**Problem 14: Volume Measurements**A punch bowl shaped like a hemisphere with a radius of 9 inches is full of punch. If we are filling cylindrical glasses that have a diameter of 3 inches and a height of 3 inches, how many glasses can we fill?**Explanation:**To solve this problem, we need to calculate the volume of the punch bowl (which is a hemisphere) and the volume of one cylindrical glass. We will then divide the total volume of the punch in the bowl by the volume of one glass to find out how many glasses we can fill.1. **Volume of the Hemisphere:** The volume $ V $ of a hemisphere is given by the formula: \[ V = \frac{2}{3} \pi r^3 \] where $ r $ is the radius of the hemisphere. For the punch bowl: \[ r = 9 \, \text{inches} \] So, \[ V_{\text{hemisphere}} = \frac{2}{3} \pi (9)^3 = \frac{2}{3} \pi (729) = 486 \pi \, \text{cubic inches} \]2. **Volume of One Cylindrical Glass:** The volume $ V $ of a cylinder is given by the formula: \[ V = \pi r^2 h \] where $ r $ is the radius of the base and $ h $ is the height of the cylinder. For the cylindrical glass: \[ r = \frac{\text{diameter}}{2} = \frac{3}{2} = 1.5 \, \text{inches} \] and \[ h = 3 \, \text{inches} \] So, \[ V_{\text{cylinder}} = \pi (1.5)^2 (3) = \pi (2.25)(3) = 6.75 \pi \, \text{cubic inches} \]3. **Number of Glasses Filled:** To find out how many cylindrical glasses we can fill, divide the volume of the punch in the bowl by the volume of one glass: \[ \frac{V_{\

Answered: Image /qna-images/answer/45f49a5e-4876-4074-b991-c34b48f23018.jpg

4. A punch bowl shaped like a hemisphere with a radius of 9 inches is full of punch. If we are filling cylindrical glasses that have a diameter of 3 inches and a height of 3 inches, how many glasses can we fill?

4. A punch bowl shaped like a hemisphere with a radius of 9 inches is full of punch. If we are filling cylindrical glasses that have a diameter of 3 inches and a height of 3 inches, how many glasses can we fill?

Algebra and Trigonometry (6th Edition)

6th Edition

ISBN:9780134463216

Author:Robert F. Blitzer

Publisher:Robert F. Blitzer

ChapterP: Prerequisites: Fundamental Concepts Of Algebra

Section: Chapter Questions

Problem 1MCCP: In Exercises 1-25, simplify the given expression or perform the indicated operation (and simplify,...

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Concept explainers

Cylinders

A cylinder is a three-dimensional solid shape with two parallel and congruent circular bases, joined by a curved surface at a fixed distance. A cylinder has an infinite curvilinear surface.

Cones

A cone is a three-dimensional solid shape having a flat base and a pointed edge at the top. The flat base of the cone tapers smoothly to form the pointed edge known as the apex. The flat base of the cone can either be circular or elliptical. A cone is drawn by joining the apex to all points on the base, using segments, lines, or half-lines, provided that the apex and the base both are in different planes.

Question

**Problem 14: Volume Measurements**

A punch bowl shaped like a hemisphere with a radius of 9 inches is full of punch. If we are filling cylindrical glasses that have a diameter of 3 inches and a height of 3 inches, how many glasses can we fill?

**Explanation:**

To solve this problem, we need to calculate the volume of the punch bowl (which is a hemisphere) and the volume of one cylindrical glass. We will then divide the total volume of the punch in the bowl by the volume of one glass to find out how many glasses we can fill.

1. **Volume of the Hemisphere:**

The volume $ V $ of a hemisphere is given by the formula:
\[
V = \frac{2}{3} \pi r^3
\]
where $ r $ is the radius of the hemisphere.

For the punch bowl:
\[
r = 9 \, \text{inches}
\]

So,
\[
V_{\text{hemisphere}} = \frac{2}{3} \pi (9)^3 = \frac{2}{3} \pi (729) = 486 \pi \, \text{cubic inches}
\]

2. **Volume of One Cylindrical Glass:**

The volume $ V $ of a cylinder is given by the formula:
\[
V = \pi r^2 h
\]
where $ r $ is the radius of the base and $ h $ is the height of the cylinder.

For the cylindrical glass:
\[
r = \frac{\text{diameter}}{2} = \frac{3}{2} = 1.5 \, \text{inches}
\]
and
\[
h = 3 \, \text{inches}
\]

So,
\[
V_{\text{cylinder}} = \pi (1.5)^2 (3) = \pi (2.25)(3) = 6.75 \pi \, \text{cubic inches}
\]

3. **Number of Glasses Filled:**

To find out how many cylindrical glasses we can fill, divide the volume of the punch in the bowl by the volume of one glass:
\[
\frac{V_{\

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