Stuck on yet another maths question - please help

Original post by knoxode

I have gotten as far as:

2(pi*r^2) + 2(pi*r*h) = (pi*r^2)h

My problem is that I am struggling to rearrange the equation.

Can somebody please help - it will be very much appreciated!

First thing to do is look for stuff that's in all three of your terms, as you can then divide through by it and make the equation simpler. Can you see two values that appear in every term?

Reply 4

Original post by sindyscape62

First thing to do is look for stuff that's in all three of your terms, as you can then divide through by it and make the equation simpler. Can you see two values that appear in every term?

Pi is in all three terms!

Reply 5

Original post by knoxode

Pi is in all three terms!

Anything else? :-)

Reply 6

Original post by simon0

Anything else? :-)

Radius?

Reply 7

Original post by knoxode

Radius?

Great!

Can you carry out what sindyscape62said and divide through by r and h?

(Note as r and h are not zero, we can divide through).

Reply 8

Original post by simon0

Great!

Can you carry out what sindyscape62said and divide through by r and h?

(Note as r and h are not zero, we can divide through).

pi (2*r^2) pi (2*r*h) = pi(r^2*h)

2r^2 2rh = r^2h

r(2r 2h) = r^2h

2r 2h = (r^2h)/r

2r = ( r^2h)/r - 2h

2r^2 = r^2h - 2h

(2r^2) /r^2 = h - 2h

(edited 6 years ago)

Reply 9

Original post by knoxode

pi (2*r^2) pi (2*r*h) = pi(r^2*h)

2r^2 2rh = r^2h

r(2r 2h) = r^2h

2r 2h = (r^2h)/r

2r = ( r^2h)/r - 2h

2r^2 = r^2h - 2h

(2r^2) /r^2 = h - 2h

Okay, a few things:

- You are missing the addition signs.

- On fourth line RHS, you can divide by r so to gain a term with "r" instead of "r^2".

- On second from last line, it appears you attempted to multiply both sides by "r" but you did not multiply the last term by r.

Start again and make "h" the subject.

-------------------------------------------------------------

So far we have:

2 \pi r h + 2 \pi r^{2} = \pi r^{2} h,

2h + 2r = rh, \quad

by dividing by

\pi

and r.

Can you re-arrange to have h as subject?

(edited 6 years ago)

Reply 10

Original post by simon0

You are missing the addition signs.

It's something with my phone - I can't put them in

Reply 11

Original post by knoxode

It's something with my phone - I can't put them in

No problem, can you take forward where I left?

Reply 12

Original post by simon0

2 \pi r h + 2 \pi r^{2} = \pi r^{2} h,

2h + 2r = rh, \quad

by dividing by

\pi

and r.

Can you re-arrange to have h as subject?

2h + 2r = rh

2h/r +2r/r = h

I am very confused.

Reply 13

Original post by knoxode

2h + 2r = rh

2h/r +2r/r = h

I am very confused.

What you've got is correct. The question want h to be on its own, so next you need to get all the terms with h in them on the same side.

Reply 14

Original post by sindyscape62

What you've got is correct. The question want h to be on its own, so next you need to get all the terms with h in them on the same side.

That is the point at which I'm stuck :/

Reply 15

Original post by knoxode

That is the point at which I'm stuck :/

So what you've currently got is:

2r+2h=rh

Can you subtract one of the terms with h in it from both sides, which would make them both on the same side?

As another example if you were given the equation:

2x+5=8x

would you know how to get x on its own in that?

Reply 16

Original post by sindyscape62

So what you've currently got is:

2r+2h=rh

Can you subtract one of the terms with h in it from both sides, which would make them both on the same side?

As another example if you were given the equation:

2x+5=8x

would you know how to get x on its own in that?

2r + 2h = rh

2h = rh - 2h

2h = h(r-2)

2h/h = r-2

Sorry, making anything the subject is one of my weak points

(edited 6 years ago)

Reply 17

medhelp

Original post by knoxode

2h + 2r = rh

2h/r +2r/r = h

I am very confused.

at 2h + 2r = rh
get h on one side
2r = rh - 2h
now try separating out h

Reply 18