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

answer: you need to organize this better it does not make sense.

step-by-step explanation: ?

there are 70 more legs than eyes at the dog show (196 – 126 = 70). dogs have two more legs than eyes, and humans have the same number of legs as eyes. thus 70 is twice the number of dogs at the show, so the number of dogs is half of 70, which is 35. to get the number of humans, subtract 35 from 63: 63 – 35 = 28.

step-by-step explanation: