Study Exponentiation

In math, exponentiation is the operation, which is written as the form of an. Where a and n is said to be base and exponent and n is any positive integer. Normally, exponentiation means that repeated multiplication. Otherwise, exponentiation an is the product of n factors of a. The exponent is usually placed as a superscript to right of base value. We are having many properties for exponentiation. Let see properties and example problems for exponentiation.


Properties - Study Exponentiation


We are having seven number of exponentiation properties that used for solving problems with exponentiation. In this properties, a, m and n are any integer values.

Product of like bases:

am an = am+n

Quotient of like bases:

`a^(m)/a^(n)` = am-n

Power to a power:

` (a^(m))^(n)` = amn

Product to a power:

(ab)m = am bm

Quotient to a power:

`(a/b)^(n)` = `a^(n)/b^(n)`

Zero exponent:

a0 = 1

Negative exponent:

a-n = `1/a^(n)` or `1/a^(-n)` = an

These are the properties that are used for exponentiation problems in study math.


Example Problems - Study Exponentiation


Example 1:

Solve 23 22.

Solution:

Given, 23 22.

This is in the form of am an, so we need to use am an = am+n property.

Here, m = 3 and n = 2 and a = 2.

Thus, 23 22 = 23+2

= 25

= 2 × 2 × 2 × 2 × 2

= 32
Hence, the answer is 23 22 = 32.

Example 2:

Shorten the following `5^5/5^3`.

Solution:

Given, `5^5/5^3` .

This is in the form of `a^m/a^n` , so we need to use `a^m/a^n` = am-n property.

Here, m = 5 and n = 3 and a = 5.

Thus, `5^5/5^3` = 55-3

= 52

= 5 × 5

= 25
Hence, the answer is `5^5/5^3` = 25.

That’s all about the study exponentiation.

Solving Change of Base Formula

The solving change of base formula is known as formulas which it permits us to rework a logarithm by means of the logs that may be is written with different base.

The change of base formula is given by,

Log a x = log b x / log b a

Here, assume that a, b and x are positive where a≠1 and b≠1.

Importance in solving change of base formula:

• Using the change of base formula we can change any base to another base. The most commonly used bases are base 10 and base e.

Log a x = log b x / log b a

• The solving change of base formula is used highly if calculators to assess a log to several base further than 10 or e.
• At the solving change of base formula having the value of x which is superior than zero.
• The log of a number to a given base is the power or an exponent to which the base must be raised in order to produce that number.

Advantages  in using change of base formula:

• Change the numeral bases, like convert  from base 2 to base 10m which is known as base conversion.
• The logarithmic change-of-base formula is applicable regularly in algebra and calculus.
• It is used for varying among the polynomial and normal bases.

solving Examples using change of base formula:


1) Solve  log 816

Solution:

By solving the change of base formula

=> log a x = log b x / log b a

log 8 16 = log 2 16 / log 2 8

=  4 / 3

2) Solve log 918

Solution:

By solving the change of base formula

=> log a x= log b x / log b a

log 9 18 = log 2 18 / log 2 9

= 4.169 / 3.169

= 1.315

3) Solve log 2 5.

Solution:

By solving the change of base formula

=> log a x= log b x / log b

log 2 5 = log 10 5 / log 102

The approximate value of the above expression is solving by,

=0 .6989 / 0.30103

=0 .3494

Practice problems in solving change of base formula :

1) Solve log 3 9 using the change of base formula.

Answer: 2

2) Solve log 10 8 using the change of base formula

Answer: 0 .9030

learning probability and odds

The probability of an incident is a proportion that tells how possible. It is that an event will take place. The numerator is the number of favorable outcomes and the

Denominator is the number of possible outcomes.

Probability:   number of flattering outcomes / number of possible outcomes

Probability is we can try to measure the chances of it occurrence

Explanation of learning probability and odds


Here learning the probability and odds

Numerical measure of the likelihood of an event to occur. If an inspection there are n possible  ways  exhaustive and mutually exclusive and out of them in m ways in the event. A occurs, then the probability of occurrence of the event. A is given by P (a) = m/n

If  in a random  sequence  of n trials of an event, M are favorable  to the event , the  probability of that event occurring is the limit of the ratio M/n, when  n is very  large , this lies between 0 and 1

P (a) = 0 means that the event can not take place.

P (a) = 1 means the event is bound to occur.

The events, A2, A3……..An are said to the mutually exclusive if ,

P(Ai∩ Aj)=0

for i=1,2,3,……n.

J=1,2,3,…….n, I  j

He events are said to be exhaustive if

P(A1)+ P(A2)+ P(A3)_.......... P(An)=1 if ,

A ∩   B  0 then

P(Aυ B)=P(A)+ P(B)

And if  A∩  B   0v then

P(A υ B)=P(A)+ P(B)-P(A ∩ B)


Learning Example of probability and odds:


Learning the probability and odds problems

1. when you toss a coin, it can fall two ways. The probability of getting a head on one roll of a coin is one chance out of two.

Solution:

P(h) means the   probability of getting a head on one toss of a  coin.

P(t) means the   probability of getting a tail on one toss of a  coin.

Step 1:  the number of favorable outcomes for head =1

Step 2:  the number of favorable outcomes for tail =1

Step 3:  number of possible outcomes    =2

Step 4:  the probability of the event for head = number of favorable outcomes / number of possible outcomes

Step 5:  the probability of the event for head= P (h) =n (e)/n(s)

Step 6:   the probability of the event for head=1/2

Answer:  1/2

Factor Tree Online

Before learning factor tree online, we'll learn important terms related with factor tree. Those terms are given below:

Factor: A factor of a number is the exact divisor of that number. In other words, a factor of the number divides the number leaving remainder 0.

We know that 15 = 1 x 15 and 15 = 3 x 5. This shows that 1, 3, 5 and 15 exactly divides 15. Therefore, 1, 3, 5 and 15 are all factors of 15.

Prime Number: Each of the numbers which has exactly two factors, namely, 1 and itself, is called a Prime Number. For example: 2, 3, 5,11, 13 etc.

Prime Factor: A factor of a given number is called a prime factor if this factor is a prime number. A prime factor cannot be factored further.Example: 2 and 3 are prime factors of 12 while 4, 6 and 12 are not.

Prime factorization : To express a given number as a product of prime factors is called Prime Factorization or complete factorization of the given number.

Example: 36 = 2 x 2 x 3 x 3, 45 = 3 x 3 x 5 etc.

Factor Tree: A diagram that shows the prime factors of a number is known as Factor tree. A factor tree contains all the prime factors of a given number.


Process to make a factor tree online


Start with the number.

• Make branches of factors - numbers that multiply to give you the original number
• Keep reducing each factor to its lowest possible prime factors.
• When we have all the prime factors identified, re-arrange them from least to greatest.
• Re-write them as exponents.

Illustration: Lets see the factor tree online preparation of 36 and 180 in the figure below:

1) Factor tree of 36:                                                                       
2. Factor tree of 180




Uses Of Factor Tree online


1. Factor Tree online is used to find the GCF or HCF

Example: Find the greatest common factor of 18 and 54.using factor tree

18
3 x 6
3 x (2x3)
The prime factors of 18 are 2, 3, and 3.

54
6 x 9
(2x3) x (3x3)
The prime factors of 54 are 2, 3, 3, and 3.

The prime factors that 18 and 54 have in common are 2, 3 and 3, so the greatest common factor is 2 x 3 x 3 = 18.

2. Factor Tree is used to find the LCM or LCD

Example: Find the LCM of 12 and 36 using factor tree

Step 1: Find the Prime factors for

12 = 2 x 2 x 3
36 = 2 x 2 x 3 x 3

Step 2: Multiply each prime factor the greatest number of times it appears in any one factorization.
The most number of times 2 appears in either factorization is twice. The most number of times 3 appears in either factorization is twice.

Thus, LCM = 2 x 2 x 3 x 3 = 36

Learn Linear Shape Functions

To learn the graphs of linear shape functions be used to resolve problems during restricted element analysis.In Lagrange functions Hermite cubic polynomials each function is unity at its individual node and zero at the further nodes. Moreover, the Lagrange shape functions summation to unity everywhere. For the Hermite polynomials H1 and H3 sum to unity.The shape functions be use to find the field variable U since recognized values at extra locations. The formula is U  = N1 U1 + N2 U2 + ...,Wherever Ni be the shape functions and Ui are identified values.

Sample problem for learn linear shape functions


The Hermite shape functions are used for beam analysis wherever together the deflection and slope of adjacent elements be required to be the similar at every node. H1 and H2 are the deflection while H2 and H4 are the slope.

D = H1 D1 + H2 S1 + H3 D2 + H4 S2
Wherever D be the deflection and Si be the slope.

In the formulas below,

L be the length
S be X - X1
r be S / L





Example problems for learn linear shape functions


Example 1:

Solving the domain of  learn linear shape function f
f (x) = sqrt (5x - 25)

Solution:
The expression function f includes a square root. The expression below the radical have to assure the condition
5x - 25 >= 0    for the function to obtain real values.
solve the linear inequality
x >= 25/5
x=5
The domain, within interval notation, is  (5 , +infinity)

Example 2:

Solving the domain of  learn linear shape function f
f (x) = sqrt (9x - 36)

Solution:
The expression function f includes a square root. The expression below the radical have to assure the condition
9x - 36 >= 0    for the function to obtain real values.
solve the linear inequality
x >= 36/9
x=4
The domain, within interval notation, is  (4 , +infinity)

Example 3:

Solving the domain of  learn linear shape function f

f (x) = sqrt (-x +9)

Solution:
The expression function f include a square root. The expression below the radical have to assure the condition
-x +9 >= 0    for the function to obtain real values.
solve the linear inequality
x >= 9/-1
x=-9
The domain, within interval notation, is (-infinity, 8).

Study about Unit Circle

In Euclidean geometry the simple shape is called circle and it consisting points in a plane which is middle from a given point called the center. Radius is the distance of the points from the center.

Unit circle:

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

If (x,y ) is the first quadrant of the unit circle, then p and q are the lengths of a right triangle whose hypotenuse has length 1. That is, from the Pythagorean Theorem satisfy the following equation with x and y,

x2 + y2 = 1.

Explanation:


1. Length of circumference:

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

c = 2`pi` r

diameter d  = 2r

r = 2 / d

c = `pi`d

2. Area enclosed the circle:

Area of the circle =  `pi ` × area of the shaded square

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

A =` pi` r2

Area of the circle interms of diameter

A  = ` pi`(d/2)2

Area =` pi` d2 / 4


Trigonometric functions on the unit circle:




In geometrical term the trigonometric functions of the angle θ can be modified to a unit circle centered at O.

If  a point of the unit circle is (x, y) , and if the origin (0, 0) to (x, y) makes an angle t from the x-axis, in the trigonometric functions like sine, cosine terms.

cos(t) =x

sin(t) = y

The equation x2 + y2 = 1 gives the relation

cos2(t) + sin2(t) = 1

Solution:

Finally from properties of circle we conclude that the circle does not have any number of sides. That is circle have 0 sides.

x and y intercept solver

Introduction to x and y intercept solver:

X-intercept:
The x-intercept is the x-value of the point where the graph of a function or relation intersect the x-axis of the coordinate system.

Y-intercept:
The y-intercept is the y-value of the point where the graph of a function or relation intersect  the y-axis of the coordinate system. (Source: Wikipedia)
In this article we are going to see the problems on  x and y intercept by using the x and y intercept solver.

Methods to find x and y intercept using solver:

X-intercept solver:

• We can find the x intercept of the line by substitute y = 0 in the line equation.
• The x intercept can be represented as (x,0)

Y-intercept solver:

• We can find the y intercept of the line by substitute x = 0 in the line equation.
• The y intercept can be represented as (y,0)

Problems on x and y intercept using intercept solver :

Problem 1:
Find the x intercept of the line 5x-3y+15 = 0

Solution:
Given, the line 5x-3y+15 = 0
To find the x intercept we need to substitute y = 0 in the equation of line.
5x-3y+15 = 0
Substitute y = 0 in the above line equation,
5x – 3(0) + 15 =0
5x – 0 + 15 = 0
5x + 15 = 0
Subtract 15 on both sides,
5x - 15 - 15 = 0 - 15
5 x = - 15
Divide by 5 on both sides,
x = - 3
The x intercept of the line is (-3,0)

Problem 2:
Find the x-intercept of the line y = x -12

Solution:

Given, the line y =x -12
To find the x intercept we need to substitute y = 0 in the equation of line.
y = x -12
Substitute y = 0 in the above line equation,
y =x -12
0 = x -12
Add 12 on both sides,
12 = x -12 + 12
12 = x
x = 12
The x intercept of the line is ( 12 , 0)

Problem 3:
Find the y-intercept of the line 3y =6x -15

Solution:
Given, the line 3y =6x -15
To find the y intercept we need to substitute x = 0 in the equation of line.
3y =6x -15
Substitute x = 0 in the above line equation,
3y =6(0) -15
3y = - 15
Divide by 3 on both sides,
y = -15/3
y =-5
The y intercept of the line is ( 0 , -5)

Problem 4:
Find the y-intercept of the line y = 2x - 10

Solution:
Given, the line y = 2x - 10
To find the y intercept we need to substitute x = 0 in the equation of line.
y = 2x - 10
Substitute x = 0 in the above line equation,
y =2(0) -10
y = - 10
The y intercept of the line is ( 0 , -10)