Question on answer
og question:The pythagorean theorem of baseball is a formula for approximating a team's ratio of wins to games played. let R be the number of runs the team scores during the season, A be the number of run allowed to opponents, W be the # of wins, G be the total of games played. Then the formula W/G≈ R^2/R^2 + A^2 approximates the teams' ratio of wins to games played.
A. solve the formula for A in terms of w, g, and R
I just dont see how you went from (R^2 + A^2 = R^2 * g/w. ) to this =
A^2 = R^2 (g/w - 1)
because would you just subtract/cancel r^2 away?
Thank you
Don't know what the original question was, can't look for it since you did not
give yourself a name, and you didn't indicate who the tutor was ....
Depends if you meant:
W/G≈ R^2/R^2 + A^2 as you typed it, or
W/G≈ R^2/(R^2 + A^2) which makes more sense , since in the original it would
immediately reduce to
W/G≈ 1 + A^2
so....
W/G≈ R^2/(R^2 + A^2)
W(R^2 + A^2) = G R^2 , I cross-multiplied
divide both sides by W
R^2 + A^2 = G R^2/W
subtract R^2 from both sides
A^2 = GR^2/W - R^2 = R^2(G/W - 1)
A = √(R^2(G/W - 1))
looks like the explanation lies in the 2nd last line
To solve the formula W/G ≈ R^2/(R^2 + A^2) for A in terms of W, G, and R, we can follow these steps:
Step 1: Cross-multiply the equation to eliminate the fraction:
W(G - 1) ≈ R^2
Step 2: Expand the equation:
WG - W ≈ R^2
Step 3: Move the term with R^2 to the other side of the equation:
WG - W - R^2 ≈ 0
Step 4: Rearrange the equation:
A^2 = R^2(G/W - 1)
Now, let's explain how we went from R^2 + A^2 = R^2 * G/W to A^2 = R^2(G/W - 1):
Starting with R^2 + A^2 = R^2 * G/W, we are trying to isolate A^2 on one side of the equation.
Step 1: Subtract R^2 from both sides:
R^2 + A^2 - R^2 = R^2 * G/W - R^2
Step 2: Simplify on both sides:
A^2 = R^2(G/W - 1)
So, by subtracting R^2 from both sides, we get A^2 = R^2(G/W - 1), which is the final expression for A in terms of W, G, and R.