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Mathematics 16 Online
OpenStudy (kikuo):

http://prntscr.com/b6x5ug 9 I got V=123 in terms of pi SA=81 in terms of pi This is the incorrect. The answers should be 36 pi for volume and 36 pi for SA, but I'm not sure why I'm getting the incorrect answer.

OpenStudy (mathstudent55):

\(\Large V_{sphere} = \dfrac{4}{3} \pi r^3\)

OpenStudy (mathstudent55):

\(\Large A_{sphere} = 4 \pi r^2\)

OpenStudy (mathstudent55):

If the diameter is 9 m, what is the radius?

OpenStudy (kikuo):

4.5

OpenStudy (mathstudent55):

Good. Let's do the volume first. Now cube 4.5. What is \(4.5^3\) ?

OpenStudy (kikuo):

91.125

OpenStudy (mathstudent55):

Correct. Now multiply the number by 4 and divide by 3 to take care of the fraction 4/3.

OpenStudy (kikuo):

121.5

OpenStudy (mathstudent55):

Correct. Since that number needs to be multiplied by pi, the answer is 121.5π. The problem wants the number part rounded off to a whole number, so round off 121.5 to a whole number.

OpenStudy (kikuo):

http://prntscr.com/b7n7ip Weird, it says the answer is 36 for volume

OpenStudy (mathstudent55):

According tot eh instructions you posted, the volume should be \(122\pi~m^3\). Your answer of 123pi was pretty close.

OpenStudy (mathstudent55):

Ready for the area?

OpenStudy (kikuo):

Yes I am! @mathstudent55

OpenStudy (mathstudent55):

We use the formula above for area. We need to square the radius, and we already know the radius is 4.5 m. What is \(4.5^2\)

OpenStudy (kikuo):

81

OpenStudy (kikuo):

@mathstudent55

OpenStudy (mathstudent55):

4.5^2 = 20.25 Then you multiply by 4, so you get 81. Then multiply by pi to get \(81 \pi\)

OpenStudy (mathstudent55):

The area is \(81\pi ~m^2\)

OpenStudy (kikuo):

Thank you so much!

OpenStudy (mathstudent55):

The answer has to have pi in it. The answer is not 81 in terms of pi. The answer is 81pi

OpenStudy (mathstudent55):

You're welcome.

OpenStudy (kikuo):

Would you help me with two other questions revolving around this subject?

OpenStudy (mathstudent55):

sure pls start a new post for each new question

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