Question

Help?

Answer 1

well if you think of this as exponential growth

\[A=Ce^{-kt}\]

Answer 2

and at time =0 you get

\[A=Ce^0=C\]

Answer 3

what is the amount at time 0

Answer 4

another name for it

Answer 5

I have no idea /:

Answer 6

alright if you have 3 gummy bears to start off (meaning no time has gone on) and then give 2 away, what was the 3 gummy bears (another name for it

Answer 7

the original amount....

Answer 8

oh, initial?

Answer 9

yes

Answer 10

Ok, that makes sense. What about the second problem?

Answer 11

yes you can look at it also like this

my example is what i saw in my Differential Class

for yours

if you let t =0
\[A(0)=4(\frac{1}{2})^0=4*1=4\]

Answer 12

for the second one you'll have to draw the graph

Answer 13

just plug in 0 1 ,2 ,3 ,4 you'll see whether it's exponentially growing (curve) or if it's a line (linear). If it's going from high to low, it's decay. If it goes from low to high it's growth

Answer 14

growth function with a linear graph?

Answer 15

not quite there is a hint with the exponent within the function... that should hint that it's an exponential graph

Answer 16

i'll do 0,1 and 2 for you but you should do 3 and 4

\[A(0)=4\]
\[A(1)=4(\frac{1}{2})^{\frac{1}{5}}=3.48\]

Answer 17

Oh..... so it's a decay function, right?

Answer 18

yep

Answer 19

I think I get how to do it now then. Thank you!

Answer 20

yep the first question is just a form type question

\[A(t)=C(g)^{kt}\]

where A(t) is the growth/decay at time t, C is the initial, k is the rate and t is time

Answer 21

second is all graphing

Answer 22

Ok I'm just going to practice some more. Thanks again!