a small rocket is shot from the edge of a cliff suppose that after t seconds the rocket is y meters above the cliff where y=30t-5t^2 what’s the greatest height the rocket reached? after how many seconds does the rocket reach its height? how far above the cliff edge is the rocket after 5 seconds? when is the rocket 40 meters above the cliff edge? where is the rocket after 7 seconds?

Greatest height: 45 meters

Time for greatest height: 3 seconds

Height after 5 seconds: 25 meters above the cliff

Time for height of 40 meters: 7.123 seconds

Height after 7 seconds: -35 meters (35 meters below the cliff)

Step-by-step explanation:

to find the maximum height, we need to calculate the derivative of y in relation to t and then find when dy/dt = 0:

dy/dt = 30 - 10t = 0

10t = 30

t = 3 seconds

In this time, the height is:

y = 30*3 - 5*3^2 = 45 meters

After 5 seconds, the height is:

y = 30*5 - 5*5^2 = 25 meters

The time for the height of 40 meters is:

40 = 30t - 5t^2

t^2 - 6t - 8 = 0

Using Bhaskara's formula, we have:

Delta = 6^2 + 4*8 = 68

sqrt(Delta) = 8.246

t1 = (6 + 8.246) / 2 = 7.123 seconds

t2 = (6 - 8.246) / 2 = -1.123 seconds (negative value for time is not valid)

So the time when the rocket reaches 40 meters is 7.123 seconds

After 7 seconds, the height is:

y = 30*7 - 5*7^2 = -35 meters

The rocket will be 35 meters below the cliff.

∠acb = 8°

step-by-step explanation:

∠xba lies on the circle and is half the measure of the intercepted arc ax

hence

measure of arc ax = 2 × 42° = 84°

∠acb is outside the circle and is

∠acb = (arcab - arcax ), hence

∠acb = 0.5 (100 - 84) = 0.5 × 16° = 8°

two data sets with the same median will not have the same varaibility

step-by-step explanation: