right cylinder with a 90 degree slice remove?

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right cylinder with a 90 degree slice remove?

Mathematics
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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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a cheese log
What do you want to know about this shape Hikari?
yea but i need to know how to do this problem

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You haven't given us a problem yet. I'll definitely help you if you can post what you want help with
that is the question(up there)
Do you need the volume, or the surface area, or what?
volume
Well that's simple Hikari. How many degrees are there in a circle? You know that it's 360 degrees. Well to find how much volume is in a right cylinder with 90 degrees removed, if you know the initial volume, you simply have to find the amount removed and subtract that from the intial value. To find how much you "cut away", you need to divide 90/360 to get the fraction of the cylinder you took away. Then you take this number and multiply it by the initial volume to get the amount you took away, and then subtract the amount you took away from the intial volume to get how much is left. Did that make sense to you?
but i have the radius is 6 and the height is 12 but can u show me the equation of this problem
Oh ok. You didn't give me the initial parameters!! Now we can move out of the theoreticals. How I would approach this problem is to envision a cylinder with r=6 and height=12. We know the volume of this (full) cylinder through the formula \[V= 2\pi r^2 * h\] We plug everything in to get a volume of \[v=864\pi\] Ok. So now we have the total volume of the cone (before we cut 0- degrees out)
Now all we have to do is find out how much we cut out of the cylinder and subtract that from the total initial volume of the cylinder We know that 90/360 = 1/4 from my previous previous explanation. So to find the area we subtracted, we multiple 1/4*864pi to get 216pi. Now we subtract the value we took from the cylinder when we cut 90 degrees out. V= 864pi - 216pi = 648 pi
oh i got thank but i am wonder if you can sent me your email so when i have my homework from geometry,you can help me
I am posting another question right now
Sure. Where's your question?
An oblique trapozidal prism. The trapozidal base has a height of 4 in. and base the measure 8 in. and 12 in. The height of the prism is 24 in.?
Hey and if you need my email, it is TheAthenian1@gmail.com
ok thank so much
So what I'm thinking is how we find the volume for any prism (nice concept to keep in mind when solving these problems). Essentially, we take the area of the base and multiply it by the height to find the volume. To find the area of the base, we can divide the trapezoid into a rectangle and a triangle (one base will have 4 in. sticking out and will form one side of the triangle).
can u show me equation please
So in this case, we have a rectangle that is 8 x 4 (think base x height) and a triangle that is (4x4)/2 (area of a triangle) If we add these two values together, we get 40 in^2 for the area of the trapezoid. Then we multiply 40 in^2 by the height 12in to get 280in^3 for the volume
In general, the volume of a prism = (area of base)(Height)
so that answer for the volume?
That's what I would think... Yes. But I have been known to be sloppy and make arithmetic mistakes in the past so it might be a good idea to check my answer as well!
hey don't worry because my mom ask someone to help me but no one did not know how to do this but u help me
Well I'm glad I could be of help! If you have any questions, feel free to email me or post here as there are some (I'm sure) very other good people who could help you as well. :)
I will sent in email because that the only thing that I can use it

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