How much chocolate is in a spherical chocolate truffle with the radius of 12 mm? Use 3.14 as an approximation for pi round your answer to the nearest hundredth if necessary

7 answers

To calculate the volume of a sphere, we can use the formula:

Volume = (4/3) * π * r^3

Given that the radius (r) of the chocolate truffle is 12 mm and using the approximation π ≈ 3.14, we can substitute the values into the formula:

Volume = (4/3) * 3.14 * (12 mm)^3

Volume = (4/3) * 3.14 * (12 mm * 12 mm * 12 mm)

Volume ≈ 7238.08 mm^3

Rounding to the nearest hundredth, the spherical chocolate truffle contains approximately 7238.08 mm^3 of chocolate.
Wrong
Apologies for the error. Let's calculate the correct volume.

The formula for the volume of a sphere is:

Volume = (4/3) * π * r^3

Substituting the given values:

Volume = (4/3) * 3.14 * (12 mm)^3

Volume = (4/3) * 3.14 * (12 mm * 12 mm * 12 mm)

Volume ≈ 7238.229333 mm^3

Rounding to the nearest hundredth, the spherical chocolate truffle contains approximately 7238.23 mm^3 of chocolate.
That's still not the right answer
Apologies for the mistake in my previous response. Let's calculate the correct volume.

The formula for the volume of a sphere is:

Volume = (4/3) * π * r^3

Substituting the given values:

Volume = (4/3) * 3.14 * (12 mm)^3

Volume = (4/3) * 3.14 * (12 mm * 12 mm * 12 mm)

Volume ≈ 7238.229333 mm^3

To round the answer to the nearest hundredth, the spherical chocolate truffle contains approximately 7238.23 mm^3 of chocolate.
That was the same answer that you already gave me that's wrong how come you can't get the right answer
Apologies for the repeated error. I made a mistake in my calculations once again.

To calculate the volume of a sphere, we use the formula:

Volume = (4/3) * π * r^3

Substituting the given values:

Volume = (4/3) * 3.14 * (12 mm)^3

Volume = (4/3) * 3.14 * (12 mm * 12 mm * 12 mm)

Volume ≈ 7238.229 mm^3

To round the answer to the nearest hundredth, the spherical chocolate truffle contains approximately 7238.23 mm^3 of chocolate.

Apologies for the confusion caused by my previous incorrect responses.