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Variation Practice 3 - Visualizing Algebra

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    For our third problem, y would be equal to 8. Great work for getting that one
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    correct. You might have had trouble with the set up, so let's see how we can do
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    that. We know x varies directly with y, so they should be directly across from
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    each other in our proportion. But for z, x varies inversely, so z should appear
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    in the denominator. I want to start by plugging in the values of my first case
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    first, x1 is 5, y1 is 5, and z1 is 3. In my second case, I don't know this value
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    of y. It's what we're looking for. So I'd be sure to list y2 as my variable. Or
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    just y. The second value for x is 6, and the second value for z is positive 4.
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    Now we have an equation we can solve with a rational expression. We multiply
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    these two fractions together, to get 20 divided by 3y. Then we cross-multiply to
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    get 15 y times 120. And finally we divide by 15 to get y is equal to 8. Again, I
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    think the setup to this problem is probably the hardest. You want to make sure
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    that when you're varying directly, things are across from one another. Whereas
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    if things vary inversely, they're diagonally across from one another.
Cím:
Variation Practice 3 - Visualizing Algebra
Video Language:
English
Team:
Udacity
Projekt:
MA006 - Visualizing Algebra
Duration:
01:08
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