Angie and kenny play online video games. angie buys 1 software package and 3 months of game play. kenny buys 1 software package and 4 months of game play. each software package cost $50. if the total cost is 240, what is the cost of one month of gameplay ?

The cost of one month of game-play =$20

Step-by-step explanation:

Let the cost of one month of gameplay be x

Then cost of game-play bought by Angie =3x(1)

and Then cost of game-play bought by Kenny=4x(2)

Cost of each software package =$50(3)

The the total cost =240= sum of costs of software bought by both of them and game-play)=50+50+3x+4x

⇒240=100+7x→(by adding like terms)

⇒140=7x⇒x=20→( dividing both sides by 7 )

∴the cost of one month of game-play =$20

Given

Angie and Kenny play online video games

Angie buys 1 software package and 3 months of game play

Kenny buys 1 software package and 4 months of game play

Each software package cost $50

total cost is 240

Find out the cost of one month of gameplay.

To proof

let us assume that the cost of one month of game play = x

As given in the question

Angie buys 1 software package and 3 months of game play

Kenny buys 1 software package and 4 months of game play

Each software package cost  = $50

Total cost =$240

Than the equation becomes

( 50 + 3x ) + ( 50 + 4x ) = 240

100 + 7x= 240

7x= 140

x = $20

The cost of one month gameplay  is $ 240.

Hence proved

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}