Fractional Powers in Math

Introduction to Fractional powers in math:

In math, when the power or exponent is a fraction number then the number said to be having the functional powers in math. The example for the fractional power is `2^ (2/3)`

Rules:

Rules for the various operations on fractional powers in math.

For multiplication:

`m^ (a/b)` *`m^ (c/d)` = `m^ ((a/b) + (c/d))`

For division:

`m^ (a/b)` /`m^(c/d)` =` m^((a/b)-(c/d))`

`(m^ (a/b))^c` = `(m^(c/b))^a`

Model Problems for Multiplication of Fractional Power in Math:

Q 1:  Find the value of the fractional power terms?

` 2^ (1/2)` * `2 ^ (3/2)`

Solution:

When we multiply the two terms with fractional powers. we need the fractional powers.

The rule is

`m^ (a/b)` *`m^ (c/d)` = `m^ ((a/b) + (c/d))`

here,

` 2^ (1/2)` * `2 ^ (3/2)`   = `2 ^(1/2+3/2).`

= `2^((1+3)/2)`

= `2^(4/2)`

= `2^2`

= 4

Q 2 : Find the value of the fractional power terms?

` 2^ (5/2)` * `2 ^ (3/2)`

Solution:

When we multiply the two terms with fractional powers.we need to add the fractional powers.

The rule is

` m^ (a/b)` *`m^ (c/d)` = `m^ ((a/b) + (c/d))`

here

`2^ (1/2)` * `2 ^ (3/2)` =` 2 ^(5/2+3/2).`

= `2^((5+3)/2)`

= `2^(8/2)`

= `2^4`

= 16

Answer is 16.

Model Problems for Division of Fractional Power in Math:

Q 1: Find the value of the fractional power terms?

`2^ (5/2)` / `2 ^ (3/2)`

Solution:

When we divide the two terms with fractional powers. we need to subtract the fractional powers

The rule is

`m^ (a/b)` /`m^ (c/d) ` = `m^ ((a/b) - (c/d))`

here

`2^ (1/2)` / `2 ^ (3/2)`   = `2 ^(5/2-3/2)` .

= `2^((5-3)/2)`

= `2^(2/2)`

= `2^1`

=  2

Answer is 2

Q 2 : Find the value of the fractional power terms?

`2^ (5/2)`   / `2 ^ (3/4)`

Solution :

When we multiply the two terms with fractional powers. we need to subtract the fractional powers

The rule is

`m^ (a/b)` /`m^ (c/d)` = `m^ ((a/b) - (c/d))`

here

`2^ (1/2)` / `2 ^ (3/2)` = `2 ^(5/2-3/4).`

= `2^((10-3)/4)`

= `2^(7/4)`

Answer : 274

Q 3 : Find the value of the `(5^ (4/5)) ^10.`

Solution:

To find the value we use

` (m^ (a/b))^c` = `(m^(c/b))^a`

`(5^ (4/5)) ^10 ` = `(5^ (10/5)) ^4.`

= `(5^ (2)) ^10.`

= `5^ (2*10)`

= `5^20`

Answer is `5^20`

Radical Form Math

Introduction to Radical form Math:

In mathematics subject there are so many chapters and so many topics are there. A radical is one of the important topics in that. A radical number is known as, which is the factor value of under the root symbol, and radical expression is containing a square root. Radical is denoted by the symbol of v (Sqrt).it has two forms, one is radical form, and other one is exponential form. Now we are going to see about radical form in math.

Radical Form Math:

Radical form is also called as Root form. it is nothing but to change the order of the exponential form into radical (root) form.

Important steps to remember how to change radical form:

Radical form into exponential form changed by to the power of exponent. For Example: 3v x = (x) 1/3 .
2.  Exponential form into radical form is change the exponent to under the radical. For example: (y) 1/2 = 2v y.

3.  Xn is called the nth root of x and is written as nv x or n Sqrt (x).

4.  If x and y are positive rational numbers and n is any rational exponent, Then xn * yn = (x * y) n. 

5.  If y = q Sqrt (x) or qv x = x 1/q then, x1/q is the exponential form and qv x or q Sqrt (x) is the radical form of x. (Where, q is the index of the radical, x is the radicand).

6.  If x and y are positive rational numbers and n is any rational exponent, Then xn * yn = (x * y) n.

Example Problems in Radical Form Math:

Radical to exponential form:

Example problem 1:

To change the given radical form into the exponential form 3v (11)

Solution:

Given: 3v (11)

= (11)1/3

The required exponential form is (11)1 / 3.

Exponential form to Radical form:

Example problem1:

To change the given exponential form into the radical form (17) - 1/2

Solution:

Given: (17) -1/2

= (17) - 1/2 = (1 / 17)1/2

The required radical form is 2v 1 / 17).

Example problem 2:

To change the given exponential form into the radical form (27 / 220) - 1/4

Solution:

Given: (27 / 220) -1/4

= (27 / 220) - 1/4 = (220 / 27)1/4

The required radical form is 4v (220 / 27).

Non Real Roots

Introduction to Non real roots:
Non real roots are the imaginary roots. A complex number is a number that consists of real part and imaginary part.   We write it as a + bi where i is the imaginary  unit.  i = v-1
If z = a + bi then a is the real part and  b is called the imaginary part.  a is denoted as Re(z) and b is denoted as Im(z)
Roots of a complex number:-
Non real roots are roots of a complex number.
A number is called the nth root of a complex number z if ?n = z  and ? = z1/n
To find the non real roots  of a complex number, we follow the following steps.
Step1 :  Write the given number in polar form.
Step2 :  Add 2kp to the argument..
Step3 :  Apply De Moivre's theorem.
Step4 :  Put k= 0,1 ...... up to infinity.

Non Real Roots-the Nth Root of Unity:

The nth root of unity:-
Take the number 1
Step1 :  1 =  1(cos 0 + i sin 0)
step 2:         1( cos (0 + 2kp) + i sin)0 + 2kp) =  cos 2kp + i sin 2kp
Step 3: nth root of unity = 11/n = (cos 2kp + i sin 2kp)1/n where k = 0,1 .......(n-1)
=  [ cos 2kp/n + i sin 2kp/n ] where k = 0,1   ... ..(n-1)
Step 4 :when k=0 the nth root is cos 0 + i son 0
when k = 1 the nth root is cos 2p/n + i sin 2p/n,...
So ? = cos 2p/n + i sin 2p/n  which is written as ei 2p/n   

Hence the nth root of unity is e0, ei2p/n, ei4p/n etc  which is written as 1,?,?2, ?3 .......... ?n-1
From the above, we note that the non real roots are in geometric progression and the sum of the non real roots  is zero.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Rationalize a Denominator and Graphing Quadratic Functions in Standard Form. I am sure they will be helpful.

Non Real Roots - Cube Root of Unity:

Cube root of unity:-
Let us find the non real roots of 11/3
Let x = 1 1/3
Then x3 = 1
x3 = cos 0 + i son 0 = cos 2kp + i sin 2kp
x  =  (cos 2kp + i sin 2kp)1/3   =   cos 2kp/3  + i sin 2kp/3  ......... k = 0, 1, 2 ......
Therefore the cube roots are cos 0 + i sin 0  = 1
cos 2p/3 + i sin 2p/3 =  ?
cos 4p/3 + i sin 4p/3 = ?2
The values of the non real roots  ? =   -1 + i v3
2
The value of the non real root  ?2 =  -1-iv3
2
When the points are plotted on garland diagram, we see them lying on the circle  of unit radius.
The roots 1,?,?2 are in geometric progression and the sum of the roots = 0                                                                 

Congruent Figures

Congruent Figures:

Two figures are said to be congruent to each other, if on placing one over the other, they exactly coincide. Congruent figures are same in size and shape both. Two circles are always congruent to each other. In the congruent triangles, the sides and the angle that coincide by superposition are called . If two line are same in size they are congruent.

In a congruent triangle corresponding sides and corresponding angles are equal. Two figures are congruence if there is an involvement with their vertices such that coupled angles are the comparable to and the comparable sides are interrelated. It is called as word like the one congruence.

SAS Property:

If two sides and the included angle of one triangle are congruent to the corresponding two sides and the included angle of another triangle, then the two triangles are congruence.

S.S.S Property: 

If the all three sides of one triangle is equal to the three sides of other triangle, both triangles are know as congruent triangles.

ASA Property:

If two angles and the included side of one triangle are congruence to the corresponding two angles and the included side of another triangle, then the two triangles are congruence.

SAA or AAS Property:

If two angles and a non- included side of one triangle are congruence to the corresponding parts of another triangle, then the triangles are congruence.

R.H.S Property:

In two triangles if one of  angles are right angles and hypotenuse and one side of first triangle is equal to hypotenuse and one side of another triangle, triangles are know as congruent triangles.

The L-L Property (The Leg – Leg Property)

If two legs of one right triangle are congruence to the corresponding legs of another right triangle, then the two triangles are congruence.

The L-AA Property (The Leg-Acute Angle Property)

If a leg and an acute angle of one right triangle are congruence to the corresponding leg and an acute angle of another. Then the two triangles are congruence.

The H-AA Property (The Hypotenuse – Acute Angle Property)

If the hypotenuse and an acute angle of one right triangle are congruence to the corresponding hypotenuse and an acute angle of another. Then the two triangles are congruence.

The H – L Property (The Hypotenuse – Leg Property)

If the hypotenuse and leg of one right triangle are congruence to the corresponding hypotenuse and leg of another, then the two triangles are congruence.

Examples of Congruent Figures:

Statement:

In SSS proof, if the corresponding sides of two triangles are proportional, then they are similar.

The proof of the above property is discussed in the form of the following solution.

Solution:

Given:    

In `Delta`ABC and `Delta`DEF, we are given that

`(AB)/(DE)` = `(BC)/(EF)` = `(AC)/(DF)`   

  
To prove:

`Delta`ABC ~ `Delta`DEF.

Construction: mark point P on DE and the point Q on DF such that DP =  AB, DQ = AC. Join P and Q.

Proof:          

Since  `(AB)/(DE)` = `(AC)/(DF)`                 (given)

`(DP)/(DE)` = `(DQ)/(EF)`                   (construction)

By the converse of Thales theorem, PQ || EF.

`anlge`DPQ = `angle`E                       (corresponding angles)

`angle`DQP = `angle`F                       (corresponding angles)

By AAA similarity ( or by AA similarity)

We get             `Delta`DPQ ~ `Delta`DEF

`(DP)/(DE)` = `(PQ)/(EF)` `rArr` `(AB)/(DE)` = `(PQ)/(EF)`                        (given)

But                  `(AB)/(DE)` = `(BC)/(EF)`                 (given)

BC = PQ and AB = DP, AC = DQ                (construction)

`Delta`ABC `~=` `Delta`DPQ                                    (SSS)

Since               `Delta`DPQ ~ `Delta`DEF

`Delta`ABC ~ `Delta`DEF.

Hence, the required result is proved.

Between, if you have problem on these topics greatest integer function, please browse expert math related websites for more help on mathematical induction.

Properties of Congruent Figures

Congruent figures have following properties:


1. The corresponding part of congruent figures are also congruent.


2. The corresponding sides lie opposite to the equal angles and corresponding angles lie opposite to the equal sides.


3. All squares can be similar but not congruent. If squares have equal size  (length and width ) they are know as congruent squares.


4. All the rectangles can be similar but Congruent rectangle are equal in size and shape.

Example : Find the missing sides of congruent triangles:

AB =  5cm

AC = 3 cm

EG = 3 Cm

EG = 4 Cm

Find BC and EF?

Sol :

Both triangles are congruent ( Given )

So All the corresponding sides are equal

BC = FG = 4cm

EF = AB = 5cm

Solve Linear Equations Graphically

There are various methods to solve linear equations like elimination method, substitution method etc. There is another method by which we can solve linear equations, called graphical method. In this method we need to draw graphs of given equations and the point of contact or the point where both the equation intersect each other becomes the solution of the linear equation. Now understand the steps to solve linear equation graphically.

Steps to Solve Linear Equation Graphically

Step1: Take first linear equation and put x = 0, you will get value of y. lets say y = a, it will give you first point (0,a).

Step2: Now put y = 0, it will give you value of x, lets say x = b, it will give you second point (b,0)

Step3: Plot both points and draw a line between them and extend this line beyond these points, This will be graph of first linear equation.

Step4: Repeat step 1,2,3 for second linear equation, you will get graph of second linear equation.

Step5: Find out the point where these lines intersect each other, lets say point is (u,v)

the solution will be x = u, y = v

Note: In case, both lines does not intersect extend both the lines to find intersecting point and once you get intersecting point, it will be required solution. Only parallel line does not intersect each other.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Online Algebra Calculator and algebra 1. I am sure they will be helpful.

Example to Solve Linear Equation Graphically

The following example will make more clear to you how to solve linear equation graphically.

Question: Solve: x + y = 1 and 2x + y = -2

Solution: x + y = 1

put x = 0, w get y = 1, so first point will be (0,1)

put y = 0, we get, x = 1, so second point will be (1,0)

2x + y = -2

put x = 0, we get, y = -2, so first point will be (0, -2)

put y = 0, we get, x = -1, so second point will be (-1, 0)

Now drawing both the lines on graph we get,

We can see that these lines cut each other at point (-3, 4)

Hence Solution will be x = -3 and y = 4.

Elementary Statistics Problems

Introduction to elementary statistics problems: 

Elementary statistics problems were done normally with data collected for specific purposes. In elementary statistics problems, we make decisions about the data by analyzing and interpreting it. There are several methods in elementary statistics for representing data graphically and in tabular form.

The method used in elementary statistics for finding a representative value for the given data are called the measure of central tendency. The three measures of central tendency used in elementary statistics problems are  Arithmetic Mean, also median and mode.

Examples on Elementary Statistics Problems

Example 1:

Find the mean deviation of the mean for given data:

8,5,15,9,13,2,6,14

Solution:

Step 1 Mean of the given data are  `barx`


`barx`  = `(8+5+15+9+13+2+6+14)/8` = `72/8` = 9

Step 2 The deviations of the respective observations from the mean x, i.e., xi– x are
8– 9,5–9,15–9,9–9,13–9,2–9,6–9,14–9      ( or )   –1,–4,6,0,4,–7,–3,5

Step 3 The absolute values of the deviations, i.e.,|xi - x |are  -1,-4,6,0,4,-7,-3,5


Step 4 The required mean deviation about the mean is

M.D. (`barx` ) = `sum` 8 i-1 |xi-x| / 8

= `(1+4+6+0+4+7+3+5)/8 ` = `30/8` = 3.75

Example 2:

Find the mean deviation of the mean for given data :

13, 4, 19, 18, 5, 10, 18, 20, 21, 8, 15, 18, 2, 3, 16, 11, 3, 1, 10, 5

Solution:

find the mean ( `barx` ) of the given data

`barx` = `1/20` `sum` 20i-1  xi = `220/20` = 11


The respective absolute values of the deviations from mean, i.e.,|x-`barx` | are

2,7,8,7,6,1,7,9,10,3,4,7,9,8,5,0,8,10,1,6

Therefore
`sum` 20i-1 |xi - `barx` | = 118

and M.D. ( `barx` ) = `118/20` = 5.9

Examples on Elementary Statistics Problems

Example 3:

Find the mean deviation of the median for the following data: 5,11,7,5,14,12,20,6,8,21,23.

Solution :

Here the total number of observations is 11 which is odd. Arranging the data into ascending order,

5 , 5 , 6 , 7 , 8 , 11 , 12 , 14 , 20 , 21 , 23

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi - M| are

6 , 6 , 5 , 4 , 3 , 0 , 1 , 3 , 9 , 10 , 12
Therefore

? 11i-1 |xi - M| = 59

M.D.(M) = `1 / 11` ? 11i-1 |xi - M| = `(1 / 11) * 59`   = 5.36

Example 4:

Find the mean deviation of  the median for the following data: 10,5, 6, 3, 12, 11, 18, 4, 7, 18, 22.
Solution:

Here the total number of observations is 11 which is odd. Arranging the data into

ascending order, we have 3 , 4 , 5 , 6 , 10 , 11 , 12 , 18 , 18 , 19 , 22

Now Median =(11+1/2) or 6th observation = 11

The absolute values of the respective deviations from the median, i.e.,|xi - M| are

8 , 7 , 6 , 5 , 1, 0 , 1, 7 , 7 , 8 , 11
Therefore
`sum` 11i-1 | - M| = 61

and  M.D.(M) = 1/1111i-1 |xi - M| =` 1/11*61 ` =5. 545. 

Lcm Prime Factorization

Introduction to LCM prime factorization:

Prime factorization is defined as the two factors of number is must be a prime number.

For example:

The prime factorization of the number x is y and z. Then the factors of the number y and z.

Y must be 1 and y similarly the factor of the number z must one and z. In this article we shall discuss about LCM prime factorization with suitable example problems

I like to share this What is LCM with you all through my article.

Example Problem for Lcm Prime Factorization:

LCM prime factorization problem: 1

find the LCM prime factorization for 24, 12.

solution:

Given that 24 and 12. Find the Factors of two numbers individually

Take the 24 and divide through small number.

`24 / 2` = 12

`12 / 2 ` = 6

`6 / 2` = 3

`3 / 3` = 1

24 = 2 * 2 * 2 * 3

`12 / 2` = 6

`6 / 2` = 3

`3 / 3` = 1

12 = 2 x 2 x 3

prime factorization of 24 = 2 x 2 x 2 x 3

prime factorization of 12 = 2 x 2 x 3

LCM of two numner 12 , 24 is 24.

So the LCM prime factorization of 24 and 12 is 2 x 2 x 2 x 3 = 24 .

LCM prime factorization problem: 2

find the LCM prime factorization for45 ,  25.

solution:

Given that 45 and 25. Find the Factors of two numbers individually

Take the 45 and divide into small number.

`45 / 3` = 15

`15 / 3` = 5

`5 / 5 ` = 1

45 = 3 x 3 x 5

`25 / 5` = 5

`5 / 5 ` = 1

25 = 5 x 5

prime factorization of 45 = 3 x 3 x 5

prime factorization of 25 = 5 x 5.

LCM of 45, 25 = 225

So LCM prime factorization of 45 and 25 is 3 x 3 x 5 x 5 = 225

Please express your views of this topic prime factorization factor tree by commenting on blog.

Example Problem for Lcm Prime Factorization:

Find the LCM prime factorization for 54, 50

solution:

Given that 54, 50. Find the Factors of two numbers individually

Take the 54 and divide throughout small number.

`54 / 2` = 27

`27 / 3` = 9

`9 / 3` = 3

`3 / 3` = 1

54 = 2 * 3 * 3 * 3

`50 / 2` = 25

`25 / 5` = 5

`5 / 5` = 1

50 = 2 x 5 x 5

prime factorization of 54 = 2 x 3 x 3 x 3

prime factorization of 50 = 2 x 5 x5

LCm of 54 and 50 = 1350

So LCM prime factorization of 54 and 50 is 2 x 3 x 3 x 3 x 5 x 5 = 1350

Example 4:

find the LCM prime factorization for 96, 18

solution:

Given that 96, 18. Find the Factors of two numbers individually

Take the 96 and divide through small number

`96 / 2` = 48

`48 / 2` = 24

`24 / 2` =12

`12 / 2` = 6

`6 / 2` = 3

`3 / 3 ` = 1

96 = 2 * 2 * 2 * 2 * 2 * 3

`18 / 2` = 9

`9 / 3` = 3

`3 / 3` = 1

18 = 2 x 3 x 3

prime factorization of 96 = 2 x 2 x 2 x 2 x 2 x 3

prime factorization of 18 = 2 x 3 x 3

LCM of 96 and 18 = 268

So LCM prime factorization of  96 and 18 is 2 x 2 x 2 x 2 x 2 x 3 x 3= 288