Study Online Exponentiation

Exponentiation should be the operation, which is written as the form of an. Where a and n is said to be base and exponent as well as n is any positive integer. In general, exponentiation means that repetitive multiplication. Otherwise, exponentiation an is the product of n factors of a. Online study should be the topic-oriented or else technical help for clarifying doubts that are convey through the computer software. Let us study properties and example problems for exponentiation.


Properties - Study Online Exponentiation

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

Product of like bases:

The product of powers with the same base means we can add the powers and keep the common base.

am an = am+n

Quotient of like bases:

To divide the powers with similar base, we can subtract the exponents and remain the common base.

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

Power to a power:

In case of raising the power to power, we need to keep same base and multiply the exponent values.

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

Product to a power:

In case of raising the product to power, we need to raise each factor to the power value.

(ab)m = am bm

Quotient to a power:

In case of raising the quotient to power, we need to raise the numerator and denominator to the power value.

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

Zero exponent:

Any number that is raised with zero power should be equivalent to ‘1’.

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 online.


Example Problems - Study Online Exponentiation


Example 1:

Solve 32 34.

Solution:

Given, 32 34.

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

Here, m = 2 and n = 4 and a = 3.

Thus, 32 34 = 32+4

= 36

= 3 × 3 × 3 × 3 × 3 × 3

= 9 × 9 × 9

= 729

Hence, the answer is 32 34 = 729.

Example 2:

Shorten the following `6^7/6^4` .

Solution:

Given, `6^7/6^4` .

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

Here, m = 7 and n = 4 and a = 6.

Thus, `6^7/6^4` = 67-3

= 64

= 6 × 6 × 6 × 6

= 36 × 36

= 1296

Hence, the answer is `6^7/6^4` = 1296.

That’s all about the study online exponentiation.

Simple Algebra Problems

Algebra is a branch of mathematics dealing with variables, equations , expressions.Here we are going to solve some of the simple algebra problems including some equations, expressions and algebraic identities.

Solve Simple algebra problems on numbers:

Ex 1: Subtract  – 4 from – 10.

Sol :
=– 10 – (– 4)

= – 10 + (additive inverse of – 4)

= – 10 + 4

= – 6 (see addition of two integers)

Ex 2: Find the product of  (a) (+ 6) × (– 5) (b) (– 12) × (+ 12) (c) (– 15) × (– 4)

Sol : 

(+ 6) × (– 5) = – 30 (positive × negative= negative)

(– 12) × (+ 12) = – 144 (negative × positive = negative) 

(– 15) × (– 4)= + 60 (negative × negative = positive) 

Ex 3: Find the sum of given algebraic expression 2x4 – 3x2 + 5x + 3 and 4x + 6x3 – 6x2 – 1.

Sol:

Using the associative and distributive property of the real numbers, we obtain (2x4 – 3x2 + 5x + 3) + (6x3 – 6x2 + 4x – 1)

= 2x4 + 6x3 – 3x2 – 6x2 + 5x + 4x + 3 – 1

= 2x4 + 6x3 – (3+6)x2 + (5+4)x + 2

= 2x4 + 6x3 – 9x2 + 9x + 2.

  The following scheme is helpful in adding two polynomials 2x4 + 0x3 – 3x2 + 5x + 3 0x4 + 6x3 – 6x2 + 4x – 1 2x4 + 6x3– 9x2 + 9x + 2 


Ex 4: Find the subtraction of given simple algebraic expression 2x3 – 3x2 – 1 from x3 + 5x2 – 4x – 6.

Sol: Using associative and distributive properties, we have (x3 + 5x2 – 4x – 6) – (2x3 – 3x2 – 1)

= x3 + 5x2 – 4x – 6 – 2x3 + 3x2 + 1

= x3 – 2x3 + 5x2 + 3x2 – 4x – 6 + 1

= (x3 – 2x3) + (5x2 + 3x2) + (–4x) + (–6+1)

= –x3 + 8x2 – 4x – 5.

The subtraction can also be performed in the following way:

Line (1): x3 + 5x2 – 4x – 6.

Line (2): 2x3 – 3x2 – 1.

Changing the sign of the polynomial in Line (2), we get Line (3): –2x3 + 3x2 + 1.

Adding the polynomial in Line (1) and Line (3), we get –x3 + 8x2 – 4x – 5 

Ex 5: Find the product of   given simple algebraic expression x3 – 2x2 – 4 and 2x2 + 3x – 1.

Sol:

(x3 – 2x2 – 4) (2x2 + 3x – 1)

= x3 (2x2 + 3x – 1) + (–2x2) (2x2 + 3x – 1) + (–4) (2x2 + 3x – 1)

= (2x5 + 3x4 – x3) + (–4x4 – 6x3 + 2x2) + (–8x2 – 12x + 4)

= 2x5 + 3x4 – x3 – 4x4 – 6x3 + 2x2 – 8x2 – 12x + 4

= 2x5 + (3x4 – 4x4) + (–x3 – 6x3) + (2x2 – 8x2) + (–12x) +4

= 2x5 – x4 – 7x3 – 6x2 – 12x + 4.


Algebra Identities to solve some simple problems :


 Algebraic identity for (x + a)(x + b)

 By using the distributive properties of numbers,

(x + a )(x + b ) = x(x + b) + a(x + b) = x2 + xb + ax + ab

= x2 + ax + bx + ab= x2 + (a + b)x + ab.

 (x + a)(x + b) ≡ x2 + (a + b)x + ab

(x – a)(x + b) ≡ x2 + (b – a)x – ab

(x + a)(x – b) ≡ x2 + (a – b)x – ab

(x – a)(x – b) ≡ x2 – (a + b)x + ab

(a + b)2 ≡ a2 + 2ab + b2

(a – b)2 ≡ a2 – 2ab + b2

(a + b)(a – b) ≡ a2 – b2

(a + b)3 ≡ a3 + 3a2b + 3ab2 + b3

(a − b)3 ≡a3−3a2b+ 3ab2−b3

a3 + b3 ≡ (a + b)3−3ab(a + b)

a3 − b3 ≡ (a −b)3+ 3ab(a−b)

Solve the following simple algebra problems using above identities:


Pro 1: Find the product:  (x + 3) (x + 5)

Sol: (x + 3) (x + 5) = x2 + (3 + 5) x + 3 × 5 = x2 + 8x + 15.


Pro 2:  Find the product:  (p + 9) (p – 2)

Sol: (p + 9) (p – 2) = p2 + (9 – 2) p – 9 × 2

= p2 + 7p – 18.


Solve simple algebra problems on the cubic identity


Algebra problems on cubic identity:

(i) (3x+2y)3 = (3x)3 + 3(3x)2(2y) + 3(3x)(2y)2 + (2y)3

= 27x3 + 3(9x2)(2y) + 3(3x)(4y2) + 8y3

= 27x3 + 54x2y + 36xy2 + 8y3.

 (ii) (2x2– 3y)3 = (2x2)3 – 3(2x2)2 (3y) + 3(2x2) (3y)2 – (3y)3

= 8x6 – 3(4x4)(3y) + 3(2x2)(9y2) – 27y3

= 8x6– 36x4y + 54x2y2 – 27y3


Pro 1: If the values of a+b and ab are 4 and 1 respectively, find the value of a3 + b3.

Solution:a3+ b3 = (a+b)3 – 3ab(a+b) = (4)3 – 3(1)(4)

= 64 – 12

= 52.


Pro 2: If a – b = 4 and ab = 2, find a3 – b3.

Solution:a3 – b3 = (a–b)3+3ab(a–b)

= (4)3+3(2)(4) =64+24

=88.

Median Test Learning

Median Definition :If the generally number of values within the case is even, next the median is the mean of the two center numbers. The median is a supportive number within cases wherever the distribution contain extremely large huge values which would or else twist the data. Represent the center value of a given set of values is called median.

To locate of values set within lower value toward higher value (rising order). If the collections of data contain an odd number of ways; the median is the middle value of the place behind arrangement the listing into rising order.


Median test learning


Method for median computation

`M=(n+1)/2`

Wherever,

M- Median,

n – Entirety number of set

But here is an odd number of values, then the median is the center value of the set.

If an even number to find median first

For example the given set{2,6,8,10}

Total number of value (n) =4

Median`M=(n+1)/2`

`M=(4+1)/2`

`M=(5)/2`

M=2.5

Consequently median must appear among second and third value.

Second value=6

Third value =8

Mean of second and third value is the Median of this set

Median`M=(14)/2`

`M=(7)/2`

M=3.5


Examples for median test learning


Example 1 for median test learning

Compute the median of this set {20, 10, 12, 40, 60,50,30}.

Solution:

First values can be arranged in order

Therefore the given set={10, 12, 20, 30, 40, 50, 60}

Here, total number of value (n) =7.

Median`M=(n+1)/2`

`M=(7+1)/2`

`M=(8)/2`

M=4

Therefore the fourth value is median that is 30.

Example 2 for median test learning

Compute the median of this set {10, 12, 13, 17, 18, and 20}

Solution:

Total number of value (n) =6

Median`M=(n+1)/2`

`M=(6+1)/2`

`M=(7)/2`

M=3.5

Consequently median must appear among Third and Fourth value.

Third value=13

Fourth value =17

Mean of third and fourth value is the Median of this set

Median`M=(13+10)/2`

`M=(30)/2`

M=15

Example 3 for median test learning
Compute the median of this set {26, 32, 36, and 55}

Solution:

Total number of value (n) =4

Median`M=(n+1)/2`

`M=(4+1)/2`

`M=(5)/2`

M=2.5

Consequently median must appear among second and third value.

Second value=32

Third value =36

Mean of second and third value is the Median of this set

Median`M=(32+36)/2`

`M=(68)/2`

M=34

Learning Weighted Median

In calculation of weighted median, the importance of all the items was considered to be equal. However, there may be situations in which all the items under considerations are not equal importance. For example, we want to find average number of marks per subject who appeared in different subjects like Mathematics, Statistics, Physics and Biology. These subjects do not have equal importance. Learning the formula to find weighted median by giving Median.


Learning Median Definition:


The arithmetic median computed by specific importance of every object is known weighted arithmetic median. To consider the every importance object, we can assume number known as weight to every object is directly proportional to its specific importance.

Weighted Arithmetic Median is computed by using following formula:

`sum` Yw = `(sum ty) / (sum t)`

Where:
Yw Stands for weighted arithmetic median.
y   Stands for values of the items and
t   Stands for weight of the item

Learning the important three methods of weighted median:

1.Arithmetic Median

2. Weighted Average

3. Average speed.

The formula for weighted average is:

Weighted Average = Sum of weighted objects / total number of objects


Learning to solve examples of Weighted median:


Ex 1:A student obtained 30, 40, 50, 60, and 35 marks in the subjects of Math, Statistics, Physics, Chemistry and Biology respectively. Assuming weights 1, 3, 5, 4, and 2 respectively for the above mentioned subjects. Find Weighted Arithmetic Mean per subject.

Solution:

Subjects	Marks Obtained       y	Weight    t	`sum` ty
Math	30	1	30
Statistics	40	3	120
Physics	50	5	250
Chemistry	60	4	240
Biology	35	2	70
Total		`sum` t = 15	`sum` y= 710

Now we will find weighted arithmetic mean as:
`sum` Yw = `( sum ty)/(sum t) = 710/ 15`   =  47.33marks/subject

Ex 2: A class of 30 students took a math test. 15 students had an average (arithmetic median) score of 90. The other students had an average score of 70. What is the average score of the whole class?

Solution:Step 1: To get the sum of weighted terms, multiply each average by the number of students that had that average and then sum them up.

90 × 15 + 70 × 15 = 1350 + 1050 = 2400

Step 2: Total number of objects = Total number of students = 30

Step 3: Using the formula

Weighted Average = Sum of Weighted objects / Total Numbers of objects.

=  `2400 / 30`

= 80.

Answer: The average score of the whole class is 80.

Learn Linear Functions in Real Life

In real life learning, linear function has single degree polynomial in its equation. The degree of the linear function always one. It cannot be increased. Linear function is  very useful in real life. In learning, linear function does not dependent any variable degrees. The linear function with the degree of one, always gives the straight line. In real life, linear function has the different variable equations. 'In learning of linear function in real life', we learn about linear equations and its degrees.

Forms of Linear function in real life


The function, which has straight line graph and that line function is known as linear function. In learning, linear functions are in three main forms.

1. Slope-Intercept Form is given by y = mx +b.

Example: y = 3x + 2

Here, slope m = 3 and y intercept b = 2

2. Point Slope Form is given by m = (y - y1) / ( x – x1)

Example: (y – 3) = 2(x – 1)

3. General Form is given by ax + by + c = 0

Example: 2x + 3y -1 = 0

Examples of set of real life linear function:

Example 1:

Solve for x and y. where, y = 3x + 4 and y = 7x.

Solution:

Plug, y = 3x + 4 in y = 7x

3x + 4 = 7x

x = 1

Now plug, x = 1 in y = 3x + 4

y = 3 + 4

y = 2

Answer:

The solutions are x = 1 and y = 2.

Example 2:

5x - y + 11 = 0 and x - y - 5 = 0, solve for x and y.

Solution:

5x - y + 11 = 0     (1)

x - y - 5 = 0         (2)

Subtract Equation (2) form Equation (1), we get,

4x + 16 = 0

4x = - 16

x = - 4

plug x = - 4 in equation (1)

- 4 – y + 13 = 0

- y + 9 = 0

y = - 9

The solutions are x = - 4 and y = - 9.


Learn linear function in real life - Practice problems:


Problem 1:

3x - y = 9 and 4x + y = 5. Solve for x, y.

A) (2, 3) B) (- 2, 3) C) (2, -3) D) (3, -2)

Answer: C

Problem 2:

2x + y = 12 and y = 3x - 3. Solve for x, y.

A) (3, 6) B) (3, - 6) C) (4, 5) D) (6, - 3)

Answer: A

Problem 3:

Solve the linear function 2x - y = 12 and y = 4x - 2

A) (- 5, 22) B) (5, - 22) C) (- 5, - 22) D) (5, 22)

Answer: C

solving online area of a circle

CIRCLE:

A line forming a closed loop, every point on which is a fixed distance from a center point.

A circle is a plane continuous figure connecting points which are equidistant from a given point, known as the center.

The word circle originates from the Latin word 'circus'. Chariot races, very popular in the Roman era were either circular or oblong and eventually the word was used to describe the shape as well.


Area of a circle - Important terms

In order to find the area of a circle, it is essential that we first understand a few important terms related to circles.

The outermost portion of the circle which separates the interior of the circle from the exterior is known as the circumference.

The center, as already mentioned, is the point inside the circle which is equidistant from all points lying on the circumference.

The distance between any point on the circumference from the center is known as the radius.

A line segment connecting any two points of lying on the circumference on the circle is called a chord.

The longest possible chord of a circle is known as a diameter. The diameter equals twice the radius.


Solving the area of a circle

Consider a circle with center P. Let us denote the radius of the circle by 'r' and the diameter by 'd'. Pi (denoted as π) is a numerical value, which is a crucial number used to calculate various attributes of circular figures. Its value up to 2 decimal points is 3.14. In terms of the radius.

we define the area (A) of the circle as:                  A = π r²

In terms of the diameter, which is twice the radius, this equation can be re-written as:

A = 0.25 π d²

The area in terms of the circumference (C) of the circle is given as:

A = C²/ 4π

Examples and exercises of area of circle.


Example 1: The radius of a circle is 3 inches. What is the area?

A = π r²

A  =  (  3.14  ) ( 3 ) 2

A = (3.14) (9)

A =  28.26 in 2

Answer  the following :

1) The diameter of a circle is 8 centimeters. What is the area?

2) The radius of a circle is 5 feet. What is the area?

Half Angle Properties

Introduction:

Let we will discuss about the half angle properties. The half angle is the angle that is half of original angle. Tha is, the product of two half angles at an edge is equal to the original(full) angle at that same edge. We will develop the properties formulas for half angle of both sine and cosine.


Half angle properties formula for sine:


We should start with formula for cosine of double angle. That is,
cos 2θ = 1 - 2sin^2 θ

Half angle properties formula - sine:

Let us consider,


Then 2θ = α and the formula becomes:


Solve for,


That is, we acquire sin(α/2) lying on the left of equation and everything else on right,



Solving of equation gives the following sine of half-angle identity:


The sign of sin α/2 represents on quadrant in which α/2 lies.
If α/2 is in the first or second quadrants then formula uses the positive case:


If α/2 is in the third or fourth quadrants, the formula uses the negative case:


Example:

Find the value of sin 45o using the sine half-angle properties.

Solution:

The sine half angle formula is ,



Therefore,

sin 45o = `sqrt((1 - cos 90 )/(2))`

= `sqrt((1 - 0 )/(2))`

= `sqrt((1)/(2))`

= 0.707


Half angle properties formula for cosine:


Half angle properties formula - cosine:

Using same process, with the similar replacement of θ = α / 2. We want to substitute into the identity,


We get,


Reverse process of the equation should be,


Addition of 1 with both sides of an equation. We get,


Making division process on both sides of equation by 2


Solving for cos(α/2), we get,


As before, the sign we need depends on quadrant.
If α/2 is in first or fourth quadrants that formula uses in positive case:


If α/2 is in the second or third quadrants, the formula uses the negative case:


Example:

Find the value of cos 115o using the cosine half-angle formula.

Solution:

We need to find cos 115o

That is, α = 230°, and so α/2 = 115°.

Therefore, cos 115o = `sqrt((1 + cos 230 )/(2))`

= `sqrt((1-0.643)/(2))`

= `sqrt((0.357)/(2))`

= 0.422

We have seen about the half angle properties.