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.

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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)

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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)

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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

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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

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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

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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.

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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


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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.

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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