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2 answers:

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$Domain:v\neq0\\\\\dfrac{2}{v}+\dfrac{2}{3}=\dfrac{v-6}{3v}\\\\\dfrac{2\cdot3}{v\cdot3}+\dfrac{2\cdot v}{3\cdot v}=\dfrac{v-6}{3v}\\\\\dfrac{6}{3v}+\dfrac{2v}{3v}=\dfrac{v-6}{3v}\\\\\dfrac{6+2v}{3v}=\dfrac{v-6}{3v}\iff6+2v=v-6\qquad\text{subtract 6 from both sides}\\\\2v=v-12\qquad\text{subtract v from both sides}\\\\\boxed{v=-12}\to\boxed{A.}$

NARA [144]3 years ago

3 0

Answer:

The answer is A, sweetie

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Y is directly proportional to square root of x<br> If y=56 when x=49 find,<br> y when x=81

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$\stackrel{\begin{array}{llll} \textit{"y" directly}\\ \textit{proportional to }\sqrt{x} \end{array}}{y = k\sqrt{x}}\qquad \textit{we know that} \begin{cases} y = 56\\ x = 49 \end{cases}\implies 56=k\sqrt{49} \\\\\\ 56=7k\implies \cfrac{56}{7}=k\implies 8=k~\hfill \boxed{y=8\sqrt{x}} \\\\\\ \textit{when x = 81, what is "y"?}\hfill y=8\sqrt{81}\implies y=8(9)\implies y=72$