It’s d, because thats the formula !! mark me brainliest so I can help more people please !!
<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>
Hi there!

We are given:
cos(7x)cos(4x) = -1 - sin(7x)sin(4x)
Begin by moving all terms with variables to one side:
cos(7x)cos(4x) + sin(7x)sin(4x) = -1
The corresponding trig identity is cos(A - B). Thus:
cos(7x - 4x) = cos(7x)cos(4x) + sin(7x)sin(4x) = -1
cos(3x) = -1
cos = -1 at π, so:
3x = π
x = π/3
We can also find another solution. Let 3π = -1:
3x = 3π
x = π
Thus, solutions on [0, 2π) are π/3 and π.
If x=1 and y=5 we get as 4product of 1+5=9
=4+5=9
9=9
lhs=rhs
I hope this helps you
90-8.5
90-40
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