Algebra Free Trial Help

Introduction to algebra free trial help:

Algebra is a branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with the geometry, analysis, topology, combinatorics, and number theory, algebra is one of the main branches of pure mathematics.(Source – Wikipedia )

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Problems in algebra free trial help


Problem 1 algebra free trial help:

Determine the equation (4x+y) (4x-3y) and x=3, y=9
Apply the x and y value.

=(4(3 )+9)(4(3)-3(9))

=(12+9)(12-27)

=(21)(- 15)

The required solution is =315

Problem 2 algebra free trial help:

Solve: -8a+24+9a-3
Solution:

-8a+24+9a-3

-8a+9a+24-3

a+21

The required solution is  a=-21

Problem 3 algebra free trial help:

Solve the given algebraic expression 3(a -4) + 6b - 3(a -b -2) + 8
Solution:

Given the algebraic expression

3(a -4) + 6b - 3(a -b -2) + 8

Multiply factors.

= 3a - 12 + 6b -3a + 3b + 6 + 8

Grouping the above terms.

The required solution is = 9b + 2

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More problems in Printable algebra 2 test


Problem 1 algebra free trial help:

Solve the equation
4(4x + 5) + (6x - 9) = -7(2x + 6) + 28

Solution:

Given the equation

4(4x + 5) + (6x - 9) = -7(2x + 6) + 28

Multiply factors

16x + 20 + 6x - 9 = -14x - 42 +28

Group like terms

22x + 11 = -14x – 14

22x + 14x = -14 – 11

Solve this equation to obtain

The required solution is X = - 25 / 36

Problem 2 algebra free trial help:

Solve       2x + 4y = 14

2x + 3y = 8

Subtract the two equations:

2x + 4y = 14

2x + 3y = 8

---------------

0 + y = 6

y = 6

Thus, y = 6. Now insert  the value of y = 6 in either one of the equations:-

2x + 4y = 14

2x + 4(6) = 14

2x + 24 = 14

2x = 14 - 24

2x = -10

x = -10/5

x = -2

Thus the required solution is, x = -2 , y = 6

Free Help with Statistics Homework

Introduction to free help with statistics homework:

In math, Statistics making effective use of numerical data between to groups of individuals or experiments. It contracts with all features of this, together with not only the group, study and understanding of such data, but also the planning of the group of data. Let us we will discuss about free help with statistics homework with some of the example homework problems.


Mean and median in Statistics:


In statistics, the mean, median, range, and mode these are main and easy terms in the.
Let us we will see the mean help with some of the homework problems.
Mean:

In free help statistics homework, the mean is one of the main important terms in statistics.
Now we will discuss how to learning the mean with help of the homework problems.
The mean is clarified about the average of the given sequence of the numbers.
Here we will locate the addition of the sequence numbers and divided by the total numbers of the given numbers.
The given numbers are 66, 55,44,55,11.
Solution:

The given numbers are 66, 55,44,55,11.
Now addition the given numbers.
= 66+55+44+55+11=231.

The total of the sequence numbers are 231.
Now divided by 5, because (Notes: 5 is the total sequence numbers)
`231 /5=46.2` .
So the mean values in this sequence numbers are 46.2.
Median:

In free help with statistics homework, the median is defines the center point of the specified numbers.
Free help with statistics home work Problems for median:
The number series are 66, 55,44,55,11.
Solution:

The given numbers are 66, 55,44,55,11.
In this given numbers, the 44 is the center value.
So the 44 is the median of the given series.

Mode and range in free help statistics homework:


Example of free help statistics homework:

Mode:

In free help statistics homework, the mode is viewed many times in the given series.
Homework problem help for mode:
The given numbers are 66, 55,44,55,11.
Solution:

The given numbers are, 66, 55,44,55,11.
Now arrange the numbers,
11,44,55,55,66.
In this given series number 55 is show frequently.
Mode in this series is 55.
Range:

The variation among highest value and the smallest value is known as the range in free help with statistics.
The given values are 66, 55,44,55,11.
Here the highest value is 55, and then the low value is 11.
So the range value is difference between the 55 and 11.
The range value is 55-11=44.
The range value of the given series is 44.

Common Multiple Solver

Introduction for common multiple solver:

Common multiple for the given set of number is also a number which divides the given set of numbers. 9,18 and 45 are the common multiple number for the given number 180. Common multiple is classified in to least common multiple and common factor. In online we can find the common multiple by using the solver. Numerous websites provides common multiple solver. Those online common multiple solvers are very interactive to the learners. This also provides the work flow.

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Internal work flow of the common multiple solver:


When we enter the sample number 180 and 360 in common multiple solver, the common multiple solver will produce the answer with the internal work flow as given below.

First, common multiple solver factor the number 180 and 360

Work Shown for 180

180 can be divided by 2: 180 = 90 * 2.

90 can be divided by 2: 90 = 45 * 2.

45 can be divided by 3: 45 = 15 * 3.

15 can be divided by 3: 15 = 5 * 3.

5 can not be divided by anything.

Work Shown for 360

360 can be divided by 2: 360 = 180 * 2.

180 can be divided by 2: 180 = 90 * 2.

90 can be divided by 2: 90 = 45 * 2.

45 can be divided by 3: 45 = 15 * 3.

15 can be divided by 3: 15 = 5 * 3.

5 can not be divided by anything.

Factors of 180: 2,2,3,3,5
Factors of 360: 2,2,2,3,3,5.
Marked common factors:
180: 2,2,3,3,5
360: 2,2,3,3,5,2
Common factors: 2,2,3,3,5
The product of the common factors is:2*2*3*3*5=180.
This number, 180, is your greatest common factor:

Greatest Common Factor( 180, 360 ) = 180.

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To find the least common multiple, we have to multiply the given number 180 and 360, then the multiplication result is divided by GCF.

Lease common multiple= (180 * 360) / 180 = 360.

There fore, the lease common multiple is 360

Completing The Square

Completing the Squares is a technique used in algebra to convert quadratic equations into a form which is sum of square of a linear equation and a constant as:

Ax^2 + bx + c = a(linear equation)2 + C
First let us see what is the expansion of (x+y)2 and (x-y)2.
(x+y)^2 = x^2 + y2 + 2xy
(x-y)^2 = x^2 + y2 - 2xy
For example (x+3)2 = x^2 + 32 +2.3x = x^2+6x+9.
(x-3)2 = x^2 + 32 - 2.3x = x^2-6x+9.

Let us see method of completing squares:

Suppose you are given a quadratic equation ax^2 + bx + c = 0 which is not a perfect square. Then follow the following steps:

1) Make the coefficient of x equal to 1 by diving the equation by its coefficient which is ‘a’ in this case:
x^2 + (b/a)x + c/a =0
2) Write the coefficient of c as multiple of 2 as:
x^2 + 2()x + (c/a) = 0…………..(1)
3) Now write the perfect square equation for which first term is x and second is coefficient of x divided by 2 which is b/2a here:
(x+(b/2a)2 )
This on expansion gives: x^2++2()x = x^2 + x +   ………..(2)
Note that in equation (1) and (2), first and second terms are equal and only difference is in constant term.
4) Now write the original equation as (x + )2 + ( - ) = 0

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Let us take an example:
x^2 + 10x + 28
Here coefficient of x is already 1 so we will come to step (2) directly. Make the coefficient of x as factor of 2. So, equation can be written as:
x^2 + 2(5)x + 28 ………….(1)
Coming to step 3 we write a perfect square equation as:
(X+5)2 which gives x^2 + 52 + 2.5x on expansion. So we have x^2 + 10x + 25 ………(2) as the expansion now.
Comparing equation 1 and equation 2 we get a difference of 3 in constant term as 28-25 = 3. So, x^2 + 10x + 25 + 3 = equation (1) = x^2 + 2(5)x + 28
Also writing the LHS of above equation in square form we get: (x+5)2 + 3 = x^2 + 10x + 28.
Completing the Square Practice is easy to do as we see in above steps.
You can also Factor by Completing the Square as in this example: x^2 – 4x – 8 = 0. as done in above  Math Completing the Square of this equation we get: (x-2)2=12.  X-2=(12)1/2 which gives x= 2+121/2.

Lines Math Terms

Introduction to lines math terms:

In day to day life, we often came across some unique math terms. Lines are one of the special math terms that falls under this category.
A line has length, but it is said to have no breadth.
A geometrical straight line extends endlessly in both the directions.
In this article of lines math terms, we are going to study various math terms related to lines.

Math terms related to lines:


Line segment -  A part or portion of a line with two end points is called a line segment.

Ray - The part of the line which has only one end point and extends endless in other direction is called a ray.

Parallel lines - The lines that lie in the same plane and never intersect are called as Parallel lines.

Perpendicular lines - If two lines lie in the same plane and intersect at right angles, they are called perpendicular lines.

Collinear points - Three or more points are said to be collinear if there is a line which contains them all.

Non-collinear points - If different points not lie on the same line, the points are said to be non collinear.

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Additional math terms related to lines:


Concurrent lines - If two or more straight lines pass through the same point, then they are called concurrent lines. The point through which the lines pass is known as point of concurrency.

Intersecting lines - If two or more lines meets at a point, then the lines are said to be intersecting lines.

Transversal -  A straight line which intersects two or more given lines at distinct point is called the transversal of the given lines.

Unique line - Two distinct points in a plane determine a unique line.

Congruent Lines - Two line segments are said to be congruent if and only if their lengths are equal.

Half-line - Let A,B,C be three points on a line L such that A lies between B and C. If we delete the points A from the line L, then each one of the remaining parts of line L is called a half-line.

Incidence Axiome - A line contains infinitely many points.The rough a given point, there pass infinitely many lines.One and only one straight line passes through the distinct points.

2 Step Equations Solver

Introduction to 2 step equation solver:

The solver uses two steps to solve the equation is called two step equation.  The main aim of the two step equation is to make the variable on one side of the equation and the constant on another side of the equation. Solving two step equations is similar to one step equation. But only the difference is two step equations have two steps rather than one step. In this article, we are going to see some of the problems to solve by substitution solver. I like to share this Variance Equation with you all through my article.


Example problems to solve by 2 step equations solver:


Example problem 1:

Find the value of x: 2 x + 10 = 20

Solution:

Step 1: Subtract 10 on both sides of the equation

2x + 10 - 10 = 20 – 10

2x = 10

Step 2: Divide by 2 on both sides of the equation

2x / 2 = 10 / 2

x = 5

So, x = 5 is the solution of the given equation.

Example problem 2:

Find the value of t: 8 t - 16 = 64

Solution:

Step 1: Add 16 on both sides of the equation

8t - 16 + 16 = 64 + 16

8t = 80

Step 2: Divide by 8 on both sides of the equation

8t / 8 = 80 / 8

t = 10

So, t = 10 is the solution of the given equation.

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Few more problems to solve by 2 step equations solver:


Example problem 3:

Find the value of m: 20 – (m / 3) = 10

Solution:

20 – (m / 3) = 10

Step 1: Subtract 20 on both sides of the equation

20 – (m / 3)  – 20 = 10 – 20

-(m / 3)  = -10

Step 2: multiply by -3 on both sides of the equation

- (m / 3) * -3 = -10 * -3

m = 30

So, m = 30 is the solution of the given equation.

Example problem 4:

Find the value of y: (y /2) + 1 = 3

Solution:

Step 1: Subtract 1 on both sides of the equation

(y /2) + 1- 1 = 3 – 1

(y /2) = 2

Step 2: multiply by 2 on both sides of the equation

(y /2) * 2 = 2* 2

y = 4.

So, y = 4 is the solution of the given equation.

Practice problems to solve by 2 step equation:

1)     Find the value of x: (x /3) + 1 = 5 (Answer: x = 12)

2)     Find the value of m: 3m + (3/5) = 5 (Answer: m =22 / 15)

3)     Find the value of x: 2 x + 10 = 12 (Answer: x = 1).

Math 150 Problems

Introduction to Math 150 Problems:

The four basic operations in math are subtraction, addition, division and multiplication. The simplifications can be performed using the help of above operations. Simplifying can be expressed in the form of algebraic expression. The solution can be obtained with decomposing numbers or else written as one’s, ten’s, hundred’s, and thousands, etc. Let us see about math 150 problems in this article.


Example Problems using Place Values in Math 150 Problems


The instructions can be used to help for solving the problems in math.

Example 1:

Decompose the numbers 150.

Solution:

Step 1:

Let us write the given number 150.

Step 2:

Split up into (1) (5) and (0).

Step 3:

1 hundred’s + 5 ten’s + 0 one’s

Step 4:

Place the digit as it is in the above step.

Put 1 in the hundred’s place, 5 in the ten’s place and 0 in the one’s place.

Example 2:

Add the given numbers.

1, 5, 0.

Solution:

Let us write the numbers one by one.

1

5

0

6

The solution for adding the given number is 6.

Example 3:

Subtract the given numbers 150.

Solution:

Subtraction can be performed from the highest number to the least number.

Step 1:

The number 1 can be subtracted from 5.

5

1

4


Step 2:

The number 0 can be subtracted from the number 4.

4

0

4

The solution for subtracting 150 is 4.

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More Examples to Practice for Math 150 Problems

Example 4:

Multiplying

75 × 2

Solution:

Multiplying first 75 × 2

75 ×

2

150

The solution for multiplying 75 × 2 is 150.

Example 5:

What solution will you obtain by adding 55 + 95?

Solution:

55

95

150

The solution for adding 95 + 55 is 150.

Example 6:

What solution will you obtain by subtracting 985 – 835?

Solution:

985

835

150

The solution for subtracting 985 – 835 is 150.