Percentage Change Calculator

A percentage change is a way to express a change in a variable. It represents the relative change between the old value and the new one.

The formula used to calculate the percentage change is

Percentage change = `((V2 - V1) / (V1)) * 100` .

V1- represents the old value

V2 - the new one.                                                              Source: - Wikipedia



some percentages as fractions:

a.       `5%=5/100=1/20` .

b.      `10%=10/100=1/10` .

c.      `25%=25/100=1/4` .

d.     `75%=75/100=3/4` .

e.      `125%=125/100=5/4` .

f.        `175%=175/100=7/4` .

g.       `(3 1/8)%=25/800=1/32` .

h.    ` (6 1/4)%=25/400=1/16` .

i.        `(8 1/3)%=25/300=1/12` .

j.        `(16 2/3)%=50/300=1/6` .

k.       `(66 2/3)%=200/300=2/3` .

l.        `(87 1/2)%=175/200=7/8` .

Calculation of Percentage:


Calculation of percentage:

The percent symbol can be treated as being equivalent to the pure number constant `1/100=0.01,`  while performing calculations with percentage.

If a number is first changed by`P% ` and then changed by `Q%` , then the net change in the number `=[P+Q+((PQ)/100)]` . Remember that any decreasing value in the formula should be taken as ‘negative’ and increasing value should be taken as ‘positive’.

Similarly, if A’s salary is `P%`  less than B’s salary, then the percentage by which B’s salary is more than A’s salary is`(100P)/(100-P)` .

If expenditure also, then percentage change in expenditure or revenue`=[P+Q+((PQ)/100)]` . Where ‘P’ is the percentage change in price and ‘Q’ is the percentage change in consumption.


Percentage change calculator - Example problems:


Percentage change calculator - Problem 1:-

Ram scored 86 runs in the cricket match on  Monday.  On Friday he scored 95 runs.  Calculate the Percentage of change?

Solution:-

Given

V2 = new value = 95 runs.

V1 = old value = 86 runs.

Percentage of change =  ?.

The formula used to calculate the percentage change is

Percentage change = `((V2 - V1) / (V1)) * 100` .

By plugging in the given values in to the formula we get

Percentage change  =` (95 - 86) / (86) * 100` .

The difference between  95 and 86 is  9.

By plugging in it to the formula we get the answer as

=`9 / 86` * 100.

The fraction 9/ 86 gives us 0.1046.

=0.1046 * 100

=10.46

The percentage of change is 10.46.

Percentage change calculator - Problem 2:-

Mary bought 40 Compact disks last month.  He bought only 30 this year.  Calculate is the percent of change.

Solution:-

Given :-

Here,

V2 = new value =  35 Compact disks

V1 = old value = 40 Compact disks

Percentage of change =?

The formula used to calculate the percentage change is

Percentage change = `((V2 - V1) / (V1)) * 100` .

V2 = new value.

V1 = old value

By plugging in the given values in to the formula we get

Percentage change =  `(35 -40) / (40)` * 100.

The difference between 35 and 40 is 5

By plugging in the given values in to the formula we get,
33333
= `5 / 40` * 100

The fraction 8/ 40 gives us  1/5.

`= 1/ 5 ` * 100.

= 20%

The percentage of change is 20 %


Percentage change calculator - Practice Problem:


Ex1 : John scored 16 runs in the cricket match on  Monday.  On Friday he scored 32 runs.  Calculate the Percentage of change?

Answer:-

The percentage of change is 50%

Ex2 : Calculate `40%`  of `625` .

Sol: `40% ` of a number `=2/5`  of the number `=2/5`  of `625=(2/5)(625)=250` .

Temperature Conversion

In this page we are going to discuss about temperature Conversion concept . Measurement is one of the important terms in day to day life. Temperature is usually measured in terms of Fahrenheit and Celsius. Temperature of is generally in terms of these two names.

Fahrenheit:

The degree Fahrenheit is usually represented as (F). Fahrenheit is names after the German scientist Gabriel Fahrenheit, who invented the Fahrenheit measurement. The zero degree in the Fahrenheit scale represents the lowest temperature recording.

Celsius:

The degree Celsius is usually represented as (C). Celsius is names after the Swedish astronomer Ander Celsius, who proposed the Celsius first. In Celsius temperature scale, water freezing point is given as 0 degrees and the boiling point of water is 100 degrees at standard atmospheric pressure.

Formula for Fahrenheit and Celsius Conversion


Fahrenheit to Celsius Conversion Formula:

The formula for converting Fahrenheit to Celsius conversion is given as,

Tc = (5/9)*(Tf-32)

Where,

Tc = temperature in degrees Celsius,

Tf = temperature in degrees



Celsius to Fahrenheit Conversion Formula:

The formula for converting Celsius to Fahrenheit conversion is given as,

Tf = (9/5)*Tc+32

Where,

Tc = temperature in degrees Celsius,

Tf = temperature in degrees Fahrenheit.


Examples on temperature Conversion

Below are the examples on fahrenheit to celsius conversion problems :

Example 1 : Convert 68 degree Fahrenheit to degree Celsius.

Solution:

The formula for converting Fahrenheit to Celsius conversion is,

Tc= (5/9)*(Tf-32)

Tc= 68

Tc= (5/9) * (68 – 32)

Tc= (5/9) * 36

Dividing 36 by 9, we get 4,

Tc= 5 * 4

Tc= 20 degree.

The answer is 20 degree Celsius

Example 2 :Convert 132 degree Fahrenheit to degree Celsius.

Solution:

The formula for converting Fahrenheit to Celsius conversion is,

Tc= (5/9)*(Tf-32)

Tf = 132

Tc= (5/9) * (132 – 32)

Tc= (5/9) * 100

Dividing 100 by 9, we get 11.11,

Tc= 5 * 11.11

Tc= 55.11 degree.

The answer is 55.11 degree Celsius.



Now see the Celsius to Fahrenheit conversion problems:

Example 1 :  Convert 50 degree Celsius to degree Fahrenheit.

Solution:

The formula for converting Celsius to Fahrenheit is given as,

Tf= (9/5)*Tc+32

Tf= (9/5) * 50 + 32

Tf= 9 * 10 + 32

Multiplying 9 and 10 we get 90,

Tf= 90 + 32

Tf= 122 degree Fahrenheit

The answer is 122 degree Fahrenheit.


Example 2 :  Convert 42 degree Celsius to degree Fahrenheit.

Solution:

The formula for converting Celsius to Fahrenheit is given as,

Tf= (9/5)*Tc+32

Tf= (9/5) * 42 + 32

Tf= 9 * 8.4 + 32

Multiplying 9 and 8.4 we get 75.6,

Tf= 75.6 + 32

Tf= 107.6 degree Fahrenheit.

The answer is 107.6 degree Fahrenheit.

Introduction to Probability and Statistics

Introduction for Probability:

Introduction for Probability is the possibility that rather will happen - how to be expected it is that some event will happen. Now and again you can measure a probability with a number: "10% chance of rain", or you know how to use words such as impossible, unlikely, and possible, even chance, likely and certain. when a coin is tossed there is a probability of getting head and tail possible outcomes for an experiment occur is sample space.

Example: "our team may won match today"

Here is a probability formula:

`P(A) = (The Number Of Ways Event A Can Occur)/(The Total Number of possibLe outcomes)`

Introduction to Statistics:

Statistics is the branch which is applied by mathematics, which deals with the scientific analysis of data. The word ‘Statistics’ is derived from Latin word ‘Status’ which means ‘political state’. In statistics introduction datas are in two types which are primary and secondary datas. Sometimes an investigator uses the primary data of another investigator collected for a different purpose. Such data are called statistics secondary data.

From the data we have learnt to calculate the measures of the central tendency like mean, median and mode. These central measures do not give us all the details about the distribution.Further descriptions of the data called measures of dispersion are necessary. According to A.L. Bowley, “Dispersion is the measure of the variation of the individual item”. That is the dispersion is used to indicate the extent to which the data is spread.

Statistics properties:

Statistics deals with three properties, which are  Mean, Median , Mode

Examples:
A cancer patient wants to identify the probability that he will survive for at least 5 years. By collecting data on survival rates of people in a similar situation, it is possible to obtain an empirical estimate of survival rates. We cannot know whether or not the patient will survive, or even know exactly what the probability of survival is. However, we can estimate the proportion of patients who survive from data.

Common Pythagorean Triples

Introduction to Pythagoras online study:

The Pythagoras Theorem is a statement relating the lengths of the sides of any right triangle.

 The theorem states that:

For any right triangle, the square of the hypotenuse
is equal to the sum of the squares of the other two sides.

Mathematically, this is written:

c^2 = a^2 + b^2

We define the side of the triangle opposite from the right angle to be the hypotenuse, c. It is the longest side of the three sides of the right triangle. The other two sides are labelled as a and b.





Pythagoras generalized the result to any right triangle. There are many different algebraic and geometric proofs of the theorem. Most of these begin with a construction of squares on a sketch of a basic right triangle. We show squares drawn on the three sides of the triangle. For a square with a side equal to a, the area is given by:

A = a * a = a2

So the Pythagorean theorem states the area c2 of the square drawn on the hypotenuse is equal to the area a2 of the square drawn on side a plus the area b2 of the square drawn on side b.

pythagoras online study-Pythagorean triplets


A knowledge of Pythagorean triplets will also help the student in working the problems at a faster pace.

 The study of these Pythagorean triples began long before the time of Pythagoras.

There are Babylonian tablets that contain lists of such triples, including quite large ones.

There are many Pythagorean triangles all of whose sides are natural numbers. The most famous has sides 3, 4,

and 5. Here are the first few examples:

32 + 42 = 52;

52 + 122 = 132;

82 + 152 = 172;

282 + 452 = 532

There are infinitely many Pythagorean triples,that is triples of natural numbers (a; b; c) satisfying the equation a2 + b2 = c2.

If we take a Pythagorean triple (a; b; c),and multiply it by some other number d, then we obtain a new Pythagorean triple

(da; db; dc). This is true because,

(da)2 + (db)2 = d2(a2 + b2) = d2c2 = (dc)2 :

Clearly these new Pythagorean triples are not very interesting. So we will concentrate our attention on triples with no common factors.They are primitive Pythagorean triples

A primitive Pythagorean triple (or PPT for short) is a triple of numbers

(a; b; c) so that a, b, and c have no common factors1 and satisfy

a2 + b2 = c2:

There are 16 primitive Pythagorean triples with c ≤ 100:

( 3 , 4 , 5 )

( 5, 12, 13)

( 7, 24, 25)

( 8, 15, 17)

( 9, 40, 41

(11, 60, 61)

(12, 35, 37)

(13, 84, 85)

(16, 63, 65)

(20, 21, 29)

(28, 45, 53)

(33, 56, 65)

(36, 77, 85)

(39, 80, 89)

(48, 55, 73)

(65, 72, 97)

 One interesting observation in a primitive  Pythagoras triple is  either a or b must be a multiple of 3.


pythagoras online study-Solved examples


The Pythagorean Theorem must work in any 90 degree triangle. This means that if you know two of the sides, you can always find the third one.

 



In the right triangle, we know that:

c^2 = 6^2 + 8^2

Simplifying the squares gives:

                                                   c2= 36 + 64

                                                  c2 = 100    

                                                   c = 10       

                                      (taking the square root of 100)



In this example, the missing side is not the long one. But the theorem still works, as long as you start with the hypotenuse:



                                                15^2 = a^2 + 9^2

Simplifying the squares gives:

                                                225 = a2 + 81

                                         225 - 81 = a2                 

                                                144 = a2        

                                                  12 = a  

                                                  a   = 12

                              (Notice that we had to rearrange the equation)

Learn Intercept Formula

Learn Intercept formula is nothing but slope intercept formula. Before going to learn intercept formula we need to know why it is called so.

It is called slope intercept form because the equation includes slope and the y-intercept. So now we know the reason why it is called slope intercept form. Now coming to actual concept.....

The general form of slope intercept form is:

y= mx^+b

Where m--> slope of the line.

b --> y-intercept.

y --> y-coordinate.

x -->x-coordinate.

Formula - learn intercept formula


Slope intercept form is the simplest of all forms as we just need to plug in the values of slope(m) and y-intercept(b).

Now how are we going to get the slope??. In problems the slope might be mentioned directly or two points through which the line passes might be given. When the second case occurs slope can be find out using formula
slope(m)= (change in x)/(change in y)

Now let us consider that two points are (x1,y1) and (x2,y2)  then slope is given by

m=(y2-y1)/(x2-x1)

We have learnt how to find slope in the slope intercept form. now the next one to be calculated is y-intercept(b).
For this we need to plug the point through which the line passes.

This point will be mentioned in the question. If slope is m and the point is (p,q) then

plugging these values in slope intercept form we get

q= m*p+b

==> b= q-mp.

Now we have slope and y-intercept substituting these values we get slope intercept form.

I think to learn intercept form is very easy  and... cool

Examples on learn intercept formula


Ex1:  What is slope intercept form of line having slope 2 and y-intercept 3?
Sol1:

Given slope (m)=2, and y-intercept (b)=3

Plug these values in slope intercept form y=mx+b

Then        y= 2*x+3

y=2x+3.

So slope intercept form of a line having slope 2 and y-intercept 3 is

y=2x+3

Ex2:  What is slope intercept form of line passing through the point (2, 3) and having a slope of 4.
Sol 2:

Here we have slope = 4 and y-intercept is not mentioned but the point through which line passes is given.

Plug m=4 and (x,y)= (2,3) in the slope intercept form y=mx+b.

Then we get

3=4*2+b.

==>b+8 =3

==>  b=3-8=-5

So y-intercept is b=-5 and we have slope as m=4.

Plugging these in slope intercept form we get

y= 4x -5.

Ex3:  Express the equation 3x+4y+5=0 in slope intercept form.
Sol 3:

The given equation is 3x+4y+5=0.

In order to convert it in to slope intercept form bring y terms on one side and the remaining terms to other side.

For this subtract with 3x and 5 on both sides

3x+4y+5-3x-5=-3x -5

By doing this 3x and 5 get cancelled and the equation becomes

4y= -3x-5

Divide both sides by 4

4y /4 = (-3x-5)/4

y=-(3/4)x  - 5/4

Thus the slope intercept form of 3x+4y+5=0 is y=-(3/4)x  - 5/4

Mean Value Theorem Derivatives

In calculus, the mean value theorem states, roughly, that given an arc of a smooth continuous (derivatives) curve, there is at least one point on that arc at which the derivative (slope) of the curve is equal (parallel) to the "average" derivative of the arc. It is used to prove theorems that make global conclusions about a function on an interval starting from local hypotheses about derivatives at points of the interval. I like to share this Anti derivative with you all through my article.

Formal Statement of Mean Value Theorem Derivatives:

Let f: [x, y] → R be a continuous function on the closed interval [x, y] , and differentiable on the open interval (x, y), where x < y. Then there exists some z in (x, y) such that,

f'(z) = `(f(y) - f(x))/(y - x)`

The mean value theorem is a generalization of Rolle's Theorem, which assumes f(x) = f (y), so that the right-hand side above is zero.

Only one needs to assume that f: [x, y] → R is continuous on [x, y] , and that for every m in (x, y) the limit,

`lim_(h ->0) (f(m + h) - f(m))/(h)`

exists as a finite number or equals + ∞ or − ∞. If finite, that limit equals f′(m). An example where this version of the theorem applies is given by the real-valued cube root function mapping m to m1/3, whose derivative tends to infinity at the origin.

Note that the theorem is false if a differentiable function is complex-valued instead of real-valued.

For example, define f(m) = eim for all real m. Then

f (2π) − f(0) = 0 = 0(2π − 0)

while, |f′(m)| = 1.

- Source Wikipedia


Proof of Mean Value Theorem Derivatives:


Let g(m) = f(m) − rm, where r is a constant, because f is continuous on the closed interval [x, y] and differentiable on the open interval (x, y), now we want to select r, so as g satisfies the conditions of the Theorem,

g (x) = g (y) `hArr` f (x) − rx =  f (y) − ry

`hArr` ry − rx = f (y) − f(x)

`hArr` r(y − x) = f (y) − f(x)

`hArr` r = `(f(y) - f(x))/(y - x)`

By Rolle's theorem, g is continuous on the closed interval [a, b] and g(a) = g(b), there is some c in (a, b) for which g′(c) = 0, and it follows from the equality g(x) = f(x) − rx that,

f' (c) = g' (c) + r = 0 + r =  `(f(b) - f(a))/(b - a)`

Algebra 2 Help and Answers

Algebra 2 help us find the unknown quantities with the help of known quantities. In algebra, we frequently use letters to represent numbers. Algebra 2 help includes real numbers, complex numbers, vectors, matrices etc. Algebra 2 is the study of the rules of relations and operations, and the constructions arising from them. An algebraic expression represents a scale that gets added or subtracted or multiplied or divided on both sides. In algebra 2 help and answers numbers are considered as constants.


Algebra 2 help and answers example Questions:


The following problems gives different answers to algebra 2 problems.

Ex 1:  Determine all real solutions to the equation

Sqrt (2 x + 13) = x - 5

Sol :   Given
sqrt (2 x + 13) = x - 5

We raise both sides to power 2.
[Sqrt (2 x + 13)] 2 = (x - 5) 2

And simplify.
2x +13 = x 2 - 10 x +25

Write the equation with right side equation to 0.
X 2 - 8 x + 12 = 0

It is a quadratic equation with 2 solutions
x = 6 and x = 2

Ex 2 :  Determine all real solutions to the equation

Sqrt (x + 8) = 12

Sol :  Given
sqrt (x + 8) = 12

We raise both sides to power two(2) in order to clear the square root.
[Sqrt (x + 8)] 2 = 12 2

And simplify
x + 8 = 144

Solve for x.
x = 136

Ex 3:   Determine all real solutions to the equation

Sqrt (x 2 – 13x+72) = 6

Sol :   Given
Sqrt (x 2 – 13x+72) = 6

We raise both sides to power 2.
[Sqrt (x 2 – 13x+72)] 2 = (6) 2

And simplify.
X 2 – 13x+72= 36

Write the equation with right side equation to 0.
X 2 - 13 x + 36 = 0

It is a quadratic equation with 2 solutions
x = 9 and  x = 4


Algebra 2 help and answers practice problems:


1) Determine all real solutions to the equation

Sqrt (x + 25) = 13

Ans : x=12

2) Determine all real solutions to the equation

Sqrt (x 2 – 11x+55) = 5

Ans : x=5 and x=6

3) Determine all real solutions to the equation

Sqrt (2 x + 24) = x

Ans : x=6 and x= - 4