, the value of the numeral with the lower value will be added to the value of the numeral of the higher value.
Rule 4: This rule is the same thing as the previous rule but the other way around. To write the number 4 in Roman, you have to subtract 1 from 5. To do that, you write a Roman 1 on the “left” hand side of a Roman 5. That way the value of the numeral with a lower value on the left will be subtracted from the value of the numeral with a higher value on the right.
Also related to this subtraction rule, you never subtract the Roman 5, 50 and 500 (V, L and D respectively) from a numeral of greater value. Meaning that the three numeral will never be written on the left hand side of a numeral with a higher value than them.
Another related topic is the subtraction of Roman 1 which is “I.” The Roman 1 can be subtracted only from the Roman 5 and 10.
Another related topic is the subtraction of the Roman 10 which is “X.” The Roman 10 can be subtracted only from the Roman 50, 100 and 1000 (L, C and M respectively).
For your convenience, here’s a list of some Roman numbers:
• 1 -> I
• 2 -> II
• 3 -> III
• 4 -> IV
• 5 -> V
• 6 -> VI
• 7 -> VII
• 8 -> VIII
• 9 -> IX
• 10 -> X
• 20 -> XX
• 30 -> XXX
• 40 -> XL
• 50 -> L
• 60 -> LX
• 70 -> LXX
• 80 -> LXXX
• 90 -> XC
• 100 -> C
These articles are used by the author in a series of mathematics courses that will teach you mathematics from the sixth standard all the way up to 12th standard. The purpose of these courses are to help us understand mathematics so that you can use them professionally well. There is a road map that we have put together about the all the courses included, where to starts, etc. To know more about that and have access to those courses, please visit mathematics page on Great IT Courses. Thank you.
Web Developer
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### Algebraic Expressions
Expressions are formed from variables and constants. Terms are added to form expressions. Expressions that contain exactly one, two and three terms are called monomials, binomials and trinomials respectively. In general, any expression containing one or more terms with non-zero coefficients (and with variables having non-negative exponents) is called a polynomial.
### Identities
An identity is an equality , which is true for all values of the variables in the equality. Four useful identities are:
1. (a+b)2 = a2 + 2ab + b2
2. (a-b)2 = a2 - 2ab + b2
3. (a+b) (a-b) = a2 - b2
4. (x+a)(x+b) = x2 + (a+b)x + ab
A quadratic equation in the variable x is of the form
ax2 + bx + c = 0, where a, b, c are real numbers and a≠0.
### Roots By Factorization Method
If the quadratic equation ax2 + bx + c = 0 can be expressed in the form (x – a)(x – b) = 0, then the roots of the equation are a and b.
The roots of a quadratic equation are given by
### Nature of Roots
1. Two distinct real roots, if b2 - 4ac > 0
2. Two equal roots, if b2 - 4ac = 0
3. No real roots, if b2 - 4ac < 0
### Sum and Product of Roots
If α and β are the roots of the quadratic equation ax2+bx+c=0. Then,
Sum of the roots = α + β = -b/a
Product of the roots = αβ = c/a
If the roots of the quadratic equation are given as α and β, the equation can be written as x2 - x(α + β) + αβ = 0.
### Maximum or Minimum Value of a Quadratic Expression
For the quadratic expression ax2 + bx + c, the minimum or minimum value is given by (4ac-b2)/4a and it occurs at -b/2a.
If a > 0, then it will be minimum value. If a < 0, then it will be maximum value.
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## Introduction: Getting Things Even
It has come to my attention that many people have difficulty figuring out how to space objects evenly other than by guess or estimate. That can lead to problems in some cases. Most people who make things or just do thing for themselves will occasionally want to evenly space objects along or within a certain distance. It may be spacing holes along a board, supports under a shelf, fence posts along a border, pickets on a fence, flower pots on a porch, or spacing trees in an orchard. Here's a mathematical way to get the spacing just right.
It helps me to draw a diagram and label the dimensions. If it helps you, do it.
## Step 1: Case 1:
CASE #1: If you want to place objects (even if they're holes) at both ends of a space and have others evenly spread between them, subtract the width (or diameter) of the objects (actually you're subtracting half the width twice, but it's the same thing), like a fence post or flower pot, from the total length. This gives you the center-to-center distance of the end objects. Then divide by the number of spaces you want (one less than the total number of objects). Alternately, you can place the two end objects wherever you want them and measure from the center of one to the center of the other. (You will get the same result if you measure from the right side of one to the right side of the other - or the left sides of both.) You can use this method to figure the spacing of hooks to hang baskets from your porch roof, install lights along your sidewalk or space plants in your garden.
As an example, if you wanted to put a number of 12 inch flower pots along a 15 foot porch (180 inches long), you would subtract 12" from 180", leaving 168". This is actually the distance from the center of a pot on one end to the center of a pot on the other end. Then divide by the number of spaces (one less than the total number of pots including the ones at each end, or one more than the number of objects between those on the ends). Let's say that you wanted a total (including the end pots) of 4 pots. That means there would be 3 spaces so divide 168" by 3 and get 56", which is the distance from the center of each pot to the center of the next one. Subtract the width of the pots (12") to get the spacing between the pots (44"). If you add up all the dimensions you will get the exact length you started from. In this case --- 12" + 44" + 12" + 44" + 12" + 44" + 12" = 180" (If you don't, you did something wrong.) If you wanted to add another pot and still keep equal spacing, divide the 168" by 4 to get 42", then subtract the size of each flower pot (12") to get a spacing of 30" between pots.
A similar method would be used to figure out the spacing of fence posts along a property line, where there has to be a post at each end.
## Step 2: Another Example of Case 1:
Another example of Case 1 is if you want to have a certain number of evenly-spaced holes in a piece of wood, metal or anything else, mark the centers of your two end holes and measure that distance. Add the size of each hole to that and divide the total by one less than the total number of holes you want (one more than the number of holes between the end holes). This will give you the center-to-center distance for perfect spacing. You can either measure out and mark each space or use a set of dividers or a compass (the kind used to make circles and arcs, not the kind for finding magnetic North) to mark your project.
## Step 3: Case 2:
If you want to evenly space objects between two existing points, measure the space you have to fill and the size of the objects you want to place in that space. Add the two together and divide by one more than the number of things you want to place between them.
As an example, say you want to put fence posts between two existing pillars. The distance between the pillars is 18'-4" (which is 220 inches) and you're using 4x4 posts (which are actually 3-1/2" x 3-1/2"). Let's also say that you don't want the posts any more than 8 feet (96") apart.
Add the width of your object to the total space you want to fill. In the example given, it would be 223-1/2". This is distance "A". Regardless of the actual size of the end posts, columns or buildings, this will give you equal spacing for posts to be added between them. See the drawing.
Add the width of your maximum spacing to the width of your object. In the example, it would be 96" + 3-1/2" or 99-1/2". This is distance "B"
Divide the distance "A" by the distance "B" to find the number of spaces you will need. Obviously, this is more important when longer distances and more spaces are involved. In the example, this would be 223.5 / 99.5 = 2.24623. Since this is more than 2 and you want equal spaces between the objects, it means that you will need more than 2 spaces (which means you'll need three, since a quarter or a half a space wouldn't be equal to the whole ones). This tells you how many spaces there need to be between objects you need to fill the distance without going over the maximum spacing you set. Since there is a space between each object (posts) and the end, you'll need one less object than spaces (in this case, two posts). If you already know how many objects you want, you can ignore this step. (Notice that if you just put in one post between the two existing posts, there would be two spaces 108.25" between them --- 220" = 108.25+3.5+108.25. This is more than the 8 foot maximum we wanted.)
Divide distance "A" by the number of spaces between objects. In our example, this is 223.5" / 3 = 74.5". This is the center-to-center distance of the posts (or flower pots on your porch, holes in your board, or pickets on your fence). The spacing between the objects is that distance less the width of your objects. From the example, 74.5" - 3.5" = 71". This is less than the 8' maximum we wanted between posts and, if you add them all up, you'll see that it comes out exactly right; 71" +3.5" + 71" + 3.5" + 71" = 220" -- the exact distance we wanted to fill between the posts.
## Step 4: Another Example of Case 2
This is also how to calculate how many pickets we need and their spacing between each post. The space from above is 71" and the pickets we intend to use measure 4-1/2" wide, so add the 4-1/2" to the 71" to get distance "A", then divide by 4-1/2 (4.5). This gives a result of 16.7777, which means that 17 spaces won't fit. Without cutting one or more of the pickets down, The most spaces which include the pickets themselves, that you can have (including the one past the post that you added) is 16. The actual number of pickets will be one less, since they will go between the posts, not covering one of them.
Now, if you divide the 75-1/2" by 16, you'll get 4.71875" (which is more than the 4.5" width of the pickets, so they will fit) then subtract the width of the pickets to get the spacing between them (4.71875 - 4.5 = 0.21875). Unless you're planning to use extremely accurate measuring devices, you're probably going to want to convert this to 16ths or 8ths, so just multiply this by 16 (or 8) to find out how many sixteenths (or eighths) are in 0.21875 (or whatever decimal you got from your measurement), which is 3.5 - 16ths. At this point, you could go to either 3/16" or 4/16" (1/4") spacing between your pickets. If you use 1/4" spacing, the last picket won't quite fit (there won't be any spacing between it and the previous picket). If you use 3/16" between them, your last picket will theoretically have 11/16" too much space between it and the post. You have to decide how you want to handle this, depending on your equipment, how fussy you are going to be, and how good your pickets are. If you use 3/16" spacing, or even 1/8" between them, and work towards the center from each post then adjust the space between the pickets for the last 5 or 6 pickets, no one will ever notice the difference... not even you.
## Step 5: Watch Out!
In reality the pickets won't be exactly 4-1/2" wide - or even straight! Wood, as purchased from a lumber yard, simply isn't all that uniform. From experience, unless you have ripped your pickets to a specific width, you're going to find at least a 1/8" variation between the narrowest and widest ones. And it will probably be more. I've also purchased many pickets that had a 1/2" or more crook or kink (side-to-side warping) in them. If you want to find out the real (average) width of your pickets, lay them out, side by side, and measure the total width without any spacing between them and divide by the number of pickets. Use that number for your calculations. If you don't, by the time you work with more than a few of them, any errors will add up and you'll find that things don't work out the way you planned.
If you want the pickets to be further apart, decrease the number of pickets used, because you want fewer, wider spaces. For example, if you used 14 pickets between posts, the center-center distance would be 75.5 / 15 = 5.03" and the spacing between pickets would be 0.5333 or about 1/2". Always remember to divide by one more than the actual number of pickets you intend to use. If you only want to use 13 pickets, each picket would be spaced 75.5 / 14 = 5.39286" from the beginning of the previous one and you would need 0.80929 or about 3/4" between pickets. Dropping to 12 pickets would increase the space between them to just over 1-1/4" (1.3076"). Just remember to measure the ACTUAL average width of your pickets the way I described above.
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Convert complex number to polar coordinates
Problem
Compute when $$x \in \mathbb{C}$$: $$x^2-4ix-5-i=0$$ and express output in polar coordinates
Attempt to solve
Solving this equation with quadratic formula:
$$x=\frac{4i \pm \sqrt{(-4i)^2-4\cdot (-5-i)}}{2}$$ $$x= \frac{4i \pm \sqrt{4(i+1)}}{2}$$ $$x = \frac{4i \pm 2\sqrt{i+1}}{2}$$ $$x = 2i \pm \sqrt{i+1}$$
I can transform cartesian complex numbers to polar with eulers formula: when $$z \in \mathbb{C}$$
$$z=re^{i\theta}$$
then: $$r=|z|=\sqrt{(\text{Re(z)})^2+(\text{Im(z)})^2}$$ $$\text{arg}(x)=\theta = \arctan{\frac{\text{Im}(z)}{\text{Re}(z)}}$$
Plugging in values after this computation would give us our complex in number in $$(r,\theta)$$ polar coordinates from $$(\text{Re},\text{Im})$$ cartesian coordinates.
Only problem is how do i convert complex number of form $$z=2i+\sqrt{i+1}$$ to polar since i don't know how to separate this into imaginary and real parts. How do you compute $$\text{Re}(z)$$ and $$\text{Im}(z)$$
• SInce $1+i=\sqrt{2}\exp\frac{\pi i}{4}$, $\sqrt{1+i}=\sqrt[4]{2}(\cos\frac{\pi i}{8}+i\sin\frac{\pi i}{8})$. You can thereby get the real and imaginary parts of $z$ and convert to polar coordinates.
– J.G.
Oct 6, 2018 at 21:10
• where does $1+i=\sqrt{2}\exp(\frac{\pi i}{4})$ come from ? @J.G.
– Tuki
Oct 6, 2018 at 21:12
• derived from $e^{i\pi}+1=0$ i suppose ?
– Tuki
Oct 6, 2018 at 21:13
Let $$a,b\in\mathbb{R}$$ so that $$\sqrt{i+1} = a+bi$$ $$i+1 = a^2 -b^2 +2abi$$
Equating real and imaginary parts, we have
$$2ab = 1$$
$$a^2 -b^2 = 1$$
Now we solve for $$(a,b)$$. \begin{align*} b &= \frac{1}{2a}\\\\ \implies \,\,\, a^2 - \left(\frac{1}{2a}\right)^2 &= 1 \\\\ a^2 &= 1 + \frac{1}{4a^2}\\\\ 4a^4 &= 4a^2 + 1\\\\ 4a^4 - 4a^2 -1 &= 0 \\\\ \end{align*}
This is a quadratic in $$a^2$$ (it's also a quadratic in $$2a^2$$, if you prefer!), so we use the quadratic formula:
$$a^2 = \frac{4 \pm \sqrt{16-4(4)(-1)}}{2(4)}$$
$$a^2 = \frac{1 \pm \sqrt{2}}{2}$$
Here we note that $$a$$ is real, so $$a^2>0$$, and we discard the negative case:
$$a^2 = \frac{1 + \sqrt{2}}{2}$$
$$a = \pm \sqrt{\frac{1 + \sqrt{2}}{2}}$$
$$b = \frac{1}{2a} = \pm \sqrt{\frac{\sqrt{2}-1}{2}}$$
This gives what you can call the principal root:
$$\sqrt{i+1} = \sqrt{\frac{1 + \sqrt{2}}{2}} + i\sqrt{\frac{\sqrt{2}-1}{2}}$$
As well as the negation of it:
$$-\sqrt{i+1} = -\sqrt{\frac{1 + \sqrt{2}}{2}} + i\left(-\sqrt{\frac{\sqrt{2}-1}{2}}\right)$$
Finally, substituting either of these into your expression $$z=2i \pm \sqrt{i+1}$$ will give you $$\text{Re}(z)$$ and $$\text{Im}(z)$$.
At that point, as you noted in your question, conversion to polar coordinates is straightforward.
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# Practice Aptitude Questions For RBI Grade B& Upcoming Exams 2017
Practice Aptitude Questions For RBI Grade – B& Upcoming Exams 2017 (Quadratic Equation):
Dear Readers, Important Practice Aptitude Questions for RBI Grade B 2017 Exam was given here with Solutions. Aspirants those who are preparing for RBI Grade – B and all other Competitive examination can use this.
Directions (Q. 1-10): In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
a) If x> y
b) If x โฅ y
c) If x< y
d) If x โค y
e) If x = y or if there is no relation between โxโ and โyโ.
1). I.15x2 – 46x + 35 = 0
II.4y2– 15y + 14 = 0
2). I.x2 + 42 = 13x
II.y=4โ1296
3).I.3x2 – 23x + 40 = 0
II.2y2– 23y + 66 = 0
4). I.x2 + 5x – 6 = 0
II.2y2 – 11y + 15 = 0
5). I.x2 + x – 2 = 0
II. y2 + 7y + 12 = 0
6). I.3x2+ 15x + 18 = 0
II.2y2 + 15y + 27 = 0
7). I.4x2+ 3x โ 1 = 0
II.6y2 โ 5y + 1 = 0
8). I.x2 + 11x + 30 = 0
II.y2 + 9y + 20 = 0
9). I.x =3โ357911
II.y = โ5041
10). I.5x + 7y = -43
II.9x โ 17y = 41
1). I.15x2 – 25x – 21x + 35 = 0
or 5x(3x – 5) – 7(3x – 5) = 0
or (5x – 7) (3x – 5) = 0
x = 7/5, 5/3
II.4y2 – 8y – 7y + 14 = 0
or 4y(y – 2) – 7(y – 2) = 0
or (4y – 7) (y – 2) = 0
y = 2, 7/4
2). I. x2+ 42 = 13x
or x2 – 13x + 42 = 0
or x2 – 7x – 6x + 42 = 0
or x(x – 7) – 6(x – 7) = 0
or (x – 6) (x – 7) = 0
x = 6, 7
II.y=4โ1296
y = 6
x โฅ y
3). I. 3x2– 23x + 40 = 0
or 3x2 – 15x – 8x + 40 = 0
or 3x(x – 5) – 8(x – 5) = 0
or (3x – 8) (x – 5) = 0
x = 5, 8/3
II.2y2 – 23y + 66 = 0
or 2y2 – 12y – 11y + 66 = 0
or 2y (y – 6) -11 (y – 6) = 0
or (y – 6)(2y – 11) = 0
y = 6, 11/ 2
x< y
4). I. x2– x + 6x -6 = 0
or x(x – 1) + 6(x – 1) = 0
or (x – 1) (x + 6) = 0
x = 1, -6
II.2y2 – 6y – 5y + 15 = 0
or 2y(y – 3) – 5(y – 3) = 0
or (y – 3) (2y – 5) = 0
y = 3, 5/2
5). I.x2+ x – 2 = 0
or x2 + 2x – x – 2 = 0
or x(x + 2) – 1(x + 2) = 0
or (x – 1) (x + 2) = 0
x = 1, – 2
II.y2 + 7y + 12 = 0
or y2 + 3y + 4y + 12 = 0
or y(y + 3) + 4(y + 3) = 0
or (y + 3) (y + 4) = 0
y = -3, -4
x> y
6). I.3x2 + 9x + 6x + 18 = 0
or 3x(x + 3) + 6(x + 3) = 0
or (x + 3)(3x + 6) = 0
x = -3, -2
II.2y2 + 6y + 9y+ 27 = 0
or 2y(y + 3) + 9(y + 3) = 0
or (2y + 9)(y + 3) = 0
y = -3, -9/2
x โฅ y
7). I.4x2 + 4x – x -1 = 0
or 4x(x+ 1)- 1(x + 1) = 0
or (4x – 1) (x + 1) = 0
x = -1, 1/4
II.6y2 – 3y – 2y + 1 = 0
or 3y(2y – 1) – 1(2y – 1) = 0
or (3y – 1) (2y – 1) = 0
y = 1/2,1/3
x< y
8). I. x2+ 11x + 30 = 0
or x(x + 5) + 6(x + 5) = 0
or (x + 5) (x + 6) = 0
x = -5, -6
II. y2 + 4y + 5y + 20 = 0
or y(y + 4) + 5(y + 4) = 0
or (y + 4) (y + 5) = 0
y = -4,-5
x โค y
9). I.x =3โ357911 = x = 71
II.y = โ5041= y = 71
x = y
10).Equation (l) ร9 & Equation (II) ร 5
45x + 63y = -387
45x – 85y = 205
– + –
148y = -592
y = -4 and x = -3
x> y
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# Dividing Rational Numbers Pre-Algebra. Vocabulary Rational Number: Any number that can be written as a fraction. Reciprocal/Multiplicative Inverse: Two.
## Presentation on theme: "Dividing Rational Numbers Pre-Algebra. Vocabulary Rational Number: Any number that can be written as a fraction. Reciprocal/Multiplicative Inverse: Two."— Presentation transcript:
Dividing Rational Numbers Pre-Algebra
Vocabulary Rational Number: Any number that can be written as a fraction. Reciprocal/Multiplicative Inverse: Two numbers whose product is 1. (Switch the numerator and the denominator.)
Example: In order to divide fractions, just remember Kentucky Chicken Fried. K – Keep the first fraction the same C – Change the division to multiplication F – Flip the second fraction (take the reciprocal)
We keep the first fraction in our problem. We change the division to multiplication. We flip the second fraction by taking the reciprocal.
Now we multiply. Since 5 and 15 share a factor of 5, we may factor out 5 from our problem. Since 24 and 36 share a factor of 12, we may factor out 12 from our problem. We multiply across horizontally. Finally, we simplify if necessary.
Example #2: This time we must change our mixed numbers into improper fractions! First we multiply 3 and 2 which yields 6. Then we add 1 to 6 and get 7. First we multiply 9 and 5 which yields 45. Then we add 4 to 45 and get 49.
Our new problem is now: We now use Keep, Change, Flip (KCF) to divide. K CF We factor. And finally, we multiply.
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# Maharashtra Board 9th Class Maths Part 1 Practice Set 4.2 Solutions Chapter 4 Ratio and Proportion
Balbharti Maharashtra State Board Class 9 Maths Solutions covers the Practice Set 4.2 Algebra 9th Class Maths Part 1 Answers Solutions Chapter 4 Ratio and Proportion.
## Practice Set 4.2 Algebra 9th Std Maths Part 1 Answers Chapter 4 Ratio and Proportion
Question 1.
Using the property $$\frac { a }{ b }$$ = $$\frac { ak }{ bk }$$, fill in the blanks by substituting proper numbers in the following.
Solution:
Question 2.
Find the following ratios.
i. The ratio of radius to circumference of the circle.
ii. The ratio of circumference of circle with radius r to its area.
iii. The ratio of diagonal of a square to its side, if the length of side is 7 cm.
iv. The lengths of sides of a rectangle are 5 cm and 3.5 cm. Find the ratio of numbers denoting its perimeter to area.
Solution:
i. Let the radius of circle be r.
then, its circumference = 2πr
Ratio of radius to circumference of the circle
The ratio of radius to circumference of the circle is 1 : 2π.
ii. Let the radius of the circle is r.
∴ circumference = 2πr and area = πr2
Ratio of circumference to the area of circle
∴ The ratio of circumference of circle with radius r to its area is 2 : r.
iii. Length of side of square = 7 cm
∴ Diagonal of square = √2 x side
= √2 x 7
= 7 √2 cm
Ratio of diagonal of a square to its side
∴ The ratio of diagonal of a square to its side is √2 : 1.
iv. Length of rectangle = (l) = 5 cm,
Breadth of rectangle = (b) = 3.5 cm
Perimeter of the rectangle = 2(l + b)
= 2(5 + 3.5)
= 2 x 8.5
= 17 cm
Area of the rectangle = l x b
= 5 x 3.5
= 17.5 cm2
Ratio of numbers denoting perimeter to the area of rectangle
∴ Ratio of numbers denoting perimeter to the area of rectangle is 34 : 35.
Question 3.
Compare the following
Solution:
Question 4.
Solve.
ABCD is a parallelogram. The ratio of ∠A and ∠B of this parallelogram is 5 : 4. FInd the measure of ∠B. [2 Marksl
Solution:
Ratio of ∠A and ∠B for given parallelogram is 5 : 4
Let the common multiple be x.
m∠A = 5x°and m∠B=4x°
Now, m∠A + m∠B = 180° …[Adjacent angles of a parallelogram arc supplementary]
∴ 5x° + 4x°= 180°
∴ 9x° = 180°
∴ x° = 20°
∴ m∠B=4x°= 4 x 20° = 80°
∴ The measure of ∠B is 800.
ii. The ratio of present ages of Albert and Salim is 5 : 9. Five years hence ratio of their ages will be 3 : 5. Find their present ages.
Solution:
The ratio of present ages of Albert and Salim is 5 : 9
Let the common multiple be x.
∴ Present age of Albert = 5x years and
Present age of Salim = 9x years
After 5 years,
Albert’s age = (5x + 5) years and
Salim’s age = (9x + 5) years
According to the given condition,
Five years hence ratio of their ages will be 3 : 5
$$\frac{5 x+5}{9 x+5}=\frac{3}{5}$$
∴ 5(5x + 5) = 3(9x + 5)
∴ 25x + 25 = 27x + 15
∴ 25 – 15 = 27 x – 25 x
∴ 10 = 2x
∴ x = 5
∴ Present age of Albert = 5x = 5 x 5 = 25 years
Present age of Salim = 9x = 9 x 5 = 45 years
∴ The present ages of Albert and Salim are 25 years and 45 years respectively.
iii. The ratio of length and breadth of a rectangle is 3 : 1, and its perimeter is 36 cm. Find the length and breadth of the rectangle.
Solution:
The ratio of length and breadth of a rectangle is 3 : 1
Let the common multiple be x.
Length of the rectangle (l) = 3x cm
and Breadth of the rectangle (b) = x cm
Given, perimeter of the rectangle = 36 cm
Since, Perimeter of the rectangle = 2(l + b)
∴ 36 = 2(3x + x)
∴ 36 = 2(4x)
∴ 36 = 8x
∴ $$x=\frac{36}{8}=\frac{9}{2}=4.5$$
Length of the rectangle = 3x = 3 x 4.5 = 13.5 cm
∴ The length of the rectangle is 13.5 cm and its breadth is 4.5 cm.
iv. The ratio of two numbers is 31 : 23 and their sum is 216. Find these numbers.
Solution:
The ratio of two numbers is 31 : 23
Let the common multiple be x.
∴ First number = 31x and
Second number = 23x
According to the given condition,
Sum of the numbers is 216
∴ 31x + 23x = 216
∴ 54x = 216
∴ x = 4
∴ First number = 31x = 31 x 4 = 124
Second number = 23x = 23 x 4 = 92
∴ The two numbers are 124 and 92.
v. If the product of two numbers is 360 and their ratio is 10 : 9, then find the numbers.
Solution:
Ratio of two numbers is 10 : 9
Let the common multiple be x.
∴ First number = 10x and
Second number = 9x
According to the given condition,
Product of two numbers is 360
∴ (10x) (9x) = 360
∴ 90x2 = 360
∴ x2 = 4
∴ x = 2 …. [Taking positive square root on both sides]
∴ First number = 10x = 10x2 = 20
Second number = 9x = 9x2 = 18
∴ The two numbers are 20 and 18.
Question 5.
If a : b = 3 : 1 and b : c = 5 : 1, then find the value of [3 Marks each]
Solution:
Given, a : b = 3 : 1
∴ $$\frac { a }{ b }$$ = $$\frac { 3 }{ 1 }$$
∴ a = 3b ….(i)
and b : c = 5 : 1
∴ $$\frac { b }{ c }$$ = $$\frac { 5 }{ 1 }$$
b = 5c …..(ii)
Substituting (ii) in (i),
we get a = 3(5c)
∴ a = 15c …(iii)
Ratio and Proportion 9th Class Practice Set 4.1 Question 6. If $$\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}$$ , then find the ratio $$\frac { a }{ b }$$.
Solution:
$$\sqrt{0.04 \times 0.4 \times a}=0.4 \times 0.04 \times \sqrt{b}$$ … [Given]
∴ 0.04 x 0.4 x a = (0.4)2 x (0.04)2 x b … [Squaring both sides]
9th Algebra Practice Set 4.2 Question 7. (x + 3) : (x + 11) = (x – 2) : (x + 1), then find the value of x.
Solution:
(x + 3) : (x + 11) = (x- 2) : (x+ 1)
$$\quad \frac{x+3}{x+11}=\frac{x-2}{x+1}$$
∴ (x + 3)(x +1) = (x – 2)(x + 11)
∴ x(x +1) + 3(x + 1) = x(x + 11) – 2(x + 11)
∴ x2 + x + 3x + 3 = x2 + 1 lx – 2x – 22
∴ x2 + 4x + 3 = x2 + 9x – 22
∴ 4x + 3 = 9x – 22
∴ 3 + 22 = 9x – 4x
∴ 25 = 5x
∴ x = 5
|
# i 1 .
Complex Numbers
Complex Number System
Definition of the Imaginary Number
i
The symbol i represents an imaginary
number with the properties:
i=
−1
i 2 = −1.
1
Definition of
−n =
−n
− 1 n = i n.
Simplifying Expressions in Terms of i
− 64
= 8i
− 100 = 10 i
− 29 = i
29
− 150 = i 150 = i 25 ⋅ 6
= 5i 6
Write the Radical Expression in Terms of i
− 100
− 25
=
10 i
5i
=2
2
The Complex Number System
Simplifying Powers of i
( )
=i
( )
= i 2 = −1
3
i13
= i4 + i
i18
= i4 + i2
i107
4
( )
= i4
26
+ i3
= −i
The Complex Number System
3
The Complex Number System
In a complex number a + bi, either a or b or
both can be 0.
If b = 0, as in 5 + 0i = 5, the number is real.
The reals are a subset of the complex numbers.
If a = 0 and b is not 0, as in 0 + 5i = 5i, the
complex number is called pure imaginary.
If a is not 0 and b is not 0, as in 3 + 2i, the
number is sometimes called imaginary.
The Complex Number System
The form a + bi is called the Standard Form for the
complex number.
Examples:
4 + 3i
!"
-3 – 5i
!#\$
#"
6
%
!&
0
&
!&
4i
&
!
(a + bi) + (c + di) = (a + c) + (b + d)i
(2 – 3i) + (4+ 5i)
= 6 + 2i
(5 + 3i) – (8 – 2i)
= -3 +5i
− 36 + − 49
= 6i + 7 i
= 13 i
4
MULTIPLICATION
(a + bi)(c + di) = ac – bd + (ad + bc)i
This is the same as “FOIL”.
Perform the Indicated Operation
(5 + 6i )( −5 − 12i ) = −25 − 60i − 30i − 72i 2
= −25 − 90i − 72i 2
= −25 − 90i − 72(−1)
= −25 − 90i + 72
= 47 − 90i
Determine the Product
7i ( 3 − 4i ) = 21i − 28i 2
21i − 28( −1) = 21i + 28
= 28 + 21i
5
The Complex Number System
The Complex Number System
The Complex Number System
6
Find the Product of the Complex
Number & Its Conjugate
5+i 3
(5 + i 3 ) (5 − i 3 )
5 + ( 3)
2
2
25 + 3 = 28
Determine the Quotient
4
5i
Rationaliz e the denominator .
4 i
⋅
5i i
=
4i
5i
2
=
4i
4i
4i
=−
=
5
−5
5( −1)
Determine the Quotient
5 − 2i
( 5 − 2i ) ( 3 + 5i )
⋅
=
3 − 5i
( 3 − 5i ) ( 3 + 5i )
=
( 5 − 2i ) ⋅ ( 3 + 5i )
32 + 52
7
Determine the Quotient
15 + 25i − 6i − 10i 2
32 + 52
15 + 25i − 6i − 10( −1)
9 + 25
15 + 25i − 6i + 10
34
=
25 + 19i
34
8
– Cards
– Cards
– Cards
– Cards
– Cards
|
Refer to our Texas Go Math Grade 1 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 1 Lesson 6.3 Answer Key Use Doubles to Add.
Essential Question
Explanation:
To get a double of a number,
we add the same number to itself.
For example, double of 2 is 2 + 2 = 4.
Explore
Draw to show the problem.
Write the number of fish.
Lesson Check
For The Teacher
• Read the following problem. There are 3 orange fish, 3 striped fish, and 1 white fish in the class fish tank. How many fish are in the fish tank?
Explanation:
There are 3 orange fish, 3 striped fish,
3 + 3 = 1
and 1 white fish in the class fish tank.
6 + 1 = 7
7 fish are in the fish tank
Math Talk
Mathematical Processes
How does knowing 3 + 3 help you solve the problem? Explain.
Explanation:
3 + 3 is known as a double fact
A number is added to itself is called a double fact
Model and Draw
Explanation:
Explanation:
6 + 7
is made as double plus one
6 + 6 + 1 = 13
Share and Show
Use to model. Make doubles. Add.
Question 1.
Explanation:
9 + 8
is made as double plus one
8 + 8 + 1 = 9
Question 2.
Explanation:
5 + 6
is made as double plus one
5 + 5 + 1 = 11
Problem Solving
Question 3.
Explanation:
7 + 8
is made as double plus one
7 + 7 + 1 = 15
Question 4.
Explanation:
5 + 4
is made as double plus one
4 + 4 + 1 = 9
Question 5.
Multi-Step Mandy has the same number of red and yellow leaves. Then she finds 1 more yellow leaf. There are 17 leaves in all. How many leaves are red? How many leaves are yellow?
_____ red leaves _____ yellow leaves
8 red leaves 9 yellow leaves
Explanation:
Total leaves are 17
8 +8 + 1 = 17
Double plus one
so, red leaves are 8 and yellow leaves are 8
8 + 1 = 9
H.O.T. Multi-Step Would you use count on or doubles to solve? Explain.
Question 6.
3 + 4
Explanation:
3 + 4 is derived as double plus one
3 + 3 + 1Â 7
Question 7.
3 + 9
Explanation:
3 + 9 = 12
started from greater number 9 and counted on
Question 8.
Leo walks in the park. He sees 3 dogs. He sees 3 cats. He sees 1 squirrel. How many animals does Leo see?
(A) 3
(B) 7
(C) 6
Explanation:
Leo walks in the park.
He sees 3 dogs. He sees 3 cats.
3 + 3 = 6
He sees 1 squirrel.
6 + 1 = 7
7 animals that Leo see
Question 9.
Analyze There are 7 birds in a tree. 8 more birds fly over. Which has the same sum as 7 + 8?
(A) 7 + 7 + 1
(B) 8 + 8 + 1
(C) 7 + 7 + 2
Explanation:
There are 7 birds in a tree. 8 more birds fly over.
7 + 7 + 1 = 15
Which has the same sum as 7 + 8
7 + 8 = 15
Question 10.
Multi-Step There are 5 white houses. There are 5 yellow houses. There is 1 brown house. How many houses
are there?
(A) 10
(B) 15
(C) 11
Explanation:
There are 5 white houses. There are 5 yellow houses.
5 + 5 = 10
There is 1 brown house.
10 + 1 = 11
11 houses are there
Question 11.
Texas Test Prep Which has the same sum as 9 + 8?
(A) 6 + 6 + 2
(B) 1 + 8 + 8
(C) 2 + 8 + 8
Explanation:
1 + 8 + 8 =17
9 + 9 = 17
Both the sums are same.
Take Home Activity
• Ask your child to show you how to use what he or she knows about doubles to help solve 6 + 5.
### Texas Go Math Grade 1 Lesson 6.3 Homework and Practice Answer Key
Question 1.
Explanation:
6 + 7
is made as double plus one
6 + 6 + 1 = 13
Question 2.
Explanation:
4 + 5
is made as double plus one
4 + 4 + 1 = 5
Problem Solving
Make doubles. Show how you find the sum.
Question 3.
There are 4 ants on the step. 5 more ants join them. How many ants are on the step?
___ + ____ + ____ = ____ ants
4 + 4 + 1 = 9
Explanation:
4 + 5 is changed as double plus one.
Texas Test Prep
Lesson Check
Question 4.
Toby finds 6 red pencils and 6 blue pencils. Then he finds 1 yellow pencil How many pencils does Toby find?
(A) 12
(B) 13
(C) 7
Explanation:
Toby finds 6 red pencils and 6 blue pencils.
6 + 6 = 12
Then he finds 1 yellow pencil
12 + 1 = 13
13Â pencils that Toby find
Question 5.
There are 8 fish in a pond. 9 more fish swim by. Which has the same sum as 8 + 9?
(A) 8 + 8 + 1
(B) 9 + 9 + 1
(C) 8 + 8 + 2
Explanation:
There are 8 fish in a pond. 9 more fish swim by.
Which has the same sum as 8 + 9
is 8 + 8 + 1 = 17
Question 6.
Multi-Step George has the same number of black shirts as white shirts. Then he gets 1 more white shirt. He has 9 shirts in all. How many shirts are white?
(A) 10
(B) 4
(C) 5
Explanation:
Total shirts are 9
9 is changed as Double plus 1
4 + 4 + 1
White and black shirts are equal
so, 4 each
1 white shirt more
4 + 1 = 5
Question 7.
Which has the same sum as 5 + 6?
(A) 5 + 5 + 1
(B) 6 + 6 + 0
(C) 5 + 6 + 1
|
# Cubic interpolation
On this page you can find explanation about (n-)cubic interpolation and implementations in Java and C++.
Anything at this page may be copied and modified.
Please contact me if you find an error.
## Cubic interpolation
If the values of a function f(x) and its derivative are known at x=0 and x=1, then the function can be interpolated on the interval [0,1] using a third degree polynomial. This is called cubic interpolation. The formula of this polynomial can be easily derived.
A third degree polynomial and its derivative:
For the green curve:
The values of the polynomial and its derivative at x=0 and x=1:
The four equations above can be rewritten to this:
And there we have our cubic interpolation formula.
Interpolation is often used to interpolate between a list of values. In that case we don't know the derivative of the function. We could simply use derivative 0 at every point, but we obtain smoother curves when we use the slope of a line between the previous and the next point as the derivative at a point. In that case the resulting polynomial is called a Catmull-Rom spline. Suppose you have the values p0, p1, p2 and p3 at respectively x=-1, x=0, x=1, and x=2. Then we can assign the values of f(0), f(1), f'(0) and f'(1) using the formulas below to interpolate between p1 and p2.
Combining the last four formulas and the preceding four, we get:
So our cubic interpolation formula becomes:
## The first and the last interval
We used the two points left of the interval and the two points right of the inverval as inputs for the interpolation function. But what if we want to interpolate between the first two or last two elements of a list? Then we have no p0 or no p3. The solution is to imagine an extra point at each end of the list. In other words, we have to make up a value for p0 and p3 when interpolating the leftmost and rightmost interval respectively. Three ways to do this are:
• Set the factor of x3 to 0 to make a quadratic polynomial:
Left: p0 = 3p1 - 3p2 + p3
Right: p3 = p0 - 3p1 + 3p2
• Repeat the first and the last point.
Left: p0 = p1
Right: p3 = p2
• Let the end point be in the middle of a line between the imaginary point and the point next to the end point.
Left: p0 = 2p1 - p2
Right: p3 = 2p2 - p1
## Bicubic interpolation
Bicubic interpolation is cubic interpolation in two dimensions. I'll only consider the case where we want to interpolate a two dimensional grid. We can use the cubic interpolation formula to construct the bicubic interpolation formula.
Suppose we have the 16 points pij, with i and j going from 0 to 3 and with pij located at (i-1, j-1). Then we can interpolate the area [0,1] x [0,1] by first interpolating the four columns and then interpolating the results in the horizontal direction. The formula becomes:
Bicubic interpolation can be used to resize images. However, this is (currently) out of the scope of this article. Please don't ask me questions about that.
## Java implementation
public class CubicInterpolator
{
public static double getValue (double[] p, double x) {
return p[1] + 0.5 * x*(p[2] - p[0] + x*(2.0*p[0] - 5.0*p[1] + 4.0*p[2] - p[3] + x*(3.0*(p[1] - p[2]) + p[3] - p[0])));
}
}
public class BicubicInterpolator extends CubicInterpolator
{
private double[] arr = new double[4];
public double getValue (double[][] p, double x, double y) {
arr[0] = getValue(p[0], y);
arr[1] = getValue(p[1], y);
arr[2] = getValue(p[2], y);
arr[3] = getValue(p[3], y);
return getValue(arr, x);
}
}
public class TricubicInterpolator extends BicubicInterpolator
{
private double[] arr = new double[4];
public double getValue (double[][][] p, double x, double y, double z) {
arr[0] = getValue(p[0], y, z);
arr[1] = getValue(p[1], y, z);
arr[2] = getValue(p[2], y, z);
arr[3] = getValue(p[3], y, z);
return getValue(arr, x);
}
}
## C++ implementation
#include <iostream>
#include <assert.h>
double cubicInterpolate (double p[4], double x) {
return p[1] + 0.5 * x*(p[2] - p[0] + x*(2.0*p[0] - 5.0*p[1] + 4.0*p[2] - p[3] + x*(3.0*(p[1] - p[2]) + p[3] - p[0])));
}
double bicubicInterpolate (double p[4][4], double x, double y) {
double arr[4];
arr[0] = cubicInterpolate(p[0], y);
arr[1] = cubicInterpolate(p[1], y);
arr[2] = cubicInterpolate(p[2], y);
arr[3] = cubicInterpolate(p[3], y);
return cubicInterpolate(arr, x);
}
double tricubicInterpolate (double p[4][4][4], double x, double y, double z) {
double arr[4];
arr[0] = bicubicInterpolate(p[0], y, z);
arr[1] = bicubicInterpolate(p[1], y, z);
arr[2] = bicubicInterpolate(p[2], y, z);
arr[3] = bicubicInterpolate(p[3], y, z);
return cubicInterpolate(arr, x);
}
double nCubicInterpolate (int n, double* p, double coordinates[]) {
assert(n > 0);
if (n == 1) {
return cubicInterpolate(p, *coordinates);
}
else {
double arr[4];
int skip = 1 << (n - 1) * 2;
arr[0] = nCubicInterpolate(n - 1, p, coordinates + 1);
arr[1] = nCubicInterpolate(n - 1, p + skip, coordinates + 1);
arr[2] = nCubicInterpolate(n - 1, p + 2*skip, coordinates + 1);
arr[3] = nCubicInterpolate(n - 1, p + 3*skip, coordinates + 1);
return cubicInterpolate(arr, *coordinates);
}
}
int main () {
// Create array
double p[4][4] = {{1,3,3,4}, {7,2,3,4}, {1,6,3,6}, {2,5,7,2}};
// Interpolate
std::cout << bicubicInterpolate(p, 0.1, 0.2) << '\n';
// Or use the nCubicInterpolate function
double co[2] = {0.1, 0.2};
std::cout << nCubicInterpolate(2, (double*) p, co) << '\n';
}
## Bicubic interpolation polynomial
This section provides an alternative way to calculate bicubic interpolation. For most purposes this way is probably less practical and efficient than the way it is done above.
We can rewrite the formula for bicubic interpolation as a multivariate polynomial:
With these values for aij, the coefficients:
In Java code we can write this as:
public class CachedBicubicInterpolator
{
private double a00, a01, a02, a03;
private double a10, a11, a12, a13;
private double a20, a21, a22, a23;
private double a30, a31, a32, a33;
public void updateCoefficients (double[][] p) {
a00 = p[1][1];
a01 = -.5*p[1][0] + .5*p[1][2];
a02 = p[1][0] - 2.5*p[1][1] + 2*p[1][2] - .5*p[1][3];
a03 = -.5*p[1][0] + 1.5*p[1][1] - 1.5*p[1][2] + .5*p[1][3];
a10 = -.5*p[0][1] + .5*p[2][1];
a11 = .25*p[0][0] - .25*p[0][2] - .25*p[2][0] + .25*p[2][2];
a12 = -.5*p[0][0] + 1.25*p[0][1] - p[0][2] + .25*p[0][3] + .5*p[2][0] - 1.25*p[2][1] + p[2][2] - .25*p[2][3];
a13 = .25*p[0][0] - .75*p[0][1] + .75*p[0][2] - .25*p[0][3] - .25*p[2][0] + .75*p[2][1] - .75*p[2][2] + .25*p[2][3];
a20 = p[0][1] - 2.5*p[1][1] + 2*p[2][1] - .5*p[3][1];
a21 = -.5*p[0][0] + .5*p[0][2] + 1.25*p[1][0] - 1.25*p[1][2] - p[2][0] + p[2][2] + .25*p[3][0] - .25*p[3][2];
a22 = p[0][0] - 2.5*p[0][1] + 2*p[0][2] - .5*p[0][3] - 2.5*p[1][0] + 6.25*p[1][1] - 5*p[1][2] + 1.25*p[1][3] + 2*p[2][0] - 5*p[2][1] + 4*p[2][2] - p[2][3] - .5*p[3][0] + 1.25*p[3][1] - p[3][2] + .25*p[3][3];
a23 = -.5*p[0][0] + 1.5*p[0][1] - 1.5*p[0][2] + .5*p[0][3] + 1.25*p[1][0] - 3.75*p[1][1] + 3.75*p[1][2] - 1.25*p[1][3] - p[2][0] + 3*p[2][1] - 3*p[2][2] + p[2][3] + .25*p[3][0] - .75*p[3][1] + .75*p[3][2] - .25*p[3][3];
a30 = -.5*p[0][1] + 1.5*p[1][1] - 1.5*p[2][1] + .5*p[3][1];
a31 = .25*p[0][0] - .25*p[0][2] - .75*p[1][0] + .75*p[1][2] + .75*p[2][0] - .75*p[2][2] - .25*p[3][0] + .25*p[3][2];
a32 = -.5*p[0][0] + 1.25*p[0][1] - p[0][2] + .25*p[0][3] + 1.5*p[1][0] - 3.75*p[1][1] + 3*p[1][2] - .75*p[1][3] - 1.5*p[2][0] + 3.75*p[2][1] - 3*p[2][2] + .75*p[2][3] + .5*p[3][0] - 1.25*p[3][1] + p[3][2] - .25*p[3][3];
a33 = .25*p[0][0] - .75*p[0][1] + .75*p[0][2] - .25*p[0][3] - .75*p[1][0] + 2.25*p[1][1] - 2.25*p[1][2] + .75*p[1][3] + .75*p[2][0] - 2.25*p[2][1] + 2.25*p[2][2] - .75*p[2][3] - .25*p[3][0] + .75*p[3][1] - .75*p[3][2] + .25*p[3][3];
}
public double getValue (double x, double y) {
double x2 = x * x;
double x3 = x2 * x;
double y2 = y * y;
double y3 = y2 * y;
return (a00 + a01 * y + a02 * y2 + a03 * y3) +
(a10 + a11 * y + a12 * y2 + a13 * y3) * x +
(a20 + a21 * y + a22 * y2 + a23 * y3) * x2 +
(a30 + a31 * y + a32 * y2 + a33 * y3) * x3;
}
}
This implementation can run faster than the implementation above when getValue is called multiple times for one call to updateCoefficients.
Similar implementations could be written for other dimensions than two.
|
## Introduction to Data Interpretation
#### Data Interpretation
Direction: Study the following histogram and answer the questions.
1. The number of persons in the age group 20 – 30 years is :
1. On the basis of given graph in question ,
Number of persons in the age group 20 – 25 years = 250
Number of persons in the age group 25 – 30 years = 250
Required answer = Number of persons in the age group 20– 25 years + Number of persons in the age group 25–30 years
##### Correct Option: B
On the basis of given graph in question ,
Number of persons in the age group 20 – 25 years = 250
Number of persons in the age group 25 – 30 years = 250
Required answer = Number of persons in the age group 20–25 years + Number of persons in the age group 25–30 years
Required answer = 250 + 150 = 400
1. The total number of persons in the age group of 15 years to 45 years is :
1. According to given histogram , we have
Number of persons in the age group of 15 - 20 years = 450
Number of persons in the age group of 20 - 25 years = 250
Number of persons in the age group of 25 - 30 years = 150
Number of persons in the age group of 30 - 35 years = 75
Number of persons in the age group of 35 - 40 years = 50
Number of persons in the age group of 40 - 45 years = 25
##### Correct Option: C
According to given histogram , we have
Number of persons in the age group of 15 - 20 years = 450
Number of persons in the age group of 20 - 25 years = 250
Number of persons in the age group of 25 - 30 years = 150
Number of persons in the age group of 30 - 35 years = 75
Number of persons in the age group of 35 - 40 years = 50
Number of persons in the age group of 40 - 45 years = 25
Total number of persons in the age group of 15 years to 45 years = 450 + 250 + 150 + 75 + 50 + 25 = 1000
Direction: The histogram shows the marks obtained by 45 students of a class. Look at the histogram and answer the questions.
1. How many students have obtained 30 or more marks but less than 40?
1. From the given bar diagram , we seen in bar of class interval 30–40
Number of students who obtained 30 or more marks but less than 40 = 3
##### Correct Option: A
From the given bar diagram , we seen in bar of class interval 30–40
Number of students who obtained 30 or more marks but less than 40 = 3
Thus , correct answer is option A .
1. How many students have obtained marks less than 10?
1. According to given bar graph , we have
Number of students who have obtained marks less than 10 = 2
##### Correct Option: A
According to given bar graph , we have
Number of students who have obtained marks less than 10 = 2
Hence , required answer is option A .
1. If the pass mark be 30, what is the percentage of successful students?
1. On the basis of given graph in question ,
Total number of students = 2 + 6 + 10 + 3 + 8 + 4 + 7 + 5 = 45
Number of failures = 2 + 6 + 10 = 18
Number of successful students = 45 – 18 = 27
∴ Required percentage = Number of successful students × 100 Total number of students
##### Correct Option: B
On the basis of given graph in question ,
Total number of students = 2 + 6 + 10 + 3 + 8 + 4 + 7 + 5 = 45
Number of failures = 2 + 6 + 10 = 18
Number of successful students = 45 – 18 = 27
∴ Required percentage = Number of successful students × 100 Total number of students
Required percentage = 27 × 100 = 60% 45
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Converting Fractions to Mixed Numbers Unit 14 > Lesson 7 of 11
You may recall the example below from a previous lesson.
Example 1
In example 1, we used circles to help us solve the problem. Now look at the next example.
Example 2: At a birthday party, there are 19 cupcakes to be shared equally among 11 guests. What part of the cupcakes will each guest get? Analysis: We need to divide 19 cupcakes by 11 equal parts. It would be time-consuming to use circles or other shapes to help us solve this problem. Therefore, we need an arithmetic method. Step 1: Look at the fraction nineteen-elevenths below. Recall that the fraction bar means to divide the numerator by the denominator. This is shown in step 2. Step 2: Step 3: Solution:
In example 2, the fraction nineteen-elevenths was converted to the mixed number one and eight-elevenths. Recall that a mixed number consists of a whole-number part and a fractional part. Let's look at some more examples of converting fractions to mixed numbers using the arithmetic method.
Example 3: Analysis: We need to divide 17 into 5 equal parts Step 1: Step 2: Answer:
Example 4: Analysis: We need to divide 37 into 10 equal parts. Step 1: Step 2: Answer:
Example 5: Analysis: We need to divide 37 into 13 equal parts. Step 1: Step 2: Answer:
In each of the examples above, we converted a fraction to a mixed number through long division of its numerator and denominator. Look at example 6 below. What is wrong with this problem?
Example 6: Analysis: In the fraction seven-eighths, the numerator is less than the denominator. Therefore, seven-eighths is a proper fraction less than 1. We know from a previous lesson that a mixed number is greater than 1. Answer: Seven-eighths cannot be written as a mixed number because it is a proper fraction.
Example 7: Can these fractions be written as mixed numbers? Explain why or why not. Analysis: In each fraction above, the numerator is equal to the denominator. Therefore, each of these fractions is an improper fraction equal to 1. But a mixed number is greater than 1. Answer: These fractions cannot be written as mixed numbers since each is an improper fraction equal to 1.
After reading examples 6 and 7, you may be wondering: Which types of fractions can be written as mixed numbers? To answer this question, let's review an important chart from a previous lesson.
Comparison of numerator and denominator Example Type of Fraction Write As If the numerator < denominator, then the fraction < 1. proper fraction proper fraction If the numerator = denominator, then the fraction = 1. improper fraction whole number If the numerator > denominator, then the fraction > 1. improper fraction mixed number
The answer to the question is: Only an improper fraction greater than 1 can be written to a mixed number.
Summary: We can convert an improper fraction greater than one to a mixed number through long division of its numerator and denominator.
### Exercises
In Exercises 1 through 5, click once in an ANSWER BOX and type in your answer; then click ENTER. After you click ENTER, a message will appear in the RESULTS BOX to indicate whether your answer is correct or incorrect. To start over, click CLEAR. Note: To write the mixed number four and two-thirds, enter 4, a space, and then 2/3 into the form.
1. Write eleven-fifths as a mixed number. ANSWER BOX: RESULTS BOX:
2. Write eleven-fourths as a mixed number. ANSWER BOX: RESULTS BOX:
3. Write thirteen-ninths as a mixed number. ANSWER BOX: RESULTS BOX:
4. On field day, there are 23 pies to share equally among 7 classes. What part of the pies will each class get? ANSWER BOX: RESULTS BOX:
5. A teacher gives her class a spelling test worth 35 points. If there are 8 words graded equally, then how many points is each word worth? ANSWER BOX: RESULTS BOX:
This lesson is by Gisele Glosser. You can find me on Google.
Lessons on Fractions Introduction to Fractions Classifying Fractions Equivalent Fractions Simplifying Fractions Comparing Fractions Ordering Fractions Converting Fractions to Mixed Numbers Converting Mixed Numbers to Fractions Practice Exercises Challenge Exercises Solutions
Related Activities Interactive Puzzles Printable Worksheets Fraction Goodies Game
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# Math Snap
## Compare the linear functions expressed by the equation, $y=-x+3$, and by data in the table. \begin{tabular}{|c|c|c|c|c|} \hline$x$ & -4 & -2 & 1 & 3 \\ \hline$y$ & 1 & -1 & -4 & -6 \\ \hline \end{tabular} Explain how to determine if these two are the same function expressed in different ways
#### STEP 1
Assumptions 1. We have a linear function expressed by the equation $y = -x + 3$. 2. We have a set of data points given in a table. 3. We need to determine if the table represents the same linear function as the equation.
#### STEP 2
First, let's write down the given linear function: $y = -x + 3$
#### STEP 3
Next, we will use the given equation to calculate the $y$-values for the $x$-values provided in the table. For $x = -4$: $y = -(-4) + 3$
#### STEP 4
Calculate the value: $y = 4 + 3 = 7$
#### STEP 5
For $x = -2$: $y = -(-2) + 3$
#### STEP 6
Calculate the value: $y = 2 + 3 = 5$
#### STEP 7
For $x = 1$: $y = -(1) + 3$
#### STEP 8
Calculate the value: $y = -1 + 3 = 2$
#### STEP 9
For $x = 3$: $y = -(3) + 3$
#### STEP 10
Calculate the value: $y = -3 + 3 = 0$
#### STEP 11
Now, let's summarize the calculated $y$-values for the given $x$-values: $\begin{array}{|c|c|} \hline x & y \\ \hline -4 & 7 \\ -2 & 5 \\ 1 & 2 \\ 3 & 0 \\ \hline \end{array}$
#### STEP 12
Compare these calculated $y$-values with the $y$-values given in the table: $\begin{array}{|c|c|c|c|c|} \hline x & -4 & -2 & 1 & 3 \\ \hline y & 1 & -1 & -4 & -6 \\ \hline \end{array}$
#### STEP 13
Notice that the $y$-values from the given table (1, -1, -4, -6) do not match the $y$-values calculated using the equation (7, 5, 2, 0).
##### SOLUTION
Since the $y$-values from the table do not match the $y$-values calculated from the equation $y = -x + 3$, we can conclude that the table does not represent the same linear function as the equation.
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# Statistics Solver
## Statistics Solver
Learn about central measures of tendency (mean, median, mode), measures of dispersion (range, variance, standard deviation, mean deviation, median deviation), and relative measures (coefficient of range, coefficient of variation) to analyze and interpret statistical data effectively.
#### Introduction to Statistics
Statistics is a branch of mathematics dealing with the collection, analysis, interpretation, presentation, and organization of data. It provides tools for making sense of complex data sets and drawing meaningful conclusions from them.
#### Central Measures of Tendency
1. Mean (Average):
• The mean is the sum of all the values divided by the number of values.
• Formula: $\text{Mean} (\bar{x}) = \frac{\sum_{i=1}^{n} x_i}{n}$
• Example: For the dataset {2, 3, 5, 7, 11}, the mean is $$\frac{2 + 3 + 5 + 7 + 11}{5} = 5.6$$.
2. Median:
• The median is the middle value when the data is ordered.
• For an odd number of observations, it is the middle value.
• For an even number of observations, it is the average of the two middle values.
• Example: For the dataset {2, 3, 5, 7, 11}, the median is 5. For {2, 3, 5, 7}, the median is $$\frac{3 + 5}{2} = 4$$.
3. Mode:
• The mode is the value that appears most frequently in the dataset.
• A dataset can have one mode, more than one mode, or no mode at all.
• Example: For the dataset {2, 3, 3, 5, 7}, the mode is 3.
#### Measures of Dispersion
Measures of dispersion describe the spread or variability of a dataset.
1. Range:
• The range is the difference between the highest and lowest values.
• Formula: $\text{Range} = \text{Max} – \text{Min}$
• Example: For the dataset {2, 3, 5, 7, 11}, the range is $$11 – 2 = 9$$.
2. Variance:
• Variance measures how far each value in the dataset is from the mean.
• Formula (for population variance): $\sigma^2 = \frac{\sum_{i=1}^{N} (x_i – \mu)^2}{N}$
• Formula (for sample variance): $s^2 = \frac{\sum_{i=1}^{n} (x_i – \bar{x})^2}{n-1}$
• Example: For the dataset {2, 3, 5, 7, 11}, the variance (sample) is $$s^2 = \frac{(2-5.6)^2 + (3-5.6)^2 + (5-5.6)^2 + (7-5.6)^2 + (11-5.6)^2}{4} = 10.3$$.
3. Standard Deviation:
• The standard deviation is the square root of the variance.
• Formula (for population): $\sigma = \sqrt{\sigma^2}$
• Formula (for sample): $s = \sqrt{s^2}$
• Example: For the dataset {2, 3, 5, 7, 11}, the standard deviation (sample) is $$s = \sqrt{10.3} \approx 3.21$$.
4. Mean Deviation:
• Mean deviation is the average of the absolute differences between each value and the mean.
• Formula: $\text{Mean Deviation} = \frac{\sum_{i=1}^{n} |x_i – \bar{x}|}{n}$
• Example: For the dataset {2, 3, 5, 7, 11}, the mean deviation is $$\frac{|2-5.6| + |3-5.6| + |5-5.6| + |7-5.6| + |11-5.6|}{5} = 2.72$$.
5. Median Deviation:
• Median deviation is the median of the absolute differences between each value and the median.
• Example: For the dataset {2, 3, 5, 7, 11}, with the median being 5, the deviations are {3, 2, 0, 2, 6}, and the median of these deviations is 2.
6. Coefficient of Range:
• The coefficient of range is a relative measure of the range.
• Formula: $\text{Coefficient of Range} = \frac{\text{Max} – \text{Min}}{\text{Max} + \text{Min}}$
• Example: For the dataset {2, 3, 5, 7, 11}, the coefficient of range is $$\frac{11 – 2}{11 + 2} = \frac{9}{13} \approx 0.69$$.
7. Coefficient of Variation:
• The coefficient of variation is the ratio of the standard deviation to the mean, expressed as a percentage.
• Formula: $\text{Coefficient of Variation} = \frac{s}{\bar{x}} \times 100\%$
• Example: For the dataset {2, 3, 5, 7, 11}, with a mean of 5.6 and a standard deviation of 3.21, the coefficient of variation is $$\frac{3.21}{5.6} \times 100\% \approx 57.32\%$$.
Understanding these statistical measures is crucial for analyzing and interpreting data effectively. These tools help summarize large datasets, identify patterns, and make informed decisions based on data.
### Summary
• Central Measures of Tendency: Mean, Median, Mode
• Measures of Dispersion: Range, Variance, Standard Deviation, Mean Deviation, Median Deviation
• Relative Measures: Coefficient of Range, Coefficient of Variation
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# How do you simplify sqrt41?
Jun 11, 2017
$\sqrt{41} \approx 6.4031242374$ is an irrational number which cannot be simplified.
#### Explanation:
$41$ is a prime number, so has no square factors.
As a result its square root cannot be simplified. It is an irrational number.
We find:
${6}^{2} = 36 < 41 < 49 = {7}^{2}$
So:
$6 < \sqrt{41} < 7$
To get a good approximation for $\sqrt{41}$ note that:
${64}^{2} = 4096$
So:
$\sqrt{41} \approx \sqrt{40.96} = 6.4 = \frac{32}{5}$
In general:
$\sqrt{{a}^{2} + b} = a + \frac{b}{2 a + \frac{b}{2 a + \frac{b}{2 a + \ldots}}}$
Putting $a = \frac{32}{5}$ and $b = \frac{1}{25}$, we find:
$\sqrt{41} = \frac{32}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \ldots}}}$
We can get rational approximations for $\sqrt{41}$ by truncating this continued fraction.
For example:
$\sqrt{41} \approx \frac{32}{5} + \frac{\frac{1}{25}}{\frac{64}{5}} = \frac{2049}{320}$
$\sqrt{41} \approx \frac{32}{5} + \frac{\frac{1}{25}}{\frac{64}{5} + \frac{\frac{1}{25}}{\frac{64}{5}}} = \frac{131168}{20485} \approx 6.4031242374$
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Integration: It’s Backwards Differentiation - Integration and Infinite Series - Calculus For Dummies
## Calculus For Dummies, 2nd Edition (2014)
### Chapter 15. Integration: It’s Backwards Differentiation
IN THIS CHAPTER
Using the area function
Getting familiar with the fundamental theorem of calculus
Finding antiderivatives
Figuring exact areas the easy way
Chapter 14 shows you the hard way to calculate the area under a function using the formal definition of integration — the limit of a Riemann sum. In this chapter, I calculate areas the easy way, taking advantage of one of the most important and amazing discoveries in mathematics — that integration (finding areas) is just differentiation in reverse. That reverse process was a great discovery, and it’s based on some difficult ideas, but before we get to that, let’s talk about a related, straightforward reverse process, namely…
Antidifferentiation
The derivative of sin x is cos x, so the antiderivative of cos x is sin x; the derivative of is , so the antiderivative of is — you just go backwards. There’s a bit more to it, but that’s the basic idea. Later in this chapter, I show you how to find areas by using antiderivatives. This is much easier than finding areas with the Riemann sum technique.
Now consider and its derivative again. The derivative of is also , as is the derivative of . Any function of the form , where C is any number, has a derivative of . So, every such function is an antiderivative of .
Definition of the indefinite integral: The indefinite integral of a function , written as , is the family of all antiderivatives of the function. For example, because the derivative of is , the indefinite integral of is , and you write
You probably recognize this integration symbol, from the discussion of the definite integral in Chapter 14. The definite integral symbol, however, contains two little numbers like that tell you to compute the area under a function between those two numbers, called the limits of integration. The naked version of the symbol, indicates an indefinite integral or an antiderivative. This chapter is all about the intimate connection between these two symbols, these two ideas.
Figure 15-1 shows the family of antiderivatives of , namely . Note that this family of curves has an infinite number of curves. They go up and down forever and are infinitely dense. The vertical gap of 2 units between each curve in Figure 15-1 is just a visual aid.
FIGURE 15-1: The family of curves All these functions have the same derivative,
Consider a few things about Figure 15-1. The top curve on the graph is ; the one below it is ; the bottom one . By the power rule, these three functions, as well as all the others in this family of functions, have a derivative of . Now, consider the slope of each of the curves where x equals 1 (see the tangent lines drawn on the curves). The derivative of each function is , so when x equals 1, the slope of each curve is , or 3. Thus, all these little tangent lines are parallel. Next, notice that all the functions in Figure 15-1 are identical except for being slid up or down (remember vertical shifts from Chapter 5?). Because they differ only by a vertical shift, the steepness at any x-value, like at , is the same for all the curves. This is the visual way to understand why each of these curves has the same derivative, and, thus, why each curve is an antiderivative of the same function.
Vocabulary, Voshmabulary: What Difference Does It Make?
In general, definitions and vocabulary are very important in mathematics, and it’s a good idea to use them correctly. But with the current topic, I’m going to be a bit lazy about precise terminology, and I hereby give you permission to do so as well.
If you’re a stickler, you should say that the indefinite integral of is and that is the family or set of all antiderivatives of (you don’t say that is the antiderivative), and you say that , for instance, is an antiderivative of . And on a test, you should definitely write . If you leave the C off, you’ll likely lose some points.
But, when discussing these matters, no one will care or be confused if you get tired of saying “+ C” after every indefinite integral and just say, for example, that the indefinite integral of is , and you can skip the indefiniteand just say that the integral of is . And instead of always talking about that family of functions business, you can just say that the antiderivative of is or that the antiderivative of is . Everyone will know what you mean. It may cost me my membership in the National Council of Teachers of Mathematics, but at least occasionally, I use this loose approach.
The Annoying Area Function
This is a tough one — gird your loins. Say you’ve got any old function, . Imagine that at some t-value, call it s, you draw a fixed vertical line. See Figure 15-2.
FIGURE 15-2: Area under f between s and x is swept out by the moving line at x.
Then you take a moveable vertical line, starting at the same point, s (“s” is for starting point), and drag it to the right. As you drag the line, you sweep out a larger and larger area under the curve. This area is a function of x, the position of the moving line. In symbols, you write
Note that t is the input variable in instead of x because x is already taken — it’s the input variable in . The subscript f in indicates that is the area function for the particular curve f or . The dt is a little increment along the t-axis — actually an infinitesimally small increment.
Here’s a simple example to make sure you’ve got a handle on how an area function works. By the way, don’t feel bad if you find this extremely hard to grasp — you’ve got lots of company. Say you’ve got the simple function, , that’s a horizontal line at . If you sweep out area beginning at , you get the following area function:
You can see that the area swept out from 3 to 4 is 10 because, in dragging the line from 3 to 4, you sweep out a rectangle with a width of 1 and a height of 10, which has an area of 1 times 10, or 10. See Figure 15-3.
FIGURE 15-3: Area under between 3 and x is swept out by the moving vertical line at x.
So, , the area swept out as you hit 4, equals 10. equals 20 because when you drag the line to 5, you’ve swept out a rectangle with a width of 2 and height of 10, which has an area of 2 times 10, or 20. equals 30, and so on.
Now, imagine that you drag the line across at a rate of one unit per second. You start at , and you hit 4 at 1 second, 5 at 2 seconds, 6 at 3 seconds, and so on. How much area are you sweeping out per second? Ten square units per second because each second you sweep out another 1-by-10 rectangle. Notice — this is huge — that because the width of each rectangle you sweep out is 1, the area of each rectangle — which is given by height times width — is the same as its height because anything times 1 equals itself. You see why this is huge in a minute. (By the way, the real rate we care about here is not area swept out per second, but, rather, area swept out per unit change on the x-axis. I explain it in terms of per second because it’s easier to think about a sweeping-out-area rate this way. And since you’re dragging the line across at one x-axis unit per one second, both rates are the same. Take your pick.)
The derivative of an area function equals the rate of area being swept out. Okay, are you sitting down? You’ve reached another one of the big Ah ha! moments in the history of mathematics. Recall that a derivative is a rate. So, because the rate at which the previous area function grows is 10 square units per second, you can say its derivative equals 10. Thus, you can write
Again, this just tells you that with each 1 unit increase in x, (the area function) goes up 10. Now here’s the critical thing: Notice that this rate or derivative of 10 is the same as the height of the original function because as you go across 1 unit, you sweep out a rectangle that’s 1 by 10, which has an area of 10, the height of the function.
And the rate works out to 10 regardless of the width of the rectangle. Imagine that you drag the vertical line from to . At a rate of one unit per second, that’ll take you 1/1,000th of a second, and you’ll sweep out a skinny rectangle with a width of 1/1,000, a height of 10, and thus an area of 10 times 1/1,000, or 1/100 square units. The rate of area being swept out would be, therefore, which equals 10 square units per second. So you see that with every small increment along the x-axis, the rate of area being swept out equals the function’s height.
This works for any function, not just horizontal lines. Look at the function and its area function that sweeps out area beginning at in Figure 15-4.
FIGURE 15-4: Area under between 2 and x is swept out by the moving vertical line at x.
Between and , grows by the area of that skinny, dark shaded “rectangle” with a width of 0.1 and a height of about 15. (As you can see, it’s not really a rectangle; it’s closer to a trapezoid, but it’s not that either because its tiny top is curving slightly. But, in the limit, as the width gets smaller and smaller, the skinny “rectangle” behaves precisely like a real rectangle.) So, to repeat, grows by the area of that dark “rectangle” which has an area extremely close to 0.1 times 15, or 1.5. That area is swept out in 0.1 seconds, so the rate of area being swept out is , or 15 square units per second, the height of the function. This idea is so important that it deserves an icon.
The sweeping out area rate equals the height. The rate of area being swept out under a curve by an area function at a given x-value is equal to the height of the curve at that x-value.
The Power and the Glory of the Fundamental Theorem of Calculus
Sound the trumpets! Now that you’ve seen the connection between the rate of growth of an area function and the height of the given curve, you’re ready for the fundamental theorem of calculus — what some say is one of the most important theorems in the history of mathematics.
The fundamental theorem of calculus: Given an area function that sweeps out area under ,
the rate at which area is being swept out is equal to the height of the original function. So, because the rate is the derivative, the derivative of the area function equals the original function:
Because you can also write the above equation as follows:
Break out the smelling salts.
Now, because the derivative of is , is by definition an antiderivative of . Check out how this works by returning to the simple function from the previous section, , and its area function, .
According to the fundamental theorem, . Thus must be an antiderivative of 10; in other words, is a function whose derivative is 10. Because any function of the form , where C is a number, has a derivative of 10, the antiderivative of 10 is . The particular number C depends on your choice of s, the point where you start sweeping out area. For a particular choice of s, the area function will be the one function (out of all the functions in the family of curves ) that crosses the x-axis at s. To figure out C, set the antiderivative equal to zero, plug the value of s into x, and solve for C.
For this function with an antiderivative of , if you start sweeping out area at, say, , then , so , and thus, , or just 10x. (Note that C does not necessarily equal s. In fact, it usually doesn’t [especially when ]. When , C often also equals 0, but not for all functions.)
Figure 15-5 shows why is the correct area function if you start sweeping out area at zero. In the top graph in the figure, the area under the curve from 0 to 3 is 30, and that’s given by . And you can see that the area from 0 to 5 is 50, which agrees with the fact that .
FIGURE 15-5: Three area functions for
If instead you start sweeping out area at and define a new area function, , then , so C equals 20 and is thus . This area function is 20 more than , which starts at , because if you start at , you’ve already swept out an area of 20 by the time you get to zero. Figure 15-5 shows why is 20 more than .
And if you start sweeping out area at , , so and the area function is . This function is 30 less than because with , you lose the 3-by-10 rectangle between 0 and 3 that has (see the bottom graph in Figure 15-5).
An area function is an antiderivative. The area swept out under the horizontal line from some number s to x, is given by an antiderivative of 10, namely where the value of C depends on where you start sweeping out area.
Now let’s look at graphs of , , and . (Note that Figure 15-5 doesn’t show the graphs of , , and . You see three graphs of the horizontal line function, ; and you see the areas swept out under by , , and , but you don’t actually see the graphs of these three area functions.) Check out Figure 15-6.
FIGURE 15-6: The actual graphs of
Figure 15-6 shows the graphs of the equations of , , and which we worked out before: , , and . (As you can see, all three are simple, lines.) The y-values of these three functions give you the areas swept out under that you see in Figure 15-5. Note that the three x-intercepts you see in Figure 15-6 are the three x-values in Figure 15-5 where sweeping out area begins.
We worked out above that and that . You can see those areas of 30 and 50 in the top graph of Figure 15-5. In Figure 15-6, you see these results on at the points and . You also saw in Figure 15-5 that was 20 more than ; you see that result in Figure 15-6 where on is 20 higher than (3, 30) on . Finally, you saw in Figure 15-5 that is 30 less than . Figure 15-6 shows that in a different way: at any x-value, the line is 30 units below the line.
A few observations. You already know from the fundamental theorem that (and the same for and ). That was explained above in terms of rates: For , , and , the rate of area being swept out under equals 10. Figure 15-6 also shows that (and the same for and ), but here you see the derivative as a slope. The slopes, of course, of all three lines equal 10. Finally, note that — like you saw in Figure 15-1 — the three lines in Figure 15-6 differ from each other only by a vertical translation. These three lines (and the infinity of all other vertically translated lines) are all members of the class of functions, , the family of antiderivatives of .
For the next example, look again at the parabola , our friend from Chapter 14 which we analyzed in terms of the sum of the areas of rectangles (Riemann sums). Flip back to Figure 14-4, and check out the shaded region under . Now you can finally compute the exact area of the shaded region the easy way.
The area function for sweeping out area under is . By the fundamental theorem, , and so is an antiderivative of . Any function of the form has a derivative of (try it), so that’s the antiderivative. For Figure 14-6, you want to sweep out area beginning at 0, so . Set the antiderivative equal to zero, plug the value of s into x, and solve for C: , so , and thus
The area swept out from 0 to 3 — which we did the hard way in Chapter 14 by computing the limit of a Riemann sum — is simply :
Piece o’ cake. That was much less work than doing it the hard way.
And after you know that the area function that starts at zero, it’s a snap to figure the area of other sections under the parabola that don’t start at zero. Say, for example, you want the area under the parabola between 2 and 3. You can compute that area by subtracting the area between 0 and 2 from the area between 0 and 3. You just figured the area between 0 and 3 — that’s 12. And the area between 0 and 2 is . So the area between 2 and 3 is , or . This subtraction method brings us to the next topic — the second version of the fundamental theorem.
The Fundamental Theorem of Calculus: Take Two
Now we finally arrive at the super-duper shortcut integration theorem that you’ll use for the rest of your natural born days — or at least till the end of your stint with calculus. This shortcut method is all you need for the integration word problems in Chapters 17 and 18.
The fundamental theorem of calculus (second version or shortcut version): Let F be any antiderivative of the function f; then
This theorem gives you the super shortcut for computing a definite integral like , the area under the parabola between 2 and 3. As I show in the previous section, you can get this area by subtracting the area between 0 and 2 from the area between 0 and 3, but to do that you need to know that the particular area function sweeping out area beginning at zero, , is (with a C value of zero).
The beauty of the shortcut theorem is that you don’t have to even use an area function like . You just find any antiderivative, , of your function, and do the subtraction, . The simplest antiderivative to use is the one where . So here’s how you use the theorem to find the area under our parabola from 2 to 3. is an antiderivative of . Then the theorem gives you:
Granted, this is the same computation I did in the previous section using the area function with , but that’s only because for the function, when s is zero, C is also zero. It’s sort of a coincidence, and it’s not true for all functions. But regardless of the function, the shortcut works, and you don’t have to worry about area functions or s or C. All you do is .
Here’s another example: What’s the area under between and ? The derivative of is , so is an antiderivative of , and thus
What could be simpler?
Areas above the curve and below the x-axis count as negative areas. Before going on, I’d be remiss if I didn’t touch on negative areas (this is virtually the same caution made at the very end of Chapter 14). Note that with the two examples above, the parabola, , and the exponential function, the areas we’re computing are under the curves and above the x-axis. These areas count as ordinary, positive areas. But, if a function goes below the x-axis, areas above the curve and below the x-axis count as negative areas. This is the case whether you’re using an area function, the first version of the fundamental theorem of calculus, or the shortcut version. Don’t worry about this for now. You see how this works in Chapter 17.
Okay, so now you’ve got the super shortcut for computing the area under a curve. And if one big shortcut wasn’t enough to make your day, Table 15-1 lists some rules about definite integrals that can make your life much easier.
TABLE 15-1 Five Easy Rules for Definite Integrals
Now that I’ve given you the shortcut version of the fundamental theorem, that doesn’t mean you’re off the hook. Below are three different ways to understand why the theorem works. This is difficult stuff — brace yourself.
Alternatively, you can skip these explanations if all you want to know is how to compute an area: forget about C and just subtract from . I include these explanations because I suspect you’re dying to learn extra math just for the love of learning — right? Other books just give you the rules; I explain why they work and the underlying principles — that’s why they pay me the big bucks.
Actually, in all seriousness, you should read at least some of this material. The fundamental theorem of calculus is one of the most important theorems in all of mathematics, so you ought to spend some time trying hard to understand what it’s all about. It’s worth the effort. Of the three explanations, the first is the easiest. But if you only want to read one or two of the three, I’d read just the third, or the second and the third. Or, you could begin with the figures accompanying the three explanations, because the figures really show you what’s going on. Finally, if you can’t digest all of this in one sitting — no worries — you can revisit it later.
Why the theorem works: Area functions explanation
One way to understand the shortcut version of the fundamental theorem is by looking at area functions. As you can see in Figure 15-7, the dark-shaded area between a and b can be figured by starting with the area between s and b, then cutting away (subtracting) the area between s and a. And it doesn’t matter whether you use 0 as the left edge of the areas or any other value of s. Do you see that you’d get the same result whether you use the graph on the left or the graph on the right?
FIGURE 15-7: Figuring the area between a and b with two different area functions.
Take a look at (see Figure 15-8). Say you want the area between 5 and 8 under the horizontal line , and you are forced to use calculus.
FIGURE 15-8: The shaded area equals 30 — well, duh, it’s a 3-by-10 rectangle.
Look back at two of the area functions for in Figure 15-5: starting at 0 (in which ) and starting at :
If you use to compute the area between 5 and 8 in Figure 15-8, you get the following:
If, on the other hand, you use to compute the same area, you get the same result:
Notice that the two 20s in the second line from the bottom cancel. Recall that all antiderivatives of are of the form . Regardless of the value of C, it cancels out as in this example. Thus, you can use any antiderivative with any value of C. For convenience, everyone just uses the antiderivative with so that you don’t mess with C at all. And the choice of s (the point where the area function begins) is irrelevant. So when you’re using the shortcut version of the fundamental theorem, and computing an area with , you’re sort of using a mystery area function with a C value of zero and an unknown starting point, s. Get it?
Why the theorem works: The integration-differentiation connection
The next explanation of the shortcut version of the fundamental theorem involves the yin/yang relationship between differentiation and integration. Check out Figure 15-9.
FIGURE 15-9: The essence of differentiation and integration in a single figure! It’s a yin/yang thing.
The figure shows a function, , and its derivative, . Look carefully at the numbers 4, 6, and 8 on both graphs. The connection between 4, 6, and 8 on the graph of f — which are the amounts of risebetween consecutive points on the curve — and 4, 6, and 8 on the graph of — which are the areas of the trapezoids under — shows the intimate relationship between integration and differentiation. Figure 15-9 is a picture worth a thousand symbols and equations, encapsulating the essence of integration in a single snapshot. It shows how the shortcut version of the fundamental theorem works because it shows that the area under between 1 and 4 equals the total rise on between and , in other words that
Note that I’ve called the two functions in Figure 15-9 and in the above equation f and to emphasize that is the derivative of . I could have instead referred to as F and referred to as f which would emphasize that is an antiderivative of . In that case you would write the above area equation in the standard way,
Either way, the meaning’s the same. I use the derivative version to point out how finding area is differentiation in reverse. Going from left to right in Figure 15-9 is differentiation: The slopes of f correspond to heights on . Going from right to left is integration: Areas under correspond to the change in height between two points on f.
Okay, here’s how it works. Imagine you’re going up along f from to . Every point along the way has a certain steepness, a slope. This slope is plotted as the y-coordinate, or height, on the graph of . The fact that goes up from to tells you that the slope of f goes up from 3 to 5 as you travel between and . This all follows from basic differentiation.
Now, as you go along f from to , the slope is constantly changing. But it turns out that because you go up a total rise of 4 as you run across 1, the average of all the slopes on f between and is , or 4. Because each of these slopes is plotted as a y-coordinate or height on , it follows that the average height of between and is also 4. Thus, between two given points, average slope on f equals average height on .
Hold on, you’re almost there. Slope equals , so when the run is 1, the slope equals the rise. For example, from to on f, the curve rises up 4 and the average slope between those points is also 4. Thus, between any two points on f whose x-coordinates differ by 1, the average slope is the rise.
The area of a trapezoid like the ones on the right in Figure 15-9 equals its width times its average height. (This is true of any other similar shape that has a bottom like a rectangle; the top can be any crooked line or funky curve you like.) So, because the width of each trapezoid is 1, and because anything times 1 is itself, the average height of each trapezoid under is its area; for instance, the area of that first trapezoid is 4 and its average height is also 4.
Are you ready for the grand finale? Here’s the whole argument in a nutshell. On f, ; going from f to , ; on , . So that gives you , and thus, finally, . And that’s what the second version of the fundamental theorem says:
These ideas are unavoidably difficult. You may have to read it two or three times for it to really sink in.
Notice that it makes no difference to the relationship between slope and area if you use any other function of the form instead of . Any parabola like or is exactly the same shape as ; it’s just been slid up or down vertically. Any such parabola rises up between and in precisely the same way as the parabola in Figure 15-9. From 1 to 2 these parabolas go over 1, up 4. From 2 to 3 they go over 1, up 6, and so on. This is why any antiderivative can be used to find area. The total area under between 1 to 4, namely 18, corresponds to the total rise on any of these parabolas from 1 to 4, namely , or 18.
At the risk of beating a dead horse, I’ve got a third explanation of the fundamental theorem for you. You might prefer it to the first two because it’s less abstract — it’s connected to simple, commonsense ideas encountered in our day-to-day world. This explanation has a lot in common with the previous one, but the ideas are presented from a different angle.
Why the theorem works: A connection to — egad! — statistics
Don’t let the title of this section put you off. I realize that many readers of this calculus book may not have studied statistics. No worries; the statistics connection I explain below involves a very simple thing covered in statistics courses, but you don’t need to know any statistics at all to understand this idea. The simple idea is the relationship between a frequency distribution graph and a cumulative frequency distribution graph (you may have run across such graphs in a newspaper or magazine). Consider Figure 15-10.
FIGURE 15-10: A frequency distribution histogram (above) and a cumulative frequency distribution histogram (below) for the annual profits of Widgets-R-Us show the connection between differentiation and integration.
The upper graph in the figure shows a frequency distribution histogram of the annual profits of Widgets-R-Us from January 1, 2001 through December 31, 2013. The rectangle marked ’07, for example, shows that the company’s profit for 2007 was \$2,000,000 (their best year during the period 2001–2013).
The lower graph in the figure is a cumulative frequency distribution histogram for the same data used for the upper graph. The difference is simply that in the cumulative graph, the height of each column shows the total profits earned since 1/1/2001. Look at the ’02 column in the lower graph and the ’01 and ’02 rectangles in the upper graph, for example. You can see that the ’02 column shows the ’02 rectangle sitting on top of the ’01 rectangle which gives that ’02 column a height equal to the total of the profits from ’01 and ’02. Got it? As you go to the right on the cumulative graph, the height of each successive column simply grows by the amount of profits earned in the corresponding single year shown in the upper graph.
Okay. So here’s the calculus connection. (Bear with me; it takes a while to walk through all this.) Look at the top rectangle of the ’08 column on the cumulative graph (let’s call that graph C for short). At that point on C, you runacross 1 year and rise up \$1,250,000, the ’08 profit you see on the frequency distribution graph (F for short). , so, since the run equals 1, the slope equals 1,250,000/1, or just 1,250,000, which is, of course, the same as the rise. Thus, the slope on C (at ’08 or any other year) can be read as a height on F for the corresponding year. (Make sure you see how this works.) Since the heights (or function values) on F are the slopes of C, F is the derivative of C. In short, F, the derivative, tells us about the slope of C.
The next idea is that since F is the derivative of C, C, by definition, is the antiderivative of F (for example, C might equal and F would equal ). Now, what does C, the antiderivative of F, tell us about F ? Imagine dragging a vertical line from left to right over F. As you sweep over the rectangles on F — year by year — the total profit you’re sweeping over is shown climbing up along C.
Look at the ’01 through ’08 rectangles on F. You can see those same rectangles climbing up stair-step fashion along C (see the rectangles labeled A, B, C, etc. on both graphs). The heights of the rectangles from F keep adding up on C as you climb up the stair-step shape. And I’ve shown how the same ’01 through ’08 rectangles that lie along the stair-step top of C can also be seen in a vertical stack at year ’08 on C. I’ve drawn the cumulative graph his way so it’s even more obvious how the heights of the rectangles add up. (Note: Most cumulative histograms are not drawn this way.)
Each rectangle on F has a base of 1 year, so, since , the area of each rectangle equals its height. So, as you stack up rectangles on C, you’re adding up the areas of those rectangles from F. For example, the height of the ’01 through ’08 stack of rectangles on C (\$8.5 million) equals the total area of the ’01 through ’08 rectangles on F. And, therefore, the heights or function values of C — which is the antiderivative of F — give you the area under the top edge of F. That’s how integration works.
Okay, we’re just about done. Now let’s go through how these two graphs explain the shortcut version of the fundamental theorem of calculus and the relationship between differentiation and integration. Look at the ’06 through ’12 rectangles on F (with the bold border). You can see those same rectangles in the bold portion of the ’12 column of C. The height of that bold stack, which shows the total profits made during those 7 years, \$7.75 million, equals the total area of the 7 rectangles in F. And to get the height of that stack on C, you simply subtract the height of the stack’s bottom edge from the height of its upper edge. That’s really all the shortcut version of the fundamental theorem says: The area under any portion of a function (like F) is given by the change in height on the function’s antiderivative (like C).
In a nutshell (keep looking at those rectangles with the bold border in both graphs), the slopes of the rectangles on C appear as heights on F. That’s differentiation. Reversing direction, you see integration: the change in heightson C shows the area under F. Voilà: Differentiation and integration are two sides of the same coin.
(Note: Mathematical purists may object to this explanation of the fundamental theorem because it involves discrete graphs (for example, the fact that the cumulative distribution histogram in Figure 15-10 goes up at one-year increments), whereas calculus is the study of smooth, continuously changing graphs (the calculus version of the cumulative distribution histogram would be a smooth curve that would show the total profits growing every millisecond — actually, in theory, every infinitesimal fraction of a second). Okay — objection noted — but the fact is that the explanation here does accurately show how integration and differentiation are related and does correctly show how the shortcut version of the fundamental theorem works. All that’s needed to turn Figure 15-10 and the accompanying explanation into standard calculus is to take everything to the limit, making the profit interval shorter and shorter and shorter: from a year to a month to a day, etc., etc. In the limit, the discrete graphs in Figure 15-10 would meld into the type of smooth graphs used in calculus. But the ideas wouldn’t change. The ideas would be exactly as explained here. This is very similar to what you saw in Chapter 14 where you first approximated the area under a curve by adding up the areas of rectangles and then were able to compute the exact area by using the limit process to narrow the widths of the rectangles till their widths became infinitesimal.)
Well, there you have it — actual explanations of why the shortcut version of the fundamental theorem works and why finding area is differentiation in reverse. If you understand only half of what I’ve just written, you’re way ahead of most students of calculus. The good news is that you probably won’t be tested on this theoretical stuff. Now let’s come back down to earth.
Finding Antiderivatives: Three Basic Techniques
I’ve been talking a lot about antiderivatives, but just how do you find them? In this section, I give you three easy techniques. Then in Chapter 16, I give you four advanced techniques. By the way, you will be tested on this stuff.
Reverse rules for antiderivatives
The easiest antiderivative rules are the ones that are the reverse of derivative rules you already know. (You can brush up on derivative rules in Chapter 10 if you need to.) These are automatic, one-step antiderivatives with the exception of the reverse power rule, which is only slightly harder.
No-brainer reverse rules
You know that the derivative of sinx is cosx, so reversing that tells you that an antiderivative of cosx is sinx. What could be simpler? But don’t forget that all functions of the form are antiderivatives of cosx. In symbols, you write
Table 15-2 lists the reverse rules for antiderivatives.
TABLE 15-2 Basic Antiderivative Formulas
The slightly more difficult reverse power rule
By the power rule for differentiation, you know that
Here’s the simple method for reversing the power rule. Use for your function. Recall that the power rule says to
1. Bring the power in front where it will multiply the rest of the derivative.
2. Reduce the power by one and simplify.
Thus, .
To reverse this process, you reverse the order of the two steps and reverse the math within each step. Here’s how that works for the above problem:
1. Increase the power by one.
The 3 becomes a 4.
2. Divide by the new power and simplify.
And thus you write .
The reverse power rule does not work for a power of negative one. The reverse power rule works for all powers (including negative and decimal powers) except for a power of negative one. Instead of using the reverse power rule, you should just memorize that the antiderivative of is (rule 3 in Table 15-2).
Test your antiderivatives by differentiating them. Especially when you’re new to antidifferentiation, it’s a good idea to test your antiderivatives by differentiating them — you can ignore the C. If you get back to your original function, you know your antiderivative is correct.
With the antiderivative you just found and the shortcut version of the fundamental theorem, you can determine the area under between, say, 1 and 2:
Guessing and checking
The guess-and-check method works when the integrand (that’s the expression after the integral symbol not counting the dx, and it’s the thing you want to antidifferentiate) is close to a function that you know the reverse rule for. For example, say you want the antiderivative of . Well, you know that the derivative of sine is cosine. Reversing that tells you that the antiderivative of cosine is sine. So you might think that the antiderivative of is . That’s your guess. Now check it by differentiating it to see if you get the original function, :
This result is very close to the original function, except for that extra coefficient of 2. In other words, the answer is 2 times as much as what you want. Because you want a result that’s half of this, just try an antiderivative that’s half of your first guess: So your new guess is . Check this second guess by differentiating it, and you get the desired result.
Here’s another example. What’s the antiderivative of ?
1. Guess the antiderivative.
This looks sort of like a power rule problem, so try the reverse power rule. The antiderivative of is by the reverse power rule, so your guess is .
2. Check your guess by differentiating it.
Your result, is three times too much, so make your second guess a third of your first guess — that’s , or .
4. Check your second guess by differentiating it.
This checks. You’re done. The antiderivative of is .
The two previous examples show that guess and check works well when the function you want to antidifferentiate has an argument like 3x or (where x is raised to the first power) instead of a plain old x. (Recall that in a function like , the 5x is called the argument.) In this case, all you have to do is tweak your guess by the reciprocal of the coefficient of x: the 3 in , for example (the 2 in has no effect on your answer). In fact, for these easy problems, you don’t really have to do any guessing and checking. You can immediately see how to tweak your guess. It becomes sort of a one-step process. If the function’s argument is more complicated than — like the in — you have to try the next method, substitution.
The substitution method
If you look back at the examples of the guess and check method in the previous section, you can see why the first guess in each case didn’t work. When you differentiate the guess, the chain rule produces an extra constant: 2 in the first example, 3 in the second. You then tweak the guesses with and to compensate for the extra constant.
Now say you want the antiderivative of and you guess that it is . Watch what happens when you differentiate to check it:
Here the chain rule produces an extra 2x — because the derivative of is 2x — but if you try to compensate for this by attaching a to your guess, it won’t work. Try it.
So, guessing and checking doesn’t work for antidifferentiating — actually no method works for this simple-looking integrand (not all functions have antiderivatives) — but your admirable attempt at differentiation here reveals a new class of functions that you can antidifferentiate. Because the derivative of is , the antiderivative of must be . This function, , is the type of function you can antidifferentiate with the substitution method.
Keep your eyes peeled for the derivative of the function’s argument. The substitution method works when the integrand contains a function and the derivative of the function’s argument — in other words, when it contains that extra thing produced by the chain rule — or something just like it except for a constant. And the integrand must not contain any other extra stuff.
The derivative of is · by the rule and the chain rule. So, the antiderivative of is . And if you were asked to find the antiderivative of , you would know that the substitution method would work because this expression contains , which is the derivative of the argument of , namely .
By now, you’re probably wondering why this is called the substitution method. I show you why in the step-by-step method below. But first, I want to point out that you don’t always have to use the step-by-step method. Assuming you understand why the antiderivative of is , you may encounter problems where you can just see the antiderivative without doing any work. But whether or not you can just see the answers to problems like that one, the substitution method is a good technique to learn because, for one thing, it has many uses in calculus and other areas of mathematics, and for another, your teacher may require that you know it and use it. Okay, so here’s how to find with substitution:
1. Set u equal to the argument of the main function.
The argument of is , so you set u equal to .
2. Take the derivative of u with respect to x.
3. Solve for dx.
4. Make the substitutions.
In , u takes the place of and takes the place of dx. So now you’ve got . The two 2xs cancel, giving you .
5. Antidifferentiate using the simple reverse rule.
6. Substitute back in for u, coming full circle.
u equals , so goes in for the u:
That’s it. So .
If the original problem had been instead of , you follow the same steps except that in Step 4, after making the substitution, you arrive at . The xs still cancel — that’s the important thing — but after canceling you get , which has that extra in it. No worries. Just pull the through the giving you . Now you finish this problem just as you did in Steps 5 and 6, except for the extra :
Because C is any old constant, is still any old constant, so you can get rid of the in front of the C. That may seem somewhat (grossly?) unmathematical, but it’s right. Thus, your final answer is . You should check this by differentiating it.
Here are a few examples of antiderivatives you can do with the substitution method so you can learn how to spot them:
·
The derivative of is , but you don’t have to pay any attention to the 3 in or the 4 in the integrand. Because the integrand contains and no other extra stuff, substitution works. Try it.
·
The integrand contains a function, , and the derivative of its argument (tanx) — which is . Because the integrand doesn’t contain any other extra stuff (except for the 10, which doesn’t matter), substitution works. Do it.
·
Because the integrand contains the derivative of namely and no other stuff except for the , substitution works. Go for it.
You can do the three problems just listed with a method that combines substitution and guess-and-check (as long as your teacher doesn’t insist that you show the six-step substitution solution). Try using this combo method to antidifferentiate the first example, . First, you confirm that the integral fits the pattern for substitution — it does, as pointed out in the first item on the checklist. This confirmation is the only part substitution plays in the combo method. Now you finish the problem with the guess-and-check method:
The antiderivative of cosine is sine, so a good guess for the antiderivative of is .
2. Check your guess by differentiating it.
Your result from Step 2, is of what you want, , so make your guess bigger (note that is the reciprocal of ). Your second guess is thus .
4. Check this second guess by differentiating it.
Finding Area with Substitution Problems
You can use the shortcut version of the fundamental theorem to calculate the area under a function that you integrate with the substitution method. You can do this in two ways. In the previous section, I use substitution, setting uequal to , to find the antiderivative of :
If you want the area under this curve from, say, 0.5 to 1, the fundamental theorem does the trick:
Another method, which amounts to the same thing, is to change the limits of integration and do the whole problem in terms of u. Refer back to the six-step solution in the section “The substitution method.” What follows is very similar, except that this time you’re doing definite integration rather than indefinite integration. Again, you want the area given by :
1. Set u equal to .
2. Take the derivative of u with respect to x.
3. Solve for dx.
4. Determine the new limits of integration.
5. Make the substitutions, including the new limits of integration, and cancel the two 2xs.
(In this problem, only one of the limits is new because when .)
6. Use the antiderivative and the fundamental theorem to get the desired area without making the switch back to .
It’s a case of six of one, half a dozen of another with the two methods; they require about the same amount of work. So you can take your pick — however, most teachers and textbooks emphasize the second method, so you probably should learn it.
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Number System Quiz-1
Maths is an important subject in CLAT,DULLB & Other Law Exams. In any of law exam, Maths carries weightage of 20 -25 % of questions. With focused practice good marks can be fetched from this section. These questions are very important in achieving your success in CLAT, DULLB and Other Law Exams..
Q1. Which fraction is the greatest among the following:
• All are same
Solution
Since what we can say by seeing 11/32, 15/47, 37/15
11/32 will be equal to 1/3 if 11/33
15/47 will be equal to 1/3 if 15/45
37/115 will be equal to 1/3 if 37/111
So, 11/32 > 1/3 Since 32 < 33
(11/32 > 11/33 = 1/3)
15/47 < 1/3, Since 47 > 45
15/47 < 15/45 = 1/3
37/115 < 1/3, Since 115 > 111
37/115 < 37/111 = 1/3
37/115 < 1/3
Hence 11/32 Will be the greatest
Q2.which fraction is the smallest of the following:
• All are same
• Solution
, Since
using same principle
Alternate Method:
When the difference of numerator and denominator of all the numbers are same, then choose the one for smallest numerator and denominator.
Therefore, 45/49 is the answer.
Q3. What is the remainder if 10400 is divided by
• 0
• 1
• 2
• 3
Solution
Last 3 digits = 400 which is divisible by 8
Q4. A man gave 1/3 of his wealth to his wife, 1/2 of remaining to his first son, 1/2 of remaining to his second son & rest of Rs. 6 lacs to his youngest son. Find out his wealth.
• 40 lacs
• 38 lacs
• 36 lacs
• 42 lacs
Solution
Let wealth be equal to w
Given to wife = w/3
Given to 1st son =
Given to 2nd son =
Youngest son = 6 lac = = w/6
Therefore, w = 36 lac
Alternate Method:
By looking at the question, we can see remaining wealth is 6 lacs and then by looking at the options only 36 and 42 are multiple of 6.
Then, check both the numbers and verify the answer.
Q5. = ?
Solution
=
Q6. The sum of the digits of a two digit number is 8; if the digit are reversed the number is decreased by 54. Find the number
• 71
• 17
• 62
• 53
Solution
x + y = 8
(10x + y) - (10y + x) = 54
9(x - y) = 54,
x - y = 6
Solving this we get, x = 7, y = 1
Alternate Method:
Work through the options
Q7. = ?
• 20
• 30
• 15
• 17
Solution
Since if a + b + c = 0, then a3 + b3 + c3 = 3abc
Q8. = 32 x 32, x = ?
• 230
• 325
• 256
• 125
Solution
= 32 x 32
(64 x 64) x x = (32)2 x (32)2
x = 16 x 16 = 256
Q9. = 0.0008, x = ?
• 30
• 28
• 25
• 40
Solution
= 0.0008, = = 5
x = 25
Q10. Find the greatest number of 4 digits which when divided by 20, 30 &15 leave 18, 28 & 13 as remainders respectively.
• 9958
• 9960
• 9985
• 9990
Solution
LCM of 20, 30 & 15 is 60 highest multiple of 60 less than 10,000 = 9960
Remainder = 18, 28, 13 or 20, 30, 15
20 – 18 = 2, 30 – 28 = 2, 15 – 13 = 2
Hence, number = 9960 - 2 = 9958
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ClearExam: Best CLAT Coaching in Delhi | Best CLAT Coaching Center | CLAT Online Coaching: quants-number-system-quiz1
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Math Home
# 1. Definition of Derivative
### Definition
Definition 1: The derivative of a function $$f$$ at a point $$a$$ is defined as $\lim_{h \rightarrow 0} \frac{f(a+h) - f(a)}{h}$
Definition 2: The derivative of a function $$f$$ at a point $$a$$ is defined as $\lim_{x \rightarrow a} \frac{f(x) - f(a)}{x - a}$
The derivative of a function $$f(x)$$ is the function $$f'(x)$$ defined by $$f'(a)$$ is equal to the derivative of $$f$$ at the point $$a$$ for every point $$a$$.
### Explanation
The derivative of $$f(x)$$ at the point $$a$$ is the slope of the tangent line of $$f$$ at $$a.$$
Consider the graph $$f(x) = \frac{x^3}{10}+1.$$
The slope between two points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is $\frac{y_2 - y_1}{x_2 - x_1}$ Given a coordinate $$x,$$ the $$y$$ value in the graph of $$f$$ is $$f(x).$$ So, the slope between two points on the graph is $\frac{f(x_2)-f(x_1)}{x_2-x_1}$ Fix the $$x$$-value $$x = 1.$$ The point on the graph is $$(1, f(1)).$$
The slope between the points $$(1,f(1))$$ and $$(2,f(2))$$ can be computed using the slope formula: \begin{align} \frac{f(2)-f(1)}{2-1} & = f(2)-f(1) \\ & = \frac{8}{10}+1-(\frac{1}{10}-1) \\ & = \frac{7}{10} \end{align}
In the above paragraph, the difference is the derivative formula without then limit when $$h = 1.$$ The slope is $\frac{f(1+1)-f(1)}{(1+1)-1}$
When $$h = \frac{1}{2},$$ we get $$x + h = 1 + \frac{1}{2} = \frac{3}{2}.$$ The slope between $$(1, f(1))$$ and $$\left(\frac{3}{2}, f\left(\frac{3}{2}\right)\right)$$ is \begin{align} \frac{f\left(\frac{3}{2}\right)-f(1)}{\frac{3}{2}-1} & = 2\left(f\left(\frac{3}{2}\right)-f(1)\right) \\ & = 2\left(\frac{27}{80}+1-\frac{1}{10}-1\right) \\ & = \frac{19}{40} \end{align}
We can find the slope between the points $$(1, f(1))$$ and $$(1+h, f(1+h))$$ for smaller and smaller $$h.$$ As $$h$$ approaches $$0,$$ the slope will approach the slope of the tangent line, which is \begin{align} \lim_{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} & = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{(1+h)^3}{10}+1-\frac{1}{10}-1\right) \\ & = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{1+3h+3h^2+h^3}{10} - \frac{1}{10} \right) \\ & = \lim_{h \rightarrow 0} \frac{1}{h}\left(\frac{3h+3h^2+h^3}{10} \right) \\ & = \lim_{h \rightarrow 0} \frac{3+3h+h^2}{10} \\ & = \frac{3}{10} \\ \end{align}
If you look at the lines point by point, you can see how the lines approach the tangent in the limit.
### Example
The derivative of $$f(x) = 4$$ is $$f'(x) = 0$$ at every point $$x.$$
For any point $$x,$$ \begin{align} \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} & = \lim_{h \rightarrow 0}\frac{4-4}{h} \\ & = \lim_{h \rightarrow 0} 0 \\ & = 0 \end{align}
Claim: If $$f(x) = c$$ for any constant $$c,$$ then $$f'(x) = 0$$ at every point $$x.$$ This can be shown by replacing $$4 - 4$$ with $$c - c.$$
The derivative of $$f(x) = x$$ is $$f'(x) = 1$$ at every point $$x.$$
For any point $$x,$$ \begin{align} \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} & = \lim_{h \rightarrow 0}\frac{x+h-x}{h} \\ & = \lim_{h \rightarrow 0} 1 \\ & = 1 \end{align}
The derivative of $$f(x) = x^2$$ is $$f'(x) = 2x$$ for every point $$x.$$
For any point $$x,$$ \begin{align} \lim_{h \rightarrow 0}\frac{f(x+h)-f(x)}{h} & = \lim_{h \rightarrow 0}\frac{(x+h)^2-x^2}{h} \\ & = \lim_{h \rightarrow 0} \frac{x^2+2xh+h^2 - x^2}{h} \\ & = \lim_{h \rightarrow 0} \frac{2xh+h^2}{h} \\ & = \lim_{h \rightarrow 0} 2x + h \\ & = 2x \end{align}
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# Proof of vector addition formula
Two vectors of lengths $a$ and $b$ make an angle $\theta$ with each other when placed tail to tail. Show that the magnitude of their resultant is :
$$r = \sqrt{ a^2 + b^2 +2ab\cos(\theta)}.$$
I understand that if we placed
the two vectors head-to-tail instead of tail-to-tail, the Law of Cosines dictates that the resultant would be:
$$\sqrt{ a^2 + b^2 -2ab\cos(\theta)}$$
However, In the situation actually described, the direction of vector $a$
has been reversed, which changes the sign of $2ab$ without
changing the sign of $a^2$. But how do I prove that mathematically?
#### Solutions Collecting From Web of "Proof of vector addition formula"
You got everything that you need.
Theorem: Given vectors $a$ and $b$ enclosing an angle $\theta$. Then the magnitude of the sum, $|a + b|$, is given by $\sqrt{ a^2 + b^2 +2ab\cos(\theta)}$.
Proof: Assuming that the Law of Cosines works for a case like the following, where $a$ and $b$ are the thick lines. The thin lines are just mirrors of the vectors.
If our vectores are defined like I just stated, this holds:
$$|a + b| = \sqrt{ a^2 + b^2 – 2ab\cos(\theta)}$$
Now we position the vector $b$ at the head of $a$. It looks like this:
$$\theta’ = \pi – \theta$$
And with $\cos(\pi – \theta) = – \cos(-\theta) = -\cos(\theta)$ we get that $-$-sign and yield the formula
$$|b – a| = \sqrt{ a^2 + b^2 + 2ab\cos(\theta)}$$
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## Do you need the quadratic formula for GRE?
(x + 3)(x + 1) = 0 Plug the correct numbers into the equation. x = -3, -1 x is the opposite of these numbers. Some quadratic equations cannot be factored and must be solved using a special formula, the “quadratic formula.” This formula is not tested on the GRE.
## Should I memorize Prime Numbers for GRE?
Know your prime numbers Prime numbers are the building blocks of any positive non-prime integer. It would certainly behoove any aspiring GRE champion to therefore memorize at least the first handful of prime numbers: 2, 3, 5, 7, 11, 13.
What are individual factors math?
Factor, in mathematics, a number or algebraic expression that divides another number or expression evenly—i.e., with no remainder. For example, 3 and 6 are factors of 12 because 12 ÷ 3 = 4 exactly and 12 ÷ 6 = 2 exactly. The other factors of 12 are 1, 2, 4, and 12.
A quadratic equation is an algebraic expression of the second degree in x. The standard form of a quadratic equation is ax2 + bx + c = 0, where a, b are the coefficients, x is the variable, and c is the constant term.
### What is a prime number over 100?
List of prime numbers to 100. 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.
### What are the factor of 100?
The numbers which we multiply to get 100 are the factors of 100. Factors of 100 are written as 1, 2, 4, 5, 10, 20, 25, 50, and 100. Factor pairs are the pairs of two numbers that, when multiplied, give the original number.
Is the sum of two prime number is always a prime number?
the sum of two prime number is always a prime number is FALSE.. The Sum of two Prime numbers 5 & 7 is 12 which is not a prime number, it is an even number.
What are the factors of 23?
The number 23 is a prime number. Being a prime number, 23 has just two factors, 1 and 23.
## Why do we use quadratic equations?
Quadratic equations are actually used in everyday life, as when calculating areas, determining a product’s profit or formulating the speed of an object. Quadratic equations refer to equations with at least one squared variable, with the most standard form being ax² + bx + c = 0.
## Do you know the rules of exponents for the GRE?
You may already know the basic rules of exponents for the GRE. These rules tell you what to do if you want to multiply or divide two exponential numbers, or raise an exponent to another power. Once you’ve memorized them, exponent problems become exponentially easier (I’m so sorry).
Can you factor out the greatest common factor?
Well notice, first of all, we can use our old trick of factoring out a greatest common factor. And, of course, a greatest common factor, both coefficients are divisible by 6, and we can factor out an x as well. So we factor out a 6x and we get x squared minus 25. Well, and of course now, that’s a difference of two squares. So it factors to this.
When do you need to factor in a problem?
Factoring is unlikely to be the thrust of an entire problem but you will need to factor as a step in a larger complicated problem. So having these skills of factoring, even factoring these really hairy expressions, is something that can help you unlock those harder Quant problems.
### What did Chelsey do to get a perfect GRE?
Chelsey always followed her heart when it came to her education. Luckily, her heart led her straight to the perfect background for GMAT and GRE teaching: she has undergraduate degrees in mathematics and history, a master’s degree in linguistics, a 790 on the GMAT, and a perfect 170/170 on the GRE.
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# How to Find Multiplicity: A Comprehensive Guide to Understanding and Solving Problems
## I. Introduction to Multiplicity
Multiplicity is a fundamental concept in mathematics that refers to the number of times a particular value appears as a solution to an equation or function. In this article, we’ll explore the basics of multiplicity, how to find it using algebra, and its applications in real-world scenarios.
### A. Brief background on the concept of multiplicity
Multiplicity has been a topic of interest in mathematics for centuries and is an essential concept in algebraic geometry, number theory, and other fields of study. The concept of multiplicity was initially introduced by the French mathematician Évariste Galois in the 19th century.
### B. Basic definitions and common examples
Multiplicity can be defined as the number of times a factor appears in the factorization of a polynomial. For example, in the polynomial (x-1)*(x-1)*(x-3), the factor (x-1) has a multiplicity of 2, while the factor (x-3) has a multiplicity of 1. Multiplicity can also refer to the number of times a particular value appears as a solution to an equation or function. For instance, the equation x^3 – 3x^2 + 3x – 1 = 0 has a root of x=1 with a multiplicity of 3.
### C. Importance of understanding multiplicity in mathematics
Understanding how to find and work with multiplicity is essential for solving equations, analyzing data sets, and modeling real-world scenarios.
## II. Finding Multiplicity Using Algebra
There are several methods for finding multiplicity, but one of the most common is through algebraic methods. Here are step-by-step instructions for finding multiplicity using algebra:
### A. Step-by-step instructions for finding multiplicity using algebra
Step 1: Factor the polynomial and identify the repeated factors.
Step 2: Take the derivative of the polynomial.
Step 3: Find the roots of the derivative.
Step 4: Determine the multiplicity of each repeated factor.
### B. Tips for approaching more complex examples
When dealing with more complex examples, it can be helpful to break down the problem into smaller parts and tackle one step at a time. Additionally, it’s essential to have a solid foundation in algebra and calculus, as they are the building blocks for understanding multiplicity.
### C. Examples of common types of problems and how to solve them
Here are a few common types of problems involving multiplicity:
Example 1: Find the multiplicity of the root x = 3 in the polynomial f(x) = (x-3)^3 * (x+2)^2
Solution: Step 1: Identify the repeated factor (x-3)^3. Step 2: Take the derivative of f(x): f'(x) = 3(x-3)^2 * (x+2)^2 + 2(x-3)^3 * 2(x+2). Step 3: Find the roots of the derivative: x = 3 and x = -2. Step 4: Determine the multiplicity of x=3: since it’s a repeated factor, the multiplicity is the number of times it appears in the polynomial, which is 3.
Example 2: Find all the roots and their multiplicities for the polynomial f(x) = x^4 – 2x^3 + x^2
Solution: Step 1: Factor the polynomial: f(x) = x^2(x^2 – 2x + 1). Step 2: Take the derivative of f(x): f'(x) = 4x^3 – 6x^2 + 2x. Step 3: Find the roots of the derivative: x=0, x=1/2, and x=3/2. Step 4: Determine the multiplicity of each root by looking at the original polynomial: x=0 has a multiplicity of 2, x=1 has a multiplicity of 1, and x=2 has a multiplicity of 0.
## III. Applications of Multiplicity
### A. Real-world applications of multiplicity in different fields
Multiplicity has several real-world applications, including analyzing financial data, predicting stock market prices, and designing computer algorithms. It’s also used in physics and engineering to model systems and predict behavior.
### B. How to use multiplicity to solve systems of equations and analyze data sets
Multiplicity is an essential tool for solving systems of equations, as it helps identify the repeated solutions and simplify the calculations. It’s also used in data analysis to identify trends, patterns, and outliers.
### C. Importance of learning more about multiplicity in different areas of study
Learning more about multiplicity can provide a solid foundation for understanding more complex mathematical concepts and pursuing advanced studies in different fields, such as algebraic geometry, number theory, and cryptography.
## IV. Practice Problems
### A. A variety of practice problems related to finding multiplicity
Here are a few practice problems related to finding multiplicity:
1. Find the multiplicity of the root x = -2 in the polynomial f(x) = (x+3)^2 * (x+4)^3 * (x-2)^4
2. Find all the roots and their multiplicities for the polynomial f(x) = x^3(x-1)^2(x+2)
3. Find the multiplicity of the root x = 5 in the polynomial f(x) = (x-5)^2 * (x^2 + 1)^2
### B. Ranging from simple to more complex problems
These practice problems range from simple to more complex problems involving multiple factors and roots.
### C. A helpful resource for students looking to test their understanding of the concept.
These practice problems are an excellent resource for students looking to test their understanding of the concept of multiplicity and prepare for exams and assignments.
## V. Advanced Topics in Multiplicity
For those looking to take their knowledge of multiplicity to the next level, here are a few advanced topics:
### A. Identification and working with higher-order multiplicities
Higher-order multiplicities involve roots that appear multiple times in a function or polynomial with exponents greater than one. It’s important to understand how to identify and work with these cases to solve complex problems.
### B. Examining more complex aspects of multiplicity
Multiplicity can be applied to a variety of mathematical concepts and scenarios, including input-output machines, differential equations, and geometric objects.
### C. Helpful for students looking to master the basics and take their knowledge to the next level
By exploring these advanced topics, students can gain a deeper understanding of multiplicity and prepare for advanced coursework and research opportunities.
## VI. Conclusion
In conclusion, understanding how to find and work with multiplicity is an essential skill for students pursuing careers in mathematics, engineering, and other STEM fields. By following the step-by-step instructions outlined in this article and practicing with the included problems, students can develop their problem-solving skills and enhance their understanding of multiplicity and its applications in real-world scenarios.
### A. Recap of key points covered in the article
Throughout this article, we’ve explored the basics of multiplicity, including its definitions, the importance of understanding it in mathematics, and how to find it using algebraic methods. We’ve also examined its applications in different fields, provided practice problems for students to master the concept, and explored advanced topics for those looking to deepen their understanding.
### B. Importance of understanding multiplicity in mathematics
Understanding multiplicity is crucial for solving equations, modeling systems, and analyzing data sets. By mastering this fundamental concept, students can develop their critical thinking skills and prepare for advanced studies and career opportunities in different fields.
### C. Final thoughts on how to improve problem-solving skills related to the concept.
By practicing with the included problems and exploring advanced topics, students can improve their problem-solving skills and develop a deeper understanding of multiplicity. It’s also essential to build a solid foundation in algebra and calculus and seek out additional resources and guidance from teachers and peers to master the concept.
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# How do you solve the quadratic equation by completing the square: x^2+2x-5=0?
Aug 1, 2015
${x}_{1 , 2} = - 1 \pm \sqrt{6}$
#### Explanation:
$\textcolor{b l u e}{{x}^{2} + \frac{b}{a} x = - \frac{c}{a}}$
To do that, add $5$ to both sides of the equation
${x}^{2} + 2 x - \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} + \textcolor{red}{\cancel{\textcolor{b l a c k}{5}}} = 5$
${x}^{2} + 2 x = 5$
Next, divide the coefficient of the $x$-term by 2, square the result. then add it to both sides of the equation.
$\frac{2}{2} = 1$, then ${1}^{2} = 1$
This will get you
${x}^{2} + 2 x + 1 = 5 + 1$
Notice that you can rewrite the left side of the equation as the square of a binomial
${x}^{2} + 2 x + 1 = {x}^{2} + 2 \cdot \left(1\right) \cdot x + {1}^{2} = {\left(x + 1\right)}^{2}$
You will now have
${\left(x + 1\right)}^{2} = 6$
Take the square root of both sides
$\sqrt{{\left(x + 1\right)}^{2}} = \sqrt{6}$
$x + 1 = \pm \sqrt{6} \implies {x}_{1 , 2} = - 1 \pm \sqrt{6} = \left\{\begin{matrix}{x}_{1} = \textcolor{g r e e n}{- 1 - \sqrt{6}} \\ {x}_{2} = \textcolor{g r e e n}{- 1 + \sqrt{6}}\end{matrix}\right.$
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# Difference between revisions of "2021 AMC 12A Problems/Problem 14"
## Problem
What is the value of $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)?$$
$\textbf{(A) }21 \qquad \textbf{(B) }100\log_5 3 \qquad \textbf{(C) }200\log_3 5 \qquad \textbf{(D) }2,200\qquad \textbf{(E) }21,000$
## Solution 1 (Condensed Explanation of Logarithmic Identities)
This equals $$\left(\sum_{k=1}^{20}k\log_5{3}\right)\left(\sum_{k=1}^{100}\log_9{25}\right)=\frac{20\cdot21}{2}\cdot\log_5{3}\cdot100\log_3{5}=\boxed{\textbf{(E) }21,000}.$$ ~JHawk0224
## Solution 2 (Detailed Explanation of Logarithmic Identities)
We will apply the following logarithmic identity: $$\log_{p^n}{q^n}=\log_{p}{q},$$ which can be proven by the Change of Base Formula: $$\log_{p^n}{q^n}=\frac{\log_{p}{q^n}}{\log_{p}{p^n}}=\frac{n\log_{p}{q}}{n}=\log_{p}{q}.$$ Now, we simplify the expressions inside the summations: \begin{align*} \log_{5^k}{{3^k}^2}&=\log_{5^k}{\left(3^k\right)^k} \\ &=k\log_{5^k}{3^k} \\ &=k\log_{5}{3}, \end{align*} and \begin{align*} \log_{9^k}{25^k}&=\log_{3^{2k}}{5^{2k}} \\ &=\log_{3}{5}. \end{align*} Using these results, we evaluate the original expression: \begin{align*} \left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)&=\left(\sum_{k=1}^{20} k\log_{5}{3}\right)\cdot\left(\sum_{k=1}^{100} \log_{3}{5}\right) \\ &= \left(\log_{5}{3}\cdot\sum_{k=1}^{20} k\right)\cdot\left(\log_{3}{5}\cdot\sum_{k=1}^{100} 1\right) \\ &= \left(\sum_{k=1}^{20} k\right)\cdot\left(\sum_{k=1}^{100} 1\right) \\ &= \frac{21\cdot20}{2}\cdot100 \\ &= \boxed{\textbf{(E) }21,000}. \end{align*} ~MRENTHUSIASM
## Solution 3 (Properties of Logarithms)
First, we can get rid of the $k$ exponents using properties of logarithms: $$\log_{5^k} 3^{k^2} = k^2 \cdot \frac{1}{k} \cdot \log_{5} 3 = k\log_{5} 3 = \log_{5} 3^k.$$ (Leaving the single $k$ in the exponent will come in handy later). Similarly, $$\log_{9^k} 25^{k} = k \cdot \frac{1}{k} \cdot \log_{9} 25 = \log_{9} 5^2.$$ Then, evaluating the first few terms in each parentheses, we can find the simplified expanded forms of each sum using the additive property of logarithms: \begin{align*} \sum_{k=1}^{20} \log_{5} 3^k &= \log_{5} 3^1 + \log_{5} 3^2 + \dots + \log_{5} 3^{20} \\ &= \log_{5} 3^{(1 + 2 + \dots + 20)} \\ &= \log_{5} 3^{\frac{20(20+1)}{2}} &&\hspace{15mm}(*) \\ &= \log_{5} 3^{210}, \\ \sum_{k=1}^{100} \log_{9} 5^2 &= \log_{9} 5^2 + \log_{9} 5^2 + \dots + \log_{9} 5^2 \\ &= \log_{9} 5^{2(100)} \\ &= \log_{9} 5^{200}. \end{align*} In $(*),$ we use the triangular numbers equation: $$1 + 2 + \dots + n = \frac{n(n+1)}{2} = \frac{20(20+1)}{2} = 210.$$ Finally, multiplying the two logarithms together, we can use the chain rule property of logarithms to simplify: $$\log_{a} b\log_{x} y = \log_{a} y\log_{x} b.$$ Thus, \begin{align*} \left(\log_{5} 3^{210}\right)\left(\log_{3^2} 5^{200}\right) &= \left(\log_{5} 5^{200}\right)\left(\log_{3^2} 3^{210}\right) \\ &= \left(\log_{5} 5^{200}\right)\left(\log_{3} 3^{105}\right) \\ &= (200)(105) \\ &= \boxed{\textbf{(E) }21,000}. \end{align*} ~Joeya (Solution)
~MRENTHUSIASM (Reformatting)
## Solution 4 (Estimations and Answer Choices)
In $\sum_{k=1}^{20} \log_{5^k} 3^{k^2},$ note that the addends are greater than $1$ for all $k\geq2.$
In $\sum_{k=1}^{100} \log_{9^k} 25^k,$ note that the addends are greater than $1$ for all $k\geq1.$
We have the inequality $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} 1\right)=19\cdot100=1,900,$$ which eliminates choices $\textbf{(A)}, \textbf{(B)},$ and $\textbf{(C)}.$ We get the answer $\boxed{\textbf{(E) }21,000}$ by either an educated guess or a continued approximation:
Since $3^3=27\approx25,$ it follows that $9^{3/2}\approx25.$ By an extremely rough underestimation, $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)\approx\left(\sum_{k=2}^{20} 1\right)\cdot\left(\sum_{k=1}^{100} \frac{3}{2}\right)=19\cdot150=2,850.$$ From here, it should be safe to guess that the answer is $\textbf{(E)}.$
As an extra guaranty, note that $\sum_{k=1}^{20} \log_{5^k} 3^{k^2} >> \sum_{k=2}^{20} 1 = 19.$ Therefore, we must have $$\left(\sum_{k=1}^{20} \log_{5^k} 3^{k^2}\right)\cdot\left(\sum_{k=1}^{100} \log_{9^k} 25^k\right)>>2,850.$$
~MRENTHUSIASM
~IceMatrix
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1. ## surds
I am having difficulty solving these surds problems
$\displaystyle (5+2 \sqrt{7})^2$
and
$\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}$
2. $\displaystyle (5+2\sqrt7)^2=25+28+20\sqrt{7}$
$\displaystyle =53+20\sqrt{7}$
$\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}=2\sqrt{2}+3\sqrt{2}-5\sqrt{2}=0$
3. Originally Posted by b3nj1d
I am having difficulty solving these surds problems
$\displaystyle (5+2 \sqrt{7})^2$
and
$\displaystyle \sqrt{8}+\sqrt{18}-\sqrt{50}$
Since we haven't learnt FOIL yet, I understand why you're having trouble with the first one.
If you have two binomials like $\displaystyle (a + b)(c + d)$, they are expanded by multiplying the Firsts, then the Outers, then the Inners, then the Lasts.
So $\displaystyle (a + b)(c + d) = ac + ad + bc + bd$.
So for your first one, you have
$\displaystyle (5 + 2\sqrt{7})^2 = (5 + 2\sqrt{7})(5 + 2\sqrt{7})$
Multiplying the firsts gives $\displaystyle 5 \times 5 = 25$.
Multiplying the outers gives $\displaystyle 5 \times 2\sqrt{7} = 10\sqrt{7}$
Multiplying the inners gives $\displaystyle 5 \times 2\sqrt{7} = 10\sqrt{7}$
Multiplying the lasts gives $\displaystyle 2\sqrt{7}\times 2\sqrt{7} = 4 \times 7 = 28$.
So $\displaystyle (5 + 2\sqrt{7})^2 = 25 + 10\sqrt{7} + 10\sqrt{7} + 28$
$\displaystyle = 53 + 20\sqrt{7}$.
For the second, you need to simplify all of the surds first. Then hopefully you'll have "like surds" that can be collected.
$\displaystyle \sqrt{8} = \sqrt{4 \times 2} = \sqrt{4} \times \sqrt{2} = 2\sqrt{2}$.
$\displaystyle \sqrt{18} = \sqrt{9 \times 2} = \sqrt{9}\times\sqrt{2} = 3\sqrt{2}$.
$\displaystyle \sqrt{50} = \sqrt{25 \times 2} = \sqrt{25}\times\sqrt{2} = 5\sqrt{2}$.
So $\displaystyle \sqrt{8} + \sqrt{18} - \sqrt{50} = 2\sqrt{2} + 3\sqrt{2} - 5\sqrt{2}$
$\displaystyle = 5\sqrt{2} - 5\sqrt{2}$
$\displaystyle = 0\sqrt{2}$
$\displaystyle = 0$.
Chookas for the exam tomorrow.
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## Elementary Algebra
{$\frac{1 - i\sqrt {31}}{8},\frac{1 + i\sqrt {31}}{8}$}
Step 1: Comparing $4x^{2}-x+2=0$ to the standard form of a quadratic equation $ax^{2}+bx+c=0$; $a=4$, $b=-1$ and $c=2$ Step 2: The quadratic formula is: $x=\frac{-b \pm \sqrt {b^{2}-4ac}}{2a}$ Step 3: Substituting the values of a, b and c in the formula: $x=\frac{-(-1) \pm \sqrt {(-1)^{2}-4(4)(2)}}{2(4)}$ Step 4: $x=\frac{1 \pm \sqrt {1-32}}{8}$ Step 5: $x=\frac{1 \pm \sqrt {-31}}{8}$ Step 6: $x=\frac{1 \pm \sqrt {-1\times31}}{8}$ Step 7: $x=\frac{1 \pm (\sqrt {-1}\times\sqrt {31})}{8}$ Step 8: $x=\frac{1 \pm (i\times \sqrt {31})}{8}$ Step 9: $x=\frac{1 \pm i\sqrt {31}}{8}$ Step 10: $x=\frac{1 - i\sqrt {31}}{8}$ or $x=\frac{1 + i\sqrt {31}}{8}$ Step 11: Therefore, the solution set is {$\frac{1 - i\sqrt {31}}{8},\frac{1 + i\sqrt {31}}{8}$}.
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# Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.1
### (v) x12 + y3 + t50
Solution:
(i) 3x2 – 4x + 15
It is a polynomial in one variable, that is, x. And all the powers of x are whole numbers.
(ii) y2 + 2√3
It is a polynomial in variable y. And all the powers of y are whole numbers.
(iii) 3√x + √2x
It is not a polynomial since the exponent of 3√x is a rational term.
(iv) x – 4/x
It is not a polynomial since the exponent of variable x is – 4/x which is not a positive term.
(v) x12 + y3 + t50
It is a three variable polynomial, where the variables are x, y and t.
### (iv) √3x – 7
Solution:
(i) 17 – 2x + 7x2
Coefficient of x2 in the above equation = 7
(ii) 9 – 12x + x3
Coefficient of x2 =0, since there is no term with x2
(iii) π/6 x2 – 3x + 4
Coefficient of x2 in the above equation = π/6
(iv) √3x – 7
Coefficient of x2 = 0, since there is no term with x2 in the above equation.
### (v) 0
Solution:
The degree is the highest possible degree of the variable in the polynomial. Now, we have
(i) Degree of the polynomial 7x3 + 4x2 – 3x + 12 is 3, since the term x3 is highest
(ii) Degree of the polynomial 12 – x + 2x3 is 3, since the term x3 is highest
(iii) Degree of the polynomial 5y – √2 is 1, since the term 5y only has the variable.
(iv) Degree of the polynomial 7 is 0, since there is no term with variable.
(v) Degree of the polynomial 0 is undefined.
### (vi) 7t4 + 4t3 + 3t – 2
Solution:
Linear polynomials have highest degree = 1. Quadratic have highest degree = 2. Cubic polynomials have highest degree = 3 and bi-quadratic as 4.
(i) x + x2 + 4: It is a quadratic polynomial since its highest possible degree is 2.
(ii) 3x – 2 : It is a linear polynomial since its highest possible degree is 1.
(iii) 2x + x2: It is a quadratic polynomial since its highest possible degree is 2.
(iv) 3y: It is a linear polynomial since its highest possible degree is 1.
(v) t2+ 1: It is a quadratic polynomial since its highest possible degree s 2.
(vi) 7t4 + 4t3 + 3t – 2: It is a bi-quadratic polynomial since its highest possible degree is 4.
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# GCD of 8, 72, 48 Calculator
Created By : Jatin Gogia
Reviewed By : Rajasekhar Valipishetty
Last Updated : Apr 06, 2023
Make use of GCD Calculator to determine the Greatest Common Divisor of 8, 72, 48 i.e. 8 largest integer that divides all the numbers equally.
GCD of 8, 72, 48 is 8
GCD(8, 72, 48) = 8
Ex: 10, 15, 20 (or) 24, 48, 96,45 (or) 78902, 89765, 12345
GCD of
GCD of numbers 8, 72, 48 is 8
GCD(8, 72, 48) = 8
### GCDof 8,72,48 is 8
Given Input numbers are 8, 72, 48
To find the GCD of numbers using factoring list out all the divisors of each number
Divisors of 8
List of positive integer divisors of 8 that divides 8 without a remainder.
1, 2, 4, 8
Divisors of 72
List of positive integer divisors of 72 that divides 72 without a remainder.
1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Divisors of 48
List of positive integer divisors of 48 that divides 48 without a remainder.
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
Greatest Common Divisior
We found the divisors of 8, 72, 48 . The biggest common divisior number is the GCD number.
So the Greatest Common Divisior 8, 72, 48 is 8.
Therefore, GCD of numbers 8, 72, 48 is 8
### Finding GCD of 8, 72, 48 using Prime Factorization
Given Input Data is 8, 72, 48
Make a list of Prime Factors of all the given numbers initially
Prime Factorization of 8 is 2 x 2 x 2
Prime Factorization of 72 is 2 x 2 x 2 x 3 x 3
Prime Factorization of 48 is 2 x 2 x 2 x 2 x 3
Highest common occurrences in the given inputs are 23
Multiplying them we get the GCD as 8
### Finding GCD of 8, 72, 48 using LCM Formula
Step1:
Let's calculate the GCD of first two numbers
The formula of GCD is GCD(a, b) = ( a x b) / LCM(a, b)
LCM(8, 72) = 72
GCD(8, 72) = ( 8 x 72 ) / 72
GCD(8, 72) = 576 / 72
GCD(8, 72) = 8
Step2:
Here we consider the GCD from the above i.e. 8 as first number and the next as 48
The formula of GCD is GCD(a, b) = ( a x b) / LCM(a, b)
LCM(8, 48) = 48
GCD(8, 48) = ( 8 x 48 ) / 48
GCD(8, 48) = 384 / 48
GCD(8, 48) = 8
GCD of 8, 72, 48 is 8
### GCD of Numbers Calculation Examples
Here are some samples of GCD of Numbers calculations.
### FAQs on GCD of numbers 8, 72, 48
1. What is the GCD of 8, 72, 48?
GCD of 8, 72, 48 is 8
2. Where do I get the detailed procedure to find GCD of 8, 72, 48?
You can find a detailed procedure to find GCD of 8, 72, 48 on our page.
3. How to find GCD of 8, 72, 48 on a calculator?
You can find the GCD of 8, 72, 48 by simply giving the inputs separated by commas and click on the calculate button to avail the Greatest Common Divisor in less time.
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# Find the first two derivatives of 2 sinx cosx.
We need to find the first derivative of 2sin(x)cos(x)
## Solution
Let us assume y = 2sin(x)cos(x)
Use the product rule:
uv’ + vu’
where u is 2sin(x) and v is cos(x)
To find first derivative:
y’ = 2sin(x)(-sin(x)) + cos(x)2cos(x)
On simplifying we get
y’ = 2cos2(x)-2sin2(x)
y’ = 2(cos2(x)-sin2(x))
We know the trignometric identity
cos(2x) = cos2(x)-sin2(x)
Hence substituting in the above equation we get,
y’ = 2cos(2x)
First derivative of 2sin(x)cos(x) is 2cos(2x)
Now will find second derivative
Take second derivative using chain rule:
y” = 2(-sin(2x)cos(2x))
Simplify:
y” = -2sin(2x)(2)
Simplify:
y” = -4sin(2x)y” = -4sin(2x)
Hence the first two derivaties of 2sin(x)cos(x) are
y’ = 2cos(2x)
y” = -4sin(2x)
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# Finding the Slope of a Line(KS3, Year 8)
homesitemaplinear equationsfinding the slope of a line
The slope is a measure of how steep the line is. Imagine a line is drawn on a graph. We can find the slope of the line.
## How to Find the Slope of a Line
Finding the slope of a line is easy.
## Question
Find the slope of the line shown below.
## 1
Find out how many units the line has gone up. In our example, the line goes up by 4 units.
## 2
Find out how many units the line has gone across. In our example, the line goes across by 2 units.
## 3
Divide the answer from Step 1 (4) by the answer from Step 2 (2).
Slope = How far up ÷ How far across
Slope = 4 ÷ 2
Slope = 2
The slope of the line is 2.
## A Formula to Find the Slope of a Line
The formula to find the slope is shown below: Let's apply the formula to the example above.
## 1
Find the change in y. Subtract the y-coordinates of the start and end of the line (or two convenient points on it.)
Change in y = 5 − 1 = 4
The change in y is 4.
## 2
Find the change in x. Subtract the x-coordinates of the start and end of the line (or two convenient points on it.)
Change in x = 2 − 0 = 2
The change in x is 2.
## 3
Divide the change in y (4) by the change in x (2).
Slope = Change in y ÷ Change in x
Slope = 4 ÷ 2
Slope = 2
The slope of the line is 2.
## Lesson Slides
The slider below gives a real example of how to find the slope of a line.
## Postive And Negative Slopes
A positive slope means the line slopes up and to the right: A negative slope means the line slopes down and to the right:
## Fractional Slope
Slope can be a fraction, such as ½ and ¾. An improper fraction is positive, but less than 1. A slope of 1 gives a 45° line. A fractional slope is less steep than this:
## Zero Slope And Undefined Slope
A line that goes straight across has zero slope: A line that goes straight across has an undefined slope:
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# 4.4: Length of a Vector
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## Learning Objectives
• Find the length of a vector and the distance between two points in $$\mathbb{R}^n$$.
• Find the corresponding unit vector to a vector in $$\mathbb{R}^n$$.
In this section, we explore what is meant by the length of a vector in $$\mathbb{R}^n$$. We develop this concept by first looking at the distance between two points in $$\mathbb{R}^n$$.
First, we will consider the concept of distance for $$\mathbb{R}$$, that is, for points in $$\mathbb{R}^1$$. Here, the distance between two points $$P$$ and $$Q$$ is given by the absolute value of their difference. We denote the distance between $$P$$ and $$Q$$ by $$d(P,Q)$$ which is defined as $d(P,Q) = \sqrt{ \left( P-Q\right) ^{2}} \label{distance1}$
Consider now the case for $$n=2$$, demonstrated by the following picture.
There are two points $$P =\left( p_{1},p_{2}\right)$$ and $$Q = \left(q_{1},q_{2}\right)$$ in the plane. The distance between these points is shown in the picture as a solid line. Notice that this line is the hypotenuse of a right triangle which is half of the rectangle shown in dotted lines. We want to find the length of this hypotenuse which will give the distance between the two points. Note the lengths of the sides of this triangle are $$\left| p_{1}-q_{1}\right|$$ and $$\left| p_{2}-q_{2}\right|$$, the absolute value of the difference in these values. Therefore, the Pythagorean Theorem implies the length of the hypotenuse (and thus the distance between $$P$$ and $$Q$$) equals $\left( \left| p_{1}-q_{1}\right| ^{2}+\left| p_{2}-q_{2}\right| ^{2}\right) ^{1/2}=\left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}\right) ^{1/2} \label{distance2}$
Now suppose $$n=3$$ and let $$P = \left( p_{1},p_{2},p_{3}\right)$$ and $$Q = \left( q_{1},q_{2},q_{3}\right)$$ be two points in $$\mathbb{R}^{3}.$$ Consider the following picture in which the solid line joins the two points and a dotted line joins the points $$\left( q_{1},q_{2},q_{3}\right)$$ and $$\left( p_{1},p_{2},q_{3}\right) .$$
Here, we need to use Pythagorean Theorem twice in order to find the length of the solid line. First, by the Pythagorean Theorem, the length of the dotted line joining $$\left( q_{1},q_{2},q_{3}\right)$$ and $$\left( p_{1},p_{2},q_{3}\right)$$ equals $\left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}\right) ^{1/2}\nonumber$ while the length of the line joining $$\left( p_{1},p_{2},q_{3}\right)$$ to $$\left( p_{1},p_{2},p_{3}\right)$$ is just $$\left| p_{3}-q_{3}\right| .$$ Therefore, by the Pythagorean Theorem again, the length of the line joining the points $$P = \left( p_{1},p_{2},p_{3}\right)$$ and $$Q = \left( q_{1},q_{2},q_{3}\right)$$ equals $\left( \left( \left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}\right) ^{1/2}\right) ^{2}+\left( p_{3}-q_{3}\right) ^{2}\right) ^{1/2}\nonumber$ $=\left( \left( p_{1}-q_{1}\right) ^{2}+\left( p_{2}-q_{2}\right) ^{2}+\left( p_{3}-q_{3}\right) ^{2}\right) ^{1/2} \label{distance3}$
This discussion motivates the following definition for the distance between points in $$\mathbb{R}^n$$.
## Definition $$\PageIndex{1}$$: Distance Between Points
Let $$P=\left( p_{1},\cdots ,p_{n}\right)$$ and $$Q=\left( q_{1},\cdots ,q_{n}\right)$$ be two points in $$\mathbb{R}^{n}$$. Then the distance between these points is defined as $\text{ distance between }P\text{ and } Q\text{ } = d( P, Q ) = \left( \sum_{k=1}^{n}\left\vert p_{k}-q_{k}\right\vert ^{2}\right) ^{1/2}\nonumber$ This is called the distance formula. We may also write $$\left\vert P - Q \right\vert$$ as the distance between $$P$$ and $$Q$$.
From the above discussion, you can see that Definition $$\PageIndex{1}$$ holds for the special cases $$n=1,2,3$$, as in Equations $$\eqref{distance1}$$, $$\eqref{distance2}$$, $$\eqref{distance3}$$. In the following example, we use Definition $$\PageIndex{1}$$ to find the distance between two points in $$\mathbb{R}^4$$.
## Example $$\PageIndex{1}$$: Distance Between Points
Find the distance between the points $$P$$ and $$Q$$ in $$\mathbb{R}^{4}$$, where $$P$$ and $$Q$$ are given by $P=\left( 1,2,-4,6\right)\nonumber$ and $Q=\left( 2,3,-1,0\right)\nonumber$
###### Solution
We will use the formula given in Definition $$\PageIndex{1}$$ to find the distance between $$P$$ and $$Q$$. Use the distance formula and write $d(P,Q)= \left( \left( 1-2\right) ^{2}+\left( 2-3\right) ^{2}+\left( -4-\left( -1\right) \right) ^{2}+\left( 6-0\right)^{2}\right) ^{\frac{1}{2}} = 47\nonumber$
Therefore, $$d( P,Q) = \sqrt{47}.$$
There are certain properties of the distance between points which are important in our study. These are outlined in the following theorem.
## Theorem $$\PageIndex{1}$$: Properties of Distance
Let $$P$$ and $$Q$$ be points in $$\mathbb{R}^n$$, and let the distance between them, $$d( P, Q)$$, be given as in Definition $$\PageIndex{1}$$. Then, the following properties hold.
• $$d( P, Q) = d( Q, P)$$
• $$d( P, Q) \geq 0$$, and equals 0 exactly when $$P = Q.$$
There are many applications of the concept of distance. For instance, given two points, we can ask what collection of points are all the same distance between the given points. This is explored in the following example.
## Example $$\PageIndex{2}$$: The Plane Between Two Points
Describe the points in $$\mathbb{R}^3$$ which are at the same distance between $$\left( 1,2,3\right)$$ and $$\left( 0,1,2\right) .$$
###### Solution
Let $$P = \left( p_1 , p_2, p_3\right)$$ be such a point. Therefore, $$P$$ is the same distance from $$\left( 1,2,3\right)$$ and $$\left( 0,1,2\right) .$$ Then byDefinition $$\PageIndex{1}$$, $\sqrt{\left( p_1 -1\right) ^{2}+\left( p_2 -2\right) ^{2}+\left( p_3-3\right) ^{2}}= \sqrt{\left( p_1 - 0 \right)^{2}+\left( p_2-1\right) ^{2}+\left( p_3-2\right) ^{2}}\nonumber$ Squaring both sides we obtain $\left( p_1 -1\right) ^{2}+\left( p_2 -2\right) ^{2}+\left( p_3 -3\right) ^{2}=p_1^{2}+\left( p_2-1\right) ^{2}+\left( p_3 -2\right) ^{2}\nonumber$ and so $\ p_1^{2}-2p_1+14+p_2^{2}-4p_2+p_3^{2}-6p_3=p_1^{2}+p_2^{2}-2p_2+5+p_3^{2}-4p_3\nonumber$ Simplifying, this becomes $-2p_1+14-4p_2-6p_3=-2p_2+5-4p_3\nonumber$ which can be written as $2p_1+2p_2+2p_3=-9 \label{distanceplane}$ Therefore, the points $$P = \left( p_1,p_2,p_3\right)$$ which are the same distance from each of the given points form a plane whose equation is given by $$\eqref{distanceplane}$$.
We can now use our understanding of the distance between two points to define what is meant by the length of a vector. Consider the following definition.
## Definition $$\PageIndex{2}$$: Length of a Vector
Let $$\vec{u} = \left[ u_{1} \cdots u_{n} \right]^T$$ be a vector in $$\mathbb{R}^n$$. Then, the length of $$\vec{u}$$, written $$\| \vec{u} \|$$ is given by $\| \vec{u} \| = \sqrt{ u_{1}^2 + \cdots + u_{n}^2}\nonumber$
This definition corresponds to Definition $$\PageIndex{1}$$, if you consider the vector $$\vec{u}$$ to have its tail at the point $$0 = \left( 0, \cdots ,0 \right)$$ and its tip at the point $$U = \left(u_1, \cdots, u_n \right)$$. Then the length of $$\vec{u}$$ is equal to the distance between $$0$$ and $$U$$, $$d(0,U)$$. In general, $$d(P,Q)=||\vec{PQ}||$$.
ConsiderExample $$\PageIndex{1}$$. ByDefinition $$\PageIndex{2}$$, we could also find the distance between $$P$$ and $$Q$$ as the length of the vector connecting them. Hence, if we were to draw a vector $$\overrightarrow{PQ}$$ with its tail at $$P$$ and its point at $$Q$$, this vector would have length equal to $$\sqrt{47}$$.
We conclude this section with a new definition for the special case of vectors of length $$1$$.
## Definition $$\PageIndex{3}$$: Unit Vector
Let $$\vec{u}$$ be a vector in $$\mathbb{R}^{n}$$. Then, we call $$\vec{u}$$ a unit vector if it has length 1, that is if $\| \vec{u} \| = 1\nonumber$
Let $$\vec{v}$$ be a vector in $$\mathbb{R}^{n}$$. Then, the vector $$\vec{u}$$ which has the same direction as $$\vec{v}$$ but length equal to $$1$$ is the corresponding unit vector of $$\vec{v}$$. This vector is given by $\vec{u} = \frac{1}{\| \vec{v} \|} \vec{v}\nonumber$
We often use the term normalize to refer to this process. When we normalize a vector, we find the corresponding unit vector of length $$1$$. Consider the following example.
## Example $$\PageIndex{3}$$: Finding a Unit Vector
Let $$\vec{v}$$ be given by $\vec{v} = \left[ \begin{array}{rrr} 1 & -3 & 4 \end{array} \right]^T\nonumber$ Find the unit vector $$\vec{u}$$ which has the same direction as $$\vec{v}$$.
###### Solution
We will use Definition $$\PageIndex{3}$$ to solve this. Therefore, we need to find the length of $$\vec{v}$$ which, by Definition $$\PageIndex{2}$$ is given by $\| \vec{v} \| = \sqrt{ v_{1}^2 + v_{2}^2+ v_{3}^2}\nonumber$ Using the corresponding values we find that \begin{aligned} \| \vec{v} \| &= \sqrt{ 1^2 + \left(-3 \right)^2 + 4^2} \\ &= \sqrt{ 1 + 9 + 16} \\ &= \sqrt{26} \end{aligned} In order to find $$\vec{u}$$, we divide $$\vec{v}$$ by $$\sqrt{26}$$. The result is \begin{aligned} \vec{u} &= \frac{1}{\| \vec{v} \|} \vec{v} \\ &= \frac{1}{\sqrt{26}} \left[ \begin{array}{rrr} 1 & -3 & 4 \end{array} \right]^T \\ &= \left[ \begin{array}{rrr} \frac{1}{\sqrt{26}} & -\frac{3}{\sqrt{26}} & \frac{4}{\sqrt{26}} \end{array} \right]^T\end{aligned}
You can verify using the Definition $$\PageIndex{1}$$ that $$\| \vec{u} \| = 1$$.
This page titled 4.4: Length of a Vector is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request.
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### MEAP Preparation - Grade 4 Mathematics1.43 Fractions (Greater Lesser or Equal)
Method When the denominators are common we simply compare the numerators. When the denominators are different, we find Least Common denominator and then compare the numerators. Example: Which is bigger 1/2 or 1/5 In the above problem the denominators are different. Denominators are made common. List out the multiples of 2 - 2,4, 6, 8,10. List out the multiples of 5 - 5, 10, 15, 20. Common multiple is 10 and hence the LCM. Multiply and divide each fraction by the LCM 1/2 x 10/10 = 5/10 1/5 x 10/10 = 2/10 The denominators are made common. Hence 1/2 = 5/10 and 1/5 = 2/10 Hence 1/2 is greater than 1/5 Answer: 1/2 Directions: Compare the following fractions. Also write at least ten examples of your own.
Name: ___________________Date:___________________
### MEAP Preparation - Grade 4 Mathematics1.43 Fractions (Greater Lesser or Equal)
Q 1: Which is bigger? 1/8 or 2/8?2/81/8 Q 2: Which is greater? 3/4 or 2/43/4They are equal.2/4 Q 3: Which is bigger? 9/8 or 8/9?9/88/9 Q 4: Which is less? 6/8 or 3/86/83/8They are equal. Q 5: Which is less? 6/8 or 3/8They are equal.6/83/8 Q 6: Which is less? 5/6 or 6/6They are equal.5/66/6 Q 7: Which is less? 5/6 or 6/65/66/6They are equal. Q 8: Which fraction is SMALLER? 10/10 19/2019/2010/10 Question 9: This question is available to subscribers only! Question 10: This question is available to subscribers only!
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<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# 8.7: Extension: Laws of Sines and Cosines
Difficulty Level: At Grade Created by: CK-12
## Learning Objectives
• Identify and use the Law of Sines and Cosines.
In this chapter, we have only applied the trigonometric ratios to right triangles. However, you can extend what we know about these ratios and derive the Law of Sines and the Law of Cosines. Both of these laws can be used with any type of triangle to find any angle or side within it. That means we can find the sine, cosine and tangent of angle that are greater than 90\begin{align*}90^\circ\end{align*}, such as the obtuse angle in an obtuse triangle.
## Law of Sines
Law of Sines: If ABC\begin{align*}\triangle ABC\end{align*} has sides of length, a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*}, then sinAa=sinBb=sinCc\begin{align*}\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}\end{align*}.
Looking at a triangle, the lengths a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*} are opposite the angles of the same letter. Let’s use the Law of Sines on a couple of examples.
We will save the proof for a later course.
Example 1: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth.
Solution: First, to find mA\begin{align*}m \angle A\end{align*}, we can use the Triangle Sum Theorem.
mA+85+38mA=180=57\begin{align*}m \angle A + 85^\circ + 38^\circ & = 180^\circ\\ m \angle A & = 57^\circ\end{align*}
Now, use the Law of Sines to set up ratios for a\begin{align*}a\end{align*} and b\begin{align*}b\end{align*}.
sin57a=sin85b=sin3812\begin{align*}\frac{\sin 57^\circ}{a} = \frac{\sin 85^\circ}{b} = \frac{\sin 38^\circ}{12}\end{align*}
sin57aasin38a=sin3812=12sin57=12sin57sin3816.4sin85b=sin3812bsin38=12sin85 b=12sin85sin3819.4\begin{align*}\frac{\sin 57^\circ}{a} & = \frac{\sin 38^\circ}{12} && \quad \frac{\sin 85^\circ}{b} = \frac{\sin 38^\circ}{12}\\ a \cdot \sin 38^\circ & = 12 \cdot \sin 57^\circ && b \cdot \sin 38^\circ = 12 \cdot \sin 85^\circ\\ a & = \frac{12 \cdot \sin 57^\circ}{\sin 38^\circ} \approx 16.4 && \qquad \quad \ \ b = \frac{12 \cdot \sin 85^\circ}{\sin 38^\circ} \approx 19.4\end{align*}
Example 2: Solve the triangle using the Law of Sines. Round decimal answers to the nearest tenth.
Solution: Set up the ratio for B\begin{align*}\angle B\end{align*} using Law of Sines.
sin952727sinBsinB=sinB16=16sin95=16sin9527sin1(16sin9527)=36.2\begin{align*}\frac{\sin 95^\circ}{27} & = \frac{\sin B}{16}\\ 27 \cdot \sin B & = 16 \cdot \sin 95^\circ\\ \sin B & = \frac{16 \cdot \sin 95^\circ}{27} \rightarrow \sin^{-1} \left ( \frac{16 \cdot \sin 95^\circ}{27} \right ) = 36.2^\circ\end{align*}
To find mC\begin{align*}m \angle C\end{align*} use the Triangle Sum Theorem. mC+95+36.2=180mC=48.8\begin{align*}m \angle C + 95 ^\circ + 36.2^\circ = 180^\circ \rightarrow m \angle C = 48.8^\circ\end{align*}
To find c\begin{align*}c\end{align*}, use the Law of Sines again. sin9527=sin48.8c\begin{align*}\frac{\sin 95^\circ}{27} = \frac{\sin 48.8^\circ}{c}\end{align*}
csin95c=27sin48.8=27sin48.8sin9520.4\begin{align*}c \cdot \sin 95^\circ & = 27 \cdot \sin 48.8^\circ\\ c & = \frac{27 \cdot \sin 48.8^\circ}{\sin 95^\circ} \approx 20.4\end{align*}
## Law of Cosines
Law of Cosines: If ABC\begin{align*}\triangle ABC\end{align*} has sides of length a,b\begin{align*}a, b\end{align*}, and c\begin{align*}c\end{align*}, then a2=b2+c22bccosA\begin{align*}a^2 = b^2 + c^2 - 2bc \cos A\end{align*}
b2c2=a2+c22ac cosB=a2+b22ab cosC\begin{align*}b^2 & = a^2 + c^2 - 2ac \ \cos B\\ c^2 & = a^2 + b^2 - 2ab \ \cos C\end{align*}
Even though there are three formulas, they are all very similar. First, notice that whatever angle is in the cosine, the opposite side is on the other side of the equal sign.
Example 3: Solve the triangle using Law of Cosines. Round your answers to the nearest hundredth.
Solution: Use the second equation to solve for B\begin{align*}\angle B\end{align*}.
b2b2b2b=262+1822(26)(18)cos26=1000936cos26=158.728812.60\begin{align*}b^2 & = 26^2 + 18^2 - 2(26)(18) \cos 26^\circ\\ b^2 & = 1000 - 936 \cos 26^\circ\\ b^2 & = 158.7288\\ b & \approx 12.60\end{align*}
To find mA\begin{align*}m \angle A\end{align*} or mC\begin{align*}m \angle C\end{align*}, you can use either the Law of Sines or Law of Cosines. Let’s use the Law of Sines.
sin2612.60=sinA1812.60sinAsinA=18sin26=18sin2612.60\begin{align*}\frac{\sin 26^\circ}{12.60} = \frac{\sin A}{18}&&\\ && 12.60 \cdot \sin A & = 18 \cdot \sin 26^\circ\\ && \sin A & = \frac{18 \cdot \sin 26^\circ}{12.60}\end{align*}
sin1(18sin2612.60)38.77\begin{align*}\sin^{-1} \left ( \frac{18 \cdot \sin 26^\circ}{12.60} \right ) \approx 38.77^\circ\end{align*} To find mC\begin{align*}m \angle C\end{align*}, use the Triangle Sum Theorem.
26+38.77+mCmC=180=115.23\begin{align*}26^\circ + 38.77^\circ + m \angle C & = 180^\circ\\ m \angle C & =115.23^\circ\end{align*}
Unlike the previous sections in this chapter, with the Laws of Sines and Cosines, we have been using values that we have found to find other values. With these two laws, you have to use values that are not given. Just keep in mind to always wait until the very last step to put anything into your calculator. This will ensure that you have the most accurate answer.
Example 4: Solve the triangle. Round your answers to the nearest hundredth.
Solution: When you are given only the sides, you have to use the Law of Cosines to find one angle and then you can use the Law of Sines to find another.
152225104310431232=222+2822(22)(28)cosA=12681232cosA=1232cosA=cosAcos1(10431232)32.16\begin{align*}15^2 & = 22^2 + 28^2 - 2(22)(28) \cos A\\ 225 & = 1268 - 1232 \cos A\\ -1043 & = -1232 \cos A\\ \frac{-1043}{-1232} & = \cos A \rightarrow \cos^{-1} \left (\frac{1043}{1232} \right ) \approx 32.16^\circ\end{align*}
Now that we have an angle and its opposite side, we can use the Law of Sines.
sin32.1615=sinB2215sinBsinB=22sin32.16=22sin32.1615\begin{align*}\frac{\sin 32.16^\circ}{15} = \frac{\sin B}{22}&&\\ &&15 \cdot \sin B & = 22 \cdot \sin 32.16^\circ\\ &&\sin B & = \frac{22 \cdot \sin 32.16^\circ}{15}\end{align*}
sin1(22sin32.1615)51.32\begin{align*}\sin^{-1} \left ( \frac{22 \cdot \sin 32.16^\circ}{15} \right ) \approx 51.32^\circ\end{align*} To find mC\begin{align*}m \angle C\end{align*}, use the Triangle Sum Theorem.
32.16+51.32+mCmC=180=96.52\begin{align*}32.16^\circ + 51.32^\circ + m \angle C &= 180^\circ\\ m \angle C &= 96.52^\circ\end{align*}
## To Summarize
Use Law of Sines when given:
• An angle and its opposite side.
• Any two angles and one side.
• Two sides and the non-included angle.
Use Law of Cosines when given:
• Two sides and the included angle.
• All three sides.
## Review Questions
Use the Law of Sines or Cosines to solve ABC\begin{align*}\triangle ABC\end{align*}. If you are not given a picture, draw one. Round all decimal answers to the nearest tenth.
1. mA=74,mB=11,BC=16\begin{align*}m \angle A = 74^\circ, m \angle B = 11^\circ, BC = 16\end{align*}
2. mA=64,AB=29,AC=34\begin{align*}m∠A = 64^\circ, AB = 29, AC = 34\end{align*}
3. mC=133,mB=25,AB=48\begin{align*}m \angle C = 133^\circ, m \angle B = 25^\circ, AB=48\end{align*}
Use the Law of Sines to solve ABC\begin{align*}\triangle ABC\end{align*} below.
1. mA=20,AB=12,BC=5\begin{align*} m\angle A = 20^\circ, AB = 12, BC = 5\end{align*}
Recall that when we learned how to prove that triangles were congruent we determined that SSA (two sides and an angle not included) did not determine a unique triangle. When we are using the Law of Sines to solve a triangle and we are given two sides and the angle not included, we may have two possible triangles. Problem 14 illustrates this.
1. Let’s say we have ABC\begin{align*} \triangle ABC \end{align*} as we did in problem 13. In problem 13 you were given two sides and the not included angle. This time, you have two angles and the side between them (ASA). Solve the triangle given that mA=20,mC=125,AC=8.4\begin{align*}m \angle A = 20^\circ, m \angle C = 125^\circ, AC = 8.4\end{align*}
2. Does the triangle that you found in problem 14 meet the requirements of the given information in problem 13? How are the two different mC\begin{align*}m \angle C\end{align*} related? Draw the two possible triangles overlapping to visualize this relationship.
It is beyond the scope of this text to determine when there will be two possible triangles, but the concept of the possibility is something worth noting at this time.
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# Fraction calculator
The calculator performs basic and advanced operations with fractions, expressions with fractions combined with integers, decimals, and mixed numbers. It also shows detailed step-by-step information about the fraction calculation procedure. Solve problems with two, three, or more fractions and numbers in one expression.
## Result:
### 3/4÷1/4 = 3/1 = 3
Spelled result in words is three.
### How do you solve fractions step by step?
1. Divide: 3/4 : 1/4 = 3/4 · 4/1 = 3 · 4/4 · 1 = 12/4 = 4 · 3 /4 · 1 = 3
Dividing two fractions is the same as multiplying the first fraction by the reciprocal value of the second fraction. The first sub-step is to find the reciprocal (reverse the numerator and denominator, reciprocal of 1/4 is 4/1) of the second fraction. Next, multiply the two numerators. Then, multiply the two denominators. In the next intermediate step, , cancel by a common factor of 4 gives 3/1.
In words - three quarters divided by one quarter = three.
#### Rules for expressions with fractions:
Fractions - use the slash “/” between the numerator and denominator, i.e., for five-hundredths, enter 5/100. If you are using mixed numbers, be sure to leave a single space between the whole and fraction part.
The slash separates the numerator (number above a fraction line) and denominator (number below).
Mixed numerals (mixed fractions or mixed numbers) write as non-zero integer separated by one space and fraction i.e., 1 2/3 (having the same sign). An example of a negative mixed fraction: -5 1/2.
Because slash is both signs for fraction line and division, we recommended use colon (:) as the operator of division fractions i.e., 1/2 : 3.
Decimals (decimal numbers) enter with a decimal point . and they are automatically converted to fractions - i.e. 1.45.
The colon : and slash / is the symbol of division. Can be used to divide mixed numbers 1 2/3 : 4 3/8 or can be used for write complex fractions i.e. 1/2 : 1/3.
An asterisk * or × is the symbol for multiplication.
Plus + is addition, minus sign - is subtraction and ()[] is mathematical parentheses.
The exponentiation/power symbol is ^ - for example: (7/8-4/5)^2 = (7/8-4/5)2
#### Examples:
subtracting fractions: 2/3 - 1/2
multiplying fractions: 7/8 * 3/9
dividing Fractions: 1/2 : 3/4
exponentiation of fraction: 3/5^3
fractional exponents: 16 ^ 1/2
adding fractions and mixed numbers: 8/5 + 6 2/7
dividing integer and fraction: 5 ÷ 1/2
complex fractions: 5/8 : 2 2/3
decimal to fraction: 0.625
Fraction to Decimal: 1/4
Fraction to Percent: 1/8 %
comparing fractions: 1/4 2/3
multiplying a fraction by a whole number: 6 * 3/4
square root of a fraction: sqrt(1/16)
reducing or simplifying the fraction (simplification) - dividing the numerator and denominator of a fraction by the same non-zero number - equivalent fraction: 4/22
expression with brackets: 1/3 * (1/2 - 3 3/8)
compound fraction: 3/4 of 5/7
fractions multiple: 2/3 of 3/5
divide to find the quotient: 3/5 ÷ 2/3
The calculator follows well-known rules for order of operations. The most common mnemonics for remembering this order of operations are:
PEMDAS - Parentheses, Exponents, Multiplication, Division, Addition, Subtraction.
BEDMAS - Brackets, Exponents, Division, Multiplication, Addition, Subtraction
BODMAS - Brackets, Of or Order, Division, Multiplication, Addition, Subtraction.
GEMDAS - Grouping Symbols - brackets (){}, Exponents, Multiplication, Division, Addition, Subtraction.
Be careful, always do multiplication and division before addition and subtraction. Some operators (+ and -) and (* and /) has the same priority and then must evaluate from left to right.
## Fractions in word problems:
• Fraction and a decimal
Write as a fraction and a decimal. One and two plus three and five hundredths
• One third 2
One third of all students in class live in a house. If here are 42 students in a class, how many of them live in house?
• Rolls
Mom bought 13 rolls. Dad ate 3.5 rolls. How many rolls left when Peter yet put two at dinner?
• Missing number
Blank +1/6 =3/2 find the missing number
• Akpan
Akpan spent 3/8 of his time in school during the week. What fraction of his time does he spend at home during the week?
• The book 4
Mr. Kinion read 3 3/4 chapters in his book on Monday. He then read 2 4/6 more chapters on Tuesday. How many chapters has he read so far?
• A class IV.C
In a class 2/5 were boys. 30 were girls. How many more girls than boys are there?
• Mixed numbers
Rewrite mixed numbers, so the fractions have the same denominator: 5 1/5 - 2 2/3
• Two thirds
Two thirds of the unknown number is 8/7. What is 7/3 of this number?
• Fruits
For the price of one pineapple, I will buy two oranges. For the price of three oranges, I will buy four apples. I will buy 6 bananas for the price of three apples. What is the price of one banana if I pay 1 euro for one pineapple?
• A pizza 2
A pizza is cut into slices that are each 1/6 of the whole. John is going to eat 1/2 of the whole pizza. How many slices will John eat?
• Land area
The land area of Asia and Africa are in a 3: 2 ratio, the European and African are is 1:3. What are the proportions of Asia, Africa, and Europe?
• Max has
Max has a 1/3 of a pound of pretzels and wants to share equal parts between himself and a friend. How much pretzels will Max's friend receive?
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Dividing Decimals
Dividing decimals can be a very tricky skill to master. After learning how to do basic division and long division, dividing decimals is a piece of cake. Instead of just placing the remainder at the top of the quotient, your child now needs to take it one step further. To refresh your memory, the remainder is the number that cannot be divided because it is too small in comparison to the divisor. When dividing decimals, you just keep moving the decimal point to the right, and add a zero to the end of the divisor until the problem can no longer be divided.
Problem:
Step 1:
The underlined portion signifies that the decimal needs to be moved over one place to the right in order to make the divisor a whole number instead of a decimal.
Step 2:
Next figure out how many times 34 goes in to 55, which is one. 34 × 1 = 34
Next bring down the 3 to continue dividing the problem.
Step 3:
Continue step 2 until the number cannot be divided.
In this case, 213 goes in to 34 six times. 34 × 6 = 204
Next bring down the 1 to continue dividing the problem.
Step 4:
Next we need to find out how many times 34 goes into 91.
In this case, 91 goes in to 34 two times. 34 × 2 = 68.
Next bring down the 8 and the end result should be 238.
Step 5:
Finally, we need to find how many times 238 goes into 34.
We have a match! 238 is a multiple of 34! 34 × 7 = 238
The quotient is 16.27.
Dividing Decimals can be a difficult concept to learn. This is why it is important to encourage your child to keep practicing outside of the classroom. Learning in class and applying it within the comforts of your own home is the most beneficial way for your child to succeed. Have them bring their class work and homework to your attention and take the time to understand your child’s pace in learning this simple concept.
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# Solving an Oblique Triangle, Part II
Last time we looked at solving triangles in the ASA, AAS, SSS, and SAS cases. We have one more case, which tends to be a little more complicated: the “ambiguous case”, SSA.
## SSA: Two sides and the angle opposite one of them
Triangles and Law of Sines
I am having problems with the Law of Sines. To be specific, I don't know how to tell if the information I'm given produces more than one triangle. Solving for one is simple. That I can do. But some of the problems in my book produce two triangles, and all the info they give to solve the problem is like 2 sides and an angle or 2 angles and a side.
For example:
Angle A = 37 degrees
Side C = 40
Side A = 28
The back of my book gives two solutions for this this problem. How do you solve it?
Problems with “2 angles and a side” (ASA or AAS) will always have one solution (unless the angles are too large); those with “2 sides and an angle” always have one solution if they are SAS. But there are cases of SSA that have no answer, or one, or two. How do you know which you have? For the moment, we’ll just examine the specific example given here.
Here is what we might draw as a (fairly accurate) sketch of this triangle:
Hi Heathe!
Using the law of sines you get:
28 40
-------- = -------
sin 37 sin C
or 28 sin C = 40 sin 37
40 sin 37
sin C = ---------
28
sin C = 0.859735747
Here we have used the Law of Sines in the form $$\frac{a}{\sin(A)} = \frac{c}{\sin(C)}$$ and solved for the sine of angle C. But, as we saw last time in working with the SSS and SAS cases, there are two angles with the same sine. That will be an issue here, as well:
To solve this equation, we use the inverse sine function. InvSin(0.859..) = 59.28 degrees. But recall that the sin of an angle and the sin of its supplement are equal. That is, sin x = sin (180-x). So, if the sin of 59.29 is 0.8597, then so is sin (180-59.29) or sin (120.71).
The triangle I drew above is the one most of us would expect, and the one we find first: the acute triangle, with a 59.29° angle at C, obtained directly from the inverse sine. The other possible answer, an obtuse triangle with a 120.71° angle at C, looks like this:
It still has the required angle and two sides, but side a has been flipped around.
So to find both solutions, we just have to remember to take the supplement of the inverse sine.
But sometimes, as we’ll be seeing, this alternative triangle might not actually exist; we need to check it:
So another possible measure for C is 120.71. We need to see if it can work. Since angle A is 37, keeping in mind that the angles of a triangle add up to 180, we can have C = 59.29 and A = 37 (and still have some angle measure left over for B) OR C = 120.71 and A = 37 and still have some angle measure left over for B.
If the sum of A and C had been greater than 180°, angle B would have turned out negative, and been rejected.
That's why there are 2 solutions in the back of your book. Now, we have to find all of the other parts of the triangle for BOTH possible angle Bs.
Finishing up, we have:
1. for the acute case, $$A = 37°$$, $$C = 59.29°$$, and $$B = 180 – (37 + 59.29) = 83.71°$$; then $$\displaystyle b = \frac{a\sin(B)}{\sin(A)} = \frac{28\sin(83.71°)}{\sin(37°)} = 46.25$$;
2. for the obtuse case, $$A = 37°$$, $$C = 120.71°$$, and $$B = 180 – (37 + 120.71) = 22.29°$$; then $$\displaystyle b = \frac{a\sin(B)}{\sin(A)} = \frac{28\sin(22.29°)}{\sin(37°)} = 17.65$$.
### Using the Law of Cosines instead
Last time, there was a mention of using the Law of Cosines instead of the Law of Sines for SSA. Since I can’t find any examples of that in our archive, let’s apply that technique to this problem.
Recall the problem:
The Law of Cosines, applied to this triangle, requires writing the unknown side b, as a variable: $$a^2 = b^2 + c^2 – 2bc\cos(A)$$ becomes $$28^2 = b^2 + 40^2 – 2(40)b\cos(37°)$$ We can rearrange this to $$b^2 – (80\cos(37°))b + 816 = 0$$
When we solve this by the quadratic formula, we get $$b = \frac{80\cos(37°) \pm \sqrt{(80\cos(37°))^2 -4(816)}}{2} = \frac{63.89\pm28.60}{2} = 46.25, 17.65$$
These are the same answers we got before. As was said last time, this is a little more work than using the Law of Sines, but the formula reminds us that there are two solutions, each of which we pursue to find the other angles.
## How many solutions are there?
To observe other things that can happen, consider this similar question from 2003:
The Ambiguous Case
I do not understand how to use the ambiguous case to determine the number of triangles that can be constructed. Is there a simple way to answer the following question:
How many triangles can be constructed if, for example, a=4, A=30, and c=12? Or a=9, b=12, and A=35?
I am confused about how to do this.
Thank you,
Les
Both examples are SSA, and we’ll look at those and yet another here.
### (a) No solution
Hi, Les.
We can start by applying the law of sines. In your first example, we get
sin(C)/c = sin(A)/a
sin(C)/12 = sin(30)/4
sin(C) = sin(30)*12/4
sin(C) = 0.5*12/4
sin(C) = 1.5
You know that the sine of any angle is between -1 and 1. There is no angle whose sine is 1.5, therefore there are no solutions in this case.
Here is what this triangle looks like if we try to draw it:
I call this the T Rex case – just imagine those tiny forelegs trying to pick up something it dropped! No matter what direction the 4-unit leg tries to move, it can’t reach the ground and make a vertex C.
And how do we recognize this case? When we find that the required sine is not in the domain of the inverse sine, so we can’t do the final step.
### (b) Two solutions
He skipped over the second example, which is the kind we’ve seen above; so let’s do it now:
We have a = 9, b = 12, and A = 35°. The Law of Sines tells us that $$\frac{\sin(B)}{12} = \frac{\sin(35°)}{9}$$ so that $$\sin(B) = \frac{12\sin(35°)}{9} = 0.764769$$ This is a valid sine, so the two possible values of B are $$\arcsin(0.764769) = 49.89°$$ and $$180° – 49.89° = 130.11°$$. Then,
1. for the acute case, $$A = 35°$$, $$B = 49.89°$$, and $$C = 180 – (35 + 49.89) = 95.11°$$; then $$\displaystyle c = \frac{a\sin(C)}{\sin(A)} = \frac{9\sin(95.11)}{\sin(35)} = 15.63$$;
2. for the obtuse case, $$A = 35°$$, $$B = 130.11°$$, and $$C = 180 – (35 + 130.11) = 14.89°$$; then $$\displaystyle c = \frac{a\sin(C)}{\sin(A)} = \frac{9\sin(14.89)}{\sin(35)} = 4.03$$.
The two triangles look like this:
### (c) One solution
But those aren’t the only things that can happen!
Let's consider another example: B = 50 deg, b = 12, c = 10. In this case
sin(C)/c = sin(B)/b
sin(C)/10 = sin(50)/12
sin(C) = sin(50)*10/12
sin(C) = 0.638370
C = 39.67 degrees
Is this the only solution? Not necessarily, because there is another angle whose sine is 0.638370, namely, 180 - 39.67 = 140.33 degrees. Can we have a triangle with angle B = 50 deg and angle C = 140.33 deg? No, because B+C = 190.33 deg, which is more than the sum of all three angles of any triangle (180 deg). Thus this case is not ambiguous: there is exactly one triangle that satisfies the conditions. The one solution has angle C = 39.67 degrees.
The work here should be familiar by now. But this time, when we check the obtuse solution with C = 140.33°, we find that the resulting triangle doesn’t exist. What does the triangle look like? Here it is:
When we flip side a to the other possible position, it is on the wrong side of A, so that the angle opposite the 12 would really be 130°, not 50°. And the angle at the new C would in fact be acute. So the second triangle is a good try, but it fails. As a result, we have only one solution this time.
(Incidentally, the SSA case is often called the “ambiguous case” because of the possibility of more than one answer; I tend to think of only the specific situation in which there are two solutions as truly ambiguous, and Doctor Rick has used the word in that sense here.)
If the law of sines gives a sine of the missing angle that is less than 1, AND both angles with this sine (the arcsine of the angle and 180 minus the arcsine) are less than 180 minus the known angle, then there are two triangles that satisfy the conditions.
This is like Les’ second problem, which we looked at above.
### Analysis
Here are some more thoughts:
You can recognize whether an SSA specification of a triangle has 0, 1 or 2 solutions without going through the law of sines. Consider the geometry of the triangle. If we have a base (b) of known length and a known angle A, the third vertex (B) must lie along a ray:
/
/
/
B /
/
D / \
/ \
B' / \ \a
/ \ \
/ \ \ \
/ a\ \ \
/ \ \\
/______________________\
A b C
We know the length of the side a, and we want to find a point (or points) B along the ray such that BC = a.
Here’s a better picture, which, as you’ll see, is in a different orientation than mine:
Consider the perpendicular CD from C to the ray. Its length is b*sin(A), and it is the shortest distance from C to the ray. If a < b*sin(A), then we know that no point on the ray will be a distance a from C; all points are farther than this. Thus the condition for no solution is
a < b*sin(A)
In the figure, I show the ambiguous case: both CB and CB' have length a. The triangle BCB' is isosceles, and CD is its altitude, so BD = B'D. This is the ambiguous case: both ABC and AB'C satisfy the conditions.
What if $$a = b \sin(A)$$? Then there is only one triangle, namely the right triangle ACD. This is a very special case.
If we increase length a, then point B moves out along the ray, but B' moves in toward A. When a = b, point B' is coincident with point A. If a > b, we no longer have an ambiguous case: only triangle ABC satisfies the conditions, because B' is in effect pushed off the ray, onto the ray in the opposite direction.
Thus, the three cases can be distinguished easily:
a < b*sin(A) no solutions
b*sin(A) < a < b two solutions
b < a one solution
In my figure, A is an acute angle. What happens if A is obtuse? It's easy to see that there is no ambiguity. If a > b, there is one solution; if a < b, there are no solutions.
Here are the two cases with an obtuse angle. First, with one solution, because a > b:
Then, with no solutions, because ab:
### Summary: The Ask Dr. Math FAQ
Now let’s look at what the FAQ says about SSA:
Case II: You are given two sides and the angle opposite one of them, say a, c, and A.
Subcase A: a < c sin(A). There is no solution.
Subcase B: a = c sin(A). There is one solution:
C = π/2, B = π/2 - A, b = c cos(A).
Subcase C: c > a > c sin(A). There are two solutions. Use the Law of Sines to find sin(C) = c sin(A)/a < 1. There are two angles C and C' = π - C having that sine, one acute and one obtuse. Then compute B = π - A - C and B' = π - A - C'. Now use the Law of Sines again to find b = a sin(B)/sin(A) and b' = a sin(B')/sin(A). The solutions are (a,b,c,A,B,C) and (a,b',c,A,B',C').
Subcase D: a >= c. There is one solution. Use the Law of Sines to find sin(C) = c sin(A)/a <= 1. Then angle C must be acute, so it can be found uniquely from sin(C). Then compute B = π - A - C. Now use the Law of Sines again to find b = a sin(B)/sin(A).
Subcase A is our case (a) above.
Subcase B is the special case where the one solution is a right triangle.
Subcase C is the usual “ambiguous” case with two solutions, our case (b).
Subcase D is our case (c) with one solution.
These don’t include the obtuse cases.
## When everything goes wrong!
Here is a question from 2002 that is more than just ambiguous:
Different Answers with Sine Rule and Cosine Rule
This is really strange. I found this problem in a textbook. A triangle ABC has measurements AB = 8.2cm, BC = 9.4cm and AC = 12.8cm and angle A = 47 degrees. Find angle B.
Method 1 - Using the Sine Rule;
sin B = AC (sin A)/BC gives B as 84.4 degrees.
Method 2 - Using the Cosine Rule;
AC^2 = BC^2 + AB^2 - 2(BC)(AB) cos B gives B as 93.1 degrees, which incidentally is the right answer.
Why does using the Sine Rule here fail to give the right answer?
Thanks for any help you can offer me.
Using the standard labeling, we have c = 8.2, a = 9.4, b = 12.8, and A = 47°.
Doctor Rick answered this, too, identifying three different issues that contribute to the confusion:
### 1. An overspecified problem
Hi, Thomas.
It's strange, isn't it? There are several things going on here at the same time, making it a bit hard to untangle.
First, the triangle is overspecified to begin with. Given just the lengths of the sides, you can use the law of cosines to find angle A. It turns out to be, not 47 degrees exactly, but 47.165 degrees.
Had you noticed that we are given not three, but four facts about the triangle? (“Overspecified” means that we are given more information than we need.) It could be called SSSA, and the SSS and the SSA or SAS are at war with one another. As a result, there is really no triangle that satisfies them all! It appears that, as he suggests, the 47 degree angle was given as an approximation, so it can’t really be used to find the answer. On the other hand, in principle you could use any three of the four givens and construct a (different) triangle you might call the “right” one!
The fact that only one angle is requested makes it easy not to check the answer; checking that the three angles add up to 180° would reveal the problem. But so does trying two methods, as Thomas did.
Here is the work for finding the “actual” value of A, assuming the three sides are exact: $$\cos(A) = \frac{b^2+c^2-a^2}{2bc} = \frac{12.8^2+8.2^2-9.4^2}{2(12.8)(8.2)} = 0.679878$$ so that $$A = \arccos(0.679878) = 47.166°$$
### 2. An amplified error
Second, since angle B is close to 90 degrees, a small error in the sine of the angle can result in a rather large error in the angle itself. In particular, when I apply the law of sines with 47 degrees for A, I get sin(B) = 0.99588589; but when I use 47.165 degrees for A, I get sin(B) = 0.99857047. That's a small error - but when I take the inverse sine of each value, I get 84.80 (I don't know why this is slightly different from your answer) and 86.936 degrees respectively.
Here is the work for these two calculations. First, using A = 47°: $$\sin(B) = \frac{b\sin(A)}{a} = \frac{12.8\sin(47)}{9.4} = 0.99588589…$$ so $$B = \arcsin(0.99588589) = 84.80°$$
Next, using A = 47.166°: $$\sin(B) = \frac{b\sin(A)}{a} = \frac{12.8\sin(47.166)}{9.4} = 0.99857231…$$ so $$B = \arcsin(0.99857231) = 86.938°$$
(I used unrounded values in my work.) In any case, an error of 1/8 of a degree in A led to an error of over 2 degrees in B.
### 3. An ambiguous SSA triangle
Now look at the corrected answer of 86.936 degrees using the law of sines. Compare it with the correct answer found using the law of cosines: 93.064 degrees. Each is 3.064 degrees away from a right angle - but in opposite directions! Does this give you an idea of what's going on?
When you use the law of cosines to find angle B, you are making use of the three sides of the triangle. Remember the SSS congruency theorem: three sides determine the shape of a triangle uniquely. This is why we are able to find a unique measure for angle B.
When you use the law of sines, you are making use of two sides and an angle. Notice how these are related: the angle is NOT between the sides. In terms of congruency theorems, we need an SSA theorem - but there isn't one. The triangle is NOT uniquely determined by these quantities.
SSA gives two possible angles, as we’ve seen before; here the correct answer is the obtuse one, not the acute one Thomas found. We are back into the realm of the ambiguity of SSA. But this time, the given sides let us choose which answer is “correct”.
Remember that we found that the sine of angle B is 0.99857047. We can't just take the inverse sine of this number: that is ONE angle with this sine, but 180 degrees minus this angle is another solution. The latter solution turns out to be correct in this case.
You see, there are two triangles that have the same angle A and sides AC and BC. The law of sines, applied without thinking, gave one of these triangles, but it was the wrong one. You must be careful in taking the inverse sine: note both solutions in the range 0 to 180 degrees, and check whether each leads to a valid solution of the triangle. If both do, you need more information (such as side AB) to tell which is correct.
Here is the actual triangle, based on the side lengths:
### Real life!
This is not your typical problem; it was given in an inherently inconsistent form. This is not good for a classroom exercise, but is not uncommon in real life! There, we typically have too much information, and some of it will be inaccurate. We have to:
1. choose which data are most reliable (in this case, the sides, which were measured more carefully);
2. avoid calculations that amplify error (such as inverse sines of ratios close to 1, which are similar to trying to intersect two lines at a shallow angle); and then
3. use the other data to check, or to choose between possibilities.
Interestingly, these are exactly the three points Doctor Rick identified in this problem.
### 1 thought on “Solving an Oblique Triangle, Part II”
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Table Charts Questions Answers of Data Interpretation includes questions on various aspects of table data which are important for all competitive exams like BANK PO, IBPS, SBI-PO, RBI, MBA, MAT, CAT, IIFT, IGNOU, SSC CGL, CBI, CPO, CLAT, CTET, NDA, CDS, Specialist Officers.
1) What was the approximate percentage increase in the sales of 55AH batteries in 1998 compared to that in 1992
• 1) 31%
• 2) 33%
• 3) 34%
• 4) 36%
Ans. C
Explanation :
Increase is = 145-108 = 37 Now we need to calculate that 37 is what percent of 108. \begin{aligned} \left(\frac{37}{108}*100 \right)\% = 34.25\% \end{aligned}
2) In the case of which battery, there was a continuous decrease in sales from 1992 to 1997
• 1) 35 AH
• 2) 4 AH
• 3) 32 AH
• 4) 7 AH
Ans. D
Explanation :
After analysing table chart, it is clear that the sales of 7AH batteries have been decreasing continuously from 1992 to 1997.
3) What is the difference in the number of 35AH batteries sold in 1993 and 1997
• 1) 39000
• 2) 40000
• 3) 43000
• 4) 49000
Ans. A
Explanation :
Required difference = [(84 - 45) * 1000] = 39000.
4) The total sales of all the seven years is the maximum for which battery
• 1) 35 AH
• 2) 4 AH
• 3) 7 AH
• 4) 32 AH
Ans. D
Explanation :
The total sales (in thousands) of all the seven years for various batteries are : a t. For 4AH = 75 + 90 + 96 + 105 + 90 + 105 + 115 = 676 For 7AH = 144 + 126 + 114 + 90 + 75 + 60 + 85 = 694 For 32AH = 114 + 102 + 75 + 150 + 135 + 165 + 160 = 901 For 35 AH = 102 + 84 + 105 + 90 + 75 + 45 + 100 = 601 For 55 AH = 108 + 126 + 135 + 75 + 90 + 120 + 145 = 799. Clearly, sales are maximum in case of 32AH batteries.
5) What is the average amount of interest per year which the company had to pay during this period
• 1) Rs. 36.66 lakhs
• 2) Rs. 36.36 lakhs
• 3) Rs. 36.26 lakhs
• 4) Rs. 36.06 lakhs
Ans. A
Explanation :
Average amount of interest paid by the Company during the given period will be \begin{aligned} \frac{23.4 + 32.5 + 41.6 + 36.4 + 49.4}{5} lakh \= 36.66 lakhs \end{aligned}
6) The total amount of bonus paid by the company during the given period is approximately what percent of the total amount of salary paid during this period
• 1) .5%
• 2) 1%
• 3) 1.5%
• 4) 2%
Ans. B
Explanation :
Here we simply need to calculate that bonus is what percent of salary. We will just sum all bonus and salary to get the percentage as below, \begin{aligned} \left(\frac{3.00 + 2.52 + 3.84 + 3.68 + 3.96}{288 + 342 + 324 + 336 + 420}*100\right)\% \= \left(\frac{17}{1710}*100\right)\% \= 1\% \text{approx.} \end{aligned}
7) Total expenditure on all these items in 1998 was approximately what percent of the total expenditure in 2002
• 1) 61%
• 2) 47%
• 3) 59%
• 4) 69%
Ans. D
Explanation :
Required percentage we can easily calculate from the above table chart. Required percentage will be, \begin{aligned} \left(\frac{288 + 98 + 3.00 + 23.4 + 83}{420 + 142 + 3.96 + 49.4 + 98}*100\right)\% \= \left(\frac{495.4}{713.36}*100\right)\% \= 69.45\% \end{aligned}
8) Calculate the total expenditure of the company over these items during the year 2000 from the table chart give
• 1) Rs. 543.44 lakhs
• 2) Rs. 544.44 lakhs
• 3) Rs. 545.44 lakhs
• 4) Rs. 546.44 lakhs
Ans. B
Explanation :
Total expenditure of the Company during 2000 = Rs. (324 + 101 + 3.84 + 41.6 + 74) lakhs = Rs. 544.44 lakhs.
9) The ratio between the total expenditure on Taxes for all the years and the total expenditure on Fuel and Transport for all the years respectively is approximatel
• 1) 4:13
• 2) 7:13
• 3) 10:13
• 4) 11:13
Ans. C
Explanation :
Required Ratio will be \begin{aligned} \left( \frac{83 + 108 + 74 + 88 + 98}{98 + 112 + 101 + 133 + 142} \right) \= \left( \frac{451}{586} \right) \= \frac{1}{1.3} = \frac{10}{13} \end{aligned}
10) In which of the following years was the number of light commercial vehicles sold approximately 25% of the number of 2-wheelers sol
• 1) 1991
• 2) 1992
• 3) 1993
• 4) 1994
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# 1375 in words
1375 in words is written as One Thousand Three Hundred and Seventy Five. 1375 represents the count or value. The article on Counting Numbers can give you an idea about count or counting. The number 1375 is used in expressions that relate to money, distance, length, year and others. Let us consider an example for 1375. ”There are One Thousand Three Hundred and Seventy Five steps to reach the temple on a mountain”.
1375 in words One Thousand Three Hundred and Seventy Five One Thousand Three Hundred and Seventy Five in Numbers 1375
## How to Write 1375 in Words?
We can convert 1375 to words using a place value chart. The number 1375 has 4 digits, so let’s make a chart that shows the place value up to 4 digits.
Thousands Hundreds Tens Ones 1 3 7 5
Thus, we can write the expanded form as:
1 × Thousand + 3 × Hundred + 7 × Ten + 5 × One
= 1 × 1000 + 3 × 100 + 7 × 10 + 5 × 1
= 1375
= One Thousand Three Hundred and Seventy Five.
1375 is the natural number that is succeeded by 1374 and preceded by 1376.
1375 in words – One Thousand Three Hundred and Seventy Five.
Is 1375 an odd number? – Yes.
Is 1375 an even number? – No.
Is 1375 a perfect square number? – No.
Is 1375 a perfect cube number? – No.
Is 1375 a prime number? – No.
Is 1375 a composite number? – Yes.
## Solved Example
1. Write the number 1375 in expanded form
Solution: 1 × 1000 + 3 × 100 + 7 × 10 + 5 × 1
We can write 1375 = 1000 +300 +70 + 5
= 1 × 1000 + 3 × 100 + 7 × 10 + 5 × 1
## Frequently Asked Questions on 1375 in words
### How to write 1375 in words?
1375 in words is written as One Thousand Three Hundred and Seventy Five.
### Is 1375 a perfect square number?
No. 1375 is not a perfect square number.
### Is 1375 a prime number?
No. 1375 is not a prime number.
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# How do you find the tension between two objects with friction?
## How do you find the tension between two objects with friction?
Account for friction.
1. Normal force (N) = 10 kg × 9.8 (acceleration from gravity) = 98 N.
2. Force from kinetic friction (Fr) = 0.5 × 98 N = 49 Newtons.
3. Force from acceleration (Fa) = 10 kg × 1 m/s2 = 10 Newtons.
4. Total tension = Fr + Fa = 49 + 10 = 59 Newtons.
## How do you find the tension in a string between two blocks with friction?
T = M1*a T = F + M2*a Both equations can be used to find the value of tension in the rope. The values of tension will be the same. The SI unit of tension force is Newton.
How does friction relate to tension?
The friction is introducing an extra force which changes the tensions on each side. As far as your question about rope stretching goes, if you anchor a rope on one side and pull, the rope will pull back, creating a tension.
### What is the formula of the tension?
Tension formula is articulated as. T=mg+ma. Where, T= tension (N or kg-m/s2) g = acceleration due to gravity (9.8 m/s2)
### How do I calculate tension?
Tension force remains a gravitational force. If the body is moving upwards then the tension will be referred to as the T = W + ma. When the body goes down, the thickness is the same as T = W – ma. T = W if the discomfort is equal to body weight.
How do you calculate tension force?
Tension Formula
1. Tension formula is articulated as. T=mg+ma.
2. Tension Formula is made use of to find the tension force acting on any object. It is useful for problems.
3. Tension Solved Examples.
4. Problem 1: A 8 Kg mass is dangling at the end of a string.
#### How do you find tension in a string physics?
The tension on an object is equal to the mass of the object x gravitational force plus/minus the mass x acceleration.
#### Does tension equal friction?
There will be kinetic friction between the surface and the block since it is in motion. This formula for tension is consistent with general observation, as the tension has to be equal to the applied force which here is the force applied plus the frictional force.
Is friction a tension?
As nouns the difference between friction and tension is that friction is the rubbing of one object or surface against another while tension is tension.
## How do you solve tension problems in physics?
How do we calculate the force of tension?
1. Draw the forces exerted on the object in question.
2. Write down Newton’s second law ( a = Σ F m ) (a=\dfrac{\Sigma F}{m}) (a=mΣF)left parenthesis, a, equals, start fraction, \Sigma, F, divided by, m, end fraction, right parenthesis for a direction in which the tension is directed.
## Is tension and force the same?
Two or more physical objects that are in contact, exert forces on each other. Based on the objects in contact we give these contact forces different names. If one of these objects in contact happens to be a string, rope, cable or spring, we call the force as tension.
What is tension force with example?
Tension is the opposite of compression force. All the objects that are present in contact with each other exert a force on each other. The best example of a tension force can be seen while pulling a rope. When a pull force is applied to the rope, a significant amount of tension gets built.
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# How do you integrate xe^(2x)dx?
Mar 7, 2015
I would use Integration by Parts:
Jun 7, 2017
$\int \setminus x {e}^{2 x} \setminus \mathrm{dx} = \frac{1}{4} {e}^{2 x} \left(2 x - 1\right) + C$
#### Explanation:
We can use the formula for Integration By Parts (IBP):
$\int \setminus u \frac{\mathrm{dv}}{\mathrm{dx}} \setminus \mathrm{dx} = u v - \int \setminus v \frac{\mathrm{du}}{\mathrm{dx}} \setminus \mathrm{dx}$, or less formally
$\text{ } \int \setminus u \setminus \mathrm{dv} = u v - \int \setminus v \setminus \mathrm{du}$
I was taught to remember the less formal rule in word; "The integral of udv equals uv minus the integral of vdu". If you struggle to remember the rule, then it may help to see that it comes a s a direct consequence of integrating the Product Rule for differentiation.
Essentially we would like to identify one function that simplifies when differentiated, and identify one that simplifies when integrated (or is at least is integrable).
So for the integrand $x {e}^{2 x}$, hopefully you can see that $x$ simplifies when differentiated and ${e}^{2 x}$ effectively remains unchanged.
Let $\left\{\begin{matrix}u & = x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} & = 1 \\ \frac{\mathrm{dv}}{\mathrm{dx}} & = {e}^{2 x} & \implies & v & = \frac{1}{2} {e}^{2 x}\end{matrix}\right.$
Then plugging into the IBP formula:
$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$
gives us
$\int \setminus \left(x\right) \left({e}^{2 x}\right) \setminus \mathrm{dx} = \left(x\right) \left(\frac{1}{2} {e}^{2 x}\right) - \int \setminus \left(\frac{1}{2} {e}^{2 x}\right) \left(1\right) \setminus \mathrm{dx}$
$\int \setminus x {e}^{2 x} \setminus \mathrm{dx} = \frac{1}{2} x {e}^{2 x} - \frac{1}{2} \int \setminus {e}^{2 x} \setminus \mathrm{dx}$
$\text{ } = \frac{1}{2} x {e}^{2 x} - \frac{1}{4} {e}^{2 x} + C$
$\text{ } = \frac{1}{4} {e}^{2 x} \left(2 x - 1\right) + C$
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# Algebra Rate Problems (upstream/downstream)
Important tips for solving Algebra Rate problems.
Algebra Rate problems are used to find the distance traveled or time required for traveling certain distance.
Important note :
For downstream ----> Rate of (boat /steamer) in still water + rate of stream
For upstream ----> Rate of (boat /steamer) in still water - rate of stream
With wind ----> Rate of the plane in still air + rate of the wind
against wind ----> Rate of the plane in still air - rate of the wind
1) A boat travels 30 km up a river in the same time it takes to travel 50 km down the same river. If the current of water is 5km/h . What is the speed of the boat in still water?
Solution :
Let speed of boat in still water = x km/h
Speed of the boat downstream = x + 5
Speed of the boat upstream = x - 5
Speed = Distance / time
So time = Distance / speed
Time taken by boat to travel 30 km upstream = 30/ (x – 5)
Time taken by boat to travel 50 km downstream = 50/ (x + 5)
Since time taken by the boat in both cases is the same ,
=> 30 (x + 5) = 50 (x -5)
=> 30 x +150 = 50x -250
=> 20x = 400
=> x = 20
Thus the speed of the boat in still water is 20 km/h.
2) A steamer goes downstream and covers the distance between two ports in 4 hrs., while it covers the same distance upstream in 5 hrs. If the speed of the stream is 2km/h , find the speed of the steamer in still water.
Solution :
Let speed of the steamer in still water = x km/h
Speed of the steamer in downstream = x +2
Speed of the steamer in upstream = x – 2
So time = Distance / speed
Distance = speed x time
D = (x +2) 4
D = (x – 2) 5
As the distance between the port is same.
4(x +2) = 5(x -2)
4x + 8 = 5x -10
-x = -18
X = 18
Speed of the the steamer in still water = 18 m/h
3) Joe traveled against the wind in a small plane for 3.2 hr. The return trip with the wind took 2.8 hr. Find the speed of the wind if the speed of the plane is still air is 170 mph.
Solution :
Let the speed of the plane in still air = x mph
Speed of the plane against the wind = 170 –x
Speed of the plane with the wind = 170 +x
Speed = Distance / time
Distance = speed x time
D = (170 –x) 3.2
D = ( 170 +x ) 2.8
As the distance is same, so
(170 –x) 3.2 = ( 170 +x ) 2.8
544 – 3.2 x = 476 + 2.8x
– 3.2 x - 2.8x = 476 – 544
-6 x = - 68
x = 68/6 = 11.33
So the speed of the wind = 11.33 mph
Linear Equation in One Variable
Linear Equation in One Variable
Algebraic Equations with Fractions
Algebra Word Problems
Algebra Rate Problems (upstream and downstream)
Algebra Age Problems
Algebra Digit Problems
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The quadratic function has maximum power of ${\displaystyle x}$ equal to ${\displaystyle 2}$:
${\displaystyle Ax^{2}+Bx+C}$.
${\displaystyle Ax^{2}+Bx+C=0\ \dots \ (1)}$, in which coefficient ${\displaystyle A}$ is non-zero.
The solution of the quadratic equation is:
${\displaystyle x={\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}\ \dots \ (2)}$
The above solution to the quadratic is well known to high-school algebra students. The proof of the solution is usually presented to students as completion of the square, not presented here. To me completion of the square seemed too complicated. To help you acquire a new respect for the quadratic, presented below are more ways to solve it.
The depressed cubic leads to a solution of the cubic and the depressed quartic leads to a solution of the quartic. This paragraph shows that the depressed quadratic leads to a solution of the quadratic. To produce the depressed function, let ${\displaystyle x={\frac {-B+t}{GA}}}$ where G is the degree of the function. For the quadratic G = 2. Let ${\displaystyle x={\frac {-B+t}{2A}}\ \dots \ (3)}$ Substitute (3) into (1) and expand: ${\displaystyle A({\frac {-B+t}{2A}})({\frac {-B+t}{2A}})+B({\frac {-B+t}{2A}})+C\ \dots \ (4)}$ ${\displaystyle (4)\ *\ 4A,\ 4AA({\frac {-B+t}{2A}})({\frac {-B+t}{2A}})+4AB({\frac {-B+t}{2A}})+4AC}$ ${\displaystyle (-B+t)(-B+t)+2B(-B+t)+4AC}$ ${\displaystyle B^{2}-2Bt+t^{2}-2B^{2}+2Bt+4AC}$ ${\displaystyle t^{2}-B^{2}+4AC}$ In the depressed quadratic above the coefficient of ${\displaystyle t^{2}}$ is ${\displaystyle 1}$ and that of ${\displaystyle t}$ is ${\displaystyle 0}$. ${\displaystyle t^{2}=B^{2}-4AC}$ ${\displaystyle t=\pm {\sqrt {B^{2}-4AC}}\ \dots \ (5)}$ Substitute (5) into (3) and the result is the solution in (2).
### p +- q
Let one value of ${\displaystyle x}$ be ${\displaystyle p+q}$ and another value of ${\displaystyle x}$ be ${\displaystyle p-q}$. Substitute these values into ${\displaystyle (1)}$ above and expand. {\displaystyle {\begin{aligned}A(p+q)(p+q)+B(p+q)+C=&\ 0\\App+A2pq+Aqq+Bp+Bq+C=&\ 0\ \dots \ (6)\\\\A(p-q)(p-q)+B(p-q)+C=&\ 0\\App-A2pq+Aqq+Bp-Bq+C=&\ 0\ \dots \ (7)\end{aligned}}} {\displaystyle {\begin{aligned}(6)-(7),\ 4Apq+2Bq=&\ 0\\2Ap+B=&\ 0\\2Ap=&\ -B\\\\p=&\ {\frac {-B}{2A}}\ \dots \ (8)\end{aligned}}} {\displaystyle {\begin{aligned}(6)+(7),\ 2App+2Aqq+2Bp+2C=&\ 0\\App+Aqq+Bp+C=&\ 0\\App+Bp+C=&\ -Aqq\ \dots \ (9)\\\\(9)*4A,\ 4AApp+4ABp+4AC=&\ -4AAqq\ \dots \ (10)\end{aligned}}} Substitute ${\displaystyle (8)}$ into ${\displaystyle (10)}$ {\displaystyle {\begin{aligned}(-B)(-B)+2B(-B)+4AC=&\ -4AAqq\\BB-2BB+4AC=&\ -4AAqq\\4AAqq=&\ BB-4AC\\\\qq=&\ {\frac {BB-4AC}{4AA}}\\\\q=&\ {\frac {\pm {\sqrt {B^{2}-4AC}}}{2A}}\ \dots \ (11)\end{aligned}}} {\displaystyle {\begin{aligned}x=&\ p+q\\\\=&\ ({\frac {-B}{2A}})+({\frac {\pm {\sqrt {B^{2}-4AC}}}{2A}})\\\\=&\ {\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}\end{aligned}}}
### By observation and elementary deduction
You have drawn a few curves by hand and suppose that each curve is symmetrical about the vertical line through a minimum point. Given ${\displaystyle f(x)=Ax^{2}+Bx+C}$ you suppose that ${\displaystyle f(p+q)=f(p-q)}$. {\displaystyle {\begin{aligned}A(p+q)(p+q)+B(p+q)+C=&\ A(p-q)(p-q)+B(p-q)+C\\\\A(pp+2pq+qq)+Bp+Bq=&\ A(pp-2pq+qq)+Bp-Bq\\\\A2pq+Bq=&\ -A2pq-Bq\\\\2A2pq+2Bq=&\ 0\\\\2q(2Ap+B)=&\ 0\\\\2Ap+B=&\ 0\\\\p=&\ {\frac {-B}{2A}}\end{aligned}}} You have found the stationary point without using calculus. Continue as per calculus below.
### By calculus
The derivative of ${\displaystyle (1)}$ is ${\displaystyle 2Ax+B}$ which equals ${\displaystyle 0}$ at a stationary point. At the stationary point ${\displaystyle x={\frac {-B}{2A}}}$. Prove that the function is symmetrical about the vertical line through ${\displaystyle x={\frac {-B}{2A}}}$. Let ${\displaystyle x={\frac {-B}{2A}}+p\ \dots \ (12)}$ Substitute (12) in (1) and expand: ${\displaystyle {\frac {4AApp+4AC-BB}{4A}}}$ Let ${\displaystyle x={\frac {-B}{2A}}-p\ \dots \ (13)}$ Substitute (13) in (1) and expand ${\displaystyle {\frac {4AApp+4AC-BB}{4A}}}$ The expansion is the same for both values of x, (12) and (13). The curve is symmetrical. If the function equals 0, then {\displaystyle {\begin{aligned}+4AApp+4AC-BB=&\ 0\\\\4AApp=&\ BB-4AC\\\\pp=&\ {\frac {BB-4AC}{4AA}}\\\\p=&\ {\frac {\pm {\sqrt {B^{2}-4AC}}}{2A}}\\\end{aligned}}} Substitute this value of p in (12) and the result is (2).
### By movement of the vertex
Begin with the basic quadratic: ${\displaystyle y=Ax^{2}}$. Move vertex from origin ${\displaystyle (0,0)}$ to ${\displaystyle (h,k)}$. ${\displaystyle y-k=A(x-h)^{2}}$. ${\displaystyle y-k=A(x^{2}-2hx+h^{2})}$. ${\displaystyle y-k=Ax^{2}-2Ahx+Ah^{2}}$. ${\displaystyle y=Ax^{2}-2Ahx+Ah^{2}+k}$. This equation is in the form of the quadratic ${\displaystyle y=Ax^{2}+Bx+C}$ where: ${\displaystyle B=-2Ah;\ C=Ah^{2}+k}$. Therefore ${\displaystyle h={\frac {-B}{2A}}}$ and ${\displaystyle h}$ is the X coordinate of the vertex in the new position. Continue as per calculus above.
### p + qi
Let ${\displaystyle x=p+qi}$ where ${\displaystyle i={\sqrt {-1}}}$ Substitute this value of ${\displaystyle x}$ into (1) and expand: {\displaystyle {\begin{aligned}A(p+qi)(p+qi)+B(p+qi)+C\\\\A(p^{2}+2pqi+(qi)^{2})+Bp+Bqi+C\\\\Ap^{2}+2Apqi-Aq^{2}+Bp+Bqi+C\end{aligned}}} Terms containing ${\displaystyle i=2Apqi+Bqi\ \dots \ (14)}$ From (14), ${\displaystyle 2Ap+B=0}$ and ${\displaystyle p={\frac {-B}{2A}}\ \dots \ (15)}$ Terms without ${\displaystyle i=Ap^{2}-Aq^{2}+Bp+C\ \dots \ (16)}$ From (15) and (16): {\displaystyle {\begin{aligned}Aq^{2}=&\ Ap^{2}+Bp+C\\\\=&\ A({\frac {-B}{2A}})({\frac {-B}{2A}})+B({\frac {-B}{2A}})+C\\\\q^{2}=&\ {\frac {B^{2}}{4AA}}-{\frac {B^{2}}{2AA}}+{\frac {C}{A}}\\\\=&\ {\frac {B^{2}}{4AA}}-{\frac {2B^{2}}{4AA}}+{\frac {4AC}{4AA}}\\\\=&\ {\frac {-B^{2}+4AC}{4AA}}\\\\=&\ {\frac {-1(B^{2}-4AC)}{4AA}}\\\\q=&\ {\frac {\pm \ i{\sqrt {B^{2}-4AC}}}{2A}}\\\\x=&\ (p)+(q)i\\\\=&\ ({\frac {-B}{2A}})\pm ({\frac {i{\sqrt {B^{2}-4AC}}}{2A}})i\\\\=&\ {\frac {-B\pm {\sqrt {B^{2}-4AC}}}{2A}}\end{aligned}}} This method shows the imaginary value ${\displaystyle i}$ coming into existence to help with intermediate calculations and then going away before the end result appears.
The quadratic function is usually defined as the familiar polynomial (1). However, the quadratic may be defined in other ways. Some are shown below:
### By three points
Figure 1: Diagram illustrating 2 quadratic curves that share 3 common points If three points on the curve are known, the familiar polynomial may be deduced. For example, if three points ${\displaystyle (-5,40),\ (4,13),\ (7,76)}$ are given and the three points satisfy ${\displaystyle y=f(x)}$, the values ${\displaystyle A,B,C}$ may be calculated. {\displaystyle {\begin{aligned}A(-5)(-5)+B(-5)+C=&\ 40\\A(4)(4)+B(4)+C=&\ 13\\A(7)(7)+B(7)+C=&\ 76\\25A-5B+C=&\ 40\ \dots \ (17)\\16A+4B+C=&\ 13\ \dots \ (18)\\49A+7B+C=&\ 76\ \dots \ (19)\\\end{aligned}}} The solution of the three equations ${\displaystyle (17),\ (18),\ (19)}$ gives the equation ${\displaystyle y=2x^{2}-x-15}$. If the three points were to satisfy ${\displaystyle x=f(y)}$, the equation would be ${\displaystyle x={\frac {2}{189}}y^{2}-{\frac {169}{189}}y+{\frac {2615}{189}}}$.
### By two points and a slope
Figure 1: Diagram illustrating quadratic curve defined by 2 points and slope at 1 point. Slope of curve at point ${\displaystyle (1,4.8)}$ is ${\displaystyle 3.2.}$ Given two points ${\displaystyle (-4,1.3)}$ and ${\displaystyle (1,4.8)}$ and the slope at ${\displaystyle (1,4.8)=3.2}$, calculate ${\displaystyle A,B,C}$. {\displaystyle {\begin{aligned}A(-4)(-4)+B(-4)+C=\ 1.3\\A(1)(1)+B(1)+C=\ 4.8\\\end{aligned}}} Slope = 2Ax + B, therefore {\displaystyle {\begin{aligned}2A(1)+B=&\ 3.2\\\\16A-4B+C=&\ 1.3\ \dots \ (20)\\A+B+C=&\ 4.8\ \dots \ (21)\\2A+B=&\ 3.2\ \dots \ (22)\\\end{aligned}}} The solution of the three equations ${\displaystyle (20),(21),(22)}$ gives the polynomial ${\displaystyle 0.5x^{2}+2.2x+2.1\ \dots \ (23)}$.
### By movement of the vertex
Begin with the basic quadratic ${\displaystyle y=x^{2}}$. If ${\displaystyle x}$ has the value ${\displaystyle p}$, then ${\displaystyle y=p^{2}}$ and the height of ${\displaystyle y}$ above the vertex = ${\displaystyle p^{2}}$. If we move the vertex to ${\displaystyle (h,k)}$, then the equation becomes ${\displaystyle (y-k)=(x-h)^{2}}$. If ${\displaystyle x}$ has the value ${\displaystyle h+p}$, then ${\displaystyle y-k=(h+p-h)^{2}=p^{2};\ y=p^{2}+k}$ and the height of ${\displaystyle y}$ above the vertex = ${\displaystyle y-k=p^{2}+k-k=p^{2}}$. The curve ${\displaystyle y=x^{2}}$ and the curve ${\displaystyle y-k=(x-h)^{2}}$ have the same shape. It's just that the vertex of the former ${\displaystyle (0,0)}$ has been moved to the vertex of the latter ${\displaystyle (h,k)}$. The latter equation expanded becomes ${\displaystyle y=x^{2}-2hx+h^{2}+k=x^{2}+(-2h)x+(h^{2}+k)}$. Consider function ${\displaystyle y=f(x)=5x^{2}+22x+21\ \dots \ (23a)}$. Therefore {\displaystyle {\begin{aligned}y=&\ ax^{2}\\\\y-k=&\ a(x-h)^{2}\\\\y-k=&\ a(x^{2}-2hx+h^{2})=ax^{2}-2ahx+ah^{2}\\\\y=&\ ax^{2}-2ahx+ah^{2}+k\\\\a=&\ 5\\\\-2ah=&\ 22;\ h={\frac {22}{-2(5)}}=-2.2\\\\ah^{2}+k=&\ 21\\\\k=&\ 21-5(-2.2)(-2.2)=-3.2\\\\\end{aligned}}} The example ${\displaystyle (23a)}$ may be expressed as {\displaystyle {\begin{aligned}y-(-3.2)=&\ 5(x-(-2.2))^{2}\\\\y+3.2=&\ 5(x+2.2)^{2}\end{aligned}}} For proof, expand: {\displaystyle {\begin{aligned}y+3.2=&\ 5(x+2.2)^{2}\\=&\ 5(x^{2}+4.4x+4.84)\\=&\ 5x^{2}+22x+24.2\\\\y=&\ 5x^{2}+22x+24.2-3.2\\=&\ 5x^{2}+22x+21\end{aligned}}}
### By compliance with the standard equation of the conic section
The quadratic function can comply with the format: ${\displaystyle Ax^{2}+Bxy+Cy^{2}+Dx+Ey+F=0.}$ (See The General Quadratic below.) For example, the function ${\displaystyle y=3x^{2}+5x-7}$ can be expressed as: ${\displaystyle (3)x^{2}+(0)xy+(0)y^{2}+(5)x+(-1)y+(-7)=0}$ or: ${\displaystyle 3x^{2}+5x-7-y=0.}$ To express a valid quadratic in this way, both ${\displaystyle A,E}$ or both ${\displaystyle C,D}$ must be non-zero.
### By a point and a straight line
The point is called the focus and the line is called the directrix. The distance from point to line is non-zero. The quadratic is the locus of a point that is equidistant from both focus and line at all times. When the quadratic is defined in this way, it is usually called a parabola.
Let the ${\displaystyle focus}$ have coordinates ${\displaystyle (p,q).}$
Let the ${\displaystyle directrix}$ have equation: ${\displaystyle y=k.}$
Let the point ${\displaystyle (x,y)}$ be equidistant from both focus and directrix.
Distance from ${\displaystyle (x,y)}$ to focus ${\displaystyle ={\sqrt {(x-p)^{2}+(y-q)^{2}}}}$.
Distance from ${\displaystyle (x,y)}$ to directrix ${\displaystyle =y-k}$.
By definition these two lengths are equal.
${\displaystyle {\sqrt {(x-p)^{2}+(y-q)^{2}}}=y-k}$
${\displaystyle x^{2}-2px+p^{2}+y^{2}-2qy+q^{2}=y^{2}-2ky+k^{2}}$
${\displaystyle (2q-2k)y=x^{2}-2px+p^{2}+q^{2}-k^{2}}$
${\displaystyle y={\frac {x^{2}}{2q-2k}}+{\frac {-2px}{2q-2k}}+{\frac {p^{2}+q^{2}-k^{2}}{2q-2k}}}$
Let this equation have the form: ${\displaystyle y=Ax^{2}+Bx+C}$
Therefore:
{\displaystyle {\begin{aligned}A&={\frac {1}{2q-2k}}\\B&={\frac {-2p}{2q-2k}}\\C&={\frac {p^{2}+q^{2}-k^{2}}{2q-2k}}\end{aligned}}}
Given ${\displaystyle A,B,C}$ calculate ${\displaystyle p,q,k}$.
${\displaystyle B={\frac {-2p}{2q-2k}}=-2pA;\ p={\frac {-B}{2A}}}$
There are two equations with two unknowns ${\displaystyle q,k:}$
{\displaystyle {\begin{aligned}A&(2q-2k)-1=0\\C&(2q-2k)-p^{2}-q^{2}+k^{2}=0\end{aligned}}}
The solutions are:
{\displaystyle {\begin{aligned}p&={\frac {-B}{2A}}\\q&={\frac {1-(B^{2}-4AC)}{4A}}\\k&={\frac {-1-(B^{2}-4AC)}{4A}}\end{aligned}}}
If the quadratic equation is expressed as ${\displaystyle y=Ax^{2}+Bx+C}$ then:
The focus is the point ${\displaystyle ({\frac {-B}{2A}},{\frac {1-(B^{2}-4AC)}{4A}})}$, and
The directrix has equation: ${\displaystyle y={\frac {-1-(B^{2}-4AC)}{4A}}}$.
The ${\displaystyle vertex}$ is exactly half-way between focus and directrix.
Vertex is the point ${\displaystyle ({\frac {-B}{2A}},{\frac {-(B^{2}-4AC)}{4A}})}$.
${\displaystyle A={\frac {1}{2q-2k}};\ 2q-2k={\frac {1}{A}};\ q-k={\frac {1}{2A}}=}$ distance from directrix to focus.
Distance from vertex to focus ${\displaystyle ={\frac {1}{4A}}}$.
If the curve has equation ${\displaystyle y=Ax^{2}}$, then the vertex is at the origin ${\displaystyle (0,0)}$.
If the focus is the point ${\displaystyle (0,q)}$, then ${\displaystyle q={\frac {1}{4A}};\ A={\frac {1}{4q}}}$ and the equation ${\displaystyle y=Ax^{2}}$ becomes ${\displaystyle y={\frac {x^{2}}{4q}}}$.
### An example with vertical focus
Figure 3: Graph of quadratic function with vertical focus ${\displaystyle 8(y+1)=(x-4)^{2}}$ showing : * vertex at (4,-1), * focus at (4,1), * directrix at y = -3. Let ${\displaystyle (p,q)=(4,1),\ k=-3.}$ Directrix has equation: ${\displaystyle y=-3}$. Focus has coordinates ${\displaystyle (4,1)}$. {\displaystyle {\begin{aligned}A&={\frac {1}{2q-2k}}={\frac {1}{2(1)-2(-3)}}={\frac {1}{2+6}}={\frac {1}{8}}\\B&={\frac {-2p}{2q-2k}}={\frac {-2(4)}{8}}=-1\\C&={\frac {p^{2}+q^{2}-k^{2}}{2q-2k}}={\frac {16+1-9}{8}}=1\end{aligned}}} This example has equation: ${\displaystyle y={\frac {1}{8}}x^{2}-x+1}$ or ${\displaystyle 8y=x^{2}-8x+8}$ or ${\displaystyle 8(y+1)=(x-4)^{2}}$. See Figure 3. Distance from vertex to focus = ${\displaystyle {\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}={\frac {8}{4}}=2.}$ Or: Vertex has coordinates ${\displaystyle (4,-1).}$ Distance from vertex to focus ${\displaystyle =2={\frac {1}{4A}};\ A={\frac {1}{8}}}$. Curve has shape of ${\displaystyle y={\frac {1}{8}}x^{2}}$ with vertex moved to ${\displaystyle (4,-1).\ y-(-1)={\frac {1}{8}}(x-4)^{2};\ 8(y+1)=(x-4)^{2}.}$
Let the ${\displaystyle focus}$ have coordinates ${\displaystyle (p,q).}$
Let the ${\displaystyle directrix}$ have equation: ${\displaystyle x=k.}$
Let the point ${\displaystyle (x,y)}$ be equidistant from both focus and directrix.
Distance from ${\displaystyle (x,y)}$ to focus ${\displaystyle ={\sqrt {(x-p)^{2}+(y-q)^{2}}}}$.
Distance from ${\displaystyle (x,y)}$ to directrix ${\displaystyle =x-k}$.
By definition these two lengths are equal.
${\displaystyle {\sqrt {(x-p)^{2}+(y-q)^{2}}}=x-k}$
${\displaystyle x^{2}-2px+p^{2}+y^{2}-2qy+q^{2}=x^{2}-2kx+k^{2}}$
${\displaystyle (2p-2k)x=y^{2}-2qy+p^{2}+q^{2}-k^{2}}$
${\displaystyle x={\frac {y^{2}}{2p-2k}}+{\frac {-2qy}{2p-2k}}+{\frac {p^{2}+q^{2}-k^{2}}{2p-2k}}}$
Let this equation have the form: ${\displaystyle x=Ay^{2}+By+C}$
Therefore:
{\displaystyle {\begin{aligned}A&={\frac {1}{2p-2k}}\\B&={\frac {-2q}{2p-2k}}\\C&={\frac {p^{2}+q^{2}-k^{2}}{2p-2k}}\end{aligned}}}
Given ${\displaystyle A,B,C}$ calculate ${\displaystyle p,q,k.}$
${\displaystyle B={\frac {-2q}{2p-2k}}=-2qA;\ q={\frac {-B}{2A}}}$
There are two equations with two unknowns ${\displaystyle p,k:}$
{\displaystyle {\begin{aligned}A&(2p-2k)-1=0\\C&(2p-2k)-p^{2}-q^{2}+k^{2}=0\end{aligned}}}
The solutions are:
{\displaystyle {\begin{aligned}p&={\frac {1-(B^{2}-4AC)}{4A}}\\q&={\frac {-B}{2A}}\\k&={\frac {-1-(B^{2}-4AC)}{4A}}\end{aligned}}}
If the quadratic equation is expressed as ${\displaystyle x=Ay^{2}+By+C}$ then:
The focus is the point ${\displaystyle ({\frac {1-(B^{2}-4AC)}{4A}},{\frac {-B}{2A}})}$, and
The directrix has equation: ${\displaystyle x={\frac {-1-(B^{2}-4AC)}{4A}}}$.
The ${\displaystyle vertex}$ is exactly half-way between focus and directrix.
Vertex is the point ${\displaystyle ({\frac {-(B^{2}-4AC)}{4A}},{\frac {-B}{2A}})}$.
${\displaystyle A={\frac {1}{2p-2k}};\ 2p-2k={\frac {1}{A}};\ p-k={\frac {1}{2A}}=}$ distance from directrix to focus.
Distance from vertex to focus ${\displaystyle ={\frac {1}{4A}}}$.
If the curve has equation ${\displaystyle x=Ay^{2}}$, then the vertex is at the origin ${\displaystyle (0,0)}$.
If the focus is the point ${\displaystyle (p,0)}$, then ${\displaystyle p={\frac {1}{4A}};\ A={\frac {1}{4p}}}$ and the equation ${\displaystyle x=Ay^{2}}$ becomes ${\displaystyle x={\frac {y^{2}}{4p}}}$.
#### An example with horizontal focus
Let ${\displaystyle (p,q)=(1,4),\ k=-3.}$
Directrix has equation: ${\displaystyle x=-3}$. Focus has coordinates ${\displaystyle (1,4)}$.
{\displaystyle {\begin{aligned}A&={\frac {1}{2p-2k}}={\frac {1}{2(1)-2(-3)}}={\frac {1}{8}}\\B&={\frac {-2q}{2p-2k}}={\frac {-2(4)}{8}}=-1\\C&={\frac {p^{2}+q^{2}-k^{2}}{2p-2k}}={\frac {1+16-9}{8}}=1\end{aligned}}}
This example has equation: ${\displaystyle x={\frac {1}{8}}y^{2}-y+1}$ or ${\displaystyle 8x=y^{2}-8y+8}$ or ${\displaystyle 8(x+1)=(y-4)^{2}}$. See Figure 4.
Distance from vertex to focus = ${\displaystyle {\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}={\frac {8}{4}}=2.}$
Given equation ${\displaystyle 8x=y^{2}-8y+8}$ calculate ${\displaystyle p,q,k}$.
Method 1. By algebra Put equation in form: ${\displaystyle x={\frac {1}{8}}y^{2}-y+1}$ where ${\displaystyle A={\frac {1}{8}},\ B=-1,\ C=1.}$ {\displaystyle {\begin{aligned}p&={\frac {1-(B^{2}-4AC)}{4A}}={\frac {1-(1-{\frac {1}{2}})}{4({\frac {1}{8}})}}={\frac {1-{\frac {1}{2}}}{\frac {1}{2}}}=1\\q&={\frac {-B}{2A}}={\frac {-(-1)}{2({\frac {1}{8}})}}={\frac {8}{2}}=4\\k&={\frac {-1-(B^{2}-4AC)}{4A}}={\frac {-1-(1-{\frac {1}{2}})}{\frac {1}{2}}}={\frac {-1{\frac {1}{2}}}{\frac {1}{2}}}=-3\end{aligned}}}
Method 2. By analytical geometry Distance from vertex to focus ${\displaystyle ={\frac {1}{4A}}={\frac {1}{4({\frac {1}{8}})}}=2.}$ Put equation in ${\displaystyle semi}$-${\displaystyle reduced}$ form: {\displaystyle {\begin{aligned}8x=y^{2}-8y+8\\(y-4)^{2}=y^{2}-8y+16\\8x=(y-4)^{2}-8\\8x+8=(y-4)^{2}\\8(x+1)=(y-4)^{2}\end{aligned}}} Vertex is point ${\displaystyle (-1,4).}$ Focus is point ${\displaystyle (-1+2,4)=(1,4)=(p,q).}$ Directrix has equation: ${\displaystyle x=-1-2=-3=k.}$
## The Parabola
See Figure 1. For simplicity values are defined as shown. Let an arbitrary point on the curve be ${\displaystyle (x,y)}$.
By definition, ${\displaystyle {\sqrt {(x-0)^{2}+(y-p)^{2}}}=y+p}$. This expression expanded gives:
${\displaystyle x^{2}-4py=0;\ y={\frac {x^{2}}{4p}}}$ and slope = ${\displaystyle {\frac {x}{2p}}}$.
If the equation of the curve is expressed as: ${\displaystyle y=Kx^{2}}$, then ${\displaystyle K={\frac {1}{4p}};\ 4Kp=1;\ p={\frac {1}{4K}}}$.
Let a straight line through the focus intersect the parabola in two points ${\displaystyle (x_{1},y_{1})}$ and ${\displaystyle (x_{2},y_{2})}$.
{\displaystyle {\begin{aligned}y=&\ {\frac {x^{2}}{4p}}=mx+p\\\\x^{2}=&\ 4pmx+4pp\\\\x^{2}-4pmx-4pp=&\ 0\\\\x=&\ {\frac {4pm\pm {\sqrt {16ppmm+16pp}}}{2}}\\\\=&\ {\frac {4pm\pm 4p{\sqrt {mm+1}}}{2}}\\\\=&\ 2pm\pm 2p{\sqrt {mm+1}}\\\\=&\ 2pm\pm 2pR\\\end{aligned}}}
where
${\displaystyle m}$ is the slope of line DB in Figure 1.
{\displaystyle {\begin{aligned}R=&\ {\sqrt {m^{2}+1}}\\\\x_{1}=&\ 2pm+2pR\\\\x_{2}=&\ 2pm-2pR\\\\y_{1}=&\ {\frac {x_{1}^{2}}{4p}}=2Rmp+2mmp+p\\\\y_{2}=&\ -2Rmp+2mmp+p\end{aligned}}}
### Characteristics of the Parabola
The parabola is a grab-bag of many interesting facts.
#### Two tangents perpendicular
Figure 2: The Parabola ${\displaystyle y={\frac {x^{2}}{4}}}$ Directrix is line ${\displaystyle y=-1}$ Tangents ${\displaystyle EDA}$ and ${\displaystyle CBA}$ intersect at point ${\displaystyle A}$ where they are perpendicular. We prove first that the tangents at ${\displaystyle (x_{1},y_{1})}$ and ${\displaystyle (x_{2},y_{2})}$ are perpendicular. {\displaystyle {\begin{aligned}s=&\ {\frac {x}{2p}}\\\\s1=&\ {\frac {x_{1}}{2p}}=m+R\\\\s2=&\ m-R\end{aligned}}} The product of ${\displaystyle s_{1}}$ and ${\displaystyle s_{2}=(m+R)(m-R)=m^{2}-R^{2}=m^{2}-(m^{2}+1)=-1}$. Therefore, the tangents (lines AB and AD in Figure 2) are perpendicular.
#### Two tangents intersect on directrix
Second, we prove that the two tangents intersect on the directrix. See Figure 2 above. Using ${\displaystyle y=mx+c}$ and ${\displaystyle c=y-mx}$: {\displaystyle {\begin{aligned}c_{1}=&\ y_{1}-(s_{1})(x_{1})\\\\=&\ 2Rmp+2mmp+p-(m+r)(2pm+2pR)\\\\=&\ -2Rmp-2mmp-p\\\\c_{2}=&\ 2Rmp-2mmp-p\end{aligned}}} The ${\displaystyle y}$ coordinate of the point of intersection satisfies both ${\displaystyle s_{1}x+c_{1}}$ and ${\displaystyle s_{2}x+c_{2}}$. Therefore, {\displaystyle {\begin{aligned}(m+R)x+(-2Rmp-2mmp-p)=&\ (m-R)x+(2Rmp-2mmp-p)\\\\x=&\ 2mp\\\end{aligned}}} ${\displaystyle 2mp}$ is the mid-point between ${\displaystyle x_{1}}$ and ${\displaystyle x_{2}}$. Point A in Figure 1 has coordinates ${\displaystyle (2mp,-p)}$ Check our work: {\displaystyle {\begin{aligned}y=s_{1}x+c_{1}=(m+R)2mp+(-2Rmp-2mmp-p)=-p\\y=s_{2}x+c_{2}=(m-R)2mp+(+2Rmp-2mmp-p)=-p\\\end{aligned}}} The tangents intersect at ${\displaystyle (2mp,-p)}$. They intersect on the directrix where ${\displaystyle y=-p}$. See Tangents perpendicular and oscillating.
#### Right angle at focus
Figure 3: Diagram illustrating right triangle ${\displaystyle GFH}$ with right angle at focus ${\displaystyle F.}$ Third, we prove that the triangle defined by the three points ${\displaystyle H\ (x1,-p),G\ (x2,-p)}$ and ${\displaystyle F\ (0,p)}$ is a right triangle. Slope of line ${\displaystyle (0,p)...(x_{1},-p)=s_{3}={\frac {p-(-p)}{0-x_{1}}}={\frac {2p}{-(2pm+2pR)}}={\frac {-1}{m+R}}}$. Slope of line ${\displaystyle (0,p)...(x_{2},-p)=s_{4}={\frac {p-(-p)}{0-x_{2}}}={\frac {2p}{-(2pm-2pR)}}={\frac {-1}{m-R}}}$. ${\displaystyle (s_{3})*(s_{4})={\frac {-1}{m+R}}\ *\ {\frac {-1}{m-R}}={\frac {1}{m^{2}-R^{2}}}={\frac {1}{-1}}=-1}$ The product of ${\displaystyle s_{3}}$ and ${\displaystyle s_{4}}$ is ${\displaystyle -1}$. Therefore the two sides ${\displaystyle HF,GF}$ are perpendicular and the triangle ${\displaystyle HFG}$ in Figure 1 is a right triangle with the right angle at ${\displaystyle F}$.
#### Two lines perpendicular
Fourth, we prove that the two lines ${\displaystyle AF,DFB}$ are perpendicular.
Point ${\displaystyle A:\ (2mp,-p)}$. Point ${\displaystyle F:(0,p)}$.
Using slope ${\displaystyle ={\frac {y_{1}-y_{2}}{x_{1}-x_{2}}}}$
Slope of line ${\displaystyle AF=s_{5}={\frac {p-(-p)}{0-2mp}}={\frac {2p}{-2mp}}={\frac {-1}{m}}}$.
Slope of line ${\displaystyle DFB=s_{6}=m.}$
${\displaystyle (s_{5})(s_{6})=-1.}$ Therefore the two lines ${\displaystyle AF,DFB}$ are perpendicular.
Because ${\displaystyle \angle AGD=\angle AFD=90^{\circ },}$ points ${\displaystyle A,G,D,F}$ are on a circle and lengths ${\displaystyle AG=AF=AH}$
In the last section we proved several points about the parabola, beginning with line ${\displaystyle DFB}$ and moving towards point ${\displaystyle A}$ on the directrix. In this section, we prove the reverse, beginning with point ${\displaystyle A}$ and moving towards line ${\displaystyle DFB}$.
Let ${\displaystyle (k,-p)}$ be any point on the directrix ${\displaystyle y=-p}$.
Using ${\displaystyle y=mx+c,-p=mk+c,\ c=-p-mk}$ and any line through ${\displaystyle (k,-p)}$ is defined as ${\displaystyle y=sx-p-sk}$ where ${\displaystyle s}$ is the slope of the line.
Let this line intersect the parabola ${\displaystyle y={\frac {x^{2}}{4p}}}$. (In Figure 1, p = 1.)
{\displaystyle {\begin{aligned}y&\ ={\frac {x^{2}}{4p}}=sx-p-sk\\\\x^{2}&\ =4psx-4pp-4psk\\\\x^{2}&\ -4psx+4pp+4psk=0\ \dots \ (24)\\\\x^{2}&\ +(-4ps)x+(4pp+4psk)=0\end{aligned}}}
The above defines the ${\displaystyle X}$ coordinate/s of any line through ${\displaystyle (k,-p)}$ that intersect/s the parabola. We are interested in the tangent, a line that intersects in only one place. Therefore, the discriminant is 0.
{\displaystyle {\begin{aligned}16ppss-4(4pp+4psk)=0\\\\16ppss-16(pp+psk)=0\\\\ppss-pks-pp=0\\\\pss-ks-p=0\\\\s=\ {\frac {k\pm {\sqrt {kk+4pp}}}{2p}}=\ {\frac {k\pm R}{2p}}\end{aligned}}}
where ${\displaystyle R={\sqrt {kk+4pp}}}$
Slope of tangent1 = ${\displaystyle s_{1}={\frac {k+R}{2p}}}$ (In Figure 1, tangent1 is the line ${\displaystyle ABC}$.)
Slope of tangent2 = ${\displaystyle s_{2}={\frac {k-R}{2p}}}$ (In Figure 1, tangent2 is the line ${\displaystyle ADE}$.)
Prove that tangent1 and tangent2 are perpendicular.
${\displaystyle s_{1}*s_{2}={\frac {k+R}{2p}}*{\frac {k-R}{2p}}={\frac {kk-RR}{4pp}}={\frac {kk-(kk+4pp)}{4pp}}={\frac {-4pp}{4pp}}=-1}$
The product of the two slopes is -1. Therefore, the two tangents are perpendicular.
From (24), we chose a value of ${\displaystyle s}$ that made the discriminant 0. Therefore
{\displaystyle {\begin{aligned}x=&\ {\frac {-B}{2A}}={\frac {4ps}{2}}=2ps\\\\x_{1}=&\ 2ps_{1}=2p*{\frac {k+R}{2p}}=k+R\\\\x_{2}=&\ k-R\\\\y=&\ {\frac {x^{2}}{4p}}\\\\y_{1}=&\ {\frac {(k+R)^{2}}{4p}}={\frac {kk+2kR+RR}{4p}}={\frac {kk+2kR+kk+4pp}{4p}}={\frac {2kk+2kR+4pp}{4p}}={\frac {kk+kR+2pp}{2p}}\\\\y_{2}=&\ {\frac {kk-kR+2pp}{2p}}\\\\m=&\ {\frac {y_{1}-y_{2}}{x_{1}-x_{2}}}={\frac {kR/p}{2R}}={\frac {k}{2p}}\end{aligned}}}
(In Figure 1, ${\displaystyle m}$ is the slope of line ${\displaystyle DFB}$. This statement agrees with ${\displaystyle k=2mp}$ proved in the last section.)
We have a line joining the two points ${\displaystyle (x1,y1),(x_{2},y_{2})}$. Calculate the intercept on the ${\displaystyle Y}$ axis.
Using ${\displaystyle y=mx+c}$,
{\displaystyle {\begin{aligned}c=&\ y-mx=y_{1}-mx_{1}\\\\=&\ {\frac {kk+kR+2pp}{2p}}-{\frac {k}{2p}}*(k+R)\\\\=&\ {\frac {kk+kR+2pp-(kk+kR)}{2p}}\\\\=&\ {\frac {2pp}{2p}}=p\end{aligned}}}
The line joining the two points ${\displaystyle (x_{1},y_{1}),\ (x_{2},y_{2})}$ passes through the focus ${\displaystyle (0,p)}$.
### Two lines parallel
Figure 2: The Parabola ${\displaystyle y={\frac {x^{2}}{4}}}$ Lines ${\displaystyle JGK,DFB}$ are parallel. Line ${\displaystyle AGH}$ divides area ${\displaystyle DGBHFD}$ into two halves equal by area. In Figure 2 tangents ${\displaystyle AB}$ and ${\displaystyle AD}$ intersect at point ${\displaystyle A}$ on the directrix. Line ${\displaystyle AGH}$ has value ${\displaystyle x=k}$. Line ${\displaystyle JGK}$ is tangent to the curve at ${\displaystyle G}$. Slope of tangent ${\displaystyle JGK={\frac {x}{2p}}={\frac {k}{2p}}}$. Slope of line ${\displaystyle DFB}$ also ${\displaystyle ={\frac {k}{2p}}}$. Therefore two lines ${\displaystyle JGK,DFB}$ are parallel.
### Area under focal chord
Area ${\displaystyle DGBHFD}$
${\displaystyle x_{1}=2pm+2pR}$ ${\displaystyle x_{2}=2pm-2pR}$ where ${\displaystyle R={\sqrt {m^{2}+1}}}$ Line ${\displaystyle DFB=mx+p}$. The integral of this value ${\displaystyle =m{\frac {x^{2}}{2}}+px}$. Area under line ${\displaystyle DFB}$ {\displaystyle {\begin{aligned}x_{1}&\\=\ \ \ \ &[m{\frac {x^{2}}{2}}+px]=8Rmmpp+4Rpp\\x_{2}&\end{aligned}}} Area under curve ${\displaystyle DGB}$ {\displaystyle {\begin{aligned}x_{1}&\\=\ \ \ \ &[{\frac {x^{3}}{12p}}]={\frac {16Rmmpp+4Rpp}{3}}\\x_{2}&\end{aligned}}} Area ${\displaystyle DGBHFD}$ {\displaystyle {\begin{aligned}=&\ 8Rmmpp+4Rpp-{\frac {16Rmmpp+4Rpp}{3}}\\\\=&\ {\frac {24Rmmpp+12Rpp-16Rmmpp-4Rpp}{3}}\\\\=&\ {\frac {8Rmmpp+8Rpp}{3}}\\\\=&\ {\frac {8Rp^{2}(m^{2}+1)}{3}}\end{aligned}}}
Area ${\displaystyle DGHD}$
Similarly it can be shown that Area ${\displaystyle DGHD={\frac {4Rp^{2}(m^{2}+1)}{3}}}$ Therefore line ${\displaystyle AGH}$ splits area ${\displaystyle DGBHFD}$ into two halves equal by area.
## Reflectivity of the Parabola
Figure 1: The Parabola ${\displaystyle y={\frac {x^{2}}{4p}}}$ Focus at point ${\displaystyle F\ (0,p)}$ Vertex at origin ${\displaystyle (0,0)}$ Directrix is line ${\displaystyle y=-p}$ By definition ${\displaystyle BF=BH}$ and ${\displaystyle DF=DG}$ In Figure 1 ${\displaystyle p=1}$ See Figure 1. ${\displaystyle \triangle GFH}$ is a right triangle and point ${\displaystyle A}$ is the midpoint of line ${\displaystyle GAH}$. ${\displaystyle \therefore \ AG=AF=AH,\ \triangle ABF}$ is congruent with ${\displaystyle \triangle ABH}$, and ${\displaystyle \angle ABF=\angle ABH}$. ${\displaystyle \angle CBJ=\angle ABH,\ \therefore \angle ABF=\angle CBJ}$. 1. Any ray of light emanating from a point source at F touches the parabola at B and is reflected away from B on a line that is always perpendicular to the directrix. ${\displaystyle \angle ABF}$ is the angle of incidence and ${\displaystyle \angle CBJ}$ is the angle of reflection. 2. The path from ${\displaystyle K}$ through focus to vertex and back to focus has length ${\displaystyle JH}$. The path from ${\displaystyle J}$ to ${\displaystyle B}$ to ${\displaystyle F}$ has length ${\displaystyle JH.\ \therefore }$ all paths to and from focus have the same length.
In Theory
1) The quadratic may be used to examine itself.
Let a quadratic equation be: ${\displaystyle y=x^{2}}$.
Let the equation of a line be: ${\displaystyle y=mx+c}$.
Let the line intersect the quadratic at ${\displaystyle (x_{1},y_{1})}$.
Therefore:
{\displaystyle {\begin{aligned}y_{1}=mx_{1}+c\\c=y_{1}-mx_{1}\\y=x^{2}=mx+c=mx+(y_{1}-mx_{1})\\x^{2}-mx-y_{1}+mx_{1}=0\end{aligned}}}
Let the line intersect the curve in exactly one place. Therefore ${\displaystyle x}$ must have exactly one value and the discriminant is ${\displaystyle 0}$.
{\displaystyle {\begin{aligned}m^{2}-4(mx_{1}-y_{1})=0\\m^{2}-4mx_{1}+4y_{1}=0\\m^{2}-4mx_{1}+4x_{1}^{2}=0\\(m-2x_{1})^{2}=0\\m=2x_{1}\end{aligned}}}
A line that touches the curve at ${\displaystyle x_{1},y_{1}}$ has slope ${\displaystyle 2x_{1}}$.
Therefore the slope of the curve at ${\displaystyle x_{1},y_{1}}$ is ${\displaystyle 2x_{1}}$. This examination of the curve has produced the slope of the curve without using calculus.
Consider the curve: ${\displaystyle y=Ax^{2}+Bx+C}$. The aim is to calculate the slope of the curve at an arbitrary point ${\displaystyle (x_{1},y_{1})}$.
{\displaystyle {\begin{aligned}Ax^{2}+Bx+C=mx+c\\Ax^{2}+Bx+C=mx+(y_{1}-mx_{1})\\Ax^{2}+Bx+C-mx-y_{1}+mx_{1}=0\\Ax^{2}+Bx-mx+C-y_{1}+mx_{1}=0\\Ax^{2}+(B-m)x+(C-y_{1}+mx_{1})=0\end{aligned}}}
If ${\displaystyle x}$ is to have exactly one value, discriminant ${\displaystyle =(B-m)^{2}-4(A)(C-y_{1}+mx_{1})=0}$.
Therefore ${\displaystyle BB-2Bm+mm-4AC+4Ay_{1}-4Amx_{1}=0}$
${\displaystyle mm-4Ax_{1}m-2Bm+BB-4AC+4Ay_{1}=0}$
${\displaystyle mm-(4Ax_{1}+2B)m+(BB-4AC+4Ay_{1})=0}$
${\displaystyle m={\frac {(4Ax_{1}+2B)\pm {\sqrt {(4Ax_{1}+2B)^{2}-4(BB-4AC+4Ay_{1})}}}{2}}}$
${\displaystyle m={\frac {(4Ax_{1}+2B)\pm {\sqrt {16AAx_{1}x_{1}+16ABx_{1}+4BB-4BB+16AC-16Ay_{1}}}}{2}}}$
${\displaystyle m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {AAx_{1}x_{1}+ABx_{1}+AC-Ay_{1}}}}{2}}}$
${\displaystyle m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {A(Ax_{1}^{2}+Bx_{1}+C-y_{1})}}}{2}}}$
${\displaystyle m={\frac {(4Ax_{1}+2B)\pm 4{\sqrt {A(0)}}}{2}}}$
${\displaystyle m=2Ax_{1}+B.}$
The slope of the curve at an arbitrary point ${\displaystyle (x_{1},y_{1})=2Ax_{1}+B}$.
2) The quadratic may be used to examine other curves, for example, the circle.
Define a circle of radius 5 at the origin:
${\displaystyle {\sqrt {x^{2}+y^{2}}}=5}$
${\displaystyle x^{2}+y^{2}=25}$
Move the circle to ${\displaystyle (8,3)}$
${\displaystyle (x-8)^{2}+(y-3)^{2}=25}$
${\displaystyle x^{2}-16x+64+y^{2}-6y+9=25}$
${\displaystyle x^{2}-16x+y^{2}-6y+48=0}$
We want to know the values of ${\displaystyle x}$ that contain the circle, that is, the values of ${\displaystyle x}$ for each of which there is only one value of ${\displaystyle y}$.
Put the equation of the circle into a quadratic in ${\displaystyle y}$.
${\displaystyle y^{2}-6y+x^{2}-16x+48=0}$
${\displaystyle A=1,\ B=-6,\ C=x^{2}-16x+48}$
There is exactly one value of ${\displaystyle y}$ if the discriminant is ${\displaystyle 0}$. Therefore
${\displaystyle (-6)(-6)-4(x^{2}-16x+48)=0}$
${\displaystyle 36-4(x^{2}-16x+48)=0}$
${\displaystyle -9+x^{2}-16x+48=0}$
${\displaystyle x^{2}-16x+39=0}$
${\displaystyle (x-3)(x-13)=0}$
${\displaystyle x_{1}=3,\ x_{2}=13}$
These values of ${\displaystyle x}$ make sense because we expect the values of ${\displaystyle x}$ to be ${\displaystyle 8\pm 5}$. This process has calculated a minimum point and a maximum point without calculus.
3) The formula remains valid for ${\displaystyle B}$ and/or ${\displaystyle C}$ equal to ${\displaystyle 0}$. Under these conditions you probably won't need the formula. For example ${\displaystyle Ax^{2}+Bx}$ can be factored by inspection as ${\displaystyle x(Ax+b)}$.
4) The quadratic can be used to solve functions of higher order.
One of the solutions of the cubic depends on the solution of a sextic in the form ${\displaystyle Ax^{6}+Bx^{3}+C=0}$. This is the quadratic ${\displaystyle AX^{2}+BX+C}$ where ${\displaystyle X=x^{3}}$.
The cubic function ${\displaystyle x^{3}+6x^{2}+13x+10}$ produces the depressed function ${\displaystyle t^{3}+9t=t(t^{2}+9)}$.
The quadratic ${\displaystyle t^{2}+9=0}$ is solved as ${\displaystyle t^{2}=-9,\ t={\sqrt {-9}}=\pm 3i}$. ${\displaystyle }$ The roots of the depressed function are ${\displaystyle t_{1}=0,\ t_{2}=3i,\ t_{3}=-3i}$.
Using ${\displaystyle x={\frac {-B+t}{3A}}}$
${\displaystyle x_{1}={\frac {-6+0}{3}}=-2}$
${\displaystyle x_{2}={\frac {-6+3i}{3}}=-2+i}$
${\displaystyle x_{3}=-2-i}$
In this example the quadratic, as part of the depressed function, simplified the solution of the cubic.
The quartic function ${\displaystyle x^{4}+4x^{3}+17x^{2}+26x+11}$ produces the depressed function ${\displaystyle t^{4}+176t^{2}-256}$ which is the quadratic ${\displaystyle T^{2}+176T-256}$ where ${\displaystyle T=t^{2}}$.
5) The quadratic appears in Newton's Laws of Motion: ${\displaystyle s=ut+{\frac {1}{2}}at^{2}}$
Graph of quadratic function ${\displaystyle y=x^{2}+4x-5}$ showing basic features : * X and Y intercepts * vertex at (-2,-9), * axis of symmetry at x = -2. Graph of quadratic function ${\displaystyle y=x^{2}-2x-3}$ showing : * X and Y intercepts in red, * vertex and axis of symmetry in blue, * focus and directrix in pink.
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# How to factorize a cubic equation?
How should I factor this polynomial: $x^3 - x^2 - 4x - 6$
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Rational root test – Martin Sleziak Dec 16 '12 at 7:14
Typically when you have a polynomial of the form $$f(x) = x^n + a_1 x^{n-1} + a_2 x^{n-2} + \cdots + a_n$$ where $a_k \in \mathbb{Z}$, to factorize it, it is a good idea to first plug in the values of the divisors of $a_n$ in $f(x)$ with the hope that it will evaluate to $0$. In your case, the polynomial is $$f(x) = x^3 - x^2 - 4x - 6$$ The divisors of $-6$ are $\{\pm1, \pm2,\pm3 \}$. We find that $f(\pm1) \neq 0$, $f(\pm2) \neq 0$ and $f(-3) \neq 0$, while $f(3) = 0$.
Hence, we have $f(x) = (x-3) (x^2 + ax + b)$. We get that $-3b = -6$ and $a-3 = -1$. Hence, we get that $b = 2$ and $a = 2$. Hence, $$f(x) = (x-3)(x^2 + 2x + 2) = (x-3)((x+1)^2 + 1) = (x-3) (x+1-i) (x+1+i)$$
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I dont see how you get a-3 = -1 ? – Ghost Dec 16 '12 at 4:58
I get $-3b = -6$ by comparing the constant term and $a-3 = -1$ by comparing the coefficient of $x^2$. – user17762 Dec 16 '12 at 4:59
@Ghost $i$ is such that $i^2 = -1$. Hence, $$(x+1)^2 + 1 = (x+1)^2 - i^2 = (x+1-i)(x+1+i)$$ since $a^2 - b^2 = (a-b)(a+b)$ – user17762 Dec 16 '12 at 5:42
Thanks :) I really appreciate it . Although, it's not possible for i^2 to be -1 . So how does that help us? Doesn't it just complicate things? – Ghost Dec 16 '12 at 5:49
Hint: the polynomial evaluates to zero when $x=3$.
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Graph the polynomial: it appears to have a zero at $x=3$ (and this appears to be the only zero). So divide your polynomial by $x-3$ (by long division) and see what you are left with.
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# Mensuration - Important Concepts and Notes
By Abhinav Gupta|Updated : August 24th, 2021
Mensuration - Important Concepts and Notes
Important Formulas on Quadrilateral and Circle
Rectangle
A four-sided shape that is made up of two pairs of parallel lines and that has four right angles; especially: a shape in which one pair of lines is longer than the other pair.
The diagonals of a rectangle bisect each other and are equal.
Area of rectangle = length x breadth = l x b
OR Area of rectangle = if one sides (l) and diagonal (d) are given.
OR Area of rectangle = if perimeter (P) and diagonal (d) are given.
Perimeter (P) of rectangle = 2 (length + breadth) = 2 (l + b).
OR Perimeter of rectangle = if one side (l) and diagonal (d) are given.
Square
A four-sided shape that is made up of four straight sides that are the same length and that has four right angles.
The diagonals of a square are equal and bisect each other at 900.
Area (a) of a square
Perimeter (P) of a square
= 4a, i.e. 4 x side
Length (d) of the diagonal of a square
Circle
A circle is a path travelled by a point which moves in such a way that its distance from a fixed point remains constant.
The fixed point is known as the centre and the fixed distance is called the radius.
(a) Circumference or perimeter of circle =
where r is radius and d is the diameter of the circle
(b) Area of circle
is circumference
(c) The radius of circle =
Sector :
A sector is a figure enclosed by two radii and an arc lying between them.
here AOB is a sector
length of arc
Area of Sector
Ring or Circular Path:
area=π(R2-r2)
Perimeter=2π(R+r)
Rhombus
Rhombus is a quadrilateral whose all sides are equal.
The diagonals of a rhombus bisect each other at 900
Area (a) of a rhombus
= a × h, i.e. base × height
Product of its diagonals
since d2
since d2
Perimeter (P) of a rhombus
= 4a, i.e. 4 x side
Where d1 and d2 are two-diagonals.
Side (a) of a rhombus
Parallelogram
A quadrilateral in which opposite sides are equal and parallel is called a parallelogram. The diagonals of a parallelogram bisect each other.
Area (a) of a parallelogram = base × altitude corresponding to the base = b × h
Area (a) of the parallelogram
where a and b are adjacent sides, d is the length of the diagonal connecting the ends of the two sides and
In a parallelogram, the sum of the squares of the diagonals = 2
(the sum of the squares of the two adjacent sides).
i.e.,
Perimeter (P) of a parallelogram
= 2 (a+b),
Where a and b are adjacent sides of the parallelogram.
Trapezium (Trapezoid)
A trapezoid is a 2-dimensional geometric figure with four sides, at least one set of which are parallel. The parallel sides are called the bases, while the other sides are called the legs. The term ‘trapezium,’ from which we got our word trapezoid has been in use in the English language since the 1500s and is from the Latin meaning ‘little table.’
Area (a) of a trapezium
1/2 x (sum of parallel sides) x perpendicular
Distance between the parallel sides
i.e.,
Where, l = b – a if b > a = a – b if a > b
And
Height (h) of the trapezium
Pathways Running across the middle of a rectangle:
X is the width of the path
Area of path= (l+b-x)x
perimeter= 2(l+b-2x)
Outer Pathways:
Area=(l+b+2x)2x
Perimeter=4(l+b+2x)
Inner Pathways:
Area=(l+b-2x)2x
Perimeter=4(l+b-2x)
Some useful Short trick:
• If there is a change of X% in defining dimensions of the 2-d figure then its perimeter will also change by X%
• If all the sides of a quadrilateral are changed by X% then its diagonal will also change by X%.
• The area of the largest triangle that can be inscribed in a semi-circle of radius r is r2.
• The number of revolution made by a circular wheel of radius r in travelling distance d is given by
number of revolution =d/2πr
• If the length and breadth of the rectangle are increased by x% and y% then the area of the rectangle will be increased by.
(x+y+xy/100)%
• If the length and breadth of a rectangle are decreased by x% and y% respectively then the area of the rectangle will decrease by:
(x+y-xy/100)%
• If the length of a rectangle is increased by x%, then its breadth will have to be decreased by (100x/100+x)% in order to maintain the same area of the rectangle.
• If each of the defining dimensions or sides of any 2-D figure is changed by x% its area changes by
x(2+x/100)%
where x=positive if increase and negative if decreases.
Thanks
Prep Smart. Stay Safe. Go BYJU'S Exam Prep.
## Important Mensuration (3D) Formulas
Cube
• s = side
• Volume: V = s^3
• Lateral surface area = 4a2
• Surface Area: S = 6s^2
• Diagonal (d) = s√3
Cuboid
• Volume of cuboid: length x breadth x width
• Total surface area = 2 ( lb + bh + hl)
Right Circular Cylinder
• Volume of Cylinder = π r^2 h
• Lateral Surface Area (LSA or CSA) = 2π r h
• Total Surface Area = TSA = 2 π r (r + h)
Hollow-Cylinder
* Volume of Hollow Cylinder = π(pie) h(r1(Square) - r2(Square))
Right Circular Cone
• l^2 = r^2 + h^2
• Volume of cone = 1/3 π r^2 h
• Curved surface area: CSA= π r l
• Total surface area = TSA = πr(r + l )
Important relation between radius, height and slant height of similar cone.
Frustum of a Cone
• h = height, s = slant height
• Volume: V = π/ 3 (r^2 + rR + R^2)h
• Surface Area: S = πs(R + r) + πr^2 + πR^2
Sphere
• Volume: V = 4/3 πr^3
• Surface Area: S = 4π^2
Hemisphere
• Volume-Hemisphere = 2/3 π r^3
• Curved surface area(CSA) = 2 π r^2
• Total surface area = TSA = 3 π r^2
Quarter-Sphere
Let 'r' is the radius of given diagram. You have to imagine this diagram, this is 1/4th part of Sphere.
Prism
• Volume = Base area x height
• Lateral Surface area = perimeter of the base x height
Pyramid
• Volume of a right pyramid = (1/3) × area of the base × height.
• Area of the lateral faces of a right pyramid = (1/2) × perimeter of the base x slant height.
• Area of whole surface of a right pyramid = area of the lateral faces + area of the base.
Important:
1.From a solid cylinder no. of maximum solid cone of same height and radius as cylinder are 3.
2.From a solid sphere, no. of maximum solid cone having height and radius equal can be made are 4.
3.From a solid hemisphere, no. of maximum solid cone having height and radius equal can be made are 2.
## CLICK HERE FOR PDF OF ALL IMPORTANT FORMULAS TO REMEMBER FOR Reasoning
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# Math in Focus Grade 1 Cumulative Review for Chapters 18 and 19 Answer Key
Go through the Math in Focus Grade 1 Workbook Answer Key Cumulative Review for Chapters 18 and 19 to finish your assignments.
## Math in Focus Grade 1 Cumulative Review for Chapters 18 and 19 Answer Key
Concepts and Skills
Count the number of groups. Count the number in each group. Then fill in the blanks.
Question 1.
6 + 6 + 6 + 6 = ___________
4 sixes = ___________
There are __________ starfishes in all.
6 + 6 + 6 + 6 = 24
4 sixes = 24
There are 24 starfishes in all.
Explanation:
Count the number of groups.
Then count the number of star fish in each group.
4 x 6 = 24
Question 2.
4 + 4 + 4 + 4 = ___________
4 fours = ___________
There are ___________ shrimps in all.
4 + 4 + 4 + 4 = 16
4 fours = 16
There are 16 shrimps in all
Explanation:
Count the number of groups.
Then count the number of shrimps in each group.
4 x 4 = 16
Look at the pictures. Then fill in the blanks.
Question 3.
___________ + ___________ + ___________ + ___________ = ___________
___________ fives = ___________
5 + 5 + 5 + 5 = 20
4 fives = 20
There are 20 cherries in all
Explanation:
Count the number of groups.
Then count the number of shrimps in each group.
4 x 4 = 16
Question 4.
There are ___________ pineapple cubes in all.
There are _________ sticks.
There are ___________ pineapple cubes in each stick.
There are 18 pineapple cubes in all.
There are 6 sticks.
There are 3 pineapple cubes in each stick.
Explanation:
Count the number of sticks.
Then count the number of pineapples in each stick.
3 x 6 = 18
Solve.
Question 5.
There are 12 sandwiches. They are shared equally by 3 children.
Each child gets ___________ sandwiches.
Each child gets 4 sandwiches.
Explanation:
There are 12 sandwiches.
They are shared equally by 3 children.
4+4+4 = 12
3 x 4 = 12
Solve.
Question 6.
There are 16 toy soldiers. Jamal packs them equally into 4 boxes.
There are ___________ toy soldiers in each box.
There are 4 toy soldiers in each box.
Explanation:
There are 16 toy soldiers.
Jamal packs them equally into 4 boxes
4+4+4+4 = 16
4 x 4 = 16
Question 7.
Timmy puts 6 pillows equally into 3 groups.
Each group has ____________ pillows.
Each group has 2 pillows
Explanation:
Timmy puts 6 pillows equally into 3 groups.
Count the number of groups.
Then count the number of shrimps in each group.
3 x 2 = 6
Question 8.
There are 9 oranges. Circle groups of 3.
There are ____________ groups of 3.
There are 3 groups of 3.
Explanation:
Count the number of groups.
Then count the number of shrimps in each group.
3 x 3 = 9
Look at the pictures. Then fill in the blanks.
Maria has 12 flowers.
Question 9.
She puts the flowers equally into 3 vases. There are flowers ____________ in each vase.
There are flowers 4 in each vase.
Explanation:
Count the number of groups.
Then count the number of shrimps in each group.
3 x 4 = 12
Question 10.
She puts 2 flowers in each vase. She needs ____________ vases.
She needs 6 vases.
Explanation:
Count the number of groups.
Then count the number of shrimps in each group.
6 x 2 = 12
Question 11.
She puts 4 flowers in each vase. She needs ____________ vases.
She needs 3 vases.
Explanation:
Count the number of groups.
Then count the number of shrimps in each group.
3 x 4 = 12
Write the value.
Question 12.
Explanation:
1 nickel = 5¢
Question 13.
3 dimes = 30¢
Explanation:
1 dime = 10¢
3 dimes = 10¢ + 10¢ + 10¢ = 30¢
Fill in the blanks.
Question 14.
Find how many of each coin are needed.
__________ dimes _________ nickels _________ pennies
2 dimes + 1 nickel + 2 pennies.
Explanation:
1 dime = 10¢
1 nickel = 5¢
1 penny = 1¢
2 dimes + 1 nickel + 2 pennies
20¢ + 5¢ + 2¢ = 17¢
Question 15.
5 nickels
Explanation:
1 quarter = 25¢
1 nickel = 5¢
1 quarter = 5 nickels.
Circle the coins to show the same value.
Question 16.
Explanation:
1 quarter = 2 dimes + 1 nickel
1 quarter = 25¢
1 dime = 10¢
1 nickel = 5¢
Circle the coins that are not dimes.
Question 17.
Explanation:
3 coins belongs to dimes,
rest of the coins are quarters, nickels and pennies.
Count on to find the value.
Question 18.
Explanation:
2 quarters + 3 dimes + 1 penny = 81¢
1 quarter = 25¢
1 dime = 10¢
1 penny = 1¢
Question 19.
25¢ + 17¢ = ____________ ¢
Explanation:
1 quarter + 1 dime + 1 nickel + 2 pennies
1 quarter = 25¢
1 dime = 10¢
1 nickel = 5¢
1 penny = 1¢
25¢ + 10¢ + 5¢ + 2¢ = 42¢
Complete.
Question 20.
45¢ – 15¢ = ____________ ¢
Explanation:
1 quarter = 25¢
1 dimes = 10¢
1 nickel = 5¢
(1 quarter + 2 dimes) – (1 dime + 1 nickel)
= (25 + 20) – (10 + 5)
= 45 – 15 = 30
Question 21.
Subtract 18¢ from 90¢
__________ – _________ = _________ ¢
Explanation:
Subtract small number form big number
90¢ – 18¢ = 72¢
Problem Solving
Solve.
Question 22.
Draw the least number of coins needed to make 64¢.
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Courses
# Chapter Notes - Fractions Class 6 Notes | EduRev
## Class 6 : Chapter Notes - Fractions Class 6 Notes | EduRev
The document Chapter Notes - Fractions Class 6 Notes | EduRev is a part of the Class 6 Course Mathematics (Maths) Class 6.
All you need of Class 6 at this link: Class 6
Fractions
A fraction is a part of a whole. A whole can be a group of objects or a single object.
Example: 3/15 is a fraction. In this, 3 is called the numerator and 15 is called the denominator.
In the figure shown here, the shaded portion is represented by 1/3, 2/6 and 4/12.
Fractions on a Number Line
Whole numbers are represented on the number line as shown here:
A fraction can be represented on the number line.
Examples:
• Consider a fraction 1/2.
1/2 is greater than 0, but less than 1.
Divide the space between 0 and 1 into two equal parts. We can show one part as the fraction 1/2
• Consider another 1/5.
1/5 fraction is greater than 0, but less than 1.
Divide the space between 0 and 1 into five equal parts. We can show the first part as 1/5 the second as 2/5 the third as 3/5 the fourth as 4/5 and the fifth part as 5/5=1.
Types of Fractions
Proper Fractions
A proper fraction is a number representing a part of a whole.
In a proper fraction, the number in the denominator shows the number of parts into which the whole is divided, while the number in the numerator shows the number of parts that have been taken.
Example: 4/20, 3/20, 10/20
Improper Fractions
A fraction in which the numerator is bigger than the denominator is called an improper fraction.
Example: 4/3, 13/8, 19/7
Mixed Fractions
A combination of a whole and a part is said to be a mixed fraction.
Example:
,
Conversion of improper fraction into mixed fraction:
An improper fraction can be expressed as mixed fraction by dividing the numerator by the denominator of the improper fraction to obtain the quotient and the remainder. Then the mixed fraction will be.
Conversion of mixed fraction into improper fraction:
A mixed fraction can be written in the form an improper fraction by writing it in the following way:
Like Fractions
Fractions with the same denominator are said to be like fractions.
Example: 4/15, 6/15, 8/15
Unlike Fractions
Fractions with different denominators are called unlike fractions.
Example: 3/15, 3/20, 9/28
Equivalent Fractions
Fractions that represent the same part of a whole are said to be equivalent fractions.
Example: 1/2 = 2/4 = 3/6 = 4/8 = 5/10 = .....
To find an equivalent fraction of a given fraction, multiply both the numerator and the denominator of the given fraction by the same number.
Simplest Form of Fraction
A fraction is said to be in its simplest form or its lowest form if its numerator and denominator have no common factor except one. The simplest form of a given fraction can also be found by dividing its numerator and denominator by its highest common factor (HCF).
Comparing Fractions
Comparing Like Fractions
In like fractions, the fraction with the greater numerator is greater.
Example: Among fractions 5/7 and 3/7, 5/7 is greater than 3/7 as 5 is greater than 3.
Two fractions are unlike fractions if they have different denominators.
Comparing Unlike Fractions
If two fractions with the same numerator but different denominators are to be compared, then the fraction with the smaller denominator is the greater of the two.
To compare unlike fractions, we first convert them into equivalent fractions. For example, to compare the following fractions ie.,
6/8 and 4/6
We find the common multiple of the denominators 6 and 8.
48 is a common multiple of 6 and 8.
24 is also a common multiple of 6 and 8.
Least Common Multiple (LCM) of 6 and 8 = 24
Hence, we can say that 6/8 is greater than 4/6
If two fractions have the same number in the denominator, then they are said to be like fractions.
Like Fractions
If two fractions have the same number in the denominator, then they are said to be like fractions.
1. Add the numerators of the fractions to get the numerator of the resultant fraction.
2. Use the common denominator of the like fractions as the denominator of the resultant fraction.
To subtract like fractions:
1. Subtract the numerators of the fractions to get the numerator of the resultant fraction.
2. Use the common denominator of the like fractions as the denominator of the resultant fraction.
Unlike Fractions
Fractions with different numbers in the denominators are said to be unlike fractions.
1. Find their equivalent fractions with the same denominator.
2. Add the numerators of the fractions to get the numerator of the resultant fraction.
3. Use the common denominator of the obtained like fraction as the denominator of the resultant fraction.
To subtract unlike fractions:
1. Find their equivalent fractions with the same denominator.
2. Subtract the numerators of the fractions to get the numerator of the resultant fraction.
Use the common denominator of the obtained like fraction as the denominator of the resultant fraction.
Addition or Subtraction of Mixed Fractions
Two mixed fractions can be added or subtracted by adding or subtracting the whole numbers of the two fractions, and then adding or subtracting the fractional parts together. Two mixed fractions can also be converted into improper fractions and then added or subtracted.
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## Mathematics (Maths) Class 6
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# How do you verify the identity tan2theta=2/(cottheta-tantheta)?
Dec 22, 2016
Rewrite $\tan \theta$ and $\cot \theta$ as sines and cosines using color(magenta)(tan theta = sintheta/costheta and cot theta = costheta/sintheta.
$\frac{\sin 2 \theta}{\cos 2 \theta} = \frac{2}{\cos \frac{\theta}{\sin} \theta - \sin \frac{\theta}{\cos} \theta}$
I would recommend you simplify the right hand side prior to expanding the left.
(sin2theta)/(cos2theta) = 2/((cos^2theta - sin^2theta)/(costhetasintheta)
$\frac{\sin 2 \theta}{\cos 2 \theta} = \frac{2 \cos \theta \sin \theta}{{\cos}^{2} \theta - {\sin}^{2} \theta}$
We know this is true because $\sin 2 \theta = 2 \sin \theta \cos \theta$ and $\cos 2 \theta$ can be written as ${\cos}^{2} \theta - {\sin}^{2} \theta$.
Practice exercises:
1. Prove the following trig identities:
a) $\frac{{\sin}^{2} \theta + {\cos}^{2} \theta + {\cot}^{2} \theta}{1 + {\tan}^{2} \theta} = {\cot}^{2} \theta$
b) $\cos \left(x + y\right) + \cos \left(x - y\right) = 2 \cos x \cos y$
c) $\csc \left(2 \alpha\right) - \cot \left(2 \alpha\right) = \tan \alpha$
Solve the following equation for x in the interval 0 ≤ x ≤ 2pi:
$\cos \left(2 x\right) = 2 {\sin}^{2} x$
Hopefully this helps, and good luck!
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# Eureka Math Precalculus Module 1 Lesson 3 Answer Key
## Engage NY Eureka Math Precalculus Module 1 Lesson 3 Answer Key
### Eureka Math Precalculus Module 1 Lesson 3 Exercise Answer Key
Opening Exercise
Recall from the previous two lessons that a linear transformation is a function f that satisfies two conditions:
(1) f(x + y) = f(x) + f(y) and (2) f(kx) = kf(x). Here, k refers to any real number, and x and y represent arbitrary elements in the domain of f.
a. Let f(x) = x2. Is f a linear transformation? Explain why or why not.
Answer:
Let x be 2 and y be 3. f(2 + 3) = f(5) = 52 = 25, but f(2) + f(3) = 22 + 32 = 4 + 9 = 13. Since these two values are different, we can conclude that f is not a linear transformation.
b. Let g(x) = $$\sqrt{x}$$. Is g a linear transformation? Explain why or why not.
Answer:
Let x be 2 and y be 3. g(2 + 3) = g(5) = $$\sqrt{5}$$, but g(2) + g(3) = $$\sqrt{2}$$ + $$\sqrt{3}$$, which is not equal to $$\sqrt{5}$$. This means that g is not a linear transformation.
### Eureka Math Precalculus Module 1 Lesson 3 Exit Ticket Answer Key
Suppose you have a linear transformation f: R → R, where f(3) = 9 and f(5) = 15.
Question 1.
Use the addition property to compute f(8) and f(13).
Answer:
f(8) = f(3 + 5) = f(3) + f(5) = 9 + 15 = 24
f(13) = f(8 + 5) = f(8) + f( 5) = 24 + 15 = 39
Question 2.
Find f(12) and f(10). Show your work.
Answer:
f(12) = f(4 ∙ 3) = 4 ∙ f(3) = 4 ∙ 9 = 36
f(10) = f(2 ∙ 5) = 2 ∙ f(5) = 2 ∙ 15 = 30
Question 3.
Find f(-3) and f(-5). Show your work.
Answer:
f(-3) = -f(3) = -9
f(-5) = -f(5) = -15
Question 4.
Find f(0). Show your work.
Answer:
f(0) = f(3 + -3) = f(3) + f(-3) = 9 + -9 = 0
Question 5.
Find a formula for f(x).
Answer:
We know that there is some number a such that f(x) = ax, and since f(3) = 9, the value of a = 3. In other words, f(x) = 3x. We can also check to see if f(5) = 15 is consistent with a = 3, which it is.
Question 6.
Draw the graph of the function y = f(x).
Answer:
### Eureka Math Precalculus Module 1 Lesson 3 Problem Set Answer Key
The first problem provides students with practice in the core skills for this lesson. The second problem is a series of exercises in which students explore concepts of linearity in the context of integer-valued functions as opposed to real¬valued functions. The third problem plays with the relation f(x + y) = f(x) + f(y), exchanging addition for multiplication in one or both expressions.
Question 1.
Suppose you have a linear transformation f: R → R, where f(2) = 1 and f(4) = 2.
a. Use the addition property to compute f(6), f(8), f(10), and f(12).
Answer:
f(6) = f(2 + 4) = f(2) + f(4) = 1 + 2 = 3
f(8) = f(2 + 6) = f(2)+ f(6) = 1 + 3 = 4
f(10) = f(4 + 6) = f(4) + f(6) = 2 + 3 = 5
f(12) = f(10 + 2) = f(10) + f(2) = 5 + 1 = 6
b. Find f(20), f(24), and f(30). Show your work.
Answer:
f(20) = f(10 ∙ 2) = 10 ∙ f(2) = 10 ∙ 1 = 10
f(24) = f(6 ∙ 4) = 6 ∙ f(4) = 6 ∙ 2 = 12
f(30) = f(15 ∙ 2) = 15 ∙ f(2) = 15 ∙ 1 = 15
c. Find f(-2), f(-4), and f(-8). Show your work.
Answer:
f(-2) = f(-1 ∙ 2) = -1 ∙ f(2) = -1 – 1 = -1
f(-4) = f(-1 ∙ 4) = -1 ∙ f(4) = -1 – 2 = -2
f(-8) = f(-2 ∙ 4) = -2 ∙ f(4) = -2 ∙ 2 = -4
d. Find a formula for f(x).
Answer:
We know there is some number a such that f(x) = ax, and since f(2) = 1, then the value of a = $$\frac{1}{2}$$.
In other words, f(x) = $$\frac{x}{2}$$. We can also check to see if f(4) = 2 is consistent with a = $$\frac{1}{2}$$ which it is.
e. Draw the graph of the function f(x).
Answer:
Question 2.
The symbol L represents the set of integers, and so g: Z → Z represents a function that takes integers as inputs and produces integers as outputs. Suppose that a function g: Z → Z satisfies g(a + b) = g(a) + g(b) for all integers a and b. Is there necessarily an integer k such that g(n) = kn for all integer inputs n?
a. Let k = g(1). Compute g(2) and g(3).
Answer:
g(2) = g(1 + 1) = g(1) + g(1) = k + k = 2k
g(3) = g(1 + 1 + 1) = g(1) + g(1) + g(1) = k + k + k = 3k
b. Let n be any positive integer. Compute g(n).
Answer:
g(n) = g(1 + …… + 1) = g(1) + ……. + g(1) = k + …… + k = nk
c. Now, consider g(0). Since g(0) = g(0 + 0), what can you conclude about g(0)?
Answer:
g(0) = g(0 + 0) = g(0) + g(0). By subtracting g(0) from both sides of the equation, we get g(0) = 0.
d. Lastly, use the fact that g(n +-n) = g(0) to learn something about g(-n), where n is any positive integer.
Answer:
g(0) = g(n + -n) = g(n) + g(-n). Since we know that g(0) = 0, we have g(n) + g(-n) = 0. This tells us that g(-n) = -g(n).
e. Use your work above to prove that g(n) = kn for every integer n. Be sure to consider the fact that n could be positive, negative, or 0.
Answer:
We showed that if n is a positive integer, then g(n) = kn, where k = g(1). Also, since k ∙ 0 = 0 and we showed that g(0) = 0, we have g(0) = k ∙ 0. Finally, if n is a negative integer, then -n is positive, which means g(-n) = k(-n) = -kn. But, since g(-n) = -g(n), we have g(n) = -g(-n) = -(-kn) = kn. Thus, in all cases, g(n) = kn.
Question 3.
In the following problems, be sure to consider all kinds of functions: polynomial, rational, trigonometric, exponential, logarithmic, etc.
a. Give an example of a function f: R → R that satisfies f(x ∙ y) = f(x) + f(y).
Answer:
Any logarithmic function works; for instance, f(x) = log(x).
b. Give an example of a function g: R → R that satisfies g(x + y) = g(x) ∙ g(y).
Answer:
Any exponential function works; for instance, g(x) = 2x.
c. Give an example of a function h: R → R that satisfies h(x ∙ y) = h(x) ∙ h(y).
Answer:
Any power of x works; for instance, h(x) = x3.
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Home >> Simple Interest >> Difference between Simple & Compound Interest >>
## Difference between Simple & Compound Interest
Amount Compound Interest Difference between Simple & Compound Interest
What is Simple Interest ?
What is Compound Interest ?
In simple interest , interest is calculated on the principal irrespective of number of years while in Compound Interest interest is added after every one year to form a new principal.
Example : Find Interest to be repaid on a loan of \$10000 for 3 years at rate 10% interest per annum ?
Solution: Here we calculate both simple interest and compound interest
Observe the following table:
1 Year Calculation
Simple Interest Compound Interest
Principal \$10000 \$10000
Interest \$1000 \$1000
2 Year Calculation
Simple Interest Compound Interest
Principal \$10000 \$11000
Interest \$1000 \$1100
3 Year Calculation
Simple Interest Compound Interest
Principal \$10000 \$12100
Interest \$1000 1210
Total Interest After 3 Years
\$3000 \$3310
You can observe that the interest paid in case of simple interest is \$3000 while interest paid in case of compound interest is \$3310.
And there is difference of \$310 in payment of interest calculated with simple interest and compound interest.
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# To Construct a Triangle when Two of its Sides and the Included Angles are given
To construct a triangle when two of its sides and the included angles are given, let’s follow the examples.
1. Draw a triangle ABC in which AB = 5 cm, BC= 7 cm, and ∠B = 75°.
Steps of Construction:
(i) Draw a line segment AB = 5 cm.
(ii) Make ∠ ABC = 75°.
(iii) Cut of BC = 7 cm.
(iv) Join AC.
Therefore, ∆ ABC is the required triangle.
2. Draw a triangle ABC in which AB = 4.5 cm, BC = 6 cm, and ∠B = 80°.
Steps of Construction:
(i) Draw a line segment AB = 4.5 cm.
(ii) Make ∠ ABC = 80°.
(iii) Cut of BC = 6 cm.
(iv) Join AC.
Therefore, ∆ ABC is the required triangle.
Triangle.
Classification of Triangle.
Properties of Triangle.
Worksheet on Triangle.
To Construct a Triangle whose Three Sides are given.
To Construct a Triangle when Two of its Sides and the included Angles are given.
To Construct a Triangle when Two of its Angles and the included Side are given.
To Construct a Right Triangle when its Hypotenuse and One Side are given.
Worksheet on Construction of Triangles.
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# Fundamental Laws of Logarithms- Part3
## Fundamental Laws of Logarithms- Part3
7. For a, n > 0 and a ≠ 1; $${{a}^{{{\log }_{a}}n}}=n$$
Proof:
Let $${{\log }_{a}}n=x$$. Then $${{a}^{x}}=n$$.
Therefore, $${{a}^{{{\log }_{a}}n}}=n$$,
Examples:
(1) $${{3}^{{{\log }_{3}}8}}=8$$
(2) $${{2}^{3{{\log }_{2}}6}}={{2}^{{{\log }_{2}}{{6}^{3}}}}={{6}^{3}}$$
(3) $${{5}^{-3{{\log }_{5}}2}}={{5}^{{{\log }_{5}}{{2}^{-3}}}}=\frac{1}{{{2}^{3}}}$$
8. $${{\log }_{{{a}^{q}}}}{{n}^{p}}=\frac{p}{q}{{\log }_{a}}n=y$$, where a, n > 0, a ≠ 1
Proof:
Let $${{\log }_{{{a}^{q}}}}{{n}^{p}}=x\ \ and\ \ {{\log }_{a}}n=y.$$then,
$${{\left( {{a}^{q}} \right)}^{x}}={{n}^{p}}\ and\ {{a}^{y}}=n$$,
$$\left( {{a}^{qx}} \right)={{n}^{p}}\ and\ {{a}^{y}}=n$$,
$$\left( {{a}^{qx}} \right)={{n}^{p}}\ and\ {{\left( {{a}^{y}} \right)}^{p}}={{n}^{p}}$$,
$${{a}^{qx}}={{a}^{yp}}$$,
qx = yp,
⇒ x = (p/q)y,
⇒ $${{\log }_{{{a}^{q}}}}{{n}^{p}}=\frac{p}{q}{{\log }_{a}}n$$.
Example: What is greater $$x={{\log }_{3}}5$$ (or) $$y={{\log }_{17}}25$$
Solution:
$$y={{\log }_{17}}25$$,
$$\frac{1}{y}={{\log }_{25}}17$$,
$$={{\log }_{{{5}^{2}}}}17$$,
$$=\frac{1}{2}{{\log }_{5}}17$$….(1),
$$x={{\log }_{3}}5$$,
$$\frac{1}{x}={{\log }_{5}}3$$….(2),
From equation (1) and (2),
$$\frac{1}{y}>\frac{1}{x}$$ ,
x > y.
9. $${{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}$$
Proof: Let $${{a}^{{{\log }_{b}}c}}=p$$$$\Rightarrow {{\log }_{b}}c={{\log }_{a}}p$$,
$$\Rightarrow \frac{\log c}{\log b}=\frac{\log p}{\log a}$$,
$$\Rightarrow \frac{\log a}{\log b}=\frac{\log p}{\log c}$$,
$${{\log }_{b}}a={{\log }_{c}}p$$,
$$p={{c}^{{{\log }_{b}}a}}\Rightarrow {{a}^{{{\log }_{b}}c}}={{c}^{{{\log }_{b}}a}}$$.
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# Maths Triangle and Its Properties part 2 (Types of Triangle) CBSE Class 7 Mathematics VII | Summary and Q&A
35.3K views
October 17, 2016
by
LearnoHub - Class 11, 12
Maths Triangle and Its Properties part 2 (Types of Triangle) CBSE Class 7 Mathematics VII
## TL;DR
This video explains the different types of triangles based on their sides and angles.
## Key Insights
• 🙃 Triangles can be classified based on their sides into scalene, isosceles, and equilateral.
• 🔺 Triangles can also be classified based on their angles into acute angled, right angled, and obtuse angled.
• 🟰 An isosceles triangle has two equal sides and two equal base angles.
• 🔺 A right angled triangle has one angle of 90°.
• 🔺 A triangle with an angle greater than 90° is called an obtuse angled triangle.
• 🔺 A triangle cannot have more than one angle of 90° or more than one obtuse angle.
• 🔺 Equilateral triangles have all sides and angles equal.
## Transcript
Hello friends this video on triangle and its properties part two is brought to you by exam fear.com no more fear from exam now there are many different types of triangles in fact the types of triangles also varies depending on what is the basis for categorizing triangles now first of all let's talk about the types of triangle based on the sides so ... Read More
### Q: What are the three types of triangles based on the length of their sides?
The three types of triangles based on the sides are scalene (with unequal sides), isosceles (with two equal sides), and equilateral (with all sides equal).
### Q: How are angles classified in triangles?
Angles in triangles can be classified as acute (less than 90°), right (one angle is 90°), or obtuse (one angle is greater than 90°).
### Q: What is an isosceles triangle?
An isosceles triangle has two equal sides and two equal angles (base angles) located on the base.
### Q: Can a triangle have two angles of 90°?
No, a triangle cannot have two angles of 90° as it would not form a closed figure. A triangle can only have three sides and three angles.
## Summary & Key Takeaways
• There are three types of triangles based on the sides: scalene, where all sides are unequal; isosceles, where two sides are equal; and equilateral, where all sides are equal.
• Triangles can also be classified based on their angles: acute angled, where all angles are less than 90°; right angled, where one angle is 90°; and obtuse angled, where one angle is greater than 90°.
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# Calculate Faster Than a Calculator: 5 Tricks for Quicker Calculations
Jul 21, 2021
Is mathematics the subject you always fear? Do you rely on the calculator to make mathematical calculations? If yes, you don’t have to anymore. There are a number of tricks that can help you get better at calculations and calculate even faster than calculators. Surprised? Well, it is true, possible, and actually works.
While making calculations, most people compromise either on speed or accuracy. However, you do not have to compromise one either of the two anymore. Here are the five best tricks that can help you calculate faster than a calculator and get accurate answers.
1. Squaring the Number
Finding the square of numbers is quite tricky at times. While squaring single-digit numbers is easy, finding the square of a double-digit or triple-digit number is complex and time-consuming. Moreover, there are also high chances that you might end up getting the wrong answers due to some calculation mistakes in between.
However, an easy way to find the squares of different numbers is to try out the simple approaches and quick tricks. It will not only save your time but also reduce your stress and help you find the correct answers. If you are willing to try out the trick to find the squares of numbers easily, here are the steps you need to follow.
Step 1: Find a base that is quite closer to the given number.
Step 2: Calculate the difference of the given number from the selected base.
Step 3: Next, add the difference you found to the given number.
Step 4: Multiply the outcome of step 3 with the base.
Step 5: Now, add the result of step 4 with the difference between the given number and the selected base. This will give you the final answer.
In order to understand the trick better, let’s take the help of an example.
Suppose you need to find the square of 79.
Step 1: Select 80 as your base number.
Step 2: Difference = 79 – 80 = -1
Step 3: Given number + Difference = 79 + (-1) = 78
Step 4: Multiplication of the result with base = 78 × 80 = 6240
Step 5: 6080 + (-1)2 = 6241
So, 6241 is the square of 79.
While adding and subtracting simple numbers is easy, it becomes difficult when fractions are involved. However, utilizing the “vertical and crosswise” approach of Vedic mathematics can make things easier for you.
Here is a simple example of how the method actually works.
Let’s take a problem: 2/4 + 1/5
In order to find the resulting numerator, cross multiply the numbers and add them.
So, for the given problem, resulting numerator = (2 x 5) + (4 x 1) = 10 + 4 = 14
In order to find the resulting denominator, simply multiply the denominators.
So, in this case, the resulting denominator = 4 x 5 = 20
3. Squaring numbers that end with 5
Many people may be familiar with this mathematics trick. However, if this is new for you, nothing to worry about; you can learn it right away. This trick is quite simple and easy. All you need to do is leave the last digit that is 5 and multiply the remaining digit with its successive number. When you get the result, put 25 in the end. This will give you the final answer.
Let’s understand this better with the help of an example.
Suppose you have to find the square of 75.
Step 1: 7 x (7+1) = 56
Step 2: 5625
So, the square of 75 is 5625.
4. Multiplying two-digit numbers by 11
When you want to get answers to multiplications in just two seconds, make sure to use this excellent trick. It is one of the popular Vedic tricks that makes calculations much easy in mathematics than you can imagine.
When you want to multiply a two-digit number by 11, first write down the digits in this pattern:
First digit [ sum of the two digits] second digit
The answer you get is the final result. This is surprisingly easy. Isn’t it? Let’s try.
In order to understand this better, let’s take an example.
Imagine that you need to find the solution of 52 x 11.
By following the pattern, you get:
5[5+2]2 = 572
So, when you multiply 52 by 11, the answer will be 572.
However, when the sum of two digits is again a two-digit number, keep the digit in the unit place and then add ‘1’ to the digit that precedes it.
48 x 11
= 4[4+8]8
= 4[12]8
= 4+1 [2]8
= 528
So, by multiplying 48 by 11, the final answer is 528.
5. Sequentially Calculating the Square of any number
When a given number is supposedly a successor of another number whose square is known to you, it is quite easy to calculate the square sequentially. In order to make this easier to understand, let’s take an example.
Given Problem = square of 141
The given number 141 is a successor of 140, which is a known square.
So, to find out the square of 141, an easy way is:
(141)2 = (140)2 + 141 + 140 = 19600 + 281 = 19881
So, 19881 is the square of the given number 141.
You can use this trick to easily find out the squares of other two-digit numbers, which are a successor of known squares.
Conclusion
Now that you know the five tricks of making calculations faster, start applying them and experience the fun of solving problems in a jiffy. Maths is actually not a very tough subject as most students think. You just need to know the right tricks to excel in the subject. Once you know the tricks, you are sure to have great fun and enjoy doing maths. Moreover, the tricks can help you save time and enhance your efficiency.
Article Posted in: How to
### Sid
Sid writes educational content periodically for Wizert and backs it up with extensive research and relevant examples. He's an avid reader and a tech enthusiast at the same time with a little bit of “Arsenal Football Club” thrown in as well. He's got more than 5 years of experience in technical content framing, digital marketing, SEO and graphic designing.
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# wikiHow to Find How Many Diagonals Are in a Polygon
Finding diagonals in a polygon is a necessary skill to develop in math. It may seem difficult at first, but is pretty simple once you learn the basic formula. A diagonal is any line segment drawn between vertices of a polygon that doesn’t include the sides of that polygon.[1] A polygon is any shape that has more than three sides. Using a very simple formula, you can calculate the number of diagonals in any polygon, whether it has 4 sides or 4,000 sides.
### Method 1 Drawing the Diagonals
1. 1
Know the names of polygons. You may need to first identify how many sides are present in the polygon. Each polygon has a prefix that indicates the number of sides it has. Here are the names of polygons with up to twenty sides:[2]
• Pentagon: 5 sides
• Hexagon: 6 sides
• Heptagon: 7 sides
• Octagon: 8 sides
• Nonagon/Enneagon: 9 sides
• Decagon: 10 sides
• Hendecagon: 11 sides
• Dodecagon: 12 sides
• Triskaidecagon/tridecagon: 13 sides
• Pendecagon: 15 sides
• Icosagon: 20 sides
• Note that a triangle has no diagonals.[3]
2. 2
Draw the polygon. If you wanted to know how many diagonals were present in a square, you would start by drawing the square. The easiest way to find diagonals and count them is to draw the polygon symmetrically, each side has the same length. It’s important to note that even if the polygon is not symmetrical, it will still have the same number of diagonals.[4]
• To draw the polygon, use a ruler and draw each side the same length, connecting all of the sides together.
• If you’re unsure what the polygon will look like, search for pictures online. For example, a stop sign is an octagon.
3. 3
Draw the diagonals. A diagonal is a line segment drawn from one corner of the shape to another, excluding the sides of the polygon.[5] Starting at one vertex of the polygon, use a ruler to draw a diagonal to every other available vertex.
• For a square, draw one line from the bottom left corner to the top right corner and another line from the bottom right corner to the top left corner.
• Draw diagonals in different colors to make them easier to count.[6]
• Note that this method gets much more difficult with polygons that have more than ten sides.
4. 4
Count the diagonals. There are two options for counting: you can count as you draw the diagonals or count them once they have been drawn. As you count each diagonal, draw a small number above the diagonal to denote that it has been counted. It is easy to lose track while counting when there are a lot of diagonals crossing each other.
• For the square, there are two diagonals: one diagonal for every two vertices.[7]
• A hexagon has 9 diagonals: there are three diagonals for every three vertices.
• A heptagon has 14 diagonals. Past the heptagon, it gets more difficult to count the diagonals because there are so many of them.
5. 5
Beware of counting a diagonal more than once. Each vertex may have multiple diagonals, but that doesn’t mean that the number of diagonals is equal to the number of vertices times the number of diagonals. Take care when counting the diagonals to count each one only once.[8]
• For example, a pentagon (5 sides) has only 5 diagonals. Each vertex has two diagonals, so if you counted each diagonal from every vertex twice, you might think there were 10 diagonals. This is incorrect because you would have counted each diagonal twice!
6. 6
Practice with some examples. Draw some other polygons and count the number of diagonals. The polygon does not have to be symmetric for this method to work. In the case of a concave polygon, you may have to draw some of the diagonals outside the actual polygon.[9]
• A hexagon has 9 diagonals.
• A heptagon has 14 diagonals.
### Method 2 Using the Diagonal Formula
1. 1
Define the formula. The formula to find the number of sides of a polygon is n(n-3)/2 where “n” equals the number of sides of the polygon.[10] Using the distributive property this can be rewritten as (n2 - 3n)/2. You may see it either way, both equations are identical.
• This equation can be used to find the number of diagonals of any polygon.
• Note that the triangle is an exception to this rule. Due to the shape of the triangle, it does not have any diagonals.[11]
2. 2
Identify the number of sides in the polygon. To use this formula, you must identify the number of sides that the polygon has. The number of sides is given in the name of the polygon, you just need to know what each name means. Here are some of common prefixes you will see in polygons:[12]
• Tetra (4), penta (5), hexa (6), hepta (7), octa (8), ennea (9), deca (10), hendeca (11), dodeca (12), trideca (13), tetradeca (14), pentadeca (15), etc.
• For very large sided polygons you may simply see it written “n-gon”, where “n” is the number of sides. For example, a 44-sided polygon would be written as 44-gon.
• If you are given a picture of the polygon, you can simply count the number of sides.
3. 3
Plug the number of sides into the equation. Once you know how many sides the polygon has, you just need to plug that number into the equation and solve. Everywhere you see “n” in the equation will be replaced with the number of sides of the polygon.[13]
• For example: A dodecagon has 12 sides.
• Write the equation: n(n-3)/2
• Plug in the variable: (12(12 - 3))/2
4. 4
Solve the equation. Finish by solving the equation using the proper order of operations. Start by solving the subtraction, then multiply, then divide. The final answer is the number of diagonals the polygon has.[14]
• For example: (12(12 – 3))/2
• Subtract: (12*9)/2
• Multiply: (108)/2
• Divide: 54
• A dodecagon has 54 diagonals.
5. 5
Practice with more examples. The more practice you have with a math concept, the better you will be at using it. Doing lots of examples will also help you memorize the formula in case you need it for a quiz, test, or exam. Remember, this formula works for a polygon of any number of sides greater than 3.
• Hexagon (6 sides): n(n-3)/2 = 6(6-3)/2 = 6*3/2 = 18/2 = 9 diagonals.
• Decagon (10 sides): n(n-3)/2 = 10(10-3)/2 = 10*7/2 = 70/2 = 35 diagonals.
• Icosagon (20 sides): n(n-3)/2 = 20(20-3)/2 = 20*17/2 = 340/2 = 170 diagonals.
• 96-gon (96 sides): 96(96-3)/2 = 96*93/2 = 8928/2 = 4464 diagonals.
## Community Q&A
Search
• How many diagonals can be drawn from one vertex of nonagon?
wikiHow Contributor
You can draw six, one for each of the vertices, except for the vertex you're drawing from, and the two adjacent vertices.
• What is the relationship between the number of sides in an icosikaipentagon and the number of diagonals?
It is the same relationship for any polygon, as expressed in the formula n(n-3)/2.
• How would we solve it if it had (n-1) sides?
wikiHow Contributor
If you had (n-1) sides you would simply plug (n-1) into the equation as the variable: n(n - 3)/2 = (n-1)((n-1) - 3)/2. Simplifying terms yields ((n-1)(n-4))/2. You can either leave it at that, or expand using FOIL: (n^2 - 5n + 4)/2.
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# Lesson Explainer: Matrix Multiplication Mathematics
In this explainer, we will learn how to identify the conditions for matrix multiplication and evaluate the product of two matrices if possible.
Let us begin by recalling scalar multiplication, which is much simpler than matrix multiplication. Scalar multiplication involves multiplying a matrix by a scalar (or a number). For example, consider the matrix
If we wanted to multiply this matrix by a scalar 2, we would multiply each of the components of the matrix by 2:
We can see that multiplying a matrix by a scalar is simple. However, it is much more complicated to multiply a matrix by another matrix. Before we can discuss how to multiply a matrix by another matrix, we need to understand when it is possible to multiply a pair of matrices.
Recall that the order of a matrix is given by
For instance, a matrix with rows and columns is said to be an matrix. In order to multiply a pair of matrices, their orders must be compatible.
### Rule: Criterion for Matrix Multiplication
Let and be matrices. To compute matrix multiplication , the number of columns in must equal the number of rows in . If is an matrix for some positive integers and , must be an matrix for some positive integer . In this case, is an matrix.
From this criterion for matrix multiplication, we can see that is not the same as for matrices and . In fact, it is possible for one of these to be defined while the other is not. For instance, say that matrices and are of order and respectively. Then, the number of columns of is equal to the number of rows of , which means that the matrix multiplication is defined. On the other hand, the number of columns of is not equal to the number of rows of . This means that the matrix multiplication is not defined. This tells us that matrix multiplication is not commutative, which means that the order of matrices in matrix multiplication cannot be changed.
In the first example, we will find the order of a matrix resulting from matrix multiplication.
### Example 1: Order of Matrices in Matrix Multiplication
Fill in the blank: If is a matrix of order and is a matrix of order , then matrix is of order .
Recall that the number of columns in matrix must be equal to the number of rows in matrix to compute matrix multiplication . We also recall that the order of a matrix is given by
Since the order of matrix is , this tells us that the number of columns in matrix is 3. The order of is , which means that matrix has 1 row and 3 columns. is the transpose of , and we know that the transpose of a matrix changes the rows of the matrix to columns of its transpose. Since the transpose of has 1 row and 3 columns, matrix must have 3 rows and 1 column. This tells us that the number of columns in matrix and the number of rows in matrix are both equal to 3, which means that matrix multiplication is valid.
Recall that the multiplication of a matrix of order by a matrix of order results in a matrix of order . In this example, we are multiplying a matrix by a matrix. This means
Hence, , , and . The order of matrix is . This is option B.
In the next example, we will find the order of a matrix being multiplied, based on the order of the product matrix as well as the order of the other matrix.
### Example 2: Order of Matrices in Matrix Multiplication
Fill in the blank: If matrix is of order and matrix is of order , then matrix is of order .
Recall that the number of columns in matrix must be equal to the number of rows in matrix to compute matrix multiplication . We also recall that the order of a matrix is given by
Since the order of matrix is , this tells us that the number of columns in matrix is 3. This number must equal the number of rows in matrix . Hence, the number of rows in matrix must be equal to 3.
We also recall that the number of rows in matrix is equal to the number of rows in matrix , and likewise the number of columns in is equal to the number of columns in matrix . We are given that matrix is of order and matrix is of order , and we can see that the numbers of rows in matrices and are the same. Since matrix has 1 column, this tells us that the number of columns in must be equal to 1.
Hence, matrix is of order . This is option D.
In the next example, we will check the criterion for matrix multiplication to determine whether the given matrix multiplication is well defined.
### Example 3: Finding the Product of Two Given Matrices
Given that determine if possible.
Recall that the number of columns in matrix must equal the number of rows in matrix to compute matrix multiplication . We also recall that the order of a matrix is given by
We can see that matrix has 2 rows and 3 columns and matrix has 2 rows and 2 columns. Since the number of columns in is not equal to the number of rows in matrix , the matrix multiplication is undefined.
In the previous examples, we considered the property of the order of matrices in matrix multiplication. Now that we know when a pair of matrices can be multiplied, let us consider how to multiply matrices. The simplest matrix multiplication is the multiplication of a row matrix by a column matrix.
### How To: Multiplying Row Matrices by Column Matrices
Let and be row and column matrices respectively. Then, matrix is well defined and is of order . The entry of this matrix is obtained by multiplying each entry in the row matrix by the corresponding entry in the column matrix and then summing all the products.
We will demonstrate this process in the next example.
### Example 4: Finding the Product of Two Given Matrices
Consider the matrices
Find , if possible.
We know that matrix multiplication is only possible if the number of columns of the first matrix matches the number of rows of the second matrix. We note that the number of columns of the first matrix and the number of rows of the second matrix are both equal to 3, so it is possible to compute matrix multiplication .
We also know that the multiplication of an matrix by an matrix results in an matrix. We can see that the orders of matrices and are and respectively. Hence, , , and . This tells us that matrix is of order .
Since matrix has only one row, it is a row matrix. Likewise, matrix is a column matrix. Recall that we can multiply a row matrix by a column matrix by multiplying each entry in matrix by the corresponding entry in the column of and summing all the products. In the following computation, we have highlighted the corresponding entries in each matrix with the same color:
We can see that the order of is , as expected.
Hence, .
In the previous example, we multiplied a row matrix by a column matrix, which resulted in a matrix with a single entry, by multiplying the th entry of the row matrix by the th entry of the column matrix and summing all the products. The resulting matrix had only one entry because the first matrix in matrix multiplication had one row and the second matrix had one column.
This process can be generalized to multiply any pair of matrices with compatible orders, where the resulting matrix may have multiple entries. To multiply a matrix with multiple rows by a matrix with multiple columns, we need to pick one row from the first matrix and one column from the second matrix. Treating the selected row and column as row and column matrices, respectively, we can multiply the row matrix and the column matrix using the method introduced earlier. We continue this process until each of the rows of the first matrix is multiplied by each of the columns of the second matrix.
### How To: Multiplying Matrices
Let and be matrices of orders and respectively. For each and , we can compute the entry in the th row and th column of matrix by multiplying the th row of by the th column of .
We know that if we multiply a matrix of order by a matrix of order , we obtain a matrix of order . This means that we need to compute this times to complete the matrix multiplication.
Let us demonstrate this process graphically by multiplying a matrix by a matrix. We know that the resulting matrix will be of order , which means that we need to multiply a row by a column 9 times. Consider the following matrices:
We can see that matrix has 2 columns and matrix has 2 rows. Hence, we can compute the product . We also know that the product will be of order . First, we can multiply the first row by the first column to find the entry in the first row and first column of matrix :
Next, we multiply the first row of by the second column of to obtain
This process continues until we have completed the matrix multiplication:
### Example 5: Finding the Product of Two Given Matrices
Given that find if possible.
We know that matrix multiplication is only possible if the number of columns of the first matrix matches the number of rows of the second matrix. We note that the number of columns of the first matrix and the number of rows of the second matrix are both equal to 3, so it is possible to compute matrix multiplication .
We also know that the multiplication of an matrix by an matrix results in an matrix. We can see that the orders of matrices and are and respectively. Hence, , , and . This tells us that matrix is of order . We need to compute the entries of this matrix.
Recall that the entry in the th row and th column of matrix is obtained by multiplying the th row of by the th column of . We begin by taking the first row of , which can be written as a column matrix . We also take the first (and only) column of , which can be written as the column matrix . Multiplying each entry in the row matrix by the corresponding entry in the column matrix and then summing all the products,
Since this is the product of the first row of and the first column of , this tells us that 7 is the entry in the first row and first column of matrix .
Next, we multiply the second row of , , by the first column of , , which leads to
Hence, 5 is the entry in the second row and first column of matrix . This leads to
In the previous example, we multiplied a pair of matrices that had compatible orders. We did this by taking the row submatrices of the first matrix and the column submatrices of the second matrix and then multiplying them together. While this is the correct process for multiplying two matrices, it is not efficient to write out the row and column matrices each time. Instead, we can abbreviate these computations by writing the product of the corresponding row and column matrices within each entry of the resulting matrix, as seen in the following formula.
### Definition: Matrix Multiplication
Let and be matrices of orders and , respectively, given by
Then, the product matrix has order and is given by where
In the next example, we will multiply a pair of matrices using this formula.
### Example 6: Finding the Product of Two Given Matrices
Consider the matrices
Find , if possible.
We know that matrix multiplication is only possible if the number of columns of the first matrix matches the number of rows of the second matrix. We note that the number of columns of the first matrix and the number of rows of the second matrix are both equal to 2, so it is possible to compute matrix multiplication .
We also know that the multiplication of an matrix by an matrix results in an matrix. We can see that the orders of matrices and are and respectively. Hence, , , and . This tells us that matrix is of order . We need to compute 9 entries of this matrix.
Recall that the entry in the th row and th column of matrix is obtained by multiplying each entry of the th row of by the corresponding entry of the th column of and then summing all the products.
For instance, to obtain the entry in the first row and first column of matrix , we need to multiply the first row matrix of , , by the first column matrix of , . This means that we multiply each entry in the row matrix by the corresponding entry in the column matrix and then sum all the products. This gives
Hence, the entry in the first row and first column of matrix is . We can continue in the same manner until we complete the matrix:
In this explainer, we discussed how to multiply two matrices of compatible orders. We note that matrix multiplication is quite different from the multiplication of two real numbers as it has a more complex structure. While it is hard to see why we need to define matrix multiplication in this manner, there is a good reason for doing this. This reason will become clearer as we learn about more advanced topics on matrices. However, we can see a glimpse of this reason when we consider a real-world example that can be solved using matrix multiplication.
In our final example, we will consider a real-world application of matrix multiplication.
### Example 7: SolvingWord Problems by Applying Operations on Matrices
The table below shows the number of different types of rooms in three hotels owned by a company. If a single room costs 160 LE per night, a double room costs 430 LE per night, and a suite costs 740 LE per night, determine the companyโs daily income when all the rooms are occupied.
HotelSingle RoomDouble RoomSuite
First Hotel457415
Second Hotel487419
Third Hotel499410
In this example, we need to determine the companyโs daily income when all the rooms are occupied. This company owns three hotels, and the number of rooms of each type is listed in the given table. If we take only the numbers in the table, we can form a matrix, which we can write as
The first column of this matrix represents the number of single rooms in each hotel. To compute the companyโs income from these rooms, we need to multiply these numbers by 160 LE, which is the cost of a single room. Similarly, we need to multiply the second column by 430 LE, which is the cost of a double room. In the same manner, the entries in the third column should be multiplied by 740 LE, which is the cost of a suite. In the end, we can obtain the total daily income by summing all the entries of the resulting matrix:
If we sum the entries in each row, we obtain
๏45ร160+74ร430+15ร74048ร160+74ร430+19ร74049ร160+94ร430+10ร740๏. (1)
The entry in each row of the matrix above tells us the daily income from each hotel. We can finish the computation by finding each entry above and adding up the entries. But let us pause here for a second to note that the matrix above can also be obtained when we multiply our original matrix by the column matrix containing the cost of each room type:
Let us consider this multiplication. We know that matrix multiplication is only possible if the number of columns of the first matrix matches the number of rows of the second matrix. We note that the number of columns of the first matrix and the number of rows of the second matrix are both equal to 3, so it is possible to compute this matrix multiplication.
We also know that we can multiply a pair of matrices by multiplying each row of the first matrix by each column of the second matrix. Let us multiply the first row of the matrix, , by the column of the second matrix, . This means that we multiply each entry in the row matrix by the corresponding entry in the column matrix and then sum all the products. This gives which is the same as the first row of the matrix given in (1). Similarly, we can see that multiplying the second and third rows of the matrix by the column matrix leads to the entries in the second and third rows of the matrix in (1). Computing each entry in (1) gives us
Adding up all the entries, we have
Hence, when all the rooms are occupied, the companyโs daily income is 159โโโ340 LE.
Let us finish by recapping a few important concepts from this explainer.
### Key Points
• Let and be matrices. To compute matrix multiplication , the number of columns in must be equal to the number of rows in . If is an matrix for some positive integers and , must be an matrix for some positive integer . In this case, is an matrix:
• Let and be row and column matrices respectively. Then, matrix is well defined and is of order . The entry of this matrix is obtained by multiplying each entry in the row matrix by the corresponding entry in the column matrix and then summing all the products.
• Let and be matrices of orders and , respectively, given by Then, the product matrix has order and is given by where
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Mathematics Empty Set , Equal Sets , Finite and Infinite Sets & Cardinality of a Finite Set
Click for Only Video
### Topics Covered
star The Empty Set
star Finite and Infinite Sets & Cardinality of a Finite Set
star Equal Sets
### The Empty Set
\color{purple}ul(✓✓) \color{purple} " DEFINITION ALERT"
A set which does not contain any element is called the "empty set" or the "null set" or the "void set".
\color{green} ✍️ The empty set is denoted by the symbol \color{blue}(Phi) or \color{blue}({ }).
color{green}(E.g.)
(i) Let A = {x : 1 < x < 2, x is a natural number}. Here is no natural number between 1 and 2. So A is the empty set.
(ii) B = {x : x^2 – 2 = 0 and x is rational number}. Then B is the empty set because the equation x^2 – 2 = 0 is not satisfied by any rational value of x.
Q 1904734658
Which of the following are examples of the null set:
(i) Set of odd natural numbers divisible by 2.
(ii) Set of even prime numbers.
(iii) { x : x is a natural numbers , x < 5 and x > 7}
(iv) { y : y is a point common to any two parallel lines}
Class 11 Exercise 1.2 Q.No. 1
Solution:
(i) Set of odd natural numbers divisible by 2 is a null set, because no odd natural number is divisible by 2.
(ii) We know that 2 is the only even prime number. Therefore, set of even prime numbers is not a null set .
(iii) { x : x in N, x < 5 and x > 7 } is a null set, because there is no natural number which is less than 5 and greater than 7 simultaneously.
(iv) { v : y is a point common to any two parallel lines} is a nuII set, because there is no point common to any two parallel lines.
### Finite and Infinite Sets & Cardinality of a Finite Set
\color{purple}ul(✓✓) \color{purple} " DEFINITION ALERT"
A set which is empty or consists of a definite number of elements is called "finite" otherwise, the set is called "infinite."
\color{green} "Consider some examples :"
(i) Let W be the set of the days of the week. Then W is finite.
(ii) Let S be the set of solutions of the equation x^2 –16 = 0. Then S is finite.
(iii) Let G be the set of points on a line. Then G is infinite.
\color{green} ✍️ \color{green} \mathbf(KEY \ POINTS)
• When we represent a set in the roster form, we write all the elements of the set within braces { }.
• It is not possible to write all the elements of an infinite set within braces { } because the numbers of elements of such a set is not finite.
• So, we represent some infinite set in the roster form by writing a few elements which clearly indicate the structure of the set followed ( or preceded ) by three dots.
"For example," {1, 2, 3 . . .} is the set of natural numbers,
\ \ \ \ \ \ \ \ \ \ \ {1, 3, 5, 7, . . .} is the set of odd natural numbers,
\ \ \ \ \ \ \ \ \ \ \ {. . .,–3, –2, –1, 0,1, 2 ,3, . . .} is the set of integers. All these sets are infinite.
\color{green} ✍️ \color{green} \mathbf(KEY \ POINTS)
• All infinite sets cannot be described in the roster form.
• For example, the set of real numbers cannot be described in this form, because the elements of this set do not follow any particular pattern.
\color{green} ✍️ \color{green} \mathbf( "Cardinality of a finite set" )
"Cardinality" of a finite Set is defined as total number of elements in a set.
Q 3017267189
State which of the following sets are finite or infinite :form.
(i)\ \ {x : x in N and (x . 1) (x .2) = 0}
(ii)\ \ {x : x in N and x^2 = 4}
(iii)\ \ {x : x in N and 2x .1 = 0}
(iv)\ \ {x : x in N and x is prime}
(v) \ \{x : x in N and x is odd}
Solution:
(i) Given set = {1, 2}. Hence, it is finite.
(ii) Given set = {2}. Hence, it is finite.
(iii) Given set = phi. Hence, it is finite.
(iv) The given set is the set of all prime numbers and since set of prime
numbers is infinite. Hence the given set is infinite
(v) Since there are infinite number of odd numbers, hence, the given set is
infinite.
Q 1914034850
Which of the following sets arc finite or infinite:
(i) The set of months of 2 year
(ii) {1, 2, 3, ... }
(iii) {1, 2, 3, ... , 99, 100}
(iv) The set of positive integers greater than 100.
(v) The set of prime numbers less than 99.
Class 11 Exercise 1.2 Q.No. 2
Solution:
(i) Finite
(ii) Infinite
(iii) Finite
(iv) Infinite
(v) Finite.
Q 1984034857
State whether each of the following set is finite or infinite:
(i) The set of lines which are parallel to the x-axis.
(ii) The set of letters in the English alphabet.
(iii) The set of numbers which are multiple of 5.
(iv) The set of animals living on the earth.
(v) The set of circles passing through the origin (0, 0).
Class 11 Exercise 1.2 Q.No. 3
Solution:
(i) Infinite
(ii) Finite
(iii) Infinite
(iv) Finite
(v) Infinite
### Equal Sets :
Given two sets A and B, if every element of A is also an element of B and if every element of B is also an element of A, then the sets A and B are said to be equal. Clearly, the two sets have exactly the same elements.
\color{purple}ul(✓✓) \color{purple} " DEFINITION ALERT"
Two sets A and B are said to be equal if they have exactly the same elements and we write A = B. Otherwise, the sets are said to be unequal and we write A ≠ B.
"We consider the following examples :"
(i) Let A = {1, 2, 3, 4} and B = {3, 1, 4, 2}. Then A = B.
(ii) Let A be the set of prime numbers less than 6 and P the set of prime factors of 30. Then A and P are equal, since 2, 3 and 5 are the only prime factors of 30 and also these are less than 6.
\color{green} ✍️ \color{green} \mathbf(KEY \ POINT)
A set does not change if one or more elements of the set are repeated.
For example, the sets A = {1, 2, 3} and B = {2, 2, 1, 3, 3} are equal, since each element of A is in B and vice-versa .
Q 3037367282
Find the pairs of equal sets, if any, give reasons:
A = {0},
B = {x : x > 15 and x < 5},
C = {x : x – 5 = 0 },
D = {x: x^2 = 25},
E = {x : x is an integral positive root of the equation x^2 – 2x –15 = 0}.
Solution:
Since 0 ∈ A and 0 does not belong to any of the sets B, C, D and E, it
follows that, A ≠ B, A ≠ C, A ≠ D, A ≠ E.
Since B = phi but none of the other sets are empty. Therefore B ≠ C, B ≠ D
and B ≠ E. Also C = {5} but –5 ∈ D, hence C ≠ D.
Since E = {5}, C = E. Further, D = {–5, 5} and E = {5}, we find that, D ≠ E.
Thus, the only pair of equal sets is C and E.
Q 3067367285
Which of the following pairs of sets are equal? Justify your answer.
(i) X, the set of letters in “"ALLOY"” and B, the set of letters in “LOYAL”.
(ii) A = {n : n ∈ Z and n^2 ≤ 4} and B = {x : x ∈ R and x^2 – 3x + 2 = 0}.
Solution:
(i) We have, X = {A, L, L, O, Y}, B = {L, O, Y, A, L}. Then X and B are
equal sets as repetition of elements in a set do not change a set. Thus,
X = {A, L, O, Y} = B
(ii) A = {–2, –1, 0, 1, 2}, B = {1, 2}. Since 0 ∈ A and 0 ∉ B, A and B are not equal sets.
Q 3037578482
Show that the set of letters needed to spell CATARACT ” and the set of letters needed to spell “ TRACT” are equal.
Solution:
Let X be the set of letters in “CATARACT”. Then
X = { C, A, T, R }
Let Y be the set of letters in “ TRACT”. Then
Y = { T, R, A, C, T } = { T, R, A, C }
Since every element in X is in Y and every element in Y is in X. It follows that X = Y.
Q 1904134958
In the following, state whether A= B or not:
(i) A = { a , b , c , d } , B = { d , c , b , a }
(ii) A= {4, 8, 12, 16} , B = {8, 4, 16, 18}
(iii) A= {2, 4, 6, 8, 10}
B = {x : x is positive even integer and x le 10}
(iv} A = {x : x is a multiple of 10}
B = { 10, 15, 20, 25, 30, ... }
Class 11 Exercise 1.2 Q.No. 4
Solution:
(i) Yes, A = B
(ii) No, A ne B
(iii) Yes, A = B
(iv) No, A ne B.
Q 1924245151
Are the following pairs of sets equal? Gives reasons.
(i) A = {2, 3}, B = {x : x is solution of x^2 + 5x + 6 = 0}
(ii) A= { x : x is a letter in the word text ('FOLLOW'})
B = {y : y is a letter in the word text ('WOLF'})
Class 11 Exercise 1.2 Q.No. 5
Solution:
(i) No, A= {2, 3}
But solution of x^2 + 5x + 6 = 0
=> x^2 + 3x + 2x + 6 = 0
=> x ( x + 3 ) + 2 (x +3) =0
=> ( x + 3 ) ( x + 2 ) = 0
=> x + 3 = 0 or x + 2 = 0
=> x = -3 or x = - 2
=> B= {-2, -3}
Hence,A ne B
(ii) Yes, A= { F, O, L, W}
and B = {W,O, L, F}
Hence, A=B
Q 1974245156
From the sets given below, select equal sets:
A= {2, 4, 8, 12},
B = {1, 2 , 3 , 4 },
C= {4, 8, 12, 14},
D = {3, 1, 4, 2},
E = {-1, 1 }, F = {0, a} ,
G = { 1 , - 1 } , H = { 0 , 1 }
Class 11 Exercise 1.2 Q.No. 6
Solution:
(i) B = D ,
(ii) E = G
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# Proof of Sum Rule of Differentiation
The sum rule of differentiation can be derived in differential calculus from first principle. $f{(x)}$ and $g{(x)}$ are two differential functions and the sum of them is written as $f{(x)}+g{(x)}$. The derivative of sum of two functions with respect to $x$ is expressed in mathematical form as follows.
$\dfrac{d}{dx}{\, \Big(f{(x)}+g{(x)}\Big)}$
### Define the derivative in limiting operation
Take, $s{(x)} = f{(x)}+g{(x)}$ and then $s{(x+\Delta x)} = f{(x+\Delta x)}+g{(x+\Delta x)}$
Now, express the derivative of the function $s{(x)}$ with respect to $x$ in limiting operation as per definition of the derivative.
$\dfrac{d}{dx}{\, \Big(s{(x)}\Big)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{s{(x+\Delta x)}-s{(x)}}{\Delta x}}$
Now, replace the values of functions $s{(x)}$ and $s{(x+\Delta x)}$
$\implies$ $\dfrac{d}{dx}{\, \Big(f{(x)}+g{(x)}\Big)}$ $\,=\,$ $\displaystyle \large \lim_{\Delta x \,\to\, 0}{\normalsize \dfrac{\Big(f{(x+\Delta x)}+g{(x+\Delta x)}\Big)-\Big(f{(x)}+g{(x)}\Big)}{\Delta x}}$
### Simplify the function
Take $\Delta x = h$ and simplify the function for deriving the derivative of sum of the functions.
$\implies$ $\dfrac{d}{dx}{\, \Big(f{(x)}+g{(x)}\Big)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{\Big(f{(x+h)}+g{(x+h)}\Big)-\Big(f{(x)}+g{(x)}\Big)}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}+g{(x+h)}-f{(x)}-g{(x)}}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}+g{(x+h)}-g{(x)}}{h}}$
$=\,\,\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \Bigg[\dfrac{f{(x+h)}-f{(x)}}{h}+\dfrac{g{(x+h)}-g{(x)}}{h}\Bigg]}$
### Evaluate the Limits of the functions
The limit of sum of functions is equal to sum of their limits as per sum rule of limits.
$\implies$ $\dfrac{d}{dx}{\, \Big(f{(x)}+g{(x)}\Big)}$ $\,=\,$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{f{(x+h)}-f{(x)}}{h}}$ $+$ $\displaystyle \large \lim_{h \,\to\, 0}{\normalsize \dfrac{g{(x+h)}-g{(x)}}{h}}$
According to the first principle of derivative, the limit of each function is the derivative of that function.
$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \Big(f{(x)}+g{(x)}\Big)}$ $\,=\,$ $\dfrac{d}{dx}{\, f{(x)}}$ $+$ $\dfrac{d}{dx}{\, g{(x)}}$
Therefore, it is proved that the derivative of sum of two functions is equal to sum of their derivatives. The derivative property is called the sum rule of differentiation. The derivative sum rule can also be used to sum of more than two terms.
$\,\,\, \therefore \,\,\,\,\,\,$ $\dfrac{d}{dx}{\, \Big(f{(x)}+g{(x)}+\ldots\Big)}$ $\,=\,$ $\dfrac{d}{dx}{\, f{(x)}}$ $+$ $\dfrac{d}{dx}{\, g{(x)}}$ $+$ $\ldots$
Thus, the sum rule of differentiation is derived mathematically in differential calculus from first principle.
Take $u = f{(x)}$, $v = g{(x)}$ and so on. The sum rule of derivatives can be written in two forms.
#### Leibniz’s notation
$(1) \,\,\,$ $\dfrac{d}{dx}{\, (u+v+w+\ldots)}$ $\,=\,$ $\dfrac{du}{dx}$ $+$ $\dfrac{dv}{dx}$ $+$ $\dfrac{dw}{dx}$ $+$ $\ldots$
#### Differentials notation
$(2) \,\,\,$ ${d}{\, (u+v+w+\ldots)}$ $\,=\,$ $du$ $+$ $dv$ $+$ $dw$ $+$ $\ldots$
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# How do you simplify (27^-⅔)/(27^-½)?
May 7, 2016
This is easier than it appears at first, the bases are both 27.
Use one of the first laws of indices. "If you are dividing and the bases are the same, subtract the indices"
Notice that $\frac{2}{3}$ is a bigger fraction than $\frac{1}{2}$
Therefore subtracting in the numerator will give a negative answer, which will need further attention, We can subtract in the denominator rather.
1/27^(-1/2 - (-2)/3 = 1/27^(-1/2 + 2/3 $\text{ } - \frac{1}{2} + \frac{2}{3}$
$\text{ } = \frac{- 3 + 4}{6}$
=$\frac{1}{27} ^ \left(\frac{1}{6}\right)$
=$\frac{1}{{3}^{3}} ^ \left(\frac{1}{6}\right)$
=1/(3^(1/2) = $\frac{1}{\sqrt{3}}$
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Question
# If the sum of the coefficients in the expansion of ${\left( {x + y} \right)^n}$ is 4096, find the greatest coefficient in the expansion.
Hint: Sum of coefficients of ${\left( {x + y} \right)^n}$ is obtained when we put $x = y = 1$. And the greatest coefficient is the coefficient of the middle term(s) in its binomial expansion.
According to the question, the sum of coefficients in the expansion of ${\left( {x + y} \right)^n}$ is 4096.
We know that the sum of coefficients is the value of the expansion if we put all the variables equal to 1. Hence here we will put $x = y = 1$. So, we have:
$\Rightarrow {\left( {1 + 1} \right)^n} = 4096, \\ \Rightarrow {2^n} = 4096, \\ \Rightarrow {2^n} = {2^{12}}, \\ \Rightarrow n = 12 \\$
Since $n = 12$, the expansion is of ${\left( {x + y} \right)^{12}}$ and it will have a total of 13 terms.
We know that the greatest coefficient is the middle term. In this case, it will be of 7th term.
The general term for binomial expansion of ${\left( {x + y} \right)^{12}}$ is:
$\Rightarrow {T_{r + 1}}{ = ^{12}}{C_r}{x^{12 - r}}.{y^r}$
For middle term (i.e. 7th term), we will put $r = 6$:
$\Rightarrow {T_7}{ = ^{12}}{C_6}{x^6}.{y^6}$
Thus the coefficient of the middle term is $^{12}{C_6} = 924$
And hence the greatest coefficient in the expansion is 924.
Note:
In the expansion of ${\left( {x + y} \right)^n}$, coefficient of the middle term is $^n{C_{\dfrac{n}{2}}}$ if $n$ is even.
But if $n$ is odd, there will be two middle terms having coefficients $^n{C_{\dfrac{{\left( {n - 1} \right)}}{2}}}$ and $^n{C_{\dfrac{{\left( {n + 1} \right)}}{2}}}$. The value of the coefficients will be the same though.
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