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Help please >_>.
I was placed in AP calc AB and have summer work but I suck at math horribly Don't know what the hell this is asking D=

Givn f(t)= C e ^-kt
1.) If f(0) = 5 what is the value of C?

2.) If f(3)= 6, What is the value of k



There are two variables but I can only plug in one, what is this witch craft ?!

If you suck at math, why are you in an AP class?

Because public schools are retarded and won't switch me out is why.....

shit finally dawned on me: I'll never do maths again

If you plug in 0 for t for the first one then k doesn't matter, c=5. You should be able to do the second one.

Can someone explain why its 5? ( Sorry haven't done math in like 3 months....) and can you do number 2 step by step please =(

Is answer to the second one: k = -1/3 ln 6/5 ?

f(t) = C e^-kt
f(0) = 5
f(0) = C e^0 = 5
f(0) = C = 5

For the 1st one.

5=Ce^(-kt) (T = 0)
5=Ce^(-k*0)
5=Ce^0
5=C (any number raised to the zeroth power equals 1)


For the 2nd one, I assume c still equals 5 (from step 1) and thus you need to solve the following equation.

6=5e^(-3k)

No, F'(t) = f(t), for the second one you have to use the first derivation of F(t) for the plugging, assuming he didn't just screw up capitalization.

They are all lower case >_>

You'll want to edit that, F(t) and f(t) are not the same thing.

Oh yeah, forgot to put my solutions, I knew I was forgetting something:
1)


2)

SM-Count wrote:You'll want to edit that, F(t) and f(t) are not the same thing.

Back to my solution?

Well the other option would be solve "6=Ce^(-3k)" while not specifying C, though you'll need to say that C can't be 0.

Yeah it says using answer from part 1 for use in the question 2

Yeah, you and azn were right, it's just plugging. I just assumed since it's calculus homework there'd be calculus XD