SRF Howework fun time >_>

[Squirt]

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 ?!

[MrJoey]

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

[Squirt]

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

[woutR]

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

[SM-Count]

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.

[Squirt]

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 =(

[aznronin]

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

[takolin]

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)

[SM-Count]

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.

[Squirt]

They are all lower case >_>

[SM-Count]

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)

[takolin]

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.

[Squirt]

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

[SM-Count]

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