The formula for finding the surface area of a cube is A=6y^2 where y is the side length. we know a=96. so we just isolate y.
96/6= 6y^2 / 6
16=y^2
sqrt (16) = sqrt(y^2)
4=y

So y is 4

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3 years ago

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the pitch, or frequency, of a vibrating string varies directly with the square root of the tension. if a string vibrates at a fr

Naily [24]

$\bf \qquad \qquad \textit{direct proportional variation}\\\\
\textit{\underline{y} varies directly with \underline{x}}\qquad \qquad y=kx\impliedby 
\begin{array}{llll}
k=constant\ of\\
\qquad variation
\end{array}\\\\
-------------------------------\\\\
\begin{array}{llll}
\stackrel{frequency}{f}\textit{ of a vibrating string varies directly}\\
\qquad \qquad \textit{with the square root of the tension }\stackrel{tension}{t}
\end{array}$

$\bf f=k\sqrt{t}\quad \textit{we also know that }
\begin{cases}
f=300\\
t=8
\end{cases}\implies 300=k\sqrt{8}
\\\\\\
\cfrac{300}{\sqrt{8}}=k\implies \cfrac{300}{\sqrt{2^2\cdot 2}}=k\implies \cfrac{300}{2\sqrt{2}}=k\implies \cfrac{150}{\sqrt{2}}=k
\\\\\\
\textit{and we can \underline{rationalize} it to }\cfrac{150\sqrt{2}}{2}\implies 75\sqrt{2}=k$

$\bf thus\qquad f=\stackrel{k}{75\sqrt{2}}\sqrt{t}\implies \boxed{f=75\sqrt{2t}}\\\\
-------------------------------\\\\
\textit{now, when t = 72, what is \underline{f}?}\qquad f=75\sqrt{2(72)}$

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3 years ago

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