Pi Fraction

Fractions :

A certain part of the whole is called as fractions. The fractions can be denoted as a/b , Where a, b are integers. We can multiply two or more fractions. There are three types of fractions in math,

1) Proper fractions

2) Improper fractions

3) Mixed fractions

Pi( pi  ):

We can express pi as an improper fraction.

pi  = 22 / 7

Here we are going to see some problems using pi fraction .


Example Problems for Pi Fraction

Problem 1:

Find the area of the circle with the radius of 5 cm. using the area formula πr2. (Note: Here substitute the fraction value for π =(22/7)

Solution:

Given radius of the circle is 5 cm.

Formula:

Area of the circle  = πr2

Substitute the given radius value and π in the given formula, we get

Area of the circle = (22/7) * (52) cm2

= 3.14 * 25 cm2

Multiplying the both the values, we get

=  78.5 cm2

Answer: The area of the circle is 78.5 cm2

Problem 2:

Find the surface area of the sphere with the radius of 12 cm. using the area formula 4πr2. (Note: Here substitute the fraction value for π =(22/7) )

Solution:

Given radius of the sphere is 12 cm.

Formula:

Surface area of the sphere = 4πr2

Substitute the given radius value and π in the given formula, we get

Surface area of the sphere = 4 * (22/7) * (122) cm2

= 4 *  3.14 * 144 cm2

Multiplying the both the values, we get

= 1808.64 cm2

Answer:

The surface area of the sphere is 1808.64 cm2

My forthcoming post is on Geometry Calculator will give you more understanding about Algebra.

Practice Problems for Pi Fraction

Practice problem 1:

Find the area of the given circle has the radius length is 7m. (Note: Here substitute the fraction value for π =(22/7) ).

Answer:

The area of the circle is 154 m2

Practice problem 2:

Find the surface area of the given sphere has the radius length is 21m. (Note: Here substitute the fraction value for π = (22/7) ).

Answer:

The area of the circle is 5544 m2

Linear Regression Line

A regression line is the straight line which gives the best fit by using the least square cocept to the given sets of data.

To get the line of regression we need to chose the sum of the squares of derivations parallel to the axis of y and minimize the same. It is called that the line of regression of y on x and it gives the best estimate of y for any given value of x.

The equation of y on x is given byy = a + bx.

To get this equation, we use the following normal equation.

sum y = na + bsumx

sum xy = asum x + bsum x 2

By solving them, we get ‘a’ and ‘b’, hence we get the regression line y on x as y = a + bx. Same principle will be applied to get the line x on y.

Now let us see few problems on linear regression line.

Example Problem on Linear Regression Line.

Ex 1: Find the regression line y on x for the following data

Table_1

Soln:

Regression Table1

Here n = 7.

The normal equations are:

sum y = an + bsum x = 56 => 7a + 56b ……….. (1)

sum xy = asum x + bsum x 2 => 469 = 56 a + 476 b ………. (2)

Therefore 8 xx (1) => 448 = 56a + 448 b

(2)=> 469 = 56a + 476 b

Simplifying them, we get: 21 = 28 b => b = 21 / 28  = 3 / 4 .

Therefore (1) => 56 = 7a + 56 (3 / 4)

=> 7a = 56 – 42 = 14 => a = 2

Therefore the equation y on x is y = 2 + 3 / 4 x.

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Example Problem on Linear Regression Line.

Ex 2: Find the regression line x on y for the following data

Table_2

soln:

Regression Table2

Here n = 9.

The normal equations are:

Σx = an + bΣy = 45 = 9a + 96b ……….. (1)

Σxy = aΣy + bΣy 2 = 536 = 96 a + 1224 b ………. (2)

Therefore 96 xx (1) = 4320 = 864a + 9216 b

9 xx (2) = 4824 = 864a + 11016b

Therefore on simplification, we get: 504 = 1800 b => b = 7 / 25 .

Therefore (1) => 45 = 9a + 96 (7 / 25)

=> 9a = 45 – 26.88 = 18.12 => a = 2.01

Therefore the equation x on y is x = 2.01 + 7 / 25 y.

Exponential Equations with Different Bases

Exponential equations with different bases mean we are going to solve the exponential functions with different bases. Normally exponential functions have the base as e. Here we are going to solve the equations other than e. We will see some example problems for exponential functions with different bases. It is better to understand how to work on the different base. If we want to solve the exponential equations we will use the logarithmic values.


Examples for Exponential Equations with Different Bases:

Solve the following equation 3x = 275x + 3

Solution:

The given equation is 3x = 275x + 3

So we get 3x = ((3)3)5x + 3

From this 3x = (3)15x + 9

From the above we can get x = 15x + 9

15x – x = -9

14x = -9

x = `-9 / 14`

Example 2 for exponential equations with different bases:

Solve the following equation 2x = 42x + 4

Solution:

The given equation is 2x = 42x + 4

So we get 2x = ((2)2)2x + 4

From this 2x = (2)4x + 8

From the above we can get x = 4x + 8

4x – x = -8

3x = -8

So x = `-8 / 3`
More Examples for Exponential Equations with Different Bases:

Example 3 for exponential equations with different bases:

Solve the following equation 5x = 125x + 3

Solution:

The given equation is 5x = 125x + 3

So we get 5x = ((5)3) x + 3

From this 5x = (5)3x + 9

From the above we can get x = 3x + 9

3x – x = -9

2x = -9

So x = `-9 / 2`

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Example 4 for exponential equations with different bases:

Solve the following equation 4x = 16x + 4

Solution:

The given equation is 4x = 16x + 4

So we get 4x = ((4)2) x + 4

From this 4x = (4)2x + 4

From the above we can get x = 2x + 4

x – 2x = 4

-x = 4

So x = 4

Logarithmic Expression Calculator

In mathematics, the logarithmic expression of an integer with a provided base is the power or exponent to which the base must be increased in order to produce the integer. For example, the logarithmic function of 1000x to base 10 is 3x, because 3 is the power to which ten must be raised to produce 1000: 103 = 1000, so log101000x  = 3x. Only positive real integers will have the real integer values; negative and complex integers hold complex logarithms.Now we are going to discuss about logarithmic expression calculator in detail.



Logarithmic Expression Calculator:

Logarithmic expression calculator is used to calculate the expression given in terms of logarithmic form. Logarithmic calculator calculates the values for the logarithmic problems. For every logarithmic function there is a value present in logarithm form. sample lograthmic calculator is given below. Here the calculate button is used to find the value of the given question. Reset button is used to clear the values and the question present in the calculator, which is useful to enter the next problem. sample calculator is given below,

Logarithmic Expression Calculator

Rules of lograthmic expression solving.

If ax = N, then loga N = x

logx mn = logx m + logx n

logx (m/n) = logx m - logx n

logx mn = n logx m

logn m = logx m/ logx n

logn m * logo n * logp o = logp m

logn m * logm n = 1, logm m = 1

logn m = `1/ (log_m n)`

logx 1 = 0 . Log 1 to any base is always 0.

these are the formulas for expression using in logarithmic expression calculator.
Exampl for Logarithmic Expression Calculator:

Exampl for logarithmic expression calculator 1:

Solve logarithmic function log4(9x)5

Solution:

Step 1: The given function is log4(9x)5

Step 2: To solve this function using log function rules

Rule 3: logb (mn) = n· logb (m)

Step 3: log4 (9x)5= 5 log4 (9x)

So the solution is log4(9x)5 = 5 log4 (9x)

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Exampl for logarithmic expression calculator 2:

Solve logarithmic function log3 `((99/2)x)`

Solution:

Step 1: The given function is log3`((99/2)x)`

Step 2: To solve this function by help of using log function rules

Rule 1:  logb (mn) = logb (m) + logb (n)

Step 3: log3 `((99/2)x)` = log3 `(99/2)`+ log3(x)

Rule 2:  logb (m/n) = logb (m) - logb (n)

log `(99/2)` = log (99) - log (2)

So the solution is log3  `((99/2)x)` = log3 (99) - log3 (2) + log3(x)

Exampl for logarithmic expression calculator 3:

Solve logarithmic functionlog2 (99/x)

Solution:

Step 1: The given function is log2 (99/x)

Step 2: To solve this function by help of using log function rules

Rule 2: logb (m/n)) = logb (m) – logb (n)

Step 3:  log2 (99/x) = log2 (99) - log2(x)

So the solution is log2 (99/x) = log2 (99) - log2(x)

Solving Simple Differential Equation

Solving simple differential equations involve the process of differentiating the algebraic function with respect to the input function. The algebraic function which is differentiable is known as differential equations. The differential equation comes under calculus category whereas to find the rate of change of the given function with respect to the input function. The following are simple example differential equations for solving.


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Simple Differential Equations Examples for Solving:

The following are the example problems with simple differential equations for solving.

Example 1:

Solve the simple differential equation.

f(k) = k2 – 4k + 8

Solution:

The given equation is

f(k) = k 2 – 4k + 8

The first derivative f ' for the algebraic function is

f '(k) = 2 k  – 4

Example 2:

Solve the simple differential equation.

f(k) = k 3 – 5 k 2  + 11k

Solution:

The given function is

f(k) = k 3 – 5 k 2  + 11k

The first derivative f ' for the algebraic function is

f '(k) = 3k 2 – 5(2 k  ) + 11

f '(k) = 3k 2 – 10 k + 11

Example 3:

Solve the simple differential equation.

f(k) = k4 – 3k 3 – 4k 2  + k

Solution:

The given function is

f(k) = k4 – 3k 3 – 4 k 2  + k

The first derivative f ' for the algebraic function is

f '(k) = 4 k 3 – 3(3k 2 ) – 4( 2 k  ) + 1

f '(k) = 4 k 3 – 9k 2  – 8 k  + 1

My forthcoming post is on Example of Hypothesis Testing and Dividing Fraction will give you more understanding about Algebra.

Example 4:

Solve the simple differential equation.

f(k) = k 5 – 6 k 3  + 11

Solution:

The given function is

f(k) = k 5 – 6 k 3  + 10

The first derivative f ' for the algebraic function is

f '(k) = 5k 4 – 6(3 k 2 )

f '(k) = 5k 4 – 18 k 2
Simple Differential Practice Equations for Solving:

1) Solve the simple differential equation.

f(k) = k 3 – 6 k 2  + 11k

Answer: f '(k) = 3k 2 – 12 k

2) Solve the simple differential equation.

f(k) = k 2 – 6 k   + 11

Answer: f '(k) = 2k – 6

About Metric Volume Units

The quantities used to find area, length, width capacities and volume of things etc are called measures. Many countries have their own system of measures. But Metric System of measures is very simple and easy to calculate. The area is measured in square unit. In metric system the volume is measured in cubic units.
Example Problems - Metric Volume Units:

The triangular prism has width 6 cm, height 9 cm and length 11 cm. find the volume of triangular prism.

Solution:

Given:

Width (w) = 6 cm

Height (h) = 9 cm

Length (l) = 11 cm

Formula:

Volume of triangular prism (V) = `1/2` (l x w x h) cubic units

= `1/2` (11 x 6 x 9)

= `1/2` (594)

= 297

Volume of triangular prism (V) = 297 cm3

2. figure out the volume trapezoidal prism whose length 11 cm, height 8cm, length of parallel sides a=7 cm and b=4cm.

Solution:

Given:

Length (l) = 11cm

Height (h) = 8cm

Parallel sides a=7cm and b=4cm

Formula:

Volume of trapezoidal prism = l x area of the base cubic units

Area of the base:

Area of the base = `1/2` x (a + b) x h

= `1/2` x (7 + 4) x 8

=`1/2` x 11 x 8

= 44 cm2

Volume of trapezoidal prism = 11x 44

= 484

Volume of trapezoidal prism = 484 cm3

3. The cylinder has the radius r = 10 feet, h=23 feet. Find the volume of cylinder.

Solution:

Given:

r=10 cm

h=23 cm

Formula:

The volume of the cylinder = π x r2 x h cubic unit

=3.14 x (10)2 x 23

The volume of the cylinder = 7222 ft3.
Example Problems - Metric Volume Units:

Cone:

4. The cone has the radius = 10 feet and height = 23 feet. Find the volume of the cone.

Solution:

Given:

Radius (r) = 10 feet

Height (h) = 23 feet

Formula:

The volume of the cone =`1/3` x π x r2  x  h

= `1/3` x 3.14 x (10)2 x 23

The volume of the cone = 2407.33 ft3

5. What is the volume rectangular prism with length 8 cm width 5 cm and height 6 cm?

Solution:

Given:

Length =8 cm

Width = 5 cm

Height = 6 cm

Formula:

Volume of rectangular solid (v) = l x w x h

= 8 x 5 x 6

= 240

Volume of rectangular solid (v) = 240 cm3

radius calculation formula

Pi is also one of the important concepts in algebra. The value of pi is 22/7 or 3.14. Using these values we can easily find the answer for complex problems in algebra. We have different kinds of problems using pi in algebra. Commonly the pi formulas are used to find area or volume of the specific shape in algebra. Here we are going to learn some problems for calculate radius.
Radius Calculation Formulas:

Perimeter of a circle = 2 * pi * radius.
Area of circle = Pi * radius 2.
Area of ellipse = pi *radius1 *radius 2.
Surface area of sphere = 4*pi* radius 2.
Surface area of cylinder = 2 * pi *radius *height.
Surface area of cone: pi* radius* side.
Surface area of torus = pi2 * (radius2 2 –radius12).
Volume of sphere = `(4)/(3)` *pi*radius 3.
Volume of ellipsoid = `(4)/(3)` *pi*radius 1*radius 2*radius3.
Volume of cylinder =pi*radius*height.
Volume of cone = `(1)/(3)` * pi* radius 2 *height.
Volume of torus = `(1)/(4)` * pi2 *(radius1 + radius2) *(radius1-radius2)2
Volume of hemisphere = `(2)/(3)` pi r3.

These are the formulas to calculate radius.
Example Problems of Radius Calculation Formula:

Example 1:

Perimeter of a circle is 40 m .calculate the radius of the circle?

Solution:

Step 1: Perimeter of a circle = 2 * pi * radius.

Step 2: We know that perimeter = 40 m.

Step 3: Plug the perimeter value in to the formula.

Step 4: 40 = 2 * 3.14 * radius.

Step 5: 40 = 6.28 *radius. (Divide using 6.28 on both the sides)

Step 6: Therefore, radius = 6.4 m.

Example 2:

Surface area of cone is 120 cm^2 and side = 7cm.Calculate the radius of the cone?

Solution:

Step 1: Surface area of cone: pi* radius* side.

Step 2: We know that surface area and side.

Step 3: Plug the surface area and side value in to the formula.

Step 4: Therefore, surface area of cone is 120 = 3.14 * radius *7.

Step 5: So, 120 = 21.98 *radius. (Divide using 21.98 on both the sides)

Step 5: Therefore radius = 5.5 cm.

These are the example problems of calculate radius using formula.

Practice problems of calculate radius using formula.

1)   Volume of cylinder = 160 m and height = 9 cm .calculate the radius of the cylinder?

2)   Suppose the area of circle is 210 m. calculate the radius of the circle?

Answer key

1)   Radius = 5.7 m

2)   Radius = 8.17 m