Suppose that w and t vary inversely and that t = when w = 4. Write a function that models the inverse variation, and find t when w = 9. (1 point) t = 1/5w; 4/45 t = 1/5w; 1/5 t = 1/20w; 1/80 t = 4/5w; 4/45
t = ??? when when w = 4?
Keep the extra "when" as a souvenir :-)
@whpalmer4 t = 1/5
@whpalmer4
We have inverse variation here, so wt = k where k is a constant. If w doubles, t is halved. We also know that t = 1/5 and w = 4. We can solve for k by simply doing \[k = (\frac{1}{5}*4) = \frac{4}{5}\] so our equation is \[wt = \frac{4}{5}\] or \[w=\frac{4}{5t}\] or \[t=\frac{4}{5w}\] For the second part, just plug in w = 9 in our final formula to get the value of t.
As a quick check, we roughly doubled w, so t should be roughly half of what it was. half of 1/5 is 1/10, and 4/45 is about 1/11, so our answer seems plausible.
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