Online Algebra ii Test

Introduction :

An online algebra ii is also one kind of  mathematics . An online algebra ii explains the main properties of algebraical expressions and relations. This online algebra ii  explains the quantity, letters and other symbols. It indicates variables, equations, objects, polynomials, and expression etc. An online algebra ii, math test II solves lessons Pre-algebra, an online algebra I, an online algebra II, and Geometry. Fundamentally , Algebra involved from the properties and processes of arithmetic which begins with the four processes: addition, subtraction, multiplication and division of numbers.


Definition on online algebra ii test:


Numerals, some literal numbers are made by algebraical symbol. Which expression denotes one number or one quantity. That is, just like the sum of 4 and 2 is one quantity, which result is 6, The sum of X and Y is one quantity, that is, X + Y. Similarly divide b,√b , a*b, a – b are algebraic expressions each of these represents one quantity or one number. Some signs and mathematical symbols are used in the algebraic expression.

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Examples of online algebra ii test


Solve the online algebra ii test equations:
Example 1 on online algebra ii test:

Problem 1:

Solve the equation:

4(-2x - 1) - (x - 2) = -3(3x + 4) + 12

Solution to Problem 1:

Given the equation

4(-2x + 1) - (x + 2) = -3(3x + 4) + 6

Multiply factors.

-8x - 4 - x - 2 = -9x - 12 +6

Group like terms.

-9x - 6 = -9x - 6

Add 9x + 6 to both sides and write the equation as follows

0 = 0

For x value will be true for all statement x and therefore all real numbers are solutions to the given equation.

Simplify the expression 3(a -2) + 3b - 3(a -b -2) + 4

Problem 2:

Given the algebraic expression

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

Solution:

Multiply factors.

= 3a - 6 + 3b -3a + 3b + 12 + 4

Group like terms.

= 6b + 10.

Problem 3:

Simplify: 6c + 4d – 5d + 7c + d = 0

Solution:

Step 1: Group together the like terms:

6c + 4d – 5d + 7c + d = 0
(6c + 7c) + (4d – 5d +d) = 0

Step 2: Then simplify:

13c   = 0

Answer: 13c = 0.

Problem 4:

Simplify: 4(c – 3) + 2 = 7

Solution:

Step 1: Remove the brackets

4c – 12 + 2 = 7

Step 2: Isolate variable a

4c = 12 – 2 + 7
4c = 17
c=17/4 = 3.4

Answer: c = 3.4

Grade 8 Math Problems

Grade 8 Math Problems

Introduction to 8th grade math:

In this article you learned how to do 8th grade math. In 8th grade math covers the following topics:

Algebra
Linear algebra
Fraction
Quadratic equation
Algebraic expression
Area of triangle and parallelogram
Polynomial
Probability
In this content we discuss about the algebraic expression and quadratic equation with suitable example problems.


Example problems on 8th grade math:


Simplify the algebraic equation.

x^2 - x = 0

Solution:

Let us take the equation is
x^2 - x = 0

Take the common factor x outside
x (x - 1) = 0

So the product x (x - 1) to be equal to zero, then we get

x = 0 or x - 1 = 0

Solve the above simple equations to obtain the solutions.
x = 0
or
x = 1

Hence the answer is x=0 and x=1

Example problems on 8th grade math:

Solve x^2-15x-100

Solution:

Let P (x) = x^2-15x-100

Step 1: The given polynomial is in the form of ax^2+bx+c =0

a = 1; b =-15; c =-100

Multiply the coefficient of x^2 and the constant term,

1*-100 =-100 (product term)

Step 2: Find the factors for the product term

-100---- > -20*5 = -100 (factors 20 and -5)

-15 ---- >-20+5 =-15 (-15 is the coefficient of x)

Step 3: Divide the coefficient of x as -15

x^2-15x-100=0

x^2+5x-20x-100=0

Step 4: Taking the common term x from the first two terms and -20 from the next two terms

x(x+5) -20(x+5) =0

(x-20) (x+5) = 0.

Now set (x+5) =0; x=-5

(x-20) =0; x=20.

The roots are x=20 and -5.

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Example problems on 8th grade math:


Solve the algebraic expression

4(x -2) + 2y - 3(x -y -4) + 5

Solution:

Given algebraic expression is
4(x -2) + 2y - 3(x -y -4) + 5

Multiplying the integer with above terms

4x - 8 + 2y -3x -3y + 12 + 5

Combine like term
4x-3x +2y -3y -8+12+5

x –y +9

Hence the answer is x-y+9

Study Online Model

Introduction to study online model:

Nowadays, studies are getting more advanced and fast paced in change. One needs to keep up his/her level to the expected level in whichever stream of education chosen by him/her. Apart from learning in school, there are various ways to get help in homework, assignments, test preparations, etc. A study online model is a way of studying online. In it you get live help on any topic you need help with or on any thing related to it, like projects, homework, etc. Thus, you get a live session in which, like a teacher teaches in the class room, you get that at home and at any time you need help; the major advantage is that one gets the full attention of the tutor, as only one student and tutor are connected through a session. I like to share this Proportions in Math with you all through my article.


Benefits of study online model:


The concept of a 'study online model' helps us to understand the theory and to solve sums relating to the topic at our own pace and ease. This is because the advantage of a 'study online model' is that we get the chance to grasp a single concept at a time and thus divide the topic to be learnt into 'basic building blocks' according to our calibre. There is no time limit to cram up the subject and one can look back at any concepts any time which may need a bit of a revision during the course of study. There are ample of rich and versatile example questions along with their solutions to refer to. A study online model has an advantage that the online content has more scope of being 'accurate', or more properly, 'updated', owing to the high speed of change on the net. This also helps those students that lag behind in studies. You can always rework and rework any part of the subject not understood properly and thus clear away any trace of hesitation with the subject to be learnt.

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The increasing scope of studying online model


From the point of view of clashing of education and daily life, a study online model is the best type of gaining education as it gives us the most flexibility which one can expect during studying. One can study at any time during the day and night and get flash backs at any concept left over or not understood properly. Thus, the 'study online model' is one of the key educational models developed in the 21st century and will play a major role in the education of all – children or adults.

A study online model helps you to develop essential skills as a person given the opportunity to critically analyse his own level can effectively correct mistakes – and furthermore, if at any time any help is needed, it is there - 24 x 7.

Online Numeric Test

Introduction to online numeric test:

Online numeric test is the process of finding appropriate solutions to the problem displayed online.  The student needs to have a clear step by step approach to solve the problems presented online fast and accurate. This lesson deals with some sample problems for online tests. I like to share this Numerical Differentiation with you all through my article.


Math - online numeric test:

In online numeric test maths there are thousands of methods to find a solution for a problem by using different methods and various formulae. the most basic operations of study online calculating are given as below:

Addition,
Multiplication,
Subtraction and
Division
Steps to do online numeric test:
The basic steps of math solving from online numeric test are

At first, perform all the operations enclosed in the brackets

Then calculate the exponents

Next perform the multiplication and division operations

At last do the addition and subtraction.

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Online numeric test - Solved Examples:


Example 1: Solve the equation y +6 = 5.

Solution:

It is easy to solve within one step by grouping like terms

y = 5 - 6

y = -1

Example 2: Solve the equation 5y + 5= 25

Solution:

It is easy to solve within one step by grouping like terms

5y + 5 = 25

5y = 25 - 5

5y = 20

Divide by 5

y = 4

Example 3: Find the value: 25 y - 10 = 5 y - 5

Solution:

It is easy to solve within one step by grouping like terms

25y - 10 = 5y - 5

25y - 5y = 10 - 5

20 y = 5

Divide by 20

y = 1/4.

The answer is y = 1/4.

Example 4: Find the sum of 1000 + 5000?

Solution:

1000 + 5000 = 6000.

Answer = 6000.

Example 5: Solve the equation y +16 = 15.

Solution:

It is easy to solve within one step by grouping like terms

y = 15 - 16

solution :y = -1

Example 6: Solve the equation 5y + 15= 45

Solution:

It is easy to solve within one step by grouping like terms

5y + 15 = 45

5y = 45 - 15

5y = 30

Divide by 5

y = 6

Example 7: Find the value: 25 y - 15 = 15 y - 5

Solution:

It is easy to solve within one step by grouping like terms

25y - 15 = 15y - 5

25y - 15y = 15 - 5

5y = 10

Divide by 5

y = 1/2.

The answer is y = 1/2.

Quadratic Equations with Square Roots

Introduction :

In mathematics, a quadratic equation is one of a polynomial equation of the second degree. The general form is

ax^2+bx+c=0

Where x represent a variable, and a, b, and c, are the constants, with a ? 0. (If a = 0, the equation becomes a linear equation.). The constants a, b, and c, are called, the quadratic coefficient, the linear coefficient and the constant term or free term. The word "quadratic" comes from quadratus; it is the Latin word for "square." Quadratic equations can also be solved by factoring, completing the square, graphing, Newton's method, and using the quadratic formula.

Source: wikipedia


Example problems on Quadratic equations with square roots:

Example 1:

Determine all real solutions to the equation

Sqrt (x + 1) = 4

Solution:

Given equation is
sqrt (x + 1) = 4

Squaring on both sides, then the above equation becomes
[sqrt (x + 1)]^2 = 4^2

Simplify the above equation
x + 1 = 16

Solve for x.
x = 15

NOTE: Because we squared both sides not including putting any conditions, extraneous solutions may be introduced, checking the solutions is necessary.

Check the left side (LS) of the given equation when x = 15

LS = sqrt (x + 1) = sqrt (15 + 1) = 4

Check the right Side (RS) of the given equation when x = 15

RS = 4

When x = 15, the left and the right sides of the given equation are equal: x = 15 is a solution to the given equation.

Example 2:

Determine all real solutions of the equation

Sqrt (3 x + 1) = x - 3

Solution:

Given equation is
sqrt ( 3 x + 1) = x - 3

Squaring on both sides, then the above equation becomes
[sqrt ( 3 x + 1) ]^2 = (x - 3)^2

Simplify the above equation
3 x + 1 = x^2 - 6 x + 9

Rewrite the above equation in factor form.
x^2 - 9 x + 8 = 0

The above form is a quadratic equation with 2 solutions
x = 8 and x = 1


Practice problems on quadratic equations with square roots:


1) Determine the real value of the given quadratic equation with square roots.

Sqrt (2 x + 15) = 5
Answer: x = 5

2) Determine the real value of the given quadratic equation with square roots.

Sqrt (4 x - 3) = x – 2

Answer: x = 7

Write Number Sentence

Number:
A number is described as a mathematical thing, which is used for calculating and counting. The system of mathematical function involves one or more numerical as input and produce its related mathematical output. This operation consists of arithmetic function for example online addition, online subtraction, online multiplication, and online division.

There are variety of numbers are used in arithmetic operation. Some of them as follow,

Natural numbers
Integers
Rational numbers
Real numbers
Complex numbers
Computable numbers
Prime number

Rules:


General rules for to write the number with sentence as follows,

Rule: 1 If the value of given number is above 9, then writing the number for it.

Examples: 1) I have 11 books.

2) There are three balls.

Rule: 2 if a line has number either above the value of 9 or below the value of 9, then writing integers for all numbers or writing in words for all numbers.

Examples: 1) They have 10 rupees and 40 paise

2) The hall contains six doors and twenty windows.

Rule: 3 the simple fractions are writing by the use of hyphens.

Examples: 1) One-half of the value is found

2) One-fourth of problems are solved.

Rule: 4 the mixed fractions are writing in numerical or writing the first word of a sentence.

Examples: 1) There are 7 ¼ water.

2) Five and one-fifth percent of acid mixed with them

Rule: 5 the maximum amount of numbers are rounding by writing in words

Examples: 1) He can earn up to five million dollars.

2) Town has 44 million people

Please express your views of this topic Solve Complex Fractions by commenting on blog.

Examples:


1) Write the following numbers in sentence: (66, 100)

Solution:

66 – Sixty six
100 – A hundred
2) Write the following fractions in sentence: (1/2, 1/4)

Solution:

1/2 – A half
1/4 – A quarter
3) Write the following roman in sentence: (X, XII)

Solution:

X – Ten
XII – Twelve
4) Write the following decimals in sentence: (6000.14, 0.003)

Solution:

6000.14 - Six thousand and fourteen hundredths
0.003 - Three thousandths
5) Write the following mixed numbers in sentence: (1 ½, 5 ¾)

Solution:

1 ½ - One-half
5 ¾ - Five and three-fourth

Online Slope Factor

Introduction :

Learning online slope factor is very simple and interesting. All linear equation has a slope factor which is attached with its x component. In general slope factor is represented by a symbol 'm'. Linear equation can be expressed as a relation between x and y with unit power. We can see linear equation in following forms:

ax + by + c = 0      ....................(1)

ax + by = c            .....................(2)

y = mx + b            ......................(3)

The equation y = mx + b is known as slope intercept form. where m is called slope factor and b is called y intercept. We can convert any linear equation in slope intercept form and then we can calculate slope factor. Slope factor is nothing but coefficient of x in the slope intercept form equation.

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eq 1 can be rewritten as: by = -ax -c   or y = (-a/b) -c/b

Therefore slope m = (-a/b)

Similarly, we can see that for eq 2 also slope m = (-a/b)


Learn to find the slope factor online

Slope or gradient of a line describes its steepness, incline, or grade. A higher slope value indicates a steeper incline.

The slope is defined as the ratio of the "rise" divided by the "run" between two points on a line, or in other words, the ratio of the altitude change to the horizontal distance between any two points on the line. Given two points (x1,y1) and (x2,y2) on a line, the slope m of the line is

m=(y2-y1)/(x2-x1)

Through differential calculus, one can calculate the slope of the tangent line to a curve at a point.

The concept of slope applies directly to grades or gradients in geography and civil. Through trigonometry, the grade m of a road is related to its angle of incline theta by

m=tan(theta).

If we have two points then find the slope  using 1st formula.

If we give any equation then we can find dy/dx=function(x) where substitute x value the we get slope.

If we have any angle then we have to use slope(m)=tan(teta) where teta is the angle.

Examples to learn online slope factor


Example1: Find the slope of 4x + 3y = 5.

Solution: Converting it into slope intercept form we get,

y = (-4/3)x + 5/3, comparing it with y = mx + b we get,

Slope m = -4/3


Example 2: Find slope of line passing through (0,0) and (2,3).

Solution: Slope m = (y2-y1)/(x2-x1)

m = (3-0)/(2-0)

or slope m = 3/2