What is the decay rate of the function y=0.5(0.83)^x
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OpenStudy (anonymous):
Please help!
OpenStudy (anonymous):
@Mertsj @abb0t @amistre64
OpenStudy (anonymous):
maybe they can help you :)
OpenStudy (anonymous):
@dan815
OpenStudy (anonymous):
lol thanks! my final is tomorrow so i'm freaking out. thanks for trying though! :)
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OpenStudy (anonymous):
@hartnn
OpenStudy (anonymous):
@mathstudent55
OpenStudy (anonymous):
well there's all my smart friends :) goodluck and your welcome
OpenStudy (amistre64):
the thing being exponentiated i believe
OpenStudy (amistre64):
think of it like an interest rate ....
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OpenStudy (anonymous):
ok...
OpenStudy (anonymous):
Like over a period of time?
OpenStudy (amistre64):
yes, "x" refers to time span
OpenStudy (anonymous):
so you make an x and y graph type thing and sort of resolve for each one?
OpenStudy (amistre64):
p0 = k(r)^0
p1 = k(r)^1
p2 = k(r)^2
the rate is the value being ^-ed
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OpenStudy (anonymous):
ok.
OpenStudy (amistre64):
if |r| > 1 it grows
if |r| < 1 it decays
OpenStudy (anonymous):
but then don't you have to find at what rate it decays at?
OpenStudy (amistre64):
it already gives you its rate ....
OpenStudy (anonymous):
on my answer sheet the teacher gaves us it says the answer isr=0.17 i just dont know how she got that.
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OpenStudy (amistre64):
y = 0.5 (0.83)^x
starting amount rate
OpenStudy (anonymous):
*is r=0.17
OpenStudy (amistre64):
let me see if im misreading this .....
OpenStudy (anonymous):
ok! :)
OpenStudy (anonymous):
thank you by the way.
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OpenStudy (amistre64):
i think i see it now
the rate of change is expressed as: a (1+r)^x
1+r = .83
r = .83 - 1
r = -.17
the negative just means its decaying instead of growing