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I have no idea how my answer is wrong. I am so certain it is right. This problem asks me to use THEIR variables to solve for density.

density = $\frac MV$

$M$ can already be used for Mass, it's $9.25$ kg in this scenario but they want VARIABLES, not numeric values. So I have no idea why they wanted to give people $9.25$ kg and $0.91$ meters.

Volume ($V$) of a sphere, using the diameter is:

$V = \frac 16\pi d^3$

(they used $d$ for diameter).

Therefore, the answer should be: Density = $\frac M{\frac 16 \pi d^3}$

What am I doing wrong? It won't accept this, it also will not accept $\frac {6M}{\pi d^3}$!

I solved for the density and it would not accept it either (using $9.25$kg and $0.91$ meters) It comes out to approximately $23.4433101635$ $kg/m^3$

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    M / ((1/6)* pi * d^3) Perhaps you need more brackets? Your expression is definitely correct. – 2017-01-14
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    It seems like your parantheses are misplaced. – 2017-01-14

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Your expression for the volume $V$ indeed correctly evaluates to:

$$V=\frac{1}{6}\pi d^3$$

Thus, using $\rho_s=\frac{M}{V}$ should give you $\rho_s=\frac{M}{\frac{1}{6}\pi d^3}=\frac{6M}{\pi d^3}$, which is the answer you have correctly obtained.

However, the issue likely has something to do with the way you typed your answer.

The way you typed it suggests the following:

$$\frac{M}{\left(\frac{1}{6}\right)} \pi d^3$$

Which is not equal to the correct expression you probably meant.

To solve this issue, just add extra parantheses:

Try the following:

M/((1/6) * pi * d^3)

(6M)/(pi*d^3)

If you want, you can let me know if these inputs work. If they do not, then I can try to provide more suggestions.

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    Phew, looks like it worked. This site was also not allowing me to use parenthesis for some reason, it would shade out the option in some scenarios, like as if it was a GPS trying to predict my address input. nerve wracking, but its over, thanks – 2017-01-14