You are jumping on a trampoline. For one jump, your height y (in feet) above the trampoline after t seconds can be represented by y= -16t^2+24t. How many seconds are you in the air?

.

Step-by-step explanation:

The person in this question is in the air whenever the height is greater than .

The graph of is a parabola opening downwards. Rearrange and find values of that would set this expression to .

.

The first factor, , suggests that would set this expression to - quite expected, since the person is on the ground right before jumping.

The second factor, , suggests that (in other words, ) would also set this expression to . Hence, this person would be once again on the ground seconds after jumping.

Hence, this person is in the air between and for a total of .

1) c 2143.6 cm

2) c 113.0 m

3) a 33.5 ft

4)   d 2 cm

5) c 20 mm

6) 180 in

7) d 120 ft

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}