Pyramid with a Trapezoid Base

A pyramid is a building where the outer surfaces are triangular and converge at a point. The base of a pyramid can be trilateral, quadrilateral, or any polygon shape, meaning that a pyramid has at least three outer surfaces (at least four faces including the base). The square pyramid, with square base and four triangular outer surfaces, is a common version. (Source: Wikipedia)

I like to share this Trapezoid Shape with you all through my article.

Pyramid with trapezoid base:
      A pyramid with a trapezoid base is called as trapezoidal pyramid.
     Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height

Example problems for pyramid with a trapezoid base

Trapezoid pyramid problem 1:
        A base length of the trapezoid pyramid is 12 cm, 14 cm and its height of the pyramid is 6 cm. Find the volume of the trapezoidal pyramid.
Solution:
     Given base lengths are b₁ = 12 cm, b₂ = 14 cm, and h = 6 cm.
Formula:
   Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height
Substitute the given values in the above formula, we get
= (1 / 3) * ((1 / 2) * (12 + 14) * 6 cm³
= (1 / 3) * 78 cm³
= 26 cm³

Answer:
 Volume of the trapezoidal pyramid is 26 cm³

Trapezoid pyramid problem 2:
        A base length of the trapezoid pyramid is 15 cm, 24 cm and its height of the pyramid is 9 cm. Find the volume of the trapezoidal pyramid.
Solution:
     Given base lengths are b₁ = 15 cm, b₂ = 24 cm, and h = 9 cm.
Formula:
Volume of trapezoidal pyramid = (1 / 3) * ((1 / 2) * (base₁ + base₂) * height
 Substitute the given values in the above formula, we get
= (1 / 3) * ((1 / 2) * (15 + 24) * 9 cm³
 = (1 / 3) * 175.5 cm³
                                               = 58.5 cm³
Answer:
 Volume of the trapezoidal pyramid is 58.5 cm³

Trapezoid pyramid problem 3:
        A base length of the trapezoid pyramid is 6 cm, 11 cm and its height of the pyramid is 5 cm. Find the area of the trapezoid.
Solution:
     Given base lengths are b₁ = 6 cm, b₂ = 11 cm, and h = 5 cm.
Formula:
Area of trapezoid = ((1 / 2) * (base₁ + base₂) * height
Substitute the given values in the above formula, we get
                                               = (1 / 2) * (6 + 11) * 5 cm³
                                               = (1 / 2) * 85 cm³
                                               = 42.5 cm²
Answer:
 Volume of the trapezoidal pyramid is 42.5 cm²

Practice problems for pyramid with a trapezoid base

Trapezoid pyramid problem 1:
        A base length of the trapezoid pyramid is 7 cm, 9 cm and its height of the pyramid is 12 cm. Find the volume of the trapezoidal pyramid.
Answer:
 Volume of the trapezoid pyramid is 32 cm³
Trapezoid pyramid problem 2:
        A base length of the trapezoid pyramid is 10 cm, 22 cm and its height of the pyramid is 7 cm. Find the volume of the trapezoidal pyramid.
Answer:

 Volume of the trapezoid pyramid is 37.33 cm³

Inverting Matrices

The inverse of a square matrix A, sometimes called a reciprocal matrix, is a matrix   such that         

Where I is the identity matrix.

A square matrix A has an inverse if the determinant . A matrix possessing an inverse is called nonsingular, or invertible.

The matrix inverse of a square matrix m may be taken in mathematics using the function inverse[m].

Inverse of 2x2 Matrix

For a 2X2  matrix

A= `[[a,b],[c,d]]`

The matrix inverse is

                  

             =  

                   =    

Inverse of 3x3 Matrix

For a 3X3 matrix

          

The matrix inverse is

A general matrix can be inverted using methods such as the gauss Jordan elimination, Gaussian elimination.

Prism Solve for h

Prism is one of the three dimensional shape. Normally the prisms faces are flat surface. Here we have two bases, one is at the top and another is at the bottom. In prisms we have two kinds of prisms; they are rectangular prism and triangular prism. When a prism bottom is rectangular shape then it’s denoted as rectangular prism. Cube and cuboids are some examples are rectangular prism. When a prism bottom is triangular shape then its said to be triangular prism. 

The diagram given below the structure of prism:

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Description to prism solve for h:

The following types are used in the prism shape they are

Volume of the regular prism is = BH cubic units,

Where B represents the base area of shape, and H represents height.

The Lateral surface area of the regular prism is given by,

              L.S.A = PH square units.

Where P represents the perimeter of shape and H represents the height of prism

Prism based on rectangular:

The area of the base is calculated by l*w

The perimeter of the base can be calculated by 2(l+w)

Therefore the total surface area is calculated by 2lw+2(l+w)H

Volume of the rectangular prism is V= L * W * H

Triangular based prism:

The area is measured as 1/2 bh

The perimeter of the base is measured by a+b+c.

So the surface are is measured by  bh+ (a+b+c) H

Circular based prism:

The area of the base is calculated as `pi` r²

The perimeter of the base is calculated by 2`pi` r

Therefore the surface area of the prism is 2`pi` r²+2`pi` rH

Let we see some basic problems regarding prisms.

Example 1:

 The dimensions of the rectangular prism is given below ,they are length is 10cm, height is 4 cm, and width is 14 cm.

Solution:

Given dimensions length = 10 cm, height = 4 cm, width = 14 cm.

Formula:

  Surface area = 2lw+2(l+w)H

apply  the values in the above formula, we get

                       = 2 *10*14+2(10+14)4

                       = 240 + 2(24)4

                      = 240 + 48*4

                     = 240 +192

                     = 432 cm²

      Volume = l * w * h  

                      = 10 * 4 * 14 cm³

                      = 560 cm³

Answer:

Surface area = 432 cm²

Volume = 560 cm³

Examples problems to solve h in prism

Example 1:

solve the h value whose prism volume is 1210 cubic units and base is 21 cm.

Solution:

The volume of the prism is V = 1210cubic units.

Base of the prism is = 21cm

                                   V= BH

apply the values in the formula we get

                              1210 = 121*H

                                     h = 1210 /121

                                     h = 10cm.

Example2:

The volume of the  rectangular prism is 360 cm³ and its length and widths are 10 ,4 cm , solve the h value

Solution:

The volume of the rectangular prism is V = 360cm³

Length of the rectangular prism is l = 10cm

Width of the rectangular prism is W = 4 cm

Apply these values in formula we get

                                     360 = 10*4*h

                                      360 = 40 *h

                                     h = 360 / 40

                                    h = 9 cm

Sample problems:

1) The volume of a regular prism is  120cm³.The base of the regular prism is 10 cm , solve its height h?

solution : height h = 12 cm.

2) The volume of the rectangular prism is 420cm3 and its length = 7 cm and width =6 cm and solve its height h?

solution: height h = 10 cm.

Conic Tutorial

The sections of a cone are  ellipses, Circles, parabolas and hyperbolas. Because these diagrams can be obtained by passing a plain through vertex of the double-napped cone. In this conic tutorial we are going to see the sections of the cone.

Circle - conic tutorial:
The circle in the section of a cone is like ellipse. When two foci are coincided to each other the circle is formed. This conic section can be formed when a circular cone is intersected with a perpendicular plane.

Sections of a cone - Parabolas - conic tutorial:

One of the section of the cone is the parabola. The points will not be on the same line but in the plain are equal from a fixed line and in a fixed point.

Solve Maxima and Minima:
The local minimum and the local maximum value must be known for finding the maxima and minima values.

Ellipse - conic tutorial:
The section of a cone is an ellipse. It is from a constant two fixed point with the sum of its focus of all the points. The concept of an ellipse makes different terminologies like focus of ellipse, loci of ellipse, etc.

Equation of ellipse:
Equation of ellipse is `(x - h)^2 / a^2 ` + `(y - k)^2 / b^2` = 1 . here real numbers are h, k, a and b.

Sections of a cone-General Equation of an Ellipse

The points of locus will be F ( p, q ) and the equation of directrix ax + by + c = 0. This is the general equation of ellipse

Theorem on Line contact with an Ellipse:
The condition for the line y = m x + c to touch the ellipse `(x^2/a^2)+(y^2/b^2) ` = 1 is that c = ±  `sqrt(a^2 . m^2 + b^2)` .

Finding the point of contact for Ellipse: 
The tangent y = m x + `sqrt(a^2 . m^2 + h^2)`  touches the ellipse `(x^2/a^2)+(y^2/b^2) `= 1, at the point `[(-a^2m/sqrt(a^2m^2+b^2) , b^2/sqrt(a^2m^2+b^2)]`

Hyperbola - conic tutorial:

A hyperbola is the section of a cone. A conic which an eccentricity greater than 1. It has two branches extending up to infinity, two foci, two directions, two vertices and two axes. ( i . e . )  transverse and conjugate and has a center which is the midpoint of the line of two foci and perpendicular conjugate axis is the line through the center and to the transverse axis.

Two Non Intersecting Lines

The two lines are concurrent means then the pencil is formed by the set of lines passing in their common point: if they are parallel, by the set of parallels to both, in the same direction means, they are non-intersecting lines. In fact, if the lines meet in the ordinary sense of the term or are parallel, the point can be at once obtained. These are called non-intersecting lines.

Example problem for two non intersecting lines:

1. Find the equation for two non intersecting lines that passes through the two points:
     Y = x – 5 and y = x + 5.
     Two non intersecting lines graph for the following equation:
     Y = x - 5
Solution:
   Now plug x values for 0, 1, 2, 3, 4
   When x = 0                                                 
   Y = (0) -5
   Y = 0-5
   Y = -5

   The co ordinate’s point is (0,-5).
   When we plug x = 1
   We get
   Y = 1 -5
   Y = -4

   The co ordinate’s point is (1, -4).
   When we plug x = 2
   We get
   Y = 2 -5
   Y = 2-5
   Y = -3

   The co ordinate’s point is (2,-3).
    When we plug x = 3
    We get
    Y = 3 -5
    Y = -2

   The co ordinate’s point is (3, -2).
    When we plug x = 4
    We get
    Y = 4-5
    Y = -1
The co ordinate’s point is (4, -1).

X	0	1	2	3	4
Y	-5	-4	-3	-2	-1

Two non intersecting lines graph for the following another equation:
   Y = x + 5
Solution:
    Now plug x values for 0, 1, 2, 3, 4
    When x = 0                                                 
    Y = 0+5
    Y = 0+5
    Y = 5

    The co ordinate’s point is (0, 5).
    When we plug x = 1
    We get
     Y = 1 +5
     Y = 6

    The co ordinate’s point is (1, 6).
     When we plug x = 2
     We get
     Y = 2 +5
     Y = 7

     The co ordinate’s point is (2,7).
     When we plug x = 3
     We get
     Y = 3 +5
     Y = 8

     The co ordinate’s point is (3, 8).
     When we plug x = 4
     We get
     Y = 4+5
     Y = 9

     The co ordinate’s point is (4, 9).

X	0	1	2	3	4
Y	5	6	7	8	9

The non intersecting lines represent in graph is,

Practice problem for two non intersecting lines:

1. Find the equation for two non intersecting lines that passes through the two points:
    Y = x – 4 and y = x + 4.
2. Find the equation for two non intersecting lines that passes through the two points:
    Y = x + 8 and y = x - 8.
3. Find the equation for two non intersecting lines that passes through the two points:

    Y = x + 2 and y = x - 2.

Trinomial Notes

Trinomial notes is a polynomial with three monomial terms. In trinomial notes, sum of three monomials is said to be  trinomial. For example, 2x5 – 3x2 + 3 is a trinomial . An Algebraic expression of the form axn is called a monomial in x, The sum of two monomials is called a binomial and the sum of three monomials is called a trinomial.  sum of the finite number of a monomials in term x is called a polynomial of x.

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How To Factor trinomial Notes:

Example 1:
Factor   trinomial notes x² + 3x − 10.

Solution.
The binomial factors will have this form:
(x   a)(x   b)
What are the factors of 10?  Let us take that they are 2 and 5:

x² + 3x − 10 = (x   2)(x   5).

We have to know how to choose the signs so that,  coefficient for middle term  will be sum of the outers plus the inners -- will be +3.                                  Choose factors , +5 and −2.

=  x² + 3x − 10

= (x − 2)(x + 5)
When 1 is the coefficient for x², the order of  factors it does not matter.

(x − 2)(x + 5) = (x + 5) (x − 2) are same factors.

Example Problems In trinomial notes:

Example 1:
Find factors for trinomial notes x² − x − 12.

Solution:
We must finding factors for 12 whose algebraic sum will be  coefficient of x is −1.
Choose −4 and + 3:
x² − x − 12 = (x − 4 )(x + 3). are facors

Example 2:

Find factors for trinomial notes    x2 + 5xy + 4y2
Solution:
x2+5xy+4y2

=x2+xy+4xy+4y2

=x(x+y)+4y(x+y)

=(x+y)(x+4y)

Example 3:

Find factors for trinomial notes    2a2 - 7ab + 6b2
Solution:
2a2-7ab+6b2
=2a2-3ab-4ab+6b2

=a(2a-3b)-2b(2a+3b)

=(2a-3b)(a-2b)

Example 4:

Find factors for trinomial notes 4x2 - 13xy + 3y2
Solution:
4x2-13xy+3y2
=4x2-xy-12xy+3y2

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

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

Acute Angles Pictures

Here we are going to see about the introduction to angles. The angle is referred as a figure which is formed by distribution of  two rays with a common point. This common point is referred as end point. The word angle is from the Latin word angulus which defines the point in corner. There are many types of angles in this we are going to see about the acute angles.

The acute angle is the type of angle which measures the angle between 0 to 90 degree and less than the 90 degree. The normal picture of the acute angle is given as,

Pictures of acute angle:
            The acute angles are referred by using the pictures which measures the angle less then 90 degree. The pictures of the acute angles are given below as examples.
Picture 1:

This is the picture of acute angle which defines that the angle measured in this picture is 48⁰ which is less than 90 degree.

Picture 2:

This is other picture of acute angle which defines that the angle measured in this picture is 28⁰ which is less than 90 degree.

Picture 3:

This is the picture of acute angle which defines that the angle measured in this picture is 37⁰ which is less than 90 degree.

Picture 4:

This is also a picture of acute angle which defines that the angle measured in this picture is 63⁰ which is less than 90 degree.

Picture 5:

This is also a picture of acute angle which defines that the angle measured in this picture is 49⁰ which is less than 90 degree.

Picture 6:

This is also a picture of acute angle which defines that the angle measured in this picture is 72⁰ which is less than 90 degree.

Example problems

Problem 1:
When the angle a-180⁰ is an acute angle. what is the value of a?
Options:
a) 240⁰
b) 300⁰
c) 290⁰
d) 320⁰

Solution:
Acute angle is the angle which measures less than 90 degree or 90 degree.
When 240⁰-180⁰ = 60⁰.
Hence the value of a is 60⁰.

Problem 2:
When the angle y - 90⁰ is an acute angle. what is the maximum value of y?
Options:
a) 170⁰
b) 200⁰
c) 100⁰
d) 540⁰

Solution:
Since the acute angle measures less than 90 degree. The maximum value of y is 170⁰.
Where 200⁰- 90⁰ = 110⁰, since it is greater than 90 degree this is not an acute angle.  
100⁰- 90⁰ = 10⁰, since it is less than 90 degree this is an acute angle but it is the minimum value.

540⁰- 90⁰ = 450⁰, since it is greater than 90 degree this is not an acute angle.  
170⁰- 90⁰ = 80⁰, since it is less than 90 degree this is an acute angle.    

Finding Volume in Math

Finding volume in math article deals with the definition of volume and the formula for the various shapes and the model problems related to volume of various shapes.

Looking out for more help on Units of Volume in algebra by visiting listed websites.

Definition of Volume:

Volume is always measured in cube.units. It is defined as the space that is occupied by the entire three-dimensional shape.

Volume formula for various shapes in math.

Volume of the cube is a3 cubic units

Here “a” is the side of the cube.

Volume of the rectangular prism is L*B*H cubic units

Here L is length, B is breadth and H is the height.

Volume of the cylinder is BH cubic units

Here B is the base area and H is the height

Volume of the pyramid is `(1/3)` BH cubic units.

Here B is the base area and H is the height

Volume of the cone is `(1/3) ` BH cubic units.

Here B is the base area and H is the height

Model problem for finding volume in math.

Problem:

1.  Finding volume of the cube when the side measures about 5 cm in length.

Solution:

Since it is a cube, the formula

Volume of the cube is a3 cubic units

Here “a” is the side of the cube.

Here a= 5cm

The volume of the cube =53

=125 cm3

The volume of the cube is125cm3

2. Finding volume of the rectangular prism when the length = 8cm, breadth = 5cm and the height is 4cm.

Solution:

Since it is a rectangular prism, the formula

Volume of the rectangular prism is L*B*H cubic units

Here L is length, B is breadth and H is the height
L= 8cm

B= 5cm

H = 4cm

Volume of the rectangular prism = (8* 5*4) cubic units.

= 160 cm3

Volume of the trectangular prism is 160cm3

3. Find the volume of the cylinder whose base area is 46m^2 and the height of the cylinder is 10m

Solution:

Base area of the cylinder is 46m2

Height of the cylinder is 10m

Formula:

Volume of the cylinder is BH cubic units

Here B is the base area and H is the height

= 46*10

= 460 m3

the volume of the cylinder is 460m3

LENGTH OF MEASURING SYSTEM

The unit of length is meter. In length of measuring system length can be measured by using various terms such as millimeter, centimeter, decimeter, meter, decameter, hectometer and kilometer.

Some of the length of measuring system

10 millimeter is equal to 1 centimeter

10 centimeter is equal to 1 decimeter

10 decimeter is equal to 1 meter

10 meter is equal to 1 decameter

10 decameter is equal to 1 hectometer

10 hectometer is equal to 1 kilometer

By knowing these values we can convert the measures of the length.

Similarly some other important conversions of length of measuring system is

1000 millimeter is equal to 1 meter

1000 meter is equal to 1 kilometer

Similarly inches, feet, yards and miles are also the measurements of length of measuring system.

CONVERSION TABLES OF LENGTH OF MEASURING SYSTEM

1 feet is equal to 12 inches

1 yard is equal to 36 inches

1 yard is equal to 3 feet

1 mile is equal to 5280 feet

We have to divide when converting a larger unit and similarly when we have to convert a smaller unit we have to multiply it.

EXAMPLES ON LENGTH OF MEASURING SYSTEM

Example 1: Convert 5000 meters to kilometer

Solution

We know that 1000 meter = 1 kilometer

So

5000 meter = 5000/1000 = 5 kilometer



Example 2: How many centimeters equal to 450 meter?

Solution:

1 meter = 100 centimeter
so

450 meter = 450 * 100

= 450000 centimeter

Example 3: Convert 6 feet to yards and inches

Solution

We know that 1 feet = 12 inches

3 feet = 1 yard

So

6 feet = 6/3 = 2 yards

6 feet = 6 * 12 = 72 inches

Example 4: Convert 5 miles into feet’s

Solution

1 mile = 5280 feet

So 5 miles = 5 * 5280

= 26400 feet’s.

PROBLEMS ON LENGTH OF MEASURING SYSTEM

Convert 5 feet to inches

Convert 20 yards into feet

Convert 100 meter to millimeter

Convert 52 kilometer to decameter

Convert 3 kilometer to millimeter

Answers:

60 inches

60 feets

100000 mm

5200 decameter

3000000 mm

Coplanar Geometry

Three or more points, lines or any other geometric shapes that lie on the common plane are knows as Coplanar.

Geometric substance lying in a same plane are said to be coplanar. In a plane three noncollinear points are some extent coplanar. Four points are coplanar, defined by them is 0, lying in the same plane. Example, any set of three points in plane are coplanar. Let us see coplanar geometry in brief.

Conditions for coplanar:

Coplanarity is corresponding to the statement that the pair of lines determined by the four points is not skew, and can be equivalently stated in vector form as

`|[x1,y1,z1,1],[x2,y2,z3,1],[x3,y3,z3,1],[x4,y4,z4,1]| = 0`

(x3 - x1).[(x2 - x1) × (x4 - x3)] = 0

The coplanar not only for four points it is also for two or three points.

An arbitrary number of n  points x1 , ..., xn can be checked for coplanarity by finding the point-plane distances of the points x4, ...,xn from the plane determined by (x1,x2,x3), and checking if they are all zero. Therefore, the points are all coplanar.

A set of n vectors v is coplanar if the nullity of the linear mapping defined by v  has dimension 1, the matrix rank of v (or equivalently, the number of its singular values) is n-1.

Parallel lines in three-dimensional space are said to be coplanar, but skew lines are not.

In this article we see the coplanar of lines on same plane. Using the line equation we find the coplanar for the three lines on a same plane.

Example problem for coplanar:

Example: prove that three lines are coplanar, equation lines are 3x + 2y = 0, 3x + 3y = 3, 2x + 2y = 2

Solution:

Given: 3x + 2y = 0 -------------(1)

3x + 3y = 3 -------------(2)

2x + 2y = 2 --------------(3)

`|[x1,y1,c],[x2,y2,c],[x3,y3,c]| = 0`

`|[3,2,0],[3,3,3],[2,2,2]| = 0`

3[(3×2) - (3×2)] - 2[(3×2) - (3×2)] +0[(3×2) - (3×2)] = 0

3(6 - 6) -2(6 - 6) -0(6 - 6) =0

0 = 0

Hence proved thus the three lines are coplanar.

To Logarithmic Table

Logarithmic Properties plays an important role in complex calculations in math. We can perform big calculations in math,physics , engineering using logarithms. It give accurate answer as the calculator. These calculations are carried out with the help of logarithmic table.We will see the logarithmic table. The ways of reading the logarithmic table and how to use it in calculations.

Any number x in standard form is written as x = m x 10^p where 1 `<=` m<10 p="">Taking log on both sides we get
log₁₀x = log₁₀(m x 10p) = log₁₀m + plog₁₀10
             = log₁₀m + p
Here p is the characteristic of log x and log₁₀m is called the mantissa of logx

How to find the logarithm of a number:

• Step 1: Write the number in the standard form.

For example 431.5 = 4.315 x 10²

• Step 2: Find the characteristic p of the logarithm.

Here p = 2

• Step 3: Find the mantissa from the table.

To find the log of 4.315 from the table. log 4.31 is 0.634473 `~~` 0.6350. We take the approximate value.

log 4.315 = p + logm = 2+0.6350 = 2.6350

Logarithmic table from 1 to 4.99

1.000	0.00000000			2.00	0.3010300	3.00	0.4771213	4.00	0.6020600
1.001	0.00043408			2.01	0.3031961	3.01	0.4785665	4.01	0.6031444
1.002	0.00086772			2.02	0.3053514	3.02	0.4800069	4.02	0.6042261
1.003	0.00130093			2.03	0.3074960	3.03	0.4814426	4.03	0.6053050
1.004	0.00173371			2.04	0.3096302	3.04	0.4828736	4.04	0.6063814
1.005	0.00216606			2.05	0.3117539	3.05	0.4842998	4.05	0.6074550
1.006	0.00259798			2.06	0.3138672	3.06	0.4857214	4.06	0.6085260
1.007	0.00302947			2.07	0.3159703	3.07	0.4871384	4.07	0.6095944
1.008	0.00346053			2.08	0.3180633	3.08	0.4885507	4.08	0.6106602
1.009	0.00389117			2.09	0.3201463	3.09	0.4899585	4.09	0.6117233
1.010	0.00432137	1.10	0.0413927	2.10	0.3222193	3.10	0.4913617	4.10	0.6127839
1.011	0.00475116	1.11	0.0453230	2.11	0.3242825	3.11	0.4927604	4.11	0.6138418
1.012	0.00518051	1.12	0.0492180	2.12	0.3263359	3.12	0.4941546	4.12	0.6148972
1.013	0.00560945	1.13	0.0530784	2.13	0.3283796	3.13	0.4955443	4.13	0.6159501
1.014	0.00603795	1.14	0.0569049	2.14	0.3304138	3.14	0.4969296	4.14	0.6170003
1.015	0.00646604	1.15	0.0606978	2.15	0.3324385	3.15	0.4983106	4.15	0.6180481
1.016	0.00689371	1.16	0.0644580	2.16	0.3344538	3.16	0.4996871	4.16	0.6190933
1.017	0.00732095	1.17	0.0681859	2.17	0.3364597	3.17	0.5010593	4.17	0.6201361
1.018	0.00774778	1.18	0.0718820	2.18	0.3384565	3.18	0.5024271	4.18	0.6211763
1.019	0.00817418	1.19	0.0755470	2.19	0.3404441	3.19	0.5037907	4.19	0.6222140
1.020	0.00860017	1.20	0.0791812	2.20	0.3424227	3.20	0.5051500	4.20	0.6232493
1.021	0.00902574	1.21	0.0827854	2.21	0.3443923	3.21	0.5065050	4.21	0.6242821
1.022	0.00945090	1.22	0.0863598	2.22	0.3463530	3.22	0.5078559	4.22	0.6253125
1.023	0.00987563	1.23	0.0899051	2.23	0.3483049	3.23	0.5092025	4.23	0.6263404
1.024	0.01029996	1.24	0.0934217	2.24	0.3502480	3.24	0.5105450	4.24	0.6273659
1.025	0.01072387	1.25	0.0969100	2.25	0.3521825	3.25	0.5118834	4.25	0.6283889
1.026	0.01114736	1.26	0.1003705	2.26	0.3541084	3.26	0.5132176	4.26	0.6294096
1.027	0.01157044	1.27	0.1038037	2.27	0.3560259	3.27	0.5145478	4.27	0.6304279
1.028	0.01199311	1.28	0.1072100	2.28	0.3579348	3.28	0.5158738	4.28	0.6314438
1.029	0.01241537	1.29	0.1105897	2.29	0.3598355	3.29	0.5171959	4.29	0.6324573
1.030	0.01283722	1.30	0.1139434	2.30	0.3617278	3.30	0.5185139	4.30	0.6334685
1.031	0.01325867	1.31	0.1172713	2.31	0.3636120	3.31	0.5198280	4.31	0.6344773
1.032	0.01367970	1.32	0.1205739	2.32	0.3654880	3.32	0.5211381	4.32	0.6354837
1.033	0.01410032	1.33	0.1238516	2.33	0.3673559	3.33	0.5224442	4.33	0.6364879
1.034	0.01452054	1.34	0.1271048	2.34	0.3692159	3.34	0.5237465	4.34	0.6374897
1.035	0.01494035	1.35	0.1303338	2.35	0.3710679	3.35	0.5250448	4.35	0.6384893
1.036	0.01535976	1.36	0.1335389	2.36	0.3729120	3.36	0.5263393	4.36	0.6394865
1.037	0.01577876	1.37	0.1367206	2.37	0.3747483	3.37	0.5276299	4.37	0.6404814
1.038	0.01619735	1.38	0.1398791	2.38	0.3765770	3.38	0.5289167	4.38	0.6414741
1.039	0.01661555	1.39	0.1430148	2.39	0.3783979	3.39	0.5301997	4.39	0.6424645
1.040	0.01703334	1.40	0.1461280	2.40	0.3802112	3.40	0.5314789	4.40	0.6434527
1.041	0.01745073	1.41	0.1492191	2.41	0.3820170	3.41	0.5327544	4.41	0.6444386
1.042	0.01786772	1.42	0.1522883	2.42	0.3838154	3.42	0.5340261	4.42	0.6454223
1.043	0.01828431	1.43	0.1553360	2.43	0.3856063	3.43	0.5352941	4.43	0.6464037
1.044	0.01870050	1.44	0.1583625	2.44	0.3873898	3.44	0.5365584	4.44	0.6473830
1.045	0.01911629	1.45	0.1613680	2.45	0.3891661	3.45	0.5378191	4.45	0.6483600
1.046	0.01953168	1.46	0.1643529	2.46	0.3909351	3.46	0.5390761	4.46	0.6493349
1.047	0.01994668	1.47	0.1673173	2.47	0.3926970	3.47	0.5403295	4.47	0.6503075
1.048	0.02036128	1.48	0.1702617	2.48	0.3944517	3.48	0.5415792	4.48	0.6512780
1.049	0.02077549	1.49	0.1731863	2.49	0.3961993	3.49	0.5428254	4.49	0.6522463
1.050	0.02118930	1.50	0.1760913	2.50	0.3979400	3.50	0.5440680	4.50	0.6532125
1.051	0.02160272	1.51	0.1789769	2.51	0.3996737	3.51	0.5453071	4.51	0.6541765
1.052	0.02201574	1.52	0.1818436	2.52	0.4014005	3.52	0.5465427	4.52	0.6551384
1.053	0.02242837	1.53	0.1846914	2.53	0.4031205	3.53	0.5477747	4.53	0.6560982
1.054	0.02284061	1.54	0.1875207	2.54	0.4048337	3.54	0.5490033	4.54	0.6570559
1.055	0.02325246	1.55	0.1903317	2.55	0.4065402	3.55	0.5502284	4.55	0.6580114
1.056	0.02366392	1.56	0.1931246	2.56	0.4082400	3.56	0.5514500	4.56	0.6589648
1.057	0.02407499	1.57	0.1958997	2.57	0.4099331	3.57	0.5526682	4.57	0.6599162
1.058	0.02448567	1.58	0.1986571	2.58	0.4116197	3.58	0.5538830	4.58	0.6608655
1.059	0.02489596	1.59	0.2013971	2.59	0.4132998	3.59	0.5550944	4.59	0.6618127
1.060	0.02530587	1.60	0.2041200	2.60	0.4149733	3.60	0.5563025	4.60	0.6627578
1.061	0.02571538	1.61	0.2068259	2.61	0.4166405	3.61	0.5575072	4.61	0.6637009
1.062	0.02612452	1.62	0.2095150	2.62	0.4183013	3.62	0.5587086	4.62	0.6646420
1.063	0.02653326	1.63	0.2121876	2.63	0.4199557	3.63	0.5599066	4.63	0.6655810
1.064	0.02694163	1.64	0.2148438	2.64	0.4216039	3.64	0.5611014	4.64	0.6665180
1.065	0.02734961	1.65	0.2174839	2.65	0.4232459	3.65	0.5622929	4.65	0.6674530
1.066	0.02775720	1.66	0.2201081	2.66	0.4248816	3.66	0.5634811	4.66	0.6683859
1.067	0.02816442	1.67	0.2227165	2.67	0.4265113	3.67	0.5646661	4.67	0.6693169
1.068	0.02857125	1.68	0.2253093	2.68	0.4281348	3.68	0.5658478	4.68	0.6702459
1.069	0.02897771	1.69	0.2278867	2.69	0.

Algebra is widely used in day to day activities watch out for my forthcoming posts on Variance Statistics Formula and Definition of least Common Multiple. I am sure they will be helpful.

Equilateral Square

Equilateral square is a geometry figure, square has four sides all the four sides are equal in length then it is said to be equilateral square. In two dimensions square we find only the area. The area is measured in terms of square units. In three dimensions, square is called cube it is also equilateral for that we find the volume only the figure has length, width and height.

Formula for equilateral square:

Area Formulas for equilateral square

The area of a square can be found by multiply the base times itself

Area of square = side × side or a2

Volume of cube =side x side x side a3



AB = BC = CD = AD

Then it is Equilateral Square

Problem for area of square:

1) Find the area of the square, length of one side id 9cm

Solution:

Given: Side a = 9cm

Area of square = a2

= 9 × 9

Area of square = 81cm2

Problems for volume of square

2) Find the volume of the cube its side is 9cm

Solution:

Given: side =9cm

Volume of cube = a3

= a × a × a

= 9 × 9 × 9

= 729cm3

3) Find the area of the square, length of one side id 4m

Solution:

Given: Side a = 4m

Area of square = a2

= 4 × 4

Area of square = 16m2

Problems for volume of square

4) Find the volume of the cube its side is 12m

Solution:

Given: side = 12m

Volume of cube = a3

= a × a × a

= 12 × 12 × 12

= 1728 m3

Examples for equilateral square:

Prove by distance formula that the square is a equilateral square:



Formula  = √(x2 - x1)2+(y2 - y1)2

A(0,0) B(0,1) C(1,1) D(1,0)

Length of AB = √(x2 - x1)2+(y2 - y1)2

AB = √(0-0)2 + (1-0)2

= √0+(1)2

= 1

Length of BC = √(x2 - x1)2+(y2 - y1)2

BC = √(1-0)2 + (1-1)2

= √(1) + (0)

= 1

Length of CD = √(x2 - x1)2+(y2 - y1)2

CD = √(1-1)2 + (0-1)2

= √(0) +(-1)

= 1

Length of AD = √(x2 - x1)2+(y2 - y1)2

AD =√ (1-0)2 + (0-0)2

= √(1) + 0

= 1

Therefore AB = BC = CD = AD =1

All the sides are equal then the square is equilateral square.

Geometry Area Perimeter

In that geometry, the area and perimeter can be calculated for two dimensional and three dimensional shapes. The quantity of surface occupied by a plane figure is called its area. The unit of area is called as square units. The length of the boundary of any closed figure is called as perimeter. The unit of perimeter is meter units. In this article we shall see about the geometry formula for finding the area and the perimeter of simple closed figures like triangle and rectangle, square, and circle. The geometry area and perimeter example problems and practice problems are given below.

Formulas:

Area of square = a2 square units

Perimeter of square = 4 * a

Area of rectangle = length * Width

Perimeter of rectangle = 2( Length + Width)

Area of circle = `Pi` r2

Circumference of circle = 2`Pi` r

Example problems - Geometry area perimeter:

Example problem 1: Find out the area and perimeter of a geometry circle with 15 cm radius.

Solution:

Given:    Radius = 15 cm

Formula:       Area of Circle = radius2

= 3.14 * 152

= 3.14 * 225

On solving this, we get

= 706.5 cm2.

The Circumference of Circle is equal to the multiplication of 2, and ∏ and the radius of the circle

Formula:              Circumference of the Circle = 2 *Π* radius

= 2 * 3.14 * 15

On solving this, we get

= 94.2 cm.

Example problem 2:

Find the geometry surface area of cube for the side is 9 inches.

Solution:

Area of cube A = 6 (side) 2 = 6 * 92

A = 6 * 81

A = 486 square inches.

Example problem 3:

Find the curved surface area, for hemisphere whose radius is 5

Solution:

Curved surface area = 2 * ∏ * r2

= 2 * 3.14 * (5)2

= 2 * 3.14 * 25

= 157

Practice problems - Geometry area perimeter:

Practice problem 1:

Find the area of  circle, For radius is 8

Answer: 200.96

Practice problem 2:

Determine the area and perimeter of rectangle, for length and width are 7meter and 5 meter respectively.

Answer: Area of rectangle = 35 m2

Perimeter of rectangle = 24 m

About Square Root

Square root of a number it can be represented by `sqrt(y)` . The square root of number is also written as exponent form `sqrt(y)` which is equal to y^1/2
                Square root of a number 25 which is equal to `sqrt(25)` =25^1/2=5.
                In this article we shall discuss about the some square roots of the numbers.

Square root table

                The following table shows the square root of the some numbers.
                Number (x)        square (x²)          Square root (x^1/2)
                1                              1                              1.000
                2                              4                              1.414
                3                              9                              1.732
                4                              16                           2.000
                5                              25                           2.236
                6                              36                           2.449
                7                              49                           2.646 
                8                              64                           2.828 
                9                              81                           3.000 
                10                           100                           3.162

Problem 1:

Find the Square root value of 150
Solution:
                Write prime factors for 150 = 2 * 3 * 5 * 5
                `sqrt(150)` = `sqrt(2 * 3 * 5 * 5)`
                                   = `sqrt(2 * 3 * 5^2)`
             `sqrt(2 * 3 * 5^2) ` we can written as (2 * 3 * 5²)^1/2
                (2 * 3 * 5²)^1/2      By using algebraic property (a*b)^m =a^m * b^m
                        (2 * 3)^1/2 * (5^2*1/2)
                5 (2 * 3)^1/2
                        Square root value of 150= 5 `sqrt(6)`

Problem 2:
Find the Square root value of 300
Solution:
                Write prime factors for 300 = 2 * 2 * 3 * 5 * 5
                `sqrt(300)` = `sqrt(2 * 2 * 3 * 5 * 5)`            
                                   = sqrt(2^2 * 3 * 5^2)
                `sqrt(2^2 * 3 * 5^2)` we can written as (2 * 3 * 5²)^1/2
                (2² * 3 * 5²)^1/2     By using algebraic property (a*b)^m =a^m * b^m
                        (3)^1/2 * (2^2*1/2 * 5^2*1/2)
                5 * 2 (3)^1/2
                        Square root value of 300=10 `sqrt(3)`

Problem 3:
Find the Square root value of 625
Solution:
                Write prime factors for 625 = 5 * 5 * 5 * 5
                `sqrt(625) ` = `sqrt(5 * 5 * 5 * 5)`   
                                   = `sqrt(5^2 * 5^2)`
                `sqrt(5^2 * 5^2)` we can written as (5² * 5²)^1/2          
                (5² * 5²)^1/2            By using algebraic property (a*b)^m =a^m * b^m
                        (5²)^1/2 * (5^2*1/2)
                5 * 5
                        Square root value of 625= 25

Problem 4:
Find the Square root value of 75
Solution:
                Write prime factors for 75 = 3 * 5 * 5
                `sqrt(75)` = `sqrt(3*5*5)`
                                   = `sqrt(3*5^2)`
                `sqrt(3*5^2)` we can written as (3*5²)^1/2 
                (3*5²)^1/2               By using algebraic property (a*b)^m =a^m * b^m
                        (3)^1/2 * (5^2*1/2)
                5 (3)^1/2
                        Square root value of 75= 5 `sqrt(3)`

Divide Complex Fractions

Complex fractions are having both the numerator and the denominator values. The complex fractions are also consists of number of rational fractions. The other names for the complex fractions are compound fractions. For example, `(1/3)/(2/3)` is called as the division complex fractions.  Complex fractions are also used for many of the arithmetic operations.

Example problem for divide complex fractions

The example problem for divide complex fractions is,

Problem 1: Divide the following complex fractions, `(2/3)/(3/8)` .

Solution:

Step 1: In the divide complex fractions, the first step is to we have to convert the given fraction into the following,

`(2/3)/(3/8)` = `(2)/(3)` `-:``(3)/(8)`

Step 2: In this step we have to multiply the fraction, so that we have to take reciprocals to the second fractions,

= `(2)/(3)` `xx` `(3)/(8)`

=`(2)/(8)`

=`(1)/(4)`

This is the required solution to the divide complex fractions.

Problem 2: Divide the improper fraction into the complex fractions, `(5 (3/4))/(2/4)` .

Solution:

Step 1: In the first step, we are convert the following into the proper complex fraction, we get,

`(23/4)/(2/4)` = `(23)/(4)` `-:` `(2)/(4)`

Step 2: In the next step, we are multiplying the complex fractions,

`(23)/(4)` `xx` `(4)/(2)`

=`(23)/(2)`

This is the required complex fractions.

Problem 3: Divide the improper fraction into the complex fractions, `(6 (2/8))/(5/8)` .

Solution:

Step 1: In the first step, we are convert the following into the proper complex fraction, we get,

`(50/8)/(5/8)` = `(50)/(8)` `-:` `(5)/(8)`

Step 2: In the next step, we are multiplying the complex fractions,

`(50)/(8)` `xx` `(8)/(5)`

=`(50)/(5)`

`This is the required solution for the complex fractions.`

Problem 4: Divide the following complex fractions, `(2/8)/(4/6)` .

Solution:

Step 1: In the divide complex fractions, the first step is to we have to convert the given fraction into the following,

`(2/8)/(4/6)` = `(2)/(8)` `-:``(4)/(6)`

Step 2: In this step we have to multiply the fraction, so that we have to take reciprocals to the second fractions,

= `(2)/(8)` `xx` `(6)/(4)`

=`(3)/(8)`

`(1)/(4)`

This is the required solution to the divide complex fractions.

Practice problem to divide complex fractions

Problem 1: Divide the following complex fractions, `(2/3)/(4/6)` .

Answer: `(2)/(3)`

Problem 2: Divide the improper fraction into the complex fractions, `(4 (2/4))/(1/4)` .

Answer: 32

Solve Stata Box Plots

Stata box plot is to construct the box and whisker plot for the given statistical records. Statistical data are very large in numbers, so for dividing and conquer the given data, understanding on the given data is necessary, and to find the relationship between the given data, studying stata box plot is more helpful.

Looking out for more help on Anova in Stata in algebra by visiting listed websites.

Solving stata box plots:

While studying stata box plots the data is to be detached. The values obtained from the given datas are Q1, Q2, and Q3, these are median values. Studying stata box plot facilitate for proportional learning of a mixture of sections of box plots.

Steps to study stata box plots:

Step 1: The first step is to arrange the given data.

Step 2: The second step is to find the median (Q2)  for the given data, if there should be two central numbers means find the mean value, it should be the median, if the count is to be odd means, median is the middle value.

Step 3: Find the values present before Q2 is called as lower quartile, the values present after Q2 is called as upper quartile.

Step 4: Find Q1 and Q3 percentiles (middle values of lower quartile and upper quartile).

Step 5: Find the subordinate and superior values.

Step 6: Mark the positions of Q1 , Q2 and Q3 percentiles in the lattice.

Step 7: Draw a box from lower to upper quartile through Q2

Step 8: Find the inter quartile range.

Step 9: Draw a line from lower to higher series.
Example to study stata box plots:

Draw a stats box plots for the given data

Cricket :    10, 20, 90, 20, -10, 80, -20, 25, 65, 85, 5

Footbal:    20, -30, 50, 60, 90, 10, 70, 65, 35, 55, 95

Solution for learning stata box plots is:

Sort out the data in ascending order.

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95

Median:

Median (Cricket)   Q2  = 20
Median (Football) Q2  = 55
Lower Median(Q1)

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95

Lower numbers are

Cricket :  :  -20, -10, 5, 10, 20

Football   :   -30, 10, 20, 35, 50

Middle (Cricket ) Q1= 5

Middle (Football) Q1= 20

Upper Median

Upper numbers are

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95

Upper three numbers are

Cricket  :  25, 65,  80, 85,  90

Football :   60, 65, 70, 90, 95
Middle (Cricket  ) Q3= 80

Middle (Football) Q3 = 70


Maximum

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95


Maximum (Cricket) = 90

Maximum (Football) =95


Minimum

Cricket :  -20, -10, 5, 10, 20, 20, 25, 65,  80, 85,  90

Football:  -30, 10, 20, 35, 50, 55, 60, 65, 70, 90, 95
Minimum (Cricket) = -20

Minimum (Football) =-30


box plot

Study stata box plots

Practice problem for learning stata box plots:

Draw stata box plot for the given set of report

Brand X      : 22, 23, 39, 27, 44, 22, 53, 32, 12, 56, 67

Brand Y      : 22, 34, 76, 45, 98, 35, 19, 49, 12, 32, 34

Multiply and Simplify

Multiplication is an one of the arithmetic operation which extending one number by another. In other words, multiplication is a product of one number to another number. Simplifying or reducing means to make a fraction as simple as possible. In other words, simplifying means both the denominator and numerator divided by the common number. For example `3/6` =`1/2` . Here 3 is a common divisor.

Example problems of multiply and simplify:

Multiplication problem 1:

Jack delivers 350 newspapers in a day. How many newspapers does he deliver in 8 days?

Solution:

We can find the count of newspapers by using the multiplication.

Jack delivers 350 newspapers per day.

Therefore delivering a newspapers in 8 days= 350*8

Here we are Multiplying the 350 and 8. Then we get the final answer.

350*8= 2800

Answer: 2800 newspapers

Multiplication problem 2:

A lodge charges $200 per night for a room. What is the cost of booking the room for 3 nights?

Solution:

We can find the count of cost by using the multiplication.

A lodge charges $200 per night for a room.

Therefore cost of booking room for 3 days = 200*3

Here we are multiplying the 200 and 3. Then we get the final answer.

200*3= 600

Answer: $600

Simplification problem 1:

Simplify the fraction `6/15`

Solution:

We can simplify the given fraction by using the following method.

In this problem, first we can find Greatest common divisor.

6, 15 = 3 is a GCF

Therefore, we are dividing by 3 on both numerator and denominator.

Then we get,

6/15= `2/5`

Answer: `2/5`

Simplification problem 2:

Simplify the fraction `4/18`

Solution:

We can simplify the given fraction by using the following method.

In this problem, first we can find Greatest common divisor.

4, 18 = 2 is a GCF

Therefore, we are dividing by 2 on both numerator and denominator.

Then we get,

`4/18` = `2/9`

Answer: `2/9`

Practice problems of simplify and multiply:

Multiply 25*56
Simplify 5/50
Simplify `14/18`

Answer: 1) 1400 2) `1/10` 3) `7/9`