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2 years ago

17=-5r what is the equation answer

Mathematics

1 answer:

alukav5142 [94]2 years ago

6 0

R=
ln(5)
−πi+ln(17)

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antiseptic1488 [7]

Answer:

$\sf -11+7\sqrt{2}$

Step-by-step explanation:

Given expression:

$\sf \dfrac{3-\sqrt{32}}{1+\sqrt{2} }$

Rewrite 32 as 16 · 2:

$\sf \implies \dfrac{3-\sqrt{16 \cdot 2}}{1+\sqrt{2} }$

Apply radical rule $\sf \sqrt{a \cdot b}=\sqrt{a}\sqrt{b}$

$\sf \implies \dfrac{3-\sqrt{16}\sqrt{2}}{1+\sqrt{2} }$

As $\sf \sqrt{16}=4$ :

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} }$

Multiply by the conjugate:

$\sf \implies \dfrac{3-4\sqrt{2}}{1+\sqrt{2} } \times \dfrac{1-\sqrt{2} }{1-\sqrt{2} }$

$\sf \implies \dfrac{(3-4\sqrt{2})(1-\sqrt{2})}{(1+\sqrt{2})(1-\sqrt{2})}$

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4\sqrt{2}\sqrt{2}}{1-\sqrt{2}+\sqrt{2}-\sqrt{2}\sqrt{2}}$

As $\sf \sqrt{2}\sqrt{2}=\sqrt{4}=2$ :

$\sf \implies \dfrac{3-3\sqrt{2}-4\sqrt{2}+4 \cdot 2}{1-\sqrt{2}+\sqrt{2}-2}$

$\sf \implies \dfrac{3-7\sqrt{2}+8}{1-2}$

$\sf \implies \dfrac{11-7\sqrt{2}}{-1}$

$\sf \implies -11+7\sqrt{2}$

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What values are excluded from the domain and range of the function f(x)=x+3/2x+5?

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