Answer:
Step-by-step explanation:
<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>
Answer:
Either A or C, I'm not quite sure tho
Step-by-step explanation:
The line is on the negative side of the graph
Hey there!
To plot the inequality y≥-6 on the coordinate plane, keep in mind that whenever you have y set equal to a constant, your graph will be a horizontal line. Therefore, your graph would be a solid (not dashed since your sign is greater than or equal to) horizontal line passing through (0,-6). Then you would shade the portion over the line to signify that the values of y must be greater than or equal to -6.
The graph below shows y≥-6: