Question

assume that the flask shown in the diagram can be modeled as a combination of a sphere and a cylinder. based on this assumption, the volume of the flask is cubic inches. if both the sphere and the cylinder are dilated by a scale factor of 2, the resulting volume would be 8 times the original volume. 100.11 50.07 35.08 20.22 reset next

Explanation:

Step1: Find radius of sphere and height - radius of cylinder

The radius of the sphere $r=\frac{4.5}{2}= 2.25$ inches, and the height of the cylinder $h = 3$ inches.

Step2: Calculate volume of original sphere

The volume formula for a sphere is $V_s=\frac{4}{3}\pi r^{3}$. Substituting $r = 2.25$, we get $V_s=\frac{4}{3}\pi(2.25)^{3}=\frac{4}{3}\pi\times11.390625 = 15.1875\pi$ cubic - inches.

Step3: Calculate volume of original cylinder

The volume formula for a cylinder is $V_c=\pi r^{2}h$. Substituting $r = 2.25$ and $h = 3$, we get $V_c=\pi(2.25)^{2}\times3=\pi\times5.0625\times3 = 15.1875\pi$ cubic - inches.

Step4: Calculate total original volume

$V_{total}=V_s + V_c=15.1875\pi+15.1875\pi = 30.375\pi\approx95.42$ cubic - inches.

Step5: Apply dilation factor

When a 3 - D shape is dilated by a scale factor $k$, the new volume $V'$ is related to the original volume $V$ by $V'=k^{3}V$. Here, $k = 2$, so the new volume $V_{new}=2^{3}\times V_{total}=8\times30.375\pi=243\pi\approx763.40$ cubic - inches.

Step6: Find ratio of new to original volume

The ratio of the new volume to the original volume is $\frac{V_{new}}{V_{total}}=8$.

Step7: Calculate new volume

The original volume $V_{total}=V_s+V_c=\frac{4}{3}\pi(\frac{4.5}{2})^{3}+\pi(\frac{4.5}{2})^{2}\times3$.
$V_{total}=\frac{4}{3}\pi\times11.390625+\pi\times5.0625\times3=15.1875\pi + 15.1875\pi=30.375\pi$ cubic - inches.
After dilation with $k = 2$, $V'=8\times30.375\pi$.
$V'=243\pi\approx763.40$ cubic - inches.
The original volume of the flask $V=\frac{4}{3}\pi(\frac{4.5}{2})^{3}+\pi(\frac{4.5}{2})^{2}\times3\approx95.42$ cubic - inches.
The new volume after dilation by a factor of 2 is $V'=8V$.
The new volume $V'=8\times( \frac{4}{3}\pi(\frac{4.5}{2})^{3}+\pi(\frac{4.5}{2})^{2}\times3)$.
$V'=8\times(\frac{4}{3}\pi\times11.390625+\pi\times5.0625\times3)$
$V'=8\times(15.1875\pi + 15.1875\pi)$
$V'=8\times30.375\pi$
$V' = 243\pi\approx763.40$ cubic - inches.
The original volume $V\approx95.42$ cubic - inches, and the new volume is 8 times the original volume.
If we calculate the original volume:
$V_{sphere}=\frac{4}{3}\pi r^{3}=\frac{4}{3}\pi(2.25)^{3}\approx53.01$ cubic - inches.
$V_{cylinder}=\pi r^{2}h=\pi(2.25)^{2}\times3\approx47.71$ cubic - inches.
$V_{total}=V_{sphere}+V_{cylinder}\approx100.72$ cubic - inches.
After dilation by a factor of 2, the new volume $V'=8V_{total}\approx805.76$ cubic - inches.
The closest value among the options (assuming the options are related to the new volume) is not clearly shown in your provided text, but the key result is that the new volume is 8 times the original volume.

Answer:

8 times