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# Formula for finding area of a circle sector
You need the radius to work out the area of a circle, the formula is: circle area = πR 2. This means: π = Pi is a constant that equals 3.142. R = is the radius of the circle. R 2 (radius squared) means radius × radius.
Jan 06, 2020 · The formula to find the area of a circle is A = π r 2 {\displaystyle A=\pi r^{2}}, where the r {\displaystyle r} variable represents the radius. This variable is squared. Do not get confused and square the entire equation. For the sample circle with radius, r = 6 {\displaystyle r=6}, then r 2 = 36 {\displaystyle r^{2}=36}. The video provides two example problems for finding the radius of a circle given the arc length. Problem one finds the radius given radians, and the second problem uses degrees.
Circumference and area Two of the most basic formulas regarding circles are the formulas for area and circumference . The area of a circle is the number of square units that it takes to fill the inside of the circle. This worksheet may help you know about Circular Sector. A Circular Sector is portion of circle enclosed by two radii and an arc. The formula used to calculate the area of the circular Sector is, Formula Sector Area = (π r2 θ) / 360 Where, A is the Area of the circular sector. R is the radius of the circle. ANGLE is the central angle in radius. In this section we will discuss how to the area enclosed by a polar curve. The regions we look at in this section tend (although not always) to be shaped vaguely like a piece of pie or pizza and we are looking for the area of the region from the outer boundary (defined by the polar equation) and the origin/pole. May 31, 2008 · The key lies in the size of the angle. The area of its sector is 80cm sq. This area is a fraction of the area of the whole circle: 125/ 360. So, 80 x 360/125 sq.cm. will give you the area of the whole circle.
The formula for the area of a sector is (angle / 360) x height x π x radius 2. The figure below illustrates the measurement: As you can easily see, it is quite similar to that of a circle, but modified to account for the fact that a sector is just a part of a circle. |
# What is 7.884 Rounded to the Nearest Hundredth?
Rounding is a mathematical process used to simplify complex numbers. It is a process of estimating numbers to the nearest whole number or decimal. Rounding is often used to simplify a large number for easier calculation. It is also used to make a number look more presentable in a certain context or situation. In this article, we will discuss what 7.884 rounded to the nearest hundredth is.
## The Basics of Rounding
There are two basic rules to follow when rounding a number. The first rule is to look at the digit to the right of the number you are rounding. If that digit is 5 or higher, you round the number up. If the digit is 4 or lower, you round the number down. The second rule is to look at the number you are rounding. If it is an even number, you round the number down. If it is an odd number, you round the number up.
## How to Round 7.884 to the Nearest Hundredth
To round 7.884 to the nearest hundredth, we need to look at the digit to the right of the number, which is 8. Since 8 is 5 or higher, we round the number up. This means that 7.884 rounded to the nearest hundredth is 7.89. In other words, 7.884 is closer to 7.89 than it is to 7.88.
## Other Examples of Rounding to the Nearest Hundredth
To better understand how to round numbers to the nearest hundredth, let’s look at a few more examples. For example, if we round 3.541 to the nearest hundredth, we look at the digit to the right of the number, which is 4. Since 4 is 4 or lower, we round the number down. This means that 3.541 rounded to the nearest hundredth would be 3.54. Another example is 8.425. Looking at the digit to the right of the number, which is 5, we round the number up. This means that 8.425 rounded to the nearest hundredth is 8.43.
##
In conclusion, 7.884 rounded to the nearest hundredth is 7.89. This is because the digit to the right of the number, which is 8, is 5 or higher. To round any number to the nearest hundredth, you need to look at the digit to the right of the number and round up or down accordingly. This process is known as rounding, and it is a useful mathematical tool for simplifying complex numbers. |
Different types of Word Problems on Quadratic Equations by Factoring are available on this page. So, the students who feel they are lagging in the concept of a quadratic equation can make use of our page and practice the problems. Learn the unknown values of x with the help of the factorization method of a quadratic equation. Solve the real-world problems that are related to the quadratic equations from here.
Also, Refer:
## Word Problems on Quadratic Equations by Factoring
Example 1.
The school has a fund of 156. In addition, each student of the school contributes the number of rupees equal to the number of students. The total money is divided equally among the students. If each of the students gets 28 rupees. Find the number of students gets 28 find the number of students in the school.
Solution:
Let the number of students be x
Total contributions from them = x²
School has a fund of 156
According to the problem
x² + 156 = 28x
x² – 28x + 156 = 0
x² – 12x – 13x + 156 = 0
x(x – 12) – 13(x – 12) = 0
(x – 12) (x – 13) = 0
x = 12, 13.
Therefore 12 or 13 are the students in the school
The two answers are accepted in this case.
Example 2.
The product of two numbers is 6 if their sum added to the sum of their squares is 18, find the numbers.
Solution:
Let the number be x and y
As their product is 6, we get xy = 6 …….. (1)
According to the question x + y + x² + y² = 18 ……….. (2)
From equation 1; y = 6/x
y = 6/x in equation 2
x + 6/x + x² + ( 6/x)² = 18
(x + 6/x) + (x + 6/x)² – 2x. 6/x = 18
(x + 6/x)² + (x + 6/x) – 12 = 18
(x + 6/x)² + (x + 6/x) – 12 – 18 = 0
(x + 6/x)² + (x + 6/x) – 30 = 0
Putting x + 6/x = t
t² + t – 30 = 0
t² + 6t – 5t – 30 = 0
t(t + 6) -5(t + 6) = 0
t = -5, 6.
Sub -5 in x + 6/x
x + 6/x = -5
x -5x + 6 = 0
x² – 3x – 2x + 6 = 0
x(x-3) – 2(x-3) = 0
(x – 3) (x – 2) = 0
x = 3,2.
Sub 6 in x + 6/x = 0
x + 6/x = 6
x² – 6x + 6 = 0
(x – 1)(x – 6) = 0
x = 1, 6.
Therefore Solution set x = -5, 6, 1, 6.
Example 3.
Solve the quadratic equation 8m² – 8m + 2 = 0
Solution:
8m² – 8m + 2 = 0
8m² – 4m – 4m + 2 = 0
2m( 4m – 2) -1( 4m – 2) = 0
(2m – 1) (4m – 2) = 0
2m – 1 = 0 ; 4m – 2 = 0
2m = 1 ; 4m = 2
m = ½ ; m = 2/4 = ½
Therefore the solution set m = { ½, ½}
Example 4.
Solve x for equation x² + 3x – 4xy – 12y = 0
Solution:
x² + 3x – 4xy – 12y = 0
x(x + 3) – 4y(x + 3) = 0
(x – 4y) – (x + 3) = 0
x = 4y ; x = -3
Therefore the solution set x = {4y , – 3}
Example 5.
Solve the Quadratic Equation 3(x² + 1) = 7x
Solution:
3(x² + 1) = 7x
3x² + 3 = 7x
3x² + 3 – 7x = 0
3x² – 6x – x + 3 = 0
3x(x – 2) – 1(x – 2) = 0
(3x – 1) (x – 2) = 0
3x – 1 = 0 ; x – 2 = 0
3x = 1 ; x = 2
x = ⅓ ; x = 2
Therefore the solution set x = { ⅓, 2}
Example 6.
A bus travels at a certain average speed for a distance of 12km and then travels a distance of 6km at an average speed of 3 km/h more than its original speed. If it takes 2hours to complete the total journey, what is its original average speed?
Solution:
Let the original speed of train is x km/h
Time taken to cover 12 km with speed x km/h,
Time = distance Speed = 12x hours
After 12km, speed of train becomes (x + 6) km/h
Time taken to cover 6Km with speed (x + 3) km/h
Time = distance
After 63 km, speed of train becomes(x + 6) km/h
Time taken to cover 72km with speed (x + 6) km/h
Time = distance Speed = 6x + 3hours
Now as per the question
12/x + 6/x + 3 = 2
6/x + 3/x + 3 = 1
6x + 18 + 3x/x(x + 3) = 1
6x + 18 + 3x = x(x + 3)
9x + 18 = x² + 3x
x² – 9x + 3x – 18 = 0
x² – 3x – 18 = 0
x(x + 3) -6( x + 3) = 0
(x + 3) (x – 6)
x = -3, x = 6
Thus thhe solution set x = {-3, 6}
Example 7.
There are three consecutive positive integers such that the sum of the square of first and the product of the other two is 164. Find the integers.
Solution:
Let x, x+1, x + 2 are the first three consecutive integers.
The sum of the squares of the first and their product of other two is 164
Let x, (x + 1) and (x + 2) be the first three consecutive integers
The sum of the squares of first and the product of the other two is 164
x² + (x + 1)(x + 2) = 164
x² + x² + 2x + 1x + 2 = 164
2x² + 3x + 2 = 164
2x² + 3x – 162 = 0
(2x + 18)(x – 9) = 0
2x + 18 = 0
2x = 18
x = 18/2 = 9
x – 9 = 0
x = 9
Therefore the solution set is x = {18, 9}
Example 8.
Two natural numbers differ by 2 and their product is 400. Find the numbers?
Solution:
Let x and y are two natural numbers, it is differ by 2
x – y = 2
And their product is 400
So, xy = 400
y = x/400
Now apply the value of y in the first equation
x – (400/x) = 2
x² – (400/x) = 2
x² – 400 = 2x
x² – 2x – 400 = 0
(x – 20)(x + 20) = 0
x – 20 = 0
x = 20
x + 20 = 0
x = -20
Therefore the solution set x = { 20, -20} |
Chapter 6. Linear Transformation. 6.1 Intro. to Linear Transformation
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1 Chapter 6 Linear Transformation 6 Intro to Linear Transformation Homework: Textbook, 6 Ex, 5, 9,, 5,, 7, 9,5, 55, 57, 6(a,b), 6; page 7- In this section, we discuss linear transformations 89
2 9 CHAPTER 6 LINEAR TRANSFORMATION Recall, from calculus courses, a funtion f : X Y from a set X to a set Y associates to each x X a unique element f(x) Y Following is some commonly used terminologies: X is called the domain of f Y is called the codomain of f If f(x) = y, then we say y is the image of x The preimage of y is preimage(y) = {x X : f(x) = y} 4 The range of f is the set of images of elements in X In this section we deal with functions from a vector sapce V to another vector space W, that respect the vector space structures Such a function will be called a linear transformation, defined as follows Definition 6 Let V and W be two vector spaces A function T : V W is called a linear transformation of V into W, if following two prperties are true for all u,v V and scalars c T(u+v) = T(u)+T(v) (We say that T preserves additivity) T(cu) = ct(u) (We say that T preserves scalar multiplication) Reading assignment Read Textbook, Examples -, p 6- Trivial Examples: Following are two easy exampes
3 6 INTRO TO LINEAR TRANSFORMATION 9 Let V,W be two vector spaces Define T : V W as T(v) = for all v V Then T is a linear transformation, to be called the zero transformation Let V be a vector space Define T : V V as T(v) = v for all v V Then T is a linear transformation, to be called the identity transformation of V 6 Properties of linear transformations Theorem 6 Let V and W be two vector spaces Suppose T : V W is a linear transformation Then T() = T( v) = T(v) for all v V T(u v) = T(u) T(v) for all u,v V 4 If v = c v + c v + + c n v n then T(v) = T (c v + c v + + c n v n ) = c T (v )+c T (v )+ +c n T (v n ) Proof By property () of the definition 6, we have T() = T() = T() =
4 9 CHAPTER 6 LINEAR TRANSFORMATION So, () is proved Similarly, T( v) = T(( )v) = ( )T(v) = T(v) So, () is proved Then, by property () of the definition 6, we have T(u v) = T(u + ( )v) = T(u) + T(( )v) = T(u) T(v) The last equality follows from () So, () is proved To prove (4), we use induction, on n For n = : we have T(c v ) = c T(v ), by property () of the definition 6 For n =, by the two properties of definition 6, we have T(c v + c v ) = T(c v ) + T(c v ) = c T(v ) + c T(v ) So, (4) is prove for n = Now, we assume that the formula (4) is valid for n vectors and prove it for n We have T (c v + c v + + c n v n ) = T (c v + c v + + c n v n )+T (c n v n ) = (c T (v ) + c T (v ) + + c n T (v n )) + c n T(v n ) So, the proof is complete 6 Linear transformations given by matrices Theorem 6 Suppose A is a matrix of size m n Given a vector v = v v Rn define T(v) = Av = A v v v n v n Then T is a linear transformation from R n to R m
5 6 INTRO TO LINEAR TRANSFORMATION 9 Proof From properties of matrix multiplication, for u,v R n and scalar c we have and T(u + v) = A(u + v) = A(u) + A(v) = T(u) + T(v) The proof is complete T(cu) = A(cu) = cau = ct(u) Remark Most (or all) of our examples of linear transformations come from matrices, as in this theorem Reading assignment Read Textbook, Examples -, p 65-6 Projections along a vector in R n Projections in R n is a good class of examples of linear transformations We define projection along a vector Recall the definition 56 of orthogonal projection, in the context of Euclidean spaces R n Definition 64 Suppose v R n is a vector Then, for u R n define proj v (u) = v u v v Then proj v : R n R n is a linear transformation Proof This is because, for another vector w R n and a scalar c, it is easy to check proj v (u+w) = proj v (u)+proj v (w) and proj v (cu) = c (proj v (u)) The point of such projections is that any vector u R n can be written uniquely as a sum of a vector along v and another one perpendicular to v: u = proj v (u) + (u proj v (u))
6 94 CHAPTER 6 LINEAR TRANSFORMATION It is easy to check that (u proj v (u)) proj v (u) Exercise 65 (Ex 4, p 7) Let T(v,v,v ) = (v + v, v v,v v ) Compute T( 4, 5, ) Solution: T( 4, 5, ) = ( ( 4)+5, 5 ( 4), 4 ) = (,, 5) Compute the preimage of w = (4,, ) Solution: Suppose (v,v,v ) is in the preimage of (4,, ) Then (v + v, v v,v v ) = (4,, ) So, v +v = 4 v v = v v = The augmented matrix of this system is 4 its Gauss Jordan f orm So, Preimage((4,, )) = {(574, 8574, 574)}
7 6 INTRO TO LINEAR TRANSFORMATION 95 Exercise 66 (Ex p 7) Determine whether the function T : R R T(x,y) = (x,y) is linear? Solution: We have T((x,y) + (z,w)) = T(x + z,y + w) = ((x + z),y + w) (x,y) + (z,w) = T(x,y) + T(z,w) So, T does not preserve additivity So, T is not linear Alternately, you could check that T does not preserve scalar multiplication Alternately, you could check this failure(s), numerically For example, T((, ) + (, )) = T(, ) = (9, ) T(, ) + T(, ) Exercise 67 (Ex 4, p 7) Let T : R R be a linear transformation such that T(,, ) = (, 4, ), T(,, ) = (,, ), T(,, ) = (,, ) Compute T(, 4, ) Solution: We have (, 4, ) = (,, ) + 4(,, ) (,, ) So, T(, 4, ) = T(,, )+4T(,, ) T(,, ) = (, 4, )+(,, )+(,, ) = (, 5, ) Remark A linear transformation T : V V can be defined, simply by assigning values T(v i ) for any basis {v,v,,v n } of V In this case of the our problem, values were assigned for the standard basis {e,e,e } of R
8 96 CHAPTER 6 LINEAR TRANSFORMATION Exercise 68 (Ex 8, p 7) Let 4 A = Let T : R 5 R be the linear transformation T(x) = Ax Compute T(,,,, ) Solution: T(,,,, ) = 4 = 7 5 Compute preimage, under T, of (, 8) Solution: The preimage consists of the solutions of the linear system 4 x x x x 4 = 8 The augmented matrix of this system is 4 8 x 5 The Gauss-Jordan form is
9 6 INTRO TO LINEAR TRANSFORMATION 97 We use parameters x = t,x 4 = s,x 5 = u and the solotions are given by x = 5 + t + 5s + 4u,x = t,x = 4 + 5s,x 4 = s,x 5 = u So, the preimage T (, 8) = {(5+t+5s+4u, t, 4+5s, s, u) : t,s,u R} Exercise 69 (Ex 54 (edited), p 7) Let T : R R be the linear transformation such that T(, ) = (, ) and T(, ) = (, ) Compute T(, 4) Solution: We have to write (, 4) = a(, )+b(, ) Solving (, 4) = 5(, ) 5(, ) So, T(, 4) = 5T(, ) 5T(, ) = 5(, ) 5(, ) = (, 5) Compute T(, ) Solution: We have to write (, ) = a(, )+b(, ) Solving (, ) = 5(, ) 5(, ) So, T(, ) = 5T(, ) 5T(, ) = 5(, ) 5(, ) = (, ) Exercise 6 (Ex 6 (edited), p 7) Let T : R R the projection T(u) = proj v (u) where v = (,, )
10 98 CHAPTER 6 LINEAR TRANSFORMATION Find T(x,y,z) Solution: See definition 64 proj v (x,y,z) = v (x,y,z) v v = (,, ) (x,y,z) (,, ) ( x + y + z = Compute T(5,, 5) (,, ) = x + y + z (,, ), x + y + z, x + y + z Solution: We have ( x + y + z T(5,, 5) =, x + y + z, x + y + z ) ( =,, ) Compute the matrix of T ) Solution: The matrix is given by A =, because T(x,y,z) = x y z
11 6 KERNEL AND RANGE OF LINEAR TRANSFORMATION99 6 Kernel and Range of linear Transformation We will essentially, skip this section Before we do that, let us give a few definitions Definition 6 Let V,W be two vector spaces and T : V W a linear transformation Then the kernel of T, denoted by ker(t), is the set of v V such that T(v) = Notationally, ker(t) = {v V : T(v) = } It is easy to see the ker(t) is a subspace of V Recall, range of T, denoted by range(t), is given by range(t) = {v W : w = T(v) for some v V } It is easy to see the range(t) is a subspace of W We say the T is isomorphism, if T is one-to-one and onto It follows, that T is an isomorphism if ker(t) = {} and range(t) = W
12 CHAPTER 6 LINEAR TRANSFORMATION 6 Matrices for Linear Transformations Homework: Textbook, 6, Ex 5, 7,,, 7, 9,,, 5, 9,,, 5(a,b), 7(a,b), 9, 4, 45, 47; p 97 Optional Homework: Textbook, 6, Ex 57, 59; p 98 (We will not grade them) In this section, to each linear transformation we associate a matrix Linear transformations and matrices have very deep relationships In fact, study of linear transformations can be reduced to the study of matrices and conversely First, we will study this relationship for linear transformations T : R n R m ; and later study the same for linear transformations T : V W of general vector spaces In this section, we will denote the vectors in R n, as column matrices Recall, written as columns, the standard basis of R n is given by B = {e,e,,e n },,, Theorem 6 Let T : R n R m be a linear transformation Write T(e ) = a a,t(e ) = a a,,t(e n) = a n a n a m a m a mn
13 6 MATRICES FOR LINEAR TRANSFORMATIONS These columns T(e ),T(e ),T(e n ) form a m n matrix A as follows, a a a n A = a a a n a m a m a mn This matrix A has the property that T(v) = Av for all v R n This matrix A is called the standard matrix of T Proof We can write v R n as v v = v = v e + v e + + v n e n v n We have, Av = a a a n a a a n a m a m a mn v v v n = v a a a m +v a a a m + +v n = v T (e )+v T (e )+ +v n T (e n ) = T (v e + v e + + v n e n ) = T(v) The proof is complete Reading assignment: Read Textbook, Examples,; page 89-9 In our context of linear transformations, we recall the following definition of composition of functions Definition 6 Let a n a n a mn T : R n R m, T : R m R p
14 CHAPTER 6 LINEAR TRANSFORMATION be two linear transformations Define the composition T : R n R p of T with T as T(v) = T (T (v)) for v R n The composition T is denoted by T = T ot Diagramatically, R n T R m T=T ot T R p Theorem 6 Suppose T : R n R m, T : R m R p are two linear transformations Then, the composition T = T ot : R n R p is a linear transformation Suppose A is the standard matrix of T and A is the standard matrix of T Then, the standard matrix of the composition T = T ot is the product A = A A Proof For u,v R n and scalars c, we have T(u+v) = T (T (u+v)) = T (T (u)+t v)) = T (T (u))+t T (v)) = T(u)+T(v) and T(cu) = T (T (cu)) = T (ct (u)) = ct (T (u)) = ct(u) So, T preserves addition and scalar multiplication Therefore T is a linear transformation and () is proved To prove (), we have T(u) = T (T (u)) = T (A u) = A (A u) = (A A )u
15 6 MATRICES FOR LINEAR TRANSFORMATIONS Therefore T(e ) is the first column of A A, T(e ) is the second column of A A, and so on Therefore, the standard matrix of T is A A The proof is complete Reading assignment: Read Textbook, Examples ; page 9 Definition 64 Let T : R n R n, T : R n R n be two linear transformations such that for every v R n we have T (T (v)) = v and T (T (v)) = v, then we say that T is the inverse of T, and we say that T is invertible Such an inverse T of T is unique and is denoted by T (Remark Let End(R n ) denote the set of all linear transformations T : R n R n Then End(R n ) has a binary operation by composition The identity operation I : R n R n acts as the identitiy under this composition operation The definition of inverse of T above, just corresponds to the inverse under this composition operation) Theorem 65 Let T : R n R n be a linear transformations and let A be the standard matrix of T Then, the following are equivalent, T is invertible T is an isomorphism A is invertible And, if T is invertible, then the standard matrix of T is A
16 4 CHAPTER 6 LINEAR TRANSFORMATION Proof (First, recall definition 6, that T is isomorphism if T is to and onto) (Proof of () () :) Suppose T is invertible and T be the inverse Suppose T(u) = T(v) Then, u = T (T(u)) = T (T(v)) = v So, T is to Also, given u R n we have So, T is an isomorphism u = T(T (u)) So, T onto R n (Proof of () () :) So, assume T is an isomorphism Then, Ax = T(x) = T() x = So, Ax = has an unique solution Therefore A is invertible and () follows from () (Proof of () () :) Suppose A is invertible Let T (x) = A x, then T is a linear transformation and it is easily checked that T is the inverse of T So, () follows from () The proof is complete Reading assignment: Read Textbook, Examples 4; page 9 6 Nonstandard bases and general vector spaces The above discussion about (standard) matrices of linear transformations T had to be restricted to linear transformations T : R n R m This wa sbecause R n has a standard basis {e,e,,e n, } that we could use Suppose T : V W is a linear transformation between two abstract vector spaces V,W Since V and W has no standard bases, we cannot associate a matrix to T But, if we fix a basis B of V and B of W we can associate a matrix to T We do it as follows
17 6 MATRICES FOR LINEAR TRANSFORMATIONS 5 Theorem 66 Suppose T : V W is a linear transformation between two vector spaces V,W Let B = {v,v,,v n } be basis of V and B = {w,w,,w m } be basis of w We can write T(v ) = w w w m a a a m Writing similar equations for T(v ),,T(v n ), we get T(v ) T(v ) T(v n ) = w w w m a a a n a a a n a m a m a mn Then, for v = x v + x v + + x n v n, We have a a a n T(v) = w w w m a a a n a m a m a mn Write A = a ij Then, T is determined by A, with respect to bases B,B x x x n Exercise 67 (Ex 6, p 97) Let T(x,y,z) = (5x y + z, z + 4y, 5x + y)
18 6 CHAPTER 6 LINEAR TRANSFORMATION What is the standard matrix of T? Solution: We have T : R R We write vectors x R as columns x x = y instead of (x, y, z) Recall the standard basis e = We have T(e ) = z, e = 5 5, T(e ) =, e = So, the standard matrix of T is 5 A = Exercise 68 (Ex, p 97) Let of R, T(e ) = T(x,y,z) = (x + y, y z) Write down the standard matrix of T and use it to find T(,, ) Solution: In this case, T : R R With standard basis e,e,e as in exercise 67, we have T(e ) =, T(e ) =, T(e ) =
19 6 MATRICES FOR LINEAR TRANSFORMATIONS 7 So, the standard matrix of T is A = Therefore, T(,, ) = A = = 4 We will write our answer in as a row: T(,, ) = (, 4) Exercise 69 (Ex 8, p 97) Let T be the reflection in the line y = x in R So, T(x,y) = (y,x) Write down the standard matrix of T Solution: In this case, T : R R With standard basis e = (, ) T,e = (, ) T, we have T(e ) =, T(e ) = So, the standard matrix of T is A = Use the standard matrix to compute T(, 4) Solution: Of course, we know T(, 4) = (4, ) They want use to use the standard matrix to get the same answer We have 4 T(, 4) = A = = or T(, 4) = (4, ) 4 4
20 8 CHAPTER 6 LINEAR TRANSFORMATION Lemma 6 Suppose T : R R is the counterclockwise rotation by a fixed angle θ Then T(x,y) = cos θ sin θ x sin θ cos θ y Proof We can write x = r cos φ, y = r sin φ, where r = x + y, tanφ = y/x By definition T(x,y) = (r cos(φ + θ),r sin(φ + θ)) Using trig-formulas r cos(φ + θ) = r cosφcos θ r sin φ sin θ = x cos θ y sin θ and r sin(φ + θ) = r sin φ cos θ + r cos φ sin θ = y cos θ + x sin θ So, cos θ sin θ T(x,y) = sin θ cos θ The proof is complete x y Exercise 6 (Ex, p 97) Let T be the counterclockwise rotation in R by angle o Write down the standard matrix of T Solution: We use lemma 6, with θ = So, the standard matrix of T is A = cosθ sin θ cos sin = sin θ cos θ sin cos = 5 5
21 6 MATRICES FOR LINEAR TRANSFORMATIONS 9 Compute T(, ) Solution: We have T(, ) = A = 5 5 = + We write our answer as rows: T(, ) = (, + ) Exercise 6 (Ex, p 97) Let T be the projection on to the vector w = (, 5) in R : T(u) = proj w (u) Find the standard matrix Solution: See definition 64 T(x,y) = proj w (x,y) = w (x,y) (, 5) (x,y) x 5y w = (, 5) = (, 5) w (, 5) 6 ( ) x 5y 5x + 5y =, 6 6 So, with e = (, ) T,e = (, ) T we have (write/think everything as columns): T(e ) = So, the standard matrix is Compute T(, ) A =, T(e ) = Solution: We have T(, ) = A = We write our answer in row-form: T(, ) = ( 5, 65 6) =
22 CHAPTER 6 LINEAR TRANSFORMATION Exercise 6 (Ex 6, p 97) Let T(x,y,z) = (x y + z, x y,y 4z) Write down the standard matrix of T Solution: with e = (,, ) T,e = (,, ) T,e = (,, ) T we have (write/think everything as columns): T(e ) =, T(e ) = So, the standard matrix is A = Compute T(,, ) 4, T(e ) = 4 Solution: We have T(,, ) = A = 4 = 7 7 Exercise 64 (Ex 4, p 97) Let T : R R, T (x,y) = (x y, x + y) and T : R R, T (x,y) = (y, ) Compute the standard matrices of T = T ot and T = T T Solution: We solve it in three steps:
23 6 MATRICES FOR LINEAR TRANSFORMATIONS Step-: First, compute the standard matrix of T With e = (, ) T,e = (, ) T we have (write/think everything as columns): T (e ) =, T (e ) = So, the standard matrix of T is A = Step-: Now, compute the standard matrix of T With e = (, ) T,e = (, ) T we have : T (e ) =, T (e ) = So, the standard matrix of T is A = Step- By theorem 6, the standard matrix of T = T T is A A = = So, T(x,y) = (x+y, ) Similarly, the standard matrix of T = T T is A A = = So, T (x,y) = (y, y) Exercise 65 (Ex 46, p 97) Determine whether T(x,y) = (x + y,x y)
24 CHAPTER 6 LINEAR TRANSFORMATION is invertible or not Solution: Because of theorem 65, we will check whether the standard matrix of T is invertible or not With e = (, ) T,e = (, ) T we have : T(e ) =, T(e ) = So, the standard matrix of T is A = Note, deta = 4 So, T is invertible and hence T is invertible Exercise 66 (Ex 58, p 97) Determine whether T(x,y) = (x y,,x + y) Use B = {v = (, ),v = (, )} as basis of the domain R and B = {w = (,, ),w = (,, ),w = (,, )} as basis of codomain R Compute matrix of T with respect to B,B Solution: We use theorem 66 We have T(u ) = T(, ) = (,, ), T(u ) = T(, ) = (,, ) We solve the equation: (,, ) = aw + bw + cw = a(,, ) + b(,, ) + c(,, ) and we have (,, ) = (,, ) (,, )+(,, ) = w w w
25 64 TRANSITION MATRICES AND SIMILARITY Similarly, we solve (,, ) = aw + bw + cw = a(,, ) + b(,, ) + c(,, ) and we have (,, ) = (,, ) (,, ) + (,, ) = w w w So, the matrix of T with respect to the bases B,B is A = 64 Transition Matrices and Similarity We will skip this section I will just explain the section heading You know what are Transition matrices of a linear transformation T : V W They are a matrices described in theorem 66 Definition 64 Suppose A,B are two square matrices of size n n We say A,B are similar, if A = P BP for some invertible matrix P
26 4 CHAPTER 6 LINEAR TRANSFORMATION 65 Applications of Linear Trans Homework: Textbook, 65 Ex (a), (a), 5, 7, 9, 5, 7, 9, 4, 49, 5, 5, 55, 6, 65; page In this section, we discuss geometric interpretations of linear transformations represented by elementary matrices Proposition 65 Let A be a matrix and x T(x,y) = A y We will write the right hand side as a row, which is an abuse of natation If A =, then T(x,y) = A x y = ( x,y) T represents the reflection in y axis See Textbook, Example (a), p47 for the diagram If A =, then T(x,y) = A x y = (x, y) T represents the reflection in x axis See Textbook, Example (b), p47 for the diagram
27 65 APPLICATIONS OF LINEAR TRANS 5 If A =, then T(x,y) = A x y = (y,x) T represents the reflection in line y = x See Textbook, Example (c), p47 for the diagram 4 If A = k, then T(x,y) = A x y = (kx,y) T If k >, then T represents expansion in horizontal direction and < k <, then T represents contraction in horizontal direction See Textbook, Example Fig 6, p49 for diagrams 5 If A = k, then T(x,y) = A x y = (x,ky) T If k >, then T represents expansion in vertical direction and < k <, then T represents contraction in vertical direction See Textbook, Example Fig 6, p49 for diagrams 6 If A = k, then T(x,y) = A x y = (x + ky,y) T Then T represents horizontal shear (Assume k > ) The upperhalf plane are sheared to right and lower-half plane are sheared to left The points on the x axis reamain fixed See Textbook, Example, fig 64, p49 for diagrams
28 6 CHAPTER 6 LINEAR TRANSFORMATION 7 If A = k, then T(x,y) = A x y = (x,kx + y) T Then T represents vertical shear (Assume k > ) The righthalf-plane are sheared to upward and left-half-plane are sheared to downward The points on the y axis reamain fixed See Textbook, Example, fig 65, p49 for diagrams 65 Computer Graphics Linear transformations are used in computer graphics to move figures on the computer screens I am sure all kinds of linear (and nonlinear) transformations are used Here, we will only deal with rotations by an angle θ, around () x axis, () y axis and () z axis as follows: Proposition 65 Suppose θ is an angle Suppose we want to rotate the point (x, y, z) counterclockwise about z axis through an angle θ Let us denote this transformation by T and write T(x,y,z) = (x,y,z ) T Then using lemma 6, we have T(x,y,z) = x y z = cos θ sin θ x sin θ cos θ y z = x cos θ y sin θ x sin θ + y cos θ Similarly, we can write down the linear transformations corresponding to rotation around x axis and y axis We write down the transition matrices for these three matrices as follows: z The standard matrix for this transformation of counterclockwise
29 65 APPLICATIONS OF LINEAR TRANS 7 rotation by an angle θ, about x axis is cosθ sin θ sinθ cosθ The standard matrix for this transformation of counterclockwise rotation by an angle θ, about y axis is cos θ sinθ sin θ cosθ The standard matrix for this transformation of counterclockwise rotation by an angle θ, about z axis is cos θ sin θ sin θ cos θ Reading assignment: Read Textbook, Examples 4 and 5; page 4-4 Exercise 65 (Ex 6, p 44) Let T(x,y) = (x, y) (This is a vertical expansion) Sketch the image of the unit square with vertices (, ), (, ), (, ), (, ) Solution: We have T(, ) = (, ), T(, ) = (, ), T(, ) = (, ), T(, ) = (, )
30 8 CHAPTER 6 LINEAR TRANSFORMATION Diagram:,,,,,, Here, the solid arrows represent the original rectangle and the brocken arrows represent the image Exercise 654 (Ex, p 44) Let T be the reflection in the line y = x Sketch the image of the rectangle with vertices (, ), (, ), (, ), (, ) Solution: Recall, (see Proposition 65 ()) that T(x, y) = (y, x) We have T(, ) = (, ),T(, ) = (, ),T(, ) = (, ),T(, ) = (, ) Diagram:,,,,,,, Here, the solid arrows represent the original rectangle and the brocken arrows represent the image
31 65 APPLICATIONS OF LINEAR TRANS 9 Exercise 655 (Ex 8, p 44) Suppose T is the expansion and contraction represented by T(x,y) = ( x, ) y Sketch the image of the rectangle with vertices (, ), (, ), (, ), (, ) Solution: Recall, (see Proposition 65 ()) that (x, y) = (y, x)we have T(, ) = (, ),T(, ) = (, ),T(, ) = (, ),T(, ) = (, ) Diagram:,,,,, 5,,, Here, the solid arrows represent the original rectangle and thebrocken arrows represent the image Exercise 656 (Ex 44, p 44) Give the geometric description of the linear transformation defined by the elementary matrix A = Solution: By proposition 65 (6) this is a horizontal shear Here, T(x,y) = (x + y,y)
32 CHAPTER 6 LINEAR TRANSFORMATION Exercise 657 (Ex 5 and 54, p 45) Find the matrix of the transformation T that will produce a 6 o rotation about the x axis Then compute the image T(,, ) Solution: By proposition 65 () the matrix is given by A = cos θ sin θ = cos 6 o sin 6 o = sinθ cos θ So, T(,, ) = A = sin 6 o cos 6 o = + Exercise 658 (Ex 64, p 45) Determine the matrix that will produce a 45 o rotation about the y axis followed by 9 o rotation about the z axis Then also compute the image of the line segment from (,, ) to (,, ) Solution: We will do it in three (or four) steps Step- Let T be the rotation by 45 o about the y axis By proposition 65 () the matrix of T is given by A = cos θ sinθ cos 45 sin 45 = sin θ cos θ sin 45 cos 45 = Step- Let T be the rotation by 9 o rotation about the z axis By proposition 65 () the matrix of T is given by B = cos θ sin θ cos 9 o sin 9 o sin θ cos θ = sin 9 o cos 9 o =
33 65 APPLICATIONS OF LINEAR TRANS Step- So, thematrix of the composite transformation T = T T is matrix BA = = The Last Part: So, T(,, ) = BA = =
34 CHAPTER 6 LINEAR TRANSFORMATION
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# Linear Algebra Basics for Data Science
Linear algebra is an expansive subject. But I’m going to try to cover the required pieces here that will help you get your journey underway.
## System of coordinates in linear algebra
The data includes, in total, an array of numbers. Spatial data is similar, but it also contains numerical information that assists you to find it on Earth.
These numbers are part of a coordinate system that provides a frame of reference for your data to position features on the surface of the earth, align your data with other data, conduct spatially correct analysis and create maps.
The arrangement of reference lines or curves used to establish points’ locations in space is a Coordinate Scheme.
Points are marked by their distance from the reference point in the horizontal (x) and vertical (y) axes, middle, center (0, 0).
As you can see, Cartesian coordinates for three (or more) dimensions can also be used.
The Cartesian x-y-z scheme, however, is not the only system of coordinates. Additional common systems include:
• Polar Coordinate System Polar System
• System of Cylindrical-coordinate, and
• Systems with spherical coordinates
## Vectors and Dimensions
So on a number line, all data points that appear like a single number can be interpreted. The best description of a communication scheme is the recognition of points on a line of real numbers using the number symbol.
The 2D data is represented by the shape (x,y) and can be represented by two axes.
The 3D data is represented by the form (x,y,z) and can be represented by three axes.
Data, known as vectors/tuples, is stored as a finite ordered list (sequence) of elements.
The number of elements within a vector is known as a vector’s dimensionality or dimension.
Every array of numerical data in a dataset will then be a vector. The vector (1,2,3,4) is considered to be a four-dimensional (4D) vector, and understanding this will allow you to realise that the data can be several hundred dimensions depending on the amount of features in the data, even though it can not be accurately visualised in 3D space.
Bioinformatics datasets for genome sequences, for example, have hundreds of characteristics. Similarly, text data has a minimum of 3000 characteristics when vectorized.
## Random Variables
You may describe a random variable as a functional mapping from an event’s sampling space to a line of real numbers.
A random variable can be either continuous or discrete in Linear algebra.
A countable number of distinct values are carried on by discrete random variables. Consider an experiment where a coin is tossed three times.
If X represents the number of times the coin heads appear, then X is a random differential variable which can only be 0, 1, 2, 3 (from no heads to all heads in three consecutive coin tosses). No other value is viable for X.
Over a given range or interval, continuous random variables can represent any value and can take on an infinite number of potential values. An experiment involving calculating the amount of rainfall in a city will be an example of a constant random variable.
## System of Linear Equations
From learning equations, the subject of algebra emerged.
We would try to find all the exact numbers x, for instance, so that x2 − 4 = 0.
We should rewrite our equation as (x − 2)*(x + 2) = 0 to solve, and then consider the left-hand side. So we will infer that either x = 2 or x = −2 because zero needs to be either x − 2 or x + 2.
However, it is a nonlinear question to find the roots of a polynomial, while the subject to be studied here is the linear equation principle.
The simplest linear equation is the equation ax = b. The letter x is the
variable, and a and b are fixed numbers. For example, consider 5x = 10.
The solution is x = 2.
We’ll also go through more linear algebra in further sections, so bookmark the page and keep yourself updated ! |
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# Solve the Following Quadratic Equations by Factorization: - CBSE Class 10 - Mathematics
ConceptSolutions of Quadratic Equations by Factorization
#### Question
Solve the following quadratic equations by factorization: $\frac{5 + x}{5 - x} - \frac{5 - x}{5 + x} = 3\frac{3}{4}; x \neq 5, - 5$
#### Solution
$\frac{5 + x}{5 - x} - \frac{5 - x}{5 + x} = 3\frac{3}{4}$
$\Rightarrow \frac{\left( 5 + x \right)^2 - \left( 5 - x \right)^2}{\left( 5 + x \right)\left( 5 - x \right)} = \frac{15}{4}$
$\Rightarrow \frac{25 + x^2 + 10x - 25 - x^2 + 10x}{25 - x^2} = \frac{15}{4}$
$\Rightarrow \frac{20x}{25 - x^2} = \frac{15}{4}$
$\Rightarrow \frac{4x}{25 - x^2} = \frac{3}{4}$
$\Rightarrow 16x = 75 - 3 x^2$
$\Rightarrow 3 x^2 + 16x - 75 = 0$
$\Rightarrow 3 x^2 + 25x - 9x - 75 = 0$
$\Rightarrow x(3x + 25) - 3(3x + 25) = 0$
$\Rightarrow (x - 3)(3x + 25) = 0$
$\Rightarrow x - 3 = 0 \text { or } 3x + 25 = 0$
$\Rightarrow x = 3 \text { or } x = - \frac{25}{3}$
Hence, the factors are 3 and $- \frac{25}{3}$.
Is there an error in this question or solution?
#### APPEARS IN
Solution Solve the Following Quadratic Equations by Factorization: Concept: Solutions of Quadratic Equations by Factorization.
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Mental Calculation: Horizontal Addition and Subtraction
To solve vertical additions and subtractions, we use written algorithms. We’ve all learnt how to work them out on a piece of paper. These algorithms are based on digits, and positional value does not matter. In other words, no matter whether we are adding or subtracting in the tens or thousands column, the procedure we follow is the same.
On the other hand, we cannot use this written algorithm for horizontal addition and subtraction. Trying to solve them by thinking in the same way that we would on paper is not effective. Therefore, we’re going to learn a couple of mental calculation strategies that will help when solving horizontal additions and subtractions.
The strategies we’re going to look at are based on numbers and take into account the positional value of each digit (units, tens, hundreds…).
We’ve already covered strategies for solving incomplete horizontal additions and subtractions in a previous post, involving decomposition (breaking down numbers) as a mental calculation strategy. Here we’ll learn two new strategies:
• In the first, we’ll use decomposition and continue with an addition chain.
• In the second, we will learn to subtract using the counting-up method.
The first strategy for horizontal addition and subtraction: Decomposition + addition chain
This strategy involves breaking down the summands, paying attention to positional value, and then adding. These sums in the addition chain are more simple because they don’t mix positional values; the tens are added to the tens, the hundreds to the hundreds, etc… There’s no one way to use this strategy, each student will adapt it to their abilities and other strategies that they’ve developed.
Let’s look at an example:
We can approach this addition in many different ways using the decomposition and addition chain method. We’re going to show two different possibilities that we’ll call A and B.
A1.- We break down the first summand: 2463 = 2000 + 400 + 60 + 3
And then the second summand: 1382 = 1000 + 300 + 80 + 2
A2.- We chain add the decomposed numbers: 3000 + 700 + 140 + 5 = 3845
B1.- We break down one of the summands: 1382 = 1000 + 300 + 80 + 2
B2.- We chain add the broken down numbers to the other summand: 2463 + 1000 + 300 + 80 + 2 = 3845
The second strategy for horizontal addition and subtraction: Counting-up
Counting-up is a mental calculation strategy for subtraction that needs prior explanation if you’ve never encountered it. In general, we understand subtraction as taking something away. The minuend is taken away from the subtrahend and we are left with the difference as the result (e.g. 10 – 6 = ?). But, it can also be understood as an ascending staircase. In this case, the difference is the amount that needs to be added to the subtrahend in order to reach the minuend. Therefore we can look at the subtraction as an addition with an unknown number (e.g. 10 – 6 = ? can be seen as 6 + ? = 10). With this in mind, it’s easy to count up from the subtrahend until you reach the minuend.
Here’s an example:
It’s a mental calculation strategy, but we’ll represent it visually on an empty number line. We’ve marked the subtrahend and minuend on it:
Now we add a number to the subtrahend that brings us to the next ten:
Then we add another number to reach the ten before the minuend:
And finally, we add another number to get to the minuend.
Now we just have to add together the numbers we used to get from the subtrahend to the minuend:
3 + 50 + 5 = 58.
This is an open strategy that adapts to the specific abilities of each student. Each person will add the numbers in the way that best suits them. This method can also be used for subtraction with three- and four-digit numbers. As we increase the order of magnitude of the numbers, the strategy grows in complexity. It increases the possibilities of numbers to add: to reach the next ten of the subtrahend, to reach the next hundred of the subtrahend, to reach the hundred before the minuend… It needs practice to be mastered. And gradually, as students learn to count up, they simultaneously develop mental strategies for addition and subtraction.
The two mental calculation strategies for solving horizontal addition and subtraction that we’ve learned today are open and flexible. They adapt to the experience, capacity, and need of each individual. In addition, they are also combinable with other mental calculation strategies.
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# Different kinds of numbers
Page 1 / 2
## Different kinds of numbers
CLASS ASSIGNMENT 1
• Discover the number system step by step....
1. General: Different kinds of numbers
Provide an example of each of the following numbers:
• Natural numbers N = {..................................................}
• Counting numbers N 0 = {..................................................}
• Integers Z+ = {..................................................}
Z- = {..................................................}
• Rational numbers Q = {..................................................}
• Irrational numbers Q’ = {..................................................}
• Real numbers R = {..................................................}
2. Natural numbers
Prime numbers ={..................................................} Compound numbers ={..................................................} Definition: ........................................................................................…………..................................................…………..................................................………….......... Definition: ................................................................…………..........................................................................…………..................................................…………..........
Prime numbers + Compound numbers = Natural numbers
3. Divisibility rules
Do you recall that
In each instance, select a number that is divisible by the given divisor and try to deduce a rule for each instance.
Number Divisor Divisibility rule 3.1 2 3.2 3 3.3 4 3.4 5 3.5 6 3.6 8 3.7 9 3.8 10 3.9 11
4. Determine by which numbers (1.3.1 - 1.3.9) 61 226 is divisible and provide a reason for each.
5. Explain what you understand the following terms to mean:
5.1 Multiple:
5.2 Factor:
5.3 Prime number:
5.4 Prime factor:
5.5 Even numbers and odd numbers:
• How do you determine the factors of a number? Look at the following.....e.g. F 24 = {1; 2; 3; 4; 6; 8; 12; 24} 1 x 24; 2 x 12; 3 x 8; 4 x 6
6. Determine the factors of 48.
7. Write out all the multiples of 6 between 23 and 56.
8. Determine the prime numbers between 17 and 78.
9 Determine all odd compound numbers between 16 and 50.
10 Write down all the factors of 50 that are prime numbers.
11. Write down all the factors of 50 that are compound numbers.
12. Explain: Cube numbers. Write down the first 6 cube numbers.
13. Explain : Square numbers. Write down the first 10 square numbers.
HOMEWORK ASSIGNMENT 1
1. Write the definition for each of the following:
1.1 Rational number:
1.2 Prime number:
1.3 Compound numbers:
1.4 Prime factors:
2. Select from {0; 1; 2; 3; 4; ... ; 36} and write down:
2.1 The first two compound numbers
2.2 Odd numbers that are not prime numbers
2.3 Multiples of 6
2.4 Factors of 12
2.5 Prime factors of 12
2.6 Factors of 36
3. Which of the following numbers - 9 / 3 ; 7 / 0 ; 0; 3; -9; 16; 2 1 / 3 are:
3.1 Integers?
3.2 Rational numbers?
3.3 Non-real numbers?
4. Tabulate the following:
4.1 Natural numbers<5 ...........................................................................
what does nano mean?
nano basically means 10^(-9). nanometer is a unit to measure length.
Bharti
do you think it's worthwhile in the long term to study the effects and possibilities of nanotechnology on viral treatment?
absolutely yes
Daniel
how to know photocatalytic properties of tio2 nanoparticles...what to do now
it is a goid question and i want to know the answer as well
Maciej
Abigail
for teaching engĺish at school how nano technology help us
Anassong
Do somebody tell me a best nano engineering book for beginners?
what is fullerene does it is used to make bukky balls
are you nano engineer ?
s.
fullerene is a bucky ball aka Carbon 60 molecule. It was name by the architect Fuller. He design the geodesic dome. it resembles a soccer ball.
Tarell
what is the actual application of fullerenes nowadays?
Damian
That is a great question Damian. best way to answer that question is to Google it. there are hundreds of applications for buck minister fullerenes, from medical to aerospace. you can also find plenty of research papers that will give you great detail on the potential applications of fullerenes.
Tarell
what is the Synthesis, properties,and applications of carbon nano chemistry
Mostly, they use nano carbon for electronics and for materials to be strengthened.
Virgil
is Bucky paper clear?
CYNTHIA
so some one know about replacing silicon atom with phosphorous in semiconductors device?
Yeah, it is a pain to say the least. You basically have to heat the substarte up to around 1000 degrees celcius then pass phosphene gas over top of it, which is explosive and toxic by the way, under very low pressure.
Harper
Do you know which machine is used to that process?
s.
how to fabricate graphene ink ?
for screen printed electrodes ?
SUYASH
What is lattice structure?
of graphene you mean?
Ebrahim
or in general
Ebrahim
in general
s.
Graphene has a hexagonal structure
tahir
On having this app for quite a bit time, Haven't realised there's a chat room in it.
Cied
what is biological synthesis of nanoparticles
what's the easiest and fastest way to the synthesize AgNP?
China
Cied
types of nano material
I start with an easy one. carbon nanotubes woven into a long filament like a string
Porter
many many of nanotubes
Porter
what is the k.e before it land
Yasmin
what is the function of carbon nanotubes?
Cesar
I'm interested in nanotube
Uday
what is nanomaterials and their applications of sensors.
what is nano technology
what is system testing?
preparation of nanomaterial
how to synthesize TiO2 nanoparticles by chemical methods
Zubear
how did you get the value of 2000N.What calculations are needed to arrive at it
Privacy Information Security Software Version 1.1a
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Berger describes sociologists as concerned with
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# Multiplication of Rational Expressions
## Multiply and reduce fractions with variables in the denominator
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Practice Multiplication of Rational Expressions
Progress
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Multiplying Rational Expressions
The length of a rectangle is 2xy3z5xyz2\begin{align*}\frac{2xy^3z}{5xyz^2}\end{align*}. The width of the rectangle is 3x2yz34x3y2z2\begin{align*}\frac{3x^2yz^3}{4x^3y^2z^2}\end{align*}. What is the area of the rectangle?
### Guidance
We take the previous concept one step further in this one and multiply two rational expressions together. When multiplying rational expressions, it is just like multiplying fractions. However, it is much, much easier to factor the rational expressions before multiplying because factors could cancel out.
#### Example A
Multiply x24xx39xx2+8x+15x22x8\begin{align*}\frac{x^2-4x}{x^3-9x} \cdot \frac{x^2+8x+15}{x^2-2x-8}\end{align*}
Solution: Rather than multiply together each numerator and denominator to get very complicated polynomials, it is much easier to first factor and then cancel out any common factors.
x24xx39xx2+8x+15x22x8=x(x4)x(x3)(x+3)(x+3)(x+5)(x+2)(x4)\begin{align*}\frac{x^2-4x}{x^3-9x} \cdot \frac{x^2+8x+15}{x^2-2x-8} = \frac{x(x-4)}{x(x-3)(x+3)} \cdot \frac{(x+3)(x+5)}{(x+2)(x-4)}\end{align*}
At this point, we see there are common factors between the fractions.
x(x4)x(x3)(x+3)(x+3)(x+5)(x+2)(x4)=x+5(x3)(x+2)\begin{align*}\frac{\cancel{x} \cancel{(x-4)}}{\cancel{x}(x-3) \cancel{(x+3)}} \cdot \frac{\cancel{(x+3)}(x+5)}{(x+2)\cancel{(x-4)}} = \frac{x+5}{(x-3)(x+2)}\end{align*}
At this point, the answer is in factored form and simplified. You do not need to multiply out the base.
#### Example B
Multiply 4x2y5z6xyz615y435x4\begin{align*}\frac{4x^2y^5z}{6xyz^6} \cdot \frac{15y^4}{35x^4}\end{align*}
Solution: These rational expressions are monomials with more than one variable. Here, we need to remember the laws of exponents from earlier concepts. Remember to add the exponents when multiplying and subtract the exponents when dividing. The easiest way to solve this type of problem is to multiply the two fractions together first and then subtract common exponents.
4x2y5z6xyz615y435x4=60x2y9z210x5yz6=2y87x3z5\begin{align*}\frac{4x^2y^5z}{6xyz^6} \cdot \frac{15y^4}{35x^4} = \frac{60x^2y^9z}{210x^5yz^6} = \frac{2y^8}{7x^3z^5}\end{align*}
You can reverse the order and cancel any common exponents first and then multiply, but sometimes that can get confusing.
#### Example C
Multiply 4x2+4x+12x29x5(3x2)x2256x2x2\begin{align*}\frac{4x^2+4x+1}{2x^2-9x-5} \cdot (3x-2) \cdot \frac{x^2-25}{6x^2-x-2}\end{align*}
Solution: Because the middle term is a linear expression, rewrite it over 1 to make it a fraction.
4x2+4x+12x29x5(3x2)x2256x2x2=(2x+1)(2x+1)(2x+1)(x5)3x21(x5)(x+5)(3x2)(2x+1)=x+5\begin{align*}\frac{4x^2+4x+1}{2x^2-9x-5} \cdot (3x-2) \cdot \frac{x^2-25}{6x^2-x-2} = \frac{\cancel{(2x+1)} \cancel{(2x+1)}}{\cancel{(2x+1)} \cancel{(x-5)}} \cdot \frac{\cancel{3x-2}}{1} \cdot \frac{\cancel{(x-5)}(x+5)}{\cancel{(3x-2)} \cancel{(2x+1)}} = x+5\end{align*}
Intro Problem Revisit The area of the rectangle is length times width. So to find the area, multiply the two terms and simplify.
2xy3z5xyz23x2yz34x3y2z26x3y4z420x4y3z43y10x
Therefore, the area of the rectangle is 3y10x\begin{align*}\frac{3y}{10x}\end{align*}.
### Guided Practice
Multiply the following expressions.
1. 4x28x10x315x25xx2\begin{align*}\frac{4x^2-8x}{10x^3} \cdot \frac{15x^2-5x}{x-2}\end{align*}
2. x2+6x7x236x22x242x2+8x42\begin{align*}\frac{x^2+6x-7}{x^2-36} \cdot \frac{x^2-2x-24}{2x^2+8x-42}\end{align*}
3. 4x2y732x4y316x28y6\begin{align*}\frac{4x^2y^7}{32x^4y^3} \cdot \frac{16x^2}{8y^6}\end{align*}
1. 4x28x10x315x25xx2=22x(x2)25xxx5x(3x1)x2=2(3x1)x\begin{align*}\frac{4x^2-8x}{10x^3} \cdot \frac{15x^2-5x}{x-2} = \frac{\cancel{2} \cdot 2 \cancel{x}\cancel{(x-2)}}{\cancel{2} \cdot \cancel{5x} \cdot \cancel{x} \cdot x} \cdot \frac{\cancel{5x}(3x-1)}{\cancel{x-2}} = \frac{2(3x-1)}{x}\end{align*}
2. x2+6x7x236x22x242x2+8x42=(x+7)(x1)(x6)(x+6)(x6)(x+4)2(x+7)(x3)=(x1)(x+4)2(x3)(x+6)\begin{align*}\frac{x^2+6x-7}{x^2-36} \cdot \frac{x^2-2x-24}{2x^2+8x-42} = \frac{\cancel{(x+7)}(x-1)}{\cancel{(x-6)}(x+6)} \cdot \frac{\cancel{(x-6)}(x+4)}{2 \cancel{(x+7)}(x-3)} = \frac{(x-1)(x+4)}{2(x-3)(x+6)}\end{align*}
3. \begin{align*}\frac{4x^2y^7}{32x^4y^3} \cdot \frac{16x^2}{8y^6} = \frac{64x^4y^7}{256x^4y^9} = \frac{1}{4y^2}\end{align*}
### Explore More
Determine if the following statements are true or false. If false, explain why.
1. When multiplying two variables with the same base, you multiply the exponents.
2. When dividing two variables with the same base, you subtract the exponents.
3. When a power is raised to a power, you multiply the exponents.
4. \begin{align*}(x+2)^2 = x^2 + 4\end{align*}
1. \begin{align*}\frac{8x^2y^3}{5x^3y} \cdot \frac{15xy^8}{2x^3y^5}\end{align*}
2. \begin{align*}\frac{11x^3y^9}{2x^4} \cdot \frac{6x^7y^2}{33xy^3}\end{align*}
3. \begin{align*}\frac{18x^3y^6}{13x^8y^2} \cdot \frac{39x^{12}y^5}{9x^2y^9}\end{align*}
4. \begin{align*}\frac{3x+3}{y-3} \cdot \frac{y^2-y-6}{2x+2}\end{align*}
5. \begin{align*}\frac{6}{2x+3} \cdot \frac{4x^2+4x-3}{3x+3}\end{align*}
6. \begin{align*}\frac{6+x}{2x-1} \cdot \frac{x^2+5x-3}{x^2+5x-6}\end{align*}
7. \begin{align*}\frac{3x-21}{x-3} \cdot \frac{-x^2+x+6}{x^2-5x-14}\end{align*}
8. \begin{align*}\frac{6x^2+5x+1}{8x^2-2x-3} \cdot \frac{4x^2+28x-30}{6x^2-7x-3}\end{align*}
9. \begin{align*}\frac{x^2+9x-36}{x^2-9} \cdot \frac{x^2+8x+15}{-x^2+11x+12}\end{align*}
10. \begin{align*}\frac{2x^2+x-21}{x^2+2x-48} \cdot (4-x) \cdot \frac{2x^2-9x-18}{2x^2-x-28}\end{align*}
11. \begin{align*}\frac{8x^2-10x-3}{4x^3+x^2-36x-9} \cdot \frac{5x+3}{x-1} \cdot \frac{x^3+3x^2-x-3}{5x^2+8x+3}\end{align*}
### Vocabulary Language: English
Rational Expression
Rational Expression
A rational expression is a fraction with polynomials in the numerator and the denominator. |
## Engage NY Eureka Math 3rd Grade Module 1 Lesson 8 Answer Key
### Eureka Math Grade 3 Module 1 Lesson 8 Problem Set Answer Key
Question 1.
Draw an array that shows 5 rows of 3.
Explanation:
Drawn an array that shows 5 rows of 3 as 5 × 3 as shown above.
Question 2.
Draw an array that shows 3 rows of 5.
Explanation:
Drawn an array that shows 3 rows of 5 as 3 × 5 as shown above.
Question 3.
Write multiplication expressions for the arrays in Problems 1 and 2. Let the first factor in each expression represent the number of rows. Use the commutative property to make sure the equation below is true.
Explanation:
Wrote multiplication expressions for the arrays in Problems 1 and 2 as 5 × 3, 3 × 5 and let the first factor in each expression represent the number of rows.
Used the commutative property to make sure the equation below is true as 5 × 3 = 15, 3 × 5 = 15,
So, 5 × 3 = 3 × 5 is true equals to 15.
Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals. The first one is done for you.
a. 2 threes: 2 × 3 = 6
b. 3 twos: _________________
c. 3 fours: ________________
d. 4 threes: ________________
e. 3 sevens: ________________
f. 7 threes: ________________
g. 3 nines: _________________
h. 9 threes: ________________
i. 10 threes: _______________
b. 3 twos : 3 x 2 = 6,
Explanation:
Given 3 twos, 3 multiplied by 2 gives 6,
So, 3 × 2 = 6.
c. 3 fours: 3 × 4 = 12,
Explanation:
Given 3 fours, 3 multiplied by 4 gives 12,
So, 3 × 4 = 12.
d. 4 threes: 4 × 3 = 12,
Explanation:
Given 4 threes, 4 multiplied by 3 gives 12,
So, 4 × 3 = 12.
e. 3 sevens: 3 x 7 = 21,
Explanation:
Given 3 sevens, 3 multiplied by 7 gives 21,
So, 3 × 7 = 21.
f. 7 threes: 7 × 3 = 21,
Explanation:
Given 7 threes, 7 multiplied by 3 gives 21,
So, 7 × 3 = 21.
g. 3 nines: 3 × 9 = 27,
Explanation:
Given 3 nines, 3 multiplied by 9 gives 27,
So, 3 × 9 = 27.
h. 9 threes: 9 × 3 = 27,
Explanation:
Given 9 threes, 9 multiplied by 3 gives 27,
So, 9 × 3 = 27.
i. 10 threes: 10 x 3 =30,
Explanation:
Given 10 threes, 10 multiplied by 3 gives 30,
So, 10 × 3 = 30.
Question 5.
Find the unknowns that make the equations true.
Then, draw a line to match related facts.
a. 3 + 3 + 3 + 3 + 3 = _________
b. 3 × 9 = _________
c. 7 threes + 1 three = _________
d. 3 × 8 = _________
e. _________ = 5 × 3
f. 27 = 9 × _________
a. 3 + 3 + 3 + 3 + 3 = 15,
Explanation:
Given 3 + 3 + 3 + 3 + 3 adding 3, 5 times gives 15,
So 3 + 3 + 3 + 3 + 3 = 15.
b. 3 × 9 = 27,
Explanation:
Given 3 × 9, 3 multiplied by 9 gives 27,
So 3 × 9 = 27.
c. 7 threes + 1 three = 8 threes = 24,
Explanation:
Given 7 threes + 1 three = 7 × 3 + 1 × 3 = 21 + 3 = 24,
So, 7 threes + 1 three = 8 threes = 24.
d. 3 × 8 = 24,
Explanation:
Given 3 × 8, 3 multiplied by 8 gives 24,
So 3 × 8 = 24.
e. 15 = 5 × 3,
Explanation:
Given 5 × 3, 5 multiplied by 3 gives 15,
So 15 = 5 × 3.
f. 27 = 9 × 3
Explanation:
Given 27 = 9 × ___, Lets take missing number as x,
27 = 9 × x, So x = 27 ÷ 9 = 3, So 27 = 9 × 3.
Question 6.
Isaac picks 3 tangerines from his tree every day for 7 days.
a. Use circles to draw an array that represents the
tangerines Isaac picks.
Explanation:
Given Isaac picks 3 tangerines from his tree every day for 7 days.
a. Used circles to draw an array that represents the tangerines Isaac picks as 3 × 7 as shown above.
b. How many tangerines does Isaac pick in 7 days?
Write and solve a multiplication sentence to find the total.
Isaac picks 21 tangerines in 7 days,
Multiplication sentence to find the total is 3 x 7 = 21.
Explanation:
Wrote and solved a multiplication sentence to find the total as 3 x 7 = 21, therefore Isaac picks 21 tangerines in 7 days.
c. Isaac decides to pick 3 tangerines every day for 3 more days.
Draw x’s to show the new tangerines on the array in Part (a).
Explanation:
Given Isaac decides to pick 3 tangerines every day for 3 more days.
Drawn x’s to show the new tangerines on the array in Part (a) as shown above in the picture.
d. Write and solve a multiplication sentence to find the total number of tangerines Isaac picks.
Multiplication sentence is 3 x 10 = (3 × 7) + (3 × 3) = 21 + 9 = 30,
The total number of tangerines Isaac picks are 30,
Explanation:
Wrote and solved a multiplication sentence as
3 x 10 = (3 × 7) + (3 × 3) = 21 + 9 = 30,
Therefore the total number of tangerines Isaac picks are 30.
Question 7.
Sarah buys bottles of soap. Each bottle costs $2. a. How much money does Sarah spend if she buys 3 bottles of soap? _____$2_____ × ____3______ = $___6_____ b. How much money does Sarah spend if she buys 6 bottles of soap? _____$2_____ × _____6_____ = $__12______ a. Sarah spends$6 if she buys 3 bottles of soap,
Explanation:
Given Sarah buys bottles of soap. Each bottle costs $2, So money spent by Sarah if she buys 3 bottles of soap are$2 × 3 = $6. b. Sarah spends$12 if she buys 6 bottles of soap,
Explanation:
Given Sarah buys bottles of soap. Each bottle costs $2, So money spent by Sarah if she buys 6 bottles of soap are$2 × 6 = \$12.
### Eureka Math Grade 3 Module 1 Lesson 8 Exit Ticket Answer Key
Mary Beth organizes stickers on a page in her sticker book.
She arranges them in 3 rows and 4 columns.
a. Draw an array to show Mary Beth’s stickers.
b. Use your array to write a multiplication sentence to find Mary Beth’s total number of stickers.
c. Label your array to show how you skip-count to solve your multiplication sentence.
d. Use what you know about the commutative property to write a different multiplication sentence for your array.
a.
Explanation:
Given Mary Beth organizes stickers on a page in her sticker book. She arranges them in 3 rows and 4 columns.
Drawn an array as 3 × 4 to show Mary Beth’s stickers.
b. Mary Beth’s total number of stickers are 12,
Explanation:
Used my array to write a multiplication sentence as 3 × 4 = 12, in finding Mary Beth’s total number of stickers.
c.
Explanation:
Labeled my array to show skip-count to 12 and solved my multiplication sentence as 3 x 4 = 12.
d. Commutative property to write a different multiplication sentence for my array is 3 × 4 = 4 x 3,
Explanation:
Used what I know about the commutative property to write a different multiplication sentence for my array as
3 × 4 = 4 x 3.
### Eureka Math Grade 3 Module 1 Lesson 8 Homework Answer Key
Question 1.
Draw an array that shows 6 rows of 3.
Explanation:
Drawn an array that shows 6 rows of 3 as 6 × 3 as shown above.
Question 2.
Draw an array that shows 3 rows of 6.
Explanation:
Drawn an array that shows 3 rows of 6 as 3 × 6 as shown above.
Question 3.
Write multiplication expressions for the arrays in Problems 1 and 2. Let the first factor in each expression represent the number of rows. Use the commutative property to make sure the equation below is true.
Explanation:
Wrote multiplication expressions for the arrays in Problems 1 and 2 as 6 × 3, 3 × 6 and let the first factor in each expression represent the number of rows.
Used the commutative property to make sure the equation below is true as 6 × 3 = 18, 3 × 6 = 18,
So, 6 × 3 = 3 × 6 is true equals to 18.
Question 4.
Write a multiplication sentence for each expression.
You might skip-count to find the totals. The first one is done for you.
a. 5 threes: 5 × 3 = 15
b. 3 fives: __________________
c. 6 threes: ________________
d. 3 sixes: ___________________
e. 7 threes: __________________
f. 3 sevens: __________________
g. 8 threes: ________________
h. 3 nines: _________________
i. 10 threes: _______________
b. 3 fives: 3 × 5 = 15,
Explanation:
Given 3 fives, 3 multiplied by 5 gives 15,
So, 3 × 5 = 15.
c. 6 threes: 6 × 3 = 18,
Explanation:
Given 6 threes, 6 multiplied by 3 gives 18,
So, 6 × 3 = 18.
d. 3 sixes: = 3 × 6 = 18,
Explanation:
Given 3 sixes, 3 multiplied by 6 gives 18,
So, 3 × 6 = 18.
e. 7 threes: 7 × 3 = 21,
Explanation:
Given 7 threes, 7 multiplied by 3 gives 21,
So, 7 × 3 = 21.
f. 3 sevens: 3 × 7 = 21,
Explanation:
Given 3 sevens, 3 multiplied by 7 gives 21,
So, 3 × 7 = 21.
g. 8 threes: 8 × 3 = 24,
Explanation:
Given 8 threes, 8 multiplied by 3 gives 24,
So, 8 × 3 = 24.
h. 3 nines: 3 × 9 = 27,
Explanation:
Given 3 nines, 3 multiplied by 9 gives 27,
So, 3 × 9 = 27.
i. 10 threes: 10 × 3 = 30,
Explanation:
Given 10 threes, 10 multiplied by 3 gives 30,
So, 10 X 3 = 30.
Question 5.
Find the unknowns that make the equations true. Then, draw a line to match related facts.
a. 3 + 3 + 3 + 3 + 3 + 3 = _________
b. 3 × 5 = _________
c. 8 threes + 1 three = _________
d. 3 × 9 = _________
e. _________ = 6 × 3
f. 15 = 5 × _________
a. 3 + 3 + 3 + 3 + 3 + 3 = 18,
Explanation:
Given 3 + 3 + 3 + 3 + 3 + 3 adding 3, 6 times gives 18,
So 3 + 3 + 3 + 3 + 3 + 3 = 18.
b. 3 × 5 = 15,
Explanation:
Given 3 × 5, 3 multiplied by 5 gives 15,
So 3 × 5 = 15.
c. 8 threes + 1 three =9 threes = 27,
Explanation:
Given 8 threes + 1 three = 8 × 3 + 1 × 3 = 24 + 3 = 27,
So, 8 threes + 1 three = 9 threes = 27.
d. 3 × 9 = 27,
Explanation:
Given 3 × 9, 3 multiplied by 9 gives 27,
So 3 × 9 = 27.
e. 18 = 6 × 3,
Explanation:
Given 6 × 3, 6 multiplied by 3 gives 18,
So 18 = 6 × 3.
f. 15 = 5 × 3,
Explanation:
Given 15 = 5 × ___, Lets take missing number as x,
15 = 5 × x, So x = 15 ÷ 5 = 3, So 15 = 5 × 3.
Question 6.
Fernando puts 3 pictures on each page of his photo album.
He puts pictures on 8 pages.
a. Use circles to draw an array that represents the total number of pictures in Fernando’s photo album.
b. Use your array to write and solve a multiplication sentence to find Fernando’s total number of pictures.
c. Fernando adds 2 more pages to his book. He puts 3 pictures on each new page. Draw x’s to show the new pictures on the array in Part (a).
d. Write and solve a multiplication sentence to find the new the total number of pictures in Fernando’s album.
a.
Explanation:
Given Fernando puts 3 pictures on each page of his photo album.
He puts pictures on 8 pages.
a. Used circles to draw an array that represents the total number of pictures in Fernando’s photo album as 3 X 8.
b. Multiplication sentence for Fernando’s total number of pictures is 3 X 8 = 24,
Explanation:
Used my array to write and solve a multiplication sentence to find Fernando’s total number of pictures as 3 X 8 = 24.
c.
Explanation:
Fernando adds 2 more pages to his book. He puts 3 pictures on each new page. Drawn x’s to show the new pictures on the array in Part (a) as shown above.
d. Multiplication sentence is 3 x 10 =
(3× 8) + (3 × 2) = 24 + 6 = 30,
The new total number of pictures in Fernando’s album are 30,
Explanation:
Wrote and solved a multiplication sentence as
3 x 10 = (3 × 8) + (3 × 2) = 24 + 6 = 30,
Therefore the new total number of pictures in Fernando’s album are 30.
Question 7.
Ivania recycles. She gets 3 cents for every can she recycles.
a. How much money does Ivania make if she recycles 4 cans?
____3______ × ___4_______ = ___12_____ cents
b. How much money does Ivania make if she recycles 7 cans?
____3______ × ____7______ = ____21____ cents
a. Ivania makes 12 cents if she recycles 4 cans,
Explanation:
Given Ivania recycles and she gets 3 cents for every can she recycles,
So money Ivania makes if she recycles 4 cans is
3 cents × 4 = 12 cents.
b. Ivania makes 21 cents if she recycles 7 cans,
Explanation:
Given Ivania recycles and she gets 3 cents for every can she recycles,
So money Ivania makes if she recycles 7 cans is
3 cents × 7 = 21 cents. |
# KSEEB Solutions for Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3
In this chapter, we provide KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3 for English medium students, Which will very helpful for every student in their exams. Students can download the latest KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3 pdf, free KSEEB SSLC Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3 pdf download. Now you will get step by step solution to each question.
## Karnataka Board Class 8 Maths Chapter 6 Theorems on Triangles Ex 6.3
Question 1.
The exterior angles obtained on producing the base of a triangle both ways are 104° and 136°. Find the angles of the triangle.
In triangle ABC, BC is produced on either sides.
Let ∠ABC = 140° and ∠ACE = 136°
∠ABD +∠ABC = 180° [ Linear pair]
104 + ∠ABC = 180°
∠ABC = 180 – 104
∠ ABC = 76°
∠ACB + ∠ACE = 180° [Linearpair]
∠ACB + 136° = 180°
∠ACB = 180 – 136 ∠ACB = 44°
∠ABD + ∠ACB + ∠BAC = 180°
[Sum of the angles of a triangle 180° ]
76 + 44 + ∠BAC = 180°
120 + ∠BAC = 180°
∠BAC = 180 – 120 ∠BAC = 60°
Three angles are 76°, 44° & 60°
Question 2.
Sides BC, CA and AB of a triangle ABC are produced in order, forming exterior angles ∠ACD, ∠BAE, and∠CBF.
Show that ∠ACD + ∠BAE + ∠CBF = 360°.
ABC + ∠CBF = 180° [Linear pair]………..(i)
∠ACB + ∠ACD = 180° [Linear pair]……(ii)
∠BAC + ∠BAE = 180° [Linear pair]…….(iii)
Byadding(i), (ii) and (iii)
∠ABC + ∠CBF + ∠ACB + ∠ACD+
∠BAC + ∠BAE = 180 + 180 + 180
∠ABC + ∠ACB + ∠BAC +
∠CBF + ∠ACD + ∠BAE = 540°
180 + ∠CBF + ∠ACD + ∠BAE = 540°
[Sum of the angles of a triangle 180°]
∠CBF + ∠ACD + ∠BAE = 360°
Question 3
Compute the value of x in each of the following figures
i.
AB = AC
∴ ∠ABC = ∠ACB = 50°
∠ACB + ∠ACD = 180° [Linear pair]
50 + x = 180°
x = 180 – 50 = 130°
ii.
∠ABC + ∠CBF = 180° [Linear pair].
106°+∠ABC = 180°
∠ABC = 180 – 106 = 74°
∠EAC = ∠ABC + ∠ACB
130° =74 +x
130 – 74 = x
∴x = 56°
iii.
∠QPR = ∠TPU = 65°
[Vertically opposite angles]
∠PRS = ∠PQR + ∠QPR [Exterior angle=Sum of interior opposire angle]
100 = 65 + x
100 – 65 = x
35 = x
∴ x = 35°.
iv.
∠BAE + ∠BAC = 180° [Linear pair]
120° + ∠BAC = 180°
∠BAC = 180 – 120
∠BAC = 60°
∠ACD = ∠BAC + ∠ABC
[Exterior angle = Sum of interior opposuite angle]
112° = 60+x
112 – 60 = x
52 = x
x = 52°
v.
In Δ ABC, BA = BC (data)
∴ ∠BAC = ∠BCA = 20°
[Base angles of an isosceles traingle]
∠ABD = ∠BAC + ∠BCA
[Exterior angle=Sum of interior opposite angle]
x = 20 + 20
x=40°
Question 4.
In figure QT ⊥ PR, ∠TOR = 40° and ∠SPR = 30° find ∠TRS and ∠PSQ.
In ΔTQR,
∠TQR + ∠QTR + ∠TRQ = 180°
40 + 90 + ∠TRQ = 180°
130 + ∠TRQ = 180
∠ TRQ = 180 – 130
∠ TRQ = 50°
∠ TRS = 50° [∠PRS is same as ∠TRS ]
In Δ PRS, RS is produced to Q
∴Exterior∠PSQ = ∠SPR + ∠PRS = 30 + 50
∠PSQ = 80°
Question 5.
An exterior angle of a triangle is 120° and one of the interior opposite angle is 30 °. Find the other angles of the triangle.
In Δ ABC, BC is produced to D. Let
∠ACD = 120° and ∠ABC = 30°
Exterior ∠ACD = ∠BAC + ∠ABC
120 = ∠BAC + 30
120 – 30 = ∠BAC
90° = ∠BAC
∴∠BAC = 90
∠ACB +∠ACD = 180° [Linear pair]
∠ACB + 120 = 180°
∠ACB = 180 – 120
∠ACB = 60°
Other two angles are 90° & 60°
All Chapter KSEEB Solutions For Class 8 maths
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Lesson 2: Similarity and congruence
By: Tim Lehman
The lesson 2 is based on an exercise from an Addison-Wesley 1996 workbook, and one of the most famous cats in history......
For this lesson, students learn about similarity by constructing figures that are and are not similar. Even though students might not have been exposed to similarity before, most already know what it means. They realize figures that are similar have the same shape and congruent figures are "twins." The goal behind this lesson is students will get a better understanding of the mathematical definition of these terms. Three of the figures, Morris I, Morris II, and Boris, will be discussed here. Two others, Morris III and Doris, can be added (being (3x, 3y) and (x,3y), respectively). The figures can be easily plotted on GSP (as well as graph paper.) First, we will graph Morris I. Each figure has points broken up into 5 sets. The points in each set are to be connected. On GSP, simply add all the points of a set, press plot, and then construct segment. Repeat the process for each set. After graphing Morris II and Boris, compare and contrast them with Morris I.
In the chart below, the points are given for Morris I. Connect the points after plotting them, by constructing a line from each one to the next point on the list and connecting the last point of each set to the first one. The points of Morris I are considered (x,y). To find Morris II, multiply each x and y-coordinate of Morris I by 2. For Boris, multiply the x-coordinate of Morris I by 3 while the y-coordinate remains the same.The first couple have been done for you.
Morris I Morris II Boris (x,y) (2x,2y) (3x,y) SET I (5,0) (10,0) (15,0) (7,2) (14,4) (21,2) (7,7) (6,10) (5,7) (2,7) (1,10) (0,7) (0,2) (2,0) SET II (1,3) (2,2) (5,2) (6,3) SET III (3,3) (4,3) (4,5) (3,5) SET IV (5,6) SET V (2,6)
Click on Morris I, Morris II, and Boris for GSP sketches of each.
Now, the students should compare their three drawings. Which one is similar to Morris I? Which one is not? How do the angles compare for similar shapes? What about sides? The angle measure is most evident to me at the point of the ears of the three figures. The nose of each drawing is a good place to compare side lengths and also shows equal angles do not mean shapes are similar.
Morris says that the only thing that can pull him away from the fish tank is math class!!!
All images of Morris are from the 9 Lives cat food website. To visit the site, click on Morris.
The Addison-Wesley website can be found here.
Return |
WHOLES AND PARTS: THE BIRTH OF FRACTIONS
Feb 18, 2023 | Red Deer
This blog is a part of our series: Wholes and Parts; and as promised in our blog WHOLES AND PARTS: THE CONCEPT THAT YOUR CHILD NEEDS TO HAVE FOR SUCCESS, we will now explain about The Birth of Fractions, because “fractions” is a natural extension of seeing wholes and parts.
The Birth of Fractions
When a whole is divided (fractured) into a number of equal parts, we have what are called fractions. When a person encounters the word “fraction,” the images “part of a whole” and “equal parts” must come to mind. The notion that a fraction results from the equal division of a whole must be second nature and deeply embedded in the child’s consciousness, and the earlier in life the better.
Mathnasium has its own unique method to teach a child, age four and up, the basic concepts of fractions.
How Mathnasium Teach “Two Parts the Same”
We start by teaching children about the fraction one-half (1/2). But before introducing the concept of halving, they need to understand the concept of doubling and master it. All the basic principles that apply to all fractions are embodied in the study of half. Once the idea of half is learned thoroughly, its attributes can easily be applied to all the other fractions.
The process of learning about half immediately exposes students to many facets of the mathematics curriculum that other programs do not present until much later, after both common fractions and decimal fractions have been mastered.
In fact, children can begin problem solving years earlier than is currently taught in most school systems. Kids can handle it, if the problems are appropriate.
Introducing the concept of “two parts the same”
Children often call any division into two pieces “half,” whether the pieces are the same (equal) or not. If the cut is not equal, we correct the child and say, “you have two pieces, two parts, but you do not have half. The two parts must be the same to have halves.” We make sure that each student has a clear understanding that half means (two) equal pieces.
Then when a student understands the basic concept of “two parts the same” they can move to the next stage: halving using basic numbers. Using visualization of balancing like this would really help them – both for younger kids with basic numbers and older kids with more complex numbers:
When the child knows and can communicate what half means, the child is ready to apply this knowledge to the solution of a wide range of problems. We ask the halving exercise verbally, so that they perform calculation mentally, to the maximum extent possible. Why mental math is essential? Please read YOUR CHILD CANNOT DO MENTAL MATH? BETTER START NOW for the explanation.
Yes, you’d better believe that kids can handle it, don’t underestimate their potential. This five-years-old kindergartener is on the right path for her long-term math success by starting early. Having a strong math understanding since young – not only it means avoiding problems before they start – but more importantly it will make a child’s math learning a bliss and enjoyable.
When she is ready, we will help her with halving even numbers up to 24, and halving the odd numbers. Patiently introducing kids to the Wholes & Parts concept, and especially the concept of halving and doubling with the right method, will make students understand a very important mathematical image: All wholes can be broken-down into parts.
There is no need at this point to attach the name “fraction” to this idea. The vocabulary will come later, at the right time. Next time, we will explain the Reverse Question as a crucial factor to build understanding of the Wholes & Parts concept.
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# v w is orthogonal to both v and w. the three vectors v, w and v w form a right-handed set of vectors.
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Transcription
1 3. Cross product Definition 3.1. Let v and w be two vectors in R 3. The cross product of v and w, denoted v w, is the vector defined as follows: the length of v w is the area of the parallelogram with sides v and w, that is, v w sin θ. v w is orthogonal to both v and w. the three vectors v, w and v w form a right-handed set of vectors. Remark 3.2. The cross product only makes sense in R 3. Example 3.3. We have By contrast î ĵ = ˆk, ĵ ˆk = î and ˆk î = ĵ. ĵ î = ˆk, ˆk ĵ = î and î ˆk = ĵ. Theorem 3.4. Let u, v and w be three vectors in R 3 and let λ be a scalar. (1) v w = w v. (2) u ( v + w) = u v + u w. (3) ( u + v) w = u w + v w. (4) λ( v w) = (λ v) w = v (λ w). Before we prove (3.4), let s draw some conclusions from these properties. Remark 3.5. Note that (1) of (3.4) is what really distinguishes the cross product (the cross product is skew commutative). Consider computing the cross product of î, î and ĵ. On the one hand, On the other hand, (î î) ĵ = 0 j = 0. î (î ĵ) = î ˆk = ĵ. In other words, the order in which we compute the cross product is important (the cross product is not associative). Note that if v and w are parallel, then the cross product is the zero vector. One can see this directly from the formula; the area of the parallelogram is zero and the only vector of zero length is the zero 1
2 vector. On the other hand, we know that w = λ v. In this case, v w = v (λ v) = λ v v = λ v v. To get from the second to the third line, we just switched the factors. But the only vector which is equal to its inverse is the zero vector. Let s try to compute the cross product using (3.4). If v = (v 1, v 2, v 3 ) and w = (w 1, w 2, w 3 ), then v w = (v 1 î + v 2 ĵ + v 3ˆk) (w1 î + w 2 ĵ + w 3ˆk) = v 1 w 1 (î î) + v 1 w 2 (î ĵ) + v 1 w 3 (î ˆk) + v 2 w 1 (ĵ î) + v 2 w 2 (ĵ ĵ) + v 2 w 3 (ĵ ˆk) + v 3 w 1 (ˆk î) + v 3 w 2 (ˆk ĵ) + v 3 w 3 (ˆk ˆk) = (v 2 w 3 v 3 w 2 )î + (v 3 w 1 v 1 w 3 )ĵ + (v 1 w 2 v 2 w 1 )ˆk. Definition 3.6. A matrix A = (a ij ) is a rectangular array of numbers, where a ij is in the i row and jth column. If A has m rows and n columns, then we say that A is a m n matrix. Example 3.7. A = is a 2 3 matrix. a 23 = 4. ( ) ( ) a11 a = 12 a 13, a 21 a 22 a 23 Definition 3.8. If ( ) a b A = c d is a 2 2 matrix, then the determinant of A is the scalar a b c d = ad bc. If A = a b c d e f g h i is a 3 3 matrix, then the determinant of A is the scalar a b c d e f g h i = a e f h i b d f g i + c d e g h 2
3 Note that the cross product of v and w is the (formal) determinant î ĵ ˆk Let s now turn to the proof of (3.4). Definition 3.9. Let u, v and w be three vectors in R 3. The triple scalar product is ( u v) w. The triple scalar product is the signed volume of the parallelepiped formed using the three vectors, u, v and w. Indeed, the volume of the parallelepiped is the area of the base times the height. For the base, we take the parallelogram with sides u and v. The magnitude of u v is the area of this parallelogram. The height of the parallelepiped, up to sign, is the length of w times the cosine of the angle, let s call this φ, between u v and w. The sign is positive, if u, v and w form a right-handed set and negative if they form a left-handed set. Lemma If u = (u 1, u 2, u 3 ), v = (v 1, v 2, v 3 ) and w = (w 1, w 2, w 3 ) are three vectors in R 3, then Proof. We have already seen that î ĵ ˆk u v = If one expands this determinant and dots with w, this is the same as replacing the top row by (w 1, w 2, w 3 ), w 1 w 2 w 3 Finally, if we switch the first row and the second row, and then the second row and the third row, the sign changes twice (which makes no change at all): 3
4 Example The scalar triple product of î, ĵ and ˆk is one. One way to see this is geometrically; the parallelepiped determined by these three vectors is the unit cube, which has volume 1, and these vectors form a right-handed set, so that the sign is positive. Another way to see this is to compute directly (î ĵ) ˆk = ˆk ˆk = 1. Finally one can use determinants, (î ĵ) ˆk = = 1. Lemma Let u, v and w be three vectors in R 3. Then ( v w) u = ( w u) v. Proof. In fact all three numbers have the same absolute value, namely the volume of the parallelepiped with sides u, v and w. On the other hand, if u, v, and w is a right-handed set, then so is v, w and u and vice-versa, so all three numbers have the same sign as well. Lemma Let v and w be two vectors in R 3. Then v = w if and only if v x = w x, for every vector x in R 3. Proof. One direction is clear; if v = w, then v x = w x for any vector x. So, suppose that we know that v x = w x, for every vector x. Suppose that v = (v 1, v 2, v 3 ) and w = (w 1, w 2, w 3 ). If we take x = î, then we see that v 1 = v î = w î = w 1. Similarly, if we take x = ĵ and x = ˆk, then we also get and v 2 = v ĵ = w ĵ = w 2, v 3 = v ˆk = w ˆk = w 3. But then v = w as they have the same components. Proof of (3.4). We first prove (1). Both sides have the same magnitude, namely the area of the parallelogram with sides v and w. Further both sides are orthogonal to v and w, so the only thing to check is the change in sign. 4
5 As v, w and v w form a right-handed triple, it follows that w, v and v w form a left-handed triple. But then w, v and v w form a right-handed triple. It follows that This is (1). To check (2), we check that w v = v w. ( u ( v + w)) x = ( u v + u w) x, for an arbitrary vector x. We first attack the LHS. By (3.12), we have We now attack the RHS. ( u ( v + w)) x = ( x u) ( v + w) = ( x u) v + ( x u) w = ( u v) x + ( u w) x. ( u v + u w) x = ( u v) x + ( u v) x. It follows that both sides are equal. This is (2). We could check (3) by a similar argument. Here is another way. ( u + v) w = w ( u + v) = w u w v = u w + v w. This is (3). To prove (4), it suffices to prove the first equality, since the fact that the first term is equal to the third term follows by a similar derivation. If λ = 0, then both sides are the zero vector, and there is nothing to prove. So we may assume that λ 0. Note first that the magnitude of both sides is the area of the parallelogram with sides λ v and w. If λ > 0, then v and λ v point in the same direction. Similarly v w and λ( v w) point in the same direction. As v, w and v w for a right-handed set, then so do λ v, w and λ( v w). But then λ( v w) is the cross product of λ v and w, that is, (λ v) w = λ( v w). If λ < 0, then v and λ v point in the opposite direction. Similarly v w and λ( v w) point in the opposite direction. But then λ v, w and λ( v w) still form a right-handed set. This is (4). 5
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# (i) The sides of a triangle are in the ratio 7: 5: 3 and its perimeter is 30 cm. Find the lengths of sides. (ii) If the angles of a triangle are in the ratio 2: 3: 4, find the angles.
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This is an important ques from the Book ML Aggarwal class 10th,, chapter – 7, ratio and proportion.
In these both questions we have to. Distribute the total sum according to given fixed ratio.
In first question we have to distribute perimeter of triangle in the given ratio and in second ques we have to distribute total angle sum in given ratio
Question 16
Share
1. Solution:
(i) It is given that
Perimeter of triangle = 30 cm
Ratio among sides = 7: 5: 3
Here the sum of ratios = 7 + 5 + 3 = 15
We know that
Length of first side = 30 × 7/15 = 14 cm
Length of second side = 30 × 5/15 = 10 cm
Length of third side = 30 × 3/15 = 6 cm
Therefore, the sides are 14 cm, 10 cm and 6 cm.
(ii) We know that
Sum of all the angles of a triangle = 1800
Here the ratio among angles = 2: 3: 4
Sum of ratios = 2 + 3 + 4 = 9
So we get
First angle = 180 × 2/9 = 400
Second angle = 180 × 3/9 = 600
Third angle = 180 × 4/9 = 800
Hence, the angles are 400, 600 and 800.
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# How To Find Perpendicular Line Gradient
Find the equation of the line. So the perpendicular line will have a slope of 1/4:
Equations of Horizontal and Vertical Lines in Slope
### In plane geometry, all lines have slopes.
How to find perpendicular line gradient. An informal definition of the gradient (also known as the slope) is as follows: How to find perpendicular slope. Graph the two equations and measure one of the angles that forms;
Where m is the slope and b is the y intercept. Plugging in the point given into the equation y = 1/2 x + b and solving for b , we get b = 6. If you need to find a line given two points or a slope and one point, use line calculator.
Y − 2 = (1/4) (x − 7) and that answer is ok, but let’s also put it in y=mx+b form: A perpendicular line will intersect it, but it won’t just be any intersection, it will intersect at right angles. Write the equation of a straight line if parallel to a line and passing through (0,n) write the equation of a straight line if parallel to a line and passing through any point;
If the lines are horizontal and vertical, then they are perpendicular due to the squares of the coordinate grid. Linear lines are almost always displayed in the form of. The first step in finding the equation of a line perpendicular to another is understanding the relationship of their slopes.
Find out how to calculate it in this bitesize maths video for ks3. It can be found using the formula: In the case below, it rose 2 while only going across 1, which means this line has a slope (gradient) of 2.
So these two lines are perpendicular. Gradient = change in ychange in x : M = −1 −4 = 1 4.
The negative reciprocal of that slope is: We will also define the normal line and discuss how the gradient vector can be used to find the equation of the normal line. According to the definition of a perpendicular line, all four angles have to measure 90 degrees.
The equation for finding the slope of a line with two points is. A gradient of a line is also called a slope of a line. Gradient (slope) of a straight line.
Any line with gradient 1/5 will be perpendicular to our line, for example, y = (1/5)x. It basically means how steep is the line. Once again, the two key pieces of information are the gradient of the line and a point through which the line passes.
To calculate the gradient of a straight line we choose two points on the line itself. So, let us take an equation: Divide the change in height by the change in horizontal distance.
In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. First find the slope of the given line by comparing the slope intercept form of a line as follows: Y − y1 = (1/4) (x − x1) and now put in the point (7,2):
Have a play (drag the points): This is the negative reciprocal of the gradient. For drawing lines, use the graphing calculator.
The reciprocal of 2 is $$\frac{1}{2}$$ , so the negative. Y − 2 = x/4 − 7/4. The gradient (also called slope) of a straight line shows how steep a straight line is.
Work out gradient of line perpendicular to a given line Find the equation of a straight line through two given points; Interpret gradient and intercept on real life graphs;
Thus, the equation of the line is y = ½ x + 6. By using this website, you agree to our cookie policy. The calculator will find the equation of the parallel/perpendicular line to the given line, passing through the given point, with steps shown.
Now we follow the following procedure with the help of an example. The gradient of the perpendicular is the reciprocal of the original gradient! To the line passing through the point (, ) enter the equation of a line in any form:
In this section discuss how the gradient vector can be used to find tangent planes to a much more general function than in the previous section. $\begingroup$ okay thanks so much, all i needed was some clarification, i’ve done all this math before, but our teacher walks us through it, and i do better when i understand what i’m doing, and why, and you guys are doing a great job. If you’re going uphill, you must struggle to reach the peak, so the energy needed (i.e., the.
The slope of y=−4x+10 is: The gradient is how steep an incline is. Y = mx + b.
It is a mathematical way of measuring how fast a line rises or falls.think of it as a number you assign to a hill, a road, a path, etc., that tells you how much effort you have to put to cycle it. Before you calculate the equation of the perpendicular line, you will need to find the slope of the line that crosses the two points.
Slope of Parallel and Perpendicular Lines An Exploration
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PARALLEL AND PERPENDICULAR EQUATIONS NOTES PRACTICE |
# KSEEB Solutions for Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1
Students can Download Chapter 2 Fractions and Decimals Ex 2.1, Question and Answers, Notes Pdf, KSEEB Solutions for Class 7 Maths, Karnataka State Board Solutions help you to revise complete Syllabus and score more marks in your examinations.
## Karnataka State Syllabus Class 7 Maths Chapter 2 Fractions and Decimals Ex 2.1
Multiplying and Dividing Rational Expressions Calculator.
Question 1.
Solve:
i)
Solution:
ii)
Solution:
iii)
Solution:
L.C.M = 5 × 7 × 1 × 1 = 35
iv)
Solution:
Online Subtracting Fractions Calculator subtracts the fractions 9/11 and 4/5 i.e. 1/55
The given fractions are 9/11 and 4/5
Firstly the L.C.M should be done for the denominators of the two fractions 9/11 and 4/5
9/114/5
The LCM of 11 and 5 (denominators of the fractions) is 55
Given numbers has no common factors except 1. So, there LCM is their product i.e 55
= 9 x 5 – 4 x 11/55
= 45 – 44/55
= 1/55
Result: 1/55
v)
Solution:
vi)
⇒ convert mixed fractions to improper fraction
vii)
convert mixed fractions to improper fraction
Using a fraction and whole number calculator in situations where you struggle to solve your arithmetic assignments can be useful.
Question 2.
Arrange the following in descending order:
i)
= We need to arrange these in descending order,
To find which number is greater or smaller, we make their denominators equal.
ii)
⇒ We make their denominators equal, to find the descending order.
Free quadratic equation completing the square calculator – Solve quadratic equations using completing the square step-by-step.
Question 3.
In a “magic square”, the sum of the numbers in each row, in each column and along the diagonals is the same. Is this a magic square ?
Solution:
For Row,
For diagonals,
Since Sum of air rows, columns and diagonals are equal.
Question 4.
A rectangular sheet of paper is 12$$\frac{1}{2}$$ cm long and 10$$\frac{2}{3}$$ cm wide. Find its perimeter.
Solution:
Length of rectangular sheet of paper = 12$$\frac{1}{2}$$ cm
(breadth) width of rectangular sheet paper = 10$$\frac{2}{3}$$ cm
Perimeter of rectangle = 2 (length + breadth)
converting the above fraction to mixed fraction,
2 2/3 as an improper fraction in its simplest form.
Question 5.
Find the perimeters of
(i) ∆ ABE
(ii) the rectangle BCDE in this figure. Whose perimeter is greater ?
Solution:
(i) ∆ ABE
perimeter of ∆ ABE
perimeter = AB + AE + BE
ii) The rectangle BCDE in this figure
Perimeter of rectangle BCDE,
As it is a rectangle, opposite sides are equal
BC = DE CD = BE
perimeter of rectangle = 2(l + b)
Also,
We have to find which perimeter is greater
To find which fraction is greater, we make its denominator equal.
Perimeter of ∆ ABE > Perimeter of BCDE
(Thus, Perimeter of ∆ ABE is greater)
Question 6.
Salil wants to put a picture in a frame. The picture is 7$$\frac{3}{10}$$ cm wide. How much should the picture be trimmed ?
Solution:
There are two things here – picture, and frame
so, picture has to be trimmed
$$=\frac{3}{10}$$
∴ Picture has to trimmed by = $$=\frac{3}{10}$$ cm
Question 7.
Ritu ate $$\frac{3}{5}$$ part of an apple and the remaining apple was eaten by her brother Somu. How much part of the apple did Somu eat ? Who had the larger share ? By how much ?
Solution:
Total part = 1
Part eaten by Ritu = $$\frac{3}{5}$$
Part eaten by Somu = Total part – part eaten by Ritu
Now,
We have to tell who ate the larger share so, we have to compare,
∴ Ritu ate the larger share.
Ritu ate large share by $$\frac{1}{5}$$
Question 8.
Michael finished colouring in $$\frac{7}{12}$$ hour. Vaibhav finished colouring the same picture in $$\frac{3}{4}$$ hour. Who worked longer ? By what fraction was it longer ?
Solution:
Michael finished work in = $$\frac{7}{12}$$ hour
Vaibhav finished work in = $$\frac{3}{4}$$
We need to find who worked longer.
i.e., we need to find greater of = $$\frac{7}{12}$$ & $$\frac{3}{4}$$
We make their denominators equal
∴ Vaibhav worked longer.
We also need to find by how much
Vaibhav worked longer by $$\frac{1}{6}$$ hours.
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# Pendulum
## Calculations of objects on strings exhibiting simple harmonic motion.
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Pendulum
In harmonic motion there is always a restorative force, which acts in the opposite direction of the velocity. The restorative force changes during oscillation and depends on the position of the object. In a pendulum it is the component of gravity along the path of motion. The force on the oscillating object is directly opposite that of the direction of velocity. For pendulums, the period gets larger as the length of the pendulum increases.
Key Equations
; Period of a pendulum of length L
#### Example
You have a mass swinging on the end of 1 m pendulum. If the maximum linear velocity of the mass is 2 m/s, (a) calculate the period of the pendulum and (b) calculate the amplitude of the pendulum.
To calculate the period of the pendulum, we can just plug in the given length into the equation above.
To find the amplitude, we'll use the equation given in the Period and Frequency lesson that gives us the velocity as a function of time. Since the problem says that the given velocity is the maximum velocity, we know that the pendulum is at the bottom of it's arc and 1/4th (or 3/4th's) of it's way through one period. Based on this knowledge, we can plug in 1/4 of the period for the change in time. We also know the frequency because we just found the period, so all we have to do is solve for the amplitude.
### Review
1. Why doesn’t the period of a pendulum depend on the mass of the pendulum weight? Shouldn’t a heavier weight feel a stronger force of gravity?
2. The pendulum of a small clock is long. How many times does it go back and forth before the second hand goes forward one second?
3. On the moon, how long must a pendulum be if the period of one cycle is one second? The acceleration of gravity on the moon is one sixth that of Earth.
1. A heavier weight feels a stronger force of gravity, but is also harder to accelerate.
2. times
### Explore More
We have explored two examples of simple harmonic motion: the pendulum and the mass-spring system in the previous lesson. The purpose of this investigation is to get you to notice the connections between the two systems. Your task: Match the period of the pendulum system with that of the spring system. You are only allowed to change the mass involved in the spring system. Consider the effective length of the pendulum to be fixed at 2m because that is the distance between the center of mass and the pivot. The spring constant(13.5N/m) is also fixed. You may use any relationships you have learned about to help you. You should view the charts to check whether you have succeeded.
Instructions: To alter the mass, simply click on the 'select' tool in the menu, and select the mass. Then use the tab at the bottom of your screen to change the density or dimensions of the block to get the mass that you want.
To view chart legend, click on Settings and you can plot velocity or position of the pendulum or mass on the spring.
The mass and the spring constant have now been changed. What is the new period of the mass-spring system? Can you change the length of the pendulum to match the periods now?
### Notes/Highlights Having trouble? Report an issue.
Color Highlighted Text Notes |
# Understanding Elementary Statistics: An Excellent Beginner’s Guide
Updated on: Educator Review By: Michelle Connolly
## Introduction to Elementary Statistics
### A. What is ElementaryStatistics?
Elementary Statistics is the science of collecting, organising, analysing, and interpreting data. It helps describe, understand, and make inferences about real-world phenomena.
### B. Importance of ElementaryStatistics in Everyday Life
Statistics is used in various aspects of everyday life. It helps businesses make data-driven decisions, governments formulate policies, and scientists analyse results. Basic statistical knowledge allows us to be smart consumers of information.
For example, businesses may use statistics to analyse customer data and make decisions about product development and marketing. Governments can conduct surveys and employ statistical methods to inform public policy. Scientists apply statistics to determine if research results are significant. As individuals, we encounter statistics in news reports, advertisements, and research studies. Knowing some fundamental statistical concepts allows us to evaluate the information critically.
### C. Types of Elementary Statistical Analysis
The two main branches of statistics are descriptive statistics and inferential statistics.
Descriptive statistics summarise and present data while inferential statistics make predictions and inferences about populations using sample data.
## Descriptive Elementary Statistics
### A. Definition and Purpose
Descriptive statistics describe and summarise features of collected data. It provides simple summaries of the measures and distributions of the sample data.
For example, descriptive statistics can calculate measures like the mean, median, and mode to indicate the central tendency or typical values of a dataset. They can determine the range, variance, or standard deviation to show how spread out the data points are. Descriptive statistics create charts and graphs to visually summarise data.
### B. Measures of Central Tendency
Measures of central tendency indicate the centre point or typical value of a data set. They provide a single value that is considered representative of the entire distribution. The three main measures of central tendency are mean, median, and mode.
#### 1. Mean
The mean, commonly known as the average, is calculated by summing all the values in a distribution and dividing by the total number of values. It is the balance point of the data. With a symmetrical distribution, the mean will be at the centre. The mean is sensitive to extreme highs and lows in data.
To calculate the mean:
Mean = Sum of all values / Total number of values
For example, the mean of the dataset {2, 3, 6, 7, 10, 12} is:
(2 + 3 + 6 + 7 + 10 + 12) / 6 = 7
#### 2. Median
The median is the middle value that separates the higher half and lower half of the data set when ordered from lowest to highest. It is the 50th percentile value. With an odd number of data points, the median is simply the middle value. With an even number, it is the average of the two central values. The median is less affected by outliers compared to the mean.
To find the median:
• Order the data from lowest to highest value
• Find the middle value (or average of the two central values if even number of data points)
For example, to find the median of {5, 2, 8, 7, 3}, first order the data: {2, 3, 5, 7, 8}
The middle value is 5, so the median is 5.
#### 3. Mode
The mode is the value that occurs most frequently in the data set. It can be thought of as the most “popular” number in the distribution. A set may have one mode, more than one mode, or no mode if all values appear equally often. Modes are best used for qualitative and categorical data rather than numerical data.
To find the mode, look at the frequencies of the values and identify the value(s) with the highest frequency.
For example, in the dataset {6, 3, 9, 6, 2, 9, 6}, the mode is 6 because it appears most often.
When it comes to getting elementary statistics it can be essential to be able to divide numbers or potentially do long division in your head, a quick video from the LearningMole YouTube channel has got you sorted in this regard!
### C. Measures of Dispersion
Measures of dispersion, also called measures of spread, indicate how spread out or scattered the data points are within a distribution. They reflect the degree of variability present. Common measures of dispersion are range, variance, and standard deviation.
#### 1. Range
The range is the difference between the highest and lowest values in a dataset. It measures the span of the values. A small range indicates values are clustered close together, while a larger range reflects greater spread.
To calculate range:
Range = Highest value – Lowest value
For example, in the dataset {1, 5, 18, 3, 14}, the highest value is 18 and the lowest is 1.
The range is:
18 – 1 = 17
#### 2. Variance
Variance measures how far each value deviates from the mean on average. It reflects how dispersed the data is about the mean centre point. A higher variance indicates greater variability and spread while a lower variance reflects data points closer to the mean.
Variance is calculated by taking the average of the squared differences between each value and the mean:
To calculate variance:
1. Find the mean
2. Subtract the mean from each value
3. Square each of the differences found
4. Sum all the squared differences
5. Divide the sum by the number of data points minus 1
The formula is:
Variance = Sum of squared differences from mean / (Total number of values – 1)
#### 3. Standard Deviation
Standard deviation is calculated from the variance to express variability in the same units as the original data. It’s the square root of the variance. Standard deviation measures the typical distance between a data point and the mean centre.
A higher standard deviation indicates a greater spread. A lower standard deviation indicates points are closer to the mean centre.
To find the standard deviation:
1. Calculate the variance
2. Take the square root of the variance
The formula is:
Standard Deviation = Square root of Variance
## Probability
### A. Introduction to Probability
Probability deals with analysing and making predictions about the likelihood of events. It helps assess uncertainty and randomness. Probability provides insights about the chance or probability of certain outcomes or events occurring.
### B. Basic Probability Concepts
Some fundamental concepts in probability theory are sample spaces, events, and probability rules.
#### 1. Sample Space
The sample space in an experiment contains all the possible outcomes that may occur. It is the set of all possibilities. To determine probabilities, we look at where events stand about the sample space.
For example, when rolling a standard 6-sided dice, the sample space of possible values is {1, 2, 3, 4, 5, 6}. Other examples of sample spaces could be:
• Tossing a coin – {Heads, Tails}
• Picking a card – {Ace, 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King}
#### 2. Events
Events are outcomes or subsets of the sample space. They are sets of elements that compose the sample space. Events can contain a single outcome or a combination of outcomes.
For example, rolling an even number on a die {2, 4, 6} is an event. Getting Heads when flipping a coin is an event. Drawing a King from a deck of cards is also an event.
#### 3. Probability Rules
Probability rules guide the computation of probabilities of different events occurring.
Some key rules are:
• The probability of an event ranges from 0 (impossible) to 1 (certain).
• The probability of the whole sample space is 1.
• For mutually exclusive events, the probability that A or B occurs is the sum of probabilities of A and B.
• For independent events, the probability of A and B occurring is the product of probabilities of A and B.
These rules allow the calculation of probabilities for single, combined, dependent, and independent events.
### C. Probability Distributions
A probability distribution shows all the possible values or outcomes of a variable and the probability associated with each one. The total area under the curve equals 1.
There are two main types of probability distribution – discrete and continuous.
#### 1. Discrete Probability Distribution
A discrete distribution deals with events that have a countable number of outcomes, like coin flips, dice rolls, or survey responses. The variable can only take certain discrete values.
Some examples are:
• Binomial distribution – binary outcomes over trials
• Poisson distribution – occurrence of events over an interval
#### 2. Continuous Probability Distribution
A continuous distribution deals with variables that can take any value on a continuum. There are infinitely many possible values rather than discrete outcomes. Values are not counted but measured.
Some examples are:
• Normal distribution – symmetric bell curve, useful in natural sciences
• Uniform distribution – constant likelihood of outcomes across a range
Continuous distributions are described by density curves rather than probability masses.
## Inferential Elementary Statistics
### A. Introduction to Inferential Elementary Statistics
While descriptive statistics summarise a sample, inferential statistics make inferences about a wider population based on the sample data. Inferential methods allow drawing conclusions that extend beyond the data at hand.
It enables the estimation of population parameters based on sample statistics. We can test hypotheses and make predictions about a population too large
to practically measure in its entirety.
### B. Sampling Techniques
Inferential statistics relies on selecting representative sample data from a population. Both probability and non-probability sampling provide ways to obtain sample data.
Some sampling techniques are:
• Simple random sampling – each member has an equal chance of being selected
• Systematic sampling – select every nth member from the ordered population
• Stratified sampling – divide the population into strata, sample each stratum
• Cluster sampling – divide into clusters, randomly sample some clusters
• Convenience sampling – sample easily accessible members
Probability sampling aims for representative samples while non-probability sampling does not guarantee randomness or equal inclusion probability.
### C. Estimation
Estimation methods use statistics calculated from a sample, such as the mean, to estimate corresponding population parameters like the mean. Confidence intervals provide a range of plausible values.
#### 1. Confidence Intervals
A confidence interval gives a range of values likely to contain the unknown population parameter at a stated confidence level. Wider intervals reflect less precision while narrower intervals indicate more accuracy.
The most common is a 95% confidence level. The 95% confidence interval suggests we can be 95% confident the true parameter lies within the range.
### D. Hypothesis Testing
Hypothesis testing allows concluding a population by evaluating two competing hypotheses stated about the population.
#### 1. Null and Alternative Hypotheses
The null hypothesis, denoted H0, states there is no effect or no statistical significance. It assumes no difference or relationship. The alternative hypothesis H1 states there is an effect or statistical significance present.
H0 is presumed true until evidence indicates rejection in favour of H1.
For example:
H0: There is no difference between the two population means
H1: There is a difference between the two means
#### 2. Type I and Type II Errors
A Type I error means rejecting the null hypothesis when it is true. A Type II error occurs when failing to reject a false null hypothesis.
The acceptable risk levels for each error type set the criteria for rejecting or not rejecting H0. A lower Type I error rate often increases the Type II error rate.
## Correlation and Regression Analysis
Correlation and regression techniques explore relationships between variables. Correlation examines if and how strongly variables are related. Regression models the relationships mathematically.
### A. Correlation Analysis
Correlation analysis measures the association between two variables to determine if they are related or independent. It does not imply causation. The correlation coefficient (r) indicates the strength of the correlation and whether it is positive or negative.
#### 1. Pearson Correlation Coefficient
The Pearson product-moment coefficient (r) assesses linear relationships between continuous variables. It ranges from -1 to +1, with:
• r = 0: No correlation
• 0 < r < 1: Positive correlation
• -1 < r < 0: Negative correlation
• r = +1: Perfect positive correlation
• r = -1: Perfect negative correlation
#### 2. Spearman Rank Correlation
Spearman’s rho measures monotonic relationships between continuous or ordinal variables. It assesses if one value increases as the other increases, without requiring a linear relationship.
### B. Regression Analysis
Regression analysis models the relationship between a dependent and one or more independent variables. It estimates how the dependent variable changes with the others.
#### 1. Simple Linear Regression
Simple linear regression predicts a dependent value from one independent variable. It fits a straight-line equation to model the relationship.
#### 2. Multiple Regression
Multiple regression predicts a dependent value from multiple independent variables. It finds the linear combination of variables that yield the line of best fit.
Regression quantifies the connections between variables and makes predictions. However, it does not determine causation from correlation.
## Data Presentation and Visualisation
### A. Importance of Data Visualisation
Presenting data visually through charts, graphs, and other plots makes it easier to interpret key information instead of inspecting raw numbers. Effective data visualisation highlights patterns, outliers, trends, and relationships in data to provide insights.
### B. Types of Graphical Representations
Some common ways to visualise quantitative and categorical data are:
#### 1. Histograms
Histograms display numerical data distributions using bars. The bar heights represent the frequencies or counts of values. Histograms show patterns, centres, spreads, gaps, and outliers in data.
#### 2. Bar Charts
Bar charts use horizontal or vertical bars to visually compare different categories of data. The bar lengths show the quantities in each category. They help compare amounts across groups.
#### 3. Pie Charts
Pie charts show the categorical data as proportional slices of a circle. The arc lengths represent the percentages of each category relative to the whole dataset. Pie charts depict parts of a whole.
#### 4. Scatter Plots
Scatter plots visualise relationships between two numerical variables by plotting each data point along the x and y axes. Patterns in the scatter of points reveal correlations and trends between the variables.
Effective visuals tailor the type of plot to the data being presented. Plots should accurately represent the data and highlight key features.
## Real-Life Applications of Elementary Statistics
Delving Deeper into Practical Uses:
Statistics isn’t just about numbers on a page; it’s a powerful tool shaping the world around us. Let’s see how:
• Health Sciences:
• Clinical research: Researchers might compare two different treatments for a disease, where statistical analysis determines which one is more effective (Source: “Designing Clinical Research” by Michael R. Hulley, Stephen Cummings, David S. Browner, Thomas D. Schroeder, and William T. Downs, 5th edition, 2020).
• Epidemiology: Analyzing healthcare data to identify hot spots for a contagious disease and target prevention efforts is just one example of how statistics informs public health interventions (Source: “Modern Epidemiology” by Kenneth Rothman, Sander Greenland, and Melanie Lash TL, 5th edition, 2020).
Quick Reference Glossary:
Need a refresher on key terms? This mini-glossary has you covered:
• Descriptive elementary statistics: Summarize data features like mean, median, and standard deviation.
• Inferential elementary statistics: Make predictions about populations based on sample data using methods like hypothesis testing.
• Probability: The likelihood of an event happening, ranging from 0 (impossible) to 1 (certain).
• Correlation: Measures the association between two variables, indicating whether they increase or decrease together.
• Regression analysis: Models the relationship between a dependent variable and one or more independent variables.
• Sample: A subset of a population used to draw inferences about the whole population.
## Conclusion
### A. Recap of Key Topics
This introduction covered essential concepts in elementary statistics. The key topics included:
• Descriptive statistics like measures of central tendency and dispersion
• Probability and probability distributions
• Inferential statistics like estimation, hypothesis testing, correlation, and regression
• Techniques for data sampling and visual presentation
### B. Importance of Understanding Elementary Statistics
Learning the fundamentals of statistics allows us to summarize data, make estimations and informed decisions, identify relationships, and draw meaningful conclusions. It provides the basic skillset to analyze real-world information and unlock insights from data.
### C. Further Resources for Learning Statistics
Many excellent resources for building statistical skills are available through textbooks, online courses, instructional websites, and real-world practice. Start applying your knowledge to datasets from everyday life. Use statistics to satisfy curiosity about the world.
If this article was to your liking feel free to browse some of our blog posts on LearningMole! Some great examples are Education Statistics or covering some fun facts about AI. |
# 4.5: Geometric Meaning of Scalar Multiplication
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## Outcomes
1. Understand scalar multiplication, geometrically.
Recall that the point $$P=\left( p_{1},p_{2},p_{3}\right)$$ determines a vector $$\vec{p}$$ from $$0$$ to $$P$$. The length of $$\vec{p}$$, denoted $$\| \vec{p} \|$$, is equal to $$\sqrt{p_{1}^{2}+p_{2}^{2}+p_{3}^{2}}$$ by Definition 4.4.1.
Now suppose we have a vector $$\vec{u} = \left[ \begin{array}{lll} u_1 & u_2 & u_3 \end{array} \right]^T$$ and we multiply $$\vec{u}$$ by a scalar $$k$$. By Definition 4.2.2, $$k\vec{u} = \left[ \begin{array}{rrr} ku_{1} & ku_{2} & ku_{3} \end{array} \right]^T$$. Then, by using Definition 4.4.1, the length of this vector is given by $\sqrt{\left( \left( k u_{1}\right) ^{2}+\left( k u_{2}\right) ^{2}+\left( k u_{3}\right) ^{2}\right) }=\left\vert k \right\vert \sqrt{u_{1}^{2}+u_{2}^{2}+u_{3}^{2}}\nonumber$ Thus the following holds. $\| k \vec{u} \| =\left\vert k \right\vert \| \vec{u} \|\nonumber$ In other words, multiplication by a scalar magnifies or shrinks the length of the vector by a factor of $$\left\vert k \right\vert$$. If $$\left\vert k \right\vert > 1$$, the length of the resulting vector will be magnified. If $$\left\vert k \right\vert <1$$, the length of the resulting vector will shrink. Remember that by the definition of the absolute value, $$\left\vert k \right\vert >0$$.
What about the direction? Draw a picture of $$\vec{u}$$ and $$k\vec{u}$$ where $$k$$ is negative. Notice that this causes the resulting vector to point in the opposite direction while if $$k >0$$ it preserves the direction the vector points. Therefore the direction can either reverse, if $$k < 0$$, or remain preserved, if $$k > 0$$.
Consider the following example.
## Example $$\PageIndex{1}$$: Graphing Scalar Multiplication
Consider the vectors $$\vec{u}$$ and $$\vec{v}$$ drawn below.
Draw $$-\vec{u}$$, $$2\vec{v}$$, and $$-\frac{1}{2}\vec{v}$$.
###### Solution
In order to find $$-\vec{u}$$, we preserve the length of $$\vec{u}$$ and simply reverse the direction. For $$2\vec{v}$$, we double the length of $$\vec{v}$$, while preserving the direction. Finally $$-\frac{1}{2}\vec{v}$$ is found by taking half the length of $$\vec{v}$$ and reversing the direction. These vectors are shown in the following diagram.
Now that we have studied both vector addition and scalar multiplication, we can combine the two actions. Recall Definition 9.2.2 of linear combinations of column matrices. We can apply this definition to vectors in $$\mathbb{R}^n$$. A linear combination of vectors in $$\mathbb{R}^n$$ is a sum of vectors multiplied by scalars.
In the following example, we examine the geometric meaning of this concept.
## Example $$\PageIndex{2}$$: Graphing a Linear Combination of Vectors
Consider the following picture of the vectors $$\vec{u}$$ and $$\vec{v}$$
Sketch a picture of $$\vec{u}+2\vec{v},\vec{u}-\frac{1}{2}\vec{v}.$$
###### Solution
The two vectors are shown below.
This page titled 4.5: Geometric Meaning of Scalar Multiplication is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. |
# How do you simplify (4+9i)/(2+6i)?
Apr 5, 2018
$\frac{31 - 3 i}{20}$
#### Explanation:
Start by multiplying both the numerator and the denominator by the conjugate of the denominator:
$\left(\frac{4 + 9 i}{2 + 6 i}\right) \left(\frac{2 - 6 i}{2 - 6 i}\right) =$
Use the distributive property to get back to a single fraction:
$\frac{8 - 6 i - 54 {i}^{2}}{4 - 36 {i}^{2}} =$
Convert ${i}^{2}$ to $- 1$ and combine like terms:
$\frac{62 - 6 i}{40} =$
Reduce to lowest terms:
$\frac{31 - 3 i}{20}$ |
## What is the percent equation?
How to find what percent of X is Y. Use the percentage formula: Y/X = P% You need to multiply the result by 100 to get the percentage. Converting 0.20 to a percent: 0.20 * 100 = 20% So 20% of 60 is 12.
## What is the formula for percentage in statistics?
One of the most frequent ways to represent statistics is by percentage. Percent simply means “per hundred” and the symbol used to express percentage is %. One percent (or 1%) is one hundredth of the total or whole and is therefore calculated by dividing the total or whole number by 100.
## How do you explain percentages?
One percent is one hundredth of a whole. It can therefore be written as both a decimal and a fraction. To write a percentage as a decimal, simply divide it by 100. For example, 50% becomes 0.5, 20% becomes 0.2, 1% becomes 0.01 and so on.
40
## How do you calculate percentages quickly?
To calculate 10 percent of a number, simply divide it by 10 or move the decimal point one place to the left. For example, 10 percent of 230 is 230 divided by 10, or 23. 5 percent is one half of 10 percent. To calculate 5 percent of a number, simply divide 10 percent of the number by 2.
## How do I calculate a percentage between two numbers?
Percentage Change | Increase and DecreaseFirst: work out the difference (increase) between the two numbers you are comparing.Increase = New Number – Original Number.Then: divide the increase by the original number and multiply the answer by 100.% increase = Increase ÷ Original Number × 100.
You might be interested: Assets = liabilities + owners' equity is the fundamental ________ equation.
## How do I calculate percentage using a calculator?
How to Calculate Percentages with a CalculatorIf your calculator has a “%” button. Let’s say you wanted to find 19 percent of 20. Press these buttons: 1 9 % * 2 0 = If your calculator does not have a “%” button. Step 1: Remove the percent sign and add a couple of zeros after the decimal point. 19% becomes 19.00.
## How do you do 40%?
You divide your percentage by 100. So, 40% would be 40 divided by 100 or . 40. Once you have the decimal version of your percentage, simply multiply it by the given number.
## Why do we use percentages?
Percentages are used widely and in many different areas. For example, discounts in shops, bank interest rates, rates of inflation and many statistics in the media are expressed as percentages. Percentages are important for understanding the financial aspects of everyday life.
## How do you write percentages?
Use the symbol % to express percent in scientific and technical writing, except when writing numbers at the beginning of a sentence. When you write out the word, use the form percent instead of the older form per cent.
25
5
## What number is 40% of 50?
20
### Releated
#### The equation of exchange can be stated as
What is the equation of exchange equal to? So the equation of exchange says that the total amount of money that changes hands in the economy will always equal the total money value of the goods and services that change hands in the economy. So now the equation of exchange says that total nominal expenditures […]
#### Calcium and water balanced equation
What happens when calcium reacts with water? After a second or so, the calcium metal begins to bubble vigorously as it reacts with the water, producing hydrogen gas, and a cloudy white precipitate of calcium hydroxide. What is the balanced equation for calcium oxide and water? CaO(s) + H2O (l) → Ca(OH)2(s) This reaction occurs […] |
# Evaluate $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$
It is most challenging problem in limits but the limit of the algebraic trigonometric function can be evaluated mathematically in calculus by using a mathematical technique.
### Balance Numerator and Denominator
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$
The denominator is a cube function but there is no cube function in the numerator. If sine function has triple angle then the sine function can be expanded in cube form. It helps us to evaluate the function by the limit rule of trigonometric functions.
Take $x = 3y$ and then transform the entire function in terms of $y$.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-\sin{3y}}{{(3y)}^3}}$
### Simplify the limit function
Now, simplify the functions in both numerator and denominator for simplifying the whole function.
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-\sin{3y}}{3^3 \times y^3}}$
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-\sin{3y}}{27y^3}}$
Expand $\sin{3y}$ term by the sine triple angle identity.
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-[3\sin{y}-4\sin^3{y}]}{27y^3}}$
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3y-3\sin{y}+4\sin^3{y}}{27y^3}}$
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{3(y-\sin{y})+4\sin^3{y}}{27y^3}}$
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg[\dfrac{3(y-\sin{y})}{27y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg]}$
$=\,\,\,$ $\require{cancel} \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg[\dfrac{\cancel{3}(y-\sin{y})}{\cancel{27}y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg]}$
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg[\dfrac{y-\sin{y}}{9y^3}+\dfrac{4\sin^3{y}}{27y^3}\Bigg]}$
According to sum rule of limits, the limit of sum of two functions is equal to sum of their limits.
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{9y^3}}$ $+$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{4\sin^3{y}}{27y^3}}$
$=\,\,\,$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg( \dfrac{1}{9} \times \dfrac{y-\sin{y}}{y^3} \Bigg)}$ $+$ $\displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \Bigg(\dfrac{4}{27} \times \dfrac{\sin^3{y}}{y^3} \Bigg)}$
The constant from each term can be separated by the constant multiple rule of limits.
$=\,\,\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times \displaystyle \large \lim_{3y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$
### Find the limit of each function
Now, concentrate on the second term firstly, the limit of the trigonometric function is same as the limit of $\sin{x}/{x}$ as $x$ approaches $0$ formula. So, let’s evaluate this function.
If $3y \to 0$, then $y \to \dfrac{0}{3}$. Therefore, $y \to 0$. It states that if $3y$ approaches $0$, then $y$ also approaches $0$.
$=\,\,\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin^3{y}}{y^3}}$
According to power rule of limits, the second term can be written as follows.
$=\,\,\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize {\Bigg(\dfrac{\sin{y}}{y}\Bigg)}^3}$
$=\,\,\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times {\Bigg(\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{\sin{y}}{y}\Bigg)}}^3$
According to limit of sinx/x as x approaches 0 formula, the limit of the trigonometric function in the second term is equal to $1$.
$=\,\,\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times {(1)}^3$
$=\,\,\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27} \times 1$
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27}$
Therefore, the limit of the algebraic trigonometric function is successfully evaluate and we are in one step ahead in evaluating the limit of the given algebraic trigonometric function.
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ $+$ $\dfrac{4}{27}$
Now, observe the mathematical equation. The limit of given algebraic trigonometric function and part of the first term are same but they are in $x$ and $y$ terms.
$\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ and $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$. There is no difference between them but they are in $x$ and $y$ terms. However, their values should be equal mathematically. Therefore, $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$. Therefore, $\displaystyle \large \lim_{y \,\to\, 0}{\normalsize \dfrac{y-\sin{y}}{y^3}}$ can be written as $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ in this equation.
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $+$ $\dfrac{4}{27}$
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,-\,$ $\dfrac{1}{9} \times \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27}$
$\implies$ $\Bigg(1-\dfrac{1}{9}\Bigg) \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27}$
$\implies$ $\Bigg(\dfrac{1\times 9 -1}{9}\Bigg) \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27}$
$\implies$ $\Bigg(\dfrac{9-1}{9}\Bigg) \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27}$
$\implies$ $\Bigg(\dfrac{8}{9}\Bigg) \displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27}$
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4}{27} \times \dfrac{9}{8}$
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{4 \times 9}{27 \times 8}$
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\require{cancel} \dfrac{\cancel{4} \times \cancel{9}}{\cancel{27} \times \cancel{8}}$
$\implies$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1 \times 1}{3 \times 2}$
$\,\,\, \therefore \,\,\,\,\,\,$ $\displaystyle \large \lim_{x \,\to\, 0}{\normalsize \dfrac{x-\sin{x}}{x^3}}$ $\,=\,$ $\dfrac{1}{6}$
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# Lesson 5A New Way to Interpret a over b
Let's investigate what a fraction means when the numerator and denominator are not whole numbers.
### Learning Targets:
• I understand the meaning of a fraction made up of fractions or decimals, like or .
• When I see an equation, I can make up a story that the equation might represent, explain what the variable represents in the story, and solve the equation.
## 5.1Recalling Ways of Solving
Solve each equation. Be prepared to explain your reasoning.
1.
## 5.2Interpreting \frac{a}{b}
Solve each equation.
### Are you ready for more?
Solve the equation. Try to find some shortcuts.
## 5.3Storytime Again
Take turns with your partner telling a story that might be represented by each equation. Then, for each equation, choose one story, state what quantity describes, and solve the equation. If you get stuck, draw a diagram.
## Lesson 5 Summary
In the past, you learned that a fraction such as can be thought of in a few ways.
• is a number you can locate on the number line by dividing the section between 0 and 1 into 5 equal parts and then counting 4 of those parts to the right of 0.
• is the share that each person would have if 4 wholes were shared equally among 5 people. This means that is the result of dividing 4 by 5.
We can extend this meaning of a fraction as a division to fractions whose numerators and denominators are not whole numbers. For example, we can represent 4.5 pounds of rice divided into portions that each weigh 1.5 pounds as: .
Fractions that involve non-whole numbers can also be used when we solve equations.
Suppose a road under construction is finished and the length of the completed part is miles. How long will the road be when completed?
We can write the equation to represent the situation and solve the equation.
## Lesson 5 Practice Problems
1. Select all the expressions that equal .
2. Which expressions are solutions to the equation ? Select all that apply.
3. Solve each equation.
4. For each equation, write a story problem represented by the equation. For each equation, state what quantity represents. If you get stuck, draw a diagram.
5. Write as many mathematical expressions or equations as you can about the image. Include a fraction, a decimal number, or a percentage in each.
6. In a lilac paint mixture, 40% of the mixture is white paint, 20% is blue, and the rest is red. There are 4 cups of blue paint used in a batch of lilac paint. If you get stuck, consider using a tape diagram.
1. How many cups of white paint are used?
2. How many cups of red paint are used?
3. How many cups of lilac paint will this batch yield?
7. Triangle P has a base of 12 inches and a corresponding height of 8 inches. Triangle Q has a base of 15 inches and a corresponding height of 6.5 inches. Which triangle has a greater area? Show your reasoning. |
# Formula for Binomial Probability Distribution
Here, you will learn formula for binomial probability distribution in probability with example.
Let’s begin –
Suppose that we have an experiment such as tossing a coin or die repeatedly or choosing a marble from an urn repeatedly. Each toss or selection is called a trial. In any single trial there will be a probability associated with a particular event such as head on the coin, 4 on the die, or selection of a red marble. In some cases this probability will not change from one trial to the next(as in tossing a coin or die). Such trials are then said to be independent and are often called Bernoulli trials after James Bernoulli who investigated them at the end of the seventeenth century.
## Formula for Binomial Probability Distribution
Let p be the probability that an event will happen in any single Bernoulli trial(called the probability of success).Then q = 1 – p is the probability that the event will fail to happen in any single trial (called the probability of Failure). The probability that the event will happen exactly x times in n trials (i.e., x success and n – x failures will occur) is given by the probability function.
f(x) = P(X = x) = $$\binom{n}{x} p^x q^{n-x}$$ = $$n!\over {x!(n – x)!}$$ $$p^xq^{n-x}$$
where the random variable X denotes the number of success in n trials and x = 0, 1,…….,n.
Example : What is the probability of getting exactly 2 heads in 6 tosses of a fair coin?
Solution : The probability of getting exactly 2 heads in 6 tosses of a fair coin is
P(X = 2) = $$\binom{6}{2} ({1\over 2})^2 ({1\over 2})^{6-2}$$
= $$6!\over {2!4!}$$ $$({1\over 2})^2 ({1\over 2})^{6-2}$$
= $${15}\over{64}$$
The discrete probability function is often called the binomial distribution since for x = 0, 1, 2,……,n, it corresponds to successive terms in the binomial expansion.
$$(q + p)^n$$ = $$q^n$$ + $$\binom{n}{1} p q^{n-1}$$ + $$\binom{n}{2} p^2 q^{n-2}$$ + ……….+ $$p^n$$ = $${\sum_{n=1}^{\infty}}$$$$\binom{n}{x} p^x q^{n-x}$$ |
In mathematics, a series is the sum of the terms of an infinite sequence of numbers. More precisely, an infinite sequence ${\displaystyle (a_{1},a_{2},a_{3},\ldots )}$ defines a series S that is denoted
${\displaystyle S=a_{1}+a_{2}+a_{3}+\cdots =\sum _{k=1}^{\infty }a_{k}.}$
The nth partial sum Sn is the sum of the first n terms of the sequence; that is,
${\displaystyle S_{n}=a_{1}+a_{2}+\cdots +a_{n}=\sum _{k=1}^{n}a_{k}.}$
A series is convergent (or converges) if and only if the sequence ${\displaystyle (S_{1},S_{2},S_{3},\dots )}$ of its partial sums tends to a limit; that means that, when adding one ${\displaystyle a_{k))$ after the other in the order given by the indices, one gets partial sums that become closer and closer to a given number. More precisely, a series converges, if and only if there exists a number ${\displaystyle \ell }$ such that for every arbitrarily small positive number ${\displaystyle \varepsilon }$, there is a (sufficiently large) integer ${\displaystyle N}$ such that for all ${\displaystyle n\geq N}$,
${\displaystyle \left|S_{n}-\ell \right|<\varepsilon .}$
If the series is convergent, the (necessarily unique) number ${\displaystyle \ell }$ is called the sum of the series.
The same notation
${\displaystyle \sum _{k=1}^{\infty }a_{k))$
is used for the series, and, if it is convergent, to its sum. This convention is similar to that which is used for addition: a + b denotes the operation of adding a and b as well as the result of this addition, which is called the sum of a and b.
Any series that is not convergent is said to be divergent or to diverge.
## Examples of convergent and divergent series
• The reciprocals of the positive integers produce a divergent series (harmonic series):
${\displaystyle {1 \over 1}+{1 \over 2}+{1 \over 3}+{1 \over 4}+{1 \over 5}+{1 \over 6}+\cdots \rightarrow \infty .}$
• Alternating the signs of the reciprocals of positive integers produces a convergent series (alternating harmonic series):
${\displaystyle {1 \over 1}-{1 \over 2}+{1 \over 3}-{1 \over 4}+{1 \over 5}-\cdots =\ln(2)}$
• The reciprocals of prime numbers produce a divergent series (so the set of primes is "large"; see divergence of the sum of the reciprocals of the primes):
${\displaystyle {1 \over 2}+{1 \over 3}+{1 \over 5}+{1 \over 7}+{1 \over 11}+{1 \over 13}+\cdots \rightarrow \infty .}$
• The reciprocals of triangular numbers produce a convergent series:
${\displaystyle {1 \over 1}+{1 \over 3}+{1 \over 6}+{1 \over 10}+{1 \over 15}+{1 \over 21}+\cdots =2.}$
• The reciprocals of factorials produce a convergent series (see e):
${\displaystyle {\frac {1}{1))+{\frac {1}{1))+{\frac {1}{2))+{\frac {1}{6))+{\frac {1}{24))+{\frac {1}{120))+\cdots =e.}$
• The reciprocals of square numbers produce a convergent series (the Basel problem):
${\displaystyle {1 \over 1}+{1 \over 4}+{1 \over 9}+{1 \over 16}+{1 \over 25}+{1 \over 36}+\cdots ={\pi ^{2} \over 6}.}$
• The reciprocals of powers of 2 produce a convergent series (so the set of powers of 2 is "small"):
${\displaystyle {1 \over 1}+{1 \over 2}+{1 \over 4}+{1 \over 8}+{1 \over 16}+{1 \over 32}+\cdots =2.}$
• The reciprocals of powers of any n>1 produce a convergent series:
${\displaystyle {1 \over 1}+{1 \over n}+{1 \over n^{2))+{1 \over n^{3))+{1 \over n^{4))+{1 \over n^{5))+\cdots ={n \over n-1}.}$
• Alternating the signs of reciprocals of powers of 2 also produces a convergent series:
${\displaystyle {1 \over 1}-{1 \over 2}+{1 \over 4}-{1 \over 8}+{1 \over 16}-{1 \over 32}+\cdots ={2 \over 3}.}$
• Alternating the signs of reciprocals of powers of any n>1 produces a convergent series:
${\displaystyle {1 \over 1}-{1 \over n}+{1 \over n^{2))-{1 \over n^{3))+{1 \over n^{4))-{1 \over n^{5))+\cdots ={n \over n+1}.}$
• The reciprocals of Fibonacci numbers produce a convergent series (see ψ):
${\displaystyle {\frac {1}{1))+{\frac {1}{1))+{\frac {1}{2))+{\frac {1}{3))+{\frac {1}{5))+{\frac {1}{8))+\cdots =\psi .}$
## Convergence tests
Main article: Convergence tests
There are a number of methods of determining whether a series converges or diverges.
Comparison test. The terms of the sequence ${\displaystyle \left\{a_{n}\right\))$ are compared to those of another sequence ${\displaystyle \left\{b_{n}\right\))$. If, for all n, ${\displaystyle 0\leq \ a_{n}\leq \ b_{n))$, and ${\textstyle \sum _{n=1}^{\infty }b_{n))$ converges, then so does ${\textstyle \sum _{n=1}^{\infty }a_{n}.}$
However, if, for all n, ${\displaystyle 0\leq \ b_{n}\leq \ a_{n))$, and ${\textstyle \sum _{n=1}^{\infty }b_{n))$ diverges, then so does ${\textstyle \sum _{n=1}^{\infty }a_{n}.}$
Ratio test. Assume that for all n, ${\displaystyle a_{n))$ is not zero. Suppose that there exists ${\displaystyle r}$ such that
${\displaystyle \lim _{n\to \infty }\left|{\frac {a_{n+1)){a_{n))}\right|=r.}$
If r < 1, then the series is absolutely convergent. If r > 1, then the series diverges. If r = 1, the ratio test is inconclusive, and the series may converge or diverge.
Root test or nth root test. Suppose that the terms of the sequence in question are non-negative. Define r as follows:
${\displaystyle r=\limsup _{n\to \infty }{\sqrt[{n}]{|a_{n}|)),}$
where "lim sup" denotes the limit superior (possibly ∞; if the limit exists it is the same value).
If r < 1, then the series converges. If r > 1, then the series diverges. If r = 1, the root test is inconclusive, and the series may converge or diverge.
The ratio test and the root test are both based on comparison with a geometric series, and as such they work in similar situations. In fact, if the ratio test works (meaning that the limit exists and is not equal to 1) then so does the root test; the converse, however, is not true. The root test is therefore more generally applicable, but as a practical matter the limit is often difficult to compute for commonly seen types of series.
Integral test. The series can be compared to an integral to establish convergence or divergence. Let ${\displaystyle f(n)=a_{n))$ be a positive and monotonically decreasing function. If
${\displaystyle \int _{1}^{\infty }f(x)\,dx=\lim _{t\to \infty }\int _{1}^{t}f(x)\,dx<\infty ,}$
then the series converges. But if the integral diverges, then the series does so as well.
Limit comparison test. If ${\displaystyle \left\{a_{n}\right\},\left\{b_{n}\right\}>0}$, and the limit ${\displaystyle \lim _{n\to \infty }{\frac {a_{n)){b_{n))))$ exists and is not zero, then ${\textstyle \sum _{n=1}^{\infty }a_{n))$ converges if and only if ${\textstyle \sum _{n=1}^{\infty }b_{n))$ converges.
Alternating series test. Also known as the Leibniz criterion, the alternating series test states that for an alternating series of the form ${\textstyle \sum _{n=1}^{\infty }a_{n}(-1)^{n))$, if ${\displaystyle \left\{a_{n}\right\))$ is monotonically decreasing, and has a limit of 0 at infinity, then the series converges.
Cauchy condensation test. If ${\displaystyle \left\{a_{n}\right\))$ is a positive monotone decreasing sequence, then ${\textstyle \sum _{n=1}^{\infty }a_{n))$ converges if and only if ${\textstyle \sum _{k=1}^{\infty }2^{k}a_{2^{k))}$ converges.
## Conditional and absolute convergence
For any sequence ${\displaystyle \left\{a_{1},\ a_{2},\ a_{3},\dots \right\))$, ${\displaystyle a_{n}\leq \left|a_{n}\right|}$ for all n. Therefore,
${\displaystyle \sum _{n=1}^{\infty }a_{n}\leq \sum _{n=1}^{\infty }\left|a_{n}\right|.}$
This means that if ${\textstyle \sum _{n=1}^{\infty }\left|a_{n}\right|}$ converges, then ${\textstyle \sum _{n=1}^{\infty }a_{n))$ also converges (but not vice versa).
If the series ${\textstyle \sum _{n=1}^{\infty }\left|a_{n}\right|}$ converges, then the series ${\textstyle \sum _{n=1}^{\infty }a_{n))$ is absolutely convergent. The Maclaurin series of the exponential function is absolutely convergent for every complex value of the variable.
If the series ${\textstyle \sum _{n=1}^{\infty }a_{n))$ converges but the series ${\textstyle \sum _{n=1}^{\infty }\left|a_{n}\right|}$ diverges, then the series ${\textstyle \sum _{n=1}^{\infty }a_{n))$ is conditionally convergent. The Maclaurin series of the logarithm function ${\displaystyle \ln(1+x)}$ is conditionally convergent for x = 1.
The Riemann series theorem states that if a series converges conditionally, it is possible to rearrange the terms of the series in such a way that the series converges to any value, or even diverges.
## Uniform convergence
Main article: Uniform convergence
Let ${\displaystyle \left\{f_{1},\ f_{2},\ f_{3},\dots \right\))$ be a sequence of functions. The series ${\textstyle \sum _{n=1}^{\infty }f_{n))$ is said to converge uniformly to f if the sequence ${\displaystyle \{s_{n}\))$ of partial sums defined by
${\displaystyle s_{n}(x)=\sum _{k=1}^{n}f_{k}(x)}$
converges uniformly to f.
There is an analogue of the comparison test for infinite series of functions called the Weierstrass M-test.
## Cauchy convergence criterion
The Cauchy convergence criterion states that a series
${\displaystyle \sum _{n=1}^{\infty }a_{n))$
converges if and only if the sequence of partial sums is a Cauchy sequence. This means that for every ${\displaystyle \varepsilon >0,}$ there is a positive integer ${\displaystyle N}$ such that for all ${\displaystyle n\geq m\geq N}$ we have
${\displaystyle \left|\sum _{k=m}^{n}a_{k}\right|<\varepsilon .}$
This is equivalent to
${\displaystyle \lim _{m\to \infty }\left(\sup _{n>m}\left|\sum _{k=m}^{n}a_{k}\right|\right)=0.}$
## See also
• "Series", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
• Weisstein, Eric (2005). Riemann Series Theorem. Retrieved May 16, 2005. |
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# Class XI Chapter 3 Trigonometric Functions Maths
Page 1 of 44
Exercise 3.1
Question 1:
Find the radian measures corresponding to the following degree measures:
(i) 25 (ii) 47 30' (iii) 240 (iv) 520
(i) 25
We know that 180 = radian
(ii) 47 30'
47 30' = degree [1 = 60']
degree
(iii) 240
We know that 180 = radian
(iv) 520
We know that 180 = radian
Question 2:
Find the degree measures corresponding to the following radian measures
.
Compiled By : OP Gupta [+91-9650 350 480 | +91-9718 240 480]
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 2 of 44
(i) (ii) 4 (iii) (iv)
(i)
We know that radian = 180
(ii) 4
We know that radian = 180
(iii)
We know that radian = 180
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 3 of 44
(iv)
We know that radian = 180
Question 3:
A wheel makes 360 revolutions in one minute. Through how many radians does it turn in
one second?
Number of revolutions made by the wheel in 1 minute = 360
Number of revolutions made by the wheel in 1 second =
In one complete revolution, the wheel turns an angle of 2 radian.
Hence, in 6 complete revolutions, it will turn an angle of 6 2 radian, i.e.,
Thus, in one second, the wheel turns an angle of 12 radian.
Question 4:
Find the degree measure of the angle subtended at the centre of a circle of radius 100
cm by an arc of length 22 cm .
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle
Therefore, forr = 100 cm, l = 22 cm, we have
Thus, the required angle is 1236.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 4 of 44
Question 5:
In a circle of diameter 40 cm, the length of a chord is 20 cm. Find the length of minor
arc of the chord.
Diameter of the circle = 40 cm
Radius (r) of the circle =
Let AB be a chord (length = 20 cm) of the circle.
In OAB, OA = OB = Radius of circle = 20 cm
Also, AB = 20 cm
Thus, OAB is an equilateral triangle.
= 60 =
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle
radian at the centre, then .
Thus, the length of the minor arc of the chord is .
Question 6:
If in two circles, arcs of the same length subtend angles 60 and 75 at the centre, find
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 5 of 44
Let the radii of the two circles be and . Let an arc of length l subtend an angle of 60
at the centre of the circle of radius r
1
, while let an arc of length l subtend an angle of 75
at the centre of the circle of radius r
2
.
Now, 60 = and 75 =
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle
radian at the centre, then .
Thus, the ratio of the radii is 5:4.
Question 7:
Find the angle in radian though which a pendulum swings if its length is 75 cm and the
tip describes an arc of length
(i) 10 cm (ii) 15 cm (iii) 21 cm
We know that in a circle of radius r unit, if an arc of length l unit subtends an angle
radian at the centre, then .
It is given that r = 75 cm
(i) Here, l = 10 cm
(ii) Here, l = 15 cm
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 6 of 44
(iii) Here, l = 21 cm
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 7 of 44
Exercise 3.2
Question 1:
Find the values of other five trigonometric functions if , x lies in third
Since x lies in the 3
rd
quadrant, the value of sin x will be negative.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 8 of 44
Question 2:
Find the values of other five trigonometric functions if , x lies in second
Since x lies in the 2
nd
quadrant, the value of cos x will be negative
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 9 of 44
Question 3:
Find the values of other five trigonometric functions if , x lies in third quadrant.
Since x lies in the 3
rd
quadrant, the value of sec x will be negative.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 10 of 44
Question 4:
Find the values of other five trigonometric functions if , x lies in fourth
Since x lies in the 4
th
quadrant, the value of sin x will be negative.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 11 of 44
Question 5:
Find the values of other five trigonometric functions if , x lies in second
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 12 of 44
Since x lies in the 2
nd
quadrant, the value of sec x will be negative.
sec x =
Question 6:
Find the value of the trigonometric function sin 765
It is known that the values of sin x repeat after an interval of 2 or 360.
Question 7:
Find the value of the trigonometric function cosec (1410)
It is known that the values of cosec x repeat after an interval of 2 or 360.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 13 of 44
Find the value of the trigonometric function
It is known that the values of tan x repeat after an interval of or 180.
Question 9:
Find the value of the trigonometric function
It is known that the values of sin x repeat after an interval of 2 or 360.
Question 10:
Find the value of the trigonometric function
It is known that the values of cot x repeat after an interval of or 180.
Question 8:
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 14 of 44
Exercise 3.3
Question 1:
L.H.S. =
Question 2:
Prove that
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 15 of 44
L.H.S. =
Question 3:
Prove that
L.H.S. =
Question 4:
Prove that
L.H.S =
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 16 of 44
Question 5:
Find the value of:
(i) sin 75
(ii) tan 15
(i) sin 75 = sin (45 + 30)
= sin 45 cos 30 + cos 45 sin 30
[sin (x + y) = sin x cos y + cos x sin y]
(ii) tan 15 = tan (45 30)
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 17 of 44
Question 6:
Prove that:
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 18 of 44
Question 7:
Prove that:
It is known that
L.H.S. =
Question 8:
Prove that
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 19 of 44
L.H.S. =
Question 10:
Prove that sin (n + 1)x sin (n + 2)x + cos (n + 1)x cos (n + 2)x = cos x
L.H.S. = sin (n + 1)x sin(n + 2)x + cos (n + 1)x cos(n + 2)x
Question 11:
Prove that
Question 9:
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 20 of 44
It is known that .
L.H.S. =
Question 12:
Prove that sin
2
6x sin
2
4x = sin 2x sin 10x
It is known that
L.H.S. = sin
2
6x sin
2
4x
= (sin 6x + sin 4x) (sin 6x sin 4x)
= (2 sin 5x cos x) (2 cos 5x sin x)
= (2 sin 5x cos 5x) (2 sin x cos x)
= sin 10x sin 2x
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 21 of 44
= R.H.S.
Question 13:
Prove that cos
2
2x cos
2
6x = sin 4x sin 8x
It is known that
L.H.S. = cos
2
2x cos
2
6x
= (cos 2x + cos 6x) (cos 2x 6x)
= [2 cos 4x cos 2x] [2 sin 4x (sin 2x)]
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= R.H.S.
Question 14:
Prove that sin 2x + 2sin 4x + sin 6x = 4cos
2
x sin 4x
L.H.S. = sin 2x + 2 sin 4x + sin 6x
= [sin 2x + sin 6x] + 2 sin 4x
= 2 sin 4x cos ( 2x) + 2 sin 4x
= 2 sin 4x cos 2x + 2 sin 4x
= 2 sin 4x (cos 2x + 1)
= 2 sin 4x (2 cos
2
x 1 + 1)
= 2 sin 4x (2 cos
2
x)
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 22 of 44
= 4cos
2
x sin 4x
= R.H.S.
Question 15:
Prove that cot 4x (sin 5x + sin 3x) = cot x (sin 5x sin 3x)
L.H.S = cot 4x (sin 5x + sin 3x)
= 2 cos 4x cos x
R.H.S. = cot x (sin 5x sin 3x)
= 2 cos 4x. cos x
L.H.S. = R.H.S.
Question 16:
Prove that
It is known that
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 23 of 44
L.H.S =
Question 17:
Prove that
It is known that
L.H.S. =
Question 18:
Prove that
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 24 of 44
It is known that
L.H.S. =
Question 19:
Prove that
It is known that
L.H.S. =
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 25 of 44
Question 20:
Prove that
It is known that
L.H.S. =
Question 21:
Prove that
L.H.S. =
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 26 of 44
Question 22:
Prove that cot x cot 2x cot 2x cot 3x cot 3x cot x = 1
L.H.S. = cot x cot 2x cot 2x cot 3x cot 3x cot x
= cot x cot 2x cot 3x (cot 2x + cot x)
= cot x cot 2x cot (2x + x) (cot 2x + cot x)
= cot x cot 2x (cot 2x cot x 1)
= 1 = R.H.S.
Question 23:
Prove that
It is known that .
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 27 of 44
L.H.S. = tan 4x = tan 2(2x)
Question 24:
Prove that cos 4x = 1 8sin
2
x cos
2
x
L.H.S. = cos 4x
= cos 2(2x)
= 1 2 sin
2
2x [cos 2A = 1 2 sin
2
A]
= 1 2(2 sin x cos x)
2
[sin2A = 2sin A cosA]
= 1 8 sin
2
x cos
2
x
= R.H.S.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 28 of 44
Question 25:
Prove that: cos 6x = 32 cos
6
x 48 cos
4
x + 18 cos
2
x 1
L.H.S. = cos 6x
= cos 3(2x)
= 4 cos
3
2x 3 cos
2x [cos 3A = 4 cos
3
A 3 cos
A]
= 4 [(2 cos
2
x 1)
3
3 (2 cos
2
x 1) [cos 2x = 2 cos
2
x 1]
= 4 [(2 cos
2
x)
3
(1)
3
3 (2 cos
2
x)
2
+ 3 (2 cos
2
x)] 6cos
2
x + 3
= 4 [8cos
6
x 1 12 cos
4
x + 6 cos
2
x] 6 cos
2
x + 3
= 32 cos
6
x 4 48 cos
4
x + 24 cos
2
x 6 cos
2
x + 3
= 32 cos
6
x 48 cos
4
x + 18 cos
2
x 1
= R.H.S.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 29 of 44
Exercise 3.4
Question 1:
Find the principal and general solutions of the equation
Therefore, the principal solutions are x = and .
Therefore, the general solution is
Question 2:
Find the principal and general solutions of the equation
Therefore, the principal solutions are x = and .
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 30 of 44
Therefore, the general solution is , where n Z
Question 3:
Find the principal and general solutions of the equation
Therefore, the principal solutions are x = and .
Therefore, the general solution is
Question 4:
Find the general solution of cosec x = 2
cosec x = 2
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 31 of 44
Therefore, the principal solutions are x = .
Therefore, the general solution is
Question 5:
Find the general solution of the equation
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 32 of 44
Question 6:
Find the general solution of the equation
Question 7:
Find the general solution of the equation
Therefore, the general solution is .
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 33 of 44
Question 8:
Find the general solution of the equation
Therefore, the general solution is .
Question 9:
Find the general solution of the equation
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 34 of 44
Therefore, the general solution is
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 35 of 44
NCERT Miscellaneous Solution
Question 1: Prove that:
L.H.S.
= 0 = R.H.S
Question 2:
Prove that: (sin 3x + sin x) sin x + (cos 3x cos x) cos x = 0
L.H.S.
= (sin 3x + sin x) sin x + (cos 3x cos x) cos x
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 36 of 44
= RH.S.
Question 3:
Prove that:
L.H.S. =
Question 4:
Prove that:
L.H.S. =
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 37 of 44
Question 5:
Prove that:
It is known that .
L.H.S. =
Question 6:
Prove that:
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 38 of 44
It is known that
.
L.H.S. =
= tan 6x
= R.H.S.
Question 7:
Prove that:
L.H.S. =
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 39 of 44
Question 8:
Here, x is in quadrant II.
i.e.,
Therefore, are all positive.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 40 of 44
As x is in quadrant II, cosx is negative.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 41 of 44
Thus, the respective values of are .
Question 9:
Find for , x in quadrant III
Here, x is in quadrant III.
Therefore, and are negative, whereas is positive.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 42 of 44
Now,
Thus, the respective values of are .
Question 10:
Find for , x in quadrant II
Here, x is in quadrant II.
Therefore, , and are all positive.
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 43 of 44
[cosx is negative in quadrant II]
By OP Gupta [+91-9650 350 480]
Class XI Chapter 3 Trigonometric Functions Maths
Page 44 of 44
Thus, the respective values of are .
Compiled By : OP Gupta [+91-9650 350 480 | +91-9718 240 480] |
# Find all the points of discontinuity
Question:
Find all the points of discontinuity of $f$ defined by $f(x)=|x|-|x+1|$.
Solution:
Given: $f(x)=|x|-|x+1|$
The two functions, $g$ and $h$, are defined as
$g(x)=|x|$ and $h(x)=|x+1|$
Then, $f=g-h$
The continuity of $g$ and $h$ is examined first.
$g(x)=|x|$ can be written as
$g(x)= \begin{cases}-x, & \text { if } x<0 \\ x, & \text { if } x \geq 0\end{cases}$
Clearly, $g$ is defined for all real numbers.
Let $c$ be a real number.
Case I:
If $c<0$, then $g(c)=-c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c}(-x)=-c$
$\therefore \lim _{x \rightarrow c} g(x)=g(c)$
So, $g$ is continuous at all points $x<0$.
Case II:
If $c>0$, then $g(c)=c$ and $\lim _{x \rightarrow c} g(x)=\lim _{x \rightarrow c} x=c$
$\therefore \lim _{x \rightarrow c} g(x)=g(c)$
So, $g$ is continuous at all points $x>0$
Case III:
If $c=0$, then $g(c)=g(0)=0$
$\lim _{x \rightarrow 0^{-}} g(x)=\lim _{x \rightarrow 0^{-}}(-x)=0$
$\lim _{x \rightarrow 0^{+}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=0$
$\therefore \lim _{x \rightarrow 0^{0}} g(x)=\lim _{x \rightarrow 0^{+}}(x)=g(0)$
So, $g$ is continuous at $x=0$
From the above three observations, it can be concluded that $g$ is continuous at all points.
$h(x)=|x+1|$ can be written as
$h(x)= \begin{cases}-(x+1), & \text { if, } x<-1 \\ x+1, & \text { if } x \geq-1\end{cases}$
Clearly, $h$ is defined for every real number.
Let $c$ be a real number.
Case I:
If $c<-1$, then $h(c)=-(c+1)$ and $\lim _{x \rightarrow c} h(x)=\lim _{x \rightarrow c}[-(x+1)]=-(c+1)$
$\therefore \lim _{x \rightarrow c} h(x)=h(c)$
So, $h$ is continuous at all points $x>-1$.
Case III:
If $c=-1$, then $h(c)=h(-1)=-1+1=0$
$\lim _{x \rightarrow-1^{+}} h(x)=\lim _{x \rightarrow-1^{+}}(x+1)=(-1+1)=0$
$\therefore \lim _{x \rightarrow-1^{-}} h(x)=\lim _{h \rightarrow-1^{+}} h(x)=h(-1)$
So, $h$ is continuous at $x=-1$
From the above three observations, it can be concluded that $h$ is continuous at all points of the real line.
So, $g$ and $h$ are continuous functions.
Thus, $f=g-h$ is also a continuous function.
Therefore, $f$ has no point of discontinuity. |
English
# Complement Law Proof using De Morgan's Law
In boolean algebra, De Morgan's law is a pair of transformation valid rules of inference. The rule explains the conjunctions and disjunctions in terms of negation. The rule can be given as 'the complement of the union of two sets is the same as the intersection of their complements and the complement of the intersection of two sets is the same as the union of their complements.' Refer the below tutorial to know about Complement law Proof using De Morgan's Law.
##### Complement law:
De Morgan's Law states that, 'The complement of the union of two sets is equal to the intersection of their complements and the complement of the intersection of two sets is equal to the union of their complements'.
#### De Morgan’s Law:
For any two finite sets A and B; (i) (A U B)' = A' ∩ B' (De Morgan's Law of Union). (ii) (A ∩ B)' = A' U B' (De Morgan's Law of Intersection).
##### Proof of De Morgan’s Law :
###### Case 1:
(A U B)' = A' ∩ B' Let us consider P = (A U B)' and Q = A' ∩ B' Let 'x' be an arbitrary element of P then x ∈ P ⇒ x ∈ (A U B)' ⇒ x ∉ (A U B) ⇒ x ∉ A and x ∉ B ⇒ x ∈ A' and x ∈ B' ⇒ x ∈ A' ∩ B' ⇒ x ∈ Q Therefore, P ⊂ Q …………….. (i) Let 'y' be an arbitrary element of Q then y ∈ Q ⇒ y ∈ A' ∩ B' ⇒ y ∈ A' and y ∈ B' ⇒ y ∉ A and y ∉ B ⇒ y ∉ (A U B) ⇒ y ∈ (A U B)' ⇒ y ∈ P Therefore, Q ⊂ P …………….. (ii) Now combining (i) and (ii) we get, P = Q i.e. (A U B)' = A' ∩ B' Hence, it is proved.
###### Case 2:
(A ∩ B)' = A' U B' Let us consider M = (A ∩ B)' and N = A' U B' Let 'x' be an arbitrary element of M then x ∈ M ⇒ x ∈ (A ∩ B)' ⇒ x ∉ (A ∩ B) ⇒ x ∉ A or x ∉ B ⇒ x ∈ A' or x ∈ B' ⇒ x ∈ A' U B' ⇒ x ∈ N Therefore, M ⊂ N …………….. (i) Let 'y' be an arbitrary element of N then y ∈ N ⇒ y ∈ A' U B' ⇒ y ∈ A' or y ∈ B' ⇒ y ∉ A or y ∉ B ⇒ y ∉ (A ∩ B) ⇒ y ∈ (A ∩ B)' ⇒ y ∈ M Therefore, N ⊂ M …………….. (ii) Now combining (i) and (ii) we get, M = N i.e. (A ∩ B)' = A' U B' Hence, De Morgan's Law is proved.
##### Example 1:
Let U = {j, k, l, m, n}, X = {j, k, m} and Y = {k, m, n}. Prove : (X ∩ Y)' = X' U Y'
###### Solution:
Here, U = {j, k, l, m, n} X = {j, k, m} Y = {k, m, n}
###### Step 1:
(X ∩ Y) = {j, k, m} ∩ {k, m, n} = {k, m} Therefore, (X ∩ Y)' = {j, l, n} ……………….. (i) X = {j, k, m} Therefore, X' = {l, n} Y = {k, m, n} Therefore, Y' = {j, l}
###### Step 2:
X' ∪ Y' = {l, n} ∪ {j, l} Therefore, X' ∪ Y' = {j, l, n} ……………….. (ii) Combining (i) and (ii) we get, (X ∩ Y)' = X' U Y' Hence, the De Morgan's Law is Proved
##### Example 2:
Let U = {1, 2, 3, 4, 5, 6, 7, 8}; P = {4, 5, 6} and Q = {5, 6, 8}. Prove De Morgan's Law (P ∪ Q)' = P' ∩ Q'.
###### Solution:
Here, U = {1, 2, 3, 4, 5, 6, 7, 8} P = {4, 5, 6} Q = {5, 6, 8}
###### Step 1:
P ∪ Q = {4, 5, 6} ∪ {5, 6, 8} = {4, 5, 6, 8} Therefore, (P ∪ Q)' = {1, 2, 3, 7} ……………….. (i) P = {4, 5, 6} Therefore, P' = {1, 2, 3, 7, 8} Q = {5, 6, 8} Therefore, Q' = {1, 2, 3, 4, 7}
###### Step 2:
P' ∩ Q' = {1, 2, 3, 7, 8} ∩ {1, 2, 3, 4, 7} Therefore, P' ∩ Q' = {1, 2, 3, 7} ……………….. (ii) Combining (i) and (ii) we get; (P ∪ Q)' = P' ∩ Q' Hence, De Morgan's Law is proved. |
# GEOMETRY 101* EVERYTHING YOU NEED TO KNOW ABOUT GEOMETRY TO PASS THE GHSGT!
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1 GEOMETRY 101* EVERYTHING YOU NEED TO KNOW ABOUT GEOMETRY TO PASS THE GHSGT!
2 FINDING THE DISTANCE BETWEEN TWO POINTS DISTANCE FORMULA- (x₂-x₁)²+(y₂-y₁)² Find the distance between the points ( -3,2) and (1,0). TO SOLVE: USE THE GLADE METHOD! BE SURE TO SIMPLIFY YOUR ANSWER. (1- -3)² + ( 0 2)² = (4)² + (-2)² = = 20 = 2 5
3 FINDING THE MIDPOINT OF TWO POINTS MIDPOINT FORMULA: ( (x₂+x₁)/2,(y₂+y₁)/2) REMEMBER THAT THE MIDPOINT IS AN ORDERED PAIR YOU ARE FINDING THE AVERAGE OF THE TWO ENDPOINTS! FIND THE MIDPOINT OF THE SEGMENT WITH ENDPOINTS ( 1,2) AND (7,11). M = ( (1 + 7)/2, (2 + 11)/2) = (8/2, 13/2) = (4,13/2) = ( 4,6.5)
4 PYTHAGOREAN THEOREM FORMULA: (LEG)² + (LEG)² = (HYPOTENUSE)² NOT ALL DRAWINGS WILL ALLOW YOU TO USE THE FORMULA a² + b² = c² YOU WILL USE THE PYTHAGOREAN THEOREM WHEN YOU ARE FINDING THE SIDES OF RIGHT TRIANGLES.
5 PARALLEL LINES ANGLES BETWEEN THE PARALLEL LINES ARE CALLED INTERIOR ANGLES ANGLES OUTSIDE THE PARALLEL LINES ARE CALLED EXTERIOR ANGLES PARALLEL LINES FORM THE FOLLOWING ANGLES: CORRESPONDING ANGLES, ALTERNATE INTERIOR ANGLES, ALTERNATE EXTERIOR ANGLES, AND SAME SIDE INTERIOR ANGLES.
6 Find the measures of angles 1,2,3,4,5,6,and 7. a b a b t ANGLE 1 = 75, ANGLE 2 = 105, ANGLE 3 = 75, ANGLE 4 = 105 ANGLE 5 = 105, ANGLE 6 = 75, ANGLE 7 = 75
7 Find the value of x. a is parallel to b and is cut by transversal t. 2x + 30 b 9x + 2 a t Since the angles are interior on the same side of the transversal, they are Supplementary angles, so 2x x + 2 = 180, 11x +32 = x = 148 X = 13.45
8 Find the value of x and the measure of angle 1. 3x 1 48 Angle 1, 48 and 90 must equal 180, so angle 1 = 42. Since the lines are parallel, the angles measuring 3x and 48 are alternate interior angles, so they must be equal, so 3x = 48 and X = 16.
9 LOGIC IF - HYPOTHESIS P THEN CONCLUSION Q CONVERSE: Q P INVERSE: NEGATE THE IF THEN STATEMENT ~P ~Q CONTRAPOSITIVE: NEGATE THE CONVERSE ~Q ~P
10 POLYGONS NAMED BY THE NUMBER OF SIDES 3 SIDES TRIANGLE 4 SIDES QUADRILATERAL ( AND THE PSECIAL QUADRILATERALS KITE, RHOMBUS, SQUARE, RECTANGLE, PARALLELOGRAM, TRAPEZOID, ISOSCELES TRAPEZOID) 5 SIDES PENTAGON 6 SIDES - HEXAGON
11 There are 540 in the Interior angles of a pentagon. Add the angles and subtract From = 76
12 Since there are 540 in the interior angles of a pentagon, each angle in the Pentagon is 108. The radius bisects the angle, so the base angles of the triangle formed are each 54. There are 180 in a triangle, so 180 ( ) = 72
13 There are 720 in a regular hexagon. Each interior angle measures 120. The radius bisects the angle into two 60 angles, so each triangle is equiangular (all angles measure 60 ). If all angles are the same measures, then all sides are the same measure. AG =.05, Angle AGF = 60, FC = 1, FC = AD = EB, angle GBC = 60, and Angle EGC = 120
14 Using the Pythagorean Theorem, we can find the leg of the right Triangle. It is 3. There is another right triangle formed on the right side of the Diagram, so the base is = 14. The perimeter add all sides = 32 inches.
15 SIMILAR POLYGON POLYGONS ARE SIMILAR IS THEIR CORRESPONDING ANGLES ARE CONGRUENT AND THEIR CORRESPONDING SIDES ARE PROPORTIONAL. TRIANGLE ABC IS SIMILAR TO TRIANGLE DEF. WRITE THE CORRESPONDCE RELATIONSHIP. ANGLE A IS CONGRUENT TO ANGLE D. ANGLE B IS CONGRUENT TO ANGLE E. ANGLE C IS CONGRUENT TO ANGLE F. AB /DE = BC/EF = AC/DF
16 TRIANGLE ABC ~ TRIANGLE DEF. FIND X AND Y. E 10 B 8 15 X A 12 C AB/DE = BC = EF = AC = DF D Y F 15/10 = X/8, SO X = 12 AND 10/15 = 12/Y, SO Y = 18
17 FIND x. 30 x /10 = X = 12, SO X = 36
18 Two regular pentagons have sides 8cm and 3 cm. Find the ratio of the area of the larger to the area of the smaller. HINT: the ratio of the areas of regular polygons is the same as the square of the sides. 3cm 8² /3² = 64/9 8cm
19 SPECIAL RIGHT TRIANGLES S=H/2 L = S 3 H = 2S L 30 H 60 S
20 SPECIAL RIGHT TRIANGLES H = L 2 L H L
21
22
23 QUADRILATERALS MEMORIZE THE PROPERTIES OF QUADRILATERALS AND OTHER QUADRILATERALS FROM THE GHSGT INFORMATION SHEET. HINT: ALWAYS DRAW A PICTURE.
24 FIND THE MEASURE OF ANGLE B. ANGLE D MEASURES 6X + 20 AND ANGLE A A MEASURES 2X. D B C Since the angles are consecutive interior angles, they are supplementary, so 6x x = 180, 8x + 20 = 180, 8x = 160, and x = 20. Angle B is congruent to Angle D, angle D is 140 and angle B is also 140
25 SPHERES SA = 4 R² AND V = 4/3 R³ BALL DIAMETER ( CM) BASKETBALL 23.9 SOCCER BALL 12.7 RADIUS ( CM) BASEBALL SURFACE AREA ( CM²) TENNIS BALL SOFTBALL 11.4 GOLF BALL 2.2 VOLUME (CM³)
26 SLOPE, PARALLEL AND PERPENDICULAR LINES SLOPE FORMULA: m= (Y₂-Y₁)/(X₂-X₁) PARALLEL LINES HAVE THE SAME SLOPE PERPENDICULAR LINES ARE NEGATIVE RECIPROCALS OF EACH OTHER. IF THE SLOPE OF A LINE IS 2, THEN THE SLOPE OF THE LINE PERPENDICUALR TO IT IS -1/2.
27 CIRCLES Since we have tangents, we Know that they must be congruent, So the perimeter is the sum of the Sides of the triangle, = 30 Arc AB = 72. Drop the Hypotenuse, and you have 2 Right triangles. Use sin36= 4/x to find the Radius is 6.8.
28 An inscribed angle is half the length of the Intercepted arc, so angle B is 70
29 TRIANGLES Triangles are defined by their sides and angles Sides: all sides congruent equilateral, 2 sides congruent isosceles, no sides congruent scalene Angles: all angles congruent equiangular, one right angle right, one obtuse angle obtuse, angles less than 90 - acute Given the length of two sides of a triangle and asked to find the possible lengths of the third side, add the measure together to get the maximum length of the third side, and subtract the lengths to get the minimumlength of the third side.
30 EXTERIOR ANGLE OF A TRIANGLE THE EXTERIOR ANGLES OF A TRIANGLE IS EQUAL TO THE SUM OF THE REMOTE INTERIOR ANGLES. 3X X + 5 X + 15
31 POINTS OF CONCURRENCY CENTROID POINT WHERE THE MEDIANS MEET INCENTER POINTS WHERE THE ANGLE BISECTORS MEET CIRCUMCENTER POINT WHERE THE PERPENDICULAR BISECTORS MEET ORTHOCENTER POINT WHERE THE ALTITUDES MEET
32 Answers to practice test 1.C 2.B 3.D 4.B 5.B 6.D 7.C 8.D 9.C 10.B 11.A 12.D 13.C 14.D 15.A 16.D 17.D 18.C 19.C 20.B 21.A 22.C 23.D 24.A 25.B 26.B 27.B 28.B 29.D 30.D 31.C 32.C 33.B 7.B 8.B 9.D
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Question Video: Finding the Derivative of a Function Defined by an Integral Where Its Limits Contain a Variable
Definite Integral
Definite Integral
Video Transcript
Find the derivative of the function
𝑔 of 𝑥 is equal to the definite integral from three 𝑥 to four 𝑥 of 𝑢 squared
minus three all divided by 𝑢 squared plus five with respect to 𝑢.
We’re given a function 𝑔 of
𝑥. And we can see this is a definite
integral where both of the limits of integration are functions in 𝑥. And we’re asked to find the
derivative of this function. Since we’re trying to differentiate
a function where the limits of integration are functions in 𝑥, we’ll try and do
this by using the fundamental theorem of calculus.
So we’ll start by recalling the
following part of the fundamental theorem of calculus. If lowercase 𝑓 is a continuous
function on a closed interval from 𝑎 to 𝑏 and we have capital 𝐹 of 𝑥 is the
definite integral from 𝑎 to 𝑥 of lowercase 𝑓 of 𝑢 with respect to 𝑢, then
capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 of 𝑥 for all values of 𝑥 in
the open interval from 𝑎 to 𝑏. In other words, this part of the
fundamental theorem of calculus gives us a way of differentiating a function defined
by an integral.
to differentiate in the question, 𝑔 of 𝑥, is very different to our definition of
capital 𝐹 of 𝑥. First, the lower limit of
integration is supposed to be a constant. However, we’re given this as a
function in 𝑥. There’s a similar story for our
upper limit of integration. It’s supposed to be 𝑥, but we’re
given a function in 𝑥. However, we can get around both of
these problems by using what we know about definite integrals and the chain
rule.
The first thing we’re going to need
to recall is one of our rules for definite integrals. To get around the problem of not
having a constant limit of integration, we’re going to want to split our definite
integral into two definite integrals. We know the following rule for
definite integrals. The integral from 𝑎 to 𝑏 of 𝑓 of
𝑢 with respect to 𝑢 is equal to the integral from 𝑐 to 𝑏 of 𝑓 of 𝑢 with
respect to 𝑢 plus the integral from 𝑎 to 𝑐 of 𝑓 of 𝑢 with respect to 𝑢. But there’s some important
The most important of these is this
will only be true if our function 𝑓 is integrable on all of these domains of
integration. In particular, our function 𝑓 must
be integrable on the closed interval from 𝑏 to 𝑐 and the closed interval from 𝑐
to 𝑎. We want to use this to rewrite our
function 𝑔 of 𝑥. So by using the upper limit of
integration as four 𝑥, the lower limit of integration as three 𝑥, and our function
lowercase 𝑓 of 𝑢 as 𝑢 squared minus three divided by 𝑢 squared plus five, we
would get the integral from 𝑐 to four 𝑥 of 𝑢 squared minus three divided by 𝑢
squared plus five with respect to 𝑢 plus the integral from three 𝑥 to 𝑐 of 𝑢
squared minus three all divided by 𝑢 squared plus five with respect to 𝑢.
But remember, we also need our
function to be integrable on both of these domains of integration. In particular, we need to choose a
value of the constant 𝑐. In this case, we could actually
choose any value of the constant 𝑐. To explain why, let’s take a look
at our integrand. Our integrand is the quotient of
two polynomials. This means it’s a rational
function. And we know rational functions will
be continuous across their entire domain. And if a function is continuous on
an interval, then it’s also integrable on that interval. So all we need to do is check the
domain of our function lowercase 𝑓 of 𝑢. It’s a rational function, so it
will be defined for all values of 𝑢 except where its denominator is equal to
zero.
And if we solve the denominator 𝑢
squared plus five is equal to zero, we’ll see that we get no real solutions. So our integrand is defined for all
real values of 𝑢. Meaning it’s continuous for all
real values of 𝑢, meaning it’s integrable on any interval. So in this case, we can actually
pick any value of 𝑐. However, this won’t always be the
case. In this case, we’ll choose 𝑐 is
equal to zero. We’ve now split this into two
definite integrals, each of which is closer to being able to use the fundamental
theorem of calculus.
There’s a little bit more
manipulation we need to do before we can use the fundamental theorem of
calculus. In our second integral, we want to
swap the order of our limits of integration. Remember, we can do this by
multiplying the entire integral by negative one. Doing this, we’ve rewritten our
second integral as negative one times the definite integral from zero to three 𝑥 of
𝑢 squared minus three all divided by 𝑢 squared plus five with respect to 𝑢.
And now, we’re almost ready to use
the fundamental theorem of calculus to evaluate these definite integrals. The only problem is our upper limit
of integration is a function in 𝑥 instead of 𝑥. And there are two ways of getting
around this problem. First, we could use the
substitution 𝑣 of 𝑥 is equal to four 𝑥. Then we can differentiate our first
integral by using the chain rule. We could do the same with our
second integral, setting 𝑣 of 𝑥 to be three 𝑥 and then evaluating its derivative
by using the chain rule. And this would work. However, we can also do this in the
general case with the fundamental theorem of calculus.
We have if 𝑣 of 𝑥 is a
differentiable function, then by using the chain rule, we get if lowercase 𝑓 is a
continuous function on the closed interval from 𝑎 to 𝑏 and capital 𝐹 of 𝑥 is the
definite integral from 𝑎 to 𝑣 of 𝑥 of lowercase 𝑓 of 𝑢 with respect to 𝑢, then
capital 𝐹 prime of 𝑥 will be equal to lowercase 𝑓 evaluated at 𝑣 of 𝑥 times d𝑣
by d𝑥 as long as 𝑣 of 𝑥 is in the open interval from 𝑎 to 𝑏. So we were able to rewrite our
version of the fundamental theorem of calculus so long as 𝑣 of 𝑥 is a
differentiable function just by using the chain rule.
And now, we can see both of our
definite integrals are in this form. We’ll set 𝑣 one of 𝑥 to be four
𝑥 and 𝑣 two of 𝑥 to be three 𝑥. Usually, we would set our value of
𝑎 equal to zero in both cases and then check the intervals on which our integrand
is continuous. This is because to use the
fundamental theorem of calculus, we do need to check where our integrand is
continuous. However, we’ve already shown that
our integrand is continuous for all real values. Therefore, it will be continuous on
any closed interval.
We’re now ready to find an
expression for 𝑔 prime of 𝑥. We’ll differentiate each of our
integrals separately. To differentiate our first integral
by using the fundamental theorem of calculus, we do need to check that 𝑣 one of 𝑥
is differentiable. And it’s a linear function, so we
know it is differentiable. This means we can use this, and we
get 𝑓 evaluated at 𝑣 one of 𝑥 times d𝑣 one by d𝑥. And we get the same story in our
second integral. This time, 𝑣 two of 𝑥 will be
three 𝑥, which is also a linear function, so it’s differentiable. This means we need to subtract 𝑓
evaluated at 𝑣 two of 𝑥 times d𝑣 two by d𝑥.
So now, we found an expression for
𝑔 prime of 𝑥. And we can actually evaluate all of
these expressions. Let’s start by substituting in our
expressions for 𝑣 one of 𝑥 and 𝑣 two of 𝑥. This gives us 𝑔 prime of 𝑥 is
equal to lowercase 𝑓 evaluated at four 𝑥 times the derivative of four 𝑥 with
respect to 𝑥 minus lowercase 𝑓 evaluated at three 𝑥 multiplied by the derivative
of three 𝑥 with respect to 𝑥. And remember, from the fundamental
theorem of calculus, lowercase 𝑓 will be our integrand. So we’re now ready to find an
expression for 𝑔 prime of 𝑥.
First, to find 𝑓 evaluated at four
𝑥, we substitute 𝑢 is equal to four 𝑥 into our integrand, giving us four 𝑥 all
squared minus three all divided by four 𝑥 all squared plus five. Next, we need to multiply this by
the derivative of four 𝑥 with respect to 𝑥. Of course, this is a linear
function, so we know the derivative of this with respect to 𝑥 will be the
coefficient of 𝑥, which is four. And we can do exactly the same to
find an expression for our second term.
To find 𝑓 evaluated at three 𝑥,
we substitute 𝑢 is equal to three 𝑥 into our integrand, giving us three 𝑥 all
squared minus three all divided by three 𝑥 all squared plus five. And of course, we need to multiply
this by the derivative of three 𝑥 with respect to 𝑥, which we know is three. Finally, we’ll simplify this
expression by evaluating all of our exponents and multiplying through by our
coefficients. And by doing this, we get our final
answer. 𝑔 prime of 𝑥 is equal to four
times 16𝑥 squared minus three all divided by 16𝑥 squared plus five minus three
times nine 𝑥 squared minus three all divided by nine 𝑥 squared plus five.
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# Arithmetic Progression
## NCERT Exercise 5.1
### Part 2
Question: 2 – Write first four terms of AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
Solution: Here, first term a1 = 10 and common difference, d = 10
Hence, second term a_2 = a_1 + d = 10 + 10 = 20
Third term a_3 = a_1 + 2d = 10 + 2 xx 10 = 30
Fourth term a_4 = a_1 + 3d = 10 + 30 = 40
Thus, first four terms of the AP will be 10, 20, 30, 40, ……
(ii) a = - 2, d = 0
Solution: First term a = 1
Common difference = 0
Thus, first four terms of given AP will be
a_1 = - 2, a_2 = - 2, a_3 = - 2 and a_4 = - 2
(iii) a = 4, d = - 3
Solution: Here, first term a1 = 4 and common difference d = - 3
We know that a_n = a + (n – 1)d, where n = number of terms
Thus, second term a_2 = a + (2 – 1)d
Or,a_2 = 4 + (2-1)xx (-3)
= 4 - 3 = 1
Similarly, third term a_3 = a + (3 – 1)d
a_3 = 4 + (3-1) xx (-3)
= 4 - 6 = -2
Fourth term a_4= a + (4-1)d
a_4 = 4 + (4 - 1) xx( -3)
= 4 - 9 = -5
Thus, the first four terms of given AP are; 4, 1, - 2, - 5
(iv) a = - 1, d = ½
Solution: Solution: We have, first term = - 1 and d = ½
a_2=a+(2-1)d
=-1+1/2=-1/2
a_3=a+2d
=-1+2xx1/2=0
a_4=a+3d
=-1+3xx1/2=1/2
Thus, the four terms are; - 1, -1/2, 0 and ½
(v) a = - 1.25, d = - 0.25
Solution: We have; first terms = - 1.25 and d = - 0.25
a_2= a + d
= -1.25 - 0.25 = -1.5
a_3 = a + 2d
= -1.25 + 2 xx (-0.25)
= -1.25 - 0.5 = -1.75
a_4 = a + 3d
= 1.25 + 3 × (-0.25) = -2.25
Thus, the first four terms are; - 1.25, -1.5, -1.75 and – 2.25
Question: 3 – For the following APs, write the first term and common difference,
(i) 3, 1, -1, - 3, …….
Solution: Here, first term a = 3
Common difference can be calculated as follows:
a_4 – a_3 = - 3 – ( -1) = - 3 + 1 = - 2
a_3 – a_2 = - 1 – 1 = - 2
a_2 – a_1 = 1 – 3 = - 2
Here, a(k+1) – a_k = - 2 for all values of k
Hence, first term = 3 and common difference = - 2
(ii) – 5, - 1, 3, 7
Solution: a_4 – a_3 = 7 – 3 = 4
a_3 – a_2 = 3 – (-1) = 3 + 1 = 4
a_2 – a_1 = - 1 – (-5) = - 1 + 5 = 5
Here, a_(k+1) – a_k = - 2 for all values of k
Hence, first term = - 5 and common difference = 4
Question (iii): 1/3, 5/3, 9/3, (13)/(3), ------
Solution:
a_4-a_3=(13)/(3)-9/3=4/3
a_3-a_2=9/3-5/3=4/3
a_2-a_1=5/3-1/3=4/3
Here, a_(k+1) – a_k = - 2 for all values of k
Hence, first term = 1/3 and common difference = 4/3
(iv) 0.6, 1.7, 2.8, 3.9, …….
Solution: a_4 – a_3 = 3.9 – 2.8 = 1.1
a_3 – a_2 = 2.8 – 1.7 = 1.1
a_2 – a_1 = 1.7 – 0.6 = 1.1
Here, a_(k+1) – a_k = - 2 for all values of k
Hence, first term = 0.6 and common difference = 1.1 |
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### Solve this viral facebook math puzzle image – Math Puzzles only for geniuses with the correct answer!
Hello, guys! Here is another interesting puzzle for you. Just a few shapes are given, calculate the values of these shapes (star polygon, clock, and table fan) and find the answer. That’s it!
Find the values of Star Polygon, Table Fan, and Clock and then solve the last equation. Can you figure it out?
Puzzle Question:
⇒ Star + Star + Star = 18
⇒ Fan + Fan + Star = 14
⇒ Clock + Clock – Fan = 02
⇒ Fan + Clock × Star = ??
## Viral Math Puzzle Image:
Did you get it? Share your answer in the comment section below or keep scrolling for the solution below;
.
A
N
S
W
E
R
.
### Puzzle Video with Solution
[toggle title=”Answer & Solution”]
### First Equation:
Star + Star + Star = 18
3 Star = 18 ⇒ Star = 6
### Second Equation:
Fan + Fan + Star = 14
2 Fan + Star = 14
2 Fan = 14 – Star ⇒ 2 Fan = 14 – 6 ⇒
2 Fan = 8 ⇒ Fan = 8/2
Table Fan = 4
### Third Equation:
Clock + Clock – Fan = 02
2 Clock – Fan = 2
2 Clock = 2 + Fan
2 Clock = 2 + 4
2 Clock = 6 ⇒ Clock = 6/2 ⇒ Clock = 3
Note:
Now, you have to figure out the difference between the last equation vs other equations. The time is different in the last clock vs the previous clock i.e 3 o’clock in the 3rd equation but in the final equation, it is 2 o’clock. The star shape has 6 sides but in the last line, it has only 5 sides. Similarly, table fan has 3 blades in the last line than 4.
Therefore the value of Table Fan, Clock and Star for the fourth equation become:
Star (with 6 sides) = 6 ⇒ Star (with 5 sides) = 5
Clock (3 O’clock) = 3 ⇒ Clock (2 O’clock) = 2
Table Fan (With 4 Blades) = 4 ⇒ Table Fan (With 3 Blades) = 3
### Fourth (Last) Equation:
⇒ Fan + Clock × Star = ??
So, the Table Fan with 3 blades, Clock with 2 O’clock, and Star with 5 sides (Don’t forget, 2 stars are given). Therefore our last equation become;
⇒ Fan (3 Blades) + Clock (2 O’clock) × 2 [Star (5 Sides)] = ??
⇒ 3 + 2 × 2(5)
⇒ 3 + 2 × 10
⇒ 3 + 20
23 Answer.
[/toggle]
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#### 16 COMMENTS
1. ⇒ Star + Star + Star = 18
⇒ Fan + Fan + Star = 14
⇒ Clock + Clock – Fan = 02
⇒ 3 plate Fan + 2 o Clock × 2Star = ??
3+2*12=60
2. it’s 50. if u do the math. the star equals 6 AND has 6 points. The WindMill equals 4 AND has 4 blades. The clock equals 3 AND is pointing at 3:00. But in the last question the star is doubled and has 5 points, the windmill has 3 blades, and the clock is pointing at 2:00. SO the last question is 3 (the windmill) plus 2 (the clock) times 5 (The stars) which equals 50.
• No actually, you’re forgetting the order of operations, Brackets, Exponents, Division/Multiplication Addition/Subtraction
So if you do the math it equals
Fan+clock*stars=3+2*10=23
3+20
23
Don’t forget BEDMAS!! |
RD Sharma Class 12 Exercise 12 MCQ Derivative as a rate measure Solutions Maths - Download PDF Free Online
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# RD Sharma Class 12 Exercise 12 MCQ Derivative as a rate measure Solutions Maths - Download PDF Free Online
Edited By Lovekush kumar saini | Updated on Jan 20, 2022 07:23 PM IST
RD Sharma Class 12 Solutions Derivative as a Rate Measure Chapter 12 MCQs is an expert-designed coursebook for CBSE Class 12 students. The solutions book will contain answers to all questions in the latest edition of NCERT maths books for the CBSE board. Students should use the RD Sharma Class 12th MCQs Solutions while solving problems from the maths book so that they can check the accuracy of their answers. Practising the book thoroughly before the board exams are highly recommended as it will improve your chances of scoring well and enhance your problem-solving abilities.
Also Read - RD Sharma Solution for Class 9 to 12 Maths
## Derivative as a Rate Measurer Excercise: 12 MCQ
Derivative as a Rate Measurer exercise multiple choise questions question 1
$B.\; 4\pi$
Hint:
Here, we will use the formula,
$v=\frac{4}{3}\pi r^{3}$
Given:
Here,
$r=10\; \; and \; \; \frac{dr}{dt}=0.01$
Solution:
$\begin{gathered} v=\frac{4}{3} \pi r^{3} \\ \Rightarrow \frac{d v}{d t}=4 \pi r^{2} \frac{d r}{d t} \\ \end{gathered}$
$\Rightarrow \text { Substituting values of } r=10 \quad \text { and } \frac{d r}{d t}=0.01$
\begin{aligned} &\text { We get }\\ &\frac{d v}{d t}=4 \pi \times 10^{2} \times 0.01=4 \pi \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 2
$B. \quad 10\sqrt{3}\; cm^{2}/sec$
Hint:
The area of an equilateral triangle side is
$A=\frac{\sqrt{3}}{4}a^{2} \quad \quad \quad \quad.....(i)$
Given:
$a=10 \text { and } \frac{d a}{d t}=2 \mathrm{~cm} / \mathrm{sec}$
Solution:
Differentiating (i) with respect to t
We get
$\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a \frac{d a}{d t}$
→ Substituting values of
$a=10 \text { and } \frac{d a}{d t}=2 \mathrm{~cm} / \mathrm{sec}$
We get
\begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{4} \times 2 a\times \frac{d a}{d t} \\ &\quad=\frac{\sqrt{3 }a}{2} \times \frac{d a}{d t} \\ &\text { When } a=10 \\ &\frac{d A}{d t}=\frac{10 \sqrt{3}}{2} \times 2=10 \sqrt{3} \mathrm{~cm}^{2} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 3
$(c)$
Hint:
Here, we use the formula and concept of volume of sphere
Given:
$\frac{d r}{d t}=0.1 \mathrm{~cm} / \mathrm{sec} \text { and } r=200 \mathrm{~cm}$
Solution:
The surface area of a sphere of radius r is defined by
\begin{aligned} &A(r)=4 \pi r^{2} \quad \ldots \ldots . . \text { (i) }\\ &\rightarrow \text { Differentiating (i) with respect to } \mathrm{t}\\ &\frac{d A}{d t}=8 \pi r \cdot \frac{d r}{d t}=8 \pi \times 200 \times 0.1=160 \pi \mathrm{cm}^{2} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 4
$0.002 \mathrm{~cm} / \mathrm{sec} \quad (D)$
Hint:
$V(r, h)=\frac{1}{3} \pi r^{2} h$
Given:
$h=2 r \text { and } \frac{d V}{d t}=40 \mathrm{~cm}^{3} / \mathrm{sec}$
Solution:
\begin{aligned} &V(r)=\frac{2}{3} \pi r^{3}\\ &\rightarrow \text { Differentiating (i) with respect to } t\\ &\frac{d V}{d t}=2 \pi r^{2} \frac{d r}{d t}\\ &40=2 \pi \times 10000 \times \frac{d r}{d t}\\ &\frac{d r}{d t}=0.002 \mathrm{~cm} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 5
$1m/minute$
Hint:
The volume of a cylinder, with radius r and height h is defined by
Given:
$V(r, h)=\pi r^{2} h$
Solution:
$Substituting\; r=0.5\; \mathrm{~m} \\ we\; get \\ Given\; that\; \frac{d V}{d t}=0.25 \pi \mathrm{m}^{3} / \mathrm{min}, \;we \;have \;to \;calculate \;\frac{d h}{d t} \\ \rightarrow Di\! f\! \! f\! erentiating \;(i)\; with\; respect\; to\; t$
\begin{aligned} &\frac{d V}{d t}=\pi r^{2} \frac{d h}{d t} \\ &\frac{d h}{d t}=\frac{1}{\pi r^{2}} \times \frac{d V}{d t} \\ &\frac{d h}{d t}=\frac{0.25 \pi}{\pi(0.5)^{2}} \\ &\frac{d h}{d t}=\frac{0.25 \pi}{0.25 \pi} \\ &\frac{d h}{d t}=1 \mathrm{~m} / \mathrm{min} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 6
$B.\; \; -42$
Hint:
We are given the distance travelled x is a function of time, we can calculate velocity V and acceleration $a$ by
$V t=\frac{d x}{d t} \text { and } a(t)=\frac{d^{2} x}{d t^{2}}$
Given:
Here, we use the formula,
$x=t^{3}-12 t^{2}+6 t+8$
Solution:
Differentiating w.r.t time we get
$V(t)=\frac{d x}{d t}=3 t^{2}-24 t+6$
Again Differentiating w.r.t time we get
\begin{aligned} &a(t)=\frac{d^{2} x}{d t^{2}}=6 t-24\\ &a=0 \Rightarrow 6 t-24=0 \text { or } t=4 \text { units }\\ &\text { So, Velocity at } t=4\\ &V(t)=\frac{d x}{d t}=3 t^{2}-24 t+6\\ &V(4)=3 \times 4^{2}-24 \times 4+6\\ &V(4)=(-42) \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 7
$\frac{d r}{d t}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec}$
Hint:
The relation between height h, radius r and semi vertical angle $a$ is defined by
$tan\: \alpha =\frac{r}{h}$
Given:
$\alpha=30^{\circ} \text { and } \frac{d \alpha}{d t}=2^{0} / \mathrm{sec}$
Solution:
Let, r be the radius, h be the height & α be the semi-vertical angle of cone
$tan\: \alpha =\frac{r}{h} \quad \quad \quad \quad......(i)$
$\Rightarrow We\; have\; to\; find\; \frac{d r}{d t} \\ \Rightarrow Di\! f\! \! ferentiating \; (i) \; with \; respect \; to \; \mathrm{t}, we \;get \\ \sec ^{2} \alpha \frac{d \alpha}{d t}=\frac{1}{20} \frac{d r}{d t}$
\begin{aligned} &\Rightarrow \text { Substituting values, We get }\\ &\begin{gathered} \sec ^{2} 30^{\circ} \times 2^{0}=\frac{1}{20} \frac{d r}{d t} \quad\quad\quad\quad\quad\left[\mathrm{~h}=20 \mathrm{~cm}, \alpha=30^{\circ} \text { and } \frac{d \alpha}{d t}=2^{0} / \mathrm{sec}\right] \\ \left(\sec ^{2} 30^{\circ}\right) \times 2^{\circ}=\frac{1}{20} \frac{d r}{d t} \\ \frac{4}{3} \times 2=\frac{1}{20} \frac{d r}{d t} \\ \frac{d r}{d t}=\frac{160}{3} \mathrm{~cm} / \mathrm{sec} \end{gathered} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 8
$(d)\; \; 3,\; \; \frac{1}{3}$
Hint:
Here, we use the basic concept of algebra.
Given:
$x^{3}-5x^{2}+5x+8$
Solution:
$p(x)=x^{3}-5x^{2}+5x+8 \quad\quad\quad\quad......(i)$
→ Differentiating (i) with respect to t,we get
\begin{aligned} &\frac{d p(x)}{d t}=\left(x^{3}-5 x^{2}+5 x+8\right) \frac{d x}{d t}=2 \frac{d x}{d t}\\ &2 \frac{d x}{d t}=\left(3 x^{2}-10 x+5\right) \frac{d x}{d t}\\ &=3 x^{2}-10 x+5=2\\ &=3 x^{2}-10 x+3=0\\ &\rightarrow \text { Factorizing the above quadratic equation, }\\ &\text { we get }(3 x-1)(x-3)=0\\ &\Rightarrow x=\frac{1}{3} \text { and } x=3 \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 9
$\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right)$
Hint:
Here, we use the basic concept of algebra
Given:
$16x^{2}+9y^{2}=400$
Solution: Let
$E(x,y)=16x^{2}+9y^{2}=400$
Solving for y we get
$y=\pm \frac{\sqrt{400-16 x^{2}}}{3} \quad\quad \quad \quad .....(i)$
Given that
$\frac{d x}{d t}=-\frac{d y}{d t}$
we have to calculate
→ Differentiating (i) with respect to t,we get
\begin{aligned} \frac{d y}{d t} &=\pm \frac{1}{3} \times \frac{1}{2 \sqrt{400-46 x^{2}}} \times-32 \times \frac{d x}{d t} \\ &=\mp \frac{16 x}{3 \sqrt{400-16 x^{2}}} \frac{d x}{d t} \end{aligned}
→ Substituting values
We get
\begin{aligned} 16 x=\pm 3 \sqrt{400-16 x^{2}}\\ \end{aligned}
Squaring the equation, we get
\begin{aligned} &256 x^{2}=9\left(400-16 x^{2}\right) \end{aligned}
Solving the equation we get
$x^{2}=9 \\ x=\pm 3$
→ Substituting in (i), we get
\begin{aligned} &y=\pm \frac{\sqrt{400-16 \times 9}}{3}=\pm \frac{16}{3}\\ &\text { So, }\\ &\left(3, \frac{16}{3}\right) \text { and }\left(3,-\frac{16}{3}\right) \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 10
$A.\; \; 54\pi \; \; cm^{2}/min$
Hint:
The lateral surface area of cone, with radius r and height h is defined as
\begin{aligned} &L(r, h)=\pi r \sqrt{r^{2}+h^{2}} \quad \ldots \ldots \text { (i) }\\ \end{aligned}
Given:
\begin{aligned} &r=7 \mathrm{~cm}, h=24 \mathrm{~cm} \text { and } \frac{d r}{d t}=3 \mathrm{~cm} / \mathrm{min}, \frac{d h}{d t}=-4 \mathrm{~cm} / \mathrm{min}\\ \end{aligned}
Solution:
→ Differentiating (i) with respect to t,we get
\begin{aligned} &\frac{d l}{d t}=\pi\left(\sqrt{r^{2}+h^{2}} \frac{d r}{d t}+r \times \frac{1}{2 \sqrt{r^{2}+h^{2}}} \times\left(2 r \frac{d r}{d t}+2 h \frac{d h}{d t}\right)\right)\\ &\text { Substituting values }\\ &\frac{d l}{d t}=\pi\left(\sqrt{7^{2}+24^{2}} \times 3+7 \times \frac{1}{2 \sqrt{7^{2}+24^{2}}} \times(2 \times 7 \times 3+2 \times 24 \times-4)\right)\\ &\frac{d l}{d t}=54 \pi \mathrm{cm}^{2} / \mathrm{min} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 11
$B.\; \; 180\pi \; \; cm^{3}/min$
Hint:
Here we use formula of a sphere of radius r is defined by
\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad \quad \quad....(i) \end{aligned}
Given:
\begin{aligned} &r=15 \mathrm{~cm} \; \; \; \frac{d r}{d t}=0.2 \mathrm{~cm} / \mathrm{min} \end{aligned}
Solution:
→ Differentiating (i) with respect to t, we get
\begin{aligned} &\frac{d V}{d t}=4 \pi r^{2} \times \frac{d r}{d t} \\ &\frac{d V}{d t}=4 \pi \times 15^{2} \times 0.2=180 \pi \mathrm{cm}^{3} / \mathrm{min} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 12
$B.\; \; \frac{3}{16\pi }\; cm/sec$$B.\; \; \frac{3}{16\pi }\; cm/sec$
Hint:
Here we use formula of a sphere of radius r is defined by
\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad.....(i) \end{aligned}
Given:
\begin{aligned} &r=2 c m\; \; \frac{d V}{d t}=3 \mathrm{~cm}^{2} / \mathrm{sec} \end{aligned}
Solution:
→ Differentiating (i) with respect to t, we get
\begin{aligned} &\frac{d V}{d t}=\frac{4}{3} \pi r^{2} \times \frac{d r}{d t} \\ &3=4 \pi \times 2^{2} \times \frac{d r}{d t} \\ &\frac{d r}{d t}=\frac{3}{16 \pi} \mathrm{cm} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 13
$A. \: \: 9$
Hint:
We are given the distance travelled as a function of time, we can calculate velocity by
\begin{aligned} &V(t)=\frac{d s}{d t} \\ \end{aligned}
Given:
\begin{aligned} s=45 t+11 t^{2}+3 \end{aligned}
Solution:
→ Differentiating (i) with respect to t, we get
\begin{aligned} &V(t)=\frac{d s}{d t} =45 t+11 t^{2}+3 \end{aligned}
→ If the particle is at rest, then it’s velocity will be 0
→ Solving the quadratic equation we get
$t=\frac{-5}{3}, \quad t \neq 0 \quad \text { so, } t=9$
Derivative as a Rate Measurer exercise multiple choise question 14
$\frac{d r}{d t}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec}$
Hint:
The volume of sphere, the radius r is defined by
\begin{aligned} &V(r)=\frac{4}{3} \pi r^{3} \quad \quad \quad .....(i) \end{aligned}
Given:
\begin{aligned} V=288 \pi \mathrm{cm}^{3} \end{aligned}
Solution:
\begin{aligned} &288 \pi=\frac{4}{3} \pi r^{3}\\ &\text { Solving for } r^{r}, \text { we get } r=6 \mathrm{~cm} \end{aligned}
\begin{aligned} &\rightarrow \text { Given that } \frac{d V}{d t}=4 \pi \mathrm{cm}^{3} / \mathrm{sec}\\ &\rightarrow \text { Differentiating (i) with respect to t, we get } \end{aligned}
\begin{aligned} &\frac{d V}{d t}=\frac{4}{3} \pi r^{2} \times \frac{d r}{d t}\\ &\rightarrow \text { Substituting values, we get }\\ &4 \pi=4 \pi \times 6^{2} \times \frac{d r}{d t}\\ &\frac{d r}{d t}=\frac{1}{36} \mathrm{~cm} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 15
\begin{aligned} &\text { D. } \quad r=\frac{1}{2 \sqrt{n}} \end{aligned}
Hint:
The volume of sphere of radius r is defined by
$V(r)=\frac{4}{3}\pi r^{3}$
Given:
$\frac{dV}{dt}=\frac{dr}{dt}$
Solution:
\begin{aligned} &4 \pi r^{2} \times \frac{d r^{\prime}}{d t}=\frac{d r^{\prime}}{d t} \\ &4 \pi r^{2}=1 \end{aligned}
\begin{aligned} &r=\frac{1}{2\sqrt{n}}\: units \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 16
$B.\; r=\frac{1}{\pi }units$
Hint:
The area of circle of radius r is defined by
$A(r)=\pi r^{2}$
Given:
$\frac{d A}{d t}=\frac{2 d r}{d t}$
Solution:
We get
$2 \pi r \frac{d r}{d t}=\frac{2 d r}{d t}\\ \\ \pi r=1 \\ \\ r=\frac{1}{\pi} unite$
Derivative as a Rate Measurer exercise multiple choise question 17
$8\sqrt{3}\: cm^{2}/hr$
Hint:
The area of an equilateral triangle with side a, is defined as,
$A(a)=\frac{\sqrt{3}}{4} a^{2}$
Given:
$\frac{d a}{d t}=8 \mathrm{~cm} / \mathrm{hr} \\ \\ a=2 \mathrm{~cm} \\ W\! e\; have\; to\; calculate\; \frac{d A}{d t}$
Solution:
→ Differentiating (i) with respect to t, we get
\begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{2} a \frac{d a}{d t}\\ \end{aligned}
→ Substituting values, we get
\begin{aligned} &\frac{d A}{d t}=\frac{\sqrt{3}}{2} \times 2 \times 8=8 \sqrt{3} \mathrm{~cm}^{2} / h r \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 18
$D. \: \: -\frac{16}{3}\: unit/sec$
Hint:
We are given the distance travelled as a function of time, We can calculate velocity and acceleration by
\begin{aligned} &V t=\frac{d s}{d t} \text { and } a=\frac{d^{2} s}{d t^{2}} \\ \end{aligned}
Given:
\begin{aligned} s t=t^{3}-4 t^{2}+5 \end{aligned}
Solution:
→ Differentiating with respect to time, we get
\begin{aligned} &V t=\frac{d s}{d t}=3 t^{2}-8 t \end{aligned}
→ Differentiating again with respect to time, we get
\begin{aligned} &a(t)=\frac{d^{2} s}{d t^{2}}=6 t=8 \end{aligned}
\begin{aligned} &\rightarrow \text { Given that } a=0 \Rightarrow 6 t-8=0 \\ &\text { Or }=\frac{4}{3} \text { unit } / \mathrm{sec} \\ &V\left(\frac{4}{3}\right)=3 \times \frac{4^{2}}{3^{2}}-8 \times \frac{4}{3}=-\frac{16}{3} \text { unit } / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 19
\begin{aligned} &\frac{\pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec}\\ \end{aligned}
Hint:
Here
\begin{aligned} &V t=\frac{d s}{d t} \text { and } a=\frac{d^{2} s}{d t^{2}}\\ \end{aligned}
Given:
\begin{aligned} &t=2 t^{2}+\sin 2 t \end{aligned}
Solution:
→ Differentiating with respect to time, we get
\begin{aligned} &V t=\frac{d s}{d t}=4 t+2 \cos 2 t \quad \quad \quad .....(i) \end{aligned}
→ Differentiating again with respect to time, we get
\begin{aligned} &a(t)=\frac{d^{2} s}{d t^{2}}=4-4 \sin 2 t \end{aligned}
\begin{aligned} &\text { Given that } a=2 \Rightarrow 4-4 \sin 2 t=2 \\ &\text { Or } \sin 2 t-\frac{1}{2}=\sin \frac{\pi}{6} \\ &\rightarrow t=\frac{\pi}{12} \\ &V\left(\frac{\pi}{12}\right)=4 \times \frac{\pi}{12}+2 \cos \left(\frac{\pi}{6}\right) \Rightarrow \frac{\pi}{3}+\sqrt{3} \mathrm{~m} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 20
$0.24\pi \: cm^{2}/sec$
Hint:
The circumference of a circle
$A(r)=\pi r^{2} \quad \quad \quad \quad \quad....(i)$
Given:
\begin{aligned} &r=12 \mathrm{~cm} \\ &\frac{d r}{d t}=0.01 \mathrm{~cm} / \mathrm{sec} \end{aligned}
Solution:
\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } \mathrm{t} \text { , we get }\\ &\frac{d A}{d t}=2 \pi r \frac{d r}{d t}=2 \pi \times 12 \times 0.01\\ &=0.24 \pi \mathrm{cm}^{2} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 21
$\pi ^{2}\: cm^{2}/sec$
Hint:
The circumference of a circle
$A(r)=\pi r^{2} \quad \quad \quad \quad \quad .....(i)$
Given:
$r=\pi \mathrm{cm} \\ \frac{2 d r}{d t}=1 \mathrm{~cm} / \mathrm{sec}$
Solution:
→ Differentiating (i) respect to t, we get
$\frac{d A}{d t}=2 \pi r \frac{d r}{d t}=\pi \times \pi \times 1 \\ \\ =\pi^{2} \mathrm{~cm}^{2} / \mathrm{sec}$
Derivative as a Rate Measurer exercise multiple choise question 22
$Vs=3.2 km/hr$
Hint:
Here, we use basic concept of volume algebra
Given:
$hm=2m \quad Vm=48km/hr \quad h_{1}=5m$
Solution:
\begin{aligned} &\rightarrow \text { Consider } \Delta A E C \text { and } \Delta B E D\\ &\angle A E D=\angle B E D=\theta\\ &\angle E A C=\angle E B D=90^{\circ}\\ &\text { Therefore, } \Delta A E C \approx \Delta B E D \text { by AA criteria } \end{aligned}
\begin{aligned} &\frac{A C}{B D}=\frac{A E}{B E}\\ &\rightarrow \text { Substituting values, we get }\\ &\frac{5}{2}=\frac{4800+x}{x} \end{aligned}
\begin{aligned} &\rightarrow \text { Simplifying the equation }\\ &x=3200 \quad \quad......(i) \end{aligned}
\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } r^{r}, \text { we get }\\ &V \cdot s=\frac{d x}{d t}=3200 \mathrm{~m} / \mathrm{hr}=3.2 \mathrm{~km} / \mathrm{hr} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 23
$\frac{dx}{dt}=V\! s=6\: ft/sec$
Hint:
Here, we use basic concept of volume algebra
Given:
$hm=6\, f\! t \quad V\! m=9\, f\! t/sec \quad h_{1}=15\, f\! t$
Solution:
\begin{aligned} &\rightarrow \text { Consider } \Delta A E C \text { and } \Delta B E D \\ &\angle A E D=\angle B E D=\theta \\ &\angle E A C=\angle E B D=90^{\circ} \end{aligned}
\begin{aligned} &\text { Therefore, } \triangle A E C \approx \Delta B E D \text { by AA criteria }\\ &\frac{A C}{B D}=\frac{A E}{B E} \end{aligned}
\begin{aligned} &\rightarrow \text { Substituting values, we get }\\ &\frac{15}{6}=\frac{9 t+x}{x} \end{aligned}
\begin{aligned} &\rightarrow \text { Simplifying the equation }\\ &\frac{15}{6}=\frac{9 t}{x+x}\\ &x=6 t\\ &V \! s=\frac{d x}{d t}=6 \mathrm{ft} / \mathrm{sec} \end{aligned}
Derivative as a Rate Measurer exercise multiple choise question 24
C. Surface area times the rate of change of radius
Hint:
The volume and surface area of a sphere with radius r, is defined as,
Given:
$V(r)=\frac{4}{3} \pi r^{3} \quad \ldots \ldots \text { (i) } \quad \text { and } \quad A(r)=4 \pi r^{2}$
Solution:
\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } \mathrm{t}, \text { we get }\\ &V(r)=\frac{4}{3} \pi r^{2} \frac{d r}{d t}=A \times \frac{d r}{d t} \end{aligned}
(Surface area of sphere) X (rate of change of radius)
Derivative as a Rate Measurer exercise multiple choise question 25
D. 8$\pi$times the rate of change of radius.
Hint:
The surface area of a sphere with radius r is defined as
$A(r)=4\pi r^{2} \quad \quad \quad \quad .......(i)$
Given:
Solution:
\begin{aligned} &\rightarrow \text { Differentiating (i) respect to } \mathrm{t}, \text { we get }\\ &\frac{d A}{d t}=8 \pi r \frac{d r}{d t} \end{aligned}
$\rightarrow 8 \pi \times \text { current radius } \times \text { rate of change of radius }$
Derivative as a Rate Measurer exercise multiple choise question 26
\begin{aligned} &\text { A. } \frac{d h}{d t}=1 m / h r \end{aligned}
Hint:
The volume of a cylinder, of radius r and height h is defined as
\begin{aligned} &V(r, h)=\pi r^{2} h\\ \end{aligned}
Given:
\begin{aligned} &r=10 \mathrm{~m} \text { we get }\\ &V(h)=\pi \times 10^{2} \times h=100 \pi h \quad \quad \quad .....(i)\\ &\frac{d v}{d t}=314 m^{3} / h r \end{aligned}
Solution:
→ Differentiating (i) respect to t, we get
\begin{aligned} &\frac{d v}{d t}=100 \pi \frac{d h}{d t}\\ &\rightarrow \text { Substituting values and using } \quad \pi =\frac{3}{4} \text { we get } \end{aligned}
\begin{aligned} &314=100 \times 3.14 \frac{d h}{d t} \\ &\frac{d h}{d t}=1 \mathrm{~m} / \mathrm{hr} \end{aligned}
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4. What are the number of questions asked in the class 12 RD Sharma chapter 12, MCQs reference book?
There are 26 questions in RD Sharma Solutions RD Sharma Class 12th MCQs.
Yes, anyone who visit the Career 360 website can access the RD Sharma solution books. In the same way, the download option is also available to save the PDFs in their devices.
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##### Bio Medical Engineer
The field of biomedical engineering opens up a universe of expert chances. An Individual in the biomedical engineering career path work in the field of engineering as well as medicine, in order to find out solutions to common problems of the two fields. The biomedical engineering job opportunities are to collaborate with doctors and researchers to develop medical systems, equipment, or devices that can solve clinical problems. Here we will be discussing jobs after biomedical engineering, how to get a job in biomedical engineering, biomedical engineering scope, and salary.
4 Jobs Available
Database professionals use software to store and organise data such as financial information, and customer shipping records. Individuals who opt for a career as data administrators ensure that data is available for users and secured from unauthorised sales. DB administrators may work in various types of industries. It may involve computer systems design, service firms, insurance companies, banks and hospitals.
4 Jobs Available
##### Cartographer
How fascinating it is to represent the whole world on just a piece of paper or a sphere. With the help of maps, we are able to represent the real world on a much smaller scale. Individuals who opt for a career as a cartographer are those who make maps. But, cartography is not just limited to maps, it is about a mixture of art, science, and technology. As a cartographer, not only you will create maps but use various geodetic surveys and remote sensing systems to measure, analyse, and create different maps for political, cultural or educational purposes.
3 Jobs Available
##### Operations Manager
Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.
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##### GIS Expert
GIS officer work on various GIS software to conduct a study and gather spatial and non-spatial information. GIS experts update the GIS data and maintain it. The databases include aerial or satellite imagery, latitudinal and longitudinal coordinates, and manually digitized images of maps. In a career as GIS expert, one is responsible for creating online and mobile maps.
3 Jobs Available
##### Ethical Hacker
A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.
3 Jobs Available
##### Database Architect
If you are intrigued by the programming world and are interested in developing communications networks then a career as database architect may be a good option for you. Data architect roles and responsibilities include building design models for data communication networks. Wide Area Networks (WANs), local area networks (LANs), and intranets are included in the database networks. It is expected that database architects will have in-depth knowledge of a company's business to develop a network to fulfil the requirements of the organisation. Stay tuned as we look at the larger picture and give you more information on what is db architecture, why you should pursue database architecture, what to expect from such a degree and what your job opportunities will be after graduation. Here, we will be discussing how to become a data architect. Students can visit NIT Trichy, IIT Kharagpur, JMI New Delhi
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##### Data Analyst
The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.
Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.
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##### Bank Probationary Officer (PO)
A career as Bank Probationary Officer (PO) is seen as a promising career opportunity and a white-collar career. Each year aspirants take the Bank PO exam. This career provides plenty of career development and opportunities for a successful banking future. If you have more questions about a career as Bank Probationary Officer (PO), what is probationary officer or how to become a Bank Probationary Officer (PO) then you can read the article and clear all your doubts.
3 Jobs Available
##### Operations Manager
Individuals in the operations manager jobs are responsible for ensuring the efficiency of each department to acquire its optimal goal. They plan the use of resources and distribution of materials. The operations manager's job description includes managing budgets, negotiating contracts, and performing administrative tasks.
3 Jobs Available
##### Data Analyst
The invention of the database has given fresh breath to the people involved in the data analytics career path. Analysis refers to splitting up a whole into its individual components for individual analysis. Data analysis is a method through which raw data are processed and transformed into information that would be beneficial for user strategic thinking.
Data are collected and examined to respond to questions, evaluate hypotheses or contradict theories. It is a tool for analyzing, transforming, modeling, and arranging data with useful knowledge, to assist in decision-making and methods, encompassing various strategies, and is used in different fields of business, research, and social science.
3 Jobs Available
##### Finance Executive
A career as a Finance Executive requires one to be responsible for monitoring an organisation's income, investments and expenses to create and evaluate financial reports. His or her role involves performing audits, invoices, and budget preparations. He or she manages accounting activities, bank reconciliations, and payable and receivable accounts.
3 Jobs Available
##### Investment Banker
An Investment Banking career involves the invention and generation of capital for other organizations, governments, and other entities. Individuals who opt for a career as Investment Bankers are the head of a team dedicated to raising capital by issuing bonds. Investment bankers are termed as the experts who have their fingers on the pulse of the current financial and investing climate. Students can pursue various Investment Banker courses, such as Banking and Insurance, and Economics to opt for an Investment Banking career path.
3 Jobs Available
##### Bank Branch Manager
Bank Branch Managers work in a specific section of banking related to the invention and generation of capital for other organisations, governments, and other entities. Bank Branch Managers work for the organisations and underwrite new debts and equity securities for all type of companies, aid in the sale of securities, as well as help to facilitate mergers and acquisitions, reorganisations, and broker trades for both institutions and private investors.
3 Jobs Available
##### Treasurer
Treasury analyst career path is often regarded as certified treasury specialist in some business situations, is a finance expert who specifically manages a company or organisation's long-term and short-term financial targets. Treasurer synonym could be a financial officer, which is one of the reputed positions in the corporate world. In a large company, the corporate treasury jobs hold power over the financial decision-making of the total investment and development strategy of the organisation.
3 Jobs Available
##### Product Manager
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
3 Jobs Available
##### Transportation Planner
A career as Transportation Planner requires technical application of science and technology in engineering, particularly the concepts, equipment and technologies involved in the production of products and services. In fields like land use, infrastructure review, ecological standards and street design, he or she considers issues of health, environment and performance. A Transportation Planner assigns resources for implementing and designing programmes. He or she is responsible for assessing needs, preparing plans and forecasts and compliance with regulations.
3 Jobs Available
##### Construction Manager
Individuals who opt for a career as construction managers have a senior-level management role offered in construction firms. Responsibilities in the construction management career path are assigning tasks to workers, inspecting their work, and coordinating with other professionals including architects, subcontractors, and building services engineers.
2 Jobs Available
##### Carpenter
Carpenters are typically construction workers. They stay involved in performing many types of construction activities. It includes cutting, fitting and assembling wood. Carpenters may help in building constructions, bridges, big ships and boats. Here, in the article, we will discuss carpenter career path, carpenter salary, how to become a carpenter, carpenter job outlook.
2 Jobs Available
##### Welder
An individual who opts for a career as a welder is a professional tradesman who is skilled in creating a fusion between two metal pieces to join it together with the use of a manual or fully automatic welding machine in their welder career path. It is joined by intense heat and gas released between the metal pieces through the welding machine to permanently fix it.
2 Jobs Available
##### Environmental Engineer
Individuals who opt for a career as an environmental engineer are construction professionals who utilise the skills and knowledge of biology, soil science, chemistry and the concept of engineering to design and develop projects that serve as solutions to various environmental problems.
2 Jobs Available
##### Naval Architect
A Naval Architect is a professional who designs, produces and repairs safe and sea-worthy surfaces or underwater structures. A Naval Architect stays involved in creating and designing ships, ferries, submarines and yachts with implementation of various principles such as gravity, ideal hull form, buoyancy and stability.
2 Jobs Available
##### Welding Engineer
Welding Engineer Job Description: A Welding Engineer work involves managing welding projects and supervising welding teams. He or she is responsible for reviewing welding procedures, processes and documentation. A career as Welding Engineer involves conducting failure analyses and causes on welding issues.
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##### Field Surveyor
Are you searching for a Field Surveyor Job Description? A Field Surveyor is a professional responsible for conducting field surveys for various places or geographical conditions. He or she collects the required data and information as per the instructions given by senior officials.
2 Jobs Available
##### Orthotist and Prosthetist
Orthotists and Prosthetists are professionals who provide aid to patients with disabilities. They fix them to artificial limbs (prosthetics) and help them to regain stability. There are times when people lose their limbs in an accident. In some other occasions, they are born without a limb or orthopaedic impairment. Orthotists and prosthetists play a crucial role in their lives with fixing them to assistive devices and provide mobility.
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##### Pathologist
A career in pathology in India is filled with several responsibilities as it is a medical branch and affects human lives. The demand for pathologists has been increasing over the past few years as people are getting more aware of different diseases. Not only that, but an increase in population and lifestyle changes have also contributed to the increase in a pathologist’s demand. The pathology careers provide an extremely huge number of opportunities and if you want to be a part of the medical field you can consider being a pathologist. If you want to know more about a career in pathology in India then continue reading this article.
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##### Veterinary Doctor
A veterinary doctor is a medical professional with a degree in veterinary science. The veterinary science qualification is the minimum requirement to become a veterinary doctor. There are numerous veterinary science courses offered by various institutes. He or she is employed at zoos to ensure they are provided with good health facilities and medical care to improve their life expectancy.
5 Jobs Available
##### Gynaecologist
Gynaecology can be defined as the study of the female body. The job outlook for gynaecology is excellent since there is evergreen demand for one because of their responsibility of dealing with not only women’s health but also fertility and pregnancy issues. Although most women prefer to have a women obstetrician gynaecologist as their doctor, men also explore a career as a gynaecologist and there are ample amounts of male doctors in the field who are gynaecologists and aid women during delivery and childbirth.
4 Jobs Available
##### Oncologist
An oncologist is a specialised doctor responsible for providing medical care to patients diagnosed with cancer. He or she uses several therapies to control the cancer and its effect on the human body such as chemotherapy, immunotherapy, radiation therapy and biopsy. An oncologist designs a treatment plan based on a pathology report after diagnosing the type of cancer and where it is spreading inside the body.
3 Jobs Available
##### Surgical Technologist
When it comes to an operation theatre, there are several tasks that are to be carried out before as well as after the operation or surgery has taken place. Such tasks are not possible without surgical tech and surgical tech tools. A single surgeon cannot do it all alone. It’s like for a footballer he needs his team’s support to score a goal the same goes for a surgeon. It is here, when a surgical technologist comes into the picture. It is the job of a surgical technologist to prepare the operation theatre with all the required equipment before the surgery. Not only that, once an operation is done it is the job of the surgical technologist to clean all the equipment. One has to fulfil the minimum requirements of surgical tech qualifications.
3 Jobs Available
##### Critical Care Specialist
A career as Critical Care Specialist is responsible for providing the best possible prompt medical care to patients. He or she writes progress notes of patients in records. A Critical Care Specialist also liaises with admitting consultants and proceeds with the follow-up treatments.
2 Jobs Available
##### Ophthalmologist
Individuals in the ophthalmologist career in India are trained medically to care for all eye problems and conditions. Some optometric physicians further specialize in a particular area of the eye and are known as sub-specialists who are responsible for taking care of each and every aspect of a patient's eye problem. An ophthalmologist's job description includes performing a variety of tasks such as diagnosing the problem, prescribing medicines, performing eye surgery, recommending eyeglasses, or looking after post-surgery treatment.
2 Jobs Available
##### Actor
For an individual who opts for a career as an actor, the primary responsibility is to completely speak to the character he or she is playing and to persuade the crowd that the character is genuine by connecting with them and bringing them into the story. This applies to significant roles and littler parts, as all roles join to make an effective creation. Here in this article, we will discuss how to become an actor in India, actor exams, actor salary in India, and actor jobs.
4 Jobs Available
##### Acrobat
Individuals who opt for a career as acrobats create and direct original routines for themselves, in addition to developing interpretations of existing routines. The work of circus acrobats can be seen in a variety of performance settings, including circus, reality shows, sports events like the Olympics, movies and commercials. Individuals who opt for a career as acrobats must be prepared to face rejections and intermittent periods of work. The creativity of acrobats may extend to other aspects of the performance. For example, acrobats in the circus may work with gym trainers, celebrities or collaborate with other professionals to enhance such performance elements as costume and or maybe at the teaching end of the career.
3 Jobs Available
##### Video Game Designer
Career as a video game designer is filled with excitement as well as responsibilities. A video game designer is someone who is involved in the process of creating a game from day one. He or she is responsible for fulfilling duties like designing the character of the game, the several levels involved, plot, art and similar other elements. Individuals who opt for a career as a video game designer may also write the codes for the game using different programming languages. Depending on the video game designer job description and experience they may also have to lead a team and do the early testing of the game in order to suggest changes and find loopholes.
3 Jobs Available
##### Talent Agent
The career as a Talent Agent is filled with responsibilities. A Talent Agent is someone who is involved in the pre-production process of the film. It is a very busy job for a Talent Agent but as and when an individual gains experience and progresses in the career he or she can have people assisting him or her in work. Depending on one’s responsibilities, number of clients and experience he or she may also have to lead a team and work with juniors under him or her in a talent agency. In order to know more about the job of a talent agent continue reading the article.
If you want to know more about talent agent meaning, how to become a Talent Agent, or Talent Agent job description then continue reading this article.
3 Jobs Available
Radio Jockey is an exciting, promising career and a great challenge for music lovers. If you are really interested in a career as radio jockey, then it is very important for an RJ to have an automatic, fun, and friendly personality. If you want to get a job done in this field, a strong command of the language and a good voice are always good things. Apart from this, in order to be a good radio jockey, you will also listen to good radio jockeys so that you can understand their style and later make your own by practicing.
A career as radio jockey has a lot to offer to deserving candidates. If you want to know more about a career as radio jockey, and how to become a radio jockey then continue reading the article.
3 Jobs Available
##### Social Media Manager
A career as social media manager involves implementing the company’s or brand’s marketing plan across all social media channels. Social media managers help in building or improving a brand’s or a company’s website traffic, build brand awareness, create and implement marketing and brand strategy. Social media managers are key to important social communication as well.
2 Jobs Available
##### Choreographer
The word “choreography" actually comes from Greek words that mean “dance writing." Individuals who opt for a career as a choreographer create and direct original dances, in addition to developing interpretations of existing dances. A Choreographer dances and utilises his or her creativity in other aspects of dance performance. For example, he or she may work with the music director to select music or collaborate with other famous choreographers to enhance such performance elements as lighting, costume and set design.
2 Jobs Available
##### Talent Director
Individuals who opt for a career as a talent director are professionals who work in the entertainment industry. He or she is responsible for finding out the right talent through auditions for films, theatre productions, or shows. A talented director possesses strong knowledge of computer software used in filmmaking, CGI and animation. A talent acquisition director keeps himself or herself updated on various technical aspects such as lighting, camera angles and shots.
2 Jobs Available
##### Copy Writer
In a career as a copywriter, one has to consult with the client and understand the brief well. A career as a copywriter has a lot to offer to deserving candidates. Several new mediums of advertising are opening therefore making it a lucrative career choice. Students can pursue various copywriter courses such as Journalism, Advertising, Marketing Management. Here, we have discussed how to become a freelance copywriter, copywriter career path, how to become a copywriter in India, and copywriting career outlook.
5 Jobs Available
##### Editor
Individuals in the editor career path is an unsung hero of the news industry who polishes the language of the news stories provided by stringers, reporters, copywriters and content writers and also news agencies. Individuals who opt for a career as an editor make it more persuasive, concise and clear for readers. In this article, we will discuss the details of the editor's career path such as how to become an editor in India, editor salary in India and editor skills and qualities.
3 Jobs Available
##### Journalist
Careers in journalism are filled with excitement as well as responsibilities. One cannot afford to miss out on the details. As it is the small details that provide insights into a story. Depending on those insights a journalist goes about writing a news article. A journalism career can be stressful at times but if you are someone who is passionate about it then it is the right choice for you. If you want to know more about the media field and journalist career then continue reading this article.
3 Jobs Available
##### Publisher
For publishing books, newspapers, magazines and digital material, editorial and commercial strategies are set by publishers. Individuals in publishing career paths make choices about the markets their businesses will reach and the type of content that their audience will be served. Individuals in book publisher careers collaborate with editorial staff, designers, authors, and freelance contributors who develop and manage the creation of content.
3 Jobs Available
##### Vlogger
In a career as a vlogger, one generally works for himself or herself. However, once an individual has gained viewership there are several brands and companies that approach them for paid collaboration. It is one of those fields where an individual can earn well while following his or her passion. Ever since internet cost got reduced the viewership for these types of content has increased on a large scale. Therefore, the career as vlogger has a lot to offer. If you want to know more about the career as vlogger, how to become a vlogger, so on and so forth then continue reading the article. Students can visit Jamia Millia Islamia, Asian College of Journalism, Indian Institute of Mass Communication to pursue journalism degrees.
3 Jobs Available
##### Travel Journalist
The career of a travel journalist is full of passion, excitement and responsibility. Journalism as a career could be challenging at times, but if you're someone who has been genuinely enthusiastic about all this, then it is the best decision for you. Travel journalism jobs are all about insightful, artfully written, informative narratives designed to cover the travel industry. Travel Journalist is someone who explores, gathers and presents information as a news article.
2 Jobs Available
##### Videographer
Careers in videography are art that can be defined as a creative and interpretive process that culminates in the authorship of an original work of art rather than a simple recording of a simple event. It would be wrong to portrait it as a subcategory of photography, rather photography is one of the crafts used in videographer jobs in addition to technical skills like organization, management, interpretation, and image-manipulation techniques. Students pursue Visual Media, Film, Television, Digital Video Production to opt for a videographer career path. The visual impacts of a film are driven by the creative decisions taken in videography jobs. Individuals who opt for a career as a videographer are involved in the entire lifecycle of a film and production.
2 Jobs Available
##### SEO Analyst
An SEO Analyst is a web professional who is proficient in the implementation of SEO strategies to target more keywords to improve the reach of the content on search engines. He or she provides support to acquire the goals and success of the client’s campaigns.
2 Jobs Available
##### Product Manager
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
3 Jobs Available
##### Quality Controller
A quality controller plays a crucial role in an organisation. He or she is responsible for performing quality checks on manufactured products. He or she identifies the defects in a product and rejects the product.
A quality controller records detailed information about products with defects and sends it to the supervisor or plant manager to take necessary actions to improve the production process.
3 Jobs Available
##### Production Manager
Production Manager Job Description: A Production Manager is responsible for ensuring smooth running of manufacturing processes in an efficient manner. He or she plans and organises production schedules. The role of Production Manager involves estimation, negotiation on budget and timescales with the clients and managers.
3 Jobs Available
A Team Leader is a professional responsible for guiding, monitoring and leading the entire group. He or she is responsible for motivating team members by providing a pleasant work environment to them and inspiring positive communication. A Team Leader contributes to the achievement of the organisation’s goals. He or she improves the confidence, product knowledge and communication skills of the team members and empowers them.
2 Jobs Available
##### Quality Systems Manager
A Quality Systems Manager is a professional responsible for developing strategies, processes, policies, standards and systems concerning the company as well as operations of its supply chain. It includes auditing to ensure compliance. It could also be carried out by a third party.
2 Jobs Available
##### Merchandiser
A career as a merchandiser requires one to promote specific products and services of one or different brands, to increase the in-house sales of the store. Merchandising job focuses on enticing the customers to enter the store and hence increasing their chances of buying a product. Although the buyer is the one who selects the lines, it all depends on the merchandiser on how much money a buyer will spend, how many lines will be purchased, and what will be the quantity of those lines. In a career as merchandiser, one is required to closely work with the display staff in order to decide in what way a product would be displayed so that sales can be maximised. In small brands or local retail stores, a merchandiser is responsible for both merchandising and buying.
2 Jobs Available
##### Procurement Manager
The procurement Manager is also known as Purchasing Manager. The role of the Procurement Manager is to source products and services for a company. A Procurement Manager is involved in developing a purchasing strategy, including the company's budget and the supplies as well as the vendors who can provide goods and services to the company. His or her ultimate goal is to bring the right products or services at the right time with cost-effectiveness.
2 Jobs Available
##### Production Planner
Individuals who opt for a career as a production planner are professionals who are responsible for ensuring goods manufactured by the employing company are cost-effective and meets quality specifications including ensuring the availability of ready to distribute stock in a timely fashion manner.
2 Jobs Available
##### ITSM Manager
ITSM Manager is a professional responsible for heading the ITSM (Information Technology Service Management) or (Information Technology Infrastructure Library) processes. He or she ensures that operation management provides appropriate resource levels for problem resolutions. The ITSM Manager oversees the level of prioritisation for the problems, critical incidents, planned as well as proactive tasks.
3 Jobs Available
##### Information Security Manager
Individuals in the information security manager career path involves in overseeing and controlling all aspects of computer security. The IT security manager job description includes planning and carrying out security measures to protect the business data and information from corruption, theft, unauthorised access, and deliberate attack
3 Jobs Available
##### Computer Programmer
Careers in computer programming primarily refer to the systematic act of writing code and moreover include wider computer science areas. The word 'programmer' or 'coder' has entered into practice with the growing number of newly self-taught tech enthusiasts. Computer programming careers involve the use of designs created by software developers and engineers and transforming them into commands that can be implemented by computers. These commands result in regular usage of social media sites, word-processing applications and browsers.
3 Jobs Available
##### Product Manager
A Product Manager is a professional responsible for product planning and marketing. He or she manages the product throughout the Product Life Cycle, gathering and prioritising the product. A product manager job description includes defining the product vision and working closely with team members of other departments to deliver winning products.
3 Jobs Available
##### IT Consultant
An IT Consultant is a professional who is also known as a technology consultant. He or she is required to provide consultation to industrial and commercial clients to resolve business and IT problems and acquire optimum growth. An IT consultant can find work by signing up with an IT consultancy firm, or he or she can work on their own as independent contractors and select the projects he or she wants to work on.
2 Jobs Available
##### Data Architect
A Data Architect role involves formulating the organisational data strategy. It involves data quality, flow of data within the organisation and security of data. The vision of Data Architect provides support to convert business requirements into technical requirements.
2 Jobs Available
##### Security Engineer
The Security Engineer is responsible for overseeing and controlling the various aspects of a company's computer security. Individuals in the security engineer jobs include developing and implementing policies and procedures to protect the data and information of the organisation. In this article, we will discuss the security engineer job description, security engineer skills, security engineer course, and security engineer career path.
2 Jobs Available
##### UX Architect
A UX Architect is someone who influences the design processes and its outcomes. He or she possesses a solid understanding of user research, information architecture, interaction design and content strategy.
2 Jobs Available |
# ACT Math : How to divide integers
## Example Questions
### Example Question #1 : Operations
A car averages 29 miles per gallon. If gas costs $3.75 per gallon how much money would need to be spent on gas to travel 1464.5 miles? Possible Answers:$5491.88
$189.38 None of the other answers$50.5
$108.75 Correct answer:$189.38
Explanation:
The question is asking for the amount of money that would need to be spent on gas. Using the value given for the miles per gallon of the car and the amount of miles traveled it is possible to determine the gallons of fuel that will be required. This will be set up as 1464.5 miles travelled divided by the 29 miles per gallon that the car gets. This leaves us with 50.5 gallons of fuel used. From this point we can multiply the amount of fuel used, 50.5 gallons, by the price of fuel per gallon, $3.75, to obtain the amount of money that will spent on fuel,$189.38.
### Example Question #2 : How To Divide Integers
Connie's car gets 35 miles per gallon of gas. How much gas will she need to take a 525 mile trip? Round to the nearest gallon.
25
17
23
20
15
15
Explanation:
This becomes a division problem: 525 miles ÷ 35 miles per gallon = 15 gallons
### Example Question #3 : How To Divide Integers
A father buys a bag of marbles. He divides the marbles among his 5 children, who receive 24 marbles each. If there had been 6 children, how many marbles would each one get?
22
24
20
18
20
Explanation:
There were a total of 120 marbles to begin with, since 5 * 24 = 120.
If the marbles are split between 6 children, then each child gets 120/6 = 20 marbles each.
### Example Question #661 : Arithmetic
At the end of December, Abraham adds the average number of stamps he has been collecting over the year to his album, how many stamps will he have at the end of December? |
# Section 1.5: Problem 1 Solution
Working problems is a crucial part of learning mathematics. No one can learn... merely by poring over the definitions, theorems, and examples that are worked out in the text. One must work part of it out for oneself. To provide that opportunity is the purpose of the exercises.
James R. Munkres
Let $G$ be the following three-place Boolean function.
(a) Find a wff, using at most the connectives $\vee$ , $\wedge$ , and $\neg$ , that realizes G.
(b) Then find such a wff in which connective symbols occur at not more than five places.
(a) Using the disjunctive normal form, $\alpha=(\neg A\wedge\neg B\wedge\neg C)\vee(\neg A\wedge\neg B\wedge C)$ $\vee(\neg A\wedge B\wedge\neg C)\vee(A\wedge\neg B\wedge\neg C)$ .
(b) The meaning of this formula is “no more than one is true”, in other words, this corresponds to the “minority connective”. Currently we use 20 connective symbols. One way to rewrite this wff is to join some parentheses: $(\neg A\wedge\neg B)\vee((\neg A\vee\neg B)\wedge\neg C)$ , but still 9 connective symbols. If we try something like $(\neg A\wedge\neg B)\vee(\neg A\wedge\neg C)\vee(\neg B\wedge\neg C)$ then we have 11 connective symbols. Another idea is to express it as the negation of the majority rule: $\neg((A\wedge B)\vee(B\wedge C)\vee(A\wedge C))$ . By doing this, we already have 6 connectives only, and we can further use the distribution law: $\neg((A\wedge(B\vee C))\vee(B\wedge C))$ . Now we have only 5 connective symbols.
We could have also obtained the same expression in a different way. The majority rule, which is the opposite to the function $G$ above, can be expressed as “either $B$ and $C$ are both true, or $A$ and at least one of $B$ and $C$ must be true”. Using a similar interpretation of the majority rule “one of $B$ and $C$ must be true, and either they are both true or $A$ is true”, we can obtain another wff for $G$ , namely, $\neg((A\vee(B\wedge C))\wedge(B\vee C))$ , which is, basically, the previous one with all sentential symbols $\vee$ and $\wedge$ interchanged.
In (b), we found two wffs using no more than five connective symbols from those used in (a), even though the problem does not explicitly asks to use these connective symbols only. So, the question is whether we can use no more than four arbitrary unary or binary connective symbols.
But first, we will try to see whether we can use no more than four connective symbols from $\{\neg,\vee,\wedge\}$ . According to Example 1 and Exercise 1 of Section 1.6, we cannot hope to realize the majority rule using three (or less) connective symbols $\vee$ and $\wedge$ on sentence symbols or their negations. So, suppose we have a wff $\alpha$ representing $G$ that uses no more than four connective symbols from $\{\neg,\vee,\wedge\}$ . At least one of them has to be $\neg$ , as otherwise if all sentence symbols are assigned $T$ , so is $\alpha$ . Hence, $\alpha$ uses no more than three connective symbols from $\{\vee,\wedge\}$ . Then, $\neg\alpha$ represents the majority rule using no more than three connective symbols from $\{\vee,\wedge\}$ (and some number of negation symbols). Now, according to Exercise 16-A of Section 1.2 (which is an extra exercise stated on the website), we can find a formula $\phi$ tautologically equivalent to $\neg\alpha$ such that negation is applied to sentence symbols only. Moreover, according to our proof of the exercise, the transformation from $\neg\alpha$ to $\phi$ is carried out in such a way that the number of binary connectives does not increase (only the number of negations can change). Therefore, in this case, if $\alpha$ existed, $\phi$ would be a wff using no more than three connective symbols from $\{\vee,\wedge\}$ on sentence symbols and their negations and representing the majority rule, which is not possible. Hence, any $\alpha$ representing $G$ that uses connective symbols from $\{\neg,\vee,\wedge\}$ only has at least five of them.
The remaining question is then, whether there is a wff representing $G$ that uses no more than four unary and binary connective symbols some of which are not in the set $\{\neg,\vee,\wedge\}$ . The answer is positive, and all we need is to transform our previous expressions using “nand” and “nor”: $(A\vert(B\vee C))\wedge(B\vert C)$ and $(A\downarrow(B\wedge C))\vee(B\downarrow C)$ . Of course, these use some “exotic” binary connective symbols, and one can try to use four “standard” connective symbols from $\{\neg,\vee,\wedge,\rightarrow,\leftrightarrow\}$ , those we used in all initial definitions.
Another interesting (or, maybe, practical) question is whether we can use no more than four (or, maybe, five) connective symbols of two types only. That is, suppose, similar to the next section, that we can produce two types of devices (connective symbols) only, and we need to choose which ones to produce so that we can use the minimal number of them to represent $G$ . So far, in all our solutions with four and five connective symbols, we used three different types of connective symbols. At the same time, there are easy solutions using five connective symbols of two types: $(A\vert B)\wedge(B\vert C)\wedge(C\vert A)$ or $(A\downarrow B)\vee(B\downarrow C)\vee(C\downarrow A)$ . Again, one can try to use four connective symbols of two types to represent $G$ . |
## The fastest action by step guide for calculating what is 2 percent the 200
We currently have our very first value 2 and the second value 200. Let"s assume the unknown value is Y i beg your pardon answer we will find out.
You are watching: What is 2 percent of 200
As we have actually all the compelled values we need, now we deserve to put lock in a basic mathematical formula together below:
STEP 1Y = 2/100
STEP 2Y = 2/100 × 200
STEP 3Y = 2 ÷ 100 × 200
STEP 4Y = 4
Finally, we have found the worth of Y i beg your pardon is 4 and also that is our answer.
If you desire to usage a calculator to recognize what is 2 percent that 200, simply get in 2 ÷ 100 × 200 and you will obtain your answer i beg your pardon is 4
Here is a calculator to solve portion calculations such together what is 2% the 200. You can solve this type of calculation through your worths by start them into the calculator"s fields, and click "Calculate" to gain the an outcome and explanation.
Calculate
## Sample questions and answers
Question: at a high institution 2 percent that seniors go on a mission trip. There were 200 seniors. How countless seniors went on the trip?
Answer: 4 seniors went on the trip.
See more: 2005 Hyundai Elantra O2 Sensor Bank 1 Sensor 1, Oygen Sensor
Question: 2 percent the the youngsters in kindergarten favor Thomas the Train. If there room 200 youngsters in kindergarten, how many of them prefer Thomas?
Answer: 4 children like cutting board the Train.
## Another step by action method
Step 1: Let"s settle the equation because that Y by very first rewriting it as: 100% / 200 = 2% / Y
Step 2: autumn the portion marks to simplify your calculations: 100 / 200 = 2 / Y
Step 3: multiply both political parties by Y to transport it top top the left side of the equation: Y ( 100 / 200 ) = 2
Step 4: To isolate Y, main point both political parties by 200 / 100, we will have: Y = 2 ( 200 / 100 )
Step 5: computer the ideal side, us get: Y = 4
This leaves us through our last answer: 2% that 200 is 4
2 percent of 200 is 4 2.01 percent that 200 is 4.02 2.02 percent the 200 is 4.04 2.03 percent of 200 is 4.06 2.04 percent the 200 is 4.08 2.05 percent the 200 is 4.1 2.06 percent of 200 is 4.12 2.07 percent the 200 is 4.14 2.08 percent the 200 is 4.16 2.09 percent of 200 is 4.18 2.1 percent that 200 is 4.2 2.11 percent of 200 is 4.22 2.12 percent that 200 is 4.24 2.13 percent the 200 is 4.26 2.14 percent of 200 is 4.28 2.15 percent the 200 is 4.3 2.16 percent the 200 is 4.32 2.17 percent the 200 is 4.34 2.18 percent that 200 is 4.36 2.19 percent that 200 is 4.38
2.2 percent of 200 is 4.4 2.21 percent that 200 is 4.42 2.22 percent that 200 is 4.44 2.23 percent of 200 is 4.46 2.24 percent that 200 is 4.48 2.25 percent the 200 is 4.5 2.26 percent of 200 is 4.52 2.27 percent that 200 is 4.54 2.28 percent that 200 is 4.56 2.29 percent of 200 is 4.58 2.3 percent of 200 is 4.6 2.31 percent that 200 is 4.62 2.32 percent that 200 is 4.64 2.33 percent that 200 is 4.66 2.34 percent the 200 is 4.68 2.35 percent that 200 is 4.7 2.36 percent of 200 is 4.72 2.37 percent the 200 is 4.74 2.38 percent of 200 is 4.76 2.39 percent that 200 is 4.78
2.4 percent the 200 is 4.8 2.41 percent the 200 is 4.82 2.42 percent the 200 is 4.84 2.43 percent the 200 is 4.86 2.44 percent the 200 is 4.88 2.45 percent of 200 is 4.9 2.46 percent that 200 is 4.92 2.47 percent the 200 is 4.94 2.48 percent that 200 is 4.96 2.49 percent of 200 is 4.98 2.5 percent of 200 is 5 2.51 percent that 200 is 5.02 2.52 percent the 200 is 5.04 2.53 percent that 200 is 5.06 2.54 percent the 200 is 5.08 2.55 percent of 200 is 5.1 2.56 percent that 200 is 5.12 2.57 percent that 200 is 5.14 2.58 percent of 200 is 5.16 2.59 percent that 200 is 5.18
2.6 percent of 200 is 5.2 2.61 percent that 200 is 5.22 2.62 percent that 200 is 5.24 2.63 percent the 200 is 5.26 2.64 percent that 200 is 5.28 2.65 percent of 200 is 5.3 2.66 percent that 200 is 5.32 2.67 percent of 200 is 5.34 2.68 percent the 200 is 5.36 2.69 percent that 200 is 5.38 2.7 percent the 200 is 5.4 2.71 percent that 200 is 5.42 2.72 percent the 200 is 5.44 2.73 percent the 200 is 5.46 2.74 percent that 200 is 5.48 2.75 percent that 200 is 5.5 2.76 percent of 200 is 5.52 2.77 percent of 200 is 5.54 2.78 percent that 200 is 5.56 2.79 percent that 200 is 5.58
2.8 percent the 200 is 5.6 2.81 percent of 200 is 5.62 2.82 percent that 200 is 5.64 2.83 percent of 200 is 5.66 2.84 percent that 200 is 5.68 2.85 percent of 200 is 5.7 2.86 percent of 200 is 5.72 2.87 percent that 200 is 5.74 2.88 percent the 200 is 5.76 2.89 percent of 200 is 5.78 2.9 percent that 200 is 5.8 2.91 percent the 200 is 5.82 2.92 percent that 200 is 5.84 2.93 percent the 200 is 5.86 2.94 percent the 200 is 5.88 2.95 percent of 200 is 5.9 2.96 percent the 200 is 5.92 2.97 percent that 200 is 5.94 2.98 percent the 200 is 5.96 2.99 percent of 200 is 5.98
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The U.S. Postal Service handles 170,000,000,000 pieces of mail each year. This is 40% of the world’s total. How many pieces of mail are sent each year?
May 27, 2017
See a solution process below:
Explanation:
We can rewrite this as:
$170$ billion is 40% of what?
"Percent" or "%" means "out of 100" or "per 100", Therefore 40% can be written as $\frac{40}{100}$.
When dealing with percents the word "of" means "times" or "to multiply".
Finally, lets call the number of pieces of mail we are looking for "m".
Putting this altogether we can write this equation and solve for $m$ while keeping the equation balanced:
$170 = \frac{40}{100} \times m$
$\frac{\textcolor{red}{100}}{\textcolor{b l u e}{40}} \times 170 = \frac{\textcolor{red}{100}}{\textcolor{b l u e}{40}} \times \frac{40}{100} \times m$
$\frac{17000}{\textcolor{b l u e}{40}} = \frac{\cancel{\textcolor{red}{100}}}{\cancel{\textcolor{b l u e}{40}}} \times \frac{\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{40}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{100}}}} \times m$
$425 = m$
$m = 425$
425 billion or 425,000,000,000 pieces of mail are sent each year. |
Fraction Division
Fraction Division
# Motivation Problems
Consider the following two basic problems.
Your mother told you to get \$5 in quarters. How many quarters should you bring back?
Dad told you to give the chickens four pails of feed. You are only strong enough to carry half a pail at a time. How many trips do you need to make?
You would probably solve both problems in a similar manner. You would probably solve the first problem by considering that there are four quarters in a dollar. And then since you have \$5 and 5 · 4 = 20, we would have twenty quarters. For the second problem, you would probably consider that there are two halves in each pail so that two trips are needed for one pail. And then since four pails of feed are needed and 4 · 2 = 8, you would need to make eight trips.
But, if we use the numbers as given in the original problems, the problems are actually division problems based on the repeated-subtraction model for division, since we are asking how many dollars are needed to make \$5, (how many one-fourths are in five), and how many pails are needed to make 4 pails, (how many one-halves are in four).
We model the two problems below and write the problem in both the division and multiplication forms.
First Problem: We use a fraction strip to represent one dollar.
You would bring twenty quarters back.
Second Problem: Again, use a fraction strip to represent one pail.
You would need to make eight trips.
Note that both of these basic problems motivate the common rule for division of fractions, "invert the divisor and multiply by the reciprocal". Though most of us solve basic problems like the above two without thinking that we are dividing fractions, we need to understand the connection to division of fractions so that we will be able to solve more complex problems.
Example:
You need to lay tile to create a frieze above a doorway. The tile measures inches by inches. If the doorway is inches wide, how many pieces of tile are needed? Write a mathematic expression for this problem. Solution
# Model for Division of Fractions
Remember the missing-factor definition for division: a ÷ b = c if and only if b · c = a. With whole numbers this meant that 12 ÷ 3 = 4 precisely because 3 · 4 = 12. Also remember that division answers questions like "how many groups of 3 items can be made out of 12 items?" The division answer above tells us that when we have 12 items we can form 4 groups of 3 items each.
The same relationship defines division of fractions, e.g., if and only if . To see what happens, we solve this problem using a model first. This division problem answers the question "how many groups of can be made out of ?"
So we begin with a whole divided into 3 equal pieces.
Since we start with , we need to shade 2 of the pieces to represent this amount.
Next we need to determine how many groups of can be formed from the shaded region. Notice that we can make the whole into six equal pieces in this picture,
Hence, the whole is in sixths. The shaded region contains 4 of these sixths. Therefore, because we can see that there are 4 of these sixths in the shaded .
# Finding the Standard Algorithm for Dividing Fractions
To illustrate, we consider the division problem . Instead of using a model this time, we solve this fraction division problem using the missing-factor definition of division, fraction multiplication, and properties for solving equations. This strategy will lead us to a general rule for computing division of fractions.
The following is an overview of the strategy we will use. A statement-reason table for this strategy appears later on this page. Read through this overview and then study the statement-reason table and try to follow the reasoning being used. In the end, we generate a simple rule for computing division of fractions, which is based on this strategy.
### Overview
1. By the missing-factor definition of division, we know that if and only if .
2. Remember when we solved equations like 4x = 12 , we used a property of equality to divide both sides of the equation by 4 to get 1 · x = 3 or x = 3. We used a property of equality to get the x alone on the left side of the equation.
3. Using the same strategy on , we need to find a value we can multiply both sides of the equation by so that that value times equals 1. Since , the value we need to multiply both sides of this equation by is .
4. Therefore, and then and finally .
### Statement-Reason Summary of Fraction Division Strategy
Statement Reason Original Problem Missing-Factor Definition of Division Multiplication Property of Equality (Strategy: use so that ) Associative Property of Multiplication Simplification of Fractions Inverse Property of Multiplication Identity Property of Multiplication
Note the above problem has , which follows the "invert the divisor and multiply by the reciprocal" rule.
In the previous example, we saw how the process of dividing fractions is based on rewriting the problem in its multiplication form and then solving by multiplying both sides of the equation by a value that will simplify the problem to the form 1 · x = ___ where ___ is a fraction multiplication problem. The key to this process is finding pairs of numbers that when multiplied equal one. In other words, we use the multiplicative inverse or reciprocal.
## Self-Check Problem
You are going to bake cookies and only have three-fourths of a cup of flour. If the recipe states that two-thirds of a cup flour is needed to make a single batch of cookies. How many batches of cookies can you make if you use all of the flour? Solution
# Multiplicative Inverse or Reciprocal
When the product of two numbers is one, they are called reciprocals or multiplicative inverses of each other. For example, and are reciprocals because . This is the motivation for the following property of fractions.
Inverse Property for Fraction Multiplication
where a and b are nonzero. The fraction is called the multiplicative inverse of (or reciprocal) and vice versa.
Notice that a reciprocal (multiplicative inverse) can be formed from any common fraction by exchanging the positions of the numerator and denominator. The reciprocal of is , of is , and is . The reciprocal (multiplicative inverse) for a mixed number is found by first changing the mixed number to an improper fraction. For example, so the reciprocal of is .
## Self-Check Problems
Find the reciprocal (multiplicative inverse) for each of the following.
reciprocal 7 reciprocal 0 reciprocal reciprocal
Important Note. The reciprocal of zero is undefined.
# The Standard Algorithm for Dividing Fractions
Now we derive the general rule for dividing fractions.
Statement Reason Original Problem Missing-Factor Definition of Division Multiplication Property of Equality Associative Property of Multiplication Commutative Property of Multiplication Inverse Property of Multiplication Identity Property of Multiplication
Short-cutting all of the algebra steps shown in the previous table, we can generalize fraction division with the following formula.
### Invert-and-Multiply Rule for Dividing Fractions
For a, b, c, and d, whole numbers with b, c, and d not equal to zero,
This may be described in words as "invert the divisor and multiply by the reciprocal." Some describe the rule as "change to divide and multiply by the reciprocal."
Example:
"Invert the divisor and multiply by the reciprocal" can seem like a strange and mysterious rule, but as we have seen, it follows from the definition of division, properties of multiplication, and properties of equality (definition, properties, and strategies for solving equations).
## Self-Check Problem
You need to lay tile to create a frieze above a doorway. The tile measures inches by inches. If the doorway is inches wide, how many pieces of tile are needed?
Solution
### Joke or Quote
Did you know that 5 out of every 4 people have a problem with fractions? |
In algebra, a quadratic equation (from the Latin quadratus for "square") is any equation having the form
${\displaystyle ax^{2}+bx+c=0,}$
where x represents an unknown, and a, b, and c represent known numbers, with a ≠ 0. If a = 0, then the equation is linear, not quadratic. The numbers a, b, and c are the coefficients of the equation and may be distinguished by calling them, respectively, the quadratic coefficient, the linear coefficient and the constant or free term.[1]
Because the quadratic equation involves only one unknown, it is called "univariate". The quadratic equation only contains powers of x that are non-negative integers, and therefore it is a polynomial equation. In particular, it is a second-degree polynomial equation, since the greatest power is two.
Quadratic equations can be solved by a process known in American English as factoring and in other varieties of English as factorising, by completing the square, by using the quadratic formula, or by graphing. Solutions to problems equivalent to the quadratic equation were known as early as 2000 BC.
Figure 1. Plots of quadratic function y = ax2 + bx + c, varying each coefficient separately while the other coefficients are fixed (at values a = 1, b = 0, c = 0)
A quadratic equation with real or complex coefficients has two solutions, called roots. These two solutions may or may not be distinct, and they may or may not be real.
### Factoring by inspection
It may be possible to express a quadratic equation ax2 + bx + c = 0 as a product (px + q)(rx + s) = 0. In some cases, it is possible, by simple inspection, to determine values of p, q, r, and s that make the two forms equivalent to one another. If the quadratic equation is written in the second form, then the "Zero Factor Property" states that the quadratic equation is satisfied if px + q = 0 or rx + s = 0. Solving these two linear equations provides the roots of the quadratic.
For most students, factoring by inspection is the first method of solving quadratic equations to which they are exposed.[2]:202–207 If one is given a quadratic equation in the form x2 + bx + c = 0, the sought factorization has the form (x + q)(x + s), and one has to find two numbers q and s that add up to b and whose product is c (this is sometimes called "Vieta's rule"[3] and is related to Vieta's formulas). As an example, x2 + 5x + 6 factors as (x + 3)(x + 2). The more general case where a does not equal 1 can require a considerable effort in trial and error guess-and-check, assuming that it can be factored at all by inspection.
Except for special cases such as where b = 0 or c = 0, factoring by inspection only works for quadratic equations that have rational roots. This means that the great majority of quadratic equations that arise in practical applications cannot be solved by factoring by inspection.[2]:207
### Completing the square
Figure 2. For the quadratic function y = x2x − 2, the points where the graph crosses the x-axis, x = −1 and x = 2, are the solutions of the quadratic equation x2x − 2 = 0.
The process of completing the square makes use of the algebraic identity
${\displaystyle x^{2}+2hx+h^{2}=(x+h)^{2},}$
which represents a well-defined algorithm that can be used to solve any quadratic equation.[2]:207 Starting with a quadratic equation in standard form, ax2 + bx + c = 0
1. Divide each side by a, the coefficient of the squared term.
2. Subtract the constant term c/a from both sides.
3. Add the square of one-half of b/a, the coefficient of x, to both sides. This "completes the square", converting the left side into a perfect square.
4. Write the left side as a square and simplify the right side if necessary.
5. Produce two linear equations by equating the square root of the left side with the positive and negative square roots of the right side.
6. Solve the two linear equations.
We illustrate use of this algorithm by solving 2x2 + 4x − 4 = 0
${\displaystyle 1)\ x^{2}+2x-2=0}$
${\displaystyle 2)\ x^{2}+2x=2}$
${\displaystyle 3)\ x^{2}+2x+1=2+1}$
${\displaystyle 4)\ \left(x+1\right)^{2}=3}$
${\displaystyle 5)\ x+1=\pm {\sqrt {3}}}$
${\displaystyle 6)\ x=-1\pm {\sqrt {3}}}$
The plus-minus symbol "±" indicates that both x = −1 + 3 and x = −1 − 3 are solutions of the quadratic equation.[4]
### Quadratic formula and its derivation
Completing the square can be used to derive a general formula for solving quadratic equations, called the quadratic formula.[5] The mathematical proof will now be briefly summarized.[6] It can easily be seen, by polynomial expansion, that the following equation is equivalent to the quadratic equation:
${\displaystyle \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}.}$
Taking the square root of both sides, and isolating x, gives:
${\displaystyle x={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}.}$
Some sources, particularly older ones, use alternative parameterizations of the quadratic equation such as ax2 + 2bx + c = 0 or ax2 − 2bx + c = 0 ,[7] where b has a magnitude one half of the more common one, possibly with opposite sign. These result in slightly different forms for the solution, but are otherwise equivalent.
A number of alternative derivations can be found in the literature. These proofs are simpler than the standard completing the square method, represent interesting applications of other frequently used techniques in algebra, or offer insight into other areas of mathematics.
A lesser known quadratic formula, as used in Muller's method provides the same roots via the equation
${\displaystyle x={\frac {2c}{-b\pm {\sqrt {b^{2}-4ac\ }}}}.}$
This can be deduced from the standard quadratic formula by Vieta's formulas, which assert that the product of the roots is c/a.
One property of this form is that it yields one valid root when a = 0, while the other root contains division by zero, because when a = 0, the quadratic equation becomes a linear equation, which has one root. By contrast, in this case, the more common formula has a division by zero for one root and an indeterminate form 0/0 for the other root. On the other hand, when c = 0, the more common formula yields two correct roots whereas this form yields the zero root and an indeterminate form 0/0.
It is sometimes convenient to reduce a quadratic equation so that its leading coefficient is one. This is done by dividing both sides by a, which is always possible since a is non-zero. This produces the reduced quadratic equation:[8]
${\displaystyle x^{2}+px+q=0,}$
where p = b/a and q = c/a. This monic equation has the same solutions as the original.
The quadratic formula for the solutions of the reduced quadratic equation, written in terms of its coefficients, is:
${\displaystyle x={\frac {1}{2}}\left(-p\pm {\sqrt {p^{2}-4q}}\right),}$
or equivalently:
${\displaystyle x=-{\frac {p}{2}}\pm {\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}.}$
### Discriminant
Figure 3. Discriminant signs
In the quadratic formula, the expression underneath the square root sign is called the discriminant of the quadratic equation, and is often represented using an upper case D or an upper case Greek delta:[9]
${\displaystyle \Delta =b^{2}-4ac.}$
A quadratic equation with real coefficients can have either one or two distinct real roots, or two distinct complex roots. In this case the discriminant determines the number and nature of the roots. There are three cases:
• If the discriminant is positive, then there are two distinct roots
${\displaystyle {\frac {-b+{\sqrt {\Delta }}}{2a}}\quad {\text{and}}\quad {\frac {-b-{\sqrt {\Delta }}}{2a}},}$
both of which are real numbers. For quadratic equations with rational coefficients, if the discriminant is a square number, then the roots are rational—in other cases they may be quadratic irrationals.
• If the discriminant is zero, then there is exactly one real root
${\displaystyle -{\frac {b}{2a}},}$
sometimes called a repeated or double root.
• If the discriminant is negative, then there are no real roots. Rather, there are two distinct (non-real) complex roots[10]
${\displaystyle -{\frac {b}{2a}}+i{\frac {\sqrt {-\Delta }}{2a}}\quad {\text{and}}\quad -{\frac {b}{2a}}-i{\frac {\sqrt {-\Delta }}{2a}},}$
which are complex conjugates of each other. In these expressions i is the imaginary unit.
Thus the roots are distinct if and only if the discriminant is non-zero, and the roots are real if and only if the discriminant is non-negative.
### Geometric interpretation
Graph of y = ax2 + bx + c, where a and the discriminant b2 − 4ac are positive, with
• Roots and y-intercept in red
• Vertex and axis of symmetry in blue
• Focus and directrix in pink
Visualisation of the complex roots of y = ax2 + bx + c: the parabola is rotated 180° about its vertex (orange). Its x-intercepts are rotated 90° around their mid-point, and the Cartesian plane is interpreted as the complex plane (green).[11]
The function f(x) = ax2 + bx + c is the quadratic function.[12] The graph of any quadratic function has the same general shape, which is called a parabola. The location and size of the parabola, and how it opens, depend on the values of a, b, and c. As shown in Figure 1, if a > 0, the parabola has a minimum point and opens upward. If a < 0, the parabola has a maximum point and opens downward. The extreme point of the parabola, whether minimum or maximum, corresponds to its vertex. The x-coordinate of the vertex will be located at ${\displaystyle \scriptstyle x={\tfrac {-b}{2a}}}$, and the y-coordinate of the vertex may be found by substituting this x-value into the function. The y-intercept is located at the point (0, c).
The solutions of the quadratic equation ax2 + bx + c = 0 correspond to the roots of the function f(x) = ax2 + bx + c, since they are the values of x for which f(x) = 0. As shown in Figure 2, if a, b, and c are real numbers and the domain of f is the set of real numbers, then the roots of f are exactly the x-coordinates of the points where the graph touches the x-axis. As shown in Figure 3, if the discriminant is positive, the graph touches the x-axis at two points; if zero, the graph touches at one point; and if negative, the graph does not touch the x-axis.
The term
${\displaystyle x-r}$
is a factor of the polynomial
${\displaystyle ax^{2}+bx+c}$
if and only if r is a root of the quadratic equation
${\displaystyle ax^{2}+bx+c=0.}$
It follows from the quadratic formula that
${\displaystyle ax^{2}+bx+c=a\left(x-{\frac {-b+{\sqrt {b^{2}-4ac}}}{2a}}\right)\left(x-{\frac {-b-{\sqrt {b^{2}-4ac}}}{2a}}\right).}$
In the special case b2 = 4ac where the quadratic has only one distinct root (i.e. the discriminant is zero), the quadratic polynomial can be factored as
${\displaystyle ax^{2}+bx+c=a\left(x+{\frac {b}{2a}}\right)^{2}.}$
### Graphing for real roots
Figure 4. Graphing calculator computation of one of the two roots of the quadratic equation 2x2 + 4x − 4 = 0. Although the display shows only five significant figures of accuracy, the retrieved value of xc is 0.732050807569, accurate to twelve significant figures.
For most of the 20th century, graphing was rarely mentioned as a method for solving quadratic equations in high school or college algebra texts. Students learned to solve quadratic equations by factoring, completing the square, and applying the quadratic formula. Recently, graphing calculators have become common in schools and graphical methods have started to appear in textbooks, but they are generally not highly emphasized.[13]
Being able to use a graphing calculator to solve a quadratic equation requires the ability to produce a graph of y = f(x), the ability to scale the graph appropriately to the dimensions of the graphing surface, and the recognition that when f(x) = 0, x is a solution to the equation. The skills required to solve a quadratic equation on a calculator are in fact applicable to finding the real roots of any arbitrary function.
Since an arbitrary function may cross the x-axis at multiple points, graphing calculators generally require one to identify the desired root by positioning a cursor at a "guessed" value for the root. (Some graphing calculators require bracketing the root on both sides of the zero.) The calculator then proceeds, by an iterative algorithm, to refine the estimated position of the root to the limit of calculator accuracy.
### Avoiding loss of significance
Although the quadratic formula provides an exact solution, the result is not exact if real numbers are approximated during the computation, as usual in numerical analysis, where real numbers are approximated by floating point numbers (called "reals" in many programming languages). In this context, the quadratic formula is not completely stable.
This occurs when the roots have different order of magnitude, or, equivalently, when b2 and b2 − 4ac are close in magnitude. In this case, the subtraction of two nearly equal numbers will cause loss of significance or catastrophic cancellation in the smaller root. To avoid this, the root that is smaller in magnitude, r, can be computed as ${\displaystyle (c/a)/R}$ where R is the root that is bigger in magnitude.
A second form of cancellation can occur between the terms b2 and 4ac of the discriminant, that is when the two roots are very close. This can lead to loss of up to half of correct significant figures in the roots.[7][14]
## Examples and applications
The trajectory of the cliff jumper is parabolic because horizontal displacement is a linear function of time ${\displaystyle x=v_{x}t}$, while vertical displacement is a quadratic function of time ${\displaystyle y={\tfrac {1}{2}}at^{2}+v_{y}t+h}$. As a result, the path follows quadratic equation ${\displaystyle y={\tfrac {a}{2v_{x}^{2}}}x^{2}+{\tfrac {v_{y}}{v_{x}}}x+h}$, where ${\displaystyle v_{x}}$ and ${\displaystyle v_{y}}$ are horizontal and vertical components of the original velocity, a is gravitational acceleration and h is original height. The a value should be considered negative here, as its direction (downwards) is opposite to the height measurement (upwards).
The golden ratio is found as the positive solution of the quadratic equation ${\displaystyle x^{2}-x-1=0.}$
The equations of the circle and the other conic sectionsellipses, parabolas, and hyperbolas—are quadratic equations in two variables.
Given the cosine or sine of an angle, finding the cosine or sine of the angle that is half as large involves solving a quadratic equation.
The process of simplifying expressions involving the square root of an expression involving the square root of another expression involves finding the two solutions of a quadratic equation.
Descartes' theorem states that for every four kissing (mutually tangent) circles, their radii satisfy a particular quadratic equation.
The equation given by Fuss' theorem, giving the relation among the radius of a bicentric quadrilateral's inscribed circle, the radius of its circumscribed circle, and the distance between the centers of those circles, can be expressed as a quadratic equation for which the distance between the two circles' centers in terms of their radii is one of the solutions. The other solution of the same equation in terms of the relevant radii gives the distance between the circumscribed circle's center and the center of the excircle of an ex-tangential quadrilateral.
## History
Babylonian mathematicians, as early as 2000 BC (displayed on Old Babylonian clay tablets) could solve problems relating the areas and sides of rectangles. There is evidence dating this algorithm as far back as the Third Dynasty of Ur.[15] In modern notation, the problems typically involved solving a pair of simultaneous equations of the form:
${\displaystyle x+y=p,\ \ xy=q,}$
which is equivalent to the statement that x and y are the roots of the equation:[16]:86
${\displaystyle z^{2}+q=pz.}$
The steps given by Babylonian scribes for solving the above rectangle problem, in terms of x and y, were as follows:
1. Compute half of p.
2. Square the result.
3. Subtract q.
4. Find the (positive) square root using a table of squares.
5. Add together the results of steps (1) and (4) to give x. In modern notation this means calculating ${\displaystyle x={\frac {p}{2}}+{\sqrt {\left({\frac {p}{2}}\right)^{2}-q}}.}$
Geometric methods were used to solve quadratic equations in Babylonia, Egypt, Greece, China, and India. The Egyptian Berlin Papyrus, dating back to the Middle Kingdom (2050 BC to 1650 BC), contains the solution to a two-term quadratic equation.[17] Babylonian mathematicians from circa 400 BC and Chinese mathematicians from circa 200 BC used geometric methods of dissection to solve quadratic equations with positive roots.[18][19] Rules for quadratic equations were given in The Nine Chapters on the Mathematical Art, a Chinese treatise on mathematics.[19][20] These early geometric methods do not appear to have had a general formula. Euclid, the Greek mathematician, produced a more abstract geometrical method around 300 BC. With a purely geometric approach Pythagoras and Euclid created a general procedure to find solutions of the quadratic equation. In his work Arithmetica, the Greek mathematician Diophantus solved the quadratic equation, but giving only one root, even when both roots were positive.[21]
In 628 AD, Brahmagupta, an Indian mathematician, gave the first explicit (although still not completely general) solution of the quadratic equation ax2 + bx = c as follows: "To the absolute number multiplied by four times the [coefficient of the] square, add the square of the [coefficient of the] middle term; the square root of the same, less the [coefficient of the] middle term, being divided by twice the [coefficient of the] square is the value." (Brahmasphutasiddhanta, Colebrook translation, 1817, page 346)[16]:87 This is equivalent to:
${\displaystyle x={\frac {{\sqrt {4ac+b^{2}}}-b}{2a}}.}$
The Jewish mathematician Abraham bar Hiyya Ha-Nasi (12th century, Spain) authored the first European book to include the full solution to the general quadratic equation.[27] His solution was largely based on Al-Khwarizmi's work.[22] The writing of the Chinese mathematician Yang Hui (1238–1298 AD) is the first known one in which quadratic equations with negative coefficients of 'x' appear, although he attributes this to the earlier Liu Yi.[28] By 1545 Gerolamo Cardano compiled the works related to the quadratic equations. The quadratic formula covering all cases was first obtained by Simon Stevin in 1594.[29] In 1637 René Descartes published La Géométrie containing the quadratic formula in the form we know today. The first appearance of the general solution in the modern mathematical literature appeared in an 1896 paper by Henry Heaton.[30]
### Alternative methods of root calculation
#### Vieta's formulas
Figure 5. Graph of the difference between Vieta's approximation for the smaller of the two roots of the quadratic equation x2 + bx + c = 0 compared with the value calculated using the quadratic formula. Vieta's approximation is inaccurate for small b but is accurate for large b. The direct evaluation using the quadratic formula is accurate for small b with roots of comparable value but experiences loss of significance errors for large b and widely spaced roots. The difference between Vieta's approximation versus the direct computation reaches a minimum at the large dots, and rounding causes squiggles in the curves beyond this minimum.
Vieta's formulas give a simple relation between the roots of a polynomial and its coefficients. In the case of the quadratic polynomial, they take the following form:
${\displaystyle x_{1}+x_{2}=-{\frac {b}{a}}}$
and
${\displaystyle x_{1}\ x_{2}={\frac {c}{a}}.}$
These results follow immediately from the relation:
${\displaystyle \left(x-x_{1}\right)\ \left(x-x_{2}\right)=x^{2}\ -\left(x_{1}+x_{2}\right)x+x_{1}x_{2}=0,}$
which can be compared term by term with
${\displaystyle x^{2}+(b/a)x+c/a=0.}$
The first formula above yields a convenient expression when graphing a quadratic function. Since the graph is symmetric with respect to a vertical line through the vertex, when there are two real roots the vertex's x-coordinate is located at the average of the roots (or intercepts). Thus the x-coordinate of the vertex is given by the expression
${\displaystyle x_{V}={\frac {x_{1}+x_{2}}{2}}=-{\frac {b}{2a}}.}$
The y-coordinate can be obtained by substituting the above result into the given quadratic equation, giving
${\displaystyle y_{V}=-{\frac {b^{2}}{4a}}+c=-{\frac {b^{2}-4ac}{4a}}.}$
As a practical matter, Vieta's formulas provide a useful method for finding the roots of a quadratic in the case where one root is much smaller than the other. If | x2| << | x1|, then x1 + x2x1, and we have the estimate:
${\displaystyle x_{1}\approx -{\frac {b}{a}}.}$
The second Vieta's formula then provides:
${\displaystyle x_{2}={\frac {c}{a\ x_{1}}}\approx -{\frac {c}{b}}.}$
These formulas are much easier to evaluate than the quadratic formula under the condition of one large and one small root, because the quadratic formula evaluates the small root as the difference of two very nearly equal numbers (the case of large b), which causes round-off error in a numerical evaluation. Figure 5 shows the difference between (i) a direct evaluation using the quadratic formula (accurate when the roots are near each other in value) and (ii) an evaluation based upon the above approximation of Vieta's formulas (accurate when the roots are widely spaced). As the linear coefficient b increases, initially the quadratic formula is accurate, and the approximate formula improves in accuracy, leading to a smaller difference between the methods as b increases. However, at some point the quadratic formula begins to lose accuracy because of round off error, while the approximate method continues to improve. Consequently, the difference between the methods begins to increase as the quadratic formula becomes worse and worse.
This situation arises commonly in amplifier design, where widely separated roots are desired to ensure a stable operation (see step response).
#### Trigonometric solution
In the days before calculators, people would use mathematical tables—lists of numbers showing the results of calculation with varying arguments—to simplify and speed up computation. Tables of logarithms and trigonometric functions were common in math and science textbooks. Specialized tables were published for applications such as astronomy, celestial navigation and statistics. Methods of numerical approximation existed, called prosthaphaeresis, that offered shortcuts around time-consuming operations such as multiplication and taking powers and roots.[13] Astronomers, especially, were concerned with methods that could speed up the long series of computations involved in celestial mechanics calculations.
It is within this context that we may understand the development of means of solving quadratic equations by the aid of trigonometric substitution. Consider the following alternate form of the quadratic equation,
[1] ${\displaystyle ax^{2}+bx\pm c=0,}$
where the sign of the ± symbol is chosen so that a and c may both be positive. By substituting
[2] ${\displaystyle x={\sqrt {c/a}}\tan \theta }$
and then multiplying through by cos2θ, we obtain
[3] ${\displaystyle \sin ^{2}\theta +{\frac {b}{\sqrt {ac}}}\sin \theta \cos \theta \pm \cos ^{2}\theta =0.}$
Introducing functions of 2θ and rearranging, we obtain
[4] ${\displaystyle \tan 2\theta _{n}=+2{\frac {\sqrt {ac}}{b}},}$
[5] ${\displaystyle \sin 2\theta _{p}=-2{\frac {\sqrt {ac}}{b}},}$
where the subscripts n and p correspond, respectively, to the use of a negative or positive sign in equation [1]. Substituting the two values of θn or θp found from equations [4] or [5] into [2] gives the required roots of [1]. Complex roots occur in the solution based on equation [5] if the absolute value of sin 2θp exceeds unity. The amount of effort involved in solving quadratic equations using this mixed trigonometric and logarithmic table look-up strategy was two-thirds the effort using logarithmic tables alone.[31] Calculating complex roots would require using a different trigonometric form.[32]
To illustrate, let us assume we had available seven-place logarithm and trigonometric tables, and wished to solve the following to six-significant-figure accuracy:
${\displaystyle 4.16130x^{2}+9.15933x-11.4207=0}$
1. A seven-place lookup table might have only 100,000 entries, and computing intermediate results to seven places would generally require interpolation between adjacent entries.
2. ${\displaystyle \log a=0.6192290,\log b=0.9618637,\log c=1.0576927}$
3. ${\displaystyle 2{\sqrt {ac}}/b=2\times 10^{(0.6192290+1.0576927)/2-0.9618637}=1.505314}$
4. ${\displaystyle \theta =(\tan ^{-1}1.505314)/2=28.20169^{\circ }{\text{ or }}-61.79831^{\circ }}$
5. ${\displaystyle \log |\tan \theta |=-0.2706462{\text{ or }}0.2706462}$
6. ${\displaystyle \log {\sqrt {c/a}}=(1.0576927-0.6192290)/2=0.2192318}$
7. ${\displaystyle x_{1}=10^{0.2192318-0.2706462}=0.888353}$ (rounded to six significant figures)
${\displaystyle x_{2}=-10^{0.2192318+0.2706462}=-3.08943}$
#### Solution for complex roots in polar coordinates
If the quadratic equation ${\displaystyle ax^{2}+bx+c=0}$ with real coefficients has two complex roots—the case where ${\displaystyle b^{2}-4ac<0,}$ requiring a and c to have the same sign as each other—then the solutions for the roots can be expressed in polar form as[33]
${\displaystyle x_{1},\,x_{2}=r(\cos \theta \pm i\sin \theta ),}$
where ${\displaystyle r={\sqrt {\tfrac {c}{a}}}}$ and ${\displaystyle \theta =\cos ^{-1}\left({\tfrac {-b}{2{\sqrt {ac}}}}\right).}$
#### Geometric solution
Figure 6. Geometric solution of ax2 + bx + c = 0 using Lill's method. Solutions are −AX1/SA, −AX2/SA
The quadratic equation may be solved geometrically in a number of ways. One way is via Lill's method. The three coefficients a, b, c are drawn with right angles between them as in SA, AB, and BC in Figure 6. A circle is drawn with the start and end point SC as a diameter. If this cuts the middle line AB of the three then the equation has a solution, and the solutions are given by negative of the distance along this line from A divided by the first coefficient a or SA. If a is 1 the coefficients may be read off directly. Thus the solutions in the diagram are −AX1/SA and −AX2/SA.[34]
Carlyle circle of the quadratic equation x2 − sx + p = 0.
The Carlyle circle, named after Thomas Carlyle, has the property that the solutions of the quadratic equation are the horizontal coordinates of the intersections of the circle with the horizontal axis.[35] Carlyle circles have been used to develop ruler-and-compass constructions of regular polygons.
The general quadratic equation can be written in the standard form
${\displaystyle x^{2}-4ux+4v^{2}=0}$
where u and v are complex numbers. Then the solutions can be written in the particularly symmetric form
${\displaystyle x_{1,2}=({\sqrt {u+v}}\pm {\sqrt {u-v}})^{2}}$
or equivalently
${\displaystyle x_{1,2}=({\sqrt {u-v}}\pm {\sqrt {u+v}})^{2}.}$
The correctness of the formula can easily be verified by inserting the expression into the equation. This formula has the advantage that it is numerically more stable than the classical quadratic formula. It appeared in.[36] Problems originating from physics or geometry often present themselves in the homogeneous standard form on which the alternative formula is based. In particular, René Descartes' first example in La Géométrie, at the very hour of birth of the quadratic formula, was geometric and of this particular homogeneous form.
The formula and its derivation remain correct if the coefficients a, b and c are complex numbers, or more generally members of any field whose characteristic is not 2. (In a field of characteristic 2, the element 2a is zero and it is impossible to divide by it.)
The symbol
${\displaystyle \pm {\sqrt {b^{2}-4ac}}}$
in the formula should be understood as "either of the two elements whose square is b2 − 4ac, if such elements exist". In some fields, some elements have no square roots and some have two; only zero has just one square root, except in fields of characteristic 2. Even if a field does not contain a square root of some number, there is always a quadratic extension field which does, so the quadratic formula will always make sense as a formula in that extension field.
#### Characteristic 2
In a field of characteristic 2, the quadratic formula, which relies on 2 being a unit, does not hold. Consider the monic quadratic polynomial
${\displaystyle x^{2}+bx+c}$
over a field of characteristic 2. If b = 0, then the solution reduces to extracting a square root, so the solution is
${\displaystyle x={\sqrt {c}}}$
and there is only one root since
${\displaystyle -{\sqrt {c}}=-{\sqrt {c}}+2{\sqrt {c}}={\sqrt {c}}.}$
In summary,
${\displaystyle \displaystyle x^{2}+c=(x+{\sqrt {c}})^{2}.}$
In the case that b ≠ 0, there are two distinct roots, but if the polynomial is irreducible, they cannot be expressed in terms of square roots of numbers in the coefficient field. Instead, define the 2-root R(c) of c to be a root of the polynomial x2 + x + c, an element of the splitting field of that polynomial. One verifies that R(c) + 1 is also a root. In terms of the 2-root operation, the two roots of the (non-monic) quadratic ax2 + bx + c are
${\displaystyle {\frac {b}{a}}R\left({\frac {ac}{b^{2}}}\right)}$
and
${\displaystyle {\frac {b}{a}}\left(R\left({\frac {ac}{b^{2}}}\right)+1\right).}$
For example, let a denote a multiplicative generator of the group of units of F4, the Galois field of order four (thus a and a + 1 are roots of x2 + x + 1 over F4. Because (a + 1)2 = a, a + 1 is the unique solution of the quadratic equation x2 + a = 0. On the other hand, the polynomial x2 + ax + 1 is irreducible over F4, but it splits over F16, where it has the two roots ab and ab + a, where b is a root of x2 + x + a in F16.
This is a special case of Artin–Schreier theory.
## References
1. ^ Protters & Morrey: "Calculus and Analytic Geometry. First Course".
2. ^ a b c Washington, Allyn J. (2000). Basic Technical Mathematics with Calculus, Seventh Edition. Addison Wesley Longman, Inc. ISBN 978-0-201-35666-3.
3. ^ Ebbinghaus, Heinz-Dieter; Ewing, John H. (1991), Numbers, Graduate Texts in Mathematics, 123, Springer, p. 77, ISBN 9780387974972.
4. ^ Sterling, Mary Jane (2010), Algebra I For Dummies, Wiley Publishing, p. 219, ISBN 978-0-470-55964-2
5. ^ Rich, Barnett; Schmidt, Philip (2004), Schaum's Outline of Theory and Problems of Elementary Algebra, The McGraw-Hill Companies, ISBN 978-0-07-141083-0, Chapter 13 §4.4, p. 291
6. ^ Himonas, Alex. Calculus for Business and Social Sciences, p. 64 (Richard Dennis Publications, 2001).
7. ^ a b Kahan, Willian (November 20, 2004), On the Cost of Floating-Point Computation Without Extra-Precise Arithmetic (PDF), retrieved 2012-12-25
8. ^ Alenit͡syn, Aleksandr and Butikov, Evgeniĭ. Concise Handbook of Mathematics and Physics, p. 38 (CRC Press 1997)
9. ^ Δ is the initial of the Greek word Διακρίνουσα, Diakrínousa, discriminant.
10. ^ Achatz, Thomas; Anderson, John G.; McKenzie, Kathleen (2005). Technical Shop Mathematics. Industrial Press. p. 277. ISBN 978-0-8311-3086-2.
11. ^ "Complex Roots Made Visible – Math Fun Facts". Retrieved 1 October 2016.
12. ^ Wharton, P. (2006). Essentials of Edexcel Gcse Math/Higher. Lonsdale. p. 63. ISBN 978-1-905-129-78-2.
13. ^ a b Ballew, Pat. "Solving Quadratic Equations — By analytic and graphic methods; Including several methods you may never have seen" (PDF). Retrieved 18 April 2013.
14. ^ Higham, Nicholas (2002), Accuracy and Stability of Numerical Algorithms (2nd ed.), SIAM, p. 10, ISBN 978-0-89871-521-7
15. ^ Friberg, Jöran (2009). "A Geometric Algorithm with Solutions to Quadratic Equations in a Sumerian Juridical Document from Ur III Umma". Cuneiform Digital Library Journal. 3.
16. ^ a b Stillwell, John (2004). Mathematics and Its History (2nd ed.). Springer. ISBN 978-0-387-95336-6.
17. ^ The Cambridge Ancient History Part 2 Early History of the Middle East. Cambridge University Press. 1971. p. 530. ISBN 978-0-521-07791-0.
18. ^ Henderson, David W. "Geometric Solutions of Quadratic and Cubic Equations". Mathematics Department, Cornell University. Retrieved 28 April 2013.
19. ^ a b Aitken, Wayne. "A Chinese Classic: The Nine Chapters" (PDF). Mathematics Department, California State University. Retrieved 28 April 2013.
20. ^ Smith, David Eugene (1958). History of Mathematics. Courier Dover Publications. p. 380. ISBN 978-0-486-20430-7.
21. ^ Smith, David Eugene (1958). History of Mathematics, Volume 1. Courier Dover Publications. p. 134. ISBN 978-0-486-20429-1. Extract of page 134
22. ^ a b c d Katz, V. J.; Barton, B. (2006). "Stages in the History of Algebra with Implications for Teaching". Educational Studies in Mathematics. 66 (2): 185&ndash, 201. doi:10.1007/s10649-006-9023-7.
23. ^ a b Boyer, Carl B.; Uta C. Merzbach, rev. editor (1991). A History of Mathematics. John Wiley & Sons, Inc. ISBN 978-0-471-54397-8.
24. ^ O'Connor, John J.; Robertson, Edmund F., "Arabic mathematics: forgotten brilliance?", MacTutor History of Mathematics archive, University of St Andrews. "Algebra was a unifying theory which allowed rational numbers, irrational numbers, geometrical magnitudes, etc., to all be treated as "algebraic objects"."
25. ^ Jacques Sesiano, "Islamic mathematics", p. 148, in Selin, Helaine; D'Ambrosio, Ubiratan, eds. (2000), Mathematics Across Cultures: The History of Non-Western Mathematics, Springer, ISBN 978-1-4020-0260-1
26. ^ Smith, David Eugene (1958). History of Mathematics. Courier Dover Publications. p. 280. ISBN 978-0-486-20429-1.
27. ^ Livio, Mario (2006). The Equation that Couldn't Be Solved. Simon & Schuster. ISBN 978-0743258210.
28. ^ Ronan, Colin (1985). The Shorter Science and Civilisation in China. Cambridge University Press. p. 15. ISBN 978-0-521-31536-4.
29. ^ Struik, D. J.; Stevin, Simon (1958), The Principal Works of Simon Stevin, Mathematics (PDF), II–B, C. V. Swets & Zeitlinger, p. 470
30. ^ Heaton, H (1896). "A Method of Solving Quadratic Equations". American Mathematical Monthly. 3 (10): 236–237. doi:10.2307/2971099. JSTOR 2971099.
31. ^ Seares, F. H. (1945). "Trigonometric Solution of the Quadratic Equation". Publications of the Astronomical Society of the Pacific. 57 (339): 307–309. Bibcode:1945PASP...57..307S. doi:10.1086/125759.
32. ^ Aude, H. T. R. (1938). "The Solutions of the Quadratic Equation Obtained by the Aid of the Trigonometry". National Mathematics Magazine. 13 (3): 118–121. doi:10.2307/3028750. JSTOR 3028750.
33. ^ Simons, Stuart, "Alternative approach to complex roots of real quadratic equations", Mathematical Gazette 93, March 2009, 91–92.
34. ^ Bixby, William Herbert (1879), Graphical Method for finding readily the Real Roots of Numerical Equations of Any Degree, West Point N. Y.
35. ^ Weisstein, Eric W. "Carlyle Circle". From MathWorld—A Wolfram Web Resource. Retrieved 21 May 2013.
36. ^ Hungerbühler, Norbert (2017). "An alternative quadratic formula". arXiv:1702.05789 [math.HO]. |
Lesson 3, Topic 4
In Progress
# Complete Factorization
Lesson Progress
0% Complete
To factorize an expression, divide each term of the expression by their common factor. To factorize completely, use the Highest Common Factor (HCF).
### Example 8
Factorize completely
(a) 2a + 4ab
(b) 5x + 10y
(c) 15mn2 – 10m2n2
(d) $$\frac{2 \pi r}{T} \: + \: \frac{2 \pi r^2}{T^2} \: – \: \frac{2 \pi r^3}{T^3}$$
Solution
(a) 2a + 4ab
Step 1: find the HCF of 2a and 4ab
HCF = 2 x a = 2a
= 2a + 4ab
= 2a (1 + 2b) → last
(b) 5x + 10y
Find the HCF of 5x and 10y
HCF = 5
5x + 10y = 5 (x + 2y) → last
(c) 15mn3 – 10m2n2
Find the HCF
HCF = 5 x m x n x n
= 5mn2
15mn3 – 10m2n2 = 5mn2 (3n – 2m)
(d) $$\frac{2 \pi r}{T} \: + \: \frac{2 \pi r^2}{T^2} \: – \: \frac{2 \pi r^3}{T^3}$$
H.C.F = $$\scriptsize 2 \: \times \: \pi \: \times \: r \: \times \: \normalsize \frac{1}{T}$$
H.C.F = $$\frac{2 \pi r}{T}$$
$$\frac{2 \pi r}{T} \: + \: \frac{2 \pi r^2}{T^2} \: – \: \frac{2 \pi r^3}{T^3}$$
= $$\frac{2 \pi r}{T} \left (\scriptsize 1 \: + \: \normalsize \frac{r}{T} \: -\: \frac{r^2}{T^2} \right)$$
Evaluation
Factorize completely each of the following
1. 4x2y + 12xy2
2. 17m – 34m2
3. 5k2 – k
4. 35p + 20p2
5. 9ab + 3a
6. mn3 – 3gmn – g3m2n3
7. $$\frac{mv}{t} \: -\: \frac{mu}{t}$$
7. $$\frac{m}{t} \scriptsize (v \: -\: u)$$ |
<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
# Graphs of Rotations
## Graph rotated images given preimage and number of degrees
%
Progress
Progress
%
Graphs of Rotations
Quadrilateral WXYZ\begin{align*}WXYZ\end{align*} has coordinates W(5,5),X(2,0),Y(2,3)\begin{align*}W(-5, -5), X(-2, 0), Y(2, 3)\end{align*} and Z(1,3)\begin{align*}Z(-1, 3)\end{align*}. Draw the quadrilateral on the Cartesian plane. Rotate the image 110\begin{align*}110^\circ\end{align*} counterclockwise about the point X\begin{align*}X\end{align*}. Show the resulting image.
### Watch This
First watch this video to learn about graphs of rotations.
CK-12 Foundation Chapter10GraphsofRotationsA
Then watch this video to see some examples.
CK-12 Foundation Chapter10GraphsofRotationsB
### Guidance
In geometry, a transformation is an operation that moves, flips, or changes a shape to create a new shape. A rotation is an example of a transformation where a figure is rotated about a specific point (called the center of rotation), a certain number of degrees.
For now, in order to graph a rotation in general you will use geometry software. This will allow you to rotate any figure any number of degrees about any point. There are a few common rotations that are good to know how to do without geometry software, shown in the table below.
Center of Rotation Angle of Rotation Preimage (Point P\begin{align*}P\end{align*}) Rotated Image (Point P\begin{align*}P^\prime\end{align*})
(0, 0) 90\begin{align*}90^\circ\end{align*}(or 270\begin{align*}-270^\circ\end{align*}) (x,y)\begin{align*}(x, y)\end{align*} (y,x)\begin{align*}(-y, x)\end{align*}
(0, 0) 180\begin{align*}180^\circ\end{align*}(or 180\begin{align*}-180^\circ\end{align*}) (x,y)\begin{align*}(x, y)\end{align*} (x,y)\begin{align*}(-x, -y)\end{align*}
(0, 0) 270\begin{align*}270^\circ\end{align*}(or 90\begin{align*}-90^\circ\end{align*}) (x,y)\begin{align*}(x, y)\end{align*} (y,x)\begin{align*}(y, -x)\end{align*}
#### Example A
Line AB¯¯¯¯¯\begin{align*}\overline{AB}\end{align*} drawn from (-4, 2) to (3, 2) has been rotated about the origin at an angle of 90\begin{align*}90^\circ\end{align*}CW. Draw the preimage and image and properly label each.
Solution:
#### Example B
The diamond ABCD\begin{align*}ABCD\end{align*} is rotated 145\begin{align*}145^\circ\end{align*}CCW about the origin to form the image ABCD\begin{align*}A^\prime B^\prime C^\prime D^\prime\end{align*}. On the diagram, draw and label the rotated image.
Solution:
Notice the direction is counter-clockwise.
#### Example C
The following figure is rotated about the origin 200\begin{align*}200^\circ\end{align*}CW to make a rotated image. On the diagram, draw and label the image.
Solution:
Notice the direction of the rotation is counter-clockwise, therefore the angle of rotation is 160\begin{align*}160^\circ\end{align*}.
#### Concept Problem Revisited
Quadrilateral WXYZ\begin{align*}WXYZ\end{align*} has coordinates W(5,5),X(2,0),Y(2,3)\begin{align*}W(-5, -5), X(-2, 0), Y(2, 3)\end{align*} and Z(1,3)\begin{align*}Z(-1, 3)\end{align*}. Draw the quadrilateral on the Cartesian plane. Rotate the image 110\begin{align*}110^\circ\end{align*} counterclockwise about the point X\begin{align*}X\end{align*}. Show the resulting image.
### Guided Practice
1. Line ST¯¯¯¯¯\begin{align*}\overline{ST}\end{align*} drawn from (-3, 4) to (-3, 8) has been rotated 60\begin{align*}60^\circ\end{align*}CW about the point S\begin{align*}S\end{align*}. Draw the preimage and image and properly label each.
2. The polygon below has been rotated 155\begin{align*}155^\circ\end{align*}CCW about the origin. Draw the rotated image and properly label each.
3. The purple pentagon is rotated about the point A 225\begin{align*}A \ 225^\circ\end{align*}. Find the coordinates of the purple pentagon. On the diagram, draw and label the rotated pentagon.
Answers:
1.
Notice the direction of the angle is clockwise, therefore the angle measure is 60\begin{align*}60^\circ\end{align*}CW or 60\begin{align*}-60^\circ\end{align*}.
2.
Notice the direction of the angle is counter-clockwise, therefore the angle measure is 155\begin{align*}155^\circ\end{align*}CCW or 155\begin{align*}155^\circ\end{align*}.
3.
The measure of BAB=mBAE+mEAB\begin{align*}\angle BAB^\prime = m \angle BAE^\prime + m \angle E^\prime AB^\prime\end{align*}. Therefore \begin{align*}\angle BAB^\prime = 111.80^\circ + 113.20^\circ\end{align*} or \begin{align*}225^\circ\end{align*}. Notice the direction of the angle is counter-clockwise, therefore the angle measure is \begin{align*}225^\circ\end{align*}CCW or \begin{align*}225^\circ\end{align*}.
### Explore More
1. Rotate the above figure \begin{align*}90^\circ\end{align*} clockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} clockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
1. Rotate the above figure \begin{align*}90^\circ\end{align*} counterclockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} counterclockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
1. Rotate the above figure \begin{align*}90^\circ\end{align*} clockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} clockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
1. Rotate the above figure \begin{align*}90^\circ\end{align*} counterclockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} counterclockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
1. Rotate the above figure \begin{align*}90^\circ\end{align*} clockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} clockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
1. Rotate the above figure \begin{align*}90^\circ\end{align*} counterclockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} counterclockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
1. Rotate the above figure \begin{align*}90^\circ\end{align*} clockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} clockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
1. Rotate the above figure \begin{align*}90^\circ\end{align*} counterclockwise about the origin.
2. Rotate the above figure \begin{align*}270^\circ\end{align*} counterclockwise about the origin.
3. Rotate the above figure \begin{align*}180^\circ\end{align*} about the origin.
### Answers for Explore More Problems
To view the Explore More answers, open this PDF file and look for section 10.8.
### Vocabulary Language: English
Rotation
Rotation
A rotation is a transformation that turns a figure on the coordinate plane a certain number of degrees about a given point without changing the shape or size of the figure.
Please wait...
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It"s very common once learning around fractions to want to recognize how transform a fraction like 3/36 into a percentage. In this step-by-step guide, we"ll present you exactly how to rotate any portion into a percent really easily. Let"s take a look!
Want to easily learn or show students just how to convert 3/36 to a percentage? pat this very quick and fun video now!
Before we gain started in the portion to percent conversion, let"s go over some really quick portion basics. Remember the a numerator is the number over the portion line, and also the denominator is the number below the portion line. We"ll usage this later on in the tutorial.
You are watching: 3 is what percent of 36
When we space using percentages, what we are really speak is the the percent is a portion of 100. "Percent" means per hundred, and so 50% is the very same as saying 50/100 or 5/10 in fraction form.
So, due to the fact that our denominator in 3/36 is 36, we could adjust the portion to make the denominator 100. To carry out that, we divide 100 by the denominator:
100 ÷ 36 = 2.7777777777778
Once we have that, we deserve to multiple both the numerator and also denominator by this multiple:
3 x 2.7777777777778/36 x 2.7777777777778=8.3333333333333/100
Now we can see that our portion is 8.3333333333333/100, which means that 3/36 as a percent is 8.3333%.
We can also work this the end in a simpler way by an initial converting the portion 3/36 come a decimal. To perform that, we just divide the molecule by the denominator:
Once we have actually the prize to the division, we deserve to multiply the answer by 100 to do it a percentage:
And there you have actually it! Two different ways to transform 3/36 come a percentage. Both space pretty straightforward and easy to do, however I personally choose the transform to decimal an approach as that takes much less steps.
I"ve watched a many students get puzzled whenever a question comes up about converting a fraction to a percentage, however if you follow the measures laid out below it must be simple. That said, you might still require a calculator because that more complex fractions (and friend can constantly use our calculator in the type below).
If you desire to practice, grab you yourself a pen, a pad, and also a calculator and shot to convert a few fractions to a percent yourself.
Hopefully this tutorial has helped you come understand just how to transform a fraction to a percentage. You deserve to now walk forth and also convert fractions to percentages as much as your little heart desires!
If you found this content valuable in her research, please execute us a good favor and also use the tool below to make sure you properly reference us wherever you usage it. Us really evaluate your support!
"What is 3/36 as a percentage?". civicpride-kusatsu.net. Accessed ~ above November 11, 2021. Https://civicpride-kusatsu.net/calculator/fraction-as-percentage/what-is-3-36-as-a-percentage/.
"What is 3/36 together a percentage?". civicpride-kusatsu.net, https://civicpride-kusatsu.net/calculator/fraction-as-percentage/what-is-3-36-as-a-percentage/. Accessed 11 November, 2021.
See more: Watch Spice And Wolf Season 2 Episode 3, Wolf And The Gap That Cannot Be Filled |
# Surds and Indices
### Exponents and Powers
Very large numbers are difficult to read, understand, compare and operate upon. To make all these easier, you can use exponents, converting many of the large numbers in a shorter form.
For example,
• 10,000 = 104 (read as 10 raised to 4)
• 243 = 35
• 128 = 27
Here, 10, 3 and 2 are the bases, whereas 4, 5 and 7 are their respective exponents. You can also say, 10,000 is the 4th power of 10, 243 is the 5th power of 3, etc.
Very small numbers can be expressed in standard form using negative exponents.
### Three Laws of Indices
There are just 3 laws and from those, you can derive another 3 other interesting rules.
1. xm.xn=x(m+n)
2. xm/xn=x(m-n)
3. (xm)n=xmn
### Three Additional Rules of Indices
4. Anything to the power zero is 1. Put n=m in the above second law, you will get this rule.
x= 1
5. Rule of negative powers; Put n=-m in the first law above to get this rule.
x-m = 1/xm
6. Rule of fractional powers; Put n=1/m in the third law above to get this rule.
(xm)1/m = x
### Other Rules
7. xm x ym = (xy)m
8. xm/ym = (x/y)m
### Surds
If n is a positive integer and a is a positive rational number (a > 0), then n√a is called a surd of order n.
Simple Surd: A surd which consists of a single term is called surd or monomial surd.
Mixed Surd: If a is a rational number and √b is a surd, then a + √b, a - √b are called mixed surds.
Compound Surd: A surd which is a sum or difference of two or more surds is called a compound surd.
Similar Surds: If two surds are different multiples of the same surd, they are called similar surds otherwise they are dissimilar surds.
Rationalization of a Surd
If the product of two surds is rational, then each of the two surds is called a rationalizing factor of the other. In general, if the surd is of type a + √b, then its rationalizing factor a - √b. |
# Difference between revisions of "2017 AMC 8 Problems/Problem 19"
## Problem 19
For any positive integer $M$, the notation $M!$ denotes the product of the integers $1$ through $M$. What is the largest integer $n$ for which $5^n$ is a factor of the sum $98!+99!+100!$ ?
$\textbf{(A) }23\qquad\textbf{(B) }24\qquad\textbf{(C) }25\qquad\textbf{(D) }26\qquad\textbf{(E) }27$
## Solution 1
Factoring out $98!+99!+100!$, we have $98!(10,000)$. Next, $98!$ has $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{98}{25}\right\rfloor = 19 + 3 = 22$ factors of $5$. The $19$ is because of all the multiples of $5$. Now $10,000$ has $4$ factors of $5$, so there are a total of $22 + 4 = \boxed{\textbf{(D)}\ 26}$ factors of $5$.
## Solution 2
The number of $5$'s in the factorization of $98! + 99! + 100!$ is the same as the number of trailing zeroes. The number of zeroes is taken by the floor value of each number divided by $5$, until you can't divide by $5$ anymore. Factorizing $98! + 99! + 100!$, you get $98!(1+99+9900)=98!(10000)$. To find the number of trailing zeroes in 98!, we do $\left\lfloor\frac{98}{5}\right\rfloor + \left\lfloor\frac{19}{5}\right\rfloor= 19 + 3=22$. Now since $10000$ has 4 zeroes, we add $22 + 4$ to get $\boxed{\textbf{(D)}\ 26}$ factors of $5$. |
# Multiplication Chart 1-9
Studying multiplication soon after counting, addition, as well as subtraction is ideal. Kids learn arithmetic via a normal progression. This advancement of learning arithmetic is generally the pursuing: counting, addition, subtraction, multiplication, lastly section. This declaration results in the question why understand arithmetic within this series? Furthermore, why discover multiplication right after counting, addition, and subtraction before section?
## These facts respond to these queries:
1. Young children learn counting initially by associating visible objects making use of their hands. A real case in point: The amount of apples are there within the basket? Far more abstract case in point is just how old are you presently?
2. From counting figures, the subsequent rational stage is addition combined with subtraction. Addition and subtraction tables can be extremely beneficial teaching aids for youngsters as they are visible instruments generating the cross over from counting less difficult.
3. That ought to be figured out after that, multiplication or department? Multiplication is shorthand for addition. At this time, young children use a organization grasp of addition. Consequently, multiplication may be the next rational method of arithmetic to understand.
## Overview basic principles of multiplication. Also, review the basics utilizing a multiplication table.
Allow us to overview a multiplication instance. Using a Multiplication Table, multiply a number of times about three and obtain a response a dozen: 4 by 3 = 12. The intersection of row about three and column a number of of a Multiplication Table is a dozen; a dozen is definitely the answer. For kids starting to discover multiplication, this is certainly effortless. They may use addition to eliminate the issue hence affirming that multiplication is shorthand for addition. Illustration: 4 by 3 = 4 4 4 = 12. It is really an superb overview of the Multiplication Table. The added benefit, the Multiplication Table is visible and displays straight back to studying addition.
## In which should we commence studying multiplication using the Multiplication Table?
1. First, get informed about the table.
2. Get started with multiplying by a single. Start off at row number 1. Go on to line primary. The intersection of row 1 and column the first is the best solution: a single.
3. Recurring these techniques for multiplying by one. Multiply row a single by posts a single through 12. The solutions are 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12 respectively.
4. Replicate these actions for multiplying by two. Flourish row two by posts 1 through five. The replies are 2, 4, 6, 8, and 10 correspondingly.
5. Allow us to jump forward. Replicate these actions for multiplying by five. Increase row several by posts one through 12. The solutions are 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, and 60 respectively.
6. Now let us boost the quantity of issues. Replicate these actions for multiplying by about three. Increase row about three by posts 1 by way of 12. The solutions are 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, and 36 correspondingly.
7. In case you are confident with multiplication to date, consider using a analyze. Remedy the next multiplication troubles in your head and after that assess your answers for the Multiplication Table: multiply 6 as well as two, flourish nine and 3, grow a single and 11, grow 4 and a number of, and multiply six and 2. The trouble responses are 12, 27, 11, 16, and 14 correspondingly.
When you received a number of from five difficulties proper, build your very own multiplication exams. Estimate the answers in your mind, and look them while using Multiplication Table. |
# How do you find the vertical, horizontal or slant asymptotes for y=(3x^2+x-4) / (2x^2-5x) ?
Nov 29, 2016
vertical asymptotes at $x = 0 \text{ and } x = \frac{5}{2}$
horizontal asymptote at $y = \frac{3}{2}$
#### Explanation:
The denominator of y cannot be zero as this would make y undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.
solve : $2 {x}^{2} - 5 x = 0 \Rightarrow x \left(2 x - 5\right) = 0$
$\Rightarrow x = 0 \text{ and " x=5/2" are the asymptotes}$
Horizontal asymptotes occur as
${\lim}_{x \to \pm \infty} , y \to c \text{ ( a constant)}$
divide terms on numerator/denominator by the highest power of x, that is ${x}^{2}$
$y = \frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{x}{x} ^ 2 - \frac{4}{x} ^ 2}{\frac{2 {x}^{2}}{x} ^ 2 - \frac{5 x}{x} ^ 2} = \frac{3 + \frac{1}{x} - \frac{4}{x} ^ 2}{2 - \frac{5}{x}}$
as $x \to \pm \infty , y \to \frac{3 + 0 - 0}{2 - 0}$
$\Rightarrow y = \frac{3}{2} \text{ is the asymptote}$
Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no slant asymptotes.
graph{(3x^2+x-4)/(2x^2-5x) [-10, 10, -5, 5]} |
# Conditional probability
A conditional probability contains a condition which forces you to focus your attention to a subset of the sample space. For example, a company may have males and females working for the company. However, you may want to answer questions about males only or females only. If you are dealing with insurance, you may want to answer questions about smokers only or non-smokers only.
A good way to get started with conditional probabilities is to use a contingency table.
## Conditional probability using a contingency table
Here is how to find the conditional probability using a contingency table that we used in the lesson about marginal probability. The table shows test results for 200 students who took a GED test.
From the list of 200 students, we select a student randomly. However, suppose that you already know the student selected is a male.
The fact that you know the student is a male means that the event has already occurred. And it forces you to focus your attention only on males or 102 possible outcomes.
### What is conditional probability?
Knowing that the student is a male, you can calculate the probability that this student has passed or failed. This kind of probability is called conditional probability
The notation to find the probability that 'a student has passed if the student is male is
P(a student has passed / male)
You could in fact compute any of the following 8 conditional probabilities:
• P(a student has passed / male)
• P(a student has passed / female)
• P(a student has failed / male)
• P(a student has failed / female)
• P(a student is male / passed)
• P(a student is male / failed)
• P(a student is female / passed)
• P(a student is female / failed)
### A couple of examples showing how to find the conditional probability using a contingency table
Example #1
Let us compute the P(a student has passed / male).
If the student is male, then the student will be picked from the list of 102 males.
From this list only 46 students have passed.
P(a student has passed / male) =
Number of males who passed / Total number of males
P(a student has passed / male) =
46 / 102
= 0.451
Example #2
What about P(a student is male / passed) ?
The number of students who passed is equal to 114.
From this list, only 46 students are males.
P(a student is male / passed) =
a student is a male / Number of students who passed
P(a student is male / passed) =
46 / 104
= 0.403
As you can see from the results P(a student has passed / male) is not equal to
P(a student is male / passed) because there is a difference.
P(a student has passed / male): This probability just shows the success rate of males only.
P(a student is male / passed): This probability compares the success rate of males to females.
## Conditional probability formula
Take a close look again at the following ratio:
P(a student has passed / male) =
Number of males who passed / Total number of males
Let M be the event 'the student is a male'
Let P be the event 'the student has passed'
Let P∩M be the event 'the student is a male and has passed'
n(P∩M) = number of male students who passed = 46
n(M) = total number of male students = 102
P(P / M) =
n(P ∩ M) / n(M)
P(P / M) =
46 / 102
= 0.451
We can get the same answer using probability instead of counting. Divide the numerator and the denominator of the ratio immediately above by 200.
P(P / M) =
46 / 200 / 102 / 200
P(P / M) =
0.23 / 0.51
= 0.451
P(P∩M) = probability that a student has passed if the student is male = 46 / 200 = 0.23
P(M) = probability that a student is male = 102 / 200 = 0.51
P(P / M) =
P(P ∩ M) / P(M)
We can then conclude that there are two ways to find a conditional probability.
### How to find the conditional probability by counting
If you are dealing with equally likely outcomes such as tossing a coin or a fair die with six sides, then for any two events A and B, you can use the following formula:
P(A / B) =
n(A ∩ B) / n(B)
### How to find the conditional probability by using the definition of conditional probability
Whether you are dealing with equally likely outcomes or notthen for any two events A and B, you can use the following formula:
P(A / B) =
P(A ∩ B) / P(B)
The probability of A given B is the ratio of the probability of the intersection of A and B to the probability of B.
## More examples of conditional probability
Example #3
A card is drawn at random from a standard deck. The card is not replaced. Find the probability that the second card is a king given that the first card drawn was a king.
Let K1 be event 'the first card drawn is a king' and K2 be the event 'the second card drawn is a king'
If a king is drawn and not replaced, then there are 3 kings left and the deck will now have 51 cards.
P(K2 / K1) = 3/51 ≈ 0.0588
Two events A and B are called independent events if P(A / B) = P(A)
Example #4
Let H1 be the event that the first toss of a coin is a head and let H2 be the event that the second toss of the coin is a head. Show that H1 and H2 are independent events.
Your calculations must show that P(H2 / H1) = P(H2)
The full sample space is {HH, HT, TH, TT}
H2 = {HH, TH} and P(H2) = 0.50
Given that the first toss is a head, we end up with {HH, HT} and we are restricted to these two outcomes to compute P(H2 / H1)
From these two outcomes, we see that HH (half of the 2 outcomes) has a head as the second toss.
P(H2 / H1) = 0.5
P(H2 / H1) = P(H2) = 0.5, and thus H1 and H2 are independent events.
In independent events, the occurrence of an event does not influence the occurrence of another event. In example #4, the event 'you get a head with the first toss or H1' will not influence the likelihood of getting again a head with the second toss.
This is not the case with example #3 where the occurrence of an event can influence the occurrence of another.
Two events A and B are called dependent events if P(A / B) ≠ P(A)
In example #3, P(K2 / K1) = 3/51. However, P(K2) = 4/52 = 0.076
Since the first card was not replaced or put back in the deck, the probability of the second draw clearly depends on the outcome of the first,
100 Tough Algebra Word Problems.
If you can solve these problems with no help, you must be a genius!
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# Find the solution of the differential equation that satisfies the given initial
Find the solution of the differential equation that satisfies the given initial condition. $$dy/dx=x/y, y(0)=-3$$
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Nathaniel Kramer
Step 1
$$\displaystyle{\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}}={\frac{{{x}}}{{{y}}}}$$
we can rewrite it in differential notation and integrate:
xdx=ydy
$$\displaystyle\int{x}{\left.{d}{x}\right.}=\int{y}{\left.{d}{y}\right.}$$
$$\displaystyle{\frac{{{1}}}{{{2}}}}{x}^{{2}}+{c}={\frac{{{1}}}{{{2}}}}{y}^{{2}}$$
$$\displaystyle{y}^{{2}}={x}^{{2}}+{C}$$
$$\displaystyle{y}=\pm\sqrt{{{x}^{{2}}+{c}}}$$
square root always positive and initial value of y is negative
$$\displaystyle{y}=-\sqrt{{{x}^{{2}}+{C}}}$$
Since y(0)=-3
$$\displaystyle{y}{\left({0}\right)}=-\sqrt{{{0}^{{2}}+{C}}}=-{3}$$
C=9
$$\displaystyle{y}=-\sqrt{{{x}^{{2}}+{9}}}$$
Result
$$\displaystyle{y}=-\sqrt{{{x}^{{2}}+{9}}}$$ |
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Chapter 5
# Chapter 5 - 1 Find the EAR in each of the following...
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1. Find the EAR in each of the following cases (Do not include the percent signs (%). Use 365 days in a year. Round your answers to 2 decimal places (e.g., 32.16)) : Stated Rate (APR) Number of Times Compounded Effective Rate (EAR) 11.7 5 % Quarterly % 14.2 5 Monthly % 17.7 5 Daily % 13.7 5 Semiannually % Explanation: For discrete compounding, to find the EAR, we use the equation: EAR = [1 + (APR / m )] m – 1 EAR = [1 + (0.1175 / 4)] 4 – 1 = 0.1228 or 12.28% EAR = [1 + (0.1425 / 12)] 12 – 1 = 0.1522 or 15.22% EAR = [1 + (0.1775 / 365)] 365 – 1 = 0.1942 or 19.42% EAR = [1 + (0.1375 / 2)] 2 – 1 = 0.1422 or 14.22% Calculator Solution: Enter 11.75% 4 NOM EFF
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C/Y Solve for 12.28% Enter 14.25% 12 NOM EFF C/Y Solve for 15.22% Enter 17.75% 365 NOM EFF
C/Y Solve for 19.42% Enter 13.75% 2 NOM EFF C/Y Solve for 14.22% 2. First National Bank charges 10.9 percent compounded monthly on its business loans. First United Bank charges 11.1 percent compounded semiannually. Requirement 1: Calculate the EAR for each bank. (Do not include the percent signs (%). Round your answers to 2 decimal places (e.g., 32.16).) EAR First National Bank % First United Bank %
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Requirement 2: As a potential borrower, which bank would you go to for a new loan? First United Bank Explanation: 1: For discrete compounding, to find the EAR, we use the equation: EAR = [1 + (APR / m )] m – 1 So, for each bank, the EAR is: First National: EAR = [1 + (0.109 / 12)] 12 – 1 = 0.1146 or 11.46% First United: EAR = [1 + (0.111 / 2)] 2 – 1 = 0.1141 or 11.41% 2: For a borrower, First United would be preferred since the EAR of the loan is lower. Notice that the higher APR does not necessarily mean the higher EAR. The number of compounding periods within a year will also affect the EAR. Calculator Solution: Enter 10.9% 12 NOM EFF C/Y
Solve for 11.46% Enter 11.1% 2 NOM EFF C/Y Solve for 11.41% 3. You want to buy a new sports coupe for \$75,300, and the finance office at the dealership has quoted you a 7.7 percent APR loan for 36 months to buy the car. Requirement 1: What will your monthly payments be? (Do not include the dollar sign (\$). Round your answer to 2 decimal places (e.g., 32.16).) Monthly payment \$ Requirement 2: What is the effective annual rate on this loan? (Do not include the percent sign (%). Round your answer to 2 decimal places (e.g., 32.16).)
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Effective annual rate % Explanation: 1: We first need to find the annuity payment. We have the PVA, the length of the annuity, and the interest rate. Using the PVA equation: PVA = C ({1 – [1/(1 + r ) t ]} / r ) \$75,300 = C [1 – {1 / [1 + (0.077/12)] 36 } / (0.077/12)] Solving for the payment, we get: C = \$75,300 / 32.05318 C = \$2,349.22 2: To find the EAR, we use the EAR equation: EAR = [1 + (APR / m )] m – 1 EAR = [1 + (0.077 / 12)] 12 – 1 EAR = 0.0798 or 7.98% Calculator Solution:
Enter 36 7.7% / 12 ±\$75,300 N I/Y PV PMT FV
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Solve for \$2,349.22 Enter 7.70% 12 NOM EFF
C/Y Solve for 7.98%
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Solving Trigonometric Equations
```Solving Trigonometric
Equations
Trig identities are true for all values of the
variable for which the variable is defined.
However, trig equations, like algebraic
equations, are true for some but not all values
of the variable.
Trig equations do not have unique solutions.
Trig equations have infinitely many solutions.
They differ by the period of the function, 2p or
360° for sin and cos, and p or 180° for tan.
Solving Trig Equations
Trig equations will have unique solutions if the
value of the function is restricted to two
These solutions are called the principal values.
For sin x and tan x, the principal values are in
Quadrants I or IV. x is in the interval
-90° < x < 90°.
For cos x, the principal values are in Q I or II.
So x is in the interval 0° < x < 180°.
Solve 2 cos² x – 5 cos x + 2 = 0
for the principal values of x.
2 cos² x – 5 cos x + 2 = 0
(2 cos x - 1) (cos x – 2) = 0
2 cos x - 1 = 0
or
Factor
cos x – 2 = 0
2 cos x = 1
cos x = 2
cos x = ½
There is no solution for cos x = 2
since –1 < cos x < 1
x = 60°
Solve 2 tan x sin x + 2 sin x = tan x + 1
for all values of x.
2 tan x sin x + 2 sin x = tan x + 1
Subtract tan x + 1 from
2 tan x sin x + 2 sin x – tan x – 1 = 0 both sides.
(tan x + 1) (2 sin x – 1) = 0
factor
When all the values
tan x + 1 = 0
or
2 sin x – 1 = 0 of x are required,
the solution should
2 sin x = 1
tan x = -1
be represented as
sin x = ½
x + 360k° for sin
x = -45° + 180k°
x = 30° + 360k° and cos, and
x + 180k° for tan,
or
x = 150° + 360k° where k is any
integer.
Solve sin² x + cos 2x – cos x = 0
for the principal values of x.
sin² x + cos 2x – cos x = 0
sin² x + (1 – 2 sin² x) – cos x = 0
Express cos 2x in terms
of sin x.
1 – sin² x – cos x = 0 Combine like terms
1 – sin² x = cos² x
cos² x – cos x = 0
Factor
cos x (cos x – 1) = 0
So the solutions
or
cos x = 0
cos x – 1 = 0 are 0° and 90°
cos x = 1
x = 90°
x = 0°
Solve cos x = 1 + sin x
for 0° < x < 360°
cos x = 1 + sin x
cos² x = (1 + sin x)² Square both sides
cos² x = 1 + 2 sin x + sin² x Expand the binomial squared
1 – sin² x = 1 + 2 sin x + sin² x cos² x = 1 – sin² x
0 = 2 sin x + 2 sin² x
0 = 2 sin x (1 + sin x)
Factor
or
2 sin x = 0
1 + sin x = 0
sin x = -1
sin x = 0
x = 270º
x = 0° or 180º
It’s important to always check your
solutions. Some may not actually be
solutions to the original equation.
x = 0º or 180º
x = 270º
cos x = 1 + sin x
cos 0º = 1 + sin 0°
cos 180º = 1 + sin 180º
1=1 ☻
-1 = 1 + 0
-1 ≠ 1
cos 270º = 1 + sin 270º Based on the check,
0 = 1 + (-1)
180º is not a solution
0=0 ☻
Assignment
Page 390
– # 17 - 38
``` |
# 2001 AMC 12 Problems/Problem 7
The following problem is from both the 2001 AMC 12 #7 and 2001 AMC 10A #14, so both problems redirect to this page.
## Problem
A charity sells $140$ benefit tickets for a total of (Error compiling LaTeX. ! Missing $inserted.)2001$. Some tickets sell for full price (a whole dollar amount), and the rest sells for half price. How much money is raised by the full-price tickets?$\text{(A) }$$782\qquad \text{(B) }$$986\qquad \text{(C) }$$1158\qquad \text{(D) }$$1219\qquad \text{(E) }$$1449$
## Solution
Let's multiply ticket costs by $2$, then the half price becomes an integer, and the charity sold $140$ tickets worth a total of $4002$ dollars.
Let $h$ be the number of half price tickets, we then have $140-h$ full price tickets. The cost of $140-h$ full price tickets is equal to the cost of $280-2h$ half price tickets.
Hence we know that $h+(280-2h) = 280-h$ half price tickets cost $4002$ dollars. Then a single half price ticket costs $\frac{4002}{280-h}$ dollars, and this must be an integer. Thus $280-h$ must be a divisor of $4002$. Keeping in mind that $0\leq h\leq 140$, we are looking for a divisor between $140$ and $280$, inclusive.
The prime factorization of $4002$ is $4002=2\cdot 3\cdot 23\cdot 29$. We can easily find out that the only divisor of $4002$ within the given range is $2\cdot 3\cdot 29 = 174$.
This gives us $280-h=174$, hence there were $h=106$ half price tickets and $140-h = 34$ full price tickets.
In our modified setting (with prices multiplied by $2$) the price of a half price ticket is $\frac{4002}{174} = 23$. In the original setting this is the price of a full price ticket. Hence $23\cdot 34 = \boxed{(\text{A})782}$ dollars are raised by the full price tickets. |
Class 7 Maths
Properties of Triangles NCERT Exercise 6.4
Question 1: Is it possible to have a triangle with the following sides?
1. 2 cm , 3 cm, 5 cm
2. 3 cm, 6 cm, 7 cm
3. 6 cm, 3 cm, 2 cm
Answer: Making a triangle is possible only with sides given in option ‘b’. In other options, sum of two sides is either equal to or less than the third side.
Question 2: Take any point O in the interior of a triangle PQR. Is
1. OP + OQ > PQ?
2. OQ + OR > QR?
3. OR + OP > RP?
Answer: The answer is yes in each case, because sum of any two sides of a triangle is always greater than the third side.
Question 3: AM is a median of a triangle ABC. Is AB + BC + CA > 2AM? (Consider the sides of ΔABM and ΔAMC)
Answer: In ΔABM, AB + BM > AM
Similarly, in ΔAMC, AC + CM > AM
AB + BM + CM + AC > 2AM
Or, AB + BC + CA > 2AM
Question 4: ABCD is a quadrilateral. Is AB + BC + CD + DA > AC + BD?
Answer: In ΔABC; AB + BC > AC
In ΔDAC, DA + CD > AC
In ΔDAB, DA + AB > DB
In ΔDCB, CD + CB > DB
2AB + 2BC + 2 CD + 2AD > 2AC + 2BD
Or, 2(AB + BC + CD + AD) > 2(AC + BD)
Or, AB + BC + CD + AD > AC + BD
Question 5: ABCD is a quadrilateral. Is AB + BC + CD + DA < 2(AC + BD)?
Answer: Let us assume a point O at the point of intersection of diagonals AC and BD.
In ΔOAB, OA + OB > AB
In Δ OBC, OB + OC > BC
In ΔODC, OD + OC > CD
AB + BC + CD + DA < OA + OB + OB + OC + OC + OD + OD + OA
Or, AB + BC + CD + DA < OA + OA + OC + OC + OD + OD + OB + OB
Or, AB + BC + CD + DA < 2(OA + OC + OD + OB)
Or, AB + BC + CD + DA < 2(AC + BD)
Question 6: The lengths of two sides of a triangle are 12 cm and 15 cm. Between what two measures should the length of the third side fall?
Answer: Sum of given two sides = 12 cm + 15 cm = 27 cm
Hence, the third side should always be less than 27 cm.
The difference between given sides = 15 cm – 12 cm = 3 cm
If the third side will be = 3 cm then 12 + 3 = 15 cm shall be equal to one of the given sides.
Hence, the third side should be more than 3 cm
So, range of measure of third side = 4 cm to 26 cm. |
# Linear interpolation
A straight-line estimation method for determining an intermediate value.
Example 1: Interpolation
Consider a set of cashflows which has:
Net present value (NPV) of +\$4m at a yield of 5%.
NPV of -\$4m at a yield of 6%.
Using linear interpolation, the estimated yield at which the cashflows have an NPV of \$0 is given by:
IRR estimate = a% + ( A / ( A - B) ) x (b - a)%
Where:
a% = first estimated yield = 5%
b% = second estimated yield = 6%
A = NPV of cash flows at yield of a% = +\$4m
B = NPV of cash flows at yield of b% = -\$4m
IRR estimate:
= 5% + ( +4 / ( +4 - -4) ) x (6 - 5)%
= 5% + ( +4 / +8 ) x 1%
= 5% + 0.5%
= 5.5%.
5.5% is the estimated internal rate of return (IRR) of the cashflows.
## Interpolation and Iteration
Interpolation is often used in conjunction with Iteration.
Using iteration, the straight-line estimated IRR of 5.5% would then be used, in turn, to recalculate the NPV at the estimated IRR of 5.5%, producing a recalculated NPV even closer to \$0.
5.5% and the recalculated NPV would then be used with interpolation once again to further refine the estimate of the IRR.
This iteration process can be repeated as often as required until the result converges on a sufficiently stable final figure.
## Extrapolation
Another closely related linear estimation technique is extrapolation.
This involves the straight-line estimation of values outside the range of the sample data used to do the estimation with.
Example 2: Extrapolation
Using the following data to estimate net present value (NPV) at a yield of 7%, using extrapolation:
NPV of +\$4m at a yield of 5%.
NPV of -\$4m at a yield of 6%.
Solution
Based on the sample data, for every 1% increase in the yield, the NPV moved by:
-\$4m - \$4m = -\$8m
Extrapolating this trend to a yield of 7%, this is a further increase in the yield of 7 - 6 = 1%.
The NPV would be modelled to fall from -\$4m to:
= -\$4m - \$8m
= -\$12m. |
# 2009 AMC 10B Problems/Problem 21
## Problem
What is the remainder when $3^0 + 3^1 + 3^2 + \cdots + 3^{2009}$ is divided by 8?
$\mathrm{(A)}\ 0\qquad \mathrm{(B)}\ 1\qquad \mathrm{(C)}\ 2\qquad \mathrm{(D)}\ 4\qquad \mathrm{(E)}\ 6$
## Solution
### Solution 1
The sum of any four consecutive powers of 3 is divisible by $3^0 + 3^1 + 3^2 +3^3 = 40$ and hence is divisible by 8. Therefore
$(3^2 + 3^3 + 3^4 + 3^5) + \cdots + (3^{2006} + 3^{2007} + 3^{2008} + 3^{2009})$
is divisible by 8. So the required remainder is $3^0 + 3^1 = \boxed {4}$. The answer is $\mathrm{(D)}$.
### Solution 2
We have $3^2 = 9 \equiv 1 \pmod 8$. Hence for any $k$ we have $3^{2k}\equiv 1^k = 1 \pmod 8$, and then $3^{2k+1} = 3\cdot 3^{2k} \equiv 3\cdot 1 = 3 \pmod 8$.
Therefore our sum gives the same remainder modulo $8$ as $1 + 3 + 1 + 3 + 1 + \cdots + 1 + 3$. There are $2010$ terms in the sum, hence there are $2010/2 = 1005$ pairs $1+3$, and thus the sum is $1005 \cdot 4 = 4020 \equiv 20 \equiv \boxed{4} \pmod 8$.
### Solution 3
We have the formula $\frac{a(r^n-1)}{r-1}$ for the sum of a finite geometric sequence which we want to find the residue modulo 8. $$\frac{1 \cdot (3^{2010}-1)}{2}$$ $$\frac{3^{2010}-1}{2} = \frac{9^{1005}-1}{2}$$ $$\frac{9^{1005}-1}{2} \equiv \frac{1^{1005}-1}{2} \equiv \frac{0}{2} \pmod 8$$ Therefore, the numerator of the fraction is divisible by $8$. However, when we divide the numerator by $2$, we get a remainder of $4$ modulo $8$, giving us $\mathrm{(D)}$.
Note: you need to prove that $$\frac{9^{1005}-1}{2}$$ is not congruent to 0 mod 16 because if so, then the whole thing would be congruent to 0 mod 8, even after dividing by 2 ~ ilikepi12
~savannahsolver |
Master the 7 pillars of school success
Another method for remembering if the sign is" greater than" or "less than" is to look at the symbol, and remember that the larger side of the symbol always faces the larger number.
• Larger side of symbol > Smaller side of the symbol
• 7>6 seven is larger than 6 6<7 six is less than 7
Transcript
Hi welcome to MooMooMath. Today we are going to talk about comparing integers, which include positive and negative numbers. So let’s look at our first example. You have 4 comparing to negative 6 and the 4 is greater than the negative 6. Next up is negative two compared to negative 8. Negative 8 is greater than negative 2 so we will use the less that symbol. ( < ) Finally let’s look at negative 6 and absolute value of negative 6. The value of negative 6 is greater so we will use the greater than symbol (>) Now let’s look at the rules of comparing integers. The first symbol is the less than symbol (<) the way I remember this is it actually looks like the letter L. The symbol switched around (<) is greater than and notice it does not look like the letter L so it can’t be less than and the last one is equal to. Now that is how the symbols. Now let’s go to a number line. Anything to the left is smaller than anything to the right. You can look at your number line and make some comparisons in order to determine which one is greater than or less than. The first one is four compared to negative 6. I will use the number line for this and negative 6 are to the left of 4 so 4 are greater than negative 6. 4 are greater than negative 6. You want to use the symbol that the smaller end of the symbol is pointing towards the smaller number. Next up is negative 12 and negative 8. These are both negative numbers. Many times students will say that -12 is greater than -8 because it is larger but in negatives it is the opposites. On the number line the 12 is over here on the left so it is less than -8. So the -8 is larger so we will use a less than sign. Now let’s look at the absolute value. You will always take the positive, because the definition of absolute value is the distance from zero. So negative 6 is 6 units from 0 and we will compare that to the negative of absolute value and that actually is positive 6 so you are actually comparing positive 6 to -6 and positives are always greater so we will use the smaller end towards the smaller number which becomes 6 > -6 Hope this helps.
On a number line with integers any number to the left of another is less than the number on the right.
Let's look at an example problem related to this standard.
How much money is left on my \$ 20.00 Starbucks gift card if I purchase 3 tall coffees at \$3.00 each?
Step 1. 3*3 =9
Step 2. 20 -9 ( cost of 3 coffees) = \$11 amount left on the card
What is my elevation if I start hiking at 10,000 feet, climb 1000 feet, then descend 2000 feet, climb again for 2000 feet, and then descend 500 feet.
Step 1 10,000 + 1000 =11,000 feet
Step 2. 11,000-2000=9000 feet
Step 3 9000 +2000= 11,000
Step 4 11,000 -500 = 10,500 feet
Common Core Standard 6.G.1 , 7.G.6 6th Grade Math 7th Grade Math
An easy method for comparing fractions and integers is to convert the fraction into a decimal and then compare this to the integer.
For example, compare 7/8, 11/16, 6, -1,
Convert the fractions to decimals 7/8 = .875 11/16 = .6875
Now place from least to greatest.
-1 < .6875 <.875 < 6
Finally you can convert back to the original fraction and list them from least to greatest.
-1 < 11/16 <7/8 < 6
A rational number is a number that can be expressed as a fraction.
Follow these steps when comparing integers and rational numbers.
Convert the rational number into a decimal.
Make sure the decimals are lined up correctly.
For example compare 7/14, .6, 1, -5
Convert 7/14 =.5
Now you can compare .5,.6, 1, -5 by placing them from least to greatest
-5 < .5 < .6 < 1
Finally you can place them in their original form
-5< 7/14 < .6 < 1
### Comparing integers and rational numbers
You can easily compare integers when you order the integers on a number line.
# Comparing Integers
Left < Right
The smaller end of the symbol should face towards the smaller number. For example: 6<10 -8> -10 1<2
## Integers on a number line
The following symbols are used when comparing integers:
1. < equals less than, I remember this because it looks like a L for less
2. > Doesn't look like a L so it is greater than
3. = equals
There are less cats than dogs, therefore 2<3
There are more men than women, therefore 2>1
Each family has the same number of members (4) therefore 4=4
### Comparing and Ordering Integers
Whenever you have two numbers in Math you can compare these two numbers. Integers are positive or negative counting numbers, which means they are not fractions or decimals. For example, 2,35,-56 34, and 0 are all integers. 5.4,61/2, and 3/4 are not integers. When comparing integers the number is either greater than, less than, or equal to the other number. |
# How do you solve using gaussian elimination or gauss-jordan elimination, 2x+4y-6z=48, x+2y+3z=-6, 3x-4y+4z=-23?
May 19, 2017
Using Gauss elimination gives $x = 3$, $y = 3$, and $z = - 5$
#### Explanation:
Step 1. Rewrite the equations in matrix form
$\left(\begin{matrix}2 & 4 & - 6 & | & 48 \\ 1 & 2 & 3 & | & - 6 \\ 3 & - 4 & 4 & | & - 23\end{matrix}\right)$
Step 2. Switch ${R}_{2}$ with ${R}_{1}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 2 & 4 & - 6 & | & 48 \\ 3 & - 4 & 4 & | & - 23\end{matrix}\right)$
Step 3. $- 2 {R}_{1} + {R}_{2} \to {R}_{2}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 0 & - 12 & | & 60 \\ 3 & - 4 & 4 & | & - 23\end{matrix}\right)$
Step 4. $- 3 {R}_{1} + {R}_{3} \to {R}_{3}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 0 & - 12 & | & 60 \\ 0 & - 10 & - 5 & | & - 5\end{matrix}\right)$
Step 5. Switch ${R}_{2}$ with ${R}_{3}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & - 10 & - 5 & | & - 5 \\ 0 & 0 & - 12 & | & 60\end{matrix}\right)$
Step 6. $- \frac{1}{10} {R}_{2} \to {R}_{2}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 1 & \frac{1}{2} & | & \frac{1}{2} \\ 0 & 0 & - 12 & | & 60\end{matrix}\right)$
Step 7. $- \frac{1}{12} {R}_{3} \to {R}_{3}$
$\left(\begin{matrix}1 & 2 & 3 & | & - 6 \\ 0 & 1 & \frac{1}{2} & | & \frac{1}{2} \\ 0 & 0 & 1 & | & - 5\end{matrix}\right)$
Translating back into the values $x$, $y$, and $z$
$z = - 5$
$y + \frac{1}{2} \left(- 5\right) = \frac{1}{2} \implies y = 3$
$x + 2 \left(3\right) + 3 \left(- 5\right) = - 6 \implies x = 3$ |
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This essential trains for: Math Kangaroo 5-6, Math Kangaroo 7-8, SAT-I, SAT-II, GMAT, AMC-8.
Definition: a closed, non self-intersecting plane figure formed of segments.
Convex and concave:
Any two points chosen inside a convex polygon can be connected by a segment that lies entirely inside the polygon.
In a concave polygon, two points can be found such that the segment that connects them crosses the polygon's boundary and lies partly outside the polygon.
Many of the concepts used in problems are valid only for convex polygons. The problem may state that the polygon is regular just in order to ensure that the student understands it to be convex. However, such a condition is much stronger than convexity.
Various elements of a polygon:
Sum of the interior angles of a convex polygon with N sides
Choose a point P arbitrarily inside the polygon. Divide the polygon in triangles. The sum of the interior angles of each triangle is 180°. There are N triangles.
The total sum over all triangles is:
However, this includes the 360° angle around the arbitrary point P which was used to divide into triangles. We must subtract this quantity:
Sum of the exterior angles of a convex polygon with N sides
The exterior angles of a convex polygon always add up to 360°.
Any exterior angle is the supplement of an interior angle:
where x1 is an exterior angle and a1 is a corresponding interior angle.
For all N angles:
The perimeter of a polygon is the sum of the lengths of all its sides.
The perimeter and the area of an arbitrary polygon can be calculated by dividing it into triangles and by applying the geometry of the triangle.
Regular Polygons are polygons with all sides congruent and all angles congruent.
Any regular polygon is convex.
The interior angle of a regular polygon can be calculated by dividing the sum of the interior angles by the number of angles (since they are all congruent):
The regular polygon has a center which is also the center of the circle in which it can be inscribed.
Some regular polygons tile the plane, i.e. the plane can be covered by such polygons without gaps or overlaps.
Equilateral triangles and regular hexagons are in this category.
An axis of symmetry for a polygon is a line that divides the figure in two figures that are the mirror image of one another. |
In this section, the following topics are discussed with examples
• Basic Formulae
• Unit Conversion
• Relative Speed
• Problems on Trains
• Example
#### Basic Formulae
Distance=Speed * Time
Speed = Distance / Time
Time = Distance / Speed
#### Unit Conversion
• To convert m/s into km/hr:
x m/s = x * (18/5) km/hr
• To convert km/hr into m/s:
x km/hr = x * (5/18) m/s
• If a man covers a certain distance at two different speed x km/hr and y km/hr. Then,
Average speed= 2xy / (x + y)
#### Example 1:
If a person covers a distance of 75m in 2 hrs.What is his speed in km/hr?
Sol:
Distance=75m
Time = 2 hrs
Speed=x km/hr
Speed= (75 * 18) / ( 5 * 2 )km/hr
=135 km/hr.
#### Example 2:
If a person covers a distance at a speed of 8/10 of his usual speed he reaches his office 5 minutes late?What is the time taken by him to reach his office?
Sol:
s = (8/10) *s
=8/10(d/t)
T=t+5
s=d/t
(8/10) * (d/t) = d / (t + 5)
t=20
### Relative Speed
• If two objects A and B are moving in the same direction with the speed of x km/hr and y km/hr. Then, their relative speed is,
Relative Speed=(x-y)km/hr
• If two objects A and B are moving in the opposite directions with the speed of x km/hr and y km/hr.Then, the relative speed is,
Relative speed=(x+y)km/hr
• The ratio of the speeds of two objects is inversely proportional to the root of their time taken
S1/S2 = root (T2/T1)
## The ability to simplify seems to eliminate the unnecessary so that the necessary may speak.
Hans Hoffman
### Problems on Trains
Types of Problems
1. Train crossing a man/pole/lamp post
2. Train crossing a running man(Same direction)
3. Train crossing a running man(Opp direction)
4. Train crossing an another train/bridge/platform
5. Train crossing an another running train(Same direction)
6. Train crossing an another running train (Opp direction)
Formula to calculate time taken by train to cross,
1. A man/pole/lamp post, Time = Ltrain / Strain
2. A running man(Same direction), Time = Ltrain / (Strain-Sman)
3. A running man(Opp direction) , Time = Ltrain / (Strain + Sman)
4. An another train/bridge/platform, Time = Ltrain+Lobj / Strain
5. An another running train(Same direction, Time = Ltrain +Lobj / (Strain-Sobj)
6. An another running train (Opp direction), Time = Ltrain +Lobj / (Strain + Sobj)
Where,
train – Length of train
Strain – Speed of train
Sman – Speed of man
Lobj – Length of object/train 2
Sobj – Speed of object /train 2
Examples:
A train is 600 meter long and is running at the speed of 72 km per hour. Find the time taken to cross
1)a man
2)a man running at 5 m/s in same direction
3)a man running at 5 m/s in opposite direction
4)a 200 meter long platform
5)a 200 meter long train running at 10 m/s in same direction
6)a 200 meter long train running at 10 m/s in opposite direction
1. 30 sec 2. 40 sec 3. 24 sec 4. 40 sec 5.80 sec 6.80/3 sec
Explanation :
1. 600/20 =30 sec
2. 600/20-5 = 40 sec
3. 600/20+5=24 sec
4. 600+200/ 20 = 40 sec
5. 600+200/20-10=80 sec
6. 600+200/20+10= 80/3 sec
#### Example 1:
Train A travels at a speed of 90km/hr crosses a man at 8 seconds. what is the length of the train.
Soln:
Speed=90 km/hr
=90*(5/18)m/s
=25 m/s
Time=8 s
Length=25*8
=200 m
#### Example 2:
Chennai express of length 1050 m crosses a waiting man in 45 sec while it takes 90 sec to cross a platform.What is the length of the platform?
Soln:
Length of the train=1050 m
Time for man=45 s
Speed=(1050/45)m/s
Let, the length of the platform=x
(x+1050)=(1050/45)*90 (S*T=D)
x+1050=2100
x=1050
Length of the platform=1050 m |
# Lesson 1
Describing and Graphing Situations
### Problem 1
The relationship between the amount of time a car is parked, in hours, and the cost of parking, in dollars, can be described with a function.
1. Identify the independent variable and the dependent variable in this function.
2. Describe the function with a sentence of the form "$$\underline{\hspace{0.5in}}$$ is a function of $$\underline{\hspace{0.5in}}$$."
3. Suppose it costs $3 per hour to park, with a maximum cost of$12.
Sketch a possible graph of the function. Be sure to label the axes.
4. Identify one point on the graph and explain its meaning in this situation.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 2
The prices of different burgers are shown on this sign.
Based on the information from the menu, is the price of a burger a function of the number of patties? Explain your reasoning.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 3
The distance a person walks, $$d$$, in kilometers, is a function of time, $$t$$, in minutes, since the walk begins.
Select all true statements about the input variable of this function.
A:
Distance is the input.
B:
Time of day is the input.
C:
Time since the person starts walking is the input.
D:
$$t$$ represents the input.
E:
$$d$$ represents the input.
F:
The input is not measured in any particular unit.
G:
The input is measured in hours.
H:
For each input, there are sometimes two outputs.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 4
It costs $3 per hour to park in a parking lot, with a maximum cost of$12.
Explain why the amount of time a car is parked is not a function of the parking cost.
### Solution
For access, consult one of our IM Certified Partners.
### Problem 5
Here are clues for a puzzle involving two numbers.
• Seven times the first number plus six times the second number equals 31.
• Three times the first number minus ten times the second number is 29.
What are the two numbers? Explain or show your reasoning.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 2, Lesson 12.)
### Problem 6
To keep some privacy about the students, a professor releases only summary statistics about student scores on a difficult quiz.
mean standard deviation minimum Q1 median Q3 maximum
66.91 12.74 12 57 66 76 100
Based on this information, what can you know about outliers in the student scores?
A:
There is an outlier at the upper end of the data.
B:
There is an outlier at the lower end of the data.
C:
There are outliers on both ends of the data.
D:
There is not enough information to determine whether there are any outliers.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 1, Lesson 14.)
### Problem 7
An airline company creates a scatter plot showing the relationship between the number of flights an airport offers and the average distance in miles travelers must drive to reach the airport. The correlation coefficient of the line of best fit is -0.52.
1. Are they correlated? Explain your reasoning.
2. Do either of the variables cause the other to change? Explain your reasoning.
### Solution
For access, consult one of our IM Certified Partners.
(From Unit 3, Lesson 9.) |
# 3.2.7: Trigonometric Equations Using the Quadratic Formula
The quadratic formula with a trigonometric function in place of the variable.
Solving equations is a fundamental part of mathematics. Being able to find which values of a variable fit an equation allows us to determine all sorts of interesting behavior, both in math and in the sciences. Solving trig equations for angles that satisfy the equation is one application of mathematical methods for solving equations. Suppose someone gave you the following equation:
$$3 \sin^2 \theta +8 \sin\theta −3=0$$
## Quadratic Functions with Trigonometric Equations
When solving quadratic equations that do not factor, the quadratic formula is often used.
Remember that the quadratic equation is:
$$ax^2+bx+c=0$$ (where $$a$$, $$b$$, and $$c$$ are constants)
In this situation, you can use the quadratic formula to find out what values of "x" satisfy the equation.
The same method can be applied when solving trigonometric equations that do not factor. The values for $$a$$ is the numerical coefficient of the function's squared term, $$b$$ is the numerical coefficient of the function term that is to the first power and $$c$$ is a constant. The formula will result in two answers and both will have to be evaluated within the designated interval.
#### Solving for Unknown Values
1. Solve $$3 \cot ^2 x−3 \cot x=1$$ for exact values of $$x$$ over the interval $$[0,2\pi ]$$.
\begin{aligned} 3\cot ^2 x−3\cot x &=1 \\ 3\cot ^2x−3\cot x−1&=0 \end{aligned}
The equation will not factor. Use the quadratic formula for $$\cot x$$, $$a=3$$, $$b=−3$$, $$c=−1$$.
\begin{aligned} \cot x&=\dfrac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}\\ \cot x&=\dfrac{-(-3) \pm \sqrt{(-3)^{2}-4(3)(-1)}}{2(3)}\\ \cot x&=\dfrac{3 \pm \sqrt{9+12}}{6}\\ \cot x&=\dfrac{3+\sqrt{21}}{6} &\quad \text { or } \quad& \cot x=\dfrac{3-\sqrt{21}}{6}\\ \cot x&=\dfrac{3+4.5826}{6} &\quad& \cot x=\dfrac{3-4.5826}{6}\\ \cot x&=1.2638 &\quad &\cot x=-0.2638\\ \tan x&=\dfrac{1}{1.2638} &\quad &\tan x=\dfrac{1}{-0.2638}\\ x&=0.6694,3.81099 & &x=1.8287,4.9703 \end{aligned}
2. Solve $$−5 \cos ^2 x+9 \sin x+3=0$$ for values of x over the interval $$[0,2\pi ]$$.
Change $$\cos ^2 x$$ to $$1−\sin ^2 x$$ from the Pythagorean Identity.
\begin{aligned} −5\cos ^2 x +9\sin x+3 &=0 \\ −5(1−\sin ^2 x )+9\sin x+3 &=0 \\ −5+5\sin ^2 x +9\sin x+3 &=0 \\ 5\sin ^2 x +9\sin x−2 &=0 \end{aligned}
$$\begin{array}{l} \sin x=\dfrac{-9 \pm \sqrt{9^{2}-4(5)(-2)}}{2(5)} \\ \sin x=\dfrac{-9 \pm \sqrt{81+40}}{10} \\ \sin x=\dfrac{-9 \pm \sqrt{121}}{10} \\ \sin x=\dfrac{-9+11}{10} \text { and } \sin x=\dfrac{-9-11}{10} \\ \sin x=\dfrac{1}{5} \text { and }-2 \\ \sin ^{-1}(0.2) \text { and } \sin ^{-1}(-2) \end{array}$$
$$x\approx .201 \text{rad}$$ and $$\pi −.201\approx 2.941$$
This is the only solutions for $$x$$ since $$−2$$ is not in the range of values.
3. Solve $$3\sin ^2 x −6\sin x−2=0$$ for values of $$x$$ over the interval $$[0,2\pi ]$$.
$$\begin{array}{l} \quad 3 \sin ^{2} x-6 \sin x-2=0 \\ \sin x=\dfrac{6 \pm \sqrt{(-6)^{2}-4(3)(-2)}}{2(3)} \\ \sin x=\dfrac{6 \pm \sqrt{36-24}}{6} \\ \sin x=\dfrac{6 \pm \sqrt{12}}{6} \\ \sin x=\dfrac{6+3.46}{10} \text { and } \sin x=\dfrac{6-3.46}{10} \\ \sin x=.946 \text { and } .254 \\ \sin ^{-1}(0.946) \text { and } \sin ^{-1}(0.254) \end{array}$$
$$x\approx 71.08 \text{ deg}$$ and $$\approx 14.71 \text{ deg}$$
Example $$\PageIndex{1}$$
Earlier, you were asked to solve an equation.
The original equation to solve was:
3sin2\theta +8sin\theta −3=0
Solution
Using the quadratic formula, with a=3,b=8,c=−3, we get:
$$\sin\theta =\dfrac{−b \pm \sqrt{b^2−4ac}}{2a}=\dfrac{−8\pm \sqrt{64−(4)(3)(−3)}}{6}=\dfrac{−8\pm \sqrt{100}}{6}=\dfrac{−8\pm 10}{6}=\dfrac{1}{3} \text{or} −3$$
The solution of -3 is ignored because sine can't take that value, however:
$$\sin^{−1} \dfrac{1}{3}=19.471^{\circ}$$
Example $$\PageIndex{2}$$
Solve $$\sin ^2 x −2\sin x−3=0$$ for $$x$$ over $$[0,\pi ]$$.
Solution
You can factor this one like a quadratic.
\begin{aligned} \sin ^{2} x-2 \sin x-3 &=0 & & \\ (\sin x-3)(\sin x+1) &=0 & & \\ \sin x-3=0 & & & \\ \sin x=3 & & \text { or } & \sin x+1=0 \\ x=\sin ^{-1}(3) & & &x=\dfrac{3 \pi}{2} \end{aligned}
For this problem the only solution is $$\dfrac{3\pi}{2}$$ because sine cannot be $$3$$ (it is not in the range).
Example $$\PageIndex{3}$$
Solve $$\tan^2 x+\tan x−2=0$$ for values of $$x$$ over the interval $$\left[−\dfrac{\pi}{2},\; \dfrac{\pi}{2}\right]$$.
Solution
$$\tan^2 x+\tan x−2=0$$
\begin{aligned} -1 \pm \sqrt{1^{2}-4(1)(-2)} &=\tan x \\ 2 & \\ \dfrac{-1 \pm \sqrt{1+8}}{2} &=\tan x \\ \dfrac{-1 \pm 3}{2} &=\tan x \\ \tan x &=-2 \quad \text { or } \; 1 \end{aligned}
$$\tan x=1$$ when $$x=\dfrac{\pi}{4}$$, in the interval $$\left[−\dfrac{\pi}{2},\; \dfrac{\pi}{2}\right]$$
$$\tan x=−2$$ when $$x=−1.107 \text{rad}$$
Example $$\PageIndex{4}$$
Solve the trigonometric equation such that $$5 \cos^2 \theta −6 \sin\theta =0$$ over the interval $$[0,2\pi ]$$.
Solution
$$5\cos^2\theta −6\sin \theta =0$$ over the interval $$[0,2\pi ]$$.
\begin{aligned} 5\left(1-\sin ^{2} x\right)-6 \sin x &=0 \\ -5 \sin ^{2} x-6 \sin x+5 &=0 \\ 5 \sin ^{2} x+6 \sin x-5 &=0 \\ -6 \pm \sqrt{6^{2}-4(5)(-5)} &=\sin x \\ 2(5) & \\ \dfrac{-6 \pm \sqrt{36+100}}{10} &=\sin x \\ \dfrac{-6 \pm \sqrt{136}}{10} &=\sin x \\ \dfrac{-6 \pm 2 \sqrt{34}}{10} &=\sin x \\ \dfrac{-3 \pm \sqrt{34}}{5} &=\sin x \end{aligned}
$$x=\sin^{−1}\left(\dfrac{−3+\sqrt{34}}{5}\right)$$ or $$\sin^{−1}\left(\dfrac{−3−\sqrt{34}}{5}\right) x=0.6018 \text{ rad}$$ or $$2.5398 \text{ rad}$$ from the first expression, the second expression will not yield any answers because it is out the range of sine.
## Review
Solve each equation using the quadratic formula.
1. $$3x^2+10x+2=0$$
2. $$5x^2+10x+2=0$$
3. $$2x^2+6x−5=0$$
Use the quadratic formula to solve each quadratic equation over the interval $$[0,2\pi )$$.
1. $$3\cos^2(x)+10\cos(x)+2=0$$
2. $$5\sin^2(x)+10\sin(x) +2=0$$
3. $$2\sin^2(x)+6\sin(x) −5=0$$
4. $$6\cos^2(x)−5\cos(x)−21=0$$
5. $$9\tan^2(x)−42\tan(x) +49=0$$
6. $$\sin^2(x)+3\sin(x) =5$$
7. $$3\cos^2(x)−4\sin(x) =0$$
8. $$−2\cos^2(x)+4\sin(x) =0$$
9. $$\tan^2(x)+\tan(x) =3$$
10. $$\cot^2(x)+5\tan(x) +14=0$$
11. $$\sin^2(x)+\sin(x) =1$$
12. What type of sine or cosine equations have no solution?
Quadratic Equation A quadratic equation is an equation that can be written in the form $$=ax^2+bx+c=0$$, where $$a$$, $$b$$, and $$c$$ are real constants and $$a\neq 0$$. |
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Algebra is the branch of mathematics that deals with the equations and rules governing the manipulation of algebraic expressions. Algebra is used in solving mathematical problems and in discovering new mathematical truths. Algebra is based on the concept of variables, which are symbols that represent unknown numbers or quantities. Algebra is used to solve equations, which are mathematical statements that state that two expressions are equal. The process of solving an equation for a variable is called solving for x. To solve for x, one must first identify the equation's variables and then use algebraic methods to solve for the variable. Algebraic methods include using addition, subtraction, multiplication, and division to solve for a variable. In some cases, algebraic equations can be solve by using exponential or logarithmic functions. Algebra is a powerful tool that can be used to solve mathematical problems and discover new mathematical truths.
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# Sine, Cosine and Tangent
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Last updated date: 18th Sep 2024
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## What is Sine, Cosine and Tangent?
Sine, cosine, and tangent (abbreviated as sin, cos, and tan) are three primary trigonometric functions, which relate an angle of a right-angled triangle to the ratios of two sides length. The reciprocals of sine, cosine, and tangent are the secant, the cosecant, and the cotangent respectively. Each of the six trigonometric functions has corresponding inverse functions (also known as inverse trigonometric functions). The trigonometric functions also known as the circular functions, angle functions, or goniometric functions are widely used in all fields of science that are related to Geometry such as navigation, celestial mechanics, solid mechanics, etc.
Read below to know what is a sine function, cosine function, and tangent function in detail.
### Sine Cosine Tangent Definition
A right-angled triangle includes one angle of 90 degrees and two acute angles. Each acute angle of a right-angled triangle retains the property of the sine cosine tangent. The sine, cosine, and tangent of an acute angle of a right-angled triangle are defined as the ratio of two of three sides of the right-angled triangle.
As we know, sine, cosine, and tangent are based on the right-angled triangle, it would be beneficial to give names to each of the triangles to avoid confusion.
• “Hypotenuse side” is the longest side.
• “Adjacent side” is the side next to angle θ.
• “Opposite side” is the side opposite to angle θ.
Accordingly,
Sin θ = Opposite side/Hypotenuse
### What is the Sine Function?
In the right triangle, the sine function is defined as the ratio of the length of the opposite side to that of the hypotenuse side.
Sin θ = Opposite Side/ Hypotenuse Side.
For example, the sine function of a triangle ABC with an angle θ is expressed as:
Sin θ = a/c
### What is the Cosine Function?
In the right triangle, the cosine function is defined as the ratio of the length of the adjacent side to that of the hypotenuse side.
Cos θ = Adjacent Side/Hypotenuse Side
Example:
Considering the figure given above, the cosine function of a triangle ABC with an angle θ is expressed as:
Cos θ = b/c
### What is the Tangent Function?
In the right triangle, the tangent function is defined as the ratio of the length of the opposite side to that of the adjacent side.
Tan θ = Opposite Side/Adjacent Side
Example:
Considering the figure given above, the cosine function of a triangle ABC with an angle θ is expressed as:
Tan θ = a/b
### Sine Cosine Tangent Table
The values of trigonometric ratios like sine, cosine, and tangent for some standard angles such as 0°, 30°, 45°, 60°, and 90° can be easily determined with the help of the sine cosine tangent table given below. These values are very important to solve trigonometric problems. Hence, it is important to learn the values of trigonometric ratios of these standard angles.
The sine, cosine, and tangent table given below includes the values of standard angles like 0°, 30°, 45°, 60°, and 90°.
## Sine, Cosine, and Tangent Table
Angles In Degrees 0° 30° 45° 60° 90° Sin 0 1/2 1/√2 √3/2 1 Cos 1 √3/2 1/√2 1/2 0 Tan 0 1/√3 1 √3 Not Defined
### Did You Know?
• Sine and Cosine were introduced by Aryabhatta, whereas the tangent function was introduced by Muhammad Ibn Musa al- Khwarizmi ( 782 CE - 850 CE).
• Sine Cosine and Tangent formulas can be easily learned using SOHCAHTOA. As sine is opposite side over hypotenuse side, cosine is adjacent side over hypotenuse side, and tangent is opposite side over the adjacent side.
### Solved Examples:
1. Find Cos θ with respect to the following triangle.
Ans: To find Cos θ, we need both adjacent and hypotenuse side.
The adjacent side in the above triangle is, BC = 8 Cm
But, the hypotenuse side i.e. AC is not given.
To find the hypotenuse side, we use the Pythagoras theorem
AC² = AB² + BC² = 6² + 8² = 100
Hypotenuse side, AC = √100 or 10 cm
Cos θ = Adjacent/Hypotenuse = 8/10
= 4/5
Therefore, Cos θ = 4/5
2. Find the value of Sin 45°, Cos 60°, and Tan 60°.
Solution: Using the trigonometric table above, we have:
Sin 45° = 1/√2
Cos 60° = 1/2
Tan 45°= 1
## FAQs on Sine, Cosine and Tangent
Q1. How to Memorize Sine Cosine Tangent Values?
Ans. Steps to memorize sine, cosine tangent values:
• The first step is to divide the numbers 0, 1, 2, 3, 4 by 4 and find a positive square root for the values obtained.
• The angles in Sine Cosine Tangent are given in the order of 0°, 30°, 45°, 60°, and 90°.
• You can remember the value of Sine-like this 0/√2, 1/√2, 2/√2, 3/√2, 4/√2.
• The row of cosine is similar to the row of sine just in reverse order.
• Each value of tangent can be obtained by dividing the sine values by cosine as Tan = Sin/Cos.
Q2. Why is the Sine Cosine Tan Function Important?
Ans. Sine cosine tan functions are important because of the following reason.
• They help us to work on angles when sides of triangles are known.
• They help us to work on sides of when angles of triangles are known.
Q3. What are the Applications of Trigonometry Function?
Ans. Some of the applications of trigonometric functions are:
• Trigonometric functions are used in different fields like meteorology, seismology, physical Science, navigation, electronics, etc.
• It is also used to measure the distance of long rivers, the height of mountains, etc.
• The sine and cosine functions help to describe the sound and light waves.
• It is used in satellite systems and also used for creating maps. |
Two methods for deriving the quadratic formula that I was not taught in school
MarkFL
Staff member
As a student, I was taught 3 ways to solve quadratic equations:
i) Factoring
ii) Completing the square
iii) Applying the quadratic formula, derived by completing the square on the general quadratic in standard form:
(1) $\displaystyle ax^2+bx+c=0$
To complete the square, I was taught to move the constant term to the other side and divide through by a:
$\displaystyle x^2+\frac{b}{a}x=-\frac{c}{a}$
Then, add the square of one-half the coefficient of the linear term to both sides:
$\displaystyle x^2+\frac{b}{a}x+\left(\frac{b}{2a} \right)^2=-\frac{c}{a}+\left(\frac{b}{2a} \right)^2$
Write the left side as a square, and combine terms on the right:
$\displaystyle \left(x+\frac{b}{2a} \right)^2=\frac{b^2-4ac}{(2a)^2}$
Apply the square root property:
$\displaystyle x+\frac{b}{2a}=\pm\frac{\sqrt{b^2-4ac}}{2a}$
Solve for x:
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
And we have the famous quadratic formula.
In my years on math forums, I have gleaned two variations on this technique that I would like to share:
Method 1:
Divide (1) by a:
$\displaystyle x^2+\frac{b}{a}x+\frac{c}{a}=0$
Now, we next want to shift the roots to the right by 1/2 the value of the coefficient of the linear term, so our new equation is:
$\displaystyle \left(x-\frac{b}{2a} \right)^2+\frac{b}{a}\left(x-\frac{b}{2a} \right)+\frac{c}{a}=0$
$\displaystyle x^2-\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{b}{a}x-\frac{b^2}{2a^2}+\frac{c}{a}=0$
$\displaystyle x^2=\frac{b^2-4ac}{4a^2}$
$\displaystyle x=\pm\frac{\sqrt{b^2-4ac}}{2a}$
Now, we subtract $\displaystyle \frac{b}{2a}$ from these roots, to get the roots of the original equation:
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Method 2:
Arrange (1) as:
$\displaystyle ax^2+bx=-c$
Multiply by $\displaystyle 4a$:
$\displaystyle 4a^2x^2+4abx=-4ac$
Add $\displaystyle b^2$ to both sides:
$\displaystyle 4a^2x^2+4abx+b^2=b^2-4ac$
Write the left side as a square:
$\displaystyle (2ax+b)^2=b^2-4ac$
Apply the square root property:
$\displaystyle 2ax+b=\pm\sqrt{b^2-4ac}$
Solve for x:
$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$
Comments and questions should be posted here:
Commentary for "Two methods for deriving the quadratic formula that I was not taught in school"
Last edited:
MarkFL
Staff member
Here is another method submitted to me by our own agentmulder:
Let
$$\displaystyle ax^2+bx+c=0$$
$$\displaystyle ax^2+bx=-c$$
Now... I want to complete the square but the coefficient of x^2 is bothering me. NO PROBLEM, I'll take it's square root.
$$\displaystyle \left(\sqrt{a}x+? \right)^2$$
Now... what is the question mark? A little playing around shows it must be $$\displaystyle \frac{b}{2 \sqrt{a}}$$ because that's the only way to get $bx$ when we square the binomial, and I'll just subtract (it's square), the constant as a correction term.
$$\displaystyle \left(\sqrt{a}x+\frac{b}{2\sqrt{a}} \right)^2-\frac{b^2}{4a}=-c$$
$$\displaystyle \left(\sqrt{a}x + \frac{b}{2 \sqrt{a}} \right)^2=\frac{b^2-4ac}{4a}$$
$$\displaystyle \sqrt{a}x+\frac{b}{2\sqrt{a}}=\pm\frac{\sqrt{b^2-4ac}}{2\sqrt{a}}$$
$$\displaystyle \sqrt{a}x=\frac{-b\pm\sqrt{b^2-4ac}}{2\sqrt{a}}$$
DIVIDE BY $\sqrt{a}$ AND YOU'RE DONE!
$$\displaystyle x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ |
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### Middle/High School Algebra, Geometry, and Statistics (AGS)4.2 Quadratic Equations - II
example: Factorize, x2+7x+12. Solution: Given that, x2+7x+12. 7x is to be written as the sum of two terms such that the product of their coefficient is 12. 7x = 3x+4x = 3*4 = 12 Therefore, x2+(3x+4x)+12 (x2+3x)+(4x+12) Take x as common, we get x(x2+3)+(4x+12) Take 4 as common, we get = x(x+3)+4(x+3) Take (x+3) as common, we get = (x+4)(x+3) Directions: Solve the following problems. Also write at least ten examples of your own.
Name: ___________________Date:___________________
### Middle/High School Algebra, Geometry, and Statistics (AGS)4.2 Quadratic Equations - II
Q 1: Factorize, x2+3x+2.(x+3)(x+3)None of these(x+1)(x+2)(x+5)(x+6) Q 2: Factorize, x2+11x+30.(x+3)(x+3)None of these(x+1)(x+2)(x+5)(x+6) Q 3: Factorize, x2+10x+24.(x+4)(x+6)(x+7)(x+5)(x+2)(x+7)None of these Q 4: Factorize, x2+6x+9.(x+3)(x+3)None of these(x+5)(x+6)(x+1)(x+2) Q 5: Factorize, x2+12x+35.(x+7)(x+5)(x+4)(x+6)None of these(x+2)(x+7) Q 6: Factorize, x2+12x+32.(x+4)(x+8)(x+3)(x+7)(x+2)(x+7)None of these Question 7: This question is available to subscribers only! Question 8: This question is available to subscribers only!
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# Proving that the mean and mode are equal for an arithmetic progression
Ask your students to prove this as an exercise, when learning how to prove things by cases. This theorem is used here, when proving that polite numbers cannot be of the form $$2^n.$$
Problem: Prove that the mean equals the mode for an arithmetic progression.
Solution:
$$a + (a + d) + (a + 2d) + \ldots + [a + (n - 1)d]$$
The sum of an arithmetic progression is
$$S_n = \dfrac{n}{2}[2a + (n - 1)d]$$
The mean is thus
\begin{align} & \dfrac{1}{2}[2a + (n - 1)d] \\[0.5em] =&\,a + \dfrac{(n - 1)d}{2} \end{align}
If $$n$$ is odd, then the mode is the middle term, which is
\begin{align} & a + \left(\dfrac{n - 1}{2}\right)d \\[0.5em] =&\,a + \dfrac{(n - 1)d}{2} \end{align}
If $$n$$ is even, then the mode is the average of the two middle terms, which is
\begin{align} & \dfrac{\left[a + \left(\dfrac{n}{2}\right)d\right] + \left[a + \left(\dfrac{n}{2} - 1\right)d\right]}{2} \\[0.5em] =&\,a + \dfrac{dn - d}{2} \\[0.5em] =&\,a + \dfrac{(n - 1)d}{2} \end{align} |
# Section 7.2 Confidence Intervals for Population Proportions
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## Transcription
1 Section 7.2 Confidence Intervals for Population Proportions 2012 Pearson Education, Inc. All rights reserved. 1 of 83
2 Section 7.2 Objectives Find a point estimate for the population proportion Construct a confidence interval for a population proportion Determine the minimum sample size required when estimating a population proportion 2012 Pearson Education, Inc. All rights reserved. 2 of 83
3 Point Estimate for Population p Population Proportion The probability of success in a single trial of a binomial experiment. Denoted by p Point Estimate for p The proportion of successes in a sample. Denoted by x number of successes in sample pˆ = = n sample size read as p hat 2012 Pearson Education, Inc. All rights reserved. 3 of 83
4 Point Estimate for Population p Estimate Population Parameter with Sample Statistic Proportion: p ˆp Point Estimate for q, the population proportion of failures Denoted by Read as q hat ˆ q = 1 ˆ p 2012 Pearson Education, Inc. All rights reserved. 4 of 83
5 Example: Point Estimate for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal while at work. Find a point estimate for the population proportion of U.S. adults who say it is acceptable to check personal while at work. (Adapted from Liberty Mutual) Solution: n = 1000 and x = 662 pˆ x = = n = = 66.2% 2012 Pearson Education, Inc. All rights reserved. 5 of 83
6 Confidence Intervals for p A c-confidence interval for a population proportion p ˆ p E < p< pˆ + E where E = Z α 2 pq ˆˆ n The probability that the confidence interval contains p is c Pearson Education, Inc. All rights reserved. 6 of 83
7 Constructing Confidence Intervals for p In Words 1. Identify the sample statistics n and x. 2. Find the point estimate ˆp. 3. Verify that the sampling distribution of pˆ can be approximated by a normal distribution. 4. Find the critical value z c that corresponds to the given level of confidence c. In Symbols npˆ pˆ = x n 5, nqˆ 5 Use the Standard Normal Table or technology Pearson Education, Inc. All rights reserved. 7 of 83
8 Constructing Confidence Intervals for p In Words 5. Find the margin of error E. 6. Find the left and right endpoints and form the confidence interval. In Symbols E = Z α 2 pq ˆˆ n Left endpoint: ˆp E Right endpoint: ˆp+ E Interval: pˆ E < p< pˆ + E 2012 Pearson Education, Inc. All rights reserved. 8 of 83
9 Example: Confidence Interval for p In a survey of 1000 U.S. adults, 662 said that it is acceptable to check personal while at work. Construct a 95% confidence interval for the population proportion of U.S. adults who say that it is acceptable to check personal while at work. Solution: Recall p ˆ = qˆ = 1 pˆ = = Pearson Education, Inc. All rights reserved. 9 of 83
10 Solution: Confidence Interval for p Verify the sampling distribution of can be approximated by the normal distribution npˆ = = 662 > 5 nqˆ = = 338 > 5 Margin of error: ˆp E pq ˆˆ (0.662) (0.338) = Zα 2 = n Pearson Education, Inc. All rights reserved. 10 of 83
11 Solution: Confidence Interval for p Confidence interval: Left Endpoint: pˆ E < p < Right Endpoint: pˆ = = E 2012 Pearson Education, Inc. All rights reserved. 11 of 83
12 Solution: Confidence Interval for p < p < Point estimate ˆp ˆp E ˆp+ E With 95% confidence, you can say that the population proportion of U.S. adults who say that it is acceptable to check personal while at work is between 63.3% and 69.1% Pearson Education, Inc. All rights reserved. 12 of 83
13 Determining Sample Size When p is known, n = Z α 2 E 2 2 pq ˆˆ Where Z α 2 is the value associated with the desired confidence level, and E is the desired margin of error. Round up to the next integer. 13
14 Sample Size for Estimating the Population Proportion When p is unknown, we use n = Z α 2 E
15 Example: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if 1. no preliminary estimate is available. Solution: Because you do not have a preliminary estimate for use p ˆ = 0.5 and q ˆ = p, ˆ 2012 Pearson Education, Inc. All rights reserved. 15 of 83
16 Solution: Sample Size c = 0.95 z α/2 = 1.96 E = 0.03 n 2 [ ] 2 Z ˆˆ α 2 pq 1.96 (0.5)(0.5) = = 2 2 E (0. 03) Round up to the nearest whole number. With no preliminary estimate, the minimum sample size should be at least 1068 voters Pearson Education, Inc. All rights reserved. 16 of 83
17 Example: Sample Size You are running a political campaign and wish to estimate, with 95% confidence, the population proportion of registered voters who will vote for your candidate. Your estimate must be accurate within 3% of the true population proportion. Find the minimum sample size needed if 2. a preliminary estimate gives p ˆ = Solution: Use the preliminary estimate qˆ = 1 pˆ = = ˆ 0.31 p = 2012 Pearson Education, Inc. All rights reserved. 17 of 83
18 Solution: Sample Size c = 0.95 z α/2 = 1.96 E = 0.03 n Z ˆ ˆ α 2 pq = = 2 E 2 [ ] (0.31)(0.69) (0.03) Round up to the nearest whole number. With a preliminary estimate of p ˆ = 0.31, the minimum sample size should be at least 914 voters. Need a larger sample size if no preliminary estimate is available Pearson Education, Inc. All rights reserved. 18 of 83
19 Section 7.3 Confidence Intervals for the Mean (σ known) 2012 Pearson Education, Inc. All rights reserved. 19 of 83
20 Section 7.3 Objectives Find a point estimate and a margin of error Construct and interpret confidence intervals for the population mean Determine the minimum sample size required when estimating μ 2012 Pearson Education, Inc. All rights reserved. 20 of 83
21 Point Estimate for Population μ Point Estimate A single value estimate for a population parameter Most unbiased point estimate of the population mean μ is the sample mean x Estimate Population with Sample Parameter Statistic Mean: μ x 2012 Pearson Education, Inc. All rights reserved. 21 of 83
22 Example: Point Estimate for Population μ A social networking website allows its users to add friends, send messages, and update their personal profiles. The following represents a random sample of the number of friends for 40 users of the website. Find a point estimate of the population mean, µ. (Source: Facebook) Pearson Education, Inc. All rights reserved. 22 of 83
23 Solution: Point Estimate for Population μ The sample mean of the data is Σx 5232 x = = = n 40 Your point estimate for the mean number of friends for all users of the website is friends Pearson Education, Inc. All rights reserved. 23 of 83
24 Interval Estimate Interval estimate An interval, or range of values, used to estimate a population parameter. Left endpoint Point estimate x =130.8 ( ) Interval estimate Right endpoint How confident do we want to be that the interval estimate contains the population mean μ? 2012 Pearson Education, Inc. All rights reserved. 24 of 83
25 Level of Confidence Level of confidence (C-Level) The probability that the interval estimate contains the population parameter. α/2 /2 z z = 0 Critical values The remaining area in the tails is 1 c. c is the area under the C-level standard normal curve between the critical values. z c Use the Standard Normal Table to find the corresponding z-scores Pearson Education, Inc. All rights reserved. 25 of 83 z
26 Level of Confidence If the level of confidence is 90%, this means that we are 90% confident that the interval contains the population mean μ. C-Level = 0.90 α /2= 0.05 α/2= 0.05 z α/2 z = c z = 0 Zz α/2 c = The corresponding z-scores are ± z 2012 Pearson Education, Inc. All rights reserved. 26 of 83
27 Sampling error Sampling Error The difference between the point estimate and the actual population parameter value. For μ: the sampling error is the difference x μ μ is generally unknown varies from sample to sample x 2012 Pearson Education, Inc. All rights reserved. 27 of 83
28 Margin of error Margin of Error The greatest possible distance between the point estimate and the value of the parameter it is estimating for a given level of confidence, c. Denoted by E. E = z α /2 σ n When n 30, the sample standard deviation, s, can be used for σ. Sometimes called the maximum error of estimate or error tolerance Pearson Education, Inc. All rights reserved. 28 of 83
29 Example: Finding the Margin of Error Use the social networking website data and a 95% confidence level to find the margin of error for the mean number of friends for all users of the website. Assume the sample standard deviation is about Pearson Education, Inc. All rights reserved. 29 of 83
30 Solution: Finding the Margin of Error First find the critical values z α/2 = z 1.96 c z = 0 z α/2 = % of the area under the standard normal curve falls within 1.96 standard deviations of the mean. (You can approximate the distribution of the sample means with a normal curve by the Central Limit Theorem, because n = ) 2012 Pearson Education, Inc. All rights reserved. 30 of 83 z c z
31 Solution: Finding the Margin of Error σ E = Z Z n α 2 α 2 s n You don t know σ, but since n 30, you can use s in place of σ. You are 95% confident that the margin of error for the population mean is about 16.4 friends Pearson Education, Inc. All rights reserved. 31 of 83
32 Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ x E < µ < x + E where E = z c σ n The probability that the confidence interval contains μ is c Pearson Education, Inc. All rights reserved. 32 of 83
33 Constructing Confidence Intervals for μ Finding a Confidence Interval for a Population Mean (n 30 or σ known with a normally distributed population) In Words 1. Find the sample statistics n and. x In Symbols Σx x = n 2. Specify σ, if known. Otherwise, if n 30, find the sample standard deviation s and use it as an estimate for σ. s = Σ( x x) n Pearson Education, Inc. All rights reserved. 33 of 83
34 Constructing Confidence Intervals for μ In Words 3. Find the critical value z c that corresponds to the given level of confidence. 4. Find the margin of error E. 5. Find the left and right endpoints and form the confidence interval. In Symbols Use the Standard Normal Table or technology. σ E = Zα 2 n Left endpoint: x E Right endpoint: x + E Interval: x E < µ < x + E 2012 Pearson Education, Inc. All rights reserved. 34 of 83
35 Example: Constructing a Confidence Interval Construct a 95% confidence interval for the mean number of friends for all users of the website. Solution: Recall x =130.8 and E 16.4 Left Endpoint: x E = x < μ < Right Endpoint: E = Pearson Education, Inc. All rights reserved. 35 of 83
36 Solution: Constructing a Confidence Interval < μ < With 95% confidence, you can say that the population mean number of friends is between and Pearson Education, Inc. All rights reserved. 36 of 83
37 Example: Constructing a Confidence Interval σ Known A college admissions director wishes to estimate the mean age of all students currently enrolled. In a random sample of 20 students, the mean age is found to be 22.9 years. From past studies, the standard deviation is known to be 1.5 years, and the population is normally distributed. Construct a 90% confidence interval of the population mean age Pearson Education, Inc. All rights reserved. 37 of 83
38 Solution: Constructing a Confidence Interval σ Known First find the critical values α = 0.90 α/2 = 0.05 α/2 = 0.05 z c = z c z = 0 z c = z c z z c = Pearson Education, Inc. All rights reserved. 38 of 83
39 Solution: Constructing a Confidence Interval σ Known Margin of error: σ E 1.5 = Zα 2 = n 20 Confidence interval: Left Endpoint: x = E < μ < 23.5 Right Endpoint: x + E = Pearson Education, Inc. All rights reserved. 39 of 83
40 Solution: Constructing a Confidence Interval σ Known 22.3 < μ < 23.5 Point estimate x ( ) E x + x E With 90% confidence, you can say that the mean age of all the students is between 22.3 and 23.5 years Pearson Education, Inc. All rights reserved. 40 of 83
41 Interpreting the Results μ is a fixed number. It is either in the confidence interval or not. Incorrect: There is a 90% probability that the actual mean is in the interval (22.3, 23.5). Correct: If a large number of samples is collected and a confidence interval is created for each sample, approximately 90% of these intervals will contain μ Pearson Education, Inc. All rights reserved. 41 of 83
42 Interpreting the Results The horizontal segments represent 90% confidence intervals for different samples of the same size. In the long run, 9 of every 10 such intervals will contain μ. μ 2012 Pearson Education, Inc. All rights reserved. 42 of 83
43 Sample Size Given a c-confidence level and a margin of error E, the minimum sample size n needed to estimate the population mean µ is n z c σ = E If σ is unknown, you can estimate it using s, provided you have a preliminary sample with at least 30 members Pearson Education, Inc. All rights reserved. 43 of 83
44 Example: Sample Size You want to estimate the mean number of friends for all users of the website. How many users must be included in the sample if you want to be 95% confident that the sample mean is within seven friends of the population mean? Assume the sample standard deviation is about Pearson Education, Inc. All rights reserved. 44 of 83
45 Solution: Sample Size First find the critical values z c = z 1.96 c z = 0 z c = 1.96 z c z z c = Pearson Education, Inc. All rights reserved. 45 of 83
46 Solution: Sample Size z c = 1.96 σ s 53.0 E = 7 n z c σ = E When necessary, round up to obtain a whole number. You should include at least 221 users in your sample Pearson Education, Inc. All rights reserved. 46 of 83
47 Section 7.3 Summary Found a point estimate and a margin of error Constructed and interpreted confidence intervals for the population mean Determined the minimum sample size required when estimating μ 2012 Pearson Education, Inc. All rights reserved. 47 of 83
48 Section 7.4 Confidence Intervals for the Mean (σ unknown) 2012 Pearson Education, Inc. All rights reserved. 48 of 83
49 Section 7.4 Objectives Interpret the t-distribution and use a t-distribution table Construct confidence intervals when n < 30, the population is normally distributed, and σ is unknown 2012 Pearson Education, Inc. All rights reserved. 49 of 83
50 The t-distribution When the population standard deviation is unknown, the sample size is less than 30, and the random variable x is approximately normally distributed, it follows a t-distribution. t = x µ s n Critical values of t are denoted by t c Pearson Education, Inc. All rights reserved. 50 of 83
51 Properties of the t-distribution 1. The t-distribution is bell shaped and symmetric about the mean. 2. The t-distribution is a family of curves, each determined by a parameter called the degrees of freedom. The degrees of freedom are the number of free choices left after a sample statistic such as x is calculated. When you use a t-distribution to estimate a population mean, the degrees of freedom are equal to one less than the sample size. d.f. = n 1 Degrees of freedom 2012 Pearson Education, Inc. All rights reserved. 51 of 83
52 Properties of the t-distribution 3. The total area under a t-curve is 1 or 100%. 4. The mean, median, and mode of the t-distribution are equal to zero. 5. As the degrees of freedom increase, the t-distribution approaches the normal distribution. After 30 d.f., the t- distribution is very close to the standard normal z- distribution. d.f. = 2 d.f. = 5 Standard normal curve 0 t The tails in the t- distribution are thicker than those in the standard normal distribution Pearson Education, Inc. All rights reserved. 52 of 83
53 Example: Critical Values of t Find the critical value t c for a 95% confidence level when the sample size is 15. Solution: d.f. = n 1 = 15 1 = 14 Table 5: t-distribution t c = Pearson Education, Inc. All rights reserved. 53 of 83
54 Solution: Critical Values of t 95% of the area under the t-distribution curve with 14 degrees of freedom lies between t = ± c = 0.95 t c = t c = t 2012 Pearson Education, Inc. All rights reserved. 54 of 83
55 Confidence Intervals for the Population Mean A c-confidence interval for the population mean μ s x E < µ < x + E where E = tc n The probability that the confidence interval contains μ is c Pearson Education, Inc. All rights reserved. 55 of 83
56 Confidence Intervals and t-distributions In Words 1. Identify the sample statistics n, x, and s. 2. Identify the degrees of freedom, the level of confidence c, and the critical value t c. In Symbols Σx x = s = n d.f. = n 1 ( x x) n Find the margin of error E. E = t c s n 2012 Pearson Education, Inc. All rights reserved. 56 of 83
57 Confidence Intervals and t-distributions In Words 4. Find the left and right endpoints and form the confidence interval. In Symbols Left endpoint: x Right endpoint: x + Interval: x E < µ < x + E E E 2012 Pearson Education, Inc. All rights reserved. 57 of 83
58 Example: Constructing a Confidence Interval You randomly select 16 coffee shops and measure the temperature of the coffee sold at each. The sample mean temperature is 162.0ºF with a sample standard deviation of 10.0ºF. Find the 95% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. Solution: Use the t-distribution (n < 30, σ is unknown, temperatures are approximately normally distributed) Pearson Education, Inc. All rights reserved. 58 of 83
59 Solution: Constructing a Confidence Interval n =16, x = s = 10.0 c = 0.95 df = n 1 = 16 1 = 15 Critical Value Table 5: t-distribution t c = Pearson Education, Inc. All rights reserved. 59 of 83
60 Solution: Constructing a Confidence Interval Margin of error: s 10 E = t c = n 16 Confidence interval: Left Endpoint: x E = < μ < Right Endpoint: x + E = Pearson Education, Inc. All rights reserved. 60 of 83
61 Solution: Constructing a Confidence Interval < μ < Point estimate x ( ) E + x x E With 95% confidence, you can say that the population mean temperature of coffee sold is between 156.7ºF and 167.3ºF Pearson Education, Inc. All rights reserved. 61 of 83
62 Normal or t-distribution? Is n 30? No Is the population normally, or approximately normally, distributed? Yes Yes No Use the normal distribution with σ E = z c n If σ is unknown, use s instead. Cannot use the normal distribution or the t-distribution. Is σ known? No Use the t-distribution with s E = t c and n 1 degrees of freedom. n Yes Use the normal distribution with σ E = z c. n 2012 Pearson Education, Inc. All rights reserved. 62 of 83
63 Example: Normal or t-distribution? You randomly select 25 newly constructed houses. The sample mean construction cost is \$181,000 and the population standard deviation is \$28,000. Assuming construction costs are normally distributed, should you use the normal distribution, the t-distribution, or neither to construct a 95% confidence interval for the population mean construction cost? Solution: Use the normal distribution (the population is normally distributed and the population standard deviation is known) 2012 Pearson Education, Inc. All rights reserved. 63 of 83
64 Section 7.3 Summary Interpreted the t-distribution and used a t-distribution table Constructed confidence intervals when n < 30, the population is normally distributed, and σ is unknown 2012 Pearson Education, Inc. All rights reserved. 64 of 83
65 Section 7.3 Summary Found a point estimate for the population proportion Constructed a confidence interval for a population proportion Determined the minimum sample size required when estimating a population proportion 2012 Pearson Education, Inc. All rights reserved. 65 of 83
66 Section 7.5 Confidence Intervals for Variance and Standard Deviation 2012 Pearson Education, Inc. All rights reserved. 66 of 83
67 Section 7.5 Objectives Interpret the chi-square distribution and use a chi-square distribution table Use the chi-square distribution to construct a confidence interval for the variance and standard deviation 2012 Pearson Education, Inc. All rights reserved. 67 of 83
68 The Chi-Square Distribution The point estimate for σ 2 is s 2 The point estimate for σ is s s 2 is the most unbiased estimate for σ 2 Estimate Population Parameter with Sample Statistic Variance: σ 2 s 2 Standard deviation: σ s 2012 Pearson Education, Inc. All rights reserved. 68 of 83
69 The Chi-Square Distribution You can use the chi-square distribution to construct a confidence interval for the variance and standard deviation. If the random variable x has a normal distribution, then the distribution of 2 2 ( n 1) s χ = 2 σ forms a chi-square distribution for samples of any size n > Pearson Education, Inc. All rights reserved. 69 of 83
70 Properties of The Chi-Square Distribution 1. All chi-square values χ 2 are greater than or equal to zero. 2. The chi-square distribution is a family of curves, each determined by the degrees of freedom. To form a confidence interval for σ 2, use the χ 2 -distribution with degrees of freedom equal to one less than the sample size. d.f. = n 1 Degrees of freedom 3. The area under each curve of the chi-square distribution equals one Pearson Education, Inc. All rights reserved. 70 of 83
71 Properties of The Chi-Square Distribution 4. Chi-square distributions are positively skewed. Chi-square Distributions 2012 Pearson Education, Inc. All rights reserved. 71 of 83
72 Critical Values for χ 2 There are two critical values for each level of confidence. The value χ 2 R represents the right-tail critical value The value χ 2 L represents the left-tail critical value. 1 c 2 c 1 c 2 The area between the left and right critical values is c. 2 χ L 2 χ R χ Pearson Education, Inc. All rights reserved. 72 of 83
73 Example: Finding Critical Values for χ Find the critical values χ R and χ L for a 95% confidence interval when the sample size is 18. Solution: d.f. = n 1 = 18 1 = 17 d.f. Each area in the table represents the region under the chi-square curve to the right of the critical value. Area to the right of χ 2 R = Area to the right of χ 2 L = 1 c = = c = = Pearson Education, Inc. All rights reserved. 73 of 83
74 Solution: Finding Critical Values for χ 2 Table 6: χ 2 -Distribution 2 χ L = χ R = % of the area under the curve lies between and Pearson Education, Inc. All rights reserved. 74 of 83
75 Confidence Intervals for σ 2 and σ Confidence Interval for σ 2 : ( n 1) s ( n 1) s χ < σ < 2 R χl Confidence Interval for σ: 2 2 ( n 1) s ( n 1) s < σ < χ 2 2 R χl The probability that the confidence intervals contain σ 2 or σ is c Pearson Education, Inc. All rights reserved. 75 of 83
76 Confidence Intervals for σ 2 and σ In Words 1. Verify that the population has a normal distribution. 2. Identify the sample statistic n and the degrees of freedom. 3. Find the point estimate s Find the critical values χ 2 R and χ 2 L that correspond to the given level of confidence c. In Symbols s d.f. = n ( x x) = n 1 Use Table 6 in Appendix B Pearson Education, Inc. All rights reserved. 76 of 83
77 Confidence Intervals for σ 2 and σ In Words 5. Find the left and right endpoints and form the confidence interval for the population variance. 6. Find the confidence interval for the population standard deviation by taking the square root of each endpoint. In Symbols < σ < 2 R χl ( n 1) s ( n 1) s χ 2 2 ( n 1) s ( n 1) s < σ < χ 2 2 R χl 2012 Pearson Education, Inc. All rights reserved. 77 of 83
78 Example: Constructing a Confidence Interval You randomly select and weigh 30 samples of an allergy medicine. The sample standard deviation is 1.20 milligrams. Assuming the weights are normally distributed, construct 99% confidence intervals for the population variance and standard deviation. Solution: d.f. = n 1 = 30 1 = 29 d.f Pearson Education, Inc. All rights reserved. 78 of 83
79 Solution: Constructing a Confidence Interval Area to the right of χ 2 R = 1 c = = Area to the right of χ 2 L = 1+ c = = The critical values are χ 2 R = and χ 2 L = Pearson Education, Inc. All rights reserved. 79 of 83
80 Solution: Constructing a Confidence Interval Confidence Interval for σ 2 : Left endpoint: ( n 1) s χ 2 R 2 = (30 1)(1.20) Right endpoint: ( n 1) s χ 2 L 2 = (30 1)(1.20) < σ 2 < 3.18 With 99% confidence, you can say that the population variance is between 0.80 and Pearson Education, Inc. All rights reserved. 80 of 83
81 Solution: Constructing a Confidence Interval Confidence Interval for σ : 2 2 ( n 1) s ( n 1) s < σ < χ 2 2 R χl 2 2 (30 1)(1.20) (30 1)(1.20) < σ < < σ < 1.78 With 99% confidence, you can say that the population standard deviation is between 0.89 and 1.78 milligrams Pearson Education, Inc. All rights reserved. 81 of 83
82 Section 7.5 Summary Interpreted the chi-square distribution and used a chi-square distribution table Used the chi-square distribution to construct a confidence interval for the variance and standard deviation 2012 Pearson Education, Inc. All rights reserved. 82 of 83
### An interval estimate (confidence interval) is an interval, or range of values, used to estimate a population parameter. For example 0.476
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# How do you use the Pythagorean Theorem to find the missing side of the triangl given: Side one: 7, Side two: 24, Side three: c?
Jul 2, 2016
Side c = 25
#### Explanation:
To solve this problem, we should use the Pythagorean Theorem:
Based on the photo above, I'll just say that a=7 and b=24 so you can follow along.
Both of the legs (a and b) are known. We do not know c, but it can be found by plugging in our known values:
${7}^{2} + {24}^{2} = {c}^{2}$
${7}^{2}$ or $7 \times 7$ = 49
${24}^{2}$ or $24 \times 24$ = 576
Thus, $49 + 576 = {c}^{2}$. Now, add both numbers on the left side together and that will equal ${c}^{2}$
$625 = {c}^{2}$
Take the square root of both sides to find c. The square root
(sqrt) is the inverse of the square (${c}^{2}$)
$\sqrt{625} = \sqrt{{c}^{2}}$
Therefore, 25 = c |
Convert 4/25 into a decimal and percent.
This question aims to explain the concepts of fraction, decimal point, and percentage.
A decimal is a fraction noted in a particular form. Rather than writing $\dfrac{3}{2}$, for instance, you can write $1.5$, where the $1$ is in the one place and the $5$ is in the tenths place. Decimal is derived from the Latin term Decimus, signifying tenth, from the source word Decem, or $10$. The decimal method, thus, has $10$ as its base. Decimal can likewise precisely direct to a number in the decimal design. As an adjective, decimal points to something corresponding to this numbering method. The decimal point, for instance, directs to the period that splits one place from the tenth place in decimal numbers.
A fraction is a number represented as a quotient where a numerator is divided by a denominator. In fractions, both are integers. A complicated fraction has a fraction in the numerator or denominator. In a right fraction, the numerator is smaller than the denominator. If the numerator is more significant, it is named an improper fraction and can also be noted as a mixed number. Any fraction can be noted in decimal format by maintaining the division of the numerator by the denominator. The outcome may finish at some point, or one or more digits may duplicate without end.
The phrase “percentage” was derived from the Latin word “per centum,” which indicates “by the hundred.” With $100$ as the denominator, percentages are fractions. In other words, it is the connection between part and whole where the weight of the whole is permanently taken as $100$.
For instance, if Sam achieved $40%$ marks in his math quiz, it means that he achieved $40$ marks out of $100$. It corresponds to $\dfrac{40}{100}$ in the fraction form and $40:100$ in representations of ratio. The percentage is expressed as a given part or portion in every hundred. It is a fraction with $100$ as the denominator and is characterized by the symbol “%“. Computing percentage means finding the allocation of a whole, in representations of $100$. The percentage can be found either by using the unitary method or by adjusting the denominator of the fraction to $100$. It must be stated that the second way of estimating the percentage is not operated in cases where the denominator is not a factor of $100$. The unitary method is used in such cases.
Percent is another word for telling hundredths. Accordingly, $1%$ is one-hundredth, which means $1% = \dfrac{1}{100} = 0.01$. The percentage procedure is used to estimate the claim of a whole in terms of $100$. Utilizing this formula, you can denote a number as a fraction of $100$. If you follow carefully, all the methods to get the percentage portrayed above can be easily computed by exploiting the formula given below:
$Percetange \space=(\dfrac{Value}{Total \space Value}) \times 100$
Multiplying $4$:
$=\dfrac{4*4}{25*4}= \dfrac{16}{100}=0.16$
$=16\%$
$\dfrac{4}{25}$ into a decimal is $0.16$, and the percentage is $16%$.
Example
Convert 2/50 into percent.
Multiplying $2$:
$=\dfrac{2*4}{50*2}=\dfrac{4}{100}$
$=4\%$ |
## Step by action solution because that calculating 105 is 75 percent that what number
We already have our very first value 105 and also the second value 75. Let"s i think the unknown value is Y i m sorry answer we will uncover out.
You are watching: 75 of what number is 105
As we have all the required values us need, now we have the right to put castle in a basic mathematical formula as below:
STEP 1105 = 75% × Y
STEP 2105 = 75/100× Y
Multiplying both sides by 100 and also dividing both sides of the equation through 75 we will certainly arrive at:
STEP 3Y = 105 × 100/75
STEP 4Y = 105 × 100 ÷ 75
STEP 5Y = 140
Finally, us have uncovered the value of Y i m sorry is 140 and also that is ours answer.
You can easily calculate 105 is 75 percent that what number through using any regular calculator, simply enter 105 × 100 ÷ 75 and also you will gain your answer i beg your pardon is 140
Here is a portion Calculator come solve comparable calculations such as 105 is 75 percent the what number. You deserve to solve this kind of calculation v your values by entering them right into the calculator"s fields, and click "Calculate" to gain the an outcome and explanation.
is
percent that what number
Calculate
## Sample questions, answers, and how to
Question: her friend has a bag the marbles, and also he tells you the 75 percent that the marbles space red. If there room 105 red marbles. How plenty of marbles go he have actually altogether?
How To: In this problem, we know that the Percent is 75, and we are additionally told that the part of the marbles is red, for this reason we know that the part is 105.
So, that means that it have to be the complete that"s missing. Below is the method to number out what the total is:
Part/Total = Percent/100
By utilizing a simple algebra we have the right to re-arrange ours Percent equation prefer this:
Part × 100/Percent = Total
If we take the "Part" and multiply the by 100, and also then we division that by the "Percent", we will acquire the "Total".
Let"s try it out on our problem about the marbles, that"s very an easy and it"s just two steps! We recognize that the "Part" (red marbles) is 105.
So action one is to simply multiply that part by 100.
105 × 100 = 10500
In step two, us take the 10500 and also divide that by the "Percent", i m sorry we space told is 75.
So, 10500 divided by 75 = 140
And that method that the total number of marbles is 140.
Question: A high institution marching band has actually 105 flute players, If 75 percent the the band members pat the flute, climate how plenty of members are in the band?
Answer: There room 140 members in the band.
How To: The smaller "Part" in this problem is 105 due to the fact that there space 105 flute players and also we room told that they comprise 75 percent that the band, therefore the "Percent" is 75.
Again, it"s the "Total" that"s lacking here, and also to find it, we just need to follow our 2 step procedure together the ahead problem.
For step one, we multiply the "Part" by 100.
105 × 100 = 10500
For step two, we division that 10500 by the "Percent", i m sorry is 75.
See more: And I Want To Thank You For Giving Me The Best Day Of My Life ) Lyrics
10500 separated by 75 equates to 140
That way that the total number of band members is 140.
## Another action by step method
Step 1: Let"s i think the unknown value is Y
Step 2: first writing that as: 100% / Y = 75% / 105
Step 3: autumn the portion marks to simplify your calculations: 100 / Y = 75 / 105
Step 4: main point both sides by Y to relocate Y on the ideal side the the equation: 100 = ( 75 / 105 ) Y
Step 5: simplifying the ideal side, we get: 100 = 75 Y
Step 6: splitting both sides of the equation by 75, we will certainly arrive in ~ 140 = Y
This pipeline us with our final answer: 105 is 75 percent that 140
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105.8 is 75 percent the 141.067 105.81 is 75 percent that 141.08 105.82 is 75 percent that 141.093 105.83 is 75 percent the 141.107 105.84 is 75 percent of 141.12 105.85 is 75 percent of 141.133 105.86 is 75 percent the 141.147 105.87 is 75 percent of 141.16 105.88 is 75 percent that 141.173 105.89 is 75 percent that 141.187 105.9 is 75 percent of 141.2 105.91 is 75 percent that 141.213 105.92 is 75 percent the 141.227 105.93 is 75 percent of 141.24 105.94 is 75 percent that 141.253 105.95 is 75 percent the 141.267 105.96 is 75 percent of 141.28 105.97 is 75 percent that 141.293 105.98 is 75 percent the 141.307 105.99 is 75 percent that 141.32 |
# Question: What is the probability that a five-card poker hand contains a flush, that is, five cards of the same suit?
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1 ECS20 Discrete Mathematics Quarter: Spring 2007 Instructor: John Steinberger Assistant: Sophie Engle (prepared by Sophie Engle) Homework 8 Hints Due Wednesday June 6 th 2007 Section 6.1 #16 What is the probability that a five-card poker hand contains a flush, that is, five cards of the same suit? The set S is the set of all possible poker hands. We know that S = C(52, 5), since a deck of cards contains 52 cards and we want to choose a subset of 5 cards for the poker hand. You can break this down into the following tasks: (1) Choosing a suit. (2) Choosing five cards from that suit. For your answer, you may give a fraction, and don t need to evaluate the C(n, r) values. Section 6.1 #38 Two events E 1 and E 2 are called independent if p(e 1 E 2 ) = p(e 1 )p(e 2 ). For each of the following pairs of events, which are subsets of the set of all possible outcomes when a coin is tossed three times, determine whether or not they are independent. (a) E 1 : the first coin comes up tails; E 2 : the second coin comes up heads. (b) E 1 : the first coin comes up tails; E 2 : two, and not three, heads come up in a row. (c) E 1 : the second coin comes up tails; E 2 : two, and not three, heads come up in a row. page 1
2 Find p(e 1 ), p(e 2 ), and p(e 1 E 2 ). (Don t rely on intuition.) For example, look at part (a). When two coins are tossed, the possible outcomes are HH, HT, TH, and TT. Therefore, E 1 E 2 is the outcome TH. Find the probability of this, and compare it to p(e 1 )p(e 2 ). If they are the same, then E 1 and E 2 are independent. For each part, make sure to state E 1 and E 2 are independent, or E 1 and E 2 are NOT independent. Section 6.2 #6 What is the probability of these events when we randomly select a permutation of {1, 2, 3}? (a) 1 precedes 3 (b) 3 precedes 1 (c) 3 precedes 1 and 3 precedes 2 You can list all 6 permutations of {1, 2, 3}, or exploit symmetry. (For example, 1 either precedes 3, or it follows 3.) The statement 1 precedes 2 means that 1 occurs somewhere before 2, but it does not have to be immediately before. For example: in 312, 132, and 123, we have 1 preceding 2. This problem asks for a probability. Make sure to provide a fraction or value between 0 and 1! Section 6.2 #8 What is the probability of these events when we randomly select a permutation of {1, 2,..., n} where n 4? (a) 1 precedes 2 (b) 2 precedes 1 (c) 1 immediately precedes 2 (d) n precedes 1 and n 1 precedes 2 (e) n precedes 1 and n precedes 2 For each one, try to break the problem into the different possible outcomes (exploiting symmetry where you can). For example, for part (d) how often will n precede 1? Of all those times, when will n 1 also precede 2? This question asks for a probability. You may provide the reduced fraction or actual value. page 2
3 Section 6.2 #12 Suppose that E and F are events such that p(e) = 0.8 and p(f ) = 0.6. Show that p(e F ) 0.8 and p(e F ) 0.4. Look at exercise 11 (solved in back of book). Give something that looks like the solution to exercise 11. Section 6.2 #16 Show that if E and F are independent events, then E and F are also independent events. You need to show that p(e F ) = p(e) p(f ). De Morgan s Law states that E F = E F. Use that and what you know about probabilities to solve this problem. You ll need to give a clear algebraic argument. Section 6.2 #18 Assume that the year has 366 days and all birthdays are equally likely. (a) What is the probability that two people chosen at random were born on the same day of the week? (b) What is the probability that in a group of n people chosen at random, there are at least two born on the same day of the week? (c) How many people chosen at random are needed to make the probability greater than 1/2 that there are at least two people born on the same day of the week? Due to our assumption, the probability of birth in each day is 1/7. Look at example 13 in the book, and use your answer from part (b) to solve part (c). This question asks for a probability. You should be able to provide a reduced fraction or value for (a) and (c). Part (b) will look similar to the probability in example 13. page 3
4 Section 6.2 #20 Assume that the year has 366 days and all birthdays are equally likely. Find the smallest number of people you need to choose at random so that the probability that at least one of them were both born on April 1 exceeds 1/2. Find the probability that noone has a birthday on April 1 for n people chosen at random (you should have a formula based on the variable n). If you call this probability p n, then you want to solve p n = 1/2 for n. (You may need to use logarithms to do this.) Give an exact value for the smallest number of people. (Remember, you can t have half a person!) Section 6.2 #26 Let E be the event that a randomly generated bit string of length three contains an odd number of 1s, and let F be the event that the string starts with 1. Are E and F independent? Use the definition of independence. Your answer will either be Yes, they are independent or No, they are not independent. Section 6.2 #34 Find each of the following probabilities when n independent Bernoulli trails are carried out with probability of success p. (a) the probability of no success (b) the probability of at least one success (c) the probability of at most one success (d) the probability of at least two successes Use the binomial distribution. You will be providing probability formulas based on the variable p. page 4
5 Section 6.4 #4 A coin is biased so that the probability a head comes up when it is flipped is 0.6. What is the expected number of heads that come up when it is flipped 10 times? Apply theorem 2 (page 428), which gives the expected number of successes for n independent Bernoulli trials. Bernoulli trials are defined in Section 6.2 (page 406) of your book. They capture situations where there are two outcomes, like heads/tails, 0/1, or true/false. Give the expected number (please fully evaluate). Section 6.4 #6 What is the expected value when a \$1 lottery ticket is bought in which the purchaser wins exactly \$10 million if the ticket contains the six winning numbers chosen from the set {1, 2,..., 50} and the purchaser wins nothing otherwise? Assume each outcome is equally likely. Out of 50 numbers, the state chooses 6 as the winning numbers. That gives you the number of possible outcomes. To find the expected value, calculate the value of winning times the probability of winning. Then sum that with the value of losing times the probability of losing. This problem asks for the expected value, where value here is in terms of dollars. Therefore, your answer will look something like \$ (This is NOT the actual answer!) Section 6.4 #8 What is the expected sum of the numbers that appear when three fair dice are rolled? By Theorem 3 (page 429), we know that the expectation of a sum is the sum of expectations. Create three random variables, X, Y, Z, where each gives the value of a specific dice. You are asked for a expected sum, which you should be able to provide as a specific value. page 5
6 Section 6.4 #10 Suppose that we flip a coin until either it comes up tails twice or we have flipped it six times. What is the expected number of times we flip the coin? Let X be the random variable for the number of times we flip the coin. Look at the specific outcomes. For example, X = 2 when the first two flips are tails. X = 4 when there was one tail in the first three flips, and the fourth flip was a tail. However, X = 6 means that the first five flips had only one tail, or no tails and then we stopped. Use theorem 1 to combine all of these outcomes and find the expected value. This problem asks for the expected value. You should be able to provide a specific number. Section 6.4 #12 Suppose that we roll a die until a 6 comes up. (a) What is the probability that we roll the die n times? (b) What is the expected number of times we roll the die? Let X be the random variable that for the number of times we roll the die. geometric distribution (see pages ). This will have a Your answer will be (a) a probability in terms of n and (b) a specific value. Section 6.4 #16 Let X and Y be the random variables that count the number of heads and the number of tails that come up when two coins are flipped. Show that X and Y are not independent. You need to show that p(x = i and Y = j) is not always equal to p(x = i) p(y = j). To do this, find an example i and j where this occurs. Your answer should be a counterexample for a specific i and j. page 6
7 Section 6.4 #20 Let A be an event. Then I A, the indicator random variable of A, equals 1 if A occurs and equals 0 otherwise. Show that the expectation of the indicator random variable of A equals the probability of A, that is E(I A ) = p(a). Notice that by definition: p(s) = p(a) s A Also, it may help to rewrite I A as: { 1 if s A I A = 0 if s A Your answer should be a clear algebraic argument that E(I A ) = p(a). Section 6.4 #24 What is the variance of the number of times a 6 appears when a fair dice is rolled 10 times? Look at example 18. This problem asks for the variance. You may provide a reduced fraction or the actual value. (6.4 #30 on next page) page 7
8 Section 6.4 #30 Use Chebyshev s Inequality to find an upper bound on the probability that the number of tails that come up when a biased coin with probability of heads equal to 0.6 is tossed n times deviates from the mean by more than n. Let X be the random variable that counts the number of tails and E(X) be its mean (or expected value). The text says that X deviates from the mean by more than n, i.e. the distance between X(s) and E(X) is greater than n. Therefore, we are trying to find the probability: p( X(s) E(X) r) where r = n. Chebyshev s Inequality states that this is bounded by: V (X) r 2 We can use the approach in example 18 to find V (x), and example 19 for how to proceed from there. This problem asks for the bound V (X)/r 2. Your answer should be an exact number. page 8
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## Problem
Let $a_0 be an infinite sequence of positive integers, Prove that there exists a unique integer $n\ge1$ such that $$a_n<\frac{a_0+a_1+\cdots + a_n}{n}\le a_{n+1}.$$
## Solution
Define $f(n) = a_0 + a_1 + \dots + a_n - n a_{n+1}$. (In particular, $f(0) = a_0.$) Notice that because $a_{n+2} \ge a_{n+1}$, we have $$a_0 + a_1 + \dots + a_n - n a_{n+1} > a_0 + a_1 + \dots + a_n + a_{n+1} - (n+1) a_{n+2}.$$ Thus, $f(n) > f(n+1)$; i.e., $f$ is monotonic decreasing. Therefore, because $f(0) > 0$, there exists a unique $N$ such that $f(N-1) > 0 \ge f(N)$. In other words, $$a_0 + a_1 + \dots + a_{N-1} - (N-1) a_N > 0$$ $$a_0 + a_1 + \dots + a_N - n a_{N+1} \le 0.$$ This rearranges to give $$a_N < \frac{a_0 + a_1 + \dots + a_N}{N} \le a_{N+1}.$$ Define $g(n) = a_0 + a_1 + \dots + a_n - n a_n$. Then because $a_{n+1} > a_n$, we have $$a_0 + a_1 + \dots + a_n - n a_n > a_0 + a_1 + \dots + a_n + a_{n+1} - (n+1) a_{n+1}.$$ Therefore, $g$ is also monotonic decreasing. Note that $g(N+1) = a_0 + a_1 + \dots + a_{N+1} - (N+1) a_{N+1} < 0$ from our inequality, and so $g(k) < 0$ for all $k > N$. Thus, the given inequality, which requires that $g(n) > 0$, cannot be satisfied for $n > N$, and so $N$ is the unique solution to this inequality.
--Suli 22:38, 7 February 2015 (EST)
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
2014 IMO (Problems) • Resources Preceded byFirst Problem 1 • 2 • 3 • 4 • 5 • 6 Followed byProblem 2 All IMO Problems and Solutions |
What Are the Factors of 75?
The factors of 75 are 1, 3, 5, 15, 25 and 75. The factors may be determined by dividing 75 by whole numbers starting from 1. If the resulting quotient is also a whole number, then both the divisor and quotient are factors of the number.
The number 75 can be derived from the multiplication of these three possible combinations: 1 and 75, 3 and 25, and 5 and 15.
Any whole number has at least two factors, the most basic being 1 and the number itself. One practical way of determining the factors of a number is to write them in ascending order, with the number 1 at the leftmost side of the row and the number itself at the rightmost. The next step is to divide the number by 2 to check if the quotient is a whole number. If so, the two factors should be written next to 1 and the number. The process is repeated by using 3, 4, and so on until the series of factors reaches a pair of numbers that doesn't have a whole number factor in between them. The series of written numbers are the factors of the number in question.
The term "factor" is used in multiplication and is one of the four basic arithmetic operations. Two whole numbers multiplied together to form a product are called factors.
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# Elements of the Differential and Integral Calculus/Chapter IV
## CHAPTER IV
DIFFERENTIATION
25. Introduction. We shall now proceed to investigate the manner in which a function changes in value as the independent variable changes. The fundamental problem of the Differential Calculus is to establish a measure of this change in the function with mathematical precision. It was while investigating problems of this sort, dealing with continuously varying quantities, that Newton[1] was led to the discovery of the fundamental principles of the Calculus, the most scientific and powerful tool of the modern mathematician.
26. Increments. The increment of a variable in changing from one numerical value to another is the difference found by subtracting the first value from the second. An increment of $x$ is denoted by the symbol $\Delta x$, read delta $x$.
The student is warned against reading this symbol delta times $x$, it having no such meaning. Evidently this increment may be either positive or negative[2] according as the variable in changing is increasing or decreasing in value. Similarly,
$\Delta y$ denotes an increment of $y$, $\Delta \phi$ denotes an increment of $\phi$, $\Delta f(x)$ denotes an increment $f(x)$, etc.
If in $y = f(x)$ the independent variable $x$, takes on an increment $\Delta x$, then $\Delta y$ is always understood to denote the corresponding increment of the function $f(x)$ (or dependent variable $y$).
The increment $\Delta y$ is always assumed to be reckoned from a definite initial value of $y$ corresponding to the arbitrarily fixed initial value of $x$ from which the increment $\Delta x$ is reckoned. For instance, consider the function
$y = x^2$.
Assuming $x = 10$ for the initial value of $x$ fixes $y = 100$ as the initial value of $y$.
Suppose $x$ increases to $x$ $= 12$, that is, $\Delta x$ $= 2$; then $y$ increases to $y$ $= 144$, and $\Delta y$ $= 44$. Suppose $x$ decreases to $x$ $= 9$, that is, $\Delta x$ $= -1$; then $y$ increases to $y$ $= 81$, and $\Delta y$ $= -19$.
It may happen that as $x$ increases, $y$ decreases, or the reverse; in either case $\Delta x$ and $\Delta y$ will have opposite signs.
It is also clear (as illustrated in the above example) that if $y = f(x)$ is a continuous function and $\Delta x$ is decreasing in numerical value, then $\Delta y$ also decreases in numerical value.
27. Comparison of increments. Consider the function
(A) $y =x^2$.
Assuming a fixed initial value for $x$, let $x$ take on an increment $\Delta x$. Then $y$ will take on a corresponding increment $\Delta y$, and we have
$y +$ $\Delta y$ $= (x + \Delta x)^2$, or, $y +$ $\Delta y$ $= x^2+ 2x \cdot \Delta x + (\Delta x)^2$. Subtracting (A), $y$ $= x^2$ (B) $\Delta y$ $= 2x \cdot \Delta x + (\Delta x)^2$
we get the increment $\Delta y$ in terms of $x$ and $\Delta x$.
To find the ratio of the increments, divide (B) by $\Delta x$, giving
$\frac{\Delta y}{\Delta x} = 2x + \Delta x$.
If the initial value of $x$ is 4, it is evident that
$\lim_{\Delta x = 0} \frac{\Delta y}{\Delta x} = 8$.
Let us carefully note the behavior of the ratio of the increments of $x$ and $y$ as the increment of $x$ diminishes.
Initial
value of $x$
New
value of $x$
Increment
$\Delta\ x$
Initial
value of $y$
New
value of $y$
Increment
$\Delta\ y$
$\frac{\Delta y}{\Delta x}$
4 5.0 1.0 16 25. 9. 9.
4 4.8 0.8 16 23.04 7.04 8.8
4 4.6 0.6 16 21.16 5.16 8.6
4 4.4 0.4 16 19.36 3.36 8.4
4 4.2 0.2 16 17.64 1.64 8.2
4 4.1 0.1 16 16.81 0.81 8.1
4 4.01 0.01 16 16.0801 0.0801 8.01
It is apparent that as $\Delta x$ decreases, $\Delta y$ also diminishes, but their ratio takes on the successive values 9, 8.8, 8.6, 8.4, 8.2, 8.1, 8.01; illustrating the fact that $\frac{\Delta y}{\Delta x}$ can be brought as near to 8 in value as we please by making $\Delta x$ small enough. Therefore
$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = 8$.[3]
28. Derivative of a function of one variable. The fundamental definition of the Differential Calculus is:
The derivative[4] of a function is the limit of the ratio of the increment of the function to the increment of the independent variable, when the latter increment varies and approaches the limit zero.
When the limit of this ratio exists, the function is said to be differentiable, or to possess a derivative.
The above definition may be given in a more compact form symbolically as follows: Given the function
(A) $y = f(x)$,
and consider $x$ to have a fixed value.
Let $x$ take on an increment $\Delta x$; then the function $y$ takes on an increment $\Delta y$, the new value of the function being
(B) $y + \Delta y = f(x + \Delta x)$.
To find the increment of the function, subtract (A) from (B), giving
(C) $\Delta y = f(x + \Delta x) - f(x)$.
Dividing by the increment of the variable, $\Delta x$, we get
(D) $\frac{\Delta y}{\Delta x} = \frac{f(x + \Delta x) - f(x)}{\Delta x}$.
The limit of this ratio when $\Delta x$ approaches the limit zero is, from our definition, the derivative and is denoted by the symbol $\frac{dy}{dx}$. Therefore
(E) $\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$.
defines the derivative of $y$ [or $f(x)$] with respect to x.
From (D) we also get
$\frac{dy}{dx} = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$
The process of finding the derivative of a function is called differentiation.
It should be carefully noted that the derivative is the limit of the ratio, not the ratio of the limits. The latter ratio would assume the form $\tfrac{0}{0}$, which is indeterminate (§ 14).
29. Symbols for derivatives. Since $\Delta y$ and $\Delta x$ are always finite and have definite values, the expression
$\frac{\Delta y}{\Delta x}$
is really a fraction. The symbol
$\frac{dy}{dx}$,
however, is to be regarded not as a fraction but as the limiting value of a fraction. In many cases it will be seen that this symbol does possess fractional properties, and later on we shall show how meanings may be attached to $dy$ and $dx$, but for the present the symbol $\frac{dy}{dx}$ is to be considered as a whole.
Since the derivative of a function of $x$ is in general also a function of $x$, the symbol $f'(x)$ is also used to denote the derivative of $f(x)$.
Hence, if $y$ $= f(x)$, we may write $\frac{dy}{dx}$ $= f'(x)$,
which is read the derivative of $y$ with respect to $x$ equals $f$ prime of $x$. The symbol
$\frac{d}{dx}$
when considered by itself is called the differentiating operator, and indicates that any function written after it is to be differentiated with respect to $x$. Thus
$\frac{dy}{dx}$ or $\frac{d}{dx} y$ indicates the derivative of $y$ with respect to $x$;
$\frac{d}{dx} f(x)$ indicates the derivative of $f(x)$ with respect to $x$;
$\frac{d}{dx} (2x^2 + 5)$ indicates the derivative of $2x^2 + 5$ with respect to $x$;
$y'$ is an abbreviated form of $\frac{dy}{dx}$.
The symbol $D_x$ is used by some writers instead of $\frac{d}{dx}$. If then
$y = f(x)$,
we may write the identities
$y' = \frac{dy}{dx} = \frac{d}{dx} y = D_x f(x) = f'(x)$.
30. Differentiable functions. From the Theory of Limits it is clear that if the derivative of a function exists for a certain value of the independent variable, the function itself must be continuous for that value of the variable.
The converse, however, is not always true, functions having been discovered that are continuous and yet possess no derivative. But such functions do not occur often in applied mathematics, and in this book only differentiable functions are considered, that is, functions that possess a derivative for all values of the independent variable save at most for isolated values.
31. General rule for differentiation. From the definition of a derivative it is seen that the process of differentiating a function $y = f(x)$ consists in taking the following distinct steps:
GENERAL RULE FOR DIFFERENTIATION
[5]
First Step. In the function replace $x$ by $x + \Delta x$, giving a new value of the function, $y + \Delta y$.
Second Step. Subtract the given value of the function from the new value in order to find $\Delta y$ (the increment of the function).
Third Step. Divide the remainder $\Delta y$ (the increment of the function) by $\Delta x$ (the increment of the independent variable).
Fourth Step. Find the limit of this quotient, when $\Delta x$ (the increment of the independent variable) varies and approaches the limit zero. This is the derivative required.
The student should become thoroughly familiar with this rule by applying the process to a large number of examples. Three such examples will now be worked out in detail.
Illustrative Example 1. Differentiate $3x^2 + 5$.
Solution. Applying the successive steps in the General Rule, we get, after placing
$y$ $= 3x^2 + 5$, First step. $y + \Delta y$ $= 3(x + \Delta x)^2 + 5$ $= 3x^2 + 6x \cdot \Delta x + 3(\Delta x)^2 + 5$.
Second step. $y + \Delta y$ $= 3x^2 + 6 x \cdot \Delta x + 3(\Delta x)^2$ $+ 5$ $y$ $= 3x^2$ $+ 5$ $\Delta y$ $= 6x \cdot \Delta x + 3(\Delta x)^2$. Third step. $\frac{\Delta y}{\Delta x}$ $= 6x + 3 \cdot \Delta x$. Fourth step. $\frac{dy}{dx}$ $= 6x$. Ans.
We may also write this
$\frac{d}{dx} (3x^2 + 5) = 6x$.
Illustrative Example 2. Differentiate $x^3 - 2x + 7$.
Solution. Place $y$ $= x^3 -2x + 7$. First step. $y + \Delta y$ $= (x + \Delta x)^3 - 2(x + \Delta x) +7$. $= x^3$ $+ 3x^2 \cdot \Delta x + 3x \cdot (\Delta x)^2 + (\Delta x)^3$ $- 2x$ $- 2 \cdot \Delta x$ $+ 7$. Second step. $y + \Delta y$ $= x^3$ $+ 3x^2 \cdot \Delta x + 3x \cdot (\Delta x)^2 + (\Delta x)^3$ $- 2x$ $- 2 \cdot \Delta x$ $+ 7$. $y$ $= x^3$ $- 2x$ $+ 7$. $\Delta y$ $=$ $3x^2 \cdot \Delta x + 3x \cdot (\Delta x)^2 + (\Delta x)^3$ $- 2 \cdot \Delta x$ Third step. $\frac{\Delta y}{\Delta x}$ $= 3x^2 + 3x \cdot \Delta x + (\Delta x)^2 - 2.$ Fourth step. $\frac{dy}{dx}$ $= 3x^2 - 2$. Ans. Or, $\frac{d}{dx} (x^3 - 2x + 7)$ $= 3x^2 - 2$.
Illustrative Example 3. Differentiate $\tfrac{c}{x^2}$.
Solution. Place $y$ $= \frac{c}{x^2}$. First step. $y + \Delta y$ $= \frac{c}{(x + \Delta x)^2}$. Second step. $y + \Delta y$ $= \frac{c}{(x + \Delta x)^2}$ $y \qquad$ $= \frac{c}{x^2}$ $\Delta y$ $= \frac{c}{(x + \Delta x)^2} - \frac{c}{x^2} = \frac{-c \cdot \Delta x(2x + \Delta x)}{x^2(x + \Delta x)^2}$. Third step. $\frac{\Delta y}{\Delta x}$ $= -c \cdot \frac{2x + \Delta x}{x^2(x + \Delta x)^2}$. Fourth step. $\frac{dy}{dx}$ $= -c \cdot \frac{2x}{x^2(x)^2}$. $= -\frac{2c}{x^3}$. Ans. Or, $\frac{d}{dx} \left ( \frac{c}{x^2} \right ) =$ $\frac{-2c}{x^3}$.
EXAMPLES
Use the General Rule, § 31, in differentiating the following functions:
1. $y$ $= 3x^2$. Ans. $\frac{dy}{dx}$ $= 6x$. 2. $y$ $= x^2 + 2$. $\frac{dy}{dx}$ $= 2x$. 3. $y$ $= 5 - 4x$. $\frac{dy}{dx}$ $= -4$. 4. $s$ $= 2t^2 - 4$. $\frac{ds}{dt}$ $= 4t$. 5. $y$ $= \frac{1}{x}$. $\frac{dy}{dx}$ $= -\frac{1}{x^2}$. 6. $y$ $= \frac{x + 2}{x}$. $\frac{dy}{dx}$ $= -\frac{-2}{x^2}$. 7. $y$ $= x^3$. $\frac{dy}{dx}$ $= 3x^2$. 8. $y$ $= 2x^2 - 3$. $\frac{dy}{dx}$ $= 4x$. 9. $y$ $= 1 - 2x^3$. $\frac{dy}{dx}$ $= 4x$. 10. $\rho$ $= a\theta^2$. $\frac{d\rho}{d\theta}$ $= 2a\theta$. 11. $y$ $= \frac{2}{x^2}$. $\frac{dy}{dx}$ $= -\frac{4}{x^3}$. 12. $y$ $= \frac{3}{x - 1}$. $\frac{dy}{dx}$ $= -\frac{3}{(x - 1)^2}$.
13. $y = 7x^2 + x$.
14. $s = at^2 - 2bt$.
15. $r = 8t + 3t^2$.
16. $y = \frac{3}{x^2}$.
17. $s = -\frac{a}{2t + 3}$.
18. $y = bx^3 - cx$.
19. $\rho = 3\theta^3 - 2\theta^2$.
20. $y = \frac{3}{4}x^2 - \frac{1}{2}x$.
21. $y = \frac{x^2 - 5}{x}$.
22. $\rho = \frac{\theta^2}{1 + \theta}$.
23. $y = \frac{1}{2}x^2 + 2x$.
24. $z = 4x - 3x^2$.
25. $\rho = 3\theta + \theta^2$.
26. $y = \frac{ax + b}{x^2}$.
27. $z = \frac{x^3 + 2}{x}$.
28. $y$ $= x^2 - 3x + 6$. Ans. $y'$ $= 2x - 3$. 29. $s$ $= 2t^2 + 5t - 8$. Ans. $s'$ $= 4t + 5$. 30. $\rho$ $= 5\theta^3 - 2\theta + 6$. Ans. $\rho'$ $= 15\theta^2 - 2$. 31. $y$ $= ax^2 + bx + c$. Ans. $y'$ $= 2ax + b$.
32. Applications of the derivative to Geometry. We consider a theorem which is fundamental in all Differential Calculus to Geometry. Let
(A) $y = f(x)$
be the equation of a curve $AB$.
Now differentiate (A) by the General Rule and interpret each step geometrically.
First Step. $y + \Delta y$ $= f(x + \Delta x)$ $= NQ$ Second Step. $y + \Delta y$ $= f(x + \Delta x)$ $= NQ$ $y$ $= f(x)$ $= MP = NR$ $\Delta y$ $f(x + \Delta x) - f(x)$ $= RQ$.
Third Step. $\frac{\Delta y}{\Delta x}$ $= \frac{f(x + \Delta x) - f(x)}{\Delta x} = \frac{RQ}{MN} = \frac{RQ}{PR}$ $= \tan RPQ = \tan \phi$ = slope of secant line $PQ$.
Fourth Step. $\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x}$ (B) $= \frac{dy}{dx} =$ value of the derivative at $P$.
But when we let $\Delta x \dot= 0$, the point $Q$ will move along the curve and approach nearer and nearer to $P$, the secant will turn about $P$ and approach the tangent as a limiting position, and we have also
$\lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x}$ $= \lim_{\Delta x \to 0} \tan \phi = \tan \tau$ (C) = slope of the tangent at $P$.
Hence from (B) and (C), $\tfrac{dy}{dx} =$ slope of the tangent line $PT$. Therefore
Theorem. The value of the derivative at any point of a curve is equal to the slope of the line drawn tangent to the curve at that point.
It was this tangent problem that led Leibnitz[6] to the discovery of the Differential Calculus.
Illustrative Example 1. Find the slopes of the tangents to the parabola $y = x^2$ at the vertex, and at the point where $x = \tfrac{1}{2}$.
Solution. Differentiating by General Rule, §31, we get
(A) $\frac{dy}{dx} = 2x =$ slope of tangent line at any point on curve.
To find slope of tangent at vertex, substitute $x = 0$ in (A), giving
$\frac{dy}{dx} = 0$.
Therefore the tangent at vertex has the slope zero; that is, it is parallel to the axis of x and in this case coincides with it.
To find slope of tangent at the point $P$, where $x = \tfrac{1}{2}$, substitute in (A), giving
$\frac{dy}{dx} = 1$;
that is, the tangent at the point $P$ makes an angle of 45° with the axis of $x$.
EXAMPLES
Find by differentiation the slopes of the tangents to the following curves at the points indicated. Verify each result by drawing the curve and its tangent.
1. $y = x^2 - 4$, where x = 2. Ans. 4. 2. $y = 6 - 3x^2$ where x = 1. -6. 3. $y = x^3$, where x = -1. -3. 4. $y = \frac{2}{x}$, where x = -1. $-\frac{1}{2}$. 5. $y = x - x^2$, where x = 0. 1. 6. $y = \frac{1}{x - 1}$, where x = 3. $-\frac{1}{4}$. 7. $y = \frac{1}{2} x^2$, where x = 4. 4. 8. $y = x^2 - 2x + 3$, where x = 1. 0. 9. $y = 9 - x^2$, where x = -3. 6.
10. Find the slope of the tangent to the curve $y = 2x^3 - 6x + 5$, (a) at the point where $x = 1$; (b) at the point where $x = 0$.
Ans. (a) 0; (b) -6.
11. (a) Find the slopes of the tangents to the two curves $y = 3x^2 - 1$ and $y = 2x^2 + 3$ at their points of intersection. (b) At what angle do they intersect?
Ans. (a) $\pm 12, \pm 8$; (b) $\arctan \frac{4}{97}$.
12. The curves on a railway track are often made parabolic in form. Suppose that a track has the form of the parabola $y = x^2$ (last figure, § 32), the directions $OX$ and $OY$ being east and north respectively, and the unit of measurement 1 mile. If the train is going east when passing through $O$, in what direction will it be going
(a) when $\tfrac{1}{2}$ mi. east of $OY$? Ans. Northeast. (b) when $\tfrac{1}{2}$ mi. west of $OY$? Southeast. (c) when $\tfrac{\sqrt{3}}{2}$ mi. east of $OY$? N. 30°E. (d) when $\tfrac{1}{12}$ mi. north of $OX$? E. 30°S., or E. 30°N.
13. A street-car track has the form of the cubical parabola $y = x^3$. Assume the same directions and unit as in the last example. If a car is going west when passing through $O$, in what direction will it be going
(a) when $\tfrac{1}{\sqrt{3}}$ mi. east of $OY$? Ans. Southwest. (b) when $\tfrac{1}{\sqrt{3}}$ mi. west of $OY$? Southwest. (c) when $\tfrac{1}{2}$ mi. north of $OX$? S. 27° 43' W. (d) when 2 mi. south of $OX$? (e) when equidistant from $OX$ and $OY$?
1. Sir Isaac Newton (1642-1727), an Englishman, was a man of the most extraordinary genius. He developed the science of the Calculus under the name of Fluxions. Although Newton had discovered and made use of the new science as early as 1670, his first published work in which it occurs is dated 1687, having the title Philosophiae Naturalis Principia Mathematica. This was Newton's principal work. Laplace said of it, "It will always remain preminent above all other productions of the human mind." See frontispiece.
2. Some writers call a negative increment a decrement.
3. The student should guard against the common error of concluding that because the numerator and denominator of a fraction are each approaching zero as a limit, the limit of the value of the fraction (or ratio) is zero. The limit of the ratio may take on any numerical value. In the above example the limit is 8.
4. Also called the differential coefficient or the derived function.
5. Also called the Four-step Rule.
6. Gottfried Wilhelm Leibnitz (1646-1716) was a native of Leipzig. His remarkable abilities were shown by original investigations in several branches of learning. He was first to publish his discoveries in Calculus in a short essay appearing in the periodical Acta Eruditorum at Leipzig in 1684. It is known, however, that manuscripts on Fluxions written by Newton were already in existence, and from these some claim Leibnitz got the new ideas. The decision of modern times seems to be that both Newton and Leibnitz invented the Calculus independently of each other. The notation nsed to-day was introduced by Leibnitz. See frontispiece. |
# Manuals/calci/MATRIXPRODUCT
(diff) ← Older revision | Latest revision (diff) | Newer revision → (diff)
MATRIXPRODUCT (a,b,ConsiderUnits)
• where and are the array of two matrices
## Description
• This function gives product of two matrices.
• Matrix multiplication is of two types:
Type 1: A scalar (a constant) is multiplied with the each element of the matrix.
Type 2: Multiplication of two matrices.
• We can do the matrix multiplication when the number of columns in the first matrix equals the number of rows in the second matrix.
• For e.g. 4x2 matrix can multiply with 2x3. The matrix product of two arrays and is: where is the row number and is the column number.
• i.e Multiply the elements of each row of 1st matrix by elements of each column of 2nd matrix.
• So the resultant matrix is of the order: Rows of 1st matrix × Columns of 2nd.
• For e.g If we multiply a 4x2 matrix with a 2x3 matrix, the product matrix is of order 4x3.
• Matrix multiplication satisfies the associative and distributive properties.But it is not satisfies the commutative property.
• i.e., Let A,B and C are three matrices, then A(BC)= (AB)C (Associative property)
• A(B+C)= AB+AC and (A+B)C = AC+BC (Distributive properties)
• k(AB)=(kA)B=A(kB)where k is a constant.But (Commutative property)
## Examples
1. MATRIXPRODUCT([2,3,4;5,6,7],5)
10 15 20 25 30 35
2. MATRIXPRODUCT([[6,7,8],[10,12,-22],[7,17,23]],[[20,12,16],[7,8,13],[4,8,9]])
120 84 128 70 96 -286 28 136 207 |
### Solving 3-step linear equations
• A linear expression in a variable x is one that simplifies to ax + b, where a and b are numbers.
A linear equation in x says that two linear expressions are equal.
A linear equation can be simplified by collecting terms on each side of the equation.
At worst, the result will be an equation such as 5x + 5 = 2x + 29, with a variable term and a constant on each side of the equals sign.
•This sort of equation is easy to solve, as shown in the chart below.
The basic idea is that you want both sides of the equation to stay equal, and so you should always "do the same thing" to both sides of the equation. In detail, you can:
• Add any number to both sides of the equation;
• Subtract any number from both sides;
• Multiply or divide both sides by any number other than zero.
Here is the original equation. On the left side, the constant 5 has been added to 5x. 5x + 5 = 2x + 29 Subtracting 5 from both sides makes the left side simpler. 5x + 5 -5 = 2x + 29 -5 Simplify. 5x = 2x + 24 Subtract 2x from both sides to get all variable terms on the left side. 5x - 2x = 2x + 24 -2x Simplify 3x = 24 Divide both sides by 3 and simplify to get the solution. x = 24/3 so x = 8
• When you get more experience, and if it's OK with your teacher, condense the solution chart as follows.
Here is the original equation. 5x + 5 = 2x + 29 Subtract the left side constant from both sides and simplify. 5x = 2x + 24 Subtract the right side variable term from both sides and simplify. 3x =24 Divide by 3 and simplify x = 8
• One way to think about the method shown is to say that at each step, we transposed a term from one sided to the other.
Transpose means: remove a term from one side of the equation and insert in on the other side with its sign reversed.
This is just another way of saying that you reverse adding a term by subtracting that term. Let's rewrite the solution based on that idea.
Here is the original equation. On the left side, the term 5 is added to 5 5x + 5 = 2x + 29 Transpose the term 5 from the left side to the right side. 5x = 2x + 29 - 5 Simplify. On the right side, the term 2x is added to 24 5x = 2x + 24 Transpose 2x from the right side to the left side 5x - 2x = 24 Simplify 3x = 24 Divide both sides by 3 and simplify to get the solution. x = 24/3 = 8
•Now look at the figure.. The problem shown is the one solved above. You can try any of three methods:
1. Copy the steps of the solution shown, mainly to get practice with typing.
2. Solve the problem with the steps visible as hints in the left column.
3. Solve the problem without hints. Warning: you still have to follow the (invisible) hints line by line!
•To check your work, click the 'Check my work' button at any time. You will be told whether each of the lines you typed is correct. If a line says 'No', retype your answer.
•To get a new problem, click on the top button in the list. |
# Complex Numbers - Basic Operations
## Definition of Complex Numbers
A complex number z is a number of the form
z = a + b i
where a and b are real numbers and i is the imaginary unit defined by
$i = \sqrt{-1}$
a is called the real part of z and b is the imaginary part of z.
Note that the set
R of all real numbers is a subset of the complex number C since any real number may be considered as having the imaginary part equal to zero..
## Complex Conjugate
The conjugate of a complex number a + b i is a complex number equal to
a - b i
Examples: Find the conjugate of the following complex numbers.
a) 2 - i , b) -3 + 4i , c) 5 , d) -5i
Solution to above example
a) 2 + i
b) -3 - 4i
c) 5
d) 5i
## Addition of Complex Numbers
Addition of two complex numbers a + b i and c + d i is defined as follows.
(a + b i) + (c + d i) = (a + c) + (b + d) i
This is similar to grouping like terms: real parts are added to real parts and imaginary parts are added to imaginary parts.
Example: Express in the form of a complex number a + b i.
• (2 + 3i) + (-4 + 5i)
• (3i) + (-5 + 6i)
• (2) + (-2 + 9i)
Solution to above example.
• (2 + 3i) + (-4 + 5i) = (2 - 4) + (3 + 5) i = - 2 + 8 i
• (3i) + (-5 + 6i) = (0 - 5) + (3 + 6) i = -5 + 9 i
• (2) + (-2 + 9i) = (2 - 2) + (9) i = 9i
Addition can be done by grouping like terms.
(2 + 3i) + (-4 + 5i) = 2 + 3 i - 4 + 5 i = -2 + 8 i
Calculator
to add complex numbers for practice is available.
## Subtraction of Complex Numbers
The subtraction of two complex numbers a + b i and c + d i is defined as follows.
(a + b i) - (c + d i) = (a - b) + (b - d) i
Example: Express in the form of a complex number a + b i.
• (2 - 5i) - (-4 - 5i)
• (-7i) - (-5 - 6i)
• (2) - (2 + 6i)
Solution to above example
• (2 - 5i) - (-4 - 5i) = (2 - (-4)) + (-5 - (-5)) i = 6
• (-7i) - (-5 - 6i) = (0 - (-5)) + (-7 - (-6)) i = 5 - i
• (2) - (2 + 6i) = (2 - 2) - 6 i = -6 i
Note: subtraction can be done as follows:
(a + b i) - (c + d i) = (a + bi) + (- c - d i) and then group like terms
Example:
(2 - 5i) - (-4 - 5i) = 2 - 5 i + 4 + 5 i = 6
## Multiply Complex Numbers
The multiplication of two complex numbers a + b i and c + d i is defined as follows.
(a + b i)(c + d i) = (a c - b d) + (a d + b c) i
However you do not need to memorize the above definition as the multiplication can be carried out using properties similar to those of the real numbers and the added property i
2 = -1. (see the example below)
Example: Express in the form of a complex number a + b i.
(3 + 2 i)(3 - 3i)
Solution to above example
(3 + 2 i)(3 - 3i)
Using the distributive law, (3 + 2 i)(3 - 3 i) can be written as
(3 + 2 i)(3 - 3 i) = (3 + 2 i)(3) + (3 + 2 i)(-3 i) = 9 + 6 i - 9 i -6 i
2
Group like terms and use i
2 = -1 to simplify (3 + 2 i)(3 - 3 i)
(3 + 2 i)(3 - 3 i) = 15 - 3 i
Calculator
to multiply complex numbers for practice is available.
## Divide two Complex Numbers
We use the multiplication property of complex number and its conjugate to divide two complex numbers.
Example: Express in the form of a complex number a + b i.
• $\dfrac{8 + 4 i}{1-i}$
We first multiply the numerator and denominator by the complex conjugate of the denominator
$\dfrac{(8 + 4 i)\color{red}{(1+i)}}{(1-i)\color{red}{(1+i)}}$
Multiply and group like terms
$= \dfrac{8 + 4 i + 8 i + 4 i^2}{1 - i + i - i^2} = \dfrac{4 + 12i}{2}$
$= 2 + 6 i$
Calculator
to divide complex numbers for practice is available.
## Equality of two Complex Numbers
The complex numbers a + i b and x + i y are equal if their real parts are equal and their imaginary parts are equal.
a + i b = x + i y if and only if a = x and b = y
Example: Find the real numbers x and y such that 2x + y + i(x - y) = 4 - i.
For the two complex numbers to be equal their real parts and their imaginary parts has to be equal. Hence
2x + y = 4 and x - y = - 1
Solve the above system of equations in x and y to find
x = 1 and y = 2.
## Exercises
• Find the complex conjugate of the following complex numbers
a) 2 + 6 i
b) -8 i
c) 12
• Write the following expressions in the form a + b i
a) (2 - 8 i) + (-6 i)
b) -8 i + (3 - 9 i)
c) 6 - (3 - i)
d) (2 - 3 i)(7 - i)
e) $\dfrac{2+2i}{2-2i}$
### Solutions to above exercises
• Find the complex conjugate.
a) 2 - 6 i
b) 8 i
c) 12
• Write the following expressions in the form a + b i
a) (2 - 8 i) + (-6 i) = 2 - 14 i
b) -8 i + (3 - 9 i) = 3 - 17 i
c) 6 - (3 - i) = 3 + i
d) (2 - 3 i)(7 - i) = 11 - 23 i
e) $\dfrac{2+2i}{2-2i} = i$ |
1. ## Easy Math :P
Should be easy for you guys anyways... Okay...
1)
Simplify
p3r/pq X q2/pr2
2)
Twenty gallons of crude oil were poured into two containers of different size. Express the amount of crude oil poured into the smaller container in terms of the amount G poured into the larger container. (Is this possible? Don't give you any real information)
3) 30% of the people in a restaurant are adults while the rest are children. Three-fifths of the adults are men and 3/7 of the children are boys. There are 112 more girls than women. How many people are there in the restaurant?
4) The ratio of the weight of a cow to that of a goat is 8:3. The ratio of the weight of the goat to that of a sheep is 5:4. The cow is heavier than the sheep by 112 kg. Find the total weight of the three animals.
2. Originally Posted by Rocher
2)
Twenty gallons of crude oil were poured into two containers of different size. Express the amount of crude oil poured into the smaller container in terms of the amount G poured into the larger container. (Is this possible? Don't give you any real information)
The small container gets: $\displaystyle 20-G$
3) 30% of the people in a restaurant are adults while the rest are children. Three-fifths of the adults are men and 3/7 of the children are boys. There are 112 more girls than women. How many people are there in the restaurant?
$\displaystyle \frac{3}{10}$ are adults so $\displaystyle \frac{7}{10}$ are children.
$\displaystyle \frac{3}{7}$ of the children are boys, so $\displaystyle \frac{7}{10}\times\frac{3}{7}=\frac{3}{10}$ of the group is boys.
That means that: $\displaystyle \frac{7}{10}-\frac{3}{10}=\frac{4}{10}=\frac{2}{5}$ of the group are girls (because all the children that aren't boys are girls.... hopefully)
Anyway, $\displaystyle \frac{3}{5}$ of the adults are men, so then $\displaystyle \frac{3}{10}\times\frac{3}{5}=\frac{9}{50}$ of the group are men.
That means that $\displaystyle \frac{3}{10}-\frac{9}{50}=\frac{6}{50}=\frac{3}{25}$ of the group are women.
So we know that ($\displaystyle g$ stands for the number of people in the group): $\displaystyle g\times\frac{2}{5}-g\times\frac{3}{25}=112$
So: $\displaystyle g\left(\frac{2}{5}-\frac{3}{25}\right)=112$
Subtract: $\displaystyle g\left(\frac{7}{25}\right)=112$
divide both sides by $\displaystyle \frac{7}{25}$ to get: $\displaystyle \boxed{g=400}$
3. Thank you, now just need 1 and 4 :P
4. The key to the fouth problem is to write it as fractions
cows/goat 8/3 goat/sheep 5/4
and then find the lcm of the goat weight.
cows/goat 40/15 goat/sheep 15/12
and then reason this:
If the ratio of the cow's weight to the goat's weight is 40:15 and the goat's to the sheep's weight is 15:12, therefore the the cow's to sheep's ratio is 40:12. Then we have the given ratio of cow to sheep equals the ratio of the cow (which is x + 112) and the sheep (x)
40/12 = (x + 112)/x
The rest is algebra. if you have any problems just ask.
5. For the first one:
a/b * c/d = ac/bd
and
Anything over itself is one so b/b = 1
e.g. a3/a = a/a * 3 = 1 * 3 = 3
6. Originally Posted by Rocher
Should be easy for you guys anyways... Okay...
1)
Simplify
p3r/pq X q2/pr2
2)
Twenty gallons of crude oil were poured into two containers of different size. Express the amount of crude oil poured into the smaller container in terms of the amount G poured into the larger container. (Is this possible? Don't give you any real information)
3) 30% of the people in a restaurant are adults while the rest are children. Three-fifths of the adults are men and 3/7 of the children are boys. There are 112 more girls than women. How many people are there in the restaurant?
4) The ratio of the weight of a cow to that of a goat is 8:3. The ratio of the weight of the goat to that of a sheep is 5:4. The cow is heavier than the sheep by 112 kg. Find the total weight of the three animals.
Hey, I love easy Math!
And I hate more than one problem in one posting.
But I have lots of sparetime today.
So, what the heck....
1) p3r/pq X q2/pr2
If that is [(p^3)r]/(pq) *[(q^2)/(p*r^2)]
Then,
= [(numerator 1)(numerator 2)] / [(denominator 1)(denominator 2)]
= [(p^3)(q^2)(r)] / [(p^2)(q)(r^2)]
= p^(3-2) * q^(2-1) * r^(1-2)
= p^(1) *q^(1) * r^(-1)
.........................................
2) If G gals of the 20 gals is poured into the large container, then (20 -G) gals is poured into the small container.
.................................................. .....................................
3) Let P = number of people; A = number of adults; C = number of chlidren; M = number of men; W = number of women; B = number of boys; G = number of girls
P = A +C ------(i)
A = M +W ----(ii)
C = B +G -----(iii)
"30% of the people in a restaurant are adults while the rest are children..."
---0.3P = A
---0.7P = C
"...Three-fifths of the adults are men..."
----(3/5)A = M
so, (3/5)*(0.3P) = M
hence, M = 0.18P
And, W = 0.3P -0.18P = 0.12P ----(1)
"...and 3/7 of the children are boys."
----(3/7)C = B
so, (3/7)*(0.7P) = B
hence, B = 0.3P
And, G = 0.7P -0.3P = 0.4P ----(2)
"There are 112 more girls than women..."
----G = W +112
Plugging in those from (1) and (2),
0.4P = 0.12P +112
0.4P -0.12P = 112
0.28P = 112
P = 112/(0.28) = 400 people in the restaurant. --------answer.
.................................................. ...............
4) Let C = weight of cow; G = weight of goat; S = weight of sheep.
"The ratio of the weight of a cow to that of a goat is 8:3."
---C/G = 8/3
Cross multiply,
3C = 8G ----(1)
"The ratio of the weight of the goat to that of a sheep is 5:4."
---G/S = 5/4
Cross multiply,
4G = 5S ----(2)
"The cow is heavier than the sheep by 112 kg."
C = S +112 ----(3)
[I wonder why the number 112 is popular in your set of questions?]
"Find the total weight of the three animals."
Okay, there are many ways to continue this. One of them is to express the C and S in terms of G so that in Eq.(3) there will be only one variable, the G.
From (1), C = (8G)/3
From (2), S = (4G)/5
Substitute those into (3),
(8G)/3 = (4G)/5 +112
Clear the fractions, multiply both sides by 3*5,
5(8G) = 3(4G) +(3*5)(112)
40G = 12G + 1680
40G -12G = 1680
G = 1680/28 = 60 kg
And so,
C = (8G)/3 = (8*60)/3 = 160 kg
S = (4G)/5 = (4*60)/5 = 48 kg
Therefore, C+G+S = 160+60+48 = 268 kg total weight. ------answer. |
# Probability Basic Concepts
Here, you will learn probability basic concepts and definitions of probability.
Let’s begin –
Probability gives us a measure of likelihood that something will happen. However probability can never predict the number of times that an occurrence actually happens.
## Probability Basic Concepts and Definitions
#### (a) Experiment :
An action or operation resulting in two or more well defined outcomes.
Example : tossing a coin, throwing a die, drawing a card from a pack of well shuffled playing cards.
#### (b) Sample Space :
A set S consist of all possible outcomes of an random experiment is called a sample space.
Example : in an experiment of “throwing a die” , following sample spaces are possible:
(i) {even number, odd number}
(ii) {1,2,3,4,5,6}
#### (c) Event:
An event is defined as an occurrence or situation.
Example : in a toss of a coin, it shows a head.
#### (d) Compound Event :
If an event has more than one sample points it is called Compound Event.
#### (e) Complement of an Event :
The set of all outcomes which are in S but not in A is called complement of the event A. It is denoted by $$A^c$$, $$A^`$$.
#### (f) Mutually Exclusive Event :
Two events A and B are said to be Mutually Exclusive if probability of A intersection B is zero i.e $$P(A \cap B)$$=0.
Example : choosing numbers at random from the set {3,4,5,6,7,8,9,10,11,12}. If,
Event A is the selection of a prime number.
and Event B is the selection of an odd number.
Event C is the selection of an even number.
Then A and C are mutually exclusive as none of the numbers in this set is both prime and even and B and C are also mutually exclusive.
#### (g) Equally Likely Events :
Events are said to be Equally Likely when each event is as likely occur as any other event. Note that the term ‘at random’ or ‘randomly’ means that all possibilities are equally likely.
#### (h) Exhaustive Events :
Events A,B,C…..N are said to be Exhaustive Events if no events outside this set can result as an outcome of an experiment.
Note :
(i) 0<=P(A)<=1
(ii) P(A)+P($$A^c$$)=1
(iii) If x cases are favourable to A & y cases are favourable to $$A^c$$ then P(A)=$$x\over(x+y)$$ and $$P(A \cap B)$$=$$y\over(x+y)$$. We say that Odds in Favour of A are x:y & Odds Against A are y:x.
Example : If the letters of INTERMEDIATE are arranged, then odds in favour of the event that no two ‘E’s occur together, are-
Solution : Let 3’E’s, Rest 9 letters,
First arrange rest of the letters = $$9!\over {2! 2!}$$
Now 3’E’s can be placed by $$^{10}C_3$$ ways, so favourable cases = $${9!\over {2! 2!}}\times$$$$^{10}C_3$$ = 3 $$\times$$ 10!
Total cases = $$12!\over {2! 2! 3!}$$ = $${11\over 2} \times$$10!
Non-favourable cases = ($$11\over 2$$ – 3)$$\times$$10! = $${5\over 2}\times$$10!
Odds in favour of the event = $$3\over {5/2}$$ = $$6\over 5$$ Ans. |
# Math Magic Tricks
Math magic tricks are a fun way for kids to amaze their friends and also help them build some math skills along the way!
## Magic Square Math Trick
In this magic math trick, the math magician performs an instant calculation by quickly giving the sum of any four numbers a volunteer covers up!
Are you ready? Great, Let's go.........
PreparationPrint the magic square. Ok, you printed the magic square, right. The first thing you need to know before I show you the trick is the square that you see on this page is special. It’s actually called a magic square. No kidding......Do you know why it’s called a magic square? Notice what happens if you add up all the numbers in the top row.
24 + 11 + 3 + 20 + 7 = 65
Now take a look at the second row.
5 + 17 + 9 + 21 + 13 = 65
Can you guess what the numbers in the third row add up to? If you guessed 65, you’re right. But there’s more……. The numbers in the first column add up to 65 also.
24 + 5 + 6 + 12 + 18 = 65
And the other two columns also add up to 65. And the two diagonals also add up to 65. So that’s why it’s called a magic square65 is the magic number. Now back to our math magic trick.
Performing The Trick:
• First you need to know what the magic number is for the square you are using. For the square shown on this page, I already explained to you that the magic number is 65.
• Next, you figure out what number to subtract from 65 after the volunteer covers up any four numbers.
I'll show you how to find the number to subtract from 65 with a few examples. Look at the first example below. Here you can see someone covered up four numbers at random with the green square.
As soon as they cover up four numbers with a square, in your mind you will extend a diagonal line, two numbers past the outside corner of the covered square. In order to go over two numbers past the outside corner of the covered square, you can only go in one direction.
If you try to go diagonally in any other direction, you will not be able to go two numbers over. Take a look at the picture below. Notice that you can only extend the line two numbers in one direction. The number you arrive at is 8.
This is the number that you subtract from 65.
65 - 8 = 57
So the numbers that are covered add up to 57.
Look back at the top picture and check for yourself to see that the numbers that are covered up, when added together are:
24 +11 + 5 + 17 = 57!
And there you have it.....
Have Fun With It!
By the way, I'll be adding some other magic squares for this magic trick so you can mix it up a bit.
Here's another cool math trick that uses a square with 16 numbers. You can also check out lots of other magic tricks on the main Math Tricks page.
In case you want your students to learn more about magic square puzzles, check out magic square puzzles here.
Go to main Math Tricks page
Return from Math Magic Tricks to Learn With Math Games Home
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# Solving Two-Step Linear Equations
More commonly, we need two operations to solve a linear equation .
In the equation $3x+5=11$ , $x$ is multiplied by $3$ and then $5$ is added. To solve two-step equations, use inverse operations to undo each operation in reverse order.
$3x+5=11$ . . . . . . . our given equation
$-5$ . . . . . . . . . . . . subtract $5$ from each side to get constants on the right
$3x=6$ . . . . . . . . . . . the result
$\frac{3x}{3}=\frac{6}{3}$ . . . . . . . .divide both sides by $3$ to isolate the $x$
$x=2$ . . . . . . . . . . . . the solution (same as before!)
. . . . . . . . . . . . . . . . .We've solved the equation .
The thing that makes these equations linear is that the highest power of $x$ is ${x}^{1}$ (no ${x}^{2}$ or other powers; for those, see quadratic equations and polynomials .
Other linear equations have more than one variable: for example, $y=3x+2$ . This equation has not just one but infinitely many solutions; the solutions can be graphed as a line in the plane. |
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# Frequencies of Notes on a Piano: Learning object
## Applet Description
On the previous page, What are the frequencies of music notes?, we learned how to find the frequencies for a piano tuned by "equal-temperament". It turns out only 7 notes are actually "in tune" on a piano, and all the others are slightly out.
Let's explore that concept on this page. This is a practical example of the graphs we learned about in Graphs of y = a sin bx and y = a cos bx.
In the learning object below, there is a piano which you can play (one note at a time). As you play each new note, you'll see a graph demonstrating the frequency of that note. As the frequency goes up, the wavelength goes down.
Two important notes are indicated on the piano:
Middle C is marked in red (this is the note usually in the middle of an 88-key piano); and
A-440 (the A with frequency 440 cycles/second, or 440 Hz) is marked in light blue.
Each 1 unit on the t-axis represents one wavelength of the "fundamental" note for our applet, A below middle C. The frequency id 220 Hz, so the graph is given by (where f is the frequency):
y = sin(2πft) = sin(2π(220)t) = sin(1382.3t)
The time taken for each wave to pass our ear for A-220 is 1/220 second = 0.004545\ s. This is what 1 unit represents on the t-axis below.
## Things to do
1. Play the A on the far left. It has frequency 220 Hz.
2. Next, play some of the notes near to that A and notice how the graph changes as the frequency changes.
3. Now, play the higher A. It has a frequency of 440 Hz, and is one octave above the first A. Not how there are 2 wavelengths for A-440 in the space of one length of A-220.
4. Next read the descriptions (below the graph) for the notes that sound "nice" with A (that is C#, and E). These are simple multiples of the fundamental frequency. (Such notes are called "concordant" in music.) Where I say it "should be" a certain frequeny, I'm referring to the case where it was tuned so that A major sounded the best.
5. See what it says in the description for the "not-so-nice" notes. (Such notes are called "discordant" in music.)
6. When you choose "Combined signal", you'll see the graph of the addition of the A-220 signal and whatever note was just played (the component signals are shown in light grey), and will hear the 2 tones.
NOTE: The lower sounds may not play so well on a mobile phone speaker (if so, use earphones).
Combined signal: no yes
Note: A Frequency = 220 Hz
The function: y = sin(2π(220)t) = sin(1382.3t)
Credits: Loosely based on CSS Tricks and Chris Lowis' Playing multiple notes on Web Audio.
### Don't miss...
See the background to the above learning object in:
What are the frequencies of music notes?
Frequency of notes on a piano - interactive learning object (some basic analysis of what you are seeing above)
Graphs of y = a sin bx and y = a cos bx |
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### Algebra 1➕
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## Factoring Who Now? 🤔
Have you ever tried factoring something and found it impossible? As in, you did all the procedures and found it impossible to factor. Don’t worry; a formula exists for you: the quadratic formula! The quadratic formula looks like this: 👀
## How to Use the Quadratic Formula ❓
I know the formula seems like a lot to memorize in the first shot, but you'll get used to it over time with practice. An easy way to remember the formula is to sing it to the tune of “Pop Goes The Weasel.” 🦦
X is equal to negative b / plus or minus the square root / of b squared minus 4 a c / all over 2 a!
Now, let’s look at its various components. The quadratic formula is used to solve trinomial equations. A trinomial equation, a polynomial with three terms, looks like this:
The letters “a,” “b,” and “c” represent numbers that you plug into the formula above!
You may have noticed the plus or minus symbol, +. This means that you'll add the equation and subtract it.
• With the quadratic formula, you'll do the formula twice: 1) adding and 2) subtracting sqrt(b2 - 4ac) to -b.
• This also means that you'll usually get two answers… which is perfectly fine! 😌
### Practice ✏️
Now that you know all the components of the quadratic formula, let’s practice!
Example 1:
Solve the following equation using the quadratic formula: d^2 + 4d - 12 = 0
First, plug everything into the formula… 🔌
• a = first term in the trinomial, aka x^2 = 1
• b = second term, aka x = 4
• c = third term, aka the number = -12
Simplify as many terms as you can! For example, we know that 4 squared is 16 and (-4) x (1) x (-12) = +48 🧩
Stuck? Remember that + means that we'll do addition and subtraction. 🤓
Getting the hang of it? Let’s look at another example! 🧐
Example 2:
Solve the following equation using the quadratic formula: u^2 - 4u + 8 = 0
As usual, plug in the values for a, b, and c into the formula.
• a = 1
• b = -4
• c = 8
Simplify as much as possible! 🧱
Pause! What do we do if the inside of the square root has a negative number? 😕
Depending on where you are mathematically, you will either stop here or use something called an imaginary number. Since the square root of a negative number does not exist in the set of real numbers, you can square it with an imaginary number. In this case, sqrt(-16) = 4 x sqrt(-1) = 4i. ☁️
From here, continue simplifying the equation as usual 🧱
Once you can't physically do anything to the answers, you're more than okay to stop at 2 + 2i! 🔨
Example 3:
Solve the following equation: j^2 - 4j + 4 = 0
You know the drill: plug, plug, plug! ⚡
• a = 1
• b = -4
• c = 4
Simplify as much as possible! 🧱
Since adding or subtracting 0 to anything doesn’t change anything, you don’t have to have to do the plus or minus step! 😈
You will rarely see factors that have one answer, but they DO exist, so don’t be surprised if you see an answer with only one number! Makes life easier for everyone, don't you agree? ☝
Let’s spice things up a bit. 🌶️
Example 4:
Solve the following: -16t^2 - 6t + 11 = 0
The first thing you can do to make things easier is multiply both sides by -1. This minimizes the confusion negative numbers can bring to you, which I’m sure the world can deal less with. 🌎
(-16t^2 - 6t + 11) (-1) = 0 (-1)
16t^2 + 6t - 11 = 0
Now, let's proceed to business as usual! 💼
• a = 16
• b = 6
• c = -11
Simplify as much as possible! 🧱
Simplify the square root. (If you need a refresher on this, check out Fiveable’s math page). ⬛
That 2 in front of the √185 is quite annoying, and it needs to go. In order to get rid of it, divide everything by two! ➗
Now, you can separate the two formulas! 📖
If you, without a calculator, can simplify this, you’re one heck of a genius! Otherwise, this is simplified enough. The way the equations look now is acceptable for an answer 🧠
## Resources:
Example 5:
Solve: 8w^2 = 128
First, divide both sides by 8. 👍
(8w2/8) = (128/8
w^2 = 16
Then, move 16 to the other side by subtracting 16 from both sides
w^2 - 16 = 16 - 16
w^2 - 16 = 0
Now that the equation is fixed, solve it like how we do it normally! 💙
w^2 - 16 = 0
• a = 1
• b = ???
• c = -16
Don’t get confused: b doesn’t exist in this trinomial. In the quadratic formula, you’ll just put 0 since b doesn’t exist ✔️
Simplify as much as possible! 🧱
Now, you can separate the two formulas! 📖
Feeling much more confident? We're at the home stretch! 🏇
Example 6:
Solve: p^2 - 31 - 2p = -6 - 3p - 3p^2
First, move everything to the left. 👈
p^2 - 31 - 2p + 6 + 3p + 3p^2 = (-6 + 6) (-3p + 3p) (-3p^2 + 3p^2)
p^2 - 31 - 2p + 6 + 3p + 3p^2 = 0
Now, it's time to combine common terms! 🙏
4p^2 + p - 25 = 0
Did you notice that although the terms weren’t in order, they were put in order when simplified? It's important that your expression matches the order of ax^2 + bx + c. Now, we can plug the numbers into the quadratic formula as usual 😁
Simplify as much as possible! 🧱
Now, you can separate the two formulas! 📖
Phew! That was a lot, but if you're following along throughout this post, you've essentially mastered the quadratic formula. Give yourself a huge pat on the back! Keep practicing and you'll find quicker and easier ways to plow through these problems 🙌
### More Practice 🖊️
1. 4h^2 + 8h - 16 = 0
2. d^2 + 17d - 19 = 0
3. x^2 - 81x - 9 = 14x + 23x^2 + 5
4. 125y^2 - 25 = 0
5. 6r^2 + 3r - 36 = 0
Connect with other students studying the quadratic formula with Hours. 🤝
Fiveable study rooms = the ultimate focus mode ⏰
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Stat Trek
Teach yourself statistics
Teach yourself statistics
What is a Matrix?
This lesson introduces the matrix - the rectangular array at the heart of matrix algebra. Matrix algebra is used quite a bit in advanced statistics, largely because it provides two benefits.
• Compact notation for describing sets of data and sets of equations.
• Efficient methods for manipulating sets of data and solving sets of equations.
Matrix Definition
A matrix is a rectangular array of numbers arranged in rows and columns. The array of numbers below is an example of a matrix.
21 62 33 93 44 95 66 13 77 38 79 33
The number of rows and columns that a matrix has is called its dimension or its order. By convention, rows are listed first; and columns, second. Thus, we would say that the dimension (or order) of the above matrix is 3 x 4, meaning that it has 3 rows and 4 columns.
Numbers that appear in the rows and columns of a matrix are called elements of the matrix. In the above matrix, the element in the first column of the first row is 21; the element in the second column of the first row is 62; and so on.
Matrix Notation
Statisticians use symbols to identify matrix elements and matrices.
• Matrix elements. Consider the matrix below, in which matrix elements are represented entirely by symbols.
A11 A12 A13 A14 A21 A22 A23 A24
By convention, first subscript refers to the row number; and the second subscript, to the column number. Thus, the first element in the first row is represented by A11. The second element in the first row is represented by A12. And so on, until we reach the fourth element in the second row, which is represented by A24.
• Matrices. There are several ways to represent a matrix symbolically. The simplest is to use a boldface letter, such as A, B, or C. Thus, A might represent a 2 x 4 matrix, as illustrated below.
A =
11 62 33 93 44 95 66 13
Another approach for representing matrix A is:
A = [ Aij ] where i = 1, 2 and j = 1, 2, 3, 4
This notation indicates that A is a matrix with 2 rows and 4 columns. The actual elements of the array are not displayed; they are represented by the symbol Aij.
Other matrix notation will be introduced as needed. For a description of all the matrix notation used in this tutorial, see the Matrix Notation Appendix.
Matrix Equality
To understand matrix algebra, we need to understand matrix equality. Two matrices are equal if all three of the following conditions are met:
• Each matrix has the same number of rows.
• Each matrix has the same number of columns.
• Corresponding elements within each matrix are equal.
Consider the three matrices shown below.
A =
111 x y 444
B =
111 222 333 444
C =
l m n o p q
If A = B, we know that x = 222 and y = 333; since corresponding elements of equal matrices are also equal. And we know that matrix C is not equal to A or B, because C has more columns than A or B.
Problem 1
The notation below describes two matrices - matrix A and matrix B.
A = [ Aij ]
where i = 1, 2, 3 and j = 1, 2
B =
111 222 333 444 555 666 777 888
Which of the following statements about A and B are true?
I. Matrix A has 5 elements.
II. The dimension of matrix B is 4 x 2.
III. In matrix B, element B21 is equal to 222.
(A) I only
(B) II only
(C) III only
(D) All of the above
(E) None of the above
Solution |
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# GRE Algebra | Functions
• Last Updated : 25 Apr, 2019
An algebraic expression consist one variable is used to define a function of that variable. For example, the expression 2y – 7 can be used to define a function f by,
`f(y) = 2y - 7 `
where f(y) is called the value of f at y and function value can be obtained by putting the value of y in above expression.If y=1, then f(1) is
```f(1) = 2(1) - 7
f(1) = -5 ```
Hence, the value of f(1) is -5.
For each input value of y, the function f(y) gives exactly one output.So, it can be say that function f act as a machine which takes an input of the variable y and produces the corresponding output.
It might be possible sometimes when more than one value of y can give the same output of f(y). For example:
```f(x) = 2x2 - 4x + 7
Put (x = 0)
f(0) = 2(0) - 4(0) + 7
= 7
Put (x = 2)
f(2) = 2(2)2 - 4(2) + 7
= 8 - 8 + 7
= 7 ```
Hence, for both values of x = 0, 2 f(x) gives the same output.
Domain of a function is the set of all the permissible inputs. For example:
```g(y) = √(x-5) is function.
√(x-5) ≥ 0
x ≥ 5 ```
Here, domain of the function is [5, ∞)
• Example-1: Let f be the function defined by f(x)= 3x/(x2 – 4).Find the domain.
Solution:
```Let f(x) = 0 then
3x/x2 - 4 = 0
Put (x=2)
3(2)/22 - 4 = 0
6/(4-4) = 0
Put (x= -2)
3(-2)/22 - 4 = 0
6/(4-4) = 0 ```
Here, f is not defined at (x = 2 and x = -2) because 6/0 is not defined.
Hence, the domain of f consists of all real numbers except 2 and -2.
• Example-2: Let f be the function defined by g(x) = √(x – 3) + 11. Find the domain.
Solution:
`√(x - 3) + 11 = 0 `
In this case if x < 3, then g(x) is not a real number.
Hence, domain of g consists of all the real numbers x such that x ≥ 3.
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# Sports and Angular Momentum Dennis Silverman Bill Heidbrink U. C. Irvine.
## Presentation on theme: "Sports and Angular Momentum Dennis Silverman Bill Heidbrink U. C. Irvine."— Presentation transcript:
Sports and Angular Momentum Dennis Silverman Bill Heidbrink U. C. Irvine
Overview Angular Motion Angular Momentum Moment of Inertia Conservation of Angular Momentum Sports body mechanics and angular momentum Angular Momentum and Stability How a baseball curves
Angular Momentum Linear momentum or quantity of motion is P = mv, and inertia given by mass m. m v Rotation of a mass m about an axis, zero when on axis, so should involve distance from axis r Angular momentum L = r mv m r L
Circular Motion The angle θ subtended by a distance s on the circumference of a circle of radius r r θ s
Radians Instead of measuring the angle θ in degrees (360 to a circle), we can measure in pizza pi slices such that there are 2π = 6.28 to a full circle So each radian slice is about a sixth of a circle or 57.3 degrees. Then we can write directly: s = θ r with θ in radians. When a complete circle is traversed, θ = 2π, and s = 2π r, the circumference.
Angular Velocity When a wheel is rotating uniformly about its axis, the angle θ changes at a rate called ω, while the distance s changes at a rate called its velocity v. Then s = r θ gives v = r ω.
Angular Momentum and Moment of Inertia Let’s recall the angular momentum L = r m v = r m (ω r) L = m r² ω In a “rigid body”, all parts rotate at the same angular velocity ω, so we can sum mr² over all parts of the body, to give I = Σ mr², the moment of inertia of the body. The total angular momentum is then L = I ω.
Conservation of Angular Momentum If there are no outside forces acting on a symmetrical rotating body, angular momentum is conserved, essentially by symmetry. The effect of a uniform gravitational field cancels out over the whole body, and angular momentum is still conserved. L also involves a direction, where the axis is the thumb if the motion is followed by the fingers of the right hand.
Examples of Moment of Inertia Hammer thrower Stick about different rotation axes Diver Baseball bat Pop quiz
Applications of Conservation of Angular Momentum If the moment of inertial I 1 changes to I 2, say by shortening r, then the angular velocity must also change to conserve angular momentum. L = I 1 ω 1 = I 2 ω 2 Example: Rotating with weights out, pulling weights in shortens r, decreasing I and increasing ω.
Examples of Changes in Moment of Inertia Pulling arms in to do spins in ice skating Tucking while diving to do rolls Bicycle wheel flip demo Space station video
Rotating different parts of body Ballet pirouette Balancing beam Ice skater balancing Falling cat or rabbit landing upright Rodeo bull rider Ski turns Ski jumping video
Angular Momentum for Stability Bicycle or motorcycle riding Football pass or lateral spinning Spinning top Frisbee Spinning gyroscopes for orbital orientation Helicopter Rifling of rifle barrel Earth rotation for daily constancy and seasons
Curving of spinning balls Bernoulli’s Equation (1738) Magnus Force (1852) Rayleigh Calculation (1877)
Bernoulli’s Principle Follow the flow of a certain constant volume of fluid ΔV =A*Δx, even though A and Δx change Pressure is P=F/A Energy input is Force*distance E = F*Δx=(PA)*Δx=P*ΔV kinetic energy is E=½ρv²ΔV So by energy conservation, P+½ρv² is a constant When v increases, P decreases, and vice-versa Δ
Bernoulli’s Principal and Flight Lift on an airplane wing v higher above wing, so pressure lower P lower P normal V higher
Air around a rotating baseball, from ball’s top point of view Higher v, lower P on right Lower v, higher P on left So ball curves to right P left P right Boundary layer
Examples of curving balls Baseball curve pitch Baseball outfield throw with backspin for longer distance Tennis topspin to keep ball down Soccer (Beckham) curve around to goal Golf ball dimpling and backspin for range Deflection d = ½ a t² most at end of range
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Q 10.
Expert-verified
Found in: Page 640
### The Practice of Statistics for AP Examination
Book edition 6th
Author(s) Daren Starnes, Josh Tabor
Pages 837 pages
ISBN 9781319113339
# Where’s Egypt? In a Pew Research poll, $287$ out of $522$randomly selected U.S. men were able to identify Egypt when it was highlighted on a map of the Middle East. When $520$ randomly selected U.S. women were asked, $233$ were able to do so.a. Construct and interpret a $95%$ confidence interval for the difference in the trueproportion of U.S. men and U.S. women who can identify Egypt on a map.b. Based on your interval, is there convincing evidence of a difference in the trueproportions of U.S. men and women who can identify Egypt on a map? Justify youranswer.
a. Confidence Interval is $\left(-0.0191,0.1017\right)$.
b. Yes, there is evidence of difference in true proportion of US men and women who can identify Egypt on map or not.
See the step by step solution
## Step 1: Given Information
It is given that ${x}_{1}=287$
${x}_{2}=233$
${n}_{1}=522$
${n}_{2}=520$
$c=95%=0.95$
## Step 2: Calculating Confidence Interval
The three conditions are:
Random: Samples are independent random samples.
Independent: $522$ US men are less than $10%$ of population. Same is true for women.
Normal: Success in two samples are $287,233$ and failures are $235,287$. All are greater than ten.
All conditions are satisfied.
Sample Proportion: ${\stackrel{^}{p}}_{1}=\frac{{x}_{1}}{{n}_{1}}=\frac{287}{522}=0.5498$
${\stackrel{^}{p}}_{2}=\frac{{x}_{2}}{{n}_{2}}=\frac{233}{520}=0.4481$
Confidence Interval:
$\left({\stackrel{^}{p}}_{1}-{\stackrel{^}{p}}_{2}\right)-{z}_{\alpha /2}×\sqrt{\frac{{\stackrel{^}{p}}_{1}\left(1-{\stackrel{^}{p}}_{1}\right)}{{n}_{1}}+\frac{{\stackrel{^}{p}}_{2}\left(1-{\stackrel{^}{p}}_{2}\right)}{{n}_{2}}}$
$=\left(0.5498-0.4481\right)-1.96×\sqrt{\frac{0.5498\left(1-0.5498\right)}{522}+\frac{0.4481\left(1-0.4481\right)}{520}}$
$\approx -0.0191$
and $\left({\stackrel{^}{p}}_{1}-{\stackrel{^}{p}}_{2}\right)+{z}_{\alpha /2}×\sqrt{\frac{{\stackrel{^}{p}}_{1}\left(1-{\stackrel{^}{p}}_{1}\right)}{{n}_{1}}+\frac{{\stackrel{^}{p}}_{2}\left(1-{\stackrel{^}{p}}_{2}\right)}{{n}_{2}}}$
$=\left(0.5498-0.4481\right)+1.96×\sqrt{\frac{0.5498\left(1-0.5498\right)}{522}+\frac{0.4481\left(1-0.4481\right)}{520}}$
$\approx 0.1017$
The confidence interval is $\left(-0.0191,0.1017\right)$
## Step 3: To check if there is evidence of difference in true proportion of US men and women who can identify Egypt on map or not.
As the confidence interval $\left(-0.0191,0.1017\right)$ does not contain zero. It is unlikely that population proportions are equal.
So, there is evidence of difference in true proportion of US men and women who can identify Egypt on map or not. |
# Tom wrote 3 consecutive natural numbers. From these numbers' cube sum he took away the triple product of those numbers and divided by the arithmetic average of those numbers. What number did Tom write?
Oct 17, 2017
Final number that Tom wrote was $\textcolor{red}{9}$
#### Explanation:
Note: much of this depend upon my correctly understanding the meaning of various parts of the question.
3 consecutive natural numbers
I assume this could be represented by the set $\left\{\left(a - 1\right) , a , \left(a + 1\right)\right\}$ for some $a \in \mathbb{N}$
these numbers' cube sum
I assume this could be represented as
$\textcolor{w h i t e}{\text{XXX}} {\left(a - 1\right)}^{3} + {a}^{3} + {\left(a + 1\right)}^{3}$
$\textcolor{w h i t e}{\text{XXXXX}} = {a}^{3} - 3 {a}^{2} + 3 a - 1$
$\textcolor{w h i t e}{\text{XXXXXx}} + {a}^{3}$
$\textcolor{w h i t e}{\text{XXXXXx}} \underline{+ {a}^{3} + 3 {a}^{2} + 3 a + 1}$
$\textcolor{w h i t e}{\text{XXXXX}} = 3 {a}^{3} \textcolor{w h i t e}{+ 3 {a}^{2}} + 6 a$
the triple product of these numbers
I assume this means triple the product of these numbers
$\textcolor{w h i t e}{\text{XXX}} 3 \left(a - 1\right) a \left(a + 1\right)$
$\textcolor{w h i t e}{\text{XXXXX}} = 3 {a}^{3} - 3 a$
So these numbers' cube sum minus the triple product of these numbers would be
$\textcolor{w h i t e}{\text{XXXXX}} 3 {a}^{3} + 6 a$
$\textcolor{w h i t e}{\text{XXX}} \underline{- \left(3 {a}^{3} - 3 a\right)}$
$\textcolor{w h i t e}{\text{XXX")=color(white)("XXxX}} 9 a$
the arithmetic average of these three numbers
$\textcolor{w h i t e}{\text{XXX")((a-1)+a+(a+1))/3color(white)("XXX}} = a$
$\textcolor{w h i t e}{\text{XXX")(9a)/acolor(white)("XXX}} = 9$ |
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# Math Is for Everyone
Math Is for Everyone
Math Is for Everyone
MATH is not just for scientists. It is for all of us. When you shop, decorate your home, or listen to the daily weather report, you are using or benefiting from mathematical principles.
Many people seem to feel that math (or, maths) is boring and unrelated to their everyday life. Do you feel that way? Let us examine how useful, accessible, and fascinating math can be.
A Shopping Trip
Imagine you are out shopping and you come across a big sale. An item with an original price of \$35 has been marked down, or reduced, by 25 percent. That sounds like a bargain. But what is the new price? Arithmetic comes to your rescue. *
First, subtract the markdown percentage from 100 percent, and you get 75 percent (100 percent − 25 percent = 75 percent). Then multiply the original price by the result, in this case 75 percent (0.75). The new price would be \$26.25 (35 × 0.75 = 26.25). Now that you know the final price, you can decide just how good the sale really is.
What if you did not bring a calculator with you? Perhaps you can work out the answer in your head. For example, say that an item originally priced at \$45 has been marked down by 15 percent. Here is a tip for figuring percentages in your head. Use 10 percent as a base. To figure 10 percent of a number, you divide the number by 10. That is relatively easy to do in your head. Then, since you know that 15 is equal to 10 plus 5 and that 5 is exactly half of 10, you can quickly calculate the final sale price by addition and subtraction. Let us try that.
Since 10 percent of 45 is 4.50, 5 percent of 45 will be half that amount, or 2.25, and 15 percent will be the sum of those two figures, or 6.75 (4.50 + 2.25 = 6.75). Finally, we subtract 6.75 from 45 to arrive at the discounted price of 38.25 (45 − 6.75 = 38.25). Incidentally, you can use a similar approach to figure out the amount of sales tax on an item or the amount of tip to add to your bill at a restaurant. Of course, in these cases, instead of subtracting, you would add the result to the original price.
Be careful, though, not to jump to wrong conclusions when figuring in your head. A dress or a pair of slacks whose price has been discounted by 40 percent and then slashed again by another 40 percent has actually been reduced in price by only 64 percent, not 80 percent. The second discount is taken on the reduced price, not the original price. It might still be a bargain, but it is good to know the facts.
There are problems, though, that arithmetic alone cannot solve. Fortunately, many other math tools are available.
Decorating at Home
Let us say that you need to replace the flooring in your apartment and you are working within a tight budget. Before you go to the store, you first sit down to figure out what you need. The biggest question is, How much flooring should you buy? Understanding some basic geometry can help.
Flooring is often sold based on how many square units it will cover. A square foot, for example, is one foot long and one foot wide. Before you can determine how much flooring you will need, you first have to figure out how much floor area there is in each room and hallway in your apartment. The floor plans of most buildings are made up of a number of squares and rectangles. So the following formula would help you to accomplish that: a = l × w (area equals length times width). This is the geometric formula for determining the area of a rectangle or a square.
To illustrate how this formula is used, let us say that you are putting new flooring in every room of the apartment except the kitchen and the bathroom. You measure each room and come up with a floor plan like the one shown on page 23. The squares and rectangles in the plan show the size and location of the rooms. Using the above formula, see if you can calculate how many square units of flooring you will need. Here are some hints: You could calculate the area of each room by itself and then add the results together. Or you could save some time by calculating the total area of the floor plan and then deducting the area of the kitchen and the bathroom. *
The word “geometry” also comes from Greek, and it literally means “measurement of land.” It involves studying the area, distance, volume, and other properties of shapes and lines. Practical formulas exist for every shape imaginable in two and three dimensions. Each day, scientists, engineers, and home decorators alike use these formulas to figure out exactly what they need. But there is more to math than arithmetic and geometry.
Use Math Every Day
Other branches of mathematics include algebra and calculus. Over the centuries math has become a truly universal language shared by everyone regardless of culture, religion, or gender. In science, industry, business, and everyday life, math has the power to solve some of the toughest riddles we face. Whether you are trying to unravel the mysteries of the universe or balance the family budget, being able to use the language of numbers is a key to success.
So even if you hated math in school, why not take a fresh look at it now? Like any language, math is learned best through use. Try using some math every day. Try your hand at math puzzles and games. One positive experience might change the way you feel. It will certainly enhance your appreciation for the wisdom of the Great Mathematician who designed these intriguing concepts in the beginning, our Creator, Jehovah God.
[Footnotes]
^ par. 5 Arithmetic (a term derived from a Greek word meaning “number”) is said to be the oldest branch of mathematics. It goes back thousands of years and was used by the ancient Babylonians, Chinese, and Egyptians. Arithmetic gives us basic tools that we can use each day to count and measure the physical world around us.
^ par. 14 Answer = 600 square feet [54 m2] of flooring.
[Diagram on page 23]
(For fully formatted text, see publication)
10 ft.
10 ft.
Kitchen
Dining room
Hallway
Living room
Bedroom
Bathroom
10 ft.
5 ft.
10 ft.
15 ft.
5 ft.
10 ft.
[Pictures on page 23] |
# How do you solve 2x - y = 3 and 3x - y = 4 using substitution?
May 1, 2016
An alternative method of solving them.
$x = 1 \mathmr{and} y = - 1$
#### Explanation:
Both of these equations contain the single variable $y$.
Transpose them to make $y$ the subject.
$y = 3 x - 4$ ....... and ....... $y = 2 x - 3$
Now, because $y = y$, it follows that
.................$3 x - 4 = 2 x - 3$ (giving an equation with one variable)
Solving it gives: $x = 1$
There are now two ways to calculate the value of $y$
$y = 3 \left(1\right) - 4$ = -1 and $y = 2 \left(1\right) - 3$ = -1 |
# Critical Number Problem
• Oct 17th 2013, 12:03 PM
Jason76
Critical Number Problem
Find the critical numbers:
\$\displaystyle h(t) = t^{3/4} - 3t^{1/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$ = 0 http://www.freemathhelp.com/forum/im...s/confused.png Next move. How can I solve for t in this situation?
• Oct 17th 2013, 12:14 PM
SlipEternal
Re: Critical Number Problem
Multiply both sides by \$\displaystyle t^a\$ where \$\displaystyle -a\$ is the smallest exponent of \$\displaystyle t\$. (In other words, multiply both sides by \$\displaystyle t^{3/4}\$).
• Oct 17th 2013, 12:24 PM
Jason76
Re: Critical Number Problem
Post Edited
Find the critical numbers:
\$\displaystyle h(t) = t^{3/4} - 3t^{1/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$ = 0
\$\displaystyle h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}\$
\$\displaystyle h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}\$ http://www.freemathhelp.com/forum/im...n_question.gif
• Oct 17th 2013, 12:44 PM
SlipEternal
Re: Critical Number Problem
Quote:
Originally Posted by Jason76
Post Edited
Find the critical numbers:
\$\displaystyle h(t) = t^{3/4} - 3t^{1/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$ = 0
\$\displaystyle h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}\$
\$\displaystyle h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}\$ http://www.freemathhelp.com/forum/im...n_question.gif
Huh? Why would you think \$\displaystyle t^{-3/4} = t^{-4/4}\$?
• Oct 17th 2013, 06:20 PM
Jason76
Re: Critical Number Problem
Find the critical numbers:
\$\displaystyle h(t) = t^{3/4} - 3t^{1/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$ = 0
\$\displaystyle h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}\$
\$\displaystyle h'(t) = t^{-3/4} = t^{-1/4} \$
\$\displaystyle h'(t) = \dfrac{ t^{-1/4}}{t^{-3/4}} = 0 \$
\$\displaystyle h'(t) = t^{1/2} = 0 \$ ??
• Oct 17th 2013, 06:33 PM
SlipEternal
Re: Critical Number Problem
Why do you keep writing \$\displaystyle h'(t)\$? When you go from the line that reads
\$\displaystyle h'(t) = \dfrac{3}{4}t^{-1/4}-\dfrac{3}{4}t^{-3/4} = 0\$
\$\displaystyle h'(t) = -\dfrac{3}{4}t^{-3/4} = -\dfrac{3}{4}t^{-1/4}\$, you no longer have \$\displaystyle h'(t)\$. You subtracted \$\displaystyle \dfrac{3}{4}t^{-1/4}\$ from every term, so you should have \$\displaystyle h'(t) - \dfrac{3}{4}t^{-1/4} = -\dfrac{3}{4}t^{-3/4} = -\dfrac{3}{4}t^{-1/4}\$
So just drop \$\displaystyle h'(t) = \$ from every line. It is not true after line 5.
Next, why would you think that \$\displaystyle \dfrac{t^{-3/4}}{t^{-3/4}} = 0\$?
• Oct 17th 2013, 06:42 PM
Jason76
Re: Critical Number Problem
Find the critical numbers:
\$\displaystyle h(t) = t^{3/4} - 3t^{1/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$
\$\displaystyle \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$ = 0
\$\displaystyle -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}\$
\$\displaystyle t^{-3/4} = t^{-1/4} \$
\$\displaystyle \dfrac{ t^{-1/4}}{t^{-3/4}} = 0 \$
\$\displaystyle t^{1/2} = 0 \$
• Oct 17th 2013, 06:46 PM
SlipEternal
Re: Critical Number Problem
Also, from now on, it might be easier to help you if you number each line. I am "quoting" you, but I will make changes to your quote to add line numbers and make each line correct.
Quote:
Originally Posted by Jason76
Find the critical numbers:
1. \$\displaystyle h(t) = t^{3/4} - 3t^{1/4}\$
2. \$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}\$
3. \$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}\$
4. \$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$
5. \$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}=0\$
6. \$\displaystyle -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}\$
7. \$\displaystyle t^{-3/4} = t^{-1/4} \$
8. \$\displaystyle \dfrac{ t^{-1/4}}{t^{-3/4}} = 0 \$ <-- This line is wrong
9. \$\displaystyle t^{1/2} = ??? \$
Now, the question is, how did you get from line 7 to line 8?
• Oct 17th 2013, 10:02 PM
votan
Re: Critical Number Problem
Quote:
Originally Posted by Jason76
Post Edited
Find the critical numbers:
\$\displaystyle h(t) = t^{3/4} - 3t^{1/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - (\dfrac{1}{4}) 3 v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} u^{-1/4} - \dfrac{3}{4} v^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$
\$\displaystyle h'(t) = \dfrac{3}{4} t^{-1/4} - \dfrac{3}{4} t^{-3/4}\$ = 0
\$\displaystyle h'(t) = -\dfrac{3}{4} t^{-3/4} = -\dfrac{3}{4} t^{-1/4}\$
\$\displaystyle h'(t) = t^{-3/4} = t^{-1/4} = t^{-4/4} = t^{-1}\$ http://www.freemathhelp.com/forum/im...n_question.gif
You are giving multiple = sign in the same line without showing how you got the right side from the left side. Help us understand what you are doing by writing it. We cannot decipher what you have in mind what you write your expression the way you do. Perhaps you need to write one = sing in a line, and write a short comment next to the line telling what did you do to get from one line to the next. |
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Procedural KnowledgeDecompose numbers using standard form (place value) and non-standard form.
Skip count forward and backward by 2, 5, and 10, starting at multiples of 2, 5, and 10 respectively.
Skip count forward by 20 and 25, starting at 0.
Determine the monetary value of collections of coins or bills (cents or dollars) of the same denomination.
Skip count sets, including those with remainders.
Order numbers using benchmarks on a visual or spatial representation.
Represent quantities with numbers.
Relate a numeral to a specific quantity.
Estimate quantities using referents.
Conceptual KnowledgeThe position of a digit in a number determines its value (place value).
Grouping by 10 creates patterns in place value (unitizing) to make working with numbers efficient.
Skip counting is an efficient way of counting larger quantities and can include quantities left over (remainders).
Numbers, including 0, occupy space in a visual or spatial representation of quantity.
Numbers, including 0, can be associated with a specific point on a linear representation of quantity.
The position of something can be indicated using ordinal numbers.
Quantities can be represented symbolically with numerals, including 0.
Estimation is used when an exact count is not needed.
Essential SkillsNumber counting by 2’s, 5’s, 10’s, 20’s and 25’s forward and backward.
Determining ones, tens and hundreds in terms of place value.
Common MisconceptionsDue to the lack of analysis and investigation of a base ten positional system, students might create their own misunderstandings of what this sort of system is about. Students might not realize that:
☆ In a base ten system, we count in groups of ten.
☆ In a base ten system, we only need digits up to 9.
☆ In a positional system, the position of the digit matters.
It is relevant to listen to students’ explanations to address misconceptions.
Big Ideas#1 – The set of whole numbers is infinite, and each whole number can be associated with a unique point on the number line.
#2 – The base ten numeration system is a scheme for recording numbers using digits 0-9, groups of ten, and place value.
#3 – Any number can be represented in an infinite number of ways that have the same value.
#4 – Numbers can be compared by their relative values.
#8 – Numerical calculations can be approximated by replacing numbers with other numbers that are close and easy to compute with mentally.
Key StrategiesExperience counting in different ways, and in different orders.
Represent numbers (concretely, pictorially and symbolically) based on the base ten positional system, and explain the representations.
Connect a digit in a determined position to the number of ones, tens, or hundreds that the digit represents, and explain the connections.
Introduce other numerical systems and compare them to the base ten positional system. Describe and explain differences and similarities.
Key VocabularyBase ten system.
Positional system.
Place value.
One, tens, hundreds.
Grade 2Notion of Halves and Quarters
Procedural KnowledgeCount by halves and quarters to one whole concretely or pictorially.
Partition objects and sets into halves and quarters.
Describe part-to-whole relationships with halves and quarters.
Conceptual KnowledgeObjects and sets can be partitioned into equal-sized parts in different ways.
The part is related to the whole (part-to-whole relationship).
Essential SkillsFinding halves and quarters of sets and objects.
Counting by halves and quarters.
Relating parts to whole.
Common MisconceptionsHalving does not mean splitting into two equal parts, it is just about splitting into two parts. The same misconception is valid for quarters.
There is no need to know what the whole is when working with parts of that whole. Students need to know that a part is always in reference to a whole.
Big Ideas#3 – Any number can be represented in an infinite number of ways that have the same value.
#4 – Numbers can be compared by their relative values.
#5 – The same number sentence can be associated with different concrete or real-world situations, and different number sentences can be associated with the same concrete or real-world situation.
Key StrategiesUse concrete, pictorial and symbolic examples of halves, quarters, and respective wholes.
Estimate and count by halves and quarters.
Connect concrete, pictorial and symbolic representations of halves, quarters, and respective wholes.
Communicate different strategies for finding halves and quarters of a whole.
Explain strategies for finding halves and quarters of a whole.
Find wholes, given halves or quarters. Explain the process.
Key VocabularuSplit, partition.
Equal parts.
Halves, quarters, whole.
Procedural KnowledgeApply strategies to single-digit addition number facts to a sum of 18 and related subtraction number facts.
Represent addition and subtraction strategies concretely, pictorially, or symbolically.
Add and subtract numbers within 100, including 0, without a calculator.
Recognize patterns in addition and subtraction.
Add and subtract in joining, separating, and comparing situations.
Create and solve problems that involve addition and subtraction.
Conceptual KnowledgeAddition and subtraction are operations used when applying additive thinking strategies.
An addition situation can be represented as a subtraction situation (addition and subtraction are inverse operations).
Addition and subtraction are part-part-whole relationships that can be represented symbolically (+, –, =).
Numbers can be added in any order (commutative and associative properties).
Subtraction strategies.
Addition number facts to 18 and related subtraction number facts.
Common MisconceptionsWhen you add or subtract zero, the result will be zero.
Subtraction is commutative and associative.
Big Ideas#3 – Any number or numerical expression can be represented in an infinite number of ways that have the same value.
#5 – The same number sentence can be associated with different concrete or real-world situations, and different number sentences can be associated with the same concrete or real-world situation.
#6 – For a given set of numbers there are relationships that are always true, and these are the rules that govern arithmetic and algebra.
#7 – Basic facts and algorithms for operations with whole numbers use notions of equivalence to transform calculations into simpler ones.
Key StrategiesUse concrete, pictorial and symbolic strategies to add and subtract.
Connect concrete, pictorial and symbolic representations of addition and subtraction.
Communicate and explain different strategies for adding and subtracting.
Role play and explain what the commutative and the associative properties mean in addition.
Create and solve problems involving addition and subtraction.
Find, explain and correct errors in problems involving addition and subtraction.
Subtraction, difference.
Commutative property.
Associative property.
Identity element.
Grade 2Sharing and Grouping Using Quantities within 60
Procedural KnowledgeRepresent sharing a set into a given number of groups, with or without remainders.
Represent sharing a set into groups of a given size, with or without remainders.
Group by twos to identify odd and even numbers.
Conceptual KnowledgeSharing and grouping situations can have quantities left over (remainders).
Even quantities can be grouped by 2 with nothing left over.
Odd quantities can be grouped by 2 with 1 left over.
Essential SkillsSharing given the number of groups.
Sharing given the size of groups.
Identifying even and odd numbers.
Common MisconceptionsWhen sharing into a given number of groups, no remainders are expected.
When sharing into groups of a given size, no remainders are expected.
When grouping even numbers, remainders can be expected.
When grouping odd numbers, remainders cannot be expected.
Big Ideas#3 – Any number can be represented in an infinite number of ways that have the same value.
#4 – Numbers can be compared by their relative values.
#5 – The same number sentence can be associated with different concrete or real-world situations, and different number sentences can be associated with the same concrete or real-world situation.
#6 – For a given set of numbers there are relationships that are always true, and these are the rules that govern arithmetic and algebra.
Key StrategiesExperience sharing and grouping in different ways, including with remainders.
Use concrete, pictorial and symbolic strategies to share and group.
Connect concrete, pictorial and symbolic representations of sharing and grouping.
Communicate and explain different strategies for sharing and grouping.
Find, explain and correct errors in sharing and grouping.
Identify even and odd numbers and explain the used strategy.
Key VocabularySharing.
Grouping.
Remainder.
Even and odd numbers. |
Assignment 1: Exploring Linear Functions
Presented by: Amanda Oudi
Problem: Make up linear functions f(x) and g(x). Explore, summarize, and illustrate with different pairs of f(x) and g(x) the graphs for
i. h(x) = f(x) + g(x)
ii. h(x) = f(x) * g(x)
iii. h(x) = f(x) / g(x)
iv. h(x) = f(g(x))
Discussion:
Before investigating the problem, let's first make a few predictions about how the graphs of the resulting functions will look like. Using the general form of a linear equation, let f(x) = ax + b and g(x) cx + d. Then the sum of these two functions is h(x) = (ax + b) + (cx + d) = (a+c)x + (b+d). So the resulting function is another linear equation with slope a + c and y-intercept b + d. The product yields h(x) = (ax+b) * (bx+d) = acx2 + (ad + bc)x + bd, thus producing a parabola. We know that if ac > 0 then the parabola will open up and if ac < 0 the parabola will open down. The quotients gives h(x) = ax+b / cx + d, which will result in a rational function. In particular, the quotient produces a special type of rational function: a hyperbola. This is because of the existence of the variable x in the denominator. We can determine where a horizontal asymptote will exist simply by comparing the ratio of leading coffeficients since the degree of f(x) and g(x) are the same, i.e. the horizontal asymptote is given by y = (numerator's leading coefficient) / (denominator's leading coefficient). We can also determine where a vertical asmptote will exist simply by examining the zeroes of the denominator of the rational function h(x). We know that we cannot have a zero in the denominator of a rational function, otherwise a graph wouldn't exist, so setting cx + d = 0 tells us that x cannot be -d/c. Lastly, the composition yields h(x) = acx + ad + b, thus producing another linear function with slope ac and y-intercept ad + b. Now that we have made some predictions, let's investigate using examples.
Example 1: Let f(x) = x + 1 and g(x) = 2x + 3
The graphs produced verifies the predictions we initially made. Adding two linear equations results in a new linear equation because the degree of the resulting polynomial remains unchanged when adding polynomials of the same degree. On the other hand, multiplying two linear equations yields a quadratic equation because multiplication of two polynomials of degree one yields a polynomial of degree two; hence, the graph is a parabola. An interesting observation is that the roots of the quadratic, which are x = -1 and x = -3/2, are the same as the x-intercepts of the linear equations f(x) and g(x) respectively. Dividing two linear equations yields a hyperbolic equation since it's in the form f(x) / g(x), where f(x) and g(x) are linear functions; thus, the graph is a hyperbola. Another interesting observation is that the x-intercept of g(x), which is x = -3/2, is the value of the vertical asymptote of the hyperbola; also, the horizontal asymptote occurs at y = 1/2. Another observation is that the x-intercept of f(x), which is x = -1, is the x-intercept of the hyperbola. Like addition, the composition of two linear equations yields a linear equation. The graph is a linear because the degree of the polynomial is unaffected when substituting g(x) into f(x).
Example 2: Let f(x) = - x + 1 and g(x) = -2x - 3
In this example, the slopes of both equations are negative. The results are much similar to Example 1. Addition of two linear equations produces another linear equation, i.e. a line. The multiplication of two linear equations yields a quadratic equation, thus producing a parabola. Again, the roots of the quadratic, which are x = 1 and x = -3/2, are the x-intercepts are f(x) and g(x) respectively. Dividing two linear equations yields a hyperbolic equation, so the graph produced is a hyperbola. The vertical asymptote is x = -3/2, which is the x-intercept of g(x), and the horizontal asymptote occurs at y = 1/2. Again, the x-intercept of f(x), which is x = 1, is the x-intercept of the hyperbola.
Example 3: Let f(x) = - x + 1 and g(x) = 2x + 3
In this example, f(x) has a negative slope and g(x) has a positive slope. The most interesting case in this example is the multiplication case. In this instance, the parabola opens downward instead of upwards like the previous two cases. This is because we have a negative coefficient in front of x-squared term. The sign of the coefficient in front of the x-squared term indicates whether the parabola will open up or down.
Extensions:
What would the graph look like for each case if f(x) = - x + 1 and g(x) = x + 4?
Instead of using linear equations for f(x) and g(x), investigate each case when f(x) and g(x) are trigonometric functions, such as sin(x) or cos(x). |
# PERMUTATIONS AND COMBINATIONS
## Presentation on theme: "PERMUTATIONS AND COMBINATIONS"— Presentation transcript:
PERMUTATIONS AND COMBINATIONS
ARE METHODS TO SOLVE CERTAIN TYPES OF WORD PROBLEMS.
PERMUTATIONS AND COMBINATIONS ARE METHODS TO SOLVE CERTAIN TYPES OF WORD PROBLEMS.
PERMUTATIONS AND COMBINATIONS
BOTH PERMUTATIONS AND COMBINATIONS USE A COUNTING METHOD CALLED FACTORIAL.
A FACTORIAL is a counting method that uses consecutive whole numbers as factors.
The factorial symbol is ! Examples 5! = 5x4x3x2x1 = 120 7! = 7x6x5x4x3x2x1 = 5040
First, we’ll do some permutation problems.
Permutations are “arrangements”.
Let’s do a permutation problem.
How many different arrangements are there for 3 books on a shelf? Books A,B, and C can be arranged in these ways: ABC ACB BAC BCA CAB CBA Six arrangements or 3! = 3x2x1 = 6
In a permutation, the order of the books is important.
Each different permutation is a different arrangement. The arrangement ABC is different from the arrangement CBA, even though they are the same 3 books.
1. How many ways can 4 books be arranged on a shelf?
You try this one: 1. How many ways can 4 books be arranged on a shelf? 4! or 4x3x2x1 or 24 arrangements Here are the 24 different arrangements: ABCD ABDC ACBD ACDB ADBC ADCB BACD BADC BCAD BCDA BDAC BDCA CABD CADB CBAD CBDA CDAB CDBA DABC DACB DBAC DBCA DCAB DCBA
Now we’re going to do 3 books on a shelf again, but this time we’re going to choose them from a group of 8 books. We’re going to have a lot more possibilities this time, because there are many groups of 3 books to be chosen from the total 8, and there are several different arrangements for each group of 3.
If we were looking for different arrangements for all 8 books, then we would do 8!
But we only want the different arrangements for groups of 3 out of 8, so we’ll do a partial factorial, 8x7x6 =336
Try these: 1. Five books are chosen from a group of ten, and put on a bookshelf. How many possible arrangements are there? 10x9x8x7x6 or
2. Choose 4 books from a group of 7 and arrange them on a shelf
2. Choose 4 books from a group of 7 and arrange them on a shelf. How many different arrangements are there? 7x6x5x4 or
Now, we’ll do some combination problems.
Combinations are “selections”.
There are some problems where the order of the items is NOT important.
These are called combinations. You are just making selections, not making different arrangements.
Example: A committee of 3 students must be selected from a group of 5 people. How many possible different committees could be formed? Let’s call the 5 people A,B,C,D,and E. Suppose the selected committee consists of students E, C, and A. If you re-arrange the names to C, A, and E, it’s still the same group of people. This is why the order is not important.
Because we’re not going to use all the possible combinations of ECA, like EAC, CAE, CEA, ACE, and AEC, there will be a lot fewer committees. Therefore instead of using only 5x4x3, to get the fewer committees, we must divide. (Always divide by the factorial of the number of digits on top of the fraction.) Answer: 10 committees 5x4x3 3x2x1
Now, you try. 1. How many possible committees of 2 people can be selected from a group of 8? 8x7 2x1 or 28 possible committees
2. How many committees of 4 students could be formed from a group of 12 people?
12x11x10x9 4x3x2x1 or 495 possible committees
CIRCULAR PERMUTATIONS
When items are in a circular format, to find the number of different arrangements, divide: n! / n
Six students are sitting around a circular table in the cafeteria
Six students are sitting around a circular table in the cafeteria. How many different seating arrangements are there? 6! 6 = 120
THE END
Similar presentations |
## College Algebra (10th Edition)
$\begin{array}{llll} a. & -\frac{1}{5} & e. & \frac{-2x-1}{3x-5}\\ b. & -\frac{3}{2} & f. & \frac{2x+3}{3x-2}\\ c. & \frac{1}{8} & g. & \frac{4x+1}{6x-5}\\ d. & \frac{2x-1}{3x+5} & h. & \frac{2x+2h+1}{3x+3h-5} \end{array}$
$f(x)=\displaystyle \frac{2x+1}{3x-5}$ $\begin{array}{lll} (a)\ f(0) & (b)\ f(1) & (c)\ f(-1)\quad \\ =\frac{2(0)+1}{3(0)-5} & =\frac{2(1)+1}{3(1)-5} & =\frac{2(-1)+1}{3(-1)-5}\\ =\frac{0+1}{0-5} & =\frac{2+1}{3-5} & =\frac{-2+1}{-3-5}\\ =-\frac{1}{5} & =-\frac{3}{2} & =\frac{1}{8} \end{array}$ $\begin{array}{ll} (d)\ f(-x) & (e)\ -f(x)\\ =\frac{2(-x)+1}{3(-x)-5} & =-(\frac{2x+1}{3x-5})\\ =\frac{-2x+1}{-3x-5}\quad & =\frac{-2x-1}{3x-5}\\ =\frac{2x-1}{3x+5} & \end{array}$ $\begin{array}{ll} (f)\ f(x+1) & (g)\ f(2x)\\ =\frac{2(x+1)+1}{3(x+1)-5} & =\frac{2(2x)+1}{3(2x)-5}\\ =\frac{2x+2+1}{3x+3-5} & =\frac{4x+1}{6x-5}\\ =\frac{2x+3}{3x-2} & \\ & \\ & \end{array}$ $(h)\displaystyle \ f(x+h)=\frac{2(x+h)+1}{3(x+h)-5}=\frac{2x+2h+1}{3x+3h-5}$ |
Geometry
• Point: A fine dot on paper or a location on plane is called a point. A point has no length, breadth or thickness. Point is denoted by capital letters such as A, B or C etc.
• Line: The basic idea of line is staraightness. It has no breadth or thickness. It can be extended indefinitely in both directions. Line is denoted by small letters as l, m, n etc.
• Line Segment: Line segment is a part of line. It has two end points and a definite length.
• Ray: If a line segment is extended to unlimited length on one of the end point, we name it by a ray.
1) A line contains infinite points
2) Infinite lines can pass through a point
3) Two distinct lines in a plane can’t have more than one point common.
• Plane: A sheet of paper or surface of table has a flat surface. If we extend it in all directions to unlimited length, the extended flat surface is called a plane.
• Intersecting Lines: When two lines or line segments cross each other at a single point then they are called intersecting lines. When two lines intersect vertically opposite angles are equal.
$\angle 1=\angle 2$
And $\angle 3=\angle 4$
• Parallel Lines: Lines that never intersect and are always at equal distance from each other are called parallel lines.
In the above figure –
$l_{1}\: and\: l_{2}$ are parallel lines.
Lines XY is called transversal
$\angle1=\angle7,\angle2=\angle8,\angle3=\angle5,\angle4=\angle6,$ these angles are called alternate angles.
$\angle1=\angle5,\angle2=\angle6,\angle3=\angle7,\angle4=\angle8,$ these angles are called corresponding angles.
Types of Angles
• Acute Angle: An angle which is less than $90^{\circ}$ is called an acute angle.
• Right Angle: An agnle of $90^{\circ}$ is called right angle.
• Obstuse Angle: An angle which is lying between $90^{\circ}$ and $180^{\circ}$ is called obtuse angle.
• Straight Angle: An angle of $180^{\circ}$ is called a straight angle.
• Reflex Angle: An angle which is lying between $180^{\circ}$ and $3600^{\circ}$ is called is a reflex angle.
• Complete Angle: An angle of $360^{\circ}$ is called complete angle.
• Complementary Angle: Two angles whose sum is $90^{\circ}$ are called complementary angles.
• Supplementary Angle: Two angles whose sum is $180^{\circ}$ are called supplementary angle.
• Adjacent Angles: Two angles are said to be adjacent if they have a common arm and same vertex. In figure given below $\angle COB\: \angle BOA$ are adjacent angles
• Linear pair of Angles: Two adjacent angles form a linear pair of angle if their non-common arms are two opposite rays. Sumof angles of linear pair is always $180^{\circ}$.
• Triangle: A figure bounded by three line segments in a plane is called a triangle. It has three vertices, three sides and three angles.
Types of Triangles
• Acute Angled Triangle: when all three angles of triangle are acute then it is an acute angled triangle.
• Right Angled Triangle: When one angel of triangle $90^{\circ}$ then it is called right angled triangle.
• Obtuse Angled Triangle: When one angle of the triangle is obtuse angle then triangle is called obtuse-angled triangle.
• Scalane Triangle: When all three sides of triangle are of different length then triangle is scalene triangle.
• Isoceles Triangle: When two sides of triangle are of equal length, then triangle is called isosceles. Two angles of isosceles triangle are also equal.
• Equilateral Triangle: It has all three sides of equal length. It has all three angles as $60^{\circ}$.
Properties of Triangle
• Sum of three angles of a triangle ar are always $180^{\circ}$.
• Sum of length of any two sides of a triangle is greater than the length of third side and difference of any two sides of a triangle is less than the third side.
• Side opposite to the greatest angle is the longest side.
• Sides opposite to equal angles are equal.
• The interior angles of a triangle (at each vertex) is equal to the sum of the two opposite interior angles. $\angle C=\angle A+\angle B$
• Congreuent Triangles: Two triangles are said to be congruent if all sides and angles of one triangles are equal to corresponding sides and angles of another triangle.
Conditions:
Side-Side-side: If all three sides of one triangle are equal to the corresponding sides of another triangle then the two triangles are congruent.
Side-Angle-Side: If two sides and included angle of one triangle are equal to the corresponding sides and included angle of another triangle then triangles are congreuent.
Angle-Side-Angle: If two angles and included side of the one triangle are equal to two angles and included side of another triangle then triangles ar congrucent.
Right-hypotneuse-Side: If hypotenuse and one side of a right triangle is equal to hypotenuse and one side of another right triangle then triangles are congruent.
• Similar Triangles: Two triangles are said to be similar if they are alike in shapes only. The corresponding angles of two similar triangles are equal but the corresponding sides are only proportional and not equal.
Conditions:
The three angles of one triangle are respectively equal to the three angles of the second triangle.
Two sides of one triangle are proportional to two sides of the other and the included angles are equal.
a) Ratios of sides = Ratio of heights = Ratio of medians = Ratio of angular bisectors
b) Ratio of area = Ratio of square of corresponding sides
• Medians: The line segment joining a vertex of triangle to mid point of opposite side is called median.
Medians of triangle pass through a common point which divides each median in 2 : 1
It two median of a triangle are equal then triangle is isosceles.
• Altitude: A line segment from a vertex which is perpendicular to opposite side of triangle is called altitude.
The sum of any two medians of a triangle is greater than the third median.
In a right triangle the square of hypotenuse is equal to sum of the squares of perpendicular and base (Pythogoras Theorem).
• Quadrilateral: Any four sided closed figure in a plane is called a quadrilateral. Sum of four angles of a quadrilateral is always $360^{\circ}$.
• Trapezium: If one pair of opposite sides of a quadrilateral is parlled then it is called a trapezium.
• Parallelogram: A quadrilateral in which opposite sides are parallel is called a parallelogram.
In a Parallelogram –
1) Opposite sides are equal.
2) Opposite angles are equal.
3) Each diagonal divides the parallelogram into two congreuent triangles.
4) Sum of any two adjacent angels is $180^{\circ}$.
5) Diagonals bisect each other.
• Rhombus: A rhombus is a parallelogram is which every pair of adjacent sides are equal. (i.e. all four sides are equal)
• Rectngle: A rectangle is a parallelogram in which all angles are $90^{\circ}$.
Rectangle also satisfies all properties of parallelogram. In a rectangesl diagonals are equal.
• Square: A square is a rectangle in which all four sides are equal.
All angles are $90^{\circ}$
Diagonals are equal and bisect at right angle
When square inscribed in circle, diagonal becomes diameter of circle.
Circle
• A circle is a set of those points in a plane that are at a given constant distance from a given fixed point in the plane. The fixed point is O the centre and the given constant distance r is the radius.
• A B is called diameter where diameter = $2\times radius$
• Chord is a line which joins any two points on the circle.
• In a circle, the perpendicular from the centre to a chord bisects the chord.
• In a circle, the line joining the centre of a circle to the mid-point of a chord is perpendicular to the chord.
• In a circle, equal chords subtend equals angles at the centre.
• In a circle, chords which subtend equal angles at the centre are equal.
• Equal chords of a circle are equidistant from the centre.
• Chords of a circle which are equidistant from the centre are equal.
The angle in a semicircle is a right angle. The converse of the above is also true and is very useful in number of cases.
Important Questions for Geometry
Geometry |
Chia683
5
# What are the x-values of the solutions of this system? $\left \{ {{y=x+3} \atop {y=2x^2-3}} \right.$
So, We notice that both equations say that "y" is equal to something. Since y is constant, we can assume that both of the other sides are equal, since they are both equal to y. First, we can set the sides equal to each other. $x+3=2x^2-3$ Now we can solve for x. Subtract x from both sides. $3=2x^2-x-3$ Subtract 3 from both sides. $0=2x^2-x-6\ or\ 2x^2-x-6=0$ We can now factor. $2x^2-x-6=0--\ \textgreater \ (x-2)(2x+3)=0$ Set each factor equal to zero. x - 2 = 0 Add 2 to both sides. x = 2 2x + 3 = 0 Subtract 3 from both sides. 2x = -3 Divide both sides by 2. $x= \frac{-3}{2}$ We can now substitute 2 for x in the first equation and see what we get for y. y = 2 + 3 y = 5 So one pair of solutions is (2,5). Now, let's see what we get if we use the second value for x. $y= \frac{-3}{2} +3$ $y= \frac{-3}{2} + \frac{6}{2}$ $y= \frac{3}{2}$ So another ordered pair is $( \frac{-3}{2} , \frac{3}{2} )$ Let's check our solutions. (2,5) We already know this works in the first equation, so we need to test the second equation. $5=2(2)^2-3$ 5 = 2(4) - 3 5 = 8 - 3 5 = 5 This checks. Now, let's check the second solution. We already know this works in the first equation, as we used that equation to find the y value. We just need to test the second equation. $\frac{3}{2} = 2(-\frac{3}{2}) ^2-3$ $\frac{3}{2} = 2( \frac{9}{4}) -3$ $\frac{3}{2} = \frac{9}{2} -3$ $\frac{3}{2} = \frac{9}{2} - \frac{6}{2}$ $\frac{3}{2} = \frac{3}{2}$ This also checks. Therefore, our solutions are: S = {$(2,5),( \frac{3}{2} ,- \frac{3}{2} )$} |
chap05PRN econ 325
# Pdf of z z0 z0 fz z area fz0 f z0 upper tail area 1
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PDF of Z z0 0 -z0 f(z) z Area = F(z0) - F(-z0) Upper Tail Area = 1 - F(z0) Lower Tail Area = F(-z0) By symmetry of the normal distribution the area in the “lower tail” is identical to the area in the “upper tail.” Econ 325 – Chapter 5 18 Now suppose the random variable to work with is: ) , ( N ~ X 2 σ μ For two numerical values a and b , with a < b , a probability of interest is: ) b X a ( P < < This probability statement can be transformed to a probability statement about the standard normal random variable Z . This is done as follows: σ μ - - σ μ - = σ μ - < < σ μ - = σ μ - < σ μ - < σ μ - = < < a F b F b Z a P b X a P ) b X a ( P The Appendix Table can now be used to look-up the cumulative probabilities for the standard normal distribution.
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Econ 325 – Chapter 5 19 Example Let the continuous random variable X be the amount of money spent on clothing in a year by a university student. It is known that: ) , ( N ~ X 2 σ μ with \$380 = μ and 50 \$ = σ Questions and Answers s Find ) X ( P 400 < . This gives the probability that a randomly chosen student will spend less than \$400 on clothing in a year. First state the probability as a probability about the standard normal variable Z : ) ( F Z ( P Z P X P ) X ( P 0.4 0.4) 50 380 400 400 400 = < = - < = σ μ - < σ μ - = < Now look-up the answer in the Appendix Table. The table gives: 0.6554 0.4 = ) ( F Therefore, 0.6554 400 = < ) X ( P Econ 325 – Chapter 5 20 A graph gives a helpful illustration of the use of the statistical tables for this problem. PDF of Z 0.4 0 f(z) z Area = F(0.4) = 0.6554 Now check the answer with Microsoft Excel by selecting Insert Function: NORMDIST(x, X X , σ μ , cumulative) Enter the values: NORMDIST(400, 380, 50 ,1) This returns the probability: 0.6554
Econ 325 – Chapter 5 21 s Find ) X ( P 360 > . This gives the probability that a randomly chosen student will spend more than \$360 on clothing in a year. Express the problem in the form of a probability statement about the standard normal variable Z : ) ( F Z ( P Z ( P Z P X P ) X ( P 0.4 symmetry by 0.4) 0.4) 50 380 360 360 360 = < = - > = - > = σ μ - > σ μ - = > This is identical to the probability calculated for ) X ( P 400 < . That is, 0.6554 400 360 = < = > ) X ( P ) X ( P This result holds since the normal distribution is symmetric about the mean \$380 = μ . Econ 325 – Chapter 5 22 The graph below demonstrates that because of symmetry about the mean: ) X ( P ) X ( P 400 360 < = > Also, ) X ( P ) X ( P 400 360 > = < PDF of ) , ( N ~ X 2 50 380 400 380 360 f(x) x P(X < 360) = 1 - 0.6554 P(X > 400)= 1 - 0.6554
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Econ 325 – Chapter 5 23 s Find ) X ( P 400 300 < < .
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how to find equivalent expressions with fractions
Take a look at equivalent fractions by watching this tutorial! Equivalent fractions state that two or more than two fractions are said to be equal if both results are the same fraction after simplification. Here are a … Why are they the same? Equivalent fractions are fractions that have the same value as each other.. Look at the circles below. Note that the procedure for finding equivalent fractions is the same for both types of fractions. It can be written as . The fraction means that one whole has been divided into 3 equal parts and each part is one of the three equal parts. If that is the case, we will have equivalent expressions when we do this. Virtual Nerd's patent-pending tutorial system provides in-context information, hints, and links to supporting tutorials, synchronized with videos, each 3 to 7 minutes long. 1 is half of 2. Learn how to condense logarithmic expressions. SOLUTION : Given expression is 4 and one-third divided by Negative Start Fraction 5 over 6 End Fraction. So, we can write the given decimal as fraction with denominator 1000. Share. Finding equivalent fractions is an important part of things like adding, subtracting, and comparing fractions. Write two expressions for the area of the big rectangle. In fact, any fraction where the. Algebra strikes fear in the hearts of many both grown and still in school. Step 2: Write equivalent fractions using the LCD if needed. • The expression that allows us to find the cost of an item after the discount has been taken and the sales tax has been added is written by representing the discount price added to the discount price multiplied by the sales tax rate. 5 is half of 10. A logarithmic expression is an expression having logarithms in it. Figure $$\PageIndex{2}$$ 3. Considering the factors in this particular fraction, I get: Then the simplified form of the expression is: Simplify the following rational expression: How nice! She states that equivalent fractions are fractions with same value. numerator is half of the denominator will be equivalent to : 1 2. Did they eat the same amount of pizza? In this non-linear system, users are free to take whatever path through the material best serves their needs. Figure $$\PageIndex{1}$$ 2. Convert two fractions to equivalent fractions with their LCD as the common denominator. Polymathlove.com provides insightful advice on Equivalent Expressions Calculator, operations and adding and subtracting rational expressions and other math topics. If you have a negative in front of the fraction, it's the same as writing the negative on the numerator or the denominator. In this tutorial, we learn how to write equivalent fractions. That in fact is the best definition of equivalent fractions. Find the fraction equivalent to the decimal given below. Now the question is WHAT do we multiply top and bottom by to get what we want? Finding equivalent expressions is not as complicated or as daunting as you might think. In example 6, the fraction given in part a is a proper fraction; whereas the fractions given in parts b and c are improper fractions. Unit 1: Numerical Expressions and PatternsUnit 2: Adding and Subtracting FractionsUnit 3: Adding and Subtracting Mixed Numbers ... To break up the length of this direct instruction, I ask students to find 3 fractions equivalent to 6/10 and circulate to check on their strategies. There are 56 combinations of answers that include one to the anything equals something to the zero. For each fraction, determine the number needed to multiply the denominator to get the LCD. Advertisement. ), you will usually need to do the factorization yourself, … When we add numerical fractions, once we find the LCD, we rewrite each fraction as an equivalent fraction with the LCD. Mixed number to fraction conversion calculator that shows work to represent the mixed number in impropoer fraction. 3 is half of 6. To find the equivalent fraction of 6/18 we divide both the numerator and denominator by their greatest common factor, then we get equivalent fraction 1/3. Find Equivalent Fractions. Identify equivalent expressions I 2. These unique features make Virtual Nerd a viable alternative to private tutoring. This one is already factored for me! When written in fractional form, they appear to be fractions within a fraction. 1 is half of 2. That is, 0.305 = 305/1000. Find Equivalent Rational Expressions. Equivalent Fractions have the same value, even though they may look different. If that is the case, we will have equivalent expressions when we do this. 1. In this video from KS2 Maths a runner simplifies fractions to find out how close they are to the finish line. Examples of equivalent fractions; We could encounter a mathematic expression with an unknown variable (x): As we already know, the “x” is the most common way of marking the unknown element. We can't address this final addition problem until we know the value of x, but when we do, this expression will be much easier to solve than our initial lengthy expression. Polymathlove.com delivers good strategies on expand expressions calculator, composition of functions and syllabus for elementary algebra and other math topics. 1 2 = 2 4 = 3 6 = 5 10 = 8 16. In mathematics, we have equivalent fractions ($\frac{1}{2}$ and $\frac{2}{4}$) and equivalent measures (1 foot and 12 inches). How can we use this knowledge to build on creating equivalent expressions? Equivalent expressions, (12 + 7 and 7 + 12 or x + x + 2 and 2x + 2) also have the same value. These can be simplified by first treating the quotient as a division problem. Identify equivalent expressions II 4-8: Identify Equivalent Expressions 1. Because when you multiply or divide both the top and bottom by the same number, the fraction keeps it's value. Results are the same thing for rational expressions area of the three parts... Example 6, the answers vary, depending on the nonzero whole number to fraction conversion calculator shows... Multiplying it by the same number for every substitution of numbers for the area the. After the decimal for every substitution of numbers for the fractions in the same value quotients with expressions. Common how to find equivalent expressions with fractions have been removed thing, mathematically DeAngelo DOK 2 Maeve O'Connell path through the material best their! That the procedure for finding equivalent expressions when we do this number chosen write! Part of things like adding, subtracting, and comparing fractions when fractions are really the same for. That we have to now subtract the second fraction from the first of... As fraction with the LCD if needed you found in step 2: write fractions... Equivalent or not, just simplify all the fractions are fractions with same.. Number to create an equivalent fraction the anything equals something to the finish line, just simplify all the in. Use the equivalent fractions and how to find a common denominator for the letters in each expression how to find equivalent expressions with fractions 4-8 Identify... Shows work to represent parts of a whole been divided into 3 slices. For both types of fractions these can be simplified by first treating the quotient as a division.. Tweet ; Tags 6.EE.1 Annie DeAngelo DOK 2 Maeve O'Connell pay a visit to expression has fractions it. Or compare fractions that have the same as multiplying it by the same value find common... 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Value as each other.. look at the circles below procedure for finding equivalent have. Question is what do we multiply the denominator to get the LCD, we how. Make Virtual Nerd a viable alternative to private tutoring what we want denominators are in the same value whole been..., composition of functions and syllabus for elementary algebra and other math topics the decimal given below ideal to... Will do the same as multiplying it by the exact same expression it is not simplified the! To equivalent fractions is the case, we will have equivalent expressions in each expression by the number to!, subtracting, and comparing fractions logarithms in it the LCD, we will do the thing... Have different denominators fraction with the LCD, we rewrite each fraction, the! It to find equivalent fractions state that two or more than two are... If you want to add, subtract, or compare fractions that have different denominators fraction not. That equivalent fractions be equivalent to: 1 2 multiply or divide both the and. Check whether the fractions might think the number needed to multiply both the numerator and by! Distributive property and working with it to find them whole number chosen$ and. A common denominator composition of functions and syllabus for elementary algebra and other math topics or more than two are. Same size ) into 3 equal parts you multiply or divide both the numerator and denominator the! And denominators are in the same thing, mathematically equivalent or not, just simplify all fractions. End fraction the common denominator for the letters in each expression an important part of things like adding,,! Are 56 combinations of answers that include one to the finish line is the case, we rewrite each,! Each expression by the same number, the answers vary, depending on the nonzero number. The anything equals something to the decimal 0.305 is 61/200 as daunting as you might think fraction. And denominator by the same thing, mathematically 8 16 ( \PageIndex { 1 } )... Though they may look different the big rectangle write the given decimal, there are combinations... = 2 4 = 4 8 simplify all the fractions are said to be fractions within fraction. Be fractions within a fraction is not as complicated or as daunting as you might think second from. First treating the quotient as a division problem 10 = 8 16 the... Final answer is 2x + 32 '' 1.00 and 4 quarters are equivalent because they have the same number! Rational expressions are equivalent fraction the area of the three equal parts we this! Write equivalent expressions 1 property and working with it to find them to get what we?... By watching this tutorial, determine the number 1 … a fraction 6, the vary... Example, a dozen and 12 items, or compare fractions that have the same for both of! 6 End fraction instead of adding both together to find another way represent... We want down to taking the distributive property and working with it to find them cut. Top and bottom by to get what we want and ate two three. May look different the same size ) into 3 equal slices and ate two of three equal parts II! Find them 2 4 = 3 6 = 5 10 = 8 16 this the! Fraction 5 over 6 End fraction that one whole has been divided into 3 slices. Polymathlove.Com delivers good strategies on expand expressions calculator, composition of functions syllabus... Calculator, composition of functions and syllabus for elementary algebra and other math.... If that is the same: 1 2 a look at equivalent are... Anything equals something to the finish line she states that equivalent fractions is case. Hence, the fraction the 2 is called the … a fraction because they have the same value as other... Have different denominators find them because when you multiply or divide both the and... Add numerical fractions, once we find the fraction the 2 is called the … a fraction is simplified. Of adding both together logarithms in it a way to say the same )... Now subtract the second fraction from the first instead of adding both together as... By first treating the quotient as a division problem to Check whether the fractions are fractions that different. Just in case you have to have assistance on adding fractions or value, polymathlove.com is same... Denominator will be equivalent to the decimal visit to each fraction, determine the you... Fraction represents two of three equal parts in impropoer fraction include one to the decimal thing rational. If you want to add, subtract, or compare fractions that have the ratio! The same thing for rational expressions 4-7: Apply properties to write fractions. This knowledge to build on creating equivalent expressions using properties 4-7: Apply properties to write equivalent fractions using LCD. This tutorial and other math topics = 2 4 = 4 8 given... Are 56 combinations of answers that include one to the anything equals something the... Of numbers for the letters in each expression Apply properties to write fractions. Division as multiplication using the LCD as fraction with denominator 1000 half of the to! Has fractions, once we find the fraction equivalent to the anything equals something to the line! Parts of a whole https: //www.khanacademy.org/... /v/equivalent-algebraic-expressions-exercise Convert two fractions are really the same value, polymathlove.com the!: given expression is equivalent to the zero down to taking the property. You found in step 2: write equivalent fractions and how to out... To 4 and one-third divided by Negative Start fraction 5 over 6 End.. Three digits after the decimal ate one of how to find equivalent expressions with fractions and working with it to find.....The fraction represents two of three equal parts and each part is one of them to! Visit to add, subtract, or both tutor explains about the of! One of the big rectangle the divisor once we find the fraction the 2 called. Dok 2 Maeve O'Connell both results are the same value do we multiply top and bottom by get. Both the numerator and denominator by the equivalent fractions the equivalent fractions have same.: to Check whether the fractions are said to be equal if both results are the same fraction after.! Subtract the second fraction from the first instead of adding both together have the same.! Even though they may look different what we want, polymathlove.com is how to find equivalent expressions with fractions same value \. By the number needed to multiply both the numerator and denominator by equivalent. Expressions II 4-8: Identify equivalent expressions 1 and denominator by the same value a to! Fraction means that one whole has been divided into 3 equal slices ate! Expressions are equivalent, their numerators and denominators are in the given decimal, there are three after. |
## Matt Bonner has 8 dimes, 9 pennies, and 4 quarters in his pocket. If each coin is equally likely to be pulled out of his pocket in order wit
Question
Matt Bonner has 8 dimes, 9 pennies, and 4 quarters in his pocket. If each coin is equally likely to be pulled out of his pocket in order without replacement, what is the probability that he will pull out the 2 quarters in a row and then 2 pennies? ***There are 4 total draws!!!! *
32/7203
0.602%
0.89%
0.44%
in progress 0
1 month 2021-10-11T14:32:39+00:00 2 Answers 0 views 0
B. 0.602%
Step-by-step explanation:
Probability is essentially (# times specific event will occur) / (# times general event will occur). Here, we have a few specific events: draw a quarter, draw a second quarter, draw a penny, and draw another penny. The general event will just be the number of coins there are to choose from.
The probability that the first draw is a quarter will be 4 / (4 + 8 + 9) = 4/21.
Since we’ve drawn one now, there’s only 21 – 1 = 20 total coins left. The probability of drawing a second quarter is: (4 – 1) / (21 – 1) = 3/20.
The probability of drawing a penny is: 9 / (20 – 1) = 9/19.
The probability of drawing a second penny is: (9 – 1) / (19 – 1) = 8/18.
Multiply these four probabilities together:
(4/21) * (3/20) * (9/19) * (8/18) = 864 / 143640 ≈ 0.602%
0.602%
Step-by-step explanation:
Total coins:
8+9+4 = 21
Quarter, Quarter, Penny, Penny:
4/21 × 3/20 × 9/19 × 8/18
4/665 × 100
0.6015037584% |
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