Angles in a parallel line

Mathematics

1 answer:

Hatshy [7]4 years ago

6 0

1.53
2.115
3.122
4.106
5.49
6.110
left to right

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Read 2 more answers

Help me solve them plsss...both of the questions :)

zaharov [31]

$\bold{\text{Answer:}\quad \dfrac{\sqrt2}{144}}$

<u>Step-by-step explanation:</u>

When you plug in 3 directly to the equation, you get 0/0

Since it is indeterminate, you have to use L'Hopital's Rule.

That means that you find the limit of the DERIVATIVE of the numerator and the DERIVATIVE of the denominator.

$\dfrac{d}{dx}(\sqrt{5+\sqrt{2x+3}}-2\sqrt2)\quad = \dfrac{1}{2\sqrt{5+\sqrt{2x+3}}(\sqrt{2x+3})}\\\\\\f'(3)\ \text{for the numerator}\ =\dfrac{1}{12\sqrt2}\\\\\\\\\dfrac{d}{dx}(x^2-9) = 2x\\\\\\f'(3) \text{for the denominator}\quad = 6\\\\\\\dfrac{f'(3)\ \text{numerator}}{f'(3)\ \text{denominator}}\quad = \dfrac{\dfrac{1}{12\sqrt2}}{6}\quad = \dfrac{1}{72\sqrt2}\quad \rightarrow \large\boxed{\dfrac{\sqrt2}{144}}$