Question

find s prime (10) for s(t)=300,000e^-.20

Answer 1

s'(t)=0 if there is no t in the equation. did you write the problem correctly?

Answer 2

here
f prime of 10 for F(x)=300,000e^-.20

Answer 3

\[f(x)=300,000e^(-0.20)\] has no variables (x's) in it, so it may as well be f(x)=5
then f'(x)=0 so f'(10)=0

Answer 4

\[find s \prime (10) for s(t)=300000e^(-.20)\]

Answer 5

once again:
s(t)=300000e^(-0.20)=A CONSTANT (no t's in the function)
the derivative of a constant is 0 so s'(t)=0 for ALL t's including t=10
so s'(10)=0

Answer 6

thats not one of the choices.. but thanks anyways

Answer 7

then you wrote the problem wrong. Or it's written wrong in you textbook. you (or they) probably forgot a t somewhere.
maybe it is supposed to be\[s(t)=300000e^(0.2t)\]
then you get \[s'(t)=60000e^(0.2t)\]\[s'(10)=60000e^2\]
is that a choice?

Answer 8

no its either
-2.72
-49123.85
13.62
-8120.12

Answer 9

sorry I forgot the - in the exponent\[s'(t)=-60000e^(-.2t)\]\[s'(10)=-60000e^(-2)=-8120.116994\] so yes, someone forgot a "t"
the answer is definitely
-8120.12

Answer 10

where are you getting -60000?

Answer 11

chain rule on derivative of f(x)=e^(-0.2t)
f'(x)=(-0.2)e^(-0.2t)
so it's 300000(-0.2)

Answer 12

wow thank you.

Answer 13

just curious, who forgot the "t", you or your book/teacher?

Answer 14

teacher. its no where on the question at all

Answer 15

what a clown. let him know how badly he messed up.

Answer 16

oh i will!