Let’s revisit the situation of making a medical supply drop from a helicopter to survivors of a natural disaster. The crate can withstand an impact of 165 ft/sec. If the helicopter has to make the drop at 450ft, then the height of the crate as a function of time is: h(t) = 450 -16t2. The crate will land 5.3 sec after the drop is initiated. Will the crate and medical supplies survive? Show/explain how you know. You need to compare the instantaneous rates of change.

When dropped from 450 feet, the crate and medical supplies will not survive the impact

Step-by-step explanation:

The impact that the crate can withstand when dropped = 165 ft,/sec

The function for the height of the crate at 450 ft. = 450 - 16·t²

The time the crate will land after drop = 5.3 sec

Therefore, we have;

Velocity, V(t) = d(h)/dt

∴ v(t) = d(450 - 16·t²)/dt = -32·t

Given that the time the cate lands after drop, t = 5.3 seconds, we have;

The velocity at 5.3 seconds after drop, v(5.3) = -32 × 5.3 = -169.6

The velocity at 5.3 seconds after drop, v(5.3) = 169.6 ft./sec. (downwards)

Therefore, given that the velocity of the crate at 5.3 seconds (169.6 ft./sec.) is higher than the velocity at which the crate can withstand an impact (165 ft./sec.) the crate and medical supplies will not survive.

8 2/3 is the answer

step-by-step explanation: just use a calculator

i believe it is c.y = 9x

rise over run. 9 units up over 1 unit right or 9/1