Question

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

Answer 1

Step-by-step explanation:

For conical part: h = 14 in & r = 9 in

For hemispherical part: r = 9 in

$V_{prop} = V_{cone} +V_{hemisphere} \\ = \frac{1}{3} \pi {r}^{2} h + \frac{2}{3} \pi {r}^{3} \\ =\frac{1}{3} \times 3.14 \times {9}^{2} \times 14 + \frac{2}{3} \times 3.14 \times {9}^{3} \ \\ = 1,186.92 + 2 \times 3.14 \times 243 \\ = 1,186.92 + 1,526.04 \\ = 2,712.96 \: {inch}^{3}$

Another way:

$V_{prop} = V_{cone} +V_{hemisphere} \\ = \frac{1}{3} \pi {r}^{2} h + \frac{2}{3} \pi {r}^{3} \\\\=\frac{1}{3} \pi {r}^{2}(h+2r)\\\\= \frac{1}{3}\times 3.14\times {9}^{2}(14+2\times 9)\\\\= \frac{1}{3}\times 3.14\times 81(14+18)\\\\= 3.14\times 27\times 32\\\\= 2,712.96\: {inch} ^3$