Volume of a Sphere Units

Introduction about volume sphere:

A sphere is a perfectly round geometrical object in three-dimensional space, such as the shape of a round ball. Like a circle in two dimensions, a perfect sphere is completely symmetrical around its center, with all points on the surface laying the same distance r from the center point. Let us see how to calculate the volume of sphere.

(Source – Wikipedia)

Examples Problems in Volume of Sphere Calculator:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic unit.

r = radius.

Volume of Sphere Units - Problems:

1. The sphere has the radius 20.3 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 20.3 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (20.3) 3

=`4/3` * 3.14 * 8365.427

Volume of the sphere (V) = 35023.25

2. The sphere has the radius 15.2 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 15.2 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (15.2) 3

=`4/3` * 3.14 * 3511.80

Volume of the sphere (V) = 14702.76 cm3

3. The sphere has the radius 11.25 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 11.25 cm

Formula:

Volume of the sphere (V) = `4/3` p r ^ 3 cubic units

= `4/3` * 3.14 * (11.25) 3

=`4/3` * 3.14 * 1423.82

Volume of the sphere (V) = 5961.09 cm3

4. The sphere has the radius 3.4 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 3.4 cm

Formula:

Volume of the sphere (V) = 4/3 p r ^ 3 cubic units

= 4/3 * 3.14 * (39.304) 3

=4/3 * 3.14 * 164.552

Volume of the sphere (V) = 164.552 cm3

5. The sphere has the radius 5.6 cm. find the volume of the sphere.

Solution:

Given:

Radius (r) = 5.6 cm

Formula:

Volume of the sphere (V) = `4/3 ` p r ^ 3 cubic units

= `4/3` * 3.14 * (5.6) 3

= `4/3` * 3.14 * 175.61

Volume of the sphere (V) = 735.22 cm3

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Practice problems in volume of sphere:

1. The sphere has the radius 8.4 cm. Find the volume of the sphere.

Answer: 2481.45408 cm3

2. The sphere has the radius 6.6 cm. Find the volume of the sphere.

Answer: 1203.64992 cm3

Solving Linear Equations Substitution

Introduction for solving linear equations using substitution:

Linear equations substitution is nothing but a process of exchanging a variable in a linear expression with its actual value. In linear algebra the substitution process plays a major role to solve the system of linear equation by substituting the value for the given variable, linear substitution is mainly used to identify a variable in an expression and to find linear relationships between the equations. Here we use the linear substitution method to solve the different types of linear equations.

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Solved Examples on Linear Equation Substitution

Ex 1:

Solve the linear equation by substitution method.

1 / (y - 1) 2 - 4 / (y - 1) + 4 = 0

Solution:

Let s = `1 / (y - 1)` and substitute this term in the given equation.

s 2 – 4s + 4 = 0

By solving the above quadratic equation, we get:
s = 2 and s=2.

Now substitute s by` 1 / (y - 1)` and solve for y
`1 / (y - 1)` = 2

1 = 2(y - 1)

1=2y - 2

3= 2y

Y = `3/2` is the solution for above equation.

Ex 2:

Solve the linear equation by substitution method.

z - 5 `sqrt (z)` = - 6

Solution:

Let s = `sqrt (z)` so that s 2 = z. Substitute z by s and `sqrt (z)` by s 2 respectively to obtain an equation in s.
s 2 - 5 s = - 6

The above equation looks like quadratic form, so rewrite the above term

s 2 - 5 s + 6 = 0

By solving the above equation we get
s = 2 or s = 3

We now substitute s by `sqrt (z)` and solve for z
`sqrt (z)` = 2 or `sqrt ( z )` = 3

z = 4 or z = 9  Is the solution for above equation.

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Practice Problems on Linear Equations Substitution for Solving:

1) Solve the linear equation by substitution method.

1 - 2 / (a - c) = 8 at c=1.

Answer:    a = 5/7.

2) Solve the linear equation by substitution method.

1 - 1 / (x - z) = -8 / (x 2 - z 2) at z=4

Answer:    x = -3.

Horizontal Line Segments

Introduction to horizontal line segments:
The line segments are distinct as the distance between two points. The line segments are make clear in another method is, the point that is joined the points of both directions. We can define a line PQ as `bar(PQ)` . Here we are going discuss about the horizontal line segment. Also we shall solve an example problem based on horizontal line segments.

Special Cases of Horizontal Line Segments:
Horizontal line segments have some special properties when compared to normal line segments,

Horizontal and vertical line segments are identified easily, that they have either x- value as similar or y value as same from both the given points of (x1, y1) (x2, y2).

In horizontal line segments, if y-value is similar then the line segment is easily identified as horizontal line segments.

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Example Problem for Horizontal Line Segments:

Plot the given horizontal line segment on the graph (8, 5) (-8, 5).

Solution:

Given: Two points to plot line segments are (8, 5) (-8, 5).

Two pairs of points are (x1, y1) (x2, y2) needed to plot the horizontal line segments in graph.

Here, from the given points x1 is 8 and y1is 5.

From the given points x2 is -8 and y2 is 5.

Now, from the given values of x co-ordinates and the values of the y co-ordinates, we can say that the values of y-coordinates are similar, so the line is horizontal line segments.

According, to the values of the x co-ordinate we have to plot the corresponding x value of the line segment and also on the y co-ordinate, plot the corresponding y value of the horizontal line segments.

These mentioned above, process is done by using graph is shown.

Thus, horizontal line segment is explained clearly and through diagrammatically is explained successfully.

Odd Square Numbers

Introduction to odd square numbers:

Odd square numbers are one of the basis for mathematics. The odd numbers are 1,3,5,7 etc. The formula for representing the odd square numbers are 2M+1, where m value is used to represent the any type of variable function. Simply the odd number can be defined as the number which are not divisible the number two. These numbers are called as the odd number.

Explanation for Odd Square Number

Odd numbers are the alternatives of the even numbers. The number that are not divisible by the number 2, 4, 6 ,8 etc are called as the odd number. For example, the numbers 3, 5, 7, 9, 11 etc are represented as the odd numbers. The diagrammatic representation of odd numbers are shown below,

The odd square numbers are represented by using the formula, (2M+1)2 . The formula can be simplified as 2(m2 + m) +1. By using the this formula the odd square number problems are solved.

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Example Problem for Odd Square Numbers

Pro 1: Find the odd square number for the given value. M = 4.

Sol :  Step 1: The formula for finding the odd square numbers is given below,

Odd square numbers = (2M+1)2 = `4M^2 + 8M + 1` = 4M(M+2) + 1

Step 2: The value of m is given. Therefore substitute the value in the formula.

Step 3: By substituting the formula we get,

= (2(4)+1)2

= 2(42 +4)+1

= 2(20)+1

= 40+1

= 41.

This is the required odd square number.

Pro 2: Find the odd square number for the given value. M = 6.

Sol :  Step 1: The formula for finding the odd square numbers is given below,

Odd square numbers = (2M+1)2 = 4(m2 + 2m) +1.

Step 2: The value of m is given. Therefore substitute the value in the formula.

Step 3: By substituting the formula we get,

= (2(6)+1)2 

= (12+1)2 = 13x13 = 169

This is the required odd square number.

Practice Problem for Odd Square Numbers

Pro 1: Find the odd square number for the given value. M = 8.

Ans : 196

Pro 2: Find the odd square number for the given value. M = 10.

Ans : 221

Slant Height Square Pyramid

Introduction to slant height of a square pyramid:

Pyramid is one of the shapes in geometry. A square pyramid is a general pyramid consist of square base. It is octahedron type.The lateral edge length and slant height, s of a right square pyramid of side length and height are In pyramid, the outer surfaces are triangular and converge at a point. The base of pyramid can be any shape like triangular, square, rectangular or of any polygon shape. Pyramid is classified into Volume of a Pyramid, square pyramid, and rectangular pyramid. All types of pyramid have three triangular faces and a base. Let’s see about basic three shapes of pyramid.

Square Pyramid Figure

Types of pyramid: 

Square pyramid
Triangular pyramid and
Rectangular pyramid.


Formula for Square Pyramid Slant Height:

Formula for finding the slant height of square pyramid:

s² = h² + (b/2)²
here s, slant height
h, height of pyramid
b, base length

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Example Problems in Square Pyramid Slant Height:

Ex 1:

The  square pyramid of height 24 cm. Find the slant height if the base edges are given as 14 cm.

Sol:
Formula for finding slant height will be  
h=24 and b=14 `b/2` =7
s² = h² +` (b/2)^2 `
s= `sqrt(24^2 + 7^2) `

= `sqrt(576 + 49)`

= `sqrt(625)`
= 25 cm

Ex 2:
The height of square pyramid 350 ft. and each side of  base is 646 ft. calculate the slant height length.

Sol:
given h=350 b=646 b/2=323
s² = h² + (b/2)²
s² = 350² + 323²
s² = 226829
s = 426.27 ft.

Ex 3:
The  square pyramid of height 32 cm. Find the slant height if the base edges are given as 12 cm.

Sol:
Formula for finding slant height will be  
h=32 and b=12 `b/2` =6
s² = h² +` (b/2)^2 `
s= `sqrt(32^2 + 6^2) `

= `sqrt(1024 + 36)`

= `sqrt(1060)`
= 32.55 cm
Ex 4:
The  square pyramid of height 16 cm. Find the slant height if the base edges are given as 4 cm.

Sol:
Formula for finding slant height will be  
h=16 and b=4 `b/2` =2
s² = h² +` (b/2)^2 `
s= `sqrt(16^2 + 4^2) `

= `sqrt(256 + 16)`

= `sqrt(272)`
= 16.5 cm

Solve Implicit Function or Relation

Introduction :

The function of implicit function is related to the variables. These two variables are given by an equation. This function has not been solved explicitly. A relation between the variables of function is said to be implicit function. For example   x^2 + y^2 = 100, y is an implicit function of x and x is an implicit function of y. In this article, we shall discuss about solve implicit function or relation.

Solve Implicit Function or Relation - Problems:

Solve implicit function or relation - problem 1:

Calculate the implicit function of x and implicit function of y in the given equation    -x^2 = - 5y  .

Solution:

Given equation is        -x^2 = - 5y.                      --------------(1)

Adding by  x^2 + 5y on both side, So we get

- x^2 + 5y +  x^2 = -5y.+ x^2 + 5y

+ 5y = + x^2                     --------------(2)

Now  divided by 5 on both sides,

`(5y)/(5)` = `( x^2) /(5)` .

Implicit function of x is                    y =  `( x^2) /(5)`.

Take equation (2)    ,         5y =  x^2 

Take square root on both sides,       `sqrt(5y)` = `sqrt(x^2)` .

` sqrt(5y)`   = x .

x = `sqrt(5y)` .

Answer:     y =  `( x^2) /(5)`.  is Implicit function of x .                       

x = `sqrt(5y)` . is Implicit function of y .                               

Solve implicit function or relation - problem 2:

The implicit function function is 10xy^2 - 5y^2  = 5. Evaluate  `(dy/dx)` .

Solution:

Given implicit function is    10xy^2 - 5y^2  = 5.

Now Find the derivative of  xy^2

`d/dx`(10xy^2)    = x 20y `(dy/dx)` + 10y^2 (1).

Find the derivative of 5y^2

`d/dx`(5y^2)    =  10y `(dy/dx)` .

Find the derivative of 5 (constant)

`d/dx`(5)    = 0.

So,                     10xy^2 - 5y^2  = 5.

20xy `(dy/dx)` + 10y^2 - 10y `(dy/dx)` . = 0

Subtract by 10 y^2 on both sides,

20xy `(dy/dx)` + 10y^2 - 10y `(dy/dx)` - 10 y^2 . = 0 - 10y^2

20xy `(dy/dx)` - 10y `(dy/dx)` .=  - 10y^2

Take `dy/dx` in common

` (dy/dx)` (20xy - 10y) = - 10y^2

Divided by (20xy - 10y) on both side so we get,

` (dy/dx)`` ((20xy - 10y)/(20xy-10y))` = `((- 10y^2)/(20xy-10y))`.

` (dy/dx)`  =  `((- 10y^2)/(20xy-10y))`.

Answer:  ` (dy/dx)`  =  `((- 10y^2)/(20xy-10y))`.

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Solve Implicit Function or Relation - Practice Problems:

Solve implicit function or relation - practice problem 1:

Find the implicit function of y in the given equation    x  = 2 - 3xy .

Answer:     Implicit function of y is               x  = `((2)/(1 + 3y))`         ..

Solve implicit function or relation - practice problem 2:

Find the implicit function of x in the given equation    3y  = 9 - 3xy .

Answer:     Implicit function of x is               y  = `((3)/(1 + x))`         ..

Basic Math Ratios

Introduction for basic math ratios:
A relation is getting form the comparison of two quantities like in some wisdom is called a ratio. Ratio will be show by the terms of fraction, that is, a:b is equal to a/b For example 25 is `1/4` of the hundred, Therefore, the ratio of 25 to 100 is `1/4` We can write it in ratio as 1:4.

Example Problems for Basic Math Ratios:
Example 1 by basic math ratios:

Find the ratio of 70 centimeters to 5 meters in its simplest form.

Solution:

5 m = 5 `xx` 100 cm

Therefore, 5 m = 500 cm

Therefore, the ratio of 70 cm to 4  m = 50cm: 500 cm

Now we have to cancel out the units. So we get

70: 500

Now we have to divide both terms by 10. So we get

7:50

Example 2 by basic math ratios:

Find the ratio of 500 m to 1.4 km in its simplest form.

Solution:

1 km = 1000 m

1.4 km = 1.4 `xx` 1000

Therefore, 1.4 km = 1400 m

Therefore, the ratio of  500 meter to 1.4 km = 500 m : 1400 m

Now we have to cancel out the units. So we get,

500:1400

Now we have to divide both terms by100. So we get

5:14

Example 3 by basic math ratios:

Find the ratio of 40 minutes to 5 hours in its simplest form.

Solution:

1 hour = 60 minutes

5 hours = 5 `xx` 60 minutes

Therefore, 5 hours = 300 minutes

Therefore, the ratio of 40 minutes to 5 hours = 40 min : 300 min

Now we have to cancel out the units. So we get

40: 300

Now we have to divide both terms by 10. So we get

4:30

Now we have to divide both terms by 2. So we get

2:15

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For Problems for Basic Math Ratios:

Problem 1 by basic math ratios:

Find the ratio of 600 milliliters to 1.8 liters in its simplest form.

Solution: 1:3

Problem 2 by basic math ratios:

Find the ratio of 700 milligrams to 5 kilograms in its simplest form.

Solution: 7:50