Arock thrown vertically upward from the surface of the moon at a velocity of 36m/sec reaches a height of s = 36t - 0.8 t^2 meters in t sec.
a. find the rock's velocity and acceleration at time t.
b. how long does it take the rock to reach its highest point?
c. how high does the rock go?
d. how long does it take the rock to reach half its maximum height?
e. how long is the rock a loft?

a. The rock's velocity is  and the acceleration is

b. It takes 22.5 seconds to reach the highest point.

c. The rock goes up to 405 m.

d. It reach half its maximum height when time is 6.59 s or 38.41 s.

e. The rock is aloft for 45 seconds.

Step-by-step explanation:

a.

The rock's velocity is the derivative of the height function

The rock's acceleration is the derivative of the velocity function

b. The rock will reach its highest point when the velocity becomes zero.

It takes 22.5 seconds to reach the highest point.

c. The rock reach its highest point when t = 22.5 s

Thus

So the rock goes up to 405 m.

d. The maximum height is 405 m. So the half of its maximum height =

To find the time it reach half its maximum height, we need to solve

For a quadratic equation of the form the solutions are

It reach half its maximum height when time is 6.59 s or 38.41 s.

e. It is aloft until s(t) = 0 again

The rock is aloft for 45 seconds.

x = 17

step-by-step explanation:

1st step: isolate x.

24 - 3x = -27

-24         -24

-3x   =   -51

2nd step: divide.

-3x     =   -51

  3             3

  x       =     17

hope this

school a:

minimum value is 2

maximum value is 22

the lower quartile is 2.5

the median is 10

the upper quartile is 15.5

school b:

minimum value is 9

maximum value is 20

the lower quartile is 12

the median is 16

the upper quartile is 18

box plot for school a isn't symmetrical. the data tails on the right

box plot for school b isn't symmetrical. the data tails on the left

hope i

-lavira: )