Question

Make A the subject of the formula.

Answer 1

\[q=\sqrt{2ghDA ^{2}}

over (s ^{2}-A ^{2})\]

Answer 2

Not sure how to put it all under the same square root, 2ghDA^2 is over (s^2-A^2) under the same root

Answer 3

nevermind i can not help, i tried sorry

Answer 4

no problem :)

Answer 5

\[q=\frac{\sqrt{2ghDA ^{2}}}{ (s ^{2}-A ^{2})}\] like this?

Answer 6

oh no hold on

Answer 7

The root above, also goes over the denom.

Answer 8

\[q=\sqrt{\frac{2ghDA^2}{s^2-A^2}}\] maybe this

Answer 9

Yeah, thats the ticket, except there are brackets around (s^2-A^2)

:)

Answer 10

\[q^2=\frac{2ghDA^2}{s^2-A^2}\] is a start

Answer 11

brackets are not needed

Answer 12

then
\[q^2(s^2-A^2)=2ghDA^2\]

Answer 13

multiply out on the left, get
\[q^2s^2-q^2A^2=2ghDA^2\] put all the stuff with A on the right get
\[q^2s^2=2ghDA^2-q^2A^2\] factor out the \(A^2\) get
\[q^2s^2=A^2(2ghD-q^2)\] divide get
\[\frac{q^2s^2}{(2ghD-q^2)}=A^2\]

Answer 14

damn damn damn i made a mistake

Answer 15

\[q^2s^2=2ghDA^2+q^2A^2\]
\[q^2s^2=A^2(2ghD+q^2)\]
\[\frac{q^2s^2}{(2ghD+q^2)}=A^2\]

Answer 16

that's better. now take the square root

Answer 17

don't forget the \(\pm\)

Answer 18

Your a genius, do you teach math?

Answer 19

\[A=\pm\frac{qs}{\sqrt{2ghD+q^2}}\]

Answer 20

i am not a genius, one day this will seem simple to you, take my word for it

Answer 21

I'm studying biomedical engineering, this is our math module.