sandy wrote:
f(t) = kt for all t, where k is a constant, and f(3)= \(\frac{1}{2}\)
Quantity A: k
Quantity B: f(1)
GIVEN: f(t) = kt for all t, where k is a constant
So, f(3) =
3kSince we're told that f(3) = \(\frac{1}{2}\), we can conclude that
3k = \(\frac{1}{2}\)
Divide both sides by 3 to get: \(k = \frac{1}{6}\)
So, we can now say that f(t) \(= \frac{1}{6}t\)
We get:
QUANTITY A: \(\frac{1}{6}\)
QUANTITY B: f(1) \(= \frac{1}{6}(1) = \frac{1}{6}\)
Answer: C
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep