Ask your own question, for FREE!
Mathematics 8 Online
OpenStudy (practice111):

Find the exponential function that satisfies the given conditions. (2 points) Initial value = 30, increasing at a rate of 13% per year f(t) = 30 ⋅ 1.13t f(t) = 13 ⋅ 1.13t f(t) = 30 ⋅ 0.13t f(t) = 30 ⋅ 13t I'm almost sure is A @imqwerty ??

OpenStudy (anonymous):

Find 13% as a decimal, and that will be your second part. Remember for decimals, you bring the decimal point from the very right over two spaces to the left.

OpenStudy (practice111):

So am i right ? @armyengineer225

OpenStudy (anonymous):

If you mean that it is the first answer, no. 13% as a decimal is 0.13. Look for the answer that has 0.13.

OpenStudy (anonymous):

that's not an increase of 13%. an increase in 13% = 113% or=1.13

OpenStudy (practice111):

so i was right? @♂

OpenStudy (anonymous):

I think so

OpenStudy (practice111):

can you help me with one more?

OpenStudy (anonymous):

give me a bit to check the graphs

OpenStudy (practice111):

Find the exponential function that satisfies the given conditions. (2 points) Initial value = 70, decreasing at a rate of 0.5% per week f(t) = 0.5⋅ 0.3t f(t) = 70 ⋅ 1.005t f(t) = 70 ⋅ 0.995t f(t) = 70 ⋅ 1.5t @♂

OpenStudy (anonymous):

c

OpenStudy (practice111):

how did you get it?

OpenStudy (practice111):

i mean the percentage

OpenStudy (anonymous):

0.5%=0.005 1-0.005=0.995

OpenStudy (practice111):

thanks :) can i ask you one more and that is it please?

OpenStudy (anonymous):

yeah

OpenStudy (practice111):

The decay of 942 mg of an isotope is described by the function A(t)= 942e-0.012t, where t is time in years. Find the amount left after 71 years. Round your answer to the nearest mg. Show all of your work for full credit.

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!