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)`