- Thread starter krisolaw
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. . . . ."y varies directly with x": y = kx

. . . . ."y varies inversely with x": y = k/x

. . . . ."y varies jointly with x and z": y = kxz

Translate the given relationship. Plug "45" in for "y" and "4" in for "x". Solve for the value of "k".

Rewrite the variation equation, this time using the value you found for "k". Plug "5" in for "y". Solve for x.

Eliz.

. . . . ."y varies directly with x": y = kx

. . . . ."y varies inversely with x": y = k/x

. . . . ."y varies jointly with x and z": y = kxz

Then what about the problem:

If y varies inversely with the

Instead of it being y = k/x wouldn't it be y = k/x^2? Thanks!

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Yes. As stated, "if y varies inversely with" whatever, then the equation is "y = k/(whatever)".krisolaw said:If y varies inversely with thesquareof x,...wouldn't it be y = k/x^2?

It is common mathematical practice to use variables for unknowns, which is what the "x" stood for. Sorry for the confusion.

Eliz.