<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>
<h2>
Answer:</h2>
<h2>
cos theta= adjacent side/ hypotenuse</h2><h2 /><h2>
cos theta= adjacent side/ hypotenusecos 36= x/8</h2><h2 /><h2>
cos theta= adjacent side/ hypotenusecos 36= x/8√5+1/4 = x/8</h2><h2 /><h2>
cos theta= adjacent side/ hypotenuse </h2>
cos 36= x/8√5+1/4 = x/8
4.7 is the approximate answer
Probably d as the the temperature outside is unpredictable and the we know the time is always going up
