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sp2606 [1]
2 years ago
5

10. What is the missing statement in the proof?

Mathematics
1 answer:
Paraphin [41]2 years ago
7 0

Answer:

∠3 ≅ ∠2.

Step-by-step explanation:

                 

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I'm assuming the limit is supposed to be

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Multiply the numerator by its conjugate, and do the same with the denominator:

\left(\sqrt{2x+3}-\sqrt{3x}\right)\left(\sqrt{2x+3}+\sqrt{3x}\right)=\left(\sqrt{2x+3}\right)^2-\left(\sqrt{3x}\right)^2=-(x-3)

so that in the limit, we have

\displaystyle\lim_{x\to3}\frac{-(x-3)}{(x^2-3x)\left(\sqrt{2x+3}+\sqrt{3x}\right)}

Factorize the first term in the denominator as

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The x-3 terms cancel, leaving you with

\displaystyle\lim_{x\to3}\frac{-1}{x\left(\sqrt{2x+3}+\sqrt{3x}\right)}

and the limand is continuous at x=3, so we can substitute it to find the limit has a value of -1/18.

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