Geophyics equation, order of operation

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To solve for b in the equation b^2 = a^2 x c^2 / 4, taking the square root yields b = (a)(c) / 2. The discussion clarifies that while algebraically there could be two solutions, only the positive value is relevant in this physical context. Additionally, the conversation notes that the original poster used excessive parentheses, which is less common in such equations. The final consensus confirms the correct formula and appreciates the assistance provided. Understanding the order of operations and simplification is key in solving such equations effectively.
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Homework Statement


solve for b given;
(a) = time (seconds)
(b) = height (meters)
(c) = Velocity (seconds)

Homework Equations


b^2 = a^2 x c^2 / 4

The Attempt at a Solution



If I take the square of the other side will I get? b = (a)(c) / 2

Thanks!
 
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Adam_9333 said:

Homework Statement


solve for b given;
(a) = time (seconds)
(b) = height (meters)
(c) = Velocity (seconds)

Homework Equations


b^2 = a^2 x c^2 / 4

The Attempt at a Solution



If I take the square of the other side will I get? b = (a)(c) / 2

Thanks!
Do you mean, if you take the square root of the other side, will you get b = ac / 2 ?
 
yes exactly
 
Adam_9333 said:

Homework Statement


(a) = time (seconds)
(b) = height (meters)
(c) = Velocity (seconds)

Homework Equations


(b)^2 = [(a)^2 x (c)^2] / 4

The Attempt at a Solution



If I take the square of the other side will I get? b = [(a)(c)] / 2

Thanks!
Much better! To solve for b take the square root of both sides, and you get b = ac/2, as you show.

If this were an algebra problem, there would be two solutions:
##b = \pm \frac{ac}{2}##
But since this is a problem of a physical nature, you're probably not concerned with the negative value for b.

BTW, you have many more parentheses than you need. More often, we get people posting here who don't use enough parentheses. Your equation can be written as b2 = a2c2/4. Having too many parentheses is not a problem, though.
 
Awesome thanks for the help, I really appreciate it!
 
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Essentially I just have this problem that I'm stuck on, on a sheet about complex numbers: Show that, for ##|r|<1,## $$1+r\cos(x)+r^2\cos(2x)+r^3\cos(3x)...=\frac{1-r\cos(x)}{1-2r\cos(x)+r^2}$$ My first thought was to express it as a geometric series, where the real part of the sum of the series would be the series you see above: $$1+re^{ix}+r^2e^{2ix}+r^3e^{3ix}...$$ The sum of this series is just: $$\frac{(re^{ix})^n-1}{re^{ix} - 1}$$ I'm having some trouble trying to figure out what to...

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