Ask your own question, for FREE!
Mathematics 12 Online
OpenStudy (anonymous):

Find the annual percent increase or decrease that y = 0.35(2.3)x models. (1 point) A.230% increase B.130% increase C.30% decrease D.65% decrease

OpenStudy (anonymous):

@hartnn help? lol

hartnn (hartnn):

u said that was last Question! lol :"P

hartnn (hartnn):

i was just kidding....

OpenStudy (anonymous):

lol i know sorry but then i just wanted to get this outta the way. I only have 4 questions this time i promise

hartnn (hartnn):

If x measures time (in years) then the value of y this year is 0.35·2.3^x, and next year it's 0.35·2.3^(x+1) so, can u calculate annual increase from this ?

OpenStudy (anonymous):

Please, please please just give me a letter? I beg you!

hartnn (hartnn):

sorry, that i cannot. its against CoC here.

hartnn (hartnn):

u just subtract 0.35·2.3^x from 0.35·2.3^(x+1)

OpenStudy (anonymous):

-_-

hartnn (hartnn):

its not difficult really.....

OpenStudy (anonymous):

is that a x sign in between 0.35 and the 2.3^x ?

hartnn (hartnn):

and i believe the Question is \(y=0.35 (0.23)^x\) right ? and yes.

hartnn (hartnn):

\(\% increase=(final-initial)*100/initial \) here, initial is =0.35·2.3^x final =0.35·2.3^(x+1)

hartnn (hartnn):

just plug in values, and u get the answer easily.

OpenStudy (anonymous):

i got it thanks. sorry i got that one an then just started doing the other ones. thank you

hartnn (hartnn):

what u got it as ?

hartnn (hartnn):

so that i can verify..

OpenStudy (anonymous):

130

OpenStudy (anonymous):

b

OpenStudy (anonymous):

130% is correct in this instance. Consider the percent increase from 0.35(2.3)^1 to 0.35(2.3)^2

hartnn (hartnn):

yes, its correct

OpenStudy (anonymous):

For growth functions, the easiest way to determine the growth rate is to consider it of the form: y = a(1 + r)^x, where r is the percent increase

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!