Problem 29 - Algebra 2

Question

Geometry The formula for the volume of a sphere is \(V=\frac{4}{3} \pi r^{3} .\) Find the radius to the nearest hundredth of a sphere with each volume. $$ 10 \mathrm{in} .^{3} $$

Step-by-Step Solution

Answer

The radius of the sphere with volume 10 cubic inches is approximately 1.34 inches.

Step 1: Rearrange the formula for the volume of a sphere

We know \(V=\frac{4}{3} \pi r^{3}\). We need to express this in terms of \(r\). Thus, after rearranging, we have \(r= \left(\frac{3V}{4\pi}\right) ^{\frac{1}{3}}\).

Step 2: Substitute the given volume

Next, substitute the given volume \(V=10\) cubic inches into the new formula. This would be \(r = \left(\frac{3*10}{4*\pi}\right)^{\frac{1}{3}}\).

Step 3: Calculate final result

Perform the calculation of the formula. This results in a radius \(r \approx 1.34\) inches, rounded to the nearest hundredth.

Key Concepts

Geometry

Geometry is an area of mathematics that deals with shapes, sizes, and properties of space. One of the three-dimensional objects widely studied in geometry is the sphere.
A sphere is perfectly round and symmetric, just like a circle in three dimensions.

• It has all its points equidistant from the center.
• The uniform distance from the center to any point on the surface is called the radius.
• Understanding the properties of a sphere helps us model real-world objects like planets, bubbles, and balls.

Being a fundamental part of geometry, spheres allow us to explore and calculate important aspects such as volume and surface area.

Sphere

Spheres are unique geometric shapes that differ significantly from other solids like cubes or cylinders.
They have no edges or vertices, only a continuous curved surface.

• The radius of a sphere is a crucial element, as it defines the size of the sphere.
• For any sphere, all radii are equal, and they are measured from the center to the surface.
• The symmetry of a sphere makes calculations involving volume and surface area simpler compared to irregular shapes.

The concepts learned through spheres can be applied to various fields like physics, engineering, and everyday problem-solving.

Radius Calculation

Calculating the radius of a sphere from its volume is a common problem in math.
This involves rearranging the volume formula to solve for the radius.

• The formula to find the radius when the volume is known is: \( r = \left( \frac{3V}{4\pi} \right)^{\frac{1}{3}} \).
• This formula comes from manipulating the standard volume formula \( V = \frac{4}{3} \pi r^3 \).
• By plugging the given volume into this rearranged formula, you can solve for the radius.

The calculation involves understanding both the algebraic manipulation and practical computation using tools like calculators.

Mathematical Formulas

Mathematical formulas serve as fundamental tools in solving problems efficiently.
They provide a clear method to relate different mathematical concepts and simplify complex problems.

• In the context of a sphere, the volume formula \( V = \frac{4}{3} \pi r^3 \) is essential in connecting the radius and volume.
• Rearranging formulas, as shown with the radius calculation, demonstrates the power of algebraic manipulation.
• Understanding the derivation and use of these formulas allows one to apply them to various real-world and theoretical problems.

Mastery of formulas and their applications not only aids in academic pursuits but also enhances logical reasoning and critical thinking skills.