text stringlengths 22 1.01M |
|---|
# How to Teach Fractions to the 3rd Grade
Save
Students are typically introduced to fractions in second grade. If you are teaching third grade students this year, start by reviewing concepts they would have learned last year, such as visually representing basic fractions, comparing simple fractions and the terms numerator and denominator. After a brief refresher, you can lead your students on a more advanced study of fractions including ordering fractions, equivalent fractions and adding and subtracting fractions. Use different teaching approaches, including demonstrations on the board, experiential activities with manipulative's, worksheets and games, so that all students are more likely to comprehend this major mathematical curriculum area.
### Things You'll Need
• Candies
• Sandwich bags
• Fraction manipulative (circles and rectangles)
• Worksheets
• Review what students would have learned last year about fractions by drawing a circle and dividing it into four equal pieces on the board. Color in one of the pieces and ask if anyone knows what fraction this represents.
• Write the correct answer, 1/4, on the board and ask students if they remember what the top number and the bottom number are called. Students should say numerator and denominator respectively.
• Pass out a sandwich bag of small candies of different colors to each student. Call out a color and ask several students what fraction of their candies are that color. Check each student to see if they have counted the total number of candies and the fraction correctly.
• Introduce the concept of equivalent fractions by passing out copies of rectangular, fraction manipulatives, such as those available on Kitchen Table's Math website.
• Ask children to color each bar a different color. Thus the whole, 1 piece would be one color, the half, 1/2 pieces would be another color, and so on.
• Demonstrate to students how to determine equivalent fractions with their rectangular manipulatives once they have cut them out. Use your own set of manipulatives or draw something similar on the board. For example, ask students how many quarter, 1/4, pieces can fit underneath one of the half, 1/2, pieces. The students should answer two pieces, meaning that one half is equivalent to two quarters -- 1/2 and 2/4 are equivalent fractions.
• Repeat this practice of determining equivalent fractions with the whole class at least 10 times; pass out a follow-up worksheet for students to work on.
• ### Other People Are Reading
• Teach students how to order fractions on a number line and to determine which fractions are worth more using the same rectangular manipulatives. For example, students can determine that 2/3s is greater than 1/2 by placing two 1/3 pieces (1/3 1/3) under one 1/2 piece. Also show students that if the numerator and the denominator are the same, the fraction always equals a whole or 1. Provide students with a follow-up worksheets.
• Teach students how to add and subtract fractions that have the same denominator. Tell them that they add or subtract the numerators and leave the denominators as is. For example one quarter plus two quarters equal three quarters: 1/4 + 2/4 = 3/4. Provide demonstrations on the board and with manipulatives and provide follow-up exercises.
• Allow students to practice the new skills they have learned through playing individual or group games. Assign 10 minutes of playing online fraction games for homework or to a student who has finished his in-class work ahead of time. Organize a fraction scavenger hunt by hiding equivalent fraction cards around the classroom or a team competition where players race to determine the answer to fraction problems.
## Tips & Warnings
• From the beginning of the fractions unit, it is a good idea to call fractions by their common names such as "one third", rather than "one over three," or "half" rather than "one over two."
## References
• Photo Credit Vintage railway distance sign or marker image by pablosphotos from Fotolia.com
Promoted By Zergnet
## Related Searches
Check It Out
### Can You Take Advantage Of Student Loan Forgiveness?
M
Is DIY in your DNA? Become part of our maker community. |
## The Geometry of Commutativity
One of the elementary observations about real numbers is that if we add the two numbers, regarding of the order, their sum is unique. We know that 4 + 3 = 3 + 4. This property is also the same when we multiply. If we have two numbers, regarding of the order, their product is also unique. For example, 6 x 3 = 3 x 6.
In general if we have real numbers a and b, we have a + b is always equal to b + a. Also, if we have real numbers c and d, c x d is always equal to d x c. These are called the commutative property of addition and multiplication, two of the axioms of algebra.
Although axioms, as we have discussed, are statements that we accept without proof, some axioms can be explained intuitively. In this post, we are going to discuss the two axioms mentioned above intuitively using geometric representations.
The geometry of a + b
One way of understanding why a + b is always equal to b + a is by representing the two numbers as lengths. For instance, we can represent the addition of integers above (4+3 = 3 + 4) as the sum of the lengths of two segments. In geometric representation, reversing the order of the numbers in the operation is just like rotating the segment 180 degrees about the center as shown below. Of course, rotating the segment way will never change its length.
Now, since that we can substitute any positive numbers to a and b we are sure that the commutative property is true for all positive numbers. » Read more
## Vectors, Parallelograms, and Commutativity
Basics of Vectors
Scalar quantities are quantities specified by magnitude. Mass, area, density are examples of scalar quantities. There are quantities that have both magnitude and direction. For example, when we say 2 kilometers east, 2 kilometers is our magnitude and east is our direction. Quantities with both magnitude and direction are called vectors.
In mathematics, vectors are usually represented by a directed line segment. The arrow of the directed segment is called its head, and the other end is called its foot.
Comparing Vectors
Two vectors are said to be equal if they have the same length and the same direction. Plainly speaking, if we look at the geometric representation of vectors, equal vectors have the “same slope and the same direction (of the arrow). In Figure 1, it is clear that there only two equal vectors – vector u and vector v.
Figure 1
The negative of a vector is the vector equal in length to a certain vector, but with opposite direction. For instance, in Figure 1, – u and u are negative of each other and vector u is also the negative of vector v. » Read more |
# Question: How Many Golf Balls Does It Take To Fill A School Bus?
## How many golf balls fit in a school bus interview question?
Divide that 2.5 cubic inches into 1.6 million and you come up with 660,000 golf balls. However, since there are seats and crap in there taking up space and also since the spherical shape of a golf ball means there will be considerable empty space between them when stacked, I’ll round down to 500,000 golf balls.
## How many footballs can fit in a bus?
74*495,000) = 366,000 Balls. Suppose a standard school bus is about 8 feet wide by 6 feet high by 20 feet long – An assumption.
## How many golf balls fit in a 747 answer?
Battison: But if we have an airliner, we do have those seats that can’t hold the golf balls. Josaphat: 56,784 cubic feet divided by 12 cubic feet. That total answer would be 4,732 golf balls.
You might be interested: Often asked: Who Invented The First School Bus?
## How many golf balls will fit in the Empire State Building?
25,780,645,161 balls. This of course disregards the irregularity of the structure and assumes the 37 million cubic feet is a volume of a simple box.
## How many balls can you fit in a limousine?
To get the total number of tennis balls that will fit into a limo, we need to divide the volume of the limo (500,000 cubic inches) by the volume of a tennis ball (4 cubic inches); 500,000/4 = 125,000. Now, we can say that 125,000 tennis ball will fit in a limousine.
## How many golf balls are on the moon?
There are two golf balls on the Moon. It is TRUE. Astronaut Alan Shepard is the fifth man to walk on the Moon and the first (and only) to have played golf there. He hit two balls during the Apollo 14 mission.
## How many golf balls can you put in a 5 gallon bucket?
The math suggests, a spherically sized bucket can contain around 340 balls, although, usually golfers use a 5-gallon bucket to carry an average of fifty to eighty balls in or out of the golf pitch.
## How many tennis balls can you fit inside a BMW?
The tennis ball fits in a regular car 28,0000/4 = 70,000. So, the total number of a tennis ball is 70,000, which easily fit in your car. How many tennis balls fit in a BMW? Some of our honorable visitors ask this question.
## How many ping pong balls are allowed on a plane?
The question is fairly satisfactorily answered: “The 747-400 has a passenger volume of 876 cubic meters, plus a cargo volume of 159 cubic meters, for a total of 1035; the balls have a diameter of 40mm; this gives about 22,870,000 ping pong balls.” The answerer notes that the weight of this many balls would prevent the
You might be interested: FAQ: When Did They Bring The School Bus Up From The Bottom Fo Lake Chelan?
## How many ping pong balls can you fit in a 747?
The volume of the tennis ball would be 47.916πcm^3. Finally, after dividing the volume of the aeroplane by that to the tennis ball we arrive at 688,705. Therefore, we can fit approximately 688,705 tennis balls in a Boeing 747.
## How many tennis balls can you fit in a Boeing 747?
Interview Answers I’d assume that 25% of the plane’s volume is taken up by seats, infrastructure (support beams etc.) and equipment, reducing the volume to 25*60*pi. This results in about 37 million tennis balls.
## How many balls can fit in a room interview questions?
Suggested answer: The average mens’ basketball has a diameter of 25cm. There are approximately 30cm in a foot. Therefore you could fit one inflated basketball in a 1 foot cubed space. Therefore you could fit 1000 inflated basketballs inside a room with a volume of 1000 ft cubed.
## How many quarters would have to be stacked to reach the Empire State Building?
A quarter is about. 05 inches thick, so there are about 240 quarters in a foot. If the Empire State Building is about 1,400 feet tall, it would take about 336,000 quarters to reach the top.
## How many balls can you fit in a Mini?
Now for the number of tennis balls which can fit in Mini Cooper. We need to divide by 4. Round up to 140,000, as it’s easier to divide by 4, so 140,000/4 is 35,000. |
NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability
In this page we have NCERT Solutions for Class 12 Maths Chapter 5: Continuity and Differentiability for EXERCISE 5.1 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Main Idea of the Exercise
A function is continuous at x = c if the function is defined at x = c and if the value of the function at x = c equals the limit of the function at x = c. If f is not continuous at c, we say f is discontinuous at c and c is called a point of discontinuity of f
Question 1
Prove that the function f(x) = 5x – 3 is continuous at x = 0, at x = – 3 and at x = 5.
Solution
At x=0
f (0) = 5(0) – 3=-3
Therefore, limit and f (0) are same. So, it is continuous
At x=-3
f(-3) = 5(-3) – 3=-18
Therefore, limit and f (-3) are same. So, it is continuous
At x=5
f(5) = 5(5) – 3=22
Therefore limit and f(5) are same. So, it is continuous
Question 2.
Examine the continuity of the function f(x) = 2x2 – 1 at x = 3.
Solution
At x=3
f(3) = 2x2 – 1 =2(9) – 1=17
Therefore limit and f(3) are same. So, it is continuous at x=3
Question 3.
Examine the following functions for continuity.
(a) f (x) = x – 5
(b) f (x) =1/(x-5) , x ≠5
(c) f (x) = (x2 -25)/ (x+5), x ≠-5
(d) f (x) = | x – 5|
Solution
a) f (x) = x – 5
The function is clearly defined at every point and f (c) =c-5 for every real number c.
Also,
Since
So, function is continuous at every real number
b) f (x) =1/(x-5) , x ≠5
The function is clearly defined at every point except x=5 and f (c) =1/(c-5) for every real number c except c=5
Since
So, function is continuous at every real number in its domain
c) f (x) = (x2 -25)/ (x+5), x ≠-5
The function is clearly defined at every point except x=-5 and f (c) = (c2 -25)/ (c+5), for every real number c except c=-5
Since
So, function is continuous at every real number in its domain
d) f (x) = | x – 5|
This function can re-written as
f(x) =
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 5 or c = 5 or c > 5
Case 1 c < 5
f (x) = 5 – x
f (c) = 5 – c
Since
Function is continuous at c< 5
Case 2 c > 5
f (x) = x – 5
f (c) = c – 5
Since
Function is continuous at c > 5
Case 3 c = 5
f (x) = x – 5
f (5) = 5 – 5=0
Since
Function is continuous at c =5
Hence, f is continuous at every real number and therefore, it is a continuous function.
Question 4.
Prove that the function f (x) = xn is continuous at x = n, where n is a positive integer.
Solution
The given function is f (x) = xn
It is evident that f is defined at all positive integers, n, and its value at n is nn
Therefore, f is continuous at n, where n is a positive integer.
Question 5.
Is the function f defined by?
continuous at x = 0? At x = 1? At x = 2?
Solution
At x = 0,
f(0) = 0.
Therefore, f is continuous at x = 0
At x = 1,
f(1) = 1.
The left-hand limit of f at x = 1 is,
The right-hand limit of f at x = 1 is,
Therefore, function is not continuous at x=1
At x = 2,
f(2) = 5.
Therefore, f is continuous at x = 2
Find all points of discontinuity of f, where f is defined by
Question 6.
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 2 or c = 2 or c > 2
It can be easily proved for c < 2 and c> 2. We have seen that in lot of previous example
For c=2
f(2) =2(2)+3=7
The left-hand limit of f at x = 2 is,
The right-hand limit of f at x = 2 is,
It is observed that the left and right hand limit of f at x = 2 do not coincide.
Therefore, f is not continuous at x = 2
Hence, x = 2 is the only point of discontinuity of f.
Question 7.
Solution
For x≤ -3
F(x) = |x| +3
Since x is negative in that, we can re-write this as
F(x) = 3-x
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < -3 or c = -3 or -3 < c < 3 or c=3 or c> 3
It can be easily proved for c < -3 , -3 < c < 3 And c> 3 We have seen that in lot of previous example, we will look at common points
For c=-3
f(-3) =3-(-3) =6
The left-hand limit of f at x = -3 is,
The right-hand limit of f at x = -3 is,
Therefore, f is continuous at x = -3
For c=3
f(3) =6(3)+2 =20
The left-hand limit of f at x = 3 is,
The right-hand limit of f at x = 3 is,
It is observed that the left and right hand limit of f at x = 3 do not coincide.
Therefore, f is not continuous at x = 3
Hence, x = 3 is the only point of discontinuity of f.
Question 8.
Solution
We know for x < 0 , |x|=-x and x> 0, |x|=x, So this function can be written as
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0
It can be easily proved for c < 0 And c> 0 We have seen that in lot of previous example, we will look at common points
For c=0
F(0) =0
The left-hand limit of f at x = 0 is,
The right-hand limit of f at x = 0 is,
It is observed that the left and right hand limit of f at x = 0 do not coincide.
Therefore, f is not continuous at x = 0
Hence, x = 0 is the only point of discontinuity of f.
Question 9.
Solution
We know for x < 0 , |x|=-x So this function can be written as
F(x) =-1 if x < 0
And
F(x) =-1 if x≥0
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0
It can be easily proved for c < 0 And c> 0 . We have seen that in lot of previous example, we will look at common points
For c=0
F(0) =0
The left-hand limit of f at x = 0 is,
The right-hand limit of f at x = 0 is,
Therefore, f is continuous at x = 0
Hence, there is no point of discontinuity of f.
Question 10.
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 1 or c = 1 or c> 1
It can be easily proved for c < 1 And c> 1 .We have seen that in lot of previous example, we will look at common points
For c=1
f(1) =1+1=2
The left-hand limit of f at x = 1 is,
The right-hand limit of f at x = 1 is,
Therefore, f is continuous at x = 1
Hence, there is no point of discontinuity of f.
Question 11.
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 2 or c = 2 or c> 2
It can be easily proved for c < 2 And c> 2 .We have seen that in lot of previous example, we will look at common points
For c=2
f(1) =x3 -3 =8-3=5
The left-hand limit of f at x = 2 is,
The right-hand limit of f at x = 2 is,
Therefore, f is continuous at x = 2
Hence, there is no point of discontinuity of f.
Question 12.
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 1 or c = 1 or c> 1
It can be easily proved for c < 1 And c> 1 . We have seen that in lot of previous example, we will look at common points
For c=1
F(1) =x10 -1 =0
The left-hand limit of f at x = 1 is,
The right-hand limit of f at x = 1 is,
Therefore, f is discontinuous at x = 1
Hence, x = 1 is the only point of discontinuity of f.
Question 13. Is the function defined by
a continuous function?
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 1 or c = 1 or c> 1
It can be easily proved for c < 1 And c> 1. We have seen that in lot of previous example, we will look at common points
For c=1
f(1) =x+5 =6
The left-hand limit of f at x = 1 is,
The right-hand limit of f at x = 1 is,
Therefore, f is discontinuous at x = 1
Hence, x = 1 is the only point of discontinuity of f.
Discuss the continuity of the function f, where f is defined by
Question 14.
Solution
The given function is defined at all points of the interval [0,10].
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, 0 ≤ c < 1 or c = 1 or 1<c < 3 or c=3 or 3 <c < 10
It can be easily proved for 0 ≤ c < 1 , 1< c < 3 , 3 <c < 10 We have seen that in lot of previous example, we will look at common points
For c=1
f(1) = 3
The left-hand limit of f at x = 1 is,
The right-hand limit of f at x = 1 is,
Therefore, f is discontinuous at x = 1
For c=3
f(3)= 5
The left-hand limit of f at x = 3 is,
The right-hand limit of f at x = 3 is,
Therefore, f is discontinuous at x = 3
Hence, f is not continuous at x = 1 and x = 3
Question 15.
Solution
This can be solved similarly above
Answer is f is not continuous only at x = 1
Question 16.
Solution
This can be solved similarly above
Answer is f is continuous at all the points
Question 17.
Find the relationship between a and b so that the function f defined by
is continuous at x = 3.
Solution
When the function is continuous at any point, it Left-hand limit, right-hand limit and function value are equal
F(3) =3a+1
So
3a+1= 3a+1= 3b+3
Or a =b + (2/3)
Question 18.
For what value of λ is the function defined by
continuous at x = 0? What about continuity at x = 1?
Solution
If f is continuous at x=0, then
f(0) =0
Since Left hand limit is not equal to right hand limit, Therefore, there is no value of λ for which the function f is continuous at x=0.
At x=1,
f(1)=4x+1=4×1+1=5
Therefore, for any values of λ, f is continuous at x=1.
Question 19.
Show that the function defined by g(x) = x – [x] is discontinuous at all integral points. Here [x] denotes the greatest integer less than or equal to x.
Solution
This function g is defined at all integral points
Let n be an integer.
Then,
g(n)=n−[n]=n−n=0
Then left hand limit of f at x=n is
Then right hand limit of f at x=n is,
So, Left hand limit and Right hand limit are not equal
Therefore, f is not continuous at x=n
Hence, g is discontinuous at all integral points.
Question 20.
Is the function defined by f(x) = x2 – sin x + 5 continuous at x = π?
Solution
At x=π, f(x)=f(π)= π2−sinπ+5=π2 −0+5=π2+5
Consider
Put x=π+h
If x→π, then it is evident that h→0
2+5
So, function is continuous at x = π
Question 21.
Discuss the continuity of the following functions:
(a) f (x) = sin x + cos x
(b) f (x) = sin x – cos x
(c) f (x) = sin x . cos x
Solution
It is known that if g and h are two continuous functions, then g+ h , g − h ,g.h are also continuous functions
.So, It has to prove first that g (x) = sin x and h (x) = cos x are continuous functions.
Let g (x) = sin x
It is evident that g (x) = sin x is defined for every real number.
g(c) =sin(c)
Let c be a real number. Put x = c + h
If x → c, then h → 0
So, g(x) is continuous function
Now
Let h (x) = cos x
It is evident that h (x) = cos x is defined for every real number.
h(c) =cos(c)
Let c be a real number. Put x = c + h
If x → c, then h → 0
So, h(x) is a continuous function
Therefore, it can be concluded that
(a) f (x) = g (x) + h (x) = sin x + cos x is a continuous function
(b) f (x) = g (x) − h (x) = sin x − cos x is a continuous function
(c) f (x) = g (x) × h (x) = sin x × cos x is a continuous function
Question 22. Discuss the continuity of the cosine, cosecant, secant and cotangent functions.
Solution
Now these functions can be rewritten as
Cosec (x) = 1/ sin (x) sin(x) ≠0
Sec (x) = 1/cos (x) cos (x) ≠0
Cot(x) = cos(x)/sin(x) sin(x) ≠0
It is known to us that if g and h are two continuous functions, then
Comparing to this, we have h(x) = cos (x) and g(x) =sin(x)
We have already proved the previous question about continuity of these function
So, we can say that
Cosec (x) is continuous at all the point except where sin(x) ≠0
i.e x ≠nπ where n ∈ Z
sec (x) is continuous at all the point except where cos(x) ≠0
i.e x ≠(2n+1)π/2 where n ∈ Z
Cot(x) is continuous at all the point except where sin(x) ≠0
i.e x ≠nπ where n ∈ Z
Question 23.
Find all points of discontinuity of f, where
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we should look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0
It can be easily proved for c < 0 And c> 0. We have seen that in lot of previous example, we will look at common points
At c=0
f(0) = 0+1=1
The left-hand limit of f at x = 0 is,
The right-hand limit of f at x = 0 is,
Therefore, f is continuous at x = 1
Hence, There is no point of discontinuity of f.
Question 24.
Determine if f defined by
is a continuous function?
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0
It can be easily proved for c < 0 And c> 0. We have seen that in lot of previous example, we will look at common points
At c=0
f(0) = 0
The left-hand limit of f at x = 0 is,
Now we know that, for x≠0
Multiplying by x2
(No change in direction as + is multiplied)
Or
So
Similarly The right-hand limit of f at x = 0 can be proved
Therefore, f is continuous at x = 0
Hence, There is no point of discontinuity of f.
Question 25.
Examine the continuity of f, where f is defined by
Solution
This function f is defined at all points of the real line.
For continuity in these type of question, we must look for the continuity at the common point and point outside the common point
Let c be a point on a real line. Then, c < 0 or c = 0 or c> 0
It can be easily proved for c < 0 And c> 0. We have seen that in lot of previous example, we will look at common points
At c=0
f(0)=-1
The left-hand limit of f at x = 0 is,
The right-hand limit of f at x = 0 is,
Therefore, f is continuous at x = 0
From the above observations, it can be concluded that f is continuous at every point of the real line.
Thus, f is a continuous function.
Find the values of k so that the function f is continuous at the indicated point in Exercises 26 to 29.
Question 26.
Solution
For the function to be continuous at x=π/2, we have
F(π/2) =3
Let x = π/2 + h
When x → π/2, then h → 0
Now k/2=3
K=6
Question 27.
Solution
For the function to be continuous at x=2, we have
F(2) =4k
The left-hand limit of f at x = 2 is,
The right-hand limit of f at x = 2 is,
Now
4k=4k=3
Or
K=3/4
Question 28.
Solution
For the function to be continuous at x=π, we have
F(π) =k π+1
The left-hand limit of f at x = π is,
The right-hand limit of f at x = π is,
Now
k π+1=-1
k=-2/ π
Question 29.
Solution
For the function to be continuous at x=5, we have
F(5) =5k+1
The left-hand limit of f at x = 5 is,
The right-hand limit of f at x = 5 is,
Now
5k+1=10
k=9/5
Question 30.
Find the values of a and b such that the function defined by
is a continuous function.
Solution
The function is defined for all the real values and it is given as continuous
It is also continuous at x=2 and x=10
Let’s check the continuity at those values
At x=2
f(2)=5
The left-hand limit of f at x = 2 is,
The right-hand limit of f at x = 2 is,
Now
2a+b=5 -(P)
At x=10
F(10)=21
The left-hand limit of f at x = 10 is,
The right-hand limit of f at x = 10 is,
Now
10a+b=21 --(Q)
From (p) and (q)
a=2 and b=1
Question 31.
Show that the function defined by f (x) = cos (x2) is a continuous function.
Solution
This function f is defined for every real number and f can be written as the composition of two functions, as,
f = g o h, where g (x) = cos x and h (x) = x2
Because g o h =g[h(x)] =cos (x2)
Now It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is
continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
So, we just need to prove that g(x) =cos(x) and h(x) =x2 are continuous function
First let us prove for g(x)=cos (x)
It is evident that g (x) = cos x is defined for every real number.
g(c) =cos(c)
Let c be a real number. Put x = c + h
If x → c, then h → 0
So, g(x) is continuous
Now h (x) = x2
It is evident that h (x) = x2 is defined for every real number.
h(c) =c2
So, h(x) is continuous
Since both g(x) and h(x) are continuous. F(x) is continuous function
Question 32.
Show that the function defined by f(x) = | cos x | is a continuous function.
Solution
This function f is defined for every real number and f can be written as the composition
of two functions, as,
f = g o h, where g (x) = |x| and h (x) = cos x
Because g o h =g[h(x)] =| cos x |
Now It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is
continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
So, we just need to prove that g(x) =|x| and h(x) =cos(x) are continuous function
First let us prove for g(x)=|x|
It can be re-written as
g(x) = -x for x < 0
=+x for x ≥0
It is evident that g (x) is defined for every real number.
For c < 0
g(c) =-c
So, it is continuous
For c = 0
g(c) =0
So, it is continuous
For c > 0
g(c) =c
So, it is continuous
let us prove for h(x)=cos (x)
It is evident that h (x) = cos x is defined for every real number.
h(c) =cos(c)
Let c be a real number. Put x = c + h
If x → c, then h → 0
Since both g(x) and h(x) are continuous. F(x) is continuous function
Question 33.
Examine that sin | x | is a continuous function.
Solution
This function f is defined for every real number and f can be written as the composition
of two functions, as,
f = g o h, where g (x) = sin(x) and h (x) = |x|
Because g o h =g[h(x)] = sin | x |
Now It is known that for real valued functions g and h, such that (g o h) is defined at c, if g is
continuous at c and if f is continuous at g (c), then (f o g) is continuous at c.
So, we just need to prove that g(x) =sin (x) and h(x) =|x| are continuous function
Let g (x) = sin x
It is evident that g (x) = sin x is defined for every real number.
g(c) =sin(c)
Let c be a real number. Put x = c + h
If x → c, then h → 0
So, g(x) is continuous function
Now let us prove for h(x)=|x|
It can be re-written as
h(x) = -x for x < 0
=+x for x ≥0
It is evident that h (x) is defined for every real number.
For c < 0
h(c) =-c
So, it is continuous
For c = 0
h(c) =0
So, it is continuous
For c > 0
h(c) =c
So, it is continuous
So h(x) is continous
Since both g(x) and h(x) are continuous. F(x) is continuous function
Question 34.
Find all the points of discontinuity of f defined by f (x) = | x | – |x + 1|.
Solution
The two functions, g and h, are defined as
g(x) =|x|
h(x) =|x+1|
Then, f = g – h
It is known that if g and h are two continuous functions, then g + h,g − h,g.h are also continuous functions.
So, we just need to prove that g(x) and h(x) is continuous
The continuity of g and h is examined first.
First let us prove for g(x)=|x|
It can be re-written as
g(x) = -x for x < 0
=+x for x ≥0
It is evident that g (x) is defined for every real number.
For c < 0
g(c) =-c
So, it is continuous
For c = 0
g(c) =0
So, it is continuous
For c > 0
g(c) =c
So, it is continuous
Now let us look
h(x)= |x+1|
It can be re-written as
h(x) = -(x+1) for x < -1
=x+1 for x ≥-1
It is evident that h (x) is defined for every real number.
For c < -1
h(c) =-(c+1)
So, it is continuous
For c = -1
h(-1) =0
So it is continuous
For c > -1
h(c) =c+1
So, it is continuous
So, both g(x) and h(x) are continuous
Therefore, f = g − h is also a continuous function.
Therefore, f has no point of discontinuity. |
Factors of 90 room the numbers, which offers the an outcome as 90 as soon as multiplied together in pair that two. The determinants of 90 deserve to be hopeful or negative, but they cannot be in the fraction or decimal form. Because that example, the pair components of 90 deserve to be (1, 90) or (-1, -90). If we multiply the pair of negative factors, such together multiplying -1 and also -90, we will get the original number 90. The factors of a number have the right to be discovered using the division method and the prime factorization method. In this article, us will learn what are the factors of 90, pair factors and the prime determinants of 90 making use of the two various methods and many addressed examples.
You are watching: What are the prime factors of 90
Table that Contents:
## What room the factors of 90?
The components of 90 are the numbers, which room multiplied in pairs resulting in an initial number 90. Together the number is an also composite number, that has countless factors various other than 1 and also 90. The components of 90 deserve to be positive or negative. Thus, the positive determinants of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Factors the 90: 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90Prime administer of 90: 3×3×5×2 or 32 × 5 × 2
## Pair components of 90
The pair of numbers which are multiplied together, resulting in an original number 90 is the pair components of 90. The positive and the an unfavorable pair components of 90 are provided below:
Positive Pair components of 90:
Positive components of 90 Positive Pair factors of 90 1 × 90 (1, 90) 2 × 45 (2, 45) 3 × 30 (3, 30) 5 × 18 (5, 18) 6 × 15 (6, 15) 9 × 10 (9, 10)
Negative Pair determinants of 90:
Negative determinants of 90 Negative Pair factors of 90 -1 × -90 (-1, -90) -2 × -45 (-2, -45) -3 × -30 (-3, -30) -5 × -18 (-5, -18) -6 × -15 (-6, -15) -9 × -10 (-9, -10)
## Factors the 90 by division Method
In the department method, the components of 90 can be found by separating 90 by different consecutive integers. If the essence divides 90 exactly without leave a remainder, then the integers space the components of 90. Now, start separating 90 by different integer numbers.
90/1 = 90 (Factor is 1 and also remainder is 0)90/2 = 45 (Factor is 2 and also remainder is 0)90/3 = 30 (Factor is 3 and remainder is 0)90/5 = 18 (Factor is 5 and remainder is 0)90/6 = 15 (Factor is 6 and also remainder is 0)90/9 = 10 (Factor is 9 and remainder is 0)90/10 = 9 (Factor is 10 and remainder is 0)90/15 = 6 (Factor is 15 and remainder is 0)90/18 = 5 (Factor is 18 and remainder is 0)90/30 = 3 (Factor is 30 and remainder is 0)90/45 = 2 (Factor is 45 and remainder is 0)90/90 = 1 (Factor is 90 and remainder is 0)
If we division 90 through numbers various other than 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45 and 90, we will acquire a remainder. Hence, the factors of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
The number 90 is a composite number. Now, the prime factors of 90 deserve to be discovered as shown below:
The an initial step is to divide the number 90 with the the smallest prime factor, i.e. 2 and also divide the output again by 2 it spins you acquire a fraction or strange number.
90 ÷ 2 = 45;
Divide 45 by 2.
45 ÷ 2 = 22.5; portion cannot be a factor.
Therefore, relocating to the following prime numbers, 3, 5, 7 and also so on.
Divide 45 through 3.
45 ÷ 3 = 15
15 ÷ 3 = 5
Now, we know 5 is a element number and it has only two factors; 1 and 5. Therefore, us cannot continue with the division method further.So, the prime determinants of 90 are 2 × 3 × 3 × 5 or 2 × 32 × 5, wherein 2, 3 and also 5 space the prime numbers.
### Sum of determinants of 90
We understand that the components of the number 90 room 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90. Now, to find the amount of the factors, include all the numbers, you will acquire a complete of 234.
Sum of factors of 90 = 1 + 2 + 3 + 5 + 6 + 9 + 10 + 15 + 18 + 30 + 45 + 90
Sum of factors of 90 = 234
If you exclude the number 90, girlfriend will obtain 144, i.e. 234 – 90 = 144
### Examples
Example 1:
Find the usual factors the 90 and 91.
Solution:
The components of 90 are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and also 90.
The determinants of 91 room 1, 7, 13, and also 91.
Therefore, the common factor of 90 and 91 is 1.
Example 2:
Find the common factors of 90 and also 89.
Solution:
Factors that 90 = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and 90.
Factors the 89 = 1 and 89.
As 89 is a prime number, the typical factor that 90 and 89 is 1 only.
Example 3:
Find the usual factors that 90 and also 100.
Solution:
The determinants of 90 room 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, and also 90.
The factors of 100 space 1, 2, 4, 5, 10, 20, 25, 50, and 100.
Hence, the typical factors that 90 and also 100 room 1, 2, 5 and 10.
See more: Direction Of Bond Polarity F Ch3 F Polar Or Non, Answered: For Each Bond, Select The Arrow That…
Links regarded Factors Factors of 24 Factors the 12 Factors of 20 Factors of 28 Factors that 21 Factors that 27 Factors of 35 Factors that 50 Factors of 40 Factors that 72
Learn an ext about factors and prime components at BYJU’S and likewise download BYJU’S – The Learning application for a much better experience and also get video lessons to study and also understand the concepts of Maths object for all the classes. |
# Word Problems on simultaneous linear equations
In this chapter we will discuss some problems related to the concept of simultaneous linear equations.
To solve the problems, you have to comprehend the statement carefully and then frame the linear equation.
After getting two linear equations, you then have to get the value of variables to get the final solution.
## Simultaneous linear equations word problems
Example 01
Twice one number minus three times a second is equal to 2, and the sum of these numbers is 11. Find the numbers.
Solution
Let the first number be x and second number be y.
(A) Let’s form the equation using the statements
First statement:
Twice the first number – three times second number equals two.
Forming the linear equation;
2x – 3y = 2.
Second statement
Sum of the two numbers is 11.
Linear equation becomes;
x + y = 11
(B) Solving the equations
Now we have two linear equations;
2x – 3y = 2
x + y = 11
Using substitution method.
Taking second equation and finding value of y.
x + y = 11
y = 11- x
Now putting this value of y in first equation.
2x – 3y = 2
2x – 3 (11 – x) = 2
2x – 33 + 3x = 2
5x = 2 + 33
5x = 35
x = 35 / 5
x = 7
Here we have got exact value of variable x.
Putting this value in any given equation.
Taking second equation
x + y = 11
7 + y = 11
y = 4
Hence, 7 & 4 are the two required numbers
Example 02
Six years hence a man’s age will be three times his son’s age, and three years ago he was nine times as old as his son. Find their present ages
Solution
Let the man’s present age be x and son’s present age be y.
(A) Forming the equations
First statement
After six years, man’s age will be three times the son age.
After 6 years;
Man age will be x + 6
Son age will be y + 6
At this point man’s age will be 6 times that of son.
(x + 6) = 3 (y + 6)
x + 6 = 3y + 18
x = 3y + 12
This is the first equation.
Second Statement
Three years ago men was nine times as old as his son
Age 3 years back;
Man age = x – 3
Son age = y – 3
At this time, men was 9 times as old as son.
x – 3 = 9(y – 3)
x – 3 = 9y -27
x – 9y = -24
This is our second equation
(B) Solving the equations
The two equations are;
x = 3y + 12
x – 9y = -24
Using substitution method.
Substitute the value x = 3y + 12 in second equation.
x – 9y = -24
3y + 12 – 9y = -24
-6y = -24 -12
-6y = -36
y = 36 / 6
y = 6 years.
Here we have got exact value of y. Put this value in first equation.
x = 3y + 12
x = 3 (6) + 12
x = 30 years
Hence, age of man is 30 years and age of son is 6 years.
Example 03
Two times a number plus ten times a second number equals twenty. Thirty times the second number plus three times the first number equals 45. Find the two numbers.
Solution
Let the first number be x and second number be y.
(A) Framing the equation
First statement
Two times a number plus ten times a second number equals twenty.
2x + 10y = 20
Second statement
Thirty times the second number plus three times the first number equals 45
3x+ 30y = 45
(B) Solving the equations
Here we have got two equations.
2x + 10y = 20
3x+ 30y = 45
We will solve the equations using substitution method.
Find the value of x in first equation.
\mathtt{2x+\ 10y\ =\ 20}\\\ \\ \mathtt{x=\frac{20-10y}{2}}\\\ \\ \mathtt{x\ =\ \frac{20}{2} -\frac{10y}{2}}\\\ \\ \mathtt{x=\ 10\ -\ 5y}
Put the value of x in 2nd equation.
3x + 30y= 45
3 (10 – 5y) + 30y = 45
30 – 15y + 30y = 45
30 + 15y = 45
15y = 15
y = 1
Here we have got exact value of y. Put this value in first equation to get the value of x.
2x + 10y = 20
2x+ 10 = 20
2x = 10
x = 5
Hence, the two numbers are 5 and 1.
Example 04
At a certain time in a deer park, the number of heads and the number of legs of deer and human visitors were counted and it was found that there were 41 heads and 136 legs. Find the number of deer and human visitors in the park.
Solution
Let number of deer be x and number of humans be y.
We know that deer has 4 legs and 1 head. So total number of deer legs and head for one deer will be 4x and x.
Total deer legs = 4x
Total deer head = x
We know that humans have 2 legs and 1 head. So count of heads and legs is given as;
Total human legs = 2y
Total human head = y
(A) Forming equations
Statement 01
There were total of 41 heads.
x + y = 41
Statement 02
There are 136 legs
4x + 2y = 136
(B) Solving equations
Here we have got two equations;
x + y = 41
4x + 2y = 136
Using substitution method for solving linear equations.
Find value of x in first equation.
x + y = 41
x = 41 – y
Substitute this value of x in 2nd equation.
4x + 2y = 136
4 (41 – y) + 2y = 136
164 -4y + 2y = 126
-2y = 136 – 164
-2y = -28
y = 14
Now we have got exact value of y. Put this value in first equation to get the value of x.
x + y = 41
x + 14 = 41
x = 27
Hence, there are 27 deer and 14 humans.
Example 5
2 shirts and 3 trousers cost $425. 3 Shirts and 2 Trousers cost$ 350. Find the price of one short and trousers.
Solution
Let price of one shirt be x and price of trouser be y.
(A) Framing equations
First statement
2 shirts and 3 trousers cost $425 2x + 3y = 425 Second statement 3 Shirts and 2 Trousers cost$ 350
3x + 2y = 350
(B) Solving equations
Here we have got two equations.
2x + 3y = 425
3x + 2y = 350
Here we will use elimination method to solve equations.
Multiply first equation by 3 and second equation by 2.
First equation
3 ( 2x + 3y = 425 )
6x + 9y = 1275
Second equation
2 ( 3x + 2y = 350)
6x + 4y = 700
Sligh both equation vertically and subtract.
Solving the equation further.
5y = 575
y = 575 / 5
y = 115
Here we got exact value of y. Put this value in first equation to get value of x.
2x + 3y = 425
2x + 3 (115) = 425
2x + 345 = 425
2x = 80
x = 40
Hence, price of shirt is 40$and price of trouser is 115$
You cannot copy content of this page |
# Which is a line of symmetry?
## Which is a line of symmetry?
A line of symmetry is a line that cuts a shape exactly in half. This means that if you were to fold the shape along the line, both halves would match exactly. Equally, if you were to place a mirror along the line, the shape would remain unchanged. A square has 4 lines of symmetry, as shown below.
## What lines of symmetry does a rhombus have?
2
How many letters have a line of symmetry?
The F and G have zero lines of symmetry. Those letters cannot be folded in half in any way with the parts matching up. The rest of the letters, A, B, C, D, and E all have only 1 line of symmetry. Notice that the A has a vertical line of symmetry, while the B, C, D, and E have a horizontal line of symmetry.
Does a square have 2 lines of symmetry?
But there’s something special about squares. They can also be divided in half diagonally. That means the square has the two lines of symmetry that all rectangles have plus two more, each diagonal. All squares have four lines of symmetry.
### What is the definition for symmetry?
1 : balanced proportions also : beauty of form arising from balanced proportions. 2 : the property of being symmetrical especially : correspondence in size, shape, and relative position of parts on opposite sides of a dividing line or median plane or about a center or axis — compare bilateral symmetry, radial symmetry.
### How many lines of symmetry is in a square?
4
How do you introduce symmetry?
Once students touch on the idea that the wings match in some way, introduce the word “symmetry.” Explain that something has symmetry if it can be split into two mirror-image halves. For example, a butterfly is symmetrical because you can fold a picture of it in half and see that both sides match.
What types of symmetry can a shape have?
There are three basic types of symmetry: rotational symmetry, reflection symmetry, and point symmetry.
## What is symmetry plan?
1 : a plane through a crystal that divides the crystal into two parts that are mirror images of each other. 2 : a vertical fore-and-aft plane that divides an airplane into symmetrical halves.
## How lines of symmetry makes monuments beautiful?
The line of symmetry of the monuments is basically the line segment which divides the monument into two geometrically equivalent segments. The architectural design on the both sides of the line of symmetry are equivalent. In this way, the line of symmetry makes monuments much more beautiful.
Which shape has no lines of symmetry?
scalene triangle
Why does a rhombus have 2 lines of symmetry?
What are the Lines of Symmetry in Rhombus? A rhombus has 2 lines of symmetry which cuts it into two identical parts. Both the lines of symmetry in a rhombus are from its diagonals. So, it can be said the rhombus lines of symmetry are its both diagonals.
### What is symmetry and its types in biology?
Types of symmetry Radial symmetry: The organism looks like a pie. This pie can be cut up into roughly identical pieces. Bilateral symmetry: There is an axis; on both sides of the axis the organism looks roughly the same. Spherical symmetry: If the organism is cut through its center, the resulting parts look the same.
### Is a square a rhombus?
A rhombus is a quadrilateral (plane figure, closed shape, four sides) with four equal-length sides and opposite sides parallel to each other. All squares are rhombuses, but not all rhombuses are squares. The opposite interior angles of rhombuses are congruent.
Does a square have 6 lines of symmetry?
There are figures and shapes that can have more than one lines of symmetry. A circle has infinite lines of symmetry. Likewise, a triangle has three lines of symmetry, while rectangle and square have four such lines which divide them into identical parts.
How do you count lines of symmetry?
When a line divides a given figure into two equal halves such that the left and right halves matches exactly then we say that the figure is symmetrical about the line.
## Do all quadrilaterals have 2 lines of symmetry?
So the square has four lines of symmetry. The rectangle has only two, as it can be folded in half horizontally or vertically: students should be encouraged to try to fold the rectangle in half diagonally to see why this does not work. The trapezoid has only a vertical line of symmetry.
## Does a parallelogram have two 90 degree angles?
Right Angles in Parallelograms Although students are taught that four-sided figures with right angles — 90 degrees — are either squares or rectangles, they are also parallelograms, but with four congruent angles instead of two pairs of two congruent angles.
What do you mean by symmetry in biology?
Symmetry, in biology, the repetition of the parts in an animal or plant in an orderly fashion. Specifically, symmetry refers to a correspondence of body parts, in size, shape, and relative position, on opposite sides of a dividing line or distributed around a central point or axis.
What shape is a rhombus?
In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. Another name is equilateral quadrilateral, since equilateral means that all of its sides are equal in length.
### How many symmetry are there?
Regular Polygons
An Equilateral Triangle (3 sides) has 3 Lines of Symmetry
A Square (4 sides) has 4 Lines of Symmetry
A Regular Pentagon (5 sides) has 5 Lines of Symmetry
A Regular Hexagon (6 sides) has 6 Lines of Symmetry
A Regular Heptagon (7 sides) has 7 Lines of Symmetry
### How do you teach symmetry?
Start out with modeling how to create symmetry. Show your child how to match the left and right sides like a mirror on either side of the vertical line. You could even grab a small mirror and place it along the center of the leaf (or object) to show him the symmetry. Next, only create one side of the symmetrical image. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
# A body rolls down a staircase of $5$steps. Each step has height $0.1m$and width $0.1m$. With what velocity will the body reach the bottom.A) $\sqrt {\dfrac{5}{2}} m/s$B) $\sqrt {\dfrac{1}{2}} m/s$C) $2\sqrt 2 m/s$D) $\dfrac{5}{{\sqrt 2 }}m/s$
Last updated date: 19th Jun 2024
Total views: 53.4k
Views today: 1.53k
Answer
Verified
53.4k+ views
Hint: Here in this question firstly we will calculate the total height and width and then we will apply equation of motion as $d = \dfrac{{g{t^2}}}{2}$ and $d = ut$ where $d$, $u$ and $t$are distance ,speed and time respectively.
Complete step by step answer:
As we know in this question we are given with us the height and width of each step as $0.1m$ and $0.1m$. So to calculate total height and width multiply given heights and widths by $5$.
So total height and width are
$h = 5 \times 0.1$
$h = 0.5$, so now for width do the same as height
$w = 5 \times 0.1$
$w = 0.5$
So now to calculate time we have $d = u \times t$,so after substituting values we get
$0.5 = u \times t$
$t = \dfrac{{0.5}}{u}$,so now substitute the value of time in $d = \dfrac{{g{t^2}}}{2}$to calculate velocity with which it has fallen on ground, so we get
$0.5 = \dfrac{{10 \times {{\left( {\dfrac{{0.5}}{u}} \right)}^2}}}{2}$
$0.5 \times 2 = \dfrac{{10 \times 0.25}}{{{u^2}}}$
${u^2} = \dfrac{{2.5}}{1}$
$u = \sqrt {\dfrac{5}{2}}$
So, The correct option is A.
Additional Information:
In such questions, an important thing that we consider is the assumption that firstly, that the horizontal velocity of the ball is constant throughout the process and air resistance or friction are not present. Secondly, we assumed the fact that the ball does not bounce back after every vertical collision with the steps, if we had not assumed the same, the question would also need to consider the time taken by the ball to rise to a specific height considering how much energy the ball loses after every collision. Proceed with the explained process if the question provides you with the needed information about energy loss after every collision with the steps.
Note: As in this question we have used initial velocity $0$ and in newton’s equations of motion is has used acceleration of gravity as g and ignoring resistance by air that is we are assuming ideal stage that there is no force acting to body other than the force of gravity and ignoring the air resistance that can also move the ball in any other side of staircase. |
# Class 9 Maths NCERT Solution for Polynomial Chapter-2 part 2
In this page we have Class 9 Maths NCERT Solution for Polynomial Chapter-2 for EXERCISE 2 . Hope you like them and do not forget to like , social share and comment at the end of the page.
Question 1:
Find the value of the polynomial P(x)= 5x-4x2 +3 at
(i) x = 0 (ii) x = −1 (iii) x = 2
Solution:
(i) P(x)= 5x-4x2 +3
P(0)= 0-0+3=3
(ii) P(x)= 5x-4x2 +3
P(-1) =-5-4+3=-6
(iii) P(x)= 5x-4x2 +3
P(2)= 10-16+3=-3
Question 2
Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 − y + 1 (ii) p(t) = 2 + t + 2t2 − t3
(iii) p(x) = x3 (iv) p(x) = (x − 1) (x + 1)
Solution:
(i) p(y) = y2 − y + 1
p(0) = (0)2 − (0) + 1 = 1
p(1) = (1)2 − (1) + 1 = 1
p(2) = (2)2 − (2) + 1 = 3
(ii) p(t) = 2 + t + 2t2 − t3
p(0) = 2 + 0 + 2 (0)2 − (0)3 = 2
p(1) = 2 + (1) + 2(1)2 − (1)3
= 2 + 1 + 2 − 1 = 4
p(2) = 2 + 2 + 2(2)2 − (2)3
= 2 + 2 + 8 − 8 = 4
(iii) p(x) = x3
p(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8
(iv) p(x) = (x − 1) (x + 1)
p(0) = (0 − 1) (0 + 1) = (− 1) (1) = − 1
p(1) = (1 − 1) (1 + 1) = 0 (2) = 0
p(2) = (2 − 1 ) (2 + 1) = 1(3) = 3
Question 3
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x =– 1/3
(ii) p(x) = 5x – π, x =4/5
(iii) p(x) = x2 – 1, x = 1, –1
(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
(v) p(x) = x2 , x = 0
(vi) p(x) = lx + m, x = – m/l
(vii) p(x) = 3x2 – 1, x =-1/√3 and 2/√3
(viii) p(x) = 2x + 1, x =1/2
Solution:
(i) p(x) = 3x + 1,x= -1/3
p(-1/3) = 3 (-1/3)+1=−1+1=0
p(-1/3) = 0 which means that -1 /3is zero of the polynomial p(x) = 3x+1.
(ii) p(x) = 5x−π, x = 4/5
p(4/5) = 5(4/5) –π = 4−π
p(4/5) ≠ 0 which means that 4/5 is not zero of the polynomial p(x) = 5x − π.
(iii) p(x) = x2−1,x = 1,− 1
p(1)=12−1=1−1=0
p(−1)=(−1)2−1=1−1=0
Both p(1) and p(−1) are equal to 0. It means that 1 and -1 are zeroes of the polynomial p(x) = x− 1.
(iv) p(x) = (x+1)(x−2),x=−1,2
p(−1) = (−1+1)(−1−2) = 0×−3 = 0
p(2) = (2+1)(2−2) = 3×0 = 0
Both p(−1) and p(2) are equal to 0. It means that -1 and 2 are zeroes of the polynomial p(x) = (x+1)(x−2).
(v) p(x) = x2,x = 0
p(0) = 02 = 0
p(0) = 0 which means that 0 is the zero of the polynomial p(x) = x2.
(vi) p(x) = lx + m,x = =-m/
p(-m/l) = l(-m/l) + m = −m + m = 0
p(-m/l) = 0 which means that (–m/l) is zero of the polynomial p(x)=lx + m
(vii) p(x) = 3x2 − 1, x =-1/√3 and 2/√3
p(-1/√3) = 3(-1/√3)2−1=3(1/3)−1=1−1=0
p(2/√3) = 3(2/√3)−1 = 3 × (4/3)−1= 4−1 = 3
p(-1/√3) = 0 which means that -1/√3 is zero of the polynomial p(x) = 3x− 1.
p(2/√3) ≠ 0 which means that 2/√3 is not zero of the polynomial p(x)=3x− 1.
(viii) p(x) = 2x + 1 ,x =1/2
p(1/2)=2 × (1/2) + 1=1+1=2
p(1/2) ≠ 0. It means that 1/2 is not zero of the polynomial p(x)=2x+1.
Question 4.
Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5 (ii) p(x) = x – 5 (iii) p(x) = 2x + 5
(iv) p(x) = 3x – 2 (v) p(x) = 3x (vi) p(x) = ax, a ≠ 0
(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
Solution:
Zero of a polynomial is that value of the variable at which the value of the polynomial is obtained as 0.
(i) p(x) = x + 5
p(x) = 0
x + 5 = 0
x = − 5
Therefore, for x = −5, the value of the polynomial is 0 and hence, x = −5 is a zero of the given polynomial.
(ii) p(x) = x − 5
p(x) = 0
x − 5 = 0
x = 5
Therefore, for x = 5, the value of the polynomial is0 and hence, x = 5 is a zero of the given polynomial.
(iii) p(x) = 2x + 5
p(x) = 0
2x + 5 = 0
2x = − 5
x = - 5/2
Therefore, for x = - 5/2 , the value of the polynomial is 0 and hence, x = - 5/2 is a zero of the given polynomial.
(iv) p(x) = 3x − 2
p(x) = 0
3x − 2 = 0
x=2/3, so 2/3 is the zero of the polynomial
(v) p(x) = 3x
p(x) = 0
3x = 0
x = 0
So x=0 is the zero of the polynomial
(vi) p(x) = ax
p(x) = 0
ax = 0
x = 0
Therefore, for x = 0, the value of the polynomial is 0 and hence, x = 0 is a zero of the given polynomial.
(vii) p(x) = cx + d
p(x) = 0
cx+ d = 0
x=-d/c |
# Proving a theorem: Changing the Order of Differentiation (large question)
• Nov 1st 2010, 01:26 PM
Runty
Proving a theorem: Changing the Order of Differentiation (large question)
This question is right out of Taylor & Mann Advanced Calculus, Third Edition, and is a REALLY tricky one. Due to how context-heavy this is, I can't write it word-for-word.
We're to prove the following theorem:
Let $f(x,y)$ and its first partial derivatives $f_1, f_2$ be defined in a neighborhood of the point $(a,b)$, and suppose that $f_1$ and $f_2$ are differentiable at that point. Then $f_{12}(a,b)=f_{21}(a,b)$.
This theorem uses part of another theorem's proof to start things off, before getting to the parts that I can't figure out.
Let $h$ be a number different from zero such that the point $(a+h,b+h)$ is inside a square having its center at $(a,b)$. We then consider the following expression:
$D=f(a+h,b+h)-f(a+h,b)-f(a,b+h)+f(a,b)$
If we introduce the function
$\phi (x)=f(x,b+h)-f(x,b)$,
we can express $D$ in the form
$D=\phi (a+h)-\phi (a)$ (*)
Now $\phi$ has the derivative
$\phi '(x)=f_1(x,b+h)-f_1(x,b)$
Hence $\phi$ is continuous, and we may apply the mean-value theorem for derivatives to (*), obtaining the following:
$D=h\phi '(a+\theta_1 h)=h(f_1(a+\theta_1 h,b+h)-f_1(a+\theta_1 h,b))$, where $0<\theta_1 <1$
That ends the part of the separate proof; now to parts which I'm in the dark about (these are from the actual question).
From the fact that $f_1$ is differentiable at $(a,b)$, one can write
$f_1(a+\theta_1 h,b+h)=f_1(a,b)+f_{11}(a,b)\theta_1 h+f{12}(a,b)h+\epsilon_1 |h|$, where $\epsilon_1\rightarrow 0$ as $h\rightarrow 0$.
Explain why this is so.
Next, go on to explain how to obtain the following expression:
$D=h^2 f_{12}(a,b)+\epsilon |h|h$ where $\epsilon\rightarrow 0$ as $h\rightarrow 0$.
Explain the derivation of the similar expression
$D=h^2f_{21}(a,b)+\epsilon '|h|h$ where $\epsilon '\rightarrow 0$ as $h\rightarrow 0$,
using the fact that $f_2$ is differentiable at $(a,b)$.
With all this, complete the proof of the theorem.
----------
This is a lot of info to work with, I know. But I can't summarize it any better than this because the question has such a high context requirement.
• Nov 2nd 2010, 04:02 AM
CaptainBlack
Quote:
Originally Posted by Runty
This question is right out of Taylor & Mann Advanced Calculus, Third Edition, and is a REALLY tricky one. Due to how context-heavy this is, I can't write it word-for-word.
We're to prove the following theorem:
Let $f(x,y)$ and its first partial derivatives $f_1, f_2$ be defined in a neighborhood of the point $(a,b)$, and suppose that $f_1$ and $f_2$ are differentiable at that point. Then $f_{12}(a,b)=f_{21}(a,b)$.
This theorem uses part of another theorem's proof to start things off, before getting to the parts that I can't figure out.
Let $h$ be a number different from zero such that the point $(a+h,b+h)$ is inside a square having its center at $(a,b)$. We then consider the following expression:
$D=f(a+h,b+h)-f(a+h,b)-f(a,b+h)+f(a,b)$
If we introduce the function
$\phi (x)=f(x,b+h)-f(x,b)$,
we can express $D$ in the form
$D=\phi (a+h)-\phi (a)$ (*)
Now $\phi$ has the derivative
$\phi '(x)=f_1(x,b+h)-f_1(x,b)$
Hence $\phi$ is continuous, and we may apply the mean-value theorem for derivatives to (*), obtaining the following:
$D=h\phi '(a+\theta_1 h)=h(f_1(a+\theta_1 h,b+h)-f_1(a+\theta_1 h,b))$, where $0<\theta_1 <1$
That ends the part of the separate proof; now to parts which I'm in the dark about (these are from the actual question).
From the fact that $f_1$ is differentiable at $(a,b)$, one can write
$f_1(a+\theta_1 h,b+h)=f_1(a,b)+f_{11}(a,b)\theta_1 h+f{12}(a,b)h+\epsilon_1 |h|$, where $\epsilon_1\rightarrow 0$ as $h\rightarrow 0$.
Explain why this is so.
Repeated application of the 1 variable Taylor theorem:
If $f(x)$ is differentiable at $a$ then:
$f(x)=f(a)+(x-a)f'(a)+ g(x-a)$
where $\lim_{h\to 0} \frac{g(h)}{h}=0$
CB
• Nov 2nd 2010, 01:07 PM
Runty
Quote:
Originally Posted by CaptainBlack
Repeated application of the 1 variable Taylor theorem:
If $f(x)$ is differentiable at $a$ then:
$f(x)=f(a)+(x-a)f'(a)+ g(x-a)$
where $\lim_{h\to 0} \frac{g(h)}{h}=0$
CB
Are you sure you mean Taylor's theorem? Checking Wikipedia, Taylor's theorem doesn't seem to relate to this.
I'm thinking more to use something akin to this:
$u(h,k)=\frac{f(x+h,y+k)-f(x,y)-f_x(x,y)h-f_y(x,y)k}{\sqrt{h^2+k^2}}$, where $\epsilon_1 |h|$ takes the place of $u(h,k)$, and so forth for other parts.
• Nov 2nd 2010, 03:23 PM
CaptainBlack
Quote:
Originally Posted by Runty
Are you sure you mean Taylor's theorem? Checking Wikipedia, Taylor's theorem doesn't seem to relate to this.
It is exactly Taylor's theorem for a once differentiable function, which informally is that for small increments on $a$ the function approximately is linear and the error decreases "faster" than the increment as the increment goes to zero.
But please ignore these posts if you feel you have a better idea.
CB
• Nov 3rd 2010, 12:02 PM
Runty
I think we're meant to use the formulations of differentiability to prove that first part. It looks to be in the same vein as this:
$f(a+h,b+k)-f(a,b)=f_1(a,b)h+f_2(a,b)k+\epsilon (|h|+|k|)$,
where $\epsilon\rightarrow 0$ as $(h,k)\rightarrow (0,0)$.
We got to Taylor's theorem, but this question comes in the textbook before that point. As such, I think it's implied that we're not meant to use Taylor's theorem, even if it is a valid choice.
• Nov 5th 2010, 12:44 PM
Runty
Okay, I got through the first hurdle; I had to show that another Theorem followed from a prior chapter (and I'm hoping that it's a valid answer), but now comes a more difficult part.
I'm not sure how I'm supposed to get from here
$f_1(a+\theta_1 h,b+h)=f_1(a,b)+f_{11}(a,b)\theta_1 h+f{12}(a,b)h+\epsilon_1 |h|$, $\epsilon_1\rightarrow 0$ as $h\rightarrow 0$
to here
$D=h^2 f_{12}(a,b)+\epsilon |h|h$, $\epsilon\rightarrow 0$ as $h\rightarrow 0$
The big problem is that epsilon statement at the end of it. I don't know how it gets there.
EDIT: I should mention this; this question is more along the lines of what is found in the following link. Click here
Unfortunately, the proof of what's in that link is NOT the correct answer for what I'm supposed to answer. It's very close, but not quite there (check the wording CAREFULLY on the theorem).
Whereas the theory in the link has the second partial derivatives defined and continuous at a point, the theorem I'm supposed to prove does not have that statement. The theorem I'm to prove says that $f_1$ and $f_2$ are differentiable, but does not say that $f_{12}$ and $f_{21}$ are defined and continuous (even if they are implied).
• Nov 7th 2010, 12:51 PM
Runty
Sorry for triple-post, but I'm still stuck on the part I listed earlier. I dunno if this will help, but I'm out of ideas. (note that there's a lot to read)
This is a theorem we're meant to use to try and answer this.
Let the function $f(x,y)$ be defined in some neighborhood of the point $(a,b)$. Let the partial derivatives $f_1,f_2,f_{12},f_{21}$ also be defined in this neighborhood, and suppose that $f_{12}$ and $f_{21}$ are continuous at $(a,b)$. Then $f_{12}(a,b)=f_{21}(a,b)$.
Note the differences between this theorem and the theorem I'm meant to prove (in the first post). The theorem above has $f_{12}$ and $f_{21}$ defined and continuous at $(a,b)$, but the theorem I'm meant to prove only says that $f_1$ and $f_2$ are differentiable at $(a,b)$. This might sound redundant, but you can clearly see there's a difference.
(much of the following is excerpted from my textbook; this proof DOES NOT answer the question)
To prove the theorem above, I work entirely inside a square having its center at $(a,b)$, and lying inside the neighborhood described earlier. Let $h$ be a number different from zero such that $(a+h,b+h)$ is inside the square. Then consider the expression
$D=f(a+h,b+h)-f(a+h,b)-f(a,b+h)+f(a,b)$
If we introduce the function
$\phi (x)=f(x,b+h)-f(x,b)$,
we can express D in the form
$D=\phi (a+h)-\phi (a)$. (*)
Now $\phi$ has the derivative
$\phi '(x)=f_1(x,b+h)-f_1(x,b)$.
Hence $\phi$ is continuous, and we may apply the mean value theorem to (*), with the result
$=h\phi '(a+\theta_1 h)=h(f_1(a+\theta_1 h,b+h)-f_1(a+\theta_1 h,b))$, where $0<\theta_1<1$. (**)
Next, let
$g(y)=f_1(a+\theta_1h,y)$.
The function g has the derivative
$g'(y)=f_{12}(a+\theta_1h,y)$.
Now we can write (**) in the form
$=h(g(b+h)-g(b))$
and apply the mean value theorem. The result is
$D=h^2 g'(b+\theta_2h)=h^2f_{12}(a+\theta_1h,b+\theta_2h)$, where $0<\theta_2<1$.
Alternatively, we could have started by expressing D in the form
$D=\psi (b+h)-\psi (b)$, where
$\psi (y)=f(a+h,y)-f(a,y)$.
This procedure would have led to the following
$D=h^2f_{21}(a+\theta_4h,b+\theta_3h)$, where $0<\theta_3<1$, $0<\theta_4<1$.
On comparing the two expressions for D, we see that
$f_{12}(a+\theta_1h,b+\theta_2h)=f_{21}(a+\theta_4h ,b+\theta_3h)$. (***)
If we now make $h\rightarrow 0$, the points at which the derivatives in (***) are evaluated both approach $(a,b)$. Hence, by the assumed continuity of $f_{12}$ and $f_{21}$, we conclude that $f_{12}(a,b)=f_{21}(a,b)$.
(Remember, this is NOT the correct answer to the theory in the starting post)
This is the theory and proof that correlate closely to what I'm meant to solve, but unfortunately using this proof would be an incorrect answer. Though the proofs start off similarly, they deviate at (**).
As to this question, this is the part I'm stuck at.
We have
$f_1(a+\theta_1 h,b+h)=f_1(a,b)+f_{11}(a,b)\theta_1 h+f{12}(a,b)h+\epsilon_1 |h|$, where $\epsilon_1\rightarrow 0$ as $h\rightarrow 0$.
I now have to explain how to obtain the following expression:
$D=h^2 f_{12} (a,b)+\epsilon |h|h$, where $\epsilon\rightarrow 0$ as $h\rightarrow 0$.
If I could just get that, I could do the same things for $f_{21}$, and hopefully complete the proof afterwards. As such, I'd greatly appreciate any help that could be provided.
• Nov 8th 2010, 11:42 AM
Runty
Again, bad interpretation has made me miss out on answering this question correctly. I have my Prof.'s answers to this, and I doubt I'll get any good mark on this. Although in all honesty, this question was ****ing ridiculous.
Here is my Prof.'s answer:
Since $f_x$ is differentiable, we have:
(*) $f_x(a+h,b+k)=f_x(a,b)+hf_{xx}(a,b)+kf_{xy}(a,b)+u( h,k)\sqrt{h^2+k^2}$
where $u\rightarrow 0$ as $(h,k)\rightarrow (0,0)$.
If we replace $h$ by $\theta_1 h$ and $k$ by $h$, we get
$f_x(a+\theta_1 h,b+h)=f_x(a,b)+\theta_1 hf_{xx}(a,b)+hf_{xy}(a,b)+u(\theta_1 h,h)\sqrt{\theta_1^2 h^2+h^2}$
so $\epsilon_1 |h|=\sqrt{\theta_1^2 h^2+h^2}u\Rightarrow \epsilon_1=\sqrt{\theta_1^2+1}u$ and $u\rightarrow 0$ as $h\rightarrow 0$ (as $(\theta_1 h,h)\rightarrow (0,0)$) so $\epsilon_1\rightarrow 0$.
Alternatively, replace $h$ by $\theta_1 h$ and take $k=0$ in (*) to get
$f_x(a+\theta_1 h,b)=f_x(a,b)+\theta_1 hf_{xx}(a,b)+0f_{xy}(a,b)+u( h,0)\sqrt{\theta_1^2 h^2}$
so $\epsilon_2=\theta_1 u\rightarrow 0$ as $h\rightarrow 0$, as before.
Substitute into
$D=h\phi '(a+\theta_1 h)=h(f_x(a+\theta_1 h,b+h)-f_x(a+\theta_1 h,b))$
to get
$D=h(hf_{xy}(a,b)+\epsilon |h|)=h^2 f_{xy} (a,b)+\epsilon |h|h$, where $\epsilon=\epsilon_1-\epsilon_2$ and so $\epsilon\rightarrow 0$ as $h\rightarrow 0$.
Everything is symmetric in $x$ and $y$, so reversing their roles in the above gives
$D=h(hf_{yx}(a,b)+\epsilon '|h|)=h^2 f_{yx} (a,b)+\epsilon '|h|h$, where $\epsilon '\rightarrow 0$ as $h\rightarrow 0$.
Then $f_{xy}(a,b)=f_{yx}(a,b)+(\epsilon '-\epsilon)|h|/h$
and $|h|/h=\pm 1, \epsilon '-\epsilon\rightarrow 0$ as $h\rightarrow 0$ so we get $f_{xy}(a,b)=f_{yx}(a,b)$ as required.
This question seriously was a pain in the neck, and I'd rather my Prof. give us a little more to work with next time, or at least be HELPFUL when I ask it of him. |
# Teachers' tricks for KS2 maths
Is your child struggling with KS2 maths? Help them learn effectively and boost their number confidence in Year 3, 4, 5 and 6 with some useful calculation strategies and tips from primary teacher Phoebe Doyle.
## Times tables tricks
Times tables are just so tricky (not to mention tedious) to learn, but they really are essential building blocks for much of later maths. Thankfully, some of the tables lend well to some useful tricks that, once learned, making knowing them easy peasy.
4 times table trick
Double the number, double again – voila!
5 times table trick
Multiply the number by 10 instead of 5, then halve your answer.
9 times table trick
Hold both hands up in front of you. Whatever you’re multiplying by (for example 9), hold down that finger (starting from your left hand, so finger 9 is your right ring finger). The fingers to the left of the bent finger are the tens (in this case 80), and the fingers to the right the units (in this case 1; 9 x 9 = 81).
12 times table trick
The answer is 10x plus 2x.
All times tables
Make up rhymes and mnemonics, for example: Eight times eight fell on the floor, picked it up it's sixty four.
## Use what you know to help with what you don’t:
Teachers tell children this from Year 1 onwards, but it’s really important to remember it, even when doing maths at secondary school (and beyond). When working with numbers there are often patterns to be found which can really help you to work out calculations.
For example: 2.30 + 5.70
• Note the number bond to 10 in the decimals, so that makes 1
• Add 1 to the two whole numbers: 2 + 5 +1 = 8
## Checking answers – does that look right?
When you’ve completed a calculation, take a moment to look at it and the answer. Ask, does that look right?
For example, if you were multiplying a whole number by 6300, the number is going to be large and, as long as the whole number is not 1, bigger than 6300. You can also help to check answers by getting into the habit of doing rough calculations prior to completing the answer and seeing if they are approximately right or by using the inverse operation (so addition to check subtraction and vice versa and multiplication to check division).
## Mental maths stress-busters
This area od maths can prove extremely stressful, as it’s often done timed. Remember:
• Focus on the numbers.
• Listen to which calculation is required.
• You can usually ignore the rest!
## Teachers’ maths test tips
1. Don’t get stressed about tests – the stress will stop your brain from working! This is your opportunity to show off and a chance for your teachers to understand what you need to learn more about.
2. Take your time, never rush – if you have some spare time at the end, go back, attempt any questions you’ve missed and check all your answers.
3. If you’re stuck on a question, move on and come back to it if you have time at the end.
4. Be aware of the time, and if needed ask how much longer you have left. This will help you make sure you don’t run out of time when you still have lots of questions to complete.
5. If you don’t understand, ask – in some tests your teacher might be able to help to read a question or to understand it, in others they won't be able to. Ask if you need help and see what they can support you with. |
# 8.8: Triangle Classification
Difficulty Level: At Grade Created by: CK-12
Estimated3 minsto complete
%
Progress
Practice Triangle Classification
MEMORY METER
This indicates how strong in your memory this concept is
Progress
Estimated3 minsto complete
%
Estimated3 minsto complete
%
MEMORY METER
This indicates how strong in your memory this concept is
Kevin is constructing a model. The model has several pieces that are shaped like triangles. He notices that triangles have a lot of different sizes and variations in appearance. He decides to make a pile for each different classification of triangles. He finds one triangle that has a \begin{align*}110^o\end{align*} angle. How should he classify the triangle?
In this concept, you will learn how to classify triangles.
### Classifying Triangles
The angles in a triangle can vary a lot in size and shape, but they always total \begin{align*}180^\circ\end{align*}. The angles are used to classify triangles. A triangle can either be acute, obtuse or right.
Acute triangles have three angles that measure less than \begin{align*}90^\circ\end{align*}. Below are a few examples of acute triangles.
Notice that each angle in the triangles above is less than \begin{align*}90^\circ\end{align*}, but the total for each triangle is still \begin{align*}180^\circ\end{align*}.
A triangle that has an obtuse angle is classified as an obtuse triangle. This means that one angle in the triangle measures more than \begin{align*}90^\circ\end{align*}. Here are three examples of obtuse triangles.
You can see that obtuse triangles have one wide angle that is greater than \begin{align*}90^\circ\end{align*}. Still, the three angles in obtuse triangles always add up to \begin{align*}180^\circ\end{align*}. Only one angle must be obtuse to make it an obtuse triangle.
The third kind of triangle is a right triangle. Right triangles have one right angle that measures exactly \begin{align*}90^\circ\end{align*}. Often, a small box in the corner tells you when an angle is a right angle. Let’s examine a few right triangles.
Once again, even with a right angle, the three angles still total \begin{align*}180^\circ\end{align*}.
One short cut is to compare the angles to \begin{align*}90^\circ\end{align*}. If an angle is exactly \begin{align*}90^\circ\end{align*}, the triangle must be a right triangle. If any angle is more than \begin{align*}90^\circ\end{align*}, the triangle must be an obtuse triangle. If there are no right or obtuse angles, the triangle must be an acute triangle.
Let's look at an example.
Label the triangle as acute, obtuse or right.
First, look at the angles and compare the angles to \begin{align*}90^o\end{align*}.
All of the angles are less than \begin{align*}90^o\end{align*}.
Next, list the classifications of triangles.
Triangles can be acute, right or obtuse.
Then, select the classification that fits the criteria.
Acute.
The answer is that the triangle is an acute triangle.
Triangles can also be classified by the lengths of their sides.
A triangle with three equal sides is an equilateral triangle. It doesn’t matter how long the sides are, as long as they are all congruent, or equal. Here are a few examples of equilateral triangles.
An isosceles triangle has two congruent sides. It doesn’t matter which two sides, any two will do. Let’s look at a few examples of isosceles triangles.
The third type of triangle is a scalene triangle. In a scalene triangle, none of the sides are congruent.
Let's look at an example.
Classify the triangle as equilateral, isosceles, or scalene.
First, examine the lengths of the sides to see if any sides are congruent.
Two sides are 7 meters long, but the third side is shorter.
Then, classify the triangle.
This triangle is an isosceles triangle.
### Examples
#### Example 1
Earlier, you were given a problem about Kevin's model.
He has one triangle that has a \begin{align*}110^o\end{align*} angle. Classify the triangle as acute, obtuse or right.
First, look at the given angle and compare it to \begin{align*}90^o\end{align*}.
The angle is larger than \begin{align*}90^o\end{align*}.
Next, list the classifications of triangles.
Triangles can be acute, right or obtuse.
Then, select the classification that fits the criteria.
Obtuse.
The answer is that the triangle is an obtuse triangle.
#### Example 2
Classify the triangle as equilateral, isosceles or scalene.
First, examine the lengths of the sides to see if any sides are congruent.
All three sides equal 4.5 inches.
Then, determine if the the triangle is equilateral, isosceles or scalene.
Equilateral.
The answer is that the triangle is equilateral.
#### Example 3
Use the angles to classify the triangle.
First, look at the angles and compare the angles to \begin{align*}90^o\end{align*}.
One of the angles is equal to \begin{align*}90^o\end{align*}.
Next, list the classifications of triangles.
Triangles can be acute, right or obtuse.
Then, select the classification that fits the criteria.
Right.
The answer is that the triangle is a right triangle.
#### Example 4
Use the angles to classify the triangle.
First, look at the angles and compare the angles to \begin{align*}90^o\end{align*}.
None of the angles are \begin{align*}90^o\end{align*} . One of the angles is larger than \begin{align*}90^o\end{align*}.
Next, list the classifications of triangles.
Triangles can be acute, right or obtuse.
Then, select the classification that fits the criteria.
Obtuse.
The answer is that the triangle is an obtuse triangle.
#### Example 5
Identify the triangle as equilateral, isosceles or scalene.
First, examine the lengths of the sides to see if any sides are congruent.
None of the sides are congruent.
Then, determine if the the triangle is equilateral, isosceles or scalene.
Scalene.
The answer is that the triangle is scalene.
### Review
Find the measure of angle \begin{align*}H\end{align*} in each figure below.
Identify each triangle as right, acute, or obtuse.
Identify each triangle as equilateral, isosceles, or scalene.
Use what you have learned to answer each question.
1. True or false. An acute triangle has three sides that are all different lengths.
2. True or false. A scalene triangle can be an acute triangle as well.
3. True or false. An isosceles triangle can also be a right triangle.
4. True or false. An equilateral triangle has three equal sides.
5. True or false. An obtuse triangle can have multiple obtuse angles.
6. True or false. A scalene triangle has three angles less than 90 degrees.
7. True or false. A triangle with a \begin{align*}100^\circ\end{align*} angle must be an obtuse triangle.
8. True or false. The angles of an equilateral triangle are also equal in measure.
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes
### Vocabulary Language: English
TermDefinition
Acute Triangle An acute triangle has three angles that each measure less than 90 degrees.
Congruent Congruent figures are identical in size, shape and measure.
Exterior angles An exterior angle is the angle formed by one side of a polygon and the extension of the adjacent side.
Interior angles Interior angles are the angles inside a figure.
Isosceles Triangle An isosceles triangle is a triangle in which exactly two sides are the same length.
Obtuse Triangle An obtuse triangle is a triangle with one angle that is greater than 90 degrees.
Right Angle A right angle is an angle equal to 90 degrees.
Scalene Triangle A scalene triangle is a triangle in which all three sides are different lengths.
Triangle A triangle is a polygon with three sides and three angles.
Equilateral A polygon is equilateral if all of its sides are the same length.
Equiangular A polygon is equiangular if all angles are the same measure.
Show Hide Details
Description
Difficulty Level:
Authors:
Tags:
Subjects: |
# Lesson 5
Normal Distributions
• Let’s investigate a specific type of distribution called a normal distribution.
### 5.1: Body Temperature
Each histogram represents a group of 500 healthy people who had their temperature taken. Three histograms represent examples of data that approximate a normal distribution and three histograms represent non-examples, or data that do not approximate a normal distribution. What do you think the elements are of a definition of normal distribution?
Examples:
Non-examples:
### 5.2: Playing a Piano
On many piano keyboards, the distance from one white key to the next is 2.39 centimeters. How many of your classmates could reach two notes that are 9 keys apart (21.5 cm) on a keyboard using only one hand?
1. Stretch your fingers apart as wide as you can and measure the farthest distance from your thumb to smallest finger. Round your measurement to the nearest tenth of a centimeter.
2. Your teacher will collect the measurements from the class. Draw a dot plot or histogram from the class data.
3. Describe the distribution you drew using terms such as: symmetric, approximately symmetric, skewed left, skewed right, approximately uniform, uniform, bell-shaped, or bimodal. Estimate the center of your distribution.
4. How would you use your distribution to determine how many people in the class can reach the two notes 9 keys apart?
### 5.3: Relative Frequency Distribution
Manufacturers of butter make sticks of butter that weigh 110 grams on average. A manufacturer suspects the machine that forms the sticks of butter may have a problem, so they weigh each stick of butter the machine produces in an hour.
The weights are grouped into intervals of 0.5 grams and are summarized in a frequency table.
weight (grams) frequency relative frequency
107–107.5 5
107.5–108 17
108–108.5 52
108.5–109 118
109–109.5 172
109.5–110 232
110–110.5 219
110.5–111 172
111–111.5 95
111.5–112 57
112–112.5 23
112.5–113 8
113–113.5 1
total 1,171
The same data are summarized in this histogram.
Although this information is useful, it might be more helpful to know the proportion of sticks of butter in each weight interval rather than the actual number of sticks in that weight interval.
1. Complete the table by dividing each frequency value by the total number of sticks of butter in the data set. Round each value to 4 decimal places.
2. A relative frequency histogram is a histogram in which the height of each bar is the relative frequency. Since the heights of the bars are found by dividing each height by the total number of sticks of butter, the shape of the distribution is the same as a regular histogram, but the labels on the $$y$$-axis are changed. Label the $$y$$-axis with the correct values for each mark.
3. The manufacturer believes they should replace the machine if more than 25% of the sticks of butter are more than 1 gram away from the intended value of 110 grams.
1. Indicate on the relative frequency histogram the bars that correspond to sticks of butter that are more than 1 gram away from the intended weight of a stick of butter.
2. Should this machine be replaced? Explain or show your reasoning.
### 5.4: The Normal Curve
These curves represent normal distributions with different means and standard deviations. What do you notice?
The equation for the curve for a normal distribution is
$$f(x) = \frac{1}{\sigma \sqrt{2\pi}} \boldcdot e^{\frac{-(x-\mu)^2}{2\sigma^2}}$$
where $$\sigma$$ is the standard deviation, $$\mu$$ is the mean, and $$e$$ is a particular number close to 2.718.
Notice the part of the function that is $$(x - \mu)$$. Using your understanding of transformations of functions, how does changing $$\mu$$ affect the graph of the function?
### Summary
A histogram shows the number of items in the data set that fall into specified intervals. A relative frequency histogram shows the proportion of the entire data set that falls into specified intervals.
For example, a study measured the handspan of 1,000 adults in centimeters. A handspan is the distance from the thumb to the smallest finger when the fingers are stretched out as much as possible. A histogram shows the number of people whose handspans are in certain intervals. The height of each bar in a histogram represents the frequency for the corresponding interval. In this example, there are 132 adults in the study whose handspans are at least 20 centimeters, but less than 20.5 centimeters.
A histogram that uses the relative frequencies shows a distribution with the same shape, but the heights of the bars represent the relative frequency for the corresponding intervals. For example, out of all the adults in the study, 13.2% have handspans that are at least 20 centimeters, but less than 20.5 centimeters (since $$\frac{132}{1,000} = 0.132$$).
In a similar way to how we can model data in a scatter plot with a line or other curve so that additional information can be estimated or predicted, it can be useful to model an approximately symmetric and bell-shaped distribution with a particular distribution called the normal distribution. A normal distribution is symmetric and bell-shaped, has an area of 1 between the $$x$$-axis and the curve, and has the $$x$$-axis as a horizontal asymptote. A normal distribution is determined entirely by the mean and standard deviation.
For the handspan data, the mean is 20.9 cm and the standard deviation is 1.41. The curve that represents the normal distribution with this same mean and standard deviation is shown on top of the relative frequency histogram for the actual data. Notice that the curve does a fairly good job of modeling the actual data in this situation, although it is not perfect.
### Glossary Entries
• normal distribution
A specific distribution in statistics whose graph is symmetric and bell-shaped, has an area of 1 between the $$x$$-axis and the graph, and has the $$x$$-axis as a horizontal asymptote.
• relative frequency histogram
A histogram where the height of each bar is the fraction of the entire data set that falls into the corresponding interval (that is, it is the relative frequency with which the data values fall into that interval). |
# 6.1: Angle Measure
From Chapter 5, we have two burning questions:
• Is there a meaning to the $t$ value?
• Is there a way to derive the terminal points for $t = \frac{\pi}{6}, \frac{\pi}{3}$ by hand?
Question 1 will be answered today. Question 2 will be answered in 6.2.
## Angle Measure
Let's first define what an angle is.
If a circle of radius 1 is drawn with the vertex of an angle at its center, then the measure of this angle in radians is the length of the arc that subtends the angle.
Picture will be drawn in class.
Since $\pi$ radians is a rotation halfway around the circle, we know that $180^\circ = \pi$ rad. Thus we have the following:
$1 \text{ rad} = \left(\frac{180}{\pi}\right)^\circ \qquad\qquad 1^\circ = \frac{\pi}{180}\text{ rad}$
• To convert degrees to radians, multiply by $\frac{\pi}{180}$.
• To convert radians to degrees, multiply by $\frac{180}{\pi}$.
• Express $60^\circ$ in radians.
• Express $\frac{\pi}{6}$ rad in degrees.
## Angles in Standard Position
You may have noticed the length of the arc is basically the $t$ value, except the initial side of the angle did not start on the positive $x$-axis. If the initial side is on the positive $x$-axis, we say the angle is in standard position.
(picture in class)
Two angles in standard position are coterminal if both the initial side and terminal sides lie on top of each other.
• Find two angles that are coterminal with the angle $\theta = 30^\circ$ in standard position.
• Find two angles that are coterminal with the angle $\theta =\frac{\pi}{3}$ in standard position.
## Length of a Circular Arc
The $t$ value in Chapter 5 is arc length in radians. We also have a method to convert $t$ into the subtended angle. What is the relationship between the arc length and the angle?
In a circle of radius $r$ the length $s$ of an arc that subtends a central angle of $\theta$ radians is $s = r\theta$
In Chapter 5 we had $r = 1$, giving $s = \theta$. This means the arc length was exactly the angle measure. Here, we are generalizing to any valid value of $r$.
• Suppose a circle has radius 10 meters. An arc is subtended by an angle of $30^\circ$. Find the arc length.
• Suppose a circle has radius 4 meters. A central angle is subtended by an arc length of 6 meters. Find the angle in radians.
## Area of a Circular Sector
In a circle of radius $r$ the area $A$ of a sector with a central angle of $\theta$ radians is $A = \frac{1}{2}r^2\theta$
Insight: Rewrite the formula as $A = \frac{\theta}{2}\cdot r^2$. What does it look like now?
Find the area of a sector of a circle with central angle $60^\circ$ if the radius of the circle is three meters.
For dinner, you tried to order one pizza with radius 9 inches. However, the pizza place ran out of pizzas that are six inches, and offered you four pizzas with radius 4 inches each for the same price. Should you take the deal?
## Circular Motion
As a point swirls around a circle, there are two different ways we can quantify how "fast" the point is moving:
• Linear speed is the rate at which the total distance traveled is changing (think the $t$ value).
• Angular speed is the rate at which the central angle $\theta$ is changing.
Suppose a point moves along a circle of radius $r$ and the ray from the center of the circle to the point traverses $\theta$ radians in time $t$. Let $s = r\theta$ (the arc length) be the distance the point travels in time $t$. Then the speed of the object can be thought of in two different ways:
Angular speed $\omega = \frac{\theta}{t}$ Linear speed $\nu = \frac{s}{t}$
A boy spins a stone in a three feet long sling at a rate of 15 revolutions every 10 seconds. Find the angular and linear velocities of the stone in seconds. |
# What is conditional probability with example?
Conditional probability: p(AB) is the probability of event A occurring, given that event B occurs. … Example: the probability that a card drawn is red (p(red) = 0.5). Another example: the probability that a card drawn is a 4 (p(four)=1/13). Joint probability: p(A and B). The probability of event A and event B occurring.
## What is conditional probability formula?
The formula for conditional probability is derived from the probability multiplication rule, P(A and B)= P(A)*P(BA). You may also see this rule as P(AB). The Union symbol () means and, as in event A happening and event B happening.
## What is the difference between conditional probability and simple probability?
Answer. P(A B) and P(AB) are very closely related. Their only difference is that the conditional probability assumes that we already know something — that B is true. … For P(AB), however, we will receive a probability between 0, if A cannot happen when B is true, and P(B), if A is always true when B is true.
Read More: Is turmeric the same as Curcuma?
## Is P AB the same as P ba?
P(AB) is the probability of A, given that B has already occurred. This is not the same as P(A)P(B.
## How do you calculate PAB?
P(A/B) Formula is given as, P(A/B) = P(AB) / P(B), where, P(A) is probability of event A happening, P(B) is the probability of event B happening and P(AB) is the probability of happening of both A and B.
## How do you calculate probability example?
For example, if the number of desired outcomes divided by the number of possible events is . 25, multiply the answer by 100 to get 25%. If you have the odds of a particular outcome in percent form, divide the percentage by 100 and then multiply it by the number of events to get the probability.
## How do you calculate PA B?
We apply P(A B) formula to calculate the probability of two independent events A and B occurring together. It is given as, P(AB) = P(A) P(B), where, P(A) is Probability of an event A and P(B) = Probability of an event B.
## What is multiplication theorem of probability?
Answer: The multiplication law states that the probability of happening of given 2 events or in different words the probability of the intersection of 2 given events is equivalent to the product achieved by finding out the product of the probability of happening of both the events.
## How do you prove conditional probability?
This probability is written P(BA), notation for the probability of B given A. In the case where events A and B are independent (where event A has no effect on the probability of event B), the conditional probability of event B given event A is simply the probability of event B, that is P(B). P(A and B)= P(A)P(BA).
## What is the difference between conditional probability and Bayes Theorem?
There are a number of differences between conditional property and Bayes theorem. … Complete answer:
Read More: Is godlike a real word?
Conditional Probability Bayes Theorem
It is used for relatively simple problems. It gives a structured formula for solving more complex problems.
## How do you report conditional probability?
Formally, the formula for the calculation of a conditional probability is written as: P(AB) = P(A B)
## What is conditional probability genetics?
Solution. Conditional Probability. p(AB) = the probability of outcome A given condition B. This is not the same as a joint probability or a simultaneous probability. It turns out that p(AB) is very easy to calculate: p(AB) = p(AB) p(B).
## What is genetic probability?
Probability is a method used to predict the likelihoods of uncertain outcomes. It is important for the field of genetics because it is used to reveal traits that are hidden in the genome by dominant alleles.
## How do you find the probability of an offspring?
Count the total number of boxes in your Punnett Square. This gives you the total number of predicted offspring. Divide the (number of occurrences of the phenotype) by (the total number of offspring). Multiply the number from step 4 by 100 to get your percent.
## What is P AUB if A and B are independent?
If A and B are independent events, then the events A and B’ are also independent. Proof: The events A and B are independent, so, P(A B)= P(A) P(B).
## What is PA or B?
If events A and B are mutually exclusive, then the probability of A or B is simply: p(A or B)= p(A) + p(B).
## What is PA or B if A and B are independent?
A and B are two events. If A and B are independent, then the probability that events A and B both occur is: p(A and B)= p(A) x p(B). In other words, the probability of A and B both occurring is the product of the probability of A and the probability of B.
## How do I calculate probability?
Divide the number of events by the number of possible outcomes.
1. Determine a single event with a single outcome. …
2. Identify the total number of outcomes that can occur. …
3. Divide the number of events by the number of possible outcomes. …
4. Determine each event you will calculate. …
5. Calculate the probability of each event.
Read More: What is difference between perfume and eau de toilette?
## Is conditional probability Independent?
A conditional probability is the probability that an event has occurred, taking into account additional information about the result of the experiment. … Two events A and B are independent if the probability P(AB) of their intersection AB is equal to the product P(A)P(B) of their individual probabilities.
## What are the 5 rules of probability?
Basic Probability Rules
• Probability Rule One (For any event A, 0 P(A) 1)
• Probability Rule Two (The sum of the probabilities of all possible outcomes is 1)
• Probability Rule Three (The Complement Rule)
• Probabilities Involving Multiple Events.
• Probability Rule Four (Addition Rule for Disjoint Events)
## How do you calculate probability and odds?
To convert from a probability to odds, divide the probability by one minus that probability. So if the probability is 10% or 0.10 , then the odds are 0.1/0.9 or ‘1 to 9’ or 0.111.
## How do you find Pa and Pb?
Formula for the probability of A and B (independent events): p(A and B)= p(A) * p(B). If the probability of one event doesn’t affect the other, you have an independent event. All you do is multiply the probability of one by the probability of another.
## How do you find P AUB given PA and PB?
If A and b are two different events then, P(A U B) = P(A) + P(B) – P(A B).
## What is P A and B in probability?
For instance P(AB) means the probability that event A occurs given event B has occurred. b. If A and B are independent – neither event influences or affects the probability that the other event occurs – then P(A and B) = P(A)*P(B). This particular rule extends to more than two independent events.
## What is addition theorem of probability?
If A and B are any two events then the probability of happening of at least one of the events is defined as P(AUB) = P(A) + P(B)- P(AB).
## What are the four rules of multiplication?
What are the rules of multiplication?
• Any number times zero is always zero. …
• Any number times one is always the same number. …
• Add a zero onto the original number when multiplying by 10. …
• The order of factors does not affect the product. …
• Products are always positive when multiplying numbers with the same signs.
## What is the formula of multiplication theorem?
The multiplication rule is a way to find the probability of two events happening at the same time (this is also one of the AP Statistics formulas). … The general multiplication rule formula is: P(A B)= P(A) P(BA) and the specific multiplication rule is P(A and B) = P(A) * P(B). |
HomeEnglishClass 11MathsChapterDifferentiation
Differentiate : <br> (i) (ax +...
# Differentiate : <br> (i) (ax + b)^(m) , (ii) (3x+5)^(6) , (iii) sqrt(ax^(2) + 2bx + c)
Updated On: 17-04-2022
Get Answer to any question, just click a photo and upload the photo
and get the answer completely free,
Watch 1000+ concepts & tricky questions explained!
(i) Let y = (ax + b)^(m). <br> Put (ax + b) = t, so that y = t^(m) and t = (ax + b). <br> :. (dy)/(dt) = mt^(m-1) and (dt)/(dx) = a. <br> So, (dy)/(dx) = ((dy)/(dt) xx (dt)/(dx)) <br> = amt^(m-1) = am (ax +b)^(m-1) [ :' t = (ax+b)]. <br> (ii) Let y = (3x+5)^(6). <br> Put (3x+5) =t, so that y = t^(6) and t = (3x+5). <br>:. (dy)/(dt) = 6t^(5) and (dt)/(dx) = 3. <br> So, (dy)/(dx) = ((dy)/(dt) xx (dt)/(dx)) <br> = 18t^(5) = 18 (3x+5)^(5), [ :' t = (3x+5)] . <br> (iii) Let y = sqrt(ax^(2) + 2bx + c). <br> Put (ax^(2) + 2bx + c) = t, so that y = sqrt(t) and t = (ax^(2) + 2bx + c). <br> :. (dy)/(dx) = 1/2 t^(-1//2) = (1)/(2sqrt(t)) and (dt)/(dx) = (2ax+2b) <br> So, (dy)/(dx) = ((dy)/(dt) xx (dt)/(dx)) = 1/(2sqrt(t)) xx 2(ax+b) <br> = ((ax + b))/(sqrt(t)) = ((ax+b))/(sqrt(ax^(2) + 2bx + c)). |
## jocelynevsq 2 years ago idenify the definite integral that represents the area of the region bounded by the graphs of y=x and y=5x-x^3
1. cherio12
try graphing them first to determine which function is the top and which is the bottom
2. jocelynevsq
ok
3. jocelynevsq
still don't get it
4. jocelynevsq
|dw:1359920225466:dw|
5. zepdrix
So ummmm yah that graph looks accurate. I always get confused by these types of problems. When the upper and lower functions switch places. But I think we can figure it out. Hmm.
6. zepdrix
Let's start by finding points of intersection.$\large y=x \qquad y=5x-x^3\qquad \qquad \rightarrow \qquad \qquad x=5x-x^3$
7. zepdrix
Which turn out to be, $$\large x=0, \quad x=-2, \quad x=2$$.
8. zepdrix
Let's look at just the right side for a moment, maybe it will make a little more sense that way.
9. zepdrix
|dw:1359921446750:dw|See how I've drawn a small rectangle? Let's find the area of that rectangle.
10. zepdrix
The area will be base x height. The base has a thickness of $$dx$$. The height is going to be the upper function minus the lower function.
11. zepdrix
$\large A=\left(height\right)(base)$$\large A=\left(5x-x^3-x\right)(dx)$See how the line y=x is BELOW the other one? So we subtract y=x from the other function to get the correct height. So this represents the area of a small rectangular slice. Now we want to Integrate (sum up) all of the rectangles between these two curves.
12. zepdrix
And we'll include our intersecting points.
13. zepdrix
$\large \int\limits_0^2 4x-x^3 \; dx$ This represents the area between the two curves on the RIGHT side. We still need to find the area of the other chunk. Notice how they switch places on the left? The line y=x is now the UPPER function! So we would subtract y=5x-x^3 from the line y=x. And our limits of integration would be a little different.
14. zepdrix
There is a little trick we could pull though c: See how they're both odd functions? They're both symmetric about the origin. So the area on the left will be the same as the area on the right. We could simply multiply this integral by 2.$\huge A_{total}\qquad =\qquad 2\int\limits\limits_0^2 4x-x^3 \; dx$
15. jocelynevsq
oh okay i see how you got that. thanks! |
HomeTren&dMastering the A^3-B^3 Formula in Mathematics
# Mastering the A^3-B^3 Formula in Mathematics
Author
Date
Category
Mathematics is often viewed as a challenging subject by many students. However, with the right tools and techniques, even complex mathematical concepts can be simplified and mastered. One such powerful formula that can help in simplifying and solving algebraic expressions is the A^3-B^3 formula. In this article, we will delve into what the A^3-B^3 formula is, how to apply it effectively, and some examples to illustrate its usage.
### Understanding the A^3-B^3 Formula
The A^3-B^3 formula is a special case of the algebraic formula for factoring the difference of cubes. It states that A^3-B^3 = (A-B) (A^2 + AB + B^2). This formula can be extremely useful when dealing with cubic expressions and simplifying them efficiently. By recognizing and using this formula, you can save time and effort in algebraic manipulations.
### Applying the A^3-B^3 Formula
To effectively apply the A^3-B^3 formula, follow these steps:
1. Identify A and B: In any given expression in the form of A^3-B^3, identify the values of A and B.
2. Substitute into the Formula: Substitute the values of A and B into the formula (A-B) (A^2 + AB + B^2).
3. Expand and Simplify: Expand the expression obtained in step 2 by multiplying the terms and simplify to get the final result.
### Example Problems
Let's consider a couple of examples to illustrate the application of the A^3-B^3 formula:
Example 1: Simplify the expression 8^3 - 2^3.
Solution:
1. Identify A = 8 and B = 2.
2. Substitute into the formula: (8-2) (8^2 + 8*2 + 2^2).
3. Simplify: 6 (64 + 16 + 4) = 6 (84) = 504.
Example 2: Factor the expression 27x^3 - 8y^3.
Solution:
1. Identify A = 3x and B = 2y.
2. Substitute into the formula: (3x-2y) (9x^2 + 6xy + 4y^2).
### Advantages of Using the A^3-B^3 Formula
• Simplification of Cubic Expressions: The formula helps in simplifying complex cubic expressions efficiently.
• Time-saving: By recognizing the pattern and applying the formula, you can save time in algebraic manipulations.
• Reduction of Errors: Using the formula reduces the chances of errors in simplification.
### Tips for Mastering the A^3-B^3 Formula
• Practice: Regular practice with different examples will enhance your understanding and proficiency in using the formula.
• Understand the Concept: Ensure you understand the concept of the A^3-B^3 formula and its application in different scenarios.
• Apply in Problem-solving: Use the formula in solving various algebraic problems to strengthen your skills.
Q1: What is the difference between the A^3-B^3 formula and the difference of squares formula?
A1: The A^3-B^3 formula is specifically for factoring the difference of cubes, while the difference of squares formula is for factoring the difference of two squares.
Q2: Can the A^3-B^3 formula be used for sum of cubes as well?
A2: No, the A^3-B^3 formula is specifically for the difference of cubes. For the sum of cubes, a different formula, A^3+B^3, is used.
Q3: Are there any real-life applications of the A^3-B^3 formula?
A3: While the formula itself may not have direct real-life applications, the concept of factoring cubic expressions is commonly used in engineering, physics, and other fields.
Q4: How can I remember the A^3-B^3 formula easily?
A4: Practice regularly and try to understand the derivation of the formula. Mnemonics or creating flashcards can also help in remembering the formula.
Q5: Is the A^3-B^3 formula limited to integer values of A and B?
A5: No, the formula can be applied to any real numbers or variables (A-B) (A^2 + AB + B^2), irrespective of whether they are integers or not.
In conclusion, mastering the A^3-B^3 formula in mathematics can significantly enhance your algebraic skills and simplification techniques. By understanding the concept, practicing regularly, and applying the formula in various problems, you can become proficient in factoring the difference of cubes efficiently and accurately. Keep practicing and exploring different examples to solidify your grasp on this powerful mathematical tool. |
# dy/dx=tan(x+y) Solve the Differential Equation [Solution]
The general solution of the differential equation dy/dx=tan(x+y) is equal to y= x -log|cos(x+y) + sin(x+y)| +C where C denotes an arbitrary constant. In this post, we will learn how to solve the differential equation dy/dx = tan(x+y).
## General Solution of dy/dx=tan(x+y)
Question: Find the general solution of $\dfrac{dy}{dx}$=tan(x+y).
Solution:
$\dfrac{dy}{dx}$=tan(x+y)
⇒ $\dfrac{dy}{dx} = \dfrac{\sin(x+y)}{\cos(x+y)}$ ...(∗)
Put x+y = v
So, 1+ $\dfrac{dy}{dx} = \dfrac{dv}{dx}$
⇒ $\dfrac{dy}{dx} = \dfrac{dv}{dx} -1$
Therefore, from (∗), we get that
$\dfrac{dv}{dx}-1 = \dfrac{\sin v}{\cos v}$
⇒ $\dfrac{dv}{dx} = \dfrac{\cos v +\sin v}{\cos v}$
⇒ $\dfrac{\cos v}{\cos v +\sin v} dv = dx$
Integrating, $\dfrac{1}{2}\int \dfrac{(\cos v +\sin v) +(\cos v -\sin v)}{\cos v +\sin v} dv = \int dx +K$
⇒ $\dfrac{1}{2} \int dv + \dfrac{1}{2} \int \dfrac{\frac{d}{dv}(\cos v +\sin v)}{\cos v +\sin v} dv = x +K$
⇒ $\dfrac{v}{2}+ \dfrac{1}{2} \log |\cos v +\sin v| = x +K$
⇒ $v+ \log |\cos v +\sin v| = 2x +C$ where C=2K.
⇒ $x+y+\log |\cos (x+y) +\sin (x+y)| = 2x +C$ as v=x+y.
⇒ $y=x-\log |\cos (x+y) +\sin (x+y)|+C$
So the solution of dy/dx=tan(x+y) is equal to y= x -log|cos(x+y) + sin(x+y)| +C where C is an integral constant.
More Differential Equations:
Solve dy/dx = x/y
Solve dy/dx = sin(x+y)
Solve dy/dx = cos(x+y)
Solve dy/dx = ex+y
How to Solve dy/dx=1+x+y+xy
## FAQs
### Q1: What is the general solution of dy/dx =tan(x+y)?
Answer: The general solution of dy/dx=tan(x+y) is given by y= x -log|cos(x+y) + sin(x+y)| +C where C is a constant. |
#### Explain solution RD Sharma class12 chapter Maxima and Minima exercise 17.5 question 28 maths.
$x=y=\frac{r}{\sqrt{2}}$
Hint: For maximum or minimum values of z, we must have $\frac{dz}{dx}=0$
Given:
$x^2+y^2=r^2$
$y = \sqrt{r^2-x^2}$ .........(1)
Solution:
Now,
$z=x+y$
$z=x+\sqrt{r^2-x^2}$ ......from (1)
\begin{aligned} &\frac{d z}{d x}=1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}} \\ &\frac{d z}{d x}=0 \\ &1+\frac{(-2 x)}{2 \sqrt{r^{2}-x^{2}}}=0 \\ &2 x=2 \sqrt{r^{2}-x^{2}} \\ &x=\sqrt{r^{2}-x^{2}} \end{aligned}
Squaring on the both sides
$x^2=r^2-x^2$
$x=\frac{r}{\sqrt{2}}$
Substituting the value of x in equation (1)
$y = \sqrt{r^2-x^2}$
\begin{aligned} &y=\sqrt{r^{2}-\left(\frac{r}{\sqrt{2}}\right)^{2}} \\ &\frac{d^{2} z}{d x^{2}}=\frac{-\sqrt{r^{2}-x^{2}}+x(-x)}{\sqrt{r^{2}-x^{2}}}=\frac{-r^{2}+x^{2}-x^{2}}{r^{2}-x^{2}} \end{aligned}
$=\frac{-r^2}{r^3}\times 2 \sqrt{2}$
$=-\frac{ 2 \sqrt{2}}{r}<0$
So z = x + y is maximum when $x=y=\frac{r}{\sqrt{2}}$ |
Making Tens
• ## Making Tens: Finding Addends That Sum to Ten
Periods: 3
Author: Susan Andrews Kunze
### Instructional Plan
Students use three different ways to find addends that sum to ten. First they make different arrangements of two number blocks that add up to ten. Then students use 2-sided counters to make and record addends that sum to 10. Lastly, students play the card game Making Tens Concentration to practice identifying addends that sum to ten.
### Preparation
For the first class period, each student will need a copy of the Making Ten Frames Activity Sheet. The teacher will need one copy of the Making Tens Overhead. This representation can also be displayed on the board. Each pair of students will need two sets of 10 snapping cubes. Each set should be a different color.
For the second period, each student will need a clean copy of the Making Tens Frames Activity Sheet, 10 two-sided counters, and a small cup.
For the third period, print a copy of the Cut Out Number Template on card stock for each student, and cut the cards apart.
### The Lesson
To begin the lesson, use 10 snapping cubes of the same color (let's say red for the sake of the directions), model the number 10 by making 2 columns of 5. Ask students, "How many cubes are there"? [10.] How many red cubes are there? [10.]. "Can you tell me how you know there are 10 cubes?" [Answers will vary.] Then introduce a second color (let's say blue for the sake of the directions). Tell the class you are replacing 2 red cubes with 2 blue cubes as you make the change. Ask, "How many red cubes are there now?" [8]. "How do you know?" [Answers will vary.] Ask, "How many blue cubes? [2] How many total cubes?" [10]. "How do you know there are 10 in all?" [Answers will vary.] Show the Making Tens Overhead to the students. Mark the overhead with "R" and "B" to show the placement of the blocks you just arranged. Write 8 + 2 = 10 on the line beneath the 10 frame. Demonstrate how the overhead could be turned 90 degrees to represent the same combination of blocks. Repeat the activity with another combination of addends to 10.
Distribute the snapping cubes and the Making Ten Frames Activity Sheet. Explain to students that they will be making different arrangements of the two colors of cubes that add to 10. They can use the activity sheet as a template for arranging the cubes in 2 columns. They will write a number sentence beneath each arrangement. Refer back to the overhead if necessary.
Observe students as they begin working, and ask them how they know they have different combinations of 10. Be aware of students who might be using an incorrect total number of cubes. If a student asks for another activity sheet, have him or her check the number sentences to see if any addends were repeated. If this occurs, it is a good opportunity to point out that, for example, 7 + 3 has the same sum as 3 + 7. Note that all combinations in this activity will inherently have the same sum. Because students are using different colored cubes, expecting students to show the commutative aspect of the combinations would be a strength for this activity.
When students are finished, have them share their ten frames with the rest of the class. Some students might not have all of the same color cubes touching, or might alternate colors. Have students discuss whether these models represent the number sentences in the same way as those that show the cubes touching. Ask students how they know it is the same or why they think it is not. Discuss the different methods students are using, why they are or are not viable, etc. If no student has used 0 as an addend, ask students how they would represent your original arrangement of 10 red cubes.
In the second period, students will use ten 2-sided counters (bean counters, disks, or pennies) to randomly make, represent, and record two addends that sum to 10 in a ten frame. Each students will receive a new copy of the Making Ten Frames Activity Sheet. On the activity sheet, above each ten frame, students label each column, depending on what counter they are using. For example, if they are using bean counters, they will label one column "Red" and the other column "White". To start, each student places the 10 objects in a cup (students should only choose two different objects), then covers, shakes, and tosses them out onto the table. Then, the student will arrange the counters to represent 2 numbers that sum to 10 and then, record each added in the corresponding column on the activity sheet. With each toss, students only record addends that are not already recorded. Be sure to point out that, for example, 2 + 5 and 5 + 2 represent the same addends. Students toss until all possible addend combinations (including 10 + 0) are tossed and recorded. As students are working, the teacher should again observe and ask questions to the students to ensure comprehension.
This activity can be done more than one time by using one or more types of two-sided manipulatives. Varying the manipulatives used when repeating this activity provides students with more opportunities to develop their visual memory of addends that sum to 10 without seeming overly repetitive.
When the teacher has determined that students have mastered the concept, this activity can be played as a game for 2 to 4 players. Each player takes a turn to toss, build, and record new sums of 10 on a new Making Ten Frames activity sheet. The first player to fill his or her recording sheet with all 6 tables of different sums of 10 wins.
In the third period, students play the card game Making Tens Concentration to practice identifying addends that sum to 10. To play, groups of 2 to 4 players make a 2 × 5 array of the cut out cards, with the numbers face down. Extra cards are placed in a pile to replace those that are removed during play. The first player flips over 2 cards from the array. If the sum is not 10, the 2 cards are replaced face-down into the array and the next player takes a turn. If the sum is 10, the player keeps the 2 cards, replaces them with cards from the extra card pile, and turns over 2 more cards in the array. Players continue to play, filling the array with extra cards until 1 player possesses all addend pairs of 10. That player is the winner.
### Assessments and Extensions
Assessment Options
1. Students may be assessed by using snapping or linking cubes to make trains that represent two numbers that sum to 10.
2. Assessment can be made by using the Making Tens: Assessment Sheet
Making Tens: Assessment Sheet
3. Students can record their combinations on large graph paper cut in the shape of a ten frame, and then glue them into a Making Tens booklet, one combination per page. Students label each number sentence and may include an appropriate fact family. To make booklets, fold and staple 3 sheets of paper together or make staple-less booklets. (In order to save time, groups of 6 can be made, where each student is responsible for one of the six pairs of addend that make 10).
4. Questioning during observation of student work can be used as a formative assessment. While doing this, make a checklist to keep track of the combinations that students know and are able to model.
Extensions
1. Play another card game: Trading for Tens. To play, each pair of students will need a deck of index cards numbered 1 - 10. To start, a player deals each player 7 cards. All of the other cards remain in the pile. Each player looks at his or her hand and puts any pair that sums to 10 in his or her pair pile. Then the first player asks another player for a card that will allow him or her to make a pair with a sum of 10. If that second player has the card, it is given to the first player. If not, the second player says "Take a card," and the first player takes a card from the pile. If the first player can use it to make a sum of 10, then he or she gets another turn. If not, the next player begins play. Play continues until all possible pairs have been made. The winner is the player with the most pairs of 10.
2. Give each student a paper bag containing 10 snapping cubes of 1 color and 10 snapping cubes of another color. Each student pulls out 10 cubes, then places those 10 cubes in 1 ten frame. Ask students who need a challenge to guess the color and the number of the remaining cubes in the bag. The student then pulls out the remaining 10 cubes and places them in a second ten frame. This provides a great illustration of the commutative property of addition for students.
3. Depending on the grade level, students may work with larger numbers, e.g. combinations to 18 or 20, using a double ten frame. This can be used in conjunction with an Illuminations app called Ten Frame.
Ten Frame
4. Introduce students to probability by dumping out ten 2-sided counters 20 times. Students will record the result every time and discuss which addend pairs occurred the most.
### Questions and Reflections
Questions for Students
1. Why are 4 + 6 and 6 + 4 considered the same pair in these activities?
[Response can be shown by changing the orientation of this pair in a ten frame or an explanation of that relationship as the commutative property of addition.]
2. What pattern can you see in the organized list of addend pairs of numbers that sum to 10?
[One number in the sequence gets larger and the other becomes smaller.]
3. Can there be more than two addends that sum to 10?
[Yes.]
4. Can you give an example of four addends that sum to 10?
[Yes, one possible answer is 1 + 2 + 3 + 4.]
5. Is there a way to keep track of the ways to make 10, ensuring that you do not repeat any?
[Yes. One way is to make one addend bigger and the other smaller by 1. Keep doing this until a repetition occurs.]
Teacher Reflection
• What were some of the ways that students demonstrated that they were actively engaged in learning by making tens?
• How did your lesson provide opportunities for differentiated instruction? Were you successful in engaging all students? If not, how could you adapt the lesson to meet the needs of students of diverse abilities?
• What aspects of classroom management worked in this lesson? What didn't work? How would you change what didn't work?
### Objectives and Standards
Learning Objectives
Students will:
• Use manipulatives to sum two numbers to make ten.
• Identify and record pairs of number addends that sum to ten.
• Develop visual patterns of numbers and addends that sum to ten. |
# Basic math glossary-P
Basic math glossary-P define words beginning with the letter P
Parallel lines: Lines that never intersect or cross one another on a plane.
Parallelogram: A quadrilateral with two pairs of equal and parallel sides.
Partial product: In a multiplication problem with at least two digits, it is the answer obtained when we multiply a digit in the first number by every digit in the other number.
Pentagon : A polygon with 5 sides.
Percent: Part per one hundred.
Percentage: The answer obtained after a number is multiplied by a percent.
Perimeter: The distance around the outside of a closed figure
Perpendicular lines: Lines that form 4 right angles when they meet or intersect.
Pi: The ratio of the perimeter of a circle to the diameter.
Pictograph: A graph that make use of pictures to display information.
Pint:. A measure of liquid capacity that is equal to 16 fluid ounces
Place value: The value of the place of a digit in a number.
Plane: A flat surface that extends forever in all directions.
Point: A location in space represented by a dot.
Polygon: A closed geometric figure with at least 3 sides.
Positive integer: A whole number bigger than 0.
Power of ten: The product of multiplying ten by itself one or more times. For example, 105 is the product of multiplying 10 by itself 5 times.
Thus,105 = 10 × 10 × 10 × 10 × 10
Prefix: In the metric system, it is a set of letters placed before a unit to form new words.
Previous balance: In consumer math, it is the amount owed before a payment is made
Prime factorization: A number written as the product of its prime factors only.
Prime number: A number than can be divided evenly only by one and the number itself.
Principal: Amount borrowed or invested.
Product:. The answer to a multiplication problem
Proper fraction: A fraction with a smaller numerator than a denominator.
Proportion: two equal ratios.
Protractor: A geometric instrument used to measure or draw angles.
Pyramid: A solid figure that has triangular sides and a base that is a polygon
## Recent Articles
1. ### Average age word problem
Aug 09, 16 01:40 PM
Average age of Dipu and Apu is 22 years. Average age of Dipu and Tipu is 24 years. Age of Dipu is 21 years. What are the ages of Apu, and tipu ? Let |
sanuluy
## Answered question
2021-01-30
Evaluate the following limits: $\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}$
### Answer & Explanation
yagombyeR
Skilled2021-01-31Added 92 answers
Given:
$\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}$
We have to evaluate the given limit
We have,
$\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}$
When we substitute y=0 in given limit we get,
$\frac{1-\mathrm{cos}\left(7\cdot 0\right)}{\left(2\cdot 0\right)}=\frac{1-1}{0}=\frac{0}{0}$
We get the indeterminate form if we substitute y=0 then we used LHospitals Rule.
LHospitals Rule :
The rule tells us that if we have an indeterminate form $\frac{0}{0}$ or $\frac{\mathrm{\infty }}{\mathrm{\infty }}$ all we need to do is differentiate the numerator and differentiate the denominator and then take the limit.
Now differentiate the numerator and differentiate the denominator with respect to y:
$⇒\underset{y\to 0}{lim}\frac{\frac{d}{dy}\left(1-cos\left(7y\right)\right)}{\frac{d}{dy}\left(2y\right)}$
$=\underset{y\to 0}{lim}\frac{0-\left(-7\mathrm{sin}\left(7y\right)\right)}{2}$
$=\underset{y\to 0}{lim}\frac{7\mathrm{sin}\left(7y\right)}{2}$
$=\frac{7}{2}\underset{y\to 0}{lim}\mathrm{sin}\left(7y\right)$
$=0$
$\left[\underset{y\to 0}{lim}\mathrm{sin}\left(7y\right)=0\right]$
Hence, $\underset{y\to 0}{lim}\frac{1-\mathrm{cos}\left(7y\right)}{2y}=0$
Do you have a similar question?
Recalculate according to your conditions!
Ask your question.
Get an expert answer.
Let our experts help you. Answer in as fast as 15 minutes.
Didn't find what you were looking for? |
# RS Aggarwal Solutions Class 9 Chapter 3 Introduction to Euclid’s Geometry
Here you can get solutions of RS Aggarwal Solutions Class 9 Chapter 3 Introduction to Euclid’s Geometry. These Solutions are part of RS Aggarwal Solutions Class 9. we have given RS Aggarwal Solutions Class 9 Chapter 3 Introduction to Euclid’s Geometry download pdf.
## RS Aggarwal Solutions Class 9 Chapter 3 Introduction to Euclid’s Geometry
### Exercise 3A
Question 1:
A theorem is a statement that requires a proof. Whereas, a basic fact which is taken for granted, without proof, is called an axiom.
Example of Theorem: Pythagoras Theorem
Example of axiom: A unique line can be drawn through any two points.
Question 2:
(i) Line segment: The straight path between two points is called a line segment.
(ii) Ray: A line segment when extended indefinitely in one direction is called a ray.
(iii) Intersecting Lines: Two lines meeting at a common point are called intersecting lines, i.e., they have a common point.
(iv) Parallel Lines: Two lines in a plane are said to be parallel, if they have no common point, i.e., they do not meet at all.
(v) Half-line: A ray without its initial point is called a half-line.
(vi) Concurrent lines: Three or more lines are said to be concurrent, if they intersect at the same point.
(vii) Collinear points: Three or more than three points are said to be collinear, if they lie on the same line.
(viii) Plane: A plane is a surface such that every point of the line joining any two points on it, lies on it.
Question 3:
(i) Six points: A,B,C,D,E,F
(ii) Five line segments: [latex]overline { EG } [/latex], [latex]overline { FH } [/latex], [latex]overline { EF } [/latex], [latex]overline { GH } [/latex], [latex]overline { MN } [/latex]
(iii) Four rays: [latex]overrightarrow { EP } [/latex], [latex]overrightarrow { GR } [/latex], [latex]overrightarrow { GB } [/latex], [latex]overrightarrow { HD } [/latex]
(iv) Four lines: [latex]overleftrightarrow { AB } [/latex], [latex]overleftrightarrow { CD } [/latex], [latex]overleftrightarrow { PQ } [/latex], [latex]overleftrightarrow { RS } [/latex]
(vi) Four collinear points: M,E,G,B
Question 4:
(i) [latex](overleftrightarrow { EF } quad overleftrightarrow { GH } ) [/latex] and their corresponding point of intersection is R.
[latex](overleftrightarrow { AB } quad overleftrightarrow { CD } ) [/latex] and their corresponding point of intersection is P.
(ii) [latex]overleftrightarrow { AB } [/latex], [latex]overleftrightarrow { EF } [/latex], [latex]overleftrightarrow { GH } [/latex] and their point of intersection is R.
(iii) Three rays are: [latex]overrightarrow { RB } [/latex], [latex]overrightarrow { RH } [/latex], [latex]overrightarrow { RG } [/latex]
(iv) Two line segments are: [latex]overline { RQ } [/latex], [latex]overline { RP } [/latex]
Question 5:
(i) An infinite number of lines can be drawn to pass through a given point.
(ii) One and only one line can pass through two given points.
(iii) Two given lines can at the most intersect at one and only one point.
(iv) [latex]overline { AB } [/latex], [latex]overline { BC } [/latex], [latex]overline { AC } [/latex]
Question 6:
(i) False
(ii) False
(iii) False
(iv) True
(v) False
(vi) True
(vii) True
(viii) True
(ix) True
(x) False
(xi) False
(xii) True
Complete RS Aggarwal Solutions Class 9
If You have any query regarding this chapter, please comment on below section our team will answer you. We Tried our best to give complete solutions so you got good marks in your exam.
If these solutions have helped you, you can also share ncertsolutionsfor.com to your friends.
Best of Luck For Your Future!! |
Courses
# Procedure - To Show that the Area of a Triangle is half the Product of the Base and the Height, Math Class 9 Notes | EduRev
## Class 9 : Procedure - To Show that the Area of a Triangle is half the Product of the Base and the Height, Math Class 9 Notes | EduRev
The document Procedure - To Show that the Area of a Triangle is half the Product of the Base and the Height, Math Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9
Materials required for real lab:
Chart paper, pencil, compass, scale, a pair of scissors, cello tape.
Procedure for Right angled triangle:
As performed in real lab:
1. Cut a right angle triangle.
2. Cut a triangle congruent to the right angle triangle.
3. Align the hypotenuse of the two triangles to obtain a rectangle.[Fig(A)]
As performed in the simulator:
1. Create a ▲ ABC by providing its three sides. For a right angled triangle with sides a, b, c where c hypotenuse, then a2 + b= c2.
2. Click on ▲ ABC to create its replica.
3. Place replicated triangle such that hypotenuse of both triangle will cover each other. Use button 'clockwise' to rotate triangle clockwise. Use button 'Anticlockwise' to rotate triangle Anticlockwise.
4. See the observation.
Observation:
We can observe that two congruent triangles aligned on hypotenuse forms a rectangle.
∴ Area of □ ABCD = 2 x Area of ▲ ABC
∴ Area of ▲ ABC = 1/2 x Area of □ ABCD
= 1/2 x [base of □ ABCD X height of □ ABCD]
= 1/2 x [BC x AB]
= 1/2 x base of triangle ABC x height of triangle ABC=1/2 x base x height
Procedure for Acute Angled Triangle:
As performed in the real lab:
1. Cut an acute angle triangle and draw the perpendicular from the vertex to the opposite side.
2. Cut a triangle congruent to it and cut this triangle along perpendicular.
3. Align the hypotenuse of these cut outs to the given triangle in order to obtain a rectangle.[Fig(B)]
As performed in simulator:
1. Create a ▲ ABC by providing its three sides. Triangle is Acute angled triangle if its square of longest side is less than to sum of products of squares of other two sides.
2. Next step is to draw perpendicular from A To line BC.
• Click on SetSquare in the 'Tools' to use it.
• Drag this set square and place at position such that point A will perpendicular to base BC.
3. Next step is to create two replica triangles of ▲ ABO and ▲ AOC respectively. Click on 'Cut Triangle' button to create these replicas.
4. Next step is to place these colored triangles at appropriate positions.
• First you have to place yellow colored triangle and then red colored triangle.
• Drag yellow colored triangle and place along with its hypotenuse to side AB of ▲ AOB which finally forms a rectangle AOBD.
• Drag Red colored triangle and place along with its hypotenuse to side AC of ▲ AOC which finally forms a rectangle AOCE.
• You can use 'clockwise' button to rotate triangle clockwise.
• You can use 'Anti-clockwise' button to rotate triangle Anti-clockwise.
5. See the observation.
Observation:
As □ DBCE is formed with ▲ ABC and 2 congruent triangles ABO and AOC.
∴ Area of □ DBCE = Area of ▲ ABC + (Area of ▲ ABO + Area of ▲ AOC)
= Area of ▲ ABC + Area of ▲ ABC
= 2 x Area of ▲ ABC
Area of ▲ ABC = 1/2 x Area of □ DBCE
= 1/2 x [base of □ DBCE X height of □ DBCE]
= 1/2 x [BC x DB] = 1/2 x [BC x AO]
= 1/2 x base of triangle ABC x height of triangle ABC = 1/2 x base x height
For Obtuse Angle Triangle:
As performed in the real lab:
1. Cut an obtuse angle triangle.
2. Cut a triangle congruent to this obtuse angle triangle.
3. Align the greatest side of the two triangles in order to obtain parallelogram.[Fig(C)]
As performed in simulator:
1. Create a ▲ ABC by providing its three side.
2. Triangle is Obtuse angle triangle if its square of longest side is greater than to sum of products of squares of other two sides.
3. Click on ▲ ABC to create its replica.
4. Place this replicated triangle such that it will forms parallelogram.
5. Use button 'clockwise' to rotate triangle clockwise.
6. Use button 'Anticlockwise' to rotate triangle Anticlockwise.
7. Place replicated triangle such that hypotenuse of both triangle will cover each other.
8. See the observation.
Observation:
You can observe that aligning these two congruent triangles forms a parallelogram.
As per property of parallelogram ▲ ABC and ▲ ADC are congruent
∴ Area of ▱ ABCD = Area of ▲ ABC + Area of ▲ ADC
= 2 x area of ▲ ABC
∴Area of ▲ ABC = 1/2 x Area of ▱ ABCD
= 1/2 x [base of ▱ ABCD X height of ▱ ABCD]
= 1/2 x [BC x height of ▲ ABC]
= 1/2 x base of ▲ ABC x height of ▲ ABC
= 1/2 x base x height
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!
## Mathematics (Maths) Class 9
190 videos|233 docs|82 tests
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; |
# How do you write the equation of a line in point slope form and slope intercept form given points (3, -8) (-2, 5)?
May 27, 2015
Given the points $\left(3 , - 8\right)$ and $\left(- 2 , 5\right)$
both the point-slope form and the slope-intercept form require that we first determine the slope.
The slope can be calculated as
$m = \frac{\Delta y}{\Delta x} = \frac{5 - \left(- 8\right)}{- 2 - 3} = - \frac{13}{5}$
Using the slope $m = - \frac{13}{5}$ and the point $\left(3 , - 8\right)$
the slope-point form ($y - {y}_{1} = m \left(x - {x}_{1}\right)$) is
$y - \left(- 8\right) = - \frac{13}{5} \left(x - 3\right)$
or
$y + 8 = - \frac{13}{5} \left(x - 3\right)$
The slope-point form can be converted into the slope-intercept form ($y = m x + b$) by some minor re-arranging of terms:
$y + 8 = - \frac{13}{5} \left(x - 3\right)$
$\rightarrow y = - \frac{13}{5} x + \frac{39}{5} - 8$
$\rightarrow y = - \frac{13}{5} x - \frac{1}{8}$ |
# How do you solve x+4=-4/x?
##### 1 Answer
Oct 11, 2015
The solution is color(blue)(x=-2
#### Explanation:
x+4=-4/color(blue)(x
$\textcolor{b l u e}{x} \cdot \left(x + 4\right) = - 4$
${x}^{2} + 4 x = - 4$
${x}^{2} + 4 x + 4 = 0$
We can Split the Middle Term of this expression to factorise it and thereby find the solution.
In this technique, if we have to factorise an expression like $a {x}^{2} + b x + c$, we need to think of 2 numbers such that:
${N}_{1} \cdot {N}_{2} = a \cdot c = 1 \cdot 4 = 4$
AND
${N}_{1} + {N}_{2} = b = 4$
After trying out a few numbers we get ${N}_{1} = 2$ and ${N}_{2} = 2$
$2 \cdot 2 = 4$, and $2 + 2 = 4$
${x}^{2} + \textcolor{b l u e}{4 x} + 4 = {x}^{2} + \textcolor{b l u e}{2 x + 2 x} + 4$
$x \left(x + 2\right) + 2 \left(x + 2\right) = 0$
$\left(x + 2\right) \left(x + 2\right) = 0$
We now equate the factor to zero to obtain the solution(both factors are equal here):
x+2=0, color(blue)(x=-2 |
# How to Find the Number of Roots by Stationary Points and Deduce for an Inverse Function Explained Quickly
The stationary points can be found by solving the derivatives to zero. Some cubic equations have one root if the stationary points are in the same sign, such as two stationary points are either above the x-axis or below the x-axis. In these cases, the cubic graph cuts the x-axis only once, so the cubic […]
# Differentiation and Displacement, Velocity and Acceleration
Distance Distance is the magnitude of the total movement from the start point or a fixed point. Displacement The displacement of a moving position relative to a fixed point. Displacement gives both the distance and direction that a particle is from a fixed point. For example, a particle moves $5$ units forwards from […]
# Maxima and Minima with Trigonometric Functions
Periodic motions can be modelled by a trigonometric equation. By differentiating these functions we are then able to solve problems relating to maxima (maximums) and minima (minimums). Remember that the following steps are used when solving a maximum or minimum problem. Step 1: Find $f^{\prime}(x)$ to obtain the gratest function. Step 2: Solve for $x$ […]
# Maximum and Minimum of Quadratics by Domain
Example 1 Find the maximum and minimum values of $y=x^2$, for $1 \le x \le 2$. From the graph, its maximum value is $4$, when $x=2$, and its minimum value is $1$, when $x=1$. Example 2 Find the maximum and minimum values of $y=x^2$, for $-2 \le x \le -1$. From the graph, its maximum […]
# How to Express the Velocity and the Acceleration as Functions of Displacement and Time
If a particle $P$ moves in a straight line and its position is given by the displacement function $x(t)$, then: the velocity of $P$ at time $t$ is given by $v(t) = x'(t)$ the acceleration of $P$ at time $t$ is given by $a(t)=v'(t)=x^{\prime \prime}(t)$ $x(0)$, $v(0)$ and $a(0)$ give the position, velocity and acceleration […]
# Motion Kinematics
Displacement Suppose an object $P$ moves along a straight line so that its position $s$ from an origin $O$ is given as some function of time $t$. We write $x=x(t)$ where $t \ge 0$. $x(t)$ is a displacement function and for any value of $t$ it gives the displacement from the origin. On the horizontal […]
# Inflection Points (Points of Inflection)
Horizontal (stationary) point of inflection (inflection point) If $x \lt a$, then $f'(x) \gt 0$ and $f^{\prime \prime}(x) \le 0 \rightarrow$ concave down. If $x = a$, then $f'(x) = 0$ and $f^{\prime \prime}(x) = 0 \rightarrow$ horizontal point inflection. If $x \gt a$, then $f'(x) \gt 0$ and $f^{\prime \prime}(x) \ge 0 \rightarrow$ concave […]
# Turning Points and Nature
A turning point of a function is a point where $f'(x)=0$. A maximum turning point is a turning point where the curve is concave up (from increasing to decreasing ) and $f^{\prime}(x)=0$ at the point. \begin{array}{|c|c|c|} \hline f^{\prime}(x) \gt 0 & f'(x) = 0 & f'(x) \lt 0 \\ \hline & \text{maximum} & \\ […]
# Increasing Functions and Decreasing Functions
Increasing and Decreasing We can determine intervals where a curve is increasing or decreasing by considering $f'(x)$ on the interval in question. $f'(x) \gt 0$: $f(x)$ is increasing $f'(x) \lt 0$: $f(x)$ is decreasing Monotone (Monotonic) Increasing or Decreasing Many functions are either increasing or decreasing for all $x \in \mathbb{R}$. These functions are called […]
# Finding the Normal Equations
A normal to a curve is a straight line passing through the point where the tangent touches the curve and is perpendicular (at right angles) to the tangent at that point. The gradient of the tangent to a curve is $m$, then the gradient of the normal is $\displaystyle -\dfrac{1}{m}$, as the product of the […] |
# How do you factor completely p^4+2p^3+2p^2-2p-3?
${p}^{4} + 2 {p}^{3} + 2 {p}^{2} - 2 p - 3 = \left(p - 1\right) \left(p + 1\right) \left({p}^{2} + 2 p + 3\right)$
When factoring polynomials, a good strategy is to use the Rational Root Theorem. When applying this theorem, it is a good idea to try substituting $\pm 1$ into the polynomial and seeing if the result is 0 because these substitutions are easy and can be done quickly. If $\pm 1$ doesn't give 0, try substituting in other factors of the constant term divided by factors of the leading term.
Looking at ${p}^{4} + 2 {p}^{3} + 2 {p}^{2} - 2 p - 3$, trying $p = 1$ gives 0 so we know $p - 1$ is a factor. Using synthetic division (or polynomial long division), we see that ${p}^{4} + 2 {p}^{3} + 2 {p}^{2} - 2 p - 3 = \left(p - 1\right) \left({p}^{3} + 3 {p}^{2} + 5 p + 3\right)$. Now trying $p = - 1$ into the cubic polynomial gives 0 so we get $\left(p - 1\right) \left({p}^{3} + 3 {p}^{2} + 5 p + 3\right) = \left(p - 1\right) \left(p + 1\right) \left({p}^{2} + 2 p + 3\right)$. Looking at the quadratic polynomial, we know it is irreducible because no factors of 3 add up to 2 and we are done. |
# Find the angles of the triangle formed by vectors vecP=5hati-3hatj+hatk, vecQ=-2hati+hatj+5hatk and vecR=9hati+5hatj+0hatk?
May 30, 2017
The triangle formed is a right isoceles triangle and $m \angle A = {90}^{\circ}$, $m \angle B = m \angle C = {45}^{\circ}$
#### Explanation:
Let the three vectors $\vec{P} = 5 \hat{i} - 3 \hat{j} + \hat{k}$, $\vec{Q} = - 2 \hat{i} + \hat{j} + 5 \hat{k}$ and $\vec{R} = 9 \hat{i} + 5 \hat{j} + 0 \hat{k}$, form a triangle with vertices $a$, $B$ and $C$ respectively.
Hence $| A B | = | \vec{Q} - \vec{P} | = | \left(- 2 - 5\right) \hat{i} + \left(1 - \left(- 3\right)\right) \hat{j} + \left(5 - 1\right) \hat{k} |$
= $| - 7 \hat{i} + 4 \hat{j} + 4 \hat{k} | = \sqrt{{\left(- 7\right)}^{2} + {4}^{2} + {4}^{2}} = \sqrt{81} = 9$
$| B C | = | \vec{R} - \vec{Q} | = | \left(9 - \left(- 2\right)\right) \hat{i} + \left(5 - 1\right) \hat{j} + \left(0 - 5\right) \hat{k} |$
= $| 11 \hat{i} + 4 \hat{j} - 5 \hat{k} | = \sqrt{{11}^{2} + {4}^{2} + {\left(- 5\right)}^{2}} = \sqrt{162} = 9 \sqrt{2}$
and $| C A | = | \vec{R} - \vec{P} | = | \left(9 - 5\right) \hat{i} + \left(5 - \left(- 3\right)\right) \hat{j} + \left(0 - 1\right) \hat{k} |$
= $| 4 \hat{i} + 8 \hat{j} - 1 \hat{k} | = \sqrt{{4}^{2} + {8}^{2} + {\left(- 1\right)}^{2}} = \sqrt{81} = 9$
It is apparent that $A {B}^{2} + C {A}^{2} = B {C}^{2}$, where $A B = C A$
Hence the triangle formed is a right isoceles triangle and $m \angle A = {90}^{\circ}$, $m \angle B = m \angle C = {45}^{\circ}$ |
# Lesson: Arithmetic and geometric patterns
3559 Views
3 Favorites
### Lesson Objective
I can recognize the difference between arithmetic and geometric patterns, and create a rule about the sequences
### Lesson Plan
Standard 6.PRA.8 - Recognize when information given in a table, graph, or formula suggests a proportional or linear relationship. 6.PRA.9 - Produce and interpret graphs that represent the relationship between two variables (x and y) in everyday situations Objective Essential Question SWBAT recognize the difference between arithmetic and geometric patterns, and create a rule about the sequences with 80% accuracy. Explain how patterns suggest a relationship. What kind of relationships do they suggest? Justify.
Lesson Agenda Agenda Item/ Time (Board Configuration) 5 E’s Learning Activities Teacher will do… Students will do … 10 minutes Daily Math Review (Warm-Up/ Do Now) Please fill in the missing item on each line 1 3 5 7 _______ 11 7 ______ 21 28 35 42 1 2 3 5 8 _______ Silently complete Do Now 5 minutes Mental Math Verbally give students the Mental Math Problem Respond verbally with an answer 5 minutes Engage Begin a monologue on patterns. Patterns such as 3, 6, 9, 12- are familiar to us since they are among the patterns you first learn as young students As you advance, you will experience number patterns again through the huge concepts of functions in math Today we will explore arithmetic and geometric patterns Engage in the conversation and give examples of simple patterns 10 minutes Explore (Conceptual Development) Provide 8 examples of sequences Complete the 8 sequence problems given to them 10 minutes Explain (Guided Practice) Explain the difference between a arithmetic pattern and geometric patterns. Arithmetic means we are either adding or subtracting in each sequence, and geometric means we are either multiplying or dividing in the pattern Actively listening and taking notes 25 minutes Extend/ Elaborate Set up 6 work stations for students to circulate through Divide the students into groups of 5 Rotate through work stations to observe patterns in different situations They are to determine the following: Is the pattern arithmetic and geometric? What is the pattern specifically doing? What will the next two numbers in the sequence be? 10 minutes Evaluate (Assessment/ Closure) Give the students the following Exit Slip: What will be the 7th number in this sequence? 1, 3, 9, 27, 8 2. The rule for a pattern is “multiply a number by 2, then add 1.” If the first number in the sequence is 0, what are the next five numbers in the sequence? 3. What rule determines the next number in this sequence? 2, 6, 18, 54, 162 Write an algebraic expression to describe the rule. Discuss their results from their findings and summarize it to turn in Homework MCAS Finish Line (6th grade) Pg. 122- Patterns CoreVocabulary Arithmetic and geometric patterns LessonDifferentiation Lower Level Learners: Find the next three terms in each sequence and give the rule. 2, 5, 8, 11 1, 10, 100 10, 9, 8, 7
### Lesson Resources
6 PRA 8 geometric arithmeticseq Combination 1,298 |
Increasing Intervals of a function
🏆Practice increasing and decreasing intervals of a function
The increasing intervals of a function
An increasing interval of a function expresses the same values of X (the interval), in which the values of the function (Y) grow parallel to the growth of the values of X to the right.
In certain cases, the increasing interval begins at the minimum point, but it does not necessarily have to be this way.
Test yourself on increasing and decreasing intervals of a function!
In what domain does the function increase?
Examples and exercises with solutions of intervals of increasing functions
Exercise #1
Determine which domain corresponds to the function described below:
The function represents the amount of fuel in a car's tank according to the distance traveled by the car.
Step-by-Step Solution
According to the definition, the amount of fuel in the car's tank will always decrease, since during the trip the car consumes fuel in order to travel.
Therefore, the domain that is suitable for this function is - always decreasing.
Always decreasing
Exercise #2
Choose the graph that best represents the following:
Temperature of lukewarm water (Y) after placing in the freezer as a function of time (X).
Step-by-Step Solution
Since the freezing point of water is below 0, the temperature of the water must drop below 0.
The graph in answer B describes a decreasing function and therefore this is the correct answer.
Exercise #3
Choose the graph that best describes the following:
The acceleration of a ball (Y) after throwing it from a building as a function of time (X).
Step-by-Step Solution
Since acceleration is dependent on time, it will be constant.
The force of gravity on Earth is constant, meaning the velocity of Earth's gravity is constant and therefore the graph will be straight.
The graph that appears in answer B satisfies this.
Exercise #4
Determine whether the function is increasing, decreasing, or constant. For each function check your answers with a graph or table:
Each number is divided by $(-1)$.
Step-by-Step Solution
The function is:
$f(x)=\frac{x}{-1}$
Let's start by assuming that x equals 0:
$f(0)=\frac{0}{-1}=0$
Now let's assume that x equals 1:
$f(1)=\frac{1}{-1}=-1$
Now let's assume that x equals 2:
$f(-1)=\frac{-1}{-1}=1$
Let's plot all the points on the function graph:
We see that we got a decreasing function.
Decreasing
Exercise #5
Determine whether the function is increasing, decreasing, or constant. For each function check your answers with a graph or table.
For each number, multiply by 0.
Step-by-Step Solution
The function is:
$f(x)=x\times0$
Let's start by assuming that x equals 0:
$f(0)=0\times0=0$
Now let's assume that x equals 1:
$f(1)=1\times0=0$
Now let's assume that x equals -1:
$f(-1)=(-1)\times0=0$
Now let's assume that x equals 2:
$f(2)=2\times0=0$
Let's plot all the points on the function's graph:
We can see that the function we obtained is a constant function. |
# Amanda Seyfried And The Stars (12/04/2019)
How will Amanda Seyfried get by on 12/04/2019 and the days ahead? Let’s use astrology to conduct a simple analysis. Note this is for entertainment purposes only – don’t get too worked up about the result. I will first work out the destiny number for Amanda Seyfried, and then something similar to the life path number, which we will calculate for today (12/04/2019). By comparing the difference of these two numbers, we may have an indication of how good their day will go, at least according to some astrology practitioners.
PATH NUMBER FOR 12/04/2019: We will take the month (12), the day (04) and the year (2019), turn each of these 3 numbers into 1 number, and add them together. How? Let’s walk through it. First, for the month, we take the current month of 12 and add the digits together: 1 + 2 = 3 (super simple). Then do the day: from 04 we do 0 + 4 = 4. Now finally, the year of 2019: 2 + 0 + 1 + 9 = 12. Now we have our three numbers, which we can add together: 3 + 4 + 12 = 19. This still isn’t a single-digit number, so we will add its digits together again: 1 + 9 = 10. This still isn’t a single-digit number, so we will add its digits together again: 1 + 0 = 1. Now we have a single-digit number: 1 is the path number for 12/04/2019.
DESTINY NUMBER FOR Amanda Seyfried: The destiny number will calculate the sum of all the letters in a name. Each letter is assigned a number per the below chart:
So for Amanda Seyfried we have the letters A (1), m (4), a (1), n (5), d (4), a (1), S (1), e (5), y (7), f (6), r (9), i (9), e (5) and d (4). Adding all of that up (yes, this can get tiring) gives 62. This still isn’t a single-digit number, so we will add its digits together again: 6 + 2 = 8. Now we have a single-digit number: 8 is the destiny number for Amanda Seyfried.
CONCLUSION: The difference between the path number for today (1) and destiny number for Amanda Seyfried (8) is 7. That is higher than the average difference between path numbers and destiny numbers (2.667), indicating that THIS IS A BAD RESULT. But don’t get too concerned! As mentioned earlier, this is just for fun. If you want a reading that people really swear by, check out your cosmic energy profile here. Go see what it says for you now – you may be absolutely amazed. It only takes 1 minute.
### Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene.
#### Latest posts by Abigale Lormen (see all)
Abigale Lormen
Abigale is a Masters in Business Administration by education. After completing her post-graduation, Abigale jumped the journalism bandwagon as a freelance journalist. Soon after that she landed a job of reporter and has been climbing the news industry ladder ever since to reach the post of editor at Tallahasseescene. |
# How do you solve 3x^2 - x < 8 - 4x?
Apr 16, 2018
$- 2.208 < x < 1.208$
#### Explanation:
First, simplify the expression:
3x^2 − x < 8 − 4x
$3 {x}^{2} + 3 x - 8 < 0$
Now we can solve for the roots of the "equality" and then check to set the directional inequality limits of the solutions.
${x}_{1} = 1.208$
${x}_{2} = - 2.208$
3x^2 − x < 8 − 4x; 3(1.208^2) − 1.208 < 8 − 4(1.208)
4.38 − 1.208 < 8 − 4.832 : $3.172 < 3.168$ (incorrect), thus the inequality value must be ${x}_{1} < 1.208$
CHECK:
3(1.20^2) − 1.20 < 8 − 4(1.20)
4.32 − 1.20 < 8 − 4.8 : $3.12 < 3.2$ Correct.
AND
3x^2 − x < 8 − 4x; 3(-2.208^2) − -2.208 < 8 − 4(-2.208)
$14.63 + 2.208 < 8 + 8.832$ : $16.838 < 16.832$ (incorrect), thus the inequality value must be ${x}_{1} > - 2.208$
CHECK:
3(-2.2^2) − -2.2 < 8 − 4(-2.2)
$14.52 + 2.20 < 8 + 8.8$ : $16.72 < 16.8$ Correct.
The range is thus:
$- 2.208 < x < 1.208$ |
# Matrices & linear systems
A matrix (plural matrices) is a rectangular array of numbers arranged in rows and columns. They have a variety of uses, but for us they are just a convenient way of staying organized while solving linear systems. Solving a system of two linear equations by elimination is easy, but it becomes much harder once you have four or five equations. Consider this linear system:
$\displaystyle x + 3 y = 4 \qquad \allowbreak\quad\allowbreak\text{and}\allowbreak\quad\allowbreak \qquad - 2 x + 3 y = 10$.
The first step is to move the constant term to the right-hand side. In this case, it’s already done. Now, we take the coefficients and we put them into an augmented matrix:
$\displaystyle \left [ \begin{array}{rr|r} 1 & 3 & 4 \\ - 2 & 3 & 10 \end{array} \right ]$.
The first two columns represent the x and y coefficients respectively; the last column, separated by a bar, represents the right-hand sides of the equations. To solve the system, we will use Gauss-Jordan elimination, which is a method of transforming the matrix to reduced row-echelon form. We want the columns left of the bar to have a diagonal of ones and zeros everywhere else. For example, if we were solving a linear system of four equations, the end result would look like this:
$\displaystyle \left [ \begin{array}{rrrr|r} 1 & 0 & 0 & 0 & x \\ 0 & 1 & 0 & 0 & y \\ 0 & 0 & 1 & 0 & z \\ 0 & 0 & 0 & 1 & w \end{array} \right ]$.
The values of x, y, z, and w would be the solution. To perform Gauss-Jordan elimination, we use elementary operations until we get to the reduced row-echelon form. There are just three operations—we can
• interchange any two rows;
• multiply one row by a nonzero scalar;
• add a multiple of one row to another.
## Example
$\displaystyle \left [ \begin{array}{rr|r} 1 & 3 & 4 \\ - 2 & 3 & 10 \end{array} \right ]$.
We can subtract the second row from the first to get
$\displaystyle \left [ \begin{array}{rr|r} 3 & 0 & - 6 \\ - 2 & 3 & 10 \end{array} \right ]$.
We can add two-thirds of the first row to the second to get
$\displaystyle \left [ \begin{array}{rr|r} 3 & 0 & - 6 \\ 0 & 3 & 6 \end{array} \right ]$.
Finally, we can divide both rows by three:
$\displaystyle \left [ \begin{array}{rr|r} 1 & 0 & - 2 \\ 0 & 1 & 2 \end{array} \right ]$.
The values $\displaystyle x = - 2$ and $\displaystyle y = 2$ are indeed the solution to this system. |
1. ## probability problem
Suppose that there are $2$ teams: $A$ and $B$. Team $A$'s pitcher throws a strike $50 \%$ of the time and a ball $50 \%$ of the time (successive pitches are independent from one another), and the pitcher never hits the batter. Knowing this, Team $B$'s manager instructs the first batter not to swing at anything. Calculate the following probabilities:
(a) The batter walks on the fourth pitch.
So $P \{\text{batter walks on fourth pitch} \} = \left(\frac{1}{2} \right)^4 = \frac{1}{16}$
(b) The batter walks on the sixth pitch (so two of the first five must be strikes).
So $P \{\text{batter walks on sixth pitch} \} = \frac{5 \cdot 4}{2^{5}} = \frac{20}{32} = \frac{5}{8}$
(c) The batter walks
So this is the same as $P \{\text{pitcher throws four balls in a row} \} = \frac{1}{16}$
(d) The first batter scores while no one is out. Not sure how to do this one.
2. In order for the batter to be walked on the sixth pitch it must be true that among the first five pitches there must be two strikes and three balls while the sixth pitch must be a ball. Here is an example $\underbrace {BSBSB}_5\boxed{B}$.
But the first five places can be rearranged in ${5 \choose 3}$ ways.
Thus the probability is ${5 \choose 3} \left( {\frac{1}{2}} \right)^6$.
3. why is it not $\frac{5!}{3!}$? Doesnt order matter?
4. also for (c), isnt the probability 1/2 from the law of large numbers?
5. Originally Posted by shilz222
why is it not $\frac{5!}{3!}$? Doesnt order matter?
How many ways are there to arrrange the string "SSBBB"?
It is not $\frac {5!} {3!}$.
For (c): How many ways are there to arrrange the string "SSBBBB"?
Each of those strings has a probiblity $\left( {\frac{1}{2}} \right)^6$
6. $\binom{5}{3}$ and $\binom{6}{4}$ respectively.
7. and for part (d) the batter scores implies that he walks (assuming he does not hit the ball) right? so how would we do this?
8. Originally Posted by shilz222
and for part (d) the batter scores implies that he walks (assuming he does not hit the ball) right? so how would we do this?
Sorry, but I know so little about sports is general and baseball in particular, that I cannot answer your question.
9. yeah what that does mean exactly? |
Trigonometry homework help
The Best Trigonometry homework help
Apps can be a great way to help students with their algebra. Let's try the best Trigonometry homework help. It is important to be able to solve expressions. This is because solving expressions is a fundamental skill in algebra. Algebra is the branch of mathematics that deals with equations and variables, and it is frequently used in physics and engineering. Many word problems can be translated into algebraic expressions, and being able to solve these expressions will allow you to solve the problem. In order to solve an expression, you need to use the order of operations. The order of operations is a set of rules that tells you the order in which to solve an equation. The order of operations is: Parentheses, Exponents, Multiplication and Division (from left to right), Addition and Subtraction (from left to right). Using the order of operations, you can solve any expression.
This can be a useful tool for solving problems in physics or engineering, where you might need to find the total amount of energy in a system, for example. There are a variety of different methods that can be used to solve series, and the choice of method will depend on the particular problem you are trying to solve. However, some of the most popular methods include the Euler-Maclaurin formula and the Ricci identity. With a little practice, you should be able to use a series solver to solve a wide range of problems.
However, more often than not, we need to solve a system of equations in order to find all of the unknown values. Fortunately, there are a variety of methods that we can use to solve systems of equations, including elimination and substitution. With a little practice, solving algebra problems can be easy and even fun!
A logarithmic equation solver is a tool that can be used to solve equations with Logarithms. Logarithmic equations often arise in settings where one is working with exponential functions. For example, if one were to take the natural log of both sides of the equation y = 2x, they would obtain the following equation: Log(y) = Log(2x). This equation can be difficult to solve without the use of a logarithmic equation solver. A logarithmic equation solver can be used to determine the value of x that satisfies this equation. In this way, a logarithmic equation solver can be a valuable tool for solving equations with Logarithms.
Differential equations are a type of mathematical equation that can be used to model many real-world situations. In general, they involve the derivative of a function with respect to one or more variables. While differential equations may seem daunting at first, there are a few key techniques that can be used to solve them. One common method is known as separation of variables. This involves breaking up the equation into two parts, one involving only the derivative and the other involving only the variable itself. Once this is done, the two parts can be solved independently and then recombined to find the solution to the original equation. Another popular method is known as integration by substitution. This approach involves substituting a new variable for the original one in such a way that the resulting equation is easier to solve. These are just a few of the many methods that can be used to solve differential equations. With practice, anyone can become proficient in this important mathematical discipline.
We cover all types of math issues
This the one of the best apps which I have ever seen for using math I have almost tried all the apps in available in Play Store it can be used offline also just I had a small complaint that it is not able to solve word problems otherwise it is the best app ever have same for solving math it will just give answer in just second not more than a second talking about the premium it should be free or it was little more much so I request to it cost must be reduced
Elisa Walker
This app is so great that I can't explain that. All the math solutions are correct and accurate. Thanks for making this app. This app helps those people like me Whose are week in math. Thank you
Octavia Perry
App that helps with math problems How to solve slope Homework help online chat Solve word problems in algebra Purple math equation solver |
# Table of 128
Created by: Team Maths - Examples.com, Last Updated: May 20, 2024
## Table of 128
The table of 128 lists the products of multiplying 128 by integers from 1 to 20. Each step increases sequentially by 128, demonstrating multiplication as repeated addition. This table helps visualize the growth pattern and is useful in educational contexts.
## What is the Multiplication Table of 128?
The multiplication table of 128 systematically lists the products of multiplying 128 by integers from 1 up to 20. Starting from 128, each entry in the table increases by 128, reflecting simple arithmetic progressions. For example, 128 multiplied by 1 equals 128, by 2 equals 256, and this pattern continues incrementally up to 128 multiplied by 20, which equals 2560. This table is a useful tool in mathematics for understanding scaling and progression in series, helping to establish a clear visual representation of multiplication as repeated addition.
## 128ᵗʰ Table
The 128th table displays sequential products of 128 with numbers 1 through 20, illustrating a linear increase in values, useful for quick multiplication reference.
## Tips for 128ᵗʰ Table
### Understand Doubling:
Since 128 is a power of 2 (2^7), multiplying it by any number can be viewed as successive doublings. This understanding can help simplify calculations and improve mental math skills.
### Use the Halving Technique:
If you know a result for a higher number, you can quickly find the result for a smaller number by halving. For example, if you know 128 x 10 = 1280, then 128 x 5 is half of 1280, which is 640.
Adding 128 repeatedly can help you calculate higher multiples without formal multiplication. This can be particularly useful for smaller multiples.
### Leverage Patterns:
Observe patterns in the digits of products for easier recall. For example, every multiple of 128 ends in an even number, and observing the cycle of last digits can provide clues for quick computation.
### Use a Calculator for Practice:
Regularly use a calculator to check your answers as you practice multiplying by 128. This reinforces learning and ensures accuracy.
### Break Down Complex Multiplications:
For higher numbers, break the multiplication into smaller parts. For example, 128 x 18 can be split into (128 x 10) + (128 x 8).
### Apply Real-Life Examples:
Apply multiplication of 128 in real scenarios such as computing memory sizes in computing, where quantities often double (128 MB, 256 MB, etc.).
## Table of 128 from 11 to 20
The table of 128 from 11 to 20 lists products of 128 multiplied by each integer from 11 through 20, demonstrating larger-scale multiplicative increments.
## How to Read 128ᵗʰ Tables?
• One time 128 is 128
• Two times 128 is 256
• Three times 128 is 384
• Four times 128 is 512
• Five times 128 is 640
• Six times 128 is 768
• Seven times 128 is 896
• Eight times 128 is 1024
• Nine times 128 is 1152
• Ten times 128 is 1280
To read the 128th multiplication table, start with 128 and multiply it sequentially by integers from 1 to 20, calculating the result for each multiplication.
## Example 1: Simple Multiplication
Question: What is 128 times 2?
Solution: To find 128 times 2, you multiply 128 by 2.
Calculation: 128 x 2 = 256
Answer: 128 times 2 equals 256.
Question: How much is 128 added to itself 3 times?
Solution: Adding 128 to itself 3 times is the same as multiplying 128 by 3.
Calculation: 128 x 3 = 384
## Example 3: Real-world Application
Question: If a jacket costs \$128, how much will 4 jackets cost?
Solution: Multiply the cost of one jacket by the number of jackets.
Calculation: 128 x 4 = 512
## Example 4: Doubling
Question: What is the double of 128?
Solution: Doubling a number is the same as multiplying it by 2.
Calculation: 128 x 2 = 256
Answer: The double of 128 is 256.
## Example 5: Multiplication by Ten
Question: What is 128 times 10?
Solution: To find 128 times 10, you multiply 128 by 10.
Calculation: 128 x 10 = 1280
Answer: 128 times 10 equals 1280.
## Example 6: Cost Analysis
Question: If one book costs \$128, how much do 7 books cost?
Solution: Multiply the cost of one book by the total number of books.
Calculation: 128 x 7 = 896
## Example 7: Division into Groups
Question: How many groups of 128 can you form with 1024 items?
Solution: Divide the total number of items by 128.
Calculation: 1024 ÷ 128 = 8
Answer: You can form 8 groups of 128 with 1024 items.
## Example 8: Multiplying by a Hundred
Question: What is 128 times 100?
Solution: To find 128 times 100, you multiply 128 by 100.
Calculation: 128 x 100 = 12800
Answer: 128 times 100 equals 12800.
## Example 9: Increment by Quarters
Question: What is 128 increased by a quarter of its value?
Solution: To increase 128 by a quarter, multiply 128 by 0.25 and add it to 128.
Calculation: 128 + (128 x 0.25) = 128 + 32 = 160
Answer: 128 increased by a quarter equals 160.
## Example 10: Triple Amount
Question: What is three times 128?
Solution: Tripling a number is the same as multiplying it by 3.
Calculation: 128 x 3 = 384
Answer: Three times 128 is 384.
Text prompt |
# Create and interpret line graphs (rational number axis)
Lesson
## Line graphs
We have seen how line graphs are ideal to see how things change over time.
As well as whole numbers, line graphs can include fractions or decimals, and this allows us to focus on things where whole numbers may not make sense.
We also look at different ways to label the axes on our graph, and how the scale may be different between the vertical axis and horizontal axis.
## Plotting a graph
When we plot a line graph with decimals or fractions, we use the same approach that we use for whole numbers, but the value of each of the increments, or ticks, on the axes may be different. Once we plot the graph, we can use it to find out some useful information.
In Video 2, let's plot a line graph to see how the black puppy is growing.
Remember!
We can have different scales on our vertical and horizontal axes.
#### Examples
##### QUESTION 1
The line graph below shows the height of a tree over four years.
1. What is the height of the tree after $2.5$2.5 years?
2. Does the tree grow more in the first or the second year?
The second year
A
The first year
B
##### QUESTION 2
A university divides its years into trimesters (one third of a year).
Roxanne is planning on enrolling at the university and uses the line graph below to see how many units of study she can complete in a given number of years.
1. How many years will it take Roxanne to complete $14$14 units of study?
2. What is the highest possible number of units of study that Roxanne can complete in $2\frac{1}{3}$213 years?
3. Roxanne needs to complete at least $18$18 units of study to complete her degree.
If she wants to complete her degree in the shortest amount of time, what is the highest number of units she can complete?
##### QUESTION 3
Luigi brings a $1.8$1.8 L bottle of water to work each day.
He records the amount of water in the bottle at the start of each hour on the line graph below.
1. How much water was in Luigi's bottle at the start of the day?
2. What time was Luigi's bottle first refilled?
$\editable{}$$\editable{}:\editable{}$$\editable{}$ PM
3. How much water did Luigi drink in the hour after first refilling his bottle?
4. How many hours were between the first and second time Luigi filled his bottle?
5. How much water did Luigi drink between the two refills?
6. How much did the amount of water in Luigi's bottle change between the start and end of the day? |
# Mixing problems
### Differential Equations
Mixing problems are an application of separable differential equations. They’re word problems that require us to create a separable differential equation based on the concentration of a substance in a tank.
Usually we’ll have a substance like salt that’s being added to a tank of water at a specific rate. At the same time, the salt water mixture is being emptied from the tank at a specific rate. We usually that the contents of the tank are always perfectly mixed, and we’re asked to model the concentration in the tank at a certain time. The formula we use to model concentration is
???\frac{dy}{dt}=C_1r_1-C_2r_2???
where
???C_1??? is the concentration of the substance being added
???r_1??? is the rate at which the substance is added
???C_2??? is the concentration of the substance being removed
???r_2??? is the rate at which the substance is removed
Once we’ve plugged everything into the mixing problem formula, we’ll need to treat it as a separable differential equation, which means that we’ll separate variables, integrate both sides of the equation, and then try to find a general solution.
Let’s look at an example.
Example
A tank contains ???1,500??? L of water and ???20??? kg of dissolved salt. Fresh water is entering the tank at ???15??? L/min (the solution stays perfectly mixed), and the solution drains at a rate of ???10??? L/min. How much salt is in the tank at ???t??? minutes and at ???10??? minutes?
???\frac{dy}{dt}=C_1r_1-C_2r_2???
In this problem, we’re interested in the concentration of salt in the tank.
???C_1=0??? kg/L because no salt is being added into the tank.
???r_1=15??? L/min because this is the rate at which water is entering the tank
???C_2=\frac{y}{1,500}??? kg/L because we’re not sure how much salt is leaving the tank, but we know the initial amount of water is ???1,500??? L
???r_2=10??? L/min because this is the rate at which the solution is leaving the tank
If we plug all these values into the formula, we get
???\frac{dy}{dt}=(0)(15)-\left(\frac{y}{1,500}\right)(10)???
???\frac{dy}{dt}=-\frac{y}{150}???
Now we’ll separate the variables.
???dy=-\frac{y}{150}\ dt???
???\frac{1}{y}\ dy=-\frac{1}{150}\ dt???
With the variables separated, we’ll integrate both sides of the equation.
???\int\frac{1}{y}\ dy=\int-\frac{1}{150}\ dt???
???\ln{|y|}+C_1=-\frac{1}{150}t+C_2???
Collect and simplify the constants.
???\ln{|y|}=-\frac{1}{150}t+C_2-C_1???
???\ln{|y|}=-\frac{1}{150}t+C???
Raise both sides to the base ???e??? in order to eliminate the natural log.
???e^{\ln{|y|}}=e^{-\frac{1}{150}t+C}???
???|y|=e^{-\frac{1}{150}t}e^C???
???e^C??? is a constant, so it can simplify to just ???C???. And we can remove the absolute value by adding ???\pm??? to the other side of the equation.
???|y|=Ce^{-\frac{1}{150}t}???
???y=\pm{C}e^{-\frac{1}{150}t}???
The ???\pm??? gets absorbed into the constant ???C??? and so the explicit equation for ???y??? is
???y=Ce^{-\frac{1}{150}t}???
We were told that initially ???20??? kg of dissolved salt existed in the tank. This is essentially the initial condition ???y(0)=20???.
???20=Ce^{-\frac{1}{150}(0)}???
???20=C(1)???
???C=20???
Plugging this back into the general solution, we get
???y(t)=20e^{-\frac{1}{150}t}???
This is the equation that models the amount of salt in the tank at ???t??? minutes. If we want to figure out how much salt is in the tank after ???5??? minutes, we just plug ???5??? in for ???t???. If we want to figure out how much salt is in the tank after ???20??? minutes, we just plug ???20??? in for ???t???.
We’ve also been asked in this problem to find the amount of salt in the tank after ???10??? minutes. Plugging ???10??? in for ???t???, we get
???y(10)=20e^{-\frac{1}{150}(10)}???
???y(10)=20e^{-\frac{1}{15}}???
???y(10)=18.7???
After ???10??? minutes, there’s ???18.7??? kg of salt in the tank. |
# Video: Calculating the Work Done by a Position Dependent Force on a Moving Particle
Consider a particle on which several forces act, one of which is known to be constant in time: π
β = (3.0π’ + 4.0π£) N. As a result, the particle moves along a straight path from a Cartesian coordinate of (0 m, 0 m) to (5.0 m, 6.0 m). What is the work done by π
β?
05:08
### Video Transcript
Consider a particle on which several forces act, one of which is known to be constant in time: π
sub one is equal to 3.0π’ plus 4.0π£ newtons. As a result, the particle moves along a straight path from a Cartesian coordinate of zero metres, zero metres to 5.0 metres, 6.0 metres. What is the work done by π
sub one?
Letβs start by highlighting some of the critical information weβve been given. First, weβre told about the force called πΉ sub one. And weβre given the expression of that force in its π₯ and π¦ or π and π components. Then, weβre told about a particle that moves under the influence of at least πΉ sub one from the location starting at the origin zero metres, zero metres moving to the point 5.0 metres, 6.0 metres in a horizontal and a vertical plane.
Weβre looking for the work done by πΉ one on this particle over the journey, and weβll represent that work using the symbol capital π. As we begin, letβs recall a definition that relates work, force, and displacement. Work, capital π, is defined as a force exerted on an object multiplied by the displacement of that object. As we look over the problem statement, we see weβve been given a force; thatβs, πΉ sub one, and weβve also been given a displacement β the path that the particle travels. We can use this information then to solve for π, the work done by πΉ one.
Now as we look at our problem, we recognize it as a two-dimensional problem. Thereβs both a horizontal and a vertical component to πΉ one and also a horizontal and a vertical component to the motion of the particle. That means we can write our total work π as the sum of the work in the horizontal and the vertical directions. Weβll call those values of work π sub π₯ for the horizontal work done and π sub π¦ for the vertical work done.
If we refer to our definition for work, we can write out equations for work in the π₯- and π¦-directions: π sub π₯ is equal to πΉ sub π₯ times π sub π₯. This is a way of saying that the work in the horizontal direction is equal to the force in the horizontal direction times the displacement in the horizontal direction. Likewise, we can write a similar equation for π sub π¦, the work done in the vertical direction.
Now that we have these two equations for work in the horizontal and vertical directions, we can begin to fill in πΉ sub π₯ and π sub π₯ and πΉ sub π¦ and π sub π¦ from the given information in our problem. Take another look at πΉ sub one, the component in the π or horizontal direction is 3.0 newtons. So we can write that in for πΉ sub π₯.
Then, as we look at the particle path, and specifically at π point, we see that its displacement in the horizontal or π₯-direction is 5.0 metres minus 0 metres, which equals 5.0 metres. Now moving on to work in the π¦ or vertical direction, we see that the component of πΉ sub one that is vertical is 4.0 newtons. And the displacement of our particle in the π¦-direction or vertical direction is 6.0 metres minus zero metres or 6.0 metres.
We can now multiply these pairs of numbers together: 3.0 newtons times 5.0 metres equals 15 newton metres. And then for π sub π¦, 4.0 newtons times 6.0 metres equals 24 newton metres.
Now, we can use these horizontal and vertical components of the total work done to solve for π, the net work done on the particle by πΉ one. To do this, we need to sum π sub π₯ and π sub π¦: 15 newton metres plus 24 newton metres equals 39 newton metres. Another name for a newton metre is a joule. This is the total amount of work done on the particle by πΉ sub one as the particle moves along this straight line path. |
# Question Video: Using Triangle Congruence Criteria to Establish Congruence Mathematics • 11th Grade
Determine whether the triangles in the given figure are congruent by applying SSS, SAS, or ASA. If they are congruent, state which of the congruence criteria proves this.
02:26
### Video Transcript
Determine whether the triangles in the given figure are congruent by applying SSS, SAS, or ASA. If they are congruent, state which of the congruence criteria proves this.
So we’re presented with two triangles, triangles 𝐴𝐵𝐶 and 𝐴 dash 𝐵 dash 𝐶 dash, and asked to determine whether or not they’re congruent. We’re also given three possible congruence criteria. Remember here, S stands for side and 𝐴 stands for angle.
Let’s look at the two diagrams and see what congruence statements we can write down. First, we see that both triangles have a length of three units. This side is 𝐴 dash 𝐵 dash in the first triangle and 𝐴𝐵 in the second, so we have the statement 𝐴 dash 𝐵 dash is equal to 𝐴𝐵. The S in brackets is used to indicate that this is a statement about the length of a side.
Next, we see that both triangles have a length of five units, its side 𝐵 dash 𝐶 dash in the first triangle and side 𝐵𝐶 in the second. So we have the statement 𝐵 dash 𝐶 dash is equal to 𝐵𝐶. And again, the S in brackets indicates that this is a statement about a side.
Finally, we see that both triangles have a length of 3.16 units, side 𝐴 dash 𝐶 dash in the first triangle and side 𝐴𝐶 in the second. So we have the statement 𝐴 dash 𝐶 dash is equal to 𝐴𝐶, and again the S in brackets to indicate it’s a statement about a side.
Now looking at the three statements we’ve made and comparing them to the congruence criteria, we can see that we do have enough information to conclude that these two triangles are congruent. The inclusion of the S’s as we wrote down our congruency statements tells us that it’s the side-side-side, SSS, congruence criteria.
And so we have our answer to the problem: yes, the two triangles are congruent, and this is due to the side-side-side congruence criteria. |
# How do you simplify the rational expression: (x-3)/(x^2-5x+6)?
Jun 8, 2018
$\frac{1}{x - 2}$
#### Explanation:
Simply the ${x}^{2} - 5 x + 6$ first:
Lets find the factors of ${x}^{2} - 5 x + 6$
$3 \times 2 = 6$ ----> adding them gives $5$ ----> we want adding it gives $- 5$
$- 3 \times - 2 = 6$ ----> adding them gives $- 5$ ---> This is the one.
Re-write the equation as follows:
${x}^{2} - 5 x + 6$
${x}^{2} - 3 x - 2 x + 6$
$x \left(x - 3\right) - 2 \left(x - 3\right)$
$\left(x - 2\right) \left(x - 3\right)$
So now we have;
$\frac{x - 3}{{x}^{2} - 5 x + 6}$ = $\frac{x - 3}{\left(x - 2\right) \left(x - 3\right)}$ = cancel(x-3)/((x-2)cancel((x-3))# = $\frac{1}{x - 2}$
Hence its simplified as:
$\frac{x - 3}{{x}^{2} - 5 x + 6}$ = $\frac{1}{x - 2}$ |
$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 10.1: Power Series and Functions
[ "article:topic", "radius of convergence", "interval of convergence", "power series", "authorname:openstax", "license:ccby" ]
$$\newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} }$$
$$\newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}}$$
A power series is a type of series with terms involving a variable. More specifically, if the variable is $$x$$, then all the terms of the series involve powers of $$x$$. As a result, a power series can be thought of as an infinite polynomial. Power series are used to represent common functions and also to define new functions. In this section we define power series and show how to determine when a power series converges and when it diverges. We also show how to represent certain functions using power series.
### Form of a Power Series
A series of the form
$\sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+\ldots ,$
where $$x$$ is a variable and the coefficients $$c_n$$ are constants, is known as a power series. The series
$1+x+x^2+\ldots =\sum_{n=0}^∞x^n$
is an example of a power series. Since this series is a geometric series with ratio $$r=|x|$$, we know that it converges if $$|x|<1$$ and diverges if $$|x|≥1.$$
Definition $$\PageIndex{1}$$: Power Series
A series of the form
$\sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+\ldots$
is a power series centered at $$x=0.$$ A series of the form
$\sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+\ldots$
is a power series centered at $$x=a$$.
To make this definition precise, we stipulate that $$x^0=1$$ and $$(x−a)^0=1$$ even when $$x=0$$ and $$x=a$$, respectively.
The series
$\sum_{n=0}^∞\dfrac{x^n}{n!}=1+x+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\ldots$
and
$\sum_{n=0}^∞n!x^n=1+x+2!x^2+3!x^3+\ldots$
are both power series centered at $$x=0.$$ The series
$\sum_{n=0}^∞\dfrac{(x−2)^n}{(n+1)3^n}=1+\dfrac{x−2}{2⋅3}+\dfrac{(x−2)^2}{3⋅3^2}+\dfrac{(x−2)^3}{4⋅3^3}+\ldots$
is a power series centered at $$x=2$$.
### Convergence of a Power Series
Since the terms in a power series involve a variable $$x$$, the series may converge for certain values of $$x$$ and diverge for other values of $$x$$. For a power series centered at $$x=a$$, the value of the series at $$x=a$$ is given by $$c_0$$. Therefore, a power series always converges at its center. Some power series converge only at that value of $$x$$. Most power series, however, converge for more than one value of $$x$$. In that case, the power series either converges for all real numbers $$x$$ or converges for all $$x$$ in a finite interval. For example, the geometric series $$\sum_{n=0}^∞x^n$$ converges for all $$x$$ in the interval $$(−1,1)$$, but diverges for all $$x$$ outside that interval. We now summarize these three possibilities for a general power series.
Note $$\PageIndex{1}$$: Convergence of a Power Series
Consider the power series $$\displaystyle \sum_{n=0}^∞c_n(x−a)^n.$$ The series satisfies exactly one of the following properties:
1. The series converges at $$x=a$$ and diverges for all $$x≠a.$$
2. The series converges for all real numbers $$x$$.
3. There exists a real number $$R>0$$ such that the series converges if $$|x−a|<R$$ and diverges if $$|x−a|>R$$. At the values $$x$$ where |x−a|=R, the series may converge or diverge.
Proof
Suppose that the power series is centered at $$a=0$$. (For a series centered at a value of a other than zero, the result follows by letting $$y=x−a$$ and considering the series
$\sum_{n=1}^∞c_ny^n. \nonumber$
We must first prove the following fact:
If there exists a real number $$d≠0$$ such that $$\sum_{n=0}^∞c_nd^n$$ converges, then the series $$\sum_{n=0}^∞c_nx^n$$ converges absolutely for all $$x$$ such that $$|x|<|d|.$$
Since $$\sum_{n=0}^∞c_nd^n$$ converges, the nth term $$c_nd^n→0$$ as $$n→∞$$. Therefore, there exists an integer $$N$$ such that $$|c_nd^n|≤1$$ for all $$n≥N.$$ Writing
$|c_nx^n|=|c_nd^n| \left|\dfrac{x}{d}\right|^n, \nonumber$
we conclude that, for all n≥N,
$|c_nx^n|≤\left|\dfrac{x}{d}\right|^n. \nonumber$
The series
$\sum_{n=N}^∞\left|\dfrac{x}{d}\right|^n \nonumber$
is a geometric series that converges if $$|\dfrac{x}{d}|<1.$$ Therefore, by the comparison test, we conclude that $$\sum_{n=N}^∞c_nx^n$$ also converges for $$|x|<|d|$$. Since we can add a finite number of terms to a convergent series, we conclude that $$\sum_{n=0}^∞c_nx^n$$ converges for $$|x|<|d|.$$
With this result, we can now prove the theorem. Consider the series
$\sum_{n=0}^∞a_nx^n \nonumber$
and let $$S$$ be the set of real numbers for which the series converges. Suppose that the set $$S={0}.$$ Then the series falls under case i.
Suppose that the set $$S$$ is the set of all real numbers. Then the series falls under case ii. Suppose that $$S≠{0}$$ and $$S$$ is not the set of real numbers. Then there exists a real number $$x*≠0$$ such that the series does not converge. Thus, the series cannot converge for any $$x$$ such that $$|x|>|x*|$$. Therefore, the set $$S$$ must be a bounded set, which means that it must have a smallest upper bound. (This fact follows from the Least Upper Bound Property for the real numbers, which is beyond the scope of this text and is covered in real analysis courses.) Call that smallest upper bound $$R$$. Since $$S≠{0}$$, the number $$R>0$$. Therefore, the series converges for all $$x$$ such that $$|x|<R,$$ and the series falls into case iii.
If a series $$\sum_{n=0}^∞c_n(x−a)^n$$ falls into case iii. of Note, then the series converges for all $$x$$ such that $$|x−a|<R$$ for some $$R>0$$, and diverges for all $$x$$ such that $$|x−a|>R$$. The series may converge or diverge at the values $$x$$ where $$|x−a|=R$$. The set of values $$x$$ for which the series $$\sum_{n=0}^∞c_n(x−a)^n$$ converges is known as the interval of convergence. Since the series diverges for all values $$x$$ where $$|x−a|>R$$, the length of the interval is $$2R$$, and therefore, the radius of the interval is $$R$$. The value $$R$$ is called the radius of convergence. For example, since the series $$\sum_{n=0}^∞x^n$$ converges for all values $$x$$ in the interval $$(−1,1)$$ and diverges for all values $$x$$ such that $$|x|≥1$$, the interval of convergence of this series is $$(−1,1)$$. Since the length of the interval is $$2$$, the radius of convergence is $$1$$.
Consider the power series $$\sum_{n=0}^∞c_n(x−a)^n$$. The set of real numbers $$x$$ where the series converges is the interval of convergence. If there exists a real number $$R>0$$ such that the series converges for $$|x−a|<R$$ and diverges for $$|x−a|>R,$$ then $$R$$ is the radius of convergence. If the series converges only at $$x=a$$, we say the radius of convergence is $$R=0$$. If the series converges for all real numbers $$x$$, we say the radius of convergence is $$R=∞$$ (Figure $$\PageIndex{1}$$).
Figure $$\PageIndex{1}$$: For a series $$\sum_{n=0}^∞c_n(x−a)^n$$ graph (a) shows a radius of convergence at $$R=0$$, graph (b) shows a radius of convergence at $$R=∞$$, and graph (c) shows a radius of convergence at $$R$$. For graph (c) we note that the series may or may not converge at the endpoints $$x=a+R$$ and $$x=a−R$$.
To determine the interval of convergence for a power series, we typically apply the ratio test. In Example $$\PageIndex{1}$$, we show the three different possibilities illustrated in Figure $$\PageIndex{1}$$.
Example $$\PageIndex{1}$$: Finding the Interval and Radius of Convergence
For each of the following series, find the interval and radius of convergence.
1. $$\displaystyle \sum_{n=0}^∞\dfrac{x^n}{n!}$$
2. $$\displaystyle \sum_{n=0}^∞n!x^n$$
3. $$\displaystyle \sum_{n=0}^∞\dfrac{(x−2)^n}{(n+1)3^n}$$
Solution
a. To check for convergence, apply the ratio test. We have
\begin{align*} ρ&=\lim_{n→∞} \left|\dfrac{\dfrac{x^{n+1}}{(n+1)!}}{\dfrac{x^n}{n!}}\right| \\[5pt] &=\lim_{n→∞} \left|\dfrac{x^{n+1}}{(n+1)!}⋅\dfrac{n!}{x^n}\right| \\[5pt] &=\lim_{n→∞}\left|\dfrac{x^{n+1}}{(n+1)⋅n!}⋅\dfrac{n!}{x^n}\right| \\[5pt] &=\lim_{n→∞}\left|\dfrac{x}{n+1}\right| \\[5pt] &=|x|\lim_{n→∞}\dfrac{1}{n+1} \\[5pt] &=0<1\end{align*}
for all values of $$x$$. Therefore, the series converges for all real numbers $$x$$. The interval of convergence is $$(−∞,∞)$$ and the radius of convergence is $$R=∞.$$
b. Apply the ratio test. For $$x≠0$$, we see that
\begin{align*} ρ&=\lim_{n→∞}\left|\dfrac{(n+1)!x^{n+1}}{n!x^n}\right| \\[5pt] &=\lim_{n→∞}|(n+1)x| \\[5pt] &=|x|\lim_{n→∞}(n+1) \\[5pt] &=∞. \end{align*}
Therefore, the series diverges for all $$x≠0$$. Since the series is centered at $$x=0$$, it must converge there, so the series converges only for $$x≠0$$. The interval of convergence is the single value $$x=0$$ and the radius of convergence is $$R=0$$.
c. In order to apply the ratio test, consider
\begin{align*} ρ&=\lim_{n→∞}\left|\dfrac{\dfrac{(x−2)^{n+1}}{(n+2)3^{n+1}}}{\dfrac{(x−2)^n}{(n+1)3^n}}\right| \\[5pt] &=\lim_{n→∞} \left|\dfrac{(x−2)^{n+1}}{(n+2)3^{n+1}}⋅\dfrac{(n+1)3^n}{(x−2)^n}\right| \\[5pt] &=\lim_{n→∞} \left|\dfrac{(x−2)(n+1)}{3(n+2)}\right|\\[5pt] &=\dfrac{|x−2|}{3}.\end{align*}
The ratio $$ρ<1$$ if $$|x−2|<3$$. Since $$|x−2|<3$$ implies that $$−3<x−2<3,$$ the series converges absolutely if $$−1<x<5$$. The ratio $$ρ>1$$ if $$|x−2|>3$$. Therefore, the series diverges if $$x<−1$$ or $$x>5$$. The ratio test is inconclusive if $$ρ=1$$. The ratio $$ρ=1$$ if and only if $$x=−1$$ or $$x=5$$. We need to test these values of $$x$$ separately. For $$x=−1$$, the series is given by
$\sum_{n=0}^∞\dfrac{(−1)^n}{n+1}=1−\dfrac{1}{2}+\dfrac{1}{3}−\dfrac{1}{4}+\ldots . \nonumber$
Since this is the alternating harmonic series, it converges. Thus, the series converges at $$x=−1$$. For $$x=5$$, the series is given by
$\sum_{n=0}^∞\dfrac{1}{n+1}=1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\ldots . \nonumber$
This is the harmonic series, which is divergent. Therefore, the power series diverges at $$x=5$$. We conclude that the interval of convergence is $$[−1,5)$$ and the radius of convergence is $$R=3$$.
Exercise $$\PageIndex{1}$$
Find the interval and radius of convergence for the series
$\sum_{n=1}^∞\dfrac{x^n}{\sqrt{n}}. \nonumber$
Hint
Apply the ratio test to check for absolute convergence.
The interval of convergence is $$[−1,1).$$ The radius of convergence is $$R=1.$$
### Representing Functions as Power Series
Being able to represent a function by an “infinite polynomial” is a powerful tool. Polynomial functions are the easiest functions to analyze, since they only involve the basic arithmetic operations of addition, subtraction, multiplication, and division. If we can represent a complicated function by an infinite polynomial, we can use the polynomial representation to differentiate or integrate it. In addition, we can use a truncated version of the polynomial expression to approximate values of the function. So, the question is, when can we represent a function by a power series?
Consider again the geometric series
$1+x+x^2+x^3+\ldots =\sum_{n=0}^∞x^n.$
Recall that the geometric series
$a+ar+ar^2+ar^3+\ldots$
converges if and only if $$|r|<1.$$ In that case, it converges to $$\dfrac{a}{1−r}$$. Therefore, if $$|x|<1$$, the series in Example $$\PageIndex{1}$$ converges to $$\dfrac{1}{1−x}$$ and we write
$1+x+x^2+x^3+\ldots =\dfrac{1}{1−x} for|x|<1.$
As a result, we are able to represent the function $$f(x)=\dfrac{1}{1−x}$$ by the power series
$1+x+x^2+x^3+\ldots when|x|<1.$
We now show graphically how this series provides a representation for the function $$f(x)=\dfrac{1}{1−x}$$ by comparing the graph of $$f$$ with the graphs of several of the partial sums of this infinite series.
Example $$\PageIndex{2}$$: Graphing a Function and Partial Sums of its Power Series
Sketch a graph of $$f(x)=\dfrac{1}{1−x}$$ and the graphs of the corresponding partial sums $$S_N(x)=\sum_{n=0}^Nx^n$$ for $$N=2,4,6$$ on the interval $$(−1,1)$$. Comment on the approximation $$S_N$$ as $$N$$ increases.
Solution
From the graph in Figure you see that as $$N$$ increases, $$S_N$$ becomes a better approximation for $$f(x)=\dfrac{1}{1−x}$$ for $$x$$ in the interval $$(−1,1)$$.
Figure $$\PageIndex{2}$$: The graph shows a function and three approximations of it by partial sums of a power series.
Exercise $$\PageIndex{2}$$
Sketch a graph of $$f(x)=\dfrac{1}{1−x^2}$$ and the corresponding partial sums $$S_N(x)=\sum_{n=0}^Nx^{2n}$$ for $$N=2,4,6$$ on the interval $$(−1,1).$$
Hint
$$S_N(x)=1+x^2+\ldots +x^{2N}=\dfrac{1−x^{2(N+1)}}{1−x^2}$$
Next we consider functions involving an expression similar to the sum of a geometric series and show how to represent these functions using power series.
Example $$\PageIndex{3}$$: Representing a Function with a Power Series
Use a power series to represent each of the following functions $$f$$. Find the interval of convergence.
1. $$f(x)=\dfrac{1}{1+x^3}$$
2. $$f(x)=\dfrac{x^2}{4−x^2}$$
Solution
a. You should recognize this function $$f$$ as the sum of a geometric series, because
$\dfrac{1}{1+x^3}=\dfrac{1}{1−(−x^3)}. \nonumber$
Using the fact that, for $$|r|<1,\dfrac{a}{1−r}$$ is the sum of the geometric series
$\sum_{n=0}^∞ar^n=a+ar+ar^2+\ldots , \nonumber$
we see that, for $$|−x^3|<1,$$
\begin{align*} \dfrac{1}{1+x^3}&=\dfrac{1}{1−(−x^3)} \\[5pt] &=\sum_{n=0}^∞(−x^3)^n \\[5pt] &=1−x^3+x^6−x^9+\ldots . \end{align*}
Since this series converges if and only if $$|−x^3|<1$$, the interval of convergence is $$(−1,1)$$, and we have
$\dfrac{1}{1+x^3}=1−x^3+x^6−x^9+\ldots for|x|<1.\nonumber$
b. This function is not in the exact form of a sum of a geometric series. However, with a little algebraic manipulation, we can relate f to a geometric series. By factoring 4 out of the two terms in the denominator, we obtain
\begin{align*} \dfrac{x^2}{4−x^2}&=\dfrac{x^2}{4(\dfrac{1−x^2}{4})}\\[5pt] &=\dfrac{x^2}{4(1−(\dfrac{x}{2})^2)}.\end{align*}
Therefore, we have
\begin{align*} \dfrac{x^2}{4−x^2}&=\dfrac{x^2}{4(1−(\dfrac{x}{2})^2)} \\[5pt] &= \dfrac{\dfrac{x^2}{4}}{1−(\dfrac{x}{2})^2} \\[5pt] &= \sum_{n=0}^∞\dfrac{x^2}{4}(\dfrac{x}{2})^{2n}. \end{align*}
The series converges as long as $$|(\dfrac{x}{2})^2|<1$$ (note that when $$|(\dfrac{x}{2})^2|=1$$ the series does not converge). Solving this inequality, we conclude that the interval of convergence is $$(−2,2)$$ and
\begin{align*} \dfrac{x^2}{4−x^2}&=\sum_{n=0}^∞\dfrac{x^{2n+2}}{4^{n+1}}\\[5pt] &=\dfrac{x^2}{4}+\dfrac{x^4}{4^2}+\dfrac{x^6}{4^3}+ \ldots \end{align*}
for $$|x|<2.$$
Exercise $$\PageIndex{3}$$
Represent the function $$f(x)=\dfrac{x^3}{2−x}$$ using a power series and find the interval of convergence.
Hint
Rewrite f in the form $$f(x)=\dfrac{g(x)}{1−h(x)}$$ for some functions $$g$$ and $$h$$.
$$\sum_{n=0}^∞\dfrac{x^{n+3}}{2^{n+1}}$$ with interval of convergence $$(−2,2)$$
In the remaining sections of this chapter, we will show ways of deriving power series representations for many other functions, and how we can make use of these representations to evaluate, differentiate, and integrate various functions.
### Key Concepts
• For a power series centered at $$x=a$$, one of the following three properties hold:
• i. The power series converges only at $$x=a$$. In this case, we say that the radius of convergence is $$R=0$$.
• ii. The power series converges for all real numbers $$x$$. In this case, we say that the radius of convergence is $$R=∞$$.
• iii. There is a real number R such that the series converges for $$|x−a|<R$$ and diverges for $$|x−a|>R$$. In this case, the radius of convergence is $$R.$$
• If a power series converges on a finite interval, the series may or may not converge at the endpoints.
• The ratio test may often be used to determine the radius of convergence.
• The geometric series $$\sum_{n=0}^∞x^n=\dfrac{1}{1−x}$$ for $$|x|<1$$ allows us to represent certain functions using geometric series.
### Key Equations
• Power series centered at $$x=0$$
$\sum_{n=0}^∞c_nx^n=c_0+c_1x+c_2x^2+\ldots nonumber$
• Power series centered at $$x=a$$
$\sum_{n=0}^∞c_n(x−a)^n=c_0+c_1(x−a)+c_2(x−a)^2+\ldots \nonumber$
### Glossary
interval of convergence
the set of real numbers $$x$$ for which a power series converges
power series
a series of the form $$\sum_{n=0}^∞c_nx^n$$ is a power series centered at $$x=0$$; a series of the form $$\sum_{n=0}^∞c_n(x−a)^n$$ is a power series centered at $$x=a$$
if there exists a real number $$R>0$$ such that a power series centered at $$x=a$$ converges for $$|x−a|<R$$ and diverges for $$|x−a|>R$$, then $$R$$ is the radius of convergence; if the power series only converges at $$x=a$$, the radius of convergence is $$R=0$$; if the power series converges for all real numbers $$x$$, the radius of convergence is $$R=∞$$ |
How to Solve a Square Root Equation
••• South_agency/E+/GettyImages
Print
The square root of a number is a value that, when multiplied by itself, gives the original number. For example, the square root of 0 is 0, the square root of 100 is 10 and the square root of 50 is 7.071. Sometimes, you can figure out, or simply recall, the square root of a number that itself is a "perfect square," which is the product of an integer multiplied by itself; as you progress through your studies, you're likely to develop a mental list of these numbers (1, 4, 9, 25, 36 . . .).
Problems involving square roots are indispensable in engineering, calculus and virtually every realm of the modern world. Although you can easily locate square root equation calculators online (see Resources for an example), solving square root equations is an important skill in algebra, because it allows you to become familiar with using radicals and work with a number of problem types outside the realm of square roots per se.
Squares and Square Roots: Basic Properties
The fact that multiplying two negative numbers together yields a positive number is important in the world of square roots because it implies that positive numbers actually have two square roots (for example, the square roots of 16 are 4 and −4, even if only the former is intuitive). Similarly, negative numbers do not have real square roots, because there is no real number that takes on a negative value when multiplied by itself. In this presentation, the negative square root of a positive number will be ignored, so that "square root of 361" can be taken as "19" rather than " −19 and 19."
Also, when trying to estimate the value of a square root when no calculator is handy, it is important to realize that functions involving squares and square roots are not linear. You'll see more on this in the section about graphs later, but as a rough example, you have already observed that the square root of 100 is 10 and the square root of 0 is 0. On sight, this might lead you to guess that the square root for 50 (which is halfway between 0 and 100) must be 5 (which is halfway between 0 and 10). But you have also already learned that the square root of 50 is 7.071.
Finally, you may have internalized the idea that multiplying two numbers together yields a number greater than itself, implying that square roots of numbers are always smaller than the original number. This is not the case! Numbers between 0 and 1 have square roots, too, and in every case, the square root is greater than the original number. This is most easily shown using fractions. For example, 16/25, or 0.64, has a perfect square in both the numerator and the denominator. This means that the square root of the fraction is the square root of its top and bottom components, which is 4/5. This is equal to 0.80, a greater number than 0.64.
Square Root Terminology
"The square root of x" is usually written using what is called a radical sign, or just a radical (√ ). Thus for any x:
\sqrt{x}
represents its square root. Flipping this around, the square of a number x is written using an exponent of 2 (x2). Exponents take superscripts on word-processing and related applications, and are also called powers. Because radical signs are not always easy to produce on demand, another way to write "the square root of x" is to use an exponent:
x^{1/2}
This in turn is part of a general scheme:
x^{(y/z)}
means "raise x to the power of y, then take the 'z' root of it." x1/2 thus means "raise x to the first power, which is simply x again, and then take the 2 root of it, or the square root." Extending this, x(5/3) means "raise x to the power of 5, then find the third root (or cube root) of the result."
Radicals can be used to represent roots other than 2, the square root. This is done by simply appending a superscript to the upper left of the radical.
\sqrt[3]{x^5}
then, represents the same number as x(5/3) from the previous paragraph does.
Most square roots are irrational numbers. This means that not only are they not nice, neat integers (e.g., 1, 2, 3, 4 . . .), but they also cannot be expressed as a neat decimal number that terminates without having to be rounded off. A rational number can be expressed as a fraction. So even though 2.75 is not an integer, it is a rational number because it is the same thing as the fraction 11/4. You were told earlier that the square root of 50 is 7.071, but this is actually rounded off from an infinite number of decimal places. The exact value of √50 is 5√2, and you'll see how this is determined soon.
Graphs of Square Root Functions
You have already seen that equations in involving squares and square roots are nonlinear. One easy way to remember this is that the graphs of the solutions of these equations are not lines. This makes sense, because if, as noted, the square of 0 is 0 and the square of 10 is 100 but the square of 5 is not 50, the graph resulting from simply squaring a number must curve its way to the correct values.
This is the case with the graph of
y = x^2
as you can see for yourself by visiting the calculator in the Resources and changing the parameters. The line passes through the point (0,0), and y does not go below 0, which you should expect because you know that x2 is never negative. You can also see that the graph is symmetrical around the y-axis, which also makes sense because every positive square root of a given number is accompanied by a negative square root of equal magnitude. Therefore, with the exception of 0, every y value on the graph of y = x2 is associated with two x-values.
Square Root Problems
One way to tackle basic square root problems by hand is to look for perfect squares "hidden" inside the problem. First, it's important to be aware of a few vital properties of squares and square roots. One of these is that, just as √x2 is simply equal to x (because the radical and the exponent cancel each other out):
\sqrt{x^2y} = x\sqrt{y}
That is, if you have a perfect square under a radical multiplying another number, you can "pull it out" and use it as a coefficient of what remains. For example, returning to the square root of 50
\sqrt{50} = \sqrt{(25)(2)} = 5\sqrt{2}
Sometimes you can wind up with a number involving square roots that is expressed as a fraction, but is still an irrational number because the denominator, the numerator or both contain a radical. In such instances, you may be asked to rationalize the denominator. For example, the number
\frac{6\sqrt{5}}{\sqrt{45}}
has a radical in both the numerator and the denominator. But after scrutinizing "45," you may recognize it as the product of 9 and 5, which means that
\sqrt{45} = \sqrt{(9)(5)} = 3\sqrt{5}
Therefore, the fraction can be written
\frac{6\sqrt{5}}{3\sqrt{5}}
The radicals cancel each other out, and you are left with 6/3 = 2. |
# How do you simplify i^10+i^2?
Dec 3, 2016
The expression simplifies to $- 2$.
#### Explanation:
${i}^{10} + {i}^{2}$
Use the exponent rule ${\left({x}^{a}\right)}^{b} = {x}^{a b}$ and change the ${i}^{10}$ term to an ${i}^{2}$ term by dividing the exponent $\textcolor{b l u e}{10}$ by $\textcolor{red}{2} = \textcolor{m a \ge n t a}{5}$.
${i}^{\textcolor{b l u e}{10}} = {\left({i}^{\textcolor{red}{2}}\right)}^{\textcolor{m a \ge n t a}{5}}$
${\left({i}^{2}\right)}^{5} + {i}^{2}$
Recall that ${i}^{2} = - 1$
${\left(- 1\right)}^{5} + \left(- 1\right)$
$- 1$ raised to an odd power is equal to $- 1$ (and btw, $- 1$ raised to an even power is equal to $1$)
$\left(- 1\right) + \left(- 1\right) = - 2$ |
# APR – APY conversion
% % < %
% % < %
= year
% % < %
= year
% % < %
= year
## Explanations
@@\mathrm{APR}@@ (Annual Percentage Rate) is the rate of simple interest for one year. If interests is given in each period, with @@N@@ periods for one year, then each period gives @@\frac{\mathrm{APR}}{N}@@ interest, always computed on the initial amount.
@@\mathrm{APY}@@ (Annual Percentage Yield) is the rate of compound interest for one year. If interests is given in each period, with @@N@@ periods for one year, then for each period the @@\frac{\mathrm{APR}}{N}@@ interest is added to the total amount, and so interests also generate interest in next periods.
By example, for a number of periods @@N = 5@@, a simple interest for the year @@\mathrm{APR} = 10\%@@ and an initial amount @@A_0 = 1000\unicode{8364}@@.
The rate for each period is @@\frac{\mathrm{APR}}{N} = \frac{10}{5}\% = 2\% = 0.02@@.
With
### simple interest
the @@5@@ periods during the year give successively:
@@\begin{array}{@{}l|l|l|l@{}} \text{Interest} & \text{Total profit} & \text{Total amount}\\ \hline & & A_0 = 1000\unicode{8364}\\ \hline I_1 = 1000\unicode{8364} \times\ 0.02 = 20\unicode{8364} & P_1 = 0\unicode{8364} +\ 20\unicode{8364} = 20\unicode{8364} & A_1 = 1000\unicode{8364}\\ I_2 = 1000\unicode{8364} \times\ 0.02 = 20\unicode{8364} & P_2 = 20\unicode{8364} +\ 20\unicode{8364} = 40\unicode{8364} & A_2 = 1000\unicode{8364}\\ I_3 = 1000\unicode{8364} \times\ 0.02 = 20\unicode{8364} & P_3 = 40\unicode{8364} +\ 20\unicode{8364} = 60\unicode{8364} & A_3 = 1000\unicode{8364}\\ I_4 = 1000\unicode{8364} \times\ 0.02 = 20\unicode{8364} & P_4 = 60\unicode{8364} +\ 20\unicode{8364} = 80\unicode{8364} & A_4 = 1000\unicode{8364}\\ I_5 = 1000\unicode{8364} \times\ 0.02 = 20\unicode{8364} & P_5 = 80\unicode{8364} +\ 20\unicode{8364} = 100\unicode{8364} & A_5 = 1000\unicode{8364} & 1\text{ year}\\ \hline & P_{10} = 200\unicode{8364} & & 2\text{ years}\\ & P_{15} = 300\unicode{8364} & & 3\text{ years}\\ & P_{20} = 400\unicode{8364} & & 4\text{ years}\\ \end{array}@@
In general,
total profit after @@k@@ periods @@P_k = A_0 + \underbrace{\frac{\mathrm{APR}}{N} + \dots + \frac{\mathrm{APR}}{N}}_{k\text{ times}} = A_0 \times \frac{\mathrm{APR}}{N} \times k@@.
Total profit after one year (i.e. @@N@@ periods) @@P_N = A_0 \times \mathrm{APR}@@.
Total profit after @@i@@ years @@P_{i N} = A_0 \times \mathrm{APR} \times i@@.
With
### compound interest
the @@5@@ periods during the year give successively:
@@\begin{array}{@{}l|l|l@{}} \text{Interest} & \text{Total amount}\\ \hline & A_0 = 1000\unicode{8364}\\ \hline I_1 = 1000\unicode{8364} \times\ 0.02 = 20\unicode{8364} & A_1 = 1000\unicode{8364} \times\ 1.02 = 1000\unicode{8364} +\ 20\unicode{8364} = 1020\unicode{8364}\\ I_2 = 1020\unicode{8364} \times\ 0.02 = 20.4\unicode{8364} & A_2 = 1020\unicode{8364} \times\ 1.02 = 1020\unicode{8364} +\ 20.4\unicode{8364} = 1040.4\unicode{8364}\\ I_3 = 1040.4\unicode{8364} \times\ 0.02 = 20.808\unicode{8364} & A_3 = 1040.4\unicode{8364} \times\ 1.02 = 1040.4\unicode{8364} +\ 20.808\unicode{8364} = 1061.208\unicode{8364}\\ I_4 = 1061.208\unicode{8364} \times\ 0.02 = 21.22416\unicode{8364} & A_4 = 1061.208\unicode{8364} \times\ 1.02 = 1061.208\unicode{8364} +\ 21.22416\unicode{8364} = 1082.43216\unicode{8364}\\ I_5 = 1082.43216\unicode{8364} \times\ 0.02 = 21.6486432\unicode{8364} & A_5 = 1082.43216\unicode{8364} \times\ 1.02 = 1082.43216\unicode{8364} +\ 21.6486432\unicode{8364} = 1104.0808032\unicode{8364} & 1\text{ year}\\ \hline & A_{10} \simeq 1218.99442 & 2\text{ years}\\ & A_{15} \simeq 1345.868338 & 3\text{ years}\\ & A_{20} \simeq 1485.947396 & 4\text{ years}\\ \end{array}@@
That gives the corresponding @@\mathrm{APY} = 0.1040808032 = 10.40808032\%@@.
In general,
total amount after @@k@@ periods @@A_k = A_0 \times \underbrace{\left(1 + \frac{\mathrm{APR}}{N}\right) \times \dots \times \left(1 + \frac{\mathrm{APR}}{N}\right)}_{k\text{ times}} = A_0 \times \left(1 + \frac{\mathrm{APR}}{N}\right)^k@@.
Total amount after one year (i.e. @@N@@ periods) @@A_N = A_0 \times \left(1 + \frac{\mathrm{APR}}{N}\right)^N = A_0 \times (1 + \mathrm{APY})@@.
Total amount after @@i@@ years @@A_{i N} = A_0 \times \left(1 + \frac{\mathrm{APR}}{N}\right)^{i N}@@.
Note that for any @@N: \mathrm{APY} \lt \lim\limits_{k\rightarrow\infty} \left(1 + \frac{\mathrm{APR}}{k}\right)^k - 1 = e^{\mathrm{APR}} - 1@@ (this is the value in the APY limit row).
### Conversion between APR and APY
For annual percentage rate @@\mathrm{APR} \in \mathbb{R}@@, annual percentage rate @@\mathrm{APY} \in \mathbb{R}@@ and the number of periods @@N \in \mathbb{N}@@.
• @@\mathrm{APY} = \left(1 + \frac{\mathrm{APR}}{N}\right)^N - 1@@
• Reciprocally @@\mathrm{APR} =\left(\sqrt[N]{1 + \mathrm{APY}} - 1\right) N@@ |
# Solving slope intercept form
In this blog post, we will explore one method of Solving slope intercept form. Let's try the best math solver.
## Solve slope intercept form
b is your solution for y and c is your solution for x. To get this answer, multiply your left side (y) by c and add that to your right side (x). This gives you your solution for x by subtracting b from both sides of the equation. This method works best when there are more than two variables in an equation. If the equation has more than two variables, you can use a calculator to simplify the equation and solve for x.
Absolute value equations are equations that have an expression with one or more variables whose values are all positive. Absolute value equations are often used to solve problems related to the measurement of length, area, or volume. In absolute value equations, the “absolute” part of the equation means that the answer is always positive, no matter what the value of the variable is. Because absolute value equations are so common, it can be helpful to learn how to solve them. Basic rules for solving absolute value equations Basic rule #1: Add negative numbers together and add positive numbers together The first step in solving any absolute value equation is to add all of the negative numbers together and then add all of the positive numbers together. For example, if you want to find the length of a rectangular room whose width is 12 feet and whose length is 16 feet, you would start by adding 12 plus (-16) and then adding 16 plus (+12). Because both of these numbers are negative, they will be added together to form a positive number.
The cosine solver iteratively solves for the cosine of a given angle. It uses a fixed value as the starting point, then iteratively increases the cosine value by each iteration until it reaches the target value. The cosine solver is an excellent tool to use when solving problems involving the cosine function. Let's take a look at an example. Say you want to find out how long it takes to drive from one location to another. You can first use a straightedge and compass to determine the distance between your starting point and destination. Then, you can plug this distance into a formula that calculates the cosine of the angle between your two points to get your driving time. This is an example of finding the exact value of something using calculus, a branch of mathematics that deals with change in quantities over time. In addition to being useful for solving problems about geometry, the cosine solver can also be used for finding accurate values of trigonometric functions such as sine and tangent . While there are many different ways to solve these problems using different formulas, one common solution method is called Simpson's rule . This method involves first calculating the ratio of opposite leg lengths and then using this ratio to calculate the hypotenuse length. By applying this step-by-step process, you can eventually reach an accurate answer for any trigonometric function
A 3x3 matrix is made up of three 3x3 matrices. To solve a 3x3 matrix, we can start by finding the least-squares solution to the following equation: With this method, a computer is used to find the linear combination of all of the three coefficients that minimizes the sum of squares.
Very easy to use, gets better with time. Math problems are very easy and comfortable to type in, even if you can't scan them with your camera. Helps you understand math problems as well as a teacher would.
### Quincie Washington
It’s the best mathematics solution app I ever seen. It’s having fantastic features like calculator but not ordinary it has all types of symbols that needed in math. It’s also had the feature of searching your own book. If your searched book is not available don't worry you can vote for your book by looking the ISBN code behind your book.
### Caroline Ross
Factoring algebra Get answers to math problems for free online Steps to solve word problems in math College algebra math answers Math help websites for algebra Derivative problem solver |
">
## Circumference Calculator
The Circumference Calculator is a helpful tool that calculates the circumference, which is the perimeter of a circle.
Desktop
Desktop
Desktop
# Circumference Calculator: Unlocking the Mysteries of Circular Measurements
## What is Circumference Calculation
Circumference is a fundamental measurement in geometry, representing the distance around a circle. In both mathematics and practical applications, understanding circumference is pivotal.
Circumference, in geometry, refers to the boundary or the perimeter of a circle or any curved geometrical shape. It plays a pivotal role in various mathematical and real-world applications. Understanding how to calculate circumference is fundamental for many fields, from basic geometry to advanced scientific calculations.
## Understanding the Circumference Formula
The formula for finding the circumference of a circle is directly linked to its diameter or radius. This formula, C = π * d (Circumference = π * diameter), is essential to grasp the relationship between these elements.
## Formula for Circumference Calculation:
The formula to calculate the circumference of a circle using the diameter ($$d$$) is given by:
$$C = \pi \times d$$
Where:
• $$C$$ = Circumference of the circle
• $$\pi$$ = Mathematical constant, approximately 3.14159
• $$d$$ = Diameter of the circle
### Example:
Let's assume we have a circle with a diameter of 10 units. Using the formula, we can calculate its circumference:
$$C = \pi \times d = \pi \times 10 = 3.14159 \times 10 = 31.4159 \, \text{units}$$
## How To Calculate Circumference
Exploring the calculations involves understanding the significance of diameter and radius in determining the circumference. We'll delve into practical examples to simplify these calculations step by step.
Finding the complete length around a circle's boundary is necessary to calculate its circumference. C = π × d is the formula for calculating the circumference, where C is the circle's circumference and d is its diameter. To put it another way, you may alternatively use the formula C = 2 × π × r, where r is the circle's radius.
For example, the circumference of a circle with a diameter of 10 units would be:
C = π × d = π × 10 = 3.14159 × 10 = 31.4159 units.
Thus, a circle with a diameter of 10 units would have a circumference of around 31.4159 units.
## Calculating Circumference Variations
While the general formula suffices for circles, variations exist when dealing with specific shapes. For instance, calculating the circumference of a circle with a given radius involves a slight adjustment to the formula: C = 2 * π * r, where r represents the radius. Similarly, the circumference of a semicircle and a sphere has its unique formulas, requiring a nuanced approach.
Calculating the circumference of different shapes involves specific variations based on their characteristics. While the general formula C = π × d works for circles, variations exist for circles with a radius, semicircles, and spheres.
For a circle with a radius (r), the formula alters slightly to C = 2 × π × r. Here, r represents the radius instead of the diameter.
A semicircle has a circumference that is half that of a full circle with the same diameter. C = π × d/2 or C = π × r, where d is the diameter and r is the radius, is still the formula.
When a sphere is involved, its circumference may be calculated using the formula C = 2 × π × r, where r is the sphere's radius.
These variations in formulas cater to different shapes, ensuring accurate calculations based on their specific attributes.
## Circumference of a Cylinder
Calculating the circumference of a cylinder requires a nuanced approach due to its unique shape. We'll uncover the method behind determining this measurement accurately.
The formula to find the circumference ($$C$$) of a cylinder with the circular base is given by:
$$C = 2 \times \pi \times r$$
Where:
• $$C$$ = Circumference of the cylinder
• $$\pi$$ = Mathematical constant, approximately 3.14159
• $$r$$ = Radius of the circular base of the cylinder
## Circumference to Diameter Chart
The relationship between the circumference ($$C$$) and diameter ($$d$$) of a circle is given by the formula:
$$C = \pi \times d$$
Where:
• $$C$$ = Circumference of the circle
• $$d$$ = Diameter of the circle
• $$\pi$$ = Mathematical constant, approximately 3.14159
Visual aids, like a circumference to diameter chart, simplify complex mathematical relationships. It provides a graphical representation, facilitating quick reference and aiding in understanding the correlation between circumference and diameter in different scenarios.
## Understanding Circumference of a Semicircle
The concept of the semicircle's circumference is intriguingly different from a full circle. We'll break down the process of calculating this measurement.
## Exploring Circumference of a Sphere
Discovering the circumference of a sphere involves a distinct formula and considerations. We'll explore this spherical calculation in depth.
## Earth's Circumference Calculation
The Earth's circumference plays a critical role in geographical measurements and navigation. Understanding how this vast measurement is calculated is fascinating.
## Conclusion
Unlocking the mysteries behind circumference calculator enhances our understanding of circular measurements. From basic formulas to real-world applications, this fundamental concept resonates across multiple spheres.
Understanding the concept of circumference and its calculation methods, from simple circles to complex shapes like spheres and cylinders, is fundamental in various fields. Tools like the diameter to circumference calculator and visual aids serve as valuable resources, simplifying mathematical complexities and finding application in real-world scenarios.
#### References:
Circumference of Circle Calculator
Circumference Formula
How should I operate the circumference calculator?
It's easy to use the circumference calculator. The calculator will instantly calculate the circumference when you enter the circle's radius or diameter
How is a circle's circumference determined?
The formula Circumference = 2 * Radius * Radius or Circumference = * Diameter, where (pi) is about 3.14159, is used to determine the circumference of a circle.
Can I use this calculator for both radius and diameter inputs?
Yes, the Circumference Calculator accommodates both radius and diameter inputs, allowing you to use whichever value is available.
Is the calculator suitable for any size of circle?
Absolutely! The Circumference Calculator efficiently calculates the circumference for circles of any size, from tiny circles to large ones.
Can I use this calculator for circular objects in real-world applications?
Yes, the Circumference Calculator is useful for real-world applications involving circular objects, such as measuring the perimeters of wheels, plates, or circular paths. |
# Four Digit Numbers – Definition, Examples | Reading and Writing 4 Digit Numbers | Number Names of Four Digit Numbers
Do you know four-digit numbers? This web page will explain the four-Digit numbers. It also includes four-digit number definition, face, and place values. You can also check the examples on reading and writing four-digit numbers, representing 4 digit numbers using an abacus for a better understanding of the concept. Get acquainted with the details like Classification of Four-Digit Numbers into Groups.
### Four-Digit Number – Definition
Four-digit numbers have four digits. In Four-Digit Numbers, the First place from the right represents one place and the second place represents tens place. The third place represents the hundred’s place. Fourth place represents thousand’s place. The smallest four-digit number is 1000. The largest four-digit number is 9999.
Four-Digit Numbers are divided into nine groups.
Four-digit numbers begin with 1000. The first group is from one thousand (1000) to one thousand nine hundred ninety-nine (1999).
The second group starts from two thousand(2000) to two thousand nine hundred ninety-nine (2999).
The third group starts from three thousand(3000) to three thousand nine hundred ninety-nine (3999).
The fourth group starts from four thousand(4000) to four thousand nine hundred ninety-nine (4999).
The fifth group starts from Five thousand(5000) to Five thousand nine hundred ninety-nine (5999).
The sixth group starts from six thousand(6000) to six thousand nine hundred ninety-nine (6999).
The seventh group starts from seven thousand(7000) to seven thousand nine hundred ninety-nine (7999).
The eighth group starts from eight thousand(8000) to eight thousand nine hundred ninety-nine (8999).
The ninth group starts from nine thousand(9000) to nine thousand nine hundred ninety-nine (9999).
If the digits from 0 to 9 are placed step by step to the right of 1000 according to the place values then four-digit numbers are formed.
Examples of four-digit numbers are 1289, 1456, 5678, 8976, 8999, etc.
In a four-digit number, the first number from the right represents the unit’s value. The second number represents tens place. The third number represents the hundred’s place. The fourth number represents the thousand’s place.
### Four Digit Numbers Examples
Example 1:
In the Four-digit number 1545 write the place values of the numbers.
Solution:
The first number 5 from the right represents the unit’s place. Here the place value of 5 is 5.
The second number 4 from the right represents the tens place. The place value of 4 is 40.
The third number 5 from the right represents the hundred’s place. The place value of 5 is 500.
The Fourth number 1 from the right represents the Thousand’s place. The place value of 1 is 1000.
Example 2:
How the four-digit number 3245 is represented in the abacus? write which digit represents which place and write the place values?
Solution:
The four-digit number 3245 is represented in the abacus is as follows:
In 3245, the first number 5 from the right represents the unit’s place. The place value of 5 is 5.
The second number 4 from the right represents the tens place. The place value of 4 is 40.
The third number 2 from the right represents the hundreds place. The place value of 2 is 200.
The fourth number 3 from the right represents the thousands place. The place value of 3 is 3000.
Example 3:
How the four-digit number 1643 is represented in the abacus? Write which number represents which place and write its place values?
Solution:
The four-digit number 1643 is represented in the abacus is as follows:
In 1643, the first number 3 from the right represents the unit’s place. The place value of 3 is 3.
The second number 4 from the right represents the tens place. The place value of 4 is 40.
The third number 6 from the right represents the hundreds place. The place value of 6 is 600.
The fourth number 1 from the right represents the thousands place. The place value of 1 is 1000.
Example 4:
In the Four-digit number 6238 write the place values of the numbers.
Solution:
The first number 8 from the right represents the unit’s place. Here the place value of 8 is 8.
The second number 3 from the right represents the tens place. The place value of 3 is 30.
The third number 2 from the right represents the hundred’s place. The place value of 2 is 200.
The Fourth number 6 from the right represents the Thousand’s place. The place value of 6 is 6000.
### FAQs on Four-Digit Numbers
1. What is a four-digit number?
Four-digit numbers have four digits. Each digit in the number represents its place value.
2. In 7893 write the place value of 9?
In 7893, 9 is in the tens place. So the place value of 9 is 90.
3. In 4326 which digit represents the hundred’s place?
In 4326,3 represents the hundreds place.
4. If 5 is in the hundreds place. What is the place value of 5?
If 5 is in the hundreds place. The place value of 5 is 500.
5. Which is the smallest four-digit number?
The smallest four-digit number is 1000.
6. Which is the largest four-digit number?
The largest four-digit number is 9999.
Scroll to Top
Scroll to Top |
Two math puzzles
What is the value of this expression?
Tags
1. The first is (1+sqrt(5))/2 by simple rearrangement and the quadratic formula.
The second is 3 by induction (which reveals a whole family of nice formulas).
2. Wesner Moise says:
First one, solve for x=sqrt(1+x)…
This gives (1+sqrt(5))/2 which I believe is the golden ratio.
3. I don’t have nice pictures like Adi but I might as well include my solutions from above.
(1) Call the value of that expression x. x=sqrt(1+sqrt(1+…)). Square both sides. Then, subtract 1 from both sides. Since the expression continues infinitely, notice now that x^2-1 is the same as x. Solve using the quadratic formula (take the positive root).
(2) I propose that sqrt(1+nsqrt(1+(n+1)sqrt(…))) is in fact just n+1. Take n=0. All of the inner terms fall away and you’re left with sqrt(1)=1=n+1.
Assume it’s true for n. n+1=sqrt(1+nsqrt(1+(n+1)sqrt(…))). Square both sides. Then, subtract 1 from both sides. Divide both sides by n. You’ve got ((n+1)^2-1)/n=sqrt(1+(n+1)sqrt(1+(n+2)sqrt(…))). RHS is the expression for n+1. LHS is (n^2+2n+1-1)/n=n+2. Formula is true for n+1. By induction, blah blah blah.
Nicolas:
Interesting, but can you detail how do you go in your induction step from n=0 to n=1?
1. golden ratio
2.
Let be: x(n) = sqrt(1 +nsqrt(1 +(n+1)sqrt(1+…
We have to calculate x(2).
By induction we will prove that: x(n) = n +1.
We have x(0) = 1 and we know that: x(n)^2 = 1 + nx(n+1).
If x(n) = n + 1, then: (n+1)^2 = 1 + n
x(n+1), so: x(n+1) = n+2, QED.
Results: x(2) = 3.
How can we arrive to the hypothesis x(n) = n+1?
We note that: x(2)^2 + x(3)^2 = (x(3) + 1)^2, so chances are that (x(2), x(3), x(3)+1) could be a Pythagorean triple (the most famous one: 3^2 + 4^2 = 5^2). If x(2) would be 3 it’s easy to observe that also x(3) would be 4, x(4) = 5, etc, so we can make the hypothesis x(n) = n+1
5. Haha, you caught me there. I stayed up way too late tonight until I was able to paper over the trouble spot with some complex analysis. Is there are a solution using purely elementary methods?
6. Smells like Ramanujan spirit 🙂
(n + 1)^2 = 1 + n((n + 1) + 1)
n + 1 = sqrt(1 + n
sqrt(1 + (n+1)sqrt(1 + (n+2)sqrt(…
7. Antimail says:
Theorem: All horses have…
8. No suggestions on putting together an elementary method solution? If the puzzle is an early one by Ramanujan, I doubt the original solution was done with complex analysis.
No complex analysis should be required, AFAIK. Hint: try to express the problem recursively.
(But an "induction" would not be useful here)
10. I did write a recursive function, but analysis is needed to prove that the sequence converges.
Define r_n(z)=sqrt(1+nz). Then, a finite approximation to some degree n is s(n,z)=r_2r_3…*r_n(z). The * there is function composition. The answer is the limit n->inf of s(n,1).
Working this morning I could replace the argument with a real-valued one, so the z can become an x.
I don’t see a purely algebraic means of doing this without a step that handwaves over the part "and then repeat this process forever".
> I did write a recursive function, but analysis is needed to prove that the sequence converges.
Correct. Well, you are definitely in the right direction…
12. I’m happy with my solution using analysis so I’m not going to push this any further. I would like to see a more classical technique if one exists. But, this was done at the very beginning of the age of rigor so perhaps handwaving was good enough at the time given that the result is indeed correct.
In essence, your solution is very close to the final proof – the only missing part is showing why "3" is the correct answer and not, say, "3.1".
I will post a complete proof soon.
14. n+p+2 = sqrt(1 + (n+p+1)(n+p+3))
n+p+1 = sqrt(1 + (n+p)
sqrt(1 + (n+p+1)(n+p+3)))
n+p = sqrt(1 + (n+p-1)
sqrt(1 + (n+p)sqrt(1 + (n+p+1)(n+p+3))))
n+1 = sqrt(1 + nsqrt(1 + (n+1)sqrt(1 +…(n+p)sqrt(1 + (n+p+1)(n+p+3) = x(n, p)
we have to prove the convergence of the x(n, p) when p->inf
15. But x(n, p) = x(n, p+1) = n + 1 because n+p+3 = sqrt(1+(n+p+2)*(n+p+4))
🙂
OK – I am going now to give one more hint to put in evidence a fundamental problem with all the approaches above.
Your formulas are correct. In fact you can write directly:
3 = sqrt(1+24)
= sqrt(1+2
sqrt(1+35))
= sqrt(1+2
sqrt(1+3sqrt(1+46)))
= sqrt(1+2sqrt(1+3sqrt(1+4sqrt(1+57))))
= sqrt(1+2sqrt(1+… sqrt(1+(n-1)(n+1))…))
… (by taking to the limit)
= sqrt(1+2sqrt(1+3sqrt(1+4…)))
But the trouble is not the convergence. The trouble is that "3" is not the only number with this property. In fact, it is easy to see that for an infinity of real numbers you can build a series which ends up with exactly the same infinite radical expression like the one above.
For example, let’s pick up:
X_1 = 4
X_2 = 1/2 * (X_1 ^ 2 – 1)
X_3 = 1/3 * (X_2 ^ 2 – 1)
X_n = 1/n * (X_(n-1) ^ 2 – 1)
It is easy to see that X_i is well-defined and monotonically growing. Then, you can write:
X_1 = sqrt(1+2X_2)
= sqrt(1+2sqrt(1+3X_3))
= sqrt(1+2sqrt(1+3sqrt(1+4X_4)))
= sqrt(1+2
sqrt(1+3sqrt(1+4sqrt(1+5X_5))))
= sqrt(1+2
sqrt(1+… sqrt(1+(n-1)X_(n-1))…))
… (by taking to the limit) …
= sqrt(1+2
sqrt(1+3sqrt(1+4…)))
Therefore:
4 = sqrt(1+2sqrt(1+3sqrt(1+4*…)))
Huh?!
So it’s not enough to proof convergence if you start with "3". You also need to _uniquely_ define what that infinite radical expression means.
17. Once you have the sequence above, it’s tedious but not impossible to prove that it converges to 3.
From the definition, s(n+1,z)=s(n,sqrt(1+(n+1)z)).
You can prove by induction that s(n,n+2)=3.
s’>0 and s”<0 (more side proofs) with respect to z so s is monotonic and concave.
So, the linear interpolate is less than the actual point.
s(n,b)>s(n,c)-(s(n,c)-s(n,a))(c-b)/(c-a)
Pick a=1,b=sqrt(1+(x+1)),c=x+2,n=x.
s(x,sqrt(1+(x+1)))>s(x,x+2)-(s(x,x+2)-s(x,1)).
s(x+1,1)>3-(3-s(x,1))
(x+2-b)/(x+1).
3-s(x+1,1)<(x+2-b)/(x+1)(3-s(x,1)).
3-s(x+1,1)<(x+2-b)/x
(3-s(x,1)).
Now we’re in a position to examine this as lim x->inf.
When x is large we get
3-s(x+1,1)<(x-1)/x(3-s(x,1)).
Take lim x->inf of both sides. Look at the RHS first.
(x-1)/x(3-s(x,1))=(1-1/x)(3-s(x,1)).
Apply the linear interpolation formula above during the limit.
(1-1/x)(1-1/(x-1))(1-1/(x-2))(3-s(C,1))
going down to some constant C when starting from finite x.
Take the log of both sides so
ln(3-s(x+1,1)) <
ln((1-1/x)
(1-1/(x-1))(1-1/(x-2))…*(3-s(C,1))) <
ln(1-1/x)+ln(1-1/(x-1))+ln(1-1/(x-2))+…
Apply ln(1+q)<=q giving
-1/x-1/(x-1)-1/(x-2)-… all as x->inf.
That’s the harmonic sequence so the limit diverges to -inf.
So, lim x->inf ln(3-s(x+1,1))=-inf.
lim x->inf 3-s(x+1,1)=0.
lim x->inf s(x,1)=3. |
Top
Conditional Probability Calculator
Top
In statistical point of view, conditional probability is nothing but the probability of an event happen provided that another event is already happened before. To find out the conditional probability, we use the online conditional probability calculator. It is denoted as $P$ $(\frac{A}{B})$ (read as probability of $A$ given $B$) if the events are dependent. If they are independent, the notation become simply $P(A)$. The formula to calculate the conditional probability when the events are dependent is given below.$P$ $(\frac{A}{B})$ = $\frac{P(A\ \cap\ B)}{P(B)}$$P$ $(\frac{B}{A})$ = $\frac{P(A\ \cap\ B)}{P(A)}$Where $P$ $(\frac{A}{B})$ is the conditional probability $P(A)$ is the probability of event $A$ $P(B)$ is the probability of event $B$
## Steps for Conditional Probability Calculator
Step 1 :
Find out the values P(A$\cap$ B) and P(A) or P(B) from the question.
Step 2 :
Determine the conditional probability using the given equations.
P(A/B)=$\frac{P(A\cap B)}{P(B)}$
P(B/A)=$\frac{P(A\cap B)}{P(A)}$
## Problems on Conditional Probability Calculator
1. ### The probability of occurrence of two events A and B is given by 0.25. The probability of occurrence of the event A and that of B is given 0.42 and 0.60 respectively. Calculate the P(A/B) and P(B/A)?
Step 1 :
P(A)=0.42, P(B)=0.60 and P(A$\cap$B)=0.25
Step 2 :
P(A/B)=$\frac{P(A\cap B)}{P(B)}$
P(A/B)=$\frac{0.25}{0.60}$=0.4166
P(B/A)=$\frac{P(A\cap B)}{P(A)}$
P(B/A)=$\frac{0.25}{0.42}$=0.5952
P(A/B)=0.4166
P(B/A)=0.5952
2. ### P(A)=0.65, P(B) and P(A$\cap$B)=0.15. Determine the values of P(A/B) and P(B/A)?
Step 1 :
From the question, it is given that
P(A)=0.65, P(B) and P(A$\cap$B)=0.15
Step 2 :
P(A/B)=$\frac{P(A\cap B)}{P(B)}$
P(A/B)=$\frac{0.15}{0.40}$=0.375
P(B/A)=$\frac{P(A\cap B)}{P(A)}$
P(B/A)=$\frac{0.15}{0.65}$=0.2307 |
# GRE Math : Algebra
## Example Questions
### Example Question #41 : Equations / Inequalities
Sally is 2 years younger than Abby
Daisy is 5 years older than Tracy
Abby is 6 years older than Tracy
A
---
Sally's age
B
---
Daisy's age
The relationship cannot be determined
Quantity A is greater
Quantity B is greater
The two quantities are equal
Quantity B is greater
Explanation:
To simplify the word problem, express the ages in terms of variables in a system of equations. Note that we want to compare S with D:
S = A – 2
D = T + 5
A = T + 6
By substituting A for T in the first equation, we can get S in terms of T, which will let us directly compare the values of S and D.
S = (T + 6) – 2 = T + 4
If D = T + 5, and S = T + 4, D must be the greater value and Daisy is one year older than Sally. B is the correct answer.
### Example Question #81 : Algebra
Quantitative Comparison
3x + 4y = 5
x – y = 6
Quantity A: x
Quantity B: y
The relationship cannot be determined from the information given.
Quantity B is greater.
The two quantities are equal.
Quantity A is greater.
Quantity A is greater.
Explanation:
First let's solve for y using the second equation, x – y = 6.
x = 6 + y. Then plug this in to the other equation.
3 (6 + y) + 4y = 5
18 + 3y + 4y = 5
18 + 7y = 5
7y = –13
y = –13/7. Now plug this value back into x = 6 + y.
x = 6 – 13/7 = 29/7. x is positive and y is negative, so clearly x is larger, so Quantity A is bigger.
### Example Question #85 : Algebra
x + y = 12 and 2xy = 6
Quantity A: x
Quantity B: y
The two quantities are equal.
The relationship cannot be determined from the information given.
Quantity A is greater.
Quantity B is greater.
The two quantities are equal.
Explanation:
Because there are two different equations for the two variables (x and y), you are able to solve for the value of each. You can rearrange the first equation to show that
y = 12 – x by subtracting x from both sides.
Then you can substitute this equation into the second equation to give you
2x – (12 – x) = 6.
This new equation can be simplified to
2x – 12 + x = 6
3x =18
x = 6
Filling this back into the first equation, we get 6 + y = 12 which means y must also equal 6. Because x and y are equal, we choose the option both quantities are equal.
### Example Question #71 : How To Find The Solution To An Equation
A theme park charges $10 for adults and$5 for kids. How many kids tickets were sold if a total of 548 tickets were sold for a total of $3750? Possible Answers: 346 269 431 157 248 Correct answer: 346 Explanation: Let c = number of kids tickets sold. Then (548 – c) adult tickets were sold. The revenue from kids tickets is$5c, and the total revenue from adult tickets is $10(548 – c). Then, 5c + 10(548 – c) = 3750 5c + 5480 – 10c = 3750 5c = 1730 c = 346. We can check to make sure that this number is correct:$5 * 346 tickets + $10 * (548 – 346) tickets =$3750 total revenue
### Example Question #87 : Algebra
Two palm trees grow next to each other in Luke's backyard. One of the trees gets sick, so Luke cuts off the top 3 feet. The other tree, however, is healthy and grows 2 feet. How tall are the two trees if the healthy tree is now 4 feet taller than the sick tree, and together they are 28 feet tall?
14 and 14 feet
12 and 16 feet
8 and 20 feet
11 and 17 feet
cannot be determined
12 and 16 feet
Explanation:
Let s stand for the sick tree and h for the healthy tree. The beginning information about cutting the sick tree and the healthy tree growing is actually not needed to solve this problem! We know that the healthy tree is 4 feet taller than the sick tree, so h = s + 4.
We also know that the two trees are 28 feet tall together, so s + h = 28. Now we can solve for the two tree heights.
Plug h = s + 4 into the second equation: (s + 4) + s = 28. Simplify and solve for h: 2s = 24 so s = 12. Then solve for h: h = s + 4 = 12 + 4 = 16.
### Example Question #71 : Algebra
Solve for z
3(z + 4)3 – 7 = 17
4
8
2
–8
–2
–2
Explanation:
1. Add 7 to both sides
3(z + 4)3 – 7 + 7= 17 + 7
3(z + 4)3 = 24
2. Divide both sides by 3
(z + 4)3 = 8
3. Take the cube root of both sides
z + 4 = 2
4. Subtract 4 from both sides
z = –2
### Example Question #89 : Algebra
Jen and Karen are travelling for the weekend. They both leave from Jen's house and meet at their destination 250 miles away. Jen drives 45mph the whole way. Karen drives 60mph but leaves a half hour after Jen. How long does it take for Karen to catch up with Jen?
She can't catch up.
Explanation:
For this type of problem, we use the formula:
When Karen catches up with Jen, their distances are equivalent. Thus,
We then make a variable for Jen's time, . Thus we know that Karen's time is (since we are working in hours).
Thus,
There's a logical shortcut you can use on "catching up" distance/rate problems. The difference between the faster (Karen at 60mph) and slower (Jen at 45mph) drivers is 15mph. Which means that every one hour, the faster driver, Karen, gains 15 miles on Jen. We know that Jen gets a 1/2 hour head start, which at 45mph means that she's 22.5 miles ahead when Karen gets started. So we can calculate the number of hours (H) of the 15mph of Karen's "catchup speed" (the difference between their speeds) it will take to make up the 22.5 mile gap:
15H = 22.5
So H = 1.5.
### Example Question #90 : Algebra
Bill and Bob are working to build toys. Bill can build toys in 6 hours. Bob can build toys in 3 hours. How long would it take Bob and Bill to build toys working together? |
# Trick to multiply with your hands with the tables of 6,7,8 and 9
We are searching data for your request:
Forums and discussions:
Manuals and reference books:
Data from registers:
Wait the end of the search in all databases.
Upon completion, a link will appear to access the found materials.
With this trick to multiply with your hands are the tables of 6, 7, 8 and 9, your children will learn more complex multiplications faster. We know the goal is to learn the multiplication tables by heart, but these math tricks can make this subject more fun.
Children too they can learn math by playingor. This trick is one such example. Pay attention and teach your children to multiply by 6, 7, 8 and 9 using only their hands.
1. The first thing you should do is number the fingers from 6 to 10. The thumbs will be number 6, the index fingers, 7, the heart, the 8, the ring fingers the 9 and the pinkies the 10.
2. To understand this trick, it is best to use examples. We start with this one: 7x8. Put the palms of your hands facing you. The first thing you should do is join finger number 7 of the left hand with finger 8 of the right hand. In this way, some fingers are left above the fingers that you put together and others are below the fingers that you put together.
3. You are going to add the fingers that you put together and those that were on top, but each finger is worth 10. So you have a total of 50.
4. And what do we do with the fingers that were left below? Multiply them! Since there were three fingers on one side and two on the other, it multiplies 3x2 and gives us 6.
5. So the result from before (50) plus this other (6) gives us 56.
6. Let's take another example: 9x7. By joining finger 9 of the left hand with finger 7 of the right, we add these fingers and those that are on top and it gives us 60.
7. Since below we have three fingers on one side and one on the other, we multiply it and give 3. The total result is ... 63!
8. One last example: 8x8. We join the two fingers number 8 of each hand. We count by ten those fingers and those that are on top. They give us 60.
9. Now we multiply the ones below: 2x2, and it gives us 4. So the final result is 64.
Remember that with this trick you can calculate the multiplications of the tables of 6,7,8 and 9 provided they are multiplied by numbers equal to or greater than 6.
You can read more articles similar to Trick to multiply with your hands with the tables of 6,7,8 and 9, in the On-site Learning category. |
$\newcommand{\dollar}{\} \DeclareMathOperator{\erf}{erf} \DeclareMathOperator{\arctanh}{arctanh} \newcommand{\lt}{<} \newcommand{\gt}{>} \newcommand{\amp}{&}$
## Section1.4Continuous Functions
###### Motivating Questions
• What does it mean to say that a function $f$ is continuous at $x = a\text{?}$
• What does it mean to say that a function $f$ is continuous everywhere?
In this section we introduce the idea of a continuous function. Many of the results in calculus require that the functions be continuous, so having a strong understanding of continuous functions will be very important.
### SubsectionBeing Continuous at a Point
Intuitively, a function is continuous if we can draw its graph without ever lifting our pencil from the page. Alternatively, we might say that the graph of a continuous function has no jumps or holes in it.
First consider the function in the left-most graph in Figure1.24. Note that $f(1)$ is not defined, which leads to the resulting hole in the graph of $f$ at $a = 1\text{.}$ If you were to draw the graph of $f$ yourself then you would need to lift your pencil when you reached $f(1)\text{.}$ We will naturally say that $f$ is not continuous at $a = 1$.
For the function $g$ in Figure1.24, we observe that while $g(1)$ is defined, the value of $g(1) = 2$ is not what you would expect. Specifically, you would expect $g(1)$ to be 3, not 2. Thus, to draw the graph of $g$ you would need to lift your pencil at $a=1\text{.}$ Again, we will say that $g$ is not continuous at $a=1\text{,}$ even though the function is defined at $a = 1\text{.}$
Finally, the function $h$ in Figure1.24 appears to be the most well-behaved of all three, since at $a = 1$ the function value is what you might expect it to be if you were to try and draw the graph of the function without lifting your pencil. In this case we would say that $h$ is continuous at $a=1\text{.}$
The above examples demonstrate a discontinuity commonly know as a removable discontinuity. This is, however, not the only way in which a function can be discontinuous. Another type of discontinuity is the so-called jump discontinuity illustrated below in Figure1.25.
A third type of discontinuity is the so-called infinite discontinuity. Infinite discontinuities exist at points where the values of a function diverge to infinity. A classic example of an infinite discontinuity is the point $x=0$ for the function $\displaystyle f(x)=\frac{1}{x}\text{;}$ you can see the behavior of the infinite discontinuity in the graph of $y=f(x)$ in Figure1.26.
### SubsectionLimits and Continuity
Using limits, we can formalize the idea of continuity. Let us start by re-examining our first example given in Figure1.24.
Consider the function in the left-most graph of Figure1.24. We noted that $f(1)$ is not defined, or $f(1)=DNE\text{,}$ thus $f$ is not continuous at $c = 1$. For the function $g\text{,}$ we observe that while $\displaystyle \lim_{x \to 1} g(x) = 3\text{,}$ the value of $g(1)$ is $2\text{,}$ and thus the limit does not equal the function value. Here, too, we will say that $g$ is not continuous, even though the function is defined at $c = 1\text{.}$ Finally, the function $h$ appears to be the most well-behaved of all three, since at $c = 1$ its limit and its function value agree. That is,
\begin{equation*} \lim_{x \to 1} h(x) = 3 = h(1)\text{.} \end{equation*}
With no hole or jump in the graph of $h$ at $c = 1\text{,}$ we say that $h$ is continuous there.
In Figure1.25 we have what we called a jump discontinuity. Note that at the location of each jump:
\begin{equation*} \lim_{x\to c^-} y \neq \lim_{x\to c^+}y, \end{equation*}
thus at each point that the graph of had a jump the $\displaystyle \lim_{x\to c} y=DNE.$
Similarly in Figure1.26 we noted that the function was discontinuous at $x=0$ again
\begin{equation*} \lim_{x\to 0^-} f(x) \neq \lim_{x\to 0^+}f(x), \end{equation*}
thus the $\displaystyle \lim_{x\to 0} f(x)=DNE.$
From these three examples we see that in order for a function to be continuous we need the limit to exist at the point. More formally, we make the following definition.
###### Continuous Function
A function $f$ is continuous at $x = c$ provided that
1. $f$ has a limit as $x \to c\text{,}$
2. $f$ is defined at $x = c\text{,}$ and
3. $\lim_{x \to c} f(x) = f(c)\text{.}$
Conditions (a) and (b) are technically contained implicitly in (c), but we state them explicitly to emphasize their individual importance. The definition says that a function is continuous at $x = c$ provided that its limit as $x \to c$ exists and equals its function value at $x = c\text{.}$ Applying this definition to the types of discontinuities we have looked we can observe the following:
• If the graph of a function has a hole at $x=a$ then $f(a)=DNE \text{.}$
• If the graph of a function has an asymptote at $x=a$ then $f(a)=DNE$ or $\displaystyle\lim_{x \to a}f(x)=DNE\text{.}$
• If the graph of a function has an jump at $x=a$ then $\displaystyle\lim_{x\to a} f(x)=DNE\text{.}$
• If the graph of a function has a hole with a value at another y-value at $x=a$ then $\displaystyle\lim_{x \to a}f(x)\neq f(a)\text{.}$
###### Continuous Function
A function is said to be continuous on an interval $[a,b]$ if the $\displaystyle \lim_{x\to c}f(x)=f(c)$ at every point $x=c$ on the interval. That is, the function has no points of discontinuity on that interval.
If a function is continuous at every point in an interval $[a,b]\text{,}$ we say the function is continuous on $[a,b]\text{.}$ If a function is continuous at every point in its domain, we simply say the function is continuous. Thus we note that continuous functions are particularly nice: to evaluate the limit of a continuous function at a point, all we need to do is evaluate the function.
For example, consider $p(x) = x^2 - 2x + 3\text{.}$ It can be proved that every polynomial is a continuous function at every real number, and thus if we would like to know $\displaystyle \lim_{x \to 2} p(x)\text{,}$ we simply compute
\begin{equation*} \lim_{x \to 2} (x^2 - 2x + 3) = 2^2 - 2 \cdot 2 + 3 = 3 = p(2)\text{.} \end{equation*}
This route of substituting an input value to evaluate a limit works whenever we know that the function being considered is continuous. Besides polynomial functions, all exponential functions and the sine and cosine functions are continuous at every point, as are many other familiar functions and combinations thereof.
###### Example1.28
Consider the function
\begin{equation*} f(x)=\frac{3x(x+4)}{x^2-16}\text{.} \end{equation*}
Are there $x$-values where the function is discontinuous? If so, how do those $x$-values fail the limit definition of continuity?
Hint
Consider the graph of $y=f(x)\text{.}$
$f(x)$ is discontinuous at $x=4, -4\text{.}$ At both points $f(4)=DNE$ and $f(-4)=DNE\text{.}$ Also $\displaystyle \lim_{x\to4}f(x)=DNE\text{.}$
Solution
We first consider a graph of $y=f(x)\text{,}$ shown below.
Visual inspection of the graph certainly indicates discontinuities. However, we can make our visual inspection more precise through a little algebra. We start by expanding the numerator and denominator of the function. Specifically, we have
\begin{equation*} f(x)=\frac{3x(x+4)}{x^2-16}=\frac{3x(x+4)}{(x+4)(x-4)}. \end{equation*}
From here we see that there are two $x$-values at which the function is undefined: at $x=-4,4\text{,}$ that is $f(x)=DNE$ at these points hence the function is not continuous.
### SubsectionPiecewise Functions
In many cases a simple function like $f(x)=x^2$ may not fully describe the behavior of a phenomenon. In some of these cases we can turn to piecewise functions to give us the tools we need.
###### Example1.32
Is the function $f(x)\text{,}$ defined below, continuous at $x=1\text{,}$ use the limit to argue your answer
\begin{equation*} f(x) = \begin{cases} x +1 \amp \text{if } x\le 1\\ 3 \amp \text{if } x\gt 1 \end{cases}. \end{equation*}
Hint
Consider the graph of $f(x)$
The function $f(x)$ is not continuous since the $\displaystyle \lim_{x\to1}f(x)=DNE\text{.}$
Solution
For a piecewise function we must examine the point where the function changes. To do so we will examine the right and left hand limits. Here we use the fact that to the left of 1, that is $x\lt1$ the function is defined as $f(x)=x+1\text{.}$ To the right of 1, that is $x\gt 1$ the function is defined as $f(x)=3\text{.}$ Thus:
\begin{equation*} \lim_{x \to 1^-} f(x)=\lim_{x \to 1^-} x+1=1+1=2\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 1^+} f(x)=\lim_{x \to 1^+} 3=3\text{.} \end{equation*}
Since the two sides are not the same
\begin{equation*} \lim_{x \to 1} f(x)=DNE\text{.} \end{equation*}
So the function $f(x)$ is not continuous at $x=1\text{.}$
The jumps that a piecewise function possesses make piecewise functions a natural place in which to explore continuity.
###### Example1.34
Is the function $h(x)\text{,}$ defined below, continuous for all values of $x\text{?}$
\begin{equation*} h(x)=\begin{cases} 4x^2 \amp \text{ if } x\leq 2 \\ 5x+6 \amp \text{ if } x\gt2 \end{cases} \end{equation*}
Hint
Consider the limit from the left and the limit from the right at $x=2\text{.}$
The function $h(x)$ is continuous on the interval $(-\infty,\infty)\text{,}$ that is for all values of $x\text{.}$
Solution
For a piecewise function we must examine the point where the function changes. To do so we will examine the right and left hand limits. Here we use the fact that to the left of 2, that is $x\lt2$ the function is defined as $h(x)=4x^2\text{.}$ To the right of 2, that is $x\gt 2$ the function is defined as $h(x)=5x+6\text{.}$ Thus:
\begin{equation*} \lim_{x \to 2^-} h(x)=\lim_{x \to 2^-} 4x^2=4(2)^2=16\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 2^+} h(x)=\lim_{x \to 2^+} 5x+6=5(2)+6=16\text{.} \end{equation*}
Since the two sides are the same
\begin{equation*} \lim_{x \to 2} h(x)=16\text{.} \end{equation*}
So the function $h(x)$ is continuous at $x=2\text{,}$ and since each piece is a polynomial this function is continuous on the interval $(-\infty,\infty)\text{,}$ that is for all values of $x\text{.}$
Not only can we ask questions about when a piecewise function is continuous but we can also ask questions about how to make a piecewise function continuous by varying parameters.
###### Example1.35
Consider the piecewise function
\begin{equation*} D(x)=\begin{cases} 4x^2-k\amp\text{ if } x \lt 2,\\ kx+1\amp\text{ if } x \geq 2.\\ \end{cases} \end{equation*}
Find the value of $k$ to make this function continuous for all $x\text{.}$
Hint
Consider the limit from the left and the limit from the right at $x=2\text{.}$
$k=5$
Solution
To determine a value of $k$ to make $D(x)$ continuous we will examine the right and left hand limits at $x=2\text{.}$ Here we use the fact that to the left of 2, that is $x\lt2$ the function is defined as $D(x)=4x^2-k\text{.}$ To the right of 2, that is $x\gt 2$ the function is defined as $D(x)=kx+1\text{.}$ Thus:
\begin{equation*} \lim_{x \to 2^-} D(x)=\lim_{x \to 2^-} 4x^2-k=4(2)^2-k=16-k\text{.} \end{equation*}
\begin{equation*} \lim_{x \to 2^+} D(x)=\lim_{x \to 2^+} kx+1=k(2)+1=2k+1\text{.} \end{equation*}
In order for $D(x)$ to be continuous need the two sides to be equal, thus set them equal and solve for $k\text{.}$
\begin{equation*} 16-k=2k+1\text{.} \end{equation*}
\begin{equation*} 15=3k\text{.} \end{equation*}
So the function $D(x)$ is continuous at $x=2$ when $k=5\text{.}$
###### Example1.36
Determine if each of the functions below is continuous at $x=2\text{.}$
1. $f(x)=\ln(3-x)\text{.}$
2. $g(x)=\frac{1}{x-2}\text{.}$
3. $h(x)=\begin{cases} x^2+1\amp\text{ if } x \neq 2,\\ 3\amp\text{ if } x=2.\\ \end{cases}$
Hint
Consider evaluating limits on each side and comparing that value to the value of the function at the point.
1. $f$ is continuous at $x=2\text{.}$
2. $g$ is not continuous at $x=2\text{.}$
3. $h$ is not continuous at $x=2\text{.}$
Solution
For each of these functions, we want to check that the limit exists at $x=2\text{,}$ the function is defined at $x=2\text{,}$ and these two values match.
1. We can examine the graph of $y=f(x)$ at $x=2$ or examine function values nearby $x=2$ on the left and right to find that $\lim_{x\to 2} f(x)=0\text{.}$ Evaluating $f(2)=\ln(3-2)=\ln(1)=0\text{.}$ Thus, $\lim_{x\to2}f(x)=f(2)\text{,}$ and $f$ is continuous at $x=2\text{.}$
2. Notice that the graph of $g$ has a vertical asymptote at $x=2\text{,}$ so $g(2)$ is undefined. Hence, $g$ is not continuous at $x=2\text{.}$
3. For values of $x$ near 2 (from the left and right), we have $h(x)$ getting close to 5. Therefore, $\lim_{x\to2}h(x)=5\text{.}$ However, $h(2)=3\text{.}$ Since $\displaystyle \lim_{x\to2}h(x)\neq h(2)\text{,}$ $h$ is not continuous at $x=2\text{.}$
### SubsectionProperties of Limits and Continuous Functions
There are several properties of limits and continuous functions that are useful to have in your toolbox. Specifically, limits and continuous functions behave well under typical mathematical operations. While these properties can be proven in detail, we proceed to only state the properties.
###### Properties of Limits
Assuming all the limits on the right-hand side exist:
• If $b$ is a constant, then $\lim\limits_{x \rightarrow c} (bf(x))=b\left(\lim\limits_{x \rightarrow c} f(x) \right)$
• $\lim\limits_{x \rightarrow c} \left( f(x)+g(x)\right)=\lim\limits_{x \rightarrow c} f(x)+\lim\limits_{x \rightarrow c}g(x)$
• $\lim\limits_{x \rightarrow c} \left( f(x) \cdot g(x)\right)=\lim\limits_{x \rightarrow c} f(x)\cdot\lim\limits_{x \rightarrow c}g(x)$
• $\lim\limits_{x \rightarrow c} \left(\frac{f(x)}{g(x)}\right)=\frac{\lim\limits_{x \rightarrow c} f(x)}{\lim\limits_{x \rightarrow c} g(x)}\text{,}$ provided $\lim\limits_{x \rightarrow c} g(x) \neq 0$
• For any constant $k\text{,}$ $\lim\limits_{x \rightarrow c} k=k$
• $\lim\limits_{x \rightarrow c} x=c$
###### Example1.37
We can use algebra to compute $\lim\limits_{x \rightarrow 1} x^2(x^3+2).$ Specifically,
\begin{align*} \lim\limits_{x \rightarrow 1} x^2(x^3+2)\amp=\left(\lim\limits_{x \rightarrow 1} x^2\right)\left(\lim\limits_{x \rightarrow 1} (x^3+2)\right)\\ \amp=\left(\lim\limits_{x \rightarrow 1} x^2\right)\left(\lim\limits_{x \rightarrow 1} x^3+\lim\limits_{x \rightarrow 1}2\right)\\ \amp=1(1+2) \\ \amp=3. \end{align*}
###### Continuity of Sums, Products, and Quotients of Functions
Suppose that $f$ and $g$ are continuous on an interval and that $b$ is a constant. Then, on that same interval,
• $bf(x)$ is continuous.
• $f(x) + g(x)$ is continuous.
• $f(x)g(x)$ is continuous.
• $\frac{f(x)}{g(x)}$ is continuous, provided $g(x) \neq 0$ on the interval.
###### Continuity of Composite Functions
If $f$ and $g$ are continuous, and if the composite function $f(g(x))$ is defined on an interval, then $f(g(x))$ is continuous on that interval.
### SubsectionSummary of Limits and Continuity
The concepts discussed in the last two sections will be important in later sections. The following is a short summary of these sections and an example that ties together the concepts of limits and continuity.
• For a function $f$ defined on an interval around a number $c\text{,}$
\begin{equation*} \lim_{x \rightarrow c} f(x)=L \end{equation*}
means that the value of $f(x)$ gets as close as we want to a number $L$ whenever $x$ is sufficiently close to $c\text{,}$ assuming the value $L$ exists.
• We define a limit from the left and a limit from the right in the same way as above, while adding the stipulation that $x\lt c$ for the left limit and $x\gt c$ for the right limit. That is, as we move $x$ sufficiently close to $c$ from the left on a number line ($x\lt c$), $f(x)$ gets as close to the limit value as we want. Similarly for the limit from the right.
• The one-sided limits help to determine if a limit exists as $x$ approaches a value $c\text{.}$ More specifically, $\lim_{x \rightarrow c} f(x)=L$ if and only if $\displaystyle \lim_{x \rightarrow c^-} f(x)=L=\lim_{x \rightarrow c^+} f(x)$
• Limits also help us determine the continuity of a function at a point $x=c\text{.}$ A function $f$ that has a limit as $x\rightarrow c\text{,}$ is defined at $x=c\text{,}$ and $\displaystyle \lim_{x\rightarrow c} f(x)=f(c)$ is continuous at $x=c\text{.}$
###### Example1.38
In this example, we take a closer look at a function whose graph we previously encountered. For convenience, this graph is reproduced below in Figure1.39.
1. At which values of $c$ does $\displaystyle \lim_{x \to c} f(x)$ not exist?
2. At which values of $c$ is $f(c)$ not defined?
3. At which values of $c$ does $f$ have a limit, but $\displaystyle \lim_{x \to c} f(x) \ne f(c)\text{?}$
4. State all values of $c$ for which $f$ is not continuous at $x = c\text{.}$
5. Which condition is stronger, and hence implies the other: $f$ has a limit at $x = c$ or $f$ is continuous at $x = c\text{?}$ Explain, and hence complete the following sentence: If $f$ at $x = c\text{,}$ then $f$ at $x = c\text{,}$ where you complete the blanks with has a limit and is continuous, using each phrase once.
Hint
1. Consider the left- and right-hand limits at each value.
2. Carefully examine places on the graph where there's an open circle.
3. Are there locations on the graph where the function has a limit but there's a hole in the graph?
4. Remember that at least one of three conditions must fail: if the function lacks a limit, if the function is undefined, or if the limit exists but does not equal the function value, then $f$ is not continuous at the point.
5. Note that the definition of being continuous requires the limit to exist.
1. $c = -2\text{;}$ $c = +2\text{.}$
2. $c = 3\text{.}$
3. $c = -1\text{;}$ $c = 3\text{.}$
4. $c=-2\text{;}$ $c = 2\text{;}$ $c = 3\text{;}$ $c = -1\text{.}$
5. If $f$ is continuous at $x = c\text{,}$ then $f$ has a limit at $x = c\text{.}$
Solution
1. $\lim_{x \to c} f(x)$ does not exist at $c = -2$ since
\begin{equation*} \lim_{x \to -2^-} f(x) = 2 \ne -1 = \lim_{x \to -2^+}\text{,} \end{equation*}
and $\lim_{x \to c} f(x)$ does not exist at $c = 2$ since $\lim_{x \to 2^+} f(x)$ does not exist due to the infinitely oscillatory behavior of $f\text{.}$
2. The only point at which $f$ is not defined is at $c = 3\text{.}$
3. At $c = -1\text{,}$ note that $\lim_{x \to -1} f(x)$ exists (and appears to have value approximately $-3.25$), but $f(-1) = 1$ and thus $\lim_{x \to -1} f(x) \ne f(-1)\text{.}$ At $c = 3\text{,}$ we have $\lim_{x \to 3} f(x) = -2.5\text{,}$ but $f(3)$ is not defined so the limit exists but does not equal the function value.
4. Based on our work in (a), (b), and (c), $f$ is not continuous at $c=-2$ and $c = 2$ because $f$ does not have a limit at those points; $f$ is not continuous at $c = 3$ since $f$ is not defined there; and $f$ is not continuous at $c = -1$ because at that point its limit does not equal its function value.
5. If $f$ is continuous at $x = c\text{,}$ then $f$ has a limit at $x = c\text{,}$ since one of the defining properties of being continuous at $x = c$ is that the function has a limit at that input value. This shows that being continuous is a stronger condition than having a limit. |
# The square of the sum of two consecutive integers is 1681. What are the integers?
Jan 8, 2016
20 and 21.
#### Explanation:
Let's say the two consecutive numbers are $a$ and $b$. We need to find an equation that we can solve to work out their values.
"The square of the sum of two consecutive integers is $1681$." That means if you add $a$ and $b$ together, then square the result, you get $1681$. As an equation we write:
${\left(a + b\right)}^{2} = 1681$
Now, there are two variables here so at first glance it looks unsolvable. But we're also told that $a$ and $b$ are consecutive, which means that $b = a + 1$!
Substituting this new information in gives us:
${\left(a + a + 1\right)}^{2} = 1681$
${\left(2 a + 1\right)}^{2} = 1681$
Next we're going to follow these steps to solve for $a$:
1) Take the square root of both sides. This will give two possible results, since both positive and negative numbers have positive squares.
2) Subtract $1$ from both sides.
3) Divide both sides by $2$.
${\left(2 a + 1\right)}^{2} = 1681$
$2 a + 1 = \sqrt{1681} = 41$
$2 a = 40$
$a = 20$
This means that $b = 21$! To check these answers, take the values $20$ and $21$ and substitute them into the original equation like this:
${\left(a + b\right)}^{2} = 1681$
${\left(20 + 21\right)}^{2} = 1681$
$1681 = 1681$
Success! |
# Selina Concise Solutions for Chapter 6 Sets Class 8 ICSE Mathematics
### Exercise 6 A
1. Write the following sets in roster (Tabular) form :
(i) A1 = {x : 2x + 3 = 11}
(ii) A2 = {x : x2 – 4x – 5 = 0}
(iii) A3 = {x : x ∈ Z, -3 ≤ x < 4}
(iv) A4 = {x : x is a two digit number and sum of digits of x is 7}
(v) A5 = {x : x = 4n, n ∈ W and n < 4}
(vi) A6 = {x : x = n/(n + 2) ; n ∈ N and n > 5}
Solution
(i) A1 = {x : 2x + 3 = 11}
∴ 2x + 3 = 11
⇒ 2x = 11 – 3
⇒ 2x = 8
⇒ x = 8/2 ⇒ x = 4
∴ Given set in roster (Tabular) form is
A1 = {4}
(ii) A2 = {x : x2 – 4x – 5 = 0}
∴ x2 – 4x – 5 = 0
⇒ x2 – 5x + x – 5 = 0
⇒ x(x – 5) + 1(x – 5) =0
⇒ (x – 5)(x + 1) = 0
∴ Either x – 5 = 0 or x + 1 = 0
⇒ x = 5 ⇒ x = -1
∴ Given set in roster (Tabular) form is A2 = {5, -1}
(iii) A3 = {x : x ∈ Z, -3 ≤ x < 4}
∵ -3 ≤ x < 4
∴ x = -3, -2, -1, 0, 1, 2, 3
∴ Given set in roster (Tabular) form is
A3 = {-3, -2, -1, 0, 1, 2, 3}
(iv) A4 = {x : x is a two digit number and sum of digits of x is 7}
∵ x is a two digit number and sum of digits of x is 7
∴ x = 16, 25, 34, 43, 52, 61, 70
∴ Given set in roster (Tabular) form is
A4 = {16, 25, 34, 43, 52, 61, 70}
(v) A5 = {x : x = 4n, n ∈ W and n < 4}
∵ x = 4n
∴ When n = 0, x = 4 × 0
⇒ x = 0
When n = 1, x = 4 × 1
⇒ x = 4
When n = 2, x = 4 × 2
⇒ x = 8
When n = 3, x = 4 × 3
⇒ x = 12
∴ Given set in roster (Tabular) form is
A5 = {0, 4, 8, 12}
(vi) A6 = {x : x = n/(n + 2); n ∈ N & n > 5}
∵ x = n/(n + 2)
∴ When n = 6, x = 6/(6 + 2)
⇒ x = 6/8
⇒ x = ¾
When n = 7, x = 7/(7 + 2)
⇒ x = 7/9
When n = 8, x = 8/(8 + 2)
⇒ x = 8/10
⇒ x = 4/5
When n = 9, x = 9/(9 + 2)
⇒ x = 9/11
∴ Given set in roster (Tabular) form is
A6 = {3/4, 7/9, 4/5, 9/11,…}
2. Write the following sets in set-builder (Rule Method) form:
(i) B1 = {6, 9, 2, 15, ...}
(ii) B2 = {11, 13, 17, 19 }
(iii) B3 = (1/3, 3/5, 5/7, 7/9, 9/11,…}
(iv) B4 = {8, 27, 64, 125, 216}
(v) B5 = {-5, -4, -3, -2, -1}
(vi) B6 = {…, -6, -3, 0, 3, 6 ….}
Solution
(i) B1 = {6, 9, 12, 15, ….}
= {x : x = 3n + 3; n ∈ N}
(ii) B2 = {11, 13, 17, 19}
= {x : x is a prime number between 10 and 20}
(iii) B3 = {1/3, 3/5, 5/7, 7/9, 9/11, …}
= {x : x = n/(n + 2), where n is an odd natural number}
(iv) B4 = {8, 27, 64, 125, 216}
= {x : x = n3; n ∈ N and 2 ≤ n ≤ 6}
(v) B5 = {-5, -4, -3, -2, -1}
= {x : x ∈ Z, -5 ≤ x ≤ -1}
(vi) B6 = {…, -6, -3, 0, 3, 6, …}
= {x : x = 3n, n ∈ Z}
3. (i) Is {1, 2, 4, 16, 64} = {x : x is a factor of 32} ? Give reason.
(ii) Is {x : x is a factor of 27} ≠ {3, 9, 27, 54}? Give reason.
(iii) Write the set of even factors of 124.
(iv) Write the set of odd factors of 72.
(v) Write the set of prime factors of 3234.
(vi) Is {x : x2 – 7x + 12 = 0} = {3, 4} ?
(vii) Is {x : x2 – 5x – 6 = 0} = {2, 3} ?
Solution
(i) No, {1, 2, 4, 16, 64} ≠ {x : x is a factor of 32}
Because 64 is not a factor of 32
(ii) Yes, { x : x is a factor of 27} + {3, 9, 27, 54}
Because 54 is not a factor of 27
(iii) 1 × 124 = 124
2 × 62 = 124
4 × 31 = 124
Factors of 124 = 1, 2, 4, 31, 62, 124
Set of even factors 124 = {2, 4, 62, 124}
(iv) 1 × 172 = 72
2 × 36 = 72
3 × 24 = 72
4 × 18 = 72
6 ×12 = 72
8 × 9 = 72
Factors of 72 = 1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72
Set of odd factors of 72 = {1, 3, 9}
(v)
3234 = 2 × 3 × 7 × 7 × 11
∴ Set of prime factors of 3234 = {2, 3, 7, 11}
(vi) x2 – 7x + 12 = 0
⇒ x2 – 4x – 3x + 12 = 0
⇒ x(x – 4) – 3(x – 4) = 0
⇒ (x – 4)(x – 3) = 0
∴ Either x – 4 = 0 or x – 3 = 0
⇒ x = 4 ⇒ x = 3
∴ {x : x2 – 7x + 12 = 0} = {3, 4} is true
(vii) x2 – 5x – 6 = 0
⇒ x2 – 6x + x – 6 = 0
⇒ x(x – 6) + 1(x – 6) = 0
⇒ (x - 6)(x + 1) = 0
∴ Either x – 6 = 0 or x + 1 = 0
i.e., x = 6 i.e., x = -1
∴ {x : x2 - 5x – 6 = 0} ≠ {2, 3}
In other words {x : x2 – 5x – 6 = 0} = {2, 3} is not true
4. Write the following sets in Roster form:
(i) The set of letters in the word ‘MEERUT’.
(ii) The set of letters in the word ‘UNIVERSAL’.
(iii) A = {x : x = y + 3, y ∈ N and y > 3}
(iv) B = {p : p ∈ W and p2 < 20}
(v) C = {x : x is composite number and 5 < x < 21}
Solution
(i) Roster form of the set of letters in the word “MEERUT” = {m, e, r, u, t}
(ii) Roster form of the set of letters in the word “UNIVERSAL” = {u, n, i, v, e, r, s, a, l}
(iii) A = {x : x = y + 3, y ∈ N and y > 3}
x = y + 3
when y = 4, x = 4 +3 = 7
when y = 5, x = 5 + 3 = 8
when y = 6, x = 6 + 3 = 9
when y = 7, x = 7 + 3 = 10
when y = 8, x = 8 + 3 = 11
∴ Roster form of the given set A = {7, 8, 9, 10, 11 ……}
(iv) B = {P : P ∈ W and p2 < 20}
When P2 = 0
P = √0 = 0
When P2 = 1
P = √1 = 1
When P2 = 4
P = √4 = 2
When P2 = 9
P = √9 = 3
When P2 = 16
P = √16 = 4
∴ Roster form of the given set B = {0, 1, 2, 3, 4}
(v) C = {x : x is composite number and 5 ≤ x ≤ 21}
5 ≤ x ≤ 21 means x = 5, 6, 7, 8, 9, 10, ……, 21
But we are given that x is a composite number
∴ x = 6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21
∴ Roster form of the given set C = {6, 8, 9, 10, 12, 14, 15, 16, 18, 20, 21}
Note: Composite numbers: The natural numbers (greater than 1), which are not prime, are called composite numbers.
5. List the elements of the following sets:
(i) {x : x2 – 2x – 3 = 0}
(ii) {x : x = 2y + 5; y ∈ N and 2 ≤ y < 6}
(iii) {x : x is a factor of 24}
(iv) {x : x ∈ Z and x2 ≤ 4}
(v) {x : 3x – 2 ≤ 10, x ∈ N}
(vi) { x : 4 – 2x > -6, x ∈ Z}
Solution
(i){x : x2 - 2x – 3 = 0}
x2 – 2x – 3 = 0
⇒ x2 – 3x + x – 3 = 0
⇒ x(x – 3) + 1(x – 3) = 0
⇒ (x + 1)(x – 3) = 0
∴ Either x – 3 = 0 or x + 1 = 0
x = 3 or x = -1
∴ Elements of the set :{x : x2 – 2x – 3 = 0} are 3 and -1
(ii) {x : x = 2y + 5; y ∈ N and 2 ≤ y < 6}
x = 2y + 5
when y = 2, x = 2 × 2 + 5 = 4 + 5 = 9
When y = 3, x = 2 × 3 + 5 = 6 + 5 = 11
When y = 4, x = 2 × 4 + 5 = 8 + 5 = 13
When y = 5, x = 2 × 5 + 5 = 10 + 5 = 15
∴ Elements of the given set {x : x = 2y + 5; y ∈ N and 2 ≤ y < 6} are 9, 11, 13, 15
(iii) {x : x is a factor of 24}
24 = 1 × 24
24 = 2 × 12
24 = 3 × 8
24 = 4 × 6
∴ Elements of the given set { x : x is a factor of 24} are 1, 2, 3, 4, 6, 8, 12, 24
(iv) {x : x ∈ Z and x2 ≤ 4}
When x2 = 4
x = ± √4 = ± 2
when x2 = 1
x = ± √1 = ±1
when x2 = 0
x = √0 = 0
∴ Elements of the given set {x : x ∈ Z and x2 ≤ 4 are +2, -2, + 1, -1, 0 or are -2, -1, 0, 1, 2
(v) {x : 3x – 2 ≤ 10, x ∈ N}
3x – 2 ≤ 10
⇒ 3x ≤ 10 + 2
⇒ 3x ≤ 12
⇒ x ≤ 12/3
≤ x ≤ 4
∴ Elements of the given set {x : 3x – 2 ≤ 10, x ∈ N} are 1, 2, 3 and 4
(vi) {x : 4 – 2x > -6, x ∈ Z}
4 – 2x > -6
⇒ -4 + 4 – 2x > -6 - 4 (Subtracting 4 from both sides)
⇒ -2x > -10
⇒ -2x + 2x + 10 > -10 + 2x + 10 [Adding 2x + 10 to both sides]
⇒ + 10 > 2x
⇒ 10/2 > x
⇒ 5 > x
∴ Elements of the given set {x : 4 – 2x > -6, x ∈ Z} are 4, 3, 2, 1, 0, -1 …
### Exercise 6 B
1. Find the cardinal number of the following sets:
(i) A1 = {-2, -1, 1, 3, 5}
(ii) A2 = {x : x ∈ N and 3 ≤ x < 7}
(iii) A3 = {p : p ∈ W 2p – 3 < 8}
(iv) A4 = {b : b ∈ Z and -7 < 3b - 1 ≤ 2}
Cardinal Number of a set: The number of elements in a set is called its Cardinal Number.
Solution
(i) A1 = {-2, -1, 1, 3, 5}
∴ Cardinal number of set A1 = 5
(ii) A2 = {x : x ∈ N and 3 ≤ x < 7}
= {3, 4, 5, 6}
∴ Cardinal number of set A2 = 4
(iii) A3 = {P : P ∈ W and 2P - 3 < 8}
2P – 3 < 8
⇒ 2P – 3 + 3 < 8 + 3
(Adding 3 to both sides)
⇒ 2P < 11
⇒ P < 11/2 (Dividing both sides by 2)
⇒ P < 5.5
∴ A3 = {0, 1, 2, 3, 4, 5}
∴ Cardinal number of set A3 = 6
(iv) A= {b : b ∈ Z and – 7 < 3b – 1 ≤ 2}
-7 < 3b – 1
⇒ - 7 + 1 < 3b – 1 + 1 (Adding 1 to both sides)
⇒ -6 < 3b
⇒ -6/3 < b (Dividing both sides by 3)
⇒ -2 < b
Again 3b - 1 ≤ 2
⇒ 3b - 1 + 1 ≤ 2 + 1 (Adding 1 to both sides)
⇒ 3b ≤ 3
⇒ b ≤ 3/3 (Dividing both sides by 3)
⇒ b ≤ 1
∴ -2 ≤ b ≤ 1
∴ Given set A4 = {-1, 0, 1}
∴ Cardinal number of set A4 = 3
2. If P = {P : P is a letter in the word “PERMANENT’}. Find n(P).
Solution
P = (P : P is a letter in the word “ PERMANENT”}
Or P = {p, e, r, m, a, n, t}
n(P) = 7
3. State, which of the following sets are finite and which are infinite:
(i) A = {x : x ∈ Z and x < 10}
(ii) B = {x : x ∈ W and 5x – 3 ≤ 20}
(iii) P = {y : y = 3x – 2, x ∈ N & x > 5}
(iv) M = {r : r = 3/n; n ∈ W and 6 < n ≤ 15}
Note: (i) A set with finite (limited) number of elements in it, is called a finite set, (ii) A set which is not finite is called an infinite set.
Solution
(i) A = {x : x ∈ Z and x < 10}
= {…., -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}
= {9, 8, 7, 6, 5, 4, 3, 2, 1, 0, -1, -2, -3, -4, ….}
∴ It is an infinite set.
(ii) B = {x : x ∈ W and 5x – 3 ≤ 20}
5x – 3 ≤ 20
⇒ 5x – 3 + 3 ≤ 20 + 3 (Adding 3 to both sides)
⇒ 5x ≤ 20 + 3
⇒ 5x ≤ 23
⇒ x ≤ 23/5 (Dividing both sides by 5)
⇒ x ≤ 4.6
∴ B = {0, 1, 2, 3, 4}
∴ It is a finite set.
(iii) P = {y : y = 3x – 2, x ∈ N and x > 5}
y = 3x – 2
When x = 6, y = 3×6 – 2 = 18 – 2 = 16
When x = 7, y = 3×7 – 2 = 21 – 2 = 19
When x = 8, y = 3 × 8 – 2 = 24 – 2 = 22
When x = 9, y = 3 × 9 – 2 = 27 – 2 = 25
∴ P = {16, 19, 22, 25, ...}
∴ It is an infinite set.
(iv) M = {r : r = 3/n; n ∈ W and 6 < n ≤ 15}
r = 3/n
when n = 7, r = 3/7
when n = 8, r = 3/8
when n = 9, r = 3/9
when n = 10, r = 3/10
when n = 11, r = 3/11
when n = 12, r = 3/12
when n = 13, r = 3/13
when n = 14, r = 3/14
when n = 15, r = 3/15
∴ M = {3/7, 3/8, 3/9, 3/10, 3/11, 3/12, 3/13, 3/14, 3/15}
∴ It is a finite set.
4. Find which of the following sets are singleton sets:
(i) The set of points of intersection of two non-parallel st. lines in the same plane.
(ii) A = {x : 7x – 3 = 11}
(iii) B = {y : 2y + 1 < 3 and y ∈ W}
Note: A set, which has only one element in it, it is called a SINGLETON or unit set.
Solution
(i) The set of points of intersection of two non-parallel st. lines in the same plane is singleton set.
(ii) A = {x : 7x – 3 = 11}
7x – 3 = 11
⇒ 7x = 11 + 3
⇒ 7x = 14
⇒ x = 14/7 = 2
∴ A = {2}
Hence, given set A is a singleton set.
(iii) B = {y : 2y + 1 < 3 and y ∈ W}
2y + 1 < 3
⇒ 2y + 1 – 1 < 3 – 1 (Subtracting 1 from both sides)
⇒ 2y < 2
⇒ y < 2/2 (Dividing both sides by 2)
⇒ y < 1
∴ B = {0}
Hence, it is singleton set.
5. Find which of the following sets are empty:
(i) The set of points of intersection of two parallel lines.
(ii) A = {x : x ∈ N and 5 < x < 6}
(iii) B = {x : x2 + 4 = 0, x ∈ N}
(iv) C = {even numbers between 7 & 11}
(v) D = {prime numbers between 7 & 11}
Note: The set, which has no element in it, is called the empty or null set.
Solution
(i) ”The set of points of intersection of two parallel lines” is an empty set because two parallel lines do not intersect anywhere.
(ii) A = { x : x ∈ N and 5 < x ≤ 6}
As 5 < x ≤ 6
∴ x = 6
∴ A = {6}
Hence, given set A is not an empty set.
(iii) B = {x : x+ 4 = 0, x ∈ N}
x2 + 4 = 0
⇒ x2 = -4
⇒ x = √-4 which is not an empty set.
(iv)
C = {Even numbers between 6 and 10}
∴ C = {8}
Hence it is not an empty set.
(v) D = {Prime numbers between 7 and 11}
Because there is no prime numbers between 7 and 11.
∴ D = { }
Hence, it is an empty set.
6. (i) Are the sets A = {4, 5, 6} and B = {x : x– 5x – 6 = 0} disjoint ?
(ii) Are the sets A = {b , c, d, e} and B = {x : x is a letter in the word ‘MASTER’} joint?
Note: (i)Two sets are said to be joint sets, if they have at least one element in common.
(ii) Two sets are said to be disjoint if they have no elements in common.
Solution
(i) A = {4, 5, 6}
B = {x : x2 – 5x – 6 = 0}
x2 – 5x – 6 = 0
⇒ x2 – 6x + x – 6= 0
⇒ (x – 6) + 1(x – 6) = 0
⇒ (x – 6)(x +1) = 0
∴ Either x – 6 = 0 Or x + 1 = 0
⇒ x = 6 or x = -1
∴ B = {6, -1}
Hence set A and set B are not disjoint because these sets have elements 6 in common.
(ii) A {b, c, d, e}
B = {x : x is a letter in the word “MASTER”}
∴ B = {m, a, s, t, e, r}
Hence set A and set B are joint because these sets have elements e in common.
7. State, whether the following pairs of sets are equivalent or not:
(i) A = {x : x ∈ N and 11 ≥ 2x – 1} and B = {y : y ∈ W and 3 ≤ y ≤ 9}
(ii) Set of intergers and set of natural numbers.
(iii) Set of whole numbers and set of multiples of 3.
(iv) P = {5, 6, 7, 8} and M = {x : x ∈ W and x < 4}
Note: Two sets are said to be equivalent, if they contain the same number of elements.
Solution
(i) A = {x : x ∈ N and 11 ≥ 2x – 1}
11 ≥ 2x – 1
⇒ 11 + 1 ≥ 2x – 1 + 1
⇒ 12 ≥ 2x
⇒ 12/2 ≥ x
⇒ 6 ≥ x
∴ A = {1, 2, 3, 4, 5, 6}
∴ n(A) = 6
B = {y : y ∈ W and 3 ≤ y ≤ 9}
∵ 3 ≤ y ≤ 9
B = {3, 4, 5, 6, 7, 8, 9}
n(B) = 7
Cardinal number of set A = 6 and cardinal number of set B = 7
Set A and Set B are not equivalent.
(ii) Set of integers has infinite number of elements. Set of natural numbers has infinite number of elements.
Set of integers and set of natural numbers are equivalent because both these sets have infinite number of elements.
(iii) Set of whole numbers, has infinite number of elements. Set of multiples of 3, has infinite number of element.
Set of whole numbers and set of multiples of 3 are equivalent because both these sets have infinite number of elements.
(iv) P = {5, 6, 7, 8}
n(P) = 4
M = {x : x ∈ W and x ≤ 4}
M = {0, 1, 2, 3, 4}
n(M) = 5
Now Cardinal number of set P = 4 and
Cardinal number of set M = 5
These sets are not equivalent.
8. State whether the following pairs of sets are equal or not:
(i) A = {2, 4, 6, 8} and
B = {2n : n ∈ N and n < 5}
(ii) M = {x : x ∈ W and x + 3 < 8} and
N = {y : y = 2n – 1, n ∈ N and n < 5}
(iii) E = {x : x2 + 8x – 9 = 0} and
F = {1, -9}
(iv) A = {x : x ∈ N, x < 3} and
B = {y : y2 – 3y + 2 = 0}
Note: Two sets are equal, if both the sets have same (identical) elements.
Solution
(i) A = {2, 4, 6, 8}
B = {2n : n ∈ N and n < 5}
When n = 1, 2n = 2 × 1 = 2
When n = 2, 2n = 2 × 2 = 4
When n = 3, 2n = 2 × 3 = 6
When n = 4, 2n = 2 × 4 = 8
∴ B = {2, 4, 6, 8}
Now we see that elements of sets A and B are the same (identical)
∴ Sets A and B are equal.
(ii) M = {x : x ∈ W and x + 3 < 8}
x + 3 < 8
⇒ x < 8 – 3
⇒ x < 5
∴ M = {0, 1, 2, 3, 4}
N = {y : y = 2n – 1, n ∈ N and n < 5}
y = 2n – 1
when n = 1, y = 2 × 1 – 1
⇒ y = 2 – 1 = 1
when n = 2, y = 2 × 2 – 1
⇒ y = 4 – 1 = 3
when n = 3, y = 2 × 3 – 1
⇒ y = 6 – 1 = 5
when n = 4, y = 2 × 4 – 1
⇒ y = 8 – 1 = 7
∴ N = {1, 3, 5, 7}
Now we see that elements of sets M and N are not the same (identical).
∴ Sets M and N are not equal.
(iii) E = {x : x + 8x – 9 = 0}
x + 8x – 9 = 0
⇒ x + 9x – x – 9 = 0
⇒ x(x + 9) – 1(x + 9) = 0
⇒ (x – 1)(x + 9) = 0
∴ Either x + 9 = 0 or x – 1 = 0
⇒ x = - 9 or x = 1
∴ E = {-9, 1}
F = {1, -9}
Now we see that elements of sets E and F are the same (identical).
∴ Sets E and F are equal.
(iv) A = {x : x ∈ N, x < 3}
= {1, 2}
B = {y : y2 - 3y + 2 = 0}
y- 3y + 2 = 0
⇒ y2 - 3y + 2 = 0
⇒ y2 - 2y - y + 2 = 0
⇒ y(y - 2) - 1(y - 2) = 0
⇒ (y - 2)(y - 1) = 0
∴ Either y - 2 = 0 or y - 1 = 0
⇒ y = 2 or y = 1
∴ B = (1, 2}
Now we see that elements of sets A and B are the same (identical).
∴ Sets A and B are equal.
9. State whether each of the following sets is a finite set or an infinite set:
(i) The set of multiples of 8.
(ii) The set of integers less than 10.
(iii) The set of whole numbers less than 12.
(iv) {x : x = 3n – 2, n ∈ W, n ≤ 8}
(v) {x : x = 3n – 2n, n ∈ Z, n ≤ 8}
(vi) {x : x = (n – 2)/(n + 1), n ∈ W)
Solution
(i) The set of multiples of 8
= {8, 16, 24, 32, ….}
It is an infinite set.
(ii) The set of integers less than 10
= {9, 8, 7, 6, 5, 4, 3, 2, 1, -1, -2, …..}
It is an infinite set.
(iii) The set of whole numbers less than 12
= {11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1. 0}
It is a finite set.
(iv) {x : x = 3n – 2, n ∈ W, n ≤ 8}
Substituting the value of n = (0, 1, 2, 3, 4, 5, 6, 7 and 8) we get
= {-2, 1, 4, 7, 10, 13, 16, 19, 22}
It is finite set.
(v) {x : x = 3n – 2, n ∈ Z, n ≤ 8}
= {22, 19, 16, 13, 10, 7, 4, 1, -2, -5, ……}
It is infinite set.
(vi) {x : x = (n – 2)/(n + 1), n ∈ W}
{-2, -1/2, 0, ¼, 2/5 ……….}
It is infinite set.
10. Answer, whether the following statements are true or false. Give reasons.
(i) The set of even natural numbers less than 21 and the set of odd natural numbers less than 21 are equivalent sets.
(ii) If E = {factors of 16} and F = {factors of 20}, then E = F.
(iii) The set A = {integers less than 20} is a finite set.
(iv) If A = {x : x is an even prime number}, then set A is empty.
(v) The set of odd prime numbers is the empty set.
(vi) The set of squares of integers and the set of whole numbers are equal sets.
(vii) In n(P) = n(M), then P → M.
(viii) If set P = set M, then n(P) = n(M).
(ix) n(A) = n(B) ⇒ A = B.
Solution
(i) Set of even natural number less than 21
= {2, 4, 6, 8, 10, 12, 14, 16, 18, 20}
∴ Cardinal Number of this set = 10
Set of odd natural numbers less than 21
= {1, 3, 5, 7, 9, 11, 13, 15, 17, 19}
∴ Cardinal number of this set = 10
Now we see that cardinal numbers of both these sets = 10
∴ “The set of even natural numbers less than 21 and the set of odd natural numbers less than 21 are equivalent sets”…… is a True statement.
(ii) E = {Factors of 16} = {1, 2, 4, 8, 16}
F = {Factors of 20}={1, 2, 4, 5, 10, 20}
Now we see that elements of set E and set F are not the same (identical)
∴ “If E = {Factors of 16} and F = {Factors of 20},
Then, E = F” ……. is a False statement.
(iii) A = {Integers less than 20}
= {19, 18, 17, 16,…0, -1, -2, -3, ….}
∴ “The set A = {Integers less than 20} is a finite set"……is a False statement.
(iv) A = {x : x is an even prime number} = {2}
∴ “If A = {x : x is an even prime number}, then set A is empty”…. is a false statement .
(v)
Set of odd prime numbers = {3, 5, 7, 11, 13, 17, 19, 23, …….}
∴ “The set of odd prime number is the empty set”…… is a False statement.
(vi) Integer Square of Integer Whole No.
∴ Set of squares of integers = {0, 1, 4, 9, 16, 25,…..}
Set of whole numbers = {0, 1, 2, 3, 4, 5, 6, 7,……}
Hence, “The set of squares of integers and the set of whole numbers are equal….. False statement.
(vii) n(P) = n(M)
It means number of elements of set P = Number of elements of set M.
∴ Sets P and M are equivalent.
∴ “If n(P) =⊇ n(M), then P ↔️ M” …… is a true Statement.
(viii) Set P = Set M
It means sets P and M are equal. Equal sets are equivalent also.
∴ Number of elements of set P = Number of elements of set M
∴ “ If set P = set M, then n(P) = n(M)”…….is a True statement.
(ix) n(A) = n(B)
⇒ Number of elements of set A = Number of elements of set B
∴ Given sets are equivalent but not equal.
∴ “ n(A) = n(B) ⇒ A = B” is a False statement.
### Exercise 6 C
1. Find all the subsets of each of the following sets:
(i) A= {5, 7}
(ii) B = {a, b, c}
(iii) C = {x : x ∈ W, x ≤2}
(iv) {p : p is a letter in the word ‘poor’}
Solution
(i) A = {5, 7}
Subsets of set A = { }, {5}, {7}, {5, 7}
(ii) B = {a, b, c}
Subsets of set B = { }, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}
(iii) C = {x : x ∈ W, x ≤ 2}
= {0, 1, 2}
∴ Subsets of set C = ɸ, {0}, {1}, {2}, {0, 1}, {0, 2}, {1, 2}, {0, 1, 2}
(iv) {P : P is a letter in the word ‘POOR’} = {p, o, r}
∴ Subsets of the given set = ɸ, {p}, {o}, {r}, {p, o}, {p, r}, {o, r}, {p, o, r}
2. If C is the set of letters in the word “cooler”, find:
(i) Set C
(ii) n(C)
(iii) Number of its subsets
(iv) Number of its proper subsets.
Note : (i) If a set has n elements, the number of its subsets = 2n
(ii) If a set has n elements, the number of its proper subsets = 2n - 1
Solution
(i) C = {c, o, l, e, r}
(ii) n(C) = 5
(iii) Number of its subsets : 25 = 2 × 2 × 2 × 2 × 2 = 32
(iv) Number of its proper subsets = 25 – 1 = 32 – 1 = 31
3. If T = {x: x is a letter in the word ‘TEETH’}, find all its subsets.
Solution
T = {t, e, h}
Subsets of set T = ɸ, {r}, {e}, {h}, {t, e}, {t, h}, {e, h}, {t, e, h}
4. Given the universal set = {-7, -3, -1, 0, 5, 6, 8, 9}, find:
(i) A = {x : x < 2}
(ii) B = {x : -4 < x < 6}
Solution
Universal set = {-7, -3, -1, 0, 5, 6, 8, 9},
(i) A = {x : x < 2} = {-7, -3, -1, 0}
(ii) B {x : -4 < x < 6} = {-3, -1, 0, 5}
5. Given the universal set = {x : x ∈ N and x < 20}, find:
(i) A = {x : x = 3p; p ∈ N}
(ii) B = {y : y – 2n + 3, n ∈ N}
(iii) C = {x : x is divisible by 4}
Solution
Universal set = {x : x ∈ N and x < 20}
= {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, …, 19}
(i) A = {x : x = 3p; p ∈ N}
x = 3p
when p = 1, x = 3 × 1 = 3
when p = 2, x = 3 × 2 = 6
when p = 3, x = 3 × 3 = 9
when p = 4, x = 3 × 4 = 12
when p = 5, x = 3 × 5 = 15
when p = 6, x = 3 × 6 = 18
∴ A = {3, 6, 9, 12, 15, 18)
(ii) B = {y : y = 2n + 3, n ∈ N}
y = 2n + 3
6. Find the proper subsets of {x : x2 – 9x – 10 = 0}
Solution
x2 - 9x – 10 = 0
⇒ x2 – 10x + x – 10 = 0
⇒ x(x – 10) + 1(x – 10) = 0
⇒ (x – 10)(x + 1) = 0
∴ Either x – 10 =0 or x + 1 = 0
x = 10
⇒ x = -1
Given set = {-1, 10}
Proper subsets of this set = ɸ, {-1}, {10}
7. Given, A = {Triangles}, B = {Isosceles triangles}, C = {Equilateral triangles}.
State whether the following are true or false. Give reasons.
(i) A ⊂ B
(ii) B ⊆ A
(iii) C ⊆ B
(iv) B ⊂A
(v) C ⊂ A
(vi) C ⊆ B ⊆ A
Solution
A = {Triangles}
B = {Isosceles triangles}
C = {Equilateral triangles}
(i) False
Since each triangle is not isosceles.
∴ A ⊂ B,
(ii) True
B ⊆ A
∵ Isosceles Δ is one of the triangles.
(iii) True
Since each equilateral triangle is isosceles also,
∴ C ⊆ B
(iv) True
B ⊂ A
∵ Isosceles Δ is one of the triangles.
(v) True
C ⊂ A
∵ Equilateral Δ is one of the triangles.
(vi) True
C ⊆ B ⊆ A
∵ Each equilateral triangle is isosceles also and each isosceles Δ is a form of triangles.
8. Given, A = {Quadrilaterals}, B = {Rectangles}, C = {Squares}, D = {rhombuses}. State, giving reasons, whether the following are true or false.
(i) B ⊂ C
(ii) D ⊂ B
(iii) C ⊆ B ⊆ A
(iv) D ⊂ A
(v) B ⊇ C
(vi) A ⊇ ⊇ D
Solution
A = {Quadrilaterals}
B = {Rectangles}
C = {Square}
D = {Rhombuses}
(i) False
B ⊂ C
∵ Rectangle is not a square also.
(ii) False
D ⊂ B
∵ Rhombus is not a rectangle also.
(iii) True
C ⊆ B ⊆ A
∵ Every square is a rectangle also and every rectangle is a quadrilateral also
(iv) True
D ⊂ A
∵ Rhombus is one of the quadrilaterals.
(v) True
⊇ C
∵ Square is a rectangle also.
(vi) False
⊇ ⊇ D
∵ Rhombus is not a rectangle also.
9. Given, universal set = {x : x ∈ N, 10 ≤ x ≤ 35}.
A = {x ∈ N: x ≤ 16} and
B = {x : x > 29} Find:
(i) A’
(ii) B’
Solution
Universal set = {x : x ∈ N, 10 ≤ x ≤ 35}
= {10, 11, 12, 13, 14, 15, ….34, 35}
A = {x ∈ N : x ≤ 16}
= {10, 11, 12, 13, 14, 15, 16}
B = {x : x > 29}
= {30, 31, 32, 33, 34, 35}
(i) A’ = {17, 18, 19, 20, 21, 22,….,33, 34, 35}
= {x : x ∈ N; 17 ≤ x ≤ 35}
(ii) B’ = {10, 11, 12, 13, 14, 15, …., 29}
= {x : x ≤ 29}
10. Given universal set = {x ∈ Z : -6 < x ≤ 6}, N = {n : n is a non – negative number} and
P = {x : x is a non-positive number}
Find : (i) N’
(ii) P’
Solution
Universal set = {x ∈ Z; - 6 < x ≤ 6}
= {-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6}
N = {n : n is a non-negative number}
= {0, 1, 2, 3, 4, 5, 6}
P = {x : x is a non-positive number}
= {-5, -4, -3, -2, -1, 0}
(i) N' = {-5, -4, -3, -2, -1}
(ii) P’ = {1, 2, 3, 4, 5, 6}
11. Let M = {letters of the word REAL} and N = {letters of the word LARE}. Write sets M and N in roster form and then state whether;
(i) M ⊆ N is true
(ii) N ⊆ M is true.
(iii) M = N is true.
Solution
M = {letters of the word REAL]
= {R, E, A, L}
And N = {letters of the word LARE}
= {L, A, R, E}
(i) M ⊆ N is true: yes
(ii) N ⊆ M is true.
(iii) M = N is true.
12. Write two sets A and B such that A ⊆ B and B ⊆ A. State the relationship between set A and B.
Solution
Let A = {letters of TALE}
B = {Letters of LATE}
Here A ⊆ B, B ⊆ A
∴ A = B
### Exercise 6 D
1. Given A = {x : x ∈ N and 3 < x ≤ 6} and B = {x : ∈ W and x < 4}. Find :
(i) Sets A and B in roster form
(ii) A ⋃ B
(iii) A ⋂ B.
(iv) A – B
(v) B – A
Solution
(i) A = (4, 5, 6)
B = (0, 1, 2, 3)
(ii) A⋃ B = {0, 1, 2, 3, 4, 5, 6}
(iii) A ⋂ B = (ɸ)
(iv) A – B = (4, 5, 6)
(v) B – A = (0, 1, 2, 3)
2. If P = {x : x ∈ W and 4 ≤ x ≤ 8}, and Q = { x : x ∈ N and x < 6}. Find:
(i) P⋃ Q and P ⋂ Q.
(ii) Is (P ⋃ Q) ⊃ (P ⋂ Q)?
Solution
(i) P = (4, 5, 6, 7, 8)
Q = (1, 2, 3, 4, 5)
P ⋃ Q = (1, 2, 3, 4, 5, 6, 7, 8)
P ⋂ Q = (4, 5)
(ii) Yes, all the element of set P ⋃ Q are contained in the set P ⋂ Q. Therefore, P ⋃ Q is a proper sunset of P ⋂ Q.
3. If A = {5, 6, 7, 8, 9}, B = {x : 3 < x < 8 and x ∈ W) and C = {x : x ≤ 5 and x ∈ N}. Find:
(i) A ⋃ B and (A ⋃ B) ⋃ C
(ii) B ⋃ C and A ⋃ (B ⋃ C)
(iii) A ⋂ B and (A ⋂ B) ⋂ C
(iv) B ⋂ C and A ⋂ (B ⋂ C)
Is (A ⋃ B) ⋃ C = A ⋃ (B ⋃ C)?
Is (A ⋂ B) ⋂ C = A ⋂ (B ⋂ C)?
Solution
A = (5, 6, 7, 8, 9)
B = (4, 5, 6, 7)
C = (1, 2, 3, 4, 5)
(i) A ⋃ B = (4, 5, 6, 7, 8, 9)
(A ⋃ B) ⋃ C = (1, 2, 3, 4, 5, 6, 7, 8, 9)
(ii) B ⋃ C (1, 2, 3, 4, 5, 6, 7)
A ⋃ (B ⋃ C) = (1, 2, 3, 4, 5, 6, 7, 8, 9)
(iii) A ⋂ B = (5, 6, 7)
(A ⋂ B) ⋂ C = (5)
(iv) B ⋂ C = (4, 5)
A ⋂ (B ⋂ C) = (5)
(v) (A ⋃ B) ⋃ C = (1, 2, 3, 4, 5, 6, 7, 8, 9)
A ⋃ (B ⋃ C) = (1, 2, 3, 4, 5, 6, 7, 8, 9)
Yes, these are equal.
(vi) (A ⋂ B) ⋂ C = A ⋂ (B ⋂ C)
{5} = {5}
Yes, these are equal.
4. Given A = {0, 1, 2, 4, 5}
B = {0, 2, 4, 6, 8} and C = {0, 3, 6, 9}. Show that
(i) A ⋃ (B ⋃ C) = (A ⋃ B) ⋃ C i. e., the union of sets is associative.
(ii) A ⋂ (B ⋂ C) = (A ⋂ B) ⋂ C i.e., the intersection of sets is associative.
Solution
A = {0, 1, 2, 4, 5}
B = {0, 2, 4, 6, 8}
C = {0, 3, 6, 9}
(i) B ⋃ C = {0, 2, 4, 6, 8} ⋃ {0, 3, 6, 9}
= {0, 2, 3, 4, 6, 8, 9}
∴ A ⋃ (B ⋃ C) = {0, 1, 2, 4, 5} ⋃ {0, 2, 3, 4, 6, 8, 9}
⇒ A ⋃ (B ⋃ C) = {0, 1, 2, 3, 4, 5, 6, 8, 9} ...(i)
A ⋃ B = {0, 1, 2, 4, 5} ⋃ {0, 2, 4, 6, 8}
= {0, 1, 2, 4, 5, 6, 8}
∴ (A ⋃ B) ⋃ C = {0, 1, 2, 4, 5, 6, 8} ⋃ {0, 3, 6, 9}
⇒ (A ⋃ B) ⋃ C = {0, 1, 2, 3, 4, 5, 6, 8, 9} ...(ii)
From I and II, we get
A ⋃ (B ⋃ C) = A ⋃ (B ⋃ C)
(ii) B ⋂ C = {0, 2, 4, 6, 8} ⋂ {0, 3, 6, 9}
= {0, 6}
Now, A ⋂ (B ⋂ C) = {1, 2, 4, 5} ⋂ {0, 6}
⇒ A ⋂ (B ⋂ C) = {0} …(i)
A ⋂ B = {0, 1, 2, 4, 5} ⋂ {0, 2, 4, 6, 8}
= {0, 2, 4}
∴ (A ⋂ B) ⋂ C = {0, 2, 4} ⋂ {0, 3, 6, 9}
⇒ A ⋂ (B ⋂ C) = {0} …(ii)
From I and II we get
A ⋂ (B ⋂ C) = (A ⋂ B) ⋂ C
5. If A = {x ∈ W : 5 < x < 10}, B = {3, 4, 5, 6, 7} and C = {x = 2n ; x ∈ N and n ≤ 4}. Find:
(i) A ⋂ (B ⋃ C)
(ii) (B ⋃ A) ⋂ (B ⋃ C)
(iii) B ⋃ (A ⋂ C)
(iv) (A ⋂ B) ⋃ (A ⋂ C)
Name the sets which are equal.
Solution
A = {x ∈ W ; 5 < x < 10}
= {6, 7, 8, 9}
B = {3, 4, 5, 6, 7}
C = {x = 2n ; n ∈ N and n ≤ 4}
x = 2n
when n = 1, x = 2 × 1= 2
when n = 2, x = 2 × 2 = 4
when n = 3, x = 2 × 3 = 6
when n = 4, x = 2 × 4 = 8
∴ C = {2, 4, 6, 8}
(i) B ⋃ C = {3, 4, 5, 6, 7} ⋃ {2, 4, 6, 8}
= {2, 3, 4, 5, 6, 7, 8}
A ⋂ (B ⋃ C) = {6, 7, 8, 9} ⋂ {2, 3, 4, 5, 6, 7, 8}
⇒ A ⋂ (B ⋃ C) = {6, 7, 8}
(ii)
B ⋃ A = {3, 4, 5, 6, 7} ⋃ {6, 7, 8, 9}
= {3, 4, 5, 6, 7, 8, 9}
(B ⋃ A) ⋃ (B ⋃ C)
= {3, 4, 5, 6, 7, 8, 9} ⋂ {2, 3, 4, 5, 6, 7, 8}
= {3, 4, 5, 6, 7, 8}
(iii) (A ⋂ C) = {6, 7, 8, 9} ⋂ {2, 4, 6, 8}
= {6, 8}
B ⋃ (A ⋂ C) = {3, 4, 5, 6, 7} ⋃ {6, 8}
= {3, 4, 5, 6, 7, 8}
(iv) (A ⋃ B) = {6, 7, 8, 9} ⋂ {3, 4, 5, 6, 7}
= {6, 7}
∴ (A ⋂ B) ⋃ (A ⋂ C) = {6, 7} ⋃ {6, 8}
= {6, 7, 8}
6. If P = {factors of 36} and Q = {factors of 48}; find:
(i) P ⋃ Q
(ii) P ⋂ Q
(iii) Q – P
(iv) P’ ⋂ Q
Solution
1 × 36 = 36, 1 × 48 = 48
2 × 18 = 36, 2 × 24 = 48
3 × 12 = 36, 3 × 16 = 48
4 × 9 = 36, 4 × 12 = 48
6 × 6 = 36, 6 × 8 = 48
∴ Factors of 36 = 1, 2, 3, 4, 6, 9, 12, 18, 36
Factors of 48 = 1, 2, 3, 4, 6, 8, 12, 16, 24, 48
P = {factors of 36}
= {1,2, 3, 4, 6, 9, 12, 18, 36}
Q = {factors of 48}
= {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
(i) P ⋃ Q = {1, 2 , 3, 4, 6, 9, 12, 18, 36} ⋃ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
= {1, 2, 3, 4, 6, 8, 9,12, 16, 18, 24, 36, 48}
(ii) P ⋂ Q = {1, 2, 3, 4, 6, 9, 12, 18, 36} ⋂ {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
= {1, 2, 3, 4, 6, 12}
(iii) Q – P = {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} – {1, 2, 3, 4, 6, 9, 12, 18, 36}
= {8, 16, 24, 48}
(iv) P’ ⋂ Q = Only Q
= Q – P
= {1, 2, 3, 4, 6, 8, 12, 16, 24, 48} – {1, 2, 3, 4, 6, 9, 12, 18, 36}
= {8, 16, 24, 48}
7. If A = {6, 7, 8, 9}, B = {4, 6, 8, 10} and C = {x : x ∈ N : 2 < x ≤ 7} find
(i) A – B
(ii) B – C
(iii) B – (A – C)
(iv) A – (B ⋃ C)
(v) B – (A ⋂ C)
(vi) B – B.
Solution
A = [6, 7, 8, 9}
B = {4, 6, 8, 10}
C = {x : x ∈ N : 2 < x ≤ 7}
= {3, 4, 5, 6, 7}
(i) A - B = {6, 7, 8, 9} – {4, 6, 8, 10}
= {7, 9}
(ii) B – C = {4, 6, 8, 10} – {3, 4, 5, 6, 7}
= {8, 10}
(iii) A – C = {6, 7, 8, 9} - {3, 4, 5, 6, 7}
= {8, 9}
B – (A – C) = {4, 6, 8, 10} – {8, 9}
= {4, 6, 10}
(iv) B ⋃ C = {4, 6, 8, 10} ⋃ {3, 4, 5, 6, 7}
= {3, 4, 5, 6, 7, 8, 10}
A – (B ⋃ C) = {6, 7, 8, 9} – {3, 4, 5, 6, 7, 8, 10}
= {9}
(v) A ⋂ C = {6, 7, 8, 9} ⋂ (3, 4, 5, 6, 7}
= {6, 7}
B – (A ⋂ C) = {4, 6, 8, 10} – {6, 7}
= {4, 8, 10}
(vi) B – B = {4, 6, 8, 10} – {4, 6, 8, 10}
= ɸ
8. If A = {1, 2, 3, 4, 5}
B = {2, 4, 6, 8}
and C = {3, 4, 5, 6}
Verify:
(i) A – (B ⋃ C) = (A – B) ⋂ (A – C)
(ii) A – (B ⋂ C) = (A – B) ⋃ (A – C)
Solution
A = {1, 2, 3, 4, 5}
B = {2, 4, 6, 8}
C = {3, 4, 5, 6}
(i) B ⋃ C = {2, 4, 6, 8} ⋃ {3, 4, 5, 6}
= {2, 3, 4, 5, 6, 8}
A – (B ⋃ C) = {1, 2, 3, 4, 5} – {2, 3, 4, 5, 6, 8}
= {1}
A – B = {1, 2, 3, 4, 5} – {2, 4, 6, 8}
= {1, 3, 5}
A – C = {1, 2, 3, 4, 5} – {3, 4, 5, 6}
= {1, 2}
∴ (A – B) ⋂ (A – C) = {1, 3, 5} – {1, 2} = {1}
∴ A – (B ⋃ C) = (A – B) ⋂ (A – C)
(ii) B ⋂ C = {2, 4, 6, 8} ⋂ {3, 4, 5, 6}
= {4, 6}
A – (B ⋂ C} = {1, 2, 3, 4, 5} – {4, 6}
= {1, 2, 3, 5}
A – B = {1, 2, 3, 4, 5} – {2, 4, 6, 8}
= {1, 3, 5}
A – C = {1, 2, 3, 4, 5} – {3, 4, 5, 6}
= {1, 2}
(A – B) ⋃ (A – C) = {1, 3, 5} ⋃ {1, 2}
= {1, 2, 3, 5}
∴ A – (B ⋂ C) = (A – B) ⋃ (A – C)
9. Given A = {x ∈ N : x < 6}, B = {3, 6, 9} and C = {x ∈ N : 2x – 5 ≤ 8}.
Show that:
(i) A ⋃ (B ⋂ C) = (A ⋃ B) ⋂ (A ⋃ C)
(ii) A ⋂ (B ⋃ C) = (A ⋂ B) ⋃ (A ⋂ C)
Solution
A = {x ∈ N : x < 6}
B = {1, 2, 3, 4, 5}
C = {x ∈ N : 2x -5 ≤ 8}
2x – 5 ≤ 8
⇒ 2x ≤ 8 + 5
⇒ 2x ≤ 13
⇒ x ≤ 13/2
⇒ x ≤ 6.5
∴ C = {1, 2, 3, 4, 5, 6}
(i) B ⋂ C = {3, 6, 9} ⋂ {1, 2, 3, 4, 5, 6}
= {3, 6}
∴ A ⋃ (B ⋂ C) = {1, 2, 3, 4, 5} ⋃ {3, 6}
= {1, 2, 3, 4, 5, 6}
A ⋃ B = {1, 2, 3, 4, 5} ⋃ {3, 6, 9}
= {1, 2, 3, 4, 5, 6, 9}
A ⋃ C = {1, 2, 3, 4, 5} ⋃ {1, 2, 3, 4, 5, 6}
= {1, 2, 3, 4, 5, 6}
∴ (A ⋃ B) ⋂ (A ⋃ C)
= {1, 2, 3, 4, 5, 6, 9} ⋂ {1, 2, 3, 4, 5, 6}
= {1, 2, 3, 4, 5, 6}
∴ A ⋃ (B ⋂ C) = (A ⋃ B) ⋂ (A ⋃ C)
(ii) B ⋃ C = {3, 6, 9} ⋃ {1, 2, 3, 4, 5, 6}
= {1, 2, 3, 4, 5, 6, 9}
A ⋂ (B ⋃ C) = {1, 2, 3, 4, 5} ⋂ {1, 2, 3, 4, 5, 6, 9}
= {1, 2, 3, 4, 5}
Now A ⋂ C = {1, 2, 3, 4, 5} ⋂ {1, 2, 3, 4, 5, 6}
= {1, 2, 3, 4, 5}
A ⋂ B = {1, 2, 3, 4, 5} ⋂ {3, 6, 9}
= {3}
∴ (A ⋂ B) ⋃ (A ⋂ C) = {3} ⋃ {1, 2, 3, 4, 5}
= {1, 2, 3, 4, 5}
∴ A ⋂ (B ⋃ C) = (A ⋂ B) ⋃ (A ⋂ C)
### Exercise 6 E
1. From the given diagram find:
(i) A ⋃ B
(ii) A’ ⋂ B
(iii) A – B
(iv) B – A
(v) (A ⋃ B)’
Solution
(i) A ⋃ B= {a, c, d, e} ⋃ {b, c, e, f}
⇒ A ⋃ B = {a, b, c, d, e, f}
(ii) A’ = {b, f, g, h}
A’ ⋂ B = {b, f, g, h} ⋂ {b, c, e, f}
⇒ A’ ⋂ B = {b, f}
(iii) A - B = {a, c, d, e} – {b, c, e, f}
⇒ A – B = {a, d}
(iv) B – A = {b, c, e, f} – {a, c, d, e}
= {b, f}
(v) A ⋃ B = {a, b, c, d, e, f}
∴ (A ⋃ B)’ = {h, g}
2. From the given diagram, find:
(i) A’
(ii) B’
(iii) A’ ⋃ B’
(iv) (A ⋃ B)’
Is A’ ⋃ B’ = (A ⋂ B)’?
Also, verify if A’ ⋂ B’ = (A ⋃ B)’.
Solution
(i) A = {1, 3, 4, 6}
∴ A’ = {2, 5, 7, 8, 9, 10}
(ii) B = {1, 2, 5}
∴ B’ = {3, 4, 6, 7, 8, 9, 10}
(iii) A’ ⋃ B’ = {2, 5, 7, 8, 9, 10} ⋃ {3, 4, 6, 7, 8, 9, 10}
= {2, 3, 4, 5, 6, 7, 8, 9, 10}
(iv) A ⋂ B = {1, 3, 4, 6} ⋂ {1, 2, 5}
= {1}
∴ (A ⋂ B)’ = {2, 3, 4, 5, 6, 7, 8, 9, 10}
From part (iii) and part (iv) we conclude
A’ ⋃ B’ = (A ⋂ B)’
Now A ⋂ B = {2, 5, 7, 8, 9, 10} ⋂ {3, 4, 6, 7, 8, 9, 10}
⇒ A’ ⋂ B’ = {7, 8, 9, 10} ...(i)
Now A ⋃ B = {1, 3, 4, 6} ⋃ {1, 2, 5}
= {1, 2, 3 4, 5, 6}
∴ (A ⋃ B’) = {7, 8, 9, 10} …(ii)
From I and II we conclude
A’ ⋃ B’ = (A ⋂ B)’
3. Use the given diagram to find:
(i) A ⋃ (B ⋂ C)
(ii) B – (A – C)
(iii) A – B.
(iv) A ⋂ B’
Is A ⋂ B’ = A – B?
Solution
(i) B ⋂ C = {d, e, f, g, h, j} ⋂ {h, i, j, k, l}
= {h, j}
∴ A ⋃ (B ⋂ C) = {a, b, c, d, g, h, i} ⋃ {h, j}
= {a, b, c, d, g, h, I, j}
(ii) A – C = {a, b, c, d, g, h, i} – {h, i, j, k, l}
= {a, b, c, d, g}
∴ B – (A – C) = {d, e, f, g, h, j} – {a, b, c, d, g}
= {e, f, h, j}
(iii) A – B = {a, b, c, d, g, h, i} – {d, e, f, g, h, i}
⇒ A – B = {a, b, c, i}
(iv) B’ = {a, b, c, i, k, l, m, n, p}
A ⋂ B’= {a, b, c, d, g, h, i} ⋂ {a, b, c, I, k, l, m, n, p}
⇒ A ⋂ B’ = {a, b, c, i} …..II
From I and II we can conclude A ⋂ B’ = A – B
4. Use the given Venn – diagram to find:
(i) B – A
(ii) A
(iii) B’
(iv) A ⋂ B
(v) A ⋃ B
Solution
(i) B – A = {1, 5} – {1, 5, 6, 7, 9}
= { }
(ii) A = {1, 5, 6, 7, 9}
(iii) B = {1, 5}
∴ B’ = {2, 3, 4, 6, 7, 8, 9, 10}
(iv) A ⋃ B = {1, 5, 6, 7, 9} ⋃ {1, 5}
= {1, 5, 6, 7, 9}
5. Draw a Venn – diagram to show the relationship between two overlapping sets A and B. Now shade the region representing:
(i) A ⋂ B
(ii) A ⋃ B
(iii) B – A
Solution
(i) A ⋂ B =
(ii) A ⋃ B =
(iii) B – A =
6. Draw a Venn –diagram to show the relationship between two sets A and B; such that A ⊆ B, Now shade the region representing:
(i) A ⋃ B
(ii) B’ ⋂ A
(iii) A ⋂ B
(iv) (A ⋃ B)’
Solution
(i) A ⋃ B =
(ii) B’ ⋂ A =
(iii) A ⋂ B =
(iv) (A ⋃ B)’
7. Two set A and B are are such that A ⋂ B = ɸ. Draw a venn-diagram to show the relationship between A and B. Shade the region representing:
(i) A ⋃ B
(ii) (A ⋃ B)’
(iii) B - A
(iv) B ⋂ A’
Solution
(i) A ⋃ B =
(ii) (A ⋃ B)’ =
(iii) B – A
(iv) B ⋂ A’ =
8. State the sets represented by the shaded portion of following venn-diagrams:
(i)
(ii)
(iii)
Solution
(i) (A ⋃ B)’
(ii) B – A or A’ ⋃ B
(iii) (B - A)’
9. In each of the given diagrams, shade the region which represents the set given underneath the diagram:
(i)
(ii)
(iii)
Solution
(i) (B – A) =
(ii) (A ⋂ B)’ =
(iii) (P ⋃ Q)’ =
10. From the given diagram, find:
(i) (A ⋃ B) – C
(ii) B – (A ⋂ C)
(iii) (B ⋂ C) ⋃ A
Verify:
A – (B ⋂ C) = (A – B) ⋃ (A – C)
Solution
(i) A ⋃ B = {a, b, c, d} ⋃ {c, d, e, g}
= {a, b, c, d, e, g}
∴ (A ⋃ B) – C = {a, b, c, d, e, g} – {b, c, e, f}
= {a, d, g}
(ii) A ⋂ C = {a, b, c, d} ⋂ {b, c, e, f}
= {b, c}
∴ B – (A ⋂ C) = {c, d, e, g} – {b, c}
= {d, e, g}
(iii) B ⋂ C = {c, e, d, g} ⋂ {b, c, e, f}
= {c, e}
11. Using the given diagram, express the following sets in the terms of A and B.
(i) {a, d}
(ii) {a, d, c, f}
(iii) {a, d, c, f, g, h}
(iv) {a, d, g, h}
(v) {g, h}
Solution
(i) {a, d} = {a, b, e, d} – {b, c, e, f}
= A – B
(ii) {a, d, c, f} = (A ⋃ B) – {b, e}
= (A ⋃ B) – (A ⋂ B)
Also {a, d, c, f} – (A – B) ⋃ (B – A)
(iii) {a, d, c, f, g, h} = (A ⋂ B)’
[∵ {b, e} = A ⋂ B
∴ (A ⋂ B)’ = {a, d, c, f, g, h}]
(iv) {a, d, g, h} = B’
[∵ {b, c, e, f} = B
∴ B’ = {a, d, g, h}]
(v) {g, h} = (A ⋃ B)’
[∴ A ⋃ B = {a, b, c, d, e, f}
∴ (A ⋃ B)’ = {g, h} |
Checkout JEE MAINS 2022 Question Paper Analysis : Checkout JEE MAINS 2022 Question Paper Analysis :
# Random Experiments
We may perform various activities in our daily existence, sometimes repeating the same actions though we get the same result every time. Suppose, in mathematics, we can directly say that the sum of all interior angles of a given quadrilateral is 360 degrees, even if we don’t know the type of quadrilateral and the measure of each internal angle. Also, we might perform several experimental activities, where the result may or may not be the same even when they are repeated under the same conditions. For example, when we toss a coin, it may turn up a tail or a head, but we are unsure which results will be obtained. These types of experiments are called random experiments.
## Random Experiment in Probability
An activity that produces a result or an outcome is called an experiment. It is an element of uncertainty as to which one of these occurs when we perform an activity or experiment. Usually, we may get a different number of outcomes from an experiment. However, when an experiment satisfies the following two conditions, it is called a random experiment.
(i) It has more than one possible outcome.
(ii) It is not possible to predict the outcome in advance.
Let’s have a look at the terms involved in random experiments which we use frequently in probability theory. Also, these terms are used to describe whether an experiment is random or not.
Terms Meaning Outcome A possible result of a random experiment is called its outcome. Example: In an experiment of throwing a die, the outcomes are 1, 2, 3, 4, 5, or 6 Sample space The set of all possible outcomes of a random experiment is called the sample space connected with that experiment and is denoted by the symbol S. Example: In an experiment of throwing a die, sample space is S = {1, 2, 3, 4, 5, 6} Sample point Each element of the sample space is called a sample point. Or Each outcome of the random experiment is also called a sample point.
## What is a Random Experiment?
Based on the definition of random experiment we can identify whether the given experiment is random or not. Go through the examples to understand what is a random experiment and what is not a random experiment.
Example 1:
Is picking a card from a well-shuffled deck of cards a random experiment?
Solution:
We know that a deck contains 52 cards, and each of these cards has an equal chance to be selected.
(i) The experiment can be repeated since we can shuffle the deck of cards every time before picking a card and there are 52 possible outcomes.
(ii) It is possible to pick any of the 52 cards, and hence the outcome is not predictable before.
Thus, the given activity satisfies the two conditions of being a random experiment.
Hence, this is a random experiment.
Example 2:
Consider the experiment of dividing 36 by 4 using a calculator. Check whether it is a random experiment or not.
Solution:
(i) This activity can be repeated under identical conditions though it has only one possible result.
(ii) The outcome is always 9, which means we can predict the outcome each time we repeat the operation.
Hence, the given activity is not a random experiment.
### Examples of Random Experiments
Below are the examples of random experiments and the corresponding sample space.
1. Tossing a coin three times
Number of possible outcomes = 8
Sample space = S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
2. Three coins are tossed simultaneously
Number of possible outcomes = 8
Sample space = S = {HHH, HHT, HTH, THH, TTH, THT, HTT, TTT}
3. Rolling a pair of dice simultaneously
Number of possible outcomes = 36
Sample space = S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
4. Throwing a die two times
Number of possible outcomes = 36
Sample space = S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}
5. Selecting a card from an urn containing 100 cards numbering from 1 to 100
Number of possible outcomes = 100
Sample space = S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10,….., 51, 52, 53, 54, 55, …., 91, 92, 93, 94, 95, 96, 97, 98, 99, 100}
6. Choosing one of the factors of 180
Number of possible outcomes = 18
Sample space = S = {1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180}
Similarly, we can write several examples which can be treated as random experiments.
### Playing Cards
Probability theory is the systematic consideration of outcomes of a random experiment. As defined above, some of the experiments include rolling a die, tossing coins, and so on. There is another experiment of playing cards. Here, a deck of cards is considered as the sample space. For example, picking a black card from a well-shuffled deck is also considered an event of the experiment, where shuffling cards is treater as the experiment of probability.
A deck contains 52 cards, 26 are black, and 16 are red.
However, these playing cards are classified into 4 suits, namely Spades, Hearts, Diamonds, and Clubs. Each of these four suits contains 13 cards.
We can also classify the playing cards into 3 categories as:
Aces: A deck contains 4 Aces, of which 1 of every suit.
Face cards: Kings, Queens, and Jacks in all four suits, also known as court cards.
Number cards: All cards from 2 to 10 in any suit are called the number cards.
• Spades and Clubs are black cards, whereas Hearts and Diamonds are red.
• 13 cards of each suit = 1 Ace + 3 face cards + 9 number cards
• The probability of drawing any card will always lie between 0 and 1.
• The number of spades, hearts, diamonds, and clubs is the same in every pack of 52 playing cards.
An example problem on picking a card from a deck is given above. |
Suggested languages for you:
Americas
Europe
|
|
# Sine and Cosine Rules
We can extend the ideas from trigonometry and the triangle rules for right-angled triangles to non-right angled triangles. We'll look at the two rules called the sine and cosine rules. We can use these rules to find unknown angles or lengths of non-right angled triangles.
## Labelling a triangle
As we see below, whenever we label a triangle, we label sides with lowercase letters and angles with uppercase letters. Opposite angles and sides are labelled with the same letter.
Example of a labelled triangle, Tom Maloy, StudySmarter Originals
## The sine rule
For a triangle with the form above, the sine rule formula is defined as:
We can also write this as:
We can interpret the sine rule like this: the ratio between the length of the side and the opposite angle is constant in any triangle. We use the sine rule when two angles and two lengths are involved. There are two situations where we will use the sine rule:
• When there are two angles and one side given, and we need to find the length of another side.
• When there are two lengths and one angle given, and we need to find another angle.
#### Example 1
Q: Find x.
A: Using the sine rule, we know that , and we can rearrange for x to get , which gives x = 23.6 (3 sf).
#### Example 2
Q: Find y.
A: Using the sine rule, we know that , which rearranges to give , thus giving y = = 54.5 ° (3.sf).
## The cosine rule
Using the same example, the cosine rule formula is defined as: a² = b² + c²-2bc · cos (A)
If we rearrange this to give A, we get: We use the cosine rule when three lengths and one angle are involved. Questions will either give three sides and we need to find an angle or give two sides and an angle, and we will need to find the length of the side.
#### Example 1
Q: Find x.
A: Using the cosine rule, we get that x² = 15² + 19² -2 · 15 · 19 · cos (40) = 149.354667 ..., which then gives x = √149.354667 ... = 12.2 (3.sf)
#### Example 2
Q: Find y
A: Using the rearranged cosine rule, we get y = = 27.7 (3.sf)
## How are the sine and cosine rules derived?
Now that we have seen what each rule is and how they work, we will look at how we get to each of them by deriving them from first principles. These may look complicated at first, but we will be using some trigonometry and Pythagoras' theorem.
### Derivation of the sine rule
Let us start with a triangle as above, but draw a line down from the top angle so that we now have two right-angled triangles, and call this line h, as shown below.
Using trigonometry, we have sin (A) = h / c , and sin (C) = h / a . We can now rearrange these for h to get h = c sin (A) and h = a sin (C) . Set these equal to each other to get c sin (A) = a sin (C) . These now rearrange to , or equivalently . We can then repeat this method with the other two angles to get the results. We then get to our desired result of
### Derivation of the cosine rule
We will now do the same with the cosine rule. We start with the same triangle, draw the same line down to create two right-angled triangles, and call this line h. We call the point that this line touches the bottom D and state that one side of the line has length x, and the other , as shown below.
By Pythagoras' theorem on the left triangle, we get x² + h² = c², and we will rearrange this to
x² = c²-h² ......... (1) Doing the same to the right triangle, we get , and we will expand this to b²-2bx + x² + h² = a² ....... ... (2) If we take the cosine of angle A, we get cos (A) = x / c, which rearranges to
.......... (3). We will now substitute (1) and (3) into (2), using (1) to get rid of the x², and (3) to replace the x in 2bx. Substituting in: b² -2b (c · cos (A)) + c²-h² + h² = a². We can then simplify this to a² = b² + c²-2bc · cos (A), which is our desired result.
## Sine and Cosine Rules - Key takeaways
• We use the sine and cosine rules when working out sides and angles on non-right-angled triangles.
• We use the sine rule when we have one unknown value and three known values from two angles and two sides.
• We use the cosine rule when we have one unknown value and three known values from one angle and three sides.
• Both the sine and cosine rules can be derived from first principles.
We use the sine and cosine rules when working out sides and angles on non-right angled triangles. We use the sine rule when we have one unknown value and three known values from two angles and two sides. We use the cosine rule when we have one unknown value and three known values from one angle and three sides.
The definitions of sine and cosine come from a right angle triangle. For any angle in a right-angled triangle, the sine of the angle is the opposite side length divided by the length of the hypotenuse, and the cosine of the angle is the adjacent side divided by the length of the hypotenuse.
For a non-right angled triangle, with uppercase letters denoting angles, and lowercase letters denoting sides, with A opposite a, and so forth, we have a²=b²+c²-2bc.cos(A)
For a non-right angled triangle, with uppercase letters denoting angles, and lowercase letters denoting sides, with A opposite a, and so forth, we have sin(A)/a=sin(B)/b=sin(C)/c
## Final Sine and Cosine Rules Quiz
Question
We have a triangle of side lengths 12, 13 and 4. Use the cosine rule to find the angle sizes in the triangle.
[Hint: Use the cosine rule three times]
66.8°, 17.8° and 95.4°
Show question
Question
A triangle has side lengths 10, 4 and x. The angle opposite x is 100°. Find the value of x.
11.4
Show question
Question
A triangle has side lengths 2, 8 and z. The angle opposite x is 18°. Find the value of x.
6.13
Show question
Question
True or false: The angle used in the cosine rule must be acute.
False
Show question
Question
A triangle has side lengths of 10, x and y, with angles of 30°, 85° and 65°. Use the sine rule to find the value of x and y. The side of length 10 is opposite the angle measuring 30°.
19.9 and 16.1
Show question
Question
A triangle has two sides of length 6 and 7, and the angle opposite the side with length 7 is 72°. Find the angle opposite the side of length 6.
54.6°
Show question
Question
A triangle has two angles measuring 67° and 33° respectively. The side opposite the angle measuring 67° has a length of 9. Find the length of the side opposite the angle measuring 33°.
5.33
Show question
Question
A triangle has two sides of length 13 and 7, and the angle opposite the side with length 13 is 144°. Find the angle opposite the side of length 7.
18.5°
Show question
Question
A triangle has two angles measuring 58° and 23° respectively. The side opposite the angle measuring 58° has a length of 15. Find the length of the side opposite the angle measuring 23°.
6.91
Show question
More about Sine and Cosine Rules
60%
of the users don't pass the Sine and Cosine Rules quiz! Will you pass the quiz?
Start Quiz
## Study Plan
Be perfectly prepared on time with an individual plan.
## Quizzes
Test your knowledge with gamified quizzes.
## Flashcards
Create and find flashcards in record time.
## Notes
Create beautiful notes faster than ever before.
## Study Sets
Have all your study materials in one place.
## Documents
Upload unlimited documents and save them online.
## Study Analytics
Identify your study strength and weaknesses.
## Weekly Goals
Set individual study goals and earn points reaching them.
## Smart Reminders
Stop procrastinating with our study reminders.
## Rewards
Earn points, unlock badges and level up while studying.
## Magic Marker
Create flashcards in notes completely automatically.
## Smart Formatting
Create the most beautiful study materials using our templates. |
# Angular Momentum - Rotation About Fixed Axis
## Introduction
Angular Momentum is a concept of rotational dynamics. We can classify motions of objects in two kinds of motions namely Translational and Rotational motions. In pure translatory motion every point of the body moves in a straight line. Each point covers the same distance in a given time interval. e.g. - A Car moving in a straight line, a man walking on the road etc. Whereas in the pure rotation, the path created by the motion of every point makes a circle whose centre lies on the axis of rotation. E.g. - earth rotating about its axis, spinning top etc.
For translational motion, in absence of external force we find the linear momentum is always conserved. We call it the principle of conservation of momentum. We used it to simplify many collision problems. One can ask, is there any such principle in Rotational motion also? The answer is-yes. In Rotational motion we have the principle of conservation of Angular momentum. Angular momentum is the angular counterpart of linear momentum. As we will see we would be able to relate it to other physical quantities like Moment of inertia, torque and linear momentum.
## Angular Momentum
Suppose a particle is rotating about an axis, then the moment linear momentum of the particle around that axis is called angular momentum of the particle.
i.e. angular momentum = linear momentum x perpendicular distance of particle
$$\mathrm{\overrightarrow{L}=\overrightarrow{p}\times \overrightarrow{r}}$$
A particle of mass m is moving with velocity v then its linear momentum L, Suppose its distance from axis of rotation is given by the radius vector $\mathrm{\overrightarrow{r}}$
### In Vector Notations Angular Momentum
$$\mathrm{\overrightarrow{L}= \overrightarrow{r}\times \overrightarrow{p}}$$
$$\mathrm{\lvert\overrightarrow{L} \rvert =rpsin\theta=mvr\:sin\theta}$$
Maschen, Angular momentum definition, marked as public domain, more details on Wikimedia Commons
The dimensions of angular momentum will be $\mathrm{[ML^2 T^{-1}]}$ and SI unit is given by $\mathrm{Kg\:m^2 s^{-1}}$.
## Angular momentum in Cartesian coordinate system
We can also write the angular momentum in cartesian coordinates explicitly. First, we write the $\mathrm{\overrightarrow{r}}$ and $\mathrm{\overrightarrow{p}}$ in cartesian forms.
$$\mathrm{\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}}$$
$$\mathrm{\overrightarrow{p}=p_x\hat{i}+p_y \hat{j}+p_z\hat{k}}$$
$$\mathrm{\overrightarrow{L}=(x\hat{i}+y \hat{j}+z\hat{k})\times (p_x\hat{i}+p_y\hat{j}+p_z\hat{k})}$$
$$\mathrm{\overrightarrow{L}_x=yp_z-zp_y}$$
$$\mathrm{\overrightarrow{L}_y=zp_x-xp_z}$$
$$\mathrm{\overrightarrow{L}_z=zp_y-yp_x}$$
## Derivation of Equation for Angular Momentum
First, we look at the analogous physical quantities of translational and rotational motion.
QuantityVelocityAccelerationForce MassMomentum
Translational$\mathrm{\overrightarrow{ u}}$
$\mathrm{\overrightarrow{a}}$$\mathrm{\overrightarrow{F}}$$\mathrm{m}$$\mathrm{\overrightarrow{p}} Rotational \mathrm{\overrightarrow{\omega}} (Angular velocity) \mathrm{\overrightarrow{a}} (Angular acceleration) \mathrm{\overrightarrow{\tau}}(torque) =\mathrm{\overrightarrow{r}\times \overrightarrow{F}}I(moment of inertia) = \mathrm{Σmr^2}$$\mathrm{\overrightarrow{L}}$(Angular momentum)
Table-1: Analogy between physical quantities of rotational and translational motion
We take a small particle of mass m having distance $\mathrm{\overrightarrow{r}}$ from the axis of rotation with linear momentum $\mathrm{\overrightarrow{p}}$.
From newton’s second law of motion, we have
$$\mathrm{\sum \overrightarrow{F}=\frac{d\overrightarrow{p}}{dt}}$$
We take cross product with r on both sides of this equation.
$$\mathrm{\overrightarrow{r}\times \sum \overrightarrow{F}=\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}}$$
We recognise that
$$\mathrm{\overrightarrow{r}\times \sum \overrightarrow{F}=\overrightarrow{\tau}}$$
Hence
$$\mathrm{\overrightarrow{\tau}=\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt} ………. (1)}$$
Now let us look at this term
$$\mathrm{\frac{d\overrightarrow{r}}{dt}\times \overrightarrow{p}}$$
since
$$\mathrm{\overrightarrow{p}=m\overrightarrow{v}\:and\:\frac{d\overrightarrow{r}}{dt}=\overrightarrow{v};}$$
$$\mathrm{\overrightarrow{v}\times \overrightarrow{p}=m(\overrightarrow{v}\times \overrightarrow{v})=0}$$
(Since for any vector $\mathrm{\overrightarrow{A}:\overrightarrow{A}\times \overrightarrow{A}=0}$)
Hence, we can add it on the both sides of equation (1)
$$\mathrm{\frac{d\overrightarrow{r}}{dt}\times \overrightarrow{p}+\overrightarrow{\tau}=\overrightarrow{r}\times \frac{d\overrightarrow{p}}{dt}+\frac{d\overrightarrow{r}}{dt}\times \overrightarrow{p}=\frac{d(\overrightarrow{r}\times \overrightarrow{p})}{dt}}$$
Hence
$$\mathrm{\overrightarrow{\tau}=\frac{d(\overrightarrow{r}\times \overrightarrow{p})}{dt}}$$
If we compare it with analogous force in translational motion
$$\mathrm{\sum F=\frac{dp}{dt}}$$
we can say that term $\mathrm{(\overrightarrow{r}\times \overrightarrow{p})}$ is some kind of momentum.
Hence, we define angular momentum
$$\mathrm{\overrightarrow{L}=(\overrightarrow{r}\times \overrightarrow{p})}$$
## Angular Momentum of Rigid Body about Fixed Axis
### Angular Momentum of System
Let, there is a system of n particles, and their individual angular momenta are given by $\mathrm{\overrightarrow{l_1},\overrightarrow{l_2},\overrightarrow{l_3},............\overrightarrow{l_n}}$ then then net angular momentum of system can be written as -
$$\mathrm{\overrightarrow{L}=\overrightarrow{l_1}+\overrightarrow{l_2}+\overrightarrow{l_3}.................+\overrightarrow{l_n}=\sum_{i}\overrightarrow{L_i}}$$
$$\mathrm{\frac{d}{dt}(\overrightarrow{L})=\sum \frac{d}{dt} \overrightarrow{l_i}=\sum_i \overrightarrow{\tau_{ext,i}}}$$
Hence net external torque on a system is equal to rate of change of Total angular momentum of system.
### For Rigids Bodies
We take the rigid body as a large system of small particles. Let the mass of each particle be $\mathrm{m_i}$ and linear velocity be $\mathrm{\overrightarrow{v_i}}$.Suppose the body is moving about the z axis with angular velocity $\mathrm{\overrightarrow{\omega_i}}$ and the particles are rotating around the z axis on the x-y plane.
Then the angular momentum of each particle
$$\mathrm{\overrightarrow{L_i}=m_i(\overrightarrow{r_i}\times \overrightarrow{v_i})}$$
As we can see in the figure, the angle between $\mathrm{\overrightarrow{r_i}}$ and $\mathrm{\overrightarrow{v_i}}$ is 90°.
$$\mathrm{\Rightarrow\:\:\overrightarrow{L _i}=m_i(\overrightarrow{r_i}\times \overrightarrow{ u_i})= m_i\:r_i\:v_i\: sin90=m_i\:r_i\:v_i}$$
Thus
$$\mathrm{\overrightarrow{L _z}=\sum \overrightarrow{L _i}=\sum m_i\:r_i\:v_i}$$
we know that
$$\mathrm{ u_i=\overrightarrow{r_i\omega}}$$
So
$$\mathrm{\overrightarrow{L_z}=\omega\sum m_i\:r_i^2=I \overrightarrow{\omega}}$$
since moment of inertia
$$\mathrm{I =\sum m_i\:r_i^2}$$
Hence for rigid body
$$\mathrm{\overrightarrow{L_z}=I \overrightarrow{\omega}}$$
i.e. The net Angular momentum of a rigid body about the z axis is equal to the product of angular momentum and moment of inertia.
We know $\mathrm{\overrightarrow{\tau}=\frac{d\overrightarrow{L}}{dt}}$
Hence
$$\mathrm{\overrightarrow{\tau}=\frac{d}{dt}(I\overrightarrow{\omega})=I\overrightarrow{\alpha}}$$
i.e. The Net torque acting on a rigid body is product of its moment of inertia and angular acceleration.
It gives us a relation between torque and angular acceleration. It is analogous to the Newton’s 2nd law
Translational motion Rotational motion $\mathrm{\overrightarrow{F}=m\overrightarrow{a}}$ $\mathrm{\overrightarrow{p}=m\overrightarrow{u}}$ $\mathrm{\overrightarrow{\tau}=I\overrightarrow{\alpha}}$ $\mathrm{\overrightarrow{L}=I\overrightarrow{\omega}}$
Table-2: Newton’s 2nd law in Rotational and Translational Motion
## Conclusion
Angular momentum is analogous to linear momentum in rotational motion. It depends upon the distance from the axis of rotation. Angular momentum has meaning only if it is defined about specified origin. It is a vector quantity always perpendicular to the plane of $\mathrm{\overrightarrow{r}}$ and $\mathrm{\overrightarrow{p}}$ The rate of change of angular momentum gives us Torque on the particle. We can use this concept to multi-particle systems also. For a rigid body it is equal to the angular velocity times the Moment of Inertia of that body. We can also show the correspondence between rotational and translational motion as summarised in Table-1.
## FAQs
Q1. What is the condition for conservation of angular momentum?
Ans. For an isolated system Angular momentum will always be conserved if external torque is zero. i.e.,$\mathrm{\overrightarrow{\tau_{ext}}=0}$
Q2. What does angular momentum physically signify?
Ans. Physically angular momentum is a measure of the turning motion of an object and Torque is a measure of the turning effect of force.
Q3. Linear momentum of a particle in circular motion is always changing, is it true for angular momentum too?
Ans. No, for a particle is circular path $\mathrm{\overrightarrow{L}=(mvr)\hat{k}}$ Hence it doesn’t change if a particle is moving about an axis passing through its centre.
Q4.Can a non-rotating body possess angular momentum?
Ans. Yes, suppose a particle is moving along some path in a straight line and if we take the axis at some point O, if O is displaced from that straight line, then we will get non zero angular momentum. Since $\mathrm{\overrightarrow{r}\times \overrightarrow{v}}$ will be non zero.
Q5. A sphere and a cylinder of same mass are rotating about the same axis with same angular velocity, which body will have larger angular momentum?
Ans. Moment of Inertia of cylinder > Moment of Inertia of sphere of same mass.
Hence Angular momentum of cylinder > Angular momentum of Sphere
Updated on: 28-Apr-2023
83 Views
##### Kickstart Your Career
Get certified by completing the course |
# MTH 111–112 Supplement
## AppendixAAnswers and Solutions to Exercises
### 1MTH 111 Supplement1.1Graph Transformations
#### Exercises
##### 1.1.1.
$$g(x)=f(-x)\text{.}$$ So, we can reflect the graph of $$y=f(x)$$ across the $$y$$-axis to obtain $$y=g(x)\text{.}$$
##### 1.1.2.
$$h(x)=-f(x)\text{.}$$ So, we can reflect the graph of $$y=f(x)$$ across the $$x$$-axis to obtain $$y=h(x)\text{.}$$
##### 1.1.3.
$$k(x)=f(x)+6\text{.}$$ So, we can shift the graph of $$y=f(x)$$ up $$6$$ units to obtain $$y=k(x)\text{.}$$
##### 1.1.4.
$$l(x)=3f(x)\text{.}$$ So, we can stretch the graph of $$y=f(x)$$ vertically by a factor of $$3$$ to obtain $$y=l(x)\text{.}$$
##### 1.1.5.
$$x$$ $$-4$$ $$-3$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$f(x)$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$\frac{1}{2}x$$ $$-1$$ $$-\frac{1}{2}$$ $$0$$ $$\frac{1}{2}$$ $$1$$ $$\frac{3}{2}$$ $$2$$ $$\frac{5}{2}$$ $$3$$ $$-2f(x)$$ $$4$$ $$2$$ $$0$$ $$-2$$ $$-4$$ $$-6$$ $$-8$$ $$-10$$ $$-12$$ $$f(x)+5$$ $$3$$ $$4$$ $$5$$ $$6$$ $$7$$ $$8$$ $$9$$ $$10$$ $$11$$ $$f(x+2)$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$ $$6$$ $$f\mathopen{}\left( \frac{1}{2}x \right)\mathclose{}$$ $$0$$ $$1$$ $$2$$ $$3$$ $$4$$ $$f(2x)$$ $$-2$$ $$0$$ $$2$$ $$4$$ $$6$$ $$f(x-3)$$ $$-2$$ $$-1$$ $$0$$ $$1$$ $$2$$ $$3$$
##### 1.1.6.
Solution.
\begin{align*} g(x) \amp = \frac{\sqrt{x-7}}{4} \\ \amp = \frac{1}{4}\sqrt{x-7} \\ \amp = \frac{1}{4}f(x-7) \end{align*}
So we can transform $$y=f(x)$$ into $$y=g(x)$$ by first shifting right $$7$$ units and then compressing vertically by a factor of $$\frac{1}{4}\text{.}$$ (There are other correct answers.)
##### 1.1.7.
Solution.
\begin{align*} g(x) \amp = \frac{2}{x}+3 \\ \amp = 2 \cdot \frac{1}{x}+3 \\ \amp = 2f(x)+3 \end{align*}
So we can transform $$y=f(x)$$ into $$y=g(x)$$ by first stretching vertically by a factor of $$2$$ and then shifting up $$3$$ units. (There are other correct answers.)
##### 1.1.8.
Solution.
\begin{align*} g(x) \amp =-4\left( \frac{1}{2}x-5 \right)^2+3 \\ \amp = -4f\mathopen{}\left( \frac{1}{2}x-5 \right)\mathclose{}+3 \\ \amp = -4f\mathopen{}\left( \frac{1}{2} (x-10) \right)\mathclose{}+3 \end{align*}
So we can transform $$y=f(x)$$ into $$y=g(x)$$ by first stretching horizontally by a factor of $$2$$ and then shifting right $$10$$ units. Then, stretching vertically by a factor of $$4$$ and reflecting across the $$x$$-axis, and finally shifting up $$3$$ units. (There are other correct answers.)
##### 1.1.9.
Solution.
\begin{align*} g(x) \amp =\frac{1}{2}\sqrt[3]{10x+30}-6 \\ \amp =\frac{1}{2}f(10x+30)-6 \\ \amp =\frac{1}{2}f\mathopen{}\left( 10( x+3 ) \right)\mathclose{} -6 \end{align*}
So we can transform $$y=f(x)$$ into $$y=g(x)$$ by first compressing horizontally by a factor of $$\frac{1}{10}$$ and then shifting left $$3$$ units. Then, compressing vertically by a factor of $$\frac{1}{2}$$ and finally shifting down $$6$$ units. (There are other correct answers.)
### 1.2Inverse Functions
#### Exercises
##### 1.2.1.
Solution.
$$m$$ is an invertible function since it is one-to-one, i.e., each output corresponds to exactly one input. Here is a table-of-values for $$m^{-1}\text{.}$$
$$x$$ $$0$$ $$5$$ $$10$$ $$15$$ $$20$$ $$m^{-1}(x)$$ $$1$$ $$2$$ $$3$$ $$4$$ $$5$$
##### 1.2.2.
Solution.
$$p$$ isn’t an invertible function since it isn’t one-to-one. Notice how the output $$0$$ corresponds to two distinct output values.
### 1.3Exponential Functions
#### Exercises
##### 1.3.1.
$$f(x)=50 \cdot 2^x$$
##### 1.3.2.
$$f(x)=4 \cdot \left( \frac{1}{2} \right)^x$$
##### 1.3.3.
$$f(x)=2 \cdot 3^x$$
##### 1.3.4.
$$f(x)=10 \cdot \left( \frac{4}{5} \right)^x$$
##### 1.3.5.
$$f(x)=5 \cdot \left( \frac{1}{5} \right)^x$$
##### 1.3.6.
$$f(x)=\frac{1}{2} \cdot \left( \frac{2}{3} \right)^x$$
### 1.4Logarithmic Functions
#### Exercises
##### 1.4.1.
$$a=\sqrt{5}$$
##### 1.4.2.
$$a=20$$
##### 1.4.3.
$$a=\sqrt{3}$$
### 2MTH 112 Supplement2.1Angles2.1.4Exercises
#### 2.1.4.1.
$$423^{\circ}$$ and $$-297^{\circ}$$ are coterminal with $$63^{\circ}\text{.}$$
#### 2.1.4.2.
$$\frac{19\pi}{9}$$ and $$-\frac{17\pi}{9}$$ are coterminal with $$\frac{\pi}{9}\text{.}$$
#### 2.1.4.3.
$$\frac{29\pi}{8}$$ and $$-\frac{3\pi}{8}$$ are coterminal with $$\frac{13\pi}{8}\text{.}$$
#### 2.1.4.4.
$$60^{\circ}$$
#### 2.1.4.5.
$$\frac{\pi}{4}$$
#### 2.1.4.6.
$$40^{\circ}$$
#### 2.1.4.7.
$$\frac{3\pi}{8}$$
#### 2.1.4.8.
$$\pi-2 \approx 1.14$$
#### 2.1.4.9.
$$\frac{\pi}{11}$$
#### 2.1.4.10.
$$20^{\circ}$$
#### 2.1.4.11.
$$\frac{\pi}{5}$$
#### 2.1.4.12.
$$80^{\circ}$$
#### 2.1.4.13.
$$243^{\circ}10' \approx 243.167^{\circ}$$
#### 2.1.4.14.
$$3^{\circ}25' \approx 3.417^{\circ}$$
#### 2.1.4.15.
$$-23^{\circ}3' = -23.05^{\circ}$$
#### 2.1.4.16.
$$75^{\circ}32'17'' \approx 75.538^{\circ}$$
#### 2.1.4.17.
$$12.4^{\circ} = 12^{\circ}24'$$
#### 2.1.4.18.
$$1.53^{\circ} = 1^{\circ}31'48''$$
#### 2.1.4.19.
$$-144.9^{\circ} = -144^{\circ}54'$$
#### 2.1.4.20.
$$0.416^{\circ} = 0^{\circ}24'57.6''$$
### 2.2Generalized Definitions of Trigonometric Functions
#### Exercises
##### 2.2.1.
\begin{align*} \cos(\theta) \amp= \frac{3}{5} \\ \sin(\theta) \amp= \frac{4}{5} \\ \tan(\theta) \amp= \frac{4}{3} \\ \sec(\theta) \amp= \frac{5}{3} \\ \csc(\theta) \amp= \frac{5}{4} \\ \cot(\theta) \amp= \frac{3}{4} \end{align*}
##### 2.2.2.
\begin{align*} \sin(\theta) \amp= -\frac{\sqrt{10}}{10} \\ \cos(\theta) \amp= -\frac{3\sqrt{10}}{10} \\ \tan(\theta) \amp= 3 \\ \sec(\theta) \amp= -\sqrt{10} \\ \csc(\theta) \amp= -\frac{\sqrt{10}}{3} \\ \cot(\theta) \amp= \frac{1}{3} \end{align*}
##### 2.2.3.
$$\mathopen{}\left( -\frac{3\sqrt{3}}{2}, -\frac{3}{2} \right)\mathclose{}$$
##### 2.2.4.
$$\mathopen{}\left( 5, 5\sqrt{3} \right)\mathclose{}$$
### 2.3Graphing Sinusoidal Functions: Phase Shift vs. Horizontal Shift
#### Exercises
##### 2.3.1.
Amplitude
$$3$$ units
Period
$$\frac{2\pi}{3}$$ units
Midline
$$\displaystyle y=0$$
Phase shift
$$\displaystyle \frac{\pi}{2}$$
Horizontal Shift
$$\frac{\pi}{6}$$ units to the right
##### 2.3.2.
Amplitude
$$1$$ unit
Period
$$\frac{\pi}{2}$$ units
Midline
$$\displaystyle y=3$$
Phase shift
$$\displaystyle -\pi$$
Horizontal Shift
$$\frac{\pi}{4}$$ units to the left
##### 2.3.3.
Amplitude
$$2$$ units
Period
$$1$$ unit
Midline
$$\displaystyle y=4$$
Phase shift
$$\displaystyle \pi$$
Horizontal Shift
$$\frac{1}{2}$$ of a unit to the right
##### 2.3.4.
Amplitude
$$4$$ units
Period
$$2$$ units
Midline
$$\displaystyle y=-2$$
Phase shift
$$\displaystyle -\frac{\pi}{4}$$
Horizontal Shift
$$\frac{1}{4}$$ of a unit to the left
##### 2.3.5.
\begin{aligned} p(x)=4 \sin \mathopen{}\left( 2 \mathopen{}\left( x-\frac{\pi}{4} \right)\mathclose{} \right)\mathclose{}-2 \\ p(x)=4 \cos \mathopen{}\left( 2 \mathopen{}\left( x-\frac{\pi}{2} \right)\mathclose{} \right)\mathclose{}-2 \end{aligned}
##### 2.3.6.
\begin{aligned}q(x)=3 \sin \mathopen{}\left( \pi \mathopen{}\left( x+\frac{1}{4} \right)\mathclose{} \right)\mathclose{}-1\\ q(x)=3 \cos \mathopen{}\left( \pi \mathopen{}\left( x-\frac{1}{4} \right)\mathclose{} \right)\mathclose{}-1\end{aligned}
### 2.4Complex Numbers and Polar Coordinates2.4.3Exercises
#### 2.4.3.1.
$$z=12e^{\frac{\pi}{3} \cdot i}$$
#### 2.4.3.2.
$$z=4e^{\frac{5\pi}{6} \cdot i}$$
#### 2.4.3.3.
$$z=10e^{-\frac{\pi}{4} \cdot i}$$
#### 2.4.3.4.
$$z=4\sqrt{3}+4i$$
#### 2.4.3.5.
$$z=-4$$
#### 2.4.3.6.
$$z=-\frac{5}{2}-\frac{5\sqrt{3}}{2} \cdot i$$
#### 2.4.3.7.
$$\sqrt{18-18\sqrt{3}\cdot i}=3\sqrt{3}-3i$$ (and the non-principal root is $$-3\sqrt{3}+3i$$)
#### 2.4.3.8.
$$\sqrt[3]{-16+16i}=2+2i$$ (and the non-principal roots are $$\left(-1-\sqrt{3}\right)+\left(-1+\sqrt{3}\right)i$$ and $$\left(-1+\sqrt{3}\right)+\left(-1-\sqrt{3}\right)i$$)
#### 2.4.3.9.
$$\sqrt{-i} = \frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}i$$ (and the non-principal root is $$-\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}i$$)
#### 2.4.3.10.
$$\sqrt[5]{-16\sqrt{3}-16i}=\sqrt{3}-i$$ (and the non-principal roots are $$2\cos\mathopen{}\left(\frac{7\pi}{30}\right)+2i\sin\mathopen{}\left(\frac{7\pi}{30}\right)\text{,}$$ $$2\cos\mathopen{}\left(\frac{19\pi}{30}\right)+2i\sin\mathopen{}\left(\frac{19\pi}{30}\right)\text{,}$$ $$2\cos\mathopen{}\left(\frac{29\pi}{30}\right)-2i\sin\mathopen{}\left(\frac{29\pi}{30}\right)\text{,}$$ and $$2\cos\mathopen{}\left(\frac{17\pi}{30}\right)-2i\sin\mathopen{}\left(\frac{17\pi}{30}\right)$$)
#### 2.4.3.11.
$$\frac{3\sqrt{3}}{2}+\frac{3}{2}i\text{,}$$ $$-3i\text{,}$$ and $$-\frac{3\sqrt{3}}{2}+\frac{3}{2}i$$
$$\frac{1}{2}+\frac{\sqrt{3}}{2}i$$ and $$-\frac{1}{2}-\frac{\sqrt{3}}{2}i$$
$$\left\{ -1, \frac{1}{2}+\frac{\sqrt{3}}{2}i, \frac{1}{2}-\frac{\sqrt{3}}{2}i \right\}$$ |
Introduction
The table below demonstrates four methods used to represent relationships.
Written out in words or stated as a problem Organized as numbers in a data table Written symbolically in an equation Arranged visually in a graph
In this lesson, we will focus on generating and interpreting relationships that are arranged visually in a graph.
People frequently use graphs to interpret relationships and make predictions. For example, business people use graphs of data to make purchasing decisions for their company. Investment bankers use graphs of data to make decisions about which stocks and bonds to buy. The media also uses graphs to communicate information to the public about the economy, population, and other socially important statistics.
Because graphs are so frequently used, it is important to understand how to correctly interpret data that is presented in graphical form.
Generating Graphs from Data
A table is an efficient way to organize numerical data. In a previous lesson, you investigated different ways to generate a graph when given a table, and different ways to generate a table when given a graph.
Consider the graph of the equation y = 0.5x – 4 and a table of data.
How can we tell if the points contained in the table actually lie on the graph of the line y = 0.5x – 4? One way to do so is to plot the points contained in the table on the graph, and test each point to see if it lies on the line.
Use a plotting applet to graph both the line and the points in the table. This tool will allow you to confirm whether or not the points in the table lie on the graph of the line.
Use the graph and table provided to determine which points lie on the graph of the line.
Use the interactive below to identify which points from the table lie on the graph of the line:
y = 0.5x – 4
For each ordered pair, place it in the Lies on Line box if the point lies on the line, or in the Does NOT Lie on Line box if the point does not lie on the line.
Practice
Use the graphing applet to determine which of the following tables contain only points that lie on the graph of the line $y=\frac{1}{3}x+3.$
Table A
x y
-9 0
-3 2
0 3
2 $3\frac{2}{3}$
6 5
Table B
x y
-9 0
-3 3
0 3
4 $4\frac{1}{3}$
6 5
Table C
x y
-9 0
-3 2
1 $3\frac{1}{3}$
3 4
9 6
Table D
x y
0 -9
2 -3
$3\frac{1}{3}$ 1
4 3
6 9
Making Predictions from Scatterplots and Graphs
In this section, you will investigate ways to make predictions about data using scatterplots and other graphs.
The table below contains data describing the population of the United States as determined by the census.
Year Year Number
(since 1900)
Population
1940 40 132122446
1950 50 152271417
1960 60 180671158
1970 70 205052174
1980 80 227224681
1990 90 249464396
2000 100 281421906
Use the graphing applet to make a scatterplot of population versus year number. Use the scatterplot that is generated to answer the questions that follow.
Click to see additional instructions for using the applet.
To better use the applet, copy and paste the data from this table, which is a simplified version of the table that was presented above.
Year Number
(since 1900)
Population
40 132122446
50 152271417
60 180671158
70 205052174
80 227224681
90 249464396
100 281421906
Question 1
What was the approximate U.S. population in 1985?
Question 2
What was the approximate U.S. population in 1965?
Question 3
If the population continues changing at the current rate, what will be the approximate population in 2010?
Practice
The graph shows the relationship between the weight of a bicycle and the maximum height of a jump off the ground.
Summary
When making predictions and justifying data from graphs, it is important to be able to represent data both in tables and in graphs. You can use the graph to visually follow trends that appear in the data, and use those trends to make predictions.
Sometimes, you are given the equation of a line and asked to confirm whether or not points lie on the line. One way to do so is to plot each point and then determine whether or not the point is on the line for the given equation.
In this case, the graph of the line 5x + 2y = 10 is shown, and the points in the table are plotted on the same grid. You can see that the point (4, 5) does not lie on the line 5x + 2y = 10.
Sometimes you are given data in a scatterplot and asked to use the scatterplot to make a prediction. To do so, you can follow the trend in the data to read either between two points or to extend the pattern beyond the data. |
Courses
Courses for Kids
Free study material
Offline Centres
More
Store
Find the distance covered by the wheel of a bus in 1000 revolutions if the diameter of the wheel is 56 cm.
Last updated date: 14th Jul 2024
Total views: 345.3k
Views today: 3.45k
Verified
345.3k+ views
Hint: Assume the given diameter of the wheel as d. Find the radius (r) of the wheel by using the relation d = 2r. Now, consider that the distance covered by the wheel in one revolution will be equal to the circumference of the wheel and find the distance covered in one revolution using the formula $C=2\pi r$ where C = circumference. Multiply both the sides with 1000 to get the answer.
Here we have been asked to find the distance covered by the wheel of a bus which takes 1000 revolutions and whose diameter is 56 cm. First let us find the radius of the wheel.
Now, we know that the diameter (d) of a circle is twice its radius (r) so using the relation d = 2r we get,
$\Rightarrow r=\dfrac{d}{2}$
Substituting the value of d = 56 cm we get,
$\Rightarrow r=\dfrac{56}{2}$
$\Rightarrow$ r = 28 cm
Now, in one revolution the wheel will cover a distance equal to its circumference, so applying the formula for the circumference (C) of a circle given as $C=2\pi r$ we get,
$\Rightarrow$ Distance covered in 1 revolution = $2\times \dfrac{22}{7}\times 28$
$\Rightarrow$ Distance covered in 1 revolution = 176 cm
Multiplying both the sides with 1000 we get,
$\Rightarrow$ Distance covered in 1000 revolution = 176000 cm
We know that 100 cm = 1m and 1000m = 1km so using these two relations we get,
$\therefore$ Distance covered in 1000 revolution = 1.76 km
Hence, the distance travelled by the wheel in 1000 revolutions is 1.76 km.
Note: Note that it is not necessary to convert the diameter of the wheel into the radius as you can remember the direct relation between the circumference and diameter given as $C=\pi d$. Although you must remember the relation d = 2r. We have taken the value of $\pi =\dfrac{22}{7}$ only to reduce our calculations because we can see that 28 can easily be cancelled by 7 present in the denominator. |
# Video: Graphing Using Derivatives
In this video, we will learn how to use derivatives to graph a function.
14:33
### Video Transcript
In this video, we’ll learn how finding the derivative can make graphing a function much more simple. Up to this stage, it’s likely we’ll have focused on particular aspects of curve sketching, such as limits, continuity, and concavity. It’s now time to combine all of this information to help us sketch graphs that reveal the full extent of the features of the various functions.
The given checklist is intended to be used like a set of guidelines when curve sketching. We won’t need to use every single point every single time. But it’s a good idea to begin by considering each of these features in turn.
The first is domain and range. It can be useful to consider the domain of the function. In other words, the set of values of 𝑥, for which the function is defined. And from this, we might consider the range, the complete set of resulting function values, though this usually develops through the other steps.
The second is intercepts. We look for both 𝑦- and 𝑥-intercepts. Remember, the 𝑦-intercept is obtained by letting 𝑥 equal to zero and solving for 𝑦, whereas the 𝑥-intercepts are obtained by setting 𝑦 equal to zero and solving for 𝑥. We look for symmetry. Are the functions even? In other words, does 𝑓 of negative 𝑥 equal 𝑓 of 𝑥? Are they odd? Does 𝑓 of negative 𝑥 equal negative 𝑓 of 𝑥? Or neither. And then we might consider asymptotes.
Remember, if the limit as 𝑥 tends to either positive or negative infinity of the function is equal to some value 𝐿, then the line 𝑦 equals 𝐿 is a horizontal asymptote of the curve. If it turns out that the limit as 𝑥 tends to infinity of the function is either positive or negative infinity, then we don’t have an asymptote. But this fact is still useful for sketching the curve.
We also recall that the line 𝑥 equals 𝑎 is a vertical asymptote if at least one of the following is true. The limit as 𝑥 tends to 𝑎 from the right of the function is either positive or negative infinity. Or the limit as 𝑥 approaches 𝑎 from the left of the function is either positive or negative infinity. The fifth point is intervals of increase or decrease.
We recall that a function is increasing when its derivative is greater than zero and decreasing when its derivative is less than zero. This tells us the shape of the graph over various intervals. We also look for local extrema. Remember, critical values occur when the derivative is equal to zero or does not exist. Then the first derivative test can tell us the nature of these critical points.
We can consider concavity and points of inflection. And this time, we evaluate the second derivative. If it’s greater than zero over some interval, then the curve is concave upward. And if it’s less than zero over some interval, the curve is concave downward. When the concavity changes, we know we have a point of inflection. Finally, we might consider the end behavior of the function. And this we usually come about as a result of the other work we’ve done.
This is, of course, a hugely extensive list. And there’ll be times where many of the features can be predicted by using a graphing calculator. But one of these will not always be accessible. We’ll now consider a number of examples which require consideration of these features.
Consider the function 𝑓 of 𝑥 equals 𝑥 minus one squared times 𝑥 plus two. Find 𝑓 prime of 𝑥. Find and classify the critical points of 𝑓. Find the intervals of increase and decrease for 𝑓. Find the limit as 𝑥 approaches infinity of 𝑓 of 𝑥.
And then there’s one more part of this question, which asks us to identify the correct graph for our function. So we’ll look at our options when we’ve completed the first four parts. So how are we going to evaluate 𝑓 prime of 𝑥?
That’s the first derivative of our function with respect to 𝑥. Well, taking a look at our function, there are a number of ways we can do this. We could distribute our parentheses and differentiate term by term. Alternatively, notice that we have a product of two functions, one of which is a composite function. So we could use the product rule alongside the general power rule.
The product rule says that the derivative of the product of two differentiable functions 𝑢 and 𝑣 is 𝑢 times d𝑣 by d𝑥 plus 𝑣 times d𝑢 by d𝑥. So let’s let 𝑢 be equal to 𝑥 minus one all squared and 𝑣 be equal to 𝑥 plus two. We’ll need to use the general power rule to find the derivative of 𝑥 minus one all squared. This is a special case of the chain rule. And it says we can multiply the entire function by the exponent, reduce the exponent by one, and then multiply all of this by the derivative of the inner function.
Well, the derivative of 𝑥 minus one is just one. So d𝑢 by d𝑥 is two times 𝑥 minus one to the power of two minus one times one. And that simplifies to two times 𝑥 minus one. d𝑣 by d𝑥 is much more straightforward. It’s simply one. So we substitute what we have into the product rule. It’s 𝑢 times d𝑣 by d𝑥 — that’s 𝑥 minus one all squared times one — plus 𝑣 times d𝑢 by d𝑥. Distributing our parentheses, and the first derivative of our function simplifies really nicely to three 𝑥 squared minus three.
In the second part of this question, we need to find and classify the critical points of our function. Remember, these occur when our first derivative is either equal to zero or does not exist. Well, we actually know that polynomial functions are all differentiable. So we’re just going to consider the case for which our first derivative is equal to zero. That is, three 𝑥 squared minus three is equal to zero.
Let’s solve for 𝑥. We divide through by one. And then we use the difference of two squares to factor 𝑥 squared minus one. It’s 𝑥 plus one times 𝑥 minus one. Now for the product of these two terms to be equal to zero, either 𝑥 plus one has to be equal to zero or 𝑥 minus one has to be equal to zero. And solving each equation for 𝑥, we obtain 𝑥 must be either negative one or 𝑥 equals one. So we have critical points where 𝑥 is either equal to positive or negative one.
Our next job is to classify these. We can perform the first derivative test and evaluate the first derivative just before and just after each critical point. Let’s add a table. We know that the first derivative is equal to zero when 𝑥 is equal to negative one or positive one. The first derivative evaluated at 𝑥 equals negative two is three times negative two squared minus three, which is nine. The first derivative evaluated when 𝑥 is equal to zero is three times zero squared minus three, which is negative three. And the first derivative evaluated at 𝑥 equals two is also nine. The function is increasing before 𝑥 equals negative one and decreasing after. So the critical point 𝑥 equals negative one is a local maximum. The opposite is true when 𝑥 equals one. So that must be our local minimum.
In the third part of this question, we’re looking to find the intervals of increase and decrease for our function. Let’s clear some space. Now this might seem awfully similar to what we just did. But at that point, we were simply considering the nature of the function at specific points. We now want to know over what intervals the function is increasing or decreasing. So we’re going to work out when the first derivative is less than zero, decreasing, or greater than zero, increasing.
We know the graph of 𝑦 equals three 𝑥 squared minus three looks a little something like this. We can see that the graph of the first derivative is greater than zero here and here. In other words, when 𝑥 is less than negative one or greater than one. The first derivative is less than zero here. That’s when 𝑥 is greater than negative one or less than one. So the intervals of increase are the open intervals from negative infinity to negative one and one to infinity. And the interval of decrease is the open interval from negative one to one.
And finally, we need to find the limit as 𝑥 approaches infinity of the function. We can do this by direct substitution. We see that as 𝑥 approaches infinity, the function itself also grows larger and larger. So the function also approaches infinity. Let’s now combine everything we’ve done to identify the graph of our function.
Which of the following is the graph of 𝑓? If we were to distribute the parentheses of our function, we’d see we have a cubic graph with a positive leading coefficient. In other words, the coefficient of 𝑥 cubed is positive. That tells us the shape of the graph will be a little something like this. We know it has critical points 𝑥 equals negative one and one and a maximum and minimum in those locations, respectively.
We can work out the value of our function at these points by substituting negative one and one in. And when we do, we see that 𝑓 of negative one is four and 𝑓 of one is zero. So we know the graph has a maximum at negative one, four and a minimum at one, zero. We also saw that it had intervals of increase over the open interval negative infinity to negative one and one to infinity and decrease over the open interval negative one to one. The only graph that satisfies all of these conditions is A. In fact, we could have very quickly spotted they could not have been D or E as these are graphs of quadratic functions. C is a cubic graph which has a negative coefficient of 𝑥 cubed. So we really only had two options.
Notice that we didn’t worry about evaluating the limits of our functions to look for asymptotes. As we know, a cubic graph doesn’t have any.
In our next example, we will consider a graph which has at least one asymptote.
Sketch the graph of 𝑓 of 𝑥 equals three 𝑥 squared over 𝑥 squared minus four.
Let’s begin by considering the domain and range of our function. We know that a function which is a quotient won’t exist at points where the denominator of the quotient is zero. So we set 𝑥 squared minus four equals zero and solve for 𝑥 to find the domain of our function. We can add four to both sides of this equation. Then we take the square root of both sides of our equation, remembering to take both the positive and negative square root of four. And we obtain that when 𝑥 squared minus four is equal to zero, 𝑥 is equal to positive or negative two. So the domain of our function is all real values, except 𝑥 is equal to positive or negative two. Now the range itself is a set of possible outputs on the function. And we can infer this through the graphing process.
Next, we’ll work out whether there are any intercepts. By setting 𝑥 equal to zero and solving for 𝑦, we’ll find the 𝑦-intercept. When we do, we obtain 𝑦 to be equal to three times zero squared over zero squared minus four, which is zero. So there’s a 𝑦-intercept at 𝑦 equals zero. Next, we set 𝑦 or 𝑓 of 𝑥 equal to zero and solve for 𝑥. That is, zero equals three 𝑥 squared over 𝑥 squared minus four.
Now for this to be true, we know that the numerator of this fraction must itself be equal to zero. And for three 𝑥 squared to be equal to zero, 𝑥 must be equal to zero. So we actually only have one intercept altogether. And that’s at the origin: zero, zero. Next, we can check for symmetry. An even function is one for which 𝑓 of negative 𝑥 equals 𝑓 of 𝑥, whereas an odd function is one for which 𝑓 of negative 𝑥 equals negative 𝑓 of 𝑥. Well, 𝑓 of negative 𝑥 is three times negative 𝑥 squared over negative 𝑥 squared minus four, which is equal to 𝑓 of 𝑥. So our function is even. And that means it’s going to be symmetrical about the 𝑦-axis.
Next, we’ll look for asymptotes. We’ll look for horizontal asymptotes by considering what happens to our function as 𝑥 approaches either positive or negative infinity. We can’t use direct substitution as when we substitute in 𝑥 equals either positive or negative infinity, we end up with infinity over infinity, which is undefined. So instead, we divide both the numerator and the denominator of our expression by 𝑥 squared. That gives us three over one minus four over 𝑥 squared.
And now we can use direct substitution. As 𝑥 approaches either positive or negative infinity, four over 𝑥 squared approaches zero. So the limit of our function is three over one, which is just three. And we see that the line 𝑦 equals three must be a horizontal asymptote.
Remember, we said that when 𝑥 is equal to positive or negative two, the denominator is zero. So we find the following limits. The limit as 𝑥 approaches two from the right of our function is infinity. And the limit as 𝑥 approaches two from the left of the function is negative infinity. And the limit as 𝑥 approaches negative two from the right of the function is negative infinity. And as it approaches negative two from the left, it’s positive infinity. So we obtain 𝑥 equals positive or negative two to be vertical asymptotes.
Let’s clear some space and look for intervals of increase and decrease. We’ll begin by using the quotient rule to find the first derivative of our function. By letting 𝑢 be equal to three 𝑥 squared and 𝑣 be equal to 𝑥 squared minus four, we obtain expressions for d𝑢 by d𝑥 and d𝑣 by d𝑥. And so the first derivative of our function is 𝑣 times d𝑢 by d𝑥 minus 𝑢 times d𝑣 by d𝑥 over 𝑣 squared. And distributing our parentheses, we find the first derivative to be negative 24𝑥 over 𝑥 squared minus four all squared.
Now since 𝑥 squared minus four all squared must be greater than zero for all 𝑥, this means the first derivative of our function must be less than zero when 𝑥 is greater than zero. So for values of 𝑥 greater than zero, we get negative 24 multiplied by a positive, which gives us a negative value. And so for 𝑥 greater than zero, the first derivative is less than zero. The opposite is true. When 𝑥 is less than zero, the first derivative is greater than zero. Since we know the function doesn’t exist when 𝑥 is equal to negative two, we can say that its intervals of increase are negative infinity to negative two and negative two to zero. And it has open intervals of decrease over zero to two and two to infinity.
Now, actually, if we look carefully, we also see that the first derivative is equal to zero when 𝑥 is equal to zero. So there’s a critical point at the point zero, zero. Coincidently, that’s also the point where our curves cross the axes.
We now have everything we need to complete our sketch. We know that there’s a horizontal asymptote at 𝑦 equals three and vertical asymptotes at 𝑥 equals negative two and 𝑥 equals two. There’s a critical point and an intercept at zero, zero. And that’s important because we see that the curve won’t be able to cross the 𝑥-axis here or here. That helps us decide where these two parts of the curve go. The curve is increasing over the open interval negative infinity to negative two and decreasing over the open interval two to infinity. And of course, we have these asymptotes. The function is also even and symmetrical about the 𝑦-axis. So this confirms we’re probably on the right track. We then use the remaining pieces of information to sketch the final part of our curve. And we’ve successfully sketched the graph of 𝑓 of 𝑥 equals three 𝑥 squared over 𝑥 squared minus four.
In this video, we’ve seen that using derivatives can help us to graph complicated functions. We have a rather extensive set of guidelines, including finding the domain or range, the intercepts, the symmetry of the curve. We look for asymptotes, intervals of increase or decrease, any local extrema, concavity, or inflection points. And we also consider the end behavior of our function. And of course, this list is extremely extensive. But we won’t need to use every single point every single time. |
# Calculus 3 : Vectors and Vector Operations
## Example Questions
### Example Question #11 : Vectors And Vector Operations
Find the angle between the two vectors.
No angle exists
Explanation:
To find the angle between two vector we use the following formula
and solve for .
Given
we find
Plugging these values in we get
To find we calculate the of both sides
and find that
### Example Question #12 : Vectors And Vector Operations
Find the approximate angle in degrees between the vectors .
Explanation:
We can compute the (acute) angle between the two vectors using the formula
Hence we have
### Example Question #13 : Vectors And Vector Operations
Find the angle in degrees between the vectors .
Explanation:
The correct answer is approximately degrees.
To find the angle between two vectors, we use the equation .
Hence we have
(This answer is small due to the fact that the two vectors nearly point in the same direction, due to and being close in value.)
### Example Question #14 : Vectors And Vector Operations
Find the acute angle in degrees between the vectors .
Explanation:
To find the angle between two vectors, we use the formula
.
So we have
### Example Question #15 : Vectors And Vector Operations
Find the angle between the following vectors (to two decimal places):
2.89
0.98
1.18
1.46
1.62
1.46
Explanation:
The dot product is defined as:
Where theta is the angle between the two vectors. Solving for theta:
To solve each component:
Putting it all together, we can solve for theta:
### Example Question #31 : Calculus 3
Find the angle between vectors and and round to the nearest degree.
Explanation:
Write the formula to find the angle between two vectors.
Evaluate each term.
Substitute the values into the equation.
### Example Question #32 : Calculus 3
Find the angle between the two vectors
Explanation:
In order to find the angle between the two vectors, we follow the formula
and solve for .
Using the vectors in the problem, we get
Simplifying we get
To solve for we find the of both sides and get
and find that
### Example Question #33 : Calculus 3
What is the angle to the nearest degree between the vectors and ?
Explanation:
In order to find the angle between the two vectors, we follow the formula
and solve for
Using the vectors in the problem, we get
Simplifying we get
To solve for
we find the
of both sides and get
and find that
### Example Question #34 : Calculus 3
Find the angle between the two vectors
and
Round to the nearest degree.
Explanation:
In order to find the angle between the two vectors, we follow the formula
and solve for
Using the vectors in the problem, we get
Simplifying we get
To solve for
we find the
of both sides and get
and find that
### Example Question #35 : Calculus 3
Find the angle between the two vectors
Explanation:
In order to find the angle between the two vectors, we follow the formula
and solve for .
Using the vectors in the problem, we get
Simplifying we get
To solve for we find the of both sides and get
and find that |
Math problem answers are solved here step-by-step to keep the explanation clear to the students. In Math-Only-Math you'll find abundant selection of all types of math questions for all the grades with the complete step-by-step solutions.
Parents and teachers can follow math-only-math to help their students to improve and polish their knowledge. Children can practice the worksheets of all the grades and on all the topics to increase their knowledge.
Various types of Math Problem Answers are solved here.
1. Mrs. Rodger got a weekly raise of $145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck. Solution: Let the 1st paycheck be x (integer). Mrs. Rodger got a weekly raise of$ 145.
So after completing the 1st week she will get $(x+145). Similarly after completing the 2nd week she will get$ (x + 145) + $145. =$ (x + 145 + 145)
= $(x + 290) So in this way end of every week her salary will increase by$ 145.
2. The value of x + x(xx) when x = 2 is:
(a) 10, (b) 16, (c) 18, (d) 36, (e) 64
Solution:
x + x(xx)
Put the value of x = 2 in the above expression we get,
2 + 2(22)
= 2 + 2(2 × 2)
= 2 + 2(4)
= 2 + 8
= 10
3. Mr. Jones sold two pipes at $1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he: (a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents Solution: 20 % profit on$ 1.20
= $20/100 × 1.20 =$ 0.20 × 1.20
= $0.24 Similarly, 20 % loss on$ 1.20
= $20/100 × 1.20 =$ 0.20 × 1.20
= $0.24 Therefore, in one pipe his profit is$ 0.24 and in the other pipe his loss is $0.24. Since both profit and loss amount is same so, it’s broke even. Answer: (a) 4. The distance light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 100 years is: (a) 587 × 108 miles, (b) 587 × 1010 miles, (c) 587 × 10-10 miles, (d) 587 × 1012 miles, (e) 587 × 10-12 miles Solution: The distance of the light travels in 100 years is: 5,870,000,000,000 × 100 miles. = 587,000,000,000,000 miles. = 587 × 1012 miles. Answer: (d) 5. A man has$ 10,000 to invest. He invests $4000 at 5 % and$ 3500 at 4 %. In order to have a yearly income of $500, he must invest the remainder at: (a) 6 % , (b) 6.1 %, (c) 6.2 %, (d) 6.3 %, (e) 6.4 % Solution: Income from$ 4000 at 5 % in one year = $4000 of 5 %. =$ 4000 × 5/100.
= $4000 × 0.05. =$ 200.
Income from $3500 at 4 % in one year =$ 3500 of 4 %.
= $3500 × 4/100. =$ 3500 × 0.04.
= $140. Total income from 4000 at 5 % and 3500 at 4 % =$ 200 + $140 =$ 340.
Remaining income amount in order to have a yearly income of $500 =$ 500 - $340. =$ 160.
Total invested amount = $4000 +$ 3500 = $7500. Remaining invest amount =$ 10000 - $7500 =$ 2500.
We know that, Interest = Principal × Rate × Time
Interest = $160, Principal =$ 2500,
Rate = r [we need to find the value of r],
Time = 1 year.
160 = 2500 × r × 1.
160 = 2500r
160/2500 = 2500r/2500 [divide both sides by 2500]
0.064 = r
r = 0.064
Change it to a percent by moving the decimal to the right two places r = 6.4 %
Therefore, he invested the remaining amount $2500 at 6.4 % in order to get$ 500 income every year.
6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was:
(a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much
Solution:
Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.
We know that
Speed = Distance/Time.
Or, Time = Distance/Speed.
So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.
And times taken to covered a distance of 300 miles on his later trip = 300/3x hr.
= 100/x hr.
So we can clearly see that his new time compared with the old time was: twice as much.
7. If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is:
(a) -10.0, (b) -0.5, (c) -0.4, (d) -0.2, (e) 10.0
Solution:
(0.2)x = 2.
Taking log on both sides
log (0.2)x = log 2.
x log (0.2) = 0.3010, [since log 2 = 0.3010].
x log (2/10) = 0.3010.
x [log 2 - log 10] = 0.3010.
x [log 2 - 1] = 0.3010,[since log 10=1].
x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].
x[-0.699] = 0.3010.
x = 0.3010/-0.699.
x = -0.4306….
x = -0.4 (nearest tenth)
8. If 102y = 25, then 10-y equals:
(a) -1/5, (b) 1/625, (c) 1/50, (d) 1/25, (e) 1/5
Solution:
102y = 25
(10y)2 = 52
10y = 5
1/10y = 1/5
10-y = 1/5
9. The fraction (5x-11)/(2x2 + x - 6) was obtained by adding the two fractions A/(x + 2) and B/(2x - 3). The values of A and B must be, respectively:
(a) 5x, -11, (b) -11, 5x, (c) -1, 3, (d) 3, -1, (e) 5, -11
Solution:
10. The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:
(a) 15, (b) 20, (c) 30, (d) 32, (e) 33
Solution:
Let the three numbers be x, y and z.
Sum of the numbers is 98.
x + y + z = 98………………(i)
The ratio of the first to the second is 2/3.
x/y = 2/3.
x = 2/3 × y.
x = 2y/3.
The ratio of the second to the third is 5/8.
y/z = 5/8.
z/y = 8/5.
z = 8/5 × y.
z = 8y/5.
Put the value of x = 2y/3 and z = 8y/5 in (i).
2y/3 + y + 8y/5 = 98
49y/15 = 98.
49y = 98 × 15.
49y = 1470.
y = 1470/49.
y = 30 .
Therefore, the second number is 30.
Unsolved Questions:
1. Fahrenheit temperature F is a linear function of Celsius temperature C. The ordered pair (0, 32) is an ordered pair of this function because 0°C is equivalent to 32°F, the freezing point of water. The ordered pair (100, 212) is also an ordered pair of this function because 100°C is equivalent to 212° F, the boiling point of water.
2. A sports field is 300 feet long. Write a formula that gives the length of x sports fields in feet. Then use this formula to determine the number of sports fields in 720 feet.
3. A recipe calls for 2 1/2 cups and I want to make 1 1/2 recipes. How many cups do I need?
4. Mario answered 30% of the questions correctly. The test contained a total of 80 questions. How many questions did Mario answer correctly? |
# Compound Interest
Imagine you put $100$ in a savings account with a yearly interest rate of $6%$ .
After one year, you have $100+6=106$ . After two years, if the interest is simple , you will have $106+6=112$ (adding $6%$ of the original principal amount each year.) But if it is compound interest , then in the second year you will earn $6%$ of the new amount:
$1.06×106=112.36$
## Yearly Compound Interest Formula
If you put $P$ dollars in a savings account with an annual interest rate $r$ , and the interest is compounded yearly, then the amount $A$ you have after $t$ years is given by the formula:
$A=P{\left(1+r\right)}^{t}$
Example:
Suppose you invest $4000$ at $7%$ interest, compounded yearly. Find the amount you have after $5$ years.
Here, $P=4000$ , $r=0.07$ , and $t=5$ . Substituting the values in the formula, we get:
$\begin{array}{l}A=4000{\left(1+0.07\right)}^{5}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx 4000\left(1.40255\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=5610.2\end{array}$
Therefore, the amount after $5$ years would be about $5610.20$ .
## General Compound Interest Formula
If interest is compounded more frequently than once a year, you get an even better deal. In this case you have to divide the interest rate by the number of periods of compounding.
If you invest $P$ dollars at an annual interest rate $r$ , compounded $n$ times a year, then the amount $A$ you have after $t$ years is given by the formula:
$A=P{\left(1+\frac{r}{n}\right)}^{nt}$
Example:
Suppose you invest $1000$ at $9%$ interest, compounded monthly. Find the amount you have after $18$ months.
Here $P=1000$ , $r=0.09$ , $n=12$ , and $t=1.5$ (since $18$ months = one and a half years).
Substituting the values, we get:
$\begin{array}{l}A=1000{\left(1+\frac{0.09}{12}\right)}^{12\left(1.5\right)}\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\approx 1000\left(1.143960\right)\\ \text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}\text{\hspace{0.17em}}=1143.960\end{array}$
Rounding to the nearest cent, you have $1143.96$ . |
Select Page
# Learning the table of nine by heart
by | Feb 10, 2020 | maths
Learning the table of nine is not difficult if you know the tricks to find the right answer. You can find the solution of every multiplication with nine by using your ten fingers or with a simple calculation trick.
## The trick with your fingers
The first trick is a counting trick and is as follows: first bend the finger with which you want to multiply nine, count from the left to come to the right finger. Then you count all fingers to the left of the bent finger. You will then find the first digit of the answer. The second digit of the solution is then the number of fingers to the right of the bent finger.
#### For example: How do you calculate 4×9?
You now bend the fourth finger from the left. This is the forefinger of your left hand. I count three fingers to the left of the forefinger and 6 fingers to the right of the forefinger. The correct number is then: 36. 4×9 = 36.
If you do not want to work with fingers, you can put ten lines on a piece of paper and then count from the left to the correct dash that you want to multiply by nine. Then you count the number of dashes left and right of this dash, and you have the answer.
## The trick with subtraction sums
Another trick is a calculation trick with subtraction sums and goes as follows: First, you calculate the first digit by subtracting the number you want to multiply by one. After that, the second number equals to the number when you deduct 10 with the number that you want to multiply with.
#### A calculation example with 6×9
The first sum is 6-1, and therefore the first digit is equal to 5. The second sum is 10-6, and this is equal to 4. The result of 6×9 is then equal to 54.
Bear in mind that these tricks only apply to the first ten numbers multiplied by 9. For higher quantities, you first have to split the amount by which you multiply by nine and then add up the answers. How this works, I explain to you in another blog.
You can find more tricks to learn the tables by heart in our fun app “the twelve times tables” With this app, you learn all the skills to learn the first twelve tables by heart quickly. This app has several educational math exercises, which you can easily exercise on your own without help.
maths
Tables & Apples |
# Analytical Reasoning Shortcuts, Examples
In ‘Analytical Reasoning’, we analyse the problems related to the counting of some geometrical figures in a given complex figures. The objective can be met by a systematic analysis of the complex figure given.
The steps involved in the analysis of the complex figure would be clear from the following examples:
## Analytical Reasoning Solved Example
Examples 1) Count the number of triangles in the problem figure.
Solution: As a first step, all the intersecting points in the problem figure are labeled as shown.
Now count the number of simplest triangles. These are ABD, BED, CBF, BEF, FIH, FHE, GDH, DHE and their number is 8.
Next count the triangles which are formed by the combination of two of these simple triangles. These are DBF, BFH, FHD and BHD and their number is 4.
Now since, the triangles formed with the combination of three or more triangles do not exist, the total number of triangles in the figure is 8 + 4 = 12. So (c) is the correct answer.
Example 2) Find the number of straight lines in the figure.
Solution: First the given problem figure is labeled as shown.
Now, count the horizontal lines, namely AC, HD and GE and their number is 3.
Now, count the vertical lines AG, BF and CE and their number is also 3.
Finally, count the six (6) slanting lines namely – BH, CG, DF, BD, AE and HF.
Thus, the total number of straight lines = 3 + 3 + 6 = 12
Example 3) Count the number of squares in the given problem figure.
Solution: As usual we first label the given problem figure as shown. Now, the squares formed with two components are-namely, ABJI, BCKJ, CDLK, IJGH, JKFG and KLEF. Their number in 6.
Similarly, a square made of four components is only 1, i.e.., CJFL
Finally, the squares composed of eight components are ACFH and BDEG. So their number is 2.
The total number of squares = 6 + 1 + 2 = 9.
Example 4) Count the number of straight lines in the given problem figure.
Solution: The given problem figure may be labeled as shown:
Now we have
(i) Three (3) horizontal lines namely, AC, HD and GE.
(ii) Three (3) slanting lines namely, AG, BF and CE. And
(iii) Six (6) slanting lines namely, BH, CG, DF, BD, AE and HF.
Total No. of lines = 3+3+6 = 12
So (a) in the correct answer.
Example 5) Find the number of parallograms in the given problem figure.
Solution: First the given figure is labeled as shown.
Now,
(i) We have 6 simplest parallelograms namely, ABKJ, BCLK, CDEL, JKHI, KLOGH, LEFG.
(ii) The parallelograms composed of two components are ACLJ, BDEK, CKFG, EFHK; LGIJ, HIAB and, BCGH and their number is 7.
(iii) The parallogram composed of four components are ACGI and BDFH and their number is 2.
(iv) The parallogram composed of six elements is ADFI, i.e.., one only.
The total number of parallelograms = 6+7+2+1 = 16 |
Edit Article
# wikiHow to Find the Reciprocal
Reciprocals are helpful in all sorts of algebraic equations. For example when you are dividing one fraction by another you multiply the first by the Reciprocal of the 2nd. You might also need reciprocals when finding equations of lines.
### Method 1 Finding the Reciprocal of a Fraction or Whole Number
1. 1
Find the reciprocal of a fraction by flipping it. The definition of "reciprocal" is simple. To find the reciprocal of any number, just calculate "1 ÷ (that number)." For a fraction, the reciprocal is just a different fraction, with the numbers "flipped" upside down (inverted).[1]
• For instance, the reciprocal of 3/4 is 4/3.
2. 2
Write the reciprocal of a whole number as a fraction. Again, the reciprocal of a number is always 1 ÷ (that number). For a whole number, write that as a fraction; there's no point in calculating it out to a decimal.
• For instance, the reciprocal of 2 is 1 ÷ 2 = 1/2.
### Method 2 Finding the Reciprocal of a Mixed Number
1. 1
Identify a mixed number. Mixed numbers are part whole number and part fraction, such as 24/5. There are two steps to finding the reciprocal of a mixed number, explained below.
2. 2
Change it to an improper fraction. Remember, the number 1 can always be written as (number)/(the same number), and fractions with the same denominator (lower number) can be added together. Here's an example with 24/5:
• 24/5
• = 1 + 1 + 4/5
• = 5/5 + 5/5 + 4/5
• = (5+5+4)/5
• = 14/5.
3. 3
Flip the fraction. Once the number is written entirely as a fraction, you can find the reciprocal just like you would with any fraction: by flipping it.
• In the example above, the reciprocal of 14/5 is 5/14.
### Method 3 Finding the Reciprocal of a Decimal
1. 1
Change it to a fraction if possible. You might recognize some common decimal numbers that can easily be turned into fractions. For instance, 0.5 = 1/2, and 0.25 = 1/4. Once in fraction form, just flip the fraction to find the reciprocal.
• For instance, the reciprocal of 0.5 is 2/1 = 2.
2. 2
Write out a division problem. If you can't change it to a fraction, calculate the reciprocal of that number as a division problem: 1 ÷ (the decimal). You can use a calculator to solve this, or continue on to the next step to solve it by hand.
• For example, you can find the reciprocal of 0.4 by calculating 1 ÷ 0.4.
3. 3
Change the division problem to use whole numbers. The first step to dividing decimals is to move the decimal point until all the numbers involved are whole numbers. As long as you move the decimal point the same number of spaces for both numbers, you'll get the correct answer.
• For example, you can take 1 ÷ 0.4 and rewrite it as 10 ÷ 4. In this case, you've moved each decimal place one space to the right, which is the same as multiplying each number by ten.
4. 4
Solve the problem using long division. Use long division techniques to calculate the reciprocal. If you calculate it for 10 ÷ 4, you'll get the answer 2.5, the reciprocal of 0.4.
## Community Q&A
Search
• What is the reciprocal of a squared number?
It is 1 divided by the squared number.
• What is the reciprocal of a+b?
1 / (a+b).
• How do I use a mathematical table to find reciprocal of a decimal number?
Some tables do have a column devoted to reciprocals, but not typically reciprocals of decimal numbers. Your best bet is simply to divide the decimal number into 1.
• How do I find the reciprocal of 4?
wikiHow Contributor
The number 4 as a fraction is 4/1. To find the reciprocal of a number, the numerator and denominator switch places, so the reciprocal of 4 is 1/4.
• How do I get the reciprocal of a mixed number?
wikiHow Contributor
Convert the mixed number into an improper fraction, then find the reciprocal as you normally would.
• Reciprocal of negative number?
The reciprocal of a negative number is the same as the reciprocal of the positive number, except it's a negative value.
• What is the reciprocal of 2.72?
wikiHow Contributor
The reciprocal of 2.72 is (1 divided by 2.72)=0.367647 to six significant figures.
• What is the reciprocal of 8?
wikiHow Contributor
1/8. Since every whole number is invisibly divided by 1 (in this case, 8/1), you just interchange the numerator and denominator, putting the denominator over the numerator and there is your reciprocal, 8/1. In other words, a reciprocal takes a number a/b and turns it into b/a.
• What is the reciprocal of 7^3 in fraction, is it (1/7)^3 or 7^-3?
Both are correct.
• What is the reciprocal of -2.45?
It is -1 / 2.45.
• What is the product of 2/n and its reciprocal?
• How do I convert a power term into a reciprocal?
200 characters left
## Tips
• A numbers negative reciprocal is the same as the regular reciprocal, multiplied by negative one.[2] For instance, the negative reciprocal of 3/4 is -4/3.
• The reciprocal is sometimes called the "multiplicative inverse."[3]
• The number 1 is its own reciprocal, since 1 ÷ 1 = 1.
• The number 0 does not have a reciprocal, since 1 ÷ 0 is undefined.[4]
## Article Info
Categories: Multiplication and Division | Fractions
In other languages:
Português: Achar o Recíproco de um Número, Deutsch: Den Kehrwert bestimmen, Español: encontrar el recíproco, Русский: найти обратное число, Français: trouver l'inverse, Italiano: Calcolare il Reciproco, Bahasa Indonesia: Mencari Resiprok atau Kebalikan
Thanks to all authors for creating a page that has been read 78,945 times. |
# Roof
Tiles are stacked in rows on the trapezoidal shaped roof.
At the ridge is 15 tiles and each subsequent row has one more tile than in the previous row.
How many tiled is covered roof if lowermost row has 37 tiles?
Result
x = 598
#### Solution:
Leave us a comment of example and its solution (i.e. if it is still somewhat unclear...):
Be the first to comment!
## Next similar examples:
1. Seats
Seats in the sport hall are organized so that each subsequent row has five more seats. First has 10 seats. How many seats are: a) in the eighth row b) in the eighteenth row
2. Sequence
Between numbers 1 and 53 insert n members of the arithmetic sequence that its sum is 702.
3. Sequence 3
Write the first 5 members of an arithmetic sequence: a4=-35, a11=-105.
4. AS sequence
In an arithmetic sequence is given the difference d = -3 and a71 = 455. a) Determine the value of a62 b) Determine the sum of 71 members.
5. Theorem prove
We want to prove the sentense: If the natural number n is divisible by six, then n is divisible by three. From what assumption we started?
6. Third member
Determine the third member of the AP if a4=93, d=7.5.
7. AP - basics
Determine first member and differentiate of the the following sequence: a3-a5=24 a4-2a5=61
8. AS - sequence
What are the first ten members of the sequence if a11=22, d=2.
9. Sequence
Write the first 7 members of an arithmetic sequence: a1=-3, d=6.
10. Sequence
Write the first 6 members of these sequence: a1 = 5 a2 = 7 an+2 = an+1 +2 an
11. Sequence 2
Write the first 5 members of an arithmetic sequence a11=-14, d=-1
12. AP - simple
Determine the first nine elements of sequence if a10 = -1 and d = 4
13. 6 terms
Find the first six terms of the sequence. a1 = 7, an = an-1 + 6
14. AP - simple
Find the first ten members of the sequence if a11 = 132, d = 3.
15. Complex number coordinates
Which coordinates show the location of -2+3i
16. Median and modus
Radka made 50 throws with a dice. The table saw fit individual dice's wall frequency: Wall Number: 1 2 3 4 5 6 frequency: 8 7 5 11 6 13 Calculate the modus and median of the wall numbers that Radka fell.
17. Trinity
How many different triads can be selected from the group 43 students? |
Homework4Solns
# Homework4Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008...
This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up to access the rest of the document.
Unformatted text preview: SHORT CALCULUS Math 16C Sec 1 Fall 2008 Homework #4 Solutions Peter Malkin Section 7.6 Question 2 Let F ( x,y,λ ) = xy- λ (2 x + y- 4). F x = y- 2 λ = 0 ⇔ λ = 1 2 y (1) F y = x- λ = 0 ⇔ λ = x (2) F λ =- (2 x + y- 4) = 0 (3) Using equation (1) and (2) gives 1 2 y = x ⇒ y = 2 x . Putting this into equation (3) gives 2 x + 2 x- 4 = 0 ⇒ x = 4 , so y = 2. Thus, f ( x,y ) = xy is maximum at the point (1 , 2), and the maximum value is f (1 , 2) = 2. Question 6 Let F ( x,y,λ ) = x 2- y 2- λ ( x- 2 y + 6). F x = 2 x- λ = 0 (4) F y =- 2 y + 2 λ = 0 (5) F λ =- ( x- 2 y + 6) = 0 (6) Equation (1) implies λ = 2 x , and substituting this into (2) gives,- 2 y + 4 x = 0, which implies y = 2 x . Substituting this into (3) gives- ( x- 4 x + 6) = 0 ⇒ x = 2 . So, y = 2 x = 4. Thus, f ( x,y ) is maximum at (2 , 4), and the maximum value is f (2 , 4) =- 12. Question 16 Let F ( x,y,λ ) = x 2- 8 x + y 2- 12 y + 48- λ ( x + y- 8). F x = 2 x- 8- λ = 0 (1) F y = 2 y- 12- λ = 0 (2) F λ =- ( x + y- 8) = 0 (3) (1)- (2) : 2 x- 2 y + 4 = 0 ⇒ x- y + 2 = 0 (4) (4) + (3) :- 2 y + 10 = 0 ⇒ y = 5 (5) Substituting y = 5 into (3) gives- ( x + 5- 8) = 0 ⇒ x = 3. So, f ( x,y ) is maximum at (3 , 5), and the maximum value is f (3 , 5) = 3 2- 8 · 3 + 5 2- 12 · 5 + 48 =- 2. 1 Question 18 Let F ( x,y,z,λ ) = x 2 y 2 z 2- λ ( x 2- y 2- z 2- 1). F x = 2 xy 2 z 2- λ 2 x = 0 ⇒ 2 x ( λ- y 2 z 2 ) = 0 ⇒ x = 0 or λ = y 2 z 2 (1) F y = 2 yx 2 z 2- λ 2 y = 0 ⇒ 2 y ( λ- x 2 z 2 ) = 0 ⇒ y = 0 or λ = x 2 z 2 (2) F z = 2 zx 2 y 2- λ 2 z = 0 ⇒ 2 z ( λ- x 2 y 2 ) = 0 ⇒ z = 0 or λ = x 2 y 2 (3) F λ =- ( x 2- y 2- z 2- 1) = 0 (4) Note that we can ignore the cases where x = 0, y = 0, or z = 0 since they do not give a maximum.= 0 since they do not give a maximum....
View Full Document
{[ snackBarMessage ]}
### Page1 / 6
Homework4Solns - SHORT CALCULUS Math 16C Sec 1 Fall 2008...
This preview shows document pages 1 - 3. Sign up to view the full document.
View Full Document
Ask a homework question - tutors are online |
## Finding Cube Root – Vedic Maths Way
This is an amazing trick which was always appreciated by the audience I have addressed in various workshops. This awe inspiring technique helps you find out the cube root of a 4 or 5 or 6 digits number mentally.
Before going further on the method to find the cube root, please make a note of the following points –
1) Cube of a 2-digit number will have at max 6 digits (99^3 = 970,299). That implies if you are given with a 6 digit number, its cube root will have 2 digits.
2) This trick works only for perfect cubes, it will not work for any arbitrary 6-digit
3) It works only for integers Read More
## Shortcut to find the Cube of a number
Very often we have to find the cube, i.e. third power of 2 digit numbers. Cubes of very large numbers are rarely used.
Cubes of all the single digits should be memorized. Find below the table of cubes of first ten natural numbers –
13 = 1, 23 = 8, 33 = 27, 43 = 64, 53 = 125,
63 = 216, 73 = 343, 83 = 512, 93 = 729, 103 = 1000
To find the cube of any 2 digit number, we have to take the following steps
First Step: The first thing we have to do is to put down the cube of the tens-digit in a row of 4 figures. The other three numbers in the row of answer should be written in a geometrical ratio in the exact proportion which is there between the digits of the given number.
Second Step: The second step is to put down, under the second and third numbers, just two times of second and third number. Then add up the two rows.
Finding the cube of 12
Or, 123 = ?
First Step: Digit in ten’s place is 1, so we write the cube of 1. And also as the ratio between 1 and 2 is 1:2, the next digits will be double the previous one. So, the first row is
1 2 4 8
Step II: In the above row our 2nd and 3rd digits (from right) are 4 and 2 respectively. So, we write down 8 and 4 below 4 and 2 respectively. Then add up the two rows.
Ex 2: 163 = ?
Soln:
Explanations: 13 (from 16) = 1. So, 1 is our first digit in the first row. Digits of 16 are in the ratio 1:6, hence our other digits should be 1×6 = 6, 6×6 = 36, 36×6 = 216. In the second row, double the 2nd and 3rd number is written. In the third row, we have to write down only one digit below each column (except under the last column which may have more than one digit). So, after putting down the unit-digit, we carry over the rest to add up with the left-hand column. Here,
i) Write down 6 of 216 and carry over 21.
ii) 36 + 72 + 21 (carried) = 129, write down 9 and carry over 12.
iii) 6 + 12 + 12 (carried) = 30, write down 0 and carry over 3.
iv) 1 + 3 (carried) = 4, write down 4. |
# 1985 AHSME Problems/Problem 28
## Problem
In $\triangle ABC$, we have $\angle C = 3\angle A$, $a = 27$ and $c = 48$. What is $b$?
$[asy] defaultpen(linewidth(0.7)+fontsize(10)); pair A=origin, B=(14,0), C=(10,6); draw(A--B--C--cycle); label("A", A, SW); label("B", B, SE); label("C", C, N); label("a", B--C, dir(B--C)*dir(-90)); label("b", A--C, dir(C--A)*dir(-90)); label("c", A--B, dir(A--B)*dir(-90));[/asy]$
$\mathrm{(A)\ } 33 \qquad \mathrm{(B) \ }35 \qquad \mathrm{(C) \ } 37 \qquad \mathrm{(D) \ } 39 \qquad \mathrm{(E) \ }\text{not uniquely determined}$
## Solution 1
Let $\angle A = x^{\circ}$, so $\angle C = 3x^{\circ}$, and thus $\angle B = \left(180-4x\right)^{\circ}$. Now let $D$ be a point on side $AB$ such that $\angle ACD = x^{\circ}$, so $\angle BCD = 3x^{\circ}-x^{\circ} = 2x^{\circ}$, which gives $$\angle CDB = 180^{\circ}-2x^{\circ}-\left(180-4x\right)^{\circ} = 2x^{\circ},$$ meaning that $\triangle CDB$ and $\triangle CDA$ are both isosceles, with $BC = BD$ and $AD = CD$. In particular, $BD = BC = 27$ and $CD = AD = AB-BD = 48-27 = 21$. Hence by Stewart's theorem on triangle $ABC$, \begin{align*}&BD \cdot AB \cdot AD + CD^2 \cdot AB = AC^2 \cdot BD + BC^2 \cdot AD \\ &\iff 27 \cdot 48 \cdot 21 + 21^2 \cdot 48 = AC^2 \cdot 27 + 27^2 \cdot 21 \\ &\iff AC^2 = \frac{27(21)(48-27) + 21^2 \cdot 48}{27} \\ &\iff AC^2 = \frac{21^2(27+48)}{27} \\ &\iff AC^2 = \frac{21^2 \cdot 25}{9} \\ &\iff AC = \frac{21 \cdot 5}{3} \qquad \text{(as } AC > 0\text{)} \\ &\iff AC = \boxed{\text{(B)} \ 35}.\end{align*}
## Solution 2
We apply the law of sines in the form $$\frac{\sin(A)}{a} = \frac{\sin(C)}{c},$$ yielding $$\frac{\sin(A)}{27} = \frac{\sin(3A)}{48} \iff 9\sin(3A) = 16\sin(A).$$
Now, the angle sum and double angle identities give \begin{align*}\sin(3A) &= \sin(2A+A) \\ &= \sin(2A)\cos(A)+\cos(2A)\sin(A) \\ &= \left(2\sin(A)\cos(A)\right)\cos(A)+\left(\cos^2(A)-\sin^2(A)\right)\sin(A) \\ &= 2\sin(A)\cos^2(A)+\sin(A)\cos^2(A)-\sin^3(A) \\ &= 3\sin(A)\left(1-\sin^2(A)\right)-\sin^3(A) \\ &\text{(using the further identity } \cos^2(\theta)+\sin^2(\theta) = 1\text{)} \\ &= 3\sin(A)-4\sin^3(A).\end{align*}
Thus our equation becomes \begin{align*}9\left(3\sin(A)-4\sin^3(A)\right) = 16\sin(A) &\iff 27\sin(A)-36\sin^3(A) = 16\sin(A) \\ &\iff 36\sin^3(A) = 11\sin(A) \\ &\iff \sin(A) = 0 \text{ or } \pm\frac{\sqrt{11}}{6}.\end{align*} Notice, however, that we must have $0^{\circ} < A < 45^{\circ}$, the latter because otherwise $A+3A \geq 180^{\circ}$, which would contradict the fact that $A$ and $3A$ are angles in a (non-degenerate) triangle. This means $\sin(A) > 0$, so the only valid solution is $$\sin(A) = \frac{\sqrt{11}}{6},$$ and the fact that $A$ is acute also means $\cos(A) > 0$, so we deduce \begin{align*}\cos(A) &= \sqrt{1-\left(\frac{\sqrt{11}}{6}\right)^2} \\ &= \sqrt{1-\frac{11}{36}} \\ &=\frac{5}{6}.\end{align*} Accordingly, using the double angle identities again, \begin{align*}\sin(4A) &= \sin(2 \cdot 2A) \\ &= 2\sin(2A)\cos(2A) \\ &= 2\left(2\sin(A)\cos(A)\right)\left(\cos^2(A)-\sin^2(A)\right) \\ & =2\left(2 \cdot \frac{\sqrt{11}}{6} \cdot\frac{5}{6}\right)\left(\left(\frac{5}{6}\right)^2-\left(\frac{\sqrt{11}}{6}\right)^2\right) \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{25-11}{36} \\ &= \frac{5\sqrt{11}}{9} \cdot \frac{7}{18} \\ &= \frac{35\sqrt{11}}{162}.\end{align*} Finally, the law of sines now gives \begin{align*}\frac{\sin(A)}{27} &= \frac{\sin(B)}{b} \\ &= \frac{\sin(180^{\circ}-3A-A)}{b} \\ &= \frac{\sin(4A)}{b} \qquad \text{(using the identity } \sin\left(180^{\circ}-\theta\right) = \sin(\theta)\text{)},\end{align*} so, substituting the above results, $$\frac{\left(\frac{\sqrt{11}}{6}\right)}{27} = \frac{\left(\frac{35\sqrt{11}}{162}\right)}{b} \iff b = \frac{6 \cdot 27 \cdot 35}{162} = \boxed{\text{(B)} \ 35}.$$ |
In this page 'Addition formula questions' you can see questions involving addition of fractions. We have also given the solution for these problems.Students can try to solve them on their own. If you have any doubts in this problem you can see this solution to get clear idea.
Adding two fractions are very easy if the denominators are same(like fractions). If the denominators are not same then we have to convert each fraction for the common denominator so that it will be easy for us to add the fractions. The following method shows how easily we can convert the given unlike fractions to like fractions.
### Difference:
If we want to add two fractions with different denominators we need to take L.C.M for the two fractions and we need to change them as same.Then only we can add the numerators.But in our method we can easily apply the formula and no need worry about taking L.C.M for the denominators.This formula will be very useful to the students who are beginner to Math.By using this formula we can easily get answer for a difficult problem too.After applying this formula we need to simplify the final answer to get the simplified form.Any way this new method will be very useful to the beginners.We have also given examples for each problem to make you understand more clear.We hope this will be very helpful to the students who are facing difficulties in doing problems in math.We have also given worksheets for each category of topics.
By practicing this worksheets you can get more knowledge.These are the most basic topics in math.Students must practice this kind of worksheets.The students who are in the elementary grade must practice this kinds of worksheets to get more knowledge in math. In this page you can get a clear idea in this topic and you have worksheets which consist of five question with clear steps has given.
Question 1:
(A) 18/56 (B) 17/49 (C) 15/48 (D) 23/39
Question 2:
(A) 18/56 (B) 19/35 (C) 19/48 (D) 14/55
Question 3:
(A) 29/58 (B) 49/51 (C) 43/77 (D) 41/56
Solution
Question 4:
(A) 29/58 (B) 39/51 (C) 37/170 (D) 41/126
Solution
Question 5:
(A) 9/10 (B) 9/11 (C) 7/17 (D) 11/12
Solution
After solving these problems in 'Addition formula questions' Students can try the worksheets whose link had been given in 'solutions' page. Parents and teachers can encourage the students to do the problems on their own. If they find any difficulty they can compare their work with solutions. Once the students complete the worksheet they will master in 'fraction-addition'. If you have any doubt you can contact us through mail, we will help you to clear your doubts. |
Select Page
Linear Algebric Equations
OUTLINE
6.1 Elementary Solution Methods
6.2 Matrix Methods for Linear Equations
6.3 Cramer’sMethod
6.4 Underdetermined Systems
6.5 Overdetermined Systems
6.6 Summary
Problems
, ..:
Linear algebraic equations such ‘~s
5x – 2y = 13
7x + 3y = 24
occur in many engineering applications. For example, electrical engineers use them to predict the power requirements for circuits; civil, mechanical, and aerospace engineers use them to design structures and machines; chemical engineers use them to compute material balances in chemical processes; and industrial
engineers apply them to design schedules and operations. The examples and homework problems in this chapter explore some of these applications. Linear algebraic equations can be solved “by hand” using pencil and paper, by calculator, or with software such as MATLAB. The choice depends on the circumstances. For equations with only two unknown variables, hand solution is easy and adequate. Some calculators can solve equation sets that have many variables. However, the greatest power and flexibility is obtained by using software.
For example, MATLAB can obtain and plot equation solutions as we vary one or more parameters. Without giving a formal definition of the term linear algebraic equations, let us simply say that their unknown variables never appear raised to a power other than unity and never appear as products, ratios, or in transcendental functions
such as In(x), e”, and cos x. The simplest linear equation is ax = b, which has the solution x = bf a if a i’ O.
In contrast, the following equations are nonlinear: which has the solutions
x = ±.J3, and
sinx = 0.5
r
which has the solutions x = 300
, 1500
, 3900
, 5100
, •••• In contrast to most nonlinear
equations, these particular nonlinear equations are easy to solve. For example,
we cannot solve the equation x +2e-x – 3 = 0 in “closed form”; that is,
we cannot express the solution as a function. We must obtain this solution numerically,
as explained in Section 3.2. The equation has two solutions: x = -0.5831
and x = 2.8887 to four decimal places.
Sets of equations are linear if all the equations are linear. They are nonlinear
if at least one of the equations is nonlinear. For example, the set
8x – 3y = 1
6x +4y = 32
is linear because both equations are linear, whereas the set
6xy – 2x = 44
. 5x – 3y =-2
is nonlinear because of the product term xy. Systematic solution methods have been developed for sets of linear equations. However, no systematic methods are available for nonlinear equations because
the nonlinear category covers such a wide range of equations. In this chapter we first review methods for solving linear equations by hand, and we use these methods to develop an understanding of the potential pitfalls that can occur when solving linear equations. Then we introduce some matrix notation that is required for use with MATLAB and that is also useful for expressing solution methods in a compact way. The conditions for the existence and uniqueness of solutions are then introduced. Methods using MATLAB are then treated in four sections: Section 6.2 covers equation sets that have unique solutions; Section 6.3 covers ‘Cramer’s method; Sections 6.4 and 6.5 explain how to determine whether a set has a unique solution, multiple solutions, or no solution at all. |
Upcoming SlideShare
×
Thanks for flagging this SlideShare!
Oops! An error has occurred.
×
Saving this for later? Get the SlideShare app to save on your phone or tablet. Read anywhere, anytime – even offline.
Standard text messaging rates apply
# Multiplication lesson
180
views
Published on
Published in: Education
0 Likes
Statistics
Notes
• Full Name
Comment goes here.
Are you sure you want to Yes No
• Be the first to comment
• Be the first to like this
Views
Total Views
180
On Slideshare
0
From Embeds
0
Number of Embeds
0
Actions
Shares
0
0
0
Likes
0
Embeds 0
No embeds
No notes for slide
### Transcript
• 1. We have added and subtracted fractions. In this lesson we will multiply fractions. When we add and subtract fractions, we count how many of the same size parts there are. When we multiply fractions, the sizes of the parts change. Consider this multiplication problem: How much is one half of one half?
• 2. How much is one-half of one-half? Using a fraction manipulative, we show one-half of a circle. To find one- half of one- half, we divide the half in half. Written out the Problem looks like this: 1 1 1 2 2 4 X = We see this equals one- fourth.
• 3. What fraction is one half of three fourths? Using a fraction manipulative, we show three fourths of a circle. To find one half of three fourths, we divide each fourth in half. Since one half of one fourth is one eighth, one half of three fourths is three eighths. We may also find one half of three fourths by multiplying. 1 3 3 2 4 8 X = EXAMPLE 1:
• 4. Multiply: We find two thirds of four fifths by multiplying. 2 4 3 5 X = EXAMPLE 2: 2 4 8 3 5 15 X = Now You’re Cooking!
• 5. 3/5 x 1/2 Now that you know how to multiply fractions, let’s try some together. 1/3 x 2/3 2/9 3/10 2/3 x 2/3 4/9 1/2 x 2/2 2/4 3/4 x 1/2 3/8 1/2 x 1/3 1/6 1/2 x 1/4 1/8
• 6. 3/5 x 1/2 Now that you know how to multiply fractions, let’s try some together. 1/3 x 2/3 2/9 3/10 2/3 x 2/3 4/9 1/2 x 2/2 2/4 3/4 x 1/2 3/8 1/2 x 1/3 1/6 1/2 x 1/4 1/8 |
# How to Convert a Fraction to a Decimal – and Why
#### (Archive Question of the Week)
Fractions are a frequent source of questions from elementary students. I plan to devote several posts to various aspects of this, from using a common denominator to add fractions, to flipping and multiplying to divide fractions, to converting between improper fractions and mixed numbers. Here, I want to look at one answer to a teacher that covers several aspects of how fractions are related to decimals.
## Why do you divide the numerator by the denominator?
Here is the question, from teacher Erin in 2001:
Dividing Fractions to find Decimals
I have some students who do not understand dividing the numerator into the denominator to find out the decimal. I cannot give them a reason why you do this except that it works. Why does this operation work to give you the decimal form of a fraction?
First, let’s check out what Erin is asking about. To convert the fraction 3/4 to a decimal, students are taught to divide 3 by 4, which gives the answer 0.75. Or, to convert 8/9 to a decimal, they divide 8 by 9 and get the answer 0.888… , a non-terminating, repeating decimal, in contrast to the terminating decimal in the first example. Why do we do this? Knowing why can help the students remember what to do; it can also help them see that math makes sense, rather than being a list of rules a teacher tells you.
Since students often get the terminology wrong here, I had to start by pointing out that Erin didn’t quite say what she meant:
Let's be careful with our terminology, first: you mean dividing the numerator BY the denominator! A lot of kids get that wrong.
We sometimes describe “6 divided by 2″ as “how many times does 2 go into 6″, and then as “divide 2 into 6″, but this can then easily get turned around into “6 divided into 2″, which is not correct. I discussed this and related issues here:
Wording Division Problems
## A fraction is a division
I continued into the heart of the matter:
The basic reason is that division is really what fractions are all about. In a very real sense, we can say that a fraction is simply a division we haven't bothered to perform yet, a division problem frozen in time.
We write "3/4" to mean "I want to divide 3 by 4, but I don't want to do the work just yet, so I can simplify the work before I finish up." That's why we use the virgule "/" or the horizontal fraction bar to signify division in more advanced math (and in computer programming languages), rather than the old-fashioned obelus.
That is, where in elementary arithmetic we write division as $$10 \div 5$$, using the symbol called an obelus, mathematicians more commonly use the horizontal bar, $$\frac{10}{5}$$ or the virgule or slash, 10/5. These show that we think of division as being the same thing as a fraction.
But why?
Let's go back to the basics of fractions to see why I can say this. What are fractions? The essence of a fraction is a division, a breaking into pieces. Take a whole object and divide it into 5 (equal) pieces; each piece is 1/5. We've divided 1 by 5, just as we divide 10 by 5 by dividing a set of 10 things into 5 parts, each of which consists of 2 objects. So 1 divided by 5 is 1/5; and 10 divided by 5 is the fraction 10/5, which simplifies to 2.
Likewise, we can divide 2 pies into 5 parts by dividing each pie into 5 parts (fifths) and taking 2 of them at a time: 2 divided by 5 is 2/5. The denominator represents the number of parts we divide each whole into (a divisor); the numerator represents a multiplier, the number of parts we have.
Let me repeat that, because it's easy to miss: 2/5 MEANS 2 divided by 5. The fraction IS a division.
Let’s do that in pictures:
(I used rectangular “cakes” instead of round “pies”, to make the comparison easier.)
So, 1 divided by 5 means dividing 1 into 5 equal parts and taking 1 of them; the result is the fraction 1/5. And 2 divided by 5 means dividing 2 into 5 equal parts, each of which is 2/5. In each case, the fraction is what you get when you divide.
At a higher level, moving away from pictures of cakes or pies, we define division as the inverse of multiplication — that is, working backward to find the missing factor that will give a known product. For example, we say that $$10 \div 5 = 2$$ because 2 is the number you can multiply by 5 to get 10: $$10 = 5 \times 2$$. This is how we check a division: Do the multiplication and see if you get the right number back. So we can check my claim that $$2 \div 5 = \frac{2}{5}$$ by multiplying: Is it true that $$2 = 5 \times \frac{2}{5}$$? Yes, it is.
One more little perspective on it:
Now look at the operations on fractions. When we multiply by a fraction, we're really dividing: 1/2 * 10 is half of ten, or ten divided by 2, or 10/2 again. Fractions mean division.
Half of something means dividing by 2; and in general, multiplying by a fraction “1 over something” means dividing by that “something”.
## To convert, we just do what it says
Now, if a fraction is a division, what does that imply?
So when we want to convert a fraction to an ordinary (decimal) number, all we're really doing is waking up a division problem that has been in suspended animation, and letting it continue: "Where was I? Oh, yeah ... 3 divided by 4 ... that's 0.75."
In other words, if we want to get 3/4 in a different form, namely a decimal, we just do the division the fraction represents, in terms of decimals.
For more on converting fractions to decimals (and more), see our FAQ:
Converting Fractions, Decimals, and Percents
This page demonstrates two ways to convert a fraction to a decimal (and also how to convert a simple decimal to a fraction):
Converting Fractions and Decimals
Also, this answer takes what I said above one step further, looking at why we use the fraction bar to represent division in algebra as well as arithmetic:
2a/5 or 2a Divided by 5?
Finally, here is a 2009 restatement of the ideas I presented above:
Why Does Division Convert a Fraction to a Decimal?
The last paragraph says it well:
Now, when we convert a fraction to a decimal, all we are doing is carrying out, in decimal form, the division that the fraction represents. That is, in my example, we've said that 3/4 means 3 divided by 4, so we go ahead and divide 3.00 by 4 and get 0.75. Since both 3/4 and 0.75 are obtained by dividing 3 by 4, they are two ways to express the same number!
This site uses Akismet to reduce spam. Learn how your comment data is processed. |
If you're seeing this message, it means we're having trouble loading external resources on our website.
If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.
Square roots of perfect squares
Learn how to find the square root of perfect squares like 25, 36, and 81.
Let's start by taking a look at an example evaluating the square root of $25$:
$\sqrt{25}=\phantom{\rule{0.167em}{0ex}}?$
Step 1: Ask, "What number squared equals $25$?"
Step 2: Notice that $5$ squared equals $25$.
${5}^{2}=5×5=25$
$\sqrt{25}=5$
Here's a question to make sure you understood:
How can we be sure that $5$ is the right answer?
Connection to a square
Finding the square root of $25$ is the same as finding the side length of a square with an area of $25$.
A square with an area of $25$ has a side length of $5$.
Practice Set 1:
Problem 1A
${4}^{2}=$
Reflection question
Which claim shows how square roots work?
Practice Set 2:
Problem 2A
$\sqrt{1}=$
Practice Set 3:
Problem 3A
$\sqrt{121}=$
Want to join the conversation?
• Can an exponent be a negative number?
• You already know that an exponent represents the number of times you have to multiply a number by itself. A negative exponent is equivalent to the inverse of the same number with a positive exponent. There is nothing special about solving a problem that includes negative exponentials.
• What happens if try to find the
the square root of an imperfect square
• Most probably it will be an irrational number in which case we can only approximate its value.
However, sometimes we can write the radicand as a fraction of two perfect squares.
Example: √5.76 = √(576∕100) = √(144∕25) = √(12²∕5²) = 12∕5 = 2.4
• I just noticed something really interesting (I think):
If I can't remember a square of some numbers (7^2 and 8^2 can be a bit tricky for me for some reason) but I remember the square number of the root that comes before it (6 and 36 in the case I'm trying to find 7^2),
I can do 36+6 to make it into 6*7, and then add a 7 to make it into a 7*7.
I tried to play around and find a rule and I think I found the formula:
n^2 + 2n + 1 = ( n + 1 )^2
(where n is the root number of the square that you do know).
if you would visualize the numbers on a grid, the n^2 is the area of the square, while the 2n+1 is the number of the additional units the is added on the side.
It's easier to see it on a times table: When looking at 25, the number diagonally next to it is 36. If you count the 'units' (the other multiplies that are on the same axes that leads toward 36 IE: 3,6,9 up to 36 on both sides) they will be the same as the 2n+1.
I tried to find a formula for a square of a root that isn't immediately follows the root I know. example: 3^2 = 9, 5^2 = ?.
unfortunately I couldn't think on one consistent formula, because there is always a need to add more and (n+1) with more additional 1s the further the number is.
does the formula I found have a name? I'm pretty sure I wasn't the first to think of that lol
• You can use the pascal's triangle I suppose
Edit1: here: https://en.wikipedia.org/wiki/Polynomial_expansion
Edit2: Sorry that is not true. That applies to expansions with higher powers, but not between different sqaures.
But maybe you can look at sum of squares where
∑x^2 = n(n+1)(2n+1)/6
• So, a perfect square is basically the answer to an exponent? (ex. 2^2(4), 12^12(144))
• it is an integer with a square root that is also an integer
• How can we use square roots in life?
• There are many ways in life you can use square roots. Some of them include a mathmatician, engineer, architect, carpenter, etc.
• It's actually a lot easier than I thought
• How do you solve square roots that can't be squared easily?
• I suggest using one of these 2 methods:
1. Long Division Method: A manual method that involves guessing, checking, and long division. It's useful for finding the square root of any number to a desired precision.
2. Newton's Method: A numerical method that uses calculus. It's similar to the Babylonian method but uses a different formula for improving the guess. It's often faster and more accurate for larger numbers or numbers with many decimal places.
(1 vote)
• How to calculate imperfect squares?
• imperfect square roots (example: 32 SQUARE)(I am going to do square uppercased because I don't know how to make a square root sign)
32 SQUARE
25 SQUARE
than the other closest
36 SQUARE
than just find the root of these numbers
5
6
it is between 5 and 6
hope this helped and may God bless your day
• How do I find the square root for bigger and not perfect numbers?
• There are a couple of ways:
1. You can approximate the value. For instance, if I have $\sqrt{70}$, I know that it'll lie between $\sqrt{64}$ and $\sqrt{81}$. As 70 is closer to 64, $\sqrt{70}$ will be closer to $\sqrt{64}$, which is 8. And if you find $\sqrt{70}$, you'll see that it is around $8.366$, which is closer to 8 than 9.
2. You can approximate the square root function with a tangent line at a nearby point. Suppose you want $\sqrt{65}$. You already know $\sqrt{64}$. So, you can first graph $y=\sqrt{x}$, then draw a tangent at $x = 64$. Find the equation of the tangent, and substitute for $x = 65$. As the tangent approximates the square root function, you'll get an rough approximation of $\sqrt{65}$.
Using this, you can get the equation of the tangent line as $y = \frac{x}{16}+4$. If I substitute $x = 65$, I get $y = 8.0625$. Now, the actual value of $\sqrt{65}$ is $8.06225$. See how good our approximation was!
3. The final method is something called a Taylor series. It's an extension of the tangent line approximation idea. If you write the Taylor series expansion of $\sqrt{x}$ at $x = 64$ to four terms, you get $8 + \frac{x-64}{16} - \frac{(x-64)^2}{4096} + \frac{(x-64)^3}{524288}$. If I substitute $x = 65$, I'll get $8 + \frac{1}{16} - \frac{1}{4096} + \frac{1}{524288} = 8.062257$, which is an even better approximation than what we got in method 2. It gets better with more terms.
I know you aren't familiar with methods 2 and 3. But hey, if you stick with Math long enough, you'll one day come back to this comment and understand everything!
• Would you do the same for a negative square root ?🤔
• You cannot find the negative square root of a number. Think of what square roots are. If we take 36 and find the square root, you'll see it's 6. Because 6 x 6 = 36. We multiply it by itself and it gives us 36.
We can't do this with negatives. Consider -36 and finding the square root of that. Which number multiplied by itself will get -36?
Well, remember the rules we have for multiplying negatives and positives.
negative x negative = positive
negative x positive = positive
positive x positive = positive.
It can't be -6, because that will give us a positive number, instead of -36.
It can't be 6, because that will also give us a positive number instead of -36.
So, negatives can't have square roots.
You'll learn in more advanced classes what we do in these cases. |
# Section 1.3 Absolute Value Equations and Inequalities
Section Objectives
1. Solve absolute value equations.
2. Solve absolute value inequalities.
### Absolute Value
The absolute value of a number, $x$, is that number's distance from zero on the number line. The absolute value of $x$ is written $|x|$, and this quantity is always either positive or zero. In fact,
• $|x| = x$ if $x$ itself is positive or zero, and
• $|x| = -x$ (think the opposite of $x$) if $x$ is negative.
Based on the definition of absolute value, it is clear that $|x| = x$ or $-x$, depending on whichever is positive. If we're not sure which is positive, we must consider both options.
For example...
If $|x|=5$, then $x$ must either be $5$ or $-5$.
### Absolute Value Equations
Suppose $P$ is positive or zero. The absolute value equation $|x|=P$ is simply a compound equation in disguise:
$|x| = P \qquad \Longleftrightarrow \qquad x=P \quad \mbox{or} \quad x = -P$
To solve an absolute value equation:
1. Use algebra to write the equation in the form $|X|=P$. In this context, $X$ and $P$ may be entire algebraic expressions, but the equation can only make sense if $P \ge 0$.
2. Solve the compound equation $X=P \mbox{ or } X =-P$.
#### Examples
Solve for $x$: $\quad |2x+4|=8$
Solve for $w$: $\quad -5|w-7|+2 = -13$
Solve for $y$: $\quad \displaystyle \left|5-\frac{2}{3} y \right| -9 = 8$
Solve for $x$: $\quad |2x-7|=0$
Solve for $x$: $\quad |2x-7|=-5$
Solve for $x$: $\quad |2x+7| = |x-1|$
### Absolute Value Inequalities
Just as we did with absolute value equations, we'll solve absolute value inequalities by rewriting them as compound inequalities. There are two cases to consider (assuming $P>0$):
• $|x| < P \qquad \Longleftrightarrow \qquad -P < x < P$
• $|x| > P \qquad \Longleftrightarrow \qquad x > P \quad \mbox{or} \quad x < -P$
These ideas should make sense if you think about $x$ representing a distance from zero.
Important idea: Be on the lookout for equations and inequalities that are never true or always true. Usually you can spot these before you even start the solution process.
#### Examples
Solve for $x$: $\quad \displaystyle \frac{|3x+2|}{4} \le 1$
Solve for $r$: $\quad -5|r-4|+12 > -28$
Solve for $x$: $\quad |3x-2| < -5$
Solve for $t$: $\quad \displaystyle \left| 3+ \frac{t}{2} \right| > 6$
Solve for $x$: $\quad |3x+7|-2 > -5$ |
Find the equation of the line perpendicular to 3x + 4y + 1= 0 and passing through (1, 1). : Kaysons Education
# Find The Equation Of The Line Perpendicular To 3x + 4y + 1= 0 And Passing Through (1, 1).
#### Video lectures
Access over 500+ hours of video lectures 24*7, covering complete syllabus for JEE preparation.
#### Online Support
Practice over 30000+ questions starting from basic level to JEE advance level.
#### National Mock Tests
Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.
#### Organized Learning
Proper planning to complete syllabus is the key to get a decent rank in JEE.
#### Test Series/Daily assignments
Give tests to analyze your progress and evaluate where you stand in terms of your JEE preparation.
## Question
### Solution
Correct option is
4x – 3y – 1 = 0
Equation of line perpendicular to 3x + 4y + 1 = 0 can be taken as
4x – 3y + k = 0.
(i) Interchange coefficient of x and y
(ii) Reverse the sign between x and y
Now, the line passes through (1, 1):
Hence required equation is:
4x – 3y – 1 = 0.
#### SIMILAR QUESTIONS
Q1
Find the area of triangle ABC with vertices A (aa2), B (bb2), C (cc2).
Q2
A straight line passes through (2, 3) and the portion of the line intercepted between the axes is bisected at this point. Find its equation
Q3
Find the slope (m), intercepts on X axis, intercept on Y axis of the line 3x+ 2y – 12 = 0. Also trace the line on XY plane.
Q4
Given the triangle A (10, 4), B(–4, 9), C(–2, 1), find the equation of median through B.
Q5
Find the equation of the straight line passing through the points (3, 3) and (7, 6). What is the length of the portion of the line intercepted between the axes of the coordinates.
Q6
Given the triangle with vertices A (–4, 9), B (10, 4), C (–2, –1). Find the equation of the altitude through A.
Q7
Find the equation of perpendicular bisector of the line joining the points (1, 1) and (2, 3).
Q8
Find the coordinates of the foot of the perpendicular from the point (2, 3) on the line y = 3x + 4.
Q9
Find the equation of the line passing through (ab) and parallel to px +qy + 1 = 0.
Q10
Find the distance of line h (x + h) + k (y + k) = 0 from the origin. |
Maths
Introduction
There are many different shapes. There might seem to be hundreds of different shapes and you could try to learn them all but
1. This is confusing
2. This is a waste of time
3. This is pointless. We need to see what they have in common and what is different about them.
It is much better to concentrate on the properties of shapes. This way you can remember a few facts and them use them to find out more about different shapes.
### Properties of shapes.
The 3 properties of shapes that we are going to look at are
1. The number of sides
2. The interior angles (the angles inside).
3. The length of the sides.
These properties help use to remember which shapes are which and why they so called (in some cases).
Let us look at shapes that have 4 sides Quadrilaterals (Quad- means 4).
Quadrilaterals ALWAYS have 4 sides
The interior angles of a quadrilateral add up to 360 degrees.
Here are the quadrilaterals you are expected to know about.
1. Square
2. Rhombus
3. Rectangle
4. Parallelogram
5. Trapezium
6. Kite
That does not seem to many does it. Let us look at the properties of quadrilaterals.
Square
A square has got 4 sides of equal length and 4 right angles (right angle = 90 degrees).
Since ALL the angles in a quadrilateral add up to 360 then 360 divided by 4 must be 90.
So if the length of one of the sides is 5 cm what is the combined length of the other 3 sides?
The answer is 15 cm because 5cm is the length of one side of a square, then each other side must be 5cm . so 5 cm + 5cm + 5 cm = 15cm.
Why do we not just say that a square has got 4 sides of equal length? Why discuss the angles?
Let us look at another quadrilateral and you will see why.
Rhombus
A Rhombus has got 4 sides of equal length and opposite sides are parallel and angles are equal.
(It looks like a square that is being pushed over). The arrows show parallel sides
What is the difference between a square and a rhombus?
The answer is in the angles. A Rhombus has got NO right angles
Now do you see why we have to look at both the side and the angle properties of a shape?
Rectangle
The rectangle (oblong) contains 4 right angles (an angle of 90˚).
It has got 2 pairs of equal sides and 4 right angles
In a rectangle, what is the same as a square and what is different?
The rectangle has got 4 right angles (each 90˚) this is the same as a square however a rectangle as not got 4 sides of equal length
Remember the rhombus and how we could squash it down? Well, if we do the same to a rectangle we get the next quadrilateral:
Parallelogram
A parallelogram is a rectangle that has been pushed over. Opposite sides are the same length and they are parallel.
What have a parallelogram and a rhombus got in common?
1. Opposite sides are parallel.
2. No angles are right angles.
Some quadrilaterals only have one set of parallel lines. Look at this next one:
Trapezium
What an odd name!
Both of these are types of trapezium. Each of them has different properties in the number of right angles. But each contains 4 sides and one pair of parallel sides.
Not all quadrilaterals have parallel sides. Here is our final member of the quadrilateral family.
Kite
A kite has got two pairs of sides next to each other that have equal length. But none of the sides are parallel.
Well, that is the last member of the quadrilateral family that you will need to know.
Time to review what you have revised. Look at the names below and see if you can remember what properties that they have. When you have thought about all six, click on check your answers and see if you were right.
1. Square
2. Rhombus
3. Rectangle
4. Parallelogram
5. Trapezium
6. Kite
### Here are the things to remember
All quadrilaterals have 4 sides
A square has got 4 sides of equal length and 4 right angles (right angle = 90 degrees).
A Rhombus has got 4 sides of equal length and opposite sides are parallel and angles are equal.
The rectangle (oblong) contains 4 right angles (an angle of 90˚). It has got 2 pairs of equal length sides
A parallelogram is a rectangle that has been pushed over. Opposite sides are the same length and they are parrallel.
A trapezium as got one pair of parallel sides.
A kite as got two pairs of sides next to each other that have equal length.
That is it for today. Well done! |
Sanderson M. Smith
UNBIASED ESTIMATORS
Consider the number set P = {2,4,6}.
If P is a population, then
Mean = m = (2+4+6)/3 = 4.
Variance = s2 = [(2-4)2 + (4-4)2 + (6-4)2]/3 = 8/3 = 2.666667.
Standard deviation = sqrt(s2) = sqrt(8/3) = s = 1.632993.
NOTE: The formula for s2 involves dividing by the population size n. In this case, n = 3.
If P is a sample, then
Sample mean = x(bar) = (2+4+6)/2 = 4.
Sample variance = s2 = [(2-4)2 + (4-4)2 + (6-4)2]/2 = 8/2 = 4.
Sample standard deviation = sqrt(s2) = sqrt(4) = s = 2.
NOTE: The formula for s2 involves dividing by n-1. In the case, n=3. Hence n-1 = 2.
Now, let's consider P to be a population.
The table below shows all possible samples of size 2 chosen from P, with replacement. There would be 3x3 = 9 samples.
Sample x(bar) for sample s2 for sample s for sample s2 for sample s for sample 2,2 2 0 0 0 0 2,4 3 2 1.414214 1 1 2,6 4 8 2.828427 4 2 4,2 3 2 1.414214 1 1 4,4 4 0 0 0 0 4,6 5 2 1.414214 1 1 6,2 4 8 2.828427 4 2 6,4 5 2 1.414214 1 1 6,6 6 0 0 0 0 Column Means 4 2.666667 1.257079 1.333333 0.888889
To summarize, we have listed all samples of size 2 (with replacement) from a population P of size 3. We have calculated statistics for each sample of size 2. Here is an important definition:
A statistic used to estimate a population parameter is unbiased if the mean of the sampling distribution of the statistic is equal to the true value of the parameter being estimated.
Note that
The mean of the sample means (4) is equal to m, the mean of the population P. This illustrates that a sample mean x(bar) is an unbiased statistic. It is sometimes stated that x(bar) is an unbiased estimator for the population parameter m .
The mean of the sample values of s2 (2.666667) is equal to s2 , the variance of the population P. This illustrates that the sample variance s2 is an unbiased statistic. It is sometimes stated that s2 is an unbiased estimator for the population variance s2.
Note carefully that the sample statistic s is not an unbiased statistic. That is, the mean of the s column in the table (1.257079) is not equal to the population parameter s = 1.632993.
Also, if you use the s2 formula for samples, the resulting statistics are not unbiased estimates for a population parameter. Note that the means for the last two columns in the table are not equal to population parameters.
In summary, the sample statistics x(bar) and s2 are unbiased estimators for the population mean m and population variance s2, respectively. |
{{ toc.signature }}
{{ 'ml-toc-proceed-mlc' | message }}
{{ 'ml-toc-proceed-tbs' | message }}
An error ocurred, try again later!
Chapter {{ article.chapter.number }}
{{ article.number }}.
# {{ article.displayTitle }}
{{ article.intro.summary }}
{{ ability.description }}
Lesson Settings & Tools
{{ 'ml-lesson-number-slides' | message : article.intro.bblockCount }} {{ 'ml-lesson-number-exercises' | message : article.intro.exerciseCount }} {{ 'ml-lesson-time-estimation' | message }}
Players typically lose and win points based on their actions in video games. The wins can be represented by natural and whole numbers but not losses. Integers, however, can represent losses. This lesson presents attributes and uses of integer numbers used in such situations.
### Catch-Up and Review
Here are a few recommended readings before getting started with this lesson.
Explore
## Measuring Temperatures
Look at the thermometer and play with it to measure different temperatures.
How can temperatures below zero be expressed?
Discussion
## Integers
Integers are whole numbers and negative numbers without fractions or decimal parts. The following are some examples of integers and non-integers.
Integers Non-integers
An integer can be positive, negative or zero. The set of all integers is denoted by
The dots indicate that the set continues to increase or decrease by one unit at a time indefinitely. An example of the use of integers is to represent temperatures below zero. One of the lowest temperatures recorded on earth was about
Discussion
## Graphing Integer Numbers
Similar to natural and whole numbers, integer numbers can be displayed on a number line. This helps to compare numbers and visualize which are greater and smaller.
Concept
## Number Line
A number line can display integer numbers by extending it on the left- and right-hand side of zero. The positive integers or natural numbers will be on the right-hand side, and the negative integers will be on the left-hand side of zero.
Recall the thermometer shown previously. A thermometer is also a number line. It can be extended down to measure positive and negative temperatures.
Pop Quiz
## Integer Numbers on a Number Line
Determine the integer represented by the point on the number line.
Example
## The Nature Scavenger Hunt Adventure
Maya is attending a teen-based summer camp. She is now taking part in a nature scavenger hunt. In this game, players are divided into teams; each has a map, a list of items to find, and tasks to complete. Each action, not only completed task, is worth points.
Find a hidden item (I)
Cheating (C)
Complete a physical challenge (P)
Solve a puzzle (S)
Take a wrong turn (W)
Run out of time before finishing a task (T)
The team that gets more points will be the winner. Maya wants a better idea of the meaning of these points to help her team win. She thinks that plotting them on a number line should help. Which of the following graphs matches Maya's?
### Hint
Identify if the points given by each action or task are negative or positive. When the number is positive, count that number of units going to the right of on the number line. On the other hand, if the number of points is negative, count that number of units to the left of
### Solution
The points for each task or action are integers. Display each on a number line and label it. Consider that finding a hidden item (I) gives points. That requires moving two units to the right of to display the points earned.
Next, cheating (C) is represented by This means that the team losses points if they cheat. Move seven units to the left of zero to represent the points given by this action.
Follow a similar method to graph the remaining points.
This graph matches option B. Now that Maya has a better understanding of the game, she can help her team to win this wonderful adventure.
Example
## Recording Temperatures Just Like Grandma
Maya is loving the summer camp. The staff even gave her what they call an explorer's box. One of the items in it is a thermometer. Her grandma used one just like it! Maya is excited to record temperatures all week wherever they go.
She created a table with the temperatures she recorded.
Temperatures Measured
She now wants to know the coldest and warmest temperature she recorded. Help Maya order these temperatures from the coldest to the warmest on a number line.
### Hint
On a number line, the smaller numbers are further left and the greater ones are further right. Plot one temperature at a time as a point on a number line. A negative temperature is on the left-hand side of A positive temperature is on the right-hand side of
### Solution
A number line helps to compare numbers. The smaller numbers are further left and the greater ones are further right. Now, because the list contains positive and negative integers, the number line must contain
The first recorded temperature is Move one unit to the left of zero and draw a point over the number line to draw this temperature.
The second recorded temperature is Plot this temperature by following a similar method. In this case, move units to the right-hand side of
The graph shows that the first temperature Maya recorded is less than the second. Now, plot the remaining temperatures following a similar reasoning.
The number line shows that the coldest recorded temperature was and the warmest was The temperatures can now be written from the coldest to the warmest.
Discussion
## Opposite Numbers
Consider the lucky number Well, perhaps the sign needs to change to get for it to be lucky. Moving on, what happens when is added to
Notice that cancels The reason for that is is the additive inverse of This property of integer numbers is worth exploring deeper!
Concept
The additive inverse of a number is another number such that their sum equals If is the additive inverse of then the following equation holds true.
Given a number, its additive inverse — also called opposite number — can be found by changing the sign. Some examples of additive inverses are listed in the table.
Pop Quiz
## Finding the Additive Inverse of Integer Numbers
The additive inverse of a number is another number such that the sum of these two numbers equals Use this information to find the additive inverse of the indicated number.
Discussion
## Comparing Distances Written as Integer Numbers
It is common to compare quantities that express distances. They help identify which quantity is further than another. Knowing which number represents a further distance is simple if the numbers are positive. What if they are negative? No fears — the absolute value comes to the rescue.
Concept
## Absolute Value
The absolute value of a number is the distance between and on the number line. It is denoted as and it is always a non-negative value.
The absolute value of a negative number is its opposite value, while the absolute value of a positive number is equal to itself.
For any integer number these two properties hold true.
Property Algebraic Representation Example
Non-negativity and
Symmetry and
The non-negative property says that the absolute value of any integer must always be greater than or equal than 0. In addition, the symmetry property means that the absolute value of a number and the absolute value of its opposite are the same.
Pop Quiz
## Comparison of Numbers
Time to identify and compare numbers. See the given statement and evaluate if it is true or false. Some statements involve absolute values.
Example
## Measuring Animals' Elevation
Near the camp is an animal rehabilitation center. There, the animals have space to roam. Maya gets the chance to visit! She takes photos of the animals and notes their elevation compared to her standing position.
Animal Elevation (In Feet)
Kingfisher feet
River otter feet
Porcupine feet
Sturgeon feet
Painted turtle feet
Use Maya's notes to identify which animal is furthest from her standing position according to elevation.
What is the animal closest to Maya's standing position?
### Hint
Use the absolute values to order the elevations of the animals from Maya's position.
### Solution
The elevations contain negative and positive integers. Apply the absolute value of the elevations to get the distances the animals are from Maya's position.
Animal Elevation (In Feet)
Kingfisher feet
River otter feet
Porcupine feet
Sturgeon feet
Painted turtle feet
The absolute value of a negative number is its additive inverse. In contrast, the absolute value of a positive number is equal to itself. Now, move units up from in a vertical number line to display the distance of the Kingfisher.
Next, the river otter's elevation is feet. Its absolute value is That means the river otter's distance is units from in the positive direction.
Use a similar reasoning to identify the remaining distances.
The graph shows that the distance of the sturgeon is the furthest. That is even with a negative elevation! On the other side of the spectrum, the painted turtle is the closest.
Closure
## The Bear Sculpture
Many campers trust Maya's math skills and entrust their money to her. Vincenzo, has already given her He then goes to the gift shop and orders a custom bear sculpture. Before paying, he runs to Maya to retrieve his Maya sees the ticket and notices that the bear sculpture costs She shows his total balance to Vincenzo.
What does this amount represent? The negative amount in Vincenzo's balance means he has a debt or money owed. The amount he is in debt is given by the absolute value of the total.
What if Vincenzo's balance has a positive amount? That means he would still have cash and credit to buy more items. What if the balance is at That means Vincenzo has no debt and no money.
Balance Meaning
Negative Debt or money owed
No debt and no credit
Positive Cash and credit
Vincenzo needs to pay that because the bear sculpture costs more than he actually has. Scouring for loose change, he finds in a secret pocket in his backpack. That bear sculpture is all his! |
# Difference between revisions of "2013 AMC 10A Problems/Problem 3"
## Problem
Square $ABCD$ has side length $10$. Point $E$ is on $\overline{BC}$, and the area of $\triangle ABE$ is $40$. What is $BE$? $[asy] pair A,B,C,D,E; A=(0,0); B=(0,50); C=(50,50); D=(50,0); E = (30,50); draw(A--B); draw(B--E); draw(E--C); draw(C--D); draw(D--A); draw(A--E); dot(A); dot(B); dot(C); dot(D); dot(E); label("A",A,SW); label("B",B,NW); label("C",C,NE); label("D",D,SE); label("E",E,N); [/asy]$
$\textbf{(A)}\ 4\qquad\textbf{(B)}\ 5\qquad\textbf{(C)}\ 6\qquad\textbf{(D)}\ 7\qquad\textbf{(E)}\ 8$
## Solution
We know that the area of $\triangle ABE$ is equal to $\frac{AB(BE)}{2}$. Plugging in $AB=10$, we get $80 = 10BE$. Dividing, we find that $BE=\boxed{\textbf{(E) }8}$
~sugar_rush |
### Lect5 - ProbStat2012
```Some Discrete Probability
Distributions
By: Prof. Gevelyn B. Itao
Probability and Statistics
Discrete Uniform Distribution
If the random variable X assumes the values x1,
x2,…,xk, with equal probabilities, then the discrete
uniform distribution is given by
Probability and Statistics
Discrete Uniform Distribution
Example 5.1: When a light bulb is selected at random from a box
that contains a 40-watt bulb, a 60-watt bulb, a 75-watt bulb, and
a 100-watt bulb, each element of the sample space
S = {40, 60, 75, 100} occurs with probability 1/4.
Therefore, we have a uniform distribution, with
Probability and Statistics
Discrete Uniform Distribution
Example 5.2: When a die is tossed, each element of the sample
space S
S = {1,2,3,4,5,6} occurs with probability 1/6.
Therefore, we have a uniform distribution, with
Probability and Statistics
Theorem 5.1
The mean and variance of the discrete uniform distribution
f (x; k) are
Probability and Statistics
Theorem 5.1
Example 5.3: referring to example 5.2, compute the mean and
variance:
Probability and Statistics
Bernoulli Process
An experiment that consists of repeated trials, each with two
possible outcomes that may be labeled success or failure.
Properties of Bernoulli Process
1. The experiment consists of n repeated trials.
2. Each trial results in 2 possible outcomes only that may be
classified as a success or failure.
3. The probability of success, denoted by p, remains constant
from trial to trial.
4. The repeated trials are independent.
Probability and Statistics
Binomial random variable
The number X of successes in n Bernoulli trials
Binomial distribution
The probability distribution of the discrete binomial random
variable.
A Bernoulli trial can result in a success with probability p and a
failure probability q = 1 – p. Then the probability distribution
of the binomial random variable X, the number of successes in
n independent trials, is
Probability and Statistics
Binomial distribution
Example 5.4: The probability that a certain kind of component
will survive a given shock test is 3/4. Find the probability that
exactly 2 of the next 4 components tested survive.
Probability and Statistics
Binomial random variable
Since p + q = 1
For P (X < r) or P (a X b), the binomial sum is provided in
tables
Probability and Statistics
Binomial random variable
Example 5.5: In In a certain city district the need for
money to buy drugs is stated as the: reason for 75% of
all thefts.
Find the probability that among the next 5 theft cases
reported in this district,
a. exactly 2 resulted from the need for money to buy
drugs;
b. at most 3 resulted from the need for money to buy
drugs.
Probability and Statistics
Binomial random variable
Example 5.6: In testing a certain kind of truck tire over a
rugged terrain, it is found that 25% of the trucks fail to
complete the test run without a blowout. Of the next 15
trucks tested, find the probability that
a. from 3 to 6 have blowouts;
b. fewer than 4 have blowouts:
c. more than 5 have blowouts.
Probability and Statistics
Binomial random variable
Example 5.7: A nationwide survey of seniors by the University
of Michigan reveals that almost 70% disapprove of daily pot
smoking, according to a report in Parade.
If 12 seniors are selected at random and asked their
opinion, find the probability that the number who disapprove
of smoking pot daily is
(a) anywhere from 7 to 9:
(b) at most 5;
(c) not less than 8.
Probability and Statistics
Multinomial Distribution
If a given trial can result in k outcomes E1, E2,…,Ek with
probabilities p1, p2,,…pk, then the probability distribution of the
random variables X1, X2, …, Xk, representing the number of
occurrences for E1, E2,…,Ek in n independent trials is
with
Probability and Statistics
Multinomial Distribution
Example 5.8: According to a genetics theory, a certain cross
of guinea pigs will result in red, black, and white offspring
in the ratio 8:4:4. Find the probability that among 8
offspring 5 will be red, 2 black, and 1 white.
Probability and Statistics
Multinomial Distribution
Example 5.9: The probabilities are 0.4, 0.2, 0.3, and 0.1,
respectively, that a delegate to a certain convention
arrived by air, bus, automobile, or train. What is the
probability that among 9 delegates randomly selected at
this convention, 3 arrived by air, 3 arrived by bus, 1
arrived by automobile, and 2 arrived by train?
.
Probability and Statistics
Hypergeometric Distribution
Similar to binomial distribution, except that it does not
require independence among trials (i.e., it can be done
without replacement)
The probability distribution of the hypergeometric random
variable X, in which the number of successes in a random
sample of size n selected from N items of which k are labeled
success and N – k labeled failure is
Probability and Statistics
Hypergeometric Distribution
Example 5.10: To avoid detection at customs, a traveler
places 6 narcotic tablets in a bottle containing 9 vitamin
pills that are similar in appearance. If the customs
official selects 3 of the tablets at random for analysis,
what is the probability that the traveler will be arrested
for illegal possession of narcotics?
Probability and Statistics
Hypergeometric Distribution
Example 5.11: A manufacturing company uses an acceptance
scheme on production items before they are shipped. The plan is
a two-stage one. Boxes of 25 are readied for shipment and a
sample of 3 is tested for defectives. If any defectives are found,
the entire box is sent back for 100% screening. If no defectives
are found, the box is shipped.
a. What is the probability that a box containing 3 defectives will
be shipped?
b. What is the probability that a box containing only 1 defective
will be sent back for screening?
Probability and Statistics
Hypergeometric Distribution
Example 5.12: A large company has an inspection system for
the batches of small compressors purchased from vendors.
A batch typically contains 15 compressors. In the inspection
system a random sample of 5 is selected and all are tested.
Suppose there arc 2 faulty compressors in the batch of 15.
(a) What is the probability that for a given sample
there will be I faulty compressor?
(b) What is the probability that inspection will discover
both faulty compressors?
Probability and Statistics
Relationship between Hypergeometric and Binomial
Distribution
If n is very small compared to N, hypergeometric distribution
approaches binomial distribution.
In other words, binomial distribution is a large population
version of hypergeometric distribution.
The quantity k/n plays the role of the binomial parameter, p
Probability and Statistics
Negative Binomial Distribution
Binomial Distribution
Number of successes is counted for a fixed number of trials
Negative Binomial Distribution
The trials are repeated until a fixed number of successes occur.
Probability and Statistics
Negative Binomial Distribution
If repeated independent trials can result in a success with
probability p and a failure with probability q = 1 – p, then the
probability distribution of the random variable X, the number
of the trial on which the kth success occurs is
Probability and Statistics
Negative Binomial Distribution
Example 5.13: Suppose the probability is 0.8 that any given
person will believe a tale about the transgressions of a
famous actress. What is the probability that
a. the sixth person to hear this tale is the fourth one to
believe it?
b. the third person to hear this tale is the first one to
believe it?
Probability and Statistics
Geometric Distribution
If repeated independent trials can result in a success with
probability p and a failure with probability q = 1 – p, then the
probability distribution of the random variable X, the number of
the trial on which the first success occurs is,
Probability and Statistics
Geometric Distribution
Example 5.14: The probability that a student pilot passes
the written test for a private pilot's license is 0.7. Find the
probability that the student will pass the test
a. on the third try;
b. before the fourth try.
Probability and Statistics
Poisson Experiments
Experiments yielding numerical values of a random variable X,
the number of outcomes occurring during a given time
interval or in a specified region
A specified region could be a line segment, an area, a volume,
or perhaps a piece of material
Probability and Statistics
Properties of Poisson Process
1. The number of outcomes occurring in one time interval or
specified region is independent of the number that occurs in
any other disjoint time interval or region of space. In other
words, Poisson process has no memory.
2. The probability that a single outcome will occur during a very
short time interval or in a small region is proportional to the
length of the time interval or the size of the region and does
not depend on the number of outcomes occurring outside
this time interval or region.
3. The probability that more than one outcome will occur in
such a short time interval or fall in such a small region is
negligible.
Probability and Statistics
Poisson Distribution
The probability distribution of the Poisson random variable X,
representing the number of outcomes occurring in a given
time interval or specified region denoted by t, is
- the average number of outcomes per unit time or region
Probability and Statistics
Poisson Distribution
Example 5.15: On average a certain intersection results in 3
traffic accidents per month. What is the probability that
for any given month at this intersection
a. exactly 5 accidents will occur?
b. less than 3 accidents will occur?
c. at least 2 accidents will occur?
Probability and Statistics
Poisson Distribution
Example 5.16: A secretary makes 2 errors per page, on
average. What is the probability that on the next page he
or she will make
(a) 4 or more errors?
(b) no errors?
Probability and Statistics
Poisson Distribution and binomial distribution
As n and p 0, and np remains constant, Binomial
Distribution approaches a Poisson Distribution where np = t
Probability and Statistics
Poisson Distribution and binomial distribution
Example 5.17: In a manufacturing process where glass products
are produced, defects or bubbles occur, occasionally rendering
the piece undesirable for marketing. It is known that, on average,
1 in every 1000 of these items produced has one or more
bubbles. What is the probability that a random sample of 8000
will yield fewer than 7 items possessing bubbles?
Some Continuous Probability
Distributions
``` |
Here is a complete list of how to add anything you may ever want to add, like whole numbers, fractions, radicals, and much much more. Please comment, rate, and ask as many questions as possible. Remember I am only an 9th grade honors student and even though I have a high math average, some things may be incorrect. Please inform me if anything is wrong. Thank You. Please comment and rate.
## Step 1: How to Add Whole Numbers
This is the most basic of the basics. All you need to do is keep counting up. For example, 7+3=10. To learn the concept, you would count up to seven, and then keep counting up three more numbers. Ex, 1,2,3,4,5,6,7 (then the next 3) 8,9,10. There are three more numbers added on to the original seven and that is basically how it works. Obviously, you cant do this when adding big numbers, but there is a better way for that. Line up the numbers vertically and, starting at the right, add together the numbers that line up. Look at the pictures (coming soon) for more detail. If there is a number 10 or greater, place the units value, left, below the line and carry the ten's digit number up and to the left. Next time you just do the same and add on the remainder from last time. If you have any questions or if I have mislabeled or spelled something wrong please leave a comment. Thank You.
TRY IT
1. 2+2
2. 5+8
3. 4+1
4. 9+0
5. 3+5
## Step 2: How to Add Integers
Adding integers is similar to adding whole numbers, but integers can be positive or negative. To add two negetive integers, you add the same way as you would add whole numbers, but place a negative sign in front of the answer. Adding positive and negative integers can get more tricky. To do that, you subtract the larger number from the ,smaller one. Then you keep the sign from the number with the greatest common value. That means the larger number without taking the negetive into account. Ex. |5| (the absolute value of 5) is equal to 5, and |-7| is equal to 7. then you get your answer. Ex, -6+3=-3. subtract 3 from 6 and keep the negative because |-6| is greater than |3|. If you have any questions or if I have mislabeled or spelled something wrong please leave a comment. Thank You.
TRY IT
1. -8+3
2. 3+(-5) (ignore the parenthesis, They just mean it is plus negative 5)
3. -6+(-2)
4. -2+1
5. 5+(-3)
## Step 3: How to Add Fractions
Adding Fractions is once again a basic concept, but occasionally it can get hard. Once you have mastered basic addition-adding up to 12 and 12-and multiplying-up to 12 and 12-you can do this. If you have a basic fraction where the denominators (the bottom part of the fraction) are equal, such as 1/2 + 1/2, you can just add the top numbers and keep the bottom the same. The reason for this is if you have half of a pizza, and your friend has half a pizza, and you put them together what do you get. A whole pizza. if the denominators are not the same, you may need to cross multiply. If you have 1/2 + 1/3 you multiply opposites. 1 x 3 and put it in the upper left. 1 x 2 and put that in the upper right. And multiply the bottoms and that becomes the denominator of both fractions. Now you can add them normally. In order to simplify fractions, you must know how to divide. Divide the numerator (top) and the denominator (bottom) by the same number. This works because 1/1, 2/2, 3/3, and even 99/99 all equal 1. If you divide by one, the number remains the same even though it appears different. If you have any questions or if I have mislabeled or spelled something wrong please leave a comment. Thank You.
TRY IT write each answer in simplest form
ex. 3/8 + 1/8 = 4/8 or 1/2
1. 2/3 + 1/3
2. 2/7 + 3/7
3. 1/2 + 1/3 (remember cross-multiplying)
4. 1/10 + 1/10
5. 2/5 + 1/5
TRY IT
1. 4*(8)+5*(8)
2. *(8)+2*(18)
3. *(44)+ *(44)
4. 2*(32)+ *(50)
5. 3*(20)+ *(45)
Whole Numbers
1. 4
2. 13
3. 5
4. 9
5. 8
Fractions
1. 1
2. 5/7
3. 5/6
4. 1/5
5. 3/5
Integers
1. -5
2. -2
3. -8
4. -1
5. 2 |
Common Core: High School - Geometry : Derive Parabola Equation: CCSS.Math.Content.HSG-GPE.A.2
Example Questions
← Previous 1
Example Question #1 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute 1 for a 10 for b and 7 for y
Now we can simplify, and solve for
Example Question #2 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and the directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute 1 for a 10 for b and 7 for y
Now we can simplify, and solve for
Example Question #3 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute 1 for a 10 for b and 7 for y
Now we can simplify, and solve for
Example Question #4 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute -10 for a 4 for b and -11 for y
Now we can simplify, and solve for
Example Question #5 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute 6 for a -9 for b and -5 for y
Now we can simplify, and solve for
Example Question #6 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute 1 for a -6 for b and -19 for y
Now we can simplify, and solve for
Example Question #7 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute -10 for a 6 for b and 15 for y
Now we can simplify, and solve for
Example Question #8 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute 7 for a 5 for b and -4 for y
Now we can simplify, and solve for
Example Question #9 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute 6 for a 8 for b and 10 for y
Now we can simplify, and solve for
Example Question #10 : Derive Parabola Equation: Ccss.Math.Content.Hsg Gpe.A.2
Find the parabolic equation, where the focus and directrix are as follows.
Explanation:
The first step to solving this problem, it to use the equation of equal distances.
Let's square each side
Now we expand each binomial
Now we can substitute -10 for a -3 for b and -4 for y
Now we can simplify, and solve for |
Properties of Scalar Product and Projection of a Vector on a Line
# Properties of Scalar Product and Projection of a Vector on a Line Video Lecture | Mathematics (Maths) Class 12 - JEE
## Mathematics (Maths) Class 12
205 videos|264 docs|139 tests
## FAQs on Properties of Scalar Product and Projection of a Vector on a Line Video Lecture - Mathematics (Maths) Class 12 - JEE
1. What is the definition of the scalar product of two vectors?
Ans. The scalar product, also known as the dot product, is an operation between two vectors that results in a scalar quantity. It is defined as the product of the magnitudes of the two vectors and the cosine of the angle between them.
2. How is the scalar product calculated?
Ans. To calculate the scalar product of two vectors, you multiply the magnitudes of the vectors by the cosine of the angle between them. This can be represented by the formula: A · B = |A| |B| cosθ, where A and B are the vectors, |A| and |B| are their magnitudes, and θ is the angle between them.
3. What are some properties of the scalar product?
Ans. The scalar product has several properties, including: - Commutative property: A · B = B · A - Distributive property: A · (B + C) = A · B + A · C - Scalar multiplication property: (kA) · B = k(A · B) = A · (kB), where k is a scalar - If A · B = 0, then A and B are orthogonal (perpendicular) to each other.
4. How can the scalar product be used to determine the angle between two vectors?
Ans. The angle between two vectors can be determined using the scalar product formula. By rearranging the formula A · B = |A| |B| cosθ, you can solve for the angle θ. Taking the inverse cosine of the scalar product divided by the product of the magnitudes gives the angle: θ = cos⁻¹(A · B / (|A| |B|)).
5. What is the projection of a vector on a line?
Ans. The projection of a vector onto a line is a vector that represents the component of the original vector that lies along the line. It is calculated by taking the scalar product of the original vector and the unit vector in the direction of the line. The formula for the projection of vector A onto a line with unit vector u is given by: projᵤA = (A · u)u.
## Mathematics (Maths) Class 12
205 videos|264 docs|139 tests
### Up next
Explore Courses for JEE exam
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev
Track your progress, build streaks, highlight & save important lessons and more!
Related Searches
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
,
; |
# What is the axis of symmetry and vertex for the graph y = 2x^2 - 2x + 5?
Vertex: $\left(0.5 , 4.5\right)$
Axis of Symmetry: $x = 0.5$
#### Explanation:
First, we have to convert $y = 2 {x}^{2} - 2 x + 5$ into vertex form, because it is currently in standard form $\left(a {x}^{2} + b x + c\right)$. To do this, we must complete the square and find the perfect square trinomial that corresponds with the equation.
First, factor the 2 out of our first two terms: $2 {x}^{2} \mathmr{and} {x}^{2}$.
This becomes $2 \left({x}^{2} - x\right) + 5$.
Now, use ${x}^{2} - x$ to complete the square, adding and subtracting ${\left(\frac{b}{2}\right)}^{2}$.
Since there is no coefficient in front of x, we can assume that it is -1 because of the sign.
${\left(\frac{- 1}{2}\right)}^{2}$ = $0.25$
$2 \left({x}^{2} - x + 0.25 - 0.25\right) + 5$
Now, we can write this as a binomial squared.
$2 \left[{\left(x - 0.5\right)}^{2} - 0.25\right] + 5$
We must multiply the -0.25 by 2 to get rid of its brackets.
This becomes $2 {\left(x - 0.5\right)}^{2} - 0.5 + 5$
Which simplifies to $2 {\left(x - 0.5\right)}^{2} + 4.5$
It's finally in vertex form! We can easily see that the vertex is $\left(0.5 , 4.5\right)$, and the axis of symmetry is simply the x coordinate of the vertex.
Vertex: $\left(0.5 , 4.5\right)$
Axis of Symmetry: $x = 0.5$
Hope this helps!
Best wishes,
A fellow highschool student |
# Difference between revisions of "2016 AMC 8 Problems/Problem 11"
Determine how many two-digit numbers satisfy the following property: when the number is added to the number obtained by reversing its digits, the sum is $132.$
$\textbf{(A) }5\qquad\textbf{(B) }7\qquad\textbf{(C) }9\qquad\textbf{(D) }11\qquad \textbf{(E) }12$
## Solution 1
We can write the two digit number in the form of $10a+b$; reverse of $10a+b$ is $10b+a$. The sum of those numbers is: $$(10a+b)+(10b+a)=132$$$$11a+11b=132$$$$a+b=12$$ We can use brute force to find order pairs $(a,b)$ such that $a+b=12$. Since $a$ and $b$ are both digits, both $a$ and $b$ have to be integers less than $10$. Thus our ordered pairs are $(3,9); (4,8); (5,7); (6,6); (7,5); (8,4); (9,3)$ or $\boxed{\textbf{(B)} 7}$ ordered pairs.
2016 AMC 8 (Problems • Answer Key • Resources) Preceded byProblem 10 Followed byProblem 12 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 All AJHSME/AMC 8 Problems and Solutions
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. |
# Distribution
View:
Sorted by:
### How to Distribute Variables
Distributing variables over the terms in an algebraic expression involves multiplication rules and the rules for exponents. When different variables are multiplied together, you can write them side by
### How to Perform Distribution in Algebra
Distributing items is an act of spreading them out equally. Algebraic distribution means to multiply each of the terms within the parentheses by another term that is outside the parentheses. To distribute
### How to Distribute Positive and Negative Numbers
When performing distribution, be aware of the sign being distributed and how that sign affects each term. Distributing a positive sign makes no difference in the signs of the terms — the signs stay the
### When to Distribute or Add First in Algebraic Distribution
When performing algebraic distribution, you get the same answer whether you distribute first or add what’s within the parentheses first. Deciding to distribute or add is a judgment call, based on the following
### How to Distribute Negative Signs with Variables
When you use negative signs and multiple variables in algebraic equations, the problems can look scary. Don’t worry, though. All you need to do is distribute the negative sign through, changing the signs
### How to Simplify Negative Exponents with Variables
Distributing with negative exponents means that you'll have fractional answers. A base that has a negative exponent can be changed to a fraction. The base and the exponent become the denominator, but the
### How to Distribute Trinomials
A trinomial, a polynomial with three terms, can be distributed over another expression. Each term in the first factor is distributed separately over the second factor, and then the entire expression is
### How to Distribute a Polynomial
Distributing a polynomial isn't hard. When distributing a polynomial over any number of other terms, you distribute each term in the first factor over all of the terms in the second factor. When the distribution
### How to Distribute with Fractional Powers or Radicals
Distribution problems with fractional powers or radicals aren't as intimidating as they look. When distributing with fractional powers or radicals, remember that exponents that are fractions work the same
### How to Distribute Binomials
When you distribute in algebra, you multiply each of the terms within the parentheses by another term that is outside the parentheses. So, when you distribute a binomial over several terms, you just apply
### How to Recognize a Perfectly Squared Binomial
Recognizing a perfectly squared binomial can make life easier. When you recognize a perfectly squared binomial, you've identified a shortcut that saves time when distributing binomials over other terms
### Finding the Sum and Difference of the Same Two Terms
When distributing binomials over other terms, knowing how to find the sum and difference of the same two terms is a handy shortcut. The sum of any two terms multiplied by the difference of the same two
### How to Find the Sum of Two Cubes
Spotting a distribution that results in the sum of two cubes is a shortcut to solving distribution problems. To recognize what distribution results in the sum of two cubes, look to see if the distribution
### How to Find the Difference of Two Cubes
An expression that results in the difference between two cubes is usually pretty hard to spot. The difference of two cubes is equal to the difference of their cube roots times a trinomial, which contains |
Question Video: Counting in Fifties to 1000 | Nagwa Question Video: Counting in Fifties to 1000 | Nagwa
# Question Video: Counting in Fifties to 1000 Mathematics
04:10
### Video Transcript
Rather than saying every single number, there’re often quicker ways to count. And this question gets us counting in fives. We do this in the first part of the question. And also counting in 50s, we do this in the second part of the question. Now, if we know how to count in fives, we can use this to help us count in 50s. So let’s look at the first part of the problem. We’re told that Michael made trays of five cookies. And if we look at each tray, we can see that they all have five cookies on them. The question tells us to find how many cookies he made. So we need to count the number of cookies there are all together.
One way to do this will be to count every single cookie, one, two, three, and so on. But because Michael has made trays of five cookies, we could skip count in fives to find the total. We can see that the first two numbers have been filled in, five, 10. Let’s continue to skip count in fives, 15, 20, 25. We skip counted in fives to find the total. And if we look at the numbers that we said as we skip counted in fives, we can see a pattern. They end in the digits five, then zero, then five again, then zero, and so on. Five, 10, 15, 20, 25. And so the number of cookies that Michael made is 25.
In the second part of the problem, we have some more objects to count. This time, we’re told about Madison who has bought boxes of 50 cookies. At this time, even if we wanted to count every single cookie, we can’t. We can’t see them. This is a good job because we’re not being asked to count every single cookie. We have to count in 50s. What do we know about the number 50? 50 is 10 lots of five. So if we know how to count in fives, we can use this to help us count in 50s. The answers will be 10 times as large.
We can see this by looking at the first two numbers that have already been completed. Instead of five, we have a number that’s 10 times as large as five, 50. And instead of 10, we have a number that’s 10 times as large as 10, 100. Because these numbers are 10 times as large, the digits have shifted one place to the left. So this has the effect of looking like we’ve added a zero onto all the numbers that we had when we counted in fives. Let’s start with 100 then and carry on skip counting in 50s, 150, 200, 250. The number of cookies that Madison bought is 250. Look at this number. It’s 10 times as large as the number of cookies that Michael made.
Both children had five groups of cookies. Michael had five trays. And Madison had five boxes. But each of them had a different number of cookies in each group. Michael had five groups of five. So we counted in fives to find the answer. Five groups of five is 25. But Madison had 10 times as many cookies in each group. She had five groups of 50. So we needed to count in 50s to find the number of cookies that she bought. Five lots of 50 equals 250.
The answers are 25 and 250.
## Join Nagwa Classes
Attend live sessions on Nagwa Classes to boost your learning with guidance and advice from an expert teacher!
• Interactive Sessions
• Chat & Messaging
• Realistic Exam Questions |
The Determinant of a Matrix
# The Determinant of a Matrix
Recall from The Invertibility of a Matrix page that a $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ is invertible if and only if $ad - bc \neq 0$. This value, $ad - bc$ is known as the Determinant for the $2 \times 2$ matrix $A$ which we define formally below.
Definition: The Determinant of the $2 \times 2$ matrix $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ denoted $\det (A)$ is $\det (A) = ad -bc$.
With this definition, we can reword the theorem from The Invertibility of a Matrix page to be that a $2 \times 2$ matrix $A$ as defined above is invertible if and only if $\det (A) \neq 0$.
Of course, the determinant quantity is defined for larger square matrices, and once again, you're invited to read much more detail on the Linear Algebra page on Math Online into computing the determinants of $n \times n$ matrices. We thus generate the following equivalent statements regarding the invertibility of a matrix.
Theorem 1: Let $A$ be an $n \times n$ matrix. Then the following statements are equivalent: a) $A$ is invertible. b) $\det A \neq 0$. c) There exists a unique solution to the system $Ax = b$. d) The trivial solution is the only solution to $Ax = 0$.
The proof that $a \implies b$ requires the reader to have knowledge about cofactor and adjoint matrices. We saw that for $2 \times 2$ matrices that:
(1)
\begin{align} \quad A^{-1} = \frac{1}{ad - bc} \begin{bmatrix} d & - b \\ -c & a \end{bmatrix} = \frac{1}{\det (A)} \begin{bmatrix} d & - b \\ -c & a \end{bmatrix} \end{align}
More generally, the matrix $\begin{bmatrix} d & - b \\ -c & a \end{bmatrix}$ is known as the Adjoint Matrix of $A$. For matrices of size larger than $2 \times 2$, we can also define an adjoint matrix and show that for any $n \times n$ matrix $A$, that:
(2)
So $A^{-1}$ exists if $\det (A) \neq 0$. |
Select Page
# Math Videos – MPM1D Unit 3 Solving Equations
#### Sec. 3.1 – Solving Simple Equations – One and Two Step Equations
Learn how to solve a simple equation involving two-steps:
Solve:
4m – 6 = 12
#### Sec. 3.2 – Solve Two-Step Equations
Learn how to solve a two-step equation:
Solve:
7x – 4 = 10
Solve:
3 + 4m + 5m = 21
#### Sec. 3.2 – How to Solve Multi-Step Equations With Variable Terms on Both Sides
Solve:
4y – 13 = -6y + 7
#### Sec. 3.2 – How to Solve Equations Involving Distributive Property
Solve:
4(k – 3) = 2 – (2k – 6)
#### Sec. 3.3 – How to Solve Equations with One Fraction
Solve:
(1/3)(x – 2) = 5
and
16 = [3(v + 7)]/2
Solve:
[3(z – 5)]/4 = 7
Solve:
3 = [2(n + 7)]/5
#### Sec. 3.3 – Solving Equations with One Fraction Application Questions
A trapezoidal backyard has an area of 100 m^2. The front and back widths are 8 m and 12 m, as shown in the diagram.
What is the length of the yard from front to back?
#### Sec. 3.4 – Modelling and Rearranging Formulas
Rearrange the following equation for ‘h’
V = p(r^2)h
#### Sec. 3.6.R – Create and Solve Equations from Word Problems
The sum of two consecutive even integers is -134. Find the numbers.
#### Sec. 3.6.R – Problem Solving Using Equations
The length of Laurie’s rectangular swimming pool is triple its width. The pool covers an area of 192 m^2.
If Laurie swims across the diagonal and back, how far does she travel? |
# Math : Probabilities
Size: px
Start display at page:
Transcription
1 Probability EP-Program - Strisuksa School - Roi-et Math : Probabilities Dr.Wattana Toutip - Department of Mathematics Khon Kaen University 200 :Wattana Toutip wattou@kku.ac.th Probability of successive events A Tree Diagram shows the probability of successive events 20.. Examples. A bag contains 5 red and 6 black marbles. Two are drawn without replacement. What is the probability that : (a) Both are red (b) They are different colors. Solution Construct the diagram as shown. The top branch corresponds to both marbles being red. Multiply the probabilities along this branch. Fig 20. (a) The Probability that both are red is The two middle branches correspond to the marbles having different colors (b) P(different colors) Four fair dice are rolled. Find the probabilities of (a) Four 6s (b) at least one 6. Solution (a) The Probability of four 6s is 6 The Probability of four 6s is (b) There will be at least one 6 unless none of the dice show a 6. This has probability The probability of at least one 6 is
2 20..2 Exercises. Two fair dice are rolled. Find the probabilities that: (a) Both are sixes (b) The total is 2 (c) The total is 7 (d) The first is greater than the second (e) A 'double' is thrown (f) At least one of the dice is a six. 2. Two card are drawn without replacement from a well-shuffled pack. Find the probabilities that : (a) The first is a heart (b) Both are heart (c) The first is a heart and the second is a spade (d) The first is a King and the second is a Queen. 3. A bag contains 5 blue are 4 green counters. Two are drawn without replacement. Find the probabilities that : (a) Both are blue (b) They are the same colour (c) There is at least one blue (d) The second is a green. 4. A sweet box contains 5 toffees, 6 liquorices and 8 chocolates. Two are drawn out. Find the probabilities that : (a) The first is a toffee and the second is a chocolate (b) At least one is a liquorice (c) Neither is a toffee. 5. Two fair dice are rolled. The score is the larger of the numbers showing. Find the probabilities that: (a) The score is (b) The score is 6 6. Two children A and B each pick at random a single digit from to 9. Find the probabilities that : (a) They pick the same number (b) A's number is larger than B's 7. To start a certain board game a die is rolled until a six is obtained. Find the probabilities that : (a) A player starts on his first roll (b) He starts on his second roll (c) He starts on his third roll (d) He has not started by his fourth roll 8. To start at darts a' double' must first be thrown. Albert has probability of throwing a 0 double, and Beatrice has probability 8. Albert throws first. Find the probabilities that : (a)both start on their first throw (b) Beatrice starts on her second throw but Albert has not started by then. 9. A fair coin is spun five times. Find the probabilities of (a) five Heads (b) at least one Head. 0. A roulette wheel has the number to. A gambler bets that a number divisible by 3 will turn up. The bet is repeated four times. Find the probabilities that the gambler (a) Wins all his bets (b) wins at least one bets.
3 . Five people take the driving test. Each has probability of passing. Find the probabilities that : (a) They all pass (b) at least one fails Exclusive and independent events. Conditional probability If two events cannot happen together, then they are exclusive. If events are exclusive then the probability that one or the other occurs is the sum of their probabilities. P A or B P A P B, provided that A and B are exclusive. Two events are independent if the truth of the one of them does not alter the probability of the other. If events A and B are independent then the probability of them both occurring is the product of their probabilities. PA & B PA PB The conditional probability of 6! 720 A given B is the probability of A, once it is known that B is true. The conditional probability of A given B is written PA B. It is obtained from the formula : PA B PB If A and B are independent then PA B PA P A & B The symbols and are sometimes used for 'and' and 'or' respectively Example. Two cards are drawn without replacement from a pack. Events A, B, C are as follows : A : the first is a heart. B : the second is a heart. C : the card is a king. Which pairs of these events are independent? Solution If A is true, then there is one fewer heart in the pack. The probability of B is 2. Hence 5 4 A and B are not independent. If A is true then C is neither more likely nor less likely than before. Its probability is still 3. Hence A and C are independent. Similarly the truth of B does not alter probability of C. The pairs A and C, B and C are independent 2. A women travels to work by bus, car or on foot with probabilities,, respectively. For each type of transport her probabilities of being late are,, arrives late one morning, find the probability that she come by bus. Solution respectively. If she
4 Here let L be the event that she is late, and B the event that she came by bus. Use the formula for conditional probability: P( B & L) PB L = PL ( ) P( B L) The probability that she came by bus is Exercises. Two cards are drawn without replacement from a pack. Events A, B,C,D are as follows : A : the first is an ace. B : the second is an 8. C : the first is red. D: the second is an spade Which pairs of events are independent? Which are exclusive? 2.Two dice are rolled. Events A, B, C, D are as follows: A : the first is an 5 B : the total is 8 C : the total is 7 D: the dice show the same number Which pairs of events are independent? Which are exclusive? 3. Two dice are rolled. Events A, B, C, D are as follows : A : the total is 7 B : the second die is a 2 C : both dice are less than 5 D: at least one die is a 6 Which pairs of events are independent? Which are exclusive? 4. Three fair coins are spun. Events A, B, C, D are as follows : A : the first coin is a head B : all the coins are heads C : there is at least one tail D: the first and last coins show the same Which pairs of events are independent? Which are exclusive? 5. With A, B, C, D as defined is Question, Find the following: (a) PB A (b) PD C 6. With A, B, C, D as defined is Question 2, Find the following: (a) PA B (b) PB D 7. With A, B, C, D as defined is Question 3, Find the following: (a) PA B (b) PC D 8. With A, B, C, D as defined is Question 4, Find the following: (a) PB A (b) PC A 9. A box contains 5 red and 6 blue marbles. Two are drawn without replacement. If the second is red find the probability that first was blue. 0. In his drawer a man has 7 left shoes and 0 right shoes. He picks two out at random. Find the probability that: (a) He has one left shoe and one right shoe (b) He has one left shoe and one right, given that the first was a left (c) The second is a left, given that he has one has one left and one right.
5 . A man travels to work by bus, car and motorcycle with probabilities 0.4, 0.5, 0. respectively. With each type of transport his chances of an accident are,, respectively. Find the probabilities that: (a) He goes by car and has an accident. (b) He does not have an accident. (c) He went by motorcycle, given that he had an accident. 2. % of the population has a certain disease. There is a test for the disease, which gives a positive response for 9 of the people with the disease, and for of the people without the 0 50 disease. A person is selected at random and tested. (a) What is the probability that the test gives a negative response? (b) If the test is positive, what is the probability that the person has the disease? (c) If the test is negative, what is the probability that the person does not have the disease? 3. An island contains two tribes; 2 3 of the population are Wache, who tell the truth with probability 0.7, and 3 are Oya, who tell the truth with probability 0.8. I meet a tribesman who tell me that he is a Wache. What is the probability that he is telling the truth? 4. In the certain town 3 4 of the voters are over 25, and they vote the Freedom Party with probability. Voters under 25 support the Freedom Party with probability 3. If a support of the Freedom Party is picked at random, what is the probability that he or she is under 25? 5. Events A and B are such that PA 0.4, PB 0.3, and Show that A and B are neither exclusive nor independent. Find PA B. 6. Events A and B are such that PA 0.3, PB 0.2, and PA or B 0.4 and B are not exclusive. Find PA & B and PA B. P A & B Show that A 7. Events A and B are such that PA 0.4, PB 0.3, and PA B 0.5. Find PA & B and PA or B Find PB A 8. Events A and B are such that PA 0.3, PB 0.5. Find PA or B and PA & B in the following cases: (a) A and B are exclusive (b) A and B are independent 20.3 Examination questions. A 2p coin a 0p are throw on a table. Event A is 'A head occurs on the 2p coin'. Event B is 'A head occurs on the 0p coin'. Event C is 'Two head or two tails obtained'. State, giving reasons, which of the following statements is (are) true and which is (are) false. (a) A and B are independent events (b) B and C are independent events. (c) A and C are mutually exclusive events
6 (d) A and BC are independent events. (e) PA B C = PA.P B.P C 2. Six balls colored yellow, green, brown, blue, pink, and black have values 2, 3, 4, 5, 6, 7 respectively. They are independent in size and placed in a box. Two balls are selected together from the box at random and the total number of point recorded (i) Find the probability that the total score is (a) 7, (b)9, (c)0, (d) greater 9, (e) odd. (ii) A game between two players, X and Y, starts with the six balls in a box. Each player in turn selects at random two balls, notes the score and then returns the balls to the box. The game is over when one of the players reaches a total score of 25 or more. (a) If X starts, calculate the probability that X wins on his second turn; (b) If Y starts, calculate the probability that Y wins on his second turn. [O ADD] 3. (a) The two electronic systems C,C2of a communications satellite operate independently and have probabilities of 0.and 0.05respectively of failing. Find the probability that (i) neither circuit fails (ii) at least one circuit fails, (iii) exactly one circuit fails (b) In a certain boxing competition all fights are either won or lost; draws are not permitted. If a boxer wins a fight then the probability that he wins his next fight is 3 ; if he loses 4 a fight the probability of him losing the next three fights is 2 3. Assuming that he won his last fight, use a tree diagram, or otherwise, to calculate the probability that of his three fights (i) he wins exactly two fights (ii) he wins at most two fights. State the most likely and least likely sequence of results for these three fights. 4. (i) The events A and B are such that PA 0.4, PB 0.45, P A B Show that the events A and B are neither mutually exclusive nor independent. (ii) A bag contains 2 red balls, 8 blue balls and 4 white balls. Three balls are taken from the bag at random and without replacement. Find the probability that all three balls are of the same colors. Find also the probability that all three balls are of different colors. 5. A box contains 25 apples, of which 20 are red and 5 are green. Of the red apples, 3 contains maggots and of the green apples, contains maggots. Two apples are chosen at random from the box. Find, in any order, (i) the probability that both apples contain maggots, (ii) the probability that both apples are red and at least one contains maggots, (iii) the probability that at least one apple contains maggots, given that both apples are red,
7 (iv) the probability that both apples are red given that at least one apple is red. 6. (a) Two digits X and Y are taken from a table of random sampling numbers. Event R is that X Y and events S is that X and Y are both less than 2. Write down (i) PR (ii) PRS (iii) PRS (iv) PR S (b) Conveyor belting for use in mines is tested for both strength and safety (the safety test is based on the amount of heat generated if the belt snaps). A testing station receives belting from three different suppliers : 30% of its tests are carried out on samples of belting from supplier A, 50% from B ; 20 % from C. From past experience the probability of failing the strength test is 0.02 for a sample from A, 0.2 from B and 0.04 from C. (i) What is the probability that a particular strength test will result in a failure? (ii) If a strength test result in a failure, What is the probability that the belting came from supplier A? (iii)what is the probability of a sample failing the safety tests given the following further information: supplier A - the probability of failing the safety tests is 0.05 and is independent of the probability of failing the strength test; supplier B % the probability of samples fail both strength and safety test supplier C exactly half the samples which fail the strength test also fail the safety test Common errors. Single probability If there are n outcomes to an experiment, then each has probability only if they are equally likely When two dice are rolled, there are possible for total score. But a total of 2 is less likely than total of 7,so neither has probability 2. Addition of probability The probability of 'A or 'B is only the sum of the probabilities if the events concerned are exclusive. In general: PA or B PA PB 3. Multiplication of probabilities The probability of A & B is only the product of the probabilities if the events concerned are independent. 4. Conditional probability independent. (a) Do not forgot to divide by PB when working out PA B (b) In the formula P A B do not assume that PA & Bis PA PB P( A& B) PB ( ) This is only true if A and B are
8 (c) Conditional probability is concerned with belief, not with cause and effect. If PA P A B. then it does not follow that B has caused A or prevented A. It may even be that B happened after A did. Solution (to exercise) (a) (e) 6 2. (a) 4 3. (a) (a) (a) (f) (b) 7 (b) 3 57 (b) (b) (b) 4 9 (c) (c) 9 7 (c) 6 (c) 5 6 (d) 5 2 (d) (d) (a) 9 (b) (a) 6 8. (a) (a) 32 (b) 5 (b) (b) 3 32 (b) 65 8 (b) (a) 8. (a) A& C, A& D, B & D indep. None exc. 2. A& C, A& D, B & D indep. B & C, C & D excl. 3. A& B indep. C& D exclusive. 4. A& Dindep. B& C excl. 5. (a) 4 (b) (c) (d) (a) 5 7. (a) 5 (b) 6 (b) 0
9 8. (a) 4 (b) (a) (b) 5 8 (c) 2. (a) (b) 5 8 (c) 2 2. (a) (b) (c) , ,0.55, (a) 0.8,0 (b) 0.65,0.5 =========================================================== References: Solomon, R.C. (997), A Level: Mathematics (4 th Edition), Great Britain, Hillman Printers(Frome) Ltd. More: (in Thai)
### Topic : ADDITION OF PROBABILITIES (MUTUALLY EXCLUSIVE EVENTS) TIME : 4 X 45 minutes
Worksheet 6 th Topic : ADDITION OF PROBABILITIES (MUTUALLY EXCLUSIVE EVENTS) TIME : 4 X 45 minutes STANDARD COMPETENCY : 1. To use the statistics rules, the rules of counting, and the characteristic of
### Module 4 Project Maths Development Team Draft (Version 2)
5 Week Modular Course in Statistics & Probability Strand 1 Module 4 Set Theory and Probability It is often said that the three basic rules of probability are: 1. Draw a picture 2. Draw a picture 3. Draw
### MEP Practice Book SA5
5 Probability 5.1 Probabilities MEP Practice Book SA5 1. Describe the probability of the following events happening, using the terms Certain Very likely Possible Very unlikely Impossible (d) (e) (f) (g)
### Chapter-wise questions. Probability. 1. Two coins are tossed simultaneously. Find the probability of getting exactly one tail.
Probability 1. Two coins are tossed simultaneously. Find the probability of getting exactly one tail. 2. 26 cards marked with English letters A to Z (one letter on each card) are shuffled well. If one
### Revision Topic 17: Probability Estimating probabilities: Relative frequency
Revision Topic 17: Probability Estimating probabilities: Relative frequency Probabilities can be estimated from experiments. The relative frequency is found using the formula: number of times event occurs.
### MEP Practice Book ES5. 1. A coin is tossed, and a die is thrown. List all the possible outcomes.
5 Probability MEP Practice Book ES5 5. Outcome of Two Events 1. A coin is tossed, and a die is thrown. List all the possible outcomes. 2. A die is thrown twice. Copy the diagram below which shows all the
### Chapter 16. Probability. For important terms and definitions refer NCERT text book. (6) NCERT text book page 386 question no.
Chapter 16 Probability For important terms and definitions refer NCERT text book. Type- I Concept : sample space (1)NCERT text book page 386 question no. 1 (*) (2) NCERT text book page 386 question no.
### Functional Skills Mathematics
Functional Skills Mathematics Level Learning Resource Probability D/L. Contents Independent Events D/L. Page - Combined Events D/L. Page - 9 West Nottinghamshire College D/L. Information Independent Events
### Q1) 6 boys and 6 girls are seated in a row. What is the probability that all the 6 gurls are together.
Required Probability = where Q1) 6 boys and 6 girls are seated in a row. What is the probability that all the 6 gurls are together. Solution: As girls are always together so they are considered as a group.
### HARDER PROBABILITY. Two events are said to be mutually exclusive if the occurrence of one excludes the occurrence of the other.
HARDER PROBABILITY MUTUALLY EXCLUSIVE EVENTS AND THE ADDITION LAW OF PROBABILITY Two events are said to be mutually exclusive if the occurrence of one excludes the occurrence of the other. Example Throwing
### Worksheets for GCSE Mathematics. Probability. mr-mathematics.com Maths Resources for Teachers. Handling Data
Worksheets for GCSE Mathematics Probability mr-mathematics.com Maths Resources for Teachers Handling Data Probability Worksheets Contents Differentiated Independent Learning Worksheets Probability Scales
### PROBABILITY.0 Concept Map Contents Page. Probability Of An Event. Probability Of Two Events. 4. Probability of Mutually Exclusive Events.4 Probability
PROGRAM DIDIK CEMERLANG AKADEMIK SPM ADDITIONAL MATHEMATICS FORM MODULE PROBABILITY PROBABILITY.0 Concept Map Contents Page. Probability Of An Event. Probability Of Two Events. 4. Probability of Mutually
### Class XII Chapter 13 Probability Maths. Exercise 13.1
Exercise 13.1 Question 1: Given that E and F are events such that P(E) = 0.6, P(F) = 0.3 and P(E F) = 0.2, find P (E F) and P(F E). It is given that P(E) = 0.6, P(F) = 0.3, and P(E F) = 0.2 Question 2:
### PROBABILITY Case of cards
WORKSHEET NO--1 PROBABILITY Case of cards WORKSHEET NO--2 Case of two die Case of coins WORKSHEET NO--3 1) Fill in the blanks: A. The probability of an impossible event is B. The probability of a sure
### PROBABILITY M.K. HOME TUITION. Mathematics Revision Guides. Level: GCSE Foundation Tier
Mathematics Revision Guides Probability Page 1 of 18 M.K. HOME TUITION Mathematics Revision Guides Level: GCSE Foundation Tier PROBABILITY Version: 2.1 Date: 08-10-2015 Mathematics Revision Guides Probability
### The Teachers Circle Mar. 20, 2012 HOW TO GAMBLE IF YOU MUST (I ll bet you \$5 that if you give me \$10, I ll give you \$20.)
The Teachers Circle Mar. 2, 22 HOW TO GAMBLE IF YOU MUST (I ll bet you \$ that if you give me \$, I ll give you \$2.) Instructor: Paul Zeitz (zeitzp@usfca.edu) Basic Laws and Definitions of Probability If
### Before giving a formal definition of probability, we explain some terms related to probability.
probability 22 INTRODUCTION In our day-to-day life, we come across statements such as: (i) It may rain today. (ii) Probably Rajesh will top his class. (iii) I doubt she will pass the test. (iv) It is unlikely
### , x {1, 2, k}, where k > 0. (a) Write down P(X = 2). (1) (b) Show that k = 3. (4) Find E(X). (2) (Total 7 marks)
1. The probability distribution of a discrete random variable X is given by 2 x P(X = x) = 14, x {1, 2, k}, where k > 0. Write down P(X = 2). (1) Show that k = 3. Find E(X). (Total 7 marks) 2. In a game
### 5.6. Independent Events. INVESTIGATE the Math. Reflecting
5.6 Independent Events YOU WILL NEED calculator EXPLORE The Fortin family has two children. Cam determines the probability that the family has two girls. Rushanna determines the probability that the family
### Lesson 3 Dependent and Independent Events
Lesson 3 Dependent and Independent Events When working with 2 separate events, we must first consider if the first event affects the second event. Situation 1 Situation 2 Drawing two cards from a deck
### Section A Calculating Probabilities & Listing Outcomes Grade F D
Name: Teacher Assessment Section A Calculating Probabilities & Listing Outcomes Grade F D 1. A fair ordinary six-sided dice is thrown once. The boxes show some of the possible outcomes. Draw a line from
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Study Guide for Test III (MATH 1630) Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the number of subsets of the set. 1) {x x is an even
### STANDARD COMPETENCY : 1. To use the statistics rules, the rules of counting, and the characteristic of probability in problem solving.
Worksheet 4 th Topic : PROBABILITY TIME : 4 X 45 minutes STANDARD COMPETENCY : 1. To use the statistics rules, the rules of counting, and the characteristic of probability in problem solving. BASIC COMPETENCY:
### 1. A factory makes calculators. Over a long period, 2 % of them are found to be faulty. A random sample of 100 calculators is tested.
1. A factory makes calculators. Over a long period, 2 % of them are found to be faulty. A random sample of 0 calculators is tested. Write down the expected number of faulty calculators in the sample. Find
### Contents 2.1 Basic Concepts of Probability Methods of Assigning Probabilities Principle of Counting - Permutation and Combination 39
CHAPTER 2 PROBABILITY Contents 2.1 Basic Concepts of Probability 38 2.2 Probability of an Event 39 2.3 Methods of Assigning Probabilities 39 2.4 Principle of Counting - Permutation and Combination 39 2.5
### Name. Is the game fair or not? Prove your answer with math. If the game is fair, play it 36 times and record the results.
Homework 5.1C You must complete table. Use math to decide if the game is fair or not. If Period the game is not fair, change the point system to make it fair. Game 1 Circle one: Fair or Not 2 six sided
### PROBABILITY. 1. Introduction. Candidates should able to:
PROBABILITY Candidates should able to: evaluate probabilities in simple cases by means of enumeration of equiprobable elementary events (e.g for the total score when two fair dice are thrown), or by calculation
### Exam III Review Problems
c Kathryn Bollinger and Benjamin Aurispa, November 10, 2011 1 Exam III Review Problems Fall 2011 Note: Not every topic is covered in this review. Please also take a look at the previous Week-in-Reviews
### Exam 2 Review F09 O Brien. Finite Mathematics Exam 2 Review
Finite Mathematics Exam Review Approximately 5 0% of the questions on Exam will come from Chapters, 4, and 5. The remaining 70 75% will come from Chapter 7. To help you prepare for the first part of the
### Mathematics 'A' level Module MS1: Statistics 1. Probability. The aims of this lesson are to enable you to. calculate and understand probability
Mathematics 'A' level Module MS1: Statistics 1 Lesson Three Aims The aims of this lesson are to enable you to calculate and understand probability apply the laws of probability in a variety of situations
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
6. Practice Problems Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Find the probability. ) A bag contains red marbles, blue marbles, and 8
### Probability - Grade 10 *
OpenStax-CNX module: m32623 1 Probability - Grade 10 * Rory Adams Free High School Science Texts Project Sarah Blyth Heather Williams This work is produced by OpenStax-CNX and licensed under the Creative
### Conditional Probability Worksheet
Conditional Probability Worksheet P( A and B) P(A B) = P( B) Exercises 3-6, compute the conditional probabilities P( AB) and P( B A ) 3. P A = 0.7, P B = 0.4, P A B = 0.25 4. P A = 0.45, P B = 0.8, P A
### Notes #45 Probability as a Fraction, Decimal, and Percent. As a result of what I learn today, I will be able to
Notes #45 Probability as a Fraction, Decimal, and Percent As a result of what I learn today, I will be able to Probabilities can be written in three ways:,, and. Probability is a of how an event is to.
### Independent Events B R Y
. Independent Events Lesson Objectives Understand independent events. Use the multiplication rule and the addition rule of probability to solve problems with independent events. Vocabulary independent
### Here are two situations involving chance:
Obstacle Courses 1. Introduction. Here are two situations involving chance: (i) Someone rolls a die three times. (People usually roll dice in pairs, so dice is more common than die, the singular form.)
### Conditional Probability Worksheet
Conditional Probability Worksheet EXAMPLE 4. Drug Testing and Conditional Probability Suppose that a company claims it has a test that is 95% effective in determining whether an athlete is using a steroid.
### LC OL Probability. ARNMaths.weebly.com. As part of Leaving Certificate Ordinary Level Math you should be able to complete the following.
A Ryan LC OL Probability ARNMaths.weebly.com Learning Outcomes As part of Leaving Certificate Ordinary Level Math you should be able to complete the following. Counting List outcomes of an experiment Apply
### ABC High School, Kathmandu, Nepal. Topic : Probability
BC High School, athmandu, Nepal Topic : Probability Grade 0 Teacher: Shyam Prasad charya. Objective of the Module: t the end of this lesson, students will be able to define and say formula of. define Mutually
### Independent and Mutually Exclusive Events
Independent and Mutually Exclusive Events By: OpenStaxCollege Independent and mutually exclusive do not mean the same thing. Independent Events Two events are independent if the following are true: P(A
### Key Concepts. Theoretical Probability. Terminology. Lesson 11-1
Key Concepts Theoretical Probability Lesson - Objective Teach students the terminology used in probability theory, and how to make calculations pertaining to experiments where all outcomes are equally
### Math 1313 Section 6.2 Definition of Probability
Math 1313 Section 6.2 Definition of Probability Probability is a measure of the likelihood that an event occurs. For example, if there is a 20% chance of rain tomorrow, that means that the probability
### STAT 311 (Spring 2016) Worksheet: W3W: Independence due: Mon. 2/1
Name: Group 1. For all groups. It is important that you understand the difference between independence and disjoint events. For each of the following situations, provide and example that is not in the
### When combined events A and B are independent:
A Resource for ree-standing Mathematics Qualifications A or B Mutually exclusive means that A and B cannot both happen at the same time. Venn Diagram showing mutually exclusive events: Aces The events
### Probability --QUESTIONS-- Principles of Math 12 - Probability Practice Exam 1
Probability --QUESTIONS-- Principles of Math - Probability Practice Exam www.math.com Principles of Math : Probability Practice Exam Use this sheet to record your answers:... 4... 4... 4.. 6. 4.. 6. 7..
### CLASSIFIED A-LEVEL PROBABILITY S1 BY: MR. AFDZAL Page 1
5 At a zoo, rides are offered on elephants, camels and jungle tractors. Ravi has money for only one ride. To decide which ride to choose, he tosses a fair coin twice. If he gets 2 heads he will go on the
### Intermediate Math Circles November 1, 2017 Probability I
Intermediate Math Circles November 1, 2017 Probability I Probability is the study of uncertain events or outcomes. Games of chance that involve rolling dice or dealing cards are one obvious area of application.
### 4. Are events C and D independent? Verify your answer with a calculation.
Honors Math 2 More Conditional Probability Name: Date: 1. A standard deck of cards has 52 cards: 26 Red cards, 26 black cards 4 suits: Hearts (red), Diamonds (red), Clubs (black), Spades (black); 13 of
### 7.1 Experiments, Sample Spaces, and Events
7.1 Experiments, Sample Spaces, and Events An experiment is an activity that has observable results. Examples: Tossing a coin, rolling dice, picking marbles out of a jar, etc. The result of an experiment
### TUESDAY, 8 NOVEMBER 2016 MORNING 1 hour 45 minutes
Surname Centre Number Candidate Number Other Names 0 GCSE NEW 3300U30- A6-3300U30- MATHEMATICS UNIT : NON-CALCULATOR INTERMEDIATE TIER TUESDAY, 8 NOVEMBER 206 MORNING hour 45 minutes For s use ADDITIONAL
### 4.1 Sample Spaces and Events
4.1 Sample Spaces and Events An experiment is an activity that has observable results. Examples: Tossing a coin, rolling dice, picking marbles out of a jar, etc. The result of an experiment is called an
### CSC/MATA67 Tutorial, Week 12
CSC/MATA67 Tutorial, Week 12 November 23, 2017 1 More counting problems A class consists of 15 students of whom 5 are prefects. Q: How many committees of 8 can be formed if each consists of a) exactly
### Unit 7 Central Tendency and Probability
Name: Block: 7.1 Central Tendency 7.2 Introduction to Probability 7.3 Independent Events 7.4 Dependent Events 7.1 Central Tendency A central tendency is a central or value in a data set. We will look at
### MTH 103 H Final Exam. 1. I study and I pass the course is an example of a. (a) conjunction (b) disjunction. (c) conditional (d) connective
MTH 103 H Final Exam Name: 1. I study and I pass the course is an example of a (a) conjunction (b) disjunction (c) conditional (d) connective 2. Which of the following is equivalent to (p q)? (a) p q (b)
### Stat210 WorkSheet#2 Chapter#2
1. When rolling a die 5 times, the number of elements of the sample space equals.(ans.=7,776) 2. If an experiment consists of throwing a die and then drawing a letter at random from the English alphabet,
### Statistics 1040 Summer 2009 Exam III
Statistics 1040 Summer 2009 Exam III 1. For the following basic probability questions. Give the RULE used in the appropriate blank (BEFORE the question), for each of the following situations, using one
### Instructions: Choose the best answer and shade in the corresponding letter on the answer sheet provided. Be sure to include your name and student ID.
Math 3201 Unit 3 Probability Test 1 Unit Test Name: Part 1 Selected Response: Instructions: Choose the best answer and shade in the corresponding letter on the answer sheet provided. Be sure to include
### Page 1 of 22. Website: Mobile:
Exercise 15.1 Question 1: Complete the following statements: (i) Probability of an event E + Probability of the event not E =. (ii) The probability of an event that cannot happen is. Such as event is called.
### Exercise Class XI Chapter 16 Probability Maths
Exercise 16.1 Question 1: Describe the sample space for the indicated experiment: A coin is tossed three times. A coin has two faces: head (H) and tail (T). When a coin is tossed three times, the total
### Probability. The MEnTe Program Math Enrichment through Technology. Title V East Los Angeles College
Probability The MEnTe Program Math Enrichment through Technology Title V East Los Angeles College 2003 East Los Angeles College. All rights reserved. Topics Introduction Empirical Probability Theoretical
### Mathacle. Name: Date:
Quiz Probability 1.) A telemarketer knows from past experience that when she makes a call, the probability that someone will answer the phone is 0.20. What is probability that the next two phone calls
### Probability. A Mathematical Model of Randomness
Probability A Mathematical Model of Randomness 1 Probability as Long Run Frequency In the eighteenth century, Compte De Buffon threw 2048 heads in 4040 coin tosses. Frequency = 2048 =.507 = 50.7% 4040
### Section Introduction to Sets
Section 1.1 - Introduction to Sets Definition: A set is a well-defined collection of objects usually denoted by uppercase letters. Definition: The elements, or members, of a set are denoted by lowercase
### Probability. Probabilty Impossibe Unlikely Equally Likely Likely Certain
PROBABILITY Probability The likelihood or chance of an event occurring If an event is IMPOSSIBLE its probability is ZERO If an event is CERTAIN its probability is ONE So all probabilities lie between 0
### PROBABILITY. Example 1 The probability of choosing a heart from a deck of cards is given by
Classical Definition of Probability PROBABILITY Probability is the measure of how likely an event is. An experiment is a situation involving chance or probability that leads to results called outcomes.
### KS3 Questions Probability. Level 3 to 5.
KS3 Questions Probability. Level 3 to 5. 1. A survey was carried out on the shoe size of 25 men. The results of the survey were as follows: 5 Complete the tally chart and frequency table for this data.
### Math June Review: Probability and Voting Procedures
Math - June Review: Probability and Voting Procedures A big box contains 7 chocolate doughnuts and honey doughnuts. A small box contains doughnuts: some are chocolate doughnuts, and the others are honey
### Section 7.3 and 7.4 Probability of Independent Events
Section 7.3 and 7.4 Probability of Independent Events Grade 7 Review Two or more events are independent when one event does not affect the outcome of the other event(s). For example, flipping a coin and
### RANDOM EXPERIMENTS AND EVENTS
Random Experiments and Events 18 RANDOM EXPERIMENTS AND EVENTS In day-to-day life we see that before commencement of a cricket match two captains go for a toss. Tossing of a coin is an activity and getting
### TEST A CHAPTER 11, PROBABILITY
TEST A CHAPTER 11, PROBABILITY 1. Two fair dice are rolled. Find the probability that the sum turning up is 9, given that the first die turns up an even number. 2. Two fair dice are rolled. Find the probability
### Contemporary Mathematics Math 1030 Sample Exam I Chapters Time Limit: 90 Minutes No Scratch Paper Calculator Allowed: Scientific
Contemporary Mathematics Math 1030 Sample Exam I Chapters 13-15 Time Limit: 90 Minutes No Scratch Paper Calculator Allowed: Scientific Name: The point value of each problem is in the left-hand margin.
### On the probability scale below mark, with a letter, the probability that the spinner will land
GCSE Exam Questions on Basic Probability. Richard has a box of toy cars. Each car is red or blue or white. 3 of the cars are red. 4 of the cars are blue. of the cars are white. Richard chooses one car
### Math 4610, Problems to be Worked in Class
Math 4610, Problems to be Worked in Class Bring this handout to class always! You will need it. If you wish to use an expanded version of this handout with space to write solutions, you can download one
### 2 A fair coin is flipped 8 times. What is the probability of getting more heads than tails? A. 1 2 B E. NOTA
For all questions, answer E. "NOTA" means none of the above answers is correct. Calculator use NO calculators will be permitted on any test other than the Statistics topic test. The word "deck" refers
### Bell Work. Warm-Up Exercises. Two six-sided dice are rolled. Find the probability of each sum or 7
Warm-Up Exercises Two six-sided dice are rolled. Find the probability of each sum. 1. 7 Bell Work 2. 5 or 7 3. You toss a coin 3 times. What is the probability of getting 3 heads? Warm-Up Notes Exercises
### Counting Methods and Probability
CHAPTER Counting Methods and Probability Many good basketball players can make 90% of their free throws. However, the likelihood of a player making several free throws in a row will be less than 90%. You
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
More 9.-9.3 Practice Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Answer the question. ) In how many ways can you answer the questions on
### STRAND: PROBABILITY Unit 2 Probability of Two or More Events
STRAND: PROAILITY Unit 2 Probability of Two or More Events TEXT Contents Section 2. Outcome of Two Events 2.2 Probability of Two Events 2. Use of Tree Diagrams 2 Probability of Two or More Events 2. Outcome
### Conditional probability 2E
Conditional probability 2E 1 a When the first token removed is red, there are 8 tokens remaining in the bag, 4 red and 4 blue. When the first token removed is blue, there are 8 tokens remaining in the
### Section 6.5 Conditional Probability
Section 6.5 Conditional Probability Example 1: An urn contains 5 green marbles and 7 black marbles. Two marbles are drawn in succession and without replacement from the urn. a) What is the probability
### The probability set-up
CHAPTER 2 The probability set-up 2.1. Introduction and basic theory We will have a sample space, denoted S (sometimes Ω) that consists of all possible outcomes. For example, if we roll two dice, the sample
### Mutually Exclusive Events Algebra 1
Name: Mutually Exclusive Events Algebra 1 Date: Mutually exclusive events are two events which have no outcomes in common. The probability that these two events would occur at the same time is zero. Exercise
### MATH 1115, Mathematics for Commerce WINTER 2011 Toby Kenney Homework Sheet 6 Model Solutions
MATH, Mathematics for Commerce WINTER 0 Toby Kenney Homework Sheet Model Solutions. A company has two machines for producing a product. The first machine produces defective products % of the time. The
### #2. A coin is tossed 40 times and lands on heads 21 times. What is the experimental probability of the coin landing on tails?
1 Pre-AP Geometry Chapter 14 Test Review Standards/Goals: A.1.f.: I can find the probability of a simple event. F.1.c.: I can use area to solve problems involving geometric probability. S.CP.1: I can define
### Probability Rules. 2) The probability, P, of any event ranges from which of the following?
Name: WORKSHEET : Date: Answer the following questions. 1) Probability of event E occurring is... P(E) = Number of ways to get E/Total number of outcomes possible in S, the sample space....if. 2) The probability,
### SHORT ANSWER. Write the word or phrase that best completes each statement or answers the question.
Math 1342 Practice Test 2 Ch 4 & 5 Name 1) Nanette must pass through three doors as she walks from her company's foyer to her office. Each of these doors may be locked or unlocked. 1) List the outcomes
### A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, OCR, Pearson-Edexcel and WJEC Eduqas. Name: Total Marks:
Probability 2 (H) A collection of 9-1 Maths GCSE Sample and Specimen questions from AQA, OCR, Pearson-Edexcel and WJEC Eduqas. Name: Total Marks: 1. Andy sometimes gets a lift to and from college. When
### KS3 Levels 3-8. Unit 3 Probability. Homework Booklet. Complete this table indicating the homework you have been set and when it is due by.
Name: Maths Group: Tutor Set: Unit 3 Probability Homework Booklet KS3 Levels 3-8 Complete this table indicating the homework you have been set and when it is due by. Date Homework Due By Handed In Please
### The probability set-up
CHAPTER The probability set-up.1. Introduction and basic theory We will have a sample space, denoted S sometimes Ω that consists of all possible outcomes. For example, if we roll two dice, the sample space
### CS1802 Week 9: Probability, Expectation, Entropy
CS02 Discrete Structures Recitation Fall 207 October 30 - November 3, 207 CS02 Week 9: Probability, Expectation, Entropy Simple Probabilities i. What is the probability that if a die is rolled five times,
### Probability and Counting Techniques
Probability and Counting Techniques Diana Pell (Multiplication Principle) Suppose that a task consists of t choices performed consecutively. Suppose that choice 1 can be performed in m 1 ways; for each
### Probability. March 06, J. Boulton MDM 4U1. P(A) = n(a) n(s) Introductory Probability
Most people think they understand odds and probability. Do you? Decision 1: Pick a card Decision 2: Switch or don't Outcomes: Make a tree diagram Do you think you understand probability? Probability Write
### PRE TEST. Math in a Cultural Context*
P grade PRE TEST Salmon Fishing: Investigations into A 6P th module in the Math in a Cultural Context* UNIVERSITY OF ALASKA FAIRBANKS Student Name: Grade: Teacher: School: Location of School: Date: *This
### Functional Skills Mathematics
Functional Skills Mathematics Level Learning Resource HD2/L. HD2/L.2 Excellence in skills development Contents HD2/L. Pages 3-6 HD2/L.2 West Nottinghamshire College 2 HD2/L. HD2/L.2 Information is the
### COMPOUND EVENTS. Judo Math Inc.
COMPOUND EVENTS Judo Math Inc. 7 th grade Statistics Discipline: Black Belt Training Order of Mastery: Compound Events 1. What are compound events? 2. Using organized Lists (7SP8) 3. Using tables (7SP8)
### MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question.
Statistics Homework Ch 5 Name MULTIPLE CHOICE. Choose the one alternative that best completes the statement or answers the question. Provide an appropriate response. 1) A coin is tossed. Find the probability
### Intermediate Math Circles November 1, 2017 Probability I. Problem Set Solutions
Intermediate Math Circles November 1, 2017 Probability I Problem Set Solutions 1. Suppose we draw one card from a well-shuffled deck. Let A be the event that we get a spade, and B be the event we get an
### Outcomes: The outcomes of this experiment are yellow, blue, red and green.
(Adapted from http://www.mathgoodies.com/) 1. Sample Space The sample space of an experiment is the set of all possible outcomes of that experiment. The sum of the probabilities of the distinct outcomes |
Paul's Online Notes
Home / Calculus II / Applications of Integrals / Arc Length
Show Mobile Notice Show All Notes Hide All Notes
Mobile Notice
You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll to see them) and some of the menu items will be cut off due to the narrow screen width.
### Section 8.1 : Arc Length
3. Determine the length of $$y = 7{\left( {6 + x} \right)^{\frac{3}{2}}}$$ , $$189 \le y \le 875$$.
Show All Steps Hide All Steps
Start Solution
Since we are not told which $$ds$$ to use we will have to decide which one to use. In this case the function is set up to use the $$ds$$ in terms of $$x$$. Note as well that if we solve the function for $$x$$ (which we’d need to do in order to use the $$ds$$ that is in terms of $$y$$) we would still have a fractional exponent and the derivative will not work out as nice once we plug it into the $$ds$$ formula.
So, let’s take the derivative of the given function and plug into the $$ds$$ formula.
$\frac{{dy}}{{dx}} = \frac{{21}}{2}{\left( {6 + x} \right)^{\frac{1}{2}}}$ \begin{align*}ds & = \sqrt {1 + {{\left[ {\frac{{dy}}{{dx}}} \right]}^2}} \,dx = \sqrt {1 + {{\left[ {\frac{{21}}{2}{{\left( {6 + x} \right)}^{\frac{1}{2}}}} \right]}^2}} \,dx = \sqrt {1 + \frac{{441}}{4}\left( {6 + x} \right)} \,dx\\ & = \sqrt {\frac{{2650}}{4} + \frac{{441}}{4}x} \,dx = \frac{1}{2}\sqrt {2650 + 441x} \,dx\end{align*}
We did a little simplification that may or may not make the integration easier. That will probably depend upon the person doing the integration and just what they find the easiest to deal with. The point is there are several forms of the $$ds$$ that we could use here. All will give the same answer.
Show Step 2
Next, we need to deal with the limits for the integral. The $$ds$$ that we choose to use in the first step has a $$dx$$ in it and that means that we’ll need $$x$$ limits for our integral. We, however, were given $$y$$ limits in the problem statement. This means we’ll need to convert those to $$x$$’s before proceeding with the integral.
To do convert these all we need to do is plug them into the function we were given in the problem statement and solve for the corresponding $$x$$. Doing this gives,
\begin{align*}y = & 189:\,\,\,\,\,\,\,\,\,189 = 7{\left( {6 + x} \right)^{\frac{3}{2}}}\,\,\,\,\, \to \,\,\,\,\,\,6 + x = {27^{\frac{2}{3}}} = 9\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,x = 3\\ y = & 875:\,\,\,\,\,\,\,\,\,875 = 7{\left( {6 + x} \right)^{\frac{3}{2}}}\,\,\,\,\, \to \,\,\,\,\,\,6 + x = {125^{\frac{2}{3}}} = 25\,\,\,\,\,\,\, \to \,\,\,\,\,\,\,x = 19\end{align*}
So, the corresponding ranges of $$x$$’s is : $$3 \le x \le 19$$.
Show Step 3
The integral giving the arc length is then,
$L = \int_{{}}^{{}}{{ds}} = \int_{3}^{{19}}{{\frac{1}{2}\sqrt {2650 + 441x} \,dx}}$ Show Step 4
Finally, all we need to do is evaluate the integral. In this case all we need to do is use a quick Calc I substitution. We’ll leave most of the integration details to you to verify.
The arc length of the curve is,
$L = \int_{3}^{{19}}{{\frac{1}{2}\sqrt {2650 + 441x} \,dx}} = \left. {\frac{1}{{1323}}{{\left( {2650 + 441x} \right)}^{\frac{3}{2}}}} \right|_3^{19} = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{{1323}}\left( {{{11029}^{\frac{3}{2}}} - {{3973}^{\frac{3}{2}}}} \right) = 686.1904}}$ |
Lesson Objectives
• Demonstrate an understanding of how to find the GCF of a polynomial
• Learn how to factor out the GCF from a polynomial
## How to Factor out the Greatest Common Factor (GCF) from a Polynomial
In our last lesson, we learned how to find the GCF or greatest common factor for a group of monomials. In this lesson, we will use that knowledge and learn how to factor out the GCF from a polynomial. Let’s begin by thinking about the distributive property:
2(3 + 5) = 2 • 3 + 2 • 5
4(3 - 5) = 4 • 3 - 4 • 5
At this point, we should know that the distributive property allows us to distribute multiplication over addition or subtraction. Since we are dealing with an equality, the reverse is also true. When we factor, we are reversing the distributive property:
Since 2 is common to both terms, we can move this outside of the parentheses:
2 • 3 + 2 • 5 = 2(3 + 5)
Since 4 is common to both terms, we can move this outside of the parentheses:
4 • 3 - 4 • 5 = 4(3 - 5)
Let's think about factoring a simple polynomial. Suppose we come across the following polynomial:
3x2 + 12x
First, let's think about the GCF of the two terms. The GCF of 3x2 and 12x is 3x. To make our next step crystal clear, let's rewrite our polynomial as:
3x • x + 3x • 4
Since 3x is common to each term, we can factor this out. This means we pull 3x out from each term and place it outside of a set of parentheses:
3x • x + 3x • 4 = 3x(x + 4)
3x2 + 12x = 3x(x + 4)
To factor out the GCF from a polynomial, we can use the following steps:
• Find the GCF of the polynomial
• Wrap the polynomial inside of parentheses
• Inside of the parentheses, pull out the GCF from each term
• We can do this by dividing each term by the GCF or we can factor each term and remove the GCF
• Write the GCF outside of the parentheses
Let's look at a few examples.
Example 1: Factor out the GCF
16x5 - 4x3 + 2x
Step 1) Find the GCF of the polynomial:
Our polynomial has terms of:
16x5, 4x3, and 2x
GCF(16x5, 4x3, 2x) = 2x
Step 2) Wrap the polynomial inside of parentheses:
(16x5 - 4x3 + 2x)
Steps 3 & 4) Factor each term and remove the GCF, write the GCF outside of the parentheses:
(2x • 8x4 - 2x • 2x2 + 2x • 1) =
2x(8x4 - 2x2 + 1)
Example 2: Factor out the GCF
25x4y2 + 50x2y2 - 10xy
Step 1) Find the GCF of the polynomial:
GCF(25x4y2, 50x2y2, 10xy) = 5xy
Step 2) Wrap the polynomial inside of parentheses:
(25x4y2 + 50x2y2 - 10xy)
Steps 3 & 4) Factor each term and remove the GCF, write the GCF outside of the parentheses:
(5xy • 5x3y + 5xy • 10xy - 5xy • 2) =
5xy(5x3y + 10xy - 2)
Example 3: Factor out the GCF
22x3y2 + 132x2y + 121x
Step 1) Find the GCF of the polynomial:
GCF(22x3y2, 132x2y, 121x) = 11x
Step 2) Wrap the polynomial inside of parentheses:
(22x3y2 + 132x2y + 121x)
Steps 3 & 4) Factor each term and remove the GCF, write the GCF outside of the parentheses:
(11x • 2x2y2 + 11x • 12xy + 11x • 11) =
11x(2x2y2 + 12xy + 11) |
## Arithmetic Series Worksheet Solution1
In the page arithmetic series worksheet solution1 you are going to see solution of each questions from the arithmetic series worksheet.
(1) Find the sum of first 75 positive integers
Solution:
To find the sum of first 75 positive integers first let us write the series
1 + 2 + 3 + ..........+ 75
Total number of terms in the series is 75 so n = 75
Sn = (n/2) (a+L)
= (75/2) (1+75)
= (75/2) (76)
= 75 x 38
= 2850
(ii) 125 natural numbers
Solution:
To find the sum of first 125 positive integers first let us write the series
1 + 2 + 3 + ..........+ 125
Total number of terms in the series is 125 so n = 125
Sn = (n/2) (a+L)
= (125/2) (1+125)
= (125/2) (126)
= 125 x 63
= 7875
(2) Find the sum of first 30 terms of an A.P whose nth term is 3 + 2 n
Solution:
nth term = 3 + 2 n
t n = 3 + 2 n
From the general term (tn) we are going to find first and last term of the arithmetic series for that first we have to apply 1 for n to get the value of first term (a) and we have to apply 30 for n to get the last term (L). Because we have only 30 terms in this series.
n = 1
t 1 = 3 + 2 (1)
t 1 = 5
a = 5
n = 30
t₃₀ = 3 + 2(30)
t₃₀ = 3 + 60
t₃₀ = 63
L = 63
Now we have to find S₃₀ for that we have to use the formula
S n = (n/2) [a+L]
S₃₀ = (30/2) [5 + 63]
= 15 [5 + 63]
= 15 [68]
S₃₀ = 1020
(3) Find the sum of each arithmetic series
(i) 38 + 35 + 32 + .......... + 2
Solution:
First we have to know that how many terms are there in the above series.
a = 38 d = t₂ - t₁ L = 2
d = 35-38
= -3
n = [(L-a)/d] + 1
= [(2-38)/(-3)] + 1
= [(-36)/(-3)] + 1
= 12 + 1
n = 13
S n = (n/2) (a+L)
= (13/2) (38 + 2)
= (13/2) (40)
= (13) (20)
= 260
These are the contents in the page arithmetic series worksheet solution1.
arithmetic series worksheet solution1 |
# How do you solve 3x^2-12x+11=0 using the quadratic function?
Jun 27, 2018
$x = \frac{6 \pm \sqrt{3}}{3}$
#### Explanation:
$x = \frac{- b = - \sqrt{{b}^{2} - 4 a c}}{2 a}$
for a quadratic taking the form
$a {x}^{2} + b x + c = 0$
So for this equation it is
$x = \frac{- \left(- 12\right) \pm \sqrt{{\left(- 12\right)}^{2} - 4 \cdot 3 \cdot 11}}{2 \cdot 3}$
$x = \frac{12 \pm \sqrt{144 - 132}}{6}$
$x = \frac{12 \pm \sqrt{4 \cdot 3}}{6}$
$x = \frac{6 \pm \sqrt{3}}{3}$
Jul 12, 2018
color(blue)(x = 2 + 1/sqrt3, 2 - 1/sqrt3= 2.5774, 1.4226
#### Explanation:
Quadratic formula to find the roots $x = \frac{- b \pm \sqrt{D}}{2 a}$ where D is the discrimination to decide whether the roots are real or imaginary.
$D = {b}^{2} - 4 a c$
Given equation is $3 {x}^{2} - 12 x + 11 = 0$
$a = 3 , b = - 12 , c = 11 , D = - {12}^{2} - \left(4 \cdot 3 \cdot 11\right) = 12$
$x = \frac{12 \pm \sqrt{12}}{6}$
$x = 2 \pm \left(\frac{1}{\sqrt{3}}\right)$
color(blue)(x = 2 + 1/sqrt3, 2 - 1/sqrt3= 2.5774, 1.4226 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.