<span>If you plug in 0, you get the indeterminate form 0/0. You can, therefore, apply L'Hopital's Rule to get the limit as h approaches 0 of e^(2+h),
which is just e^2.
</span><span><span><span>[e^(<span>2+h) </span></span>− <span>e^2]/</span></span>h </span>= [<span><span><span>e^2</span>(<span>e^h</span>−1)]/</span>h
</span><span>so in the limit, as h goes to 0, you'll notice that the numerator and denominator each go to zero (e^h goes to 1, and so e^h-1 goes to zero). This means the form is 'indeterminate' (here, 0/0), so we may use L'Hoptial's rule:
</span><span>
=<span>e^2</span></span>
The answer is C.
SA = bh + (s1 + s2 + s3)H
SA = (6.8)(5.1) + (6.8 + 5.1 + 8.5)(2.5) = 85.68
Answer:
55
Step-by-step explanation:
2−5+1+8
11−5
i add 8 because n is 6 so add 2 it is 8 so i put it in the problem and got 55
have a nice day and sorry if wrong T_T
K^2 - 100h^2
(k - 10h)(k + 10h)
