When a baseball is hit by a batter, the height of the ball, h(t), at time t, , is determined by the equation If t is in seconds, for which interval of time is the height of the ball greater than or equal to 52 feet?

t ≤ 1

Step-by-step explanation:

Complete question:

When a baseball is hit by a batter, the height of the ball, h(t), at time t, t=0 is determined by the equation h(t)=-16t^2+64t^2+4. If t is in seconds, for which interval of time is the height of the ball greater than or equal to 52 feet?

If the height of the ball is greater than or equal to 52, the equation becomes;

h(t)=-16t^2+64t^2+4

-16t^2+64t^2+4≥ 48

-16t²+64t + 4 - 52≥0

16t²-64t² +48≥0

-48t²+48≥0

Subtract 48 from both sides

-48t²+48-48≥0-48

-48t²≥-48

t²≤-48/-48

t²≤ 1

t ≤ 1

Hence the required interval is t ≤ 1

step-by-step explanation:

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}