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

Wrong

That's still not the right answer

That was the same answer that you already gave me that's wrong how come you can't get the right answer

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.

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.

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.

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.