Category: Arithmetic
"Published in Newark, California, USA"
Express 0.454545454545454545454545454545....... into its equivalent fraction.
Solution:
The above decimal is a repeating decimal because the digit, 45 is repeating endlessly. It is usually happened if a fraction cannot be converted into basic fraction. Basic fraction is a fraction whose denominator is a multiple of 10. If the denominator is not a multiple of 10 but the multiples of its factors like 2 and 5 are given, then you can convert it into the multiples of 10. For example, if the denominator is 4, then you need to multiply both the numerator and the denominator by 25 so that the denominator will be 100.
How about if the denominator is not a multiple of 10 and the factors are not the factors of 10? Well, there's no way to convert it into the multiples of 10. For example if the denominator is 35, then the factors are 5 and 7. Since 7 is not a factor of 10, then there's no way to convert the denominator into the multiples of 10. If you will do the manual division by dividing the numerator by denominator, you will notice that there's no end in division because the remainder is always the same and the digits of the quotient are repeating endlessly that's why the given decimal is called a repeating decimal.
Let's go back to the problem. How do you convert the repeating decimal into a fraction? Well, there's a way to convert the repeating decimal into a fraction. Here's the procedure and solution
Let x = 0.454545454545454545454545454545..........
100 x = 45.45454545454545454545454545454545......
Subtract x from 100 x, we have
100 x = 45.4545454545454545454545454545454545......
- x = -0.454545454545454545454545454545454......
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99 x = 45
Divide both sides of the equation by 99
As a rule in Mathematics that all fractions must be simplified into a lowest term. Factor 45 and 99 as follows
Their Greatest Common Factor (GCF) is 9. Cancel 9 on both sides of the fraction. Therefore, the answer is

This website will show the principles of solving Math problems in Arithmetic, Algebra, Plane Geometry, Solid Geometry, Analytic Geometry, Trigonometry, Differential Calculus, Integral Calculus, Statistics, Differential Equations, Physics, Mechanics, Strength of Materials, and Chemical Engineering Math that we are using anywhere in everyday life. This website is also about the derivation of common formulas and equations. (Founded on September 28, 2012 in Newark, California, USA)
Tuesday, December 4, 2012
Monday, December 3, 2012
Solving Trigonometric Equations
Category: Trigonometry, Algebra
"Published in Newark, California, USA"
Find the values of x in the range from 0º to 360º for
Sin x + Sin 2x + Sin 3x = 0
Solution:
Consider the given equation
Sin x + Sin 2x + Sin 3x = 0
Apply the Sum and Difference of Two Angles Formula and Double Angle Formula for the above equation
Sin x + 2 Sin x Cos x + Sin (x + 2x) = 0
Sin x + 2 Sin x Cos x + Sin x Cos 2x + Cos x Sin 2x = 0
Sin x + 2 Sin x Cos x + Sin x (Cos2 x - Sin2 x) + Cos x (2 Sin x Cos x) = 0
Sin x + 2 Sin x Cos x + Sin x Cos2 x - Sin3 x + 2 Sin x Cos2 x = 0
Sin x + 2 Sin x Cos x + 3 Sin x Cos2 x - Sin3 x = 0
Take out their common factor which is Sin x, we have
Sin x (1 + 2 Cos x + 3 Cos2 x - Sin2 x) = 0
Sin x [1 + 2 Cos x + 3 Cos2 x - (1 - Cos2 x)] = 0
Sin x (1 + 2 Cos x + 3 Cos2 x - 1 + Cos2 x) = 0
Sin x (2 Cos x + 4 Cos2 x) = 0
(Sin x )(2 Cos x)(1 + 2 Cos x) = 0
Equate each factor in zero
For Sin x = 0
x = Sin-1 0
x = 0º, 180º, 360º
For 2 Cos x = 0
Cos x = 0
x = Cos-1 0
x = 90º, 270º
For 1 + 2 Cos x = 0
2 Cos x = -1
Cos x = - ½
x = Cos-1 -½
x = 120º, 240º
Therefore,
x = 0º, 90º, 120º, 180º, 240º, 270º, and 360º
"Published in Newark, California, USA"
Find the values of x in the range from 0º to 360º for
Sin x + Sin 2x + Sin 3x = 0
Solution:
Consider the given equation
Sin x + Sin 2x + Sin 3x = 0
Apply the Sum and Difference of Two Angles Formula and Double Angle Formula for the above equation
Sin x + 2 Sin x Cos x + Sin (x + 2x) = 0
Sin x + 2 Sin x Cos x + Sin x Cos 2x + Cos x Sin 2x = 0
Sin x + 2 Sin x Cos x + Sin x (Cos2 x - Sin2 x) + Cos x (2 Sin x Cos x) = 0
Sin x + 2 Sin x Cos x + Sin x Cos2 x - Sin3 x + 2 Sin x Cos2 x = 0
Sin x + 2 Sin x Cos x + 3 Sin x Cos2 x - Sin3 x = 0
Take out their common factor which is Sin x, we have
Sin x (1 + 2 Cos x + 3 Cos2 x - Sin2 x) = 0
Sin x [1 + 2 Cos x + 3 Cos2 x - (1 - Cos2 x)] = 0
Sin x (1 + 2 Cos x + 3 Cos2 x - 1 + Cos2 x) = 0
Sin x (2 Cos x + 4 Cos2 x) = 0
(Sin x )(2 Cos x)(1 + 2 Cos x) = 0
Equate each factor in zero
For Sin x = 0
x = Sin-1 0
x = 0º, 180º, 360º
For 2 Cos x = 0
Cos x = 0
x = Cos-1 0
x = 90º, 270º
For 1 + 2 Cos x = 0
2 Cos x = -1
Cos x = - ½
x = Cos-1 -½
x = 120º, 240º
Therefore,
x = 0º, 90º, 120º, 180º, 240º, 270º, and 360º
Sunday, December 2, 2012
Frustum - Right Circular Cone
Category: Solid Geometry, Plane Geometry
"Published in Newark, California, USA"
The frustum of a right circular cone has a slant height of 9 ft. , and the radii of the bases are 5 ft. and 7 ft. Find the lateral area and the total area. What is the altitude of this frustum? Find the altitude of the cone that was remove to leave this frustum. Find the volume of the frustum.
Solution:
To illustrate the problem, you can draw the figure as follows
Since the slant height and the radii of the bases are given, we can get the lateral area, area of the bases, and the total area of the frustum.
Let's get the circumference of the top base of the frustum as follows
For the bottom base of the frustum
Therefore, the lateral area of the frustum is
Let's get the area of the top base of the frustum as follows
For the bottom base of the frustum
Therefore, the total area of the frustum is
Since the altitude of the frustum is not given in the problem, we have to find the altitude of the frustum with the use of the vertical section of the frustum. The vertical section of the frustum is a trapezoid.
From the figure, we can get the altitude of a trapezoid which is also the altitude of a frustum. As you notice that the end portion of a trapezoid is a right triangle. We can use the Pythagorean Theorem to solve for the altitude of the frustum as follows
In order to get the altitude of the cone that was removed to leave this frustum, we have use the formula for similar triangles as follows
Finally, we can get the volume of the frustum as follows
"Published in Newark, California, USA"
The frustum of a right circular cone has a slant height of 9 ft. , and the radii of the bases are 5 ft. and 7 ft. Find the lateral area and the total area. What is the altitude of this frustum? Find the altitude of the cone that was remove to leave this frustum. Find the volume of the frustum.
Solution:
To illustrate the problem, you can draw the figure as follows
![]() |
Photo by Math Principles in Everyday Life |
Since the slant height and the radii of the bases are given, we can get the lateral area, area of the bases, and the total area of the frustum.
Let's get the circumference of the top base of the frustum as follows
For the bottom base of the frustum
Therefore, the lateral area of the frustum is
Let's get the area of the top base of the frustum as follows
For the bottom base of the frustum
Therefore, the total area of the frustum is
Since the altitude of the frustum is not given in the problem, we have to find the altitude of the frustum with the use of the vertical section of the frustum. The vertical section of the frustum is a trapezoid.
![]() |
Photo by Math Principles in Everyday Life |
From the figure, we can get the altitude of a trapezoid which is also the altitude of a frustum. As you notice that the end portion of a trapezoid is a right triangle. We can use the Pythagorean Theorem to solve for the altitude of the frustum as follows
In order to get the altitude of the cone that was removed to leave this frustum, we have use the formula for similar triangles as follows
Finally, we can get the volume of the frustum as follows
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