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Coming from the volume of a wire's equation where r is the cross-section diagonal, and l is the lenght: $$V=πlr^2$$

$$6*10^{-3}= π*61*r^2$$

I rearranged the equation, getting

$$(6*10^{-3})/(6.1*10)*π =r^2 $$

From here I typed the left side of the equation into my calculator, under the square root sign, and got $$r = 5.6*10^{-3}$$

Is there a practical way this can be solved on paper, without using the calculator? Because I got numbers like 0.0005 etc. and its pretty frustrating to work with those.

P.S: I would like to ask a question about math, without any example, just a theory question, under which tag should I ask that?

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  • $\begingroup$ @zz20s sorry, its meant to be multiplication, edited. $\endgroup$ – JohnFire Jan 20 '16 at 15:15
  • $\begingroup$ $\LaTeX$ hint: to get multicharacter exponents, enclose them in braces. So 10^{-3} will give $10^{-3}$ $\endgroup$ – Ross Millikan Jan 20 '16 at 15:17
  • $\begingroup$ If the scientific notation bothers you, you can fix the number of digits after the comma with most calculators. This avoids scientific notation. You can also take the reciproque ($1$ divided by the number) and write down $\frac{1}{number}$. This would avoid scientific notation in this case. Calculating square roots by hand is more annyoing than the scientific notation. $\endgroup$ – Peter Jan 20 '16 at 15:18
  • $\begingroup$ @RossMillikan Thank you :) $\endgroup$ – JohnFire Jan 20 '16 at 15:25
  • $\begingroup$ @Peter I Thank you,, that might be easier for me :) $\endgroup$ – JohnFire Jan 20 '16 at 15:26
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I don't know where the subtraction of $3$ comes from in the third equation, so I will ignore it. You can use scientific notation to avoid the $0.00$ part, so your equation becomes $r^2=\frac 6{6.1\pi}\cdot 10^{-2}$. You can compute the fraction to get about $0.313092$, so you have $r^2=31.3092 \cdot 10^{-4}$ The square root of $10^{-4}$ is $10^{-2}$ so you can apply that in the end. The only hard part is taking the square root by hand. You can do that with estimation, Newton's method, or the old digit extraction technique.

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  • $\begingroup$ That was supposed to be 10^-3 sorry $\endgroup$ – JohnFire Jan 20 '16 at 15:25
  • $\begingroup$ The 6/6.1π * 10^-2 looks easier, thanks :) $\endgroup$ – JohnFire Jan 20 '16 at 15:26

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