Derivatives Related Rates

Derivative related rates is the normal derivatives whereas the differentiation is carried out with respect to the time function t. The Differentiation of a given function with respect to time is called related rates derivatives. The related rates derivatives are also one of the parts of calculus which deals with calculating the rate of change of function with respect to time. The following are the example problems for related rates derivatives.

Related Rates Example Problems:

Example 1:

Find the related rate derivatives for the given function.

f(t) = t 4 – 18t + 16

Solution:

The given function is

f(t) = t 4 – 18t + 16

The first derivative f ' is given by

f '(t) = 4 t 3 – 18

Example 2:

Find the related rate derivatives for the given function.

f(t) = t 5 – 6 t 3  + 11

Solution:

The given function is

f(t) = t 5 – 6 t 3  + 10

The first derivative f ' is given by

f '(t) = 5t 4 – 6(3 t 2 )

f '(t) = 5t 4 – 18 t 2

Example 3:

Find the related rate derivatives for the given function.

f(t) = t2 – 4t + 8

Solution:

The given function is

f(t) = t 2 – 4t + 8

Differentiate the above equation with respect to t.

f '(t) = 2 t  – 4

Example 4:

Find the related rate derivatives for the given function.

f(t) = t 3 – 5 t 2  + 11t

Solution:

The given function is

f(t) = t 3 – 5 t 2  + 11t

The first derivative f ' is given by

f '(t) = 3t 2 – 5(2 t  ) + 11

f '(t) = 3t 2 – 10 t + 11

Example 5:

Find the related rate derivatives for the given function.

f(t) = t4 – 3t 3 – 4t 2  + t

Solution:

The given function is

f(t) = t4 – 3t 3 – 4 t 2  + t

The first derivative f ' is given by

f '(t) = 4 t 3 – 3(3t 2 ) – 4( 2 t  ) + 1

f '(t) = 4 t 3 – 9t 2  – 8 t  + 1

Related Rates Practice Problems:

1) Find the related rate derivatives for the given function.

f(t) = t 3 – 6 t 2  + 11t

Answer: f '(t) = 3t 2 – 12 t

2) Find the related rate derivatives for the given function.

f(t) = t 2 – 6 t   + 11

Answer: f '(t) = 2t – 6

Construct Probability Table

Construct Probability table is defined as an equation or table with its probability happenings. Usually, probability table is for constant occurrence for much number of data produces results. For construct probability table function we learn about discrete random variable, if a random variable obtains only a finite or a countable number of values, it is known as discrete random variable. To construct Probability table it contains probability mass function and moments.

Classification of Construct Probability Table:

Probability Mass Function:

The statistical description of discrete probability table function p (x)  functions and satisfies the following properties:

(1.) The probability which x can take a particular value x is p (x)

That is P(x = x) = p (x) = px.

(2). P(x) is a non – negative for every real x.

(3). The amount of p (x) over all likely values of x is one. Which is Σpi = 1 here j denotes that x and pi is the probability at x = xi

Moments:

A probable value of a function of a chance variable x is used for calculating the moments. Let us see the two types of moments.

(i) Moments about the origin.
(ii) Moments about the mean that are known as central moments.

Example Problem for Construct Probability Table:

A box has 4 green and 3 black pens. Construct a probability table distribution of number of black pens in 3 draws one by one from the box. (i) With replacement

Solution:

(i) With replacement

Let x be the approximate variable of drawing number of black pens in three draws.

X will takes the values 0,1,2,3.

P (Black pen) = `3 / 7 ` = P (B)

P (Not Black pen) = `4 / 7` = P (G)

Hence P(X = 0) = P (GGG) = `4/ 7` * `4 / 7` * `4 / 7` = `64 / 343`

P(X = 1) = P (BGG) + P (GBG) + P (GGB)

=  ( `3 / 7` * `3 / 7` * `4 / 7` )  + ( `4 / 7` * `3 / 7`   * `4 / 7` ) +  ( `4 / 7` * `3 / 7`   * `3 / 7` )

= 3 * (`48 / 343` )

= `144 / 343 `

P(X = 2) = P (BBG) + P (BGB) + P (GBB)

= (`3 / 7` * `3 / 7` * `4 / 7` ) + (`3 / 7` * `4 / 7` * `3 / 7` ) + (`4 / 7` * `3 / 7` * `3 / 7` )

= 3 * ( `3 / 7` ) * ( `3 / 7` ) * ( `4 / 7` )

= 3 * (`36 / 343` )

= `108 / 343`

P(X = 3) = P (BBB) = ( `3 / 7` ) * ( `3 / 7` ) * ( `3 / 7` )

= `27 / 343`

Therefore the wanted probability table distribution is

x	0	1	2	3
p (x =x)	`64 / 43`	`144 / 343`	`108 / 343`	`27 / 343`

Example 2:

Find out the probability mass function, and the collective distribution function for obtaining ‘3’s while two dice are thrown.

Solution:

2 dice are thrown. Let x be the approximate variable of obtaining number of ‘3’s. Hence x can take the values 0, 1, 2.

P (no ‘3’) = P (X = 0) =` 25 / 36`

P (one ‘3’) = P (X = 1) = `10 / 36`

P (two ‘3’s) = P (X = 2) = `1 / 36`

Obtained Probability mass function is

x	0	1	2
P (x = x)	`25 / 36`	`10 / 36`	`1 / 36`

Limit Point of a Sequence

In this article, we will discuss the limit point of a sequence. A set of numbers said to be a limit point of a sequence. It has two types of sequences.

1. Arithmetic sequence and

2. Geometric sequence.

Arithmetic sequence means that, the sequence of a numbers such that the difference between two consecutive members of the sequence is a constant. Geometric sequence means that, the sequence of a numbers such that the ratio between two consecutive members of the sequence is a constant. The limit point of a sequence formulas and example problems are given below.

Formulas and Example Problems for Limit Point of a Sequence

Sequences formulas are given below.

Formula for arithmetic sequence:

nth term of the sequence : an = a1 + (n - 1)d

Series of the sequence: sn = `(n(a_1 + a_n))/2 `

Formula for geometric sequence:

nth term of the sequence: an = a1 * rn-1

Series of the sequence: sn = `(a_1(1-r^n))/(1 - r)`


Example problem 1:

Find the 11th term of the given series 11, 12, 13, 14, 15,......

Solution:

First term of the series, a1 = 11

Difference of two consecutive terms, d = 12 - 11 = 1

n = 11

The formula to find the nth term of an arithmetic series, `a_n = a_1 + (n-1)d`

So, the 11th term of the series  11, 12, 13, 14, 15,... = 11 + (11 - 1) 1

= 11 + 10 * 1

= 11 + 10

After simplify this, we get

= 21

So, the 11th term of the sequence 11, 12, 13, 14, 15,... is 21.

More Example Problems for Limit Point of a Sequence

Example problem 2:

Find out the 5th term of a geometric sequence if a1 = 70 and the common ratio (C.R) r = 2

Solution:

Use the formula `a_n = a_1 * r^(n-1)` that gives the nth term to find `a_5` as follows

`a_5 = a_1 * r^(5-1)`

= 70 * (2)4

= 70 * 16

After simplify this, we get

= 1120.

The 5th term of a geometric sequence is 1120.

The above examples are helpful to study of limit point of a sequence.

Parts of Quadrilateral

Quadrilateral is a type of polygon with four sides and four vertices's or corners. The quadrilaterals are whichever convex or concave. The entire convex quadrilateral covers the plane by continual revolving around the midpoints of their ends.

Parts of Quadrilateral

• Sides
• Vertices's
• Interior angles
• Diagonals
• Adjacent sides
• Opposite angles
• Opposite sides
• Consecutive angles

Parts of Quadrilateral-side

The quadrilateral is of many types, special cases are tangent quadrilaterals and cyclic quadrilateral. The sides of the tangent quadrilateral are opposite and have equal length, in cyclic quadrilateral the product of the opposite sides are the same, are called a harmonic quadrilateral.

Vertices's

Vertex is the two side’s meets at the end point. They are four vertices's on a quadrilateral. The consecutive vertices's are same at the endpoints.

Diagonal

The Line segment that joins two vertices's in opposite. In Parallelogram the diagonal bisect each other. In Rectangle the diagonals are congruent. In kite, one diagonal is perpendicular bisector of other, and the diagonal bisects a pair of opposite angles. In Rhombus, the diagonals bisect the angles and are perpendicular bisector of each other; the diagonals divide the rhombus into four congruent right triangles.

Interior angle of quadrilateral:

Interior angle of quadrilateral  is the set of all points in its plane which lie in between both the arms. The addition of the interior angles of a non-crossed quadrilateral is 360.  In a crossed quadrilateral, the addition of the one side of  interior angles equals the addition of the interior angles on the other side.

Adjacent sides

Consecutive sides or adjacent sides have a common endpoint. In kite two dissimilar pairs of adjacent sides are similar.

Opposite angles

Opposite angles are angles whose vertices's are not successive.

Opposite sides:

Opposite sides of a quadrilateral are sides that do not have a general endpoint. In parallelogram and in rectangle the opposite sides are parallel and  congruent.

Consecutive angles

In parallelogram, any pair of consecutive angles are supplementary.In Rhombus two sides are congruent

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

Volume Divided by Area

Volume is how much three-dimensional space a substance (solid, liquid, gas, or plasma) or shape occupies or contains, often quantified numerically using the SI derived unit, the cubic meter.

Area is a quantity expressing the two-dimensional size of a defined part of a surface, typically a region bounded by a closed curve. The surface area of a 3-dimensional solid is the total area of the exposed surface. (Source: From Wikipedia).

Use of Volume Divided by Area:

The term volume divided by area is used to find the a dimension of a solid.

For example,

In a sphere, volume divided by area gives the radius of the sphere.

Volume of sphere = `4/3 pi r^3`

Surface area of sphere = `4 pi r^2`

Volume divided by area = `(4/3 pi r^3)/(4 pi r^2)`

= `r/3`

In a cube, volume divided by area gives the side of the cube,

Volume of cube = a^3

Surface area of cube = 6a^2

Volume divide by area = `(a^3)/(6a^2)`

= `a/6`

In a cylinder, volume divided by area gives the radius of the cylinder

Volume of cylinder = `pi r^2 h`

Surface area of cylinder = `2 pi r h`

Volume divided by area = `(pi r^2 h)/(2 pi r h)`

= `r/2`

Example Problems for Volume Divided by Area:

Here, we are going to see some example problems to find the dimensions of solid shapes using volume divided by area.

Example 1

Find the side of a cube whose volume is 125 cubic meter and the surface area is 150 square meter.

Solution

For a cube, Volume divide by area = `a/6` = `125/150`

`a/6` = `5/6`

a = 6

So, the side of the given cube is 5 meters.

Example 2

Find the radius of a sphere whose volume is 2393.88 cubic centimeter, and the surface area is 865.26 square centimeter.

Solution

We know that, the volume divided by area in a sphere gives `r/3`

Here, volume of the sphere = 2393.88 cubic centimeter

Surface area of the sphere = 865.26 square centimeter

Volume divided by area; `r/3` = `(2393.88)/(865.26)`

r = `(2393.88)/(865.26)` * 3

r = 8.3

So, the radius of the given sphere is 8.3 centimeter.

Example 3

The volume and lateral surface area of a cylinder are 565.2 cubic feet and 188.4 square feet respectively. Find the radius of the cylinder.

Solution

We know that, the volume divided by area in a sphere gives `r/2`

Volume of the cylinder = 565.2 cubic feet

Lateral area of cylinder = 188.4 square feet

Volume divided by area; `r/2` = `(565.2)/(188.4)`

r = `(565.2)/(188.4)` * 2

r = 6

So, the radius of the cylinder is 6 feet.

Solving Trigonometric Examples

Answering the trigonometric examples is nothing but we are solving the trigonometric functions and trigonometric identities. Here we will take trigonometric functions and equations. Trigonometric examples contain trigonometric functions. In trigonometric model we will solve the Sin, Cos, Tan Identities. We can find the angles from these identities. We will solve some trigonometric examples.

Explanation for Solving Trigonometric Examples:

Ex 1:     Solve the following trigonometric equation Cos4A – Sinn 2A =0

Sol :      Cos 4A – sin2A =0

2Sin2 (2A) + Sin (2A) – 1 = 0

Here we can use quadratic formula to find A value. Let us take any variable equal to Sin 2A

Let us take y = Sin 2A

2y2 + y – 1 =0

2y2+2y – y – 1 = 0

2y(y + 1) – (y + 1) = 0

If we factor this we will get two values for y.

(y + 1)(2y – 1) = 0

Now y + 1 = 0 2y – 1 = 0

Now plug y = Sin2A

Sin 2A + 1 = 0                                      2Sin2A – 1 =0

Sin 2A = -1                                          2Sin 2A = 1

2A = Sin-1 (-1)                                       Sin 2A =

2A = 270                                             2A = Sin-1

2A = 30

A = 135                                                      A = 15

From this we will get two value for A.

Practice Problem for Solving Trigonometric Examples:

Ex 2:          Solve the assessment of the following trigonometric identity Sin 75 - Cos 15

Sol :           Sin 75 – Cos 15

Here we have to use sum and variation formula to find the value os Sin 75 - Cos 15

Sin (45 + 30) – Cos (45 - 30)

Sin (A + B) = Sin A Cos B + Cos A Sin B

Cos (A - B) = Cos A Cos B + Sin A Sin B

Here A = 45

B = 30

Sin (45 + 30) = Sin45.Cos30 + Cos45Sin30

= 0.7071 * 0.8660 + 0.7071 * (0.5)

= 0.6123 + 0.3536

Sin 75 = 0.9659

Cos (45 – 30) = Cos45Cos30 + Sin45Sin30

=0.7071 * 0.8660 + 0.7071 * (0.5)

=0.6123 + 0.3536

Cos 15 = 0.9659

Now plug the values in the equation is

Sin 75 – Cos15 = 0.9659 – 0.9659

Sin 75 – Cos15 = 0

Ex 3:               Solve for x Sin x = 0.5, Cos x = 0.8660

Sol :             (I)  Given    Sin x = 0.5

x = Sin-1 (0.5)

x = 30o

(II)    Cos x = 0.8660

x = Cos-1 (0.8660)

x = 300

So from this angle x =30o. Here we use the opposite trigonometric functions to find the value of x.

Solving Simple Linear Equations

Solving simple linear equation involves the process of the solving basic linear equations in simple method. Linear equations come under the category of linear algebra whereas linear algebra is defined as the process of calculating system of unknown variables with the help of known things. Simple linear equations have relations with the families of vectors called vector or linear spaces. The following are the simple linear equations examples for solving.

Simple Linear Equations Examples for Solving:

Example 1:

Solve the simple linear equation to find the variable value.

-2(h - 1) – 4h - 1 = 3(h + 2) – 4h

Solution:

Given expression is
-2(h - 1) – 4h - 1 = 3(h + 2) – 4h

Multiplying the integer terms
-2h + 2 – 4h - 1 = 3h + 6 – 4h

Grouping the above terms
-6h + 1 = -h + 6

Subtract 1 on both sides
-6h + 1 - 1 = -h + 6 -1

Group the above terms
-6h = -h + 5

Add h on both sides
-6h + h = h + 5 -h

Group the above terms
-5h = 5

Multiply -1/5 on both sides
h = - 5/5

h = - 1 is the solution.

Example 2:

Solve the simple linear equation to find the variable value.

-4(h + 3) = h + 9

Solution:

Given expression is
-4(h + 3) = h + 9

Multiplying the integer terms
-4h - 12 = h + 9

Add 12 on both sides
-4h - 12 + 12 = h + 9 + 12

Grouping the above terms
-4h = h + 21

Subtract h on both sides
-4h - h = h + 21 -h

Grouping the above perms
-5h = 21

Multiply -1/5 on both sides
h = -21/5

h = -21/5 is the solution.

Simple Linear Equations Practice Problems for Solving:

1) Solve the simple linear equation to find the variable value.

-3(h - 2) – 2h - 3 = 2(h + 1) – 4h

Answer:  h = -1/3 is the solution.

2) Solve the simple linear equation to find the variable value.

-2(h + 3) = h - 1

Answer:   h = - 5/3 is the solution.

Explanation for Multiples and Factors of an Integers

Introduction to multiples and factors of an integers:
                        In math, the natural numbers are form the integer and another name of number is an integer. The factor is divisor of a given number. This divisor is divides the given integer without any remainder. The factors may be two or more in an integer. The multiple is a one quantity in multiplication. Now we are going to see about sum of factors of a number.

Explanation for Multiples and Factors of an Integers

Multiples of an integers:

                          If we multiply the one integer with another integer means that first integer is called as multiples. For example, x. y = x is multiple of y that is 2 x 3 = integer 2 is a multiple of integer 3.

Properties of multiple:

• The integer zero is common multiple of all integers.
• If we multiply any integer with 1 means that any integer is multiple of 1.

Factors of integers:

                             The factors are referred as divisors and each integer contains more than one factors. Types of factors are,

• Prime factors – It define the prime numbers (only two divisors).
• Composite factors – It define the composite number (two or more factors).

More about Multiples and Factors of an Integers

Example problems for multiples and factors of an integers:

Problem 1: Find out the multiples of given number.

8

Answer:

The given integer is 8.

The multiples of given integer is 8 x 1 = 8

                                                     8 x 2 = 16

                                                     8 x 3 = 24

                                                     8 x 4 = 32

Therefore, the multiples of an integer 8 are 8, 16, 24, 32…….

Problem 2: Find out the factors of given integer.

42

Answer:

The given integer is 42.

Factors of an integer 42 are 1, 2, 3, 6, 7, 14, 21, and 42.

Example problems for multiples and factors of an integers:

1. Determine the factors of an integer 20.

Answer: Factors are 1, 2, 4, 5, 10 and 20.

2. List the multiples of an integer 11.

Answer: The multiples of an integer are 11, 22, 33, 44….

Fraction least to Greatest

Introduction to fraction least to greatest:

   A fraction number value is one part of the whole number value in decimals.  A Fraction number value consisting of a two division in the number. The one part is top place of a number value is called as a numerator value. Another part is bottom place of the number value is called as a denominator value. That is numerator value of the fraction number divided by a denominator value of the fraction number. In this article we shall discuss about fraction least to greatest.

Sample Problem for Fraction least to Greatest:

This problem shows proper fraction for least to greatest:

Problem 1:

Find the value of given fraction numbers `(2)/(5)` + `(3)/(5)`

Solution:

   In the proper fraction a denominator values are same. So we directly add or subtract the numerator value.

Step 1: In the denominator values are same.

Step 2: Add the numerator values and put over the same denominator values.

       `(2)/(5)` + `(3)/(5)` = `(2 + 3)/(5)`

                 = `(5)/(5)`

Step 3: Now we simplify the fraction values.

               = 1

Problem 2:

Find the value of given fraction numbers `(2)/(9)` + `(7)/(9)`

Solution:

   In the proper fraction a denominator values are same. So we directly add or subtract the numerator value.

Step 1 Here  denominator values are same.

Step 2: Add the numerator values and put over the same denominator values.

      `(2)/(9)` + `(7)/(9)`   = `(2 + 7)/(9)`

                   = `(9)/(9)`  

Step 3: Now we simplify the fraction values.

               = 1

Improper fraction problem for least to greatest:

Problem 1:

Find the value of given fraction numbers `(11)/(7)` + `(13)/(5)`

Solution:

   In the improper fraction a denominator values are not same. So we don’t directly add or subtract the numerator values. So, we take least common multiplier for the denominator values.

Step 1: In the denominator values are not same. So, we take LCM for denominator

Step 2: multiply the numerator and denominator values for common multiplier.

The LCM value of the given fraction denominator value is 33

       `(11)/(7)` * `(5)/(5)` = `(55)/(35)`

        `(13)/(5)` * `(7)/(7)`   = `(91)/(35)`

 Step 3: Now we add the numerator values.

     `(11)/(7)` + `(13)/(5)` = `(11 + 13)/(35)`

                    = `(24)/(35)`

Practice Problem for Fraction least to Greatest:

Problem 1:

Find the value of given fraction numbers `(5)/(8)` + `(7)/(8)`

               Answer: `(4)/(3)`

Problem 2:

Find the value of given fraction numbers `(5)/(6)` - `(11)/(6)`

             Answer: -1

Mixed Word Problems

Introduction:

                A mixed word problems contain different types of word problems. The types of word problems are age problem, percent problems, quadratic problems, etc, to find the solution for the word problems first step is to analyze the word problem. After analyzing the problem use the appropriate method to find the answer.

Example on Mixed Word Problems

Example 1- Mixed word problems:

Kevin is two times old as john. The sum of their age is 30. Find the age of Kevin and john.

Solution:

Given Kevin is two times old as john.

So K = 2J `=>` 1

The sum of their age is 30

So, K+J = 30 `=>` 2

Now substitute the equation 1 in equation 2

2J + J = 30.

3J = 30

Now divide by 3 on both sides

3J/3 = 30/3

J = 10.

Substitute the value of J in equation 1 to find the value of K.

K= 2 (10)

K = 20.

Kevin’s age is 20 and John’s age is 10.

Example 2 - Mixed word problems:

What is 8 percent of 100?

Solution:

Given, what is 8 percent of 100

Let us take unknown as x, is is same as =, of refers ()

So what is 8 percent of 100

x = 8 %( 100)

8% = 8 /100

So the above expression would be

x = 8/100(100)

x= 8.

So 8 is 8 percent of 100.

More Example on Mixed Word Problems:

Example 3 - Mixed word problems:

The area of the rectangle is 24 cm² and the perimeter of the rectangle is 20cm

Solution:

Given, Area of the rectangle = 24cm² and the perimeter of the rectangle is 20cm.

The formula to find the area of the rectangle is length * width

The formula to find the perimeter of the rectangle is 2*(length + width)

Let us take l as length and w as width.

l*w =24

Divide by w on both sides

l= 24/w`=>` 1

2(l+w) =20

Divide by 2 on both sides

l+w =10`=>`

24/w + w =10

24+w² =10w

w² -10w +24 =0

w² – 6w-4w +24 =0

Take w as common from first two terms

w(w-6)-4w+24=0

Take -4 as common from last two terms

w (w-6)-4(w-6)=0

(w-6)(w-4)=0

w=6 or w=4

Now substitute w=6 and w=4 in equation 1 to find the corresponding length

When w=6`=>` l=24/6 `=>` l=4

When w=4`=>` l=24/4`=>` l=6

The length of the rectangle =6 and the width of the rectangle =4.