NASA launches a rocket at t = 0 seconds. Its height, in meters above sea level, as a function of time is given by h(t) = – 4.9t2 + 139t + 185,
Assuming that the rocket will splash down into the ocean, at what time does splashdown occur?
The rocket splashes down after
seconds.
How high above sea level does the rocket get at its peak?
The rocket peaks at
meters above sea level.

Step-by-step explanation:

y = ax² + bx + c

D = b² - 4ac

= ( - b ± √D ) ÷ 2a

= -

~~~~~~~~~~~~~

h(t) = - 4.9t² + 139t + 185

- 4.9t² + 139t + 185 = 0

D = 139² + 4(- 4.9)(185) = 22947

= ( - 139 ± √22947) ÷ 2(- 4.9)

≈ 29.641 ≈ 30 seconds

< 0

= - 4.9( )² + 139( ) + 185 ≈ 1170.7653 ≈ 1171 meters

what is the approximate angle of acb if abc is 25cm and the length of segment bc is 15

step-by-step explanation:

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