Multiply Formula

In this article we are going see about how to multiply the two variables and numbers. Multiplication is the process of find the product of the given values. We use the distributive formula for finding the multiplication values. Multiplication formulas are reducing the man power and easy for calculation. Now in this article we solve some problems using multiplying formula.

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Distributive formula:

a * (b + c) = (a * b) + (a * c)

x * (y * z) = (x * y * z)
Example Problems for Multiply Formula

Multiply formula example problem 1:

Multiply the given two function f(x) = (2x + 4) and g(x) = (6x - 2)

Solution:

Given functions are f(x) = (2x + 4) and g(x) = (6x - 2)

Add each and every variables of the given two functions, we get

f(x) * g(x) = (2x + 4) * (6x - 2)

Using distributive property,

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

Expand the above values, we get

= 12x2 - 4x + 24x - 8

After simplification, we get

f(x) * g(x) = 12x2 + 20x - 8

Answer:

The final multiplication value of the given function is 12x2 + 20x - 8

Multiply formula example problem 2:

Multiply the given two function f(x) = (10x + 3) and g(x) = (x - 3)

Solution:

Given functions are f(x) = (10x + 3) and g(x) = (x - 3)

Add each and every variables of the given two functions, we get

f(x) * g(x) = (10x + 3) * (x - 3)

Using distributive property,

= 10x (x - 3) + 3 (x - 3)

Expand the above values, we get

= 10x2 - 30x + 3x - 9

After simplification, we get

f(x) * g(x) = 10x2 - 27x - 9

Answer:

The final multiplication value of the given function is 10x2 - 27x - 9

I am planning to write more post on Log Calculator and Example of Histogram. Keep checking my blog.

Multiply Formula Example Problem 3:

Multiply the given two function f(x) = (3x2 + 1) and g(x) = (x + 2)

Solution:

Given functions are f(x) = (3x2 + 1) and g(x) = (x + 2)

Add each and every variables of the given two functions, we get

f(x) * g(x) = (3x2 + 1) * (x + 2)

Using distributive property,

= 3x2 (x + 2) + 1 (x + 2)

Expand the above values, we get

= 3x3 + 6x2 + x + 2

After simplification, we get

f(x) * g(x) = 3x3 + 6x2 + x + 2

Answer:

The final multiplication value of the given function is 3x3 + 6x2 + x + 2

Volume of Three Dimensional Shapes

Volume is how much three-dimensional space a substance or shape occupies or contains. Three dimensional mathematical shapes are also assigned volumes. Volumes of some simple shapes, can be easily calculated using arithmetic formulas. One-dimensional figures (such as lines) and two-dimensional shapes (such as squares) are assigned zero volume in the three-dimensional space. (Source: From Wikipedia).

Here we are going to see the formulas to find the volume of three dimensional shapes and example problems.
Formulas to Find the Volume of three Dimensional Shapes

Here we are going to see some arithmetic formulas to find the volume of simple three dimensional shapes such as cube, cone, cylinder, and sphere.

Volume of cube = a3 cubic units. Where, a is the side of the cube.
Volume of cone = `1/3 pi r^2 h` cubic units. Where, r and h are the radius and height of the cone.
Volume of cylinder = `pi r^2 h` cubic units. Where, r and h are the radius and height of the cylinder.
Volume of sphere = `4/3 pi r^3` cubic units. Where, r is the radius of the sphere.

Example Problems to Find the Volume of three Dimensional Shapes

Example 1

Find the volume of a three dimensional shape with all sides equal to 3 feet.

Solution

A three dimensional shape with equal sides is a cube.

Volume of a cube = `a^3` cubic units

= 33

= 3 * 3 * 3

= 27

So, the volume of the given three dimensional shape is 27 cubic feet.

Example 2

Find the volume of a sphere, whose radius is 5 m.

Solution

Volume of a sphere = `4/3 pi r^3` cubic units

= `4/3` * 3.14 * `5^3`

= `4/3` * 3.14 * 5 * 5 * 5

= 523.33

The volume of the given sphere is 523.33 cubic meter.

Example 3

Find the volume of a cylinder with radius 3 cm and height 4 cm.

Solution

Volume of a cylinder = `pi r^2 h` cubic units

= 3.14 * 3 * 3 * 4

= 113.04

So, the volume of the given cylinder is 113.04 cubic cm.

Algebra is widely used in day to day activities watch out for my forthcoming posts on equation of line and Distance From a Point to a Line. I am sure they will be helpful.

Example 4

Find the volume of the cone, whose radius is 3 cm and height is 4 cm.

Solution

Volume of a cone = `1/3 pi r^2 h` cubic units

= `1/3` * 3.14 * 3 * 3 * 4

= 3.14 * 3 * 4

= 37.68

So, the volume of the given cone is 37.68 cubic cm.

Depreciation Value Formula

Depreciation is the reduction in the value of the asset year by year due to wear and tear. Depreciation can be calculated using the straight line depreciation method or the accelerated depreciation method.

The straight line method calculates the depreciation  by spreading the cost evenly over the life period of the fixed asset.
Accelerated depreciation method calculates the depreciation by expensing  a large part of the cost at the beginning of the life of the fixed asset.



Formula for Straight line depreciation method =  Cost / life.

Ex :- The cost of a machine is $ 100.  It is expected to last for 4 years. Calculate the annual depreciation.

Sol:  Cost of the machine = $ 100

Expected time         =       4

Therefore annual depreciation is 100 / 4 = $25

Every year $25 would be expensed as depreciation value.

In this method the salvage value is not taken into account.
Depreciation Value:

Another method used is called accelerated depreciation  method  or declining balance depreciation method. It uses a factor based on the life of the asset. The factor is the percentage of the asset that would be depreciated each year under straight line depreciation times the accelerator.

Let us take the cost of the machine that is $100. Under this system we double the depreciation period a 50% from 25%. This is called double declining balance

Ex:  Cost of a machine is $100.  Its life period is 4 years.  What is the depreciation factor?

Sol: The depreciation factor is calculated  by doubling  as 200% = 2 * (1/4) = 0.50

So the calculation runs like this:-

Year              Depreciable basis   Depreciation calculation    Depreciation expenses  Accumulated depreciation

1                     $100                       100 * 0.5                              50                                  50

2                     $  50                        50 * 0.5                              25                                   75

3                     $  25                        25 * 0.5                              12.50                              87.50

4                     $  12.50                 12.5 * 0.5                                6.25                              93.75

So over the  four years the depreciation has been $ 93.75

The salvage value is $100 - $93.75 = $6.25

This depreciation value formula is adopted by most of the manufacturing units where there is considerable wear and tear.

It is also used by  transport system such as lorries, goods carrier where the wear and tear is considerable.
Formula of Depreciation Value :

There is yet another formula for finding the depreciation value. It is  written as  A= P( 1 - i)n

A = the depreciated amount : P = the present value ;  i = rate of depreciation and n = number of years.

It is also written as FV = PV ( 1 - i)n  where FV is future value ; PV = present value ; i = rate of depreciation and n = number of years.

We use the subtraction sign (1 - i) because the value goes down.

Let us do a problem using this formula.

Ex: A machine depreciates  in value each year at the rate of 10% of its value at the beginning of a year.  The machine was purchased for $10,000.  Obtain its value at the end of 10th year.

Sol:

Present value = PV = $10,000

rate of depreciation =10% = 0.1

Number of years = 10                                                                                                                          _                               _

So   FV = 10,000 ( 1 -0.1)10  = 10,000 (0.9)10  = log  0f 10,000 + 10 log 0.9 = 4.0000 + 10 x  1.9542 = 4.0000 + 1.5420

=  3.5420

Antilog of  3.5420 = 3483

Hence the depreciated value of the machine whose purchase price was $10,000  at the depreciation rate of 10% for 10 years becomes $ 3,483

Ans = $ 3,483

Independent Random Vectors

The basic needs of vectors are to provide the relationship between the magnitude and direction. Let us take any two random vectors respectively s and t. If the directions of these random vectors are in the different direction and different magnitude we can say that these two vectors are independent random vectors.  If any of two vectors will be in the different places, we can say that the two are independent.

Independent Random Vectors:

Let us take {X1, X2, . . .  Xn} are the set of n random variables.

This is often convenient to consider these random variables set at a single object x = {X1, X2, . . .  Xn}. This will be called random vectors.  If it is particularly true true when this set of random variables submitted to the linear transformations.

If a random variable is described using its probability distribution the random vector is always described using its joint probability distribution of the n random variables which will make the random vector.
Expectation of Independent Random Vectors:

By the definition of independent random vectors the expectation is the vector whose components are the expectation of the independent random vectors.

If we make the vector

E(x) =` [[E[x_1]],[E[x_2]],[[....]],[E[x_n]]] `

Where the mean μ = E[x] and this is the center of gravity of the joint probability distribution of the random variables {X1, X2, . . .  Xn}

Linearity of the independent random vectors:

If the random vectors are immediately verified that the expectation is linear

If x and y are two independent random vectors, and if A and B are two constant matrices:

Then E [Ax + By] = AE[x] + BE[y]

And if b is a constant vector:

E [Ax + b] = AE[x] + b

My forthcoming post is on First Order Differential Equation and Definite Integrals will give you more understanding about Algebra.


Variance of the independent random vector:

Definition of the variance of the random vector and the random variable are equal. If we denote μ and the vector E[x], then by definition of the variance:

Var(x) = E[(x – μ)(x – μ)']

Estimating Fraction Calculator

Fractions:

A fraction is a number that can represent part of a whole. The earliest fractions were reciprocals of integers: ancient symbols representing one part of two, one part of three, one part of four, and so on. A much later development was the common or "vulgar" fractions which are still used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.

Calculator:

A calculator is a small (often pocket-sized), usually inexpensive electronic device used to perform the basic operations of arithmetic.(source : Wikipedia).

estimating fractions calculator

By giving input fraction we can get the output as per our operation. Let us see some problems on estimating fractions calculator.
Problems on Estimating Fractions Calculator :

1. Addition of fractions:

To add a fraction we need the common denominator.  We can add the fractions with same denominator.

Example 1:

Estimation the addition of fractions  3/4 + 6/4

Solution:

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

= 9/4

Example 2:

Estimate the addition of fractions 5/7 + 9/6

Solution:

To add a fraction we need to make common denominator,

5/7 + 9/6 = (5 * 6) / (7 * 6) + ( 9 *7) / ( 6 * 7)

= 30/42 + 49 /42

= ( 30 + 49 ) / 42

= 79 / 42

I am planning to write more post on Surface Area of a Cone Formula and Rectangular Prism Volume. Keep checking my blog.

2.Estimating subtraction of fraction:

Example 1:

Estimating the subtraction for the following fractions 5/4 - 2/4

Solution:

Given , 5/4 - 2/4

In the above fractions , the denominator is equal.

So we can subtract the numerator of the fractions  just like integers and keep the denominator,

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

= 3/4

Answer: 5/4 - 2/4 = 3 / 4

Example 2:

Estimating the subtraction for the following fractions 5/4 - 9/8

Solution:

Given,  5/4 - 9/8

Both fractions are has different denominator,

So we need to find the lcd for 4 and 8.

Multiple of 8 = 8 , 16 ,24 ,.....

Multiple of 4 = 4 ,8, 12 ,16 ,....

The least common multiple of 4 and 8 is 8.

Multiply 5/4 by 2 on both numerator and denominator ,

(5*2) / ( 4 * 2) = 10 /8

Now we can add the fractions,

5/4  - 9 / 8 = (10 /8) - (9 / 8)

= (10- 9 ) /8

= 1 / 8

Answer : 5/4  - 11/8 = 1/ 8
Problems on Estimating Fractions Calculator :

3. Estimating multiplication of  fractions:

Just like integers we can multiply the fractions.

Example:

Multiply 3/8 * 24 / 27

Solution:

Given, 3/8 * 24 / 27

3/8 * 24 / 27 = (3 * 24) / ( 8 *27)

= 72 / 216

Divide by 72 on both numerator and denominator,

= 1/3

4.Estimating division of fractions:

Example :

Divide the fractions 14/18 ÷ 7/9

Solution:

Given, 14/18 ÷ 7/9

14/18 -> Dividend

7/9 -> Divisor

Take a reciprocal of the divisor,

Reciprocal of 7/9 = 9/7

Multiply the reciprocal of divisor by the dividend.

14/18 * 9/7 = (14 * 9) / ( 18 * 7)

= 126 / 126

= 1

Answer: 14/18 ÷ 7/9 = 1

Solving for a Specific Variable

Algebra includes all the concepts like variables, constants, expressions, exponents, equation and etc . Variable is one of the main terms in math. Variables do not change the meaning of the expressions. Generally algebra expression includes variables. Commonly variables can be represented using alphabets. Here is the example, 2y^2+4y+2.Here we are going to learn about solving for a specific variables.

I like to share this Different Types of Variables with you all through my article.

Simple Example Problems of Solving for a Specific Variable:

Example 1:

A= bc then solve for b .

Solution:

Step 1: divide using c on both the sides .

Step 2:So, A/c = (bc)/c . (c term will be cancelled )

Step 3:therefore, the value of b is A/c .

Example 2:

P= 2l+2w Solve for w.

Solution :

Step 1: the given question is p= 2l+2w .

Step 2: p-2l =2l-21+2w.(Subtract 2l on both the sides )

Step 3: When we simplify we get p-2l=2w.

Step 4: (p-2l)/2 =(2w)/2 (Divide using 2 on both the sides)

Step 5: Therefore the value of w is (p-21)/2 .

These are the simple examples of solving  for a specific variables.
Some more Examples of Solving for a Specific Variables.

Example 3:

Q=(c+d)/2  then solve d.

Solution :

Step 1: The given question is q= (c+d)/2  .

Step 2: Multiply 2 on both the sides. So, 2q=(c+d) /2   xx 2 .

Step 3: When we simplify we get 2q= c+d.

Step 4: Subtract c on both the sides . So 2q-c=c-c+d.

Step 5: Therefore, the value d is 2q-c

Example 4:

V= (3k)/t  then solve t .

Solution :

Step 1: Multiply t on both the sides Vt = (3k)/t  xx t

Step 2: When we simplify we get Vt = 3k.

Step 3: Divide using V on both the sides. So (Vt)/V =(3k)/V

Step 4: Therefore, the value of t = (3k)/v .

I am planning to write more post on prime factorization of 72 and how to do standard deviation. Keep checking my blog.

Example 5:

Q =3a+5ac then solve a.

Solution:

Step 1: The given question is q= 3a+5ac .

Step 2: Here a is a common term. Take the common term as outside.

Step 3: Now we have q= a(3+5c). (divide using 3+5c on both the sides).

Step 4: (q)/(3+5c) = (a(3+5c))/(3+5c) .

Step 5: When we simplify we get (q)/(3+5c) = a. Therefore the answer is (q)/(3+5c) =a.

These are the example problems of solving for a specific variables.