Average Annual Growth Formula

The average annual growth rate formula can usually be calculated with the help of  the simple interest as well as the compound interest formula. It is calculated in virtue of the principal amount especially the unpaid principal amount.

Formula for calcualting the average annual growth rate is given by

` I = P*N*R`

Where P = principal amount

and R = rate of interest

Simple interest for finding Average annual growth:

Here we are using Simple interest Formula for finding the average annual growth,

The simple interest formula is given by I=prt

In this We have to calculate the annual growth rate ,  then we have to divide it by the number of years given , Finally we got the annual growth rate.

 

Example problem :

we have  $ 2000 which are to invest with a given rate of 3% for a 4 year period time.

So , Here P value = 2000 , R = 3 and T = 4.

By the formula of simple interest I =PRT

By plugging in the values we have ,

= 2000 * `3 /100` * 4

= `(2000*3*4)/100`

=20*3*4

=240

The interest amount for the period of 4 years = 240

So ,  average annual growth rate is` 240 / 4` => 60 (Answer)

Compound interest method for finding Average annual growth:

This is an alternate method for finding the average annual rate.

The formula is given by :

In the case , if  interest is compounded once a year:

A = `P(1 + r)^n`

In the case , if interest is compounded q times a year:

A = `P(1 + r/q)^(nq)`

Here P is the principal amount, r is the annual rate of interest and n is the number of years

A is to find the money accumulated after n years, including interest.

 

Example problem:

The karthic deposit amount is 2000 the rate of interest is 2% and the 3 years to find the compound interes.

Here P=2000 , r = 2/100 and n = 3years

A is how much money you’ve accumulated after 3 years, including interest.

If the interest is compounded once a year:

A = P(1 + r)ⁿ

=2000(1+0.02)³

=2000(1.02)³

            =2000 x1.0612

=2122.416

the compound interest of the given amount is =2122.416

If the interest is compounded 2 times a year:

A = P(1 + `r/q)` ^nq

q = is the number of times to year

==2000(1+0.02)^3*2

=2000(1.02)⁶

            =2000 x1.126

=2252.32

The compound interest of the given amount is =2252.32

The average annual growth rate is 2122.142 – 2000 =122.142.

Answer :  The average annual growth value is 122.142

Increasing Proportionally To y

Increasing proportionally to y is a direct proportion. It is mostly done with two variables. If one of the variable increases then the other variable also increases. It works according to the law of directly proportional. The general form of direct proportion is y = k x where k is a constant, x and y are variables. x increases when y increases.

Example problems – Increasing proportionally to y:

 

What is the value of x when y = 10 and k is 2.

Solution:

The general form of direct proportion is y = k * x

Given, y= 10 and k =2.

Now substitute the value of 10 and 2 in y = k * x to find the value of x.

10= 2 * x

Divide by 2 on both sides.

10/2 = 2x/2

5 = x.

The value of x increasing proportionally to y.

What is the value of x when y = 3 and k is 1/4.

Solution:

The general form of direct proportion is y = k * x

Given, y= 3 and k =1/ 4.

Now substitute the value of 10 and 2 in y = k * x to find the value of x.

3= 1/4 * x

Multiply by 4 on both sides.

4 * 3 = 4 * 1/ 4 * x

12 = x

The value of x increasing proportionally to y.

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More example problems:

 

Check whether the value of x increasing proportionally to y when k is 2.

Solution:

The general form of direct proportion is y = k x.

Given, k = 2.

Let us take the value of y are -3, -2, -1, 0, 1, 2, 3.

When y = -3`=>` -3 = 2 * x`=>` x=-3/2 `=>` x= -1.5.

When y = -2 `=>` -2 = 2 * x`=>` x =-2/2`=>` x = -1.

When y = -1 `=>` -1 = 2 * x`=>` x =-1/2`=>` x= -0.5.

When y =  0 `=>` 0  = 2 * x `=>` x = 0   `=>` x= 0.

When y = +1 `=>` +1 = 2 * x `=>` x =1/2`=>` x= 0.5.

When y = +2 `=>` +2 = 2 * x `=>` x =2/2`=>` x= 1.

When y = +3 `=>` +3 = 2 * x `=>` x =3/2`=>` x= 1.5.

In this when the value of y increases the value of x also increasing proportionally to y.

Proof Of Identity Requirements

Learning proof of identity requirements:

In mathematics, the proof of identity requirements is easy to derive. In general we can derive many kind of identity proofs, like Algebra identity proof, trigonometric identity proof. Here we can see some of trigonometric identity proof requirements. To study trigonometry, identities are the basic concepts.Let us see some of the basic identity proofs.

 

Proof of trigonometric identity requirements:

 

Now we have to prove the identity requirement sin² `theta` + cos²`theta` = 1

The proof of the identity requirement is as follows,

Let `theta` be the acute angle.The vertex `theta` is taken as the origin and the initial arm `theta` is taken as positive x-axis.

Let RS be drawn perpendicular to x-axis.

Consider OS= x , RS= y and OR =r

Using Pythagoras identity in the right triangle ORS we get,  x²+y²=r²

Now Divide r² on both sides,

`(x^2)/(r^2)` +`(y^2)/(r^2)` = `(r^2)/(r^2)`

`(x^2)/(r^2)`   +  `(y^2)/(r^2)` = 1

We know that sin `theta` = `(y)/(r)`   and   cos `theta` = `(x)/(r)`

Therefore, ( cos `theta` )² + ( sin `theta` )² = 1

 Hence the indentity ( cos `theta` )² + ( sin `theta` )² = 1 is proved.

`(1)/(sin^2 theta)`

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Another identity proof requirements:

 

Using the identity we can prove the 1 + tan²θ = sec² θ

Proof:

Consider the identity proof ( cos `theta` )² + ( sin `theta` )² = 1

   Divide both sides by  ( cos `theta` )2

`(cos^2 theta +sin^2 theta)/(cos^2 theta)`     = `1/(cos^2 theta)`

`(cos^2 theta)/(cos^2 theta) + `` (sin^2 theta)/(cos^2 theta)``(1)/(cos^2 theta)`

1   +  ` (sin^2 theta)/(cos^2 theta)`    =    `(1)/(cos^2 theta)`

1 +(tan θ)2 = (sec θ)2

Hence 1 + tan²θ = sec² θ is proved

Now consider the identity reqirement  ( cos `theta` )² + ( sin `theta` )² = 1

Proof:

Dividing both sides of (1) by sin²θ, we get

`(cos^2 theta)/(sin^2 theta) + ``(sin^2 theta)/(sin^2 theta)``(1)/(sin^2 theta)`

cot²θ + 1 = cosec²θ

Hence Proved.

Hence the three identies are based on the Pythagoras identity.

Using the  identity (1),
(i) sin²θ = (sin²θ + cos²θ)− cos²θ = 1 – cos²θ.
(ii) cos²θ = (cos²θ + sin²θ) – sin²θ = 1 – sin²θ

Using the identity (2),

(i) tan²θ = (1 + tan²θ) – 1= sec²θ − 1
(ii) sec²θ − tan²θ = (1 + tan²θ) – tan²θ = 1.

Using the identity (3),

(i) cot²θ = (1 + cot²θ) – 1 = cosec²θ − 1
(ii) cosec²θ − cot²θ = (1 + cot²θ) – cot²θ = 1.

These are identity requirements using in trigonometry.

Parallel Tangent Line

If we have two functions, f(x) and g(x), then we have to find the parallel tangent lines for an x-value on both functions. Let the line could not be parallel line and tangent at the same time, so we just took the derivative of both the parallel line and tangent line, and made them equal, without paying concentration to the parallel part of the functions f(x) and g(x).

                           

 

Example 1:

 

Find all points on the graph of y = x ⁴ − 2x² where the tangent line is parallel to the x axis.

Solution:

• Lines that are parallel to the x axis have slope = 0. The slope of a tangent line to the graph of y = x ⁴ − 2x² is given by the first derivative y ‘.

  y ‘ = 4x ³ – 4x
  

• We now find all values of x for which y ‘ = 0.

  4x ³ – 4x = 0
  

• To obtain the solutions for x, solve the above equation 4x ³ – 4x = 0.

  x = 0, x = 1, and x = – 1
  

• Find the y coordinates by plugging the values of x coordinates in, y = x ⁴ – 2x²

  for x = 0 , y = 0,

  for x = 1 , y = –1, and

  for x = –1 , y = –1
  

• The tangent lines are parallel to the x axis at the points, (0, 0), (–1, –1), and (1, –1).

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Example 2:

 

Find a, and b so that the graph of y = ax³ + bx is tangent to the line y = –3x + 4 at x = 1.

Solution:

• We should find out two algebraic equations in order to find a, and b. Hence the position of tangent is on the graph of y = ax³ + bx and y = –3x + 4, at x = 1 we have,

  a(1)³ + b(1) = –3(1) + 4

• Simplify to write an equation in a and b

  a + b = 1

•  –3 is the slope for the tangent line, which is also equal to the first derivative y ‘ of y = ax³ + bx at x = 1,

  y ‘ = 3ax² + x = –3 at x = 1.
  

• The above gives a second equation in a and b

  3a + b = –3.

• To find the values of a and b, solve the system of equations a + b = 1 and 3a + b = –3,

  a = –2 and b = 3.

Inverse Problems

Introduction :

•  Inverse problem deals with the linear function determined by expressing the independent variable of another function in terms of the dependent variable which is then regarded as an independent variable. Thus, for example, the trigonometric function gives rise to the inverse trigonometric functions. The introduction problems are explained below.

• For example, consider inverse function is the formula that changes the Celsius temperature to Fahrenheit temperature is the formula that converts Fahrenheit to Celsius.  When a mathematical process is introduced, one of the most essential questions is how to invert it. Let us discuss about some examples of inverse problems.

Inverse functions example problems:

Example 1:

Solve the inverse problem that the function f is a function with inverse f ^-1. Function h is defined as h(y) = A*f(y – h) + k where A, k and h are constants. Calculate the inverse function of h in terms of f ^-1, A, k and h.

Solution:

h ^-1(y) = f ^-1((y – k) / A) + h

Example 2:

Solve the inverse function f(y) = √(y – 4) + 3.

a) Calculate the inverse of f.

b) Calculate the range of f ^-1.

Solution:

a)      f ^-1(y) = (y – 3)³ + 4 , y≥3

b)      [4 , +infinity) : it is the domain of f

Example 3:

Solve the inverse function f is a function with inverse f ^-1. Function h be defined by h(y) = f(y)  + k where k is a constant. Calculate the inverse function of h in terms of f ^-1 and k

Solution:

h ^-1(y) = f ^-1(y – k)

Example 4:

Solve the inverse function h(y) = (y – 1) / (-y + 3).

a) Calculate the inverse of h.

b) Calculate the range of h.

Solution:

a)      h ^-1(y) = (-3y – 1) / (y + 1)

b)      (-infinity , -1) U (-1 , +infinity) : it is the domain of h^-1

Inverse functions practice problems:

1)Solve the inverse function f(z) = √(z – 8) + 9.

a) Calculate the inverse of f.

b) Calculate the range of f ^-1.

Answer:  a) f ^-1(z) = (z – 9)³ + 8 , z≥8

                  b) [8, +infinity): it is the domain of f

2)   Solve the inverse function h(z) = (z – 5)/ (-z + 8).

a)  Calculate the inverse of h.

b) Calculate the range of h.

Answer:  a) h ^-1(z) = (-8z – 1) / (z – 5)

                  b) (-infinity, -5) U (-5, +infinity): it is the domain of h^-1

Formula For Speed Distance

The average speed of an object in an interval of time is the distance traveled by the object divided by the duration of the interval.

The relation between speed, distance and time is given by the formula,

Speed=distance/time

(i.e.) Distance=speed*time

Now, we are going to see some of the problems in speed, distance and time. From these problems, we can get clear view about speed, distance and time.

 

Problems on Speed and distance:

 

Example problem 1:

If a car is travelling at 30 mph and we want to figure out how far it will go in 3 hours.

Solution:

The relation between speed, distance and time is given by the formula,

Distance=speed*time

Distance=30 mph*3 hours.

Distance=90 miles.

I am planning to write more post on free online math solver with steps with example,

Rationalising Denominator

Introduction :

A rational equation is various functions to recognize how to be written as the part of two polynomial functions. Algebra is the sharing of arithmetic about the rules of method and relations, and the shape and suggestion occur from them, with expressions, equations moreover algebra structures. The divisions of algebra standard basic algebra are frequently branch of the curriculum in derived knowledge and establish the proposal of variables instead of numbers.

 

Rationalising denominator:

 

Statements place on these variables are function using the rules of method that relate to numbers. This recognizes how to be completed for a range of reasons, with equation solving. Algebra is radically broader than necessary and when exact rules of method are used and when operations are expanding for equipment other than numbers.

 

 

Examples for Rationalising denominator:

 

Example 1:

How to solve rationalising the denominator  `(sqrt(x)+7/(sqrt(x))`

Solution:

Step 1: the given expression is `(sqrt(x)+7)/(sqrt(x))`

Step 2: to multiply

`(sqrt(x)+7)/(sqrt(x))((sqrt(x))/(sqrt(x)))`

Step 3:  `(x+7sqrt(x))/(x)`

So the solution is    `(x+7sqrt(x))/(x)`

Example 2:

How to solve rationalising the denominator  `sqrt(8)/(sqrt(5))`

Solution:

Step 1: the given expression is  `sqrt(8)/(sqrt(5))`

Step 2: to multiply

`sqrt(8)/(sqrt(5))(sqrt(5)/sqrt(5))`

Step 3:  `sqrt(40)/(5)`

So the solution is   `(2sqrt(10))/(5)`

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Example 3:

How to solve rationalising the denominator  `(3sqrt(x))/(5sqrt(x))`

Solution:

Step 1: the given expression is   `(3sqrt(x))/(5sqrt(x))`

Step 2:  to multiply

`(3sqrt(x))/(5sqrt(x))((5sqrt(x))/(5sqrt(x)))`

Step 3:      `(15x)/(25x)`

Step 4:    So the solution is `3/5`

Example 4:

How to solve rationalising the denominator  `4/sqrt(xy)`

Solution:

Step 1: the given expression is    `4/sqrt(xy)`

Step 2: to multiply

`4/sqrt(xy)(sqrt(xy)/sqrt(xy))`

Step 3:   `((4sqrt(xy))/(xy))`

So the solution is     `((4sqrt(xy))/(xy))`

Example 5:

How to solve rationalising the denominator   `14/(sqrt(7)-sqrt(5))`

Solution:

Step 1: the given expression is  `14/(sqrt(7)-sqrt(5))`

Step 2: `to multiply the`

`14/(sqrt(7)-sqrt(5))((sqrt(7)+sqrt(5))/(sqrt(7)+sqrt(5)))`

Step 3:  `(14(sqrt(7)+sqrt(5)))/(7-5)`

Step 4:   `(14(sqrt(7)+sqrt(5)))/(2)`

So the solution is     `(7(sqrt(7)+sqrt(5)))`

Example 6:

How to solve rationalising the denominator  `(8)/sqrt(x)`

Solution:

Step 1: the given expression is  `(8)/sqrt(x)`

Step 2: `to multiply the`

`(8)/sqrt(x)(sqrt(x)/sqrt(x))`

Step 3: `(8sqrt(x))/x`

So the solution is    `(8sqrt(x))/x`

Example 7:

How to solve rationalising the denominator   `10/(sqrt(x)+sqrt(y))`

Solution:

Step 1: the given expression is  `10/(sqrt(x)+sqrt(y))`

Step 2: `to multiply the`

`10/(sqrt(x)+sqrt(y))((sqrt(x)-sqrt(y))/(sqrt(x)-sqrt(y)))`

Step 3:  `((10)(sqrt(x)-sqrt(y)))/(x-y))`

So the solution is    `((10)(sqrt(x)-sqrt(y)))/(x-y)`

 

 

Preparation For Graphs

Graphs are holistic representations of mathematical associations. The three types generally used are the a) line graph, b) bar graph, and c) pie graph. A drawing that specifies a connection, often practical, between two sets of values as a set of points having coordinates represented by the connection, also called plot. A diagram which is used to monitoring data, in particular one viewing the association between more than one variable.

Preparation of Types of graphs:

 

Preparation of Types of graphs is an essential one. The types of graphs are as follows:

1) Line Graphs:

The line graph is the only one three type that represents two proportions of quantity, one on the perpendicular axis and one on the parallel. All point on the graph specifies a value in mutually proportions.

2) Bar Graph:

A bar graph can also match up to the relative size of several items or the parts of an entire in relative to that whole.

3) Pie Graph:

Pie graphs are save almost only for viewing the parts of an entire relative to one other. A familiar example is the “healthy diet concept,” presentation the percentage of dairy products, meat, vegetables, sea food, grains, and fruit that should be addicted for best healthiness.

Advantages:

1. Numerical or comparative associations in sets of data can be rapidly and exactly perceived by the viewer.

2. A graph can be constructed with very small try and/or creative talent.

3. Graphs can be done in special colors or shadings to comprise a third dimension or excellence.

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Preparation for graphs through examples:

 

Preparation for graphs through examples are explained below:

1) Line graph:

Draw the line graph for the following equation Y = 3x – 1.

2) Bar Graph:

Draw the bar graph for the following data.

Months	Jan	Feb	Mar	Apr	May	June
Profit value	150	100	125	-25	-170	25
Sales Value	450	370	475	350	125	425

 

Disadvantages or Limitations for graph:

1. Care must be taken that pictorial representation doesn’t oversimplify and, therefore, Inaccurately depict complex data.

2. Planning and preparation take time and special materials such as graph paper, colored pens, straightedge, poster board.

 

Finding Nth Term Of a Series

What is series?

In context with numbers , a series is nothing but arrangements of numbers in some order. In series one number will have some relationship with next number.

consider the series

  2, 4 ,6 ,8 …

The dots at the end signifies that the series have no end. Each number in the sequence is called a term. Here in the above series, first term is  ” 2 “, second term is ” 4 “ and so on. In the given series above, we can say that it is a series is multiply of 2. Therefore the nth term will be ,

n^th  term = 2n

similarly n^th  term in 7n  where n=100  =  7 * 100  = 700

EXAMPLE TO FIND THE Nth TERM IN A SERIES

1)7, 12, 17, 22, 27,…?

The numbers are increasing by 5 each time, but the formula is not simply nth term = 5n because that gives, 5,10,15,20,25,…

But compare these two sequences and you can see that the first one is always 2 more than the second one, so the sequence 7,12,17,22,27,… has the formula nth term = 5n+2

This can be checked, eg. 3rd term = 5×3+2 = 17 which is correct

And the formula can be used to work out, for example, the 40th term. 40th term = 5×40+2 = 202

2)

Find the 11th term of the sequence 4, 7, 10,…

a. Notice that 3 is being added to all numbers, so the first term should be 3, then 6, then 9, etc.

b. You need to add 1 to 3 to equal 4.

c. So the expression is 3n + 1.

d. The answer is 3(11) + 1 or 34.

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3) Find the 8th term of the sequence -2, 0, 2,…

a. In this case ‘a’ is 2. So normally we would count: 2,4,6,etc. So the first term should be 2.

b. If we subtract 4 from 2 we get -2.

c. So the expression is 2n – 4.

d. The answer is 2(8) – 4 or 12.

Area under a Curve

The integral calculus has many applications. An important one of such applications is determining the
area under a normal curve. We mean a normal curve to stress that the curve should be smooth and
without breaks. In other words, the function represented by the graph should be continuous.

Before we study about area under a curve let us brush up our basic knowledge of integration. An integral

in the form f(x)dx tells us the overall summation of the products of value of the function f(x) and the

infinitesimal value dx of the variable.

∫f(x)dx

Look at the above diagram. let dx be an infinitesimal width of the variable at any point and f(x) be the
value of the function at that point. The product f(x)dx is nothing but the tiny area shaded. (The width is
exaggerated for explanation purposes). Suppose we calculate all such areas and sum them up, it must

be equal to the integral

not practicable to do such calculation. In fact, only for this reason these types of integrals are called as
indefinite integrals.

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Now suppose we restrict our study to a limited interval like between a and b shown in the diagram, then
the value of the integral of the function between x =a and x = b is same as the sum of areas of infinite
number of rectangles between those points which effectively is the area below the curve between the
same points. This concept is reversely applied to calculate area under curve. That is the area of the
region acdb in the above diagram is nothing but,.

This concept of definite integral is a boon to calculate the area of odd shapes. For example how do you
find the area of a parabolic segment of height h and width w? The standard function for the shape of
parabola is f(x) = x2. So the required area A is w*h – .= w*h – (w3/12) = wh – (w/3)(w2/4) = wh – (w/3)(h),
since at x = w/2, y = h = (w2/4). So A = (2/3)(w*h). That is, (2/3) times the width and height.

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The concept of finding area under a curve has leads to an immaculate help in certain cases of evaluating
difficult integrals. In many cases it may not be possible to evaluate some integrals due to the complexity.
But it can be evaluated between the desired limits by graphing the function of the integrand. The area of
the region below the curve between the given limits can be calculated by a number of methods and hence
the value of the definite integral can be estimated.

∫ f(x)dx. But since the domain or the end limits of the variable is indefinite, it is