Answer: The expression that represents Meg's finishing time in June is "y - 10".
The problem started with Meg running in April. She had a time in April and we called it "y".
Now, Meg ran again in June. In June, she did 10 seconds faster. So it makes since that we need to subtract 10 from her April time, "y". Therefore, the expression is simply "y - 10".
4.7 because if you plug it in- (2,1) and (6,6) it is 4.7 units
Hello from MrBillDoesMath!
Answer:
(g o f)(0) cannot be evaluated
Discussion:
(g o f)(0) = g ( f(0)) if it exists but f(0) is not defined (as division by zero is not defined) so
(g o f)(0) cannot be evaluated
Thank you,
MrB
Answer:
3
Step-by-step explanation:
lim(t→∞) [t ln(1 + 3/t) ]
If we evaluate the limit, we get:
∞ ln(1 + 3/∞)
∞ ln(1 + 0)
∞ 0
This is undetermined. To apply L'Hopital's rule, we need to rewrite this so the limit evaluates to ∞/∞ or 0/0.
lim(t→∞) [t ln(1 + 3/t) ]
lim(t→∞) [ln(1 + 3/t) / (1/t)]
This evaluates to 0/0. We can simplify a little with u substitution:
lim(u→0) [ln(1 + 3u) / u]
Applying L'Hopital's rule:
lim(u→0) [1/(1 + 3u) × 3 / 1]
lim(u→0) [3 / (1 + 3u)]
3 / (1 + 0)
3
