GIVEN: f(t) = kt for all t, where k is a constant
So, f(3) = 3k

Since 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
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The exercise gives a function f(t)=kt.

And ask you to compare f(1) with k

Replacing t=1 in the function

f(1)=kx1=k

Answer: Quantities are equal (independent of the numeric value of k)

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