Question

Can I please get some help with this problem? I will fan and medal, the most helpful answer.

Answer 1

A cylinder has a radius of 5x+2 and a height of 2x+8. Which polynomial on standard form best describes the total volume of the cylinder? \[ V=\pi r^2 h\]

Answer 2

\[A.) 100 \pi x^4+880 \pi x^3 + 2256 \pi x^2+1408 \pi x +256 \pi\]
\[B.) 50 \pi x^3 +240 \pi x^2 + 168 \pi x + 32 \pi\]
\[C.) 20 \pi x^3 +168 \pi x^2 + 384 \pi x +128 \pi\]
\[D.) 10 \pi x^2 + 44 \pi x +16 \pi\]

Answer 3

I am sorry...I do not know this...geometry is not my best subject...again, I am sorry

Answer 4

:/ meh its okay do u know anyone who can help

Answer 5

@Zarkon ...can you help

Answer 6

Thanks @texaschic101
@beccaboo333 can you help me please or point me in the right direction

Answer 7

@SolomonZelman ...help please

Answer 8

either @robtobey or @mathmale can help with this. >.< I don't quite understand it. sorry.

Answer 9

:/ meh its okay i just need help on it soon i have to turn it in soon

Answer 10

the area of the cylinder is the circumference times the height.

Answer 11

π(5x+2)^2 is the circumference

2x+8 is the height

So, it would be

(2x+8)π(5x+2)^2 (Or (2x+8)π(5x+2)(5x+2) )

All you need to do is just to expand :)

Answer 12

not the circumference, the area. My bad...

Answer 13

But the numbers are same

Answer 14

its okay and okay so would it be \[\pi (5x+2)^2\times(2x+8)\]

Answer 15

but how do i simplify it to get the answer?

Answer 16

Yeah, but in expanded form

Answer 17

But the numbers are same

Answer 18

\[V=\pi r^2h \]\[V=\pi (2 x+8) (5 x+2)^2 \]\[V=\pi (2 x+8) \left(25 x^2+20 x+4\right) \]\[V=\pi \left(50 x^3+240 x^2+168 x+32\right) \]

Answer 19

\[\huge\color{blue}{ π(5x+2)^2 \times (2x+8) }\]
lets brake this up,
\[\color{green}{ (5x+2)^2 = (5x+2)(5x+2)=25x^2 +20x+4 }\]
\[So~~~now:~~~~~~\color{blue}{ π(25x^2+20x+4) (2x+8) }\]\[\color{green}{ (25x^2+20x+4) (2x+8) =50x^3+240x^2 +168x+32 }\]
\[\color{green}{ 50πx^3+240πx^2 +168πx+32π }\]

Answer 20

WOW, got the exact same thing :)

Answer 21

Thanks I should be able to give both of you a medal

Answer 22

I was about to put up a pic, but got disconnected :(

Answer 23

Yeah, unfortunately it's only 1 medal per question, but ...

Good luck with the rest !

Answer 24

it otay you guys a re very helpful

Answer 25

nice to hear that :)

Answer 26

:)