Binomial Theory

In algebra, a binomial theory is nothing but the study of binomial. The binomial is defined as the polynomial with sum of two monomial terms. Some of the binomial term is given as,The binomial also consist of distributions, coefficient, variables etc.some of the binomial terms are given as,

3x + 1 = 0

4x² + 5 = 0

9x³ + 15 = 0

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).



Properties of binomial:

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

A sum of binomial coefficients of even term is equal to a sum of binomial coefficients of odd terms, and it is equal to

2n-1

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

Examples

Example1: find the value of 5²- 3² and prove it?

Solution:

The difference of binomial a2 − b2 is the product of two other binomials:

a2 − b2 = (a + b) (a − b).

52 − 32 = (5 + 3) (5 − 3).

= (8)(2)

= 16.      ---------- (1)

Proof:

a2 − b2 = 52 − 32

= 25 – 9

= 16       ---------- (2)

From 1 and 2 it proved.

Example 2: find the value of (3x + 1) and (2x + 3)?

Solution:

The product of a pair of linear binomials (ax + b) and (cx + d) is:

(ax + b)(cx + d) = acx2 + adx + bcx + bd.

(3x + 1)(2x + 3) = 3*2x2 + 3*3x + 1*2x + 1*3.

= 6x2 + 9x + 2x + 3.

Therefore, (3x + 1) (2x + 3) = 6x2 + 9x + 2x + 3

Example 3: Prove that (a + b) ³ = 8?

Solution:

From the property of binomial,

A sum of binomial coefficients of an exponent (a + b) n is equal to 2 n.

(1 + 1) n = 2 n

(a + b) ³ = (1 + 1) 3 = 2 3

= 8

Hence it is proved.

Algebra is widely used in day to day activities watch out for my forthcoming posts on taylor series cos and cbse previous year question papers class 12. I am sure they will be helpful.

Practice problem:

Problem1: find the value of 9²- 4² and prove it?

Answer is 65

Problem2: find the value of (2x + 1) and (x + 3)?

Answer is 2x2 + 6x + 1x + 3.

Solving Sums Integrals

Integration is an important concept in mathematics and, together with differentiation, is one of the two main operations in calculus. The term integral may also refer to the notion of antiderivative, a function F whose derivative is the given function f. In this case it is called an indefinite integral. Integral can be classified as definite and indefinite integral. (Source: Wikipedia)

I like to share this Table of Derivatives and Integrals with you all through my article.

Examples problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function ∫ (234x2 + 132x4 - 5x) dx.

Solution:

Given ∫ (234x2 + 132x4 - 5x) dx

Integrate the given function with respect to x, we get

∫ (234x2 + 132x4 - 5x) dx = ∫ 234x2 dx + ∫ 132x4 dx - ∫ 5x dx.

= 234 (x3 / 3) + 132 (x5 / 5) - 5 (x2 / 2) + c.

= 78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Answer:

The final solution is  78x3 + `(132 / 5)` x5 - `(5 / 2)` x2 + c.

Solving sums integral problem 2`:`

Find the value of the integration

`int_2^5(x^6)dx`

Solution:

Integrate the given function with respect to x, we get

`int_2^5(x^6)dx`  = `(x^7 / 7)`52

Substitute the lower and upper limits, we get

= `((5^7 / 7) - (2^7/ 7))`

= `((78125 / 7) - (128 / 7))`

= `(77997 / 7)`

Answer:

The final answer is `(77997 / 7)`

Solving sums integral problem 3:

Integration using algebraic rational function ∫ 7dx / (17x + 37)

Solution:

Using integrable function method,

Given function is ∫ `(7dx) / (17x + 37)`

Formula:

∫ [L / (ax + c)] dx = (L / a) log (ax + c)

From given, L = 7, a = 17, and c = 37

Integrate the given equation with respect to x, we get

= `(7 / 17)` log (17x + 37)

Answer:

The final answer is `(7 / 17)` log (17x + 37).

Practice problems for solving sums integrals

Solving sums integral problem 1:

Integrate the given function using integrable function ∫ `(14 / (11x + 12))` dx

Answer:

The final answer is `(14 / 11)` log (11x +12)

Solving sums integral problem 2:

Integrate the given function using integrable function ∫ `(15 / (21x + 92))` dx

Answer:

The final answer is `(15 / 21)` log (21x + 92)

My forthcoming post is on cbse class 10 syllabus will give you more understanding about Algebra.

Solving sums integral problem 3:

Integrate the given function ∫ (7.9x2 - 12.9x) dx

Answer:

The final answer is `(7.9 / 3)` x3 - `(12.9 / 2)` x2

Difference of Two Means

Mean is defined as the average for the total number of values given in the data set. It is the sum made between the given data set and it is divided it by the total number of values given in the data set. Tis is called as the mean for the given data set. This can also be called as arithametic mean or sample mean. The difference between two mean is nothing but the taking difference for the two sets of data (mu_1 , mu_2 ) and calculating the difference for that mean.

Difference of two mean  = mu_1 - mu_2

Where as

mu_1 is the mean value for the first set of data.
mu_2 is the mean value for the second set of data.

Mean is calculated by the way,
mu = (Sum of all the values given) / (Total Number of values)


Steps for calculating the difference of two means:

Get the two sets of data for calculating the mean.
Measure the sum for the first set of data and divide it by the total number of data's given. This is the mean for the first data set
Measure the sum for the second set of data and divide it by the total number of data's given. This is the mean for the second data set.
Now measure the difference of the two means from the mean calculated.



Difference of two means - Example Problems:

Difference of two means - Problem 1:

Find the difference of two means from the given two data set. 2, 3, 4, 5, 6, 7 and 1, 2, 3, 4, 5, 6

Solution:

Mean for the first set of the data given

mu_1 = (2+3+4+5+6+7) / 6

= 27 / 6

=  4.5

Mean for the second set of the data given

mu_2 = (1+2+3+4+5+6) / 6

=  21 / 6

mu_2    = 3.5

Difference of the two means is given by

Difference of two mean = mu_1 - mu_2

= 4.5 - 3.5

Difference of two mean = 1

Difference of two means - Problem 2:

Find the difference of two means from the given two data set. 24, 23, 24, 25, 26, 27 and 11, 13, 13, 14, 15, 16

Solution:

Mean for the first set of the data given

mu_1 = (24+23+24+25+26+27) / 6

= 149 / 6

=  24.8333333

Mean for the second set of the data given

mu_2 = (11+13+13+14+15+16) / 6

=  82 / 6

mu_2    = 13.6666667

Algebra is widely used in day to day activities watch out for my forthcoming posts on icse syllabus and cbse books. I am sure they will be helpful.

Difference of the two means is given by

Difference of two mean = mu_1 - mu_2

= 24.8333333 - 13.6666667

Difference of two mean  = 11.1666666

Is a Cube a Polygon

Cube is not a polygon,because cube is a three dimensional shaped figure .But polygon is a two dimensional object. Generally polygon must be flat, plane figure and it’s made up of line segment. Here we are going to study about cube and polygon shape and its example problems.


Shape of cube:

Cube is a regular solid three-dimensional figure it has six square faces .it has 12 edges of equal length and 8 vertices it is otherwise called as regular hexahedron.

Shape of polygon:

It will be triangle, quadrilateral, pentagon, hexagon, octagon, heptagon…etc. All are two dimensional shapes.

Example problems for cube:

Example: 1

Find the volume of the cube with side length 6 meter.

Solution:

We know that volume of the cube is a 3

Given a = 6

Therefore a3 = (6)3

= 6 * 6 * 6

= 216 meter cube

Example: 2

Find the surface area of the cube each side length of a cube is 16 feet.

Solution:

We know that surface area of cube is,

A = 6a2

Here the given  a = 16 feet

Substitute the a value in the above formula we get

A= 6*162

162 = 16*16 = 256

Therefore the area of the cube is

= 6*256

=1536 feet square

The surface area of the given cube is 1536 feet square.

Example problems for polygon:

Example: 1

Find the perimeter of the polygon which is hexagon shape with side length is 9 feet

Solution:

We know that perimeter of the hexagon is =6*a

a represents the side length

Therefore perimeter = 6 *9

= 54

Perimeter of the given hexagon is 54 feet

My forthcoming post is on syllabus of class x cbse and tamil nadu text book will give you more understanding about Algebra.

Example: 2

Find the perimeter of the regular pentagon with side length is 8 meter

Solution:

We know that pentagon is one of a polygon and formula for finding the perimeter is 5 *a

= 5 * 8

= 40

Perimeter of the pentagon is = 40 meter

Unit of Measurement for Volume

The volume is defined as the space occupied by any object in three dimensions. There are various units for the measurement of the volume of the objects. The most commonly used is the standard international unit of measurement, the cubic meter unit. But there are other units of volume measurement. In this article we will see them in detail.


Unit of measurement for volume:

The various units of volume measurement are related with the standard international unit of volume measurement the cubic meter. With the relations the various units can also be related to one other. The various units of volume measurement and their relations are,

1 cubic meter = 1000 L = 264.2 gallons

1 cubic meter = 35.31 ft3 = 1.308 yd3

1 gallon = 0.1337 ft3 = 3.785 L

1 cubic feet = 7.481 gallons = 0.0283 m3

1 cubic yard = 27 ft3 = 202 gallons = 0.7646 m3 = 764.6 L

1 imperial barrel = 163.7 L = 6.10 m3

The above relations can be used for the conversion between the various units above by relating with each other.

Example problems on unit of measurement for volume:

1. A container volume is measured to be 15000 liters. Convert the volume into m3 and gallons.

Solution:

1 cubic meter = 1000 L

1 liter = `1/1000` m3

15000 liter = `15000*(1/1000)` m3

15000 liter = 15 m3

1000 liter = 264.2 gallons

15000 liters = 15*264.2 gallons

15000 liters = 3963 gallons

2. A water tank can store 20.5 m3 of water. Convert the volume into yd3 and gallons.

Solution:

1 cubic meter = 1.308 yd3

20.5 cubic meters = 20.5 * 1.308 yd3

20.5 cubic meters = 26.8 yd3

1 cubic meter = 264.2 gallons

20.5 cubic meter = 20.5*264.2 gallons

20.5 cubic meter = 5416 gallons

Algebra is widely used in day to day activities watch out for my forthcoming posts on Multiplying Mixed Number Fractions and polynomial function degree. I am sure they will be helpful.

3. Convert 1.35 Barrel into cubic meters.

Solution:

1 imperial barrel = 6.10 m3

1.35 imperial barrel = 1.35*6.1 m3

1.35 imperial barrel = 8.23 m3
Practice problems on unit of measurement for volume:

1. Convert the volume of 560 gallons into ft3 and liters.

Answer: 74.9 ft3 and 2119.6 L

2. Convert the volume of 2.5 yd3 into liters.

Answer: 1911.5

Median Frequency Table

Median is defined as one of the most important topic in mathematics. Mainly it is used to find the middle values. The values are given in the frequency table. By using the table we can find the median. Both the even numbers and the odd numbers, the median can be find. In this article, we are going to find the calculation of median from the frequency table.

I like to share this Median in Statistics with you all through my article.

Explanation to median frequency table

The explanation to median frequency table are given below the following,

Median for odd values:

Median = `(n + 1)/2`,      if the n value is odd.

Median for even values:

Median = `(n/2)+1` , if the n value is even.

Example problems to median frequency table

Problem 1: Find median for the following frequency table,
Values 2 3 4 5 6
Frequency 3
4 5
6
7

Solution:

Step 1: Given:

Values = 2, 3, 4, 5, 6

Frequency = 3, 4, 5, 6, 7

Step 2: Find:

Values = 2 + 3 + 4 + 5 + 6

= 20 ( Its a even function)

Step 3: Formula:

Median = `(n/2)+1` , if the n value is even.

Step 4: Solve:

Median = `(n/2)+1`

= `(20/2) + 1`

= 10 + 1

= 11

Therefore, the median is in the position of 11.

Step 5: To find position:

Add values and the frequencies, we get,
Values 2 3 4 5 6
Frequency 3 4 5 6 7
Position 2 + 3 = 5 5 + 4 = 9 5 + 9 = 14

Since the frequency is at 11 position, it will be between the 9 and the 14 position, So, 4 is the median value.

Result: Median = 4

Problem 2: Find median for the following frequency table,
Values 1 3 4 5 6
Frequency 2
4 6
8
10

Solution:

Step 1: Given:

Values = 1, 3, 4, 5, 6

Frequency = 2, 4, 6, 8, 10

Step 2: Find:

Values = 1 + 3 + 4 + 5 + 6

= 19 ( Its a odd function)

Step 3: Formula:

Median = `(n + 1)/2`,  if the n value is odd.

Step 4: Solve:

Median = `(n+1)/2`

= `(19 + 1)/2`

= `20/2`

= 10

Therefore, the median is in the position of 10.

Step 5: To find position:

Add values and the frequencies, we get,
Values 1 3 4 5 6
Frequency 2 4 6 8 10
Position 2 + 1 =3 4 + 3 =7 7 + 6 =13

Since the frequency is at 10 position, it will be between the 7 and the 13 position, So, 4 is the median value.

Result: Median = 4


My forthcoming post is on Divide Polynomials and Regular Convex Polygon will give you more understanding about Algebra.

Practice problems to median frequency table

Problem 1: Find median for the following frequency table,
Values 3
5 7 9 10
Frequency 5
6
8
8
9

Answer: 7

Problem 2: Find median for the following frequency table,
Values 3
6
9
12
15
Frequency 2
4
6
8
10

Answer: 9

Study Definition of Subset

Online gives the definition of subset as the elements of subset are contained by another set. By studying the definition of subset we can understand that subsets are part of another set which is used a symbol `sube`. Online gives a clear definition of subsets which helps to study problems easily.

Explanation to study definition of subset:

The definition of subset is as follows.

A set X is called as a subset of a set Y if some of or all the elements of X are existing in the set Y which can be denoted as X `sube` Y. We can also write the set as the set y is a superset of set X and denoted as Y `supe` X. A empty set is also taken as a subset for any kind of set.

Representation to study definition of subset:

The elements in X are existing in the set Y called X is the subset of Y.

Example problems to study definition of subset:

Example: 1

Write the subset relation for the following sets.

A = {m, n, o} , B = {o, p} and C = {m, n, o, p, q, r}

Solution:

Given sets are,

A = {m, n, o}

B = {o, p}

C = {m, n, o, p, q, r}

A and B:

The elements in A is not in B as well as the elements in B are not in A.

A and C:

The elements in A are in C. So, A is called as subset for the set C = {m, n, o, p, q, r} which can be represented as A `sube` C.

B and C:

The elements in B are in C. So, B is called as subset for the set C = {m, n, o, p, q, r} which can be represented as B `sube` C.

Example: 2

Say whether the set P = {12, 14, 15} is a subset for a set Q = {11, 12, 13, 14, 15}.

Solution:

Given sets are,

P = {12, 14, 15}

Q = {11, 12, 13, 14, 15}

The elements in P are in Q. So, P is called as subset for the set Q = {11, 12, 13, 14, 15} which can be represented as P` sube ` Q.

My forthcoming post is on Decimals Place value and pre algebra online will give you more understanding about Algebra.

Practice problems to study definition of subset:

Problem: 1

Write the subset relation for the sets C = {9, 7} and D = {9, 8, 7, 6}

Answer: C `sube` D

Problem: 2

Write the subset relation for the sets S = { } and P = {4, 5, 10}

Answer: S `sube` P

Preparation for Subset

In mathematics subsets are the terms used in set theory. The preparation depends on the elements in the sets. For the preparation for subset we have the derive a set by having elements of another set. The subset preparation uses the symbol `sube` .

For example, C `sube` A denotes C is subset of the set A and A `supe` C denotes A is a superset of C.

Understanding subset of a set is always challenging for me but thanks to all math help websites to help me out. 

Explanation to preparation for subset:

The preparation for subset is as follows.
Every set has a subset which is derived from the set. This subset may have all elements from the given set and it may be an empty set.

For example, set V = {l, p m , n}. Some of the possible subset for the given set V are { } , {l, p, m , n} , {l}, {l , p} , { m, n} , {p, m, n} etc. In this the set V is called as superset.

Example problems to preparation for subset:

Example: 1

Prepare the subsets of a set C = {2, 3}

Solution:
Given: C = {2, 3}

Subsets has the elements from the given set C = {2, 3} such as { }, {2}, {3}, {2, 3}

Example: 2

Which of the following is true for the sets A = {1, 2, 3, 4} B = {3, 4}?

a) B `supe` A

b) A `sube` B

c) A = B

d) B` sube` A

Solution:
Given A = {1, 2, 3, 4} B = {3, 4}
        B has the elements 3, 4 which is in A and has the elements 1, 2, 3, 4 where 1, 2 are not in B.So, B is a subset of A and A is a super set of B. (B `sube` A)
Answer: d

Algebra is widely used in day to day activities watch out for my forthcoming posts on Sum of Exterior Angles Formula and Multiplication Fractions. I am sure they will be helpful.

Practice problems to preparation fo subset:

Problem: 1

Which of the following is true for the sets X = {a, c, b, m, l, i} Y = {a, b, l}?

a) Y `supe` X

b) Y `sube` X

c) Y = X

d) X `sube` Y

Answer: b
Example: 2

Which of the following is true for the sets P = {p, m} Q = {p, q, r, s}?

a) P `supe` Q

b) P `sube` Q

c) P = Q

d) Q `sube` P

Answer: b

Height of Trapezoid

Geometry deals with shapes, structures, lines, planes and angle’s. Geometry learning is also known as architectural learning. Basic shapes of geometry are square, triangle, rectangle, parallelogram, trapezoid etc. Trapezoid is one of the basic shapes in geometry. Trapezoid is a quadrilateral which has 4 sides. The total internal angle of the trapezoid is 360 degree. In trapezoid, one pair of opposite sides is parallel.

Formula for finding the height of the Trapezoid

The formula for finding the area of pyramid is given as,

Area of pyramid = h (b1 + b2)/2

Where,

h = height of the pyramid,

b1, b2 = bases of the trapezoid.

From the given area formula we can find the height of the trapezoid when the area of trapezoid is given,

A = h (b1 + b2)/2

2A = h (b1 + b2)

h = 2A/b1+ b2



Problems on height of pyramid:

Example 1:

Find the area of the trapezoid, whose bases are 10 cm and 12 cm, height, is 6 cm.

Solution:

Formula for finding the area of the trapezoid is,

Area of pyramid = h (b1 + b2)/2

= 6 (10 + 12) / 2

= 6 (22) / 2

= 3 * 22

= 66 cm2.

The answer is 66cm2.



Example 2:

Find the height of the trapezoid, whose bases are 8 cm and 12 cm, area, is 120 cm2.

Solution:

Formula for finding the area of the trapezoid is,

Area of pyramid = h (b1 + b2)/2

120 = h (8 + 12)/2

120 * 2 = h (20)

240 =   h*20

Divide 20 on both sides,

240/20 = 20*h/20

12 = h

The height is 12 cm.

I am planning to write more post on Statistics Quartiles and Calculate Geometric Mean. Keep checking my blog.

Example 3:

Find the height of the trapezoid, whose bases are 6 cm and 4 cm, area, is 100 cm2.

Solution:

Formula for finding the area of the trapezoid is,

Area of pyramid = h (b1 + b2)/2

100 = h (6 + 4)/2

100 * 2 = h (10)

200 =   h*10

Divide 10 on both sides,

200/10 = 10*h/10

20 = h

The height is 20 cm.

Least Common Multiple of 3 and 6

In mathematics, the least common multiple of two rational numbers a and b is the smallest positive rational number that is an integer multiple of both a and b. Since it is a multiple, it can be divided by a and b without a remainder. If either a or b is 0, so that there is no such positive integer, then LCM(a, b) is defined to be zero. (Source: From Wikipedia).

Least common multiple of two numbers can be found by their multiples. Here we are going to learn how to find the least common multiple of two or more numbers.

Least common multiple of 3 and 6

The least common multiple of 3 and 6 can be found by finding the multiples 3 and 6.

The list of multiples of 3 and 6 are given below

The multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60.

The multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60.

Here 6, 12, 18, 24, 30, 36, 42, 48, 54, 60 are the common factors, among those 6 is the lowest common number.

So, 6 is the lowest common multiple of 3 and 6.

Example problems for least common multiple

Example 1

Find the least common multiple of 3 and 16

Solution

The least common multiple of 3 and 16 can be found by finding the multiples 3 and 16.

The list of multiples of 3 and 16 are given below

The multiples of 3 = 3, 6, 9, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 54, 57, 60.

The multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160.

Here 48 is the lowest common number. So, 48 is the lowest common multiple of 3 and 16.

Example 2

Find the least common multiple of 13 and 6

Solution

The least common multiple of 13 and 6 can be found by finding the multiples 13 and 6.

The list of multiples of 13 and 6 are given below

The multiples of 13 = 13, 26, 39, 52, 65, 78, 91, 104, 117, 130, 143.

The multiples of 6 = 6, 12, 18, 24, 30, 36, 42, 48, 54, 60, 72, 78, 84.

Here 78 is the lowest common number. So, 78 is the lowest common multiple of 13 and 6.

My forthcoming post is on Positive Correlation Graph and percentage formulas will give you more understanding about Algebra.

Example 3

Find the least common multiple of 13 and 16

Solution

The least common multiple of 13 and 16 can be found by finding the multiples 13 and 16.

The list of multiples of 13 and 16 are given below

The multiples of 13 = 13, 26, 39, 52, 65, 78, 91, 104, 117, 130, 143, 156, 169, 182, 195, 208.

The multiples of 16 = 16, 32, 48, 64, 80, 96, 112, 128, 144, 160, 176, 192, 208.

Here 208 is the lowest common number. So, 208 is the lowest common multiple of 13 and 16.

Beginner Multiplication

Multiplication (x) is the arithmetical operation of calculating one value by another value. It is a one kind of essential operations in basic arithmetic (the others operations are addition, subtraction and division). Since the outcome of calculating by whole numbers can be thinking of as including of a few number of copies of the original, whole-number products larger than 1 can be calculated by frequent addition.

Types of multiplication for beginner:

• Multiplication of variable by exponent

• Multiplication of fraction

• Multiplication of unlike signs ( positive(+),negative (-))


Understanding Multiplication Fractions is always challenging for me but thanks to all math help websites to help me out.

Multiplying Variables with Exponents:

Multiplication of exponent:

1) Exponent of 0:

If the exponent is 0 specifies you are not multiplying by anything and the answer is 1

For example, a0 = 1

x0 = 1

2) Exponent of 1:

If the exponent is 1 specifies you are multiplying the variable with 1. (Example x1 = x)

Rules for multiplication terms on fractions:

• First multiple the value of numerator.

• Then multiple the value of denominator.

• Lastly decrease the fraction (if required).

Example:

`2/5` ×` 3/4`

Step1 Multiply the numerators:

`3 / 4` × `2 / 5` = > 3 × 2 / 4 x 5 = `6/20`

Step2 Multiply the denominators:

` 3/4` ×`2/5` = 3×2 / 4 x 5 = `6/20`

Step3 Lastly decrease the fraction

Therefore solution is `3/10` .

My forthcoming post is on Population Versus Sample and The Prime Numbers will give you more understanding about Algebra.

Multiplication of unlike signs for beginner:

• Positive(+) × Positive(+) = Positive(+)

Ex:  6 × 2   =12
• Positive (+) × Negative (-) = Negative (-).

Ex:   6 × (-2) = -12
• Negative (-) × Positive (+) = Negative (-).

Ex: (-6) × 2 = -12
• Negative(-) × Negative(-) = Positive(+).

Ex: (-6) × (-2)=12

Example:

Multiply by (a + 4) (a -5)

Step 1: multiply by y in the second factor

a (a-5) =  a2- 5a

Step 2: multiply by 5 in the second factor

4(a - 5) = 4a-20

Step 3: add step 1 and step 2

(a + 4) (a - 5) = a2- 5a + 4a - 20

= a2 -1a - 20

Basic multiplication problems for beginner:

1) Multiply the values i) 23 × 2 ii) 4 × (6 × 3) = (4 × 6) ×3

Solution:

i) 23 × 2 = 46

ii)    4 × (6 × 3) = (4 × 6) ×3

4 × 18 = 24 × 3

72 = 72.

2) Multiply the values i) 4 × (2 + 6)   ii) 6 × (5 × 3) = (6 × 5)×3

Solution:

i) 4 × (2 + 6) = 4 × 2 + 4 × 6

= 8 + 24

= 32.

ii)6 × (5 × 3) = (6 × 5) × 3

6 × 15 = 30 × 3

90 = 90.

Isosceles Triangle Hypotenuse

When the two sides of the triangle are said to be equal then the triangle is called as isosceles triangle. When all the sides of the triangle are equal then the triangle is called as the equilateral triangle and when no sides of the triangle are equal then it is said to be scalene triangle. In an isosceles triangle only the sides but also the two angles are said to be equal.
                                                               
Here we will see about the isosceles triangle hypotenuse.

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Isosceles triangle hypotenuse.

The hypotenuse side of the isosceles triangle is the sides that are found opposite to that of the right angle. According to the right angle theorem the height of the isosceles triangle is given by
                         h = √(b² – ¼ a²)
Thus the area of an isosceles triangle is
                        A = ½ ah
                            = ½ a √(b² – ¼ a²)
                            = ½ a² √((b²/a²) – (¼))
An isosceles triangle is also called as the triangle with two congruent sides. The angles which are opposite to these congruent sides are called as the base angles and the angles found between those sides are called as the vertex angle of the isosceles triangle.
Any equilateral triangle can be an isosceles triangle but no isosceles triangle is an equilateral triangle.                           

Properties of the isosceles triangle base:

• According to the right angle theorem the sum of the squares of the hypotenuse side is equal to the sum of the squares of the other two sides. The hypotenuse side is the side which is opposite to the right angle

• The side which is not equal to the other sides of a triangle is called as the base of the isosceles triangle.

• The base angles of the isosceles triangle are found to be equal.

• When the third angle of the isosceles triangle is a right angle then it is called as the right isosceles triangle.

• The perpendicular distance from the base to the vertex of an isosceles triangle is called as the altitude of the isosceles triangle.

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Externally Tangent Circles

Two intersect circles in a single point is known as tangent circles. It can be divided into two types of tangency: internal and external. By using the tangent circles many problems and constructions in geometry are solved. This type of problems have real-life applications. The followings are some real-time applications: trilateration and maximizing the use of materials.

                                             

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Illustration:

A tangent that is common to two circles and does not intersect the segment joining the centers of the circles is called Common External Tangent. A common tangent can be any one of the following:External tangent or Internal tangent.

                                                   
          From the above figure, we can see that line PQ is the common external tangent.

                                               
          From the above figure, we can see that line AB is the common external tangent.

Theorem for Externally Tangent Circles:

Theorem:If a line is tangent to a circle, it is perpendicular to the radius drawn to the point of tangency.
                                                       
Solution:
            If:        Line  AB is a tangent
                       D is point of tangent.
            Then:   OD ┴ AB

Example problems:

Example 1: Find the common external tangent from the below figure.

                                             
Solution:
            A tangent that is common to two circles and does not intersect the segment joining the centers of the circles is called common external tangent.
            From the given figure, the line CD touches both the circles on the same side of the line is the external tangent to the circle.
            Therefore, CD is the common external tangent line.

Example 2: Find the common external tangent from the below figure.
                                                                                    
Solution:
            The given figure is concentric circles. Concentric circles having the same center and not tangent.

Example 3: Find the common external tangent from the below figure.
                                                     
Solution:
          From the above figure, we can identify that given is two complete separate circles.
          The given figure have two external tangents.

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Example 4: Find the equation for tangent lines y=3x³ from the point x= 2.
Solution :
Given, y=3x³………….(1)
Step 1: To find the first derivative of y

                y ' = 3x ² ………..(2)
Step  2: To plug x =2 into y ' to find the slope at x.
                y ' = 3(2) ²
                y= 12
               Slope of tangent line =12
Step 3:To plug x=2 into y to find the y value of the tangent point
              y =3(2)³
              y=24 
              Hence tangent point (2,24)
Step 4: To plug the slope=12 and point (2,24) using the point-slope equation to find the equation     for the tangent line.
              y- y₁ =m (x-x₁)
              y- 24 =12 (x-2)
              y =12x- 24 + 24
              y=12x
Example 5: Find the slope of tangent line which is passing through the point (2, 0) and the curve y = 2x – x³
Solution:
            Step 1: Find the derivative y¹
                                y = 2x – x³
                                y¹ = 2 -3x²
            Step 2: To find the slope of tangent line,
                         Given point (2, 0) --- > (x, y)
                         y¹ = 2 -3(2)² ---- > 2 – 3(4) --- > 2 -12 = -10
                         Slope of the tangent line = -10         

Practice problems:

1. Determine the equation of the tangent line passing through the point (− 1, − 2) and having slope4 /7
Answer:   4x − 7y − 10 = 0
2. Determine the equation of the line with the slope 3 and y-intercept 4.
Answer: y = 3x + 4

Power Factor Conversion

To signify a lot of the numbers we have used scientific notation. For example the number 7500 is identified as 7.5E3. Here the letter “E" can be read as "times ten to the power of". Every unit is written out in full before its contraction is used. Some units have superscripts in them. In power factor conversion includes the watt to joule, foot pound to force per second, kilogram to meter per second, horse power to watt.

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Example problems of power factor conversion:

Power factor conversion Problem 1:

Convert watt to joule per second.

25 W

30 W

Solution:

We can convert watt to joules per second by using the following method.

1 watt = 1 joule per second

The initial unit is 25 W and 30 W

Here conversion factor is 1 W/ 1 joule/sec

That is, `25*` `1/1` = 25

`30*` `1/1` =30

Therefore the final unit is 25 and 30 units

Answer: 25 joule/sec and 30 joule/sec

Power factor conversion Problem 2:

Convert watt to joule per minute.

55

Solution:

We can convert watt to joule per minute by using the following method.

1 watt = 60 joule per minute

The initial unit is 55 W

Here conversion factor is 1 watt = 60 joule per minute.

That is, `55*` `60/1` =3300

Therefore the final unit is 3300 units

Answer: 3300 joule/minute

Power factor conversion Problem 3:

Convert watt to horsepower.

25

Solution:

We can convert watt to horsepower by using the following method.

1 watt = 0.001341022 horsepower

The initial unit is 25 watt

Here conversion factor is `(0.001341022 hp)/(1 watt)`

That is, 25*0.001341022= 0.033525552 horsepower

Therefore the final unit is 0.033525552 horsepower

Answer: 0.033525552 horsepower

Practice problems of power factor conversion:

1. Convert watt to joule per second: 75

2. Convert watt to joule per minute: 22

3. Convert watt to horsepower: 45

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Answer:

75 joule per second
1320 joule per minute

3.  0.060345994 horsepower

Prime and Rate

Prime: Definition of prime number is a number which cannot divide by any other numbers. That is ,only the number itself and 1 can divide the prime number.
        Rate: Definition of rate is the ratio between the distance and time. Definition of rate is always in the form of fraction.
Let us see the definition of prime and rate.

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Definition of prime number:

Definition of prime number:

If a number is not allowed to divide by any number then it is called as prime number. Other wise it is called as composite number.

Example: 133

This number can not be divide by any number. Only 1 and 133 can divide it. So, 133 is considered as prime number.

Example:1

Find the prime number among the following numbers.

a. 155

b. 242

c. 141

d. 339

Ans :141

Example: 2

Find the prime number among the following numbers.

 a.122

b.223

c.144

d.166

Ans : 223

Definition of rate:

Definition of rate:
  Formula for finding a rate is,

Rate=Distance/Time

        Rate has a units depends on the units of distance and time. If the distance has a unit in km and the time is in hr then the rate should be written as km/hr.
Example problems for finding a rate:
Example:1

Find the rate of scooter. The distance travelled by a scooter is 89 km and the time taken for that is 3 hours.

Solution:
Given,

Distance = 89km

Time=3 hr

We know the rate formula is Rate=distance/time

Rate=`89/3`km/hr.

Therefore the rate of the given scooter is `89/3` km/hr.

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Example:2

Find the time taken by a cycle. The distance travelled by a cycle is 24 km and the rate is `12/6`km/hr.

Solution:
Given,

Distance = 24km

Rate=`12/6`km/hr

We know the rate formula is Rate=distance/time

Time= Distance/time

         =`24/(12/6)`

        =`144/12`

       =12 hours.

Therefore the time taken by a cycle to travel 24 km is 12 hours.

Fraction Conversion Table

Fraction:
A fraction is a number that can represent part of a whole. The earliest fractions are reciprocals of integers: ancient symbols represent one part of two, one part of three, one part of four, and so on. A much later development was the common fractions which are used today (½, ⅝, ¾, etc.) and which consist of a numerator and a denominator.

Fraction can be converted into to forms as shown below,

• Fraction conversion into decimal
• Fraction conversion into percentage.

Through this article you can learn about fraction conversion table, how to make conversion of fraction into decimal and how to make conversion of fraction into percentage.

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Example for fraction conversion table:

Fraction conversion table:

         We can express the conversion of fraction in terms of table format.This table is known as fraction conversion table.The following are the two examples for fraction conversion table.

                       

Fig(i) Fraction conversion table 1                                      Fig(ii) Fraction conversion table 2

The fractions that contains the denominator as a  power of  10 called as decimal fractions.
Ex:
0.1 = 1/10
0.01 = 1/100
0.001 = 1/1000

Problems on fraction conversion table:

Pro 1: Convert the fraction 45 / 90
Sol:
Given, 45 / 90 into decimal.
To find the decimal,divide the numerator by the denominator ,
             _________
         90 )  45      (
45 cannot be divided by 90.
So Multiply 45 by 10 and add one decimal place in quotient,
             _________
         90 )  450     ( 0.5
                450     
                 0        
So the decimal of fraction 45/90 is 0.5
Ans: 45/90 = 0.5
Pro 2: Convert the fraction 76 / 43 into decimal.
Sol:
Given, 76 / 43 as decimal
To find the decimal,divide the numerator by the denominator ,
             _________
         43)  76     ( 1.76
               43     
                33      
33 cannot be divide by 43. So multiply 33 by 10 and one decimal place in quotient,
            _________
         43)  76     ( 1.7
               43     
               330    
               301   
                 290
                 258
                   38
The decimal for 76 / 43 = 1.76
Ans:76 / 43 = 1.76

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Pro 3: Convert the fraction 6 / 5 into decimal.
Sol:
Given, 6 / 5 as decimal
To find the decimal,divide the numerator by the denominator ,
           _________
        5 )  6     ( 1
              5     
              1   
1 cannot be divide by 5. So multiply 1 by 10 and one decimal place in quotient,
            _________
           5 )  6     ( 1.2
                 5     
                 10
                 10  
                   0 
The decimal for 6/5 = 1.2
Ans: 6/5 = 1.2

Venn Diagram Example of Disjoint

Let use see about venn diagram example of disjoint.Venn diagrams also called as the set diagram.These are diagrams that show all theoretically possible logical relations between finite groups of set. Venn diagrams were considered approximately 1880 by John Venn. They are used to teach elementary set theory, and show simple set relationships in logic, probability and statistics.


venn diagram example of disjoint - Notations


Curly  braces - {...} - are used to phrase.
These braces can be used in various ways.
For example:

• List the elements of a set: {-3, -2, -1, 0, 1, 2, 3,4}

• Describe the elements of a set: {integers between -3 and 3 inclusive}

• Use an identifier (the letter x for example) to symbolize a typical element, a '|' symbol to stand for the axiom such that', and then the rule or rules that the identifier must follow: {x | x is an integer and |x| < 5}

The Greek letter ∈ is used as follows:

• ∈ means 'is an element of ...'. For example: 3 ∈ {positive integer}

• ∉ means 'is not an element of ...'. For example: Washington DC ∉ {European capital cities}

• The set is a finite: {British citizens}

                      Or

•  infinite: {6, 12, 21, 24, 35, ...}

Sets are usually be represented using upper case letters: A, B,X,Z ...

venn diagram example of disjoint - Example

The following is the diagram representation of disjoint in venn diagram.
Two sets are equally exclusive also called disjoint. If do not have any elements in common and need not together contain the universal set.
The following venn diagram represents the disjoint sets.

Set of all elements of A is also known as difference of set A-B, which do not go to B. In the set planner form, the difference set is :   

 

Example problem for disjoint set.
A={2,3,4,1,8,9}   B={2,3,4,1,10,12} What is the A-B and B-A?
Solution:
A-B=?
Given A={2,3,4,1,8,9}
            B={2,3,4,1,9,10,12}
Here all elements of A  an available in B except 9.
So the A-B is 9.
B-A=?
Here all elements of A  an available in A except 12.
So the B-A is 12.

Line Intercept Sampling

In this article, we will discuss about the line intercept sampling. Sampling means method of selecting sample. The line intercepts have two types, one is x intercept and next one is y intercept. X intercept means that, the point crosses the x -coordinates or axis and y intercept means point crosses the y- axis of the line. The slope intercepts form of line y = mx + b, where m is slope of the line, and b - the y intercept. Let us learn about the line intercept sampling example problems are given below.

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Example problems for line intercept sampling:

Example problem 1:

Find the x intercept of the line, 11.5x + 13.5y = 27

Solution:

The slope intercept form of line y = mx + b, where m is the slope of the line

Given equation is in the form of ax + by = c. To find the x- intercept Plug y = 0 in the equation

Here x intercept, so y = 0

11.5x + 13.5(0) = 27

11.5x = 27

Divide by 11.5 on both sides.

` (11.5x)/(11.5)` = ` (27)/(11.5)`

After simplify this, we get

x = 2.34

x intercept = (2.34, 0)

Example problem 2:

Find the y intercept of the line, 4x + 24y = 72

Solution:

The slope intercept form of line y = mx + b, where m is the slope of the line

Given equation is in the form of ax + by = c. To find the y- intercept Plug x = 0 in the equation

Here y intercept, so x = 0

4(0) + 24y = 72

0 + 24y = 72

24y = 72

Divide by 24 on both sides.

`(24y)/(24)` = ` (72)/(24)`

After simplify this, we get

y = 3

y intercept = (0, 3).

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More example problems for line intercept sampling:

Example problem 3:

Find the y intercept of the line, 5x + 30y = 120

Solution:

The slope intercept form of line y = mx + b, where m is the slope of the line

Given equation is in the form of ax + by = c. To find the y- intercept Plug x = 0 in the equation

Here y intercept, so x = 0

5(0) + 30y = 120

0 + 30y = 120

30y = 120

Divide by 30 on both sides.

`(30y)/(30)` = ` (120)/(30)`

After simplify this, we get

y = 4

y intercept = (0, 4)

The above examples are helpful to learn of line intercept sampling.

Alternating Current Generator

The household electricity that we used is based on the principle of Alternating Current. One advantage of Alternating Current is that it does not induce fatal shock. Also it has an added advantage of easy voltage amplification. In this article we shall discuss the Alternating Current Generator which is often abbreviated as A.C. Generator.

Understanding Alternating Exterior Angles is always challenging for me but thanks to all math help websites to help me out. 

Introduction to A.C. generator

The phenomenon of electromagnetic induction has been technologically exploited in many ways. An exceptionally important application is the generation of alternating current (A.C.).

Principle of ac generator

One method to induce an emf or current in a loop is through a change in the loop orientation or a change in it’s effective area. As the coil rotates in a magnetic field B , the effective area of the loop (the face perpendicular to the field) is AcosӨ, where Ө is the angle between A and B. This method of producing a flux change is the principle of operation of a simple ac generator.

A. C. Generator : Construction and Working

It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil called the armature is mechanically rotated in the uniform magnetic field by some external means . The rotation  of the coil causes the magnetic flux through it to change, so an emf is induced in the coil . The ends of the coil are connected to an external circuit by means of slip rings and brushes.

Working
When the coil is rotated with the constant angular speed of w , the angle Ө between the magnetic field vector B and the area vector A of the coil at any instant t is Ө=wt(assuming Ө=0⁰at t=0). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and the flux at any time is ф_B=BAcosӨ=BAcoswt From faraday’s law, the induced emf for the rotating coil of N turns is then,  E=-N×dф_B/dt = -NBA×dcoswt∕dt thus, the instantaneous value of the emf is  E= NBAwsinwt.Where NBAw is the maximum value of the emf,which occurs when sinwt= +1 or -1. If we denote NBAw as E₀, then E=E₀sinwt since the value of the sine function varies between +1 or -1, the sign, or polarity of the emf changes with time. 

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The emf has its extremum value when Ө=90⁰ or Ө=270, as the change of flux is greatest at these points. The direction current changes periodically and therefore the current is called alternating current.
The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine.

Polynomials gcf Calculator

Polynomials gcf calculator is one of the interesting topics in mathematics. It is the process of performing greatest common factor for the given polynomial expression. It is the sums of a finite number of monomials are called as polynomial. Polynomial has more than one term and it has a constant value for the given each term, for that variable power of integral is raised to more than two.

Example for Polynomial expression is a2 – 26a – 28.

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Definition of Greatest common factor calculator:

Greatest common factor (gcf):

Greatest common factor is defined as the process of the highest number which divides more than two numbers or terms exactly.

Steps to find the Greatest common factor:

Step 1:

Given polynomial expression can be arranged in the order of powers

Step 2:

Each term in the given expression can be factored.

Step 3:

Find the common factors in each terms

Step 4:

Take the greatest common factor

Step 5:

Simplify the each term.

Example problem for polynomials gcf calculator:

Some example problems for polynomials gcf calculator are,

Example 1:

Find the gcf for the given Polynomial expression 3x2 – 9x

Solution:

Step 1:

3x2 – 9x

Step 2:

3 . x . x – 9 . x

Step 3:

In the given expression, common factors in the each term is 3x

3 . x . x – 3 . 3 . x

Step 4:

Take the common term outside

3x ( x – 3)

Step 5:

Solution to the given polynomial gcf is 3x (x – 3)

Example 2:

Find the gcf for the given Polynomial expression 5x2y5 – 20x4y3

Solution:

Step 1:

6x2y5 – 24x4y3

Step 2:

6 . x . x . y . y . y . y . y – 24 . x . x . x .x . y . y . y

Step 3:

In the given expression, common factors in the each term is 6x2y3

6 . x . x . y . y . y – 6 . 4 . x . x . x . x . y . y . y

Step 4:

Take the common term outside

6x2y3 ( y2 – 4x2)

Step 5:

Solution to the given polynomial gcf is 6x2y3 ( y2 – 4x2)

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Example 3:

Find the gcf for the given Polynomial expression 15x3y4 – 45x5y6

Solution:

Step 1:

15x3y4 – 45x5y6

Step 2:

15 . x . x . x . y . y . y . y  – 45 . x . x . x .x . x . y . y . y . y . y . y

Step 3:

In the given expression, common factors in the each term is 15x3y4

15 . x . x . x . y . y . y . y – 15 . 3 . x . x . x . x . x . y . y . y . y . y . y

Step 4:

Take the common term outside

15x3y4 ( y – 3x2  y2)

Step 5:

Solution to the given polynomial gcf is 15x3y4( y – 3x2  y2 )

How to add Trinomials

Trinomials, a function is in the structure of ax2+bx+c =0 (where a≠0, b, c are constants). Quadratic function or quadratic equation also called  trinomials. An algebraic expression which has 3 terms known as trinomials. The product of two binomials gives a trinomial and there are two solutions for the given trinomial. To sum trinomial first it needs to combine the liked terms and then solve it. Since, the trinomial is a combination of two terms and the sum of trinomials also has three terms.

Examples problem for add trinomials:

1. Add trinomials for given function is 2x + 3y + 4z and 6x + 4y +2z.

   Solution:
   Given functions is,
    2x + 3y + 4z and 6x + 4y +2z,

Step 1:
    Write the given trinomials as,
                       = 2x + 3y + 4z + 6x + 4y +2z

Step 2:
    Then combine the terms like as,
                       = 2x + 6x + 3y + 4y + 4z + 2z

Step 3:
    Then add the given trinomials, and we get the answer,
             Answer = 8x +7y + 6z.

2. Add trinomials for given function is 8x - 2y - 9z and 3x + y +2z.

    Solution:
    Given functions is,
     8x - 2y - 9z and 3x + y +2z,

Step 1:
    Write the given trinomials as,
                       = 8x - 2y - 9z + 3x + y +2z

Step 2:
    Then combine the terms like as,
                   = 8x + 3x - 2y + y - 9z + 2z

Step 3:
    Then add the given trinomials, and we get the answer,
            Answer = 11x - y - 7z.

Examples problem for add trinomials:

3. Add trinomials for given function is 3x² + 6y² + 4z² and x² + 5y² + 6z².
    Solution:
    Given functions is,
    3x² + 6y² + 4z² and x² + 5y² + 6z²,
Step 1:
    Write the given trinomials as,
                       = 3x² + 6y² + 4z² + x² + 5y² + 6z²
Step 2:
    Then combine the terms like as,
                       = 3x² + x² + 6y² +5y² + 4z² + 6z².
Step 3:
    Then add the given trinomials, and we get the answer,
            Answer = 4x² + 11y² + 10z².

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4. Add trinomials for given function is x + y + 4z and 6x + 4y +2z.
    Solution:
    Given functions is,
    x + y + 4z and 6x + 4y +2z,
Step 1:
    Write the given trinomials as,
                        = x + y + 4z + 6x + 4y +2z
Step 2:
    Then combine the terms like as,
                        = x + 6x + y + 4y + 4z + 2z
Step 3:
    Then add the given trinomials, and we get the answer,
             Answer = 7x + 5y + 6z.

Sine Cosine Cotangent

Sine:

          A trigonometry functions of an angle. The sine of an angle theta shortened as sin theta In a right angled triangle is the ratio of the side opposite angle to the hypotenuse. This definition applies only of angles between 0 to 90 (0 and `pi/2 ` radians).

Sin `theta` = Opposite / hypotenuse  = `(BC)/(AC)`

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Cosine:

A trigonometry function of an angle.The cosine of an angle `theta` abbreviated as cos `theta` In a right angled triangle is the ratio of the side adjacent to the hypotenuse.

Cos `theta` = Adjacent / hypotenuse = `(BC)/(AC)`

cotangent:

A trigonometry function of an angle.The cotangent of an angle `theta` ( cot  `theta` ) in a right angled triangle is the ratio of the side adjacent to it to the opposite side.
Cot `theta` = adjacent / opposite = `(BC)/(AB)`

Example problem for sin

Find the measure of the length of other sides and also find the sin function values for the given right angle triangle.

we want to find the length of side c, the hypotenuse.

Here, we know that side a has a length of 8 and side b has a length of 6.
To find the length of side c, we can use the Pythagorean Theorem which says that c²=a²+b², or
Substitute in that a=8 and b=6, we find that:

c = √ (( 8²) + (6²))

  = √ (64 + 36)

  = √ 100

c = 10 m

So the value of x is found as x = 10 m
Now we have to find the value of `theta` . we can use the sin function to find the value of `theta`
Sin `theta`     = Opposite / hypotenuse  
          = 6/10
          = 0.6
Sin `theta`  = 0.6
  `theta` = sin^-1 (0.6)
`theta`  = 37<sup>o


Algebra is widely used in day to day activities watch out for my forthcoming posts on Statistics Hypothesis Testing and Integers Number Line. I am sure they will be helpful.</sup>

Example problem for cosine, cotangent:

Find the cosine and cotangent function of the given right angled triangle.

Solution:
Here we have to find the cosine and cotangent of the given right angled triangle
Cosine`theta` = Adjacent / hypotenuse
               =  4 /  5

= 0.8

cos `theta`  = 0.8
`theta` = cos^-1 (0.8)

      = 36^o
Cotangent `theta` = adjacent / opposite =  4/3

= 1.33

cot`theta`  = 1.33
`theta` = cot^-1 (1.33)