# Mathematics

## qlt_task5_0612.docx

A. Joe, a newly hired employee of a News Agency, needs a new cell phone for his job. His company gives him an allowance of $300.00 for his communication line. It is up to him to consume this allowance or save some of it as long as his communication with his company will never be affected.

Now, he meets up with a sales agent of a network service provider who has offered Joe two options: Plan 250 and Plan 300. Plan 250 will require Joe to pay $250 monthly which offers 20 Hours Voice Messaging and unlimited free Text Messages consumable. While the Plan 300, will require Joe to pay $275 monthly with unlimited call and text. Plan 250 charges the rate of voice messaging @ $10/hour when the consumables are used up. Joe was offered these two options since he is expected to use 20 or up to 30 hours of voice messaging. Investigate what Plan should Joe choose and show him his options and limits.

B.

Let y be the amount to be paid monthly and x be the hours used in making voice calls.

Plan 250: y

This is because at x 20 the charge is fixed at $250 otherwise the rate of $10 is added per hour.

Plan 300: y=300

250 + 10(x-20) = 300

( Point where the two Plans will be equal (25, 300), after this point it will be better for Joe to choose Plan 300 )250 + 10x – 200 = 300, distributing 10 to (x-20)

10x = 300 + 200 – 250, transposing all constants to the right side

10x = 250, simplifying

x = 25, dividing both sides by 10 @ x = 25, Plan 250 will pay $300

therefore the solution set is at (25, 300)

_

PLAN 250 10 Hours 20 Hours 30 Hours 250 250 350 PLAN 300 10 Hours 20 Hours 30 Hours 300 300 300 Column1 10 Hours 20 Hours 30 Hours

## revised_rqlt_task_2_V1.docx

y = -(2/3)x + 30

this form follows the slope-intercept form:

y = mx + b, where b is the y-intercept.

b = 30, therefore y – intercept is 30

(0, 30)

to find the x – intercept set y = 0

0 = -(2/3)x + 30, transpose term with x to left side

(2/3)x = 30, multiply 3/2 both sides

( y-intercept (0, 30) this also represent the 3 rd story window (0, 30) )x = 45, x – intercept is at (45, 0)

( x-intercept (45, 0) this also represent the point on the ground where the beam hit (45, 0) )@ x = -30 we solve for y:

y = – (2/3)(30) + 30

y = -20 + 30

y = 10 (height of the beam @ 30 ft

away from the building as

shown also in the graph)

( Note: every division line in the y – axis is incremented by 10 units, and 5 units for every line division on x – axis. )

The first quadrant of the graph is the relevant part for

this problem. This graph maybe a good representation of

the problem since a third-story window maybe realistically

represented by a height of 30-feet and the beam following

a line path y=-(2/3)x + 30, will hit the ground 45 feet away from the building.

## revised_rqlt_task_3_v1.docx

y = 20x + 400

Plan B:

y = 10x + 600

solving for amount when Plan A = Plan B we equate equation 1 and 2

20x + 400 = 10x + 600, transpose all terms with x on the left

20x – 10x = 600 – 400, simplifying

10x = 200

x = 20

@ x = 20 we get:

Plan A: y = 20(20) + 400 = $800 or

( Solution (20, 800) )Plan B: y = 10(20) + 600 = $800

( Solution (20, 800) )Both Plans will have an equal amount of $800 after 20 months (x = 20 months)

( @ 14 months Plan B has a higher amount of savings. )

( y – axis (amount of money in $100) )

( 900 ) ( 800 )

( Solution (20, 800) )

( @ 14 months Plan B has a higher amount of savings. ) ( 700 ) ( 600 ) ( y – intercept (0, 600) )

( 500 )

( x = 14 months ) ( 400 ) ( y – intercept (0, 400) )

( x = 23 months ) ( Solution (20, 800) ) ( 300 )

( 200 )

( 100 )

( x – axis (number of months) )

( 2 0 ) ( 15 ) ( 5 ) ( 1 ) ( 10 )

Legend:

Plan A

Plan B The first quadrant is the relevant quadrant on this.