Can someone help me with this 1 math problem. I posted a picture of the question.

Step-by-step explanation:
For conical part: h = 14 in & r = 9 in
For hemispherical part: r = 9 in
[tex]V_{prop} = V_{cone} +V_{hemisphere} \\ = \frac{1}{3} \pi {r}^{2} h + \frac{2}{3} \pi {r}^{3} \\ =\frac{1}{3} \times 3.14 \times {9}^{2} \times 14 + \frac{2}{3} \times 3.14 \times {9}^{3} \ \\ = 1,186.92 + 2 \times 3.14 \times 243 \\ = 1,186.92 + 1,526.04 \\ = 2,712.96 \: {inch}^{3} \\ [/tex]
Another way:
[tex]V_{prop} = V_{cone} +V_{hemisphere} \\ = \frac{1}{3} \pi {r}^{2} h + \frac{2}{3} \pi {r}^{3} \\\\
=\frac{1}{3} \pi {r}^{2}(h+2r)\\\\
= \frac{1}{3}\times 3.14\times {9}^{2}(14+2\times 9)\\\\
= \frac{1}{3}\times 3.14\times 81(14+18)
\\\\
= 3.14\times 27\times 32\\\\
= 2,712.96\: {inch} ^3 [/tex]