floridastudent wrote:
Hi can someone elaborate how you made that final jump from (w/z) = (x/y)^3 into (w/z) =(w/x)^3
In the problem statement it saysthat:
\(\frac{w}{x} =\frac{x}{y}\)
So you only have to substitute both values.
A faster way to solve it is to play with values:
w=8, x=4, y=2,z=1
\(\frac{8}{4} =\frac{4}{2} =\frac{2}{1}\)
\(\frac{w}{z} =\frac{8}{1} = 8\)
A)\(\frac{w}{x} =\frac{8}{4} ≠ 8\)
B)\(\frac{w}{x^3} =\frac{8}{4^3} =\frac{8}{64} ≠ 8\)
C)\(\frac{w^2}{x^2} =\frac{8^2}{4^2} =\frac{64}{16} = 4 ≠ 8\)
D)\(\frac{w^2}{x^3} =\frac{8^2}{4^3} =\frac{64}{64} = 1 ≠ 8\)
E)\(\frac{w^3}{x^3} =\frac{8^3}{4^3} =\frac{2^9}{2^6} = 8\)
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Statistics : Posted by andreagonzalez2k • on 24 Oct 2023, 19:50 • Replies 4 • Views 509
