Question

I will give brainliest to anyone who gives me a correct answer

Answer 1

Answer:

C

Step-by-step explanation:

Let's check the first condition [y=110, t=0] for all the choices and eliminate:

A)

$y = 110e^{7t} - 80\\y = 110e^{0} - 80\\y=30$

NO

B)

$y = 110 - 80e^{-5t}\\y = 110 - 80e^{0}\\y=30$

NO

C)

$y = 30 + 80e^{-2t}\\y = 30 + 80e^{0}\\y=110$

YES

D)

$y = 110 - 80e^{2t}\\y = 110 - 80e^{0}\\y=30$

NO

E)

$y = 30 - 80e^{-4t}\\y = 30 - 80e^{0}\\y=-50$

NO

Note: e^0 = 1

Only C is satisfied. But let's make sure the 2nd condition applies as well.

$y = 30 + 80e^{-2t}\\y = 30 + 80e^{-\infty}\\y=30+\frac{80}{e^{\infty}}\\y>>30$

Hence, y approaches 30 as t goes to infinity. 2nd condition is satisfied as well.

Answer is C