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Sagot :
To solve for the radius [tex]\( r \)[/tex] of a sphere given its volume [tex]\( V \)[/tex], we start with the formula for the volume of a sphere:
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
Given that the volume [tex]\( V \)[/tex] is 435 cubic inches, and using [tex]\( \pi = 3.14 \)[/tex], we can substitute these values into the equation:
[tex]\[ 435 = \frac{4}{3} \times 3.14 \times r^3 \][/tex]
First, we isolate [tex]\( r^3 \)[/tex]. To do this, we will multiply both sides of the equation by the reciprocal of [tex]\( \frac{4}{3} \times 3.14 \)[/tex]:
[tex]\[ r^3 = \frac{435 \times 3}{4 \times 3.14} \][/tex]
Now let's calculate the value inside the fraction:
[tex]\[ r^3 = \frac{1305}{12.56} \][/tex]
By performing the division, we get:
[tex]\[ r^3 \approx 103.90 \][/tex]
Now, to solve for [tex]\( r \)[/tex], we need to take the cube root of 103.90:
[tex]\[ r \approx 103.90^{\frac{1}{3}} \][/tex]
Using a calculator or approximating the cube root, we find:
[tex]\[ r \approx 4.70 \][/tex]
To the nearest tenth of an inch, the radius [tex]\( r \)[/tex] of the basketball is:
[tex]\[ r \approx 4.7 \text{ inches} \][/tex]
Therefore, the correct answer is:
D. 4.7 inches
[tex]\[ V = \frac{4}{3} \pi r^3 \][/tex]
Given that the volume [tex]\( V \)[/tex] is 435 cubic inches, and using [tex]\( \pi = 3.14 \)[/tex], we can substitute these values into the equation:
[tex]\[ 435 = \frac{4}{3} \times 3.14 \times r^3 \][/tex]
First, we isolate [tex]\( r^3 \)[/tex]. To do this, we will multiply both sides of the equation by the reciprocal of [tex]\( \frac{4}{3} \times 3.14 \)[/tex]:
[tex]\[ r^3 = \frac{435 \times 3}{4 \times 3.14} \][/tex]
Now let's calculate the value inside the fraction:
[tex]\[ r^3 = \frac{1305}{12.56} \][/tex]
By performing the division, we get:
[tex]\[ r^3 \approx 103.90 \][/tex]
Now, to solve for [tex]\( r \)[/tex], we need to take the cube root of 103.90:
[tex]\[ r \approx 103.90^{\frac{1}{3}} \][/tex]
Using a calculator or approximating the cube root, we find:
[tex]\[ r \approx 4.70 \][/tex]
To the nearest tenth of an inch, the radius [tex]\( r \)[/tex] of the basketball is:
[tex]\[ r \approx 4.7 \text{ inches} \][/tex]
Therefore, the correct answer is:
D. 4.7 inches
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