We need to make $1.15\times10^5$ coins with a thickness of $2.0\times10^{-3}m(2mm)$ from a $0.12m^3$ block of silver. I'm trying to calculate what diameter the coins will end up being.
Did I set up the equation correctly where I'd get my answer from solving for D? I tried to draw it out to help understand what needs to be done. $0.12m^3 = (1.15\times10^5) \times ((2.0\times10^{-3}m)\times \frac{\pi D}{2})$
My solution: $0.12m^3 = (1.15\times10^5) \times ((2.0\times10^{-3}m)\times \pi r^2)$ $\frac{0.12}{(1.15\times10^5)} = (2.0\times10^{-3})\times \pi r^2$ $\pi r^2 = \frac{(0.12)}{(2.0\times10^{-3})(1.15\times10^5)}$ $r^2 = \frac{(0.12)}{\pi(2.0\times10^{-3})(1.15\times10^5)}$ $ 2r = 2\sqrt{\frac{(0.12)}{\pi(2.0\times10^{-3})(1.15\times10^5)}} = 0.025774m$ $ D = 2.5774cm $