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Variation Practice 2 - Visualizing Algebra

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    And in this problem, a would equal 3. Way to go if you got that one correct.
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    Since a varies inversely with b, we want to make sure that we put the b quantity
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    in the denominator. But for c, a varies directly with it. So we want to put that
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    straight across in the numerator. This is perhaps the trickiest part about
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    setting up variation problems. And you want to make sure you get it right each
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    time. Remember, with inverse relationships. They'll appear in opposite
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    positions, whereas with direct relationships, the variables will appear directly
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    across from one another. So in my first case, a is 12, b is three, and c is
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    eight. So I plug in those values. In the second case, we don't know the value of
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    a, so I leave that written as a in the denominator. But I do know that b2 is
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    equal to six. And c2 is equal to 4. These are the values of b and c in our
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    second case. Now that we have this equation, we multiply these two fractions
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    together to get 48 divided by 12. We cross multiply to get 144 equals 48a, and
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    finally we divide both sides by 48 to get a is equal to 3.
Cím:
Variation Practice 2 - Visualizing Algebra
Video Language:
English
Team:
Udacity
Projekt:
MA006 - Visualizing Algebra
Duration:
01:04
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