Aball is dropped from a height of 12 feet returns to a height that is 1/2 of the height from which it fell. the ball continue to bounce half the height of the previous bounce each time. how far will the ball have traveled when it hits the ground the fifth time

Step-by-step explanation:

I'm not sure if you're asking the total amount of feet it will have traveled in 5 bounces, or if you're asking how far the ball will travel on its 5th bounce.  I think it's the latter you want answered, so here goes.

This is an exponential equation, of the form

 where a is the initial height and b is the decay factor.  Our initial height is 12 feet, and since it comes up 1/2 of its previous height each time it bounces, b = 1/2.  Our equation, then, is

If you want to find out how far it drops on the fifth bounce, sub in a 5 for x and solve for y:

which gives you a height of .375 feet which is 3/8 of a foot.

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}