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: $$0.12m^3 = (1.15\times10^5) \times ((2.0\times10^{-3}m)\times \pi r^2)$$ $$\frac{0.12}{(1.15\times10^5)} = (2.0\times10^{-3})\times \pi r^2$$ $$\pi r^2 = \frac{(0.12)}{(2.0\times10^{-3})(1.15\times10^5)}$$ $$r^2 = \frac{(0.12)}{\pi(2.0\times10^{-3})(1.15\times10^5)}$$ $$ 2r = 2\sqrt{\frac{(0.12)}{\pi(2.0\times10^{-3})(1.15\times10^5)}} = 0.025774m$$ $$ D = 2.5774cm $$