Answer:
r*4, m+15, 21/k, m-16, 17/g, p+24 . tell me if this is wrong i may have another answer
Step-by-step explanation:
Answer:
1/14
Step-by-step explanation:
There is only 1 common multiple of 4 and 6 between 1 and 14.
So the probability is:
2x-1=5
2(3)-1=5
6-1=5
5=5
x=3
<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>
