Question

Please Help! Just equations!

Answer 1

Ae^(-Bt)
When t=0, the rate is 0.1, so A=0.1
When t=100, e^(-Bt)=1/2 and B=ln(2)/100
Then just put the given t and you get the rate at that time.
To get the total oil that leaked in a certain time, just integrate that, because integration is the opposite of rate of change that is a derivative.

Answer 2

it says its not correct..

Answer 3

What is the correct answer?

Answer 4

idk

Answer 5

can you write out the formula? i got 0.1e^ln2/100

Answer 6

That is the formula I used.

Answer 7

its not correct

Answer 8

Oh, but there is a minus in the exponent. But do you have the correct answer with you?

Answer 9

no the answer is online. when i get it right i know. idk what it is now

Answer 10

I cheked my answer again and I really think that is correct, Maybe I'm missing something.
Is that the equation you put in the field?\[R(t)=0.1e^{-\frac{\ln 2}{100}}\]

Answer 11

yes i am putting that in and its not correct

Answer 12

Oh, I wrote it wrong, forgot the t in the exponent.
Well, I would think that the answer key is wrong, but apart from that I can't help you.

Answer 13

where does the t go?

Answer 14

\[R(t)=0.1e^{-\frac{\ln 2}{100}t}\]

Answer 15

ok that was correct. now for the second part do i take the integral?