Re-Arrange Equation?

Question

please can someone help me re-arrange the equation here :

http://upload.wikimedia.org/math/a/f/8/af83d4c94e8b16be266c820267b42463.png

to make H the subject ?

Answer 1

s = SQR (r^2 + h^2)
s^2 = r^2 + h^2
h^2 = s^2-r^2
h = SQR(s^2 - r^2)

Answer 2

i'm assuming that once you're saying "re-arranging", you propose shifting words to unravel for the unknown , or on your case, problem 'x'. #a million....x - 5 = 6v .....you want to re-set up words such that 'x' will equivalent something. the purely something you've is the different words contained in the equation. So let's flow each and everything however the 'x' to the right so as that 'x' will equivalent in spite of is on the different fringe of the equation. to that end, you want to flow (-5y) to the right side. you want (-5y) to vanish from the left, so in case you further (+5y) to the left, it would want to be 0, and all you've left on the left is 'x'. so that you've sucsessfully remoted 'x' with information from itself on the left. Now, save in recommendations this rule: once you do something to at least one fringe of an equation, you should do the same to the different side. save in recommendations you further (+5y) to the left, so that you also ought to operate it to the right side. So now you've an equation that you'll be certain for'x'. ( x = 6v +5y) problem #2 is largely the same. Subtract v squared from the left side and also from the right side. ............note: in case you do not have THE SUPERSCRIPT '2' to point ............THAT A time period IS SQUARED, the widely USED pc ............image to point A RAISED skill OF ANY cost IS THE ........... image (^). So v squared might want to be written v^2. so that you presently have 'x' remoted on the left and (-v^2) is on the right to provide you x = w - v^2. problem #3 is carried out the same way. Subtract 6v from each and each and every side. problem #4 is a touch distinct. There aren't any words to move, however the 'x' time period is squared and with the intention to unravel for 'x' you should take the sq. root of x^2 meaning you may favor to also take the sq. root of the time period (2Y) on the right. The sq. root of x^2 is ± x and the sq. root of (2y) is (2y) less than the unconventional signal ?. problem #5 is like #4 except you even ought to move (upload) the (-v) to each and each and every side.

Answer 3

h^2 +r^2 =s^2(sqyare both sides)
h^2=s^2 -r^2
h=sqrt(s^2 -r^2)

Answer 4

s = ( r ^ 2 + h ^ 2 ) ^ (1/2)

s ^ 2 = r ^ 2 + h ^ 2

s ^ 2 - r ^ 2 = h ^ 2

*****

(s+r) * (s - r ) = h ^ 2

[(s+r) * (s - r )] ^ (1/2) = h

or

(s ^ 2 - r ^ 2) ^ (1/2) = h

Answer 5

kindly state your question clearly

Answer 6

s=root(r^2+h^2)
s^2=r^2+h^2
s^2-r^2=h^2
h^2=s^2-r^2
h=root(s^2-r^2)