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.