Examples of Functions

Introduction to examples of functions:

A function is a special type of relation. In a function, no two ordered pairs can have the same initial element and a different second element. That is, for a function, corresponding to each first element of the ordered pairs, there must be a different second element. i.e. In a function we cannot have ordered pairs of the form (a1, b1) and (a2, b2) with a1 = a2 and b1 ? b2.

Types of functions:(examples of functions)


1. Onto function

2. One-to-one function:

3. Identity function:

4. Constant function

5. Linear function:

6. Trigonometrical functions:


Explanation of Types of functions: examples of functions


1. Onto function

If the range of a function is equal to the co-domain then the function is

called an onto function. Otherwise it is called an into function.


Definition: A function f is onto if to each element b in the co-domain, there is

at least one element a in the domain such that b = f(a)

Example for Onto function


2. One-to-one function:

A function is said to be one-to-one if each element of the range is

associated with exactly one element of the domain.

i.e. two different elements in the domain (A) have different images in the

co-domain (B).



Note: (1) A function is said to be injective if it is one-to-one.

(2) It is said to be injective if it is both one-to-one and onto.

Example for one to one function:


3. Identity function:

A function f from a set A to the same set A is said to be an identity

function if f(x) = x for all x e A i.e. f : A ? A is defined by f(x) = x for all

x e A. Identity function is denoted by IA or simply I. Therefore I(x) = x always.

Example of Identity function:

f(x) = x

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4. Constant function:

If the range of a function is a singleton set then the function is called a

constant function.

Example for constant function:

If

f(x) = 8

then

f (13) = 8

Here f is called the constant function. Whatever comes in to f, the number 8 comes out.


5. Linear function:

If a function f : R ? R is defined in the form f(x) = ax + b then the function

is called a linear function. Here a and b are constants.


6. Trigonometrical functions:

In Trigonometry, we have two types of functions.

(1) Circular functions (2)Hyperbolic functions.

We will discuss circular functions only. The circular functions are

Examples of trigonometrical functions are:

(a) f(x) = sinx (b) f(x) = cos x (c) f(x) = tan x

(d) f(x) = secx (e) f(x) = cosecx (f) f(x) = cotx

Automatic Math Solutions

Introduction to automatic math solutions:
Automatic math solution deals with solving math problems in online. Only through online we get automatic math solution for the given problem. We get solution for any kind of math problems automatically. The following are some of the example problems which give the solution for the given question automatically. Here mathematics solution includes calculus problem, linear algebra problem and algebra problem. Some of the solved problems are given below. I like to share this Combinations Calculator with you all through my article.


Automatic math solutions examples:

Example 1:

Find the y value from the equation

(y - 4)(y + 8) = - y + 4

Solution:

Given equation with slight changes in the right hand side

(y - 4) (y + 8) = - (y - 4)

Make the above equation with right side equal to zero.

(y - 4)(y + 8) + (y - 4) = 0

Take (y – 4) as common factor.

(y - 4)(y + 8 + 1) = 0

Simplify the above equation
y - 4 = 0

y + 9 = 0

To obtain
y = 4
or
y = 9

Example 2:

Solve the algebraic expression

4(x -1) + 2y - 5(x -y -4) + 5

Solution:

Given algebraic expression is

4(x -1) + 2y - 5(x -y -4) + 5

Multiplying the integer terms

= 4x - 4 + 2y -5x + 5y + 20 + 5

Grouping the above terms

= -x + 7y + 21

Example 3:

Solve the Expression     -2(c - 3) – 6c - 1 = 8(c +2) – 2c

Solution:

Given equation is
-2(c - 3) – 6c - 1 = 8(c + 2) – 2c

Multiplying the integer term
-2c + 6 – 6c - 1 = 8c + 16 – 2c

Grouping the above terms
-8c + 5 = 6c + 16

Subtract 6c + 5 on both sides
-8c + 5 – 6c -16 = 6c +16 -6c-16

Grouping the above terms
-14c = 11

C = -11 / 14

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Automatic math solutions practice problems:

1) Find the a value from the equation

(a - 3)(a + 4) = - a + 3

Answer: a = 3 or a = -4

2) Find the x value from the given function using differentiation.

f(x) = x 4 - 108x + 100

Answer: x = 3 or x = -3

Sample Mean Formula

Introduction to Mean:

Mean is one of the more common term in statistics and it's easy to compute. The average is calculated by arranging the values from the set in a particular way and computing a single number as being the average of the set. The mean may different with median, mode or range. For example, mean income is twisted upwards by a small number of people with very large incomes, so that the majority has an income lower than the mean.


Mean Formula

In general sample mean (A.M) or average of n observations x1, x2, …, xn is defined to be the number x such that the sum of the deviations of the observations from x is 0. That is, the arithmetic mean x of n observations x1, x2, …, xn is given by the equation

(x1 ? x) +(x2 ? x) + ... +(xn ? x) = 0

Hence sample mean formula = x1+x2+x3+…….xn / n

sample mean formula= sum of elements / number of elements


Sample Mean Example

Following steps used to calculate the sample mean value

Step1: find the sum of the numbers

Step2: Calculate the total numbers

Step3: Using the formula finding the mean



Example:

The salaries of its nine employees:

The CEO makes $100,000 per year,

Two managers make $50,000 per year,

Four factory workers make $15,000 each, and

Two trainees make $9,000 per year.

So add 100000+50,000 + 50000+15,000 + 15000+15,000 + 15000 + 9,000 + 9000(all the values in the set of data) which gives 278, 000. Then divide that total by 9 (value in the set of data).

That gives you the mean, which is $30,889.

Not a bad average salary. But be careful when using this number. After all, only three of the nine workers at xyz Co. make that much money.


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Practice Problem:

Using the mean formula find the mean value

The salaries of its ten employees:

The CEO makes $100,000 per year,

Two managers make $50,000 per year,

Four factory workers make $15,000 each, and

Three trainees make $9,000 per year.

Answer: 28700

how to do math proportion

Introduction to math proportion:

Algebra is that separation of mathematics in which calculation are made by using any arbitrary characters to stand for the quantities or things considered. Which are used to correspond to numbers is often given the name literal numbers or simply literals. Since the literal numbers are used to represent numbers,

In our daily life, by learning ratio and proportion many a times we compare two quantities of the same type. Thus, in convinced situations, comparison by division makes better sense than comparison by taking the difference. The comparison by division is the Ratio. We denote ratio-using symbol ‘:’. If two ratios are equal, we state that they are in proportion and use the symbol ‘:’ or ‘=’ to equate the two ratios.


Math proportion – Definition and types:

Definition of proportion:

If two ratios are not equal, then we state that they are not in proportion. In a statement of learning proportion, the four quantities involved when taken in order are known as respective terms. First and fourth terms are known as extreme terms. Second and third terms are known as middle terms.

a: b = c : d

There are two types of propositions,

1. Simple proposition:

A propositions consisting of just one subject and one predicate is called a simple proposition.

Example: The following are simple proposition

1. Ram is blind.

2. The flower is not red.

2.  Compound proposition

A proposition consisting of two or more simple propositions in the form of a single sentence is called a compound proposition.

Example: The following are compound propositions,

Quadrilateral ABCD is a square and each side of this quadrilateral is 4cm long.


Math proportion – Example problems:


Problem 1:

Are the ratios 60g: 30g and 32 kg: 48 kg in proportion?

Solution:

60 g: 30 g =60/30 = 2:1

= 2:1

32 kg: 56 kg = 4 / 7 = 4:7

= 4: 7           So, 60: 30 = 32: 56.

Therefore, the ratios 60 g: 30 g and 32 kg: 56 kg are in proportion,

i.e. 60 : 30 :: 32 : 56.

The middle terms in this are 30, 32 and the extreme terms are 60, 56.

Problem 2:

If 4: 5 = 8: X is a proportion, find the missing term.

Solution:

Product of extremes = 4 × x

Product of means = 5 × 8 = 40.

Since it is a proportion, 4 × x = 40

Divide both sides by 4

(4 × x) / 4 = 40 / 4

x =40 / 4 = 10.

X = 10

Problem 3:
The income and Savings of a family are in the ratio 5: 2. If the income of the family is Rs.5, 800. Find how much is being saved.

Solution:

Let the savings be Rs. x.

The proportion is 5: 2 = 5800: x

(Income: Saving) = (Income: Saving)

5x = 2 × 5800

5x / 5 = (2 × 5800) / 5

x = 11600 / 5

x = 2320

The Savings = Rs.2320.

grade 5 math probability

Introduction to grade 5 math probability:

Probability is a way of expressing knowledge or belief that an event will occur or has occurred. In mathematics the concept has been given an exact meaning in probability theory that is used extensively in such areas of study as mathematics, statistics, finance, gambling, and science. In this article we shall discuss about grade 5 probability problems.(Source: wikipedia) I like to share this Probability of Independent Events with you all through my article.


Grade 5 math probability example problems

Here we are going to discuss grade 5 probability problems with detailed solutions.

Problem:

An unbiased die is rolled once at a time find the probability of shown the number 6 in dice.

Solution:

A dice is rolled once at a time, the possible chance of sample space is S= {1, 2, 3, 4, 5, 6} and, therefore, the sample space n(S)=6..

Consider E1 = event of rolling a dice getting the number 6

E1= {6} and, therefore, n (E1) =1.

Therefore probability (getting number 6) = P (E1) = n (E1) / n(S) = `1 / 6`

Example:

An unbiased die is rolled once at a time. Find the probability of getting a number greater than 2.

Solution:

A dice is rolled once at a time, the possible chance of sample space is S= {1, 2, 3, 4, 5, 6} and, therefore, the sample space n(S)=6..

Let E1 = event of rolling a dice getting number greater than 2 Then,

E1= {3, 4, 5, 6} and, therefore, n (E1) =4.

Therefore Probability of getting a number rolling dice greater than 2 = P (E1) = n(E1) / n(S) = `4 / 6` = `2 / 3` .

Example:

An unbiased die is rolled once at a time. Find the probability of getting a number less than 3.

Solution:

A dice is rolled once at a time, the possible chance of sample space is S= {1, 2, 3, 4, 5, 6} and, therefore, the sample space n(S)=6..

Let E1 = event of rolling a dice getting number less than 3 Then,

E1= {2 , 1} and, therefore, n (E1) =2.

Therefore Probability of getting a number rolling dice less than 3 = P (E1) = n(E1) / n(S) = `2 / 6` = `1 / 3` .

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Grade 5 math probability example problems


Problem:

An unbiased die is rolled once at a time find the probability of shown the number 1 in dice.

Answer:

`1/6.`

Problem:

An unbiased die is rolled once at a time. Find the probability of getting a number greater than 4.

Answer:

`1/3`

Y Intercept Slope Form

Introduction on y intercept slope form:

If the line passes through the points A (x1, y1) and B (x2, y2) and then its slope is,

Slope (m) = y2-y1/x2-x1.

The slope is defined as the relation between the perpendicular line distance and flat distance between any two points. The slope is denoted by letter ‘m’.

m=y/x where y and x distances between the two points

The point slope equation of a line is,

Y=mx+b

Where x and y is the points, m is the slope and b is the Y-intercept

The Y intercepts are taking place when horizontal values are zero.

Example Problem on Y Intercept Slope Form:

Example 1

Given point (5, 2) and the slope is 4. Find the equation of this line and the Y intercept.

Solution:

Point (5, 2)? (x1, y1)

Slope m= 4.

Equation of the slope point form is

y-y1 = m (x-x1)

y-2=4(x-5)

y-2=4x-20 (using distribute property)

y-2+2=4x-20+2 (Add 2 on both side)

y=4x-18.

Slope m=4. Y intercepts b=-18.

The line of the equation is y=4x-18

Example 2:

Given point (7, 3) and (-2, 5). Find the equation of this line and the Y intercept.

Solution:

Points (7, 3) ? (x1, y1)

(-2, 5) --- > (X2, y2)

To find the equation of line is,

y-y1 = m (x-x1)

Step 1: Take points (7, 3) (-2, 5)

Slope m = y2-y1/ x2-x1

m=5-3/-2-7 ? 2/-9

Step 2: Now take slope m=-2/9 and point (-2, 5)

y-y1 = m (x-x1)

y-5=-2/9(x-(-2))

Step 3: Simplification

y-5=-2/9x+4/9 (Using distributive property)

y-5+5=-2/9x+0.44+5 (Add 5 on both side)

y=-2/9x+5.44.

The equation is in slope intercept form

Slope m=-2/9 and Y intercept b=5.44

The line of equation is y=-2/9x+5.44

some more Examples on Y Intercept Slope Form:

Example3

Find the Y intercepts of the function f(x) =3.5x+25.

Solution:

To find the Y intercept, plug x=0

F (0) = 3.5(0)+25

=25

This means the Y intercepts taking place at (0, 25)

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Example 4:

Find the Y intercepts of the function f(x) =1.2x+3.5

Solution:

To find the Y intercept, plug x=0

F(0)=1.2(0)+3.5

=3.5

This means the Y intercepts taking place at (0, 3.5)

Practice Root Mean Square

Introduction for root mean square practice:

Generally root mean square is the main topic for statistics and mathematics subject. It is known as well as quadratic mean. The root mean square is a learning topic of the statistical measure of the degree of the variable quantity; it is helpful for while positive and negative integers are required for mean value. Here in this article we are going to explain about root mean square, and solving root mean square example and practice problems.

Practice Root Mean Square:

The root mean square values are the set of values.  It is the average mean value of the squares of the unique (original) values. Root mean square abbreviated from is RMS or rms.

The square root of arithmetic mean is square values of the casual variables. In other terms, we can explain that the root mean square is a statistics evaluate of the degree of the different quantity. It can be evaluate for the serious of discrete ideals or for a constant varying function.

Root mean square formula:

Root mean square =` sqrt (((x1^2) + (x2^2) + (x3^2)+ .....+ (xn^2))/ N)`

Where as, x = individual value

N = number of total values.

This formula is very significant for practice problems solving in root mean square.

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Practice Problems from Root Mean Square:

Practice problem 1: Solve the root mean square value of (-6, 10, 15, 21, -12, 16).

Solution:

Step 1: to count the total number of values (N),

Here N = 6.

Step 2: Square all the values,

36, 100, 225, 441, 144, 256

Step 3: Take the average of all the square values,

`(36+100+225+441+144+256)/6`

= `1202/6`

= `200.33`

Step 4: To take the square root of the average values,

Rms = `sqrt (200.33)`

= 14.1539

Answer is ~ 14.15.

Practice problem 2: To solving the following numbers and find the root mean square values (-15, 12, 50, 8, -32).

Solution:

Here the total numbers are, N = 5.

Square all the values (225, 144, 2500, 64, and 1024)

Take the average of the square values.

`(225 +14 4 + 2500 + 64 + 1024) / 5`

= `3957 /5.`

= `791.4`

Take the square root of average values.

Rms = `sqrt (791.4)`

= `28.131`

Answer is = ~28.1.

Home work practice problems of root mean square:

Problem 1: To find the Rms value of (-6, 12, 24, 36, 18, -12) the above values.

Answer is: 20.49

Problem 2: To Solve the values and fined the Rms value of (-12, 24, -10, 20, -8, 16)

Answer is: 16.02

Those all above explanations and example problems make clear and know how to solving the roots mean square practice problems.