Interest Expense Equation

Introduction :

In this article we will discuss about the interest expensive equation. The “  Interest expense equation ” gives the period of the interest rate. Interest rate is not always charged yearly. It will be charged every 6 months, Monthly, some times Daily!. The “ Interest rate” contains two types,  simple interest rate and Second compound interest rate. Here compound interest only denoted by the interest expense equation because compound interest gives the expensive interest amount. In summary, to calculate the  simple interest rate, just find out the interest rate for one period and then multifly the number of periods. To calculate the compound  interest rate or interest expense equation, first find out the interest rate for the first period, add its total  and then calculate the interest rate for the next period, and so on..,

Formula

Formula for interest expense Equation:

Formula for simple interest:

simple interest SI = pnr

where 

p  =  principal amount
n  =  period
r  =  interest rate


Formula for compound interest or interest expense equation:

compound interest  CI   =   P (1  +`r/100` )n

Where

p  =  principal amount
n  =  period
r  =  interest rate

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Examples

Example For interest expense equation:

Ricky  paid $2000 at 10%  in finance then after 4 years what amount he will get? .   Finance using simple interest and compound interest method. But ricky expect interest expense method.

Solution:

Here,

Principal amount  p = $2000

Period        n  =  4 years

Interest  rate        r =  10%

Simple interest SI = pnr

SI =2000*4*10/100 

=  800

Simple interest    =  800

Compound interest or interest expense CI   =   P (1  +` r/100`   )n

CI = 2000( 1+`10/100` )4

=  2000(`11/100` )4

=  2000( `11/10*11/10*11/10*11/10` )

=  2928

CI   =  $2928     

Total amount in simple interest  = 2000+800

Simple interest amount is   =$2800

But the compound interest amout is $2928 .

Interest expensively gives the compound interest.

So after 4 years Ricky got $2928.

Associative Property in Math

Introduction :

Algebra holds many laws. The followings are the basic algebraic laws. They are Associative law, distributive law and commutative law. In math associative property states that changes in group do not change the answer means the moving of brackets does not affect the answer. In math associative law consists of two types of operations. They are addition and multiplications.

Associative Property:

Associativeproperty:

In math associative property states that changes in group do not change the answer means the moving of brackets does not affect the answer. Associative law consists of two types of operations.

Two types of associative property as follows

Addition
Multiplication


Addition:

The addition of numbers is same. In math the associative property for addition says that grouping of numbers does not change the result or answer.

x + (y + z) = (x + y) +z

Example:

4 + (6 +3) = (4 + 6) +3

Multiplication:

The product of more numbers is same. Associative property for multiplication says that grouping of numbers does not change the result or answer.

x * (y * z) = (x * y) * z

Example:

7 * (2 * 8) = (7 *2) *8

Associative property is true for addition and multiplication but fails for both subtraction and division.

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Example Sums for Associative Property:

Example sums for associative property:

Example 1:

5 + (8 + 2) = (5 + 8) +2

5 * (8 * 2) = (5 * 8) * 2

Through the above problem prove both the associative property of addition and the associative property of multiplication

Associative property of addition:

Proof:

5 + (8 + 2) = (5 + 8) +2

LHS:

5 + (8 + 2)

= 5 + (10)

= 15

RHS:

(5 + 8) +2

=13 +2

=15

Left hand side equal to right hand side so the associative property of addition is proved

Associative property of multiplication:

Proof:

5 * (8 * 2) = (5 * 8) * 2

LHS:

5 * (8 * 2)

= 5 * (16)

= 80

RHS:

(5 * 8) * 2

= (40) * 2

= 80

Left hand side equal to right hand side so the associative property multiplication is proved

Example 2:

7 + (3 + 5) = (7 + 3) +5

Through the above problem prove the associative addition property.

Associative property of addition:

Proof:

7 + (3 + 5) = (7 + 3) +5

LHS:

7 + (3 + 5)

=7 + (8)

=15

RHS:

(7 + 3) +5

= (10) +5

=15

Left hand side equal to right hand side so the associative property of addition is proved

Lateral Area of a Cylinder Calculator

Introduction to lateral area of the cylinder calculator:

A cylinder is one of the most basic curvilinear geometric shapes, the surface formed by the points at a fixed distance from a given straight line, the axis of the cylinder. The solid enclosed by this surface and by two planes perpendicular to the axis is also called a cylinder. Lateral area of the cylinder is calculated by the following formula,                                                              (Source from Wikipedia)

Lateral area of the cylinder = 2 `pi` r h

Here

r is the radius of the cylinder

h is the height of the cylinder

Here we are going t o study about how to find the lateral area of the cylinder using calculator and its example problems.

Lateral Area of Cylinder Calculator:

The above figure represents the lateral area of cylinder calculator .First we have to enter the radius and height of the cylinder .then we have to press the calculate button we get the answer.

Example problems:

Example: 1

Find the lateral area of the cylinder with radius is 7 meter and height of the cylinder is 9 meter.

Solution:

We know the formula for lateral area of the cylinder = 2 `pi ` r h

Here r = 7 meter, h = 9 meter `pi` = 3.14

Substitute this value into the formula we get

Lateral area = 2 * 3.14 * 7 * 9

= 395.64

Therefore the lateral area of the cylinder is 395.64 square meter

Example: 2

Find the lateral area of the cylinder with radius is 11 inches and height of the cylinder is 13 inches.

Solution:

The formula for lateral area of the cylinder = 2` pi` r h

Here r = 11 inches, h = 13 inches, `pi` = 3.14

Lateral area = 2 * 3.14 * 11*13

Simplify the above we get

= 898.04

Therefore the lateral area of the cylinder is 898.04 square inches

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Lateral Area of Cylinder Calculator - Practice Problems

Find the lateral area of the cylinder with radius is 12 inches and height of the cylinder is 9 inches.
Answer = 678.58 square inches

Find the lateral area of the cylinder with radius is 15 meter and height of the cylinder is 14 meter.
Answer = 1319.46 square meter

Dividing Rational Numbers

Introduction to Dividing Rational Numbers

The conditions for dividing positive and negative integers also apply to division with rational numbers.

Keep in mind that to divide by nonzero number, multiply by the reciprocal of that number.

Dividing Rational Numbers – Examples:

Dividing Rational Numbers – Example 1:

`(-39b +65)/13`

Step 1:

The Factor bar Indicates division:

`(-39b +65)/13 ` = -39b + 65 `-: ` 13

Step 2:

Multiply by 1/13 the reciprocal of 13.

= -39b+65(`1/13` )

Step 3:

Distributive Property

`(-39b)(1/13)` + `65(1/13)`

Answer:

-3b +5

Dividing Rational Numbers – Example 2:

Simplify:

(-51 + 34c)/-17

Step 1:

The Factor bar Indicates division:

= (-51+34c) `-:` (-17)

Step 2:

Multiply by` 1/17` the reciprocal of 17.

= (-51 + 34C). (`- 1/17` )

Step 3:

Distributive Property

= -51`(-1/17)` + (34C)`(-1/17)`

Answer:

= 3 -2c

Dividing Rational Numbers – Example 3:

Cricket pitch perimeter of a square cricket pitch is 36 feet. Find the length of one side of the rectangle

To find the length of one side, divide the perimeter by the number of sides.

Same signs -> Positive quotient

36` -:` 4 =9

Answer: The length of one side is 9 feet.

Dividing Rational Numbers – Example 4:

Solve:

`-56/27` `-:` ` -14/9`

Step 1:

Reciprocal of` -14/9 ` is `-9/14`

= `(-56/27) xx(-9/14)`

Step 2:

Multiply with reciprocated number

= `(56 xx 9 )/ (27 xx 14)`

Step 3:

Simplify

=`(4xx1)/(3xx1)`

Answer:

=` 4/3`

Dividing Rational Numbers – Example 5:

Solve: 

`35/32` `-: ``-21/16`

Step 1:

Reciprocal of` -21/16` is` -16/21`

`-35/32 xx 16/21`

Step 2:

Multiply with reciprocated number

= `(35xx16)/(-32 xx 21)`

Step 3:

Simplify

=`(5xx1)/(-2xx3)`

Answer:

= `-5/6`

Dividing Rational Numbers – Example 6:

`(-39b +65)/12`

Step 1:

The Factor bar Indicates division:

`(-39b +65)/12` = -39b + 65 `-: ` 12

Step 2:

Multiply by `1/12` the reciprocal of 12.

= -39b+65`(1/12)`

Step 3:

Distributive Property

`(-39b)(1/12) + 65(1/12)`

Answer:

`(-39b)/12 + (65/12)`

Dividing Rational Numbers – Example 7:

Cricket the perimeter of a square baseball diamond is 400 feet. Find the length of one side of the circle

To find the length of one side, divide the perimeter by the number of sides.

Same signs -> Positive quotient

400` -:` 4 =100

Answer: The length of one side is 100 feet.

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Dividing Rational Numbers – Practice Problems:

Solve

`(-40b +66)/12`

Answer

`((-10b)/3) + (33/6)`

Solve Inconsistent System of Equations

Introduction:

Algebra is a section in math, which comprises of infinite number of operations on equations, polynomials, radicals, rational numbers, logarithms, etc. Solving a set of linear equation is also an element of algebra. While solving, if there is no solution it is called as inconsistent set of equations else they are said to be consistent.

Examples to Solve Inconsistent System of Equations:
1. Find whether the following set of equation is consistent or inconsistent?

2x+3y = 1

4x+6y =2

Solution:

Elimination method:

In this method we try to eliminate one variable either ‘x’ or ‘y’ and then find out the remaining variable’s value.

2x+3y = 1-----------> (1)

4x+6y =2 -----------> (2)

Multiply equation (1) with ‘2’and subtract it from (2),

Therefore, 2(2x+3y – 1) =0

4x+6y -2 =0

4x+6y-1 – (4x+6y -2) =0’

4x +6y-1-4x-6y+2 =0,

4x-4x =0 similarly 6y-6y =0,

Therefore, -1+2=0

Since both the ‘x’ and ‘y’ terms have vanished, it is not possible to get a solution.

Hence, this set of equations is inconsistent.

2. Find whether the following set of equation is consistent or inconsistent?

x+4y = 1

3x+12y =2

Solution:

Substitution method:

x+3y = 1-----------> (1)

3x+9y =2 -----------> (2)

Solve the first equation for ‘x’

Subtract by ‘3y’ on both sides,

Therefore x+3y -3y =1-3y,

Since 3y-3y =0, x =1-3y,

Substitute x =1-3y in (2)

3(1-3y) + (9y) =2,

3-9y+9y =2,

3=2.

Since both the ‘x’ and ‘y’ terms have vanished, it is not possible to get a solution.

Hence, this set of equations is inconsistent.

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Solve Practice Problems: Inconsistant System of Equations

1. Check whether the system of equations are inconsistent?
x +y=1
x +y=2
2. Solve and comment on the system solution,
5x +3y =1
15x+9y =4
3. Check whether the set of equations are inconsistent?
3x+2y=5
4x+6y=3
4. Solve and comment on the system solution,
2x+2y=5
4x+6y=6
5. Solve and comment on the solution,
7x+2y=5
5x+6y=7
6. Solve and comment on the system  solution?
8x+6y=5
8x+6y=3
7. Solve and comment on the system solution?
x -y=1
x -y=5

Correct Scientific Notation

Introduction to correct scientific notation: 

Generally, we approach across extremely big as well as extremely little numbers. Writing this number such as extremely difficult. The use of correct scientific notation helps us to state them in a suitable form. Exponential notation is also identified as scientific conversion which is nothing but writing the numbers in the shape of m x 10 n.

Correct Scientific Notation Terms:

The correct scientific notation help terms are given below that,

Coefficient

Exponent

In the term m x 10 n, m is the co-efficient and n is the exponent.  The co-efficient term m can be of some real number and the exponent n have to be an integer. The coefficient expression is as well recognized as mantissa.

Example: 

3. 4 x 108

3.4 = coefficient or mantissa

8 = exponent.

Steps to find correct scientific notation:

Steps to find correct scientific notation is given below that:

The correct scientific notations have one number to the left of the decimal point.

We surround to set the decimal point subsequent the first digit.

How many places the decimal position was stimulated point out through the power of ten (10).

A fact in correct Scientific notation is writing as an ordinary form:

X×10y  (where X is the coefficient that is in among 1 and 10. And ‘y’ is the exponents)

Coefficient ×10exponents

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Examples for Correct Scientific Notation:

Example 1:

4876 change into correct scientific notation.

Solution:

Specified that 6581

Specified integer is not a big number, other than it is easy to convert scientific conversion.

Initial note down 6.581this is not equivalent to specified integer

(6.581)(1000) = 6581

At the present 1000 = 10 3          

Following that it can be note down in scientific conversion = 6.581 x 10 3

Example 2:

654123897 change into correct scientific notation.

Solution:

Specified that 654123897

Specified integer is a large number, other than it is easy to convert scientific conversion.

Initial note down 6.54123897 this is not equivalent to specified integer

Estimate how many digits available subsequent the decimal point in previous step

There are 8 digits subsequent the decimal point so you be able to place 8 zero

6.54123897 (100000000)

So the conversion determination be 10 8

The scientific conversion can be note down as 6.54 x 10 8

Example 3:

0.00753 change into correct scientific notation.

Solution:

Specified that 0.00753

Specified integer is not a large number, other than it is simple to convert scientific conversion.

Initial note down the integer as 7.53

Present are 3 digit subsequent the number is there so we can write 10 -3

The scientific conversion can be note down as 7.53 x 10 -3

Practice problems for correct scientific notation:

Problem 1:

91000000 change into correct scientific notation.

Answer: 9.1 x 107

Problem 2:

591000 change into correct scientific notation.

Answer: 5.91 x 105

Problem 3:

35759182 change into correct scientific notation. 

Answer: 3.57 x 107

Sector of a Circle

Introduction

The two radii of the central angle θ and the formation of the arc in between the points on the circle which forms the sector of a circle .The sector of a circle is nothing but the calculation of a part of the whole area of the circle. The angles in the circle which can be measured in degrees or radians.

Let us take the angle θ of the circle to be the central angle and the total area circle is given as πr2.In this the 2π is the angle for the whole part of the circle.

Formula for the Sector of Circle:

Formula:  The area of a sector of circle:

Area = `(theta pi R^2)/360`

Where, C is the central angle in degrees,

R is the radius of the circle

π is Pi, approximately 3.142

The formula is obtained by taking the area of a whole circle, and then a fraction of it which depends on the central angle is the area of sector. In the sector of the circle, if the central angle is 90°, then the sector of the circle will have some portion which is equal to one quarter of the whole circle..

When the central angle is in radians, then the formula for area of the sector can be given as,

Area=R2C/2

Where, C is the central angle in radians and

R is the radius of the circle of which the sector is part.          

Example for Sector of Circle:

Ex 1: The sector of a circle consists of radius 8 cm and angle 40°.Find the area of the sector of a circle and also the major part of the circle.

Sol:

Area of sector = `theta/360` × πr2

= `40/360`  × 3.14 × 64

= 22.34 cm2

Thus, the area of the major circle = area of circle – area of minor circle.

= 50.2 – 22.34

= 27.86 cm2

Ex 2: If the radius of the circle is equal to the value of 12 inches and the central angle is 1500.  Solve the area of the sector of a circle.

Sol:

The given values are r = 12 inches and `theta` = 1500

Area:

Area of a sector is solved through the following formula,

A=` (theta)/(360^0)` x π r2

That is A= `(150^0) / (360^0) ` x 3.14 x 12 x 12

Simplifying this we have to get

A= 10800 square inches.

This is the area of the arc of a sector.

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Ex 3: If the radius of the circle is  5 cm and the central angle is  1470.  Solve the area of the sector of a circle.

Sol:

The given values are r = 5 cm and theta = 1470

Area:

Area of a sector is solved through the following formula,

A= `(theta)/(360^0)` x π r2

That is A= `(147^0) / (360^0)` x 3.14 x 5 x 5

Simplifying this we have to get

A= 1837.5 square cm.

This is the area of the arc of a sector.