12) A new amusement park decides to offer its visitors a choice between two plans. With Plan 1 they pay $6 to enter the park and they pay $3.50 for each ride. With Plan 2 they pay $12 to enter the park and $1.50 for each ride. Write an
equation then find the number of rides when the cost will be the same. What will that cost be?

9514 1404 393

Step-by-step explanation:

The plan 1 cost for x rides will be ...

  y = 6 + 3.50x

The plan 2 cost for x rides will be ...

  y = 12 + 1.50x

These costs will be equal when ...

  y = y

  6 + 3.50x = 12 + 1.50x . . . . . substitute for y

  2.00x = 6 . . . . . . . . . . . . . subtract 6+1.50x

  x = 3 . . . . . . . . . . . . . . . divide by 2

For 3 rides, the cost will be the same.

__

For 3 rides the cost is ...

  y = 6 + 3.50(3) = 6 + 10.50 = 16.50

The cost for 3 rides will be $16.50 for either plan.

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}