math
posted by brittany .
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.
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