Raynor2i
2022-08-01
Answered

z is jointly proportional to ${x}^{3}$ and ${y}^{2}$. If z=172 when x=6 and y=6, find z when x=4 and y=3

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Tristan Pittman

Answered 2022-08-02
Author has **14** answers

$z=k{x}^{3}{y}^{2}$

If so, we plug in the original data to find k, then so from there thus we have

$172=k({6}^{2})({6}^{3})$

which given us

k=0.02212

Now we plug this in to the top equation and plug in the new x and y.

$z=(0.02212)\cdot ({4}^{3})\cdot ({3}^{2})=12.74$

If so, we plug in the original data to find k, then so from there thus we have

$172=k({6}^{2})({6}^{3})$

which given us

k=0.02212

Now we plug this in to the top equation and plug in the new x and y.

$z=(0.02212)\cdot ({4}^{3})\cdot ({3}^{2})=12.74$

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Solve ${2}^{x}={x}^{2}$

I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log:

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And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple…

I've been asked to solve this and I've tried a few things but I have trouble eliminating $x$. I first tried taking the natural log:

$x\mathrm{ln}\left(2\right)=2\mathrm{ln}\left(x\right)$

$\frac{\mathrm{ln}\left(2\right)}{2}}={\displaystyle \frac{\mathrm{ln}\left(x\right)}{x}$

I don't know what to do from here so I decided to try another method:

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And then I get stuck here, I'm all out of ideas. My guess is I've overlooked something simple…

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