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:

I like to share this Acute Angled Triangle with you all through my article.

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