Respuesta :

The answer is 170.09 in³ ≈ 170.1 in³
 
The volume (V) of the figure is the sum of the volume of a cylinder (V1) and half of the volume of the sphere (V2):
V = V1 + 1/2V2

V1 = π * r² * h
π = 3.14
r = 5/2 = 2.5 in
h = 7 in
V1 = 3.14 * 2.5² * 7
V1 = 137.38 in³

V2 = 4/3 * π * r³
π = 3.14
r = 5/2 = 2.5 in
V2 = 4/3 * 3.14 * 2.5³
V2 = 65.42 in³

Hence:
V = V1 + 1/2 V2
V = 137.38 in³ + 1/2 * 65.42 in³
V = 170.09 in³ ≈ 170.1 in³

Hello!

The figure is made up of a cylinder and a hemisphere. To the nearest whole number, what is the approximate volume of this figure? Use 3.14 to approximate π

Data: (Cylinder)

h (height) = 7 in

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

Adopting: [tex]\pi \approx 3.14[/tex]

V (volume) = ?

Solving: (Cylinder volume)

[tex]V = \pi *r^2*h[/tex]

[tex]V = 3.14 *2.5^2*7[/tex]

[tex]V = 3.14*6.25*7[/tex]

[tex]V = 137.375 \to \boxed{V_{cylinder} \approx 137.38\:in^3}[/tex]

Note: Now, let's find the volume of a hemisphere.

Data: (hemisphere volume)

V (volume) = ?

r (radius) = 2.5 in (The diameter is 5 being twice the radius)

Adopting: [tex]\pi \approx 3.14[/tex]

If: We know that the volume of a sphere is [tex]V = 4* \pi * \dfrac{r^3}{3}[/tex] , but we have a hemisphere, so the formula will be half the volume of the hemisphere [tex]V = \dfrac{1}{2}* 4* \pi * \dfrac{r^3}{3} \to \boxed{V = 2* \pi * \dfrac{r^3}{3}}[/tex]

Formula: (Volume of the hemisphere)

[tex]V = 2* \pi * \dfrac{r^3}{3}[/tex]

Solving:

[tex]V = 2* \pi * \dfrac{r^3}{3}[/tex]

[tex]V = 2*3.14 * \dfrac{2.5^3}{3}[/tex]

[tex]V = 2*3.14 * \dfrac{15.625}{3}[/tex]

[tex]V = \dfrac{98.125}{3}[/tex]

[tex]\boxed{ V_{hemisphere} \approx 32.70\:in^3}[/tex]

Now, to find the total volume of the figure, add the values: (cylinder volume + hemisphere volume)

Volume of the figure = cylinder volume + hemisphere volume

Volume of the figure = 137.38 in³ + 32.70 in³

[tex]Volume\:of\:the\:figure =170.08 \to \boxed{\boxed{\boxed{Volume\:of\:the\:figure = 170\:in^3}}}\end{array}}\qquad\quad\checkmark[/tex]

_______________________

I Hope this helps, greetings ... Dexteright02! =)

ACCESS MORE