Find the volume of half a sphere with the radius of 10.
Do you know the formula for volume of a sphere? It's \(\large V= \frac{4}{3}\pi r^3\) where: r = radius Now, for the volume of the half sphere, do as it says, cut it in half, which means, divide the volume of a sphere by 2.
So what would you get as the answer?
:) What did \(\huge YOU\) get?
something in the 2000's. which i know is wrong
isn't just 3.14 times radius^3 divide by two
Yes, you're right. I got: \(2094.4\)
As long as you plug everything in correctly into your calculator, you're answer should be correct. You're simply plugging your given value of \(10^3\) which is same thing as 1000
Okay well I moved onto a different problem. Using the quadratic formula, solve x^2+5x-14=0. & approximate to the nearest tenth. I got -4,7 but it says Im wrong. Am I?
Yes, you are wrong. One of your values IS close but is actually supposed to be -7. Anyways, here is the quadratic formula for when you have: \(ax^2+bx+c\) it is: \[\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\]
I don't know if it's displaying properly on your screen, but the root should be all over like this: \(\sqrt{b^2-4ac}\)
If you plug in everything: \[\frac{ -5 \pm \sqrt{(5)^2 - 4(-14)(1)} }{ 2(1) }\]
abbot is that volume formula right for spherical coords?
Thats how I have it.
I give up
why wud u give up >_> u just need to volume of a spher with radius 10 and divide by 2 -.-
@dan815 please do not spam in other people questions. And yes, in sperical coordinates, I think that's how you would prove the volume of a sphere..but that's irrelevant here obviously.
@amanda_deason you need to plug in the given values into your calculator, you should get: \[\frac{ -5 \pm \sqrt{81} }{ 2 }\]
@abb0t it's just frustrating because this program keeps saying im getting it wrong when its actually right. i just wish that i could have all the answers because im supposed to graduate tomorrow and now i cant.
do u wanna learn complete the square? thats where quadratic formula comes from
I think you forgot to divide by \(\huge 2\) because if you notice, \(-5+9 = 4\) for the top and \(-5 - 9 = -14\) You SHOULD get: x = 2 and x = -7
thats what i put and i still got it wrong.
You put \(\large x = 2\) and \(\large x = -7\) and it says you're wrong?
then put X=-7 and X = 2 :)
If that is the case, you need to discuss that with your school because those are the solutions to your quadratic function.
okay thank you.
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