A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet? A. 3 B. t<4
C. t>4
D. 3>t>4

3 < t < 4

Step-by-step explanation:

Given the following information :

Initial Velocity of projectile = 112 ft/s

Height (h) above the ground after t seconds :

h = –16t2 + 112t

To calculate the time when h exceeds 192 Feets (h >192)

That is ;

-16t^2 + 112t = > 192

-16t^2 + 112t - 192 = 0

Divide through by - 16

t^2 - 7t + 12 = 0

Factorizing

t^2 - 3t - 4t + 12 =0

t(t - 3) - 4(t - 3) = 0

(t-3) = 0 or (t-4) =0

t = 3 or t=4

Therefore, t exist between 3 and 4 for height in excess of 192ft

–16(3)^2 + 112(3) = 192 feets

–16(4)^2 + 112(4) = 192 feets

3 < t < 4

question 1 is 8, question 2 is 9, question 3 is 24

step-by-step explanation:

now, the numbers they give you at the top are the key. from there, you just have to plug and chug! let's try 1, 2, and 3!

key: a=3, b=5, c=6

1. a+5

remember, a=3, so we plug that in to get 3+5=8

2. 15-c

c=6, so we plug that in to get 15-6=9

3. 4b

b=6, so we plug it in to get 4*6=24

time to try the rest on your own! if you put the answers below in a comment, i'll check them for you!

45

step-by-step explanation:

to find a percentage, add a decimal in front and multiply it by the other number.

.25 times 180= 45