Solve Basic Math Practice

Introduction to solving basic math practice:

The basic math practice consists of the operation like addition, subtraction, multiplication, and division. In this basic math practice all the operation are easy to solve. We are living in a mathematical world. Whenever we want to purchase, choose an insurance or health plan, we rely on math understanding. The mathematical thinking introduces the new problem solving technique increased dramatically.math problem can be solve usually a rough sheet. but now all are having the system and internet connections so we can able to see in to the computer.Please express your views of this topic Derivative Division Rule by commenting on blog.


Solve addition for basic math practice:


To solve addition for basic math practice has the numbers for addition operation it can be solve by using the following operation.

Example problem 1:

Solve 4+4=?

Given that 4 + 4

4+4=8

Example problem 2:

Solve 4+3+4=?

Given that 4 + 3+4

=7+4 (4+3=7)

=11

Solve subtraction for basic math practice:

To solve subtraction for basic math practice has the numbers for subtraction operation it can be solve by using the following operation.

Example problem 1:

Solve 5 -2=?

Given that 5 -2

=5-2

=3

Example problem 2:

Solve 9 -4 -3=?

Given that 9 -4 -3

=5 - 3 (9-4=5)

=2

Solve multiplication for basic math practice:

To solve multiplication for basic math practice has the numbers for multiplication operation it can be solve by using the following operation.

Example problem 1:

Solve 2*4=?

Given that 2*4

=2*4

=8

Example problem 2:

Solve 2*3*4=?

Given that 2*3*4

=6*4 (2*3=6)

=24

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Solve division for basic math practice


To solve division for basic math practice has the numbers for division operation it can be solve by using the following operation.

Example problem 1:

Solve 3/3=?

Given that 3/3

=3/3

=1

Example problem 2:

Solve 7/3=?

Given that 7/3

=7/3

If the denominator is lesser than numerator

=7/3           (6/3=2)

After that we multiply the numerator* (10)

=7/3 =2.33

=2.33

Solve practice problem

Addition

5+7

8+2

6+3

Subtraction

3-2

5-5

8-9

Multiplication

2*1

2*2

2*3



Division

6/6

5/1

20/2

Answers

Addition

12

10

9

Subtraction

1

0

-1

Multiplication

2

4

6

Division

1

5

10

Basic Statistics Problems Tutor

Introduction to Basic Statistics Problems Tutor

Statistics is the proper science of making effective use of numerical data connecting to groups of individuals or experiments. It deals by means of all aspects of this, counting not only the collection, analysis and interpretation of such information, except the planning of the gathering of data, in terms of the design of surveys and experiments. Now we will study the basic problems given by tutor. Let us see about the basic statistics problems.


Basic Statistics Example Problems Tutor


The following examples problems are used to learn statistics basic. It is the examples are given by the tutor.

Example 1

Find out the mean, median, mode, range of the following given numbers in statistics?

4,8,12,16,19,22,27.

Solution

The given numbers are 4,8,12,16,19,22,27.

Mean

Mean is the average of the given number. First find the total value of the given numbers.

Sum of the given numbers are = 4+8+12+16+19+22+27.

= 108.

The total value is divided by 7 (7 is the total given numbers) = 108/ 7

= 15.4

Median

Middle element of the given series is known as median.

The number series is 4,8,12,16,19,22,27.

The middle element of the above series is 16.

So the value of median is 16.

Mode

Mode is a copy value of the given series. Here no copy value.

Therefore the mode value is empty.

Range

Range is the subtraction of the values form least to high of the given number series.

Range = 27-4

= 23.


Example 2


Solve the mean, median, range of the following in basic statistics?

12,17,19,24,29.

Solution

The given numbers are 12,17,19,24,29.

Mean

Mean is the average of the given number. Find the total value of the given numbers.

Sum of the given numbers are = 12+17+19+24+29

=101.

Total values are divided by 5 (Here 5 is the total given numbers) = 101/5

= 20.2.

Median

Middle element of the given series is known as median.

The number series is 12,17,19,24,29.

The middle element of the above series is 19.

So the median value is 19.

Range

Subtract the minimum value from the topmost value of the series.

Range =29-12

=17.

These examples are the basic statistics problems given by tutor.

Basic Math Definitions Solve Online

Introduction to Mathematical definition solving online:

The basic mathematical definitions are which very useful in our daily life. For every incident we use mathematics in one or the other kind. The mathematical definitions are like addition, subtraction, multiplication, division, average, place value, fractions and so on. There are many basic technique is used to solve math problems with some example problems and practice problems. The student can view and solve math definitions from any point of the world using online. So now let us see the mathematical definitions solving in this article.I like to share this Find the Average Rate of Change with you all through my article.

Addition
Subtraction
Multiplication
Division
Fraction
Place value
Average

Basic math definitions solving online


Solving Addition online: Addition is the sum of two or sometimes more numbers and symbol for addition is ”+”.

For example: Add 33 and 52?

Sol : Given data 33 ,52

= 33 + 52

= 85

Answer: 85

Solving Subtraction online: Subtraction is the difference between two numbers and symbol for subtraction is “-”

For example:  Sub 20 from 40?

Sol:      Given data 20, 40

= 40 - 20

= 20

Answer: 20

Solving Multiplication online: Multiplication is a Product of two or more numbers or it is also called as repeated addition. The symbol for the multiplication are “x, (), * and also .”

For example:  Multiply the numbers 3 from 3

Sol:      Given 3, 3

= 3 x 3

= 9

Answer is: 9

Solving Division online: Division is a arithmetic operation and its operation is to Split of numbers into equal parts. The symbol for division is ( / )

For example: divide 2 from 20

Sol:   Given 20, 2

Divide 20 by 2

= 20 / 2

= 10

Answer is: 5

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Solving Fraction online: Fraction is a part of a whole number.

For example: fractions 1/3, 4/5

Solving Place value online: place value is a digit value in a number.

For example: Find the place value of 2 in this number 321?

the value of 2 in the given number is tens place 2x10.

Solving Average online: Average is the sum of total numbers divided by to total count of those numbers

For example: find the average of these numbers 2, 2, 2, 2,

Sum of the above numbers are 8

Total numbers are 4

So average of these numbers becomes 8/4 = 2

Study About Basic Geometry Test

Introduction to study about basic geometry test:

Geometry is the one of the fundamental concept in mathematics. In this section we will see the study about basic geometry test, which is include several set of test questions along with answer keys. And also geometry test helps the students to recognize the geometry concepts with neat and clear. It also help students to self approximation their ability to perform test in geometry. Let us see the study about basic geometry test.

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Test questions for study about basic geometry test:


Test question 1: The measure of an angle is 45 degree. Find the measure of a complementary angle.

Test question 2: The measure of an angle is 32 degree. Find the measure of a supplementary angle.

Test question 3: A triangle has angle measurements of 100 degree, 60 degree, and 20 degree. Find the kind of triangle is it.

Test question 4: A triangle has angle measurements of 110 degree, 40 degree, and 30 degree. Find the kind of triangle is it.

Test question 5: Do calculation to find the area of the triangle whose base is 10 inches and height 14 inches.

Test question 6: The area of the triangle is 30 sq cm. The base of the triangle is 5 cm. Do the calculation for find the length of the triangle.

Test question 7: The diameter of the circle is 50 cm. Do the calculation for find the radius of the circle.

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Answer keys for study about basic geometry test:


Answer key 1: The complementary angle of 45 degree is 45 degree.

Answer key 2: The supplementary angle of 32 degree is 148 degree.

Answer key 3: Measurement of the triangle is obtuse.

Answer key 4: Measurement of the triangle is scalene.

Answer key 5: Area of the triangle is 70 sq inches.

Answer key 6: Length of the triangle is 12 cm.

Answer key 7: Radius of the circle is 25 cm.

Introduction to Basic Algebra

Introduction to basic algebra:

Basic algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorial, and number theory, basic algebra is one of the main branches of pure mathematics.

Source: Wikipedia


Basic algebra rules:


Commutative Property of Addition.

a + b = b + a

Example:

5+6=6+5

Commutative Property of Multiplication.

a * b = b * a

Example:

7*3 = 3*7

Associative Property of Addition.

(a + b) + c = a + (b + c)

Example:

(5+8) + 2=5 + (8+2)

Associative Property of Multiplication.

(a * b) * c = a * (b * c)

Example:

(5*9) * 4=5 * (9*4)



Distributive Properties of Addition over Multiplication.

a * (b + c) = a * b + a * c

(a + b) * c = a * c + b * c

Example:

4 * (5+7) =4*5 + 4*7

(2+7) * 9=2 * 9+7 * 9

The reciprocal of a non-zero real number a is `1/a`.

a*(`1/a`) = 1

Example:

reciprocal of 7 is `1/7` and 7*(`1/7`) = 1

The additive inverse of a is -a.

a + (-a) = 0

Example:

additive inverse of -7 is -(-7) = 7 and - 7+(7) = 0

The additive identity is 0.

a + 0 = 0 + a = a

The multiplicative identity is 1.

a * 1 = 1 * a = a

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More about basic algebra:


Basic algebra expressions:

An expression in algebra 1 is a grouping of numbers, variables, constants, operators and parentheses or grouping brackets that are stated as an entity. An equation has an expression on each area of the equals sign. Some expressions are prepared of sub-expressions. Much of algebra concern simplifies expressions to facilitate the solution of equations.

Steps for simplify basic algebra expressions:

Step 1: assembly the terms containing the same variable collectively in algebra 1 expressions.

Step 2: execute the operation inside the parentheses for the variable and other.

Step 3: revise the expressions and simplifying the algebra expressions.

Step 4: To verify the equation, if there is capable to simplify the expression, then recur the step 1 to 4.

Introduction for basic algebra equation:

An algebra equation is mathematical statements that assert the equality of two expressions. Basic 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 basic algebra equation is in mathematical identities, assertions that are true independent of the values of any variables contained within them.

Laws of addition in basic algebra:

For the addition of positive and negative number, the follow rules, established in the First Course, apply make easy to solve algebra equation:

A number represented by a letter is called a literal number, and any number expression in which more than numbers of symbols are variable is called a literal expression.

Laws for addition in basic algebra:

1)  To add two numbers with similar signs, find the sum of their total values, and prefix the sign common to both.

2)  To add two numbers with different signs, find the diversity of their total values, and prefix the sign of the one with the larger total value.

How to Solve Basic Integers

Introduction to basic integers:
In mathematics, basic integers formed by natural numbers and including zero (0). The natural numbers like as, (1, 2, 3…) and 0 not including the natural numbers. Integers formed with positive numbers and negative numbers. Negative numbers are like this symbol, (-1, -2, -3 …). Negative numbers not including zero. Fractions and decimals are all not integers. Solve basic integers used to find the sum of basic integers. Here we are going to solve about  basic integers.


Examples problem for solve basic integers:


Rule 1: Sum of two or more negative integers is a negative integer.

Problem 1:

Solve the sum of basic integers, given integers is -6 and -2.

Solution:

Given integers is -6 and -2.

= -6 + -2

= -8

Sum of the integers is -8.

Problem 2:

Solve the sum of basic integers, given integers is -12 and -52.

Solution:

Given integers is -12 and -52.

= -12 + -52

= -74

Sum of the integers is -74.

Problem 3:

Solve the sum of basic integers, given integers is -16 and -12.

Solution:

Given integers is -16 and -12.

= -16 + -12

= -28

Sum of the integers is -28.

Rule 2: Sum of two or more positive integers is a positive integer.

Problem 4:

Solve the sum of basic integers, given integers is +6 and +6.

Solution:

Given integers is +6 and +6.

= +6 + +6

= +12

Sum of the integers is +12.

Problem 5:

Solve the sum of basic integers, given integers is +20 and +20.

Solution:

Given integers is +20 and +20.

= +20 + +20

= +40

Sum of the integers is +40.

Problem 6:

Solve the sum of basic integers, given integers is +14 and +12.

Solution:

Given integers is +14 and +12.

= +14 + +12

= +26

Sum of the integers is +26.

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Practice problems for solve basic integers:


1. Solve the sum of basic integers, given integers is -4 and -5.

Answer is -9.

2. Solve the sum of basic integers, given integers is -8 and -12.

Answer is -20.

3. Solve the sum of basic integers, given integers is -50 and -15.

Answer is -65.

4. Solve the sum of basic integers, given integers is +5 and +2.

Answer is +7.

5. Solve the sum of basic integers, given integers is +25 and +22.

Answer is +47.

6. Solve the sum of basic integers, given integers is +11 and +11.

Answer is +22.

Basic Probability Calculations

Introduction to basic probability calculations:
The basic probability calculations are very helpful to the students. Calculations are the best tool to study what you want without disturbing the time. Calculation can be more interesting and interact than books and lecturing. Probability is result of examine Probability is number of possible actions divide by total quantity of proceedings. If the event arises it represents like one, or else not occurs it must symbolize zero. Now we will see basic probability calculations.


Example problems for basic probability calculations:


Example 1:- Using basic probability calculations

In school there are 100 students available. In those students, there are 50 students are science group, 25 students are maths group and 25 students are the English group. Find the probability if,

i) Science group students are leaving first from the school.

ii) English group students are leaving second from the school.

Solution:

Given,

Total students n(S)=100

Science group students n(A)=50

Maths group students n(B)=25

English group students n(C)=25

i) Let P(A) is the event of the science group students are leaving first.

So the P(A)=`(n(A))/(n(S))`

=`50/100`

=1/2

ii)Let P(B) is the event of the English group students are leaving first.

P(B)=`(n(B))/(n(S))`

=`25/100`

=1/4.

Example 2:- Using basic probability calculations

The bag has the 60 kerchiefs. In those kerchiefs, there are 15 is the blue color, 20 is the yellow color and 25 is the white color. If John select the white color kerchief means solve the probability?

Solution:

First list the given information,

Total kerchiefs n(S)=60

Blue colour kerchiefs n(A)= 15

Yellow colour kerchiefs n(B)= 20

White colour kerchiefs n(C) = 25

Assume P(E) is the probability for john select the white colour kerchief.

So P(E)=`(n(C ))/(n(S))`

=`25/60`

=`5/12` .

In the above way we can solve the problems.

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Example 3:- Using basic probability calculations


Two coins are spinning. What is the probability for if two head are receiving?

Answer

Two coins flipped means the feasible outcomes are {(H,T),(H,H),(T,H),(T,T)}

So the total numbers of outcomes are 4.

We get the two head possibility is 1.

Probability = 1/4.

Example 4:- Using basic probability calculations

From the following scores are 41,13,46,27,79,25,15. We can find the probability for above 25 scores? 2.

Solution:

Probability for finding above 25 scores:

Above 25 scores=41, 46, 27, 79.

In the given scores the above 35scores probability values are (41, 46, 27, 79)

Probability for above 25 scores=41+46+27+79

=193.

Total scores values = (41+13+46+27+79+25+15) =246.

Probability=above 25/Total given scores.

Total scores=246

Probability for above 25 scores=`193/246` .

=`193/246` .

Formulas for finding probabilities for below 25 scores are,

Basic Geometry Concepts

Introduction to basic geometry concepts:

In basic geometry concepts, the geometry is all on the subject of shapes. Geometry was enormously major to prehistoric societies and was used for surveying, astronomy, direction-finding, and structure. Geometry is the study of angles and triangles, outer limits, area and quantity. It differs from algebra in that one develops a reasonable formation where mathematical relations are prove and functional.

Basic Geometry Terms:

Some of the basic terms in geometry are line, rays, angles, and plane.
Now we will discuss about basic concepts in geometry.
Line segment:

In these basic geometry concepts, a line segment is a straight line up segment which is division of the straight line among two points.
To organize a line section, one can inscribe AB.
The points on each surface of the line division are referred to as the finish points.
Ray:

In basic geometry concepts, a ray is the element of the line which consists of the arranged point and locates of all points on one elevation of the closing stages point.
Angle:

In basic geometry concepts, angle can be discrete as two rays or two line segments having a common finale position.
The endpoint becomes predictable as the vertex.
Angles come about while two rays accumulate or mix up at the matching endpoint.

Plane:

A plane is often instead of by a blackboard, communication board, a plane of a box or the peak of a table.
These 'plane' surfaces are damaged to connect any two or added points in a straight line.
A plane is a parallel frontage.

Types of Angles:

An angle is different as wherever two emission or two-line segment join mutually at a general endpoint is known as vertex.
In basic geometry concepts having some types of angles. these are,
Acute angle.
Right angle.
Obtuse angle.
Acute angle:

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An acute angle procedure a lesser of that 90°
Right Angle:

A right angle connections correctly 90°
Obtuse Angle:

An obtuse angle measures not required than 90° but less than 180°

Math Factors of 59

Introduction to math factors of 59:

Math factors are the numbers that we do product in show to get another number:

For example the number 6 has the factors 1, 2 and 3 as shown as follows,

6 =1 x 2 x 3

In this article we are going to learn how to factorize the numbers and all about the math factors of 59.


Math factors of 59:


In order to get the factors we have to divide with the numbers starting with the number 1.

Let us do the math factors of 59.

Step 1: Divide the number 59 by 1 we get 59.

Step 2: We can’t divide 59 further as it is a prime and it gives the result with the decimal

Point.

Step 3: Finally the factors of 59 are 1 and 59.

That is 59 = 1 x 59.

That is 59 = 59 x 1.

Note: prime number is a number which is divided by 1 and the number itself.


Example problems- Math factors of 59:


The following examples will make you clear about how to factorize the numbers in math.

Example 1:

Find the factors of 159

Solution:

Step 1: Divide the number 159 by 1 we get 159.

Step 2: As it is a odd number we cant divide it by 2 therefore it is better to go with 3

Step 3: Divide 159 by 3 :

159/3 = `(3*53)/3`

= 53 (where 3 cancels each other)

Step 4: As 53 includes in the above step it is also a factor of 159.

Finally the factors of 159 are 1, 3, 53 and 159.

That is 159 = 1 x 159;

That is 159 = 3 x 53;

That is 159 = 53 x3;

That is 159 = 159 x 1.

Example 2:

Find the factors of 259

Solution:

Step 1: Divide the number 259 by 1 we get 259.

Step 2: As it is a odd number we cant divide it by 2 therefore it is better to go with 3

Step 3: Dividing by 3 is also not possible so Divide 159 by 7 :

259/7 = `(7*37)/7`

= 37 (where 7 cancels each other)

Step 4: As 37 includes in the above step it is also a factor of 259.

Finally the factors of 259 are 1, 7, 37 and 259.

That is 259 = 1 x 259;

That is 259 = 7 x 37;

That is 259 = 37 x 7;

That is 259 = 259 x 1.

Example 3:

Find the factors of 359

Solution:

Step 1: Divide the number 359 by 1 we get 359.

Step 2: We can’t divide 359 further as it is a prime and it gives the result with the decimal

Point.

Step 3: Finally the factors of 359 are 1 and 359.

That is 359 = 1 x 359.

That is 359 = 359 x 1.

Practice problems- Math factors of 59:

Problem 1:

Find the factors of 559.

Solution:

The factors of 559 are 1 , 13 ,43 , and 559.

Problem 2:

Find the factors of 659.

Solution:

The factors of 659 are 1 and 659

Problem 3:

Find the factors of 759.

Solution:

The factors of 759 are 1 , 3 , 11 , 23 , 33 , 69 , 253 , 759.

Basic Notes on Algebra

Introduction to basic notes on algebra:

Algebra is defined as the major part of mathematics which deals with the study of laws of the operations and those concepts which involves various polynomials, algebraic structures and the equations. There are various kinds of algebra which includes elementary algebra, abstract algebra, etc.

In this article we are going to see about the basic notes of algebra which includes various laws and some example problems.

Basic notes on the laws of algebra:

laws of algebra:

There are various laws that are used in algebra for solving the equations and various algebraic structures.

The first basic notes on algebra are addition which involves the commutative law.

It states that a + b = b + a

Example:  5+6 = 6 + 5

11 = 11

The commutative law can also be applied for multiplication.

The basic law for multiplication is given by

a * b = b *a

Example:    5* 6 = 6* 5

30 = 30

We can also use the associative law for both the addition and the multiplication.

The basic notes for the associative law in algebra for addition is given by

(a + b) + c = a+ (b + c)

Example: (1+2) + 3 = 1 + (2+3)

3 + 3 = 1 + 5

6 = 6

The associative law for multiplication is given by

(a * b) * c = a*(b * c)


Example: (1*2) * 3 = 1* (2*3)

2 * 3 = 1 * 6

6 = 6

The next important basic notes of algebra include the distributive law.

(a * b) + c = (a*b) + (a * c)

Example: (1 * 2) + 3 = (1 *2) + (1 * 3)

2 + 3 = 2 + 3

5 = 5

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Notes on basic rules of algebra:

The additive inverse of a is given by –a for any constant or a variable.

Example :a + (-a) = 0 and 5 + (-5) = 0

The multiplicative inverse of a is 1/a for any constant or a variable.

Example: a * 1/a = 1 and 5 * 1/5 = 1

In case of addition the identity element is zero for any constant or variable.

Example: a + 0 = a and 5 + 0 = 5

In case of multiplication the identity element is one for any constant or variable.

Example: a * 1 = 1 and 5 * 1 = 5.

Do Basic Division

Introduction to do basic division:

Division is one of the arithmetic operations. Division is the inverse form of multiplication. The term division is sometimes said to be sharing. Division is denoted by the symbol `-:`. For example: When M is divided by N.

M/N = O

M – Dividend.
N – Divisor.
O – Quotient.
For example:

The value 120 divided by12.

120/12 = 10

In this,

120 – Dividend.

12 – Divisor.

10 – Quotient.


Do basic division


In this division there are divided, divisor, remainder and quotient will appear in the basic way. It is very easy to solve. This division is very easily understandable by the kids. Let see some examples of the basic division and do some practice problems for this division method.

Example problems for basic division:

Calculate the value 245 is divided by 15.

Solution:

Steps to solve the given problem

First step:

To find the number of times do 15 go into the 240?

Second step:

24 is divided by 15

24/15 = 1

The remainder is 9

Third step:

Then the remainder is put in front of the value 0.

90 is divided by 15

90/15 = 6.

The remainder is said to be zero (0).

Solution:

15) 240 (16

15

_______

90

90

_______

0

________

Divisor = 15

Dividend = 240

Quotient = 16

Remainder = 0

Another example for basic division:

The value 418 is divided by 16.

Dividend = 418.

Divisor = 16.

Solution:

16) 418 (24

32

_________

99

90

__________

9

_________

Answer:

Quotient = 24

Remainder = 9.

Another example for basic division:

The value 16724 is divided by 12.

The dividend = 16724

The divisor = 12.

Solution:

12) 16732 (1311

12

_______

47

36

_______

13

12

_________

12

12

_________

0

__________

Answer:

Quotient = 1311

Remainder = 0.


Some practice problems for the basic division

Problem 1:

Find out value 450 is divided by 5.

Answer:

The correct answer is 90.

Problem 2:

Find the value of 43978 is divided by 14.

Answer:

The correct answer is

Quotient = 3141

Remainder = 4

Problem 3:

Find the value of 12450 is divided by 10

Answer:

The correct answer is

Quotient = 1245

Remainder = 0.

Problem 4:

Find the value of 29726 is divided by 2.

Answer:

The correct answer is

Quotient = 14863

Remainder = 0

Area Math 4th Grade

Introduction to 4th grade math area:

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The surface area of a 3-dimensional solid is the total area of the exposed surface, such as the sum of the areas of the exposed sides of a polyhedron. Area is an important invariant in the differential geometry of surfaces. In this article we shall discuss about 4th grade math area problem. (Source: Wikipedia)

4th grade math area example problem

Here we are going to discuss some 4th grade math area problems with detailed solutions.

Example 1:

Find the area of the rectangle, the rectangle base value 23m and height of rectangle is 10 m.

Solution:

Let base b = 23 and rectangle height h = 10.

Then the area of the rectangle formula = b × h

The area of the rectangle = 23 × 10

= 230 sq. meters.

Therefore the area of rectangle = 230 sq. meters.

Example 2:

Find the area of the circle, the circle radius value 12.

Solution:

Area of the circle formula = pi r2

Pi = 22/7 or 3.14

The area of the circle = 22/7× 12x12

= 452

Answer:

Therefore the area of circle = 452

Example 3:

Find the area of the square; the square one side of length value is 22units.

Solution:

Formula for finding area of square = a2 of (Side) 2

The side value of the area = 22

Therefore (22) 2

That is 22 * 22 = 484

Area of the square = 484units.

Example 4:

Find the area of triangle, the triangle base value=20cm and the height of the triangle = 12cm

Solution:

Area of the triangle formula = `1/2` b x h

Height of the triangle = 12

Base of the triangle = 20

Therefore area of triangle = `1/2` x 20 x 12

The area of triangle = 120cm

Example 5:

Find the area of the rectangle, the rectangle base value 22m and height of rectangle is 15 m.

Solution:

Let base b = 22 and rectangle height h = 15.

Then the area of the rectangle formula = b × h

The area of the rectangle = 22 × 15

= 330 sq. meters.

Therefore the area of rectangle = 330 sq. meters.

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4th grade math area practice problem

Problem 1:

Find the area of the rectangle, the rectangle base value 20m and height of rectangle is 10 m.

Answer:

The area of rectangle = 200 sq. meters.

Problem 2:

Find the area of the square; the square one side of length value is 20 units.

Answer:

Area of square = 400 units

Practice Test for Basic Math

Introduction to mathematics:

Mathematics is the study of quantity, structure, space, and change. Mathematicians seek out patterns, formulate new conjectures, and establish truth by rigorous deduction from appropriately chosen axioms and definitions. There is debate over whether mathematical objects such as numbers and points exist naturally or are human creations. (Source: Wikipedia)

Example problems of practice test for basic math

Basic math test example problem 1:

Add the given two values 35 and 84.

Solution:

Given numbers are 35 and 84

(35 + 84) = 35
84  ( + )
____
119
____

Answer:

The final answer is 119

Basic math test example problem 2:

Find the area of the triangle with the base length is 20 m and its height is 12 m.

Solution:

Given base length (b) = 20 m and height (h) = 12 m

Area of the triangle = `(1 / 2)` * (base length) * (height)

= `(1 / 2)` * 20 m * 12 m

= 120 m^2

Answer:

The final answer is 120 m^2

Basic math test example problem 3:

The sum of the two numbers is 23. Smaller number is three less than that of larger number. Find out the two number values.

Solution:

Let us consider,

x is the larger number and y is the smaller number

Given, sum of the two numbers is 23

x + y = 23 ------- (1)

Smaller number is three less than larger number, we get

y = x - 3 --------- (2)

Substitute equation 2 in equation 1, we get

2x - 3 = 23

After simplifying, we get

x = 13

Substitute the value of x in equation 2, we get

y = 10

Finally, that two numbers are x = 13, y = 10

Answer:

The final answer is x = 13, y = `10`

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Practice problems of practice test for basic math

Basic math test practice problem 1:

Subtract 45 and 34

Answer:

The final answer is 11

Basic math test practice problem 2:

Multiply the given values 31 and 10

Answer:

The final answer is 310

Basic math test practice problem 3:

Solve: x = 2x - 7

Answer:

The final answer is x = 7

Basic math test practice problem 4:

Solve:

3y - 10 = 2

Answer:

The final answer is y = 4

Basic Algebra Calculations

Introduction to basic algebra calculations:

Basic algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. The part of basic algebra calculations called elementary algebra calculations is often part of the curriculum in secondary education and introduces the concept of variables representing numbers. The variables are manipulated using the rules of operations that apply to numbers, such as addition. (Source: Wikipedia).

Examples for basic algebra calculations:

Example 1 for basic algebra calculations:

Find the value for x in (x+2) +(x-22) +(x+45) =0.

Solution:

The given expression is(x+2) +(x-22) +(x+45) =0.

In this above expression first add the constants separately and then add the variable x.

(x+2) +(x-22) +(x+45) = (x+ x+ x) + (2-22+45)

(x+ x+ x) + (2-22+45) =3x+25

3x+ 25=0

3x =-25

x=-25/3

The value for x in (x+2) +(x-22) +(x+45) =0 is -25/3.

Example 2 for basic algebra calculations:

Find the value for the quadratic equation x ^2+6x +8=0.

Solution:

The given quadratic equation is x ^2+6x +8=0.

We have to find the roots for the above quadratic equation.

This can be solved by the factoring by middle term.

A=co-efficient 0f x ^2=1

B= co- efficient of x=6

C= constant=8

We split the 6 as (2+4), and then only we get 2 x4 =8.

x ^2+6x +8= x ^2+2x+4x +8

Take x as common in first two terms and 4 as common in next two terms.

x ^2+6x +8= x(x+2)+4(x+2)

x ^2+6x +8= (x+4) (x+2)

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

x+4 =0 and x+2 =0

In the equation x+4 =0, subtract 4 on both sides.

x+4 =0

x+4-4 =0-4

x=-4

In the equation x+2 =0, subtract 2 on both sides.

x+2 =0

x+2-2 =0-2

x=-2

The roots for the equation x ^2+6x +8=0 is x=-2,-4.

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Practice problem for basic algebra calculations:

Find the value of x for the equation (x+12)+( x-22)=0
Answer: x=5.

Find the roots for the quadratic equation x ^2+ 11x+30=0.
Answer: x=-5,-6.

Basic Proportionality Theorem

Introduction of basic proportionality theorem:
The Basic proportionality theorem is the theorem associated to triangle's property. This theorem says that if a line intersects two sides of one triangle and is parallel to the third side, then it divides the first two sides proportionally. That’s the Basic Proportionality Theorem shows the relation between the triangle and a straight line which bisects the two sides and parallel to another side.

Explanatory of basic proportionality theorem:

If a line is drawn parallel to the triangle, thus the line intersect two sides of the triangle parallel to the other (third) side in distinct points, then it cuts(divide) the two sides proportionally in the same ratio. This is known as proportionality theorem. It is otherwise called as Thales theorem.

Now we are going to prove the theorem and illustrative the statements

Given: A triangle LMN, S and T are any other two points on PQ and PR respectively such that ST parallel to QR.

To prove:

LS / SM = PT / TN

Construction:

Join MT and NS. From TU perpendicular to LM.

Proof:

In triangle LMN

ST is parallel to MN

And

If LS/MS = k (Where k is constant)

Thus LT / NT is also equal to k

Or we can say

LS / MS = LT / NT

We are given LMN,

ST is parallel to MN.

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Basic proportionality theorem:

Proportionality Theorem says that, A line parallel to one side of a triangle divides the other two sides into parts of equal proportion.

DATA:

In triangle ABC, a line PQ drawn parallel to BC cuts AB and AC at P and Q respectively.

To prove:

AP/ PB = AQ /QC

Construction:

Let the point P divide AB in the ratio of l: m where l and m are natural numbers. Divide AP into 'l' and PB into 'm' equal parts. Through all these points on AB, draw lines parallel to BC to cut AC.

Proof:

Statement                                                                           Reason

1. AP is cut in l equal parts and lines are drawn through                    Construction

these points parallel to BC.

2. Therefore, AQ is cut l equal parts.                                                        Intercept  theorem

3. Similarly,                                                                                                    Statement 1

QC is cut in m equal parts.

4. AP / PB = 1 /m                                                                                          Construction

5. AQ / QC = 1/ m                                                                                         Statement 2 and 3

6. Therefore AP / PB = AQ / QC                                                                 Statement 4 and 5.

Simplify Monomials

An algebraic expression consisting of only one term is called a monomial. It can be a product/quotient of variable and numerals or only variables or even constant terms.

Some of the examples are 3x^2, -p/2, 4xy, 63, -7z etc. Operations addition and subtraction cannot be used to represent a mono form as they are used when there are more than one term in an algebraic expression also monomials cannot have negative exponents. For instance 5-y, 7x+3, ab-3 are not mono forms.

The sum of the exponents of all the variables would be the degree of the same.
The degree of such a constant term is zero. To simplify monomials we would be using law of exponents, multiplication of mono forms and division of mono forms accordingly wherever required.

The order of operations is also followed while solving
The combination of these expressions which are given inside the parenthesis have to be simplified first.
If we have a power of an exponential algebraic expression then law of exponents is used. For example, given an expression (3a2b3)2 can be simplified by using the law of multiplying the exponents [(am)n=amn] giving 3a4b6

Multiplication of Monomials:

To multiply the given expression the exponents of the like terms are to be added using the law of exponents, am.an=am+n
Example: Given an expression 3x2y4. 4xy3 then we simplify applying the law of exponents am.an=am+n, we get 3.4 x2+1y4+3= 12x3y7
Example: -3(m2n).(-4mn3)
Coefficients are multiplied first and then the variable part
(-3.-4)(m2.m1)(n1.n3)=(12)(m2+1)(n1+3)= 12m3n4

Dividing monomials:

While dividing exponential expressions, the exponents of like terms are subtracted
Example: 9x3y4/3x2y this expression can be simplified applying the law of exponents am/an=am-n
We get, 3x3-2y4-1= 3xy3

Let us now learn the steps involved in solving monomials with the help of few examples as given below,
Example: -18a^3b^2c^5/3ab^2c
Here first the numeral part is simplified, -18/3= -6
Each of the like terms are simplified using the law of exponents,
a^3/a= a^3-1= a^2
b^2/b^2 = b^2-2=b0=1
c^5/c= c^5-1=c^4
Writing all the terms together we get, -6a^2c^4 which is the simplified form

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Example: Simplify (3p^2q3/3p) . (9p^2q^2/q^5)
First step would be to multiply the numerators and add the exponents of like terms
Numerator= (3.9)(p^2+2.q3+2) = 27p4q5
Then multiply the denominators and add the exponents of like terms if any
Denominator= (3.1)(p1q^5)  [when the exponent is not given it is considered as 1]
Now divide the coefficients of the expression if possible
27/3 = 9
Next step would be to subtract the exponents of like terms
P4-1q5-5=p3.q0=p3.1=p3
The final answer is (3p^2q^3/3p).(9p^2q^2/q^5)= 9p^3

Basic Algebra For

Introduction of Basic Algebra For:
Algebra is the branch of mathematics concerning the study of the rules of operations and relations, and the constructions and concepts arising from them, including terms, polynomials, equations and algebraic structures. Together with geometry, analysis, topology, combinatorics, and number theory, algebra is one of the main branches of pure mathematics. The part of algebra called elementary algebra.

Source – Wikipedia.

Basic Algebra for Addition and Subtraction:

The followings are some of the examples for basic algebra of addition and subtraction.

Addition of fractions:  `4/9 ` +` 4/9` .

Solution:

From the fraction denominators are same.

By adding the numerator and same as in the denominator:

= `4/9` +` 4/9`

=` (4 +4)/9`

=` 8/9`

By Simplify and adding the fraction:

= `8/9`

= `4/3`

Subtraction of fractions: ` 2/6` - `4/6` .

Solution:

From the fraction denominators are same.

By subtract the numerator and same as in the denominator:

= `2/6 ` - `4/6`

= `(2- 4)/ 6`

=` -2/6`

By Simplify and subtract the fraction:

=` -2/6`

= `-1/3`

Algebraic expression of addition: 10x + 7y + 3x + 5a.

Solution:

The given expressions are 10x + 7y + 3x + 5a

The like terms in expression are 10x and -3x. We cannot do anything the 7y or 5a.

Then group the terms and add.

= (10x + 3x) + 5a + 7y

= 5a + 13x + 7y

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Basic Algebra for Multiplication and Division:

The followings are some of the examples for basic algebra of multiplication and division.

Multiply the integers: 5 × 9.

Solution:

Given integers are (5) × (9)

= |5| × |9|

= 5 × 9

= 45 is the solution for the given integers.

Multiply the integers: 4 × (-5).

Solution:

Given integers are 4 × (-5)

= 4 × -5

= 4 × -5

= -20 is the solution for the given integers.

Divide: 8a^2 – 40 by 8

Solution:

The given polynomials are 8a^2 – 40 ÷ 8

= `(8a^2-40)/8`

By dividing them we get

= `(8a^2)/ 8` * `-40/8`

= a^2 - 5

Divide: 12a^2 – 24 by 4

Solution:

The given polynomials are 12a^2 – 24 ÷ 4

= `(12a^2-24)/4`

By dividing them we get

= `(12a^2)/4` *` -24/4`

= 3a^2 - 6

Grade 7 Math Powers

Introduction of grade 7 math powers:

Grade 7 math powers are nothing but the repeated multiplication of the particular value upto the numbers in the powers. The grade 7 math powers are made to have the multiplication, addition, subtraction etc. The grade 7 math powers are the way which shows the powers of the particular term leads to the multiple of the particular value. The math power saves the time for the multiplication process.

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Grade 7 math powers:


The grade 7 math power is involved in the number sense. The powers that involves in the number sense that gives the power values on the particular term. The grade 7 math powers are made to express through the scientific notations such as power values. The math powers are made to have the range of the numerical values of the particular value.

The grade 7 math powers are simple to compute the answers let us see the simple examples for the grade 7 math powers. The multiplication of the same factor is mentioned as the power of the terms. The math powers are done along the way which has the products of the factor which represents on the right top of the factor.


Example problems for grade 7 math powers:


Example 1:

Calculate 2 power 1. That is `2^(1)` = ?

Solution:

`2^(1)` = 2.

Repeated multiplication of two upto one time.

Example 2:

Calculate 2 power 2. That is `2^(2)` = ?

Solution:

`2^(2)` = 2 * 2 = 4.

Repeated multiplication of two upto two times.

Example 3:

Calculate 2 power 3. That is `2^(3)` = ?

Solution:

`2^(3)` = 2 * 2 * 2 = 4 * 2 = 8.

Repeated multiplication of two upto two times.

These are the example problems on grade 7 math powers.

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Practice problems for grade 7 math powers:


Practice problem help you to get thorough in the concept of powers.

Problem 1:

Calculate 2 power 7. That is calculate `2^(7)`

Answer:

128

Problem 2:

Calculate 2 power 8. That is calculate `2^(8)`

Answer:

256.

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Help in 7th Grade Math

Introduction to Help in 7th grade math:

In this article we discuss about help in 7th grade math solving problems. In this article we are going to discuss for 7th grade topics for help. In everyday life we are using mathematical concepts often. 7th grade math covers following chapters

Numbers
Measures
Algebra
Geometry
Handling data.
The some help in 7th grade solving problems are given below.

Example problems for help in 7th grade math:

Example 1:

Addition of integers: ( + 5) + ( + 7) =?

Solution:

The sum of positive integers is a positive integer obtained by adding the two integers.

(+5) + (+7) = +12

Solution is 12

Example 2:

Subtraction of integers:

(+5) – (+3) =?

Solution:

To subtract an integer from another integer, add the additive inverse of the second number to the first number.

(+5) – (+3) = 5 – 3 = 2

Solution is 2

Example 3:

Multiplication of integers:

(+5) * (+8) =?

Solution:

The product of two positive integers is a positive integer.

(+5) * (+8) = 5 * 8 = 40

Solution is 40

Example 4:

Division of integers:

(+10)/(+5) =?

Solution:

Positive integer / positive integer = positive value

(+10)/ (+5) = 10/5 = 2

Solution is 2

Example 5:

John inverse RS 6000 for 60 months. Hear received the simple interest of RS 1500. Find the rate of interest.

Solution:

One year = 12 months

Five year = 60 months

Principal, p = RS 6000

Number of years, n = 5

Simple interest, i = RS 1500

Rate of interest, r =?

Rate of interest, r = 100 * i/ p * n

100 * 1500 / 6000 * 5 = 5

Rate of interest 5 %

Example 6:

Find the area and perimeter of a rectangle with length 7.5 cm and breadth 4.4cm.

Solution:

Given length = 7.5 cm

Breadth = 4.4 cm

Area of a rectangle = l * b square unit

= 7.5 * 4.4 square unit

= 33

Area = 33 cm2

Perimeter of rectangle = 2 (l * b) unit

= 2 ( 7.5 * 4.4 )

= 2 * 11.9

= 23.8

Perimeter = 23.8 cm

Example 7:

Solve 4x + 5 = 25

Solution:

Given expression 4x + 5 = 25

Subtract 5 on both sides

4x + 5 – 5 = 25 – 5

4x = 20

Divide 4 on both sides

4x/4 = 20/ 4

x = 5

Solution is 5

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Practice problems and solutions for help in 7th grade math:


Problem 1: Addition of integer (+4) + (+2) =?

Solution: 6

Problem 2: Subtraction of integer (+8) – (+3) =?

Solution: 5

Problem 3: Multiplication of integer (+5) * (+5) =?

Solution: 25

Problem 4: Division of integer (+8) / (+4) =?

Solution: 2

Problem 5: Find the area of a rectangle with length 7 cm and breadth 4cm.

Solution: 28cm2

Problem 6: Solve 5x + 2 = 37

Solution: 7

How To Do Basic Long Division

INTRODUCTION FOR HOW TO DO BASIC LONG DIVISION:

Division is the arithmetical operation for finding how many times a number is in another number. This is a one kind of operations in basic arithmetic. Division is nothing but the repeated reduction.

For example: 24 ÷ 12 = 2. Here 24 is the dividend, 12 is the divisor, and 2 is the quotient.  Here we are going to see how we are going to divide a number using long division method.

The two definitions :

the number that is to be divided into is called as the dividend
The number which divides the dividend is called as the divisor


The steps followed to do long division are:

Division , Multiplication, Subtraction, and bring down.

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Example problem to show how to do basic long division:


Ex 1: Solve 3264 ÷ 2.

2)3265(1632
2
12
12
6
6
5
4
1
Steps to show how to do basic long division.

Step – 1: In the given problem the dividend is 3264 and the divisor is 2

Step – 2: Now we are going to divide 2364 by 2, we take the first number of the divisor. Since 3 is less than 2 we can stop with 3

Step – 3: The divisor 2 is multiplied with the quotient 1 and gives the result or 2, so when we subtract 3 with 2 we get the remainder of 1.

Step – 4: The remainder we have is 1; one will not go with 2 so we bring down 2.

Step – 5: Now 2 is multiplied with 6 to get 12. When 12 subtracted with 12 we get 0 so we bring 6 down.

Step – 6: When 2 multiplied with 3 we will get 6 so when 6 subtracted with 6 we will get 0. Now we bring 5 down

Step – 7: When two multiplied with two we will get four. And the remainder is one.

Step – 8: we cannot proceed further. So we stop with this. The remainder is 1 and the quotient is 1632. This is how we do basic long division.

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Practice problems for how to do basic long division:


Pro -1: solve 564 by 4

Pro -2: solve 840 by 10.

Ans 1) 141 and 2) 84