List of Math Terms

Introduction to list of math terms:
List of math terms glossary of basic mathematics words from the level of kindergarten through college level. In many cases one or more definition is given. These lists of math terms are listed in order of difficulty, as younger students need only the earlier definitions. These lists are used to clear up the confusions of students so they could do duties related to math. This article listed the various list of math terms.

Names of List of Math Terms:

Algorithm - A step-by-step procedure of problem solving for solving computational mathematical problems.

Angle – Angle is formed by two rays starting at the same point.

Arc - A section of the circumference in the circle.

Area - The space measured in square units that any two-dimensional shape.

Arithmetic - A part of mathematics concerned with the four operations of positive numbers.

Base - The bottom of a figure, solid or three-dimensional objects

Bell Curve - The shape of the graph is indicates the normal distribution.

Binomial - A polynomial equation with two terms are joined by a plus or minus sign.

Centimeter - Metric units of measurement. It measure of length.

Circumference - The complete distance around the circle.

Chord - The segment which joins any two points on a circle.

Coefficient - A constant term with a variable.

Decagon - A polygon has ten sides, ten angles and ten straight lines.

Degree – it is unit of angle.

End Point - The point at which a line ends.

Equilateral – four side shape and all sides are equal.

Even Number - A number is divided or divisible by 2.

Exponent - The number is gives reference to the repeated multiplication required.

Some more List of Math Terms:

Face - The shape is joined by the edges on a three-dimensional object.

Factoring - The process of breaking numbers are down into all of their factors.

Graph Theory - A part of mathematics focusing on the properties of a variety of graphs.

Hexagon - A six side polygon and it has six angles.

Integers - Whole numbers, positive numbers, negative numbers including zero.

Isosceles - A polygon has two equal length sides.

Kilometer – it is the one of unit of measure. It is equals 1000 meters.

Like Terms - Terms have the same variable and different coefficient and the same exponents/degrees.

Mean - The mean is the like as the average.

Mixed Numbers - whole number with a fraction.

Obtuse Angle - An angle has a measure greater than 90°.

Quadrilateral - A four side polygon/shape.

Range - The difference between the high value and the low value.

Scalene Triangle - A triangle has 3 unequal sides.

Uniform – All are same.

Volume – one of the units of measure.

X-Axis - The horizontal line or axis in a plane.

Y-Axis - The vertical line axis in a plane.

Domain of Exponential Functions

Introduction to domain of exponential functions:

The exponential function in mathematics is defined as ex, where e is an integer. For example let us denote ex is an exponential function, Let us assume the value of x is zero, x = 0 then the solution is in the form of e0 = 1. Provided e.g. shows general concept of the exponential function. Here we are going to discuss about what is domain and the exponential function together as domain of exponential functions.

Domain of Exponential Functions:

Domain is the set of point or atlest a single point which refers an open connected space. We can say the domain in other words as group of all possible values from an independent variable of a function. The domain and range of the function is lies between (-infinity) to (+infinity).

First we have to start with the fundamental functions of the exponential function of base a,

f (x) = ax , a > 0 and also not equal to 1.

Then the value of the domain functions f is the set of all real numbers.

The range of f is the interval (0 , +infinity).

The given function have a horizontal asymptote which given by y = 0, then the function f has a y intercept at (0, 1).

When the value of the given function f is increased, then the value of a is greater than 1

When the value of the given function f is decreased, then the value of a is lesser than 1.

Example for Domain of Exponential Functions:

Example for domain of exponential functions 1:

Find out the domain of the given function f(x) = ln of (`sqrt (x^2 - 9x + 18)` )

Solution:

Step 1: Given exponential function is f(x) = ln of (`sqrt (x^2 - 9x + 18)` )

Step 2: When we take any value of the 'ln' then it must be a positive value. So:

`sqrt(x^2 - 9x - 18)` > 0

Therefore, `x^2` - 9x - 18 > 0

`x^2 ` - 3x - 6x + 18 > 0

x(x - 6) - 3 (x - 6) > 0

(x - 6)(x - 3) > 0

So either x - 6 and x - 3 are both positive, or x - 6 and x - 3 are both negative. So either x < 6 or x > 3. That's the domain. Having problem with help solve math problem keep reading my upcoming posts, i will try to help you.

Example for domain of exponential functions 2: Determine the domain of the function f(x) = log10 { 1 - log [`x^2` - 4x + 13]}

Solution:

Step 1: The given function is f(x) = log10 { 1 - log [`x^2` - 4x + 13]}

Step 2: Let there are two function present in log. Let we take outer log as nesting log and inner log as nested log.

Step 3: First nesting log is simplified as,

1 - log (`x^2` - 4x + 13)

log (`x^2` - 4x + 13)  < 1

log10 (`x^2` - 4x + 13) < log1010

x2 - 4x + 13 < 10

x2 - 4x + 3 < 0

x - 3x - x + 3 < 0

(x - 3)(x - 1) < 0

x `in` (3, 1)

For nested log function,

`x^2` - 3x + 12 > 0

Here, from the given function squared terms and their co efficient are positive and also domain is > 0. There fore the inequality is true for every real numbers.

Practice Math Facts Online

Practice math facts online – Introduction:

In algebra basic arithmetic operation (addition, subtraction, multiplication, and division) generally used in day to day life. Addition defined as adding the two numbers. Subtraction (-) defined as the inverse of addition. Multiplication defined as the product of numbers. Division can be considered as repeated subtraction.

In online, students can learn about math facts. Through online, students can have interactive sessions with tutors. Online is one of the efficient tools for one to one learning. In these articles we are going to see about practice math facts online. Understanding What is a Line Segment is always challenging for me but thanks to all math help websites to help me out.

Practice Math Facts Online – Addition and Subtraction Rules and Examples:

Practice math facts online - Addition rule with examples:

Positive + Positive = Positive: 7 + 7 = 14

Negative + Negative = Negative: (- 5) + (- 6) = - 11

Addition of a negative and a positive integer: Use the sign of the larger number and subtract, examples are,

(- 5) + 4 = -1

7 + (-2) = 5

Practice problems:

1.  (-9) + 2 =?

Answer: (-7)

2.  5 + (-1) =?

Answer: 4

Practice math facts online - Subtracting rule with examples:

Negative - Positive = Negative: (- 4) - 2 = -4 + (-2) = -6

Positive - Negative = Positive + Positive = Positive: 4 - (-4) = 4 + 4 = 8

Negative - Negative = Negative + Positive = Use the sign of the larger number and subtract (Change double negatives to a non negative)

(-5) - (-2) = (-5) + 2 = -3

(-2) - (-4) = (-2) + 4 = 2

Practice problems:

1   (-9) – 2 =?

Answer: (-11)

2.   5 - (-1) =?

Answer: 6

Is this topic Equation of a Line Standard Form hard for you? Watch out for my coming posts.

Practice Math Facts Online – Multiplication and Division Examples:

Practice math facts online - Multiplication rules:

Rule1: Multiplication of two integers by the related signs resolve be positive sign

a) Positive x positive = positive

b) Negative x negative = positive

Rule2: Multiplication of two integers by the unlike signs will be negative

a) Positive x negative = negative

b) Negative x positive = negative

Example problems:

1. Positive x positive = positive

Example: 2 * 4 = 8

2. Negative x negative = positive

Example: (-2) * (-6) = (12)

3. Positive x negative = negative

Example: 6 * (-3) = (-18)

4. Negative x positive = negative

Example: (-4) * 3 = (-12)

Practice problems:

1.  (-7) * 2 =?

Answer: (-14)

2.  (-5) * (-2) =?

Answer: 10

Practice math facts online - Division rules and example problems:

Rule 1: Division of two integers by the related signs resolve be positive sign

a) Positive ÷ positive = positive

b) Negative ÷ negative = positive

Rule 2:  Division of two integers by the unlike signs will be negative

a) Positive ÷ negative = negative

b) Negative ÷ positive = negative

Example problems:

1. Positive ÷ positive = positive

Example: 9 ÷ 3 = 3

2. Negative ÷ negative = positive

Example: (-14 ÷ (-2 = 7

3. Positive ÷ negative = negative

Example: 10 ÷ (-2) = (-5)

4. Negative ÷ positive = negative

Example: (-15) ÷ (3) = (-5)

Practice problems:

1.  (-9) ÷ 3 =?

Answer: (-3)

2.  25 ÷ (-5) =?

Answer: (-5)

Standard Deviation Z Value

Introduction for standard deviation z value:

In this article we shall discuss about the standard deviation z value. Standard deviation is the part of measurement of the z-value,standard scores are also called as z-values, z-scores, normal scores.The z-value be simply define the population parameter, because in standardized testing; but one just contain a sample set, next the similar calculation by sample mean also illustration standard deviation yield the Student's t-statistic.

Standard Deviation Z Value

Z value:

The z value formula is

` Z=(x- mu)/ sigma`

where

x is a raw value to be standardized

μ is the mean of the population

σ is the standard deviation of the population

In standard deviation, z value be the number of standard deviations to a value, x, be above or else below the mean.Value of x is fewer than the mean- z value negative.The value of x is more than the mean- z value positive and value of x equals the mean - z value zero.

Let us see some of the examples of standard deviation z value. Having problem with Finding the least Common Denominator keep reading my upcoming posts, i will try to help you.

Examples for Standard Deviation Z Value

Example 1:

Find the Z value to the raw data of 14 coming from the mean 8 and the standard deviation 3.

Solution:

The formula for the Z value is

` Z = (x-mu)/sigma`

Where, x =14, µ = 8 and σ = 3

`Z = (14-8)/3`

`Z = 6 / 3`

Z  = 2

The Z value is 2

Example 2:

The Z value  3 was observed from the outcome of the normal distribution with mean 16 and the standard deviation 6 and evaluate the raw data.

Solution:

The  Z value  formula is

`Z =(x-mu)/sigma`

Where, Z= 3, µ = 16 and σ = 6.

2 = `(x-16)/6`

6(2)=x-16

12=x-16

x = 16 + 12

x = 28.

The raw data for the Z value 3, mean 16 and standard deviation 6 is 28.

These are the examples of standard deviation z value.

Practice examples for standard deviation z value:

1.Measure the Z value to the raw data of 18 coming from the mean 6 and the standard deviation 3.

Answer : 4

2.Calculate  the Z value to the raw data of 8 coming from the mean 4 and the standard deviation 2.

Answer :  2

Area Perimeter Rectangles

Introduction to area perimeter of rectangles:

In mathematics, the rectangle is a polygon of 4 sides, it's called quadrilateral. The rectangle is one of the quadrilateral with it has the four right angle.

Area of rectangles:

The sum of surface occupied by a plane rectangle is labeled the area of rectangle. The units of the rectangle area are square units.

Perimeter of rectangles:

The length of the boundary of any rectangle is labeled the perimeter of rectangle. The units of perimeter of rectangle are units.



Formulas for Area and Perimeter of Rectangles:

Area and perimeter of a rectangle:

For a rectangle of length = l and breadth or width = b we have:

Area = (l x b) square units.

Length = `(Area)/(width)` and Width =`(Area)/(Leng th or Width)`

Perimeter = 2(l + b) units.

Diagonal =`sqrt(l^2 + b^2)` units.I like to share this free 8th grade math problems with you all through my article.

Examples for Area and Perimeter of Rectangles:

Example 1:

Determine the area and perimeter of rectangles of the length is 23 cm and width is 50 cm?

Solution:

Given:

length( l) = 23 cm.

breath or width( b) = 50 cm.

To find area of the rectangles:

area A = (l x b) square units.

= (23 x 50) square cm.

=  1150 cm2 or square cm.

Therefore, Area of Rectangles = 1150 square cm.

To find perimeter of the rectangles:

Perimeter = 2(l + b) units.

= 2(23 + 50) cm.

= 2(73) cm.

= 146 cm.

Therefore, Perimeter of Rectangles = 146cm.

Example 2:

Find the area and perimeter of rectangles, if length is 12 inches and width or breadth is 5 inches?

Solution:

Given:

Length = 12 inches.

Width or breadth = 5 inches.

To find area of rectangle:

Area = (l x b) square units.

= (12 x 5) square inches.

= 60 square inches.

Therefore, Area of Rectangles = 60 square inches.

To find the perimeter of rectangle:

Perimeter = 2(l + b) units.

= 2(12 + 5) inches

= 2(17) inches.

= 34 inches.

Therefore, Perimeter of Rectangles = 34 inches.

Function Machines in Math

Introduction to function machines in math:

In this article function machines in math, we will discuss about basic arithmetic operation in the mathematics. Function machine should perform the following operation like addition, subtraction, multiplication and division of single and two digit number. And also half the number, double the number, addition of half and single digit number, subtraction of half and single digit number etc. Let us see some problems for function machines in math.

Single Step Performance - Function Machines in Math

Consider the variable x and y as a single digit number

1.Double the number 2x

2.Half the number` x/2`

3.Adding of single digit number  x+y

4.Subtracting of single digit number  x-y (Here the x is a largest number and y is a smallest number)

5.Multipling of single digit number  x(y)

6.Dividing of single digit number ` x/y`

Consider the variable x and y as a two digit number

7.Adding of double digit number  x+y

8.Subtracting of double digit number  x-y (Here the x is a largest number and y is a smallest number)

Two step performance -  Function machines in math

1.Double the number x and add a single digit number 2x+y

2.Double the number x and subtract a single digit number 2x-y

3.Half and add a single digit number `(x/2)+y`

4.Half and subtract a single digit number `(x/2)-y`

5.Add half of the number with itself `(x/2)+x`

Worked Example - Function Machines in Math

Perform the function math operation to the single and two digit number.

Single step performance

Consider the following  single digit number x=5 and  y=2

1.Double the number 2x=2(5)=10

2.Half the number x/2=5/2=2.5

3.Adding of single digit number  x+y=5+2=7

4.Subtracting of single digit number  x-y=5-2=3

5.Multipling of single digit number  x(y) = 5(2)=10

6.Dividing of single digit number  `x/y=5/2=2.5`

Consider the following two digit number x=20 and y=15

7.Adding of double digit number  x+y=20+15=35

8.Subtracting of double digit number  x-y =20-15=5

Two step performance

1.Double the number x and add a single digit number 2x+y => 2(5)=10+y=10+5=15

2.Double the number x and subtract a single digit number 2x-y => => 2(5)=10+y=10+5=15

3.Half and add a single digit number `(x/2)+y =gt 5/2=2.5+y=2.5+5=7.5`

4.Half and subtract a single digit number `(x/2)-y =gt 5/2=2.5+y=2.5-5=-2.5`

5.Add half of the number with itself`(x/2)+x =gt 5/2=2.5+x=2.5+5=7.5 `

Example of Cubic Function

Introduction of cubic function:

A Cubic function is a small different from a quadratic function. A Cubic functions have a 3 x intercept, The cubic function refer to as 3 degrees. The example of a cubic function is y=(x-1)(x+3)(x-4). it has 3 x intercepts which loaded on (1,0)(-3,0)(4,0).A cubic function is one of the functions which is formed as, F(x) =ax3+bx2+cx+dwhere a- nonzero (or) say polinomial of degree three. Quadratic function is derivation for cubic function. Also, a intergral for a cubic function.By ƒ(x) = 0 and assuming a ≠ 0 gives the cubic formula of the form:ax3+bx2+cx+d=0Coefficient a, b, c, d are real numbers. However, most of theory is also legal if they belong to field of characteristic other than 2 or 3.

Example Problems on Cubic Function:

Roots of a cubic function:

Each cubic equation with real coefficients have at least one solution x among the real numbers; this is a consequence of the Intermediate value theorem. We are able to differentiate several likely cases using the discriminant.

`Delta=18abcd-4b^3d+b^2c^2-4ac^3-36a^2d^2`

The next cases require to be measured

If Δ > 0, the equation have three distinct real roots.
If Δ = 0, the equation has a multiple root along with all its roots are real.
If Δ < 0, t the equation have one real root along with two non real complex conjugate roots.


Let us see some examples of cubic function:
Example 1:

Solving the factors of the cubic of the equation  x3-3x2-25x+75.

Solution:

The given equation is x3-3x2-25x+75.

This form as ax3 + bx2 + cx + d

So,  (x3-3x2) + (-25x+75)

Take a common variable:

=x2(x-3) -25(x-3)

=( x2-25) (x-3)

Here x2 – 25 in the form of a2 + b2 = (a + b) (a - b)

So, x2 – 25 = (x + 5)(x - 5)

=(x-5)(x+5)(x-3)

Answer: The solutions are 5,-5,3

Example 2:

Solving the factors of the cubic of the equation 6x3-36x2 = -54x

Solution:

This equation can be written as 6x3-36x2 + 54x=0.

This form as ax3 + bx2 + cx + d

So, 6x(x2-6x+9) =0.

Here x2 – 6x + 9 in the form of Ax2 + Bx + C so we find the factor.I like to share this Free math problem solver with you all through my article.

6x(x-3)(x-3) =0.

x=0, x=3, x=3.

Answer: The solutions are 0, 3, and 3.

Example 3:

Solving the factors of the cubic  of the equation x3-4x2-100x+400.

Solution:

The given cubic equation is (x3-4x2) + (-100x+400)

=x2(x-4) -100(x-4)

=( x2-100) (x-4)

=(x-10)(x+5)(x-3)

Answer : The solutins are 5,-5,3

Elementary Math Practice

Introduction to elementary math practice:

In elementary math the children are mostly study about arithmetics. Arithmetic or arithmetics is the oldest and most elementary branch of mathematics, used by almost everyone. It involves the study of quantity, especially as the result of combining numbers. Professional mathematicians sometimes use the term arithmetic when referring to more advanced results related to number theory, but this should not be confused with elementary school arithmetic. Let us see elementary math practice. (Source: Wikipedia)

Example Problems for Elementary Math Practice:

Example 1: Find the prime number from the following: 14, 17 and 60

Solution:

Given 14, 17 and 60

14 is divisible by 1, 2, 7, 14.

17 is divisible by 1, 17.

60 is divisible by 1, 2, 3, 4, 5, 6, 10, 12, 15, 20, 30, 60

From the given 17 is a prime number.

Answer: 17 is a prime number and 14, 60 are composite numbers.

Example 2: Multiply the following: `12/5 * 3/4`

Solution:

Given `12/5 * 3/4`

Multiply the denominators and numerators individually ,

`12/5 * 3/4 = (12*3) / (5*4) `

=  ` (36/20)`

=  ` 9/5 `(divide by 4)

Answer: `12/5 * 3/4 = 9/5.`

Example 3: Solve for `x/2` = 10.

Solution:

Given `x/2` = 10

x = 10 * 2 (multiply by 2 on both sides)

x=20.

Answer: x=20.

Practice Problems for Elementary Math Practice:

Problem 1: Add the followings: 3458 + 0.24 + 457 + 0.97

Answer: 3916.21

Problem 2: Find the common multiples of 6, 10 and 15.

Answer: Common multiples of 6, 10 and 15 are 30, 60, 90 …..

Problem 3: Find the summation of fraction numbers `2/5` , `3/5` and `10/5` .

Answer: Summation of fraction numbers` 2/5` , `3/5` and `10/5` = `18/5` .

Problem 4: Solve for z: (2z + 9) – (z – 10) =2z.

Answer: z=19.

Problem 5: Solve for k: k+11 =25.

Answer: k = 14.

Problem 6: Solve for s: 2s + 4 =10.

Answer: s=3.

Problem 7: Divide 455 ÷ 5

Answer: 91.

Pure Math Explained

Introduction to pure math explained:

The pure mathematics is defined as combinations of algebra, geometry, topology and number theory and analysis. Pure mathematics looks at the boundary of math and pure reason. It has been explained as "that part of math activity that is done without explicit or immediate consideration of direct application," although what is "pure" in one era often becomes applied later. In this article we will solve geometry and algebra examples for pure math explained.

Examples – Pure Math Explained:

Let us we will solve the example problems for pure math explained.

Now we are going to solve the example problem in geometry using line and circle shape for pure math explained.

Example 1:

The slope of the geometric line which passes through (3.3, 17.2) and (9.8, 21.5), find the slope value of this line?

Solution:

We know that the slope of a line can be found as m =`(Y2-Y1)/(X2-X1)`

Here, x1 = 3.3 x2 = 9.8 y1=17.2 y2 =21.5

m = `(21.5 - 17.2)/(9.8-3.3)`

= `4.3/6.5`

m =0.66

So the slope of the line is found to be 0.66.

Example 2:

What is the area of the circle if r=1.8cm?

Solution:

Formula = `pi` r2

= 3.14*1.8*1.8

= 10.17cm2

Example for Pure Math Explained:
Let us we will solve the example problem in algebra for pure math explained.

Problem 3:

Solve the given polynomial equations.

7x2 + 5 + 6x + 3x2 + 2x + 4 + 8x

Solution:

Step 1:

First we have to mingle terms x2

7x2 + 3x2=10x2

Step 2:

Now combine the terms x

6x + 2x + 8x = 16x

Step 3:

Then join the constants terms

5 + 4 =9

Step 4:

Finally, combine all the terms

10x2 + 16x +9

So, the final answer is 10x2 + 16x +9

Example 4:

Enlarge the following using identities, (6m)2-(15n)2

Solution:

Step 1:

Given, (6m)2-(15n)2

Take, (6m)2 - (15n)2, for this we need to use the identity, a2 - b2 = (a+b) (a-b)

Here, a = 6m and b = 15n.

Step 2:

As a result, (6m)2 -(15n)2 = (6m+15n) (6m-15n)

These are example problems for pure math explained.

That’s all about pure math explained.

The Sum of Twice a Number and Seven

The sum of twice a number and seven :

In general ,twice means a number multiplied by 2 .the sum of twice a number means a number multiplied by 2 +7.in other words,Sum of twice a number and seven means ,the product of  number by 2  plus seven.

For example : Let us consider the unknown number x,y,z, The expression should becomes 2x+7,2y+7,2z+7

Example Problems to Find Sum of Twice a Number and Seven :
Example 1:

The sum of twice a number and seven is 19.Find the number ?

Solution:

The sum of twice a number and 7 is 19.

Let us consider the unknown number is x,

Write this sentence into equation,we get,

2x+7 =19

Subtract 7 on both sides, 2x+7-7=19-7

2x=12

Divide both sides by 2,x=`12/2`

x=6

Therefore the unknown number x=6

Example 2:

The sum of twice a number and seven is 17.Find the number ?

Solution:

The sum of twice a number and 7 is 17.

Let us consider the unknown number is x,

Write this sentence into equation,we get,

2x+7 =17

Subtract 7 on both sides, 2x+7-7=17-7

2x=10

Divide both sides by 2,x=`10/2`

x=5

Therefore the unknown number x=5

Example 3:

The sum of twice a number and seven is 53.Find the number ?

Solution:

The sum of twice a number and 7 is 53Let us consider the unknown number is x,

Write this sentence into equation,we get,

2x+7 =53

Subtract 7 on both sides, 2x+7-7=53-7

2x=46

Divide both sides by 2,x=`46/2`

x=23

Therefore the unknown number x=23

Example 4:

The sum of twice a number and seven is 121.Find the number ?

Solution:

The sum of twice a number and 7 is 121

Let us consider the unknown number is x,

Write this sentence into equation,we get,

2x+7 =121

Subtract 7 on both sides, 2x+7-7=121-7

2x=114

Divide both sides by 2,x=`114/2`

x=57

Therefore the unknown number x=57

Please express your views of this topic What is a Composite Number? by commenting on blog.

Practice Problem to Find the Sum of Twice a Number and Seven:

1)The sum of twice a number and seven is 13.Find the number ?

Answer:3

2)The sum of twice a number and seven is 25.Find the number ?

Answer:9

Primary Solutions Math

Introduction to primary solutions math:

In this article, we are going to learn about,

"What is the definition of Mathematics"

"Primary math problems with solutions"

Definition of Mathematics:
Mathematics is a group of related sciences, including algebra, geometry, and calculus, concerned with the study of number, quantity, shape, and space and their interrelationships by using a specialized notation

Primary Solutions Math:

Here are provided some examples of primary math along with the solutions.

Ex 1:

Cost of 3 pencils is 45 cents and that of 5 pencils is 75 cents. What is the proportion?

Sol:

Ratio of the two quantities = 3 : 5

Ratio of their costs = 45 : 75

Therefore, the proportion is 3 : 5 = 45 : 75

Ex 2:

Find the slope of given line 2x+3y=0

Sol:

Given line is: 2x+3y=0

Slope =-x/y

Y==3

X=2

M=slope=-2/3

Ex 3:

Solve 2x + 12 = 0

Sol:

2x + 12 = 0

Put all the variables aside and value on other side.

We get, 2x = -12

Now to get x value divide both sides with 2

2x/2 = -12/2

x = -6

Ex 4:

Is 6 a composite or a prime?

Sol:

1 x 6 = 6

2 x 3 = 6

6 can be divided with 1 and 6 and also with 2 and 3. So 6 is a composite number.

Ex 5:

Write 9 table.

Sol:

Given: 9 table

9 × 1 = 9 (9)

9 × 2 = 18 (9+9=18)

9 × 3 = 27 (9+9+9=27)

9 × 4 = 36 (9+9+9+9=36)

9 × 5 = 45 (9+9+9+9+9=45)

9× 6 = 54 (9+9+9+9+9+9=54)

9 × 7 = 63(9+9+9+9+9+9+9=63)

9 × 8 = 72 (9+9+9+9+9+9+9+9=72)

9 × 9 = 81(9+9+9+9+9+9+9+9+9=81)

9x10=90(9+9+9+9+9+9+9+9+9+9=90)

Ex 6:

A die is rolled on a desk; what is the probability to get number 2 face.

Sol:

Given data:

Total number of out comes = 6 (A die has 6 faces)

Favorable outcome = 1 (only one face is having number 2)

P (E) = Number of favorable outcomes/Total possible outcomes.

Probability to get 2 face is = 1/6.

Ex 7:

Write the place value of  748   

Sol:

Place value of 8 = 8 x ones = 8
Place value of 4 = 4 x tens = 40
Place value of 7 = 7 x hundreds = 700

Ex 8:

Simplify x + y+2x-3y=0

Sol:

Given expression,

x + y+2x-3y=0

Add and subtract the like terms we get ,

X+2x+y-3y=0

3x-2y=0

Front End Estimation Math

Introduction to front end estimation math:

The front end estimation is typically produces the earlier estimation of the addition or the difference than the answer formed through adding or subtracting rounded numbers for the estimation in the mathematics. It may give the closer estimated values for the actual estimation for the given problem.

In the front end estimation of math, one takes the two maximum digits of a number, and then put zeros on the other position values. This is supposed to create the earlier estimation rather than rounding and then addition numbers. In this article we shall discuss the front end estimation in the math.

How to Calculate an Addition by front End Estimation Math:

Add the digits of the two maximum place values

Plug in zeros for the other place values

Ex 1:  5596 + 3845 is estimated to be 9300 through front end estimation math(i.e. 5500 + 3800).

Ex 2: 5596 + 855 is estimated to be 6300 through front end estimation math(i.e. 5500 + 800).

Examples for front End Estimation:

Ex 1: 77 + 53 = 80 + 50 = 130

Sol: Make a note of that we are only adding the left the majority numbers (8 and 5)



31 + 68 = 30 + 70 = 100



77 - 52 = 80 - 50 = 30



88 - 42 = 90 - 40 = 50.



But the numbers contains the three digits; round it to the nearby hundred places earlier than adding the left the majority digits or numbers

Ex 2:    264 + 590 = 300 + 600 = 900.



Sol: Make the note of that the front end estimation for the every other numbers apart from the left the majority digits (3 and 6) are equivalent to zero and we only added the left the majority digits as stated before.



335 + 555 = 300 + 600 = 900
666 - 354 = 700 - 400 = 300
832 - 678 = 800 - 700 = 100.


But the numbers contains the four digits, so round it to the nearby thousand place earlier than adding the left the majority digits or numbers



Ex 3:  3354 + 2677 = 3000 + 3000 = 6000.



Sol: Make a note of that the other numbers apart from the left the majority digits (3 and 3) are equivalent to zero and we only added the left the majority digits as stated before

6687 - 3765 = 7000 - 4000 = 3000

4245 + 2897 = 4000 + 3000 = 7000

8501 - 5508 = 9000 - 6000 = 3000.

But the numbers contains the five digits; round it to the nearby ten thousand places before adding the left the majority digits or numbers

Ex 4:  59974 - 36799 = 60000 - 40000 = 20000.

Sol: Make a note of that the other numbers apart from the left the majority digits (6 and 4) are equivalent to zero and we only subtracted the left the majority digits as stated before

93113 + 48893 = 90000 + 50000 = 140000

Determining Sample Space

Introduction to determining sample space:

The probability separation be capable of be described regarding the sample space. The probability may be describing more requisites as, experiment, outcome, sample space, events. The probability is a one of dimension from the set of events. The many algorithms can be used to determining the sample spaces. The simply to determining the probable outcomes is straightforward represented as sample spaces. The probability triple is a further name of the sample space. The probability process can be classified three types like sample space, events and function of event.

Determining Sample Space-definition:

Sample space is defined as; the amount of probable outcomes is known as determining sample space. The determining sample space is helped to calculate the whole space, so it is normally represented as measure space.  The sample space can be specified as ?, and it is one of the chances of non- empty set.

Some of the results can be followed through the experiment so it is simply called as deterministic experiments. From the experiments, set of possible outcome is called as random experiment. The put of probable outcomes of a random research is known as determining sample space

Determining Sample Space-examples:

Problem 1:

Determining the sample space, when three coins are tossed randomly.

Solution:

Basically the coins contain two sample outcomes like head or tail. Here, we have to toss three coins randomly. The possible outcomes are,

The required sample space = {HHH,HHT,HTH,THH,THT,TTH,HTT,TTT}

Checking:

The number of coins (n) = 3.

Normal formula for sample space = 2number of entity.

The sample space       = 23.

So, the sample space = {HHH,HHT,HTH,THH,THT,TTH,HTT,TTT}.

Understanding Least Common Multiple Finder is always challenging for me but thanks to all math help websites to help me out.

Problem 2:

Determining the sample space, where the two dice throw same time. Find the equal pair of dice.

Solution:

The each dice has been providing the six possible outcomes.

So, the two dice has been providing 36 possible outcomes.

The possible outcomes of two dice = { (1,1), (1,2),(1,3), (1,4),(1,5),(1,6),

(2,1),(2,2),(2,3) ,(2,4), (2,5),(2,6),

(3,1),(3,2),(3,3),(3,4), (3,5),(3,6),

(4,1),(4,2),(4,3),(4,4),(4,5),(4,6),

(5,1), (5,2),(5,3),(5,4),(5,5),(5,6),

(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

These are the possible outcomes by throwing two dice.

From the given problem, we have to find the pair of outcomes.

The pair of sample space = { (1,1),(2,2),(3,3),(4,4),(5,5),(6,6)}.

This is the sample spaces for throwing two dice at the same time.

Solve Calculus Help Integrals

Introduction for solve calculus help integrals:

Calculus is divisions in mathematics. It contains limits, functions, derivatives, integrals, and infinite series. Calculus is divided into two major parts, differential and integral calculus, which are associated by the basic theorem of calculus. Calculus is the learning of vary, in the equal way that geometry is the learning of shape and algebra is the learning of procedure and their application to solving equations.A math tutorial is one technique of shifting knowledge and may be used as a piece of learning. More interactive and detailed than a book; a tutorial look for to tutor by example and provide the information to complete a certain task.I like to share this Triple Integrals with you all through my article.

3 important topics involved in Calculus are,

Limits

Derivatives

Integrals

Solve Calculus Help Integrals - Examples

Solve calculus help integrals - Example 1:

Integrate the given equation with respect to x, we get

`int ` x7 dx = `(x^7 +1)/(7 + 1) ` + c

= `(x^8)/8 ` + c

Answer:

The final answer is  ` (x^8)/8` + C

Solve calculus help integrals - Example 2:

I am planning to write more post on Integration by Part, Fundamental Theorem of Calculus Examples. Keep checking my blog.

Evaluate:

`int` e(5x+3) dx

Solution:

Put (5x+3) =t so that 5 dx = dt or dx =`1/5` dt.

= `int` e(5x+3) dx =`1/5` et + C

= `1/5` e (5x+3) + C

Solve calculus help integrals - Example 3:

Integrate the given function int 2 sin 4x dx

Solution:

Integrate the given function with respect to x, we get

`int` 2 sin 4x dx = 2 `int` sin 4x dx

= 2 `xx`  - cos `(4x)/4`

= - cos `(4x)/4`

Answer:

The final answer is - cos` (4x)/4`

Solve Calculus Help Integrals - more Examples:

Solve calculus help integrals - Example 1:

Integrate int arcsin 2x dx

Solution:

U = arcsin 2x and du = dx = 1 dx

du =` 1/sqrt (1-(2x)^2) 2 dx = 2/sqrt(1-4x^2) dx ` and  v=x..

Therefore,

int arcsin 2xdx = x arcsin 2x – `int x 2/(sqrt(1-4x^2))` dx  = x arcsin 2x – 2` int x/(sqrt(1-4x^2))` dx.

Now apply u-substitution.

Let     U = 1-4x2

du = -6x dx,

`(-1/6)` du = x dx.

`int `  arcsin 2x dx = x arcsin 2x – 2` int` `x/sqrt(1-4x^2)` dx

= x arcsin 2x-2 `int 1/(sqrt(1-4x^2))` x dx

= x arcsin 2x-2 `int 1/sqrt(u) (-1/6)` du

= x arcsin 2x + `(1/3) int 1/sqrt(u)` du

= x arcsin 2x +` (1/3) int ` u-1/2 du

= x arcsin 2x+ `(1/3) (u^(1/2))/(1/2)` + C

= x arcsin 2x + `1/3(1-4x2)^(1/2)` + C

= x arcsin 2x +` (1/3) sqrt(1-4x^2)` + C.

Solve calculus help integrals - Example 2:

Evaluate:

1) `int` cos 2x dx

2) `int` e(4x+3) dx

Solutions:

1)    Put 2x =t so that 2 dx    = dt or dx =`1/2` dt.

= `int` cos 2x dx =`1/2` sin t + C

= `1/2` sin 2x + C

2)    Put (4x+3) =t so that 5 dx = dt or dx =`1/4` dt.

= `int` e(4x+3) dx =`1/4` et + C

= `1/4` e (4x+3) + C

Rules in Terminating Decimals

Introduction to rules in terminating decimals:
Since our numbering system decimals are fraction having their denominators as 10 or powers of 10, it is called a decimal system. Dec em in Latin means ten.

Here we will learn about numbers that are less than 1.

The numbers written in decimal form are called decimal numbers (or) simply decimals.
A decimal has two parts:
(1) whole number part and
(2) decimal part

Example:

In 473.619, the whole number part = 437 and the decimal part = .619

Rules in Terminating Decimals- Basic Definitions

Decimal point:

The dot that we use to separate the whole number part from the decimal part is called the decimal point.

Decimal places:

In a decimal, the places occupied by the digits after the decimal point are called decimal places.

For examples: 1) 7.37 has two decimal places.

2) 2.176 has three decimal places.

Decimal fractions:

A fraction whose denominator is 10 or a power of 10 is called decimal fractions

To round off a decimal:

Step 1: Look for the number of decimal point we would like to keep.

Step 2: Check the digit after the decimal point we would like to keep.

Step 3: Drop the digit if it is less than 5.

Step 4: If the digit is 5 or more, add 1 to the previous number before dropping

the extra digit

I am planning to write more post on What is an Irrational Number?, List of Prime Numbers to 100. Keep checking my blog.

Rules in Terminating Decimals-some Definitions

Definition of Terminating Decimal:

The word “terminate” means “end”. A decimal that ends is a terminating decimal. Inother words, if a fraction can be converted into a decimal completely, then such a decimal number is called a terminating decimal.

In the process of converting a fraction into a decimal by the division method, if we obtain a zero remainder after a certain number of steps, then the decimal obtained is a terminating decimal.

example for rules in terminating decimals:

Consider the following fractions namely 3/4 5/8, 1/8, 3/2,.by using convert fractions to decimals

Non-terminating decimals:

If a fraction cannot be converted into a decimal completely, then the division does not come to an end. Such a decimal number is called a Non-terminating decimal.

Example: ? = 22/7 = 3.412 . . . is a non-terminating decimal.

Recurring decimals:

Non-terminating decimals in which some digits are repeated continuously are called recurring decimals.

Example: 1.09090909. . . is a recurring decimal. It can be written as 1. 09 with ‘bar’ over the first and last digits of the repeated numbers.

Surface Integrals of Vector Fields

Introduction to Surface integrals of vector fields over parameterized surface:

The surface integral of vector field F on R3 over a surface S described in parametric form by r : D ? R3 is given by the number :

?? S F`*` dS = ?? DF(r(u,v)) `*` (ru `xx` rv) du dv

Surface integrals of vector fields for graphs:

Consider the vector field F = Pi + Qj + Rk and the surface S given by the graph z = f(x, y) on a domain D. Then the surface integral of vector field F over the surface S is :

`intint` S F`*` dS = ?? S F`*` ndS = ?? D(-Pfx - Qfy + R)dx dy

Example for Surface Integrals of Vector Fields

2?  ?

Let S be the unit sphere given by the parameterization:

x = cos ? sin f , y = sin ? sin f , z = cos f  for 0 = ? = 2p , 0 = f = p

Find the surface integral of vector field F(x, y, z) = xi + yj + zk  over the surface S.

Solution:

Here we have r(?, f) = ( cos ? sin f,  sin ? sin f,  cos f)

Hence,

r? = (-sin ? sin f, cos ? sin f, 0)

rf = (cos ? cos f,  sin ? cos f,  -sin f)

r? × rf = (-sin2 f cos ?, -sin2 f sin ?, -sin f cos f)

Next we evaluate,

F  · (r? × rf) = -sin f

Hence,

?? S F· dS = ?? DF(r(?, f)) · (r? × rf) d ? d ?

= $\int_{0}^{2\Pi }\int_{0}^{\Pi } -sin \varphi  d\varphi d\theta$

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Exercise on Surface Integrals of Vector Fields

Consider the vector field F  = x2i + y2j + zk. Let S be the surface given by the graph of the function z = x + y + 1 over the rectangle 0=x=1 , 0=y=1. Evaluate ?? S F`*` ndS

Hint: Here z = f(x, y) = x+y+1. Hence find fx and fy and plug in the formula given above for surface integrals of vector fields for graphs.

Conclusion on Surface Integrals of Vector Fields:

Surface integrals of vector fields are also referred to as flux across S. In other words, it is the volume of fluid moving with a velocity described by the vector field F(x,y,z) crossing the surface S per unit time.

Add Radical Numbers

Introduction to add radical numbers:

A square root of a number is said to be a radical number. Which is obtained by closing the numbers under add, subtract, multiply, and root extraction. A radical expression is containing a square root.  Radical is the (Sqrt) v symbol. That is used to denote square root, otherwise it is called as nth root. Here we have to go for add radical numbers, and how to work out addition of radical numbers.

Add Radical Numbers:

Add (Adding) radical expressions, which are containing radicals. Adding radicals are simple, when similar terms.

Similar terms:

If the radicals are similar terms means it has meet with following conditions.

The radical both have same index.
Both two radicals under the root sign is have identical quantities.
Any radicals, which are outside is same value.
Different terms:

If the radicals are different (not same) terms means it has meet with following conditions:

The radicals both are having same indexes.
The quantities which are under the radicals are different (not same), but both radicals need to simplify.
The outside variables are different (Not as same).

I am planning to write more post on Significant Figures Practice, Ratios and Proportions Word Problems. Keep checking my blog.

Example Problems for Add Radical Numbers:

Example 1:

Solve: v125 + v45. Add the radical numbers and find the answer.

Given: v125 + v45

First we have to simplify it. (125 = 25 * 5, and 45 = 9 * 5)

= v (25*5) + v (9*5)             (take common factor outside)

= 5v5 + 3v5

= 8v5.

Answer is: 8v5

Example 2:

Solve: v256 + v196. Add the radical numbers and find the answer.

Given: v256 + v196

Simplify the given terms, and we get (256 = 16* 16 and 196 = 14 * 14)

= v 16 * 16 + v 14 * 14                       (take common terms outside)

= 16 + 14

= 30

Answer is: 30

Example 3:

Solve: 2av3a3b3 + 3bv27a5b. Add the radical equation, and find the answer.

Solution:

Given: 2av3a3b3 + 3bv27a5b

Simplify it, and write it as separate,

= 2av3a2. a. b2.b + 3bv9.3a2. a2.a.b   (take common factors as outside) and write,

= 2a2bv3ab + 9a2bv3ab

Then we combine the co-efficient terms and we get 11a2bv3ab.

Answer is: 11a2bv3ab.

Double Integration by Parts

Introduction for double integration by parts:
The integration process represents the inverse process of the differentiation in progress. If `dy/dx` =h(x) here y is the function then the double integrals `int` `int`h(x) dx dy= y. There are two types of integration namely indefinite integral (without  limits) and definite integral (with limits). If we consider u = f(x), v = h(x), then the product rule in its simplest form is: `intint u dv/dx dxdy = int [uv - int v (du)/dx dx]dy`

In this article we shall discuss about the double integration by parts with the examples for the solution.

Examples to Explain "double Integration by Parts "

Double integration by parts on this function  `int int 5x^3 ln x dxdy` ,solve for the answer after calculating the integration on the first step by the integration by parts.
Solution:

First we take inner integral on the functions   `int 5x^3 ln x dx`

It is solved under the integration by parts as the first step for the prosecution of the steps.

The logarithmic form is exist in the given integration function.

Take ln x as u and `x^3 dx` as dv

u =5 `ln x`                   dv =`x^3 dx`

`(du)/dx = 5/x`                   v = `int x^3 dx`    

du = `5dx/x`                  v = `x^4/4 `      

We have  `int u (dv)/dx dx = uv - int v du/dx dx`

0r `int u dv = uv - int v du`

`int 5x^3 ln x dx`  = 5 ln x . `x^4/4 - ` `int x^4/4 5dx/x`             

=  5 ln x . `x^4/4 - ` `5/4int x^3 dx` 

=  5 ln x . `x^4/4 - `   `5/4 (1/4 x^4) `           

= `5 ln x (x^4)/4``-(5x^4)/16`          

This is the result obtained under the integration by parts for the inner integral.

Operate it with the outer integral functions by,

`int int 5x^3 ln x dxdy` =  `int [5 ln x (x^4)/4-(5x^4)/16 ] dy`         

=  `int` 5 ln x . `x^4/4 dy ` ` - int (5x^4)/16 dy` 

= `5y ln x (x^4)/4``-(5yx^4)/16 + c `          

This is the result obtained under the double integration by parts.

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Problems to Explain "double Integration by Parts "

Double integration by parts on this function  `int int x log x dx dy` ,solve for the answer after calculating the integration on the first step by the integration by parts.
Solution:

First we take inner integral on the functions   `int x log x dx`

It is solved under the integration by parts as the first step for the prosecution of the steps.

`int x log xdx` = `int ( log x) (x dx)`

The logarithmic function on the x function with another function x is not integrated directly ,so we take it as

u = logx       and        dv = x dx

? du = 1/x dx              v = `x^2/2`           

? `int x log x dx` = (logx) `x^2/2`  -  `int (x^2/2) (1/x dx)`                        

= `x^2/2`  (logx) `-1/2` `int x dx`       

? `int x log x dx` =  `x^2/2`  (logx) - `1/4 x^2 `

This is the result obtained under the integration by parts for the inner integral.

Operate it with the outer integral functions by,

`int int x log x dx dy`  =    `int [x^2/2(log x) - 1/4 x^2 ] dy`           

=    `int [x^2/2(log x)]dy - int [1/4 x^2 ] dy`

=       `x^2y/2(log x) - 1/4 x^2y dy + c`              

This is the result obtained under the double integration by parts.

Addition Fraction Calculator

Introduction to addition fraction calculator
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. much later development still used today (½, ?, ¾, etc.) and which consist of a numerator and a denominator, the numerator representing a number of equal parts and the denominator telling how many of those parts make up a whole. Mathematical calculators are used to perform mathematical operations. With the help of fraction addition calculator we can easily adding the fraction.  In this article of addition fraction calculator we are going to discuss about adding the fraction using calculator.

Calculation of Addition of Fractions.

Addition Fractions with same denominators:

A fraction contains two numbers, the top number value is called the numerator, and the bottom number value is called the denominator.

numerator
denominator

To add two fractions the denominator have the same value, add the numerators that is top value and put that sum over the common denominator value.

Addition Fractions with different denominators:

Find the Least Common multiplication (LCM) of the fractions

Add the numerators values (top value) of the fractions

Simplifying the fraction value

Addition two mixed numbers the fractions have the same denominator:

Adding the numerators value (top value) of the two fractions

Place that sum of top value over the common denominator (top value).

If the fraction is improper fraction (numerator larger than or equal to the denominator) then we have to convert it to a mixed number

Adding the integer part of the two mixed numbers

I am planning to write more post on real numbers, solving inequalities. Keep checking my blog.

Fraction Addition Calculator Examples

Example for adding fraction with same denominator:

Find `(4/12)+(5/12)`

Adding top values

=`(9)/(12)`

Example for adding fraction with different denominator:

Find the sum of 2/3 and 4/6`(2/3)+(5/6)`

= `(2)/(3)` =`(5)/(6)`

=`(2*2)/(3*2)`  + `(5*1)/(6*1)`

`(4)/(6)`+  `(5)/(6)`

`(9)/(6)`

= 1 `(3)/(6)`

Example for adding fraction with different denominator:

Adding 2 `(3/4)+3(1/2)`

= 2 `3/4` = `11/4`

=3 `1/2` = `7/12`

=`11/4` +`14/4` =`25/4`

convert int to mixed fraction `25/4` = 6 `1/4`

Semi Circle Math Definitions

Introduction of Semi circle math definitions:

Another name of the semi circle is half of the  circle. The diameter of a circle is the distance from a one point on the sphere to an end point pi Radians, and is the distance from one point on a sphere to another.

If r is the semi circle radius, then d = 2r. d is the diameter of the semi circle.

Example for Semi Circle in Math Definitions:

Semi circle- Example 1 in math definitions:

Find the diameter of a semi circle given that its radius is 5

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 5 = 10

Semi circle- Example 2 in math definitions:

Find the diameter of a semi circle known that area of the semi circle is 153.86

Solution:

Step 1:

Find the radius of the semi circle

`( pi . r^2)/2` = 153.86

`(r^2)/2` = `153.86 / 3.14`

`(r^2)/2`   = 49

r2 = 49 x 2  = 9.899



Step 2:

Then calculate the diameter of the semi circle,

Diameter = 2 * r

=  2 * 9.899

= 19.79800

Therefore, the diameter of the semi circle is 19.798.

Semi circle- Example 3 in math definitions:

Find the diameter of a semi circle known that its radius is 6

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 6 = 12

Semi circle- Example 4 in math definitions:

Find the diameter of a semi circle known that area of the semi circle is 133.04

Solution:

Step 1:

Find the radius of the semi circle

` (pi . r^2)/2` = 113.04

`(r^2)/2` = `113.04 / 3.14`

r2  = 36 x 2 = 72

r = 8.485

Step 2:

Then calculate the diameter of the semi circle,

Diameter = 2 * r

=  2 *8.485

= 16.97

Therefore, the diameter of the semi circle is 16.97.

Semi circle- Example 5 in math definitions:

Find the diameter of a semi circle known that its radius is 8

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 8 = 16

Semi circle- Example 6 in math definitions:

Find the diameter of a semi circle known that its radius is 21

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 21 = 42

Semi circle- Example 7 in math definitions:

Find the diameter of a semi circle known that area of the circle is 1384.74

Solution:

Step 1:

Find the radius of the circle

`(pi . r^2)/2` = 1384.74

r2/2 = 1384.74 / 3.14

r2  = 441 x 2

r2 =  882

r   = 29.698



Step 2:

Then calculate the diameter of the circle,

Diameter = 2 * r

=  2 * 29.698

= 59.396

Therefore, the diameter of the semi circle is 59.396.

Semi circle- Example 8 in math definitions:

Find the diameter of a semi circle known that its radius is 32

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 32 = 64

Semi circle- Example 9 in math definitions:

Find the diameter of a semi circle known that its radius is42

Solution:

The diameter is always multiply by 2 the length of the radius,

Diameter = 2 x 42 = 84

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Practice Problems for Semi Circle in Math Definitions:

Practice Problem 1:

Find the diameter of a semi circle known that its radius is100

Answer:

200

Practice Problem 2:

Find the diameter of a semi circle known that area of the semi circle is 530.66

Answer:

36.768

Word Problems about Prime Factorization

Introduction :

In Mathematics, the basic theorem of arithmetic states that each composite number can be factorized uniquely into a product of prime factors. A number is a prime number if its only whole number factors are 1 and itself. 7 are primes because its only factors are 1 and 7. If a number is not primes, it is called a composite number.

Because 4 have factors of 2 and 2, 4 is a composite number. The number 1 is not considered a prime number. Therefore, it is not included in the following list of prime numbers less than 50

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47

Word Problems about Prime Factorizations – Definition and Example Problems:
Word problems about prime factorizations – Definition:

A whole number greater than 1 with precisely two factors, itself and 1, is a prime number. A whole number greater than 1 with above 2 factors is a composite number. The numbers 0 and 1 are neither prime nor composite: 0 has infinite factor, and 1 only has one factor, itself.

The number 2 is the only even prime number. A number expressed as a produce of factors that are all prime is called the prime factorizations of the number. For example, the prime factorizations of 12 are 2 x 2 x 3. There are two main ways of finding the primes of a number: dividing and splitting.

Word problems about prime factorizations - Example problems:

Word problem 1:          

Jim purchased some oranges to distribute to her friends. If she gave 3 or 5 or 6 oranges to each of her friends she will be left with 100 oranges. What is the least number of oranges she purchased?

Solution:

The L.C.M of 3, 5, 6 = 30
100 + 30 = 130
The least number of oranges she purchased is 130.
Solution: 130 oranges.

Word problem 2:

George wished to buy some crackers. If he gave 2 crackers or 3 crackers or 5 crackers to each of his friends, he is left with no crackers. What is the least number of crackers he must purchase?

Solution:

George wished to buy some crackers.

The L.C.M of 2, 4, 5 = 20

The least number of crackers he purchased is 20.

Solution: 20 crackers.

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Word Problems about Prime Factorizations – Practice Problems:
Word problem 1:

Jane purchased a few oranges to distribute to her friends. If she gave 4 or 5 or 6 oranges to each of her friends she will be left with 15 oranges. What is the least number of oranges she purchased?

Answer: 75 oranges she purchased.

Word problem 2:

Three clocks ring once at the same time. After that, the first clock rings after every 3 hours, the second after every 7 hours, and third after every 5 hours. After how many hours will they again ring together?

Answer: 105 hours will they again ring together.

Word problem 3:

What is the smallest number of students required so that they may be uniformly arranged in rows of 12, 14 or 20?

Answer: 252 arranged students

Prime Number Square

Introduction to prime number square:

Prime numbers:

In arithmetic, a prime number is normal numerals that have accurately two separate normal numeral divisors: 1 and itself. There are infinitely a lot of prime facts. The first twenty-five prime numbers are:

2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97.

Square numbers:

In arithmetic, a square numeral, also called a perfect square, is an numeral that is the square of an digit; in other terms, it is the multiple of a number of numeral with itself. For example, 4 is a square numeral, as it can be written as 2 × 2. Square facts are non-negative. Example: 22 = 4

Example for Prime Number Square:
Locate the prime number square on first five prime numbers. 

Solution:

First five prime numbers are 2, 3, 5, 7, and 11.

Prime number square is given below that:

2 is a Prime number. So, 22 = 4

3 is a Prime number. So, 32 = 9

5 is a Prime number. So, 52 = 25

7 is a Prime number. So, 72 = 49

11 is a Prime number. So, 112 = 121

The first five prime numbers square is 4, 9, 25, 49, and 121.

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More Examples for Prime Number Square:
Example 1:

Locate the prime number square on 31, 37, 41, 43, and 47. 

Solution:

Prime number square is given below that:

31 is a Prime number. So, 312 = 31 x 31 = 961

37 is a Prime number. So, 372 = 37 x 37 = 1369

41 is a Prime number. So, 412 = 41 x 41 = 1681

43 is a Prime number. So, 432 = 43 x 43 = 1849

47 is a Prime number. So, 472 = 47 x 47 = 2209

The prime numbers square is 961, 1369, 1681, 1849, and 2209.

Example 2:

Locate the prime number square on first five prime numbers. 

Solution:

First five prime numbers are 13, 17, 19, 23, and 29.

Prime number square is given below that:

13 is a Prime number. So, 132 = 169

17 is a Prime number. So, 172 = 289

19 is a Prime number. So, 192 = 361

23 is a Prime number. So, 232 = 529

29 is a Prime number. So, 292 = 841

The first five prime numbers square is 169, 289, 361, 529, and 841.

Basic Number Theory

Introduction to basic Number theory:

Let us see about the topic is basic number theory is said to be as the complete theory of numbers in which we can study about basic process of math like addition, subtraction, multiplication and division in step by step method. The basic number theory is the basic theory of mathematics. We shall prepare some example of explain basic number theory in the below articles.

Types of Number Theory:

Theron are four types of basic number theory are mostly used in the basic number theory; these types are classified according to the symbol of the number. It will be shown as below,

Addition Number Theory
Subtraction  Number Theory
Multiplying  Number Theory
Division Number Theory

Explanation with Basic Number Theory

Addition Number Theory:

Numbers are used for including process in daily life. The math symbol for addition is plus or +. The math symbol of plus operation is used for adding two or more quantities into single sum of quantities. The two numbers are added jointly to obtain the solution for the sum of two numbers.

For example:

203 + 15 = 218, here, 203 and 15 are addends and 218 are called as sum.

Subtraction Number Theory:

Subtraction number theory get leave in such situations where present is a loss or reduction of somewhat as a result of subtracting a number from several extra number. Subtraction is getting gone part from a total.

For example:

119 – 9 = 110, here, 119 and 9 are minus are called as subtraction.

Multiplying Number Theory:

Here we are leaving to see the basic number theory using multiplication, usually multiplication is called as the constant addition, understand if we are combine the similar 7 groups with 7 objects in each group means it will be written in multiplication format of 7*7=49, we get the same answer in addition also, it will be shown below 7+7+7+7+7 = 49. 

For example:

12 * 7 = 84

Division Number Theory:

In arithmetic divisor is distinct as “the numeral that you are available to divide by”. Divisor of an exacting numeral is also known as the divisor of a given digit.

Dividend /Divisor = Quotient.

For example:

The given number is 60/10

Answer is 6

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Example Problems for Basic Number Theory:

Example 1:

Adding 81 and 4

Solution:

Let us write the given problem as 81 + 4.

Add the number 81 and 4 we get the sum as,

81 + 4 = 85

Therefore, the solution for adding 81 + 4 is 85. 

Example 2:

There are 163 students playing in a ground. 21 of them are getting ready to class room. How many students will be left?

Solution:

Total number of students = 163

Number of students go to class room = 21

Number of students left = 163 - 21

= 142

Example 3

Multiplying two numbers 53 and 29

Solution:

The given two numbers 53 and 29.

We need to find the product of two numbers

By multiplying 53 and 29

We get 1537

So the answer is 1537

Problem 4:

Divide 800 / 10

Solution:

80(quotient)

----

10)  800

800 -

---------

0(remainder)     

---------

Solving Polynomial Equations in Factored Form

Introduction to polynomial equations in factored form:
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. (Source: Wikipedia)

Generally polynomial equations can be written as, x^2 - 3x + 9. Polynomial equations in factored form can be written as

(x - 2) (x - 3).

Example Problems for Solving Polynomial Equations in Factored Form
Solving polynomial equations in factored form example problem 1:

Write the given polynomial expression x^2 - 17x + 16 in factored form.

Solution:

Given polynomial expression is x^2 - 17x + 16

First factorize the given expression, we get

(x^2 - 17x + 16) = (x^2 - 16x - x + 16)

Grouping the first two terms and second two terms, we get

= (x^2 - 16x) - (x - 16)

= x (x - 16) - 1 (x - 16)

= (x - 16) (x - 1)


The factors of the given polynomial expression is (x - 16) and (x - 1)

Answer:

The final answer is (x - 16) and (x - 1)

Solving polynomial equations in factored form example problem 2:

Write the given polynomial expression x^2 + 27x + 140 in factored form.

Solution:

Given polynomial expression is x^2 + 27x + 140

First factorize the given expression, we get

(x^2 + 27x + 140) = (x^2 + 20x + 7x + 140)

Grouping the first two terms and second two terms, we get

= (x^2 + 20x) + (7x + 140)

= x (x + 20) + 7 (x + 20)

= (x + 20) (x + 7)


The factors of the given polynomial expression is (x + 20) and (x + 7)

Answer:

The final answer is (x + 20) and (x + 7)

Solving polynomial equations in factored form example problem 3:

Write the given polynomial expression x^2 + 17x - 434 in factored form.

Solution:

Given polynomial expression is x^2 + 17x - 434

First factorize the given expression, we get

(x^2 + 17x - 434) = (x^2 + 31x - 14x - 434)

Grouping the first two terms and second two terms, we get

= (x^2 + 31x) - (14x + 434)

= x (x + 31) - 14 (x + 31)

= (x + 31) (x - 14)


The factors of the given polynomial expression is (x + 31) and (x - 14)

Answer:

The final answer is (x + 31) and (x - 14)

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Practice Problems for Solving Polynomial Equations in Factored Form
Solving polynomial equations in factored form practice problem 1:

Write the given polynomial expression x^2 + 4x - 32 in factored form.

Answer:

The final answer is (x + 8) and (x - 4)

Solving polynomial equations in factored form practice problem 2:

Write the given polynomial expression x^2 - 7x - 260 in factored form.

Answer:

The final answer is (x - 20) and (x + 13)

Solving polynomial equations in factored form practice problem 3:

Write the given polynomial expression x^2 + 21x + 110 in factored form.

Answer:

The final answer is (x + 10) and (x + 11)

Fourth Grade Math Expressions

Introduction to fourth grade math expressions:

Fourth grade math expressions involve the process of solving basic math expressions using arithmetic symbols. The algebraic expressions with arithmetic symbols are referred as fourth grade math expressions. For fourth grade students, the simple math expressions include addition, subtraction, division and multiplication. The elementary math expressions are solved for fourth grade students. The following are the example math expressions with detailed solutions for fourth graders.

Fourth Grade Math Expressions Examples:

Fourth grade math expressions are divided into four sections as Basic Math Operation, Fractions, Integers Operations, Combining like terms of Variables. The example problems are discussed below with step by step detailed solution for fourth graders. The solved problems are under arithmetic categories.

Basic Math Operations:

Fourth grade math expressions mainly cover addition, subtraction, multiplication, and division problems

1) Solve, 3(4+3) – 9 + 3(4)

Solution:
= 3(7) – 9 + 12
= 21 + 3
= 24

2) Solve, 5 + `8/4` * 5 + 5

Solution:
= 5 + 2 * 5 + 5
= 5 + 10 + 5
= 20

3) Solve, 6^2 + 7^2  

Solution:

= (6 * 6) + (7 * 7)
= 36 + 49
= 85

Fractions:

Fraction is one of the parts of fourth grade math expressions. Simplify the following (Write in mixed number if any of the answer is an improper fraction)

1) Solve` (2/3) + (1/5)`

Solution:

Take LCM as 15
= `(10/15) + (3/15)`

Add the above terms.
` = (10+3) / 15 `

` = 13/15`

2) Solve` (3/5) - (1/6)`

Solution:

Take LCM as 30
` = (18/30) - (5/30)`

Add the above terms.
=` (18 - 5) / 30`


=` 13/30`

Integer Operations:

Integer operation deals with performing math operations with integers such as positive and negative integers.

1) Solve, 11(–7)
= – 77

2) Solve, –81 / –3
= –27 / –1
= 27

3) Solve, –14 – (–3)
= –14 + 3
= –11

Variables – Combining Like Terms:

Variables plays major role in fourth grade math. Simplifying variables by combining like terms helps to calculate the value of the variable.

1) Solve, 5x + 3 – 3x
= 5x – 3x + 3
= 2x + 3

2) Solve, y + 2y – 2 + 5
= 3y + 3


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Fourth Grade Math Expressions Practice Problems:

The following are the fourth grade math problems for self practicing.

1) Simplify the math expressions.

4^2 + (2 * 4) – 4

Solution: 20

2) Simplify the math expressions.

(–2)2 + 8 – 10 – 6

Solution: - 4

3) Simplify the math expressions.

8x + 4y – 8 + 4y – 2x

Solution: 6x + 8y – 8

Rational Expressions

Expression is a finite combination of symbols that are well formed and rational expressions is that where the numerator and the denominator or both of them are polynomials. You can practice rational expressions problems with expert and highly qualified tutor vista tutors. Our tutors help you out to learn the concept of rational expressions and have a gaining ground over the topic.

It includes the concepts of polynomial fractions. Get help with rational expressions online and gain valuable math learning.


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Introduction:

Polynomial fractions are declaring the rational expression, usual fractions you can do with rational expressions. When dealing with rational expression, you will frequently need to estimate the appearance, and it can be useful to know which values would cause division by zero, so you can pass up these x-values. Ratio of two polynomials is declaring the rational expression.

Definition:

A rational number is a few number that can be printed in the form a/b, there are a is specified the integer and b is also specified the integers and b ? 0. It is needed to declare exclude 0 because the fraction specified the fraction and division by zero is undefined.

Rules

Each problem by factoring everything you can.
Retain information that, even with all the difficult looking functions, a rational expression is just a fraction: you control them using all the rules of fractions that you are common with.
Two rational expressions same to both other. It is known as the rational equation. Rational expression goal is specified the solve for x, it is locate the x value that create the equation is true.

Simplifying Rational Expressions
Simplifying rational expressions becomes easy with little help. Following is a detail explanation of rational expression examples. A rational expression is more than a fraction in which the numerator and/or the denominator are polynomials.

x/52

The x is specified the numerator and 52 is specified the denominator
The 52 is specified denominator, it is a constant, the expression is known as for all real number values of x.

102/x

The 102 is called the numerator and x is called the denominator
Denominator is represents the x is a variable, expression is approximate specified x=0

Basic Method

`(a)/(b)=``(ad)/(bd)`

Additional method

`(a)/(b)+(c)/(b)=(a+c)/(b)`

subtraction method

`(a)/(b)-(c)/(b)=(a-c)/(b)`

Multiplication method

`(a)/(b).(c)/(d)=(ac)/(bd)`

Division method

`(a)/(b)-:(c)/(b)=(ad)/(bc)`

Example

1.  `(10)/(30)`+ `(5)/(20)`

= `(20+15)/(60)`

=  `(35)/(60)`

=   `(7)/(12)`

2.     `(60x^(3))/(80x^(2))`   =   `(10x^(2).6x)/(10x^(2).8)`

=    `(10x^(2).6x)/(10x^(2).8)`

=     `(6x)/(8)`

=   `(3x)/(4)`

3.     `(40)/(60)`       = `(4.10)/(6.10)`         

=  `(4)/(6)`

4.  `(x-5)^(2)`

`x^(2)+25-10x-` `x^(2)+11x`

x+25

5.   `(8)/(6)` -`(5)/(8)`

=   `(32-15)/(24)`

=`(17)/(24)`

6.  `x^(2)+8x = -16`

`x^(2)+16+8x =0`   

`(x+4)^(2)=0`

`x=-4`

Practice problem:

1.`(8x+16)/(4y)` =      `(2x+4)/(y)`

2.  `(4x+2)/(4)`    =   x +  `(1)/(2)`

Consistent System of Equations

Introduction to consistent system of equations:

                  Algebra is a subdivision in math, which comprises of vast operations on equations, polynomials, radicals, rational numbers, logarithms, etc. consistent system of equations are contains variable equations which is to be solved. While solving, if the there is at least one solution it is called as consistent system of equations. If there is more than one solution then they are said to be undetermined system of equations.

     
Consistent System of Equations-example Problem:

The goal here is to solve
3x+2y=5
4x+6y=3 for the variables x and y.
Let's start by solving 3x+2y = 5 for the variable x.
Move the 2y to the right hand side by subtracting 2y from both sides, like this:
Now, the equation system reads:
3x = 5-2y
To isolate the x, we have to divide both sides of the equation by the other variables
         around the x on the left side of the equation.
The last step is to divide both sides of the equation by 3 like this:
3x ÷ 3 = x
To divide 5-2y by 3
Divide each term in 5-2y by 3 term by term.
The solution to your equation is:
x =5/3-2/3y
Next, let's solve 4x+6y = 3 for the variable y.
Move the 4x to the right hand side by subtracting 4x from both sides, like this:
From the left hand side:
4x - 4x = 0
The answer is 6y
From the right hand side:
The answer is 3-4x
Now, the equation reads:
6y = 3-4x
To isolate the y, we have to divide both sides of the equation by the other variables
     around the y on the left side of the equation.
The last step is to divide both sides of the equation by 6 like this:
Divide each term in 3-4x by 6 term by term.
The solution to your equation is:
y =1/2-2/3x
Now, plug the earlier result, x=5/3-2/3y, in for x everywhere it occurs in
y=1/2-2/3x.
This gives y=1/2-2/3(5/3-2/3y). Now all we have to do is solve this for y,to have our first solution.
1/2-2/3*(5/3-2/3*y) evaluates to ½ - 10/9 + 4y/9
Y - 4y/9 = -11/18
5y/9 =-11/18
The last step is to divide both sides of the equation by 5/9 like this:
y = - 11/18 * 9/5
The solution to the equation is:
y = -11/10
Lastly, to find the solution for x, we plug this answer for y into the earlier result that
x=5/3-2/3y.
This gives x=5/3-2/3(-11/10).
On, simplifying this.
x= 12/5,
So, the solutions to your equations are:
x= 12/5
y= -11/10
Since this set of equations has one solution, they are consistent system in nature.

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Consistent System of Equations-practice Problem:

1. Check whether the following set of equation is consistent or inconsistent?
2x+3y=1
3x+2y=2
[Ans: x =4/5, y = -1/5]

Simplifying Algebra Equation

Introduction to simplifying algebra equation:

        An algebra equation is a mathematical statement that asserts the equality of two expressions. An algebra equations consist of the expressions that are to be equal on opposite sides of an equal sign, as in

                x+3=5

        One use of an algebra equations is in mathematical identities, assertions that are true independent of the values of any variables contained within them. For example, for any given value of x it is true that

                x(x-1)=x^2-x

         However, an algebra equations can also be correct for certain values of the variables.

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Problems on Simplifying Algebra Equation:
 Solve equation of the form ax+b=c

              In solve equation of the form ax+b=c, the goal is to rewrite the equation in the form variable=constant. This needs applying both the addition and the multiplication property of equations.

           Solve:`(2)/(5)` x-3=7

  Solution:

                        `(2)/(5)` x-3=-7

                        `(2)/(5)` x-3+3=-7+3       add 3 to each side of the equation.

                        `(2)/(5)`x=-4                  Simplify.

                        `(5)/(2)`*`(2)/(5)`x=`(5)/(2)`(-4)            Multiply each side of the equation by the coefficient 5/2

                        x=-10                    Simplify. Now the equation is in the form variable=constant.

Example 2:

Solve: 3x-7=-5

Solution:

            3x-7=-5

            3x-7+7=-5+7                      Add 7 to each side of the equation.

            3x=2                                 Simplify.

            `(3x)/(3)` =`(2)/(3)`                               Divide each side of the equation by 3.

            x=`(2)/(3)`                                 Simplify. Now the equation is in the form variable=constant.

            The solution is `(2)/(3)`.            Write the solution.



Example 3:

            Solve: `(1)/(2)`=x+`(2)/(3)`

Solution:

            `(1)/(2)`=x+`(2)/(3)`

            6(`(1)/(2)` )=6(x+`(2)/(3)`)           Multiply both side of the equation by 6, the LCM of the denominators.

            3=6(x)+6(`(2)/(3)`)           Simplifying equation. Use the Distributive Property on the right side of the equation.

            3=6x+4                 Note that multiplying both side of the equation by the LCM of the denominators eliminated by the fractions.

            3-4=6x+4-4             Add -4 to each side of the equation.

            -1=6x                     Divide each side of the equation by 6.

            x=-`(1)/(6)`                       Simplify. Now the equation is in the form variable=constant.

Practices Problems on Simplifying Algebra Equation:
Problem 1                  

           Solve: 4x-1=5

           Answer: x=1

Problem 2:

            Solve:  `(1)/(2)`=x+`(1)/(3)`

             Answer: -`(1)/(2)`