Arrange Formulas: Isolate Variable Easily

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To isolate a variable in a formula, such as v, one must rearrange the equation by moving all terms involving v to one side and constants to the other. This process may involve operations like addition, subtraction, multiplication, division, or taking square roots, depending on the complexity of the formula. For example, starting with the equation 1/2mv^2 = mgh, one can eliminate the fraction by multiplying both sides by 2, resulting in mv^2 = 2mgh. Next, dividing by mass cancels out m, leading to v^2 = 2gh, and finally taking the square root gives v = sqrt(2gh). Understanding these steps is crucial for effectively isolating variables in various equations.
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How do you rearrange a formula to isolate a particlur variable.
 
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What do you mean ? Give an example.

Daniel.
 
Specfic

Like they give a forumula and they ask to isolate like v for example.
 
That is a very vague question I am afraid, as there are many methods you may have to use, depending on what kind of formula. It sometimes is even impossible.

Assuming you mean something simple, basically if we want to isolate v, we take all the things with a v to one side of the formula, and everything else on the other side. If there are common factors, cancel them out. If it is still not plain v, say v^2, then we cancel that out with a square root, we've got v. But that's not a very good explanation. Give us an example and we'll tell you how to do that one.
 
To isolate a particular variable you must add, subtract, square root, power, divide or multiply variables around. Trying to cancel what's on the left or on the right.

Say you have

1/2mv^2 = mgh

To get velocity by itself. We first times each side by 2. This gets rid of the half.

mv^2 = 2mgh

Then we divide by mass and the masses cancel out.

v^2 = 2gh

Then we square root each side because it is v^2, and to get a powered number/variable by itself we must square root it.

v = sqrt(2gh)
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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