Exponentiation Tutoring

This article is a online exponentiation tutoring.

Generally the exponentiation is used in all sciences including the mathematical, physical, chemical or even in biological sciences but it is assumed or preferably called a mathematical operation which generally written as an it is clear from this notation that the exponentiation involves two numbers one is called the base i.e. a and other is called the exponent i.e. n. When n is any positive integer then the exponentiation simply means the repeated multiplication of the number ‘a’ to itself n times which are written as:

 

Generally, exponent is the superscript to the base. Exponentiation above will be read as a raised to the power n or exponent of n, or a to the n. A few exponents have their pronunciation: for instance, a2 is read as a square and a3 as a cube.

Exponentiation comes to use in economics or biology or chemistry or physics or computer science to calculate compound interest or population growth.


Example problems on exponentiation tutoring:


Positive integer exponents:

a2 is pronounced as a square because any square with side a has area equal to a2. The expression a3 is referred to as the cube of a since the volume of a cube with side length a is a3.

So 32 is called as "three squared", and 23 is "two cubed".

The numbers with positive integer exponents may be defined by the initial condition

a1 = a

thus in general the recurrence relation

an+1 = a·an.


Exponents of one and zero:


If a is not zero and n an m are two positive integer exponents such that the exponent (n – m) is also positive then we can say that



If we consider a special case in which the exponent’s n and m are equal then we can wrote the equality as

 

Thus we lead to the following rule: Any number with raised to the power 1 is the number itself.

Solving Quadratic Formula

Quadratic Equation:

An equation which has one or more terms are squared but no higher power in terms, having the syntax, ax2+bx+c =0 where a represents the numerical coefficient of x2, b represents the numerical coefficient of x, and c represents the constant numerical term.

Types of quadratic equation

Pure quadratic equation:

The numerical coefficient cannot be zero. If b=0 then the quadratic equation is called as a ‘pure’ quadratic equation

Complete quadratic equation:

If the equation having x and x2 terms such an equation is called a ‘complete’ quadratic equation. The constant numerical term ‘c’ may or may not be zero in a complete quadratic equation. Example, x2 + 5x + 6 = 0 and 2x2 - 5x = 0 are complete quadratic equations.


Quadratic Equation Formula


The quadratic equation has the solutions ax2+bx+c =0

x =√(b2-4ac)/2a

Consider the general quadratic equation

ax2+bx+c =0

With a`!=` 0. First divide both sides of the equation by a to get

x2+b/a x + c/a =0

which leads to

x2+ b/a x = - c/a

Next complete the square by adding ((b)/(2a) )2to both sides

X2+ ((b)/(a) )x+((b)/(2a) )2 = -((c)/(a) )+ ((b)/(2a) )2

(x+(b)/(2a) )2=-((c)/(a) ) + ((b^2)/(4a^2) )

(x+(b)/(2a) )2 = (b^2-4ac)/(4a^2)

Finally we take the square root of both sides:

x+(b)/(2a) = +-(sqrt(b^2-4ac))/(2a)

or

x =-(b)/(2a)  +-(sqrt(b^2-4ac))/(2a)

The final form of Quadratic Formula is

x =-b+-sqrt(b^2-4ac)/(2a)

The two roots of the equation is

``          -b-(sqrt(b^2-4ac))/(2a)

-b+(sqrt(b^2-4ac))/(2a)


Example Problem on solving Quadratic formula


Example:

Find the roots of the equation by quadratic formula method, x2-10x+25=0

Solution:

Step 1:  From the equation, a = 1, b = - 10 and c = 25.

Step 2:  To Find X:
plug-in the values in the formula below
x = ``

Step 3:  We get the roots, x = ``
x = 5 and x = 5
which means x1 = 5 and x2 = 5.

Here x = 5 is root of the equation.

learning problems in probability

Introduction:

Learning Probability is also one part of mathematics subject. Probability disturbed with investigation of approximate phenomena. Learning the fundamental things of the probability is random variables, stochastic processes, and events. Mathematical abstraction of the non-deterministic events or calculated quantities that may moreover be single occurrences or evolve over time in an apparently random fashion. While a coin toss or roll of a die is a approximate event.

Let us some example problems in probability with solved solutions.

Learning example problems in probability:

Learning Example problem 1:

Three dice are rolled once. What is the chance that the sum of the expression numbers on the three dice is greater than 15?

Solution:

In probability problems while three dice are rolled,

the trial space S = {(1,1,1), (1,1,2), (1,1,3) ...(6,6,6)}.

S contains 6 × 6 × 6 = 216 outcomes.

Let A be the event of receiving the sum of appearance numbers greater than 15.

A = { (4,6,6), (6,4,6), (6,6,4), (5,5,6), (5,6,5), (6,5,5), (5,6,6), (6,5,6),

(6,6,5), (6,6,6)}.

N (S) = 216 and n (A) = 10.

Therefore P (A) equal to n (A) / n(S)

= 10/ 216

=5 / 108.


Probability Problem 2:


Learning Example problem 2:

In a vehicle Stand there are 100 vehicles, 60 of which are cars, 30 are mini truck and the rest are Lorries. If each vehicle is uniformly likely to leave, find out the probability in this problem.

a) Mini truck departure first.

b) Lorry departure first.

c) Car leaving second if either a lorry or mini truck had gone first.

Solution:

a) Let S be the model space and A be the event of a mini truck leaving first.

n (S) = 100

n (A) = 30

Probability of a mini truck departure first:

P (A) = 30 / 100 = 3 / 10.

b) Let B be the event of a lorry departure first.

n (B) = 100 – 60 – 30 = 10

Probability of a lorry departure first:

P (B) = 10 / 100 = 1 / 10.

c) If moreover a lorry or mini truck had gone first, then there would be 99 vehicles totally left there, 60 of which are cars. Let T be the trial space and C be the occurrence of a car departure.

n(T) = 99

n(C) = 60

Probability of a car departure after a lorry or mini truck has gone:

P(C) = 60 / 99 = 20 / 33.

Learning Probability Density Functions

The Learning Probability density functions are known as Gaussian functions. A probability density function of a continues random variable is defined as a function that describes the virtual probability for that random variable to occur at a given point within the using space. The Learning of probability density functions is same for the probability distribution functions. In this article we see the learning  the definition of probability density functions and some example problem for the probability density functions.

Definition of Learning the Probability Density Functions:


Probability density function is defined as continues random variable functions f(x). this functions satisfies the following properties.

Probability functions limits between a and b

P(a≤x≤b) = ∫ a to b  f(x) . dx

A probability density function has only real for the real value.

F(x) ≥ 0

Integral of the probability density functions is 1.

∫ -oo to oo f(x) dx  = 1

Learning the fundamental properties of probability density functions:

F(x) is continues random functions

P(a<= x <b) =  P(a < x<b) = P( a< x <= b)

= P(a<=X<b) =

Where f is probability density distribution functions

In the differentional functional of the probability density functions, we have

P(x<X<=x+dx) = F(x+dx)-F(x) = dF(x)=F'(x)dx = f(x)dx

Where

X is Known as probability differential functions.


Learning the some example problems for the probability density functions:


Example 1:

If the probability density function of a random variable is given by f(x) = H (x – x^3);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x^2/2 – x^4/4] limits 0 to1 = 1

H [ 1/2- 1/4] = 1

1 After simplify, We get

H= 4

Example-2

If the probability density function of a random variable is given by f(x) = H (1– x^5);  0 < x <1 .find the H.

Solution:

Since ,

F(x) is the probability density functions means,

F(x)= ∫ a to b f(x) dx

Substitute the values,

H [ x – x^6/6] limits 0 to1 = 1

H [ 1- 1/6] = 1

1 After simplify, We get

H= 6/5 = 1.2

Practice problem of  learning the probability density functions:

Problem -1:

A continuous random variable X follows the probability law, f(x) =H x (x – x^2 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 12

Problem -2:

A continuous random variable X follows the probability law, f(x) =H x^2 (x– x^4 ), 0 < x < 1and 0 for elsewhere. Find k

Answer: H = 9.33

Learn Operations on Functions

Generally a function is represented in terms of  f(x).

Functions behave exactly as one would expect with regard to the four basic operations of algebra (addition, subtraction, multiplication, and division). When functions are combined by these operations, though, the domain of the new combined function is only the elements that were shared by the domains of the original functions.

Operations on functions :

As we add, subtract, multiply the numbers, we can add , subtract, multiply, divide the functions. There is one operation more called the composition of functions.

Let us learn through examples, that would be easier to understand.

Consider the two functions,

f(x) = x +2

g (x) = x2 -5


Operations on functions : Addition


f(x) = x +2

g (x) = x^2 -5

(f+g) (x) = x +2 +x2 -5

= x2+x -5 +2

= x2 + x -3

(f+g) (2) = 22 +2 -3 = 6 -3 =3


Operations on functions : Subtraction


f(x) = x +2

g (x) = x^2 -5

(f-g) (x) = x+2 – (x2 -5)

= x2+x +2 -5

=x2 +x -3

(f-g)(2) = 2 +2 - 22 -5 = -5


Operations on functions : Multiplication


f(x) = x +2

g (x) = x^2 -5

(f)(g)(x) = (x+2) (x2-5)

= x3 -5x +2x2 -10

= x3 +2x2 -5x -10

(f)(g)(2) = (2+2) (22-5) = -4


Operations on functions : Division


f(x) = x +2

g (x) = x2 -5

g(x) /f(x) =

=  [x2 -5] / [x+2]

g (2) / f (2 ) = (22 - 5)/ (2+2)

=(4 -5)/ (2+2) = -1 / 4


Operations on functions : Composition of Function


f(x) = x +2

g (x) = x^2 -5

(f o g)(x) read as f of g of x, we can say  that first, we find g(x) and then we find f(x) for the result that we got from g(x)

The concept is simple. First, the value of g at x is taken, and then the value of f at that value is taken

g(x) = x2 -5

(f o g)(x)= f (x2-5) = x2 -5 +2 = x2 -3

(f o g)(2) =

= g(2) =22 – 5 = -1

= f(-1) = -1+2 =1

Trigonometric Functions Unit Circle

Unit circle:

In mathematics, a unit circle is a circle with a radius of one. Frequently, except in trigonometry, the unit circle is the circle of radius one centered at the origin (0, 0) in the Cartesian coordinate system in the Euclidean plane. The unit circle is denoted as s1; the generalization to higher dimensions is the unit sphere.

Trigonometric functions on the unit circle:

All of the trigonometric functions of the angle θ can be modified by geometrically in terms of a unit circle centered at O.

The trigonometric functions of cosine and sine can be defined on the unit circle as follows. If (x, y) is a point of the unit circle, and if the ray from the origin (0, 0) to (x, y) makes an angle t from the positive x-axis, (where counterclockwise turning is positive), then

cos(t) = x

sin(t) = y

The equation x2 + y2 = 1 gives the relation

cos2(t) + sin2(t) = 1

The unit circle also operated that sine and cosine are periodic functions, with the identities

cos t = cos(2πk + t)

sin t = sin(2πk + t)

For any integer k

Area & Circumference:

1. Length of circumference:

The circumference’s length is related to the radius (r) by

c = 2*π*r

For unit circle circumference is 2*(because r = 1).

2. Area enclosed:

Area of the circle = π × area of the shaded square

Main article: Area of a disk

The area of the circle is π multiplied with the radius squared:

A = π *r2

For unit circle area is


Application:


Circle group:

The complex numbers can be pointed with points in the Euclidean plane, namely the number a + bi is identified with the point (a, b). Under this identification, the unit circle is a group under multiplication, called the circle group. It has main applications in mathematics and science.

Application of the Unit Circle:

• It is used to understand the relationship between the unit circle and the trigonometric (sine and cosine) functions.

• Because of this reason it is used to solve a real-world application.

Multiplication Solver

Multiplication (symbol "×") solver is the mathematical operation of scaling one number by another. It is one of the four basic operations in a elementary arithmetic (the others being addition, subtraction and division) (Source.Wikipedia). Multiplication solver  there are two jobs such that one of them can be completed in m ways, and when it has been completed in any one of these m ways, second job can be completed in n ways; then the two jobs in succession can be completed in m × n ways.

Examples for multiplication solver:


Rules for multiplication solver:

•    The multiplication solver of two positive integers is a positive integer.

•    The multiplication solver of a positive integer and the negative integer is a negative integer.

•    The multiplication solver of two negative integers is a positive integer.

Example 1:

Multiply 283 by 104

Solution:

283 × 104 means 104 times 283

or 100 times 283 + four time 283

or 28300 + [4* 283]

or 29432

∴ 283 × 104 = 29432 [multiplication rules: The product of two positive integers is a positive integer.]

Example 2:

Multiply 750 × 98

Solution:

750 × 98 means 98 times 750

or 100 times 750 – two time 750

or 75000 – 1500

or 73500

∴ 750 × 98 = 73500      [multiplication rules: The product of a positive integer and the negative integer is a negative integer.]

Example 3:

Find the value of 10 × 5 + 12 × 5 + 14 × 5

Solution:

10 × 5 + 12 × 5 + 14 × 5 means

10 times 5 + 12 times 5 + 14 times 5

or 36 times 5

or 36 × 5

or 180

∴ The value = 180


Understanding Example 4:

Multiply 843 by 55 by ordinary method

Solution:

8 4 3 ×

5 5
..............
4215
...................
4 6 3 6 5              [multiplication rules: The product of two positive integers is a positive integer.]
.......................

Example 5:

Find the product of (a) (+ 6) × (– 6) (b) (– 13) × (+ 13) (c) (– 15) × (– 4)

Solution:

(a) (+ 6) × (– 6) = – 36 (positive × negative = negative)

(b) (– 13) × (+ 13) = – 169 (negative × positive = negatives)

(c) (– 15) × (– 5) = + 75 (negative × negative = positive)

Example 5:

Find the product of (a) (– 4) × (+ 5) × (+ 2) (b) (– 2) × (– 5) × (+ 10) (c) (– 4) × (– 2) × (– 6)

Solution:

(a) (– 4) × (+ 5) × (+ 2) (b) (– 2) × (– 5) × (+ 10) (c) (– 4) × (– 2) × (– 6)

= (– 20) × (+ 2) = (+ 10) × (+ 10) = (+ 8) × (– 6) [The product of a positive integer and the negative integer is a negative integer. multiplication rules]

= – 40 = + 100 = – 48

Example 6 :

Multiply  2 and 3

Solution:

By repeated addition we get the following:

2 × 4 = 2 + 2 + 2+2 = 8.

Practice problem for multiplication solver:


1.Multiply  2 and 5

Answer:10

2.Multiply  100 and 108

Answer:10800