Parallelogram with Four Equal Sides

In geometry, a parallelogram is a quadrilateral with the two pairs of parallel sides. In Euclidean Geometry, the opposites or facing sides of a parallelograms are of equal length and the opposite angles of a parallelogram are of equal measure. The congruences of opposite sides and opposite angles are a direct consequence of the Euclidean Parallel Postulate and neither condition can be proven without appealing to the Euclidean Parallel Postulate or one of its equivalent formulations. The three-dimensional counterparts of a parallelogram is a parallelepiped.(Source.Wikipedia)

Parallelogram with Equal Sides are Square and Rhombus:

Parallelograms:

Area of parallelograms A = b *  h   sq. units



where h is the perpendicular height of the parallelogram.

b is the base length of the parallelogram.

Examples for Parallelogram with Equal Sides:

Example 1:

Find the area of the rhombus whose base is 150cm and the perpendicular height will be equal to 150cm.

Given:

H=150cm

B=150cm

Solution:

Area = b*h

= 150*150

= 22500cm2

Example 2:

Find the Area of a Parallelogram with a base of 12 centimeters and a height of 15 centimeters.

Solution:

A=b*h

A= (12 cm) · (15 cm)

A= 180 cm2

Example 3:

Find the area of a parallelogram with a base of 8 inches and a height of 14 inches.

Solution:

A=b*h

A= (8 in) · (14 in)

A= 112 in2

Example 4:

The area of a parallelogram is 30 square centimeters and the base is 60 centimeters. Find the height.

Solution:

A=b*h

30 cm2 = (60 cm) · h

30 cm2 ÷ (60 cm) = h

h= 0.5 cm

Example 5:

The area of a parallelogram is 100 square centimeters and the base is 50 centimeters. Find the height.

Solution:

A=b*h

100 cm2 = (50 cm) · h

100 cm2 ÷ (50 cm) = h

h= 2 cm.

These are the examples on parallelogram with four equal sides.

Fundamental Theorem of Integral Calculus

In this section we are going to discuss about the fundamental theorem of integral calculus concept. The development of short form the fundamental theorem of integral calculus corresponding to over screening the significant concept and problem in using calculus theorems are referred as review calculus. This article helps to improve the knowledge for using fundamental theorem of integral calculus problem and below the problems are helping toll for the exam. Fundamental theorem of integral calculus problem solutions also shows below. The fundamental theorem of integral calculus handled the differentiation, integration and inverse operations are process here now.

Important of Fundamental Theorem of Integral Calculus:-

Let` f(x)` is a continuous function on the closed interval `[a, b]` .

Let the area function `A(x)` be defined by `A(x) = int_a^xf(x)dx ` for` xgt=a`

Then `A'(x) = f(x)` for all `X in [a,b]`

Let` f(x)` be a continuous function defined on an interval `[a,b]` .

`If intf(x)dx = F(x) then int_a^a f(x)dx = [F(x)]^b_a`

`=F(b) - F (a)` is called the definite integral or `f (x) ` among the limits` a` and `b` .

This declaration is also known as fundamental theorem of calculus.

We identify `b` the upper limit of `x` and a the lower limit.

If in place of `F(x)` we take` F(x) +c` as the value of the integral, we have

`int_a^b f(x)dx = [F(x) + C ]^b_a`

`= [F (b) + c] - [F (a) + c]`

`= F (b) + c - F (a) - c`

`= F (b) - F (a)`

Therefore, the value of a definite integral is unique. It does not depend on the constant c and hence in the evaluation of a definite integral the constant of integration does not play any role.

Let `int f(x) dx = F(x) +C`

Then `int_a^b f(x)dx = F(b)-F(a)`

Note down:-

From the above two theorem, we infer the following

`intf(x)dx` =(Anti derivative of the function `f(x) ` at `b` ) - (Anti derivative of the function `f(x) ` at `a` )

The fundamental theorem of integral calculus gives you an idea about a close relationship between differentiation and integration. These theorems give an exchange method evaluating definite integral, without calculating the limit of a sum.


Example on Fundamental Theorem of Integral Calculus:-

Evaluate the definite integral of the following

`int^(pi/4)_0(3sec^2 x + x^2 + 3)`

Solution:

`int_0^(pi/4) (3sec^2 x + x^2 + 3)`

`= 3 int_0^(pi/4)sec^2 xdx + int_0^(pi/4)x^2dx + int_0^(pi/4)3dx`

`= 3[tanx] ^ (pi/4) _0 + [(x^3)/ (3)] ^ (pi/4) _0 + [3x] ^ (pi/4) _0`

`= 3 (tan (pi/4) - tan0) + 1/3 (pi/4) ^3 - 0+ [(pi/2) - 0]`

`= 3 + (pi^3)/192+ (pi)/ (2).`

Cumulative Probability

In probability theory the cumulative distribution function (CDF) or just distribution function, completely describes the probability distribution of a real valued random variable X. Cumulative distribution functions are used to specify the distribution of multivariate random variables.

For every real number x, the CDF of real valued random variable is X its given by
                              f(x)=  P[X <= x]

where the function right hand side is represents the probability of  the random variable X takes on the value less than or equal to x. The probability of  X is lies in the interval (a, b) is therefore F_X (b) − F_X (a).

Cumulative Probability Example:

Consider a coin flip experiment. If we flip a coin two times, we might ask that what is the probability that the coin flips would result in one or fewer heads? The answer would be a cumulative probability. It would be the probability when that the coin is flip results in zero heads plus the probability that the coin flip results in one head. Thus, the cumulative probability would equal:

P(X < 1) = P(X = 0) + P(X = 1) = 0.25 + 0.50 = 0.75

The table below shows that the both of the probabilities and the cumulative probabilities associated with this experiment.

Number of heads              Probability       Cumulative Probability

0                                            0.25                                0.25

1                                             0.50                                0.75

2                                             0.25                                1.00

 Assume we have a random variable X. Cumulative probabilities that are provide for each value x, the probability of a result less than or equal to X, P[X <= x].

Example:

Here's the probability distribution and for a discrete random variable X
X                  f(x)
1                   0.1
2                   0.2
3                    0.4
4                    0.3
The cumulative distribution function tables and for each value x = 1, 2, 3, 4, the probability of a result less than or equal

For example:
* P[ X <= 1 ] = 0.1

* P[ X <= 2 ] = 0.1 + 0.2 = 0.3

* P[ X <= 3 ] = 0.1 + 0.2 + 0.4 = 0.7

* P[ X <= 4 ] = 0.1 + 0.2 + 0.4 + 0.3 = 1

 These probabilities can be tabled

 X                    P[X <= x]
 1                        0.1
 2                        0.3
 3                        0.7
 4                        1.0

Decrease in Percentages

To understand this concept, let us start with an agenda or a plan to go further which will make us stay organized all through the explanation. The explanation will begin with the introduction to the concept followed by the introduction to the concept of percent-ages, which will be followed by the change in the same and then the polarity or the direction of the change will be seen, then what do we with just finding out the solution the next step will obviously be the interpretation part which requires the complete understanding of the concept.

The concept which will be dealt is the same given in the header. The change can always be an increase or a Percentage Decrease; it can be either way round. So it is necessary for us to learn both the method of calculations, actually both require the same formula which is the Percent Decrease Formula, the only difference will be the final answer, that too in the interpretation part, since obviously without understanding the number one can never say if it is an increase or a decrease in the change occurred. Before starting with, one has to be clear with the concepts of pct to solve the questions easily without much strain.

Let us now try to answer the question of how to Calculate Percentage Decrease. Answering this question is never a great task, it is all about the numbers we get for solving, and one can easily solve such type of questions, if one understands the problem and interprets the Percent Decrease. Let us consider an example to get a better understand Calculating Percent Decrease of the numbers whose initial value that is the original value is 50 and the final value is 25; the answer for this question is 50 – 25 / 50 * 100 = 50%, the answer shows us that there is an 50 % decrease in the final value from the original value and that shows the decrease.

The fact is that anyone can solve such problems of increase or decrease, but the interpretation part can only be done by the person who completely understood the concept. The final interpretation shows us the direction of change that is it shows whether the change is positive or negative and increase or decrease in the number in terms of percentages which makes the concept complete with the final interpretation.

Rational Equations Practice

An equation which has rational expression is known as rational equation. A fraction is also termed as a rational equation. By certain steps we can solve the rational equations. We can practice many problems through online. We are going to see some problems to practice the rational equations.

Example: `1/x` + `5/x` = `7/9` is a rational expression called rational equation.

Explanation to Rational Equations Practice

The following steps are helpful for solving the Rational Expressions.

Step 1:

If possible factor the denominator of every rational expression term in rational equation.

Step 2:

Be sure that the rational expressions in rational equation has same denominator. If not, make it by taking LCM for all rational expression.

Step 3:

Now cancelling all the common terms and simplifying for final answer.

Example Problems to Rational Equations Practice:

Example: 1

Solve: `1/6` = `x/8`

Solution:

Given,

`1/6` = `x/8`

Step 1:

1 × 8 = 6 x

Step 2:

6 x = 8

x = `8/6`

x = `4/3`

Answer: x = `4/3`

Example: 2

Solve:  `x/8` + `5/8` = `7/8`

Solution:

Given: `x/8` + `5/8` = `7/8`

Step 1:

The fractions `x/8` , `5/8` and `7/8` having a common factor 8 in their denominator.

Step 2:

`x/8` + `5/8` = `7/8`

`(x + 5)/8` = `7/8`

Step 3:

Cancel a common factor 8 on both sides of above rational equation.

x + 5 = 7

x + 5 - 5 = 7 - 5

x = 2

Answer: x = 2


Example: 3

Solve: `6/x` + `8/(x + 3)` = `(10)/(x^2 + 3x)`

Solution:

Given,on: `6/x` + `8/(x + 3)` = `(10)/(x^2 + 3x)`

Step 1:

Factor the last rational expression `10/(x^2 + 3x)` = `10/(x(x +3))`

Step 2:

LCM of the all the terms in given rational equation is x(x + 3)

`6/x` x `(x + 3)/(x + 3)` = `(6(x + 3)) /(x(x + 3))`

`8/(x+ 3) ` x `x/x` = `(8x)/(x(x+3))`

`(10)/(x^2 +3x)` x `1/1` = `(10)/(x(x + 3))`

Step 3:

Now the rational equation is,

`"(6(x + 3))/(x(x+3))` + `(8x)/(x(x+3))` = `(10)/(x(x+3))`

`(6x + 18 + 8x)/(x(x+3))` = `(10)/(x(x+3))`

Step 4:

By cancelling the common x(x + 3) we get,

6x + 18 + 8x = 10

14x + 18 = 10

14x = 10 - 18

14x = -8

x = -`8/14`

x = `-4/7`

Answer: x = `-4/7`.

Practice Problem to Rational Equations Practice:

Problem: 1

Solve: `6/x` = `5/6`

Answer: x = `36/5`

Problem: 2

Solve the rational equation, `(x -9)/5` = `21/9`

Answer: x = `62/3`

Decimals least to Greatest

We have learnt the fractions in the lower class mathematics. Now we shall study about special fractions whose denominators are 10, 100, 1000 etc. These fractions are called decimal fractions. Decimals least to greatest is also called as decimals in ascending order (Lower value to higher value)

Let us these fractions in a new way. That is

1/10 is understand writing one-tenth and is correspond to in decimal fraction as 0.1

1/100 is understand writing as one-hundredth and is correspond to in decimal fraction as 0.01

1/1000 is understand writing as one-thousandth and is correspond to in decimal fraction as 0.001

Steps for Calculating Decimals least to Greatest:

Step 1: Look the whole number first, and then look at the tenth, then the hundredth place and so on.

Step 2: Find out which number is smallest

Step 3: Arrange the decimals from least to greatest

For example, arrange the following numbers from least to greatest

0.56, 0.102, 0. 272, 0.5

Ans:  0.102, 0.272, 0.5, 0.56

Examples on Decimals least to Greastest:

Let us see some examples of decimal least to greatest:

Ex 1:  Find order of the decimals from least to greatest

5.6, 2.5, 1.2, 6.5, 0.6 and 0.2

Sol :   Step 1: First look at the tenth, and then the hundredth place so on. Next find the compare the two decimal numbers.

6.5 > 5.6

5.6 > 2.5

2.5 > 1.2

1.2 > 0.6

0.6 > 0.2

Step 2:  Arrange the decimals from least to greatest.

0.2, 0.6, 1.2, 2.5, 5.6 and 6.5

The above values are the arranged values from least to greatest.

Ex  2:   Find order of the decimals from least to greatest

5.7, 57, 0.71, 0.26

Sol :  Step 1: First look at the tenth, and then the hundredth place so on. Next find the compare the two decimal numbers.

56 > 5.7

5.7 > 0.71

0.71 > 0.26

Step 2:  Arrange the decimals from least to greatest.

0.26, 0.71, 5.7, 57

The above values are the arranged values from least to greatest.

Value Weighted Average

Arithmetic average is an average in which is defined as the summation of all the given elements divided by the total number of elements. In this all the values get same weighted.

Ex: Find the arithmetic average of 7, 16, 20, 57 and 60.


Arithmetic average = `("sum of all the elements")/("total number of elements")`

sum of all the given elements = 7 + 16 + 20 + 57+ 60 = 160

total number of elements = 5

Arithmetic average = `(160)/(5)`  = 32

so  arithmetic average = 32

In the above example we weighted to all the values.

Weighted average is same as that of arithmetic average but the only difference is that each of the element have assigned different weighted in the data given. The above example changes like that

Ex: Find the weighted  average of 7, 16, 20, 57 and 60, whose occurrences (weighted) values are 4, 3, 6, 2 and 1.

Solution is given after definition and formula

Definition of Weighted Average Value

Weighted average is defined as the summation of the element values multiplied with the occurrences (allocated weighted) which is divided by the summation of the occurrences values. This is called Value Weighted Average.

Formula for the weighted average is given by,

where

`barx_(w) ->`weighted average value

`w_(i) ->`allocated weighted value(occurrences) for the given element.

`x_(i) ->`element value

Steps to calculate weighted average value:

Step 1: Calculate the summation of the element values multiplied with the occurrences (allocated weighted).

Step 2: Calculate the summation of the occurrences values.

Step 3: Now division the value obtain by step 1 with step2 and get weighted average value.

Examples on Weighted Average Value

Ex 1: Find the weighted average of 7, 16, 20, 57 and 60 whose occurrences (weighted) values are 4, 3, 6, 2 and 1.

Sol :    Formula for the weighted average is

Step 1:     `sum_(n=1)^5(w_(i)x_(i))` =  (4) (7) + (3) (16) + (6) (20) + (2) (57) + (1) (60)

= 370

Step 2:         `sum_(n=1)^5 (x_(i))`   =  4 + 3 + 6 + 2 + 1

= 16

Step 3:     weighted average value = `barx_(w)` =   `(370)/(16)`

= 23.125                    Ans

Ex 2: Find the weighted average of 7, 2, 8 and 10, whose occurrences (weighted) values are 3, 2, 1 and 1.

Sol : Formula for the weighted average is

Step 1:   `sum_(n=1)^4(w_(i)x_(i))`  (3) (7) + (2) (2) + (1) (8) + (1) (10)

= 43

Step 2:      `sum_(n=1)^4(w_(i))`  3 + 2 + 1 + 1

= 7

Step 3:       weighted average value = `barx_(w)`   =  `(43)/(7)`

Ans =   6.142