Tuesday, May 19, 2009

A problem on Simplifying Algebraic Fractions

Fractions is a basic section of math where student get to learn how to do fractions, types of fractions and how to simplify fractions involving algebraic terms or numeric terms as shown in the below problem.


Topic : Simplifying Algebraic Fractions

Below problem is an algebra question which explains how to do fraction.

Problem : Show the steps that you would take to solve the following algebraically




What potential solution did you obtain? Explain why this is not a solution.

Solution :










=> 3 * 2 = 2x - 3x + 9

=> 6 = - x + 9

=> - x + 9 = 6

=> - x = 6 - 9

=> -x = - 3

=> x = 3

This is not a solution because at x = 3, the denominators of the terms at the right hand side and the left hand side become ‘zero’. So, x = 3 is only a potential solution and not a solution.

Friday, May 15, 2009

Using Shell or Disc Method to Find Volume of the Solid

Disc method and Shell(cylinder) method of integration are the two different methods of finding volume of solid of a revolution, using rectangular coordination system the functions are defined in terms of x in the below problem.

Topic : Disc or Cylinder Method of Finding Volume of the Sphere.
Problem : Use the disc or shell method to find the volume of the solid generated by revolving the regions bounded by the graphs of the equations about the specified line y = 8.
y=x3 y=0, x=2
Solution :
Here the solid is rotated along x-axis
y = x3 => x = (y)1/3

or x = y1/3

when y = 8, x = (8)1/3 = (23)1/3 = 23*1/3 = 2

So a = 0, b = 2

Volume of a Solid by rotating about y = 8 is given by:

V = 2π02(y)1/3 . y dy

= 2π02(y)4/3 dy

= 2π[y7/3/(7/3)0]2

= 2π[(2)7/3/(7/3)- (0)7/3/(7/3)]

= 2π[((27)1/3/(7/3)]

= 2π 3√(128)/(7/3)

= 6.3√(128)π/7

= 4.32π

Sunday, May 3, 2009

Word problem on Geometry construction - Ellipse

Here is a word problem on geometrical construction of geometrical figure called Ellipse, where maximum distance between two points is calculated.

Topic : Ellipse
Geometry Construction

Focus point, elliptical orbit and major axis are few important terms used in this Ellipse concept.


Question : Hally's comet has an elliptical orbit with the sun at one focus and a major axis of 1,636,484,848 miles. T
he closest the comet comes to the sun is 54,004,000miles. What is the maximum distance from the comet to the sun?

Solution :














So SC = 54,004,000miles

maximum distance from the comet to the sun will be DS

That is when comet comes at D there is a maximum distance from sun


So SD = CD - SC
where CD is the major axis = 1,636,484,848 miles


SD = 1,636,484,848 miles - 54,004,000miles

SD = 1582480848 miles


So the maximum distance is 1,582,480,848 miles


Hope the attributes involved in ellipse concept are understood.

Thursday, April 9, 2009

Briefly Explain How to Find Sum of Geometric Progression

Topic : Geometric Progression

Question : Explanation to Find How to Find Sum of terms in Geometric Progression

Solution :


Monday, April 6, 2009

Question to Prove Rectangle Inscribed the Circle Properties

Topic : Rectangle Inscribed the Circle

Question : If RTCE is a rectangle, then show that RC is the diameter of the circle.









Solution :
Since ∟RTC = 90º
So measure of arc REC = 180
mREC = 180º
This shows that the arc REC is half of the circle as the total measure of circle is 180º X 2 = 360º

That means arc REC forms a semi circle
Hence RC passes through the center of the circle or we can say that RC is the diameter of the circle.
Hence proved.

Tuesday, March 31, 2009

Question to Draw Box Plot for a Collection of Data

Topic : Box Plot

Question : In a survey of the cafeteria food at Metropiolis Middle School, 50 students were asked to rate how well they liked the lunches on a scale of 1 to 10, with 1 being the lowest rating and 10 being the highest rating. The box plot was made from the collected data.
What percent of the students in the sample rated the cafeteria food between 5.75 and 9


Solution :








First Quartile = Q1 = 5.75
That is 25% of students lie between 3 and 5.75
Second Quartile = Q2 = 6.5
That is 50% of students lie between 3 and 6.5
Third Quartile = Q3 = 8
That is 75% of the students lie between 3 and 8
100% of students lie between 3 and 9
So the % of students who lie between 5.75 and 9 = 100% - 25% = 75%
and Answer is 75%





Monday, March 23, 2009

Question on Finding the area of a Field

Topic : Area of Rectangle


Question : A farmer wants to use 1000 meters of fencing to enclose 3 sides of a rectangular field. What dimensions should be used to maximize the area of the field, and what is the maximum area of the field?


Solution :