Category: Mathematics

Cube and Cube Root of numbers can be found easily using the simplest and quickest methods. Check the complete details of the How to Find the Cube and Cube Root of a Number? We have covered everything like the definition of cube and Cubes Relation with Cube Numbers, Perfect Cube, etc. For better understanding, we even jotted the solved examples explained in detail.

• Cube
• To Find if the Given Number is a Perfect Cube
• Cube Root
• Method for Finding the Cube of a Two-Digit Number
• Table of Cube Roots
• Worksheet on Cube
• Worksheet on Cube and Cube Root
• Worksheet on Cube Root

Cube

The cube of a number is calculated by multiplying a number itself by 3 times. If you consider a number n, then the cube of a number n is n. Here, n is the natural number.

Example:

1, 8, 27 are the cube number of the numbers 1, 2, and 3 respectively.

Cube of 9 = 9 × 9 × 9 = 729
Cube of 8 = 8 × 8 × 8 = 512
Cube of 6 = 6 × 6 × 6 = 216

Cubes Relation with Cube Numbers

In mathematics, a cube is defined as a solid figure where all edges are of the same sizes and each edge is perpendicular to other edges.

Example:

If you take cubes of 4 units, then you can form a bigger cube of 64 units. Or else, if you take cubes of 3 units, then you can form a bigger cube of 27 units.

Perfect Cube Cube Numbers

The product of the three same numbers will give you a cube of a number (perfect cube).

Example:

The cube of a number 2 is 2 × 2 × 2 = 8.
8 is a perfect cube.

Properties of Cube Numbers

1. The cube of an even number is always an even number.

Example:
(i) Find the cube of a number 2?
2 × 2 × 2 = 8
8 is an even number.
(ii) Find the cube of a number 4?
4 × 4 × 4 = 64
64 is an even number.
(iii) Find the cube of a number 6?
6 × 6 × 6 = 216

2. The cube of an odd number is always an odd number.

Example:
(i) Find the cube of a number 3?
3 × 3 × 3 = 27
27 is an odd number.
(ii) Find the cube of a number 5?
5 × 5 × 5 = 125
125 is an odd number.
(ii) Find the cube of a number 7?
7 × 7 × 7 = 343
343 is an odd number.

Units Digits in Cube Numbers

If a number is even or odd, its cube is even or odd respective to the given number. The cube of a unit’s digit always shows the below results.

(i) Cube of 1 = 1 × 1 × 1 = 1;
The Units Digits of Cube of 1 is 1.
(ii) Cube of 2 = 2 × 2 × 2 = 8
The Units Digits of Cube of 2 is 8.
(iii) Cube of 3 = 3 × 3 × 3 = 27
The Units Digits of Cube of 3 is 7.
(iv) Cube of 4 = 4 × 4 × 4 = 64
The Units Digits of Cube of 4 is 4.
(v) Cube of 5 = 5 × 5 × 5 = 125
The Units Digits of Cube of 5 is 5.
(vi) Cube of 6 = 6 × 6 × 6 = 216
The Units Digits of Cube of 6 is 6.
(vii) Cube of 7 = 7 × 7 × 7 = 343
The Units Digits of Cube of 7 is 3.
(viii) Cube of 8 = 8 × 8 × 8 = 512
The Units Digits of Cube of 8 is 2.
(ix) Cube of 9 = 9 × 9 × 9 = 729
The Units Digits of Cube of 9 is 9.

Cube roots

Cube Root of a Number is the inverse of finding the cube of a number. If the cube of a number 3 is 27, then the cube root of 27 is 3.

How to Find the Cube Root of a Number by Prime Factorisation Method?

The Prime Factorisation of any Number Cube Root can be calculated by grouping the triplets of the same numbers. Multiply the numbers by taking each one from each triplet to provide you the Cube Root of a Number.

Example:
Cube Root of 216 = 2 × 2 × 2 × 3 × 3 × 3 = 2 × 3 = 6
6 is the cube root of 216.

FAQs on Cube and Cube Roots

1. Find the cube of 3.4?
The cube of a number can be calculated by multiplying it three times.
Cube of 3.4 = 3.4 x 3.4 x 3.4 = 39.304

2. Is 288 a perfect cube? If not, find the smallest natural number by which 288 should be multiplied so that the product is a perfect cube.
The prime factorization of 288 is
288 = 2 x 2 x 2 x 6 x 6
Since we can see number 6 cannot be paired in a group of three. Therefore, 288 is not a perfect cube.
To make it a perfect cube, we have to multiply the 6 by the original number.
Thus, 2 x 2 x 2 x 6 x 6 x 6 = 1728, which is a perfect cube.
Hence, the smallest natural number which should be multiplied to 288 to make a perfect cube is 6.

3: Find the smallest number by which 256 must be divided to obtain a perfect cube.
The prime factorization of 256 is
256 = 2×2×2×2×2×2×4
Now, if we group the factors in triplets of equal factors,
256 = (2×2×2)×(2×2×2)×4
Here, 4 cannot be grouped into triples of equal factors.
Therefore, we will divide 256 by 4 to get a perfect cube.

4. Michael makes a cuboid of plasticine of sides 3 cm, 2 cm, 3 cm. How many such cuboids will he need to form a cube?
Given that the sides of the cube are 3 cm, 2 cm, and 3 cm.
Therefore, volume of cube = 3×2×3 = 18
The prime factorization of 18 = 3×2×3
Here, 2, 3, and 3 cannot be grouped into triplets of equal factors.
Therefore, we will multiply 18 by 2×2×3 = 12 to get a perfect square.
Hence, 12 cuboids are needed.

Know whether zero falls under Rational Numbers or not and the statements supporting it here. Yes, Zero is a Rational Number and you will have clarity on it by the end. As we can write the Integer 0 in any of the below forms.

For instance, 0/1, 0/-1, 0/2, 0/-2, 0/3, 0/-3, 0/4, 0/-4 …..

In other words, we can express as 0 = 0/b where b is a non zero integer.

Thus, you can write 0 as a/b = 0 where a is 0 and the denominator b is a non- zero integer.

Therefore, 0 is a Rational Number.

Examples

(i) 0/6

0/6 is a rational number as we have the denominator non- zero integer.

(ii) 0/-2

0/-2 is a rational number since -2 is an integer and is non zero.

(iii) 0/10

0/10 is a rational number since we have 12 in the denominator which is a non zero integer.

Thus, the above instances prove that 0 is a Rational Number.

You will no longer feel drawing 2 or more linear equations on a graph difficult anymore with our article. Get a detailed procedure to represent simultaneous equations graphically easily. Furthermore, have a look at the example questions provided below to get clarity on the topic and solve related problems.

To solve a pair of simultaneous equations graphically, we first draw two equations on the graph. Those two equations form straight lines intersecting each other at a common point. This common point gives the solution of the pair of simultaneous equations.

How to Solve Simultaneous Linear Equations Graphically?

Let us take two first-order linear equations

p₁x + q₁y + r₁ = 0

p₂x + q₂y + r₂ = 0

Draw these two lines on a coordinate graph. These lines always form a straight line on the graph. Suppose L₁ represent the graph of p₁x + q₁y + r₁ = 0 and L₂ represent the graph of p₂x + q₂y + r₂ = 0. By solving the simultaneous equations graphically we will get three possible solutions. They are as follows.

1. When the two lines L₁, L₂ interest at a single point

• Here, two lines meet at a point called p (x, y).
• So, the obtained point x coordinate, y coordinate is the unique solution of the given linear equations.
• This system is called independent.

2. When two lines L₁, L₂ are coincident

• Here, two equations represent the single line.
• Therefore, the equations have infinitely many solutions.
• This system is called dependent.

3. When two lines L₁, L₂ are parallel to each other

• The two equations have no common solutions.
• This system is called inconsistent.

Examples of Simultaneous Equations Graphically

1. Solve this system of equations by graphing: y = x + 1 and x + y = 5

Solution:

Given linear equations are y = x + 1 and x + y = 5

On a graph paper to solve simultaneous equations graphically, draw a horizontal line X’OX and a vertical line YOY’ as the x-axis and y-axis respectively.

The first equation y = x + 1 is in the slope intercept form y = mx + c [m = 1, c = 1]

Now apply the trial and error method to get the 3 pairs of points (x, y) which satisfy the equation y = x + 1.

If the value of x = -1 then y = -1 + 1 = 0

If the value of x = 0 then y = 0 + 1 = 1

If the value of x = 1 then y = 1 + 1 = 2

If the value of x = 2 then y = 2 + 1 = 3

Arrange these values of the linear equation y = x + 1 in the table

x	-1	0	1	2
y	0	1	2	3

Plot the points of the equation y = x + 1; A (-1, 0), B (0, 1), C (1, 2), D (2, 3) on the graph.

Join the points A, B, C, and D to get the line equation AD.

So, the line AD represents the equation y = x + 1.

Convert the second equation x + y = 5 into the slope-intercept form.

y = 5 – x [ here, m =-1, c = 5]

Apply the trial and error method to get the points that satisfy the given equation.

If the value of x = -1 then y = 5 – (-1) = 5 + 1 = 6

If the value of x = 0 then y = 5 – 0 = 5

If the value of x = 1 then y = 5 – 1 = 4

If the value of x = 2 then y = 5 – 2 = 3

Arrange the values of the linear equation in the table.

x	-1	0	1	2
y	6	5	4	3

Plot the points P (-1, 6), Q (0, 5), R (1, 4), S (2, 3) on the graph

Join the points to get a straight line of x + y = 5

We get two straight lines intersecting each other at (2, 3).

Therefore, x = 2 and y = 3 is the solution of the given system of equation.

2. Solve graphically the system if linear equation y = 2x = 4 and y + 2x = 1.

Solution:

Given two linear equations are y + 2x = 4 and y + 2x = 1

Convert the first equation y + 2x = 4 into the slope intercept form.

y = 4 – 2x [ here m = -2, c = 4]

Apply the trial and error method to get the points on the line.

If the value of x = -1 then y = 4 – 2 (-1) = 4 + 2 = 6

If the value of x = 0 then y = 4 – 2 (0) = 4 – 0 = 4

If the value of x = 1 then y = 4 – 2 (1) = 4 – 2 = 2

If the value of x = 2 then y = 4 – 2 (2) = 4 – 4 = 0

Arrange these values of the linear equation y = 4 – 2x in the table.

x	-1	0	1	2
y	6	4	2	0

Now plot the points A (-1, 6), B (0, 4), C (1, 2), D (2, 0) on the graph.

Join the points to get the linear equation y + 2x = 4.

Convert the second equation y + 2x = 1 into the slope intercept form.

y = 1 – 2x [m = -2, c = 1]

Apply the trial and error method to get the points on the equation.

When the value of x = -1 then y = 1 – 2(-1) = 1 + 2 = 3

When the value of x = 0 then y = 1 – 2(0) = 1 – 0 = 1

When the value of x = 1 then y = 1 – 2(1) = 1 – 2 = -1

When the value of x = 2 then y = 1 – 2(2) = 1 – 4 = -3

Arrange these values of the equation in the table.

x	-1	0	1	2
y	3	1	-1	-3

Plot the points P (-1, 3), Q (0, 1), R (1, -1), S (2, -3) on the graph.

Join the points to form a linear equation y + 2x = 1

We get from the graph that two straight lines are parallel to each other.

Therefore, the given system of equations has no solutions.

3. Solve this system of equations by graphing: x – 2y = 2 and 2y + 2 = x.

Solution:

Given linear equations are x – 2y = 2 and 2y + 2 = x.

Convert the first equation into y = mx + c form

y = x / 2 – 1 [ here m = 1/2, c = -1]

Apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation x – 2y = 2.

If the value of x = -1 then y = -1/2 – 1 = -3/2

If the value of x = 0 then y = -0/2 – 1 = -1

If the value of x = 1 then y = 1/2 – 1 = -1/2

Arrange these values in the table

x	-1	0	1
y	-3/2	-1	-1/2

Now plot the points of the equation A (-1, -3/2) B (0, -1), C (1, -1/2)

Join the points A, B, C to get the graph line x – 2y = 2

Convert the second equation 2y + 2 = x into the slope-intercept form.

y = ½ (x – 2) [ m = 1/2, c = -1]

Apply the trial and error method to find 3 pairs of values of (x, y) which satisfy the given equation 2y + 2 = x.

If the value of x = -1 then y = ½ (-1 – 2) = -3/2

If the value of x = 0 then y = ½ (0 – 2) = -1

If the value of x = 1 then y = ½ (1 – 2) = -1/2

Arrange these values in the table

x	-1	0	1
y	-3/2	-1	-1/2

Now plot the points of the equation 2y + 2 = x; A (-1, -3/2) B (0, -1), C (1, -1/2) on the graph paper.

Join the points of A, B, and C; to get the graph line AC.

Thus line AC is the graph of 2y + 2 = x.

We find that the two straight lines coincide.

Therefore, the given system of equations has an infinite number of solutions.

Do you want to calculate the highest common factor of polynomials by using the division method? If yes then stay on this page. Here we are providing the detailed step by step procedure to find the G.C.F of polynomials by division method. Along with the steps you can also check some solved example questions from the below sections.

How to find the Highest Common Factor of Polynomials by Division Method?

We are using the division method to find the G.C.F of polynomials when the polynomials have the highest factor and it is difficult to compute the HCF. Then follow the steps and instructions provided below and get the answer easily.

• Let us take two polynomials f(x), g(x).
• Divide the polynomials f(x) / g(x) to get f(x) = g(x) * q(x) + r(x). Here the degree of g(x) > degree of r(x).
• If the remainder r(x) is zer0, then g(x) is the highest common factor of polynomials.
• If the remainder is not equal to zero, then again divide g(x) by r(x) to obtain g(x) = r(x) * q(x) + r1(x). Here if r1(x) is zero then required H.C.F is r(x).
• If it is not zero, then continue the process until we get zero as a remainder.

Solved Examples on GCF of Polynomials

Example 1.

Find the H.C.F of x⁴ + 4x³ + x – 10 and x² + 3x – 5 by using the division method?

Solution:

Given polynomials are f(x) = x⁴ + 4x³ + x – 10, g(x) = x² + 3x – 5

Arranging the polynomials in the descending order f(x) = 1x⁴ + 4x³ + 0x² + x – 10

By using the division method,

The remainder is zero.

So, the required H.C.F of x⁴ + 4x³ + x – 10 and x² + 3x – 5 is x² + 3x – 5.

Example 2.

Calculate the greatest common factor of 8x³ – 10x² – x + 3 and x – 1 by using the division method?

Solution:

Given two polynomials are 8x³ – 10x² – x + 3 and x – 1

Let us take f(x) = 8x³ – 10x² – x + 3, g(x) = x – 1

Divide f(x) by g(x)

The remainder is zero.

So, the required H.C.F of 8x³ – 10x² – x + 3 and x – 1 is x – 1.

Example 3.

Find the HCF of the following pairs of polynomials using the division algorithm

2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12?

Solution:

Given two polynomials are 2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12

Let f(x) = 2 x³ + 2 x² + 2 x + 2 , g(x) = 6 x³ + 12 x² + 6 x + 12

Take 2 common from the first polynomial.

f(x) = 2(x³ + x² + x + 1)

Take 6 common from the second polynomial.

g(x) = 6(x³ + 2x² + x + 2)

Divide f(x) by g(x)

 

The remainder is not zero. So, we have to repeat this long division once again.

Then divide g(x) by r1(x). i.e x³ + 2x² + x + 2 by x² + 1

The remainder is zero.

The H.C.F of 2, 6 is 2.

So, required H.C.F of 2 x³ + 2 x² + 2 x + 2 , 6 x³ + 12 x² + 6 x + 12 is 2 ( x² + 1).

Disjoint Sets are the Sets whose intersection with each other results in a Null Set. In Set Theory if two or more sets have no common elements between them then the Intersection of Sets is an Empty Set or Null Set. This kind of Sets is Called Disjoint Sets. Go through the entire article to know about Disjoint Sets Definition, Disjoint Sets using Venn Diagram, Pairwise Disjoint Sets, etc.

Disjoint Sets Definition

Two Sets are said to be disjoint if they have no common elements between them. If elements in two sets are common then they are said to be Non-Disjoint Sets. Condition for Disjointness is just that intersection of the entire collection needs to empty.

For Example, A = { 4, 5, 6} B = { 7, 8, 9} then A and B are said to be Disjoint Sets since they have no common elements between them. Some Sets can have null set as Intersection without being Disjoint.

Disjoint Sets using Venn Diagram

Two sets A and B are disjoint sets if the intersection of two sets A and B is either a null set or an empty set. In other words, we can say the Intersection of Sets is Empty.

i.e. A ∩ B = ϕ

Pairwise Disjoint Sets

Definition of Disjoint Sets can be proceeded to any group of sets. Collection of Sets is said to be pairwise disjoint if it has any two sets disjoint in the collection. These are also called as Mutually Disjoint Sets.

Consider P is a set of any collection of Sets and A and B. i.e. A, B ∈ P. Then, P is known as pairwise disjoint if and only if A ≠ B. Therefore, A ∩ B = ϕ.

Example:

{ {7}, {3, 4}, {5, 6, 8} }

Solved Examples on Disjoint Sets Venn Diagrams

1. Determine whether the following Venn Diagram represents Disjoint Sets or not?

Solution:

S = {2, 4, 6 8} T = {1, 3, 5, 7}

Sets S, T doesn’t have any common elements between them. Thus, Sets S, T are said to be Disjoint.

2. If A = { 1, 2, 3, 4, 5, 6, 7, 8, 9} B = {14, 15, 16, 17, 18, 19}. Check whether the following are Disjoint Sets are not?

Solution:

Given Sets are A = { 1, 2, 3, 4, 5, 6, 7, 8, 9} B = {14, 15, 16, 17, 18, 19}

Sets A, B doesn’t have any common elements between them. Thus, Sets A, B are said to be Disjoint.

Learn How to Calculate Compound Interest when Interest is Compounded Half-Yearly. Computation of Compound Interest by the growing principal can be complicated. Check out Solved Examples explaining the step by step process for finding the compound interest when compounded half-yearly. To help you better understand we have given the Compound Interest Formula when Interest Rate is Compounded Half-Yearly.

How to find Compound Interest when Interest is Compounded Half-Yearly?

If the rate of interest is annual and interest is compounded half-yearly then the annual interest rate is halved(r/2) and the number of years is doubled i.e. 2n. The Formula to Calculate the Compound Interest when Interest Rate is Compounded Half Yearly is given by

Let Principal = P, Rate of Interest = r/2 %, time = 2n, Amount = A, Compound Interest = CI then

A = P(1+r/2/100)²ⁿ

In the Case of the Half-Yearly Compounding, Rate Interest is divided by 2 and the number of years is multiplied by 2.

CI = A – P

= P(1+r/2/100)²ⁿ – P

=P{(1+r/2/100)²ⁿ-1}

If any three of the terms are given the fourth one can be found easily.

Problems on Compound Interest when Interest is Compounded Half-Yearly

1. Find the amount and the compound interest on $12,000 at 8 % per annum for 3 1/2 years if the interest is compounded half-yearly?

Solution:

Given Principal = $12, 000

r = 8% per annum

rate of interest half-yearly = 8/2 %

= 4%

n = 3 1/2

= 7/2 years

when compounded half yearly multiply by 2 i.e. 2n

= 7/2*2

= 7

We know Amount A = P(1+r/100)ⁿ

= 12,000(1+4/100)⁷

=12,000(1+0.04)⁷

= 12,000(1.04)⁷

A = Rs. 15791

We know CI = A – P

= 15791 – 12,000

= Rs. 3791

2. Find the compound interest on Rs 5000 for 3/2 years at 5% per annum, interest is payable half-yearly?

Solution:

A = P(1+r/100)ⁿ

P = 5000

n = 3/2

2n = 3/2*2

= 3

r = 5%

A = 5000(1+5/100)³

A = 5000(1.05)³

A = Rs. 5788

CI = A – P

= 5788 – 5000

= Rs. 788

Every Integer is a Rational Number but a Rational Number need not be an Integer. Check out the statements, examples supporting whether or not All Rational Numbers are Integers.

We know 1 = 1/1, 2 = 2/1, 3 = 3/1 ……..

Also, -1 = -1/1, -2 = -2/1, -3 = -3/1 ……..

You can also express integer a in the form of a/1 which is also a Rational Number.

Hence, every integer is clearly a Rational Number.

Clearly, 5/2,-4/3, 3/7, etc. are all Rational Numbers but not Integers.

Therefore, every integer is a Rational Number but a Rational Number need not be an Integer. Check out the following sections and get a complete idea of the statement.

Determine whether the following Rational Numbers are Integers or not

(i) 3/5

3/5 is not an Integer and we can’t express it other than a fraction form or decimal value.

(ii) 6/3

6/3 is an integer. On simplifying 6/3 to its lowest form we get 6/3 = 2/1 which is an integer.

(iii) -3/-3

-3/-3 is an integer. On reducing -3/-3 to its reduced form we get -1/-1 =1 which is an integer.

(iv) -13/2

-13/2 is not an integer and we can’t express it other than a fraction form or decimal value.

(v) -36/9

-36/9 is an integer as we get the reduced form -36/9=-4 which is an integer.

(vi) 47/-9

47/-9 is not an integer and we can’t express it other than fraction form or decimal value.

(vii) -70/-20

-70/-20 is not an integer and we can’t express it other than fraction form or decimal value.

(viii) 1000/-10

1000/-10 is an integer as we get 1000/-10 = -100 on reducing to its lowest form and -100 is an integer.

From the above instances, we can conclude that Not Every Rational Number is an Integer.

https://www.youtube.com/watch?v=9yvtLN_24G0

Let us learn in detail how to arrange Rational Numbers in Ascending Order. Have a look at the general method to arrange the Rational Numbers in Increasing Order. To help you get a better idea of the concept we even listed the solved examples provided step by step.

Procedure to arrange from Smallest to Largest Rational Numbers

Go through the below-listed guidelines in order to arrange Rational Numbers from smallest to largest. They are along the lines

Step 1: Express the given rational number in terms of a positive denominator.

Step 2: Determine the Least Common Multiple of the positive denominators obtained.

Step 3: Express each rational number with the LCM acquired as the common denominator.

Step 4: The number which has the smaller numerator is the smaller rational number.

Refer to our free online Polynomial in Ascending Order Calculator & reorder the given polynomial in small to big within a fraction of seconds.

Solved Examples for Rational Numbers in Ascending Order

1.  Write the following rational numbers in Ascending Order -3/5, -1/5, -2/5

Solution:

Since all the numbers have a common denominator the one with a smaller numerator is the smaller rational number. However, when it comes to negative numbers the higher one is the smaller one.

Therefore arranging the given rational numbers we get -3/5, -2/5, -1/5

2.  Arrange the rational numbers 1/2, -2/9, -4/3 in Ascending Order?

Solution:

Find the LCM of the denominators 2, 9, 3

LCM of 2, 9, 3 is 18

Express the given rational numbers with the LCM in terms of common denominator.

1/ 2= 1*9/2*9 = 9/18

-2/9 = -2*2/9*2 = -4/18

-4/3 = -4*6/3*6 = -24/18

Check the numerators of all the rational numbers expressed with a common denominator.

Since -24 is less than the other two we can arrange the given rational numbers in Ascending Order.

-4/3, -2/9, 1/2 is the Ascending Order of Given Rational Numbers.

3. Arrange the Rational Numbers 5/8, 4/-6, 3/5 in Ascending Order?

Solution:

Firstly, express the rational numbers with positive denominators by multiplying with -1

4/-6 = 4*(-1)/-6*(-1) = -4/6

So, find the LCM of the denominators 8, 6, 5

LCM of 8, 6, 5 is 120

5/8 = 5*15/8*15 = 75/120

-4/6 = -4*20/6*20 = -80/120

3/5 = 3*24/5*24 = 72/120

Check the numerator of the rational numbers having common denominators.

since -80 is the smallest that itself is the smallest rational number.

Therefore, 4/-6, 3/5, 5/8 are in Ascending Order.

Also Read: Program to Read a Number n and Compute n+nn+nnn in C++ and Python

BODMAS RULE:

We must remember the word VBODMAS in solving sums on simplification. These letters stand for vinculum, bracket, of, division, multiplication, addition and subtraction respectively.

The sums on simplification must be solved in that order i.e., first solve vinculum followed by bracket and so on until the sum is solved.

V	Vinculum means bar as (-)
B	Bracket- () {} and then [ ]
O	of
D	Division [÷]
M	Multiplication [x]
A	Addition [+]
S	Subtraction [-]

In simplifying an expression first of all bar must be removed. After removing the bar, the brackets must be removed, strictly in the order ( ), { }and [ ].

After removing the brackets, we must use the following operations strictly in the order: (i) of (ii) Division (iii) Multiplication (iv) Addition (v) Subtraction

		Example 1	Example 2
B	Brackets	5 × ( 7 – 3) = 5 × 4 = 20	24 ÷ ( 8 – 5 ) = 24 ÷ 3 = 8
O	Orders	7 + 22 = 7 + 4 = 11	10 – 32 = 10 – 9 = 1
D	Divide	9 + 12 ÷ 3 = 9 + 4 = 13	10 – 6 ÷ 2 = 10 – 3 = 7
M	Multiply	10 ÷ 2× 3 = 10 + 6 = 16	9 – 4 × 2 = 9 – 8 = 1
A	Add	4 × 3 + 5 = 12 + 5 = 17	2 × 7 + 8 = 14 + 8 = 22
S	Subtract	10 ÷ 2 – 2 = 5 – 2 = 3	9 ÷ 3 – 1 = 3 – 1 = 2

Ordering Mathematical Operations

B	O	D	M	A	S
Brackets [ ], { }, ( )	Orders x2, √x	Divide ÷	Multiply ×	Addition +	Subtract ×

Use of brackets and the BODMAS rule

Let us consider an example to illustrate the use of brackets.
Rima bought 35 chocolates and ate 5 of them. She distributed the remaining chocolates equally among 6 of her friends. How many chocolates did she give to each of them?

In this problem we have to subtract 5 chocolates that Rima ate, from 35 chocolates she had, before dividing them among 6 of her friends. So we have to first perform the operation of subtraction and then do division. In such cases, we use brackets around the part that has to be done first, that is
 (35 – 5) ÷ 6 (First solve bracket, i.e., 35 – 5 = 30)
= 30 ÷ 6      (Division: 30 ÷ 6 = 5)
= 5
Consider another example.

Example 1: Solve 2 of 3 × (5 + 2).
Solution: 2 of 3 × (5 + 2)
= 2 of 3 × 7  (First bracket: 5 + 2 = 7)
= 6 × 7          (Of: 2 of 3 = 2 × 3 = 6)
= 42              (Multiplication: 6 × 7 = 42)
Hence, when problems involve brackets, of, ×, ÷, +, and – then
 To make it easy to remember this order, we remember the word BODMAS, where B stands for brackets, O for of’, D for division, M for multiplication, A for addition, and S for subtraction. This is called the ‘BODMAS‘ rule.
Sometimes numerical expressions may involve different types of brackets. These brackets are

• Vinculum or bar —
• Parentheses or small brackets ( )
• Braces or curly brackets { }
• Square brackets or big brackets [ ]

We simplify expressions by starting with the innermost bracket. Usually the vinculum is the innermost bracket, next is the parentheses, then the braces, and finally the square brackets. Let us now consider some examples.

Example 2: Simplify \(25-[20-\{10-(7-\overline{5-3})\}]\).
Solution:

Example 3: Simplify [72 -12 ÷ 3 of 2] + (18 – 6) ÷ 4.
Solution:

RULES FOR SIMPLIFICATION

1. Order of operation: The use of brackets take us to a new order of operation. The operation inside the brackets comes before the ODMAS. There are different types of brackets already mentioned here.

2. If there is no sign between a number and the bracket, then it is implied that the operation to be performed is multiplication.
Examples

3. When there is ‘+’ sign before a bracket, you can simply remove the bracket.

Examples:

4. When there is a ‘–’ sign before a bracket, then all signs within the bracket change while removing the bracket.

Examples:

Ex: 1 :
Simplify:

Solution:

Exercise
1.

2.

3.

4.

Answers

1.  c     2. a      3. d      4. c

2. Application of the formula, (a + b)² = a² + b² + 2ab
Ex: 1.
Simplify 0.46 x 0.46 + 0.54 x 0.54 + 0.92 x 0.54
Solution:
We have the expression
0. 46 x 0.46 + 0.54 x 0.54 + 0.92 x 0.54
(0.46)² +(0.54)² + 2. x 0.46 x 0.54
If we suppose a = 0.46 and b = 0.54, then
= a² + b² + 2ab = (a + b)²
= (0.46 + 0.54)2 = (1.00)² = 1
∴ Answer = 1.

If you are searching for help on how to get the coordinates of a point you have arrived at the correct place. Check out the complete details about a point x coordinate, y coordinate in the following sections of this page. Get to know what is meant by coordinates and how to find the coordinates of a point in a coordinate geometry along with few examples to get a better idea of the concept.

Coordinates of a Point Definition

Coordinates are nothing but a set of real numbers that represent the exact position of a point in a cartesian plane, Generally, a two-dimensional coordinate plane has two axes namely the x-axis, y-axis. The two coordinates of a point are abscissa or x coordinate, ordinate or y coordinate.

Steps to Find Coordinates of a Point in a Plane

In order to find the coordinates of a point, you must have a coordinate plane. The concept of finding the point coordinates is quite easy.

• As a first step, check out the coordinate graph having both horizontal and vertical lines.
• Find in which quadrant the given point falls and based on that you will get the signs of coordinates of a point.
• Draw two dotted perpendicular lines from the point to the x-axis, y-axis.
• Measure the distance between the dotted line meeting point on the x-axis and consider it as x coordinate.
• Likewise, measure the distance from the origin to the dotted line meeting point on the y-axis.
• Take it as y coordinate.
• Then, write your point as (x_coordinate_value, y_coordinate_value).

Point Coordinates

In any two dimensional plane, each point has two coordinates. The first coordinate is called the x coordinate which is measured along the x-axis. The second coordinate of the point is called the y coordinate which is measured along the y-axis. More details about the coordinates are mentioned below.

Abscissa: It is defined as the first number of a point which is present before the comma and after the opening parenthesis. It is also known as the x coordinate. Its value can be positive or negative. If the abscissa is zero, then we call it a point on the y-axis. Usually, the x coordinate value is measured along the x-axis.

Ordinate: The second number of an ordered pair is called the ordinate. It is also called the y coordinate. It is available after the comma and before the closing parenthesis, it can be negative or positive. If the y coordinate value is zero in a point, then it is called the point on the x-axis. Generally, its value is measured along the y-axis.

Solved Examples

1. In the adjoining figure, XOX’ and YOY’ are the co-ordinate axes. Find out the coordinates of points P, Q, R, S. And also write x coordinate, y coordinate of each point?

Solution:

To find the point P:

Point P is the fourth quadrant. So its abscissa is positive, the ordinate is negative.

The perpendicular distance of the point from the x-axis is 1 unit, y-axis also 1 unit.

So, x coordinate is 1, y coordinate is -1.

Therefore, point P (1, -1)

To locate the point Q:

As Q is in the first quadrant, its coordinates are positive.

The perpendicular distance between point and origin on the x-axis is 2 units, on the y-axis is 4 units.

So, abscissa is 2, ordinate is 4

Therefore, Q (2, 4)

To locate point R:

The point R is the second quadrant. So, the R x coordinate is negative and the y coordinate is positive.

The perpendicular distance of the point on the x-axis is 5 units, the y-axis is 3 units.

X coordinate is -5, y coordinate is 3.

Therefore R (-5, 3)

To find the point S:

The point S is located in the fourth quadrant.

The perpendicular distance of the point from the x-axis is 6 units and it is positive.

The perpendicular distance along the y-axis is 3 units and it is negative.

So, abscissa is 6, ordinate is -3

Therefore S (6, -3)

Example 2.

Find the coordinates of A, B from the following figure?

Solution:

To locate point A:

Point A is on the x-axis.

So, the y coordinate is 0.

The point is -4 units away from the origin.

So the x coordinate is -4.

Therefore, A (-4, 0).

To find point B:

It is located in the third quadrant. So both coordinates are negative.

The distance from the point to the x-axis is 7 units, the y-axis is 2 units.

Therefore, Point B (-7, -2).