A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find out
the time at which the rocket will reach its max, to the nearest 100th a second.
y = -16x^2 +100x+ 72

30005

yes ok                                            

(a) {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …}

(b) $59 398

step-by-step explanation:

(a) the annual sequence of profits

aₙ = a₁rⁿ⁻¹

the model for the sequence of profits is {a₁, a₁r, a₁r², … a₁rⁿ⁻¹, …} .

(b) profit after 15 yr

  a₁ = $ 30 000

  r = 1.05

a₁₅ = 30 000 × 1.05¹⁴

a₁₅ = 30 000 × 1.979 93

a₁₅ = $59 398

the profit in year 15 will be $59 398.

the answer is option a you would get more money.it would be $250.

step-by-step explanation:

for option b 200/2=100=50% 100/2=50=25% 50/5=10=5% it would be $220