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.

Please express your views of this topic Multiply Fraction by commenting on blog.

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.

Solve Interval Notation

The interval notation denotes the set of real datas with the property that any data that deception among in the set is also has in the set. For instance, the group of all the datas x satisfying 0 ≤ x ≤1 this is an range which has 0 and 1, as well as all numerical values among 0 and 1.

We can represent the inequality result by solving the interval notation.
Solving Interval Notation:

We can express the inequality outcomes by solving the interval notation

The symbols used in solving the interval notation are,

( - is the “not included” symbol or “open” symbol

[ - is the “included” symbol or “closed” symbol

Solving Open Interval:

(p, q)  is written as p ≤ x ≤ q  here the termination points are not included.

open interval

Solving Closed Interval:

[p, q]  is written as p < x < q  here the termination points are included.

closed interval

Solving Half-Open Interval:

(p, q]  is written p < x
half-open interval

Solving Half-Closed Interval:

[p, q) is taken as p < x
half-closed interval

Solving Non-ending Interval:

( p , ∞) is taken as x > p where p is not incorporated and infinity is always defined as being "open".

non-ending interval

Solving Non-ending Interval:

( -∞ , q ] is interpreted as x < q here q is incorporated and again, infinity is always defined as being "open".

non-ending interval1

I am planning to write more post on hex to decimal conversion and cbse nic sample papers. Keep checking my blog.

More about Solving Interval Notation:

If we accomplish the favored set of outcomes we can use the concoction of interval notations. For example define the interval of all values except 9.

As an inequality  x<9 or="or" x="x">9

In an interval notation ( -∞ ,9) U (9,∞).

Scalar Line Integrals

Scalar line integral is a definite integral it will be taken over a surface and integrated, so the scalar line integral is defined as the sum of all points in the surface.  Let x: [a, b] gives R3 be a path of class C and f: X subeR3 gives R be a continuous function contains the image of x.  The scalar line integral of f along x is

int_a^bf(x(t)) || x'(t) dt

Notation is usually written as int_x f ds.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Definite Integrals and Anti derivative. I am sure they will be helpful.

General form of scalar line integral:

The scalar function of line integral  is

if F = F1 i + F2 j + F3 k

r = x i + y j + z k

So int_cF. dr = int_c (f_(1)dx + f_(2) dy + f_(3) dz)

Scalar line integral and parametrizations

If y is a reparametrization of x. Then

If y is orientation-preserving, then int_y F.ds =  int_x F.ds

If y is orientation-preserving, then  int_y F.ds  = -int_x F.ds


When the path is parametrized by length of arc, the natural analog of the integral has done in 1 dimension. The integral of a scalar function f along a curve r(s) is simply int f (r(s)) ds

Applications:

F is a force acting upon a the particular particle so the particle moves along a curve C in  sample space and r be the position vector of the  given particle at a point on C. Then work done by the given particle at C is F.dr and the total work is done by F along a curve C is given by the line integral  int_c F. dr



Having problem with Convergent and Divergent Boundaries keep reading my upcoming posts, i will try to help you.


Scalar Line Integrals don’t depend on parameterizations.
Scalar Line Integrals-example Problems

Evaluate I = int_e f(x, y, z) ds where f(x, y, z) = = x2 – y2 – 1 + z and e is the helix parametrized by c(t) = (cos t, sin t, t)  [o <= t <=pi]. 3

Solution:

I = int_alpha^beta f(c(t)) || c'(t) || dt

See that f (c (t)) = cos2t + sin2t - 1

Also  c' (t) = (-sin t, cos t, 1) =    sqrt(-sin t^(2) + cos^(2) t + 1^(2))

See that f(c (t)) = cos2t + sin2t - 1

I = int_0^(3pi)sqrt(2) t dt

= sqrt(2) [(1)/(2)  t^(2) ][]_0^(3pi)

= (sqrt(2))/(2) 9pi^(2).

The correct answer for scalar line integral is = (sqrt(2))/(2) 9pi^(2).

Practice Problem:

Evaluate int sin 6x cos 3x dx


Answer: -1/2 [(cos 9x / 9) + (cos 3x / 3)] + c.

Define Natural Logarithm

There are three mathematical quantity related to the function  ea = x ,  Here the quantity " x " is said to be natural lagarithm of the number " a ". And the quantity " e " is said to be the base of the log and last one  is x which is power of the natural logarithm .The value of natural logarithm is given by as follow:

logex =a.

We can state it as above .

To show the formula

logex =a. and  ea = x represents the same we can take some examples as .

loge 10 = 2.3025

And the      e2.3025 = 10,so both formula exists.
Graph of Natural Logarithm:


Examples on Natural Logarithms:

Addition rule –


logex + logey   = logexy

Subtraction rule –

logex + logey   = logex/y
Solved problem :

Ex 1: Solve loge2 + loge4

Sol: Assume base as e. so  log 2 +log 4 = log 8

Ex 2: Solve loge4 - loge2

Sol: Base is e than log 4 -log 2 = log 2

Practice questions:

Que 1 : Change the following from exponential form to natural log form

e2.3025 =10
e1 =e

Ans : a. is  loge10 =2.3025

b.    is   normal form  as   logee = 1

Que 2: For log 23+ log 3 = log x then x=?,where all base is e.

Ans: 69

Que 3:For log 24 – log 4 =log x ,where all base is e.What is the value of x?

Ans: 6

Simple Logarithms

In mathematics, the logarithms of a number to a given base is the power or exponent to which the base must be raised in order to produce that number. For example, the logarithm of 100000 to base 10 is 5, because 5 is the power to which ten must be raised to produce 100000: 105 = 100000, so log10100000  = 5. Only positive real number  have real number logarithms; negative and complex numbers have complex logarithms.

I like to share this Simplifying Logarithms with you all through my article.

Simple logarithms are simple step produced by the problem.
Simple Logarithms Rules:

Let us see some of the simple steps that used to solve the logarithims.

Product rule: If a, p and q are positive numbers and a ?1, then

loga(pq) = logap +logaq

Quotient rule: If p, q and a are positive numbers and a ? 1, then,

log a(p/q) = log ap –loga q

Power rule: If a and q are positive numbers, a ? 1 and m is a real number, then

logapq =qlogap

Change of base rule: If p, q and a are positive numbers and p ? 1, a ? 1, then

Log pq = logap* logqa

Reciprocal rule: If p and q is the positive numbers other than 1, then

Log pq =1 / log pq
Examples Simple Logarithms:

Example 1:

Reduce: 22log3 27 + 22log3 729      (ii)75 log5 8 +75  log5 5/1000

Solution:

(i) Since the expressions is a sum of two logarithms and the bases are equal, we can apply the product rule

(i) 22log3 27 +4log3 729 =22 [log 3 (27*729)]

=22[ log3 (33*36)]

= 22log3 39 =22* 9 log33 = 22*9=198

(ii)               75 log58+75log5(`5/1000` )  =75 log5 (`(8*5)/1000` )

=75 log 5(`1/25` )

=75log 5(`1/ 5^2` )

=75 log5(5-2)

= -2*75 log55 = -150*1= -150



My forthcoming post is on prime factorization method, and  formula for conditional probability will give you more understanding about Algebra.


Example 2:

solve: 89log7 98-89log714

Solution:

89log7 98-89log714 =89log 7(`98/14` )

=89 log77 =89*1= 89

Example 3:

Solve .log650x – log6(4x+1)

Solution:

Using quotient law, we can write the equations as log6 (50x / 4x+1) Changing into exponential form, we get

(50x /4x+1) = 60

50x = 4x + 1

46x=1

x=`1/46`

polynomials

In mathematics, a polynomial is an expression of finite length constructed from variables (also known as indeterminates) and constants, using only the operations of addition, subtraction, multiplication, and non-negative, whole-number exponents.

Rational function:

In mathematics, a rational function is any function which can be written as the ratio of two polynomial functions. It can be  written as((x - 2) / (x + 3))

(Source: Wikipedia)
Example Problems for Polynomials Rational Expressions

Polynomials rational expressions example problem 1:

Simplifying the given rational expressions ((5x + 15) / (10x + 40))

Solution:

Given rational expression is ((5x + 15) / (10x + 40))

Take 5 as common in the numerator value, we get

= ((5(x + 3)) / (10x + 40))

Take 10 as common in the denominator value, we get

= ((5(x + 3)) / (10(x + 4)))

Divide the both numerator and denominator value by 5, we get

= ((x + 3) / (2 (x + 4)))          

Answer:

The final answer is ((x + 3) / (2(x + 4)))

Polynomials rational expressions example problem 2:

Simplifying the given rational expressions ((4x + 12) / (2x + 84))

Solution:

Given rational expression is ((4x + 12) / (2x + 84))

Take 4 as common in the numerator value, we get

= ((4(x + 3)) / (2x + 84))

Take 2 as common in the denominator value, we get

= ((4(x + 3)) / (2(x + 42)))

Divide the both numerator and denominator value by 2 , we get

= ((2(x + 3)) / (x + 42))          

Answer:

The final answer is ((2(x + 3)) / (x + 42))

Algebra is widely used in day to day activities watch out for my forthcoming posts on Sphere Definition and Hemisphere Definition. I am sure they will be helpful.

Polynomials rational expressions example problem 3:

Simplifying the given rational expressions ((3x + 12) / (12x + 84))

Solution:

Given rational expression is ((3x + 12) / (12x + 84))

Take 3 as common in the numerator value, we get

= ((3(x + 4)) / (12x + 84))

Take 12 as common in the denominator value, we get

= ((3(x + 4)) / (12(x + 7)))

Divide the both numerator and denominator value by 3, we get

= ((x + 4) / (4 (x + 7)))          

Answer:

The final answer is ((x + 4) / (4(x + 7)))

Polynomials rational expressions example problem 4:

Simplifying the given rational expressions ((13x + 13) / (13x + 26))

Solution:

Given rational expression is ((13x + 13) / (13x + 26))

Take 13 as common in the numerator value, we get

= ((13(x + 1)) / (13x + 26))

Take 13 as common in the denominator value, we get

= ((13(x + 1)) / (13(x + 2)))

Divide the both numerator and denominator value by 13, we get

= ((x + 1) / (x + 2))          

Answer:

The final answer is ((x + 1) / (x + 2))
Practice Problems for Polynomials Rational Expressions

Polynomials rational expressions practice problem 1:

Simplifying the given rational expressions ((5x + 75) / (15x + 45))

Answer:

The final answer is ((x + 15) / (3(x + 3)))

Polynomials rational expressions practice problem 2:

Simplifying the given rational expressions ((3x + 30) / (6x + 54))

Answer:

Given rational expression is ((x + 10) / (2(x + 9)))

Polynomials rational expressions practice problem 3:

Simplifying the given rational expressions ((2x + 30) / (4x + 56))

Answer:

Given rational expression is ((x + 15) / (2(x + 14)))

Having problem with math solver algebra 1 keep reading my upcoming posts, i will try to help you.

List Of Rational Numbers

Let us study about list of rational numbers. The basic of mathematics is said to be as clear knowledge of numbers and its operations.
The math includes various forms of numbers as integers, whole numbers, natural numbers, real rational and irrational numbers.
The rational numbers are defined as the form of fractional numbers that are with the two simple integers. Some of the examples for list of rational numbers are below.

List of Rational Numbers:

List of rational numbers – example 1:

Find out the rational numbers that are in the following number series (1, `sqrt(2)` , `1/2` , 0.3333…, `-5/2` , `7/4` , `sqrt(9)` , `55/10` , 4.44232…) and list them in the order that smallest to the largest rational numbers?

Solution:

The number series given is as follows: (1, `sqrt(2)` , `1/2` , 0.3333…, `-5/2` , `7/4` , `sqrt(9)` , `55/10` , 4.44232…)
All the possible rational numbers that present in the given number series is found to be as follows:
(1, `1/2` , `-5/2` , `7/4` , `sqrt(9)` and `55/10` )
Therefore the rational numbers in the order from the smallest to the largest numbers are as follows:
(`-5/2` , `1/2` , 1, `7/4` , `sqrt(9)` , `55/10` )
(-2.5, 0.5, 1, 1.75, 3, 5.5)

List of rational numbers – example 2:

Find out the rational numbers that are in the following number series (`1/5` , `sqrt(16)` , 0.0222…, `3/2` , `8/4` , `5/100` , 4.2, 343.234…) and list them in the order that smallest to the largest rational numbers?

Solution:

The number series given is as follows: (`1/5` , `sqrt(16)` , 0.0222…, `3/2` , `8/4` , `5/100` , 4.2, 343.234…)
All the possible rational numbers that present in the given number series is found to be as follows:
    (`1/5` , `sqrt(16)` , `3/2` , `8/4` , `5/100` , 4.2)
    Therefore the rational numbers in the order from the smallest to the largest numbers are as follows:
    (`5/100` , `1/5` , `3/2` , `8/4,` `sqrt(16)` , 4.2)
     (0.05, 1.2, 1.5, 2, 4, 4.2)

I am planning to write more post on algebra 2 help free, word problems algebra 2. Keep checking my blog.

List of rational numbers – exercises:

Find out the rational numbers that are in the following number series (`-11/2` , `sqrt(6)` , `sqrt(4)`, 0.02, `7/3` , `28/4` , `1/1000` , 43.234…) and list them in the order that smallest to the largest rational numbers?(Answer:-5.2, 0.001, 0.02, `sqrt(4)` , `28/4` )
    Is the number 23.3434…. is a rational number? (Answer: no)
    Is the number `sqrt(25)` is a rational number? (Answer: yes)

Stepwise Calculation of Standard Deviation

In Statistics a branch of Mathematics, Standard Deviation is the mean of mean. It is the measure which helps us to know how the data is spread out.  It is denoted by the Greek symbol sigma. Formula for Standard-Deviation is given by square root of variance, variance is defined as the average of the squared differences from the mean given by the formula sigma^2 = 1/(n-1)[summation(i=1 to n)(xi – x bar)^2]. So, the Formula Standard Deviation is given as sigma = sqrt{ [summation(i=1 to n)(xi – x bar)^2]/(n-1)}
Equation for Standard Deviation is given by sigma = sqrt[summation(k=1 to n)(xk – x bar)^2/(n-1)]
 n=total number of values, x bar = mean of the data, xk= each of the data values

How to find Standard Deviation for a given data? The following are the steps involved to find Standard-Deviation of a given data:
1. Mean of the given data is calculated
2. Then the deviations are calculated
3. Find the square of these deviations
4. Find the sum of the squares of the deviations
5. Divide the sum by one less than the total numbers in the data
6. Finally find the square root of the value got from the step 5, this result is the standard-deviation
Let us consider an example to understand how to find standard-deviation
Given data, 91, 23, 47, 62, 76, 38, 82, 29

Step1: First we find the average of the given data, [91+23+47+62+76+38+28+82+29]/8= 59.5
So, the mean =  59.5
Step2: The deviations are calculated by subtracting the mean from each given data value.
(91-59.5), (23-59.5), (47-59.5), (62-59.5), (76-59.5), (38 – 59.5), (82-59.5), (29 – 59.5)
      So, the deviations are, 31.5, -36.5, -12.5, 2.5, 16.5, -21.5, 22.5, -30.5
Step3: Squares of the above deviations are,
992.25, 1332.25, 156.25, 6.25, 272.25, 462.25, 506.25, 930.25
Step4: Sum of the squares of the deviations is 4658
Step5: Divide the sum 4658 with (n-1) = one less than total number of values = (8-1) = 7 which gives,
             4658/7 = 665.43
Step6: Standard-Deviation is the square root of the value got in the step5,
Standard-deviation, sigma = sqrt(665.43) = 25.79

Calculate Standard Deviation of the given data,  7, 9, 12, 6, 4, 13, 21, 14, 22, 16
Total Number of values = n = 10
]
Step1: Mean of the given data, (7+9+12+6+4+13+21+14+22+16)/10 = 124/10 = 12.4
Step2: Deviations are calculated by subtracting the mean from each of the data values which are,
            -5.4, -3.4, -0.4,-6.4, -8.4, 0.6, 8.6, 1.6, 9.6, 3.6
Step3: Squares of the deviations are, 29.16, 11.56, 0.16, 40.96, 70.56, 0.36, 73.96, 2.56, 92.16, 12.96
Step4: Sum of the squares of the deviations is 334.4
Step5: divide the sum with (n-1) = (10-1= 9) which gives 334.4/9 = 37.15
Step6: Standard-deviation, sigma = sqrt(37.15) = 6.095

Functions in math

What is a mathematical function? A function is a technical term used to define relation between variables. Let us more understand what is a Math Function? A variable y is called a function of a variable if for every value of x there is a definite value of y. example  of mathematical functions y = x^2.x is called the independent variable as it takes any arbitrary assigned value whereas y is called the dependent variable as its value depends upon the value of x. The set of all possible value of the independent variable in a function is called the domain of the function and the set of values of the dependent variable is called the range of the function.

Let us more understand about functions math. We can classify functions in math as follow:
1. Into functions
A function f : X -> Y is called an into function if there is atleast one element in Y which has no pre-image in X. A function is an into function if its range is a proper subset of codomain.
2. Onto functions(Surjection)
A function f : X -> Y is called an onto function if every element in Y has atleast one pre-image in X i.e. every element of Y is image of some element of X under f. i.e. for every y belongs to Y, there exists an element x in X such that
f(x) = y.
3. One-one function (Injection)
A function f : X -> Y is called a one-one function if no element of X has more than one image in Y. In other words, if the images of distinct elements in X under f are distinct i.e. for every x1, x2 belongs to X.
4. Many-one function
A function f : X -> Y is called a many-one function if two or more elements of X have the same image in Y.
5. One-one into function
When a function is both 1-1 and into function, then it is called one-one into function. It satisfies the following properties:
(i) No two elements of the domain have the same image.
(ii) There is atleast one element in codomain which is not the image of any element of the domain.
6. One-one onto function
It is a function which is both 1-1 and onto. It satisfies the following properties:
(i) No two elements of the domain have the same image.
(ii) It is one-one i.e. f(x) = f(y)
(iii) It is onto i.e. for all y belongs to Y, there exists x belongs to X such that f(x) = y.
7. Many one into function
It is a function which is both many-one and into. It has the following properties:
(i) There are atleast two elements of the domain which correspond to the same element of the codomain.
(ii) There is atleast one element of the codomain which is not the image of any element of the domain.
8. Many-one onto function
It is a function which is both many-one and onto. It satisfies the following properties:
(i) There are atleast two elements of the domain which correspond to the same element of the codomain.
(ii) Every element of the codomain corresponds to some element of the domain.
9. Identity function
A function f : X -> X such that f(x) = x for all x belongs to X is called the identity function. In an identity function each element of the domain corresponds to itself.
10. Constant function
A function f in which all elements of X are associated with the same element of Y is called a constant function.
11. Equal function
Two functions f and g are said to be equal if their domains are same and f(x) = g(x) for all x. If f and g are equal we write them as f = g.

Know more about Continuity definition and function in calculus. If you have problem in solving calculus problems leave a comment with the problem. I will try to solve.

First order linear differential equation

First order differential equations: A first order differential equation is a relation dy/dx = f(x, y).....(1)in which f(x, y) is a function of two variables defined on a region in the xy-plane. A solution of the equation (1) is a differential function y = y(x) defined on an interval of x-values such that d/dx y(x) = f(x, y(x)) on that interval.
The initial condition that y(x0) = y0 amounts to requiring the solution curve y = y(x) to pass through the point (x0, y0).The general form of a first order and first degree differential equation is f(x, y, dy/dx) =0....(i).we know that the tangent of the direction of a curve in Cartesian rectangular coordinates at any point is given by dy/dx .

so the equation in (i)  can be known as the equation which establish the relationship between the coordinates of a point and the slope of the tangent ie dy/dx to the integral curve at a point .Solving the differential equation  given by (i) means finding those curves for which the direction of tangent at each point coincides with the direction of the field.All the curves represented by the general solution when takes together will give the locus of the differential equation .Since there is one arbitrary constant in the general  solution of the equation of the first order. The locus of the equation can be said to be made up of single infinity of curve.

First Order Linear Differential Equation A first order differential equation that can be written in the form  dy/dx + P(x) y = Q(x),where P and Q are functions of x, is a linear first order equation. The above equation is the standard form.  Let us understand First Order Differential Equation Solver using some example. Suppose we have the equation dy/dx = 1 – y/x is a first order differential equation in which f(x, y) = 1 – (y/x).
now let us take one more example show that the function y = 1/x + x/2 is a solution of the initial value problem dy/dx = 1 – y/x,y(2) = 3/2 . The given function satisfies the initial condition because y(2) = (1/x + x/2)x = 2= ½ + 2/2 = 3/2To show that it satisfies the differential equation, we show that the two sides of the equation agree when we substitute (1/x) + (x/2) for y.On the left: dy/dx = d/dx(1/x + x/2) = -1/x^2 + ½.On the right: 1 – y/x = 1 – 1/x(1/x + x/2) = 1 – 1/x^2 – ½ = -1/x^2 + ½.The function y = (1/x) + (x/2) satisfies both the differential equation and the initial condition, which is what we needed to show.

What is a Mode in Math?

Math teacher, Ms Grace assigned the following number of problems for homework on 8 different days; 10, 12, 8, 10, 7, 10, 8, 10. Let us arrange the data in the increasing order, 7,8,8,10,10,10,10,12. What do you observe in the data? 10 is the number which is seen most number of times. This number 10, which occurs most often, is called the Mode in Math or Mode Math. So, Mode Definition Math would be, the Mode of a set of data is the value in the set that occurs most often (most number of times)

By definition, Statistics Mode or statistical mode is the value that occurs most frequently in the data set. For example, what would be the mode of the data set, 12, 11, 13, 9, 11, 8, 6, 11. Let us first arrange the data in the ascending order. That gives us, 6, 8, 9, 11, 11, 11, 12, 13. As you can see the value or the number 11, occurs most number of times than any other number in the data and hence 11 is the mode of the given data set.

To learn the steps involved in finding mode of a given set, let us consider an example. Let us assume that there is a basket ball match going on and the scores of the game are as listed below. Let us determine the mode of the scores.

Scores: 19, 6, 3, 22, 19, 9

Step1: list in the scores in the ascending order
3, 6, 9, 19, 19, 22
Step2: identify the number which is occurs most often, 19
  Hence,  19 is the mode of the score

So, the steps involved in finding mode are, first we need to order the list in ascending order and then identify the number or value which occurs most often that will give us mode.

Note: If there is no number which occurs most often, then we can say that the given data has no mode
On a cold winter day in December, the temperature of 7 cities in North America is recorded in Fahrenheit. How do we find the mode of these temperatures?  Temperatures recorded:  -9,-12, 0, -6,      - 10, -3, 5
To find the mean, the first step would be to order the given temperatures in ascending order and then identify the temperature which occurs most often
-12, -10, -9, -6,-3, 0, 5
From the above ordered list, we can see that there is no value or number which occurs most often. So, we can conclude that there is no mode for the given temperatures.

This article gives basic information about Online Statistics Tutoring. Next article will cover more Statistics concept and its problems and many more. Please share your comments.

Definition of Absolute Value

Definition of Absolute Value
The absolute value is defined as the distance of ‘a’ from zero and is denoted by the symbol |a|. The absolute value only tells how far from zero and not in which direction is the location of a.

From the above figure, the absolute value of |3| = 3 since it is 3 units from the right side of the zero, and |-3|= 3 since it is 3 unit from the left side of the Zero. So the absolute unit should not have the negative sign, and is always positive or zero.

Absolute value equation

To solve absolute value equations, first split the equation into two equations. Then solve the equation to get the solution of absolute value equation.
Example 1:
|x| = 7
X = 7 x = -7
The solutions are {7, -7}
Example2:
|3x-4| = 5
3x-4  = 5 3x-4 = -5
3x = 9 3x = -1
X = 3 x = -1/3
The solutions are {1/3, 3}


Absolute Value Inequality
To solve absolute value inequalities first isolate the one side on the inequality symbol. Then write the first equation with out absolute sign and solve the inequalities and write second equation without absolute sign, reverse the inequalities and then solve the problem. The absolute value should be greater than any negative number and it should not be less than a negative number, sincethe absolute value should be always positive.
Example1:  (greater than)
|x-20| > 5
x-20> 5 X-20 < -5
x> 25 x < 15
The solution is 15 > x > 25
Example2: (less then or equal to)
|X-3| = 4
X-3 = 4 X-3 = -4
X = 7 X = -1
The solution is -1 = x = 4

Absolute value of a complex number
The definition for absolute value ofa real number is not directly generalized for definition for complex numbers, since the complex numbers are not ordered. But the geometric interpretation of the absolute value of a real number is as its distance from ‘0’ to be generalized.

The definition for absolute value ofa complex number is, it is a distance in the complex plane from the origin ‘0’ by using of Pythagorean Theorem. The absolute value of the difference of the complex number is equal to the difference between them.
The definition for absolute value ofa complex number is,
Z = x + iy
Where,
Z - Absolute value or modulus
x, y -  Real numbers
|z| = v(x^2+y^2 )
In above equation, the absolute value of a real number is x, when the complex part becomes 0.
In polar form the complex number z is expressed like z = re i?
If the absolute value is (r = 0, ?-real)
|z| = r
So, the absolute value of a complex number in the complex analogue equation is,

The properties of absolute value of a complex number are as same as absolute values of a real numbers.

Give example how to solve the absolute value equations and inequalities
Example 1: absolute value equation
|x - 7| = |2x-2|
Write in to two equations with out absolute symbol.
X-7 = 2x-2 x-7 = -(2x-2)
X-2x = -2+7 x+2x = 7 + 2
-x = 5 3x = 9
X = -5 x = 9
The solutions are { 9, -5}
Example 2: Absolute value inequalities
|2x-3| > 5
Write in to two inequalities with out absolute symbol.

2x-3 > 5 2x-3 < - 5
2x > 8 2x< -2
X < 4 x < -1
The solution is -1 > x < 4

Equation of a line which is passing through two points

Question :-

Find the equation of the line which is passing through two points (-3,7)(5,-1)

Answer:-

We have to use the point formula to find the equation of the line,this is much similar to midpoint formula

y-y1     x-x1
------ = ------
y2-y1     x2-x1

We have 2 points

( -3 , 7 )  and ( 5 , -1 )
x1  y1          x2  y2

So the equation is

y-7      x-(-3)
------ = ------
-1-7      5-(-3)

y-7      x+3
------ = ------
-8        8

We can further simplify it by cross multiplication.which is a part of indices maths

similarly we can find all points having an x-coordinate of 2 whose distance from the point 2 1 is 5

Statistics: Mean

Mean Math Or Mean Statistics
In Mathematics in the branch of Statistics, the expression for the mathematical mean of a statistical distribution is the mathematical average of all the terms in the data. To calculate this, we add up the values of the terms given and divide the sum by the number of terms in the data. This expression is also called the Arithmetic Mean.
Example: Find the Arithmetic Mean of the following data 5, 5,10,10,15,15,20,30
Solution: Arithmetic Mean = Regular average = sum of the values of the terms/number of terms
Sum of the values of the terms =5+5+10+10+15+15+20+30=110
Number of terms = 8
Mean = 110/8 = 13.75
Sample Mean
The sample mean in statistics branch of Mathematics is the sum of all observed outcomes from the sample divided by the total number of events. It is denoted by the symbol x with a (bar) above it. The formula used to compute the sample mean is as follows:
X (bar)= (1/n) (x1+x2+x3………xn)
 If we consider the sampling of some non-indigenous weed in a land of five acres in Springfield and came up with the counts of this weeds in that area as 38, 56, 84,105,116
We calculate sample mean for the above sampling by adding the weed counts and divide by the number of samples, 5
(38+56+84+105+116)/5= 79.8
So, we get the sample mean of non-indigenous weeds = 79.8
Weighted Mean
While computing Arithmetic Mean all the terms or values have equal importance. But at times we come across some situations where all the terms or values do not have equal importance. For instance, when we compute the average number of marks per subject, as per the different subjects like English, Science, Mathematics, Social Sciences; each subject has different levels of importance.
So, weighted mean is the arithmetic mean calculated by considering the relative importance (weight) of each term.
Each item is assigned a weight in proportion to its relative importance
Formula to compute Weighted Mean: xw (bar) = sigma(wx)/sigma(w)
(here x =value of the items  w= weight of the item)
Example:  A student scored 50, 70, 80 in English, Science and Math respectively and assume the weights of each subject to be 4,5,6  respectively. Find the weighted arithmetic mean of each subject.
Solution: let us tabulate the given data
Subjects              Marks obtained          weight              wx
English                50                          4                  150
Science                       60                          5                  300
Math    80                          6                  480
         Sigma(w)=15  sigma(wx)=930
Arithmetic Weighted mean = 930/15= 62 marks/subject
Short-cut Method
A short-cut method of calculating the arithmetic mean is based on the property of arithmetic average. In this method the deviations (D) of the items from an assumed mean are first calculated and then multiplied with their respective frequencies (f). Then, the total of these products [sigma(fD)] is divided by the total frequencies [sigma(f)]and added to the assumed mean(A). The figure we get is the actual arithmetic average or the Arithmetic Mean.
Formula used in the short-cut method of calculating the arithmetic mean:
X(bar) = A + sigma(fD)/sigma(f)

Operations on Decimals

Decimals:
Numbers representing fractions without having any numerator or denominator is called a decimal number.
Example: 0.567, 0.2, 0.49
Decimal Chart:
The value of a number in a decimal is decided by its place value.
0.345
3 represents three tenths = 3/10
4 represents four hundredths = 4/100
5 represents five thousandths = 5/1000

In engineering calculations, fractions are used extensively. As we know that, we can always express a fraction in terms of its corresponding decimal number. We have a ready reckoner that gives us corresponding converted values of a fraction.
1/64   = .015625 1/32   = .03125 3/64  = .046875 1/16 = .0625
5/64   = .078125 3/32   = .09375 7/64  = .109375 1/8  = .125
9/64   = .140625 5/32   = .15625 11/64 = .171875 3/16 = .1875
13/64   = .203125 7/32   = .21875 15/64 = .234375 1/4  = .25
17/64   = .265625 9/32   = .28125 19/64 = .296875 5/16 = .3125
21/64   = .328125 11/32  = .34375 23/64 = .359375 3/8  = .375
25/64   = .390625 13/32  = .40625 27/64 = .421875 7/16 = .4375
29/64   = .453125 15/32  = .46875 31/64 = .484375 1/2  = .5
33/64   = .515625 17/32  = .53125 35/64 = .546875 9/16 = .5625
37/64   = .578125 19/32  = .59375 39/64 = .609375 5/8  = .625
41/64   = .640625 21/32  = .65625 43/64 = .671875 11/16= .6875
45/64   = .703125 23/32  = .71875 47/64 = .734375 3/4  = .75
49/64   = .765625 25/32  = .78125 51/64 = .796875 13/16= .8125
53/64   = .828125 27/32  = .84375 55/64 = .859375 7/8  = .875
57/64   = .890625 29/32  = .90625 59/64 = .921875 15/16= .9375
61/64   = .953125 31/32  = .96875 63/64 = .984375

Multiplying decimals
In multiplication of decimal numbers, we follow two steps:
Step 1: Multiply the numbers without taking into consideration of decimal points
Step 2: Place the decimal point starting from right and moving as many digits as the sum of the number of digits in decimal part in both numbers towards left.
Example:


Dividing decimals:
We have two steps to follow.
Step 1: Convert both numerator and denominator into fractions
Step 2:  Follow the rules of division of fractions.
Example: 14.4 ?0.12
Step 1: 14.4 = 144/10 and 0.12 = 12/100
Step 2: 144/ 10 ? 12/100 = 144/10 ?100/12 = 120
Subtracting decimals:
To subtract decimals one from the other we follow
Step 1: Write the numbers one below the other with decimal point coinciding
Step 2: Add zeros to make length of the numbers same
Step 3: Follow the laws of subtraction to subtract
Example:
0.56- 0.287
0.560
0.287
-------
0.213
Terminating decimals:
The word terminate means end. A decimal number with definite number of digits (ending) is called terminating decimal.
Example: 0.495. There are three digits after the decimal point.
All terminating decimals can be represented as a rational number.
Dewey Decimal System:
In public libraries, books are catalogued according to the subjects. Dewey Decimal System (DDC) is the most widely used system of classification for organizing books in libraries. In this system, integer part of the number gives us the specific subject and decimal part gives us the specific part of the subject.
For example:
In DDC, 512.12 represents a book on Algebra, 815.4 represents a book on 16th century Italian poetry.

Exponents and their rules

Fractional exponent is also called as a rational exponent which is in the form of a fraction like ½ (square root), ⅓ (cube root), ¼ (fourth root) etc.

A fractional exponent with 1/n is the n-th root of the given base.

(a)^ ⅓ = ³√a

A fractional exponent with m/n, we need to take the m-th power of the base and then find the n-th root or vice versa

(a) ^⅔ = ³√a²

The basic exponential rules are as follows: 

Exponents and tLaw of exponents:

1. Any number except zero when raised to zero equals 1
2. Any number raised to 1 equals itself
3. A number with negative power equals its reciprocal with a positive power
4. When we divide terms with same bases, the powers are subtracted
5. When we multiply terms with same bases, the powers are added
6. When a term is raised to a power with a whole power, the powers are multiplied.
7. When product of terms are raised to a power, each term is raised to that power

Subtracting exponents:

When we divide terms with same bases, the exponents are subtracted

The basic exponential rules are as follows: 

For example,

= 3³

Exponent rules:

1. When we multiply terms with same bases, the powers are added

2. When we divide terms with same bases, the powers are subtracted

3. Any number except zero when raised to zero equals 1

4. Any number raised to 1 equals itself

5. When a term is raised to a power with a whole power, the powers are multiplied.

6. When product of terms are raised to a power, each term is raised to that power

7. A number with negative power equals its reciprocal with a positive power