# math

posted by
**brittany** on
.

Solve for the indicated variable

A=1/2 bh Solve for b

2A = bh, so

b = 2A/h

ok thanks so on this example

C=2pie r

Solve for R

Would the answer be

2C/pie=r ????

ok thanks so on this example

C=2pie r

Solve for R

Would the answer be

2C/pie=r ????

**Divide both sides by everything we DON'T want. That is 2 & pi on the right side.
C = 2pi r
so
C/(2 pi) = (2 pi r)/2 pi
Now on the right the 2's cancel and the pi's cancel leaving just r.
C/(2 pi) = r
** I hope this helps.

yes that helped

I did that on my next problem which is

A=4pi r^2 Solve for r

I divided everything on both sides.

A/(4 pi)= 4pi r^2/(4 pi)

the 4 pi cancels on the right

A/(4 pi)= r^2

What do I do with the r^2 so I can get r by its self???

A/(4 pi)= r^2

What do I do with the r^2 so I can get r by its self???

**Take the square root of both sides.**

sqrt[A/(4 pi)] = sqrt r^2 = r.

But note that you can take the square root of 4, which is 2, and pull it out of the sqrt part like so,

1/2[sqrt A/pi] = r. Remember, too, that sqrt 2 is +/- 2 and not just 2 for +2 x +2 = 4 and -2 x -2 = 4. Likewise, sqrt r^2 is +r or -r.

sqrt[A/(4 pi)] = sqrt r^2 = r.

But note that you can take the square root of 4, which is 2, and pull it out of the sqrt part like so,

1/2[sqrt A/pi] = r. Remember, too, that sqrt 2 is +/- 2 and not just 2 for +2 x +2 = 4 and -2 x -2 = 4. Likewise, sqrt r^2 is +r or -r.