1. ## rearranging formula

Hi All;
I need to make a2 the subject of a1 sqrt 2gh/(a1/a2)-1 = q

firat divide bt a1 giving sqrt 2gh/(a1/a2)-1 = q/a1

square both sides giving 2gh/(a1/a2)-1 = qsq / a1sq

add one to both sides giving 2gh/(a1/a2) = qsq / a1sq + 1

stuck here.

Thanks.

2. Divide by a2?

Please use parentheses to clarify meaning. In this case, a1 sqrt(2gh/(a1/a2)-1) = q would have been helpful.
Please don't use implicit multiplication. Use an effective symbol. In this case, a1 * sqrt(2gh/(a1/a2)-1) = q would have been helpful.
Use "^" to indicate exponentiation. 2gh/(a1/a2)-1 = (q^2) / (a1^2) = (q/a2)^2

3. Dont understand divide by a2?

4. I am troubled by this...

$\displaystyle \frac{2gh}{\frac{a1}{a2}} = 2gh\cdot\frac{a2}{a1}$

Okay, you talked me out of it. Instead of that, multiply by $\displaystyle \frac{a1}{2gh}$

Please scroll down to question 6 and there answer is at the bottom of the page.
I think there answer is wrong just let me know if it is.

Thanks.

6. Originally Posted by anthonye
Hi All;
I need to make a2 the subject of a1 sqrt 2gh/(a1/a2)-1 = q

firat divide bt a1 giving sqrt 2gh/(a1/a2)-1 = q/a1

square both sides giving 2gh/(a1/a2)-1 = qsq / a1sq

add one to both sides giving 2gh/(a1/a2) = qsq / a1sq + 1

stuck here.

Thanks.
The givn problem is
A1*sqrt{2gh/[(A1/A2)^2 - 1]} = q
Now simplify.

7. Dont understand I'm having trouble making a2 the subject.

8. q^2/A1^2 = 2gh/[(A1/A2 )^2 - 1]

(A1/A2)^2 - 1 = 2gh*A1^2/q^2

(A1/A2)^2 = 1 + 2gh*A1^2/q^2

A1^2/A2^2 = [q^2 + 2gh*A1^2]/q^2

A2^2 = q^2A1^2/[q^2 + 2gh*A1^2]

9. Why in the fourth step does q^2 appear in numerator and denominator.

10. Let A1 = A and A2 = B; then Sarigama's (which is correct) becomes:
B^2 = q^2 A^2 / (q^2 + 2gh * A^2)
So complete this way:
B = qA / SQRT(q^2 + 2gh * A^2)

You need practice with these:
try and solve this for n:
q = a * SQRT[m / (n - 1)]

11. Originally Posted by anthonye
Why in the fourth step does q^2 appear in numerator and denominator.
Example: if x = 1 + a/y, then x = (y + a) / y
That's quite basic: are you attending math classes?

12. Hello, anthonye!

$\text{Solve for }A_2\!:\;\;A_1\sqrt{\frac{2gh}{(\frac{A_1}{A_2})^2 - 1}}\;=\;q$

$\text{Square both sides: }\;A_1^2\cdot\frac{2gh}{(\frac{A_1}{A_2})^2 - 1} \;=\;q^2$

$\text{Cross-multiply: }\;2A_1^2gh \;=\;q^2\left[\left(\frac{A_1}{A_2}\right)^2 - 1\right]$

. . . . . . . . . . . . $2A_1^2gh \;=\;q^2\left(\frac{A_1}{A_2}\right)^2 - q^2$

. . . . . . . . .$2A_1^2gh + q^2 \;=\;q^2\left(\frac{A_1}{A_2}\right)^2$

. . . . . . . . $\frac{2A_1^2gh + q^2}{q^2} \;=\;\left(\frac{A_1}{A_2}\right)^2$

. . . . . . . . $\frac{2A_1^2gh + q^2}{q^2} \;=\;\frac{A_1^2}{A_2^2}$

. . . . . . . . . . . . . . $A_2^2 \;=\;\frac{A_2^2q^2}{2A_1^2gh + q^2}$

. . . . . . . . . . . . . .$A_2 \;=\;\pm \sqrt{\frac{A_1^2q^2}{2A_1^2gh + q^2}}$

. . . . . . . . . . . . . .$A_2 \;=\;\pm\frac{A_1q}{\sqrt{2A_1^2gh + q^2}}$

13. Ok so 1/y is equal to its reciprocal y/1 which is just y I take it thats correct.

Because y is unknown yes.

14. Ok Wilmer I get n=ma^2 - 1/ mq^2

15. Originally Posted by anthonye
Ok Wilmer I get n=ma^2 - 1/ mq^2
NOOOooooo.......PLUS you forgot brackets: n = (ma^2 - 1) / mq^2 ;
VERY important; hope you know why...

We have:
q = a * SQRT[m / (n - 1)]

Divide by a:
q / a = SQRT[m / (n - 1)]

Square both sides:
q^2 / a^2 = m / (n - 1) ; still with me?

Cross multiply:
q^2(n - 1) = ma^2

Expand:
nq^2 - q^2 = ma^2

Isolate the n term:
nq^2 = ma^2 + q^2

Divide by q^2:
n = (ma^2 + q^2) / q^2 : dig them brackets on left !!

Amen!

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