Evaluate [10/y + 13] -3, when y=5

Mathematics

2 answers:

notsponge [240]3 years ago

8 0

Here you go, I hope this helps

Oliga [24]3 years ago

5 0

Answer:

(10/y+13)-3

(10/5+13)-3

(2+13)-3

15-3

I hope this will help you

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The maximum weight that a rectangular beam can support varies jointly as its width and the square of its height and inversely as

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$\bf \qquad \qquad \textit{joint compound proportional variation} \\\\ \begin{array}{llll} \textit{\underline{y} varies directly with \underline{x}}\\ \textit{and inversely with \underline{z}} \end{array}\implies y=\cfrac{kx}{z}\impliedby \begin{array}{llll} k=constant\ of\\ \qquad variation \end{array} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \stackrel{\textit{W varies jointly as its width, squared heght and inversely with length}}{W=\cfrac{wh^2}{L}}$

$\bf \textit{now we also know that }~~ \begin{cases} w=\frac{1}{4}\\ h=\frac{1}{3}\\ L=12\\ W=30 \end{cases}\implies 30=\cfrac{~~k\frac{1}{4}\left( \frac{1}{3} \right)^2~~}{12} \\\\\\ 360=\cfrac{k}{36}\implies 12960=k~\hfill \boxed{W=\cfrac{12960wh^2}{L}}$

$\bf \textit{when } \begin{cases} w=\frac{1}{4}\\ h=\frac{1}{2}\\ L=12 \end{cases}\textit{ what is \underline{W}?}\qquad W=\cfrac{12960\left( \frac{1}{4} \right)\left(\frac{1}{2} \right)^2}{12} \\\\\\ W=1080\left( \cfrac{1}{4} \right)\left( \cfrac{1}{4} \right)\implies W=1080\cdot \cfrac{1}{16}\implies W=\cfrac{135}{2}\implies W=67\frac{1}{2}$