Solving Monotonicity and Concavity

A function is said to be rising on an interval [a, b] = I if f (z1) < f (z2) whenever z1 < z2 for z1, z2 `in` I. A function is said to be decreasing on an interval [a, b] = I if f (z1) > f (z2) whenever z1 < z2 for z1, z2 `in` I. A function which is strictly increasing or strictly decreasing on an interval is said to be monotonic on that interval. A function is said to be curved in up on an interval [a, b] if f ’’ (z) > 0 on [a, b]. A function is said to be curved in down on an interval [a, b] if f ’’ (z) < 0 on [a, b]. A point (c, f(c)) on the graph of y = f (z) is called an inflection point if the concavity changes at the point. (I.E., it is curved in up on some interval (a, c) and curved in down on some interval (c, b) or vice versa.)

Solving monotonicity and concavity - Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f ‘(x) = `(x^5)/20 + (x^4)/12 - (3x^3)/3 - 10`

f ’(x) = `(x^4)/4 + (x^3)/3 - 3(x^2)`

f ‘’ (x) = x3 + x2 – 3x

= x(x2 + x - 3)

f ‘‘(x) = 0

x = 0, -2.3, 1.3

Inflection pts: x= -2.3, 0, 1.3

Curved in up: (-2.3,0), (1.3,infinity)

Curved in down: (-oo,-2.3), (0,1.3)

Solving monotonicity and concavity

Solving monotonicity and concavity - More Examples:

Solving monotonicity and concavity - Example 1:

Find inflection points & determine concavity for f (x)

f (x) = `(x^5)/20 + (x^4)/12 - (x^3)/3 +10`

f ‘ (x) = `(x^4)/4 + (x^3)/3 - x^2`

f ‘’ (x) = x3 + x2 – 2x = x(x2 + x - 2)

= x (x+2) (x-1)

f ‘‘ (x) = 0

x = 0, -2, 1

f ’’(-5) < 0, f’’(-1) > 0, f ’’(.5) < 0, f ’’(10) > 0

Inflection pts: x=-2,0,1

Curved in up: (-2,0), (1,`oo` )

Curved in down: (`-oo` .,-2), (0,1)

Solving monotonicity and concavity

Trinomial Multiplication

In algebraic expression consisting of only one term is called a monomial, two terms is called a binomial and when there are three terms separated by an addition or a subtraction operation is called a trinomial. ‘Tri’ in the word trinomial means three and hence the name. It is also referred to as a polynomial as ‘poly’ means more than two terms. For example, a+3b-c consists of three terms a, b and c and hence a trinomial. In 3b, 3 is called the coefficient of b. In a given polynomial we come across like terms and unlike terms.

Like terms are the terms which have the same variable or literal but a different coefficient. For example, (3b, -7b);(5x2, 12x2);(-4xy, 7xy) are some of the like terms. Unlike terms as the name suggests are the terms which have different variables. For example, (7x, 8y); (4ab, -4ac); (2y2, -2x2) etc are some of the unlike terms.

It is not necessary that there would be only two like or unlike terms, it depends on the number of terms in the given polynomial. Now let us learn as to How do you multiply trinomials. There are two methods in which we can multiply trinomials, one is the horizontal method the other is the vertical method.

The steps to be followed in a Horizontal method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
Solution: First and second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term

Arrange the two polynomials horizontally
(4x2-3x+5) X (x2+5x-3)
Now distribute each of the terms of the first trinomial with each of the terms of the second trinomial
=4x2(x2+5x-3)- 3x(x2+5x-3) +5(x2+5x-3)
=4x4+20x3-12x2 - 3x3-15x2+9x +5x2+25x-15

Combine the like terms and simplify
=4x4+x3(20-3)+x2(-12-15+5)+ x(9+25) – 15
= 4x4+17x3-22x2+34x-15

The steps to be followed in a Vertical method of multiplication are as follows:
Example: Multiply (4x2-3x+5)(x2+5x-3)
The first and the second terms are chosen irrespective of the order
Let (4x2-3x+5) be the first term and (x2+5x-3) be the second term
Arrange the two polynomials in a vertical form
4x2-3x+5
X  x2+5x-3

Working from right towards left each term of the lower trinomial is multiplied with each of the terms of the upper trinomials. Then the products are written underneath the second trinomial in three rows in the order of the degree of each term and then the like terms simplified as shown below
4x2-3x+5
X  x2+5x-3
-12x2+ 9x – 15
+20x3-15x2 + 25x
4x4 - 3x3  + 5x2                      .
4x4+17x3-22x2+ 34x-15 is the final product!

Solve Explicit Differentiation

In calculus,Explicit is a function which the independent variable. The function f explicitly is to provide a preparation for determining the output of the given function y in terms of the input value x: y = f(x). Derivative of an explicit function is called as explicit differentiation. for example  y = x3 + 5. The process of finding the differentiation of the independent variable in an explicit function by differentiating each term separately, by expressing the derivative of the independent variable as a symbol, and by solving the resulting expression for the symbol.

Solve explicit differentiation problems:

Let us see some problems and its helps to solve an explicit differentiation.

Solve explicit differentiation problem 1:

Find the differentiation of given explicit function   y = x2 - 15x + 3.

Solution:

Given explicit function is  y = x2 - 15x + 3.

Differentiation of explicit function is   dy/dx  = d/dx ( x2 - 15x + 3)

Separate the each term, so, we get

= d/dx (x2) - d/dx (15x) + d/dx (3)

= d/dx (x2) - 15d/dx(x) + d/dx (3).

= 2 x(2-1) - 15 + 0

= 2x - 15

The differentiation of an explicit function is  2x - 15

Solve explicit differentiation problem 2:

Find the differentiation of an explicit function  x2 +  cot x = - y

Solution:

Given explicit function is  x2 +  cot x = - y

Multiply by (-1) on both sides,  y = - x2 -  cot x

Differentiation of an explicit function,

y =   - x2 -  cot x

dy/dx = d/dx -x2   - d/dx (cot x) .

=  - 2x - (-cosec2 x) .

= - 2x + cosec2x

The Differentiation of an explicit function is  - 2x + cosec2x

Solve explicit differentiation problem 3:

Find the differentiation of given explicit function  2x2 + y2 = 1

Solution:

Given explicit function is  2x2 + y2 = 1

Subtract 2x2 on both sides.we get,

2x2 + y2 - 2x2 = 1 - 2x2

y 2 = 1 - 2x2

Take square root on both sides, we get

 sqrt (y^2) =  +- sqrt (1 - 2x^2)

y = +- sqrt (1 - 2x^2) .

Differentiate the function,  Let u = 1 - 2x2            and           y = sqrt u

(du)/(dx) = - 4x                              dy/(du) = 1/(2sqrtu)

So,   dy/dx =  ((dy)/(du)) ((du)/(dx)) .

=  1/(2sqrtu) . (-4x )

=  (- 2x) / sqrtu .

Substitute u = y, So we get

= - (2x)/y   .

The differentiation of an explicit function is  - (2x)/y    .

Substitution Geometry

Geometry is a module of mathematics, which involves the study of shapes, line equation, angles problem, dimensions, relative position of figures etc.  The term ‘Geometry’ means study of properties. A point is used to represent a position in space. A plane to be a surface extending infinitely in all directions such that all points lying on the line joining any two points on the surface. Substitution geometry problems are given below.

Example problems for substitution geometry :

1. Find out the geometry equation of straight line passing through the points 2x + y = 8 and 3x - 2y + 7 = 0 and parallel to 4x+ y - 11 = 0
Solution:
Let (x1, y1) be the intersection lines
2x1 +  y1 =  8       …  (1)
3x1 - 2y1 = - 7   …   (2)


(1) × 2 ? 4x1 + 2y1 = 16      …    (3)
(2) + (3) ? x1 =9/7 `=>` y1 =38/7   (x1, y1) =( 9/7 ,38/7)


The straight line parallel to 4x + y - 11 = 0 is of the form 4x + y + k = 0
But it passes through (9/7 ,38/7)


36/7 +38/7 + k = 0 ? k = -74/7
4x + y -74/7 = 0
28x + 7y - 74 = 0 this is the equation of straight line.


2. For what values of ‘a’, the three straight lines 3x + y + 2 = 0, 2x - y + 3 = 0and x + a y - 3 = 0 are concurrent?

Solution:

Let (x1, y1) be the point of concurrency. This point satisfies the first two equations.
3x1 + y1 + 2 = 0 … (1)
2x1 - y1 + 3 = 0 … (2)

Solving (1) and (2) By using substitution method, we get (- 1, 1) as the point of intersection. Since it is a point of concurrency, it lies on x + a y - 3 =0
- 1 + a - 3 = 0
a -4 = 0

a = 4

Practice problems for substitution geometry:

1. Find the point of intersection of the straight lines 5x + 4y - 13 = 0 and 3x + y - 5 = 0

Ans: The point of intersection is (1, 2)

2. Find the geometry equation of straight- line perpendicular to the straight line 3x + 4y + 28 = 0 and passing through the  point (- 1, 4).

Ans: 4x - 3y + 16 = 0

3. Find the equation of straight line passing through the intersection of straight lines 2x + y = 8 and 3x - y = 2 and through the point (2, - 3).

Ans: x = 2

Proportionality Problems

In mathematics, proportionality indicates that two variables are related in a linear manner. If one number doubles in size, so does the other; if one of the variables diminishes to 1/10 of its former value, so does the other.The symbol for proportionality resemble a stretched-out, lowercase Greek letter alpha . When this symbol appear that two quantities or variables, it is read "is proportional to" or "varies in direct proportion with." Thus, the expression x alpha y is read " x is proportional to y " or " x varies in direct proportion with y ." In this condition, as long as x and y do not attain values of zero, the quotient x / y is always equal to the same value k , which is called the proportionality constant. (source: wiki)

Example proportionality problems:

Proportionality problem 1:

Find a if a/4= 3/2.

Proportionality  solution:

By using property 1:

a/4=3/2

(a) (2) = (4) (3)

2a = 12

Dividing both side by 2

a=6

Therefore the value of a=6

Proportionality problem 2:

Is 6: 4 = 3: 2 a proportion?

No. If this were a proportion, Property 1 would produce

(6) (2) = (4) (3)

12   = 12, this is true. It is a proportion.

Practice proportionality problems:

Practice proportionality problem 1:

Janette's car uses 9 gallons to go 200 miles.

a) How many gallons will she use to go 400 miles?

B) How many gallons will she use to go 600 miles?

c) How far can she drive with 36 gallons?

(a)18

(b) 27

(c)800 miles

Practice proportionality problem 2:

Carlos makes $25 in 4 hours.

a) How much will he make in 2 hours?

b) How much will he make in 20 hours?

c) How much will he make in 22 hours?

(a)$12.50

(b)$125

(c)$137.50


practice proportionality problem 3:

Which is a better value?

a) 15 ounces for $ 9.25 or 30 ounces for $18.00?

b) 15 ounces for $ 9.25 or 5 ounces for $3.29?

Answer:  ( a )30 ounces for $18.

( b )15 ounces for $9.25.

practice proportionality problem 4:

6 bottles cost $7.00.

a) How much will 18 bottles cost?

b) How much will 15 bottles cost?

c) How much will 27 bottles cost?

d) How many bottles can you buy with $10.50?

Answer:   ( a ) $21

( b ) $17.50

( c ) $31.50

( d ) 9

Geometry Area and Volume

Geometry” Earth-measuring" is an part of the mathematics concerned with questions of size, shape, relative position of figures, and the properties of space. Geometry is one of the oldest sciences. Initially a body of the  practical knowledge concerning lengths, areas, and volumes. And now we can see about the problems in geometry area and volume.

Geometry area and volume problem 1:

Pro 1 :Find the volume of cylinder with the radius 8 cm and the height 12 cm.

Solution:We can find the volume of  an cylinder by using the following formula:

Volume of cylinder V=πr2h

Substitute the values of r and h into the above formula. Than, we get

V=π*82*12

=3.14*64*12

=2411.52 cm3

Pro 2 :Find the volume of sphere with the radius is 12 cm.

Solution:We can find the volume of  the sphere by using the following formula

Volume of sphere V= (4/3) πr3

Substitute the value of radius into the above formula. Then we get,

V= (4/3) *3.14*123

= 1.333*3.14*144

= 602.72 cm3

Ans: 602.72 cm3

Problem 3:Find the amount of pyramid with the base 8.2 mt and height 10.2 mt.

Solution:We can find the volume of  the pyramid by using the following formula

Volume of pyramid V= (1/3) b h

Substitute the values of  the base and height into the above formula. Then we get,

V= (1/3)*8.2*10.2

=0.333*8.2*10.2

=27.8521 mt3

Ans: 27.8521 mt3

Geometry - Find the volume of shapes when area is given

Find the volume of the right prism whose area of the base is 550 cm2 and height is 38cm



Solution:Given that area of the base, A = 550 cm2 and height (h) of the prism = 38 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 20900

Volume of right prism = 20900 cm3

Find the volume of the right prism whose area of the base is 450 cm2 and height is 34cm

Solution:Given that area of the base, A = 450 cm2 and height (h) of the prism = 34 cm

Volume of the right prism = area of the base * height cu.units

= A * h

= 15300

Tutoring About Correlation Coefficient

Concept of correlation:

Correlation is a method of studying the relationship between the two variables. In statistical analysis we come across the study of two variables wherein the change in the value of one variable produces a change in the value of other variable. In that case we say that the variables are correlated or there is a correlation between the two variables.

The formula for the correlation coefficient r can be expressed in the form,

r = `(sum (X - barX) ( Y - barY))/(sqrt(sum (X - barX)^2) sqrt(sum(Y - barY)^2))`

It is conventionally taken as x = X - X and y = Y - Y and hence we write

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

The above formula is expressed in terms of deviations of the variables from their means. Instead, if the actual values of the observations are taken then the formula can be written as,

r= `(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2) sqrt(NsumY^2 - (sum Y)^2))`

Instead of the deviations from their means, the deviations are measured from the value A and B for X and Y variables by taking dx = X - A, dy = Y - B, the correlation coefficient r is given by,

r = `(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2) sqrt(Nsumdy^2 - (sumdy)^2))`

Tutoring about formulas for calculating correlation coefficient:

r = `(sum xy)/(sqrt(sum x^2) sqrt(sum y^2))`

This formula is used when deviations are measured from their mean.
r= `"(N sum XY - sum X sum Y)/(sqrt(N sum X^2 - (sumX)^2)(sqrt(N sum Y^2 - (sumY)^2))) `

This formula is used if no assumed average is taken for x and y series
r = `"(N sumdxdy - sumdx.sumdy)/(sqrt(Nsumdx^2- (sumdx)^2)(sqrtNsumdy^2 - (sumdy)^2)) `

This formula is applied when deviations for x and y series are taken from some assumed values.


Tutoring on problems on correlation coefficient::

Calculate the correlation coefficient between x and y from the following data:
x 1 3 5 8 9 10
y 3 4 8 10 12 11


Solution:
x y x - `barx` y - `bary` (x - `barx` )2 (y - `bary` )2 (x-`barx` )(y - `bary` )
1 3 -5 -5 25 25 25
3 4 -3 -4 9 16 12
5 8 -1 0 1 0 0
8 10 2 2 4 4 4
9 12 3 4 9 16 12
10 11 4 3 16 9 12
36 48 0 0 64 70 65


`barx` = `(sum x)/(n)` = 36/6 = 6

`bary` = `(sum y)/(n)` = 48/6 = 8

r = `(sum(x- barx)(y - bary))/(sqrt(sum(x-barx)^2)sqrt(sum(y-bary)^2))`

= `(sum xy)/(sqrt(sumx^2)sqrt(sumy^2))`

= `(65)/(sqrt(64)sqrt(70))` = 0.97

Limit of a Function

Students can study about Limit of a Function here. Consider the function f(x). Let the independent variable x take values near a given constant a. Then f(x) takes a corresponding set of values. Suppose that when x is close to a, the values of f(x) are close to some constant. Suppose f(x) can be made to differ arbitrarily small from A by taking values of x that are sufficiently close to a but not equal to a and that is true for all such values of x. Then f(x) is said to approach limit A as x approaches a.

If the function f(x) approaches a constant A when x approaches a in whatever manner without assuming the value a, A is said to be the limit of f(x) as x approaches a. Thus we write lim_(x->a) f(x) = A

Find the Limit of a Function

Students can learn to Find the Limit of a Function if they know what Functions are and how they behave at the given limits.

A function may approach two different limits. One where the variable approaches its limit through values larger than the limit and the other where the variable approaches its limit through values smaller than the limit. In such a case the limit is not defined but the right and left hand limit exists.

The right hand limit of a function is the value of the function approaches when the variable approaches its limit from the right. Here, we write lim_(x->a^+) f(x) = A+

The left hand limit of a function is the value of the function approaches when the variable approaches its limit from the left.

here, we write lim_(x->a^-) f(x) = A-

The limit of a function exists if and only if the left hand limit = right hand limit.

In that case, lim_(x->a^+) f(x) = lim_(x->a^-) f(x) = f(x)

Properties of limit of a function

The following are some of the properties of limits which are useful in evaluating the limit of a function.

1. lim_(x->a) k = k ( k is a constant)

2. lim_(x->a) [f(x) ± g(x)] =  lim_(x->a) f(x) ± lim_(x->a) g(x)

3. lim_(x->oo) [f(x).g(x)] = lim_(x->oo) f(x) . lim_(x->oo) g(x)

4. lim_(x->a) (f(x))/(g(x)) = (lim_(x->a)f(x))/(lim_(x->a)g(x))

5. lim_(x->a) [f(x)]n =  [ lim_(x->a) f(x)]n

Standard limit theorems:

1. lim_(x->a) (x^n- a^n)/(x - a) = nan-1

2. lim_(x->0) (e^x-1)/(x) = 1

3. lim_(x->0) (sin x)/(x) = 1

4. lim_(x->0) (1 + (1)/(n))^(n) = e



Solved Examples

Example 1: Evaluate the right hand limit of the function

f(x) = {│x – 4│/x – 4, x ≠ 4, 0 x = 4

at x = 4

Sol: (RHL of f(x) at x = 4)

= lim f(x) = lim f(4 + h) = lim │4+ h – 4│4 + h – 4

x→4+      h→0             h→0

= lim │h│/h = lim h/h = lim 1 = 1

h→0            h→0      h→0

Example 2: Let f be the function given by f(x) = x2 – a2/x – a, x ≠ a

Using (in , δ) definition show that lim f(x) = 2a

x → 0

Sol: Let in > 0 be given. In order to show that

lim f(xi) = 2a

x → a

We have to show that that for any given in > 0, there exists a number δ >0 such that

│f(x) – 2a│< in whenever 0 < │x – a│< δ

If x ≠ a, then │f(x) – 2a│= │x3 – a2/x – a│

= │(x + a) – 2a│                                         [... x ≠ a]

= │x – a│

... │f(x) – 2a│< in , if │x – a│< in

Choosing a number δ such that 0 < δ < in , we have

│f(x) – 2a│< in when whenever 0 < │x – a│< δ

Hence    lim f(x) = 2a

x → 0

Standard Deviation in Statistics

Statistics is a division of applied mathematics which contracts with the particular interpolation of data. The term ‘Statistics’ has been taken from the Latin name ‘Status’ in which it defines ‘political state’. This statistics mainly used to measure the arithmetic mean, median, mode and standard deviation. This measurement gives idea about where the data points are centered. Let us discuss about standard deviation with some example problems.

Evaluation of Standard Deviation in statistics:

Mean:

In Statistics, Mean is defined as the average of the given total numbers, i.e., total number of data divided by the number of data set given.

Formula for finding mean,

barx   = (sum(x))/(n)

Standard Deviation:

In Statistics, Standard Deviation is the measure of describing squared mean difference variability and spread of the Data set in the given total numbers. It is used to take the measurement of taking square root and average of numbers in the Data set.

Formula for standard deviation,

S =  sqrt ((sum(x - barx)^2 )/ (n-1))

Here Standard Deviation is calculated by using the mean Value  barx

Example Problems to find standard deviation in statistics :

Problem 1:

Here are 4 measurements   66, 45, 67, 45, 34, 56, 78 and 57. Calculate statistics standard deviation for the given measurements

Sol:

Mean: Calculate the mean the using the formula,

barx   = (sum(x)) / n

barx   =( 66+45+67+45+34+56+78+57) / 8

= 448 / 8

barx   = 56

Standard Deviation,

S = sqrt((sum(x-barx)^2) / (n-1) )

S = sqrt((( 66-56)^2+(45-56)^2+(67-56)^2+(45-56)^2+(34-56)^2+(56-56)^2+(78-56)^2+(57-56)^2) / (8-1))

= sqrt((100+121+121+121+484+0+484+1)/7)

=  sqrt( 1432 / 7)

S = sqrt(204.571429)

Standard Deviation S =  14.3028469

Problem 2:

Find the Statistics Standard deviation of the given Data 16, 17, 18, 20 and 24

Sol:

Mean: Calculate the mean using the formula,

barx   = (sum(x)) / n

barx  = ( 16 + 17 + 18 + 20 + 24 ) / 5

barx = 95 / 5

barx = 19

X               x-barx              (x-barx )^2

6               16 - 19 = -3           9

7               17 - 19 = -2           4

8               18 - 19 = -1           1

9               20 - 19 =  1           1

10              21 - 19 =  2           4

Sum of the  (x-barx)^2

9+4+1+1+4 = 19

Standard Deviation: Calculate the standard deviation

S =   sqrt((sum(x- barx)^2 ) / (n-1))

S = sqrt( ( 9+4+1+1+4) / 4 )

S = sqrt (19 / 4 )

S = sqrt(4.75 )

S = 2.17944947

Practice Problems for Statistics Standard Deviation:

1. Find the statistics standard deviation for the following given data.37, 56, 54, 54, 26, 67, 12, 65 and 34.

Answer:                     Mean = 45

Standard Deviation = 18.714967272213

2. Calculate the statistics standard deviation for the following. 77, 56, 33, 87, 90, 23, 67, 80 and 99.

Answer:                    Mean = 68

Standard Deviation = 26.043233286211

Study Online Second Derivatives

Study online second derivatives involves the process differentiating the given polynomial function twice with respect to the given variables whereas all the process is clearly explained with the help of online. Generally the second derivative is discussed in calculus whereas it is mainly helps to find the rate of change of the given function with respect to the change in the input. The following are the solved example problems with detailed step by step solution for second derivatives study discussed in online.

Example 1:

Determine the second derivative from the polynomial.

f(b) = 5b 2 +5b 4  + 12

Solution:

The given function is

f(b) = 5b 2 +5b 4  + 12

The above function is differentiated with respect to b to find the first derivative

f '(b) = 5(2b  )+5(4 b 3 ) + 0

By solving above terms

f '(b) = 10b +20b3

The above function is again differentiated with respect to b to find second derivative

f ''(b) =  10(1 ) +20(3b2)

f ''(b) =  10 + 60b2 is the answer.

Example 2:

Determine the second derivative from the polynomial.

f(b) = 4b4 +5b 5 +6b 6  + 2b

Solution:

The given function is

f(b) = 4b4 +5b 5 +6b 6  + 2b

The above function is differentiated with respect to b to find the first derivative

f '(b) = 4(4b 3 )+5(5b 4 ) +6( 6b 5) +2

By solving above terms

f '(b) = 16b 3 +25b 4 +36 b 5 + 2

The above function is again differentiated with respect to b to find second derivative

f ''(b)= 16(3b 2) +25(4b 3) +36 (5b 4)

f ''(b)= 48b 2 +100b 3 +180b 4 is the answer.

Example 3:

Determine the second derivative from the polynomial.

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

Solution:

The given equation is

f(b) = 2b6 + 2 b5 + 3 b4 + 3b

The above function is differentiated with respect to b to find the first derivative

f '(b) =  2(6b 5)  +2 (5 b4 ) +3(4 b3) + 3

By solving above terms

f '(b) =  12b 5  +  10b4  + 12 b3 – 3

The above function is again differentiated with respect to b to find second derivative

f ''(b) =  12(5b 4 ) – 10(4b3)  + 12(3b2)

f ''(b) =  60b 4 – 40b3 +36b2  is the answer.

Online second derivatives practice problems for study:

1) Determine the second derivative from the polynomial.

f(b) = b 3 + b 4 + b 5

Answer: f ''(b) = 6b +12b2+ 20b 3

2) Determine the second derivative from the polynomial.

f(b) = 2b 3+3b5 + 4b 6

Answer: f ''(b) = 12b + 60b3 + 120 b 4

Solving Negative Number

Definition:

Negative number is defined as the number which indicate by negative sign or minus ('-') . Negative number is less then zero and placed left to zero.

Ex: ...-5,-4,-3,-2,-1,0,1,2,3,4...

Comparison of Positive and Negative:

For each negative number , there is a positive number that is its opposite . Here we can write the opposite of negative number with a positive of same number or plus sign used In front of the number and call these numbers are positive numbers. Ex : 1,2,3 ,.....positive numbers are grater than zero. Similarly, the opposite of any positive number is a negative number .

Ex:    1,2,3 is -1,-2,-3.

Solving examples for negative numbers:

Zero cannot be taken as a negative number or positive number.
For every positive number x, there exists a negative number y such that x + y = 0
Positive number is denoted as plus ('+')sign and negative number is denoted as minus sign('-').
Example of negative number:-2,-43,-34 and example for positive number is 2,43,34.
negative and positive number may be written as mixed numbers or fraction numbers.

The equal fraction of negative numbers are given bellow:

(-3)/7,3/(-7),-(3/7) and -3/7 .

The equal mixed numbers are given bellow.

-2/5,-(2/5)       (-4)/9, 4/-(9) and -4/9

More about solving negative numbers:

Solving Addition of Negative Numbers:

To add the negative numbers which consist of minus sign. To provide the answer of addition of negative number.

Solving examples for addition negative numbers:

-12+(-6)=?

Solution:

-12+(-6)= -18.

Here the values of  -12 and -6are 12 and 6 adding the smaller from the larger gives -12+(-6)= -18, and since the larger  value was 12, so we can give result the same sign as -12, '-' so -12+(-6)= -18.

Example:

(-6) + (-6) = ?

Here the absolute values of -6 and -6 are 6 and 6.  Adding the smaller from the larger gives -6 - 6 = 12 ,  but here both has same value . In this case sign is the matter, here 12 and -12 are the not same and then -6 and -6 are same numbers. The property of all same number sum is 12. The addition of two number up to zero are called as additive inverses.

Multiplying Negative Numbers:

Solving example for multiplying negative numbers :

Product of negative number ,here we can take the product of their values.

(-3.3) × (-5) = ?


(-3.3) × (-5) = (-3.3) × (-5)

= (-3.3) × (-5)

=  +16.5.
Dividing Negative Numbers

Solving example for dividing negative number:

To divide two negative numbers, here we can divide the value of the first by the value of the second.

(-1.6) ÷ (-4) = (-1.6) ÷ (-4)

= (-1.6) ÷ (-4)

=  -0.4

Geometry Without Common Vertices

In geometry, some figures have common vertices. Mostly the geometric figures are without common vertices. If a triangle, quadrilateral and some geometric figures in a graph are lying without common vertices are refered same. In graph there are four quadrants in that some vertices are fall on common vertices, with out common vertices points of are plotted.

Geometric figure without common vertices

Without common vertices find the distance of two points

In geometry the triangle has three vertices; the vertices are not common vertices. The common vertices are formed only if two triangles are in same point without three common vertices. The distance between two un common vertices are find out by using the coordinates of the vertices (x1,x2) and (x2,x2) of the vertices.

Distance between two vertices = √(x2-x1)2 +(y2-y1)2

If the geometry figure having the common vertices in a graph



Examples for without common vertices

Distance of a vertices are find using distance formula:

Examples for distance between two vertices:

Ex 1:   Find the distance formed by without common vertices, Vertices A(4,5), B(7,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=4    X2=7    Y1=5    Y2=4

AB = √(7-4)2+(4-5)2

= √(3)2+ (-1)2

= √9+1

= √10 units

Ex 2 :  Find the distance formed by without common vertices, Vertices A(3,2), B(5,4)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=5    Y1=2    Y2=4

AB = √(5-3)2+(4-2)2

= √(2)2+ (2)2

= √4+4

= √8 units

Ex 3 :      Find the distance formed by without common vertices, Vertices A(6,4), B(10,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=6    X2=10    Y1=4    Y2=8

AB = √(10-6)2+(8-4)2

= √(4)2+ (4)2

= √16+16

= √32 units

Ex 4:    Find the distance formed by without common vertices, Vertices A(3,3), B(4,8)

Sol :            Distance AB = √(x2-x1)2 +(y2-y1)2

X1=3    X2=4    Y1=3    Y2=8

AB = √(4-3)2+(8-3)2

= √(1)2+ (5)2

= √1+25

= √26 units

Practice problems:

Q 1   Find the distance of two vertices A(1,1) B(1,2)   Answer: √1 units

Q 2   Find the distance of two vertices A(2,2) B(1,1)   Answer: √2 units

Straight Line Equation

The equation of straight line is generally written as    y = mx + b

Graphical Representation

where, m= slope or gradient of the straight line equation

b = the y-intercept

Suppose we want to find equation of a straight line that passes through a known point and has a known slope. Let (x ,y) represent the co-ordinates of any point on the line and let (x1 ,y1) represent the co-ordinates of other point. The slope of the straight line equation is given as,

m = (y-y_1)/(x-x_1)

After finding the slope m as we are given the co-ordinates of the point (x1 ,y1) in the equation    y= mx + b Then, the constant 'b' can also be found so that finally the straight line equation is obtained.

Other Forms of straight Line Equations

Different Forms of Straight Line Equations:

There are many other forms of Straight Line Equations.

1. Straight Line Equation through two points:

The line through two different points ( x1 ,y1) and ( x2 ,y2) is given by

y-y1  = [(y_2-y_1)/(x_2-x_1)]  . (x - x1)

2. Straight Line Equation in general form:

A straight line is defined by a linear equation as

Ax+By+C=0     where A , B are not both 0

3. Straight Line Equation in Intercept-intercept form:

Let us consider that a straight line intersects x-axis at (a , 0) and y-axis at (0 , b). Then it is defined by equation

(x/a) +(y/b) = 1

4. Straight Line Equation in Point-Slope Form:

The equation of straight line through the point (a , b) with slope m is

y = m ( x - a ) + b

Solved Examples on equations of straight line

Ex:1 Find the equation of a line passing through (2 , 3) and having a slope of 3?

sol: step 1: Compare the given point(2 , 3) with the general point (x1 , y1)

Now x1=2  and  y1=3

step 2: we have the straight line equation as  y-y1=m(x-x1)

now substitute the given point in the above equation

y-3=3(x-2)

y-3=3x-6

3x-y=6-3

3x-y-3=0

The straight line equation is 3x-y-3=0

Ex:2 Find out the straight line equation of the line passing through the points (1,2) and (2,4).

Sol: Given points are compared with (x1 , y1) and (x2 , y2) and substitute the points in the equation

y-y1  = [(y_2-y_1)/(x_2-x_1) ] . (x-x1)

y-2  =  [(4-2)/(2-1)] . (x-1)

y-2 = 2(x-1)

y-2 = 2x-2

2x - y = 0

The straight line equation is 2x-y=0.

Practice Problems on equations of straight line

Pro:1 Find the equation of the line passing through the points (-3,4) and (4,-2)

Ans: we have      

m = (y_2-y_1)/(x_2-x_1)

=(-2-4)/(4+3)

=(-6/7)

let (x,y) be compared to (-3,4)

y-4 = (-6/7) . (x-(-3))

7(y-4) = -6(x+3)

7y-28 = -6x-18

7y-28 = -6x-18

7y+6x = 10

Pro:2 Write equation of line having points and slopes as follows;

P(3,3) , m=-2

P(-2,-1) , m=1/3

P1(2,2) and P2(-4,-1)

y-intercept = 2 , m=3

Ans: The answers to the above given practice problems are

y+2x=11

3y-x=-1

2y-x=2

y-3x=2

Hyperbola Axis

The locus of a point whose distance from a point rest at center shows an unaltered ratio, higher than one to its length from the fixed line is known as hyperbola. In this article, you can learn about the axis of hyperbola, general equation of the hyperbola and definitions regarding hyperbola.

Axis of Hyperbola:

Transverse Axis:

Transverse axis can be defined as the line segment, which joins the vertices of the hyperbola. 2a is calculated as the difference between two vertices. The equation of the transverse axis is y = 0. This is showing that y-coordinate is zero.

Conjugate Axis:

Conjugate axis can be defined as the line segment, which joins the y-coordinates of the hyperbola. The distance between the two coordinates are 2a (i.e., the length of the Conjugate axis is 2a.). x = 0 is considered as the equation of conjugate axis. Therefore the x-coordinates of Conjugate axis is zero.

General Equation of the Hyperbola:

Some important points to be considered:

Fixed point is represented as F.
Fixed line as l.
Eccentricity as e, then it should be greater 1.
Moving point is represented as P(x,y).

Steps:

Fixed point F is plotted and also the fixed line ‘l’ is drawn.
Perpendicular (FZ) is dropped from F to l.
As next step drop PM which is perpendicular one from P to l.
Plot the points A, A’ which divides FZ internally and externally in the ratio e : 1            
respectively.
Take AA’ = 2a and treat it as x-axis.
Draw a perpendicular bisector of AA’ and treat it as y-axis.
Consider C as the origin, then the known points are C(0,0), A(a,0) and A’(-a,0).
The general equation of the hyperbola x^2/a^2 -y^2/b^2 = 1.

Definitions Regarding Hyperbola:

Focus:

The fixed point F is known as the focus of the hyperbola.

Directrix:

The directrix is nothing but the fixed line. Then the directrix equation is given by x = a/e.

Centre:

The centre of the hyperbola is a point, at which the transverse and conjugate axes intersect and they are represented as ‘C’.

Vertices:

The vertices of the hyperbola are the points, where the curve and its transverse axis meet. The vertices are A(a,0) and A’(-a,0).

Studying Standard Deviation Examples

The standard deviation of a statistical population, a data set, or a probability distribution is the square root of its variance. Standard deviation is a widely used measure of the variability or dispersion, being algebraically more tractable though practically less robust than the expected deviation or average absolute deviation. It shows how much variation there is from the "average" (mean) (or expected/budgeted value). It helps detect tampering of data. Examples for studying standard deviation is given below.

Formula for studying standard deviation examples:

Formula for studying standard deviation examples are defined below:

Measurement of standard deviation is prepared by taking square root for the addition of mean difference with the certain data divided by the total number of values subtracted by one. The following formula for standard deviation as shows given below.

s=v?(X-M)2/n-1

Here S = Sum of values

X = Individual value

M = Mean of total all value

N = Sample size (Total number of values)

Variance:

Variance = s2

Steps for calculating Standard Deviation examples:

• Step 1: calculate the average for given n numbers using the formula this is called mean of given numbers

• Step 2: Find distance between each given numbers in the Data set from the calculated average value. This is called  "deviation" from the mean value.

• Step 3: Take the Square of each deviation value found from mean. This is squared deviation from mean.

• Step 4: Calculate the sum for all the squared standard deviations.

• Step 5: Now apply the Standard Deviation formula and find standard deviation formula. It will be the square root of variance.

Examples for Studying Standard deviation:

Examples for Studying Standard deviation are as follows:

Pro 1:   Here are 4 measurements 4, 6, 7, 9 and 10Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
4+6+7+9+ 10
x =   ------------------------
5-1

= 36/4

= 9

Standard Deviation,
v (4-9)2 + (6-9)2 + (7-9)2 + (9-9)2 + (10-9)2
S=  -------------------------------------------------------------
5-1

= v 81 /4

= v 20.25

=  4.5

Standard Deviation  S = 2.12132

Pro 2:  Here are 4 measurements 10, 20, 30, 40 and 50 Calculate the Standard Deviation

Sol :     Mean: Calculate the average for the given values. To find the mean.
10+20+30+40+60
x = ---------------------------
5-1

= 160/4

= 40

Standard Deviation,
v (10-40)2 + (20-40)2 + (30-40)2 + (40-40)2 + (60-40)2
S=  -----------------------------------------------------------------------------
5-1

= v 1600 /4

= v 400

S = 20

Standard Deviation S = 20

Absolutely Continuous Function

DEFINITION

Let [p,q] be the close bounded interval of C. Then a function f:[p,q]→ R will be an absolutely continuous function on [p,q], if for any δ>0 there will be a ε>0 such that the certain conditions which are mentioned below holds good

If (p1q1)..............(pnqn) is a collection which is finite with disjoint open intervals in [p,q] such that

Σni=1 (qi-pi) < ε

and

Σni=1 |f(qi)-f(pi)| < δ



EQUIVALENT DEFINITIONS

The condition on a real-valued function  "  f  "  on the compact interval [ p, q ] are equivalent if

1) f is an absolutely continuous function

2) ' f ' has a derivative f1 almost everywhere which is a Lebesgue integral and

f ( x ) = f ( p ) +  ∫x a  f1  ( c ) dt

for all x on [ p , q ]

3)There exists a Lebesgue integrable function such that g on [ p , q ] such that f ( x ) = f ( p ) + ∫x p  g ( c ) dt

for all x on [ p , q ]

If these conditions are satisfied by the function the definitely g = f' almost everywhere

Properties of the absolute continuous function

PROPERTIES OF ABSOLUTELY CONTINUOUS FUNCTION

1. The sum and difference of two absolute continuous function are also absolutely continuous. The products of two absolute continuous function defined on the bounded interval will also be a absolute continuous function.

2. If an absolutely absolute continuous function is defined on a bounded closed interval and is nowhere zero then its reciprocal is also an absolutely continuous function.

3. Every absolutely continuous function is an uniformly continuous function.

4. If  f: [ p , q ] → R is absolutely continuous, then it will be a function of the bounded variation on [ p,q ]

Learning Geometric Probability

Numerical measure of the likelihood of an event to occur is called as Probability. The probability should be a range in between 0 and 1.In this case we say probability is 0. If the event is certain to occur, we say probability is likely 1. The probabilities involved in the geometric problem that also called as geometric probability. This geometry probability may be a circle or any polygon from the geometric. It involves the length, area and volume of any one of the geometric shapes.

The definition of probability of an event has shown in below that is depend on their outcomes from the possibilities



Number of successful outcomes

Probability  =        _____________________________

Total number of possible outcomes.

learning geometric probability example problem in triangle:

The figure shows a triangle divided into sectors of different colours. Find the probability of angle for blue sector and orange sector?

learning geometric probability

Solution:

We have to find the Probability for blue color:

Step 1:

Total angle of triangle is 180 degree

Step: 2

Probability finding blue color sector= The blue sector triangle angle/ The total angle of  entire triangle

The blue sector triangle angle=30 degree

=30/360

=1/12

Step 3:

Next we have to find Probability for orange color:

Probability finding orange color sector= The orange sector triangle angle/ The total angle of entire triangle

The orange sector triangle angle=45

=45/360

=9/72

learning geometric probability example in Rectangle:

Example 2:

A rectangle has four sides and that corner having each 4 ball. Find the probability of each corner having 4 balls.?

Solution:

Here the Rectangle has the five corners such as A, B, C, D

We have to find the probability here,

Probability = Number of successful outcomes / total number of possible outcome

Probability  of each corner having the balls=4/4=1

Example 3:

A triangle with area 15 cm2 is inscribed in a circle with radius 3 Cm.Find the probability that a ball thrown fall into the triangle?

Solution:

We have to find the area of the circle and area of the triangle is given. From this we can find the probability of a ball that fall into the triangle easily.

Area of the circle = `Pi` r2

Radius  = 3. So Area of the circle = 3.14 * 3 * 3 = 28.26 cm2.

Area of the triangle  = 15 cm2.

So the probability  = 15 / 28.26 = 0.53

Answer is 0.53.

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How to solve a percent problem

In mathematics, a percentage is a way of expressing a number as a fraction of 100 (per cent meaning "per hundred" in French). It is often denoted using the percent sign, "%", or the abbreviation "pct". For example, 45% (read as "forty-five percent") is equal to 45 / 100, or 0.45. Let us see how to solve percent problem.                                                                                                                                - Source from Wikipedia

Formula – How to solve percent problem:

To solve the percent problem convert the fraction else a decimal to a percent and then multiply it by 100:

To get the percent value multiply the fraction into hundred. For example

` 2 / 4` (100) = `2 / 4 ` * 100 = 0.5 * 100 = 50%

To convert the percent value to fraction divide it by hundred.

25% = `25 / 100 ` = `1 / 4`

Example: How to solve percent problem:

Problem 1:

What is the decimal value of 72%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `72 / 100` = 0.72.

Therefore the decimal value of 72% is 0.72.

Problem 2:

What is the decimal value of 86.3%?

Solution:

To obtain the decimal value we have to divide the given number by 100. Which is `(86.3) / 100` = 0.863.

Therefore the decimal value of 86.3% is 0.863.

Problem 3:

What is the decimal value 18% of 60?

Solution:

To find 18 percentages divide 18 by 100 and then multiply it by 60.

= `18 / 100` * 60 = 10.8

Hence 18% of 60 = 10.8

Problem 4:

What is the decimal value for 15.5% of 150?

Solution:

To find 15.5 percent divide 15.5 by 100 and then multiply it by 150.

=` (15.5) / 100` * 150 = 23.25

Hence 15.5% of 150 = 23.25

Problem 5:

The book cost is 5.00 dollars. Now it was 15% increased. What will be the new cost?

Solution:

Multiply the cost by 15%. That is `15 / 100` * 5.00.

= `15 / 100` * 5.00 = 0.75

Add this decimal value with original cost (5.00+0.75= 5.75)

Therefore the original cost of the book is 5.75 dollars.

Problem 6:

In a box there are 85 chocolates. Suppose if a boy took out 60% of chocolates, how many chocolates did he left in the box?

Solution:

The boy took out the chocolates from the box is 60% of 85 or else 60 / 100 × 85

`60 / 100` × 85 = 51%

So the box has 85 chocolates and the boy took out 51 chocolates. Hence the number of chocolates the boy left in the box is 85 – 51 = 34.
Therefore the boy left 34 chocolates in the box.

Problem 7:

A theatre contains 530 chairs. 470 of them are occupied. What percentages of the chairs not occupied?

Solution:

Number of chairs                        = 530

Number of chairs occupied       = 470

Number of chairs not occupied = 530 – 470 = 60

Number of chairs not occupied percentage = `60 / 530` × 100 = 11.32%
Practice problems – How to solve percent problem:

Problem 1:

What is the decimal value of 27%?

Solution:

27% = 0.27

Problem 2:

What is the decimal value 8% of 75?

Solution:

8% of 75 = 6

What is the Associative Property

The addition or multiplication of a collection of integer is the same separately from of how the numbers are grouped. A binary operation * is assumed to be associative if any three elements a, b, c of a set a *(b * c) = (a * b) * c. Multiplication of real integers are associative but the division of real integers are not assosiative. And addition is associative but subtraction is not.

What is the Associative Properties

What is the associative property?..The associative property is a property which always involve 3 or more numbers in calculations. The parenthesis indicates the terms that are consider one unit. The gathering (Associative Property) are within the parenthesis. Hence, the figures are 'associated' together. In multiplication, the result is always the same regardless of their combination.

There were two properties are involved in what is the associative property,

They were, Associative property in addition

Associative property in multiplication


Examples for what is the Associative Property (Addition and Multiplication)

Example for what is the Associative property in addition:

Let a, b, c be any real numbers, then

A + (B + C) = (A +B) + C

Let A = 5, B =9, C = 8

L.H.S = 5 + (9 + 8) = 5 + (17) = 22

R.H.S = (5 + 9) + 8 = (14) + 8 = 22

L.H.S = R.H.S

More Examples for addition:

When we change the groupings of addends, the sum does not change:

(8 + 5) + 9 = 22               (or)

8 + (5 + 9)  = 22

(9 + 1) +8 = 18                 (or)

9 + (1 + 8) = 18
Just consider that when the grouping of addends vary, the sum leftovers the same.

Example for what is the Associative property in multiplication:

Let A, B, C be any real numbers, then

A * (B * C) = (A *B) * C

Let, A = 5, B =9, C = 8

L.H.S = 5 * (9 * 8) = 5 * (72) = 360

R.H.S = (5 * 9) * 8 = (45) * 8 = 360

L.H.S = R.H.S

More Examples for Multiplication:

When we change the groupings of factors, the product does not change:

(1 x 9) x 5 = 45          or

1 x (9 x 5) = 45

(2 * 5) * 10 = 100      or

2 * (5 * 10) = 100

Just remember that when the grouping of factors changes, the product remains the same.

Simple Math Multiplication

In mathematics multiplication is used to multiple the numeric values or numbers. Multiplication can also be visualize as counting objects set in a rectangle or as result the area of a four-sided figure whose sides have given lengths. The area of a four-sided figure is not depend on which side is calculated first which illustrates that the order in which values are multiply as one does not matter

Multiplication examples:

Example 1:12 X 8

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 8 = 8.

Step 2: Multiply this 8 by 10 giving 80.

Step 3: Now multiply the 8 by the 2 of twelve giving 16. Add this to 80 giving 84.

Therefore 8 X 12 = 96

Example 2: 18 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 18. So 1 X 8 =1 8.

Step 2: Multiply this 18 by 10 giving 180.

Step 3: Now multiply the 18 by the 2 of twelve giving 36. Add this to 180 giving 216.

Therefore 18 X 12 = 216

Example 3 :12 X 9

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So 1 X 9 = 9.

Step 2: Multiply this 9 by 10 giving 90.

Step 3: Now multiply the 9 by the 2 of twelve giving 18. Add this to 70 giving 108.

Therefore 9 X 12 = 108

Sample examples:

Example 4: 19 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 9 =1 9.

Step 2: Multiply this 19 by 10 giving 190.

Step 3: Now multiply the 19 by the 2 of twelve giving 34. Add this to 170 giving 228.

Therefore 19 X 12 = 228

Example 5 :23 X 7

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 7. So, 1 X 7 = 7.

Step 2: Multiply this 7 by 10 giving 70.

Step 3: Now multiply the 7 by the 2 of twelve giving 14. Add this to 70 giving 161.

Therefore 7 X 23 = 161

Example 6: 20 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 20. So 2 X 10 = 20.

Step 2: Multiply this 20 by 10 giving 170.

Step 3: Now multiply the 17 by the 2 of twelve giving 34. Add this to 170 giving 240.

Therefore 17 X 12 = 240

Example 7 :12 X 10

Step 1: Multiply the 1 of the 12 by the

number we are multiplying by, in this case 7. So 1 X 10 = 10.

Step 2: Multiply this 12 by 10 giving 120.

Step 3: Now multiply the 10 by the 2 of twelve giving 20. Add this to 70 giving 84.

Therefore 12 X 10 = 120.

Example 8: 30 X 12

Step 1: Multiply the 1 of the 12 by the
number we are multiplying by, in this case 17. So 1 X 30 =30.

Step 2: Multiply this 30 by 12 giving 360.

Step 3: Now multiply the 30 by the 2 of twelve giving 60. Add this to 170 giving 360.

Therefore 17 X 12 = 360.

Complex Analysis Problems

Complex analysis is the functions of the complex numbers. Complex numbers have the real and also the imaginary parts. The complex numbers can be represented as x +i y, where x denoted as the real part and y denoted as the imaginary part. The complex numbers comprises the addition of two complex numbers, subtraction of two complex numbers, multiplication of two complex numbers and also the division of two complex numbers. The conception of the complex numbers is the reflection of the fact. This article has the functions of the complex numbers.

Examples for learn complex analysis problems:

Example 1 to learn complex analysis problems:

Compute the value for the complex number f (x) = (75+42i) + (90+72i).

Solution:

The given complex number is f (x) = (75+42i) + (90+72i).

Step 1: (75+42i) + (90+72i) = (75+90) + (42i + 72i)

Step 2: (75+42i) + (90+72i) = 165 + (42i + 72i)

Step 3: (75+42i) + (90+72i) = 165 +114i

The value for the complex number f (x) = (75+42i) + (90+72i) is f (x) = 165 +114i.

Example 2 to learn complex analysis problems:

Resolve the value for the complex numbers f (x) = (13+i) x (11-i).

Solution:

The given complex numbers are f (x) = (13+i) x (11-i).

Step 1: (13+i) x (11-i) = 13(11 - i) + i (11 -i)

Step 2: (13+i) x (11-i) = (143- 13 i) + (11i - i2)

Step 3: (13+i) x (11-i) = 143 - 13i + 11i - (-1) (where i2= -1)

Step 4: (13+i) x (11-i) = (143+1) + (-13i +11i)

Step 5: (13+i) x (11-i) = 144 - 2i

The value for the complex number f (x) = (13+i) x (11-i) is f (x) = 144 - 2i.

Example 3 to learn complex analysis problems:

Calculate the value for the complex number f (x) = (180+53i) - (82+ 7i).

Solution:

The given complex number is f (x) = (180+53i) - (82+ 7i).

Step 1: (180+53i) - (82+ 7i) = (180-82) + (53i - 7i)

Step 2: (180+53i) - (82+ 7i) = 98 +46i

The value for the complex number f (x) = (180+53i) - (82+ 7i) is f (x) = 98 +46i.

Practice problem for learn complex analysis problems:

Compute the value for the complex numbers f (x) = (45+29i)-(2+7i).

Answer: f (x) = 43- 22i

Predict the value for the complex numbers f (x) = (145+88i) + (33+39i).

Answer: f (x) = 178+127i

Compute the value for the complex numbers f (x) = (5+3i) x (2+i).

Answer: f (x) = 7+11i

Calculate Ratio Math

In mathematics, a ratio expresses the magnitude of quantities relative to each other. Specifically, the ratio of two quantities indicates how many times the first quantity is contained in the second and may be expressed algebraically as their quotient.

Example:

For every Spoon of sugar, you need 2 spoons of flour (1:2)

Source: Wikipedia

Definition- calculate ratio math:

The ration of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the terms of the ratio.

Concept - calculate ratio math:

The numeric ratio of two numbers r and s(s≠0) is the section of the numbers. The numbers r and s are called the conditions of the numeric ratio.

Types of ratio- calculate ratio math:

Compounded ratio in math.

Duplicate ratio in math.

Triplicate ratio in math.

Define compounded ratio in math:

Regular format for compounded numeric ratio is

p/q *r/s=(pr)/(qs)

Example for compounded ratio in math:

How to calculate ratios: 7/5*3/3

Solution:

=7/5*3/3

=(7*3)/(5*3)

=21/15

=7/5 or 7:5

Define duplicate ratio in math:

Regular format for duplicate ratio is

(r/s)*(r/s)=(r^2)/(s^2)

Example for duplicate ratio in math:

How to calculate ratios: 6/5*6/5

Solution:

=6/5*6/5

=(6^2)/(5^2)

=36/25 or 36:25

Define triplicate ratio in math:

Regular format for triplicates ratio is

r/sr/sr/s=(r^3)/(s^3)

Example for triplicate ratios in math:

How to calculate ratios: 5/7 * 5/7 * 5/7

Solution:

=(5*5*5)/(7*7*7)

=(5^3)/(7^3) or 5^3:7^3

=125/343   or 125:343

More about ratios- calculate ratio math:

Define Inverse ratio in math:

It is often want to estimate the numbers of a ratio in the inverse order. To perform this, we simply switch over the numerator and the denominator. Therefore, the inverse of 21:7 is 7:21. When the terms of a ratio are switch over are called  the INVERSE NUMERIC RATIO results.

Example problem for inverse ratios in math:

In math, how to calculate inverse numeric ratios:  7/21*7/2

Solution:

=7/21*7/2    (7 Divided by both the numerator and denominator)

=(7*7)/(21*2)

=49/42

=7/6 or 7:6

Answer is 7/6 or 7:6

Function Examples Math

Let f be a function from X to Y.  Then it have to satisfy the following two conditions.

(i) No two different ordered pairs in f have the same first element.

(ii) All the elements of X will occur as the first elements in f.

Now let us discuss few examples that satisfies these conditions.


Example problems on functions:

Ex 1: Check whether the following ordered pair represents a function:

Here X = {1, 2, 3, 4}

Y = {a, b, c}

(i) {(1, a), (2, b), (3, c), (4, c)}.

(ii) {(1, a), (2, c), (3, b)}.

Sol: Given: (i) {(1, a), (2, b), (3, c), (4, c)}

Here all the first values are from x.  All the first values are different.  Therefore it is a function.

(ii) {(1, a), (2, c), (3, b)}.

Here 4 of X is not there in the ordered pair.  Therefore, this cannot be a function.

Ex 2: Let X = {7, 8} and Y = {1, 2}. Which one of the following is not a function of

f from X to Y.

(i) {(7, 1), (8, 2), (7, 2)}

(ii) {(7, 1), (8, 2)}

(iii) {(7, 1), (8, 1)}

Sol: (i) {(7, 1), (8, 2), (7, 2)}.  This is not a function from X to Y.  Because 7 is repeated in the first element.

(ii) {(7, 1), (8, 2)}.  This satisfies the definition of the function. Hence it is a function.

(iii) {(7, 1), (8, 1)}. This satisfies the definition of the function.  Hence it is a function.


More example problems on functions:

Ex 3: Let X = { 1,2,3,4,5,6,7,8,9} and  Y = {3,6,9}

R be a relation “is three times of”, find R on X.

Can this be a function {(3, 1), (6, 2), (9, 3)}?

Sol: R = {(3, 1), (6, 2), (9, 3)}.

Since here {(3, 1), (6, 2), (9, 3)} the first values are different, this can be a function from Y to X.

Continuous Probability Distribution

In statististics if we have the finite number of set and the probability of the set is called Discrete probability distribution. For an example for a cars on a road, deaths by cancer, tosses until a die shows a first 6. If we measure anything it is called as continuous probability distribution function. For an example we can to say the electric voltage, rainfall, and hardness of steel. In both cases we can determined the distribution by the following distribution function

F(x) = P(X ≤ x)

This is the probability that X will assume any value not exceeding x.

Continuous random variables and probability distribution:

Discrete random variables appear in experiments which is having finite set (defectives in a production, days of sun shines in Chicago, customers standing in a line etc.). Continuous random variables is appear in the experiments which we can't count but we can measure (length of screws, voltage in a power etc.). If we are find the the value probability for  the continuous  variable then it is called continuous probability distribution. By the definition of a random variable of X and its for the distribution are of continuous type will be defined as the following integral.

F(x) = `int_-oo^xf(v)dv`

it is called density of the distribution, is non negative, and is continuous perhaps except for finitely many x-values.

The continuous random variables are simpler than the discrete ones with respect to intervals. Indeed  in the continuous case the continuous probability distribution of the four probabilities corresponding to a < X ≤ b, a < X < b, a ≤ X < b, a ≤ X ≤ b with any fixed a and b (> a) are all same.

Sample problem for probability distribution:

Continuous probability distribution problem 1:

Let X have the density function f(x) = .75(1 - x2) if  -1 ≤ x ≤ 1 and zero otherwise. Find the value of distribution function. Find the continuous probability distribution of P(-1/2 ≤ X ≤ 1/2). Find x such that P(X ≤ x) = 0.95.

Solution:

F(x) = `int_-1^x`(1 - v2) dv= 0.5 + 0.75 x2 -.25x3              if  -1 ≤ x ≤ 1,

And F(x) = 1 if x > 1

P(-1/2 ≤ x ≤ 1/2) = F(1/2)  - F(-1/2) =  `int_(-1/2)^(1/2)`(0.5 + 0.75 x2 -.25x3) dx =  68.75%

Because for continuous probability distribution    P(-1/2 ≤ x ≤ 1/2) =    P(-1/2 < x ≤ 1/2)

Continuous probability distribution P(X = x) = F(x) = 0.5 + 0.75x -1.25x3  = 0.95

If we solve this we will get x = 0.73

Domain and Range of Trigonometric Functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant   Since tan ?=sin?/cos?  and sec?=1/cos?

,cos? being in denominator tan? and sec?  is not defined when cos? iszero

that is ? being odd multiple of ?/2 R- (2n + 1)/2| n`in`  Z

that is these are not defined at odd multiple of pi /2

and range of these functions is set of  all real numbers

Domain of cotangent and cosecant  since cot?= cos?/sin? and cosec?=1/sin?

sin? being in the denominator ,cot ? and cosec? is not defined when sin? is zero

that is  ? being multiple of ?

therefore for cot and cosec to be defined multiple of ? are dicarded

and range of these functions is all real numbers except interval  [-1,1]

Trignometric functions domain and range

Domain and Range of Trigonometric Functions




Explanation for domain and range of some trignometric functions

Domain of sine and cosine is set of all real numbers means it is defined for all values which is real

and domain of these function is [-1,1] means that whatever we  is put from domain ,we always get value with in it

 Domain of Tangent and Secant is R- {`pi` (2n + 1)/2| n `in`  Z } that is these are not defined at odd multiple of `pi`/2

and range of these functions is all values except (-1,1) interval

same can be explained  for cosecant and cotangent.

Learn Divisibility Tests

Divisibility tests means that to discover a number is divisible by a particular number (divisor) without perform the division operation. The rules shown below are used to translate a given number into a normally smaller number while preserve divisibility by the divisor. There are divisibility test for numbers in any radix, and they are all dissimilar, we represent rules simply for decimal numbers below. In this article we shall learn about divisibility tests rules with examples.

Divisibility tests rules

To learn a number is divisible by 2:

The number is end with even number (that is 0,2,4,6) the number are divisible by 2.

To learn a number is divisible by 3:

The sum of the digit is multiple of 3. The given number is divisible by 3.

To learn a number is divisible by 4:

The numbers are formed by the final pairs of digits is divisible by 4.

Example:

4672 => 72 ÷ 4 = 18

So 4672 ÷ 4 = 1168

A number is divisible by 8:

The numbers are formed by the final three digits are equally divisible by 8.

Example:

53,104 => 104 ÷ 8 = 13

So 53,104 ÷ 8 = 6638

To learn a number is divisible by 9:

The sum of the digits is a multiple of 9.

Example: 3,726

3,726 => 3 + 7 + 2 + 6 = 18

Since 18 = 9 × 2,

Then 3,726 ÷ 9 = 414

To learn a number is divisible by 6:

The numbers satisfy the rule for 2 and 3; that is, first it must be an even number, then a digit sum is a multiple of 3.

To learn a number is divisible by 12:

The numbers satisfy the rules for 3 and 4; that is, the digit sum is a multiple of 3, and its final digit pair is a multiple of 4.

Divisible tests practice problem

Problem1:

Check whether the given number 4652 is divisible by 4

Answer:

4652 => 52 ÷ 4 = 18

So 4672 ÷ 4 = 1163 therefore the given number is divisible by 4.

Problem2:

Check whether the given number 53112 is divisible by 8

Answer:

53,112 => 112 ÷ 8 = 14

So 53,112 ÷ 8 = 6639 therefore the given number is divisible by 8.

Vertical Stretch and Compression

Vertical stretch and compression mean transforming the graph based on the scale factors. Here we will see how we are performing the stretching and compression of graph using the scale factor. Stretching the graph is nothing but we are transforming the graph away from axis. Compression of the graph means squeezing the given graph towards the axis. We will see some example problems foe vertical stretch and compression of the graphs.
Vertical Stretch:

Vertical stretch is nothing but the stretching the graph away from x – axis. If the given function is f(x) then the vertical stretch of the given function is y = a f(x). Where 0 < a < 1

Example for vertical stretch:

Graph the following function and its vertical stretch. Where f(x) = x2 – 1.6x and the vertical stretch scale factor of the function f (x) is 0.4.

Solution:

Given function f(x) = x2 – 1.6x

We can write the given functions like y = x2 – 1.6x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the given function

We have to plug y = 0

So 0 = x2 – 1.6x

x (x – 1.6) = 0

x = 0 and x = 1.6

So the x intercept point is (0, 0) and (1.6, 0)

For y – intercept of the given function

We have to plug x = 0

So y = (x)2 – 1.6(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical stretch function.

The vertical stretch function is

f(x) = a f (x)

So y = 0.4 (x2 – 1.6x)

If we graph both the function we will get the following graph

vertical stretch - stretch graph
Vertical Compression:

Vertical compression is nothing but the squeezing the graph towards the x – axis. If the given function is f(x) then the vertical compression of the given function is y = af(x). Where a > 1

Example for vertical compression:

Graph the following function and its vertical compression. Where f(x) = x2 – 3.5x and the vertical compression scale factor of the function f (x) is 1.25

Solution:

The given function f(x) = x2 – 3.5x

We can write the given functions like y = x2 – 3.5x

If we want to graph the function we have to find the x and y intercept of the original function.

For x – intercept of the original function

We have to plug y = 0

So 0 = x2 – 3.5x

x (x – 3.5) = 0

x = 0 and x = 3.5

So the x intercept point is (0, 0) and (3.5, 0)

For y – intercept of the original function

We have to plug x = 0

So y = (x)2 – 3.5(0) = 0

So y intercept point is (0, 0)

Now we have to graph the vertical compression function.

The vertical compression function is

f(x) = a f (x)

So y = 1.25 (x2 – 3.5x)

Solve Vector Spherical Coordinates

This article is to solve vector spherical coordinates. Solve vector spherical coordinates is nothing but solving problems under spherical coordinate with the help of the tutors. Tutor vista tutor have many highly qualified tutors in online. A quantity which has magnitude and direction is a vector. There are three methods  to describe a vector.  Specific lengths, direction, angles, projections or components are the methods. Below we can see about solve vector spherical coordinates.

Solve Vector Spherical Coordinates

From this system any point as the point of intersection of the spherical surface (radius r = constant) conical surface (theta phitheta and phi by dr, rd theta and dphi .The element sides are dr,rd theta and rsin theta dphi . = constant). A differential volume element is obtained in spherical coordinate. This happens by increasing r value,

The differential length dl = sqrt(dr^2+(rd theta)^2+(rsin theta dphi)^2)

The differential areas ds = dr.rd theta = rdrd theta

= dr.rsin theta dphi = r sin theta dphi dr

= rd theta. rsin theta dphi = r2sin2 theta d theta dphi

The differential volume dv = dr.rd theta.rsin theta dphi

dv = r2sin theta dphi dr.

So spherical co ordinates = (r, theta, phi)
Solve Vector Spherical Coordinates

A vector in cartesian co-ordinate system can be converted to spherical coordinate.

Conversion of  cartesian to spherical system

The Cartesian co-ordinates is (x,y,z). This cartesian coordinate is converted into spherical co-ordinates ( r, theta, phi )

Given                                  Transform

x                                     r = sqrt (x2+y2+z2)

y                                     theta = cos-1(z/(sqrt(x2+y2+z2))) = cos-1(z/r)

z                                    phi = tan-1(y/x)

Problem 1: Convert the cartesian coordinates x = 2, y = 1, z = 3 into spherical co-ordinates.

Given                        Transform

x = 2                           r = sqrt (x2+y2+z2)

= sqrt (4+1+9)

= sqrt (14) = 3.74

y = 1                            cos-1(z/r) =  cos-1(3/sqrt(14)) =  36.7°

z = 3                           phi = tan-1(y/x) tan-1(1/2)26.56°

Spherical co- ordinates are (3.74, 36.7°, 26.56°)